CBSE Class 8

These NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.2

Question 1.
Find the product of the following pairs of monomials
(i) 4, 7p
(ii) -4p, 7p
(iii) -4p, 7pq
(iv) 4p³, – 3p
(v) 4p, 0
Solution:
(i) 4 × 7p
= (4 × 7)p
= 28p

(ii) -4p × 7p
= {(-4) × 7} × (p × p)
= (-28) × p²
= -28p²

(iii) -4p × 7pq
= {(-4) × 7} × {p × (pq)}
= -28 × (p × p × q)
= -28p²q

(iv) 4p³ × -3p
= {4 × (-3)} × (p³ × p)
= -12 × (p⁴)
= -12p⁴

(v) 4p × 0
= (4 × 0) × p
= 0 × p
= 0

Question 2.
Find the areas of rectangles, with the following pairs of monomials as their lengths and breadths respectively.
(p, q); (10m, 5n); (20x², 5y²); (4x, 3x²); (3mn, 4np)
Solution:
(i) (p, q)
Area of the rectangle = length × breadth
= p x q
= pq

(ii) (10m, 5n)
Area of the rectangle = length × breadth
= 10m × 5n
= (10 × 5) × (m × n)
= 50 × mn
= 50mn

(iii) 20x², 5y²
Area of the rectangle = length × breadth
= 20x² × 5y²
= (20 × 5) × (x² × y²)
= 100 × x²y²
= 100x²y²

(iv) (4x, 3x²)
Area of the rectangle = length × breadth
= 4x × 3x²
= (4 × 3) × (x × x²)
= 12 × x³
= 12x³

(v) (3mn, 4np)
Area of the rectangle = length × breadth
= 3mn × 4np
= (3 × 4) × (mn × np)
= 12 × m × (n × n) × p
= 12 × m × n² × p
= 12mn²p

Question 3.
Complete the table products.

Solution:

Question 4.
Obtain the volume of rectangular boxes with the following length, breadth and height respectively.
(i) 5a, 3a², 7a4
(ii) 2p, 4q, 8r
(iii) xy, 2x²y, 2xy²
(iv) a, 2b, 3c
Solution:
(i) 5a, 3a², 7a⁴
Volume of the rectangular box = length × breadth × height
= (5a) × (3a²) × (7a⁴)
= (5 × 3 × 7) × (a × a² × a⁴)
= 105a⁷

(ii) 2p, 4q, 8r
Volume of the rectangular box = Length × Breadth × Height
= 2p × 4q × 8r
= (2 × 4 × 8) × (p × q × r)
= 64pqr

(iii) xy; 2x²y; 2xy²
Volume of the rectangular box = length × breadth × height
= xy × 2x²y × 2xy²
= (1 × 2 × 2) × (x × x² × x) × (y × y × y²)
= 4 × x⁴ × y⁴
= 4x⁴y⁴

(iv) a, 2b, 3c
Volume of the rectangular box = length × breadth × height
= a × 2b × 3c
= (1 × 2 × 3) × (a × b × c)
= 6abc

Question 5.
Obtain the product of
(i) xy, yz, zx
(ii) a, -a², a³
(iii) 2, 4y, 8y², 16y³
(iv) a, 2b, 3c, 6abc
(v) m, -mn, mnp
Solution:
(i) xy, yz, zx
(xy) × (yz) × (zx)
= (x × x) × (y × y) × (z × z)
= x² × y² × z²
= x²y²z²

(ii) a; -a²; a³
(a) × (-a²) × (a³)
= -(a × a² × a³)
= -a⁶

(iii) 2, 4y, 8y², 16y³
(2) × (4y) × (8y²) × (16y³)
= (2 × 4 × 8 × 16) × (y × y² × y³)
= 1024y⁶

(iv) a, 2b, 3c, 6abc
(a) × (2b) × (3c) × (6abc)
= (2 × 3 × 6) × (a × a) × (b × b) × (c × c)
= 36 × a² × b² × c²
= 36a²b²c²

(v) m, -mn, mnp
(m) × (-mn) × (mnp)
= -1 × (m × m × m) × (n × n) × p
= -1 × m³ × n² × p
= -m³n²p

These NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.1

Question 1.
Identify the terms, their coefficients for each of the following expressions.
(i) 5xyz² – 3zy
(ii) 1 + x + x²
(iii) 4x²y² – 4x²y²z² + z²
(iv) 3 – pq + qr – rp
(v) \(\frac{x}{2}+\frac{y}{2}\) – xy
(vi) 0.3a – 0.6ab + 0.5b
Solution:
(i) 5xyz² – 3zy
Terms are 5xyz² and -3zy
Coefficients of 5xyz² is 5
Coefficients of-3zy is -3

(ii) 1 + x + x²
Terms are x², x, and 1
Coefficient of x² is 1
coefficient of x is 1
and constant term is 1

(iii) 4x²y² – 4x²y²z² + z²
Terms are 4x²y², -4x²y²z² and z²
Coefficient of x²y² is 4
Coefficient of -4x²y²z² is -4
Coefficient of z² is 1

(iv) 3 – pq + qr – rp
Terms are 3, -pq, qr and -rp
Coefficient of 3 is 3.
Coefficient of -pq is -1.
Coefficient of qr is 1 and coefficient of -rp is -1

(v) \(\frac{x}{2}+\frac{y}{2}\) – xy
Terms are \(\frac{x}{2}, \frac{y}{2}\) and -xy
Coefficient of \(\frac{\mathrm{x}}{2}\) is \(\frac{1}{2}\)
Coefficient of \(\frac{\mathrm{x}}{2}\) is \(\frac{1}{2}\)
and coefficient of -xy is -1

(vi) 0.3a – 0.6ab + 0.5b
Terms are 0.3a, -0.6ab and 0.5b
Coefficient of 0.3a is 0.3
Coefficient of -0.6ab is -0.6
Coefficient of 0.5b is 0.5

Question 2.
Classify the following polynomial as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?
x + y, 1000, x + x² + x³ + x⁴, 7 + y + 5x, 2y – 3y², 2y – 3y² + 4y³, 5x – 4y + 3xy, 4z – 15z², ab + bc + cd + da, pqr, p²q + pq², 2p + 2q
Solution:
Monomials: 100pqr
Binomials: x + y; 2y – 3y²; 4z – 15z²; p²q + pq²; 2p + 2q
Trinomials: 7 + y + 5x; 2y – 3y² + 4y³; 5x – 4y + 3xy
Polynomials that do not fit in these categories: x + x² + x³ + x⁴ and ab + bc + cd + da
(Since the above polynomials has four terms)

Question 3.
Add the following.
(i) ab – bc, bc – ca, ca – ab
(ii) a – b + ab, b – c + bc, c – a + ac
(iii) 2p²q² – 3pq + 4, 5 + 7pq – 3p²q²
(iv) l² + m², m² + n², n² + l², 2lm + 2mn + 2nl
Solution:
(i) ab – bc; bc – ca; ca – ab

(ii) a – b + ab; b – c + bc; c – a + ac

(iii) 2p²q² – 3pq + 4; 5 + 7pq – 3p²q²

(iv) l² + m²; m² + n²; n² + l²; 2lm + 2mn + 2nl

= 2(l² + m² + n² + lm + mn + nl)

Question 4.
(a) Subtract 4a – 7ab + 3b + 12 from 12a – 9ab + 5b – 3
(b) Subtract 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + 10xyz
(c) Subtract 4p²q – 3pq + 5pq² – 8p + 7q – 10 from 18 – 3p – 11q + 5pq – 2pq² + 5p²q
Solution:

These NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities InText Questions and Answers are prepared by our highly skilled subject experts.

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

Try These (Page No. 119)

Question 1.
In a primary school, the parents were asked about the number of hours they spend per day in 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 following 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.
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 man than 1\(\frac{1}{2}\) hours?

Solution:
(i) Since, 30% of total surveyed parents helped their children for “\(\frac{1}{2}\) hours to 1\(\frac{1}{2}\) hours”. And 90 parents helped their children for “\(\frac{1}{2}\) hours to 1\(\frac{1}{2}\) hours”.
30% [ surveyed parents] = 90
or \(\frac{30}{100}\) × [surveyed parents] = 90
or surveyed parents = 90 × 100/30
= 3 × 100
= 300

(ii) Since, 50% of surveyed parents did not help their children.
Number of parents who did not help = 50% of surveyed parents
= 50% of 300
= 50/100 × 300
= 150

(iii) Since, 20% of surveyed parents help their children for more than 1\(\frac{1}{2}\) hours, i.e. 20% of surveyed parents help for more than 1\(\frac{1}{2}\) hours.
Number of parents who helped for more than 1\(\frac{1}{2}\) hours = 20% of 300
= 20/100 × 300
= 20 × 3
= 60
Note: ‘of’ means multiplication

