Question

The cost of 13 cups and 16 spoons is Rs.296,while the cost of 16 cups and 13 spoons is Rs.284. Find the cost of 2 cups and 5 spoons.\[\]

Answer 1

Hint: We assume the cost of 1 cup is $x$ rupees and the cost of 1 spoon is $y$ rupees. We form the equation $13x+16y=296$ from the given data cost of 13 cups and 16 spoons is Rs.296. We also form the equation $16x+13y=284$ from the given data cost of 16 cups and 13 spoons is Rs.284. We solve the equations by method elimination to get $x,y.$\[\]

Complete step by step answer:
Let us assume the cost of 1 cup is $x$ rupees and the cost of 1 spoon is $y$ rupees. We are given that the cost of 13 cups and 16 spoons is 296 rupees. So we have
\[13x+16y=296....(1)\]
We are also given the question that the cost of 16 cups and 13 spoons is 284 rupees. So we have
\[16x+13y=284....(2)\]
We see the two given equations are linear equations with two unknowns $x$ and $y$. We solve the equation by first eliminating $x$ from both the equations. In order to do that , we first multiply 16 with equation (1) and 13 with equation (2). We get
\[\begin{align}
  & 208x+256y=4736 \\
 & 208x+169y=3692 \\
\end{align}\]
Now we subtract equation (2) from equation (1) and get ,
\[\begin{align}
  & 87y=1044 \\
 & \Rightarrow y=\dfrac{1044}{87}=12 \\
\end{align}\]
We put the value of $y$ in equation (2) in order to obtain the value of $x$. We get ,
\[\begin{align}
  & 16x+13\times 12=284 \\
 & \Rightarrow 16x=284-156=128 \\
 & \Rightarrow x=\dfrac{128}{16}=8 \\
\end{align}\]
So the cost of 1 cup is $x=8$ rupees and the cost of 1 spoon is $y=12$ rupees. We are asked to find in the question the cost of 2 cups and 5 spoons. So we have the cost in rupees as
\[5x+2y=5\times 12+2\times 8=60+16=76\]

Note: We note that when we are given a pair of linear equations say with two unknowns ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0,{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0$ we can directly find unknowns as $x=\dfrac{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}},y=\dfrac{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}$ subjected to condition that ${{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}$ is non-zero.