BITS PILANI Test 9 – Answer Key & Explanatory Answers

BITS PILANI Test - 9 – Answer Key & Explanatory Answers 1.ABCD is a cyclic quadrilateral.∠ DBA = 50° and ∠ADB = 33°. then the measure off ∠BCD is 1) 83° 2) 80° 3) 75° 4) 60° 2.A shop keeper earns a profit of 12% on selling a book at 10% discount on the printed price. The ratio of cost price to the printed price of the book is 1) 45:56 2) 50: 61 3) 99: 125 4) None of these 3.The number of pupils of a class is 55. The ratio of the number of male pupils to the number of female pupils is 5: 6. The number of female pupils is 1) 11 2) 25 3) 30 4) 35 4.5% more is gained by selling a watch for Rs. 350 than by selling it for Rs. 340. The cost price of the watch is 1) Rs. 110 2) Rs. 140 3) Rs. 200 4) Rs. 250 5.If 60% of the students in a school are boys and number of girls is 812, how many boys are there in the school? 1) 1128 2) 1218 3) 1821 4) 1281 6.It takes eight hours for a 600 km journey, if 120 km is done by train and the rest by car. It takes 20 minutes more, if 200 km is done by train and the rest by car. The ratio of the speed of the train to that of the car is: 1) 3:5 2) 3:4 3) 4: 3 4) 4: 5 7.If m + n = 1, then the value of m3 + n3 + 3mn is equal to 1) 0 2) 1 3) 2

4) 3

8.The maximum number of common tangents that can be drawn to two disjoint circles is 1) 1 2) 2 3) 4 4) Infinitely many 9.If cos4θ - sin4θ = 1/3, then the value of tan2θ is 1) 1/2 2) 1/3 3) 1/4

4) 1/5

10.If a perfect square, not divisible by 6, be divided by 6, the remainder will be 1) 1, 3 or 5 2) 1, 2 or 5 3) 1, 3 or 4 4) 1, 2 or 4 11.Integers from 100 to 99999 are written consecutively from left to right as shown: 100101102 ... 9999899999. If xn denotes the nth digit from the left, then which is the least possible value of n for which xn-2 = xn-1 = xn= xn+1 = xn+2 = 4? (1)1035 (2) 1033 (3) 16480 (4) 93143 (5) None of these Solution: Given series: 100101102103…99910001001…9999899999. We have to get the least possible value of ‘n’. Here, by inspection, it can be found out that the first time five 4’s occur consecutively is when the numbers 444 and 445 are lined up next to each other (443444445446 …). Here, the required nth digit is the last digit of the number 444. Starting from 100, including 444, there are (444 – 100 + 1) = 345 three digit numbers. Hence, the required nth digit = 345 × 3 = 1035th digit. Hence, the correct answer is option 1. 12.The number N in base 7 has 3 digits. When it is written in base 9, the digits are reversed. Which are the following digits is one of the digits of N? (1) 4 (2) 2 (3) 0 (4) 6 (5) None of these Solution: Let N be denoted by abc. In base 10, the value would be c×70 + b×71 + a×72 = c + 7b + 49a. In base 9, the number would be denoted by cba and its value in base 10 would be a×90 + b×91 + c×92 = a + 9b + 81c. Thus, c + 7b + 49a = a + 9b + 81c. → 48a = 2b + 80c → 24a = b + 40c. c = 3a/5 – b/40 We know that a, b, c are positive integers that are less than 7. So 3a/5 and b/40 should also be integers. © 2017 ETHNUS

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BITS PILANI Test - 9 – Answer Key & Explanatory Answers So, ‘a’ can only take the value 5 and ‘b’ can only take the value 0. So, c = 3 – 0 = 3. So, the number N is 503. Hence, the correct answer is option 3. 13. What is the remainder when 444 is divided by 15? (1)14 (2) 12 (3) 5

(4) 1

(5) 3

Solution: Let N = 444 = (42)22 = (16)22. By remainder theorem, when f(x) is divided by (x-a), the remainder is given by f(a). Here, let f(x) = xn = (16)22, a = 1, then (x-a) = (16 – 1) = 15. Thus, remainder = f(a) = (1)22 = 1. Hence, the correct answer is option 4. Alternately, 41 when divided by 15, the remainder is 4 42 when divided by 15, the remainder is 1 43 when divided by 15, the remainder is 4 and so on. Since it follows a pattern, 4x when divided by 15 will leave a remainder of 4, when x is odd. 4x when divided by 15 will leave a remainder of 1, when x is even. Hence, the correct answer is option 4. 14.If log45 125 = x, then find the value of log15 81? (1) 4×(3-x)/(3+x) (2) 4×(3+x)/(3-x)

