ans: 500 Tk.
Let, the capital be x Tk.
$\frac{2}{3}$ rd of his capital be $\frac{2x}{3}$ TK
$\frac{1}{6}$ th of his capital be $\frac{x}{6}$
Rest of the capital be = $(x-\frac{2x}{3}-\frac{x}{6})$
= $\frac{6x-4x-x}{6}$ = $\frac{x}{6}$
According to the question.
$(\frac{2x}{3}\times \frac{3}{100}\times 1)$ + $(\frac{x}{6}\times \frac{6}{100}\times 1)$ = 25
=> $\frac{x}{50}+\frac{x}{50}+\frac{x}{50}$ = 25
=> $\frac{2x+x+2x}{100}$ = 25
=> 5x =100×25
=> $\frac{25\times 100}{5}$
x = 500 TK
ans: 35 hours.
The time required for pipe A alone take to fill the tank = 2x $\times$ 2 = 4x hours
C alone can do in 1 hour =$\frac{1}{x}$ part of work
B alone can do in 1 hour =$\frac{1}{2x}$ part of work
A alone can do in 1 hour = $\frac{1}{4x}$part of work
Now, (A + B + C) together can do in 5 hour total 1 work
A + B + C together can do in 1 hour = $\frac{1}{5}$ part of work
According to the question, $\frac{1}{x}$+$\frac{1}{2x}$+$\frac{1}{4x}$ = $\frac{1}{5}$
=> x = $\frac{35}{4}$
The time required for pipe A alone take to fill the tank = 4x = 4 x $\frac{35}{4}$ = 35 hours
Alternative: Suppose pipe A alone takes x hours to fill the tank.
We know, speed and time have inverse relationship and as the pipe C is twice as fast as B and B is twice as fast as A
So, pipes B and C will take $\frac{x}{2}$ and $\frac{x}{4}$ hours respectively to fill the tank.
=> $\frac{1}{x}$+$\frac{2}{x}$+$\frac{4}{x}$ =$\frac{1}{5}$
=> $\frac{7}{x}$ = $\frac{1}{5}$
= x = 35
MCQ Shortcut: A,B,C এই তিনজনের একত্রে কাজ = $\frac{ABC}{AB+BC+CA}$
=> 5 = $\frac{(4X)(2X)(X)}{(4x\times 2x)+(2x\times x)(x\times 4x)}$
=> 5 = $\frac{8x}{14}$
=> x = $\frac{35}{4}$
A = 4x = 4 x $\frac{35}{4}$ = 35
50 years and 20 years.
Let, 10 years ago they were 4x and x
so, their present age 4x+10, x+10
After 10 years they will, 4x+10+10: x+10+10= 2:1
=> $\frac{4x+20}{x+20}=\frac{2}{1}$
=> 4x+20 = 2x+40
=> x= 10
So, father’s age now = 4 x 10+10 = 50 years and son is now = 10+10 = 20 years
ans: 12$\frac{1}{2}$
Let work done by 1 man on 1 day be I, one boy in 1 day be y
Given, two men and three boys can do a piece of work in ten days
Work done by 2 men and 3 boys in 1 day = $\frac{1}{10}$
=> 2x + 3y = $\frac{1}{10}$…………….(i)
Given, 3 men and 2 boys can do the same piece of work in 8 days
Work done by 3 men and 2 boys in 1 day = $\frac{1}{8}$
=> 3x+2y = $\frac{1}{8}$………………………..(ii)
Now, to get values of x and y, we need to solve (i) and (ii)
(i) x 3 => 6x + 9y = $\frac{3}{10}$ ………….(iii)
(ii) x 2 => 6x+ 4y = $\frac{1}{4}$………….(iv)
(iii) – (iv ) => 5y = $\frac{14}{20}$
y = $\frac{1}{100}$
Using this value of y in (i)
=> 2x + 3 $\times \frac{1}{100}$ = $\frac{1}{10}$
=> 2x + $\frac{3}{100}$ = $\frac{1}{100}$
=> 2x = $\frac{7}{100}$
x = $\frac{7}{200}$
Work done by 2 men and 1 boy in 1 day = 2x + y
= 2$\times$$\frac{7}{200}$+$\frac{1}{100}$ = $\frac{8}{100}$ = $\frac{2}{25}$
Therefore, required number of days to complete the work if 2 men and 1 boy work = $\frac{25}{2}$ = 12$\frac{1}{2}$
