Class 8

Students can Download Maths Chapter 2 Algebra Ex 2.3 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 2 Chapter 2 Algebra Ex 2.3

Question 1.
Fill in the blanks:

Question (i)
X – axis and Y – axis intersect at ……..
Answer:
Origin (0, 0)

Question (ii)
The coordinates of the point in third quadrant are always ……….
Answer:
negatives

Question (iii)
(0, -5) point lies on ………. axis.
Answer:
y – axis.

Question (iv)
The x coordinate is always ……… on the y – axis.
Answer:
Zero

Question (v)
Coordinates are the same for a line parallel to Y – axis.
Answer:
X.

Question 2.
Say True or False:

Question (i)
(-10, 20) lies in the second quadrant.
Answer:
True
Hint:
(-10, 20)
x = – 10,
y = 20
∴ (- 10, 20) lies in second quadrant – True

Question (ii)
(-9, 0) lies on the x – axis.
Answer:
True
Hint:
(-9, 0) on x – axis, Y – coordinate is always zero.
∴ (-9, 0) lies on x – axis – True

Question (iii)
The coordinates of the origin are (1, 1).
Answer:
False
Hint:
Coordinate of origin is (0, 0), not (1, 1). Hence – False

Question 3.
Find the quadrants without plotting the points on a graph sheet.
(3, -4), (5, 7), (2, 0), (- 3, – 5), (4, – 3), (- 7, 2), (- 8, 0), (0,10), (- 9, 50).
Solution:

• If x & y coordinate are positive – I quad
• If x is positive, y is negative – IV quad
• If x is negative, y is positive – II quad
• If both are negative, then – III quad

Question 4.
Plot the following points in a graph sheet.
A(5, 2), B(- 7, – 3), C(- 2, 4), D(- 1, – 1), E(0, – 5), F(2, 0), G(7, – 4), H(- 4, 0), 1(2,3), J(8, – 4) K (0, 7).
Solution:

Question 5.
Use the grid graph to determine the coordinates where each figure is located.

a) Star ……..
b) Bird ……..
c) Red Circle ……..
d) Diamond ……..
e) Triangle ……..
f) Ant ……..
g) Mango ……..
h) Housefly ……..
i) Medal ……..
j) Spider ……..
Solution:
a) Star (3, 2)
b) bird (-2, 0)
c) Red Circle (-2, -2)
d) Diamond (-2, 2)
e) Triangle (-1, -1)
f) Ant (3, -1)
g) Mango (0, 2)
h) Housefly (2, 0)
i) Medal (-3, 3)
j) Spider (0, -2)

Students can Download Maths Chapter 2 Algebra Ex 2.2 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 2 Chapter 2 Algebra Ex 2.2

Question 1.
Fill in the blanks:

Question (i)
The solution of the equation ax + b = 0 is ………
Answer:
– \(\frac{b}{a}\)
Solution:
ax + b = 0
ax = – b
∴ x = – \(\frac{b}{a}\)

Question (ii)
If a and b are positive integers then the solution of the equation ax = b has to be always ………
Answer:
positive
Hint:
Since a & b are positive integers,
The solution to the equation ax = b is x = – \(\frac{b}{a}\) is also positive.

Question (iii)
One-sixth of a number when subtracted from the number itself gives 25. The number is ……..
Answer:
30
Hint:
Let the number be x.
As per question, when one sixth of number is subtracted from itself it gives 25
x – \(\frac{x}{6}\) = 25
∴ \(\frac{6x-x}{6}\) = 25
∴ \(\frac{5x}{6}\) = 25
∴ x = \(\frac{25×6}{3}\) = 5 x 6 = 30

Question (iv)
If the angles of a triangle are in the ratio 2 : 3 : 4 then the difference between the greatest and the smallest angle is
Answer:
40°
Hint:
Given angles are in the ratio 2 : 3 : 4
Let the angles be 2x, 3x & 4x
Since sum of the angles of a triangle is 180°,
We get
2x + 3x + 4x = 180
∴ 9x = 180
∴ X = \(\frac{180}{9}\) = 20°
∴ The angles are 2x = 2 x 20 = 40°
3x = 3 x 20 = 60°
4x = 4 x 20 = 80°
∴ Difference between greatest & smallest angle is
80° – 40° = 40°

Question (v)
In an equation a + b = 23. The value of a is 14 then the value of b is ……..
Answer:
b = 9
Hint:
Given equation is a + b = 23
a = 14
14 + b = 23
b = 23 – 14 = 9
b = 9

Question 2.
Say True or False

Question (i)
“Sum of a number and two times that number is 48” can be written as y + 2y = 48
Answer:
True
Hint:
Let the number be ‘y’
Sum of number & two times that number is 48
Can be written as y + 2y = 48 – True

Question (ii)
5(3x + 2) = 3(5x – 7) is a linear equation in one variable.
Answer:
True
Hint:
5 (3x + 2) = 3 (5x – 7) is a linear equation in one variable – ‘x’ – True

Question (iii)
x = 25 is the solution of one third of a number is less than 10 the original number.
Answer:
False
Hint:
One third of number is 10 less than original number.
Let number be ‘x’ Therefore let us frame the equation
\(\frac{x}{3}\) = x – 10
∴ x = 3x – 30
3x – x = 30
2x = 30
x = 15 is the solution

Question 3.
One number is seven times another. If their difference is 18, find the numbers.
Solution:
Let the numbers be x & y
Given that one number is 7 times the other & that the difference is 18.
Let x = 7y
also, x – y = 18 (given)
Substituting for x in the above
We get 7y – y = 18
∴ 6y = 18
y = \(\frac{18}{6}\) = 3
x = 7y = 7 x 3 = 21
The number are 3 & 21

Question 4.
The sum of three consecutive odd numbers is 75. Which is the largest among them?
Solution:
Given sum of three consecutive odd numbers is 75
Odd numbers are 1, 3, 5,1,9, 11, 13,……..
∴ The difference between 2 consecutive odd numbers is always 2. or in other words, if one odd number is x, the next odd number would be x + 2 and the next number would be x + 2 + 2 = x + 4
i.e x + 4
Since sum of 3 consecutive odd nos is 75
∴ x + x + 2 + x + 4 = 75
3x + 6 = 75 ⇒ 3x = 75 – 6
3x = 69
x = \(\frac{69}{3}\) = 23
The odd numbers are 23, 23 + 2, 23 + 4
i.e 23, 25, 27
∴ Largest number is 27.

Question 5.
The length of a rectangle is \(\frac{1}{3}\)rd of its breadth. If its perimeter is 64 m, then find the length and breadth of the rectangle.
Solution:

Let length & breadth of rectangle be l and ‘6’ respectively
Given that length is \(\frac{1}{3}\) of breadth,
∴ l = \(\frac{1}{3}\) x b ⇒ l = \(\frac{b}{3}\) ⇒ b = 3l …..(1)
Also given that perimeter is 64 m
Perimeter = 2 x (l + b)
2 x l + 2 x b = 64
Substituting for vahie of b from (1), we get
2l + 2(3l) = 64
∴ 2l + 6l = 64
8l = 64
∴ l = \(\frac{64}{8}\) = 8 m
b = 3l = 3 x 8 = 24 m
length l = 8 m & breadth h = 24 m

Question 6.
A total of 90 currency notes, consisting only of X5 and ?10 denominations, amount to ₹ 500. Find the number of notes in each denomination.
Solution:
Let the number of ₹ 5 notes be ‘x’
And number of ₹ 10 notes be ly ’
Total numbers of notes is x + y = 90 (given)
The total value of the notes is 500 rupees.
Value of one ₹ 5 rupee note is 5
Value of x ₹ 5 rupee notes is 5 × x = 5x
Value ofy ₹ 10 rupee notes is 10 × y = 10y
∴ The total value is 5x + 10y which is 500
we have 2 equations:
x + y = 90
5x + 1oy = 500
Multiplying both sides of (1) by 5, we get
5 × x + 5 x y = 90 x 5
5x + 5y = 450
Subtracting (3) from (2), we get

∴ y = \(\frac{50}{5}\) = 10
Substitute y = 10 in equation (1)
x + y = 90 ⇒ x + 10 = 90
⇒ x = 90 – 10 ⇒ x = 80
There are ₹ 5 denominations are 80 numbers and ₹ 10 denominations are 10 numbers

Question 7.
At present, Thenmozhi’s age is 5 years more than that of Murali’s age. Five years ago, the ratio of Thenmozhi’s age to Murali’s age was 3:2. Find their present ages.
Solution:
Let present ages ofThenmozhi & Murali be l’ & ‘m’
Given that at present
Thenmozhi’s age is 5 years more than Murali
∴ t = m + 5
5 years ago, Thenmozhi’s age would be t – 5
& Murali’s age would be m – 5
Ratio of their ages is given as 3 : 2
∴ \(\frac{t-5}{m-5}\) = \(\frac{3}{2}\) [∴ By cross multiplication]

2(t – 5) = 3(m – 5)
2 x t – 2 x 5 = 3 x m – 3 x 5 ⇒ 2t – 10 = 3m – 15
Substituting for t from (1)
2(m + 5) – 10 = 3m – 15
2m + 10 – 10 = 3m – 15
2m = 3m – 15
3m – 2m = 15
m = 15
t = m + 5 = 15 + 5 = 20
∴ Present ages of Thenmozhi & Murali are 20 & 15

Question 8.
A number consists of two digits whose sum is 9. If 27 is subtracted from the original number, its digits are interchanged. Find the original number.
Solution:
Let the units/digit of a number be ‘u’ & tens digit of the number be ‘t’
Given that sum of it’s digits is 9
∴ t + u = 9 ….(1)
If 27 is subtracted from original number, the digits are interchanged
The number is written as 10t + u
[Understand: Suppose a 2 digit number is 21,
it can be written as 2 x 10 + 1
∴ 32 = 3 x 10 + 2
45 = 4 x 10 + 5
tu = t x 10 + u = 10t + u]
Given that when 27 is subtracted, digits interchange
10t + u – 27 = 10u + t (number with interchanged digits)
∴ By transposition & bringing like variables together
10t – t + 10u =27
∴ 9t – 9u = 27
Dividing by ‘9’ throughout, we get
\(\frac{9t}{9}\) – \(\frac{9u}{9}\) = \(\frac{27}{9}\) ⇒ t – u = 3 ….(2)
Solving (1) & (2)

t = \(\frac{12}{2}\) = 6
∴ u = 3
t = 6 substitute in (1)
t + u = 9
⇒ 6 + u = 9
⇒ u = 9 – 6 = 3
Hence the number is 63.

