Class 8

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 9 Algebraic Expressions and Identities Ex 9.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3

1. Carry out the multiplication of the expressions in each of the following pairs:

Question (i)
4p, q + r
Solution:
= 4p × (q + r)
= (4p × q) + (4p × r)
= 4pq + 4pr

Question (ii)
ab, a – b
Solution:
= ab × (a-b)
= (ab × a) – (ab × b)
= a²b – ab²

Question (iii)
a + b, 7a²b²
Solution:
= (a + b) × 7a²b²
= (a × 7a²b²) + (b × 7a²b²)
= 7 a³b² + 7a²b³

Question (iv)
a² – 9, 4a
Solution:
= (a² – 9) × 4a
= (a² × 4a) – (9 × 4a)
= 4a³ – 36a

Question (v)
pq + qr + rp, 0
Solution:
= (pq + qr + rp) × 0
= 0

2. Complete the table:

First expression	Second expression	Product
1. a	b + c + d	……………
2. x + y- 5	5xy	……………
3. P	6p2 – 7p + 5	…………..
4. 4p2q- q2	p2 – q2	…………..
5. a + b + c	abc	………….

Solution:
(i) a × (b + c + d)
= (a × b) + (a × c) + (a × d)
= ab + ac + ad

(ii) (x + y – 5) × 5xy
= (x × 5xy) + (y × 5xy) + [(- 5) × 5xy]
= 5x²y + 5xy² – 25xy

(iii) p × (6p² – 7p + 5)
= (p × 6p²) + [p × (- 7p)] + (p × 5)
= 6p³ – 7p² + 5p

(iv) 4p²q² × (p² – q²)
= (4p²q² × p²) + [4p²q² × (-q²)]
= 4p⁴q² – 4p²q⁴

(v) (a + b + c) × abc
= (a × abc) + (b × abc) + (c × abc)
= a²bc + ab²c + abc²

3. Find the product:

Question (i)
(a²) × (2a²²) × (4a²⁶)
Solution:
= (1 × 2 × 4) × a² × a²² × a²⁶
= 8 × a⁵⁰
= 8a⁵⁰

Question (ii)
(\(\frac {2}{3}\)xy) × (\(\frac {-9}{10}\)x²y²)
Solution:
= \(\frac {2}{3}\) × (\(\frac {-9}{10}\)) × xy × x²y²
= \(\frac {-2}{3}\) × \(\frac {9}{10}\) × x³y³
= \(\frac {-3}{5}\)x³y³

Question (iii)
(\(\frac {-10}{3}\)pq³) × (\(\frac {6}{5}\)p³q)
Solution:
= [(\(\frac {-10}{3}\)) × \(\frac {6}{5}\)] × pq³ × p³q
= – \(\frac {10}{3}\) × \(\frac {6}{5}\) × p⁴q⁴
= – 4p⁴q⁴

Question (iv)
x × x² × x³ × x⁴
Solution:
= (1 × 1 × 1 × 1) × x × x² × x³ × x⁴
= (1) × x¹⁰
= x¹⁰

4.

Question (a)
Simplify 3x(4x – 5) + 3 and find its value for
(i) x = 3
(ii) x = \(\frac {1}{2}\)
Solution:
3x (4x- 5) + 3
= (3x × 4x) – (3x × 5) + 3
= 12x¹⁰ – 15x + 3

(i) When x = 3, then
12x² – 15x + 3
= 12 (3)² – 15(3) + 3
= 12 (9) – 15 (3) + 3
= 108 -45 + 3
= 111 – 45
= 66

(ii) x = \(\frac {1}{2}\), then
12x² – 15x + 3
= 12(\(\frac {1}{2}\))² – 15(\(\frac {1}{2}\)) + 3
= 12(\(\frac {1}{4}\)) – 15(\(\frac {1}{2}\)) + 3
= 3 – \(\frac {15}{2}\) + 3
= 6 – \(\frac {15}{2}\)
= \(\frac{12-15}{2}\)
= \(\frac {-3}{2}\)

Question (b)
Simplify a (a2 + a + 1) + 5 and find its values for
(i) a = 0 (ii) a = 1 (iii) a = (-1)
Solution:
a (a² + a + 1) + 5
= (a × a²) + (a × a) + (a × 1) + 5
= a³ + a² + a + 5

(i) When a = 0, then
a³ + a² + a + 5
= (-1)³ + (0)³ + (0) + 5
= 0 + 0 + 0 + 5
= 5

(ii) a = 1, then
a³ + a² + a + 5
= (1)³ + (1)² + (1) + 5
= 1 + 1 + 1 + 5
= 8

(iii) a = (-1), then
a³ + a² + a + 5
= (-1)³ + (-1)² + (-1) + 5
= (-1) + (1) + (-1) + 5
= 6 – 2 = 4

5.

