Sequences and Series Test 16

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Sequences and Series Test 16

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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

Observe the given pattern
$4\times 0+1=01$
$4\times1+2=06$
$4\times 2+3=11$
$4\times 3+4=16$
Then the value of $4\times8+9$ is -----

A
$43$

B
$42$

C
$41$

D
$40$

Solution

If we carefully observe the pattern
in first row it shows four multiply with zero addition one equals one
same for other pattern too
so answer of $4\times 8+9 = 41$
Question 3
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 4
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 5
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 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

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 8
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 9
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 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 $^6P_3$

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Sequences and Series Test 16

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Sequences and Series Test 16

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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

434343

424242

414141

404040

Solution

Question 3

Combinations

Exponential functions

Logarithmic functions

Permutations

Solution

Question 4

111

36C17 ^{36}C_{17} 36C17​

219\dfrac {2}{19}192​

36C18 ^{36}C_{18} 36C18​

Solution

Question 5

20

18

16

14

Solution

Question 6

aaa

bbb

ccc

000

Solution

Question 7

181818

808080

10P8{^1}{^0}P_810P8​

10C8{^1}{^0}C_810C8​

Solution

Question 8

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

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

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

None of these

Solution

Question 9

141414

282828

888

666

Solution

Question 10

15P3{^1}{^5}P_315P3​

4353634^3 5^3 6^3435363

4P35P36P3{^4}P_3 {^5}P_3 {^6}P_34P3​5P3​6P3​

180180180

Solution

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$$

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 )

$$

$43$

$42$

$41$

$40$

$1$

$^{36}C_{17}$

$\dfrac {2}{19}$

$^{36}C_{18}$

$a$

$b$

$c$

$0$

$18$

$80$

${^1}{^0}P_8$

${^1}{^0}C_8$

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

$14$

$28$

$8$

$6$

${^1}{^5}P_3$

$4^3 5^3 6^3$

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

$180$