Author name: Bhagya

Haryana State Board HBSE 8th Class Maths Solutions Chapter 10 Visualizing Solid Shapes Ex 10.2 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 10 Visualizing Solid Shapes Exercise 10.2

Question 1.
Look at the given map of a city.

Answer the following:
(a) Colour the maps 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 cf 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 lo 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:
Student fill this map with colours.

Question 2.
Draw a map of your class room using proper scale and symbols for different objects.
Solution:
Try yourself.

Question 3.
Draw a map of your school compound using proper scale and symbols for various features like play ground main building, garden etc.
Solution:
Try yourself.

Question 4.
Draw a map giving instructions to your friend so that she reaches your house without any difficulty.
Solution:
Try yourself.

Haryana State Board HBSE 8th Class Maths Solutions Chapter 10 Visualizing Solid Shapes Ex 10.1 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 10 Visualizing Solid Shapes Exercise 10.1

Question 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:

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


Solution:

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


Solution:

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


Solution:

Haryana State Board HBSE 8th Class Maths Solutions Chapter 9 Algebraic Expressions and Identities InText Questions and Answers.

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

Try These (Page 138)

Question 1.
Give five examples of expressions containing one variable and five examples of expression containing two variables.
Solution:
One variable :
(i) x + 3,
(ii) x + 4,
(iii) y + 3,
(iv) 2 + 5,
(v) p + 2.

Two variables :
(i) 2xy + 3,
(ii) 3xy + 4,
(iii) 5xy + 3,
(iv) zx + 9,
(v) 2pq + 3.

Question 2.
Show on the number line x, x – 4, 2x + 1, 3x – 2.
Solution:

Identify the coefficient of each term in the expression
x²y – 10x²y + 5xy – 20
Coefficient of x²y = 1
Coefficient of -10x²y = -10
Coefficient of 5xy = 5.

Question 3.
Classify the following polynomials as monomials, binomials, trinomials
-z + 5, x + y + z, y + z + 100, ab – ac, 17.
Solution:
Monomials : 17
Binomials : -z + 5, ab – ac
Trinomials : x + y + z, y + z + 100.

Question 4.
Construct:
(a) 3 binomials with only x as a variable;
(b) 3 binomials with x and y as variables;
(c) 3 monomials withx and y as variables;
(d) 2 polynomials with 4 or more terms.
Solution:
(a) x + 2, 2x + 5, 3x + 1
(b) 2xy + 3, 3xy + 7, 4xy + 2
(c) xy, x²y², 4x³y³ .
(d) a + b + c + d, 3x + 2y + z + 5

Try These (Page 139)

Question 1.
Write two terms which are like
(i) 7xy,
(ii) 4mn²,
(iii) 2l.
Solution:
(i) (7xy + 2); (14xy + 3)
(ii) (4mn² – 1); (6mn² + 2)
(iii) (2l + 5); (4l + 3).

Try These (Page 142)

Question 1.
Can you think of two more such situations, where we may need to multiply algebraic expressions ?
[Hint: (i) Think of Speed and Time, (ii) Think of interest to be paid, the principal and the rate of simple interest etc.]
Solution:
(i) Speed = \(\frac{Distance}{Time}\)
s = \(\frac{d}{t}\)
⇒ d = s × t

(ii) Simple Interest
= \(\frac{Principal × Rate × Time}{100}\)
⇒ S.I. = \(\frac{P × R × T}{100}\)
⇒ P × R × T = S.I. × 100

Try These (Page 143)

Question 1.
Find 4x × 5y × 7z.
First find 4x × 5y and multiply it by 7r; 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 × 5y × 7z = (4x + by) × 8z
= (20xy) × 7z
= 140 xyz.

Try These (Page 144)

Question 1.
Find the product:
(i) 2x(3x + 5xy)
(ii) a²(2ab – 5c).
Solution:
(i) 2x(3x + 5xy)
= (2 × 3)x² + (2 × 5)x²y
= 6x² + 10x²y.

(ii) a²(2ab – 5c) = 2a³b – 5a²c.

Try These (Page 145)

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

Try These (i) (Page 149)

Question 1.
Put -6 in place of 6 in Identity (I). Do you get Identity (II) ?
Solution:
Identity I
⇒ (a + b)² = a² = a² + 2ab + b²
If b = -b.
⇒ [a + (-b)]² = a² + 2a(-b) + (-b)²
⇒ (a – b)² = a² – 2ab + b²
Hence Identity-II verified.

Try These (ii) (Page 149)

Question 1.
Verify identity (IV), for a = 2, b = 3, x = 5.
Solution:
∵ (x + a) (x + b) = x² + (a + b)x + ab
If, a = 2, b = 3, x = 5
Then (x + a) (x + b)
= (5)² + (2 + 3) × 5 + 2 × 3
= 25 + 25 + 6 = 56.

Question 2.
Consider, the special case of identity (IV) with a = b, what do you get ? Is it related to identity (I) ?
Solution:
If a = b,
Then, (x + a) (x + b) = x² + (b + b) x + b × b
= x² + 2bx + b².

Question 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:
If a = -c and b = -c
Then, (x + a) (x + b)
= x² + [(-c) + (-c)] x + (-c × -c)
= x² – 2cx + c².

Question 4.
Consider the special case of identity (IV) with b = -a. What/ do you get ? Is it related to identity (III) ?
Solution:
If b = -a,
Then, (x + a) (x + b)
= x² + [a + (-a)]x + a × (-a)
= x² + (a – a) x – a²
= x² + 0 – a²
= x² – a²
= (x + a) (x – a).

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

Haryana Board 8th Class Maths Solutions Chapter 9 Algebraic Expressions and Identities Exercise 9.5

Question 1.
Use a suitable identity to get each of the following products :
(i) (x + 3) (x + 3)
(ii) (2y + 5) (2y + 5)
(iii) (2a – 7) (2a – 7)
(iv) (3a – \(\frac{1}{2}\))(3a – \(\frac{1}{2}\))
(v) (1.1m – 0.4) (1.1m + 0.4)
(vi) (a² + b²) (-a² + b²)
(vii) (6x- 7) (6x + 7)
(viii) (-a + c) (-a + c)
(ix)
(x) (7a – 9b) (7a – 9b).
Solution:
(i) ∵ (x + 9) (x + b)
= x² + (a + b) x + ab
∴ (x + 3) (x + 3) = x² + (3 + 3)x +3 × 3
= x² + 6x + 9.

