Mathematical Skills Test - 1

In 1 st container, acid : water \(=2: 7\)

In 2nd container, acid: water \(=4: 5\)

Ratio of mixture \(=2: 5\)

Let " \(x\) " be the acid content of the New mixture Using the principal of allegation

\( \Rightarrow\frac{[(\frac{4 }{ 9})-x] }{[x-(\frac{2 }{ 9})]}=\frac{2 }{ 5} \)

\( \Rightarrow \frac{4 }{ 9-x}=\frac{2 x }{ 5}-\frac{4 }{ 45 }\)

\( \Rightarrow \frac{7 x }{ 5}=\frac{4 }{ 9}+\frac{4 }{ 45} \)

\( \Rightarrow \frac{7 x }{ 5}=\frac{24 }{ 45 }\)

\( \Rightarrow x=\frac{8 }{ 21}\)

Ratio of Acid : water in the new mixture \(=8:(21-8)\)

Ratio of Acid : water in the new mixture \(=8: 13\)

\(\therefore\) The required ratio is \(8: 13\).

Given,

The price of an article is first increased by 20% and later decreased by 25% due to a decrease in sales.

Net percentage change \(=x-y-\frac{x \times y}{100}\)

\(=20-25-\frac{25 \times 20}{100}\)

\(=20-25-5\)

\(=-10 \%\)

Given,

The sum becomes \(9\) times in \(2\) years.

Let \(P\) be the principal.

\(\therefore A=9P\)

As we know,

\(A=P\left(1+\frac{r}{100}\right)^{t}\)

\(\therefore 9 P=P\left(1+\frac{r}{100}\right)^{2}\)

\(\Rightarrow 9=\left(1+\frac{r}{100}\right)^{2}\)

\(\Rightarrow \sqrt{9}=1+\frac{r}{100}\)

\(\Rightarrow 3=1+\frac{r}{100}\)

\(\Rightarrow 3-1=\frac{r}{100}\)

\(\Rightarrow 2=\frac{r}{100}\)

\(\Rightarrow r=200 \%\)

\(\therefore\) The rate of interest is \(200 \%\).

Given:

Sum = Rs. 16000, Time = 1 year, Rate = 30%

Formula:

$\mathrm{A}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{T}}$

C.I = A - P

Where, A = Amount, P = Principal, T = Time, C.I = Compound interest and R = Rate of interest

For compounded half yearly,

R $=\frac{30\%}{2}=15\%$

T = 2 × 1 = 2 years

According to the question,

$\mathrm{A}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{T}}$

$=16000{\left(1+\frac{15}{100}\right)}^2$

$=16000{\left(1+\frac{3}{20}\right)}^2$

$=16000{\left(\frac{23}{20}\right)}^2$

$=16000\left(\frac{529}{400}\right)$

⇒ 40 × 529 = 21160

Now, C.I = 21160 - 16000

= 5160

∴ The compound interest is Rs. 5160.

Given:

The selling price of each chair = Rs. 800, Profit = 20%, Loss = 20%

Formula:

Profit % = $\left(\frac{\mathrm{S}.\mathrm{P}-\mathrm{C}.\mathrm{P}}{\mathrm{C}.\mathrm{P}}\right)\times 100$

Loss % = $\left(\frac{\mathrm{C}.\mathrm{P}-\mathrm{S}.\mathrm{P}}{\mathrm{C}.\mathrm{P}}\right)\times 100$

Where, C.P = Cost Price and S.P = Selling Price

According to the question,

C.P of the first chair is:

$\mathrm{C}.\mathrm{P}\times \frac{120}{100}=800$

⇒ C.P = 666.66

and C.P of second chair is:

$\mathrm{C}.\mathrm{P}\times \frac{80}{100}=800$

⇒ C.P = 1000

The total C.P of two chairs are:

= 1000 + 666.66

= 1666.66

The total S.P of two chairs are:

= 800 + 800

= 1600

So that, the S.P is less than the C.P

⇒ Loss % = $\left(\frac{1666.66-1600}{1666.66}\right)\times 100$

$=\frac{66.66\times 100}{1666.66}$

= 3.99% in approx 4%.

∴ The total loss percentage is 4%.

Given:

Speed of boat \(= 20\) km/h

Speed of stream \(= 2\) km/h

We know that:

Distance \(=\) Speed \(\times\) Time

Downstream speed \(=\) Speed of boat \(+\) Speed of stream

Downstream speed \(= 20 + 2 = 22\) km/h

\(\therefore\) Required time \(=\frac{198}{22}=9\) Hr

Let the sum of money be \(P\).

