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:

Write the quantities of Q, R, S in terms of T, then use both the total item count (250) and the total revenue (250 x 60) to form two equations in P's quantity and T.

Answer 1

The Correct Option is B

A faster route through this kind of weighted-average problem is to work with how far each product's price sits above or below the overall average, instead of writing out full revenue totals. Since the average price is Rs. 60, look at each product's price minus 60.

1. Price deviations from the average of Rs. 60: P is $100-60 = 40$ above average, Q is $50-60 = -10$ (below), R is $40-60=-20$ (below), S is $60-60=0$ (exactly average), T is $60-60=0$ (exactly average).
2. Because the overall average is fixed at 60, the total of (quantity times deviation) across all five products must add up to exactly $0$, the extra earned above average must exactly cancel the shortfall below average.
3. Let quantity of T $=t$, so Q $=4t$, R $=3t$, S $=2t$, and let quantity of P $=p$. Since S and T both have a deviation of $0$, they contribute nothing to this balance, and only P, Q, R matter: $40p + (-10)(4t) + (-20)(3t) = 0$.
4. Simplify: $40p - 40t - 60t = 0$, so $40p = 100t$, giving $p = 2.5t$.

Now use the total count equation: $p + 4t+3t+2t+t = 250$, so $p+10t=250$. Substitute $p=2.5t$: $2.5t+10t=250$, so $12.5t=250$, giving $t=20$. Then $p=2.5(20)=50$.

Let's summarize:

• In a weighted-average problem, the sum of (quantity times deviation-from-average) across all groups always equals zero, a shortcut that skips computing full totals.
• Combining this deviation equation with the simple total-count equation solves for both unknowns quickly.

The quantity of product P sold that day works out to 50 units, matching option (B).