Question

Alfred and Rani both picked different two digit numbers. If you multiply Alfred’s number by 5 and double Rani’s number, the sum is 300. If you double Alfred’s number and multiply Rani’s number by 3, the sum of the two numbers is 252. What is the sum of their two numbers?
(a) 96
(b) 112
(c) 128
(d) 144

Answer 1

Hint: To solve the given question, we will first assume that the number picked by Alfred is x and the number picked by Rani is y. Now, we will multiply x by 5 and multiply y by 2 and then add both the numbers. We will equate its sum to 300. We will get a linear equation in two variables. Similarly, we will find another linear equation in two variables and then we will solve these equations by the method of elimination. After solving, we will get the values of x and y. Then we will add x and y to get the sum of the numbers.

Complete step by step solution:
To start with, we will assume that the number picked by Alfred is x and the number picked by Rani is y. Now, when we multiply Alfred’s number by 5, we get 5x and when we double Rani’s number we get 2y. Now it is given in the question that after multiplication, they are added and their sum is 300. Thus, we will get,
\[5x+2y=300......\left( i \right)\]
Now, when Alfred’s number is doubled, we get 2x and when Rani’s number is multiplied by 3, we get 3y. Now, it is given in the question that after multiplication, they are added and their sum is 252. Thus, we will get the following linear equation.
\[2x+3y=252.....\left( ii \right)\]
Now, we have got two linear equations in two variables. We will solve these equations with the help of 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. In our case, we will try to remove x. For this, we will multiply (i) with 2 and (ii) with 5. Thus, we will get the new equations as
\[2\left( 5x+2y \right)=2\left( 300 \right)\]
\[\Rightarrow 10x+4y=600.....\left( iii \right)\]
\[5\left( 2x+3y \right)=5\left( 252 \right)\]
\[\Rightarrow 10x+15y=1260.....\left( iv \right)\]
Now, we will subtract (iii) from (iv). Thus, we will get,
\[\left( 10x+15y \right)-\left( 10x+4y \right)=1260-600\]
\[\Rightarrow 0+11y=660\]
\[\Rightarrow 11y=660\]
\[\Rightarrow y=\dfrac{660}{11}\]
\[\Rightarrow y=60.....\left( v \right)\]
Now, we will put the value of y from (v) to (i). Thus, we will get,
\[\Rightarrow 5x+2\left( 60 \right)=300\]
\[\Rightarrow 5x+120=300\]
\[\Rightarrow 5x=180\]
\[\Rightarrow x=\dfrac{180}{5}\]
\[\Rightarrow x=36\]
Now, we have to find the sum of the numbers picked by them. Thus, we have,
Sum = x + y
\[\Rightarrow \text{Sum}=60+36\]
\[\Rightarrow \text{Sum}=96\]
Hence, option (a) is the right answer

Note: We can also solve the linear equations obtained while solving the question by substitution method and find the values of x and y. From equation (i), we have,
\[5x+2y=300\]
\[\Rightarrow 2y=300-5x\]
\[\Rightarrow y=\dfrac{300-5x}{2}\]
\[\Rightarrow y=150-\dfrac{5x}{2}\]
Now, we will put this value of y in (ii). Thus, we will get,
\[2x+3\left( 150-\dfrac{5x}{2} \right)=252\]
\[\Rightarrow 2x+450-\dfrac{15x}{2}=252\]
\[\Rightarrow \dfrac{-11x}{2}=252-450\]
\[\Rightarrow \dfrac{-11x}{2}=198\]
\[\Rightarrow x=\dfrac{198\times 2}{11}\]
\[\Rightarrow x=36\]
On putting this value of x in y, we get,
\[\Rightarrow y=150-\dfrac{5\left( 36 \right)}{2}\]
\[\Rightarrow y=150-90\]
\[\Rightarrow y=60\]
Therefore sum = x + y = 36 + 60 = 90.