Premier Bank Ltd. || TJO (24-01-2020) || 2020

Let, the remaining amount would be invested at x% interest.

Here, remaining amount = 100000 - 40000 - 35000 = 25000 Tk.

According to question,

( 40000 $\times$ 5%)+(35000$\times$4%)+(2500$\times$ x%)=5000

$$\begin{gathered} =\frac{40000\times 5}{100}+\frac{35000\times 4}{100}+\frac{25000\times x}{100}=5000 \\ =\frac{200000+140000+25000x}{100}=5000 \\ =340000+25000x=500000 \\ =25000x=160000 \\ \therefore x=\frac{160000}{25000}=6.4\% \end{gathered}$$

M can do in 5 days =$\frac{5}{40}$=$\frac{1}{8}$ Part of the work

Remaining work = (1 - $\frac{1}{8}$) = ($\frac{8-1}{8}$) = $\frac{7}{8}$ Part

Now, N does $\frac{7}{8}$ Part of the work in = 42 days

$\therefore$1 or complete part of the work done by N in = ($\frac{42\times 8}{7}$)=48 days

Now, (M+N)'s 1 day's work = ($\frac{1}{40}+\frac{1}{48}$)= $\left(\frac{88}{40\times 48}\right)=\frac{11}{240} part$

They together do $\frac{11}{240}$ part of the work in =1 day's

$\therefore$They I or complete part of the work in = $\frac{240}{11}$ = 21 $\frac{9}{11}$ days

Let, x be the price of one shirt,

y be the price of one pair of trousers and z be the price of one hat.

According to the question,

4x + 4y + 2z = 560……….(i)

9x + 9y + 6z = 1, 290………(ii)

3x + 3y + 2z = 430……………….(ii)              [Divide all terms of equation (ii) by 3]

Now, we have [(i)-(iii))]

$$\begin{gathered} \frac{4x + 4y + 2z = 560....................\left(i\right) \\ 3x + 3y + 2z = 430...................\left(iii\right)}{\therefore x + y = 130......................\left(iv\right)} \\ Again, from equation \left(iii\right) \\ 3x + 3y + 2z = 430 \\ = 3(x + y) + 2z = 430 \\ Putting value from equation \left(iv\right)] \\ 3 \times 130 + 2z = 430 \\ \therefore z = 20 \\ \therefore x + y + z = 130 + 20 = 150 ans. \end{gathered}$$