Straight Lines Test 45

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Straight Lines Test 45

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Weekly Quiz Competition

Question 1
1 / -0

In a test there were n questions. In the test $$\displaystyle 2^{n-i}$$ students gave wrong answers to i questions where $$\displaystyle i=1,2,3...,n$$. If the total number of wrong answers given is 2047 then n is

A
12

B
11

C
10

D
none of these

Solution

Total number of wrong answers $$= 2^{n-1}+2^{n-2}+...+2^{2}+2+1$$
$$=2^{n-1}+2^{n-2}+...+2^{4}+2^{3}+2^{2}+2+1$$
$$\Rightarrow 2^{n}-1=2043$$ [Using formula for sum of G.P]
$$ \Rightarrow 2^{n}=2048$$
$$ \Rightarrow 2^{n}=2^{11}$$
$$\Rightarrow n = 11$$
Hence, option 'B' is correct.
Question 2
1 / -0

If $$^nC_3=^nC_{13}$$, then $$^{20}C_n$$ is.

A
$$1825$$

B
$$3801$$

C
$$4845$$

D
$$300$$

Solution

We know that $$ { { n }_{ C } }_{ r }= { { n }_{ C } }_{n- r }$$

Given, $$ { { n }_{ C } }_{3}= { { n }_{ C } }_{13} $$
$$ => r = 3 $$ and $$ n -r = 13 $$
$$ => n - 3 = 13 $$
$$ => n = 16 $$

Also, $$ { { n }_{ C } }_{ r }=\dfrac { n! }{ r!(n-r)! } $$
So, $$ { { 20 }_{ C } }_{ 16 }=\dfrac { 20! }{ 16!(20-16)! } = \dfrac { 20! }{ 16! \times 4! } = \dfrac { 20 \times 19 \times 18 \times 17 \times 16! }{ 16! \times 4 \times 3 \times 2 } = 4845 $$
Question 3
1 / -0

The sides of a quadrilateral are all positive integers and three of them are $$5, 10, 20.$$ How many possible value are there for the fourth side?

A
$$29$$

B
$$31$$

C
$$32$$

D
$$34$$

Solution
Question 4
1 / -0

Directions For Questions

The area of a triangle whose vertices are $$\displaystyle \left( { x }_{ 1 },{ y }_{ 1 } \right) ,\left( { x }_{ 2 },{ y }_{ 2 } \right) $$ and $$\displaystyle \left( { x }_{ 3 },{ y }_{ 3 } \right) $$ is given by $$\displaystyle \Delta =\frac { 1 }{ 2 } \left| { x }_{ 1 }\left( { y }_{ 2 },{ y }_{ 3 } \right) +{ x }_{ 2 }\left( { y }_{ 3 },{ y }_{ 1 } \right) +{ x }_{ 3 }\left( { y }_{ 1 },{ y }_{ 2 } \right) \right| $$. The points $$\displaystyle \left( { x }_{ 1 },{ y }_{ 1 } \right) ,\left( { x }_{ 2 },{ y }_{ 2 } \right) $$ and $$\displaystyle \left( { x }_{ 3 },{ y }_{ 3 } \right) $$ are collinear of $$\displaystyle \Delta =0$$

...view full instructions

Determine the area of the triangle whose vertices are $$\displaystyle \left( \frac { 1 }{ 2 } ,\frac { -1 }{ 2 } \right) ,\left( 2,\frac { -1 }{ 2 } \right) $$ and $$\displaystyle \left( 2,\frac { \sqrt { 3 } -1 }{ 2 } \right) $$.

A
$$\displaystyle 3\sqrt { 3 } $$

B
$$\displaystyle \frac { 3\sqrt { 3 } }{ 8 } $$

C
$$\displaystyle 12\sqrt { 3 } $$

D
$$\displaystyle \frac { 3\sqrt { 3 } }{ 4 } $$

Solution
Question 5
1 / -0

A college offers $$7$$ courses in the morning and $$5$$ courses in the evening. Find the number of ways a student can select exactly one course either in the morning or in the evening.

