Class 7 Maths Chapter 1 Question Answers - Integers

Q1. Verify  a - (-b)=a + b  for the following values of  a   and  b
(a) a = 25, b = 12
Ans: Substituting value of  a and  b in given equation
a - (-b)=a + b
25 - (-12)=25 + 12
25 + 12 = 25 + 12
37=37
Hence, verified.

(b) a=113,b=16
Ans: Substituting value of  a  and  b in given equation
a - (-b) = a + b
113 - (-16) = 113 + 16
113 + 16 = 113 + 16
129 = 129

Q2. Use ,  or  =  sign for the below statements to make it true
(a) (-9) + (-28) ____  (-9) - (-28)
Ans: Solving both sides-
(-9) + (-28)=−37
(-9) - (-28)=19
Thus,  (-9) + (-28) < (-9) - (-28)

(b)  25 + (-14) - 18 ____   25 + (-14) - (-18)
Ans: Solving both sides
25 + (−14) − 18 = 11 − 18
=- 7
25 + (-14) - (-18) = 11 + 18
= 29
Thus,  25 + (−14) − 18 < 25 + (-14) - (-18).

Q3. Write down a pair of integers for the following 
(a) Sum gives  -9
Ans:  A pair of integers that gives sum  -9 is  (−6,−3).

(b) Difference gives  -11
Ans:  A pair of integers that gives sum  -11 is  (−14,3).

Q4.  (a) Write a positive and negative integer whose sum is  -4 .
Ans: (4,−8)is a positive and negative integer whose sum is  -4 .

(b) Write a negative integer and a positive integer whose difference is  -2.
Ans:  (−1,1) is a positive and negative integer whose sum is  -2 .

Q5. Fill in the blanks
(a) (-4) + (-11) = (-11)+ _ _ _
Ans: (-4) + (-11) = (-11)+ _ _ _
⇒ (-4) + (-11)+11
⇒ −4
Thus,  (-4) + (-11) = (-11) + -4

(b) [22 + (-9)] + (-2)=22 + [ _ _ _ _ +(-2)]
Ans: [22 + (-9)] + (-2) = 22 + [ _ _ _ _ + (-2)]
⇒ 13-2=22+[ _ _ _ _ +(-2)]
⇒ 11-22=[ _ _ _ _ +(-2)]
⇒ -11= _ _ _ _ +(-2)
⇒ -11+2
⇒ -9
Thus, [22 + (-9)] + (-2) = 22 + [-9 + (-2)]

Q6. Verify  7 × [(22) + (-9)] = [(7) × 22]+[7 × -9]
Ans:  On solving both sides
7 ×  [(22) + (-9)] = [(7) × 22]+[7 × -9]
7 × [13] = 154−63
91= 91
Hence, verified.

Q7.Find the product of
(a) 63 × 0 × -7
Ans: The product of  63 × 0 × -7 is  0.

(b) 5 × (-3) × -2
Ans: So, 5 × (−3) × −2 = 5 × 6
=3 0
The product of  5×(−3)×−2 is  30.

Q8. (a)  -2 × ____ = 14
Ans: So,  -2 × ____ = 14
⇒ 14 / - 2
⇒ -7
Hence,  -2 × 7 =14 .

(b)  ____ × -8 = -32
Ans:  So,  _ _ _ × -8 = -32
⇒ −32 / −8
⇒ 4
Hence, 4 × -8 = -32

Q9. Evaluate:
(a) -39 ÷ 13
Ans:  −39÷13
⇒ −39 / 13
⇒ −3
Hence,  −39 ÷ 13 = −3

(b) -64 ÷ [-8 × -8]
Ans:  −64 ÷ [−8×−8]
⇒ −64 / [−8×−8]
⇒ −64 / 64
⇒ −1
Hence,  −64 ÷ [−8×−8] = −1

Q10. Write two pairs of integers such that  a ÷ b = -5
Ans:The two pairs of integers such that  a ÷ b=-5 are:
> (10,−2)
> (−70,14)

Q11.  Manvita deposits Rs.  5000 in her bank account after two days. She withdraws Rs.  3748 from it. If the amount deposited is a positive integer. How will you represent the amount withdrawn and also find the balance amount in the account?
Ans: The amount withdrawn should always be represented as a negative integer.
Thus, it would be  −3748
Since, Total balance  =   Amount deposited  −  Amount withdrawn
Therefore,
Total balance  = 5000−3748
= Rs. 1250
Hence, the amount withdrawn would be negative integer i.e.,  −3748 and the balance amount in the account is  Rs. 1252.

Q12. In a game Mishala scored  20, -40, 10  and Meera scored  -40, 10, 20. Who scored more and can we add scores (integers) in any order?
Ans: Since, Mishala scored  20,-40,10
Therefore, total score of Mishala is
= 20 - 40 + 10
= -20 + 10
= -10
And since, Meera scored  -40, 10 20.
Therefore, total score of Meera is
= -40 + 10 + 20
= -20 + 10
= -10
Hence, both scored the same points in a game but in a different order.
Yes, we can add integers in any order.

