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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