CBSE Class 8

These NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers InText Questions

NCERT Intext Question Page No. 194

Question 1.
Find the multiplicative inverse of the following:
(i) 2^-4
(ii) 10⁵
(iii) 7^-2
(iv) 5^-3
(v) 10^-100
Answer:
(i) The multiplicative inverse of 2^-4 = 2⁴
(ii) The multiplicative inverse of 10^-5 = 10⁵
(iii) The multiplicative inverse of 7^-2 = 7²
(iv) The multiplicative inverse of 5^-3 = 5³
(v) The multiplicative inverse of 10^-100 = 10¹⁰⁰

Question 2.
Expand the following numbers using exponents.
(i) 1025.63
(ii) 1256.249
Answer:

NCERT Intext Question Page No. 195

Question 1.
Simplify and write in exponential form.
(i) (-2)^-3 x (-2)^-4
(ii) p³ x p^-10
(iii) 3² x 3^-5 x 3⁶
Answer:
(i) (-2)^-3 x (-2)^-4 = (-2)^3-4
[a^m x aⁿ = a^m+n]
= (-2)^-7 = \(\frac{1}{(-2)^{7}}\)

(ii) p³ x p^-10 = p^{3 + (-10)} = p^{3 – 10} = p⁷ or \(\frac{1}{\mathrm{p}^{7}}\)

(iii) 3² x 3^-5 x 3⁶ = 3^{2 + (- 5) + 6} = 3^{2 – 5 + 6} = 3^8-5 = 3³

NCERT Intext Question Page No. 199

Question 1.
Write the following numbers in standard form.
(i) 0.000000564
(ii) 0.0000021
(iii) 15240000
Answer:
(i) 0.000000564 = \(\frac{564}{1000000000}=\frac{564}{10^{9}}\)
= \(\frac{5.64 \times 10^{2}}{10^{9}}\) = 5.64 x 10^-7

(ii) 0.0000021 = \(\frac{21}{10000000}=\frac{21}{10^{7}}\)
= \(\frac{2.1 \times 10}{10^{7}}\) = 2.1 x 10^1-7 = 2.1 x 10^-6

(iii) 15240000 = 1524 x 10000
= 1.524 x 1000 x 10000 = 1.524 x 10⁷
∴ 15240000 = 1.524 x 10⁷

Question 2.
Write all the facts give in the standard form.
Answer:
A number is said to be in the standard form when it is written as k x 10ⁿ, Where 1 ≤ k < 10 and ‘n’ is an integer. A number expressed as the product of a number between 1 and 10 and an integral power of 10.
Example: Compare the size of a red blood cell which is 0.000007m to that of a plant cell which is 0.0000129m.
Size of red blood cell = 0.000007 m
= \(\frac{7}{1000000}\) x 10^-6 m
Size of the plant cell = \(\frac{7}{10000000} \mathrm{~m}=\frac{1.29 \times 10^{2}}{10000000} \mathrm{~m}\)
= 1.29 x 10^-5 m
Now = \(\frac{7 \times 10^{-6}}{1.29 \times 10^{-5}}=\frac{7 \times 10^{-1}}{1.29}\)
= \(\frac{7 \times 10^{-1}}{1.3}=\frac{0.7}{1.3}=\frac{1}{2}\) (approx)
Thus, the size of a red blood cell is half of the plant cell size.

These NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers Ex 12.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers Exercise 12.2

Question 1.
Express the following numbers in standard form.
(i) 0.0000000000085
(ii) 0.00000000000942
(iii) 6020000000000000
(iv) 0.00000000837
(v) 31860000000
Answer:
(i) 0.0000000000085 = \(\frac{85}{10^{13}}=\frac{8.5 \times 10}{10^{13}}\)
= 8.5 × 10 × 10^-13 = 8.5 × 10^-12

(ii) 0.00000000000942 = \(\frac{942}{10^{14}}\)
= \(\frac{9.42 \times 100}{10^{14}}=\frac{9.42 \times 10^{2}}{10^{14}}\)
= 9.42 × 10² × 10^-14 = 9.42 × 10^-12

(iii) 6020000000000000 = 602 × 10¹³
= 6.02 × 10² × 10¹³ = 6.02 × 10¹⁵

(iv) 0.00000000837 = \(\frac{837}{10^{11}}=\frac{8.37 \times 10^{2}}{10^{11}}\)
= 8.37 × 10^{2 -11} = 8.37 × 10^-9

