Class 7

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 1 Knowing Our Numbers Ex 1.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.3

1. Estimate each of the following using general rule:

Question (a)
837 + 987
Solution:
While rounding off to hundreds place
837 + 987 = 800 + 1000
= 1800

Question (b)
783 – 427
Solution:
While rounding off to hundreds place
783 – 427 = 800 – 400
= 400

Question (c)
1391 + 2783
Solution:
(i) While rounding off to thousands place
1391 + 2783 = 1000 + 3000
= 4000
(ii) While rounding off to hundreds place
1391 + 2783 = 1400 + 2800
= 4000

Question (d)
28292 – 21496.
Solution:
While rounding off to ten thousands place.
28292 – 21496 = 30000 – 20000
= 10000

2. Estimate the product using general rule:

Question (a)
898 × 785
Solution:
898 rounds off to hundreds place = 900
785 rounds off to hundreds place = 800
Estimated product = 900 × 800
= 720000

Question (b)
9 × 795
Solution:
9 rounding off to tens place = 10
795 rounding off to tens place = 800
Estimated product = 10 × 800
= 8000

Question (c)
(c) 87 × 317
Solution:
87 rounded off to hundreds place = 100
317 rounded off to hundreds place = 300
Estimated product = 90 × 300
= 27000

Question (d)
9250 × 29
Solution:
9250 rounds off to thousands place = 9000
29 rounds off to tens place = 30
Estimated product = 9000 × 30
= 270000

3. Estimate by rounding off to nearest hundred:

Question (a)
439 + 334 + 4317
Solution:
439 rounds off to nearest hundreds = 400
334 rounds off to nearest hundreds = 300
4317 rounds off to nearest hundreds = 4300
Estimated sum = 400 + 300 + 4300 = 5000

Question (b)
108734 – 47599.
Solution:
108734 rounds off to nearest hundreds = 108700
47599 rounds off to nearest hundreds = – 47600
Estimated difference = 61100

4. Estimate by rounding off to nearest tens:

Question (a)
439 + 334 + 4317
Solution:
439 + 334 + 4317
439 rounds off to nearest tens = 440
334 rounds off to nearest tens = + 330
4317 rounds off to nearest tens = + 4320
Estimated sum = 5090

Question (b)
108734 – 47599
Solution:
108734 rounds off to nearest tens = 108730
47599 rounds off to nearest tens = – 47600
Estimated difference = 61130

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 1 Knowing Our Numbers Ex 1.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.2

1. Convert the following measurements as directed:

Question (a)
5 km into metre
Solution:
1 km= 1000 m
∴ 5 km = 5 × 1000 m
= 5000 m

Question (b)
35 kilometre into metre
Solution:
1 km = 1000 m
∴ 35 km = 35 × 1000 m
= 35000 m

Question (c)
2000 milligram into gram
Solution:
1000 mg = 1 gm
∴ 2000 mg = \(\frac {1}{1000}\) × 2000 gm
= 2 gm

Question (d)
500 decigram into gram
Solution:
10 decigram = 1 gm
∴ 500 decigram = \(\frac {1}{10}\) × 500 gm
= 50 gm

Question (e)
2000 millilitre into litre
Solution:
1000 ml = 1 litre
∴ 2000 ml = \(\frac {1}{1000}\) × 2000 litre
= 2 litre

Question (f)
12 kilolitre into litre
Solution:
1 kilolitre = 1000 litre
∴ 12 kilolitre = 12 × 1000 litres
= 12000 litres

2. In an election, the successful candidate registered 6317 votes whereas his nearest rival could attain only 3761 votes. By what margin did the successful candidates defeat his rival?
Solution:
Votes attained by successful candidate = 6317
Votes attained by nearest rival = 3761
Difference between their votes = 6317 – 3761 = 2556
Successful candidate defeat his rival by 2556 votes.

3. A monthly magazine having 37 pages is published on 20th day of each month. This month 23791 copies were printed. Tell us how many pages were printed in all?
Solution:
Number of pages in one copy = 37
Number of pages in 23791 copies
= 23791 × 37
= 880267

4. A shopkeeper has 37 reams. One ream contain 480 pages and he wants to make quires of all these sheets to sell in retail. One quire of sheets contain 24 sheets. How many quires will be made?
Solution:
Number of pages in one ream = 480
Number of pages in 37 reams = 37 × 480
= 17760

Number of quires in 24 sheets = 1
Number of quires in 17760 sheets
= \(\frac {1}{24}\) × 17760
= 740 quires

5. Veerpal serves milk to the guests in glasses of capacity 250 ml each. Suppose that the glasses are filled to capacity and there was 5 litre milk that got consumed. How many guests were served with milk?
Solution:
Total quantity of milk consumed = 5 litre
= 5 × 1000 ml
= 5000 ml
Capacity of the glass = 250 ml
Number of glasses served = 5000 ÷ 250 = 20
Now

The milk is served in 20 glasses

6. A box of medicine contain 2,00,000 tablets each weighing 20 mg. What is the total weight of tablets inbox?
Solution:
Weight of each tablet = 20 mg
Weight of 2,00,000 tablets
= 2,00,000 × 20 mg
= 40,00,000 mg
= \(\frac {40,00,000}{1000}\) g
= 4000 g
= \(\frac {4000}{1000}\) = 4 kg
Hence, total weight of tablets is 4 kg

7. A bookstore sold books worth Rupees Two lakh eighty-five thousand eight hundred ninety-one in the first week of June. They sold books worth Rupees Four lakh seven hundred sixty-eight in the second week of June. How much was the total sale for two weeks together?
Solution:
Worth of books sold in first week = Rupees Two lakh eighty-five thousand eight hundred ninety one only.
= ₹ 2,85,891
Worth of books sold in second week = Rupees Four lakh seven hundred sixty-eight = ₹ 4,00,768.
Total sale for two weeks

8. A famous cricket player has so far scored 6978 runs in test matches. He wishes to complete 10,000 runs. How many more runs he need?
Solution:
The number of runs player wishes to complete = 10,000
The number of runs he scored = – 6,978
The number of more runs he needed = 3,022.
He needed 3,022 more runs

9. Surinder has ₹ 78592 with him. He placed an order for purchasing 39 radio sets at ₹ 1234 each. How much money will remain with him after the purchase?
Solution:
Cost of one radio set = ₹ 1234
Cost of 39 radio sets = 39 × ₹ 1234
= ₹ 48126

Total money Surinder has = ₹ 78592
Cost of 39 radio sets = – ₹ 48126
Money remained with him = ₹ 30466

10. A vessel has 3 litre 650 ml of curd. In how many glasses each of 25 ml capacity can it be distributed?
Solution:
Total quantity of curd = 3 l 650 ml
= 3 × 1000 ml + 650 ml
= 3000 ml + 650 ml
= 3650 ml
Capacity of one glass = 25 ml
Number of glasses distributed
= 3650 ÷ 25 = 146

∴ The curd can be distributed in 146 glasses.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 1 Knowing Our Numbers Ex 1.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.1

1. Write the smallest and the greatest number:

Question (a)
30900, 30594, 30945, 30495
(b) 10092, 10029, 10209, 10920.
Solution:
(a) All the given numbers are: 30900, 30594, 30945, 30495 are five-digit numbers. Let us examine digits on extreme left side of each number. First digit and second digit of all the numbers are same.

