KSEEB Solutions for Class 8 Maths Chapter 8 Comparing Quantities InText Questions

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KSEEB Solutions for Class 8 Maths Chapter 8 Comparing Quantities InText Questions

Try These (Page 119)

Question 1.
In a primary school, the parents were asked about the number of hours they spend per day helping their children to do homework. There were 90 parents who helped for \(\frac{1}{2}\) hour to 1\(\frac{1}{2}\) hours. The distribution of parents according to the time for which, they said they helped is given in the adjoining figure; 20% helped for more than 1\(\frac{1}{2}\) hours per day; 30% helped for \(\frac{1}{2}\) hour to 1\(\frac{1}{2}\) hours; 50% did not help at all.
KSEEB Solutions for Class 8 Maths Chapter 8 Comparing Quantities InText Questions Q1
Using this, answer the following:
(i) How many parents were surveyed?
(ii) How many said that they did not help?
(iii) How many said that they helped for more than 1\(\frac{1}{2}\) hours?
Solution:
(i) Let the total number of parents = x
Parents who helped \(\frac{1}{2}\) hour to 1\(\frac{1}{2}\) hours = 90
The percentage of these parents = 30%
∴ 30% of x = 90
⇒ \(\frac{30}{100} \times x\) = 90
⇒ x = \(\frac{90 \times 100}{30}\)
∴ x = 300
∴ Number of parents = 300

(ii) Percentage of parents who did not help = 50%
Number of parents who did not help = 50% of 300
= \(\frac{50}{100}\) × 300
= 150

KSEEB Solutions for Class 8 Maths Chapter 8 Comparing Quantities InText Questions

(iii) Percentage of parents who helped more than 1\(\frac{1}{2}\) hours = 20%
Number of such parents = 20% of 300
= \(\frac{20}{100}\) × 300
= 60

Try These (Page 121)

Question 1.
A shop gives a 20% discount. What would the sale price of each of these be?
(a) A dress marked at Rs. 120
(b) A pair of shoes marked at Rs. 750
(c) A bag marked at Rs. 250
Solution:
(a) Marked price of a dress = Rs. 120
Discount = 20%
Discount = 20% of 120
= \(\frac{20}{100} \times 120\)
= Rs. 24
S.P. = M.P – Discount
= 120 – 24
= Rs. 96

(b) Marked price of a pair of shoes = Rs. 750
Discount = 20% of 750
= \(\frac{20}{100} \times 750\)
= Rs. 150
∴ S.P. = M.P – Discount
= 750 – 150
= Rs. 600

(c) Marked price of a bag = Rs. 250
Discount = 20% of 250
= \(\frac{20}{100} \times 250\)
= Rs. 50
∴ S.P. = M.P – Discount
= 250 – 50
= Rs. 200

KSEEB Solutions for Class 8 Maths Chapter 8 Comparing Quantities InText Questions

Question 2.
A table marked at Rs. 15,000 is available for Rs. 14,400. Find the discount given and the discount percent.
Solution:
Marked price of a table = Rs. 15,000
Sale price of a table = Rs. 14,400
Discount = M.P.- S.P
= 15,000 – 14,400
= Rs. 600
Discount % = \(\frac{\text { Discount }}{\text { M.P }} \times 100\)
= \(\frac{600}{15000} \times 100\)
= 4%
∴ Discount = Rs. 600 and Discount percent = 4

Question 3.
An almirah is sold at Rs. 5,225 after allowing a discount of 5%. Find its marked price.
Solution:
Let Marked price = Rs. x
Sale price of a table = Rs. 5225
Discount = 5%
S.P. = M.P. – Discount
⇒ 5225 = x – 5% of x
⇒ 5225 = x – \(\frac{5}{100} \times x\)
⇒ 5225 = \(\frac{100 x-5 x}{100}\)
⇒ 5225 × 100 = 95x
⇒ x = \(\frac{5225 \times 100}{95}\)
⇒ x = 5500
∴ M.P. = Rs. 5500

Try These (Page 123)

Question 1.
Find selling price (SP) if a profit of 5% is made on
(a) a cycle of Rs. 700 with Rs. 50 as overhead charges.
(b) a lawnmower bought at Rs. 1150 with Rs. 50 as transportation charges.
(c) a fan bought for Rs. 560 and expenses of Rs. 40 made on its repairs.
Solution:
(a) Cost price of a cycle = Rs. 700
Overhead charges = Rs. 50
Total cost price = 700 + 50 = Rs. 750
Profit percent = 5%
Profit = 5% of total C.P.
= \(\frac{5}{100} \times 750\)
= Rs. 37.50
∴ S.P. = C.P. + Profit
= 750 + 37.50
= Rs. 787.50

