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Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 11 Mensuration Ex 11.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.1

1. A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area?

Question (a)

Solution:
Side of a square field = 60 m
∴ Perimeter of a square field = 4 × side
= 4 × 60 = 240 m
Area of a square field = (side)²
= (60)²
= 60 × 60
= 3600m²

Question (b)

Solution:
Perimeter of a rectangular field = Perimeter of square field
∴ Perimeter of a rectangular held = 240
∴ 2 (length + breadth) = 240
∴ 2 (80 + breadth) = 240
∴ 80 + breadth = \(\frac {240}{2}\)
∴ 80 + breadth =120
∴ breadth = 120 – 80
∴ breadth = 40
Breadth of rectangular field = 40 m
∴ Area of rectangular field = length × breadth
= (80 × 40)
= 3200 m²
Area of square field > Area of rectangular field
Thus, area of square field (a) is larger.

2. Mrs. Kaushik has a square plot with the measurement as shown g in the figure. She wants to construct a house in the middle of the plot. A garden is developed around the house. Find the total cost of developing a garden around the house at the rate of ₹ 55 per m².

Solution:
Side of the square plot = 25 m
∴ Area of the square plot = (side)²
= (25 × 25) m²
= 625 m²
In square plot, a rectangular-shaped house is to be constructed.
∴ Area of the constructed house
= length × breadth
= (20 × 15) m²
= 300 m²
∴ Area of the garden = Area of square plot – Area of constructed house
= 625 – 300 = 325 m²
Cost of developing garden of 1 m² is ₹ 55
∴ Cost of developing garden of 325 m²
= ₹ (55 × 325)
= ₹ 17,875
Thus, total cost of developing garden is ₹ 17,875.

3. The shape of a garden is rectangular in the middle and semicircular at the ends as shown in the diagram. Find the area and the perimeter of this garden [Length of rectangle is 20 – (3.5+ 3.5) metres].

Solution:
[Note: Here 2 semicircles at the ends of a rectangular garden makes a whole circle. So first find area of a circle and then area of a rectangle. Sum of these two areas is total area. Follow same pattern to find perimeter too. For perimeter of a garden, take only length as rectangle is between two semicircles. Diameter of a circle = Breadth of a rectangle = 7 m]
For semicircle:
∴ Radius = \(\frac{\text { diameter }}{\text { 2 }}\) = \(\frac {7}{2}\)m
Area of circle = πr²
Area of a semicircle = \(\frac {1}{2}\)πr²
∴ Area of 2 semicircles = 2(\(\frac {1}{2}\)πr²)
= \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\)m²
= 38.5 m²
Circumference of two semicircles = 2πr
= 2 × \(\frac {22}{7}\) × \(\frac {7}{2}\)
= 22 m

For rectangle:
length = 20 – (3.5 + 3.5) = 20 – 7 = 13 m
breadth = 7 m
Area of the rectangle = length × breadth
= 13 × 7 = 91 m²
Perimeter of the rectangle
= 2 (length × breadth)
= 2 (13 + 0)
= 2 × 13 = 26 m
∴ Total area of the garden = (38.5 + 91) m²
= 129.5 m²
∴ Perimeter of the garden = (22 + 26) m
= 48 m
Thus, area of the garden is 129.5 m² and the perimeter is 48 m.

4. A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m²? (If required you can split the tiles in whatever way you want to fill up the corners.)
Solution:
[Note : To find number of tiles, divide the area of the floor by area of a tile. Let us do it in a simple way. Unit of floor area and tile area should be same.] Here, tile is parallelogram shaped.
So it’s area = base × corresponding height
Area of a floor = 1080 m²
Base of a tile = 24 cm = \(\frac {24}{100}\) m
Corresponding height of a tile = 10 cm = \(\frac {10}{100}\) m
Number of tiles = \(\frac{\text { Area of a floor }}{\text { Area of a title }}\)
= \(\frac{1080}{\frac{24}{100} \times \frac{10}{100}}\)
= \(\frac{1080 \times 100 \times 100}{24 \times 10}\)
= 45,000
Thus, 45,000 tiles are required to cover the given floor.

