Integers Class 7 Exercise 1.1 |
NCERT Answers for Class 7 Maths Chapter 1 Integers Exercise 1.3
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Ex 1.3 Class 7 Maths Question 1. Find each of the following products:
(a) 3 × (–1)
(b) (–1) × 225
(c) (–21) × (–30)
(d) (–316) × (–1)
(e) (–15) × 0 × (–18)
(f) (–12) × (–11) × (10)
(g) 9 × (–3) × (– 6)
(h) (–18) × (–5) × (– 4)
(i) (–1) × (–2) × (–3) × 4
(j) (–3) × (–6) × (–2) × (–1)
Answer
(a) 3 × (–1)
By the rule of Multiplication of integers,
= 3 × (-1)
= -3 … [Since (+ × – = -)]
(b) (–1) × 225
By the rule of Multiplication of integers,
= (-1) × 225
= -225 … [Since (- × + = -)]
(c) (–21) × (–30)
By the rule of Multiplication of integers,
= (-21) × (-30)
= 630 … [Since (- × – = +)]
(d) (–316) × (–1)
By the rule of Multiplication of integers,
= (-316) × (-1)
= 316 … [Since (- × – = +)]
(e) (–15) × 0 × (–18)
By the rule of Multiplication of integers,
= (–15) × 0 × (–18)
= 0
Since, any integer is multiplied with zero and the answer is zero itself.
(f) (–12) × (–11) × (10)
By the rule of Multiplication of integers,
= (–12) × (-11) × (10)
First multiply the two numbers having same sign,
= 132 × 10 … [Since (- × – = +)]
= 1320
(g) 9 × (–3) × (– 6)
By the rule of Multiplication of integers,
= 9 × (-3) × (-6)
First multiply the two numbers having same sign,
= 9 × 18 … [Since (- × – = +)]
= 162
(h) (–18) × (–5) × (– 4)
By the rule of Multiplication of integers,
= (-18) × (-5) × (-4)
First multiply the two numbers having same sign,
= 90 × -4 … [Since (- × – = +)]
= – 360 … [Since (+ × – = -)]
(i) (–1) × (–2) × (–3) × 4
By the rule of Multiplication of integers,
= [(–1) × (–2)] × [(–3) × 4]
= 2 × (-12) … [Since (- × – = +), (- × + = -)]
= – 24
(j) (–3) × (–6) × (–2) × (–1)
By the rule of Multiplication of integers,
= [(–3) × (–6)] × [(–2) × (–1)]
First multiply the two numbers having same sign,
= 18 × 2 … [Since (- × – = +)
= 36
Ex 1.3 Class 7 Maths Question 2. Verify the following:
(a) 18 × [7 + (–3)] = [18 × 7] + [18 × (–3)]
(b) (–21) × [(– 4) + (– 6)] = [(–21) × (– 4)] + [(–21) × (– 6)]
Answer
From the given equation,
Let us consider the Left Hand Side (LHS) first = 18 × [7 + (–3)]
= 18 × [7 – 3]
= 18 × 4
= 72
Now, consider the Right Hand Side (RHS) = [18 × 7] + [18 × (–3)]
= [126] + [-54]
= 126 – 54
= 72
By comparing LHS and RHS,
72 = 72
LHS = RHS
Hence, the given equation is verified.
(b) (–21) × [(– 4) + (– 6)] = [(–21) × (– 4)] + [(–21) × (– 6)]
From the given equation,
Let us consider the Left Hand Side (LHS) first = (–21) × [(– 4) + (– 6)]
= (-21) × [-4 – 6]
= (-21) × [-10]
= 210
Now, consider the Right Hand Side (RHS) = [(–21) × (– 4)] + [(–21) × (– 6)]
= [84] + [126]
= 210
By comparing LHS and RHS,
210 = 210
LHS = RHS
Hence, the given equation is verified.
Ex 1.3 Class 7 Maths Question 3.
(i) For any integer a, what is (–1) × a equal to?
(ii) Determine the integer whose product with (–1) is
(a) –22 (b) 37 (c) 0
Answer
(i) For any integer a, what is (–1) × a equal to?
= (-1) × a = -a
Because, when we multiplied any integer a with -1, then we get additive inverse of that integer.
