# Class 8 RD Sharma Solutions- Chapter 9 Linear Equation In One Variable – Exercise 9.4 | Set 2

### Chapter 9 Linear Equation In One Variable – Exercise 9.4 | Set 1

**Question 14. I am currently 5 times as old as my son. In 6 years time I will be three times as old as he will be then. What are our ages now?**

**Solution:**

Let assume that present son’s age will be x years

Present father’s age will be 5x years

Son’s age after 6 years will be (x + 6) years

Fathers’ age after 6 years will be (5x + 6) years

5x + 6 = 3(x + 6)

5x + 6 = 3x + 18

5x – 3x = 18 – 6

2x = 12

x = 12/2= 6

Now ,present son’s age will be x = 6years

And Present father’s age will be 5x = 5(6) = 30years

**Question 15. I have Rs 1000 in ten and five rupee notes. If the number of ten rupee notes that I have is ten more than the number of five rupee notes, how many notes do I have in each denomination?**

**Solution:**

Let us assume that the number of five rupee notes be x

Number of ten rupee notes will be (x + 10)

Amount due to five rupee notes will be = 5 × x = 5x

Amount due to ten rupee notes will be = 10 (x + 10) = 10x + 100

The total amount = Rs 1000

5x + 10x +100 = 1000

15x = 900

x = 900/15 = 60

The number of five rupee notes are x = 60

Number of ten rupee notes are x + 10 = 60+10 = 70

**Question 16. At a party, colas, squash and fruit juice were offered to guests. A fourth of the guests drank colas, a third drank squash, two fifths drank fruit juice and just three did not drink anything. How many guests were in all?**

**Solution:**

Let assume that number of guests will be x

Let The given details are Number of guests who drank colas are x/4

Let Number of guests who drank squash are x/3

Let Number of guests who drank fruit juice are 2x/5

Let Number of guests who did not drink anything are 3

x/4 + x/3 + 2x/5 + 3 = x

By take LCM for 4, 3 and 5 is 60

(15x+20x+24x-60x)/60 = -3

By doing cross-multiply,

(15x+20x+24x-60x) = -3(60)

-x = -180

x = 180

The total number of guests in all were 180.

**Question 17. There are 180 multiple choice questions in a test. If a candidate gets 4 marks for every correct answer and for every unattempted or wrongly answered question one mark is deducted from the total score of correct answers. If a candidate scored 450 marks in the test, how many questions did he answer correctly?**

**Solution:**

Let assume that number of correct answers be x

Number of questions answered wrong are (180 – x) (let)

Total score when answered right = 4x (let)

Marks deducted when answered wrong = 1(180 – x) = 180 – x

4x – (180 – x) = 450

4x – 180 + x = 450

5x = 450 + 180

5x = 630

x = 630/5 = 126

**Question 18. A labourer is engaged for 20 days on the condition that he will receive Rs 60 for each day, he works and he will be fined Rs 5 for each day, he is absent. If he receives Rs 745 in all for how many days he remained absent?**

**Solution:**

Let us assume that number of absent days as x

Let the number of present days is (20 – x)

Wage for one day work will be Rs 60

Fine for absent day will be Rs 5

60(20 – x) – 5x = 745

1200 – 60x – 5x = 744

-65x = 744-1200

-65x = -456

x = -456/-65 = 7

**Question 19. Ravish has three boxes whose total weight is 60 ½ Kg. Box B weighs 3 ½ kg more than box A and box C weighs 5 1/3 kg more than box B. Find the weight of box A.**

**Solution:**

The given details are total weight of three boxes is 60 ½ kg = 121/2 kg

Let assume that the weight of box A is x kg

Weight of box B is x + 7/2 kg

Weight of box C is x + 7/2 + 16/3 kg

x + x + 7/2 + x + 7/2 + 16/3 = 121/2

3x = 121/2 – 7/2 – 7/2 – 16/3

Now take LCM for 2 and 3 is 6

3x = (363 – 21 – 21 – 32)/6

3x = 289/6

x = 289/18

**Question 20. The numerator of a rational number is 3 less than the denominator. If the denominator is increased by 5 and the numerator by 2, we get the rational number 1/2. Find the rational number.**

**Solution:**

Let assume that denominator be x

The numerator be (x – 3)

We know that formula

Fraction = numerator/denominator

= (x – 3)/x

when Numerator is increased by 2

Denominator is increased by 5, then fraction is ½

(x – 3 + 2)/(x + 5) = 1/2

(x – 1)/(x + 5) = 1/2

By doing cross-multiplication

2(x – 1) = x + 5

2x – 2 = x + 5

2x – x = 2 + 5

x = 7

Denominator is x = 7, numerator is (x – 3) = 7 – 3 = 4

And the fraction is numerator/denominator = 4/7

**Question 21. In a rational number, twice the numerator is 2 more than the denominator If 3 is added to each, the numerator and the denominator. The new fraction is 2/3. Find the original number.**

**Solution:**

Let us assume the numerator be x

The denominator be (2x – 2)

According formula

Fraction = numerator/denominator

= x / (2x – 2)

So, the numerator and denominator are increased by 3, then fraction is 2/3

(x + 3)/(2x – 2 + 3) = 2/3

(x + 3)/(2x + 1) = 2/3

By doing cross-multiplying

3(x + 3) = 2(2x + 1)

3x + 9 = 4x + 2

3x – 4x = 2 – 9

-x = -7

x = 7

The numerator is x = 7, denominator is (2x – 2) = (2(7) – 2) = 14-2 = 12

And the fraction is numerator/denominator = 7/12

**Question 22. The distance between two stations is 340 km. Two trains start simultaneously from these stations on parallel tracks to cross each other. The speed of one of them is greater than that of the other by 5 km/hr. If the distance between the two trains after 2 hours of their start is 30 km, find the speed of each train.**

**Solution:**

Let assume the speed of one train be x km/hr.

