CA Foundation 2026 · Paper 3 · Quantitative AptitudeQ38 · 1 mark↻ Appears in 2 of 6 yearsOfficial key verified
The mean of five observations is 28. Among the five observations, three observations are 10, 23 and 62. The difference between the remaining two observations is 13. Then the remaining two observations are ______.
Figure for question 38
Single correct — pick one, then checkICAI format
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

The last two observations sum to 4545 and differ by 1313, giving 2929 and 1616.

Step 1 — Find the total sum

Mean =28= 28 over 55 observations, so

∑x=28×5=140\sum x = 28 \times 5 = 140

Step 2 — Subtract the known observations

The three given values total 10+23+62=9510 + 23 + 62 = 95, so the remaining two satisfy

x+y=140−95=45x + y = 140 - 95 = 45

Step 3 — Use the difference and solve

Given x−y=13x - y = 13, add the two equations:

2x=45+13=58  ⇒  x=29,y=45−29=162x = 45 + 13 = 58 \;\Rightarrow\; x = 29,\qquad y = 45 - 29 = 16

Check: 29+16=4529 + 16 = 45 ✓ and 29−16=1329 - 16 = 13 ✓.

Why the other options are wrong: (A) 30+17=4730+17=47, (B) 38+15=5338+15=53, (D) 35+22=5735+22=57 — none sum to the required 4545, so only 2929 and 1616 fit both conditions. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.

← Back to the Paper 3 · Quantitative Aptitude 2026 paper