Class 7 Exam  >  Class 7 Notes  >  Mathematics (Maths) Class 7  >  NCERT Solutions: Comparing Quantities

NCERT Solutions for Class 7 Maths - Comparing Quantities- 1

Exercise 7.1

Q1: Convert the given fractional numbers to percent: 
(a) 1/8
(b) 5/4
(c) 3/40
(d) 2/7

Ans: To convert the given fraction into percent we multiply the fraction by 100 and put the percent (%) Sign :

NCERT Solutions for Class 7 Maths - Comparing Quantities- 1
NCERT Solutions for Class 7 Maths - Comparing Quantities- 1


Q2: Convert the given decimal fractions to per cents: 
(a) 0.65
(b) 2.1

(c) 0.02
(d) 12.35

Ans:  To convert the decimals into percent, first convert them into fractions and then multiply by 100 and put a percent sign.
(a) 0.65:
0.65 can be written as 65/100.
To convert it to a percentage, we multiply it by 100.
(65/100) x 100 = 65%
Therefore, 0.65 is equal to 65%.

(b) 2.1:
2.1 remains the same.
To convert it to a percentage, we multiply it by 100.
2.1 x 100 = 210%
Therefore, 2.1 is equal to 210%.

(c) 0.02:
0.02 can be written as 2/100.
To convert it to a percentage, we multiply it by 100.
(2/100) x 100 = 2%
Therefore, 0.02 is equal to 2%.

(d) 12.35:
12.35 remains the same.
To convert it to a percentage, we multiply it by 100.
12.35 x 100 = 1235%
Therefore, 12.35 is equal to 1235%.


Q3: Estimate what part of the figures is coloured and hence find the percent which is coloured. 

NCERT Solutions for Class 7 Maths - Comparing Quantities- 1

Ans: (i) We can observe that only 1 part is coloured out of 4 parts.
Therefore, Coloured part = 1/4
 ∴ Percent of coloured part = NCERT Solutions for Class 7 Maths - Comparing Quantities- 1NCERT Solutions for Class 7 Maths - Comparing Quantities- 1

(ii) We can observe that only 3 parts is coloured out of 5 parts.
Therefore, Coloured part = 3/5
∴ Percent of coloured part NCERT Solutions for Class 7 Maths - Comparing Quantities- 1NCERT Solutions for Class 7 Maths - Comparing Quantities- 1

(iii) We can observe that only 3 parts is coloured out of 8 parts.
Therefore, Coloured part = 3/8
∴ Percent of coloured partNCERT Solutions for Class 7 Maths - Comparing Quantities- 1NCERT Solutions for Class 7 Maths - Comparing Quantities- 1
= 37.5%


Q4: Find: 
(a) 15% of 250 
(b) 1% of 1 hour 
(c) 20% of 2500 
(d) 75% of 1 kg 
Ans: 
(a) 15% of 250 NCERT Solutions for Class 7 Maths - Comparing Quantities- 1

(b) We know that, 1 hour = 60 minutes = 60 x 60 seconds = 3600 seconds
1% of 1 hours = 1% of 60 minutes = 1% of (60 )Minutes

NCERT Solutions for Class 7 Maths - Comparing Quantities- 1

(c) 20% of Rs 2500 NCERT Solutions for Class 7 Maths - Comparing Quantities- 1

(d) 75% of 1 kg = 75% of 1000 g NCERT Solutions for Class 7 Maths - Comparing Quantities- 1


Q5: Find the whole quantity if: 
(a) 5% of it is 600 
(b) 12% of it is ₹1080
(c) 40% of it is 500 km 
(d) 70% of it is 14 minutes 
(e) 8% of it is 40 litres 
Ans: Let the whole quantity be x in given questions:

NCERT Solutions for Class 7 Maths - Comparing Quantities- 1
NCERT Solutions for Class 7 Maths - Comparing Quantities- 1
NCERT Solutions for Class 7 Maths - Comparing Quantities- 1
NCERT Solutions for Class 7 Maths - Comparing Quantities- 1
NCERT Solutions for Class 7 Maths - Comparing Quantities- 1

Q6: Convert given per cents to decimal fractions and also to fractions in simplest forms: 
(a) 25% 
(b) 150% 
(c) 20% 
(d) 5% 
Ans: 
NCERT Solutions for Class 7 Maths - Comparing Quantities- 1

 
Q7: In a city, 30% are females, 40% are males and remaining are children. What percent are children? 
Ans: Given Percentage of females = 30%
Percentage of males = 40%
Total percentage of females and males = 30 + 40 = 70%
Percentage of children = Total percentage - Percentage of males and females
100% - 70% = 30%
Hence, 30% are children.


