Wipro Questions with solutions LCM and HCF Part 1

 Wipro Questions with solutions LCM and HCF

1. The least perfect square, which is divisible by each of 20, 35, and 65 is

a: 3314434

b: 3341200

c: 3312400

d: 4331444

Option C

Explanation:

L.C.M. of 20, 35, 65 = 1820

Now, 1820= 5 x 4 x 7 x 13

To make it a perfect square, it must be multiplied by 5 x 4 x 7 x 13

So, required number =5 x 4 x 7 x 13 x 5 x 4 x 7 x 13

= 52 x 42 x 72 x 132 = 3312400

 

2. The HCF and LCM of the two numbers are 13 and 455 respectively. If one of the      numbers lies between 75 and 125, then, that number is: 

a: 78

b: 91

c: 10

d: 11

Option b i.e 91

LCM×HCF=Product of two numbers

This question is one of the easiest ones, if you still cannot find how to do it, just comment on the page we will share the step-wise solutions.

 

3. Ram got a question in his maths test that,If LCM of two number is 369, HCF of two numbers is 27 and one number is 9, then find other

a: 1390

b: 1212

c: 1107

d: 1502

Answer:1107

Explanation:

For any this type of question, remember

Product of two numbers = Product of their HCF and LCM

So Other number = (369*27)/9 = 1107

Easy trick is

We all know

Product of two numbers = Product of their HCF and LCM

So, 369*27 = 9*X

therefore X = (369 x 27) / 9

X = 1107

 

4. Find the greatest number which on dividing 1654 and 2153, leaves a remainder of   24 and 23 respectively?

a: 15

b: 12

c: 10

d: 13

Answer:10

In this type of question, it's obvious we need to calculate the HCF, the trick is

Required Number = HCF of ( 1654 - 24 ) and HCF of ( 2153 - 23 )

therefore HCF of 1630 and 2130 is 10

 

5. Which greatest possible length can be used to measure exactly 20 meters 15 cm, 21 meters 55 cm, and 12 meters 65 cm?

a: 8

b: 5

c: 4

d: 2 

Answer: 5

We need to find out the HCF for the given length.

20 meter 15 cm = 2015 cm.

21 meter 55 cm = 2155 cm.

12 meter 65 cm = 1265 cm.

HCF of 2015, 2155, and 1265 is 5.


wipro questions with solutions
Wipro Questions with solutions LCM and HCF  Part - 1


 


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