Let the four consecutive numbers be represented as $x$, $x+1$, $x+2$, and $x+3$.
The problem states that twice the first, three times the second, four times the third, and five times the fourth sum up to 236. This can be written as an equation:
$2x + 3(x+1) + 4(x+2) + 5(x+3) = 236$
Since the first number ($x$) is 15, the four consecutive numbers are:
The four consecutive numbers are 15, 16, 17, and 18.
Verification: $2(15) + 3(16) + 4(17) + 5(18) = 30 + 48 + 68 + 90 = 236$. The condition is satisfied.
The correct option is 15, 16, 17, 18.
The sum of three fractions A, B, and C, A > B > C, is \(\frac{121}{60}\) . When C is divided by B, the resulting fraction is \(\frac{9}{10}\) , which exceeds A by \(\frac{3}{20}\) . What is the difference between B and C?
7 is added to a certain number and the sum is multiplied by 5. The product is then divided by 3 and 4 is subtracted from the quotient. If the result comes to 16, then what is the original number?
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