Study MaterialsImportant QuestionsComparing Quantities Class 8 Extra Questions Maths Chapter 8

Comparing Quantities Class 8 Extra Questions Maths Chapter 8

Comparing Quantities Class 8 Extra Questions Maths Chapter 8

Extra Questions for Class 8 Maths Chapter 8 Comparing Quantities

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    Comparing Quantities Class 8 Extra Questions Very Short Answer Type

    Question 1.
    Express the following in decimal form:
    (a) 12%
    (b) 25%
    Solution:
    (a) 12% = \(\frac { 12 }{ 100 }\) = 0.12
    (b) 25% = \(\frac { 25 }{ 100 }\) = 0.25

    Question 2.
    Evaluate the following:
    (a) 20% of 400
    (b) 12\(\frac { 1 }{ 2 }\)% of 625
    Solution:
    Comparing Quantities NCERT Extra Questions for Class 8 Maths Q2

    Question 3.
    If 20% of x is 25, then find x.
    Solution:
    20% of x = 25
    Comparing Quantities NCERT Extra Questions for Class 8 Maths Q3
    Hence x = 125

    Question 4.
    Express the following as a fraction
    (a) 35%
    (b) 64%
    Solution:
    Comparing Quantities NCERT Extra Questions for Class 8 Maths Q4

    Question 5.
    Express the following into per cent
    (а) 1\(\frac { 3 }{ 5 }\)
    (b) 2 : 5
    Solution:
    Comparing Quantities NCERT Extra Questions for Class 8 Maths Q5

    Question 6.
    There are 24% of boys in a school. If the number of girls is 456, find the total number of students in the school.
    Solution:
    Let the total number of students be 100.
    Number of boys = 24% of 100 = \(\frac { 24 }{ 100 }\) × 100 = 24
    Number of girls = 100 – 24 = 76
    ⇒ If number of girls is 76, then total number of students = 100
    ⇒ If Number of girls is 1, then total number of students = \(\frac { 100 }{ 76 }\)
    If Number of girls is 456, then total number of students = \(\frac { 100\times 456 }{ 76 }\) = 600
    Hence, the total number of students in the school = 600

    Question 7.
    The cost of 15 articles is equal to the selling price of 12 articles. Find the profit per cent.
    Solution:
    Let CP of 15 articles be ₹ 100
    CP of 1 article = ₹ \(\frac { 100 }{ 15 }\)
    SP of 12 articles = ₹ 100
    SP fo 1 article = ₹ \(\frac { 100 }{ 12 }\)
    SP > CP
    Comparing Quantities NCERT Extra Questions for Class 8 Maths Q7
    Comparing Quantities NCERT Extra Questions for Class 8 Maths Q7.1
    Hence, profit = 25%

    Question 8.
    An article is marked at ₹ 940. If it is sold for ₹ 799, then find the discount per cent.
    Solution:
    MP = ₹ 940
    SP = ₹ 799
    Discount = MP – SP = 940 – 799 = ₹ 141
    Comparing Quantities NCERT Extra Questions for Class 8 Maths Q8
    Hence, discount = 15%

    Question 9.
    A watch was bought for ₹ 2,700 including 8% VAT. Find its price before the VAT was added.
    Solution:
    Cost of watch including VAT = ₹ 2,700
    Let the initial cost of the watch be ₹ 100
    VAT = 8% of ₹ 100 = ₹ 8
    Cost of watch including VAT = ₹ 100 + ₹ 8 = ₹ 108
    If cost including VAT is ₹ 108, then its initial cost = ₹ 100
    If cost including VAT is ₹ 1, then its initial cost = ₹ \(\frac { 100 }{ 108 }\)
    If cost including VAT is ₹ 2,700, then its initial cost = ₹ \(\frac { 100 }{ 108 }\) × 2700 = ₹ 2500
    Hence, the required cost = ₹ 2,500

    Question 10.
    Find the amount if ₹ 2,000 is invested for 2 years at 4% p.a. compounded annually.
    Solution:
    Comparing Quantities NCERT Extra Questions for Class 8 Maths Q10
    Comparing Quantities NCERT Extra Questions for Class 8 Maths Q10.1

    Comparing Quantities Class 8 Extra Questions Short Answer Tpye

    Question 11.
    A number is increased by 20% and then it is decreased by 20%. Find the net increase or decrease per cent. (NCERT Exemplar)
    Solution:
    Let the number be 100
    20% increase = \(\frac { 20 }{ 100 }\) × 100 = 20
    Increased value = 100 + 20 = 120
    Now it is decreased by 20%
    Decreased value = 120 – \(\frac { 120 }{ 100 }\) × 20 = 120 – 24 = 96
    Net decrease = 100 – 96 = 4
    Decrease per cent = \(\frac { 4 }{ 100 }\) × 100 = 4%
    Hence, the net decrease per cent = 4%

