# Class 10 RD Sharma Solutions – Chapter 8 Quadratic Equations – Exercise 8.13

### Question 1. A piece of cloth costs â‚¹ 35. If the piece were 4 m longer and each meter costs â‚¹ 1 less, the cost would remain unchanged. How long is the piece?

**Solution:**

Let us considered the length of piece of cloth = x m

Given: The total cost = â‚¹ 35

So, the cost of 1 m cloth = â‚¹ 35/x

According to the question,

(x + 4)(35/x – 1) = 35

â‡’ 35 – x + (140/x) – 4 = 35

â‡’ -x + (140/x) + 31 – 35 = 0

â‡’ -x + (140/x) – 4 = 0

â‡’ -x

^{2 }+ 140 – 4x = 0â‡’ x

^{2 }+ 4x – 140 = 0â‡’ x

^{2 }+ 14x – 10x – 140 = 0â‡’ x (x + 14) -10 (x + 14) = 0

â‡’ (x + 14) (x – 10) = 0

x = 10 or x = -14

Here the value of x = -14 is negative which is not possible

So, the length of piece of cloth = 10 m.

### Question 2. Some students planned a picnic. The budget for food was â‚¹ 480. But eight of these failed to go and thus the cost of food for each member increased by â‚¹ 10. How many students attended the picnic?

**Solution:**

Let us considered the number of students = x

Given: The total budget = â‚¹ 480

So, the share of each student = â‚¹ 480/x

According to the question,

(480/x – 8) – (480/x) = 10

â‡’

â‡’

â‡’ 10x

^{2 }– 80x – 3840 = 0â‡’ x

^{2 }– 8x – 384 = 0â‡’ x

^{2 }+ 16x – 24x – 384 = 0â‡’ x (x + 16) – 24 (x + 16) = 0

â‡’ (x + 16) (x – 24) = 0

x = 24, or x = -16

Here the value of x = -16 is negative which is not possible

So, the number of students = 24.

The total number of students attend the picnic = 24 – 8 = 16

### Question 3. A dealer sells an article for â‚¹ 24 and gains as much percent as the cost price of the article. Find the cost price of the article.

**Solution:**

Let cost price of the article = â‚¹ x

Selling price = â‚¹ 24

Gain = x %

According to the Question,

S.P.= C.P. Ã— (100 + Gain%)/100

24 = x(100 + x)/100

â‡’ 2400 = 100x + x

^{2}â‡’ x

^{2}+ 100x – 2400 = 0â‡’ x

^{2}– 20x +120x – 2400 = 0â‡’ x(x – 20) + 120(x – 20) = 0

â‡’ (x – 20)(x + 120) = 0

x = 20 or x = -120

Here the value of x = -120 is negative which is not possible

Therefore, the cost price of the article = â‚¹20

### Question 4. Out of a group of swans, 7/2 times the square root of the total number are playing on the share of a pond. The two remaining ones are swinging in water. Find the total number of swans.

**Solution:**

Let us considered the total number of swans = x

According to the question,

7/2(âˆšx) + 2 = x

â‡’ 7âˆšx = 2x – 4

On squaring both sides, we get

â‡’ 49x = 4x + 16 – 16x

â‡’ 4x

^{2}– 65x + 16 = 0â‡’ 4x

^{2}– 64x – x + 16 = 0â‡’ 4x(x – 16) – (x â€“ 16) = 0

â‡’ (4x – 1)(x – 16) = 0

â‡’ x =1/4 or x = 16

Since, number of swans is a natural number we can neglect the solution of a = 1/4

Hence, the total number of swans is 16.

### Question 5. If the list price of a toy is reduced by â‚¹ 2, a person can buy 2 toys mope for â‚¹ 360. Find the original price of the toy.

**Solution:**

Let the original price of the toy = x

The number of toys he can buy at the original price for â‚¹ 360 = 360/x

According to the question,

â‡’ 360x = (x – 2)(360 + 2x)

â‡’ 360x = 360x + 2x

^{2}– 720 – 4xâ‡’ x

^{2}– 2x – 360 = 0â‡’ x

^{2}– 20x + 18x – 360 = 0â‡’ x(x – 20) + 18(x – 20) = 0

â‡’ (x + 18)(x – 20) = 0

â‡’ x + 18 = 0 or x – 20 = 0

â‡’ x = -18 or x = 20

As, the price canâ€™t be negative, x = -18 is neglected.

Thus, the original price of the toy is â‚¹ 20.

### Question 6. â‚¹ 9000 were divided equally among a certain number of persons. Had there been 20 more persons, each would have got â‚¹ 160 less. Find the original number of persons.

**Solution:**

Letâ€™s consider the original number of people = a

Amount which each receives when a person are present = 9000/a

According to the question,

â‡’ 9000a = (9000 â€“ 160a)(a + 20)

â‡’ 9000a = 9000a + 180000 â€“ 160a

^{2}â€“ 3200aâ‡’ a

^{2}+ 20a â€“ 1125 = 0â‡’ a

^{2}+ 45a â€“ 25a â€“ 1125 = 0â‡’ a(a + 45) â€“ 25(a + 45) = 0

â‡’ (a â€“ 25)(a + 45) = 0

â‡’ a = 25 or a = -45 (Number of people can never be negative,

so we can neglect this value)

Therefore, the original number of people = 25.

### Question 7. Some students planned a picnic. The budget for food was â‚¹ 500. But 5 of them failed to go and thus the cost of food for each number increased by â‚¹ 5. How many students attended the picnic?

