# CBSE Important Questions Class 7 Maths Chapter 1

**Important Questions Class 7 Mathematics Chapter 1 – Integers**

Mathematics is an important subject that we need in our daily life too. Students must solve questions to clear their concepts and boost their confidence. The first chapter of Class 7 Mathematics under CBSE curriculum is integers.

Students have learned integers in their previous class. In this chapter, they will learn how to put the integers on the number line, their properties, and the addition and multiplication of integers. It is a very important chapter. Students must practice the textbook exercise and questions from other sources to build their concepts.

Extramarks is a leading company that provides a wide range of study materials related to CBSE and NCERT. Our experts have made the Important Questions Class 7 Mathematics Chapter 1 to help students in regular practice. They collected the questions from different sources such as the textbook exercises, CBSE sample papers, CBSE past years’ question papers and important reference books. They have solved the questions too. Hence, the question series will help students increase their exam marks.

Extramarks is a leading company that helps students by providing all the important study materials related to CBSE and NCERT. You may register on our official website and download these study materials. You will find the CBSE syllabus, NCERT textbooks, CBSE past years’ question papers, CBSE sample papers, CBSE revision notes, CBSE extra questions, NCERT solutions, NCERT important questions, vital formulas and many more.

**Important Questions Class 7 Mathematics Chapter 1 – With Solutions**

The experts of Extramarks have made this question series so that students can solve the questions daily. They collected the questions from the textbook exercises, CBSE sample papers and important reference books. They have included a few questions from the past years’ question papers so that students may have an idea regarding questions in exams. Experienced professionals have further checked the answers to ensure the best quality of the content. Thus, the Important Questions Class 7 Mathematics Chapter 1 will help students to score better in exams. The questions are-

**Question 1.** Following number line given below shows the temperature present in degree celsius at different places on a particular day.

Image Source: Internet / NCERT Textbook

(i) Observe the number line and write down the temperature of the places marked on it.

**Answer 1:-**

By observing the above number line, we can find out the temperature of the cities as follows,

The temperature in the city of Lahulspiti is -8°C.

The temperature in the city of Srinagar is -2°C

The temperature in the city of Shimla is 5°C.

The temperature in the city of Ooty is 14°C.

The temperature in the city of Bengaluru is 22°C.

(ii) What is the temperature difference between the hottest and the coldest places among the cities stated above?

Answer:-

From the above number line, we can observe that,

The temperature at the given hottest place, that is, Bengaluru, is 22°C.

The temperature at the given coldest place, that is, Lahulspiti, is -8°C

The temperature difference between the hottest and the coldest place is given as = 22°C – (-8°C)

= 22°C + 8°C

= 30° Celsius

Hence, the total temperature difference between the hottest and the coldest place is 30oC.

(iii) What is the temperature difference between the cities of Lahulspiti and Srinagar?

Answer:-

From the above-given number line,

The temperature in the city of Lahulspiti is -8°C.

The temperature in the city of Srinagar is -2°C

∴The temperature difference between the cities Lahulspiti and Srinagar is = -2oC – (8oC)

= – 2°C + 8°C

= 6°C

(iv) Can we say that the temperature of Srinagar and Shimla taken together is less than the temperature present at Shimla? Is it also less than the temperature present at Srinagar?

Answer:-

From the above-given number line,

The temperature in the city of Srinagar =-2°C

The temperature in the city of Shimla = 5°C

The temperature of the cities Srinagar and Shimla taken together becomes = – 2°C + 5°C

= 3° degree C

5°C > 3°C

Hence, the temperature of the cities Srinagar and Shimla taken together is indeed less than the temperature present at Shimla.

Then,

3° > -2°

And No, the temperature of the cities Srinagar and Shimla taken together is not less than the temperature of the city Srinagar.

**Question 2.** Mohan deposits ₹ 2,000 in his bank account and then withdraws ₹ 1,642 from it the following day. Now, if the withdrawal of the amount from the account is represented by a negative integer, then how will you represent the total amount deposited? Also, Find the balance in Mohan’s account after the withdrawal.

**Answer 2**

Withdrawal of these amounts from the account is represented by a negative integer.

Then, the deposit of the amount to the account is represented by a positive integer.

