AMCAT Previous Years Questions on LCM and HCF



Topics
Sub- Topics
Expected Ques

Basic Mathematics


       6 - 8 Questions


Applied Mathematics



      8 - 10 Questions


Engineering Mathematics


       8 - 10 Questions




Ques. If the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to:
A. 11/120
B. 55/501
C. 11/60
D. 12/55

Correct Op: A
As we know that LCM * HCF = Product of two Numbers Thus a + b = 55 and ab = 5 x 120 = 600 Thus 1/a + 1/b = (b + a)/(ab) = 55/600 = 11/120

Question 2
The least perfect square, which is divisible by each of 21, 36 and 66 is
A. 213414
B. 213424
C. 213434
D. 213444

Correct Op: D
Option D Explanation: L.C.M. of 21, 36, 66 = 2772 Now, 2772 = 2 x 2 x 3 x 3 x 7 x 11 To make it a perfect square, it must be multiplied by 7 x 11. So, required number = 2 x 2 x 3 x 3 x 7 x 7 x 11 x 11 = 21344

Question 3
The LCM of three different numbers is 1024. Which one of the following can never be there HCF?
A. 8
B. 32
C. 124
D. 256

Correct Op: C
1024 is divisible by all numbers except 124

Question 4
If LCM of two number is 693, HCF of two numbers is 11 and one number is 99, then find other
A. 12
B. 45
C. 77
D. 34

Correct Op: C
Answer: Option C Explanation: For any this type of question, remember Product of two numbers = Product of their HCF and LCM So Other number = (693*11)/99 = 77

Question 5
Which greatest possible length can be used to measure exactly 15 meter 75 cm, 11 meter 25 cm and 7 meter 65 cm
A. 45
B. 255
C. 55
D. 36

Correct Op: B 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

Question 6
Find the greatest number which on dividing 1661 and 2045, leaves a remainder of 10 and 13 respectively?
A. 91
B. 127
C. 137
D. 140

Correct Op: B
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

Question 7
Six bells commence tolling together and toll at the intervals of 2,4,6,8,10,12 seconds respectively. In 60 minutes how many times they will toll together?
A. 16
B. 15
C. 31
D. 30

Correct Op: C
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

Question 8
There are three numbers, these are co-prime to each other are such that the product of the first two is 551 and that of the last two is 1073. What will be the sum of three numbers :
A. 67
B. 85
C. 129
D. 87

Correct Op: C
As given the questions these numbers are co primes, so there is only 1 as their common factor. It is also given that two products have the middle number in common. So, middle number = H.C.F. of 551 and 1073 = 29; So first number is : 551/29 = 19 Third number = 1073/29 = 37 So sum of these numbers is = (19 + 29 + 37) = 85

Question 9
Find the smallest 4 digit number which is divisible by 18, 24, 32?
A. 1216
B. 1512
C. 1152
D. 1680

Correct Op: C
LCM of 18, 24 and 32: 18 = 2 x 3 x 3 24 = 2 x 2 x 2 x 3 32 = 2 x 2 x 2 x 2 x 2 LCM = 2 x 2 x 2 x 2 x 2 x 3 x 3 = 288 The smallest four digit number that is divisible by 288: 288 x 3 = 864 288 x 4 = 1152 Since 288 is divisible by 18, 24 and 32, 1152 is also divisible by all these numbers. Therefore, 1152 is the smallest four digit number divisible by 18, 24 and 32.

Question 10
What is the smallest number that leaves a remainder of 4 when divided by 3, 4, 5, and 6?
A. 56
B. 128
C. 64
D. 32

Correct Op: C
Take the lcm of 3,4,5,6. 3=3×1 4=2×2 5=5×1 6=2×3 LCM=2×2×3×5=60 2. Therefore 60 is the number that is completely divisible by these numbers. But we have to find the smallest number that leaves remainder 2. Answer : 60+2=64

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