Mathematical Skills Test - 3

Given:

The selling price of the two articles is 2 : 3

First article sold at 20% profit

And second at a loss of 20%

Formula:

Profit% or Loss% = $\left(\frac{\mathrm{Proft} \mathrm{or} \mathrm{Loss}}{\mathrm{Cost} \mathrm{Price}}\right)\times 100$

Let the selling price be 2x and 3x.

First article sold at 20% profit

Selling price 120% = 2x

Cost price = $2\mathrm{x}\times \frac{100}{120}=\frac{5\mathrm{x}}{3}$

Second article sold at 80%

Cost price = $3\mathrm{x}\times \frac{100}{80}=\frac{15\mathrm{x}}{4}$

Total cost price = $\frac{5\mathrm{x}}{3}+\frac{15\mathrm{x}}{4}$

$=\frac{65\mathrm{x}}{12}$

Total selling price = 2x + 3x = 5x

Cost price > Selling price

So, there is loss of

$\frac{65\mathrm{x}}{12}-5\mathrm{x}=\frac{5\mathrm{x}}{12}$

Loss% $=\frac{{\displaystyle \frac{5\mathrm{x}}{12}}}{{\displaystyle \frac{65}{12}}}\times 100$

$=\frac{100}{13}\%$ Loss

∴ There will be loss of $\frac{100}{13}\%$.

Given,

Principal \((P)=\) Rs. \(1725\)

Time \((T)=4\) years

Simple Interest \(=\) Rs. \(897\)

Let \(R\) be the rate of interest.

As we know,

Simple Interest \(=\frac{P \times R \times T}{100}\)

\(\Rightarrow 897=\frac{1725 \times R \times 4}{100}\)

\(\Rightarrow \frac{897 \times 100}{1725 \times 4}=R\)

\(\Rightarrow \frac{89700}{6900}=R\)

\(\therefore R= 13 \%\)

So, Rate of interest is \(13 \%\).

Let A and B invested Rs. ‘x’ and Rs. ‘y’ respectively

∵ A invested Rs. 20000 after 6 months

⇒ A’s total capital investment = 6x + 6 × (x + 20000) = 12x + 120000

∵ B invested Rs. 25000 after 6 months

⇒ B’s total capital investment = 6y + 6 × (y + 25000) = 12y + 150000

∵ C invested Rs. 20000 after 6 months

⇒ C’s total capital investment = 6 × 40000 + 6 × 60000 = 240000 + 360000 = 600000

∵ Ratio of capital investments = Ratio of profit shares

Considering ratios of A and C,

$\Rightarrow \frac{12\mathrm{x}+120000}{600000}=\frac{3600}{3000}$

$\Rightarrow \frac{\mathrm{x}+10000}{50000}=\frac{6}{5}$

⇒ 5x + 50000 = 300000

⇒ x = $\frac{250000}{5}$ = Rs. 50000

Considering ratios of B and C,

$\Rightarrow \frac{12\mathrm{y}+150000}{600000}=\frac{4050}{3000}$

⇒ 12y + 150000 = 810000

⇒ y = $\frac{660000}{12}$ = Rs. 55000

∴ Option (D) satisfies the given blanks.

Given:

A person makes a profit = 20%

Formula:

Profit% = $\frac{\mathrm{Profit}}{\mathrm{Cost} \mathrm{Price}}\times 100$

Let the person bought an article for Rs. 100.

And since he gains 20% he sold it for Rs. 120

Now let's move to the second part of the question,

He purchased the article at 20% less

New Cost Price = $100\times \frac{100-20}{100}$

New Cost Price = Rs. 80

The new selling price is

New Selling price = $\frac{120\left(100+10\right)}{100}$

New Selling price = Rs. 132

New profit percent = $\left(\frac{132-80}{80}\right)\times 100$

= 65%

∴ The new profit per cent is 65.

Let \(A=\) Tripti and \(B=\) Pragati

Probability of solving the problem by \({A}, {P}({A})=\frac{1}{2}\)

Probability of solving the problem by \({B}, {P}({B})=\frac{1}{3}\) 

Since the problem is solved independently by \({A}\) and \({B}\), 

\(\therefore {P}({AB})={P}({A}) .{P}({B})=\frac{1}{2} \times \frac{1}{3}=\frac{1}{6}\)

\({P}({A}^{\prime})=1-{P}({A})=1-\frac{1}{2}=\frac{1}{2}\)

\({P}({B}^{\prime})={1}-{P}({B})=1-\frac{1}{3}=\frac{2}{3}\)

Probability that the problem is solved \(={P}({A} \cup {B})\)

\(={P}({A})+{P}({B})-{P}({A B})\)

\(=\frac{1}{2}+\frac{1}{3}-\frac{1}{6}\)

\(=\frac{4}{6}\)

\(=\frac{2}{3}\)

Given:

Pipe A can fill a tank in 24 hours.

