NCERT Solutions Class 7 Maths Chapter 1 - Integers

What is an Integer?

The word ‘Integer’ is derived from the Latin word intact or whole. So, integers are always the whole number, which consists of positive or negative numbers or zero, which is simply called a combination of negative and whole numbers. Integers will never be fractional or decimal numbers.  Usually, the set of integers is denoted by Z.

Eg, Z = { -5, -2, 0, 3, 9}

Addition and Subtraction of Integers

Learning the addition and subtraction of Integers is most important to make simple calculations in day-to-day life for performing simple calculations. Like calculating the pocket money spent by you in a day. Calculating average among integers. 

Multiplication of Integers

Multiplication is a simple one for many of you. But, while multiplying the integers it is important to note a sign of the numbers. This will mainly help while simplifying an equation.

Multiplication of a Positive and Negative Integer

Multiplication of a positive and negative integer will always leave the answer as a negative integer.

Multiplication of Two Negative Integers

Multiplication of two negative integers will always result in a positive integer. Should keep in the note of this topic while simplifying the quadratic equation.

Properties of Multiplication of Integers

1. Multiplication of two positive integers leaves a positive result

2. Multiplication of two negative integers leaves a positive result 

3. Multiplication of a positive integer and a negative integer leaves a negative result 

4. Multiplication of zero with any number is zero.

Division of Integers

Properties of Division of Integers

1. Division of two positive integers leaves a positive result.

2. The division of two negative integers leaves a positive result. 

3. The division of a positive integer and a negative integer leaves a negative result. 

4. The division of zero by any number is zero. 

5. The division of any number by zero is infinite.

Access NCERT Solutions for Maths Chapter 1 – Integers

Exercise 1.1

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 1.2

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.

Exercise 1.2

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

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.

Class 10 Maths Chapter 1: Exercise Breakdown

Conclusion

The NCERT Solutions for Class 7 Maths Chapter 1 Integers, provided by Vedantu, are essential for understanding the basics of integers. These solutions help students grasp concepts like addition, subtraction, multiplication, and division of integers, as well as their properties. It's important to focus on solving the examples and practice problems to reinforce these concepts. The step-by-step explanations make complex topics easier to understand, which is crucial for building a strong foundation in mathematics. For best results, regularly practice the exercises and review the key concepts highlighted in these solutions.

Other Related Links for CBSE Class 7 Maths Chapter 1

NCERT Solutions for Class 7 Maths - Chapter-wise List

Given below are the chapter-wise NCERT Solutions for Class 7 Maths. These solutions are provided by the Maths experts at Vedantu in a detailed manner. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.