NCERT Solutions for Class 8 Math Chapter 2 - Microorganisms- Friend and Foe

Answer:

x – 2 = 7 [Transposing]
$\Rightarrow$ x = 7 + 2
$\Rightarrow$ x = 9

Answer:

y + 3 = 10
$\Rightarrow$ y = 10 – 3 [Transposing]
$\Rightarrow$ y = 7

Answer:

6 = z + 2
$\Rightarrow$ 6 – 2 = z [Transposing]
$\Rightarrow$ 4 = z

Answer:

Answer:

6x = 12

Answer:

Answer:

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Answer:

7x – 9 = 16
$\Rightarrow$ 7x = 16 + 9 (Transposing)
$\Rightarrow$ 7x = 25

Answer:

14y – 8 = 13
$\Rightarrow$ 14y = 13 + 8 (Transposing)
$\Rightarrow$ 14y = 21

Answer:

17 + 6p = 9
$\Rightarrow$ 6p = 9 – 17 (Transposing)
$\Rightarrow$ 6p = –8

Answer:

Answer:

Answer:

Perimeter of the pool = 154 m
Let breadth of the pool = x m
$\therefore$ Length of the pool = (2x + 2) m
$\therefore$ Perimeter = 2(Length + Breadth)
$\Rightarrow$ 154 = 2[x + 2x + 2]
$\Rightarrow$ 154 = 2[3x + 2]
$\Rightarrow$ 154 = 6x + 4
$\Rightarrow$ 6x = 154 – 4 $\Rightarrow$ 6x = 150

= 25 m
$\therefore$ Breadth of the pool = 25 m
Length of the pool = (2x + 2)
= 2(25) + 2 = 50 + 2 = 52

Answer:

Answer:

Let one number = x
Other number = x + 15
According to question,
x + x + 15 = 95
$\Rightarrow$ x + x = 95 – 15
$\Rightarrow$ 2x = 80

$\therefore$ One number = x = 40
Other number = x + 15 = 40 + 15 = 55.

Answer:

Since, two numbers are in the ratio 5 : 3
$\therefore$ Let first number = 5x
and second number = 3x
According to question,
5x – 3x = 18
2x = 18

$\therefore$ First number = 5x = 5 × 9 = 45
and second number = 3x = 3 × 9 = 27

Answer:

Let first integer = x
Second consecutive integer = x + 1
Third consecutive integer = x + 2
According to question,
x + x + 1 + x + 2 = 51
$\Rightarrow$ 3x + 3 = 51
$\Rightarrow$ 3x = 51 – 3
$\Rightarrow$ 3x = 48

$\therefore$ First integer = x = 16
second integer = x + 1 = 16 + 1 = 17
and third integer = x + 2 = 16 + 2 = 18

Answer:

Let first multiple of 8 is 8x.
Second consecutive multiple = 8x + 8
And third consecutive multiple = (8x + 8) + 8
= 8x + 16
According to question,
8x + 8x + 8 + 8x + 16 = 888
$\Rightarrow$ 24x + 24 = 888
$\Rightarrow$ 24x = 888 – 24
$\Rightarrow$ 24x = 864

$\therefore$ First multiple of 8 = 8x = 8 × 36 = 288
Second multiple of 8 = 8x + 8 = 8 × 36 + 8 = 288 + 8 = 296 And third multiple of 8 = 8x + 16 = 8 × 36 + 16 = 288 + 16 = 304

Answer:

Let three consecutive integers in increasing order
are x, x + 1 and x + 2
$\therefore$ According to question,
2x + 3 (x + 1) + 4 (x + 2) = 74
$\Rightarrow$ 2x + 3x + 3 + 4x + 8 = 74
$\Rightarrow$ 9x + 11 = 74
$\Rightarrow$ 9x = 74 – 11
$\Rightarrow$ 9x = 63

$\therefore$ First integer = x = 7 Second integer = x + 1 = 7 + 1 = 8 And third integer = x + 2 = 7 + 2 = 9

Answer:

Let present age of Rahul = 5x
and present age of Haroon = 7x
Four years later;
Age of Rahul = 5x + 4
Age of Haroon = 7x + 4
According to question,
(5x + 4) + (7x + 4) = 56
$\Rightarrow$ 5x + 4 + 7x + 4 = 56
$\Rightarrow$ 12x + 8 = 56
$\Rightarrow$ 12x = 56 – 8
$\Rightarrow$ 12x = 48