Let Speed of other train be (x + 5) km/hr.

Total distance between two stations = 340 km

As we know the formula

Distance = speed × time

Distance covered by one train in 2 hrs. = x × 2 = 2x km

Distance covered by other train in 2 hrs. = 2(x + 5) = (2x + 10) km

Distance between the trains is 30 km

2x + 2x + 10 + 30 = 340

4x + 40 = 340

4x = 340 – 40

4x = 300

x = 300/4 = 75

The speed of one train is 75 km/hr.

Speed of other train is (x + 5) = 75 + 5 = 80 km/hr.

**Question 23. A steamer goes downstream from one point another in 9 hours. It covers the same distance upstream in 10 hours. If the speed of the stream be 1 km/hr., find the speed of the steamer in still water and the distance between the ports.**

**Solution:**

Let us assume the speed of steamer be x km/hr.

Speed of stream = 1 km/hr. (According to question)

Let Downstream speed = (x + 1) km/hr.

Let Upstream speed = (x – 1) km/hr.

According to formula

Distance = speed × time

= (x + 1) × 9

= (x – 1) × 10

9x + 9 = 10x – 10

9x – 10x = -10 -9

-x = -19

x = 19 km/hr.

The speed of the steamer in still water 19 km/hr.

Distance between the ports is 9(x + 1) = 9(19+1) = 9(20) = 180 km.

**Question 24. Bhagwanti inherited Rs 12000.00. She invested part of it as 10% and the rest at 12%. Her annual income from these investments is Rs 1280.00 How much did she invest at each rate?**

**Solution:**

At rate of 10% Let the investment be Rs x (let)

At the rate of 12% the investment will be Rs (12000 – x) (let)

At 10% of rate the annual income will be x × (10/100) = 10x/100 (let)

At 12% of rate the annual income will be (12000 – x) × 12/100 = (144000 – 12x)/100

Total investment = 1280

10x/100 + (144000 – 12x)/100 = 1280

(10x + 144000 – 12x)/100 = 1280

(144000 – 2x)/100 = 1280

By doing cross-multiply

144000 – 2x = 1280(100)

-2x = 128000 – 144000

-2x = -16000

x = -16000/-2= 8000

At 10% of rate she invested Rs 8000 and at 12% of rate she invested Rs (12000 – x) = Rs (12000 – 8000) = Rs 4000

**Question 25. The length of a rectangle exceeds its breadth by 9 cm. If length and breadth are each increased by 3 cm, the area of the new rectangle will be 84 cm2 more than that of the given rectangle. Find the length and breadth of the given rectangle.**

**Solution:**

Let assume that breadth of the rectangle be x meter

Length of the rectangle be (x + 9) meter (let)

Area of the rectangle length × breadth = x(x +9) m2 (As we know the formula)

When length and breadth increased by 3cm

New length will be x + 9 + 3 = x + 12

New breadth will be x + 3

Area =(x + 12) (x + 3) = x (x + 9) + 84

x2 + 15x + 36 = x2 + 9x + 84

15x – 9x = 84 – 36

6x = 48

x = 48/6 = 8

Length of the rectangle (x + 9) = (8 + 9) = 17cm

breadth of the rectangle is 8cm.

**Question 26. The sum of the ages of Anup and his father is 100. When Anup is as old as his father now, he will be five times as old as his son Anuj is now. Anuj will be eight years older than Anup is now, when Anup is as old as his father. What are their ages now?**

**Solution:**

Let assume that age of Anup be x years

Age of Anup’s father will be (100 – x) years

The age of Anuj will be (100-x)/5 years

Anup is as old as his father after (100 – 2x) years

Anuj’s age = present age of his father (Anup) + 8

Present age of Anuj + 100 – 2x = Present age of Anup + 8

(100 – x)/5 + (100 – 2x) = x + 8

(100-x)/5 – 3x = 8 – 100

(100 – x – 15x)/5 = -92

By doing cross-multiplication,

100 – 16x = -460

-16x = -460 – 100

-16x = -560

x = -560/-16 = 35

Present age of Anup is 35 years then, Age of Anup’s father will be (100-x) = 100-35 = 65 years

The age of Anuj is (100-x)/5 = (100 – 35)/5 = 65/5 = 13 years

**Question 27. A lady went shopping and spent half of what she had on buying hankies and gave a rupee to a beggar waiting outside the shop. She spent half of what was left on lunch and followed that up with a two rupee tip. She spent half of the remaining amount on a book and three rupees on bus fare. When she reached home, she found that she had exactly one rupee left. How much money did she start with?**

**Solution:**

Let assume that amount lady had be Rs x

Amount spent for hankies and given to beggar is x/2 + 1 (let)

Remaining amount is x – (x/2 + 1) = x/2 – 1 = (x-2)/2 (let)

Amount spent for lunch (x-2)/2×1/2 = (x-2)/4

Amount given as tip is Rs 2

Remaining amount after lunch = (x-2)/2 – (x-2)/4 – 2 = (2x – 4 – x + 2 – 8)/4 = (x – 10)/4

Amounts spent for books =1/2 × (x-10)/4 = (x-10)/8

Bus fare is Rs 3

Amount left = (x-10)/4 – (x-10)/8 – 3 = (2x – 20 – x + 10 – 24)/8 = (x-34)/8

According to the question we know that the amount left = Rs 1

(x-34)/8 = 1

By cross-multiplication

x – 34 = 8

x = 8 + 34 = 42

The lady started with Rs. 42

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