Question

Q83. There are 45 children in a classroom. After distributing the sweets equally among the children 10 sweets remain. If the number
had been 50, after equal distribution 25 sweets would have remained. What would be the least number of sweets that satisfy th
conditions?
O (A) 725
O (B) 695
O (C) 775
(D) 595
O
(E) 325
17%​

Answer 1

Given : There are 45 children in a classroom. After distributing the sweets equally among the children 10 sweets remain

If the number  had been 50, after equal distribution 25 sweets would have remained.

To Find :   What would be the least number of sweets that satisfy the conditions?

(A) 725

(B) 695

(C) 775

(D) 595

(E) 325

Solution:

After distributing the sweets equally among the 45 children 10 sweets remain.

=> Sweets = 45A  + 10

After distributing the sweets equally among the 50 children 25 sweets remain.

=> Sweets = 50B  + 25

45A  + 10 = 50B  + 25

=> 9A  + 2  = 10B  + 5

=> 9A  = 10B  +  3

A = 7  , B = 6   satisfy thus

Sweets = 50B  + 25  = 50*6 + 25  =  325

325 = 45 x 7 + 10

325  = 50 x 6 + 25

least number of sweets that satisfy the conditions  = 325

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