Category: Class 7

Students can Download Chapter 2 Fractions and Decimals Ex 2.7, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.7

Question 1.
Find:
i) 0.4 ÷ 2
Solution:

ii) 0.35 ÷ 5
Solution:

iii) 2.48 ÷ 4
Solution:

iv) 65.4 ÷ 6
Solution:

v) 651.2 ÷ 4
Solution:

vi) 14.49 ÷ 7
Solution:

vii) 3.96 ÷ 4
Solution:

viii) 0.80 ÷ 5
Solution:

Question 2.
i) 4.8 ÷ 10
Solution:

ii) 52.5 ÷ 10
Solution:

iii) 0.7 ÷ 10
Solution:

iv) 33.1 ÷ 10
Solution:

v) 272.23 ÷ 10
Solution:

vi) 0.56 ÷ 10
Solution:

vii) 03.97 ÷ 10
Solution:

Question 3.
i) 2.7 ÷ 100
Solution:

ii) 0.3 ÷ 100
Solution:

iii) 0.78 ÷ 100
Solution:

iv) 432.6 ÷ 100
Solution:

v) 23.6 ÷ 100
Solution:

vi) 98.53 ÷ 100
Solution:

Question 4.
Find :
i) 7.9 ÷ 1000
Solution:

ii) 26.3 ÷ 1000
Solution:

iii) 38.53 ÷ 1000
Solution:

iv) 128.9 ÷ 1000
Solution:

v) 0.5 ÷ 1000
Solution:

Question 5.
Find:
i) 7 ÷ 3.5
Solution:

ii) 36 ÷ 2
Solution:

iii) 3.25 ÷ 0.5
Solution:

iv) 30.94 ÷ 0.7
Solution:

v) 0.5 ÷ 0.25
Solution:

vi) 7.75 ÷ 0.25
Solution:

vii) 76.5 ÷ 0.15
Solution:

viii) 37.8 ÷ 1.4
Solution:

ix) 2.73 ÷ 1.3
Solution:

Question 6.
A vehicle covers a distance of 43.2 km in 2.4 litres of petrol. How much distance will it cover in one litre of petrol ?
Solution:
Distance covered in 2.4 liters petrol = 43.2km
Distance covered in 1 liters

(∴ vehicle covers 18 km in 1 litre of petrol)

Students can Download Chapter 1 Integers Ex 1.4, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 1 Integers Ex 1.4

Question 1.
Evaluate each of the following
a) (-30) ÷ 10
Solution:

b) 50 ÷ (-5)
Solution:

c) (-36) ÷ (-9)
Solution:

= 4

d) (-49) ÷ (49)
Solution:

= -1

e) 13 ÷ [(-2) + 1]
Solution:
= 13 ÷ (-2 + 10)
= 13 ÷ (-1)
\(=\frac{13}{-1}\)
= -13

f) 0 ÷ (-12)

= 0

g) (-31) ÷ [(-30) + (-1)]
= (-31) ÷ [-30 – 1]
= (-31) ÷ (-31)

= 1

h) [(-36) ÷ 12] ÷ 3
Solution:


= -1

i) [(-6) + 5] ÷ [(- 2) + 1]
=[-6 + 5] ÷ [-2 + 1]
= (-1) ÷ (-1)

= 1

Question 2.
Verify that a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c) for each of the following values of a, b and c.
a) a = 12, b = -4, c = 2
Solution:
a) a = 12, b = -4, c = 2
To verify a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c)

∴ LHS ≠ RHS
hence verified

b) a = (-10), b = 1, c = 1

∴ LHS ≠ RHS
hence verified

Question 3.
Fill in the blanks :

a) 369 ÷ ___ = 369
369 ÷ 1 = 369
Let number be x

x = 1

b) (-75) ÷ ___ = -1
(-75) ÷ 75 = -1
Let number be x

x = 75

c) (-206) ÷ ___ = 1
(-206) ÷ -206 = 1
Let number be x

x = -206

d) -87 ÷ ___ = 87
-87 ÷ -1 = 87
let the number be x

x = -1

e) ___ ÷ 1 = -87
87 ÷ 1 = -87
Let the number be x

1 × -87 = x
-87 = x
x = -87

f) ___ ÷ 48 = -1
48 ÷ 48 = -1
Let number be x

– 1 × 48 = x
-48 = x
∴ x = -48

g) 20 ÷ ___ = -2
20 ÷ -10 = (-2)

x = -10

h) __ ÷ (4) = -3
-12 ÷ (4) = -3
Let number be x

x = -3 × 4
x=-12

Question 4.
Write five pairs of integers (a, b) such that a – b = -3. One such pair is (6, -2) because 6 ÷ (-2) = (-3)
Solution:
a ÷ b = -3

∴ The other five pairs are (-6, 2), (-3, 1), (-9, 3), (-12, 4), (-15, 5)

Question 5.
The temperature at 12 noon was 10°C above zero. If it decreases at the rate of 2°C per hour until midnight, at what time would the temperature be 8°C below zero? What would be the temperature at mid-night ?
Solution:
Temperature at 12pm = 10°C
Temperature decreases at 2°C per hour
we need to find, time when temperature is 8°C below zero, ie., -8°C
∴ Total decrease = Final temperature – initial temperature = -8°C -10°C
= -8 – 10 = -18°C
Now,
Temperature decrease in 1 hour × Total number of hours = Total decrease

So, temperature becomes -8°C = 9 hours
∴ Temperature is -8°C at 9PM
Now,
We need to find temperature at mid – night temperature at 12AM = Temperature at 12PM
Temperature change in 12 hours
Temperature change in 1 hour = -2°C
Temperature change in 12 hour = -2 × 12°C = -24°C
Now,
Temperature at 12AM – Temperature at 12PM = Temperature change in 12 hours. Temperature at 12AM – 10°C = -24°C
Temperature at 12AM = – 24°C + 10°C
Temperature at 12 AM = -14°C
∴ Temperature at Mid – night = -14°C

