NCERT Solutions for Class 7 Maths Chapter 1 - Integers

Exercise 1.1

1. Following number line shows the temperature in degree Celsius $\left( ^{\circ }\text{C} \right)$ at different places on a particular day:

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

Ans: First find the distance between two marks on the number line. Observe that there is total five partitions between every two consecutive numbers written say $0,5$. Hence, one partition will represent ${{1}^{\circ }}C$.

Therefore, the temperature of the places marked on it is:

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

Ans: From the table we can see that the temperature of the hottest place is Bangalore which is ${{22}^{\circ }}\text{C}$ and the temperature of the coldest place Lahulspiti $-{{8}^{\circ }}\text{C}$ and their difference will be 

Difference = Temperature of Bangalore – Temperature of Lahulspiti 

Difference $=22-\left( -8 \right)=22\text{+}8\text{=3}{{\text{0}}^{\circ }}\text{C}$.

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

Ans: From the table we can see that the temperature of the Srinagar is $-{{2}^{\circ }}\text{C}$ and the temperature of Lahulspiti is $-{{8}^{\circ }}\text{C}$ and their difference will be 

Difference = Temperature of Srinagar – Temperature of Lahulspiti 

Difference \[=-2-\left( -8 \right)=-2+8={{6}^{\circ }}\text{C}\].

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?

Ans: From the table we can see that the temperature of the Srinagar is $-{{2}^{\circ }}\text{C}$ and the temperature of Shimla is ${{5}^{\circ }}\text{C}$ and their sum will be 

Sum = Temperature of Shimla – Temperature of Srinagar

Sum \[=5+\left( -2 \right)=5-2={{3}^{\circ }}\text{C}\].  …..(1)

Temperature of Shimla $={{5}^{\circ }}\text{C}$.  …..(2)

From (1) and (2), the temperature of Srinagar and Shimla taken together is less than the temperature at Shimla.

Temperature of Srinagar $=-{{2}^{\circ }}\text{C}\,$.  …..(3)

From (1) and (3), the temperature of Srinagar and Shimla taken together is not less than the temperature at 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?

Ans: Jack’s scores in five successive rounds are $25,-5,-10,15\text{ and }10$. Then the total marks got by Jack will be the sum of all these marks i.e., 

$25+\left( -5 \right)+\left( -10 \right)+15+10=25-5-10+15+10$ 

$\Rightarrow 25-15+5+10=35$

Therefore, Jack scored \[35\] marks in the quiz.

3. At Srinagar temperature was \[-{{5}^{\circ }}\text{C}\] on Monday and then it dropped by ${{2}^{\circ }}\text{C}$ on Tuesday. What was the temperature of Srinagar on Tuesday? On Wednesday, it rose by${{4}^{\circ }}\text{C}$. What was the temperature on this day?

Ans: It is given that on Monday, temperature at Srinagar is $-{{5}^{\circ }}\text{C}$. On Tuesday, temperature dropped by ${{2}^{\circ }}\text{C}$ i.e., the temperature becomes $2$ less than Monday.

$\therefore $  Temperature on Tuesday$=-{{5}^{\circ }}\text{C}-{{2}^{\circ }}\text{C}=-{{7}^{\circ }}\text{C}$  ….. (1)

On Wednesday, temperature rose up by ${{4}^{\circ }}\text{C}$ i.e., the temperature becomes $4$ more than Tuesday. Hence from (1),

$\therefore $ Temperature on Wednesday $=-{{7}^{\circ }}\text{C}\,\text{+}\,{{\text{4}}^{\circ }}\text{C}=-{{3}^{\circ }}\text{C}$ 

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

Ans: Given, height of a place above the sea level is $5000\text{ m}$ and a floating submarine is $1200\text{ m}$ below the sea level. Let us denote the distance above the sea level by $+$ sign and the distance below the sea level by  $-$ sign.

Hence, Distance of plane from sea level $=5000m$ 

Distance of submarine from sea level $=-1200m$ 

$\therefore $  The vertical distance between the plane and the submarine is $5000-\left( -1200 \right)=6200\text{ m}$.

Thus, the vertical distance between the plane and the submarine is $6200\text{ m}$.

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 accounts after the withdrawal?

Ans: Deposit and Withdrawal are opposite of each other. Therefore, if we are representing withdrawal by $-$, we should represent deposit by $+$.

Given, Deposit amount $=$ ₹$2,000$ and Withdrawal amount $=$ ₹$1,642$. In integer representation, deposit $=+2000$ and withdrawal $=-1642$. 

$\therefore $  Balance$=2,000-1,642$ $=$ ₹$358$

Therefore, the balance in Mohan’s account after 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?

Ans: East and West are opposite of each other. Therefore, if we are representing east by $+$, we should represent west by $-$.

Let $A$ be the point of origin i.e., $A$ will lie at $0$ on the number line.

Given, 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.

Plotting the given information on the number line we get,

Distance from $A$ to $B$$=20\text{ km}$ 

Distance from $B$ to final point $=30\text{ km}$ 

Distance from $A$ to final point $=20+\left( -30 \right)=-10\,\text{km}$.

Therefore, Rita is $10$ km west from her initial position $A$ and will be represented by the integer $-10$.

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

Ans: In a magic square each row, column and diagonal have the same sum. So let us check whether the given square is a magic square or not.

Sum of row $1$ $=5+\left( -1 \right)+\left( -4 \right)=0$  …..(1) 

Sum of row $2$ $=\left( -5 \right)+\left( -2 \right)+7=-7+7=0$  …..(2) 

Sum of row $3=$ $0+3+\left( -3 \right)=3-3=0$  …..(3) 

Sum of column $1=$ $5+\left( -5 \right)+0=5-5=0$  …..(4) 

Sum of column $2=$ $\left( -1 \right)+\left( -2 \right)+3=-3+3=0$  …..(5) 

Sum of column $3=$ $\left( -4 \right)+7+\left( -3 \right)=7-7=0$  …..(6) 

Sum of diagonal $1=$ $5+\left( -2 \right)+\left( -3 \right)=5-5=0$  …..(7) 

Sum of diagonal $2=$ $\left( -4 \right)+\left( -2 \right)+0=-6$  …..(8) 

From (1) to (8) we can conclude that this box is not a magic square because all the sums are not equal.

Ans: In a magic square each row, column and diagonal have the same sum. So let us check whether the given square is a magic square or not.

