Class 8

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 9 Algebraic Expressions and Identities Ex 9.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.2

1. Find the product of the following pairs of monomials:

Question (i)
4, 7p
Solution:
= 4 × 7p
= 4 × 7 × p
= 28p

Question (ii)
– 4p, 7p
Solution:
= – 4p × 7p
= (-4 × 7) × p × p
= – 28p²

Question (iii)
– 4p, 7pq
Solution:
= (- 4p) × 7pq
= – 4 × 7 × p × pq
= – 28p²q

Question (iv)
4p³, – 3p
Solution:
= 4p³ × (- 3p)
= 4 × (- 3) × p³ × p
= – 12p⁴

Question (v)
4p, 0
Solution:
= 4p × 0
= 0

2. Find the areas of rectangles with the following pairs of monomials as their s lengths and breadths respectively:
(p, q); (10m, 5n); (20x², 5y²); (4x, 3x²); (3mn, 4np)
Solution:
Area of a rectangle = length × breadth
(i) length = p, breadth = q
∴ Area = length × breadth
= p × q = pq unit²

(ii) length = 10m, breadth = 5n
∴ Area = length × breadth
= 10m × 5n
= 50mn unit²

(iii) length = 20x², breadth = 5y²
∴ Area = length × breadth
= 20x² × 5y²
= 100x²y² unit²

(iv) length = 4x, breadth 3x²
∴ Area = length × breadth
= 4x × 3x²
= 12x³ unit²

(v) length = 3mn, breadth = 4np
∴ Area = length × breadth
= 3mn × 4np
= 12 mn²p unit²

3. Complete the table of products:
Solution:

First monomial → Second monomial ↓	2x	-5 y	3x2	-4xy	7x2y	– 9x2y2
2x	4x2	-10xy	6x3	-8x2y	14x3y	– 18x3y2
– 5 y	– 10xy	25y2	– 15x2y	20xy2	– 35x2y2	45x2y3
3x2	6x3	– 15x2y	9x4	– 12x3y	21x4y	– 27x4y2
– 4xy	– 8x2y	20xy2	-12x3y	16x2y2	– 28x3y2	36x3y3
7x2y	14x3y	– 35x2y2	21x4y	– 28x3y2	49x4y2	– 63x4y3
– 9x2y2	– 18x3y2	45x2y3	– 27x4y2	36x3y3	– 63x4y3	81x4y4

4. Obtain the volume of rectangular boxes with the following length, breadth and height respectively:

Question (i)
5a, 3a², 7a⁴
Solution:
Volume of rectangular box = length × breadth × height
length = 5a, breadth = 3a², height = 7a⁴
∴ Volume = length × breadth × height
= 5a × 3a² × 7a⁴
= (5 × 3 × 7) × a × a² × a⁴
= 105a⁷ cubic unit

Question (ii)
2p, 4q, 8r
Solution:
length = 2p, breadth = 4q, height = 8r
∴ Volume = length × breadth × height
= 2p × 4q × 8r
= (2 × 4 × 8) × p × q × r
= 64pqr cubic unit

Question (iii)
xy, 2x²y, 2xy²
Solution:
length = xy, breadth = 2x²y, height = 2 xy²
∴ Volume = length × breadth × height
= xy × 2x²y × 2xy²
= (1 × 2 × 2) × xy × x²y × xy²
= 4x⁴y⁴ cubic unit

Question (iv)
a, 2b, 3c
Solution:
length = a, breadth = 2b, height = 3c
∴ Volume = length × breadth × height
= a × 2b × 3c
= (1 × 2 × 3) × a × b × c
= 6abc cubic unit

5. Obtain the product of:

Question (i)
xy, yz, zx
Solution:
= xy × yz × zx
= [x × y) × (y × z) × (z × x)
= x × x × y × y × z × z
= x²y²z²

Question (ii)
a, -a², a³
Solution:
= (a) × (- a²) × (a³)
= – a × a² × a³
= – a⁶

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

Question (iv)
a, 2b, 3c, 6abc
Solution:
= a × 2b × 3c × 6abc
= 1 × 2 × 3 × 6 × a × b × c × abc
= 36 a²b²c²

Question (v)
m, – mn, mnp
Solution:
= (m) × (- mn) × (mnp)
= – (m × mn × mnp)
= – m³n²p

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 7 Cubes and Cube Roots InText Questions and Answers.

PSEB 8th Class Maths Solutions Chapter 7 Cubes and Cube Roots InText Questions

Try These (Textbook Page No. 111)

Find the one’s digit of the cube of each of the following numbers:
(i) 3331
(ii) 8888
(iii) 149
(iv) 1005
(v) 1024
(vi) 77
(vii) 5022
(viii) 53
Solution:

Sl. No.	Number	Number ending in	Units place digit of the cube
(i)	3331	1	1
(ii)	8888	8	2
(iii)	149	9	9
(iv)	1005	5	5
(v)	1024	4	4
(vi)	77	7	3
(vii)	5022	2	8
(viii)	53	3	7

Some interesting patterns: (Textbook Page No. 111)

Observe the following pattern of sums of odd numbers.

Try These (Textbook Page No. 111)

1. Express the following numbers as the sum of odd numbers using the above pattern ?
(a) 6³
(b) 8³
(c) 7³
Solution:
From above pattern, we can conclude n³ = [n(n – 1) + 1] + [n(n – 1) + 3] + [n(n – 1) + 5]… + n terms
(a) 6³
Here, n = 6, n- 1=5
6 (6 – 1) + 1 → 6 × 5 + 1 → 31

OR
= [6(6 – 1) + 1] + [6(6 – 1) + 3] + [6(6 – 1) + 5] + [6(6 – 1) + 7] + [6(6 – 1) + 9] + [6(6 – 1) + 11]
= (6 × 5 + 1) + (6 × 5 + 3) + (6 × 5 + 5) + (6 × 5 + 7) + (6 × 5 + 9) + (6 × 5 + 11)
= (30 + 1) + (30 + 3) + (30 + 5) + (30 + 7) + (30 + 9) + (30 + 11)
= 31 +33 + 35 + 37 + 39 + 41
= 216

(b) 8³
Here, n = 8, n – 1 = 7
8 (8 – 1) + 1 → 8 × 7 + 1 → 57

OR
= [8(8 – 1) + 1] + (8(8 – 1) + 3] + [8(8 – 1) + 5] + [8(8 – 1) + 7] + [8(8 – 1) + 9] + [8(8 – 1) + 11] + [8(8 – 1) + 13] + [8(8 – 1) + 15]
= (8 × 7 + 1) + (8 × 7 + 3) + (8 × 7 + 5) + (8 × 7 + 7) + (8 × 7 + 9) + (8 × 7 + 11) + (8 × 7 + 13) + (8 × 7 + 15)
= (56 + 1) + (56 + 3) + (56 + 5) + (56 + 7) + (56 + 9) + (56 + 11) + (56 + 13) + (56 + 15)
= 57 + 59 + 61 + 63 + 65 + 67 + 69 + 71
= 512