From the above question,

The total amount that is deposited in the bank account by the Mohan = ₹ 2000

The total amount that is withdrawn from the bank account by the Mohan is = – ₹ 1642

Final Balance in Mohan’s account after the withdrawal = amount deposited + amount is withdrawn

= ₹ 2000 + (-₹ 1642)

= ₹ 2000 – ₹ 1642

= ₹ 358

Hence, the total balance in Mohan’s account after the withdrawal is ₹ 358

**Question 3.** In the following quiz, positive marks are given for every correct answer and negative marks are given for each incorrect answer. If Jack’s scores in the quiz for five successive rounds were 25, – 5, – 10, 10, and 15 so, what was his total at the end?

**Answer 3:-**

From the above question,

Jack’s scores in the five successive rounds are 25, -5, -10, 15 and 10

Hence, Their total score of Jack at the end will be = 25 + (-5) + (-10) + 15 + 10

= 25 – 5 – 10 + 15 + 10

= 50 – 15

= 35 marks

∴ Now, Jack’s total score at the end is 35.

**Question 4. **In the city of Srinagar, temperature was – 5°C on Monday, and then it dropped by two °C on Tuesday. What was the temperature of the city of Srinagar on Tuesday? On Wednesday, the temperature rose by 4°C. What was the temperature on this day?

**Answer 4:-**

From the above question,

The temperature on Monday at Srinagar is = -5C

The temperature on Tuesday at the city of Srinagar is dropped by 2C = Temperature on Monday – 2C

= -5C – 2C

= -7 celsius

The temperature on Wednesday at the city Srinagar rose by 4C = Temperature on Tuesday + 4C.

= -7C + 4C

= -3 celsius

Thus, the temperature on days Tuesday and Wednesday was found to be -7C and -3C, respectively.

**Question 5.** In a magic square, every row, column and diagonal has the same sum. Check which of these following is a magic square.

Image Source: Internet / NCERT Textbook

**Answer 5:-**

Firstly we consider the square (i)

Now By adding these numbers in each of the rows, we get,

= 5 + (- 1) + (- 4) equals to 5 – 1 – 4 = 5 – 5 = 0

= -5 + (-2) + 7 equals to – 5 – 2 + 7 = -7 + 7 = 0

= 0 + 3 + (-3) = 3 – 3 = 0

By adding these numbers in every column we receive,

= 5 + (- 5) + 0 is equal to 5 – 5 = 0

= (-1) + (-2) + 3 equals to -1 – 2 + 3 = -3 + 3 = 0

= -4 + 7 + (-3) equals to -4 + 7 – 3 = -7 + 7 = 0

By adding these numbers in diagonals, we receive,

= 5 + (-2) + (-3) is equal to 5 – 2 – 3 = 5 – 5 = 0

= -4 + (-2) + 0 is equal to – 4 – 2 = -6

Because the sum of one diagonal is not always equal to zero,

Hence, (i) is not a magic square.

Now, we should consider the square (ii)

By adding these numbers to each rows we receive,

= 1 + (-10) + 0 is equal to 1 – 10 + 0 = -9

= (-4) + (-3) + (-2) equal to -4 – 3 – 2 = -9

= (-6) + 4 + (-7) becomes equal to -6 + 4 – 7 = -13 + 4 = -9

By adding these numbers in each column we receive,

= 1 + (-4) + (-6) equals to 1 – 4 – 6 = 1 – 10 = -9

= (-10) + (-3) + 4 equals to -10 – 3 + 4 = -13 + 4

= 0 + (-2) + (-7) equals to 0 – 2 – 7 = -9

By adding these numbers in diagonals, we receive,

= 1 + (-3) + (-7) equals to 1 – 3 – 7 = 1 – 10 = -9

= 0 + (-3) + (-6) equal to 0 – 3 – 6 = -9

Hence This (ii) square is a magic square because the sum of each row, each column and the diagonal becomes equal to -9 (negative).

**Question 6.** Verify a – (– b) is equal to a + b for the following values of alphabets a and b.

(i) a = 21, b = 18

**Answer 6:-**

From the above question,

a = 21 and b = 18

So To verify a – (- b) is equal to a + b

Let us take the Left Hand Side (LHS) = a – (- b)

= 21 – (- 18)

= 21 + 18

= 39

Now, lets take Right Hand Side (RHS) = a + b

= 21 + 18

= 39

By comparing both the LHS and the RHS.

LHS = RHS

39 = 39

Hence, the value of a and b are verified.

