SSC CGL 2024 Tier-I · 12 September 2024, Shift 2 · Quantitative Aptitude · Question 20 of 100

Text as printed on a candidate's response sheet for this shift; answer per SSC's tentative key as marked on that sheet. Solution last reviewed 6 September 2026.

Where this paper and its answers come from

SSC does not publish CGL question papers or a public answer key. The questions and options here are reproduced as printed on a candidate's own response sheet for this shift — the Commission's text, unchanged. Each answer is SSC's tentative key as marked on that sheet; SSC allowed challenges to it and never publishes a final key. The solutions are written by UnlockIAS.

In a garden, there are 6 daisy plants the first year. Each year, a gardener adds 3 new daisy plants. He has 26 jasmine plants the first year and loses 2 each year. When will the number of daisy plants equal to the number of jasmine plants after the first year?

(a)6 years
(b)4 years
(c)5 years
(d)2 years

Answer

Why

Correct — B. Let n be the number of years after the first year and write each plant count as a straight line.

Daisies = 6 + 3n
Jasmines = 26 − 2n

Set them equal:
6 + 3n = 26 − 2n
5n = 20
n = 4

Check: daisies 6 + 12 = 18, jasmines 26 − 8 = 18, equal after 4 years → option (b).

Why the others are wrong

(a)
6 years — At n = 6 the daisies stand at 6 + 18 = 24 and the jasmines at 26 − 12 = 14. The counts crossed two years earlier and the gap has already reversed.
(c)
5 years — 5 is the year number, not the years elapsed. Counting the first year itself as one of the n years shifts everything by one; at n = 5 the daisies are 21 against 16 jasmines.
(d)
2 years — At n = 2 the counts are 12 and 22, still 10 apart. The starting gap is 26 − 6 = 20 and it closes by 5 a year, so it takes 20 ÷ 5 = 4 years.

Concept

Two quantities that change by a fixed amount each year are two arithmetic sequences, and the question of when they become equal is one linear equation.

Write each as start + (rate × n), watching the sign: +3 for a gain, −2 for a loss. Solving 6 + 3n = 26 − 2n gives n = 4.

A faster route uses the gap. The counts begin 26 − 6 = 20 apart, and the gap shrinks by 3 + 2 = 5 every year, so they meet after 20 ÷ 5 = 4 years.

The question says after the first year, so n counts years elapsed from the first-year figures of 6 and 26. Reading n as the serial number of the year gives 5 instead.

Key facts

Daisies follow 6 + 3n and jasmines follow 26 − 2n, where n is years after the first year.
The counts are equal when 6 + 3n = 26 − 2n, which gives n = 4.
At n = 4 both stand at 18 plants.
Gap method: an initial gap of 20 closing at 5 a year meets in 4 years.

Study next

Arithmetic progressions and the nth term a + (n − 1)dSolving a linear equation in one variable by transpositionRelative-rate problems where a gap closes at the sum of two rates

Common traps

Counting the first year as one of the n years and answering 5.
Adding 2 to the jasmine count instead of subtracting it.
Dividing the starting gap by one rate alone, 20 ÷ 3 or 20 ÷ 2, instead of by their sum.

SSC dresses a fixed yearly change as plants, salary or stock. The salary version, where a fixed annual increment connects two known years, is asked on 11 Sep 2024, 09:00, Quant Q.20.

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