(c) 7³
Here, n = 7, n – 1 = 6
7 × 6 + 1 → 42 + 1 → 43

OR
= [7(7 – 1) + 1] + [7(7 – 1) + 3] + [7(7 – 1) + 5] + [7(7 – 1) + 7] + [7(7 – 1) + 9] + [7(7 – 1) + 11] + [7(7 – 1) + 13]
= (7 × 6 + 1) + (7 × 6 + 3) + (7 × 6 + 5) + (7 × 6 + 7) + (7 × 6 + 9) + (7 × 6 + 11) + (7 × 6 + 13)
= (42 + 1) + (42 + 3) + (42 + 5) + (42 + 7) + (42 + 9) + (42 + 11) + (42 + 13)
= 43 + 45 + 47 + 49 + 51 + 53 + 55
= 343

Consider the following pattern:
2³ – 1³ = 1 + 2 × 1 × 3
3³ – 2³ = 1 + 3 × 2 × 3
4³ – 3³ = 1 + 4 × 3 × 3
Using the above pattern, find the value of the following:
(i) 7³ – 6³
(ii) 12³– 11³
(iii) 20³ – 19³
(iv) 51³ – 50³
Solution:
From above pattern, we can conclude
n³ – (n – 1)³ = 1 + n × (n – 1) × 3
(i) 7³ – 6³ = 1 + 7 × 6 × 3
= 1 + 126
= 127

(ii) 12³ – 11³ = 1 + 12 × 11 × 3
= 1 + 396
= 397

(iii) 20³ – 19³ = 1 + 20 × 19 × 3
= 1 + 1140
= 1141

(iv) 51³ – 50³ = 1 + 51 × 50 × 3
= 1 + 7650
= 7651

Try These (Textbook Page No. 112)

1. Which of the following are perfect cubes?

Question (1).
400
Solution:
\(\begin{array}{l|l}
2 & 400 \\
\hline 2 & 200 \\
\hline 2 & 100 \\
\hline 2 & 50 \\
\hline 5 & 25 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)
400 = 2 × 2 × 2 × 2 × 5 × 5
Here, the prime factors 2 and 5 do not appear in triples.
∴ 2 × 5 × 5 is left over.
∴ 400 is not a perfect cube.

Question (2).
3375
Solution:
\(\begin{array}{l|l}
3 & 3375 \\
\hline 3 & 1125 \\
\hline 3 & 375 \\
\hline 5 & 125 \\
\hline 5 & 25 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)
3375 = 3 × 3 × 3 × 5 × 5 × 5
Here, the prime factors 3 and 5 appear in triples.
No factor is left over.
∴ 3375 is a perfect cube.
3375 = 3³ × 5³

Question (3).
8000
Solution:
\(\begin{array}{l|l}
2 & 8000 \\
\hline 2 & 4000 \\
\hline 2 & 2000 \\
\hline 2 & 1000 \\
\hline 2 & 500 \\
\hline 2 & 250 \\
\hline 5 & 125 \\
\hline 5 & 25 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)
8000 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 × 5
Here, the prime factors 2 and 5 appear in triples.
No factor is left over.
∴ 8000 is a perfect cube.
8000 = 2³ × 2³ × 5³

Question (4).
15625
Solution:
\(\begin{array}{l|l}
5 & 15625 \\
\hline 5 & 3125 \\
\hline 5 & 625 \\
\hline 5 & 125 \\
\hline 5 & 25 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)
15625 = 5 × 5 × 5 × 5 × 5 × 5
Here, the prime factor 5 appear in triples.
No factor is left over.
∴ 15625 is a perfect cube.
15625 = 5³ × 5³

Question (5).
9000
Solution:
\(\begin{array}{l|l}
2 & 9000 \\
\hline 2 & 4500 \\
\hline 2 & 2250 \\
\hline 3 & 1125 \\
\hline 3 & 375 \\
\hline 5 & 125 \\
\hline 5 & 25 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)
9000 = 2 × 2 × 2 × 3 × 3 × 5 × 5 × 5
Here, among the prime factors 2 and 5 appear in triples but 3 does not appear in triple.
3 × 3 is left over.
∴ 9000 is not a perfect cube.

Question (6).
6859
Solution:
\(\begin{array}{l|l}
19 & 6859 \\
\hline 19 & 361 \\
\hline 19 & 19 \\
\hline & 1
\end{array}\)
6859 = 19 × 19 × 19
Here, the prime factor 19 appears in triple.
No factor is left over.
∴ 6859 is a perfect cube.
6859 = 193

Question (7).
2025
Solution:
\(\begin{array}{l|l}
3 & 2025 \\
\hline 3 & 675 \\
\hline 3 & 225 \\
\hline 3 & 75 \\
\hline 5 & 25 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)
2025 = 3 × 3 × 3 × 3 × 5 × 5
Here, the prime factor 3 appears in triple, but 3 × 5 × 5 is left over.
∴ 2025 is not a perfect cube.

Question (8).
10648
Solution:
\(\begin{array}{r|l}
2 & 10648 \\
\hline 2 & 5324 \\
\hline 2 & 2662 \\
\hline 11 & 1331 \\
\hline 11 & 121 \\
\hline 11 & 11 \\
\hline & 1
\end{array}\)
10648 = 2 × 2 × 2 × 11 × 11 × 11
Here, the prime factors 2 and 11 appear in triples.
No factor is left over.
∴ 10648 is a perfect cube.
10648 = 2³ × 11³

Think, Discuss and Write (Textbook Page No. 113)

1. Check which of the following are perfect cubes:
(i) 2700
(ii) 16000
(iii) 64000
(iv) 900
(v) 125000
(vi) 36000
(vii) 21600
(viii) 10000
(ix) 27000000
(x) 1000
What pattern do you observe in these perfect cubes ?
Solution:
(i) 2700
The number is ending with two zeros. If a number ends with three zeros or a multiple of 3 zeros, it may be a perfect cube.
∴ 2700 is not a perfect cube.

(ii) 16000
The number is ending with three zeros.
So it may be a perfect cube.
But, 16 is not a perfect cube.
∴ 16000 is not a perfect cube.

(iii) 64000
The number is ending with three zeros.
So it may be a perfect cube.
64 is a perfect cube. (∵ 4³ = 64)
∴ 64000 is a perfect cube.

(iv) 900
The number is ending with two zeros.
So it is not a perfect cube.
∴ 900 is not a perfect cube.

(v) 125000
The number is ending with three zeros.
So it may be a perfect cube.
125 is a perfect cube. (∵ 5³ = 125)
∴ 125000 is a perfect cube.

(vi) 36000
The number is ending with three zeros.
So it may be a perfect cube.
But, 36 is not a perfect cube.
∴ 36000 is not a perfect cube.

(vii) 21600
The number is ending with two zeros.
So it is not a perfect cube.
∴ 21600 is not a perfect cube.

(viii) 10000
The number is ending with four zeros.
So it is not a perfect cube.
∴ 10000 is not a perfect cube.

(ix) 27000000
The number is ending with six zeros.
So it may be a perfect cube.
27 is a perfect cube. (∵ 3³ = 27)
∴ 27000000 is a perfect cube.

(x) 1000
The number is ending with three zeros.
So it may be a perfect cube.
1 is a perfect cube, (∵ 1³ = 1)
∴ 1000 is a perfect cube.

