Question

The expenses of a party numbering 43 were Rs 229; if each man paid Rs 10, each woman paid Rs 5 and each child Rs 2, how many were there of each?
a. 25,5,13 respectively
b. 35,8,0 respectively
c. 21,12,10 respectively
d. 16,5,22 respectively

Answer 1

Hint: The above problem can be solved by assuming the number of men or women or children present in the party to derive the total no. of people present and the money spent by men can be calculated by multiplying the number of men with the amount one man gives and similarly for the women and children.
Then we get the two equations, upon solving those two equations we get the number of men, women and children present at the party.

Complete step by step solution: Party numbering = 43; which means that a total of 43 people were present at the party; Now the number of men, women and children are not known to us; so we can assume some variable to denote the number of men, women and children present at the party.
Let the number of men be x;
Let the total number of women be ‘y’; Let the total number of children be ‘z’.
According to the question;
Total number of people at party = 43;
$\Rightarrow x+y+z=43$ (1)
Now, Money paid by man = Rs 10
Money paid by woman = Rs 5
Money paid by each child = Rs2
So, Money paid by 1 man = Rs 10
So, Money paid by 2 man = Rs 10 × 2 = Rs 20
∴ Money paid by ‘x’ men = Rs 10 × x = Rs 10x
Similarly, money paid by 1 woman = Rs 5
⇒ Money paid by 2 women = Rs 5 × 2 = Rs 10
∴ Money paid by ‘y’ women = Rs 5 × y = Rs 5y
Similarly, money paid by 1 child = Rs 2
⇒ Money paid by 2 children = Rs 2 × 2 = Rs 4
∴ Money paid by ‘z’ children = Rs 2 × z = Rs 2z
So, Total money for party = Rs 229
⇒ 10x + 5y + 2z = 229 (2)
Let the number of women be 5 because in two of the options the number of women is mentioned as 5,and so we tried the hit and trial method, and hence we get;
In equation (1);
$\begin{align}
  & x+y+z=43 \\
 & \Rightarrow x+5+z=43 \\
 & \Rightarrow x+z=43-5 \\
 & \Rightarrow x+z=38 \\
\end{align}$ (3)
And, In equation (2);
\[\begin{align}
  & 10x+5y+2z=229 \\
 & \Rightarrow 10x+\left( 5\times 5 \right)+2z=229 \\
 & \Rightarrow 10x+25+2z=229 \\
 & \Rightarrow 10x+2z=229-25 \\
 & \Rightarrow 10x+2z=204 \\
 & \\
\end{align}\] (4)
Multiplying equation (3) with ‘2’;
$\Rightarrow x+z=38\xrightarrow[{}]{}$(3) × 2
\[\Rightarrow 2(x+z)=38\times 2\]
$\Rightarrow 2x+2z=76$ (5)
and, eqn (4) $10x+2z=204$
Subtracting eqn (5) from eqn (4)
\[\underline{\begin{matrix}
   10x & + & 2z & = & 204 \\
   2x & + & 2z & = & 204 \\
   - & {} & - & {} & - \\
\end{matrix}}\]
$8x=128$
Putting value of ‘x’ in equation (3);
$x+z=38$
$\Rightarrow 16+z=38$
$\Rightarrow $
So, \[\begin{align}
  & x=16 \\
 & y=5 \\
 & z=22 \\
\end{align}\]
Hence (x, y, z) (16, 5, 22) is a solution to the above problem

Note: The core rule of algebra is the number of unknowns should be equal to the number of equations. then only we’ll be able to find the unique solution. If the number of unknowns is greater then the number of equations then we need to assume some initial value as we did in this question.