Try These (Page No. 121)

Question 2.
A shop gives a 20% discount. What would the sale price of each of these be?
(a) A dress marked at ₹ 120
(b) A pair of shoes marked at ₹ 750
(c) A bag marked at ₹ 250
Solution:
(a) Marked price of the dress = ₹ 120
Discount rate = 20%
Discount = 20% of ₹ 120
= ₹ \(\frac{20}{100}\) × 120
= ₹ 2 × 12
= ₹ 24
Sale price of the dress = [Marked Price] – [Discount]
= ₹ 120 – ₹ 24
= ₹ 96

(b) Marked price of the pair of shoes = ₹ 750
Discount rate = 20%
Discount = 20% of ₹ 750
= ₹ \(\frac{20}{100}\) × 750
= ₹ 2 × 75
= ₹ 150
Now, Sale price of the pair of shoes = [Marked Price] – [Discount]
= ₹ 750 – ₹ 150
= ₹ 600

(c) Marked price of the bag = ₹ 250
Discount rate = 20%
Discount = 20% of ₹ 250
= ₹ \(\frac{20}{100}\) × 250
= ₹ 2 × 25
= ₹ 50
Sale price of the bag = [Marked price] – [ Discount]
= ₹ 250 – ₹ 50
= ₹ 200

Question 3.
A table marked at ₹ 15,000 is available for ₹ 14,000. Find the discount given and the discount percent.
Solution:
The marked price of the table = ₹ 15000
Sale price of the table = ₹ 14400
Discount = [Marked Price] – [Sale price]
= ₹ 15000 – ₹ 14400
= ₹ 600
Discount percent = Discount/Marked Price × 100

Question 4.
An almirah is sold at ₹ 5,225 after allowing a discount of 5%. Find its marked price.
Solution:
Sale price of the almirah = ₹ 5255
Discount rate = 5%
Since, Discount = 5% of marked price
Marked Price – Discount = Sale price
or [Marked Price] – \(\frac{5}{100}\) × [Marked Price] = Sale price
or [Marked Price] × \(\frac{95}{100}\) = ₹ 5255
or [Marked Price] = ₹ 5255 × \(\frac{95}{100}\) = ₹ 5500

Try These (Page No. 123)

Question 5.
Find selling price (SP) if a profit of 5% is made on
(a) a cycle of ₹ 700 with ₹ 50 as overhead charges.
(b) a lawnmower bought at ₹ 1150 with ₹ 50 as transportation charges.
(c) a fan bought for ₹ 560 and expenses of ₹ 40 made on its repairs.
Solution:
(a) Total cost price = ₹ 700 + ₹ 50 (overhead expenses) = ₹ 750
profit = 5% of ₹ 750
= ₹ \(\frac{5}{100}\) × ₹ 780
= ₹ 5 × \(\frac{15}{100}\)
= ₹ \(\frac{75}{100}\)
= ₹ 37.50
Now, SP = CP + profit
= ₹ 750 + ₹ 37.50
= ₹ 787.50

(b) Total cost price = ₹ 1150 + ₹ 50 (Overhead expenses) = ₹ 1200
Profit = 5% of 1200
= \(\frac{5}{100}\) × ₹ 1200
= ₹ 5 × 12
= ₹ 60
Now, SP = CP + Profit
= ₹ 1200 + ₹ 60
= ₹ 1260

(c) Total cost price = ₹ 560 + ₹ 40 (Overhead expenses) = ₹ 600
Profit = 5% of ₹ 600 = \(\frac{5}{100}\) × ₹ 600
= ₹ 5 × 6
= ₹ 30
Now, SP = CP + profit
= ₹ 600 + ₹ 30
= ₹ 630

Try These (Page No. 123)

Question 6.
A shopkeeper bought two TV sets at ₹ 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:
(a) Marked price of the dress = ₹ 120
Discount rate = 20%
Discount = 20% of ₹ 120
= ₹ \(\frac{20}{100}\) × 120
= ₹ 2 × 12
= ₹ 24
Sale price of the dress = [Marked Price] – [Discount]
= ₹ 120 – ₹ 24
= ₹ 96

