SCERT AP 7th Class Maths Solutions Pdf Chapter 1 Integers Unit Exercise Questions and Answers.
AP State Syllabus 7th Class Maths Solutions 1st Lesson Integers Unit Exercise
Question 1.
 Calculate the following.
 (i) 8 × (-1)
 Answer:
 8 × (-1)
 We know, a × (- b) = – (a × b)
 = – (8 × 1) = – 8
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(ii) (- 2) × 175
 Answer:
 (-2) × 175
 We know, (- a) × b = – (a × b)
 = – (2 × 175) = – 350
(iii) (- 3) × (-40)
 Answer:
 (- 3) × (-40)
 We know, (- a) × (-b) = (a × b)
 = (- 3) × (- 40)
 = 3 × 40 = 120
(iv) (- 24) × (- 7)
 Answer:
 (- 24) × (- 7)
 We know, (- a) × (- b) = (a × b)
 = 24 × 7 = 168
(v) (- 7) ÷ (-1)
 Answer:
 (- 7) ÷ (-1)
 We know, (- a) + (- b) = a ÷ b
 = (- 7) ÷ (- 1)
 = 7 ÷ 1 = 7
(vi) (- 12) ÷ (+ 6)
 Answer:
 (- 12) ÷ ( + 6)
 We know, (- a) ÷ (- b) = – (a ÷ b)
 = (- 12) ÷ 6 = – 2
(vii) (- 49) ÷ (-7 )
 Answer:
 (- 49) ÷ (-7)
 We know, (- a) ÷ (- b) = a ÷ b
 = (- 49) ÷ (- 7)
 = 49 ÷ 7 = 7
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(viii) (+ 63) ÷ (- 9)
 Answer:
 (+ 63) ÷ (- 9)
 We know, a ÷ (- b) = – (a ÷ b)
 = 63 ÷ (-9) = – (63 ÷ 9) = – 7
Question 2.
 Replace the blank with an integer to make it a true statement.
 (i) (- 7) × _______ = 21
 Answer:
 (- 7) ×   x    = 21
 × = 21 ÷ (-7)
 We know, a ÷ (-b) = – (a ÷ b)
 x = – (21 ÷ 7)
 ∴ x = – 3
(ii) 7 × _______= – 42
 Answer:
 7 ×   x   = – 42
 x = (- 42) ÷ 7
 We know, (- a) ÷ b = – (a ÷ b)
 x = – (42 ÷ 7)
 ∴ x = – 6
(iii) ________ × (-9) = – 72
 Answer:
   x    × (- 9) = – 72
 x = (- 72) ÷ (- 9)
 We know, (- a) ÷ (-b) = (a ÷ b)
 x = (72 ÷ 9)
 ∴ x = 8
(iv) ________ × (- 11) = 132
 Answer:
   x    × (- 11) = 132
 × = 132 ÷ (-11)
 We know, a ÷ (- b) = – (a ÷ b)
 x = – (132 ÷ 11)
 ∴ × = – 12
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(v) (- 25) ÷ ________ = 1
 Answer:
 (- 25) ÷   x    = 1
 (- 25) = 1 × x
 x = (- 25) ÷ 1
 We know, (- a) ÷ b = – (a ÷ b)
 x = – (25 ÷ 1)
 ∴ x = – 25
(vi) 42 ÷ ________ = – 6
 Answer:
 42 ÷   x    = – 6
 42 = (- 6) × x
 x = 42 ÷ (- 6)
 We know, a ÷ (- b) = – (a ÷ b)
 x = – (42 ÷ 6)
 ∴ x = – 7
(vii) ______ ÷ 4 (- 15) = 6
 Answer:
   x    × 4 (- 15) = 6
 x = 6 × (- 15)
 We know, a × (-b) = – (a × b)
 x = – (6 × 15)
 ∴ x = – 90
(viii) ________ ÷ (- 9) = 16
 Answer:
   x    ÷ (- 9) = 16
 x = 16 × (- 9)
 We know, a × (- b) = – (a × b)
 x = – (16 × 9)
 ∴ x = – 144
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Question 3.
