Differentiation Test 24

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Differentiation Test 24

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Weekly Quiz Competition

Question 1
1 / -0

There is no symmetry in symmetry in legs of Fig

A

B

C

D

E

Solution

There is no symmetry in legs of Fig.(e)
Question 2
1 / -0

The area of a triangle with the vertices $$(1, 2), (3, 4)$$ and $$(5, 6)$$, is ____ square units.

A
$$0$$

B
$$1$$

C
$$2$$

D
$$5$$

Solution

Area of the triangle with three vertices $$(x_1,y_1)$$, $$(x_2,y_2)$$ and $$(x_3,y_3)$$ is

$$A=\dfrac { 1 }{ 2 } \left| { y }_{ 1 }({ x }_{ 2 }-{ x }_{ 3 })+{ y }_{ 2 }({ x }_{ 3 }-{ x }_{ 1 })+{ y }_{ 3 }({ x }_{ 1 }-{ x }_{ 2 }) \right|$$

Therefore, the area of triangle with vertices $$(1,2)$$, $$(3,4)$$ and $$(5,6)$$ is:

$$A=\dfrac { 1 }{ 2 } \left| 2(3-5)+4(5-1)+6(1-3) \right| =\dfrac { 1 }{ 2 } \left| -4+16-12 \right| =\dfrac { 1 }{ 2 } \left| 16-16 \right| =\dfrac { 1 }{ 2 } \left| 0 \right| =0$$

Hence, the area of the triangle is $$0$$.
Question 3
1 / -0

$$L, M$$ and $$N$$ are the midpoints of the sides $$BC, CA$$ and $$AB$$ respectively of triangle $$ABC$$. If the vertices are $$A(3,-4), B(5,-2)$$ and $$C(1,3)$$ the area of $$\displaystyle \triangle LMN$$ is ____ square units.

A
2.25

B
2.5

C
2.75

D
3

Solution

Area of triangle having vertices $$(x_1,y_1), (x_2,y_2)$$ and $$(x_3,y_3)$$ is given by
$$ = \dfrac{1}{2} \times [ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ] $$

Let $$(x_1,y_1)$$ $$=(3,-4)$$, $$(x_2,y_2)$$ $$=(5,-2)$$ and $$(x_3,y_3)$$ $$=(1,3)$$

Area of $$\displaystyle \Delta ABC=\dfrac{1}{2}\left [ 3\left ( -2-3 \right )+5\left ( 3+4 \right )+1\left ( -4+2 \right ) \right ]$$

$$=\dfrac{1}{2}\left ( -15+35-2 \right )$$

$$=9$$
$$\displaystyle \therefore $$ Area of $$\displaystyle \Delta LMN=\frac{1}{4}\times 9=2.25$$ square units.
Question 4
1 / -0

- T - CN - P - NT - C

A
PNTCT

B
TCCTN

C
NPTCP

D
NPCPT

Solution

The series is N T P C / N T P C / N T P C
Question 5
1 / -0

If coordinates of P,Q, and R are (3,6),(-1,3) and (2,-1) respectively. Then area of $$\displaystyle \triangle PQR$$ is ____ square units.

A
12

B
12.5

C
13

D
14

Solution

Area of triangle having vertices $$(x_1,y_1), (x_2,y_2)$$ and $$(x_3,y_3)$$ is given by
$$ = \dfrac{1}{2} \times [ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ] $$
Area of $$\displaystyle \Delta PQR$$ is given by
$$\displaystyle =\frac{1}{2}\left [ 3\left ( 3+1 \right )-1\left ( -1-6 \right )+2\left ( 6-3 \right ) \right ]$$
$$\displaystyle =\frac{1}{2}\left ( 12+7+6 \right )=12.5$$ square units.
Question 6
1 / -0

The area of the triangle formed by the points $$(2, 6), (10, 0)$$ and $$(0, k)$$ is zero square units. Find the value of $$k.$$

A
$$\dfrac {15}{2}$$

B
$$\dfrac 32$$

C
$$\dfrac 72$$

D
$$\dfrac {13}{2}$$

Solution

The area of the triangle is:

$$A=\dfrac { 1 }{ 2 } \left| { y }_{ 1 }({ x }_{ 2 }-{ x }_{ 3 })+{ y }_{ 2 }({ x }_{ 3 }-{ x }_{ 1 })+{ y }_{ 3 }({ x }_{ 1 }-{ x }_{ 2 }) \right|$$

