The sum of four numbers is 64. If you add 3 t | vedclass

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(C) Let the four numbers be $a, b, c,$ and $d.$According to the problem,$a+3 = b-3 = 3c = d/3 = k.$From this,we can express each variable in terms of

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$32$

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(C) Let the four numbers be $a, b, c,$ and $d.$According to the problem,$a+3 = b-3 = 3c = d/3 = k.$From this,we can express each variable in terms of $k$:$a = k-3$$b = k+3$$c = k/3$$d = 3k$Given that the sum of the numbers is $64$:$(k-3) + (k+3) + (k/3) + 3k = 64$$5k + k/3 = 64$$(15k + k)/3 = 64$$16k/3 = 64$$k = (64 	imes 3) / 16 = 12.$Now,substitute $k=12$ to find the numbers:$a = 12-3 = 9$$b = 12+3 = 15$$c = 12/3 = 4$$d = 3 	imes 12 = 36.$The numbers are $9, 15, 4,$ and $36.$The largest number is $36$ and the smallest is $4.$The difference is $36 - 4 = 32.$

The sum of four numbers is $64.$ If you add $3$ to the first number,$3$ is subtracted from the second number,the third is multiplied by $3,$ and the fourth is divided by $3,$ then all the results are equal. What is the difference between the largest and the smallest of the original numbers?

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