Samacheer Class 7 Maths - Number System Intext Questions

Plotting the points on the number line, we get

The numbers are placed in an increasing order from left to right.
∴ Ascending order: -10 < -6 < -5 < 0 < 2 < 4 < 10

Separating positive and the negative integers, we get -27, -4, -22, -9, -35
Arranging the numbers in descending order -4 > -9 > -22 > -27 > -35
The positive numbers are 19,12,47, 3
Arranging in descending order, we get 47 > 19 > 12 > 3
0 stands in the middle.
∴ Descending order: 47 > 19 > 12 > 3 > 0 > -4 > -9 > -22 > -27 > -35
(Try This Text Book Page No. 3)

(i) (-4) + (+3)
To find the sum of (-4) and (+3), we start at zero facing positive direction continuing in the same direction and move 4 units backward to represent (-4).
Since the operation is addition we maintain the same direction and move three units forward to represent (+3)
We land at -1
So (-4) + (+3) = -1

(ii) (-4) + (-3)
From zero move 4 steps backward to represent (-4)
From the same direction again move 3 units backward to represent (-3)
We land at -7 So (-4) + (-3) = -7

(iii) (+4) + (-3)
We start at zero facing positive direction and move 4 steps forward to represent (+4) Since the operation is addition we maintain the same direction and move three units backward to represent (-3).
We land at +1.
So (+4) + (-3) = +1

(Properties of Addition Textbook Page No. 6)

(i) [(-2) + (-9)] + 6 = (-2) + [(-9) + 6]
[(-2) + (-9)] + 6 = (-11) + 6 = -5
Also (-2) + [(-9) + 6] = (-2) + (-3) = -5
Both the cases the sum is -5.
∴ – [(-2) + (-9)] + 6 = (-2) + [(-9) + 6]
(ii) [7 + (-8)] +(-5) = 7 + [(-8) + (-5)]
Here [7 + (-8)] + (-5) = (-1) + (-5) = -6
Also 7 + [(-8) + (-5)] = 7 + (-13) = 7 – 13 = -6
In both the cases the sum is -6.
∴ [7 + (-8)] + (-5) = 7 + [(-8) + (-5)]
(iii) [(-11) + 5] + (-14) = (-11) + [5 + (-14)]
Here [(-11) + 5] + (-14) = (-6) + (-14) = (-20)
(-11) + [5 + (-14)] = (-11) +(-9) = (-20)
In both the cases the sum is -20.
∴ [(-11) + 5] + (-14) = (-11) + [5 + (-14)]
(iv) (-5) + [(-32) + (-2)] = [(-5) + (-32)] + (-2)
(-5) + [(-32) + (-2)] = (-5) + (-34) = -39
Also [(-5) + (-32)] + (-2) = (-37) + (-2) = -39
In both the cases the sum is -39.
∴ (-5)+ [(-32) +(-2)] = [(-5)+ (-32)] +(-2)

We start at zero facing positive direction move 4 units backward to represent (-4). Then turn towards negative side and move 3 units forward.

We reach -7.
∴ (-4) – (+3) = -7.
(ii) (-4) – (-3)
We start at zero facing positive direction. Move 4 units backward to represent -4. Then turn towards the negative side and move 3 units backwards.

We reach at-1.
∴ (-4) – (-3) = -1.

(-6) – (-2) = -6 + (Additive inverse of-2)
= -6 + (+2) = -4
Also (-6)+ 2 = -4
∴ (-6) – (-2) = (-6) + 2
(ii) 35 – (-7) and 35 + 7.
35 – (-7) = 35 + (Additive inverse of -7) = 35 + (+7) = 42
Also 35 + 7 = 42 ; 35 – (-7) =35 + 7
(iii) 26 – (+10) and 26 + (-10)
26 – (+10) = 26 + (Additive inverse of +10) = 26 + (-10) = 16
Also 26 + (-10) = 16; 26 – (+10) = 26 + (-10)

(i) -10 – 8 = -18 & -10 + 8 = – 2
(ii) (-20) + 10 = -10 & (-20) – (-10) = -10
(iii) -70 – 50 = (-70) + (-50) = -20
(iv) 100 – (+100) = 0 & 100 – (-100) = 100 + (+100) = 200
(v) -50 – 30 = -50 + (-30) = -80 Also -100 + 20 = – 80
(Try These Text book Page No. 14)

