Straight Lines Test 21

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Straight Lines Test 21

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Weekly Quiz Competition

Question 1
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 2
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 3
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 4
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 $$45.$$
Question 5
1 / -0

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

A
$$\dfrac 12$$

B
$$1$$

C
$$2$$

D
$$\dfrac 32$$

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:
$$A=\dfrac { 1 }{ 2 } \left| 0(2-5)+3(5-0)+8(0-2) \right| =\dfrac { 1 }{ 2 } \left| 0+15-16 \right| =\dfrac { 1 }{ 2 } \left| -1 \right| =\dfrac { 1 }{ 2 }$$
Hence, the area of the triangle is $$\dfrac { 1 }{ 2 }$$ square units.
Question 6
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 7
1 / -0

In how many ways can 3 diamond cards be drawn simultaneously from a pack of cards?

A
$$78$$

B
$$1716$$

C
$$286$$

D
$$13$$

Solution

Pack of cards $$52$$
Diamond cards are $$13$$
To select $$3$$ cards
Total number of choices =$$^{ 13 }C_{ 3 }$$
$$= \dfrac { 13\times 12\times 11 }{ 3\times 2\times 1 } \\ = 143\times 2\\ = 286$$
Hence $$286$$ ways are possible and is a correct answer.
Question 8
1 / -0

Find the area of $$\Delta ABC$$, in which $$A = (2, 1), B = (3, 4)$$ and $$C = (-3, -2).$$

A
$$3$$ square units

B
$$4$$ square units

C
$$6$$ square units

D
$$12$$ square units

Solution

Area of triangle with vertices $$(x_1,y_1)$$, $$(x_2,y_2)$$ and $$(x_3,y_3)$$ is:
$$A=\dfrac { 1 }{ 2 } \left| x_{ 1 }(y_{ 2 }-y_{ 3 })+x_{ 2 }(y_{ 3 }-y_{ 1 })+x_{ 3 }(y_{ 1 }-y_{ 2 }) \right|$$
Therefore, the area of triangle with vertices $$(2,1)$$, $$(3,4)$$ and $$(-3,-2)$$ is:
$$A=\displaystyle \frac { 1 }{ 2 } \left| 2(4-(-2))+3(-2-1)-3(1-4) \right| $$
$$=\dfrac { 1 }{ 2 } \left| 2(6)+3(-3)-3(-3) \right| $$
$$=\dfrac { 1 }{ 2 } \left| 12-9+9 \right|$$
$$ =\dfrac { 1 }{ 2 } \left| 12 \right| =6$$ square units.
Question 9
1 / -0

A researcher plans to identify each participant in a $$174$$ certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are $$12$$ participants,and each participant is to receive a different code?

A
$$4$$

B
$$5$$

C
$$6$$

D
$$7$$

E
$$8$$

Solution

The number of single letter code possible are $$n$$ and no of distinct codes $$={^n{C}_{2}}$$
So, $${^n{C}_{2}}+n\ge 12$$
$$\dfrac{n(n-1)}{2}+n\ge 12$$
$$n(n+1)\ge 24$$
$$n$$ min $$=5$$ as least number is asked.
Hence the answer is $$5.$$
Question 10
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 $$where$$

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Re-Attempt Weekly Quiz Competition

Straight Lines Test 21

Result

Straight Lines Test 21

Score

-

Rank

-

SHARING IS CARING

Question 1

12

12.5

13

14

Solution

Question 2

1/2

1

2

3/2

Solution

Question 3

70

75

65

80

Solution

Question 4

$$14$$

$$4$$

$$40$$

$$45$$

Solution

Question 5

$$\dfrac 12$$

$$1$$

$$2$$

$$\dfrac 32$$

Solution

Question 6

$$0$$

$$1$$

$$2$$

$$5$$

Solution

Question 7

$$78$$

$$1716$$

$$286$$

$$13$$

Solution

Question 8

$$3$$ square units

$$4$$ square units

$$6$$ square units

$$12$$ square units

Solution

Question 9

$$4$$

$$5$$

$$6$$

$$7$$

$$8$$

Solution

Question 10

+

$$ \displaystyle \div $$

?

@

Solution

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