Question

1. dean besteedt een vijfde van het geld in zijn portefeuille aan kledij en dan een vijfde van wat overblijft aan voeding. in totaal heeft hij 72 euro uitgegeven. hoeveel euro zat er in het begin in zijn portefeuille?

a. 100
b. 125
c. 150
d. 175
e. 200

Explanation:

Step1: Define the initial amount

Let the initial amount of money in the portfolio be \( x \) euros.

Step2: Calculate money spent on clothing

Dean spends \( \frac{1}{5}x \) on clothing. The remaining money is \( x - \frac{1}{5}x=\frac{4}{5}x \).

Step3: Calculate money spent on food

He then spends \( \frac{1}{5} \) of the remaining on food. So the amount spent on food is \( \frac{1}{5}\times\frac{4}{5}x = \frac{4}{25}x \).

Step4: Total amount spent

The total amount spent is the sum of money spent on clothing and food, which is \( \frac{1}{5}x+\frac{4}{25}x \). We know this total is 72 euros.
First, simplify \( \frac{1}{5}x+\frac{4}{25}x \):
\( \frac{1}{5}x=\frac{5}{25}x \), so \( \frac{5}{25}x+\frac{4}{25}x=\frac{9}{25}x \).

Step5: Solve for \( x \)

We have the equation \( \frac{9}{25}x = 72 \).
To solve for \( x \), multiply both sides by \( \frac{25}{9} \):
\( x=72\times\frac{25}{9} \)
\( 72\div9 = 8 \), so \( x = 8\times25=200 \)? Wait, no, wait, let's check again. Wait, maybe I made a mistake. Wait, let's re - calculate:

Wait, \( \frac{9}{25}x=72 \)

\( x = 72\times\frac{25}{9}\)

\( 72\div9 = 8 \), \( 8\times25 = 200 \). But wait, the options have E as 200. But let's check the calculation again.

Wait, initial amount \( x \)

Clothing: \( \frac{1}{5}x \), remaining: \( x-\frac{1}{5}x=\frac{4}{5}x \)

Food: \( \frac{1}{5}\times\frac{4}{5}x=\frac{4}{25}x \)

Total spent: \( \frac{1}{5}x+\frac{4}{25}x=\frac{5 + 4}{25}x=\frac{9}{25}x \)

Set \( \frac{9}{25}x = 72 \)

\( x=72\times\frac{25}{9}=200 \)

Wait, but let's verify with \( x = 200 \)

Clothing: \( \frac{1}{5}\times200 = 40 \)

Remaining: \( 200 - 40=160 \)

Food: \( \frac{1}{5}\times160 = 32 \)

Total spent: \( 40 + 32=72 \), which matches. So the initial amount is 200 euros.

Answer:

E. 200