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Kevin and Danny had \$200 altogether. Kevin gave 2/7 of his money to Danny. As a result, Danny had 4 times as much money as Kevin. How much did Danny have at first?

Oct 3, 2021
edited by Guest  Oct 3, 2021

#1
+1

Let initially Kevin had x dollars and Danny had y dollars. The initial sum of money is \$200 therefore,

x + y = 200     (1)

Kevin gave 2/7 of his money to Danny. So, amount of money Kevin gave to Danny is 2x/7. The Danny total money was y + 2x/7

dollars and Kevin's money was x - 2x/7 = 5x/7 dollars.

The amount of money Danny had = y + 2x/7 (2)

After taking money from the Kevin the Danny had 4 times as much money as Kevin therefore the Danny had 4 (5x/7) dollars.

The amount of money Danny had = 4(5x/7) (3)

Combining both (2) and (3)

It gives,

y + 2x/7 = 4(5x/7)

Solve the equation for x

2x/7 = 20x/7 - y

2x = 20x - y ⋅ 7

x = 7y/18

Substitute x = 7x/18 in (1) and solve for y.

7y/18 + y = 200

7y/18 ⋅ 18 + y ⋅ 18=200 ⋅ 18

25y=3600

y=144

The Danny had \$144 at first.

Oct 4, 2021
#2
+36420
+1

k + d = 200       or   k= 200 - d

Then     k - 2/7 k = 1/4   (d + 2/7 k)

5/7 k - 2/28 k = 1/4 d

18/28 k = 1/4 d                 sub in the red equation

18/28 (200 - d) = 1/4 d

d = 144

Oct 4, 2021