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A sum of Rs 3.75 was paid in 25 paise, 10 paise and 5 paise coins. The number of 10 paise coins are 4 times the number of 25 paise coins and twice of the number of 5 paise coins. How many were there of each type?

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Hint: In this particular question use the concept that in one rupee there are 100 paise, then assume any variable be the number of 25 paise coins so the number of 10 paise coins is 4 times the number of 25 paise coins i.e. 4 times the assumed variable so use these concepts to reach the solution of the question.

Complete step by step answer:
Given data:
Sum of Rs = 3.75
Was paid in 25 paise, 10 paise and 5 paise coins.
Let the number of 25 paise coins be X.
And let the number of 5 paise coins be Y.
And it is given that 10 paise coins are 4 times the number of 25 paise coins.
So the number of 10 paise coins = 4 times the number of 25 paise coins.
So the number of 10 paise coins = 4X......... (1)
And it is also that 10 paise coins are twice the number of 5 paise coins.
So the number of 10 paise coins = 2Y............ (2)
Now from equation (1) and (2) we have,
4X = 2Y
Therefore, Y = 2X.
So the number of 5 paise coins are 2X, 10 paise coins are 4X, and 25 paise coins are X.
Now as we know that in one rupee there are 100 paise.
So in 3.75 rupees there are 100 (3.75) = 375 paise.
Now the sum of the product of respective paise with the respective number of coins is equal to the required paise which is paid.
$ \Rightarrow \left( {5 \times 2X} \right) + \left( {10 \times 4X} \right) + \left( {25 \times X} \right) = 375$
$ \Rightarrow 10X + 40X + 25X = 375$
$ \Rightarrow 75X = 375$
$ \Rightarrow X = \dfrac{{375}}{{75}} = 5$

So the number of 25 paise coins = 5, the number of 10 paise coins = 4X = 4(5) = 20, and the number of 5 paise coins = 2X = 2(5) = 10.

Note: Whenever we face such types of questions the key concept we have to remember is that the sum of the product of respective paise with the respective number of coins is equal to the required paise which is paid, so first find out the number of coins in terms of one variable as above, then substitute this value in the above described formula as above substituted and simplify we will get the required answer.
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