Question

The table lists the unit selling price of five products P, Q, R, S, and T. On a particular day, 250 items were sold with the average selling price of Rs. 60. The following observations were made:
(i) The quantity of S sold was twice that of T.
(ii) The quantity of R sold was thrice that of T.
(iii) The quantity of Q sold was four times that of T.

Product	P	Q	R	S	T
Unit selling price (Rs.)	100	50	40	60	60

What is the quantity of product P sold on that day?

• A. 40
• B. 50
• C. 60
• D. 70

Hint:

Use the ratio clues to write everyone's quantity in terms of T's quantity, then solve total-quantity and total-revenue equations together.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Instead of writing two separate equations, we can use the idea of deviation from the average price. Every item's price minus the average price of Rs. 60, weighted by how many of that item were sold, must add up to zero, since the average is exactly 60.

Step 2: Key Formula or Approach:
For each product, find (unit price minus 60) and multiply by its quantity. The sum of these weighted deviations across all five products must equal zero:
\[ \sum (\text{price}_i - 60)\times \text{quantity}_i = 0 \]

Step 3: Detailed Explanation:
Let quantity of T be $t$, so Q $=4t$, R $=3t$, S $=2t$, and P $=p$ (unknown).
Deviations from 60: P deviation $=100-60=40$, Q deviation $=50-60=-10$, R deviation $=40-60=-20$, S deviation $=60-60=0$, T deviation $=60-60=0$.
Applying the zero-sum rule:
\[ 40p + (-10)(4t) + (-20)(3t) + 0(2t) + 0(t) = 0 \]
\[ 40p - 40t - 60t = 0 \]
\[ 40p = 100t \]
\[ p = 2.5t \]
Now use the total quantity equation, $p + 4t+3t+2t+t = 250$, that is, $p + 10t = 250$.
Substitute $p = 2.5t$:
\[ 2.5t + 10t = 250 \]
\[ 12.5t = 250 \]
\[ t = 20 \]
So $p = 2.5(20) = 50$.

Step 4: Final Answer:
The quantity of product P sold on that day is 50, which matches option (B). This deviation method avoids computing the full revenue total of Rs. 15000 directly, since the zero-sum-of-deviations trick already encodes the average condition. \[ \boxed{p = 50} \]