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# XAT LCM and HCF Questions PDF [Most Important]

Download LCM and HCF Questions for XAT PDF – XAT LCM and HCF questions pdf by Cracku. Practice XAT solved LCM and HCF Questions paper tests and These sample questions are to analyse the XAT exam pattern. Top 15 very Important LCM and HCF Questions for XAT based on asked questions in previous exam papers.  The XAT question papers contain actual questions asked with answers, and solutions.

Question 1: 0ne-eighth of a number is 17.25. What will 73% of the number be ?

a) 100.74

b) 138.00

c) 96.42

d) 82.66

e) None of these

Solution:

Let the number be $8x$

Acc to ques,

=> $\frac{1}{8} * 8x = 17.25$

=> $x = 17.25$

$\therefore$ 73 % of the number = $\frac{73}{100} * 8x$

= 0.73 * 8 * 17.25 = 100.74

Question 2: ‘ A’ ,’B’ and ‘C’ are three consecutive even integers such that four times ‘A’ is equal to three times ‘C’. What is the value of B’?

a) 12

b) 10

c) 16

d) 14

e) None of these

Solution:

let A be 2x ,B be 2x +2 and C be 2x +4

Given that 4A = 3C

4(2x) = 3 (2x +4)

8x = 6x +12

2x = 12

x = 6

B = 2*6 + 2 = 14

Question 3: What is the LCM of the following fractions? 3/11, 2/5, 1/9

a) 0

b) 1

c) 2

d) 3

e) 6

Solution:

The LCM of the fractions = LCM of the numerators/ HCF of the denominators
LCM of the numerators = LCM of 3, 2 and 1 = 6
HCF of the denominators = HCF of 11, 5 and 9 = 1
So, the required LCM = 6/1 = 6

Question 4: What is the LCM of the following fractions? ⅔, 4/7, 1/4

a) 1

b) 2

c) 4

d) 7

e) 8

Solution:

LCM of fractions = LCM of the numerators/ HCF of the denominators.
LCM of 2, 4 and 1 is 4
HCF of 3, 7 and 4 is 1
So, the LCM of the given fractions is 4/1 = 4

Question 5: The LCM of two numbers is 360, and their HCF is 15. If one of the numbers is 45, what is the value of the remainder when the other number is divided by 7?

a) 1

b) 2

c) 3

d) 4

e) 0

Solution:

let the second number be x.
We know that LCM*HCF = Product of the two numbers
So 360*15 = x *45
=> x = $\frac{360*15}{45}$
=> x = 120
Now 120 on being divided by 7, leaves a remainder of 1.

Question 6: What is the greatest number which divides 1070 and 1265 and leaves remainders 3 and 4 respectively?

a) 91

b) 93

c) 95

d) 97

e) 101

Solution:

The required number is the HCF of (1070 – 3, 1265 – 4) = HCF of (1067, 1261)
1067 = 11*97 and 1261 = 13*97
So, HCF = 97

Question 7: The sum of two numbers is 65 and the HCF of the two numbers is 5. If the LCM of the two numbers is 180, what is the sum of the reciprocals of the two numbers?

a) 11/13

b) 11/625

c) 13/625

d) 11/180

e) 13/180

Solution:

Let the numbers be ha and hb. HCF of the numbers = h and LCM of the numbers = hab.
Sum of reciprocals = (a+b)/hab
h(a+b) = 65
h = 5
(a+b) = 13
hab = LCM = 180
So, sum of reciprocals = 13/180

Question 8: The ratio of two numbers is 5:6. The product of the HCF and LCM of the two numbers is 120. What is the HCF of the two numbers?

a) 2

b) 4

c) 6

d) 8

e) 1

Solution:

Let the two numbers be 5h and 6h where h is the HCF of the numbers. Product of the numbers = Product of LCM and HCF.
=> $30 h^2$ = 120 => h = 2
So, the HCF of the two numbers is 2

Question 9: What is the smallest number which when multiplied by 10 is exactly divisible by 12, 18, 24 and 32?

a) 288

b) 144

c) 1440

d) 552

e) None of the above

Solution:

The LCM of the numbers is 288. So, multiples of 288 are divisible by all the numbers. LCM of 288 and 10 = 1440. So, the required number = 144

Question 10: In a church, the first bell rings at intervals of 2 seconds, the second bell rings at intervals of 4 seconds and so on till the fifth bell, which rings at intervals of 10 seconds. How many times do all the bells ring together in a half-an-hour period?

a) 60

b) 45

c) 120

d) 15

e) 30

Solution:

The LCM of 2, 4, 6, 8 and 10 is 120. So, the bells rings together once every 2 minutes. So, they ring together 15 times in a half-an-hour period.

Question 11: The largest number that divides 43, 91 and 183 and leaves the same remainder in each case is

a) 16

b) 8

c) 9

d) 13

e) 4

Solution:

The required number is the HCF of (91 – 43), (183 – 91) and (183 – 43) = HCF of (48, 92, 140) = 4

Question 12: The LCM of two numbers x and y is 20. Find the number of unordered pairs (x,y)?

a) 8

b) 9

c) 7

d) 10

e) None of the above