SEBA Class 7 Mathematics Chapter 8 Comparing Quantities

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SEBA Class 7 Mathematics Chapter 8 Comparing Quantities

Also, you can read the SCERT book online in these sections Solutions by Expert Teachers as per SCERT (CBSE) Book guidelines. SEBA Class 7 General Maths Textual Question Answer. These solutions are part of SCERT All Subject Solutions. Here we have given SEBA Class 7 Mathematics Chapter 8 Comparing Quantities Solutions for All Subject, You can practice these here.

Comparing Quantities

Chapter – 8

PART – I

Exercise – 8.1

1. Find the ratios–

(i) 5 km and 500 m.

Ans: = 5 km/500 m

= 5 × 100 m/500 m

= 10/1

= 10:1

(ii) 6 week and 21 days.

Ans: 6 weeks/21 days

= 6 × 7 days/21 days

= 2/1

= 2:1

(iii) 30 days and 210 hours.

Ans: = 30 days/210 hours

= 30 × 24 hours/210 hours

= 24/7

= 24:7

(iv) 12 meter and 100 centimeter.

Ans: 12 m/100 centimeter

= 30 × 100 centimeter/100 centimeter

= 12/1

= 12:1

(v) 7 years and 18 months.

Ans: 7 years/18 months

= 7 × 12 months/18 months

= 84/18

= 14/3 

2. Examine whether the following ratios are equivalent or not-

(i) 4:5 and 2:3

Ans: 4:5 and 2:3

4/5 = (4 × 3)/(5 × 3) = 12/15

2/3 = (2 × 5)/(3 × 5) = 10/15

∴ 4/5 > 2/3

∴ Not Equivalent.

(ii) 7:15 and 21:45

Ans: 7:15 and 21:45

7/15 = (7 × 3)/(15 × 3) = 21/45

21/45 = (21 × 1)(/45 × 1) = 21/45

∴ 7/15 = 21/45

∴ 7:15 and 21:45 are equivalent.

(iii) 6:8 and 54:72

Ans: 6:8 and 54:72

6/8 = (6 × 9)/(8 × 9) = 54/72

54/72 = (45 × 1)/(72 × 1) = 54/72

∴ 6/8 = 54/72

∴ 6:8 and 54:72 are equivalent.

(iv) 13:14 and 14:13

Ans: 13:14 and 14:13

13/14 = (13 × 13)/(14 × 13) = 169/182

14/13 = (14 × 14) / (13 × 14)= 196/182

13/14 ≠ 14/13 and 13/14<14/13

∴ 13:14 and 14:13 are not equivalent.

3. The price of three geometry boxes is 33. How many geometry boxes can be bought with 286?

Ans: Number of geometry boxes with Rs. 33 is 3

∴ Number of geometry boxes with Rs. 1 is 3/33

∴ Number of geometry boxes with Rs. 286 is

4. Kalpana has bought 10 balls with Rs. 160 and Jaymati has bought 8 balls with Rs. 128. Find if both of them have purchased at an equal price?

Ans: Number of balls has brought with Rs. 160 = 10

∴ Number of balls has brought with Rs. 1 = 10/160

= 1/16  

= 1:16

Again, number of balls has brought with Rs. 128 = 8

∴ Number of balls has brought with Rs. 1 = 8/128

= 1/16

= 1:16

Again, They have purchased at an equal price.

5. Jenelia needs to pay 66,000/- as house rent for the whole year. If she wants to pay house rent every three months then how much will she need to pay each time?

Ans: For 12 months Jenelia needs to pay Rs. 66,000/- as a rent

∴ For 1 month Jenelia needs to pay Rs. 66,000/12 as house rent

∴ For 3 months Jenelia needs to pay Rs. 66,000/12 × 3 as house rent 

= Rs. 16,500

Exercise – 8.2

1. Express the following fraction in percentage–

(i) 3/5

Ans: 3/5 × 100% 

= 3 × 20% 

= 60%

(ii) 7/12

Ans: 7/12 × 100% 

= 175/3% = 58.33%

(iii) 15/32

Ans: 15/32 × 100% 

= 375/8% 

= 46.88%

(iv) 91/175

Ans: 

