# Class 8

## Samacheer Kalvi 8th Books Solutions Guide

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## Samacheer Kalvi 8th Social Science Book Answers Solutions Guide

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## Samacheer Kalvi 8th Science Book Answers Solutions Guide

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## Samacheer Kalvi 8th Tamil Book Answers Solutions Guide

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## Samacheer Kalvi 8th English Book Back Answers Solutions Guide

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## Samacheer Kalvi 8th Maths Book Answers Solutions Guide

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### Samacheer Kalvi 8th Maths Book Back Answers

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Samacheer Kalvi 8th Maths Book Solutions Chapter 2 Measurements

Tamil Nadu State Board 8th Std Maths Guide Chapter 5 Geometry

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TN 8th Standard Maths Text Book Samacheer Kalvi Chapter 7 Information Processing

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8th Samacheer Kalvi Maths Guide Chapter 1 எண்கள்

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Samacheer Kalvi 8th Maths Book Back Answers (Old Syllabus)

Samacheer Kalvi 8th Maths Book Solutions Term 1

Samacheer Kalvi 8th Maths Book Solutions Chapter 1 Rational Numbers

Samacheer Kalvi 8th Maths Book Answers Chapter 2 Measurements

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 4 Geometry

8th Standard Maths Text Book Samacheer Kalvi Chapter 5 Information Processing

Samacheer Kalvi 8th Maths Solutions Book Solutions Term 2

Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics

8th Standard Samacheer Kalvi Maths Solutions Chapter 2 Algebra

Tamilnadu Board Class 8 Maths Chapter 3 Geometry

Tamil Nadu State Board 8th Std Maths Chapter 4 Information Processing

Samacheer Kalvi 8th Maths Book Solutions Term 3

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Samacheer Kalvi 8th Maths Book Answers Chapter 2 Life Mathematics

Tamilnadu Board 8th Maths Solutions Term 3 Chapter 4 Statistics

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## Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.4

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## Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.4

Miscellaneous Practice Problems

Question 1.
Two gates are fitted at the entrance of a library. To open the gates easily, a wheel is fixed at 6 feet istance from the wall ito which the gate is fixed. If one of the gates is opened to 90°, find the distance moved by the wheel (π = 3.14).

Solution:
Let A be the position of the wall AC be the gate in initial position and AB be position when it is moved 90°.
Now the arc length BC gives the distance moved by the wheel.
Length of the arc
= $$\frac{\theta}{360^{\circ}}$$ × 2πr units
= $$\frac{90^{\circ}}{360^{\circ}}$$ × 2 × 3.14 × 6 feets
= 3.14 × 3 feets
= 9.42 feets
∴ Distance moved by the wheel = 9.42 feets.

Question 2.
With his usual speed, if a person covers a circular track of radius 150 ra in 9 minutes, find the distance that he covers in 3 minutes (π = 3.14).
Solution:
Radius of the circular track = 150m
Distance covers in 9 minutes = Perimeter of the circle = 2 × π × r units
Distance covered in 9 min = 2 × 3.14 × 150 m

Distance he covers in 3 min = 314 m

Question 3.
Find the area of the house drawing given in the figure.

Solution:
Area of the house = Area of a square of side 6 cm + Area of a rectangle with l = 8cm, h = 6 cm + Area of a ∆ with b = 6 cm and h = 4 cm + Area of a parallelogram with b = 8 cm, h = 4 cm
= (side × side) + (l × b) + ($$\frac{1}{2}$$ × b × h) + 6h cm2
= (6 × 6) + (8 × 6) + ($$\frac{1}{2}$$ × 6 × 4) + (8 × 4) cm2
= 36 + 48 + 12+ 32 cm2
Required Area = 128 cm2

Question 4.
Draw the top, front and side view of the following solid shapes.

Solution:

Question 5.
Draw the net for the cube of side 4 cm in a graph sheet.
Solution:

Challenging Problems

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Question 6.
Guna has fixed a single door of 3 feet wide in his room whereas Nathan has fixed a double door, each 1 $$\frac{1}{2}$$ feet wide in his room. From the closed state, if each of the single and double doors can open up to 120°, whose door requires a minimum area?

