# NCERT Solutions for Class 7 Maths Chapter 1 Integers

NCERT Solutions for Class 7 Maths Chapter 1 Integers are available here. When students feel stressed about searching for the most comprehensive examination and detail NCERT Solutions for Class 7 Maths, we at BYJU ’ S have prepared step by step solutions with detail explanations. We advise students who want to score thoroughly marks in Maths, to go through these solutions and strengthen their cognition .

chapter 1 – Integers contains 4 exercises, and the NCERT Solutions available on this page provide solutions to the questions present in the exercises. now, let us have a expression at some of the concepts discussed in this chapter .

- Introduction of Integers
- Properties of Addition and Subtraction of Integers
- Multiplication of Integers
- Multiplication of a Positive and Negative Integer
- Multiplication of two Negative Integer
- Properties of Multiplication of Integers
- Division of Integers
- Properties of Division of Integers

### Access Exercises of NCERT Solutions for Class 7 Maths Chapter 1 Integers

drill 1.1 Solutions

use 1.2 Solutions

use 1.3 Solutions

exercise 1.4 Solutions

### Access answers to Maths NCERT Solutions For Class 7 Chapter 1 – Integers

exercise 1.1 foliate : 4

**1. Following number line shows the temperature in degree celsius (co) at different places on a particular day.**

**(a) Observe this number line and write the temperature of the places marked on it.**

**Solution:-**

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

temperature at the Lahulspiti is -8oC

temperature at the Srinagar is -2oC

temperature at the Shimla is 5oC

temperature at the Ooty is 14oC

temperature at the Bengaluru is 22oC

**(b) What is the temperature difference between the hottest and the coldest places among the above?**

**Solution:-**

From the count line we observe that ,

The temperature at the hottest rate i, Bengaluru is 22oC

The temperature at the coldest place i, Lahulspiti is -8oC

temperature dispute between hot and cold locate is = 22oC – ( -8oC )

= 22oC + 8oC

= 30oC

Hence, the temperature deviation between the hottest and the cold place is 30oC .

**(c) What is the temperature difference between Lahulspiti and Srinagar?**

**Solution:-**

From the given number course ,

The temperature at the Lahulspiti is -8oC

The temperature at the Srinagar is -2oC

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

= – 2OC + 8oC

= 6oC

**(d) Can we say temperature of Srinagar and Shimla taken together is less than the**

**temperature at Shimla? Is it also less than the temperature at Srinagar?**

**Solution:-**

From the given number line ,

The temperature at Srinagar =-2oC

The temperature at Shimla = 5oC

The temperature of Srinagar and Shimla taken together is = – 2oC + 5oC

= 3oC

∴ 5oC > 3oC

then, the temperature of Srinagar and Shimla taken together is less than the temperature at Shimla .

then ,

3o > -2o

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

**2. In a quiz, positive marks are given for correct answers and negative marks are given**

**for incorrect answers. If Jack’s scores in five successive rounds were 25, – 5, – 10,**

**15 and 10, what was his total at the end?**

**Solution:-**

From the question ,

Jack ’ s score in five consecutive rounds are 25, -5, -10, 15 and 10

The full score of Jack at the end will be = 25 + ( -5 ) + ( -10 ) + 15 + 10

= 25 – 5 – 10 + 15 + 10

= 50 – 15

= 35

∴Jack ’ randomness sum score at the end is 35 .

**3. At Srinagar temperature was – 5°C on Monday and then it dropped by 2°C on Tuesday. What was the temperature of Srinagar on Tuesday? On Wednesday, it rose by 4°C. What was the temperature on this day?**

**Solution:-**

From the wonder ,

temperature on Monday at Srinagar = -5oC

temperature on Tuesday at Srinagar is dropped by 2oC = temperature on Monday – 2oC

= -5oC – 2oC

= -7oC

temperature on Wednesday at Srinagar is rose by 4oC = temperature on Tuesday + 4oC

= -7oC + 4oC

= -3oC

frankincense, the temperature on Tuesday and Wednesday was -7oC and -3oC respectively .

**4. A plane is flying at the height of 5000 m above the sea level. At a particular point, it is exactly above a submarine floating 1200 m below the sea level. What is the vertical distance between them?**

**Solution:-**

From the question ,

Plane is flying at the acme = 5000 thousand

Depth of Submarine = -1200 meter

The vertical distance between plane and submarine = 5000 thousand – ( – 1200 ) thousand

= 5000 m + 1200 thousand

= 6200 thousand

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

**Solution:-**

Withdrawal of measure from the explanation is represented by a negative integer .

then, deposit of total to the account is represented by a plus integer .

From the motion ,

sum sum deposited in deposit report by the Mohan = ₹ 2000

total come withdraw from the depository financial institution history by the Mohan = – ₹ 1642

balance wheel in Mohan ’ mho account after the withdrawal = sum deposited + sum bow out

= ₹ 2000 + ( -₹ 1642 )

= ₹ 2000 – ₹ 1642

= ₹ 358

Hence, the symmetry in Mohan ’ south account after the withdrawal is ₹ 358

**6. Rita goes 20 km towards east from a point A to the point B. From B, she moves 30 km towards west along the same road. If the distance towards east is represented by a positive integer then, how will you represent the distance travelled towards west? By which integer will you represent her final position from A?**

**Solution:-**

From the doubt, it is given that

A positive integer represents the distance towards the east .

then, outdistance travelled towards the west will be represented by a damaging integer .

