SEBA SCERT Solutions for Class 7 Maths Chapter 1 Integers

SEBA SCERT Solutions for Class 7 Maths Chapter 1 Integers

 

SEBA SCERT Solutions for Class 7 Maths Chapter 1 Exercise 1.1 Integers in English and Assamese Medium modified and updated for academic year 2024-25. All the question-answers and solutions are revised on the basis of new syllabus and latest SEBA SCERT textbooks issued for curriculum 2024-25.

 


All the question answers with solutions are done according to latest SEBA SCERT Books for 2024-25.

 

SEBA SCERT Solutions for Class 7 Maths Chapter 1 Integers Exercise 1.1

 

Class: 7-

Mathematics

Chapter: 1

Exercise 1.1

Session:

SEBA 2024 - 25

Medium:

English, Assamese & Hindi

 

 

Class 7 Maths Solutions: All the solutions can be viewed by clicking on the Solutions Link below.

Integers
Solutions Link
Exercise 1.1  Click Here
Exercise 1.2  Click Here
Exercise 1.3  Click Here

follow us on

Instagram
Facebook
Subscribe Our YouTube Channel

 

 

SEBA SCERT Solutions for Class 7 Maths Chapter 1 Integers Exercise 1.1

 

 

Exercise -1.1

      1.    How many integers are there in between 5 and (-13)?

Solution:




Now, starting from 4 and counting down to -12, we have:

4, 3, 2, 1, 0, -1, -2, -3, -4, -5, -6, -7, -8, -9, -10, -11, -12

There are a total of 17 integers between 5 and -13.

 

      2.    Write greatest and smallest integers in between 13 and (-13)?

Solution:


The greatest integer between 13 and -13 is 12, and the smallest integer between them is -12.

 

      3.    Plot the following integers on number line

-6,4, -10,5, -1

 Solution:



 


      4.    Write 5 negative integers which are greater than -15.

Solution:

Here are five negative integers greater than -15:

 

-14, -13, -12, -11, -10

 

      5.    Mention if true or false-

(i)              Positive integers are called as Natural Number.

(ii)            All the integers are whole number.

(iii)         Number line is extended to infinity on the both sides of '0'

(iv)          '0' and negative integers form the collection of whole number.

(v)            If a + b = 0, then one of them is additive inverse of other and vice versa.

Solution:

(i)              True - Positive integers are indeed a subset of natural numbers.

(ii)            False

(iii)         True - The number line extends infinitely in both directions from zero.

(iv)          False - '0' and non-negative integers (including positive integers) form the collection of whole numbers. Negative integers are not included.

(v)            True - If 𝑎 + 𝑏 = 0 then 𝑏 is the additive inverse of 𝑎 and vice versa. This property is fundamental to the concept of additive inverses.

 

      6.    Write a pair of integers –

(i)              Whose sum is – 3            (ii) Whose difference is – 5

(iii)         Whose sum is 0               (iv) Whose difference is 2

Solution:

(i)              Integers whose sum is -3:

Let's call the integers x and y. So, x + y = -3. One possible pair of integers that satisfy this equation is (-5, 2), because -5 + 2 = -3.

 

(ii)            Integers whose difference is 5:

Again, let's call the integers x and y. So, x - y = 5. One possible pair of integers that satisfy this equation is (6, 1), because 6 - 1 = 5.

 

(iii)         Integers whose sum is 0:

For integers x and y, if x + y = 0, then x = -y. One possible pair of integers that satisfy this equation is (3, -3), because 3 + (-3) = 0.

 

(iv)          Integers whose difference is 0:

For integers x and y, if x - y = 0, then x = y. One possible pair of integers that satisfy this equation is (4, 4), because 4 - 4 = 0.

 

      7.    Write a pair of negative integers whose subtraction is 6.

Solution:

let's call the negative integers x and y. So, x - y = 6.

One possible pair of negative integers that satisfy this equation is (-3, -9), because (-3) - (-9) = -3 + 9 = 6.

 

      8.     Find the integers a and b such that (i) a + b is positive (ii) a ≠ b (iii) a – b = 0

Solution:

let's find integers a and b that satisfy the given conditions:

 

(i)              a + b is positive:

For a + b to be positive, both a and b cannot be negative.

 

(ii)             a ≠ b:

This condition ensures that the integers are different.

 

(iii)          a - b = 0:

This means that a and b are equal.

 

Combining these conditions:

 

We need two integers that are positive, not equal to each other, and equal when subtracted. One such pair of integers is (4, 4). Here, a + b = 4 + 4 = 8, which is positive, and a - b = 4 - 4 = 0. These integers meet all the given conditions.

 

 

      9.    Fill in the boxes

(i)              (–15) + (–4) = (–4) +          

(ii)                      + {(–7) + 8} = {5 + (–7)} +8

(iii)         (– 23) +       = – 23 = (–23) +         

(iv)          (–19) +            = (–27)

(v)            x + 12 = 0 then x =           

Solution:

Let's fill in the boxes:

(i)              (-15) + (-4) = (-4) + (-15)

(ii)            5 + {(–7) + 8} = {5 + (–7)} +8

(iii)         (-23) + 0 = -23 = (-23) + 0

(iv)          (-19) + (-8) = (-27)

(v)            x + 12 = 0 then x = -12

  

      10.A man moved 14 kilometers towards East from the position A. But another man moved 6 Kilometers towards West from the position A. What is the distance between them?

