Sequences and Series Test 33

Result

Sequences and Series Test 33

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Weekly Quiz Competition

Question 1
1 / -0

Find the missing factor among the given figure ?

A
$$4$$

B
$$6$$

C
$$8$$

D
$$12$$

Solution
Question 2
1 / -0

A box contains 5 pairs of shoes. If 4 shoes are selected, then the number of ways in which exactly one pair of shoes obtained is :

A
120

B
140

C
160

D
180

Solution

We have,

A box contains pair of shoes $$=5$$

Selected pair of shoes $$=4$$

Then,

Exactly one pair of shoes obtained is

$$ {{=}^{5}}{{P}_{4}} $$

$$ =\dfrac{5!}{\left( 5-4 \right)!} $$

$$ =\dfrac{5!}{1!}=5! $$

$$ =5\times 4\times 3\times 2\times 1 $$

$$ =120 $$
Hence, this is the answer.
Question 3
1 / -0

There are n locks and n matching keys. If all the locks and keys are to be perfectly matched, find the maximum number of trails required to open a lock.

A
$$^{n}C_{2}$$

B
$$\sum _{ k=2 }^{ n }{ \left( k+2 \right) } $$

C
$$\dfrac{n(n+1)}{2}$$

D
$$^{n+1}C_{2}$$

Solution
Question 4
1 / -0

Find the area of the triangle formed by the mid points of sides of the triangle whose vertices are $$(2, 1)$$, $$(-2, 3)$$, $$(4, -3)$$.

A
$$1.5$$ sq. units

B
$$3$$ sq. units

C
$$6$$ sq. units

D
$$20$$ sq. units
Question 5
1 / -0

The number of intersection points of diagonals of $$2009$$ sides polygon, which lie inside the polygon.

A
$$^{2009}C_4$$

B
$$^{2009}C_2$$

C
$$^{2008}C_4$$

D
$$^{2008}C_2$$

Solution

Choosing any $$4$$ vertices make $$1$$ intersection

$$\therefore \ $$ No. of intersection points of diagonals $$=^nC_4$$

Here $$n=2009$$, No. of points $$=^{2009}C_4$$

$$\therefore \ $$ Option $$A$$ is correct.
Question 6
1 / -0

The midpoints of the sides of a triangle are $$\left(1,1\right),\left(4,3\right)$$ and $$\left(3,5\right)$$. The area of the triangle is ___ square units.

A
$$14$$

B
$$16$$

C
$$18$$

D
$$20$$

Solution
Question 7
1 / -0

Ten persons, amongst whom are $$A$$,$$B$$ and $$C$$ to speak at a function. The number of ways in which it can be done if $$A$$ wants to speak before $$B$$ AND $$B$$ wants to speak before $$C$$ is

A
$$\dfrac{{10!}}{6}$$

B
$$3!$$ $$7!$$

C
$$^{10}{P_3}.7!$$

D
none of these

Solution
Question 8
1 / -0

Let $$p=\dfrac{1}{1\times2}+\dfrac{1}{3\times4}+\dfrac{5}{1\times6}+.......+\dfrac{1}{2013\times2014}$$ and $$Q=\dfrac{1}{1008\times2014}+\dfrac{1}{1009\times2013}+.........+\dfrac{1}{2014\times1008}$$
then $$\dfrac{P}{Q}=$$

A
$$2013$$

B
$$2014$$

C
$$1511$$

D
$$2$$
Question 9
1 / -0

The number of ways in which a mixed doubles tennis game can be arranged between 10 players consisting of 6 men and 4 women is .

A
180

B
90

C
48

D
12

Solution

Number of ways to select two men out of six men
$$ = ^{6}C_{2} $$
Number of ways to select two women out of four
women $$ = ^{4}C_{2} $$
Number of ways of selecting the players for the
mixed double $$ = ^{6}C_{2}\times ^{4}C_{2} = 15\times 16 = 90 $$
Let $$ M_{1},M_{2},W_{1}\& W_{2} $$ are selected players for we
mixed double tennis game
if $$ M_{1} $$ chooses $$W_{1}$$ then $$ M_{2}$$ has $$W_{2}$$ as the partner or if
$$ M_{1}$$ chooses $$ W_{2}$$, lean $$M_{2}$$ has $$ W_{1}$$ as the partner
$$ \therefore $$ There are $$2$$ choice for the teams.
Thus, Number of ways in which mixed double
tennis game be arranged $$ 90\times 2 = 180 $$
Question 10
1 / -0

solve that :-

A
$$14$$

B
$$18$$

C
$$11$$

D
$$26$$

Solution

Given,

Image $$A$$

$$4+2+5+3=14 \times 2=28$$

Image $$B$$

$$7+5+4+3=19 \times 2=38$$

$$\therefore$$ Image $$C$$

$$2+1+3+7=13 \times 2=26$$