\(\therefore\) Amount \(A=3 P\)

Simple Interest \(=A-P=3P-P=2 P\)

Rate of interest \(=8 \%\)

Let time be \(T\) years.

As we know,

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

\(\therefore 2P= \frac{P \times 8 \times T}{100}\)

\(\Rightarrow 2P=\frac{8PT}{100}\)

\(\Rightarrow \frac {2P \times 100}{8P} =T\)

\(\therefore T=25\)

After \(25\) years, the sum of money at simple interest will be thrice itself at the rate of \(8 \%\) per annum.

Given:

Cost Price of 50 Coca-Cola Bottles = Selling Price of 40 Coca-Cola Bottles.

Formula:

Profit% = $\left(\frac{\mathrm{SP}–\mathrm{CP}}{\mathrm{CP}}\times 100\right)$

Loss% = $\left(\frac{\mathrm{CP}–\mathrm{SP}}{\mathrm{CP}}\right)$ × 100

where,

CP = Cost Price & SP = Selling Price

According to the question,

Cost Price of 50 Coca-Cola Bottles = Selling Price of 40 Coca-Cola Bottles

⇒ CP × 50 = SP × 40

$\Rightarrow \frac{\mathrm{SP}}{\mathrm{CP}}=\frac{50}{40}$

$\Rightarrow \frac{\mathrm{SP}}{\mathrm{CP}}=\frac{5}{4}$

Here, $\frac{\mathrm{SP}}{\mathrm{CP}}>1$.

This means SP > CP and there is a profit in the transaction.

Let the SP be 5x and CP be 4x.

So, Profit% = $\left(\frac{\mathrm{SP}–\mathrm{CP}}{\mathrm{CP}}\right)\times 100\mathrm{\%}$

$=\frac{5\mathrm{x}–4\mathrm{x}}{4\mathrm{x}}\times 100\mathrm{\%}$

$=\left(\frac{1\mathrm{x}}{4\mathrm{x}}\right)\times 100\%$

$\Rightarrow \left(\frac{1}{4}\right)\times 100\%$

= 25%

∴ The Profit percentage is 25%.

We know that,

\(\tan \theta=\frac{\text { Opposite side }}{\text { Adjacent side }}\)

Given,

From the top of a lighthouse, \(70 \) m high with its base at sea level, the angle of depression of a boat is \(15^{\circ}\).

Let the base be '\(x\)'.

\(\tan \theta=\frac{70}{{x}}\)

\(\Rightarrow \tan 15^{\circ}=\frac{70}{{x}}\quad\dots(1)\)

\(\Rightarrow \tan \left(45^{\circ}-30^{\circ}\right)=\frac{70}{x}\)

We know that,

\(\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \cdot \tan B}\)

\(\Rightarrow \tan \left(45^{\circ}-30^{\circ}\right)=\frac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ} \cdot \tan 30^{\circ}}\)

\(\Rightarrow \tan \left(15^{\circ}\right)=\frac{1-\frac{1}{\sqrt{3}}}{1+1. \frac{1}{\sqrt{3}}}\)

\(\Rightarrow \tan \left(15^{\circ}\right)=\frac{\sqrt{3}-1}{\sqrt{3}+1}\)

By Rationalization, we get

\( \tan 15^{\circ}=2-\sqrt{3}\)

From equation \((1)\), we get

\( \frac{70}{{x}}=2-\sqrt{3}\)

\(\Rightarrow {x}=\frac{70}{2-\sqrt{3}}\)

\(\Rightarrow  {x}=\frac{70}{2-\sqrt{3}} \times \frac{2+\sqrt{3}}{2+\sqrt{3}}\)

\(\Rightarrow {x}=70(2+\sqrt{3})\) m

It is given that,

The ratio of one-rupee and fifty paise = 13 : 11

The ratio of coins = 13(1) : 11(2) = 13 : 22   [∵ 1 rupee coin = 2 × 50 paise coin]

Let the proportionality constant be ‘x’.

The one-rupee coins = 13x

The fifty paise coins = 22x

Therefore, according to the question,

⇒ 13x + 22x = 210

⇒ 35x = 210

⇒ x = 6

Therefore, the number of one-rupee coins = 13x = 13 × 6 = 78