Question

At a certain fast food restaurant, Amit can buy 3 burgers, 7 shakes and one order of fries for Rs. 120. At the same time, it would cost Rs. 164.50 for 4 burgers, 10 shakes and one order of fries. How much would it cost for an ordinary meal of one burger, one shake and one order of fires?
A.Rs. 31
B.Rs. 41
C.Rs. 21
D.Cannot be determined

Answer 1

Hint: The given question can be solved by forming two different equations based on the costs of burger, shakes and fries that is, with 3 variables. Then subtracting two equations to get another equation. From forming these equations we will be able to solve by substituting the third one in one of the two equations. So that we can find the cost of one burger, one shake and one order of fries.

Complete step-by-step answer:
Let cost of burger = b, cost of shake = s and cost of fries = f
Given, Cost of 3 burgers, 7 shakes and one order of fries = Rs. 120
Now, $3{\text{b + 7s + f = 120 }}...{\text{(1)}}$
And Cost of 4 burgers, 10 shakes and one order of fries = Rs. 164.50
And $4{\text{b + 10s + f = 164}}{\text{.50 }}...{\text{(2)}}$
Subtract the two equations we get
$\begin{gathered}
  {\text{b + 3s = 44}}{\text{.5}} \\
  \therefore {\text{ b = 44}}{\text{.5 - 3s }}...{\text{(3)}} \\
\end{gathered} $
We have got three equations, we can substitute (3) in (1)
Now, $3{\text{b + 7s + f = 120}}$
This equation can be written as
Or ${\text{2b + b + 7s + f = 120}}$
Now in this equation we now substitute ${\text{b = 44}}{\text{.5 - 3s}}$ in $2{\text{b}}$
That is,
${\text{2}}\left( {44.5{\text{ - 3s}}} \right){\text{ + b + 7s + f = 120 }}...{\text{from (3)}}$
$89 - 6{\text{s + b + 7s + f = 120}}$
$\therefore {\text{ b + s + f = 31}}$
That is, Cost of one burger + Cost of one shake + Cost of one order of fries = Rs. 31
Thus, the cost of one burger, one shake and one fries is Rs. 31.

Note: Here in the solution there is no necessity to find the individual cost of burger, shake and fires. We just need the sum. There is a different approach for this by not substitution. We can find the cost by subtracting the equation (3) two times from the (1) equation or 3 times from the (2) equation we will get b+s+f = Rs. 31.