Mathematics Test

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Mathematics Test - 61

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Weekly Quiz Competition

Question 1
1 / -0

In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is:

A
21/46

B
25/117

C
1/50

D
3/25

Solution

Let \(S\) be the sample space and \(E\) be the event of selecting 1 girl and 2 boys Then, \(n(S)=\) Number ways of selecting 3 students out of 25
\[
\begin{array}{l}
={ }^{25} C_{3} \\
=\frac{(25 \times 24 \times 23)}{(3 \times 2 \times 1)} \\
=2300 \\
n(E)=\left({ }^{10} C_{1} \times{ }^{15} C_{2}\right) \\
=\left[10 \times \frac{(15 \times 14)}{(2 \times 1)}\right] \\
=1050 \\
\therefore P(E)=\frac{n(E)}{n(S)}=\frac{1050}{2300} \\
=\frac{21}{46}
\end{array}
\]
Question 2
1 / -0

A question paper has two parts, A and B, each containing 10 questions. If a student has to choose 8 from part A and 5 from part B, in how many ways can he choose the questions?

A
11340

B
12750

C
40

D
320

Solution

There 10 questions in part A out of which 8 question can be chosen as \(={ }^{10} \mathrm{C}_{8}\)
Similarly, 5 questions can be chosen from 10 questions of Part B as \(=^{10} \mathrm{C}_{5}\)
Hence, total number of ways,
\(={ }^{10} \mathrm{C}_{8} \times{ }^{10} \mathrm{C}_{5}\)
\(=\frac{10 !}{2 ! \times 8 !} \times \frac{10 !}{5 ! \times 5}\)
\(=\left\{10 \times \frac{9}{2}\right\} \times\left\{\frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1}\right\}\)
\(=11340\)
Question 3
1 / -0

Two pipes A and B together can fill a cistern in 4 hours. Had they been opened separately, then B would have taken 6 hours more than A to fill the cistern. How much time will be taken by A to fill the cistern separately?

A
1 hour

B
2 hours

C
6 hours

D
8 hours

Solution

Let the cistern be filled by pipe \(A\) alone in \(x\) hours. Then, pipe \(B\) will fill it in \((x+6)\) hours \(\therefore \frac{1}{x}+\frac{1}{(x+6)}=\frac{1}{4}\)
\(\Rightarrow \frac{x+6+x}{x(x+6)}=\frac{1}{4}\)
\(\Rightarrow x^{2}-2 x-24=0\)
\(\Rightarrow(x-6)(x+4)=0\)
\(\Rightarrow x=6 . \text { [neglecting the negative value of } x]\)
Question 4
1 / -0

A bacteria gives birth to two new bacteria in each second and the life span of each bacteria is 5 seconds. The process of the reproduction is continuous until the death of the bacteria. initially there is one newly born bacteria at time t = 0, the find the total number of live bacteria just after 10 seconds :

A
3¹⁰/2

B
3¹⁰- 2¹⁰

C
243 × (3⁵-1)

D
None of these

Solution

Total number of bacteria after 10 seconds,
\(=3^{10}-3^{5}\)
\(=3^{5} \times\left(3^{5}-1\right)\)
\(=243 \times\left(3^{5}-1\right)\)
since, just after 10 seconds all the bacterias (i.e. \(3^{5}\) ) are dead after living 5 seconds each.
Question 5
1 / -0

f(x)=x²−3x, then the points at which f(x)=f′(x)are

A
1, 3

B
1, ? 3

C
? 1, 3

D
None of these

Solution

\(f(x)=x^{2}-3 x\) and \(f^{\prime}(x)=2 x-3 \quad\) But
\(f(x)=f^{\prime}(x) \Rightarrow x^{2}-3 x=2 x-3\) p \(x^{2}-5 x+3=0\)
\(\therefore x=\frac{5 \pm \sqrt{25-12}}{2}=\frac{5 \pm \sqrt{13}}{2}\)
Question 6
1 / -0

If y=cos(sinx²),then at x=√π2,dy/dx=

A
? 2

B
2

C
−2√π/2

D
0

Solution

since \(\frac{d y}{d x}=-\sin \left(\sin x^{2}\right) \cdot \cos x^{2} \cdot 2 x\)
Therefore, at \(x=\sqrt{\frac{\pi}{2}}, \cos x^{2}=\cos \frac{\pi}{2}=0\) p \(\frac{d y}{d x}=0\)
Question 7
1 / -0

The average marks of four subjects is 120. If 33 was misread as 13 during the calculation, what will be the correct average?

