Mathematics Test

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Mathematics Test - 18

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Weekly Quiz Competition

Question 1
1 / -0

The probability that A speaks truth is4/5, while this probability for B is3/4. The probability that they contradict each other when asked to speak on a fact is

A
3/23

B
1/20

C
1/5

D
14/40

Solution

In short the event described here is $\mathrm{P}(\mathrm{A})^{\prime}$ P(B) $+\mathrm{P}(\mathrm{A}) \mathrm{P}(\mathrm{B})^{\prime}$
where $P(A)$ and $P(B)$ are the probabilities of $A$ and $B$ speaking the truth
respectively.
\[
P(E)=P(A)^{\prime} P(B)+P(A) P(B)^{\prime}=\frac{1}{5} \times \frac{3}{4}+\frac{4}{5} \times \frac{1}{4}
\]
$P(E)=\frac{3}{20}+\frac{4}{20}=\frac{7}{20}=\frac{14}{40}$
Question 2
1 / -0

The minimum number of elements that must be added to the relation R= {(1,2)(2,3)} on the set of natural numbers so that it is an equivalence is

A
4

B
7

C
6

D
5

Solution

For $R$ to be equivalence relation, it should be reflexive, symmetric and transitive Now, $R$ is reflexive $\Rightarrow(x, x) \in R$ for all $x \in\{1,2,3\}$
$\Rightarrow(1,1),(2,2),(3,3) \in R$
$R$ is symmetric $\Rightarrow(1,2),(2,3) \in R$
$\Rightarrow(2,1),(3,2) \in R$
$R$ is transitive
$\Rightarrow(1,2),(2,3) \in R$
$\Rightarrow(1,3) \in R$
Also (3,1)$\in R$ as $R$ is symmetric. So, the total number of elements is 9
Hence, minimum 7 elements must be added.
Question 3
1 / -0

If the circles x² + y² = 9 and x² + y² − 2αx + 2y + 1 = 0 touch each other internally, then α=

A
± 4/3

B
1

C
4/3

D
-4/3

Solution

Circle $1: x^{2}+y^{2}=3^{2}$
radius is $r_{1}=3$ and centre is (0,0) Circle $2: x^{2}+y^{2}-2 \alpha x+(\alpha)^{2}+2 \cdot 1 \cdot y+1-(\alpha)^{2}+1-1=0$
$(x-\alpha)^{2}+(y+1)^{2}-1-(\alpha)^{2}+1=0$
$(x-\alpha)^{2}+(y+1)^{2}=(\alpha)^{2}$
radius is $r_{2}=\alpha$ and centre is $(\alpha,-1)$
Distance between 2 centres is diff between their radii
$r_{2}-r_{1}=C_{1} C_{2}$
$C_{1} C_{2}=\sqrt{(\alpha-0)^{2}+(-1-0)^{2}}=\sqrt{\alpha^{2}+1^{2}}$
$\Rightarrow \alpha-3=\sqrt{\alpha^{2}+1^{2}}$
$\Rightarrow(\alpha-3)^{2}=\alpha^{2}+1^{2}$
$\Rightarrow \alpha^{2}+9-6 \alpha=\alpha^{2}+1$
$\Rightarrow 6 \alpha=8$
$\Rightarrow \alpha=\frac{4}{3}$
Question 4
1 / -0

The value of $(0.2)^{\log _{\sqrt{5}}\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots .10 n\right)}$ is

A
4

B
1/4

C
2

D
1/2

Solution

Let's calculate the sum of infinite series $\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots \ldots$
Let's consider it to be $S$
The series is a GP with first term $(a)=\frac{1}{4}$ and common ratio $(r)=\frac{1}{2}$
$\therefore S=\frac{a}{1-r}$
$\Rightarrow S=\frac{1}{2}$
To find $\left(\frac{1}{5}\right)^{\log _{\sqrt{3}} \frac{1}{2}}$
By Properties of log, we can write it as $\left(\frac{1}{2}\right)^{2 \log _{5} \frac{1}{5}}$
$=\left(\frac{1}{2}\right)^{-2 \log _{5} 5}$
$=\left(\frac{1}{2}\right)^{-2}$
$=2^{2}=4$
Hence, option A is correct.
Question 5
1 / -0

If one root of the equation ax² + bx + c = 0 be n times the other root, then

A
na²= bc(n + 1)²

B
nb² = ac(n + 1)²

C
nc² = ab(n + 1)²

D
nb³ = ac(n + 1)²

Solution

Let the roots be $\alpha$ and $n \alpha$
Sum of roots, $\alpha+n \alpha=-\frac{b}{a} \Rightarrow \alpha=-\frac{b}{a(n+1)} \quad \ldots \ldots(i)$
and product, $\alpha . n . \alpha=\frac{c}{a} \Rightarrow \alpha^{2}=\frac{c}{n a} \quad \ldots \ldots .$ (ii)
From (i) and (ii), we get
\[
\begin{array}{l}
\Rightarrow\left[-\frac{b}{a(n+1)}\right]^{2}=\frac{c}{n a} \Rightarrow \frac{b^{2}}{a^{2}(n+1)^{2}}=\frac{c}{n a} \\
\Rightarrow n b^{2}=a c(n+1)^{2}
\end{array}
\]
Note : Students should remember this question as a fact.
Question 6
1 / -0

If n(A) + n(B) = m, then the number of possible bijections from A to B is

A
(m/2)!

