Download Tamil Nadu 8th Standard Maths question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.

Questions:

Generic 3

8 x 3 = 24

Use Ceasar Cipher table set + 4 and to try to solve the given secret sentence.
fvieo mr gshiw ger fi xvmgoc

Relation between Principal and Simple Interest:
A bank gives 10% 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 compute
(i) The annual interest obtainable for investment of Rs.450;
(ii) The amount a Senior citizen has to invest to get an annual simple interest of Rs.80.

Graph the equation y = x + 1.
Begin by choosing a couple of values for x and y. It will firstly help to see
(i) what happens to y when x is zero and
(ii) what happens to x when y is zero.
After this we can go on to find one or two more values.
Let us find at least two more ordered pairs. For easy graphing, let us avoid fractional answers. We shall make suitable guesses.

Given that one pair of new born rabbits they produce a new pair each month and from the second month, each new pair can breed themselves. Find how many pairs of rabbits are bred from one pair in a year, and find the relationship between the number of months and the number of pairs of rabbits by tabulation (a pair means (a male and a female)).

Construct a parallelogram BEAR with BE = 7 cm, BA = 7.5 cm and ㄥBEA = 80⁰. Also find its area.

Construct a parallelogram DUCK with DC = 8 cm, UK = 6 cm and ㄥDOU = 110⁰. Also find its area.

The value of a motor cycle 2 years ago was Rs.70000. It depreciates at the rate of 4% p.a. Find its present value.

When a number is decreased by 25% it becomes 120. Find the number.

Generic 2

12 x 2 = 24

Do the given problems by repeated subtraction method and verify the result.

10.

Using repeated subtracting method find HCF of the following:
280 and 420

11.

Frame Additive cipher table (key = 4).

12.

The following is a table of values connecting the radii of a few circles and their circumferences (Taking \(\pi =\frac { 22 }{ 7 } \)) Illustrate the relation with a graph and find
(i) The radius when the circumference is 242 units.
(ii) The circumference when the radius is 24.5 units.

Radius (r)	Circumference (2πr)
7	2 x \(\frac { 22 }{ 7 } \) x 7=44
14	2 x \(\frac { 22 }{ 7 } \) x 14=88
21	2 x \(\frac { 22 }{ 7 } \) x 21=132
28	2 x \(\frac { 22 }{ 7 } \) x 28=176
35	2 x \(\frac { 22 }{ 7 } \) x 35=220
42	2 x \(\frac { 22 }{ 7 } \) x 42=264
49	2 x \(\frac { 22 }{ 7 } \) x 49=308

13.

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

14.

I. Construct the following parallelograms with the given measurements and find their area.
1. ARTS, AR = 6 cm, RT = 5 cm and ㄥART = 70⁰. .
2. CAMP, CA = 6 cm, AP = 8 cm and CP = 5.5 cm.
3. EARN, ER = 10 cm, AN = 7 cm and ㄥEOA = 110° where \(\overset { \_ \_ }{ ER } \) and \(\overset { \_ \_ }{ AN } \) intersect at O.
4. GAIN, GA = 7.5 cm, GI = 9 cm and ㄥGAI = 100⁰.

15.

Find x and m: (i)\(\frac { 3x-2 }{ 4 } =\frac { (x-3) }{ 5 } =-1\)
(ii) \(\frac { m+9 }{ 3m+15 } =\frac { 5 }{ 3 } \)

16.

I. Construct the following trapeziums with the given measures and also find their area.
1. AIMS with \(\overset { \_ \_ }{ AI } \) || \(\overset { \_ \_ }{ SM } \), AI = 6 cm, IM = 5 cm, AM = 9 cm and MS = 6.5 cm.
2. CUTE with \(\overset { \_ \_ }{ CD } \) || \(\overset { \_ \_ }{ ET } \), CU = 7 cm, ㄥUCE = 80⁰ CE = 6 cm and TE = 5 cm..
3. ARMY with \(\overset { \_ \_ }{ AR } \) || \(\overset { \_ \_ }{ YM } \), AR = 7 cm, RM = 6.5 cm ㄥRAY = 100⁰ and ㄥARM = 60⁰
4. CITY with \(\overset { \_ \_ }{ CI } \) || \(\overset { \_ \_ }{ YT } \), CI = 7 cm, IT = 5.5 cm, TY = 4 cm and YC = 6 cm.