5. An ant is moving around a few food pieces of different shapes scattered on the floor. For which food piece would the ant have to take a longer round? Remember, circumference of a circle can be obtained by using the expression c = 2 πr, where r is the radius of the circle.

Question (a)

Solution:
Here, the shape is semi-circular.
Diameter = 2.8 cm
Radius = \(\frac{\text { Diameter }}{2}=\frac{2.8}{2}\) = 1.4 cm
Circumference of a semicircle = πr
Perimeter of the given figure
= πr + diameter
= (\(\frac {22}{7}\) × 1.4) + 2.8
= 4.4 + 2.8
= 7.2 cm

Question (b)

Solution:
Here, given shape is semicircular at one side, (radius = \(\frac {2.8}{2}\) = 1.4 cm)
So perimeter of semicircular region (circumference) = πr
= \(\frac {22}{7}\) × 1.4
= \(\frac {22}{7}\) × \(\frac {14}{10}\)
= 4.4 cm … (i)
Perimeter of the other portion
= breadth + length + breadth
= (1.5 + 2.8 + 1.5) cm
= 5.8 cm … (ii)
∴ Perimeter of the given figure
= (4.4 + 5.8) cm [from (i) and (ii)]
= 10.2 cm

Question (c)

Solution:
Perimeter of a given part
(semi circular circumference) = πr
= \(\frac {22}{7}\) × 1.4
= 4.4 cm
∴ Perimeter of the given figure
= (4.4 + 2 + 2) cm
= 8.4 cm
Thus, 7.2 cm < 8.4 cm < 10.2 cm.
Thus, the ant would has to take a longer round for food piece (b), as it has a larger perimeter.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter Statistics Ex 14.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.2

Question 1.
The following table shows the ages of the patients admitted in a hospital during a year:

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Solution:
For mode:
In the given data, Maximum frequency is 23 and it corresponds to the class interval 35 – 45
Modal class = 35 – 45
So, l = 35; f₁ = 23; f₀ = 21; f₂ = 14 and h = 10
Using fonnula, Mode l = l + \(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\) × h
Mode = 35 + \(\left[\frac{23-21}{2(23)-21-14}\right]\) × 10
= 35 + \(\frac{2}{46-35}\) × 10
= 35 + \(\frac{20}{11}\) = 35 + 1.8 = 36.8.

For Mean:

From above data,
Assumed mean (a) = 30
Width of class (h) = 10
∴ \(\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}=\frac{43}{80}\) = 0.5375
Using formula, Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)
\(\bar{X}\) = 30 + 10 (0.5375)
= 30 + 5.375 = 35.375 = 35.37
Hence, mode of given data is 36.8 years and mean of the given data is 35.37 years. Also, it is clear from above discussion that average age of a patient admitted in the hospital is 35.37 years and maximum number of patients admitted in the hospital are of age 36.8 years.

Question 2.
The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:

Determine the ¡nodal lifetimes of the components.
Solution:
In the given data.
Maximum frequency is 61 and it corresponds to the class interval 60 – 80.
∴ Model class = 60 – 80
So, l = 60; f₁ = 61 ; f₀ = 52; f₂ = 38 and h = 20
Using formula, Mode = l + \(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\) × h

Mode = 60 + \(\left(\frac{61-52}{2(61)-52-38}\right)\) × 20

= 60 + \(\frac{9}{122-52-38}\) × 20

= 60 + \(\frac{9}{32}\) × 20

= 60 + \(\frac{180}{32}\)

= 60 + 5.625 = 65.625
Hence, modal Lifetimes of the components is 65.625 hours.

