Question

Amar has only 3 denominations of coins i.e., 1, 2, and '5. If he has a total of 40 coins and the total amount of 130, the maximum
number of 5 coins he can have, is
A) 24
B) 22
C) 23
D) 21​

Answer 1

Answer:

22

let's consider 1 rupee coin as X

2 rupee as y

5 rupee as z

totally 40 coins in a bag so,x+y+z=40

similarly,1x+2y+5z= 140

by simplify above equation we get y+4z = 90

here they are given option

first consider 24 as z because they are given the option based on maximum number of 5 rupee coins

then option 1,. 5*24=120

coins remain==> 14 coins

amount remaining==>20 rupees only so this option gets eliminate

opt 2. ,5*22=110

coins remain=18

amount remain=30

therefore,12*2=24 rupee

remaining amount and coin ia 6 rupee and 6 coins respectively

then,6*1=6 rupee

Result:

1 rupee coin= 6

2 rupee coin=12

5 rupee coin=22