Aryabhatta National Maths Competition - 2021: Complete Practice Questions Set (CPQS)

Aryabhatta National Maths Competition 2021 Complete Practice Questions Set (CPQS) (For Group 1)

Chain Rule

Rule of three The method of finding the 4th proportional when the other three are given is called Simple Proportional or Rule of three.

Compound Proportional Repeated use of three is called Compound Proportional.

Direct Proportional Two quantities are said to vary directly if on the increase ( or decrease) of the one , the other increase (or decrease) to the same extent. EXAMPLES 1. More articles , more cost. 2. More men, more work. Indirect Proportional Two quantities are said to vary indirectly if on the increase ( or decrease) of the one , the other decrease (or increase ) to the same extent.

EXAMPLES 1. Less time to finish a work , more persons at work. 2. More speed , less is the time taken.

Sample Question Paper on Chain Rule Q 1 - If 15 dolls cost Rs 35, what do 39 dolls cost? A - Rs 71 B - Rs 81 C - Rs 91 D - Rs 101

Answer - C Explanation Let the required cost be Rs x. More dolls, more cost (direct) ∴ 15 : 39:: 35 : x ⇒15 *x = (39 *35) ⇒ x= (39 *35)/15 = 91. ∴ Cost of 39 dolls = Rs 91 Q 2 - If 36 men can do a piece of work in 25 days, in how many days will 15 men do it? A - 30 B - 40 C - 50 D - 60

Answer - D Explanation Let the required number of days be x. Less men, more days (indirect) ∴ 15: 36:: 25 : x ⇒ 15 *x = (36 * 25) ⇒ x= (36 *25 ) / 15 = 60. ∴ Required number of days = 60. Q 3 - If 20 men can build a wall 112m long in 6 days, what length of a similar wall can be built by 25 men in days? A - 40m B - 50m C - 60m D - 70m

Answer - D Explanation Let the required length be x metres. More men, more length built (direct) Less days, less length built (direct) Men 20 : 25 :: 112 : x Days 6:3 ∴ (20 * 6 *x ) = ( 25 * 3 *112) ⇒ x= (25 * 3 * 112) / (20 * 6 ) = 70. Required length 70m.

Q 4 - If 8 men working 9 hours a days can built a wall 18m long, 2 m broad and 12m high in 10 days, how many men will be required to build a wall 32m long , 3m broad and 9m high by working 6 hours a days, in 8 days? A - 20 B - 30 C - 40 D - 50

Answer - B Explanation let the required number of men be x. More length, more men (Direct) More breadth, more men (Direct) Less height, less men (Direct) Less hours per day, more men (Indirect) Less days, more men (Indirect) Length 18:32 Breadth 2:3 Height 12:9 :: 8 : x Hrs / Day 6: 9 Days 8:10 ∴ ( 18 * 2 * 12 * 6 * 8 * x) = ( 32 * 3 * 9 * 9 * 10) ⇒ x= 32*3*9*9*10 / 18*2*12*6*8 =30.

Q 5 - A contract was to be completed in 56 days and 104 men were set to works, Each working 8 hours per days. After 30 days , 2/5 of the work is completed. How many additional men may be employed so that the work may be completed in time, each man now working 9 hours a day? A - 36 B - 46 C - 56 D - 66

Answer - C Explanation Remaining work = (1- 2/5 ) = 3/5 , Remaining period = (56 – 30) =26 days. Let the additional men employed be x. More work , more men (direct) More days , less men (indirect) More hrs/ day, less men (indirect) Work 2/5 : 3/5 Days 26: 30 :: 104 : (104 + x) Hrs/ day 9:8 ∴ 2/5 *26 * 9 * (104 +x ) = 3/5 *30 * 8 * 104 ⇒ (104 +x) = 3 * 30 * 8 * 104 / 2* 26 * 9 = 160 ⇒ x = (160 – 104) = 56. Additional men to be employed = 56. Q 6 - 5 men or 9 women can do a piece of work in 19 days. In how many days will 3 men and 6 women do it? A - 12 B - 13

C - 14 D - 15

Answer - D Explanation 9 women = 5 men ⇒ 1 women = 5/9 men ⇒ 6 women = (5/9 * 6) men = 10/3 men. 3 men +6 women = (3+ 10/3 ) men = 19/3 men. Let the required number of days be x. More men, less days 19/3 : 5 :: 19 : x ⇒ 19 /3 *x = (5 *19) ⇒ x= (5* 19 * 3 /19 ) = 15. ∴ Required number of days = 15. Q 7 - 8 women can complete the work in 10 days and 10 children take 16 days to complete the same work. How many days will 10 women and 12 children take to complete the work ? A-8 B-7 C-6 D-5

Answer - D Explanation 1 women can complete the work in (10 * 8) days= 80 days.

1 child can complete the work in (16 * 10) days= 160 days. 1 women 1 days work = 1/80 , 1 child 1 days work= 1/160. (10 women +12 children) 1 days work = ( 10 * 1/80 +12 * 1/160) = ( 1/8 + 3/40 ) = 8/ 40 = 1 /5. ∴ 10 women and 12 children will finish the work in 5 days. Q 8 - If 6 engines consume 15 metric tonnes of coal when each is running 9 hours a days , how much coal will be required for 8 engines, each running 12 hours a days, it being given that 3 engines of former type consume as much as 4 engines of latter type? A - 17 tonnes B - 18 tonnes C - 19 tonnes D - 20 tonnes

Answer - D Explanation Let the required quantity of coal consumed be x tones. More engines, more coal consumption (direct) More hours, more coal consumption (direct) Less rate of consumption, less consumption (direct) Engines 6:8 Working Hrs 9:12 :: 15 : x Rate of consumption 1/3 : 1/4 ∴ (6 * 9 * 1/3 * x) = (8 * 12 * 1/4 * 15 ⇒ 18x = 360 ⇒ x = 20. Quantity of coal consumed = 20 tonnes.

Q 9 - If 22.5 m of a uniform rod weighs 85.5 kg , what will be the weight of 6m of the same rod? A - 22.8 kg B - 25.6 kg C - 26.5 kg D - 28kg

Answer - A Explanation Let the required weight be x kg. Less length, less weight (direct) 22.5: 6 :: 85.5 :x ⇒ 22.5x = (6 * 85.5) ⇒ x= (6 * 85.5) / 22.5 = (6 *885 / 225) = 22.8 kg. Required weight = 22.8 kg.

Q 10 - On a scale of map 1.5cm represents 24km. If the distance between two points on the map is 76.5 cm, the distance between these points is: A - 1071 km B - 1224 km C - 1377 km D - None of these

Answer - B Explanation Let the actual distance be x km. More distance on the map, more is actual distance (direct) 1.5 : 76.5 :: 24 : x ⇒ 1.5x = (76.5 * 24) ⇒ x = (76.5 * 24) / 1.5 = 1224 km. Required distance= 1224km. Q 11 - 6 dozen eggs are bought for Rs 48. How much will 132 eggs cost? A - Rs 78 B - Rs 80 C - Rs 82 D - Rs 88

Answer - D Explanation Let the required cost be Rs x. More eggs, more cost ( direct) 72: 132 :: 48 : x ⇒ 72 x = (132 * 48) ⇒ x= (132 *48) / 72 = 88. ∴ Required cost = Rs 88. Q 12 - In a race, Raghu cover 5 km in 20 minutes, how much distance will he cover in 50 minutes? A - 10.5 km B - 12 km

C - 12.5 km D - 13 km Answer - C Explanation Let the required distance be x km. More time , more distance covered ( direct) 20: 50: :: 5 : x ⇒ 20x = (50 * 5 ) ⇒ x= (50 * 5) / 20 = 12.5 km. Required distance = 12.5 km. Q 13 - A certain number of men can finish a piece of work in 100 days. If there were 10 men less, it would take 10 days more for the work to be finished. How many men were there originally? A - 75 B - 82 C - 100 D - 110

Answer - D Explanation Originally let there be x men. Less men, More days

(Indirect Proportion)

Therefore, (x-10) : x :: 100 :110

=> (x - 10) * 110 = x * 100 => x= 110 Q 14 - If 7 spiders make 7 webs in 7 days, then 1 spider will make 1 web in how many days? A-1 B-3 C-7 D - 14

Answer - C Explanation Let the required number days be x. Less spiders, More days (Indirect Proportion) Less webs, Less days (Direct Proportion) Spiders1:7Webs7:1}⋮⋮ 7:xSpiders1:7Webs7:1⋮⋮ 7:x => 1 * 7 * x = 7 * 1 * 7 => x= 7 Q 15 - 2 men and 7 boys can do a piece of work in 14 days; 3 men and 8 boys can do the same in 11 days. Then, 8 men and 6 boys can do three times the amount of this work in A - 18 Days

B - 21 Days C - 24 Days D - 30 Days

Answer - B Explanation

(2 x 14) men +(7 x 14) boys = (3 x 11) men + (8 x 11) boys =>5 men= 10 boys => 1man= 2 boys Therefore, (2 men+ 7 boys) = (2 x 2 +7) boys = 11 boys ( 8 men + 6 boys) = (8 x 2 +6) boys = 22 boys. Let the required number of days be x. More boys , Less days

(Indirect proportion)

More work , More days (Direct proportion) Boys22:11Work1 : 3}⋮⋮ 14:xBoys22:11Work1 : 3⋮⋮ 14:x Therefore, (22 * 1 * x) = (11 * 3 * 14) => x = 21 Hence, the required number of days = 21

Q 16 - 3 pumps working 8 hours a day, can empty a tank in 2 days. How many hours a day must 4 pumps work to empty the tank in 1 day? A-9 B - 10 C - 11 D - 12

Answer - D Explanation

Let the required no of working hours per day be x. More pumps , Less working hours per day (Indirect Proportion) Less days, More working hours per day

(Indirect Proportion)

Pumps4 : 3Days1 : 2}⋮⋮ 8:xPumps4 : 3Days1 : 2⋮⋮ 8:x => (4 * 1 * x) = (3 * 2 * 8) => x=12

Q 17 - 36 men can complete a piece of work in 18 days. In how many days will 27 men complete the same work ? A - 24 Days B - 28 Days C - 34 Days D - 35 Days

Answer - A Explanation Less Men, means more Days {Indirect Proportion} Let the number of days be x then, 27 : 36 :: 18 : x [Please pay attention, we have written 27 : 36 rather than 36 : 27, in indirect proportion, if you get it then chain rule is clear to you :)] =>27x = 36 * 18 => x = 24 So 24 days will be required to get work done by 27 men.

Q 18 - Some persons can do a piece of work in 12 days. Two times the number of such persons will do half of that work in A - 6 Days B - 12 Days C - 4 Days D - 3 Days

Answer - D Explanation

Let x men can do the in 12 days and the required number of days be z More men, Less days

[Indirect Proportion]

Less work, Less days

[Direct Proportion ]

men2x:xwork1:12} :: 12 : zmen2x:xwork1:12 :: 12 : z ∴(2x×1×z)=(x×12×12)∴2x×1×z=x×12×12 ⇒z=3 Days

Q 19 - If 36 men can do a piece of work in 25 hours, in how mwny hours will15 men do it? A - 40 B - 50 C - 60 D - 70

Answer - C Explanation Let the required no of hours be x. Then Less men , More hours (Indirct Proportion) Therefore, 15:36 ::25:x => (15 x X)=(36 x 25) => x = 60 Hence, 15 men can do it in 60 hours. Q 20 - If 18 binders bind 900 books in 10 days , How many binders will be required to bind 660 books in 12 days? A - 22 B - 14 C - 13

D - 11

Answer - D Explanation Let the required no of binders be X. Less books , Less binders

(Direct Proportion)

More days, Less binders

(Indirect proportion)

Books900:660Days12 : 10}⋮⋮ 18:xBooks900:660Days12 : 10⋮⋮ 18:x Therefore, (900 * 12 * x) = (660 * 10 * 18) => x = 11

Percentage

Percentage Percent means many hundredths.Example: z% is z percent which means z hundredths. It will be written as: z% = z⁄100 ⁄q as percent: (p⁄q x 100)%

p

Commodity If the price of a commodity increases by R%, then the reduction in consumption so as not to increase the expenditure is: [R⁄(100 + R)x 100]% If the price of a commodity decreases by R%, then the increase in consumption so as not to decrease the expenditure is: [R⁄(100 - R)x 100]%

Population The population of a city is P and let it increases at the rate of R% per annum: Population after t years: P(1 + R⁄100)t Population t years ago: P⁄(1 + R⁄100)t

Depreciation Let V be the present value of machine. Suppose it depreciates at the rate of R% per annum: Machine's value after t years:P(1 - R⁄100)t Machine's value t years ago: P⁄(1 - R⁄100)t



If P is R% more than Q, then Q is less than P by how many percent? [R⁄(100 + R)x 100]%



If P is R% more than Q, then Q is more than P by how many percent? [R⁄(100 - R)x 100]%

Sample Question Paper on Percentage Q 1 - What is fraction equivalent of 32%. A - 6/30 B - 8/25 C - 7/50 D - 11:10

Answer - B Explanation 32% = 32/100 = 8/25.