(v) 21/40

Ans: 21/40 × 100%

= 105/2%

= 52.5%

2. Express the decimal fraction in percentage–

(i) 0.8

Ans: 0.8 × 100%

= 8/10 × 100%

= 80%

(ii) 5.75

Ans: 5.75 × 100% 

= 575/100 × 100%

= 575%

(iii) 0.08

Ans: 0.08 × 100% 

= 8/100 × 100%

= 8%

(iv) 32.1

Ans: 32.1 × 100% 

= 321/10 × 100%

= 3210%

(v) 0.004

Ans: 0.004 × 100% 

= 4/4000 × 100%

= 4/10%

= 0.4%

3. Express in fraction–

(i) 20%

Ans: 20% 

= 20/100 

= 1/5

(ii) 32%

Ans: 32% 

= 32/100 

= 8/25

(iii) 0.5%

Ans: 0.5%

= 5/100

= 5/10 × 100

= 1/200

(iv) 7.25%

Ans: 7.25% 

= 725/100 × 100

= 29/400

(v) 180%

Ans: 180%

= 180/100

= 1 4/5

4. Convert the percentages into decimal fraction and express the fraction in lowest form of the following–

(i) 18%

Ans: 18% 

= 18/100 

= 18

Again, 18/100

= 9/50

(ii) 0.25%

Ans: 0.25%

= 25/100 × 100

= 25/10000

= 0.0025

Again, 25/100 × 100

= 1/400

(iii) 60%

Ans: 60%

= 60/100

= 0.60

Again, 60/100

= 3/5

(iv) 42.5%

Ans: 42.5% 

= 425/10 × 100 

= 0.425

Again, 425/10 × 100 

= 17/40

(v) 56%

Ans: 56% 

= 56/100 

= 0.56

Again, 56/100 = 14/25

5. Express in percentage–

(i) 4:10

Ans: 4:10

= 4/10 × 100% 

= 40%

(ii) 11:20

Ans: 11:20

= 11/20 × 100% 

= 55%

(iii) 19:50

Ans: 19:50

= 19/50 × 100% 

= 38%

(iv) 3:25

Ans: 3:25

= 3/25 × 100% 

= 12%

(v) 9:4

Ans: 9:4

= 9/4 × 100% 

= 225%

6. Express in ratio–

(i) 72%

Ans: 72% 

= 72/100 

= 18/25 

= 18:25

(ii) 15/4%

Ans: 15/4% 

= 15/4 × 100

= 3/80 

= 3:80

(iii) 0.14%

Ans: 0.14%

= 14/100 × 100

= 7/5000

= 7:5000

(iv) 6 2/5%

Ans: 6 2/5%

= 32/5 × 100

= 8/125

= 8:125

(v) 4.6%

Ans: 4.6%

= 46/10 × 100

= 23/500

= 23:500

7. Express the coloured portion of the following figures in percentage–

(i)

Ans: = 1/4 × 100%

= 25%

Ans: = 1/2/3 × 100%

= 1/6 × 100%

= 50/3%

= 16 ⅔%

Ans: 12/40 × 100%

= 30%

8. Express in percentage–

(i) 8 hours of day.

Ans: 8 hours of day

= 8/24 hours 

= 1/3 × 100%

= 33 ⅓%

(ii) 72 person out of 80.

Ans: 72 person out of 80

= 72/80 × 100% 

= 90%

(iii) 25 paisa of Rs. 5.

Ans: 25 paisa of Rs. 5 

= 25/5paisa/Rupees

= 25/5 × 1/100

= 1/20 × 100

= 5%

9. Determine the value–

(i) 20% of 50

Ans: 20% of 50

= 50 × 20/100

= 10

(ii) 30% of Rs 600

Ans: 30% of Rs 600

= Rs. (600 × 30/100)

= Rs. 108

(iii) 20% of 5 km.

Ans: 20% of 5 km

= 5 × 1000 × 20/100 m

= 100 m

= 1 km

(iv) 30% of 1 hours

Ans: 30% of 1 hours

= 60 × 30/100 minutes

= 18 minutes.

10. If value of 20% of a number is 50, determine the number.

Ans: Let the number be x

A/Q, 20% of x = 50

⇒ 20 × x/100 = 50

⇒ x = 250

∴ The reqd. Number = 250.

11. 13% of which number is 91?

Ans: Let the number be x

A/Q, 13% of x = 91

⇒ 13x/100 = 91

⇒ 13x = 91 × 100

⇒ x =  91 × 100/13 = 700

∴ The reqd. Number is 700.

12. What percentage of 40 is 16?

Ans: Let the percentage be x%

∴ x% of 40 = 16

⇒ 40 × x/100 = 16

⇒ 2x/5 = 16

⇒ 2x = 16 × 5/2 = 40%

∴ The reqd. Number is 40%

13. Express ’30 out of 40′ and ’40 out of 50′, both in percentage. Which is greater?

Ans: 

∴ 40 out of 50 is greater than 30 out of 40.

14. The area of a garden is 400 square meter. The garden is arranged in the following manner.

(i) 30% space of the garden is reserved for flowers. Find the area of the space for flowers.

Ans: The area of garden = 400 sq. m.

30% space of the garden is reserved for flowers.

∴ Area of the space for flowers

= 30% of 400 sq.m.

= 30 × 400/100 sq. m.

= 120 sq.m.

(ii) The area of a pond in the garden is 80 square meter. What percentage of the garden is covered by the pond?

Ans: The area of pond in the garden is 80 sq.m.

∴ The reqd. percentage = 80/400 × 100

= 20%

Exercise – 8.3

1. Find the profit or loss in the following transactions. Also find profit percent or loss percent in each case.

(a) A ball bought for Rs. 300 and sold for Rs. 450.