Then by observing the third and fourth digits from left side we conclude that
Smallest number = 30495
Greatest number = 30945

Question (b)
10092, 10029, 10209, 10920.
Solution:
All the given numbers are: 10092, 10029, 10209, 10920 are five digit numbers. Let us examine digits on extreme left side of each number. First digit and second digit from left of all the numbers are same.

Then by observing third and fourth digits from left we conclude that
Smallest number = 10029
Greatest number = 10920

2. Arrange the numbers in ascending order:

Question (a)
6089, 6098, 5231, 3953
Solution:
Ascending order is:
3953, 5231, 6089, 6098

Question (b)
49905, 6073, 58904, 7392
Solution:
Ascending order is:
6073, 7392, 49905, 58904

Question (c)
9801, 25751, 36501, 38802.
Solution:
Ascending order is:
9801, 25751, 36501, 38802

3. Arrange the numbers in descending order:

Question (a)
75003, 20051, 7600, 60632
Solution:
Descending order is:
75003, 60632, 20051, 7600

Question (b)
2934, 2834, 667, 3289
Solution:
Descending order is:
3289, 2934, 2834, 667

Question (c)
1971, 45321, 88715, 92547.
Solution:
Descending order is:
92547, 88715, 45321, 1971.

4. Use the given digits without repetition and make the greatest and smallest 4 digit number:

Question (a)
6, 4, 3, 2
Solution:
6432, 2346

Question (b)
9, 7, 0, 3
Solution:
9730, 3079

Question (c)
5, 4, 0, 3
Solution:
5430, 3045

Question (d)
3, 2, 7, 1.
Solution:
1321, 1237.

5. Using any one digit twice make the greatest and the smallest 4 digit number:

Question (i)
(a) 2, 3,7
(b) 5,0,3
(c) 2, 3, 0
(d) 1, 3, 4
(e) 2, 5, 8
(f) 1, 2, 3
Solution:
(a) 7732, 2237
(b) 5530, 3005
(c) 3320, 2003
(d) 4431, 1134
(e) 8852, 2258
(f) 3321, 1123

6. Read the following numbers using place value chart:

Question (i)
(a) 638975
(b) 84321
(c) 29061058
(d) 60003608.
Solution:
Place Value Chart:

	C	TL	L	TTh	Th	H	T	O
(a)			6	3	8	9	7	5
(b)				8	4	3	2	1
(c)	2	9	0	6	1	0	5	8
(d)	6	0	0	0	3	6	0	8

(a) Six lakh thirty-eight thousand nine hundred seventy-five
(b) Eighty-four thousand three hundred twenty-one
(c) Two crore ninety lakh sixty one thousand fifty-eight
(d) Six crore three thousand six hundred eight.

7. Insert commas suitably and write the names according to Indian System of Numeration:

Question (a)
98606873
Solution:
9,86,06,873
Nine crore eighty-six lakh six thousand eight hundred seventy-three.

Question (b)
7635172
Solution:
76,35,172
Seventy-six lakh thirty-five thousand one hundred seventy-two.

Question (c)
89700057
Solution:
8,97,00,057
Eight crore ninety-seven lakh fifty-seven.

Question (d)
89322602
Solution:
8,93,22,602
Eight crore ninety-three lakh twenty-two thousand six hundred two.

Question (e)
4503217
Solution:
45,03,217
Forty-five lakh three thousand two hundred seventeen.

Question (f)
90032045.
Solution:
9,00,32,045
Nine crore thirty-two thousand forty-five.

8. Insert commas suitably and write the names according to International System of Numeration:

Question (a)
89832081
Solution:
89,832,081
Eighty-nine million eight hundred thirty-two thousand eighty-one.

Question (b)
6543374
Solution:
6,543,374
Six million five hundred fourty three thousand three hundred seventy-four.

Question (c)
88976306
Solution:
88,976,306
Eighty-eight million nine hundred seventy-six thousand three hundred six.

Question (d)
9860001
Solution:
9,860,001
Nine million eight hundred sixty thousand one.

Question (e)
90032045
Solution:
90,032,045
Ninety million thirty-two thousand forty-five.

Question (f)
4503217
Solution:
4,503,217
Four million five hundred three thousand two hundred seventeen.

9. Write the number names as numerals:

Question (a)
Seven lakh fifty-four thousand
Solution:
7,54,000

Question (b)
Nine crore fifty-three lakh seventy-four thousand five hundred twenty-three.
Solution:
9,53,74,523

Question (c)
Six hundred forty-seven thousand five hundred twenty-five.
Solution:
647,525

Question (d)
Seventy-two million three hundred thirty-two thousand one hundred twelve.
Solution:
72,332,112

Question (e)
Fifty-eight million four hundred twenty-three thousand two hundred two.
Solution:
58,423,202

Question (f)
Twenty-three lakh thirty thousand ten.
Solution:
23,30,010.

10. How many eight-digit numbers are there in all?
Solution:
Largest eight-digit number is 99999999.
Largest seven-digit number is 9999999.
Total number of eight digit numbers = Largest eight digit – Largest seven digit number
= 99999999 – 9999999
= 90000000

11. Fill in the blanks:

Question (i)
(a) 1 Lakh = ten thousand
(b) 1 Million = hundred thousand
(c) 1 Crore = ten lakh
(d) 1 Crore = million
(e) 1 Million = lakh.
Solution:
(a) Ten
(b) Ten
(c) Ten
(d) Ten
(e) Ten

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 7 Congruence of Triangles Ex 7.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.1

1. Identify the pairs of congruent figures and write the congruence in symbolic form.

Question (i).

Answer:
In figure
Radius of circle C₁ = 2 cm
Radius of circle C₂ = 1.5 cm
As radius of circle C₁ ≠ Radius of circle C₂
∴ Circle C₁ is not congruent to circle C₂

Question (ii).

Answer:
In figure
Length of line segment AB = 6 cm
Length of line segment MN = 7 cm
As length of line segment AB ≠ Length of line segment MN
∴ AB is not congruent to MN

Question (iii).

Answer:
In ΔXYZ and ΔPQR
XY = PQ, YZ = PR, XZ = QR
So, ΔXYZ and ΔPQR have the same size and shape
∴ ΔXYZ ≅ ΔQPR

Question (iv).

Answer:
In figure ΔABC and ΔDEF do not have the same size and shape
∴ ΔABC and ΔDEF are do not have the same size and shape.
∴ ΔABC and ΔDEF are not-congruent.

Question (v).

Answer:

Question (vi).

Answer:

2. If ΔPQR as ΔOMN under the correspondence PQR ↔ OMN, write all the corresponding congruent parts of the triangle.
Solution:
For better understanding of the correspondence, let us draw a diagram of given correspondence.