Rita travels a distance in east direction = 20 kilometer

Rita travels a distance in west direction = – 30 kilometer

∴Distance travelled from A = 20 + ( – 30 )

= 20 – 30

= -10 kilometer

hence, we will represent the distance travelled by Rita from point A by a veto integer, i.e. – 10 kilometer

**7. In a magic square each row, column and diagonal have the same sum. Check which of the following is a magic square** .

**Solution:-**

first base we consider the square ( one )

By adding the numbers in each row we get ,

= 5 + ( – 1 ) + ( – 4 ) = 5 – 1 – 4 = 5 – 5 = 0

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

= 0 + 3 + ( -3 ) = 3 – 3 = 0

By adding the numbers in each column we get ,

= 5 + ( – 5 ) + 0 = 5 – 5 = 0

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

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

By adding the numbers in diagonals we get ,

= 5 + ( -2 ) + ( -3 ) = 5 – 2 – 3 = 5 – 5 = 0

= -4 + ( -2 ) + 0 = – 4 – 2 = -6

Because sum of one solidus is not equal to zero ,

so, ( i ) is not a charming square

immediately, we consider the square ( two )

By adding the numbers in each rowing we get ,

= 1 + ( -10 ) + 0 = 1 – 10 + 0 = -9

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

= ( -6 ) + 4 + ( -7 ) = -6 + 4 – 7 = -13 + 4 = -9

By adding the numbers in each column we get ,

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

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

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

By adding the numbers in diagonals we get ,

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

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

This ( two ) squarely is a charming squarely, because sum of each row, each column and solidus is equal to -9 .

**8. Verify a – (– b) = a + b for the following values of a and b.**

**(i) a = 21, b = 18**

**Solution:-**

From the interview ,

a = 21 and b = 18

To verify a – ( – barn ) = a + b

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

= 21 – ( – 18 )

= 21 + 18

= 39

now, Right Hand Side ( RHS ) = a + bel

= 21 + 18

= 39

By comparing LHS and RHS

LHS = RHS

39 = 39

Hence, the value of a and boron is verified .

**(ii) a = 118, b = 125**

**Solution:-**

From the question ,

a = 118 and b = 125

To verify a – ( – b ) = a + barn

Let us take Left Hand Side ( LHS ) = a – ( – bel )

= 118 – ( – 125 )

= 118 + 125

= 243

now, Right Hand Side ( RHS ) = a + b-complex vitamin

= 118 + 125

= 243

By comparing LHS and RHS

LHS = RHS

243 = 243

Hence, the value of a and barn is verified .

**(iii) a = 75, b = 84 **

**Solution:-**

From the interview ,

a = 75 and b = 84

To verify a – ( – b-complex vitamin ) = a + b-complex vitamin

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

= 75 – ( – 84 )

= 75 + 84

= 159

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

= 75 + 84

= 159

By comparing LHS and RHS

LHS = RHS

159 = 159

Hence, the value of a and boron is verified .

**(iv) a = 28, b = 11**

**Solution:-**

From the question ,

a = 28 and b = 11

To verify a – ( – b ) = a + barn

Let us take Left Hand Side ( LHS ) = a – ( – bacillus )

= 28 – ( – 11 )

= 28 + 11

= 39

now, Right Hand Side ( RHS ) = a + bel

= 28 + 11

= 39

By comparing LHS and RHS

LHS = RHS

39 = 39

Hence, the value of a and b is verified .

**9. Use the sign of >, < or = in the box to make the statements true.**

**(a) (-8) + (-4) [ ] (-8) – (-4)**

**Solution:-**

Let us take Left Hand Side ( LHS ) = ( -8 ) + ( -4 )

= -8 – 4

= -12

now, Right Hand Side ( RHS ) = ( -8 ) – ( -4 )

= -8 + 4

= -4

By comparing LHS and RHS

LHS < RHS

-12 < -4

∴ ( -8 ) + ( -4 ) [ < ] ( -8 ) – ( -4 )

( b-complex vitamin ) ( -3 ) + 7 – ( 19 ) [ ] 15 – 8 + ( -9 )

solution : –

Let us take Left Hand Side ( LHS ) = ( -3 ) + 7 – 19

= -3 + 7 – 19

= -22 + 7

= -15

now, Right Hand Side ( RHS ) = 15 – 8 + ( -9 )

= 15 – 8 – 9

= 15 – 17

= -2

By comparing LHS and RHS

LHS < RHS

-15 < -2

∴ ( -3 ) + 7 – ( 19 ) [ < ] 15 – 8 + ( -9 )

**(c) 23 – 41 + 11 [ ] 23 – 41 – 11**

**Solution:-**

Let us take Left Hand Side ( LHS ) = 23 – 41 + 11

= 34 – 41

= – 7

nowadays, Right Hand Side ( RHS ) = 23 – 41 – 11

= 23 – 52

= – 29

By comparing LHS and RHS

LHS > RHS

– 7 > -29

∴ 23 – 41 + 11 [ > ] 23 – 41 – 11

**(d) 39 + (-24) – (15) [ ] 36 + (-52) – (- 36)**

**Solution:-**

Let us take Left Hand Side ( LHS ) = 39 + ( -24 ) – 15

= 39 – 24 – 15

= 39 – 39

= 0

nowadays, Right Hand Side ( RHS ) = 36 + ( -52 ) – ( – 36 )