Solution:
(a) Width of the door that Guna fixed = 3 feet.
When the door is open the radius of the sector = 3 feet
Angle covered = 120°
∴ Area required to open the door = $$\frac{120^{\circ}}{360^{\circ}}$$ × πr2 = $$\frac{120^{\circ}}{360^{\circ}}$$ × π × 3 × 3 = 37π feet2

(b) Width of the double doors that Nathan fixed = 1$$\frac{1}{2}$$ feet.
Angle described to open = 120°
Area required to open = 2 × Area of the sector
= 2 × $$\frac{120^{\circ}}{360^{\circ}} \times \pi \times \frac{3}{2} \times \frac{3}{2} \text { feets }^{2}=\frac{3 \pi}{2}$$ feet2
= $$\frac{1}{2}$$ (3π) feet2
∴ The double door requires the minimum area.

Question 7.
In a rectangular field which measures 15 m × 8m, cows are tied with a rope of length 3m at four corners of the field and also at the centre. Find the area of the field where none of the cow can graze. (π = 3.14).
Solution:
Area of the field where none of the cow can graze = Area of the rectangle – [Area of 4 quadrant circles] – Area of a circle

Area of the rectangle = l × b units2
= 15 × 8 m2 = 120 m2
Area of 4 quadrant circles = 4 × $$\frac{1}{4}$$ πr2 units
Radius of the circle = 3 m
Area of 4 quadrant circles = 4 × $$\frac{1}{4}$$ × 3.14 × 3 × 3 = 28.26m2
Area of the circle at the middle = πr2 units
= 3.14 × 3 × 3m2 = 28.26m2
∴ Area where none of the cows can graze
= [120 – 28.26 – 28.26]m2 = 120 – 56.52 m2 = 63.48m2

Question 8.
Three identical coins, each of diameter 6 cm are placed as shown. Find the area of the shaded region between the coins, (π = 3.14) ( $$\sqrt{3}$$ = 1.732)

Solution:
Given diameter of the coins = 6 cm
∴ Radius of the coins = $$\frac{6}{2}$$ = 3 cm
Area of the shaded region = Area of equilateral triangle – Area of 3 sectors of angle 60°
Area of the equilateral triangle

∴ Area of the shaded region = 15.588 – 14.13 cm2 = 1.458 cm2
Required area 1.458 cm2 (approximately)

Question 9.
Using graph sheet, draw the net for the cuboid whose length is 5cm, breadth is 4cm and height is 3cm and also find its area.
Solution:
Net for the cuboid is:

One of the possible nets for a cuboid of length = 5 cm, breadth = 4 cm, height = 3 cm is given above
Area of the cuboid
= 20 cm2 + 15 cm2 + 20 cm2 + 15 cm2 + 12 cm2 + 12 cm2 = 94 cm2
Using formula,
Surface area of a cuboid
= 2 (lb + bh + lh) unit2
= 2(5 × 4 + 4 × 3 + 5 × 3) cm2
= 2(20 + 12 + 15) cm2
= 94 cm2

Question 10.
Using Euler’s formula, find the unknowns.

Solution:
Euler’s formula is given by F + V- E = 2
(i) V = 6, E = 14
By Euler’s formula
= F + 6 – 14 = 2
F = 2 + 14 – 6
F = 10

(ii) F = 8, E = 10
By Euler’s formula
= 8 + V – 10 = 2
V = 2 – 8 + 10
V = 4

(iii) F = 20, V = 10
By Euler’s formula
= 20 + 10 – E = 2
30 – E = 2
E = 30 – 2
E = 28
Tabulating the required unknowns

## Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 4 Information Processing Ex 4.2

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## Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 4 Information Processing Ex 4.2

The factors of 70 are 1, 2, 5, 7, 10, 14, 35, and 70.

Question 1.

Question (i)
Common prime factors of 30 and 250 are
(a) 2 x 5
(b) 3 x 5
(c) 2 x 3 x 5
(d) 5 x 5
(a) 2 x 5
Hint:
Prime factors of 30 are 2 x 3 x 5
Prime factors of 250 are 5 x 5 x 5 x 2
∴ Common prime factors are 2 x 5

The Prime Factorization of 84 is 22 × 3 × 7.

Question (ii)
Common prime factors of 36, 60 and 72 are
(a) 2 x 2
(b) 2 x 3
(c) 3 x 3
(d) 3 x 2 x 2
(d) 3 x 2 x 2
Hint:
Prime factors of 36 are 2 x 2 x 3 x 3
Prime factors of 60 are 2 x 2 x 3 x 5
Prime factors of 72 are 2 x 2 x 2 x 3 x 3
∴ Common prime factors are 2 x 2 x 3

Question (iii)
Two numbers are said to be co-prime numbers if their HCF is –
(a) 2
(b) 3
(c) 0
(d) 1
(d) 1

What is the greatest common factor of 12 and 18? Prealgebra Greatest Common Factor.