Question (a)
Add : p(p – q), q(q – r) and r(r – p)
Solution:
[p(p-q)] + [q(q-r)] + [r(r-p)]
= p² – pq + q² – qr + r² – rp
= p² + q² + r² – pq – qr – rp

Question (b)
Add : 2x(z – x – y) and 2y(z – y – x)
Solution:
[2x (z – x – y)] + [2y (z – y – x)]
= 2xz – 2x² – 2xy + 2yz – 2y² – 2xy
= – 2x² – 2y² – 2xy – 2xy + 2yz + 2xz
= – 2x² – 2y² – 4xy + 2yz + 2xz

Question (c)
Subtract : 31(l – 4m + 5n) from 4l(10n – 3m + 2l)
Solution:
[4l(10n – 3m + 21)] – [3l(l – 4m + 5n)]
= [40ln – 12lm + 8l²] – [3l² – 12lm + 15ln]
= 40ln – 12lm + 8l² – 3l² + 12lm – 15ln
= 40ln – 15ln – 12lm + 12lm + 8l² – 3l²
= 25ln + 0lm + 5l²
= 25ln + 5l²

Question (d)
Subtract : 3a(a + b + c) – 2b(a – b + c) from 4c(- a + b + c)
Solution:
[4c (- a + b + c)] – [3a (a + b + c) – 2b (a – b + c)]
= [- 4ac + 4bc + 4c²]
– [3a² + 3ab + 3ac – 2ab + 2b² – 2bc]
= – 4ac + 4be + 4c² – 3a² – 3ab – 3ac + 2ab – 2b² + 2bc
= – 3a² – 2b² + 4c² – 3ab + 2ab + 4bc + 2bc – 4ac – 3ac
= – 3a² – 2b² + 4c² – ab + 6bc – 7ac

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 9 Algebraic Expressions and Identities InText Questions and Answers.

PSEB 8th Class Maths Solutions Chapter 9 Algebraic Expressions and Identities InText Questions

Try These: [Textbook Page No. 139]

1. Write two terms which are like:

Question (i)
7xy
Solution:
14xy, 21xy

Question (ii)
4mn²
Solution:
8mn² , – 11mn²

Question (iii)
21
Solution:
– 5l, 9l

Try These : [Textbook Page No. 142]

1. Can you think of two more such situations, where we may need to multiply algebraic expressions?
Solution:
1. Aarush purchased x notebooks and y pens. If cost of a notebook and a pen is same ₹ z, what amount has he to pay? → ₹ z (x + y)
2. Shailja wants to spread a carpet in her room having length (l + 5) m and breadth (b – 2) m. Find the area of the carpet. → (l + 5) (b – 2) m²

Try These: [Textbook Page No. 143]

1. Find 4x × 5y × 7z.
Solution:
4x × 5y × 7z = 4 × 5 × 7 × x × y × z
= 140 xyz

2. Find 4x × 5y × 7z. First find (4x × 5y) and multiply it by 7z; or first find (5y × 7z) and multiply it by 4x. Is the result the same? What do you observe? Does the order in which you carry out the multiplication matter?
Solution:
(4x × 5 y) = 4 × 5 × x × y
= 20xy
Now, 20xy × 7z = 20 × 7 × xy × z
= 140xyz … (i)
Also, (5y × 7z) = 5 × 7 × y × z = 35 yz
Now, 35yz × 4x = 35 × 4 × yz × x
= 140xyz … (ii)
Yes, the result is same.
We can conclude that product remains same if we change order of the terms.

3. Complete the table for area of a rectangle with given length and breadth.
Solution:

length	breadth	area
3x	5y	3x × 5y = 15xy
9y	4y2	9y × 4y2 = 36y3
4ab	5bc	4ab × 5be = 20ab2c
2l2m	3lm2	2l2m × 3lm2 = 6l3m3

Try These : [Textbook Page No. 144]

1. Find the product:

Question (i)
2x (3x + 5xy)
Solution:
= (2x × 3x) + (2x × 5xy)
= 6x² + 10x²y

Question (ii)
a² (2ab – 5c)
Solution:
= (a² × 2ab) – (a² × 5c)
= 2a³b – 5a²c

Try These: [Textbook Page No. 145]

1. Find the product: (4p² + 5p + 7) × 3p
Solution:
(4p² + 5p + 7) × 3p
= (4p² × 3p) + (5p × 3p) + (7 × 3p)
= 12p³ + 15p² + 21p

Try These : [Textbook Page No. 149]

1. Put -b in place of b in Identity (I). Do you get Identity (II)?
Solution:
Identity (I): (a + b)² = a² + 2ab + b²
Let us put (- b) instead of b [a + (- b)]²
= a² + 2a (- b) + (- b)²
∴ (a – b)²
= a² – 2ab + b²
Identity (II): (a – b)² = a² – 2ab + b²
Yes, we get Identity (II).

Try These : [Textbook Page No. 149]

1. Verify Identity (IV), for a = 2, b = 3, x = 5.
Solution:
Identity (IV):
(x + a) (x + b) = x² + (a + b) x + ab
Substitute a = 2, b = 3 and x = 5
LHS
= (x + a) (x + b)
= (5 + 2) (5 + 3)
= (7)(8)
= 56

RHS
= x² + (a + b) x + ab
= (5)² + (2 + 3) × 5 + (2 × 3)
= 25 + (5) × 5 + (6)
= 25 + 25 + 6 = 56
∴ LHS = RHS
∴ The given identity is true for the given values.