Q 2 - What is fraction equivalent of 160%. A - 8/5 B - 9/5 C - 6/7 D - 6/23

Answer - A Explanation

160% = 160/100 = 8/5

Q 3 - What is decimal equivalent of 0.8%. A - 0.008 B - 0.08 C - 0.8 D - 0.0008

Answer - A Explanation 0.8% = 0.8/100 = 0.008.

Q 4 - What is decimal equivalent of 18%. A - 0.0018 B - 0.18 C - 18 D - 0.018

Answer - B Explanation 18% = 18/100 = 0.18

Q 5 - What is decimal equivalent of 5%. A - 0.0005 B - 0.005 C - 0.05 D - 0.5

Answer - C Explanation 5% = 5/100 = 0.05

Q 6 - What is decimal equivalent of 126%. A - 1.26 B - 126 C - 12.6 D - 1260

Answer - A Explanation 126% = 126/100 = 1.26

Q 7 - What is decimal equivalent of 0.06%. A - 0.6 B - 0.06 C - 0.006 D - 0.0006

Answer - D Explanation 0.06% = 0.06/100 = 0.0006.

Q 8 - What is 3/4 as per cent? A - 45 B - 55 C - 65 D - 75

Answer - D Explanation 3/4= (3/4*100)% = 75%

Q 9 - 45% of 280 + 28% of 450 = ?. A - 352 B - 252 C - 452 D - 552

Answer - B Explanation 45% of 280+28% of 450 = (450/100*280)(28/100*450) = (126+126) = 252

Q 10 - Which is largest in 50/3%, 2/15, 0.18, 3/7 ? A - 0.18 B - 3/7 C - 2/15 D - 50/3%

Answer - B Explanation 162/3% = 50/(3*100) = 1/6 = 0.166, 2/15 =0.133, 3rd number = 0.18, 3/7 = 0.42. Clearly, 3/7 is largest.

Q 11 - 65% of a number is 21 less than 4/5 th of that number. Find the number. A - 140 B - 130 C - 120 D - 110

Answer - A Explanation Let the number be x. Then, (4/5 x)-(65% of x)= 21 => 4x/5-65x/100=21 => 4x/5-13x/20=21 => 16x-13x=420 =>3x=420 ?x=140. Required number = 140.

Q 12 - What per cent is 120 of 90? A - 400/3% B - 400/6% C - 200/3% D - 200/6%

Answer - A Explanation Required % = (120/90*100)%= 400/3%

Q 13 - What percent is 5gm of 1kg? A - 0.15% B - 0.05% C - 0.25% D - 0.35%

Answer - B Explanation Required % = (5/1000*100)%= 1/2%=0.05%

Q 14 - What per cent is 120 ml of 3.5 liters? A - 30/7% B - 20/7% C - 10/7% D - 1/7%

Answer - A Explanation Required % = (150/3500*100)%= 30/7%

Q 15 - If A's pay is 20% more than that of B, then what number of per penny is B's compensation not as much as that of A? A - 50/3% B - 60/3% C - 70/3% D - 80/3%

Answer - A Explanation B's salary is less than that of A by is less than that of a by {R/((100+R) )*100}% = {20/((120) )*100}% = 50/3%

Q 16 - If A's salary is 25% not as much as that of B, then what number of per penny is B's compensation more than that of A? A - 10/3% B - 100/3% C - 20/3 % D - 200/3 %

Answer - B Explanation B's salary is more than that of A by {R/((100-R) )*100}%

= {25/((100-25) )*100}% = (25/75*100)=100/3%

Q 17 - If the cost of tea is expanded by 20%, by what amount of percent must the utilization of tea be lessened so as not to build the consumption? A - 50/3% B - 100/3% C - 20/3% D - 200/3%

Answer - A Explanation Reduction % in consumption = {R/((100+R) )*100}%(20/120*100)% = 50/3%

Q 18 - If the cost of sugar falls by 10%, by how much per penny should a householder expand its utilization, so as not to diminish its use on sugar? A - 100/9 B - 10/9 C - 100/3 D - 10/3

Answer - A Explanation Increase % in consumption = {R/((100-R) )*100}% = {10/((100-10) )*100} = 100/9%

Q 19 - The tax on a commodity is diminished by 20% and its consumption increases by 15%. Find the effect on revenue. A - 8% increase B - 8% decrease C - 10% decrease D - 10% increase

Answer - B Explanation Let originally revenue obtained by Rs. x. New revenue = (Consumption * Tax) = (115% of 80% of Rs. x) = Rs(115/100*80/100*x)=Rs.92x/100 = 92% of the original. Hence, the revenue is decreased by 8%.

Q 20 - The population of a town is 176400. It increases annually at the rate of 5% p.a. What will be its population after 2 years? A - 194481 B - 294481 C - 394481 D - 494481

Answer - A Explanation Population after 2 years = {176400 * (1+5/100)2 } 176400* 21/20*21/20=194481.

Q 21 - 88% of 370 + 24% of 210 - ? = 118 A - 258 B - 298 C - 320 D - 18 Answer : A Explanation Let the unknown value is p 88% of 370 + 24% of 210 - p = 118

p = 325.60 + 50.40 - 118 = 258

Q 22 - If 120 is 20% of a number, then 120% of that number will be? A - 120 B - 620 C - 720 D - 20 Answer : C Explanation let z be the number. Then, 20% of z = 120 i.e (20⁄100 x z) = 120 Therefore, z = 600 120% of z = (120⁄100 x 600) = 720

Q 23 - A batsman scored 110 runs which included 3 boundaries and 8 sixes. What percent of his total score did he make by running between the wickets? A - 5⁄11% B - 45% C - 11 5⁄45% D - 45 5⁄11%

Answer : D Explanation Number of runs made by running = 110 - (3 x 4 + 8 x 6) = 50 Therefore, percentage = (50⁄100 x 100)% = 45 5⁄11%

Q 24 - 10% of the voters did not cast their votes in an election between two candidates. 10% of the votes polled were found invalid. The successful candidate for 54% of the valid votes and won by a majority of 1620 votes. Find the number of voters enrolled on the voters list? A - 27000 B - 32000 C - 25000 D - 28000 Answer : C Explanation Let z be the total number of voters. Votes polled = 90% of z Valid votes = (90% of (90% of z)) Therefore, 54% of [90% of (90% of z)] - 46% of [90% of (90% of z)] = 1620 → 8% of [90% of (90% of z)] = 1620 The number of voters enrolled were z = 25000

Q 25 - A sum of Rs 817 is divided among A, B and C such that A receives 25% more than B and B receives 25% less than C. What is the A share in the amount? A - 228 B - 247 C - 285 D - 304 Answer : C Explanation Let B receives Rs x ∴ A's Share = Rs 5x/4 C's Share = Rs 4x/3 According to question, x + 5x/4 + 4x/3 = 817 Or, (12x + 15x + 16x)/12 = 817 Or, x = (817*12)/43 = Rs 228 A's Share = 5x/4 = (5*228)/4= Rs 285

Q 26 - The length of a rectangle is increased by 50%. By what percent would the width have to be decreased to maintain the same area? A - 75/3% B - 78/3% C - 93/3% D - 100/3%

Answer : D Explanation Let length =100 m, breadth = 100m New length =150, new breadth = x m Then 150 * x =100*100 Or x =10000/150=200/3 Decrease in breadth = (100-200/3) =100/3%

Q 27 - Water tax is increased by 30% but its consumption is decreased by 20%. Then, the increase or decreased in the expenditure of the money is A - no change B - 5% increase C - 4% decrease D - 4 % increase Answer : D Explanation Let tax =Rs.100 and Consumption = 100 units Original expenditure = Rs.100*100=Rs.10000 New expenditure=Rs.(130*80) =Rs. 10400 Increase in expenditure by 4 %

Q 28 - The population of a town is 176400. It increases annually at the rate of 5% p.a. What was it 2 years ago? A - 160000 B - 260000 C - 360000 D - 460000 Answer : A Explanation Population 2 years ago = 176400/(1+5/100)2 =(176400*20/21*20/21)=160000.

Q 29 - ? % of 24 = 0.72 A-7 B-6 C-3 D-4 Answer : C Explanation Let x% of 24 = 0.72. Then, (x/100*24)=0.72= 72/100=>24x=72 =>x=3.

Q 30 - (43% of 2750) � (38% of 2990) = ? A - 43.6 B - 46.3 C - 44.7 D - 49.3 Answer : B Explanation (43% of 2750) � (38% of 2990) = 1182.5 - 1136.2 = 46.3

Speed & Distance

Formulas Speed = Distance⁄Time Time = Distance⁄Speed P km/hr = (P x 5⁄18)m/sec P m/sec = (P x 18⁄5)km/hr If the ratio of the speeds of P and Q is p:q, then the ratio of the times taken by them to cover the same distance is ⁄ : ⁄ or q : p

1 p 1 q

The average speed of the journey is (2pq⁄p+q)km/hr if a man covers a certain distance at p km/hr and an equal distance at q km/hr

Sample Question Paper on Speed & Distance

Q 1 - What is meters/sec for 54 km/hr? A - 15 m/sec B - 20 m/sec C - 25 m/sec D - 30 m/sec

Answer - A Explanation 54 km/hr = (54*5/18) m/sec = 15 m/sec.

Q 2 - What is km/hr for 16 m/sec? A - 53.6km/hr B - 55.6km/hr C - 57.6km/hr D - 59.6km/hr

Answer - C

Explanation 16 m/sec = (16*18/5)km/hr = 288/5km/hr =57.6 km/hr.

Q 3 - Anita can cover a sure separation in 1 hr 24min by covering two-third of the separation at 4km/hr and the rest at5 km/hr. Find the aggregate separation. A - 3 kms B - 4 kms C - 5 kms D - 6 kms

Answer - D Explanation Let the aggregate separation be x km. Then, 2/3x/4+1/3x/5=7/5 => x/6+x/15=7/5 => 5x+2x=42 => 7x=42 => x=6. ∴ Total separation = 6 kms Q 4 - A man strolls from his home to the railroad station. On the off chance that he strolls at 5 km/hr, he misses a train by 7 minutes. However, on the off chance that he strolls at 6km/hr, he achieves the station 5 minute before the flight of the train. Discover the separation secured by him to achieve the station.

A - 5 kms B - 6 kms C - 7 kms D - 8 kms

Answer - B Explanation Let the required separation be x km. At that point, X/5 - x/6 = 12/60 (distinction between two time interims is 12 min.) => x/5 - x/6 = 1/5 => 6x-5x=6 => x= 6 Required separation = 6 kms

Q 5 - Strolling at 7/8 of its typical velocity, a train is 10 minutes past the point of no return. Locate its standard time to cover the trip? A - 60 min B - 70 min C - 80 min D - 90 min

Answer - B Explanation New speed = 7/8 of its standard velocity

New time taken = 8/7 of the standard time. (8/7 of the standard time)- (common time) = 10 min. => 1/7 of the standard time = 10 min => usual time = 70 min.

