Originally he had 20 apples.
Let, the number of apples be x
The man sells apples to the first customer =$\frac{x}{2}+1=\frac{x+2}{2}$
Remaining Apples = x-$\frac{x+2}{2}=\frac{2x-x-2}{2}=\frac{x-2}{2}$
He sells apples to the second customer = $\frac{1}{3}(\frac{x-2}{2})+1=\frac{x-2}{6}+1=\frac{x-2+6}{6}=\frac{x+4}{6}$
Remaining apples = $\frac{x-2}{2}-\frac{x-10}{15}=\frac{5(x-5)-(x+10)}{15}=\frac{5x-25-10}{15}$ =$\frac{4x-35}{15}$
According to the question,
$\frac{4x-35}{15}$=3
=> 4x-35=45
=> 4x = 45+35 =80
so, x= $\frac{80}{4}$=20
500
Let, x passengers were boarded at station A.
After getting down 10% passengers, the number of passengers = x-10% of x = x-$\frac{10x}{100}$=x-$\frac{x}{10}$ =$\frac{9x}{10}$
After getting in 100 passengers, the number of passengers at station B = $\frac{9x}{10}$+100= $\frac{9x+1000}{10}$
After getting down 50% passengers, the number of passengers = $\frac{9x+1000}{10}$-50% of ($\frac{9x+1000}{10}$)
= $\frac{9x+1000}{10}-\frac{50}{100}(\frac{9x+1000}{10})$
= $\frac{9x+1000}{10}-\frac{1}{2}(\frac{9x+1000}{10})=\frac{1}{2}(\frac{9x+100}{10})$
Afer getting in 25 passenger, the number of passengers at station c
= $\frac{1}{2}(\frac{9x+1000}{10})+25=\frac{9x+1000}{20}+25=\frac{9x+1500}{20}$
After getting down 50% passengers, the number of passengers = $\frac{9x+1500}{20}-50% of \frac{9×1500}{20}$
= $\frac{9x+1500}{20}-\frac{50}{100}(\frac{9x+1500}{20})$-$\frac{1}{2}(\frac{9x+1500}{20})$
= $\frac{1}{2}(\frac{9x+1500}{20})=\frac{9x+1500}{40}$
= The number of passengers at station D will be, $\frac{9x+1500}{40}+50=200$
=> $\frac{9x+1500}{40}+50=200-50$
=> $\frac{9x+1500}{40}$ =150
=> 9x+1,500 =6,000
=>9x=6,000-1,500 = 4,500
so, x=$\frac{4500}{9}$ = 500
500 passengers were boarded at station A.
The cost of the cow is TK.53.061.22 & The cost of the ox is TK.13,061.22
Let. the cost price of Cow be X TK
And the cost price of Ox be Y TK
According to the questions.
120%x+l25%y = 80,000
=> $\frac{120X}{100}+\frac{125y}{100}=80,000$
=> 120x + 125y = 80,000 x 100
=> 24x + 25 = 16,00,000 …………….(i)
Again. 125%x + 120%y = 82.000
=> $\frac{125x}{100}+\frac{120y}{100}$=82,000
=> 125x + 120y = 82,000 x 100
so, 25x + 24y = 16,40,000 ……………………….(ii)
(i) x 25 – (ii) x 24
600x + 625y = 4,00,00,000
600x+ 576y = 3,93,60,000
=> 49y = 6,40,000
y = 13,061,22
putting the valu of y =13,061,22 in equation (i) we get
24x=16,00,000-3,26,530.61
=> 24x = 12,73,469.4
so, x=53,061.22
The cost of the cow is TK.53,061,22 & The cost of the ox is TK.13,061.22