We are given that the area of the triangle is $$0$$ and the points are $$(2,6)$$, $$(10,0)$$ and $$(0,k)$$, therefore,

$$\Rightarrow 0=\dfrac { 1 }{ 2 } \left| 6(10-0)+0(0-2)+k(2-10) \right| \\ \Rightarrow 0=\dfrac { 1 }{ 2 } \left| 60-0-8k \right| \\ \Rightarrow 0=\dfrac { 1 }{ 2 } \left| 60-8k \right| $$

$$\Rightarrow 60-8k=0\\ \\ \Rightarrow 8k=60\\ \Rightarrow k=\dfrac { 60 }{ 8 } =\dfrac { 15 }{ 2 }$$

Hence, $$k=\dfrac { 15 }{ 2 }$$.
Question 7
1 / -0

In a certain language '$$ \displaystyle +\div ? $$' means 'where are you' , '$$ \displaystyle @-\div $$' means ' we are here' , and '$$ \displaystyle +@\times $$' means ' you come here' What is the code for ' Where' ?

A
+

B
$$ \displaystyle \div $$

C
?

D
@

Solution

s.no code sentence
1. $$ \displaystyle '+\div ?' $$ Where are you
2. $$ \displaystyle '@-\div ' $$ We are here
3. $$ \displaystyle '+@\times $$ you come here

As we can see that where is only in sentence $$1$$ [Where is the word for which we have to find the code ] Therefore we need to gather the codes for $$are$$ & $$you$$ to find out the code for where sentence 1 & 2 have the word $$are$$ in common and the symbol $$ \displaystyle \div $$ in common Therefore $$ \displaystyle \div $$ is the symbol for $$are$$
Sentence 1 & 3 have the word you in common and the symbol + in common therefore + stands for $$you$$

thus, leaving only ?, which stands for $$where$$
Question 8
1 / -0

Area of a triangle whose vertices are (0, 0), (2, 3) , (5, 8) is _______ square units.

A
1/2

B
1

C
2

D
3/2

Solution

Area of a triangle with vertices $$({ x }_{ 1 },{ y }_{ 1 })$$ ; $$({ x }_{ 2 },{ y }_{ 2 })$$ and $$({ x }_{ 3 },{ y }_{ 3 })$$ is $$ \left| \frac { { x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right| $$

Hence, substituting the points $$({ x }_{ 1 },{ y }_{ 1 }) = (0,0) $$ ; $$({ x }_{ 2 },{ y }_{ 2 }) = (2,3) $$ and $$({ x }_{ 3 },{ y }_{ 3 }) = (5,8)$$ in the area formula, we get

Area of triangle ABC $$ = \left| \frac { (0)(3-8)+(2)(8-0)+5(0-3) }{ 2 } \right| = \left| \frac { 16 -15 }{ 2 } \right| = \frac {1}{2} units $$
Question 9
1 / -0

Image

A
70

B
75

C
65

D
80

Solution

Suppose, the apple is $$x$$ and the pear is $$y$$
Given in the question,

$$2x+y=230 $$ eq—1

And,
$$y-x=5$$ eq—2

So, from —2
$$y= 5+x$$ eq—3

Putting eq—3 in eq—1

$$2x+5+x=230$$

$$3x+5=230$$

$$3x=225$$

$$x=\dfrac{225}{3}$$

$$x=75$$

Now from eq—3

$$y=5+75$$

$$y=80$$

So the value of a pear is $$80$$

Hence Option D is the correct answer.
Question 10
1 / -0

In Hyderabad there are 5 routes to Begumpet from Kukatpally and 9 routes to Dilsukhnagar from Begumpet In how many ways can a person travel from Kukatpally to Dilsukhnagar via Begumpet?

A
$$14$$

B
$$4$$

C
$$40$$

D
$$45$$

Solution

This is an implication of AND principal so multiplication shall be done.
So, number of ways $$=5\times9$$
$$=45$$
Hence, the answer is $$45.$$