Think, Discuss and Write (Textbook Page No. 115)

1. State true or false for any integer m, m² < m³. Why ?
Solution:
It seems true, but not always true.
m × m = m² and m × m × m = m³
∴ m² < m³
e.g. if m = 1
∴ m² = 1² = 1 and m³ = 1³ = 1
∴ m² ≮  m³, but m² = m³
If m = (- 1)
∴ m² = (- 1)² = 1 and m³ = (- 1)³ = (- 1)
∴ m² ≮  m³, but m2 > m³
So the above statement is not always true.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 8 Comparing Quantities Ex 8.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.3

1. Calculate the amount and compound interest on:

Question (a)
₹ 10,800 for 3 years at 12\(\frac {1}{2}\) per annum compounded annually.
Solution:
Here, P = ₹ 10,800;
R = 12\(\frac {1}{2}\) % = \(\frac {25}{2}\) %;
T = 3 years; ∴ n = 3

Amount = ₹ 15,377.34
Compound’interest = Amount – Principal
= ₹ (15377.34 – 10800)
= ₹ 4577.34

Question (b)
₹ 18,000 for 2\(\frac {1}{2}\) years at 10% per annum compounded annually.
Solution:
Here, P = ₹ 18,000; R = 10 %;
T = 2\(\frac {1}{2}\) years; ∴ n = 2 + \(\frac {1}{2}\)

Amount = ₹ 22,869
Compoimd interest = Amount – Principal
= ₹ (22869 – 18000)
= ₹ 4869

Question (c)
₹ 62,500 for 1\(\frac {1}{2}\) years at 8% per annum compounded half yearly.
Solution:
Here, the interest is compounded half-yearly.
Here, P = ₹ 62,500; R = \(\frac {8}{2}\) = 4 %
T = 1\(\frac {1}{2}\) years ∴ n = \(\frac {3}{2}\) × 2 = 3

Amount = ₹ 70,304
Compound interest = Amount – Principal
= ₹ (70304 – 62500)
= ₹ 7804

Question (d)
₹ 8000 for 1 year at 9 % per annum compounded half-yearly.
(You could use the year-by-year calculation using SI formula to verify.)
Solution:
Here, the interest is compounded half-yearly.
Here, P = ₹ 8000; R = \(\frac {9}{2}\) %;
T = 1 year ∴ n = 2

Amount = ₹ 8736.20
Compound interest = Amount – Principal
= ₹ (8736.20 – 8000)
= ₹ 736.20
[Note : By finding simple interest also we can calculate.)
SI = \(\frac {PRT}{100}\)
= \(\frac{8000 \times 9 \times 1}{2 \times 100}\)
= ₹ 376.20
Thus, total interest of 1 year
= ₹ (360 + 376.20)
= ₹ 736.20

Question (e)
₹ 10,000 for 1 year 8% per annum compounded half yearly.
Solution:
Here, the interest is compounded half-yearly.
Here, P = ₹ 10,000; R = \(\frac {8}{2}\) = 4 %;
T = 1 year ∴ n = 1 × 2 = 2

Amount = ₹ 10,816
Compound interest = Amount – Principal
= ₹ (10816 – 10000)
= ₹ 816

2. Kamala borrowed ₹ 26,400 from a Bank to buy a scooter at a rate of 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:
[Note: Here, find amount after 2 years by compound interest. This amount is principal for \(\frac {4}{12}\) year. For this \(\frac {4}{12}\) year, find simple interest.)
Here, P = ₹ 26,400; R = 15%;
T = 2 years ∴ n = 2

Now, ₹ 34,914 will be principal to find interest of 4 months.
SI = \(\frac {PRT}{100}\)
= \(\frac{34914 \times 15 \times 4}{100 \times 12}\)
= \(\frac{174570}{100}\)
= ₹ 174570
= ₹ 1745.70
Amount = ₹ (34914 + 1745.70)
= ₹ 36,659.70
Thus, Kamala Mil have to pay ₹ 36,659.70 to clear the loan.

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:
Here, P = ₹ 12,500; R = 12%; T = 3 years
SI = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
= \(\frac{12500 \times 12 \times 3}{100}\)
= 125 × 12 × 3
= ₹ 4500
Simple interest = ₹ 4500

For Radha:
Here, P = ₹ 12,500; R = 10%;
T = 3 years ∴ n = 3


Amount = ₹ 16,637.50
Compound interest = Amount – Principal
= ₹ (16637.50 – 12500)
= ₹ 4137.50
Fabina has to pay ₹ 4500 as interest and Radha has to pay ₹ 4137.50 as interest.
∴ Fabina has to pay more interest.
Difference in interest = ₹ (4500 – 4137.50)
= ₹ 362.50
Thus, Fabina has to pay ₹ 362.50 more than Radha as interest.

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:
For Simple Interest:
Here, P = ₹ 12,000; R = 6 %; T = 2 years
SI = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
= \(\frac{12000 \times 6 \times 2}{100}\)
= 120 × 6 × 2
= ₹ 1440
SI = ₹ 1440

For Compound Interest:
Here, P = ₹ 12,000, R = 6 %
T = 2 years ∴ n = 2

= ₹ 13,483.20
Amount = ₹ 13,483.20
CI = A – P
= ₹ (13483.20 – 12000)
= ₹ 1483.20
∴ Extra amount to be paid
= ₹ (1483.20 – 1440)
= ₹ 43.20
Thus, I have to pay ₹ 43.20 as extra amount.

5. Vasudevan invested ₹ 60,000 at an interest rate of 12% per annum compounded half-yearly. What amount would he get.

Question (i)
after 6 months?
Solution:
Interest after 6 months:
Here, P = ₹ 60,000; R = \(\frac {12}{2}\) = 6%;
T = 6 months ∴ n = 1

∴ Amount = ₹ 63,600

Question (ii)
after 1 year?
Solution:
After 1 year:
∴ Amount = ₹ 67,416
Here, P = ₹ 60,000; R = \(\frac {12}{2}\) = 6 %;
T = 1 year ∴ n = 2

∴ Amount = ₹ 67, 146
Thus, Vasudevan will get ₹ 63,600 after 6 months and ₹ 67,416 after 1 year.

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

Question (i)
compounded annually.
Solution:
Calculation of CI annually:
Here, P = ₹ 80,000; R = 10 %;
T = 1\(\frac {1}{2}\) year

For 1st year:
R = 10% and n = 1

∴ Amount of CI after 1 year = ₹ 88,000
Now, calculate simple interest of ₹ 88,000 for 6 months.
P = ₹ 88,000; R = 10 %;
T = 6 months = \(\frac {1}{2}\) year
∴ Interest = \(\frac{P \times R \times T}{100}\)
= \(\frac{88000 \times 10 \times 1}{100 \times 2}\)
= 4400
∴ Interest of 6 months = ₹ 4400
Thus, A = P + I
= ₹ (88000 + 4400)
= ₹ 92,400
Thus, according to CI, Arif has to pay ₹ 92,400

Question (ii)
compounded half yearly.
Solution:
If interest is compounded half yearly
Here, P = ₹ 88000, R = \(\frac {10}{2}\) = 5%
T = 1\(\frac {1}{2}\) years ∴ n = 3

∴ Amount = ₹ 92,610
Thus, according to half-yearly CI, Arif has to pay ₹ 92,610
∴ Difference = ₹ (92,610 – 92,400)
= ₹ 210

7. Maria invested ₹ 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find

Question (i)
The amount credited against her name at the end of the second year,
Solution:
(i) Here, P = ₹ 8000, R = 5 %,
T = 2 years ∴ n = 2


Thus, amount credited against Marlas name at the end of second year is ₹ 8820.