(i) 15 – 12 = 3 & 12-15 = 12 +(-15) = -3

(ii) -21 – 32 = (-21) + (-32) = -53
Also -32 – (-21) = (-32) + (+21) = -11 ; -53 < -11

Consider the numbers 1,2 and 3. Now (1 – 2) – 3 = -1 – 3 = -4
Also 1 – (2 – 3) = 1 – (-1) = 1 + 1 = 2
∴ (1 – 2) – ≠ 1 – (2 – 3)
∴ Associative property is not true for subtraction of integers.
Exercise 1.3
Multiplication of Integers
(Try These Textbook Page No. 16)

We know that
(i) product of two positive integers is positive
(ii) product of two negative integers is
(ii) product of two negative integers is positive
(iii) product of integers with opposite sign is negative.
∴ The table will be as follows:

Here 18 × (-5) = -90 Also (-5) × 18 = -90
∴ 18 × (-5)= (-5) × 18
(ii) 31 × (-6) and (-6) × 31
Here 31 × (-6) = -186 Also (-6) × 31 =-186
∴ 31 × (-6)= (-6) × 31
(iii) 4 × 51 and 51 × 4
Here 4 × 51 = 204 Also 51 × 4 = 204
∴ 4 × 51 = 51 × 4

LHS = (-20) × (13 × 4) = (-20) × 52 = -1040
RHS = [(-20) × 13] × 4 = (-260) × 4 = -1040
LHS = RHS
∴ (-20) × (13 × 4) = [(-20) × 13] × 4
(ii) [(-50) × (-2)] × (-3) = (-50) × [(-2) × (-3)]
LHS = [(-50) × (-2)] × (-3) = 100 × (-3) = -300
RHS = (-50) × [(-2) × (-3)] = (-50) × 6 =-300
LHS = RHS
∴ [(-50) × (-2)] × (-3) = (-50) × [-2) × (-3)]
(iii) [(-4) × (-3)] × (-5) = (-4) × [(-3) × (-5)]
LHS = [(-4) × (-3)] × (-5) = 12 × (-5) = -60
RHS = (-4) × [(-3) × (-5)] = (-4) × 15 = -60
LHS = RHS
∴ [(-4) × (-3)] × (-5) = (-4) × [(-3) × (-5)]
(Try These Textbook Page No. 19)

(-6) × (4 + (-5)) = (-6) × (-1) = 6 .
((-6) × 4) + ((-6) × (-5)) = (-24)+ 30 = 6
Hence (-6) × (4 + (-5)) = ((-6) × 4) + ((-6) × (-5))
(ii) (-3) × [2 + (-8)] and [(-3) × 2] + [(-3) × 8]
(-3) × [2 + (-8)] = (-3) × (-6) = 18
Also [(-3) × 2] + [(-3) × 8] = (-6)+ (-24) = -30
(-3) × [2 + (-8)] ≠ [(-3) × 2] + [(-3) × 8]

LHS = (-5) × [(-76) + 8] = (-5) × (-68)
= +340
RHS = [(-5) × (-76)] + [(-5) × 8]
= +380 + (-40) = +380 – 40
= +340
LHS = RHS
∴ (-5) × [(-76) + 8] = [(-5) × (-76)] + [(-5) × 8]
(ii) (42 × 7) + [42 × (-3)]
LHS = 42 × [7 + (-3)]
= 168
RHS = (42 × 7) + [42 × (-3)] = 294 – 126
= 168
LHS = RHS
∴ 42 × [7 + (-3)] = (42 × 7) + [42 × (-3)]
(iii) [(-3) × (-4)] + [(-3) × (-5)]
LHS = (-3) × [(-4) + (-5)] = (-3) × (-9)
= +27
RHS = [(-3) × (-4)] + [(-3) × (-5)] = 12 + 15 = 27
LHS = RHS
∴ (-3) × [(-4) + (-5)] = [(-3) × (-4)] + [(-3) × (-5)]
(iv) 103 × 25 = (100 + 3) × 25 = (100 × 25) + (3 × 25)
First consider 103 × 25 = 2575
Now (100 + 3) × 25 = 103 × 25 = 2575
Also (100 × 25) + (3 × 25) = 2500 + 75
= 2575
∴ All the three are same. 103 × 25 = (100 + 3) × 25 = (100 × 25) +(3 × 25)
Exercise 1.4
Division of Integers
(Try These Text book Page No. 22)

Starting at zero on the number line facing positive direction and move 8 steps forward reaching 8.

