A two-digit number is such that the sum of its digits is 11. When the digits are reversed, the new number formed is 27 greater than the original number. Find the number.

• Kwanza, fanya kutoa ndani ya mabano: $1\frac{1}{2} - \frac{1}{3}$. Badilisha $1\frac{1}{2}$ kuwa sehemu isiyofaa: $\frac{3}{2}$. $\frac{3}{2} - \frac{1}{3} = \frac{9}{6} - \frac{2}{6} = \frac{7}{6}$.
• Kisha, zidisha matokeo kwa $\frac{3}{4}$: $\frac{7}{6} \times \frac{3}{4} = \frac{21}{24}$.
• Rahisisha sehemu: $\frac{21}{24} = \frac{7}{8}$. Hivyo, nambari ya juu ni $\frac{7}{8}$.

• Kwanza, zidisha: $\frac{1}{2} \text{ of } 2\frac{1}{2}$. Badilisha $2\frac{1}{2}$ kuwa sehemu isiyofaa: $\frac{5}{2}$. $\frac{1}{2} \times \frac{5}{2} = \frac{5}{4}$.
• Kisha, jumlisha matokeo na $1\frac{1}{4}$. Badilisha $1\frac{1}{4}$ kuwa sehemu isiyofaa: $\frac{5}{4}$. $\frac{5}{4} + \frac{5}{4} = \frac{10}{4}$.
• Rahisisha sehemu: $\frac{10}{4} = \frac{5}{2}$. Hivyo, nambari ya chini ni $\frac{5}{2}$.

Rahisisha jibu: $\frac{14}{40} = \frac{7}{20}$.
Jibu ni: $\boxed{\mathbf{\frac{7}{20}}}$

Acha namba ya asili iwe $10t + u$, ambapo $t$ ni tarakimu ya makumi na $u$ ni tarakimu ya mamoja.
Kutoka taarifa ya kwanza: $t + u = 11$ (Mlinganyo 1).
Namba mpya baada ya kubadilisha tarakimu ni $10u + t$.
Kutoka taarifa ya pili: $10u + t = (10t + u) + 27$ (Mlinganyo 2).

$10u + t = 10t + u + 27$
$10u - u + t - 10t = 27$
$9u - 9t = 27$
Gawanya pande zote mbili kwa 9:
$u - t = 3$ (Mlinganyo 3).

Tuna milinganyo miwili:

1. $t + u = 11$
2. $u - t = 3$ Jumlisha Mlinganyo 1 na Mlinganyo 3: $(t + u) + (u - t) = 11 + 3$ $2u = 14$ $u = 7$.

Badilisha $u = 7$ katika Mlinganyo 1:
$t + 7 = 11$
$t = 11 - 7$
$t = 4$.

Namba ya asili ni $10t + u = 10(4) + 7 = 40 + 7 = 47$.
Jibu ni: $\boxed{\mathbf{47}}$