Q 6 - Hitesh covers a sure separation via auto driving at 70 km/hr and returns back to the beginning stage riding on a bike at 55km/hr. locate his normal velocity for the entire trip? A - 62.6 km/hr B - 61.6 km/hr C - 60.6 km/hr D - 59.6 km/hr

Answer - B Explanation Normal velocity = 2xy/(x+y) km/hr = (2*70*55)/ (70+55) km/hr = (2*70*55)/125 km/hr = 308/5 km/hr = 61.6 km/hr

Q 7 - The separation between two stations A and B is 450 km. A train begins at 4 pm from A and moves towards B at a normal velocity of 60 km/hr. Another train begins from B at 3.20 p.m and moves towards A at a normal velocity of 80 km/hr. How a long way from A will the two train s meet and what time? A - 6:50 pm B - 5:50 pm C - 4:50 pm

D - 3:50 pm

Answer - A Explanation Assume two trains meet at x km from A (time taken by B to cover (450-x) km-(time taken by A to cover x km) = 40/60 => (450-x)/80 - x/60 = 40/60 ?3 (450-x) - 4x = 160 => 7x=1190 ∴ x = 170 In this way the two trains meet at a separation of 170 km from A. Time taken by A to cover 170 km = 170/60 = 2hrs 50 min. Along these lines, the two trains meet at 6:50 pm

Q 8 - A man cycles from A to B, a separation of 21 km in 1 hr 40 min. The street from A is level for 13 km and afterward it is tough to B. The man's normal rate on level is 15 km/hr. Locate his normal tough pace? A - 10 km/hr B - 11 km/hr C - 12 km/hr D - 13 km/hr

Answer - A Explanation Let the normal tough rate be x km/hr. at that point, 13/15 + 8/x = 5/3 =>8/x= (5/3-13/15) = 12/15 = 4/5

=>x = (8*5)/4 = 10 ∴ Normal tough rate = 10 km/hr Q 9 - A hoodlum is spotted by a policeman from a separation of 100 meters. At the point when the policeman begins the pursuit, the criminal likewise begin s running. In the event that the pace of the criminal be 8km/hr and that of the policeman 10 km/hr, how far the hoodlum will have keep running before he is overwhelmed? A - 200 m B - 300 m C - 400 m D - 500 m

Answer - C Explanation Relative pace of the policeman = (10-8) km/hr = 2 km/hr Time taken by policeman to cover 100 m = (100/1000*1/2) hr = 1/20 hr In 1/20 hr, the cheat covers a separation of (8*1/20) km= (2/5 km) = (2/5*1000) m = 400 m

Q 10 - I walk a sure separation and ride back setting aside an aggregate time of 37 minutes. I could walk both routes in 55 minutes. To what extent would it take me to ride both ways? A - 18 min B - 19 min

C - 20 min D - 21 min

Answer - B Explanation Let the given separation be x km. At that point, (Time taken to walk X km)+ (time taken to ride x km) = 37 min. => (time taken to walk 2X km) + (time taken to ride 2x km) = 74 min. => 55 min + (time taken to ride 2 x km) = 74 => time taken to ride 2x km = 19 min.

Q 11 - A and B are two stations 390 km separated. A train begins from An at 10 am and ventures towards B at 65 kmph. Another train begins from B at 11 am and towards A at 35 kmph. AT what time they meet? A - 2.15 pm B - 3.15 pm C - 4.15 pm D - 5.15 pm

Answer - A Explanation Assume they meet x hours after 10 am then, (separation moved by first in x hrs.) + (separation moved by second in (x-1) hrs) = 390

∴ 65x+ 35(x-1) = 390 => 100x =425 => x= 17/4 In this way, they meet 4 hrs 15 min. after 10 am at 2.15 pm

Q 12 - A products train leaves a station at a sure time and at a settled rate. Following 6 hours, an express prepare leaves the same station and moves in the same bearing at a uniform velocity of 90 kmph. This train makes up for lost time the merchandise train in 4 hours. Discover the velocity of the merchandise train. A - 33 kmph B - 34 kmph C - 35 kmph D - 36 kmph

Answer - B Explanation Let the velocity of the merchandise train be x kmph. Separation secured by products train in 10 hours = separation secured by express prepare in 4 hours ∴ 10x =4*90 or x = 36 Along these lines, rate of products train = 36 kmph

Q 13 - A man finishes 30 km of a voyage at 6km/hr and the staying 40km of the venture in 5 hr.His normal pace for the entire voyage is: A - 70/11 km/hr

B - 7 km/hr C - 15/2 km D - 8 km/hr Answer : B Explanation Total distance = (30+40)km= 70 km Total time taken = (30/6 +5) hrs =10 hrs Average speed = 70/10 km/hr = 7 km/hr

Q 14 - A is twice as quick as B and B is thrice as quick as C. The excursion secured by C in 42 min. will be secured by A in A - 7 min. B - 14 min. C - 28 min. D - 63 min. Answer : A Explanation Let c speed be x meters/min. Then, B speed=3x meters /min and A speed =6x meters/ min. Ratio of speed of A and C =ratio of times taken by C and A 6x:x=42:ymin⇒6x/x=42/y⇒y=42/6min=7 min.

Q 15 - If an understudy strolls from his home to class at 5km/hrs, he is late by 30 min. However, on the off chance that he strolls at 6 km/hr. he is late by 5 min. just. The separation of his school from his home is: A - 2.5 km B - 3.6 km C - 5.5 km D - 12.5 km Answer : D Explanation Let the required distance be x km. then, x/5 - x/6 = 25/7 (difference between two times is 25 min.) ⇒ 12x- 10 x = 25 ⇒2x = 25 ⇒ x= 25/2 km = 12.5 km Q 16 - A train secured a separation at a uniform velocity. On the off chance that the train had been 6 km/hr speedier, it would have taken 4 hours not exactly the booked time, and if the train were slower by 6 km/hr, the train would have taken 6 hours more than the planned time. The length of the trip is? A - 700 km B - 720 km C - 740 km D - 760 km Answer : B

Explanation Let the required distance be x km and uniform speed by y km/hr x/y - x/(y+6) = 4 ...(a) x/(y-6) - x/y = 6 ...(b) ⇒xy+6x-xy = 4y (y+6) and xy - xy +6x = 6y (y-6) ⇒4y2+24 y - 6x = 0 and 6y2- 36y - 6x = 0 ⇒2y2- 60 y = 0 ⇒ 2y (y-30) = 0 ⇒y = 30 ∴x/30 - x/36 = 4 ⇒6x- 5x = 720 ⇒x = 720 km Q 17 - A star is 8.1* 10ⁱ3km far from the earth. Assume light goes at the pace of 3.0* 10⁵ km for every second. To what extent will it take light from star to achieve the earth? A - 7.5 * 103 hrs B - 7.5 * 10⁴ hrs C - 2.7 * 10ⁱ⁰ sec D - 2.7 * 10ⁱⁱ sec Answer : B Explanation (3*10⁵) km is covered in 1 sec. (8.1 * 10ⁱ3) km is covered in (1/3 *10⁵* 8.1*10ⁱ3) sec = (2.7 *10⁸*1/60*1/60) hrs = (2.7*10⁶)/36 hrs= (2.7 *100*10⁴)/36 hrs = (7.5 *10⁴) hrs.

Q 18 - A constable is 114 m behind a thief. The constable runs 21 m and the thief15 m in a moment. In what the reality of the situation will become obvious eventually constable catch the criminal? A - 16 min B - 17 min C - 18 min D - 19 min Answer : D Explanation (21-15) m i.e.6m is covered in 1 min. 114m will be covered in (1/6*114) min=19 min.

Q 19 - Renu began cycling along the limits of a square field ABCD from corner point A. after thirty minutes, he came to the corner point C, slantingly inverse to A. In the event that his rate was 8 km/hr, the zone of the field is: A - 64 sq. km B - 8 sq. km C - 4 sq. km D - cannot be resolved Answer : B Explanation

Length of diagonal = (8*1/2 )km = 4 km Area of the field = (1/2 *4*4) sq. km = 8 sq. km

Q 20 - The proportion between the rates of strolling of A and B is 2:3. In the event that the time taken by B to cover a sure separation is 36 minutes, the time taken by A to cover that much separation is A - 24 min B - 54 min C - 48 min D - none of these Answer : B Explanation Ratio of time taken = 1/2:1/3 = 3:2 Time taken by B = 36 min. let the time taken by A be x min. ∴x/36 = 3/2 ⇒x = (3*36/2) min. = 54 min Q 21 - The proportion between the rates of going of An and B is 2:3 and in this manner A takes 10 minute more than the time taken by B to achieve a destination. In the event that A had strolled at twofold speed, he would have secured the separation in A - 30 min B - 25 min

C - 20 min D - 15 min Answer : D Explanation Ratio of time taken by A and B = 1/2: 1/3 Suppose B takes x min. Then, A takes (x+10) min. ∴(x+10): x= 1/2: 1/3 = 3:2 ⇒ x+10/x = 3/2 ⇒ 2x+20 = 3x ⇒x = 20 Thus B takes 20 min. and A takes 30 min. AT double speed A would covers it in 15 min.

Q 22 - A agriculturist voyaged a separation of 61 km in 9 hours. He voyaged halfway by walking at 4 km/hr and incompletely on bike at 9 km/hr. The separation went by walking is: A - 14 km B - 15 km C - 16 km D - 17 km Answer : C Explanation Let the distance travelled on foot be x km Then, distance covered on bicycle = (61-x) km

∴x/4 + 61-x/9 =9 ⇒ 9x+4(61-x)= 324 ⇒ 5x = (324-244)= 80 ⇒x = 16 Distance covered on foot = 16 km

AVERAGE

Average = (Sum of Observations⁄Number of Observations) Average Speed Suppose a man covers a certain distance at m kmph and an equal distance at y kmph. Then,:

Average speed (during the whole journey) = (2xy⁄(x + y))kmph

Sample Question Paper on AVERAGE Q 1 - The average of 20 numbers is zero. Of them, at the most, how many may be greater than zero? A - 19 B - 10 C-0 D-1

Answer - A Explanation Average of 20 numbers = 0. Therefore Sum of 20 numbers = (0 x 20) = 0 It is quite possible that 19 of these numbers may be positive and if their sum is a, then 20th number is (-a).

Q 2 - Find the average of all the numbers between 6 and 34 which are divisible by 5? A - 30 B - 24

C - 20 D -18

Answer - C Explanation Average = (10 + 15 + 20 + 25 + 30)⁄5 = 100⁄5 = 20.

Q 3 - The average of first five multiples of 3 is? A - 15 B - 12 C-3 D-9

Answer - D Explanation Average = 3(1 + 2 + 3 + 4 + 5)⁄5 = 45⁄5 = 9

Q 4 - The average of first nine prime numbers is? A - 10 B - 111⁄9 C-9 D - 112⁄9

Answer - B Explanation Average = (2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23)⁄9 = 100⁄9 = 111⁄9

Q 5 - A student was asked to find the arithmetic mean of the numbers 3, 11, 7, 9, 15, 13, 8, 19, 17, 21, 14, and z? A-3 B-7 C - 17 D - 31

Answer - B Explanation Clearly, we have (3 + 11 + 7 + 9 + 15 + 13 + 8 + 19 + 17 + 21 + 14 + z)⁄12 = 12 or 137 + z = 144 or z = 144 - 137 = 7.

Q 6 - If the mean of 5 observation z, z + 2, z + 4, z + 6 and z + 8 is 11, then the mean of the last three observation is? A - 11 B - 13 C - 15 D - 17

Answer - B Explanation we have : (z + (z + 2) + (z + 4) + (z + 6) + (z + 8))⁄5 = 11 or 5z + 20 = 55 or z = 7. So the numbers are 7, 9, 11, 13, 15. therefore required mean = (11 + 13 + 15⁄3) = 39⁄3 = 13.

Q 7 - The average of the two-digit numbers, which remain the same when the digits interchange their positions, is? A - 55 B - 33 C - 44 D - 66

Answer - A Explanation Average = (11 + 22 + 33 + 44 + 55 + 66 + 77 + 88 + 99⁄9) ((11 + 99) + (22 + 88) + (33 + 77) + (44 + 66) +55⁄9) (4 x 110 + 55⁄9) (495⁄9) = 55

Q 8 - The average of a non-zero number and its square is 5 times the number. The number is? A-9 B - 17 C - 29 D - 295

Answer - A Explanation Let the number be z. then, z+z2 ⁄2 = 5z 2 = z - 9z = 0 z (z - 9) = 0 z = 0 or z = 9 so the number is 9.

