Chapter 9.7, Problem 32P

To determine

To solve: the problems about two-digit numbers.

Answer to Problem 32P

The two- decimal number is $0.36$ .

Explanation of Solution

Given information:

A two-decimal place number between 0 and 1 such that the sum of its digits is 9 and such that if the digits are reversed the number is increased by 0.27.

Formula used:

Digit problems are based on our decimal system of numeration.

Step1. The problem asks for the original number.

Step2. Let t = the tenth digit .

Let u = the hundredth digit .

Step3. Use the facts of the problem to write two equations.

Step4. Solve the equations.

Calculation:

Consider ,

Let t = the tenth digit .

Let u = the hundredth digit .

	Tenth	Hundredth	Value
Original number	t	u	$0.1t+0.01u$
Number reversed	u	t	$0.1u+0.01t$

According to the question,

Sum of digits is 9.

  $=t+u=9\text{ (i)}$

Also,

  $=0.1u+0.01t=0.1t+0.01u+0.27\text{ (ii)}$

  $\begin{array}{l}\text{Multiplying by }100\text{ on both the sides}\\ =10u+t=10t+u+27\\ =10u-u+t-10t=27\\ =9u-9t=27\\ =\frac{9(u-t)}{9}=\frac{27}{9}\\ =u-t=3\\ =u=3+t\text{ (iii)}\end{array}$

Putting value of u in equation (i) we get,

  $\begin{array}{l}=t+3+t=9\\ =2t+3=9\\ =2t=9-3\\ =\frac{2t}{2}=\frac{6}{2}\\ =t=3\end{array}$

Putting value of t in equation (iii) we get,

  $\begin{array}{l}=u=3+3\\ =u=6\end{array}$

Thus, the two- decimal number is $0.36$ .