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Question 1 

The students of Vidya Mandir Secondary school, Jalpaiguri went for an excursion. It was seen that a boat travels at 14.5 km per hour when it goes along with the stream. By the time he starts returning the speed of the river doubled than its original value due to a sudden storm. The speed of the boat is 7 km per hour when it goes against the river stream. What is the speed of boat per hour? a) 10 km b) 8 km c) 12 km d) 7 km Answer : c) 12 km

Solution : Let the speed of the boat be B and the speed of stream be S. Equation for travel along with Stream : B + S = 14.5 ----> eq 1 During the Equation travel against the stream, of =the doubled.We are using Therefore, for travel againstthe thespeed stream B stream - 2S = 7temporarily ----> eq 2 (NOTE: 2S instead of S in the equation as speed of the stream has temporarily doubled when he travelled against the stream) eq 1 - eq 2 => 3S = 7.5 or S = 2.5 kph Substitute S = 2.5 in eq 1 B = 14.5 - 2.5 = 12 Question 2 

A boat man was driving a boat during a cyclonic storm which time 222 the speed the hours river was considerably high than normal times. He found thatduring the boat travelled km inof three when he was driving along with the river. But when he drove the boat against the river the boat travelled 100 km in two hours. What is the speed of the river per hour during cyclonic storm? a) 10 km b) 12 km c) 8 km d) 9 km Answer : b) 12 km

Solution : Let the eq boat B + the S =speed 222/3of ----> 1 be B and the speed of the river be S. B - S = 100/2 ----> eq 2 Adding equations 1 and 2 we get : 2B = 300 + 444 / 6 2B = 744/6 2B = 124 B = 62

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Substituting B = 62 in eq 2 we get, S = B - 100/2 = 62 - 50 = 12 Question 3 

Maruti Cruisers sailed km along with the river the in six hours. theCruiser boat hadtravelled to returnat to the starting point and132 it started returning against river and Suddenly this time the 128 km in eight hours. By how much percentage Cruiser's speed exceeds the speed of the River ? a) 533.33% b) 444.44% c) 267.67% d) none of these. Answer : a) 533.33%

Solution : Speed of boat along with the river per hour = 22 km (132 divided by 6) Speed boat against B + S =of22 ----> eq 1 the river per hour = 16 km (128 divided by 8) B –  S = 16 ----> eq 2 Adding equations 1 and 2, we get : B+S-B+S = 22 -16 = 6 =2S S = 3 km/hour Substituting S= 3, in eq 1 we get : B + 3 = 22 B = 22 - 3 = 19 km /hour Cruiser's speed exceeds the speed of the River By : Cruiser's Speed - River Speed / River Speed x 100 = B - S / S X 100% = 19 - 3/ 3 X 100 = 16/3 X 100 = 533.33% Question 4 

In a radio Station, the songs database contains songs from 2 Tamil movies, 1 Hindi movie and 1 Telugu movie. Each movie has 5 songs.They can play only 3 songs.If they want to play 3 tamil songs then what will be the probability? a) 10C3 / 20C3 b) 5C3 / 15C3 c) 5C3 / 20C3 d) 20C3 / 10C3 Answer : a) 10C3 / 20C3

Solution : Since moviethey has have 5 songs, the total number of songs is 20. Out ofeach 20 songs, to play 3 songs. Therefore, number of ways of selecting 3 songs = 20C3.  Number of ways of selecting 3 Tamil songs = 10C3. Probability of playing three tamil songs = Number of ways of selecting 3 Tamil songs / Number of ways of selecting 3 songs (any language) = 10C3 / 20C3.

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Question 5 

There are two cricket teams namely Team A and Team B. Each has 15 players (including substitutes). Now a new team of playing 11 has to be formed by pulling players from both the teams. In how many ways 11 players can be selected such that at least 5 players are picked from each team? a) 15C5 x 15C6 b) 2(15C5 x 15C6) c) (15C5)2 d) 15C6 Answer : b) 2(15C5 x 15C6)

Solution : Since at least 5 players from each team should be pulled up the possible ways of selection are 5 players from Team A and 6 players from Team B (OR) 6 players from team A and 5 players from Team B. Case I : Therefore, the number of way of selecting 5 players from Team A AND 6 players from Team B = 15C5 x 15C6. Case II : Similarly, the number of way of selecting 6 players from Team A AND 5 players from Team B = 15C6 x 15C5.

Hence, the number of ways of selecting 11 players such that at least 5 players in each team = Case I OR  Case II = 15C5 x 15C6 + 15C6 x 15C5 = 2(15C5 x 15C6) P.S : In probability equations like those employed in the above question, AND's need to be replaced by 'x' (multiply) and OR needs to be replaced by '+' (addition). Question 6 

In a College, there are 10 students in Maths department, 12 Students in Computer Science department and 8 students in Physics department. A recruiting company wants to select 4 students out of these 30 students with at least 1 student from each department. In how many ways the company can achieve this ? a) 12960 b) 12690 c) 19260 d) 16920 Answer : 

Solution: Since there should be at least 1 student from each department, the possible ways of selection are 1 Student from Maths, 1 student from computer science and 2 students from physics (OR) 1 Student from Maths, 2 students from computer science and 1 student from physics (OR) 2 Students from Maths, 1 student from computer science and 1 student from physics.

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Note : Applying a very similar logic as that of the previous question, Therefore, the number of ways of selecting 4 students such that at least 1 student from each department = 10C1 x 12C1 x 8C2 + 10C1 x 12C2 x 8C1 + 10C2 x 12C1 x 8C1 = (10 x 12 x 8 x 7)/2 + (10 x 12 x 11 x 8)/2 + (10 x 9 x 12 x 8)/2 = (10 x 12 x 8)(7+11+9)/2

= = 10 480xx12 27x 8 (27/2) = 12960 Question 7 

Priya wanted to mail 120 messages to her friend Banu. She mailed 1 message on the first day, 2 messages on the second day, 3 messages on the third day and so on. How many days she required to send all messages to Banu? a)16 days b) 17 days c) 15 days d) 14 days. Answer : c) 15 days.

Solution: Total number of messages = 120 On the 1st day, Priya mailed 1 message to Banu. On the 2nd day, she mailed 2 messages to Banu and so on..... Let X denote the number of messages send on the Xth day. Therefore, 1st day messages + 2 nd day messages + ....... + Xth day messages = 120. 1 + 2 + 3 + ...... + X = 120 X / 2 = 120 X2( +XX+=1)240 2 X  + X - 240 = 0 By factoring the above eqn. we get X 2 - 15 X + 16 X - 240 = 0 ( the middle term is obtained by the multiplicants of last term i.e 15 x 16 =240 and the subtracted value is 1 which is the middle term) X(X - 15) + 16 ( X - 15) = 0 or (X + 16)(X-15) X = 15 0r X = -16 X = 15 ( since number of days cannot be negative). So Priya required 15 days to send all messages. Question 8 

An IT company conducted interview for B.E Students. On the first day, they selected one student. On the second day, they selected 8 students and on the third day, they selected 27 students and so on. How many students will be selected if they conduct interview for 10 days? a) 6050 b) 2530 c) 3025 d) 6025

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Answer : c) 3025

Solution: On the first day, they selected 1 student. On day, they selected students.and so on. On the the second third day, they selected 278students Since they conduct interview for 10 days, the number of students selected on the 10th day is 103 = 1000 students. Total number of students selected = No. of students selected on 1 st day + No. of students selected on 2nd day + ..... + No. of students selected on 10 th day. Therefore, Total number of students selected = 1 + 8 + 27 + ..... + 1000. = 13 + 23 + 33 + .... + 103. By using the formula, 13 + 23 + 33 + .... + n 3 = (n (n+1) / 2)2, we get

2

 Number of students/ 4selected = (10 (10+1) / 2) . = (10 *11*10*11) = 3025. Hence, the company has selected 3025 students in 10 days. Question 9 

Jeeva has a story book of 2047 pages. He read 1 page on the first day, 2 pages on the second day, 4 pages on the third day, 8 pages on the fourth day and so on. How many days it took for Jeeva to complete the book? a)10 days b) 12 days c) 11days d) 13 days Answer : c) 11 days

Solution: Total number of Pages = 2047 On the 1st day, Jeeva read 1 page. On the 2nd day, he read 2 pages On the 3rd day, he read 4 pages and so on..... Let X denote the number of pages read on the X th day. Therefore, 1 + 2 + 4 + .... + X = 2047 Then by using the formula, the sum of X numbers in G.P = a ( r X -1)/(r-1) where a –  First term & r –  Common ratio, we get 1 (2x - 1) / (2 - 1) = 2047 2X-1 = 2047 2X = 2048

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2X = 211  X = 11. Question 10 

Trisha talked over certain important for nearly onecoffee. hour. After thatdrinks they went Coffee and DayPriya shop —  Trisha had cold coffee and issues Priya asked for hot Both the wereto charged at the same rate of Rs.75 per cup. Trisha started walking at 6 kmph towards eastern direction and Priya started walking at 5 kmph towards western direction. They started at 11 am. At what time will they be 44 km apart ? a) 3 am b) 2 pm c) 3pm d) 4 pm Answer : c) 3 pm

Solution : Trisha and Priya started walking in eastern and western direction respectively at 11 am. Trisha walked at 6 kmph and Priya started walking at 5 kmph. So the relative speed is 11 kmph. Time required to be 44 Kms apart when the relative speed is 11 Km/hr = Distance Between Them / Relative Speed = 44/11 = 4. They will be 44 km apart after four hours from the time of starting. i.e. 3 pm. (candidates must  be alert to not to mark a) 3 am as answer) Question 11 

Sundaramurthy can row at 8.5 kmph in still water and he finds it takes him thrice as long to row up as to row down the river. Find the speed of river in kmph. a) 3.25 km/hr b) 5.25 km/hr c) 2.75 km/hr d) 4.25 km/hr Answer : d) 4.25 km/hr

Solution : Let the speed of the river be c. His speed when rowing down = 8.5 + c His speed when rowing up = 8.5 - c In the question it has "it takes himupthrice as long rowtaken up as for to row river". This means thebeen timegiven takenthat for him to row is 3 times thetotime himdown to rowthe down. In other words, his overall speed when rowing down will be 3 times greater than his overall speed when rowing up. Let the time taken for upstream be Tup and for downstream be Tdown. Tup = 3 x Tdown Tdown / Tup = 1 / 3

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Speed Up = Distance / Time Up Speed Down = Distance / Time Down Speed Up / Speed Down = Time Down / Time Up 8.5 - c / 8.5 + c = 1 / 3 8.5 8.5 + + cc = = (8.5 25.5 -- c) 3c3 c + 3c = 25.5 –  8.5 = 17 4c = 17 c = 4.25 km/hr Question 12 

In a software company employees arranged for a picnic. At the planning stage it was found that  per employee share of the total cost would be same as the number of enlisted employees. However, on the day of picnic the number of employees who were present was 20 less than the enlisted number. Consequently, per employee share upemployee by 20%. What was the number of employees who had enlisted initially? How much didshot each pay finally? a) 120, Rs.172.80 b) 100, Rs.172.80 c) 120, Rs.144 d) none of these. Answer : c) 120, Rs.144

Solution : Let the number of employees initially be n From the question, the Cost per Employee = n (as it is given that "employee share of the total cost would same as the of enlisted employees") Hence total be cost = Cost pernumber Employee x Number of Employees = n x n = n2  Twenty employees did not turn up and hence those participating in picnic = (n –  20). Hence the cost of n2 has to be shared by the present (n - 20) employees. Total cost per employee will be n2 / (n-20) This is to be given to be 20% more than the original figure. So, n2 / (n-20) = (120/100) n n2 = 1.20 n (n-20) Dividing by n on both sides, n = 1.2 (n - 20) n 24==1.2n 0.2n- 24 120 = n Initial amount planned to be paid by each employee = n = Rs.120 Final and actual amount paid by each employee = n 2 / (n-20) = 14400/100 = Rs. 144. Question 13 

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Vijay, a shop owner was running a shop near IEFS Engineering College, Warangal. The shop is  patronised by students, teachers, other staff members of the College besides to tourist public. One day Rehamana a tourist bought a new wrist watch from Vijay for Rs. 2480. Rehamana was not having sufficient money in cash and then issued a cheque for Rs.2800. Vijay paid the difference amount to Rehamana . Vijay gave the cheque to his cousin brother and took money him. Vijay’s cousinunpaid. found that there no balancethe in cheque the bank account from the cheque was returned Vijay hadwas to reimburse amount to of hisRehamana cousin andand take back the cheque. The cost price of wrist watch was Rs. 2120. What is the gain or loss made  by Vijay in this transaction?

a) Gain of Rs. 2440 b) loss of Rs. 2440 c) gain of Rs.5240 d) Loss of Rs. 5240 Answer : b) loss of Rs. 2440/.

