NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.4 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.4.

Board CBSE
Textbook NCERT
Class Class 7
Subject Maths
Chapter Chapter 1
Chapter Name Integers
Exercise Ex 1.4
Number of Questions Solved 7
Category NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 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].

Solution:

(a) (- 30) + 10 = – 3
(b) 50 +(-5) = – 10
(c) (-36) +(-9) = 4
(d) (- 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 = (- 3) + 3 = – 1
(i) [(- 6) + 5] + [(- 2) + 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 = l.
Solution:
(a) a + (b + c) = 12 ÷ [(- 4) + 2] = 12 + (- 2) = – 6
(a ÷ b) + (a ÷ c) = 12 ÷ (- 4) + 12 ÷ 2 = -3 + 6 = 3
So, a + (b + c) ≠ (a + b) + (a + c)

(b) a ÷ (b + c) = (- 10) + (1 + 1) = (- 10) + 2 = – 5
a ÷ b + a ÷ c = (- 10) ÷ 1 + (- 10) ÷ 1 = (- 10) + (- 10) = – 20
So, a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c).

Question 3.
Fill in the blanks:

(a) 369 ÷ …….. = 369
(b) -75 ÷ …….. = – 1
(c) (- 206) ÷ ……. = 1
(d) -87 ÷ …….. = 87
(e) ……. ÷ 1 = -87
(f) ……. ÷ 48 = -1
(g) 20 ÷ …… = -2
(h) …… ÷ (4) = – 3.

Solution:

(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.

Question 4.
Write five pairs of integers (a, b) such that a + b = -3. One such pair is (6, -2) because 6 +(-2) = (-3).
Solution:
Five pairs of integers (a, b) such that a + b = -3 are (- 6, 2), (-9, 3), (12,- 4), (21, -7), (-24, 8)
Note: We may write many such pairs of integers.

Question 5.
The temperature at 12 noon was 10°C above zero. If it decreases at the rate of 2°C per hour until mid-night, at what time would the temperature be 8°C degrees below zero? What would be the temperature at mid night?
Solution:
Difference in temperatures +10 °C and -8
= [10 – (- 8)] °C = (10 + 8)° C = 18 °C
Decrease in temperature in one hour = 2°C
Number of hours taken to have temperature 8 °C below zero \(=\frac { Total\quad decrease }{ Decrease\quad in\quad one\quad hour } \)
\(=\frac { 18 }{ 2 }\)
So, at 9 P.M., the temperature will be 8 °C below zero
Temperature at mid-night = 10 °C – (2 x 12) °C
= 10°C – 24 °C = -14 °C

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?
Solution:
(i) Let ‘x’ be the number of incorrect questions attempted by Radhika.
According to the question, we get
(+ 3) × 12 + x × (-2) = 20
⇒ 36 – 2x = 20
⇒ 2x = 36 – 20
⇒ x = \(\frac { 16 }{ 2 } \) = 8
Therefore, Radhika attempted 8 incorrect questions.

(ii) Let ‘x’ be the number of incorrect question attempted by Mohini.
According to the question, we get
(+ 3) × 7 + x × (- 2) = – 5
⇒ 21 – 2x = -5
⇒ 2x = 21 + 5
⇒ x = \(\frac { 26 }{ 2 } \) = 13
Therefore, Mohini attempted 13 incorrect questions.

Question 7.
An elevator descends into a mine shaft at the rate of 6m/min. If the descent starts from 10 m above the ground level, how long will it take to reach – 350 m.
Solution:
Difference in heights at two positions = 10 m – (-350 m) = 360 m
Rate of descent = 6 m/minute
∴ Time taken \(=\left( 360 \right) \div \left( 6 \right)\) minutes = 60 minutes = 1 hour
Hence, the elevator will take 1 hour to reach = 350 m.

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