Let Sunilâ€™s age be x; Anilâ€™s age be y. then,
x - y = 4, xy = 672.
(x + y)^{2} = (x â€“ y)^{2} + 4xy.
(x + y)^{2} = 4^{2} + 4 x 672 => 16 + 2688.
(x + y)^{2} = 2704 = x + y = 52.
x |
+ |
= |
52 |
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x |
- | = |
4 |
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2x |
= |
56 |
2x = 56; x = $\displaystyle \frac{56}{2} = 28.$
x = 28 => Sunil age = 28, Anil age = 52 â€“ 28 = 24.
Let number of sons be x, then grandsons = x (x â€“ 1).
Age of Kala = x + x^{2} â€“ x = between 80 and 90 or x^{2} = between 80 and 90.
The perfect square that lies between 80 and 90 is 81, then age of Kala must be 81.
Note: in such type of questions the age lies between the given ranges of age.
Ex: if range is 60 and 70 the age must be 64 years. which is a perfect square between 60 and 70.
Suchaâ€™s age 6 years, then Meenaâ€™s age = 6 x 3 = 18.
After 5 years Meenas age = 18 + 5 = 23 years.
Let age of Jaya and Amit age 15 years ago be 8x : 9x.
Then present ages = $\displaystyle \frac{8\text{x} \ + 15}{9\text{x} \ + 15} = \frac{3}{4}$ => 32x + 60 = 27x + 45.
5x = 15 => x = 3.
Age of Jaya 15 years ago was 8x = 8 x 3 = 24, Amit = 9x = 9 x 3 = 27.
Present age of Jaya = 24 + 15 = 39; Amit = 27 + 15 = 42.
We can find from the options.
(X + Y) â€“ (Y + Z) = 6 => X â€“ Z = 6 years.
Sum of ages of son and father 8 years ago = 31 x 2 = 62.
Present sum of ages = (31 + 8) x 2 = 78.
Father = 78 $\displaystyle \times \frac{2}{3}$ = 52.
Son = 78 $\displaystyle \times \frac{1}{3}$ = 26.
Sum of ages = 18 x 18 = 324.
New average = 18.8 years.
Sum of ages including 12 new students = (18 + 12) x 18.8 = 564.
Sum of ages of 12 new students = 564 â€“ 324 = 240.
Average age = $\displaystyle \frac{240}{12}$ = 20 years.
From the options we can find, with the help of trial and error method.
From the options we can find, with the help of trial and error method.
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