(ii) a = 118, b = 125

Answer:-

From the above question,

a = 118 and b = 125

To verify this a – (- b) = a + b

Let us take the Left Hand Side (LHS) = a – (- b)

= 118 – (- 125)

= 118 + 125

= 243

Now, take the Right Hand Side (RHS) = a + b

= 118 + 125

= 243

By comparing both the LHS and the RHS

LHS = RHS

243 = 243

Hence, the values of a and b are verified.

(iii) a = 75, b = 84

Answer:-

From the above question,

a = 75 and b = 84

To verify that the a – (- b) = a + b

Let us take the Left Hand Side (LHS) = a – (- b)

= 75 – (- 84)

= 75 + 84

= 159

Now, the Right Hand Side (RHS) = a + b

= 75 + 84

= 159

By comparing both LHS and RHS, we find that,

LHS = RHS

159 = 159

Hence, the value of a and b is verified as.

(iv) a = 28, b = 11

Answer:-

From the above question,

a = 28 and b = 11

To verify that a – (- b) = a + b

Let us now take Left Hand Side (LHS) = a – (- b)

= 28 – (- 11)

= 28 + 11

= 39

Now, Right Hand Side (RHS) = a + b

= 28 + 11

= 39

By comparing both the LHS and the RHS

LHS = RHS

39 = 39

Hence, the value of a and b are verified.

**Question 7**. A water tank has stepped inside it. A monkey is sitting on the utter topmost step (which is the first step). The water level is present at the ninth step.

(i) He jumps three steps down the stairs and then successively jumps back two steps upwards. In how many jumps will the Monkey reach the following water level?

**Answer 7:-**

Let us consider the steps moved down are represented by a positive integer, and then the steps moved up are represented by a negative integer.

Initially, the Monkey is sitting on the topmost step, which is the first step.

In the 1st jump monkey will be at the step = 1 + 3 = 4 steps

In the 2nd jump monkey will be at the step = 4 + (-2) = 4 – 2 = 2 steps

In the 3rd jump monkey will be at the step = 2 + 3 = 5 steps

In the 4th jump monkey will be at the step = 5 + (-2) = 5 – 2 = 3 steps

In the 5th jump monkey will be at the step = 3 + 3 = 6 steps

In the 6th jump monkey will be at the step = 6 + (-2) = 6 – 2 = 4 steps

In the 7th jump monkey will be at the step = 4 + 3 = 7 steps

In the 8th jump monkey will be at the step = 7 + (-2) = 7 – 2 = 5 steps

In the 9th jump monkey will be at the step = 5 + 3 = 8 steps

In the 10th jump monkey will be at the step = 8 + (-2) = 8 – 2 = 6 steps

In the 11th jump monkey will be at the step = 6 + 3 = 9 steps

∴Monkey took a total of 11 jumps (i.e., 9th step) to reach the water level.

(ii) After drinking water, the Monkey wants to go back. For this, the Monkey jumps four steps up and then successively jumps back two steps down in his every move. In how many total jumps will he reach back to the top step?

Answer:-

Let us consider the steps moved down are represented by the positive integers, and then the steps moved up are represented by the negative integers.

Initially, the Monkey is sitting on the ninth step, i.e., at the water level.

In the 1st jump monkey will be at the step = 9 + (-4) = 9 – 4 = 5 steps

In the 2nd jump monkey will be at the step = 5 + 2 = 7 steps

In the 3rd jump monkey will be at the step = 7 + (-4) = 7 – 4 = 3 steps

In the 4th jump monkey will be at the step = 3 + 2 = 5 steps

In the 5th jump monkey will be at the step = 5 + (-4) = 5 – 4 = 1 step

∴ Hence the Monkey took five jumps to reach back to the top step, i.e., the first step.

**Question 8.** Fill in the blanks to make the following statements true:

(i) (–5) + (– 8) = (– 8) + (…………)

**Answer 8:-**

Let us assume that the missing integer is x,

So,

= (–5) + (– 8) which equals to (– 8) + (x)

= – 5 – 8 = – 8 + x

= – 13 = – 8 + x

By sending – 8 from the RHS to the LHS, it becomes 8,

= – 13 + 8 = x

= x = – 5

Now substitute the x value in the place of the blank place present,

(–5) + (– 8) = (– 8) + (- 5) … [This following equation is present in the form of the Commutative law of Addition]

(ii) –53 + ………… = –53

Answer:-

Let us assume that the missing integer is x,

So,

= –53 + x = –53

By sending – 53 from the LHS to the RHS, it becomes 53,

= x = -53 + 53

= x = 0

Now substitute the following x value in the blank place,

= –53 + 0 = –53 … [This equation is present in the form of Closure property of Addition]

