Suppose B was there in the business for $x$ months. Then
A : B $=76000×12:57000×x$
Therefore,
$76000×12:57000×x=2:1$
$76×12:57x=2:1$
$76×12×1=57x×2$
$76×4=19x×2$
$4×4=x×2$
$x=8$
Hence B was there in the business for 8 months, or joined after 12-8 = 4 months
P : Q : R = $120000:135000:150000$
$=120:135:150=24:27:30=8:9:10$
Share of P
$=56700×\dfrac{8}{27}=2100×8=16800$
Share of Q
$=56700×\dfrac{9}{27}=2100×9= 18900$
Share of R
$=56700×\dfrac{10}{27}=2100×10=21000$
Let maximum marks of the examination $=x$
Marks that Arun got $=30%$ of $x$ $=\dfrac{30x}{100}$
Given that Arun failed by 10 marks.
=> pass mark = $\dfrac{30x}{100}+10~~\cdots(1)$
Marks that Sujith got $=40%$ of $x$ $=\dfrac{40x}{100}$
Given that Sujith got 15 marks more than the passing marks.
=> pass mark $=\dfrac{40x}{100}-15~~\cdots(2)$
From $(1)$ and $(2)$,
$\dfrac{30x}{100}+10=\dfrac{40x}{100}-15\dfrac{10x}{100}=25\dfrac{x}{10}=25x=10×25=250$
pass mark
$=\dfrac{30x}{100}+10$
=$\dfrac{30×250}{100}+10$
=75+10=85
Given that $\dfrac{Q}{P}=\dfrac{P}{P+Q}\cdots(1)$
Since Q can be written as a certain percentage of P, we can assume that $Q=kP$
Hence $(1)$ becomes
$\dfrac{kP}{P}=\dfrac{P}{P + kP}$
$\Rightarrow k=\dfrac{1}{1+k}\\\Rightarrow k(k+1)=1~~\cdots(2)$
Q as a percentage of P
$=\dfrac{Q}{P}×100\\=\dfrac{kP}{P}×100=100k\%~~\cdots(3)$
From here we have two approaches.
Approach 1 - trial and error method
Here we use the values given in the choices to find out the answer.
Take 50% from the given choice.
If 50% is the answer,
$100k=50 \Rightarrow k=\dfrac{50}{100}=\dfrac{1}{2}$
But if we substitute $k=\dfrac{1}{2}$ in $(2),$
$k(k+1)=\dfrac{1}{2}\left(\dfrac{1}{2}+ 1\right)$ $=\dfrac{1}{2}×\dfrac{3}{2}=\dfrac{3}{4} \neq 1$
Now Take another choice, say 62%
If 62% is the answer,
$100k=62 \Rightarrow k=\dfrac{62}{100}$
If we substitute $k=\dfrac{62}{100}$ in $(2),$
$k(k+1)=\dfrac{62}{100}\left(\dfrac{62}{100}+ 1\right)$ $= \dfrac{62}{100}×\dfrac{162}{100}=\dfrac{10044}{10000}\approx 1$
Hence 62% is the answer.
Approach 2
lets solve the quadratic equation $(2)$
$k(k+1)=1\\k^2+k-1=0\\k=\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}=\dfrac{-1 \pm \sqrt{1^2-\left[4×1×(-1)\right]}}{2×1}=\dfrac{-1 \pm \sqrt{5}}{2}\\= \dfrac{-1 \pm 2.24}{2}\\ = \dfrac{1.24}{2}\text{ or }\dfrac{-3.24}{2}\\= 0.62$
avoiding -ve value
From $(3)$, Q as a percentage of P
$=100k\%=(100×.62)\%=62\%$
Let the labeled price = $x$
SP = 704
Initial Discount = 20%
Price after initial discount = $x \times \dfrac{80}{100}$
Additional discount = 12%
Price after additional discount= $x \times \dfrac{80}{100}\times \dfrac{88}{100}$
But Price after additional discount = SP = 704
$\Rightarrow x \times \dfrac{80}{100}\times \dfrac{88}{100}$ = 704
$\Rightarrow x \times \dfrac{4}{5}\times \dfrac{22}{25} $= 704
$\Rightarrow x = 704 \times \dfrac{25}{22}\times \dfrac{5}{4} = 176 \times \dfrac{25}{22}\times 5 $
= $8 \times 25 \times 5 = 40 \times 25$ = 1000
Let C.P. be Rs. $ x $.
Then,$ \dfrac{1920 - x}{x} \times 100$ =$ \dfrac{x - 1280}{x} \times 100$
$\Rightarrow$ 1920 - $ x $ = $ x $ - 1280
$\Rightarrow$ 2$ x $ = 3200
$\Rightarrow x $ = 1600
$\therefore$ Required S.P. = 125% of Rs. 1600 = Rs.$ \left(\dfrac{125}{100} \times 1600\right) $= Rs 2000.
Let investment in each case be Rs. $(143 \times 117)$.
| Income in 1st case = Rs.$ \left(\dfrac{11}{143} \times 143 \times 117\right) $= Rs. 1287. |
| Income in 2nd case = Rs.$ \left(\dfrac{39}{4 \times 117} \times 143 \times 117\right) $= Rs. 1394.25 |
| Clearly, 9$ \dfrac{3}{4} $% stock at 117 is better. |
Face value of each share = Rs.20
Market value of each share = Rs.25
Number of shares = 12500
Amount required to purchase the shares = 12500 × 25 = 312500
Mohan further sells the shares at a premium of Rs. 11 each
ie, Mohan further sells the shares at Rs.(20+11) = Rs.31 per share
Total amount he gets by selling all the shares = 12500 × 31 = 387500
His gain = 387500 - 312500 = Rs.75000
Let the duration of the flight be $ x $ hours.
| Then,$ \dfrac{600}{x} $-$ \dfrac{600}{x + (1/2)} $= 200 |
| $\Rightarrow \dfrac{600}{x} $-$ \dfrac{1200}{2x + 1} $= 200 |
$\Rightarrow x \left(2x + 1\right)$ = 3
$\Rightarrow$ 2$ x $2 + $ x $ - 3 = 0
$\Rightarrow$ $\left(2x + 3\right)\left(x- 1\right)$ = 0
$\Rightarrow x $ = 1 hr. [neglecting the -ve value of $ x $]
Time from 12 p.m. on Monday to 2 p.m. on the following Monday = 7 days 2 hours = 170 hours.
$\therefore$ The watch gains$ \left(2 + 4\dfrac{4}{5} \right) $min.or$ \dfrac{34}{5} $min. in 170 hrs.
Now,$ \dfrac{34}{5} $min. are gained in 170 hrs.
$\therefore$ 2 min. are gained in$ \left(170 \times\dfrac{5}{34} \times 2\right) $hrs= 50 hrs.
$\therefore$ Watch is correct 2 days 2 hrs. after 12 p.m. on Monday i.e., it will be correct at 2 p.m. on Wednesday.