ans: A Profit of 1,317,077 Tk. B Profit of 91,385 Tk. & C Profit of 68,538 Tk.
According to the question,
2A = BB = 4C
Let, 2A = BB = 4C = 12x ———– (i)
(I) =>2A = 12x => A = 6x
=> 3B = 12x => B = 4x
=> 4C = 12x => C = 3x
A:B:C=6x:4x:3x=6:4:3
Sum of the ratio=6+4+3 =13
As, profit is distributed with the same ratio of investment
Profit of
A=$\frac{6}{13}\times$2,97,000=1,37,077 TK
B=$\frac{4}{13}\times$2,97,000=91,385
C=$\frac{3}{13}\times$2,97,000=68538
ans: 1.5 hour
Here, Distance, D = 600km
Let, Time = t & Speed = s
According to the question,
=>D = st
=>600 = st
=>s=$\frac{600}{t}$
From 2″” condition,
=>(s-200)$(t+\frac{1}{2})$=600
=>($\frac{600}{t}$-200)$(t+\frac{1}{2})$=600
=>($\frac{600-200}{t}$)($\frac{2t+1}{2})$)=600
=>(600 + 200t) (2t + 1) = 600
=>1200t + 600 – 400$t^{2}$ – 200t=1200t
=>600 – 400$t^{2}$ – 200t= 0
=>400$t^{2}$ + 200t- 600 = 0
=>4$t^{2}$ +2t -6 =0
=>2$t^{2}$ + t – 3 = o
=>(2t+3)(t-1)=0
=>t = l and t= -$\frac{3}{2}$
As, t cannot be negative,
So, t = 1 hour
Duration of the journey = Normal time + Extra/Delay
= 1h + 0.5h = 1.5 hour
ans: 46 min.
Here, at 36 min A fill up = 1 tank
At 1 min A fill up =$\frac{1}{36}$tank
Similarly At 1 min B fill up =$\frac{1}{45}$tank
At 1 min (A + B) fill up =$(\frac{1}{36}+\frac{1}{45})=\frac{1}{20})$tank
Similarly 7 min A + B fill up = $\frac{7}{20}$tank
Remaining tank = 1 – $\frac{7}{20}=\frac{13}{20}$tank
Now, C empty at 1 min =$\frac{1}{30}$tank
At’ 1 (A+B)+C Fill up = $\frac{7}{20}-\frac{1}{30}=\frac{1}{60}$ tank
Now, $\frac{1}{60}$ tank fill in = 1 min
1 tank fill in =1x $\frac{60}{1}$ min
$\frac{13}{20}$ tank fill in= 1x $\frac{60}{1}x\frac{13}{20}$= 39 min
Total time = 7 + 39 = 46 min.