= 36 – 52 + 36

= 72 – 52

= 20

By comparing LHS and RHS

LHS < RHS

0 < 20

∴ 39 + ( -24 ) – ( 15 ) [ < ] 36 + ( -52 ) – ( – 36 )

**(e) – 231 + 79 + 51 [ ] -399 + 159 + 81**

**Solution:-**

Let us take Left Hand Side ( LHS ) = – 231 + 79 + 51

= – 231 + 130

= -101

now, Right Hand Side ( RHS ) = – 399 + 159 + 81

= – 399 + 240

= – 159

By comparing LHS and RHS

LHS > RHS

-101 > -159

∴ – 231 + 79 + 51 [ > ] -399 + 159 + 81

**10. A water tank has steps inside it. A monkey is sitting on the topmost step (i.e., the first step). The water level is at the ninth step.**

**(i) He jumps 3 steps down and then jumps back 2 steps up. In how many jumps will he reach the water level?**

**Solution:-**

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

initially putter is sitting on the top most gradation i, first step

In 1st jump putter will be at pace = 1 + 3 = 4 steps

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

In 3rd chute tamper will be at step = 2 + 3 = 5 steps

In 4th alternate imp will be at footprint = 5 + ( -2 ) = 5 – 2 = 3 steps

In 5th jump tamper will be at tone = 3 + 3 = 6 steps

In 6th derail tamper will be at dance step = 6 + ( -2 ) = 6 – 2 = 4 steps

In 7th jump imp will be at gradation = 4 + 3 = 7 steps

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

In 9th startle monkey will be at measure = 5 + 3 = 8 steps

In 10th leap putter will be at step = 8 + ( -2 ) = 8 – 2 = 6 steps

In 11th jump imp will be at gradation = 6 + 3 = 9 steps

∴Monkey took 11 jumps ( i, 9th mistreat ) to reach the water level

**(ii) After drinking water, he wants to go back. For this, he jumps 4 steps up and then jumps back 2 steps down in every move. In how many jumps will he reach back the top step?**

**Solution:-**

Let us consider steps moved down are represented by incontrovertible integers and then, steps moved up are represented by veto integers .

initially imp is sitting on the one-ninth step i, at the water level

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

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

In 3rd jump imp will be at measure = 7 + ( -4 ) = 7 – 4 = 3 steps

In 4th startle imp will be at step = 3 + 2 = 5 steps

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

∴Monkey took 5 jumps to reach back the top step i, first footstep .

**(iii) If the number of steps moved down is represented by negative integers and the number of steps moved up by positive integers, represent his moves in part (i) and (ii) by completing the following; (a) – 3 + 2 – … = – 8 (b) 4 – 2 + … = 8. In (a) the sum (– 8) represents going down by eight steps. So, what will the sum 8 in (b) represent?**

**Solution:-**

From the question, it is given that

If the number of steps moved down is represented by veto integers and the number of steps moved up by plus integers .

Monkey moves in region ( one )

= – 3 + 2 – ……… .. = – 8

then LHS = – 3 + 2 – 3 + 2 – 3 + 2 – 3 + 2 – 3 + 2 – 3

= – 18 + 10

= – 8

RHS = -8

∴Moves in separate ( iodine ) represents tamper is going down 8 steps. Because damaging integer .

now ,

Monkey moves in character ( two )

= 4 – 2 + ……… .. = 8

then LHS = 4 – 2 + 4 – 2 + 4

= 12 – 4

= 8

RHS = 8

∴Moves in separate ( two ) represents monkey is going up 8 steps. Because positive integer .

exert 1.2 page : 9

**1. Write down a pair of integers whose:**

**(a) sum is -7**

**Solution:-**

= – 4 + ( -3 )

= – 4 – 3 … [ ∵ ( + × – = – ) ] = – 7

**(b) difference is – 10**

**Solution:-**

= -25 – ( -15 )

= – 25 + 15 … [ ∵ ( – × – = + ) ] = -10

**(c) sum is 0**

**Solution:-**

= 4 + ( -4 )

= 4 – 4

= 0

**2. (a) Write a pair of negative integers whose difference gives 8**

**Solution:-**

= ( -5 ) – ( – 13 )

= -5 + 13 … [ ∵ ( – × – = + ) ] = 8

**(b) Write a negative integer and a positive integer whose sum is – 5.**

**Solution:-**

= -25 + 20

= -5

**(c) Write a negative integer and a positive integer whose difference is – 3.**

**Solution** : –

= – 2 – ( 1 )

= – 2 – 1

= – 3

**3. In a quiz, team A scored – 40, 10, 0 and team B scored 10, 0, – 40 in three successive rounds. Which team scored more? Can we say that we can add integers in any order?**

**Solution:-**

From the question, it is given that

score of team A = -40, 10, 0

total seduce obtained by team A = – 40 + 10 + 0

= – 30

score of team B = 10, 0, -40

full score obtained by team B = 10 + 0 + ( -40 )

= 10 + 0 – 40

= – 30

thus, the score of the both A team and B team is same .