Solution:

Given,

A man moved 14 kilometers towards the East from position A, and the other moved 6 kilometers towards the West from the same position.

 

Since they moved in opposite directions, we can subtract their distances to find the total distance between them.

 

Distance between them = Distance traveled by the first man + Distance traveled by the second man

= 14 km + 6 km

= 20 km

 

So, the distance between them is 20 kilometers.

 

      11.A man has a deposited ₹ 35, and another man has a debt of ₹ 40. How much rich is first man compared to second man?

Solution:

Given,

First man's wealth = ₹ 35

Second man's debt = ₹ 40

 

To compare their wealth, we subtract the second man's debt from the first man's deposit:

 

Richness of first man compared to second man = Amount deposited by first man - Debt of second man

= ₹ 35 - ₹ 40

= -₹ 5

 

Since the result is negative, it means the first man is not richer than the second man; rather, the second man has a debt of ₹ 5 more than what the first man deposited. So, the first man is ₹ 5 poorer compared to the second man.

 

      12.On a certain Tuesday temperature of Guwahati at 5 am in the morning was 25°C. But temperature was increased by 8°C at 2 pm in the afternoon and at 10 pm the temperature was decreased by 3°C. On Wednesday at 12 noon again temperature was increased by 5°C.What was the temperature at 12 noon on Wednesday?

Solution:

Given,

 

Tuesday 5 am temperature = 25°C

Increased by 8°C at 2 pm

Decreased by 3°C at 10 pm

Increased by 5°C at 12 noon on Wednesday

Let's calculate the changes:

 

From Tuesday 5 am to Tuesday 2 pm: Increase of 8°C

Temperature at 2 pm = 25°C + 8°C = 33°C

Again, From Tuesday 2 pm to Tuesday 10 pm: Decrease of 3°C

Temperature at 10 pm = 33°C - 3°C = 30°C

Again, From Tuesday 10 pm to Wednesday 12 noon: Increase of 5°C

Temperature at 12 noon on Wednesday = 30°C + 5°C = 35°C

So, the temperature at 12 noon on Wednesday was 35°C.

 

      13. Anuradha had deposited ₹ 3200 in the bank and next day she had withdrawn ₹ 2,540. How much money left in the account of Anuradha after withdrawal?

Solution:

 

Anuradha Initial deposit = ₹ 3200

next day she Withdrawal = ₹ 2540

 

Amount left in the account of Anuradha = Initial deposit - Withdrawal

= ₹ 3200 - ₹ 2540

= ₹ 660

 

So, after the withdrawal, Anuradha has ₹ 660 left in her account.

 

     14.Sum of two numbers is –5. If one of the number is 18 then what is the other number?

Solution:

Let' the other number x.

 

Given,

One number = 18

Sum of the two numbers = -5

 

So, we can write the equation:

18 + x = -5

 

To find the value of x, we need to isolate it on one side of the equation,

x = -5 - 18

x = -23

 

So, the other number is -23.

 

      15.What should be added with –23 to get ‘0’?

Solution:

We need to subtract -23 from 0.

 

0 - (-23) = 0 + 23 = 23

 

So, to get 0, you should add 23 to -23.

 

      16.Sum of two integers is – 48. If one of the number is –20 then what is the other number?

Solution:

Let's call the other number x.

 

Given,

One number = -20

Sum of the two numbers = -48

 

So, we can write the equation.

-20 + x = -48

 

To find the value of x, we need to isolate it on one side of the equation.

x = -48 + 20

x = -28

 

So, the other number is -28.

 

 

      17.Evaluate using number line:

(i)              (+5) – (+3) (iii) (–6) – (+5)

(ii)            (+6) + (–5) (iv) (–8) + (–3)

Solution:

(i)              (+5) - (+3)

Start at positive 5 on the number line and move 3 units to the left (since it's subtraction):

 

+5−(+3) =+2

 

(ii)            (+6) + (-5)


Start at positive 6 on the number line and move 5 units to the left (since it's addition of a negative number):




+6+(−5)=+1

(iii)         (-6) - (+5):


Start at negative 6 on the number line and move 5 units to the right (since it's subtraction of a positive number):

 



-6 – 5= -11

(iv)          (-8) + (-3):

Start at negative 8 on the number line and move 3 units to the left (since it's addition of negative numbers):

 


−8+(−3) =−11

 

        18.Find whether of the statements are true or false:

(i)              (–6) + 23 + (–2) = (–2) + (–6) + 23

(ii)            (16–15) + (–7) = 16– {15+(–7)}

(iii)         Natural numbers are closed under subtraction.

(iv)          Of the two numbers 0 and – 670, – 670 is greater.

(v)            With respect to subtraction of integers commutative property and associative property do not hold.

Solution:

(i)              True. Addition is commutative, so the order of the terms doesn't affect the result.

 

(ii)            False.

 

(iii)         False. Natural numbers are not closed under subtraction. For example, subtracting a larger natural number from a smaller one may result in a negative integer, which is not a natural number.

 

(iv)          False

 

(v)            True


 

Digital Pipal Academy Verified By

Sudev Chandra Das(B.Sc. Mathematics)

 

 

More Resources For Class 10 Solutions in Assamese Medium




    More Resources For Class 9 Solutions in Assamese Medium

Previous Post Next Post