(ii) Determine the integer whose product with (–1) is
(a) –22
Now, multiply -22 with (-1), we get
= -22 × (-1)
= 22
Because, when we multiplied integer -22 with -1, then we get additive inverse of that integer.
(b) 37
Now, multiply 37 with (-1), we get
= 37 × (-1)
= -37
Because, when we multiplied integer 37 with -1, then we get additive inverse of that integer.
(c) 0
Now, multiply 0 with (-1), we get
= 0 × (-1)
= 0
Because, the product of negative integers and zero give zero only.
Ex 1.3 Class 7 Maths Question 4. Starting from (–1) × 5, write various products showing some pattern to show (–1) × (–1) = 1.
Answer
The various products are,
= -1 × 5 = -5
= -1 × 4 = -4
= -1 × 3 = -3
= -1 × 2 = -2
= -1 × 1 = -1
= -1 × 0 = 0
= -1 × -1 = 1
We concluded that the product of one negative integer and one positive integer is negative integer. Then, the product of two negative integers is a positive integer.
Ex 1.3 Class 7 Maths Question 5. Find the product, using suitable properties:
(a) 26 × (– 48) + (– 48) × (–36)
(b) 8 × 53 × (–125)
(c) 15 × (–25) × (– 4) × (–10)
(d) (– 41) × 102
(e) 625 × (–35) + (– 625) × 65
(f) 7 × (50 – 2)
(g) (–17) × (–29)
(h) (–57) × (–19) + 57
(a) 26 × (– 48) + (– 48) × (–36)
Answer
The given equation is in the form of Distributive law of Multiplication over Addition.
= a × (b + c) = (a × b) + (a × c)
Let, a = -48, b = 26, c = -36
Now,
= 26 × (– 48) + (– 48) × (–36)
= -48 × (26 + (-36)
= -48 × (26 – 36)
= -48 × (-10)
= 480 … [Since (- × – = +)
(b) 8 × 53 × (–125)
The given equation is in the form of Commutative law of Multiplication.
= a × b = b × a
Then,
= 8 × [53 × (-125)]
= 8 × [(-125) × 53]
= [8 × (-125)] × 53
= [-1000] × 53
= – 53000
(c) 15 × (–25) × (– 4) × (–10)
The given equation is in the form of Commutative law of Multiplication.
= a × b = b × a
Then,
= 15 × [(–25) × (– 4)] × (–10)
= 15 × [100] × (–10)
= 15 × [-1000]
= – 15000
(d) (– 41) × 102
The given equation is in the form of Distributive law of Multiplication over Addition.
= a × (b + c) = (a × b) + (a × c)
= (-41) × (100 + 2)
= (-41) × 100 + (-41) × 2
= – 4100 – 82
= – 4182
(e) 625 × (–35) + (– 625) × 65
The given equation is in the form of Distributive law of Multiplication over Addition.
= a × (b + c) = (a × b) + (a × c)
= 625 × [(-35) + (-65)]
= 625 × [-100]
= – 62500
(f) 7 × (50 – 2)
The given equation is in the form of Distributive law of Multiplication over Subtraction.
= a × (b – c) = (a × b) – (a × c)
= (7 × 50) – (7 × 2)
= 350 – 14
= 336
(g) (–17) × (–29)
The given equation is in the form of Distributive law of Multiplication over Addition.
= a × (b + c) = (a × b) + (a × c)
= (-17) × [-30 + 1]
= [(-17) × (-30)] + [(-17) × 1]
= [510] + [-17]
= 493
(h) (–57) × (–19) + 57
The given equation is in the form of Distributive law of Multiplication over Addition.
= a × (b + c) = (a × b) + (a × c)
= (57 × 19) + (57 × 1)
= 57 [19 + 1]
= 57 × 20
= 1140
Ex 1.3 Class 7 Maths 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?
Answer
From the question, it is given that
Let us take the lowered temperature as negative,
Initial temperature = 40^{o}C
Change in temperature per hour = -5^{o}C
Change in temperature after 10 hours = (-5) × 10 = -50^{o}C
Therefore, The final room temperature after 10 hours of freezing process = 40^{o}C + (-50^{o}C)
= -10^{o}C
Ex 1.3 Class 7 Maths 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 questions not attempted.
(i) Mohan gets four correct and six incorrect answers. What is his score?
