(a) The present ages of a father and his son are in the ratio 10
(a) The present ages of a father and his son are in the ratio 10 : 3. If the son is 15 years old now, in how many years will the ratio of their ages be 2 : 1?
(b) The arithmetic mean of x, y and z is 6 while that of x, y, z, l, u, v and w is 9. Calculate the arithmetic mean of l, u, v and w.
Explanation
(a) Present ages of a father and his son are in the ratio 10 : 3.
Let the sum of their ages be x.
The son's age = 15 years = \(\frac{3}{13} \times x = 15\)
\(x = \frac{15 \times 13}{3} = 65\)
\(\therefore\) The Father's present age = 65 - 15 = 50 years.
In z years time, the ratio of their ages = 2 : 1
\(\frac{50 + z}{15 + z} = \frac{2}{1}\)
\(\implies 50 + z = 2(15 + z)\)
\(50 + z = 30 + 2z \implies 50 - 30 = 2z - z\)
\(20 = z\)
Therefore, in 20 years' time, the ratio of their age will be 2 : 1.
(b) \(\frac{x + y + z}{3} = 6 \implies x + y + z = 18 .... (1)\)
\(\frac{x + y + z + l + u + v + w}{7} = 9 \implies x + y + z + l + u + v + w = 63 .... (2)\)
Putting (1) into (2), we have
\(18 + l + u + v + w = 63 \)
\(\therefore l + u + v + w = 63 - 18 = 45\)
Mean of l, u, v and w = \(\frac{45}{4} = 11.25\).