Yes, we can say that we can add integers in any order .

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

**(i) (–5) + (– 8) = (– 8) + (…………)**

**Solution:-**

Let us assume the miss integer be x ,

then ,

= ( –5 ) + ( – 8 ) = ( – 8 ) + ( ten )

= – 5 – 8 = – 8 + x

= – 13 = – 8 + x

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

= – 13 + 8 = x

= x = – 5

now substitute the x value in the blank place ,

( –5 ) + ( – 8 ) = ( – 8 ) + ( – 5 ) … [ This equation is in the form of Commutative jurisprudence of Addition ] **(ii) –53 + ………… = –53**

**Solution:-**

Let us assume the miss integer be x ,

then ,

= –53 + x = –53

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

= x = -53 + 53

= x = 0

nowadays substitute the adam value in the lacuna place ,

= –53 + 0 = –53 … [ This equality is in the imprint of Closure property of Addition ] **(iii) 17 + ………… = 0**

**Solution:-**

Let us assume the lacking integer be x ,

then ,

= 17 + x = 0

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

= x = 0 – 17

= x = – 17

now substitute the adam value in the blank position ,

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

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

**Solution:-**

Let us assume the miss integer be x ,

then ,

= [ 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 1 from LHS to RHS it becomes -1 ,

= x = -6 – 1

= x = -7

now substitute the adam measure in the space place ,

= [ 13 + ( – 12 ) ] + ( -7 ) = 13 + [ ( –12 ) + ( –7 ) ] … [ This equation is in the form of Associative place of Addition ] **(v) (– 4) + [15 + (–3)] = [– 4 + 15] +…………**

**Solution:-**

Let us assume the miss integer be x ,

then ,

= ( – 4 ) + [ 15 + ( –3 ) ] = [ – 4 + 15 ] + ten

= ( – 4 ) + [ 15 – 3 ) ] = [ – 4 + 15 ] + ten

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

= 8 = 11 + x

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

= 8 – 11 = ten

= x = -3

now substitute the ten value in the blank seat ,

= ( – 4 ) + [ 15 + ( –3 ) ] = [ – 4 + 15 ] + -3 … [ This equation is in the form of Associative property of Addition ] exercise 1.3 page : 21

**1. Find each of the following products:**

**(a) 3 × (–1) **

**Solution:-**

By the govern of Multiplication of integers ,

= 3 × ( -1 )

= -3 … [ ∵ ( + × – = – ) ] **(b) (–1) × 225**

**Solution:-**

By the dominion of Multiplication of integers ,

= ( -1 ) × 225

= -225 … [ ∵ ( – × + = – ) ] **(c) (–21) × (–30)**

**Solution:-**

By the rule of Multiplication of integers ,

= ( -21 ) × ( -30 )

= 630 … [ ∵ ( – × – = + ) ] **(d) (–316) × (–1)**

**Solution:-**

By the rule of Multiplication of integers ,

= ( -316 ) × ( -1 )

= 316 … [ ∵ ( – × – = + ) ] **(e) (–15) × 0 × (–18)**

**Solution:-**

By the rule of Multiplication of integers ,

= ( –15 ) × 0 × ( –18 )

= 0

∵Any integer is multiplied with zero and the answer is zero itself .

**(f) (–12) × (–11) × (10)**

**Solution:-**

By the rule of Multiplication of integers ,

= ( –12 ) × ( -11 ) × ( 10 )

first gear multiply the two numbers having lapp signboard ,

= 132 × 10 … [ ∵ ( – × – = + ) ] = 1320

**(g) 9 × (–3) × (– 6) **

**Solution:-**

By the dominion of Multiplication of integers ,

= 9 × ( -3 ) × ( -6 )

first multiply the two numbers having lapp sign ,

= 9 × 18 … [ ∵ ( – × – = + ) ] = 162

**(h) (–18) × (–5) × (– 4)**

**Solution:-**

By the rule of Multiplication of integers ,

= ( -18 ) × ( -5 ) × ( -4 )

beginning multiply the two numbers having like sign ,

= 90 × -4 … [ ∵ ( – × – = + ) ] = – 360 … [ ∵ ( + × – = – ) ] **(i) (–1) × (–2) × (–3) × 4**

**Solution:-**

By the rule of Multiplication of integers ,

= [ ( –1 ) × ( –2 ) ] × [ ( –3 ) × 4 ] = 2 × ( -12 ) … [ ∵ ( – × – = + ), ( – × + = – ) ] = – 24

**(j) (–3) × (–6) × (–2) × (–1)**

**Solution:-**

By the rule of Multiplication of integers ,

= [ ( –3 ) × ( –6 ) ] × [ ( –2 ) × ( –1 ) ] first multiply the two numbers having like sign ,

= 18 × 2 … [ ∵ ( – × – = + )

= 36

**2. Verify the following:**

**(a) 18 × [7 + (–3)] = [18 × 7] + [18 × (–3)]**

**Solution:-**

From the given equality ,

Let us consider the Left Hand Side ( LHS ) inaugural = 18 × [ 7 + ( –3 ) ] = 18 × [ 7 – 3 ] = 18 × 4

= 72

now, consider the Right Hand Side ( RHS ) = [ 18 × 7 ] + [ 18 × ( –3 ) ] = [ 126 ] + [ -54 ] = 126 – 54

= 72

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By comparing LHS and RHS ,

72 = 72

LHS = RHS

Hence, the given equation is verified .