(iii) 17 + ………… = 0

Answer:-

Let us assume that the missing integer is x,

So,

= 17 + x = 0

By sending 17 from the LHS to the RHS, it becomes -17,

= x = 0 – 17

= x = – 17

Now substitute this x value in the blank place,

= 17 + (-17) = 0 … [This equation is present in the form of Closure property of Addition]

= 17 – 17 = 0

(iv) [13 + (– 12)] + (…………) = 13 + [(–12) + (–7)]

Answer:-

Let us assume that the missing integer is x,

So,

= [13 + (– 12)] + (x) = 13 + [(–12) + (–7)]

= [13 – 12] + (x) = 13 + [–12 –7]

= [1] + (x) = 13 + [-19]

= 1 + (x) = 13 – 19

= 1 + (x) = -6

By sending one from the LHS to the RHS, it becomes -1,

= x = -6 – 1

= x = -7

Now substitute the following x value in the blank place value,

= [13 + (– 12)] + (-7) equals to 13 + [(–12) + (–7)] … [This equation is present in the form of the Associative Property of Addition]

(v) (– 4) + [15 + (–3)] equals to [– 4 + 15] +…………

Answer:-

Let us assume that the missing integer is x,

So,

= (– 4) + [15 + (–3)] is equal to [– 4 + 15] + x

= (– 4) + [15 – 3)] equals to [– 4 + 15] + x

= (-4) + [12] = [11] + x

= 8 = 11 + x

Now, By sending 11 from the RHS to the LHS, it becomes -11,

= 8 – 11 = x

= x = -3

Now substitute the x value in the place of the blank place,

= (– 4) + [15 + (–3)] equals to [– 4 + 15] + -3 … [The following equation is in the form of the Associative property of the Addition]

**Question 9.** Find the product using the suitable properties:

(i) 26 × (– 48) + (– 48) × (–36)

**Answer 9:-**

This given equation is in the form of the Distributive law of the Multiplication property over Addition.

= a × (b + c) becomes equal to (a × b) + (a × c)

Let, a = -48, b = 26, c = -36

So,

= 26 × (– 48) + (– 48) × (–36)

= -48 × (26 + (-36)

= -48 × (26 – 36)

= -48 × (-10)

= 480 … [∵ (- × – = +)

(ii) 8 × 53 × (–125)

Answer:-

The given equation is present in the form of the Commutative law of Multiplication.

= a × b = b × a

Then,

= 8 × [53 × (-125)]

= 8 × [(-125) × 53]

= [8 × (-125)] × 53

= [-1000] × 53

= – 53000

(iii) 15 × (–25) × (– 4) × (–10)

Answer:-

This given equation is in the form of the Commutative law of the Multiplication property.

= a × b = b × a

So,

= 15 × [(–25) × (– 4)] × (–10)

= 15 × [100] × (–10)

= 15 × [-1000]

= – 15000

(iv) (– 41) × 102

Answer:-

This given equation is in the form of a Distributive law of the Multiplication property over Addition.

= a × (b + c) = (a × b) + (a × c)

= (-41) × (100 + 2)

= (-41) × 100 + (-41) × 2

= – 4100 – 82

= – 4182

(v) 625 × (–35) + (– 625) × 65

Answer:-

This given equation is in the form of the Distributive law of Multiplication over Addition.

= a × (b + c) = (a × b) + (a × c)

= 625 × [(-35) + (-65)]

= 625 × [-100]

= – 62500

**Question 10. **A certain freezing process requires that the room temperature be lowered from 40°C at the rate of 5°C every hour. What will be the final room temperature 10 hours after the actual process begins?

**Answer 10:-**

From the above question, it is given that

Let us take the lowered temperature as a negative integer,

Initial temperature will be= 40oC

Change in temperature per hour is = -5oC

Change in temperature after 10 hours will be = (-5) × 10 = -50oC

∴The final room temperature after the 10 hours of freezing process = 40oC + (-50oC)

= -10oC

**Question 11.** In a class test containing about ten questions, five marks are awarded for each correct answer and (–2) marks are awarded for every incorrect answer and 0 for questions which are not attempted.

(i) Mohan gets four correct answers and six incorrect answers on his test. What is his total score?