Sum of row $1=$ $1+\left( -10 \right)+0=1-10=-9$  …..(1) 

Sum of row $2=$ $\left( -4 \right)+\left( -3 \right)+\left( -2 \right)=-7-2=-9$  …..(2) 

Sum of row $3=$ $\left( -6 \right)+4+\left( -7 \right)=-2-7=-9$  …..(3) 

Sum of column $1=$ $1+\left( -10 \right)+0=1-10=-9$  …..(4) 

Sum of column $2=$ $\left( -10 \right)+\left( -3 \right)+4=-13+4=-9$  …..(5) 

Sum of column $3=$ $0+\left( -2 \right)+\left( -7 \right)=0-9=-9$  …..(6) 

Sum of diagonal $1=$ $1+\left( -3 \right)+\left( -7 \right)=1-10=-9$  …..(7) 

Sum of diagonal $2=$ $0+\left( -3 \right)+\left( -6 \right)=-9$  …..(8) 

From (1) to (8) we can conclude that this box is magic square because all the sums are equal.

8. Verify $a-\left( -b \right)=a+b$ for the following values of $a\,\text{and }b$:

i. Given, $a=21,b=18$

Ans: Given, $a=21,b=18$. We have to verify $a-\left( -b \right)=a+b$.

LHS $=21-\left( -18 \right)=21+18=39$  ….. (1) 

RHS $=a+b=21+18=39$  ….. (2)

From (1) and (2), since LHS $=$RHS therefore, it is verified that $a-\left( -b \right)=a+b$.

ii) Given, $a=118,b=125$

Ans: Given, $a=118,b=125$. We have to verify $a-\left( -b \right)=a+b$.

LHS $=118-\left( -125 \right)=118+125=243$  ….. (1) 

RHS $=118+125=243$  ….. (2)

From (1) and (2), since LHS $=$RHS therefore, it is verified that $a-\left( -b \right)=a+b$.

iii) Given, $a=75,b=84$

Ans: Given, $a=75,b=84$. We have to verify $a-\left( -b \right)=a+b$.

LHS $=75-\left( -84 \right)=75+84=159$  ….. (1) 

RHS $=a+b=75+84=159$  ….. (2)

From (1) and (2), since LHS $=$RHS therefore, it is verified that $a-\left( -b \right)=a+b$.

iv) Given, $a=28,b=11$

Ans: Given, $a=28,b=11$. We have to verify $a-\left( -b \right)=a+b$.

LHS $=28-\left( -11 \right)=28+11=39$  ….. (1) 

RHS $=28+11=39$  ….. (2)

From (1) and (2), since LHS $=$RHS therefore, it is verified that $a-\left( -b \right)=a+b$.

9. Use the sign of $>,<\,\,\text{or}\,=$ in the blank to make the statements true:

a) $\left( -8 \right)+\left( -4 \right)\,\_\_\left( -8 \right)-\left( -4 \right)$.

Ans: $\left( -8 \right)+\left( -4 \right)=-12$ and $\left( -8 \right)-\left( -4 \right)=-4$ 

$\Rightarrow \text{ }-12\,\,<\,\,-4$.

$\Rightarrow \left( -8 \right)+\left( -4 \right)\,<\,\,\left( -8 \right)-\left( -4 \right)\,$

b) $\left( -3 \right)+7-\left( 19 \right)\_\_15-8+\left( -9 \right)$ 

Ans: $\left( -3 \right)+7-\left( 19 \right)\,\,=-15$ and $15-8+\left( -9 \right)=-2$ 

$\Rightarrow -15\,\,<-2$ 

$\Rightarrow \left( -3 \right)+7-\left( 19 \right)\,<15-8+\left( -9 \right)$

c) $23-41+11\_\_23-41-11$ 

Ans: $23-41+11=-7$ and $23-41-11=-29$ 

$\Rightarrow -7>-29$ 

$\Rightarrow 23-41+11>23-41-11$

d) $39+\left( -24 \right)-\left( 15 \right)\_\_36+\left( -52 \right)-\left( -36 \right)$ 

Ans: $39+\left( -24 \right)-\left( 15 \right)\,=0$ and $36+\left( -52 \right)-\left( -36 \right)=20$ 

$\Rightarrow 0<20$

$\Rightarrow 39+\left( -24 \right)-\left( 15 \right)<36+\left( -52 \right)-\left( -36 \right)$ 

e) $\left( -231 \right)+79+51\_\_\left( -399 \right)+159+81$ 

Ans: \[\left( -231 \right)+79+51=-101\] and \[\left( -399 \right)+159+81=-159\] 

\[\Rightarrow -101>-159\]

\[\Rightarrow \left( -231 \right)+79+51>\left( -399 \right)+159+81\]

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 jump back $2$ steps up. In how many jumps will he reach the water level?

Ans: Given that the monkey jumps $3$ steps down and jumps back $2$ steps up. We can represent the path of the monkey as:

First jump $=1+3=4$ steps

Second jump $=4-2=2$ steps

Third jump $=2+3=5$ steps 

Fourth jump $=5-2=3$ steps 

Fifth jump $=3+3=6$ steps

Sixth jump $=6-2=4$ steps

Seventh jump $=4+3=7$ steps

Eighth jump $=7-2=5$ steps

Ninth jump $=5+3=8$ steps 

Tenth jump $=8-2=6$ steps

Eleventh jump $=6+3=9$ steps 

Therefore, the monkey will reach ninth steps in $11$ jumps.

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?

Ans: Given that the monkey jumps four steps up and them jumps down $2$ steps down. We can represent the path of the monkey as:

First jump $=9-4=5$ steps

Second jump $=5+2=7$ steps

Third jump $=7-4=3$ steps 

Fourth jump $=3+2=5$ steps 

Fifth jump $=5-4=1$ steps

Therefore, the monkey reaches back on the first step in fifth jump.

iii) If the number of steps moved down is represented by negative integers and the number of steps move up by positive integers, represent his moves in part (i) and (ii) by completing the following: 

1. $-3+2-..........=-8$

2. $4-2+..........=8$ 

In (a) the sum (−8) represent going down by eight steps. So, what will the sum 8 in (b) represent? 

Ans:  (a) $-3+2-3+2-3+2-3+2-3+2-3+2-3+2-3+2=-8$ 

(b)  $4-2+4-2+4-2+4-2=8$ 

Thus, sum $8$ in (b) represents going up by eight steps.

Exercise 1.2

1. Write down a pair of integers whose:

a) sum is $-7$ 

Ans: The pair of integers whose sum is $-7$ is $\left( -4,-3 \right)$ i.e., $-4+\left( -3 \right)=-7$.

b) difference is $-10$

Ans: The pair of integers whose difference is $-10$ is $\left( -3,7 \right)$ i.e., $-3-7=-10$

c) sum is $0$

Ans: The pair of integers whose sum is $0$ is $\left( -30,30 \right)$ i.e., $-30+30=0$.

