Sequences and Series Test 32

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Sequences and Series Test 32

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Weekly Quiz Competition

Question 1
1 / -0

The unit's place digit in $$(1446)^{4n + 3}$$ is

A
$$n$$

B
$$0$$

C
$$6$$

D
None of these

Solution
Question 2
1 / -0

From 0 to 9 , four digited numbers can be formed such that
the digits are in ascending order is

A
$${}^{10}{P_4}$$

B
$${}^{10}{C_4}$$

C
$${}^{10}{P_4} - {}^9{P_3}$$

D
$${}^{10}{C_4} - {}^9{C_3}$$

Solution

Selection of$$ 4$$ digits out of$$ 10$$ (including 0)=$$ ^{10}C_4$$
Question 3
1 / -0

If $$p:q:r=1:2:3\;then$$ $$\sqrt {5{p^2} + {q^2} + {r^2}} $$ is equal to:

A
4r

B
5p

C
2q

D
5

Solution

Consider the problem

Given, $$p:q:r=1:2:3$$

Therefore, $$\frac{p}{1} = \frac{q}{2} = \frac{r}{4} = k$$ (let)

So, $$p = k,q = 2k,r = 4k$$

Therefore,

$$\begin{array}{l} \sqrt { 5{ p^{ 2 } }+{ q^{ 2 } }+{ r^{ 2 } } } \\ =\sqrt { 5{ k^{ 2 } }+4{ k^{ 2 } }+16{ k^{ 2 } } } \\ =\sqrt { 25{ k^{ 2 } } } \\ 5k=5p \end{array}$$

Hence, Option $$B$$ is the correct answer which is $$5p$$
Question 4
1 / -0

$$7$$ boys and $$8$$ girls have to sit in a row on $$15$$ chairs numbered from $$1$$ to $$15$$ then?

A
Number of ways boys and girls sit alternately is $$8!7!$$

B
Number of ways boys and girls sit alternately is $$2(8!7!)$$

C
The number of ways in which first and fifteenth chair are occupied by boys and between any two boys an even number of girls sit is $$^9C_4$$ $$8!7!$$

D
The number of ways in which first and last seat are occupied by boys and between any two boys an even number of girls sit is $$(2{^9C_4}8!7!)$$.

Solution

$$\begin{matrix} G-denotes\, \, Girls \\ X-denotes\, \, boys \\ \Rightarrow G\times G\times G\times G\times G\times G\times G\times G \\ \Rightarrow Number\, \, of\, \, ways\, \, of\, Girls\, \, is\, \, 8!\, \, alternate \\ \Rightarrow Number\, \, of\, \, ways\, \, of\, boys\, \, is\, \, 7! \\ \, \, \, \, \, \, \, after\, \, late \\ \Rightarrow Number\, \, of\, \, boys\, \, and\, \, girls\, \, sit\, \, alternate\, \, =8!7! \\ When\, \, first\, \, and\, \, fifteen\, \, chair\, \, are\, \, fixed\, \, and\, \\ \, first\, \, chair\, \, \times G\times G\times G\times G\times G\times G\times G\times fifteen\, \, chair\, \, between\, \, any\, \, two\, \, boys \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \underline { 9{ C_{ 4 } }8!7! } \, \, \, Ans. \\ \end{matrix}$$
Question 5
1 / -0

The three digits in 10! are:

A
$$800$$

B
$$700$$

C
$$500$$

D
$$600$$

Solution

As we know that $$10!=1\times 2\times 3\times 4\times 5\times 6\times 7\times 8\times 9\times 10$$
$$10!=10^2(3\times 4\times 6\times 7\times 8\times 9)$$
$$10!=10^2(3^4\times 2^6\times 7)$$
$$10!=10^2(81\times 64\times 7)$$
The last three digit in $$10!$$ is $$800$$
The correct option is A.
Question 6
1 / -0

Replace the question mark (?) with the correct option.

A
$$2$$

B
$$7$$

C
$$4$$

D
$$9$$

Solution
Question 7
1 / -0

A is a set containing n elements. A subset P of A is chosen. The set A is reconstructed by replacing the element of P.A subset Q of A is again chosen. The number of way of choosing P and Q so that P Q =$$\phi $$ is :-

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

B
$$2^{n}$$

C
$$2^{n}-1$$

D
$$3^{n}$$

Solution

Let A = $${a_{1},a_{2},a_{3},....a_{n}}$$. For $$a_{1}$$ $$\epsilon $$ For $$a_{1} \epsilon $$A we have following choices:
(i) $$a_{1} \epsilon $$ P and $$a_{1} \epsilon $$ Q
(ii) $$a_{1} \epsilon $$P $$\notin $$
(iii) $$_{1}\notin P and a_{1}\epsilon Q$$
(Iv)$$_{1}\notin P and a_{1}\notin Q$$
Out of these only (Ii), and (iii) and (iv) imply $$a_{1}\notin P\cap Q $$ therefore, the number of ways in which none of $$a_{1},a_{2}.....a_{n}$$ belong $$P\cap Q$$ is $$3^{n}$$.
Question 8
1 / -0

Find: $$6,25,62,123,(?),341$$

A
$$216$$

B
$$214$$

C
$$215$$

D
$$218$$

Solution

$$6=2^{3}-2=6$$
$$25=3^{3}-2=25$$
$$62=4^{3}-2=62$$
$$123=5^{3}-2=123$$
$$214=6^{3}-2=214$$
$$341=7^3-2=341$$

Missing term is $$214$$
Question 9
1 / -0

Let $$5 < n_1 < n_2 < n_3 < n_4$$ be integers such that $$n_1+n_2+n_3+n_4=35$$. The number of such distinct arrangements $$(n_1, n_2, n_3, n_4)$$.

A
$$^{38}C_3$$

B
$$^8C_3$$

C
$$5$$

D
$$6$$

Solution

$${ n }_{ 1 }+{ n }_{ 2 }+{ n }_{ 3 }+.....+{ n }_{ k }=R$$
for this arrangement,
No. of arrangement or distinct arrangements are $${ R+K-1 }_{ { C }_{ K-1 } }$$
So, for $${ n }_{ 1 }+{ n }_{ 2 }+{ n }_{ 3 }+{ n }_{ 4 }=35$$
No. of arrangements $$={ 35+4-1 }_{ { C }_{ 4-1 } }={ 38 }_{ { C }_{ 3 } }$$
Question 10
1 / -0

Select the missing number from the given responses ?

A
$$15$$

B
$$10$$

C
$$12$$

D
$$17$$

Solution