In this page quadratic equation solution26 we are going to see solution of the word problems of the topic quadratic equation.

**Question 33**

An aeroplane traveled a distance of 400 km at an average speed of x km/hr. On the return journey,the speed was increased by 40 km/hr. Write down an expression for the time taken for (i) the onward journey and (ii) the return journey. If the journey took 30 minutes less than onward journey,write down an equation in x and its value.

**Solution:**

Let “x” be average speed of aeroplane

On the return journey,the speed was increased by 40 km/hr

So “x + 40” be the speed of aeroplane

Distance to be covered = 400 km

Let T1 be the time taken for onward journey in the speed of x km/hr

Let T2 be the time taken for downward journey to cover the same distance 400 km at the speed of (x + 40) km/hr

Time = Distance /speed

T1 = 400/x

T2 = 400/(x + 40)

T1 - T2 = 30 minutes

[400/x] - [400/(x + 40)] = 30/60

400[(1/x) - 1/(x + 40)] = 1/2

400[(x + 40 - x)/x(x + 40)] = 1/2

400[40/(x² + 40 x)] = 1/2

16000 (2) = (x² + 40 x)

x² + 40 x - 32000 = 0

x² + 160 x - 100 x - 32000 = 0

x(x + 160)- 100 (x + 160) = 0

(x - 100) (x + 160) = 0

x - 100 = 0 x + 160 = 0

x = 100 x = -160

Here x represents the speed of the areoplane. So we should not take the negative value - 160 for x.

Speed of the aeroplane = 48 km/hr

Increased speed = (x + 40) = (48 + 40) = 88 km/hr

**Verification:**

Time = Distance/speed

Time taken by the aeroplane to cover 400 km

for onward journey = 400/100

= 4 hours

= 4 x 60 = 240 minutes

Time taken by the aeroplane to cover 400 km

for downward journey = 400/88

= 4.54 hours

= 272 minutes

difference of time taken = 272 - 240 = 30 minutes

quadratic equation solution26

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