Permutations and Combinations Test 22

Result

Permutations and Combinations Test 22

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

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

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 3
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 4
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 5
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 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

Directions For Questions

In a certain code, ' il be pee ' means 'roses are blue', silk hee means ' red flowers' and ' pee mit hee' means ' flowers are vegctables'

...view full instructions

How is 'red written in that code ?

A
hee

B
silk

C
be

D
cannot be determined

Solution

Type Direct coding (Jumbled fashion)
S.no. code sentence
1. il be pee roses are blue
2. silk hee red flowers
3. pee mit hee Flowers are Vegetables
common word in sentences 1 & 3 $$ \displaystyle \rightarrow $$ 'are' and code $$ \displaystyle \rightarrow $$ 'pee'
common word in sentences 2 & 3 $$ \displaystyle \rightarrow $$ flowers and code $$ \displaystyle \rightarrow $$ 'hee'
Therefore
Codes words
pee are
hee flowers
silk red
niit vegetables
il roses/ blue
be blue./ rose
Question 10
1 / -0

Directions For Questions

In a certain code, ' il be pee ' means 'roses are blue', silk hee means ' red flowers' and ' pee mit hee' means ' flowers are vegctables'

...view full instructions

How is 'roses' written in that code?

A
il

B
pee

C
be

D
cannot be determined

Solution

Type Direct coding (Jumbled fashion)
S.no. code sentence
1. il be pee roses are blue
2. silk hee red flowers
3. pee mit hee Flowers are Vegetables
common word in sentences 1 & 3 $$ \displaystyle \rightarrow $$ 'are' and code $$ \displaystyle \rightarrow $$ 'pee'
common word in sentences 2 & 3 $$ \displaystyle \rightarrow $$ flowers and code $$ \displaystyle \rightarrow $$ 'hee'
Therefore
Codes words
pee are
hee flowers
silk red
niit vegetables
il roses/ blue
be blue./ rose