17.

Rithika buys an LED TV which has a 25 inches screen. If its height is 7 inches, how wide is the screen? Her TV cabinet is 20 inches wide. Will the TV fit into the cabinet? Why?

18.

Some articles are bought at 2 for Rs.15 and sold at 3 for Rs. 25. Find the gain percentage

19.

If the numerator of a fraction is increased by 50% and the denominator is decreased by 20%, then it becomes \(\frac { 3 }{ 5 } \). Find the original fraction.

20.

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.
(i) How many members are there in the sports club?
(ii) How many play cricket, volleyball and badminton?

Multiple Choice Question

5 x 1 = 5

21.

There are four groups of letters in each set. Three of these sets are a like in some way while one is different. Find the one which is different.

(a)

H K N Q

(b)

I L O R

(c)

J M P S

(d)

A D G J

22.

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°

23.

(a) \(\frac { x }{ 2 } \) = 10 (i) x = 4
(b) 20 = 6x – 4 (ii) x = 1
(c) 2x – 5 = 3 – x (iii) x = 20
(d) 7x – 4 – 8x = 20 (iv) x = \(\frac { 8 }{ 3 } \)
(e) \(\frac { 4 }{ 11 } \)- x = \(\frac { -7 }{ 11 } \) (v) x = –24

(a)

(i), (ii), (iv), (iii), (v)

(b)

(iii), (iv), (i), (ii), (v)

(c)

(iii), (i), (iv), (v), (ii)

(d)

(iii), (i), (v), (iv), (ii)

24.

The sides of a right angled triangle are in the ratio 5: 12: 13 and its perimeter is 120 units then, the sides are ______________.

(a)

25, 36, 59

(b)

10, 24, 26

(c)

36, 39, 45

(d)

20, 48, 52

25.

What is the marked price of a hat which is bought for Rs. 210 at 16% discount?

(a)

Rs. 243

(b)

Rs. 176

(c)

Rs. 230

(d)

Rs. 250

Fill in the blanks

5 x 1 = 5

26.

H X R V M X V

27.

The value of x in the equation x + 5 = 12 is _____________

28.

The compound interest on Rs.5000 at 12% p.a for 2 years compounded annually is ____________.

29.

If the sides of a triangle are in the ratio 5: 12: 13 then, it is ________.

30.

If 'l' and ‘m’ are the legs and 'n' is the hypotenuse of a right angled triangle then, l² = ________.

True or False

5 x 1 = 5

31.

y = −9 x not passes through the origin.

(a)

True

(b)

False

32.

The coordinates of the origin are (1,1).

(a)

True

(b)

False

33.

The compound interest on Rs.16000 for 9 months at 20% p.a, compounded quarterly is Rs.2522.

(a)

True

(b)

False

34.

One of the legs of a right angled triangle PQR having ㄥR = 90⁰ is PQ.

(a)

True

(b)

False

35.

In a right angled triangle, the hypotenuse is the greatest side.

(a)

True

(b)

False

Generic 5

3 x 5 = 15

36.

Construct a trapezium DESK in which \(\overset { \_ \_ }{ DE } \) is parallel to \(\overset { \_ \_ }{ KS } \), DE = 8 cm, ES = 5.5 cm, KS = 5 cm and KD = 6 cm. Find also its area.

37.

Construct a trapezium CARD in which \(\overset { \_ }{ CA } \) is parallel to \(\overset { \_ }{ DR} \), CA = 9 cm, ㄥCAR = 70, AR = 6 cm and CD = 7 cm. Also find its area.

38.

Construct a trapezium BOAT in which \(\overset { - }{ BO } \) is parallel to \(\overset { - }{ TA } \), BO = 7 cm, OA = 6 cm, BA = 10 cm and TA = 6 cm. Also find its area.