A
$$35$$

B
$$12$$

C
$$40$$

D
$$30$$

Solution

$$7$$ Courses in morning
$$5$$ courses in evening
Total number of courses $$=12$$
Selecting any one of the course
Number of ways $$^{ 12 }{ C }_{ 1 }$$
$$=\cfrac { 12! }{ 1!\times 1! } $$
$$=12$$ Ways
Therefore total ways $$=12$$
Question 6
1 / -0

If $$m$$ denotes the number of $$5$$ digit numbers if each successive digits are in their descending order of magnitude and $$n$$ is the corresponding figure. When the digits and in their ascending order of magnitude then $$(m-n)$$ has the value

A
$$ { _{ }^{ 9 }{ C } }_{ 4 }$$

B
$${ _{ }^{ 9 }{ C } }_{ 5 }$$

C
$${ _{ }^{ 10 }{ C } }_{ 3 }$$

D
$$ { _{ }^{ 9 }{ C } }_{ 3 }$$

Solution

For $$m$$,

First we select any $$5$$ digits from $$0,1,2,...,9$$

Number of ways $$=10C_5$$

Now after selection there is only $$1$$ way to arrange these selected digits, i.e., in descending order.

Therefore $$m=10C_5 \times 1=10C_5$$

For $$n$$,

First we select any $$5$$ digits from $$1,2,...,9$$

We can't select zero as then it would be the first digit and then the number of won't be a $$5-$$digit one.

Therefore number of ways $$=9C_5$$

$$\Rightarrow n=9C_5 \times 1=9C_5$$

$$m-n=10C_5-9C_5$$

We know that, $$nC_r+nC_{r-1}=(n+1)C_r$$

$$\Rightarrow (n+1)C_r-nC_r=nC_{r-1}$$

$$\therefore 10C_5-9C_5=9C_4$$

Hence $$m-n=9C_4$$
Question 7
1 / -0

Numbers can be classified into two categories,depending on their divisible conditions.
They are (i) Even numbers $$(2p) \vee p \epsilon N$$ (ii) odd numbers $$(2p + 1) \vee p \epsilon N$$
a. $$a_1, a_2 ...... a_{2013}$$ are integers, not necessarily distinct.
$$x = (-1)^{a_1}+(-1)^{a_2}+.....+(-1)^{a_{1006}}$$
$$y = (-1)^{a_{1007}}+(-1)^{a_{1008}}+......+(-1)^{a_{2013}}$$

Then which of the following is true?

A
$$(-1)^x = 1; (-1)^y = 1$$

B
$$(-1)^x = 1; (-1)^y = -1$$

C
$$(-1)^x = -1; (-1)^y = 1$$

D
$$(-1)^x = -1; (-1)^y = -1$$

Solution
Question 8
1 / -0

If $$ { _{ }^{ n }{ C } }_{ 4 },{ _{ }^{ n }{ C } }_{ 5 }$$ and $$ { _{ }^{ n }{ C } }_{ 6 }$$ are in AP, then $$n$$ is

A
$$7$$ or $$14$$

B
$$7$$

C
$$14$$

D
$$14$$ or $$21$$

Solution

Since, $${ _{ }^{ n }{ C } }_{ 4 },{ _{ }^{ n }{ C } }_{ 5 }$$ and $$ { _{ }^{ n }{ C } }_{ 6 }$$ are in AP
$$\therefore \quad 2{ _{ }^{ n }{ C } }_{ 5 }={ _{ }^{ n }{ C } }_{ 4 }+{ _{ }^{ n }{ C } }_{ 6 }$$
$$\Rightarrow 2\times \cfrac { n! }{ 5!(n-5)! } =\cfrac { n! }{ 4!(n-4)! } +\cfrac { n! }{ 6!(n-6)! } $$
$$\Rightarrow \cfrac { 2 }{ 5(n-5) } =\cfrac { 1 }{ (n-4)(n-5) } +\cfrac { 1 }{ 30 } $$
$$ \Rightarrow \cfrac { 2 }{ 5(n-5) } =\cfrac { 30+{ n }^{ 2 }-9n+50 }{ 30(n-4)(n-5) } $$
$$\Rightarrow 12(n-4)={ n }^{ 2 }-9n+50$$
$$\Rightarrow { n }^{ 2 }-21n+98=0$$
$$\Rightarrow (n-14)(n-7)=0\quad $$
$$\Rightarrow n=7,14$$
Question 9
1 / -0