(ii) ∵ (a + b)² = a² + 2ab + b²
∴ (2y + 5) (2y + 5) = (2y + 5)²
= (2y)² + 2 × 2y × 5 + (5)²
= 4y² + 20y + 25.

(iii) (2a – 7) (2a – 7)
= (2a-7)²
[∵ (a – b)² = a² – 2ab + b²]
= (2a)² – 2 × 2a × 7 + (7)²
= 4a² – 28a+ 49.

(iv) (3a – \(\frac{1}{2}\))(3a – \(\frac{1}{2}\)) = (3a – \(\frac{1}{2}\))²
= (3a)² – 2 × 3a × \(\frac{1}{2}\) + (\(\frac{1}{2}\))²
= 9a² – 3a + \(\frac{1}{4}\)

(v) (1.1m -0.4) (1.1m + 0.4)
∵ (a + b) (a – b) = a² – b²
∴ (1.1m + 0.4) (1.1m – 0.4)
= (1.1m)² – (0.4)²
= 1.21m² – 0.16.

(vi) (a² + b²) (-a² + b²) = (b² + a²) (b² – a²)
= (b²)² – (a²)² – b⁴ – a⁴.

(vii) (6x – 7) (6x + 7) = (6x)² – (7)²
= 36x² – 49.

(viii) (c -a) (c- a) = (c – a)²
= c² – 2ca, + a²

(x) (7a – 9b) (7a – 9b)
= (7a – 9b)²
= (7a)² – 2 x 7a x 9b + (9b)²
= 49a² – 126ab + 81b².

Question 2.
Use the identity (x + a) (x + b) = x² + (a + b) x + ab to find the following products :
(i) (x + 3) (x + 7)
(ii) (4x + 5) (4x + 1)
(iii) (4x – 5)(4x – 1)
(iv) (4x + 5)(4x – 1)
(v) (2x + 5y) (2x + 3y)
(vi) (2a² + 9) (2a² + 5)
(vii) (xyz – 4) (xyz – 2).
Solution:
(i) (x + 3) (x + 7)
= x² + (3 + 7)x + 3 × 7
= x² + 10x + 21.

(iii) (4x + 5) (4x + 1)
= (4x)² + (5 + 1)x + 5 × 1
= 16x² + 6x + 5.

(iii) (4x – 5) (4x – 1)
= (4x)² + [-5 + (-1)]x + (-5x – 1)
= 16x² – 6x + 5.

(iv) (4x + 5) (4x – 1)
= (4x)² + [5 + (-1)]x + [5 × (-1)]
= 16x² + 4x – 5.

(v) (2x + 5y) (2x + 3y)
= (2x)² + (5y + 3y)x + 5y × 3y
= 4x² + 8xy + 15y².

(vi) (2a² + 9) (2a² + 5)
= (2a²)² + (9 + 5)2a² + 9 × 5
= 4a⁴ + 28a² + 45.

(vii) (xyz – 4) (xyz – 2)
= (xyz)² + [(-4) + (-2)] xyz + (-4) x (-2)
= x²y²z² – 6xyz + 8.

Question 3.
Find the following squares by using the identities :
(i) (b – 7)²
(ii) (xy + 3z)²
(iii) (6x² – 5y)²
(iv)
(v) (0.4p – 0.5q)²
(vi) (2xy + 5y)²
Solution:
(i) b – 7)² = b² – 2 x b x 7 + 7²
[∵ (a – b)² = a² – 2ab + b²]
= b² – 14 b + 49.

(ii) (xy + 3z)² = (xy)² + 2 × xy × 3z + (3z)²
[∵ (a + b)² = a² + 2ab + b²]
= x²y² + 6xyz² + 9z².

(iii) (6x² – 5y)² = (6x²)² – 2 × 6x² × 5y + (5y)²
= 36x⁴ – 60x²y + 25y².

(iv)

(v) (0.4p – 0.5q)² = (0.4p)² – 2 × 0.4p × 0.5q + (0.5q)²
= 0.16p² – 0.40pq + 0.25q².

(vi) (2xy + 5y)² = (2xy)² + 2 × 2xy × 5y + (5y)²
= 4x²y² + 20xy² + 25y².

Question 4.
Simplify :
(i) (a² – b²)²
(ii) (2x + 5)² – (2x – 5)²
(iii) (7m – 8n)² + (7m + 8n)²
(iv) (4m + 5n)² + (5m + 4n)²
(v) (2.5p – 1.5q)² – (1.5p – 2.5q)²
(vi) (ab + bc)² – 2ab²c
(vii) (m² – n²m)² + 2m³n².
Solution:
(i) (a² – b²)² = (a²)² – 2 × a² × b² + (b²)²
= a⁴ – 2a²b² + b⁴.

(ii) (2x + 5)² – (2x – 5)²
= [(2x + 5) + (2x – 5)] [(2x + 5) – (2x – 5)]
= (2x + 5 + 2x – 5) (2x + 5 – 2x + 5)
= 4x × 10 = 40x.

(iii) (7m – 8n)² + (7m + 8n)²
= (7m)² – 2 × 7m × 8n + (8n)² + (7m)² + 2 × 7m × 8n + (8n)²
= 2 × (7m)² + 2 × (8n)².
= 2 × 49m² + 2 × 64n²
= 98m² + 128n².

(iv) (4m + 5n)² + (5m + 4n)²
= (4m)² + 2 × 4m × 5n + (5n)² + (5m)² + 2 × 5m × 4n + (4n)²
= 16m² + 40mn + 25n² + 25m² + 40mn + 16n²
= 41m² + 41n² + 80mn

(v) (2.5p – 1.5q)² – (1.5p – 2.5q)²
= [(2.5p)² – 2 × 2.5p × 1.5q + (1.5q)²] – [(1.5p)² – 2 × 1.5p × 2.5q + (2.5q)²]
= 6.25p² – 0.75pq + 2.25q² – 2.25p² + 0.75pq + 6.25q²
= 12.50p².

(vi) (ab + bc)² – 2ab²c
= (ab)² + 2 × ab × bc + (bc)² – 2ab²c
= a²b² + 2ab²c + b²c² – 2ab²c
= a²b² + b²c².

(vii) (m² – n²m)² + 2m³n²
= (m²)² – 2 × m² × n²m + (n²m)² + 2m³n²
= m⁴ – 2m³n² + n⁴m² + 2m³n² = m⁴ + n⁴m².

Question 5.
Show that:
(i) (3x + 7)² – 84x = (3x – 7)²
(ii) (9p – 5q)² + 180pq = (9p + 5q)²
(iii) (4pq + 3q)² – (4pq – 3q)² = 48pq²
(iv) (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0.
Solution:
(i) (3x + 7)² – 84= (3x – 7)²
L.H.S. = (3²)² + 2 × 3x × 7 + (7)² – 84x
= 9x² + 42x + 49 – 84x
= 9x² – 42x + 49 = (3x – 7)²
= R.H.S. Hence proved.

(ii) (9p – 5q)² + 180pg = (9q + 5q)²
L.H.S. = (9p)² + 2 × 9p × 5q + (5q)² + 180pg
= (9p)² + 2 × 9pq × 5q + (5q)²
= (9p + 5q)²
= R.H.S. Hence proved.

(iii) (4pq + 3q)² – (4pq – 3q)² = 48pq²
L.H.S. = [(4pg + 3q) + (4pq – 3q)]
[(4pq + 3q) – (4pq – 3q)]
= (4pq + 3q + 4pq – 3g)
(4pq + 3q – 4pq + 3q)
= 8pq × 6q = 48pq²
= R.H.S. Hence proved.

(iv) (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0
L.H.S. = a² – b² + b² – c² + c² – a² = 0
= R.H.S Hence proved.

Question 6.
Using identities, evaluate.
(i) (71)²
(ii) (99)²
(iii) (102)²
(iv) (998)²
(v) (5.2)²
(vi) 297 × 303
(vii) 78 × 82
(viii) (8.9)²
(ix) 10.5 × 9.5
Solution:
(i) (70 + 1)² = (70)² + 2 × 70 × 1 + (1)²
= 4900 + 140 + 1
= 5041.

(ii) (99)² = (100 – 1)²
= (100)² – 2 × 100 × 1 + (1)²
= 10000 – 200 + 1 = 9801

(iii) (102)² = (100 + 2)²
= (100)² + 2 × 100 × 2 + (2)²
= 10000 + 400 + 4 = 10402.

(iv) (998)² = (1000 – 2)²
= (1000)² – 2 × 1000 × 2 + (2)²
= 1000000 – 4000 + 4
= 996004.

(v) (5.2)² = (5 + 0.2)²
= (5)² + 2 × 5 × 0.2 + (0.2)²
= 25 + 2.0 + 0.04 = 27.04.

(vi) 297 × 303 = (300 – 3) (300 + 3)
= (300)² – (3)²
= 90000 – 9
= 89991

(vii) 78 × 82 = (80 – 2) (80 + 2)
= (80)² – (2)²
= 6400 – 4
= 6396.

(viii) (8.9)² = (9 – 0.1)²
= (9)² – 2 × 9 × 0.1 + (0.1)²
= 81 – 1.8 + 0.01
= 79.21.

(ix) (10.5) × (9.5) = (10 + 0.5) (10 – 0.5)
= (10)² – (0.5)²
= 100 – 0.25
= 99.75.

Question 7.
Using a² – b² = (a + b) (a – b), find
(i) (51)² – (49)²
(ii) (1.02)² – (0.98)²
(iii) (153)² – (147)²
(iv) (12.1)² – (7.9)²
Solution:
(i) (51 + 49) (51 – 49) = 100 × 2 = 200.
(ii) (1.02)² – (0.98)²
= (1.02 + 0.98) (1.02 – 0.98)
= 2.00 × 0.04 = 0.08.

(iii) (153)² – (147)²
= (153 + 147) (153 – 147)
= 200 × 6 = 1200.

(iv) (12.1)² – (7.9)² = (12.1 + 7.9) (12.1 – 7.9)
= 20.0 × 4.2 = 84.0 = 84.

Question 8.
Using
(x + a) (x + b) = x² + (a + b) x + ab, find
(i) 103 × 104
(ii) 5.1 × 5.2
(iii) 103 × 98
(iv) 9.7 × 9.8.
Solution:
(i) (100 + 3) (100 + 4)
= (100)² + (3 + 4) × 100 + 3 × 4
= 10000 + 700 + 12 = 10712.

(ii) (5 + 0.1) (5 + 0.2)
= 5² + (0.1 + 0.2) × 5 + (0.1 × 0.2)
= 25 + 0.02 × 5 + 0.02
= 25 + 0.10 + 0.02
= 25.12.

(iii) (100 + 3) (100 – 2)
= (100)² + [3 + (-2)] × 100 + 3 × (-2)
= 10000 + 100 – 6
= 10094.

(iv) 9.7 × 9.8 = (10 – 0.3) (10 – 0.2)
= (10)² + (-0.3 + (-o.2)] × 10 + (-0.3) × (-0.2)
= 100 – 5 – 0.06
= 94.94.

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

Haryana Board 8th Class Maths Solutions Chapter 9 Algebraic Expressions and Identities Exercise 9.4

Question 1.
Multiply the binomials.
(i) (2x + 5) and (4x- 3)
(ii) (y – 8) and (3y – 4)
(iii) (2.5l – 0.5 m) and (2.5l + 0.5 m)
(iv) (a + 3b) and (x + 5)
(v) (2pq + 3q²) and 3(pq – 2q²)
(vi)
Solution:
(i) (2x + 5) × (4x – 3)
= 8x² – 6x + 20x – 15
= 8x² + 14x – 15.

(ii) (y – 8) × (3y – 4) = 3y² – 4y – 24y + 32
= 3y² – 28y + 32.

(iii) (2.5l – 0.5m) × (2.5l + 0.5m)
= 6.25l² + 1.25lm – 1.25lm + 0.25m²
= 6.25l² + 0.25m².

(iv) (a + 3b) × (x,+ 5)
= ax + 5a + 3bx + 15b
= ax + 3bx + 5a + 15b.

(v) (2pq + 3q²) × (3pq – 2q²)
= 6p²q² – 4pq³ + 9pq³ – 6q⁴
= 6p²a² + 5pq³ – 6q⁴.

(vi)

Question 2.
Find the product.
(i) (5 – 2x)(3 + x)
(ii) (x + 7y)(7x – y)
(iii) (a² + b) (a + b²)
(iv) (p² – q²) (2p + q).
Solution:
(i) (5 – 2x) × (3 + x)
= 15 + 5x – 6x – 2x²
= 15 – x – 2x².

(ii) (x + 7y) × (7x -y)
= 7x² – xy + 49xy – 7y²
= 7x² – 7y² + 48xy.

(iii) (a² + b) × (a + b²)
= a³ + a²b² + ab + b³
= a³ + b³ + a²b² + ab.

(iv) (p² – q²) × (2p + q)
= 2p³ + p²q – 2pq² – q³
= 2p³ – q³ + p²q – 2pq².

Question 3.
Simplify :
(i) (x² – 5) (x + 5) + 25
(ii) (a² + 5) (b³ + 3) + 5
(iii) (t + s²) (t² – s)
(iv) (a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd)
(v) (x +y) (2x +y) + (x + 2y) (x – y)
(vi) (x + y) (x² – xy + y²)
(vii) (1.5x – 4y) (1.5x + 4y + 3)- 4.5x + 12y
(viii) (a + b + c) (a + b – c).
Solution:
(i) (x² – 5) × (x + 5) + 25
= x³ + 5x² – 5x – 25 + 25
= x³ + 5x² – 5x.

(ii) (a² + 5) × (b³ + 3) + 5
= a²b³ + 3a² + 5b² + 15 + 5
= a²b³ + 3a² + 5b³ + 20.

(iii) (t + s²) (t² – s)
= t³ – st + s²t² – s³
= t³ – s³ + s² + t² – st.

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

(v) (x + y) × (2x + y) + (x + 2y) x (x – y)
= 2x² + xy + 2xy + y² + x² – xy + 2xy – 2y²
= 3x² – y² + 4xy.

(vi) (x + y) × (x² – xy + y²)
= x³ – x²y + xy² + x²y – xy² + y³
= x³ + y³.

(vii) (1.5x – 4y) × (1.5x + 4y + 3) – 4.5x + 12y
= 2.25x² + 60xy + 4.5x – 60xy – 16y² – 12y – 4.5x + 12y
= 2.25x².

(viii) (a + b + c) × (a + b – c)
= a² + ab – ac + ab + b² – bc + ac + bc – c²
= a² + b² – c² + 2ab.

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

Haryana Board 8th Class Maths Solutions Chapter 9 Algebraic Expressions and Identities Exercise 9.3

Question 1.
Carry out the multiplication of the expressions in each of the following pairs :
(i) 4p, q + r
(ii) ab, a – b
(iii) a + b, 7a²b²
(iv) a² – 9, 4a
(v) pq + qr + rp, 0.
Solution:
(i) 4p(q + r) = 4pq + 4pr.
(ii) ab × (a – b) = a²b – ab².
(iii) (a + b) × (7a²b²) = 7a²b² + 7a²b³.
(iv) (a² – 9) × 4a = 4a³ – 36a.
(v) (pq – qr + rp) × 0 = 0

Question 2.
Complete the table.

Solution:
(i) ab + ac, + ad
(ii) 5x²y + 5xy² – 25xy
(iii) 6p³ – 7p² + 5p
(iv) 4p⁴q² – 4p²q⁴
(v) a²bc + ab²c + abc².

Question 3.
Find the product.
(i) (a²) × (2a²²) × (4a²⁶)

(iv) x × x² × x³ × x⁴.
Solution:
(i) a² × 2 × a²² × 4 × a²⁶ = 8a⁵⁰.

(iv) x × x² × x³ × x⁴ = x¹⁰.

Question 4.
(a) Simplify 3x (4x – 5) + 3 and find its values for
(i) x = 3
(ii) x = \(\frac{1}{2}\)

(b) Simplify a (a² + a + 1) + 5 and find its value for
(i) a = 0
(ii) a = 1
(iii) a = -1.
Solution:
(a) 3x (4x – 5) + 3 = 12x² – 15x + 3
(i) If x = 3, 12 × (3)² – 15 × 3 + 3
= 12 × 9 – 45 + 3
= 108 – 42 = 66.

(ii) If x = \(\frac{1}{2}\) 12 × (\(\frac{1}{2}\))² – 15 × \(\frac{1}{2}\) + 3
= 12 × \(\frac{1}{4}\) – \(\frac{15}{2}\) + 3
= 6 – \(\frac{15}{2}\) = \(\frac{12-15}{2}\) = \(\frac{-3}{2}\)

(b) a(a² + a + 1) + 5 = a³ + a² + a + 5
(i) If a = 0
a³ + a² + a + 5 = 0³ + 0² + 0 + 5 = 5.

(ii) If a = 1,
a³ + a² + a + 5 = (1)³ + 1² + 1 + 5
= 1 + 1 + 1 + 5 = 8.

(iii) If a = 1
a³ + a² + a + 5 = (-1)³ + (-1)² + (-1) + 5
= 1 + 1 – 1 + 5 = 6.

Question 5.
(a) Add : p(p – q), q(q – r) and r(r – p)
(b) Add : 2x (z – x – y) and 2y (z – y- x)
(c) Subtract : 3l(l – 4m + 5n) from 4l(10n – 3m + 2l)
(d) Subtract: 3a(a + b + c) – 2b(a – b + c) from 4c(-a + b + c).
Solution:
(a) Add : (p² – pq) + (q² – qr) + (r² – rp)
= p² + q² + r² – pq – qr – rp.

(b) Add : (2xz – 2x² – 2xy) + (2yz – 2y² – 2xy)
= 2xz – 2x² – 2xy + 2yz – 2y² – 2xy
= -2x² – 2y² – 4xy + 2xz + 2yz.

(c) Subtract : (40ln – 12lm + 8l²) – (3l² – 12lm + 15ln)
= 40ln – 12lm + 8l² – 3l² + 12lm – 15ln
= 5l² + 25ln.

(d) Subtract : (-4ca + 4bc + 4c²) – (3a² + 3ab + 3ac – 2ab + 2b² – 2bc)
= (-4ca + 4bc + 4c²) – (3a² + 2b² + ab — 2bc + 3ac)
= -4ac + 45c + 4c² – 3a² – 2b² – ab + 25c – 3ac
= -3a² – 2b² + 4c² – ab + 6bc – 7ac.

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

Haryana Board 8th Class Maths Solutions Chapter 9 Algebraic Expressions and Identities Exercise 9.2

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

(ii) -4p × 7p = (-4 × 7) × (p × p)
= -28p².

(iii) -4p × 7pq = (-4 × 7) × (p × pq).
= -28p²q.

(iv) 4p3, -3p = [(4) × (-3)] × (p3 × p)
= -12 p⁴.

(v) 4p × 0 = (4 × 0) × p = 0 × p = 0.

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

(ii) If l = 10m, b = 5n
Area of rectangle = 10m × bn
= 50mn sq. unit.

(iii) If l = 20x², b =-5y²
Area of rectangle = 20x² × 5y²
= 100x²y² sq. unit

(iv) If l = 4x, b = 3x²
Area of rectangle = 4x × 3x² = 12x³sq. unit

(v) If l = 3mn, b = 4np
Area of rectangle = 3mn × 4np
= 12mn²p sq. unit.

Question 3.
Complete the table of products.

Solution:
Complete the table of products as shown in Table :

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

(i) l = 5a, b = 3a², h = 7a⁴
Volume of rectangular box
= l × b × h cubic unit
= 5a × 3a² × 7a⁴
= (5 × 3 × 7) × (a × a2 × a4)
= 105a⁷ cubic unit.

(ii) l = 2p, b = 4q, h = 8r
v = l × b × h, = 2p × 4q × 8r
= (2 × 4 × 8) × (p × q × r)
= 64pqr.

(iii) l = xy, b = 2x²y, h = 2xy²
v = xy × 2x²y × 2xy²
= (1 × 2 × 2) × (xy × x²y × xy²)
= 4x⁴y⁴.

(iv) l = a, b = 2b, h = 3c
v = l × b × h = a × 2b × 3c
= (1 × 2 × 3) (a × b × c) = 6abc.

Question 5.
Obtain the product of
(i) xy, yz, zx
(ii) a, -a², a³
(iii) 2, 4y, 8y², 16y³
(iv) a, 2b, 3c, 6abc
(v) m, -mn, mnp.
Solution:
(i) xy × yz × zx = x²y²z².
(ii) a × (-a²) × (a³) = -a⁶.

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

(iv) a × 2b × 3c × 6abc
= (2 × 3 × 6) x (a x b × c × abc)
= 36a²b²c².

(v) mx (-mn) × mnp = -m³n²p.

Haryana State Board HBSE 8th Class Maths Solutions Chapter 9 Algebraic Expressions and Identities Ex 9.1 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 9 Algebraic Expressions and Identities Exercise 9.1

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

(ii) There are three terms; 1, x and x²
Coefficient of x = 1
Coefficient of x² = 1

(iii) There are three terms; 4x²y², -4x²y²z², z².
Coefficient of 4x²y² = 4
Coefficient of -4x²y²z² = -4
Coefficient of z² = 1

(iv) There are four terms; 3, -pq, qr and – rp.
Coefficient of -pq = -1
Coefficient of qr = 1
Coefficient of -rp = -1.

(v) There are three terms; \(\frac{x}{2}\) , \(\frac{y}{2}\), -xy
Coefficient of \(\frac{x}{2}\) = \(\frac{1}{2}\)
Coefficient of \(\frac{y}{2}\) = \(\frac{1}{2}\)
Coefficient of-xy = -1

(vi) There are three terms; 0.3a, -0.6ab, 0.5b.
Coefficient of 0.3a = 0.3
Coefficient of-0.6a6 = -0.6 .
Coefficient of 0.56 = 0.5.

Question 2.
Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories ?.
x + y, 1000, x + x² + x³ + x⁴, 7 + y + 5x, 2y – 3y², 2y – 3y² + 4y³, 5x – 4y + 3xy, 4z – 15z², ab + bc + cd + da, pqr, p²q + pq², 2p + 2q.
Solution:
Monomial = 1000, par
Binomial = x + y, 2y – 3y², 4z – 15z², p²q + pq², 2p + 2q
Trinomial = 7 + y + 5x, 2y – 3y² + 4y³, 5x – 4y + 3xy
Polynomial = x + x² + x³ + x⁴, ab + bc + cd + da,

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

Question 4.
(a) Subtract 4a – 7ab + 36 + 12 from 12a – 9ab + 5b – 3
(6) Subtract 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + 10xyz
(c) Subtract 4p²q – 3pq + 5pq² – 8p + 7q – 10 from 18 – 3p – 11q + 5pq – 2pq² + bp²q.
Solution:
(a)

Haryana State Board HBSE 8th Class Maths Solutions Chapter 8 Comparing Quantities InText Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 8 Comparing Quantities InText Questions

Try These (Page 119)

Question 1.
In a primary school, the parents were asked about the number of hours they spend per day in helping their children to do homework. There were 90 parent who helped for \(\frac{1}{2}\) hour to 1\(\frac{1}{2}\) hours. The distribution of parents according to time for which, they said they helped is given in the adjoining figure; 20% helped for more than 1\(\frac{1}{2}\) hours per day; 30% helped for \(\frac{1}{2}\) hour to 1\(\frac{1}{2}\) hours; 50% did not help at all. It

Using this answer the following :
(i) How many parents were surveyed ?
(ii) How many said that they did not help ?
(iii) How many said that they helped for more than 1\(\frac{1}{2}\) hours ?
Solution:
(i) Let total no. of parents be x 30% of x helped for \(\frac{1}{2}\) hour to 1\(\frac{1}{2}\) hour
So 30% of x = 90
\(\frac{30}{100}\) × x = 90 or x = \(\frac{90 \times 100}{30}\) = 300
Number of parents = 300.

(ii) 50% parents did not help,
So number of parents did not help
= 50% of 300 = \(\frac{50}{100}\) × 300 = 150

(iii) 20% parents helped for more than 1\(\frac{1}{2}\) hours
So, number of such parents
= 20% of 300 = \(\frac{20}{100}\) × 300 = 60

Try These (Page 121)

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

(b) 20% discount means
20% discount on M.P.
⇒ Selling price = 100 – 20
= 80% on M.P.
∴ Sale price of shoe of MP Rs. 750
= 80% of 750
= \(\frac{80}{100}\) × 750 = 600
∴ Sale price of shoe = Rs. 600.

(c) 20% discount means
Selling price = 80% of M.P.
∴ Selling price of bag of M.P. Rs. 250
= 80% of 250
= \(\frac{80}{100}\) × 250 = 200
∴ Sale price of bag = Rs. 200.

Question 2.
A table marked at Rs. 15,000 is available for Rs. 14,400. Find the discount given and the discount per cent.
Solution:
M.P. = Rs. 15000
S.P. = 1440
Discount = ?
Discount % = ?
Discount = M.P. – S.P.
= 15000 – 14400 = 600
Discount % = \(\frac{Discount}{M.P.}\) × 100
= \(\frac{600}{15000}\) × 100 = 4%
Discount – Rs. 600, discount % = 4.

Question 3.
An almirah is sold at Rs. 5,225 after allowing a discount of 5%. Find its marked price.
Solution:
M.P. = ?
S.P. = Rs. 5,225
Discount % = 5 %

Try These (Page 123)

Question 1.
Find selling price (S.P.) if a profit of 5% is made on
(a) a cycle of Rs. 700 with Rs. 50 as overhead charges.
(b) a lawn mower bought at Rs. 1150 with Rs. 50 as transportation charges.
(c) a fan bought for Rs. 560 and expenses of Rs. 40 made on its repairs.
Solution:
(a) Cycle of worth Rs. 700
Overhead charges on cycle = 50
∴ C.P. of cycle = 700 + 50 = 750
Profit % = 5
∴ Profit of Rs. 5 on C.P. of Rs. 100
⇒ S.P. = (100 + 5)% of C.P.
∴ S.P. of cycle of C.P. Rs. 750 = 105% of 750
= \(\frac{105}{100}\) × 750 = 787.50
∴ S.P. of cycle = Rs. 787.50.

(b) Lawn mower bought in Rs. 1150
Transportation charge = Rs. 50
∴ C.P. of lawn mower = 1150 + 50 = 1200
5% profit on lawn mower is made
∴ S.P. = 105% of C.P.
= \(\frac{105}{100}\) × 1200 = 1260
∴ Sale price of lawn mower = Rs. 1260.

(c) Fan bought in Rs. 560
Expenses on its repairs = Rs. 40
∴ C.P. of fan = 560 + 40 = 600
5% profit on fan is made
∴ S.P. of fan = 105% of C.P. of fan
= \(\frac{105}{100}\) × 600 = 630
S.P. of fan = Rs. 630.

Question 2.
A shopkeeper bought two T.V. sets at Rs. 10,000 each. He sold one at a profit of 10% and the other at a loss of 10%. Find whether he made an overall profit or loss.
Solution:
Overall C.P. of each T.V. set = Rs. 10,000
One is sold at a profit of 10%
⇒ If C.P. is Rs. 100, S.P. is Rs. 110
Therefore, when C.P. is Rs. 10,000,
Then, S.P. = Rs. \(\frac{110}{100}\) × 10000 = 11000
In second case, T.V. at a loss of 10%
⇒ If C.P. is Rs. 100, S.P. is Rs. 90
Therefore, when C.P. is Rs. 10,000
Then, S.P. = Rs. \(\frac{90}{100}\) × 10000 = 9000
Total C.P. of two sets of T.V.
= Rs. 10000 + Rs. 10000
= Rs. 20,000
Total S.P. of two sets of T.V.
= Rs. 11000 + Rs. 9000
= Rs. 20,000
Since Total S.P. = Total C,P.
⇒ Neither profit nor loss.

Try These (Page 124-125)

Question 1.
Find the buying price of each of the following when 5% S.T. is added on the purchase of
(a) A towel at Rs. 50.
(b) Two bars of soap at Rs. 35 each.
(c) 5 kg of flour at Rs. 15 per kg.
Solution:
(a) S.T. of Rs 50 = 50 × \(\frac{10}{100}\) = Rs. 5
Bill amount = Cost of towel + sales tax
= 50 + 5 = Rs. 55.

(b) S.T. of Rs. 35 = 35 × \(\frac{10}{100}\) = Rs. 3.50
∴ Bill amount = cost of two bars of soap + S.T.
= 70 + 7 = Rs. 77.

(c) Cost of 5 kg flour = 5 × 15 = Rs. 75
Sales tax = Rs. 75 × \(\frac{10}{100}\)
= Rs. 7.5
Bill amount = Cost of flour + S.T.
= 75 + 7.5
= Rs. 82.50.

Question 2.
If 8% VAT is included in the prices, find the original price of
(a) A TV bought for Rs. 13,500.
(b) A shampoo bottle bought for Rs. 180.
Solution:
(a) 8% VAT mean an article of Rs. 100
will costs Rs. 100 + 8 = Rs. 108
Therefore, when price including VAT is Rs. 108, original price = Rs. 100
∴ When price including VAT is Rs. 13500
Original price = \(\frac{100}{108}\) × 13500 = 12500
∴ Original price of T.V. = Rs. 12500.

(b) 8% VAT included in the price
⇒ When price including VAT is Rs. 108, original price = Rs. 100
∴ When price including VAT is Rs. 180, original price
= \(\frac{100}{108}\) × 180 = 166.67
∴ Original price of shampoo = Rs. 166.67.

Try These (Page 126)

Question 1.
Find interest and amount to be paid on Rs. 15000 at 5% per annum after 2 years.
Solution:
P = Rs. 15000, R = 5% per annum, T = 2 year, I = ?, A = ?
I = \(\frac{P \times R \times T}{100}\) × \(\frac{5}{100}\) × 2 = Rs. 1500
A = P + I
= Rs. 15000 + Rs. 1500
= Rs. 16500.

Try These (Page 129)

Question 1.
Find C.I. on a sum of Rs. 8000 for 2 years at 5% per annum compounded annually.
Solution:
P = Rs. 8000, R = 5% per annum, T = 2 year, C.I. = ?

= 20 × 21 × 21 = Rs. 8820
C.I. = A – P
= Rs. 8820 – Rs. 8000
= Rs. 820.

Try These (Page 130)

Question 1.
Find the time period and rate for each.
1. A sum taken for 1\(\frac{1}{2}\) years at 8% per annum is compounded half yearly.
2. A sum taken for 2 years at 4% per annum compounded half yearly.
Solution:
1. If the time period yearly and the interest is compounded half yearly, then the time, T will be T × 2
∴ T = 1\(\frac{1}{2}\) year
= \(\frac{3}{2}\) × 2 = 3 years.
and, if the rate of interest is R% per annum and the interest is compounded half yearly, then the rate of interest will be \(\frac{R}{2}\) % per half hear.
R = 8% = \(\frac{R}{2}\)
= 4% per half year.

2. If the time period yearly and the interest is compounded half yearly then the time, T will be T × 2
∴ T = 2 years = 2 × 2 = 4 years
and, if the rate of interest is R% per annum and the interest is compounded half yearly, then the rate of interest will be \(\frac{R}{2}\) % per half year.
∴ R = 4% = \(\frac{4}{2}\)%
= 2% per half year.

Try These (Page 131)

Question 1.
Find the amount to be paid
(i) At the end of 2 years on Rs. 2,400 at ty5% per annum compounded annually.
(ii) At the end of 1 year on Rs. 1,800 at 8% per annum compounded quarterly.
Solution:
(i) A = ?, T = 2 years, P = Rs. 2400, R = 5% per annum compounded annually.

= 6 × 441 = Rs. 2646
∴ Amount = Rs. 2646.

(ii) R = 8% per annum compounded quarterly = 2% per quarterly
T = 1 year = 4 quarter
P = Rs. 1800, A = ?

= 1800 × \(\frac{51}{50}\) × \(\frac{51}{50}\)
= Rs. 1872.7
Amount = Rs. 1872.72.

Try These (Page 133)

Question 1.
A machinery worth Rs. 10,500 depreciated by 5%. Find its value after one year.
Solution:
P = Rs. 10500, R = 5%, T = 1 year
Cost of machinery after one year
= P(1 – \(\frac{R}{100}\))¹ = 10500(1 – \(\frac{5}{100}\))¹
= 10500(1 – \(\frac{1}{20}\))¹
= 10500 \(\frac{19}{20}\)
= 525 × 19 = Rs. 9975.

Question 2.
Find the population of a city after 2 years, which is at present 12 lakh, if the rate of increase is 4%.
Solution:
T = 2 year, present population, P = 12 lakh
R = 4%, A = ?
= 1200000 (1 + \(\frac{4}{100}\))²
= 1200000(1 + \(\frac{1}{25}\))²
= 1200000 × \(\frac{26}{25}\) × \(\frac{26}{25}\)
= 1297920
∴ Population after 2 years
= 1297920.

Haryana State Board HBSE 8th Class Maths Solutions Chapter 8 Comparing Quantities Ex 8.3 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 8 Comparing Quantities Exercise 8.3

Question 1.
Calculate the amount and compound interest on :
(a) Rs. 10,800 for 3 years at 12\(\frac{1}{2}\)% per annum compounded annually.
(b) Rs. 18,000 for 2\(\frac{1}{2}\) years at 10% per annum compounded annually.
(c) Rs. 62,500 for 1\(\frac{1}{2}\) years at 8% per annum compounded half yearly.
(d) Rs. 8,000 for 1 year at 9% per annum compounded half yearly. (You could use the year by year calculation using S.I. formula to verify).
(e) Rs. 10,000 for 1 year at 8% per annum compounded half yearly.
Solution:
(a) A = ?, C.I. = ?, P = Rs. 10,800,
T = 3 years, R = 12 \(\frac{1}{2}\)% per annum
A = P(1 + \(\frac{R}{100}\))^T
= 1080o(1 + \(\frac{25}{2}\) × \(\frac{1}{100}\))³
= 10800(\(\frac{17}{16}\))³
= 10800 × \(\frac{17}{16}\) × \(\frac{17}{16}\) × \(\frac{17}{16}\)
= Rs. 12954.20
C.I. = A – P
= Rs. 1295.4.20 – Rs. 10800
= Rs. 2154.20.

(b) P = Rs. 18000, T = 2\(\frac{1}{2}\) years, R = 10% per annum compounded annually, A = ?, C.I. = ?
A = P(1 + \(\frac{R}{100}\))^T
= 18000(1 + \(\frac{10}{100}\))^{2\(\frac{1}{2}\)}
So amount for 2 years is given by
A = Rs. 18000(1 + \(\frac{10}{100}\))²
= 18000(1 + \(\frac{1}{10}\))²
= 18000 × \(\frac{11}{10}\) × \(\frac{11}{10}\)
= 180 × 121 = Rs. 21780
Rs. 21780 would act as principal for next \(\frac{1}{2}\) year. We find S.I. on Rs. 21780 for \(\frac{1}{2}\) year
S.I = Rs. \(\frac{21780 \times \frac{1}{2} \times 10}{100}\)
= Rs. 1089
Interest for two years
= Rs. 21780 – Rs. 18000
= Rs. 3780
Interest for next \(\frac{1}{2}\) year = Rs. 1089
Therefore, total compound interest
= Rs. 3780 + Rs. 1089
= Rs. 4869
A = P + C.I.
= Rs. 18000 + Rs. 4869
= Rs. 22869.

(c) P = Rs. 62500, T = 1\(\frac{1}{2}\) years = 3 half years, R = 8% per annum = 4% per half yearly, A = ?, C.I. = ?

= Rs 70304
C.I. = A – P
= Rs. 70304 – Rs. 62500
= Rs. 7804.

(d) P = Rs. 8000, T = 1 year = 2 half years, R = 9% per annum = \(\frac{9}{2}\) % per half year, A = ?, C.I. = ?

= Rs. 8736.20
C.I. = A – P
= Rs. 8736.20 – Rs. 8000
= Rs. 736.20.

(e) P = Rs. 10000, T = 1 year = 2 half years, R = 8% per annum = 4% per half year, A = ?, C.I. = ?

= Rs. 10816
C.I. = A – P
= Rs. 10816 – Rs. 10000
= Rs. 816.

Question 2.
Kamla borrowed Rs. 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 S.I. on the 2nd year amount for 4/12 years]
Solution:
P = Rs. 26400, R = 15% per annum, T = 2 years and 4 months, A = ?


= 66 × 23 × 23
= Rs. 34914
Rs. 34914 would act as principal for next 4 months i.e. \(\frac{1}{3}\) year
So S.I. = P × R × T
= 34914 × \(\frac{15}{100}\) × \(\frac{1}{3}\)
= Rs. 1745.70
Amount paid by Kamla after 2 years
= Rs. 34914 + Rs. 1745.70
= Rs. 36659.70.

Question 3.
Fabina borrows Rs. 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:
In case of Fabina :
P = Rs. 12500,
R = 12% per annum, T = 3 years
S.I = P × R × T
= 12500 × \(\frac{12}{100}\) × 3 = RS. 4500.
In case of Radha :
P = Rs. 12500,
R = 10% per annum,
T = 3 years, C.I. = ?

= Rs. 16637.50
C.I. = A – P
= Rs. 16637.50 – Rs. 12500
= Rs. 4137.50
It is obvious that Fabina would pay more interest by Rs. 4500 – Rs. 4137.50 = Rs. 362.50 So,
Fabina paid Rs. 362.50 more than Radha.

Question 4.
I borrowed Rs. 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:
P = Rs. 12000, R = 6% per annum, T = 2 years, S.I. = ?, C.I. = ?
In case of simple interest:
S.I = P × R × T
= 12000 × \(\frac{6}{100}\) × 2
= Rs. 1440
In case of compound interest:

= Rs. 13483.20
So, Amount = Rs. 13483.20
∴ C.I. = A – P
= Rs. 13483.20 – Rs. 12000
= Rs. 1483.20
It is obvious that C.I. > S.I
So extra amount paid in case of C.I. is
= Rs. 1483.20 – Rs. 1440
= Rs. 43.20.

Question 5.
Vasudevan invested Rs. 60,000 at an interest rate of 12% per annum compounded half yearly. What amount would he get
(i) after 6 months ?
(ii) after 1 year ?
Solution:
P = Rs. 60000,
R = 12% per annum compounded half yearly
= 6% half yearly
A₆ = ?
A₁₂ = ?
T₆ = 6 months = 1 half year
T₁₂ = 1 year = 2 half years

(i) Amount after 6 months

= Rs. 67416.
∴ A₁₂ = Rs. 67416.

Question 6.
Arif took a loan of Rs. 80,000 from a bank. If the rate of interest is 10% per annum, find the difference in amounts he
would be paying after 1\(\frac{1}{2}\) years if the interest is
(i) Compounded annually.
(ii) Compounded half yearly.
Solution:
P = Rs. 80000,
R = 10% per annum,
T = 1\(\frac{1}{2}\) years
= 5% half yearly = 3 half years
In 1st Case : When amount calculated annual compounding

Rs. 88000 would be principal for half year to calculate S.I. for \(\frac{1}{2}\) year
S.I. = P × R × T
= 88000 × \(\frac{10}{100}\) × \(\frac{1}{2}\)
= Rs. 4400
∴ C.I. for 1\(\frac{1}{2}\) on given principal
= (Rs. 88000 – Rs. 80000) + Rs. 4400
= Rs. 12400
So, A = P + C.I.
= Rs. 80000 + Rs. 12400
= Rs. 92400
In 2nd Case : When amount calculated half yearly compounding

= Rs. 92610
Difference in amount in both the cases
= Rs. 92610 – Rs. 92400
= Rs. 210.

Question 7.
Maria invested Rs. 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find
(i) The amount credited against her name at the end of the second year.
(ii) The interest for the 3rd year.
Solution:
(i) P = Rs. 8000, R = 5% per annum, T = 2 year, T = 3 year, A₂ = ?, A₃ = ?
Amount after 2 years,

Amount credited after 2 years = Rs. 8820.
Amount after 3 years,

= Rs. 9261
∴ Interest for 3rd year = A₃ – A₂
= Rs. 9261 – Rs. 8820
= Rs. 441.

Question 8.
Find the amount and the compound interest on Rs. 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:
P = Rs. 10000
T = 1\(\frac{1}{2}\) years = 3 half years
R = 10% per annum = 5% half yearly
A_h =?, A_a = ?

= Rs. 11576.25
So, Amount = Rs. 11576.25
∴ C.I. = A – P
= Rs. 11576.25 – Rs. 10000
= Rs. 1576.25
If n = 1\(\frac{1}{2}\) = \(\frac{3}{2}\) years
A_a = P(1 + \(\frac{R}{100}\))^T

= 1000 × 1.05 = Rs. 11550
∴ C.I. = A_a – P = 11550 – 10000
= Rs. 1550.
or, Rs. 11000 would be principal for \(\frac{1}{2}\) year to calculate S.I. for \(\frac{1}{2}\) year.
S.I. = P × R × T
= 11000 × \(\frac{10}{100}\) × \(\frac{1}{2}\) = Rs. 550
∴ C.I. = (Rs. 11000 – Rs. 10000) + Rs. 550 = Rs. 1550
It is obvious that C.I. calculated half yearly is greater than that when calculated yearly.
Extra amount of interest = Rs. 1576.25 – Rs. 1500 = Rs. 76.25.

Question 9.
Find the amount which Ram will get on Rs. 4096, if have gave it for 18 months at 12\(\frac{1}{2}\) % per annum, interest being compounded half yearly.
Solution:
P = Rs. 4096,
T = 18 months = 3 half years,
R = 12\(\frac{1}{2}\) % per annum
= \(\frac{25}{4}\) % per half year
A = ?

= Rs. 4913

Question 10.
The population of a place increased to 54,000 in 2003 at a rate of 5% per annum
(i) find the population in 2001.
(ii) what would be its population in 2005 ?
Solution:
(i) P = 54000, R = 5% per annum, A = ?, T = 2 years
Population before 2 years

(ii) For population in 2005 i.e. 2 years after 2003
P = 54000, T = 2 years, R = 5% per annum, A = ?

∴ Population in 2005 be 59535.

Question 11.
In a laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5,06,000.
Solution:
Initial count of bacteria P = 506000, T = 2 hour
Rate of increasing bacteria R = 2.5% per hour, A = ?

∴ Count of bacteria after 2 hour = 531616.

Question 12.
A scooter was bought at Rs. 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.
Solution:
P = Rs. 42000, R = 8% per annum, T = 1 year, A = ?