Solved 2024 Specimen Paper ICSE Class 10 Mathematics — Question 7

(a) Let the three positive numbers be $\dfrac{a}{r}, a, ar$.

Given,

Product of the numbers is 3375.

$\Rightarrow \dfrac{a}{r} \times a \times ar = 3375 \\[1em] \Rightarrow a^3 = 3375 \\[1em] \Rightarrow a^3 = (15)^3 \\[1em] \Rightarrow a = 15.$

(b) Given,

Result of the product of first and second number added to the product of second and third number is 750.

$\therefore \dfrac{a}{r} \times a + a \times ar = 750 \\[1em] \Rightarrow \dfrac{a^2}{r} + a^2r = 750 \\[1em] \Rightarrow a^2\Big(\dfrac{1}{r} + r\Big) = 750 \\[1em] \Rightarrow 15^2\Big(\dfrac{1}{r} + r\Big) = 750 \\[1em] \Rightarrow \Big(\dfrac{1 + r^2}{r}\Big) = \dfrac{750}{15^2} \\[1em] \Rightarrow \Big(\dfrac{1 + r^2}{r}\Big) = \dfrac{10}{3} \\[1em] \Rightarrow 3(1 + r^2) = 10r \\[1em] \Rightarrow 3 + 3r^2 = 10r \\[1em] \Rightarrow 3r^2 - 10r + 3 = 0 \\[1em] \Rightarrow 3r^2 - 9r - r + 3 = 0 \\[1em] \Rightarrow 3r(r - 3) - 1(r - 3) = 0 \\[1em] \Rightarrow (3r - 1)(r - 3) = 0 \\[1em] \Rightarrow 3r - 1 = 0 \text{ or } r - 3 = 0 \\[1em] \Rightarrow 3r = 1 \text{ or } r = 3 \\[1em] \Rightarrow r = \dfrac{1}{3} \text{ or } r = 3.$

Let r = $\dfrac{1}{3}$

Numbers : $\dfrac{a}{r}, a, ar$

= $\dfrac{15}{\dfrac{1}{3}}, 15, 15 \times \dfrac{1}{3}$

= 45, 15, 5.

Let r = 3

Numbers : $\dfrac{a}{r}, a, ar$

= $\dfrac{15}{3}, 15, 15 \times 3$

= 5, 15, 45.

Hence, numbers are 5, 15 and 45 or 45, 15 and 5.