2. Consider, the special case of Identity (IV) with a = b, what do you get ? Is it related to Identity (I)?
Solution:
When a = b (∴ Take y for both)
(x + a) (x + b) = x² + (a + b) x + ab
Substitute a = y and b = y
(x + y)(x + y) = x² + (y + y)x + (y × y)
= x² + (2y) x + (y × y) = x² + 2xy + y²
∴ Yes, it is the same as Identity ( I).

3. Consider, the special case of Identity (IV) with a = -c and b = -c. What do you get ? Is it related to Identity (II) ?
Solution:
Identity (IV):
(x + a)(x + b) = x² + (a + b) x + ab
Substitute (- c) instead of a and (- c) instead of b,
(x – c) (x – c)
= x² + [(-c) + (-c)]x + [(-c) × (-c)]
= x² + [- 2c] x + (c²)
= x² – 2cx + c²
∴ Yes, it is the same as Identity (II).

4. Consider the special case of Identity (IV) with b = – a. What do you get ? Is it related to Identity (III)?
Solution :
Identity (IV):
(x + a) (x + b) = x² + (a + b) x + ab
Substitute (-a) instead of b,
(x + a) (x – a)
= x² + [a + (- a)] x + [a × (- a)]
= x² + (a – a) x + [- a²]
= x² + (0) x – a²
= x² – a²
∴ Yes, it is the same as Identity (III).

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 7 Cubes and Cube Roots Ex 7.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots Ex 7.1

1. Which of the following numbers are not perfect cubes ?

Question (i).
216
Solution:
216
\(\begin{array}{l|l}
2 & 216 \\
\hline 2 & 108 \\
\hline 2 & 54 \\
\hline 3 & 27 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)
216 = 2 × 2× 2 × 3 × 3 × 3
Here, the prime factors 2 and 3 appear in a group of three (triples).
∴ 216 is a perfect cube.
216 = 2³ × 3³

Question (ii).
128
Solution:
\(\begin{array}{l|l}
2 & 128 \\
\hline 2 & 64 \\
\hline 2 & 32 \\
\hline 2 & 16 \\
\hline 2 & 8 \\
\hline 2 & 4 \\
\hline 2 & 2 \\
\hline & 1
\end{array}\)
128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
Here, the prime factor 2 does not appear in a group of three.
∴ 128 is not a perfect cube.

Question (iii).
1000
Solution:
\(\begin{array}{l|l}
2 & 1000 \\
\hline 2 & 500 \\
\hline 2 & 250 \\
\hline 5 & 125 \\
\hline 5 & 25 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)
1000 = 2 × 2 × 2 × 5 × 5 × 5
Here, the prime factors 2 and 5 appear in a group of three.
∴ 1000 is a perfect cube.
1000 = 2³ × 5³

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

Question (v).
46656
Solution:
\(\begin{array}{l|l}
2 & 46656 \\
\hline 2 & 23328 \\
\hline 2 & 11664 \\
\hline 2 & 5832 \\
\hline 2 & 2916 \\
\hline 2 & 1458 \\
\hline 3 & 729 \\
\hline 3 & 243 \\
\hline 3 & 81 \\
\hline 3 & 27 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)
46656 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3
Here, the prime factors 2 and 3 appear in a group of three.
∴ 46656 is a perfect cube.
46656 = 2³ × 2³ × 3³ × 3³

2. Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube:

Question (i).
243
Solution:
\(\begin{array}{l|l}
3 & 243 \\
\hline 3 & 81 \\
\hline 3 & 27 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)
243 = 3 × 3 × 3 × 3 × 3
Here, the prime factor 3 does not appear in a group of three.
∴ 243 is not a perfect cube.
(243) × 3 = (3 × 3 × 3 × 3 × 3) × 3
∴ 729 = 3³ × 3³
Thus, 3 is the smallest number by which 243 must be multiplied to get a perfect cube.

Question (ii).
256
Solution:
\(\begin{array}{l|l}
2 & 256 \\
\hline 2 & 128 \\
\hline 2 & 64 \\
\hline 2 & 32 \\
\hline 2 & 16 \\
\hline 2 & 8 \\
\hline 2 & 4 \\
\hline 2 & 2 \\
\hline & 1
\end{array}\)
256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
Here, the prime factor 2 does not appear in a group of three.
∴ 256 is not a perfect cube.
(256) × 2
= (2 × 2 × 2 × 2 × 2 × 2 × 2 × 2) × 2
∴ 512 = 2³ × 2³ × 2³
Thus, 2 is the smallest number by which 256 must be multiplied to get a perfect cube.