(3)(3-x)/(3+x)

(4) (3+x)/(3-x)

(5) None of these

Solution: Given: log45 125 = x = (log 125)/(log 45) = (log 53)/(log (32×5)) = (3 log 5)/(2 log 3 + log 5). → 2 x log 3 + x log 5 = 3 log 5 → 2 x log 3 = (3 – x) log 5. → log 5 = (2 x log 3)/ (3 – x) (1) Now, log1581 = log15 34 = (4 log 3)/log 15 = (4 log 3)/(log 3+log 5) Using (1) here, log1581 = (4 log 3)/(log 3+((2 x log 3)/ (3 – x))) = (4)/(1 + (2 x/(3 – x))) = (4) ×(3 – x)/(3 – x + 2 x) → log1581 = (4) ×(3 – x)/(3 + x) Hence, the correct answer is option 1. 15.Find the volume of the largest box that can be made, by cutting equal squares out of the corners of a piece of rectangular sheet measuring 24 cm by 15 cm and then folding upwards (to make a box without the top face)? (1) 584 cm3 (2) 484 cm3 (3) 216 cm3 (4) 384 cm3 (5) None of these Solution: Given, the width is 15 cm and the length of the rectangular sheet is 24 cm. Let the side of each square that is cut out of the 4 corners be denoted by ‘x’. Then, the new width of the box = (15 – 2x) and the new length = (24 – 2x) and the height = x. Let volume of the box = V = (15 – 2x) (24 – 2x) (x) = 360x – 78x2 + 4x3. To maximize the volume, set dV/dx = 0. dV/dx = 360 – 156x + 12x2 = 0. → x2 - 13 x + 30 = 0 → (x – 3)(x – 10) = 0 Thus, we get x = 3 or 10, here x = 10 is an inappropriate value as the width of the box = (15 – 2x). Hence, x = 3. So, the maximum volume = (15 – 2×3) (24 – 2×3) (3) = 486 cm3. Hence, the correct answer is option 5. 16.Working at full efficiency, Raja alone can complete a new development project in 8 days. But each successive day, his efficiency goes down to half of his previous day. Starting with full efficiency on day 1, he worked alone for 3 days and then on day 4, Rani joined him and thereafter, Raja started working at full efficiency until the end of the project. Now working together, they both completed the remaining part of the project in 5 more days. In all, if they were paid Rs 1440, what is Rani’s share? (Assume equal number of working hours for both on all days). (1) 480 (2) 180 (3) 225 (4) 288 (5) 360 © 2017 ETHNUS