Question 6.
In a class test (+3) marks are given for every correct answer and (-2) marks are given for every incorrect answer and no marks for not attempting any question.
Solution:
Marks given for correct answer = +3
Marks given for incorrect answer = -2
Marks given for not attempting = 0
i) Radhika scored 20 marks. If she has got 12 correct answers, how many questions has she attempted incorrectly ?
Solution:
Total marks scored = 20
Number of correct answer = 12
Marks of for correct answer = 12 × 3 = 36
Also
Marks for incorrect answer = (-2) × Number of incorrect answer
Marks for not attempt answer = 0 × Number of not attempt = 0
Now,
Total marks = marks for correct answer + marks for incorrect answer + marks for not attempt
20 = 36 + (-2) × No of incorrect answers + 0
20 – 36 = (-2) × No of incorrect answers
-16 = (-2) × No of incorrect answer
\(\frac{-16}{-2}\) = No of incorrect answers
8 = No of incorrect answers
So, Radhika attempted 8 questions incorrectly

ii) Mohini scores -5 marks in this test, thought she has got 7 correct answers. How many questions has she attempted incorrectly?
Solution:
Total marks = – 5
No of correct answers = 7
Marks for correct answers = 7 × 3 = 21
Also,
Marks for incorrect answer = (- 2) × No. of incorrect answers
Marks for not attempt answer = 0 × No of not attempt answers
Now,
Total marks = marks for correct answer + marks for incorrect answer + marks for not attempt
– 5 = 21 + (- 2) × No. of incorrect answers + 0
– 5 – 21 = (- 2) × No. of incorrect answers
– 26 = – 2 × No. of incorrect answers
\(\frac{-26}{-2}\) = No. of incorrect answers
13 = No, of incorrect answers.
∴ Mohini attempted 13 questions incorrectly

Question 7.
An elevator descends into a mine shaft at the rate of 6m/min. If the descent starts from 10m above the ground level, how long will it take to reach -350 m.
Solution:
Initial position = 10m final position = -350 m
Total distance elevator has to travel = initial position – final position
= 10 – (- 350)
= 10 + 350 = 360 m
Now,
Elevator descends at rate 6m/min
∴ Elevator travels 6m in = 1 minute
Elevator travels 1 m in: = \(\frac{1}{6}\)
Elevator travels 360m in = \(\frac{1}{6}\) × 360 minutes
= 60 minutes
= 1 hour
∴ It will take the elevator 1 hour.

Students can Download Chapter 4 Simple Equations Ex 4.3, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 4 Simple Equations Ex 4.3

Question 1.
Solve the following equations :

Solution:

Substitute the value of y = 8 in the given equation

∴ LHS = RHS (verified)

b) 5t + 28 = 10
Solution:
The given equation is 5t + 28 = 10
Transposing 28 from LHS to RHS
5t = 10 – 28 [on transposing + 28 becomes -28]
5t = -18
Dividing by 5 both the sides

Substitute the value of t = \(\frac{-18}{5}\) in the given equation
5t + 28 = 10

-18 + 28 = 10
10 = 10
∴ LHS = RHS (verified)

c)

Solution:

Transposing + 3 from LHS to RHS

Multiply by 5 both the sides

Substitute the value of a = -5 in the given equation.

-1 + 3 = 2
2 = 2
∴ LHS = RHS (verified)

d)

Solution:

Transposing +7 from LHS to RHS


Substitute the value of q = -8 in the given equation.

-2 + 7 = 5
5 = 5
∴ LHS = RHS (verified)

e)

Solution:

Multiply by 2 both the sides

Substitute the value of x = -4 in the given equation.

∴ LHS = RHS (verified)

f)

Solution:

Multiply by 2 both the sides


Divide by 5 both the sides

∴ LHS = RHS (verified)

g)
Solution:

Divide by 7 both the sides

Substitute the value of m = \(\frac{1}{2}\) in the given equation


13 = 13
∴ LHS = RHS (verified)

h) 6z + 10 = -2
Solution:
The given equation is 6z + 10 = -2
Transposing + 10 from LHS to RHS
6z = -2 – 10(on Transposing + 10 becomes – 10)
6z = -12
Divide by 6 both the sides

Substitute the value of z = -2 in the given equation
6z + 10 = -2
6(-2) + 10 = -2
-12 + 10 = -2
-2 = -2
∴ LHS = RHS (verified)

i)

Solution:

multiply by 2 both the sides

Divide by 3 both the sides

Substitute the value of l = \(4 / 9\) in the given equation.


∴ LHS = RHS (verified)

j)

Solution:

Transpose -5 from LHS to RHS

Multiply by 3 both the sides

Divide by 2 both the sides

Substitute the value of b = 12 in the given equation

3 = 3
∴ LHS = RHS (verified)

Question 2.
Solve the Mowing equations :
a) 2(x+4) = 12
Solution:
The given equation is 2(x + 4) = 12
Divide by 2 both the sides

x + 4 = 6
Transpose +4 from LHS to RHS x = 6 – 4 (on transposing 4 becomes -4)
∴ x = 2
Substitute the value of x = 2 in the given equation
2(x + 4) = 12
2(2 + 4) = 12
2(6) = 12
12 = 12
∴ LHS = RHS (verified)

b) 3(n – 5) = 21
The given equation is 3 (n – 5) = 21
Divide by 3 both the sides

∴ n – 5 = 7
Transpose -5 from LHS to RHS
n = 7 + 5 [on transposing -5 becomes +5]
∴ n = 12
Substitute the value of n = 12 in the given equation
3(n – 5) = 21
3(12 – 5) = 21
3(7) = 21
21 = 21
LHS = RHS (verified)

c) 3 (n – 5) = -21
The given equation is 3 (n – 5) = -21
Divide by 3 both the sides 3