Answer:-

From the above question,

Marks awarded for one correct answer is = 5

Hence,

The total marks awarded for his four correct answers are = four × 5 = 20 marks.

Marks awarded for 1 wrong answer = -2 (negative)

Hence,

Total marks awarded for 6 wrong answers is = 6 × -2 = -12

∴Total score obtained by Mohan = 20 + (-12)

= 20 – 12

= 8

(ii) Reshma gets five correct answers and similarly five incorrect answers; what is her total score?

Answer:-

From the above question,

Marks awarded for one correct answer is = 5

Hence,

Total marks awarded for 5 correct answer becomes = 5 × 5 = 25

Marks awarded for one wrong answer is = -2

Hence,

Total marks awarded for 5 wrong answer becomes = 5 × -2 = -10

∴Total score obtained by Reshma is = 25 + (-10)

= 25 – 10

= 15

(iii) Heena gets two correct answers and five incorrect answers out of the seven questions she attempts. What is her final score?

Answer:-

From the above question,

Marks awarded for one correct answer is = 5

Hence,

Total marks awarded for 2 correct answer is = 2 × 5 = 10

Marks awarded for one wrong answer is = -2

Hence,

Total marks awarded for 5 wrong answer becomes = 5 × -2 = -10

Marks awarded for the questions which are not attempted is = 0

∴Total score obtained by Heena is = 10 + (-10)

= 10 – 10

= 0

**Question 12.** A cement company earns a profit of around ₹ 8 per bag of white cement that is sold and simultaneously a loss of ₹ 5 per bag of grey cement that is sold.

(i) The company sells 3,000 bags of white cement and 5,000 bags of grey cement in a month. What is its profit or loss?

Answer:-

We denote profit by a positive integer and loss by a negative integer,

So From the above question,

The Cement company earns a profit on selling one bag of white cement = ₹ 8 per bag.

So,

The cement company earns a total profit on selling 3000 bags of white cement = 3000 × ₹ 8

= ₹ 24000

And also the,

Loss on selling one bag of grey cement is = – ₹ 5 per bag.

Hence,

Loss on selling the 5000 bags of the grey cement = 5000 × – ₹ 5

= – ₹ 25000

Total loss or profit earned by these cement companies is = profit + loss.

= 24000 + (-25000)

= – ₹1000

Hence, a loss of ₹ 1000 will be incurred by the company.

(ii) What is the number of white cement bags that must sell to have neither a profit nor loss if the total number of grey bags sold is 6,400 bags?

Answer:-

We denote the profit as a positive integer and the loss as a negative integer,

From the above question,

The cement company earns the profit on selling one bag of white cement as = ₹ 8 per bag.

Now Let the number of white cement bags present be x.

Then,

The cement company earns a profit on selling these x bags of white cement as = (x) × ₹ 8

= ₹ 8x

Loss on selling one bag of grey cement becomes = – ₹ 5 per bag.

Then,

Loss on selling 6400 bags of grey cement becomes = 6400 × – ₹ 5

= – ₹ 32000

According to the above question,

Company to have neither profit nor loss, must sell,

= Profit + loss = 0

= 8x + (-32000) =0

By sending -32000 from the LHS to the RHS, it becomes 32000

= 8x = 32000

= x = 32000/8

= x = 4000

Hence, the 4000 bags of white cement should sell to have neither profit nor loss.

**Question 13. **Evaluate each of the following:

(i) (–30) ÷ 10

Answer:-

= (–30) ÷ 10

= – 3

When we divide the negative integer by a positive integer, we first divide them as whole numbers and then put the minus sign (-) before the quotient.

(ii) 50 ÷ (–5)

Answer:-

= (50) ÷ (-5)

= – 10

When we divide the positive integer by a negative integer, we first divide them as whole numbers and then apply the minus sign (-) before the quotient.

(iii) (–36) ÷ (–9)

Answer:-

= (-36) ÷ (-9)

= 4

When we divide the negative integer by a similar negative integer, we first divide these as whole numbers and then put the positive sign (+) before the quotient.

(iv) (– 49) ÷ (49)

Answer:-

= (–49) ÷ 49

= – 1

When we divide the negative integer by a positive integer, we first divide these as whole numbers and then put the minus sign (-) before the quotient.

(e) 13 ÷ [(–2) + 1]

Answer:-

= 13 ÷ [(–2) + 1]

= 13 ÷ (-1)

= – 13

When we divide the positive integer by a negative integer, we first divide these as whole numbers and then put the minus sign (-) before the quotient.