Question 9.
The denominator of a fraction exceeds its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, we get \(\frac{3}{2}\). Find the original fraction.
Solution:
Let the numerator & denominator be ‘n’ & ‘d’
Given that denominator exceeds numerator by 8
∴ d = n + 8
If numerator increased by 17 & denominator decreased by 1,
it becomes (n + 17) & (d – 1), fraction is \(\frac{3}{2}\).
i.e = \(\frac{n+17}{d-1}\) = \(\frac{3}{2}\) by cross multiplying, we get
\(\frac{n+17}{d-1}\) = \(\frac{3}{2}\)
2(n + 17) = 3(d – 1)
2n + 2 x 17 = 3d – 3

∴ 34 + 3 = 3d – 2n
∴ 3d – 2n = 37
Substituting eqn. (1) in (2), we get,
3 x (n + 8) – 2n = 37
3n + 3 x 8 – 2n = 37

∴ n = 37 – 24 = 13
d = n + 8 = 13 + 8 = 21
The fraction is \(\frac{n}{d}\) = \(\frac{13}{21}\)

Question 10.
If a train runs at 60 km/hr it reaches its destination late by 15 minutes. But, if it runs at 85 kmph it is late by only 4 minutes. Find the distance to be covered by the train.
Solution:
Let the distance to be covered by train be ‘d’
Using the formula, time take (t) = \(\frac{Distance}{Speed}\)
Case 1:
If speed = 60 km/h
The time taken is 15 minutes more than usual (t + \(\frac{15}{60}\))
Let usual time taken be ‘t’ hrs.
Caution:
Since speed is given in km/hr, we should take care to maintain all units such as time should be in hour and distance should be in km.
Given that in case 1, it takes 15 min. more
15 m = \(\frac{15}{60}\) hr = \(\frac{1}{4}\) hr.
∴ Substituting in formula
\(\frac{Distance}{Speed}\) = time
∴ \(\frac{d}{60}\) = t + \(\frac{1}{4}\)
Since usually it takes ‘t’ hr, but when running at 60 km/h, it takes 15 min (\(\frac{1}{4}\)) extra.
Multiplying by 60 on both sides
d = 60 x t + 60 x \(\frac{1}{4}\) = 60t + 15 …..(1)
Case 2:
Speed is given as 85 km/h
Time taken is only 4 min (\(\frac{4}{60}\) hr) more than usual time
time taken = (t + \(\frac{1}{15}\)) hr. (\(\frac{4}{60}\) = \(\frac{1}{15}\))
Using the formula,
\(\frac{Distance}{Speed}\) = time
∴ \(\frac{d}{85}\) = t + \(\frac{1}{15}\)
Multiplying by 85 on both sides
\(\frac{d}{85}\) x 85 = 85 x t + 85 x \(\frac{1}{15}\)
∴ d = 85 t + \(\frac{17}{3}\)
From (1) & (2), we will solve for ‘t’
Equating & eliminating ‘d we get

By transposing, we get
15 – \(\frac{17}{3}\) = 85t – 60t
\(\frac{45-17}{3}\) = 25t
∴ 25t = \(\frac{28}{3}\)
∴ t = \(\frac{28}{3×25}\) = \(\frac{28}{75}\) hr (\(\frac{28}{75}\) x 60 = 22.4 min)
Substituting this value of ‘t’ in eqn. (1), we get
d = 60t+ 15 = 60 x \(\frac{28}{75}\) + 15 = \(\frac{1680}{75}\) + 15 = 22.4 + 15
= 37.4 km

Objective Type Questions

Question 11.
Sum of a number and its half is 30 then the number is ……..
(a) 15
(b) 20
(c) 25
(d) 40
Answer:
(b) 20
Hint:
Let number be V
half of number is \(\frac{x}{2}\)
Sum of number and it’s half is given by
x + \(\frac{x}{2}\) = 30 [Multiplying by 2 on both sides]
2x + x = 30 x 2
3x = 60
x = \(\frac{60}{3}\) = 20

Question 12.
The exterior angle of a triangle is 120° and one of its interior opposite angle 58°, then the other opposite interior angle is ………
(a) 62°
(b) 72°
(c) 78°
(d) 68°
Answer:
(a) 62°

Hint:
As per property of ∆, exterior angle is equals to sum of interior opposite angles Let the other interior angle to be found be ‘x’
∴ x + 58 = 120°
∴ x = 120 – 58 = 62°

Question 13.
What sum of money will earn ₹ 500 as simple interest in 1 year at 5% per annum?
(a) 50000
(b) 30000
(c) 10000
(d) 5000
Answer:
(c) 10000
Hint:
Let sum of money be ‘P’
Time period (n) is given as 1 yr.
Rate of simple interest (r) is given as 5% p.a
∴ As per formula for simple interest.
S.I = \(\frac{Pxrxn}{100}\) = \(\frac{px5x1}{100}\)
P x 5 x 1 = 500 x 100
∴ P = \(\frac{500×100}{5}\) = 100 x 100 = 10,000

Question 14.
The product of LCM and HCF of two numbers is 24. If one of the number is 6, then the other number is ………
(a) 6
(b) 2
(c) 4
(d) 8
Answer:
(c) 4
Hint:
Product of LCM & HCF of 2 numbers is always product of the numbers. [this is property]
Product of LCM & HCF is given as 24.
∴ Product of the 2 nos. is 24
Given one number is 6.
Let other number be ‘x’
∴ 6 × x = 24
∴ x = \(\frac{24}{6}\) = 4

Students can Download Maths Chapter 1 Life Mathematics Ex 1.4 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 2 Chapter 1 Life Mathematics Ex 1.4

Miscellaneous Practise Problems

Question 1.
Nanda’s marks in 3 Math tests Tl, T2 and T3 were 38 out of 40,27 out of 30 and 48 out of 50. In which test did he do well? Find his overall percentage in all the 3 tests.
Solution:
Nanda’s marks are as follows
In Test 1 → T₁ \(\frac{38}{40}\)
Test 2 → T₂ \(\frac{27}{30}\)
Test 3 → T₃ \(\frac{48}{50}\)
For percentage, multiply by 100
T₁ = \(\frac{38}{40}\) x 100 = 95%
T₂ = \(\frac{27}{30}\) x 100 = 90%
T₃ = \(\frac{48}{50}\) x 100 = 96%
Hence, he has scored highest percentage in test 3
∴ He is done well in T3
Overall percentage is the average of the 3 percentages.
i.e \(\frac{90+95+96}{3}\) = \(\frac{281}{50}\) = 93\(\frac{2}{3}\)%

Question 2.
Sultana bought the following things from a general store. Calculate the total bill amount to be paid by her.
(i) Medicines costing ₹ 800 with GST @ 5% ………..

(ii) Cosmetics costing ₹ 650 with GST @ 12% ………….

(iii) Cereals costing ₹ 900 with GST @ 0% …………

(iv) Sunglass costing ₹ 1750 with GST @ 18 % ………..

(v) Air Conditioner costing ₹ 28500 with GST @ 28% ………..

Solution:

(i) Medicine : bill amount is 800(1 + \(\frac{5}{100}\)) = 800 x \(\frac{105}{100}\) = 840
(ii) Cosmetics: Bill amount is 650(1 + \(\frac{12}{100}\)) = 650 x \(\frac{112}{100}\) = 728
(iii) Cereals : Bill amount is 900(1 + \(\frac{0}{100}\)) = 900
(iv) Sunglass: bill amount is 1750(1 + \(\frac{18}{100}\)) = 1750 x \(\frac{118}{100}\) = 2065
(v) AC : Bill amount is 28500(1 + \(\frac{28}{100}\)) = 28500 x \(\frac{128}{100}\) = 36480
∴ Total bill amount = 840 + 728 + 900 + 2065 + 36480
= ₹ 41,013 (total bill amount)

Question 3.
P’s income is 25% more than that of Q. By what percentage is Q’s income less than P’s?
Solution:
Let Q’s income be 100.
P’s income is 25% more than that of Q
∴ P’s income = 100 + \(\frac{25}{100}\) x 100 = 125
Q’s income is 25 less than that of P
In percentage terms, Q’s income is less than P’s with respect to P’s income is
\(\frac{P-Q}{P}\) x 100 = \(\frac{125-100}{125}\) x 100 = \(\frac{25}{100}\) x 100 = 20%

Question 4.
Gopi sold a laptop at 12% gain. If it had been sold for ₹ 1200 more, the gain would have been 20%. Find the cost price of the laptop.

Solution:
Let the cost price of the laptop be ‘x’
Gain = 12%

If the selling price was 1200 more
i.e \(\frac{112}{100}\)x +1200, the gain is 20%
i.e new selling price = x(1 + \(\frac{2}{100}\))
= \(\frac{112}{100}\)x + 1200 = x(100 + \(\frac{20}{100}\)) = \(\frac{112}{100}\)x
∴ 1200 = \(\frac{120}{100}\)x = \(\frac{112}{100}\)x = \(\frac{8}{100}\)x
∴ x = \(\frac{120×100}{8}\) = 15000
Cost price of the laptop is ₹ 15000

Question 5.
Vaidegi sold two sarees for ₹ 2200 each. On one she gains 10% and on the other she loses 12%. Calculate her gain or loss percentage in the sales.

Solution:
Saree 1:
The selling price is ₹ 2200, let cost price be CP₁, gain is 10%
Cost price ₹ Using the formula

Saree 2:
The selling price is 2200, let cost price be CP₂, loss is given as 12%. We need to find CP₂ using the formula as before,

Cost price of both together is CP₁ + CP₂
= 2000 + 2500 = 4500 ……(1)
Selling price of both together is 2 x 2200 = 4400 …….(2)
Since net selling price is less than net cost price, there is a loss.
loss% = \(\frac{loss}{cost price}\) x 100
Loss = Net cost price – Net selling price
(1) – (2) = 4500 – 4400 = 100
∴ loss% = \(\frac{100}{4500}\) x 100 = \(\frac{100}{45}\) = \(\frac{20}{9}\) = 2\(\frac{2}{9}\)%
= 2\(\frac{2}{9}\)% loss

Question 6.
A sum of money becomes ₹ 18000 in 2 years and ₹ 40500 in 4 years on compound interest. Find the sum.
Solution:
Let the sum of money be ‘P’
Given that sum ‘P’ becomes 18000 in 2yrs.
i.e Amount (A) = P(1 + \(\frac{r}{100}\))ⁿ
18000 = P(1 + \(\frac{r}{100}\))²
(1 + \(\frac{r}{100}\))² = \(\frac{18000}{P}\) ⇒ (1 + \(\frac{r}{100}\))⁴ = (\(\frac{18000}{P}\))² ……(1)
Also given that sum ‘P’ becomes46500 in 4 yrs.
i.e Amount (A) = P(1 + \(\frac{r}{100}\))ⁿ
40500 = P(1 + \(\frac{r}{100}\))⁴
Substituting (1) in (2), we get

Question 7.
Find the difference in the compound interest on ₹ 62500 for 1\(\frac{1}{2}\) years at 8% p.a compounded annually and when compounded half-yearly.
Solution:
Case 1:
P = ₹ 62500
n = 1\(\frac{1}{2}\)yrs. (a\(\frac{b}{c}\)) formula
r = 8% Compound annully
CI = A – P

CI = A – P = 70200 – 62500 = 7700 ……(1)
CAse 2:
P = ₹ 62500
n = 1\(\frac{1}{2}\)yrs.
r = 8% p.a when compound half yearly
r = 4% compound half yearly

= 70304 – 62500 = ₹ 7804
Difference between case 1 & case 2 is (2) – (1)
∴ (2) – (1) = 7804 – 7700 = ₹ 104

Challenging Problems

Question 8.
If the first number is 20% less than the second number and the second number is 25% more than 100, then find the first number.
Solution:
Second number is 25% more than 100
∴ 2nd number is 100 + \(\frac{25}{100}\) x 100= 125
First number is 20% less than second no.