$\therefore$ Present age of Rahul = 5x = 5 × 4 = 20
And present age of Haroon = 7x = 7 × 4 = 28

Answer:

Let the number of boys in the class = 7x
and the number of girls in the class = 5x
According to question,
7x = 5x + 8
$\Rightarrow$ 7x – 5x = 8
$\Rightarrow$ 2x = 8

$\therefore$ Number of boys = 7x = 7 × 4 = 28
and number of girls = 5x = 5 × 4 = 20
$\therefore$ Total class strength = 28 + 20 = 48 students

Answer:

Let the age of Baichung = x years
$\therefore$ Age of Baichung’s father = (x + 29) years
Since, sum of ages of all three is 135 years.
$\therefore$ Age of Baichung’s grandfather
= 135 – (x + x + 29)
= 135 – (2x + 29)
= 135 – 2x – 29
= 106 – 2x ...(1)
According to question,
Age of Baichung’s grandfather
= (x + 29) + 26
= x + 55 ...(2)
$\therefore$ From (1) and (2), we have:
106 – 2x = x + 55
106 – 55 = x + 2x
51 = 3x

Answer:

Let the present age of Ravi = x years
After 15 years, age of Ravi = (x + 15) years
According to question,
x + 15 = 4x
$\Rightarrow$ 4x – x = 15
$\Rightarrow$ 3x = 15

$\therefore$ Present age of Ravi = 5 years

Answer:

Answer:

Let the number of ₹100 notes = 2x
Number of ₹50 notes = 3x
and number of ₹10 notes = 5x
$\therefore$ Total money she had
= 2x × 100 + 3x × 50 + 5x × 10
= 200x + 150x + 50x
= 400x
$\therefore$ According to question,
400x = 4,00,000

$\therefore$ Number of ₹100 notes = 2x = 2 × 1000 = 2000
Number of ₹50 notes = 3x = 3 × 1000 = 3000
Number of ₹10 notes = 5x = 5 × 1000 = 5000

Answer:

Total money I had = ₹300
Total number of coins = 160
Let the number of ₹5 coins = x
$\therefore$ The number of ₹2 coins = 3x
$\therefore$ The number of ₹1 coin = 160 – (x + 3x)
= 160 – 4x
$\therefore$ Total money I had
= 5 × x + 2 × 3x + 1 × (160 – 4x)
= 5x + 6x + 160 – 4x
= 160 + 7x
According to question,
160 + 7x = 300
$\Rightarrow$ 7x = 300 – 160
$\Rightarrow$ 7x = 140

$\therefore$ Number of ₹5 coins = x = 20
Number of ₹2 coins = 3x = 3 × 20 = 60
Number of ₹1 coin = 160 – 4x
= 160 – 4 × 20
= 160 – 80 = 80

Answer:

Total prize money = ₹3000
Prize given to a winner = ₹100
Prize given to a participant = ₹25
Total number of participants = 63
Let number of winners = x
$\therefore$ Number of participants who does
not win = (63 – x)
$\therefore$ Prize money distributed to winners
= ₹100 × x = ₹100x
Prize money distributed to participants who
does not win = ₹25 (63 – x)
$\therefore$ Total prize money = ₹100 x + ₹25 (63 – x)
$\therefore$ We have,
100x + 25 (63 – x) = 3000
$\Rightarrow$ 100x + 1575 – 25x = 3000
$\Rightarrow$ 75x = 3000 – 1575
$\Rightarrow$ 75x = 1425

$\therefore$ Number of winners = x = 19

Answer:

3x = 2x + 18
$\Rightarrow$ 3x – 2x = 18
$\Rightarrow$ x = 18
Check:
3x = 2x + 18
$\Rightarrow$ 3(18) = 2(18) + 18
$\Rightarrow$ 54 = 36 + 18
$\Rightarrow$ 54 = 54
$\therefore$ LHS = RHS

Answer:

