Mathematical Skills Test - 4

Given:

P = 2000, R = 10%, Time = 4 year

Formula:

The given identity is \(\sin ^{-1}\left(\frac{2 \mathrm{a}}{1+\mathrm{a}^{2}}\right)+\sin ^{-1}\left(\frac{2 \mathrm{~b}}{1+\mathrm{b}^{2}}\right)=2 \tan ^{-1} \mathrm{n}\).

Let \(\mathrm{a}=\tan \mathrm{y}_{1}\) and \(\mathrm{b}=\tan \mathrm{y}_{2}\).

Therefore, the given equation becomes:

\(\sin ^{-1}\left(\frac{2 \mathrm{a}}{1+\mathrm{a}^{2}}\right)+\sin ^{-1}\left(\frac{2 \mathrm{~b}}{1+\mathrm{b}^{2}}\right)=2 \tan ^{-1} \mathrm{n}\)

\(\sin ^{-1}\left(\frac{2\left(\tan \mathrm{y}_{1}\right)}{1+\left(\tan \mathrm{y}_{1}\right)^{2}}\right)+\sin ^{-1}\left(\frac{2\left(\tan \mathrm{y}_{2}\right)}{1+\left(\tan \mathrm{y}_{2}\right)^{2}}\right)=2 \tan ^{-1} \mathrm{n}\)

\(\sin ^{-1}\left(\sin 2 \mathrm{y}_{1}\right)+\sin ^{-1}\left(\sin 2 \mathrm{y}_{2}\right)=2 \tan ^{-1} \mathrm{n}\)

\(2 \mathrm{y}_{1}+2 \mathrm{y}_{2}=2 \tan ^{-1} \mathrm{n}\)

\(\mathrm{y}_{1}+\mathrm{y}_{2}=\tan ^{-1} \mathrm{n}\)

\(\tan \left(\mathrm{y}_{1}+\mathrm{y}_{2}\right)=\mathrm{n}\)

\(\frac{\tan \mathrm{y}_{1}+\tan \mathrm{y}_{2}}{1-\tan \mathrm{y}_{1} \tan \mathrm{y}_{2}}=\mathrm{n}\)

\(\frac{\mathrm{a}+\mathrm{b}}{1-\mathrm{ab}}=\mathrm{n}\)

Therefore, \(\mathrm{n}=\frac{\mathrm{a}+\mathrm{b}}{1-\mathrm{ab}}\)

Let the time period of their investments of Ajit, Anand, and Ayush be \(2x:y:x\) respectively.

We know that,

Profit \(=\)Capital \(\times\) Time

Profit will be shared in the ratio \(6x:4y:5x\)

Given that,

\((\frac{5 x}{(11 x+4 y)})\times51000=15000\)

\(\Rightarrow \frac{5 x}{(11 x+4 y)}=\frac{5}{17}\)

\(\Rightarrow 85 x=55 x+20 y\)

\(\Rightarrow 30 x=20 y\)

\(\Rightarrow 3 x=2 y\)

Let \(x=2 k\) and \(y=3 k\)

\(\therefore\) Required ratio \(=2 x: y: x=4 k: 3 k: 2 k=4: 3: 2\)

The probability that both the tickets may show even numbers = P(A). P(B).

Where, A: event that the ticket drawn with even number

B: event that another even number ticket is drawn without replacing the first one

Given, A bag contains 17 tickets numbered from 1 to 17.

Even numbers in 1 to 17 = 2, 4, 6, 8, 10, 12, 14, 16.

There are 8 tickets with an even number.

A: event that the ticket was drawn with an even number

$\mathrm{P}\left(\mathrm{A}\right)=\frac{8}{17}$

Now, there are 16 tickets with 7 even number of tickets.

B: event that another even number ticket is drawn without replacing the first one

$\mathrm{P}\left(\mathrm{B}\right)=\frac{7}{16}$

The probability that both the tickets may show even numbers = P(A). P(B)

$=\frac{8}{17}.\frac{7}{16}$

$=\frac{7}{34}$

So, A bag contains 17 tickets numbered from 1 to 17. A ticket is drawn at random, then another ticket is drawn without replacing the first one. The probability that both the tickets may show even numbers by $\frac{7}{34}$.

