Question

In: Math

Four numbers have a sum of 9900. The second exceeds the first by 1/7 of the...

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.

Expert Solution

Let the first number = x

Second number = x + x/7 = 8x/7

Third number = x + 8x/7 + 300 = 15x/7 + 300

The fourth number = x + 8x/7 + + 15x/7 + 300 + 300

= 30x/7 + 600

The total sum of the numbers = 9900

or, x + 8x/7 + 15x/7 + 300 + 30x/7 + 600 = 9900

or,(7x + 8x + 15x + 30x)/7 = 9900 - 900

or, 60x/7 = 9000

or, 60x = 63000

or, x = 1050

So the first number = 1050

Second number = 1050 * 8/7 = 1200

Third number = 15x/7 + 300

= (15 * 1050)/7 + 300

= 2550

Fourth number = 30x/7 + 600

= (30 * 1050)/7 + 600

= 5100

milcah answered 2 years ago

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