Class 7 Maths Chapter 1
Class 7 Maths Chapter 1 Exercise 1.1 Solutions, Here are the NCERT Solutions for Class 7 Maths Exercise 1.1 Chapter 1 Integers in Simple PDF. This exercise from NCERT Solutions for Class 7 Maths Chapter 1 covers topics related to Introduction to Integers.
We know that integers make up a large collection of numbers, including whole numbers and negative numbers. Our expert teachers have designed these solutions in a precise, comprehensive, and detailed manner.
Class 7 Maths Chapter 1 Exercise 1.1 Solutions, Know more about these topics by solving the NCERT Solutions for Class 7 Maths Chapter 1 integer questions with the help of solutions provided here.
Class 7 Maths Chapter 1 Exercise 1.1 Solutions
Class 7 Maths Chapter 1 Question 1.
The following number line shows the temperature in degree celsius (co) at different places on a particular day.
(a) Observe this number line and write the temperature of the places marked on it.
Solution:-
According to Questions, we can find the temperature of the cities as,
The temperature in Lahulspiti is -8oC
The temperature in Srinagar is -2oC
The temperature in Shimla is 5oC
The temperature in Ooty is 14oC
The temperature in Bangalore is 22oC
(b) What is the temperature difference between the hottest and the coldest places among the above?
Solution:-
According to Questions, we see that,
The temperature of the hottest place i.e. Bangalore is 22oC
The temperature of the coldest place i.e. Lahulspiti is -8oC
The temperature difference between the hottest and the coldest place = 22oC – (-8oC)
= 22oC + 8oC
= 30oC
Therefore, the temperature difference between the hottest and the coldest place is 30oC.
(c) What is the temperature difference between Lahulspriti and Srinagar?
Solution:-
According to Questions, we see that,
The temperature of Lahulspiti is -8oC
The temperature in Srinagar is -2oC
Temperature difference between Lahulspiti and Srinagar = -2oC – (-8oC)
= – 2oC + 8oC
= 6oC
(d) Can we say the temperature of Srinagar and Shimla have taken together is less than the temperature at Shimla? Is it also less than the temperature at Srinagar?
Solution:-
According to Questions, we see that,
Temperature of Srinagar = -2°C
Temperature of Shimla = 5°C
∴ Temperature of the above cities taken together
= -2°C + 5°C = 3°C
Temperature of Shimla = 5°C
Hence, the temperature of Srinagar and Shimla have taken together is less than that of Shimla by 2°C.
i.e., (5°C – 3°C) = 2°C
Again,
3o > -2o
No, the temperature of Srinagar and Shimla have taken together is not less than the temperature of Srinagar.
Class 7 Maths Chapter 1 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, what was his total at the end?
Solution:-
According to Questions, we see that,
Jack’s scores in five consecutive rounds are 25, -5, -10, 15, and 10.
Finally Jack’s total score will be = 25 + (-5) + (-10) + 15 + 10
= 25 – 5 – 10 + 15 + 10
= 50 – 15
= 35
In the end, Jack’s overall score is 35.
Class 7 Maths Chapter 1 Question3.
At Srinagar 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?
Solution:
According to Questions, we see that,
The temperature on Monday at Srinagar= -5°C
Temperature on Tuesday at Srinagar is dropped by 2°C = – 2°C
Temperature on Tuesday = -5°C – 2°C = -7°C
On Wednesday, the temperature increased by 4°C (degrees Celsius).
Temperature on Wednesday
= -7°C + 4°C = -3°C
Hence, the required temperature on Tuesday = -7°C
and temperature on Wednesday = -3°C
Class 7 Maths Chapter 1 Question 4.
A plane is flying at a height of 5000 m above sea level. At a particular point, it is exactly above a submarine flowing 1200 m below sea level. What is the vertical distance between them?
Solution:
According to Questions,
Height of the flying plane = 5000 m
Depth of the submarine = -1200 m
∴ Distance between them
= + 5000 m – (-1200 m)
= 5000 m + 1200 m = 6200 m
Hence, the vertical distance = 6200 m
Class 7 Maths Chapter 1 Question 5.