Question (ii)
The interest for the 3rd year.
Solution:
To find the interest for the 3rd year:
P = ₹ 8820, R = 5%, T = 1 years
SI = \(\frac{\mathrm{PRT}}{100}=\frac{8820 \times 5 \times 1}{100}\) = 441
The interest for 3rd year is ₹ 441
OR
The interest for the 3rd year:
Here, P = ₹ 8000, R = 5 %, T = 3 years ∴ n = 3

∴ At the end of 3rd year ₹ 9261 will be credited against Maria’s name.
∴ Interest of the 3rd year = Amount of 3 years – Amount of 2 years
= ₹ (9261 – 8820)
= ₹ 441

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:
(i) Here, interest is compounded half-yearly.
Here, P = ₹ 10,000; R = \(\frac {10}{2}\) = 5 %;
T = 1\(\frac {1}{2}\) years ∴ n = 3

Amount = ₹ 11,576.25
CI = A – P
= ₹ (11576.25 – 10000)
= ₹ 1576.25

(ii) Here, interest is compounded yearly. CI for 1 year:
Here, P = ₹ 10,000; R = 10%;
T = 1 year ∴ n = 1

Amount at the end of 1 year = ₹ 11,000
∴ CI = A – P
= ₹ (11000 – 10000)
= ₹ 1000
Now, calculate SI for 6 months.
Here, P = ₹ 11,000; R = 10%; T = \(\frac {1}{2}\) year
SI = \(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}=\frac{11000 \times 10 \times 1}{100 \times 2}\)
= 550
∴ Total interest of 1\(\frac {1}{2}\) years = ₹ (1000 + 550)
= ₹ 1550
After comparing (i) and (ii), we can conclude ₹ 1576.25 > ₹ 1550
∴ Yes, the interest is more if compounded half-yearly.

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:
Here, interest is compounded half-yearly.
Here, P = ₹ 4096, R = 12\(\frac{1}{2} \times \frac{1}{2}=\frac{25}{4}\)
T = 18 months = 1\(\frac {1}{2}\) years ∴ n = 3

Amount = ₹ 4913
Thus, Ram will get ₹ 4913 at the end of period.

10. The population of a place increased to 54,000 in 2003 at a rate of 5% per annum

Question (i)
find the population in 2001.
Solution:
[Note: In 1st case, we have to find P as population of 2003 is given. From that we have to find population of 2001.]
Population in 2003 = 54,000
Here, A = 54,000; R = 5%; T = 2 years ∴ n = 2

∴ P = 48979.59 (approx)
P = 48980 (approx)
Thus, the population in 2001 is 48,980.

Question (ii)
What would be its population in 2005?
Solution:
Here, P = 54,000; R = 5 %;
T = 2 years ∴ n = 2

Thus, the population in 2005 is 59,535.

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:
Initial count of bacteria = 5,06,000
Rate of increasing = 2.5% per hour
Bacteria count after 2 hours
Here, P = 5,06,000, R = 2.5 % = \(\frac {5}{2}\)%;
T = 2 hours ∴ n = 2

A = 531616 (approx)
Thus, the number of bacteria count after 2 hours will be 5,31,616 (approx).

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:
CP of a scooter = ₹ 42,000
Here, P = ₹ 42,000; R = 8 %; T = 1 ∴ n = 1
R = – 8 % (as depreciation)

Thus, the value of a scooter after 1 year will be ₹ 38,640.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 9 Algebraic Expressions and Identities Ex 9.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4

1. Multiply the binomials:

Question (i)
(2x + 5) and (4x – 3)
Solution:
= (2x + 5)(4x – 3)
= 2x(4x – 3) + 5 (4x – 3)
= 8x² – 6x + 20x – 15
= 8x² + 14x – 15

Question (ii)
(y – 8) and (3y – 4)
Solution:
= (y -8) (3y – 4)
= y (3y – 4) – 8 (3y – 4)
= 3y² – 4y – 24y + 32
= 3y² – 28y + 32

Question (iii)
(2.51 – 0.5m) and (2.51 + 0.5m)
Solution:
= (2.5l – 0.5m) (2.5l + 0.5m)
= 2.5l(2.5l + 0.5m) – 0.5m (2.5l + 0.5m)
= 6.25l² + 1.25lm – 1.25lm – 0.25m²
= 6.25l² – 0.25m²

Question (iv)
(a + 3b) and (x + 5)
Solution:
= (a + 3b) (x + 5)
= a (x + 5) + 3b (x + 5)
= ax + 5a + 3bx + 15b

Question (v)
(2pq + 3q²) and (3pq – 2q²)
Solution:
= (2pq + 3q²) (3pq – 2q²)
= 2pq (3pq – 2q²) + 3q² (3pq – 2q²)
= 6p²q² – 4pq³ + 9pq³ – 6q⁴
= 6p²q² + 5pq³ – 6q⁴

Question (vi)
(\(\frac {3}{4}\)a² + 3b²) and 4 (a² – \(\frac {2}{3}\)b²)
Solution:
= (\(\frac {3}{4}\)a² + 3b²) (4a² – \(\frac {8}{3}\)b²)
= \(\frac {3}{4}\)a⁴ (4a² – \(\frac {8}{3}\)b²) + 3b² (4a² – \(\frac {8}{3}\)b²)
= 3a₄ – 2a²b² + 12a²b² – 8b₄
= 3a⁴ + 10a²b² – 8b⁴

2. Find the product:

Question (i)
(5 – 2x) (3 + x)
Solution:
= 5 (3 + x) – 2x (3 + x)
= 15 + 5x – 6x – 2x²
= 15 – x – 2x²

Question (ii)
(x + 7y) (7x – y)
Solution:
= x(7x – y) + 7y (7x – y)
= 7x² – xy + 49xy – 7y²
= 7x² + 48xy – 7y²

Question (iii)
(a² + b) (a + b²)
Solution:
= a² (a + b²) + b (a + b²)
= a³ + a²b² + ab + b³

Question (iv)
(p² – q²) (2p + q)
Solution:
= p²(2p + q) – q²(2p + q)
= 2p³ + p²q – 2pq² – q³

3. Simplify:

Question (i)
(x² – 5) (x + 5) + 25
Solution:
= x² (x + 5) – 5 (x + 5) + 25
= x³ + 5x² – 5x – 25 + 25
= x³ + 5x² – 5x

Question (ii)
(a² + 5) (b³ + 3) + 5
Solution:
= a²(b³ + 3) + 5 (b³ + 3) + 5
= a²b³ + 3a² + 5b³ + 15 + 5
= a²b³ + 3a² + 5b³ + 20

Question (iii)
(t + s²)(t² – s)
Solution:
= t (t² – s) + s² (t² – s)
= t³ – st + s²t² – s³

Question (iv)
(a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd)
Solution:
= a(c – d) + b(c – d) + a(c + d) – b(c + d) + 2 (ac + bd)
= ac – ad + bc – bd + ac + ad – bc – bd + 2ac + 2bd
= ac + ac + 2ac – ad + ad + bc – bc – bd – bd + 2 bd
= 4 ac

