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NCERT Solutions for Class 7 Maths Chapter 1 Integers All Exercise
Here are the NCERT solutions for Class 7 Maths Chapter 1 Integers, designed according to the CBSE syllabus. Our experts at Infinity Learn have carefully created these NCERT solutions to provide clear and detailed explanations for students aiming to do well in their math exams. We recommend students to explore these class 7 NCERT solutions to improve their understanding and performance in mathematics.
Class 7 Maths Chapter 1 Integers includes four detailed exercises, which are exercise 1.1, exercise 1.2, exercise 1.3, and exercise 1.4 and the solutions provided here cover all the questions from these exercises. These Class 7 maths solutions are aimed at helping students understand the concept of integers and their operations, which are essential for the Class 7 Maths curriculum. We also have class 7 Maths chapter 1 Integers exemplar solutions, students can build a solid foundation in mathematics and prepare effectively for their exams.
Let’s take a closer look at some of the concepts covered in this chapter.
- Integers are introduced.
- Integer Addition and Subtraction Properties
- Integer Multiplication is a technique for multiplying integers.
- Positive and Negative Integer Multiplication
- Two Negative Integers are multiplied. Integer Multiplication’s Characteristics
- Integers are divided into two groups.
- Division of Integers Properties
Also Check: CBSE Class 7 Maths Syllabus
NCERT Solutions for Class 7 Maths Chapter 1 Integers PDF Download
Below is the NCERT solutions for class 7 maths chapter 1 in pdf format to download which cover all 4 exercises.
NCERT Solutions for Class 7 Maths Chapter 1 Integers
- NCERT Solutions for Class 7 Maths Ex 1.1
- NCERT Solutions for Class 7 Maths Ex 1.2
- NCERT Solutions for Class 7 Maths Ex 1.3
- NCERT Solutions for Class 7 Maths Ex 1.4
Access NCERT Solutions for Class 7 Maths Chapter 1 All Exercise
Below is the NCERT solutions for class 7 Maths chapter 1 Integers exercise 1.1, exercise 1.2, exercise 1.3, and exercise 1.4.
NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.1
These are easy solved Class 7 Maths chapter 1 exercise 1.1 solutions
Q 1: Following number line shows the temperature in degree Celsius )(ºC)at different places on a particular day:
- Observe this number line and write the temperature of the places marked on it.
- What is the temperature difference between the hottest and the coldest places among the above?
- What is the temperature difference between Lazulite and Srinagar?
- 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
Sol. (a)
Places | Temperature |
---|---|
Bangalore | +14°C |
Ooty | -2°C |
Shimla | -8°C |
Srinagar | +5°C |
Lahulspiti | +22°C |
(b) Temperature difference between the hottest and the coldest places
= Temperature of Bangalore − Temperature of Lazulite
= 22º C (−8º C) = 22º C+8 C = 30º C
(c) Temperature between Lazulite and Srinagar
= Temperature of Srinagar − Temperature of Lazulite
= −2º C− (−8º C) = −2º C+8º C = 6º C
(d) Temperature of Srinagar and Shimla together
Temperature of Srinagar + Temperature of Shimla = −2°C + 5°C = 3°C
Temperature at Shimla = 5º C
Temperature of Srinagar = -2º C
Therefore, we can say that the temperature of Srinagar and Shimla together is less than the temperature at Shimla but not less than the temperature at Srinagar.
Q 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, what was his total at the end?
Sol. Total at the end = 25 − 5 − 10 + 15 + 10 = 35
Q 3: At Srinagar, the temperature was −5°C on Monday and then it dropped by 2°C on Tuesday. What was the temperature of Srinagar on Tuesday? On Wednesday, it rose by 4°C. What was the temperature on this day?
Sol. Temperature at Srinagar on Monday = −5°C. Drop in temperature on Tuesday = 2°C. Temperature at Srinagar on Tuesday = −5°C − 2°C = −7°C. Rise in temperature on Wednesday = 4°C. Temperature at Srinagar on Wednesday = −7°C + 4°C = −3°C.
Q 4: A plane is flying at the height of 5000m above sea level. At a particular point, it is exactly above a submarine floating 1200m below sea level. What is the vertical distance between them?
Sol. Vertical distance between the plane and submarine = 5000m + 1200m = 6200m
Q 5: Mohan deposits ₹2000 in his bank account and withdraws ₹1642 from it the next day. If the 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.
