Question

The total cost of three prizes is Rs. 2550. If the value of the second prize is $\dfrac{3}{4}th$ of the first and the value of the 3rd prize is $\dfrac{1}{2}$ of the second prize. Find the value of the first prize.
A. Rs. 900
B. Rs. 1500
C. Rs. 1200
D. Rs. 450

Answer 1

Hint: In this, we have to form a linear equation in one variable by assuming the value of first prize and then applying the conditions given to us.
A linear equation is of the form $ax+b=0$, where a and b are constants and x is a variable.

Complete step-by-step answer:
Given that the total cost of three prizes is Rs. 2550. We assume that the cost of first prize is x.
Then, according the given condition,
Value of second prize = $\dfrac{3}{4}th$ of the first prize = $\dfrac{3}{4}\times x =\dfrac{3x}{4}$
Also, value of third prize = $\dfrac{1}{2}$ of the second prize = $\dfrac{1}{2}\times $\Rightarrow$ \dfrac{3x}{4}=\dfrac{3x}{8}$
Now, since, the total value of all the three prizes is Rs. 2550. Thus,
$\Rightarrow$ $x+\dfrac{3x}{4}+\dfrac{3x}{8}=2550$
Taking LCM in denominator, we get,
$\Rightarrow$ $\dfrac{8x+6x+3x}{8}=2550$
Cross multiplying, we get,
$\Rightarrow$ $8x+6x+3x=8\times 2550$
Solving this, we get,
$\Rightarrow$ $17x=20400$
$\implies x =\dfrac{20400}{17}=1200$
Thus, the value of first prize is Rs. 1200.
Hence, option C is correct.

Note: A linear equation in one variable can be solved by simplifying the terms. A linear equation in one variable has a unique solution, i.e. every linear equation has only and only one solution.