Question (v)
(x + y) (2x + y) + (x + 2y) (x – y)
Solution:
= x (2x + y) + y (2x + y) +x(x – y) + 2y (x – y)
= 2x² + xy + 2xy + y² + x² – xy + 2xy – 2y²
= 2x² + x² + xy + 2xy – xy + 2xy + y² – 2y²
= 3x² + 4xy – y²

Question (vi)
(x + y) (x² – xy + y²)
Solution:
= x (x² – xy + y²) + y (x² – xy + y²)
= x³ – x²y + xy² + x²y – xy² + y³
= x³ – x²y + x²y + xy² – xy² + y³
= x³ + y³

Question (vii)
(1.5x – 4y) (1.5x + 4y + 3) – 4.5x + 12y
Solution:
= 1.5x (1.5x + 4y + 3) – 4y (1.5x + 4y + 3) – 4.5x + 12y
= 2.25x² + 6xy + 4.5x – 6xy – 16y² – 12y – 4.5x + 12y
= 2.25x² + 6xy – 6xy + 4.5x – 4.5x – 16y² – 12y + 12y
= 2.25x² – 16y²

Question (viii)
(a + b + c) (a + b – c)
Solution:
= a (a + b – c) + b (a + b – c) + c (a + b – c)
= a² + ab – ac + ab + b² – bc + ac + bc – c²
= a² + ab + ab – ac + ac + b² – bc + bc – c²
= a² + 2ab + b² – c²
= a² + b² – c² + 2 ab

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 8 Comparing Quantities Ex 8.2 Textbook Exercise Questions and Answers.

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

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 be ₹ 100.
After 10% increase, the new salary = ₹ (100 + 10)
= ₹ 110
If his new salary is ₹ 110,
then the original salary = ₹ 100
If new salary is ₹ 1,54,000,
then original salary = (\(\frac {100}{110}\) × 1,54,000)
= 100 × 1400
= ₹ 140000
Thus, his original salary was ₹ 1,40,000.

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:
Number of people went to the zoo on Sunday = 845
Number of people went to the zoo on Monday = 169
∴ Decrease in the number of people visiting the zoo on Monday = (845 – 169) = 676
Percentage decrease = \(\left(\frac{\text { Decrease }}{\text { Original number }} \times 100\right) \%\)
= (\(\frac {676}{845}\) × 100) %
= 80 %
Thus, 80% decrease in the number of people visiting the zoo on Monday.

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
∴ Cost price of 1 article = ₹ \(\frac {2400}{80}\) = ₹ 30
Profit = 16%
∴ Profit on 1 article = ₹ \(\left(\frac{16}{100} \times 30\right)\)
= ₹ 4.80
∴ Selling price of 1 article
= Cost price + Profit
= ₹ 30 + ₹ 4.80
= ₹ 34.80
Thus, selling price of one article is ₹ 34.80.

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:
Cost price of an article = ₹ 15,500
Repair charge (overhead charge) = ₹ 450
∴ Total cost price = Cost price of an article + Overhead expenses
= ₹ (15500 + 450) = ₹ 15,950
Profit per cent = 15%
Amount of profit = 15% of ₹ 15,950
= ₹ \(\left(\frac{15}{100} \times 15950\right)\)
= ₹ \(\left(\frac{23925}{10}\right)\)
= ₹ 2392.50
∴ Selling price = Total cost + Profit
= ₹ (15950 + 2392.50)
= ₹ 18342.50
Thus, the selling price of an article is ₹ 18,342.50.

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 TV. Find the gain or loss per cent on the whole transaction.
Solution:
(i) Cost price of a VCR = ₹ 8000,
Loss per cent = 4 %
∴ Loss amount = 4% of cost
= ₹ \(\left(\frac{4}{100} \times 8000\right)\)
= ₹ (4 × 80)
= ₹ 320
∴ Selling price = Cost price – Loss
= ₹ (8000 – 320)
= ₹ 7680

(ii) Cost price of a TV = ₹ 8000
Profit per cent = 8%
∴ Profit amount = 8 % of cost
= ₹ \(\left(\frac{8}{100} \times 8000\right)\)
= ₹ (8 × 80)
= ₹ 640
∴ Selling price = Cost price + Profit
= ₹ (8000 + 640)
= ₹ 8640
Total cost price of a VCR and TV = ₹ (8000 + 8000)
= ₹ 16,000
Total selling price of a VCR and TV = ₹ (7680 + 8640)
= ₹ 16,320
SP > CR
∴ profit = SP – CP
= ₹ (16,320 – 16,000)
= ₹ 320
∴ Profit per cent = \(\left(\frac{\text { Profit }}{\text { Cost price }} \times 100\right) \%\)
= \(\left(\frac{320}{16000} \times 100\right) \%\)
= 2%
Thus, there is 2% profit on the whole transaction.

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:
MP of a pair of jeans = ₹ 1450
MP of one shirt = ₹ 850
∴ MP of two shirts = ₹ (2 × 850)
= ₹ 1700
∴ Total MP of three items = ₹ (1450 + 1700)
= ₹ 3150
Discount per cent = 10%
∴ Amount of discount = 10% of total cost
= ₹ \(\left(\frac{10}{100} \times 3150\right)\)
= ₹ 315
Bill amount = MP – Discount
= ₹ (3150 – 315)
= ₹ 2835
Thus, customer would have to pay ₹ 2835 for a pair of jeans and two shirts.

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. (Hint: Find CP of each)
Solution:
Let CP of 1st buffalo be ₹ x
Gain (Profit) per cent = 5%
Amount of profit = 5 % of CP
= ₹ \(\left(\frac{5}{100} \times x\right)\)
= ₹ \(\frac{5 x}{100}\)
∴ SP = CP + Profit
= ₹ \(\left(x+\frac{5 x}{100}\right)\)
= ₹ \(\left(\frac{100 x+5 x}{100}\right)\)
= ₹ \(\frac{105 x}{100}\)
But, SP of a buffalo = ₹ 20,000 (Given)
∴ \(\frac{105 x}{100}\) = ₹ 20,000
∴ x = ₹ \(\left(\frac{20000 \times 100}{105}\right)\)
= ₹ 19047.62

Let the cost price of 2nd buffalo = ₹ y
Loss per cent = 10%
Amount of loss = 10% of CP
= ₹ \(\left(\frac{10}{100} \times y\right)\)
= ₹ \(\frac{10 y}{100}\)
SP = CP – Loss
= ₹ \(\left(y-\frac{10 y}{100}\right)\)
= ₹ \(\left(\frac{100 y-10 y}{100}\right)\)
= ₹ \(\frac{90 y}{100}\)
But SP of a buffalo = ₹ 20,000 (Given)
∴ \(\frac{90 y}{100}\) = ₹ 20,000
∴ y = ₹ \(\left(\frac{20000 \times 100}{90}\right)\)
= ₹ 22222.22

Total CP of both buffaloes = ₹ (x + y)
= ₹ (19047.62 + 22222.22)
= ₹ 41,269.84

Total SP of both buffaloes = ₹ (20000 + 20000)
= ₹ 40,000
∴ SP < CP
Amount of loss = CP – SP
= ₹ (41269.84 – 40000)
= ₹ 1269.84
Thus, there is overall loss of ₹ 1269.84.