The correspondence is PQR → OMN
This means vertices P ↔ O, Q ↔ M, R ↔ N
Sides : PQ ↔ OM, QR ↔ MN, RP ↔ NO
and Angles: ∠PQR ↔ ∠OMN, ∠QRP ↔ ∠MNO, ∠RPQ ↔ ∠NOM

3. Draw any two pairs of congruent triangles.
Solution:
Two pairs of congruent triangles are :
(i) Draw a ΔABC in which AB = 5 cm, BC = 4 cm and CA = 6 cm.
Draw another ΔPQR in which PQ = 6 cm, QR = 5 cm and RP = 4 cm shown in the following figure.


Make a trace copy of ΔABC using a tracig paper and superinpose it on ΔPQR, where C falls on P, A falls on Q and B falls on R. We observe that ΔABC will corr. ΔPQR.
∴ ΔABC ≅ ΔQRP

(ii) Draw a ΔXYZ in which XY = 5 cm, YZ = 6 cm and ZX = 3 cm. Draw another ΔLMN in which LM = 5 cm, MN = 6 cm and NL = 3 cm. Since both ΔXYZ and ΔLMN have the same size and shape.

4. If ΔABC ≅ ΔZYX, write the parts of ΔZYX that correspond to.
(i) ∠B
(ii) CA
(iii) AB
(iv) ∠C
Solution:
First of all we draw a diagram of given correspondence.

The correspondence is ABC ↔ ZYX.
This means A ↔ Z, B ↔ Y, C ↔ X
Therefore
(i) ∠B = ∠Y
(ii) CA = XZ
(iii) AB = ZY
(iv) ∠C = ∠X

5. Multiple Choice Questions :

Question (i).
If ΔABC as ΔXYZ under the correspondence ABC ↔ XYZ. Then
(a) ∠A = ∠Z
(b) ∠X = ∠B
(c) ∠A = ∠X
(d) ∠C = ∠X.
Answer:
(c) ∠A = ∠X

Question (ii).
Two line segments are congruent if,
(a) They are parallel
(b) They intersect each other
(c) They are part of same line
(d) They are of equal length.
Answer:
(d) They are of equal length.

Question (iii).
Two triangles ΔABC and ΔLMN are congruent AB = LM, BC = MN. If AC = 5 cm then LN is :
(a) 3 cm
(b) 15 cm
(c) 5 cm
(d) Can’t find.
Answer:
(c) 5 cm

6. Two right angles are always congruent. (True/False)
Answer:
True

7. Two opposite sides of a rectangle are always congruent. (True/False)
Answer:
True

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 6 Triangles Integers MCQ Questions with Answers.

PSEB 7th Class Maths Chapter 6 Triangles Integers MCQ Questions

Multiple Choice Questions :

Question 1.
The closed curve made up of three lines segments is called :
(a) Quadrilateral
(b) Triangle
(c) Rectangle
(d) Square
Answer:
(b) Triangle

Question 2.
The number of median a triangle has, is:
(a) Two
(b) One
(c) Three
(d) Four
Answer:
(c) Three

Question 3.
What is the number of altitudes of a triangle ?
(a) Three
(b) Four
(c) One
(d) Two
Answer:
(a) Three

Question 4.
In the following fig. value of exterior angle x is :

(a) 50°
(b) 70°
(c) 120°
(d) 60°
Answer:
(c) 120°

Question 5.
In the following fig. the value of unknown interior angle x is :

(a) 50°
(b) 115°
(c) 65°
(d) 130°
Answer:
(c) 65°

Fill in the blanks :

Question 1.
A triangle can have ……………. medians.
Answer:
Three

Question 2.
An exterior angle of a triangle is equal to ……………. of its interior opposite angles.
Answer:
Sum

Question 3.
A triangle is a ……………. figure.
Answer:
Closed

Question 4.
Sum of the angles of a triangle is …………….
Answer:
180°

Question 5.
The point of concurrence of the medians of a triangle is called …………….
Answer:
Centroid

Write True or False :

Question 1.
A triangle can have three altitudes. (True/False)
Answer:
True

Question 2.
There are three angle bisectors in a triangle. (True/False)
Answer:
True

Question 3.
A triangle is possible with angle 60°, 70°, 80°. (True/False)
Answer:
False

Question 4.
The sum of all the interior angles of a triangle is 180°. (True/False)
Answer:
True

Question 5.
Measure of each interior angle of equilateral tree triangle to 60°. (True/False)
Answer:
True

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 6 Triangles Ex 6.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 6 Triangles Ex 6.2

1. State, if a triangle is possible with the following angles.

Question (a).
35°, 70°, 65°
Answer:
No Reason :
Sum of three angles
= 35° + 70° + 65° = 170°
But, we know that sum of angles of a triangle is always 180°
∴ A triangle cannot have angles 35°, 70° and 90°.

Question (b).
70°, 50°, 60°
Answer:
Yes
Reason :
Sum of three angles
= 70° + 50° + 60°
= 180°
By angle sum property
∴ A triangle can have angles 70°, 50° and 60°.

Question (c).
90°, 80°, 20°
Answer:
No
Reason :
Sum of three angles
= 90° + 80° + 20°
=190°
But, we know that sum of angles of a triangle is always 180°
(Angle sum properly)
∴ A triangle cannot have angles 90°, 80° and 20°.

Question (d).
60°, 60°, 60°
Answer:
Yes
Reason :
Sum of three angles
= 60° + 60° + 60°
= 180°
by angle sum property.
∴ A triangle can have angles 60°, 60° and 60°.

Question (e).
90°, 90°, 90°
Answer:
No Reason :
Sum of three angles
= 90° + 90° + 90°
= 270°
But, we know that sum of angles of a triangle is always 180° (Angle sum properly)
∴ A triangle cannot have angles 90°, 90° and 90°.

2. Find the value of x in the following figures :

Question (i).

Answer:
By angle sum property of a triangle
x + 53° + 60° = 180°
x + 113° = 180°
x = 180° – 113°
x = 67°

Question (ii).

Answer:
By angle sum property of a triangle
90° + x + 42° = 180°
132° + x = 180°
x = 180° – 132°
x = 48°

Question (iii).

Answer:
By angle sum property of a triangle
x + x + 70° = 180°
2x + 70° = 180°
2x = 180° – 70°
2x = 110°
x = \(\frac{110^{\circ}}{2}\)
x = 55°

Question (iv).

Answer:
By angle sum property of a triangle
x + 3x + 2x = 180°
6x = 180°
x = \(\frac{180^{\circ}}{6}\)
x = 30°

Question (v).

Answer:
By angle sum property of a triangle
x + x + x = 180°
3x = 180°
x = \(\frac{180^{\circ}}{3}\)
x = 60°

Question (vi).

Answer:
By angle sum property of a triangle
x – 5° + 60° + x + 5° = 180°
2x + 60° = 180°
2x = 180° – 60°
2x = 120°
x = \(\frac{120^{\circ}}{2}\)
x = 60°

3. Find the values of x and y in the following figures :

Question (i).

Answer:
Since in ΔABC, BC is produced to D
∴ 60° + x = 110°
(By exterior angle property)
x = 110°- 60°
x = 50° ………. (1)
Now, in ΔABC
60° + x + y = 180°
(By angle sum property of triangle)
60° + 50° + y = 180° [(By using (1)]
110° + y = 180°
y = 180° – 110°
y = 70°
Hence, x = 50°,
y = 70°

Question (ii).

