KSEEB Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.2

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KSEEB Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.2

Question 1.
A man got a 10% increase in his salary. If his new salary is Rs. 1,54,000, find his original salary.
Solution:
Let Initial salary = Rs. 100
Increase % = 10%
Amount of increase = 10% of 100
= \(\frac{10}{100} \times 100\)
= Rs. 10
New salary after increase = 100 + 10 = Rs. 110
If new salary Rs. 110 then initial salary = Rs. 100
If new salary Rs. 1 then initial salary = \(\frac{100}{110}\)
If new salary is Rs. 1,54,000 then initial salary = \(\frac{100}{110}\) × 1,54,000 = Rs. 1,40,000

Question 2.
On Sunday 845 people went to the Zoo. On Monday only 169 people went. What is the percent decrease in the people visiting the Zoo on Monday?
Solution:
On Sunday No. of people visited zoo on Sunday = 845
On Monday No. of people visited zoo on Monday = 169
Decrease in No. of people visited zoo on Monday = 845 – 169 = 676
% Decrease on Monday = \(\frac{\text { Decrease }}{\text { Visitor on Sunday }} \times 100\)
= \(\frac{676}{845} \times 100\)
= 80%

KSEEB Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.2

Question 3.
A shopkeeper buys 80 articles for Rs. 2,400 and sells them for a profit of 16%. Find the selling price of one article.
Solution:
C.P. of 80 articles = Rs. 2400
Profit % = 16%
Amount of profit = 16% of 2400
= \(\frac{16}{100} \times 2400\)
= Rs. 384
S.P. = C.P. + profit
= 2400 + 384
= Rs. 2784

Question 4.
The cost of an article was Rs. 15,500, Rs. 450 were spent on its repairs. If it is sold for a profit of 15%, find the selling price of the article.
Solution:
C.P. of an article = Rs. 15,500
Repair = Rs. 450
Total C.P. = 15,500 + 450 = Rs. 15,950
Profit % = 15%
Amount of profit = 15% of total C.P.
= \(\frac{15}{100} \times 15950\)
= Rs. 2392.50
S.P. = C.P. + profit
= 15950 + 2392.50
= Rs. 18342.50

KSEEB Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.2

Question 5.
A VCR and T.V. were bought for Rs. 8,000 each. The shopkeeper made loss of 4% on the VCR and a profit of 8% on the T.V. Find the gain or loss percent on the whole transaction.
Solution:
C.P. of V.C.R. = Rs. 8,000
C.P. of T.V. = Rs. 8,000
Loss on V.C.R. = 4%
Profit on T.V. = 8%
Amount of loss on V.C.R. = 4% of C.P.
= \(\frac{4}{100} \times 8000\)
= Rs. 320
S.P. = C.P. – loss
S.P. of V.C.R. = Rs. 8000 – 320 = Rs. 7680
Amount of profit on T.V. = 8% of C.P.
= \(\frac{8}{100} \times 8000\)
= Rs. 640
S.P. of T.V. = 8000 + 640 = Rs. 8640
∴ S.P. of V.C.R. and T.V. = 7680 + 8640 = Rs. 16320
∴ C.P. of V.C.R. and T.V.= 16000
Profit = S.P. – C.P.
= 16320 – 16000
= Rs. 320
% of profit = \(\frac{\text { Total Profit }}{\text { Total C.P. }} \times 100\)
= \(\frac{320}{16000} \times 100\)
= 2%

Question 6.
During a sale, a shop offered a discount of 10% on the marked prices of all the items. What would a customer have to pay for a pair of jeans marked at Rs. 1450 and two shirts marked at Rs. 850 each?
Solution:
(a) Marked price of pair of jeans = Rs. 1450
Marked price of two shirts = 2 × 850 = Rs. 1700
Total price M.P. = 1450 + 1700 = Rs. 3150
Discount % = 10%
Amount of discount = 10% of total M.P.
= \(\frac{8}{100} \times 3150\)
= Rs. 315
Price after discount = M.P. – Discount
= 3150 – 315
= Rs. 2835

KSEEB Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.2

Question 7.
A milkman sold two of his buffaloes for Rs. 20,000 each. On one he made a gain of 5% and on the other a loss of 10%. Find his overall gain or loss. (Hint: Find CP of each)
Solution:
S.P. of each buffaloes = Rs. 20,000
Loss on one = 10%
Profit on other = 5%
Let C.P. of one buffaloe = Rs. x
Loss = 10%
Amount of loss = \(\frac{10}{100} x\) = Rs. \(\frac{x}{10}\)
S.P. = C.P. – Loss
= x – \(\frac{x}{10}\)
= \(\frac{10 x-x}{10}\)
= \(\frac{9 x}{10}\)
\(\frac{9 x}{10}\) = 20000
x = \(\frac{2000}{9} \times 10\)
= Rs. 22222.22 (Approx.)
Amount of profit on other buffaloe = \(\frac{5}{100} \times x\) = Rs. \(\frac{x}{20}\)
S.P. = C.P. + profit
x + 5% of x S.P. = x + \(\frac{x}{20}\)
= x + \(\frac{x}{20}\)
= \(\frac{20 x+x}{20}\)
= \(\frac{21 x}{20}\)
Now, \(\frac{21 x}{20}\) = 20000
x = \(\frac{20000 \times 20}{21}\) = 19047.62
Total C.P. = 19047.62 + 22222.22 = Rs. 41269.84
S.P. = 20,000 + 20,000 = Rs. 40,000
Loss = C.P. – S.P. = 41269.84 – 40000 = Rs. 1269.84

Question 8.
The price of a TV is Rs. 13,000. The sales tax charged on it is at the rate of 12%. Find the amount that Vinod will have to pay if he buys it.
Solution:
Marked price = 13,000
Sales Tax = 12%
Amount of S.T. = 12% of 13,000
= \(\frac{12}{100} \times 13,000\)
= Rs. 1,560
Amount paid = 13000 + 1560 = Rs. 14,560

KSEEB Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.2

Question 9.
Arun bought a pair of skates at a sale where the discount given was 20%. If the amount he pays is Rs. 1,600, find the marked price.
Solution:
S.P. = M.P. – Discount
Let the marked price = Rs. x
Discount % = 20%
Amount of discount = 20% of x
= \(\frac{20}{100} \times x\)
S.P. = Rs. 1600
1600 = x – \(\frac{20 x}{100}\)
1600 = \(\frac{100 x-20 x}{100}\) = \(\frac{80 x}{100}\)
x = \(\frac{1600 \times 100}{80}\) = 2000
Hence marked price = Rs. 2,000

Question 10.
I purchased a hair-dryer for Rs. 5,400 including 8% VAT. Find the price before VAT was added.
Solution:
S.P. including VAT = Rs. 5400
VAT = 8%
Let marked price = Rs. x
Amount of VAT = \(\frac{8 x}{100}\)
S.P. = M.P. + VAT
= \(\frac{100 x+8 x}{100}\)
5400 = x + \(\frac{8 x}{100}\)
5400 = \(\frac{108 x}{100}\)
\(\frac{5400 \times 100}{108}\) = x
5000 = x
∴ M.P. = Rs. 5000

KSEEB Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.2

Question 11.
An article was purchased for ₹ 1239 including GST of 18%. Find the price of the article before GST was added?
Solution:
Price of Article = ₹ 1239
GST = 18%
= 18% of 1239
= \(\frac {18}{100}\) × 1239
= Rs. 223
Price of Article before GST was added = 1239 – 223 = Rs. 1016