2. Find the integers in the below questions:

a) Write a pair of negative integers whose difference gives $8$.

Ans: The pair of negative integers whose difference is $8$ is $\left( -1,-9 \right)$ i.e., $-1-\left( -9 \right)=-1+9=8$

b) Write a negative integer and a positive integer whose sum is $-5$. 

Ans: The pair of negative and positive integers whose sum is $-5$ is $\left( -9,4 \right)$ i.e., $\left( -9 \right)+4=-5$.

c) Write a negative integer and a positive integer whose difference is $-3$.

Ans: The pair of negative and positive integers whose difference is $-3$ is $\left( -1,2 \right)$ i.e., $\left( -1 \right)-2=-1-2=-3$.

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

Ans: Given that the team $A$ scored $-40,10,0$. Therefore, total score of the team $A$ $=-40+10+0=-30$.

Given that the team $B$ scored $10,0,-40$.Therefore, total score of Team $B$ $=10+0+\left( -40 \right)=-30$. 

Therefore, scores of both teams are same and we can add integers in any order due to commutative property.

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

i) $\left( -5 \right)+\left( -8 \right)=\left( -8 \right)+\left( ...... \right)$ 

Ans: Using Commutative Property we can fill the blank as $-5$. 

$\therefore $ $\left( -5 \right)+\left( -8 \right)=\left( -8 \right)+\left( -5 \right)$.

ii) $-53+\left( ...... \right)=-53$

Ans: Using Zero additive property we can fill the blank as $0$ .

$\therefore $ \[-53+0=-53\] 

iii) $17+\left( ...... \right)=0$ 

Ans: Using Zero Additive identity we can fill the blank as $-17$ .

$\therefore $ $17+\left( -17 \right)=0$ 

iv) $\left[ 13+\left( -12 \right) \right]+\left( ...... \right)=13+\left[ \left( -12 \right)+\left( -7 \right) \right]$ 

Ans: Using Associative property we can fill the blank as $-7$.

$\therefore $ $\left[ 13+\left( -12 \right) \right]+\left( -7 \right)=13+\left[ \left( -12 \right)+\left( -7 \right) \right]$ 

v) $\left( -4 \right)+\left[ 15+\left( -3 \right) \right]=\left[ -4+15 \right]+\left( ...... \right)$

Ans: Using Associative property we can fill the blank as $-3$.

$\therefore $ $\left( -4 \right)+\left[ 15+\left( -3 \right) \right]=\left[ -4+15 \right]+\left( -3 \right)$ 

Exercise 1.3

1. Find each of the following products:

a) $3\,\,\times \,\,\left( -1 \right)$

Ans: While multiplying a negative integer and a positive integer, multiply them as whole numbers and then put a minus sign $\left( - \right)$ before the product i.e., 

$3\times \left( -1 \right)=-3$

b) $\left( -1 \right)\,\,\times \,\,225$

Ans: While multiplying a negative integer and a positive integer, multiply them as whole numbers and then put a minus sign $\left( - \right)$ before the product i.e., 

$\left( -1 \right)\times 225=-225$

c) $\left( -21 \right)\,\,\times \,\,\left( -30 \right)$

Ans: While multiplying two negative integers, multiply them as whole numbers and then put a plus sign $\left( + \right)$ before the product i.e., 

$\left( -21 \right)\times \left( -30 \right)=630$

d) $\left( -316 \right)\,\,\times \,\,\left( -1 \right)$

Ans: While multiplying two negative integers, multiply them as whole numbers and then put a plus sign $\left( + \right)$ before the product i.e., 

$\left( -316 \right)\times \left( -1 \right)=316$

e) $\left( -15 \right)\,\,\times \,\,0\,\,\times \,\,\left( -30 \right)$

Ans: While multiplying two negative integers, multiply them as whole numbers and then put a plus sign $\left( + \right)$ before the product i.e., 

$\left( -15 \right)\times \,0\times \left( -18 \right)=0$

f) $\left( -12 \right)\,\,\times \,\,\left( -11 \right)\,\,\times \,\,\left( 10 \right)$

Ans: While multiplying two negative integers, multiply them as whole numbers and then put a plus sign $\left( + \right)$ before the product i.e., 

$\left[ \left( -12 \right)\times \left( -11 \right) \right]\times \left( 10 \right)=132\times 10=1320$

g) $9\,\,\times \,\,\left( -3 \right)\,\,\times \,\,\left( -6 \right)$

Ans: While multiplying two negative integers, multiply them as whole numbers and then put a plus sign $\left( + \right)$ before the product i.e., 

$9\times \left[ \left( -3 \right)\times \left( -6 \right) \right]=9\times 18=162$

h) $\left( -18 \right)\,\,\times \,\,\left( -5 \right)\,\,\times \,\,\left( -4 \right)$

Ans: While multiplying two negative integers, multiply them as whole numbers and then put a plus sign $\left( + \right)$ before the product i.e., $\left[ \left( -18 \right)\times \left( -5 \right) \right]\times \left( -4 \right)=90\times \left( -4 \right)$   ….. (1)

While multiplying a negative integer and a positive integer, multiply them as whole numbers and then put a minus sign $\left( - \right)$ before the product i.e., from (1), 

$\left[ \left( -18 \right)\times \left( -5 \right) \right]\times \left( -4 \right)=-360$

i) $\left( -1 \right)\,\,\times \,\,\left( -2 \right)\,\,\times \,\,\left( -3 \right)\,\,\times \,\,4$

Ans: While multiplying two negative integers, multiply them as whole numbers and then put a plus sign $\left( + \right)$ before the product and while multiplying a negative integer and a positive integer, multiply them as whole numbers and then put a minus sign $\left( - \right)$ before the product i.e., 

$\left[ \left( -1 \right)\times \left( -2 \right) \right]\times \left[ \left( -3 \right)\times 4 \right]=2\times \left( -12 \right)=-24$

j) $\left( -3 \right)\,\,\times \,\,\left( -6 \right)\,\,\times \,\,\left( 2 \right)\,\,\times \,\,\left( -1 \right)$

Ans: While multiplying two negative integers, multiply them as whole numbers and then put a plus sign $\left( + \right)$ before the product and while multiplying a negative integer and a positive integer, multiply them as whole numbers and then put a minus sign $\left( - \right)$ before the product i.e., 

$\left[ \left( -3 \right)\times \left( -6 \right) \right]\times \left[ \left( 2 \right)\times \left( -1 \right) \right]=\left( 18 \right)\times \left( -2 \right)=-36$

2. Verify the following:

a) $18\,\,\times \,\,\left[ 7+\left( -3 \right) \right]=\left[ 18\times 7 \right]+\left[ 18\times \left( -3 \right) \right]$

Ans: Given expression, $18\times \left[ 7+\left( -3 \right) \right]=\left[ 18\times 7 \right]+\left[ 18\times \left( -3 \right) \right]$.

