Sequences and Series Test 20

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Sequences and Series Test 20

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Question 1
1 / -0

If the area of a triangle is $$68 $$ sq. units and the vertices are $$(6, 7), (-4, 1)$$ and $$(a, -9) $$ then the value of $$a$$ is

A
1

B
2

C
3

D
4

Solution

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

}_{ 2 })$$ and $$({ x }_{ 3 },{ y }_{ 3 })$$ is:
$$A= \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 $$({ x }_{ 1 },{ y }_{ 1 }) = (6,7) $$ ; $$({ x }_{ 2

},{ y }_{ 2 }) = (-4,1) $$ and $$({ x }_{ 3 },{ y }_{ 3 }) = (a,-9)$$ in the formula for area, we get:

Area of triangle $$ = \left| \dfrac { (6)(1+9)+(-4)(-9-7) + a(7-1) }{ 2 }

\right| = 68 $$
$$ \left| \dfrac { 60 + 64 + 6a }{ 2 } \right| = 68 $$
$$ \dfrac{124 + 6a}{2} = 68 $$
$$ 124 + 6a = 136 $$
$$ 6a = 12 $$
$$ \implies a = 2 $$
Question 2
1 / -0

Find the missing letters
A, B, D, G, .....

A
L

B
M

C
J

D
K

Solution

Pattern is $$\displaystyle ^{1}A,^{1}A+1,^{2}B+2,^{4}D+3,^{7}G+4,^{11}K$$
$$\displaystyle \therefore $$ Missing letter = K
Question 3
1 / -0

Find the missing letters
M, N, O, L, R, I, V, ........

A
H

B
K

C
E

D
S

Solution

Pattern is combination of two patterns M, O, R, V, A, .... and N, L, I, H, .......
$$\displaystyle ^{13}M,^{14}N,^{15}O,^{11}L,^{18}R,^{9}I,^{22}V,^{8}H,$$ .....
$$\displaystyle \therefore $$ Missing letter = H
Question 4
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 5
1 / -0

YEB, CFI, DHL, ....

A
QOL

B
QGL

C
TOL

D
QNL

E
none of these

Solution

First letter of each term is -2 steps; Second of each term +1, +2, +3, +4 steps forward Third letter is alternatively moved +2, +3 steps
Question 6
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 7
1 / -0

There are 8 buses running from Kota to Jaipur and 10 buses running from Jaipur to Delhi. In how many ways a Person can travel from Kota to Delhi via Jaipur by bus?

A
80

B
8

C
10

D
160

Solution

Let $$E_1$$ be the event of travelling from Kota to Jaipur & $$E_2$$ be the event of travelling from Jaipur to Delhi by the person.
$$E_1$$ can happen in 8 ways and $$E_2$$ can happen in 10 ways.
Since both the events $$E_1$$ and $$E_2$$ are to be happened 12 in order, simultaneously,the number of ways $$=8\times10=80$$
Question 8
1 / -0

The area of triangle formed by $$(0, 0), (0, a)$$ and $$(b, 0)$$ is .......... .

A
$$ab$$

B
$$\displaystyle \frac{ab}{2}$$

C
$$\left | \displaystyle \frac{ab}{2} \right |$$

D
$$\left | ab \right |$$

Solution

The area of a triangle formed by joining the points $$(x_1, y_1)$$, $$(x_2, y_2)$$ and $$(x_3, y_3)$$ is

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

Therefore, the area of a triangle formed by joining the points $$(0, 0)$$, $$(0, a)$$ and $$(b, 0)$$ is:

$$A=\dfrac { 1 }{ 2 } \bigg| 0(0-b)+a(b-0)+0(0-0) \bigg| $$

$$=\dfrac { 1 }{ 2 } \bigg| 0+ab \bigg| $$

$$=\bigg| \dfrac { ab }{ 2 } \bigg|$$

Hence, the area of the triangle is $$\left| \dfrac { ab }{ 2 } \right|$$
Question 9
1 / -0

The midpoints of the sides of triangle $$ABC$$ are $$ (-1,-2), (6,1)$$ and $$(3,5) $$. The area of $$\displaystyle \triangle ABC$$ is ____ square units.

A
$$72$$

B
$$73$$

C
$$74$$

D
$$75$$

Solution

Area of triangle having vertices $$(x_1,y_1), (x_2,y_2)$$ and $$(x_3,y_3)$$ is given by
Area $$ = \dfrac{1}{2} \times [ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ] $$
Area of triangle formed with midpoints
$$\displaystyle =\frac{1}{2}\left [ -1\left ( 1-5 \right )+6\left ( 5+2 \right )+3\left ( -2-1 \right ) \right ]$$
$$\displaystyle =\frac{37}{2}$$
$$\displaystyle \therefore $$ Area of $$\displaystyle \Delta ABC=4\times \frac{37}{2}=74$$ square units.
$$[\because$$ Area of triangle formed by joining mid-points of the sides of given triangle is $$\left(\dfrac{1}{4}\right)^{th}$$ of the area of original triangle $$]$$
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 $$45.$$

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Sequences and Series Test 20

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Sequences and Series Test 20

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Rank

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SHARING IS CARING

Question 1

1

2

3

4

Solution

Question 2

L

M

J

K

Solution

Question 3

H

K

E

S

Solution

Question 4

2.25

2.5

2.75

3

Solution

Question 5

QOL

QGL

TOL

QNL

none of these

Solution

Question 6

PNTCT

TCCTN

NPTCP

NPCPT

Solution

Question 7

80

8

10

160

Solution

Question 8

$$ab$$

$$\displaystyle \frac{ab}{2}$$

$$\left | \displaystyle \frac{ab}{2} \right |$$

$$\left | ab \right |$$

Solution

Question 9

$$72$$

$$73$$

$$74$$

$$75$$

Solution

Question 10

$$14$$

$$4$$

$$40$$

$$45$$

Solution

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