KSEEB Solutions for Class 7 Maths Chapter 1 Integers Ex 1.3

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

KSEEB Solutions

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

KSEEB Solutions

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)

KSEEB Solutions

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

KSEEB Solutions

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
KSEEB Solutions for Class 7 Maths Chapter 1 Integers Ex 1.3 30
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

Leave a Comment