Additional Tk. 56,000 is needed to make income equal.
Let, he has to invest more Tk. x at 10% interest
Then his total investment will be = (24,000 + x) Tk.
Now at 7.5%, the amount of interest on Tk. 24,000 = (24000$\times \frac{7.5}{100}$) TK
At 10%, the amount of interest on Tk. x =(x$\times \frac{10}{100}$) TK
At 9.25%, the amount of interest on Tk. (24,000 + x) = (24,000 + x)$\times \frac{9.25}{100}$)
=> $\frac{180000}{100}$+$\frac{10x}{100}$ = $\frac{222000+9.25x}{100}$
=> 1,80,000 + 10x = 2,22,000 + 9.25
=> 10x – 9.25x = 2,22,000 -1,80,000
=> 0.75x = 42,000
=> x = $\frac{42000}{0.75}$
x = 56,000
Profit 50 Tk. & Cost price 350 Tk.
Let, the sale price of the product is Tk. x
& the cost price of the product is Tk. y
7% of sale price is Tk. 7% of x = $\frac{7x}{100}$
8% of cost price is Tk. 8% of y = $\frac{8y}{100}$
According to the question, $\frac{7x}{100}$ = $\frac{8y}{100}$
According to the question, $\frac{7x}{100}$ = $\frac{8y}{100}$
=> 7x = 8y
=> x = $\frac{8y}{7}$……….(i)
Again, 9% of sale price is Tk. 9% of x = $\frac{9x}{100}$
10% of cost pr1ce is Tk. 10% of y = $\frac{10y}{100}$
Aecordin to the qestion , $\frac{9x}{100}= \frac{10y}{100}$+1
=> $\frac{9x}{100}$ = $\frac{10y+100}{100}$
=> 9x = 10y+100
=> 2y = 700
y = 350
Now putting the value of y in equation (i), we get
x = $\frac{8\times 350}{7}$
=> x = 400
Cost price of the product is Tk. 350 & the sale price of the products is Tk. 400
Profit = 400 -350 = 50 Tk.