(b) Marked price of the pair of shoes = ₹ 750
Discount rate = 20%
Discount = 20% of ₹ 750
= ₹ \(\frac{20}{100}\) × 750
= ₹ 2 × 75
= ₹ 150
Now, Sale price of the pair of shoes = [Marked Price] – [Discount]
= ₹ 750 – ₹ 150
= ₹ 600

(c) Marked price of the bag = ₹ 250
Discount rate = 20%
Discount = 20% of ₹ 250
= ₹ \(\frac{20}{100}\) × 250
= ₹ 2 × 25
= ₹ 50
Sale price of the bag = [Marked price] – [Discount]
= ₹ 250 – ₹ 50
= ₹ 200

Try These (Page No. 126)

Question 7.
Find the interest and amount to be paid on ₹ 15000 at 5% per annum after 2 years.
Solution:
Here P = ₹ 15000, R% = 5%, T = 2 years
Simple Interest = \(\frac{\text { PRT }}{100}\)
= \(\frac{15000 \times 5 \times 2}{100}\)
= ₹ 1500
Amount = ₹ 15000 + ₹ 1500 = ₹ 16,500

Try These (Page No. 129)

Question 8.
Find Cl on a sum of ₹ 8000 for 2 years at 5% per annum compounded annually.
Solution:
We have
P = ₹ 8000, R = 55 p.a., T = 2 years
∴ \(A=P\left[1+\frac{R}{100}\right]^{n}\)
= ₹ 8000 \(\left[1+\frac{5}{100}\right]^{2}\)
= ₹ \(8000\left[\frac{21}{20}\right]^{2}\)
= ₹ \(8000 \times \frac{21}{20} \times \frac{21}{20}\)
= ₹ (20 × 21 × 21)
= ₹ 8820
Now, compound interest = A – P
= ₹ 8820 – ₹ 8000
= ₹ 820

Try These (Page No. 130)

Question 9.
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. We have interest rate 8% per annum 1\(\frac{1}{2}\) year.
It is compounded half-yearly.
Time period (n) = 2(1\(\frac{1}{2}\)) = 3 half yearly
Rate (R) = \(\frac{1}{2}\) (8%) = 4% per half yearly.

2. We have an interest rate of 4% per annum for the year
Time period (n) = 2(2) = 4 half-yearly
Rate (R) = \(\frac{1}{2}\) (4%) = 2% per half yearly.

Try These (Page No. 131)

Question 10.
Find the amount to be paid
1. At the end of 2 years on ₹ 2,400 at 5% per annum compounded annually.
2. At the end of 1 year on ₹ 1,800 at 8% per annum compounded quarterly.
Solution:
1. We have:
P = ₹ 2400, R = 5% p.a., T = 2 years
∴ Interest is compounded annually i.e, n = 2

2. Here, interest compounded quarterly.
R = 8%, p.a. = \(\frac{8}{4}\), i.e., 2% per quarter
T = 1 year = 4 × 1, i.e., 4 quarters on
n = 4

Try These (Page No. 133)

Question 11.
A machinery worth ₹ 10,500 depreciated by 5%. Find its value after one year?
Solution:
Here, P = 10, 500, R = 5% p.a.
T = 1 year, n = 1
∴ A = \(\mathrm{P}\left[1+\frac{5}{100}\right]^{1}\)
[∴ Depreciation is there, r = -5%]
= ₹ \(10500\left[1+\frac{5}{100}\right]^{1}\)
= ₹ 10500 × \(\frac{19}{20}\)
= ₹ 525 × 19
= ₹ 9975
Thus, machinery value after 1 year = ₹ 9975
Note: For depreciation, we use the formula as A = \(\mathrm{P}\left[1-\frac{\mathrm{R}}{100}\right]^{\mathrm{n}}\)

Question 12.
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, P = 12 lakh
Rate of increase, R = 4% p.a.
Time, T = 2 years
n = 2
Population after 2 years = \(\mathrm{P}\left[1+\frac{\mathrm{R}}{100}\right]^{\mathrm{n}}\)
= 12 lakh \(\left[1+\frac{4}{100}\right]^{2}\)
= 12 lakh \(\left[\frac{26}{25}\right]^{2}\)
= 12,00, 000 × \(\frac{26}{25} \times \frac{26}{25}\)
= 1920 × 26 × 26
= 12,97,920
Thus, the population of the town will be 12,97,920 after 2 years.

These NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Exercise 8.3

Question 1.
Calculate the amount and compound interest on
(a) ₹ 10,800 for 3 years at 12\(\frac{1}{2}\)% per annum compounded annually.
(b) ₹ 18,000 for 2\(\frac{1}{2}\) years at 10% per annum compounded annually.
(c) ₹ 62,500 for 1\(\frac{1}{2}\) years at 8% per annum compounded half yearly.
(d) ₹ 8,000 for 1 year at 9% per annum compounded half yearly.
(You could use the year by year calculation using SI formula to verify)
(e) ₹ 10,000 for 1 year at 8% per annum compounded half-yearly.
Solution:
(a) Here P = ₹ 10800, n = 3 years

Compound interest = ₹ 15377.34 – ₹ 10,800 = ₹ 4577.34
Amount = ₹ 15377.34
Compound interest = ₹ 4577.34

(b) HereP = ₹ 18,000, n = 2\(\frac{1}{2}\), R = 10% p.a
For first 2 years

CI = A – P = 21,780 – 18000 = ₹ 3780
For the last \(\frac{1}{2}\) year, principal become ₹ 21,780
So, Interest = \(\frac{21780 \times 10 \times 1}{2 \times 100}\) = ₹ 1089
∴ Total interest = 3780 + 1089 = ₹ 4869

(c) Here P = ₹ 62,500, R = \(\frac{8}{2}\) = 4% per half year
n = \(\frac{3}{2}\) × 2 = 3 years

A = ₹ 4 × 26 × 26 × 26 = ₹ 70304
Compound interest = Amount – Principal
= ₹ 70304 – ₹ 62500
= ₹ 7804

(d) Here P = ₹ 8000, n = 2 (2 half year),
R = \(\frac{9}{2}\) % (half yearly)

Compound interest = ₹ 8736.2 – ₹ 8000 = ₹ 736.20

(e) Here P = ₹ 10,000; n = 2 (2 half year);
R = \(\frac{8}{2}\) % = 4% (half yearly)

Compound Interest = ₹ 10816 – ₹ 10,000 = ₹ 816

Question 2.
Kamala borrowed ₹ 26,400 from a Bank to buy a scooter at a rate 15% p.a. compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan?
(Hint: Find A for 2 years with interest is compounded yearly and then find SI on the 2nd year amount for \(\frac{4}{12}\) years).
Solution:
Here P = ₹ 26,400, R% = 15% p.a, n = 2 years

Simple interest on ₹ 34914 at 15% p.a for 4 months (\(\frac{1}{3}\) years)
= ₹ \(\frac{34914 \times 15 \times 1}{100 \times 3}\)
= ₹ 1745.70
Requiredamount = ₹ 34914 + ₹ 1745.70 = ₹ 36,659.70

Question 3.
Fabina borrows ₹ 12,500 at 12% per annum for 3 years at simple interest and Radha borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much?
Solution:
For Fabina,
P = ₹ 12,500, R% = 12%, T = 3 years
Simple interest = \(\frac{\text { PRT }}{100}\)
= ₹ \(\frac{12500 \times 12 \times 3}{100}\)
= ₹ 4500
For Radha,
P = ₹ 12500, R = 10% p.a. n = 3 years

Compound interest = Amount – Principal
= ₹ 16,637.50 – ₹ 12500
= ₹ 4137.50
Difference between C.I and S.I = ₹ 4500 – ₹ 4137.50 = ₹ 362.5
Fabina pays more by ₹ 362.50

Question 4.
I borrowed ₹ 12,000 from Jamshed at 6% per annum simple interest for 2 years. Had I borrowed this sum at 6% per annum compound interest, what extra amount would I have to pay?
Solution:
Simple interest
Here P = ₹ 12000, R% = 6%, T = 2 years
I = \(\frac{\text { PRT }}{100}\)
= ₹ \(\frac{12000 \times 6 \times 2}{100}\)
= ₹ 1440
Compound interest
Here P = ₹ 12000, R = 6%, n = 2 years

C.I. = ₹ 13,483.20 – ₹ 12000 = ₹ 1483.20
∴ Excess amount = ₹ 1483.20 – ₹ 1440 = ₹ 43.20
I would have to pay to him an excess amount of ₹ 43.20.

