SCERT AP 7th Class Maths Solutions Pdf Chapter 1 Integers Ex 1.2 Textbook Exercise Questions and Answers.

## AP State Syllabus 7th Class Maths Solutions 1st Lesson Integers Exercise 1.2

Question 1.
Calculate the following,
(i) (- 96) ÷ 16
(- a) ÷ b = – $$\frac{a}{b}$$
= $$\frac{-96}{16}$$ = – 6 (ii) 98 ÷ (- 49)
a ÷ (- b) = – $$\frac{a}{b}$$
= $$\frac{98}{-49}$$ = – 2

(iii) (- 51) ÷ 17
(- a) ÷ b = – $$\frac{a}{b}$$
= $$\frac{-51}{17}$$ = – 3

(iv) 38 ÷ (- 19)
a ÷ (- b) = – $$\frac{a}{b}$$
= $$\frac{38}{-19}$$ = – 2

(v) (- 80) ÷ 20
(- a) ÷ b = – $$\frac{a}{b}$$
= $$\frac{-80}{20}$$ = – 4

(vi) (- 150) ÷ (- 25)
(- a) ÷ (- b) = – $$\frac{-a}{-b}$$ = $$\frac{a}{b}$$
= $$\frac{-150}{-25}$$ = $$\frac{150}{25}$$ = – 6

(vii) (- 600) ÷ 60
(- a) ÷ b = – $$\frac{a}{b}$$
= $$\frac{-600}{60}$$
= – 10

(viii) (-54) ÷ 9
(- a) ÷ b = – $$\frac{a}{b}$$
= $$\frac{- 54}{9}$$
= – 6

(ix) 130 ÷ 65
a ÷ b = $$\frac{a}{b}$$
= $$\frac{130}{65}$$
= 2 (x) (- 315) ÷ ( -315)
(- a) ÷ (- b) = $$\frac{-a}{-b}$$ = $$\frac{a}{b}$$
= $$\frac{-315}{-315}$$
= $$\frac{315}{315}$$
= 1

Question 2.
The product of two integers is – 165. If one number is – 15, find the other integer.
Given one number a = – 15
Other number b = ?
Product of two numbers = (- 165)
a × b = – 165
(- 15) × b = – 165
b = (- 165) × (- 15)
Other number b = 11

Question 3.
Because of Covid-19 a company lockdown for 6 months and got loss of ₹1,32,000 in the year 2020. Find the average loss of each month. Given loss of 6 months = ₹ 1,32,000
Average loss of each month = 1,32,000 ÷ 6
Average loss of each month = ₹ 22,000 Question 4.
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 ?
The temperature at 12 noon = 10°C above zero.
The temperature is decreasing at 2°C per hour 8°C below zero = – 8°C
So, total temperature falls 2°C in every one hour.
Therefore, to decrease 18°C, time taken
= 18/2 = 9 hrs.
Present time is 12 noon, so, the time when the temperature at 12 midnight will be
= (- 8° C) + (- 2° C × 3 hrs.)
= (- 8° C) + (- 6° C)
= 160
So, at 12 midnight the temperature will be – 14°C

Question 5.
A green grocer earns a profit of ₹ 7 per kg of tomato and got loss of ₹ 4 per kg of brinjal by selling. On Monday he gets neither profit nor loss,.if he sold 68 kgs of tomato. How many kgs of brinjal did he sell ? Given profit per kg of tomato = ₹ 7
Loss per kg of brinjal = ₹ 4
Weight of tomatoes sold = 68 kgs
Let number of kgs of tomato be x kg.
Number of kgs of brinjals be y kg.
7x – 4y = 0
7 × 68 – 4y = 0
– 4y = – 7 × 68
y = $$\frac{-7 \times 68}{-4}$$
∴ Weight of brinjals sold = + 7 × 17
= + 119 kgs. Question 6.
In a test, each correct answer carry + 3 marks and incorrect answer – 1 mark, Sona attempted all the questions and scored + 20 marks though she got 10 correct answers.
(i) How many incorrect answers did she attempt ?
(ii) How many questions were given in the test ?
Given marks for every correct answer = + 3
Given marks for every incorrect answer = – 1
No. of correct answer questions = 10
Total marks gained on correct answer questions = 10 × (+3) = + 30

(i) No. of incorrect answer questions = x
Total marks gained on incorrect answer questions = y × (-1) = – y.
Marks scored by Sona
⇒ + 30 + (- y) = + 20
⇒ – y = + 20 – 30
⇒ – y = – 10
⇒ y = 10
∴ No. of incorrect answers Sona attempts =10

(ii) Number of questions given in the test = No. of correct answer questions + No. of incorrect answer questions.
= 10 + 10 = 20 questions. Question 7.
Write 5 pairs of integers (a, b) such that a ÷ b = -4.
(Ex : (12, – 3) because 12 + (-3) = -4) 