Match the following

5 x 1 = 5

39.

multiplication

40.

subtraction

41.

mathematics

42.

7x – 4 – 8x = 20

43.

20 = 6x – 4

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Answers Key:

Generic 3

8 x 3 = 24

Let us make Ceasar Cipher table first. Here, we have to set to + 4 table. For that, we have to start letter e to set as A, f as B … likewise d as Z. Now, the + 4 Ceasar Cipher table looks like

Plain Text

Cipher Text

The given plain text is
fvieo mr gshiw ger fi xvmgoc
To crack this secret code, follow the steps given below.
Step 1: Using Ceasar Cipher table, let us first match the most repeated letters. This will help us to progress faster.
fvieo mr gshiw ger fi xvmgoc

Step 2: Then, let us find remaining letters to complete the code.

Thus, the secret sentence is, BREAK IN CODES CAN BE TRICKY

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

Deposit(in.Rs)	100	200	300	500	1000
Anaual S.I(in.Rs)	10	20	30	50	100

Deposit	Interest
100	\(\frac { 100\times 1\times 10 }{ 100 } \)=10
200	\(\frac { 200\times 1\times 10 }{ 100 } \)=20
300	\(\frac { 300\times 1\times 10 }{ 100 } \)=30
500	\(\frac { 500\times 1\times 10 }{ 100 } \)=50
1000	\(\frac { 1000\times 1\times 10 }{ 100 } \)=100

These are the points which are 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:
(i) Corresponding to Rs.300 on the x-axis, we get the interest as Rs.30 on the y-axis.
(ii) Corresponding to Rs.70 on the y-axis, we get the deposit as Rs.700 on the x-axis.

x	-2	-1	0	1	2
y	-1	0	1	2	3

x	y = x+1
−2 −1 0 1 2	−2+1 = –1 −1+1 = 0 0+1 = 1 1+1 = 2 2+1 = 3

The above figure clearly forms the sequence is 1, 1, 2, 3, 5, 8... Here, we find the pattern in which each number is in the Fibonacci sequence, obtained by adding together with previous two. Going on like this to find subsequent numbers at the twelfth month, we will get 144 pairs of rabbits. In the other words, twelfth Fibonacci number is 144.

Number of months	1	2	3	4	5	6	7	8	9	10	11	12
Number of pairs of rabbits	1	1	2	3	5	8	13	21	34	55	89	144

Given:
BE = 7 cm, BA = 7.5 cm and ㄥBEA = 80⁰

Steps:
1. Draw a line segment BE = 7 cm.
2. Make an angle ㄥBEX =80° at E on \(\overset { \_ \_ }{ BE } \).
3. With B as centre, draw an arc of radius 7.5 cm cutting \(\overset { \_ \_ }{ EX } \) at A and Join BA.
4. With B as centre, draw an arc of radius equal to the length of \(\overset { \_ \_ }{ AE } \).
5. With A as centre, draw an arc of radius 7 cm. Let both arcs cut at R.
6. Join BR and AR.
7. BEAR is the required parallelogram.
Calculation of area:
Area of the parallelogram BEAR = bh sq. units
= 7 x 4.1 = 28.7 sq. cm

Given:
DC = 8 cm, UK = 6 cm and ㄥDOU = 110⁰

Steps:
1. Draw a line segment DC = 8 cm.
2. Mark O the midpoint of \(\overset { \_ \_ }{ DC } \).
3. Draw a line \(\overset { \_ \_ }{ XY } \) through O which makes ㄥDOY = 110⁰.
4. With O as centre and 3 cm as radius draw two arcs on \(\overset { \_ \_ }{XY } \)on either sides of \(\overset { \_ \_ }{DC } \). Let the arcs cut \(\overset { \_ \_ }{ OX } \)at K and \(\overset { \_ \_ }{OY } \) at U
5. Join \(\overset { \_ \_ }{DU } \) , \(\overset { \_ \_ }{UC } \), \(\overset { \_ \_ }{CK } \) and \(\overset { \_ \_ }{KD } \).
6. DUCK is the required parallelogram.
Calculation of Area:
Area of the parallelogram DUCK = bh sq.units
= 5.8 x 3.9 = 22.62sq.cm

