Straight Lines Test 44

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Straight Lines Test 44

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Weekly Quiz Competition

Question 1
1 / -0

If $$A$$ and $$B$$ are two points having co-ordinates $$(3,4)$$ and $$(5,-2)$$ respectively and $$P$$ is a point such that $$PA=PB$$ and area of triangle $$PAB=10$$ square units, then the co-ordinates of $$P$$ are

A
$$(7,4)$$ or $$(13,2)$$

B
$$(7,2)$$ or $$(1,0)$$

C
$$(2,7)$$ or $$(4,13)$$

D
$$None\ of\ these$$

Solution

Question 2

1 / -0

The letters of word 'ZENITH' are written in all positive ways. If all these words are written in the order of a dictionary, then the rank of the word 'ZENITH' is

$$716$$

$$692$$

$$698$$

$$616$$

Solution

The total number of words is $$6! = 720$$. Let us write the letters of word ZENITH alphabetically, i.e, EHINTZ.

For ZENITH word starts with	Word starting with	Number of words
$$Z$$	$$E$$	$$5!$$
	$$H$$	$$5!$$
	$$I$$	$$5!$$
	$$N$$	$$5!$$
	$$T$$	$$5!$$
ZEN	ZEH	$$3!$$
	ZEI	$$3!$$
ZENI	ZENH	$$2$$
ZENIT	ZENIH	$$1$$
	Total number of words before ZERNITH	$$615$$

Hence, there are $$615$$ words before ZENITH, so the rank of ZENITH is $$616$$.

Question 3
1 / -0

The next number in the pattern 62, 37, 12 ____ is

A
25

B
13

C
0

D
-13

Solution

Option (b) is correct; here, the given series has a common difference of +25.
$$\begin{array}{l}\Rightarrow-62+25=-37,-37+25=-12 \\\therefore-12+25=13\end{array}$$
Question 4
1 / -0

the area of a triangle with vertices $$ (-3, 0),(3,0) $$ and $$ (0, k) $$ is $$ 9 $$ sq units. then the value of $$ k $$ will be

A
$$ 9 $$

B
$$ 3 $$

C
$$ -9 $$

D
$$ 6 $$

Solution

We know that , area of a triangle with vertices $$ (a_1,y_1),(x_2,y_2) $$ and $$ (x_3,y_3) $$ is given by
$$ \Delta = \frac {1}{2} \left| \begin{matrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{matrix} \right| $$
$$ \therefore \Delta = \frac {1}{2} \left| \begin{matrix} -3 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & k & 1 \end{matrix} \right| $$
Expanding along $$ R_1 $$
$$ 9 = \frac {1}{2} [ -3 ( -k) - 0 +1 ( 3k) ] $$
$$ \Rightarrow 18 = 3k + 3k = 6 k $$
$$ \therefore K = \frac {18 }{6} = 3 $$
Question 5
1 / -0

The number of distinct rational numbers x such that $$\displaystyle 0 < x < 1$$ and $$\displaystyle x = \frac{p}{q}r$$, where $$\displaystyle p,q \epsilon \left \{ 1,2,3,4,5,6 \right \}$$,is

A
$$15$$

B
$$13$$

C
$$12$$

D
$$11$$

Solution

Since total number of digits $$=6$$
So to get a number of a number of the form $$\dfrac { p }{ q } $$ we will have to choose two numbers out of the 6 numbers.
There is only one way to arrange them since the number $$x$$ has to satisfy 0 Hence no. $$x = \left( \begin{matrix} 6 \\ 2 \end{matrix} \right) \times 1=15$$, but some numbers have to be removed because they are repeated.
They are :
$$\dfrac { 2 }{ 4 } =\dfrac { 1 }{ 2 } \\ \dfrac { 4 }{ 6 } =\dfrac { 2 }{ 3 } \\ \dfrac { 2 }{ 6 } =\dfrac { 1 }{ 3 } \\ \dfrac { 3 }{ 6 } =\dfrac { 1 }{ 2 } $$
Thus the number of removed numbers$$=4$$
Therefore, total no. of distinct $$x = 15-4=11$$
Hence, option 'D' is correct.
Question 6
1 / -0