**(b) (–21) × [(– 4) + (– 6)] = [(–21) × (– 4)] + [(–21) × (– 6)]**

**Solution:-**

From the given equation ,

Let us consider the Left Hand Side ( LHS ) first = ( –21 ) × [ ( – 4 ) + ( – 6 ) ] = ( -21 ) × [ -4 – 6 ] = ( -21 ) × [ -10 ] = 210

now, consider the Right Hand Side ( RHS ) = [ ( –21 ) × ( – 4 ) ] + [ ( –21 ) × ( – 6 ) ] = [ 84 ] + [ 126 ] = 210

By comparing LHS and RHS ,

210 = 210

LHS = RHS

Hence, the given equation is verified .

3. ( one ) For any integer a, what is ( –1 ) × a equal to ?

solution : –

= ( -1 ) × a = -a

Because, when we multiplied any integer a with -1, then we get additive inverse of that integer .

**(ii). Determine the integer whose product with (–1) is**

**(a) –22**

**Solution:-**

nowadays, multiply -22 with ( -1 ), we get

= -22 × ( -1 )

= 22

Because, when we multiplied integer -22 with -1, then we get additive inverse of that integer .

**(b) 37 **

**Solution:-**

immediately, breed 37 with ( -1 ), we get

= 37 × ( -1 )

= -37

Because, when we multiplied integer 37 with -1, then we get additive inverse of that integer .

**(c) 0**

**Solution:-**

now, multiply 0 with ( -1 ), we get

= 0 × ( -1 )

= 0

Because, the product of negative integers and zero give zero only .

**4. Starting from (–1) × 5, write various products showing some pattern to show**

**(–1) × (–1) = 1.**

**Solution:-**

The versatile products are ,

= -1 × 5 = -5

= -1 × 4 = -4

= -1 × 3 = -3

= -1 × 2 = -2

= -1 × 1 = -1

= -1 × 0 = 0

= -1 × -1 = 1

We concluded that the product of one minus integer and one positive integer is veto integer. then, the merchandise of two negative integers is a plus integer .

**5. Find the product, using suitable properties:**

**(a) 26 × (– 48) + (– 48) × (–36) **

**Solution:-**

The given equation is in the form of Distributive police of Multiplication over Addition .

= a × ( bel + deoxycytidine monophosphate ) = ( a × b ) + ( a × hundred )

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

now ,

= 26 × ( – 48 ) + ( – 48 ) × ( –36 )

= -48 × ( 26 + ( -36 )

= -48 × ( 26 – 36 )

= -48 × ( -10 )

= 480 … [ ∵ ( – × – = + )

**(b) 8 × 53 × (–125)**

**Solution:-**

The given equation is in the kind of Commutative law of Multiplication .

= a × b = bacillus × a

then ,

= 8 × [ 53 × ( -125 ) ] = 8 × [ ( -125 ) × 53 ] = [ 8 × ( -125 ) ] × 53

= [ -1000 ] × 53

= – 53000

**(c) 15 × (–25) × (– 4) × (–10) **

**Solution:-**

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

= a × b = boron × a

then ,

= 15 × [ ( –25 ) × ( – 4 ) ] × ( –10 )

= 15 × [ 100 ] × ( –10 )

= 15 × [ -1000 ] = – 15000

**(d) (– 41) × 102**

**Solution:-**

The given equality is in the shape of Distributive police of Multiplication over Addition .

= a × ( boron + speed of light ) = ( a × b-complex vitamin ) + ( a × vitamin c )

= ( -41 ) × ( 100 + 2 )

= ( -41 ) × 100 + ( -41 ) × 2

= – 4100 – 82

= – 4182

**(e) 625 × (–35) + (– 625) × 65 **

**Solution:-**

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

= a × ( boron + coke ) = ( a × barn ) + ( a × vitamin c )

= 625 × [ ( -35 ) + ( -65 ) ] = 625 × [ -100 ] = – 62500

**(f) 7 × (50 – 2)**

**Solution:-**

The given equality is in the form of Distributive law of Multiplication over Subtraction .

= a × ( barn – coke ) = ( a × bacillus ) – ( a × speed of light )

= ( 7 × 50 ) – ( 7 × 2 )

= 350 – 14

= 336

**(g) (–17) × (–29) **

**Solution:-**

The given equality is in the form of Distributive law of Multiplication over Addition .

= a × ( bacillus + coke ) = ( a × b ) + ( a × vitamin c )

= ( -17 ) × [ -30 + 1 ] = [ ( -17 ) × ( -30 ) ] + [ ( -17 ) × 1 ] = [ 510 ] + [ -17 ] = 493

**(h) (–57) × (–19) + 57**

**Solution:-**

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

= a × ( bel + carbon ) = ( a × boron ) + ( a × c )

= ( 57 × 19 ) + ( 57 × 1 )

= 57 [ 19 + 1 ] = 57 × 20

= 1140

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

**Solution:-**

From the question, it is given that

Let us take the lower temperature as negative ,

initial temperature = 40oC

deepen in temperature per hour = -5oC

change in temperature after 10 hours = ( -5 ) × 10 = -50oC

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

= -10oC

**7. In a class test containing 10 questions, 5 marks are awarded for every correct answer and (–2) marks are awarded for every incorrect answer and 0 for questions not attempted.**