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BITS PILANI Test - 9 – Answer Key & Explanatory Answers Solution: Let one day’s work (full efficiency) of Raja and Rani be denoted by a and b respectively and the total work be 1 unit. Raja’s 1st day work = 1/8 unit. (full efficiency) Raja’s 2nd day work = 1/(8×2) = 1/16 unit. Raja’s 3rd day work = 1/(16×2) = 1/32 unit. Thus, in the first 3 days, Raja completes (1/8 + 1/16 + 1/32) = (7/32) units. So, the work remaining = (1 – 7/32) = 25/32 units. Rani joins Raja on the 4th day and they both work together for 5 days and complete the remaining work of (25/32) units. Hence, working together, in one day, they complete (5/32) units. Hence, Rani’s one day work = (5/32) – (1/8) = 1/32 units. Now, the total work done by Raja = (7/32) + (5×4/32) = 27/32 units and the total work done by Rani = (1/32) × 5 = 5/32 units. Hence, the total amount should be divided in the ratio of 27:5. Thus, Rani’s share = (5/32) × 1440 = Rs 225. Hence, the correct answer is option 3. Alternately, The one day wage of Raja = 1440/8 = Rs. 180. Raja worked 6 days with full of his efficiency and 1 day with half of his efficiency and 1 day at 1/4th his full efficiency. The wages of Raja = 6 x 180 + 90 + 45 The wages Rani = 1440 – [6 x 180 + 90 + 45] = Rs. 225 Hence, the correct answer is option 3. 17.Given that: f(x) = √x, f 1(x) = e^(2logef(x)) and so on such that: f n(x) = e^(2loge[f n-1(x)]). What is the product of the following series: f(x) × f 1(x) × f 2(x) × ... f n(x)? (1) 2n loge x (2) 1 (3) x(2n) (4) e(2nx) (5) None of these Solution: Given that: f(x) = √x = x^(1/2). f 1(x) = e^(2logef(x)) = e^(logef(x)) × e^(logef(x)) = f(x) × f(x) (Since e^(loge x) = x). Therefore, f 1(x) = (f(x))^2 = x f 2(x) = (f 1(x))^2 = x2 f 3(x) = (f 2(x))^2 = x4 f 4(x) = (f 3(x))^2 = x8 f n(x) = (f n-1(x))^2 = x2(n-1) Now, the product : f(x) × f 1(x) × f 2(x) × ... f n(x) = x^(1/2) × x × x2 × x4 × x2 × x4 × x8 × … x2(n-1) = x^((1/2)+1+2+4+8+…+2(n-1)) = x^(2n – ½). Hence, the correct answer is option 5. 18.A right-angled triangle has sides of distinct, integral values, which are in Arithmetic Progression. If at least one side of the triangle is a multiple of ten, then how many distinct sets of values can be found out for the sides of the given triangle, given that the perimeter of the triangle is less than 125 units? (1) 2 (2) 6 (3) 4 (4) 8 (5) 5 Solution: Let the sides of the right-angled triangle be a, b and c. Given the perimeter is f(x) = a + b + c < 125. Since it is a right-angled triangle, the 3 sides form a Pythagorean triplet. The only triplet that is in Arithmetic Progression is 3, 4, 5. Hence, the sides are in the ratio 3:4:5. Therefore, the sides can be taken to be equal to: 3k, 4k and 5k, where k is a positive integer. So now, the sum of the sides = 12k < 125 → 0 < k ≤ 10. For k = 2, 4, 6, 8, 10; c = 5k, is a multiple of 10. For k = 5; b = 4k, is a multiple of 10. Thus, we have 6 different sets of values for the sides of the triangle which satisfy all the given conditions. Hence, the correct answer is option 2. 19.A basket of articles contain a supply of balls of three different colours. In how many different ways can you choose six balls from the basket? (1) 39 (2) 36 (3) 42 (4) 18 (5) 21 © 2017 ETHNUS

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BITS PILANI Test - 9 – Answer Key & Explanatory Answers Solution: Consider the following cases possible: (a) All 6 balls are of the same colour (C1) = 3C1 = 3 cases. (Corresponding to the 3 different colours) (b) Cases with balls of 2 different colours (C1, C2) and cases with balls of all 3 different colours (C1, C2, C3): The number of ways we can select the balls is given by: (Any of the 3 colours for the ball of color 1) (Any of the remaining 2 colours for the other balls) = 3C1 2C1 = 32 = 6 ways. In this, we also have 6 different ball-colour combinations: (i) 5 Balls C1 and 1 ball C2 (ii) 4 Balls C1 and 2 balls C2 (iii) 3 Balls C1and 3 balls C2 (iv) 4 Balls C1, 1 ball C2, 1 ball C3 (v) 3 Balls C1, 2 balls C2, 1 ball C3 (vi) 2 Balls C1, 2 balls C2, 2 balls C3 The number of cases = 6 * 6 = 36 cases. Thus, adding all the above possible cases, we get the total number of ways = 3 + 36 = 39 ways. Hence, the correct answer is option 1. 20. A bus and a train leave for Mysore from Bangalore at 11 am and 1 pm on the same day respectively and both reach the destination at the same time. Travelling at the same speeds as before, if the train had left for Bangalore from Mysore and the bus had left for Mysore from Bangalore at the same time, they would have meet each other after 80 minutes. What is the ratio of the speed of the bus to the train? (1)2:1 (2) 1:1 (3) 1:2 (4) 3:4 (5) Cannot be determined uniquely Solution: Let the time taken (in hrs) by the bus to reach Mysore from Bangalore be denoted by t. Then, the time taken (in hrs) by the train to reach Mysore from Bangalore would be (t – 2) hrs. Also, let the distance between Bangalore and Mysore be ‘D’. Speed of the bus is then given by Sb = D/t and the speed of the train is given by St = D/(t – 2). When travelling in opposite directions, the relative speed would be the sum of the speeds of the bus and the train. Hence, relative speed = (D/t) + (D/(t – 2)). Time taken to meet when travelling in opposite directions = Total distance/Relative speed = D/[(D/t)+(D/(t – 2)) = 80 mins = 4/3 hrs. → (t)(t – 2)/(2t – 2) = 4/3. Solving this equation, we get t = 4 hrs. So, the bus takes 4 hrs to complete a journey while the train takes (4 – 2) = 2 hrs to complete the same journey. Hence, the ratio of their speeds = 2:4 = 1:2. Hence, the correct answer is option 3. 21. For the following questions Find the odd word/letters/number from the given alternatives. 1) Time 2) Skill 3) Interest