Then we move 12 steps
backward to represent -12 –
and reach at -4.
∴ 8 + (-12) = -4
(ii) (-3) and (-5) using number line.
Starting at zero on the number line facing positive direction and move 3 steps backward reaching-3.

Then we move 5 steps backward to represent -5 and reach -8.
∴ (-3) + (-5) = -8
(iii) (-100) + (-10)
(-100) + (-10) = -100 – 10 = -110
(iv) 20 + (-72)
20 + (-72) = 20 – 72 = -52
(v) 82 + (-75)
82 + (-75) = 82 – 75 = 7
(vi) -48 + (-15)
-48 + (-15) = -48 – 15 = -63
(vii) -225 + (-63)
-225 + (-63) = -225 – 63 = -288

For each incorrect question the score = -1
In paper I, score for 25 incorrect questions – 25 × (-1) = -25
In paper II, for 13 incorrect question the score = 13 × (-1) = -13
The total marks get reduced = (-25) + (-13) = -38
-38 marks will be reduced.

Total score of team A = [(+30) + (-20)] + 0 = (+10) + 0 = 10
Total score of team B = [(-20) + 0] + (+30)
= -20 + 30 = +10
Score of team A = Score of team B.
Yes, we say that we can add integers in any order.

First we take (11 + 7) + 10 = 18 + 10 = 28
Now 11 + (7 + 10) = 11 + 17 = 28
In both the cases the sum is 28. ∴ (11 + 7) + 10 = 11 + (7 + 10)
This property is known as associative property of integers under addition.

(iv) 10 pm
: ’Temperature at 12 noon = 18° above zero = +18°
Rate of decrease per hour = -3°
Temperature 12° below zero = -12°
-12 is 30 units to the left of+18°
Time at which it reach -12° = \(\frac{30}{3}\) = 10 h
10 hrs after 12 noon = 10 pm

(i) -9 +(-5) + 6
(i) -9 + (-5) + 6 = -14 + 6 = -8
(ii) 8 + (-12) + 6 = -4 + 6 = – 2
(iii) -4 + 2 + 10 = -2 + 10 = 8
(iv) 10 + (-4) + 8 = 6 + 8 = 14

We start at zero facing positive direction. Move 3 units backward to represent (-3). Then turn towards the negative side and move 4 units backwards. We reach+1.

∴ (-3) – (-4) = +1
(ii) 7 – (-10) using number line

We start at zero facing positive direction. Move 7 units forward to represent (+7). Then turn towards the negative side and move 10 units backwards.
We reach +17
∴ 1 – (-10) = +17
(iii) 35 – (-64)
35 – (-64) = 35+ (Additive inverse of-64) = 35 + (+64) = 99
∴ 35 – (-64) = 99
(iv) -200 – (+100)
-200 – (+100) = -200 + (Additive inverse of+100) = -200 + (-100) = -300
-200 – (+100) = -300

Total pencils Kabilan had = 10
No. of pencils given to Senthil = 2
No. of pencils given to Karthick = 3.
Now number of pencils left with Kabilan = 10 – 2 – 3 = 8 – 3 = 5
Number of pencils got from his father = 6
No. total pencils Kabilan had = 5 + 6 = 11
Number of pencils given to his sister = 8
Number of pencils left with Kabilan = 11 – 8 = 3

Initially the lift will be in the ground floor representing ‘0’
It goes to 5 floors down ⇒ -5
Then it moves 10 floors up +10.
Now the lift will be = 0 – 5 + 10 = -5 + 10
= 5 th floor (above the ground floor)

According to the problem = -17 + A number = -19
The number = -19 + 17 = -2
∴ -2 should be added to -17 to get -19

Subtracting -12 from -47, we get
-47 – (-12) = -47 + (Additive inverse of-12)
= -47 + (+12) = -35
But the students answer is -30.
So he is not correct.
Objective Type Questions

(i) 16 is an even interger.
If negative integers are multiplied even number of times, the product is a positive integer.
∴ 16 times a negative integer is a positive integer.
(ii) 29 times negative integer.
If negative integers are multiplied odd number of times, the product is a negative integer. 29 is odd.
∴ 29 times negative integers is a negative integer.