KSEEB Solutions for Class 8 Maths Chapter 8 Comparing Quantities InText Questions

(b) C.P. of lawn mower = Rs. 1150
Transportation charges = Rs. 50
Total C.P. = 1150 + 50 = Rs. 1200
Profit percent = 5%
Profit = 5% of total C.P.
= \(\frac{5}{100} \times 1200\)
= Rs. 60
∴ S.P. = C.P. + Profit
S.P. = 1200 + 60 = Rs. 1260

(c) C.P. of a fan = Rs. 560
Repair charges = Rs. 40
Total C.P. = 560 + 40 = Rs. 600
Profit % = 5%
∴ Profit = 5% of total C.P.
= \(\frac{5}{100} \times 600\)
= Rs. 30
∴ S.P. = C.P. + Profit
= 600 + 30
= Rs. 630

Try These (Page 123)

Question 1.
A shopkeeper bought two TV sets at Rs. 10,000 each. He sold one at a profit of 10% and the other at a loss of 10%. Find whether he made an overall profit or loss.
Solution:
C.P. of T.V. set = Rs. 10,000
Profit = 10%
∴ Profit = 10% of 10,000
= \(\frac{10}{100} \times 10,000\)
= Rs. 1000
S.P. = C.P. + Profit
= 10,000 + 1,000
= Rs. 11,000
Loss % on another T.V. set = 10%
Loss = 10% of C.P.
= \(\frac{10}{100} \times 10000\)
= Rs. 1000
∴ S.P. = C.P. – Loss
= 10,000 – 1,000
= Rs. 9,000
Total C.P. of both T.V. sets
= 10,000 + 10,000
= Rs. 20,000
Total S.P. of both T.V. sets = 11,000 + 9000
= Rs. 20,000
Hence C.P. = S.P.
∴ No profit or loss.

KSEEB Solutions for Class 8 Maths Chapter 8 Comparing Quantities InText Questions

Try These (Pages 124 – 125)

Question 1.
Find the buying price of each of the following when 5% S.T. is added on the purchase of
(a) A towel at Rs. 50
(b) Two bars of soap at Rs. 35 each
(c) 5 kg of flour at Rs. 15 per kg.
Solution:
(a) Price of a towel = Rs. 50
S.T. = 5%
Amount of S.T. = 5% of price of towel
= \(\frac{5}{100} \times 50\)
= Rs. 2.50
∴ Buying price = Price + S.T.
= 50 + 2.50
= Rs. 52.50

(b) Price of two bars of soap = 35 × 2 = Rs. 70
S.T. = 5%
∴ Amount of S.T. = 5% of price of soap
= \(\frac{5}{100} \times 70\)
= Rs. 3.50
∴ Total price = Price + S.T.
= 70 + 3.50
= Rs. 73.50

(c) Price of 1 kg of flour = Rs. 15
Price of 5 kg of flour = 15 × 5 = Rs. 75
S.T. = 5%
∴ Amount of S.T. = 5% of price of flour
= \(\frac{5}{100} \times 75\)
= \(\frac{15}{4}\)
∴ Total price of flour = Price + S.T.
= 75 + 3.75
= Rs. 78.75

KSEEB Solutions for Class 8 Maths Chapter 8 Comparing Quantities InText Questions

Question 2.
If 8% VAT is included in the price, find the original price of
(a) A TV bought for Rs. 13,500
(b) A shampoo bottle bought for Rs. 180.
Solution:
(a) Let marked price = Rs. 100
V.A.T. = 8%
Amount of V.A.T. = 8% of 100
= \(\frac{8}{100} \times 100\)
= Rs. 8
Total price = 100 + 8 = Rs.108
∴ Price of T.V. included V.A.T. = 13,500
Original price = \(\frac{100}{108} \times 13500\) = Rs. 12,500

(b) Let marked price = Rs. 100
V.A.T. = 8%
Amount of V.A.T. = 8% of 100
= \(\frac{8}{100} \times 100\)
= Rs. 8
Total price = 100 + 8 = Rs. 108
∴ Price of Shampoo = Rs. 180
Original price = \(\frac{100}{108} \times 180\) = Rs. 166.67