Question 5.
Vasudevan invested ₹ 60,000 at an interest rate of 12% per annum compounded half yearly. What amount would he get
(i) after 6 months?
(ii) after 1 year?
Solution:
(i) After 6 months
Here, P = ₹ 60,000, R = 12% per annum = \(\frac {1}{2}\) × 12 = 6%, n = 1 (one half year)

He would get ₹ 63,600 after 6 months

(ii) After one year
Here P = ₹ 60,000, R = \(\frac{12}{2}\) = 6% per half year
n = (1 × 2) = 2 half years

Hence, he would get ₹ 67,416 after one year

Question 6.
Arif took a loan of ₹ 80,000 from a bank. If the rate of interest is 10% per annum, find the difference in amounts he would be paying after 1\(\frac{1}{2}\) years if the interest is
(i) compounded annually
(ii) compounded half-yearly
Solution:
(i) Compounded annually
Here P = ₹ 80,000, R = 10% p.a, n = 1 year

Total interest = ₹ 8000 + ₹ 4400 = ₹ 12400
∴ Required amount = ₹ 80000 + ₹ 12400 = ₹ 92400

(ii) Compound half yearly
Here P = ₹ 80,000
R = \(\frac{10}{2}\)% = 5% per half year
n = 1\(\frac{1}{2}\) × 2 years = 3 half years

= \(\frac{80000 \times 105 \times 105 \times 105}{100 \times 100 \times 100}\)
= ₹ 92610
∴ The required amount = ₹ 92610
Difference in amount = ₹ 92610 – ₹ 92400 = ₹ 210
Hence, difference in amount is ₹ 210.

Question 7.
Maria invested ₹ 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find
(i) The amount credited against her name at the end of the second year.
(ii) The interest for the 3rd year.
Solution:
(i) Here P = ₹ 8000, R = 5%, n = 2 years

The amount credited at the end of 2nd year = ₹ 8820

(ii) Here P = ₹ 8820, R = 5%, n = 1 year
Note: To find the interest for the 3rd year
(Find the S.I. for one year)
I = \(\frac{\text { PRT }}{100}\)
= \(\frac{8820 \times 5 \times 1}{100}\)
= ₹ 441
∴ Interest for the 3rd year is ₹ 441

Question 8.
Find the amount and the compound interest on ₹ 10,000 for 1\(\frac{1}{2}\) years at 10% per annum, compounded half-yearly. Would this interest be more than the interest he would get if it was compounded annually?
Solution:
If compounded half-yearly
Here P = ₹ 10,000
R = \(\frac{10}{2}\) = 5% per half year

C.I = A – P
= ₹ 11576.25 – ₹ 10,000
= ₹ 1576.25
₹ 1576.25 is the required C.I, when compounded half-yearly.

If compounded annually
Here, P = ₹ 10,000, R = 10%, n = 1 year

Total compound Interest = ₹ (11000 – 10,000) + 550
= ₹ 1000 + ₹ 550
= ₹ 1550
When compounded half-yearly, the compound interest = ₹ 1576.25
When compounded annually, the compound interest = ₹ 1550
Hence, the interest when compounded half-yearly would be more than the interest when compounded annually.

Question 9.
Find the amount which Ram will get on ₹ 4096, if he gave it for 18 months at 12\(\frac{1}{2}\)% per annum, interest being compounded half-yearly.
Solution:

Hence, the required amount ₹ 4913.

Question 10.
The population of a place increased to 54,000 in 2003 at a rate of 5% per annum
(i) find the population in 2001.
(ii) what would be its population in 2005?
Solution:
Let the population in 2001 be ‘x’
R = 5%, n = 2years (2003 – 2001)
Population in 2003 is 54,000

= 48979.59
= 48980 (approx).
Hence, the population in 2001 was about 48980

(ii) Here P = 54,000, R = 5% p.a., n = 2 years (2003 – 2005 = 2)

Hence, the population in 2005 would be 59,535.

Question 11.
In a Laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5,06,000.
Solution:
Here P = 5,06,000, R = 2.5% per hour, n = 2 hours

= 531,616.25
= 531616 (approx).
Number of bacteria at the end of 2 hours = 531616 (approx.)