(v) 31860000000 = 3186 × 10⁷
= 3.186 × 10³ × 10⁷
= 3.186 × 10³ + 10³⁺⁷
= 3.186 × 10¹⁰

Question 2.
Express the following numbers in usual form.
(i) 3.02 × 10^-6
(ii) 4.5 × 10⁴
(iii)3 × 10^-8
(iv) 1.0001 × 10⁹
(v) 5.8 × 10¹²
(vi) 3.61492 × 10⁶
Answer:
(i) 3.02 × 10^-6 = 302 x 10^-2 × 10^-6
= 302 × 10^-8 = 0.00000302

(ii) 4.5 × 10⁴ = \(\frac{45}{10}\) × 10000 = 45000

(iii) 3 × 10^-5 = \(\frac{3}{10^{8}}\) = 0.00000003

(iv) 1.0001 × 10⁹ = \(\frac{10001 \times 10^{9}}{10^{4}}\)
= 10001 × 10^9-4 = 10001 × 10⁵
= 1000100000

(v) 5.8 × 10¹² = \(\frac{58}{10}\) × 10¹² = 58 × 10^12-1
= 58 × 10^12-1 = 58 × 10¹¹
= 5800000000000

(vi) 3.61492 x 10⁶ = \(\frac{361492}{10^{5}}\) × 10⁶
= 361492 × 10^6-5 = 361492 × 10
= 36149200

Question 3.
Express the number appearing in the following statements in standard form.
(i) 1 micron is equal to \(\frac{1}{1000000}\) m.
(ii) Charge of an electron is 0.000,000,000,000,000,000,16 coulomb.
(iii) Size of a bacteria is 0.0000005 m
(iv) Size of a plant cell is 0.00001275 m
(v) Thickness of a thick paper is 0.07 mm
Answer:
(i) \(\frac{1}{1000000}=\frac{1}{10^{6}}\) = 1 × 10^-6
1 micron = 1 × 10^-6

(ii) 0.000,000,000,000,000,000,16 = \(\frac{16}{10^{20}}\)
= \(\frac{1.6 \times 10}{10^{20}}\) = 1.6 × 10^1-20 = 1.6 × 10^-19
.’. Charge on an electron is 1.6 × 10^-19 coulomb.

(iii) 0.0000005 = \(\frac{5}{10^{7}}\) = 5 × 10^-7
Size of a bacteria = 5 × 10^-7 m

(iv) 0.00001275 = \(\frac{1275}{10^{8}}=\frac{1.275 \times 10^{3}}{10^{8}}\)
= 1.275 × 10^3-8 = 1.275 × 10^-5
Size of a plant cell is 1.275 × 10^-5m

(v) 0.07 = \(\frac{7}{10^{2}}\) = 7 × 10^-2
.’. Thickness of a thick paper is 7 × 10^-2mm

Question 4.
In a stack, there are 5 books each of thickness 20 mm and 5 paper sheets each of thickness 0.016 mm. What is the total thickness of the stack.
Answer:
Thickness of one book = 20 mm
Thickness of 5 books = 5 × 20 mm = 100 mm
Again, Thickness of one paper sheet = 0.016 mm
.’. Thickness of 5 paper sheets = 5× 0.016 mm
= 0.080 m
Total thickness =100 mm + 0.080 mm = 100.08 mm
\(\frac{10008}{10^{2}}=\frac{1.0008 \times 10^{4}}{10^{2}}\)
= 1.0008 × 10² mm

These NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers Ex 12.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers Exercise 12.1

Question 1.
Evaluate
(i) 3^-2
(ii) (-4)^-2
(iii) \(\left(\frac{1}{2}\right)^{-5}\)
Answer:
(i) 3^-2 = \(\frac{1}{3^{2}}=\frac{1}{9}\)
(ii) (-4)^-2 = \(\frac{1}{(-4)^{2}}=\frac{1}{16}\)

Question 2.
Simplify and express the result in power notation with positive exponent.
(i) (-4)5 ÷ (-4)⁸
(ii) \(\left(\frac{1}{2^{3}}\right)^{2}\)
(iii) (-3)⁴ × \(\left(\frac{5}{3}\right)^{4}\)
(iv) (3^-7 ÷ 3¹⁰) × 3^-5
(v) 2^-3 × (-7)-3
Answer:

