# Class 10 RD Sharma Solutions – Chapter 8 Quadratic Equations – Exercise 8.8

• Last Updated : 11 Dec, 2020

### Question 1. If the speed of a boat in still water is 8 Km/hr. It can go 15 Km upstream and 22 Km downstream in 5 hours. Find the speed of the stream.

Solution:

Let the speed of stream be x Km/hr,

Then, speed downstream = (8 + x) Km/hr

speed upstream = (8 – x) Km/hr

Since, speed = distance / time

Time taken by the boat to 15 Km upstream = 15/(8 – x) hr

Time taken by the boat to 22 Km downstream = 22/(8 + x)hr

According to the question : boat returns to the same point in 5 hr so,

⇒ 15/(8-x) + 22/(8 + x) = 5

⇒ 15(8+x) + 22(8-x) = 5 (8+x) (8-x)

⇒ 120 + 15x + 176 -22x = 5(82 – x2)

⇒ 296 -7x = 5(64 -x2 )

⇒ 296 -7x = 320 – 5x2

⇒ 5x2 -7x -320 +296 = 0

⇒ 5x2 -7x -24 = 0

⇒ 5x2 -15x +8x -24 = 0                   [by using factorization method]

⇒ 5x (x-3)+ 8(x-3) = 0

⇒ (5x+8) (x-3) = 0

So, the values of x are x = 3, x = -8/5

Since, the speed of stream can never be negative so, x = -8/5 will be neglected

Hence, the speed of stream is 3 Km/hr.

### Question 2. A train travelling at a uniform speed for 360 Km, would have taken 48 minutes less to travel the same distance if it’s speed were 5 Km/hr more. Find the original speed of the train.

Solution:

Total distance covered = 360 Km

Let the usual speed of train be x Km/hr

Then, the increased speed of train = (x+5) Km/hr

Since, speed = distance / time

Time taken by the train under usual speed to cover 360 Km = 360/x hr

Time taken by the train under increased speed to cover 360 Km = 360/ (x+5) hr

According, to the question it takes 48 minutes less to travel the same distance:

[48 min in hours = 48/60 = 4/5 hr]

⇒ 360/x – 360/(x+5) = 48/60

⇒ [360(x + 5) – 360 (x) ] / (x+5) (x) = 4/5

⇒ (360x + 1800- 360x) /( x2 + 5x ) = 4/5

⇒ 1800 = 4/5 (x2 + 5x)

⇒ 1800 × 5 / 4 = x2+5x

⇒ 2250 = x2+ 5x

⇒ x2 + 5x – 2250 = 0

⇒ x2 + 50x – 45x -2250 = 0       [by using factorization method]

⇒ x (x+50) -45(x +50) = 0

⇒ (x – 45) (x+ 50) = 0

So, the values of x are x = 45, x = – 50

Since, the speed can never be negative so, x = -50 will be neglected

Hence, the original speed of the train is 45 Km/hr.

### Question 3. A fast train takes one hour less than a slow train for a journey of 200 Km. If the speed of the slow train is 10 Km/hr less than that of the fast train, find the speed of the two trains.

Solution:

Total journey covered = 200 Km

Let the speed of fast train be x Km/hr

Then, speed of slow train = (x – 10) Km/hr

Since, speed = distance / time

Time taken by the fast train to cover 200 Km = 200/x hr

Time taken by the slow train to cover 200 Km = 200/(x – 10) hr

According to question, fast train takes one hour less than a slow train for a journey

⇒ 200/(x-10) – 200/x = 1

⇒ 200(x) – 200(x-10) = (x-10)(x)

⇒ 200x – 200x + 2000 = x2 – 10x

⇒ x2 -10x -2000 = 0

⇒ x2 – 50x +  40x – 2000 = 0                              [by using factorization method]

⇒ x (x – 50) + 40 (x – 50) = 0

⇒ (x + 40) (x – 50) = 0

So, the values of x are x = – 40 , x = 50

Since, the speed can never be negative so, x = -40 will be neglected

Hence, the speed of fast train is 50 Km/hr and speed of slow train is (50 – 10) Km/hr which is 40 Km/hr.

### Question 4. A passenger train takes one hour less for a journey of 150 km if its speed is increased by 5 km/hr from its usual speed. Find the usual speed of the train.

