Shobo's mother's present age is six times Shobo's present age. Shobo's age five years from now will be one third of this mother's present age. What are their present ages?

Asked by Sakshi | 1 year ago |  94

##### Solution :-

Let Shobo's age be x years. Therefore, his mother's age will be 6x years.

According to the given question,

After 5 year, shobo's age = $$\frac{shobo's\, mother's\,present\,age}{3}$$

x + 5 = $$\frac{6x}{3}$$

x + 5 = 2x

Transposing x to R.H.S, we obtain

5 = 2x - x

5 = x

6x = 6 × 5 = 30

Therefore, the present ages of Shobo and Shobo's mother will be 5 years and 30 years respectively.

Answered by Aaryan | 1 year ago

### Related Questions

#### Solve the inequations and graph their solutions on a number line  – 1 < (x / 2) + 1 ≤ 3, x ε l

Solve the inequations and graph their solutions on a number line  – 1 < ($$\dfrac{x}{2}$$) + 1 ≤ 3, x ε l

#### Solve the inequations and graph their solutions on a number line – 4 ≤ 4x < 14, x ε N

Solve the inequations and graph their solutions on a number line – 4 ≤ 4x < 14, x ε N

#### Solve (x / 3) + (1 / 4) < (x / 6) + (1 / 2), x ε W. Also represent its solution on the number line.

Solve ($$\dfrac{x}{3}$$) + ($$\dfrac{1}{4}$$) < ($$\dfrac{x}{6}$$) + ($$\dfrac{1}{2}$$), x ε W. Also represent its solution on the number line.

If the replacement set is {-3, -2, -1, 0, 1, 2, 3}, solve the inequation $$\dfrac {(3x – 1) }{2} < 2$$. Represent its solution on the number line.
Solve the inequations ($$\dfrac{3}{2}$$) – ($$\dfrac{x}{2}$$) > – 1, x ε N