10th Standard Mathematics All Chapter Creative Questions-I-2019-2020

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Questions:

Generic 2

7 x 2 = 14

Find the depth of a cylindrical tank of radius 28 m, if its capacity is equal to that of a rectangular tank of size 28 m x 16 m x 11 m.

In figure OA· OB = OC·OD
Show that \(\angle A=\angle C\ and\ \angle B=\angle D\)

Find the standard deviation for the following data. 5, 10, 15, 20, 25. And also find the new S.D. if three is added to each value.

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 an arrow .

Solve the following system of linear equations in three variables.
x + y + z = 6; 2x + 3y + 4z = 20;
3x + 2y + 5z = 22

Show that any positive odd integer is of the form 4q + 1 or 4q + 3, where q is some integer.

If A (-5, 7), B (-4, -5), C (-1, -6) and D (4, 5) are the vertices of a quadrilateral, find the area of the quadrilateral ABCD.

Generic 5

16 x 5 = 80

If ATB=90^o then prove that
\(\sqrt { \frac { tanA\quad tanB+tanA\quad cotB }{ sinA\quad secB } } -\frac { { Sin }^{ 2 }A }{ { Cos }^{ 2 }A } =tanA\)

In figure 0 is any point inside a rectangle ABCD. Prove that OB² + OD²= OA²+ OC²

10.

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.

11.

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

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.

What is the ratio of the volume of a cylinder, a cone, and a sphere. If each has the same diameter and same height?

15.

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

16.

A two digit number is such that the product of its digits is 18, when 63 is subtracted from the number, the digits interchange their places. Find the number.

17.

The sum of two numbers is 15. If the sum of their reciprocals is \(\frac{3}{10}\), find the numbers.

18.

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

19.

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

Find the value of f(3),

20.

Find a relation between x and y if the points (x, y) (1, 2) and (7, 0) are collinear.

21.

Let f = {(2, 7); (3, 4), (7, 9), (-1, 6), (0, 2), (5,3)} be a function from A = {-1,0, 2, 3, 5, 7} to B = {2, 3, 4, 6, 7, 9}. Is this
(i) an one-one function
(ii) an onto function,
(iii) both one and onto function?

22.

Which of the following list of numbers form an AP? If they form an AP, write the next two terms:
4,10,16, 22, ...

23.

Find the coordinates at the points of trisection (i.e. points dividing in three equal parts) of the line segment joining the points A(2, -2) and B(-7, 4).

Multiple Choice Question

10 x 1 = 10

24.

Kamalam went to play a lucky draw contest 135 tickets of the lucky draw were sold. If the probability of Kamalam winning is \(\frac { 1 }{ 9 } \), then the number of tickets bought by kamalam is ____________

(a)

(b)

(c)

(d)

25.

The four vertices of a quadrilateral are (1, 2), (5, -6), (7, -4) and (k, -2) taken in order. If the area of quadrilateral is zero then find the value of k.

(a)

(b)

-2

(c)

(d)

26.

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)

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

27.

In figure \(\angle OAB={ 60 }^{ o }\) and OA = 6cm then radius of the circle is ____________

(a)

\(\frac { 3 }{ 2 } \sqrt { 3 } cm\)

(b)

2 cm

(c)

\(3\sqrt { 3 } cm\)

(d)

\(2\sqrt { 3 } cm\)

28.

The square root of 4m²- 24m + 36 is ___________

(a)

4(m-3)

(b)

2(m-3)

(c)

(2m-3)²

(d)

(m-3)

29.

Given a₁= -1, \(a=\frac { { a }_{ n } }{ n+2 } \), then a₄ is ____________

(a)

\(-\frac { 1 }{ 20 } \)

(b)

\(-\frac { 1 }{ 4 } \)

(c)

\(-\frac { 1 }{ 840 } \)

(d)

\(-\frac { 1 }{ 120 } \)

30.

If \(f(x)=\frac { 1 }{ x } \), and \(g(x)=\frac { 1 }{ { x }^{ 3 } } \) then f o g o(y), is ________

(a)

\(\frac { 1 }{ { y }^{ 8 } } \)

(b)

\(\frac { 1 }{ { y }^{ 6 } } \)

(c)

\(\frac { 1 }{ { y }^{ 4 } } \)

(d)

\(\frac { 1 }{ { y }^{ 3 } } \)

31.

which of the following is true?

(a)

0 ≤ p(∈) ≤ 1

(b)

p(∈) > 1

(c)

p(∈) < 0

(d)

\(-\frac { 1 }{ 2 } \ge P(\in )\le \frac { 1 }{ 2 } \)

32.

