A collection of 58 coins, consisting of ..... (elementary algebra)

Question

A collection of 58 coins, consisting of 25p, 50p, Re. 1 and Rs. 2 coins are in a bag. The Re. 1 coins number 5 times that of the 25p coins. The Rs. 2 coins are double the number of Re. 1 coins, and thrice the number of the 50p coins. If the total value of the coins is Rs. 80/75, how many coins of each kind are there??

Answer 1

Let R1, R2, H, Q represent numbers of Re1, Rs 2, 50p, 25p coins. Assuming there is a coin for 1 rupee (Re1) and a separate coin for 2 rupees (Rs2), R1+R2+H+Q=58; R1+2R2+0.50H+0.25Q=80/75=80.75 (for convenience). R1=5Q. R2=2R1=10Q=3H. H=10Q/3

5Q+10Q+H+Q=58; 16Q+H=58=16Q+10Q/3; 58Q/3=58, so Q=3. 2R1=10Q=30, so R1=15. R2=2R1=30. H=10Q/3=10. Therefore we have R1=15, R2=30, H=10 and Q=3.

CHECK: Add up the numbers of coins: 15+30+10+3=58 OK. Add up value: 15+60+5+0.75=80.75=Rs80/75.

 

 

Answer 2

our equations are d + q = 22, (1) 10d + 25q = 325. (2) where d is the number of dimes and q is the number of quarters. Express d = 22-q from (1) and substitute it into (1). You will get a single equation for the unknown q 10*(22-q) + 25q = 325. Simplify it: 220 - 10q + 25q = 325 ---> 15q = 325 -220 ---> 15q = 105 ---> q= = 7. Answer. The coin bank contains 7 quarters and 22-7 = 15 dimes.