NCERT Solutions for Class 7 Maths Chapter 1 Integers (EX 1.4) Exercise 1.4

Q.1

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

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.2

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

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.3

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

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.4

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

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.5

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

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.6

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

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.7

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

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}