## AMCAT Previous Year Questions 2018

NUMBER  SYSTEM

Rs. 1200 is lent out at 5% per annum simple interest for 3 years. Find the amount after 3 years.

1. Rs. 1380
2. Rs. 1290
3. Rs. 1470
4. Rs.1200

Explanation:

A = P+I

A = 1200+(PTR/100)

A = [1200+(1200*5*3/100)]

A = Rs. 1380.

Basu  started a business investing Rs 9000. After five months, Sameer joined with a capital of Rs 8000. If at the end of the year, they earn a profit of Rs. 6970, then what will be the share of Sameer in the profit ?

1. Rs 2380
2. Rs 2300
3. Rs 2280
4. Rs 2260

Explanation:

Basu invested for 12 months and Sameer invested for 7 months.

So Kamal: Sameer = (9000*12):(8000*7)

= 108 : 56

= 27 : 14 ,

Sameer Ratio in profit will be =(6970*14/41)=Rs 2380.

` `

Two trains for Mumbai leave Delhi at 6 am and 6.45 am and travel at 100 kmph and 136 kmph respectively. How many kilometers from Delhi will the two trains be together:

1. 262.4 km
2. 260 km
3. 283.33 km
4. 275 km

Explanation:

Difference in time of departure between two trains = 45 min. = 45/60 hour = 3/4 hour.

Let the distance be x km from Delhi where the two trains will be together.

Time taken to cover x km with speed 136 kmph be t hour

and time taken to cover x km with speed 100 kmph (As the train take 45 mins. more) be

(t+3/4)

= ((4t+3)/4);

Now,

100*(4t+3)/4 = 136t;

Or, 25(4t+3) = 136t;

Or, 100t+75 = 136t;

Or, 36t = 75;

Or, t = 75/36 = 2.08 hours;

Then, distance x km = 136*2.084 = 283.33 km.

A sum was invested at simple interest at a certain interest for 2 years. It would have fetched Rs. 60 more had it been invested at 2% higher rate. What was the sum?

1. Rs. 1500
2. Rs. 1300
3. Rs. 2500
4. Rs. 1000

Explanation:

Let the rate be R at which Principal P is invested for 2 years.

(Interest at Rate (R+2)) % – (interest at rate R %) = Rs. 60;

(P*2*(R+2))/100 – (P*2*R)/100 = 60;

(2PR+4P-2PR)/100 = 60;

4P = 60*100;

Or, P = 60*100/4;

Hence, P = Rs. 1500

Bhargab  has a certain average for 9 innings. In the tenth innings, he scores 100 runs thereby increasing his average by 8 runs. His new average is:

1. 20
2. 21
3. 28
4. 32

Explanation:

Let Bhargab”s average be x for 9 innings. So, Ajit scored 9x run in 9 innings.

In the 10th inning, he scored 100 runs then average became (x+8). And he scored (x+8)*10 runs in 10 innings.

Now,

=>9x+100 = 10*(x+8)

Or, 9x+100 = 10x+80

Or, x = 100-80

Or, x = 20

New average = (x+8) = 28 runs.

A box has 210 coins of denominations one-rupee and fifty paise only. The ratio of their respective values is 13:11. The number of one-rupee coin is

1. 65
2. 66
3. 77   D. 78

Explanation:

Respective ratio of the number of coins;

= 13:11*2 = 13:22

Hence, Number of 1 rupee coins;

= 13*210/(13+22) = 78.

A man completes a certain journey by a car. If he covered 30% of the distance at the speed of 20kmph. 60% of the distance at 40km/h and the remaining of the distance at 10 kmph, his average speed is:

1. 25 km/h
2. 28 km/h
3. 30 km/h
4. 33 km/h

Explanation:

If  total distance be 100 km.

Average speed = total distance covered/ time taken;

= 100/[(30/20)+(60/40)+(10/10)];

= 100/[(3/2)+(3/2)+(1)];

= 100/[(3+3+2)/2]

= (100*2)/8

= 25 kmph

In how many ways can 8 Indians and, 4 American and 4 Englishmen can be seated in a row so that all person of the same nationality sit together?