Question 3.
The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:

Solution:
For Mode: In the given data.
Maximum frequency is 40, and it corresponds to the class interval 1500 – 2000.
∴ Model class = 1500 – 2000
So, l = 1500; f₁ = 40; f₀ = 24; f₂ = 33 and h = 500
Using formula, Mode = l + \(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\) × h

= 1500 + \(\left\{\frac{40-24}{2(40)-24-33}\right\}\) × 500

= 1500 + \(\left\{\frac{16}{80-24-33}\right\}\) × 500

= 1500 + \(\frac{16 \times 500}{23}\)

= 1500 + \(\frac{8000}{23}\) = 1500 + 347.83 = 1847.83

For Mean:

From above data,

Assumed Mean (a) = 2750
Length of width (h) = 500
\(\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}=-\frac{35}{200}\) = – 0.175
Using formula, Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)
\(\overline{\mathbf{X}}\) = 2750 + 500 (- 0.175)
= 2750 – 87.50 = 2662.50
Hence, the modal monthly expenditure of family is 1847.83 and the mean monthly expenditure is 2662.50.

Question 4.
The following distribution gives the slate-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

Solution:
For Mode:
In the given data,
Maximum frequency is 10 and it corresponds to the class interval is 30 – 35.
∴ Modal class = 30 – 35.
So, l = 30; f₁ = 10; f₀ = 9; f₂ = 3 and h = 5
using formula, Mode = l + \(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\) × h
Mode = 30 + \(\left(\frac{10-9}{2(10)-9-3}\right)\) × 5
= 30 + \(\frac{1}{20-12}\) × 5
= 30 + \(\frac{5}{8}\) = 30 + 0.625 = 30.625 = 30.63 (approx.)

For mean:

From above data, Assumed Mean (a) = 32.5
Width of class (h) = 5
∴ \(\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}=-\frac{23}{35}\) = – 0.65

Using formula, Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)
\(\overline{\mathbf{X}}\) = 32.5 + 5 (- 0.65)
= 32..5 – 3.25 = 29.25 (approx.)
Hence, mode and mean of given data is 30.63 and 29.25. Also, from above discussion, it clear that states/U.T. have student per teacher is 30.63 and on average, this ratio is 29.25.

Question 5.
The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches

Find the mode of the data.
Solution:
In the given data,
Maximum frequency is 18 and it corresponds to the class interval 4000 – 5000.
∴ Modal class = 4000 – 5000
So, l = 4000; f₁ = 18; f₀ = 4; f₂ = 9 and h = 1000
Using formula, Mode = l + \(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\) × h

Mode = 4000 + \(\left(\frac{18-4}{2(18)-4-9}\right)\) × 1000

= 4000 + \(\frac{14}{36-13}\) × 1000

= 4000 + \(\frac{14000}{23}\) = 4000 + 608.6956

= 4000 + 608.7 = 4608.7 (approx.)
Hence, mode of the given data is 4608.7.

Question 6.
A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised It in the table given below. Find the mode of the data:

Solution:
In the given data,
Maximum frequency is 20 and it corresponds to the class interval 40 – 50
∴ Modal Class = 40 – 50
So, l = 40; f₁ = 20; f₀ = 12; f₂= 11 and h = 10
Using formula, Mode = l + \(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\) × h

Mode = 40 + \(\left(\frac{20-12}{2(20)-12-11}\right)\) × 10

= 40 + \(\frac{8}{40-23}\) × 10

= 40 + \(\frac{80}{17}\) = 40 + 4.70588

= 40 + 4.7 = 44.7 (approx.)
Hence, mode of the given data is 44.7 cars.

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 6th Class English Book Solutions English Vocabulary Antonyms Exercise Questions and Answers, Notes.

PSEB 6th Class English Vocabulary Antonyms

An Antonym is a word which is opposite in meaning (विपरीतार्थक) to another word; as-

Word – Antonym
absent – present
active – passive
above – below
accept – reject
before – after
bitter – sweet

blunt – sharp
bold – timid
beautiful – ugly
bright – dim
cheap – costly
clean – dirty
clever – stupid
dark – bright
defeat – victory
difficult – easy
death – life
early – late
empty – full
enemy – friend
far – near
foolish – wise
fresh – stale
good – bad
great – small
happy – sad
high – low
hot – cold
in – out
import – export
increase – decrease
joy – sorrow
Word – Antonym
junior – senior
kind – cruel
lend – borrow
light – heavy
love – hate
long – short
old – young/new
oral – written
night – day
peace – war
poor – rich
profit – loss
right – wrong
shallow – deep
slow – fast
stale – fresh

strong – weak
summer – winter
thick – thin
top – bottom
tall – short
up – down
warm – cool
wide – narrow
wise – silly

Punjab State Board PSEB 6th Class English Book Solutions English Vocabulary Sounds of Animals Exercise Questions and Answers, Notes.