Transpose -5 from LHS to RHS n = -7 + 5 (on transposing -5 becomes +5)
n = -2
Substitute the value n = -2 in the given equation
3 (n – 5) = -21 .
3(-2 – 5) = -21
3(-7) = -21
-21 = -21
∴ LHS = RHS (verified)

d) -4(2 + x) = 8
Solution:
The given equation is -4 (2 + x) = 8
Divide by -4 both the sides

2 + x = -2
Transpose 2 from LHS to RHS x = -2 -2 [on transposing 2 becomes -2]
∴ x = -4
Substitute the value of x = -4 in the given equation
-4(2 + x) = 8
-4(2 – 4) = 8
-4 (-2) = 8
8 = 8
∴ LHS = RHS (verified)

e) 4 (2 – x) = 8
The given equation is 4 (2 – x) = 8
Divide by 4 both the sides

2 – x = 2
Transpose +2 from LHS to RHS
-x = 2 – 2 (on transposing 2 becomes -2)
-x = 0
multiply by -1 both the sides
-x × -1 = 0 × -1 x = 0
Substituting the value of x = 0 in the given equation
4(2 – x) = 8
4(2 – 0) = 8
8 – 0 = 8
8 = 8
∴ LHS = RHS (verified)

Question 3.
Solve the following equations :
a) 4 = 5 (p – 2)
Solution:
The given equation is 4 = 5 (p – 2)
5(p – 2) = 4 [LHS and RHS are inter-changed the equation remains the same]
Divide by 5 both sides


Transpose -2 from LHS to RHS

Substituting the value of p = \(\frac{14}{5}\) in the given equation

∴ LHS = RHS (verified)

b) -4 = 5 (p – 2)
Solution:
The given equation is – 4 = 5 (p – 2)
[LHS & RHS are interchanged the equation remains same]
5 (p – 2) = -4
Divide by 5 both the sides

Transpose -2 from LHS to RHS

Substitute the value of P = 6/5 in the given equation
5(p – 2) = -4

∴ LHS = RHS (verified)

c) 16 = 4 + 3 (t + 2)
Solution:
The given equation is 16 = 4 + 3 (t + 2)
4 + 3 (t + 2) = 16 [LHS and RHS are interchanged the equation remains the same.] 4 + 3t + 6 = 16
3t + 10 = 16
3t = 16 – 10 [ transposing + 10 from LHS to RHS, it becomes -10]
3t = 6
t = 6/3 = 2 ∴ t = 2
Substitute the value of t = 2 in the given equation
16 = 4 + 3 (t +2)
16 = 4 + 3t + 6
16 = 10 + 3(2)
16 = 10 + 6
16 = 16
∴ LHS = RHS (verified)

d) 4 + 5 (p – 1) = 34
Solution:
The given equation is 4 + 5 (p – 1) = 34
Transpose 4 from LHS to RHS
5(p – 1) = 34 – 4 (on transposing 4 becomes -4)
5(p – 1) = 30
Divide by 5 both sides

p – 1 = 6
Transposing -1 from LHS to RHS p = 6 + 1 [on transposing -1 becomes 1]
p = 7
Substitute the value of p = 7 in the given equation
4 + 5(p – 1) = 34
4 + 5p – 5 = 34
4 + 5 (7) – 5 = 34
4 + 35 – 5 = 34
34 = 34
∴ LHS = RHS (verified)

e) 0 = 16 + 4(m – 6)
Solution:
The given equation is 0 = 16 + 4 (m – 6)
16 + 4 (m – 6) = 0 [ LHS ami RHS are interchanged the equation remains the same] Transpose 16 from LHS to RHS .
4(m – 6) = 0 – 16 [on transposing 16 becomes -16]
4(m – 6) = -16
Divide by 4 both sides

m – 6 = – 4
Transpose – 6 from LHS to RHS
m = – 4 + 6 [on transposing – 6 becomes + 6]
m = 2
Substitute the value m = 2 in the given equation
16 + 4(m – 6) = 0
16 + 4(2 – 6) = 0
16 + 4(-4) = 0
16 – 16 = 0
0 = 0
∴ LHS = RHS (verified)

Question 4.
a) Construct 3 equations starting with x = 2
Solution:
x = 2
multiply by 3 both sides
3(x) = 2(3)
3x = 6
Add 6 to both sides
3x + 6 = 6 + 6
equation 1 3x + 6 = 12
x = 2
multiply by 12 both sides
12(x) = 2(12)
12x = 24
Subtract 10 from both sides
12x – 10 = 24 – 10
equation 2 12x – 10 = 14
multiply by 15 both the sides
x (15) = 2(15)
15x = 30
Divide by 2 both the sides

b) Construct 3 equations starting with x = -2
Solution:
x = -2
multiply by 5 both sides
5(x) = -2 (5)
5x = -10

x = -2
multiply by 8 both the sides
8(x) = -2 × 8
8x = -16
Divide by 3 both the sides

x = -2
multiply by 9 both the sides
9(x) = -2 × 9
9x = -18
Subtract 7 from both the sides 9x-l – -18 – 7
3. 9x – 7 = -25 —– 3. Equation

Students can Download Chapter 2 Fractions and Decimals Ex 2.6, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.6

Question 1.
Find :
i) 0.2 × 6
Solution:
0.2 × 6 = 1.2

ii) 8 × 4.6
Solution:
8 × 4.6 = 36.8

iii) 2.71 × 5
Solution:
2.71 × 5 = 13.55

iv) 20.1 × 4
Solution:
20.1 × 4 = 80.4

v) 0.05 × 7
Solution:
0.05 × 7 = 0.35

vi) 211.02 × 4
Solution:
211.02 × 4 = 844.08

vii) 2 × 0.86
Solution:
2 × 0.86 = 1.72

Question 2.
Find the area of a rectangle whose length is 5.7 cm and breadth is 3 cm.
Solution:
Length = 5.7 cm
Breadth = 3 cm
∴ Area of the rectangle = l × b
= 5.7 × 3
= 17.1 sq. cms.