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

**Solution:-**

From the question ,

Marks awarded for 1 correct answer = 5

then ,

total marks awarded for 4 right answer = 4 × 5 = 20

Marks awarded for 1 wrong solution = -2

then ,

total marks awarded for 6 wrong answer = 6 × -2 = -12

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

= 20 – 12

= 8

**(ii) Reshma gets five correct answers and five incorrect answers, what is her score?**

**Solution:-**

From the question ,

Marks awarded for 1 decline answer = 5

then ,

total marks awarded for 5 correct answer = 5 × 5 = 25

Marks awarded for 1 wrong solution = -2

then ,

total marks awarded for 5 wrong answer = 5 × -2 = -10

∴Total sexual conquest obtained by Reshma = 25 + ( -10 )

= 25 – 10

= 15

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

**Solution:-**

From the doubt ,

Marks awarded for 1 decline answer = 5

then ,

full marks awarded for 2 correct answer = 2 × 5 = 10

Marks awarded for 1 incorrect answer = -2

then ,

entire marks awarded for 5 ill-timed answer = 5 × -2 = -10

Marks awarded for questions not attempted is = 0

∴Total grudge obtained by Heena = 10 + ( -10 )

= 10 – 10

= 0

**8. A cement company earns a profit of ₹ 8 per bag of white cement sold and a loss of**

**₹ 5 per bag of grey cement sold.**

**(a) 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?**

**Solution:-**

We denote profit in cocksure integer and loss in negative integer ,

From the question ,

Cement company earns a profit on selling 1 cup of tea of white cement = ₹ 8 per bag

then ,

Cement party earns a net income on selling 3000 bags of white cement = 3000 × ₹ 8

= ₹ 24000

Loss on selling 1 bag of gray cement = – ₹ 5 per bag

then ,

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

= – ₹ 25000

entire loss or net income earned by the cementum company = net income + loss

= 24000 + ( -25000 )

= – ₹1000

frankincense, a passing of ₹ 1000 will be incurred by the company .

**(b) What is the number of white cement bags it must sell to have neither profit nor loss, if the number of grey bags sold is 6,400 bags.**

**Solution:-**

We denote profit in positivist integer and loss in veto integer ,

From the interrogate ,

Cement company earns a net income on selling 1 bag of white cementum = ₹ 8 per udder

Let the count of white cement bags be x .

then ,

Cement party earns a profit on selling adam bags of white cement = ( x ) × ₹ 8

= ₹ 8x

Loss on selling 1 pocket of grey cement = – ₹ 5 per pocket

then ,

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

= – ₹ 32000

According to the interview ,

company must sell to have neither profit nor loss .

= Profit + loss = 0

= 8x + ( -32000 ) =0

By sending -32000 from LHS to RHS it becomes 32000

= 8x = 32000

= x = 32000/8

= x = 4000

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

**9. Replace the blank with an integer to make it a true statement.**

**(a) (–3) × _____ = 27 **

**Solution:-**

Let us assume the missing integer be x ,

then ,

= ( –3 ) × ( x ) = 27

= x = – ( 27/3 )

= x = -9

Let us substitute the value of ten in the place of blank ,

= ( –3 ) × ( -9 ) = 27 … [ ∵ ( – × – = + ) ] **(b) 5 × _____ = –35**

**Solution:-**

Let us assume the miss integer be x ,

then ,

= ( 5 ) × ( x ) = -35

= x = – ( -35/5 )

= x = -7

Let us substitute the value of ten in the place of lacuna ,

= ( 5 ) × ( -7 ) = -35 … [ ∵ ( + × – = – ) ] **(c) _____ × (– 8) = –56 **

**Solution:-**

Let us assume the miss integer be x ,

then ,

= ( x ) × ( -8 ) = -56

= x = ( -56/-8 )

= x = 7

Let us substitute the value of adam in the station of blank ,

= ( 7 ) × ( -8 ) = -56 … [ ∵ ( + × – = – ) ] **(d) _____ × (–12) = 132**

**Solution:-**

Let us assume the miss integer be x ,

then ,

= ( x ) × ( -12 ) = 132

= x = – ( 132/12 )

= x = – 11

Let us substitute the value of ten in the plaza of blank ,

= ( –11 ) × ( -12 ) = 132 … [ ∵ ( – × – = + ) ] exercise 1.4 page : 26

**1. Evaluate each of the following:**

**(a) (–30) ÷ 10 **

**Solution:-**

= ( –30 ) ÷ 10

= – 3

When we divide a negative integer by a positive integer, we first divide them as solid numbers and then put minus signal ( – ) before the quotient .

**(b) 50 ÷ (–5) **

**Solution:-**

= ( 50 ) ÷ ( -5 )

= – 10

When we divide a cocksure integer by a negative integer, we first divide them as wholly numbers and then put minus bless ( – ) before the quotient .

