Question

If P is x% more than Q but y% less than R, then find the ratio of Q to R.

• A. (100 – y):(100 + x)
• B. (100 + y):(100 – x)
• C. (100 – x):(100 + y)
• D. (100 + x):(100 – y)

Answer 1

The correct option is A (100 – y):(100 + x)

$P=Q(1+\frac{x}{100})....\left(1\right)$

$andP=R(1-\frac{y}{100})....\left(2\right)$

$Dividing\left(1\right)by\left(2\right),wegetQ:R=(1-\frac{y}{100}):(1+\frac{x}{100})=(100-y):(100+x)$

Alternate method:
We can assume any arbitrary values for x and y.

Let x = 10 and y = 20.
Also, let Q = 100q
So, P = 100q + 10% of 100q = 110q
Let R = 100r
So, P = 100r – 20r = 80r
So, 110 q = 80r
⇒ Ratio of q to r = 8:11
Now, putting x = 10 and y = 20 in A, we can see that it is coming out to be 8:11.
Hence, option A is the correct answer.