Sol. Amount deposited = +₹2000. Balance in Mohan’s account after the withdrawal = ₹2000 − ₹1642 = ₹358.
Q 6: Rita goes 20km towards east from a point A to the point B. From B, she moves 30km towards west along the same road. If the distance towards the east is represented by positive integers, then how will you represent the distance travelled towards west? By which integer will you represent her final position from A?
Sol. The distance towards the west = −30km. Her final position from A = 20km − 30km = −10km.
Q:7. In a magic square each row, column and diagonal have the same sum. Check which of the following is a magic square.
Sol.
(i)
Taking rows
5+(−1)+(−4) = 5−5 = 0
(−5)+(−2)+7 = −7+7 = 0
0+3+(−3) = 3−3= 0
Taking columns
5+(−5)+0 = 5−5 = 0
(−1)+(−2)+3= −3+3= 0
(−4)+7+(−3) = 7−7 = 0
Taking diagonals
5+(−2)+(−3) = 5−5 = 0
(−4)+(−2)+0 = −6
This box is not a magic square because all the sums are not equal.
(ii)
Taking rows
1+(−10)+0 =1−10 = −9
(−4)+(−3)+(−2) = −7−2 = −9
(−6)+4+(−7) = −2−7 = −9
Taking columns
1+(−4)+(−6) =1−10 = −9
(−10)+(−3)+4 = −13+4 = −9
0+(−2)+(−7) = 0−9 = −9
Taking diagonals
1+(−3)+(−7) =1−10 = −9
0+(−3)+(−6) = −9
This box is a magic square because all the sums are equal.
Q 8: Verify a−(−b) = a + b for the following values of a and b.
(i) a = 21,b =18
(ii) a =118,b =125
(iii) a = 75,b = 84
(iv)a = 28,b =11.
Sol.
(i) Given: a = 21,b =18
We have a −(−b) = a +b
Putting the values in L.H.S. = a −(−b) = 21−(−18) = 21+18 = 39
Putting the values in R.H.S. = a + b = 21+19 = 39
Since, L.H.S. =R.H.S Hence, verified.
(ii) Given: a =118,b =125
We have a −(−b) = a +b
Putting the values in L.H.S. = a −(−b) =118−(−125) =118+125 = 243
Putting the values in R.H.S. = a + b =118 +125 = 243
Since, L.H.S. =R.H. S Hence, verified.
(iii) Given: a = 75, b = 84
We have a – (-b) = a + b
Putting the values in L.H.S. = a −(−b) = 75−(−84) = 75+84 =159
Putting the values in R.H.S. = a + b = 75 +84 =159
Since, L.H.S. =R.H. S Hence, verified.
(iv)
Given: a = 28, b = 11
We have a – (-b ) = a+b
Putting the values in L.H.S. = a −(−b) = 28−(−11) = 28+11= 39
Putting the values in R.H.S. = a + b = 28 +11 = 39
Since, L.H.S.=R.H. S Hence, verified.
Q 9: Use the sign of >, < or = in the box to make the statements true.
(a) (−8)+(−4) ………. (−8)−(−4)
(b) (−3)+7−(19) ………..15−8+(−9)
(c) 23− 41+11……….. 23− 41−11
(d) 39+(−24)−(15) ………. 36+(−52)−(−36)
(e) (−231)+79+51………. (−399)+159+81
Sol.
