SEBA SCERT Solutions for Class 7
Maths Chapter 1 Integers Exercise 1.3
SEBA SCERT Solutions for Class 7 Maths Chapter 1 Exercise 1.3 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.
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SEBA SCERT Solutions for Class 7 Maths Chapter 1 Integers Exercise 1.3
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Class: 7- |
Mathematics |
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Chapter: 1 |
Exercise 1.3 |
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Session: |
SEBA 2024 - 25 |
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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 |
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SEBA SCERT Solutions for Class 7 Maths Chapter 1 Integers Exercise 1.3
Exercise -1.3
1. Find the quotient
(i) 15÷(−5)
(ii) (−60) ÷10
(iii) (−54)÷( −6)
(iv) 0÷(−15)
(v) (−61) ÷{(−60)+( −1)}
(vi) {(−72) ÷(−6)} ÷(−3)
Solution
(i) 15÷(−5)
`=\frac{15}{(-5)}`
`=-3`
(ii) (−60) ÷10
`=\frac{(-60)}{10}`
`=-6`
(iii) (−54)÷( −6)
`=\frac{(-54)}{(-6)}`
`=9`
(iv) 0÷(−15)
` =\frac0{(-15)}`
`=0`
(v) (−61) ÷{(−60)+( −1)}
`=(-61)\div(-61)`
`=\frac{(-61)}{(-61)}`
`=1`
(vi) {(−72) ÷(−6)} ÷(−3)
`=\frac{(-72)}{(-6)}\div(-3)`
`=12\div(-3)`
`=\frac{12}{(-3)}`
`=-4`
2. Fill in the blanks
(i) (−600) ÷25 =________
(ii) {(−4) ×18} ÷ = 12
(iii) ÷(5−6) = −20
(iv) (−123) ÷(−1) = ________
Solution
(i) (−600) ÷25 =__−24__
(ii) {(−4) ×18} ÷ −6 = 12
(iii) 20 ÷(5−6) = −20
(iv) (−123) ÷(−1) = __123
3. (i) If a ÷ (−7) = 8, then find the value of integer ‘a’.
(ii) If 125 ÷ b = −5, then find the value of integer ‘b’.
Solution
a ÷ (−7) = 8
`\Rightarrow\frac a{(-7)}=8`
Multiplying both sides by −7
`\Rightarrow\frac a{(-7)}\times(-7)=8\times(-7)`
`\Rightarrow a=-56`
So, the value of 𝑎 is −56.
(ii) 125 ÷ b = −5
`\Rightarrow\frac{125}b=-5`
Multiplying both sides by 𝑏
`\Rightarrow\frac{125}b\times b=-5\times b`
`\Rightarrow125=-5b`
Now, dividing both sides by −5
`\Rightarrow\frac{125}{-5}=\frac{-5b}{-5}`
`\Rightarrow-25=b`
`\Rightarrow b=-25`
So, the value of b is −25
4. Write three pairs of integers a,b such that a÷b = −5
Solution
Pair (i): a = −10, b = 2
`\Rightarrow\frac{-10}2=-5`
Pair (ii): a = 15, b = −3
`\Rightarrow\frac{15}{-3}=-5`
Pair (iii): a = −25, b = 5
`\Rightarrow\frac{-25}5=-5`
Thus, the three pairs of integers a and b are
(i) (−10,2)
(ii) (15, −3)
(iii) (−25,5)
5. In a class test of a school 20 questions were given. For each correct answer 5 marks is awarded and for each wrong answer (-2) is awarded.
(i) A student answered all the questions. But her 10 answers were correct. How much marks did she score?
(ii) Other student could answer only 5 answers correctly. What was the score by the student?
Solution
(i) Given,
Total questions = 20
Correct Answers = 10
For each correct answer, 5 marks are awarded,
∴ 10×5 = 50
For each incorrect answer, -2 marks are awarded,
∴ 10×(−2)=−20
∴Total marks scored = 50+(−20)
= 30
(ii) Given,
Total questions = 20
Correct Answers = 5
Incorrect Answers = 20−5 = 15
For each correct answer, 5 marks are awarded
5×5=25
For each incorrect answer, -2 marks are awarded
15×(−2)=−30
Total marks scored = 25+(−30)
= −5
6. In an examination for each correct answer 5 marks is awarded and for each wrong answer (-2) is awarded.
(i) Sumon answered all the questions. But his 16 answers were correct and obtained 64.
(ii) Jaya answered all the questions. But she could answer only 6 questions correctly and obtained (-6). How many questions were answered incorrectly by them.
Solution
(i)Given
Suman obtained = 64 marks
His total correct answers = 16
Marks awarded for correct answers = 16×5=80
Let x be the number of incorrect answers
We know the total marks Sumon obtained include the penalty for incorrect answers
`80+(-2\times x)=64`
`\Rightarrow80-2x=64`
`\Rightarrow-2x=64-80`
`\Rightarrow-2x=-16`
`\Rightarrow x=\frac{-16}{-2}`
`\Rightarrow x=8`
So, Sumon answered 8 questions incorrectly
(ii) Jaya obtained = −6
Her Total correct answers = 6
Marks awarded for correct answers = 6×5=30
Let x be the number of incorrect answers.
We know the total marks Jaya obtained include the penalty for incorrect answers
`30+(-2\times x)=-6`
`\Rightarrow30-2x=-6`
`\Rightarrow-2x=-6-30`
`\Rightarrow-2x=-36`
`\Rightarrow x=\frac{-36}{-2}`
`\Rightarrow x=18`
So, Jaya answered 18 questions incorrectly
7. A rubber company makes a profit of 15 per bag of rubber. Loss incured on every waste bag of rubber is of ₹ 8.
(i) The Company sold 1500 bags of good rubber and 500 bags of waste rubber. How much profit or loss incurred by the company?
(ii) If a the company sold 750 bags of waste rubber then how many bags of good rubber are to be sold so that company incured neither profit nor loss?
Solution
(i) Given,
Profit per bag of good rubber = ₹15
Loss per bag of waste rubber = ₹8
Total profit from good rubber bags,
1500×15= ₹22500
Total loss from waste rubber bags,
500×8= ₹4000
Net profit = Total Profit−Total loss
= 22500−4000
= ₹18500
So, the company incurred a profit of ₹18,500.
(ii) Given,
Number of waste rubber bags sold = 750
Loss per waste rubber bag= ₹8
Total loss from waste rubber bags = 750×8
= 6000₹
Since The total profit from good rubber bags must equal the total loss from waste rubber bags.
Let x be the number of good rubber bags needed to be sold.
Profit per good rubber bag = ₹15
Total profit from good rubber bags.
`15x=6000`
`\Rightarrow x=\frac{6000}{15}`
`\Rightarrow x=400`
So, the company needs to sell 400 bags of good rubber to incur neither profit nor loss