Answer:
In ΔPQR,
∠P + ∠Q + ∠R = 180°
60° + 40° + x = 180°
(By angle sum property of triangle)
100° + x = 180°
x = 180° – 100°
x = 80°
Now,in ΔPQR, QR is produced
∴ y = 60° + 40°
(By exterior angle property)
y = 100°
Hence, x = 80°,
y = 100°

Question (iii).

Answer:
∠ACB = ∠ECD
∴ x = 80°….(1)
(vertically opposite angles)
∠ACD + ∠ECD = 180° (Linear pair)
∴ ∠ACD + 80° = 180° [by using (1)]
∠ACD = 180°- 80°
∠ACD = 100° ….(2)
In ΔABC, BC is produced to D
∴ x + y = ∠ACD
(By exterior angle property)
80° + y = 100°
(By using (1) and (2))
y = 100° – 80°
y = 20°
Hence, x = 80° and y = 20°

Question (iv).

Answer:
By angle sum property of a triangle
∠L + ∠M + ∠N = 180°
y + 90° + y = 180°
2 y + 90° = 180°
2y = 180° – 90°
2y = 90°
y = \(\frac{90^{\circ}}{2}\)
y = 45°

Question (v).

Answer:
∠ABC = ∠HBI
(Vertically opposite angles)
∴ y = x …(1)
∠BAC = ∠GAF
(Vertically opposite angles)
∴ ∠BAC = x ….(2)
∠ACB = ∠EFD
(Vertically opp. angles)
∠ACB = x …(3)
Now, in ΔABC
∠BAC + ∠ABC + ∠ACB = 180°
(By angle sum property of triangle)
x + x + x = 180°
[by using (1), (2) and (3)]
3x = 180°
x = \(\frac{180^{\circ}}{3}\)
x = 60°
y = x
= 60° (by using (1) and (4))
Hence, x = 60°, y = 60°

Question (vi).

Answer:
In ΔPQR, QR is produced to S,
∴ 2x – 5° = 50° + x + 5°
(By exterior angle property.)
2x – 5°= 55° + x
2x – x = 55° + 5°
x = 60° ….(i)
Now, by angle sum property of a ΔPQR
50° + x + 5° + y = 180°
55° + 60° + y = 180°
115°+ y = 180°
y = 180° – 115°
y = 65°
Hence, x = 60° and y = 65°

4. The angles of a triangle are in the ratio 5:6:7. Find the measure of each of the angles.
Solution:
Let the measure of the given angles be
(5x)°, (6x)°, (7x)°
By angle sum property of a triangle
(5x)° + (6x)° + (7x)° = 180°
(18x)° = 180°
x = \(\frac{180^{\circ}}{18}\)
x = 10
Required angles
= (5 × 10)°, (6 × 10)°, (7 × 10)°
= 50°, 60°, 70°

5. One angle of a triangle is 60°. The other two angles are in the ratio 4 : 8. Find the angles.
Solution:
One angle of triangle = 60°
Let the other two angles be (4x)° and (8x)°
By angle sum property of a triangle
60° + (4x)° + (8x)° = 180°
60° + (12x)° = 180°
(12x)° = 180° – 60°
(12x)° = 120°
x = \(\frac{120^{\circ}}{12}\)
x = 10
Required angles = (4x)°, (8x)°
(4 × 10)°, (8 × 10)°
= 40°, 80°

6. In a triangle ABC, ZB = 50°, ∠C = 62°. Find ∠A.
Solution:
In a ΔABC, ∠B = 50°, ∠C = 62°
By angle sum property of a triangle
∠A + ∠B + ∠C = 180°
∠A + 50° + 62° = 180°
∠A + 112° = 180°
∠A = 180° – 112°
∠A = 68°

7. In a right angled triangle two acute angles are in the ratio 2 : 3. Find the angles.
Solution:
In a right angle triangle one angle = 90°
Let the other two angles be (2x)°, (3x)°
By angle sum property of a triangle.
90° + (2x)° + (3x)° = 180°
90° + (5x)° = 180°
(5x)° = 180° – 90°
(5x)° = 90°
x = \(\frac{90^{\circ}}{5}\)
x = 18
Required angles = (2x)°, (3x)°
= (2 × 18)°, (3 × 18)°
= 36°, 54°

8. Three angles of a triangle are (2x + 20)°, (x + 30)° and (2x – 10)°. Find the angles.
Solution:
Since, we know that the sum of angles of a triangle is always 180°
∴ (2x + 20)° + (x + 30)° + (2x – 10)° = 180°
(5x + 40)° = 180°
(5x)° = 180° – 40°
(5x)° = 140°
x = \(\frac{140^{\circ}}{5}\)
x = 28
Required angles
= (2x + 20)°, (2x + 30)° and (2x – 10)°
= (2 × 28 + 20)°, (28 + 30)° and (2 × 28 -10)°
= (56 + 20)°, (58)° and (56 – 10)°
= 76°, 58° and 46°

9. Multiple choice questions :

Question (i).
A triangle can have two …………….
(a) Acute angles
(b) Obtuse angles
(c) Right angles
(d) None of these.
Answer:
(a) Acute angles

Question (ii).
A triangle is possible with measure of angles
(a) 30°, 40°, 100°
(b) 60°, 60°, 70°
(c) 60°, 50°, 70°
(d) 90°, 89°, 92°
Answer:
(c) 60°, 50°, 70°

Question (iii).
One of the equal angles of an isosceles triangle is 45° then its third angle is
(a) 45°
(b) 60°
(c) 100°
(d) 90°
Answer:
(d) 90°

Question (iv).
The number of obtuse angles that a triangle can have
(a) 2
(b) 1
(c) 3
(d) 4.
Answer:
(b) 1

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 6 Triangles Ex 6.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 6 Triangles Ex 6.3

1. Find the length of the unknown side in each of following figures

Question (i).

Answer:
Take a = 3 cm, b = 4 cm and unknown side = c
By Pythagoras Theorem
c² = a² + b²
c² = (3)² + (4)²
c² = 9 + 16
c² = 25
∴ c = \(\sqrt{25}\)
c = 5
Thus, the length of unknown side = 5 cm.

Question (ii).

Answer:
Take a = 15 cm, b = 20 cm
By Pythagoras Theorem
c² = a² + b²
∴ c² = (15)² + (20)²
c² = 225 + 400
c² = 625
∴ c = \(\sqrt{625}\)
c = 25
Thus, the length of unknown side = 5 cm

2. Which of the following can be the sides of a right triangle ?
(i) 4 cm, 5 cm, 7 cm
(ii) 1.5 cm, 2 cm, 2.5 cm
(iii) 2 cm, 2 cm, 5 cm
In the case of right angled triangles, identify the right angles.
Solutions:
(i) Let in ΔABC, the longest side is AB = 7 cm

(BC)² + (AC)²
= (4)² + (5)²
= 16 + 25 = 41
(BC)² + (AC)² = 41
Also AB² = (7)² = 49
Since AB² ≠ (BC)² + (AC)²
∴ The triangles with the given sides is not a right triangle.

