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?
Show Hint
Write one equation for the total number of items (250) and one for the total revenue (250 times Rs. 60), using T as the base quantity.
Step 1: Use the deviation-from-mean shortcut.
For a weighted average, the sum of (price minus average) times quantity must equal zero. Here the average price is $60$, so
\[ \sum (\text{price}_i - 60)\times \text{quantity}_i = 0 \]
Step 2: Work out each deviation.
$P: 100-60=40$, $Q: 50-60=-10$, $R: 40-60=-20$, $S: 60-60=0$, $T: 60-60=0$. Only $P$, $Q$ and $R$ contribute, since $S$ and $T$ sit exactly at the average price.
Step 3: Bring in the quantities.
Let $T=t$. Then $Q=4t$ and $R=3t$. Let $P=p$. The deviation equation becomes
\[ 40p + (-10)(4t) + (-20)(3t) = 0 \]
\[ 40p - 40t - 60t = 0 \]
\[ 40p = 100t \]
\[ p = 2.5t \]
Step 4: Use the total count.
All five quantities add to $250$:
\[ p + 4t + 3t + 2t + t = 250 \]
\[ p + 10t = 250 \]
Step 5: Solve for t and p.
Substitute $p=2.5t$:
\[ 2.5t + 10t = 250 \]
\[ 12.5t = 250 \]
\[ t = 20 \]
So $p = 2.5\times20 = 50$.
\[ \boxed{50} \]
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