If S₁ denotes the total surface area at a sphere of radius ૪ and S₂ denotes the total surface area of a cylinder of base radius ૪ and height 2r, then ___________

(a)

S₁= S₂

(b)

S₁> S₂

(c)

S₁< S₂

(d)

S₁= 2S₂

33.

Sin(45^o+ θ ) - cos(45^o- θ) is equal to ___________

(a)

2cos θ

(b)

(c)

2sin θ

(d)

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Answers Key:

Generic 2

7 x 2 = 14

Volume of the cylindrical tank = Volume of the rectangle tank
πr²h = 28 x 16 x 11 m³

\(h=\frac { 16\times 11 }{ 88 } =2m\)

OA· OB = OC . OD (Given)

so, \(\cfrac { OA }{ OC } =\cfrac { OD }{ OB } \)
Also we have
(vertically opposite angles) ...(2)
From (1) and (2)
\(\Delta AOD\sim \Delta COB\)( SAS similarity criterion)
So

(corresponding angles of similar triangles)

x	d' = \(\frac { x-15 }{ 5 } \)	d'²
5	-2	4
10	-1	1
15	0	0
20	1	1
25	2	4
	Σd = 0	Σd'²= 10

\(\bar { x } =\frac { \Sigma x }{ n } =\frac { 75 }{ 5 } \)=15
d'=\(\frac { x-\bar { x } }{ c } =\frac { x-A }{ c } \)
A is assumed mean c is common factor.
Here A= 15, C = 5
σ =\(\sqrt { \left( \frac { \Sigma d'^{ 2 } }{ n } \right) -\left( \frac { \Sigma d' }{ n } \right) ^{ 2 } } \) x c
=\(\sqrt { \frac { 10 }{ 5 } -0 } \) x c
=\(\sqrt { 2 } \) x 5
= 5\(\sqrt { 2 } \)
If 3 is added to each value, we get 8, 13, 18,23, 28 as new values.

x	d' = \(\frac { x-18 }{ 5 } \)	d'²
8	-2	4
13	-1	1
18	0	0
23	1	1
28	2	4
	Σd' = 10	Σd'²= 10

∴ σ =\(\sqrt { \left( \frac { \Sigma d'^{ 2 } }{ n } \right) -\left( \frac { \Sigma d' }{ n } \right) ^{ 2 } } \) x c
=\(\sqrt { \frac { 10 }{ 5 } -0 } \) x 5
=\(\sqrt { 2 } \) x 5
= 5\(\sqrt { 2 } \)
S.D. doesn't change when a number is added or subtracted to the values.

An arrow diagram

x + y + z = 6 ....(1)
2x + 3y + 4z = 20 ...(2)
3x + 2y + 5z = 22 ....(3)

Sub. z = 3 in (5) ⇒ y - 2(3) =-4
y=2
Sub. y = 2, z = 3 in (1), we get
x+2+3=6
x=1
x= 1,y = 2, z = 3

Let us start with taking a, where a is a +ve odd integer.
We apply the division algorithm with 'a' and 'b' = 4.
Since 0 ≤ r < 4, the possible remainders are 0,1,2,3.
That is, a can be 4q, or 4q + 1, or 4q + 2 or 4q + 3, where 1 is the quotient. However, since a is odd, a cannot be 4q or 4q + 2 (since they are both divisible by 2).
Any odd integer is of the form 4q + 1 or 4q + 3

By joining B to D, you will get too triangles ABD and BCD.
Now, the area of ΔABD
=\(\frac { 1 }{ 2 } \)[ -5(-5 - 5) + (-4)(5 -7) + 4(7 + 5)]
=\(\frac { 1 }{ 2 } \)(50 + 8 + 48)
=\(\frac { 106 }{ 2 } \) = 53 square units.
Also, the area of ΔBCD
=\(\frac { 1 }{ 2 } \) = [-4(-6 - 5) - 1(5 + 5) + 4(-5 + 6)]
=\(\frac { 1 }{ 2 } \)(44 - 10 + 4)
= 19 square units.
So, the area of quadrilateral ABCD
= 53 + 19 = 72 square units.

