10th Standard Mathematics Creative Test Model

Download Tamil Nadu 10th 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 2

7 x 2 = 14

If the radii of the circular ends of a conical bucket which is 45 cm high are 28 cm and 7 cm, find the capacity of the bucket. (Use π = \(\frac{22}{7}\))

In figure if PQ || RS Prove that \(\Delta POQ\sim \Delta SOQ\)

Find the standard deviation of 30, 80, 60, 70, 20, 40, 50 using the direct method.

Let A = {0, 1, 2, 3} and B = {1, 3, 5, 7, 9} be two sets. Let f: A \(\rightarrow\)B be a function given by f(x) = 2x + 1. Represent this function as a graph.

Using quadratic formula solve the following equations.9x²-9(a+b)x+(2a²+5ab+2b²)=0

Prove that \(\sqrt { 3 } \) is irrational

Find a relation between x and y such that the point (x, y) is equidistant from the points (7, 1) and (3, 5).

Generic 5

16 x 5 = 80

A ladder is placed against a wall such that its foot is at a distance of 2.5 m from the wall and its top reaches a window 6 m above the ground. Find the length of the ladder.

From a point on a bridge across a river, the angles of depression of the banks on opposite sides at the river are 30° and 45°, respectively. If the bridge is at a height at 3 m from the banks, find the width at the river.

10.

In \(AD\bot BC\) prove that AB²+ CD² = BD²+ AC².

11.

If sin (A - B) = \(\frac12\), cos (A + B) = \(\frac12\), 0^o < A + ≤ 90°, A > B, find A and B.

12.

A wooden article was made by scooping out a hemisphere from each end of a cylinder as shown in figure. If the height of the cylinder is 10 cm and its base is of radius 3.5 cm find the total surface area of the article.

13.

S.D. of a data is 2102, mean is 36.6, then find its C.V.

14.

Find the number of coins, 1.5 cm is diameter and 0.2 cm thick, to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.

15.

Find the co-efficient of variation for the following data: 16, 13, 17,21, 18.

16.

Find two consecutive natural numbers whose product is 20.

17.

A chess board contains 64 equal squares and the area of each square is 6.25 cm², A border round the board is 2 cm wide.

18.

How many terms of the AP: 24, 21, 18, ... must be taken so that their sum is 78?

19.

Which of the following list of numbers form an AP? If they form an AP, write the next two terms:
1,-1,-3, -5, ...

20.

A function f: (1,6) \(\rightarrow\)R is defined as follows:

Find the value of f(3),

21.

Let A = {1, 2, 3, 4, 5}, B = N and f: A \(\rightarrow\)B be defined by f(x) = x². Find the range of f. Identify the type of function.

22.

Find the value of k if the points A(2, 3), B(4, k) and (6, -3) are collinear.

23.

Find the area of a triangle vertices are(1, -1), (-4, 6) and (-3, -5).

Multiple Choice Question

10 x 1 = 10

24.

The range of first 10 prime number is ___________

(a)

(b)

(c)

(d)

25.

It S₁ denotes the total surface area of a sphere of radius r and S₂ denotes the total surface area of a cylinder of base radius r and height 2r, then ___________

(a)

S₁= S₂

(b)

S₁> S₂

(c)

S₁< S₂

(d)

S₁= 2S₂

26.

The y-intercept of the line 3x - 4y + 8 = 0 is ___________

(a)

\(-\frac { 8 }{ 3 } \)

(b)

\(\frac { 8 }{ 3 } \)

(c)

(d)

\(\frac { 1 }{ 2 } \)

27.

Two concentric circles if radii a and b where a>b are given. The length of the chord of the circle which touches the smaller circle is ____________

(a)

\(\sqrt { { a }^{ 2 }-{ b }^{ 2 } } \)

(b)

\(\sqrt { { a }^{ 2 }-{ b }^{ 2 } } \)

(c)

\(\sqrt { { a }^{ 2 }+{ b }^{ 2 } } \)

(d)

\(2\sqrt { { a }^{ 2 }+{ b }^{ 2 } } \)

28.

The ratio of the areas of two similar triangles is equal to ____________

(a)

The ratio of their corresponding sides

(b)

The cube of the ratio of their corresponding sides

(c)

The ratio of their corresponding attitudes

(d)

The square of the ratio of their corresponding sides

29.

