NCERT Class 7 Maths Chapter 1 Integers Exercise 1.3 Solutions
In this page we have provided solutions of the Exercise 1.3 of NCERT Class 7 Maths Chapter 1 Integers. These solutions are made by our team of expert teachers. Practice these solutions carefully for a better understanding of the topic which will help in scoring good marks in the examination.
Examples of integers are {….-5, -4, -3, -2, -1, 0, +1, +2, + 3, +4, +5….}.
Publishing Organisation |
NCERT |
Class |
7 |
Subject |
Mathematics |
Chapter |
1: Integers |
Exercise |
1.3 |
EXERCISE 1.3
(1) Evaluate each of the following:
(a) (–30) ÷ 10
= – 30/ 10 = -3
(b) 50 ÷ (–5)
= 50/ (- 5) = – 10
(c) (–36) ÷ (–9)
= – 36 / (- 9) = 4
(d) (– 49) ÷ (49)
= – 49/ 49 = – 1
(e) 13 ÷ [(–2) + 1]
= 13/ (- 1) = – 13
(f) 0 ÷ (–12) = 0
(g) (–31) ÷ [(–30) + (–1)]
= – 31 / (- 31) = 1
(h) [(–36) ÷ 12] ÷ 3
= (- 36/ 12) ÷ 3
= -3 / 3 = – 1
(i) [(– 6) + 5)] ÷ [(–2) + 1]
= -1 / – 1 = 1
(2) Verify that a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c) for each of the following values of a, b and c.
(a) a = 12, b = – 4, c = 2
Solution: L. H. S. = a ÷ (b + c)
= 12 ÷ (- 4 + 2)
= 12/ (- 2) = – 6
- H. S. = (a ÷ b) + (a ÷ c)
= {12 ÷ (- 4)} + (12/ 2)
= {12/(- 4)} + 6
= – 3 + 6 = 3
∴ L. H. S. ≠ R. H. S. (proved)
(b) a = (–10), b = 1, c = 1
Solution: L. H. S. = a ÷ (b + c)
= (-10) / (1 + 1)
= – 10 / 2 = – 5
- H. S. = (a ÷ b) + (a ÷ c)
= {(- 10)/ 1} + {(- 10)/ 1}
= – 10 + (- 10) = – 20
(3) Fill in the blanks:
(a) 369 ÷ 1 = 369
(b) (–75) ÷ 75 = –1
(c) (–206) ÷ (- 206) = 1
(d) – 87 ÷ (- 1) = 87
(e) (- 87) ÷ 1 = – 87
(f) (- 48) ÷ 48 = –1
(g) 20 ÷ (- 10) = –2
(h) (- 12) ÷ (4) = –3
(4) Write five pairs of integers (a, b) such that a ÷ b = –3. One such pair is (6, –2) because 6 ÷ (–2) = (–3).
Solution: (i) (3, – 1) because 3 ÷ (- 1) = – 3
(ii) (- 3, 1) because (- 3) ÷ 1 = – 3
(iii) (- 6, 2) because (- 6) ÷ 2 = – 3
(iv) (- 9, 3) because (-9) ÷ 3 = – 3
(v) (9, – 3) because 9 ÷ (- 3) = – 3
(5) The temperature at 12 noon was 10°C above zero. If it decreases at the rate of 2°C per hour until midnight, at what time would the temperature be 8°C below zero? What would be the temperature at mid-night?
Solution: Initial temperature at 12 noon = 10^{o} C
Change the temperature per hour = – 2^{o} C
Temperature at 1:00 pm = {10 + (- 2)}^{o} C = 8^{o }C
Temperature at 2:00 pm = {8 + (- 2)}^{o} C = 6^{o} C
Temperature at 3:00 pm = {6 + (- 2)}^{o} C = 4^{o} C
Temperature at 4:00 pm = {4 + (- 2)}^{o} C = 2^{o} C
Temperature at 5:00 pm = {2 + (- 2)}^{o }C = 0^{o} C
Temperature at 6:00 pm = {0 + (- 2)}^{o} C = – 2^{o} C
Temperature at 7:00 pm = {- 2 + (- 2)}^{o }C = – 4^{o} C
Temperature at 8:00 pm = {- 4 + (- 2)}^{o} C = – 6^{o} C
Temperature at 9:00 pm = {- 6 + (- 2)}^{o} C = – 8^{o} C
Hence, the temperature will be 8^{o} C below zero at 9:00 pm.
(6) In a class test (+ 3) marks are given for every correct answer and (–2) marks are given for every incorrect answer and no marks for not attempting any question.
(i) Radhika scored 20 marks. If she has got 12 correct answers, how many questions has she attempted incorrectly?
Solution: Let the number of incorrect answer be x.
∴ (12 × 3) + {x × (- 2)} = 20
or, 36 + (- 2x) = 20
or, – 2x = 20 – 36
or, – 2x = – 16
or, x = 8
(ii) Mohini scores –5 marks in this test, though she has got 7 correct answers. How many questions has she attempted incorrectly?
Solution: Let the number of incorrect answer be x.
∴ (7 × 3) + {x × (- 2) = – 5
or, 21 + (- 2x) = – 5
or, – 2x = – 5 – 21
or, – 2x = – 26
or, x = 13
(7) An elevator descends into a mine shaft at the rate of 6 m/min. If the descent starts from 10 m above the ground level, how long will it take to reach – 350 m.
Solution: Initial height = 10 m
Final depth = – 350 m
Total distance descends by elevator = (- 350) – 10 = – 360 m
The time taken by elevator = – 360 / (- 6) = 60 minutes = 1 hour
See Also NCERT Class 7 Math Solution