1st number is 100

Question 9.
A shopkeeper gives two successive discounts on an article whose marked price is ? 180 and selling price is ? 108. Find the first discount percent if the second discount is 25%.
Solution:
Marked price is given as ? 180
Let 1^st discount be d₁% = ? (to find)
2nd discount be d₂% = 25%
Selling price is 108 (given)
Price after 1^st discount = 180(1 – \(\frac { d_{ 1 } }{ 100 } \)) = P₁
Price after 2^nd discount = P₁(1 – \(\frac { d_{ 2 } }{ 100 } \)) = 108
Substituting for P₁ from (1), we get

∴ 1 – \(\frac { d_{ 1 } }{ 100 } \) = \(\frac{4}{5}\)
∴ \(\frac { d_{ 1 } }{ 100 } \) = 1 – \(\frac{4}{5}\) = \(\frac{1}{5}\)
∴ d₁ = \(\frac{1}{5}\) x 100 = 20%

Question 10.
A man bought an article on 30% discount and sold it at 40% more than the marked price. Find the profit made by him.
Solution:
Let marked price be ‘P’
Discounted price = P(1 – \(\frac{d}{100}\)) where d is the discount %
∴ Discounted price = P(1 – \(\frac{30}{100}\)) = \(\frac{70}{100}\)P → this is the cost price.
Selling price = 40% more than marked price

Question 11.
Find the rate of compound interest at which a principal becomes 1.69 times itself in 2 years.
Solution:
Let principal be ‘P’
Amount is given to be 1.69 times principal
i.e 1.69 P
Time period is 2yrs. = (n)
Rate of interest = r = ? (required)
Applying the formula,
Amount = Principal (1 + \(\frac{r}{100}\))ⁿ
Substituting, 1.69 P = P (1 + \(\frac{r}{100}\))²
∴ (1 + \(\frac{r}{100}\))² = \(\frac{1.69P}{P}\)
Taking square root on both sides, we get
\(\sqrt{1.69}\) = 1 + \(\frac{r}{100}\)
∴ 1 + \(\frac{r}{100}\) = 1.3
∴ \(\frac{r}{100}\) = 1.3
r = 30%
∴ rate of compound interst is 30%

Question 12.
The simple interest on a certain principal for 3 years at 10% p.a is ₹ 300. Find the compound interest accrued in 3 years.
Solution:
Let principal be ‘P’
Rate of interest is 10% p.a (r)
Time period (n) = 3 yrs.
Formula for simple interest
∴ P = \(\frac{300×100}{10×3}\) = 1000
Compound Interest = CI = A – P
Amount (A) = P (1 + \(\frac{r}{100}\))ⁿ
= 1000(1 + \(\frac{10}{100}\))³ = 1000 x (\(\frac{10o+10}{100}\))³
= 1000 x \(\frac{110}{100}\) x \(\frac{110}{100}\) x \(\frac{110}{100}\) = 1331
Compound Interest (Cl) = Amount – Principal
= 1331 – 1000 = 331
Compound Interest = ₹ 331

Students can Download Maths Chapter 2 Algebra Additional 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 2 Chapter 2 Algebra Additional Questions

Question 1.
A bus runs constantly at a speed of 60 km/hr. Draw a time-distance graph for the situation. Also find the

1. Time taken to cover 360 km.
2. Distance covered in 4\(\frac{1}{2}\) hours.

Solution:
Given the bus runs constantly at a speed of 60 km/hr.
i.e. for 1 hour = 60 km
2 hours = 2 x 60 = 120 km
3 hours = 3 x 60 = 180 km
We can tabulate as below,

Take a suitable scale:

1. Mark the number of hours on the x – axis
2. Mark the distance in km on the y – axis
3. Plot the points (1, 60), (2,120), (3, 180), (4, 240) and (5, 300)
4. Join the points and get a straight line from the graph.
   • Time taken to cover 360 km is 6 hours.
   • Distance covered in 4\(\frac{1}{2}\) hours is

Question 2.
A company dealing with finance gives 15% simple interest on deposits made by senior citizens. Illustrate by a graph the relation between the deposit and the interest gained. Use the graph to complete the following.

1. The annual interest obtainable for investment of ₹ 450
2. The amount a senior citizen has to invest to get an annual simple interest of ₹ 100.

Solution:
Using the formula for calculating the simple interest, the following table of values is prepared.


These are the points to be plotted in the graph sheet, let us take the deposits along x – axis and annual simple interest along y – axis.
We choose the scale as follows.
Then we plot the points and draw the straight line.
From the graph we find.

1. Corresponding to ₹ 450 on the x – axis, we get the interest as ₹ 67.5 on the y – axis.
2. Corresponding to ₹ 100 on the y – axis we get the deposit as ₹ 666 on the x – axis.

Question 3.
The following table gives the quantity of petrol and its cost

Plot the graph.
Solution:

1. Take a suitable scale on both the axes. Here we take on the x – axis
   1 cm = 1 litre, on the y axis 1 cm = 70 rupees.
2. Mark number of litres of petrol along the x – axis
3. Mark the cost of petrol along the y – axis.
4. Plot the points (1, 70), (2, 140), (3, 210), (4, 280), (5, 350), (6, 420) and (7,490)
5. Join the points.

The graph can help us to estimate few more things also.

Students can Download Maths Chapter 1 Life Mathematics Additional 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.

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

Additional Questions and Answers

Question 1.
Fill in the blanks

Question (a)
Percent means ……….
Answer:
Per hundred or out of hundred.

Question (b)
Percent is useful in ………
Answer:
Comparing quantities easily

Question (c)
The formula to find the increased quantity ………
Answer:
I = (1 + \(\frac{x}{100}\))

Question (d)
The formula to find the decreased quantity ………
Answer:
D = (1 + \(\frac{x}{100}\))

Question (e)
Gain or profit % ……..
Answer:
(\(\frac{Profit}{C.P}\) x 100)%

Question (f)
Loss % = ………..
Answer:
(\(\frac{Loss}{C.P}\) x 100)%

Question (g)
S.P = ………. (if gain % is given)
Answer:

Question (h)
C.P = ……… (if gain % is given)
Answer:

Question (f)
S.P = ………. (if loss % is given)
Answer:

Question (h)
C.P = ………. (if loss % is given)
Answer:

Question (k)
Selling price = Marked price – …………
Answer:
Discount

Question (l)
Cost price = Cost price + ……….
Answer:
Over head expenses

Question 2.
If y% of ₹ 1000 is 600, find the value ofy.
Solution:
y% of 1000 = 600
\(\frac{y}{100}\) x 1000 = 600
y = \(\frac{600}{10}\)
y = 60

Question 3.
A number when decreased by 10% becomes 900. Then find the number.
Solution:
Let the number be ‘x’
Given x – \(\frac{10}{100}\)x = 900
\(\frac{100x-10x}{100}\) = 900
\(\frac{90x}{100}\) = 900
x = \(\frac{900×100}{90}\) = 1000

Question 4.
If the population in a city has increased from 5,00,000 to 7,00,000 in a year, find the percentage increase in population.
Solution:
Increase in population = 7,00,000 – 5,00,000 = 2,00,000
Percentage increase in population = \(\frac{2,00,000}{5,00,000}\) x 100 = 40%

Question 5.
If the selling price of a refrigerator is equal to \(\frac{10}{8}\) of its cost price, then find the gain/ profit percent.
Solution:
Let the C.P of the refrigerator be x
S.P = \(\frac{10}{8}\)x
Profit = S.P – C.P = \(\frac{10}{8}\)x – x = \(\frac{10x-8x}{8}\) = \(\frac{2x}{8}\)

Question 6.
Karnan bought a dishwasher for ₹ 32,300 and paid ₹ 2700 for its transportation. Then he sold it for X 38,500. Find his gain or loss percent.
Solution:
Total C.P of the dishwasher = C.P + Overhead expenses.
= ₹ 32300 + ₹ 2700 = ₹ 35000
S.P = ₹ 38500
Therefore, we find S.P > C.P

Question 7.
The value of a car 2 years ago was ₹ 1,40,000. It depreciates at the rate of 4% p.a. Find its present value.
Solution:
Depreciated value = P(1 + \(\frac{r}{100}\))ⁿ = 1,40,000(1 – \(\frac{4}{100}\))²
= 1,40,000(\(\frac{96}{100}\)) x (\(\frac{96}{100}\)) = ₹ 1,29,024

Question 8.
Find the difference in C.I and S.I for P = ₹ 10,000, r = 4% p.a, n = 2 years.
Solution:
C.I – S.I = P(\(\frac{r}{100}\))² = 10,000(\(\frac{4}{100}\))²
= 10,000 x \(\frac{4}{100}\) x \(\frac{4}{100}\) = ₹ 16

Question 9.
Find the C.I for the given Principal = ₹ 8,000, r = 5% p.a, n = 2 years
Amount A = P(1 + \(\frac{r}{100}\))ⁿ = 8000(1 + \(\frac{5}{100}\))²
= 8000 x \(\frac{105}{100}\) x \(\frac{105}{100}\)
= 8000 x \(\frac{21}{20}\) x \(\frac{21}{20}\)
A = ₹ 8820
Cl = A – P = 8820 – 8000 = 820

Question 10.
Find the S.I for the principal P = ₹ 16,000, r = 5% p.a, n = 3 years
Solution:
P = ₹ 16,000, n= 3 years, r = 5%
S.I = \(\frac{Pnr}{100}\) = \(\frac{16000x3x5}{100}\) = ₹ 2400

Students can Download Maths Chapter 1 Life Mathematics Ex 1.3 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 2 Chapter 1 Life Mathematics Ex 1.3

Question 1.
Fill in the blanks

Question (i)
The compound interest on ₹ 5000 at 12% p.a for 2 years compounded annually is ………..
Answer:
₹ 1272
Hint:
Compound Interest (Cl) formula is
Cl = Amount – Principal
Amount = A (1 + \(\frac{r}{100}\))ⁿ = 5000 (1 + \(\frac{12}{100}\))²
= 5000 (1 + \(\frac{112}{100}\))² = 6272
∴ Cl = 6272 – 5000 = ₹ 1272

Question (ii)
The compound interest on ₹ 8000 at 10% p.a for 1 year, compounded half yearly is …………
Answer:
₹ 820
Hint:
Compound interest (CI) = Amount – Principal
∴ Amount = P (1 + \(\frac{r}{100}\))²ⁿ [2n as it is compounded half yearly]
r = 10% p.a, for half yearly r = 1 + \(\frac{10}{2}\) = 5
A = 8000 (1 + \(\frac{5}{100}\))^2×1 = 8000 x (\(\frac{105}{100}\))² = 8820
CI = Amount – principal
= 8820 – 8000 = ₹ 820

Question (iii)
The annual rate of growth in population of a town is 10%. If its present population is 26620, the population 3 years ago was ………..
Answer:
₹ 20,000
Hint:
Rate of growth of population r = 10%; Present population = 26620
Let population 3 years ago be x
∴ Applying the formula for population growth which is similar to compound interest,
26620 = x (1 + \(\frac{r}{100}\))³
∴ 26620 = x (1 + \(\frac{10}{100}\))³ = x (\(\frac{110}{100}\))³
∴ x = 26620 x (\(\frac{110}{100}\))³
= ₹ 20,000
The population 3 years ago was ₹ 20,000

Questions (iv)
The amount if the compound interest is calculated quarterly, is found using the formula ………….
Answer:
A = P (1 + \(\frac{r}{400}\))⁴ⁿ
Hint:
Quarterly means 4 times in a year.
∴ The formula for compound interest is
A = P (1 + \(\frac{r}{400}\))⁴ⁿ

Question (v)
The difference between the S.I and C.I for 2 years for a principal of ₹ 5000 at the rate of interest 8% p.a is …….
Answer:
₹ 32
Hint:
Difference between S.I & C.I is given by the formula
CI – SI = P (\(\frac{r}{100}\))²
Principal (P) = 5000, r = 8% p.a
∴ CI – SI = 5000 (\(\frac{8}{100}\))² = 5000 x \(\frac{8}{100}\) x \(\frac{8}{100}\) = ₹ 32