(a) (-8) + (-4) [ ] (-8) – (-4)
Solution:-
Let us take Left Hand Side (LHS) = (-8) + (-4)
= -8 – 4
= -12
Now, Right Hand Side (RHS) = (-8) – (-4)
= -8 + 4
= -4
By comparing LHS and RHS,
LHS < RHS
-12 < -4
∴ (-8) + (-4) [<] (-8) – (-4)
(b) (-3) + 7 – (19) [ ] 15 – 8 + (-9)
Solution:-
Let us take Left Hand Side (LHS) = (-3) + 7 – 19
= -3 + 7 – 19
= -22 + 7
= -15
Now, Right Hand Side (RHS) = 15 – 8 + (-9)
= 15 – 8 – 9
= 15 – 17
= -2
By comparing LHS and RHS,
LHS < RHS
-15 < -2
∴ (-3) + 7 – (19) [<] 15 – 8 + (-9)
(c) 23 – 41 + 11 [ ] 23 – 41 – 11
Solution:-
Let us take Left Hand Side (LHS) = 23 – 41 + 11
= 34 – 41
= – 7
Now, Right Hand Side (RHS) = 23 – 41 – 11
= 23 – 52
= – 29
By comparing LHS and RHS,
LHS > RHS
– 7 > -29
∴ 23 – 41 + 11 [>] 23 – 41 – 11
(d) 39 + (-24) – (15) [ ] 36 + (-52) – (- 36)
Solution:-
Let us take Left Hand Side (LHS) = 39 + (-24) – 15
= 39 – 24 – 15
= 39 – 39
= 0
Now, Right Hand Side (RHS) = 36 + (-52) – (- 36)
= 36 – 52 + 36
= 72 – 52
= 20
By comparing LHS and RHS,
LHS < RHS
0 < 20
∴ 39 + (-24) – (15) [<] 36 + (-52) – (- 36)
(e) – 231 + 79 + 51 [ ] -399 + 159 + 81
Solution:-
Let us take Left Hand Side (LHS) = – 231 + 79 + 51
= – 231 + 130
= -101
Now, Right Hand Side (RHS) = – 399 + 159 + 81
= – 399 + 240
= – 159
By comparing LHS and RHS,
LHS > RHS
-101 > -159
∴ – 231 + 79 + 51 [>] -399 + 159 + 81
Q 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 with every move. In how many jumps will he reach back to 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 parts (i) and (ii) by completing the following: (a) – 3 + 2 – … = – 8 (b) 4 – 2 + … = 8. In (a), the sum (– 8) represents going down by eight steps. So, what will the sum 8 in (b) represent?
Sol:
(i) He jumps 3 steps down and jumps back 2 steps up. Following number ray shows the
jumps of monkey:
First jump =1+3 = 4steps
Second jump = 4−2 = 2steps
Third jump= 2+3 = 5steps
Fourth jump = 5−2 = 3steps
Fifth jump = 3+3 = 6steps
Sixth jump = 6−2 = 4steps
Seventh jump= 4+3 = 7steps
Eight jump = 7 −2 = 5steps
Ninth jump= 5+3 = 8steps
Tenth jump = 8−2 = 6steps
Eleventh jump = 6+3 = 9steps
He will reach ninth steps in 11 jumps.
(ii) He jumps four steps and then jumps down 2 steps. Following number ray shows the
jumps of monkey:
Thus, monkey reach bank on the first step in fifth jump.
(iii)
(a) −3+ 2 − 3+ 2 − 3+ 2 − 3+ 2 − 3+ 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.
- Chapter 2 Fractions and Decimals
- Chapter 3 Data Handling
- Chapter 4 Simple Equations
- Chapter 5 Lines and Angles
- Chapter 6 The Triangles and Its Properties
- Chapter 7 Congruence of Triangles
- Chapter 8 Comparing Quantities
- Chapter 9 Rational Numbers
- Chapter 10 Practical Geometry
- Chapter 11 Perimeter and Area
- Chapter 12 Algebraic Expressions
- Chapter 13 Exponents and Powers
- Chapter 14 Symmetry
- Chapter 15 Visualizing Solid Shapes
NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.2
These are easy solved Class 7 Maths chapter 1 exercise 1.2 solutions
Q1: Write down a pair of integers whose:
(a) sum is -7
(b) difference is 10
(c) sum is 0.
Sol.
(a) (-15 ) and 8
(b) 15 and 25.
(c) ( 49)− and 49
Q2:
(a) Write a pair of negative integers and a positive integer whose sum is −5
(b) Write a negative integer and a positive integer whose difference is -3.
Sol.
(a) (−10) and (-18)
(b) (−10) and 5
Q 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?
Sol.
Total score of team A
= −40+10+0 = (−40)+10
= (−40−10) = −30
Total score of Team B
=10+0+(−40) =10+(−40)
= −(40−10) = −30
So, both the teams scored equally. Yes! We can say that can add integers in any order.
Yes, we can add integers in any order due to commutative property.
Q 4: Fill in the blanks to make the following statements true:
(i) (−5)+(−8) = (−8)+(…..)
(ii) −53+….. = −53
(iii)17 +….. = 0(iv)[ 13+(12)] +(…..) =13+ [(12)+(−7)]
(v) (−4)+ [15+(−3)] = [−4+15]+…… .
Sol.