As given, the ratio between the speed of the boat in still water to the speed of the stream is 5: 2.

Let the speed of the boat in still water and speed of the stream be \(5x\) and \(2x\) respectively.

According to the question,

\(\frac{224}{4}=5 x+2 x\)

\(\Rightarrow x=\frac{224}{7} \times 4\)

\(\Rightarrow x=\frac{32}{4}=8\)

Required difference \(=5 x-2 x\)

\(=3 x\)

\(=24 \) km/hr

Given:

The efficiency of Varun is \(\frac{1}{5}\) more of Karan.

The efficiency of Varun = Efficiency of Karan +\(\frac{1}{5}\) × Efficiency of Karan

\(\frac{\text{(Efficiency of Varun)}}{\text{(Efficiency of Karan)}} = \frac{6}{5}\)

The total efficiency of Varun and Karan = 11

Work = Efficiency × Time

Total work = 11 × 30 = 330

Number of days required to complete the whole work by Karan

\(⇒ \frac{330}{5} = 66\) days

∴ Karan can complete the whole work in 66 days.

Cost price of the article = Rs. 120

Selling price = 120 × 1.1 = Rs. 132

Checking with the help of options:

By option (A):

Cost price = 120 - 12 = 108

Selling price = 132 + 5 = 137

Profit percentage =$\frac{137-108}{108}$ × 100 = 26.85%

So, it does not satisfy the blanks.

By option (B):

Cost price = 120 - 20 = 100

Selling price = 132 + 10 = 142

Profit percentage =$\frac{142-100}{100}$ × 100 = 42%

So, it does not satisfy the blanks.

By option (C):

Cost price = 120 – 12 = 108

Selling price = 132 + 3 = 135

Profit percentage =$\frac{135-108}{108}$ × 100 = 25%

So, it satisfies the blanks.

By option (D):

Cost price = 120 – 15 = 105

Selling price = 132 + 10 = 142

Profit percentage =$\frac{142-105}{105}$ × 100 = 35.2%

So, it does not satisfy the blanks.

Given,

Number of days by Rakesh to complete the project = 10 days

Number of days taken by Ravi to complete a project = 15 days

∵ Rakesh takes 10 days to complete a project.

∴ Work done by Rakesh in a day \(= \frac{1}{10}\)

∵ Ravi takes 15 days to complete a project.

∴ Work done by Ravi in a day \(= \frac{1}{15}\)

Let Anas takes X days to complete the project.

Work done by Rakesh in 4days + work done by Rakesh and Ravi in 2 days + work done by all three persons in \(\frac{16}{13}\) days = 1

\(\Rightarrow \frac{4 }{10}+2 \times(\frac{1}{10}+\frac{1}{15})+\frac{16}{13} \times(\frac{1}{10}+\frac{1}{15}+\frac{1}{ X})=1\)

\(\Rightarrow \frac{2}{ 5}+2 \times(\frac{5}{30})+\frac{16}{ 13} \times(\frac{5 }{30}+\frac{1}{X})=1\)

\(\Rightarrow \frac{1}{3}+\frac{16}{13} \times(\frac{1}{6}+\frac{1}{ X})=1-\frac{2 }{5}\)

\(\Rightarrow \frac{16}{ 13} \times(\frac{1}{ 6}+\frac{1}{ X})=\frac{3}{5}\)

\(\Rightarrow \frac{16}{13} \times(\frac{1}{ 6}+\frac{1}{X})=\frac{3}{ 5}-\frac{1}{3}\)

\(\Rightarrow \frac{1}{6}+\frac{1}{X}=\frac{13}{16} \times \frac{4 }{15}\)

\(\Rightarrow \frac{1}{X}=\frac{13}{ 60}-\frac{1}{6}\)

\(\Rightarrow \frac{1}{X}=\frac{1}{20}\)

\(\Rightarrow X=20\)

∴ Number of days taken by Anas to complete the project = 20 days.