Two classrooms A and B having capacity of $$25$$ and $$(n-25)$$ seats respectively. $$A_n$$ denotes the number of possible seating arrangements of room $$'A'$$, when 'n' students are to be seated in these rooms, starting from room $$'A'$$ which is to be filled up to its capacity. If $$A_n-A_{n-1}=25!(^{49}C_{25})$$ then 'n' equals:

A
$$50$$

B
$$48$$

C
$$49$$

D
$$51$$

Solution

Given $$A_n=nC_{25} \cdot 25!$$

$$A_{n-1}={n-1}C_{25} \cdot 25!$$

Hence $$nC_{25} \cdot 25! - (n-1)C_{25} \cdot 25!=25! 49C_{25}$$

$$\Rightarrow (n-1)C_{25}+(n-1)C_{24}-(n-1)C_{25}=49C_{24}$$

$$\Rightarrow n-1=49$$

$$\Rightarrow n=50$$
Question 10
1 / -0

The line $$\displaystyle 3x+2y=24$$ meets x-axis at A and y-axis at B. The perpendicular bisector of $$\displaystyle \overline { AB } $$ meets the line through (0, -1) and parallel to x-axis at C. Find the area of $$\displaystyle \Delta ABC$$.

A
85 $$\displaystyle { units }^{ 2 }$$

B
87 $$\displaystyle { units }^{ 2 }$$

C
90 $$\displaystyle { units }^{ 2 }$$

D
91 $$\displaystyle { units }^{ 2 }$$

Solution

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Re-Attempt Weekly Quiz Competition

Straight Lines Test 45

Result

Straight Lines Test 45

Score

-

Rank

-

SHARING IS CARING

Question 1

12

11

10

none of these

Solution

Question 2

$$1825$$

$$3801$$

$$4845$$

$$300$$

Solution

Question 3

$$29$$

$$31$$

$$32$$

$$34$$

Solution

Question 4

$$\displaystyle 3\sqrt { 3 } $$

$$\displaystyle \frac { 3\sqrt { 3 } }{ 8 } $$

$$\displaystyle 12\sqrt { 3 } $$

$$\displaystyle \frac { 3\sqrt { 3 } }{ 4 } $$

Solution

Question 5

$$35$$

$$12$$

$$40$$

$$30$$

Solution

Question 6

$$ { _{ }^{ 9 }{ C } }_{ 4 }$$

$${ _{ }^{ 9 }{ C } }_{ 5 }$$

$${ _{ }^{ 10 }{ C } }_{ 3 }$$

$$ { _{ }^{ 9 }{ C } }_{ 3 }$$

Solution

Question 7

$$(-1)^x = 1; (-1)^y = 1$$

$$(-1)^x = 1; (-1)^y = -1$$

$$(-1)^x = -1; (-1)^y = 1$$

$$(-1)^x = -1; (-1)^y = -1$$

Solution

Question 8

$$7$$ or $$14$$

$$7$$

$$14$$

$$14$$ or $$21$$

Solution

Question 9

$$50$$

$$48$$

$$49$$

$$51$$

Solution

Question 10

85 $$\displaystyle { units }^{ 2 }$$

87 $$\displaystyle { units }^{ 2 }$$

90 $$\displaystyle { units }^{ 2 }$$

91 $$\displaystyle { units }^{ 2 }$$

Solution

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