(ii) Reshma gets five correct answers and five incorrect answers, what is her score?
(iii) Heena gets two correct and five incorrect answers out of seven questions she attempts. What is her score?
Answer
From the question,
Marks awarded for 1 correct answer = 5
Then, Total marks awarded for 4 correct answer = 4 × 5 = 20
Marks awarded for 1 wrong answer = -2
Then, Total marks awarded for 6 wrong answer = 6 × -2 = -12
Therefore, Total score obtained by Mohan = 20 + (-12)
= 20 – 12
= 8
(ii) Reshma gets five correct answers and five incorrect answers, what is her score?
From the question,
Marks awarded for 1 correct answer = 5
Then, Total marks awarded for 5 correct answer = 5 × 5 = 25
Marks awarded for 1 wrong answer = -2
Then, Total marks awarded for 5 wrong answer = 5 × -2 = -10
Therefore, Total score obtained by Reshma = 25 + (-10)
= 25 – 10
= 15
(iii) Heena gets two correct and five incorrect answers out of seven questions she attempts. What is her score?
From the question,
Marks awarded for 1 correct answer = 5
Then, Total marks awarded for 2 correct answer = 2 × 5 = 10
Marks awarded for 1 wrong answer = -2
Then, Total marks awarded for 5 wrong answer = 5 × -2 = -10
Marks awarded for questions not attempted is = 0
Therefore, Total score obtained by Heena = 10 + (-10)
= 10 – 10
= 0
Ex 1.3 Class 7 Maths 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?
(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.
(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?
We denote profit in positive integer and loss in negative integer,
From the question,
Cement company earns a profit on selling 1 bag of white cement = ₹ 8 per bag
Then, Cement company earns a profit on selling 3000 bags of white cement = 3000 × ₹ 8
= ₹ 24000
Loss on selling 1 bag of grey cement = – ₹ 5 per bag
Then, Loss on selling 5000 bags of grey cement = 5000 × – ₹ 5
= – ₹ 25000
Total loss or profit earned by the cement company = profit + loss
= 24000 + (-25000)
= – ₹1000
Thus, a loss of ₹ 1000 will be incurred by the company.
(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.
We denote profit in positive integer and loss in negative integer,
From the question,
Cement company earns a profit on selling 1 bag of white cement = ₹ 8 per bag
Let the number of white cement bags be x.
Then, Cement company earns a profit on selling x bags of white cement = (x) × ₹ 8
= ₹ 8x
Loss on selling 1 bag of grey cement = – ₹ 5 per bag
Then,
Loss on selling 6400 bags of grey cement = 6400 × – ₹ 5
= – ₹ 32000
According to the question,
Company must sell to have neither profit nor loss.
= Profit + loss = 0
= 8x + (-32000) =0
By sending -32000 from LHS to RHS it becomes 32000
= 8x = 32000
= x = 32000/8
= x = 4000
Hence, the 4000 bags of white cement have neither profit nor loss.
Ex 1.3 Class 7 Maths Question 9. Replace the blank with an integer to make it a true statement.
(a) (–3) × _____ = 27
(b) 5 × _____ = –35
(c) _____ × (– 8) = –56
(d) _____ × (–12) = 132
(a) (–3) × _____ = 27
Let us assume the missing integer be x,
Then,
= (–3) × (x) = 27
= x = – (27/3)
= x = -9
Let us substitute the value of x in the place of blank,
= (–3) × (-9) = 27 … [Since (- × – = +)]
(b) 5 × _____ = –35
Let us assume the missing integer be x,
Then,
= (5) × (x) = -35
= x = – (-35/5)
= x = -7
Let us substitute the value of x in the place of blank,
= (5) × (-7) = -35 … [Since (+ × – = -)]
(c) _____ × (– 8) = –56
Let us assume the missing integer be x,
Then,
= (x) × (-8) = -56
= x = (-56/-8)
= x = 7
Let us substitute the value of x in the place of blank,
= (7) × (-8) = -56 … [Since (+ × – = -)]
(d) _____ × (–12) = 132
Let us assume the missing integer be x,
Then,
= (x) × (-12) = 132
= x = – (132/12)
= x = – 11
Let us substitute the value of x in the place of blank,
= (–11) × (-12) = 132 … [Since (- × – = +)]
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