Question

A sum was divided among P, Q and R. R got double than P who got double than Q. If the difference between the shares of Q and R is Rs. 3675.00, then the sum in rupee is
(a) 4900
(b) 8575
(c) 11025
(d) 7350

Answer 1

Hint: To solve the above question, we will first assume that the total sum of the money is S. Then we will assume P got ‘x’ rupees, Q got ‘y’ rupees and R got ‘z’ rupees. We will then derive two equations with the help of the relation given in the question. The third equation will be derived by the relation of y and z given in the question, i.e. the difference between y and z is Rs. 3675.00. Finally, we will derive the relation between S, x, y and z according to the fact that the sum of x, y and z is S. We will solve these 4 equations in 4 variables with the help of the substitution method and find out the value of S.

Complete step by step solution:
To start with, we will assume that the total sum of money is S. Out of S, P got ‘x’ rupees, Q got ‘y’ rupees and R got ‘z’ rupees. So, we have the following equation
\[S=x+y+z......\left( i \right)\]
It is given in the question that R got double than P i.e. z is twice the x. Thus, we will get,
\[z=2x....\left( ii \right)\]
Another information given in the question is that P got double than Q, i.e. x is twice y. Thus, we will get,
\[x=2y....\left( iii \right)\]
It is given that the difference between z and y is Rs. 3675.00. Thus, we have,
\[z-y=3675....\left( iv \right)\]
Now, we have got four equations in four variables. We will solve these equations by the method of substitution. For this, we will put the value of x from (iii) to (ii). Thus, we will get,
\[\Rightarrow z=2\left( 2y \right)\]
\[\Rightarrow z=4y.....\left( v \right)\]
Now, we will substitute the value of z from (v) to (iv). Thus, we will get,
\[\Rightarrow 4y-y=3675\]
\[\Rightarrow 3y=3675\]
\[\Rightarrow y=1225\text{ Rupees}......\left( vi \right)\]
Now, we will put this value of y in (iii). Thus, we will get,
\[x=2\left( 1225\text{ Rupees} \right)\]
\[\Rightarrow x=2450\text{ Rupees}......\left( vii \right)\]
Now, we will put this value of x in (ii). Thus, we will get,
\[z=2\left( \text{2450 Rupees} \right)\]
\[\Rightarrow z=4900\text{ Rupees}......\left( viii \right)\]
Now, we will put the value of x, y and z from (vi), (vii) and (viii) to (i). Thus, we will get,
\[\Rightarrow S=2450\text{ Rupees}+1225\text{ Rupees}+4900\text{ Rupees}\]
\[\Rightarrow S=\left( 2450+1225+4900 \right)\text{ Rupees}\]
\[\Rightarrow S=8575\text{ Rupees}\]
Hence, option (b) is the right answer.

Note: We can solve equations (i), (ii), (iii) and (iv) by elimination method also. For this, we will multiply (iv) by 2 and subtract from (iii). Thus, we will get,
\[\Rightarrow x-2\left( z-y \right)=2y-2\left( 3675 \right)\]
\[\Rightarrow x-2z+2y=2y-7350\]
\[\Rightarrow x-2z=-7350.....\left( 1 \right)\]
Now, we will multiply (1) with 2 and add in (ii). Thus, we will get,
\[2\left( x-2z \right)+z=-2\left( 7350 \right)+2x\]
\[\Rightarrow 2x-4z+z=-14700+2x\]
\[\Rightarrow 3z=14700\]
\[\Rightarrow z=4900\text{ Rupees}\]
Now, we will put the value of z in (1). Thus, we will get,
\[\Rightarrow x-2\left( 4900 \right)=-7350\]
\[\Rightarrow x=9800-7350\]
\[\Rightarrow x=2450\text{ Rupees}\]
Now, we will put the value of x in (iii). Thus, we will get y = 1250. So, the sum will be
\[S=x+y+z\]
\[\Rightarrow S=\left( 2450+1225+4900 \right)\text{ Rupees}\]
\[\Rightarrow S=8575\text{ Rupees}\]