1. 3! 4! 8! 4!
2. 3! 8!
3. 4! 4!
4. 8! 4! 4!

Correct Answer: 3! 4! 8! 4!

Explanation:

Taking all person of same nationality as one person, then we will have only three people.

These three person can be arranged themselves in 3! Ways.

8 Indians can be arranged themselves in 8! Way.

4 American can be arranged themselves in 4! Ways.

4 Englishman can be arranged themselves in 4! Ways.

Hence, required number of ways = 3!*8!*4!*4! Ways.

A train 125 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 10 seconds. The speed of the train is:

1. 45 km/hr
2. 50 km/hr
3. 54 km/hr
4. 55 km/hr

Explanation:

Speed of train relative to man=(125/10)m/sec

=(25/2)m/sec

=((25/2)*(18/5))km/hr

=45km/hr

Let the speed of the train be x km/hr. Then, relative speed = (x – 5) km/hr.

∴ x – 5 = 45 ⇒ x = 50 km/hr.

A man’s speed with the current is 15 km/hr and the speed of the current is 2.5 km/hr. The man’s speed against the current is:

1. 8.5 km/hr
2. 9 km/hr
3. 10 km/hr
4. 12.5 km/hr

Explanation:

Man’s rate in still water = (15 – 2.5) km/hr = 12.5 km/hr.Man’s rate against the current = (12.5 – 2.5) km/hr = 10 km/hr.

The speed of the train going from Nagpur to Allahabad is 100 km/h while when coming back from Allahabad to Nagpur, its speed is 150 km/h. find the average speed during whole journey.

1. 125 km/hr
2. 75 km/hr
3. 135 km/hr  D. 120 km/hr

Explanation:

Average speed,

= (2*x*y)/(x+y)

= (2*100*150)/(100+150)

= (200*150)/250

= 120 km/hr.

A boatman goes 2 km against the current of the stream in 1 hour and goes 1 km along the current in 10 minutes. How long will it take to go 5 km in stationary water?

1. 40 minutes
2. 1 hour
3. 1 hr 15 min
4. 1 hr 30 min

Correct Answer: 1 hr 15 min

Explanation:

Rate downstream=((1/10)*60)km/hr=6km/hr

rate upstream=2km/hr

Speed in still water=1/2(6+2)km/hr=4km/hr

requre time=(5/4)hrs=1hr15min

The average of the first five multiples of 9 is:

1. 20
2. 27
3. 28
4. 30

Explanation:

Required average = (total sum of multiple of 9)/5

= (9+18+27+36+45)/5

= 27

# D)46 minutes and 12 seconds

Explanation:

L.C.M. of 252, 308 and 198 = 2772.So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.

Find the largest number which divides 62, 132 and 237 to leave the same remainder in each case

A.22

B.35

C.48

D.52.

Explanation:

NO =  H.C.F. of (132 – 62), (237 – 132) and (237 – 62)

=  H.C.F. of 70, 105 and 175 = 35.

Find the largest number of four digits exactly divisible by 12,15,18 and 27.

A)6720

B)7720

C)8720

D)9720

Explanation: The Largest number of four digits is 9999. Required number must be divisible by L.C.M. of 12,15,18,27 i.e. 540.On dividing 9999 by 540,we get 279 as remainder .Required number = (9999-279) = 9720.

Find the smallest number of five digits exactly divisible by 16,24,36 and 54.

# D)10568

Explanation: Smallest number of five digits is 10000.Required number must be divisible by L.C.M. of 16,24,36,54 i.e 432,On dividing 10000 by 432,we get 64 as remainder.Required number = 10000 +( 432 – 64 ) = 10368.

Find the least number which when divided by 20,25,35 and 40 leaves remainders 14,19,29 and 34 respectively.

A)1194

B)1294

C)1394

D)1494

Explanation: Here,(20-14) = 6,(25 – 19)=6,(35-29)=6 and (40-34)=6.