A
122

B
120

C
125

D
121

Solution

Correct average
=120+(33−13/4)
=120+5
=125
Solve while reading method:
Average given is 120.
Difference of 33 and 13 is 20.
That means 20 must be added to total.
Then average of is and so must be added to average, i.e.
Correct average = 120 + 5 = 125
Question 8
1 / -0

If tan(π/2−θ/2)=√the value of cosθ is?

A
0

B
1/√2

C
1/2

D
1

Solution

\(\tan \left(\frac{\pi}{2}-\frac{\theta}{2}\right)=\sqrt{3}\)
\(\Rightarrow \tan \left(90^{\circ}-\frac{\theta}{2}\right)=\sqrt{3}\left[\pi=180^{\circ}\right]\)
\(\Rightarrow \cot \frac{\theta}{2}=\sqrt{3}\)
\(\Rightarrow \cot \frac{\theta}{2}=\cot 30^{\circ}\)
\(\Rightarrow \frac{\theta}{2}=30^{\circ}\)
\(\Rightarrow \theta=60^{\circ}\)
\(\Rightarrow \cos 60^{\circ}=\frac{1}{2}\)
Question 9
1 / -0

If log₁₀2 = 0.3010, the value of log₁₀80 is:

A
1.6020

B
1.9030

C
3.9030

D
None of these

Solution

\(\log _{10} 80\)
\(=\log _{10}(8 \times 10)\)
\(=\log _{10} 8+\log _{10} 10\)
\(=\log _{10}\left(2^{3}\right)+1\)
\(=3 \log _{10} 2+1\)
\(=(3 \times 0.3010)+1\)
1.9030
Question 10
1 / -0

IF a + b = 1, c + d = 1 and a - b = d/c,then the value of c²+ d² = ?

A
a/b

B
b/a

C
-1

D
1

Solution

\(a+b=1\)
\(c+d=1\)
\(a-b=\frac{d}{c}\)
or \(\frac{1}{a-b}=\frac{c}{d}\)
\(\Rightarrow \frac{a+b}{a-b}\)
\(=\frac{c}{d}\)
\((\because a+b=1)\)
By \(\mathrm{C} \& \mathrm{D}\) rule
\(\begin{aligned} \Rightarrow & \frac{(a+b)-(a-b)}{(a+b)-(a-b)} \\ &=\frac{c+d}{c-d} \\ \Rightarrow & \frac{2 a}{2 b}=\frac{c+d}{c-d} \\ \Rightarrow \frac{a}{b} &=\frac{c+d}{c-d} \end{aligned}\)
Now multiply \& divide by \((c+d)\)
\(\frac{a}{b}=\frac{(c+d)}{(c-d)}\)
\(\times \frac{(c+d)}{(c+d)}\)
\(=\frac{(c+d)^{2}}{c^{2}-d^{2}}\)
\(\Rightarrow \frac{a}{b}=\frac{(c+d)^{2}}{\left(c^{2}-d^{2}\right)}\)
\(\Rightarrow c+d=1\)
\(\Rightarrow \frac{a}{b}=\frac{1}{c^{2}-d^{2}}\)
\(\Rightarrow c^{2}-d^{2}=\frac{b}{c}\)

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Re-Attempt Weekly Quiz Competition

Mathematics Test - 61

Result

Mathematics Test - 61

Score

-

Rank

-

SHARING IS CARING

Question 1

21/46

25/117

1/50

3/25

Solution

Question 2

11340

12750

40

320

Solution

Question 3

1 hour

2 hours

6 hours

8 hours

Solution

Question 4

310/2

310 - 210

243 × (35 -1)

None of these

Solution

Question 5

1, 3

1, ? 3

? 1, 3

None of these

Solution

Question 6

? 2

2

−2√π/2

0

Solution

Question 7

122

120

125

121

Solution

Question 8

0

1/√2

1/2

1

Solution

Question 9

1.6020

1.9030

3.9030

None of these

Solution

Question 10

a/b

b/a

-1

1

Solution

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3¹⁰/2

3¹⁰- 2¹⁰

243 × (3⁵-1)