8. The price of a TV is ₹ 13,000. The GST charged on it is at the rate of 12%. Find the amount that Vinod will have to pay if he buys it.
Solution:
Price of a TV = ₹ 13,000
GST per cent = 12%
∴ Amount of GST = 12% of price
= ₹ \(\left(\frac{12}{100} \times 13,000\right)\)
= ₹ 1560
∴ Total amount = Price of a TV + GST
= ₹ (13,000 + 1560)
= ₹ 14,560
Thus, Vinod will have to pay ₹ 14,560.

9. Aran bought a pair of skates at a sale where the discount given was 20%. If the amount he pays is ₹ 1600, find the marked price.
Solution:
Let the MP of a pair of skates be ₹ x.
Discount per cent = 20%
∴ Amount of discount = 20% of MP
= ₹ \(\left(\frac{20}{100} \times x\right)\)
= ₹ \(\frac{20 x}{100}\)
∴ Selling price = MP – Discount
= ₹ \(\left(x-\frac{20 x}{100}\right)\)
= ₹ \(\left(\frac{100 x-20 x}{100}\right)\)
= ₹ \(\frac{80 x}{100}\)
= ₹ \(\frac{4}{5} x\)
But, SP of a pair of skates = ₹ 1600 (Given).
∴ \(\frac{4}{5} x\) = 1600
∴ x = ₹ \(\left(\frac{1600 \times 5}{4}\right)\)
∴ x = ₹ 2000
Thus, the marked price of a pair of skates is ₹ 2000.

10. I purchased a hair dryer for ₹ 5400 including 18% GST. Find the price before GST was added.
Solution:
Cost price of hair dryer with GST = ₹ 5400
GST per cent = 18%
Let the price of a hair dryer before GST ) was added be ₹ x.
∴ Amount of GST = 18% of x
= ₹ \(\left(\frac{18}{100} \times x\right)\)
= ₹ \(\frac{18 x}{100}\)
∴ Price after adding GST = ₹ \(\left(x+\frac{18}{100} x\right)\)
= ₹ \(\left(\frac{100 x+18 x}{100}\right)\)
= ₹ \(\frac{118 x}{100}\)
But, price after adding GST = ₹ 5400 (Given)
∴ \(\frac{118 x}{100}\) = 5400
∴ x = ₹ \(\left(\frac{5400 \times 100}{118}\right)\)
= ₹ 4576.27
Thus, the price of a hair dryer before adding GST is ₹ 4576.27.

11. An article was purchased for ₹ 1239 including GST of 18%. Find the price of the article before GST was added.
Solution:
Such type of sums can be done by two methods.
One method:
Let the price of an article before adding GST be ₹ 100.
GST = 18%
∴ price including GST = ₹ (100 + 18)
= ₹ 118
If price including GST is ₹ 118,
then price before adding GST = ₹ 100
∴ if price including GST is ₹ 1239,
then price before adding GST = ₹ \(\left(\frac{1239}{118} \times 100\right)\)
= ₹ 1050
Thus, the price of an article before adding GST was ₹ 1050.

Another method:
Let the price of an article before adding GST be ₹ x.
GST = 18%
Amount of GST = 18% of ₹ x
= ₹ \(\left(\frac{18}{100} \times x\right)\)
= ₹ \(\frac{18 x}{100}\)
Price after adding GST = ₹ \(\left(x+\frac{18 x}{100}\right)\)
= ₹ \(\left(\frac{118 x}{100}\right)\)
But, the price of an article = ₹ 1239 (Given)
∴ \(\frac{118 x}{100}\) = 1239
∴ x = ₹ \(\left(\frac{1239 \times 100}{118}\right)\)
= ₹ 1050
Thus, the price of an article before adding GST was ₹ 1050.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 7 Cubes and Cube Roots Ex 7.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots Ex 7.2

1. Find the cube root of each of the following numbers by prime factorisation method.

Question (i).
64
Solution:
\(\begin{array}{l|l}
2 & 64 \\
\hline 2 & 32 \\
\hline 2 & 16 \\
\hline 2 & 8 \\
\hline 2 & 4 \\
\hline 2 & 2 \\
\hline & 1
\end{array}\)
By prime factorisation,
64 = 2 × 2 × 2 × 2 × 2 × 2
∴ \(\sqrt[3]{64}\) = 2 × 2
= 4
Thus, cube root of 64 is 4.

Question (ii).
512
Solution:
\(\begin{array}{l|l}
2 & 512 \\
\hline 2 & 256 \\
\hline 2 & 128 \\
\hline 2 & 64 \\
\hline 2 & 32 \\
\hline 2 & 16 \\
\hline 2 & 8 \\
\hline 2 & 4 \\
\hline 2 & 2 \\
\hline & 1
\end{array}\)
By prime factorisation,
512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
∴ \(\sqrt[3]{512}\) = 2 × 2 × 2
= 8
Thus, cube root of 512 is 8.

Question (iii).
10648
Solution:
\(\begin{array}{r|l}
2 & 10648 \\
\hline 2 & 5324 \\
\hline 2 & 2662 \\
\hline 11 & 1331 \\
\hline 11 & 121 \\
\hline 11 & 11 \\
\hline & 1
\end{array}\)
By prime factorisation,
10648 = 2 × 2 × 2 × 11 × 11 × 11
∴ \(\sqrt[3]{10648}\) = 2 × 11
= 22
Thus, cube root of 10648 is 22.

Question (iv).
27000
Solution:
\(\begin{array}{l|l}
2 & 27000 \\
\hline 2 & 13500 \\
\hline 2 & 6750 \\
\hline 3 & 3375 \\
\hline 3 & 1125 \\
\hline 3 & 375 \\
\hline 5 & 125 \\
\hline 5 & 25 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)
By prime factorisation
27000 = 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5 × 5
∴ \(\sqrt[3]{27000}\) = 2 × 3 × 5
= 30
Thus, cube root of 27000 is 30.

Question (v).
15625
Solution:
\(\begin{array}{l|l}
5 & 15625 \\
\hline 5 & 3125 \\
\hline 5 & 625 \\
\hline 5 & 125 \\
\hline 5 & 25 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)
By prime factorisation,
15625 = 5 × 5 × 5 × 5 × 5 × 5
∴ \(\sqrt[3]{15625}\) = 5 × 5
= 25
Thus, cube root of 15625 is 25.

Question (vi).
13824
Solution:
\(\begin{array}{l|l}
2 & 13824 \\
\hline 2 & 6912 \\
\hline 2 & 3456 \\
\hline 2 & 1728 \\
\hline 2 & 864 \\
\hline 2 & 432 \\
\hline 2 & 216 \\
\hline 2 & 108 \\
\hline 2 & 54 \\
\hline 3 & 27 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)
By prime factorisation,
13824 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
∴ \(\sqrt[3]{13824}\) = 2 × 2 × 2 × 3
= 24

Question (vii).
110592
Solution:
\(\begin{array}{l|l}
2 & 110592 \\
\hline 2 & 55296 \\
\hline 2 & 27648 \\
\hline 2 & 13824 \\
\hline 2 & 6912 \\
\hline 2 & 3456 \\
\hline 2 & 1728 \\
\hline 2 & 864 \\
\hline 2 & 432 \\
\hline 2 & 216 \\
\hline 2 & 108 \\
\hline 2 & 54 \\
\hline 3 & 27 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)
By prime factorisation,
110592 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
∴ \(\sqrt[3]{110592}\) = 2 × 2 × 2 × 2 × 3
= 48
Thus, cube root of 110592 is 48.