PSEB 6th Class English Vocabulary Sounds of Animals

Animal – Sound
Ass – Brays
Bee – hum
Cat/Kitten – mew
Chicks – cheep
Cows – low, moo

Crows – caw
Cuckoos – cuckoo
Cock – Crow
Deers – bell
Dogs – barks
Dolphins – click
Donkeys – hee-haw
Doves – coo
Ducks – quack
Eagles – scream
Elephants – trumpet, roar
Fly – buzz
Foxes – bark, yelp.
Frogs – croak
Geese – cackle/quack
Grasshoppers – chirp
Goat – baah
Hare – squeak
Horse – neigh
Lions – roar, growl
Monkey – scream, chatter
Owl – hoot
Parrot – talk
Sparrow – chirrup, chirp
Snakes – hiss

Sheep – bleat
Tiger – roar, growl
Tortoise – grunt.

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 7th Class English Book Solutions English Message Writing Exercise Questions and Answers, Notes.

PSEB 7th Class English Message Writing

Type – I

1. Read the following telephonic conversation between Kavita and Karan. Karan will not be able to meet Avik. He leaves a message for him. Write this message by using not more than 50 words.

Kavita : Hello ! Hello ! I am Kavita from Indore. Can I speak to Avik? I am his sister.
Karan : Hello ! Kavita I am Karan, Avik’s colleagues. Avik is on leave today. Can I take a message ?
Kavita : Yes, Karan, I am coming to Mumbai tomorrow. Ask him to pick me up at the
airport. I have an interview for the post of Scientist at NPL on the day after tomorrow.
Karan : Which flight are you coming on ?
Kavita : It is the Jetline flight which arrives there at 7.15 p.m. I am bringing along with me that big box which contains his books. I hope it won’t be any trouble for you coming.
Karan : Not at all, I will leave a message on his table. Okay, Kavita.
Kavita : Thank you, Karan. Bye.

Message

Avik

Today, there was a telephonic message from your sister, Kavita. She is coming to Mumbai tomorrow as she has some interview here. She is coming by the Jetlnte flight which will arrive here at 7.15 p.m. She says she will be bringing with her a big box containing your books. Please pick her up at the airport.

Karan

2. Read the following telephone conversation which took place when Suresh was staying with his uncle. Write the message from Suresh to his maid, using not more than 50 words.

Seshu : Hello ! Hello ! This is Sheshu from Lucknow. Can I speak to Mrs. Rao, please ? I am a friend of his son, Madhav.
Suresh : This is Suresh Rao. My uncle is not here at the moment. We heard about the earthquake. Is Madhav all right ?
Seshu : Yes, yes. He’s okay now. He had a bad fall during the earthquake and he broke his left leg. It was a multiple fracture, but there’s nothing to worry about now.
Suresh : Is he in hospital ?
Sheshu : Yes, he’s at the Tata Memorial Hospital here. Would you please inform his family ?
Suresh : Of course I will.

Message

Dear Uncle

There was a telephonic call for you from one Mrs. Sheshu. He is our Madhav’s friend, from Lucknow. He had a fall during the earthquake and he broke his leg. He got a multiple fracture and has been in the Tata Memorial Hospital there. But he added that there was nothing to worry.

Suresh

3. Here is telephonic talk between Gurbani and Jaspreet. Gurbani give her a message , Write the message on behalf of Gurmeet not more than 50 words.

Gurbani : Hi ! Gurmeet.
Jaspreet : Sorry, I’m not Gurmeet. I’m her elder sister Jaspreet. Can I know who is calling ?
Gurbani : I Gurbani, her friend, Is it not her contact number ? I have some urgent message for her.
Jaspreet : Yes, it is but she has gone to the market to by some fruit and her mobile, is with me.
Gurbani : Would you please convey my message to her ? .
Jaspreet : But she is not coming back for about two hours. I am also going.to the hospital to see one of our neighbours. Would you gave me the message. I’ll leave it on her table before I go.
Gurbani : Sure ! Please note our family is going to Hazoor Sahib on Sunday. She can accompnay as if her parents allow. It will be a good company for me.
Jaspreet : All right, Don’t worry. The message will reach her.
Gurbani : Thank you very much.

