Permutations and Combinations Test 16

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Permutations and Combinations Test 16

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Weekly Quiz Competition

Question 1
1 / -0

A point $$(a, b)$$ is called a good point if both $$a$$ and $$b$$ are integers. Number of good points on the curve $$xy$$ $$=$$ $$225$$ are

A
20

B
18

C
16

D
14

Solution

The order pair $$(x, y)$$ satisfying $$xy=225$$ are $$(1, 225), (3, 75) (5, 45), (9, 25), (15, 15)$$. Order can be changed in the first four pairs and both $$x$$ and $$y$$ can be negative also, so the no. of pairs $$=2(2\times 4+1)=18$$
Question 2
1 / -0

In a class there are $$10$$ boys and $$8$$ girls. The teacher wants to select either a boy or a girl to represent the class in a function. The number of ways the teacher can make this selection.

A
$$18$$

B
$$80$$

C
$${^1}{^0}P_8$$

D
$${^1}{^0}C_8$$

Solution

There are $$10$$ boys and $$8$$ girls in a class.
The teacher wants to select just one to represent the class.
One boy out of $$10$$ can be selected in $$10$$ ways.
Similarly, one girl out of $$8$$ can be selected in $$8$$ ways.
So, if he has to select either a boy OR a girl, he can do it in total $$10+8$$ number of ways. i.e $$18$$
Question 3
1 / -0

The number of ways in which 7 persons can be arranged around a circle is:

A
$$360$$

B
$$720$$

C
$$5040$$

D
$$1440$$

Solution

$$no$$ of ways of arranging n persons in a circle is $$(n-1)!=(7-1)=6!=720$$
Question 4
1 / -0

The sum of the value of the digits at the tens place of all the numbers formed with the help of $$3, 4, 5, 6$$ taken all at a time is

A
$$1080$$

B
$$4320$$

C
$$360$$

D
$$180$$

Solution

Total number of numbers formed with the digits $$3,4,5,6$$ is $$4!$$ $$=$$ $$24$$
If a digit is fixed in tens place, number of numbers formed will be $$6$$
i.e for every digit in tens place there will be $$6$$ numbers formed
Now, sum of the digits $$ = (6\times3)+(6\times4)+(6\times5)+(6\times6) $$ $$ = 108 $$.
Now its value in tens place $$= 108\times10=1080 $$.
Question 5
1 / -0

Three Men have $$4$$ coats $$5$$ waist Coats, and $$6$$ caps. The number of ways they can wear them is

A
$${^1}{^5}P_3$$

B
$$4^3 5^3 6^3$$

C
$${^4}P_3 {^5}P_3 {^6}P_3$$

D
$$180$$

Solution

Coats $$\rightarrow ^4P_3$$
4 3 2
Waist Coats $$\rightarrow ^5P_3$$
5 4 3
Caps $$\rightarrow ^6P_3$$
6 5 4
Total no. of ways of wearing them= $$ ^4P_3$$ x $$ ^5P_3$$ x $$ ^6P_3$$
Question 6
1 / -0

The points $$A (2, 9), B (a, 5)$$ and $$C (5, 5) $$ are the vertices of a triangle $$ABC$$ right angled at $$B$$. Find the values of $$a$$ and hence the area of $$\Delta $$ $$ABC$$.