Question 2.
Using repeated division method find HCF of the following:

1. 455 and 26
2. 392 and 256
3. 6765 and 610
4. 184, 230 and 276

1. 455 and 26 divisor
Solution:

• Step 1: The larger number should be dividend = 455 & smaller number should be divisor = 26
• Step 2: After 1st division, the remainder becomes new divisor & the previous divisor
• becomes next dividend.
• Step 3: This is done till remainder is zero.
• Step 4: The last divisor is the HCF

The HCF is 13.

2. 392 and 256
256 is smaller, so it is the 1st divisor

∴ HCF = 8

3. 6765 and 610

∴ HCF = 5

4. 184, 230 and 276
First let us take 184 & 230

∴ 46 is the Hi of 184, and 230.
Now the HCF of the first two numbers is the dividend for the third number.

HCF of 184, 230 & 276 is 46

Question 3.
Using repeated subtracting method find HCF of the following:

1. 42 and 70
2. 36 and 80
3. 280 and 420
4. 1014 and 654

1. 42 and 70
Solution:
Let number be m & n
m > n
We do m – n & the result of subtraction becomes new ‘m’. if m becomes less than n, we do n – m and then assign the result as n. We should do this till m = n. When m = n then ‘m’ is the HCF.
42 and 70 now m = 70, n = 42
70 – 42 = 28, now m = 42, n = 28
42 – 28 = 14, now m = 28, n = 14
28 – 14= 14, now m = 14, n = 14
we stop here as m = n
∴ HCF of 42 & 70 is 14

2. 36 and 80
Solution:
36 and 80 m = 80, n = 36
80 – 36 = 44, now n = 44, m = 36
Since n > m, we should do n m
44 – 36 = 8, now n = 8, m = 36
36 – 8 = 28
Similarly, processing, proceeding, we do repeated subtraction
till m = n
28 – 8 = 20
20 – 8 = 12
12 – 8 = 4
8 – 4 = 4 now m = n = 4
∴ HCF is 4

3. 280 and 420
Solution:
Let m = 420, n = 280
m – n = 420 – 280 = 140
now m = 280, n = 140
m – n = 280 – 140 = 140
now m = n = 140
∴ HCF is 140

4. 1014 and 654
Solution:
Let m = 1014, n = 654
m – n = 1014 – 654 = 360
now m = 654, n = 360
m – n = 654 – 360 = 294
now m = 360, n = 294
m – n = 360 – 294 = 66
now m = 294, n = 66
m – n = 294 – 66 = 228
now m = 66, n = 228
n – m = 228 – 66 = 162
now m = 162, n = 66
∴ m – n = 162 – 66 = 96
n – m = 96 – 66 = 30
Similarly 66 – 30 = 36
36 – 30 = 6
30 – 6 = 24
24 – 6 = 18
18 – 6 = 12
12 – 6 = 6
now m = n
∴ HCF of 1014 and 654 is 6

Question 4.
Do the given problems in repeated subtraction method
1. 56 and 12

2. 320,120 and 95

1. 56 and 12
Solution:
56 & 12
Let n = 56 & n = 12
m – n = 56 – 12 = 44
now m = 44, n = 12
m – n = 44 – 12 = 32
m – n = 32 – 12 = 20
m – n = 20 – 12 = 8
n – m = 12 = 8 = 4
m – n = 8 – 44. now m = n
HCF of 56 & 12 is 4

2. 320, 120 and 95
Let us take 320 & 120 first m = 320, n = 120
m – n = 320 – 120 = 200
m = 200, n = 120
∴ m – n = 200 – 120 = 80
120 – 80 = 40
80 – 40 = 40
∴ m = w = 40 → HCF of 320, 120
Now let us find HCF of 40 & 95
m = 95, n = 40
∴ m – n = 95 – 40 = 55
55 – 40 = 15
40 – 15 = 25
25 – 15 = 10
15 – 10 = 5
HCF of 40 & 95 is 5 10 – 5 = 5
HCF of 320, 120 & 95 is 5

Question 5.
On a school trip, 56 girls and 98 boys went to Kanyakumari. They were divided into as many groups as possible so that there were equal numbers of girls and boys in each group. Find the largest group possible? (To find the HCF using repeated division method)
Solution:
56 girls & 98 boys. Now, we need to find HCF of 56 & 98
Using repeated division method. So, we first divide. 98 by 56

14.