Question 12.
A scooter was bought at ₹ 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.
Solution:
Here P = ₹ 42,000 R = 8% p.a, n = 1 year
When the value depreciated

Value of the scooter after one year = ₹ 38,640

These NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Exercise 8.2

Question 1.
A man got a 10% increase in his salary. If his new salary is ₹ 1,54,000, find his original salary.
Solution:
Let the original salary of a man be ₹ x.
Increase in salary = 10% = \(\frac{10}{100}\) × x = \(\frac{x}{10}\)
New salary = ₹ 1,54,000
x + \(\frac{x}{10}\) = 154000
\(\frac{10 x+x}{10}\) = 154000
11x = 154000 × 10
x = \(\frac{154000 \times 10}{11}\) = ₹ 1,40,000
∴ Original salary = ₹ 1,40,000

Question 2.
On Sunday 845 people went to the zoo. On Monday only 169 people went. What is the per cent decrease in the people visiting the zoo on Monday?
Solution:
Decrease Value = 845 – 169 = 676
Decrease % = \(\frac{\text { Decrease Value }}{\text { Original Value }} \times 100\)
= \(\frac{676}{845}\) × 100 = 80%
Decrease % = 80%

Question 3.
A shopkeeper buys 80 articles for ₹ 2400 and sells them for a profit of 16%. Find the selling price of one article.
Solution:
Cost price of 80 articles = ₹ 2400
Profit = \(\frac{16}{100}\) × 2400 = ₹ 384
Selling price of 80 articles = ₹ 2400 + 384 = ₹ 2784
Selling price of one article = ₹ \(\frac{2784}{80}\) = ₹ 34.80

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

Question 5.
A VCR and TV were bought for ₹ 8000 each. The shopkeeper made a loss of 4% on the VCR and a profit of 8% on the T. V. Find the gain or loss per cent on the whole transaction.
Solution:
Total C.P. of VCR and T.V. = ₹ 2 × 8000 = ₹ 16000
Loss on VCR = \(\frac{4}{100}\) × 8000 = ₹ 320
Selling price of the VCR = ₹ 8000 – 320 = ₹ 7680
Profit on T.V. = \(\frac{8}{100}\) × 8000 = ₹ 640
Selling price of the T.V. = ₹ 8000 + ₹ 640 = ₹ 8640
Total selling price = ₹ 7680 + ₹ 8640 = ₹ 16,320
Combined Profit = ₹ 16320 – ₹ 16000 = ₹ 320
Gain % = \(\frac{\text { Gain }}{\text { Total C.P. }} \times 100\)
= \(\frac{320}{16000}\) × 100
= 2%
Profit % on the whole = 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 ₹ 1450 and two shirts marked at ₹ 850 each?
Solution:
Marked price of a pairs of jeans = ₹ 1450
Discount = 10% of M.P
= \(\frac{10}{100}\) × 1450
= ₹ 145
Sale price of a pair of jeans = ₹ 1450 – ₹ 145 = ₹ 1305
Marked price of two shirts = ₹ 850 × 2 = ₹ 1700
Discount = 10% of M.P
= \(\frac{10}{100}\) × 1700
= ₹ 170
Sale price of two shirts = ₹ 1700 – 170 = ₹ 1530
∴ Total amount paid by the customer = ₹ 1305 + ₹ 1530 = ₹ 2835
Amount paid by the customer = ₹ 2835

Question 7.
A milkman sold two of his buffaloes for ₹ 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. (Find CP of each)
Solution:
Selling price of 1st buffalo = ₹ 20,000
gain% = 5%
Cost price = \(\left(\frac{100}{100+\text { Profit } \%}\right) \times \mathrm{S} . \mathrm{P}\)

Total selling price = ₹ 2 × 20,000 = ₹ 40,000
Loss = Total C.P. – Total S.P
= ₹ 41269.84 – ₹ 40,000
= ₹ 1269.84

Question 8.
The price of a TV is ₹ 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:
Cost price of the T.V. = ₹ 13,000
Sales tax = 12% of C.P
= \(\frac{12}{100}\) × 13000
= ₹ 1560
Bill amount = ₹ 1560 + ₹ 13,000 = ₹ 14,560
∴ Vinod has to pay ₹ 14,560 to buy a TV.

Question 9.
Arun bought a pair of skates at a sale where the discount given was 20%. If the amount he pays is ₹ 1,600. Find the marked price.
Solution:
Let the marked price be ₹ ‘x’
Discount = 20% of M.P
= \(\frac{20}{100}\) × x
= \(\frac{\mathrm{x}}{5}\)
Amount paid by Arun = ₹ 1600
x – \(\frac{\mathrm{x}}{5}\) = 1600
\(\frac{5 x-x}{5}=1600\)
x = \(\frac{1600 \times 5}{4}\) = ₹ 2000
∴ Amount paid by Arun = ₹ 2,000