Depreciated value = P\({ \left( 1-\frac { r }{ 100 } \right) }^{ n }\)
= 70000\({ \left( 1-\frac { 4 }{ 100 } \right) }^{ 2 }\)
= 70000 x \(\frac { 96 }{ 100 } \times \frac { 96 }{ 100 } \)
= Rs.64512

Let the number be x.
x-\(\frac { 25 }{ 100 } x\) = 120
\(\frac { 100x-25x }{ 100 } \) = 120
\(\frac { 75x }{ 100 } \) = 120
x = \(\frac { 120\times 100 }{ 75 } \)
x = 160

Generic 2

12 x 2 = 24

320,120 and 95
First find,the HCF of 320 and 120.
There m = 320 and n = 120, Here m > n.
320 - 120 = 200
200 - 120 = 80
120 - 80 = 40
80 - 40 = 40
40 - 40 = 0
The HCF of 320 and 120 is 40.
Now find the HCF of 40 and 95.
95 - 40 = 55
55 - 40 = 15
40 - 15 = 25
25 -15 = 10
15 - 10 = 5
10 - 5 = 5
5 - 5 =0
The HCF of 320,120 and 95 is 5.

10.

140

11.

Plain Text	A	B	C	D	E	F	G	H	I	J	K	L	M	N	O	P	Q	R	S	T	U	V	W	X	Y	Z
CipherText	04	05	06	07	08	09	10	11	12	13	14	15	16	17	18	19	20	21	22	23	24	25	00	01	02	03

12.

13.

l = 8 cm, b = 24 cm

14.

1. AR = 6 cm; RT = 5 cm and
Rough diagram

Steps:
1. Diaw a liqe segment AR = 6 cm.
2. Make an angle o at R on AR
3. With R as centre, draw an arc of radius 5 cm cutting RX at T.
4. With A and T as centres draw arcs of radii 5 cm and 5 cm respectively. Let them cut at S.
5. ]oin AS and TS
6. ARTS is the required parallelogram
Calculation of area:
Area of the Parallelogram ARTS
= bh sq. units
= 6 x 4.2 = 28.2 sq. cm
2. Given: CA = 6 cm; AP = 8 cm and CP = 5.5 cm.
Rough diagram

Steps:
1. Draw a line segment CA = 6 cm.
2. With C and A as centres, draw an arc of radius 5.5 cm and 8 cm respectively. Let them cut at p.
3. ]oin CP and AP.
4. With A and P as centres draw an arcs of radius 5.5 cm and 6 cm respectively. Let them cut at M.
5. Join AM and PM.
6. CAMP is the required parallelogram.
Calculation of area:
Area of the Parallelogram CAMP '
= bh sq. units
= 6 x 5.4 = 32.4 sq. cm
3. Given: ER = 10cm ; AN = 7 cm and o
Rough diagram

Steps:
1. Draw a line segment ER = 10 cm.
2. Mark O the mid point of an angle 4. With O as centre and 3.5 cm arcs on XY on either side of .gn. tet the arcs cut OX at N and OY at A.
5. EARN is the required parallelogram.
Calculation of area:
Area of the Parallelogram EARN
= bh sq. units
= 7 x 4.8 = 33.6 sq. cm
4. Given: GA,=7.5 cm; GI = 9 cm and o
Rough diagram

Steps:
1. Draw a line segment GA = 7.5 cm.
2. Make an angle IGAI = 100^oat A on GA.
3. With G as centre, draw an arc of radius 9 cm cutting AX at I and join GI.
4. With G as centre drawer an arc of radius equal to the length of AI.
5. With I as centre, draw an arc of radius 7.5 cm. Let both arcs cut at N.
6. Join IN and GN.
7. GAIN is the required parallelogram
Calculation of area:
Area of the Parallelogram GAIN
= bh sq. units .
= 7.5 x 3.7 = 27.75 sq. cm

15.

