Sequences and Series Test 39

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Sequences and Series Test 39

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Question 1
1 / -0

The number of seven letter words that can be formed by using the letters of the word $$SUCCESS$$ that the two $$C$$ are together but no two $$S$$ are together is

A
$$24$$

B
$$18$$

C
$$54$$

D
none of these

Solution

using the letters of the word $$SUCCESS$$ that the two $$C$$ are together but no two $$S$$ are together
let two C's be 1 unit
$$\therefore $$ no. of ways $$=\frac{6!}{4!}=30$$
We have to put 1 letter between one S
So, No of ways$$=2\times 3!=12$$
No. of ways=30-12=18
Question 2
1 / -0

Find the missing number :

A
$$125$$

B
$$90$$

C
$$105$$

D
$$225$$

Solution

$$\textbf{Step 1: Observe the numbers opposite to each other}$$
$$ \text{We can see that the number opposite to 315 is 21. Also, } \dfrac{315}{21}=15$$
$$\text{Hence, we can say the missing number has at least one factor of 15 }$$
$$\text{We see the missing number is opposite to 15}$$
$$\text{Also, } 15\times15=225$$
$$\textbf{Step 2: Verification}$$
$$\text{Let the missing number be x.}$$
$$\text{ATQ, }$$
$$\Rightarrow 21+15+x=261$$
$$\Rightarrow x=261-21-15$$
$$x=225$$
$$\text{Hence, it is verified that x=225}$$
$$\textbf{Thus, the missing number is D 225.}$$
Question 3
1 / -0

Find the missing number.

A
$$74$$

B
$$62$$

C
$$91$$

D
$$97$$

Solution

$$ \textbf{ Step 1: Observing the difference between the adjacent numbers}$$

$$ \text{If we notice the adjacent numbers, we see } 4\times 2 -1 =7$$

$$ \text{Similarly, } 7 \times 2-1=13$$

$$ \text{Similarly, } 13\times2 -1=25$$

$$ \text{Similarly, } 25\times2-1=49$$

$$ \textbf{Step 2: Calculating the missing number}$$

$$ \text{Thus, looking at the above observations, we can conclude the missing number will be:}$$

$$ \Rightarrow 49\times 2-1=97$$

$$\textbf{Thus, the missing number is D 97}$$
Question 4
1 / -0

The value of $$^{47}C_{4}+\displaystyle \sum _{ j=1 }^{ 5 }\ ^{ \left( 52-j \right) } { C }_{ 3 }$$ is

A
$$^{47}C_{5}$$

B
$$^{52}C_{5}$$

C
$$^{52}C_{4}$$

D
$$^{52}C_{3}$$

Solution

We have,
$$\begin{array}{l} ^{ 47 }{ C_{ 4 } }{ +^{ 51 } }{ C_{ 3 } }{ +^{ 50 } }{ C_{ 3 } }{ +^{ 49 } }{ C_{ 3 } }{ +^{ 48 } }{ C_{ 3 } }{ +^{ 47 } }{ C_{ 3 } } \\ { =^{ n } }{ C_{ r } }{ +^{ n } }{ C_{ r-1 } }{ =^{ n+1 } }{ C_{ r } } \\ { =^{ 48 } }{ C_{ 4 } }{ +^{ 48 } }{ C_{ 3 } }{ +^{ 49 } }{ C_{ 3 } }{ +^{ 50 } }{ C_{ 3 } }{ +^{ 51 } }{ C_{ 3 } } \\ { =^{ 49 } }{ C_{ 4 } }{ +^{ 49 } }{ C_{ 3 } }{ +^{ 50 } }{ C_{ 3 } }{ +^{ 51 } }{ C_{ 3 } } \end{array}$$
Similarly,
$$=^{52}{C_4}$$.
Question 5
1 / -0

$$496 : 204 : : 329 : ?$$

A
$$90$$

B
$$110$$

C
$$115$$

D
$$135$$
Question 6
1 / -0

A committee of $$4$$ persons is to be formed from $$2$$ ladies, $$2$$ old men and $$4$$ young men such that it includes at least $$1$$ lady. at least $$1$$ old man and at most $$2$$ young men. Then the total number of ways in which this committee can be formed is :

A
$$40$$

B
$$41$$

C
$$16$$

D
$$32$$

Solution
Question 7
1 / -0

Find the sum of first $$15$$ terms of the sequence whose $${n}^{th}$$ term is $$3+4n$$.

A
$$525$$

B
$$425$$

C
$$495$$

D
$$None\,of\,thes$$

Solution

$$n^{th}$$ term = 3+4 n
$$1^{st}$$ term = 3+4(1)=7
$$2^{nd}$$ term = 3+4(2)=11
$$3^{rd}$$ term 3+4(3)=15
It is an AP with a=7, d=4
$$S_{15}=\frac{15}{2}[2(7)+14(4)]$$
$$= 15[7+28]$$
$$= 525$$
Question 8
1 / -0

Find $$x$$, if $$\dfrac {1}{4!}-\dfrac {1}{x}=\dfrac {1}{5!}$$.

A
$$5$$

B
$$4$$

C
$$30$$

D
$$None$$

Solution
Question 9
1 / -0

solve

A
$$225$$

B
$$175$$

C
$$335$$

D
$$350$$

Solution

Pattens is as follows:-

$$ (20 - 9)^{2} = 121 , $$ $$(24 - 11)^{2} = 169 $$

So $$ (28 - 13)^{2} = 15^{2} = 225$$

Hence missing place = 225
Question 10
1 / -0

insert the missing number: 5,8,12,17,23,___,38

A
$$29$$

B
$$30$$

C
$$32$$

D
$$25$$

Solution

$$5,8,12,17,23,......38$$
$$5+3=8;8+4=12,12+5=17,17+6=23$$
$$23+7=30,30+8=30$$
$$\therefore 30$$