(i) (−5)+(−8) = (−8)+(−5) [Commutative property]
(ii) -53 + 0 = -53 [Zero additive property]
(iii)17+(−17) = 0 [Additive identity]
(iv) [13+(12)] +(−7) =13+ [(12)+(−7)] [Associative property]
(v) (−4) + [15+(−3)] = [−4+15]+(−3) [Associative property]
NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.3
These are easy solved Class 7 Maths chapter 1 exercise 1.3 solutions
1. Find each of the following products:
(a) 3 × (–1)
(b) (–1) × 225
(c) (–21) × (–30)
(d) (–316) × (–1)
(e) (–15) × 0 × (–18)
(f) (–12) × (–11) × (10)
(g) 9 × (–3) × (– 6)
(h) (–18) × (–5) × (– 4)
(i) (–1) × (–2) × (–3) × 4
(j) (–3) × (–6) × (–2) × (–1)
Sol.
(a) 3 × (–1)
= 3 × (-1)
= -3 … [∵ (+ × – = -)]
(b) (–1) × 225
= (-1) × 225
= -225 … [∵ (- × + = -)]
(c) (–21) × (–30)
= (-21) × (-30)
= 630 … [∵ (- × – = +)]
(d) (–316) × (–1)
= (-316) × (-1)
= 316 … [∵ (- × – = +)]
(e) (–15) × 0 × (–18)
= (–15) × 0 × (–18)
= 0
(f) (–12) × (–11) × (10)
= 132 × 10 … [∵ (- × – = +)]
= 1320
(g) 9 × (–3) × (– 6)
= 9 × 18 … [∵ (- × – = +)]
= 162
(h) (–18) × (–5) × (– 4)
= 90 × -4 … [∵ (- × – = +)]
= – 360 … [∵ (+ × – = -)]
(i) (–1) × (–2) × (–3) × 4
= 2 × (-12) … [∵ (- × – = +), (- × + = -)]
= – 24
(j) (–3) × (–6) × (–2) × (–1)
= 18 × 2 … [∵ (- × – = +)
= 36
Q 2: Verify the following:
(a) 18 × [7 + (–3)] = [18 × 7] + [18 × (–3)]
(b) (–21) × [(– 4) + (– 6)] = [(–21) × (– 4)] + [(–21) × (– 6)]
Sol.
a)
Left Hand Side (LHS) first = 18 × [7 + (–3)]
= 18 × [7 – 3]
= 18 × 4
= 72
Right Hand Side (RHS) = [18 × 7] + [18 × (–3)]
= [126] + [-54]
= 126 – 54
= 72
comparing LHS and RHS
72 = 72
LHS = RHS
b)
Left Hand Side (LHS) first = (–21) × [(– 4) + (– 6)]
= (-21) × [-4 – 6]
= (-21) × [-10]
= 210
Right Hand Side (RHS) = [(–21) × (– 4)] + [(–21) × (– 6)]
= [84] + [126]
= 210
comparing LHS and RHS
210 = 210
LHS = RHS
Q 3:
(i) For any integer a, what is (–1) × a equal to?
(ii). Determine the integer whose product with (–1) is
(a) –22
(b) 37
Sol.
i) (−1) × a = −a , where a is an integer.
ii)
(a) (−1) × 22 = 22
(b) (−1) × 37 = −37
(c) (−1) × 0 = 0
Hence (c) 0 is the required integer.
4. Starting from (–1) × 5, write various products showing some pattern to show
(–1) × (–1) = 1.
Sol.
= -1 × 5 = -5
= -1 × 4 = -4
= -1 × 3 = -3
= -1 × 2 = -2
= -1 × 1 = -1
= -1 × 0 = 0
= -1 × -1 = 1
Thus, we can conclude that this pattern shows the product of one negative integer and one positive integer is negative whereas the product of two negative integers is a positive integer.
5. Find the product using suitable properties:
(a) 26 × (– 48) + (– 48) × (–36)
(b) 8 × 53 × (–125)
(c) 15 × (–25) × (– 4) × (–10)
(d) (– 41) × 102
(e) 625 × (–35) + (– 625) × 65
(f) 7 × (50 – 2)
(g) (–17) × (–29)
(h) (–57) × (–19) + 57
Sol.