Question 3.
Find the value of
(i) (3° + 4^-1) × 2²
(ii) (2^-1 × 4^-1) ÷ 2^-2
(iii) \(\left(\frac{1}{2}\right)^{-2}+\left(\frac{1}{3}\right)^{-2}+\left(\frac{1}{4}\right)^{-4}\)
(iv) (3^-1 + ^-1 + 5^-1)0
(v) \(\left\{\left(\frac{-2}{3}\right)^{-2}\right\}^{2}\)
Answer:
(i) (3° + 4^-1) × 2²
= (1 + \(\frac{1}{4}\) ) × 2²
[(i) a⁰ = 1 (ii) a^-m = \(\frac{1}{a^{m}}\) ]
= \(\left(\frac{4+1}{4}\right)\) × 4 = \(\frac{5}{4}\) × 4 = 5

Question 4.
Evaluate:
(i) \(\frac{8^{-1} \times 5^{3}}{2^{-4}}\)
(ii) (5^-1 × 2^-1) × 6^-1
Answer:
\(\frac{8^{-1} \times 5^{3}}{2^{-4}}=\frac{5^{3} \times 2^{4}}{8^{1}}=\frac{(5 \times 5 \times 5) \times 2^{4}}{2^{3}}\)
= 125 × 2^4-3 = 125 × 2¹ = 250

(ii) (5^-1 × 2^-1) × 6^-1 = \(\left(\frac{1}{5} \times \frac{1}{2}\right) \times \frac{1}{6}\)
[a^-m = \(\frac{1}{\mathrm{a}^{\mathrm{m}}}\) ]
= \(\frac{1}{10} \times \frac{1}{6}=\frac{1}{60}\)

Question 5.
Find the value of m for which 5m ÷ 5 3 = 55.
Answer:
5^m ÷ 5^-3 = 5⁵
5^m ÷ \(\frac{1}{5^{3}}\) = 5⁵
5^m × 5³ = 5⁵
5^m+3 = 5⁵ (a^m × aⁿ = a^m+n)
∴ m + 3 = 5 (since the bases are equal, the exponents are equal)
m = 5 – 3
m = 2
The value of m = 2.

Question 6.
Evaluate
(i) \(\left\{\left(\frac{1}{3}\right)^{-1}-\left(\frac{1}{4}\right)^{-1}\right\}^{-1}\)
(ii) \(\left(\frac{5}{8}\right)^{-7} \times\left(\frac{8}{5}\right)^{-4}\)
Answer:

Question 7.
Simplify:
(i) \(\frac{25 \times \mathrm{t}^{-4}}{5^{-3} \times 10 \times \mathrm{t}^{-8}}(\mathrm{t} \neq 0)\)
(ii) \(\frac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}}\)
Answer:

These NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions InText Questions

NCERT Intext Question Page No. 204

Question 1.
Observe the following tables and find if x and y are directly proportional.

Answer:

i. e. each ratio is the same
.’. x and y are directly proportional

i. e. all the ratios are not the same x
and y are not directly proportional.

i.e. all the ratios are not the same
x and y are not directly proportional.

Question 2.
Principal = ₹ 1000, Rate = 8% per annum. Fill in the following table and find which type of interest (simple or compound) change is in direct proportion with time period.

Answer:
Case of Simple Interest
[P = ₹ 1000, r = 8% p.a.]

In each case the ratio is the same.
The simple interest changes in direct proportion with time period.
Case of Compound Interest [P = ₹ 1000, r = 8% p.a.]

\(\frac{\mathrm{CI}}{\mathrm{T}}\) is not the same in each case.
∴ The compound interest does not change in direct proportion with time period.

NCERT Intext Question Page No. 211

Question 1.
Observe the following tables and find which pair of variables (here x and y) are in the inverse proportion.

Answer:
(i) x₁y₁ = 50 x 5 = 250
x₂y₂ = 40 x 6 = 240
x₁y₁ ≠ x₂y₂
Hence, x and y are not in inverse proportion.