The number of rectangles that can be obtained by joining four of the twelve vertices of a $$12$$ sided regular polygon is

A
$$66$$

B
$$30$$

C
$$24$$

D
$$15$$

Solution

The first vertex can be choosed in $$12$$ ways and diagonally opposite to it is $$1$$ vertex. Now for $$3rd$$ vertex we have $$10$$ choices and for $$4th$$ $$1.$$
However, each rectangle is counted $$8$$ times.
$$\therefore$$ No. of ways $$=\dfrac{12\times1\times10\times1}{8}$$ $$=15$$ ways.
Hence, the answer is $$15.$$
Question 7
1 / -0

Three vertices are chosen randomly from the seven vertices of a regular $$7$$ -sided polygon. The probability that they form the vertices of an isosceles triangle is

A
$$\displaystyle \frac{1}{7}$$

B
$$\displaystyle \frac{1}{3}$$

C
$$\displaystyle \frac{3}{7}$$

D
$$\displaystyle \frac{3}{5}$$

Solution

A regular 3 sided polygon is nothing else but a heptagon for creating isosceles triangle we need to choose adjacent sides only.
No. of $$\triangle 's$$ formed $$=7{ { C }_{ 3 } }$$
While number of isosceles triangle formed $$=$$ No. of points $$\times $$ points available $$=7\times 3.$$
$$\Rightarrow$$ So, probability $$=\dfrac { 7\times 3 }{ 7{ { C }_{ 3 } } } = \dfrac { 21 }{ 35 } =\dfrac { 3 }{ 5 } .$$
Hence, the answer is $$\dfrac { 3 }{ 5 }.$$
Question 8
1 / -0

The number of ways in which three numbers in A.P. can be seleced from the set of first n natural number if n is odd is

A
$$ \displaystyle \frac{n\left ( n-2 \right )}{4} $$

B
$$ \displaystyle \frac{n\left ( n-1 \right )^2}{4} $$

C
$$ \displaystyle \frac{\left ( n-1 \right )^2}{4} $$

D
None of these

Solution

In order to solve this question, we must observe the number of ways in which we can select the first term of the required A.P. for different values of the common difference($$r$$) starting from $$r=1$$ given that there are only $$3$$ terms before $$n$$.
For $$r=1$$, the number of ways in which we can select the first term of the A.P. $$=n-2$$
For $$r=2$$, the number of ways to select the first term $$=n-4$$
For $$r=3$$, the number of ways to select the first term $$=n-6$$
Now we see a pattern emerging, we also realize from this that $$r<=\dfrac{n-1}{2}$$ for an A.P. with 3 terms to exist in the given interval.
$$\therefore$$ The final answer $$=n-2+n-4+n-6+...+5+3+1$$
Now we use the formula to find the sum of an A.P. which is $$S_n=\dfrac{n}{2}[a_1+a_n]$$
$$\therefore$$ Answer $$=\dfrac{n-1}{4}[1+n-2]=\dfrac{(n-1)^2}{4}$$
Question 9
1 / -0

If the difference of the number of arrangements of three things from a certain number of dissimilar things and the number of selections of the same number of things from them exceeds $$ 100$$, then the least number of dissimilar things is

A
$$8$$

B
$$6$$

C
$$5$$

D
$$7$$

Solution

Here,
$$^nP_3-^nC_3 > 100$$

or $$\dfrac {n!}{(n-3)!}-\dfrac {n!}{3!(n-3)!} > 100$$

or $$\dfrac {5}{6}n(n-1)(n-2) > 100$$

or $$n(n-1)(n-2) > 100$$

or $$n(n-1)(n-2) > 6\times 5\times 4$$

or $$n=7, 8, .....$$
Question 10
1 / -0

If four points are $$A(6,3),B(-3,5),C(4,-2)$$ and $$P(x,y),$$ then the ratio of the areas of $$\triangle PBC$$ and $$\triangle ABC$$ is:

A
$$\displaystyle \frac { x+y-2 }{ 7 } $$

B
$$\displaystyle \frac { x-y-2 }{ 7 } $$

C
$$\displaystyle \frac { x-y+2 }{ 2 } $$

D
$$\displaystyle \frac { x+y+2 }{ 2 } $$

Solution

Given: Coordinates of points $$\displaystyle A\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 6,3 \right) ,B\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( -3,5 \right) ,C\left( { x }_{ 3, }{ y }_{ 3 } \right) =\left( 4,-2 \right) $$ and $$\displaystyle P\left( x,y \right) .$$

We know that the area of:
$$\displaystyle\triangle PBC=\frac { 1 }{ 2 } \left[ x\left( { y }_{ 2 }-{ y }_{ 3 } \right) +{ x }_{ 3 }\left( y-{ y }_{ 2 } \right) +{ x }_{ 2 }\left( { y }_{ 3 }-y \right) \right] $$

$$\displaystyle=\frac { 1 }{ 2 } \left[ x\left( 5+2 \right) +4\left( y-5 \right) -3\left( -2-y \right) \right] $$ $$\displaystyle=\frac { 1 }{ 2 } \left[ 7x+7y-14 \right] $$

Similarly, the area of

$$\displaystyle\triangle ABC=\frac { 1 }{ 2 } \left[ { x }_{ 1 }\left( { y }_{ 2 }-{ y }_{ 3 } \right) +{ x }_{ 2 }\left( { y }_{ 3 }-{ y }_{ 1 } \right) \right] +{ x }_{ 3 }\left( { y }_{ 1 }-{ y }_{ 2 } \right) $$

$$\displaystyle=\dfrac{1}{2}[6\left( 5+2 \right) -3\left( -2-3 \right) +4\left( 3-5 \right)] =\dfrac{49}{2}$$

Therefore, the ratio of the areas of $$\triangle PAB$$ and $$\triangle ABC$$

$$\displaystyle=\frac { 7x+7y-14 }{ 49 } =\frac { 7\left( x+y-2 \right) }{ 49 } =\frac { x+y-2 }{ 7 } $$

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Re-Attempt Weekly Quiz Competition

Straight Lines Test 44

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Straight Lines Test 44

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SHARING IS CARING

Question 1

$$(7,4)$$ or $$(13,2)$$

$$(7,2)$$ or $$(1,0)$$

$$(2,7)$$ or $$(4,13)$$

$$None\ of\ these$$

Solution

Question 2

$$716$$

$$692$$

$$698$$

$$616$$

Solution

Question 3

25

13

0

-13

Solution

Question 4

$$ 9 $$

$$ 3 $$

$$ -9 $$

$$ 6 $$

Solution

Question 5

$$15$$

$$13$$

$$12$$

$$11$$

Solution

Question 6

$$66$$

$$30$$

$$24$$

$$15$$

Solution

Question 7

$$\displaystyle \frac{1}{7}$$

$$\displaystyle \frac{1}{3}$$

$$\displaystyle \frac{3}{7}$$

$$\displaystyle \frac{3}{5}$$

Solution

Question 8

$$ \displaystyle \frac{n\left ( n-2 \right )}{4} $$

$$ \displaystyle \frac{n\left ( n-1 \right )^2}{4} $$

$$ \displaystyle \frac{\left ( n-1 \right )^2}{4} $$

None of these

Solution

Question 9

$$8$$

$$6$$

$$5$$

$$7$$

Solution

Question 10

$$\displaystyle \frac { x+y-2 }{ 7 } $$

$$\displaystyle \frac { x-y-2 }{ 7 } $$

$$\displaystyle \frac { x-y+2 }{ 2 } $$

$$\displaystyle \frac { x+y+2 }{ 2 } $$

Solution

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