Question (iii).
72
Solution:
\(\begin{array}{l|l}
2 & 72 \\
\hline 2 & 36 \\
\hline 2 & 18 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)
72 = 2 × 2 × 2 × 3 × 3
Here, the prime factor 3 does not appear in a group of three.
∴ 72 is not a perfect cube.
(72) × 3
= (2 × 2 × 2 × 3 × 3) × 3
∴ 216 = 2³ × 3³
Thus, 3 is the smallest number by which 72 must be multiplied to get a perfect cube.

Question (iv).
675
Solution:
\(\begin{array}{l|l}
3 & 675 \\
\hline 3 & 225 \\
\hline 3 & 75 \\
\hline 5 & 25 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)
∴ 675 = 3 × 3 × 3 × 5 × 5
Here, the prime factor 5 does not appear in a group of three.
∴ 675 is not a perfect cube.
(675) × 5
= (3 × 3 × 3 × 5 × 5) × 5
∴ 3375 = 3³ × 5³
Thus, 5 is the smallest number by which 675 must be multiplied to get a perfect cube.

Question (v).
100
Solution:
\(\begin{array}{l|l}
2 & 100 \\
\hline 2 & 50 \\
\hline 5 & 25 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)
100 = 2 × 2 × 5 × 5
Here, the prime factors 2 and 5 do not appear in a group of three.
∴ 100 is not a perfect cube.
(100) × 2 × 5 = (2 × 2 × 5 × 5) × 2 × 5
∴ 1000 = 2³ × 5³
Thus, 10 is the smallest number by which 100 must be multiplied to get a perfect cube.

3. Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube:

Question (i).
81
Solution:
\(\begin{array}{l|l}
3 & 81 \\
\hline 3 & 27 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)
81 = 3 × 3 × 3 × 3
Here, the prime factor 3 is left over after making triples of 3.
∴ 81 is not a perfect cube.
(81) ÷ 3 = (3 × 3 × 3 × 3) ÷ 3
∴ 27 = 3 × 3 × 3 = 33
Thus, 3 is the smallest number by which 81 must be divided to get a perfect cube.

Question (ii).
128
Solution:
\(\begin{array}{l|l}
2 & 128 \\
\hline 2 & 64 \\
\hline 2 & 32 \\
\hline 2 & 16 \\
\hline 2 & 8 \\
\hline 2 & 4 \\
\hline 2 & 2 \\
\hline & 1
\end{array}\)
128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
Here, the prime factor 2 is left over after making triples of 2.
∴ 128 is not a perfect cube.
(128) ÷ 2
= (2 × 2 × 2 × 2 × 2 × 2 × 2) ÷ 2
∴ 64 = 2 × 2 × 2 × 2 × 2 × 2
= 2³ × 2³
Thus, 2 is the smallest number by which 128 must be divided to get a perfect cube.

Question (iii).
135
Solution:
\(\begin{array}{l|l}
3 & 135 \\
\hline 3 & 45 \\
\hline 3 & 15 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)
135 = 3 × 3 × 3 × 5
Here, the prime factor 5 is left over after making triples of 3.
∴ 135 is not a perfect cube.
(135) ÷ 5 = (3 × 3 × 3 × 5) ÷ 5
∴ 27 = 3 × 3 × 3
= 3³
Thus, 5 is the smallest number by which 135 must be divided to get a perfect cube.

Question (iv).
192
Solution:
\(\begin{array}{l|l}
2 & 192 \\
\hline 2 & 96 \\
\hline 2 & 48 \\
\hline 2 & 24 \\
\hline 2 & 12 \\
\hline 2 & 6 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)
192 = 2 × 2 × 2 × 2 × 2 × 2 × 3
Here, the prime factor 3 is left over after making triples of 2.
∴ 192 is not a perfect cube.
(192) ÷ 3
= (2 × 2 × 2 × 2 × 2 × 2 × 3) ÷ 3
∴ 64 = 2 × 2 × 2 × 2 × 2 × 2
= 2³ × 2³
Thus, 3 is the smallest number by which 192 must be divided to get a perfect cube.

Question (v).
704
Solution:
\(\begin{array}{l|l}
2 & 704 \\
\hline 2 & 352 \\
\hline 2 & 176 \\
\hline 2 & 88 \\
\hline 2 & 44 \\
\hline 2 & 22 \\
\hline 11 & 11 \\
\hline & 1
\end{array}\)
704 = 2 × 2 × 2 × 2 × 2 × 2 × 11
Here, the prime factor 11 is left over after making triples of 2.
∴ 704 is not a perfect cube.
(704) ÷ 11
= (2 × 2 × 2 × 2 × 2 × 2 × 11) ÷ 11
∴ 64 = 2 × 2 × 2 × 2 × 2 × 2
= 2³ × 2³
Thus, 11 is the smallest number by which 704 must be divided to get a perfect cube.

4. Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube ?
Solution:
Sides of the cuboid are 5 cm, 2 cm and 5 cm (Given).
∴ Volume of the cuboid = l × b × h
= 5 cm × 2 cm × 5 cm
To form it as a cube, its dimensions should be in the group of triplets.
Here, the prime factors 2 and 5 are not in group of three.
∴ Volume of the required cube
= (5 cm × 5 cm × 2 cm) × 5 cm × 2 cm × 2 cm
= (5³ × 2³) cm³
Thus, Parikshit needed 5 × 2 × 2 = 20 cuboids to form a cube.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 9 Algebraic Expressions and Identities Ex 9.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.5

1. Use a suitable identity to get each of the following products:

Question (i)
(x + 3) (x + 3)
Solution:
= (x + 3)²
= (x)² + 2(x)(3) + (3)²
[∵ (a + b)² = a² + 2ab + b²]
= x² + 6x + 9

Question (ii)
(2y + 5) (2y + 5)
Solution:
= (2y + 5)²
= (2y)² + 2 (2y)(5) + (5)²
[∵ (a – b)² = a² – 2ab + b²]
= 4y² + 20y + 25

Question (iii)
(2a – 7) (2a – 7)
Solution:
= (2a – 7)²
= (2a)² – 2(2a)(7) + (7)²
[∵ (a – b)² = a² – 2ab + b²]
= 4a² – 28a + 49

Question (iv)
(3a – \(\frac {1}{2}\))(3a – \(\frac {1}{2}\))
Solution:
= (3a – \(\frac {1}{2}\))²
= (3a)² – 2(3a)(\(\frac {1}{2}\)) + (\(\frac {1}{2}\))²
[∵ (a – b)² = a² – 2ab + b²]
= 9a² – 3a + \(\frac {1}{4}\)

Question (v)
(1.1m – 0.4) (1.1m + 0.4)
Solution:
= (1.1m)² – (0.4)²
[∵ (a + b) (a – b) = a² – b²]
= 1.21m² – 0.16

Question (vi)
(a² + b²) (-a² + b²)
Solution:
= (b² + a²) (b² – a²)
= (b²)² – (a²)²
[∵ (a + b) (a – b) = a² – b²]
= b⁴ – a⁴

Question (vii)
(6x – 7) (6x + 7)
Solution:
= (6x)² – (7)²
[∵ (a + b) (a – b) = a² – b²]
= 36x² – 49

Question (viii)
(-a + c) (-a + c)
Solution:
= (-a + c)²
= (-a)² + 2 (-a) (c) + (c)²
[∵ (a + b)² = a² + 2ab + b²]
= a² – 2ac + c²

Question (ix)
(\(\frac{x}{2}+\frac{3 y}{4}\))
Solution:
= (\(\frac{x}{2}+\frac{3 y}{4}\))²
= (\(\frac {x}{2}\))² + 2(\(\frac {x}{2}\))(\(\frac {3y}{4}\)) + (\(\frac {3y}{4}\))²
[∵ (a + b)² = a² + 2ab + b²]
= \(\frac{x^{2}}{4}+\frac{3 x y}{4}+\frac{9 y^{2}}{16}\)

Question (x)
(7a – 9b) (7a – 9b)
Solution:
= (7a – 9b)²
= (7a)² – 2(7a)(9b) + (9b)²
[∵ (a – b)² = a² – 2ab + bsup>2]
= 49a² – 126ab + 81b²

2. Use the identity (x + a) (x + b) = x² + (a + b) x + ab to find the following products:

Question (i)
(x + 3) (x + 7)
Solution:
Identity : (x + a) (x + b) = x² + (a + b) x + ab
= (x)² + (3 + 7)x + (3) (7)
= x² + (10) x + 21
= x² + 10x + 21

Question (ii)
(4x + 5) (4x + 1)
Solution:
= (4x)² + (5 + 1) 4x + (5)(1)
= 16x² + (6) 4x + 5
= 16x² + 24x + 5

Question (iii)
(4x – 5) (4x – 1)
Solution:
= (4x)² + (- 5 – 1) 4x + (- 5) (- 1)
= 16x² + (- 6) 4x + 5
= 16x² – 24x + 5

Question (iv)
(4x + 5) (4x- 1)
Solution:
= (4x)² + (5 – 1) 4x + (5) (- 1)
= 16x² + (4) 4x – 5
= 16x² + 16x – 5

Question (v)
(2x + 5y) (2x + 3y)
Solution:
= (2x)² + (5y + 3y) 2x + (5y) (3y)
= 4x² + (8y) 2x + 15y²
= 4x² + 16xy + 15y²

Question (vi)
(2a² + 9) (2a² + 5)
Solution:
= (2a²)² + (9 + 5) 2a² + (9)(5)
= 4a⁴ + (14)2a² + 45
= 4a⁴ + 28a² + 45

Question (vii)
(xyz – 4) (xyz – 2)
Solution:
= (xyz)² + (- 4 – 2) xyz + (- 4)(- 2)
= x²y²z² + (- 6) xyz + 8
= x²y²z² – 6xyz + 8

3. Find the following squares by using the identities:

Question (i)
(b – 7)²
Solution:
= (b)² – 2 (b)(7) + (7)²
[∵ (a – b)² = a² – 2ab + b²]
= b² – 14 b + 49

Question (ii)
(xy + 3z)²
Solution:
= (xy)² + 2 (xy)(3z) + (3z)²
[∵ (a + b)² = a² + 2ab + b²]
= x²y² + 6xyz + 9z²

