Class 8

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 12 Exponents and Powers InText Questions and Answers.

PSEB 8th Class Maths Solutions Chapter 12 Exponents and Powers InText Questions

Try These : [Textbook Page No. 194]

1. Find the multiplicative inverse of the following:

Question (i)
2^-4
Solution:
Multiplicative inverse of 2^-4 = 2⁴

Question (ii)
10^-5
Solution:
Multiplicative inverse of 10^-5 = 10⁵

Question (iii)
7^-2
Solution:
Multiplicative inverse of 7^-2 = 7²

Question (iv)
5^-3
Solution:
Multiplicative inverse of 5^-3 = 5³

Question (v)
10^-100
Solution:
Multiplicative inverse of 10^-100 =10¹⁰⁰

Try These : [Textbook Page No. 194]

1. Expand the following numbers using exponents:
(i) 1025.63
(ii) 1256.249
Solution:

Number	Expanded form
(i) 1025.63	(1 × 1000) + (0 × 100) + (2 × 10) + (5 × 1) + (6 × \(\frac {1}{10}\)) + (3 × \(\frac {1}{100}\)) OR (1 × 103) + (2 × 101) + (5 × 100) + (6 × 10-1) + (3 × 10-2)
(ii) 1256.249	(1 × 1000) + (2 × 100) + (5 × 10) + (6 × 1) + (2 × \(\frac {1}{10}\))+ (4 x \(\frac {1}{100}\)) + (9 × \(\frac {1}{1000}\)) OR (1 × 103) + (2 × 102) + (5 × 101) + (6 × 100) + (2 × 10-1) + (4 × 10-2) + (9 × 10-3)

Try These : [Textbook Page No. 195]

1. Simplify and write in exponential form:

Question (i)
(- 2)^-3 × (- 2)^-4
Solution:
= (-2)^-3+(-4)
= (-2)^-3-4
= (-2)^-7 or \(\frac{1}{(-2)^{7}}\)

Question (ii)
p³ × p^-10
Solution:
= p^3+(-10)
= p^3-10
= (p)^-7 or \(\frac{1}{(p)^{7}}\)

Question (iii)
3² × 3^-5 × 3⁶
Solution:
= 3^2+(-5)+6
= 3^2-5+6
= 3^2+6-5
= 3³

Try These : [Textbook Page No. 199]

1. Write the following numbers in standard form:

Question (i)
0.000000564
Solution:
= \(\frac{564}{1000000000}\)
(The decimal point is shifted to nine places to the right.)
= \(\frac{5.64}{10^{9}}\)
= \(\frac{5.64}{10^{7}}\)
= 5.64 × 10^-7
∴ 0.000000564 = 5.64 × 10^-7

Question (ii)
0.0000021
Solution:
\(\frac{21}{10000000}\)
= \(\frac{2.1 \times 10}{10000000}\)
= \(\frac{2.1}{10^{6}}\)
= 2.1 × 10^-6
∴ 0.0000021 = 2.1 × 10^-6

Question (iii)
21600000
Solution:
= 216 × 100000
= 216 × 10⁵
= 2.16 × 10² × 10⁵
= 2.16 × 10⁷
∴ 21600000 = 2.16 × 10⁷

Question (iv)
15240000
Solution:
= 1524 × 10000
= 1.524 × 1000 × 10000
= 1.524 × 10³ × 10⁴
= 1.524 × 10⁷
∴ 15240000 = 1.524 × 10⁷

2. Write all the facts given in the standard form. Observe the following facts: [Textbook Page No. 198 ]

Question 1.
The distance from the Earth to the Sun is 150,600,000,000 m.
Solution:
150,600,000,000 m
= 1.506 × 10¹¹ m

Question 2.
The speed of light is 300,000,000 m/sec.
Solution:
300,000,000 m / sec
= 3 × 10⁸m/sec

Question 3.
Thickness of Class VII Mathematics book is 20 mm.
Solution:
20 = 2 × 10¹ mm

Question 4.
The average diameter of a Red Blood Cell is 0.000007 m.
Solution:
0.000007 = 7 × 10^-6 m

Question 5.
The thickness of human hair is in the range of 0.005 cm to 0.01 cm.
Solution:
0.005 = 5 × 10^-3cm and
0.01 = 1 × 10^-2 cm

Question 6.
The distance of moon from the Earth is 384,467,000 m (approx).
Solution:
384,467,000 = 3.84467 × 10⁸ m

Question 7.
The size of a plant cell is 0.00001275 m.
Solution:
0.00001275 = 1.275 × 10^-5m

Question 8.
Average radius of the Sun is 695000 km.
Solution:
695000 = 6.95 × 10⁵ km

Question 9.
Mass of propellant in a space shuttle solid rocket booster is 503600 kg.
Solution:
503600 = 5.036 × 10⁵ kg

Question 10.
Thickness of a piece of paper is 0.0016 cm.
Solution:
0.0016 = 1.6 × 10^-3 cm

Question 11.
Diameter of a wire on a computer chip is 0.000003 m.
Solution:
0.000003 = 3 × 10^-6 cm

Question 12.
The height of Mount Everest is 8848 m.
Solution:
8848 = 8.848 × 10³ m

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 4 Practical Geometry InText Questions and Answers.

PSEB 8th Class Maths Solutions Chapter 4 Practical Geometry InText Questions

Think, Discuss and Write (Textbook Page No. 58)

Question 1.
Arshad has five measurements of a quadrilateral ABCD. These are AB = 5 cm, ∠A = 50°, AC = 4 cm, BD = 5 cm and AD = 6 cm. Can he construct a unique quadrilateral ? Give reasons for your answer.
Solution:
No, the quadrilateral ABCD cannot be constructed with the given combination of measurements. If the length of side BC or DC is given, then only □ ABCD can be constructed.

Think, Discuss and Write (Textbook Page No. 60)

Question (i).
We saw that 5 measurements of a quadrilateral can determine a quadrilateral uniquely. Do you think any five measurements of the quadrilateral can do this ?
Solution:
No, any 5 measurements can’t determine a s quadrilateral. To construct a quadrilateral, specific combination of measurements should be needed such as :

• Four sides and one diagonal
• Four sides and one angle
• Three sides and two diagonals
• Two adjacent sides and three angles
• Three sides and two included angles
• Some special properties should be given,

Question (ii).
Can you draw a parallelogram BATS where BA = 5 cm, AT = 6 cm and AS = 6.5 cm? Why?
Solution:

Here, to draw a parallelogram BATS, BA = 5 cm, AT = 6 cm and AS = 6.5 cm are given. In parallelogram length of opposite sides are equal. So ST = AB = 5 cm and SB = AT = 6 cm. First we can draw A ASB where SB = 6 cm, AB = 5 cm and AS = 6.5 cm. Then draw ΔATS where AT = 6 cm, ST = 5 cm.
Thus, we can draw a parallelogram from given measurements.