Required number = (L.C.M. of 20,25,35,40) – 6 =1394.

A, B and C start running around a circular stadium and complete one round in 27 s, 9 s and 36 s, respectively. In how much time will they meet again at the starting point?

A)1 minute 48 seconds

B.2 minute 48 seconds

C.3 minute 48 seconds

D.4 minute 48 seconds

Explanation:

LCM of 27, 9 and 36 = 108

So they will meet again at the starting point after 108 sec.

i.e., 1 min 48 sec.

Three friends Raju , Ramesh and Sunil start running around a circular stadium and complete a single round in 24 s, 36 s and 40 s, respectively. After how many minutes will they meet against at the sta.

A)5 minutes

B.6 minutes

C.7 minutes

D.8 minutes

Explanation:

24 = 3 × 2 × 2 × 2 = 3 × 2³

36 = 3 × 3 × 2 × 2 = 3² × 2²

40 = 2 × 2 × 2 × 5 = 5¹ × 23

LCM of 24, 36 and 40 = 3² × 2³ × 5

= 9 × 8 × 5 = 360

Hence, they will meet  after 360 s, i.e., 6 min

The least number which should be added to 2497 so that sum is divisible by 5, 6, 4, 3?

A)21

B)23

C)25

D)27

Explanation:LCM of 5,6,4,3 is 60.

On dividing 2497 by 60 we get 37 as remainder.

Therefore number to be added is 60 – 37 =23.

Six bells commence tolling together and toll at the intervals of 2,4,6,8,10,12 seconds resp. In 60 minutes how many times they will toll together.

A)17

B)21

C)24

D)31

Explanation:

LCM of 2-4-6-8-10-12 is 120 seconds, that is 2 minutes.

Now 60/2 = 30

Adding one bell at the starting it will 30+1 = 31

An electronic device makes a beep after every 60 sec. Another device makes a beep after every 62 sec. They beeped together at 10 a.m. The time when they will next make a beep together at the earliest,

A)9 : 31 am

B)10 : 31 am

C)11 : 31 am

D)12 : 31 am

Explanation:

L.C.M. of 60 and 62 seconds is 1860 seconds

1860/60 = 31 minutes

If the product of two numbers is 84942 and their H.C.F. is 33, find their L.C.M.

A)2474

B.2574

C.2674

D.2774

Explanation:

HCF * LCM = 84942, because

Product of two numbers = Product of HCF and LCM

LCM = 84942/33 = 2574

Find the greatest number which on dividing 1661 and 2045, leaves a reminder of 10 and 13 respectively

A)127

B)129

C)131

D)133.

Explanation:

In this type of question, its obvious we need to calculate the HCF, trick is

HCF of (1661 – 10) and (2045 -13)

= HCF (1651, 2032) = 127

Which greatest possible length can be used to measure exactly 15 meter 75 cm, 11 meter 25 cm and 7 meter 65 cm

A)255 cm

B)265 cm

C)275 cm

D)285 cm

Explanation:

Convert first all terms into cm.

i.e. 1575 cm, 1125cm, 765cm.

Now whenever we need to calculate this type of question, we need to find the HCF.

HCF of above terms is 255.

The product of two numbers is 2025 and their HCF is 15 their LCM is:

A)2040

B)2010

C)135

D)150

Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ?

A)4

B)10

C)15

D)16

Explanation:

L.C.M. of 2, 4, 6, 8, 10, 12 is 120. So, the bells will toll together after every 120 seconds(2 minutes). In 30 minutes, they will toll together 30 + 1 = 16 times. 2

Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is

A)40

B)80

C)120

D)200

Explanation:

Let the numbers be 3x, 4x and 5x.

Then, their L.C.M. = 60x.So, 60x = 2400 or x = 40.

The numbers are (3 x 40), (4 x 40) and (5 x 40).

Hence, required H.C.F. = 40.

The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:

A)101

B)107

C)111

D)185

Explanation:

Let the numbers be 37a and 37b.

Then, 37a x 37b = 4107

ab = 3.

Now, co-primes with product 3 are (1, 3).

