Mathematics Test

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Mathematics Test - 66

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Weekly Quiz Competition

Question 1
1 / -0

If tan 9°=p/q, then the value of $\frac{s e c^{2} 81 °}{1 + c o t^{2} 81 °}$ is ?

A
p/q

B
1

C
p²/q²

D
q²/p²

Solution

$\tan 9^{\circ}=\frac{p}{q}$
$\Rightarrow \frac{\sec ^{2} 81^{\circ}}{1+\cot ^{2} 81^{\circ}}$
$\Rightarrow \frac{\sec ^{2} 81^{\circ}}{\operatorname{cosec}^{2} 81^{\circ}}$
$\Rightarrow \frac{1}{\cos ^{2} 81^{\circ}}$
$\Rightarrow \sin ^{2} 81^{\circ}$
$\Rightarrow \tan ^{2} 81^{\circ}$
$\Rightarrow \tan ^{2}\left(90^{\circ}-9^{\circ}\right)$
$\Rightarrow \cot ^{2} 9^{\circ}$
$\Rightarrow \frac{q^{2}}{p^{2}}$
Question 2
1 / -0

Two pipes can fill a tank in 20 and 24 minutes respectively and a waste pipe can empty 3 gallons per minute. All the three pipes working together can fill the tank in 15 minutes. The capacity of the tank is:

A
60 gallons

B
100 gallons

C
120 gallons

D
180 gallons

Solution

Work done by the waste pipe in 1 minute $=\frac{1}{15}-\left(\frac{1}{20}+\frac{1}{24}\right)$
$=\left(\frac{1}{15}-\frac{11}{120}\right)$
$=-\frac{1}{40} \quad[-v e \text { sign means emptying }]$
$\therefore$ Volume of $\frac{1}{40}$ part $=3$ gallons Volume of whole
$=(3 \times 40)$ gallons
$=120$ gallons
Question 3
1 / -0

If the fifth term of a GP is 81 and first term is 16, what will be the 4th term of the GP?

A
36

B
18

C
54

D
24

Solution

$5^{t h}$ term of $\mathrm{GP}$
$=a r^{5-1}$
$=16 \times r^{4}$
$=81$
$\mathrm{Or}, r=\left(\frac{81}{16}\right)^{\frac{1}{4}}=\frac{3}{2}$
$4^{t h}$ term of $\mathrm{GP}$
$=a r^{4-1}$
$=16 \times\left(\frac{3}{2}\right)^{3}$
$=54$
Question 4
1 / -0

If log₁₀ 2 = 0.3010, then log₂ 10 is equal to:

A
699/301

B
1000/301

C
0.3010

D
0.6990

Solution

$\log _{2} 10$
$=\frac{1}{\log _{10} 2}$
$=\frac{1}{0.3010}$
$=\frac{10000}{3010}$
$=\frac{1000}{301}$
Question 5
1 / -0

In a boat there are 8 men whose average weight is increased by 1 kg when 1 man of 60 kg is replaced by a new man. What is weight of new comer?

A
70

B
66

C
68

D
69

Solution

Member in group × age increased = difference of replacement
Or, 8 × 1 = new comer - man going out
Or, new comer = 8 + 60;
Or, new comer = 68 years.
Question 6
1 / -0

If 7sin²θ + 3cos²θ = 4, (0° ≤ θ ≤ 90°), then the value of θ is?

A
π/2

B
π/3

C
π/6

D
π/4

Solution

\begin{array}{l}
7 \sin ^{2} \theta \\
+3 \cos ^{2} \theta=4 \\
\Rightarrow 7 \sin ^{2} \theta \\
+3\left(1-\sin ^{2} \theta\right) \\
=4 \\
\Rightarrow 7 \sin ^{2} \theta+3 \\
-3 \sin ^{2} \theta=4 \\
\Rightarrow 4 \sin ^{2} \theta=1 \\
\Rightarrow \sin ^{2} \theta=\frac{1}{4} \\
\Rightarrow \sin \theta=\frac{1}{2} \\
=\sin 30^{\circ} \\
\theta=30^{\circ} \\
=\frac{\pi}{6} \\
{\left[\because \pi^{c}=180^{\circ}\right]}
\end{array}
Question 7
1 / -0