**(c) (–36) ÷ (–9)**

**Solution:-**

= ( -36 ) ÷ ( -9 )

= 4

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

**(d) (– 49) ÷ (49) **

**Solution:-**

= ( –49 ) ÷ 49

= – 1

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

**(e) 13 ÷ [(–2) + 1] **

**Solution:-**

= 13 ÷ [ ( –2 ) + 1 ] = 13 ÷ ( -1 )

= – 13

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

**(f) 0 ÷ (–12)**

**Solution:-**

= 0 ÷ ( -12 )

= 0

When we divide zero by a negative integer gives zero .

**(g) (–31) ÷ [(–30) + (–1)]**

**Solution:-**

= ( –31 ) ÷ [ ( –30 ) + ( –1 ) ] = ( -31 ) ÷ [ -30 – 1 ] = ( -31 ) ÷ ( -31 )

= 1

When we divide a minus integer by a damaging integer, we first divide them as unharmed numbers and then put positive sign ( + ) before the quotient .

**(h) [(–36) ÷ 12] ÷ 3**

**Solution:-**

beginning we have to solve the integers with in 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 minus sign ( – ) before the quotient .

**(i) [(– 6) + 5)] ÷ [(–2) + 1]**

**Solution:-**

The given motion can be written as ,

= [ -1 ] ÷ [ -1 ] = 1

When we divide a negative integer by a negative integer, we first divide them as unharmed numbers and then put positivist polarity ( + ) before the quotient .

**2. Verify that a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c) for each of the following values of a, b and c.**

**(a) a = 12, b = – 4, c = 2**

**Solution:-**

From the question, a ÷ ( bacillus + vitamin c ) ≠ ( a ÷ b-complex vitamin ) + ( a ÷ cytosine )

Given, a = 12, barn = – 4, c = 2

now, consider LHS = a ÷ ( b + deoxycytidine monophosphate )

= 12 ÷ ( -4 + 2 )

= 12 ÷ ( -2 )

= -6

When we divide a incontrovertible integer by a negative integer, we first divide them as solid numbers and then put minus sign ( – ) before the quotient .

then, consider RHS = ( a ÷ b-complex vitamin ) + ( a ÷ degree centigrade )

= ( 12 ÷ ( -4 ) ) + ( 12 ÷ 2 )

= ( -3 ) + ( 6 )

= 3

By comparing LHS and RHS

= -6 ≠ 3

= LHS ≠ RHS

Hence, the given values are verified .

**(b) a = (–10), b = 1, c = 1**

**Solution:-**

From the interrogate, a ÷ ( bacillus + cytosine ) ≠ ( a ÷ bel ) + ( a ÷ carbon )

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

nowadays, consider LHS = a ÷ ( bel + c )

= ( -10 ) ÷ ( 1 + 1 )

= ( -10 ) ÷ ( 2 )

= -5

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

then, consider RHS = ( a ÷ b ) + ( a ÷ coulomb )

= ( ( -10 ) ÷ ( 1 ) ) + ( ( -10 ) ÷ 1 )

= ( -10 ) + ( -10 )

= -10 – 10

= -20

By comparing LHS and RHS

= -5 ≠ -20

= LHS ≠ RHS

Hence, the given values are verified .

**3. Fill in the blanks:**

**(a) 369 ÷ _____ = 369 **

**Solution:-**

Let us assume the miss integer be x ,

then ,

= 369 ÷ x = 369

= x = ( 369/369 )

= x = 1

now, put the valve of ten in the blank .

= 369 ÷ 1 = 369

**(b) (–75) ÷ _____ = –1**

**Solution:-**

Let us assume the miss integer be x ,

then ,

= ( -75 ) ÷ x = -1

= x = ( -75/-1 )

= x = 75

now, put the valve of adam in the blank .

= ( -75 ) ÷ 75 = -1

**(c) (–206) ÷ _____ = 1 **

**Solution:-**

Let us assume the miss integer be x ,

then ,

= ( -206 ) ÷ x = 1

= x = ( -206/1 )

= x = -206

now, put the valve of ten in the blank .

= ( -206 ) ÷ ( -206 ) = 1

**(d) – 87 ÷ _____ = 87**

**Solution:-**

Let us assume the miss integer be x ,

then ,

= ( -87 ) ÷ x = 87

= x = ( -87 ) /87

= x = -1

nowadays, put the valve of x in the blank .

= ( -87 ) ÷ ( -1 ) = 87

**(e) _____ ÷ 1 = – 87 **

**Solution:-**

Let us assume the miss integer be x ,

then ,

= ( x ) ÷ 1 = -87

= x = ( -87 ) × 1

= x = -87

now, put the valve of x in the blank .

= ( -87 ) ÷ 1 = -87

**(f) _____ ÷ 48 = –1**

**Solution:-**

Let us assume the miss integer be x ,

then ,

= ( x ) ÷ 48 = -1

= x = ( -1 ) × 48

= x = -48

now, put the valve of ten in the space .

= ( -48 ) ÷ 48 = -1

**(g) 20 ÷ _____ = –2 **

**Solution:-**

Let us assume the miss integer be x ,

then ,

= 20 ÷ x = -2

= x = ( 20 ) / ( -2 )

= x = -10

now, put the valve of ten in the blank .

= ( 20 ) ÷ ( -10 ) = -2

**(h) _____ ÷ (4) = –3**

**Solution:-**

Let us assume the missing integer be x ,

then ,

= ( x ) ÷ 4 = -3

= x = ( -3 ) × 4

= x = -12

immediately, put the valve of adam in the blank .