Q 9 - The average of 7 consecutive number is 20. The largest of these numbers is? A - 20 B - 22 C - 23 D - 24

Answer - C Explanation Let the number be z, z + 1, z + 2, z + 3 ,z + 4, z + 5 ,z + 6. then, (z + (z + 1) + (z + 2) + (z + 3) + (z + 4) + (z + 5) + (z + 6)) ⁄7 = 20 7z + 21 = 140 or 7z = 119 or z = 17 Largest number = z + 6 = 17 + 6 = 23

Q 10 - The average of five consecutive odd numbers is 61. What is the difference between the highest and lowest numbers? A-9 B-8 C - 10 D - 11

Answer - B Explanation Let the number be z, z + 2, z + 4, z + 6 and z + 8. then, (z + (z + 2) + (z + 4) + (z + 6) + (z + 8))⁄5 = 61 5z + 20 = 305 or z = 57 so the required number is = (57 + 8) - 57 = 8

Q 11 - The sum of three consecutive odd numbers is 38 more than the average of these numbers. What is the first of these numbers? A - 17 B - 13 C - 19 D - none

Answer - A Explanation Let the number be z, z + 2, and z + 4. then, (z + z + 2 + z + 4) - (z + z + 2 + z + 4)⁄3 = 38 2(3z + 6) = 114 or 6z = 102 or z = 17.

Q 12 - The average age of the boys in a class is 16 years and that of the girls is 15 years. The average age for the whole class is A - 15 years B - 15.5 years C - 16 years D - Cannot be computed with the given information

Answer - D Explanation Clearly to find the average we ought to know the number of boys, girls or students in the class neither of which is given. So, data is inadequate.

Q 13 - The average annual income (in Rs.) of certain agricultural workers is S and that of other workers is T. The number of agriculture workers is 11 times that of other workers. Then the average monthly income (in Rs.) of all the workers is? A - S + T⁄2 B - 11S + T⁄12 C - 1 +⁄11ST D - S + 11T⁄2

Answer - B Explanation Let the number of other workers be z. then, number of agricultural workers = 11z Total number of workers = 12z Therefore Average monthly salary = S x 11z + T x z⁄12z = 11S + T⁄12

Q 14 - A family consists of grandparents, parents and three grandchildren. The average age of the grandparents is 67 years, that of the parent is 35 years and that of the grandchildren is 6 years. What is the average age of the family? A - 284⁄7 B - 315⁄7 C - 321⁄7

D - none

Answer - B Explanation Required average = (67 x 2 + 35 x 2 + 6 x 3⁄2 + 2 + 3) = (134 + 70 + 18⁄7) = 222⁄7 = 315⁄7

Q 15 - A library has an average of 510 visitors on Sundays and 240 on other days. The average number of visiters per day in a month of 30 days begining with a Sunday is? A - 276 B - 280 C - 285 D - 250

Answer - C Explanation Since the month begins with a sunday, so there will be five sundays in the month Therefore Required average = (510 x 5 + 240 x 25⁄30) = 8550⁄30 = 285

Q 16 - If the average marks of three batches of 55, 60 and 45 students respectively is 50, 55 and 60, then the average marks of all the students is? A - 53.33 B - 54.68 C - 55 D - none

Answer - B Explanation Required average = (55 x 50 + 60 x 55 + 45 x 60⁄55 + 60 + 45) (2750 + 3300 + 2700⁄160) = (8750⁄160) = 54.68

Q 17 - The average weight of 16 boys in a class is 50.25 kgs and that of the remaining 8 boys is 45.15 kgs. Find the average weight of all the boys in the class? A - 48.55 B - 49.25 C - 45 D - 47

Answer - A Explanation Required average = (50.25 x 16 + 45.15 x 8⁄16 + 8) (804 + 361.20⁄24) = (1165.20⁄24) = 48.55

Q 18 - A car owner buys petrol at Rs. 7.50, Rs. 8 and Rs. 8.50 per litre for three successive years. What approximately is the average cost per litre of petrol if he spends Rs. 4000 each year? A - 7.98 B-8 C - 8.50 D-9

Answer - A Explanation Total quantity of petrol consumed in 3 years. = (4000⁄7.50 + 4000⁄8 + 4000⁄8.50) litres = 4000 2⁄15 + 1⁄8 2⁄17 = 76700⁄51 litres Total amount spent = Rs. (3 x 4000) = Rs 12000 Therefore Average cost = Rs. (12000 x 51⁄76700) = Rs. 6120⁄767 = Rs. 7.98.

Q 19 - The average of six numbers is z and the average of three of these is y. If the average of the remaining three is w, then? A - 2z = 2y + 2w B - z = 2y + 2w C-z=y+w D - 2z = y + w

Answer - D Explanation Clearly, we have: z = 3y + 3w⁄6 or 2z = y + w.

Q 20 - Out of 9 persons, 8 persons spent Rs. 30 each for their meals. The ninth one spent Rs. 20 more than the average expenditure of all the nine. The total money spent by all of them was? A - 290 B - 260 C - 292.50 D - 400.50

Answer - C Explanation Let the average expenditure be Rs z then, 9z = 8 x 30 + (z + 20) or 9z = z + 260 or 8z = 260 or z = 32.50. Therefore total money spent = 9z = Rs. (9 x 32.50) = Rs. 292.50.

Number Systems

In Decimal number system, there are ten symbols namely 0,1,2,3,4,5,6,7,8 and 9 called digits. A number is denoted by group of these digits called as numerals.

Face Value Face value of a digit in a numeral is value of the digit itself. For example in 321, face value of 1 is 1, face value of 2 is 2 and face value of 3 is 3.

Place Value Place value of a digit in a numeral is value of the digit multiplied by 10n where n starts from 0. For example in 321: 

Place value of 1 = 1 x 100 = 1 x 1 = 1



Place value of 2 = 2 x 101 = 2 x 10 = 20



Place value of 3 = 3 x 102 = 3 x 100 = 300 0th position digit is called unit digit and is the most commonly used topic in aptitude tests.

Types of Numbers 1. Natural Numbers - n > 0 where n is counting number; [1,2,3...] 2. Whole Numbers - n ≥ 0 where n is counting number; [0,1,2,3...]. 0 is the only whole number which is not a natural number. Every natural number is a whole number.

3. Integers - n ≥ 0 or n ≤ 0 where n is counting number;...,-3,-2,-1,0,1,2,3... are integers. o

Positive Integers - n > 0; [1,2,3...]

o

Negative Integers - n < 0; [-1,-2,-3...]

o

Non-Positive Integers - n ≤ 0; [0,-1,-2,-3...]

o

Non-Negative Integers - n ≥ 0; [0,1,2,3...]

0 is neither positive nor negative integer. 4. Even Numbers - n / 2 = 0 where n is counting number; [0,2,4,...] 5. Odd Numbers - n / 2 ≠ 0 where n is counting number; [1,3,5,...] 6. Prime Numbers - Numbers which is divisible by themselves only apart from 1. 1 is not a prime number. To test a number p to be prime, find a whole number k such that k > √p. Get all prime numbers less than or equal to k and divide p with each of these prime numbers. If no number divides p exactly then p is a prime number otherwise it is not a prime number. Example: 191 is prime number or not? Solution: Step 1 - 14 > √191 Step 2 - Prime numbers less than 14 are 2,3,5,7,11 and 13.

Step 3 - 191 is not divisible by any above prime number. Result - 191 is a prime number. Example: 187 is prime number or not? Solution: Step 1 - 14 > √187 Step 2 - Prime numbers less than 14 are 2,3,5,7,11 and 13. Step 3 - 187 is divisible by 11. Result - 187 is not a prime number. 7. Composite Numbers - Non-prime numbers > 1. For example, 4,6,8,9 etc. 1 is neither a prime number nor a composite number. 2 is the only even prime number. 8. Co-Primes Numbers - Two natural numbers are co-primes if their H.C.F. is 1. For example, (2,3), (4,5) are co-primes. Divisibility Following are tips to check divisibility of numbers. 1. Divisibility by 2 - A number is divisible by 2 if its unit digit is 0,2,4,6 or 8. Example: 64578 is divisible by 2 or not? Solution: Step 1 - Unit digit is 8. Result - 64578 is divisible by 2. Example: 64575 is divisible by 2 or not? Solution: Step 1 - Unit digit is 5.

Result - 64575 is not divisible by 2. 2. Divisibility by 3 - A number is divisible by 3 if sum of its digits is completely divisible by 3. Example: 64578 is divisible by 3 or not? Solution: Step 1 - Sum of its digits is 6 + 4 + 5 + 7 + 8 = 30 which is divisible by 3. Result - 64578 is divisible by 3. Example: 64576 is divisible by 3 or not? Solution: Step 1 - Sum of its digits is 6 + 4 + 5 + 7 + 6 = 28 which is not divisible by 3. Result - 64576 is not divisible by 3. 3. Divisibility by 4 - A number is divisible by 4 if number formed using its last two digits is completely divisible by 4. Example: 64578 is divisible by 4 or not? Solution: Step 1 - number formed using its last two digits is 78 which is not divisible by 4. Result - 64578 is not divisible by 4. Example: 64580 is divisible by 4 or not? Solution: Step 1 - number formed using its last two digits is 80 which is divisible by 4. Result - 64580 is divisible by 4. 4. Divisibility by 5 - A number is divisible by 5 if its unit digit is 0 or 5.

Example: 64578 is divisible by 5 or not? Solution: Step 1 - Unit digit is 8. Result - 64578 is not divisible by 5. Example: 64575 is divisible by 5 or not? Solution: Step 1 - Unit digit is 5. Result - 64575 is divisible by 5. 5. Divisibility by 6 - A number is divisible by 6 if the number is divisible by both 2 and 3. Example: 64578 is divisible by 6 or not? Solution: Step 1 - Unit digit is 8. Number is divisible by 2. Step 2 - Sum of its digits is 6 + 4 + 5 + 7 + 8 = 30 which is divisible by 3. Result - 64578 is divisible by 6. Example: 64576 is divisible by 6 or not? Solution: Step 1 - Unit digit is 8. Number is divisible by 2. Step 2 - Sum of its digits is 6 + 4 + 5 + 7 + 6 = 28 which is not divisible by 3. Result - 64576 is not divisible by 6. 6. Divisibility by 8 - A number is divisible by 8 if number formed using its last three digits is completely divisible by 8. Example: 64578 is divisible by 8 or not? Solution: Step 1 - number formed using its last three digits is 578

which is not divisible by 8. Result - 64578 is not divisible by 8. Example: 64576 is divisible by 8 or not? Solution: Step 1 - number formed using its last three digits is 576 which is divisible by 8. Result - 64576 is divisible by 8. 7. Divisibility by 9 - A number is divisible by 9 if sum of its digits is completely divisible by 9. Example: 64579 is divisible by 9 or not? Solution: Step 1 - Sum of its digits is 6 + 4 + 5 + 7 + 9 = 31 which is not divisible by 9. Result - 64579 is not divisible by 9. Example: 64575 is divisible by 9 or not? Solution: Step 1 - Sum of its digits is 6 + 4 + 5 + 7 + 5 = 27 which is divisible by 9. Result - 64575 is divisible by 9. 8. Divisibility by 10 - A number is divisible by 10 if its unit digit is 0. Example: 64575 is divisible by 10 or not? Solution: Step 1 - Unit digit is 5. Result - 64578 is not divisible by 10. Example: 64570 is divisible by 10 or not? Solution:

Step 1 - Unit digit is 0. Result - 64570 is divisible by 10. 9. Divisibility by 11 - A number is divisible by 11 if difference between sum of digits at odd places and sum of digits at even places is either 0 or is divisible by 11. Example: 64575 is divisible by 11 or not? Solution: Step 1 - difference between sum of digits at odd places and sum of digits at even places = (6+5+5) - (4+7) = 5 which is not divisible by 11. Result - 64575 is not divisible by 11. Example: 64075 is divisible by 11 or not? Solution: Step 1 - difference between sum of digits at odd places and sum of digits at even places = (6+0+5) - (4+7) = 0. Result - 64075 is divisible by 11. Tips on Division 1. If a number n is divisible by two co-primes numbers a, b then n is divisible by ab. 2. (a-b) always divides (an - bn) if n is a natural number. 3. (a+b) always divides (an - bn) if n is an even number. 4. (a+b) always divides (an + bn) if n is an odd number. Division Algorithm When a number is divided by another number then

Dividend = (Divisor x Quotient) + Reminder Series Following are formulaes for basic number series: 1. (1+2+3+...+n) = (1/2)n(n+1) 2. (12+22+32+...+n2) = (1/6)n(n+1)(2n+1) 3. (13+23+33+...+n3) = (1/4)n2(n+1)2

Basic Formulaes These are the basic formulae: (a + b)2 = a2 + b2 + 2ab (a - b)2 = a2 + b2 - 2ab (a + b)2 - (a - b)2 = 4ab (a + b)2 + (a - b)2 = 2(a2 + b2) (a2 - b2) = (a + b)(a - b) (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) (a3 + b3) = (a + b)(a2 - ab + b2) (a3 - b3) = (a - b)(a2 + ab + b2) (a3 + b3 + c3 - 3abc) = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)

Sample Question Paper on Number Systems Q 1 - Which of the following is a prime number? A - 187 B - 811 C - 341 D - 437

Answer - B Explanation Step 1. Find a whole number k such that k2 > n for each number. 142 > 187. 302 > 811. 192 > 341. 212 > 437. Step 2. Get all prime numbers which are < k 14 - 2 , 3, 5, 7, 11, 13 30 - 2 , 3, 5, 7, 11, 13, 17, 19, 23, 29 19 - 2 , 3, 5, 7, 11, 13, 17 21 - 2 , 3, 5, 7, 11, 13, 17, 19 Step 3. Check divisiblity of each number

with prime numbers which are < k. 187 is divisible by 11. 811 is not divisible by any prime number. 341 is divisible by 11. 437 is divisible by 19. Result: 811 is the prime number.