If log₁₀5+log₁₀(5x+1)= log₁₀(x+5)+1,then x is equal to :

A
1

B
3

C
5

D
10

Solution

$\Rightarrow \log _{10} 5+\log _{10}(5 x+1)=\log _{10}(x+5)+1$
$\Rightarrow \log _{10} 5+\log _{10}(5 x+1)=\log _{10}(x+5)+1$
$\Rightarrow \log _{10}[5(5 x+1)]=\log _{10}[10(x+5)]$
$\Rightarrow 5(5 x+1)=10(x+5)$
$\Rightarrow 5 x+1=2 x+10$
$\Rightarrow 3 x=9$
$\Rightarrow x=3$
Question 8
1 / -0

From a pack of 52 cards, two cards are drawn together at random. What is the probability of both the cards being kings?

A
1/15

B
25/57

C
35/256

D
1/221

Solution

Let S be the sample space Then, $n(S)={ }^{52} C_{2}$
$=\frac{(52 \times 51)}{(2 \times 1)}$
$=1326$
Let $E=$ event of getting 2 kings out of 4 $\therefore n(E)={ }^{4} C_{2}=\frac{(4 \times 3)}{(2 \times 1)}=6$
$\therefore P(E)=\frac{n(E)}{n(S)}$
$=\frac{6}{1326}$
$=\frac{1}{221}$
Question 9
1 / -0

If a² + b² + 4c² = 2(a + b - 2c) - 3 and a, b, c are real, then the value of (a²+ b² + c²) is?

A
3

B
3(1/4)

C
2

D
2(1/4)

Solution

$a^{2}+b^{2}+4 c^{2}=2(a+b-2 c)-3$
$\Rightarrow a^{2}+b^{2}+4 c^{2}-2 a-2 b+4 c+3=0$
$\Rightarrow a^{2}-2 a+1+b^{2}-2 b+1+4 c^{2}+4 c+1=0$
$\Rightarrow(a-1)^{2}+(b-1)^{2}+(2 c+1)^{2}=0$
$\therefore a-1=0 a=1$
$b-1=0 b=1$
$2 c+1=0 c=\frac{1}{2}$
$\therefore a^{2}+b^{2}+c^{2}$
$\Rightarrow 1+1+\frac{1}{4}$
$\Rightarrow 2+\frac{1}{4}$
$\Rightarrow \frac{9}{4}$
$\Rightarrow 2 \frac{1}{4}$
Question 10
1 / -0

The number of positive integers which can be formed by using any number of digits from 0, 1, 2, 3, 4, 5 without repetition.

A
1200

B
1500

C
1600

D
1630

Solution

One digit positive numbers = 5
Two digit positive numbers = 25
Three digit positive numbers = 100
4 digit positive numbers = 300
5 digit positive numbers = 600
Six digit positive numbers = 600
Total positive numbers,
= 5 + 25 + 100 + 300 + 600 + 600
= 1630

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Re-Attempt Weekly Quiz Competition

Mathematics Test - 66

Result

Mathematics Test - 66

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SHARING IS CARING

Question 1

p/q

1

p2/q2

q2/p2

Solution

Question 2

60 gallons

100 gallons

120 gallons

180 gallons

Solution

Question 3

36

18

54

24

Solution

Question 4

699/301

1000/301

0.3010

0.6990

Solution

Question 5

70

66

68

69

Solution

Question 6

π/2

π/3

π/6

π/4

Solution

Question 7

1

3

5

10

Solution

Question 8

1/15

25/57

35/256

1/221

Solution

Question 9

3

3(1/4)

2

2(1/4)

Solution

Question 10

1200

1500

1600

1630

Solution

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p²/q²

q²/p²