Permutations and Combinations Test 21

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Permutations and Combinations Test 21

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Weekly Quiz Competition

Question 1
1 / -0

Directions For Questions

In this type of questions certain numbers are given, out of which all except one are alike in some manner while one is different, and this number is the answer.

...view full instructions

Choose the one which is different from other

A
119

B
136

C
147

D
153

Solution

Only 119 has different factors 7 and 17 and no factor is repeated
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

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 5
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 6
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 7
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 8
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 9
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 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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Permutations and Combinations Test 21

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Permutations and Combinations Test 21

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SHARING IS CARING

Question 1

119

136

147

153

Solution

Question 2

L

M

J

K

Solution

Question 3

H

K

E

S

Solution

Question 4

$$72$$

$$73$$

$$74$$

$$75$$

Solution

Question 5

2.25

2.5

2.75

3

Solution

Question 6

QOL

QGL

TOL

QNL

none of these

Solution

Question 7

PNTCT

TCCTN

NPTCP

NPCPT

Solution

Question 8

12

12.5

13

14

Solution

Question 9

$$ab$$

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

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

$$\left | ab \right |$$

Solution

Question 10

$$14$$

$$4$$

$$40$$

$$45$$

Solution

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