(a) 26x(-48)+(-48)x(-36)
⇒(-48)×[26+(-36)] =(-48)x(-10)
⇒ 480
(b) 8x53x(-125)
⇒ 53×[8x(-125)] ⇒ 53×(-1000) -53000
(c) 15x(-25)x(-4)x(-10)
⇒ 15x[(-25)x(-4)x(-10)] ⇒ 15×(-1000) =-15000
(d) (-41)x(102)
⇒-41x(100+2) [(-41)×100][(-41)x2] ⇒-4100+(-82) ⇒-4182
(e) 625x(-35)+(-625)×65
⇒ 625x[(-35)+(-65)]
⇒ 625×[-100] ⇒-62500
(f) 7×(50-2)
⇒7×50-7×2
⇒350-14-336
(g) (-17)x(-29)
⇒(-17)×[(30)+1]
⇒(-17)x(30)+(-17)x1
⇒ 510+(-17)
(h) (-57)x(-19)+57
⇒(-57)x(-19)+57×1
⇒57×19+57×1
⇒57×[19+1]
⇒57×20=1140
Q6: A certain freezing process requires that room temperature be lowered from 40°C at the rate of 5°C every hour. What will be the room temperature 10 hours after the process begins?
Sol.
Present room temperature = 40^{o}C
Decreasing the temperature every hour = 5^{o}C
Room temperature after 10 hours = 40ºC +10 x (- 5^{o}C)
= 40^{o}C + (-50^{o}C)
= -10^{o}C
Q 7: In a class test containing 10 questions, 5 marks are awarded for every correct answer and (–2) 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?
Sol.
(i) Mohan gets marks for four correct questions = 4×5 = 20
He gets marks for six incorrect questions=6x(-2)=-12
Therefore, total scores of Mohan = (4×5)+[6x(-2)]
=20-12=8
Thus, Mohan gets 8 marks in a class test.
(ii) Reshma gets marks for five correct questions= 5×5 = 25
She gets marks for five incorrect questions=5x(-2)=-10
Therefore, total score of Resham = 25+(-10)=15
Thus, Reshma gets 15 marks in a class test.
(iii) Heena gets marks for two correct questions=2×5=10
She gets marks for five incorrect questions=5x(-2)=-10
Therefore, total score of Resham=10+(-10)=0
Thus, Reshma gets O marks in a class test.
Q 8: A cement company earns a profit of ₹ 8 per bag of white cement sold and a loss of
₹ 5 per bag of grey cement sold.
(a) The company sells 3,000 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?
Given: Profit of 1 bag of white cement = `8
And Loss of 1 bag of grey cement=`5
(a) Profit on selling 3000 bags of white cement=3000×8=`24,000Loss of selling
5000 bags of grey cement=5000×5=25,000
Since Profit<Loss
Therefore, his total loss on selling the grey cement bags = Loss – Profit
=25,000-24,000 = `1000
Thus, he has lost of `1,000 on selling the grey cement bags.
(b) Let the number of bags of white cement be x.
According to question, Loss-Profit
5×6,400=X × 8
5×6400
X = 5 × 6400 / 8
= 4000bags
Thus, he must sell 4000 white cement bags to have neither profit nor loss.
Q 9: Replace the blank with an integer to make it a true statement.
(a) (–3) × _____ = 27
(b) 5 × _____ = –35
(c) _____ × (– 8) = –56
(d) _____ × (–12) = 132
Sol.
(a) (-3)x(-9)=27
(c) 7×(-8)=-56
(b) 5x……=-35
(d)(-12)=132
(b) 5x(-7)=-35
(d) (-11)x(-12)=132
NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.4
These are easy solved Class 7 Maths chapter 1 exercise 1.4 solutions
1. Evaluate each of the following.
(a) (–30) ÷ 10
(b) 50 ÷ (–5)
(c) (–36) ÷ (–9)
(d) (– 49) ÷ (49)
(e) 13 ÷ [(–2) + 1]
(f) 0 ÷ (–12)
(g) (–31) ÷ [(–30) + (–1)]
(h) [(–36) ÷ 12] ÷ 3
(i) [(– 6) + 5)] ÷ [(–2) + 1]
Sol.
