NCERT Solutions Class 7 Mathematics Chapter 1

Q.1

$\begin{array}{l}\mathrm{Following}\mathrm{number}\mathrm{line}\mathrm{shows}\mathrm{the}\mathrm{temperature}\mathrm{in}\mathrm{degree}\\ \mathrm{celsius}(°\mathrm{C})\mathrm{at}\mathrm{different}\mathrm{places}\mathrm{on}\mathrm{a}\mathrm{particular}\mathrm{day}\\ \left(\text{a}\right)\text{Observe this number line and write the}\\ \text{temperature of the places marked on it}.\\ \left(\text{b}\right)\text{What is the temperature difference between the}\\ \text{hottest and the coldest places among the above}?\\ \left(\text{c}\right)\text{What is the temperature difference between Lahulspiti}\\ \text{and Srinagar}?\\ \left(\text{d}\right)\text{Can we say temperature of Srinagar and Shimla taken}\\ \text{together is less than the temperature at Shimla}?\\ \text{Is it also less than the temperature at Srinagar}?\end{array}$

Ans.

\begin{array}{l}\text{(a)}\text{By observing the number line, the temprature of the}\\ \text{places marked are as follows:}\\ \text{Lahulspiti:}-8°C\\ \text{Srinagar:}-2°C\\ \text{Shimla:}5°C\\ \text{Ooty:}14°C\\ \text{Banglore:}22°C\\ \\ \text{(b)}\text{The hottest place is Banglore with temprature}22°C\\ \text{and coldest place is Lahulspiti}\text{with temprature}-8°C\text{.}\\ \text{So},\text{the temperature difference between the hottest}\\ \text{and the coldest places is}\\ \text{=22}°\text{C}-\left(-8°\text{C}\right)\\ =22°\text{C}+8°\text{C}\\ =\boxed{30°\text{C}}\\ \\ (c)\text{The temperature difference between Lahulspiti and}\\ \text{Srinagar}\\ =-2°\text{C}-\left(-8°\text{C}\right)\\ =-2°\text{C}+8°\text{C}\\ =\boxed{6°\text{C}}\\ \\ (d)\text{The temprature of Srinagar and Shimla are}-2°C\\ \text{and}5°C.\\ \text{So, together their temprature would be}\\ -2°C+5°C=3°C\text{, which is less than temprature of Shimla}\\ \text{Thus, temprature of Srinagar and Shimla taken together}\\ \text{is less than the temprature at Shimla}\text{.}\\ \text{But, it is not less than temprature at Srinagar}\text{.}\end{array}

Q.2

$\begin{array}{l}\mathrm{In}\mathrm{a}\mathrm{quiz},\mathrm{positive}\mathrm{marks}\mathrm{are}\mathrm{given}\mathrm{for}\mathrm{correct}\mathrm{answers}\\ \mathrm{and}\mathrm{negative}\mathrm{marks}\mathrm{are}\mathrm{given}\mathrm{for}\mathrm{in}\mathrm{correct}\mathrm{answers}.\\ \mathrm{If}\mathrm{Jack}’\mathrm{s}\mathrm{scores}\mathrm{in}\mathrm{five}\mathrm{successive}\mathrm{rounds}\mathrm{were}\\ 25,–5,–10,15\mathrm{and}10,\mathrm{what}\mathrm{was}\mathrm{his}\mathrm{total}\mathrm{at}\mathrm{the}\mathrm{end}?\end{array}$

Ans.

\begin{array}{l}\text{Jack’s scores in five successive rounds were 25,}-5,-10,\\ \text{15 and and 10}\text{.}\\ \text{His total at the end would be}\\ \text{=25+}\left(-5\right)+\left(-10\right)+15+10\\ =50-15\\ =\boxed{35}\end{array}

Q.3

$\begin{array}{l}\text{At Srinagar temperature was-5°C on Monday and then}\\ \text{it dropped by2°C on Tuesday. What was the temperature}\\ \text{of Srinagar on Tuesday? On Wednesday, it rose by 4°C.}\\ \text{What was the temperature on this day?}\end{array}$

Ans.

$\begin{array}{l}\text{Temprature on Monday=}-5°C\\ \text{Since, temperature dropped by}2°C\text{on Tuesday}\text{.}\\ \text{So, the temperature of Srinagar on Tuesday was}\\ =-5°C-2°C\\ =\boxed{-7°C}\\ \text{On Wednesday, temperature rose by}4°C,\text{so temperature of}\\ \text{Srinagar on Wednesday was}\\ =-7°C+4°C\\ =\boxed{-3°C}\\ \text{Thus, Temperature on Tuesday and Wednesday was}\\ \boxed{-7°C\text{and}-3°C}\text{respectively}\text{.}\end{array}$

Q.4

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

Ans.

\begin{array}{l}\text{Height of the plane = 5000 m}\\ \text{Depth of the submarine}=-1200\text{m}\\ \text{So, the distance between plane and submarine}\\ \text{=5000 m}-\left(-1200\text{m}\right)\\ =5000\text{m+1200 m}\\ \text{=}\boxed{\text{6200 m}}\\ \text{Thus, the vertical distance between them is 6200 m}\text{.}\end{array}

Q.5

$\begin{array}{l}\text{Mohan deposits Rs 2,000 in his bank account and}\\ \text{withdraws Rs 1,642 from it, the next day. If withdrawal}\\ \text{of amount from the account is represented by a negative}\\ \text{integer, then how will you represent the amount deposited?}\\ \text{Find the balance in Mohan’s account after the withdraw.}\end{array}$

Ans.

\begin{array}{l}\text{Since, withdrawal of amount from the account is represented}\\ \text{by a negative integer, so we take amount deposited as a positive}\\ \text{integer}\\ \text{Amount deposited = Rs 2000}\\ \text{Amount withdrawn}=-\text{Rs 1642}\\ \text{Balance left in Mohan’s account}\\ \text{= Rs 2000}-\text{Rs 1642}\\ \text{=}\boxed{\text{Rs 358}}\\ \text{Thus, the balance in Mohan’s account after withdrawal}\\ \text{is Rs 358}\text{.}\end{array}