(i) x = −2
(ii) m = −4

16.

Given:
AI = 6 cm,
IM = 5 cm,
AM = 9 cm,
MS = 6.5 cm.
Rough Diagram

Steps:
1. Draw aline segment AI = 6 cm.
2. With A and I as centres draw arcs of radius
9 cm and 5 cm respectively and let them cut at M.
3. Join AM and IM.
4. Draw MX parallel to AI.
5. With M as centre, draw an arc of radius6.5 cm cutting MX at S.
6. |oin AS. AIMS is the required trapezium
Calculation of Area:
Area of the trapezium AIMS
\(=1 / 2 \times h \times(a+b) \)
\(=1 / 2 \times 4.8 \times(6+6.5) \)
\(=1 / 2 \times 4.8 \times 12.5=30 \mathrm{sq} . \mathrm{cm} . \)
2. Given:
CU = 7cm; o
CE = 6cm; TE = 5cm.
Rough Diagram

Steps:
1. Draw a line segment CU = 7cm
2. Constract an angle o at c
3. With C as centre, draw an arc of radius 6 cm cutting CX at E.
4. Draw EY parallel to CU.
5. With E as centre, draw an arc of radius 5 cm cutting EY at T.
6. Join TLI CUTE is the required trapezium.
Calculation of Area:
Area of the trapezium CUTE
\(=1 / 2 \times \mathrm{h} \times(\mathrm{a}+\mathrm{b}) \)
\(=1 / 2 \times 6 \times(5+7) \)
\(=1 / 2 \times 6 \times 12=36 \text { sq. } \mathrm{cm} . \)
3. Given:
AR = 7 cm; RM = 6.5 cm
ZRAY = 100^o; ZARM = 60^o
Rough Diagram

Steps: -
1. Draw a line segment AR 7 cm.
2. Construct an angle o at R. S.
3. With R as centre draw an arc of radius 6.5 cm cutting RX at M.
4. Draw MZ parallel to AR.
5. Construct an angle ZRAY = 100^o A cutting MZ at Y.
6. ARMY is the required trapezium
Calculation of Area:
Area of the trapezium ARMY
\(=1 / 2 \times h \times(a+b) \)
\(=1 / 2 \times 5.6 \times(7+7) \)
\(=1 / 2 \times 5.6 \times 14=39.2 \text { sq. cm } \)
4. Given:
CI = 7 cm ; IT = 5.5 cm
TY = 4 cm ; YC = 6cm.
Rough Diagram

Steps:
1..Draw a line segment CI = 7 cm.
2. Mark the point A on CI such that CA = 4 cm.
3. With A and I as centres, draw arcs of radii 6 cm and S.S cm. Let them cut at T. join AT and IT.
4. With C and T as centres, draw arcs of radii 6 cm and 4 cm respectively. Let them cut at Y. Join TY and CY.
5. CITY is the required trapezium.
Calculation of Area:
Area of the trapezium CITY
\(=1 / 2 \times h \times(a+b) \)
\(=1 / 2 \times 6 \times(4+4) \)
\(=1 / 2 \times 6 \times 8=24 \text { sq. } \mathrm{cm} . \)

17.

Let x be the wide of the screen.
From the figure,
x²+ 2²= 25²
x² + 49 = 625
x² = 625 - 49
= 576
\(x=\sqrt{576}\)
= 24
The wide of the screen is 24 inches
The TV cabinet wide is 21 inches. It is not fit for the TV which has the wide screen 24 inches

18.

Let x be the total number of articles
\(\therefore \ \text { C.P }=\frac{15}{2} \times x \text { and S.P }=\frac{25}{3} \times x\)
Gain = S.P-C.P
\(=\frac{25}{3} x-\frac{15}{2} x\)
\(=\frac{50 x-45 x}{6}=\frac{5 x}{6}\)
Gain percentage \( =\frac{6}{\frac{15 x}{2}} \times 100 \)
\(=\frac{5 x}{6} \times \frac{2}{15 x} \times 100=\frac{100}{9} \)
\(=11 \frac{1}{9} \% \)

19.