Question 2.
Say True or False

Question (i)
Depreciation value is calculated by the formula P (1 – \(\frac{r}{100}\))ⁿ
Answer:
True
Hint:
Depreciation formula is P (1 – \(\frac{r}{100}\))ⁿ

Question (ii)
If the present population of a city is P and it increases at the rate of r % p.a, then the population n years ago would be P (1 + \(\frac{r}{100}\))ⁿ.
Answer:
False
Hint:
Let the population ‘n’ yrs ago be ‘x’
Present population (P) = x × (1 + \(\frac{r}{100}\))ⁿ
x = \(\frac { P }{ (1+\frac { r }{ 100 } )^{ n } } \)

Question (iii)
The present value of a machine is ₹ 16800. It depreciates @25% p.a. Its worth after 2 years is ₹ 9450.
Answer:
True
Hint:
Present value of machine = ₹ 16800
Depreciation rate (n) = 25%
Value after 2 years = P (1 – \(\frac{r}{100}\))ⁿ = 16800 (1 – \(\frac{25}{100}\))²
= 16800 x (1 – \(\frac{1}{4}\))² = 16800 x \(\frac{3}{4}\) x \(\frac{3}{4}\) = 9450

Question (iv)
The time taken for ₹ 1000 to become ₹ 1331 @20% p.a compounded annually is 3 years.
Answer:
False
Principal money = 1000
rate of interest Amount = 20%
Amount = 1331, applying in formula we get
A = (1 + \(\frac{r}{100}\))ⁿ
1331 = 1000(1 + \(\frac{r}{100}\))ⁿ
∴ \(\frac{1331}{1000}\) = (1 – \(\frac{1}{5}\))ⁿ
\(\frac{1331}{1000}\) = (\(\frac{6}{5}\))ⁿ
∴ n ≠ 3 (False)

Question (v)
The compound interest on ₹ 16000 for 9 months @20% p.a, compounded quarterly is ₹ 2522.
Answer:
True
Hint:
Principal (P) = 16000
n = 9 months = \(\frac{9}{12}\) years
r = 20% p.a
For compounding quarterly, we have to use below formula.
Amount (A) = P x (1 + \(\frac{r}{100}\))⁴ⁿ
Since quarterly we have to divide r by 4
r = \(\frac{20}{4}\)

∴ Interest = A – P = 18522 – 16000 = 2522 (True)

Question 3.
Find the compound interest on ₹ 3200 at 2.5% p.a for 2 years, compounded annually.
Solution:
Principal (P) = ₹ 3200
r = 2.5% p.a
n = 2 years comp, annually
∴ Amount (A) = (1 + \(\frac{r}{100}\))ⁿ = (1 + \(\frac{2.5}{100}\))²
= 3200 x (1.025)² = 3362
Compound interest (Cl) = Amount – Principal = 3362 – 3200 = ₹ 162

Question 4.
Find the compound interest for 2\(\frac{1}{2}\) years on ₹ 4000 at 10% p.a if the interest is compounded yearly.
Solution:
Principal (P) = ₹ 4000
r = 10% p.a
Compounded yearly n = 2\(\frac{1}{2}\) years. Since it is of the form a\(\frac{b}{c}\) years

= 4000x 1.1 x 1.1 x 1.05 = 5082
∴ Cl = Amount – Principal = 5082 – 4000 = 1082

Question 5.
Magesh invested ₹ 5000 at 12% p.a for one year. If the interest is compounded half yearly, find the amount he gets at the end of the year.
Solution:
Principal (P) = ₹ 5000
Interest compounded half yearly
r = 12% p.a = \(\frac{12}{2}\) = 6% for half yearly
t = 1 yr.
Since compounded half yearly, the formula to be used is
Amount A = P (1 + \(\frac{r}{100}\))²ⁿ
A = 5000 (1 + \(\frac{6}{100}\))^2×1 = 5000 x (\(\frac{106}{100}\))² = ₹ 5618

Question 6.
At what time will a sum of ₹ 3000 will amount to ₹ 3993 at 10% p.a compounded annually?
Solution:
Amount A = ₹ 3993
Principal = ₹ 3000
r = 10% p.a
n = ?

Question 7.
A principal becomes ₹ 2028 in 2 years at 4% p.a compound interest. Find the Principal.
Solution:
n = 2 years
r = rate of interest = 4% p.a
Amount A = ₹ 2028
Amount (A) = P (1 + \(\frac{r}{100}\))ⁿ
2028 = P (1 + \(\frac{4}{100}\))ⁿ
2028 = P (\(\frac{r}{100}\))²
∴ P = \(\frac{2028x100x100}{104×104}\) = ₹ 1875

Question 8.
At what rate percentage p.a will ₹ 5625 amount to ₹ 6084 in 2 years at compound interest?
Solution:
Principal (P) = ₹ 5625
Amount (A) = ₹ 6084
n = 2 years
r = ?
Amount (A) = P (1 + \(\frac{r}{100}\))ⁿ [Applying in formula]
6084 = 5625 (1 + \(\frac{r}{100}\))²
(1 + \(\frac{r}{100}\))² = \(\frac{6084}{5625}\)
Taking square root on both sides, we get
1 + \(\frac{r}{100}\) = \(\frac{78}{75}\)
\(\frac{r}{100}\) = \(\frac{78}{75}\) – 1 = \(\frac{3}{75}\) = \(\frac{1}{25}\)
∴ r = \(\frac{1}{25}\) x 100 = 4%

Question 9.
In how many years will ₹ 3375 amount to ₹ 4096 at 13\(\frac{1}{3}\)% p.a where interest is compounded half-yearly?
Solution:
Principal = ₹ 3375
Amount = ₹ 4096
r = 13\(\frac{1}{3}\)% p.a = \(\frac{40}{3}\)% p.a
Compounded half yearly r = \(\frac { \frac { 40 }{ 3 } }{ 2 } \) = \(\frac{2}{3}\)
Let no. of years be n
for compounding half yearly, formula is

Question 10.
Find the C.I on ₹ 15000 for 3 years if the rates of interest is 15%, 20% and 25% for I, II and III years respectively.
Solution:
Principal (P) = ₹ 15000
rate of interest 1 (a) = 15% for year I
rate of interest 2 (b) = 20% for year II
rate of interest 3 (c) = 25% for year III
Formula for amount when rate of interest is different for different years is

Compound Interest (Cl) = A – P = 25,875 – 15,000 = 10,875
Cl = ₹ 10,875

Question 11.
The present height of a tree is 847 cm. Find its height two years ago, if it increases at 10 % p.a.

Solution:
Present height of tree = 847 cm
Present height = ‘h’
n = 2 yrs
rate of growth = 10% p.a
Applying in formula, we get

∴ Original height of tree = 70 cm

Question 12.
Find the difference between the C.I and the S.I on ₹ 5000 for 1 year at 2% p.a, if the interest is compounded half yearly.
Solution:
Principal (P) = ₹ 5000
time period (n) = 1 yr.
Rate of interest (r) = 2% p.a
for half yearly r = 1%
Difference between Cl & SI is given by the formula
CI – SI = P (\(\frac{r}{100}\))²ⁿ [for half yearly compounding]
CI – SI = P (\(\frac{1}{100}\))^2×1
= 5000 x \(\frac{1}{100}\) x \(\frac{1}{100}\) = ₹ 0.50

Question 13.
What is the difference in simple interest and compound interest on 115000 for 2 years at 6% p.a compounded annually.
Solution:
Principal (P) = ₹ 15,000
Time period (n) = 2 yrs.
Rate of interest (r) = 6% p.a compounded annually
Difference between CI and SI given by
CI – SI = P (\(\frac{r}{100}\))ⁿ = 15000 (\(\frac{6}{100}\))²
= 15000 x \(\frac{6}{100}\) x \(\frac{6}{100}\)
= ₹ 54

Question 14.
Find the rate of interest if the difference between the C.I and S.I on ₹ 8000 compounded annually for 2 years is ₹ 20.
Solution:
Principal (P) = ₹ 8000
time period (n) = 2 yrs.
rate of interest (r) = ?
Difference between Cl & SI is given by the formula
CI – SI = P (1 + \(\frac{r}{100}\))ⁿ
Difference between Cl & SI is given as 20
∴ 20 = 8000 x (\(\frac{r}{100}\))²
∴ (\(\frac{r}{100}\))² = \(\frac{20}{8000}\) = \(\frac{1}{400}\)
Taking square root on both sides
\(\frac{r}{100}\) = \(\sqrt { \frac { 1 }{ 400 } } \) = \(\frac{1}{20}\)
∴ r = \(\frac{100}{20}\)

Question 15.
Find the principal if the difference between C.I and S.I on it at 15% p.a for 3 years is ₹ 1134.
Solution:
Rate of interest (r) = 15% p.a
time period (n) = 3 years
Difference between Cl & SI is given as 1134
Principal = ? → required to find
Simple Interest SI = \(\frac{Pnr}{100}\)
Compound Interest CI = P (1 + i)ⁿ – P
Cl – SI = P [(1 + i)ⁿ – 1 – \(\frac{nr}{100}\)]

Objective Type Questions

Question 16.
The number of conversion periods, if the interest on a principal is compounded every two months is ……………
(a) 2
(b) 4
(c) 6
(d) 12
Answer:
(c) 6
Hint:
Conversion period is the time period after which the interest is added to the principal. If principal is compounded every two months then in a year, there will be 6\(\frac{12}{2}\) conversation periods.

Question 17.
The time taken for ₹ 4400 to become ₹ 4851 at 10%, compounded half yearly is
(a) 6 months
(b) 1 year
(c) 1\(\frac{1}{2}\) years
(d) 2years
Answer:
(b) 1 year
Hint:
Principal = ₹ 4400
Amount = ₹ 4851
Rate of interest = 10% p.a
for half yearly, divide by 2,
r = \(\frac{10}{2}\) = 5 %
Compounded half yearly, so the formula is
A = P (1 + \(\frac{r}{100}\))²ⁿ
Substuting in the above formula, we get

Taking square root on both sides, we get
(\(\frac{21}{20}\))²ⁿ = (\(\frac{21}{20}\))²
Equating power on both sides
∴ 2n = 2,
n = 1

Question 18.
The cost of a machine is ₹ 18000 and it depreciates at 16\(\frac{2}{3}\)% annually. Its value after 2 years will be ………..
(a) ₹ 12000
(b) ₹ 12500
(c) ₹ 15000
(d) ₹ 16500
Answer:
(b) ₹ 12500
Hint:
Cost of machine = ₹ 18000
Depreciation rate = 16\(\frac{2}{3}\)% = \(\frac{50}{3}\)% p.a
time period = 2 years
∴ As per depreciation formula,
Depriciated value = Original value (1 – \(\frac{r}{100}\))ⁿ
Substituting in above formula, we get

Question 19.
The sum which amounts to ₹ 2662 at 10% p.a in 3 years compounded yearly is ………..
(a) ₹ 2000
(b) ₹ 1800
(c) ₹ 1500
(d) ₹ 2500
Answer:
(a) ₹ 2000
Hint:
Amount = ₹ 2662
rate of interest = 10% p.a
Time period = 3 yrs. Compounded yearly
Principal (P) → required to find?
Applying formula A = P (1 + \(\frac{r}{100}\))ⁿ