(ii) x₁y₁ = 100 x 60 = 6000
x₂y₂ = 200 x 30 = 6000
x₃y₃ = 300 x 20 = 6000
x₄y₄ = 400 x 15 = 6000

x₁y₁ = x₂y₂ = x₃y₃ = x₄y₄
Hence x and y are in inverse proportion.
(iii) x₁y₁ = 90 x 10 = 900
x₂y₂ = 60 x 15 = 900
x₃y₃ = 45 x 20 = 900
x₄y₄ = 30 x 25 = 750
x₁y₁ = x₂y₂ = x₃y₃ ≠ x₄y₄
.’. x and y are not in inverse proportion.
Note: (1) When two quantities x and y are in direct proportion (or vary directly), they are also written is x ∝ y.
(2) When two quantities x and y are in inverse proportion (or very inversely), they are also written as x ∝ \(\frac{1}{y}\).

These NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Exercise 11.3

Question 1.
There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make?


Answer:
(a) Here, l = 60 cm, b = 40 cm; h = 50 cm Total surface area of the cuboid
= 2 (lb + bh + hl)
= 2 [(60 × 40) + (40 × 50) + (50 × 60)] cm²
= 2 [2400 + 2000 + 3000] cm²
= 2 (7400) cm² = 14800 cm²

(b) Here, l = 50 cm, b = 50 cm, h = 50 cm
Total surface area of the cuboid
= 2 (lb + bh + lh)
= 2 [(50 × 50)+ (50 × 50)+ (50 × 50)] cm²
= 2 [2500 + 2500 + 2500] cm²
= 2 × 7500 cm² = 15000 cm²
Since, the total surface area of the second (b) is greater:
∴ Cuboid (a) will required lesser material.

Question 2.
A suitcase with measures 80 cm × 48 cm × 24 cm is to be covered with a tarpaulin cloth. How many metres of tarpaulin of width 96 cm is required to cover 100 such suitcases?
Answer:
Here, length = 80 cm, breadth = 48 cm, height = 24 cm
Total surface are a of asuitcase = 2(lb+bh+lh)
= 2[80 × 48 + 48 × 24 + 24 × 80]
= 2 [3840 + 1152 + 1920]cm²
= 2[6912] cm² = 13824 cm²
Required tarpaulin for 100 suitcases = 13824 x 100 cm²
Width of the tarpaulin = 96 cm
Length, tarpaulin required = \(\frac{13824 \times 100}{96}\)cm
[1 m = 100 cm]
= \(\frac{13824 \times 100}{96 \times 100}\)m = \(\frac{13824}{96}\)m = 144 m
The required length of tarpaulin for 100 suitcases = 144 m

Question 3.
Find the side of a cube whose surface area is 600 cm2
Answer:
Let the side of a cube be ‘x’ cm
Total surface area of the cube = 600 cm²
6 × x² = 600
x² = \(\frac{600}{6}\)
x² = 100
x² = 10²
x = 10
The required side of the cube =10 cm.

Question 4.
Rukshar painted the outside of the cabinet of measure 1m × 2m × 1.5m. How much surface area did she cover if she painted all except the bottom of the cabinet.

Answer:
Total surface area of the cabinet = 2 × [lb + bh + lh] sq m
Area not to be painted = Bottom of the cabinet = lb
Area to be painted = T.S.A – lb
= 2 (lb + bh + lh) – lb
= 2[1 × 2 + 2 × 1.5 + 1.5 × 1]-(1 × 2)m²
= 2(2 + 3 + 1.5) -2 m²
= 2 × 6.5 – 2 m²
= 13 – 2 m²
= 11 m²
Area to be painted = 11 m².

Question 5.
Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth and height of 15 m, 10 m and 7 m respectively. From each can of paint, 100 m² of area is painted. How many cans of paint will she need to paint the room?
Answer:
Here l = 15 m, b = 10 m and h = 7 m.
Area to the painted = Area of four walls + Area of ceiling.
= 2 (bh + hl) + lb
2 [10 × 7 +7 × 15] + 15 × 10 m²
= 2 [70 + 105] + 150 m²
= 2(175) + 150 m²
= 350 + 150 m² = 500 m²
= Number of cans needed = \(\frac{\text { Area to be painted }}{\text { Area painted by } 1 \text { can }}\) = \(\frac{500}{100}\) = 5
Number of cans needed = 5.

Question 6.
Describe how the two figures at the right are alike and how they are different. Which box has larger lateral surface area?