**Solution:**

Let us considered the number of students = x

Given: The total budget = â‚¹ 500

So, the share of each student = â‚¹ 500/x

The number of students failed to go = 5

According to the question,

â‡’ 5x

^{2}– 25x – 2500 = 0â‡’ x

^{2}– 5x – 500 = 0â‡’ x

^{2}– 25x + 20x – 500 = 0â‡’ x(x – 25) + 20(x – 25) = 0

â‡’ (x – 25)(x + 20) = 0

â‡’ x – 25 = 0 or x + 20 = 0

â‡’ x = 25 or x = -20

Here the value of x = -20 is negative which is not possible

So, the number of students = 25

### Question 8. A pole has to be erected at a point on the boundary of a. circular park of diameter 13 meters in such a way that the difference, of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 meters. Is it the possible to do so? If yes, at what distances from the two gates Should the pole be erected?

**Solution:**

In the given circle, AB(Diameter) = 13 m

Let us considered P be the pole in the circle.

So, PB = x m and PA= (x + 7) m

In the triangle APB

AB

^{2 }= AP^{2}+ PB^{2}â‡’ 13

^{2}= (x + 7)^{2}+ (x)^{2}â‡’ 169 = x

^{2}+ 49 +14x + x^{2}â‡’ 2x

^{2}+ 14x -120 = 0â‡’ x

^{2}+7x – 60 = 0â‡’ x

^{2}+12x – 5x – 60 = 0â‡’ x(x + 12) – 5(x + 12) = 0

â‡’ (x + 12)(x – 5) = 0

x = -12 or x = 5

Here the value of x = -12 is negative which is not possible

So the value valid of x = 5

Hence, PB = 5 m and PA = 5 + 7 = 12 m.

### Question 9. In a class test, the sum of the marks obtained by P in Mathematics and Science is 28. Had he got 3 marks more in Mathematics and 4 marks less in Science. The product of their marks, would have been 180. Find his marks in the two subjects.

**Solution:**

Given: The sum of marks obtained by P in Mathematics and Science = 28

So let us considered the marks in Mathematics = x

and the marks in Science = 28 â€“ x

According to the question

(x + 3) (28 â€“ x â€“ 4) = 180

â‡’ (x + 3) (24 â€“ x) = 180

â‡’ 24x â€“ xÂ² + 72 â€“ 3x = 180

â‡’ 21x â€“ xÂ² + 72 â€“ 180 = 0

â‡’ â€“ xÂ² + 21x â€“ 108 = 0

â‡’ xÂ² â€“ 21x + 108 = 0

â‡’ xÂ² â€“ 9x â€“ 12x + 108 = 0

â‡’ x (x â€“ 9) â€“ 12 (x â€“ 9) â€“ 0

â‡’ (x â€“ 9)(x â€“ 12) = 0

x = 9 or x = 12

So, if we take x = 9 then the marks in

Mathematics = 9 and marks in Science = 19

So, if we take x = 12 then the marks in

Mathematics = 12 and marks in English = 16

### Question 10. In a class test, the sum of Shefaliâ€™s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of her marks would have been 210. Find her marks in two subjects.

**Solution:**

Given: Sum of Shefali’s marks in Mathematics and English = 30

So let us considered the marks in Mathematics = x

and the marks in English = 30 â€“ x

According to the question

(x + 2) (30 â€“ x â€“ 3) = 210

â‡’ (x + 2) (27 â€“ x) = 210

â‡’ 27x â€“ xÂ² + 54 â€“ 2x â€“ 210 = 0

â‡’ â€“ xÂ² + 25x â€“ 156 = 0

â‡’ xÂ² â€“ 25x + 156 = 0

â‡’ xÂ² â€“ 12x â€“ 13x +156 = 0

â‡’ x (x â€“ 12) â€“ 13 (x â€“ 12) = 0

â‡’ (x â€“ 12) (x â€“ 13) = 0

x = 13 or x = 12

So, if we take x = 13 then the marks in

Mathematics = 12 and marks in English = 18

So, if we take x = 12 then the marks in

Mathematics = 13 and marks in English = 17

### Question 11. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was â‚¹ 90, And the number of articles produced and the cost of each article.

**Solution:**

Given: Total cost of production on that day = â‚¹90

So, Let us considered the number of articles = x

Hence the price of each article = 2x + 3

According to the question,

x (2x + 3) = 90

â‡’ 2xÂ² + 3x – 90 = 0

â‡’ 2xÂ² -12x + 15x – 90 = 0

â‡’ 2x (x – 6) + 15 (x – 6) = 0

â‡’ (x – 6) (2x + 15) = 0

If x – 6 = 0

So, x = 6

If 2x + 15 = 0

So x = -15/2

Here the value of x is negative which is not possible

So the value valid of x = 6

Hence the number of articles = 6

And price of each article = 2x + 3 = 2 x 6 + 3 = 12 + 3 = 15

### Question 12. At t minutes past 2 pm the time needed by the minutes hand and a clock to show 3 pm was found to be 3 minutes less than t^{2}/4 minutes. Find t.

**Solution:**

As we already know that, the time between 2 pm to 3 pm = 1 h = 60 minutes

Given: At t minutes past 2 pm, the time needed by the minutes hand of a clock

to show 3 pm was found to be 3 minutes less than t

^{2}/4 minutes.Find: the value of t

So,

â‡’ 4t + tÂ² – 12 = 240

â‡’ tÂ² + 4t – 252 = 0

â‡’ tÂ² + 18t – 14t – 252 = 0

â‡’ t (t + 18) – 14 (t + 18) = 0

â‡’ (t + 18) (t – 14) = 0

As we know that time cannot be negative, so t â‰ -18

Hence, t = 14 min

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