Solution : Transaction I (involving Vijay and Rehamana) Rehamana gave a cheque for Rs. 2800 for the watch worth Rs. 2480. The difference amount = 2800 - 2480 = Rs. 320 was given by Vijay to Rehamana. Transaction II (check realization by Vijay with the help of his brother) Vijay when tried to realize the cheque, the cheque bounced. Though he got the money in advance from his brother, at the end, he had to return that money as cheque was not valid. Since he had returned the money that he got from his brother, his net gain or loss in transacton II (with his  brother is 0). Summing Up :

Total Vijay Loss In I + Loss in Transaction II + Cost Price of Watch (as he hasLoss nowTo given the=watch to Transaction Rehamana which he cannot collect back) = 320 + 0 + 2120 = Rs. 2440 Question 14 

A leading seller of Vegetables in Koyambedu converted a loss of 20% into a profit of 25% when the selling price of potatoes was increased by Rs.225 per tonne. . Find the cost price of potato per tonne? a) Rs.600 b) Rs. 800 c) Rs. 500 d) None of these. Answer : c) Rs. 500

Solution : Let the cost price per tonne = CP. Initial Selling Price SP1 when he was making a loss of 20% = SP1 = 80% of CP = .8CP

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Selling Price SP2 when he was making a profit of 25% = SP2 = 125% of CP = 1.25CP Increase in selling price = SP2 - SP1 = 225 1.25 CP - 0.80 CP = 225 0.45CP = 225 CP = Rs. 500 Question 15 

A toy vendor has four toys each costing Rs.100 with him. He plans to make an overall profit of 25% by spreading the profit equally across all the toys. There comes a buyer who wants three toys but at a discount of 5% on the SP. He accepts and sells three toys at a discount of 5% from the marked selling price. Now, if another buyer comes to his shop, by how much percentage the selling price needs to be increased from the cost price so that still he manages to make an overall  profit of 25% with all 4 toys. a) 43.25% b) 43.75% c) 43.55% d) 43.65% Answer : b) 43.75%

Solution : CP of each of the toys - Rs. 100. According to his initial plan, he would had thought to sell each of the toys at Rs. 125 so that he would make a net profit of 25% by selling all the toys. But the first buyer wants a discount of 5% on the selling price of Rs. 125. Therefore, his discount per toy = 5/100 x 125 = Rs. 6.25. Thus he by has selling threeup toys a discount 6.25 he the would had price lost 3of x 6.25 = 18.75.  Now to make foratthis amount of byRs. increasing selling the 4th toy. Therefore his new selling price on the 4th toy = Rs. 125 (his planned selling price) + 18.75 = 143.75 Therefore his new marked price 143.75 is greater than the cost price of Rs.100 by, (143.75 - 100) / 100 % = 43.75%. Question 16 

There are nine different toppings in Merry Brown Pizza shop. They have to make pizzas with different toppings. In how many ways can the shop keeper make pizzas with 2 different toppings? a) 36 b) 48 c) 72 d) none of these. Answer : a) 36

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Solution : The question deals with choosing 2 toppings from 9 available toppings REGARDLESS of the ORDER in which we pick. In cases, where the order of choice doesn't matter we need to use Combination Formula of nCr (formula to choose r options from n available options regardless of order). of ways to choose 2 different toppings from available 9 toppings = nCr = 9C2 = 9 X 8 /  Number 1 x 2 = 36 Question 17 

In a toy shop there are seven different varieties of toys. The shopkeeper wants to make a selection of different toys of 3 different varieties? a) 28 b) 35 c) 56 d) none of these. Answer : b) 35

Solution : This question is similar to the 1st question.  Number of ways to select 3 varieties from 7 available varieties = nCr = 7C3 = 7 x 6 x 5 / 1 x 2 x 3 = 35 Question 18 

In how many different ways can a group of 4 boys and 2 girls be made out of a total of 8 boys and 4 girls? a) 240 b) 360 c) 280 d) 420 Answer : d) 420

Solution :  N1 = Number of ways to choose 4 boys from 8 boys = nCr = 8C4  N2 = Number of ways to choose 2 girls from 4 girls = nCr = 4C2  Number of ways to choose 4 boys and 2 girls out of a total of 8 boys and 4 girls = N1 X N2 = 8C4 x 4C2 = 70 x 6 = 420 Question 19 

A typist was appointed in Correct technology services and she was asked to prepare 23 letters addressed to different people and prepare 23 covers addressed to them. She inserts these letters at random into the envelopes. (one letter per envelope). What is the probability that exactly one letter is inserted to the wrong envelope?

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a) 7/23 b) 9/23 c) 0 d) 11/23 Answer : c) 0.

Solution : There are 23 letters and 23 covers with addresses. By mistake if the typist puts one of the letters in wrong address, there is no way but another letter has to be put in wrong cover. Hence, there is NO CHANCE that ONLY ONE of the letter is inserted in wrong envelope. Hence probability of such occurrence is 0. For example, let the letters be L1, L2, L3..... L23 and let the respective covers with addresses be C1,C2,C3..... C23. Lets say, if L2 is placed in C4 instead of C2 then L4 has left with no option but to be placed in C2. Hence, if one letter is misplaced, another letter has to be misplaced. Hence probability of ONLY ONE misplacement is zero as chances of ONLY ONE misplacement is zero. Question 20 

Ansar can do a work of making a sofa set in 40 hours and with the help of Nayeem, he can complete the same work in 24 hours. If they get Rs.1200 for the work, then what could be  Nayeem’s share?  a) Rs. 360 b)Rs.420 c)Rs.480 d) Rs.540 Answer : c) Rs.480

Solution : Ansar can do in one hour 1/40 of the sofa set work Both Ansar and Nayeem perform in one hour 1/24 of the sofa set work.  Nayeem’s one hour work = One hour work of Both Ansar and Nayeem - One hour work of Ansar = 1/24 - 1/40 = 1/60 So, Nayeem will finish the work in 60 hours. Ansar’s revenue share / Nayeem’s revenue share = Time taken for Ansar to complete the work / Time taken for Nayeem to complete the work = 60 / 40 = 3 / 2 Ansar’s revenue share : Nayeem’s revenue share = 3 : 2 Let Ansar's share be 3X and let Nayeem's share be 2X. Therefore 3X + 2X = 1200 (Total revenue of both) Or X = 1200/5 = 240. Therefore Nayeem’s share = 2X = 2 x (240) = Rs. 480  

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Question 21 

8 men and 12 women together can complete construction of a house in 8 days. 6 men and 14 women can complete the construction of the same house in 10 days. Suppose 20 women alone work for construction of this house, how many days will they take to complete the same? a) 36 days b)40 days c)32 days d)24 days Answer : b) 40 days

Solution : Work done by 8 men and 12 women for 8 days is equal to Work done by 6 men and 14 women in 10 days. Let m be the amount of work done by one man and let w be the amount of work done by one woman. Then (8m = (6m 14 w w) 10 64 m++12 96w) w=8 60 m ++140 64 m - 60 m = 140 w - 96 w 4 m = 44 w 1 m = 11 w Construction of the house requires (8 x 11w + 12 w)8 = 800 women days. So if the construction is to be done by 20 women only It will take - 800/20 = 40 days. Question 22 

Bharath and Rajani together can complete a piece of work in 12 days and Rajani and Kamal together in 15 days. If Bharath is twice as good a workman as Kamal, then in how many days will Rajani along complete the same work? a) 30 b)24 c)25 d) 20 Answer : d) 20

Solution : Assume Bharath completes Rajani completes in R days in B days Kamal completes in K days Then 1/B + 1/R = 1/12 -->eq 1 (Bharath and Rajani take 12 days to complete the work) 1/R + 1/K = 1/15 --> eq 2 eq 1 - eq 2 gives, 1/B - 1/K = 1/12 - 1/15 = 1/60 --> eq 3 From the question it is given that Bharath is twice as efficient as Kamal. That is, if Bharath takes B days Kamal will take 2B days to complete the work. Therefore eq 3 becomes,

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1/B - 1/2B = 1/60 1/2B = 1/60 B = 30 days 1/R = 1/12 - 1/30 = 3/60 = 1/20 So Rajani will take 20 days to complete the work. Question 23 

Adhvaith can do a certain work in 30 days. Kashyap can do same work in 25 days. Adhvaith started the work and worked for 9 days. Kashyap came and joined to do the work from the 10th day. How many more days would they have taken together to complete the work? a)10 3/11 days b)11 2/11 days c) 9 6/11 days d) 8 2/11 days Answer : c) 9 6/11 days.

Solution: Adhvaith can do 1/30 of the work in one day In 9 days he would have completed - 9 x 1/30 = 3/10 of the work Balance work = 1 - 3/10 = 7/10 Kashyap can do 1/25 of the work in one day Work that can be done by Adhvaith and Kashyap in one day = (1/30 + 1/25) = 11/150 of the work. So Adhvaith and Kashyap can complete 7/10 of the work in 7/10 x 150/11 = 105/11 = 9 6/11 days. Question 24 

A private limited company entrusts works to 20 men, working 12 hours a day. This group can complete the work in 24 days. The company now wants to entrust twice the work to 60 men working 4 hours a day. Assume that 2 men of the first group do as much work in one hour as 3 men of second group do in 1 ½ hours. How many number of days will the second group of men take to complete this work? a) 108 days b)120 days c)124 days d)81 days Answer : a)108 days

Solution : Let efficiency of men in I group be E1 and that of second group be E2. Ratios of efficiency of men in I group to that of II group can be found by using the formula,

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E1/E2 = Time taken by men in II group to do certain amount of work / Time taken by men in I group to do the same amount of work as that of men in II goup = (3 x 1.5) : (2 x 1) = 4.5 : 2  Now, M1D1T1E1W2 = M2D2T2E2W1 --> 1 (where M1 = number of men in I group, M2 = number of men in II group. D1 = number of days required to complete work by group I, D2 = number of days required to complete work by group II. T1 = working hours per day by group I. T2 = working hours per day by group II. w2 = amount of work by group II, w1 = amount of work by group I.) Since we are to calculate the time taken by group II to complete twice the amount of work as that of group I, W2 = 2 x W1. We had earlier calculated E1/E2 = 4.5/2. Also the =question we can infer that, M1 =from 20, M2 60. T1 = 12, T2 = 4. D1 = 24 and D2 is what we need to find. Substituting all the values in eq 1, we can find D2 as follows. D2 = (20 X 24 X 12 X 4.5 X 2)/(60 X 4 X 2 X 1) = 108 days. Question 25 

Three persons Manmohan, Anna And Sushma working together, can do a job in X hours. When working alone, Manmohan needs an additional six hours to do the job; Anna, working alone needs an additional hour and Sushma working along needs X additional hours. What is the value of X? a)2/3 b)3/2 c)11/12 d)2 Answer : a)2/3

Solution : In this type of problems where answers cannot be easily found out using equations, it is advisable to go from the given answer choices. Based on information given one hour work done by all the three together = 1/ X+6 + 1/ X+1 + 1/2X = 1/X X is not known. Using the data given 1/ (2/3) +6 + 1/(2/3) +1 + 1/(4/3) = 1/ ( 2/3)

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This comes out correctly. Whereas other values given in b), c) and d) do not get the result  properly. Hence a) is correct. Question 26 

If 40%percentage of the people read newspaper X, 50% read newspaper Y and 10% read both the papers. What of the people read neither newspaper? a) 10%. b) 15%. c) 20%. d) 25%. Answer : c) 20%.

Reason :

40% read newspaper X 50% read newspaper Y and out of this 10% read both X and Y That means X alone is read by 30%, Y alone is read by 40% Both are read by 10% People who read X alone + People who read Y alone + People who read both X and Y = 30% + 40% + 10% = 80% This means the balance 20% don't read either of the newspapers. Question 27 

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In a metropolitan city 25% neither watch TV news nor read newspaper, 35% read a newspaper and 20% read a newspaper and watch the news on television also. What percent of people watch Television? a)55% b) 60% c) 50% d) 40% Answer : b) 60%.

Reason :

Let X% of people watch television. People who read Newspaper –  35% People who watch TV as well as read newspaper –  20% From Venn diagram, we can infer that people who watch Television alone = (X-20)% People who read Newspaper alone = 35% - 20% = 15% People who neither watch TV nor newspaper is given as 25%. This means, people who watch either TV or Newspaper or Both = 100 - 25 = 75% Also this 75% will be equal to People who watch TV alone + People who read Newspaper alone + People who watch TV and read Newspaper Or 75% = (X-20)% + 15% + 20% Or X = 60%

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Question 28 

Out of the total of 200 musicians in Rehman Music club, 10% can play all the three instrumentsguitar, violin and flute. The number of musicians who can play any two and only two of the above instruments is 70. The number of musicians who can play the guitar alone is 80. What is the total number of those who can play violin alone or flute alone? a) 20 b) 30 c) 40 d) 50 Answer : b) 30

Reason :

 Number of people who can play all the three instruments = 10% of 200 = (10/100)200 = 20 (people in pink colored area) Musicians who can play any two and only two of the three instruments - 70 (Sum of people in grey colored areas)  Number of musicians who can play Guitar alone = 80  Number of musicians who can play violin alone or flute alone = Sum of people in blue areas = 200 -170 = 30

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Question 29 

In a college of arts and science 96 students are seated in rows and columns in such a way that the number of students in each row is 50 % more than the number of students in each column. How many students are there in each row? a) 7 b) 9 c) 12 d) 11 Answer : c) 12

Reason : (These kinds of questions can be answered in two ways. Either by applying pure mathematical calculations or by applying pure logic considering the options given. But logical answering may not yield correct answers all the time as some questions may require mathematical solving for accurate answers when options are close to each other.) In each row there are 50 % more students than that of the columns. This means the number of students in each row should be divisible by 2 (Because 50 % of any number can be found by dividing it by 2.) Of the given choices c) alone satisfies this condition (As simple as that!) Question 30 

A secondary school student scored 31 marks in Science, 39 marks in mathematics, 28 marks in Hindi, 26 marks in Social studies and 36 marks in English. The maximum marks a student can score in each subject is 80. How much percentage did the student get in this examination? a ) 30% b) 50% c) 40% d) 44% Answer : c). 40%