Question (iii).
Can you draw a rhombus ZEAL where ZE = 3.5 cm, diagonal EL = 5 cm ? Why?
Solution:

Here, to draw a rhombus ZEAL, ZE = 3.5 cm and EL = 5 cm are given. All of rhombus are equal to each other. So ZE = EA = AL = LZ = 3.5 cm. Moreover, diagonal EL = 5 cm Is given.
So We know all necessary measurements to draw rhombus. Yes, we can draw a rhombus ZEAL.

Question (iv).
A student attempted to draw a quadrilateral PLAY where PL = 3 cm, LA = 4 cm, AY = 4.5 cm, PY = 2 cm and LY = 6 cm, but could not draw it. What is the reason?
[Hint: Discuss it using a rough sketch. ]
Solution :
Here, to draw a quadrilateral PLAY,
PL = 3 cm, LA = 4 cm, AY = 4.5 cm, PY = 2 cm and LY = 6 cm are given.

Now, let us look at measurements of
ΔPLY. PL + PY = 3 cm + 2 cm = 5 cm while YL = 6 cm.
We know that the sum of the lengths of any two sides of triangle is always greater than the length of the third side.
So point P cannot be determined even after constructing ΔLAY. Thus, a student failed s due to this reason.

Think, Discuss and Write (Textbook Page No. 62)

Question 1.
In the above example, can we draw the quadrilateral by drawing ΔABD first and then find the fourth point C ?
Solution:
Since, the measurement of AB is not given, we cannot draw ΔABD, so question does not arise to find the foruth point C.

Thus, we cannot draw the quadrilateral

Question 2.
Can you construct a quadrilateral PQRS with PQ = 3 cm, RS = 3 cm, PS = 7.5 cm, PR = 8 cm and SQ = 4 cm ? Justify your answer.
Solution:
No, the quadrilateral PQRS cannot be constructed as in ΔQSP, SQ + PQ ≯ SP

Think, Discuss and Write (Textbook Page No. 64)

Question 1.
Can you construct the above quadrilateral MIST if we have 100° at M instead of 75° ?
Solution:
Yes. The quadrilateral MIST can be constructed with ∠M = 100° instead of 75°.
[Note : Only size of quadrilateral is changed.]

Question 2.
Can you construct the quadrilateral PLAN if PL = 6 cm, LA = 9.5 cm, ∠P = 75°, ∠L = 150° and ∠A = 140° ?
[Hint: Recall angle sum property.]
Solution:
Here, ∠P + ∠L + ∠A + ∠N
= 75° + 150° + 140° + ∠N
= 365° + ∠N
∴ Construction of quadrilateral PLAN is not possible as according to angle sum property. The sum of all the angles of a quadrilateral is 360°. Here, 365° + N > 360°.

Question 3.
In a parallelogram, the lengths of adjacent sides are known. Do we still need measures of the angles to construct as in the example above?
Solution :
In a parallelogram, the opposite sides are parallel and of equal length. Here, we know the lengths of adjacent sides, so measures of the angles are not needed.
If we know the length of a diagonal the quadrilateral can be drawn.

Think, Discuss and Write (Textbook Page No. 66)

Question 1.
In the above example, we first drew BC. Instead, what could have been be the other starting points?
Solution :
Instead of drawing BC, we can start with \(\overline{\mathrm{AB}}\) or \(\overline{\mathrm{CD}}\).

Question 2.
We used some five measurements to draw quadrilaterals so far. Can there be different sets of five measurements (other than seen so far) to draw a quadrilateral ? The following problems may help you in answering the question.
(i) Quadrilateral ABCD with AB = 5 cm, BC = 5.5 cm, CD = 4 cm, AD = 6 cm and ∠B = 80°.
(ii) Quadrilateral PQRS with PQ = 4.5 cm, ∠P = 70°, ∠Q = 100°, ∠R = 80° and ∠S = 110°.
Construct a few more examples of your own to find sufficiency/insufficiency of the data for construction of a quadrilateral.
Solution:
(i) Here, four sides and one angle are given. So given data is sufficient to construct quadrilateral ABCD.
(ii) We cannot locate the points R and S with the help of given data. So given data is insufficient to construct quadrilateral PQRS.

Few examples of sufficient data to construct quadrilaterals :

• Quadrilateral PQRS in which RS = 6 cm, QR = 5 cm, PQ = 5 cm, ∠Q = 135°, ∠R = 90°. (three sides and two angles)
• Quadrilateral ABCD in which AB = 5 cm, BC = 4 cm, ∠B = 60°, ∠A = 90° and ∠C = 135°. (two sides and three angles)

Try these (Textbook Page No. 67)

Question 1.
How will you construct a rectangle PQRS if you know only the lengths PQ and QR?
Solution:

Each angle of a rectangle is a right angle. Rectangle has opposite sides of equal lengths. Here, PQ is given, So PQ = RS.
ΔPQR can be drawn using PQ, QR and ∠Q = 90°.
ΔQRS can be drawn using QR, RS and ∠R = 90°.
Thus, the required rectangle PQRS can be constructed.

Question 2.
Construct the kite EASY if AY = 8 cm, EY = 4 cm and SY = 6 cm. Which properties of the kite did you use in the process ?
Solution:

Diagonals intersect each other at right angle. ? In kite pairs of consecutive sides are of equal lengths.
While constructing, E cannot be located.
∴ Kite EASY cannot be constructed
(∵ For Δ EYA, the sum of lengths of two sides EY + EA (4 + 4) is not greater than length of third side AY (8).

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 12 Exponents and Powers Ex 12.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 12 Exponents and Powers Ex 12.2

1. Express the following numbers in standard form:

Question (i)
0.0000000000085
Solution:
= \(\frac{85}{10000000000000}\)
= \(\frac{85}{10^{13}}\)
= \(\frac{8.5 \times 10}{10^{13}}\)
= 8.5 × 10 × 10^-13
= 8.5 × 10^-12

Question (ii)
0.00000000000942
Solution:
= \(\frac{942}{100000000000000}\)
= \(\frac{942}{10^{14}}\)
= \(\frac{9.42 \times 10^{2}}{10^{14}}\)
= 9.42 × 10² × 10^-14
= 9.42 × 10^-12

Question (iii)
6020000000000000
Solution:
= 602 × 10000000000000
= 602 × 10¹³
= 6.02 × 10² × 10¹³
= 6.02 × 10¹⁵

Question (iv)
0.00000000837
Solution:
= \(\frac{837}{100000000000}\)
= \(\frac{837}{10^{11}}\)
= \(\frac{8.37 \times 10^{2}}{10^{11}}\)
= 8.37 × 10² × 10^-11
= 8.37 × 10^2+(-11)
= 8.37 × 10^-9

Question (v)
31860000000
Solution:
= 3186 × 10000000
= 3186 × 10⁷
= 3.186 × 10³ × 10⁷
= 3.186 × 10^{3 + 7}
= 3.186 × 10¹⁰

2. Express the following numbers in usual form:

Question (i)
3.02 × 10^-6
Solution:
= 302 × 10^-2 × 10^-6
= 302 × 10^-8
= \(\frac{302}{100000000}\)
= 0.00000302

