These NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.4 Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 7 Maths Chapter 1 Integers Exercise 1.4
Question 1.
Evaluate each of the following:
(a) (-30) ÷ 10
(b) 50 ÷ (-5)
(c) (-36) ÷ (-9)
(d) (-49) ÷ (49)
(e) 13 ÷ [(-2) + 1]
(f) 0 ÷ (-12)
(g) [(-31) ÷ [(-30) + (-1)]
(h) [(-36) ÷ 12] ÷ 3
(i) = (-6 + 5) ÷ [(-2 + 1)]
Answer:
(a) (-30) ÷ 10 = – (30 ÷ 10) = – (3) = -3
(b) 50 ÷ (-5) = – (50 ÷ 5) = – (10) = -10
(c) (-36) ÷ (- 9) = + (36 ÷ 9) = + 4
(d) (-49) ÷ 49 = – (49 ÷ 49) = – (1) = -1
(e) 13 ÷ [(-2) + 1] = 13 ÷ (-2 + 1)
= 13 ÷ (-1) = – (13 ÷ 1) = -13
(f) 0 ÷ (-12) = – (0 ÷ 12) = 0
(g) (-31) ÷ [(-30) + (-1) = (-31) ÷ [-30 -1]
-31 ÷ (-31) = + (31 ÷ 31) = + 1
(h) [(-36) ÷ 12] ÷ 3 = [-36 ÷ 12] ÷ 3
= – (36 ÷ 12) ÷ 3 = – (3) ÷ 3
= -3 ÷ 3 = – 1
(i) [(-6) + 5] ÷ [(-2) + 1]
= (-6 + 5) ÷ (-2 + 1) = – 1 ÷ (-1)
= + (1 ÷ 1) = 1
Question 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
(b) a = (-10), b = 1, c = 1
Answer:
(a) Given a = 12, b = – 4 and c = 2
L.H.S = a + (b + c)
= 12 ÷ (-4 + 2) = 12 ÷ (- 2)
= – [12 ÷ 2] = – 6
R.H.S = (a ÷ b) + (a ÷ c)
= (12 + -4) + [12 ÷ 2]
= -(12 ÷ 4) + (6) = -3 + 6 = 3
∴ L.H.S ≠ R.H.S
(b) Here a = – 10, b = 1, c = 1
L.H.S = a ÷ (b + c)
= (-10) ÷ (1 + 1) = -10 ÷ 2
= -(10 ÷ 2) = -5
R.H.S = (a ÷ b) + (a ÷ c)
= (-10 ÷ 1) + (-10 ÷ 1)
= -(10 ÷ 1) + (-10) ÷ 1
= -10 -10 = -20
L.H.S ≠ R.H.S
Question 3.
Fill in the blanks :
Answer:
(a) 369 ÷ 1 = 369 (a ÷ 1 = a)
(b) (-75) ÷ 75 = -1 (-a ÷ a = -1)
(c) (-206) ÷ 206 = 1 (- a) ÷ (-a) = 1
(d) -87 ÷ -1 = 87 (-a ÷ – 1 = a)
(e) -87 ÷ = -87 [(-a) ÷ 1 = -a]
(f) -48 ÷ 48 = -1 (-a ÷ a = -1)
(g) 20 ÷ -10 = -2
(h) -12 ÷ 4 = -3
Question 4.
Write five pairs of integers (a, b) such that a ÷ b = -3. One such pair is (6, -2) because 6 ÷ (-2) = (-3).
Answer:
Five pairs of integers (a, b) such that a ÷ b = -3
(1) (3, -1)
(2) (-3, 1)
(3) (9, -3)
(4) (-9, 3)
(5) (-12, 4)
Question 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?
Answer:
Temperature at 12 noon = + 10°C
Rate of change in temperature = -2°C per hour
Number of hours from 12 noon to midnight = 12
∴ Change in temperature in 12 hours = 12° × (-2) = -24°C
Temperatureat mid-night= 10°C + (-24°C)
(ie 12 hours after 12 noon) = -14°C
Temperature difference between + 10°C and -8°C = + 10°C – (-8°C)
= 10°C + 8°C = 18°C
Time taken = 18 ÷ 2 = 9 hours
Temperature change of 18°C will take place in 9 hours from 12 noon
Time after 9 hours from 12 noon = 9 P.M.
Question 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?
(ii) Mohini scores -5 marks in this
test, though she has got 7 correct answers. How many questions has she attempted incorrectly?
Answer:
(i) Let ‘x’ be the number of incorrect question attempted by Radhika
According to the given question, we get
(+3) × 12 + x × (-2) = 20
36 – 2x = 20
2x = 36 – 20
2x = 16
x = 8
∴ Radhika attempted 8 incorrect questions.
(ii) Let ‘x’ be the number of incorrect question attempted by Mohini
According to the given question, we get
(+3) × 7 + x × (-2) = -5
21 – 2x = -5
2x = 21 + 5
2x = 26
x = \(\frac{26}{2}\) = 13
∴ Mohini attempted 13 incorrect questions.
Question 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?
Answer:
Present position of the elevator is at 10 m above the ground level;
Distance moved by the elevator below the ground level = 350 m
∴ Total distance moved by the elevator = 350 m + 10 m = 360 m
∴ The rate of descend = 6 m/min
Time taken = \(\frac{360}{6}\) minutes = 60 minutes 6 = 1 hour
Required time = 1 hour