CAT 2008 LRDI Question 6 · DI - Tables & Graphs

Let the number of men subscribers in 2003 be m, and the number of women subscribers in 2003 be w.

It is given that the women subscribers increase at the rate of 10% per year and the men increase at the rate of 5% per year.

∴ Number of men subscribers in 2007 = m(1 + 0.05)⁷ ≈ m[1 + ⁷C₁ (0.05) + ⁷C₂ (0.05)² + ⁷C₃ (0.05)³]

≈ m(1 + 0.35 + 0.0525 + 0.0043)

≈ 1.4m

Number of women subscribers in 2007 = w(1 + 0.1)⁷ ≈ w [1 + ⁷C₁(0.1) + ⁷C₂ (0.1)² + ⁷C₃ (0.1)³]

≈ w$\left[1+7(0.1)+\frac{7\times 6}{2}{(0.1)}^2+\frac{7\times 6\times 5}{2\times 3}{(0.1)}^3\right]$

≈ w(1 + 0.7 + 0.21 + 0.035)

≈ 1.95w

Now, it is given that in 2003, men are 60% of the total European subscribers. So, women are 40% of the subscribers. Let the total European subscribers be P. Then,

m = 0.6P and

w = 0.4P

∴ The total number of subscribers in 2003 = m + w = 0.6P + 0.4P = P

And the total number of subscribers in 2007 = 1.4m + 1.95w

= 1.4(0.6P) + 1.95(0.4P)

= 0.84P + 0.78P = 1.62P

∴ Percentage growth of subscribers between '03 and '07 = $\frac{(1.62)P-P}{P}$ × 100 = 62%

Hence, option (a).