(f) 0 ÷ (–12)

Answer:-

= 0 ÷ (-12)

= 0

When we divide zero by a negative integer, it gives zero.

(g) (–31) ÷ [(–30) + (–1)]

Answers:-

= (–31) ÷ [(–30) + (–1)]

= (-31) ÷ [-30 – 1]

= (-31) ÷ (-31)

= 1

When we divide the negative integer by a negative integer, we first divide these as whole numbers and then put the positive sign (+) before the quotient.

(h) [(–36) ÷ 12] ÷ 3

Answer:-

First, we have to solve these integers within the bracket,

= [(–36) ÷ 12]

= (–36) ÷ 12

= – 3

Then,

= (-3) ÷ 3

= -1

When we divide a negative integer by a positive integer, we first divide them as whole numbers and then put the minus sign (-) before the quotient.

(i) [(– 6) + 5)] ÷ [(–2) + 1]

Answer:-

The given question can be written as,

= [-1] ÷ [-1]

= 1

When we divide the negative integer by a negative integer, we first divide these as whole numbers and then put the positive sign (+) before the quotient.

**Question 14. **Verify that a ÷ (b + c) is not equal to (a ÷ b) + (a ÷ c) for each of the following symbols of a, b and c.

(i) a = 12, b = – 4, c = 2

Answer:-

From the above question, a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c)

Given, a = 12, b = – 4 (negative), c = 2

Now, consider that the LHS = a ÷ (b + c)

= 12 ÷ (-4 + 2)

= 12 ÷ (-2)

= -6

When we divide a following positive integer by any of the negative integers, we first divide them as a whole number and then put the minus sign (-) before their quotient.

Then, consider that the RHS is equal to = (a ÷ b) + (a ÷ c)

= (12 ÷ (-4)) + (12 ÷ 2)

= (-3) + (6)

= 3

By comparing the LHS and RHS, we get,

= -6 ≠ 3

= LHS ≠ RHS

Hence, the given values have been verified.

(ii) a = (–10), b = 1, c = 1

Answers:-

From the above question, a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c)

Given, a = (-10), b = 1, c = 1

Now, consider that the LHS = a ÷ (b + c)

= (-10) ÷ (1 + 1)

= (-10) ÷ (2)

= -5

When we divide a negative integer by any other positive integer, we first divide them as a whole number and then put the minus sign (-) before the quotient.

Then, consider RHS = (a ÷ b) + (a ÷ c)

= ((-10) ÷ (1)) + ((-10) ÷ 1)

= (-10) + (-10)

= -10 – 10

= -20

By comparing LHS and RHS

= -5 ≠ -20

= LHS ≠ RHS

Hence, the given values are verified.

Question. Fill in the following blanks:

(a) 369 ÷ _____ = 369

Answer:-

Let us assume that the missing integer is x,

Now,

= 369 ÷ x = 369

= x = (369/369)

= x = 1

Hence, put the valve of x in the blank place.

= 369 ÷ 1 = 369

(b) (–75) ÷ _____ = –1

Answer:-

Let us assume that the missing integer is x,

Hence,

= (-75) ÷ x = -1

= x = (-75/-1)

= x = 75

Now, put the above valve of x in the blank place.

= (-75) ÷ 75 = -1

(c) (–206) ÷ _____ = 1

Answer:-

Let us assume that the missing integer is x,

So,

= (-206) ÷ x = 1

= x = (-206/1)

= x = -206

Now, put the above valve of x in the blank place.

= (-206) ÷ (-206) = 1

(d) – 87 ÷ _____ = 87

Answer:-

Let us assume that the missing integer is x,

So,

= (-87) ÷ x = 87

= x = (-87)/87

= x = -1

Now, put the above valve of x in the blank place.

= (-87) ÷ (-1) = 87

(e) _____ ÷ 1 = – 87

Answer:-

Let us assume that the missing integer is x,

Now,

= (x) ÷ 1 = -87

= x = (-87) × 1

= x = -87

So, put the valve of x in the blank.

= (-87) ÷ 1 = -87

(f) _____ ÷ 48 = –1

Answer:-

Let us assume that the missing integer is x,

So,

= (x) ÷ 48 = -1

= x = (-1) × 48

= x = -48

Now, put the above valve of x in the following blank.