Question 20.
The difference between simple and compound interest on a certain sum of money for 2 years at 2% p.a is ₹ 1 the sum of money is ……….
(a) ₹ 2000
(b) ₹ 1500
(c) ₹ 3000
(d) ₹ 2500
Answer:
(d) ₹ 2500
Difference between Cl and SI is given as Re 1
Time period (n) = 2 yrs.
Rate of interest (r) = 2% p.a
Formula for difference is
CI – SI = P x (1 + \(\frac{r}{100}\))ⁿ
Substituting the values in above formula, we get
1 = P x (\(\frac{2}{100}\))²
∴ P = 1 x (\(\frac{100}{2}\))²
= 1 x (50)² = ₹ 2500

Students can Download Maths Chapter 2 Algebra Ex 2.1 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 2 Chapter 2 Algebra Ex 2.1

Question 1.
Fill in the blanks

Question (i)
The value of x in the equation x +5 12 ¡s ……….
Answer:
7
Hint:
Given,
x + 5 = 12
x = 12 – 5 = 7 (by transposition method)
Value of x is 7

Question (ii)
The value ofy in the equation y – 9 = (-5) + 7 is ……….
Answer:
11
Hint:
Given,
y – 9 = (- 5) + 7
y – 9 = 7 – 5 (re-arranging)
y – 9 = 2
∴ y = 2 + 9 = 11 (by transposition method)

Question (iii)
The value of m in the equation 8m = 56 is ………
Answer:
7
Hint:
Given,
8m = 56
Divided by 8 on both sides
\(\frac{8xm}{8}\) = \(\frac{56}{8}\)
∴ m = 7

Question (iv)
The value ofp in the equation \(\frac{2p}{3}\) = 10 is ……….
Answer:
1
Hint:
Given,
\(\frac{2p}{3}\) = 10
Multiplying by 3 on both sides,

∴ p = 15

Question (v)
The linear equation in one variable has ……… Solution.
Answer:
one.

Question 2.
Say True or False.

Question (i)
The shifting of a number from one side of an equation to other is called transposition.
Answer:
True

Question (ii)
Linear equation in one variable has only one variable with power 2.
Answer:
False
[Linear equation in one variable has only one variable with power one – correct statement]

Question 3.
Match the following :

(A) (i), (ii), (iv), (iii), (v)
(B) (iii), (iv), (i), (ii), (v)
(C) (iii), (i), (iv), (v), (ii)
(D) (iii), (i), (v), (iv), (ii)
Answer:
(C) (iii),(i), (iv), (v), (ii)
Hint:
a. \(\frac{x}{2}\) = 10,
multiplying by 2 on both sides, we get
\(\frac{x}{2}\) x 2 = 10 x 2 ⇒ x = 20

b. 20 = 6x – 4
by transposition ⇒ 20 + 4 = 6x
6x = 24
dividing by 6 on both sides,
\(\frac{6x}{6}\) = \(\frac{24}{6}\) ⇒ x = 4

c. 2x – 5 = 3 – x
By transposing the variable ‘x’, we get
2x – 5 + x = 3
by transposing – 5 to other side,
2x + x = 3 + 5
∴ 3x = 8

∴ x = \(\frac{8}{3}\)

d. 7x – 4 – 8x = 20
by transposing – 4 to other side,
7x – 8x = 20 + 4
– x = 24
∴ x = – 24
\(\frac{4}{11}\) – x = \(\frac{-7}{11}\)
Transposing \(\frac{4}{11}\) to other side,
– x = \(\frac{-7}{11}\)\(\frac{-4}{11}\) = \(\frac{-7-4}{11}\) = \(\frac{-11}{11}\) = – 1
∴ – x = – 1 ⇒ x = 1

Question 4.
Find x:

Question (i)
\(\frac{2x}{3}\) – 4 = \(\frac{10}{3}\)
Solution:
Transposing -4 to other side, it becomes +4
∴ \(\frac{2x}{3}\) = \(\frac{10}{3}\) + 4
Taking LCM & adding,
\(\frac{2x}{3}\) = \(\frac{10}{3}\) + \(\frac{4}{1}\) = \(\frac{10+12}{3}\) = \(\frac{22}{3}\)
\(\frac{2x}{3}\) = \(\frac{22}{3}\)
Multiplying by 3 on both sides

⇒ 2x = 22
dividing by 2 on both sides,
We get \(\frac{2x}{2}\) = \(\frac{22}{2}\)
∴ x = 11

Question (ii)
y + \(\frac{1}{6}\) – 3y = \(\frac{2}{3}\)
Solution:
Transposing \(\frac{1}{6}\) to the other side,
y – 3y = \(\frac{2}{3}\) – \(\frac{1}{6}\)
Taking LCM,
– 2y = \(\frac{2}{3}\) – \(\frac{1}{6}\) = \(\frac{2×2-1}{6}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\)
∴ – 2y = \(\frac{1}{2}\) ⇒ 2y = – \(\frac{1}{2}\)
dividing by 2 or both sides.

Question (iii)
\(\frac{1}{3}\) – \(\frac{x}{3}\) = \(\frac{7x}{12}\) + \(\frac{5}{4}\)
Transposing \(\frac{-x}{3}\) to the other side, it becomes + \(\frac{x}{3}\)
∴ \(\frac{1}{3}\) = \(\frac{7x}{12}\) + \(\frac{5}{4}\) + \(\frac{x}{3}\)
Transposing \(\frac{5}{4}\) to the other side, it becomes \(\frac{-5}{4}\)
\(\frac{1}{3}\) + \(\frac{5}{4}\) = \(\frac{7x}{12}\) + \(\frac{x}{3}\)
Multiply by 12 throughout
[we look at the denominators 3,4, 12, 3 and take the LCM, which is 12]

4 – 15 = 7x + x × 4
-11 = 7x + 4x
11x = – 11
x = -1

Question 5.
Find x:

Question (i)
-3(4x + 9) = 21
Solution:

Expanding the bracket,
-3 × 4x + (-3) × 9 = 21
-12x + (-27) = 21
-12x – 27 = 21
Transposing – 27 to other side, it becomes +27
-12x = 21 + 27 = 48
12x = 48 ⇒ 12x = -48
Dividing by 12 on both sides

⇒ x = – 4

Question (ii)
20 – 2 ( 5 – p) = 8
Solution:

Expanding the bracket,
20 – 2 x 5 – 2 x (-p) = 8
20 – 10 + 2 + p = 8 (-2 x -P = 2p)
10 + 2p = 8 transporting 10 to other side
2P = 8 – 10 = -2
∴ 2p = -2
∴ p = -1

Question (iii)
(7x – 5) – 4(2 + 5x) = 10(2 – x)
Solution:

Expanding the brackets,
7x – 5 – 4 × 2 – 4 × 5x = 10 × 2 + 10 × (-x)
7x – 5 – 8 – 20x = 20 – 10x
7x – 13 – 20x = 20 – 10x
Transposing 10x & -13, we get
7x – 13 – 20x + 10x = 20
7x – 20x + 10x = 20 + 13,
Simplifying,
-3x = 33
∴ 3x = -33
x = \(\frac{-33}{3}\) = -11
x = -11

Question 6.
Find x and m:

Question (i)
\(\frac{3x-2}{4}\) – \(\frac{(x-3)}{5}\) = -1
Solution:
\(\frac{3x-2}{4}\) – \(\frac{(x-3)}{5}\)
Taking LCM on LHS, [LCM of 4 & 5 is 20]

∴ 11x + 2 = -20
∴ 11x = – 20 – 2 = – 22
x = \(\frac{-22}{11}\) = -2
x = -2

Question (ii)
\(\frac{m+9}{3m+15}\) = \(\frac{5}{3}\)
Solution:
\(\frac{m+9}{3m+15}\) = \(\frac{5}{3}\)
Cross multiplying, we get

∴ (m + 9) x 3 = 5 x (3m + 15)
m x 3 + 9 x 3 = 5 x 3m + 5 x 15

Transporting 3m & 75, we get
27 – 75 = 15m – 3m
-48 = 12m

⇒ m = -4

Students can Download Maths Chapter 1 Life Mathematics 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 2 Chapter 1 Life Mathematics Ex 1.2

Question 1.
Fill in the blanks:

Question (i)
Loss or gain percent is always calculated on the ………
Answer:
Cost price

Question (ii)
A mobile phone is sold for ? 8400 at a gain of 20%. The cost price of the mobile phone is …………
Answer:
₹ 7000
Hint:
Let cost price of mobile be ₹ x. Given that selling price is ? 8400 and gain is 20%

Question (iii)
An article is sold for ₹ 555 at a loss of 7\(\frac{1}{2}\)% the cost price qfthe article is ……….
Answer:
₹ 600
Hint:
Given selling price is ₹ 555 & loss is 7\(\frac{1}{2}\)%
as per formula,

Question (iv)
The marked price of a mixer grinder is ₹ 4500 is sold for ₹ 4140 after discount. The rate of discount is ……..
Answer:
8%
Hint:
Marked price is X 4500. Discounted price in ₹ 4140
∴ Discount = Marked price – Discounted price = 4500 – 4140 – 360

Question (v)
The total bill amount of a shirt costing ₹ 575 and a T-shirt costing ₹ 325 with GST of 5% is ………..
Answer:
₹ 945
Hint:
Cost price of shirt = ₹ 575 (CP)
GST = 5%

= 575 x (\(\frac { (100+5) }{ 100 } \)) = 575 x \(\frac { 105 }{ 100 } \)
= ₹ 603.75
Cost price of T-shirt = ₹ 325 (CP)
GST = 5%

= 325 x (\(\frac { (100+5) }{ 100 } \))
Total bill amount = ₹ 603.75 + ₹ 341.25 = ₹ 945

Question 2.
If selling an article for ₹ 820 causes 10% loss on the selling price, find its cost price.
Solution:
Given that selling price (SP) = ₹ 820
Loss % = 10 %

Question 3.
If the profit earned on selling an article for ₹ 810 is the same as loss on selling it for ₹ 530, then find the cost price of the article.
Solution:
Case 1: Profit = Selling price (SP) – Cost price (CP)
Case 2: Loss = Cost price (CP) – Selling price (SP)
Given that profit of case 1 = loss of case 2
∴ P = 810 – CP
L = CP – 530
Since profit (P) = loss (L)
810 – CP = CP – 530
∴ 2CP = 810 + 530 = 1340 ⇒ C.P = \(\frac{1340}{2}\)
∴ CP = 670

Question 4.
Some articles are bought at 2 for ₹ 15 and sold at 3 for ₹ 25. Find the gain percentage.
Solution:
Let cost price of one article be C.P
Given that 2 are bought for ₹ 15
∴ 2 x CP = 15 ⇒ CP = \(\frac{15}{2}\)
Let selling price of one article be SP
Given that 3 are sold for ₹ 25
∴ 3 x CP = 25 ⇒ SP = \(\frac{25}{3}\)
∴ Gain = SP – CP = \(\frac{25}{3}\) – \(\frac{15}{2}\) = \(\frac{50-45}{6}\) = \(\frac{5}{6}\)

Question 5.
If the selling price of 10 rulers is the same as the cost price of 15 rulers, then find the gain percentage.
Solution:
Let cost price of one ruler be x
Given that selling price (SP) of 10 rulers
i.e., same as cost price (CP) of 15 rulers
∴ SP of 10 rulers = 15 × x = 15x
SP of 1 ruler = \(\frac{15x}{10}\) = 1.5x
∴ Gain = SP of 1 ruler – CP of ruler = 1.5x -x = 0.5x

Question 6.
By selling a speaker for ₹ 768, a man loses 20%. In order to gain 20% how much should he sell the speaker?

