Chapter 1 Integers
Exercise
1.1
Ex 1.1 Class 7 Maths
Question 1.
Following number line shows the temperature in degree Celsius (°C) at different
places on a particular day.
(a) Observe this number line
and write the temperature of the places marked on it.
(b) What is the temperature difference between the hottest and the coldest
places among the above?
(c) What is the temperature difference between Lahulspriti and Srinagar?
(d) 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?
Solution:
(a) From the given number line, we observe the following temperatures.
Cities |
Temperature |
Lahulspriti |
-8°C |
Srinagar |
-2°C |
Shimla |
5°C |
Ooty |
14°C |
Bengaluru |
22°C |
(b) The temperature of the hottest place = 22°C
The temperature of the coldest place = -8°C
Difference = 22°C – (-8°C)
= 22°C + 8°C = 30°C
(c) Temperature of Lahulspriti = -8°C
Temperature of Srinagar = -2°C
∴ Difference = -2°C – (-8°C)
= -2°C + 8°C = 6°C
(d) 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 taken together is less than that
of Shimla by 2°C.
i.e., (5°C – 3°C) = 2°C
Ex 1.1 Class 7 Maths
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:
Given scores are 25, -5, -10, 15, 10
Marks given for correct answers
= 25 + 15 + 10 = 50
Marks given for incorrect answers
= (-5) + (-10) = -15
∴ Total marks given at the end
= 50 + (-15) = 50 – 15 = 35
Ex 1.1 Class 7 Maths
Question 3.
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:
Initial temperature of Srinagar on Monday = -5°C
Temperature on Tuesday = -5°C – 2°C = -7°C
Temperature was increased by 4°C on Wednesday.
∴ Temperature on Wednesday
= -7°C + 4°C = -3°C
Hence, the required temperature on Tuesday = -7°C
and the temperature on Wednesday = -3°C
Ex 1.1 Class 7 Maths
Question 4.
A plane is flying at the height of 5000 m above the sea level. At a particular
point, it is exactly above a submarine flowing 1200 m below the sea level. What
is the vertical distance between them?
Solution:
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
Ex 1.1 Class 7 Maths
Question 5.
Mohan deposits ₹ 2,000 in a bank account and withdraws ₹ 1,642 from it, the
next day. If 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.
Solution:
The deposited amount will be represented by a positive integer i.e., ₹ 2000.
Amount withdrawn = ₹ 1,642
∴ Balance in the account
= ₹ 2,000 – ₹ 1,642 = ₹ 358
Hence, the balance in Mohan’s account after the withdrawal
= ₹ 358
Ex 1.1 Class 7 Maths
Question 6.
Rita goes 20 km towards east from a point A to the point B. From B, she moves
30 km towards west along the same road. If the distance towards east is
represented by a positive integer, then how will you represent her final
position from A?
Solution:
Distances travelled towards east from point A will be represented by positive
integer i.e. +20 km.
Distance travelled towards the west from point B will be represented by
negative integer, i.e., —30 km.
Final position of Rita from A
= 20 km – 30 km = – 10 km
Hence, the required position of Rita will be presented by a negative number,
i.e., -10.
Ex 1.1 Class 7 Maths
Question 7.
In a magic square each row, column and the diagonal have the same sum. Check
which of the following is a magic square?
Solution:
(i) Row one R_{1} = 5 + (-1) + (—4)
=5 – 1 – 4 = 5 – 5 = 0
Row two R_{2} = (-5) + (-2) + 7
= -5 – 2 + 7 = -7 + 7 = 0
Row three R_{3} = 0 + 3 + (-3)
= 0 + 3- 3 = 0
Column one C_{1}t = 5 + (-5) + 0
= 5 – 5 + 0 = 0
Column two C_{2} = (-1) + (-2) + (3)
=-1 – 2 + 3 = -3 + 3 = 0
Column three C_{3} = (-4) + 7 + (-3)
= -4 + 7 – 3 = 7 – 7 = 0
Diagonal d_{12} = 5 + (-2) + (-3)
= 5 – 2- 3 = 5 – 5 = 0
Diagonal d_{2} = (-4) + (-2) + 0
= -4 – 2 + 0 = -6 + 0 = -6
Here, the sum of the integers of diagonal d2 is different from the others.
Hence, it is not a magic square.
(ii) Row one R_{1} = 1 + (-10) + 0
= 1 – 10 + 0 = -9
Row two R_{2} = (-4) + (-3) + (-2)
= -4 – 3 – 2 = -9
Row three R_{3} = (-6) + (4) + (-7)
= -6 + 4 – 7 = -9
Column one C_{3} = 1 + (-4) + (-6)
= 1 – 4 – 6 = -9
Column two C_{2} = (-10) + (-3) + 4
= -10 – 3 + 4 = -9
Column three C_{3} = 0 + (-2) + (-7)
= 0 – 2 -7 = -9
Diagonal d_{1} = 1 + (-3) + (-7)
= 1 – 3 – 7 = 1 – 10 = -9
Diagonal d_{2} = 0 + (-3) + (-6)
= 0 – 3- 6 = -9
Here, sum of the integers column wise, row wise and diagonally is same i.e. -9.