Solution:

Total journey covered = 150 Km

Let, the usual speed of train be x Km/hr

the increased speed of train = (x + 5) Km/hr

Since, speed = distance / time

Time taken by the train under usual speed to cover 150 Km = 150/ x hr

Time taken by the train under usual speed to cover 150 Km = 150/ (x+5) hr

According to the question, passenger train takes one hour less if its speed is increased :

⇒ 150 / x – 150 / (x+5) = 1

⇒ 150(x +5) – 150(x) = (x+5)(x)

⇒ 150x – 750 – 150x = x2 + 5x

⇒ x2 + 5x +750 = 0                                 [by using factorization method]

⇒ x2 +30x – 25x +750 = 0

⇒ x (x+30) – 25 (x+30) = 0

⇒ (x – 25) (x + 30) = 0

So, the values of x are x = 25, x = – 30

Since, the speed can never be negative so, x = -30 will be neglected

Hence, the usual speed of the train is 25 Km/hr.

### Question 5. The time taken by a person to cover 150 km was 2.5 hrs more than the time taken in the return journey. If he returned at a speed of 10 km/hr more than the speed of going, what was the speed per hour in each direction?

Solution:

Total distance covered = 150 Km

Let,the speed of the person while going be x Km/hr

Then the speed while returning = (x + 10) Km/hr

Since, speed = distance / time

Time taken by the person while going = 150/ x hr

Time taken by the person while returning = 150/ (x+10) hr

According to the question, time taken by a person to cover 150 km was 2.5 hrs more than the time taken in the return journey:

⇒ 150/x – 150/ (x + 10) = 2.5 = 5/2

⇒ 150 (x + 10) – 150x = 5/2(x+10)(x)

⇒ 150x + 1500 – 150x = 5/2 (x2 + 10x)

⇒ 1500 × 2 = 5x2 + 50x

⇒ 3000=5x2 + 50x

⇒ 5x2 + 50x – 3000 = 0     [divide the equation by 5]

⇒ x2 + 10x – 600 = 0      [by using factorization method]

⇒ x2 + 30x – 20x -600 = 0

⇒ x (x+30) -20 (x +30) = 0

⇒ (x – 20) (x + 30) = 0

So, the values of x are x = 20, x = – 30

Since, the speed can never be negative so, x = -30 will be neglected

Hence, the usual speed of the train is 20 Km/hr.

### Question 6. A plane left 40 minutes late due to bad weather and in order to reach its destination, 1600 Km away in time, it had to increase its speed by 400 Km/hr from its usual speed. Find the usual speed of the plane.

Solution:

Total distance covered = 1600 Km

Let,the usual speed of plane while be x Km/hr

Then the increased speed = (x + 400) Km/hr

Since, speed = distance / time

Time taken by the plane for usual speed = 1600/ x hr

Time taken by the plane for increased speed = 1600/ (x+400) hr

According to the question, plane left 40 minutes late :      [40 minutes in hours be 40/60 = 2/3]

⇒ 1600/x – 1600/ (x + 400) = 2/3

⇒ 1600 (x + 400) – 1600x = 2/3 (x+400) (x)

⇒ 1600x + 640000 – 1600x = 2/3 (x2 + 400x)

⇒ 640000 × 3 = 2x2 + 800x

⇒ 1920000=2x2 + 800x

⇒ 2x2 + 800x – 1920000 = 0     [divide the equation by 2]

⇒ x2 + 400x – 960000 = 0      [by using factorization method]

⇒ x2 + 1200x – 800x -960000 = 0

⇒ x (x+1200) -800 (x +1200) = 0

⇒ (x – 800) (x + 1200) = 0

So, the values of x are x = 800, x = -1200

Since, the speed can never be negative so, x = -1200 will be neglected

Hence, the usual speed of the plane is 800 Km/hr.

### Question 7. An Aeroplan takes 1 hour less for a journey of 1200 Km if its speed is increased by 100 Km/hr from its usual speed. Find its usual speed.

Solution:

Total distance covered = 1200 Km

Let, the usual speed of Aeroplan while be x Km/hr

Then the increased speed = (x + 100) Km/hr

Since, speed = distance / time

Time taken by the plane for usual speed = 1200/ x hr

Time taken by the plane for increased speed = 1200/ (x+100) hr

According to the question, Aeroplane takes 1 hour less for a journey :

⇒ 1200/x – 1200/ (x + 100) = 1

⇒ 1200 (x + 100) – 1200x = (x+100) (x)

⇒ 1200x + 120000 – 1200x = (x2 + 100x)

⇒ 120000  = x2 + 100x

⇒ x2 + 100x – 120000 = 0

⇒ x2 + 100x – 120000 = 0             [by using factorization method]

⇒ x2+ 400x – 300x -120000 = 0

⇒ x (x+400) -300 (x +400) = 0

⇒ (x – 300) (x + 400) = 0

So, the values of x are x = 300, x = -400

Since, the speed can never be negative so, x = -400 will be neglected

Hence, the usual speed of the Aeroplan is 300 Km/hr.