B
m²

C
m!

D
2m

Solution

A bijection from $A$ to $B$ is a function which maps every element of $A$ to uniqu element of $B$ i.e. injective. $\Longrightarrow n(B) \geq n(A)$
Also, it ensures that every element of $B$ is an image of some element of $A$ $\Longrightarrow n(A) \geq n(B)$
$\therefore n(A)=n(B)$
$\Longrightarrow n(A)=\frac{m}{2}=n(B)$
Let $A=\left\{a_{1}, a_{2}, \ldots, a_{\frac{m}{2}}\right\}$ and
$B=\left\{b_{1}, b_{2}, \ldots \ldots, b_{\frac{m}{2}}\right\}$
Let $f: A \rightarrow B$ defined by $f\left(a_{i}\right)=b_{i}$ is a bijection. Any and all images of some fixed $a_{i}$ appears in at least one such $f$. And each $f$ unique for each permutation $\left(b_{1}, b_{2}, \ldots b_{\frac{m}{2}}\right)$ Hence, the number of functions is exactly equal to the number of such permutations, which is $\left(\frac{m}{2}\right) !$
Question 7
1 / -0

A
m

B
a

C
a - m

D
a + m

Solution
Question 8
1 / -0

If the $7^{t h}$ term of a harmonic progression is 8 and the $8^{t h}$ term is 7, then its $15^{t h}$ term is

A
16

B
14

C
27/14

D
56/15

Solution

Obviously, $7^{t h}$ term of corresponding A.P is $\frac{1}{8}$ and $8^{t h}$ term will be $\frac{1}{7} a+6 d=\frac{1}{8}$ and $a+7 d=\frac{1}{7}$
Solving these, we get $\mathrm{d}=\frac{1}{56}$ and $\mathrm{a}=\frac{1}{56}$
Therefore $15^{t h}$ term of this A.P.
$=\frac{1}{56}+14 \times \frac{1}{56}=\frac{15}{56}$
Hence the required $15^{t h}$ term of the H.P. is $\frac{56}{15}$
Question 9
1 / -0

Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house is

A
2/9

B
2/18

C
3/8

D
1/20

Solution

For a particular house being selected Probability $=\frac{1}{3}$ Prob(all the persons apply for the same house) $=\left(\frac{1}{3} \times \frac{1}{3} \times \frac{1}{3}\right) \times 3=\frac{1}{9}$ or $\frac{2}{18}$
Question 10
1 / -0

In a town of 10,000 families it was found that 40% family buy newspaper A, 20% buy newspaper B and 10% families buy newspaper C, 5% families buy A and B, 3% buy B and C and 4% buy A and C. If 2% families buy all the three newspapers, then number of families which buy A only is

A
3100

B
3300

C
2900

D
1400

Solution

n(A) = 40% of 10,000 = 4,000
n(B) = 20% of 10,000 = 2,000
n(C) = 10% of 10,000 = 1,000
n(A∩ B) = 5% of 10,000 = 500
n(B∩ C) = 3% of 10,000 = 300
n(C∩ A) = 4% of 10,000 = 400
n(A∩ B∩ C) = 2% of 10,000 = 200
We want to find the number of families which buy only A= n(A) - [n(A ∩ B) + n(A ∩ C) - n(A ∩ B∩ C)]
=4000 - [500 + 400 - 200] = 4000 - 700 = 3300

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Re-Attempt Weekly Quiz Competition

Mathematics Test - 18

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Mathematics Test - 18

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SHARING IS CARING

Question 1

3/23

1/20

1/5

14/40

Solution

Question 2

4

7

6

5

Solution

Question 3

± 4/3

1

4/3

-4/3

Solution

Question 4

4

1/4

2

1/2

Solution

Question 5

na2 = bc(n + 1)2

nb2 = ac(n + 1)2

nc2 = ab(n + 1)2

nb3 = ac(n + 1)2

Solution

Question 6

(m/2)!

m2

m!

2m

Solution

Question 7

m

a

a - m

a + m

Solution

Question 8

16

14

27/14

56/15

Solution

Question 9

2/9

2/18

3/8

1/20

Solution

Question 10

3100

3300

2900

1400

Solution

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na²= bc(n + 1)²

nb² = ac(n + 1)²

nc² = ab(n + 1)²

nb³ = ac(n + 1)²

m²