There r 36 coins of 25 Paise 10 Paise 5 Paise of rs 3
Find no.of 25 Paise coins
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Let there be m of 25p coins, and p number of 5paise.
Then there are n = 36 - m - p number of 10p coins.
Assume that there is at least 1 of each type of coin.
Their total value in paise is :
25 m + 10 n + 5 p = 300 paise
25 m + 10 (36 - m - p) + 5 p = 300
5 p = 15 m + 60
p = 3 m + 12
As m >= 1, p >= 15 and p is a multiple of 3. m , n , p are all integers.
m = 1 p = 15 n = 36 - 15 - 1 = 20
m = 2 p = 18 n = 16
m = 3 p = 21 n = 12
m = 4 p = 24 n = 8
m = 5 p = 27 n = 4
So there are five combinations of coins possible.
Then there are n = 36 - m - p number of 10p coins.
Assume that there is at least 1 of each type of coin.
Their total value in paise is :
25 m + 10 n + 5 p = 300 paise
25 m + 10 (36 - m - p) + 5 p = 300
5 p = 15 m + 60
p = 3 m + 12
As m >= 1, p >= 15 and p is a multiple of 3. m , n , p are all integers.
m = 1 p = 15 n = 36 - 15 - 1 = 20
m = 2 p = 18 n = 16
m = 3 p = 21 n = 12
m = 4 p = 24 n = 8
m = 5 p = 27 n = 4
So there are five combinations of coins possible.
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