4) Knowledge

22. For the following questions Find the odd word/letters/number from the given alternatives. 1) CPA 2) REB 3) QUD

4) AOT

23. For the following questions Find the odd word/letters/number pair from the given alternatives. 1) 15 - 21 2) 32 - 41 3) 22 - 27

4) 31 - 35

24.Arrange the following words as per order in the dictionary and then choose the one which comes last. 1. Qualify 2.Quarter 3.Quarrel 4.Quaver 1) Qualify 2) Quarter 3) Quarrel 4) Quaver 25.A series is given, with one term missing. Choose the correct alternative from the given ones that will complete the series. DF, GJ, KM, NQ, RT, ? 1) EI 2) UX 3) UV 4) XY © 2017 ETHNUS

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BITS PILANI Test - 9 – Answer Key & Explanatory Answers 26.A series is given, with one term missing. Choose the correct alternative from the given ones that will complete the series. 2, 7, 14, 23, 34, ? 1) 47 2) 39 3) 42 4) 46 27.A and B are standing at place "P". They start moving in the opposite directions at the speed of 5 Kmph and 4 Kmph respectively. What will be the distance between them after 3 hours 1) 3 kms 2) 21 kms 3) 18 kms 4) 27 kms 28.If Usha is taller than Nisha; Nisha is taller than Asha; Alka is taller then Usha. Harsha is shorter than Asha; then who among them is the tallest? 1) Usha 2) Alka 3) Nisha 4) Asha 29.From the given alternative words, select the word which cannot be formed using the letters of the given word: DHARAMSALA 1) MASALA 2) ARAMANA 3) RAMA 4) SAHARA 30.If in a code GONE is written as ILPB then how may CRIB be written in that code? 1) EUKY 2) EKUY 3) EYUK 4) EOKY Directions for questions 31 to 35: Refer the data given below to answer the following questions. Eight places, Banaswadi, Jayanagar, M.G.Road, Indira nagar, Koramangala, K.R.Puram, Basavangudi and Richmond town in Bangalore are connected by a network of eleven roads. There are eight roads that allow only one-way traffic while there are three roads that allow two-way traffic. From Banaswadi there are three roads, one to Jayanagar, one to Indira nagar and the other to Basavangudi. From K.R.Puram, there are two roads one to Koramangala and the other to Jayanagar. There are two roads from Basavangudi – one to Richmond town and the other to Indira nagar. There are two roads from Jayanagar, one to M.G.Road and the other to K.R.Puram. There is a road to Jayanagar and also a road to Richmond town from M.G.Road. There is a road from Koramangala to Indira nagar and a road from Indira nagar to Jayanagar. There is a road to M.G.Road from Richmond town. There are no other roads, connecting any two places apart from the routes that are mentioned above. Solutions for questions 31 to 35: Refer the following network. Banaswadi

Basavangudi

Jayanagar Koramanga

K.R.P

Indira nagar Richmond town

M.G.Road

31.Which of the following places must be visited, if a person wants to go from Banaswadi to K.R.Puram? (1) Basavangudi (2) Indira nagar (3) Jayanagar (4) M.G.Road (5)Richmond town Solution: K.R.Puram can only be reached by Jayanagar. Hence, the correct answer is option 3. 32.If a person wants to go from Banaswadi to Koramangala, then in how many ways can he go without crossing any place twice? (1) 4 (2) 2 (3) 5 (4) 3 (5) 6 Solution: The ways are Banaswadi-Jayanagar-K.R.Puram-Koramangala Banaswadi-Indira nagar-Jayanagar-K.R.Puram-Koramangala Banaswadi-Basavangudi-Indira nagar-Jayanagar-K.R.Puram-Koramangala Banaswadi-Basavangudi-Richmondtown-M.G.Road-Jayanagar-K.R.Puram-Koramangala Hence, the correct answer is option 1.