Answer:
Similarity: Both have the same height
Difference: Cylinder has curved and circular surfaces.
but in cube, all faces are identical squares
Lateral surface area of the cylinder
= 2πrh = 2 × \(\frac{22}{7}\) × \(\frac{7}{2}\) × 7 cm²
= 7 × 22 cm² =154 cm²
Lateral surface area of the cube = 4l2 sq.m
= 4 × 7 × 7 cm² = 196 cm²

Question 7.
A closed cylindrical tank of radius 7 m and height 3 m is made from a sheet of metal. How much sheet of metal is required?
Answer:
Here r = 7m, h = 3m
Total surface area of the cylinder = 2 π r (r + h)sq. units
= 2 × \(\frac{22}{7}\) × 7(7 + 3) = 44 × 10 m² = 440 m²
Metal required for the tank = 440 m².

Question 8.
The lateral surface area of a hollow cylinder is 4224 cm². It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of rectangular sheet?
Answer:
Lateral surface area of the cylinder = 4224 cm²
Let the length of the rectangular sheet be l cm
Width of the sheet = 33 cm.
Area of the rectangular sheet = L.S.A of the cylinder
l × 33 = 4224
l = \(\frac{4224}{33}\) cm = 128 cm
Perimeter of the rectangular sheet = 2 (l + b)
= 2 (128 + 33) cm = 2 × 161 cm = 322 cm

Question 9.
A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is 84 cm and length is 1m.

Answer:
Radius of the roller = \(\frac{84}{2}\) = 42cm
Length of the roller = 1 m = 100 cm
Lateral surface area = 2 π rh sq unit
[road roller is a cylinder]
= 2 × \(\frac{22}{7}\) × 42 × 100 cm²
= 44 × 6 × 100 cm² = 26400 cm²
Area levelled by 750 revolutions
= 750 × 26400 cm²
=\(\frac{750 \times 26400}{100 \times 100}\)m² = \(\frac{75 \times 264}{10}\) = 1980 m²

Question 10.
A company packages its milk powder in cylindrical container whose base has a diameter of 14 cm and height 20 cm. Company places a label around the surface of the container (as shown in the figure). If the label is placed 2 cm from top and bottom, what is the area of the label.

Answer:
Radius of the label = \(\frac{14}{2}\) = 7 cm
Height of the label = [20 – (2 + 2)] cm
= (20 – 4) cm = 16 cm
Area of the label = Lateral surface area of the cylinder.
= 2πrh = 2 × \(\frac{22}{7}\) × 7 × 16
= 44 × 16 cm² = 704 cm²

These NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions Exercise 13.1

Question 1.
Following are the car parking charges near a railway station upto

Check if the parking charges are in direct proportion to the parking time.
Answer:

Since \(\frac{15}{1} \neq \frac{25}{2} \neq \frac{35}{3} \neq \frac{15}{2}\)
∴ The parking charges are not in direct proportion with the parking time.

Question 2.
A mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base. In the following table, find the parts of base that need to be added.

Answer:
Let the number of parts of red pigment be x₁, x₂, x₃, x₄ , x₅ and the number of parts of
the base be y₁, y₂, y₃, y₄, y₅
As the number of parts of red pigment increases, number of parts of the base also increases in the same ratio It is a direct proportion

The required parts of bases are 32, 56, 96 and 160.

Question 3.
In question 2 above, if 1 part of a red pigment requires 75 mL of base, how much red pigment should we mix with 1800 mL of base?
Answer:
We have \(\frac{\mathrm{x}_{1}}{\mathrm{y}_{1}}=\frac{1}{8}=\frac{\mathrm{x}_{2}}{\mathrm{y}_{2}}\)
Here x₁ = 1, y₁ = 75 and y₂ = 1800
∴ \(\frac{1}{75}=\frac{x_{2}}{1800}\)
x₂ = \(\frac{1 \times 1800}{75}\) = 24
Thus, the required no. of red pigments = 24.