Q13. Find the product with suitable properties for the following-
(a) 16 × (-34) + (-34) × (-18)
Ans: Given
16 × (-34) + (-34) × (-18)
By distributive property-
 a × b + a × c = a[b + c]
Thus,
= -34[16 - 18]
= -34 × -2
= 68
Hence,  16 × (-34)+(-34) × (-18) = 68 .

(b) 23 × -36 × 10
Ans: Given
23×−36×10
By commutative property-
(a × b) × c = a × (b × c)
Thus,
= 23 × [−36 × 10]
= 23 × −360
= −8280

Q14.  A fruit merchant earns a profit of Rs. 6 per bag of orange sold and a loss of Rs. 4 per bag of grapes sold.
(a) Merchant sells  1800 bags of orange and  2500 bags of grapes. What is the profit or loss?
Ans: Since, profit is denoted by a positive integer and a loss is denoted by a positive integer.
Therefore, profit earned by selling  1 bag of orange is Rs.  6
Profit earned by selling  1800 bags or orange is
6 × 1800
= Rs. 10,800
Loss incurred by selling  1 bag of grapes is Rs.  −4
Loss incurred by selling  2500 bags of grapes is
= −4 × 2500
= 10,000
Total profit or loss earned  = Profit  + Loss
=10,800+10,000
=800
Hence, a profit of Rs. 800 will be earned by a merchant.

(b) What is the number of bags of oranges to be sold to have neither profit nor loss if the number of grapes bags are sold is  900 bags?
Ans: Since profit is denoted by a positive integer and a loss is denoted by a positive integer.
Therefore, Loss incurred while selling  1 bag of grapes  = -Rs.4
Loss incurred while selling 900 bags of grapes be
= −4 × 900
= −3600
Let the number of bags of oranges to be sold  = x
Profit earned when  1 bag of orange is sold  = Rs.6
Profit earned while selling x bags of orange  = 6x
Condition for no profit, no loss
Profit earned  +  Loss incurred  =0
6x - 3600 = 0
6x = 3600
x = 36006
x = 600
Hence, to have neither profit nor loss 600 number of bags of oranges to be sold.

Q15. Verify that a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c) for each of the following values of  a , b and  c.

(a) a = 8, b = 4, c = 2
Ans: For equation a ÷ (b+c) ≠ (a ÷ b) + (a ÷ c)
L.H.S  = a ÷ (b+c)
= 8 ÷ (-4 + 2)
= 8 ÷ (-2)
= -4
R.H.S  =(a ÷ b) + (a ÷ c)
= (8 ÷ -4) + (8 ÷ 2)
= -2 + 4
= 2
Hence,  L.H.S≠R.H.S
Thus, a ÷ (b+c) ≠ (a ÷ b) + (a ÷ c) for  a = 8,b = 4,c = 2

(b) a = -15, b = 2,c = 1
Ans: For equation a ÷ (b+c)≠(a ÷ b)+(a ÷ c)
L.H.S  = a ÷ (b+c)
 =-15 ÷ (2+1)
=-15 ÷ 3
R.H.S  =(a ÷ b) + (a ÷ c)
= (-15 ÷ 2) + (-15 ÷ 1)
= -7.5 + (-15)
= -22.5
Hence,  L.H.S≠R.H.S
Thus, a ÷ (b+c) ≠ (a ÷ b) + (a ÷ c) for  a = -15, b = 2, c= 1.

Q16.  In a CET Examination  (+2) marks are given for every current answer and  (-0.5)  marks are given for every wrong answer and  0 for non-attempting any question.
(a) Likitha scores  30 marks. If she got  20 correct answers, how many questions she has attempted incorrectly?
Ans: Marks obtained for  1  correct answer  = +2
Marks obtained for  1 wrong answer  = -0.5
So, Marks scored by Likitha =  30
Marks obtained by  20 correct answers = 20 × 2 = 40
Marks obtained for incorrect answer  = Total score  −   Marks obtained by  20 correct answer
= 30 − 40
= −10
Marks obtained for  1  wrong answer = −0.5
The number of incorrect answers = −10 − 0.5
= 20
Hence, she attempted  20 questions wrongly.

(b) Saara scores  -4  marks if she got  3 correct answers. How many were incorrect?
Ans: Marks obtained for  1  correct answer  = +2
Marks obtained for  1  wrong answer  = -0.5
So, Marks scored by Saara  = −4
Marks obtained for 3 correct answers = 3 × 2 = 6
Marks obtained for incorrect answers  = Total score  −  Marks obtained for  3  correct answer
= −4 − 6 = −10
Marks obtained for 1 wrong answer = −0.5
The number of incorrect questions  = −10 − 0.5
= 20
Hence,  20 questions were incorrect.