CSAT Solved Papers/ 2023/Q44

2023 CSAT — Q44

Quant Arithmetic & numeracy 2.5 marks Medium

A principal PP becomes QQ in 11 year when compounded half-yearly with R%R\% annual rate of interest. If the same principal PP becomes QQ in 11 year when compounded annually with S%S\% annual rate of interest, then which one of the following is correct?

  1. A R = S
  2. B R > S
  3. C R < S Answer
  4. D R ≤ S

Worked rationale

Both schemes turn the same PP into the same QQ in one year, so their growth factors are equal:

(1+R200)2=1+S100.\left(1 + \frac{R}{200}\right)^{2} = 1 + \frac{S}{100}.

Expand the left side:

1+R100+R240000=1+S100.1 + \frac{R}{100} + \frac{R^{2}}{40000} = 1 + \frac{S}{100}.

Cancel the 11 and multiply by 100100:

S100=R100+R240000  ⇒  S=R+R2400.\frac{S}{100} = \frac{R}{100} + \frac{R^{2}}{40000} \;\Rightarrow\; S = R + \frac{R^{2}}{400}.

Since R>0R > 0, the extra term R2400>0\dfrac{R^{2}}{400} > 0, so S>RS > R, i.e.

R<S.R < S.

Answer: (c) R < S.

Why the other options miss

  • A
    solved the wrong question: assumes nominal rates match when final amounts match, ignoring that more frequent compounding earns more, so a lower nominal rate suffices.
  • B
    the proportion is backwards: inverts the effect of compounding frequency — thinks the half-yearly scheme needs a higher rate to reach the same QQ.
  • D
    missed a case: allows equality, but R=SR = S is impossible for R>0R > 0 (the R2/400R^2/400 term is strictly positive), so the relation is strict.

Specialist insight

The principle: for the same final amount, more frequent compounding requires a lower nominal rate, because half-yearly compounding has an effective annual rate (1+R200)2−1=R100+R240000>R100\left(1+\frac{R}{200}\right)^2 - 1 = \frac{R}{100} + \frac{R^2}{40000} > \frac{R}{100}. Equating effective rates gives S=R+R2/400S = R + R^2/400, so S>RS > R strictly. The "≤\le" decoy tempts a cautious reader, but the strict positivity of R2/400R^2/400 rules out equality — answer-craft reward goes to the strict inequality (c).

The trap, in one line

Equal amounts ⇒(1+R/200)2=1+S/100⇒S=R+R2/400>R\Rightarrow (1+R/200)^2 = 1+S/100 \Rightarrow S = R + R^2/400 > R, so R<SR < S (strict) == (c).

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