(a) (–30) ÷ 10
= -30 / 10
= -3
(b) 50 ÷ (–5)
= 50 / -5
= -10
(c) (–36) ÷ (–9)
= -36 / -9
= 4
(d) (–49) ÷ (49)
= -49 / 49
= -1
(e) 13 ÷ [(–2) + 1]
= 13 / (-2 + 1)
= 13 / -1
= -13
(f) 0 ÷ (–12)
= 0 / -12
= 0
(g) (–31) ÷ [(–30) + (–1)]
= -31 / (-30 – 1)
= -31 / -31
= 1
(h) [(–36) ÷ 12] ÷ 3
= (-36 / 12) / 3
= -3 / 3
= -1
(i) [(–6) + 5)] ÷ [(–2) + 1]
= (-6 + 5) / (-2 + 1)
= -1 / -1
= 1
2. Verify that a ÷ (b + c) ≠ (a ÷ b) + (a ÷ 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
Sol.
(a)
Given:
a÷(b+c)+(a÷b)+(a÷c)
a=12, b=-4, c=2
Putting the given values in L.H.S. = 12÷(−4+2)
= 12 ÷ (-2) = 12 x (-1/2) = -12/2 = 6
Putting the given values in R.H. S=[12÷(−4)]+(12÷2)
=(12 x -1/4) + 6=- -3 + 6 = 3
Since,
L.H.S. ≠ R.H.S
Hence verified.
(b) Given:
a÷(b+c)+(a÷b)+(a÷c)
a=-10, b=1, c=1
Putting the given values in L.H.S. = −10÷(1+1)
=-10÷(2)=-5
Putting the given values in R.H. S=[−10÷1]+[−10÷1]
=-10-10-20
Since,
L.H.S. ≠ R.H.S
Hence verified.
3. Fill in the blanks:
(a) 369 ÷ _____ = 369
(b) (–75) ÷ _____ = –1
(c) (–206) ÷ _____ = 1
(d) – 87 ÷ _____ = 87
(e) _____ ÷ 1 = – 87
(f) _____ ÷ 48 = –1
(g) 20 ÷ _____ = –2
(h) _____ ÷ (4) = –3
Sol.
(a) 369 ÷ 1 = 369
(b) (-75) ÷ 75 = -1
(c) (-206) ÷ (-206) = 1
(d) -87 ÷ (-1) = 87
(e)-87 ÷ 1 = -87
(f) (-48) ÷ 48 = -1
(g) 20 ÷ (-10) = -2
(h) (-12) ÷ (4) = -3
Q 4: Write five pairs of integers (a, b) such that a ÷ b = -3. One such pair is (6, -2) because 6 + (-2) = -3.
Solution:
(a) (24, -8) because 24 ÷ (-8) = -3
(b) (-12, 4) because (-12) ÷ 4 = -3
(c) (15, -5) because 15 ÷ (-5) = -3
(d) (18, -6) because 18 ÷ (-6) = -3
(e) (60, -20) because 60 ÷ (-20) = -3
Q 5: The temperature at 12 noon was 10°C above zero. If it decreases at the rate of 2°C per hour until midnight, at what time would the temperature be 8°C below zero? What would be the temperature at midnight?
The temperature at 12 noon was 10°C above zero. If it decreases at the rate of 2°C per hour until midnight, at what time would the temperature be 8°C below zero? What would be the temperature at midnight?
Solution:
Temperature at 12 noon was 10°C above zero i.e. +10°C
Rate of decrease in temperature per hour = 2°C
Number of hours from 12 noon to midnight = 12
∴ Change in temperature in 12 hours
= 12 × (-2°C) = -24°C
∴ Temperature at midnight
= +10°C + (-24°C) = -14°C
Hence, the required temperature at midnight =-14°C
Difference in temperature between + 10°C and -8°C
= +10°C – (-8°C) = +10°C + 8°C = 18°C
Number of hours required = 18∘C /2∘C = 9 hours
∴ Time after 9 hours from 12 noon = 9 pm.
Q 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?
Sol.
Given that:
+3 marks are given for each correct answer. (-2) marks are given for each incorrect answer. Zero marks for not attempted questions.
(i) Marks obtained by Radhika for 12 correct answers = (+3) × 12 = 36
Marks obtained by Radhika for correct answers = 12 × 3 = 36
Total marks obtained by Radhika = 20
∴ Marks obtained by Radhika for incorrect answers = 20 – 36 = -16
Number of incorrect answers =(−16)÷(−2)=(−16)(−2)=8
Hence, the required number of incorrect answers = 8
(ii) Marks scored by Mohini = -5
Number of correct answers = 7
∴ Marks obtained by Mohini for 7 correct answers = 7 × (+3) – 21
Marks obtained for incorrect answers = -5 – 21 = (-26)
∴ Number of incorrect answers = (-26) ÷ (-2) = 13
Hence, the required number of incorrect answers – 13.
