Questions are taken verbatim from the NCERT textbook; answers were grounded against the chapter's content during generation. Items needing review are marked.
Your Progress - Chapter 10% complete
1Exercise 1.11 questions
Q.1-41. Write down a pair of integers whose: (a) sum is -7 (b) difference is -10 (c) sum is 0
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.
3. In a quiz, team A scored -40, 10, 0 and team B scored 10, 0, -40 in three successive rounds. Which team scored more? Can we say that we can add integers in any order?
4. Fill in the blanks to make the following statements true: (i) $(-5)+(-8)=(-8)+(\ldots)$ (ii) $-53+\ldots=-53$ (iii) $17+\ldots=0$ (iv) $[13+(-12)]+(\ldots)=13+[(-12)+(-7)]$ (v) $(-4)+[15+(-3)]=[-4+15]+\ldots$v
SolutionFor Q3, Team A total $=-40+10+0=-30$ and Team B total $=10+0-40=-30$. Q4 uses commutative, identity, inverse and associative properties of addition of integers.
Answer:1. One possible set of answers is: (a) $-10, 3$ (b) $-6, 4$ because $-6-4=-10$ (c) $-3, 3$.
2. One possible set is: (a) $-2, -10$ because $-2-(-10)=8$ (b) $-6, 1$ (c) $-1, 2$ because $-1-2=-3$.
3. Both teams scored the same total, $-30$. Yes, integers can be added in any order.
4. (i) $-5$ (ii) $0$ (iii) $-17$ (iv) $-7$ (v) $-3$.
2Exercise 1.21 questions
Q.1-41. Find each of the following products: (a) $3\times(-1)$ (b) $(-1)\times225$ (c) $(-21)\times(-30)$ (d) $(-316)\times(-1)$ (e) $(-15)\times0\times(-18)$ (f) $(-12)\times(-11)\times10$ (g) $9\times(-3)\times(-6)$ (h) $(-18)\times(-5)\times(-4)$ (i) $(-1)\times(-2)\times(-3)\times4$ (j) $(-3)\times(-6)\times(-2)\times(-1)$
2. Verify the following: (a) $18\times[7+(-3)]=[18\times7]+[18\times(-3)]$ (b) $(-21)\times[(-4)+(-6)]=[(-21)\times(-4)]+[(-21)\times(-6)]$
3. (i) For any integer $a$, what is $(-1)\times a$ equal to? (ii) Determine the integer whose product with $(-1)$ is (a) $-22$ (b) $37$ (c) $0$
4. Starting from $(-1)\times5$, write various products showing some pattern to show $(-1)\times(-1)=1$.v
SolutionUse integer sign rules: one negative factor gives a negative product, two negative factors give a positive product, and any product with 0 is 0.
Answer:1. (a) $-3$ (b) $-225$ (c) $630$ (d) $316$ (e) $0$ (f) $1320$ (g) $162$ (h) $-360$ (i) $-24$ (j) $36$.
2. (a) LHS $=18\times4=72$ and RHS $=126-54=72$. (b) LHS $=(-21)\times(-10)=210$ and RHS $=84+126=210$.
3. (i) $(-1)\times a=-a$ (ii) (a) $22$ (b) $-37$ (c) $0$.
4. A pattern is $-1\times5=-5$, $-1\times4=-4$, $-1\times3=-3$, $-1\times2=-2$, $-1\times1=-1$, $-1\times0=0$, so continuing the pattern gives $-1\times(-1)=1$.
3Exercise 1.31 questions
Q.1-71. Evaluate each of the following: (a) $(-30)\div10$ (b) $50\div(-5)$ (c) $(-36)\div(-9)$ (d) $(-49)\div49$ (e) $13\div[(-2)+1]$ (f) $0\div(-12)$ (g) $(-31)\div[(-30)+(-1)]$ (h) $[(-36)\div12]\div3$ (i) $[(-6)+5]\div[(-2)+1]$
2. Verify that $a\div(b+c)\ne(a\div b)+(a\div c)$ for each of the following values of $a$, $b$ and $c$: (a) $a=12,b=-4,c=2$ (b) $a=-10,b=1,c=1$
3. Fill in the blanks: (a) $369\div\ldots=369$ (b) $(-75)\div\ldots=-1$ (c) $(-206)\div\ldots=1$ (d) $-87\div\ldots=87$ (e) $\ldots\div1=-87$ (f) $\ldots\div48=-1$ (g) $20\div\ldots=-2$ (h) $\ldots\div4=-3$
4. Write five pairs of integers $(a,b)$ such that $a\div b=-3$. One such pair is $(6,-2)$ because $6\div(-2)=-3$.
5. The temperature at 12 noon was $10^\circ C$ above zero. If it decreases at the rate of $2^\circ C$ per hour until midnight, at what time would the temperature be $8^\circ C$ below zero? What would be the temperature at mid-night?
6. In a class test $(+3)$ marks are given for every correct answer and $(-2)$ 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?
7. An elevator descends into a mine shaft at the rate of 6 m/min. If the descent starts from 10 m above the ground level, how long will it take to reach -350 m.v
SolutionQ5: drop needed from $10^\circ C$ to $-8^\circ C$ is $18^\circ C$, at $2^\circ C$ per hour, so time $=9$ hours after noon. By midnight the drop is $24^\circ C$, so temperature is $10-24=-14^\circ C$. Q7: total descent $=10-(-350)=360$ m; time $=360/6=60$ min.
Answer:1. (a) $-3$ (b) $-10$ (c) $4$ (d) $-1$ (e) $-13$ (f) $0$ (g) $1$ (h) $-1$ (i) $1$.
2. (a) $12\div(-2)=-6$, while $12\div(-4)+12\div2=-3+6=3$, so they are not equal. (b) $-10\div2=-5$, while $-10\div1+(-10)\div1=-20$, so they are not equal.
3. (a) $1$ (b) $75$ (c) $-206$ (d) $-1$ (e) $-87$ (f) $-48$ (g) $-10$ (h) $-12$.
4. Examples: $(-6,2)$, $(-12,4)$, $(12,-4)$, $(9,-3)$, $(-9,3)$.
5. The temperature is $-8^\circ C$ at 9 p.m.; at midnight it is $-14^\circ C$.
6. (i) 8 incorrect answers (ii) 13 incorrect answers.
7. It will take 1 hour.