Q8: Out of 15,000 voters in a constituency, 60% voted. Find the percentage of voters who did not vote. Can you now find how many actually did not vote? 
Ans: Total voters = 15,000
Percentage of voted candidates = 60%
Percentage of not voted candidates = 100 - 60 = 40%
Actual candidates, who did not vote = 40% of 15000 NCERT Solutions for Class 7 Maths - Comparing Quantities- 1
Hence, 6,000 candidates did not vote.


Q9: Meeta saves Rs 4000 from her salary. If this is 10% of her salary. What is her salary? 
Ans: Let Meera's total salary be Rs x.
Now, 10% of salary = Rs 4000
⇒ 10% of x = Rs 400
⇒ (10/100) * x = Rs 4000
⇒ x = (4000 * 100)/10 = Rs 40,000
Hence, Meera's salary is Rs 40,000.


Q10: A local cricket team played 20 matches in one season. It won 25% of them. How many matches did they win? 
Ans: Number of matches played by cricket team = 20
Percentage of won matches = 25%
Total matches won by them = 25% of 20
NCERT Solutions for Class 7 Maths - Comparing Quantities- 1
Hence, they won 5 matches.

Exercise 7.2

Q1: Tell what is the profit or loss in the following transactions. Also find profit per cent or loss per cent in each case.

(a) Gardening shears bought for ₹250 and sold for ₹325.

  • Profit/Loss: ₹325 - ₹250 = ₹75 (Profit)
  • Profit per cent: 75250×100=30%

(b) A refrigerator bought for ₹12,000 and sold at ₹13,500.

  • Profit/Loss: ₹13,500 - ₹12,000 = ₹1,500 (Profit)
  • Profit per cent: 150012000×100=12.5%\frac{1500}{12000} \times 100 = 12.5\%

(c) A cupboard bought for ₹2,500 and sold at ₹3,000.

  • Profit/Loss: ₹3,000 - ₹2,500 = ₹500 (Profit)
  • Profit per cent: 5002500×100=20%\frac{500}{2500} \times 100 = 20\%

(d) A skirt bought for ₹250 and sold at ₹150.

  • Profit/Loss: ₹150 - ₹250 = -₹100 (Loss)
  • Loss per cent: 100250×100=40%\frac{100}{250} \times 100 = 40\%

Q2: Convert each part of the ratio to a percentage:

(a) 3: 1

  • Total parts = 3 + 1 = 4
  • Percentage of first part: 34×100=75%\frac{3}{4} \times 100 = 75\
  • Percentage of second part: 14×100=25%\frac{1}{4} \times 100 = 25\%41×100=25%

(b) 2 : 3: 5

  • Total parts = 2 + 3 + 5 = 10
  • Percentage of first part: 210×100=20%\frac{2}{10} \times 100 = 20\%102×100=20%
  • Percentage of second part: 310×100=30%\frac{3}{10} \times 100 = 30\%
  • Percentage of third part: 510×100=50%\frac{5}{10} \times 100 = 50\%

(c) 1: 4

  • Total parts = 1 + 4 = 5
  • Percentage of first part: 15×100=20%\frac{1}{5} \times 100 = 20\%
  • Percentage of second part: 45×100=80%\frac{4}{5} \times 100 = 80\%

(d) 1: 2: 5

  • Total parts = 1 + 2 + 5 = 8
  • Percentage of first part: 18×100=12.5%\frac{1}{8} \times 100 = 12.5\%
  • Percentage of second part: 28×100=25%\frac{2}{8} \times 100 = 25\%
  • Percentage of third part: 58×100=62.5%\frac{5}{8} \times 100 = 62.5\%

Q3: The population of a city decreased from 25,000 to 24,500. Find the percentage decrease.

  • Decrease in population: 25,000 - 24,500 = 500
  • Percentage decrease:NCERT Solutions for Class 7 Maths - Comparing Quantities- 1

Q4: Arun bought a car for ₹3,50,000. The next year, the price went up to ₹3,70,000. What was the Percentage of price increase?

