Sum of ages of mother and daughter is 60 years. 5 years ago, the ratio of their age were 4:1. What will be the ratio of their age after 10 years?

(Age)

Basin is filled by Ain 5 hours

$\therefore$In 1 hour, $\frac{1}{5}$part of Basin is filled

Similarly by ‘ B’ in 1 hour, $\frac{1}{8}$ part of Basin is filled

The both tap can fill in 1 hour =$\frac{1}{5}+\frac{1}{8}=\frac{8+5}{40}=\frac{13}{40}$ part of Basin

As, the leak can empty the fill Basin in 20 hours

$\therefore$ In 1 hour, the leak can empty $\frac{1}{20}$ part of Basin.

$\therefore$The amount of water remain in the Basin in 1 hour=$\frac{13}{40}-\frac{1}{20}=\frac{13-2}{40}=\frac{11}{40}$ part of Basin

Thus, while two taps are working,

$\frac{11}{40}$ of the Basin is filled in = 1 hour

$\therefore$ 1 or full part of the Basin is filled n = $\frac{1\times 40}{11}=\frac{40}{11}=3\frac{7}{11}$ hour

Let, Radius of the garden with road is = R meter

& Radius of the garden without road is = r meter

$\therefore$Outer circumstance of the garden 2πR meter

$\therefore$Inner circumstance of the garden 2πr meter

Here, width of the road (R-r) meter

According to question,

2πR - 2πr = 88

= (R - r) $\times$ 2π= 88

$\therefore$R-r=$\frac{88}{2\mathrm{\pi }}=\frac{88}{{\displaystyle \frac{2\times 22}{7}}}=\frac{88}{1}\times \frac{7}{2\times 22}=14 meters$

$\therefore$Width of the road is 14 meters.

x number of person were supposed to travel

$\therefore$Per head cost =$\frac{5500}{x}$ taka

But, due to absence, new per head cost will be $\frac{5500}{x-5}$ taka

According to question,

$$\begin{gathered} \frac{5500}{x}\times 2=\frac{5500}{x-5} \\ =\frac{2}{x}=\frac{1}{x-5} [Dividing both sides by 5500] \\ =2x-10=x \\ =2x-x=10 \\ \therefore x=10 \\ \therefore Number of person who were availed the service=10-5=5 \end{gathered}$$

$$\begin{gathered} Given that, a+\frac{1}{a}=3 \\ L.H.S= a^4+\frac{1}{a^4}={\left(a^2\right)}^2+{\left(\frac{1}{a^2}\right)}^2=(a^2+\frac{1}{a^2})-2\times a^2\times \frac{1}{a^2}[\because a^2+b^2={(a+b)}^2-2ab] \\ ={(a^2+\frac{1}{a^2})}^2-2={\{{(a-\frac{1}{a})}^2+2\times a\times \frac{1}{a}\}}^2-2 \\ ={\{{\left(3\right)}^2+2\}}^2-2 \\ ={(9+2)}^2-2 \\ ={11}^2-2 \\ =121-2 \\ =119 \\ \therefore L.H.S.=R.H.S.\left(Proved\right) \end{gathered}$$

Let, number of total bench is x

$\therefore$Number of students = 5x - (4 $\times$ 5) = 5x - 20

In second case, number of students will bc = 3x + 20

According to the question,

5x - 20 = 3x + 20

= 5x - 3x = 20 + 20

= 2x = 40

$\therefore$ $x=\frac{40}{2}=20$

$\therefore$Number of benches is 20.

Larger number is x & smaller number is y.

According to question,

4x + 3y =71........(i)

and 2x = 5y + 3 (ii)

From equation (ii)

2x = 5y + 3

=2x-5y=3…………(iii)

Multiplying equation (iii) by 2 and subtracting it from equation (i), we get

$$\begin{gathered} \frac{4x+3y=71 \\ 4x-10y=6 \\ - + -}{13y=65} \\ =y=\frac{65}{13}=5 \end{gathered}$$

Putting (y = 5) in equation (ii) we get,

2x = 5y + 3

=2x = (5 $\times$ 5) + 3 = 25 + 3 = 28

$\therefore$x=$\frac{28}{2}=14$

$\therefore$Larger number 14 and smaller number 5.