Sequences and Series Test 18

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Sequences and Series Test 18

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Weekly Quiz Competition

Question 1
1 / -0

In a combination, the ordering of the selected objects is immaterial whereas in a permutation, the ordering is essential.

A
True

B
False

C
Either

D
Neither

Solution

Combination of few things .
The order of selected things is not specified .
Combination is basically selection hence ordering in immaterial .
Permtation of few things the order of selected things is specified .
Permutation basically means selecting and arranging .
Hence ordering is essential .
Question 2
1 / -0

There are $$6$$ boxes numbered $$1, 2 ....... 6$$. Each box is to be filled up either with a red or a green ball in such a way that at least $$1$$ box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is:

A
$$5$$

B
$$21$$

C
$$33$$

D
$$60$$

Solution

(B) The number of ways in which 1 green ball can be put $$=6$$ . The number of ways in which two green balls can be put such that the boxes are consecutive
$$=5$$ $$(i.e., (1, 2),(2, 3),(3, 4),(4, 5),(5, 6))$$

Similarly, the number of ways in which three green balls can be put
$$=4( i.e. (1, 2, 3),(2, 3, 4),(3, 4, 5),(4, 5, 6))$$
$$\cdots \cdots \cdots \cdots \cdots $$ and so on.
$$\therefore $$ Total number of ways of doing this
$$=6+5+4+3+2+1=21$$
Question 3
1 / -0

Rajdhani Express going from Bombay to Delhi stops at five inter-mediate stations, $$10$$ passengers enter the train during the journey with $$10$$ different ticket of two classes. The number of different sets of tickets they may have is

A
$$^{15}C_{10}$$

B
$$^{20}C_{10}$$

C
$$^{30}C_{10}$$

D
None of these

Solution

For a particular class, the total number of different tickets from first intermediate station is $$5.$$
Similarly, number of different tickets from second intermediate station is $$4.$$
So the total number of different tickets is $$5+4+3+2+1=15$$.
And same number of tickets for another class is equal to total number of different tickets,
which is equal to $$30$$ and number of selection is $$^{30}C_{10}$$.
Question 4
1 / -0

Find the area of the triangle whose vertices are $$(3,2), \ (-2, -3)$$ and $$(2,3)$$.

A
$$8$$ sq.unit

B
$$7$$ sq.unit

C
$$6$$ sq.unit

D
$$5$$ sq.unit

Solution

Area of a triangle with vertices $$({ x }_{ 1 },{ y }_{ 1 });$$ $$({ x }_{ 2 },y_2)$$ and $$({ x }_{ 3 },{ y }_{ 3 })$$
$$ = \left |\dfrac{ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3 (y_1 - y_2) } {2} \right |$$

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

Area = $$\left | \dfrac {3(-3-3) +(-2)(3-2) + 2 (2-(-3)) }{2}\right | = 5 $$ sq. unit
Question 5
1 / -0

Find the area of the right-angled triangle whose vertices are $$(2, -2)$$ , $$(-2, 1)$$ and $$(5, 2).$$

A
$$\displaystyle 5\sqrt{2}$$ sq. units

B
$$\displaystyle \dfrac{25}{2}$$ sq. units

C
$$\displaystyle 15\sqrt{2}$$ sq. units

D
$$10$$ sq. units

Solution

Let the points be $$ A(2,-2), B(-2,1), C(5,2) $$.

Distance between two points $$ \left( { x }_{ 1 },{ y }_{ 1} \right) $$ and $$ \left( { x }_{ 2 },{ y }_{ 2 } \right) $$ can be calculated using the formula $$ \sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( { y }_{ 2 }-{ y }_{ 1 } \right) ^{ 2 } } $$
Hence, Length of side AB $$ = \sqrt { \left(-2-2 \right) ^{ 2 }+\left(1 + 2\right) ^{ 2 } } = \sqrt { 16+ 9 } = \sqrt { 25 } = 5 $$
Length of side BC $$ = \sqrt {\left(5 + 2\right) ^{ 2 }+\left(2-1\right) ^{ 2 } } = \sqrt { 49 + 1 } = \sqrt { 50} $$
Length of side AC $$ = \sqrt { \left(5-2 \right) ^{ 2 }+\left(2+ 2\right) ^{ 2 } } = \sqrt { 9+16 } = \sqrt { 25 } = 5 $$
Since, $$ {(\sqrt { 50 }) }^{2} = { 5 }^{2} + { 5 }^{2} $$,
$$ \Rightarrow {BC}^{2} ={AB}^{2} + {AC}^{2} $$
Hence, the triangle has a right angle at $$A$$, with $$AB$$ and $$AC$$ as base and height.
So, area of the triangle $$ABC = \cfrac {1}{2} \times base \times height = \cfrac {1}{2} \times 5 \times 5= \cfrac {25}{2}$$ sq units.
Question 6
1 / -0