Question (ii)
4.5 × 10⁴
Solution:
= \(\frac {45}{10}\) × 10000
= 45 × 1000
= 45000

Question (iii)
3 × 10^-8
Solution:
= \(\frac{3}{100000000}\)
= 0.00000003

Question (iv)
1.0001 × 10⁹
Solution:
= \(\frac{10001}{10000}\) × 1000000000
= 10001 × 100000
= 1000100000

Question (v)
5.8 × 10¹²
Solution:
= \(\frac {58}{10}\) × 100000000000
= 58 × 10000000000
= 5800000000000

Question (vi)
3.61492 × 10⁶
Solution:
= \(\frac{361492}{100000}\) × 1000000
= 361492 x 10
= 3614920

3. Express the number appearing in the following statements in standard form:

Question (i)
1 micron is equal to \(\frac{1}{1000000}\) m.
Solution:
1 micron = \(\frac{1}{1000000}\)m
= \(\frac{1}{10^{6}}\)m
∴ 1 micron = 1 × 10^-6 m

Question (ii)
Charge of an electron is 0.000,000,000,000,000,000,16 coulomb.
Solution:
Charge of an electron
= 0.000,000,000,000,000,000,16 coulomb
= \(\frac{16}{100000000000000000000}\) coulomb
= \(\frac{1.6 \times 10}{10^{20}}\) coulomb
= \(\frac{1.6}{10^{19}}\) coulomb
= 1.6 × 10^-19 coulomb
∴ Charge of an electron = 1.6 × 10^-19 coulomb

Question (iii)
Size of a bacteria is 0.0000005 m.
Solution:
Size of a bacteria = 0.0000005 m
= \(\frac{5}{10000000}\)m
= \(\frac{5}{10^{7}}\)m
= 5 × 10^-7 m
∴ Size of a bacteria = 5 × 10^-7 m

Question (iv)
Size of a plant cell is 0.00001275 m.
Solution:
Size of a plant cell = 0.00001275 m
= \(\frac{1275}{100000000}\)m
= \(\frac{1275}{10^{8}}\)m
= \(\frac{1.275 \times 10^{3}}{10^{8}}\) m
= 1.275 × 10^3-8
= 1.275 × 10^-5
∴ Size of a plant cell
= 1.275 × 10^-5 m

Question (v)
Thickness of a thick paper is 0.07 mm.
Solution:
Thickness of a thick paper = 0.07 mm
= \(\frac {7}{100}\) mm
= \(\frac{7}{10^{2}}\) mm
= 7 × 10^-2 mm
∴ Thickness of a thick paper
= 7 × 10^-2 mm

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.
Solution:
Thickness of a book = 20 mm
∴ Thickness of 5 books = (5 × 20) mm
= 100 mm
Thickness of a paper sheet = 0.016 mm
∴ Thickness of 5 paper sheets
= (5 × 0.016) mm
= 0.08 mm
∴ Total thickness of a stack = Thickness of books + Thickness of paper sheets
= (100 + 0.08) mm
= 100.08 mm
In standard form 100.08 is written as 1.0008 × 10².
Thus, the total thickness of the stack is 1.0008 × 10² mm.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 12 Exponents and Powers Ex 12.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 12 Exponents and Powers Ex 12.1

1. Evaluate:

Question (i)
3^-2
Solution:
= \(\frac{1}{3^{2}}\)
= \(\frac{1}{3 \times 3}\)
= \(\frac {1}{9}\)

Question (ii)
(-4)^-2
Solution:
= \(\frac{1}{(-4)^{2}}\)
= \(\frac{1}{(-4) \times(-4)}\)
= \(\frac {1}{9}\)

Question (iii)
(\(\frac {1}{2}\))^-5
Solution:
= \(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\)
= \(\frac{1}{\frac{1}{32}}\)
= 32

2. Simplify and express the result in power notation with positive exponent:

Question (i)
(-4)⁵ ÷ (-4)⁸
Solution:
= (-4)^{5 – 8} (∵ a^m ÷ aⁿ = a^m-n)
= (-4)^{– 3}
= \(\frac{1}{(-4)^{3}}\)

Question (ii)
\(\left(\frac{1}{2^{3}}\right)\)²
Solution:
= \(\frac{(1)^{2}}{\left(2^{3}\right)^{2}}\) [∵ \(\left(\frac{a}{b}\right)^{m}=\frac{a^{m}}{b^{m}}\)]
= \(\frac{1}{2^{3 \times 2}}\) [∵ (a^m)ⁿ = a^mn
= \(\frac{1}{2^{6}}\)

Question (iii)
(-3)⁴ × (\(\frac {5}{3}\))⁴
Solution:
= [(-3) × \(\frac {5}{3}\)]⁴ [∵ a^m × b^m = (ab)^m]
= [(-1) × 5]⁴
= (-1)⁴ × 5⁴
= 1 × 5⁴
= 5⁴

Question (iv)
(3^-7 ÷ 3^-10) × 3^-5
Solution:
= (3^(-7)-(-10)) × 3^-5 (∵ a^m ÷ aⁿ = a^m-n)
= (3^-7+10 × 3^-5
= 3³ × 3^-5
= 3³ + (-5) (∵ a^m × aⁿ = a^m+n)
= 3^-2
= \(\frac{1}{3^{2}}\)

Question (v)
2^-3 × (-7)^-3
Solution:
= [2 × (-7)]^-3 [∵ a^m × b^m = (ab)^m
= (-14)^-3
= \(\frac{1}{(-14)^{3}}\)

3. Find the value of:

Question (i)
(3⁰ + 4^-1) × 2²
Solution:
= (1 + \(\frac {1}{4}\)) × 2²
= (\(1 \frac{1}{4}\)) × 2²
= (\(\frac {5}{4}\)) × 4
= 5

Question (ii)
(2^{– 1} × 4^-1) ÷ 2²
Solution:
= (\(\frac {1}{2}\) × \(\frac {1}{4}\)) ÷ \(\frac{1}{2^{2}}\)
= (\(\frac {1}{8}\)) ÷ \(\frac {1}{4}\)
= \(\frac {1}{8}\) × \(\frac {4}{1}\)
= \(\frac {1}{2}\)

Question (iii)
(\(\frac {1}{2}\))^{– 2} + (\(\frac {1}{3}\))^{– 2} + (\(\frac {1}{4}\))^{– 2}
Solution:
=\(\frac{1}{\left(\frac{1}{2}\right)^{2}}+\frac{1}{\left(\frac{1}{3}\right)^{2}}+\frac{1}{\left(\frac{1}{4}\right)^{2}}\)
= \(\frac{1}{\frac{1}{4}}+\frac{1}{\frac{1}{9}}+\frac{1}{\frac{1}{16}}\)
= 4 + 9 + 16
= 29

Question (iv)
(3^{– 1} + 4^{– 1} + 5^{– 1})⁰
Solution:
(3^-1 + 4^-1 + 5^-1)⁰
∴ [3^-1 + 4^-1 + 5^-1]⁰ = 1
[∵ a⁰ = 1]