Question (viii).
46656
Solution:
\(\begin{array}{l|l}
2 & 46656 \\
\hline 2 & 23328 \\
\hline 2 & 11664 \\
\hline 2 & 5832 \\
\hline 2 & 2916 \\
\hline 2 & 1458 \\
\hline 3 & 729 \\
\hline 3 & 243 \\
\hline 3 & 81 \\
\hline 3 & 27 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)
By prime factorisation,
46656 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3
∴ \(\sqrt[3]{46656}\) = 2 × 2 × 3 × 3
= 36
Thus, cube root of 46656 is 36.

Question (ix).
175616
Solution:
\(\begin{array}{l|l}
2 & 175616 \\
\hline 2 & 87808 \\
\hline 2 & 43904 \\
\hline 2 & 21952 \\
\hline 2 & 10976 \\
\hline 2 & 5488 \\
\hline 2 & 2744 \\
\hline 2 & 1372 \\
\hline 2 & 686 \\
\hline 7 & 343 \\
\hline 7 & 49 \\
\hline 7 & 7 \\
\hline & 1
\end{array}\)
By prime factorisation,
175616 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 7 × 7 × 7
∴ \(\sqrt[3]{175616}\) = 2 × 2 × 2 × 7
= 56
Thus, cube root of 175616 is 56.

Question (x).
91125
Solution:
\(\begin{array}{l|l}
3 & 91125 \\
\hline 3 & 30375 \\
\hline 3 & 10125 \\
\hline 3 & 3375 \\
\hline 3 & 1125 \\
\hline 3 & 375 \\
\hline 5 & 125 \\
\hline 5 & 25 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)
By prime factorisation,
91125 = 3 × 3 × 3 × 3 × 3 × 3 × 5 × 5 × 5
∴ \(\sqrt[3]{91125}\) =3 × 3 × 5
= 45
Thus, cube root of 91125 is 45.

2. State true or false.

Question (i).
Cube of any odd number is even.
Solution:
False, cube of any odd number is odd.

Question (ii).
A perfect cube does not end with two zeros.
Solution:
True, a perfect cube ending with zero will always have a triplet of zeros.

Question (iii).
If square of a number ends with 5, then its cube ends with 25.
Solution:
False, let us understand with an example.
15² = 15 × 15 = 225
15³ = 15 × 15 × 15 = 3375

Question (iv).
There is no perfect cube which ends with 8.
Solution:
False, let us understand with an example.
8 = 2³, 1728 = 12³

Question (v).
The cube of a two-digit number may be a three-digit number.
Solution:
False, let us take an example.
The smallest two-digit number is 10.
10³ = 1000
which is a four-digit number and not a three-digit number.

Question (vi).
The cube of a two-digit number may have seven or more digits.
Solution:
False, let us take an example.
The greatest two-digit number is 99.
99³ = 970299
which is a six-digit number.

Question (vii).
The cube of a single digit number may be a single digit number.
Solution:
True, let us take examples.
1³ = 1 and 2³ = 8

3. You are told that 1,331 is a perfect cube. Can you guess without factorisation what is its cube root? Similarly, guess the cube roots of 4913, 12167, 32768.
Solution:
Yes, we can guess without prime factorisation.
1331 : Separate given number into two groups.
1331 → 1 and 331
331 → Units place digit of 331 is 1.
∴ Unit place digit of cube root of 1331 = 1 (∵ 1³ = 1)
1 → 1³ = 1 and 2³ = 8
∴ Tens digit of cube root of 1331 = 1
Thus, the cube root of 1331 is 11.

(i) 4913 : Separate given number into two groups.
4913 → 4 and 913
913 → Units place digit of 913 is 3.
∴ Unit digit of cube root of 4913 = 7 (∵ 7³ = 343)
4 → 1³ = 1 and 2³ = 8
1 < 4 < 8 (∴ 1³ < 4 < 2³)
∴ The tens digit of cube root of 4913 = 1
Thus, the cube root of 4913 is 17.

(ii) 12167 : Separate given number into two groups.
12167 → 12 and 167
167 → Units place digit of 167 is 7.
∴ Unit digit of cube root of 12167 = 3
(∵ 3³ = 27)
12 → 2³ = 8 and 3³ = 27
8 < 12 < 27 (∵ 2³ < 12 < 3³)
∴ Tens place digit of cube root of 12167 = 2
Thus, the cube root of 12167 is 23.

(iii) 32768 : Separate given number into two groups.
32768 → 32 and 768
768 → Units place digit of 768 is 8.
∴ Unit digit of cube root of 32768 = 2 (∵ 2³ = 8)
32 → 3³ = 27 and 4³ = 64
27 < 32 < 64 (∵ 3³ < 32 < 4³)
∴ The tens digit of cube root of 32768 = 3
Thus, the cube root of 32768 is 32.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 10 Visualising Solid Shapes InText Questions and Answers.

PSEB 8th Class Maths Solutions Chapter 10 Visualising Solid Shapes InText Questions

Do This : [Textbook Page No. 153 – 154]

1. Match the following: (First one is done for you.)

Solution:
(A) → (ii) → (c),
(B) → (iii) → (g),
(C) → (i) → (b),
(D) → (iv) → ( h ),
(E) → (v) → (f),
(F) → (vii) → (d),
(G) → (vi ) → (e),
(H) → (viii) → (a).

Do This : [Textbook Page No. 154 – 155]

1. Match the following pictures (objects) with their shapes:

Solution:
(i) → (c),
(ii) → (d),
(iii) → (e),
(iv) → (b),
(v) → (a).

Do This : [Textbook Page No. 162-163]

1. Look at the following map of a city:

(a) Colour the map as follows: Blue – Water, Red – Fire Station, Orange – Library, Yellow – Schools, Green -Park, Pink – Community Centre, Purple – Hospital, Brown – Cemetery.
(b) Mark a Green ‘X’ at the intersection of 2nd street and Danim street. A Black ‘Y’ where the river meets the third street. A red ‘Z’ at the intersection of main street and 1st street.
(c) In magenta colour, draw a short street route from the college to the lake.

2. Draw a map of the route from your house to your school showing important landmarks.
[Note: Friends, do activity 1 and 2 by yourself. Enjoy drawing and colouring.]

Do This : [Textbook Page No. 165-166]

1. Tabulate the number of faces, edges and vertices for the following polyhedrons: (Here ‘V’ stands for number of vertices, ‘F’ stands for number of faces and ‘E’ stands for number of edges.)

Solid	F	V	E	F + V	E + 2
Cuboid
Triangular pyramid
Triangular prism
Pyramid with square base
Prism with square base

What do you infer from the last two columns ? In each case, do you find F + V = E + 2, i.e., F + V – E = 2? This relationship is called Euler’s formula. Infact this formula is true for any polyhedron.
Solution:

Solid	F	V	E	F + V	E + 2
Cuboid	6	8	12	6 + 8 =14	12 + 2 = 14
Triangular pyramid	4	4	6	4 + 4 = 8	6 + 2 = 8
Triangular prism	5	6	9	5 + 6 =11	9 + 2 =11
Pyramid with square base	5	5	8	5 + 5 = 10	8 + 2 = 10
Prism with square base	6	8	12	6 + 8 = 14	12 + 2 = 14

Yes, in each case Euler’s formula F + V = E + 2 or F + V – E = 2 is true.