Question (iii)
(6x² – 5y)²
Solution:
= (6x²)² – 2 (6x²) (5y) + (5y)²
[∵ (a – b)² = a² – 2ab + b²]
= 36x⁴ – 60x²y + 25y²

Question (iv)
(\(\frac {2}{3}\)m + \(\frac {3}{2}\)n)²
Solution:
= (\(\frac {2}{3}\)m)² + 2(\(\frac {2}{3}\)m)(\(\frac {3}{2}\)n) + (\(\frac {3}{2}\)n)²
[∵ (a + b)² = a² + 2ab + b²]
= \(\frac {4}{9}\)m² + 2mn + \(\frac {9}{4}\)n²

Question (v)
(0.4p – 0.5q)²
Solution:
= (0.4p)² – 2 (0.4p)(0.5q) + (0.5q)²
[∵ (a – b)² = a² – 2ab + b²]
= 0.16p² – 0.4pq + 0.25q²

Question (vi)
(2xy + 5y)²
Solution:
= (2xy)² + 2 (2xy)(5y) + (5y)²
[∵ (a + b)² = a² + 2ab + b²]
= 4x²y² + 20xy² + 25 y²

4. Simplify:

Question (i)
(a² – b²)²
Solution:
= (a²)² – 2(a²)(b²) + (b²)²
= a⁴ – 2a²b² + b⁴

Question (ii)
(2x + 5)² – (2x – 5)²
Solution:
= [(2x)² + 2(2x)(5) + (5)²] – [(2x)² – 2 (2x)(5) + (5)²]
= [4x² + 20x + 25] – [4x² – 20x + 25]
= 4x² + 20x + 25 – 4x² + 20x – 25
= 4x² – 4x² + 20x + 20x + 25 – 25
= 40x

Question (iii)
(7m – 8n)² + (7m + 8n)²
Solution:
= [(7m)² – 2(7m)(8n) + (8n)²] + [(7m)² + 2 (7m)(8n) + (8n)²]
= [49m² – 112mn + 64n²] + [49m² + 112mn + 64n²]
= 49m² – 112mn + 64n² + 49m² + 112mn + 64n²
= 49m² + 49m² – 112mn + 112mn + 64n² + 64n²
= 98m² + 128n²

Question (iv)
(4m + 5n)² + (5m + 4n)²
Solution:
= [(4m)² + 2 (4m)(5n) + (5n)²] + [(5m)² + 2 (5m)(4n) + (4n)²]
= 16m² + 40mn + 25n² + 25m² + 40mn + 16n²
= 16m² + 25m² + 40mn + 40 mn + 25n² + 16n²
= 41m² + 80mn + 41n²

Question (v)
(2.5p – 1.5q)² – (1.5p – 2.5q)²
Solution:
= [(2.5p)² – 2 (2.5p)(1.5q) + (1.5q)²] – [(1.5p)² – 2 (1.5p)(2.5q) + (2.5q)²]
= [6.25p² – 7.5pq + 2.25q²] – [2.25p² – 7.5pq + 6.25q²]
= 6.25p² – 7.5pq + 2.25q² – 2.25p² + 7.5pq – 6.25q²
= 6.25p² – 2.25p² – 7.5pq + 7.5pq + 2.25q² – 6.25q²
= 4p² – 4q²

Question (vi)
(ab + bc)² – 2ab²c
Solution:
= [(ab)² + 2 (ab)(bc) + (bc)²] – 2ab²c
= a²b² + 2ab²c + b²c² – 2ab²c
= a²b² + 2ab²c – 2ab²c + b²c²
= a²b² + b²c²

Question (vii)
(m² – n²m)² + 2m³n²
Solution:
= [(m²)² – 2 (m²)(n²m)² + (n²m)²] + 2m³n²
= m⁴ – 2 m³n² + n⁴m² + 2 m³ n²
= m⁴ – 2 m³n² + 2 m³n² + n⁴m²
= m⁴ + m²n⁴

5. Show that:

Question (i)
(3x + 7)² – 84x = (3x – 7)²
Solution:
LHS = (3x + 7)² – 84x
= (3x)² + 2(3x)(7) + (7)² – 84x
= 9x² + 42x + 49 – 84x
= 9x² + 42x – 84x + 49
= 9x² – 42x + 49

RHS = (3x – 7)²
= (3x)² – 2(3x)(7) + (7)²
= 9x² – 42x + 49
Thus, LHS = RHS
∴ (3x + 7)² – 84x = (3x – 7)²

Question (ii)
(9p – 5q)² + 180pq = (9p + 5q)²
Solution:
LHS = (9p – 5q)² + 180pq
= (9p)² – 2(9p)(5q) + (5q)² + 180pq
= 81p² – 90pq + 25q² + 180pq
= 81p² – 90pq + 180pq + 25q²
= 81p² + 90pq + 25q²

RHS = (9p + 5q)²
= (9p)² + 2(9p)(5q) + (5q)²
= 81p² + 90pq + 25q²
Thus, LHS = RHS
∴ (9p – 5q)² + 180pq = (9p + 5q)²