Q 2 - Which of the following is the output of 6894 x 99 ? A - 685506 B - 682506 C - 683506 D - 684506

Answer - B Explanation 6894 x 99 = 6894 x (100 - 1) = 6894 x 100 - 6894 x 1 = 689400 - 6894 = 682506

Q 3 - Which of the following is the output of 685798 x 125 ? A - 85724750 B - 82257509 C - 82259503 D - 82247608

Answer - A Explanation 685798 x 125 = 685798 x 53 = 685798 x (10/2)3 = (685798 x 103) / 23 = 685798000 / 8 = 85724750

Q 4 - Which of the following is the output of 43986 x 625 ? A - 27491450 B - 27491350 C - 27491250 D - 27491750

Answer - C Explanation 43986 x 625 = 43986 x 54 = 43986 x (10/2)4 = (43986 x 104) / 24 = 439860000 / 16 = 27491250

Q 5 - Which of the following is the output of 869 x 738 + 869 x 262 ? A - 262000 B - 738000 C - 969000 D - 869000

Answer - D Explanation 869 x 738 + 869 x 262 = 869 x (738 + 262) = 869 x 1000 = 869000

Q 6 - Which of the following is the output of 936 x 587 - 936 x 487 ? A - 93600 B - 58700 C - 48700 D - 100

Answer - A Explanation 936 x 587 - 936 x 487 = 936 x (587 - 487) = 936 x 100 = 93600

Q 7 - Which of the following is the output of 1496 x 1496 ? A - 3338016 B - 2238016 C - 2248016 D - 2258016

Answer - B Explanation 1496 x 1496 = 14962 = (1500-4)2 = 15002 + 42 - 2 x 1500 x 4 = 2250000 + 16 - 12000 = 2238016 We've used following formula here: (a-b)2 = a2 + b2 - 2ab.

Q 8 - Which of the following is the output of 1607 x 1607 ? A - 2581449 B - 2583449 C - 2582449 D - 2584449

Answer - C Explanation 1607 x 1607 = 16072 = (1600+7)2

= 16002 + 72 + 2 x 1600 x 7 = 2560000 + 49 +22400 = 2582449 We've used following formula here: (a+b)2 = a2 + b2 + 2ab.

Q 9 - Which of the following is the output of 596 x 596 - 104 x 104 ? A - 377700 B - 366600 C - 355500 D - 344400

Answer - D Explanation 596 x 596 - 104 x 104 = 5962 - 1042 = (596 + 104) x (596 - 104) = 700 x 492 = 344400 We've used following formula here: a2 - b2 = (a + b)(a - b).

Q 10 - Which of the following is the output of 57 x 57 + 43 x 43 + 2 x 57 x 43 ? A - 10000 B - 5700 C - 4300 D - 1000

Answer - A Explanation 57 x 57 + 43 x 43 + 2 x 57 x 43 = (57 + 43)2 = (100)2 = 10000 We've used following formula here: (a + b)2 = a2 + b2 + 2ab.

Q 11 - Which of the following is the output of 93 x 93 + 73 x 73 - 2 x 93 x 73 ? A - 200 B - 400 C - 300 D - 100

Answer - B Explanation 93 x 93 + 73 x 73 - 2 x 93 x 73 = (93 - 73)2 = (20)2 = 400 We've used following formula here: (a - b)2 = a2 + b2 - 2ab.

Q 12 - Which of the following is the output of (578 x 578 x 578 + 432 x 432 x 432) / (578 x 578 - 578 x 432 + 432 x 432) ? A - 2000 B - 4000 C - 3000 D - 1000

Answer - D Explanation (578 x 578 x 578 + 432 x 432 x 432) / (578 x 578 - 578 x 432 + 432 x 432) Let's have a = 578, b = 432 Now expression is (a3 + b3) / (a2 - ab + b2)

=a+b = 578 + 432 = 1000 We've used following formula here: a3 + b3 = (a + b)(a2 - ab + b2).

Q 13 - Which of the following is the output of (141 x 141 x 141 - 58 x 58 x 58) / (141 x 141 + 141 x 58 + 58 x 58) ? A - 83 B - 100 C - 90 D - 73

Answer - A Explanation (141 x 141 x 141 - 58 x 58 x 58) / (141 x 141 + 141 x 58 + 58 x 58) Let's have a = 141, b = 58 Now expression is (a3 - b3) / (a2 + ab + b2) =a-b = 141 - 58 = 83 We've used following formula here:

a3 - b3 = (a - b)(a2 + ab + b2).

Q 14 - Which of the following is the output of 213 x 213 + 187 x 187 ? A - 50338 B - 80338 C - 90338 D - 70338

Answer - B Explanation 213 x 213 + 187 x 187 Let's have a = 213, b = 187 Now expression is a2 + b2 Using following formula, (a + b)2 + (a - b)2 = 2 x (a2 + b2) 2 x ( 213 x 213 + 187 x 187) = (213 + 187)2 + (213 - 187)2 2 x ( 213 x 213 + 187 x 187) = 4002 + 262 2 x ( 213 x 213 + 187 x 187) = 160000 + 676 213 x 213 + 187 x 187 = 160676 / 2 = 80338

Q 15 - Which of the following is the output of ((637 + 478)2 - (637 - 478)2) /(637 x 478) ? A-8 B-6 C-4 D -24

Answer - C Explanation ((637 + 478)2 - (637 - 478)2)/(637 x 478) Let's have a = 637, b = 478 Now expression is ((a + b)2 - (a - b)2) / ab = (a2 + b2 + 2ab - (a2 + b2 - 2ab)) / ab = (a2 + b2 + 2ab - a2 - b2 + 2ab) / ab = 4ab / ab =4 We've used following formulae here: (a + b)2 = a2 + b2 + 2ab. (a - b)2 = a2 + b2 - 2ab.

Q 16 - Which of the following is the output of ((964 + 578)2 + (964 - 578)2) /(964 x 964 + 578 x 578) ? A-4 B-6 C-8 D-2

Answer - D Explanation ((964 + 578)2 + (964 - 578)2) /(964 x 964 + 578 x 578) Let's have a = 964, b = 578 Now expression is ((a + b)2 + (a - b)2) / (a2 + b2) = (a2 + b2 + 2ab + (a2 + b2 - 2ab)) / (a2 + b2) = (a2 + b2 + 2ab + a2 + b2 - 2ab) / (a2 + b2) = 2(a2 + b2) / (a2 + b2) =2 We've used following formulae here: (a + b)2 = a2 + b2 + 2ab. (a - b)2 = a2 + b2 - 2ab.

Q 17 - On dividing a number by 342, 47 is the remainder. What will be remainder if same number is divided by 18? A - 11 B-6 C-8 D-2

Answer - A Explanation Let's quotient is a and given number be b. b = 342a + 47 = (18 x 19)a + 36 + 11 = (18 x 19)a + (18 x 2) + 11 = 18 x (19a + 2) + 11 Thus, if same number is divided by 18, remainder will be 11. We've used following formulae here: Dividend = (Divisor x Quotient) + Reminder

Q 18 - What will be unit digit in (3157)754? A-8 B-9

C-7 D-6

Answer - B Explanation unit digit in (3157)754 = unit digit in (7)754 = unit digit in (74)188 x 72 = unit digit in (1 x 49) =9 Thus Unit digit in (3157)754 is 9. We've used following formulae here: Unit digit in 71 = 7 Unit digit in 72 = 9 Unit digit in 73 = 3 Unit digit in 74 = 1 Unit digit in 75 = 7 Unit digit in 76 = 9 Unit digit in 77 = 3 Unit digit in 78 = 1 So pattern is 7-9-3-1. This pattern works for all numbers. So Unit digit in ((7)4)n) will be 1.

Q 19 - What will be unit digit in 658 x 539 x 436 x 312? A-8 B-9 C-4 D-6

Answer - C Explanation Multiply unit digits of each number. Unit digit in 658 x 539 x 436 x 312 = Unit digit in 8 x 9 x 6 x 2. = Unit digit in 864. = 4.

Q 20 - What will be unit digit in 357 x 641 x 763? A-8 B-9 C-4 D-6

Answer - C Explanation 357 = (34)14 x 3 So Unit digit in 357 = Unit digit in 1 x 3 =3 641 = (64)10 x 6 So Unit digit in 641 = Unit digit in 6 x 6 =6 763 = (74)15 x 73 So Unit digit in 761 = Unit digit in 1 x 343 =3 So Unit digit in 357 x 641 x 763 = Unit digit in 3 x 6 x 3 =4 We've used following formulae here: Unit digit in 34 = 1 Unit digit in 64 = 6 Unit digit in 74 = 1 So Unit digit - in ((3)4)n) will be 1. - in ((6)4)n) will be 6. - in ((7)4)n) will be 1.

Time & Work

1. If A can do a piece of work in n days, then A's 1 day work = 1/n. 2. If A's 1 day work = 1/n, then A can finish the work in n days. 3. If A is twice as good a workmen as B, then o

Ratio of work done by A and B in the same time = 2:1

o

Ratio of time taken by A and B in doing the same work = 1:2

Sample Question Paper on Time & Work Q 1 - A can do a bit of work in 8 days, which B alone can do in 10 days in how long . In how long both cooperating can do it? A - 40/9 days B - 41/9 days C - 42/9 days D - 43/9 days

Answer - A Explanation A's 1 day work= 1/8, B`s 1 day work = 1/10 ∴ (A+B) 1 day work = (1/8+1/10) = 9/40 Both cooperating can complete it in 40/9 days. Q 2 - A and B together can dive a trench in 12 days, which an alone can dive in 30 days. In how long B alone can burrow it? A - 18 days B - 19 days

C - 20 days D - 21 days

Answer - C Explanation (A+B)'s 1 day work = 1/12, A's 1 day work =1/30 ∴ B's 1 day work = (1/12-1/30) = 3/60 = 1/20 Henceforth, B alone can dive the trench in 20 days. Q 3 - A can do a bit of work in 25 days which B can complete in 20 days. Both together labor for 5 days and afterward A leaves off. How long will B take to complete the remaining work? A - 7 days B - 8 days C - 9 days D - 11 days

Answer - D Explanation (A+B)'s 5 days work = 5(1/25+1/20) = (5*9/100) = 9/20 Remaining work = (1-9/20) = 11/20

1/20 work is finished by B in 1 day 11/20 work is finished by B in (1*20*11/20) = 11 days

Q 4 - A and B can do a bit of work in 12 days. B and C can do it in 15 days while C and A can do it in 20 days. In how long will they complete it cooperating? Additionally, in how long can A alone do it? A - 10 days, 30 days. B - 15 days, 20 days. C - 20 days, 40 days. D - 10 days, 50 days.