 Write all the possible pairs of integers that give a product of – 50.
 Answer:
| a | b | a × b = – 50 | 
| 41 | (-50) | 41 × (-50) = – 50 | 
| (-D | (+50) | (-1) × (50) = – 50 | 
| +2 | (-25) | 42 × (-25) = – 50 | 
| (-2) | (+25) | (-2) × (25) = – 50 | 
| +5 | (-10) | 45 × (-10) = – 50 | 
| (-5) | (+10) | (-5) × (10) = – 50 | 
| 450 | (-D | 50 × (-1) = – 50 | 
| (-50) | (1) | (-50) × (1) = – 50 | 
Question 4.
 Sarikar, a fruit vendor sells 100 kg of oranges and 75 kg of pomegranates. If he makes a profit of ₹ 11 per one kg of pomegranates and loss of ₹ 8 per one kg oranges, what will be his overall profit or loss ?
 
 Answer:
 Given
 Profit on 1 kg of pomegranates = ₹ 11
 Profit on 75 kg of pomegranates
 = 75 × 11
 = ₹ 825
 Loss on 1 kg of oranges = ₹ 8
 Loss on 100 kg of oranges
 = 100 × 8
 = ₹ 800
 Profit is greater than loss.
 So, Sankar will get profit.
 Overall profit = ₹ 825 – ₹ 800
 = ₹ 25
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Question 5.
 Bhargavi lost 5700 calories in the month of June using yoga. If the calory loss is uniform, calculate the loss of calories per day ?
 
 Answer:
 Given number of calories Bhargavi lost in the month of June = 5700 calories June month has 30 days.
 So, number of calories lost in 30 days = 5700
 Number of calories lost in 1 day = 5700 ÷ 30
 Number of calories lost per day = 190 Calories.
Question 6.
 Simplify 625 × (-35) + 625 × 30 using suitable law.
 Answer:
 625 × (-35) + 625 × 30
 Multiplication distributes over addition of integers.
 We know, a × b + a × c = a(b + c)
 = 625 [(- 35) + 30]
 = 625 (- 35 + 30)
 = 625 (- 5)
 We know, a × (- b) = – (a × b)
 = – (625 × 5)
 = – 3125
Question 7.
 Simplify the following using BODMAS.
 (i) 12 – 36 ÷ 3
 Answer:
 12 – 36 = 3 (Division)
 = 12 – 12 (Subtraction)
 = 0
(ii) 6 × (-7) + (- 3) ÷ 3
 Answer:
 6 × (- 7) + (- 3) ÷ 3
 We know, (- a) ÷ b = – (a ÷ b)
 = 6 × (- 7) – 3 ÷ 3 (Division)
 We know, a × (-b) = – (a × b)
 = – (6 × 7) – 1 (Multiplication)
 = – 42 – 1 (Subtraction)
 = – 43
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(iii) 38 – {35 – (36 – \(\overline{34-37}\))}
 Answer:
 38 – {35 – (36 – \(\overline{34-37}\))} (Vinculum)
 = 38 – {35 – (36 – (-3)}
 = 38 – {35 – (36 + 3)} (Simple bracket)
 = 38 – {35 – 39} (Curly bracket)
 = 38 – (-4)
 = 38 + 4 (Addition)
 = 42
Question 8.
 Write the absolute values of following numbers.
 (i) – 700
 Answer:
 We know, |- x| = x
 |- 700| = 700
(ii) 150
 Answer:
 We know, | x | = + x
 So, |150| = 150
(iii) – 150
 Answer:
 We know, | – x | = x
 So, |- 150| = 150
(iv) – 35
 Answer:
 We know, | – x | = x
 So, |- 35| = 35
(v) If p < 10, then |p – 10|
 Answer:
 We know, if × < a, then |x – a | = a – x
 So, |p – 10| = 10 – p
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(vi) If y > 7, then |7 – y|
 Answer:
 We know, if x > a, then |a – x| = x – a
 So, |7 – y | = y – 7