Generic 5

16 x 5 = 80

Through O, draw PQIIBC so that P lies on AB and Q lies on DC
Now, PQ II BC
\(PQ\bot AB\quad PQ\bot OC\)
\(\left( \because \angle B={ 90 }^{ 0 }and\angle C={ 90 }^{ 0 } \right) \)
So, \(\angle BPQ={ 90 }^{ 0 }\quad \angle CQP={ 90 }^{ 0 }\)
Therefore BPQC and APQD are both rectangles. Now from \(\Delta OPB\)
OB²= BP²+ OP²
Similarly from \(\Delta OQD\)
OD² = OQ² + DQ²
From \(\Delta OQC\)
OC² = OQ² + CQ²
\(\Delta OAP\) we have
OA² = AP² +OP²
Adding (1) and (2)
OB² + OD² = BP² + OP² + OQ² + DQ²
(As BP = CQ and DQ = AP)
= CQ² + OP²+ OQ² + AP²
= CQ² + OQ² + OP² + AP²
= OC²+ OA
[From (3) and (4)]

10.

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

11.

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

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.

Volume of a cylinder \(=\pi { r }^{ 2 }h\)
Volume of a cone \(=\frac { 1 }{ 3 } \pi { r }^{ 2 }h\)
Volume of a sphere \(=\frac { 4 }{ 3 } \pi { r }^{ 3 }\)
Their ratio V₁ : V₂ : V₃

\(h:\frac { h }{ 3 } :\frac { 4r }{ 3 } \)
3h: h: 4r
3h : h : 2(2r) (where 2r = h)
∴ V₁ : V₂ : V₃ = 3: 1 : 2

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 the tens digits be x. Then the uints digits=\(\frac{18}{x}\)
∴ Number =10x+\(\frac{18}{x}\)
and number obtained by interchanging the digits =10x+\(\frac{18}{x}\)
\(\\ \therefore \left( 10x+\frac { 18 }{ x } \right) -\left( 10\times \frac { 18 }{ x } +x \right) =63\)
\(\Rightarrow 10x+\frac { 18 }{ x } -\frac { 180 }{ x } -x=0\)
\(\Rightarrow 9x-\frac { 162 }{ x } -63=0\)
⇒ 9x²-63x-162=0
⇒ x²-7x-18=0
⇒ (x-9)(x+2)=0⇒ x=9,-2
But a digit can never be (-ve), so x = 9.
So, the required number \(=10\times 9+\frac { 18 }{ 9 } =92\)

17.

Let the numbers be ∝, β
Sum of the roots = ∝ + β = 15 ...(1)
\(\frac { 1 }{ \alpha } +\frac { 1 }{ \beta } =\frac { 3 }{ 10 } \quad \quad \quad ...(2)\)
\(\\ \frac { +\alpha }{ \alpha \beta } =\frac { 3 }{ 10 } \)
10(∝+ β)= 3∝β ....(3)
30∝β=10x15=150
Products of the roots =∝β=50 ....(4)
∴ From (1) & (4), we have
x²-15x+50=0
(x-10)(x-5)=0⇒x=10,5

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.

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

20.

If A(-2, -1), B(a, 0), C(4, b) and D(1, 2) are the vertices of a parallelogram, find the values of a and b.
We know that the diagonals of a parallelogram bisect each other. Therefore the co-ordinates of the midpoint of AC are same as the co-ordinates of the mid-point of BD. i.e.
\(\left( \frac { -2+4 }{ 2 } ,\frac { -1+b }{ 2 } \right) =\left( \frac { a+1 }{ 2 } ,\frac { 0+2 }{ 2 } \right) \)
⇒ \(\left( 1,\frac { b-1 }{ 2 } \right) =\left( \frac { a+1 }{ 2 } ,1 \right) \)
⇒ \(\frac { a+1 }{ 2 } \) = 1 ⇒ a + 1 = 2 ⇒ a = 1
⇒ \(\frac { b-1 }{ 2 } \) = 1 ⇒ b - 1 = 2 ⇒ b = 3

21.

It is both one-one and onto function

All the elements in A have their separate images in B. All the elements in B have their preimage in A. Therefore it is one-one and onto function.

22.

4, 10, 16,22, ...
We have a² - a² = 10 - 4 = 6
a³ - a² = 16 -10 = 6
ஃ It is an A.P. with common difference 6
ஃ The next two terms are,

23.

Let P and Q be the points of trisection at AB.
i.e., AP = PQ = QB

Therefore, P divides AB internally in the ratio 1:2. Therefore, the coordinates at P, by applying the section formula, are
\(\left[ \frac { 1(-7)+2(2) }{ 1+2 } ,\frac { 1(7)+2(-2) }{ 1+2 } \right] \) i.e., (-1,10)
Now, Q also divides AB internally in the ratio 2:1, so, the coordinates at Q are
\(\left[ \frac { 2(-7)+1(2) }{ 2+1 } ,\frac { 2(4)+(-2) }{ 2+1 } \right] \) i.e., (-4,2)
Therefore, the coordinates at the points at trisection of the line segment joining A and B are (-1, 0) and (-4, 2).