Combinations of previous exercises, concepts are put into this exercise to test the students. This ensures a student can face any problem involving quadratic equations. The Selina Solutions for Class 10 Maths prepared by expert faculty team at BYJUâ€™S has all the solutions to problems to help students strengthen their problem-solving skills. Solutions to these other miscellaneous problems in this exercise are available in the Concise Selina Solutions for Class 10 Maths Chapter 6 Solving (simple) Problems (Based On Quadratic Equations) Exercise 6(E) PDF given below.

## Selina Solutions Concise Maths Class 10 Chapter 6 Solving (simple) Problems (Based On Quadratic Equations) Exercise 6(E) Download PDF

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### Access Selina Solutions Concise Maths Class 10 Chapter 6 Solving (simple) Problems (Based On Quadratic Equations) Exercise 6(E)

**1. The distance by road between two towns A and B is 216 km, and by rail it is 208 km. A car travels at a speed of x km/hr and the train travels at a speed which is 16 km/hr faster than the car. Calculate:**

**(i) the time taken by the car to reach town B from A, in terms of x;**

**(ii) the time taken by the train to reach town B from A, in terms of x.**

**(iii) If the train takes 2 hours less than the car, to reach town B, obtain an equation in x and solve it.**

**(iv) Hence, find the speed of the train.**

**Solution: **

Given,

Speed of car = x km/hr

Speed of train = (x + 16) km/hr

And, we know that

Time = Distance/ Speed

(i) Time taken by the car to reach town B from town A = 216/x hrs

(ii) Time taken by the train to reach town B from A = 208/(x + 16) hrs

(iii) According to the question, we have

4x + 1728 = x^{2} + 16x

x^{2}Â + 12x â€“ 1728 = 0

x^{2}Â + 48x â€“ 36x â€“ 1728 = 0

x(x + 48) â€“ 36(x + 48) = 0

(x + 48) (x – 36) = 0

x = -48, 36

As speed cannot be negative,

x = 36

(iv) Therefore, the speed of the train is (x + 16) = (36 + 16)km/hr = 52 km/h

**2. A trader buys x articles for a total cost of Rs 600.**

**(i) Write down the cost of one article in terms of x.**

**If the cost per article were Rs 5 more, the number of articles that can be bought for Rs 600 would be four less.**

**(ii) Write down the equation in x for the above situation and solve it for x.**

**Solution:**

We have,

Number of articles = x

And, the total cost of articles = Rs 600

Then,

(i) Cost of one article = Rs 600/x

(ii) From the question we have,

x^{2} â€“ 4x â€“ 480 = 0

x^{2} â€“ 24x â€“ 20x â€“ 480 = 0

x(x – 24) + 20(x – 24) = 0

(x – 24) (x + 20) = 0

x = 24 or -20

As the number of articles cannot be negative, x = 24.

**3. A hotel bill for a number of people for overnight stay is Rs 4800. If there were 4 people more, the bill each person had to pay, would have reduced by Rs 200. Find the number of people staying overnight. **

**Solution: **

Letâ€™s assume the number of people staying overnight as x.

Given, total hotel bill = Rs 4800

So, hotel bill for each person = Rs 4800/x

Then, according to the question

x^{2} + 4x â€“ 96 = 0

x^{2} + 12x â€“ 8x â€“ 96 = 0

x(x +12) â€“ 8(x + 12) = 0

(x – 8) (x + 12) = 0

So, x = 8 or -12

As, the number of people cannot be negative. We take x = 8.

Therefore, the number of people staying overnight is 8.

**4. An aeroplane travelled a distance of 400 km at an average speed of x km/hr. On the return journey, the speed was increased by 40 km/hr. Write down an expression for the time taken for:**

**(i) the onward journey;**

**(ii) the return journey.**

**If the return journey took 30 minutes less than the onward journey, write down an equation in x and find its value.**

**Solution:**

Given,

Distance = 400 km

Average speed of the aeroplane = x km/hr

And, speed while returning = (x + 40) km/hr

We know that,

Time = Distance/ Speed

(i) Time taken for onward journey = 400/x hrs

(ii) Time take for return journey = 400/(x + 40) hrs

Then according to the question,

x^{2}Â + 40x â€“ 32000 = 0

x^{2}Â + 200x â€“ 160x â€“ 32000 = 0

x(x + 200) â€“ 160(x + 200) = 0

(x + 200) (x â€“ 160) = 0

So, x = -200 or 160

As the speed cannot be negative, x = 160 is only valid.

