Integers

integers class 7 selina

Step by Step solutions of Concise Mathematics ICSE Class-7 Maths chapter 1- Integers by Selina is provided

Table of Contents

Exercise: 1-D

Q1: The sum of two integers is -15. If one of them is 9, find the other.

Step 1: Let the other number be x.
Step 2: According to the question:
x + 9 = -15
Step 3: Subtract 9 from both sides:
x = -15 – 9
x = -24
Answer: -24


Q2: The difference between integers x and -6 is -5. Find the values of x.

Option 1:
Step 1: According to the question:
x – (-6) = -5
Step 2: Simplify the expression:
x + 6 = -5
Step 3: Subtract 6 from both sides:
x = -5 – 6 = -11
Answer: -11
Option 2:
Step 1: According to the question:
(-6) – x = -5
Step 2: Simplify the expression:
-6 – x = -5
Step 3: Subtract 6 from both sides:
x = -6 + 5 = -1
Answer: -1


Q3: The sum of two integers is 28. If one integer is -45, find the other.

Step 1: Let the other integer be x.
x + (-45) = 28
Step 2: Add 45 to both sides:
x = 28 + 45 = 73
Answer: 73


Q4: The sum of two integers is -56. If one integer is 42, find the other.

Step 1: Let the other number be x.
x – 42 = -56
Step 2: Subtract 42 from both sides:
x = -56 + 42 = -14
Answer: -14


Q5: The difference between an integer x and (-9) is 6. Find all possible values of x.

Option 1:
Step 1: x – (-9) = 6
x + 9 = 6
Step 2: Subtract 9 from both sides:
x = 6 – 9 = -3
Answer: -3
Option 2:
Step 1: (-9) – x = 6
-9 – x = 6
Step 2: Subtract 9 from both sides:
x = (-9) – 6 = -15
Answer: -15


Q6: Write all the integers between -15 and 15, which are divisible by 2 and 3.

Step 1: We need numbers divisible by both 2 and 3 ⇒ LCM = 6
Step 2: Integers divisible by 6 between -15 and 15: -12, -6, 0, 6, 12
Answer: -12, -6, 0, 6, 12


Q7: Write all the integers between -5 and 5, which are divisible by 2 or 3.

Step 1: Integers between -5 and 5: -4, -3, -2, -1, 0, 1, 2, 3, 4
Step 2: Divisible by 2 or 3: -4, -3, -2, 0, 2, 3, 4
Answer: -4, -3, -2, 0, 2, 3, 4


Q8: Find the result of subtracting the sum of all integers between 20 and 30 from the sum of all integers from 20 to 30.

Step 1: Integers from 20 to 30 (inclusive): 20, 21, …, 30
Total = 11 numbers, Sum = (n/2)(first + last) = (11/2)(20 + 30) = (11/2)(50) = 275
Step 2: Integers *between* 20 and 30: 21 to 29
Sum = (9/2)(21 + 29) = (9/2)(50) = 225
Step 3: Required result = 275 – 225 = 50
Answer: 50


Q9: Add the product of (-13) and (-17) to the quotient of (-187) and 11.

Step 1: Product = (-13) × (-17) = 221
Step 2: Quotient = (-187) ÷ 11 = -17
Step 3: Required sum = 221 + (-17) = 204
Answer: 204


Q10: The product of two integers is -180. If one of them is 12, find the other.

Step 1: Let other number be x.
12 × x = -180
Step 2: x = -180 ÷ 12 = -15
Answer: -15


Q11:

i. A number changes from -20 to 30. What is the increase or decrease in the number?
Change = 30 – (-20) = 30 + 20 = 50
Answer: Increase of 50

ii. A number changes from 40 to -30. What is the increase or decrease in the number?
Change = -30 – 40 = -70
Answer: Decrease of 70

previous
next

Share the Post:

Leave a Comment

Your email address will not be published. Required fields are marked *

Related Posts​

  • Type casting in Java
    The process of converting the value of one data type to another data type is known as typecasting.
  • Identities
    Step by Step solutions of Test Yourself Concise Mathematics ICSE Class-8 Maths chapter 12- Identities by Selina is provided.

Join Our Newsletter

Name
Email
The form has been submitted successfully!
There has been some error while submitting the form. Please verify all form fields again.

Scroll to Top