Question (v)
\(\left\{\left(\frac{-2}{3}\right)^{-2}\right\}\)²
Solution:
= \(\left(\frac{-2}{3}\right)^{(-2) \times 2}\) [∵ (a^m)^m = a^mn]
= (\(\frac {-2}{3}\))^-4
= \(\frac{(-2)^{-4}}{(3)^{-4}}\) [∵ \(\left(\frac{a}{b}\right)^{m}=\frac{a^{m}}{b^{m}}\)]
= \(\frac{3^{4}}{(-2)^{4}}\)
= \(\frac{3 \times 3 \times 3 \times 3}{(-2) \times(-2) \times(-2) \times(-2)}\)
= \(\frac {81}{16}\)

4. Evaluate:

Question (i)
\(\frac{8^{-1} \times 5^{3}}{2^{-4}}\)
Solution:
= \(\frac{2^{4} \times 5^{3}}{8}\)
= \(\frac{2^{4} \times 5^{3}}{2^{3}}\)
= 2^4-3 × 5³
= 2 × 125
= 250

Question (ii)
(5^-1 × 2^-1) × 6^-1
Solution:
= (\(\frac {1}{5}\) × \(\frac {1}{2}\)) × \(\frac {1}{6}\)
= (\(\frac {1}{10}\)) × \(\frac {1}{6}\)
= \(\frac {1}{60}\)

Another method:
= 5^-1 × 2^-1 × 6^-1
= (5 × 2 × 6)^-1
= (60)^-1
= \(\frac {1}{60}\)

5. Find the value of m for which 5^m ÷ 5^-3 = 5⁵.
Solution:
∴ 5^m-(-3) = 5⁵.
∴ 5^{m + 3} = 5⁵
∴ m + 3 = 5 (∵ a^m = aⁿ then m = n)
∴ m = 5 – 3
∴ m = 2
Thus, value of m is 2.

6. Evaluate:

Question (i)
\(\left\{\left(\frac{1}{3}\right)^{-1}-\left(\frac{1}{4}\right)^{-1}\right\}^{-1}\)
Solution:
= {\(\frac{3}{1}-\frac{4}{1}\)}^-1 (∵ a^-m = \(\frac {1}{a^m}\))
= {3 – 4}^-1
= {-1}^-1
= \(\frac {1}{-1}\)
= -1

Question (ii)
(\(\frac {5}{8}\))^-7 × (\(\frac {8}{5}\))^-4
Solution:
(\(\frac {5}{8}\))^-7 × (\(\frac {5}{8}\))⁴
(∵ a^-m = \(\frac{1}{a^{m}}\))
= (\(\frac {5}{8}\))^-7+4 (∵ a^m × aⁿ = a^m+n)
= (\(\frac {5}{8}\))^-3
= (\(\frac {8}{5}\))³
= \(\frac{8 \times 8 \times 8}{5 \times 5 \times 5}\)
= \(\frac {512}{125}\)

7. Simplify:

Question (i)
\(\frac{25 \times t^{-4}}{5^{-3} \times 10 \times t^{-8}}\) (t ≠ 0)
Solution:

Question (ii)
\(\frac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}}\)
Solution:

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 4 Practical Geometry Ex 4.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.5

Draw the following.

Question 1.
The square READ with RE = 5.1 cm.
Solution:

Steps of construction :

• Draw a line segment RE = 5.1 cm.
• At E, draw \(\overrightarrow{\mathrm{EM}}\), such that ∠REM = 90°.
• With E as centre and radius =5.1 cm, draw an arc intersecting \(\overrightarrow{\mathrm{EM}}\) at A.
• With R as centre and radius = 5.1 cm, draw an arc.
• With A as centre and radius = 5.1 cm, draw an arc which intersect the previous arc at D.
• Draw \(\overline{\mathrm{RD}}\) and \(\overline{\mathrm{AD}}\).

Thus, READ is the required square.

Question 2.
A rhombus whose diagonals are 5.2 cm and 6.4 cm long.
Solution:

[Note : The diagonals of a rhombus bisect each other at right angle.]
Here, in rhombus XYZW, YW 6.4 cm.
∴ OW = OY = 3.2 cm

Steps of construction:

• Draw a line segment XZ = 5.2 cm.
• Draw \(\overleftrightarrow{\mathrm{AB}}\), the perpendicular bisector of \(\overline{\mathrm{XZ}}\), which intersect \(\overline{\mathrm{XZ}}\) at O.
• With O as centre and radius = 3.2 cm, draw an arc intersecting \(\overleftrightarrow{\mathrm{AB}}\) above \(\overline{\mathrm{XZ}}\) at W.
• With O as centre and radius = 3.2 cm, draw another arc which intersects \(\overleftrightarrow{\mathrm{AB}}\) below \(\overline{\mathrm{XZ}}\) at Y.
• Draw \(\overline{\mathrm{XY}}\), \(\overline{\mathrm{YZ}}\), \(\overline{\mathrm{ZW}}\) and \(\overline{\mathrm{XW}}\).

Thus, XYZW is the required rhombus.

Question 3.
A rectangle with adjacent sides of lengths 5 cm and 4 cm.
Solution:

[Note: Each angle of a rectangle is a right angle.]
Steps of construction :

• Draw a line segment AB = 5 cm.
• At A, draw \(\overrightarrow{\mathrm{AX}}\), such that ∠XAB = 90°.
• With A as centre and radius = 4 cm, draw an arc intersecting \(\overrightarrow{\mathrm{AX}}\) at D.
• With B as centre and radius = 4 cm, draw an arc.
• With D as centre and radius = 5 cm, draw an arc which intersects the previous arc at C.
• Draw \(\overline{\mathrm{BC}}\) and \(\overline{\mathrm{CD}}\).

Thus, ABCD is the required rectangle.

Question 4.
A parallelogram OKAY where OK = 5.5 cm and KA = 4.2 cm. Is it unique ?
Solution:

A parallelogram cannot be constructed as sufficient measurements are not given. It is not unique as angles may vary in parallelogram drawn by given measurements.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 4 Practical Geometry Ex 4.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.4

1. Construct the following quadrilaterals:

Question (i).
Quadrilateral DEAR
DE = 4 cm
EA = 5 cm
AR = 4.5 cm
∠E = 60°
∠A = 90°
Solution:

Steps of construction:

• Draw a line segment DE = 4 cm.
• At E, draw \(\overrightarrow{\mathrm{EM}}\) such that ∠DEM = 60°.
• With E as centre and radius = 5 cm, draw an arc intersecting \(\overrightarrow{\mathrm{EM}}\) at A.
• At A, draw \(\overrightarrow{\mathrm{AN}}\) such that ∠EAN = 90°.
• With A as centre and radius = 4.5 cm, draw an arc intersecting \(\overrightarrow{\mathrm{AN}}\) at R.
• Draw \(\overline{\mathrm{DR}}\).

Thus, DEAR is the required quadrilateral.