If \(2A+3B=\left[ \begin{matrix} 2 & -1 & 4 \\ 3 & 2 & 5 \end{matrix} \right] \) and \(A+2B=\left[ \begin{matrix} 5 & 0 & 3 \\ 1 & 6 & 2 \end{matrix} \right] \) then B = [hint: B = (A+2B)-(2+3B)]

(a)

\(\left[ \begin{matrix} 8 & -1 & -2 \\ -1 & 10 & -1 \end{matrix} \right] \)

(b)

\(\left[ \begin{matrix} 8 & -1 & 2 \\ -1 & 10 & -1 \end{matrix} \right] \)

(c)

\(\left[ \begin{matrix} 8 & 1 & 2 \\ -1 & 10 & -1 \end{matrix} \right] \)

(d)

\(\left[ \begin{matrix} 8 & 1 & 2 \\ 1 & 10 & 1 \end{matrix} \right] \)

30.

Sum of first n terms of the series \(\sqrt { 2 } +\sqrt { 8 } +\sqrt { 18 } +...\) is ____________

(a)

\(\frac { n(n+1) }{ 2 } \)

(b)

\(\sqrt { n } \)

(c)

\(\frac { n\left( n+1 \right) }{ \sqrt { 2 } } \)

(d)

31.

The function t which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined Fahrenheit degree is 95, then the value of C \(t(C)=\frac { 9c }{ 5 } +32\) is ___________

(a)

(b)

(c)

(d)

32.

A cylinder 10 cone and have there are of a equal base and have the same height. what is the ratio of there volumes?

(a)

3:1:2

(b)

3:2:1

(c)

1:2:3

(d)

1:3:2

33.

The value of the expression \(\left[ \frac { { sin }^{ 2 }{ 22 }^{ o }+{ sin }^{ 2 }{ 68 }^{ o } }{ { cos }^{ 2 }{ 22 }^{ 0 }+{ cos }^{ 2 }{ 68 }^{ 0 } } +{ sin }^{ 2 }{ 63 }^{ o+ }{ cos }63^{ 0 }{ sin27 }^{ 0 } \right] \)is ___________

(a)

(b)

(c)

(d)

Visit QB365-SLMS - https://www.qb365.in/slms/

Answers Key:

Generic 2

7 x 2 = 14

Clearly bucket forms frustum of a cone such that the radii of its circular ends are r₁= 28 cm, r₂= 7 cm, h = 45 cm
Capacity of the bucket = volume of the frustum
\(\Rightarrow \frac { 1 }{ 3 } \times \pi h[{ r }_{ 1 }^{ 2 }+{ r }_{ 2 }^{ 2 }+{ r }_{ 1 }{ r }_{ 2 }]\)
\(\Rightarrow \frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times 45[{ 28 }^{ 2 }+{ 7 }^{ 2 }+28\times 7)]\)
= 22 x 15 x (28 x 4 + 7 + 28)
\(\Rightarrow 330\times 147{ cm }^{ 2 }\Rightarrow 48510{ cm }^{ 2 }\)

PQ II RS

Also

\(\angle \therefore \Delta POQ\sim \Delta SOR\) (AAA similarity criterion)

x	x²
30	900
80	6400
60	3600
70	4900
20	400
40	1600
50	2500
Σx = 350	Σx²= 20300

σ =\(\sqrt { \frac { \Sigma x^{ 2 } }{ n } -\left( \frac { \Sigma x }{ n } \right) ^{ 2 } } \)
=\(\\ \sqrt { \frac { 20300 }{ 7 } -\left( \frac { 350 }{ 7 } \right) ^{ 2 } } \)
=\(\sqrt { 400 } \) = 20

A Graph f = {(x, f(x) / x \(\epsilon \) A}
{(0, 1), (1, 3), (2, 5), (3, 7)}

9x²-9(a+b)x+(2a²+5ab+2b²)=0
Comparing this with ax² + bx + c = O.
a =9
b = -9(a + b)
c = (2a² + 5ab + 2b²)
∴ ∆=B²-4AC
⇒ 81(a+b)²-36(2a²+5ab+2b²)
⇒ 9a² + 9b² - 18ab
⇒ 9(a - b)2> 0
∴ the roots are real and given by
\(\beta =\frac { -B-\sqrt { D } }{ 2A } =\frac { 9(a+b)+3(a-b) }{ 18 } \)
\(=\frac { 12a+6b }{ 18 } =\frac { 2a+b }{ 3 } \)
\(\beta =\frac { -B-\sqrt { D } }{ 2A } =\frac { 9(a+b)+3(a-b) }{ 18 } \)
\(=\frac { 6a+12b }{ 18 } =\frac { a+2b }{ 3 } \)