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BITS PILANI Test - 9 – Answer Key & Explanatory Answers 33.Which of the following two places have exactly two routes, such that no route is common and no place is crossed more than once? (1) Banaswadi to Koramangala (2) Banaswadi to Richmond town (3) Basavangudi to Jayanagar (4) Jayanagar to K.R.Puram (5) None of these Solution: The routes to Basavangudi to Jayanagar are 1. Basavangudi-Indira nagar-Jayanagar 2. Basavangudi-Richmond town-M.G.Road-Jayanagar Hence, the correct answer is option 3. 34.How many places will be in between if one goes from Banaswadi to Koramangala by the fewest number of intervenings enroute? (1) 1 (2) 2 (3) 3 (4) 4 (5) 5 Solution: The shortest route is Banaswadi-Jayanagar-K.R.Puram-Koramangala Hence, the correct answer is option 2. 35.How many places will be crossed if one goes from Basavangudi to M.G.Road by the longest route possible without crossing any place twice? (Include the destination and starting point) (1) 1 (2) 2 (3) 3 (4) 4 (5) 5 Solution: The ways are Basavangudi -Indira nagar -- Jayanagar-M.G.Road Basavangudi-Richmond town- M.G.Road Hence, the correct answer is option 4. Directions for questions 36 to 40: Refer the data below to answer the following questions The following are the details of different steps involved in constructing a fly-over in Bangalore. The total number of activities to be completed are 12. Please note that in any month, one or more activities are in progress. Work/Activity M J K O F H N B D X P S

Duration (In Months) 2 1 1 2 4 6 3 4 1 1 2 2

Other works/activities that should be completed before the work J M, J M O J F, H, X S N B P

For example, the works that should be before the start of the work B are F, H and X.

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BITS PILANI Test - 9 – Answer Key & Explanatory Answers Explanations for questions 36 to 40: The following table gives the minimum possible time for the entire activities to be completed. Month 1st 2nd 3rd 4th 5th 6th 7th 8th 9th 10th

Activity Activities Month (in progress) completed M, J M,K,N N,O,F N,O,F F,H,X F,H H H H H

J M,K N,O X F

Activity (in progress)

11th 12th 13th 14th 15th 16th 17th 18th 19th

B B B B P P S S D

Activity completed

B P S D

H

36.What can be the maximum time gap between the start of the work H and the start of the work S? (1) 3 months (2) 9 months (3) 15 months (4) 21 months

(5) 17 months

Solution: Before H, the activities that are to be completed = M, J and O. The maximum number of activities that can be performed between H and S (including H) are H, K, F, N, B, X and P. So, the maximum time gap = 6 + 1 + 4 + 3 + 4 + 1 + 2 = 21 months. Hence, the correct answer is option 4. 37.What is the least possible time gap between the start of the work M and the start of work X? (1) 1 month (2) 2 months (3) 4 months (4) 6 months

(5) None of these

Solution: One only activity that is to be completed before X is N. Assuming that N is completed before M, the least time gap between M and X is 0 months. Hence, the correct answer is option 5. 38.If only one activity can be undertaken at any given time, then what is the minimum possible time in which all the above mentioned works can be completed? (1) 19 months (2) 26 months (3) 27 months (4) 29 months (5) 24 months Solution: The answer for this question is nothing but the sum of the time taken for the activities M to S. So, the time required = 2 + 1 + 1 + 2 + 4 + 6 + 3 + 4 + 1 + 1 + 2 + 2 = 29 months Hence, the correct answer is option 4. 39.If the flyover construction is to be finished at the earliest, then the maximum time by which the work X can start is (1) 8th month (2) 9th month (3) 10th month (4) 11th month (5) 12th month Solution: To start the activity B, the activity X has to be completed. Referring to the table above, B starts at 11th month. Since only one month is required for X to be completed, the latest time that X could start is 10th month. Hence, the correct answer is option 3. 40.The earliest time by which the work B can start is (1) 6th month (2) 8th month

(3) 10th month

(4) 11th month

(5) 9th month

Solution: Referring to the table above, the earliest time that the activity B could start is the 11th month. Hence, the correct answer is option 4. © 2017 ETHNUS

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