A
$$a = -1$$ , Area $$=$$ $$15$$ sq. unit

B
$$a = 2$$ , Area $$=$$ $$6$$ sq. unit

C
$$a = 0 $$ , Area $$=$$ $$8$$ sq. unit

D
$$a = 1$$ , Area $$=$$ $$12$$ sq. unit

Solution

Given that the vertices of the $$\Delta ABC$$ are $$A\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 2,9 \right) , B\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( a,5 \right) $$ and $$C\left( { x }_{ 3 },{ y }_{ 3 } \right) =\left( 5,5 \right) and \angle B={ 90 }^{ o }$$.
So, $$AC$$ is the hypotenuse.
$$ \therefore$$ by Pythagoras theorem, we have
$$ { AB }^{ 2 }+{ BC }^{ 2 }={ AC }^{ 2 }$$ .........(i)
Now, by distance formula
$$d=\sqrt { { \left( { x }_{ 1 }-{ x }_{ 2 } \right) }^{ 2 }+{ \left( { y }_{ 1 }-y_{ 2 } \right) }^{ 2 } } $$
So, $$AB=\sqrt { { \left( { x }_{ 1 }-{ x }_{ 2 } \right) }^{ 2 }+{ \left( { y }_{ 1 }-y_{ 2 } \right) }^{ 2 } } =\sqrt { { \left( 2-a \right) }^{ 2 }+{ \left( 9-5 \right) }^{ 2 } } =\sqrt { { a }^{ 2 }-4a+20 } $$,
$$BC=\sqrt { { \left( { x }_{ 3 }-{ x }_{ 2 } \right) }^{ 2 }+{ \left( { y }_{ 3 }-y_{ 2 } \right) }^{ 2 } } =\sqrt { { \left( 5-a \right) }^{ 2 }+{ \left( 5-5 \right) }^{ 2 } } =\sqrt { { a }^{ 2 }-10a+25 } $$ and
$$AC=\sqrt { { \left( { x }_{ 3 }-{ x }_{ 1 } \right) }^{ 2 }+{ \left( { y }_{ 3 }-y_{ 1 } \right) }^{ 2 } } =\sqrt { { \left( 5-2 \right) }^{ 2 }+{ \left( 5-9 \right) }^{ 2 } } $$ units $$=\sqrt { 25 } $$ units.
$$ \therefore$$ by (i), we get
$${ a }^{ 2 }-4a+20{ +a }^{ 2 }-10a+25=25$$
$$ \Rightarrow { a }^{ 2 }-7a+10=0$$
$$ \Rightarrow a=5,2$$
We reject $$a=5$$, since $$B$$ and $$C$$ will coincide in that case and $$\Delta ABC$$ will collapse.
So, $$a=2. i.e. AB=\sqrt { { a }^{ 2 }-4a+20 } =\sqrt { 4-8+20 } 3$$ units $$=43$$ units
$$ BC=\sqrt { { a }^{ 2 }-10a+25 } =\sqrt { 4-20+25 } 3$$ units $$=3$$ units.
Now,
We know that the area of the triangle whose vertices are $$\displaystyle (x_{1},y_{1}),(x_{2},y_{2}),$$ and $$(x_{3},y_{3})$$ is $$\cfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1}) +x_{3}(y_{1}-y_{2})\right |$$
Therefore, required area is $$6$$ square units.
Question 7
1 / -0

Find the area of triangle having vertices are $$A (3, 1), B (12, 2)$$ and $$C (0, 2)$$.

A
$$4$$

B
$$6$$

C
$$12$$

D
$$18$$

Solution

The given points are $$A\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 3,1 \right) , B\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( 12,2 \right) \& C\left( { x }_{ 3 },{ y }_{ 3 } \right) =\left( 0,2 \right)$$
Let us obtain $$ar\Delta ABC$$ by applying the area formula.
$$ ar\Delta =\dfrac { 1 }{ 2 } \left\{ { x }_{ 1 }\left( { y }_{ 2 }-{ y }_{ 3 } \right) +{ x }_{ 2 }\left( { y }_{ 3 }-{ y }_{ 1 } \right) +{ x }_{ 3 }\left( { y }_{ 1 }-{ y }_{ 2 } \right) \right\}$$
$$\Rightarrow ar\Delta =\dfrac { 1 }{ 2 } \left\{ 3\left( 2-2 \right) +12\left( 2-1 \right) +0\left( 1-2 \right) \right\}$$ units $$=6$$ units
$$\therefore ar\Delta =6$$ units
Hence, option B.
Question 8
1 / -0

The area of triangle ABC (in sq. units) is :

