A 45-litre mixture of kerosene and diesel contains kerosene and diesel in the ratio 4:5. How much diesel must be added to make the ratio 4:6?
- (a)3 L
- (b)5 L
- (c)6 L
- (d)7.5 L
Answer
Why
Correct — B. Only diesel is added, so the kerosene stays fixed. Split the 45 L first.
One ratio part: 45 ÷ (4 + 5) = 5 L
Kerosene = 4 × 5 = 20 L, diesel = 5 × 5 = 25 L
Diesel needed for 4 : 6 = 20 × 6⁄4 = 30 L
Diesel to add = 30 − 25 = 5 L → option (b)
Why the others are wrong
- (a)3 L — 3 L brings diesel to 28 L, and 20 : 28 = 5 : 7, not 4 : 6. Diesel has to reach 30 L.
- (c)6 L — 6 L brings diesel to 31 L, so kerosene to diesel becomes 20 : 31, just past the 20 : 30 that 4 : 6 requires.
- (d)7.5 L — 7.5 L brings diesel to 32.5 L: 20 : 32.5 = 8 : 13, which overshoots the target 20 : 30.
Concept
When only one liquid is added to a mixture, the other liquid does not change, so it anchors the new ratio.
Find the fixed quantity first (20 L of kerosene), then read off what the new ratio asks of the other: kerosene : diesel = 4 : 6 means diesel = 6⁄4 of the kerosene.
In ratio parts: kerosene stays at 4 parts while diesel goes from 5 to 6, so add 1 part = 5 L.
The target 4 : 6 is 2 : 3 in lowest terms. Left as 4 : 6, it lines up with 4 : 5, since kerosene is 4 parts in both.
Key facts
- In a ratio a : b with total T, one part = T ÷ (a + b).
- Adding only one component leaves the other unchanged, so work from the unchanged one.
- 4 : 6 and 2 : 3 are the same ratio.
Study next
Common traps
- Applying 4 : 6 to the old 45 L: that puts kerosene at 18 L, but no kerosene was removed.
- Answering with the new diesel total, 30 L, instead of the amount added, 30 − 25 = 5 L.
Adding one liquid to reach a new ratio also appears at 12 Sep 2025, 09:00, Quant Q.15, where 40 L of juice and water at 5 : 3 needs 22.5 L of water to reach 2 : 3.
23 Sep 2024, 09:00, Quant Q.20 sets the target as a percentage: 8 L of water takes a 32 L milk-water mix at 5 : 3 to 50% water.
Related PYQs
No directly related past PYQ was found.