Let the numerator be x and the denominator be y.
The original fraction = x/y
Given the numerator of a fraction is increased by 50%
\( \therefore \mathrm{Nr} =x+50 \% \text { of } x \)
\(=x+\frac{50}{100} x=x+\frac{1}{2} x \\ \mathrm{Nr} =\frac{3}{2} x \)
Also given the denominator is decreased by 20 %
\( \text { Dr } =y-20 \% \text { of } y \)
\(=y-\frac{20}{100} y=y-\frac{1}{5} y=\frac{4}{5} y \)
By given data
\( \frac{\frac{3}{2} x}{\frac{4}{5} y}=\frac{3}{5} \Rightarrow \frac{3}{2} x \times \frac{5}{4 y}=\frac{3}{5} \)
\(\frac{x}{y}=\frac{8}{25} \)
The original fraction is 8/25

20.

(i) 500
ii) Cricket -45,Volleyball-42, Badminton -33

Multiple Choice Question

5 x 1 = 5

21.

(d)
A D G J

22.

(a)
62°

23.

24.

(d)
20, 48, 52

25.

(d)
Rs. 250

Fill in the blanks

5 x 1 = 5

26.

()
SCIENCE

27.

()
x = 7

28.

()
1272

29.

()
right angled triangle

30.

()
n² − m²

True or False

5 x 1 = 5

31.

(b)
False

32.

(b)
False

33.

(a)
True

34.

(b)
False

35.

(a)
True

Generic 5

3 x 5 = 15

36.

Given:
DE = 8 cm, ES = 5.5 cm, KS = 5 cm, KD = 6 cm and \(\overset { \_ \_ }{ DE } \) || \(\overset { \_ \_ }{ KS } \)

Steps:
1. Draw a line segment DE = 8cm.
2. Mark the point A on DE such that DA = 5 cm.
3. With A and E as centres, draw arcs of radii 6 cm and 5.5 cm respectively. Let them cut at S. Join AS and ES.
4. With D and S as centres, draw arcs of radii 6 cm and 5 cm respectively. Let them cut at K. Join DK and KS.
5. DESK is the required trapezium.
Calculation of area:
Area of the trapezium DESK = \(\frac { 1 }{ 2 } \) x h x (a + b) sq. units
= \(\frac { 1 }{ 2 } \) x 5.5 x (8 + 5) = 35.75 sq. cm

37.

Given:
CA = 9 cm, ㄥCAR = 70⁰ AR = 6 cm, and CD = 7 cm and \(\overset { \_ }{ CA } \) || \(\overset { \_ }{ DR } \)

Steps:
1. Draw a line segment CA = 9 cm.
2. Construct an angle ㄥCAX = 70⁰at A.
3. With A as centre, draw an arc of radius 6 cm cutting AX at R.
4. Draw RY parallel to CA.
5. With C as centre, draw an arc of radius 7 cm cutting RY at D.
6. Join CD. CARD is the required trapezium.
Calculation of area:
Area of the trapezium CARD = \(\frac { 1 }{ 2 } \) x h x(a + b) sq.units
= \(\frac { 1 }{ 2 } \) x 5.6 x (9 + 11) = 56 sq. cm

38.

Given:
BO = 7cm, OA = 6cm, BA = 10cm,
TA = 6 cm and \(\overset { - }{ BO } \) || \(\overset { - }{ TA } \)

Steps:
1. Draw a line segment BO = 7 cm.
2. With B and O as centres, draw arcs of radii 10cm and 6cm respectively and let them cut at A.
3. Join BA and OA.
4. Draw AX parallel to BO
5. With A as centre, draw an arc of radius 6cm cutting AX at T.
6. Join BT. BOAT is the required trapezium.
Calculation of area:
Area of the trapezium BOAT = \(\frac { 1 }{ 2 } \) x h x (a+b) sq units
= \(\frac { 1 }{ 2 } \) x 5.9 x (7+6) = 38.35 sq. cm