Hence, (ii) is a magic square.
Ex 1.1 Class 7 Maths
Question 8.
Verify a – (-b) = a + b for the following values of a and 6.
(i) a = 21, b = 18
(ii) a = 118, b = 125
(iii) a = 75, b = 84
(iv) a= 28, 6 = 11
Solution:
(i) a – (-b) = a + b
LHS = 21 – (-18) = 21 + 18 = 39
RHS = 21 + 18 = 39
LHS = RHS Hence, verified.
(ii) a – (-b) = a + b
LHS = 118 – (-125) = 118 + 125 = 243
RHS = 118 + 125 = 243
LHS = RHS Hence, verified.
(iii) a – (-b) = a + b
LHS = 75 – (-84) = 75 + 84 = 159
RHS = 75 + 84 = 159
LHS = RHS Hence, verified.
(iv) a – (-b) = a + b
LHS = 28 – (-11) = 28 + 11 = 39
RHS = 28 + 11 = 28 + 11 = 39
LHS = RHS Hence, verified.
Ex 1.1 Class 7 Maths
Question 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
Solution:
(a) (-8) + (-4) □ (-8) – (-4)
LHS = (-8) + (-4) = -8 – 4 = – 12
RHS = (-8) – (-4) = -8 + 4 = -4
Here – 12 < -4
Hence, (-8) + (-4) [<] (-8) – (-4)
(b) (-3) + 7 – (19) □ 15 – 8 + (-9)
LHS = (-3) + 7 – (19) =-3 + 7-19
= -3 – 19 + 7
= -22 + 1 = -15
RHS = 15 – 8 + (-9)
= 15-8-9
= 15 – 17 = -2
Here -15 < -2
Hence, (-3) + 7 – (19) [<] 15 – 8 + (-9)
(c) 23 – 41 + 11 □ 23 – 41 – 11
LHS = 23 – 41 + 11 = 23 + 11 – 41 = 34 – 41 = -7
RHS = 23 – 41 – 11 = 23 – 52 = -29 Here, -7 > -29
Hence, 23 – 41 + 11 [>] 23 – 41 – 11
(d) 39 + (-24) – (15) □ 36 + (-52) – (-36)
LHS = 39 + (-24) – (15)
= 39 – 24 – 15
= 39 – 39 = 0
RHS = 36 + (-52) – (-36) = 36 – 52 + 36
= 36 + 36 – 52
= 72 – 52 = 20
Here 0 < 20
Hence, 39 + (-24) – (15) [<] 36 + (-52) – (-36)
(e) -231 + 79 + 51 □ -399 + 159 + 81
LHS = -231 + 79 + 51 = -231 + 130 = -101
RHS = -399 + 159 + 81 = -399 + 240 = -159
Here, -101 > -159
Hence, -231 + 79 + 51 [>] -399 + 159 + 81
Ex 1.1 Class 7 Maths
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?
(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?
(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 move in part (t) 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:
(i) The position of monkey after the
1^{st} jump J_{1} is at 4^{th} step ↓
2^{nd} jump J_{2} is at 2^{nd} step ↑
3^{rd} jump J_{3} is at 5^{th} step ↓
4^{th} jump J_{4} is at 3^{rd} step ↑
5^{th} jump J_{5} is at 6^{th} step ↓
6^{th} jump J_{6} is at 4^{th} step ↑
7^{th} jump J_{7} is at 7^{th} step ↓
8^{th} jump J_{8} is at 5^{th} step ↑
9^{th} jump J_{9} is at 8^{th} step ↓
10^{th} jump J_{10} is at 6^{th} step ↑
11^{th} jump J_{11} is at9^{th} step ↓ (Water level)
Hence the required number of jumps = 11.
(ii) Monkey’s position after the
1^{st} jump J_{1} is at 5^{th} step ↑
2^{nd} jump J_{2} is at 7^{th} step ↓
3^{rd} jump J_{3} is at 3^{rd} step ↑
4^{th} jump J_{4} is at 5^{th} step ↓
5^{th} jump J_{5} is at 1^{st} step ↑
Hence, the required number of jumps = 5.
(iii) According to the given conditions we have the
following tables
Jumps |
J_{1} |
J_{2} |
J_{3} |
J_{4} |
J_{5} |
J_{6} |
J_{7} |
J_{8} |
J_{9} |
J_{10} |
J_{11} |
Number of steps |
-3 |
+2 |
-3 |
+2 |
-3 |
+2 |
-3 |
+2 |
-3 |
+2 |
-3 |
Therefore (a) Total number of steps
=-3 + 2 – 3 + 2 – 3 + 2 – 3 + 2 – 3 + 2 – 3
= -8 which represents the monkey goes down by 8 steps.
In case (ii), we get
Jumps |
J_{1} |
J_{2} |
J_{3} |
J_{4} |
J_{5} |
Number of steps |
+
4 |
-2 |
+4 |
-2 |
+4 |
Therefore (b) Total number of steps.
= +4 – 2 + 4 – 2 + 4 = 8
Here, the monkey is going up by 8 steps.