Simplifying the given expression by first solving the square brackets.

While multiplying a negative integer and a positive integer, multiply them as whole numbers and then put a minus sign $\left( - \right)$ before the product.

$\Rightarrow \,\,18\times \left[ 4 \right]=\left[ 126 \right]+\left[ -54 \right]$ 

$\Rightarrow \,\,72=72$

$\Rightarrow \,\,\text{L}\text{.H}\text{.S}\text{.}=\text{R}\text{.H}\text{.S}\text{.}$

Hence verified.

b) $\left( -21 \right)\times \left[ \left( -4 \right)+\left( -6 \right) \right]=\left[ \left( -21 \right)\times \left( -4 \right) \right]+\left[ \left( -21 \right)\times \left( -6 \right) \right]$ 

Ans: Given expression, $\left( -21 \right)\times \left[ \left( -4 \right)+\left( -6 \right) \right]=\left[ \left( -21 \right)\times \left( -4 \right) \right]+\left[ \left( -21 \right)\times \left( -6 \right) \right]$

Simplifying the given expression by first solving the square brackets.

While multiplying two negative integers, multiply them as whole numbers and then put a plus sign $\left( + \right)$ before the product

$\Rightarrow \,\,\left( -21 \right)\times \left( -10 \right)=84+126$

$\Rightarrow \,\,210=210$

$\Rightarrow \,\,\text{L}\text{.H}\text{.S}\text{.}=\text{R}\text{.H}\text{.S}\text{.}$                                                                              

Hence verified.

3. Solve the following:

i)  For any integer $a$, what is $\left( -1 \right)\,\,\times \,\,a$ equal to?

Ans: $\left( -1 \right)\times a=-a,\,\text{ where }a\text{ is an integer}\text{.}$

ii) Determine the integer whose product with $\left( -1 \right)$ is:

a) $-22$

Ans: The integer whose product with $-1$ is \[-22\] is $22$ i.e., $\left( -1 \right)\times \left( 22 \right)=-22$. 

b) $37$ 

Ans: The integer whose product with $-1$ is \[37\] is $-37$ i.e., $\left( -1 \right)\times \left( -37 \right)=37$.

c) $0$ 

Ans: The integer whose product with $-1$ is \[0\] is $0$ i.e., $-1\times 0=0$.

4. Starting from $\left( -1 \right)\,\,\times \,\,5$ write various products showing some patterns to show $\left( -1 \right)\,\,\times \,\,\left( -1 \right)=1$.

Ans: Consider the product, $\left( -1 \right)\times 5=-5$

Also, $\left( -1 \right)\times 4=-4$, $\left( -1 \right)\times 3=-3$, $\left( -1 \right)\times 2=-2$, $\left( -1 \right)\times 1=-1$, etc.

Thus, we can observe that the product of one negative integer and one positive integer is negative integer.

Similarly, $\left( -1 \right)\times \left( -1 \right)=1$ i.e., the product of two negative integers is a positive integer.

5. Find the product, using suitable properties:

a) $26\,\,\times \,\,\left( -48 \right)+\left( -48 \right)\,\,\times \,\,\left( -36 \right)$

Ans: $26\times \left( -48 \right)+\left( -48 \right)\times \left( -36 \right)$ …..(1)

Taking $-48$ common from (1) using distributive property we get,

 $\Rightarrow \,\,\left( -48 \right)\times \left[ 26+\left( -36 \right) \right]$

$\Rightarrow \,\,\left( -48 \right)\times \left( -10 \right)$

$\Rightarrow \,\,480$

b) $8\,\,\times \,\,53\,\,\times \,\,\left( -125 \right)$

Ans: $8\times 53\times \left( -125 \right)$ …..(1)

Using Commutative property on (1) we get,

$\Rightarrow \,\,53\times \left[ 8\times \left( -125 \right) \right]$

\[\Rightarrow \,\,53\times \left( -1000 \right)\]

$\Rightarrow \,\,-53000$

c) $15\,\,\times \,\,\left( -25 \right)\times \left( -4 \right)\,\,\times \,\,\left( -10 \right)$

Ans: $15\times \left( -25 \right)\times \left( -4 \right)\times \left( -10 \right)$ 

$\Rightarrow \,\,15\times \left[ \left( -25 \right)\times \left( -4 \right)\times \left( -10 \right) \right]$   …..(1)

Using Commutative property on (1) we get,

\[\Rightarrow \,\,15\times \left( -4 \right)\times \left[ -25\times -10 \right]\]

\[\Rightarrow \,\,15\times \left( -4 \right)\times 250\] 

$\Rightarrow \,\,-15000$

d) $\left( -41 \right)\,\,\times \,\,\left( 102 \right)$

Ans: $\left( -41 \right)\times \left( 102 \right)$

$\Rightarrow \,\,-41\times \left[ 100+2 \right]$    …..(1)          

Using distributive property on (1) we get,

$\Rightarrow \,\,\left[ \left( -41 \right)\times 100 \right]+\left[ \left( -41 \right)\times 2 \right]$

$\Rightarrow \,\,-4100+\left( -82 \right)$

$\Rightarrow \,\,-4182$ 

e) $625\,\,\times \,\,\left( -35 \right)+\left( -625 \right)\,\,\times \,\,65$

Ans: $625\times \left( -35 \right)+\left( -625 \right)\times 65$ …..(1)

Taking $625$ common from (1) using distributive property we get,

$\Rightarrow \,\,625\times \left[ \left( -35 \right)+\left( -65 \right) \right]$

$\Rightarrow \,\,625\times \left( -100 \right)$

$\Rightarrow \,\,-62500$

f) $7\,\,\times \,\,\left( 50-2 \right)$

Ans: $7\times \left( 50-2 \right)$ ….. (1)

Using distributive property on (1) we get,

$\Rightarrow \,\,7\times 50-7\times 2$  

$\Rightarrow \,\,350-14=336$

g) $\left( -17 \right)\,\,\times \,\,\left( -29 \right)$

Ans: $\left( -17 \right)\times \left( -29 \right)$ 

$\Rightarrow \,\,\left( -17 \right)\times \left[ \left( -30 \right)+1 \right]$  …..(1)

Using distributive property on (1) we get,

$\Rightarrow \,\,\left( -17 \right)\times \left( -30 \right)+\left( -17 \right)\times 1$

$\Rightarrow \,\,510+\left( -17 \right)$       

$\Rightarrow \,\,493$

h) $\left( -57 \right)\,\,\times \,\,\left( -19 \right)+57$

Ans: $\left( -57 \right)\times \left( -19 \right)+57$ 

$\Rightarrow \,\,\left( -57 \right)\times \left( -19 \right)+57\times 1$

$\Rightarrow \,\,57\times 19+57\times 1$        …..(1)

Taking $57$ common from (1) using distributive property we get,

$\Rightarrow \,\,57\times \left( 19+1 \right)$ 

$\Rightarrow \,\,57\times 20=1140$

6. A certain freezing process requires that room temperature be lowered from ${{40}^{\circ }}\text{C}$ at the rate of ${{5}^{\circ }}\text{C}$ every hour. What will be the room temperature $10$ hours after the process begins?

Ans: It is given that the present temperature of the room is ${{40}^{\circ }}\text{C}$. Also, the temperature is decreasing the temperature every hour by ${{5}^{\circ }}\text{C}$.

Therefore, temperature after $1$ hour is ${{40}^{\circ }}-{{5}^{\circ }}={{35}^{\circ }}C$.

Temperature after $2$ hours is ${{40}^{\circ }}-2\times \left( {{5}^{\circ }} \right)={{30}^{\circ }}C$. 

Temperature after $3$ hours is ${{40}^{\circ }}-3\times \left( {{5}^{\circ }} \right)={{25}^{\circ }}C$. 

Similarly, temperature after $n$ hours is ${{\left[ 40-n\left( 5 \right) \right]}^{\circ }}C$.    ….. (1)

Form (1), room temperature after 10 hours \[={{\left[ 40-10\times \left( 5 \right) \right]}^{\circ }}C\]

$\Rightarrow {{\left[ 40-50 \right]}^{\circ }}\text{C}=-{{10}^{\circ }}\text{C}$

Therefore, the room temperature $10$ hours after the process begins is $-{{10}^{\circ }}\text{C}$.

7. In a class test containing $10$ questions, $5$ marks are awarded for every correct answer and $\left( -2 \right)$ 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? 

Ans: Given that out of $10$ questions, Mohan gets $4$ correct and $6$ incorrect answers. 

Marks for $1$ correct answer $=5$ 

Marks for $4$ correct answers $=4\times 5=20$  …..(1)

Marks for $1$ wrong answer $=-2$ 

Marks for $6$ wrong answers $=6\times \left( -2 \right)=-12$  …..(2)

Therefore from (1) and (2), total scores of Mohan$=\left( 20 \right)+\left( -12 \right)=8$.    

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

Ans: Given that out of $10$ questions, Reshma gets $5$ correct and $5$ incorrect answers. 

Marks for $1$ correct answer $=5$ 

Marks for $5$ correct answers $=5\times 5=25$  …..(1)

Marks for $1$ wrong answer $=-2$ 

Marks for $5$ wrong answers $=5\times \left( -2 \right)=-10$  …..(2)

Therefore from (1) and (2), total scores of Resham $=\left( 25 \right)+\left( -10 \right)=15$.    

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

Ans: Given that out of $10$ questions, Heena gets $2$ correct, $5$ incorrect and $3$ un-attempted.

Marks for $1$ correct answer $=5$ 

Marks for $2$ correct answers $=2\times 5=10$  …..(1)

Marks for $1$ wrong answer $=-2$ 

Marks for $5$ wrong answers $=5\times \left( -2 \right)=-10$  …..(2)

Marks for $1$ un-attempt question $=0$ 

Marks for $3$ un-attempt questions $=0\times 3=0$  …..(3)

Therefore from (1), (2) and (3), total scores of Heena $=\left( 10 \right)+\left( -10 \right)+0=0$.

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?

Ans: Given that profit of 1 bag of white cement $=$ ₹$8$

Therefore, Profit on selling $3000$ bags of white cement

\[=3000\times 8=\] ₹$24,000$ ….. (1) 

And loss of 1 bag of grey cement $=$ ₹$5$.

Therefore, Loss on selling $5000$ bags of grey cement

\[=5000\times 5=\] ₹$25,000$ …..(2)

Let us denote profit by  $+$ and loss by $-$. Then from (1) and (2), 

Total selling $=24000+\left( -25000 \right)=-1000$.

Therefore, his total loss on selling cement bags is ₹$1,000$. 

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.

Ans: Let the number of bags of white cement bags sold be $x$.

Given that profit of 1 bag of white cement $=$ ₹$8$

Therefore, Profit on selling $x$ bags of white cement ₹\[8x\] ….. (1) 

And loss of 1 bag of grey cement $=$ ₹$5$.

Therefore, Loss on selling $6400$ bags of grey cement 

\[=6400\times 5=\] ₹$32,000$ …..(2)

To have neither profit nor loss, from (1) and (2) we get,

$8x=32000$

$\Rightarrow \,\,x=\dfrac{32000}{8}\text{ }$ 

$\therefore x=4000$ 

Thus, he must sell $4000$ white cement bags to have neither profit nor loss.

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

a) $\left( -3 \right)\,\,\times \,\,\_\_\_\_\_=27$$$

Ans: Let the number in the blank be $x$, then 

\[\left( -3 \right)\times x=27\] 

$\Rightarrow x=\dfrac{27}{-3}$ 

$\Rightarrow x=-9$ 

\[\therefore \left( -3 \right)\times \underline{\left( -9 \right)}=27\]

b) $5\,\,\times \,\_\_\_\_\_=-35$$$

Ans: Let the number in the blank be $x$, then

\[5\times x=-35\text{ }\] 

$\Rightarrow x=\dfrac{-35}{5}$ 

$\Rightarrow x=-7$ 

\[\therefore 5\times \underline{\left( -7 \right)}=-35\text{ }\]

c) $\_\_\_\_\_\times \left( -8 \right)=-56$$$

Ans: Let the number in the blank be $x$, then

$x\times \left( -8 \right)=-56$ 

$\Rightarrow x=\dfrac{-56}{-8}$ 

$\Rightarrow x=7$ 

$\therefore \underline{7}\times \left( -8 \right)=-56$

d) $\_\_\_\_\_\times \left( -12 \right)=132$$$

Ans: Let the number in the blank be $x$, then

\[x\times \left( -12 \right)=132\] 

\[\Rightarrow x=\dfrac{132}{-12}\] 

$\Rightarrow x=-11$ 

\[\therefore \underline{\left( -11 \right)}\times \left( -12 \right)=132\]

Exercise 1.4

1. Evaluate each of the following:

a) \[\left( -30 \right)\div 10\]

Ans: While dividing a negative integer by a positive integer, divide them as whole numbers and then put a minus sign $\left( - \right)$ before the quotient. Therefore, \[\left( -30 \right)\div \text{10}=-\dfrac{30}{10}=-3\].

b) \[50\div \left( -5 \right)\]

Ans: While dividing a positive integer by a negative integer, divide them as whole numbers and then put a minus sign $\left( - \right)$ before the quotient. Therefore, \[50\div \left( -5 \right)=-\dfrac{50}{5}=-10\].

c) \[\left( -36 \right)\div \left( -9 \right)\]

Ans: While dividing a positive integer by a positive integer, divide them as whole numbers and then put a plus sign $\left( + \right)$ before the quotient. Therefore, \[\left( -36 \right)\div \left( -9 \right)=\dfrac{36}{9}=4\].

d) \[\left( -49 \right)\div 49\]

Ans: While dividing a negative integer by a positive integer, divide them as whole numbers and then put a minus sign $\left( - \right)$ before the quotient. Therefore, \[\left( -49 \right)\div 49=-\dfrac{49}{49}=-1\].

e) $13\div \left[ \left( -2 \right)+1 \right]$

Ans: Simplifying the given expression, \[13\div \left[ \left( -\text{2} \right)+1 \right]=13\div \left( -1 \right)\] ….(1)

While dividing a positive integer by a negative integer, divide them as whole numbers and then put a minus sign $\left( - \right)$ before the quotient. Therefore from (1), \[13\div \left( -1 \right)=-13\].

f) $0\div \left( -12 \right)$

Ans: While dividing $0$ by any integer, the quotient is $0$. Therefore  \[0\div \left( -12 \right)=0\].

g) $\left( -31 \right)\div \left[ \left( -30 \right)\div \left( -1 \right) \right]$

Ans: While dividing a positive integer by a positive integer, divide them as whole numbers and then put a plus sign $\left( + \right)$ before the quotient. Therefore, \[\left( -30 \right)\div \left( -1 \right)=30\].  …..(1)

Hence from (1), \[\left( -31 \right)\div \left[ \left( -30 \right)\div \left( -1 \right) \right]=\left( -31 \right)\div \left( 30 \right)\] ….. (2)

While dividing a negative integer by a positive integer, divide them as whole numbers and then put a minus sign $\left( - \right)$ before the quotient. Therefore from (2), \[\left( -31 \right)\div \left[ \left( -30 \right)\div \left( -1 \right) \right]=-\dfrac{31}{30}\].

h) $\left[ \left( -36 \right)\div 12 \right]\div 3$

Ans: While dividing a negative integer by a positive integer, divide them as whole numbers and then put a minus sign $\left( - \right)$ before the quotient. Therefore, $\left( -36 \right)\div 12=-\dfrac{36}{12}=-3$  …..(1)

While dividing a negative integer by a positive integer, divide them as whole numbers and then put a minus sign $\left( - \right)$ before the quotient. Therefore from (1), $\left[ \left( -36 \right)\div 12 \right]\div 3=\left( -3 \right)\div 3=-1$.

i) $\left[ \left( -6 \right)+5 \right]\div \left[ \left( -2 \right)+1 \right]$

Ans: Simplifying the given expression, $\left[ \left( -6 \right)+5 \right]\div \left[ \left( -2 \right)+1 \right]=\left( -1 \right)\div \left( -1 \right)$ ….(1)

While dividing a negative integer by a negative integer, divide them as whole numbers and then put a plus sign $\left( + \right)$ before the quotient. Therefore from (1), $\left[ \left( -6 \right)+5 \right]\div \left[ \left( -2 \right)+1 \right]=1$.

2. Verify that \[a\div \left( b+c \right)\ne \left( a\div b \right)+\left( a\div c \right)\]  for each of the following values of $a,b\text{ and }c.$

a) \[a=12,b=-4,c=2\]

Ans: Given, $a=12,b=-4,c=2$. We have to verify $a\div \left( b+c \right)\ne \left( a\div b \right)+\left( a\div c \right)$.

Substituting the given values of $a,b,c$ in LHS we get,

L.H.S.$=12\div \left( -4+2 \right)=12\div \left( -2 \right)$ ….. (1)

While dividing a positive integer by a negative integer, divide them as whole numbers and then put a minus sign $\left( - \right)$ before the quotient. Therefore from (1), \[12\div \left( -4+2 \right)=-6\].   …..(2)  

Substituting the given values of $a,b,c$ in RHS we get,

R.H.S.$=\left[ 12\div \left( -4 \right) \right]\text{+}\left( 12\div 2 \right)\text{ }$ …..(3)

While dividing a negative integer by a positive integer, divide them as whole numbers and then put a minus sign $\left( - \right)$ before the quotient. Therefore, from (3), $\left[ 12\div \left( -4 \right) \right]\text{+}\left( 12\div 2 \right)=\left( -3 \right)\text{+}6\text{ }$

$\Rightarrow RHS=3$ …..(4)

From (2) and (4), we can conclude that $\,\text{L}\text{.H}\text{.S}\text{.}\ne \text{R}\text{.H}\text{.S}\text{.}$ Hence verified.     

b) $a=\left( -10 \right),b=1,c=1$

Ans: Given, $a=-10,b=1,c=1$. We have to verify $a\div \left( b+c \right)\ne \left( a\div b \right)+\left( a\div c \right)$.

Substituting the given values of $a,b,c$ in LHS we get,

L.H.S.$=-10\div \left( 1+1 \right)=-10\div \left( 2 \right)$ ….. (1)

While dividing a negative integer by a positive integer, divide them as whole numbers and then put a minus sign $\left( - \right)$ before the quotient. Therefore from (1), \[-10\div \left( 1+1 \right)=-5\].   …..(2)  

Substituting the given values of $a,b,c$ in RHS we get,

R.H.S.$=\left[ -10\div 1 \right]\text{+}\left( -10\div 1 \right)\text{ }$ …..(3)

While dividing a negative integer by a positive integer, divide them as whole numbers and then put a minus sign $\left( - \right)$ before the quotient. Therefore, from (3), $\left[ -10\div 1 \right]\text{+}\left( -10\div 1 \right)\text{ }=\left( -10 \right)\text{+}\left( -10 \right)\text{ }$

$\Rightarrow RHS=-20$ …..(4)

From (2) and (4), we can conclude that $\,\text{L}\text{.H}\text{.S}\text{.}\ne \text{R}\text{.H}\text{.S}\text{.}$ Hence verified.