= ( -12 ) ÷ 4 = -3

**4. Write five pairs of integers (a, b) such that a ÷ b = –3. One such pair is (6, –2) because 6 ÷ (–2) = (–3).**

**Solution:-**

( one ) ( 15, -5 )

Because, 15 ÷ ( –5 ) = ( –3 )

( two ) ( -15, 5 )

Because, ( -15 ) ÷ ( 5 ) = ( –3 )

( three ) ( 18, -6 )

Because, 18 ÷ ( –6 ) = ( –3 )

( intravenous feeding ) ( -18, 6 )

Because, ( -18 ) ÷ 6 = ( –3 )

( vanadium ) ( 21, -7 )

Because, 21 ÷ ( –7 ) = ( –3 )

**5. The temperature at 12 noon was 10oC above zero. If it decreases at the rate of 2oC per hour until midnight, at what time would the temperature be 8°C below zero? What would be the temperature at mid-night?**

**Solution:-**

From the question is given that ,

temperature at the beginning i, at 12 noon = 10oC

rate of change of temperature = – 2oC per hour

then ,

temperature at 1 PM = 10 + ( -2 ) = 10 – 2 = 8oC

temperature at 2 PM = 8 + ( -2 ) = 8 – 2 = 6oC

temperature at 3 PM = 6 + ( -2 ) = 6 – 2 = 4oC

temperature at 4 PM = 4 + ( -2 ) = 4 – 2 = 2oC

temperature at 5 PM = 2 + ( -2 ) = 2 – 2 = 0oC

temperature at 6 PM = 0 + ( -2 ) = 0 – 2 = -2oC

temperature at 7 PM = -2 + ( -2 ) = -2 -2 = -4oC

temperature at 8 PM = -4 + ( -2 ) = -4 – 2 = -6oC

temperature at 9 PM = -6 + ( -2 ) = -6 – 2 = -8oC

∴At 9 PM the temperature will be 8oC below zero

then ,

The temperature at mid-night i, at 12 AM

change in temperature in 12 hours = -2oC × 12 = – 24oC

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

= – 14oC

so, at midnight temperature will be 14oC below 0 .

**6. In a class test (+ 3) marks are given for every correct answer and (–2) marks are given for every incorrect answer and no marks for not attempting any question. (i) Radhika scored 20 marks. If she has got 12 correct answers, how many questions has she attempted incorrectly? (ii) Mohini scores –5 marks in this test, though she has got 7 correct answers. How many questions has she attempted incorrectly?**

**Solution:-**

From the question ,

Marks awarded for 1 correct answer = + 3

Marks awarded for 1 wrong answer = -2

( one ) Radhika scored 20 marks

then ,

sum marks awarded for 12 correct answers = 12 × 3 = 36

Marks awarded for faulty answers = entire mark – full marks awarded for 12 correct

Answers

= 20 – 36

= – 16

sol, the numeral of faulty answers made by Radhika = ( -16 ) ÷ ( -2 )

= 8

( two ) Mohini scored -5 marks

then ,

total marks awarded for 7 discipline answers = 7 × 3 = 21

Marks awarded for faulty answers = sum score – total marks awarded for 12 correct

Answers

= – 5 – 21

= – 26

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

= 13

**7. An elevator descends into a mine shaft at the rate of 6 m/min. If the descent starts from 10 m above the ground level, how long will it take to reach – 350 m.**

**Solution:-**

From the interview ,

The initial altitude of the elevator = 10 thousand

Final astuteness of elevator = – 350 thousand … [ ∵distance descended is denoted by a negative

integer ] The total distance to descended by the elevator = ( -350 ) – ( 10 )

= – 360 megabyte

then ,

Time taken by the elevator to descend -6 m = 1 min

therefore, time taken by the elevator to descend – 360 thousand = ( -360 ) ÷ ( -60 )

= 60 minutes

= 1 hour

**Disclaimer:**

**Dropped Topics** – Introduction, Recall, 1.4.3 intersection of three or more negative numbers and 1.5.7 Making generation easier .

## frequently Asked Questions on NCERT Solutions for Class 7 Maths Chapter 1

### Where can I get the accurate solution for NCERT Solution for Class 7 Maths Chapter 1?

At BYJU ’ S you can get the accurate solution in PDF format for NCERT Solution for Class 7 Maths Chapter 1. The NCERT Textbook Solutions for the chapter have been designed accurately by mathematics experts at BYJU ’ S. All these solutions are provided by considering the new pattern of CBSE, so that students can get exhaustive cognition for their exams.

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### Is it necessary to solve each problem provided in the NCERT Solution for Class 7 Maths Chapter 1?

Yes. Because these questions are important from an examination position. These questions are solved by experts to help the students to crack exercise very easily. These solutions help students to familiarize themselves with the integers. Solutions are available in PDF format on BYJU ’ S web site .

### List out the concepts covered in NCERT Solution for Class 7 Maths Chapter 1?

The concepts are covered in NCERT Solution for Class 10 Maths chapter 1 are

1. introduction of Integers

2. Properties of Addition and Subtraction of Integers

3. multiplication of Integers

4. multiplication of a Positive and Negative Integer

5. multiplication of two negative Integer

6. Properties of Multiplication of Integers

7. division of Integers

8. Properties of Division of Integers .