Observe the given multiples of 37.
$${37\times3=111}$$
$${37\times 6 =222}$$
$${37\times9=333}$$
$${37\times12=444}$$-------------------------------
Find the product of $${37\times27}$$

A
$$999$$

B
Greatest $$3$$ digit number

C
Either A or B

D
Smallest $$3$$ digit number

Solution

$${37\times3=37\times(3\times1)=111}$$
$${37\times 6=37\times(3\times2)=222}$$
$${37\times9=37\times(3\times3)=333}$$

$${37\times27=37\times(3\times9)=999}$$
Question 7
1 / -0

Consider the points $$A( a, b + c)$$, $$B(b, c + a)$$, and $$C(c, a +b)$$ be the vertices of $$\bigtriangleup$$ABC. The area of $$\bigtriangleup$$ABC is:

A
$$2(a^2 + b^2 +c^2)$$

B
$$a^2 + b^2 +c^2$$

C
$$2(ab + bc + ca)$$

D
None of these

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 $$\cfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1}) +x_{3}(y_{1}-y_{2})\right |$$
Since, vertices are $$A(a, b + c),~ B(b, c + a),$$ and $$ C(c, a + b)$$
$$\therefore$$ Area $$ \bigtriangleup ABC = \dfrac{1}{2}|a(c + a) -b(b + c) + b(a + b) - c(c + a) +c(b + c) -a(a + b)|=0$$ .
Hence option 'D' is correct.
Question 8
1 / -0

$$(3, 1), (-3, 2)$$ and $$\displaystyle (0,2-\sqrt{3})$$ are the vertices of __________ triangle of area ___________.

A
an isosceles, $$81$$ sq. units

B
a scalene, $$\cfrac {-3+6\sqrt{3}}{2}$$ sq. units

C
an equilateral, $$\displaystyle 9\sqrt{3}$$ sq. units

D
a right angled, $$81$$ sq. units

Solution

Distance between two points $$= \sqrt { \left( { x }_{ 2 }-{x }_{ 1 } \right) ^{ 2 }+\left( { y }_{ 2 }-{ y }_{ 1 } \right) ^{ 2 } } $$
Distance between the points $$A (3,1) $$ and $$B (-3,2) = \sqrt { \left( -3-3 \right) ^{ 2 }+\left( 2 - 1 \right) ^{ 2 } } = \sqrt { 36 + 1 } = \sqrt { 37 }$$
Distance between the points $$B(-3,2) $$ and $$C (0,2-\sqrt {3}) = \sqrt { \left( 0 + 3 \right) ^{ 2 }+\left(2 - \sqrt {3} - 2\right) ^{ 2 } } = \sqrt { 9 + 3 } = \sqrt { 12 } $$
Distance between the points $$A(3,1) $$ and $$ C(0,2-\sqrt {3})$$
$$ = \sqrt { \left( 0-3\right) ^{ 2 }+\left( 2-\sqrt {3} - 1\right) ^{ 2 } } = \sqrt { 9 + 1 + 3 -2\sqrt {3} } = \sqrt { 13 - 2\sqrt {3} } $$
Since the length of the sides between all vertices are different, they are the vertices of a scalene triangle.
Area of a triangle $$= \left| \cfrac { {x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right| $$

Area of $$\triangle ABC= \left| \cfrac { (3)(2-2+\sqrt {3})+(-3)(2-\sqrt {3} - 1)+0(1-2) }{ 2 } \right| $$
$$ = \left| \cfrac { 3\sqrt {3} -3 + 3\sqrt {3} }{ 2 } \right| $$
$$ = \cfrac{-3 + 6 \sqrt {3}}{2}$$ sq. units
Question 9
1 / -0

Find the area of the triangle formed by joining the midpoints of the sides of the triangle whose vertices are $$(2,2)$$, $$(4,4)$$ and $$(2,6)$$.

A

5

B
3

C
1

D
0

Solution

Let $$ A(2,2), B(4,4) $$ and $$ C(2,6) $$ be the vertices of a given triangle ABC.