Think, Discuss and Write : [Textbook Page No. 166]

1. What happens to F, V and E if some parts are sliced off from a solid ? (To start with, you may take a plasticine cube, cut a corner off and investigate.)
Solution:

Suppose ABCDEFGH is a cube. It has 6 faces, 8 vertices and 12 edges.
V = 8, F = 6 and E = 12
∴ V + F – E
= 8 + 6 – 12 = 2
So the Euler’s formula is verified.

Now, suppose we sliced off ∆ PQR from this cube, then F = 6 + 1 = 7
V = 8 – 1 + 3
= 7 + 3 = 10
E = 12 + 3 = 15
∴ V + F – E = 10 + 7 – 15 = 2
∴ Here also Euler’s formula is verified. Thus, if some parts are sliced off a solid, then the number of vertices, edges and faces will be changed but still the Euler’s formula is verified.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 10 Visualising Solid Shapes Ex 10.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.3

1. Can a polyhedron have for its faces:

Question (i)
3 triangles?
Solution:
[Note: Polyhedron is a solid, which is made by polygonal regions and these polygonal regions are called its faces.]
No, a polyhedron cannot have 3 triangles for its faces because it have atleast 4 faces.

Question (ii)
4 triangles?
Solution:
Yes, a polyhedron can have 4 triangles for its faces as triangular pyramid.

Question (iii)
a square and four triangles?
Solution:
Yes, a polyhedron can have a square and four triangles for its faces as a pyramid with square base.

2 Is it possible to have a polyhedron with any given number of faces? (Hint: Think of a pyramid.)
Solution:
Yes, it can be possible only if the number of faces is greater than or equal to 4, because a polyhedron has atleast 4 faces.

3. Which are prisms among the following?

Solution:
Note: A prism is a polyhedron whose base and top are congruent polygons and lateral faces are rectangles.
(i) No, a nail is not a prism because its base and top are not congruent polygons.
(ii) Yes, an unsharpened pencil is a prism because its base and top are congruent polygons and faces are rectangles.
(iii) No, a table weight is not a prism because its top and base are not congruent polygons.
(iv) Yes, box is a prism because its base and top are congruent polygons and lateral faces are rectangles.

4.

Question (i)
How are prisms and cylinders alike?
Solution:
Top and base of a prism and of a cylinder are congruent and parallel to each other and a prism becomes a cylinder, if the number of sides of its base becomes larger and larger.

Question (ii)
How are pyramids and cones alike?
Solution:
The pyramids and cones are alike because their lateral faces meet at a vertex. Also, a pyramid becomes a cone if the number of sides of its base becomes larger and larger.

5. Is a square prism same as a cube? Explain.
Solution:
No, a square prism is not same as a cube. It can be a cuboid also.

6. Verify Euler’s formula for these solids:

Solution:
(i) For figure (i)
F = 7, V= 10 and E = 15
∴ F + V = 7 + 10 = 17
Now, F + V – E = 17 – 15 = 2
Thus, F + V – E = 2
Hence, Euler’s formula is verified.

(ii) For figure (ii)
F = 9, V = 9 and E = 16
∴ F + V = 9 + 9 = 18
Now, F + V – E = 18 – 16 = 2
Thus, F + V – E = 2
Hence, Euler’s formula is verified.

7. Using Euler’s formula find the unknown:

	(i)	(ii)	(iii)
Faces	?	5	20
Vertices	6	?	12
Edges	12	9	?

Solution:
(i) Here, F = ?, V = 6 and E = 12
Now, F + V – E = 2 (∵ Euler’s formula)
∴ F + 6 – 12 = 2
∴ F – 6 = 2
∴ F = 2 + 6
∴ F = 8

(ii) Here, F = 5, V = ? and E = 9
Now, F + V – E = 2 (∵ Euler’s formula)
∴ 5 + V – 9 = 2
∴ V – 4 = 2
∴ V = 2 + 4
∴ V = 6

(iii) Here, F = 20, V = 12 and E = ?
Now, F + V – E = 2 (∵ Euler’s formula)
∴ 20 + 12 – E = 2
∴ 32 – E = 2
∴ – E = 2 – 32
∴ – E = – 30
∴ E = 30

8. Can a polyhedron have 10 faces, 20 edges and 15 vertices?
Solution:
Here, F = 10, E = 20 and V = 15
By Euler’s formula, F + V – E = 2
LHS = F + V – E
= 10 + 15 – 20
= 25 – 20 = 5
Thus, F + V – E ≠ 2
Hence, such a polyhedron is not possible.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 10 Visualising Solid Shapes Ex 10.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.2

1. Look at the given map of a city:

Answer the following:
(a) Colour the map as follows: Blue – Water, Red – Fire Station, Orange – Library, Yellow – Schools, Green – Park, Pink – College, Purple – Hospital, Brown – Cemetery.
(b) Mark a Green ‘X’ at the intersection of Road ‘C’ and Nehru Road, Green *Y’ at the intersection of Gandhi Road and Road A.
(c) In red, draw a short street route from Library to the bus depot.
(d) Which is further east, the city park or the market?
(e) Which is further south, the primary school or the Sr. Secondary School?
Solution:
[Note: Enjoy (a) to (c) by doing oneself. (d) city park (e) Sr. Secondary School]

2. Draw a map of your classroom using proper scale and symbols for different objects.

3. Draw a map of your school compound using proper scale and symbols for various features like playground, main building, garden, etc.

4. Draw a map giving instructions to your friend so that she reaches your house without any difficulty.

[Note: Enjoy while doing 2, 3, 4 by yourself.]

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 10 Visualising Solid Shapes Ex 10.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.1

1. For each of the given solid, the two views are given. Match for each solid the corresponding top and front views: The first one is done for you.


Solution:
(a) → (iii) → (iv)
(b) → (i) → (v), (c) → (iv) → (ii)
(d) → (v) → (iii), (e) → (ii) → (i)

2. For each of the given solid, the three views are given. Identify for each solid the corresponding top, front and side views.


Solution:
(a)

Object	(i)	(ii)	(iii)
View	Front view	Side view	Top view

(b)

Object	(i)	(ii)	(iii)
View	Side view	Front view	Top view

(c)

Object	(i)	(ii)	(iii)
View	Front view	Side view	Top view

(d)

Object	(i)	(ii)	(iii)
View	Front view	Side view	Top view

3. For each given solid, identify the top view, front view and side view:


Solution:
(a)

Object	(i)	(ii)	(iii)
View	Top view	Front view	Side view

(b)

Object	(i)	(ii)	(iii)
View	Side view	Front view	Top view

(c)

Object	(i)	(ii)	(iii)
View	Top view	Side view	Front view

(d)

Object	(i)	(ii)	(iii)
View	Side view	Front view	Top view

(e)

Object	(i)	(ii)	(iii)
View	Front view	Top view	Side view

4. Draw the front view, side view and top view of the given objects:

Solution: