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S.chand publication New Learning Composite mathematics solution of class 8 Chapter 12 Mensuration Exercise 12C

Exercise 12C

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Question 1

Find the surface area and lateral area of each cuboid / cube. All lengths are in centimetres.

(a)

Surface Area=6×(side)²

=6×(10)²

=6×100=600unit²

Lateral Area=4×(side)²

=4×(10)²

=4×100

=400 unit²

(b)

S.A=2(l×b + b×h + h×l)

=2(10×6 + 6×4 + 4×10)

=2(60+24+40)

=2×124

=248cm²

L.A=2h(l+b)

=2×4×(10+6)

=8×(16)

=128cm²

(c)

S.A=6×(side)²

=6×(3)²

=6×9

=54cm²

L.A=4×(side)²

=4×(3)²

=4×9

=36cm²

(d)

S.A=2(l×b + b×h + h×l)

=2(15×15 + 15×12 + 12×15)

=2(225+180+180)

=2×585

=1170 cm²

L.A=2h(l+b)

=2×12(15+15)

=2×12×30

=720 cm²

(e)

Sol :

S.A=2(l×b + b×h + h×l)

=2(17×11+11×12 +12×17)

=2×(187+132+204)

=2×(523)

=1046 cm²

L.A=2h(l+b)

=2×12×(17+11)

=24×28

=672 cm²

(f)

S.A=6×(side)²

=6×(4.5)²

=6×20.25

=121.50 cm²

L.A=4×(side)²

=4×(4.5)²

=4×20.25

=81 cm²

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Question 2

Find the surface area of a cuboid, given

(a) Length=8 cm, Breadth=7 cm, Height=12 cm

(b) Length=60 mm, Breadth=40 mm, Height=55 mm

Sol :

(a)

L=8cm , B=7cm, H=12cm

∴S.A={2(8×7)}+{2(7×12)}+{2(8×12)}

=(2×56)+(2×84)+(2×96)

=112+168+192=472cm²

(b)

L=60cm , B=40cm , H=55cm

∴S.A={2(60×55)}+{2(40×55)}+{2(60×40)}
=(2×3300)+(2×2500)+(2×2400)
=6600+4800+4800=15800cm²

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Question 3

Find the surface area of a cube with side length

(a) 6 cm

S.A={2(6×6)}+{2(6×6)}+{2(6×6)}

=(2×36)+(2×36)+(2×36)

=72+72+72=216cm²

(b)

=(2×324)+(2×324)+(2×324)

=648+648+648=1944cm²

(c)

S.A={2(1.7×1.7)}+{2(1.7×1.7)}+{2(1.7×1.7)}

=(2×2.89)+(2×2.89)+(2×2.89)

=5.78+5.78+5.78=17.34cm²

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Question 4

Find the surface area of a wooden cube of side length 30cm

Sol :

Cube side=30cm

S.A={2(30×30)}+{2(30×30)}+{2(30×30)}

=(2×900)+(2×900)+(2×900)

=1800+1800+1800=5400cm²

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Question 5

A cornflakes box measures 30 cm by 24 cm by 8 cm. Calculate the surface area of the box

Sol :

L=24cm, B=8cm ,H=30cm

S.A={2(24×8)}+{2(24×30)}+{2(30×8)}

={2×192}+{2×720}+{2×240}

=384+1440+480=2304cm²

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Question 6

Find the missing dimension in the figure with the given surface area.

Sol :

B=5cm ,H=12cm ,L=?

S.A={2(x×5)}+{2(x×12)}+{2(5×12)}

={5x×2}+{24x}+{2×60}

=10x+24x+120=3x+120

ATQ,

S.A=256=34x+120

34x=256-120=136

$x=\frac{136}{34}$=4cm

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Question 7

A cuboid is 3 cm by 4 cm by 7 cm. Explain whether doubling all of the dimensions would double the surface area?

Sol :

L=4cm, B=3cm ,H=7cm

S.A₁={2(4×3)}+{2(4×7)}+{2(3×7)}

={2×12}+{2×28}+{2×21}

=24+56+42=122cm²

If doubling all of dimensions

∴L=4×2=8cm ,B=3×2=6cm ,H=7×2=14cm

∴S.A₂={2(8×6)}+{2(6×14)}+{2(8×14)}

=(2×48)+(2×84)+(2×112)

=96+168+224=488

∴S.A₂=488

S.A₁=122

∴S.A₂=122×4=S.A₁×4

∴No, It will be 4 times

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Question 8

The net consists of 6 equal squares with side length 8 cm

(a) Name the solid it represents.

(b) Find the surface ares of the solid.

Sol :

(a) The solid shape is cube

(b) S.A={2(8×8)}+{2(8×8)}+{2(8×8)}

={2×64}+{2×64}+{2×64}=128+128+128=384cm²

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Question 9

A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long. 25 cm wide and 25 cm high.

(a) What is the area of glass?

(b) How much tape is needed for all the 12 edges?

Sol :

(a)

S.A.={2(30×25)}+{2(30×25)}+{2(25×25)}

=(2×750)+(2×750)+(2×625)

=1500+1500+1250=4250cm²

(b)

Length of top for all 12 edges=Perimeter of the top+Four vertical edge

={2(30+25)}+{25×4}

=(2×55)+(100)

=110+100=210cm

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Question 10

Two cubes each of 10 cm edge are joined end to end. Find the surface area of the resulting cuboid.

Sol :

S.A={2(20×10)}+{2(20×10)}+{2(10×10)}

=(2×200)+(2×200)+(2×100)

=400+400+200=1000cm²

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Question 11

Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth and height of is 15 m, 10 m and 7 m respectively. From each can of paint, $100 \mathrm{~m}^{2}$ of area is painted. Hone many cans of paint will he need to paint the entire room?

Sol :

S.A={2(15×7)}+{2(10×7)}+{2(15×10)}

=(107×2)+(70×2)+(150)

=210+140+150=500cm²

∴Cans of paint needed$=\frac{500}{100}=5$

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Question 12

A cuboidal tin box opened at the top has dimensions 20cm×16cm×14cm. What is the total area of metal duct sheet required to make 10 such boxes?

Sol :

S.A=2(l×b + b×h + h×l)

=2(20×16+16×14+14×20)

=2×(320+224+280)

=2×824

=1648 cm²

∴1 box need 1648 cm²sheet

∴10 box need 1648×10=16480cm²sheet

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Question 13

Find the area to be painted in the given block with a cylindrical hole. Given that length is 15 cm, width is 12 cm, height is 20 cm and radius of the holes is 2.8 cm.

Sol :

S.A=2(l×b + b×h + h×l)

=2(15×12 + 12×20 + 20×15)

=2(180+240+300)

=2×720

=1440 cm²

∴Area of hole=πr²

$=\frac{22}{7}\times (2.8)^2$

$=\frac{22}{7}\times \frac{28}{10}\times \frac{28}{10}$

$=\frac{2464}{100}$

=24.64

∴Area to be painted=Total surface of the block-(2×Area of one circular hole)

=1440-(24.64×2)

=1440-49.28

=1390.72cm²

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Question 14

The area of the four walls in a room is $120 \mathrm{~m}^{2}$. The length is twice the breadth. If the height of the room is 4m , find the area of the floor.

Sol :

Let breadth of room =x m

Length of room =2x m

Height of the rm=4 m

The area of the four walls=2(bh+lh)=120

2(x×4+2x×4)=120

2(12x)=120

$x=\frac{120}{4}$=5 m

Breadth=x=5 m ,

Length=2x=2×5=10 m

Area of floor=l×b=10×5=50 m²

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Question 15

A cube of edge 5 cm is cut into cubes, each of edge 1 cm. Find the ratio of the total surface area of one of small cubes to that of the large cube.

Sol :

Edge of large cube =5 cm

Area of large cube $=6(a)^{2}=6 \times(5)^{2}$

=6×25=150 cm²

Edge of small cube=1 cm

Area of small cube$=6 \times(a)^{2}=6 \times(1)^{2}$

$=6 \times 1=6$

Area of small cube : Area of large cube $=\frac{6}{150}=\frac{1}{25}$

1 : 25

S.chand publication New Learning Composite mathematics solution of class 8 Chapter 6 Factorisation of Algebraic Expressions Exercise 6C

Exercise 6C

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Question 1

ax² + by² + bx² + ay²

Sol :

=ax²+by²+bx²+ay²

=x²(a+b)+y²(a+b)

=(a+b)(x²+y²)

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Question 2

my + nz + mz + ny

Sol :

=my+mz+nz+ny

=m(y+z)+n(y+z)

=(y+z)(m+n)

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Question 3

ab – 5c + 5a – bc

Sol :

=ab+5a+5c-bc

=ab-bc-5c+5a

=b(a-c)+5(a-c)

=(a-c)(b+5)

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Question 4

x² + ab – ax – bx

Sol :

=x²+ab–ax–bx

=x(x-a)-b(x-a)

=(x-a)(x-b)

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Question 5

ab² – bc² – ab + c²

Sol :

=ab(b-1)-c²(b-c)

=(b-1)(ab-c²)

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Question 6

2xw – 3 – 6x + w

Sol :

=2xw+w-3-6x

=w(2x+1)-3(2x+1)

=(2x+1)(w-3)

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Question 7

54x + 10y – 12xy – 45

Sol :

=54x-45+10y-12xy

=9(6x-5)-2y(6x-5)

=(9-2y)(6x-5)

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Question 8

m³ + 4m + 5m² + 20

Sol :

=m³+5m²+4m+20

=m²(m+5)+4(m+5)

=(m²+4)(m+5)

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Question 9

2axy² + 10x + 3ay² + 15

Sol :

=2axy²+10x+3ay²+15

=ay²(2x+3)+5(2x+3)

=(ay²+5)(2x+3)

S.chand publication New Learning Composite mathematics solution of class 8 Chapter 2 Exponents Exercise 2A

Exercise 2A

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Question 1

Evaluate

(a) 9⁰

Sol : 1

(b) $\left(\frac{1}{3}\right)^0$

Sol : 1

(d) (395.008)⁰

Sol : 1

(e) $\left(\frac{3xy}{ab}-z\right)^0$

Sol : 1

(f) $\frac{1}{a^0 + b^0 + c^0}$

Sol : $=\frac{1}{3}$

(g) (8⁰ – 2⁰)x 25⁰

Sol :

=(1-1)×1

=0×1=0

(h) (35⁰ + 2⁰ + 3⁰) ÷ 3

Sol : 1

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Question 2

Simplifying giving the answer as a fraction:
(a) 6^-1

Sol : $=\frac{1}{6}$

(b) 10^-1

Sol : $=\frac{1}{10}$

Sol : $=\frac{1}{4^3}=\frac{1}{64}$

(d) 2^-7

Sol :

$=\frac{1}{2^7}=\frac{1}{128}$

(e) 3^-3

Sol :

$=\frac{1}{3^3}=\frac{1}{27}$

(f) p^-1

Sol :

$=\frac{1}{p}$

(g) (x^-2)²

Sol :

$=\frac{1}{x^4}$

(h) (10a)^-2

Sol :

$=\frac{1}{100a^2}$

(i) (x+y)^-2

Sol :

$=\frac{1}{(x+y)^2}$

(j) (m⁵)^-3

Sol :

$=\frac{1}{m^{15}}$

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Question 3

Find the reciprocal of the following.

(a) (3)^-2

Sol :

$=\frac{1}{3^2}=\frac{1}{9}$

(b) (-2)^-5

Sol :

$=\frac{1}{(-2)^5}=-\frac{1}{32}$

$=\left(\frac{5}{2}\right)^3=\frac{125}{8}$

(d) $\left(\frac{2}{3}\right)^{-4}$

Sol :

$=\left(\frac{3}{2}\right)^4=\frac{81}{16}$

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Question 4

Simplify and express the result in power notation with positive exponent.
(a) (-2)^-4 x (-2)^-6

Sol : $=\frac{1}{-2^4}\times \frac{1}{-2^6}=\frac{1}{2^{10}}$

(b) 3⁷ x 3^-9 x 3⁶

Sol : $=\frac{1}{3^7}\times \frac{1}{3^9}\times \frac{1}{3^{6}}$

=3⁴

Sol : $=3^7 \times \left(\frac{5}{3}\right)^7=3^7 \times \frac{5^7}{3^{7}}$

=5⁷

(d) 2^-7 x [(2)^-3]^-4

Sol : $=\frac{1}{2^7}\times 2^{12}$

=2⁵

(f) (4^-8 ÷ 4^-10) x 4^-5

Sol : $=\frac{4^{-8}}{4^{-10}}\times 4^{-5}$ $=\frac{1}{4^2}\times 4^{-5}=4^{-3}=\frac{1}{4^3}$

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Question 5

Evaluate:
(a) (8⁰ + 5^-1)×$\left(\frac{1}{5}\right)^{-2}$

Sol :

$=\left(1+\frac{1}{5}\right)\times (5^{-1})^{-2}$

$=\frac{5+1}{5}\times 5^2$

$=\frac{6}{5}\times 5^2$

=6×5=30

(b) (30⁰+ 15⁰)×$\left(\frac{1}{2}\right)^{-3}$

Sol :

=(1+1)×(2^-1)^-3=2×2³

=2⁴=16

Sol :

$=\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}\right)^0 \div \left(\frac{9}{-1}\right)^2$
$=1\div \frac{81}{1}=\frac{1}{81}$

(d) $\left[\left(\frac{1}{3}\right)^{-2}+\left(\frac{1}{4}\right)^{-2}+\left(\frac{1}{6}\right)^{-2}\right]$

Sol :

=3²+4²+6²

=9+16+36

=25+36=61

(e) $\left\{\left(-\frac{4}{5}\right)^{-2}\right\}^{-2}$

Sol :

$=\left\{\left(-\frac{5}{4}\right)\right\}^{-2}=\left(-\frac{5}{4}\right)^{-4}$

$=\left(\frac{4}{5}\right)^4=\frac{256}{625}$

(f) $9^{-1}\times \frac{4^3}{3^{-4}}$

Sol :

$=\frac{(3^2)^{-1}\times 4^3}{3^{-4}}=\frac{3^{-2}\times 4^3}{3^{-4}}$

$=\frac{4^3}{3^{-2}}$

=4³×3²=576

(g) $(5^{-1}\times 4^{-1})\times \left(\frac{3}{2}\right)^{-2}$

Sol :

$=\left(\frac{1}{5}\times \frac{1}{4}\right)\times \left(\frac{2}{3}\right)^2$

$=\frac{1}{20}\times \frac{2^2}{3^2}$

$=\frac{1}{20}\times \frac{4}{9}=\frac{1}{45}$

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Question 6

Evaluate:

(a) $\left[\left(\frac{1}{4}\right)^{-1}-\left(\frac{1}{5}\right)^{-1}\right]+1$
Sol :

=(4-5)+1

=-1+1=0

(b) $\left[\left(\frac{11}{12}\right)^{-10} \div \left(\frac{11}{12}\right)^{-12}\right]\times 1\frac{23}{121}$

Sol :

$=\left[\left(\frac{12}{11}\right)^{10}\div \left(\frac{12}{11}\right)^{12}\right]\times \frac{144}{121}$

$=\left(\frac{12}{11}\right)^{-2}\times \left(\frac{12}{11}\right)^{2}$

$=\frac{(11)^2}{(12)^2}\times \frac{(12)^2}{(11)^2}$

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Question 7

Find the value of m for which, 7^2m ÷ 7^-4 = 7^-6

Sol :

$=\frac{7^{2m}}{7^{-4}}=$7^-6

⇒7^2m=7^-6×7^-4=7^-10

⇒(7²)^m=(7²)^-5

⇒m=-5

S.chand publication New Learning Composite mathematics solution of class 8 Chapter 1 Rational numbers Exercise 1F

Exercise 1F

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Question 1

A group of friends hike $5\frac{3}{4}$ km, stops for lunch, and then hike another $3\frac{1}{5}$ km. How far did they hike?

Sol :

Before lunch hiking distance $=5\frac{3}{4}$km $=\frac{23}{4}$ km

After lunch hiking distance $=3\frac{1}{5}=\frac{16}{5}$ km

Total distance$=\frac{23}{4}+\frac{16}{5}=\frac{115+64}{20}$

$=8\frac{19}{20}$km

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Question 2

Ms Joshi walks her dog $\frac{4}{5}$ km each day. What is the total distance that Ms Joshi walks her dog in 6 days?

Sol :

Ms Joshi walks her dog $\frac{4}{5}$ km each day

Total distance that Ms Joshi walks her dog in 6 days$=6\times \frac{4}{5}=\frac{24}{5}$ $=4\frac{4}{5}$km

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Question 3

Priya completes $\frac{1}{20}$ of her painting each day. How much of her painting does she complete 5 days?

Sol :

Priya completes $\frac{1}{20}$ of her painting each day

Priya completes a portion of painting in 5 day $=5\times \frac{1}{20}=\frac{1}{4}$

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Question 4

An oxygen tank contained $219\frac{2}{3}$ L of oxygen before $32\frac{1}{3}$ L were used. If the tank can hold $245\frac{3}{8}$ L, how much space in the tank is unused?

Sol :

An oxygen tank contained $219\frac{2}{3}=\frac{659}{3}$ L

Oxygen used$=32\frac{1}{3}=\frac{97}{3}$ L

Capacity of tank$245\frac{3}{8}=\frac{1963}{8}$ L

Space unused in the tank $=\frac{1963}{8}-\left[\frac{659}{3}-\frac{97}{3}\right]$

$=\frac{1963}{8}-\frac{562}{3}=\frac{5883-4432}{24}$

$=\frac{1393}{24}=58\frac{1}{24}$ L

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Question 5

A water pipe has an outside diameter of 3.5 cm and a wall thickness of $\frac{25}{32}$ cm. What is the inside diameter of the pipe?

Sol :

Let ,

Inner diameter(d)=x

Outer diameter(D)=3.5cm

Thickness(t)$=\frac{257}{32}$cm

$t=\frac{3.5-d}{2}$

or $\frac{25}{32}=\frac{3.5-d}{2}$

or 3.5-d$=\frac{25\times 2}{32}$

or -d$=\frac{25}{16}-\frac{35}{10}$

or $d=\frac{35}{10}-\frac{25}{16}=\frac{560-250}{160}$

$=\frac{310}{160}=1\frac{15}{16}$cm

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Question 6

Find the perimeter and area of a rectangular piece of land measuring $12\frac{1}{2}$m by $5\frac{1}{5}$ m. How much area of the land is left unused if 4 small square flower beds of side length 1 m are made at the center?

Sol :

Length of rectangle(L)$=12\frac{1}{2}$cm $=\frac{25}{2}$cm

Breadth of rectangle (B)$=5\frac{1}{5}$cm $=\frac{26}{5}$ cm

Perimeter of rectangle=2(L+B)

$=2\left(\frac{25}{2}+\frac{26}{5}\right)$

$=2\left(\frac{125+52}{10}\right)=2\times \frac{177}{10}$

$=\frac{177}{5}=35\frac{2}{5}$cm

∴Area of rectangle(L×B)$=\left(\frac{25}{2}\times \frac{26}{5}\right)$

=65cm²

∴The flowered area=1²=1cm²

∴4 flowered bed area=4×1=4cm²

∴Area of the land is left unused=(65-4)=61cm²

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Question 7

Out of a certain sum of money a boy spends $\frac{3}{5}$, and then $\frac{1}{4}$ of the remainder. He has Rs 15 left. How much amount had he at first?

