# HCF And LCM - MCQ 3

## 20 Questions MCQ Test Quantitative Techniques for CLAT | HCF And LCM - MCQ 3

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Attempt HCF And LCM - MCQ 3 | 20 questions in 15 minutes | Mock test for Quant preparation | Free important questions MCQ to study Quantitative Techniques for CLAT for Quant Exam | Download free PDF with solutions
QUESTION: 1

### The least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is

Solution:

L.C.M of 5, 6, 7, 8 = 840

Therefore, Required Number is of the form 840k+3.

Least value of k for which (840k+3) is divisible by 9 is k = 2

Therefore, Required  Number = (840 x 2+3)=1683

QUESTION: 2

### Find the 4-digit smallest number which when divided by 12, 15, 25, 30 leaves no remainder?

Solution:

LCM of 12, 15, 25 and 30 is 300
least number of 4-digit divided by 300 is 1200

QUESTION: 3

### Find the least number which when divided by 12, 27 and 35 leaves 6 as a remainder?

Solution:

Number = LCM (12, 17, 35) + 6 = 3780 + 6 = 3786

QUESTION: 4

The HCF and LCM of two numbers is 84 and 840 respectively. If the first number is 168, find the second one

Solution:

LCM*HCF = a*b
840*84 = 168*b, b = 420

QUESTION: 5

Find the last number which when divided by 6, 8, 15 and 30 leaves remainder 2, 4, 11 and 26 respectively?

Solution:

LCM (6, 8, 15, 30) – 4 = 120 – 4 = 116

QUESTION: 6

HCF and LCM of two numbers is 5 and 275 respectively and the sum of these two numbers is 80. Find the sum of the reciprocals of these numbers

Solution:

a*b = 5*275 and a+b = 80
(a+b)/(a*b) = 80/(5*275) = 16/275

QUESTION: 7

Three buckets contains balloons filled with water. First bucket contains 243 balloons. Second contains 304 balloons and last bucket contains 127 balloons. Find the largest number of balloons that can be given equally to the children such that 3, 4 and 7 balloons are left in first, second and third bucket respectively?

Solution:

HCF (240, 300, 120) = 60

QUESTION: 8

Riya, Anil and Rishi start running around a circular stadium and complete one round in 15s, 12s and 21s respectively. In how much time will they meet again at the starting point?

Solution:

LCM (15, 12, 21) = 420 second = 7 minutes

QUESTION: 9

In a college all the students are made to stand in four rows. 4 rows contains 12, 8, 22, 30 students respectively. Find the least number of students in the college?

Solution:

LCM (12, 8, 22, 30) = 3960

QUESTION: 10

Find the greatest number that will divide 427 and 900 leaving the remainders 3 and 8 respectively?

Solution:

HCF (427 – 3, 900 – 8) (424, 892) = 4

QUESTION: 11

What is the HCF of 3/5, 7/10, 2/15, 6/21?

Solution:

HCF will be HCF of numerators/LCM of denominators
So HCF = HCF of 3,7,2,6)/(LCM of 5,10, 15, 21)
= 1/210

QUESTION: 12

The HCF and LCM of two numbers is 84 and 840 respectively. If the first number is 168, find the second one.

Solution:

Product of two numbers = HCF * LCM
So 2nd number = 84*840/168

QUESTION: 13

Sum of 2 numbers is 128 and their HCF is 8. How many numbers of pairs of numbers will satisfy this condition?

Solution:

Since HCF = 8 is highest common factor among those numbers
So, let first no = 8x, 2nd number = 8y
So 8x + 8y = 128
x + y = 16
the co-prime numbers which sum up 16 are (1,15), (3,13), (5,11), and (7,9) *co-prime numbers are those which do no have any factor in common. These pairs are taken here because in HCF highest common factor is taken, so the remaining multiples x and y must have no common factors.
Since there are 4 pairs of co-prime numbers here, there will be 4 pairs of numbers satisfying given conditions. These are (8*1, 8*15), (8*3, 8*13) , (8*5, 8*11) , (8*7, 8*9)

QUESTION: 14

Find the least number exactly divisible by 10, 15, 18 and 30.

Solution:

It will be the LCM of these numbers.

QUESTION: 15

Find the least number which when divided by15, 21, 24 and 32 leaves the same remainder 2 in each case.

Solution:

It will be the LCM of these numbers + 2
So 3360 + 2

QUESTION: 16

Find the least number which when divided by 10, 15, 18 and 30 leaves remainders 6, 11, 14 and 26 respectively.

Solution:

Since 10-6 = 4, 15-11 = 4, 18-14 = 4, 30-26 = 4
So answer will be the LCM of these numbers – 4
So 90 – 4

QUESTION: 17

Find the smallest number of 4 digits which is exactly divisible by 12, 15, 21 and 30.

Solution:

Least 4 digit number = 1000
LCM of (12, 15, 21, 30) = 420
On dividing 420 by 1000, leaves remainder = 160
So answer = 1000 + (420 – 160)

QUESTION: 18

Find the least number which when divided by 2, 5, 9 and 12 leaves a remainder 3 but leaves no remainder when same number is divided by 11.

Solution:

LCM(2, 5, 9, 12) = 180
So the number will be somewhat = 180x + 3
This number leaves no remainder when divided by 11, so x = 2 to get (180x+3) fully divided by 11
So number = 180*2 + 3

QUESTION: 19

HCF and LCM of two numbers is 5 and 275 respectively and the sum of these two numbers is 80. Find the sum of the reciprocals of these numbers.

Solution:

x + y = 80, xy = 5*275 = 1375
sum of reciprocals = 1/x + 1/y = (x+y)/xy = 80/1375

QUESTION: 20

The HCF and LCM of two numbers is 70 and 1050 respectively. If the first number is 210, find the second one.

Solution:

Produvt of two numbers = HCF * LCM
So 2nd number = 70*1050/210

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