Mathematics Test

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Mathematics Test - 54

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Weekly Quiz Competition

Question 1
1 / -0

If sec²70^∘− cot²20^∘/2(cosec²59^∘− tan²31^∘) = 2/m, then m is equal to?

A
2

B
3

C
4

D
1

Solution

$\frac{\sec ^{2} 70^{\circ}-\cot ^{2} 20^{\circ}}{2\left(\operatorname{cosec}^{2} 59^{\circ}-\tan ^{2} 31^{\circ}\right)}=\frac{2}{m}$
$\Rightarrow \frac{\sec ^{2} 70^{\circ}-\cot ^{2}\left(90^{\circ}-70^{\circ}\right)}{2\left(\operatorname{cosec}^{2} 59^{\circ}-\tan ^{2}\left(90^{\circ}-59^{\circ}\right)\right)}=\frac{2}{m}$
$\Rightarrow \frac{\sec ^{2} 70^{\circ}-\tan ^{2} 70^{\circ}}{2\left(\operatorname{cosec}^{2} 59^{\circ}-\cot ^{2} 59^{\circ}\right)}=\frac{2}{m}$
$\Rightarrow \frac{1}{2}=\frac{2}{m}\left[\sec ^{2} \theta-\tan ^{2} \theta=1\right]$
$\left.\operatorname{cosec}^{2} \theta-\cot ^{2} \theta=1\right)$
$\Rightarrow m=2 \times 2$
$\Rightarrow m=4$
Question 2
1 / -0

Pipes A and B can fill a tank in 5 and 6 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in:

A
$1 \frac{13}{17}$

B
$2 \frac{8}{11}$

C
$3 \frac{9}{17}$

D
$4 \frac{1}{2}$

Solution

Net part filled in 1 hour = (1/5+1/6−1/12) =17/60
∴The tank will be full in 60/17 hours
i.e.,3(9/17) hours
Question 3
1 / -0

Find the 15th term of the sequence 20, 15, 10 . . . . .

A
-45

B
-55

C
-50

D
0

Solution

$\begin{aligned} 15^{\text {th }} \text { term }=& a+14 d=20+14 \times(-5) \\ &=20-70 \\ &=-50 \end{aligned}$
Question 4
1 / -0

If log 2 = 0.3010 and log 3 = 0.4771, the value of log5 512 is:

A
2.870

B
2.967

C
3.876

D
3.912

Solution

$\log _{5} 512$
$=\frac{\log 512}{\log 5}$
$=\frac{\log 2^{9}}{\log \left(\frac{10}{2}\right)}$
$=\frac{9 \log 2}{\log 10-\log 2}$
$=\frac{(9 \times 0.3010)}{1-0.3010}$
$=\frac{2.709}{0.699}$
$=\frac{2709}{699}$
$=3.876$
Question 5
1 / -0

Ajit has a certain average for 9 innings. In the tenth innings, he scores 100 runs thereby increasing his average by 8 runs. His new average is:

A
20

B
21

C
28

D
32

Solution

Let Ajit's average be x for 9 innings. So, Ajit scored 9x run in 9 innings.
In the 10^th inning, he scored 100 runs then average became (x+8). And he scored (x + 8) × 10 runs in 10 innings.
Now,
⇒9x+100=10×(x+8)
or,9x+100=10x+80
or,x=100−80
or,x=20
Newaverage=(x+8)
=28runs
Question 6
1 / -0

sinθ + cosθ/sinθ − cosθ = 3,then the value of sin⁴θ − cos⁴θ is?