Question 3.
Find :
i) 1.3 × 10
Solution:
1.3 × 10 = 13.0

ii) 36.8 × 10
Solution:
36.8 × 10 = 368.0

iii) 153.7 × 10
Solution:
153.7 × 10 = 1537

iv) 168.07 × 10
Solution:
168.07 × 10 = 1680.7

v) 31.1 × 100
Solution:
31.1 × 100 = 311.00

vi) 156.1 × 100
156.1 × 100 = 15610.0

vii) 3.62 × 100
3.62 × 100 = 362.00

viii) 43.07 × 100
43.07 × 100 = 4307.00

ix) 0.5 × 10
0.5 × 10 = 5.0

x) 0.08 × 10
0.08 × 10 = 0.80

xi) 0.9 × 100
0.9 × 100 = 90.0

xii) 0.03 × 1000
0.03 × 1000 = 30.00

Question 4.
A wheeler covers a distance of 55.3 km in one litre of petrol. How much distance will it cover in 10 litres of petrol?
Solution:
Distance covered by a two-wheeler in one litre petrol = 55.3 kms
∴ Distance covered by a two-wheeler in 10 1 petrol = 55.3 × 10 = 553.0 kms.

Question 5.
Find :
i) 2.5 × 0.3
Solution:
2.5 × 0.3 = 0.75

ii) 0.1 × 51.7
Solution:
0.1 × 51.7 = 5.17

iii) 0.2 × 316.8
Solution:
0.2 × 316.8 = 63.36

iv) 1.3 × 3.1
Solution:
1.3 × 3.1 = 14.031

v) 0.5 × 0.05
0.5 × 0.05 = 0.025

vi) 11.2 × 0.15
Solution:
11.2 × 0.15 = 1 .680

vii) 1.07 × 0.02
Solution:
1.07 × 0.02 = 0.0214

viii) 10.05 × 1.05
Solution:
10.05 × 1.05 = 10.5525

ix) 101.01 × 0.01
Solution:
101.01 × 0.01 = 1.01011

x) 100.01 × 1.1
Solution:
100.01 × 1.1 = 110.011

Students can Download Chapter 6 The Triangles and Its Properties Ex 6.5, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.5

Question 1.
PQR is a triangle, right-angled at P. If PQ = 10cm and PR = 24 cm, find QR.
Solution:

PQR is a right angled triangle. ∠RPQ = 90°
∴ QR² = PQ² + PR² (By pythagoras property)
= 10² + 24²
= 100 + 576
∴ QR =  = 26 cms

Question 2.
ABC is a triangle, right-angled at C. If AB = 25 cm and AC = 7 cm, find BC.
Solution:

ABC is a right-angled triangle.
∠ACB = 90°
∴ AB² = AC² + BC² (By pythagoras property)
25² = 7² + BC²
BC² = 25² – 7² = 625 – 49
∴ BC = \(\sqrt{576}\) = 24
∴ BC = 24 cms

Question 3.
A 15m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.
Solution:

The distance of the foot of the ladder from the wall is ‘a’ m.
Then
a² + 12² = 12² (pythagoras property)
a² = 15² – 12²
a² = 225 – 144
a² = 81
∴ a = 
∴ The distance of the foot of the ladder from the wall be 9m.

Question 4.
Which of the following can be the sides of a right triangle?
Solution:
In the case of right-angled triangles, identify the right angles.
i) 2.5 cm, 6.5 cm, 6 cm
Solution:
The greatest side is hypotenuse = 6.5 cm
(2.5)² + (6)² = (6.5)² (According to Pythagoras property)
6.25 + 36 = 42.25
42.25 = 42.25
∴ This is a right-angled triangle. The given length can be the sides of a right-angled triangle. The right angle between the lengths of 2.5 cm and 6 cm.

ii) 2 cm, 2 cm, 5 cm
The greatest side is 5 cm
(2)² + (2)² = (5)² (According to pythagoras property)
4 + 4 ≠ 25
∴ It is not the lengths of right angled triangle.

iii) 1.5 cm, 2 cm, 2.5 cms
The greatest side is 2.5 cms.
According to pythagoras property
(2.5)² = (1.5)² + (2)²
6.25 = 2.25 + 4
6.25 = 6.25
It is a right angled trianges, the given lengths are sides of the right angled triangles.
∴ The right angle between the lengths of 1.5 cms and 2 cms.

Question 5.
A Tree is broken at a height of 5m from the ground and its top touches the ground at a distance of 12 m from the base tree. Find the original height of the tree.
Solution:

According to the given statement the sketch we get is right angled triangle
RQ² = PR² + PQ²
= 5² + 12²
= 25 + 144 = 169
∴ RQ =  = 13m
∴ The original height of the tree is PR + RS = 5 + 13 = 18 m

Question 6.
Angles Q and R of a ∆ PQR are 25° and 65°.
Write which of the following is true :
Solution:

∠p = 90°
∴ According to Pythagoras property

ii) PQ² + RP² = QR²
PQ² + RP² = QR² is true

Question 7.
Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.
Solution:

PQRS is a rectangle.
According to the figure, PQS is a right-angled triangle.
∴ QS² = PQ² + PS²
41² = 40² + PS²
∴ PS² = 42² + 40²
= 1681 – 1600
PS =  = 9 cms
The perimeter of the rectangle = 2 (PQ + PS)
= 2 (40 + 9) = 2 (49)
= 98 cms

Question 8.
The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.
Solution:

PQRS is a rhombus. The diagonals are PQ & RS.
PQ = 30 cms RS = 16 cms
According to fig In Rhombus the diagonals bisect
Each other. ∴ PQ & RS bisect at O.
PO = OQ SO = OR

