A stationery shop ordered 6 dozen red pens and some additional dozen blue pens. The price of the red pens per dozen was three times that of the blue ones. When the order was delivered, it was found that the number of dozens of the two colours had been interchanged. This increased the bill by 40%. Find the ratio of the number of dozens of red pens to the number of dozens of blue pens in the original order.

Correct — B. Let blue cost ₹p a dozen, so red costs ₹3p, and let x dozen blue be ordered.

Original bill: 6 × 3p + x × p = (18 + x)p
Delivered bill: x × 3p + 6 × p = (3x + 6)p

40% increase: 3x + 6 = 1.4(18 + x)
Expand: 3x + 6 = 25.2 + 1.4x
Collect terms: 1.6x = 19.2, so x = 12

Red : blue = 6 : 12 = 1 : 2 → option (b)
Check: the bills are 30p and 42p, and 42 ÷ 30 = 1.4

Why the others are wrong

• (a)
  1:1 — 1 : 1 means 6 dozen of each, so swapping the counts changes nothing and the bill rises by 0%, not 40%.
• (c)
  2:3 — 2 : 3 means 9 dozen blue. The bill would go from 27p to 33p, a rise of about 22%, short of the stated 40%.
• (d)
  3:4 — 3 : 4 means 8 dozen blue. The bill would go from 26p to 30p, a rise of about 15%, not 40%.

Swapping the counts of a dear item and a cheap one changes the bill only through the difference in counts. More of the cheap item was ordered, so the swap moves those extra dozens onto the dearer price and the bill rises.

Write both bills in terms of one price p, apply the percentage change, and p cancels, leaving one equation in the unknown count.

Key facts

• A 40% increase means new bill = 1.4 × old bill.
• 6 dozen red at 3p and 12 dozen blue at p cost 18p + 12p = 30p.
• After the swap, 12 dozen red and 6 dozen blue cost 36p + 6p = 42p.

Common traps

• Swapping the prices instead of the counts: each colour keeps its price, only the dozens change.
• Giving blue : red (2 : 1) when the question asks for red : blue.

No directly related past PYQ was found.