Question

A bag contains a number of 10paise, 20paise, 25paise coins in the ratio \[7:4:3\]. If the money value comes to be Rs. 90, the number of 25paise coins in the bag is:

Answer 1

Hint: In this question, let us assume some variables for all the three types of coins present in the bag. Then from the given proportion write them in terms of a common constant. Now, equate the sum of the coins in the bag to 90 which gives the value of constant and so all the assumed variables.

Complete step-by-step answer:
Now, from the given conditions in the question there are 10paise, 20paise, 25paise coins in the bag
Let us assume that number of 10paise coins as x, number of 20paise coins as y and number of 25paise coins as z
Now, given in the question that these coins are in the ratio \[7:4:3\]which gives
\[\Rightarrow x:y:z=7:4:3\]
Now, let us assume a proportionality constant as a then we get,
\[\begin{align}
  & \Rightarrow x=7a \\
 & \Rightarrow y=4a \\
 & \Rightarrow z=3a \\
\end{align}\]
Now, as given in the question that the sum of coins in the bag is Rs. 90
As we already know that the value of 100paise is equal to Rs. 1 we can get the relation that
\[\Rightarrow 1paise=Rs.\dfrac{1}{100}\]
Now, let us convert the given types of coins into rupees using the above relation
\[\Rightarrow 10paise=Rs.\dfrac{10}{100}\]
\[\Rightarrow 20paise=Rs.\dfrac{20}{100}\]
\[\Rightarrow 25paise=Rs.\dfrac{25}{100}\]
Now, as the sum of the coins is Rs.90 we get,
\[\Rightarrow \dfrac{10}{100}\times x+\dfrac{20}{100}\times y+\dfrac{25}{100}\times z=90\]
Now, this can be further written in the simplified form as
\[\Rightarrow \dfrac{x}{10}+\dfrac{y}{5}+\dfrac{z}{4}=90\]
Let us now substitute the respective values in terms of constant assumed
\[\Rightarrow x=7a,y=4a,z=3a\]
Now, on substituting these values in the above equation we get,
\[\Rightarrow \dfrac{7a}{10}+\dfrac{4a}{5}+\dfrac{3a}{4}=90\]
Now, on taking L.C.M of the denominators it can be further written as
\[\Rightarrow \dfrac{14a+16a+15a}{20}=90\]
Now, on multiplying with 12 on both sides and simplifying further we get,
\[\Rightarrow 45a=90\times 20\]
Now, on cancelling the common terms on both sides we get,
\[\Rightarrow a=2\times 20\]
\[\therefore a=40\]
Now, as we know the relation between z and a we get,
\[\Rightarrow z=3a\]
\[\begin{align}
  & \Rightarrow z=3\times 40 \\
 & \therefore z=120 \\
\end{align}\]
Hence, there are 120 25paise coins in the bag

Note: Instead of assuming a constant of proportionality we can also solve it by writing x and y in terms of z and then substituting them in the equation gives the value of z directly. Both the methods give the same result.
It is important to note that when equating the sum to 90 we need to divide all the value of paise by 100 as the sum is in rupees. When substituting the values and simplifying we should not neglect any of the terms because it changes the result.