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