  • Increase in price: ₹3,70,000 - ₹3,50,000 = ₹20,000
  • Percentage increase: NCERT Solutions for Class 7 Maths - Comparing Quantities- 1

Q5: I buy a T.V. for ₹10,000 and sell it at a profit of 20%. How much money do I get for it?

  • Profit: 20% of ₹10,000 NCERT Solutions for Class 7 Maths - Comparing Quantities- 1
  • Selling price: ₹10,000 + ₹2,000 = ₹12,000

Q6: Juhi sells a washing machine for ₹13,500. She loses 20% in the bargain. What was the price at which she bought it?

Ans: Loss: 20% of Cost Price (CP)
Selling Price (SP) = ₹13,500
Therefore, SP = CP - 20% of CP
NCERT Solutions for Class 7 Maths - Comparing Quantities- 1

Q7: (i) Chalk contains calcium, carbon and oxygen in the ratio 10:3:12. Find the percentage of carbon in chalk.
Ans: Total parts = 10 + 3 + 12 = 25

Percentage of carbon:
NCERT Solutions for Class 7 Maths - Comparing Quantities- 1

(ii) If in a stick of chalk, carbon is 3g, what is the weight of the chalk stick?
Ans: Carbon percentage: 12%
Therefore, 12%  of weight of chalk = 3g 
Let the weight of the chalk be W
NCERT Solutions for Class 7 Maths - Comparing Quantities- 1

Q8: Amina buys a book for ₹275 and sells it at a loss of 15%. How much does she sell it for?

  • Loss: 15% of ₹275
    NCERT Solutions for Class 7 Maths - Comparing Quantities- 1
  • Selling price: ₹275 - ₹41.25 = ₹233.75

Q9: Find the amount to be paid at the end of 3 years in each case:

(a) Principal = ₹1,200 at 12% p.a.

  • Simple Interest: 1200×12×3/100=432\frac{1200 \times 12 \times 3}{100} = 
  • Amount: ₹1,200 + ₹432 = ₹1,632

(b) Principal = ₹7,500 at 5% p.a.

  • Simple Interest: 7500×5×3/100=1,125\frac{7500 \times 5 \times 3}{100} = ₹1,125
  • Amount: ₹7,500 + ₹1,125 = ₹8,625

Q10: What rate gives ₹280 as interest on a sum of ₹56,000 in 2 years?

Ans: Interest (I): ₹280
Principal (P): ₹56,000
Time (T): 2 years
Using the formula for Simple Interest: I=P×R×T/100
NCERT Solutions for Class 7 Maths - Comparing Quantities- 1

Q11: If Meena gives an interest of ₹45 for one year at 9% rate p.a., what is the sum she has borrowed?
Ans:  Interest (I): ₹45
Rate (R): 9%
Time (T): 1 year
Using the formula for Simple Interest: I = P x R x T/100
NCERT Solutions for Class 7 Maths - Comparing Quantities- 1\frac{P \times R \times T}{100} = ₹500

The document NCERT Solutions for Class 7 Maths - Comparing Quantities- 1 is a part of the Class 7 Course Mathematics (Maths) Class 7.
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FAQs on NCERT Solutions for Class 7 Maths - Comparing Quantities- 1

1. What are comparing quantities in mathematics?
Ans.Comparing quantities involves evaluating two or more amounts to determine their relationship. This can include finding percentages, ratios, or using fractions to see how one quantity relates to another.
2. How do you calculate the percentage increase or decrease?
Ans.To calculate the percentage increase, subtract the original value from the new value, divide the result by the original value, and then multiply by 100. For percentage decrease, the process is similar: subtract the new value from the original value, divide by the original value, and multiply by 100.
3. What is the formula for finding the ratio of two quantities?
Ans.The formula for finding the ratio of two quantities is to divide one quantity by the other. This can be expressed in the form of a:b, where 'a' is the first quantity and 'b' is the second quantity.
4. How can we apply comparisons in real-life situations?
Ans.Comparisons can be applied in various real-life situations such as budgeting money, comparing prices while shopping, or assessing discounts during sales. It helps individuals make informed decisions based on quantitative analysis.
5. What are some common mistakes to avoid when comparing quantities?
Ans.Common mistakes include not using the same units for comparison, forgetting to convert percentages into decimals or vice versa, and miscalculating ratios by not simplifying them properly. Always double-check calculations to avoid errors.
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