Question 6.
Kalai wants to cut identical squares as big as she can, from a piece of paper measuring 168 mm and by 196 mm. What is the length of the side of the biggest square? (To find HCF using repeated subtraction method)
Solution:
Sides are 168 & 196
To find HCF of 168 & 196, we are to use repeated subtraction method.
m = 196, n = 168
m – n = 196 -168 = 28 now n = 28, m = 168
m – n = 168 – 28 = 140 now m = 140, n = = 28
m – n = 140 – 28 = 112 now m = 112, n = 28
m – n = 112 -28 = 84 now m = 84, n = 28
m – n = 84 – 28 = = 56 now m = 56, n = 28
m – n = 56 – 28 = 28
∴ HCF is 28
∴ Length of biggest square is 28

## Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Ex 1.1

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## Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 1 Life Mathematics Ex 1.1

Question 1.
Fill in the blanks:

Question (i)
If 30% of x is 150, then x is ………….
500
Hint:
Given 30% of x is 150

Question (ii)
2 minutes is ………… % to an hour.
3$$\frac{1}{3}$$%
Hint:
Let 2 min be x% of an hour
and 1 hr = 60 min
x% = $$\frac{2}{60}$$ x 100 = $$\frac{200}{3}$$ = $$\frac{10}{3}$$ = 3$$\frac{1}{3}$$
x = 3$$\frac{1}{3}$$

Question (iii)
If x % of x = 25, then x = …………
50
Hint:
Given that x % of x is 25
∴ $$\frac{x}{100}$$ = 25
x2 = 25 x 100 = 2500
∴ x = $$\sqrt{2500}$$ = 50

Question (iv)
In a school of 1400 students, there are 420 girls. The percentage of boys in the school is ………….
70
Hint:
Given total number of students in school = 1400
Number of girls in school = 420
Number of boys in school = 1400 – 420 = 980

% of boys = 70%

Question (v)
0.5252 is ………….. %.
52.52%
Hint:
Given a number, and to express as a percentage, we need to multiply by 100
∴ to express 0.5252 as percentage, we should multiply by 100
∴ 0.5252 x 100 = 52.52%

The Percentage Difference Calculator (% difference calculator) will find the percent difference between two positive numbers greater than 0.

Question 2.
Rewrite each underlined portion using percentage language.

Question (i)
One half of the cake is distributed to the children.
50% of the cake is distributed to the children
Hint:
One half is nothing but $$\frac{1}{2}$$ as percentage, we need to multiply by 100
$$\frac{1}{2}$$ x 100 = 50%

Question (ii)
Aparna scored 7.5 points out of 10 in a competition.
Aparna scored 75% in a competition
Hint:
7.5 points out of 10 is $$\frac{7.5}{10}$$ = 0.75
For percentage, we need to multiply by 100
∴ We get 0.75 x 100 = 75%

Question (iii)
The statue was made of pure silver.
96% students participated in sports
Hint:
Pure silver means there are no other metals
so, 100 out of 100 parts is made of silver = $$\frac{100}{100}$$
∴ to express as percentage, $$\frac{100}{100}$$ x 100% = 100%

Question (iv)
48 out of 50 students participated in sports
96% students participated in sports
Hint:
48 out of 50 students in fraction form is $$\frac{48}{50}$$.
As a percentage, we need to multiply by 100

Question (v)
Only 2 persons out of 3 will be selected in the interview
Only 66$$\frac{2}{3}$$% persons will be selected in the interview.
Hint:
2 out of 3 in fraction form is $$\frac{2}{3}$$ to express as percentage, we need to multiply by 100
$$\frac{2}{3}$$ x 100 = $$\frac{200}{3}$$ = 66$$\frac{2}{3}$$%

Online percentage off calculator tool makes the calculation faster, and it displays the percentage off of the product in a fraction of seconds.