5t – 3 = 3t – 5
$\Rightarrow$ 5t – 3t = – 5 + 3
$\Rightarrow$ 2t = – 2

Check:
5t – 3 = 3t – 5
$\Rightarrow$ 5 (– 1) – 3 = 3 (– 1) – 5
$\Rightarrow$ – 5 – 3 = – 3 – 5
$\Rightarrow$ – 8 = – 8
$\therefore$ LHS = RHS

Answer:

5x + 9 = 5 + 3x
$\Rightarrow$ 5x – 3x = 5 – 9
$\Rightarrow$ 2x = – 4

$\Rightarrow$ x = – 2
Check:
5x + 9 = 5 + 3x
$\Rightarrow$ 5 (– 2) + 9 = 5 + 3 (– 2)
$\Rightarrow$ – 10 + 9 = 5 – 6
$\Rightarrow$ – 1 = – 1
$\therefore$ LHS = RHS

Answer:

4z + 3 = 6 + 2z
$\Rightarrow$ 4z – 2z = 6 – 3
$\Rightarrow$ 2z = 3

Check:
4z + 3 = 6 + 2z

Answer:

2x – 1 = 14 – x
$\Rightarrow$ 2x + x = 14 + 1
$\Rightarrow$ 3x = 15

Check:
2x – 1 = 14 – x
$\Rightarrow$ 2(5) – 1 = 14 – 5
$\Rightarrow$ 10 – 1 = 9
$\Rightarrow$ 9 = 9
$\therefore$ LHS = RHS

Answer:

8x + 4 = 3(x – 1) + 7
$\Rightarrow$ 8x + 4 = 3x – 3 + 7
$\Rightarrow$ 8x – 3x = 4 – 4
$\Rightarrow$ 5x = 0

Check:
8x + 4 = 3(x – 1) + 7
$\Rightarrow$ 8(0) + 4 = 3(0 – 1) + 7
$\Rightarrow$ 0 + 4 = – 3 + 7
$\Rightarrow$ 4 = 4
$\therefore$ LHS = RHS

Answer:

$\Rightarrow$ 5x = 4x + 40
$\Rightarrow$ 5x – 4x = 40
x = 40
Check:

$\Rightarrow$ 40 = 4(10)
$\Rightarrow$ 40 = 40
$\therefore$ LHS = RHS

Answer:

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Answer:

Let the number thought by Amina = x
Now, according to statement,

$\Rightarrow$ 8x – 20 = 3x
$\Rightarrow$ 8x – 3x = 20
$\Rightarrow$ 5x = 20

$\therefore$ The number thought by Amina = 4

Answer:

Let first positive number = x
Then, another positive number = 5x
When 21 is added to both the numbers, then the
numbers becomes twice.
First new number = x + 21
And another new number = 5x + 21
Now,
According to question,
5x + 21 = 2(x + 21)
$\Rightarrow$ 5x + 21 = 2x + 42
$\Rightarrow$ 5x – 2x = 42 – 21
$\Rightarrow$ 3x = 21

$\therefore$ The first number = 7
and another number = 5x = 5(7) = 35

Answer:

Sum of digits = 9
Let digit at unit’s place = x
$\Rightarrow$ Digit at tens place = 9 – x
$\Rightarrow$ Number = 10 × (Digit at tens place) + Digit
at unit’s place
= 10(9 – x) + x
= 90 – 10x + x
= 90 – 9x
Now, the number obtained by interchanging the
digits = 10x + (9 – x)
= 10x + 9 – x
= 9x + 9
$\therefore$ According to question,
9x + 9 = (90 – 9x) + 27
$\Rightarrow$ 9x + 9 = 90 – 9x + 27
$\Rightarrow$ 9x + 9x = 117 – 9
$\Rightarrow$ 18x = 108

$\therefore$ Number = 90 – 9x
= 90 – 9 (6)
= 90 – 54 = 36

Answer:

Let unit’s digit = x
and ten’s digit = 3x
$\therefore$ The number = 10 (Digit at tens place)
+ (Digit at unit’s place)
= 10(3x) + x
= 30x + x
= 31x
Number obtained by interchanging the digits
= 10(x) + 3x
= 10x + 3x = 13x
Now,
According to question,
31x + 13x = 88
$\Rightarrow$ 44x = 88

$\therefore$ The number = 31x = 31(2) = 62

Answer:

Let the present age of Shobo = x years
$\therefore$ The present age of Shobo’s mother = 6x years
After 5 years,
Age of Shobo = (x + 5) years
Age of Shobo’s mother = (6x + 5) years
According to question,