**5. Rs 6500 was divided equally among a certain number of persons. Had there been 15 persons more, each would have got Rs 30 less. Find the original number of persons.**

**Solution: **

Letâ€™s take the original number of persons to be x.

Total money which was divided = Rs 6500

Each personâ€™s share = Rs 6500/x

Then, according to the question

x^{2}Â + 15x â€“ 3250 = 0

x^{2}Â + 65x â€“ 50x â€“ 3250 = 0

x(x + 65) – 50(x + 65) = 0

(x + 65) (x â€“ 50) = 0

So, x = -65 or 50

As, the number of persons cannot be negative. x = 50

Therefore, the original number of persons are 50.

**6. A plane left 30 minutes later than the schedule time and in order to reach its destination 1500 km away in time, it has to increase its speed by 250 km/hr from its usual speed. Find its usual speed.**

**Solution: **

Letâ€™s consider the usual speed of the plane to be x km/hr

The distance to travel = 1500km

We know that,

Time = Distance/ Speed

Then according to the question, we have

x^{2} + 250x â€“ 750000 = 0

x^{2} + 1000x â€“ 750x â€“ 750000 = 0

x(x + 1000) â€“ 750(x + 1000) = 0

(x + 1000) (x – 750) = 0

So, x = -1000 or 750

As, speed cannot be negative. We take x = 750 as the solution.

Therefore, the usual speed of the plan is 750km/hr.

**7. Two trains leave a railway station at the same time. The first train travels due west and the second train due north. The first train travels 5 km/hr faster than the second train. If after 2 hours, they are 50 km apart, find the speed of each train.**

**Solution: **

Let take the speed of the second train to be x km/hr.

Then, the speed of the first train is (x + 5) km/hr

Let O be the position of the railway station from which the two trains leave.

Distance travelled by the first train in 2 hours = OA = speed x time = 2(x + 5) km

Distance travelled by the second train in 2 hours = OB = speed x time = 2x km

By Pythagoras Theorem, we have

AB^{2} = OA^{2} + OB^{2}

(50)^{2} = [2(x + 5)]^{2} + (2x)^{2}

2500 = 4(x^{2} + 10x + 25) + 4x^{2}

2500 = 8x^{2} + 40x + 100

x^{2} + 5x â€“ 300 = 0

x^{2} + 20x â€“ 15x â€“ 300 = 0

(x + 20) (x – 15) = 0

So, x = -20 or x = 15

As x cannot be negative, we have x = 15

Thus, the speed of the second train is 15 km/hr and the speed of the first train is 20 km/hr.

**8. The sum S of first n even natural numbers is given by the relation S = n(n + 1). Find n, if the sum is 420.**

**Solution: **

Given relation, S = n(n + 1)

And, S = 420

So, n(n + 1) = 420

n^{2}Â + n – 420 = 0

n^{2}Â + 21n – 20n – 420 = 0

n(n + 21) – 20(n + 21) = 0

(n + 21) (n – 20) = 0

n = -21, 20

As, n cannot be negative.

Therefore, n = 20.

**9. The sum of the ages of a father and his son is 45 years. Five years ago, the product of their ages (in years) was 124. Determine their present ages.**

**Solution: **

Letâ€™s assume the present ages of father and his son to be x years and (45 – x) years respectively.

So five years ago,

Father’s age = (x – 5) years

Son’s age = (45 – x – 5) years = (40 – x) years

From the question, the below equation can be formed

(x – 5) (40 – x) = 124

40x – x^{2}Â – 200 + 5x = 124

x^{2}Â – 45x +324 = 0

x^{2}Â – 36x – 9x +324 = 0

x(x – 36) – 9(x – 36) = 0

(x – 36) (x – 9) = 0

x = 36, 9

So, if x = 9,

The father’s age = 9 years and the son’s age = (45 – x) = 36 years

This is not possible.

Hence, x = 36

Therefore,

The father’s age = 36 years

The son’s age = (45 – 36) years = 9 years