Solution:
Selling price (SP) of speaker = ₹ 768
Loss % = 20%
as per formula

For gain of 20%, we should now calculate the selling price

= 960 (\(\frac{100+20}{100}\)) = 960 x \(\frac{120}{100}\)
= 96 x 12 = ₹ 1152

Question 7.
A man sold two gas stoves for ₹ 8400 each. He sold one at a gain of 20% and the other at a loss of 20%. Find his gain or loss % in the whole transaction.

Solution:
Let the CP of gas stove 1 be x and gas stove 2 be y
Given that selling price (SP) for both is the same = ₹ 8400
Case i:
First gastove: Cost price (CP) = x
Selling Price (SP) = ₹ 8400
Gain % = 20%

Cost price of first gas stove = ₹ 7000

Case 2:
Second gastove: Cost price (CP) = y
Selling price (SP) = ₹ 8400
loss % = 20%

∴ Cost price of second stove = ₹ 10,500
From (1) and (2),
Total cost price = Cost of stove 1 + Cost of stove 2
= 7000+ 10500 = 17,500
Total selling price = SP of stove 1 + SP of stove 2
= 8400 + 8400 = 16,800
Now, we find that total selling price is less than total cost price, therefore it is a loss
∴ Loss = CP – SP= 17,500 – 16,800 = 700
Loss % = \(\frac{loss}{CP}\) x 100 = \(\frac{700}{17500}\) x 100 = 4%
Loss % = 4 %

Question 8.
Find the unknowns x, y and z

Solution:
(i) Book marked price = ₹ 225 discount = 8%

∴ 225 x (\(\frac { (100-8) }{ 100 } \))
= 225 x (\(\frac { 92 }{ 100 } \)) = ₹ 207

(ii) LED TV selling price = 11970 discount = 5%,

∴ 11970 = y x \(\frac { (100-d%) }{ 100 } \)
∴ y = \(\frac { 11970×100 }{ 95 } \) = 126 x 100 = ₹ 12,600

(iii) Digital clock marked price (MP) = ₹ 750, MP = ₹ 12,600
Selling price (SP) = ₹ 615, Discount = z

∴ 615 = 750 x \(\frac { (100-z) }{ 100 } \)
∴ (100 – z) = \(\frac { 615×100 }{ 700 } \)
100 – z = 82
∴ z = 100 – 82
Discouont = 18%

Question 9.
Find the total bill amount for the data below.

Solution:

For bill amount, we should apply GST on the discounted value of the items.

Total bill amount = Bill amount of School bag + Stationary + Cosmetics + Hair drier
= 532 + 252 + 1357 + 2304
= ₹ 4,445

Question 10.
A shopkeeper buys goods at \(\frac { 4 }{ 5 } \) of its marked price and sells them at \(\frac { 4 }{ 5 } \) of the marked price find his profit percentage.
Solution:
Let marked price be MP
Given that he buys good at \(\frac { 4 }{ 5 } \) of marked price
∴ CP (cost price) = \(\frac { 4 }{ 5 } \) MP
Given that selling price (SP) = \(\frac { 7 }{ 5 } \) x MP
∴ Profit = Selling price – Cost price = \(\frac { 7 }{ 5 } \) MP – \(\frac { 4 }{ 5 } \) MP = \(\frac { 3 }{ 5 } \)  MP

Question 11.
A branded AC has a marked price of ₹ 37250. There are 2 options given for the customer.

1. Selling Price is ₹ 37250 along with attractive gifts worth ₹ 3000 (or)
2. Discount of 8% but no free gifts. Which offer is better?

Marked price of AC = ₹ 37,250
Option 1:
Selling price = ₹ 37250 & gifts worth ₹ 3000
∴ Net gain for customer = ₹ 3000 as there is no discount on AC

Option 2:
Discount of 8%, but no gift

= 37250 x \(\frac { (100-8) }{ 100 } \) = 37250 x 0.92 = 34270
∴ Savings for customer = 37250 – 34270 = ₹ 2980
Therefore, the customer gets 3000 gift in option 1 where as he is able to save only ₹ 2980 in option 2. Therefore, option 1 is better.

Question 12.
If a mattress is marked for ₹ 7500 and is available at two successive discount of 10% and 20%, find the amount to be paid by the customer.
Solution:
Marked price of mattress = ₹ 7500
Discount d₁ = 10%
Discount d₂ = 20%

= 7500 x \(\frac { (100-10) }{ 100 } \) = 7500 x \(\frac { 90 }{ 100 } \) = 6750

= 6750 x \(\frac { (100-20) }{ 100 } \) = ₹ 5400

Objective Type Questions

Question 13.
A fruit vendor sells fruits for ₹ 200 gaining ₹ 40. His gain percentage is –
(a) 20%
(b) 22%
(c) 25%
(d) 16\(\frac { 2 }{ 3 } \)
Answer:
(c) 25%
Hint:
Selling price = ₹ 200
Gain = 40
∴ CP = Selling price – gain = 200 – 40 = 160
Gain % = \(\frac { Gain }{ CP } \) x 100 = \(\frac { 40 }{ 160 } \) x 100 = 25%

Question 14.
By selling a flower pot for ₹ 528, a woman gains 20%. At what price should she sell it to gain 25%?
(a) ₹ 500
(b) ₹ 550
(c) ₹ 553
(d) ₹ 573
Answer:
(b) ₹ 550
Hint:
If selling price (SP) = ₹ 528
Gain % = 20%
∴ CP = ?

∴ 528 = CP x \(\frac { 100+20 }{ 100 } \)
∴ CP = \(\frac { 528×100 }{ 120 } \)
If gain % = 25 %, Selling ?

= 440 x \(\frac { (100+gain%) }{ 100 } \) = 440 x \(\frac { 125 }{ 100 } \)
= ₹ 550

Question 15.
A man buys an article for ₹ 150 and makes overhead expenses which are 12% of the cost price. At what price must he sell it to gain 5%?
(a) ₹ 180
(b) ₹ 168
(c) ₹ 176.40
(d) ₹ 85
Answer:
(c) ₹ 176.40
Hint:
Cost price of article = ₹ 150
Over head expenses = 12% of cost price = \(\frac { 12 }{ 100 } \) x 150 = ₹ 85
∴ Effective cost of article = 150 + 18 = ₹ 168
Now, to gain 5%, he has to sell at

Question 16.
The price of a hat is ₹ 210. What was the marked price of the hat if it is bought at 16% discount?
(a) ₹ 243
(b) ₹ 176
(c) ₹ 230
(d) ₹ 250
Answer:
(d) ₹ 250
Hint:
Let marked price be MP
Discounted price = ₹ 210
Rate of discount = 16%
As per formula:

Question 17.
The single discount which is equivalent to two successive discount of 20% and 25% is –
(a) 40%
(b) 45%
(c) 5%
(d) 22.5%
Answer:
(a) 40%
Let marked price be MP, after discount 1 of 20%,

After discount 2 of 25%,

Comparing with formula, we get
∴ This is equivalent to a single discount of 40%

Students can Download Maths Chapter 1 Life Mathematics Intext 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 2 Chapter 1 Life Mathematics Intext Questions

Exercise 1.1
Try This (Text book Page no. 1)

Question 1.
Find the indicated percentage value of the given numbers
Solution:

Think (Text book page no. 6)

Question 1.
An increase from 200 to 600 ¡s clearly a 200%increase. Isn’t it? (check!). With a lot of pride, the traffic police commiuêener of a city reported that the accidents had decreased by 200% ¡n one year. He cameupwith this number stating that the accidents had gone down from 600 last year to 2this year. Is the decrease from 600 to 200, the same 200% as above? Justify.
Solution:
Increase from original value 200 to 600

Decrease from original value 600 to 200

here original value is 600
% decrease = \(\frac{600-200}{600}\) x 100 = \(\frac{400}{600}\) x 100 = 66.67% decrease
Increase from 200 → 600 and % decrease from 600 → 200 are not the same

Try This (Text book page no. 7)

Question 1.
What percent of a day is 10 hours?
Solution:
In a day, there are 24 hours
10 hrs out of 24 hrs is \(\frac{10}{24}\)
As a percentage, we need to multiply by 100
∴ Percentage = \(\frac{10}{24}\) x 100 = 41.67%

Question 2.
Divide ₹ 350 among P, Q and R such that P gets 50% of what Q gets and Q gets 50% of what R gets.
Solution:
Let R get x, Q gets 50% of what R gets
∴ Q gets = \(\frac{50}{100}\) × x = \(\frac{x}{2}\)
P gets 50% of what Q gets
∴ P gets = \(\frac{50}{100}\) x \(\frac{x}{2}\) = \(\frac{x}{4}\)
Since 350 is divided among the three
∴ 350 = x + \(\frac{x}{2}\) + \(\frac{x}{4}\)
350 = \(\frac{4x+2x+x}{4}\) = \(\frac{7x}{4}\) = 350
x = \(\frac{350×4}{7}\) = 200
Q gets = \(\frac{x}{2}\) = \(\frac{200}{2}\) = 100,
P gets = \(\frac{x}{4}\) = \(\frac{200}{4}\) = 50
∴ P = 50
Q = 100
R = 200

Exercise 1.2
Think (Text book Page No. 13)

Question 1.
A shopkeeper marks the price of a marker board 15% above the cost price and then allows a discount of 15% on the marked price. Does he gain or lose in the transaction?
Solution:
Let cost price of marker board be 100
CP = 100 Marks it 15% above CP
∴ Marked price = MP = \(\frac{15}{100}\) x CP + CP = \(\frac{15}{100}\) 100 + 100 = 15 + 100 = 115
Discount % = 15%

∴ He sells it 97.75 which is less than his cost price. Therefore he loses
Loss = 97.75 – 100 = – 2.25

Try This (Text book Page No. 14)

Question 1.
By selling 5 articles, a man gains the cost price of 1 article. Find his gain percentage.
Solution:
Let cost price of article be C.P. Let S.P of 1 articles be SP by selling 5 articles at SP he makes a gain of cost price of one article.
Gain on 1 article = SP – CP; ⇒ Gain% = \(\frac{SP-CP}{CP}\) x 100
Gain on 2 articles = 2 x (SP – CP)
Gain on 5 articles = 5 x (SP – CP)
Given than gain on 5 articles is CP of 1 article
∴ 5(SP – CP) = CP
\(\frac{SP-CP}{CP}\) = \(\frac{1}{5}\)
Gain percentage \(\frac{SP-CP}{CP}\) x 100 = \(\frac{1}{5}\) x 100 % = 20%

Question 2.
By selling 8 articles, a shopkeeper gains the selling price of 3 articles. Find his gain percentage.
Solution:
Let cost price of 1 article be CP. Let selling price of 1 article be SP.
Gain on 1 article = SP – CP
Gain on 8 articles = 8 x (SP – CP)
Given that gain on 8 articles in selling price of 3 articles
8(SP – CP) = 3 x SP
∴ \(\frac{SP-CP}{CP}\) = \(\frac{3}{8}\)
∴ \(\frac{SP}{SP-CP}\) = \(\frac{8}{3}\)
Subtracting 1 on both sides
\(\frac{SP}{SP-CP}\) – 1 = \(\frac{8}{3}\) – 1 = \(\frac{SP-(SP-CP)}{SP-CP}\)= \(\frac{8-3}{3}\)
\(\frac{SP-SP-CP}{SP-CP}\) = \(\frac{5}{3}\) ⇒ \(\frac{CP}{SP-CP}\) = \(\frac{5}{3}\)
\(\frac{SP-CP}{CP}\) = \(\frac{3}{5}\) (taking reciprocals on both sides)
Gain% \(\frac{SP-CP}{CP}\) x 100 = \(\frac{3}{5}\) x 100 = 3 x 20 = 60%

Question 3.
If the C.P of 20 articles is equal to the S.P of 15 articles, find the profit or loss percentage.
Solution:
Given CP of 20 article = SP of 15 articles
∴ SP of 15 articles = CP of 20 articles
∴ SP of 1 article = \(\frac{1}{15}\) x CP of 20 articles
SP = \(\frac{1}{15}\) x 20 CP = \(\frac{20}{15}\) CP = \(\frac{4}{3}\) CP
∴ SP = \(\frac{4}{3}\) CP ⇒ SP is greater than CP
It is a profit.

