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Change due = $ 0.56

If A denotes every $0.10 and B denotes every $0.01, then

Possible combinations using (A) $0.10 and (B) $0.01 are as follows:

1A + 46B --> total 47 coins
2A + 36B --> total 38 coins
3A + 26B --> total 29 coins --> Correct answer!
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Bunuel
Randall purchased a shirt for $19.44 using a $20 bill. If his correct change was returned in only dimes ($0.10) and pennies ($0.01), how many coins could Randall have received?

(A) 9
(B) 21
(C) 29
(D) 37
(E) 44

Randall received a change of $0.56.

Let \(d\) be the # of dimes & \(p\) be the # of pennies Randall got in return.

Hence we have, \(0.10d + 0.01p = 0.56\)

Multiplying by 100 on both sides, we get

\(10d + p = 56\), hence (d,p), that satisfy the equation are (1,46), (2,36), (3,26), (4,16), (5,6)

# of coins received by Randall can be 47 or 38 or 29 or 20 or 11.

Answer C.

Thanks,
GyM
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D + P = Total Coins
5 + 6 = 9
4 + 16 = 20
3 + 26 = 29 Only Direct Hit on the options

2 + 36 = 37
1 + 46 = 47
0 + 56 = 56
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