17
No. of researchers in team A = $70\frac{35}{100}$
= 24.5 = 25 (approx.)
And No. of researchers in team B = $70\frac{65}{100}$
= 45.5= 45.(approx.)
Now, No. of researchers who prefer team A = $70\frac{60}{100}$ = 42.
And No. of researchers who prefer team B = $70\frac{40}{100}$ = 28.
No. of researchers prefer team A but not assigned to team A = 42 — 25 = 17
And No. of researchers prefer team B but not assigned to team B = 45 — 28 = 17.
The least possible number of researchers not to be assigned to the preferred team = 17.
$3\frac{1}{3}$ hours (Machine B)
Machine A takes 5 hours to produce = 1 lot
Machine A takes 1 hours to produce = $\frac{1}{5}$ lot
And machine B takes x hours to produce = 1 lot
Machine B takes it hours to produce = $\frac{1}{x}$ lot
Together they take 1 hour to produce = $\frac{1}{5}$ + $\frac{1}{x}$
= $\frac{x+5}{5x}$ Portion of lot
Again, they take 2 hours to complete the job
They take 1 hour to complete $\frac{1}{2}$ job
So, $\frac{x+5}{5x}$ = $\frac{1}{2}$
=> x = $3\frac{1}{3}$ hours
TK 312,500
Let, in 2004 no. of pairs of the shoes was 10 and price per pair was Tk. 10
In 2005, at 20% decreasing of pair of shoes = (10 – $10\frac{20}{100}$) = 8 pair
And at 20% increasrng of price per pair = (10 + $10\frac{20}{100}$) = 12 Tk.
In 2004 total revenue was = 10 x 10 = 100 Tk.
And in 2005 total revenue was = 8 x 12 = 96 Tk.
In 2005 revenue Tk. 96 then revenue was = Tk. 100 in 2004
In 2005 revenue Tk. I then revenue was = Tk. $\frac{100}{96}$ in 2004
In 2005 revenue Tk. 300,000 then revenue was = Tk. $\frac{100×300000}{96}$ = Tk. 312,500 in 2004