Reason : Student's total marks in all the subjects = 160. This is in five subjects. Each subject carries 80 marks . Total maximum marks = 400 Hence, his percentage in the examination = (Student's total marks in all the subjects / Total maximum marks) X 100 160/400 x 100 = 40%

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Question 31 

In a class of certain number of students, Kamal’s rank is tenth from the top. Ram is seven ranks  below Kamal. Also Ram ranks 30th from bottom. What is the strength of the class.

a)47 b)57 c)28 d)46 Answer : a). 47

Reason : Kamal is tenth rank from the top. Ram is seven ranks below Kamal. This means he is 17th rank from the top. Let the number of students be x. Ram's rank from bottom = Strength - Ram's Rank from top Or 30 = x - 17 Or x = 47 Hence, strength of the class = 47. Question 32 

In a class of 50 students 26% students play only cricket, 18% students play only badminton, 10% students play only football. 20% students play only badminton and cricket, 12% students play only cricket and football and 8% students play only football and badminton, 6% students play all the three games. Totally how many students play cricket? a)32 b) 24 c) 14 d)12 Answer : a)32

Reason : Below Venn diagram represents percentage of students playing Cricket, Badminton and Football

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  You will find cricket is played by 26% + 20% + 6% + 12% = 64% 64% of 50 = 50 x 64/100 = 32. Question 33 

Consider three people A,B and C. Let A and B can finish a job in 21 days, B and C in 14 days and A and C in 28 days. Who will take the least time when working independently ? Options : 1) A 2) B 3) C 4) Can't be determined Answer 1 

Correct answer is B Consider WA, WB and WC be the work done per day by A,B and C respectively. Then WA + WB = 1/21 -- eq 1 WB + WC = 1/14 -- eq 2 WA + WC = 1/28 -- eq 3 Eq 2 - Eq 3 will give WB - WA = 1/14 - 1/28 = 1/28 -- eq 4

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Eq 1 + Eq 4 will give 2WB = 1/21 + 1/28 = 7/84 WB = 7/168 Sub value of WB in eq 1, we get WA = 1/21 - 7/168 = 1/168. Sub value of WA in eq 3, we get wc = 1/28 - WA = 1/28 - 1/168 = 5/168 Since WB (work done by B per day) is greater when compared to WA and WB clearly B will be able to the maximum work on any given day and hence he should consume least amount of time when working independently. Question 34 

Consider two postmen A and B respectively. A is young and can deliver 20 parcels in 3 hours while B is older than A and can deliver only 15 parcels in 4 hours. If the total number of parcels to deliver is 60, how long they will take working together. a. 121/12 hours b. 144/36 hours c. 144/25 hours d. 121/25 hours Answer 2 

Correct ans is option c. 144/25 hours. A can deliver 20 parcels in 3 hours. Hence for 1 hour he can deliver 20/3 parcels. B can deliver 15 parcel in 4 hours. Hence for 1 hour B needs 15/4 parcels. When A and B work together, for 1 hour they can deliver, 20/3 + 15/4 parcles = 80 + 45 /12 = 125/12 parcels. Hence to deliver 60 parcels they would require : 60 X 12/125 = 720/125 = 144/25 hours Question 35 

Consider a courier company A which can deliver 100 parcels in 5 days with 5 men working for 8 hours a day. Consider another courier company B where every employee is equally effecient as that of company B. Company B is short of one man when compared to A and has a policy of asking its workers to work only for 6 hours a day. How long (in days) company B will take to deliver 100 parcels. Options : a. 8.3 b. 24 c. 12 d 6.6

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Answer 

Correct answer is a. 8.3 days Total amount of work W = N x D X W where N = number of men, D = number of days, W = amount of work per day Applying the above formula for company A we get, Work done by company A to deliver 100 parcels = 5 X 5 X 8 = 200 -- eq 1 Work done by company B to deliver 100 parcels = 4 X D x 6 = 24D -- eq 2 Since the work to be done is same in both the cases, eq 1 = eq2 or 200 = 24D or D = 8.3 Question 36 

In software applications, especially in JAVA based applications, how design patterns help after a  problem/requirement is identified ? Answer 1 Design pattern can be treated as a way or template to solve a problem depending upon the nature of the problem/requirement. Though they themselves cannot solve a problem, they provide greater insights into the best possible ways to solve them. Question 37 

Say two programs essentially dealing with solving same problem through similar algorithms. Let A be written using pointers and B be written using arrays. Which one will be memory efficient and which will be more readable. Answer 2 A will be more memory efficient than B and B will be more readable than A. Reason : generally  pointers are more memory efficient when used by an experienced programmer. And arrays are generally easily readable and can be easily understood by even beginner level programmer. Question 38 

What is the integrity constraint in SQL which when used ensures that values in a column of a table have a corresponding member on another column of another table.

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Answer 3 Question 39 

Tell one good programming requirement that illustrates the fact that C is a strongly typed language ? Answer 1 C involves variable declarations before their usage. This makes it clear that C is strongly typed. Question 40 

Consider a weather report application in C where any change in temperature by +- 2 degrees will  be constantly reported to the system and the application should store the times of the day when these changes took place. Which is a better data structure to be used in this scenario - an array or a linked list ? Answer 2 Linked List would handle dynamically growing data very well when compared to arrays. Hence it would be a better choice. Question 41 

Can one use arrays to implement stacks or queues over linked lists ? Is it feasible ? Answer 3 Thought linked lists are commonly used for implementing stacks or queues, with good  programming logic, arrays can also be used to implement those. However, as you would expect, linked lists are far more efficient in this scenario. Question 42

1) Unnikrishnan is a famous singer from South India. One should never miss to hear his mellifluous melodies. Question : Pick an appropriate synonym for mellifluous. a) Sad b) Nostalgic c) Pleasant d) Priceless Answer: Based on the choices, the appropriate synonym would be c) Pleasant. Question 43

2) One should know when to be evasive. It is the way the prudent overcome difficulties due to overwork. Question : Pick an appropriate synonym for evasive.

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a) leave the place b) avoid commitment c) struggle d) fight Answer: Evasive means to avoid commitment so that one would not be held responsible for an incomplete work. Hence correct answer is b) avoid commitment . Question 44

3) John alluded in a way that he did not like Annie's friendship. Question : Pick an appropriate synonym for allude. a) Argued b) Refer Indirectly c) Fought d) Shouted Answer: Correct synonym for allude from the options is "Refer Indirectly".

Question 45 A bacteria doubles itself each single day. It totally takes 15 days for the bacteria to fill a test tube. Find an approximate number of days for the bacteria to fill 1/3 of the jar. Answer: Though the question looks like a tough one, it is actually a simple question. If it takes 15 days for the bacteria to fill the entire test tube, on 14th day it would had filled half of the tube, on 13th day it would had filled a quarter of the tube and so on. Now 1/3th comes somewhere in middle  between quarter and half filling of the tube. Hence the bacteria would need somewhere between 13 to 14 days to fill 1/3th of the jar. Question 46 Pick the odd man out from the following : a) elation b) frenzy c) enthusiasm d) despair Answer: "despair" would be the obvious answer. All the other options are synonyms with ecstasy while despair is a clear antonym of ecstasy.You should have reasonably good vocabulary to tackle any  placement question of this kind. Question 47 Ram travels by car from city A to city B heading towards north in 4 hours. From there he travels west to City C in 3 hours. Say his average speed is same during both the courses. Now from C he travels back to A in shortest path possible. How much time he would had taken to reach A from C. Answer: Again this is a very simple question which appears difficult. The entire course of Ram takes the form of a right angled triangle. Hence the time taken for him to travel back to A would be sqrt(3^2 + 4^2) = sqrt(25) = 5 hours.

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Questions 48 

Below are few sample questions compiled on the basis of latest recruitment papers of the company. These are technical questions and have been solved for you. You can expect questions of similar nature in the placement tests. Expansion of CDMA ? a) Code Division Multiple Access  b) Cell Data Multiple Access c) Code Data Multiple Access d) None of the above. Solution: The above aptitude question is from communication systems. However this question is a basic one whichshould remember. The answer is a)Code Division Multiple Access. You could expect everyone similar questions like expansion of GSM, GPS etc in wipro placement papers. Which of the following layers are not in TCP / IP? a) Application Layer  b) Network Layer c) Bridging Layer d) Data Link Layer Solution: Answer is c) Bridging Layer. You can always expect one or two questions from computer networks in Wipro papers. What will printf(++i) in C will do? a) Print i and then increment i by 1  b) Increment i by 1 and then print i c) i and ithen i by ii d) Print Increment by iincrement and then print Solution: This is one common question which you can expect in Wipro Papers. These kinds of questions dealing with  basics of C statements are common in other company placemente papers as well. This includes

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TCS, Accenture, HCL, HP etc. What is the function of assembler? a) Converts binary code  b) Converts assembly high levelto language code into assembly code c) Interprets and executes assembly code d) Noe of the above Answer a) Assembly to binary code   Questions 49 

1) My friend collects antique stamps. She purchased two, but found that she needed to raise money urgently. So she sold them for Rs. 8000 each. On one she made 20% and on the other she lost 20%. How much did she gain or lose in the entire transaction? Answer She lost Rs 666.67

2) 12 members were present at a board meeting. Each member shook hands with all of the other members before & after the meeting.How many hand shakes were there? Answer 132

3) An emergency vehicle at a speed oftime 50 miles per20 hour.How vehicle travel on the returntravels trip if10 themiles round-trip travel is to be minutes?fast must the Answer 75 mil es per hour

4) All of the students at a college are majoring in psychology, business, or both. 73% of the students are psychology majors, & 62% are business majors. If there are 200 students, how many of them are majoring in both psychology & business? Answer 70 students ar e majori ng i n both, psychol ogy & busin ess

Questions 50 

1) Grass in lawn grows equally thickand in a uniform rate. It akes 24 days for 70 cows and 60 for 30 cows . How many cows can eat away the same in 96 days.? An s : 18 or 19   

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2) There is a certain four digit number whose fourth digit is twise the first digit.Third digit is three more than second digit.Sum of the first and fourth digits twise the third number.What was that number ? An s : 2034 and 4368   

3) A theif steals half the total no of loaves of bread plus 1/2 loaf from a backery. A second theif steals half the remaing no of loaves plus 1/2 loaf and so on. After the 5th theif has stolen there are no more loaves left in the backery. What was the total no of loaves did the  backery have at the biggining? An s: 31   

4) A person needs 6 steps to cover a distance of one slab. if he increases his foot length(step length) by 3 inches he needs only 5 steps to cover the slabs length. what is the length of the each slab? An s : 31 in ches   

1) There is one lily in the pond on 1st june. There are two in the pond on 2nd june . There are four on 3rd june and so on. The pond is full with lilies by the end of the june. i)On which date the  pond is half full.ans. 29th. --the june has 30 days) (ii)if we start with 2 lilies on 1st june when will be the pond be full with lilies? Ans: 29th ju ne   

2) A lorry starts from Banglore to Mysore at 6.00 A.M,7.00am.8.00 am.....10 pm. Similarly one another starts from Mysore to Banglore at 6.00 am,7.00 am, 8.00 am.....10.00pm. A lorry takes 9 hours to travel from Mysore andcross vice versa. (i) A lorry which hasBanglore started atto6.00 am will how many lorries. (ii)A lorry which had started at 6.00pm will cross how many lorries. Ans: (i)10 lor r ies (ii )14 lor ri es

3) There is an element which triplicates in every hour. Each of these 3 items inturn reproduce exactly 3 other items. If a single compund is kept in a container at noon and the container is full  by midnight. After how many hours is the container 1/3 full? Ans:11:00pm   

4) A person goes to a bank and Quotes x Rs and y paise on a cheque. The cashier misreads it and gives y Rs and x paise. The man comes out and donates 5 paise to a begger. Now, the man has exactly double the amount he has quoted on the cheque. An s: 31 Rs. and 63 paise.

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Questions 51 

1) A 2 mb PCM (Pulse Code Modulation) has a)32 voice channels  b)30 voice channels & 1 signalling channel c)31 chanels &which 1 signalling channel d)32 voice channels out of 30 voice channels, 1 signalling channel and 1 syncronization channel. An swer is :31 voice chanels & 1 signall in g channel   

2) Word alignment is a)alligning the address to the next word boundary of the machine  b)alligning to even boundary c)alligning to word boundary d)none of the above The corr ect answer is : al li gni ng the addr ess to the next wor d boun dary of th e machin e   

3) To send a packet data using datagram, connection will be established.. a)before data transmission  b)connection is not establshed before data transmission c)no connection required d)none of the above The corr ect answer is : no connection requir ed   

4) The status of the kernel is? a)task  b)process c)not defined d)none of the above Th e corr ect asnwer is : Pr ocess

Questions 52 

1) Explain the use of the WHERE clause. An swer : I t dir ects SQL to extr act data f rom r ows where the value of the colu mn is the same as the cur r ent value of the WH ERE clause vari able. 

2) What technique is used to retrieve data from more than one table in a single SQL statement? An swer :T he Join statement combines data fr om mor e that two tables   

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3) What is a foreign key? An swer : I t identif ies a related row i n anoth er table and establi shes a logical r elati onshi p between rows in two tables. 

4) What are the reasons for adding an index to a table ? An swer : To i ncr ease looku p perf ormance. F or a uni que key to guarantee the uni queness of the valu es f or a colu mn on a table.

Questions 53 

1) What is PL/SQL ? An swer : A pr ogramming language available in Or acle. It is a way of wr apping SQL wi th 3GL constru cts such as I F ..TH EN ..EL SE, WH I L E etc. 

2) What is locking and what is it used for ? An swer: L ocking i s a mechani sm to restr ict access to data. I t prevents one user updati ng data whil st another user is looking at i t. 