Q.6

$\begin{array}{l}\text{Rita goes 20 km towards east from a point A to the point B.}\\ \text{From B, she moves 30 km towards west along the same road.}\\ \text{If the distance towards east is represented by a positive}\\ \text{integer then, how will you represent the distance travelled}\\ \text{towards west? By which integer will you represent her final}\\ \text{position from A?}\end{array}$

Ans.

\begin{array}{l}\text{Here, the distance towards east is represented by a positive}\\ \text{integer and the distance travelled towards west will be}\\ \text{represented by a negative integer}\text{.}\\ \text{So, distance travelled in east direction = 20 km}\\ \text{distance travelled in west direction =}-\text{30 km}\\ \text{Distance travelled from A = 20 km+}\left(-30\text{km}\right)\\ =-10\text{km}\\ \text{Thus, Rita’s distance travelled from point A}\text{}\text{will be represented}\\ \text{by a negative integer}\left(-10\text{km}\right)\text{.}\\ \text{Rita is in west direction}\text{.}\end{array}

Q.7

$\begin{array}{l}\text{In a magic square each row, column and diagonal have the}\\ \text{same sum. Check which of the following is a magic square.}\end{array}$

Ans.

\begin{array}{l}\text{In a magic square, each row, column and diagonal have}\\ \text{the same sum}\text{.}\\ \text{So, in square (i), every row and column sum up to 0}\text{.}\\ \text{However sum of one of its diagonal is not zero}\text{.}\\ -4-2=-6\ne \text{0}\\ \text{So, (i) is not a magic square}\text{.}\\ \text{Similarly, in square (ii) every row, column and diagonal}\\ \text{sum up to}-\text{9}\text{.}\\ \text{Thus, (ii) is a magic square}\text{.}\end{array}

Q.8

$\begin{array}{l}\text{Verifya}-\left(-\text{b}\right)\text{\hspace{0.33em}=\hspace{0.33em}a\hspace{0.33em}+\hspace{0.33em}b for the following values of a and b.}\\ \left(\text{i}\right)\text{\hspace{0.17em}\hspace{0.33em}a\hspace{0.33em}=\hspace{0.33em}21,b\hspace{0.33em}=\hspace{0.33em}18\hspace{0.33em}}\left(\text{ii}\right)\text{\hspace{0.33em}\hspace{0.17em}a\hspace{0.33em}=\hspace{0.33em}118,\hspace{0.33em}b\hspace{0.33em}=\hspace{0.33em}125}\\ \left(\text{iii}\right)\text{\hspace{0.33em}a\hspace{0.33em}=\hspace{0.33em}75,b\hspace{0.33em}=\hspace{0.33em}84\hspace{0.33em}}\left(\text{iv}\right)\text{\hspace{0.33em}a\hspace{0.33em}=\hspace{0.33em}28,\hspace{0.33em}b\hspace{0.33em}=\hspace{0.33em}11}\end{array}$

Ans.

$\begin{array}{l}\left(\mathrm{i}\right)\text{\hspace{0.17em}a=21, b=18}\\ \mathrm{a}-(-\mathrm{b})=21-(-18)=21+18=39\\ \mathrm{a}+\mathrm{b}=21+18=39\\ \mathrm{Thus},\mathrm{a}-(-\mathrm{b})=\mathrm{a}+\mathrm{b}\\ \left(\mathrm{ii}\right)\text{\hspace{0.17em}a=118, b=125}\\ \mathrm{a}-(-\mathrm{b})=118-(-125)=118+125=243\\ \mathrm{a}+\mathrm{b}=118+125=243\\ \mathrm{Thus},\mathrm{a}-(-\mathrm{b})=\mathrm{a}+\mathrm{b}\\ \left(\mathrm{iii}\right)\text{\hspace{0.17em}a=75, b=84}\\ \mathrm{a}-(-\mathrm{b})=75-(-84)=75+84=159\\ \mathrm{a}+\mathrm{b}=75+84=39\\ \mathrm{Thus},\mathrm{a}-(-\mathrm{b})=\mathrm{a}+\mathrm{b}\\ \left(\mathrm{iv}\right)\text{\hspace{0.17em}a=28, b=11}\\ \mathrm{a}-(-\mathrm{b})=28-(-11)=28+11=39\\ \mathrm{a}+\mathrm{b}=28+11=39\\ \mathrm{Thus},\mathrm{a}-(-\mathrm{b})=\mathrm{a}+\mathrm{b}\end{array}$

Q.9

$\begin{array}{l}\text{Use the sign of\hspace{0.33em}>,<\hspace{0.33em}or\hspace{0.33em}=\hspace{0.33em}in the box to make the statement}\\ \text{true.}\\ \left(\text{a}\right)\left(-\text{8}\right)\text{+}\left(-\text{4}\right)\boxed{}\left(-\text{8}\right)-\left(-\text{4}\right)\\ \left(b\right)(–3)+7–\left(19\right)\boxed{}15–8+(–9)\\ \left(c\right)23–41+11\boxed{}23–41–11\\ \left(d\right)39+(–24)–\left(15\right)\boxed{}36+(–52)–(–36)\\ \left(e\right)–231+79+51\boxed{}–399+159+81\end{array}$

Ans.

\begin{array}{l}\text{(a) (}-\text{8) + (}-\text{4)}\boxed{<}\text{(}-\text{8)}-\text{(}-\text{4)}\\ \text{(b) (}-\text{3) +7}-\text{(19)}\boxed{<}\text{15}-\text{8+(}-\text{9)}\\ \text{(c) 23}-\text{41+11}\boxed{>}23-41-11\\ \text{(d) 39+(}-\text{24)}-\text{(15)}\boxed{<}36+\text{(}-\text{52)}-\text{(}-\text{36)}\\ \text{(e)}-\text{231+79+51}\boxed{>}-\text{399 + 159 + 81}\end{array}

