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.
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:
- [9, (- 3)]
- [12, (- 4)]
- [15, (-5)]
- [-9, 3]
- [-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 mid-night, at what time would the temperature be 8°C degrees below zero? What would be the temperature at mid night?
Solution:
Temperature at 12 noon = + 10°C
Temperature at 1 P.M.= 10°C – 2°C = 8°C
Temperature at 2 P.M. = 8°C – 2°C = 6°C
Temperature at 3 P.M. = 6°C – 2°C = 4°C
Temperature at 4 P.M. = 4°C – 2°C = 2°C
Temperature at 5 P.M. = 2°C – 2°C = 0°C
Temperature at 6 P.M. = 0°C – 2°C = – 2°C
Temperature at 7 P.M. = – 2°C – 2°C = – 4°C
Temperature at 8 P.M. = – 4°C – 2°C = – 6°C
Temperature at 9 P.M. = – 6°C – 2°C = – 8°C
So, the temperature was 8°C below zero at 9 P.M.
Temperature at 10 P.M. = – 8°C – 2°C = – 10°C
Temperature at 11 P.M. = – 10°C – 2°C = – 12°C
Temperature at 12 P.M. = – 12°C – 2°C = – 14°C
So, the temperature at mid-night was 14 degrees below zero.
Aliter:
Difference between the temperatures 10°C above zero and 8°C below zero
= 10°C – (- 8°C)
= 10°C + 8°C = 18°C
Time for this degrees
= 18 ÷ 2 = 9 hours
So, the temperature was 8°C below zero at 9 P.M.
Difference between the times 12 noon and mid-night = 12 hours
Degrees in temperature in 12 hours = 12 × 2 = 24°C
So, temperature at mid night
= 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 question 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:
Let ‘x’ minutes be the time taken to reach – 350 m.
According to the question, we get
x = \(\frac { (-10)+(-350) }{ 6 } \) = \(\frac { -360 }{ 6 } \) = -60
Negative sign indicates descent.
∴ x = 60 minutes = 1 hour
We hope the NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.4, help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.4, drop a comment below and we will get back to you at the earliest.