3) What types of locking are available with Oracle? An swer : Row locki ng and table locking. 

4) What happens if a tablespace clause is left off of a primary key constraint clause? An swer : Th is resul ts in the in dex th at i s automaticall y generated bein g placed in the user s def aul t tablespace. Since this wil l usual ly be the same tablespace as the table is bein g created in, th is can cause serious performance problems.

Questions 54 

1) A garrison of 3300 men has provisions for 32 days, when given at a rate of 850 grams per head. At the end of 7 days a reinforcement arrives and it was found that now the provisions will last 8 days less, when given at the rate of 825 grams per head.How, many more men can it feed? An s. 1700 men.  

2) From 5 different green balls, four different blue balls and three different red balls, how many combinations of balls can be chosen taking at least one green and one blue ball?

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An s. 3720. 

3) Three pipes, A, B, & C are attached to a tank. A & B can fill it in 20 & 30 minutes respectively while C can empty it in 15 minutes. If A, B & C are kept open successively for 1 minute each, how soon will the tank be filled? An s. 167 mi nu tes. 

4) A person walking 5/6 of his usual rate is 40 minutes late. What is his usual time? An s. 3 hour s 20 mi nutes. 

Question 55 

The students of Vidya Mandir Secondary school, Jalpaiguri went for an excursion. It was seen that a boatthe travels per hour when alongvalue withdue the to stream. By the timeThe he starts returning speedatof14.5 the km river doubled thanititsgoes original a sudden storm. speed of the boat is 7 km per hour when it goes against the river stream. What is the speed of boat per hour? a) 10 km b) 8 km c) 12 km d) 7 km Answer : c) 12 km

Solution : Let the speed of thealong boat be B and the :speed be S.eq 1 Equation for travel with Stream B + Sof=stream 14.5 ----> During the travel against the stream, the speed of the stream temporarily doubled. Therefore, Equation for travel against the stream = B - 2S = 7 ----> eq 2 (NOTE: We are using 2S instead of S in the equation as speed of the stream has temporarily doubled when he travelled against the stream) eq 1 - eq 2 => 3S = 7.5 or S = 2.5 kph Substitute S = 2.5 in eq 1 B = 14.5 - 2.5 = 12 Question 56 

A boat man was driving a boat during a cyclonic storm during which time the speed of the river was considerably high than normal times. He found that the boat travelled 222 km in three hours when he was driving along with the river. But when he drove the boat against the river the boat travelled 100 km in two hours. What is the speed of the river per hour during cyclonic storm? a) 10 km b) 12 km c) 8 km d) 9 km

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Answer : b) 12 km

Solution : Let the speed of the boat be B and the speed of the river be S. B 222/3----> ---->eq eq21 B -+SS==100/2 Adding equations 1 and 2 we get : 2B = 300 + 444 / 6 2B = 744/6 2B = 124 B = 62 Substituting B = 62 in eq 2 we get, S = B - 100/2 = 62 - 50 = 12 Question 57 

Maruti Cruisers sailed km along with the river the in six hours. theCruiser boat had to returnat to the starting point and132 it started returning against river and Suddenly this time the travelled 128 km in eight hours. By how much percentage Cruiser's speed exceeds the speed of the River ? a) 533.33% b) 444.44% c) 267.67% d) none of these. Answer : a) 533.33%

Solution : Speed of boat along with the river per hour = 22 km (132 divided by 6) Speed boat against B + S =of22 ----> eq 1 the river per hour = 16 km (128 divided by 8) B –  S = 16 ----> eq 2 Adding equations 1 and 2, we get : B+S-B+S = 22 -16 = 6 =2S S = 3 km/hour Substituting S= 3, in eq 1 we get : B + 3 = 22 B = 22 - 3 = 19 km /hour Cruiser's speed exceeds the speed of the River By : Cruiser's Speed - River Speed / River Speed x 100 = B - S / S X 100% = 19 - 3/ 3 X 100 = 16/3 X 100 = 533.33% Question 58

1) The minute hand of a clock overtakes the hour hand at intervals of 64 minutes of correct time. How much a day does the clock gain or lose? An s. 32 8/11 mi nu tes. 

2) Solve for x and y: 1/x - 1/y = 1/3, 1/x2 + 1/y2 = 5/9.

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An s. x = 3/2 or -3 and y = 3 or -3/2. 

3) Daal is now being sold at Rs. 20 a kg. During last month its rate was Rs. 16 per kg. By how much percent should a family reduce its consumption so as to keep the expenditure fixed? An s. 20 %. 

4) Find the least value of 3x + 4y if x2y3 = 6. Ans. 10. Question 59 

1) What is an appropriate synonym for Bifid ? a) Divided  b) Divided in two c) Timid An swer is Di vided   

2) Find the antonym for gaurish . a) Cheap  b) Flashy c) Costly Answer is Cheap   

3) Choose an appropriate antonym for the word deliberate. a) Unintended  b) Targeted c) Focussed An swer is Uni ntended   

4) Which of the following is an appropriate synonym for the word Debauch ? a) Demoralize  b) Encourage c) Cultivate An swer is Demor ali ze

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Question 60 

1)Find the sum of the first 14 terms for a sequence starting with 2, ending with 120 and common difference 2 a)845  b)854 c)800 An swer i s 854   

2) Find the 10th element in the series 5,15,35,45..... a)85  b)95 c)Cannot be determined Answer is Cannot be determined 

3) Find the difference between last and last but one term in the Sequence 1, 9, 17, 25… which has 40 terms in total a)8  b)16 c)24 Answer is 8 

4) Find the 7th term in the series 4,8,16… 

a)512  b)256 c)64 Answer is 256 Question 61 

Here are some worked sample questions from WIPRO Placement Papers 1) In 8085 microprocessor READY signal does.which of the following is incorrect statements [a] It is input to the microprocessor [b] It sequences the instructions An swer: I t sequences the in str ucti ons   

2) int zap(int n) {

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  if(n 1). 1. On the first day 1 medal and 1/6 of the remaining m - 1 medals were awarded. 2. On the second day 2 medals and 1/6 of the now remaining medals was awarded; and so on. 3. On the nth and last day, the remaining n medals were awarded. How many days did the contest last, and how many medals were awarded altogether? a) 6 days , 36 medals b) 7 days, 49 medals c) 5 days, 25 medals d) none of these. Answer : c) 5 days , 25 medals.

Solution : For this question, we have to go by options. Only for option c) all the conditions in questions will be met as given below. (We recommend you to try with other options, to prove that they are wrong.) On day 1: Medals awarded = (1 + 24/6) = 5 : Remaining 20 medals On day 2: Medals awarded = (2 + 18/6) = 5 : Remaining 15 medals On day 3: Medals awarded = (3 + 12/6) = 5 : Remaining 10 medals On day 4: Medals awarded = (4 + 6/6) = 5 : Remaining 5 medals On day 5: Medals awarded = 5

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Question 124 

 Arun got a new mobile number. The first five digits are 99445. Find the next 5 digits, if the fourth digit from the right is twice the last digit from the left, the second digit from the left is two greater than the second digit from the right, the sum of last two digit from the left is same as the middle digit, the second digit from left is one less than third digit from the right and the second digit from the left is three times the first digit from the left? a) 26734 b) 26752 c) 26743 d) 24752 Answer : c) 26743

Solution : Let the five digit be abcde. From the question, we can infer that b = 2e, b = d+2, c = d+e, c = b+1 ( b = c - 1)and b = 3a. Substitute c = d + e for c in c = b + 1 we get , d+e=b+1 Now Substitute b = 2e we get, d + e = 2e + 1 d - e = 1 ----------- eqn 1 Substitute b = 2e in b = 2 + d we get, 2e - d = 2 ------------ eqn 2 By Solving eqn 1 & 2 we get e = 3 Substituting e value we get, d = 4, b = 6 , a = 3, c = 7 Hence the remaining 5 digits are 26743 Question 125 

 Age of Kamal is a two digit number. If the sum of age and the reverse of his age is 110 and if the ten's digit is 4 less than the unit digit, then the age of Kamal is? a) 73 b) 28 c) 82 d) 37 Answer : d) 37

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Solution : Let the age of Kamal be 10a + b Then the reversed age is 10b + a Given that 10a + b + 10b + a = 110 11a + 11b = 110 ----- (1)  Also given that a = b – 4 ---- (2) Solving (1) and (2), we get a = 3 and b = 7 So, the age of Kamal is 37 Question 126 

If a five digit number 526ab is divisible by 40, then find a and b. a) 8, 0 b) 1, 6 c) 4, 0 d) 2, 0 Answer : c) 4, 0

Solution : L.C.M of 40 = 5 x 8 Since 526ab is divisible by 5 and 8, b = 0  And since 526a0 is divisible by 8, the last three digits must be divisible by 8. i.e. 6a0 must be divisible by 8. This will happen only when a = 4. Hence, the last two digits are 4 and 0. Question 127 

In a conference hall, 56 members are seated by the following arrangements. After the first women, one man seated. After the second women, two men are seated. After the third women, three men are seated and so on. How many numbers of men are seated in the second half of the arrangements? a) 27 b) 25 c) 24 d) 26 Answer : b) 25

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Solution : The seating arrangement will be

From the above diagram, we can see that the second half arrangement is : MMMMMMMWMMMMMMMMWMMMMMMMMMWM Here, there are 25 men in the above arrangement. Question 128 

In a bank, there are two counters namely counter1 and counter2. They will call the customers by a token number. If counter1 calls the token numbers from 40 in descending order and the counter2 calls only the token numbers which are prime numbers in ascending order. What token number will they call out at the same time if they were calling in the same speed? a) 29 b) 31 c) 23 d) never call out the same number Answer : d) never call out the same number

Solution : Both counters will call in the following manner

So, both will never call out the same token number at a time. Question 129 

If the numbers from 1 to 100 which are exactly divisible by 4 and 6 are arranged in descending order, maximum number being on the top, which would come at the seventh place from the top? a) 24 b) 12 c) 36 d) 18 Answer :a) 24

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Solution : From 1 to 100, the number which is divisible by 4 are 100, 96, 92, 88, 84, 80, 76, 72, 68, 64, 60, 56, 52, 48, 44, 40, 36, 32, 28, 24, 20, 16, 12, 8 and 4. From 1 to 100, the numbers which is divisible by 6 are 96, 90, 84, 78, 72, 66, 60, 54, 48, 42, 36, 30, 24, 18, 12 and 6. Therefore, the numbers which are divisible by 4 and 6 are 96, 84, 72, 60, 48, 36, 24 and 12. Hence, the seventh place from the top is 24. Question 130 

If Jeya celebrated her 18th birthday on 28 February 2009, Saturday. Then what day will come on her 25th birthday? a)Monday b) Sunday c) Friday d) Saturday Answer : b) Sunday.

Solution: Jeya Celebrated her 18th birthday on 28 February 2009, Saturday. th

Then her 19  birthday will be on 28 February 2010, Sunday. (This is because 2010 is not a leap year and hence contains 365 days. 365 divided by 7 gives remainder 1 which means any date on 2009 corresponds to the same date on 2010 advanced by one day.) Her 20th birthday will be on 28 February 2011, Monday. Her 21st birthday will be on 28 February 2012, Tuesday. Her 22nd birthday will be on 28 February 2013, Thursday. (Since, 2012 is a leap year hence contains 366 days. 366 divided by 7 gives remainder 2 which means any date on 2013 corresponds to the same date on 2012 advanced by two days.) Her 23rd birthday will be on 28 February 2014, Friday. Her 24th birthday will be on 28 February 2015, Saturday. Her 25th birthday will be on 28 February 2016, Sunday. th

So, her 25  birthday will be on Sunday.

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Question 131 

Ravi went to the hospital on Monday. Doctors made some check up and they asked to come three days before the two days after the next day of day after tomorrow. When will Ravi go to Hospital? a) Friday b) Saturday c) Tuesday d) Wednesday Answer : d) Wednesday

Solution: Ravi went to hospital on Monday. Then, three days before Monday is Friday and the two days after Friday is Sunday.  And the next day of Sunday is Monday and the day after tomorrow of Monday is Wednesday. So, Ravi will go to hospital on Wednesday. Question 132 

The Calendar for the year 2013 is the same as for the year? a) 2020 b) 2018 c) 2017 d) 2019 Answer : d) 2019

Solution: For calendar to repeat exactly, the dates and days have to match perfectly. Consider 2014 :  Any date on 2014 will correspond to same date on 2013 advanced by one day. (same logic used in first question.) For example if Jan 1 is Tuesday on 2013, then Jan 1 will be Wednesday on 2014. Year Advanced

2014 days

2015 1

2016 (leap year) 1

2

2017

2018 1

2019 1

1

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In 2019 the total number of advancements will be 1 + 1 + 1 + 2 + 1 + 1 = 7 . Any week has seven days. Hence advancement of 7 days also means the days are going to be the same for any dates. That is if 1 st Jan on 2013 is Tuesday, then 1st Jan on 2019 will also be Tuesday. Hence the calendar for the year 2013 and 2019 is the same. Question 133 

Narain Karthikeyan‟s friend participated in a cycling feat and he drove through the lap at the rate of 6 kmph, 12 kmph, 18 kmph and 24 kmph. What is the average speed of his friend‟s cycling ?  a) 15 kmph b) 18 kmph c) 14.25 kmph d) 11.52 kmph Answer : d) 11.52 kmph

Solution :  Assume the length of the lap is 72 kmph (LCM 6,12,18 and 24) Then his friend will take 12 hours, 6 hours, 4 hours and 3 hours respectively at the rate of 6,12,18 and 24 kmph. Total time taken 25 hours. Total distance 288 km 288/25 = 11.52 kmph (We will get this same answer irrespective of whatever we assume the lap length. However, taking the LCM of the speeds has made the solution much easier.) Question 134 

Krishnakumar bought an acid bottle quite evaporative in nature. He bought acid containing 24 litres of concentrated acid. He paid at the rate of Rs.2452 per litre since he could not get them at a cheaper price. On the first day 1/8 of the acid got evaporated. On the second day 2/3 of the balance got evaporated. He forgot to close the container and on the third day 1/7 of the balance acid got evaporated . How many litres of acid is left in the container? a) 12 litres b)6 litres c)3 litres d) 18 litres Answer : b) 6 litres.

