Question

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.

Answer 1

Answer:

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 2

Answer:

1050, 1200, 2550 and  5100.

Step-by-step explanation:

If the first number is x then the second is  x + 1/7x.

The third =  x + x + 1/7x + 300 and the fourth = x + x + 1/7x + x + x + 1/7x+ 300 + 300.

So our equation is:-

8x +  4 * 1/7x  + 900 = 9900

60/7 = 9000

60x = 7*9000

x = 1050.

Therefore the 4 numbers are 1050,  *1/7*1050 = 1200, 2550 and  5100.