= (-48) ÷ 48 = -1

**Question 15. **The temperature at 12 noon was 10 degrees C above zero. If it decreases at the rate of 2C per hour until midnight, at what time would the temperature be eight °C below zero? Also, What would be the temperature at midnight?

Answer:-

From the above question, it is given that,

The temperature at the beginning, which is, at 12 noon, is = 10C

The rate of change of temperature becomes = – 2C per hour.

Then,

Temperature present at 1 PM = 10 + (-2) = 10 – 2 = 8° C

Temperature present at 2 PM = 8 + (-2) = 8 – 2 = 6° C

Temperature present at 3 PM = 6 + (-2) = 6 – 2 = 4°C

Temperature present at 4 PM = 4 + (-2) = 4 – 2 = 2°C

Temperature present at 5 PM = 2 + (-2) = 2 – 2 = 0°C

Temperature present at 6 PM = 0 + (-2) = 0 – 2 = -2°C

Temperature present at 7 PM = -2 + (-2) = -2 -2 = -4°C

Temperature present at 8 PM = -4 + (-2) = -4 – 2 = -6°C

Temperature present at 9 PM = -6 + (-2) = -6 – 2 = -8°C

∴At 9 PM, the temperature will be 8° C below zero.

Then,

The temperature at mid-night which is at 12 AM

Change in the temperature in every 12 hours = -2°C × 12 = – 24°C

So, at midnight the temperature will be = 10 + (-24)

= – 14°C

At midnight the temperature will be 14°C below 0.

**Question 16. **In the following class test, (+ 3) marks are given for every correct answer, (–2) marks are given for every the incorrect answer and no marks are given for not attempting any question.

(i) Radhika scored 20 marks. If she has got around 12 correct answers, then how many questions has she attempted that are incorrect?

(ii) Mohini scores –5 (negative) marks on this test, and though she has got seven correct answers. How many questions has she attempted incorrectly?

Answer:-

From the above question,

Marks awarded for 1 correct answer is = + 3

Marks awarded for one wrong answer is = -2

(i) Radhika, in the test, scored 20 marks

So,

Total marks awarded for every 12 correct answers is = 12 × 3 = 36

Marks awarded for every incorrect answer = Total score – Total marks awarded for 12 correct questions.

Answers

= 20 – 36

= – 16

So, the number of incorrect answers done by Radhika = (-16) ÷ (-2)

= 8

(ii) Mohini scored a total of -5 marks

Then,

Total marks awarded for her 7 correct answers is = 7 × 3 = 21

Marks awarded for her incorrect answers = Total score – Total marks awarded for the 12 correct answers.

= – 5 – 21

= – 26

Hence, the number of incorrect answers made by Mohini = (-26) ÷ (-2)

= 13.

**Question 17.** An elevator descends down into a mine shaft at the rate of 6 m per min. If the descent starts from 10 meters above the ground level, how much time will it take to reach – 350 m?

Answers :-

From the above question,

The initial height of the elevator becomes = 10 m

Final depth of elevator is = – 350 m … [the distance descended is denoted by a negative integer]

The total distance to descend by the elevator becomes = (-350) – (10)

= – 360 m

Hence,

Time taken by the elevator to descend (negative) -6 m is = 1 min

So, the total time taken by the elevator to descend – 360 m becomes = (-360) ÷ (-60)

= 60 minutes

### = 1 hour

**Benefits of Solving Important Questions Class 7 Mathematics Chapter 1**

Practice is the key to success. The practice habit is very important for students because it will help them in many ways. It will help them to score better in exams. Apart from this, practice will clear doubts, generate interest in the subject matter, and strengthen the concepts. Thus, students must practice sums regularly to improve their exam preparation. The Important Questions Class 7 Mathematics Chapter 1 will help students in many ways. These are-

- The experts have collated the questions from various sources. They have accumulated the questions from the textbook exercises, CBSE sample papers, CBSE past years’ question papers and important reference books. Thus, students will find all the vital questions In this article, and they can solve the questions regularly. Thus, students don’t have to search for questions in different books, but they will find them here. Thus, Chapter 1, Class 7 Mathematics Important Questions includes all the important concepts.
- The experts have not only collated the questions but also provided the solutions. They have given a step-by-step solution for each chapter to help students. Experienced professionals have further checked the answers. Thus, we have ensured the best quality of content for the students. They can follow the solutions and check their answers with the experts’ answers. So, the Mathematics Class 7 Chapter 1 Important Questions will help students to clarify their doubts, boost their confidence and build their concepts.
- The subject matter experts of Extramarks understand the student’s needs. They have built the question series to help students with their exam preparation. They have collected all the vital questions so students can find them in a single article. Sometimes, students need more than the textbook. Hence, they can follow the Class 7 Mathematics Chapter 1 Important Questions because they will find chapter-wise questions for each subject. Regular practice will strengthen their ideas, and they can solve any question that comes in exams. Thus, the question series will help them to score better in exams.

