A bus conductor's purse has 171 coins in all, made up of ₹10, ₹5, ₹ 1, and ₹2 coins. The number of ₹10 coins is 1 4/5 times the number of ₹2 coins while the number of ₹1 coins 2/3 is times the number of ₹2 coins. If the number of 5 coins is half the number of ₹1 coins, find how many ₹10,₹ 5, ₹ 1, and ₹2 coins are there in the bus conductor's purse.
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Answer:
10₹=81
2₹=45
1₹=30
5₹=15
Step-by-step explanation:
suppose, coin of 10₹=x
coin of 2₹=y
coin of 1₹=z
coin of 5₹=a
according to question,
x=(1 4/5)y=9/5y
z=(2/3)y
a=z/2= (2/6)y(after putting the value of z)
since, x+y+z+a=171, putting the value of x, z and a and taking y as common
(9/5+1+2/3+2/6)y=171
(19/5)y=171
on solving y=45
x=(9/5)*45=81
z=(2/3)*45=30
a=30/2=15
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