Question (ii).
Quadrilateral TRUE
TR = 3.5 cm
RU = 3 cm
UE = 4 cm
∠R = 75°
∠U = 120°
Solution:

Steps of construction :

• Draw a line segment TR = 3.5 cm.
• At R, draw a \(\overrightarrow{\mathrm{RM}}\) such that ∠TRM = 75°.
• With R as centre and radius = 3 cm, draw an arc intersecting \(\overrightarrow{\mathrm{RM}}\) at U.
• At U, draw \(\overrightarrow{\mathrm{UN}}\) such that ∠RUN = 120°.
• With U as centre and radius = 4 cm, draw an arc intersecting \(\overrightarrow{\mathrm{UN}}\) at E.
• Draw \(\overline{\mathrm{ET}}\).

Thus, TRUE is the required quadrilateral.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 4 Practical Geometry Ex 4.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.3

1. Construct the following quadrilaterals:

Question (i).
Quadrilateral MORE
MO = 6 cm
OR = 4.5 cm
∠M = 60°
∠O = 105°
∠R = 105°
Solution:

• Draw a line segment MO = 6 cm.
• At M, draw \(\overrightarrow{\mathrm{MA}}\), such that ∠OMA = 60°
• At O, draw \(\overrightarrow{\mathrm{OB}}\) such that ∠MOB = 105°.
• With O as centre and radius = 4.5 cm, draw an arc intersecting \(\overrightarrow{\mathrm{OB}}\) at R.
• At R, draw \(\overrightarrow{\mathrm{RC}}\) such that ∠ORC = 105°.
• Locate E at intersection of \(\overrightarrow{\mathrm{RC}}\) and \(\overrightarrow{\mathrm{MA}}\).

Thus, MORE is the required quadrilateral.

Question (ii).
Quadrilateral PLAN
PL = 4 cm
LA = 6.5 cm
∠P = 90°
∠A = 110°
∠N = 85°
Solution:

[Note : In □ PLAN, m∠P = 90°, m∠A = 110° and m∠N = 85°)
∴ m∠L = 360°- (m∠P + m∠A + m∠N)
= 360° – (90° + 110° + 85°)
= 360° – 285°
= 75°

Steps of construction:

• Draw a line segment AL = 6.5 cm.
• At A, draw \(\overrightarrow{\mathrm{AX}}\) such that ∠XAL = 110°. (Use protractor)
• At L, draw \(\overrightarrow{\mathrm{LY}}\) such that ∠YLA = 75°. (Use protractor)
• With L as centre and radius = 4 cm, draw an arc intersecting \(\overrightarrow{\mathrm{LY}}\) at P.
• At P, draw \(\overrightarrow{\mathrm{PZ}}\) such that ∠ZPL = 90°. (∵ ∠ ZPY = 90°)
• Locate N at intersection of \(\overrightarrow{\mathrm{AX}}\) and \(\overrightarrow{\mathrm{PZ}}\).

Thus, PLAN is the required quadrilateral.

Question (iii).
Parallelogram HEAR
HE = 5 cm
EA = 6 cm
∠R = 85°
Solution:

[Note : □ HEAR is a parallelogram.
Opposite sides of parallelogram are of equal lengths.]
HE = 5 cm, ∴ AR = 5 cm, EA = 6 cm, ∴ HR = 6 cm
Adjacent angles of a parallelogram arc supplementary.
m∠R = 85° (given)
∴ m∠H = 180° – 85° = 95°
Opposite angles of a parallelogram are of equal measures,
m ∠ R = 85°
∴ m ∠ E = 85°

Steps of construction:

• Draw a line segment HE = 5 cm.
• At H, draw \(\overrightarrow{\mathrm{HX}}\), such that ∠ XHE = 95°. (Use protractor)
• With H as centre and radius = 6 cm, draw an arc intersecting \(\overrightarrow{\mathrm{HX}}\) at R.
• At E, draw \(\overrightarrow{\mathrm{EY}}\) such that ∠ HEY = 85°. (Use protractor)
• With E as centre and radius = 6 cm, draw an arc intersecting \(\overrightarrow{\mathrm{EY}}\) at A.
• Draw \(\overline{\mathrm{AR}}\).

Thus, HEAR is the required parallelogram.

Question (iv).
Rectangle OKAY
OK = 7 cm
KA = 5 cm
Solution:

[Note: Here, OKAY is a rectangle. Opposite sides of a rectangle are of equal lengths.]
OK = 7 cm, ∴ AY = 7 cm and KA = 5 cm, ∴ OY = 5 cm
Moreover, all angles of a rectangle are right angles.

Steps of construction:

• Draw a line segment OK = 7 cm.
• At O, draw \(\overrightarrow{\mathrm{OM}}\) such that ∠ MOK = 90°.
• With O as centre and radius = 5 cm, draw an arc intersecting \(\overrightarrow{\mathrm{OM}}\) at Y.
• At K, draw \(\overrightarrow{\mathrm{KN}}\) such that ∠ NKO = 90°.
• With K as centre and radius = 5 cm, draw an arc intersecting \(\overrightarrow{\mathrm{KN}}\) at A.
• Draw \(\overline{\mathrm{AY}}\).

Thus, OKAY is the required rectangle.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 13 Direct and Inverse Proportions InText Questions and Answers.

PSEB 8th Class Maths Solutions Chapter 13 Direct and Inverse Proportions InText Questions

Try These : [Textbook Page No. 204]

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

Question (i)

X	20	17	14	11	8	5	2
y	40	34	28	22	16	10	4

Solution:
\(\begin{aligned}
&\frac{20}{40}=\frac{1}{2}, \frac{17}{34}=\frac{1}{2}, \frac{14}{28}=\frac{1}{2}, \frac{11}{22}=\frac{1}{2}, \frac{8}{16}=\frac{1}{2} \\
&\frac{5}{10}=\frac{1}{2}, \frac{2}{4}=\frac{1}{2}
\end{aligned}\)
The value of \(\frac{\text {x}}{\text {y}}\) is same for different values of x and y. So these values x and y are directly proportional.

Question (ii)

X	6	10	14	18	22	26	30
y	4	8	12	16	20	24	28

Solution:
\(\begin{aligned}
&\frac{6}{4}=\frac{3}{2}, \frac{10}{8}=\frac{5}{4}, \frac{14}{12}=\frac{7}{6}, \frac{18}{16}=\frac{9}{8}, \frac{22}{20}=\frac{11}{10}, \\
&\frac{26}{24}=\frac{13}{12}, \frac{30}{28}=\frac{15}{14}
\end{aligned}\)
The values of \(\frac{\text {x}}{\text {y}}\) are different for different values of x and y respectively. So these values of x and y are not directly proportional.

Question (iii)

X	5	8	12	15	18	20
y	15	24	36	60	72	100

Solution:
The values of \(\frac{\text {x}}{\text {y}}\) are different for different values of x and y respectively.
So these values of x and y are not directly proportional.

