Sequences and Series Test 45

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Sequences and Series Test 45

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Weekly Quiz Competition

Question 1
1 / -0

Total number of ways of selecting two numbers from the set $${1,2,3,...90}$$ so that their sum is divisible by $$3$$ is

A
$$885$$

B
$$1335$$

C
$$1770$$

D
$$3670$$

Solution
Question 2
1 / -0

The number of $$n$$ digit numbers which consists of the digits $$1$$ & $$2$$ only if each digits is to be used atleast once, is equal to $$510$$ then $$n$$ is equal to

A
$$7$$

B
$$8$$

C
$$9$$

D
$$10$$
Question 3
1 / -0

Fill in the missing values.
J V R B
Z 18 24 22 14
L 11 17 15 7
? 9 15 13 5
T 15 21 ? 11

A
X and $$29$$

B
P and $$15$$

C
H and $$19$$

D
F and $$12$$

Solution
Question 4
1 / -0

Find the missing number, if same rule is followed in all the three figures.

A
12

B
16

C
14

D
9

Solution

REF.Image.
Given

If we consider down two number of fig(i)

then their product is $$ \Rightarrow 4\times 16 = 64 $$

Now if we take square root then =$$ \sqrt{64}$$

$$ = 8 \to$$upper are we are getting.

similarly from fig (iii)

down number product $$ \Rightarrow 12 \times27 = 324$$

take square roots $$ = \sqrt{324} = 18 \to$$upper number.

So, similarly from fig(ii)

Down product = $$ 18 \times 8 = 144 $$

so square root = $$ \sqrt{144} = 12 \rightarrow $$ option A
Question 5
1 / -0

The value of $$\displaystyle\sum^{10}_{r=0}(r)$$ $$^{20}C_r$$ is equal to?

A
$$20(2^{18}+{^{19}}C_{10})$$

B
$$10(2^{18}+{^{19}}C_{10})$$

C
$$20(2^{18}+{^{19}}C_{11})$$

D
$$10(2^{18}+{^{19}}C_{11})$$

Solution

$$\begin{matrix} \sum _{ r=0 }^{ 10 }(r){ \, ^{ 20 } }{ C_{ r } } \\ \Rightarrow 0{ +^{ 20 } }{ C_{ 1 } }+{ 2^{ 20 } }{ C_{ 2 } }+{ 3^{ 20 } }{ C_{ 3 } }+............+{ 10^{ 20 } }{ C_{ 10 } } \\ \Rightarrow 20+20\times 19+\dfrac { { 20\times 19\times 18 } }{ 2 } +.........+10\times \dfrac { { 20! } }{ { 10!\times 10! } } \\ \Rightarrow 20\left[ { 1+19+\dfrac { { 19\times 18 } }{ 2 } +\dfrac { { 19\times 18\times 17 } }{ 6 } +...........+\dfrac { { 10\times 19! } }{ { 10!\times 10! } } } \right] \\ \Rightarrow 20\left[ { 1+19+\dfrac { { 19\times 18 } }{ 2 } +\dfrac { { 19\times 18\times 17 } }{ 6 } +.........+\dfrac { { 19! } }{ { 9!\times 10! } } } \right] \\ \, \, \, \, \, \, 20\left[ { ^{ 19 }{ C_{ 0 } }{ +^{ 19 } }{ C_{ 1 } }{ +^{ 19 } }{ C_{ 2 } }{ +^{ 19 } }{ C_{ 3 } }+..........{ +^{ 19 } }{ C_{ 10 } } } \right] \, \, \, \, Ans. \\ \end{matrix}$$
Question 6
1 / -0

Find the missing number in the given figure?

A
$$25$$

B
$$361$$

C
$$196$$

D
$$144$$

Solution
Question 7
1 / -0

A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is

A
140

B
196

C
280

D
346

Solution

There would be two cases,

(1) When 4 is selected from first five and rest 6 from remaining 8.
This is done in $$^{5}C_{4}\times ^{8}C_{6}$$ ways.

(2) When all 5 are selected from first five and rest five from remaining 8.
This can be done in $$^{5}C_{5}\times ^{8}C_{5}$$ ways.

$$\therefore $$ A total of $$^{5}C_{4}\times ^{8}C_{6}+^{5}C_{5}\times ^{8}C_{5}$$

$$=5\times \cfrac{8!}{6!2!}+\cfrac{8!}{3!5!}$$

$$=\cfrac{5\times 8\times 7}{2}+\cfrac{8\times 7\times 8}{3\times 2}$$

$$=140+56=196\,ways$$
Question 8
1 / -0

If $$\displaystyle\sum^m_{k=1}(k^2+1)k!=1999(2000!)$$, then m is?

A
$$1999$$

B
$$2000$$

C
$$2001$$

D
$$2002$$

Solution
Question 9
1 / -0

If $$24 + 37 = 7,12 + 18 = 3$$, then 54+21=?

A
11

B
9

C
12

D
3

Solution
Question 10
1 / -0

In the following number series,one number is wrong.find out the ?

1,2,8,33,149,765,4626

A
$$33$$

B
$$8$$

C
$$149$$

D
$$0$$

Solution

1,2,8,33,149,765,4626
Carefully analyzing the series, we can see the following pattern
$$1\times 1+(1)^{2}=2$$
$$2\times 2+(2)^{2}=8$$
$$8\times 3+(3)^{2}=33$$
$$33\times 4+(4)^{2}=148$$
$$148\times 5+(5)^{2}=765$$
$$765\times 6+(6)^{2}=4626$$
So, we can clearly see that 149 is the wrong entry in the series.