Considering the ∆ POR
PR² = PO² + OR²
= 15² + 8²
= 225 + 64 = 289
∴ PR = \(\sqrt{289}\) = 17cms
∴ The perimeter of the rhombus = 4 × side PR
= 4 × 17 = 68 cms

Students can Download Chapter 2 Fractions and Decimals Ex 2.5, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.5

Question 1.
Which is greater ?
i) 0.5 or 0.05
Solution:
0.5 > 0.05 (∵ 0.5 = 0.50)

ii) 0.7 or 0.5
Solution:
0.7 > 0.5

iii) 7 or 0.7
Solution:
7 > 0.7 (∵ 7 = 7.0)

iv) 1.37 or 1.49
Solution:
1.49 > 1.37

v) 2.03 or 2.30
Solution:
2.30 > 2.03

vi) 0.8 or 0.88
Solution:
0.88 > 0.8 (∵ 0.8 = 80)

Question 2.
Express as rupees using decimals :
i) 7 paise
Solution:
7 paise = Rs. 0.07

ii) 7 rupees 7 paise
Solution:
7 rupees 7 paise Rs. 7.07

iii) 77 rupees 77paise
Solution:
77 rupees 77 paise = ₹ 77.77

iv) 50 paise
Solution:
50 paise = 0.50

v) 235 paise
Solution:
235 paise = 2.35

Question 3.
i) Express 5cm in metre and kilometre
Solution:
5 cm = 0.05m = 0.00005km.

ii) Express 35mm in cm, m and km
Solution:
35mm = 3.5cm = 0.000035km

Question 4.
Express in kg :
i) 200g
Solution:
200g = 0.200kg = 0.2 kg

ii) 3470g
Solution:
3470g = 3.470kg

iii) 4 kg 8 g
Solution:
4kg 8g = 4.008 kg

Question 5.
Write the following decimal numbers in the expanded form :
i) 20.03
Solution:

ii) 2.03
Solution:

iii) 200.03
Solution:

iv) 2.034
Solution:

Question 6.
Write the place value of 2 in the following decimal numbers :

i) 2.56
Solution:
2.56 place value of 2 in the decimal number 2.56 = 2 ones place)

ii) 21.37
Solution:
21.37 place value of 2 in the decimal number 21.37 = 2 × 10 = 20. (2 is in tens place)

iii) 10.25
Solution:
10.25 = palce value of 2 in the decimal

iv) 9.42
Solution:
9.42 = place value of 2 in the decimal number

(2 is in hundredths place)

v) 63.352
Solution:
63.352 = place value of 2 in the decimal number

Question 7.
Dinesh went from place A to place B and from 12.7 km from C. Ayub went from place A to place D and from there to place C. D is 9.3 km from A and C is 11.8 km from D. Who travelles more and by how much ?
Solution:
Distance travelled from A to B = 7.5kms
Distance travelled from B to C = 12.7kms
Total distance travelled from A to B & B to C = 20.2 kms.
Ayub
Distance travelled from A to D = 9.3 kms
Distance travelled from D to C = 11.8 kms
Total distance travelled from A to D & D to C = 21.1 kms
∴ Ayub travelled more distance than Dinesh by = 21.1 – 20.2 = 0.9 kms.

Question 8.
Shyama bought 5 kg 300g apples and 3 kg 250 g mangoes. Sarala bought 4 kg of 800 g oranges and 4 kg 150g bananas. Who bought more fruits?
Solution:
Shyama:
Weight of Apples bought = 5.300kgs
Weight of Mangoes bought = 3.250kgs
Total weight of fruits = 8.550kgs

Sarala:
Weight of oranges bought = 4.800kgs
Weight of Bananas bought = 4.150kgs
Total weight of fruits = 8.950kgs
8.950 >8.550
∴ Sarala bought more fruits than Shyama.

Question 9.
How much less is 28 km than 42.6 km?
Solution:

∴ 28 km is less than 42.6 km by 14.6 kms.

Students can Download Chapter 2 Fractions and Decimals Ex 2.4, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.4

Question 1.
Find:
i)

Solution:

ii)

Solution:

iii)

Solution:

iv)

Solution:

v)

Solution:

vi)

Solution:

Question 2.
Find the reciprocal of each of the following fractions. Classify the reciprocals as proper fractions, improper fractions and whole numbers.
Solution:
Whole number ➝ Number starting from 0
⇒ 0, 1, 2, 3, 4, 5, 6, ..
Proper fraction ➝ where Numerator < Denominator

Improper fraction ➝ where Numerator < Denominator

here Numerator > Denominator
∴ It is improper fraction

ii) \(\frac{5}{8}\)
Solution:

here, Numerator > Denominator
∴ It is improper fraction

iii) \(\frac{9}{7}\)
Solution:

here Numerator > Denominator
∴ It is improper fraction

iv)

Solution:

here, Numerator > Denominator
∴ It is improper fraction

v) \(\frac{12}{7}\)
Solution:

here, Numerator > Denominator
∴ It is improper fraction

vi) \(\frac{1}{8}\)
Solution:

∴ It is a whole number

Question 3.
Find:
i)

Solution:

ii)

Solution:

iii)

Solution:

iv)

Solution:

v)

Solution:

vi)

Solution:

Question 4.
Find
i)

Solution:

ii)

Solution:

iii)

Solution:

iv)

Solution:

v)

Solution:

vi)

Solution:

vii)

Solution:

viii)

Solution:

Students can Download Chapter 4 Simple Equations Ex 4.2, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 4 Simple Equations Ex 4.2

Question 1.
Given first the step you will use to separate the variable and then solve the equation:
a) x – 1 = 0
Solution:
The given equation = x – 1 = 0
Add 1 to both the sides