Answer - A Explanation (A+B)'s 1 day work = 1/12, (B+C)'s 1 day work = 1/15, (C+A)'S 1 day work = 1/20 Including: 2(A+B+C)'s 1 day work = (1/12+ 1/15+ 1/20)= 12/60 = 1/5 ∴ (A+B+C) `s 1 day work = (1/2 *1/5) = 1/10 ∴ working together they can complete the work in 10 days. A's 1 day work = (1/10-1/15) = 1/30, B`s 1 day work = (1/10-1/20) = 1/20 C's 1 day work = (1/10-1/12) = 1/60 ∴ A alone can take the necessary steps in 30 days.

Q 5 - A can fabricate a divider in 30 days , while B alone can assemble it in 40 days, If they construct it together and get an installment of RS. 7000, what B's offer? A - 2000 B - 3000 C - 4000 D - 6500

Answer - B Explanation A's 1 days work = 1/30, B's 1 day work = 1/40, Proportion of their shares = 1/30:1/40 = 4:3 B's offer = (7000*3/7) = Rs. 3000

Q 6 - A can do a bit of work in 10 days while B alone can do it in 15 days. They cooperate for 5 days and whatever remains of the work is finished by C in 2 days. On the off chance that they get Rs. 4500 for the entire work, by what means if they partition the cash? A - Rs 1250, Rs 1200, Rs 550 B - Rs 2250, Rs 1500, Rs 750

C - Rs 1050, Rs 1000, Rs 500 D - Rs 650, Rs 700, Rs 500

Answer - B Explanation (A+B)'s 5 days work = 5(1/10+ 1/15)= (5* 1/6)= 5/6 Remaining work = (1-5/6) = 1/6 C's 2 days work = 1/6 (A's 5 day work): (B's 5 day work): (C's 2 days work) = 5/10: 5/15: 1/6 = 15: 10:5 = 3:2:1 A's offer = (4500*3/6) = Rs. 2250 B's offer = (4500*2/6) = Rs. 1500 C's share= (4500*1/6) = Rs. 750

Q 7 - A, B and C can do a piece of work in 24 days, 30 days and 40 days respectively. They began the work together but C left 4 days before the completion of the work. In how many days was the work completed? A - 11 Days B - 12 Days C - 13 Days D - 14 Days

Answer: A) 11 days Explanation One day's work of A, B and C = (1/24 + 1/30 + 1/40) = 1/10. C leaves 4 days before completion of the work, which means only A and B work during the last 4 days. Work done by A and B together in the last 4 days = 4 (1/24 + 1/30) = 3/10. Remaining Work = 7/10, which was done by A,B and C in the initial number of days. Number of days required for this initial work = 7 days. Thus, the total numbers of days required = 4 + 7 = 11 days.

Q 8 - P can complete a work in 12 days working 8 hours a day.Q can complete the same work in 8 days working 10 hours a day. If both p and Q work together,working 8 hours a day,in how many days can they complete the work? A - 60/11 B - 61/11 C - 71/11 D - 72/11

Answer: A) 60/11 Explanation P can complete the work in (12 x 8) hrs = 96 hrs Q can complete the work in (8 x 10) hrs=80 hrs Therefore, P's 1 hour work=1/96 and Q's 1 hour work= 1/80 (P+Q)'s 1 hour's work =(1/96) + (1/80) = 11/480. So both P and Q will finish the work in 480/11 hrs Therefore, Number of days of 8 hours each = (480/11) x (1/8) = 60/11

Q 9 - A and B can do a piece of work in 30 days , while B and C can do the same work in 24 days and C and A in 20 days . They all work together for 10 days when B and C leave. How many days more will A take to finish the work? A - 18 Days B - 24 Days C - 30 Days D - 36 Days

Answer: A) 18 days Explanation 2(A+B+C)'s 1 day work = 1/30 + 1/24 + 1/20 = 1/8 =>(A+B+C)'s 1 day's work= 1/16

work done by A,B and C in 10 days=10/16 = 5/8 Remaining work= 3/8 A's 1 day's work= (1 / 16−1 / 24)=1 / 48 Now, 1/48 work is done by A in 1 day. So, 3/8 work will be done by A in =48 x (3/8) = 18 days

Q 10 - An air conditioner can cool the hall in 40 miutes while another takes 45 minutes to cool under similar conditions. If both air conditioners are switched on at same instance then how long will it take to cool the room approximately ? A -18 minutes B - 19 minutes C - 22 minutes D - 24 minutes

Answer: C) 22 minutes Explanation Let the two conditioners be A and B 'A' cools at 40min

'B' at 45min Together = (a x b)/(a + b) = (45 x 40)/(45 + 40) = 45 x 40/85 = 21.1764 = 22 min (approx).

Q 11 - A works twice as fast as B.If B can complete a work in 18 days independently,the number of days in which A and B can together finish the work is: A - 4 days B - 6 days C - 8 days D - 10 days Answer: B) 6 days Explanation Ratio of rates of working of A and B =2:1. So, ratio of times taken =1:2 Therefore, A's 1 day's work=1/9

B's 1 day's work=1/18 (A+B)'s 1 day's work= 1/9 + 1/18 = 1/6 so, A and B together can finish the work in 6 days

Q 12 - A can do a certain work in the same time in which B and C together can do it.If A and B together could do it in 20 days and C alone in 60 days ,then B alone could do it in: A - 20 days B - 40 days C - 50 days D - 60 days

Answer: D) 60 days Explanation

(A+B)'s 1 day's work=1/20 C's 1 day work=1/60 (A+B+C)'s 1 day's work= 1/20 + 1/60 = 1/15 Also A's 1 day's work =(B+C)'s 1 day's work

Therefore, we get: 2 x (A's 1 day 's work)=1/15 =>A's 1 day's work=1/30 Therefore, B's 1 day's work= 1/20 - 1/30 = 1/60 So, B alone could do the work in 60 days.

Q 13 - A is thrice efficient as B and C is twice as efficient as B. what is the ratio of number of days taken by A,B and C, when they work individually? A - 2:6:3 B - 2:3:6 C -1:2:3 D - 3:1:2 Answer: A) 2:6:3 Explanation

A : B : C Ratio of efficiency

3

: 1 : 2

Ratio of No.of days

1/3 : 1/1 : 1/2

or

2 : 6 : 3

Hence A is correct.

Q 14 - 4 men can repair a road in 7 hours. How many men are required to repair the road in 2 hours ? A - 17 men B - 14 men C - 13 men D - 16 men Answer: B) 14 men Explanation

M x T / W = Constant where, M= Men (no. of men) T= Time taken W= Work load So, here we apply M1 x T1/ W1 = M2 x T2 / W2 Given that, M1 = 4 men, T1 = 7 hours ; T2 = 2 hours, we have to find M2 =?

Note that here, W1 = W2 = 1 road, ie. equal work load. Clearly, substituting in the above equation we get, M2 = 14 men.

Q 15 - A,B,C together can do a piece of work in 10 days.All the three started workingat it together and after 4 days,A left.Then,B and C together completed the work in 10 more days.In how many days can complete a work alone ? A - 25 B - 24 C - 23 D - 21 Answer: A) 25 Explanation

(A+B+C) do 1 work in 10 days. So (A+B+C)'s 1 day work=1/10 and as they work together for 4 days so workdone by them in 4 days=4/10=2/5 Remaining work=1-2/5=3/5 (B+C) take 10 more days to complete 3/5 work. So( B+C)'s 1 day work=3/50 Now A'S 1 day work=(A+B+C)'s 1 day work - (B+C)'s 1 day work=1/10-3/50=1/25

A does 1/25 work in in 1 day Therefore 1 work in 25 days.

Q 16 - (x-2) men can do a piece of work in x days and (x+7) men can do 75% of the same work in (x-10)days. Then in how many days can (x+10) men finish the work? A - 27 Days B - 12 Days C - 25 Days D - 18 Days Answer: B) 12 days Explanation 34×(x−2)x=(x+7)(x−10)34×(x-2)x=(x+7)(x-10) ⇒x2−6x−280 =0⇒x2-6x-280 =0 => x= 20 and x=-14

so, the acceptable values is x=20 Therefore, Total work =(x-2)x = 18 x 20 =360 unit Now 360 = 30 x k

=> k=12 days

Q 17 - 3 men, 4 women and 6 children can complete a work in 7 days. A woman does double the work a man does and a child does half the work a man does. How many women alone can complete this work in 7 days ? A-6 B-9 C-5 D-7 Answer: D) 7 Explanation Let 1 woman's 1 day work = x. Then, 1 man's 1 day work = x/2 and 1 child's 1 day work x/4. So, (3x/2 + 4x + + 6x/4) = 1/7 28x/4 = 1/7 => x = 1/49 1 woman alone can complete the work in 49 days. So, to complete the work in 7 days, number of women required = 49/7 = 7.

Q 18 - A, B and C can complete a piece of work in 24,6 and 12 days respectively.Working together, they will complete the same work in: A - 1/24 days B - 7/24 days C - 24/7 days D - 4 Days Answer: C) 24/7 days Explanation (A+B+C)'s 1 day's work = (1/24 + 1/6 + 1/12) = 7/24 so, A,B and C together will complete the work in 24/7 days.

Q 19 - A and B together can do a piece of work in 40 days. A having worked for 20 days, B finishes the remaining work alone in 60 days. In How many days shall B finish the whole work alone? A - 60 B - 70 C - 80 D - 90

Answer: C) 80 Explanation Let A's 1 day's work=x and B's 1 day's work=y Then x+y = 1/40 and 20x+60y=1 Solving these two equations , we get : x= 1/80 and y= 1/80 Therefore B's 1 day work = 1/80 Hence,B alone shall finish the whole work in 80 days

Q 20 - A is 30% more efficient than B. How much time will they, working together, take to complete a job which A alone could have done in 23 days ? A - 9 Days B - 11 Days C - 13 Days D - 15 Days

Answer: C) 13 days Explanation

Ratio of times taken by A and B = 100 : 130 = 10 : 13. Suppose B takes x days to do the work. Then, 10 : 13 :: 23 : x

=>

x = ( 23 x 13/10 )

=>

x = 299 /10.

A's 1 day's work = 1/23 ; B's 1 day's work = 10/299 . (A + B)'s 1 day's work = ( 1/23 + 10/299 ) = 23/299 = 113 . Therefore, A and B together can complete the work in 13 days.

