Quadratic Equations Word Problems on Distance, Speed and Time Class 10 Concise Exe-6C ICSE Maths Selina Solutions Ch-6. In this article you would learn how to solve problems / questions of quadratic equations based on Distance, Speed and Time including C.P. and S.P. Visit official Website CISCE for detail information about ICSE Board Class-10 Mathematics.
Quadratic Equations Word Problems on Distance, Speed and Time Class 10
Concise Exe-6C ICSE Maths Selina Solutions Ch-6
| Board | ICSE |
| Publications | Selina |
| Subject | Maths |
| Class | 10th |
| Chapter-6 | Solving (Simple) Problems (Based on Quadratic Equations) |
| Writer | R.K. Bansal |
| Exe-6C | Problems based on Distance, Speed and Time and on C.P. and S.P. |
Quadratic Equations Word Problems on Distance, Speed and Time Class 10
Concise Exe-6C ICSE Maths Selina Solutions Ch-6
Que-1: The speed of an ordinary train is x km per hr and that of an express train is (x + 25) km per hr.
(i) Find the time taken by each train to cover 300 km.
(ii) If the ordinary train takes 2 hrs more than the express train; calculate speed of the express train.
Sol: (i) Speed of ordinary train = x km/hr
Speed of express train = (x + 25) km/hr
Distance = 300 km
We know
Time = Distance/Speed
∴ Time taken by ordinary train to cover 300 km = 300/𝑥 hrs
Time taken by express train to cover 300 km = 300/(𝑥+25) hrs
(ii) Given that the ordinary train takes 2 hours more than the express train to cover the distance.
Therefore,
(300/𝑥) − {300/(𝑥+25)} = 2
(300𝑥+7500−300𝑥)/{𝑥(𝑥+25)} = 2
7500 = 2𝑥² + 50𝑥
2𝑥² + 50𝑥 − 7500 = 0
𝑥² + 25𝑥 − 3750 = 0
𝑥² +75𝑥 −50𝑥 −3750 =0
𝑥(𝑥+75) −50(𝑥+75) =0
(𝑥+75)(𝑥−50) =0
x = –75, 50
But, speed cannot be negative.
So, x = 50.
∴ Speed of the express train = (x + 25) km/hr = 75 km/hr.
Que-2: If the speed of a car is increased by 10 km per hr, it takes 18 minutes less to cover a distance of 36 km. Find the speed of the car.
Sol: Let the speed of the car be x km/hr.
Distance = 36 km
Time taken to cover a distance of 36 km = 36/x hrs
(Time = Distance/Speed)
New speed of the car = (x + 10) km/hr
New time taken by the car to cover a distance of 36 km = 36/(𝑥+10) hrs
From the given information, we have:
(36/𝑥) − {36/(𝑥+10)} = 18/60
{36(𝑥+10)−36𝑥}/{𝑥(𝑥+10)} = 3/10
(36𝑥+360−36𝑥)/{𝑥(𝑥+10)} = 3/10
360/{𝑥(𝑥+10)} = 3/10
3x2 + 30x = 3600
3x2 + 30x – 3600 = 0
x2 + 10x – 1200 = 0 …(Dividing by 3)
x2 + 40x – 30x – 1200 = 0
x(x + 40) – 30(x + 40) = 0
(x + 40)(x – 30) = 0
x = – 40, 30
But speed cannot be negative.
So, x = 30.
Hence, the original speed of the car is 30 km/hr.
Que-3: If the speed of an aeroplane is reduced by 40 km/hr, it takes 20 minutes more to cover 1200 km. Find the speed of the aeroplane.
Sol: Let the original speed of the aeroplane be x km/hr.
Time taken to cover a distance of 1200 km = 1200/x hrs
(Time = Distance/Speed)
Let the new speed of the aeroplane be (x – 40) km/hr
Time taken to cover a distance of 1200 km = 1200/(𝑥−40) hrs
From the given information, we have
{1200/(𝑥−40)} − (20/60) = 1200/𝑥
{1200/(𝑥−40)} −(1200/𝑥) = 20/60
{1200𝑥−1200𝑥+48000}/{𝑥(𝑥−40)} = 1/3
x(x – 40) = 48000 × 3
x2 – 40x – 144000 = 0
x2 – 400x + 360x – 144000 = 0
x(x – 400) + 360(x – 400) = 0
(x – 400)(x + 360) = 0
x = 400, – 360
But, speed cannot be negative.
So, x = 400.
Thus, the original speed of the aeroplane is 400 km/hr.
Que-4: A car covers a distance of 400 km at a certain speed. Had the speed been 12 km/h more, the time taken for the journey would have been 1 hour 40 minutes less. Find the original speed of the car.
Sol: Let the original speed of the car be x
km/hr,
so, Time taken by car = 400/𝑥 hrs.
Again, Speed = (x + 12) km/hr
Time taken by car = 400/(𝑥+12)
so, (400/𝑥) − {400/(𝑥+12)} = 1hr + (40/60)
{400(𝑥+12−𝑥)}/{𝑥(𝑥+12)} = 5/3
4800/(𝑥²+12𝑥) = 5/3
⇒ 5 (x2 + 12x) = 14,400
⇒ x2 + 12x – 2,880 = 0
⇒ x2 + 60x – 48x – 2,880 = 0
⇒ x (x + 60) – 48 (x + 60) = 0
⇒ (x + 60) (x – 48) = 0
Either, x + 60 = 0
x = -60 …(Neglect, Speed can’t be negative)
or
x – 48 = 0
x = 48
⇒ Original speed of the car is 48 km/hr.
