Straight Lines Test 37

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Straight Lines Test 37

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Weekly Quiz Competition

Question 1
1 / -0

In the following question, groups of letters are given. Find the odd one out.

A
$$\mathrm {EfgiL}$$

B
$$\mathrm {RcjkY}$$

C
$$\mathrm {DacsO}$$

D
$$\mathrm {QuegP}$$

Solution
Question 2
1 / -0

In how many ways atleast one horse and atleast one dog can be selected out of eight horses and seven dogs.

A
$${2}^{15}-2$$

B
$${2}^{15}-1$$

C
$$({2}^{8}-1)({2}^{7}-1)$$

D
$${ _{ }^{ 15 }{ C } }_{ 2 }$$

Solution

As we can either select or not select any horse; total ways for horses$$=2^r$$ ways to be excluded are when no horse is selected, which is only one way, therefore, favourable ways for horses$$=2^8-1$$
similarly, favourable ways for dogs$$=2^7-1$$
$$\Rightarrow$$ Total ways$$=(2^8-1)(2^7-1)$$.
Question 3
1 / -0

The total number of ways of arranging the letters $$AAABBBCCDEF$$ in a row such that letters $$C$$ are separated from one another is

A
$$277200$$

B
$$138600$$

C
$$453600$$

D
none of these

Solution
Question 4
1 / -0

The number of ways in which the letters of the word $$"ARRANGE"$$ can be permuted such that $$R's$$ occur together is

A
$$\dfrac{!7}{!2!2}$$

B
$$6!$$

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

D
none of these

Solution
Question 5
1 / -0

$$\begin{bmatrix} 12 & 47 & 21 \\ 10 & 52 & 4 \\ 64 & ? & 24 \end{bmatrix}$$

A
$$40$$

B
$$83$$

C
$$62$$

D
$$16$$

Solution

$$\begin{bmatrix} 12 & 47 & 21\\ 10 & 52 & 4\\ 64 & ? & 24\end{bmatrix}$$
We can see that dividing third number in each row with second digit of middle number and multiplying with first digit gives first member.
Similarly, out of $$4$$ options, dividing $$24$$ by $$3$$ and multiplying by $$8$$ gives $$64$$.
$$\Rightarrow$$ Option $$(B)$$.
Question 6
1 / -0

If $$64a^2+36b^2=400, ab=4 $$ , then $$8a+6b$$ is:

A
26

B
28

C
22

D
None of the above

Solution

Given $$64a^2+36b^2=400,\quad ab=4$$
We know
$$(a+b)^2=a^2+b^2+2ab$$
therefore,
$$(8a+6b)^2=(8a)^2+(6b)^2+2(8a)(6a)$$
$$=64a^2+36b^2+96ab$$
$$=400+96\times4$$
$$=400+384$$
$$(8a+6b)^2=784$$
$$8a+6b=\sqrt{784}$$
$$8a+6b=28$$
Question 7
1 / -0

The number of positive integral solutions of the equation $$x _ { 1 } x _ { 2 } x _ { 3 } x _ { 4 } x _ { 5 } = 1050$$ is

A
1800

B
1600

C
1400

D
1875

Solution

We have,
$${x_1}{x_2}{x_3}{x_4}{x_5} = 1050 = 2 \times 3 \times {5^2} \times 7$$
Now,
$$2,3$$ and $$7$$ can be put in any boxes $${x_1},{x_2},{x_3},{x_4}$$ and $${x_5}$$.
Also $$5,5$$ can be distributed in $$5$$ boxes in
$$^{2 + 5 - 1}{C_{5 - 1}}{ = ^6}{C_4} = 15$$Ways.
So total no. of positive integral solutions $$=15 \times5 \times5 \times5$$
$$=1875$$
Option $$D$$ is correct answer.
Question 8
1 / -0

If the area of triangle formed by the points $$(2a,b),(a+b,2b+a)$$ and $$(2b,2a)$$ be $$\lambda$$ then the area of the triangle whose vertices are $$(a+b,a-b), (3b-a,b+3a)$$ and $$(3a-b,3b-a)$$ will be

A
$$\dfrac{3}{2} \lambda$$

B
$$3\lambda$$

C
$$4\lambda$$

D
$$none\ of\ these$$

Solution

Area of $$ \triangle \,= \dfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2}) \right |$$
Given area of triangle whose vertices are $$ (2a,b), (a+b,2b+a), (2b,2a)$$ is $$ \lambda $$
$$\therefore \lambda = \dfrac{1}{2}\left | 2a(2b+a-2a)+(a+b)(2a-b)+2b(b-2b-a) \right | $$
$$ \lambda = \dfrac{1}{2}\left | 4ab+2a^{2}-4a^{2}+2a^{2}-ab+2ab-b^{2}+2b^{2}-4b^{2}-2ab \right |$$
$$ \lambda = \dfrac{1}{2}\left | 3ab - 3b^{2} \right | $$ __ (1)
Now area of triangle whose vertices are $$ (a+b,a-b), (3b-a,b+3a), (3a-b,3b-a)$$ is given by
Area $$ = \dfrac{1}{2}\left | (a+b)(b+3a-3b+a)+(3b-a)(3b-a-a+b)+(3a-b)(a-b-b-3a) \right |$$
$$ = \dfrac{1}{2}\left | (a+b) (-2b+4a)+(3b-a)(4b-2a)+(3a-b)(-2a-2b) \right |$$
$$ = \dfrac{1}{2}\left | -2ab+4a^{2}-2b^{2}+4ab+12b^{2}-6ab-4ab+2a^{2}-6a^{2}-6ab+2ab+2b^{2} \right |$$
$$ = \dfrac{1}{2}\left | 12b^{2}-12ab \right |$$
$$ = \dfrac{4}{2}\left | -(3ab-3b^{2}) \right |$$
$$ = 4.\dfrac{1}{2}\left | 3ab-3b^{2} \right |$$
$$ = 4 \lambda .$$
Question 9
1 / -0

If $$(1+x+x^2)^n=\displaystyle\sum^{2n}_{r=0}a_rx^r$$, then $$a_0a_{2r}-a_1a_{2r+1}+a_2a_{2r+2}-....=?$$

A
$$a_r$$

B
$$a_{n-2r}$$

C
$$a_{n+r}$$

D
$$a_{2r}$$
Question 10
1 / -0

If repetitions are not allowed, the number of numbers consisting of $$4$$ digits and divisible by $$5$$ and formed out of $$0,1,2,3,4,5,6$$ is

A
$$220$$

B
$$240$$

C
$$370$$

D
$$588$$

Solution

$$\rightarrow$$ For divisibility by $$5$$, unit's place should be $$0$$ or $$5$$
with $$0$$, ----$$0$$
$$6$$ choices $$\Rightarrow ^6C_3\times 3!$$
$$=\dfrac{6!}{3!3!}\times 3!=6.5.4$$
$$=120$$
with $$5$$, ----$$5$$
$$6$$ choices $$=120$$
$$^5C_2\times 2!=10\times 2=20$$
$$\Rightarrow 120+120-20=220$$
$$\rightarrow (A)$$