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Solution : In this problem rate at which the acid was bought is given only to distract the attention of the reader. On the first day 1/8 got evaporated means for the second day balance left is 7/8. On the second day 2/3 of the balance got evaporated means 7/8 x 2/3 = 7/12 got evaporated Balance acid left out = 7/8 - 7/12 = 7/24 On the third day 1/7 got evaporated means what is left in the container is 6/7 of the balance at the end of second day. Quantity of acid left in the container = 7/24 x 6/7 = 1/4 1/4 x 24 = 6 litres Question 135 

Ramesh and Rahul were making joint-study for their semester examinations and the lights went off. It was around 10.00 pm . They lighted two uniform candles of equal length but one thicker than the other. The thick candle is supposed to last six hours and the thin one two hours less. When they finally went to sleep, the thick candle was thrice as long as the thin one. Approximately at what time did they go to sleep? a) 1.00 am b) 1:24 am c) 11.00 pm d) none of these. Answer : b) 1:24 am

Solution :  Assume that the initial length of both the candle was L and they studied for X hours. Consider thick candle :

In 6 hours the entire length would be burnt. Therefore in 1 hour L/6th of the candle would be burnt. Similarly in X hours, XL/6 th of the candle would be burnt.  Applying same logic to thin candle, length of thin candle burnt in X hours = XL/4  After X hours, total thick candle remaining = L - XL/6  After X hours, total thin candle remaining = L - XL/4

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 Also, it is given that the thick candle was thrice as long as the thin one when they finally went to sleep. (L - XL/6) = 3(L - XL/4) (6 - X)/6 =(12 - 3X)/4 24 - 4X = 72 - 18X X = 3.4 (approximately) = 3 hours and 24 minutes (as .4 hours = 4/10 x 60 = 24 minutes) So they slept 3 hours and 24 minutes after 10.00 p.m i.e. at 1.24 am. Question 136 

 Anantharama Iyer and Gopalakrishna Gandhi – two friends were meeting after a long time. Gandhi told he has two kids and their age is less than 10. The sum of the squares of their age is 130.Anantharama Iyer told that he has three kids and first two of them are of the age of Gandhi‟s two children and the sum of the squares of their age is 139. Wh at is the age of the third child of Anantharama Iyer? a) 4 b)5 c)3 d)cannot be determined Answer : c) 3.

Solution : The age of the two kids of Anantharama Iyer is same as that of Gandhi‟s children. The sum of the squares of age of Gandhi‟s children is 130.  The sum of squares of age of  Anantharama Iyer‟s children is 139. That means the square of the age of the third child is 9. Age is 3 years. Question 137 

Skilled Men, skilled women and unskilled men are employed to do a road construction work in the proportion of 1:2:3 and their wages in the proportion of 6:3:2. When 50 skilled men are employed total daily wages of all the hands amount to Rs. 45000. Find the total weekly wages to be paid to a skilled man, a skilled woman and an unskilled man. a) Rs.3850 b) Rs.4850 c)Rs.5850 d)Rs.6850

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Answer : a) Rs. 3850

Solution :

Number of men, women and unskilled men are 1: 2: 3. But the skilled men are 50 in number. Therefore number of skilled women workers = 50 x 2 = 100 and the number of unskilled men = 50 x 3 = 150 Their wages are in the ratio 6:3:2.  Assuming a man gets paid Rs.6, woman will get Rs.3 and an unskilled man will get Rs. 2 Total wages paid to all the employees - 50 x 6 + 100 x 3 + 150 x 2 = Rs.900 If the total wages are Rs. 900, a skilled man gets Rs.6 If the total wages are Rs.45000, a skilled man gets (45000 /900 ) x 6 = Rs. 300 Weekly wage of a skilled man = daily wage x 7 = 300 x 7 = Rs.2100 The ratio of wages between men and women is 6:3. i.e Each woman will earn half as that of a skilled man. Total weekly wages of a woman = 2100/2 = Rs.1050 The ratio of wages between skilled men and unskilled men is 6:2. i.e Each unskilled man will earn 1/3rd of that of a skilled man. Total weekly wages of an unskilled man = 2100/3 = Rs.700 Total weekly wages of a man, a woman and an unskilled man = 2100 + 1050 + 700 = Rs.3850 Question 138 

Fill the Sequence DI,MR,VA, ? a) EJ b) SZ c) BK d) FR Answer : a) EJ

Solution : In the sequence M is the 9th letter from D and V is the 9th letter from M. So the next letter will be the 9th character from V i.e E.

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In DI, I is the 5th letter from D. In MR, R is the 5th letter from M and in VA and so on. The fifth character from E will be J. So the next sequence letters will be EJ

Question 139 

In a certain code RELATED to written as EFUBKDQ. How is RETAINS written in that code? a) SDQBTOJ b) JOTBQDS C) JOTBSDQ d) TOJBSDQ Answer : d) TOJBSDQ

Reason : Inspecting how RELATED was written as EFUBKDQ, we can easily find that, RELATED – is first written as DETALER DETALER becomes EFUBKDQ (+1,+1,+1, +1,-1,-1,-1  – alphabetically)   Applying the same logic to RETAINS we get, RETAINS is first written as SNIATER SNIATER becomes TOJBSDQ (applying the above rule)

Question 140 

If AEGI is encoded as CH how CEDH will be encoded. a) GF b) AC c) DF d) DI Answer : c) DF

Reason : Inspecting the encoding of AEGI : Pairing two-two letters we have AE GI. The middle letter between A and E is C and the middle letter between G and I is H. Similarly applying the same logic to ACDF we get Pairing two-two letters, we have CE DH

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The middle letter between C and E is D while the middle letter between D and H is F Hence the answer is DF. Question 141 

Six people – A,B,C,D, E and F are standing in a straight line facing north not necessarily in the same order. B is standing to the right of D. A is standing fourth to the left of F and F is not standing on the extreme end of the line. D is standing second to the left of B. In the above arrangement, which of the following pair represents the people standing at the extreme ends of the line? a) DF b) AC c) BC d) AF

Answer : b) AC

Reason : Based on information given seating arrangement is as follows: north  A D E B F C south Question 142 

 Aravamudhan gives Babu as many rupees as Babu has and Chiranjeevi as many rupees as Chiranjeevi has. Similarly, Babu then gives Aravamudhan and Chiranjeevi as many rupees as each of them has. Chiranjeevi, similarly gives Aravamudhan and Babu as many rupees as each of them has. Now finally each of them has Rs.32, with how many rupees does Aravamudhan start? a)52 b)56 c)60 d)62 Answer : a)52

Solution :

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We have to take a bottom up (reverse) approach to tackle these kinds of problems. Below are the steps to solve this problem. Aravamudhan

Babu

Chiranjeevi

Finally Before Chiranjeevi gives

32 16

32 16

8

56

52

28

Before Babu gives Before Aravamudhan gives

32 64 32 16

Hence, Aravamudhan start with Rs 52. Question 143 

Fifty one books are arranged from left to right in order of increasing prices. The price of each book differs by Rs. 2 from that of each adjacent book. For the price of the book at the extreme right a buyer can buy the middle book and an adjacent one. Then, which of the following statements is correct ? a) The adjacent book referred to is at the left of the middle magazine. b) The middle book sells for Rs. 56 c) The most expensive book sells for Rs.116 d) None of the above statements is correct. Answer : a) The adjacent book referred to is at the left of the middle magazine.

Solution : Let the price of the first book be B Price of the last book (51st) will be B + (2 x 50) = B + 100 Price of the middle book = B + (2 x 25) = B + 50 Price of the book to the left of middle book = Price of the middle book  – 2 = B + 48 But, from the question, we can infer that, Price of the middle book + Price of the book to the left of middle book = Price of the last book. B + 50 + B + 48 = B + 100 B=2

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Instead if the buyer buys the middle book and the book adjacent right side B + 50 + B + 52 = B + 100 B = -2 (not possible) Cost of middle book = 2 + 50 = 52 Price of the Most expensive book (i.e last book) = B + 100 = 2 + 100 = Rs.102. Hence a) is correct. Question 144 

 A group of six friends are sitting around a hexagonal table, each one at one corner of the hexagon. (facing the centre inside) Raman is sitting opposite to Ramesh. Jolly is sitting next to Seema. Neeta is sitting opposite to Seema, but not next to Ram. Amit has a person sitting between Ramesh and himself. If Neeta sits to the right of amit, then who is sitting to the left of Amit? a)Ramesh b)Raman c)Jolly d) Seema Answer : b) Raman

Solution :

Based on the information provided, seating pattern will be drawn as above Clockwise = AMIT – RAMAN - SEEMA - JOLLY- RAMESH - NEETA Question 145 

Statement : Some watches are hens. All hens are wall clocks.

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C o n c l u s i o n s   :

I. All watches are wall clocks II. some wall clocks are hens. O p t i o n s . 

a) Only I can be true always b) Only II can be true always c) Both I and II can be true always d) Both I and II cannot be true always Answer 1 

Option b is the correct answer. Reason : Conclusion I cannot be true as only some watches could be wall clocks. However, conclusion II is true always as some wall clocks have to be hens. Question 146 

 All animals are monkeys. All monkeys are reptiles. C o n c l u s i o n s   :

I. All reptiles are animals. II. All animals are reptiles. O p t i o n s . 

a) Only I can be true always b) Only II can be true always c) Both I and II can be true always d) Both I and II cannot be true always Answer 2 

Option b is the right answer.

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Reason : Though all animals are reptiles from the two statements, there is no indication that all reptiles are monkeys (and in turn animals). Hence option I cannot be true always. But second conclusion is always true which is straightforward to interpret. Question 147 

Some Red Boxes are Green Boxes. All Red Boxes are Yellow Boxes. C o n c l u s i o n s   :

I. Some Yellow Boxes are Green Boxes. II. All Green Boxes are Red Boxes. O p t i o n s . 

a) Only I can be true always b) Only II can be true always c) Both I and II can be true always d) Both I and II cannot be true always Answer 3 

Option a is the right answer. Question 148 

1) What could be a possible arrangement of 10 points in space so that five lines could be drawn with a condition that every line has four points on it.  Answer : This is a question which you would had come across in several puzzle books and previous questions. Answer is a "star". You could try drawing a star and count the number of intersections which should be 10. 2) Ram has 30 Rupees with him. He has two Rupee notes. One of the rupee notes is not a 20 Rupee note. Then find out the notes he has with him ?

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 Answer: These kinds of questions are the simplest of the simpler ones. But one could tend to miss out due to nervousness.Here is the answer : One of the rupee notes is not a 20 Rupee note. So the other one can be a 20 Rupee :). And the first one should be a 10 Rupee note. 3) In a dark room with no equipments other than a clock, you are given a pendulum. How can you measure your height now ?  Answer: This is a bit difficult one. Time period of pendulum is given by 2 X 3.14 X (sqrt(height of rope/gravity)). Now, gravity = g = 9.18. Using stop clock you can find the Time Period of pendulum.(average time for one oscillation). Using the formula for time period you can measure the height of the rope. Then using the rope you can measure your own height. Question 

1) An elephant fell into a well of depth 20mts. Everyday it climbs 4 mts but at the end falls by 2 mts. Find the number of days it would take for the elephant to get out of the well ? Solution : Since it climbs 4 mts and falls by 2 mts everyday, effectively it climbs 2 mts everyday. In 8 days it climbs 16mts. Then on 9th day it climbs by 4 mts and can get out without having to fall back. Hence it takes 9 days for the elephant to be safe. Question 149 

2) Ramu was driving to a city full of mathematicians. There was a distance board reading the number of Kms he has to travel further. It stated that the distance when divided by 2,3,4,5 and 6 leaves a remainder 1. But when divided by 11 it leaves a remainder 0. Help Ramu in knowing the exact distance. Solution : LCM of 2,3,4,5 and 6 is 60. LCM and its multiples are perfectly divisible by all the numbers 2,3,4,5 and 6.

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i.e, 60,120,180... are perfectly divisible by 2,3,4,5 and 6. Then 61,121,181.... would leave a remainder 1 when divided by any of the numbers 2,3,4,5 or 6.  Among these 121 is perfectly divisible by 11. Hence Ramu has to travel 121 Kms to reach the city of mathematicians. Question 150 

3) Brothers Ramu ,Raju and Rahul are given ropes of equal lengths. They are given a chance by their ailing father that they can own the land that they manage to fence around in father's common land. Ramu fences in the shape of a triangle, Raju fences in the shape of a circle and Rahul fences in a star shaped manner. At the end, who will manage to acquire most land ?? Solution: Raju would be the clear winner. This is because for a given perimeter, circle takes the maximum area. Question 

Hard work --------- with smart preparation can help a long way in passing your exams. a) Coupled b) Attached c) Adjoined with d) None  Answer: a) Coupled Explanation & Inference: Coupled is the work that fits well the given blank. Though "attached" and "adjoined with" look almost similar, you could easily make out that "Coupled" is the correct answer by reading the sentence filled with each of the options.

