Roger had 8,550 pesos more than Peter.If 5/8 of Peter's money is equal to 2/7 of Rogers money,?

Question

Roger had 8,550 pesos more than Peter.If 5/8 of Peter's money is equal to 2/7 of Rogers money,

a. how much did Peter have?
b. how much did they have altogether?

Answer 1

R = P +8550
5/8 P = 2/7R
Substitute like this
P = 8/5 x 2/7 R
P = 16/35 R
So R = 16/35R + 8550
17/35 R = 8550
R= 17603 pesos for Roger
P = 17603 -8550 = 9053 pesos for Peter
Total is 26656 pesos

Answer 2

You have two unknowns, but enough information to set up the needed 2 equations.

Roger's pesos = R
Peter's pesos = P
"Roger had 8,550 pesos more than Peter"
R = 8500+P
"5/8 of Peter's money is equal to 2/7 of Rogers money"
5/8 P = 2/7 R

Take the second equation. I'll isolate R. (7/2) (5/8 P) = R
35/16 P = R

Now, substitute the new value for R into the first equation:
35/16 P = 8550 + P (16/16) [16/16 = 1, so that future +/- calculations can occur.]
Subtract the P on the right, do so also from the left.

(35/16 P) - (16/16 P) = 19/16 P = 8550
P = 8550 (16/19)
P = 7200

Now, back to the first equation:
R = 8550 + 7200 = 15750

You can add the two together for part "b."

Answer 3

Hmmm... Well if we let R = Roger's money and P = Peter's money, we can set up a few equations here. We know that:

R = P + 8550

AND

5/8*P = 2/7*R

So let's solve for a variable and plug it into the other equations. Since R is already solved (isolated), we can plug R = P + 8550 into the second equation, and get this:

5/8*P = 2/7*(P + 8550)

Now solve for P:

5/8*P = 2/7*P + 17100/7 <-----Distribute
35/56*P = 16/56*P + 17100/7 <-----Least Common Denom.
19/56*P = 17100 <------Subtract
19*P = 136800 <-----Multiply both sides by 56
P = 7200 <----- Divide both sides by 19

Ok so now we know how much Peter had. So Since we know what:

R = P + 5880, then R = 7200 + 8550, so R = 15750!

Final answer: Peter has 7200 pesos, Roger has 15750 pesos.

To CHECK your answer completely, try these two numbers in your second equation: 5/8*P = 2/7*R:

5/8*7200 = 2/7*15750
4500 = 4500

So it checks out, and we got the right answer!

Answer 4

R - Roger.
P - Peter.
x - Common value.


R = x + 8550
P = x
------------------------
5/8 x = 2/7(x + 8550) (x by 5/8)
8/5 * 5/8 x = 2/7(x + 8550)
x = 16/35(x + 8550)
x = 16/35x + 3908∙671 429...
x - 16/35x = 3908∙671 429...
0∙542 857 142...x = 3908∙671 429... (÷ by 0∙542 857 142...)
x = 7200

(a).
P = x
P = 7200 pesos.

(b).
R = x + 8550
R = 7200 + 8550
R = 15,750 pesos.

Together ?
R + P = 7,200 + 15,750
R + P = 22,950 pesos.

Answer 5

x = Roger's money, x - P8,550.00 = Peter's money

a. How much did Peter have?
2/7x = 5/8(x - P8,550.00)
2/7x = 5/8x - P5,343.75
5/8x - 2/7x = P5,343.75
35/56x - 16/56x = P5,343.75
19/56x = P5,343.75
x = P15,750.00

Answer: Roger's money is P15,750.00

b. How much did they have altogether?
= P15,750.00 + (P15,750.00 - $8,550.00)
= P15,750.00 + P7,200.00
= P22,950.00

Answer: Altogether they have P22,950.00.

Proof:
5/8(P15,750.00 - P8,550.00) = 2/7 * P15,750.00
5/8 * P7,200.00 = P4,500.00
P4,500.00 = P4,500.00

Answer 6

x = number of pesos Peter has
y = number of pesos Roger has
y = x + 8550
5/8 times x = 2/7 times (x + 8550)
56(5/8 x) = 56 (2/7) (x + 8550)
35 x = 16(x + 8550)
35 x = 16x + 16(8550)
19x = 16(8550)
19x/19 = 16(8550)/19
x = 16(450)
x = 7200
y = 8550 + 7200
y = 15750

Answer 7

R = Roger
P = Peter

R = P + 8550

5/8 * P = 2/7 * R
R = 35/16 * P

35/16 * P = P + 8550
35/16 * P - P = 8550
19 / 16 * P = 8550
P = 7200

R = 7200 + 8550 = 15750

TOTAL = R + P = 15750 + 7200 = 22950

Answer 8

Peter has 7200 pesos.

They have 22950 pesos altogether.

Answer 9

1526.78 pesos
10076.78 pesos all together