Consider (8 – 13) × 7 = (-5) × 7 = -35
Now 8 – (13 × 7) = 8 – 91 = -83
∴ (8 – 13) × 7 ≠ 8 – (13 × 7)
(ii) [(-6) – (+8)] × (-4) and (-6) – [8 × (-4)]
[(-6) – (+8)] × (-4) = [(-6) + (-8)] × (-4) = (-14) × (-4) = +56
Now (-6) – [8 × (-4)] = (-6) – (-32)
= (-6) + (+32) = +26
∴ [(-6) – (+8)] × (-4) ≠ (-6) – [8 × (-4)]
(iii) 3 × [(-4) + (-10)] and [3 × (-4) + 3 × (-10)]
Consider 3 × [(-4) + (-10)] = 3 × -14 = -42
Now [3 × (-4) + 3 × (-10)] = (-12) + (-30) = -42
Here 3 × [(-4) + (-10)] = [3 × (-4) + 3 × (-10)]
It is the distributive property of multiplication over addition.

Factor of 50 are 1, 2, 5, 10, 25, 50.
Possible pairs of integers that gives product -50:
(-1 × 50), (1 × (-50)), (-2 × 25), (2 × (-25)), (-5 × 10), (5 × (-10))
Objective Type Questions

(iv) (-6) × (+5)
Hint:
(i) -20 + (10) = -10
(ii) 60 – 30 = 30
(iii) 10 + 20 = 30
(iv) (-6) × (+5) = – 30

Given the product of two integers = -135
One of them = -15
∴ -15 × Another number = -135
Other number = \(\frac{-135}{-15}\) = 9
∴ The other number = 9.

The elevator’s position = 15 m above ground level = +15 m
It should reach = -250 m
The distance to be travelled = 15 – (-250) m = 15 + (+250) m 265 m
Time taken to descend 5 m = 1 min
∴ Time required to descend 265 m = \(\frac{24}{8}\) = 53 min

Temperature in the first day = -5°C
Temperature in the next day = 9°C
∴ Increase in temperature = 9°C – (-5°C)
= 9°C + (+5°C) = 14°C

(ii) 6 protons and 6 electrons → (+6) + (-6) = 0
(iii) 9 protons and 12 electrons → (+9) + (-12) = 9-12 = -3 ⇒ 3 electrons \(\ominus\ominus\ominus\)
(iv) 4 protons and 8 electrons → (+4) + (-8) = +4 – 8 = -4 ⇒ 4 electrons \(\ominus\ominus \ominus\ominus\)
(v) 7 protons and 6 electrons → (+7) + (-6) = +1 = 1 proton \(\oplus\)

(i) -275°C = (-275 + 273)K = -2K
(ii) 45°C = (45 + 273)K = 318 K
(iii) -400°C = (-400 + 273)K = -127 K
(iv) -273°C = (-273 + 273) K = 0K

(i) Initial balance of student’s account = ₹ 690
Deposited amount = ₹ 485 (+)
∴ Amount left in the account = ₹ 690 + ₹ 485 = ₹ 1175
(ii) Balance in the account = ₹ 1175
Amount withdrawn = ₹ 500 (-)
Amount left = ₹ 1175 – ₹ 500 = ₹ 675
(iii) Balance in the account = ₹ 675
Amount withdrawn = ₹ 350 (-)
Amount left = ₹ 675 – ₹ 350 = ₹ 325
(iv) Balance in the account = ₹ 325
Amount deposited = ₹ 89(+)
Amount left = ₹ 325 + ₹ 89 = ₹ 414
(v) Balance in the account = ₹ 414
Amount withdrawn = ₹ 300 (-)
Amount left = ₹ 414 – ₹ 300 = ₹ 114

Total pages lost – 35
One day work = 5 page 35
35 pages = \(\frac{35}{5}\) = 7 days work
∴ 7 day’s work he lost.
(ii) Number of characters in four pages = 1800
Number of characters in one page = \(\frac{1800}{4}\) = 450
∴ Number of characters in 35 pages = 450 × 35 = 15,750 characters
(iii) Payment for one page = ₹ 250
∴ Payment for 35 pages = ₹ 250 × ₹ 35 = ₹ 8,750
(iv) Number of pages recreated a day = 7
∴ To recreate 35 pages day’s needed = \(\frac{35}{7}\) = 5 days
(v) Payment of Kavimaan = ₹ 100 per page
∴ for 35 pages payment = ₹ 100 × 35 = ₹ 3,500

According to the problem {[(I + 2) × 5] – 10} ÷ 4 = 15
{[(I + 2) × 5] – 10} = 15 × 4 = 60
I + 2 = \(\frac{70}{5}\) = 14
(I + 2) × 5 = 60 + 10 = 70
I = 14 – 2 ; I = 12

Number of apples Kamatchi sold = 30
Profit per apple = ₹ 8(+)
∴ Profit for 30 apples = 30 × 8 = ₹ 240
Number of pomegranates sold 50
Loss per pomegranate = ₹ 5(-)
Loss on selling 50 pomegranates = 50 × (-5) = ₹ -250
Overall loss = -250 + 240 = ₹ -10
i.e. loss ₹ 10.

