Question

determine the input
a number went into this machine and 38 came out.
what number went in?
(image shows a machine with increase by 55% and decrease by 35% steps)

Explanation:

Step1: Define the input number

Let the input number be \( x \).

Step2: Apply the first operation (increase by 55%)

After increasing by 55%, the number becomes \( x + 0.55x = 1.55x \).

Step3: Apply the second operation (decrease by 35%)

Decreasing \( 1.55x \) by 35% means we take \( (1 - 0.35) \times 1.55x = 0.65 \times 1.55x \).

Step4: Set up the equation

We know the output is 38, so \( 0.65 \times 1.55x = 38 \). First, calculate \( 0.65 \times 1.55 = 1.0075 \)? Wait, no, wait. Wait, maybe the order is: increase by 55% then decrease by 35%? Wait, no, let's re - examine. Wait, the machine: first, the number goes into "Increase by 55%", then into "Decrease by 35%", and then 38 comes out. So the first transformation: \( x \) becomes \( x(1 + 0.55)=1.55x \). Then, decrease that result by 35%: \( 1.55x(1 - 0.35)=1.55x\times0.65 \). Let's compute \( 1.55\times0.65 = \frac{155}{100}\times\frac{65}{100}=\frac{155\times65}{10000}=\frac{10075}{10000} = 1.0075 \)? Wait, that can't be right. Wait, maybe I made a mistake. Wait, no, wait, maybe the operations are: increase by 55% and then decrease by 35%? Wait, but let's check the calculation again. Wait, \( 1.55\times0.65=(1 + 0.55)\times(1 - 0.35)=1\times1-1\times0.35 + 0.55\times1-0.55\times0.35=1 - 0.35+0.55 - 0.1925=1.0075 \). But if the output is 38, then \( 1.0075x = 38 \), so \( x=\frac{38}{1.0075}\approx37.72 \)? Wait, that seems odd. Wait, maybe the order is decrease by 35% first then increase by 55%? No, the machine has "Increase by 55%" first, then "Decrease by 35%". Wait, maybe I messed up the percentage operations. Let's re - express:

When we increase a number by 55%, the multiplier is \( 1 + 55\%=1.55 \). When we decrease a number by 35%, the multiplier is \( 1-35\% = 0.65 \). So the overall transformation is \( x\times1.55\times0.65=38 \).

First, calculate \( 1.55\times0.65=\frac{155}{100}\times\frac{65}{100}=\frac{155\times65}{10000}=\frac{(150 + 5)\times65}{10000}=\frac{150\times65+5\times65}{10000}=\frac{9750 + 325}{10000}=\frac{10075}{10000}=1.0075 \)

Then, \( x=\frac{38}{1.0075}=\frac{38\times10000}{10075}=\frac{380000}{10075}=\frac{380000\div25}{10075\div25}=\frac{15200}{403}\approx37.72 \). Wait, that seems incorrect. Wait, maybe the operations are: the number is first increased by 55% and then decreased by 35%, but maybe I have the order wrong. Wait, maybe the machine is: the number goes into "Increase by 55%", then into "Decrease by 35%", so the formula is \( x(1 + 0.55)(1 - 0.35)=38 \). Let's compute \( (1 + 0.55)(1 - 0.35)=1.55\times0.65 = 1.0075 \), so \( x=\frac{38}{1.0075}\approx37.72 \). But this seems strange. Wait, maybe I made a mistake in the percentage application. Wait, another way: Let's work backwards. Let the number after the first operation (increase by 55%) be \( y \). Then, when we decrease \( y \) by 35%, we get 38. So, \( y(1 - 0.35)=38 \), so \( y=\frac{38}{0.65}=\frac{38\times100}{65}=\frac{3800}{65}=\frac{760}{13}\approx58.46 \). Then, \( y \) is the result of increasing \( x \) by 55%, so \( x(1 + 0.55)=y \), so \( x=\frac{y}{1.55}=\frac{\frac{760}{13}}{1.55}=\frac{760}{13\times1.55}=\frac{760}{20.15}\approx37.72 \). Wait, but this is the same as before. But let's check with \( x = 40 \): increase by 55%: \( 40\times1.55 = 62 \), decrease by 35%: \( 62\times0.65 = 40.3 \). Not 38. If \( x = 38 \): increase by 55%: \( 38\times1.55 = 58.9 \), decrease by 35%: \( 58.9\times0.65 = 38.285 \). Close to 38. If \( x = 37.7 \): increase by 55%: \( 37.7\times1.55=37.7\times(1 + 0.55)=37.7+37.7\times0.55 = 37.7+20.735 = 58.435 \). Decrease by 35%: \( 58.435\times0.65 = 58.435-58.435\times0.35=58.435 - 20.45225 = 37.98275\approx38 \). So the input number is approximately 37.72. But let's do the exact calculation.

From the equation \( x\times1.55\times0.65 = 38 \)

\( x=\frac{38}{1.55\times0.65}=\frac{38}{\frac{155}{100}\times\frac{65}{100}}=\frac{38\times100\times100}{155\times65}=\frac{380000}{10075}=\frac{380000\div25}{10075\div25}=\frac{15200}{403}\approx37.72 \)

Wait, but maybe the operations are in the reverse order? Like decrease by 35% first then increase by 55%? Let's try that. Let \( x \) be the input. Decrease by 35%: \( x(1 - 0.35)=0.65x \). Then increase by 55%: \( 0.65x(1 + 0.55)=0.65x\times1.55 = 1.0075x \), same as before. So regardless of the order (since multiplication is commutative), the result is the same. So the input number \( x=\frac{38}{1.55\times0.65}=\frac{38}{1.0075}\approx37.72 \). But let's check with a calculator: \( 1.55\times0.65 = 1.0075 \), \( 38\div1.0075 = 37.717\approx37.72 \). But maybe we made a mistake in the operation order. Wait, the problem says "A number went into this machine and 38 came out". The machine has two operations: "Increase by 55%" and "Decrease by 35%". Let's assume the order is: first, increase by 55%, then decrease by 35%. So the formula is correct.

Answer:

The input number is approximately \( \boldsymbol{37.72} \) (if we want an exact fraction, \( \frac{15200}{403}\approx37.72 \))