Question 10.
I purchased a hair-dryer for ₹ 5400 including 8% VAT. Find the price before VAT was added.
Solution:
Let the original price of a hair-dryer be ‘x’
VAT = 8% of original price = \(\frac{8}{100}\) × x = \(\frac{2 \mathrm{x}}{25}\)
Bill amount = ₹ 5400
x + \(\frac{2 \mathrm{x}}{25}\) = 5400
\(\frac{25 x+2 x}{25}\) = 5400
\(\frac{27 \mathrm{x}}{25}\) = 5400
x = \(\frac{5400 \times 25}{27}\) = ₹ 5000
∴ Original price of the hair dryer = ₹ 5000

Question 11.
An article was purchased for ₹ 1239 including GST of 18%. Find the price of the article before GST was added?
Solution:
Cost with GST included = ₹ 1239
Cost with out GST = x rupees
GST = 18%
If we take this into equation
x + ((18 ÷ 100 ) × x) = 1239
Cost before GST + GST = Cost with GST
⇒ x + ( 9x ÷ 50 ) = 1239
⇒ (50x + 9x) ÷ 50 = 1239
⇒ 59x ÷ 50 = 1239
⇒ 59x = 1239 × 50
⇒ 59x = 61950
⇒ x = 61950 ÷ 59
⇒ x = 1050
Price before GST = ₹ 1050

These NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Exercise 8.1

Question 1.
Find the ratio of the following:
(a) Speed of a cycle 15km per hour to the speed of scooter 30 km per hour.
(b) 5 m to 10 km
(c) 50 paise to ₹ 5
Solution:
(a) Speed of the cycle = 15 km per hour.
Speed of the scooter = 30 km per hour.
∴ Ratio = Speed of the cycle : Speed of the scooter
= 15 : 30
= 1 : 2 (Divided by 15)

(b) 10 km = 10 × 1000 m = 10000 m
Ratio = 5 m : 10 km
= 5 : 10000
= 1 : 2000

(c) ₹ 5 = 5 × 100 paise = 500 paise
Ratio = 50 paise : ₹ 5
= 50 : 500
= 1 : 10

Question 2.
Convert the following ratios to percentages.
(a) 3 : 4
(b) 2 : 3
Solution:
(a) 3 : 4 = \(\frac{3}{4}\)
= \(\frac{3}{4}\) × 100%
= 75%

(b) 2 : 3 = \(\frac{2}{3}\)
= \(\frac{2}{3}\) × 100
= \(\frac{200}{3}\)
= 66\(\frac{2}{3}\) %

Question 3.
72% of 25 students are good at mathematics. How many are not good in mathematics.
Solution:
Total number of students = 25
Number of students good in mathematics = 72% of 25
= \(\frac{72}{100}\) × 25
= 18
Number of students who are not good in mathematics = 25 – 18 = 7
Percentage of students not good in mathematics = \(\frac{7}{25}\) × 100
= 7 × 4
= 28%
or = \(\frac{28}{100}\) × 25 = 7 students are not good in Mathematics.

Question 4.
A football team won 10 matches out of the total number of matches they played. If their win percentage was 40, then how many matches did they play in all?
Solution:
Number of matches won by the team = 10
Let the total number of matches be ‘x’
The team won 40% the total number matches.
∴ \(\frac{40}{100}\) × x = 10
x = \(\frac{10 \times 100}{40}\) = 25
∴ Total number of matches played = 25.

Question 5.
If Chameli had ₹ 600 left after spending 75% of her money, how much did she have in the beginning?
Solution:
Amount left with Chameli = (100 – 75)% = 25%
Let the total amount with Chameli be ₹ ‘x’
∴ 25% of total money = ₹ 600
\(\frac{25}{100}\) × x = 600
x = \(\frac{600 \times 100}{25}\) = ₹ 2400
Amount in the beginning = ₹ 2400

Question 6.
If 60% of people in a city like a cricket 30% like football and the remaining like other games, then what percent of the people like other games? If the total number of people is 50 lakh, find the exact number who like each type of game.
Solution:
People who like cricket = 60%
People who like football = 30%
People who like other games = 100 – (60 + 30)
= 100 – 90
= 10%
Total number of people = 50,00,000
No. of people who like cricket = \(\frac{60}{100}\) × 50,00,000 = 30,00,000
No. of people who like football = \(\frac{30}{100}\) × 50,00,000 = 15,00,000
No. of people who like other games = \(\frac{10}{100}\) × 50,00,000 = 5,00,000