Try These (Page 126)

Question 1.
Find the interest and amount to be paid on Rs. 15000 at 5% per annum after 2 years.
Solution:
P = Rs. 15,000
R = 5% (p.a)
T = 2 years
S.I. = \(\frac{\mathbf{P} \times \mathbf{R} \times \mathbf{T}}{100}\)
= \(\frac{15000 \times 5 \times 2}{100}\)
= Rs. 1500
Amount = S.I. + P
= 15000 + 1500
= Rs. 16,500

KSEEB Solutions for Class 8 Maths Chapter 8 Comparing Quantities InText Questions

Try These (Page 129)

Question 2.
Find Cl on a sum of Rs. 8000 for 2 years at 5% per annum compounded annually.
Solution:
P = Rs. 8000
Time = 2 years
Rate = 5%
A = \(\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{T}}\)
= \(8000\left(1+\frac{5}{100}\right)^{2}\)
= \(8000\left(1+\frac{1}{20}\right)^{2}\)
= \(8000 \times\left(\frac{21}{20}\right)^{2}\)
= \(8000 \times \frac{21}{20} \times \frac{21}{20}\)
= 20 × 21 × 21
= 20 × 441
= Rs. 8820
C.I. = A – P
= 8820 – 8000
= Rs. 820

Try These (Page 130)

Question 1.
Find the time period and rate for each.
1. A sum taken for 1\(\frac{1}{2}\) years at 8% per annum is compounded half-yearly.
2. A sum taken for 2 years at 4% per annum compounded half-yearly.
Solution:
1. Time = \(\frac{3}{2}\) years
= \(\frac{3}{2}\) × 2
= 3 (half years)
Rate = 8 (p.a) = \(\frac{8}{2}\) = 4 (half yearly)

KSEEB Solutions for Class 8 Maths Chapter 8 Comparing Quantities InText Questions

2. Time = 2 years
= 2 × 2
= 4 (half years)
Rate = 4% (p.a)
= \(\frac{4}{2}\)
= 2% (half yearly)

Try These (Page 131)

Question 1.
Find the amount to be paid
1. At the end of 2 years on Rs. 2,400 at 5% per annum compounded annually.
2. At the end of 1 year on Rs. 1,800 at 8% per annum compounded quarterly.
Solution:
1. P = Rs. 2400
R = 5% (p.a)
T = 2
A = \(P\left(1+\frac{R}{100}\right)^{T}\)
= \(2400\left(1+\frac{5}{100}\right)^{2}\)
= \(2400 \times \frac{21}{20} \times \frac{21}{20}\)
= 6 × 21 × 21
= 6 × 441
= Rs. 2646

2. P = Rs. 1800
R = 8% (p.a), or (\(\frac{8}{4}\) = 2 % quarterly)
Time = 1 year = 4 quarters
A = \(P\left(1+\frac{R}{100}\right)^{T}\)
= \(1800\left(1+\frac{2}{100}\right)^{4}\)
= \(1800 \times \frac{51}{50} \times \frac{51}{50} \times \frac{51}{50} \times \frac{51}{50}\)
= \(\frac{121773618}{62500}\)
= Rs. 1948.38

KSEEB Solutions for Class 8 Maths Chapter 8 Comparing Quantities InText Questions

Try These (Page 133)

Question 1.
Machinery worth Rs. 10,500 depreciated by 5%. Find its value after one year.
Solution:
P = Rs. 10,500
R = 5% (p.a)
Time = 1 year
Machineis depreciated
A = \(P\left(1-\frac{R}{100}\right)^{T}\)
= \(10,500\left(1-\frac{5}{100}\right)^{1}\)
= \(10,500 \times \frac{95}{100}\)
= 105 × 95
= Rs. 9975

KSEEB Solutions for Class 8 Maths Chapter 8 Comparing Quantities InText Questions

Question 2.
Find the population of a city after 2 years, which is at present 12 lakh, if the rate of increase is 4%.
Solution:
Present population = 12 lakh
Time = 2 years
Rate of increase in population = 4% (p.a.)
Population after 2 years = 12,00,000 \(\left(1+\frac{4}{100}\right)^{2}\)
= 12,00,000 × \(\frac{26}{25} \times \frac{26}{25}\)
= 1297920