(ii) Let in ΔABC, the longest side is AB = 2.5 cm

(AB)² = (2.5)² = 6.25 ….(1)
(BC)² + (AC)²
= (1.5)² + (2)²
= 2.25 + 4
= 6.25
∴ (BC)² + (AC)² = 6.25 ….(2)
From (1) and (2)
(AB)² = (BC)² + (AC)²
Therefore, the given triangle is a right triangle.
The angle opposite to the longest side is right angle.

(iii) Let in ΔABC, the longest side is AB = 5 cm

(AB)² = (5)²
(AB)² = 25 ….(1)
(BC)² + (AC)² = (2)² + (2)²
(BC)² + (AC)² = 4 + 4
(BC)² + (AC)² = 8 …..(2)
From (1) and (2)
(AB)² ≠ (BC)² + (AC)²
Therefore the triangle whose sides are 5 cm, 2 cm and 2 cm is not a right triangle.

3. Find the area and the perimeter of the rectangle whose length is 15 cm and the length of one diagonal is 17 cm.
Solution:
Let ABCD be a rectangle with length AB = 15 cm and diagonal AC = 17 cm.
In ΔABC, ∠B = 90° (Each angle of a rectangle)

By Pythagoras Theorem,
(AC)² = (AB)² + (BC)²
(17)² = (15)² + (BC)²
289 = 225 + (BC)²
(BC)² = 289 – 225 = 64
BC = 8 cm
Area of rectangle ABCD
= AB × BC
= 15 cm × 8 cm
= 120 cm²
Perimeter of rectangle ABCD = 2(AB + BC)
= 2(15 cm + 8 cm)
= 2(23 cm) = 46 cm²

4. A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance, find the distance of the foot of the ladder from the wall.

Solution:
Let AB be the ladder and BC be the distance of the foot of the ladder from the wall then AB = 15 m and AC = 12 m

By Pythagoras Theorem,
AB² = BC² + AC²
(15)² = BC² + (12)²
225 = BC² + 144
BC² = 225 – 144
BC² = 81
BC =9
Hence the distance of the foot of the ladder from the wall is 9 m.

5. The side of a rhombus is 5 cm. If the length of one of the diagonals of the rhombus is 8 cm, then find the length of the other diagonal.
Solution:
Let ABCD be a rhombus with side AB = 5 cm and diagonal AC = 8 cm
Let diagonal AC and BD bisect each other at O.
Then OA = OC = \(\frac {8}{2}\) cm = 4 cm
The diagonals of a rhombus bisect each other at right angle

∴ In right angled ΔAOB,
AO = 4 cm, AB = 5 cm
By Pythagoras Theorem,
OA² + OB² = AB²
(4)² + OB² = (5)²
16 + OB² = 25
OB² = 25 – 16 = 9
OB = 3 cm
Diagonal BD = 2 × OB = 2 × 3cm = 6cm
Therefore other diagonal of rhombus = 6 cm.

6. A right triangle is isosceles. If the square of the hypotenuse is 50 m, what is length of each of its sides ?
Solution:
Let ΔABC is a right isosceles triangle in which (AC)² = 50 m and AB = AC

∴ By Pythagoras Theorem
AB² + BC² = AC²
∴ AB² + AB² = AC²
2AB² = 50
AB² = 25
AB = 5
Therefore length of each equal side = 5m.

7. ΔABC is a triangle right angled at C if AC = 8 cm and BC = 6 cm, find AB.
Solution:
In right angled triangle ABC right angle at C
AC = 8 cm and BC = 6 cm

By Pythagoras Theorem
AB² = AC² + BC²
AB² = (8)² + (6)²
AB² = 100
AB = 10 cm.

8. State whether the following triplets are Pythagorean or not.

Question (i).
(5, 7, 12)
Solution:
Let a = 5, b = 1, c = 12
∴ c² = (12)² = 144
a² + b² = (5)² + (7)²
= 25 + 49 = 74
∴ a² + b² ≠ c²
∴ (5, 7, 12) is not a pythagorean triplet.

Question (ii).
(3, 4, 5)
Solution:
Let a = 3, b = 4, c = 5
∴ a² + b² = (3)² + (4)²
= 9 + 16 = 25
c² = (5)² = 25
∴ c² = a² + b²
∴ (3, 4, 5) is a pythagorean triplet

Question (iii).
(8, 9, 10)
Solution:
Let a = 8, b = 9, c = 10
∴ a² + b² = (8)² + (9)²
= 64 + 81 = 145
c² = (10)² = 100
c² ≠ a² + b²
Therefore (8, 9, 10) is not a pythagorean triplet.

Question (iv).
(5, 12, 13)
Solution:
Let a = 5, b = 12, c = 13
∴ a² + b² = (5)² + (12)²
= 25 + 144 = 169
c² = (13)² = 169
a² + b² ≠ c²
Therefore (5, 12, 13) is a pythagorean triplet.

9. Multiple Choice Questions :

Question (i).
In a ΔABC, if ∠A = 40° and ∠B = 55° then ∠C is
(a) 75°
(b) 80°
(c) 95°
(d) 85°
Answer:
(d) 85°

Question (ii).
If the angles of a triangle are 35°, 35° and 110°, then it is
(a) an isosceles triangle
(b) an equilateral triangle
(c) a scalene triangle
(d) right angled triangle
Answer:
(a) an isosceles triangle

Question (iii).
A triangle can have two
(a) right angles
(b) obtuse angles
(c) acute angles
(d) straight angles
Answer:
(c) acute angles

Question (iv).
A triangle whose angles measure 35°, 55° and 90° is
(a) acute angled
(b) right angled
(c) obtuse angled
(d) isosceles
Answer:
(b) right angled

Question (v).
A triangle is not possible whose angles measure
(a) 40°, 65°, 75°
(b) 50°, 56°, 74°
(c) 72°, 63°, 45°
(d) 67°, 42°, 81°
Answer:
(d) 67°, 42°, 81°

Question (vi).
A triangle is not possible with sides of lengths (in cm)
(a) 6, 4, 10
(b) 5, 3, 7
(c) 7, 8, 9
(d) 3.6, 5.4, 8
Answer:
(a) 6, 4, 10

Question (vii).
In a right angled triangle, the length of two legs are 6 cm and 8 cm. The length of the hypotenuse is
(a) 14 cm
(b) 10 cm
(c) 11 cm
(d) 12 cm
Answer:
(b) 10 cm

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 4 Simple Equations Ex 4.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.4

Question 1.
If 7 is added to five times a number, the result is 57. Find the number.
Solution:
Let the required number = x
Five times the number = 5x
7 added to five times the number = 5x + 7
According to the problem
5x + 7 = 57
5x = 57 – 7
5x = 50
x = \(\frac {50}{5}\)
So, x = 10
Hence the required number is 10.

Question 2.
9 decreased from four times a number yields 43. Find the number.
Solution:
Let the required number = x
Four times the number = 4x
9 decreased from four times the number = 4x – 9
According to the problem
4x – 9 = 43
4x = 43 + 9
4x = 52
x = \(\frac {52}{4}\)
x = 13
Hence, the required number is 13.