Q.10

$\begin{array}{l}\text{A\hspace{0.33em}water tank has steps inside it. A monkey is sitting on}\\ \text{the topmost step}\left(\text{i.e., the first step}\right)\text{. The water level is}\\ \text{at the ninth step.}\\ \left(\text{i}\right)\text{He jumps 3 steps down and then jumps back 2 steps}\\ \text{up.In how many jumps will he reach the water level?}\\ \left(\text{ii}\right)\text{After drinking water, he wants to go back. For this,}\\ \text{he jumps 4 steps up and then jumps back 2}\\ \text{steps down in every move. In how many jumps will he}\\ \text{reach back the top step?}\\ \left(\text{iii}\right)\text{If the number of steps moved down is represented}\\ \text{by negative integers and the number of steps moved up}\\ \text{by positive integers, represent his moves in part}\left(\text{i}\right)\text{and}\\ \left(\text{ii}\right)\text{by completing the following;}\\ \left(\text{a}\right)\text{– 3 + 2 –}\mathrm{\dots }\text{=- 8}\\ \left(\text{b}\right)\text{4 – 2 +}\mathrm{\dots }\text{= 8.}\\ \text{In}\left(\text{a}\right)\text{the sum}\left(\text{– 8}\right)\text{represents going down by eight}\\ \text{steps. So, what will the sum 8 in}\left(\text{b}\right)\text{represent?}\end{array}$

Ans.

\begin{array}{l}\text{Consider the steps moved down be represented by positive}\\ \text{integers and steps moved up be represented by negative}\\ \text{integers}\text{.}\\ (i)\text{Initially, the monkey was at step = 1}\\ \text{After 1st jump, monkey will be at step = 1+3=4}\\ \text{After 2nd jump, monkey will be at step = 4+}\left(-2\right)\text{=2}\\ \text{After 3rd jump, monkey will be at step = 2+3=5}\\ \text{After 4th jump, monkey will be at step = 5+}\left(-2\right)\text{=3}\\ \text{After 5th jump, monkey will be at step = 3+3=6}\\ \text{After 6th jump, monkey will be at step = 6+}\left(-2\right)\text{=4}\\ \text{After 7th jump, monkey will be at step = 4+3=7}\\ \text{After 8th jump, monkey will be at step = 7+}\left(-2\right)\text{=5}\\ \text{After 9th jump, monkey will be at step = 5+3=8}\\ \text{After 10th jump, monkey will be at step = 8+}\left(-2\right)\text{=6}\\ \text{After 11th jump, monkey will be at step = 6+3=9}\\ \text{Clearly, the monkey will be at water level (i}\text{.e}\text{., 9th step)}\\ \text{after 11 jumps}\text{.}\\ \\ \text{(ii)}\\ \text{Initiall, the monkey was at step = 9}\\ \text{After 1st jump, monkey will be at step = 9+}\left(-4\right)\text{=5}\\ \text{After 2nd jump, monkey will be at step = 5+2=7}\\ \text{After 3rd jump, monkey will be at step = 7+}\left(-4\right)\text{=3}\\ \text{After 4th jump, monkey will be at step = 3+2=5}\\ \text{After 5th jump, monkey will be at step = 5+}\left(-4\right)\text{=1}\\ \text{Clearly,}\text{the will reach back at the top step after 5 jumps}\text{.}\\ \text{(iii) If steps moved down are represented by a negative}\\ \text{integers and steps moved up are represented bya positive}\\ \text{integers, then his moves will be as follows:}\\ \text{Moves in part (i):}\\ -\text{3+2}-\text{3+2}-\text{3+2}-\text{3+2}-\text{3+2}-\text{3=}-\text{8}\\ \text{Moves in part (ii):}\\ \text{4}-\text{2+4}-\text{2+4=8}\\ \text{Moves in part (ii) represent goin up 8 steps}\text{.}\end{array}

Q.11

$\begin{array}{l}\text{Write down a pair of integers whose:}\\ \left(\text{a}\right)\text{\hspace{0.33em}sumis -7}\\ \left(\text{b}\right)\text{\hspace{0.33em}differenceis -10}\\ \left(\text{c}\right)\text{\hspace{0.33em}sumis 0}\end{array}$

Ans.

\begin{array}{l}\text{(a)}-\text{8+(+1) =}-\text{8+1=}-\text{7}\\ \text{So, the pair is (}-\text{8, 1)}\\ \text{(b)}-\text{12}-\text{(}-\text{2)=}-\text{12+2=}-\text{10}\\ \text{So, the pair is (}-\text{12,}-\text{2)}\\ \text{(c) 5+(}-\text{5)=5}-\text{5=0}\\ \text{So, the pair is (}-\text{5,}-\text{5)}\end{array}

Q.12 

$\begin{array}{l}\begin{array}{l}\left(\text{a}\right)\text{\hspace{0.33em}Write\hspace{0.33em}a\hspace{0.33em}pair\hspace{0.33em}of\hspace{0.33em}negative\hspace{0.33em}integers\hspace{0.33em}whose\hspace{0.33em}difference}\\ \text{gives\hspace{0.33em}8.}\\ \left(\text{b}\right)\text{\hspace{0.33em}Write\hspace{0.33em}an\hspace{0.33em}egative\hspace{0.33em}integer\hspace{0.33em}and a\hspace{0.33em}positive\hspace{0.33em}integer\hspace{0.33em}whose}\end{array}\\ \begin{array}{l}\text{sumis\hspace{0.33em}-5.}\\ \left(\text{c}\right)\text{\hspace{0.33em}Write\hspace{0.33em}an\hspace{0.33em}egative\hspace{0.33em}integer\hspace{0.33em}and\hspace{0.33em}a\hspace{0.33em}positive\hspace{0.33em}integer\hspace{0.33em}whose}\\ \text{differenceis\hspace{0.33em}-3.}\end{array}\end{array}$

Ans.

\begin{array}{l}\text{(a) To write a pair of negative integers whose difference is 8}\\ \text{So,}-\text{2}-\text{(}-\text{10)=}-\text{2+10=8}\\ \text{So, pair of negative}\mathrm{int}\text{egers is}\boxed{(-2,-10)}.\\ (b)\text{To write a negative integer and a positive integer whose}\\ \text{sum is}-\text{5}\text{.}\\ -\text{8+3=}-\text{5}\\ \text{So, a negative integer and a positive}\mathrm{int}\text{eger}\text{is}\boxed{(-8,3)}.\\ (c)\text{To write a negative integer and a positive integer whose}\\ \text{difference is}-\text{3}\\ -2-\text{(+1)=}-\text{2}-\text{1=}-\text{3}\\ \text{So, a negative integer and a positive}\mathrm{int}\text{eger}\text{is}\boxed{(-2,1)}.\end{array}

