Differentiation Test 14

Result

Differentiation Test 14

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

If P (n, n) denotes the number of permutations of n different things taken all at a time then P (n, n ) is also identical to:

A
$$

P ( n - 1 , n - 1 )

$$

B
$$

P ( n , n - 1 )

$$

C
$$

\mathrm { r } ! \mathrm { P } ( \mathrm { n } , \mathrm { n } - \mathrm { r } )

$$

D
$$

( n - r ) \cdot P ( n , r )

$$

Solution
Question 2
1 / -0

When we realize a specific implementation of a pancake algorithm, every move when we find the greatest of the sized array and flipping can be modeled through ____________.

A
Combinations

B
Exponential functions

C
Logarithmic functions

D
Permutations

Solution

Unlike a traditional sorting algorithm, which attempts to sort with the fewest comparisons possible, the goal is to sort the sequence in as few reversals as possible.
Here when we flipping the array or stack, we have to take utmost priority to preserve the order of the list so that that sorting doesnt become invalid. Hence we use permutations, we are ensuring that order matters.
Question 3
1 / -0

The value of $$\displaystyle E = \frac { (1+17)(1+\frac { 17 }{ 2 } )(1+\frac { 17 }{ 3 } )......(1+\frac { 17 }{ 19 } ) }{ (1+19)(1+\frac { 19 }{ 2 } )(1+\frac { 19 }{ 3 } ).....(1+\frac { 19 }{ 17 } ) } $$ is,

A
$$1$$

B
$$ ^{36}C_{17} $$

C
$$\dfrac {2}{19}$$

D
$$ ^{36}C_{18} $$

Solution

$$\displaystyle E = \frac { (1+17)(1+\frac { 17 }{ 2 } )(1+\frac { 17 }{3 } )......(1+\frac { 17 }{ 19 } ) }{ (1+19)(1+\frac { 19 }{ 2 } )(1+\frac { 19 }{ 3 } ).....(1+\frac { 19 }{ 17 } ) } $$

$$\quad =\displaystyle \frac{\displaystyle \frac{18\cdot 19\cdot 20.............36}{1\cdot 2\cdot 3\cdot 4...............19}}{\displaystyle \frac{20\cdot 21\cdot 22.............36}{1\cdot 2\cdot 3\cdot 4...............17}}$$

$$ \ \ \ =\displaystyle \frac{18\cdot 19\cdot 20.............36}{1\cdot 2\cdot 3\cdot 4...............19}\times \frac{1\cdot 2\cdot 3\cdot 4...............17}{20\cdot 21\cdot 22.............36}$$

$$\quad = \displaystyle \frac{18\cdot 19\cdot 20.............36}{20\cdot 21\cdot 22.............36}\times \frac{1\cdot 2\cdot 3\cdot 4...............17}{1\cdot 2\cdot 3\cdot 4...............19}$$

$$\displaystyle \ \ \ =\frac{18\cdot 19}{1}\times \frac{1}{18\cdot 19}=1$$
Question 4
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 5
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 6
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 7
1 / -0

Plot $$(3, 0), (5, 0)$$ and $$(0, 4) $$ on cartesian plane. Name the figure formed by joining these points and find its area.

A
Figure formed is a Rectangle and Area $$= 52$$ sq. unit

B
Figure formed is a triangle and Area $$= 4$$ sq. unit

C
Figure formed is a square and Area $$= 40$$ sq. unit

D
None of these

Solution

The figure is a triangle.
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 $$= \dfrac{1}{2} [3(0-4)+5(4-0)+0] = 4$$ square units.
Question 8
1 / -0

The area of a triangle with vertices $$A (3, 0), B (7, 0)$$ and $$C (8, 4)$$ is

A
$$14$$

B
$$28$$

C
$$8$$

D
$$6$$

Solution

The area of the $$ \Delta ABC$$ with vertices
$$ A\left( { x }_{ 1 },{ y }_{ 1 } \right)=(3,0), B\left( { x }_{ 2 },{ y }_{ 2 } \right)=(7,0)\ \&\ C\left( { x }_{ 3 },{ y }_{ 3 } \right)=(8,4)$$ is
$$ Ar.\Delta ABC=\dfrac { 1 }{ 2 } \left[ { x }_{ 1 }\left( { y }_{ 2 }-{ y }_{ 3 } \right) +{ x }_{ 2 }\left( { y }_{ 3 }-{ y }_{ 1 } \right) +{ x }_{ 3 }\left( { y }_{ 1 }-{ y }_{ 2 } \right) \right]$$
$$=\dfrac { 1 }{ 2 } \left\{ 3\left( 0-4 \right) +7\left( 4-0 \right) +8\left( 0-0 \right) \right\}$$ sq.units $$=8$$ sq. units.
Hence, option C.
Question 9
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 10
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 $$ ^6P_3$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Differentiation Test 14

Result

Differentiation Test 14

Score

-

Rank

-

SHARING IS CARING

Question 1

$$P ( n - 1 , n - 1 )$$

$$P ( n , n - 1 )$$

$$\mathrm { r } ! \mathrm { P } ( \mathrm { n } , \mathrm { n } - \mathrm { r } )$$

$$( n - r ) \cdot P ( n , r )$$

Solution

Question 2

Combinations

Exponential functions

Logarithmic functions

Permutations

Solution

Question 3

$$1$$

$$ ^{36}C_{17} $$

$$\dfrac {2}{19}$$

$$ ^{36}C_{18} $$

Solution

Question 4

20

18

16

14

Solution

Question 5

$$18$$

$$80$$

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

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

Solution

Question 6

$$a$$

$$b$$

$$c$$

$$0$$

Solution

Question 7

Figure formed is a Rectangle and Area $$= 52$$ sq. unit

Figure formed is a triangle and Area $$= 4$$ sq. unit

Figure formed is a square and Area $$= 40$$ sq. unit

None of these

Solution

Question 8

$$14$$

$$28$$

$$8$$

$$6$$

Solution

Question 9

$$1080$$

$$4320$$

$$360$$

$$180$$

Solution

Question 10

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

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

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

$$180$$

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.

$$

P ( n - 1 , n - 1 )

$$

$$

P ( n , n - 1 )

$$

$$

\mathrm { r } ! \mathrm { P } ( \mathrm { n } , \mathrm { n } - \mathrm { r } )

$$

$$

( n - r ) \cdot P ( n , r )

$$