Mohan deposits Rs 2,000 in his bank account and withdraws Rs 1,642 from it, the next day. If the withdrawal of the 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:-
According to the Questions,
Total amount deposited by Mohan in the bank account = Rs 2000
The total amount was withdrawn by Mohan from the bank account = – Rs1642
Balance after withdrawal in Mohan’s account = Amount deposited + Amount withdrawn
= Rs 2000 + (- Rs 1642)
= Rs2000 – Rs1642
= Rs 358
Hence, the balance in Mohan’s account after withdrawal is Rs 358.
Class 7 Maths Chapter 1 Question 6.
Rita goes 20 km towards the east from point A to point B. From B, she moves 30 km towards the west along the same road. If the distance towards the east is represented by a positive integer, then how will you represent her final position from A?
Solution:-
According to the Questions,
A positive integer indicates the distance to the east.
Then, the distance traveled west will be represented by a negative integer.
Distance covered by Rita in east direction = 20 km
Distance covered by Rita in west direction = – 30 km
Distance covered by A = 20 + (- 30)
= 20 – 30
= -10 km
Hence, the required position of Rita will be presented by a negative number, i.e., -10.
Class 7 Maths Chapter 1 Question 7.
In a magic square each row, column, and diagonal have the same sum. Check which of the following is a magic square.
Solution:-
First, we consider square (i)
Adding the numbers in each row, we get,
= 5 + (-1) + (-4) = 5 – 1 – 4 = 5 – 5 = 0
= -5 + (-2) + 7 = – 5 – 2 + 7 = -7 + 7 = 0
= 0 + 3 + (-3) = 3 – 3 = 0
Adding the numbers in each column, we get,
= 5 + (- 5) + 0 = 5 – 5 = 0
= (-1) + (-2) + 3 = -1 – 2 + 3 = -3 + 3 = 0
= -4 + 7 + (-3) = -4 + 7 – 3 = -7 + 7 = 0
Adding the numbers in the diagonals, we get,
= 5 + (-2) + (-3) = 5 – 2 – 3 = 5 – 5 = 0
= -4 + (-2) + 0 = – 4 – 2 = -6
Since the sum of a diagonal is not equal to zero,
Hence, (i) there is no magic square
Now, let us consider square (ii).
Adding the numbers in each row, we get,
= 1 + (-10) + 0 = 1 – 10 + 0 = -9
= (-4) + (-3) + (-2) = -4 – 3 – 2 = -9
= (-6) + 4 + (-7) = -6 + 4 – 7 = -13 + 4 = -9
Adding the numbers in each column, we get,
= 1 + (-4) + (-6) = 1 – 4 – 6 = 1 – 10 = -9
= (-10) + (-3) + 4 = -10 – 3 + 4 = -13 + 4= -9
= 0 + (-2) + (-7) = 0 – 2 – 7 = -9
Adding the numbers in the diagonals, we get,
= 1 + (-3) + (-7) = 1 – 3 – 7 = 1 – 10 = -9
= 0 + (-3) + (-6) = 0 – 3 – 6 = -9
This (ii) square is a magic square because the sum of each row, each column, and diagonal is equal to -9
Class 7 Maths Chapter 1 Question 8.
Verify a – (– b) = a + b for the following values of a and b.
(i) a = 21, b = 18
Solution:
a – (-b) = a + b
LHS = 21 – (-18) = 21 + 18 = 39
RHS = 21 + 18 = 39
LHS = RHS
39 = 39
LHS = RHS Hence, verified.
(ii) a = 118, b = 125
Solution:
a – (-b) = a + b
LHS = 118 – (-125) = 118 + 125 = 243
RHS = 118 + 125 = 243
LHS = RHS
243 = 243
LHS = RHS Hence, verified.
(iii) a = 75, b = 84
Solution:
a – (-b) = a + b
LHS = 75 – (-84) = 75 + 84 = 159
RHS = 75 + 84 = 159
LHS = RHS
159 = 159
LHS = RHS Hence, verified.