Profit & Loss

1. Cost Price, (c.p.) = The price, at which an article is purchased, is called its cost price. 2. Selling price (s.p) = The price, at which an article is sold, is called its selling price. 3. Profit or Gain = (S.P) - (C.P) 4. Loss = (C.P) - (S.P) 5. Gain or Loss is always reckoned on C.P. Formulae 1. Gain% = (Gain*100) / C.P 2. Loss% = (Loss*100) / C.P 3. S.P = (100+ Gain %) /100 * (C.P) 4. S.P = (100-Loss %) / 100 * (C.P) 5. C.P = 100 / (100 + Gain %)* (S.P) 6. C.P = 100 / (100-Loss %)* (S.P) Important cases 1. If an article is sold at a profit of say, 20%, then SP = 120% of CP. 2. If an article is sold at a loss of say, 20%, then SP = 80% of CP. 3. When a person sells two similar items, one at a gain of say x% and the other at a loss of say x%. then the seller always incurs a loss given by: Loss% = (x/10)2 4. If a seller sells his goods at cost price but uses false weights, then Gain% = [Error/(True value - Error) * 100]%

Sample Question Paper on Profit & Loss Q 1 - A vendor bought 6 oranges for Re 10 and sold them at 4 for Re 6. Find his loss or gain percent. A - 8% gain B - 10% gain C - 8% loss D - 10% loss

Answer - D Explanation Suppose, number of oranges bought = LCM of 6 and 4 = 12 ∴CP = Re (10/6 * 12) = Re 20 and SP = Re (6/4 * 12) = Re 18 ∴Loss% = (2/20 * 100)% = 10%

Q 2 - By selling 33 meters of cloth, one gains the selling price of 11 meters. Find the gain percent. A - 50% B - 45% C - 40% D - 60%

Answer - A Explanation (SP of 33m) - (CP of 33m) = Gain = SP of 11m ∴ SP of 22m = CP of 33m Let CP of each meter be Re 1. Then, CP of 22m = Re 22. Hence SP of 22m = Re 33. ∴ %Gain = 11/22 * 100 = 50% Q 3 - Pure ghee costs Re 100 per kg. A shopkeeper mixes vegetable oil costing Re 50 per kg and sells the mixture at Re 96 per kg, making a profit of 20%. In what ratio does he mix the pure ghee with the vegetable oil. A - 3:2 B - 2:3 C - 4:3 D - 3:1

Answer - A Explanation Mean Cost price = Re (100/120)*96 = Re 80 per kg Apply rule of allegation, there4; Required ratio = 30:20 = 3:2 Q 4 - The CP of 25 articles is equal to SP of 20 articles. Find the loss or gain percent. A - 35%

B - 30% C - 25% D - None of these

Answer - C Explanation Let the CP of each article = Re 1. Then CP of 20 articles = Re 20. SP of 20 articles = CP of 25 articles = Re 25. ∴ Gain% = (5/20)*100% = 25% Q 5 - A shopkeeper bought 80 kg of sugar at Re 13.50/kg and mixed it with 120 kg sugar at Re 16/kg. If he is to make a profit of 16% what rate should he sell the sugar to his customers? A - Re 12/kg B - Re 15.25/kg C - Re 17/kg D - Re 17.40/kg

Answer - D Explanation CP of 200 kg of mixture = Re (80 * 13.50) + (120 * 16) = Re 3000 SP = 116% of Re 3000 = Re (116/100)*3000 = Re 3480 ∴Rate of SP = Re 3480/200 = Re 17.40/kg

Q 6 - A man bought cookies at 3 for a rupee. How many for a rupee should he sell to make a profit a 50%. A-1 B-2 C - 1.5 D - None of these

Answer - B Explanation CP of 3 cookies = Re 1 SP of 3 cookies = 150% of Re 1 = 3/2 For Re 3/2, the man sells 3 cookies. Hence for Re 1, number of cookies sold = 3*2/3 = 2 Q 7 - Anil buys a calculator for Re 600 and sells it to Vikash at 10% profit. Vikash sells it to Chandan for 5 % profit. Chandan after using it for certain time, sells it to Dinesh at a loss of 20%. For how much Chandan sell the calculator to Dinesh. A - Re 550.50 B - Re 564.40 C - Re 554.40 D - None of these

Answer - C Explanation SP for Chandan = 600 * (110/100) * (105/100) * (80/100)

= 600 * 924/1000 = Re 554.40 Q 8 - An article is sold by X to Y at a loss of 20%, Y to Z at a gain of 15%, Z to W at a loss of 5% and W to V at a profit of 10%. If v had to pay Re 500, how much X paid for it? A - Re 520.07 B - Re 490.07 C - Re 510.07 D - Re 530.07

Answer - A Explanation CP for X = 500 * (100/80) * (100/115) * (100/95) * (100/110) = 500 * 10000/9614 = Re 520.07 Q 9 - A vendor when could not find buyers for his vegetable at Re 10/kg, reduced the rate to Re 8.10 per kg but uses a faulty weight of 900 gm in place of 1 kg weight. Find the percent change in the actual price or loss. A - 8% B - 8.10% C - 9% D - 10%

Answer - D Explanation After the price was reduced, 900 gm now costs Re 8.10. Hence 1000gm will cost (1000/900)*8.10 = Re 9 % change in actual price or loss = [(10 - 9)/10]*100% = 10% Q 10 - A trader marks the SP of an object at a profit of 20%. Considering the demand o the object, he further increases the price by 10%. Find the final profit %. A - 35% B - 31% C - 32% D - 25%

Answer - C Explanation Let the CP = Re 100 ∴ SP = 100 * (120/100) * (110/100) = Re 132 Final profit = (132 - 100)*100% = 32%

Q 11 - An article when sold for Re 4600 makes a 15% profit. Find the profit or loss % if it was sold for Re 3600. A - 10% gain B - 11% loss C - 10% loss D - 11% gain

Answer - C Explanation CP = 4600 * (100/115) =Re 4000 Loss% = [(4000-3600)/4000]*100% = 10% Q 12 - A seller sells a watch at 5% loss. If he had bought it at 20% more and sold it for Re 115 less, he would have incurred a loss of 40%. Find the cost price of the watch. A - Re 500 B - Re 5000 C - Re 550 D - Re 450

Answer - A Explanation Assume CP = x

Selling price at the first case = (95/100)x Selling price at the second case = (60/100)*(120/100)x = (7200/10000)x As per question, (95/100)x - (7200/10000)x = 150 Or, x = Re 500 Q 13 - When a man sold an article for Re 540, he made a loss of 10%. At what price should he sell it, so that he incurs a loss of only 5%. A - Re 550 B - Re 525 C - Re 575 D - Re 570

Answer - D Explanation CP = 540*(100/90) = Re 600 New SP = 600*(95/100) = Re 570 Q 14 - Ram sells chocolates at a profit of 20% for Re 60. What will be the percentage loss or gain if he reduces the price to Re 55 due to less demand. A - 11% B - -11%

C - 10% D - -10%

Answer - C Explanation CP = 60*(100/120) = Re 50 New SP = Re 55 Gain% = (5/50)*100 =10% Q 15 - A shopkeeper buys rice for Re 1600. He had to sell 1/4th at a loss of 20%. If he is to make an overall gain of 10%, what percentage of profit he needs to make out of the remaining stock of rice? A - 20% B - 25% C - 15% D - 18%

Answer - A Explanation CP of 1/4th of the stock = 1600/4 = Re 400 SP of 1/4th of the stock = 400*(80/100) =Re 320 In order to make a profit of 10% on total CP, the SP should be:

SP = 1600*(110/100) = Re 1760 ∴The SP for the remaining 3/4th of the stock should be Re 1760 - Re 320 = Re 1440. Cost Price of the 3/4th of stock = Re 1600 - Re 400 = Re 1200. ∴%Gain = [(1440-1200)/1200*100}] = (240/1200)*100 = 20% Q 16 - A 10% hike in the price of wheat forces a person to purchase 2 kg less for Re 110. Find the new and the original price of the wheat. A - Re 10/kg B - Re 5/kg C - Re 6/kg D - Re 8/kg

Answer - B Explanation 10% of Re 110 = Re 11 Cost of 2 kg of wheat at new price = Re 11 So, cost of 1 kg of wheat at new price = Re 5.50 = Re 11/2 Original Price = (11/2)*(100/110) = Re 5 per kg

Q 17 - 10 kg of rice costs as much as 20 kg of wheat, 25 kg of wheat costs as much as 2kg of tea, 5 kg of tea costs as much as 25kg of sugar. Find the cost of 6 kg of sugar if 14 kg of rice costs Re 32. A - Re 50 B - Re 55 C - Re 60 D - Re 65

Answer - C Explanation 4 kg of rice costs Re 32 ∴10kg of rice will cost = (32/4)*10 = Re 80

20 kg of wheat costs Re 80. ∴25kg of wheat costs = (80/20)*25 = Re 100 2kg of tea costs Re 100 ∴5 kg of tea costs = (100/2)*5 = Re 250

25kg of sugar costs Re 250. ∴6 kg of sugar costs = (250/25)*6 = Re 60 Q 18 - A fruit seller sells bananas at a profit of 20%. If he increases the selling price of each banana by 25 paisa, he earns a profit of 45%. Find the initial selling price of each banana and also its cost price. A - SP = Re1.20, CP = Re 1 B - SP = Re1.50, CP = Re 1

C - SP = Re1.20, CP = Re 1.10 D - None of the above.

Answer - A Explanation Let CP = x paisa. Initial SP = x*(120/100) paisa As per question, 120x/100 + 25 = (145/100)x or, 145x/100 - 120x/100 = 25 or, 25x/100 = 25 or, x = 100 paisa CP = 100 paisa or Re 1. Initial SP = 120 paisa or Re 1.20. Q 19 - A man sold two plots for Re 8 lakhs each. One on he earns a profit of 16% and the other he loses 16%. How much does he loss or gain in the whole transaction? A - 2.5% loss B - 3% gain C - 2.56% loss D - 3.56% loss

Answer - C Explanation

Applying direct formula, %loss = (16/10)2% = 64/25% = 2.56% Q 20 - An uneducated retailer marks all his goods at 50% above the cost price and thinking that he will still make 25% profit, offers a discount of 25% on the marked price. What is his actual profit on the sales? A - 10% B - 12.50% C - 11.50% D - 12%

Answer - B Explanation Let CP = Re 100. The, marked price, MP = Re 150 SP = 75% of Re 150 = Re 112.50 ∴ Gain% = 12.50% Q 21 - A book shop offers a book at an increase of 10%. On the off chance that he had purchased it at 4% less and sold it for Rs. 6 more, he would have picked up 75/4 %. The expense cost of the book is: A - 130 B - 140 C - 150 D - 160

Answer : C Explanation Let the C.P. be Rs x.Then, S.P. =Rs (110/100 * x) =Rs 11x/10. New C.P. = 96% of Rs x=Rs (96/100 * x) =Rs 24x/25. New S.P. =Rs (11x/10+6). ∴ (11x/10+6) =475/4% of 24x/25⇒11x+60/10=475/400*24x/25=57x/50 ⇒550x+3000=570x ⇒20x=3000⇒x=150. ∴ C.P. =Rs 150. Q 22 - Mohan purchased 20 feasting tables for rs. 12000 and sold them at a benefit equivalent to the offering cost of 4 eating tables. The offering cost of every eating table is: A - 700 B - 750 C - 725 D - 775 Answer : B Explanation C.P of each table = rs. (12000/20) = rs. 600 (S.P of 20 tables)- (C.P of 20% table) = profit = S.P of 4 tables ⇒S.P of 16 tables = C.P of 20 tables= 1200 ⇒ S.P of 1 table = (12000/16) = Rs. 750

Q 23 - A man offers an article at 12.5% misfortune. Had he sold it for Rs. 103.60 more, he could have picked up 6%. What is the C.P of the articles? A - 278.60 B - 350 C - 432 D - 560 Answer : D Explanation Let the C.P be Rs. x. Then , (106/100 *x) - (87.5/100*x) =103.60 ⇒ 106x- 87.5x= 103.60 ⇒ 18.5x =10360, x= 103600/185 = 560 ∴ C.P= Rs. 560 Q 24 - A shipper has 1000kg of sugar, a portion of which he offers at 8% benefit what's more, and the rest at 18% benefit. He increases 14% in general. The amount sold at 18% benefit is: A - 560 kg B - 600 kg C - 400 kg D - 640 kg Answer : B Explanation Let the sugar sold at *5 gain be x kg. Then, sugar sold at 18% gain = (1000-x) kg

Let the C.P Of sugar be Rs. Y per Kg. Total C.P = Rs.(1000y) (108/100*xy)+ 118/100 (1000-x)y = 114/100 *1000y ⇒108xy + 118000y -118xy = 114000y ⇒ 10x =4000 ⇒ x= 400 Quantity sold at 18% gain= (1000-400) = 600 kg Q 25 - The profit earned after selling an article for Re 625 is the same as the loss incurred after selling the article for Re 435. What is the cost price of the article? A - Re 530 B - Re 520 C - Re 550 D - None of these Answer : A Explanation Let profit = loss = Re x As per question, 625 ? x = 435 + x Or, 2x = 190 Or, x = Re 95 ∴ CP = Re (435 + 95) = Re 530

Q 26 - A shopkeeper expects a gain of 17.5% on his cost price. If in a week, his sale was of Rs. 235, what was his profit? A - 35 B - 32.5 C - 30 D - 27.5 Answer : A Explanation C.P. = Rs 100/( 117.5) x 235=Rs.200 Profit = 235-200=Rs. 35 Q 27 - The aggregate expense cost of two watches is R. 840. One is sold at a benefit of 16% and the other at lost 12%. There is no misfortune or addition in the entire exchange. The expense cost of the watch on which the businessperson additions, is: A - 360 B - 370 C - 380 D - 390 Answer : A Explanation Let their C.P be rs. x and Rs. (840-x) resp. Then, (116/100*x) + 88/100*(840-x) ⇒ 29x/25+ (22 (840-x)/25) = 840 ⇒ (29x-22x) + 18480 =21000 ⇒ 7x =2520 ∴ Required C.P = Rs. 360

Q 28 - Which of the accompanying is the most reduced proportion? A - 7:13 B - 17:25 C - 7:15 D - 15:23 Answer : C Explanation 7/13 =0.538, 17/25 =0.68, 7/15= 0.466, 15/23 =0.652 Clearly, 7/15 is the smallest. Q 29 - A businessperson denotes his merchandise to gain35%. In any case, he permits 10% markdown for money installment. His benefit percent is: A - 27/2% B - 43/2% C - 25% D - 63/2% Answer : B Explanation Let the C.P be Rs.100. Then, M.p= rs. 135 S.P = 90% of rs. 135= Rs. (90/100*135)= rs. 121.50 Profit% = (121.50-100)% =21.50% =43/2%

Q 30 - On a Rs. 10000 installment arrange a man has decision between three progressive rebates of 10%, 10% and 30% and three progressive rebates of 40%, 5% and 5%. By picking the better one, he can spare: A - Rs. 200 B - Rs. 255 C - Rs. 400 D - Rs. 433 Answer : B Explanation 1st payment= 90% 0f 90% of 70% of rs. 10000 = rs. (90/100*90/100* 70/100*10000) =5670 2nd. Payment = 60% of 95% of 95% of 10000. = (60/100*95/100*95/100*10000)= 5415 By choosing the better one he can save rs . (5670-5415) = 255

Time & Calendar

1.

