# Quadratic equation class 10 word problem with solution

Here, you will get all types of quadratic equation class 10 word problems with the solution. So, first try to solve it, on your own and then see the solution.

## Problem-based on numbers

Q.1 The sum of two numbers is 15.if the sum of their reciprocal is \frac{3}{10} :find the numbers

Solution: Let the required numbers are x and 15-x .Then,

It is given that ,sum of their reciprocal is \frac{3}{10}

\frac{1}{x}+\frac{1}{15-x}=\frac{3}{10}

\frac{15-x+x}{x(15-x)}=\frac{3}{10}

\frac{15}{x(15-x)}=\frac{3}{10}

150=3x(15-x)

150=45x-3x²

x²-15x+50=0

x²-10x-5x+50=0

x(x-10)-5(x-10)=0

(x-10)(x-5)=0

(x-10) =0 or (x-5)=0

x=10         or       x=5

Hence both numbers are 10 and 5

Q.2 A two-digit number is such that the product of the digits is 14.When 45 is added to the number, then the digits are reversed. Find the number

Solution:

Let the ten’s digit be x .Then ,unit digit will be \frac{14}{x}.

Number becomes=10x+\frac{14}{x}.

Required number when digit get reversed =10×\frac{14}{x}+x

According to question

10x+\frac{14}{x}+45=10×\frac{14}{x}+x

9x²+45x-126=0

x²+5x-14=0

x²+7x-2x-14=0

(x+7)(x-2)=0

(x+7)=0 or (x-2)=0

x=-7     or x=2

Required number =10×2+\frac{14}{2}=27

## Speed and time word problem

Q.1 A train travels a distance of 300km at a constant speed. If the speed of the train is increased by 5km an hour, the journey would have taken 2 hours less. Find the original speed of the train

Solution :

Let the original speed of the train be x km/hr

Time taken by the train to cover 300km =\frac{300}{x}

If the speed of the train increased by 5km/hr ,then time taken to cover same distance=\frac{300}{x+5}

According to question,

\frac{300}{x}\frac{300}{x+5}=2

\frac{300x+1500-300x}{x²+5x}=2

2x²+10x=1500

x²+5x-750=0

x²+30x-25x-750=0

x(x+30)-25(x+30)=0

(x+30)(x-25)=0

(x+30)=0   or  (x-25)=0

x=-30          or x=25

X cannot be negative

So, original speed of the train is 25km/hr

Q.2 The speed of a boat in still water is 8km/hr. It can go 15km upstream and 22km downstream in 5hours. Find the speed of the stream

Solution: Let the speed of the stream=xkm/hr

speed of boat during downstream=(8+x)km/hr

Speed of boat during upstream=(8-x)km/hr

Time taken for going 15 upstream =\frac{15}{8-x}km/hr

Time taken for going 22 downstream=\frac{22}{8+x}km/hr

It is given that ,total time taken during this journey is 5 hour

So,

\frac{15}{8-x}km/hr+\frac{22}{8+x}km/hr=5

\frac{296-7x}{64-x²}=5

296-7x=320-5x²

5x²-7x-320+296=0

5x²-7x-24=0

(x-3)(5x+8)=0

(x-3)=0     or    (5x+8)=0

x=3             or              x=\frac{-8}{5}     this is not possible

So ,speed of the stream=5km/hr

## Age,triangle and work based  problem

Q.1 The sum of ages of a father and his son is 45 years. Five years ago, the product of their ages was 124 years . Determine their present ages

Solution: Let the present age of the father is x years. Then,

Age of the son =(45-x) years

Five years ago,

Father’s age =(x-5) years

Son’s age =(45-x-5) years

It is given that five years ago ,the product of their ages was 124

(x-5)(40-x)=124

40x-x²-200+5x=124

x²-45x+324=0

x(x-36)-9(x-36)=0

(x-36)(x-9)=0

(x-9)=0    or (x-36)=0

The value of x are 9 and 36

Case:I When x=36 ,we have

Father’s present age =36 years

Son’s present age =9 years

Whereas Case II is not possible

So, the present age of the son and father is 9 year and 45 years respectively

Q.2 The hypotenuse of the right-angled triangle is 6 meters more than twice the shortest side.If the third side is 2 meters less than the hypotenuse, find the sides of the triangle

Solution: Let the length of the shortest of the triangle = x meters

Hypotenuse = (2x+6) metres

And ,The third side =(2x+6-2) metres =(2x+4) metres

Applying Pythagoras theorem

(2x+6)²=x²+(2x+4)²

x²-8x-20= 0

x²-10x+2x-20=0

x(x-10)+2(x-10)=0

(x-10) (x+2)=0

Hence Length of the shortest side of the triangle = 10 meters

Length of the third side = 24 meters

Length of the hypotenuse = 26 meters

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