Que-5: A girl goes to her friend’s house, which is at a distance of 12 km. She covers half of the distance at a speed of x km/hr and the remaining distance at a speed of (x + 2) km/hr. If she takes 2 hrs 30 minutes to cover the whole distance, find ‘x’.
Sol: We know
Time = Distance/Speed
Given, the girl covers a distance of 6 km at a speed x km/hr.
Time taken to cover first 6 km = 6𝑥
Also, the girl covers the remaining 6 km distance at a speed (x + 2) km/hr.
Time taken to cover next 6 km = 6/(𝑥+2)
Total time taken to cover the whole distance = 2 hrs 30 mins
= 2*(30/60)
= 2*(1/2) = 5/2 hrs
∴ (6/𝑥) + {6/(𝑥+2)} = 5/2
(6𝑥+12+6𝑥)/{𝑥(𝑥+2)} = 5/2
(12+12𝑥)/(𝑥²+2𝑥) = 5/2
24 + 24x = 5x2 + 10x
5x2 – 14x – 24 = 0
5x2 – 20x + 6x – 24 = 0
5x(x – 4) + 6(x – 4) = 0
(5x + 6)(x – 4) = 0
𝑥 =−6/5, 4
Since, speed cannot be negative.
Therefore, x = 4.
Que-6: A car made a run of 390 km in ‘x’ hours. If the speed had been 4 km/hour more, it would have taken 2 hours less for the journey. Find ‘x’.
Sol: Let the original speed of the car be y km/hr
We know
Speed = Distance/Time
∴ 𝑦 =390/𝑥
⇒𝑥 = 390/𝑦 …(1)
New speed of the car = (y + 4) km/hr
New time taken by the car to cover 390 km = 390/(𝑦+4)
From the given information,
(390/𝑦) − {390/(𝑦+4)} = 2
(390𝑦+1560−390𝑦)/{𝑦(𝑦+4)} = 2
780/(𝑦²+4𝑦) = 1
y2 + 4y – 780 = 0
y2 + 30y – 26y – 780 = 0
y(y + 30) – 26(y + 30) = 0
(y + 30)(y – 26) = 0
y = –30, 26
Since, time cannot be negative, so y = 26
From (1), we have
𝑥 = 390/𝑦 = 390/26 = 15
Que-7: A goods train leaves a station at 6 p.m., followed by an express train which leaved at 8 p.m. and travels 20 km/hour faster than the goods train. The express train arrives at a station, 1040 km away, 36 minutes before the goods train. Assuming that the speeds of both the train remain constant between the two stations; calculate their speeds.
Sol: Let the speed of goods train be x km/hr.
So, the speed of express train will be (x + 20) km/hr.
Distance = 1040 km
We know
Time = Distance/Speed
Time taken by good train to cover a distance of 1040 km = 1040/𝑥 hrs
Time taken by express train to cover a distance of 1040 km = 1040/(𝑥+20) hrs
It is given that the express train arrives at a station 36 minutes before the goods train. Also the express train leaves the station 2 hours after the goods train. This means that the express train arrives at the station ((36/60)+2)hrs = 13/5hrs before the good train.
Therefore, we have
(1040/𝑥) − {1040/(𝑥+20)} = 13/5
(1040𝑥+20800−1040𝑥)/{𝑥(𝑥+20)} = 13/5
20800/(𝑥²+20𝑥) = 13/5
1600𝑥² + 20𝑥 = 1/5
x2 + 20x – 8000 = 0
(x – 80)(x + 100) = 0
x = 80, –100
Since, the speed cannot be negative.
So, x = 80.
Thus, the speed of goods train is 80 km/hr and the speed of express train is 100 km/hr.
Que-8: A man bought an article for Rs x and sold it for Rs 16. If his loss was x per cent, find the cost price of the article.
Sol: C.P. of the article = Rs. x
S.P. of the article = Rs. 16
Loss = Rs. (x – 16)
We know:
Loss% = (Loss/C.P.) ×100
∴ 𝑥 = {(𝑥−16)/𝑥} × 100
x2 = 100x – 1600
x2 – 100x + 1600 = 0
(x – 80)(x – 20) = 0
x = 80, 20
Thus, the cost price of the article is Rs. 20 or Rs. 80
Que-9: A trader bought an article for Rs x and sold it for Rs 52, thereby making a profit of (x – 10) per cent on his outlay. Calculate the cost price.
Sol: C.P. of the article = Rs. x
S.P. of the article = Rs. 52
Profit = Rs. (52 – x)
We know:
Profit% = (Profit/C.P.) × 100
∴ 𝑥 −10 = {(52−𝑥)/𝑥} ×100
x2 – 10x = 5200 – 100x
x2 + 90x – 5200 = 0
(x + 130)(x – 40) = 0
x = –130, 40
Since, C.P. cannot be negative.
So, x = 40.
Thus, the cost price of the article is Rs. 40
Que-10: By selling a chair for Rs 75, Mohan gained as much per cent as its cost. Calculate the cost of the chair.
Sol: Let the C.P. of the chair be Rs. x
S.P. of chair = Rs. 75
Profit = Rs. (75 – x)
We know:
Profit% = (Profit/C.P.) ×100
∴ 𝑥 = {(75−𝑥)/𝑥} × 100
x2 = 7500 – 100x
x2 + 100x – 7500 = 0
(x + 150)(x – 50) = 0
x = –150, 50
But, C.P. cannot be negative.
So, x = 50.
Hence, the cost of the chair is Rs. 50
— : End of Quadratic Equations Word Problems on Distance Speed and Time Class 10 Concise Exe-6C Selina Solutions Ch-6 Practice Problems. :–
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