Q.13

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

Ans.

\begin{array}{l}\text{Sum of team A scored}=-40+10+0=-30\\ \text{Sum of team B scored}=10+0-40=-30\\ \text{Both teams scored equal}\text{.}\\ \text{Yes, we can add integers in any order}\text{.}\end{array}

Q.14

$\begin{array}{l}\mathrm{Fill}\mathrm{in}\mathrm{the}\mathrm{blanks}\mathrm{to}\mathrm{make}\mathrm{the}\mathrm{following}\mathrm{statements}\mathrm{true}:\\ \left(\mathrm{i}\right)\mathrm{}(–5)+(.\dots \dots \dots ..)=(–8)+(.\dots \dots \dots ..)\\ \left(\mathrm{ii}\right)–53+.\dots \dots \dots ..=–53\\ \left(\mathrm{iii}\right)17+.\dots \dots \dots ..=0\\ \left(\mathrm{iv}\right)[13+(–12)]+(.\dots \dots \dots ..)=.\dots \dots \dots ..+[(–12)+(–7)]\\ \left(\mathrm{v}\right)(–4)+[.\dots \dots \dots ..+(–3)]=[.\dots \dots \dots ..+15]+.\dots \dots \dots ..\end{array}$

Ans.

\begin{array}{l}\left(\text{i}\right)\left(-\text{5}\right)+\left(\boxed{-8}\right)=\left(-\text{8}\right)+\left(\boxed{-5}\right)\\ \left(\text{ii}\right)-\text{53}+\boxed{0}=-\text{53}\\ \left(\text{iii}\right)\text{17}+\boxed{-17}=0\\ \left(\text{iv}\right)\left[\text{13}+\left(-\text{12}\right)\right]+\left(\boxed{-7}\right)=\boxed{13}+\left[\left(-\text{12}\right)+\left(-\text{7}\right)\right]\\ \left(\text{v}\right)\left(-\text{4}\right)+\left[\boxed{15}+\left(-\text{3}\right)\right]=\left[\boxed{-4}+\text{15}\right]+\boxed{-3}\end{array}

Q.15

$\begin{array}{l}\mathrm{Find}\mathrm{each}\mathrm{of}\mathrm{the}\mathrm{following}\mathrm{products}:\\ \left(\mathrm{a}\right)3\times (–1)\\ \left(\mathrm{b}\right)(–1)\times 225\\ \left(\mathrm{c}\right)(–21)\times (–30)\\ \left(\mathrm{d}\right)(–316)\times (–1)\\ \left(\mathrm{e}\right)(–15)\times 0\times (–18)\\ \left(\mathrm{f}\right)(–12)\times (–11)\times \left(10\right)\\ \left(\mathrm{g}\right)9\times (–3)\times (–6)\\ \left(\mathrm{h}\right)(–18)\times (–5)\times (–4)\\ \left(\mathrm{i}\right)(–1)\times (–2)\times (–3)\times 4\\ \left(\mathrm{j}\right)(–3)\times (–6)\times (–2)\times (–1)\end{array}$

Ans.

\begin{array}{l}\left(\text{a}\right)\text{3}\times \left(-\text{1}\right)=\boxed{-3}\\ \left(\text{b}\right)\left(-\text{1}\right)\times \text{225}=\boxed{-225}\\ \left(\text{c}\right)\left(-\text{21}\right)\times \left(-\text{3}0\right)=\boxed{630}\\ \left(\text{d}\right)\left(-\text{316}\right)\times \left(-\text{1}\right)=\boxed{316}\\ \left(\text{e}\right)\left(-\text{15}\right)\times 0\times \left(-\text{18}\right)=\boxed{0}\\ \left(\text{f}\right)\left(-\text{12}\right)\times \left(-\text{11}\right)\times \left(\text{1}0\right)=\boxed{1320}\\ \left(\text{g}\right)\text{9}\times \left(-\text{3}\right)\times \left(-\text{6}\right)=\boxed{162}\\ \left(\text{h}\right)\left(-\text{18}\right)\times \left(-\text{5}\right)\times \left(-\text{4}\right)=\boxed{-360}\\ \left(\text{i}\right)\left(-\text{1}\right)\times \left(-\text{2}\right)\times \left(-\text{3}\right)\times \text{4}=\boxed{-24}\\ \left(\text{j}\right)\left(-\text{3}\right)\times \left(-\text{6}\right)\times \left(-\text{2}\right)\times \left(-\text{1}\right)=\boxed{36}\end{array}

Q.16

$\begin{array}{l}\mathrm{Verify}\mathrm{the}\mathrm{following}:\\ \left(\mathrm{a}\right)18\times [7+(–3)]=[18\times 7]+[18\times (–3)]\\ \left(\mathrm{b}\right)(–21)\times [(–4)+(–6)]=[(–21)\times (–4)]+[(–21)\times (–6)]\end{array}$

Ans.

\begin{array}{l}\left(\text{a}\right)\\ \text{18}\times \left[\text{7}+\left(–\text{3}\right)\right]=18\times \left[4\right]=72\\ \text{and}\\ \left[\text{18}\times \text{7}\right]+\left[\text{18}\times \left(–\text{3}\right)\right]=\left[126\right]+\left[-56\right]=70\\ \text{So},\boxed{\text{18}\times \left[\text{7}+\left(–\text{3}\right)\right]\ne \left[\text{18}\times \text{7}\right]+\left[\text{18}\times \left(–\text{3}\right)\right]}\\ \left(\text{b}\right)\\ \left(-\text{21}\right)\times \left[\left(-\text{4}\right)+\left(-\text{6}\right)\right]=\left(-21\right)\times \left[-10\right]=210\\ \text{and}\\ \left[\left(-\text{21}\right)\times \left(-\text{4}\right)\right]+\left[\left(-\text{21}\right)\times \left(-\text{6}\right)\right]=\left[84\right]+\left[126\right]=210\\ \text{So},\\ \boxed{\left(–\text{21}\right)\times \left[\left(–\text{4}\right)+\left(–\text{6}\right)\right]=\left[\left(–\text{21}\right)\times \left(–\text{4}\right)\right]+\left[\left(–\text{21}\right)\times \left(–\text{6}\right)\right]}\end{array}