Question 3.
48 is 32% of what number?
Solution:
Let the number required to be found be ‘x’
Given that 32% of x is 48

Question 4.
A bank pays ₹ 240 as interest for 2 years for a sum of ? 3000 deposited as savings. Find the rate of interest given by the bank.
Solution:
The formula for simple interest is given by
Interest (I) = $$\frac{PxRxT}{100}$$ 100
Where P is principal amount; R is rate of interest in percentage; t is time period Substituting given values in formula, we get
P = 3000, t = 2 yrs, I = Rs 240

Question 5.
A Welfare Association has a sports club where 30% of the members play cricket, 28% play volleyball, 22% play badminton and the rest play indoor games. If 30 member play indoor games.

1. How many members are there in the sports club?
2. How many play cricket, volleyball and badminton?

Solution:
Given data
Members who play Cricket = 30% …..(1)
Members who play Vollyball = 28% …..(2)
Members who play Batminton = 22% ……(3)
Members who play outdoor games = (1) + (2) + (3) = 30
Members who play indoor games = 100 – 80 = 20%
Also given 30 members play indoor games
Let total no of members be ‘x’
∴ $$\frac{30}{x}$$ = 20
∴ x = $$\frac{30×100}{20}$$ = 15

1. Total number of members is 150

2. Members who play Cricket = 30% of total = $$\frac{30}{100}$$ x 150

Question 6.
What is 25% of 30% of 400?
Solution:
Required to find 25% of 30% of 400

Question 7.
If the difference between 75% of a number and 60% of the same number is 82.5, then find 20% of the number.
Solution:
Given that 75% of number less 60% of number is 82.5. Let the number be ‘x’
∴ $$\frac{75}{100}$$ × x = 82.5
∴ 0.75 – 0.6.60x = 82.5
∴ 0.15x = 82.5
∴ x = $$\frac{82.5}{0.15}$$ = $$\frac{8250}{15}$$ = 550
Required to find 20% of number ie 20% of x.

Question 8.
If a car is sold for ₹ 2,00,000 from its original price of? 3,00,000 find the percentage for decrease in the value of the car.

Solution:
Original price of car = ₹ 3,00,000
actual selling price of car = ₹ 2,00,000
Decrease in amount from original = 3,00,000 – 2,00,000 = 1,00,000

Question 9.
A number when increased by 18% gives 236. Find the number.
Solution:
Let the number be x. Given that when it is increased by 18%, we get 236.
x + $$\frac{18}{100}$$ = 236
$$\frac{100x+18x}{100}$$ = 236
$$\frac{118}{100}$$x = 236
The number = x = $$\frac{236×100}{118}$$ = 200

Question 10.
A number when decreased by 20% gives 80. Find the number.
Solution:
Let the number be x. Given that when it is increased by 20% we get 80.

Question 11.
A number is increased by 25% and then decreased by 20%. Find the change in that number.
Solution:
Method 1:
Let the number be x. First it is increased by 25%

Method 2:
[to understand, let us assume that number is 100]
So, first when we increase by 25%, we get

We get back 100 ⇒ No change

Question 12.
If the numerator of a fraction is increased by 25% and the denominator is increased by 10%, it becomes $$\frac{2}{5}$$. Find the original fraction.
Solution:
Let the fraction be $$\frac{N}{D}$$ where N is numerator & D is denominator
It is given that numerator is include by 25%

Question 13.
A fruit vendor bought some mangoes of which 10% were rotten. He sold 33 $$\frac{1}{3}$$% of the rest. Find the total number of mangoes bought by him initially, if he still has 240 mangoes with him.

Solution:
Let the number of mangoes bought by fruit seller initially be x.
Given that 10% of mangoes were rotten
∴ Number of rotten mangoes = $$\frac{10}{100}$$ × x
Number of good mangoes = x – no. of rotten mangoes
= x – $$\frac{10}{100}$$x = $$\frac{100x-10x}{100}$$ = $$\frac{90}{100}$$x …(1)
Number of mangoes sold = 33$$\frac{1}{3}$$% of good mangoes = $$\frac{100}{3}$$%
∴ Mangoes sold = $$\frac{100}{3}$$ x $$\frac{90}{100}$$x × $$\frac{1}{100}$$ = $$\frac{30}{100}$$x
Number of mangoes remaining = No. of good mangoes – No. of mangoes sold
From (1) and (2)

