Permutations and Combinations Test 18

Result

Permutations and Combinations Test 18

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

A graph may be defined as a set of points connected by lines called edges. Every edge connects a pair of points. Thus, a triangle is a graph with 3 edges and 3 points. The degree of a point is the number of edges connected to it. For example, a triangle is agraph with three points of degree 2 each. Consider a graph with 12 points. It is possible to reach any point from any other point through a sequence of edges. The number of edges "e" in the graph must satisfy the condition

A
$11 \leq e \leq 66$

B
$10 \leq e \leq 66$

C
$11 \leq e \leq 65$

D
$0 \leq e \leq 11$

Solution

(A) Since every edge connects a pair of points, the given 12 points have to be joined using lines. We may have minimum number of edges if all the 12 points are collinear.
No. of edges in this particular case
$=12-1=11$
Maximum number of edges are possible when all the 12 points are non-collinear. In this particular case number of different straight lines that can be formed using 12 points which is equal to $^12C_{2}$
$=\frac{12\times 11}{2}=66$
Therefore, following inequality holds for "e"
$11 \leq e \leq 66$
Question 2
1 / -0

In the following questions, the numbers are arranged in a particular order or pattern. Choose the missing number from the given alternatives.3, 9, 17, 27, _ ,

A
33

B
37

C
39

D
40

Solution

$3 + 6 = 9$
$9 + 8 = 17$
$17 + 10 = 27$
$27 + 12 = 39$
Question 3
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 4
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 5
1 / -0

Seven girls are to dance in a circle. In how many different ways can they stand on the circumference of the circle?

A
$120$

B
$640$

C
$720$

D
$840$

Solution

Since circular permutations of $n$ objects is $(n-1)!$
Hence, the number of ways is $\quad (7-1)!=6!=720$
Question 6
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 7
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 8
1 / -0

The area of the triangle whose vertices are $A(1,1), B(7, 3)$ and $C(12, 2)$ is

A
$25$ square units

B
$8$ square units

C
$16$ square units

D
$12$ square units

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 $\cfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1}) +x_{3}(y_{1}-y_{2})\right |$
$\therefore$
Area $= \dfrac{1}{2} [1(1)+ 7(1)+ 12(-2)] = -8$ , but area can not be negative
Hence, area = 8 square 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

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.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Permutations and Combinations Test 18

Result

Permutations and Combinations Test 18

Score

-

Rank

-

SHARING IS CARING

Question 1

11≤e≤6611 \leq e \leq 6611≤e≤66

10≤e≤6610 \leq e \leq 6610≤e≤66

11≤e≤6511 \leq e \leq 6511≤e≤65

0≤e≤110 \leq e \leq 110≤e≤11

Solution

Question 2

33

37

39

40

Solution

Question 3

555

212121

333333

606060

Solution

Question 4

15C10^{15}C_{10}15C10​

20C10^{20}C_{10}20C10​

30C10^{30}C_{10}30C10​

None of these

Solution

Question 5

120120120

640640640

720720720

840840840

Solution

Question 6

888 sq.unit

777 sq.unit

666 sq.unit

555 sq.unit

Solution

Question 7

999999999

Greatest 333 digit number

Either A or B

Smallest 333 digit number

Solution

Question 8

252525 square units

888 square units

161616 square units

121212 square units

Solution

Question 9

5

3

1

0

Solution

Question 10

2(a2+b2+c2)2(a^2 + b^2 +c^2)2(a2+b2+c2)

a2+b2+c2a^2 + b^2 +c^2a2+b2+c2

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

None of these

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.

$11 \leq e \leq 66$

$10 \leq e \leq 66$

$11 \leq e \leq 65$

$0 \leq e \leq 11$

$5$

$21$

$33$

$60$

$^{15}C_{10}$

$^{20}C_{10}$

$^{30}C_{10}$

$120$

$640$

$720$

$840$

$8$ sq.unit

$7$ sq.unit

$6$ sq.unit

$5$ sq.unit

$999$

Greatest $3$ digit number

Smallest $3$ digit number

$25$ square units

$8$ square units

$16$ square units

$12$ square units

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

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

$2(ab + bc + ca)$