Q.17

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{For}\mathrm{any}\mathrm{integera},\mathrm{what}\mathrm{is}(–1)\times \mathrm{a}\mathrm{equal}\mathrm{to}?\\ \left(\mathrm{ii}\right)\mathrm{Determine}\mathrm{the}\mathrm{integer}\mathrm{whose}\mathrm{product}\mathrm{with}(–1)\mathrm{is}\\ \left(\mathrm{a}\right)–22\left(\mathrm{b}\right)\mathrm{}37\left(\mathrm{c}\right)0\end{array}$

Ans.

\begin{array}{l}\left(\text{i}\right)\left(-\text{1}\right)\times \text{a=}-\text{a}\\ \text{(ii) (a)}\boxed{22}\times \left(-1\right)=-22\\ \text{(b)}\boxed{-\text{37}}\times \left(-1\right)=37\\ \text{(c)}\boxed{\text{0}}\times \left(-1\right)=0\end{array}

Q.18

$\begin{array}{l}\mathrm{Starting}\mathrm{from}(–1)\times 5,\mathrm{write}\mathrm{various}\mathrm{products}\mathrm{showing}\mathrm{some}\\ \mathrm{pattern}\mathrm{to}\mathrm{show}(–1)\times (–1)=1\end{array}$

Ans.

\begin{array}{l}-1\times 5=-5\\ -1\times 4=-4=-5+1\\ -1\times 3=-3=-4+1\\ -1\times 2=-2=-3+1\\ -1\times 1=-1=-2+1\\ -1\times 0=0=-1+1\\ \text{Thus,}\boxed{-\text{1}\times \left(-1\right)=0+1=1}\end{array}

Q.19 

\begin{array}{l}Findtheproduct,usingsuitableproperties:\\ \left(a\right)26\times \left(–48\right)+\left(–48\right)\times \left(–36\right)\left(b\right)8\times 53\times \left(–125\right)\\ \left(c\right)15\times \left(–25\right)\times \left(–4\right)\times \left(–10\right)\left(d\right)\left(–41\right)\times 102\\ \left(e\right)625\times \left(–35\right)+\left(–625\right)\times 65\left(f\right)7\times \left(50–2\right)\\ \left(g\right)\left(–17\right)\times \left(–29\right)\left(h\right)\left(–57\right)\times \left(–19\right)+57\end{array}

Ans.

\begin{array}{l}\left(\text{a}\right)\text{26}\times \left(-\text{48}\right)+\left(-\text{48}\right)\times \left(-\text{36}\right)\\ =\left(-48\right)\times 26+\left(-48\right)\times \left(-36\right)\\ =\left(-48\right)\left[26-36\right]\\ =\left(-48\right)\left[-10\right]\\ =\boxed{480}\\ \left(\text{b}\right)\text{8}\times \text{53}\times \left(-\text{125}\right)\\ =424\times \left(-125\right)\\ =\boxed{53000}\\ \left(\text{c}\right)\text{15}\times \left(-\text{25}\right)\times \left(-\text{4}\right)\times \left(-\text{1}0\right)\\ =\left(15\times -25\right)\times \left(-4\times -10\right)\\ =\left(-375\right)\times 40\\ =\boxed{-15000}\\ \left(\text{d}\right)\left(-\text{41}\right)\times \text{1}0\text{2}\\ =\boxed{-\text{4182}}\\ \left(\text{e}\right)\text{625}\times \left(-\text{35}\right)+\left(-\text{625}\right)\times \text{65}\\ =-21875+\left(-40625\right)\\ =\boxed{-62500}\\ \\ \left(\text{f}\right)\text{7}\times \left(\text{5}0-\text{2}\right)\\ =7\times \left(48\right)\\ =\boxed{336}\\ \left(\text{g}\right)\left(-\text{17}\right)\times \left(-\text{29}\right)\\ =\boxed{493}\\ \left(\text{h}\right)\left(-\text{57}\right)\times \left(-\text{19}\right)+\text{57}\\ =1083+57\\ =\boxed{1140}\end{array}

Q.20

$\begin{array}{l}\left(\text{i}\right)\text{For any integer a, what is}\left(\text{–1}\right)\text{× a equal to?}\\ \left(\text{ii}\right)\text{Determine the integer whose product with}\left(\text{–1}\right)\text{is}\\ \left(\text{a}\right)\text{–22}\left(\text{b}\right)\text{37}\left(\text{c}\right)\text{0}\end{array}$

Ans.

\begin{array}{l}\left(\text{i}\right)\left(-\text{1}\right)\times \text{a=}-\text{a}\\ \text{(ii) (a)}\boxed{22}\times \left(-1\right)=-22\\ \text{(b)}\boxed{-\text{37}}\times \left(-1\right)=37\\ \text{(c)}\boxed{\text{0}}\times \left(-1\right)=0\end{array}

Q.21

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

Ans.

\begin{array}{l}-1\times 5=-5\\ -1\times 4=-4=-5+1\\ -1\times 3=-3=-4+1\\ -1\times 2=-2=-3+1\\ -1\times 1=-1=-2+1\\ -1\times 0=0=-1+1\\ \text{Thus ,}\boxed{-\text{1}\times \left(-1\right)=0+1=1}\end{array}

Q.22

$\begin{array}{l}\text{Find the product, using suitable properties:}\\ \left(\text{a}\right)\text{26 ×}\left(\text{– 48}\right)\text{+}\left(\text{– 48}\right)\text{×}\left(\text{–36}\right)\left(\text{b}\right)\text{8 × 53 ×}\left(\text{–125}\right)\\ \left(\text{c}\right)\text{15 ×}\left(\text{–25}\right)\text{×}\left(\text{– 4}\right)\text{×}\left(\text{–10}\right)\left(\text{d}\right)\left(\text{– 41}\right)\text{× 102}\\ \left(\text{e}\right)\text{625 ×}\left(\text{–35}\right)\text{+}\left(\text{– 625}\right)\text{× 65}\left(\text{f}\right)\text{7 ×}\left(\text{50 – 2}\right)\\ \left(\text{g}\right)\left(\text{–17}\right)\text{×}\left(\text{–29}\right)\left(\text{h}\right)\left(\text{–57}\right)\text{×}\left(\text{–19}\right)\text{+ 57}\end{array}$

