AMCAT Sample Questions Quants Solutions 21- 30

Ques 21 : Choose the correct answer.
In how many ways can a number 6084 be written as a product of two different factors ?
Option 1 : 27 Option 2 : 26 Option 3 : 13 Option 4 : 14

Detailed Solution-
6084 = 2^2 * 3^2 * 13^2 = so, total number of factors are (a+1)(b+1)(c+1)= 3*3*3 =27 factors. For, each two factors we can get one way to write the number as a product of two different factors.But as the two factors are to be distinct it should be (27-1)/2 = 13 ways

Ques 22 : Choose the correct answer.
What is the smallest four-digit number which when divided by 6, leaves a remainder of 5 and when divided by 5 leaves a remainder of 3?
Option 1 : 1043 Option 2 : 1073 Option 3 : 1103 Option 4 : None of these


Detailed Solution-
1000 is the smallest 4 digit no.
The first one completely divisible by 6 is 1002
Then 1008 …
In the range of those leaving a remainder 5 on being divided by 6 are…1007,1013,1019…
Next case, divided by 5,remainder 3…
1003,1008,1013…
1013 is the common term so the answer is 1013

Ques 23 : Choose the correct answer.
P is an integer. P>883. If P-7 is a multiple of 11, then the largest number that will always divide (P+4) (P+15) is:
Option 1 : 11 Option 2 : 121 Option 3 : 242 Option 4 : None of these

Detailed Solution-
Given P is an integer>883.
P-7 is a multiple of 11=>there exist a positive integer a such that
P-7=11 a=>P=11 a+7
(P+4)(P+15)=(11 a+7+4)(11 a+7+15)
 =(11 a+11)(11 a+22)
 =121(a+1)(a+2)
As a is a positive integer therefore (a+1)(a+2) is divisible by 2.Hence (P+4)(P+15) is divisible by 121*2=242

Ques 24 : Choose the correct answer.
Let C be a positive integer such that C + 7 is divisible by 5. The smallest positive integer n (>2) such that C + n2 is divisible by 5 is:
Option 1 : 4 Option 2 : 5 Option 3 : 3 Option 4 : Does not exist

Ques 25 : Choose the correct answer.
Four bells begin to toll together and then each one at intervals of 6 s, 7 s, 8 s and 9 s respectively. The number of times they will toll together in the next 2 hr is:
Option 1 : 14 times Option 2 : 15 times Option 3 : 13 times Option 4 : 11 times

Detailed Solution-
Lcm of 6,7,8,9=504 sec
in 2 hrs =3600*2 sec
so no. times they will ring =3600*2/504=14 times

Ques 26 : Choose the correct answer.
The product of two numbers is 16200. If their LCM is 216, find their HCF.
Option 1 : 75 Option 2 : 70 Option 3 : 80 Option 4 : Data inconsistent

Detailed Solution-
The product of two numbers is 16200. If their LCM is 216, find their HCF

Ques 27 : Choose the correct answer.
There are four prime numbers written in ascending order of magnitude. The product of first three is 385 and that of last three is 1001. Find the first number.
Option 1 : 5 Option 2 : 7 Option 3 : 11 Option 4 : 17

Detailed Solution-
two middle numbers are common in these 2 products whose product is equal to HCF of 385 and 1001.
HCF = 77
largest number = 1001/77 =13
smallest number = 385/77 =5

Ques 28 : Choose the correct answer.
M and N are two distinct natural numbers. HCF and LCM of M and N are K and L respectively. A is also a natural number, which of the following relations is not possible?
Option 1 : K*L=A Option 2 : K*A=L Option 3 : L*A=K Option 4 : None of these

Detailed Solution-
Hcf cannot be larger than lcm.

Ques 29 : Choose the correct answer.
On dividing a number by 999,the quotient is 366 and the remainder is 103.The number is:
Option 1 : 364724 Option 2 : 365387 Option 3 : 365737 Option 4 : 366757

Detailed Solution
Required number = 366 x 999 + 103
=366x(1000 - 1) + 103
= 366000 - 366 + 103 = 365737.Detailed Solution-
Hcf cannot be larger than lcm.


Ques 30 : Choose the correct answer.
The difference between two numbers is 1365.When the larger number is divided by the smaller one ,the quotient is 6 and the remainder is 15.The smaller number is:
Option 1 : 240 Option 2 : 270 Option 3 : 295 Option 4 : 360

let the smaller no be x
thenlarger is x+1365
now acc to the Questions,
x+1365=6*x+15
=>x=270 which is the smaller no.

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