Question

sanjiv has ruppes3.10 in10 paisa and 25 paisa coins.if the 10 paisa coins exceed the number of 25 paisa coins by three, how many coins of each does he have!

Answer 1

number of 10 paisa coin = x
number of 25 paisa coin =y
A/q,
x - y = 3 
⇒x = 3 +y
and, 10x + 25y = 310
⇒ 10 ( 3 +y) + 25y = 310
⇒30 +10y + 25y = 310
⇒35 y= 280
⇒y =8 and x = 11
there are 11 coins of 10 paisa and 8 coins of 25 paisa

Answer 2

Let x be the number of 10 paise coins and y be the number of 25 paise coins.
given,
10x+25y=310                     [3.10=310 paise]
divide the eqn by 5,
we get,
2x+5y=62-------------------(1)
then, x-y=3
multiply the eqn by 2,we get,
2x-2y=6---------------------(2)
(1)-(2)======>7y=56
thus, y=8.
putting y=8 in (1)
2x=62-40
    =22
thus x=11.
ans:there are 11 10 paise coins and 8 25 paise coins