So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111).

Greater number = 111.

The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is:

A)216

B)389

C)443

D)548

Explanation:

Required number = (L.C.M. of 12, 15, 20, 54) + 8

= 540 + 8

= 548.

The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one of the numbers is 275, then the other is:

A)208

B)308

C)408

D)508

Explanation:

Other number = (11 x 7700/275) = 308.

The ratio of two numbers is 3 : 4 and their H.C.F. is 4. Their L.C.M. is:

A)48

B)36

C)24

D)18

Explanation:

Let the numbers be 3x and 4x. Then, their H.C.F. = x. So, x = 4.

numbers 12 and 16.

L.C.M. of 12 and 16 = 48.

The smallest number which when diminished by 7, is divisible 12, 16, 18, 21 and 28 is:

A)268

B)432

C)689

D)1015

Explanation:

number = (L.C.M. of 12,16, 18, 21, 28) + 7

= 1008 + 7

= 1015

The H.C.F. of two numbers is 11 and their L.C.M. is 693. If one of the numbers is 77,find the other.

A)79

B)89

C)99

D)109

Explanation:

Other number = 11 X 693   = 99

Find the greatest possible length which can be used to measure exactly the lengths 4 m 95 cm, 9 m and 16 m 65 cm.

A)25 cm

B)35 cm

C)45 cm

D)55 cm

Explanation:

Required length = H.C.F. of 495 cm, 900 cm and 1665 cm.

495 = 3² x 5 x 11, 900 = 2² x 3² x 5², 1665 = 3² x 5 x 37.

H.C.F. = 32 x 5 = 45.

Hence, required length = 45 cm.

Average:

In the first 10 overs of a cricket game, the run rate was only 3.2. What should be the run rate in the remaining 40 overs to reach the target of 282 runs?

A)6.25

B)5.5

C)7.4

D)5

Explanation:

Runs IN first 10 overs = 10 × 3.2 = 32

Total runs = 282

Remaining  = 282 – 32 = 250

Remaining overs = 40

Run rate  = 25040=6.25

 The average of 20 numbers is zero. Of them, How many of them may be greater than zero, at the most?
1. 1
2. 20
3. 0
4. 19

Explanation:

Average of 20 numbers = 0

=> Sum of 20 numbers20=0Sum of 20 numbers20=0

The average weight of A, B and C is 45 kg. If the average weight of A and B be 40 kg and that of B and C be 43 kg, what is the weight of B?

A)31kg

B)28kg

C)40kg

D)45kg

Explanation:

Let the weight of A, B and C are a,b and c respectively.

Average weight of A,B and C = 45

a + b + c = 45 × 3 = 135 — equation(1)

Average weight of A and B = 40

a + b = 40 × 2 = 80 — equation(2)

Average weight of B and C = 43

b + c = 43 × 2 = 86 — equation(3)

equation(2) + equation(3) – equation(1)

=> a + b + b + c – (a + b + c) = 80 + 86 – 135

=> b = 80 + 86 -135 = 166 – 135 = 31

Weight of B = 31 Kg

The average weight of 8 person’s increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg. What is the weight of the new person?

A)85kg

B)60kg

C)75kg

D)23kg

Explanation:

Total increase in weight = 8 × 2.5 = 20

` `

If xx is the weight of the new person, total increase in weight = x−65x−65

=> 20 = xx – 65

=> xx = 20 + 65 = 85

There are two divisions A and B of a class, consisting of 36 and 44 students respectively. If the average weight of divisions A is 40 kg and that of division b is 35 kg. What is the average weight of the whole class?

A)38.25

B)37.25

C)38.5

D)37

Explanation:

Total weight of students in division A = 36 × 40

Total weight of students in division B = 44 × 35

Total students = 36 + 44 = 80

Average weight of the whole class

=(36×40)+(44×35)80=(9×40)+(11×35)20=(9×8)+(11×7)4=72+774=1494=37.25

A batsman makes a score of 87 runs in the 17th inning and thus increases his averages by 3. What is his average after 17th inning?