Question (iii)
(\(\frac {4}{3}\)m – \(\frac {3}{4}\)n)² + 2mn = \(\frac {16}{9}\)m² + \(\frac {9}{16}\)n²
Solution:
LHS = [\(\frac {4}{3}\)m – \(\frac {3}{4}\)n]² + 2mn
= [(\(\frac {4}{3}\)m)² – 2(\(\frac {4}{3}\)m)(\(\frac {3}{4}\)n) + (\(\frac {3}{4}\)n)²] + 2mn
= \(\frac {16}{9}\)m² – 2mn + \(\frac {9}{16}\)n² + 2mn
= \(\frac {16}{9}\)m² – 2mn + 2mn + \(\frac {9}{16}\)n²
= \(\frac {16}{9}\)m² + \(\frac {9}{16}\)n² = RHS
Thus, LHS = RHS
∴ (\(\frac {4}{3}\)m – \(\frac {3}{4}\)n)² + 2mn = \(\frac {16}{9}\)m² + \(\frac {9}{16}\)n²

Question (iv)
(4pq + 3q)² – (4pq – 3q)² = 48pq²
Solution:
LHS = (4pq + 3q)² – (4pq – 3q)²
= [(4pq)² + 2 (4pq)(3q) + (3q)²] – [(4pq)² – 2 (4pq)(3q) + (3q)²]
= [16p²q² + 24pq² + 9q²] – [16p²q² – 24pq² + 9q²]
= 16p²q² + 24pq² + 9q² – 16p²q² + 24pq² – 9q²
= 16p²q² – 16p²q² + 24pq² + 24pq² + 9q² – 9q²
= 48pq² = RHS
Thus, LHS = RHS
∴ (4pq + 3q)² – (4pq – 3q)² = 48pq²

Question (v)
(a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0
Solution:
LHS = (a – b)(a + b) + (b – c)(b + c) + (c – a) (c + a)
= (a² – b²) + (b² – c²) + (c² – a²)
= a² – b² + b² – c² + c² – a²
= a² – a² + b² – b² + c² – c²
= 0 = RHS
Thus, LHS = RHS
∴ (a -b)(a + b) + (b- c) (b + c) + (c – a) (c + a) = 0

6. Using identities, evaluate:

Question (i)
71²
Solution:
= (70 + 1)²
= (70)² + 2 (70)(1) + (1)²
[∵ (a + b)² = a² + 2ab + b²]
= 4900 + 140 + 1
= 5041

Question (ii)
99²
Solution:
= (100 – 1)2
= (100)² – 2(100)(1) + (1)²
[∵ (a – b)² – a² – 2ab + b²]
= 10000 – 200 + 1
= 9801

Question (iii)
102²
Solution:
= (100 + 2)²
= (100)² + 2 (100)(2) + (2)²
[∵ (a + b)² = a² + 2ab + b²]
= 10000 + 400 + 4
= 10404

Question (iv)
998²
Solution:
= (1000 – 2)²
= (1000)² – 2 (1000)(2) + (2)²
[∵ (a – b)² = a² – 2ab + b²]
= 1000000 – 4000 + 4
= 996004

Question (v)
5.2²
Solution:
= (5 + 0.2)²
= (5)² + 2 (5)(0.2) + (0.2)²
[∵ (a + b)² = a² + 2ab + b²]
= 25 + 2 + 0.04
= 27 + 0.04
= 27.04

Question (vi)
297 × 303
Solution:
= (300 – 3) × (300 + 3)
= (300)² – (3)²
[∵ (a – b)(a + b) = a² – b²]
= 90000 – 9
= 89991

Question (vii)
78 × 82
Solution:
= (80 – 2) × (80 + 2)
= (80)² – (2)²
[∵ (a – b)(a + b) = a² – b²]
= 6400 – 4
= 6396

Question (viii)
8.9²
Solution:
= (9 – 0.1)²
= (9)² – 2(9)(0.1) + (0.1)²
[∵ (a – b)² = a² – 2ab + b²]
= 81 – 1.8 + 0.01
= 81.01 – 1.8
= 79.21

Question (ix)
10.5 × 9.5
Solution:
= (10 + 0.5) × (10 – 0.5)
= (10)² – (0.5)²
[∵ (a + b)(a – b) = a² – b²]
= 100 – 0.25
= 99.75

7. Using a² – b² = (a + b) (a – b), find:

Question (i)
51² – 49²
Solution:
= (51 + 49) (51 – 49)
= (100) × (2)
= 200

Question (ii)
(1.02)² – (0.98)²
Solution:
= (1.02 + 0.98) (1.02 – 0.98)
= (2.0) × (0.04)
= 0.08

Question (iii)
153² – 147²
Solution:
= (153 + 147) (153 – 147)
= (300) × (6)
= 1800