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

Time Period	1 year	2 years	3 years
Simple Interest (in ₹)
Compound Interest (in ₹)

Solution:
Simple interest : SI = \(\frac{P \times R \times T}{100}\)
For calculation:
P = ₹ 1000, R = 8 %, T = …………….

Time (T) →	1 year: T = 1	2 years : T = 2	3 years : T = 3
Simple interest SI = \(\frac{P \times R \times T}{100}\)	₹ \(\frac{1000 \times 8 \times 1}{100}\) = ₹ 80	₹ \(\frac{1000 \times 8 \times 2}{100}\) = ₹ 160	₹ \(\frac{1000 \times 8 \times 3}{100}\) = ₹ 240
\(\frac{\text { SI }}{\text { T }}\)	\(\frac {80}{1}\) = 80	\(\frac {160}{2}\) = 80	\(\frac {240}{3}\) = 80

Here, the ratio of simple interest with time period is same for every year.
Hence, simple interest changes in direct proportion with time period.
Compound interest:
For calculation:
P = ₹ 1000, R = 8 %, T = ……………

Time →	1 year : n = 1
A = P(1 + \(\frac {R}{100}\))n CI = A – P	A = 1000(1 + \(\frac {8}{100}\))1 = 1000 × \(\frac {108}{100}\) = 1080 ∴ CI = 1080 – 1000 = ₹ 80
\(\frac{\text { CI }}{\text { T }}\)	\(\frac {80}{1}\)

Time →	2 years : n = 2
A = P(1 + \(\frac {R}{100}\))n CI = A – P	A = 1000(1 + \(\frac {8}{100}\))2 = 1000 × \(\frac {108}{100}\) × \(\frac {108}{100}\) = ₹ 1166.40 ∴ CI = ₹ 1166.40 – 1000 = ₹ 166.40
\(\frac{\text { CI }}{\text { T }}\)	\(\frac {166.40}{2}\)

Time →	3 years : n = 3
A = P(1 + \(\frac {R}{100}\))n CI = A – P	A = 1000(1 + \(\frac {8}{100}\))3 = 1000 × \(\frac {108}{100}\) × \(\frac {108}{100}\) × \(\frac {108}{100}\) = ₹ 1259.712 ∴ CI = ₹ 1259.712 – ₹ 1000 = ₹ 259.712
\(\frac{\text { CI }}{\text { T }}\)	\(\frac {259.712}{3}\)

Here, the ratio of CI and T is not same.
Thus, compound interest is not proportional with time period.

Think, Discuss and Write: [Textbook Page No. 204]

1. If we fix time period and the rate of interest, simple interest changes proportionally with principal. Would there be a similar relationship for compound interest? Why?
Solution:
Time period (T) and rate of interest (R) are fixed, then
Simple interest = \(\frac {PRT}{100}\) = P × Constant
So simple interest depends on principal. Simple interest changes proportionally with principal.
Now, compound interest = P(1 + \(\frac {R}{100}\))^T – P
i.e., A – P
= P [(1 + \(\frac {R}{100}\)^T – 1]
= P × Constant
So compound interest depends on principal.
If principal increases or decreases, then compound interest will also increases or decreases.
Thus, compound interest changes with principal.

Think, Discuss and Write : [Textbook Page No. 209]

1. Take a few problems discussed so far under ‘direct variation’. Do you think that they can be solved by ‘unitary method’?
Solution:
Yes, each problem can be solved by unitary method.
e.g. Question 4 of Exercise: 13.1
Number of bottles filled in 6 hours = 840
∴ The number of bottles filled in 1 hour = \(\frac {840}{6}\) = 140
The number of bottles filled in 5 hours = 140 × 5 = 700
Thus, 700 bottles will it fill in five hours.

Try These : [Textbook Page No. 211]

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

Question (i)

X	50	40	30	20
y	5	6	7	8

Solution:
x₁ = 50 and y₁ = 5
∴ x₁y₁ = 50 × 5
∴ x₁y₁ = 250

x₂ = 40 and y₂ = 6
∴ x₂y₂ = 40 × 6
∴ x₂y₂ = 240

x₃ = 30 and y₃ = 7
∴ x₃y₃ = 30 × 7
∴ x₃y₃ = 210

x₄ = 20 and y₄ = 8
∴ x₄y₄ = 20 × 8
∴ x₄y₄ = 160
Now 250 ≠ 240 ≠ 210 ≠ 160
∴ x₁y₁ ≠ x₂y₂ ≠ x₃y₃ ≠ x₄y₄
∴ x and y are not in inverse proportion.

Question (ii)

X	100	200	300	400
y	60	30	20	15

Solution:
x₁ = 100 and y₁ = 60
∴ x₁y₁ = 100 × 60
∴ x₁y₁ = 6000

x₂ = 200 and y₂ = 30
∴ x₂y₂ = 200 × 30
∴ x₂y₂ = 6000

x₃ = 300 and y₃ = 20
∴ x₃y₃ = 300 × 20
∴ x₃y₃ = 6000

x₄ = 400 and y₄ = 15
∴ x₄y₄ = 400 × 15
∴ x₄y₄ = 6000
Now x₁y₁ = x₂y₂ = x₃y₃ = x₄y₄
∴ x and y are in inverse proportion.

Question (iii)

X	90	60	45	30	20	5
y	10	15	20	25	30	35

Solution:
x₁ = 90 and y₁ = 10
∴ x₁y₁ = 90 × 10
∴ x₁y₁ = 900

x₂ = 60 and y₂ = 15
∴ x₂y₂ = 60 × 15
∴ x₂y₂ = 900

x₃ = 45 and y₃ = 20
∴ x₃y₃ = 45 × 20
∴ x₃y₃ = 900

x₄ = 30 and y₄ = 25
∴ x₄y₄ = 30 × 25
∴ x₄y₄ = 750

x₅ = 20 and y₅ = 30
∴ x₅y₅ = 20 × 30
∴ x₅y₅ = 600

x₆ = 30 and y₆ = 35
∴ x₆y₆ = 5 × 35
∴ x₆y₆ = 175

Now x₁y₁ = x₂y₂ = x₃y₃ ≠ x₄y₄ ≠ x₅y₅ ≠ x₆y₆
∴ x and y are in inverse proportion.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 13 Direct and Inverse Proportions Ex 13.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.2

1. Which of the following are in inverse proportion?

Question (i)
The number of workers on a job and the time to complete the job.
Solution:
If the number of workers on a job increases, then time to complete the job decreases. So, it is the case of inverse proportion.

Question (ii)
The time taken for a journey and the distance travelled in a uniform speed.
Solution:
Here, the speed is uniform.
∴ The time taken for a journey is directly proportional to the speed. So, it is not the case of inverse proportion.

Question (iii)
Area of cultivated land and the crop harvested.
Solution:
For more area of cultivated land, more crops would be harvested. So, it is not the case of inverse proportion.

Question (iv)
The time taken for a fixed journey and the speed of the vehicle.
Solution:
If speed of vehicle is more, then time to cover the fixed journey would be less. So, it is the case of inverse proportion.

Question (v)
The population of a country and the area of land per person.
Solution:
For more population, less area per person would be in the country. So, it is a case of inverse proportion.