Exercise 1.3
Try This (Text book Page No. 23)

Question 1.
Find the principal which gives ₹ 420 as C.I @ 20% p.a compounded half yearly for one year.
Solution:
CI = ₹420
Rate = ₹ 20% p.a
Principle = ₹ [required to find] Time period (n) = 1 year.
However, let us value of r to be 20% p.a so for half yearly, r is \(\frac{20}{2}\) = 10%
Formula for Amount (A) when compounded half yearly is

Question 2.
The price of a laptop depreciates @ 4% p.a. If its present price is ₹ 24,000, find its price after 3 years.
Solution:
Let original price of laptop be ‘P’, Rate of depreciation is 4% p.a,
Present price is ₹ 24,000 (D).
Formula for depreciation is

Price after 3 years from now is
P(1 – \(\frac{4}{100}\))ⁿ⁺³ = ? ⇒ (1 – \(\frac{4}{100}\))ⁿ (1 – \(\frac{4}{100}\))³
From (1)
24,000 x (1 – \(\frac{4}{100}\))³
24,000 x \(\frac{96}{100}\) x \(\frac{96}{100}\) x \(\frac{96}{100}\) = 21233.66

Actvity I (text book Page No. 23)

Question 1.
Mukunthan invests 30,000/- for 3 months in a bank which gives C.I at the rate of 12% compounded monthly. A private company offers his S.l at the rate of 12% p.a What is the difference in the interests received by Mukunthan? Do by traditional method and verify your answer by calculator.
Solution:
Principal = 30,000
Time period = 3 months
In Bank rate of interest for CI = 12% compounded monthly
∴ A = (1 + \(\frac{r}{100}\))ⁿ = 30,000(1 + \(\frac{12}{100}\))³
30,000 x \(\frac{112}{100}\) x \(\frac{112}{100}\) x \(\frac{112}{100}\) = 42147.84
∴ CI = A – P = 42147.84 – 30,000
CI = 12147.84
In private company,
Rate of single Interest SI = 12% p.a
So, for 3 months, i.e \(\frac{3}{12}\) = \(\frac{1}{4}\) year,

∴ Difference in interest = CI – SI = 12,147.84 – 900 = 11247.84

8th Standard English Guide Term 1 Unit 2 Prose Friendship Book Back Answers

Students can Download English Lesson 2 Friendship Questions and Answers, Summary, Activity, Notes, Samacheer Kalvi 8th English Book Solutions Guide Pdf  helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Friendship 8th Standard English 2nd Lesson Question and Answer

Read And Understand

A. Choose the correct answer.

Question 1.
Vetri went to Asif ‘s _______
(a) home
(b) office
(c) room
Answer:
(b) office

Question 2.
Vetri came to Chennai to visit his _______
(a) father
(b) friend
(c) brother
Answer:
(b) friend

Question 3.
Asif saw his friend through the _______
(a) camera
(b) window
(c) glass
Answer:
(a) camera

B. Choose correct synonyms for the italic word.

Question 1.
Vetri constructed a bungalow.
(a) design
(b) build
(c) foundation
(d) destroy
Answer:
(b) build

Question 2.
The brothers started a business, separately.
(a) apart
(b) alone
(c) united
(d) combined
Answer:
(a) apart

Question 3.
I am living in the outskirts of the village.
(a) border
(b) outpost
(c) center
(d) region
Answer:
(b) outpost

Question 4.
Asif quarreled with his friend.
(a) fight
(b) differ
(c) peace
(d) fun
Answer:
(a) fight

Question 5.
He stood astounded.
(a) happy
(b) surprised
(c) shocked
(d) excited
Answer:
(c) shocked

C. Choose correct Antonyms for the italic word.

Question 1.
The wife replied angrily.
(a) calm
(b) annoyed
(c) irritate
Answer:
(a) calm

Question 2.
The vegetables look fresh.
(a) rotten
(b) dull
(c) new
Answer:
(a) rotten

Question 3.
Vetri had a strong will to start a new business.
(a) desire
(b) thin
(c) weak
Answer:
(c) weak

Question 4.
Vetri was surprised by his friend.
(a) unsurprised
(b) expected
(c) shocked
Answer:
unsurprised

Question 5.
He spoke nervously
(a) Scared
(b) confident
(c) anxious
(b) confident
Answer:
(b) confident

D. Answer the following questions in one or two words.

Question 1.
What was the name of Vetri’s company?
Answer:
The name of his company was “Vetri Constructions”.

Question 2.
Why did he sell his properties?
Answer:
He sold his properties to pay his loans.

Question 3.
Which was the home town of Vetri and Asif?
Answer:
The home town of Vetri and Asif was Keelakudi near Coimbatore.

Question 4.
When was the school established?
Answer:
The school was on the outskirts of Keelakudi village. It was in middle school.

Question 5.
When did Vetri receive a call from Asif’s office?
Answer:
Vetri received a call from Asif’s office two days later.

E. Answer the following questions in 100 words.

Question 1.
How Vetri lost his properties?
Answer:
Vetri was once a successful businessman in Coimbatore. His Vetri Construction was a leading construction company. After his father’s death, his brothers demanded to split the wealth, as they wanted to start their business separately. From then on, Vetri found it difficult to establish his business. He took loans to run his company, but he could not pay the loan. So he sold all his properties and paid the loans. His family then moved to a very small house. He found a job and started to lead a normal life.

Question 2.
What happened when Vetri met Asif?
Answer:
Vetri went straight to Asifs office to meet him. When he reached Asifs office, Asif saw Vetri through CCTV camera came out and pat told him on his back. There they discussed their school days and the fun they had. After that Asif took Vetri to his home for lunch. Vetri was surprised to see that everyone knows him. He stayed till evening and Asif dropped him in the railway station.

Question 3.
How did Asif show his friendship?
Answer:
Asif was a true friend of Vetri. When Vetri came to his office, he came to receive him, as he saw through the CCTV camera. He gave Vetri a pat on his back. Vetri was speechless on seeing him. He took to his cabin. They spoke about their school days and the fun they had. They discussed their business. Asif took Vetri to his home for lunch. Vetri was surprised to see that everyone knows him. Vetri stayed there till evening. Asif dropped him at the railway station. After two days, he invited Vetri to his office and assigned him a project.

Vocabulary

Compound Words

A. Match the following compound words and write them:
Answer:

First word	Second word	New word
match	mark	matchbox
air	cut	airport
blood	port	blood bank
Pop	gum	popcorn
sky	bank	sky blue
hair	blue	hair cut
book	corn	book mark
chewing	box.	chewing gum

B. Choose the best answer to make a compound word.

Question 1.
Which can be placed after ‘soft’?
(a) play
(b) ware
(c) run
(d) cycle
Answer:
(b) software

Question 2.
Which can be placed before ‘light’?
(a) try
(b) sun
(c) horse
(d) cat
Answer:
(b) sunlight

Question 3.
Which can be placed after ‘safe’?
(a) chair
(b) guard
(c) shop
(d) van
Answer:
(b) safeguard

Question 4.
Which can be placed after ‘blue’?
(a) cane
(b) print
(c) see
(d) land
Answer:
(b) blueprint

Question 5.
Which can be placed after ‘water’?
(a) food
(b) stick
(c) fall
(d) out
Answer:
(c) waterfall

Singular, Plural

Write the plural form of the given words :
Answer:

No.	Singular	Plural
1.	food	food
2.	radius	radii
3.	governor-general	governors-general
4.	syllabus	syllabi
5.	datum	data
6.	commander-in-chief	commanders-in-chief
7.	thesis	theses
8.	forum	fora
9.	cattle	cattle
10.	genius	geniuses / genii

Listening
Listen carefully to the passage given in the QR code and answer the following questions.

Questions:

Question 1.
Whose speech is it?
Answer:
It is Rahim’s speech.

Question 2.
What did Rahul engrave?
Answer:
He engraved in his heart that his friend Rahim had helped him.

Question 3.
Who is lucky?
Answer:
Rahim is lucky.

Question 4.
Who bagged the ‘all-rounder’ award?
Answer:
Rahul bagged an all-rounder medal in school.

Question 5.
Whose birthday party is it?
Answer:
It is Rahul’s birthday party.

Writing
Arrange the picture in order by writing the numbers 1,2,3 and 4 in the given boxes and write this familiar story in about 100 words.
Make use of the words given below.

(thirsty, village, pitcher, disappointment, pebbles, water level )

Answer:
One hot day, a thirsty crow in a village, searched for water everywhere. At last, it found a pitcher. But there was a disappointment for the crow, as the water level in the pitcher was very low. The crow got an idea. It saw some pebbles nearby. It started putting pebbles in the pitcher. Slowly, the water level raised. The crow drank the water and flew away happily.

Grammar

Degrees Of Comparison

a. Fill in the blanks:
Answer:

Positive	Comparative	Superlative
tall	taller	tallest
smart	smarter	smartest
large	larger	largest
more	more	most
late	Later/latter	latest (last)

Let’s compare two things.

Question 1.
Which is faster a train or a plane?
Answer:
plane is faster than a train.

Question 2.
Which is cheaper gold or silver?
Answer:
Silver is cheaper than gold.

Question 3.
Which is larger, city, or village?
Answer:
A city is larger than a village.

Question 4.
Which is bigger, a sea or an ocean?
Answer:
An ocean is bigger than the sea.

Question 5.
Which is taller, a giraffe or a camel?
Answer:
A giraffe is taller than a camel.

Let’s compare three things.

EG: Donkey, horse, and elephant (strong)
A donkey is strong.
The horse is stronger than a donkey.
An elephant is the strongest.

Question 1.
Town – city – village (quiet)
Answer:
A city is quiet.
Town is quieter than the city.
The village is the quietest.

Question 2.
Istanbul – Moscow – London (populated)
Answer:
Istanbul is populated.
Moscow is more populated than Istanbul.
London is the most populated.

Question 3.
Windy weather – warm weather – rainy weather (good)
Answer:
Windy weather is good.
Warm weather is better than windy weather.
Rainy weather is the best.

Question 4.
Ocean – river – lake (deep)
Answer:
River is deep.
Lake is deeper than the river.
Ocean is the deepest.

Question 5.
The USA – Russia – Spain (large)
Answer:
Hitl Spain is a large country.
The USA is larger than Spain.
Russia is the largest country.

Question 6.
The Mahanadi – The Cauvery – The Ganga (long)
Answer:
The cauvery is a long river.
The Mahanadi is longer than the Cauvery.
Ganga is the longest river in our country.