(iv) a= 28, b= 11
Solution:
a – (-b) = a + b
LHS = 28 – (-11) = 28 + 11 = 39
RHS = 28 + 11 = 28 + 11 = 39
LHS = RHS
39 = 39
LHS = RHS Hence, verified.
Class 7 Maths Chapter 1 Solution
Class 7 Maths Chapter 1 Question 9.
Use a sign of >, < or = in the box to make the statements true.
(a) (-8) + (-4) [ ] (-8) – (-4)
Solution:-
Let us take left hand side (LHS) = (-8) + (-4)
= -8 – 4
= -12
Now, Right Hand (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 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 consider Left Hand Side (LHS) = 23 – 41 + 11
= 34 – 41
= – 7
Now, Right Hand (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 consider Left Hand Side (LHS) = 39 + (-24) – 15
= 39 – 24 – 15
= 39 – 39
= 0
Now, on the 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 side (RHS) = – 399 + 159 + 81
= – 399 + 240
= – 159
By comparing LHS and RHS
LHS > RHS
-101> -159
– 231 + 79 + 51 [>] -399 + 159 + 81
Class 7 Maths Chapter 1 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?
Solution:-
Let us consider that the steps to move down are represented by positive integers and then the steps to move up are represented by negative integers.
Initially, the monkey is sitting on the top of the ladder i.e. the first step
In the first jump the monkey will be on the step = 1 + 3 = 4 steps
In the second jump the monkey will be on the step = 4 + (-2) = 4 – 2 = 2 steps
In the third jump the monkey will be on step = 2 + 3 = 5
In the fourth jump the monkey will be on the step = 5 + (-2) = 5 – 2 = 3 steps
In 5th jump monkey will be at step = 3 + 3 = 6 steps
In 6th jump monkey will be at step = 6 + (-2) = 6 – 2 = 4 steps
In 7th jump the monkey will be on step = 4 + 3 = 7 steps
In 8th jump the monkey will be on the step = 7 + (-2) = 7 – 2 = 5 steps
In 9th jump the monkey will be on step = 5 + 3 = 8 steps
In 10th jump the monkey will be on the step = 8 + (-2) = 8 – 2 = 6 steps
In the 11th jump, the monkey will be on step = 6 + 3 = 9 steps
The monkey took 11 jumps (i.e., the 9th step) to 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?
Solution:-
Let us consider that the steps to move down are represented by positive integers and then the steps to move up are represented by negative integers.
Initially, the monkey is sitting on the ninth step i.e. on the water level.
In the first jump the monkey will be on the step = 9 + (-4) = 9 – 4 = 5 steps
In the second jump the monkey will be on step = 5 + 2 = 7 steps
In the third jump the monkey will be on the step = 7 + (-4) = 7 – 4 = 3 steps
In the fourth jump the monkey will be on the step = 3 + 2 = 5 steps
In 5th jump the monkey will be on step = 5 + (-4) = 5 – 4 = 1 step
The monkey made 5 jumps to reach back to the top step i.e. the first 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 (– 8) represents going down by eight steps. So, what will the sum 8 in (b) represent?
Solution:-
It is given from the question that
If the number of steps to be taken down is represented by negative integers and the number of steps is represented by positive integers.
A monkey walks in (i)
= – 3 + 2 – ……….. = – 8
Then LHS = – 3 + 2 – 3 + 2 – 3 + 2 – 3 + 2 – 3 + 2 – 3
= – 18 + 10
= – 8
RHS = -8
In part (i) the monkey is going down 8 steps. Because of a negative integer.
Now,
A monkey walks in part (ii)
= 4 – 2 + ……….. = 8
Then LHS = 4 – 2 + 4 – 2 + 4
= 12 – 4
= 8
RHS = 8
Move-in part (ii) Monkey is going up 8 steps. Because of positive integers.
NCERT Solutions for Class 7 Maths Chapter 1 PDF Download
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