2. In a Television game show, the prize money of ₹ 1,00,000 is to be divided equally amongst the winners. Complete the following table and find whether the prize money given to an individual winner is directly or inversely proportional to the number of winners?

Number of winners	1	2	4	5	8	10	20
Prize for each winner (in ₹)	1,00,000	50,000	…	…	…	…	…

Solution:
Here, more the number of winners, less is the prize money for each winner.
∴ This is a case of inverse proportion.

Now, the table is as follows:

Number of winners	1	2	4	5	8	10	20
Prize for each winner (in ₹)	1,00,000	50,000	25,000	20,000	12,500	10,000	5,000

3. Rehman is making a wheel using spokes. He wants to fix equal spokes in such a way that the angles between any pair of consecutive spokes are equal. Help him by completing the following table:

Number of spokes	4	6	8	10	12
Angle between a pair of consecutive spokes	90°	60°	…	…	…

Question (i)
Are the number of spokes and the angles formed between the pairs of consecutive spokes in inverse proportion?
Solution:
Here, more the number of spokes, less the measure of angle between a pair of consecutive spokes.
∴ This is a case of inverse proportion.
Here,

Now, the table is as follows:

Number of spokes	4	6	8	10	12
Angle between a pair of consecutive spokes
	90°	60°	45°	36°	30°

Question (ii)
Calculate the angle between a pair of consecutive spokes on a wheel with 15 spokes.
Solution:
Let the measure of angle be x°.
∴ 15 × x° = 4 × 90°
∴ x° = \(\frac{4 \times 90^{\circ}}{15}\) = 24°
Hence, this angle should be 24°.

Question (iii)
How many spokes would be needed, if the angle between a pair of consecutive spokes is 40°?
Solution:
Let the number of spokes be n.
∴ n × 40° = 4 × 90°
∴ n = \(\frac{4 \times 90^{\circ}}{40^{\circ}}\) = 9
Thus, 9 spokes would be needed.

4. If a box of sweets is divided among 24 children, they will get 5 sweets each. How many would each get, if the number of the children is reduced by 4?
Solution:

Number of children x	Number of sweets y
x1 = 24	y1 = 5
x2 = 24 – 4 = 20	y2 = (?)

Here, if the number of children decreases, then the number of sweets received by each child will increase.
∴ This is a case of inverse proportion.
x₁ × y₁ = x₂ × y₂
∴ 24 × 5 = 20 × y₂
∴ y₂ = \(\frac{24 \times 5}{20}\) = 6
Thus, each child will get 6 sweets.

5. A farmer has enough food to feed 20 animals in his cattle for 6 days. How long would the food last if there were 10 more animals in his cattle?
Solution:

Number of animals x	Number of days y
x1 = 20	y1 = 6
x2 = 20 + 10 = 20	y2 = (?)

Here, the number of animals increases, so the number of days to feed them will decrease.
∴ This is a case of inverse proportion.
∴ x₁ × y₁ = x₂ × y₂
∴ 20 × 6 = 30 × y₂
∴ y₂ = \(\frac{20 \times 6}{30}\) = 4
Thus, the food will last for 4 days.

6. A contractor estimates that 3 persons could rewire Jasminder’s house in 4 days. If, he uses 4 persons instead of three, how long should they take to complete the job?
Solution:

Number of persons x	Number of days y
x1 = 42	y1 = 63
x2 = (?)	y1 = 63

Here, more the number of persons, less will be the time required to complete the job.
∴ This is a case of inverse proportion.
∴ x₁ × y₁ = x₂ × y₂
∴ 3 × 4 = 4 × y₂
∴ y₂ = \(\frac{3 \times 4}{4}\) = 3
Thus, 3 days will be required to complete the job.

7. A batch of bottles were packed in 25 boxes with 12 bottles in each box. If the same batch is packed using 20 bottles in each box, how many boxes would be filled?

Solution:

Number of bottles in a box x	Number of boxes y
x1 = 12	y1 = 25
x2 = 20	y2 = (?)

Here, more the number of bottles in a box, less would be the number of boxes.
∴ This is a case of inverse proportion.
∴ x₁ × y₁ = x₂ × y₂
∴ 3 × 4 = 4 × y₂
∴ y₂ = \(\frac{3 \times 4}{4}\) = 3
Thus, 15 boxes would be filled.

8. A factory requires 42 machines to produce a given number of articles in 63 days. How many machines would be required to produce the same number of articles in 54 days?
Solution:

Number of machines x	Number of days y
x1 = 42	y1 = 63
x2 = (?)	y2 = 54

Here, if the number of days will be less, the number of machines required will be more.
∴ This is a case of inverse proportion.
∴ x₁ × y₁ = x₂ × y₂
∴ 42 × 63 = x₂ × 54
∴ x₂ = \(\frac{42 \times 63}{54}\) = 49
Thus, 49 machines would be required.

9. A car takes 2 hours to reach a destination by travelling at the speed of 60 km/h. How long will it take when the car travels at the speed of 80 km/h?
Solution:

Speed (km/h) x	Number of hours y
x1 = 60	y1 = 2
x2 = 80	y2 = (?)

Here, if the speed of car increases, then the time taken to cover the same distance will decrease.
∴ This is a case of inverse proportion.
∴ x₁ × y₁ = x₂ × y₂
∴ 60 × 2 = 80 × y₂
∴ y₂ = \(\frac{60 \times 2}{80}=\frac{3}{2}=1 \frac{1}{2}\) h
Thus, car would take 1\(\frac {1}{2}\) hours.

10. Two persons could fit new windows in a house in 3 days.

Question (i)
One of the persons fell ill before the work started. How long would the job take now?
Solution:

Number of persons x	Number of days y
x1 = 2	y1 = 3
x2 = 2 – 1 = 1	y2 = (?)

Here, less the number of persons, more would be the number of days to complete the job.
∴ This is a case of inverse proportion.
∴ x₁ × y₁ = x₂ × y₂
∴ 2 × 2 = 1 × y₂
∴ y₂ = \(\frac{2 \times 3}{1}\) = 6
Thus, it would take 6 days to complete the job.

Question (ii)
How many persons would be needed to fit the windows in one day?
Solution:

Number of days x	Number of persons y
x1 = 3	y1 = 2
x2 = 1	y2 = (?)

Here, less the number of days, more will be the number of persons needed.
∴ This is a case of inverse proportion.
∴ y₂ = ? and x₂ = 1
∴ x₁ × y₁ = x₂ × y₂
∴ 3 × 2 = 1 × y₂
∴ y₂ = \(\frac{3 \times 2}{1}\) = 6
Thus, 6 persons would be needed.

11. A school has 8 periods a day each of 45 minutes duration. How long would each period be, if the school has 9 periods a day, assuming the number of school hours to be the same?
Solution:

Number of periods x	Length of each period (in minute) y
x1 = 8	y1 = 45
x2 = 9	y2 = (?)

