One of the faster workers can do thejob in 50 days.
Let, each of the faster workers can do the job in x days
Slower worker can do the job in 2x days.
3 workers in one day can do = $\frac{1}{x}$+$\frac{1}{x}$+$\frac{1}{2x}$ portion
= $\frac{5}{2x}$ portion
3 workers In 20 days can do $\frac{5}{2x}$ x 20 — $\frac{50}{x}$ Portlon
$\frac{50}{x}$ =1
x = 50
Question 2:
$20\sqrt{3}$ ft
$\Delta$ABD is a equal side triangle
$\angle $ADB = 60°
$\angle $BAD = 60°
$\angle $DBC = 30° & $\angle $BDC = 120°
$\angle $DCB = 30°
in $\Delta$BDC, $\angle $DBC = $\angle $DCB
BD = DC = 20
AC=AD+DC=20+20=40
ABC is a right triangle, AC is hypotenuse
AC² = AB²+ BC²
BC² = AC² — A8² = 40²-20² = 1200
BC = $\sqrt{400×3} =20\sqrt{3}$
$31\frac{1}{3}$ %
As property P.1 is 20% less than purchase price,
Purchase ofP1= 100000 x $31\frac{100\%}{80\%}$
= Tk. 125000
Loss = Tk. 25000
Mr X made profit from P2 = Tk. 25000
The purchase price of P2 was = Tk. 100000 — Tk. 25000
= T k. 75000
Profit in P2 was =$31\frac{25000}{75000}x100$ = $31\frac{1}{3}$ %
Question 4:
$2\pm \sqrt{10}$
$\frac{2}{x-2}+\frac{3}{x+3}=1$
=> $\frac{2x+6+3x-6}{(x-2)(x+3)}=1$
=> $\frac{5x}{x^{2}+x-6}=1$
=> $x^{2}+x-6 = 5x$
=> $x^{2}-4x-6 = 0$
=> $x = \frac{-(-4)\pm \sqrt{(-4)^{2}-4x1x(-6)}}{2×1}$
=> x = $2\pm \sqrt{10}$
$9\frac{1}{3}$
In first 8 hours he gets = Tk. 8 x 8 =TK 64
He gets from ever time = Tk. (80 – 64) = Tk. 16
Overtime (excess of 8 hours duty) rate = Tk. $8\frac{3}{2}$ = Tk. 12 per hour
Overtime =$\frac{16}{12}$ =$\frac{4}{3}$ hours
Total time he works = (8 +$\frac{4}{3}$) hours = $9\frac{1}{3}$ hours
Question 6:
$\frac{9}{254}$
80
Let, Number of older children = x
Number of younger children = (200 – x)
x x 2 + 3(200 —x) = 480
2x+600—3x=480
x = 120
Number of younger children = (200 — 120) = 80
10 years
Here,
R = 5?
T =?
l= 15000Tk.
P = 30000Tk.
A = 45000Tk.
Let,
Interest be, I = 3x
and investment be, P = 6x
Amount be, A = 3x + 6x = 9::
According to the question,
9x = 45000
x = $\frac{45000}{9}$ = 5000 Tk
Interest, 1 = 3x = 3 x 5000 = 15000Tk
Investment, P = 6x = 6 x 5000 = 30000Tk
We know,
T = $\frac{100xI}{PR}$
=$\frac{100×15000}{30000×5}$ =10
10
Let, the age of C = x
The age ofB = 2::
The age ofA =2x+ 2
According to the question,
$\frac{x+2x+2x+2}{3}=9$
=> 5x+ 2 =27
=> 5x=27—2
=> 5x=25
=> x=5
Now, Age of Be will be = 2x years
=2×5 =10
