Straight Lines Test 31

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Straight Lines Test 31

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Weekly Quiz Competition

Question 1
1 / -0

How many different $$4$$-person committees can be chosen form the $$100$$ members of the Senate ?

A
$$25$$

B
$$400$$

C
$$3,921,225$$

D
$$94,109,400$$

Solution

Total number of members $$=100$$.
Members has to be selected $$=4$$.
$$\therefore$$ different committees $$=^{100}C_{4}=\large{\frac{100!}{4!\times96!}}$$ $$=3,921,225$$.
Question 2
1 / -0

The number of ways in which we can select 5 letters of the word INTERNATIONAL is equal to

A
$$200$$

B
$$220$$

C
$$242$$

D
$$256$$

Solution

We have thirteen letters.
$$2l, 3N, 2T, 2A$$ and $$(E, O, L,R)8$$ (types)
We can select 5(five) letters in the following manner:

1. All different $${^8C}_5 = \dfrac{8.7.5}{1.2.3} = 56$$ ways

2. 2 alike, 3 different $${^4C}_1 . {^7C}_3 = 435 = 140$$ ways (we have 4 sets of alike letters)

3. 3 alike, 2 different $${^1C}_1 . {^7C}_2 = 1.21 = 21$$ ways (we have only one set of 3 alike)

4. 3 alike and 2 alike $${^1C}_1 . {^3C}_1 = 1.3 = 3$$ ways

5.Two sets of alike and one different
$${^4C}_2 . {^6C}_1 = 6.6 = 36$$ ways

$$\therefore$$ Total number of selection is
$$56 + 140 + 21 + 3 + 36 = 256$$ ways $$\Longrightarrow$$ (d)
Question 3
1 / -0

The area of the triangle with vertices at $$(-4, 1), (1, 2)(4, -3)$$ is

A
$$14$$

B
$$16$$

C
$$15$$

D
None of these

Solution
Question 4
1 / -0

The area of the triangle whose vertices are (3,8), (-4,2) and (5,-1) is

A
$$75 \ sq. units$$

B
$$37.5 \ sq. units$$

C
$$45 \ sq. units$$

D
$$22.5 \ sq. units$$

Solution

Let $$A(3,8), B(-4,2),C(5,-1)$$ be the vertices of the given $$\triangle ABC$$.

Then,

$$(x_{1}=3,y_{1}=8),(x_{2}=-4,y_{2}=2),(x_{3}=5,y_{3}=-1)$$

Area of $$\triangle ABC$$ = $$\dfrac{1}{2}|[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]|$$

$$=\dfrac{1}{2}|3[2-(-1)]-4(-1-8)+5(8-2)|$$

$$=\dfrac{1}{2}|9+36+30|=\dfrac{75}{2}=37.5 \ sq. units$$
Question 5
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Statement I: The function f(x) in the figure is differentiable at x = a
Statement II: The function f(x) continuous at x = a

A
Both Statement I and Statement II are true and the Statement II is the correct explanation of the Statement I.

B
Both Statement I and Statement II are true and the Statement II is not the correct explanation of the Statement I.

C
Statement I is true but Statement II is false

D
Statement I is false but Statement II is true.

Solution

$$f(x)$$ is continuous at as it can be observed from the figure there is no discontinuous at $$x=a$$ but coming to differentiability $$f(x)$$ is not differentiable at $$x=a$$ since the curve is sharp[i.e the left slope and right slope are not equal]
therefore the answer is statement I is false but statement II is true
Question 6
1 / -0

If $$(1+3+5+....+p)+(1+3+5+....+q)=(1+3+5+.....+r),$$ where each set of parentheses contains the sum of consecutive odd integers as shown, the smallest possible value of $$p+q+r$$, (where $$P>6$$ ) is :

A
$$12$$

B
$$21$$

C
$$27$$

D
$$24$$

Solution

$$ 1+3+5+...+2n-1 = n^{2} $$
no. of terms $$ = \dfrac{n+1}{2} $$
According to question we can write
$$\left (\dfrac{p+1}{2}\right)^{2}+\left(\dfrac{q+1}{2}\right)^{2}\pm \left(\dfrac{r+1}{2}\right)^{2} $$
It forms Pythagoras triplet Thus for min value
take 3, 4 and 5 as our Pythagoras triplet
$$ \therefore \dfrac{p+1}{2} = 4 $$ $$ \dfrac{q+1}{2} = 3 $$ $$ \dfrac{r+1}{2} = 5 $$
$$ p = 7 $$ $$ q = 5 $$ $$ r = 9 $$
$$ p + q + r = 21 $$
Question 7
1 / -0

The total number of non-negative integer n satisfying the
equation $$ {n^2} = p + q\,and\,{n^3} = {p^2} + {q^2}$$, where p and q are
integers, is

A
$$1$$

B
$$2$$

C
$$3$$

D
infinite

Solution
Question 8
1 / -0

If $$\dfrac { ^{ n }{ P }_{ r-1 } }{ a } =\dfrac { ^{ n }{ P }_{ r } }{ b } =\dfrac { ^{ n }{ P }_{ r+1 } }{ c } $$, then which of the following holds good:

A
$$c^{2}=a(b+c)$$

B
$$a^{2}=c(a+b)$$

C
$$b^{2}=c(a-b)$$

D
$$\dfrac {1}{a}+\dfrac {1}{b}+\dfrac {1}{c}=1$$

Solution

$$\dfrac{^{n} {P}_{r-1}}{a} = \dfrac{^{n}{P}_{r}}{b} = \dfrac{^{n}{P}_{r+1}}{c} $$

$$\implies \dfrac{n!}{(n - r+ 1)!a} = \dfrac{n!}{(n-r)!b} = \dfrac{n!}{(n-r-1)!c} $$

$$\implies \dfrac{n!}{(n - r+ 1)!a} = \dfrac{n!}{(n-r)(n-r+1)!b} = \dfrac{n!}{(n-r-1)(n-r)(n-r+1)!c} $$

$$\implies \dfrac{1}{a} = \dfrac{1}{(n-r)b} = \dfrac{1}{(n-r-1)(n-r)c} $$

$$\implies n-r = \dfrac{a}{b}$$ and $$n-r-1 = \dfrac{b}{c}$$

$$\implies n-r - 1 = \dfrac{a}{b} - 1 $$

$$\implies \dfrac{b}{c} = \dfrac{a}{b} - 1$$

$$\implies \dfrac{b}{c} - \dfrac{a}{b} = - 1$$

$$\implies b^2 - ac = - bc $$

$$\implies b^2 = ac - bc$$

$$\therefore b^2 = c(a - b)$$ (Ans)
Question 9
1 / -0

Pam likes the numbers 1689 and 6891. Knowing this, which pair of numbers will she like among the ones below?

A
1981 and 1891

B
19 and 91

C
190 and 160

D
1198911 and 1168611

Solution
Question 10
1 / -0

The number of permutations of the letters of the word $$HONOLULU$$ taken $$4$$ at a time is

A
$$ 354$$

B
$$ 314$$

C
$$ 210$$

D
$$ 124$$

Solution

$$H-1$$
$$N-1$$
$$O-2$$
$$L-2$$
$$U-2$$
Number of permutation $$=\dfrac {n!}{r_1\times r_2\times....\times r_n} =\dfrac { 8! }{ { (2! })^{ 3 }4! } =210$$