Question 14.
A student gets 31% marks in an examination but fails by 12 marks. If the pass percentage is 35%, find the maximum marks of the examination.
Solution:
Let the maximum marks in the exam be ‘x’. Pass percentage is given as 35%
∴ Pass mark = $$\frac{35}{100}$$ × x = $$\frac{35}{100}$$x
Student gets 31 % marks = $$\frac{31}{100}$$ × x = $$\frac{31}{100}$$x
But student fails by 12 marks → meaning his mark is 12 less than pass mark.
∴ $$\frac{31}{100}$$x = $$\frac{35}{100}$$x – 12
∴ $$\frac{35}{100}$$x – $$\frac{31}{100}$$x = 12
∴ $$\frac{35x-31x}{100}$$ = 12 ⇒ $$\frac{4x}{100}$$ = 12
∴ $$\frac{12×100}{4}$$ = 300

Question 15.
The ratio of boys and girls in a class is 5:3. If 16% of boys and 8% of girls failed in an examination, then find the percentage of passed students.
Solution:
Let number of boys be ‘6’ and number of girls be ‘g’
Ratio of boys and girls is given as 5:3
b:g = 5:3 ⇒ $$\frac{b}{g}$$ = $$\frac{5}{3}$$ …..(A)
Failure in boys = 16% = $$\frac{16}{100}$$ x b = $$\frac{16b}{100}$$
Failure in girls = 8% = $$\frac{8}{100}$$ x g = $$\frac{8g}{100}$$
Pass in boys = 100 – 16% = 84% $$\frac{84}{100}$$b …..(1)
Pass in girls = 100 – 8% = 92% = $$\frac{92}{100}$$g
From A, we have $$\frac{b}{g}$$ = $$\frac{5}{3}$$, = adding 1 on both sides, we get
$$\frac{b}{g}$$ + 1 = $$\frac{5}{3}$$ + 1
$$\frac{b+g}{g}$$ = $$\frac{5+3}{3}$$ = $$\frac{8}{3}$$
∴ g = $$\frac{3}{8}$$(b + g)
Similarly b = $$\frac{5}{8}$$(b + g)
Total pass = Pass in girls + Pass in boys
= (1) + (2) = $$\frac{84}{100}$$b + $$\frac{92}{100}$$g

Objective Type Questions

Question 16.
12% of 250 litres is the same as ………. of 150 liters
(a) 10%
(b) 15%
(c) 20%
(d) 30%
(c) 20%
Hint:
12% of 250 = $$\frac{12}{100}$$ x 250 = 30 lit
Percentage : $$\frac{30}{150}$$ x 100 = 20%

Question 17.
If three candidates A, B and C is 3 in a school election got 153, 245 and 102 votes respectively, the percentage of votes for the winner is ……..
(a) 48%
(b) 49%
(c) 50%
(d) 45%
(b) 49%
Hint:
Candidate 1: 153
Candidate 2: 245 – winner [as maximum votes]
Candidate 3: 102
Total votes = 1 + 2 + 3 = 153 + 245 + 102 = 500

Question 18.
15% of 25% of 10000 = ………..
(a) 375
(b) 400
(c) 425
(d) 475
(a) 375
Hint:
15% of 25% of 10000 is
First let us do 25% of 10,000, which is

Next 15% of the above is $$\frac{15}{100}$$ x 2500 = 375

Question 19.
When 60 is subtracted from 60% of a number to give 60, the number is –
(a) 60
(b) 100
(c) 150
(d) 200
(d) 200
Hint:
Let the number be ‘x’.
60% of the number is $$\frac{60}{100}$$ × x = $$\frac{60x}{100}$$
Given that when 60 is subtracted from 60%, we get 60
i.e $$\frac{60}{100}$$x – 60 = 60
∴ $$\frac{60}{100}$$x = 60 + 60 = 120
∴ x = $$\frac{120×100}{60}$$ = 200

Question 20.
If 48% of 48 = 64% of x, then x =
(a) 64
(b) 56
(c) 42
(d) 36
(d) 36
Hint:
Given that 48% of 48 48 = 64% of x

## Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 1 Numbers Ex 1.2

Students can Download Maths Chapter 1 Numbers Ex 1.2 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

## Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 1 Numbers Ex 1.2

Question 1.
Fill in the blanks:
(i) If a number has 5 or 6 digits in it then, its square root will have………digits.
(ii) The value of 180 lies between integers………and……….
(iii) $$\sqrt{10}$$ × $$\sqrt{6}$$ × $$\sqrt{15}$$ =……………
(iv) $$\frac{\sqrt{300}{\sqrt{192}}$$ =…………….
(v) $$\sqrt{65.61}$$ =…………….
Solution:
(i) 3
(ii) 13, 14
(iii) 30
(iv) $$\frac{5}{4}$$
(v) 8.1

Polynomial roots calculator.