### Question 8. A train travels at a certain average speed for a distance 63 Km and then travels a distance of 72 Km at an average speed of 6 Km/hr more than the original speed. If it takes 3 hours to complete total journey, what is its original average speed?

Solution:

Let the average speed of train be x Km/hr for distance of 63 Km.

the average speed of train for distance of 72 Km = (x + 6) Km/ hr

Since, speed = distance / time

Time taken by train to cover 63 Km with speed x Km/hr = 63 / x hr

Time taken by train to cover 72 Km with speed (x+6) Km/hr = 72 / (x+6) hr

According, to question train takes 3 hours to complete total journey:

⇒ 63 / x + 72 / (x+6) = 3

⇒ 63 ( x +6) + 72(x) = 3 (x) (x+6)

⇒ 63x + 378 + 72x = 3x2 + 18x

⇒ 135x + 378 -18x = 3x2

⇒ 117x + 378 = 3x2

⇒ 39x + 126 = x2                         [by using factorization method]

⇒ x2 – 39x – 126 = 0

⇒ x2 -42x + 3x -126 = 0

⇒ x(x – 42) + 3 ( x – 42) = 0

⇒ (x+3) (x-42) = 0

So, the values of x are x = 42, x = -3

Since, the speed can never be negative so, x = -3 will be neglected

Hence, the average speed of the train is 42 Km/hr.

### Question 9. A train covers a distance of 90 Km at a uniform speed. Had the speed been 15 Km/hr more, it would have taken 30 minutes less for the journey. Find the original speed of the train.

Solution:

Total distance covered = 90 Km

Let the original speed of train be x Km/hr

Then, the increased speed of train = (x+15) Km/hr

Since, speed = distance / time

Time taken by the train under usual speed to cover 90 Km = 90/x hr

Time taken by the train under increased speed to cover 90 Km = 90/ (x+15) hr

According, to the question it takes 30 minutes less to travel the same distance:

[ 30 min in hours = 30/60 = 1/2 hr ]

⇒ 90/x – 90/(x+15) = 1/2

⇒ [90(x + 15) – 90 (x) ] / (x+15) (x) = 1/2

⇒ (90x + 1350- 90x) /( x2 + 15x ) = 1/2

⇒ 1350 = 1/2 ( x2 + 15x)

⇒ 1350 × 2 = x2+15x

⇒ 2700 = x2+ 15x

⇒ x2 + 15x – 2700 = 0

⇒ x2 + 60x – 45x – 2700 = 0       [by using factorization method]

⇒ x (x+60) -45(x +60) = 0

⇒ (x – 45) ( x+ 60) = 0

So, the values of x are x = 45, x = – 60

Since, the speed can never be negative so, x = -60 will be neglected

Hence, the original speed of the train is 45 Km/hr.

### Question 10. A train travels 360 Km at a uniform speed. If the speed had been 5 Km/hr more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Solution:

Total distance covered = 360 Km

Let the uniform speed of train be x Km/hr

Then, the increased speed of train = (x+5) Km/hr

Since, speed = distance / time

Time taken by the train under usual speed to cover 360 Km = 360/x hr

Time taken by the train under increased speed to cover 360 Km = 360/ (x+5) hr

According, to the question it takes 1 hour less to travel the same distance:

⇒ 360/x – 360/(x+5) = 1

⇒ [360(x + 5) – 360 (x) ] / (x+5) (x) = 1

⇒ (360x + 1800- 360x) /( x2 + 5x ) = 1

⇒ 1800 =  ( x2 + 5x)

⇒ 1800  = x2+5x

⇒ x2 + 5x – 1800 = 0

⇒ x2 + 45x – 40x -1800 = 0       [by using factorization method]

⇒ x (x+45) -40(x +45) = 0

⇒ (x – 40) ( x+ 45) = 0

So, the values of x are x = -45, x =  40

Since, the speed can never be negative so, x = -45 will be neglected

Hence, the original speed of the train is 40 Km/hr.

### Question 11. 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/hr more than that of the passenger train, find the average speeds of the two trains.

Solution:

Total distance between Mysore and Bangalore = 132 Km

Let, the average speed of the passenger train be x Km/hr

Then speed of express train = (x + 11) Km/hr

Since, speed = distance / time

Time taken by the  passenger train  = 132/x hr

Time taken by express train = 132 / (x+11) hr

According, to the question, express train takes 1 hour less than a passenger train to travel :

⇒ 132/x – 132/(x+11) = 1

⇒ 132(x+11) – 132 (x) = (x+11) (x)

⇒ 132x +  1452 – 132x = x2 + 11x

⇒ x2 +11x – 1452 = 0                          [by using factorization method]

⇒ x2 -33x + 44x -1452 = 0

⇒ x (x – 33) + 44 ( x – 33) = 0

⇒ (x+44) (x -33) = 0

So, the values of x are x = -44, x =  33

Since, the speed can never be negative so, x = -44 will be neglected

Hence, the speed of the passenger train is 33 Km/hr and the speed of express train is 44 Km/hr.