Extramarks is a leading company that provides all the important study materials related to CBSE and NCERT. You can download these after registering on our official website. We provide CBSE syllabus, CBSE past years, question papers, CBSE sample papers, NCERT books, NCERT exemplar, NCERT solutions, important questions, CBSE revision notes, CBSE extra questions, vital formulas and many more. Like the Important Questions Class 7 Mathematics Chapter 1, you will also find important questions for other chapters. The links to study materials are given below-

- NCERT books
- Important questions
- CBSE Revision Notes
- CBSE syllabus
- CBSE sample papers
- CBSE past years’ question papers
- Important formulas
- CBSE extra questions

**Q.1 Which one of the following statements is false?**

**1. For any two positive integers a and b, a ÷ (–b) = – a ÷ b, where b ≠ 0.**

**2. The commutativity, associativity and distributivity of integers help to make calculations simpler.**

**3. The product of three integers does not depend upon the grouping of integers.**

**4. Division is closed for integers.**

**Marks: ****1**

**Ans**

Option 4.

**Explanation**

Division is not closed for integers.

For example:

2 ÷ 6 =

$\frac{1}{3}$

is not an integer.

**Q.2 Which one of the following is false?**

**Marks:**1

**1. **Sum of integers a and b is an integer.

**2. **a + b = b + a, for all integers a and b

**3. **a – b = b – a, for all integers a and b

**4. **a + (b + c) = (a + b) + c, for all integers a, b and c

**Ans **Option3

**Explanation**

a – b = b – a, for all integers a and b is false.

For example, 2 – 4 = – 2 and 4 – 2 = 2

Thus, 2 – 4 ≠ 4 – 2

**Q.3 What is the difference between a temperature of 7º C above zero and a temperature of 3º C below zero?**

**1. 10º C**

**2. 7º C**

**3. 4º C**

**4. 3º C**

**Marks:**1

**Ans **Option 1.

**Explanation**

Difference between a temperature of 7º C above zero and a temperature of 3º C below zero

= 7º C – (– 3º C)

= 7º C + 3º C

= 10º C

**Q.4 A plane is flying at the height of 8750 m above sea level. At a particular point, it is exactly above a submarine floating 1340 m below sea level. What is the vertical distance between them?**

**Marks:**2

**Ans**

Height of the plane above sea level = 8750 m

Distance of submarine below sea level = – 1340 m

Vertical distance = 8750 m – (– 1340 m)

= 8750 m + 1340 m

= 10,090 m

**Q.5 A man walks 22 m towards east and then 17 m towards west. The position of the man with respect to his starting point is ______________.**

**1.5 m towards west**

**2.5 m towards east**

**3.39 m towards east**

**4.39 m towards west**

**Marks:**1

**Ans **Option 2.

**Explanation**

Let 22 m towards east be represented by +22, then –17 m represents 17 m towards west.

On adding, +22 – 17 = +5 (positive)

The position of the man with respect to his starting point = 5 m towards east

##### FAQs (Frequently Asked Questions)

## 1. Is Class 7 Mathematics Chapter 1 easy?

Class 7 Mathematics Chapter 1 under CBSE curriculum is about integers. Students will study the properties of integers, how to add and multiply integers and how to put them on the number line. The concepts may be new to them, but they have studied integers in Class 6. They can easily understand the concepts if they follow the textbook seriously. The chapter is relatively easy. Students can take help from the Important Questions Class 7 Mathematics Chapter 1 to solve questions from the chapter.

## 2. How can the Important Questions Class 7 Mathematics Chapter 1 help students?

The experts of Extramarks have made the question series after taking help from several sources. They have collated the questions from the textbook exercise, CBSE sample papers, important reference books and NCERT exemplar. They have included questions from CBSE past years’ question papers too. Apart from this, they have solved the questions for students, and experienced professionals have further checked the answers. Thus, the Important Questions Class 7 Mathematics Chapter 1 will help the students to practice the sums regularly. It will boost their confidence and increase their marks in exams.