Question 151 

The anti riot force was asked not to resort to lathi charge but to -------- restraint till a final command. a) Exercise b) Commit c) Undertake d) None

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 Answer: a) Exercise Explanation & Inference: This question is a much simpler one when compared to the first question. Other than the word "Exercise" all the other words can potentially alter the meaning of the sentence. Question 152 

-------- of parents is a most important factor when deciding partner in Indian weddings. a) Agreement b) Consent c) Selection d) None  Answer : b) Consent Explanation & Interference: Consent is the correct answer. "Consent" gives the apt meaning for the sentence. "Agreement" is a formal term while "Selection" should not have followed by "of" (instead "by" should had come). Question 153 

Ramu is a ------ singer. a) Skilled b) Bright c) Clever d) None  Answer: a) Skilled. Explanation & Interference: Bright and Clever do not fit well for the context of the sentence. Question 154 

1) A man has Ten Horses and nine stables as shown here. [][][][][][][][][] The man wants to fit Ten Horses into nine stables. How can he fit Ten horses into nine stables? Answer T h e a n s w e r i s s i m p l e . It s a y s t h e m a n w a n t s t o f i t " T e n H o r s e s " i n t o n i n e s t a b l e s . T h e r e ar e n i n e l e tt e r s i n t h e p h r a s e " T e n H o r s e s " . S o y o u c a n p u t o n e l e t t e r ea c h in all nine s tables. [T] [E] [N] [H] [O] [R] [S] [E] [S]   

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2) To move a Safe, two cylindrical steel bars 7 inches in diameter are used as rollers.How far will the safe have moved forward when the rollers have made one revolution? Answer The safe must have mo ved 22 inches forward.

3) If a rook and a bishop of a standard chess set are randomly placed on a chessboard, what is the probability that one is attacking the other? Note that both are different colored pieces. Answer T h e p r o b a b i l i t y o f e i t h e r t h e R o o k o r t h e B i s h o p a t t a c k i n g t h e o t h e r i s 0 .3 61 1

4) If three babies are born every second of the day, then how many babies will be born in the year 2001? Answer 9,46,08,000 ba bi es   

Question 155 

1) A man is visited by his 7 friends in the following pattern.The first friend visited every day.The second visited every second day.The third visited every third day and so on upto 7th friend on every 7th day.When will they all meet together. A n s : 4 2 0   

2) Every man dances with 3 women and every woman dances with 3 men.And also 2 men have 2 women in common in the party. How many people attended the party? A n s : 8   

3) Two girls go to a shopping. In the first shop they spend half of what they had in their purse + 2 rupees.Next half of remaining + 5 rs.Next half Of whatremained.The remaining 5 rs spent on coffee.How much did they start with?

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A n s : 6 4. 

4) A man spent 1/5 of his life as young boy...,1/7,1/12, and then 5 years upto his sons birth his son was chosen alderman 4 years ago when he was half the age of the man now. Find the mans age. ( refer shakuntala devi - puzzles to puzzle u- problem from that - we dont remember exact figures ). A n s : 8 4y r s    

Question 156 

1) I have 3 rs. and the types of stamps are 2,7,10,15,20 parse. I should buy 6 each of 2 types and 5 each of remaining 3 types exactly.What should be the types of stamps in the 5 each lot and 6 each lot? Ans : 5 each lo t: 2,7,15 6 each lot : 10,20

2) A,B,C,D,E denote diff digits.* means multiply. AB * CD = EEE (CD * E)-AB = CC .Then find the digits. An s: A,B ,C,D,E = 3,7,1,2,4   

3) There r some chickens and some chicken field. If we sell 75 chickens Then the feed will last 20 days longer.If we buy 100 chickens more, then the feed will get over before 15 days.What is the present number of chickens? A n s : 3 00   

4) What is the next number in the series:  A : 3,6,13,26,33,66,__ A n s : 7 3   

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  Question 157 

1) Let S be a Set of some positive integral numbers ; with an average of 47 ; and containing the number 83 . The numbers may or maynot be distinct .However ; when the number 83 is removed ; the Avg drops to 46 .What is the largest number that can be possibly contained in that Set ? Solution: L e t S b e t h e s u m o f t h a t s e t o f n p o s i t i v e in t e g e r s . S/n = 47 (S - 83)/(n - 1) = 46 Solvin g th e above 2 equation s, we get S = 1739; n = 37. T h i s s e t o f 3 7 p o s i t i v e i n t e g e r s c o n t a i n 8 3 . To g e t t o t h e e x p e c t e d a n s w e r , w e h a v e t o s u p p o s e t h a t 3 5 o f t h e r e m a i n i n g 3 6 i n t e g e r s h a s a v a l u e o f 1 e a c h ( l ea s t +ve integ er). Thus , the largest p oss ible interg er in th e set = 1739 - 83 - 35*1 = 1621   

2) Sum of squares of two numbers 'x' and 'y' is less than or equal to 100 . How many sets of integer solutions of 'x', 'y' is possible ? Solutions: x = 0, |y| 21 so lut ion s |x| = 1, |y| 38 so lut ion s |x| = 2, |y| 38 so lut ion s |x| = 3, |y| 38 so lut ion s |x| = 4, |y| 38 so lut ion s |x| = 5, |y| 34 so lut ion s |x| = 6, |y| 34 so lut ion s |x| = 7, |y| 30 so lut ion s |x| = 8, |y| 26 so lut ion s |x| = 9, |y| 18 so lut ion s |x| = 10, y = 0 -> 2 so lut ion s   

Total 317 solutions

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3) There is a unique number of which the square and the cube together use all ciphers from 0 up to 9 exactly once. Which number is this? Solution: The num ber is 69. 69^ 2=4761 and 69^ 3=328509.. 

4) You are standing next to a well, and you have two jugs. One jug has a content of 3 litres and the other one has a content of 5 liters. How can you get just 4 liters of water using only these two jugs? Solution: F i ll 3 l i t e r j u g p o u r t o 5 l i t er j u g Fill again 3 liter jug and add to 5 liter jug then 1 liter will be there in 3 liter jug P o u r a l l w a t e r o u t s i d e f r o m 5 l i t er j u g Fill 1 liter water from 3 l i t er j u g t o 5 l i t e r j u g F i ll 3 l i t e r j u g a n d a d d t o 5 l i t er j u g m a k i n g i t 4 l i t er s o f w a t e r . 

Question 158 

1) Eight friends Harsha, Fakis, Balaji, Eswar, Dhinesh, Chandra, Geetha, and Ahmed are sitting in a circle facing the center. Balaji is sitting between Geetha and Dhinesh. Harsha is third to the left of Balaji and second to the right of Ahmed. Chandra is sitting between Ahmed and Geetha and Balaji and Eshwar are not sitting opposite to each other. Who is third to the left of Dhinesh? Answer: Ahmed

Fakis

Chandra Geetha   

Harsha Eswar

Balaji Dhinesh

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2) A fast typist can type some matter in 2 hours and a slow typist can type the same in 3 hours. If both type combinely, in how much time will they finish? A n s w e r : 1 h r 12 m i n Explanation : The fast typist's wo rk do ne in 1 hr = 1/2 T h e s l o w t y p i s t ' s w o r k d o n e i n 1 h r = 1 /3 If they wo rk co mb inely, work d on e in 1 hr = 1/2+1/3 = 5/6 So, the wo rk w ill be com pleted in 6/5 hou rs. i.e., 1+1/5 hou rs = 1hr 12 min   

3) 3)1, ? , 9, 7, 7, 9, 13, 10, 9, 1, 4, 9, 16, 16, ..... What number comes at '?' at the second place from the start ? Answer: THE SERIES IS THE SUM OF THE DIGITS IN THE SQUARE OF NA TURAL NUMBERS : 1^ 2 - 1 - 1 2^ 2 - 4 - 4 3^ 2 - 9 - 9 4^ 2 - 16 - 1+6 - 7 5^ 2 - 25 - 2+5 - 7 ............ A n s w e r i s 4.  

4) Today is 4.11.09. keeping that figure 41109 in mind, i have arrived at the following sequence : 2, 1, 9, 5, ? which of the following four numbers suits '?' a) 7 b) 65 c) 4563 d) 262145 Answer is: 41109 1^ 4 + 1 = 2 1^ 1 + 0 = 1 0^ 1 + 9 = 9

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9^ 0 + 4 = 5 So n ext is 4^ 9 + 1 = 262145   

Question 159 

1) In a race Andrew beats Jim. Jack is not the last. Dennis loses to both Jack and Lucia. Jim beats Jack. Who won the race? ANSWER:   

The possible positions in the race are,  Andrew

Andrew

Jim

Jim

Jack Lucia

Lucia Jack

Dennis

Dennis

In either case Andrew is the winner. 2) Find the digits X,Y,Z XXXX YYYY + ZZZZ ----------------YXXXZ ----------------ANSWER: x+y+z = z ==> x+y = 10 max value for y = 1 hence x = 1 also , (carry )1+x+y+z = x ==> 1+y+z = 10 henc e z=8 X

Y

Z

9

1

8   

3) A 1 k.m. long wire is held by n poles. If one pole is removed, the length of the gap becomes 12/3m. What is the number of poles initially?

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ANSWER: Let the no. of poles originally = n After taking away one pole (n-1)*12/3 = 1000 n = 251   

4) A + B + C +D = D + E + F + G = G + H + I =17. IF A = 4 WHAT ARE THE VALUES OF D AND G. EACH LETTER TAKEN ONLY ONE OF THE DIGIT FROM 1 TO 9. A ns w er: A = 4 ,B = 2, C =6, D = 5, E = 3, F = 8, G = 1,H = 7, I = 9.

Question 160 

1) The L.C.M. (Lowest (or Least) Common Multiple) of two numbers is 45 times to their H.C.F (Highest Common Factor). If one of the numbers is 125 and sum of L.C.M. and H.C.F. is 1150, the other number is: a) 215 b) 220 c) 225 d) 235 Solut ion: Let L .C.M. be l and H.C.F. be h. Then l = 45h l + h =1150 This g ives h =25 and l = 1125 Thus the s econ d n um ber = (25 * 1125)/125 = 225   

2) Six bells commence tolling together and toll at intervals 2,4,6,8,10 and 12 seconds respectively. In 30 minutes how many times they toll together. a) 4 b) 10 c) 15 d) 16 Solu tio n: L.C.M. of 2,4,6,8,10,12 = 120

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So they to ll togeth er after every 120 secon ds i.e. 2 minu tes. So in 30 minu tes they toll to gether 30/2 + 1 = 16 time   

3) The value of (0.625 * 0.0729 * 28.9)/(0.0017 * 0.025 * 8.1) is a) 0.3825 b) 3.825 c) 38.25 d) 382.5 S o l u t i o n : T h e s u m o f d e c i m a l p l ac e s i n n u m e r at o r a n d d e n o m i n a t o r b e i n g t h e s a m e , d ec i m a l p o i n t c a n b e r e m o v e d (625*729 *289) / (17*25*81) = 3825   

4) Linear equations are part of all math aptitude tests. They can be age related problems, upstream, downstream problems etc I am three times as old as my son. Five years later I shall be two and a half times as old as my son. What is my age? Solution: L e t m y a g e b e x y e a r s a n d a g e o f m y s o n b e y y e a rs . (i) x = 3y Five years later (ii) x+5 = 5/2(y+5) From (i) and (ii) y=15 and x=45   

Question 161 

1) Can you find out what day of the week was January 12, 1979? A n s . F r id a y . 

2) Find x+2y (i). x+y=10 (ii). 2x+4y=20

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A n s : ( b )   

3) Is angle BAC is a right angle

(i) AB=2BC (2) BC=1.5AC A n s : ( e)   

4) Is x greater than y (i) x=2k (ii) k=2y A n s : ( e)   

Question 162 

1) If x and y are the two digits f the number 653xy such that this number is divisible by 80, then x+y is equal to: (i) 2 (ii) 3 (iii) 4 (iv) 6  Answer is : Since 653xy is div isible by 2 as well as by 5, so y = 0 Now 653x0 is divisible by 8 so 3x0 is also d ivisible by 8. B y h i t a n d t r i a l x = 6 an d x + y = 6   

2) The smallest number which when diminished by 3 is divisible by 21,28,36 and 45 is... (i) 869 (ii) 859

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(iii) 4320 (iv) 427 Answer is The required number = l.c.m. of (21,28,36 ,45)+3=1263  

3) If 1.5x=0.04y then the value of (y-x)/(y+x) is (i) 730/77 (ii) 73/77 (iii) 7.3/77 (iv) None Answer is: x/y = 0.04/1.5 = 2/75 So (y-x)/(y+x) = (1 - x/y)/(1 + x/y) = (1 - 2/75)/ (1 + 2/75) = 73/77  

4) The average age of a class is 15.8 years. The average age of boys in the class is 16.4 years while that of girls is 15.4 years. What is the ratio of boys to girls in the class? (i) 1:2 (ii) 3:4 (iii) 3:5 (iv) None of these A n s w e r i s :    Let the ratio b e k:1. Then k*16.4+1*15.4 + (k+1)*15.8 (16.4-15.8)k = 15.8 - 15.4 k=0.4/0.6 = 2/3 so required ratio = 2:3

Question 163 

1) Which is the smallest addressable unit of memory? a) 1 bit b) 1 byte c) 1 word Ans wer is 1 bit   

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2) What is the size of the memory unit, WORD ? a) CPU specific b) 1 byte c) 2 bytes A n s w e r i s C P U s p e c i f i c   

3) How many bytes make one Octet ? a) 2 b) 1 c) 8 A n s w e r i s 1   

4) Which layer of OSI is responsible for Modulation and Demodulation? a) Physical layer b) Data link layer c) Network layer

A n s w e r i s P h y s i c a l l a y e r   

Question164  1) Two people enter a race in whick you run to a point and back. Person A runs 20 mph to and from the point. Person B runs to the point going 10 mph and 30 mph going back. Who came in first? Answer Per son A came in f ir st.