Question 3.
If one-fifth of a number minus 4 gives 3, find the number.
Solution:
Let the required number = x
One fifth of the number = \(\frac {1}{5}\)x
One fifth of the number minus 4 = \(\frac {1}{5}\)x – 4
According to problem
\(\frac {1}{5}\)x – 4 = 3
\(\frac {1}{5}\)x = 3 + 4
\(\frac {1}{5}\)x = 7
x = 35
Hence the required number is 35.

Question 4.
In a class of 35 students, the number of girls is two-fifth the number of boys. Find the number of girls in the class.
Solution:
Let the number of boys = x
∴ number of girls = \(\frac {2}{5}\)x
Total number of students = 35
x + \(\frac {2}{5}\)x =35
\(\frac{5 x+2 x}{5}\) = 35
7x = 5 × 35
x = \(\frac{5 \times 35}{7}\)
x = 25
Therefore number of boys = 25
Number of girls = 35 – 25 = 10.

Question 5.
Sham’s father’s age is 5 years more than three times Sham’s age. Find Sham’s age, if his father is 44 years old.
Solution:
Let Sham’s age = x years
Then Sham’s father age = 3x + 5
But Sham’s fathers age = 44
According to question
3x + 5 = 44
3x = 44 – 5
3x = 39
Dividing both sides by 3
\(\frac{3 x}{3}=\frac{39}{3}\)
or x = 13
Hence Sham’s age is 13 years.

Question 6.
In an isosceles triangle the base angles are equal, the vertex angle is 40°. What are the base angles of the triangle ? (Remember, the sum of three angles of a triangle is 180°)
Solution:
Let each base angle of an isosceles triangle = x (in degrees)
Vertex angle = 40°
The sum of angles of a triangle = 180°
∴ x + x + 40° = 180°
2x = 180° – 40°
2x = 140°
Divide both sides by 2
\(\frac{2 x}{2}-\frac{140^{\circ}}{2}\)
Or x = 70°
Each equal angle is of 70°

Question 7.
Irfan says that he has 7 marbles more than five times the marbles Pannit has. Irfan has 37 marbles. How many marbles does Pannit have ?
Solution:
Let marbles Parmit has = x
Marbles Irfan has = 5x + 7
But Marbles Irfan has = 37
∴ 5x + 7 = 37
5x = 37 – 7
5x = 30
x = \(\frac {30}{5}\) = 6
Therefore Parmit has 6 marbles.

Question 8.
The length of a rectangle is 3 units more than its breadth and the perimeter is 22 units. Find the breadth and length of a rectangle.
Solution:
Let breadth of rectangle (l)
= x units
∵ length of rectangle (b) = (x + 3) units
∴ Perimeter of rectangle = 2(l + b)
= 2 (x + x + 3) units
= 2(2x + 3) units
According to the question
Perimeter = 22 units
2 (2x + 3) =22
\(\frac{2(2 x+3)}{2}=\frac{22}{2}\)
2x + 3 = 11
2x = 11 – 3
or 2x = 8
Dividing both sides by 2 we get
\(\frac{2 x}{2}=\frac{8}{2}\)
x = 4
∴ breadth = 4 units
Length = (4 + 3) units
= 7 units

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 4 Simple Equations Ex 4.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3

1. Solve each of the following equation.

Question (i).
6x + 10 = – 2
Answer:
Given equation is 6x + 10 = – 2
Transposing + 10 from L.H.S to R.H.S
we get
6x = -2 – 10
or 6x = -12
Dividing both sides by 6, we get
\(\frac{6 x}{6}=\frac{-12}{6}\)
or x = – 2, which is the required solution.

To check Put x = – 2 in the LHS of the equation 6x + 10 = – 2
L.H.S. = 6x + 10
= 6 × -2 + 10
= -12 + 10
= – 2 = R.H.S.
∴ L.H.S. = R.H.S.

Question (ii).
2y – 3 = 2
Answer:
Given equation is 2y – 3 = 2
Transposing – 3 from L.H.S. to R.H.S,
we get
2y = 2 + 3
or 2y = 5
Dividing both sides by 2, we get:
\(\frac{2 y}{2}=\frac{5}{2}\)
or y = \(\frac {5}{2}\), which is the required solution

To check. Put y = \(\frac {5}{2}\) in the L.H.S of the equation 2y – 3 = 2
L.H.S = 2y – 3 = 2 × \(\frac {5}{2}\) – 3
= 5 – 3 = 2 = R.H.S.
∴ L.H.S. = R.H.S.

Question (iii).
\(\frac{a}{5}\) + 3 = 2
Answer:
Given equation is \(\frac{a}{5}\) + 3 = 2
Transposing + 3 from L.H.S to R.H.S., we get
\(\frac{a}{5}\) = 2 – 3
or \(\frac{a}{5}\) = -1
Multiplying both sides, by 5, we get
5 × \(\frac{a}{5}\) = 5 × – 1
or a = – 5, which is the required solution.

To Check: Put a = – 5 in the L.H.S of the equation
\(\frac{a}{5}\) + 3 = 2,
L.H.S. = \(\frac{a}{5}\) + 3
= \(\frac {-5}{5}\) + 3
= – 1 + 3
= 2 = R.H.S.
∴ L.H.S. = R.H.S.

Question (iv).
\(\frac{3 x}{2}=\frac{2}{3}\)
Answer:
Given equation is \(\frac{3 x}{2}=\frac{2}{3}\)
Multiplying both sides by 2, we get
2 × \(\frac{3 x}{2}\) = 2 × \(\frac {2}{3}\)
or 3x = \(\frac {4}{3}\)
Dividing both sides by 3 we get
\(\frac{3 x}{3}=\frac{4}{3} \times \frac{1}{3}\)
or x = \(\frac {4}{9}\), which is the required solution.

To Check. Put x = \(\frac {4}{9}\) in the L.H.S. of equation \(\frac{3 x}{2}=\frac{2}{3}\)
L.H.S. = \(\frac{3 x}{2}=\frac{3}{2} \times \frac{4}{9}\) = \(\frac {2}{3}\) = R.H.S.
∴L.H.S. = R.H.S.

Question (v).
\(\frac {5}{2}\)x = -5
Answer:
Given equation is \(\frac {5}{2}\) x = – 5
Multiplying both sides by 2, we get
2 × \(\frac {5}{2}\) x = 2 × – 5
or 5x = – 10
Dividing both sides by 5, we get
\(\frac{5 x}{5}=\frac{-10}{5}\)
or x = – 2, which is the required solution.

To Check. Put x = – 2 in L.H.S. of the equation \(\frac {5}{2}\)x = – 5
L.H.S. = \(\frac {5}{2}\)x = \(\frac {5}{2}\) × -2
= – 5 = R.H.S.
∴ L.H.S. = R.H.S.

Question (vi).
2x + \(\frac {5}{2}\) = \(\frac {37}{2}\)
Answer:
Given equation is 2x + \(\frac {5}{2}\) = \(\frac {37}{2}\)
Subtract \(\frac {5}{2}\) from both sides, we get
2x + \(\frac {5}{2}\) – \(\frac {5}{2}\)
= \(\frac {37}{2}\) – \(\frac {5}{2}\)
or 2x = \(\frac{37-5}{2}\)
or 2x = \(\frac {32}{2}\)
or 2x = 16
Dividing both sides by 2, we get
\(\frac{2 x}{2}=\frac{16}{2}\)
or x = 8, which is the required solution.

