Question

A bag contains Rs. 11,400 in the form of currency notes of Rs. 100, Rs. 50 and Rs. 10 in the ratio 3:4:10. How many notes are there of Rs. 50?
(a) 76
(b) 72
(c) 56
(d) 48

Answer 1

Hint: To solve the above question, we will first assume that there are ‘x’ notes of Rs. 100, ‘y’ notes of Rs. 50 and ‘z’ notes of Rs. 10. We will develop the first linear equation in x, y and z with the help of the fact that the sum of 100x, 50y and 10z is Rs. 11400. Then we will develop another two relations with the help of the ratio given in the question. Then, we will solve these three equations with the help of the elimination method and finally, we will find the value of y which is the required answer.

Complete step by step solution:
To start with we will assume that there are ‘x’ notes of Rs. 100, ‘y’ notes of Rs. 50 and ‘z’ notes of Rs. 10. Now, we know that the value of one Rs. 100 note is 100 so the value of x Rs. 100 notes will be 100x. Similarly, we can say that the value of y Rs. 50 notes will be 50y and the value of z Rs. 10 notes will be 10z. Now, it is given that the total value of money in the bag is 11400. Thus, we will get,
\[100x+50y+10z=11400\]
We will divide the above equation by 10. Thus, we will get,
\[\Rightarrow 10x+5y+z=1140....\left( i \right)\]
Now, it is given that the ratio of x, y and z is 3:4:10. Thus, we have,
\[x:y:z=3:4:10\]
From the above relation, we can say that,
\[\dfrac{x}{y}=\dfrac{3}{4}\]
\[\Rightarrow 4x=3y\]
\[\Rightarrow 4x-3y=0.....\left( ii \right)\]
Similarly, we have,
\[\dfrac{y}{z}=\dfrac{4}{10}\]
\[\Rightarrow 10y=4z\]
\[\Rightarrow 5y=2z.....\left( iii \right)\]
Now, we have three equations in three variables. We will solve these equations by the elimination method. In the elimination method, we multiply one equation with a constant and another equation with another constant and then add or subtract these equations to eliminate one variable. For this, first, we will multiply (iii) with \[\dfrac{1}{2}\] and then add in (i). Thus, we will get,
\[\dfrac{1}{2}\left( 5y \right)+10x+5y+z=1140+\dfrac{1}{2}\left( 2z \right)\]
\[\Rightarrow \dfrac{5y}{2}+10x+5y+z=1140+z\]
\[\Rightarrow \dfrac{15y}{2}+10x=1140.....\left( iv \right)\]
Now, we will multiply equation (ii) with \[\dfrac{5}{2}\] and then we will subtract it from (i). Thus, we will get,
\[\left( 10x+5y+z \right)-\dfrac{5}{2}\left( 4x-3y \right)=1140-\dfrac{5}{2}\left( 0 \right)\]
\[\Rightarrow 10x+5y+z-10x+\dfrac{15y}{2}=1140\]
\[\Rightarrow \dfrac{25y}{2}+z=1140\]
\[\Rightarrow 25y+2z=2280.....\left( v \right)\]
Now, we will add (iii) and (v). Thus, we will get,
\[\left( 25y+2z \right)+\left( 5y \right)=2280+2z\]
\[\Rightarrow 30y+2z=2280+2z\]
\[\Rightarrow 30y=2280\]
\[\Rightarrow y=\dfrac{2280}{30}\]
\[\Rightarrow y=76\]
Thus, there are 76 notes of Rs. 50 in that bag.
Hence, option (a) is the right answer.

Note: We can solve the equations (i), (ii) and (iii) by substitution method. For this, we will write x and z in terms of y from (ii) and (iii) and then we will substitute them in y. Thus, from (ii), we have,
\[4x-3y=0\]
\[\Rightarrow 4x=3y\]
\[\Rightarrow x=\dfrac{3y}{4}.....\left( 1 \right)\]
Similarly, from (iii), we have,
\[\Rightarrow z=\dfrac{5y}{2}....\left( 2 \right)\]
From (1), (2) and (i), we have,
\[10\left( \dfrac{3y}{4} \right)+5y+\dfrac{5y}{2}=1140\]
\[\Rightarrow \dfrac{15y}{2}+5y+\dfrac{5y}{2}=1140\]
\[\Rightarrow 15y=1140\]
\[\Rightarrow y=76\]