A
1/5

B
3/5

C
2/5

D
4/5

Solution

\begin{aligned}
&\frac{\sin \theta+\cos \theta}{\sin \theta-\cos \theta}=3\\
&\sin \theta+\cos \theta=3 \sin \theta-3 \cos \theta\\
&\sin \theta=2 \cos \theta\\
&\text { i. } e . \frac{\sin \theta}{\cos \theta}=\frac{2}{1}\\
&\tan \theta=\frac{2}{1}=\frac{P}{B}\\
&\text { we know that, } H=\sqrt{P^{2}+B^{2}}=\sqrt{5}\\
&\sin \theta=\frac{2}{\sqrt{5}} \text { and } \cos \theta=\frac{1}{\sqrt{5}}\\
&\therefore \sin ^{4} \theta-\cos ^{4} \theta\\
&\Rightarrow\left(\sin ^{2} \theta+\cos ^{2} \theta\right)\left(\sin ^{2} \theta-\cos ^{2} \theta\right)\\
&\Rightarrow 1 \times\left\{\left(\frac{2}{\sqrt{5}}\right)^{2}-\left(\frac{1}{\sqrt{5}}\right)^{2}\right\}\\
&\Rightarrow \frac{4}{5}-\frac{1}{5}=\frac{3}{5}
\end{aligned}
Question 7
1 / -0

If log 27 = 1.431, then the value of log 9 is:

A
0.934

B
0.945

C
0.954

D
0.958

Solution

$\log 27=1.431$
$\Rightarrow \log \left(3^{3}\right)=1.431$
$\Rightarrow 3 \log 3=1.431$
$\Rightarrow \log 3=0.477$
$\therefore \log 9=\log \left(3^{2}\right)$
$=2 \log 3$
$=(2 \times 0.477)$
$=0.954$
Question 8
1 / -0

Three unbiased coins are tossed. What is the probability of getting at most two heads?

A
3/4

B
1/4

C
3/8

D
7/8

Solution

Getting at most Two heads means 0 to 2 but not more than 2 Here $\mathrm{S}=\{\mathrm{T} \mathrm{T} \mathrm{T}, \mathrm{TTH}, \mathrm{THT}, \mathrm{HTT}, \mathrm{THH}, \mathrm{HTH}, \mathrm{HHT}, \mathrm{HHH}\}$
Let $E=$ event of getting at most two heads Then $E=\{T T T, T T H, T H T, H T T, T H H, H T H, H H T\}$
$\therefore P(E)=\frac{n(E)}{n(S)}=\frac{7}{8}$
Question 9
1 / -0

If $x^{2}-3 x+1=0,$ then the value of $x^{2}+x+\frac{1}{x}+\frac{1}{x^{2}}$ is?

A
32

B
16

C
23

D
-23

Solution

$a^{2}+b^{2}=5 a b$
$\Rightarrow \frac{a^{2}}{a b}+\frac{b^{2}}{a b}=5$
$\Rightarrow \frac{a}{b}+\frac{b}{a}=5$
Squaring the both sides $\Rightarrow\left(\frac{a}{b}\right)^{2}+\left(\frac{b}{a}\right)^{2}+2 \times \frac{a}{b} \times \frac{b}{a}=25$
$\Rightarrow \frac{a^{2}}{b^{2}}+\frac{b^{2}}{a^{2}}=25-2$
$\Rightarrow \frac{a^{2}}{b^{2}}+\frac{b^{2}}{a^{2}}=23$
Question 10
1 / -0

A two member committee comprising of one male and one female member is to be constitute out of five males and three females. Amongst the females. Ms. A refuses to be a member of the committee in which Mr. B is taken as the member. In how many different ways can the committee be constituted ?

A
11

B
12

C
13

D
14

Solution

⁵C1 × ³C1 - 1
= 15 - 1
= 14

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Re-Attempt Weekly Quiz Competition

Mathematics Test - 54

Result

Mathematics Test - 54

Score

-

Rank

-

SHARING IS CARING

Question 1

2

3

4

1

Solution

Question 2

11317

2811

3917

412

Solution

Question 3

-45

-55

-50

0

Solution

Question 4

2.870

2.967

3.876

3.912

Solution

Question 5

20

21

28

32

Solution

Question 6

1/5

3/5

2/5

4/5

Solution

Question 7

0.934

0.945

0.954

0.958

Solution

Question 8

3/4

1/4

3/8

7/8

Solution

Question 9

32

16

23

-23

Solution

Question 10

11

12

13

14

Solution

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$1 \frac{13}{17}$

$2 \frac{8}{11}$

$3 \frac{9}{17}$

$4 \frac{1}{2}$