Question

1.Which is the largest number which divides 38 and 125 ,leaving remainders 5 and 4 respectively ?

2.To enhance the reading skills of grade X students, the school nominates you and two of your friends to set up a class library. There are two sections- section A and section B of grade X. There are 35 students in section A and 40 students in section B.


What is the minimum number of books you will acquire for the class library, so that they can be distributed equally among students of Section A or Section B?

Answer 1

Question:- 1

Which is the largest number which divides 38 and 125 ,leaving remainders 5 and 4 respectively ?

Answer :- 1

Let largest number be 'x' which divides 38 and 125 ,leaving remainders 5 and 4 respectively.

It implies,

• 38 - 5 is divisible by x. i.e. 33 is divisible by x.

• 125 - 4 is divisible by x. i.e. 121 is divisible by x.

It means,

• x = HCF(33, 121)

So,

Prime factorization of 33 = 3 × 11

Prime factorization of 121 = 11 × 11

So,

• HCF( 33, 121 ) = 11

Hence,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \blue{\underbrace\red{ \boxed{ \bf \:Largest \: number, \: x = 11}}}$

Question :- 2

To enhance the reading skills of grade X students, the school nominates you and two of your friends to set up a class library. There are two sections- section A and section B of grade X. There are 35 students in section A and 40 students in section B. What is the minimum number of books you will acquire for the class library, so that they can be distributed equally among students of Section A or Section B?

Answer :- 2

Since equal number of books are to distributed among the students of Section A and Section B of grade 10.

So,

• The number of books should be a multiple of 35 and 40.

Hence,

• The required number of books = LCM( 35, 40 ).

Prime factorization of 35 = 5 × 7

Prime factorization of 40 = 2 × 2 × 2 × 5

So,

• LCM( 35, 40 ) = 2 × 2 × 2 × 5 × 7 = 280

Hence,

$\: \: \: \: \blue{\underbrace\red{ \boxed{ \bf \:Required \: number\: of \: books\: = \: 280}}}$

Additional Information :-

1. Let a and b are two positive integers, then

• a × b = HCF (a, b) × LCM (a, b)

2. HCF always divides LCM