Question 7.
Chennai – Hyderabad – Bangaluru (modern)
Answer:
Chennai is a modern city.
Hyderabad is more modern than Chennai.
Bengaluru is the most modern city.

Friendship Additional Questions

I. Choose the correct Synonyms for the italic word

Question 1.
After his father’s death, his brothers demanded to split the wealth.
(a) break up
(b) brush
(c) open
(d) close
Answer:
(a) break up

Question 2.
But he always long to start a new construction company.
(a) fabrication
(b) devastation
(c) real estate
(d) building formation
Answer:
(d) building formation

Question 3.
He told his wife about his decision.
(a) hesitation
(b) delay
(c) judgment
(a) Abnormal
Answer:
(c) judgment

Question 4.
They lived in the beautiful village of Keelakudi.
(a) ugly
(b) awful
(c) vile
(d) Pretty
Answer:
(d) Pretty

Question 5.
Their friendship grew stronger with time.
(a) weaker
(b) slender
(c) firmer
(d) Punier
Answer:
(c) firmer

Question 6.
Asif consoled him with chocolate.
(a) comforted
(b) distressed
(c) troubled
(d) annoyed
Answer:
(a) comforted

Question 7.
I don’t know if you will get an appointment.
(a) employment
(b) dismissal
(c) arrangement for meeting
(d) discharge
Answer:
(c) arrangement for meeting

Question 8.
Vetri was dumbfounded.
(a) speechless
(b) mindful
(c) conscious
(d) Casual
Answer:
(a) speechless

Question 9.
Our MD Mr. Asif has assigned a project to you.
(a) refused
(b) cancelled
(c) given
(d) relieved
Answer:
(c) given

Question 10.
The security was astounded.
(a) annoyed
(b) irked
(c) upset
(d) surprised
Answer:
(d) surprised

II. Choose the correct Antonyms for the italic word.

Question 1.
Vetri was once a successful businessman in Coimbatore.
(a) powerful
(b) unsuccessful
(c) wonderful
(d) awful
Answer:
(b) unsuccessful

Question 2.
Vetri found it difficult to establish his business.
(a) abolish
(b) setup
(c) deal
(d) offer
Answer:
(a) abolish

Question 3.
He started to lead a normal life.
(a) usual
(b) same
(c) ordinary
(d) abnormal
Answer:
(d) abnormal

Question 4.
His brothers demanded to split the wealth.
(a) offered
(b) searched
(c) rushed
(d) ordered
Answer:
(a) offered

Question 5.
One day he discussed with his wife.
(a) ignored
(b) considered
(c) examined
(d) argued
Answer:
(a) ignored

Question 6.
To everyone’s surprise, they continued to be good friends.
(a) lingered
(b) extended
(c) discontinued
(d) lasted
Answer:
(a) lingered

Question 7.
One day there was a quarrel between Vetri and Asif’s families.
(a) row
(b) fight
(c) contest
(d) agreement
Answer:
(d) agreement

Question 8.
The results and marks never affected their friendship.
(a) moved
(b) stirred
(c) unaffected
(d) pretended
Answer:
(c) unaffected

Question 9.
Two days later, Vetri received a call.
(a) paid
(b) gave
(c) lived
(d) took
Answer:
(b) gave

Question 10.
Vetri told his wife everything in detail.
(a) same
(b) anything
(c) nothing
(d) all
Answer:
(c) nothing

III. Choose the Correct Answer [MCQ]

Question 1.
Vetri’s constructions was once a ________ construction company.
(a) small
(b) leading
(c) dull
(d) vague
Answer:
(b) leading

Question 2.
Everything went well, until his ________ died
(a) father
(b) sister
(c) uncle
(d) brother
Answer:
(a) father

Question 3.
They were always together in learning and ________
(a) writing
(b) reading
(c) studying
(d) playing
Answer:
(d) playing

Question 4.
They also helped others with their ________
(a) homework
(b) lessons
(c) projects
(d) assignment
Answer:
(b) lessons

Question 5.
The security was ________
(a) surprised
(b) astounded
(c) arrogant
(d) bold
Answer:
(b) astounded

Question 6.
They spoke about their school days and the ________ they had.
(a) boring
(b) lethargic
(c) fun
(d) indolent
Answer:
(c) fun

Question 7.
Then they discussed about their ________
(a) business
(b) children
(c) relatives
(d) plans
Answer:
(a) business

Question 8.
Asif dropped him in the ________ station.
(a) police
(b) railway
(c) bus
(d) fire
Answer:
(b) railway

IV. Short Questions with Answers.

Question 1.
What did Vetri’s Constructions construct? ‘
Answer:
Vetri’s constructions constructed many shopping complexes, houses, and a few apartments in and around Coimbatore.

Question 2.
What happened after his father’s death?
Answer:
After his father’s death, his brothers demanded to split the wealth, as they wanted to start their business separately.

Question 3.
Whom did Vetri decide to meet? Why?
Answer:
Vetri decided to meet his friend Asif in Chennai for help.

Question 4.
How strong was their friendship?
Answer:
Vetri and Asif studied in the same school at Keelakudi village. They were always together in learning and playing. They continued to be good friends till their X Std. After that, Asif had to move to Chennai.

Question 5.
Who stopped Vetri at the gate of Asif’s office?
Answer:
The security stopped Vetri at the gate and asked who does he wants to meet.

Question 6.
How did Asif know that Vetri had come to his office?
Answer:
Asif saw him through the CCTV camera and came to receive him.

V. Paragraph Questions with Answers.

Question 1.
How did Vetri and Asif’s friendship start?
Answer:
Their friendship started on the first day of school. When Vetri’s parents dropped him at the school, he started crying. Asif consoled him with chocolate and asked him not to cry. From that day, they stayed together, played together, and even exchanged their food. Their friendship grew stronger with time. They were always good in their studies and helped each other in lessons. There was always a healthy competition between them. Surprisingly, the exams, results, and marks never affected their friendship. Their friendship continued till the tenth standard. Then Asif moved to Chennai

Friendship Grammer Additional

Degrees of Comparison

I. Fill in the blanks.

Question 1.

Positive	Comparative	Superlative
big
	costlier
		hardest
short
	simpler
		tallest
long
	faster

Answer:

Positive	Comparative	Superlative
big	bigger	biggest
costly	costlier	costliest
hard	harder	hardest
short	shorter	shortest
simple	simpler	simplest
tall	taller	tallest
long	longer	longest
fast	faster	fastest

Question 2.

Positive	Comparative	Superlative
beautiful
	more different
		most enjoyable
effectively
	more delicious
		most useful
honest		most honest
	more qualified

Answer:

Positive	Comparative	Superlative
beautiful	more beautiful	most beautiful
different	more different	most different
enjoyable	more enjoyable	most enjoyable
effectively	more effectively	most effectively
delicious	more delicious	most delicious
useful	more useful	most useful
honest	more honest	most honest
qualified	more qualified	most qualified

Question 3.

Adjectives	Comparative	Superlative
bad
	drier
far (place)		farthest
far (time)		furthest
	better
late (time)		latest
late (order)	latter
	less
		most
safe
simple	simpler

Answer:

Adjectives	Comparative	Superlative
bad	worse	worst
dry	drier	driest
far (place)	farther	farthest
far (time)	further	furthest
good	better	best
late (time)	later	latest
late (order)	latter	last
little	less	least
many	more	most
safe	safer	safest
simple	simpler	simplest

II. Change the following sentences into comparative degree sentences.

Question 1.
India is not as large as China.
Answer:
China is larger than India.

Question 2.
Water is not as light as air.
Answer:
Air is lighter than water.

Question 3.
My brother is not as tall as I am.
Answer:
I am taller than my brother.

Question 4.
Ravi is not as intelligent as Magesh.
Answer:
Magesh is more intelligent than Ravi.

Question 5.
No other problem facing our country is as serious as unemployment.
Answer:
Unemployment is more serious than any other problem facing our country.

III. Change the following sentences into Superlative degree sentences.

Question 1.
Chandrasekar is richer than any other man in this town.
Answer:
Chandrasekar is the richest man in this town.

Question 2.
Vijay is taller than any other student in this class.
Answer:
Vijay is the tallest student in this class.

Question 3.
Shakespeare is greater than any other English poet.
Answer:
Shakespeare is the greatest of English Poets.

Question 4.
Greenland is larger than any other island.
Answer:
Greenland is the largest island.

Question 5.
Anita is prettier than any other girl in my class.
Answer:
Anita is the prettiest girl in my class.

Friendship Summary

Section – I

This is a story about the Friendship between Vetri and Asif. Vetri was a successful businessman in Coimbatore. His Vetri construction company constructed many shopping complexes, houses, and a few apartments in and around Coimbatore. After his father’s death, he was forced to split the wealth to his brothers. From then on, Vetri found it difficult to establish his business. He could not pay the loans taken to run his company. So he sold all his properties and paid the loans. He moved to a very small house with his family.

He found a job and led a normal life. But he desired to start a company again. No one was ready to lend him money. One day, his wife suggested him to ask his friend Asif for help. But Vetri hesitated to ask him for help, as he had not seen him for a long time. After deep thought, he decided to meet his friend Asif who was in Chennai.

Fill in the blanks :

1. _________ constructions was once a leading company
2. He took a_________to run his company.
3. Vetri’s friend is_________

Answers:

1. Vetri
2. loan
3. Asif

Section – II

Vetri boarded the train to Chennai. His memory went back to his school days. They lived in the beautiful village of Keelakudi near Coimbatore. Vetri and Asif studied in a school in the outskirts of the village Keelakudi. The teachers and the students would never forget Vetri and Asif and their friendship. They were always together in learning and playing.

Their friendship started on the first day of school. On the first day, when Vetri started crying, Asif consoled him with chocolate and told him not to cry. After that day, they stayed together played together, and even exchanged their food. They were always good in their studies and helped each other in lessons.

One day, there was a quarrel between Vetri and Asif’s families. But this did not affect their friendship. Their exams, results and marks never affected their friendship. After their tenth standard, Asif moved to Chennai and Vetri settled in Coimbatore.

Say True or False.

1. Keelakudi was a native of Vetri.
2. The school was a middle school.
3. Vetri and Asif were good in their studies.
4. Vetri never visited Chennai.
5. Asif was a businessman.

Answer:

1. True
2. True
3. True
4. True
5. True

Section – III

When Vetri went to Asif’s office, he was stopped by the security. Vetri told him that he wanted to meet his friend Asif. He directed Vetri to the Receptionist. When Vetri was asking the receptionist whether he could meet Asif his friend, Vetri got a pat on his back. It was Asif his friend. Asif saw him through the CCTV camera and came down to receive him. Vetri was speechless and apologized to Asif that he did not get a chance to visit Chennai till then. He said that since he had come to attend his friend’s wedding he thought that he could meet him.

They spoke about their school days, the fun they had. They discussed their business. Asif took Vetri to his home to lunch. Vetri was surprised to see that everyone knows him. In the evening, Asif dropped him at the railway station. Vetri reached home and told his wife everything in detail. Two days later, he received a call from Asif’s office stating that Asif, their M.D. had assigned a project to Vetri.

Samacheer Kalvi 8th English Prose

• The Nose-Jewel Book Back Answers
• Hobby – Turns A Successful Career Book Back Answers
• Sir Isaac Newton – The Ingenious Scientist Book Back Answers
• My Reminiscence Book Back Answers
• Being Safe Book Back Answers
• Friendship Book Back Answers
• Cyber Safety Book Back Answers