Let us assume the opposite, (1) \(\sqrt { 3 } \) is irrational.
Hence \(\sqrt { 3 } =\frac { p }{ q } \)
Where p and q (q ≠ 0) are co-prime (no common factor other than 1)
Hence, \(\sqrt { 3 } =\frac { p }{ q } \)
\(\sqrt { 3 } \)q = p
Squaring both side
\({ (\sqrt { 3 }q ) }^{ 2 }={ p }^{ 2 }\)
3q² = p²
\({ q }^{ 2 }=\frac { p }{ 3 } \)
Hence, 3 divides p²So 3 divides p also .....(1)
Hence we can say
\(\frac{p}{3}\) = c where c is some integer
s, p =p²
Putting p = 3c
3q²= (3c)²
3q²= 9c²
q2 = \(\frac13\) x 9c²
q² = 3c²
\(\frac{9^2}{3}\) = c²
Hence 3 divides q²
So, 3 divides q also ...(2)
By (1) and (2) 3 divides both p and q
By contradiction \(\sqrt { 3 } \) is irrational.

Let P(x, y) be equidistant from the points A (7, 1) and B (3, 5).
We are given that AP = BP. So, AP² = BP²
(x - 7)² + (y - 1)²= (x - 3)² + (y - 5)²
x²- 14x + 49 + y - 2y + 1 = x²- 6x + 9 + y -10y + 25
x - y = 2
Which is the required relation.

Generic 5

16 x 5 = 80

Let AB be the ladder and CA be the wall with . the window at A.
Also, BC = 2.5 m and CA = 6 m
From Pythagoras theorem
AB² = BC²+ CA²
= (2.5)² + (6)²
= 42.25
AB = 6.5
Thus, length at the ladder is 6.5 m.

A and B represent points on the bank on opposite sides at the river, so that AB is the width of the river. P is a point on the bridge at a height of 3m i.e., DP = 3 m. We are interested to determine the width at the river which is the length at the side AB of the ΔAPB.

Now, AB = AD + DB
In right ΔAPD,
ㄥA = 30^o
So, tan 30° = \(\frac { 1 }{ \sqrt { 3 } } =\frac { 3 }{ AD } \)
i.e., \(\frac { 1 }{ \sqrt { 3 } } =\frac { 3 }{ AD } \) (or) AD = 3\(\sqrt { 3 } \)
Also, in right ΔPBD,
B = 45^o
So, BD = PD = 3m
Now, AB = BD + AD
= 3 + 3\(\sqrt { 3 } \) = 3(1+\(\sqrt { 3 } \))m
Therefore, the width at the river is 3(+\(\sqrt { 3 } \))m

10.

From \(\Delta ADC\) we have
AC²= AD² + CD²....(1)
(Pythagoras theorem)
From \(\Delta ADB\) we have
AB² = AD² + BD² ...(2)
(Pythagoras theorem)
Subtracting (1) from (2) we have,
AB² - AC² = BD² - CD²
AB² + CD² = BD² + AC²

11.

Since, sin(A - B) = \(\frac12\), ∴ A-B = 30° ..... (1)
Also, since cos (A + B) = \(\frac12\)
∴ A + B = 60° ...(2)
Solving (1) and (2)
A - B + A + B = 30^o + 60^o
2A = 90^o
A = 45^o
We get,
A = 45° and B = 15°

12.

Radius of the cylinder be r
Height of the cylinder be h
Total surface area of the article
= CSA of cylinder + CSA of 2 hemispheres
= 2ㅠrh + 2πr²= 2πr(h + 2r)
\(=2\times \frac { 22 }{ 7 } \times 3.5\times (10+2\times 3.5)\)
= 22 x 17 = 374 cm²

13.

σ = 21.2, \(\bar { x } \) = 36.6
CV = \(\frac { \sigma }{ \bar { x } } \) x 100 = \(\frac { 21.2 }{ 36.6 } \) x 100 = 57.92%

14.