    Question 12.
    Two candidates Raman and Rajan contested an election. Raman gets 46% of the valid votes and is#defeated by 1600 votes. Find the total number of valid votes cast in the election.
    Solution:
    Let the total number of valid votes be 100
    Number of votes got by Raman = 46% of 100 = \(\frac { 46 }{ 100 }\) × 100 = 46
    Number of votes got by Rajan = 100 – 46 = 54
    Difference between the votes = 54 – 46 = 8
    8% of Valid votes = 1,600
    ⇒ \(\frac { 8 }{ 100 }\) × Valid votes = 1,600
    ⇒ Valid votes = \(\frac { 1600\times 100 }{ 8 }\) = 20,000
    Hence, the total number of valid votes = 20,000

    Question 13.
    A man whose income is ₹ 57,600 a year spends ₹ 43,200 a year. What percentage of his income does he save?
    Solution:
    Annual income of a man = ₹ 57,600
    Amount spent by him in the year = ₹ 43,200
    Net amount saved by him = ₹ 57,600 – ₹ 43,200 = ₹ 14,400
    Percentage of his annual saving Saving = \(\frac { Saving }{ Income }\) × 100
    = \(\frac { 14400 }{ 57600 }\) × 100
    = 25%
    Hence, the saving percentage = 25%

    Question 14.
    A CD player was purchased for ₹ 3,200 and ₹ 560 were spent on its repairs. It was then sold at a gain of 12\(\frac { 1 }{ 2 }\) %. How much did the seller receive?
    Solution:
    Cost price of the CD player = ₹ 3,200
    Amount spent on its repairing = ₹ 560
    Net cost price = ₹ 3,200 + ₹ 560 = ₹ 3,760
    Comparing Quantities NCERT Extra Questions for Class 8 Maths Q14
    Hence, the required amount = ₹ 4,230

    Question 15.
    A car is marked at ₹ 3,00,000. The dealer allows successive discounts of 6%, 4% and 2\(\frac { 1 }{ 2 }\) % on it. What is the net selling price of it?
    Solution:
    Marked price of the car = ₹ 3,00,000
    Net selling price after the successive discounts
    Comparing Quantities NCERT Extra Questions for Class 8 Maths Q15
    Hence, the net selling price = ₹ 2,63,952

    Question 16.
    Ramesh bought a shirt for ₹ 336, including 12% ST and a tie for ₹ 110 including 10% ST. Find the list price (without sales tax) of the shirt and the tie together.
    Solution:
    List price of the shirt = \(\frac { 110 }{ 112 }\) × 336 = ₹ 300
    List price of the tie = \(\frac { 100 }{ 110 }\) × 110 = ₹ 100
    List price of both together = ₹ 300 + ₹ 100 = ₹ 400

    Question 17.
    Find the amount of ₹ 6,250 at 8% pa compounded annually for 2 years. Also, find the compound interest.
    Solution:
    Comparing Quantities NCERT Extra Questions for Class 8 Maths Q17

    Question 18.
    Find the compound interest on ₹ 31,250 at 12% pa for 12\(\frac { 1 }{ 2 }\) years.
    Solution:
    Comparing Quantities NCERT Extra Questions for Class 8 Maths Q18

    Question 19.
    Vishakha offers a discount of 20% on all the items at her shop and still makes a profit of 12%. What is the cost price of an article marked at ₹ 280? (NCERT Exemplar)
    Solution:
    Marked Price = ₹ 280
    Discount = 20% of ₹ 280
    = \(\frac { 1 }{ 2 }\) × 280 = ₹ 56
    So selling price = ₹ (280 – 56) = ₹ 224
    Let the cost price be ₹ 100
    Profit = 12% of ₹ 100 = ₹ 12
    So selling price = ₹ (100 + 12) = ₹ 112
    If the selling price is ₹ 112, cost price = ₹ 100
    If the selling price is ₹ 224, cost price = ₹ (\(\frac { 100 }{ 112 }\) × 224) = ₹ 200

    Question 20.
    Find the compound interest on ₹ 48,000 for one year at 8% per annum when compounded half yearly. (NCERT Exemplar)
    Solution:
    Principal (P) = ₹ 48,000
    Rate (R) = 8% p.a.
    Time (n) = 1 year
    Interest is compounded half yearly
    Comparing Quantities NCERT Extra Questions for Class 8 Maths Q20
    Therefore Compound Interest = A – P = ₹ (519,16.80 – 48,000) = ₹ 3,916.80.
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