Pipe A and B together can fill the tank in 36 hours.

Let the capacity of the tank be 1.

Efficiency of Pipe A = $\frac{1}{A}$

Efficiency of Pipe B = $-\left(\frac{1}{B}\right)$

A negative sign denotes an outlet.

Efficiency of Pipe A and Pipe B together,

$\frac{1}{A} + \frac{1}{B} = \frac{1}{total work}$

⇒ $\frac{1}{24} - \frac{1}{B} = \frac{1}{36}$

⇒ $\frac{(3 - 2)}{72} = \frac{1}{B}$

⇒ $\frac{1}{72} = \frac{1}{B}$

⇒ B = 72

∴ Pipe B independently will empty the tank in 72 hours.

Given: 

\(\mathrm{P}(\mathrm{x}, \mathrm{y})\) is equidistant from points \(\mathrm{A}(3,4)\) and \(\mathrm{B}(5,3)\)

As we know that, distance between two points \(\mathrm{A}\) and \(\mathrm{B}\) is given by:

\(|A B|=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)

\(|P A|=\sqrt{(x-3)^{2}+(y-4)^{2}}\) ..(1)

\(|P B|=\sqrt{(x-5)^{2}+(y-3)^{2}}\) ..(2)

\(\because|\mathrm{PA}|=\mid \mathrm{PB}\)

\(\Rightarrow \sqrt{(x-3)^{2}+(y-4)^{2}}=\sqrt{(x-5)^{2}+(y-3)^{2}}\)

By squaring both the sides we get,

\((x-3)^{2}+(y-4)^{2}=(x-5)^{2}+(y-3)^{2}\)

\(\left(x^{2}+9-6 x\right)+\left(y^{2}+16-8 y\right)=\left(x^{2}+25-10 x\right)+\left(y^{2}+9-6 y\right)\)

\(4 x-16=2 y-7\)

\( 4 x-2 y=9\)

Suppose the first pipe was closed after \(x\) hr.

Then, the first pipe was functional for \(x\) hours, while second pipe was functional for \(16\) hours.

\(\Rightarrow \frac{x}{24}+\frac{16}{32}=1\)

\(\Rightarrow \frac{x}{24}=1-\frac{1}{2}=\frac{1}{2}\)

\(\Rightarrow x=12\) hr

So, first pipe must be closed after \(12\) hours to fill the tank in exactly \(16\) hours.

If the event \(\mathrm{B}\) occurs does not change the probability that event \(\mathrm{A}\) occurs, and event \(\mathrm{A}\) and \(\mathrm{B}\) are the independent event then

\(\Rightarrow \mathrm{P}(\mathrm{A} \mid \mathrm{B})=\mathrm{P}(\mathrm{A})\)

According to bayes's theorem It states the relation between the probabilities of \(\mathrm{A}\) and \(\mathrm{B,~ P(A)}\) and \(\mathrm{P(B)}\) and the conditional probabilities of \(\mathrm{A}\) gives \(\mathrm{B}\) and \(\mathrm{B}\) gives \(\mathrm{A}\).

\(\Rightarrow \mathrm{P}(\mathrm{A} \mid \mathrm{B})\) and \(\mathrm{P}(\mathrm{B} \mid \mathrm{A})\)

\(\Rightarrow \mathrm{P(A} \mid \mathrm{B})=\frac{\mathrm{P(B} \mid \mathrm{A}) \cdot \mathrm{P(A)}}{\mathrm{P(B)}}\)

\(\Rightarrow \mathrm{P}(\mathrm{B} \mid \mathrm{A})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}\)

\(\Rightarrow \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})\)

\(\Rightarrow \mathrm{P}(\mathrm{A} \mid \mathrm{B})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \cdot \mathrm{P}(\mathrm{A})}{\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})}\)

\(\Rightarrow \mathrm{P}(\mathrm{A} \mid \mathrm{B})=\frac{\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B}) \cdot \mathrm{P}(\mathrm{A})}{ \mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})}\)

\(\Rightarrow \mathrm{P}(\mathrm{A} \mid \mathrm{B})=\mathrm{P}(\mathrm{A})\)

\(\therefore\) The value of \(\mathrm{P(A} \mid \mathrm{B})\) is \(\mathrm{P(A)}\).

In an examination it is required to get 290 of the aggregate marks to pass. A student gets 203 marks and is declared failed by 12% of total marks.

Let's maximum aggregate marks \(=x\)

\(12 \%\) of \(203+x\) \(=290\)

\(12 \%\) of \(x\) \(=290-203\)

\( x=\frac{87 \times 100}{12}\)

\(x=725 \)