No. of coins required \(=\frac{Volume\ of\ the\ cylinder}{Volume\ of\ 1\ coin}\)
\(=\frac { \pi { r }_{ 1 }^{ 2 }{ h }_{ 1 } }{ \pi { r }_{ 2 }^{ 2 }{ h }_{ 2 } } =\frac { \pi \times \frac { 42 }{ 20 } \times \frac { 45 }{ 20 } \times 10 }{ \pi \times \frac { 15 }{ 20 } \times \frac { 15 }{ 20 } \times \frac { 2 }{ 10 } } \)

= 450

15.

Mean \(\bar { x } \) = \(\frac { 16+13+17+21+18 }{ 5 } =\frac { 85 }{ 5 } \) = 17

x	d = x - 17	d²
16	-1	1
13	-4	16
17	0	0
21	4	16
18	1	1
	Σd = 0	Σd²= 34

σ =\(\sqrt { \frac { \Sigma d^{ 2 } }{ n } } =\sqrt { \frac { 34 }{ 5 } } =\sqrt { 638 } \)
σ = 2.61
Co-efficient of variation
CV = \(\frac { \sigma }{ \bar { x } } \) x 100 = \(\frac { 2.61 }{ 17 } \) x 100
= 15.35%

16.

Let a natural number be x.
The next number = x + 1
x (x + 1) = 20
X²+x-20 = 0
(x + 5)(x - 4) = 0
x=-5,4
∴ x=4
(∵ x≠-5, x is a natural number
The next number = 4 + 1= 5
Two consecutive numbers are 4,5.

17.

Let the length of the side of the chess board be x cm. Then
Area of 64 squares = (x - 4)²
(x - 4)² = 64 x 6.25
⇒ x²-8x+ 16=400
⇒X²- 8x - 384 =0
⇒ X²- 24x + 16x - 384 = 0
⇒(x - 24)(x + 16) = 0
⇒ x=24 cm.

18.

Here a = 24, d = 21-24 = -3, S_n = 78. We need to find n.
We know that,
S_n= \(\frac{n}{2}\) (2a + (n - 1)d)
78 = \(\frac{n}{2}\) 48 + (n -1)(-3))
78 = \(\frac{n}{2}\) 2(51-3n)
or 3n²- 51n + 156 = 0
n² -17n + 52 = 0
(n - 4)(n - 13) = 0
n = 4 or 13
The number of terms are 4 or 13.

19.

1,-1,-3, -5, ...
t₂ - t₁= -1 - 1 = -2
t₃ - t₂ = -3 - (-1)= -2
t₄ - t₃ = -5 - (-3) = -2
The given list of numbers form an A.P with the common difference -2.
The next two terms are (-5 + (-2)) = -7, -7 + (-2) = -9.

20.

f(3) =. 2x - 1
= 2(3) - 1 = 6 - 1 = 5

21.

A = {1,2,3,4,5}
B = {1,2,3,4, ... }
f: A \(\rightarrow\)B, f(x) = x²
\(\therefore\) f(1) = 1² = 1
f(2) = 2²= 4
f(3) = 3² = 9
f(4) = 4²= 16
f(5) = 5²= 25
\(\therefore\) Range of f = {1, 4, 9, 16, 25}
Elements in A have been connected with different elements in B. Therefore it is one-one function. But not all the elements in B have preimages in A. Therefore it is not on-to function.

22.

Since the given points are collinear, the area a the triangle formed by them must be 0, i.e.,
= \(\frac { 1 }{ 2 } \) [2(k + 3) + 4(-3 - 3) + 6(3 - k)] = 0
= \(\frac { 1 }{ 2 } \) [-4k] = 0
k = 0
Area of ΔABC
= \(\frac { 1 }{ 2 } \)[2(0 + 3) + 4(-3 - 3) + 6(3 - 0)] = 0

23.

The area of the triangle formed by the vertices A(1, -1), B(-4, 6) and C(-3, -5), by using the formula above, is given by
= \(\frac { 1 }{ 2 } \)[1(6 + 5) +(-4) (-5 + 1) + (-3)(-1 - 6)]
= \(\frac { 1 }{ 2 } \)[11 + 16 + 21] = 24 square units.

Multiple Choice Question

10 x 1 = 10

24.

25.