Ans.

\begin{array}{l}\left(\text{a}\right)\text{26}\times \left(-\text{48}\right)+\left(-\text{48}\right)\times \left(-\text{36}\right)\\ =\left(-48\right)\times 26+\left(-48\right)\times \left(-36\right)\\ =\left(-48\right)\left[26-36\right]\\ =\left(-48\right)\left[-10\right]\\ =\boxed{480}\\ \left(\text{b}\right)\text{8}\times \text{53}\times \left(-\text{125}\right)\\ =424\times \left(-125\right)\\ =\boxed{53000}\\ \left(\text{c}\right)\text{15}\times \left(-\text{25}\right)\times \left(-\text{4}\right)\times \left(-\text{1}0\right)\\ =\left(15\times -25\right)\times \left(-4\times -10\right)\\ =\left(-375\right)\times 40\\ =\boxed{-15000}\\ \left(\text{d}\right)\left(-\text{41}\right)\times \text{1}0\text{2}\\ =\boxed{-\text{4182}}\\ \left(\text{e}\right)\text{625}\times \left(-\text{35}\right)+\left(-\text{625}\right)\times \text{65}\\ =-21875+\left(-40625\right)\\ =\boxed{-62500}\\ \left(\text{f}\right)\text{7}\times \left(\text{5}0-\text{2}\right)\\ =7\times \left(48\right)\\ =\boxed{336}\\ \left(\text{g}\right)\left(-\text{17}\right)\times \left(-\text{29}\right)\\ =\boxed{493}\\ \left(\text{h}\right)\left(-\text{57}\right)\times \left(-\text{19}\right)+\text{57}\\ =1083+57\\ =\boxed{1140}\end{array}

Q.23

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

Ans.

\begin{array}{l}\text{Given initial temprature}=40°C\\ \text{Change in temprature per hour}=-5°C\\ \text{Change in temprature after 10 hours}=-5°C\times 10=-50°C\\ \text{Final temprature}=40°C-50°C=\boxed{-10°C}\end{array}

Q.24

$\begin{array}{l}\text{In a class test containing 10 questions, 5 marks are}\\ \text{awarded for every correct answer and}\left(\text{-2}\right)\text{marks are}\\ \text{awarded forevery incorrect answer and 0 for questions}\\ \text{not attempted.}\\ \left(\text{i}\right)\text{Mohan gets four correct and six incorrect answers.}\\ \text{What is his score?}\\ \left(\text{ii}\right)\text{Reshma gets five correct answers and five incorrect}\\ \text{answers, what is her score?}\\ \left(\text{iii}\right)\text{Heena gets two correct and five incorrect answers}\\ \text{out of seven questions she attempts. What is her score?}\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ \text{Marks given for 1 correct answer}=5\\ \text{Marks given for 4 correct answer}=5\times 4=20\\ \text{Marks given for 1 wrong answer}=-2\\ \text{Marks given for 6 wrong answer}=-2\times 6=-12\\ \text{Score obtained by Mohan}=20-12=\boxed{8}\\ \text{(ii)}\\ \text{Marks given for 1 correct answer}=5\\ \text{Marks given for 5 correct answer}=5\times 5=25\\ \text{Marks given for 1 wrong answer}=-2\\ \text{Marks given for 5 wrong answer}=-2\times 5=-10\\ \text{Scored obtained by Reshma}=25-10=\boxed{15}\\ \text{(iii) Similarly,}\\ \text{Marks given for 2 correct answer}=5\times 2\\ \text{Marks given for 5 correct answer}=-2\times 5=-10\\ \text{Score btained by Heena}=10–10=\boxed{0}\end{array}$

Q.25

$\begin{array}{l}\text{A cement company earns a profit of ₹ 8 per bag of white}\\ \text{cement sold and a loss of ₹ 5 per bag of grey cement sold.}\end{array}$

$\begin{array}{l}\left(\text{a}\right)\text{The company sells 3,000 bags of white cement and 5,000}\\ \text{bags of grey cement in a month. What is its profit or loss?}\\ \left(\text{b}\right)\text{What is the number of white cement bags it must sell to}\\ \text{have neither profit nor loss, if the number of grey bags sold}\\ \text{is 6,400 bags.}\end{array}$

Ans.

\begin{array}{l}(a)\\ \text{Profit earned while selling 1 bag of white cement}=₹8\\ \text{Profit earned while selling 3000 bag of white cement}\\ =₹8\times 3000=24000\\ \text{Loss incurred while selling 1 bag of grey cement}=-₹5\\ \text{Loss incurred while selling 5000 bag of grey cement}\\ =-₹5\times 5000=-₹25000\\ \text{Total profit/loss earned = Profit+loss}\\ =₹24000-₹25000=-₹1000\\ \text{Thus, there will be a loss of}₹\text{1000 to the company}\text{.}\\ \text{(b)}\\ \text{Loss incurred while selling 1 bag of grey cement}=-₹5\\ \text{Loss incurred while selling 6400 bag of grey cement}\\ =-₹5\times 6400=-₹32000\\ \text{Let the number of white bag to be sold be}x\text{.}\\ \text{Profit earned selling 1 bag of white cement}=₹8\\ \text{Profit earned selling x bag of white cement}=₹8x\\ \text{When there is no profit or no loss, we have}\\ \text{Profit earned + Loss incurred}=0\\ 8x–32000=0\\ 8x=32000\\ x=\frac{32000}{8}\\ =\boxed{4000}\\ \text{Thus},4000\text{bags of white cement should be sold}\text{.}\end{array}

Q.26

$\begin{array}{l}\text{Replace the blank with an integer to make it a true}\\ \text{statement.}\\ \left(\text{a}\right)\left(-\text{3}\right)\text{× \_\_\_\_\_ = 27}\\ \left(\text{b}\right)\text{5 × \_\_\_\_\_ = -35}\\ \left(\text{c}\right)\text{\_\_\_\_\_ ×}\left(-\text{8}\right)\text{=}-\text{56}\\ \left(\text{d}\right)\text{\_\_\_\_\_ ×}\left(-\text{12}\right)\text{= 132}\end{array}$

Ans.