2) Mark ate half of a pizza on Monday. He ate half of what was left on Tuesday and so on. He followed this pattern for one week. How much of the pizza would he have eaten during the week? Answer M ark woul d have ate 127/128 (99.22%) of the pizza dur in g the week.  

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3) In the General meeting of "Friends Club", Sameer said, "The repairs to the Club will come to a total of Rs 3120 and I propose that this amount should be met by the members, each paying an equal amount." The proposal was immediately agreed. However, four members of the Club chose to resign, leaving the remaining members to pay an extra Rs 26 each.How many members did the Club originally have? Answer The Clu b ori ginal ly had 24 member s.

4) If you look at a clock and the time is 3:15.What is the angle between the hour and the minute hands? ( The answer to this is not zero!) Answer 7.5 degrees   

Question 165 1) Mrs. Watsherface had a garage sale. A custmer named Gina bought an old lamp and a rug. She paid a total of $5.25 for everything. The rug cost 25 cents more than the lamp. How much did each cost? Answer Th e lamp cost $ 2.50 and the rug cost $ 2.75   

2) SlowRun Express runs between Bangalore and Mumbai, For the up as well as the down  journey, the train leaves the starting station at 10:00 PM everyday and reaches the destination at 11:30 PM after three days.Mr. Haani once travelled by SlowRun Express from Mumbai to Bangalore. How many SlowRun Express did he cross during his journey? Answer M r. H aani cr ossed 7 SlowRun E xpresses dur in g hi s jou rn ey.

3) In a certain year, the number of girls who graduated from City High School was twice the number of boys. If 3/4 of the girls and 5/6 of the boys went to college immediately after graduation, what fraction of the graduates that year went to college immediately after graduation? Answer Assume numberofof graduated fromCity CityH H ighSchool School Thereforthat e, number girboys ls graduated fr om igh == 2*BB I t i s given that 3/4 of the gir ls and 5/6 of the boys went to coll ege immediately after graduation. H ence, total students went to coll ege = (3/4)(2* B) + (5/6)(B) = B * (3/2 + 5/6) = (7/3)B

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F r action of the graduates that year went to coll ege immediately after gr aduation = [(7/3)B] / [3* B] = 7/9 Th erefor e, the answer is 7/9   

4) mule and a donkey wereThe carrying sacks on their mule started If complaining thatAhis load was too heavy. donkeyfull said to him "Whybacks.The are you complaining? you gave me one of your sacks I'd have double what you have and if I give you one of my sacks we'd have an even amount."How many sacks were each of them carrying? Give the minimal possible answer. Answer Th e mul e was carr yin g 5 sacks and th e donkey was carr yin g 7 sacks.

Question 166  1) There are 6561 balls out of them 1 is heavy.Find the min. no. of times the balls have to be weighed for finding out the haevy ball? An s. 8   

2) If i walk with 30 miles/hr i reach 1 hour before and if i walk with 20 miles/hr i reach 1 hour late.Find the distance between 2 points and the exact time of reaching destination is 11 am then find the speed with which it walks. An s: 120mi les and 24 mi les/hr  

3) No. of animals is 11 more than the no. of birds. If the no. of birds were the no. of animals and no.legs of animals wereby theone no.fifth of birds( interchanging animals andwere birds.), the total no. of get reduced (1/5).ie., How many no. ofno.s birdsofand animals there? An s: bir ds:11,anim als:22   

4) Complete the series:5, 20, 24, 6, 2, 8, ? An s: 12 (as 5* 4=20, 20+4=24, 24/4=6, 6-4=2, 2* 4=8, 8+4=12). 

Question167  1) Which one of the following conditions must p, q and r satisfy so that the following system of linear simultaneous equations has at least one solution, such that p + q + r not=0? x+ 2y –  3z = p 2x + 6y –  11z = q x –  2y + 7z = r

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a) 5p – 2q –  r = 0  b) 5p + 2q + r = 0 c) 5p + 2q –  r = 0 d) 5p –  2q + r = 0 Solution: I t i s given th at , if we consider the fi r st option, and mu lti ply the fi rst equati on by 5, second by –  2 and thir d by – 1, we see that the coef fici ents of x , y and z all add up-to zer o. Thus, 5p –  2q  –  r = 0. No other option satisf ies thi s. 

2) In a 4000 meter race around a circular stadium having a circumference of 1000 meters, the fastest runner and the slowest runner reach the same point at the end of the 5th minute, for the first time after the start of the race. All the runners have the same staring point and each runner maintains a uniform speed throughout the race. If the fastest runner runs at twice the speed of the slowest runner, what is the time taken by the fastest runner to finish the race? a) 20  b) 15 min min c) 10 min d) 5 min Solution: Th e rati o of the speeds of the fastest and the slowest r un ners is 2 : 1. H ence they shou ld meet at onl y one point on the cir cumf erence i.e. the startin g point (A s the dif ference in the r atio i n r educed for m i s 1). For the two of them to meet f or t he fi rst ti me, the faster shoul d have compl eted one complete r oun d over the slower one. Since the two of th em meet f or the f ir st ti me after 5 mi n, the faster one should h ave completed 2 r ounds (i.e. 2000 m) and the slower one shou the ld have d. min (i.e.. 1000 m) in this ti me. Th us, the faster one woul d complete racecompleted (i.e. 4000 1 m)roun in 10

3) Dealer sold a radio at a loss of 2.5 %. Had he sold for Rs. 100 more, he would have gained 7 %.In order to gain 12 %, he should sell it for: a) 850  b) 1080 c) 925 d) 1120 Solution: L et c.p be c let s.p be s s = 0.975c H ad he sold f or Rs. 100 more, he woul d have gained 7 % (100+0.975c-c)/c* 100=7 (100-0.025c)* 100 = 7c 10000-2.5c=7c

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c=1000 Th e answer shoul d be 1120. 

4) At the end of the year 1998, shepard bought nine dozen goats.Henceforth, every year he had P% of goat at the beginning of the year and sold q% at the end of the year where p>0 and q>0.If the shepard nine dozen goat at end of the year 2002, after making the sales for the year, which of thehad following is true?? a) p=q  b) p>q c) pq.   th e per centage sold.T he onl y way, the number of goats wi ll

Question168 

1) Who among the following won the men?s singles title of the French Open 2004 ? (a) Guillermo Coria (b) Roger Federer (c) Andy Roddick (d) Gaston Gaudio An s: Gaston Gaudi o   

2) The Lingaraja Temple built during the medieval period is at (a) Bhubaneswar (b) Khajuraho (c) Madurai (d) Mount Abu An s: Bh ubaneswar  

3) During the Mughal period, which one of the following were the first to come to India as traders ? (a) Portuguese (b) Dutch

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(c) Danish (d) English An s: Portu guese   

4) ? Who among the following Delhi Sultans is known for introducing market control mechanism (a) Iltutmish (b) Balban (c) Alauddin Khalji (d) Firoze Tughlaq Ans: A lauddin Kh alji   

Question169  1) Which of the following cricketers holds the world record of maximum number of sixes in Tests ? (a) Chris Carins (New Zealand) (b) Viv Richards (West Indies) (c) Sachin Tendulkar (India) (d) Wasim Akram (Pakistan) An s:Ch r is Carins (New Zealand)   

2) Who among the following has been appointed the new Chief Justice of India ? (a) Justice Rajendra Babu (b) Justice V. N. Khare (c) Justice R. C. Lahoti (d) None of these An s: Justice R. C. Lahoti  

3) Who among the following sports persons got the honour of lighting the Olympic flame at the Major Dhyan Chand Stadium in New Delhi recently ? (a) Anjali Bhagwat (b) Abhinav Bindra (c) Viswanathan Anand (d) K. M. Beenamol Ans: Anj ali B hagwat   

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4) Who among the following has been appointed new chairman of the National Commission for Farmers ? (a) Ajit Singh (b) K. C. Pant (c) M. S. Swaminathan (d) Dr. Sharad Pawar An s: Dr . M . S. Swaminathan   

Question 170  1) 1.A boy walks 40 km north and then walks 50km towards east. Then turns right and walks 30km and then turns right and walks 50km.Now what is the distance between this point and the starting place? An swer is 10km   

2) A boy walks some distance towards north and then turns right, walks some distance. After walking some time he turns left and walks some distance and then walks some dist and then he walks at an angle of 45degree towards right and then turns to his left and walks. Towards Which direction was he walking finally? An swer is Nor th-west 3) A man walks 10km towards west. Then which of the combinations will take him to the original starting position? a) left, left, right  b)left, right, right c)right, left, right d)right, right, right Answer is right, right, right 

4) My flight takes of at 2am from a place at 18N 10E and landed 10 Hrs later at a place with coordinates 36N70W. What is the local time when my plane landed? a)6:00 am  b) 6:40am c) 7:40 d) 7:00 e) 8:00 Answer is The destination place is 80 degree west to the starting place. Hence the time difference between these two places is 5 hour 20 min. (=24hr*80/360). When the flight landed, the time at the starting place is 12 noon (2 AM + 10 hours). Hence, the time at the destination place is 12 noon - 5:20 hours = 6: 40 AM  

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1) A contractor agreeing to finish a work in 150 days, employed 75 men each working 8 hours daily. After 90 days, only 2/7 of the work was completed. Increasing the number of men by a)150 mens  b)200 mens c)100 mens An swer i s 150 mens   

2) A dishonest shopkeeper professes to sell pulses at the cost price, but he uses a false weight of 950gm. for a kg. His gain is? a)6.2%  b)5.3% c)3.9% An swer i s 5.3% 

3) A student divided a number by 2/3 when he required to multiply by 3/2. Calculate the  percentage of error in his result. a)0%  b)2% c)1% An swer i s 0% 

4) What is the sum of the first 25 natural odd numbers? a)575  b)600 c)625 An swer i s 625   

Question 171  1) It was calculated that 75 men could complete a piece of work in 20 days. When work was scheduled to commence, it was found necessary to send 25 men to another project. How much longer will it take to complete the work? a)27 days  b)30 days c)25 days An swer i s 30 days   

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2) Five boys were climbing a hill. J was following H. R was just ahead of G. K was between G & H. They were climbing up in a column. Who was the second? a)R  b)K c)G d)H e)J An swer is G   

3) Five farmers have 7, 9, 11, 13 & 14 apple trees, respectively in their orchards. Last year, each of them discovered that every tree in their own orchard bore exactly the same number of apples. Further, if the third farmer gives one apple to the first, and the fifth gives three to each of the second and the fourth, they would all have exactly the same number of apples. What were the yields per tree in the orchards of the third and fourth farmers? a)13 & 9 apples per tree.  b)11 & 10 apples per tree. c)11 & 9 apples per tree. Answer i s 11 & 9 apples per tr ee.  4) A man bought a horse and a cart. If he sold the horse at 10 % loss and the cart at 20 % gain, he would not lose anything; but if he sold the horse at 5% loss and the cart at 5% gain, he would lose Rs. 10 in the bargain. The amount paid by him was Rs? a)Cost price  b)Cost price of of horse horse == Rs. Rs. 100 400 & & the the cost cost price price of of cart cart == 200. 200. c)Cost price of horse = Rs. 500 & the cost price of cart = 300. An swer i s Cost pr ice of h orse = Rs. 400 & the cost price of car t = 200. 

Question 172  1) A building with height D shadows up to G. A Neighbour building with what height shadows C feet.? a) CD/G  b) C/DG c) CG/G d) DG/C An swer is CD/G   

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2)A person was fined for exceeding the speed limit by 10mph. Another person was also fined for exceeding the speed limit by twice the same. If the second person was travelling at speed of 35mph . Findthe speed limit ? a) 25 mph  b) 15 mph mph c) 28 An swer i s 15 mph   

3) A bus started from the bus stand at 8 Am and after staying 30 minutes at a destination return  back to the bus stand. The Destination is 27 miles from the bus stand . The Speed of the bus is 18mph . In the return journey the bus travels with 50% fast speed. At what time it is return to the  bus stand? a) 1.30 p.m.  b) 1.45 p.m. p.m. c) 1.00 An swer i s 1.00 p.m. 

4) Two lemons cost 10 cents. Then one and a half dozen cost a) 60 cents  b) 55 cents c) 90 cents An swer i s 90 cents   

Question 173 1) A car is filled with 4.5 gallons of fuel for a round trip . Car is taken 1/4 more than in going than com ing up. What is the fuel consumed in coming up ? a) 2.5 gallon  b) 2.25 gallon c) 2.0 gallon An swer is 2.0 gall on   

2) A work is done by two people in 24 minutes. One of them alone can do it in 40 minutes. How much time will the other person wiil take to complete it?