Checking
x – 1 = 0
put the value of x = 1
1 – 1 = 0
∴ LHS = RHS (checked)

b) x + 1 = 0
The given equation = x + 1 = 0
Subtract 1 from both the sides

Verification
x+ 1 = 0
substitute the value of x = -1 then x + 1 = 0
(-1) + 1 = 0
0 = 0
LHS = RHS (verified)

c) x – 1 = 5
The given equation is x – 1 = 5
Add 1 to both sides

Substitute the value in the equation x = 6.
x – 1 = 5
6 – 1 = 5
5 = 5
∴ LHS = RHS (verified)

d) x + 6 = 2
The given equation is x + 6 = 2
Subtract 6 from both the sides

Substitute the value of x = -4 in the given equation,
x + 6 = 2
-4 + 6 = 2
2 = 2
∴ LHS = RHS (verified)

e) y – 4 = -7
The given equation is y – 4 = -7
Add 4 to both the sides

substitute the value of y = -3 in the given equation
y – 4 = -7
-3 – 4 = -7
-7 = -7
∴ LHS = RHS (verified)

f) y – 4 = 4
Solution:
The given equation is y – 4 = 4
Add 4 to both the sides

substitute the value of y = 8 in the given equation
y – 4 = 4
8 – 4 = 4
4 = 4
∴ LHS = RHS (verified)

g) y + 4 = 4
Solution:
The given equation is y + 4 = 4
subtract 4 from both sides

substitute the value of y = 0 in the given equation.
y + 4 = 4
0 + 4 = 4
4 = 4
∴ LHS = RHS (verified)

h) y + 4 = -4
Solution:
The given equation is y + 4 = -4
subtract 4 from both the sides

y = -8 (verified)
substitute the value of y = -8 in the given equation
y + 4 = -4
-8 + 4 = -4
-4 = -4
∴ LHS = RHS (verified)

Question 2.
Give first the step you will use to separate the variable and then solve the equation:

a) 3l = 42
Solution:
The given equation is 3l = 42
Divide by 3 both the sides

Substitute the value of l = 14 in the given equation
3l = 42
3(14) = 42
42 = 42
∴ LHS = RHS (verified)

b) \(\frac{\mathbf{b}}{2}\) = 6
Solution:
The given equation is \(\frac{\mathbf{b}}{2}\) = 6
Multiplied by 2 both the sides b

substitute the value of b = 12 in the given equation

∴ LHS = RHS (verified)

c)
\(\frac{\mathbf{p}}{7}\) = 4
Solution:

Multiplied by 7 to both the sides

Substitute the value of p = 28 in the given equation

∴ LHS = RHS (verified)

d) 4x = 25
Solution:
The given equation is 4 x = 25
Divide both sides by 4

substitue the value of x = \(\frac{25}{4}\) in the given equation
4x = 25

∴ LHS = RHS (verified)

e) 8y = 36
Solution:
The given equation is 8y = 36
Divide both sides by 8

Substitute the equation by y = \(\frac{9}{2}\) in the given equation
8y = 36

∴ LHS = RHS

f)

Solution:

Substitute the equation by z = \(\frac{15}{4}\) in the given equation

∴ LHS = RHS (verified)

g)

Solution:

Substitute the value of a = \(\frac{7}{3}\) in the given equation

∴ LHS = RHS (verified)

h) 20t = -10
Solution:
The given equation is 20 t = -10
Divide by 20 to both sides


LHS = RHS (verified)

Question 3.
Given the steps you will use to separate the variable and then solve the equation:
a) 3n – 2 = 46
Solution:
The given equation is 3n – 2 = 46
Add 2 to both sides 3n – 2 + 2 = 46+ 2
3n = 48
Divide the equation by 3

∴ n = 16 This is required solution.
substitute the value of n = 16 in the given equation
3n – 2 = 46
3(16) – 2 = 46
48 – 2 = 46
46 = 46
∴ LHS = RHS (verified)

b) 5m + 7 = 17
The given equation is 5m + 7 = 17
Subtract 7 from both sides

5m = 10
Divide the equation by 5 both the sides

Substitute the value of m = 2 in the given equation
5m + 7 = 17
5(2) + 7 =17
10 + 7 = 17
∴ LHS = RHS (verified)

c) \(\frac{20 p}{3}\) = 40
Solution:

Substitute the value of p = 6 in the given equation. .

40 = 40
∴ LHS = RHS (verified)

d)

Solution:

Multiply by 10 to both the sides

Divide by 3 to both sides

Substitute the value of p = 20 in the given equation

Question 4.
Solve the following equations :
a) 10p = 100
Solution:
The given equation is 10p = 100
Divide by 10 both the sides

Substitute the value of p in the given equation
10p = 100
10 x 10 = 100
100 = 100
∴ LHS = RHS (verified)

b) 10p + 10 = 100
Solution:
The given equation is 10p + 10 = 100
subtract 10 from both the sides

Divide by 10 both the sides

Substitute the value of p = 9 in the given equation
10p + 10 = 100
10 × 9 + 10 = 100
90 + 10 = 100
100 = 100
LHS = RHS (verified)

c) \(\frac{\mathbf{p}}{4}\) = 5
Solution:

5 = 5
∴ LHS = RHS (verified)

d) \(\frac{-p}{3}\) = 5
The given equation is \(\frac{-p}{3}\) = 5
The equation is multiplied by -3 both the sides

Substitute the value of p = -15 in the given equation

∴ LHS = RHS (verified)

e)

Solution:

Substitute the value p = 8 in the given equation

f) 3s = -9
The given equation is 3s = -9
Divide the equation by 3 both the sides

Substitute the value of s = -3 in the given equation
3s = -9
3(-3)=-9
-9 = -9
∴ LHS = RHS (verified)

g) 3s + 12 = 0
Solution:
The given equation is 3s + 12 = 0
subtract 12 from both the sides