Message

22.06.2020
Dear Gurmeet,

There was a call from your friend Gurbani in your absence. Their family is going to Hazoor Sahib on Sunday. You can accompany them if our parents permit. She will be feeling good in your company. Talk to her for confirmation.

Jaspreet.

4. Read the conversation between Mrs. Singh and the Principal of G.S.S. Modern School. Write the message on behalf of the Principal that he will send to the Preeti’s class-teacher.

Mrs. Singh : Hello ! Is that G.S.S. Modem School ?
Principal : Yes, what do you want ?
Mrs. Singh : I would like to speak to the school Principal.
Principal : yes, speaking. What can I do for you.
Mrs. Singh : My daughter, Preeti is a student of VII A of your school. Today was the last day for the payment of her school fee. I have deposited it the school account.
Principal : What is the problem in that ? It was your duty.
Mrs. Singh : Madam, she was worried about fine before she left for school and looked sad. Send this message to her in class, to make her tension free.
Principal : Of course ! The message will be sent to her through her class teacher.
Mrs. Singh : Thank you, Madam.

Message

12.05.2020
Dear Preeti

You looked worried and sad before you left for school. It was natural because your school fee was not paid and you could be fined or punished in some other way for that. Now, don’t take any tension as your fee has been deposited.

Mummy

5. Read the following telephonic conversation between Ravinder and Ranjit about to leave for her coaching soon and his mother is not at home for the moment. Write this message on behalf of Ranjit.

Ravinder : Hello, is that Amit ?
Ranjit : No I am her elder brother, Ranjit speaking May I know who is speaking ?
Ravinder : I am Ravinder speaking. I wanted to speak to your brother for an important message.
Ranjit : He is not at home now. He has just gone to visit one of two friends. I am also leaving for my office. If there is some message I shall give him.
Ravinder : OK, then Please tell Amit all about me. We are to play a friendly cricket match tomorrow morining. I am one of the players in his team front. I will not be able to take part in it as I am suddenly fell ill and the doctor has advised me complete rest. He can take any other player with him.

Message

23 March, 2020
Dear Amit

In your absence there was a telephone from your friend, Ravinder. He is one of the players of your cricket team for the tomorrow match.

But he will not be able to come, as he has suddenly fallen ill and the doctor has advised him complete rest. You can take any other friend with you.

Ranjit

6. Here is telephonic talk between Gurbani and Jaspreet Gurmeet gives her message for her brother. Write the message on behalf of Gurmeet not more than 50 words.

Gurmeet : Hi ! Gurmeet.
Jaspreet : Sorry, I’m not Gurmeet. I am his elder sister, Jaspreet. Can I know who is calling ?
Gurbani : I am Gurbani, Bedi, her tutor. Is it not his contact number ? I have some urgent message for her.
Jaspreet : Yes it is. But she has gone to the market to buy fruit and her mobile is with me.
Gurbani : Would you please, convey my message to him
Jaspreet : But he is not coming back for about two hours, I’m also going to the hospital to see one of our neighbours. Would you give me the message. I’ll leave it on his table before I go.
Gurbani : Sure ! Please note this. I shall not be able ‘to come for coaching as I have sprained my ankle. Therefore she should not write for me in the evening.
Jaspreet : All right. Don’t worry the message will reach him.
Gurbani : Thank you very much.

Message

April 10, 2020
Dear Sarbjit

There was a call from you tutor when you were not at home. She will not be coming today for coaching because she has sprained her ankle. Therefore don’t wait for her in the evening.

Mother/Mummy

7. Read the following telephone conversation between kamal and Hardeep from a hospital. Hardeep wants to talk to kamlesh sharma but she is at present not at home. She will be back after an hour. Thinking yourself as Kamal and using the telephone conversation as the subject, write a message to Kamlesh Sharma in not more than 50 words.