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?
Starting position of mine shaft is 10 m above the ground but it moves in opposite direction so it travels the distance (−350)m below the ground
So total distance covered by mine shaft =10m−(−350)m+350 = 360m
Now, time taken to cover a distance of 6 m by it =1 min
So, time taken to cover a distance of 1 m by it = 1 / 6 min
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Class 7 Maths Chapter 1 Integers Overview
In what ways may we use integers in our everyday lives?
Aside from math class, integers are used in a variety of ways in our everyday lives. In all domains, we can utilize integers to calculate the efficiency of positive and negative numbers.
We come encounter integers in a variety of scenarios throughout our daily lives:
Profit, income, increase, rise, high, north, east, above depositing, ascending, and so forth are all determined using positive numbers.
Negative integers are employed to calculate numbers such as loss, expense, reduction, fall, low, south, west, below, withdrawing, sliding, and so on.
For example:
Position A is located on a mountain 4680 meters above sea level, while point B is located in a mine 765 meters below sea level. What is the length of the distance between points A and B?
In this case, we’ll use a point O at sea level as an illustration.
Then,
OA = + 4680m and OB = – 765m.
Distance between A and B = OA + OB
= +4680 + 765 m
= (4680 + 765) m
= 5445 m
Integers’ Characteristics
Numbers for addition and multiplication through patterns are among the features of integers. They also take into account the complete numbers. Integers are used to express general communicative and associative features.
Facts
Natural numbers are the counting numbers 1, 2, 3, 4, 5, and so on, whereas whole numbers are the set of natural numbers plus zero, such as 0, 1, 2, 3, 4, 5, and so on.
On a number line, negative integers are represented by points to the left of zero, and positive integers are represented by points to the right of zero.
For integers to the left of zero and positive integers to the right of zero, the integer 0 serves as an additive identity.
0 is neither a positive nor a negative number.
The numerical value of an integer, independent of its sign, is its absolute value. | a | denotes the absolute value of an integer a.
Number Line
Natural numbers, whole numbers, positive integers, and negative integers are all represented on a number line. To determine numerical operations, the identities are marked at equal intervals on a line. Number lines are significant because they reflect numbers that we encounter on a regular basis.
Steps of Drawing a Number Line
- Draw any length of the straight line.
- To divide the drawn line into the needed number, mark points at equal intervals on it.
- Any of the locations noted on the line in step 2 should be marked as 0.
- Mark the positive integers + 1, + 2, + 3, and so on on the right-hand side of the line, starting at 0. Similarly, starting at 0, mark the negative integers -1, -2, -3, and so on on the left side.
- The numbers continue up to infinity, as indicated by the arrowheads on both sides of the drawn line.
Other Resources for Class 7 | |
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FAQs on NCERT Solutions for Class 7 Maths Chapter 1 Integers
Where can I find an exact NCERT Solution for Class 7 Maths Chapter 1 solution?
At Infinity learn, you can find the correct NCERT Solution for Class 7 Maths Chapter 1 in PDF format. Infinity learns mathematics experts meticulously created the NCERT Textbook Solutions for the chapter. All of these solutions are based on the new CBSE pattern, ensuring that students have a complete understanding of the material for their exams.
Is it necessary to complete all of the problems in Chapter 1 of the NCERT Solution for Class 7 Maths?
Yes. Because these questions are crucial in terms of the exam. Experts have answered these questions to assist students in easily completing the task. These answers assist students in becoming more familiar with integers. On Infinity learn'S website, solutions are provided in PDF format.
In 7th grade, what are integers?
The first chapter of the NCERT Class 7 Maths textbook is Integers. Integers are a collection of natural numbers such as -4, -3, -2, -1, 0, 1, 2, 3, 4, and so on. They have a wide range of real-world applications. In daily calculations and transactions, both positive and negative integers play a significant role. To do well in Math in higher courses, it is necessary to learn basic topics such as integers.
What is the name of Chapter 1 of Maths Class 7?
Chapter 1 of Maths Class 7 in the NCERT syllabus is called Integers. It introduces students to the concept of integers and their properties.
Is 0 an integer?
Yes, 0 is an integer. In mathematics, integers are a set of numbers that include all positive whole numbers, their negative counterparts, and zero.