Scheme A offers R% p.a. compound interest, compounded annually. Find the value…

2020

Scheme A offers R% p.a. compound interest, compounded annually. Find the value of R.
Statement I – The interest received on Rs. (2/5)P at simple interest for 11 years at the rate of (R/4)% p.a. is half of the interest when Rs. P is invested in Scheme A for 2 years.
Statement II – Rs. X is invested in Scheme A for two years and it amounts to Rs. 3600. If Rs. (16/25)X is invested at the rate of (R – 10)% on compound interest annually for two years, then it amounts to Rs. 1936.
Statement III – When Rs. Q is invested in Scheme A for two years and for three years, the difference between the interests is Rs. 1152.

  1. A.

    Statement (I) alone is sufficient to answer the question

  2. B.

    Statement (II) alone is sufficient to answer the question

  3. C.

    All the three statements taken together are necessary to answer the question

  4. D.

    Either statement (I) alone or statements (II) and (III) together are sufficient to answer the question

  5. E.

    Either statement (I) alone or statement (II) alone is sufficient to answer the question

Show answer & explanation

Correct answer: E

Concept

This is a data-sufficiency question: the task is not to compute a single number but to judge, for each statement independently, whether the data it carries is enough to pin down R uniquely. The two governing tools are the simple-interest formula SI = P·T·r/100 and the compound-interest amount A = P(1 + r/100)n. Because Scheme A is described as offering an R% rate of interest, R is a genuine positive interest rate, so R = 0 (a no-interest scheme) is not an admissible value. A statement is sufficient only when the unknown principal cancels out, leaving one admissible value of R; if a statement still contains an extra unknown alongside R, it cannot fix R by itself.

Applying it to each statement

Statement I.

  1. SI on Rs. (2/5)P for 11 years at (R/4)% = (2/5)P · 11 · (R/4)/100 = 11PR/1000.

  2. CI on Rs. P in Scheme A for 2 years = P[(1 + R/100)2 − 1] = P(200R + R²)/10000.

  3. Given SI is half of that CI: 11PR/1000 = ½ · P(200R + R²)/10000.

  4. P cancels. Multiply through by 20000: 220R = 200R + R², so R² − 20R = 0, giving R(R − 20) = 0.

  5. The roots are R = 0 and R = 20; R = 0 is rejected because a 0% scheme carries no interest and contradicts the stem, leaving the single admissible value R = 20. The principal vanished and only one valid R survives — so this statement is sufficient on its own.

Statement II.

  1. X(1 + R/100)2 = 3600 (Scheme A, 2 years).

  2. (16/25)X(1 + (R − 10)/100)2 = 1936.

  3. Divide the second amount by the first to cancel X: (16/25)·[(1 + (R − 10)/100)/(1 + R/100)]² = 1936/3600.

  4. So [(1 + (R − 10)/100)/(1 + R/100)]² = 1936/2304 = (44/48)² = (11/12)².

  5. Take the positive root: (1 + (R − 10)/100)/(1 + R/100) = 11/12. Let a = 1 + R/100; then (a − 0.1)/a = 11/12, so 12a − 1.2 = 11a, giving a = 1.2 and R = 20 — a single value, with no spurious root. So this statement is also sufficient on its own.

Statement III.

  1. Interest difference = Q(1 + R/100)3 − Q(1 + R/100)2 = Q(1 + R/100)²·(R/100) = 1152.

  2. This is a single equation in two unknowns, Q and R. The principal Q does not cancel, so R cannot be determined — this statement is not sufficient by itself.

Cross-check and conclusion

  • Statement I → R = 20 alone (R = 0 inadmissible); Statement II → R = 20 alone with no extra root; Statement III → cannot fix R alone.

  • Plugging R = 20 back: in I, 11·20/1000 = 0.22 and ½·(200·20 + 400)/10000 = ½·0.44 = 0.22 ✓; in II, with a = 1.2 the ratio is (1.1/1.2) = 11/12 ✓.

  • Therefore either the first statement alone or the second statement alone is sufficient, and the third adds nothing on its own.

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