Let D, E, and F be the midpoints of AB, BC and CA respectively.

Mid point of two points $$ { (x }_{ 1 },{ y }_{ 1 }) $$ and $$ { (x }_{ 2 },{ y }_{

2 }) $$ is calculated by the formula $$ \left( \dfrac { { x }_{ 1 }+{ x

}_{ 2 } }{ 2 } ,\dfrac { { y }_{ 1 }+y_{ 2 } }{ 2 } \right) $$

Using this formula,the coordinates of D, E, and F are given as $$\displaystyle D \left (\frac{2+4}{2},\frac {2+4}{2} \right ), E\left (\frac {4+2}{2},\frac {4+6}{2}\right ) $$ and $$ F\left (\dfrac {2+2}{2},\dfrac {2+6}{2} \right ) $$

i.e., $$ D(3,3), E(3,5) and F(2,4) $$

Area of a triangle with vertices $$({ x }_{ 1 },{ y }_{ 1 })$$ ; $$({ x }_{ 2 },{ y

}_{ 2 })$$ and $$({ x }_{ 3 },{ y }_{ 3 })$$ is $$ \left| \dfrac { {

x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{

3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right| $$

Hence, substituting the points $$({ x }_{ 1 },{ y }_{ 1 }) = (3,3) $$ ; $$({ x

}_{ 2 },{ y }_{ 2 }) = (3,5) $$ and $$({ x }_{ 3 },{ y }_{ 3 }) = (2,4)$$

in the area formula, we get

Area of triangle DEF $$ = \left| \dfrac { 3(5-4)+(3)(4-3)+2(3-5) }{ 2

} \right| = \left| \dfrac { 3 +3 -4 }{ 2 } \right| =

\dfrac {2}{2} = 1 \ sq \ units $$
Question 10
1 / -0

Find the area of the triangle whose vertices are $$(a, b + c), (a, b - c)$$ and $$(-a, c)$$.

A
$$2ac$$

B
$$2bc$$

C
$$b(a + c)$$

D
$$c (a - b)$$

Solution

Area of a triangle $$ (A) =\left| \cfrac { {x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right| $$

Hence, substituting the points $$({ x }_{ 1 },{ y }_{ 1 }) = (a,b+c) $$ , $$({ x }_{ 2

},{ y }_{ 2 }) = (a,b-c) $$ and $$({ x }_{ 3 },{ y }_{ 3 }) = (-a,c) $$

$$ A=\left| \cfrac { a(b-c-c)+a(c-b-c)-a(b+c-b+c) }{ 2 } \right| $$
$$=\left| \cfrac { a(b-2c)+a(-b)-a(2c) }{ 2 } \right|$$
$$ =\left| \cfrac { ab-2ac-ab-2ac }{ 2 } \right| $$
$$=\left| \cfrac { -4ac }{ 2 } \right| $$
$$=2ac$$ square units

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Sequences and Series Test 18

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Sequences and Series Test 18

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SHARING IS CARING

Question 1

True

False

Either

Neither

Solution

Question 2

$$5$$

$$21$$

$$33$$

$$60$$

Solution

Question 3

$$^{15}C_{10}$$

$$^{20}C_{10}$$

$$^{30}C_{10}$$

None of these

Solution

Question 4

$$8$$ sq.unit

$$7$$ sq.unit

$$6$$ sq.unit

$$5$$ sq.unit

Solution

Question 5

$$\displaystyle 5\sqrt{2}$$ sq. units

$$\displaystyle \dfrac{25}{2}$$ sq. units

$$\displaystyle 15\sqrt{2}$$ sq. units

$$10$$ sq. units

Solution

Question 6

$$999$$

Greatest $$3$$ digit number

Either A or B

Smallest $$3$$ digit number

Solution

Question 7

$$2(a^2 + b^2 +c^2)$$

$$a^2 + b^2 +c^2$$

$$2(ab + bc + ca)$$

None of these

Solution

Question 8

an isosceles, $$81$$ sq. units

a scalene, $$\cfrac {-3+6\sqrt{3}}{2}$$ sq. units

an equilateral, $$\displaystyle 9\sqrt{3}$$ sq. units

a right angled, $$81$$ sq. units

Solution

Question 9

5

3

1

0

Solution

Question 10

$$2ac$$

$$2bc$$

$$b(a + c)$$

$$c (a - b)$$

Solution

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