Question

A is greater than B by 1/3rd the sum of A and B.if B is increased by 40,it becomes greater than twice A by 10.find A,B

Answer 1

Answer :

A is greater than B by 1/3rd the sum of A and B.

=> (A - B) = 1/3 (A + B)
=> 3×(A - B) = A + B
=> 3A - 3B = A + B
=> 3A - A - 3B - B = 0
=> 2A - 4B = 0
=> A - 2B = 0 -----------(1)


Now,if B is increased by 40,it becomes greater than twice A by 10.

=> (B+40) - (2A) = 10
=> -2A + B = 10-40
=> -2A + B = -30
=> 2A - B = 30 ---------(2)

Now,lets multiply equation (2) by 2,
4A - 2B = 60 ----------(3)


Now,lets subtract equation (3) by (1),

(A - 2B) - (4A - 2B) = 0-60
=> -3A -60
=> A = 20

Lets put value of A in equation (1),
A - 2B = 0
=> 20 - 2B = 0
=> 2B = 20
=> B = 10

● So the value of A is 20
&
● value of B is 10.


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