Divide by 3 both the sides

Substitute the value of s = -4 in the given equation
3s = 12 = 0
3(-4) = 12 = 0
-12 + 12 = 0
0 = 0
∴ LHS = RHS

h) 3s = 0
Solution:
The given equation is 3s = 0
Divide by 3 both the sides

Substitute the value of s = 0 in the given equation
3s = 0
3(0) = 0
0 = 0
∴ LHS = RHS (verified)

i) 2q = 6
Solution:
The given equation is 2q = 6
Divide the equation by 2 both the sides

substitute the value of q = 3 in the given equation
2q = 6
2(3) = 6 6 = 6
∴ LHS = RHS (verified)

j) 2q – 6 = 0
Solution:
The given equation is 2q – 6 = 0
Add 6 to both the sides

Substitute the value of q = 3 in the given equation
2q – 6 = 0
2(3) – 6 = 0
6 – 6 = 0
0 = 0
LHS = RHS (verified)

k) 2q + 6 = 0
Solution:
The given equation is 2q + 6 = 0
Subtract 6 from both the sides

Divide by 2 both the sides by 2

Substitute the value of q = -3 in the given equation
2q + 6 = 0
2(-3) + 6 = 0
-6 + 6 = 0
0 = 0
∴ LHS = RHS (verified)

l) 2q + 6 = 12
The given equation is 2q + 6 = 12
Subtract 6 from both the sides

Divide by 2 both the sides

Students can Download Chapter 2 Fractions and Decimals Ex 2.3, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.3

Question 1.
Find :

Solution:

ii)

Question 2.
Multiply and reduce to lowest form (if possible)
i)

Solution:

ii)

Solution:

iii)

Solution:

iv)

Solution:

v)

Solution:

vi)

Solution:

vii)

Solution:

Question 3.
Multiply the following fractions:

i)

Solution:

ii)

Solution:

iii)

Solution:

iv)

Solution:

v)

Solution:

vi)

Solution:

vii)

Solution:

Question 4.
Which is greater :
i)

Solution:

Now,
we have to find greater of

To compare fractions, we make denominator equal.

ii)

Solution:

Now,
we have to find greater of \(\frac{3}{7}\) & \(\frac{2}{7}\)
Since denominator is same, number with greater numerator is greater.i.e., \(\frac{3}{7}>\frac{2}{7}\)
∴ Thus \(\frac{1}{2}\) of \(\frac{6}{7}\) is greater

Question 5.
Saili plants 4 saplings in a row, in her garden. The distance between two adjacent saplings is \(\frac{3}{4}\)m. Find the distance between the first and the last sapling.
Solution:
Saili plants 4 saplings, in a row.
The distance between each sapling = \(\frac{3}{4}\) m

Question 6.
Lipika reads a book for \(1 \frac{3}{4}\) everyday. She reads the entire book in 6 days. How many hours in all were required by her to read the book ?
Solution:
Number of hours read everyday = \(1 \frac{3}{4}\) hours
Number of days to read whole book = 6 days
Therefore,

Question 7.
A car runs 16 km using 1 litre of petrol. How much distance will it cover using \(2 \frac{3}{4}\) litres of petrol.
Solution:
Distance of car runs in 1 litre = 16km

∴ Car will run 44 km in \(2 \frac{3}{4}\) litre of petrol.

Question 8.
a) i) Provide the number in the box □, such that \(\frac{2}{3} \times \square=\frac{10}{30}\)
Solution:

ii) The simplest form of the number obtained in □ is ___
Solution:

b) i) Provide the number in the box □, such that \(\frac{3}{5} \times \square=\frac{24}{75}\)
Solution:

b) ii) The simplest form of the number obtained in □ is ___.
Solution:

it cannot be simplified further

Students can Download Chapter 1 Integers Ex 1.3, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 1 Integers Ex 1.3

Question 1.
Find each of the following products
a) 3 × (-1)
3 × (-1) = -3

b) (-1) × 225
(-1) × 225 = – 225

c) (-21) × (-30)
(-21) × (-30) = 630

d) (-316) × (-1)
(-316) × (-1) = 316

e) (-15) × 0 × (-18)
(-15) × 0 × (-18)

f) (-12) × (-11) × (10)
(-12) × (-11) × (10) = 1320

g) 9 × (-3) × (-6)
9 × (-3) × (-6) = 162

h) (-18) × (-5) × (-4)
(-18) × (-5) × (-4) = – 360

i) (-1) × (2) × (-3) × 4
(-1) × (-2) × (-3) × 4 = -24

j) (-3) × (-6) × (-2) × (-1)
(-3) × (-6) × (-2) × (-1) = 36

Question 2.
Verify the following :
a) 18 × [7 + (-3)] = [18 × 7] + [18 × (-3)]
Solution:
18 × [7 – 3] = [126] + [18 × -3]
18 × [4] = 126 + [-54]
72 = 72 (verified)

b) (-21) × [(-4) + (-6)] = [(-21) × (-4)] + [(-21) × (-6)]
Solution:
-21 × [-4 + -6] = [-21 × -4] + [-21 × -6]
-21 × -10 = [84]+[126]
+ 210 = 210
LHS = RHS (verified)

Question 3.
i) For any integer a, what is (-1) × a equal to?
-1 × a = -a

ii) Determine the integer whose product with (-1) is
a) -22
-1 × -22 = +22
b) 37
-1 × 37 = -37
c) 0
-1 × 0 = 0

Question 4.
Starting from (-1) × 5, write various products showing some pattern to show (-1) × (-1) = 1.
Solution:
(-1) × 5 = -5
(-1) × 4 = -4 [= (-5) + 1]
(-1) × 3 = -3 [= (-4) + 1]
(-1) × 2 = -2 [= (-3) + 1]
(-1) × 1 = -1 [= (-2) + 1]
(-1) × 0 = 0 [= (-1) + 1]
(-1) × (-1) = 1 [= 0 + 1]