Sol :

Total money=x

Money spend $=\frac{3x}{5}$

Remainder$=\frac{1}{4}\left(x-\frac{3x}{5}\right)$

ATQ,

$x-\left[\frac{3x}{5}+\frac{1}{4}\left(x-\frac{3x}{5}\right)\right]=15$

or $x-\left[\frac{3x}{x}+\frac{1}{4}\left(\frac{5x-3x}{5}\right)\right]=15$

or $x-\left[\frac{3x}{5}+\frac{2x}{20}\right]=15$

or $x-\left[\frac{12x+2x}{20}\right]=15$

or $x-\frac{14x}{20}=15$

or $\frac{20x-14x}{20}=15$

or 6x=15×20

or $x=\frac{515\times 20}{6}=50$

Ans 50

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Question 8

$10\frac{1}{2}$ tonnes of sand are to be shared between a number of builders. One of them receives $\frac{4}{7}$ of the total and the remaining sand is shared by 3 builders. How much does the fourth builder receive and how much sand does each of the other 3 builders receive?

Sol :

Total sand $=10\frac{1}{2}=\frac{21}{2}$ tonnes

Fourth builder get$=\frac{21}{2}\times \frac{4}{7}$=6 tonnes

Rest sand$=\frac{21}{2}-6=\frac{21-12}{2}=\frac{9}{2}$

Distributing into 3$=\frac{9}{2\times 3}=\frac{3}{2}$ tonnes

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Question 9

A man aged 27 marries a woman aged 24, he dies at the age of 81 and she dies at the age of 91. For what fraction of his life is the man married? For what fraction of her life is the woman a widow?

Sol :

Man married life duration$=\frac{81-27}{81}=\frac{2}{3}$

Woman widow life duration$=\frac{91-(54+24)}{91}=\frac{91-48}{91}$ $=\frac{13}{91}=\frac{1}{7}$

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Question 10

Shalini has to cut out circles a diameter $1\frac{1}{4}$ cm from an aluminium strip of dimensions $8\frac{3}{4}$ cm by $1\frac{1}{4}$ cm. How many full circles can Shalini cut? Also calculate the wastage of the aluminium strip.

Sol :

Circle diameter(d)$=1\frac{1}{4}$cm

Strip length$=8\frac{3}{4}=\frac{35}{4}$cm

Strip breadth$=1\frac{1}{4}=\frac{5}{4}$

∴Number of full circles which can be cut from the strip$=\frac{35}{4}\div \frac{5}{4}$

$=\frac{35}{4}\times \frac{4}{5}$

$r=\dfrac{\frac{5}{4}}{2}=\frac{5}{8}$

Area of circle=πr²=$\pi\times \frac{5}{8}\times \frac{5}{8}$

$=\frac{22}{7} \times \frac{5}{8}\times \frac{5}{8}$cm²

Area of 7 circles $=7\times\frac{22}{7} \times \frac{5}{8}\times \frac{5}{8}$

$=\frac{275}{32}$cm²

Area of strip$=\frac{35}{4}\times \frac{5}{4}=\frac{175}{16}$cm²

∴Wastage=(Area of strip-Area of 7circles)

$=\frac{175}{16}-\frac{275}{32}$

$=\frac{175\times 2-275}{32}=\frac{350-275}{32}$

$=\frac{75}{32}=2\frac{11}{32}$ cm²

S.chand publication New Learning Composite mathematics solution of class 8 Chapter 12 Mensuration Exercise 12C

Exercise 12C

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

S.chand publication New Learning Composite mathematics solution of class 8 Chapter 6 Factorisation of Algebraic Expressions Exercise 6C

Exercise 6C

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

S.chand publication New Learning Composite mathematics solution of class 8 Chapter 2 Exponents Exercise 2A

Exercise 2A

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

S.chand publication New Learning Composite mathematics solution of class 8 Chapter 1 Rational numbers Exercise 1F

Exercise 1F

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

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