AMCAT Quantitative Ability Previous Papers-14

Ques. Find the largest number which divides 62,132,237 to leave the same reminder
30
32
35
45

Answer: Option C

Explanation:

Trick is HCF of (237-132), (132-62), (237-62)
= HCF of (70,105,175) = 35

Ques. Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:

A.    4
B.    5
C.    6
D.    8

N = H.C.F. of (4665 – 1305), (6905 – 4665) and (6905 – 1305)

= H.C.F. of 3360, 2240 and 5600 = 1120.

Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4

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

Answer: Option C

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.

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
Answer: Option A

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 least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:

A.    3
B.    13
C.    23
D.    33
Answer: Option C

Explanation:
L.C.M. of 5, 6, 4 and 3 = 60.
On dividing 2497 by 60, the remainder is 37.
Number to be added = (60 – 37) = 23.

A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and c in 198 seconds, all starting at the same point. After what time will they again at the starting point ?

A.    26 minutes and 18 seconds
B.    42 minutes and 36 seconds
C.    45 minutes
D.    46 minutes and 12 seconds
Answer: Option D

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.

Ques.     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.
15
16
30
31

Answer: Option D

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

Ques. 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, is
10:28 am
10:30 am
10:31 am
None of above

Answer: Option C

Explanation:
L.C.M. of 60 and 62 seconds is 1860 seconds
1860/60 = 31 minutes
They will beep together at 10:31 a.m.
Sometimes questions on red lights blinking comes in exam, which can be solved in the same way

Ques. Find the greatest number that will divide 400, 435 and 541 leaving 9, 10 and 14 as remainders respectively
19
17
13
9

Answer: Option B
Explanation:
Answer will be HCF of (400-9, 435-10, 541-14)
HCF of (391, 425, 527) = 17