### Question 12. An Aeroplan left 50 minutes later than its scheduled time, and in order to reach the destination, 1250 Km away, in time, it had to increase its speed by 250 Km/hr from its usual speed. Find its usual speed.

Solution:

Total distance covered = 1250 Km

Let the usual speed of Aeroplan be x Km/hr

Then the speed of Aeroplan  = (x + 250) Km/hr

Since, speed = distance / time

Time taken by Aeroplan for usual speed = 1250/x hr

Time taken by Aeroplan for increased speed = 1250/(x+250) hr

According to question, Aeroplane left 50 minutes later than its scheduled time:

[50 minutes in hours = 50/60 = 5/6]

⇒ 1250/x – 1250/(x+250) = 5/6

⇒ 1250 (x+250) – 1250 (x)= 5/6 (x+250)(x)

⇒ 1250 x + 312500 – 1250x = 5/6 (x2 – 250x)

⇒ 312500 x 6/5 = x2 – 250x

⇒ 375000 = x2 -250x

⇒ x2 – 250x – 375000 = 0                      [by using factorization method]

⇒ x2 -500x + 750x -375000 = 0

⇒ x(x – 500) + 750 ( x – 500) = 0

⇒ (x + 750) (x-500) = 0

So, the values of x are x = -750, x =  500.

Since, the speed can never be negative so, x = -750 will be neglected

Hence, the usual speed of the Aeroplan is 500 Km/hr.

### Question 13. While boarding an Aeroplan, a passenger got hurt. The pilot showing promptness and concern, made arrangements to hospitalize the injured and so the plane started late by 30 minutes to reach the destination, 1500 Km away, in time, the pilot increased the speed by 100 Km/hr. Find the original speed/hour of the plane.

Solution:

Total distance to be travelled  : 1500 Km

Let the original speed of the plan be x Km/hr

Then, the increased speed of the plan = (x+100) Km/hr

Since, speed = distance / time

Time taken by the plan in original speed = 1500/ x hr

Time taken by the plan in  increased speed = 1500/ (x + 100) hr

According to question, plane started late by 30 minutes:

[30 minutes in hours is 30/60 hours = 1/2 hr]

⇒ 1500/x – 1500/(x+100) = 1/2

⇒ 1500(x+100) – 1500 (x) = 1/2 (x) (x+100)

⇒ 1500x +150000 – 1500x = 1/2 (x2 + 100x )

⇒ 150000 x 2 = x2 + 100x

⇒ 300000 = x2 + 100x

⇒ x2 + 100x – 300000 = 0              [by using factorization method]

⇒ x2 + 600x – 500x -300000 = 0

⇒ x (x +600) – 500( x+ 600) = 0

⇒ (x – 500) (x+600) = 0

So, the values of x are x = -600, x = 500.

Since, the speed can never be negative so, x = -600 will be neglected

Hence, the original speed of the Aeroplan is 500 Km/hr.

### Question 14. A motorboat whose speed in still water is 18 Km/hr, takes 1 hour more to go 24 Km upstream than to return downstream to the same spot. Find the speed of the stream.

Solution:

Total distance covered = 24 Km

Speed of the boat in still water is = 18 Km/hr

Let, the usual speed of the stream be x Km/hr

Speed of the boat upstream = speed of the boat in still water – speed of the stream = (18 – x) Km/hr

Speed of the boat downstream = speed of the boat in still water + speed of the stream = (18 + x) Km/hr

Since, speed = distance / time

Time taken by boat for upstream = 24/(18 – x) hr

Time taken by boat for downstream = 24/(18 + x) hr

According to  the question, motorboat takes 1 hour more to return downstream to the same spot:

⇒ 24 / (18 – x) – 24 / (18 + x) = 1

⇒ 24 (18 + x) – 24(18 – x) = (18 – x)(18 + x)

⇒ 432 + 24x – 432 + 24x = 182 – x2

⇒ 48x = 324 – x2

⇒ x2 + 48x – 324 = 0                       [by using factorization method]

⇒ x2 + 54x – 6x – 324 = 0

⇒ x (x+ 54) – 6(x + 54) = 0

⇒ (x – 6) (x+54) = 0

So, the values of x are x = -600, x = 500.

Since, the speed can never be negative so, x = -600 will be neglected

Hence, the original speed of the Aeroplane is 500 Km/hr.

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