$\Rightarrow$ x + 5 = 2x
$\Rightarrow$ 2x – x = 5 $\Rightarrow$ x = 5
$\therefore$ Present age of Shobo = 5 years
Present age of Shobo’s mother
= 6x = 6(5) = 30 years

Answer:

Let length of plot = 11x
and breadth of plot = 4x
$\therefore$ Perimeter of rectangular plot = 2 (l + b)
= 2(11x + 4x)
= 2(15x) = 30x
It is given that cost of fencing the plot at the rate
of ₹ 100 per metre is ₹ 75000.
$\therefore$ 30x × 100 = 75000
$\Rightarrow$ 3000x = 75000

$\Rightarrow$ x = 25
$\therefore$ Length of rectangular plot = 11x = 11 × 25 = 275 m
Breadth of rectangular plot = 4x = 4 × 25 = 100 m

Answer:

Let the total material purchased for shirts = 3x
metre and the total material purchased for
trousers = 2x metre
Cost price of shirt material = ₹50 per metre
Cost price of trouser material = ₹90 per metre
Profit = 12%
Profit on shirt material = 12% of ₹50

Answer:

Let the number of deer in the herd = x
Number of deer grazing in the field

Answer:

Let the age of grand-daughter = x years
$\therefore$ Age of grandfather = 10x years
Also, it is given that,
10x = x + 54
$\Rightarrow$ 10x – x = 54
$\Rightarrow$ 9x = 54 $\Rightarrow$ x = 6 years
$\therefore$ Present age of grand-daughter = x = 6 years
Present age of grandfahter = 10x
= 10(6) = 60 years

Answer:

Let age of Aman’s son = x years
$\therefore$ Age of Aman = 3x years
Ten years ago, age of Aman’s son = (x – 10) years
Ten years Ago, age of Aman = (3x – 10) years
$\therefore$ According to question,
3x – 10 = 5(x – 10)
$\Rightarrow$ 3x – 10 = 5x – 50
$\Rightarrow$ 3x – 5x = – 50 + 10
$\Rightarrow$ – 2x = – 40 $\Rightarrow$ x = 20
$\therefore$ Age of Aman’s son = 20 years
Age of Aman = 3x = 3(20) = 60 years

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Answer:

We have,
3(t – 3) = 5(2t + 1)
$\Rightarrow$ 3t – 9 = 10t + 5
$\Rightarrow$ 3t – 10t = 5 + 9
$\Rightarrow$ – 7t = 14
$\Rightarrow$ t = – 2

Answer:

We have,
$\Rightarrow$ 15(y – 4) – 2(y – 9) + 5(y + 6) = 0
$\Rightarrow$ 15y – 60 – 2y + 18 + 5y + 30 = 0
$\Rightarrow$ 15y – 2y + 5y = 60 – 18 – 30
$\Rightarrow$ 18y = 12

Answer:

We have,
3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
$\Rightarrow$ 15z – 21 – 18z + 22 = 32z – 52 – 17
$\Rightarrow$ – 3z – 32z = – 69 + 21 – 22
$\Rightarrow$ – 35z = – 70 $\Rightarrow$ z = 2

Answer:

We have,
0.25(4f – 3) = 0.05(10f – 9)
$\Rightarrow$ 1.00f – 0.75 = 0.5f – 0.45
$\Rightarrow$ 1.0f – 0.5f = –0.45 + 0.75
$\Rightarrow$ 0.5f = 0.30

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Answer:

Let age of Hari = 5x
and age of Harry = 7x
After four years,
Age of Hari = 5x + 4
Age of Harry = 7x + 4
According to question,

Multiplying both sides by (7x + 4), we have

$\Rightarrow$ 4(5x + 4) = 3(7x + 4)
$\Rightarrow$ 20x + 16 = 21x + 12
$\Rightarrow$ 20x – 21x = 12 – 16
$\Rightarrow$ – x = – 4
$\Rightarrow$ x = 4
$\therefore$ Age of Hari = 5x = 5 (4) = 20 years
Age of Harry = 7x = 7 (4) = 28 years

Answer:

Let the numerator = x
$\therefore$ Denominator = x + 8
According to question,