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# Class ten maths Quadratic Equations NCERT Solution

#### TEXTBOOK’S EXERCISE 4.3

6. The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.

Sol. Let shorter side of the rectangular field = AD = x m

Longer side of the rectangular field = AB

= (x + 30)m

Diagonal of rectangular field = DB = (x + 60) m

Using Pythagoras Theorem in ΔDAB,

we have, (DB)2 = (AD)2 + (AB)2

⇒ (x + 60)2 = (x)2 + (x + 30)2

⇒ x2 + 3600 + 120x = x2 + x2 + 900 + 60x.

⇒ x2 + 3600 + 120x – 2x2 – 900 – 60x = 0

⇒ –x2 + 60x + 2700 = 0

⇒ x2 – 60x – 2700 = 0

Using the quadratic formula, we get

Since a side cannot be negative, so we will take the positive value.

So, x = 90

Hence, shorter side of rectangular field = 90 m

∴ Longer side of a rectangular field

= (90 + 30) m = 120 m

7. The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.

Sol. Let larger number = x. Smaller number = y

As per condition :

x2 – y2 = 180 …(i)                                        (∵ x > y)

Also, y2 = 8x …(ii)

From eqn. (i) and (ii), we have x2 – 8x = 180

⇒ x2 – 8x – 180 = 0

Using the quadratic formula, we get

When, x = –10, then from eqn. (ii),

y2 = 8(–10)

= – 80, which is impossible.

When, x = 18, then from Eqn. (ii),

y2 = 8(18) = 144

Hence, the required numbers are 18 and 12 or 18 and –12.

8. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Sol. Let the speed of the train = x km/hour

Distance covered by the train = 360 km

Time = Distance /Speed = 360 /x hours

Increased speed of the train = (x + 5) km/hour

∴ Time is taken by the train with increased speed

As per condition:

⇒ 1800 = x2 + 5x

⇒ x2 + 5x – 1800 = 0

Using the quadratic formula, we get

or 80/2 and −90/2 or 40 and – 45.

Since speed cannot be negative. So, we will take a positive value.

So, x = 40.

Hence, speed of train = 40 km/hour.

9. Two water taps together can fill a tank in 9 3/8 hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Sol. Let the time taken by larger tap = x hours

Time taken by smaller tap = (x + 10) hours

A portion of a tank filled in one hour by the larger tap = 1/x

A portion of a tank filled in one hour by the smaller tap

∴ A portion of a tank filled in one hour by both taps

But, two taps together can fill the tank in

hours, i.e., in 75 8 hours.

Now, a portion of a tank filled by two taps together in one hour = 8/75 …(ii)

From Eqn. (i) and Eqn. (ii), we have

⇒ 75(2x + 10) = 8 (x2 + 10x)

⇒ 150x + 750 = 8x2 + 80x

⇒ 8x2 + 80x – 150x – 750 = 0

⇒ 8x2 – 70x – 750 = 0

⇒ 4x2 – 35x – 375 = 0

Using the quadratic formula, we get

or 120/ 8 and −50/ 8 or 15 and −25/ 4

Since x is the time, it cannot be negative. Therefore, x = 15.

Hence, larger water tap will fill the tank separately in 15 hours.

Smaller water tap will fill the tank separately in (15 + 10) = 25 hours.

10. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/hour more than that of the passenger train, find the average speed of the two trains.

Sol. Let speed of passenger train = x km/hour

So, speed of express train = (x + 11) km/hour

Distance between Mysore and Bangalore = 132 km

Time taken by passenger train = 132/ x hour

Time is taken by express train

As per condition :

⇒ x2 + 11x – 1452 = 0

Using the quadratic formula, we get

So, the roots of the equation are

Or 66/2 and 88/2 or 33 and – 44.

Since x is the speed of the train, it cannot be negative.

Hence, speed of passenger train = 33 km/hour

Speed of express train = (33 +11) = 44 km/hour.

11. Some of the areas of two squares are 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.

Sol. Let the length of the side of one square = x m

∴ Area of square = x2 m2

Perimeter of square = 4x m

Let the length of the side of another square = y m

∴ Area of that square = y2 m2

The perimeter of that square = 4y m

As per condition,

x2 + y2 = 468 …(i)

And 4x – 4y = 24

⇒ 4(x – y) = 24

⇒ x – y = 6 …(ii) From (i) and (ii), we have

Using the quadratic formula, we get

⇒ y = 12 and – 18

As length cannot be negative. So, we will take a positive value.

So, y = 12

From (ii), x = 6 + 12 = 18

Hence, sides of two squares are 12 m and 18 m.

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