A)39

B)35

C)42

1. D) 40.5

Explanation:

Let the average after 17 innings = x

Total runs scored in 17 innings = 17x

Average after 16 innings = (x-3)

Total runs scored in 16 innings = 16(x-3)

Total runs scored in 16 innings + 87 = Total runs scored in 17 innings

=> 16(x-3) + 87 = 17x

=> 16x – 48 + 87 = 17x

=> x = 39

The average age of a husband and his wife was 23 years at the time of their marriage. After five years they have a one year old child. What is the average age of the family ?

A)21 years

B)20 years

C)18 years

D)19 years

Explanation

Total age of husband and wife (at the time of their marriage) = 2 × 23 = 46

Total age of husband and wife after 5 years + Age of the 1 year old child

= 46 + 5 + 5 + 1 = 57

Average age of the family = 573573 = 19

Suresh drives his car to a place 150 km away at an average speed of 50 km/hr and returns at 30 km/hr. What is his average speed for the whole journey ?

A)32.5 km/hr.

B)35 km/hr.

C)37.5 km/hr

D)40 km/hr

Explanation:

If a car covers a certain distance at x kmph and an equal distance at y kmph. Then,

average speed of the whole journey = 2xyx+y2xyx+y kmph.

Therefore, average speed

=2×50×3050+30=2×50×3080=2×50×38=50×34=25×32=752=37.5

The average score of a cricketer for ten matches is 38.9 runs. If the average for the first six matches is 42, what is the average for the last four matches?

A)34.25

B)36.4

C)40.2

D)32.25

Explanation:

Total runs scored in 10 matches = 10 × 38.9

Total runs scored in first 6 matches = 6 × 42

Total runs scored in the last 4 matches = 10 × 38.9 – 6 × 42

Average of the runs scored in the last 4 matches = 10×38.9−6×42410×38.9−6×424

=389−2524=1374=34.25

There are two sections A and B of a class, consisting of 36 and 44 students’ respectively.If the average weight of section A is 40kg and that of section B is 35kg, find the average weight of the whole class

A)30Kg

B)37.25Kg

C)36Kg

D)44Kg

Explanation:

Total weight of (36+44=80) Students =(36×40+44×35) kg = 2980 kg

Therefore average weight of the whole class =(298080) kg

Therefore average weight = 37.25kg

The speed of the train going from A to B is 100 km/h while when coming back from B to A, its speed is 150 km/h. find the average speed during whole journey

A)125 km/hr

1. 75 km/hr
2. 135 km/hr
3. 120 km/hr

Explanation:Average speed,

= (2*x*y)/(x+y)

= (2*100*150)/(100+150)

= (200*150)/250

= 120 km/hr.

The average age of three boys is 15 years. If their ages are in ratio 3:5:7, the age of the youngest boy is

1. A) 21 years
2. B) 18 years
3. C) 15 years
4. D) 9 years

Explanation:

Sum of ages of three boys = 45 years

Now, (3x+5x+7x) = 45

Or, 15x = 45

Or, x = 3

So, age of youngest boy = 3x = 3*3 = 9 years.

In a boat there are 8 men whose average weight is increased by 1 kg when 1 man of 60 kg is replaced by a new man. What is weight of new comer?E. 12 years

1. A) 70
2. B) 66
3. C) 68
4. D) 69

Explanation: 8*1 = new comer – man going out

Or, new comer = 8+60;

Or, new comer = 68 years.

The average of 25 results is 18. The average of first 12 of those is 14 and the average of last 12 is 17. What is the 13th result?

A)74

B)75

C)69

D)78

Explanation:

Sum of 1st 12 results = 12*14

Sum of last 12 results = 12*17

13th result = x (let)

Now,

12*14+12*17+x = 25*18

Or, x = 78.

A train covers the first 16 km at a speed of 20 km per hour another 20 km at 40 km per hour and the last 10 km at 15 km per hour. Find the average speed for the entire journey.

A)24 km

B)26 km

C)21 km

D)23(23/59) km