Here, the number of periods is more, then the length of each period will be less.
∴ This is a case of inverse proportion.
∴ x₁ × y₁ = x₂ × y₂
∴ 8 × 45 = 9 × y₂
∴ y₂ = \(\frac{8 \times 45}{9}\) = 40
Thus, each period would be of 40 minutes.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 13 Direct and Inverse Proportions Ex 13.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.1

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.
Solution:
Here, ratio of parking charges and parking time are as follow:

Parking time	Parking charges	Parking charge / Parking time
4 hours	₹ 60	\(\frac{60}{4}=\frac{15}{1}\)
8 hours	₹ 100	\(\frac{100}{8}=\frac{25}{2}\)
12 hours	₹ 140	\(\frac{140}{12}=\frac{35}{3}\)
24 hours	₹ 180	\(\frac{180}{24}=\frac{15}{2}\)

Here, \(\frac {15}{1}\) ≠ \(\frac {25}{2}\) ≠ \(\frac {35}{3}\) ≠ \(\frac {15}{2}\)
Thus, the parking charges are not in direct proportion to the parking time.

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:
Solution:

Parts of red pigment	1	4	7	12	20
Parts of base	8	…	…	…	…

If parts of red pigment are x₁, x₂, x₃, x₄ and x₅ respectively, then parts of base are y₁, y₂, y₃, y₄ and y₅ respectively. Here, it is clear that mixture preparation is in direct proportion.

Thus, the table is

Parts of red pigment	1	4	7	12	20
Parts of base	8	32	56	96	160

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?
Solution:
See as per question 2 –
x₁ = 1, y₁ = 75, x₂ = ? and yx₂ = 1800
\(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)
∴ \(\frac{1}{75}=\frac{x_{2}}{1800}\)
∴ x₂ = \(\frac{1 \times 1800}{75}\)
∴ x₂ = 24
Thus, 24 ml of red pigment should be mixed with 1800 ml of base.

4. A machine in a soft drink factory fills 840 bottles in six hours. How many bottles will it fill in five hours?
Solution:
Let the number of bottles filled by machine in 5 h be x.

Number of hours (x)	6	5
Number of bottles filled (y)	840	?

Here, as the number of hours decreases, the number of bottles filled will also decrease.
∴ It is case of direct proportion.
Here, x₁ = 6, y₁ = 840, x₂ = 5 and y₂ = ?
\(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)
∴ \(\frac{6}{840}=\frac{5}{y_{2}}\)
∴ y₂ = \(\frac{5 \times 840}{6}\)
∴ y₂ = 700
Thus, 700 bottles will be filled in 5 hours.

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?
Solution:

Enlargement in picture of bacteria	Length (cm)
50,000 times enlarged (x1)	5 (y1)
1 (x2)	? (y2)

Here, length of bacteria increases as picture of bacteria enlarges.
∴ It is case of, direct proportion.

Hence, the actual length of bacteria is 10^-4 cm.
Now, the photograph is enlarged 20,000 times.

Enlargement in picture of bacteria	Length (cm)
50,000 times enlarged (x1)	5 (y1)
20,000 times enlarged (x1)	? (y2)

∴ \(\frac{50,000}{5}=\frac{20,000}{y_{2}}\)
∴ y₂ = \(\frac{20,000 \times 5}{50000}\)
∴ y₂ = 2
Thus, its enlarged length would be 2 cm.

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?
Solution:

–	Actual ship	Model ship
Length of the ship x	28 m	?
Height of mast y	12m	9 cm

This is a case of direct proportion.
x₁ = 28, y₁ = 12, x₂ = ?, y₂ = 9
\(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)
∴ \(\frac{28}{12}=\frac{x_{2}}{9}\)
∴ x₂ = \(\frac{28 \times 9}{12}\)
∴ x₂ = 21
Thus, the length of model ship is 21 cm.

7. Suppose 2 kg of sugar contains 9 × 10⁶ crystals. How many sugar crystals are there in

Question (i)
5 kg of sugar?
Solution:

Weight of sugar (kg) x	Number of sugar crystals y
x1 = 2	y1 = 9 × 106
x2 = 5	y2 = (?)

This is a case of direct proportion.

Thus, there are 2.25 × 10⁷ crystals of sugar in 5 kg of sugar.

Question (ii)
1.2kg of sugar?
Solution:

Weight of sugar (kg) x	Number of sugar crystals y
x1 = 2	y1 = 9 × 106
x2 = 1.2	y2 = (?)

Here. x₁ = 2, x₁ = 9 × 10⁶, x² = 1.2, y² = ?
\(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)
∴ \(\frac{2}{9 \times 10^{6}}=\frac{1.2}{y_{2}}\)
∴ y₂ = \(\frac{1.2 \times 9 \times 10^{6}}{2}\)
∴ y₂ = 0.6 × 9 × 10⁶
∴ y₂ = 5.4 × 10⁶
Thus, there are 5.4 × 10⁶ crystals of sugar in 1.2 kg of sugar.

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?
Solution:

Actual distance (km) x	Distance on the map (cm) y
x1 = 18	y1 = 1
x2 = 72	y2 = (?)

This is a case of direct proportional.
Here, x₁ = 18 km, y₁ = 1 cm, x₂ = 72 km, y₂ = ?
\(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)
∴ \(\frac{18}{1}=\frac{72}{y_{2}}\)
∴ y₂ = \(\frac{72 \times 1}{18}\)
∴ y₂ = 4
Thus, the distance covered by her on the map is 4 cm.

9. A 5 m 60 cm high vertical pole casts a shadow 3 m 20 cm long. Find at the same time

Question (i)
the length of the shadow cast by another pole 10 m 50 cm high
Solution:

Height of vertical pole x	Length of shadow y
x1 = 5 m 60 cm = 560 cm	y1 = 3 m 20 cm = 320 cm
x2 = 10 m 50 cm = 1050 cm	y2 = (?)

This is a case of direct proportionality.

Thus, the length of the shadow cast by another pole is 6 m.

Question (ii)
the height of a pole which casts a shadow 5 m long.
Solution:

Height of vertical pole x	Length of shadow y
x1 = 560 cm	y1 = 320 cm
x2 = (?)	y2 = 5 m = 500 cm

x₁ = 560, y₁ = 320, x₂ = ?, y₂ = 500
\(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)
∴ \(\frac{560}{320}=\frac{x_{2}}{500}\)
∴ x₂ = \(\frac{560 \times 500}{320}\)
∴ x₂ = 875 cm
∴ x₂ = 8.75 cm
Thus, the height of the pole is 8,75 m.

10. A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?
Solution:

Distance (km) x	Time (minute) y
x1= 14	y1 = 25
x2 = (?)	y2 = 5 hours = 300

This is a case of direct proportion.
∴ Here, x₁ = 14, y₁ = 25, x₂ = ?, y₂ = 300
\(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)
∴ \(\frac{14}{25}=\frac{x_{2}}{300}\)
∴ x₂ = \(\frac{14 \times 300}{25}\)
∴ x₂ = 168
Thus, loaded truck can travel 168 km in 5 h.