General Test - 3

For 1st Six months A, B, C all three have equal share of Rs. 10000

⇒ As after 6 months A & C Withdraw Rs. 5000/- each from the business hence for next 6 months the Share in business ⇒ A = 5000, B = 10000, C = 5000 Respectively

So, the ratio Becomes:

⇒ A : B : C = (10000 × 6 + 5000 × 6) : (10000 × 12) : (10000 × 6 + 5000 × 6)

⇒ A : B : C = 90000 : 120000 : 90000 = 3 : 4 : 3

⇒ A's Share in Profit $=\frac{3}{3+4+3}\times 100000$

$=\frac{3}{10}\times 100000$

$=30,000$

 ⇒ B's Share in Profit $=\frac{4}{3+4+3}\times 100000$

$=\frac{4}{10}\times 100000$

$=40,000$

⇒ C's Share in Profit $=\frac{3}{3+4+3}\times 100000$

$=\frac{3}{10}\times 100000$

$=30,000$

So, profit of A and C = 60000

Given:

The rate of interest for \(1^{\text {st }} 2\) years is \(8 \%\)

For the next 3 years it is \(10 \%\)

For the period beyond 5 years it is \(12.5 \%\)

Principal \(=\operatorname{Rs} 20 \mathrm{~L}\)

Amoun paid \(=\mathrm{Rs} ~36.7 \mathrm{~L}\)

Amount \(=P+S I\)

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

Where \(\mathrm{P} \rightarrow\) Principal, \(\mathrm{R} \rightarrow\) rate of interest, \(\mathrm{T} \rightarrow\) time

Total \(\mathrm{SI}=\mathrm{A}-\mathrm{P}=36.7 \mathrm{~L}-2 \mathrm{OL}\)

Total \(\mathrm{SI}=16.7 \mathrm{~L}\)

SI for first 2 years \(=20 \mathrm{~L} \times 2 \times \frac{8}{100}=3.2 \mathrm{~L}\)

SI for next 3 years \(=20 \mathrm{~L} \times 3 \times \frac{10}{100}=6 \mathrm{~L}\)

So, total SI for the first 5 years \(=9.2 \mathrm{~L}\)

Then, the rest of the interest is obtained at the rate of \(12.5 \%\)

Remaining interest \(=16.7 \mathrm{~L}-9.2 \mathrm{~L}=7.5 \mathrm{~L}\)

SI for next \(\mathrm{N}\) years \(=20 \mathrm{~L} \times \mathrm{N} \times 12.5 \%=7.5 \mathrm{~L}\)

\(\mathrm{N}=7.5 \mathrm{~L} \times \frac8{20} \mathrm{~L} \quad(12.5 \% \rightarrow \frac18 \mathrm{in}\) fraction \()\)

\(N=3\)

That is, total years \(=2+3+3=8\) years.

Given-

A certain sum amounts to $27$ times of itself in $3$ years.

Let the sum $=Rs.P$

Amount $A=Rs.27P$

Time $T=3$ years

According to the formula-

$A=P{(1+\frac{R}{100})}^T$   where $R$ is rate percent per annum

$\Rightarrow 27P=P{(1+\frac{R}{100})}^3$

$\Rightarrow 27={(1+\frac{R}{100})}^3$

$\Rightarrow 3^3={(1+\frac{R}{100})}^3$

On equating the bases,

$\Rightarrow 3=1+\frac{R}{100}$

$\Rightarrow \frac{R}{100}=2$

$\Rightarrow R=200\%$

Let the M.P. is Rs. \(x\).

\(\Rightarrow\) S.P. \(=x-8 \%\) of \(x\) \(=\frac{92 x}{100}\)

C.P. \(=75 \%\) of selling price

\(\Rightarrow 75 \% \times \frac{92 x}{100}=0.69 x\)

S.P. = C.P. + Profit

\(\Rightarrow 0.92 x=0.69 x+\) Profit \(\Rightarrow 0.23 x=\) Profit

\(\therefore\) Required percent \(=\frac{Profit}{C.P.} \times 100\)

\(=[(\frac{0.23 x}{0.69 x}) \times 100]\) \(= 33 \frac{1}{3} \%\)

Cost Price of 5 eggs \(=\) Rs. 4

Cost Price of 1 egg \(({CP})=\) Rs. \(\frac{4}{5}\)

Selling Price pf 4 eggs \(=\) Rs. 5

Selling Price of 1 egg \(({SP})=\) Rs. \(\frac{5}{4}\)

Profit \(\%=\frac{({SP}-{CP})}{ {CP}}\times 100\)

Profit \(\%=\frac{(\frac{5 }{4}-\frac{4}{5})}{\frac{4}{5}} \times 100\)

Profit \(\%=\frac{225}{4} \%\) \(=56 \frac{1}{4} \%\)

Quantity of sugar \(=6\times\frac{4}{100}=\frac{6}{25}\)

Quantity of water remaining after evaporation \(=6-1=5\)

Percentage of sugar in remaining solution \(=\frac{\frac{6}{25}}{5} \times 100\) \(=4.8 \%\)

∴ The required result is \(4.8\%\).

Suppose, \(4 x=2880\) \(\Rightarrow x=720\)

Total amount being spent on food, clothes and rent

\(=2 x+4 x+5 x=11 x\) \(=11×720\) \(=\) Rs \(7920\)

Now, the amount being spent is \(90\%\) of his income as he saves \(10\%\).

\(\therefore 90 \%\) of income \(=7920\) \(\Rightarrow\) Income \(=\)\(\frac{7920}{0.9}\)\(=\) Rs \(8800\)

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.

\(\frac{547.527}{0.0082}=x \Rightarrow(\frac{547527}{82}) \times(\frac{10000}{1000})=x \)

\(\Rightarrow(\frac{547527}{82}) \times 10=x \Rightarrow(\frac{547527}{82})=\frac{x}{10}\)

Given, \(1+\frac{1}{2+\frac{1}{3+\frac{5}{6}}}\)

\( =1+\frac{1}{2+\frac{1}{\frac{18+5}{6}}} \)

\(=1+\frac{1}{2+\frac{6}{23}}  =1+\frac{1}{\frac{46+6}{23}} \)

\(=1+\frac{1}{\frac{52}{23}}  =1+\frac{23}{52} \)

\( =\frac{52+23}{52} =\frac{75}{52}\)