Four numbers have a sum of 9900. the second exceeds the first by 1/7 of the first. the third exceeds the sum of the first two by 300. the fourth exceeds the sum of the first three by 300. find the four numbers.

Numbers are 1050, 1200, 2550, 5100.

Step-by-step explanation:

Let the four numbers are a, b, c, d.

Since second exceeds the first by 1/7 of the first.

b = a + (a/7) = 8a/7

Third exceeds the sum of first two by 200.

c = a + b + 300 = a + (8a/7) + 300 = (15a/7) + 300

Fourth exceeds the sum of the first three by 300

d = a + b + c + 300 = a + (8a/7) + (15a/7) + 300 + 300 = a + (23a/7) + 600

d = (30a/7)+600

And sum of four numbers is 9900

a + b + c + d = 9900(1)

By putting the values of b, c, d in the equation (1)

a + (8a/7) + (15a/7) + 300 + (30a/7) + 600 = 9900

a + (8a/7) + (15a/7) + (30a/7) = 9900 - 300 - 600 = 9000

a + [(8a + 15a + 30a)/7] = 9000

a + (53a/7) = 9000

60a/7 = 9000

a = 9000×7/60 = 1050

Now we put the values of a and get the numbers

b = 8a/7 = 8×1050/7 = 8×150 = 1200

c = (15a/7) + 300 = (15×1050)/7 + 300 = 15×150 + 300 = 2250 + 300 = 2550

and finally from equation 1

1050 + 1200 + 2550 + d = 9900

d + 4800 = 9900

d = 9900 - 4800 = 5100

Therefore the numbers are 1050, 1200, 2550 & 5100.

answer: i literally just answered this question but they want me to do it again

step-by-step explanation:

i think the answer is d im not sure