Water level fall per week = -3 inches
∴ Water level decrease for 6 weeks = 6 ₹ (-3) = 18 inches
∴ decrease of 18 inches of water level.

Years in BCC (BCE) are taken as negative integers.

Buddha was bom in -563
and died in -483
So he was alive in 500 BC (BCE)
Life time = -483 – (-563) = -483 + 563 = +80
Buddha’s life time = 80 years.
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(i) LHS = (11 + 7) + 10 = 18 + 10 = 28
RHS = 11 + (7 + 10)
= 11 + (17) = 28
LHS = RHS
∴ (11 + 7) + 10 = 11 + (7 + 10)
(ii) LHS = (8 – 13) × 7 = -5 × 7 = -35
RHS = 8 – (13 × 7) = 8 – 91 = -83
LHS ≠ RHS
∴ (8 – 13) × 7 ≠ 8 – (13 × 7)
(iii) LHS = [(-6) – (+8)] × (-4) = [(-6) + (-8)] × (-4) = (-14) × (-4) = +56
RHS = (-6) – [8 × (-4)] = -6 – (-32)
= -6 + (+32) = +26
LHS ≠ RHS
∴ [(-6) – (+8)] × (-4) ≠ (-6) – [8 × (-4)]
(iv) LHS = 3 × [(-4) + (-10)] = 3 × (-14) = -42
RHS = [3 × (-4) + 3 × (-10)] = (-12) + (-30) = -42
LHS = RHS
3 × [(-4) + (-10)] = [3 × (-4) + 3 × (-10)]

Initial bank balance = ₹ 5000 ; Total deposits: January : ₹ 2000 ; March : ₹ 1000
Total deposits upto March = ₹ 5000 + ₹ 2000 + ₹ 1000 = ₹ 8000
Amount withdrawn: February : ₹ 700 (-)
March : ₹ 500 (-)
∴ Total amount withdrawn = (-700) + (-500) ₹ -1200
Net bank balance = ₹ 8000 – ₹ 1200 = ₹ 6800

Amount increases for x every year = ₹ 10.
Price ofx in 2018 = ₹ 50 ; Price of x in 2019 = ₹ 50 + ₹ 10 = ₹ 60
Price of x in 2020 = ₹ 60 + ₹ 10 = ₹ 70 Amount decreases for y per year = ₹ 15
Price of y in 2018 = ₹ 90
Price of y in 2019 = ₹ 90 – ₹ 15 = ₹ 75
Price of y in 2020 = ₹ 75 – ₹ 15 = ₹ 60
Here 70 > 60. Item x will costlier in year 2020.

Given

(i) My name LEENA → 12 + 5 + 5 + 0 + 1 = 23
(ii) SUCCESS → (-5) + (-7) + 3 + 3 + 5 + (-5) + (-5)
= -12 + 6 + 5 + (-10) = -6 + 5 + (-10) = (-1) + (-10)
= -11

Water used for one day = 100 litres.
Water used for 10 days = 100 × 10 = 1000 litres.
After 10 days water left in the tank = 2000 litres
Initially amount of water will be = 2000 + 1000 = 3000 litres

The water in the well is at 20th step.
For each jump the dog goes low 4 steps. 5
∴ Number of jumps the dog to reach the water = \(\frac{20}{4}\) = 5 jumps

Position of submarine = 650 feet below sea level = -650 feet
Again the depth it descends = 200 feet below = – 200 feet
∴ Position of submarine = (-650) + (-200) = -850 feet
The submarine will be 850 feet below the sea level.

Column total = Row total = diagonal total
∴ 1 + y + (-6) = (-10) + (-3) + 4
y + (- 5) = -13 + 4
y = -9 + 5
y = -4
So 1 + (-10) + x = y + (-3) + (-2)
-9 + x = (-4) + (-3) + (-2)
-9 + x = -9
x = -9 + 9
x = 0
Now x + (-2) + z = (-10) + (-3) + 4
0 + (-2) + z = (-13) + 4
-2 + z = -9
z = -9 + 2 = -7
z = -7
∴ x = 0, y = -4, z = -7
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