\begin{array}{l}\left(\text{a}\right)\left(\text{–3}\right)\text{×}\underset{\_}{\boxed{\text{–9}}}\text{= 27}\\ \left(\text{b}\right)\text{5 ×}\underset{\_}{\boxed{\text{–7}}}\text{= –35}\\ \left(\text{c}\right)\underset{\_}{\boxed{\text{7}}}\text{×}\left(\text{–8}\right)\text{= –56}\\ \left(\text{d}\right)\underset{\_}{\boxed{\text{–11}}}\text{×}\left(\text{–12}\right)\text{= 132}\end{array}

Q.27

$\begin{array}{l}\text{Evaluate each of the following:}\\ \left(\mathrm{a}\right)(–30)÷10\\ \left(\mathrm{b}\right)50÷(–5)\\ \left(\mathrm{c}\right)(–36)÷(–9)\\ \left(\mathrm{d}\right)(–49)÷\left(49\right)\\ \left(\mathrm{e}\right)13÷[(–2)+1]\\ \left(\mathrm{f}\right)0÷(–12)\\ \left(\mathrm{g}\right)(–31)÷[(–30)+(–1)]\\ \left(\mathrm{h}\right)[(–36)÷12]÷3\\ \left(\mathrm{i}\right)[(–6)+5)]÷[(–2)+1]\end{array}$

Ans.

\begin{array}{l}\left(\text{a}\right)\left(-\text{3}0\right)÷\text{1}0=\boxed{-3}\\ \left(\text{b}\right)\text{5}0÷\left(-\text{5}\right)=\boxed{-10}\\ \left(\text{c}\right)\left(-\text{36}\right)÷\left(-\text{9}\right)=\boxed{4}\\ \left(\text{d}\right)\left(-\text{49}\right)÷\left(\text{49}\right)=\boxed{-1}\\ \left(\text{e}\right)\text{13}÷\left[\left(-\text{2}\right)+\text{1}\right]=13÷\left[-1\right]=\boxed{-13}\\ \left(\text{f}\right)0÷\left(-\text{12}\right)=\boxed{0}\\ \left(\text{g}\right)\left(-\text{31}\right)÷\left[\left(-\text{3}0\right)+\left(-\text{1}\right)\right]=\left(-31\right)÷\left[-31\right]=\boxed{1}\\ \left(\text{h}\right)\left[\left(-\text{36}\right)÷\text{12}\right]÷\text{3=}\left[-3\right]÷3=\boxed{-1}\\ \left(\text{i}\right)\left[\left(-\text{6}\right)+\text{5}\right)\left]÷\right[\left(-\text{2}\right)+\text{1}]=\left[-1\right]÷\left[-1\right]=\boxed{1}\end{array}

Q.28

$\begin{array}{l}\text{Verify that a÷}\left(\text{b+c}\right)\text{¹}\left(\text{a÷b}\right)\text{+}\left(\text{a÷c}\right)\text{\hspace{0.33em}for each of the}\\ \text{following.}\\ \text{values of a, b and c.}\\ \text{(a) a=12, b=-4, c=2}\\ \text{(b) a=(-10), b=1, c=1}\end{array}$

Ans.

$\begin{array}{l}\left(\mathrm{a}\right)\mathrm{a}=12,\mathrm{b}=-4,\mathrm{c}=2\\ \mathrm{a}÷(\mathrm{b}+\mathrm{c})=12÷(-4+2)\\ =12÷(-2)\\ =-6\\ \text{and}\\ (\mathrm{a}÷\mathrm{b})+(\mathrm{a}÷\mathrm{c})=(12÷(-4))+(12÷2)\\ =(-3)+6=3\\ \mathrm{So},\boxed{\mathrm{a}÷(\mathrm{b}+\mathrm{c})\ne (\mathrm{a}÷\mathrm{b})+(\mathrm{a}÷\mathrm{c})}\\ \left(\mathrm{b}\right)\\ \mathrm{a}=-10,\mathrm{b}=1,\mathrm{c}=1\\ \mathrm{a}÷(\mathrm{b}+\mathrm{c})=-10÷(1+1)\\ =-10÷\left(2\right)\\ =-5\\ \text{and}\\ (\mathrm{a}÷\mathrm{b})+(\mathrm{a}÷\mathrm{c})=(-10÷1)+(-10÷1)\\ =(-10)-10=-20\\ \mathrm{So},\boxed{\mathrm{a}÷(\mathrm{b}+\mathrm{c})\ne (\mathrm{a}÷\mathrm{b})+(\mathrm{a}÷\mathrm{c})}\end{array}$

Q.29

$\begin{array}{l}\text{Fill in the blanks:}\\ \left(\text{a}\right)\text{\hspace{0.33em}369 ÷ \_\_\_\_\_ = 369}\\ \left(\text{b}\right)\left(\text{–75}\right)\text{÷ \_\_\_\_\_ = –1}\\ \left(\text{c}\right)\left(\text{–206}\right)\text{÷ \_\_\_\_\_ = 1}\\ \left(\text{d}\right)\text{\hspace{0.33em}– 87 ÷ \_\_\_\_\_ = 87}\\ \left(\text{e}\right)\text{\hspace{0.33em}\_\_\_\_\_ ÷ 1 = – 87}\\ \left(\text{f}\right)\text{\hspace{0.33em}\_\_\_\_\_ ÷ 48 = –1}\\ \left(\text{g}\right)\text{\hspace{0.33em}20 ÷ \_\_\_\_\_ = –2}\\ \left(\text{h}\right)\text{\hspace{0.33em}\_\_\_\_\_ ÷}\left(\text{4}\right)\text{= –3}\end{array}$

Ans.

\begin{array}{l}\left(\text{a}\right)\text{369}÷\underset{\_}{\boxed{1}}=\text{369}\\ \left(\text{b}\right)\left(-\text{75}\right)÷\underset{\_}{\boxed{75}}=-\text{1}\\ \left(\text{c}\right)\left(-\text{2}0\text{6}\right)÷\underset{\_}{\boxed{-206}}=\text{1}\\ \left(\text{d}\right)-\text{87}÷\underset{\_}{\boxed{-1}}=\text{87}\\ \left(\text{e}\right)\underset{\_}{\boxed{-87}}÷\text{1}=-\text{87}\\ \left(\text{f}\right)\underset{\_}{\boxed{-48}}÷\text{48}=-\text{1}\\ \left(\text{g}\right)\text{2}0÷\underset{\_}{\underset{\_}{\boxed{-10}}}=-\text{2}\\ \left(\text{h}\right)\underset{\_}{\boxed{-12}}÷\left(\text{4}\right)=-\text{3}\end{array}