Question (iv)
12.1² – 7.9²
Solution:
= (12.1 + 7.9) (12.1 – 7.9)
= 20 × 4.2
= 84

8. Using (x + a)(x + b) = x² + (a + b) x + ab, find:

Question (i)
103 × 104
Solution:
= (100 + 3) × (100 + 4)
= (100)² + (3 + 4) × 100 + (3)(4)
= 10000 + 700 + 12
=10712

Question (ii)
5.1 × 5.2
Solution:
= (5 + 0.1) (5 + 0.2)
= (5)² + (0.1 + 0.2) × 5 + (0.1)(0.2)
= 25 + (0.3) × 5 + 0.02
= 25 + 1.5 + 0.02
= 26.52

Question (iii)
103 × 98
Solution:
= (100 + 3) (100-2)
= (100)² + (3 – 2) 100 + (3)(-2)
= 10000 + 100 – 6
= 10094

Question (iv)
9.7 × 9.8
Solution:
= (10 – 0.3) (10 – 0.2)
= (10)² + [(-0.3) + (-0.2)] 10 + (-0.3) (-0.2)
= 100 + [-0.5] × 10 + 0.06
= 100 – 5 + 0.06
= 95.06

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 8 Comparing Quantities Ex 8.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.1

1. Find the ratio of the following.

Question (a).
Speed of a cycle 15 km per hour to the speed of scooter 30 km per hour.
Solution:
Speed of a cycle = 15 km/h
Speed of a scooter = 30 km / h
∴ Ratio of the speed of a cycle to the speed of a scooter
= \(\frac{15 \mathrm{~km} / \mathrm{h}}{30 \mathrm{~km} / \mathrm{h}}\)
= \(\frac {1}{2}\)
= 1 : 2

Question (b).
5 m to 10 km
Solution:
[Note : Unit of both quantities should be same.]
1 km = 1000 m
∴ 10 km = 10 × 1000 m
= 10,000 m
∴ Ratio of 5 m to 10 km = \(\frac{5 \mathrm{~m}}{10 \mathrm{~km}}\)
= \(\frac{5 \mathrm{~m}}{10000 \mathrm{~m}}\)
= \(\frac{1}{2000}\)
= 1 : 2000

Question (c).
50 paise to ₹ 5
Solution:
[Note : Unit of both quantities should be same.]
₹ 1 = 100 paise
∴ ₹ 5 = 500 paise
∴ Ratio of 50 paise to ₹ 5 = \(\frac{50 \text { paise }}{₹ 5}\)
= \(\frac{50 \text { paise }}{500 \text { paise }}\)
= \(\frac{1}{10}\)
= 1 : 10

2. Convert the following ratios to percentages.

Question (a).
3 : 4
Solution:
Given ratio = 3 : 4
∴ Percentage = (\(\frac{3}{4}\) × 100) %
= (3 × 25) %
= 75 %

Question (b).
2 : 3
Solution:
Given ratio = 2 : 3
∴ Percentage = (\(\frac{2}{3}\) × 100) %
= (\(\frac{200}{3}\)) %
= 66 \(\frac{2}{3}\)%

3. 72% of 25 students are interested in Mathematics. How many are not interested in Mathematics?
Solution:
Total number of students = 25
Students interested in Mathematics = 72%
∴ Students who are not interested in Mathematics = (100 – 72) %
= 28 %
Number of students who are not interested in Mathematics = 28% of 25
= \(\frac{28}{100}\) × 25
= \(\frac{28}{4}\)
= 7
Thus, 7 students are not interested in Mathematics.

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 football team = 10
Let x matches be played by the team.
∴ 40% of x = 10
∴ \(\frac{40}{100}\) × x = 10
∴ x = \(\frac{10 \times 100}{40}\)
= 25
Thus, the football team played 25 matches in all.

5. If Chameli had ₹ 600 left after spending 75% of her money, how much did she have in the beginning?
Solution:
Let Chameli had in the beginning ₹ x
Percentage of money spent by Chameli = 75 %
Percentage of money left with Chameli = (100 – 75)%
= 25%
But money left = ₹ 600 (Given)
∴ 25% of x = 600
∴ \(\frac{25}{100}\) × x = 600
∴ x = \(\frac{600 \times 100}{25}\)
∴ x = 2400
Thus, Chameli had ₹ 2400 in the beginning.

6. If 60% people in a city like cricket, 30% like football and the remaining like other games, then what per cent 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:
Percentage of people who like cricket = 60 %
Percentage of people who like football = 30 %
∴ Percentage of people who like other games = [ 100 – (60 + 30)]%
= (100 – 90)%
= 10 %
Total number of people = 50,00,000 (Given)
Now,
People who like cricket
= 60% of 50,00,000
= \(\frac {1}{2}\) × 50,00,000
= 60 × 50000
= 3000000
= 30 lakh

People who like football
= 30% of 5000000
= \(\frac {30}{100}\) × 5000000
= 30 × 50000
= 1500000
= 15 lakh

People who like other games
= 10% of 5000000
= \(\frac {10}{100}\) × 5000000
= 500000
= 5 lakh
Thus, number of people who like
cricket = 30 lakh,
football = 15 lakh
and other games = 5 lakh

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.