Question 5.
End the product, using suitable properties:
a) 26 × (-48) + (-48) × (-36)
= 26 × (-48)+ (-36) × (-48)
= [26 + (-36)] × (-48) ∵ Distributive Property = (-10) × (-48)
= + 480

b) 8 × 53 × (-125)
= 8 × (-125) × 53
= [8 × (-125)] × 53
= [-(8 × 125)] × 53 ∵ a × (-b) = -(a × b)
= (-1000) × 53 = -(1000 x 53) ∵ (-a) × (b) = -(a × b)
= -53000

c) 15 × (-25) × (-4) × (-10)
= 15 × (-25) × (-10) × (-4)
∵ commutativity of multiplication
= 15 × (-10) × (-25) × (-4)
= [15 × (-10)] × [(-25) × (-4)]
= [-(15 × 10)] × [25 × 4] (∵ a × (-b) = -(a × b)
= -150 × 100 (-a) × (-b) = a × b
= -15000

d) (-41) × 102
= -(41 × 102) ∵ (-a) × b = -(a × b)
=-[41 × (100 + 2)]
= -[41 × 100 + 41 × 2]
Distributivity of multiplication over addition.
= -[4100 + 82]
= -4182

e) 625 × (-35) + (-625) × 65
= 625 × (-35) + 625 × (-65) ∵ (-a) × b = a × (-b)
= 625 × [(-35) + (-65)] ∵ Distributivity of multiplication over addition
= 625 × (-100)
= -(625 × 100)

f) 7 × (50 – 2)
= 7 × 50 – 7 × 2 ∵ Distributivity of multiplication over subtraction
= 350 – 14 = 336

g) (-17) × (-29)
= 17 × 29 ( ∵ (-a) × (-b) = a × b)
= 17 × (30 – 1)
= 17 × 30 – 17 × 1
∴ Distributivity of multiplication over subtraction
= 510 – 17
= 493

h) (-57) × (-19) + 57
= 57 × 19 + 57 ∵ (-a) × (-b) = a × b
= 57 × 19 + 57 × 1 ∵ (a × 1 = a)
= 57 × (19 + 1)
∴ Distributivity of multiplication over subtraction.
= 57 × 20 = 1140

Question 6.
A certain freezing process requires that room temperature be lowered from 40°C at the rate of 5°C every hour. What will be the room temperature 10 hours after the process begins?
Solution:
Initial temperature = 40°C
Room temperature decreases at 5°C per hour
Temperature change in 1 hour = – 5°C
Temperature change in 10 hour = – 5 × 10°C = – 50°C
Room temperature after 10 hours = Initial temperature + Temperature change in 10 hour
= 40 + (- 50)
= 40 – 50
∴ Room temperature after 10 hours = – 10°C

Question 7.
In a class test containing 10 questions, 5 marks are awarded for every correct answer, and (-2) marks are awarded for every incorrect answer and 0 for question not attempted.
i) Mohan gets four correct and six incorrect answers. What is his score?
Solution:
Total questions = 10
Marks for correct answer = + 5
Marks for incorrect answer = – 2
Marks for no attempt = 0
Mohan marks for 4 correct answers = 4 × 5 = 20
Mohan marks for 6 incorrect answers= 6 × (- 2) = 12
∴ Total marks = 20 + (- 12) = 20 – 12 = 8
Total marks scored by Mohan is = 8

ii) Reshma gets five correct answers and five incorrect answers, what is her score?
Solution:
Reshma marks for 5 correct answers = 5 × 5
= 25
Reshma marks for 5 incorrect answers
= 5 × (-2) = -10
∴ Total marks = 25 + (-10)
= 25 – 10
= 15
∴ Total marks scored by Reshma = 15

iii) Heena gets two correct and five incorrect answers out of seven questions she attempts. What is her score ?
Solution:
Heena marks for 2 correct answers = 2 × 5 = 10
Heena marks for 5 incorrect answers = 5 × (- 2) = -10
∴ Heena total marks = 10 + (- 10) = 10 – 10 = 0
∴ Total marks scored by Heena = 0

Question 8.
A cement company earns a profit of 8 per bag of white cement sold and a loss of 5 per bag of grey cement sold.
a) The company sells 3,000 bags of white cement and 5,000 bags of grey cement in a month. What is its profit or loss?
Solution:
Profit on white cement = Rs + 8
Loss on grey cement = Rs – 5
∴ Profit on 1 bag of white cement = Rs 8
Profit on 3,000 bags of white cement
= Rs 8 × 3000 = Rs 24,000
Loss on 1 bag cement = Rs – 5
Loss on 5000 bags grey cement
= Rs -5 × 5000 = Rs – 25000
Total profit / loss = Total profit + Total loss
= 24000 + (- 2500)
= 24000 – 25000
∴ Total loss = – 1000

b) What is the number of white cement bags it must sell to have neither profit nor loss, if the number of grey bags sold is 6,400 bags.
Solution:
Loss on 1 bag grey cement bag = Rs -5
Loss on 6400 bag grey cement =Rs -5 × 6400
= Rs -32000
Let number of bags of white cement = n
profit on 1 bag white cement = Rs 8
Profit on n bags white cement = Rs 8n
∴ Total profit / loss = 0
Total profit + Total loss = 0
8n + (-32000) = 0
8n – 32000 = 0
⇒ 8n = 32000

n = 4000
∴ Number of white cement bags = n = 4000

Question 9.
Replace the blank with integer to make it a true statement.
a) (-3) × ___ = 27
(-3) × -9 = 27
WKT,
3 × 9 = 27
and -1 × -1 = 1
∴ (-3) × (-9) = 27

b) 5 × ___ = -35
5 × -7 = -35
WKT,
5 × 7 = 35 and 1 × -1 = -1
∴ 5 × (-7) = -35

c) __ × (-8) = -56
7 × (-8) = -56
WKT,
7 × 8 = 56
and 1 × -1 = -1
∴ 7 × (-8) = -56

d) __ × (-12) = 132
-11 × (-12)= 132
WKT,
11 × 12 = 132 and -1 × -1 = 1
∴ (-11) × (-12)= 132