To Check. Put x = 8 in the L.H.S. of the equation 2x + \(\frac {5}{2}\) = \(\frac {37}{2}\)
L.H.S. = 2x + \(\frac {5}{2}\)
= 2 × 8 + \(\frac {5}{2}\)
= 16 + \(\frac {5}{2}\)
= \(\frac{32+5}{2}\)
= \(\frac {37}{2}\) = R.H.S.
∴ L.H.S. = R.H.S.

2. Solve the following equation

Question (i).
5 (x + 1) = 25
Answer:
Given equation is 5 (x + 1) = 25
Dividing both sides by 5 we get
\(\frac{5(x+1)}{5}=\frac{25}{5}\)
or x + 1 = 5
Transposing 1 from L.H.S. to R.H.S. we get
x = 5 – 1
or x = 4, which is the required solution.

To Check. Put x = 4 in the L.H.S. of the equation 5 (x + 1) = 25
L.H.S. = 5 (x + 1)
= 5 (4 + 1)
= 5 (5)
= 25 = R.H.S.
∴ L.H.S. = R.H.S.

Question (ii).
2 (3x – 1) = 10
Answer:
Given equation is 2 (3x – 1) = 10
Dividing both sides by 2, we get
\(\frac{2(3 x-1)}{2}=\frac{10}{2}\)
or 3x – 1 = 5
Transposing – 1 from L.H.S. to R.H.S we get
3x = 5 + 1
3x = 6
Dividing both sides by 3, we get \(\frac{3 x}{3}=\frac{6}{3}\)
or x = 2, which is the required solution.

To Check. Put x = 2, in the L.H.S. of the equation 2 (3x – 1) = 10
L.H.S. = 2 (3x – 1) = 10
L.H.S = 2 (3x – 1) = 2 (3 × 2 – 1)
= 2 (6 – 1)
= 2 × 5
= 10 = R.H.S.
∴L.H.S. = R.H.S.

Question (iii).
4 (2 – x) = 8
Answer:
Given equation is 4 (2 – x) = 8
Dividing both sides by 4, we get
\(\frac{4(2-x)}{4}=\frac{8}{4}\)
or 2 – x= 2
Transposing 2 from L.H.S. to R.H.S. we get
-x = 2 – 2
or – x = 0
Multiplying both sides by – 1, we get
-x × – 1 = x – 1
or x = 0, which is the required solution.

To Check. Put x = 0 in the L.H.S. of the equation 4 (2 – x) = 8
L.H.S. = 4 (2 – x) = 4 (2 – 0)
= 4 × 2
= 8 = R.H.S.
∴ L.H.S. = R.H.S.

Question (iv).
– 4 (2 + x) = 8.
Answer:
Given equation is – 4 (2 + x) = 8
Dividing both sides by – 4, we get
\(\frac{-4(2+x)}{-4}=\frac{8}{-4}\)
Transposing 2 from L.H.S. to R.H.S. we get :
x = – 2 – 2
or x = – 4, which is the required solution

To Check. Put x = – 4 in the L.H.S. of equation – 4 (2 + x) = 8
L.H.S. = – 4 (2 + x) = – 4 [2 + (- 4)]
= – 4 (2 – 4)
= – 4 (- 2)
= 8 = R.H.S.
∴ L.H.S. = R.H.S.

3. Solve the following equations :

Question (i).
4 = 5 (x – 2)
Answer:
Given equation is 4 = 5 (x – 2)
or 4 = 5x – 10
Transposing 5x to L.H.S. and 4 to R.H.S.,
we get
– 5x = – 4 – 10
or – 5x = – 14
Dividing both sides by – 5, we get
\(\frac{-5 x}{-5}=\frac{-14}{-5}\)
or, x = \(\frac {14}{5}\), which is the required solution.

To Check. Put x = \(\frac {14}{5}\) in the R.H.S. of the equation 4 = 5 (x – 2)
R.H.S. = 5 (x – 2) = 5\(\left(\frac{14}{5}-2\right)\)
= 5\(\left(\frac{14-10}{5}\right)\)
= 5 \(\left(\frac{4}{5}\right)\)
= 4 = L.H.S.
∴ L.H.S. = R.H.S.

Question (ii).
– 4 = 5 (x – 2)
Answer:
Given equation is – 4 = 5 (x – 2)
or – 4 = 5x – 10
Transposing -4 to R.H.S and 5x to L.H.S
we get
-5x = 4 – 10 or -5x = -6
Dividing both sides by – 5 we get
\(\frac{-5 x}{-5}=\frac{-6}{-5}\)
or x = \(\frac {6}{5}\), which is the required solution.

To Check. Put x = \(\frac {6}{5}\) in the R.H.S. of the equation – 4 = 5 (x – 2)
L.H.S. = 5 (x – 2)
= 5\(\left(\frac{6}{5}-2\right)\)
= 5\(\left(\frac{6-10}{5}\right)\)
= 5\(\left(\frac{-4}{5}\right)\)
= -4 = L.H.S.
L.H.S. = R.H.S.

Question (iii).
4 + 5 (p – 1) = 34
Answer:
Given equation is 4 + 5(p – 1) = 34
Transposing 4 to R.H.S. we get
5(p – 1) = 34 – 4
5(p – 1) = 30
Dividing both sides, by 5, we get
\(\frac{5(p-1)}{5}=\frac{30}{5}\)
p – 1=6
Transposing -1 to R.H.S. we get
p = 6 + 1
p = 7 which is the required solution.

To Check : Put p = 7 in L.H.S. of the equation 4 + 5 (p – 1) = 34
L.H.S. = 4 + 5 (p – 1)
= 4 + 5 (7 – 1)
= 4 + 5 (6)
= 4 + 30
= 34 = R.H.S.
∴ L.H.S. = R.H.S.

Question (iv).
6y – 1 = 2y + 1.
Answer:
Given equation is 6y – 1 = 2y + 1
Transposing – 1 to R.H.S. and 2y to L.H.S,
we get
6y – 2y = 1 + 1
or 4y = 2 or y = \(\frac {2}{4}\)
or y = \(\frac {1}{2}\), which is the required solution.

To Check Put y = \(\frac {1}{2}\) in both L.H.S. and R.H.S. of the equation
6y – 1 = 2y + 1
L.H.S. = 6y – 1 = 6 × \(\frac {1}{2}\) – 1 = 3 – 1 = 2
R.H.S. = 2y + 1 = 2 × \(\frac {1}{2}\) + 1 = 1 + 1 = 2.
∴ L.H.S. = R.H.S.

4.

Question (i).
Construct 3 equations starting with x = 2
Answer:
First Equation.
(i) Start with x = 2
Multiplying both sides by 10
10x = 20
Adding 2 to both sides
10x + 2 = 20 + 2
or 10x + 2 = 22
This has resulted in an equation.

Second Equation. Start with x = 2
Divide both sides by 5
∴ \(\frac{x}{5}=\frac{2}{5}\)
This has resulted in an equation.

Third Equation. Start with x = 2
Multiply both sides by 5, we get
5x = 5 × 2
or 5x = 10
Subtracting 3 from both sides, we get
5x – 4 = 10 – 3
or 5x – 3 = 7
This has resulted in an equation.

Question (ii).
Construct 3 equation starting with x = – 2
Answer:
First Equation. Start with x = – 2
Multiplying both sides with 3, we get
3x = – 6
This has resulted in an equation

Second Equation. Start with x = – 2
Multiplying both sides with 3, we get 3x = -6
Adding 7 to both sides, we get 3x + 7
= -6 + 7 or 3x + 7 = 1
This has resulted in an equation.

Third Equation. Start with x = – 2
Multiplying both side with 2 we get 3x = – 6
Adding 10 to both sides we get
3x+ 10 = -6 + 10
or 3x + 10 = 4
This has resulted in an equation.

Multiple Choice Questions :

5. If 7x + 4 = 39, then x is equal to :
(a) 6
(b) -4
(c) 5
(d) 8
Answer:
(c) 5

6. If 8m – 8 = 56 then m is equal to :
(a) -4
(b) -2
(c) -14
(d) 8
Answer:
(d) 8

7. Which of the following number satisfies the equation – 6 + x = -18 ?
(a) 10
(b) – 13
(c) – 12
(d) – 16.
Answer:
(a) 10

8. If \(\frac{x}{2}\) = 14, then the value of 2x + 6 is equal to :
(a) 62
(b) -64
(c) 16
(d) -62.
Answer:
(a) 62

9. If 3 subtracted from twice a number is 5, then the number is :
(a) -4
(b) -2
(c) 2
(d) 4
Answer:
(d) 4

10. If 5 added to thrice an integer is – 7, then the integer is :
(a) – 6
(b) – 5
(c) -4
(d) 4
Answer:
(c) -4

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 4 Simple Equations Ex 4.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2

1. Write the first step that you will use to separate the variable and then solve the equation.

Question (i).
x + 1 = 0
Answer:
Given equation x + 1 = 0
Subtracting 1 from both sides, we get
x + 1 – 1 = -1
or x = – 1

Question (ii).
x – 1 = 5
Answer:
Given equation is x – 1 = 5
Adding 1 to both sides we get
x – 1 + 1 = 5 + 1
or x = 6
Thus x = 6 is the solution of the given equation

Question (iii).
x + 6 = 2
Answer:
Given equation is x + 6 = 2
Subtracting 6 from both sides, we get:
x + 6 – 6 = 2 – 6
or x = – 4
Thus, x = – 4 is the solution of the given equation.

Question (iv).
y + 4 = 4
Answer:
Given equation is y + 4 = 4
Subtracting 4 from both sides we get
y + 4 – 4 = 4 – 4
or y = 0
Thus, y = 0 is the solution of the given equation.

Question (v).
y – 3 = 3
Answer:
Given equation is y – 3 = 3
Adding 3 to both sides we get
y – 3 + 3 = 3 + 3
or y = 6
Thus, y = 6 is the solution of the given equation.

2. Write the first step that you will use to separate the variable and then sotye the equation :

Question (i).
3x = 15
Answer:
Given equation is 3x = 15
Dividing both sides by 3 we get
\(\frac{3 x}{3}=\frac{15}{3}\)
or x = 5

Question (ii).
\(\frac{P}{7}\) = 4
Answer:
Given equation is \(\frac{P}{7}\) = 4
Multiplying both sides by 7, we get
7 × \(\frac{P}{7}\) = 7 × 4
or p = 28
Thus, p = 28 is the solution of the given equation.

Question (iii).
8y = 36
Answer:
Given equation is 8y = 36
Dividing both sides by 8, we get
\(\frac{8 y}{8}=\frac{36}{8}\)
or y = \(\frac {9}{2}\)

Question (iv).
20x = – 10
Answer:
Given equation is
20x = – 10
Dividing both sides by 20
\(\frac{20 x}{20}=\frac{-10}{20}\)
or x = \(\frac {-1}{2}\)

3. Give the steps you will use to separate the variable and then solve the equation.

Question (i).
5x + 7 = 17
Answer:
Given equation is 5x + 7 = 17
Subtracting 7 from both sides, we get
5x + 7 – 7 = 17 – 7
or 5x = 10
Dividing both sides by 5, we get
\(\frac{5 x}{5}=\frac{10}{5}\)
or x = 2

Question (ii).
\(\frac{20 x}{3}\) = 40
Answer:
Given equation is \(\frac{20 x}{3}\) = 40
Multiplying both sides by 3, we get
3 × \(\frac{20 x}{3}\) = 3 × 40
or 20x = 3 × 40
Dividing both sides by 20, we get
\(\frac{20 x}{20}\) = \(\frac{3 \times 40}{20}\)
or x = 6

Question (iii).
3p – 2 = 46
Answer:
Given equation is 3p – 2 = 46
Adding 2 to both sides, we get
3p – 2 + 2 = 46 + 2
or 3 p = 48
Dividing both sides by 3, we get:
\(\frac{3 p}{3}=\frac{48}{3}\)
or p = 16

4. Solve the following equations :

Question (i).
10x + 10 = 100
Answer:
Given equation is 10x + 10 = 100
Subtracting 10 from both sides, we get
10x + 10 – 10 = 100 – 10
or 10x = 90
Dividing both sides by 10, we get
\(\frac{10 x}{10}=\frac{90}{10}\)
or x = 9
Thus x = 9 is the solution of the given equation.

Question (ii).
\(\frac{-p}{3}\) = 5
Answer:
Given equation is \(\frac{-p}{3}\) = 5
Multiplying both sides by – 3, we get
– 3 × \(\frac{-p}{3}\) = -3 × 5
or p = -15
Thus p = – 15 is the solution of the given equation.

Question (iii).
3x + 12 = 0
Answer:
Given equation is 3x + 12 = 0
Subtracting 12 from both sides, we get
3x + 12 – 12 = – 12
or 3x = – 12
Dividing both sides by 3, we get
\(\frac{3 x}{3}=\frac{-12}{3}\)
or x = -4
Thus x = – 4 is the solution of the given equation.

Question (iv).
2q – 6 = 0
Answer:
The given equation is 2q – 6 = 0
Adding 6 to both sides, we get
2q – 6 + 6 = 0 + 6
or 2q = 6
Dividing both sides by 2, we get
\(\frac{2 q}{2}=\frac{6}{2}\)
or q = 3
Thus, q = 3 is the solution of the given equation.

Question (v).
3p = 0
Answer:
The given equation is 3p = 0
Dividing both sides by 3, we get
\(\frac{3 p}{3}=\frac{0}{3}\)
or p = 0
Thus, p = 0 is the solution of the given equation.

Question (vi).
3s = -9
Answer:
The given equation is
3s = -9
Dividing both sides by 3, we get
\(\frac{3 s}{3}=-\frac{9}{3}\)
or s = – 3
Thus, s = – 3 is the solution of the given equation.