Let the quantity be \( x \).
According to the information given in the problem, we can set up the following equation:
\[ x + \frac{2}{3}x + \frac{1}{4}x + \frac{1}{5}x = 37 \]
Now, we will find a common denominator for the fractions. The least common multiple of 3, 4, and 5 is 60. We will express each term with a denominator of 60:
\[ x = \frac{60}{60}x \] \[ \frac{2}{3}x = \frac{40}{60}x \] \[ \frac{1}{4}x = \frac{15}{60}x \] \[ \frac{1}{5}x = \frac{12}{60}x \]
Now, we can rewrite the equation:
\[ \frac{60}{60}x + \frac{40}{60}x + \frac{15}{60}x + \frac{12}{60}x = 37 \]
Combining the fractions on the left side:
\[ \frac{60 + 40 + 15 + 12}{60}x = 37 \]
Calculating the numerator:
\[ 60 + 40 + 15 + 12 = 127 \]
So, we have:
\[ \frac{127}{60}x = 37 \]
Now, to isolate \( x \), multiply both sides by \( \frac{60}{127} \):
\[ x = 37 \times \frac{60}{127} \]
Calculating this value:
\[ x = \frac{2220}{127} \approx 17.4418 \]
To express the quantity as a fraction or decimal:
Calculating \( 37 \times 60 \):
\[ 37 \times 60 = 2220 \]
Now divide by 127:
\[ x \approx 17.4418 \]
Thus, the exact value of the quantity is:
\[ x = \frac{2220}{127} \]
The approximate value is \( x \approx 17.44 \). However, if you need the exact fraction, you can leave it as \( \frac{2220}{127} \).