A
$$15$$ sq. units

B
$$10$$ sq. units

C
$$7.5$$ sq. units

D
$$2.5$$ sq. units

Solution

The area of a triangle is given as:
Area $$=\dfrac { 1 }{ 2 } \left[ { x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }) \right] $$
the given points are $$ A(1,3),~B(-1,0) $$and $$C(4,0)$$
by Substituting ,
$$\text{Area} =\dfrac { 1 }{ 2 } \left[ 1(0-0)+(-1)(0-3)+4(3-0) \right] \\ =\dfrac { 1 }{ 2 } \left[ 3+12 \right] =\dfrac { 15 }{ 2 } =7.5$$
$$\therefore$$ Area of the given triangle ABC is $$7.5$$ sq. units
Question 9
1 / -0

The area of a triangle with vertices $$(a, b + c), (b, c + a)$$ and $$(c, a + b)$$ is

A
$$a$$

B
$$b$$

C
$$c$$

D
$$0$$

Solution

The area of $$\triangle ABC$$ with vertices $$A\equiv(x_1, y_1)$$, $$B\equiv(x_2, y_2)$$ and $$C\equiv(x_3, y_3)$$ is given as,
$$A(\triangle ABC) = \left|\dfrac12[x_1(y_3-y_2) +x_2(y_1 - y_3) + x_3(y_2-y_1) ]\right|$$
In this problem,
$$A(\triangle ABC) = \left|\dfrac12[a(a+b-(c+a)) +b(b+c - (a+b)) + c(c+a-(b+c)) ]\right|$$
$$\therefore A(\triangle ABC) = \left|\dfrac12[ab-ac + bc-ba + ca-cb]\right|$$
$$\therefore A(\triangle ABC) = 0$$
Hence, the correct Option is D.
Question 10
1 / -0

Area of the triangle formed by the points P(-1.5, 3), Q(6, -2) and R(-3, 4) is 0.

A
True

B
False

C
Ambiguous

D
Data insufficient

Solution

We know that the area of the triangle whose vertices are $$\displaystyle (x_{1},y_{1}),(x_{2},y_{2}),$$ and $$(x_{3},y_{3})$$ is $$\cfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1}) +x_{3}(y_{1}-y_{2})\right |$$
Given $$P(-1.5, 3), Q(6, -2)$$ and $$R(-3, 4)$$
Therefore, area is given by
$$= \frac{1}{2}\times[(-1.5)(-2-4) + 6(4-3)+(-3)(3+2)]$$
$$= \frac{1}{2}\times(9 +6-15)$$
$$= 0$$

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Re-Attempt Weekly Quiz Competition

Permutations and Combinations Test 16

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Permutations and Combinations Test 16

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SHARING IS CARING

Question 1

20

18

16

14

Solution

Question 2

$$18$$

$$80$$

$${^1}{^0}P_8$$

$${^1}{^0}C_8$$

Solution

Question 3

$$360$$

$$720$$

$$5040$$

$$1440$$

Solution

Question 4

$$1080$$

$$4320$$

$$360$$

$$180$$

Solution

Question 5

$${^1}{^5}P_3$$

$$4^3 5^3 6^3$$

$${^4}P_3 {^5}P_3 {^6}P_3$$

$$180$$

Solution

Question 6

$$a = -1$$ , Area $$=$$ $$15$$ sq. unit

$$a = 2$$ , Area $$=$$ $$6$$ sq. unit

$$a = 0 $$ , Area $$=$$ $$8$$ sq. unit

$$a = 1$$ , Area $$=$$ $$12$$ sq. unit

Solution

Question 7

$$4$$

$$6$$

$$12$$

$$18$$

Solution

Question 8

$$15$$ sq. units

$$10$$ sq. units

$$7.5$$ sq. units

$$2.5$$ sq. units

Solution

Question 9

$$a$$

$$b$$

$$c$$

$$0$$

Solution

Question 10

True

False

Ambiguous

Data insufficient

Solution

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