Ans: Profit = Selling price – Cost price

= Rs. (450 – 300)

= Rs. 150

∴ Profit percentage = Profit/Cost Price × 100%

= 150/300 × 100%

= 50%

Profit = Rs. 150 and Profit percentage 50%

(b) A TV bought for 10,000 and sold for Rs. 15,500.

Ans: Profit = Selling price – Cost price

= Rs. (15,500 – 10,000)

= Rs. 5,500, Profit percent = 50%

∴ Profit percent = 5,500/10,000 × 100%

= 55%

Profit = Rs. 5,500, Profit percent = 55%

(c) A cupboard bought for Rs. 5,000 and sold for Rs. 3,500.

Ans: Loss = Rs. (5,000 – 3500)

= Rs. 1500

∴ Loss percent = 1500/5000 × 100%

= 30%

Loss = Rs. 1500 and Loss percent = 30%

(d) A shirt bought for Rs. 400 and sold for Rs. 280.

Ans: Loss = Rs. (400 – 280)

= Rs. 120

∴ Loss percent = 120/400 × 100%

= 30%

Loss = Rs. 120, Loss percent = 30%

2. Population of an area decreases from 45,000 to 42,000. Calculate percentage of decrease.

Ans: Population of an area from 45,000 to 42,000

∴ Decreases population = 45,000 – 42,000

= 3,000

∴ Percentage of decrease = 3,000/45,000 × 100%

= 20/3%

= 6 ⅔%

3. A book bought for Rs. 250 and sold for Rs. 190. What is the profit or loss in percentage?

Ans: Loss = Rs. (250 – 190)

= Rs. 60

∴ Loss percentage = 60/250 × 100

= 24%

4. Rohan has bought a TV for Rs. 10,500 and sold if at a profit of 30% Calculate the selling price of the TV.

Ans: Let, 

The cost price = Rs. 100

∴ Selling price = Rs. (100 + 30)

= Rs. 130

∴ If C. P. is Rs. 100

S.P. = Rs. 130

∴ If C.P. is Rs.1 S.P. = Rs. 130/100

∴ If C.P. is Rs. 10,500 S.P. = Rs. 130/100 × 10500

= 13,650

Selling price of the TV = Rs. 13,650

5. Sumona has bought a car for Rs. 2,50,000 and sold for Rs. 2,30,000. What is the loss or profit in percentage?

Ans: Loss = Rs. (2,50,000 – 2,30,000)

= 20,000

∴ Loss percentage = 20000/250000 × 100%

= 8%

6. A motor cycle has been sold for Rs. 23,000 at a profit of 15%. What is the cost price of the motor cycle?

Ans: Let, C. P. = Rs. 100

∴ S.P. = Rs. (100 + 15)

= Rs. 115

∴ If S.P. is Rs. 115, C.P. = 100

∴ If S.P. is Rs. 1, C.P. = Rs. 100/115

∴ If S.P. is Rs. 23,000 C.P.

= Rs.100 × 23,000/115

= 20,000

∴ Cost price = 20,000

7. Ratan has bought an almirah for Rs. 6,250 and sold at a loss of 24%. What is the selling price of the almirah?

Ans: Let the C.P. = Rs. 100

Now, S.P. = Rs. (100 – 24)

= Rs. 76

∴ Selling price = 76 × 6250/100

= Rs. (38 × 125)

= Rs. 4750

8. Determine the interest of Rs. 5000 at 8% rate of interest per annum for 3 years.

Ans: Here, P = Rs. 5000

Rate of interest (R) = 8,

Time (T) = 3 years

∴ The reqd. S.I. = P × R × T/100

= Rs. 5000 × 8 × 3/100

= (50 × 24)

= Rs. 1200

9. Dalimi pays Rs. 120 as interest for one year at a rate of interest of 5%. How much money did she borrow?

Ans: Rs. 5 as the interest on Rs. 100

∴ Rs. 1 as the interest on Rs. 100/5

∴ Rs. 120 as the interest on Rs. 100 × 120/5

= Rs. 2400

She borrowed = Rs. 2400.

10. Find out the interest of Rs. 1,25,000 at 10% rate of interest for a period of 1 year and 6 months.

Ans: Let, the principal (P) = Rs. 1,25,000

Rate of interest ( R) = 10%

And time (T) = 1 year and 6 months

= 1 6/12 years

= 3/2 years

∴ Interest (I) = P × R × T/100

= 1,25,000 × 10 × 3 × 2 / 100

= Rs. (6250 × 3)

= Rs. 18,750

11. Find out the interest and total amount of Rs. 1500 at rate of interest of Rs. 12 per annum for 2 years and 9 months.

Ans: Principal (P) = Rs. 1500

Rate of interest (R) = 12

Time (T) = 2 years and 9 months

= 2 9/12 years

= 11/4 years

∴ Interest = P × R × T/100

= Rs. 1500 × 12 × 11 × 4 /100

= Rs.15 × 3 × 11

= Rs. 495

Now, interest = Rs. 495

And Amount = I + P

= Rs. (495 + 1500)

= Rs. 1995