Rahim has 36 Tk.
Let, the amount of Rahim be Tk.x
The amount of Karim be Tk. x
Rumana has =$(X\times \frac{1}{2})=\frac{X}{2}$
Amena has = $\frac{X}{2}$x$\frac{1}{2}$=$\frac{X}{4}$
According to the question,
x+x+$\frac{X}{2}$+$\frac{X}{4}$+1=100
=> 2x+$\frac{X}{2}$+$\frac{X}{4}$+100-1
=> $\frac{8x+2x+x}{4}$=99
=> 11x=99×4
=> x=$\frac{99×4}{11}$
x=36
Cost price is 30.000 tk
Let, the cost price be x Tk.
Marked price = 120% of x =$\frac{120x}{100}=\frac{6x}{5}$
At 10% discount on Tk. $\frac{6x}{5}$ selling price will be
= ($\frac{6x}{5}$-10% of $\frac{6x}{5}$) = ($\frac{6x}{5}-\frac{10}{100}X\frac{6x}{5}$)=($\frac{6x}{5}$-$\frac{3x}{25}$)=$\frac{30x-3x}{25}$=$\frac{27x}{25}$
Profit = ($\frac{27x}{25}$-X)=$\frac{27x-25x}{25}$=$\frac{2x}{25}$ tk
According to question, $\frac{2x}{25}$ =2,400
x=$\frac{2400×25}{2}$ = 30,000
Total cost is Tk. 14,000
Given that, the length of garden is 60 meter
And the breadth of garden is 20 meter
The area of the garden = (60 x 20)= 1,200m$^{2}$
Now, length without path = [60 – (5 x 2)] meter = (60 -10) meter = 50 meter
And breadth without path = [20 -(5 x 2)] meter =(20 – 10) meter =10 meter
The area of the garden without path = (50 x 10) = 500m$^{2}$
The area of the path is = (1,200 – 500)m$^{2}$= 700m$^{2}$
The amount of Tk. need to cover the path with grass is (20 X 700) = 14,000 Tk’
Total cost is Tk. 14,000.
Question 4:
x=15
$\frac{10}{2x-5}+\frac{1}{x+5}=\frac{18}{3x-5}$
=> $\frac{10}{2x-5}+\frac{1}{x+5}=\frac{15+3}{3x-5}$
=>$\frac{10}{2x-5}+\frac{1}{x+5}=\frac{15}{3x-5}+\frac{3x}{3x-5}$
=> $\frac{10}{2x-5}-\frac{15}{3x-5}=\frac{3}{3x-5}-\frac{1}{x+5}$
=> $\frac{10(3x-5)-15(2x-5)}{(2x-5)(3x-5)}=\frac{3(x+5)-1(3x-5)}{(3x-5)(x+5)}$
=> $\frac{25}{(2x-5)(3x-5)}=\frac{20}{(3x-5)(x+5)}$
=> $\frac{5}{2x-5}=\frac{4}{x+5}$
=> 4(2x-5) = 5(x+5)
=> 8x-20 = 5x+25
=> 8x-5x=25 +20
=> 3x=45
so, x=15
