1 the sum of the ages of a man and his son is 45 years 5 years ago,the product - Maths - Quadratic Equations

Question

1 the sum of the ages of a man and his son is 45 years 5 years
ago,the product of their ages was 4 times the man's age at that
time
2 7 yrs ago varun's age was 5 times the square of swati's age 3
yrs hence her age will be two fifth of varun's age find der
ages
3 the speed of a boat in still water is 8km/hr it can go 15km
upstream and 22km downstream in 5hrs find da speed of the
stream
4 a fast train takes 1 hour less than a slow train for a journey
of 200km if the speed of the slow train is 10km/hr less than that
of the fast train find the speed of the 2 trains
(pleez help in the distance sums im not understanding
which value will come 1st n which sine 2 put)

Lalit Mehra · Accepted Answer

Dear Student!

Let the present age of Swati be x years.

7 years ago,

Varun's age = 5 × (Swati's age)²

∴ Varun's age = 5 × (x – 7)²

Present age of Varun = 5 (x – 7)² + 7

3 years hence,

Swati's age = $\frac{2}{5} \times Varun's age$

$∴ (x + 3) = \frac{2}{5} [{5 (x - 7)^{2} + 7} + 3] \Rightarrow (x + 3) = \frac{2}{5} [5 (x - 7)^{2} + 10] \Rightarrow 5 (x + 3) = 2 [5 (x^{2} + 49 - 14 x) + 10] \Rightarrow 5 x + 15 = 10 (x^{2} + 49 - 14 x) + 20 \Rightarrow 5 x + 15 = 10 x^{2} - 140 x + 490 + 20 \Rightarrow 10 x^{2} - 145 x + 495 = 0 \Rightarrow 2 x^{2} - 29 x + 99 = 0 \Rightarrow 2 x^{2} - 18 x - 11 x + 99 = 0 \Rightarrow 2 x (x - 9) - 11 (x - 9) = 0$

$\Rightarrow (2 x - 11) (x - 9) = 0 \Rightarrow 2 x - 11 = 0 or x - 9 = 0 \Rightarrow x = \frac{11}{2} or x = 9 ∴ x = 9 (∵ x cannot be \frac{11}{2})$

Present age of Swati = 9 years

Present age of Varun = $5 (x - 7)^{2} + 7 = 5 (9 - 7)^{2} + 7 = 5 \times 4 + 7 = 20 + 7 = 27$

3.

Let the speed of the stream be x km/hr.

Speed of the boat upstream = Speed of boat in still water – Speed of the stream

∴ Speed of the boat upstream = (8 – x) km/hr

Speed of the boat downstream = Speed of boat in still water + Speed of the stream

∴ Speed of the boat downstream = (8 + x) km/hr

Given, Time for upstream journey + Time for downstream journey = 5 hrs

$∴ \frac{Distance covered upstream}{Speed of the boat upstream} + \frac{Distance covered downstream}{Speed of the boat downstream} = 5 hrs \Rightarrow \frac{15 km}{(18 - x) km/hr} + \frac{22 km}{(8 + x) km/hr} = 5 hrs \Rightarrow \frac{15}{8 - x} + \frac{22}{8 + x} = 5 \Rightarrow \frac{15 (8 + x) + 22 (8 - x)}{(8 - x) (8 + x)} = 5 \Rightarrow \frac{120 + 15 x + 176 - 22 x}{64 - x^{2}} = 5 \Rightarrow 296 - 7 x = 320 - 5 x^{2} \Rightarrow 5 x^{2} - 7 x - 24 = 0 \Rightarrow 5 x^{2} - 15 x + 8 x - 24 = 0 \Rightarrow 5 x (x - 3) + 8 (x - 3) = 0$

$\Rightarrow (x - 3) (5 x + 8) = 0 \Rightarrow x - 3 = 0 or 5 x + 8 = 0 \Rightarrow x = 3 or x = - \frac{8}{5} ∴ x = 3 (∵ S peed cannot be negative)$

Thus, speed of the stream is 3 km/hr.

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