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a) 45 min  b) 50 min c) 60 min An swer is 60 min   

3)Low temperature at the night in a city is more than 1/2 high as higher temperature are 100. Then What is low temperature? a) 40 degree  b) 29 degree c) 35 degree An swer i s 40 degree   

4) A sales person multiplied by a number and get the answer 3. Instead of that number divided  by What is the answer she actually has to get.? a) 1/5  b) 1/3 c) 1/4 An swer is 1/3   

Question 174 

The perimeter of a square and a rectangle is the same. If the rectangle is 12 cms. by 10 cms., then  by what percentage is the area of the square more than that of the rectangle? a) 1 b) 3 c) 5/6 d) 1/2 Answer : c) 5/6

Reason : Given that Perimeters of square and rectangle are the same. Perimeter of rectangle of size 12 cms and 10 cms is 2(Length + Breadth) = 2(12+10) = 44 Perimeter of square = 44. (4 x side of square) Length of each side of the square = 44/4 =11 Area of the square = Side 2 = 11 x 11 = 121 sq. meter Area of the rectangle = Length X Breadth = 12 x 10 = 120 sq. meter

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Percentage Increase of area of the square to that of the rectangle = (Area of Square - Area of Rectangle) / Area of Rectangle X 100 % = (121 - 120)/120 X 100 % = 5/6 % Question 175 

Capacity cylindrical vessel 25,872 litres. height of the cylinder is 200% more than the radiusof ofaits base, what is theisarea of the baseIfinthe square cms? a) 336 b) 1232 c) 616 d) cannot be determined Answer : c)616

Reason : Let the radius be R and height be H Height is 200% more than its radius. This means H = R + 200% R = R (1 + 200/100) = 3 R 2

Volume of the cylinder (formula) = π R  H

It has been mentioned that volume of the cylinder is 25,872 cm3  22/7 R 2H = 25,872 cm3  Substituting H = 3R we get R = 14 cm 2

2

Area of the base = πR   = 22/7 x 142 = 616 cm   Question 176 

If the radius of a circle is increased by 12%, then the area of the circle a) decreases by 25.44%  b) increases by 25.44% c) no change in area d) decreases by 12% Answer : b) increases by 25.44%

Reason : 2

Area of circle πr   where r is the radius

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Assume initially the radius was 10. Then area becomes 22/7x10x10 = 314.28 --> eq 1  Now if the radius is increased by 12%, it will become 10+(12/100)10 = 10+1.2=11.2  New area = 22/7 x 11.2x 11.2 = 394.24 --> eq 2 Increase in area in our case = eq 2 - eq 1 = 79.96 Percentage increase in area when radius is increased by 12% = Increase in Area / Original Area X 100% = 79.96 / 314.28 X 100 =25.44% Question 177 

Pointing to the woman in the picture, Govind said, ― Her mother has only one grand child whose mother is my wife‖. How is the woman in the picture related to Govind?   a) Sister b) Wife c) cousin d) data inadequate Answer : b) Wife

Woman in the picture is told as  –  Her mother has one only one grand child. Mother of that grand child is Govind’s wife. So woman in the picture is wife of Govind.   Question 178 

Anand is the uncle of Bimala, who is the daughter of Cauvery and Cauvery is the daughter in law of Palani. How is Anand related to Palani? a) Son b) son in law c) Brother d) newphew Answer : a) Son

Palani’s daughter in law is Cauvery.Cauvery’s daughter is Bimala. Anand is uncle of Bimala. Cauvery and Anand are siblings. Therfore Anand is son of Palani. Question 179 

Pointing to a man, Lakshmi says, ―This man’s son’s sister is my mother -in-law‖. How is the Lakshmi’s husband related to the man?  

a) son b) son-in-law c) grandson d)nephew Answer : c) grandson

Man’s son’s sister = Man’s daughter. Man’s daughter is the mother of Lakshmi’s husband. 

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Question 180 

A cylinder of diameter 14 and height 7 is converted into a cone of radius 6. Now, what could be the percentage height of the new shape ? a. 308% b. 32% c. 2150% d. 215% Answer: a. 308%

Reason: Volume of cylinder = pi X r 2 X h = pi X 49 X 7 = 343pi Volume of the cone = Volume of cylinder = 343pi -> eq 1 Volume of cone is given by formula = 1/3 (pi X r 2 X h) = 36/3(pi)(h) = 12(pi)(h)--> eq 2 (here h is the height of the cone) From eq 1 and eq 2 we get 343pi = 12(pi)(h) Therefore, height of the cone = 343/12 Percentage increase in height = ((height of the cone - height of the cylinder )/ height of the cylinder) X 100% = (343/12 - 7)/7 X 100 = 259/84 X 100 = 308% approximately. Question 181 

Find the largest number which gives out reminders 2 and 5 when used to divide 80 and 122. a. 19 b. 17 c. 13 d. infinite number of solutions Answer: c. 13.

Reason: This is a very simple question where finding out the HCF of 80 - 2 and 122 - 5 will give the answer. The HCF of 78 and 117 is 13

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Hint : In case you are confused about HCF, LCM stuff, you can very well do a trial and error with options which is not a difficult task but could consume a little more time than the direct method. Question 182 

Find the ratio of the volumes of a sphere and cone of same radius. Also the height of the cone is twice its radius. a. 3 : 1 b. 4 : 3 c. 4 : 1 d. 2 : 1 Answer : d. 2 : 1

Reason: This is again a simple question that can be answered by knowing formulas. Volume of a cone = 1/3 pi r 2 h where r is the radius and h is the height of the cone.  Now, h = 2r as per the question. Hence volume of the cone = 2/3 pi r 3  Volume of a sphere = (4/3) pi r3 where r is the radius which is the same as that of the cone. Ratio of volume of sphere to that of cone = (4/3) pi r 3 / 2/3 pi r 3 = 2 : 1 Question 183 

Find the missing number in the sequence 504 , _ , 990, 1320, 1716 a. 720 b. 1000 c. 721 d. 790 Answer : a. 720

The given sequence is of the form n3-n where n starts from 8. Hence the blank should be filled by 93-9 which is 720. Question 184 

Consider a group of working men. If 1/10th of the workers are absent then work of each person would be increased by. a. 20% b. 22.22% c. 10% d. 11.11% Answer 1 

Answer is d. 11.11%

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Let us assume 100 units of work is divided among 10 men. When no one is absent, each member have to do 100/10 = 10 units of the work With 1/10th of men absent, there would be 9 men present.  Now, if 100 units of work is to be done by the 9 men present, each should do 100/9 units of work. Extra work required from each member when only 9 men are present = 100/9 - 10 = 10/9 units. Percentage of increase of work for each member = (Extra work required from each member when only 9 men are present)/ (Original amount of work when everyone is present) X 100 % = ((10/9)/10)X 100% = 11.11% Question 185 

Consider the following statements, All members of team A are good tall guys with an average height of 6 feet. Average of the height of members of team B is lesser than that of team A. Which could be true among the below options. a) Every member of team B is shorter than each member of team A.  b) Few members of team B could be over 6 feet. Answer 2 

Answer is Option b) Few members of team B could be over 6 feet. Reason : The statement says that there is a possibility that few members of team B can actually  be taller than few members of team A. Question 186 

John is supposed to walk from his house to park every morning. One morning, he is in real hurry and wants to save at least 1/3rd of the time. By how much percentage he should increase his speed. a. 100% b. 33% c. 66% d. 50% Answer 3 

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Answer is d. 50% Let the distance between his house and park be 100 metres. Lets assume he takes 30 mins daily. Hence his speed would be 100/30. One day, if he wishes to save 1/3rd of 30 minutes, that is 10 minutes, he should cover the distance in 20 minutes in which case his speed would be 100/20. Required increment in speed = 100/20 - 100/30 Percentage of increase in speed = (Required increment in speed / Original speed) X 100 % = ((100/20 - 100/30) / 100/30 )X100 % = 50%. Question 187

Consider three statements. P isofshorter than R, is Q true. is taller than R, S is taller than Q. Based on the the following above statements, state which the following a) S is the tallest of all b) R is taller than S c) None of the above Answer 1 

Option a) S is the tallest of all, is the correct answer. The second option is not true because S is taller than Q who in turn is taller than R. Question 189

Ravi describes his relationship with Raju as follows. "Raju is the eldest son of his grand father's only daughter in law". So, how Raju is related to Ravi? a) elder brother b) brother in law c) younger brother Answer 2 

Option a) is the right answer. Ravi's grandfather's only daughter in law has to be Ravi's mother. Hence, the eldest son of Ravi's mother should be his elder brother. Question 190

Say a C union has two members one requiring 2 bytes of memory and the other requiring 4 bytes of memory? What would be the memory allocated to this data type ? a) 4 bytes b) 2 bytes c) 6 bytes Answer 3 

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Option a) 4 bytes is the right answer. Union memory allocation is based on the member that occupies the maximum data size. Hence the answer would be 4 bytes.

Question 191

Consider the following three statements. P is shorter than R, Q is taller than R, S is taller than Q. Based on the above statements, state which of the following is true. a) S is the tallest of all b) R is taller than S c) None of the above Answer 1 

Option a) S is the tallest of all, is the correct answer. The second option is not true because S is taller than Q who in turn is taller than R. Question192

Ravi describes his relationship with Raju as follows. "Raju is the eldest son of his grand father's only daughter in law". So, how Raju is related to Ravi? a) elder brother b) brother in law c) younger brother Answer 2 

Option a) is the right answer. Ravi's grandfather's only daughter in law has to be Ravi's mother. Hence, the eldest son of Ravi's mother should be his elder brother. Question 193

Say a C union has two members one requiring 2 bytes of memory and the other requiring 4 bytes of memory? What would be the memory allocated to this data type ? a) 4 bytes b) 2 bytes c) 6 bytes Answer 3 

Option a)the 4 bytes is the right answer. Union memorywould allocation based on the member that occupies maximum data size. Hence the answer be 4 is bytes. Question 195

1) Total number of boys and girls in a class is A. The total number of girls subtracted from the total number of boys is B. Now please express the portion of girls to the total strength in terms of 'A' and 'B'.

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Let x denote the number of boys. Let y denote the number of girls. A = x + y -> equation 1 B = x - y -> equation 2 Subtracting equation 2 from equation 1 we will get y = (A - B)/2 -> equation 3 Ratio of total number of girls to the total strength will be y / x + y.  Now substituting A for x + y as per equation 1 and (A - B)/2 for y as per equation 3 we will get the answer as (A - B)/(2*(A + B)) Question 196

2) Find the ratio of weights of 1 gram of cotton to 1.5 gram of rubber if their densities are in the ratio of 1 : 2. This is a very simple question. Since the ratio of weights is the question, one need not care about their densities. Hence 1 gram of cotton : 1.5 gram of rubber will be in the ratio 1/1.5 = 2/3 (as simple as that :)) Question 197

3) Ravi had got twice as much as marks as Ramu. His teacher had made him a promise that, for every mark he scores above Ramu, he would be awarded 50% of those marks as bonus. Find the ratio of his bonus marks to the total marks of Ravi and Ramu. Lets say Ramu has got x marks. Then Ravi would had got 2x marks initially. But his teacher has promised a bonus of 50% of marks from the extra marks he got more than Ravi. Hence he would be awarded x/2 marks as bonus. Total marks of Ravi = 2x + x/2 = 5x/2. Total marks of both the students = Total marks of Ramu + Total marks of Ravi = x + 5x/2 = 7x/2. Ratio of Ravi's bonus mark to the Total marks of Ravi and Ramu would be (x/2) divided by (7x/2) = 1/7. Question 198  1) A person with some money spends 1/3 for cloths, 1/5 of the remaining for food and 1/4 of the remaining for travel. He is left with Rs 100/How much did he have with him in the begining?

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Answer Rs. 250/-

2) Grass in lawn grows equally thick and in a uniform rate. It takes 24 days for 70 cows and 60 days for 30 cows to eat the whole of the grass. How many cows are needed to eat the grass in 96 days? Answer 20 cows

3) There is a safe with a 5 digit number as the key. The 4th digit is 4 greater than the second digit, while the 3rd digit is 3 less than the 2nd digit. The 1st digit is thrice the last digit. There are 3 pairs whose sum is 11. Find the number. Answer 65292

4) Consider a number 235, where last digit is the sum of first two digits i.e. 2 + 3 = 5.How many such 3-digit numbers are there? Answer There are 45 dif f er ent 3-di git numbers. 

Question 199  1) There are 4 parties A,B,C,D. Ram told that either A or B will win. Shyam told C will never win. Hari told either B or C or D will win. Only one of them was Correct. Which party won? Ans: C   

2) A colck takes 33 seconds to complete the pendulum sound when it is 12:00 noon. How long one can hear the pendulum sound it is 6:00 a.m. ie the the differrence between 1st sound and last sound. Ans: 15 Secs

3) There are 111 players participating in a singles tennis tournament. The player who is loosing will be out of the tournament. For each and every match, One new ball is taken. Find the no. of balls required for the entire tournament. Ans:110

5)  I have got some money in my bag .(which is stolen after shopping I and II). I spent 10% of my money for shopping (I) For second time,10% of the remaining money is spent for shopping (II) The total bill amount=Rs.18. Find the amount which will be remaining in the bag?

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6)  An swer: Rs.81    Question 200  1) Three cards are drawn at random from an ordinary pack of cards.Find the probability that they will consist king, a queen and of anaace. An s. 64/2210. 

2) A number of cats got together and decided to kill between them 999919 mice. Every cat killed an equal number of mice. Each cat killed more mice than there were cats. How many cats do you think there were ? An s. 991. 

3) If Log2 x - 5 Log x + 6 = 0, then what would the value / values of x be? An s. x = e2 or e3. 

4) The square of a two digit number is divided by half the number.After 36 is added to the quotient, this sum is then divided by 2. The digits of the resulting number are the same as those in the original number, but they are in reverse order. The ten's place of the original number is equal to twice the difference between its digits. What is the number? An s. 46   

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