AMCAT Sample Questions Quants Solutions 11-20

Ques 11 : Choose the correct answer.
The L.C.M. of two numbers is 4800 and their G.C.M. is 160. If one of the numbers is 480, then the other number is:

Option 1 : 1600 
Option 2 : 1800
Option 3 : 2200
Option 4 : 2600
Option 5 : None of these

Answer :     C
Explanation :
Let n be the required number.
Since, HCF x LCM = product of two numbers.
=>4800 x 160 = 480 x n.
=>n=(4800 x 160)/480.
=>n=1600.

Ques 12 : Choose the correct answer.
The L.C.M. of two numbers is 140. If their ratio is 2:5, then the numbers are:

Option 1 : 28,70 
Option 2 : 28,7
Option 3 : 8,70
Option 4 : 8,40
Option 5 : None of these

LCM = 140
Let HCF be x
Then the 2 nos. ll be 2x nd 5x
Then Prod. of 2 nos. = LCM*HCF
2x * 5x = 140x
10x = 140
x = 14
2 nos. ll be 2*14 nd 5*14 => 28, 70

Ans : 28, 70

Ques 13 : Choose the correct answer.
If a number is exactly divisible by 85, then what will be the remainder when the same number is divided by 17?

Option 1 : 3
Option 2 : 1
Option 3 : 4
Option 4 : 0
zero 17*5=85 so remainder =0

Ques 14 : Choose the correct answer.
The least perfect square number which is exactly divisible by 3, 4, 7, 10 and 12 is:

Option 1 : 8100
Option 2 : 17600
Option 3 : 44100 
Option 4 : None of these

LCM (3, 4, 5, 6 ,8) = 120
120 = 2 x 2 x 2 x 3 x 5.
As 2,3 and 5 are not in pair in LCM’s factor so we need to multiply 120 by 5 and 3,2 to make it a perfect square.
120 x 2 x 5 x 3 = 3,600.
∴ 3600 is the least perfect square divisible by 3, 4, 5, 6 and 8.

Ques 15 : Choose the correct answer.
(xn+yn) is divisible by (x-y):

Option 1 : for all values of n
Option 2 : only for even values of n
Option 3 : only for odd values of n
Option 4 : for no values of n

Ques 16 : Choose the correct answer.
The greatest number that will divide 63, 138 and 228 so as to leave the same remainder in each case:
Option 1 : 15 
Option 2 : 20
Option 3 : 35
Option 4 : 40

HCF of ((138-63),(228-138),(228-63))
HCF of (75,90,165)
75=3*5*5
90=2*3*3*5
165=3*5*11
ans=3*5=15

Ques 17 : Choose the correct answer.
Find the largest number, smaller than the smallest four-digit number, which when divided by 4,5,6 and 7 leaves a remainder 2 in each case.
Option 1 : 422
Option 2 : 842 
Option 3 : 12723
Option 4 : None of these

Take LCM of 4,5,6,7. It is 420
BUt the no must leave remainder 2 in each case, so the no is of the form: 420k + 2.
The smallest 4-digit no is 1000. So keeping k=0,1,2,3....
We get that the largest no smaller than the smallest 4 -digit no is 842

Ques 18 : Choose the correct answer.
What is the highest power of 5 that divides 90 x 80 x 70 x 60 x 50 x 40 x 30 x 20 x 10?
Option 1 : 10 
Option 2 : 12
Option 3 : 14
Option 4 : None of these
This number can be factorized into (18*5)*(16*5)*(14*5)*(12*5)*(2*5*5)*(8*5)*(6*5)*(4*5)*(2*5)

Total number of 5’s in this factor=10 is the highest power of 5 that divides this number.

Ques 19 : Choose the correct answer.
If a and b are natural numbers and a-b is divisible by 3, then a3-b3 is divisible by:
Option 1 : 3 but not by 9 
Option 2 : 9
Option 3 : 6
Option 4 : 27

If a − b is divisible by 3, then a − b = 3k, for some integer k
(a − b)² = (3k)²
a² − 2ab + b² = 9k²
a³ − b³ = (a−b) (a² + ab + b²)
. . . . . = (a−b) (a² − 2ab + b² + 3ab)
. . . . . = 3k (9k + 3ab)
. . . . . = 3k * 3 (3k + ab)
. . . . . = 9 k(3k+ab)
Since k(3k+ab) is an integer, then 9k(3k+ab) is divisible by 9
Answer:  9

Ques 20 : Choose the correct answer.
What is the greatest positive power of 5 that divides 30! exactly?
Option 1 : 5
Option 2 : 6
Option 3 : 7
Option 4 : 8

5,10,15,20,25 & 30(When we write 30! as 1*2*3*4*5........*28*29*30(

Are having 7 , 5s as factors.

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