Hardeep : Is that Kamlesh Sharma ?
Kamal : No, we are her tanents.
Hardeep : May I speak to Mrs Sharma ? I have to talk to her urgently.
Kamal : She is not here at present. She has gone out and will return after an hour.
Hardeep : Then, would you give a message to her as soon as she returns ?
Kamal : Yes, of course. But I am also going to my friend’s home for his birthday party this evening. However, you needn’t worry I shall leave your message on table for her. What’s it ?
Hardeep : Kindly tell her that her grandmother has met with an accident and she is at present, in the civil hospital here in Khanna. She has broken her leg and is in plaster. Now she is feeling easy. Please tell kamlesh sharma to reach the hospital at her earliest with her husband.
Kamal : O.M.G. Please worry not. I shall leave the message for her before I leave Hardeep. Thank you very much.
Kamal : By, Who speaks on the other side ?
Hardeep : I am her neighbour, Hardeep Sodhi.

Message

May 5, 2020.
Dear Sharma Aunty

There was a telephone call for you from Khanna. I am sorry to inform that your grandmother has broken her leg and is in the Civil Hospital there. Don’t worry she is feeling easy now though she is still in plaster. Reach the hospital with your husband at your earliest.

Kamal

8. Read the following telephonic conversation between Ravinder and Shilpa. Shilpa is about to leave for her coaching within five minutes and her sister is not at home for the moment. Write this message conveyed on behalf of Shilpa.

Mrs. Ravinder : Hello, is that Jaspreet ?
Shilpa : No, I am her sister, Shilpa speaking. May I know who speaks on the other side ?
Mrs. Ravinder : I am Mrs. Ravinder, her friend Sonum’s mother speaking from bus stand. I wanted to speak to your sister urgently.
Shilpa : She is not at home now aunty. She has just gone to her college if there is some message I shall give her.
Mrs. Ravinder : Ok, then please note down I have returned from Amritsar. I have brought some holy books for her. I wanted her to collect the packet at bus stand as I am already late for home. However, she can collect it from there in the evening today urgently. I will not be availabe for the night as we have to attend a marriage party.
Shilpa : You needn’t worry aunty. As I am also going to the Gurudwara, I shall leave this message on table before leave.
Mrs. Ravinder : Thank you very much.

Message

March 5, 2020
Didi

There was a telephone from Mrs. Ravinder, Sonum’s mother. She has bought some holy books for you from Amritsar. Collect the packet from her home in the evening today, urgenlty as she will not be available at night as they are going to attend a marriage party.

Shilpa.

TYPE – II

1. You want to send a message to your niece on her birthday as you are unable to attend it because of an urgent meeting in the office. Write the S.M.S. is not more than 50 words.

May 15, 2020
Dear Vani

Many many happy returns of the day. Stay blessed and in high spirits. Don’t mind my absence as I am unable to attend your birthday party due to an urgent meeting in the office. You will soon receive a lovely gift from me through courier.

Your loving uncle
Gurnarn

2. You were to attend the marriage of your friend but you suddenly fell ill that night. Send your friend an S.M.S. informing your friend about your disability to reach and giving him congratulations and expressing your good wishes for the wedding couple.

Dear Madhur

Congratulations on the wedding of your elder brother. Please pardon my absence as I am unable to attend the marriage ceremony because of sudden illness. I had got my briefcase ready to take a bus, but I was forced it lie in bed. Let me convey my hearty wishes for the happy life of the wedding couple.

Sharan

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 10th Class Maths Book Solutions Chapter Statistics Ex 14.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.1

Question 1.
A survey was conducted by a group of students as a part of their enviroment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Which method did you use for finding the mean, and why?
Solution:
Since, number of plants and houses are small in their values; so we must use direct method

Mean X = \(\overline{\mathrm{X}}=\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}\)

\(\frac{162}{20}\) = 8.1

Hence, mean number of plants per house is 8.1.

Question 2.
Consider the following distribution of daily wages of 50 workers of a factory.

Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution:

From given data,
Assumed Mean (a) = 150
and width of the class (h) = 20
∴ \(\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}\)
= \(\frac{-12}{50}\) = – 0.24

Using formula, Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)
= 150 + (20) (- 0.24)
= 150 – 4.8 = 145.2
Hence,mean daily wages of the workers of factory is ₹ 145.20.

Question 3.
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is 18. Find the missing frequency f.

Solution;

From the above data,
Assumed mean (a) = 18
Using formula, Mean \((\overline{\mathrm{X}})=a+\frac{\sum f_{i} d_{i}}{\Sigma f_{i}}\)

\(\bar{X}=18+\frac{2 f-40}{44+f}\)
But. Mean of data \((\bar{x})\) = 18 …(Given)
∴ 18 = 18 + \(\frac{2 f-40}{44+f}\)
or \(\frac{2 f-40}{44+f}\) = 18 – 18 = 0
or 2f – 40 = 0
or 2f = 40
or f = \(\frac{40}{2}\) = 20
Hence, missing frequency f is 20.

Question 4.
Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method.

Solution:

From above data,
Assumed Mean (a) = 75.5
Width of Class (h) = 3
∴ \(\bar{u}=\frac{\sum f_{i} u_{i}}{\sum f_{i}}\)
= \(\frac{4}{30}\) (approx.)
Using formula,
Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)
= 75.5 + 3 (0.13) = 75.5 + 0.39
\(\overline{\mathrm{X}}\) = 75.89
Hence, mean heart beats per minute for women is 75.89.

Question 5.
In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number mangoes. The following was the distribution of mangoes according to the number of boxes.

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
Solution:
Since, values of number of mamgoes and number of boxes are large numerically. So, we must use step-deviation method.

From above data,
Assumed Mean (a) = 57
Width of class (h) = 3
∴ \(\bar{u}=\frac{\sum f_{i} u_{i}}{\sum f_{i}}\)

\(\bar{u}=\frac{25}{400}\) = 0.0625

Using formula, Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)
\(\bar{X}\) = 57 + 3(0.0625)
= 57 + 0.1875 = 57.1875 = 57.19.
Hence, mean number of mangoes kept in a packing box is 57.19.

Question 6.
The table below shows the daily expenditure on food of 25 households in a locality.

Find the mean daily expenditure on food by a suitable method.

Solution:

From above data,
Assumed mean (a) = 225
Width of class (h) = 50
∴ \(\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}\)

\(\bar{u}=-\frac{7}{25}\) = 0.28
Using formula,
Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)
\(\bar{X}\) = 225 + 50 (- 0.28)
\(\bar{X}\) = 225 – 14
\(\bar{X}\) = 211
Hence, mean daily expenditure on food is ₹ 211.

Question 7.
To find out the concentration of SO₂ in the air (in parts per million, e., ppm), the data was collected for 30 localities in a certain city and is presented below:

Find the mean concentration of SO₂ in the air.
Solution:

From above data,
Assumed mean (a) = 0.10
Width of class (h) = 0.04
∴ \(\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}\)

\(\bar{u}=-\frac{1}{30}\) = – 0.33(approx.)
Using formula, Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)
\(\bar{X}\) = 0.10 + 0.04 (- 0.33)
= 0.10 – 0.0013 = 0.0987 (approx.)
Hence, mean concentration of SO₂ in the air is 0.0987 ppm.

Question 8.
A class teacher has the folloing absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Solution:

From above data,
Assumed Mean (a) = 17
Using formula, Mean(\(\bar{X}\)) = a + \(\frac{\Sigma f_{i} d_{i}}{\Sigma f_{i}}\)
\(\bar{X}\) = 17 + \(-\frac{181}{40}\)
= 17 – 4.52 = 12.48.
Hence, mean 12.48 number of days a student was absent.

Question 9.
The following table gives the literacy rate (in percentage) of 35 citIes. Find the mean literacy rate.

Solution:

From the above data,
Assumed Mean (a) = 70
Width of class (h) = 10
∴ \(\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}=\frac{-2}{35}\) = – 0.057

Using formula, Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)
\(\bar{X}\) = 70 + 10 (- 0.057)
= 70 – 0.57 = 69.43
Hence, mean literacy rate is 69.43%.

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: