Unit 1: Integers

# Question: 1

Following number line shows the temperature in degree celsius ( $°\text{C}$ ) at different places on a particular day.

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

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

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

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

a.    The temperature of the places marked on the given number line is given in the following table:

 Places Temperature Bengaluru $22°\text{C}$ Ooty $14°\text{C}$ Shimla $5°\text{C}$ Srinagar $-2°\text{C}$ Lahulspiti $-8°\text{C}$

b.   The temperature of the hottest place (Bengaluru) $=22°\text{C}$

The temperature of the coldest place (Lahulspiti) $=-8°\text{C}$

Difference $=22°\text{C}–\left(-8°\text{C}\right)$

$\begin{array}{l}=22°\text{C}+8°\text{C}\\ =30°\text{C}\end{array}$

c.    The temperature of Srinagar $=-2°\text{C}$

The temperature of Lahulspiti $=-8°\text{C}$

Difference $=-2°\text{C}-\left(-8°\text{C}\right)$

$\begin{array}{l}=-2°\text{C+}8°\text{C}\\ =6°\text{C}\end{array}$

d.   Sum of the temperatures of Srinagar and Shimla $=5°\text{C}+\left(-2°\text{C}\right)$

$\begin{array}{l}=5°\text{C}-2°\text{C}\\ =3°\text{C}\end{array}$

The temperature at Shimla $=5°\text{C}$

$3°\text{C}<5°\text{C}$

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

Now, temperature of Srinagar $=-2°\text{C}$

$3°\text{C}>-2°\text{C}$

So, temperature of $3°\text{C}$ is not less than the temperature at Srinagar.

# Question: 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 $10\text{,}$ what was his total at the end?

## Solution

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

Total marks got by Jack $=25+\left(-5\right)+\left(-10\right)+15+10$

$\begin{array}{l}=25-15+25\\ =35\end{array}$

Thus, $35$ marks are obtained by Jack in a quiz.

# Question: 3

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

## Solution

On Monday, temperature at Srinagar $=-5°\text{C}$

On Tuesday, temperature dropped by $=2°\text{C}$

$\therefore$ Temperature on Tuesday $=-5°\text{C}-2°\text{C}=-7°\text{C}$

On Wednesday, temperature rose up by $=4°\text{C}$

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

Thus, temperature on Tuesday and Wednesday was $-7°\text{C}$ and $-3°\text{C}$ respectively.

# Question: 4

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

## Solution

Height of a plane above the sea level

Depth of a submarine below the sea level

$\therefore$ The vertical distance between the plane and the submarine

Thus, the vertical distance between the plane and the submarine is

# Question: 5

Mohan deposits $Rs\text{\hspace{0.17em}}2,000$ in his bank account and withdraws $Rs\text{\hspace{0.17em}}1,642$ 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

Amount deposited $=\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}2,000$ and Amount withdraw $=\text{\hspace{0.17em}}\text{\hspace{0.17em}}-Rs\text{\hspace{0.17em}}1,642$

$\therefore$ Balance $=\text{\hspace{0.17em}}2,000-1,642=\text{\hspace{0.17em}}\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}358$

Thus, the balance in Mohan’s account after withdrawal is $Rs\text{\hspace{0.17em}}358.$

# Question: 6

Rita goes  towards east from a point A to the point B. From B, she moves  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

According to the question distance towards east is represented by a positive integer. So, the distance opposite of east, that is west direction is represented by a negative integer.

On the given number line:

Distance from A to B

Distance from B to C

Thus, final position of Rita from A

# Question: 7

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

 $5$ $-1$ $-4$ $1$ $-10$ $0$ $-5$ $-2$ $7$ $-4$ $-3$ $-2$ $0$ $3$ $-3$ $-6$ $4$ $-7$

(i)                                 (ii)

## Solution

(i)             Sum of numbers in each row:
$5+\left(-1\right)+\left(-4\right)=5-5=0$

$\begin{array}{l}\left(-5\right)+\left(-2\right)+7=-7+7=0\\ 0+3+\left(-3\right)=3-3=0\end{array}$

Sum of numbers in each column:
$5+\left(-5\right)+0=5-5=0$

$\begin{array}{l}\left(-1\right)+\left(-2\right)+3=-3+3=0\\ \left(-4\right)+7+\left(-3\right)=7-7=0\end{array}$

Sum of numbers along each diagonal:
$5+\left(-2\right)+\left(-3\right)=5-5=0$

$\left(-4\right)+\left(-2\right)+0=-6\ne 0$
As the sum of numbers along a diagonal is not zero, the box is not a magic square.

(ii)          Sum of numbers in each row:
$1+\left(-10\right)+0=1-10=-9$

$\begin{array}{l}\left(-4\right)+\left(-3\right)+\left(-2\right)=-7-2=-9\\ \left(-6\right)+4+\left(-7\right)=-2-7=-9\end{array}$

Sum of numbers in each column:
$1+\left(-4\right)+\left(-6\right)=1-10=-9$

$\begin{array}{l}\left(-10\right)+\left(-3\right)+4=-13+4=-9\\ 0+\left(-2\right)+\left(-7\right)=0-9=-9\end{array}$

Sum of numbers along each diagonal: $1+\left(-3\right)+\left(-7\right)=1-10=-9$

$0+\left(-3\right)+\left(-6\right)=-9$

The given square is a magic square as its each row, column and diagonal have the same sum.

# Question: 8

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

(i)

(ii)

(iii)

(iv)

## Solution

(i)             Given:

We have, $a-\left(-b\right)=a+b$

Put the values in L.H.S.

$a-\left(-b\right)=21-\left(-18\right)$

$=21+18=39$

$=39$

Put the values in R.H.S.
$a+b=21+19$ $=39$

L.H.S. $=$ R.H.S.

Hence, verified.

(ii)          Given:

We have, $a-\left(-b\right)=a+b$

Put the values in L.H.S. $a-\left(-b\right)=118-\left(-125\right)$

$=118+125$

$=243$

Put the values in R.H.S.
$a+b=118+125$

$=243$

Thus, L.H.S. $=$ R.H.S.

Hence, verified.

(iii)      Given:

We have, $a-\left(-b\right)=a+b$

Put the values in L.H.S. $a-\left(-b\right)=75-\left(-84\right)$

$=75+84$

$=159$

Put the values in R.H.S.
$a+b=75+84$

$=159$

Thus, L.H.S. $=$ R.H.S.

Hence, verified.

(iv)       Given:

We have, $a-\left(-b\right)=a+b$

Put the values in L.H.S. $a-\left(-b\right)=28-\left(-11\right)$

$=28+11$

$=39$

Putting the values in R.H.S.
$a+b=28+11$

$=39$

Thus, L.H.S. $=$ R.H.S.

Hence, verified.

# Question: 9

Use the sign of  in the box to make the statements true.

a.

b.

c.

d.

e.

## Solution

a.

b.

c.

d.

e.

# Question: 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?

(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?

(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 $\left(-8\right)$ represents going down by eight steps. So, what will the sum $8$ in (b) represent?

## Solution

(i)             He jumps $3$ steps down and then jumps back $2$ steps up. Following number line shows the jumps of monkey:

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,

He will reach ninth steps in $11$ jumps.

(ii)          Monkey jumps four steps up and then jumps down $2$ steps. Following number line shows the jumps of monkey:

The money reaches back on the first step in fifth jump.

(iii)      (a) $-3+2-3+2-3+2-3+2-3$

$+\text{\hspace{0.17em}}2-3+2-3+2-3+2=-8$

(b) $4-2+4-2+4-2+4-2=8$

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

# Question: 1

Write down a pair of integers whose:

a.    sum is $-7$

b.   difference is $-10$

c.    sum is $0$

## Solution

a.     A pair of integers whose sum is  $-5+\left(-2\right)=-7$

b.    A pair of integers whose difference is  $\left(-2\right)-\left(8\right)=-2-8=-10$

c.    A pair of integers whose sum is  $-5+5=0$

# Question: 2

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

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

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

## Solution

a.    $-2-\left(-10\right)=-2+10=8$

b.   $\left(-7\right)+2=-5$

c.    $\left(-2\right)-\left(1\right)=-2-1=-3$

# Question: 3

In a quiz, team A scored $-40,\text{\hspace{0.17em}}\text{\hspace{0.17em}}10,\text{\hspace{0.17em}}\text{\hspace{0.17em}}0$ and team B scored $10,\text{\hspace{0.17em}}\text{\hspace{0.17em}}0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}-40$ in three successive rounds. Which team scored more? Can we say that we can add integers in any order?

## Solution

Team A scored $-40,\text{\hspace{0.17em}}\text{\hspace{0.17em}}10,\text{\hspace{0.17em}}\text{\hspace{0.17em}}0.$

Total score of Team A $=-40+10+0=-30$

Team B scored $10,\text{\hspace{0.17em}}\text{\hspace{0.17em}}0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}-40.$

Total score of Team B $=10+0+\left(-40\right)=10+0-40=-30$

Thus, scores of both the teams are same.

Yes, we can add integers in any order as per commutative property.

# Question: 4

Fill in the blanks to make the following statements true:

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

(ii)     $-53+_______=-53$

(iii) $17+_______=0$

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

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

## Solution

(i)        $\left(-5\right)+\left(\underset{_}{-8}\right)=\left(-8\right)+\left(\underset{_}{-5}\right)$

(ii)     $-53+\underset{_}{0}=-53$

(iii) $17+\underset{_}{\left(-17\right)}=0$

(iv)  $\left[13+\left(-12\right)\right]+\left(\underset{_}{-7}\right)=13+\left[\left(-12\right)+\left(-7\right)\right]$

(v)     $\left(-4\right)+\left[15+\left(-3\right)\right]=\left[-4+15\right]+\left(\underset{_}{-3}\right)$

# Question: 1

Find each of the following products:

a.    $3×\left(-1\right)$

b.   $\left(-1\right)×225$

c.    $\left(-21\right)×\left(-30\right)$

d.   $\left(-316\right)×\left(-1\right)$

e.    $\left(-15\right)×0×\left(-18\right)$

f.     $\left(-12\right)×\left(-11\right)×\left(10\right)$

g.   $9×\left(-3\right)×\left(-6\right)$

h.   $\left(-18\right)×\left(-5\right)×\left(-4\right)$

i.      $\left(-1\right)×\left(-2\right)×\left(-3\right)×4$

j.      $\left(-3\right)×\left(-6\right)×\left(-2\right)×\left(-1\right)$

## Solution

a.    $3×\left(-1\right)=-3$

b.   $\left(-1\right)×225=-225$

c.    $\left(-21\right)×\left(-30\right)=630$

d.   $\left(-316\right)×\left(-1\right)=316$

e.    $\left(-15\right)×0×\left(-18\right)=0$

f.     $\left(-12\right)×\left(-11\right)×\left(10\right)=132×10=1320$

g.   $9×\left(-3\right)×\left(-6\right)=9×18=162$

h.   $\left(-18\right)×\left(-5\right)×\left(-4\right)=90×\left(-4\right)=-360$

i.      $\left(-1\right)×\left(-2\right)×\left(-3\right)×4=\left(-6×4\right)=-24$

j.      $\left(-3\right)×\left(-6\right)×\left(-2\right)×\left(-1\right)=\left(18\right)×\left(2\right)=36$

# Question: 2

Verify the following:

a.    $18×\left[7+\left(-3\right)\right]=\left[18×7\right]+\left[18×\left(-3\right)\right]$

b.   $\left(-21\right)×\left[\left(-4\right)+\left(-6\right)\right]$

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

## Solution

a.    $18×\left[7+\left(-3\right)\right]=\left[18×7\right]+\left[18×\left(-3\right)\right]$

$\begin{array}{l}⇒18×4=126+\left(-54\right)\\ ⇒72=72\end{array}$

$⇒$ $\text{L}.\text{H}.\text{S}.=\text{R}.\text{H}.\text{S}.$

Hence verified.

b.   $\left(-21\right)×\left[\left(-4\right)+\left(-6\right)\right]$

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

$\begin{array}{l}⇒\left(-21\right)×\left(-10\right)=84+126\\ ⇒210=210\end{array}$

$⇒$ $\text{L}.\text{H}.\text{S}.=\text{R}.\text{H}.\text{S}.$

Hence verified.

# Question: 3

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

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

(a) $-22$

(b)                   $37$

(c) $0$

## Solution

(i)        $\left(-1\right)×a=-a\text{,}$ where $a$ is an integer.

(ii)

(a) $\left(-1\right)×\left(22\right)=-22$

(b)                   $\left(-1\right)×\left(-37\right)=37$

(c) $\left(-1\right)×0=0$

# Question: 4

Starting from $\left(-1\right)×5\text{,}$ write various products showing some pattern to show $\left(-1\right)×\left(-1\right)=1.$

## Solution

$\begin{array}{l}\left(-1\right)×5=-5\\ \left(-1\right)×4=-4\\ \left(-1\right)×3=-3\\ \left(-1\right)×2=-2\\ \left(-1\right)×1=-1\end{array}$

$\left(-1\right)×\left(-1\right)=1$

This pattern shows the product of a negative integer and a positive integer is a negative integer, whereas the product two negative integers is a positive integer.

# Question: 5

Find the product, using suitable properties:

a.    $26×\left(-48\right)+\left(-48\right)×\left(-36\right)$

b.   $8×53×\left(-125\right)$

c.    $15×\left(-25\right)×\left(-4\right)×\left(-10\right)$

d.   $\left(-41\right)×102$

e.    $625×\left(-35\right)+\left(-625\right)×\left(65\right)$

f.     $7×\left(50-2\right)$

g.   $\left(-17\right)×\left(-29\right)$

h.   $\left(-57\right)×\left(-19\right)+57$

## Solution

a.    $26×\left(-48\right)+\left(-48\right)×\left(-36\right)$

$=\left(-48\right)×\left[26+\left(-36\right)\right]$       [Distributive property]

$\begin{array}{l}=\left(-48\right)×\left(-10\right)\\ =480\end{array}$

b.   $8×53×\left(-125\right)$

$=53×\left[8×\left(-125\right)\right]$           [Associative property]

$\begin{array}{l}=53×\left(-1000\right)\\ =-53000\end{array}$

c.    $15×\left(-25\right)×\left(-4\right)×\left(-10\right)$

$=15×\left[\left(-25\right)×\left(-4\right)×\left(-10\right)\right]$
[Commutative property]

$\begin{array}{l}=15×\left(-1000\right)\\ =-15000\end{array}$

d.   $\left(-41\right)×102$

$=-41×\left[100+2\right]$            [Distributive property]

e.    $\begin{array}{l}\\ \\ =\left[\left(-41\right)×100\right]+\left[\left(-41\right)×2\right]\\ =-4100+\left(-82\right)\\ =-4182\end{array}$$625×\left(-35\right)+\left(-625\right)×\left(65\right)$

$=625×\left[\left(-35\right)+\left(-65\right)\right]$    [Distributive property]

$\begin{array}{l}=625×\left(-100\right)\\ =-62500\end{array}$

f.     $7×\left(50-2\right)$

$=7×50-7×2$                  [Distributive property]

$\begin{array}{l}=350-14\\ =336\end{array}$

g.   $\left(-17\right)×\left(-29\right)$

$\begin{array}{l}=\left(-17\right)×\left(-30+1\right)\\ =\left(-17\right)×\left(-30\right)+\left(-17\right)×1\\ =510+\left(-17\right)\\ =493\end{array}$  [Distributive property]

h.   $\left(-57\right)×\left(-19\right)+57$

$=\left(-57\right)×\left(-19\right)+57×1$

$\begin{array}{l}=57×19+57×1\\ =57×\left(19+1\right)\\ =57×20\\ =1140\end{array}$              [Distributive property]

# Question: 6

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

## Solution

Present room temperature $=40°\text{C}$

Decrease in the temperature every hour $=5°\text{C}$
Change in the temperature every hour $=-5°\text{C}$

Room temperature after $10$ hours $=40°\text{C}+10×\left(-5°\text{C}\right)$

$=40°\text{C}-50°\text{C}=-10°\text{C}$

# Question: 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?

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

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

## Solution

(i)             Marks for four correct question $=4×5=20$

Marks for six incorrect question $=6×\left(-2\right)=-12$

Therefore, total score

$\begin{array}{l}=\left(4×5\right)+\left[6×\left(-2\right)\right]\\ =20-12\\ =8\end{array}$

Thus, Mohan gets $8$ marks in a class test.

(ii)         Marks for five correct question $=5×5=25$

Marks for five incorrect question $=5×\left(-2\right)=-10$

Therefore, total score

$\begin{array}{l}=25+\left(-10\right)\\ =15\end{array}$

Thus, Reshma gets $15$ marks in a class test.

(iii)      Marks for two correct question $=2×5=10$ Marks for five incorrect question $=5×\left(-2\right)=-10$

Therefore, total score

$\begin{array}{l}=10+\left(-10\right)\\ =0\end{array}$

Thus, Reshma gets $0$ marks in a class test.

# Question: 8

A cement company earns a profit of $Rs\text{\hspace{0.17em}}8$ per bag of white cement sold and a loss of $Rs\text{\hspace{0.17em}}5$ per bag of grey cement sold.

a.    The company sells $3000$ bags of white cement and $5,000$ bags of grey cement in a month. What is its profit or loss?

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

Given, profit of $1$ bag of white cement $=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}8$

Loss of $1$ bag of grey cement $=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}5$

a.    Profit on selling $3000$ bags of white cement $=3000×\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}8=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}24,000$

Loss on selling $5000$ bag of grey cement $=5000×\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}5=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}25,000$

Since, Profit $<$ Loss

Therefore, total loss on selling grey cement bags $=$ Loss $-$ Profit

$\begin{array}{l}=\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}25,000-Rs\text{\hspace{0.17em}}24,000\\ =\text{\hspace{0.17em}}Rs\text{\hspace{0.17em}}1,000\end{array}$

b.   Let the number of bags of white cement be $x\text{.}$

According to the question, Loss $=$ Profit

$x=\frac{5×6400}{8}=4000$ bags

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

# Question: 9

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

a.    $\left(-3\right)×_______=27$

b.   $5×_______=-35$

c.    $_______×\left(-8\right)=-56$

d.   $_______×\left(-12\right)=132$

## Solution

a.    $\left(-3\right)×\underset{_}{\left(-9\right)}=27$

b.   $5×\underset{_}{\left(-7\right)}=-35$

c.    $\underset{_}{7}×\left(-8\right)=-56$

d.   $\underset{_}{\left(-11\right)}×\left(-12\right)=132$

# Question: 1

Evaluate each of the following:

a.    $\left(-30\right)÷10$

b.   $50÷\left(-5\right)$

c.    $\left(-36\right)÷\left(-9\right)$

d.   $\left(-49\right)÷\left(49\right)$

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

f.     $0÷\left(-12\right)$

g.   $\left(-31\right)÷\left[\left(-30\right)+\left(-1\right)\right]$

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

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

## Solution

a.    $\left(-30\right)÷10$

$\begin{array}{l}=\left(-30\right)×\frac{1}{10}\\ =\frac{-30×1}{10}\\ =-3\end{array}$

b.   $50÷\left(-5\right)$

$\begin{array}{l}=50×\left(\frac{-1}{5}\right)\\ =\frac{50×\left(-1\right)}{5}\\ =-10\end{array}$

c.    $\left(-36\right)÷\left(-9\right)$

$\begin{array}{l}=\left(-36\right)×\left(\frac{-1}{9}\right)\\ =\frac{\left(-36\right)×\left(-1\right)}{9}\\ =\frac{36}{9}\\ =4\end{array}$

d.   $\left(-49\right)÷\left(49\right)$

$\begin{array}{l}=\left(-49\right)×\frac{1}{49}\\ =\frac{-49}{49}\\ =-1\end{array}$

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

$\begin{array}{l}=13÷\left(-1\right)\\ =13×\left(\frac{-1}{1}\right)\\ =-13\end{array}$

f.     $0÷\left(-12\right)$

$\begin{array}{l}=0×\left(\frac{-1}{12}\right)\\ =\frac{0}{12}\\ =0\end{array}$

g.   $\left(-31\right)÷\left[\left(-30\right)+\left(-1\right)\right]$

$\begin{array}{l}=\left(-31\right)÷\left(-30-1\right)\\ =\left(-31\right)÷\left(-31\right)\\ =\left(-31\right)×\left(\frac{-1}{31}\right)\\ =\frac{31}{31}\\ =1\end{array}$

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

$\begin{array}{l}=\left[\left(-36\right)×\frac{1}{12}\right]×\frac{1}{3}\\ =\left(\frac{-36}{12}\right)×\frac{1}{3}\\ =\left(-3\right)×\frac{1}{3}\\ =\frac{-3}{3}\\ =-1\end{array}$

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

$\begin{array}{l}=\left(-6+5\right)÷\left(-2+1\right)\\ =\left(-1\right)÷\left(-1\right)\\ =\left(-1\right)×\frac{\left(-1\right)}{1}\\ =1\end{array}$

# Question: 2

Verify that $a÷\left(b+c\right)\ne \left(a÷b\right)+\left(a÷c\right)$ for each of the following values of $a,b$ and $c.$

a.    $a=12,\text{\hspace{0.17em}}\text{\hspace{0.17em}}b=-4,\text{\hspace{0.17em}}\text{\hspace{0.17em}}c=2$

b.   $a=\left(-10\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}b=1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}c=1$

## Solution

a.    Given: $a÷\left(b+c\right)\ne \left(a÷b\right)+\left(a÷c\right)$ and $a=12,\text{\hspace{0.17em}}\text{\hspace{0.17em}}b=-4,\text{\hspace{0.17em}}\text{\hspace{0.17em}}c=2$

On putting the given value in L.H.S. L.H.S. $=12÷\left(-4+2\right)$

$\begin{array}{l}=12÷\left(-2\right)\\ =12÷\left(\frac{-1}{2}\right)\\ =\frac{-12}{2}\\ =-6\end{array}$

On putting the given value in R.H.S. R.H.S. $=\left[12÷\left(-4\right)\right]+\left(12÷2\right)$

$\begin{array}{l}=\left(12×\frac{-1}{4}\right)+6\\ =-3+6\\ =3\end{array}$

Thus, L.H.S. $\ne$ R.H.S.

Hence verified.

b.   Given: $a÷\left(b+c\right)\ne \left(a÷b\right)+\left(a÷c\right)$ and $a=\left(-10\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}b=1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}c=1$
L.H.S. $=-10÷\left(1+1\right)$

$\begin{array}{l}=-10÷\left(2\right)\\ =-5\end{array}$

R.H.S. $=\left[-10÷1\right]+\left(-10÷1\right)$

$\begin{array}{l}=-10-10\\ =-20\end{array}$

Thus, L.H.S. $\ne$ R.H.S. Hence verified.

# Question: 3

Fill in the blanks:

a.    $369÷_______=\left(369\right)$

b.   $\left(-75\right)÷_______=-1$

c.    $\left(-206\right)÷_______=1$

d.   $-87÷_______=87$

e.    $_______÷1=-87$

f.     $_______÷48=-1$

g.   $20÷_______=-2$

h.   $_______÷\left(4\right)=-3$

## Solution

a.    $369÷\underset{_}{1}=\left(369\right)$

b.   $\left(-75\right)÷\underset{_}{75}=-1$

c.    $\left(-206\right)÷\underset{_}{-206}=1$

d.   $-87÷\underset{_}{-1}=87$

e.    $\underset{_}{\left(-87\right)}÷1=-87$

f.     $\underset{_}{\left(-48\right)}÷48=-1$

g.   $20÷\underset{_}{\left(-10\right)}=-2$

h.   $\underset{_}{\left(-12\right)}÷\left(4\right)=-3$

# Question: 4

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

## Solution

(i)       $\left(-6\right)÷2=-3$

(ii)        $9÷\left(-3\right)=-3$

(iii)       $12÷\left(-4\right)=-3$

(iv)       $\left(-9\right)÷3=-3$

(v)       $\left(-15\right)÷5=-3$

So, five pairs of required integers are:
(–6, 2), (9, –3), (12, –4), (–9, 3) and (–15, 5)

# Question: 5

The temperature at $12$ noon was $10°\text{C}$ above zero. If it decreases at the rate of $2°\text{C}$ per hour until midnight, at what time would the temperature be $8°\text{C}$ below zero?

What would be the temperature at mid-night?

## Solution

Following number line represents the temperature.

The temperature decreases
The temperature decreases

$10°\text{C}-\left(-8°\text{C}\right)=18°\text{C}$

The temperature decreases

Total time = .

Thus, at $9$ pm the temperature would be $8°\text{C}$ below $0°\text{C}\text{.}$

Time between mid-noon and mid-night $=12$ hours

Temperature at mid-night $=10°\text{C}-2°×12=-14°\text{C}$

# Question: 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?

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

## Solution

(i)             Marks given for one correct answer $=3$

Marks given for $12$ correct answers $=3×12=36$

Radhika scored $20$ marks.

Therefore, marks obtained for incorrect answers $=20-36=-16$

Now, marks given for one incorrect answer $=-2$

Therefore, number of incorrect answers $=\left(-16\right)÷\left(-2\right)=8$

Thus, Radhika has attempted $8$ questions incorrectly.

(ii)          Marks given for seven correct answers $=3×7=21$

Mohini’s score $=-5$

Marks obtained for incorrect answers $=-5-21=-26$

Now, marks given for one incorrect answer $=-2$

Therefore, marks given for incorrect answers $=\left(-26\right)÷\left(-2\right)=13$

Thus, Mohini has attempted $13$ questions incorrectly.

# Question: 7

An elevator descends into a mine shaft at the rate of  If the descent starts from  above the ground level, how long will it take to reach

## Solution

Starting position of mine shaft is  above the ground but it moves in opposite direction so it travels the distance  below the ground.

So total distance covered by mine shaft

Now, time taken to cover a distance of  $=1$ minute

So, time taken to cover a distance of  $=\frac{1}{6}$ minute

Therefore, time taken to cover a distance of

[  ]

Thus, in one hour the mine shaft reaches $-350$ m below the ground.