Question 4.
A machine in a soft drink factory fills 840 bottles in six hours. How many bottles will it fill in five hours?
Answer:

More the number of hours, more the number of bottles that would be filled. Thus, given quantities vary directly.
∴ \(\frac{840}{x}=\frac{6}{5}\)
x = \(\frac{840 \times 5}{6}\) = 140 × 5 = 700
∴ The required number of bottles = 700

Question 5.
A photograph of a bacteria enlarged 50,000 times attains a length of 5 cm as shown in the diagram. What is the actual length of the bacteria? If the photograph is enlarged 20,000 times only, what would be its enlarged length?
Answer:
Actual length = \(\frac{5}{50000}=\frac{1}{10000}\) = 10^-4 cm

The length increases with an increment in the number of times the photograph enlarged.
∴ It is a case of direct proportion.
\(\frac{50000}{20000}=\frac{5}{x}\)
50,000 × x = 5 × 20,000
x = \(\frac{5 \times 20,000}{50,000}\) = 2
∴The enlarged length = 2 cm.

Question 6.
In a model of a ship, the mast is 9 cm high, while the mast of the actual ship is 12 m high. If the length of the ship is 28 m, how long is the model ship?
Answer:
Let the required length of the model ship be ‘x’ cm

Since, more the length of the ship, more would be the length of its mast.
∴ It is a direct variation \(\frac{28}{x}=\frac{12}{9}\)
12 × x = 28 × 9
x = \(\frac{28 \times 9}{12}=\frac{7 \times 9}{3}\) = 7 × 3 = 21
The required length of the model = 21 cm.

Question 7.
Suppose 2 kg of sugar contains 9 x 106 crystals. How many sugar crystals are there in (i) 5 kg of sugar? (ii) 1.2 kg of sugar?
Answer:
(i) Let the required number of sugar crystals be Y.
Since more the amount of sugar, more would be the number of sugar crystals.
∴ It is a direct variation
\(\frac{2}{5}=\frac{9 \times 10^{6}}{\mathrm{x}}\)
2x = 5 × 9 × 10⁶
x = \(\frac{5 \times 9 \times 10^{6}}{2}=\frac{45 \times 10^{6}}{2}\) = 22.5 × 10⁶
= 2.25 × 10 × 10⁶
= 2.25 × 10⁷
∴ Required number of sugar crystals = 2.25 x 10⁷

(ii) Let the number of sugar crystals in a sugar be ‘y’
It is a direct variation
\(\frac{2}{1.2}=\frac{9 \times 10^{6}}{\mathrm{y}}\)
2y = 9 × 10⁶ × 1.2
y = \(\frac{9 \times 10^{6} \times 1.2}{2}\) = 9 × 10⁶ × 0.6
= 5.4 × 10⁶
The required number of sugar crystals = 5.4 × 10⁶

Question 8.
Rashmi has a road map with a scale of 1 cm representing 18 km. She drives on a road for 72 km. What would be her distance covered in the map?
Answer:
Let the required distance covered in the map be ‘x’ cm.

It is a direct variation
\(\frac{18}{72}=\frac{1}{x}\)
18 × x = 72 × 1
x = \(\frac{72 \times 1}{18}\) = 4 × 1 = 4
∴ The required distance on the map is 4 cm

Question 9.
A 5 m 60 cm high vertical pole casts a shadow 3 m 20 cm long. Find at the same time
(i) the length of the shadow cast by another pole 10 m 50 cm high
(ii) the height of a pole which casts a shadow 5 m long.
Answer:
Let the required length of shadow be ‘x’ cm

As the height of the pole increases, the length of the shadow also increases in the same ratio
It is a direct variation.
\(\frac{560}{1050}=\frac{320}{\mathrm{x}}\)
560 × x = 1050 × 320
x = \(\frac{1050 \times 320}{560}\) = 600 cm
∴ Required length of the shadow = 600 cm (6 m)

(ii) Let the required height of the pole be y’ cm

It is a direct variation.
\(\frac{560}{y}=\frac{320}{500}\)
y × 320 = 560 × 500
y = \(\frac{560 \times 500}{320}\) = 125 × 7 = 875
7 320
∴ The height of the pole = 875 cm (or) 8 m 75 cm.

Question 10.
A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how for can it travel in 5 hours?
Answer:
Let the distance travelled in 5 hours be ‘x’ km.

If the time increases, the distance also increases.
∴ It is a direct variation.
\(\frac{14}{x}=\frac{25}{300}\)
25 × x = 300 × 14
x = \(\frac{300 \times 14}{25}\) = 12 × 14 = 168
∴ The required distance =168 km.