Sequences and Series Test 42

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Sequences and Series Test 42

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Weekly Quiz Competition

Question 1
1 / -0

If AM of two numbers a and 17 is 15. Then a is ___

A
15

B
13

C
17

D
None of the above

Solution
Question 2
1 / -0

Arithmetic mean of two numbers a+d and a-d is

A
$$a$$

B
$$b$$

C
Cannot be determined

D
None of the above

Solution
Question 3
1 / -0

Find the sum of Arithmetic means of $$3 , 9$$ and $$12 , 8$$.

A
$$6$$

B
$$9$$

C
$$20$$

D
$$16$$

Solution
Question 4
1 / -0

The arithmetic mean of $$4$$ and $$14$$ is

A
$$9$$

B
$$18$$

C
$$10$$

D
$$4$$

Solution
Question 5
1 / -0

Insert an arithmetic mean between $$7$$ and $$21$$

A
$$10$$

B
$$14$$

C
$$28$$

D
$$30$$
Question 6
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 7
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 8
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 9
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 10
1 / -0

Let $$A$$ be the set of all $$3 \times 3$$ symmetric matrices all of whose entries are either $$0$$ or $$1$$. Five of these entries are $$1$$ and four of them are $$0$$.
The number of matrices in $$A$$ is

A
$$12$$

B
$$6$$

C
$$9$$

D
$$3$$

Solution

If two zeros are the entries in the diagonal, then
$$^{3}\mathrm{C}_{2}\times^{3}\mathrm{C}_{1}$$
If all the entries in the principle diagonal is 1, then
$$^{3}\mathrm{C}_{1}$$
$$\Rightarrow$$ Total matrix $$= 12.$$