Odd days: In a given period, the quantity of days more than the complete weeks are called Odd days.

2.

Leap Year: Every year divisible by 4 in a leap year. But not all century years are leap years. Only those century years which are divisible by 400 are leap years and other century years are ordinary years. As an example, 1100, 1300, 1400, 1500, 1700 are ordinary years but 1200, 1600, 2000 are leap years. So every 4th century is a leap year.

3.

Ordinary Year:A non-leap year is an ordinary year. A conventional year has 365 days. A leap year has 366 days.

4.

Counting of odd days:  1 customary year = 365 days (52 weeks+ 1 day) ∴ 1 customary year has 1 odd day.

 1 jump year =366 days = (52 weeks+2 days) ∴ 1 jump year has 2 odd days.

 100 year =76 normal year +24 jump year = (76*1+24*2) odd days =124 odd days = (17 weeks +5 days) =5 odd days. 5.

No. of odd days in 100 years = 5

6.

No. of odd days in 200 years = (5*2) =3 odd days.

7.

No. of odd days in 300 years = (5*3) =1 odd day.

8.

No. of odd days in 400 years = (5*4+1) = 0 odd days.

9.

Every one of 800 years, 1200 years, 1600 years, 2000 years and so on has 0 odd days.

10.

Odd days related to days of the week: No. of odd days

0

1

2

3

4

5

6

Days

Sun

Mon

Tue

Wed

Thu

Fri

Sat

11.  February: 28 days(ordinary year) gives '0' odd days, 29 days (leap year) gives '1' odd day.  Jan, March, May, July, Aug, Oct and Dec have 31 days each and therefore give '3' odd days.  April, June, Sep and Nov each have 30 days and therefore give '2' odd days.

Sample Question Paper on Time & Calendar Q 1 - What was the day of the week on 15th June, 1776? A - Sunday B - Saturday C - Thursday D - None of these

Answer - B Explanation 15th June 1776 = (1775 years + Period from 01.01.1776 to 15.06.1776) Counting of odd days: No of odd days in 1600 years = 0 No of odd days in 100 years = 5 75 years = 18 leap years + 57 ordinary years = 18*2 + 57*1 = 36 + 57 = 93 odd days = 13 weeks + 2 odd days = 2 odd days ∴ 1775 years have (0+5+2) = 7 odd days = 0 odd days. Jan to May = (31+29+31+30+31) = 152 days

Add 15 days of June. = 152 + 15 = 167 days = 23 weeks + 6 days = 6 odd days. ∴ Total number of odd days = 0 + 6 = 6 odd days. Hence 15.06.1776 was Saturday. Q 2 - January 15, 1997 was a Wednesday. What day of the week was on Jan 5, 2000? A - Wednesday B - Thursday C - Friday D - Saturday

Answer - A Explanation 1997, 1998 and 1999 are not leap years. 1998 and 1999 has 2 odd days. No of days remaining in 1997 = 365 - 15 = 350 = 50 weeks of 0 odd days. 05.01.2000 = 5 odd days. Total no of odd days = 2 + 0 + 5 = 7

7 days from Wednesday is Wednesday. ∴ Jan 5, 2000 was also Wednesday. Q 3 - The calendar for the year 2007 will be the same for the year: A - 2018 B - 2017 C - 2016 D - 2014

Answer - A Explanation We will count the no of odd days from the year 2007 onwards to get the sum equal to 0 odd days. Year

2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017

Odd day 1

2

1

1

1

2

1

1

Sum = 14 odd days = 0 odd days Calendar for the year 2018 will be the same for the year 2007.

1

2

1

Q 4 - Will date-book for the year 2003 serve for the year 2014? A - no B - yes

Answer - B Explanation We must have same day on 1.1.2003 and 1.1.2014. Along these lines, number of odd days somewhere around 31.12.2002 and 31.12.2013 must be 0. This period has 3 jump years and 8 common years. Number of odd days = (3*2+8*1) =14=0 odd days. ∴ Calendar for the year 2003 will serve for the year 2014. Q 5 - What was the week's day on fifteenth august, 1947? A - Sunday B - Monday C - Friday D - Thursday

Answer - C Explanation fifteenth Aug.1947 =(1946 years +period from 1.1.1947 to 15.8.1947)

Odd days in 1600 years =0 Odd days in 300 years = (5*3) =15 =1946 years = (11 jump years+35 customary years) = (11*2 +35*1) odd days= 57 days = (8 weeks +1 day) = 1 odd day ∴ odd days in 1946 years= (0+1+1) =2

Jan + Feb. + March + April + May + June + July + Aug (31 + 28 +31+ 30 + 31 +30+31+15) = 227 days 227 days = (32 weeks +3 days) = 3 odd days. Aggregate no. of odd days = (2+3) = 5 Consequently the required day is Friday.

Q 6 - If Feb 12th,1986 falls on Wednesday then Jan 1st,1987 falls on which day? A - Wednesday B - Tuesday C - Thursday D - Friday

Answer - C Explanation

First,we count the number of odd days for the left over days in the given period. Here,given period is 12.2.1986 to 1.1.1987 Feb Mar Apr May June July Aug Sept Oct Nov Dec Jan 16 31 30 31 30 31 31 30 31 30 31 1 (left days) 2 + 3 + 2 + 3 + 2 + 3 + 3 + 2 + 3 + 2 + 3 + 1(odd days) = 1 odd day So,given day Wednesday + 1 = Thursday is the required result.

Q 7 - The year next to 2005 will have the same calendar as that of the year 2005? A - 2016 B - 2022 C - 2011 D - None

Answer - C

Explanation

NOTE :

Repetition of leap year ===> Add +28 to the Given Year. Repetition of non leap year Step 1 : Add +11 to the Given Year. If Result is a leap year, Go to step 2. Step 2: Add +6 to the Given Year. Solution : Given Year is 2005, Which is a non leap year. Step 1 : Add +11 to the given year (i.e 2005 + 11) = 2016, Which is a leap year. Step 2 : Add +6 to the given year (i.e 2005 + 6) = 2011 Therfore, The calendar for the year 2005 will be same for the year 2011

Q 8 - If today is Saturday, what will be the day 350 days from now ? A - Saturday B - Friday C - Sunday D - Monday

Answer - A

Explanation 350 days = (350/7=50 weeks) i.e No odd days, So it will be a Saturday

Q 9 - Given that on 9th August 2016 is Saturday. What was the day on 9th August 1616 ? A - Saturday B - Sunday C - Friday D - Monday

Answer - A Explanation

We know that, After every 400 years, the same day occurs. Thus, if 9th August 2016 is Saturday, before 400 years i.e., on 9th August 1616 has to be Saturday.

Q 10 - Which year has 366 days? A - 1900

B - 1200 C - 2500 D - 1700

Answer - B Explanation

NOTE: When a century year leaves a remiander 0, when divided by 400 then it is a leap year (366 days). So, 1200 has 366 days.

Q 11 - It was Sunday on Jan 1, 2006. What was the day of the week Jan 1, 2010 ? A - Sunday B - Friday C - Monday D - Tuesday

Answer: B) Friday Explanation

Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days. On 1st Jan 2006, it was Sunday. Thus, On 1st Jan, 2010 it is Friday.

Q 12 - What is 90 days from today? Hints : Today is 20th January 2017, Sunday

A - 18th April, Friday B - 20th April, Saturday C - 21th April, Sunday D - 19th April, Saturday

Answer: B) 20th April, Saturday Explanation

Given Today is 20th January 2017, Sunday In january, we have 31 days

February - 28 days (Non leap year) March - 31 days April - 30 days => Remaining days => 31 - 20 = 11 in Jan + 28 in Feb + 31 in Mar = 11 + 28 + 31 = 70 days More 20 days to complete 90 days => upto 20th April Therefore, after 90 days from today i.e, 20th Jan 2017 is 20th Apr 2017. Now, the day of the week will be 90/7 => Remainder '6' As the day starts with '0' on sunday 6 => Saturday. Required day is 20th April, Saturday.

Q 13 - On 19th June, 1984 Monday falls. What day of the week was it on 19th June, 1985? A - Sunday

B - Monday C- Tuesday D - Wednesday Answer: C) Tuesday Explanation

The year 1985 is an ordinary year. So, it has 1 odd day. So, the day on 19th June, 1985 will be 1 day after the day on 19th June, 1984. But, 19th June,1984 is Monday S0, 19th June, 1985 is Tuesday.

Q 14 - The calendar of year 1939 is same as which year? A - 1943 B - 1964 C - 1950 D - 1956

Answer: C) 1950 Explanation Given year 1939, when divided by 4 leaves a remainder of 3.

NOTE: When remainder is 3, 11 is added to the given year to get the result. So, 1939 + 11 = 1950 Q 15 - How many seconds in 10 years? A - 31523500 sec B - 315360000 sec C - 315423000 sec D - 315354000 sec

Answer: B) 315360000 sec Explanation

We know that, 1 year = 365 days 1 day = 24 hours 1 hour = 60 minutes 1 minute = 60 seconds. Then, 1 year = 365 x 24 x 60 x 60 seconds.

= 8760 x 3600 1 year = 31536000 seconds. Hence, 10 years = 31536000 x 10 = 315360000 seconds. Q 16 - What will be the day of the week 15th August, 2010? A - Sunday B - Monday C - Tuesday D - Friday

Answer: A)Sunday Explanation

Q 17 - What is two weeks from today? A - Same Day B - Previous Day C - Next Day D - None

Answer: A) same day Explanation We know that the day repeats every 7 days, 14 days, 21 days,... So if today is Monday, after 7 days it is again Monday, after 14 days again it is Monday. Hence, after 2 weeks i.e, 14 days the day repeats and is the same day.

Q 18 - Which century year is a leap year? A - 1900 B - 1700 C - 1600 D - 1100

Answer: C) 1600 Explanation

NOTE: For an century year,it is divided by 400,when leaves a remainder 0 then it is a leap year. So, 1600

Q 19 - How much does a clock lose per day, if its hands coincide ever 65 minutes ? A - 1440/143 min B - 184/13 min C - 1425/18 min D - 541/9 min

Answer: A) Explanation:

55 min. spaces are covered in 60 min. 60 min. spaces are covered in ((60/55) x 60) min = 720/11 min. Loss in 65 min. = (720/11 - 65 ) = 5/11 Loss in 24 hrs = (5/11 x 1/65 x 24 x 60)min = 1440/143 min.

Q 20 - How many days in 4 years? A - 1460 B - 1461 C - 1462 D - 1459

Answer: B) 1461 Explanation:

Days in 4 years => Let the first year is Normal year i.e, its not Leap year. A Leap Years occurs once for every 4 years. 4 years => 365 + 365 + 365 + 366(Leap year) 4 years => 730 + 731 = 1461 Therefore, Number of Days in 4 Years = 1461 Days.