3. Fill in the blanks:

a) \[369\div \_\_\_\_\_=369\]

Ans: While dividing any integer by $1$, the quotient is the original integer. Therefore, \[369\div \underline{1}=369\].

b) \[\left( -75 \right)\div \_\_\_\_\_=\left( -1 \right)\]

Ans: While dividing any integer by itself, the quotient is $1$. Also, while dividing a negative integer by a positive integer, divide them as whole numbers and then put a minus sign $\left( - \right)$ before the quotient. Therefore, \[\left( -75 \right)\div \underline{75}=\left( -1 \right)\]

c) \[\left( -206 \right)\div \_\_\_\_\_=1\]

Ans: While dividing any integer by itself, the quotient is $1$. Also, while dividing a negative integer by a negative integer, divide them as whole numbers and then put a plus sign $\left( + \right)$ before the quotient. Therefore, $\left( -206 \right)\div \underline{\left( -206 \right)}=1$

d) \[\left( -87 \right)\div \_\_\_\_\_=87\]

Ans: While dividing any integer by $1$, the quotient is the original integer. Also, while dividing a negative integer by a negative integer, divide them as whole numbers and then put a plus sign $\left( + \right)$ before the quotient. Therefore, $\left( -87 \right)\div \underline{\left( -1 \right)}=87$.

e) $\_\_\_\_\_\div 1=-87$

Ans: While dividing any integer by $1$, the quotient is the original integer. Therefore, $\underline{\left( -87 \right)}\div 1=-87$.

f) $\_\_\_\_\_\div 48=-1$

Ans: While dividing any integer by $1$, the quotient is the original integer. Also, while dividing a negative integer by a positive integer, divide them as whole numbers and then put a minus sign $\left( - \right)$ before the quotient. Therefore, $\underline{\left( -48 \right)}\div 48=-1$

g) \[20\div \_\_\_\_\_=-2\]

Ans: While dividing a positive integer by a negative integer, divide them as whole numbers and then put a minus sign $\left( - \right)$ before the quotient. Therefore, $20\div \underline{\left( -10 \right)}=-2$

h) $\_\_\_\_\_\div \left( 4 \right)=-3$

Ans: While dividing a negative integer by a positive integer, divide them as whole numbers and then put a minus sign $\left( - \right)$ before the quotient.$\left( -12 \right)\div \left( 4 \right)=-3$

4. Write five pairs of integers $\left( a,b \right)$ such that $a\div b=-3$. One such pair is $\left( 6,-2 \right)$ because $6\div \left( -2 \right)=\left( -3 \right)$.

Ans: Five pair of integers $\left( a,b \right)$ such that $a\div b=-3$ are:

i) The pair of integers $\left( -9,3 \right)$ is such that $\left( -9 \right)\div 3=-3$.

ii) The pair of integers $\left( -15,5 \right)$ is such that $\left( -15 \right)\div 5=-3$.

iii) The pair of integers $\left( 12,-4 \right)$ is such that $12\div \left( -4 \right)=-3$.

iv) The pair of integers $\left( -3,1 \right)$ is such that $\left( -3 \right)\div 1=-3$.

v) The pair of integers $\left( 21,-7 \right)$ is such that $21\div \left( -7 \right)=-3$.  

5. The temperature at noon was ${{10}^{\circ }}\text{C}$ above zero. If it decreases at the rate of ${{2}^{\circ }}\text{C}$ per hour until mid-night, at what time would the temperature be ${{8}^{\circ }}\text{C}$ below zero? What would be the temperature at mid-night?

Ans: Given that the temperature at $12$ noon is \[{{10}^{\circ }}C\] above the zero i.e., $+{{10}^{\circ }}C$. Now, the temperature decreases ${{2}^{\circ }}C$ each hour until midnight. Therefore, following number line represents the temperature path:

Temperature at $12$ noon $={{10}^{\circ }}C$.

Temperature at $1PM$ $={{\left( 10-2 \right)}^{\circ }}C={{8}^{\circ }}C$.

Temperature at $2PM$ $={{\left( 8-2 \right)}^{\circ }}C={{6}^{\circ }}C$.

Temperature at $3PM$ $={{\left( 6-2 \right)}^{\circ }}C={{4}^{\circ }}C$.

Therefore, temperature at $nPM$ $={{\left( 10-2n \right)}^{\circ }}C$.

According to the question, $\left( 10-2n \right)=-8$ ….(1)

Solving (1) by rearranging terms we get, 

$10+8=2n$

$\Rightarrow n=9$ 

Therefore, at $9\,\text{pm}$ the temperature would be ${{8}^{\circ }}\text{C}\,\,\text{below}\,\,{{0}^{\circ }}\text{C}$.

6. In a class test $\left( +3 \right)$ marks are given for every correct answer and $\left( -2 \right)$ 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?

Ans: Given that Radhika scored $20$ marks and she has got $12$ correct answers.

Let the number of incorrect answers be $x$ then, 

Marks given for one correct answer $=3$

Marks given for $12$ correct answers $=3\times 12=36$ ….. (1)

Marks given for one wrong answer $=-2$

Marks given for $x$ wrong answers $=-2x$ ….. (2)

Also, Radhika scored $20$ marks. Hence from (1) and (2),

$20=36+\left( -2x \right)$ ….. (3)

Solving (3) by rearranging terms we get, 

$2x=16$

$\Rightarrow x=8$ 

Therefore, Radhika has attempted $8$ incorrect questions.

ii) Mohini scores $\left( -5 \right)$ marks in this test, though she has got $7$ correct answers. How many questions has she attempted incorrectly?

Ans: Given that Mohini scored $-5$ marks and she has got $7$ correct answers.

Let the number of incorrect answers be $x$ then, 

Marks given for one correct answer $=3$

Marks given for $7$ correct answers $=3\times 7=21$ ….. (1)

Marks given for one wrong answer $=-2$

Marks given for $x$ wrong answers $=-2x$ ….. (2)

Also, Mohini scored $-5$ marks. Hence from (1) and (2),

$-5=21+\left( -2x \right)$ ….. (3)

Solving (3) by rearranging terms we get, 

$2x=26$

$\Rightarrow x=13$ 

Therefore, Mohini has attempted $13$ incorrect questions.

7. An elevator descends into a mine shaft at the rate of $6\text{ m/min}$ . If the descent starts from \[10\] above the ground level, how long will it take to reach $\text{-350 m}$?

Ans: Given that the starting position of mine shaft is $10\,\,\text{m}$ above the ground.

And its destination is $350m$ below the ground.

Let us denote the distance above the ground by $+$ sign and the distance below the ground by $-$ sign. Therefore, starting from $10m$ it has to go $-350m$. 

Total distance covered by mine shaft $=10\,\text{m}-\left( -350 \right)\text{m}=10+350=360\,\text{m}$…..(1)

Now, it is given that elevator takes $1\text{ }\min $ to cover the distance of $6m$ i.e., 

Time taken to cover a distance of $6\,\text{m}$$=1$ minute.    …..(2) 

Hence from (2),

Time taken to cover a distance of $1\,\text{m}$ $=\dfrac{1}{6}$ minute.   …..(3)

Hence from (1) and (3), time taken to cover a distance of $360\,\text{m}$ 

$=\,\dfrac{1}{6}\times 360=60$ minutes $=1$ hour.

Therefore, in $1$ hour the mine shaft reaches $-350$ below the ground.

NCERT Solutions for Class 7 Chapter 1 Maths Integers – Free PDF Download

Integers : The set of natural numbers like ……., -4, -3, -2, -1, 0, 1, 2, 3, 4 ….. are called integers.

How are Integers applicable in our real life?

Integers are applied in many ways in our real-life besides math class. We can use integers for calculating the efficiency of positive and negative numbers in all fields. 

In our day-to-day life, we come across many situations where integers are used:

i. Positive integers are used to determine profit, income, increase, rise, high, north, east, above depositing, climbing and so on. 

ii. Negative integers are used to determine quantities like loss, expenditure, decrease, fall, low, south, west, below, withdrawing, sliding and so on. 

Example: A point A is on a mountain which is 4680 m above sea level and a point B is in a mine which is 765 m below sea-level. What is the distance between A and B?

In this example, we can consider a point O at the sea level. 

Then, height OA = + 4680m;

Height OB = - 765m.

Distance between A and B  = [OA] + [OB]

            = {[ + 4680 ]  + [ -765]} m

            = (4680 + 765) m = 5445 m

Benefits of Vedantu NCERT Solutions:

• Comprehensive and Accurate Solutions: Our NCERT solutions are curated by subject experts in adherence to the latest CBSE guidelines. They provide accurate and reliable solutions to all the questions and exercises in the textbook, ensuring a comprehensive understanding of the concepts.

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NCERT Solutions for Class 7 Maths

• Chapter 2 - Fractions and Decimals

• Chapter 3 - Data Handling

• Chapter 4 - Simple Equations

• Chapter 5 - Lines and Angles

• Chapter 6 - The Triangle and Its Properties

• Chapter 7 - Congruence of Triangles

• Chapter 8 - Comparing Quantities

• Chapter 9 - Rational Numbers

• Chapter 10 - Practical Geometry

• Chapter 11 - Perimeter and Area

• Chapter 12 - Algebraic Expressions

• Chapter 13 - Exponents and Powers

• Chapter 14 - Symmetry

• Chapter 15 - Visualising Solid Shapes
  

Students can also refer to the following study material for Chapter 1 of Class 7 Maths:

• Chapter 1 Integers: Important Questions

• Chapter 1 Integers: Revision Notes

• Chapter 1 Integers: Formulas

• Chapter 1 Integers: RD Sharma Solutions

• Chapter 1 Integers: NCERT Exemplar Solutions

Properties of Integers

The properties of integers include numbers for addition and multiplication through patterns. They also include the whole numbers as well. Integers involve expressing communicative and associative properties in a general form.

Facts

• The counting numbers 1, 2, 3, 4, 5, …….. and so on are called Natural Numbers, whereas the set of natural numbers together with zero like  0, 1, 2, 3, 4, 5, ……. and so on are called whole numbers.

• On a number line, we represent the negative integers by the points to the left of zero and positive integers by the points to the right of zero.

• The integer 0 is an additive identity for integers by the points to the left of zero and positive integers by the points to the right of zero. 

• The integer 0 is neither positive nor negative.

• The absolute value of an integer is its numerical value of the integer regardless of its sign. The absolute value of an integer a is denoted by | a |.

Example:

 Absolute value of 8 i.e. | 8 | = 8

    Absolute value of –8 i.e. | -8 | = 8

• If ‘a’ and ‘b’ are integers, then (a + b), (a - b) and (a x b) are also integers. Integers are used for addition, subtraction and multiplication. 

• If ‘a’ and ‘b’ are integers then a x b = b x a and a + b = b + a. Multiplication is a commutative property for integers. 

• For any three integers a, b and c, we have:

1. Addition is associative property for integers. a + ( b + c ) = ( a + b ) + c = c + ( b + a )

2. Multiplication is associative property for integers. a x ( b x c ) = ( a x b ) x c = c x ( b x a)

• For any three integers a, b and c, we have:

1. a x (b + c) = (a x b) + ( a x c)

2. a x (b-c) = (a x b) – (a x c)

• 1 is the multiplicative identity for integers and 0 is the identity under addition. 

Ex: a + 0 = a = 0 + a and a x 1= a =1 x a

• If the integers are of like signs, their product is positive.

• When we add two positive or negative integers with like signs, we add their numerical values and assign the sign of the numbers added with the sum.

Ex: 6 + 5 = 11

            - 6 + ( - 5 ) = - 6  - 5 = -11

• If the integers have unlike signs, their product is negative.   a x –b = - ab

Ex:  5 x – 4 = -20

           - 5 x 4 = -20 

• When we add two integers with unlike signs then we take the difference of their numerical values and assign the sign of the integer with the greater numerical value. 

Ex: 2 + ( - 8 ) = - 2

            8 + ( - 2 ) = 2

• The product of an integer and 0 is always 0.

• If the dividend and divisor are integers of like signs, then the quotient is a positive integer. 

• If the dividend and divisor are integers of unlike signs, then the quotient is a negative integer.

• Division by 0 is not possible. However, 0 divided by any integer (except 0) is equal to 0.

Important Note

• 0 is neither positive nor negative.

• The + sign is not written before a positive number.

• ½  and 0.5 are not integers because they are not whole numbers. 

• Negative numbers are usually placed in brackets to avoid confusion arising due to two signs in evaluations. 

          Ex: 4 + ( - 2 ) = -2

Number Line

A number line represents natural numbers, whole numbers, positive integers and negative integers. The identities are marked at equal intervals on a line to determine numerical operations. Number lines are important because they represent numbers that are used in our daily life. 

Steps for drawing a number line

1. Draw a straight line of any length.

2. Mark points at equal intervals on the drawn line to divide it into the required number.

3. Mark any one of the points, marked on the line in step 2, as 0.

4. Starting from 0 and on the right-hand side of the line, mark the positive numbers + 1, + 2, + 3, and so on. Similarly, starting from 0 on the left side of mark the negative integers -1, -2, -3, and so on.

5. The arrowheads on both the sides of the drawn line indicate that the numbers continue up to infinity.