Q.30

$\begin{array}{l}\text{Write five pairs of integers}\left(\text{a,b}\right)\text{such that a÷b\hspace{0.33em}=}-\text{3. One}\\ \text{such pair is}\left(\text{6,\hspace{0.33em}}-\text{2}\right)\text{\hspace{0.33em}because 6÷}\left(-\text{2}\right)\text{\hspace{0.33em}=}\left(-\text{3}\right)\text{.}\end{array}$

Ans.

\begin{array}{l}\text{Five pair of integers are:}\\ \boxed{\left(3,-1\right),\left(-3,1\right),\left(9,-3\right),\left(-9,3\right)\text{and}\left(12,-4\right)}\end{array}

Q.31

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

Ans.

\begin{array}{l}\text{Initial temprature}=10°C\\ \text{Change in temprature per hour}=-2°C\\ \text{Temprature at 1:00 PM=10}°C+\left(-2°C\right)=8°C\\ \text{Temprature at 2:00 PM=8}°C+\left(-2°C\right)=6°C\\ \text{Temprature at 3:00 PM=6}°C+\left(-2°C\right)=4°C\\ \text{Temprature at 4:00 PM}=4°C+\left(-2°C\right)=2°C\\ \text{Temprature at 5:00 PM}=2°C+\left(-2°C\right)=0°C\\ \text{Temprature at 6:00 PM}=0°C+\left(-2°C\right)=-2°C\\ \text{Temprature at 7:00 PM}=-2°C+\left(-2°C\right)=-4°C\\ \text{Temprature at 8:00 PM}=-4°C+\left(-2°C\right)=-6°C\\ \text{Temprature at 9:00 PM}=-6°C+\left(-2°C\right)=-8°C\\ \text{Thus, the temprature will be}8°C\text{below zero at 9:00PM}\\ \text{It will take 12 hours to be midnight (i}\text{.e}\text{. 12:00 AM) after 12:00}\\ \text{noon}\text{.}\\ \text{Change in temprature in 12 hours}=-\text{2}°C\times 12=-24°C\\ \text{At midnight the temprature would be}=\text{10+}\left(-24\right)=-14°C\\ \text{Thus,}\boxed{\text{the temprature at midnight will be 14}°C\text{below}0}.\end{array}

Q.32

$\begin{array}{l}\text{In a class test}\left(\text{+ 3}\right)\text{marks are given for every correct}\\ \text{answer and}\left(\text{-2}\right)\text{marks are given for every incorrect}\\ \text{answer and no marks for not attempting any question.}\\ \left(\text{i}\right)\text{Radhika scored 20 marks. If she has got 12 correct}\\ \text{answers, how many questions has she attempted}\\ \text{incorrectly?}\\ \left(\text{ii}\right)\text{Mohini scores -5 marks in this test, though she has}\\ \text{got 7 correct answers. How many questions has she}\\ \text{attempted incorrectly?}\\ \left(\text{iii}\right)\text{Rakesh scores 18 marks by attempting 16 questions.}\\ \text{How many questions has he attempted\hspace{0.33em}correctly and how}\\ \text{many has he attempted incorrectly?}\end{array}$

Ans.

\begin{array}{l}\text{Marks obtained for 1 right answer}=+3\\ \text{Marks obtained for 1 wrong answer}=-2\\ (i)\\ \text{Marks obtained by Radhika=20}\\ \text{Marks obtained for 12 correct answers=12}\times \text{3=36}\\ \text{Marks obtained for incorrect answer=Total score-Marks}\\ \text{obtained for 12 correct answers}\\ \text{=20-36=–16}\\ \text{Marks obtained for 1 wrong answer=-2}\\ \text{Thus, number of incorrect answer=}\left(-16\right)÷\left(-2\right)=\boxed{8}\\ \text{Thus, Radhika attempted 8 questions wrongly}\text{.}\\ \text{(ii)}\\ \text{Marks scored by Mohini=-5}\\ \text{Marks obtained for incorrect answers}\\ \text{=Total answer-Marks obtained for 12 incorrect answer}\\ =-5-21=-26\\ \text{Marks obtained for 1 wrong answer=-2}\\ \text{Thus, number of incorrect answer=}\left(-26\right)÷\left(-2\right)=\boxed{13}\\ \text{Therefore, Mohini attempted 13 questions wrongly}\text{.}\\ \text{(iii)}\\ \text{Total marks scored by Rakesh =18}\\ \text{Number of questions attempted =16}\\ \left(\text{Number of correct answers}\right)\left(3\right)\\ +\left(\text{Number of incorrect answers}\right)\left(-2\right)=18\\ \Rightarrow \left(\text{Number of correct answers}\right)\left(5\right)+\left(-32\right)=18\\ \Rightarrow \left(\text{Number of correct answers}\right)=\frac{50}{5}=10\\ So,\left(\text{Number of incorrect answers}\right)=16-10=6\\ \text{Thus, Total number of correct and incorrect answers scored}\\ \text{by Rakesh is 10 and 6 respectively}\text{.}\end{array}

Q.33

$\begin{array}{l}\text{An elevator descends into a mine shaft at the rate}\\ \text{of 6 m/min. If the descent startsfrom 10 m above}\\ \text{the ground level,how long will it take to reach – 350\hspace{0.33em}m.}\end{array}$

Ans.

\begin{array}{l}\text{Distance descended is denoted by a negative integer}\text{.}\\ \text{Initial height}=+10\text{m}\\ \text{Final depth}=-350\text{m}\\ \text{Total distance to be descended by the elevator}=\left(-350\right)-\left(+10\right)=-360\text{m}\\ \text{Time}\text{taken by the elevator to be descend}-\text{6 m}=\text{1 min}\\ \text{Thus, time taken by the elevator to descend}-\text{360 m =}\left(-360\right)÷\left(-6\right)=\boxed{60\text{minutes}=1\text{hour}}\end{array}