RBSE Solutions for Class 7 Maths Chapter 1 Integers Ex 1.4

Rajasthan Board RBSE Solutions for Class 7 Maths Chapter 1 Integers Ex 1.4 Textbook Exercise Questions and Answers.

RBSE Class 7 Maths Solutions Chapter 1 Integers Ex 1.4

Question 1.
Evaluate each of the following :
(a) (- 30) ÷ 10
Answer:
(- 30) ÷ 10 = - (30 ÷ 10) = - 3

(b) 50 ÷ (- 5)
Answer:
50 ÷ (- 5) = - (50 ÷ 5) = - 10

(c) (- 36) ÷ (- 9)
Answer:
(- 36) ÷ (- 9) = 36 ÷ 9 = 4

(d) (- 49) ÷ 49
Answer:
(- 49) ÷ (49) = - (49 ÷ 49) = - 1

(e) 13 ÷ [(- 2) + 1]
Answer:
13 ÷ [(- 2) + 11 = 13 ÷ (- 1) = - (13 ÷ 1) = - 13

(f) 0 ÷ (- 12)
Answer:
0 ÷ (- 12) = 0

(g) (- 31) ÷ [(- 30) + (- 1)]
Answer:
(-31) ÷ [(- 30)+ (-1)1 = (- 31) ÷ (- 31) = 31 ÷ 31 = 1

(h) [(- 36) ÷ 12] ÷ 3
Answer:
[(- 36) ÷ 12] ÷ 3 = (- 3) ÷ 3 = - (3 ÷ 3) = - 1

(i) [(- 6) + 5] ÷ [(- 2) + 1]
Answer:
[(- 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
Answer:
LHS = a ÷ (b + c)
= 12 ÷ (- 4 + 2)
= 12 ÷ (- 2) = - 6
RHS = (a ÷ b) + (a ÷ c)
= (12 ÷ - 4) + (12 ÷ 2)
= - 3 + 6 = 3
∴ LHS ≠ RHS

(b) a = (- 10), b = 1, c = 1
Answer:
LHS = a ÷ (b + c)
= - 10 ÷ (1 + 1) .
= - 10 ÷ 2 = - 5
RHS = (a ÷ b) + (a ÷ c)
= (- 10 ÷ 1) + (- 10 ÷ 1)
= - 10 + (- 10) = - 20
∴ LHS ≠ RHS

Question 3.
Fill in the blanks:
(a) 369 ÷ = 369
Answer:
1

(b) (- 75) ÷ = - 1
Answer:
75

(c) (- 206) ÷ _= 1
Answer:
- 206

(d) - 87 ÷ = 87
Answer:
- 1

(e) _______ + 1 = - 87
Answer:
- 87

(f) _______ + 48 = - 1
Answer:
- 48

(g) 20 ÷ _________ = - 2
Answer:
- 10

(h) ________ ÷ (4) = - 3
Answer:
- 12

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:
(a) 12 ÷ (- 4) = - 3
(b) - 9 ÷ 3 = - 3
(c) 45 ÷ (-15) = - 3
(d) (- 18) ÷ 6 = - 3
(e) - 21 ÷ 7 = - 3

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
Required temperature = - 8°C
Required decrease in temperature
= 10°C - (- 8°C) = 10°C + 8°C = 18°C
No. of hrs. for 2°C decrease = 1
No. of hrs. for 1°C decrease = \(\frac{1}{2}\)
No. of hrs. for 18°C decrease = \(\frac{1}{2}\) × 18 = 9
So, the temperature will be 8°C below zero at (12 noon + 9 hrs.) = 9 pm
No. of hrs. from 12 noon to midnight = 12
Decrease in temperature in 12 hrs. = 12 × 2 = 24°C
∴ Temperature at midnight = 10°C - 24°C = - 14°C
So, the temperature at midnight would be - 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 question.
(i) Radhika scored 20 marks. If she has got 12 correct answers, how many questions has she attemped 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) Marks given for one correct answer = 3
Marks given for 12 correct answers = 12 × 3 = 36
Radhika1 s score = 20
∴ Marks given for incorrect answers = 20 - 36 = -16
Marks for 1 incorrect answer = - 2
∴ No. of incorrect answers = (- 16) ÷ (- 2) = 8
So, Radhika attempted 8 questions incorrectly.

(ii) Marks given for 7 correct answers = 3 × 7 = 21
Mohini's score = - 5
Marks given for incorrect answer = - 5 - 21 = - 26
Marks for 1 incorrect answer = - 2
∴ No. of incorrect answers = (- 26) ÷ (- 2) = 13
So, Mohini attempted 13 questions incorrectly.

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:
Starting position of shaft = 10 m
Final position of shaft = - 350 m
Total distance to be covered = 10 m - (- 350 m) = 360 m
Time taken to cover 6 m = 1 min
Time taken to cover 1 m = \(\frac{1}{6}\) min
Time taken to cover 360 m = \(\frac{1}{6}\) × 360 min = 60 min
So, the elevator will take 60 min or 1 hr. to reach = 350 m.