Straight Lines Test 34

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Straight Lines Test 34

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Weekly Quiz Competition

Question 1
1 / -0

A question paper on mathematics consists of twelve questions divided into three parts A, B and C, each containing four questions. In how many ways can an examinee answer five questions, selecting atleast one from each part.

A
$$624$$

B
$$208$$

C
$$2304$$

D
None

Solution

Total no. of ways
$$A$$ $$B$$ $$ C$$ $$A$$ $$B$$ $$ C$$
$$=(^4C_1\times \ ^4C_1 \times \ ^4C_3)\times 3+ (^4C_1\, ^4C_2\, ^4C_2)\times 3$$ ($$\therefore$$ multiple by 3 as for each A,B,C different combination)
$$=192+432$$
$$=624$$
Question 2
1 / -0

The number of zeroes, at the end of $$50!$$, is

A
$$5$$

B
$$10$$

C
$$11$$

D
$$12$$

Solution

Solution-
50!$$\rightarrow $$ Number of zeroes of the end
Zeroes can be obtained by multiples of 10
as multiples of 5 and 2.
5 zeroes there are many powers of 2 in 50!
we need to find power of 5 in 50!
Power $$ = 1(5)+1(15)+2(25)+1(35)+1(45)+1(50)$$
$$ = 7 $$
7 zeroes caused by 5, 15, 25, 35, 45, 50
Total zeroes = 5+7 = 12
Question 3
1 / -0

Garlands are formed using 6 red roses and 6 yellow roses of different sizes. The number of arrangements in garland which have red roses and yellow roses come alternately is

A
$$5!\times 6!$$

B
$$6! \times 6!$$

C
$$\dfrac{5!}{2!}\times 6!$$

D
$$2(6!\times 6!)$$

Solution
Question 4
1 / -0

If $$f(x)=1-x+x^2-x^3+....-x^{15}+x^{16}+x^{17}$$, then the coefficient of $$x^2$$ in $$f(x-1)$$ is?

A
$$826$$

B
$$816$$

C
$$822$$

D
None of these

Solution

$$f\left( x \right) = 1 - \left( {x - 1} \right) + {\left( {x - 1} \right)^2} - {\left( {x - 1} \right)^3} + {\left( {x - 1} \right)^4}.... - {\left( {x - 1} \right)^{15}} + {\left( {x - 1} \right)^{16}} + {\left( {x - 1} \right)^{17}}$$
The coefficient of $${x^2}$$ by using binomial expansion
$$^2{c_0}{x^2}{ + ^3}{c_1}{x^2}{ + ^4}{c_2}{x^2}{ + ^5}{c_3}{x^2}{ + ^6}{c_4}{x^2}{ + ^7}{c_5}{x^2}{ + ^8}{c_6}{x^2}{ + ^9}{c_7}{x^2}$$
$${ + ^{10}}{c_8}{x^2}{ - ^{11}}{c_9}{x^2}{ + ^{12}}{c_{10}}{x^2}{ - ^{13}}{c_{11}}{x^2}{ + ^{14}}{c_{12}}{x^2}{ - ^{15}}{c_{13}}{x^2}{ + ^{16}}{c_{14}}{x^2}{ - ^{17}}{c_{15}}{x^2}$$
$$ \Rightarrow {x^2}\left( {1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 + 45 + 55 + 66 + 78 + 91 + 105 + 120 + 136} \right)$$
= 372 + 444 = 816
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

Area of the triangle formed by $$(x_{1,}y_{1}),(x_{2},y_{2}), (3y_{2},(-2y_{1}))$$

A
$$ \dfrac{1}{2}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{1}}-3{{y}_{2}}^{2} \right] $$

B
$$ \dfrac{1}{2}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{1}}-3{{y}_{2}}^{3} \right] $$

C
$$ \dfrac{2}{3}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{1}}-3{{y}_{2}}^{2} \right] $$

D
$$ =\dfrac{1}{2}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{3}} \right] $$

Solution

We have,

$$ A\left( {{x}_{1}},{{y}_{1}} \right)=\left( {{x}_{1}},{{y}_{1}} \right) $$

$$ B\left( {{x}_{2}},{{y}_{2}} \right)=\left( {{x}_{2}},{{y}_{2}} \right) $$

$$ C\left( {{x}_{3}},{{y}_{3}} \right)=\left( 3{{y}_{2}},-2{{y}_{1}} \right) $$

We know that the area of triangle is

$$ Area\,of\,\Delta ABC=\dfrac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right] $$

$$ =\dfrac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}+2{{y}_{1}} \right)+{{x}_{2}}\left( -2{{y}_{1}}-{{y}_{1}} \right)+3{{y}_{2}}\left( {{y}_{1}}-{{y}_{2}} \right) \right] $$

$$ =\dfrac{1}{2}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{1}}-3{{y}_{2}}^{2} \right] $$
Hence, the is the answer
Question 7
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 8
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 9
1 / -0

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

A
$$2$$

B
$$7$$

C
$$4$$

D
$$9$$

Solution
Question 10
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 $$3^{n}$$.

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Straight Lines Test 34

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Straight Lines Test 34

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Question 1

$$624$$

$$208$$

$$2304$$

None

Solution

Question 2

$$5$$

$$10$$

$$11$$

$$12$$

Solution

Question 3

$$5!\times 6!$$

$$6! \times 6!$$

$$\dfrac{5!}{2!}\times 6!$$

$$2(6!\times 6!)$$

Solution

Question 4

$$826$$

$$816$$

$$822$$

None of these

Solution

Question 5

$$800$$

$$700$$

$$500$$

$$600$$

Solution

Question 6

$$ \dfrac{1}{2}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{1}}-3{{y}_{2}}^{2} \right] $$

$$ \dfrac{1}{2}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{1}}-3{{y}_{2}}^{3} \right] $$

$$ \dfrac{2}{3}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{1}}-3{{y}_{2}}^{2} \right] $$

$$ =\dfrac{1}{2}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{3}} \right] $$

Solution

Question 7

4r

5p

2q

5

Solution

Question 8

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

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

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 $$^9C_4$$ $$8!7!$$

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

Question 9

$$2$$

$$7$$

$$4$$

$$9$$

Solution

Question 10

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

$$2^{n}$$

$$2^{n}-1$$

$$3^{n}$$

Solution

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