Question 2.
Estimate the value of the following square roots to the nearest whole number:
(i) $$\sqrt{440}$$
(ii) $$\sqrt{800}$$
(iii) $$\sqrt{1020}$$
Solution:
(i) We have 20² = 400
21² = 441
∴ $$\sqrt{440}$$  $$\widetilde { – }$$ 21

(ii) We have 28² = 784
29² = 841
∴ $$\sqrt{800}$$ $$\widetilde { – }$$ 28

(iii) We have 31² = 961
32² = 1024
∴ $$\sqrt{1020}$$ $$\widetilde { – }$$ = 32

Question 3.
Find the least number that must be added to 1300 so as to get a perfect square. Also find the square root of the perfect square.
Solution:
We work out the process of finding square root by long division method.
The given number is 1300

So we have 36² < 1300 < 37²
Also 1300 is (469 – 400) = 69 less than 37². So if we add 69 to 1300 it will be perfect square. Hence the required, least number is 69 and the perfect square number is 1300 + 69 = 1369

∴ $$\sqrt{1369}$$ = 37

Question 4.
Find the least number that must be subtracted to 6412 so as to get a perfect square. Also find the square root of the perfect square.
Solution:
Let us work out the process of finding the square root of 6412 by long division method.

The remainder in the last step is 12. Is if 12 be subtracted from the given number the remainder will be zero and the new number will be a perfect square.
∴ The required number is 12.
The square number is 6412 – 12 = 6400
Also $$\sqrt{6400}$$ = 80

Question 5.
Find the square root by long division method.
(i) 17956
(ii) 11025
(iii) 6889
(iv) 1764
(v) 418609
Solution:

Question 6.
Find the square root of the following decimal numbers:
(i) 2.89
(ii) 1.96
(iii) 67.24
(iv) 31.36
(v) 2.0164
(vi) 13.9876
Solution:

Question 7.
Find the square root of each of the following fractions:
(i) $$\frac{144}{225}$$
(ii) 7$$\frac{18}{49}$$
(iii) 6$$\frac{1}{4}$$
(iv) 4$$\frac{25}{36}$$
Solution:

Question 8.
Say True or False:
(i) $$\frac{\sqrt{32}}{\sqrt{8}}=2$$
(ii) $$\sqrt{\frac{625}{1024}}=\frac{25}{32}$$
(iii) $$\sqrt{28}{7}=2\sqrt{7}$$
(iv) $$\sqrt{225}{64}=\sqrt{289}$$
(v) $$\sqrt{1 \frac{400}{441}}=1 \frac{20}{21}$$
Solution:
(i) true
(ii) true
(iii) false
(iv) false
(v) false

Objective Type Questions

Question 9.
$$\sqrt{48}$$ is approximately equal to
(a) 5
(b) 6
(c) 7
(d) 8
Solution:
(c) 7
Hint:
$$\sqrt{49}$$ = 7

Question 10.
$$\sqrt{128}$$ – $$\sqrt{98}$$ + $$\sqrt{18}$$ =
(a) $$\sqrt{2}$$
(b) $$\sqrt{8}$$
(c) $$\sqrt{48}$$
(d) $$\sqrt{32}$$
Solution:
(d) $$\sqrt{32}$$
Hint:
$$\sqrt{128}-\sqrt{98}+\sqrt{18}=8 \sqrt{2}-7 \sqrt{2}+3 \sqrt{2}=4 \sqrt{2}=\sqrt{32}$$

Question 11.
$$\sqrt{22+\sqrt{7+\sqrt{4}}}=$$
(a) $$\sqrt{25}$$
(b) $$\sqrt{33}$$
(c) $$\sqrt{31}$$
(d) $$\sqrt{29}$$
Solution:
(a) $$\sqrt{25}$$
Hint:
$$\sqrt{22+\sqrt{7+\sqrt{4}}}=\sqrt{22+\sqrt{7+2}}=\sqrt{22+3}=\sqrt{25}$$

Question 12.
The number of digits in the square root of 123454321 is
(a) 4
(b) 5
(c) 6
(d) 7
Solution:
(b) 5
Hint:
$$=\frac{n+1}{2}=\frac{10}{2}=5$$