MHB Raju had 5 times as much money as Ann. How much did Raju have at first?

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Raju and Ann saved a total of $1560, with Raju initially having five times as much money as Ann after giving her $50. The equations established were R + A = 1560 and R - 50 = 5(A + 50). By substituting A with (1560 - R) in the second equation, it was determined that R = 1350. Thus, Raju initially had $1350 before giving Ann $50. The calculations confirmed the solution and clarified any confusion regarding the math involved.
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Raju and Ann saved a total of 1560 dollars. After Raju gave Ann 50 dollars, Raju had 5 times as much money as Ann. How much did Raju have at first?Number of dollars saved = R
Number of dollars saved = A

1) We know Raju and Ann saved 1560 dollars

R + A = 1560

2) (i got confuse at this part) Raju gave Ann 50 dollars, Raju had 5 times as much money as Ann

R = 5A (not sure if this is the correct way to do it)So then I did => 5A + A = 1560. So A = 260.

Then, I put it back to R = 5A, and then I added 50. So the answer is R = 1350.
 
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After Raju gave Ann 50 dollars Raju had \(R-50\) and Ann had \(A+50\), so we want:

$$R-50=5(A+50)$$

Now, as you observed:

$$R+A=1560\implies A=1560-R$$

And so we can now substitute into the first equation to get an equation in \(R\) only:

$$R-50=5((1560-R)+50)$$

Solving this equation, you will indeed find \(R=1350\).
 
MarkFL said:
After Raju gave Ann 50 dollars Raju had \(R-50\) and Ann had \(A+50\), so we want:

$$R-50=5(A+50)$$

Thank you for this.

When I went back to this question, I had a feeling that I was making math up. Thanks again.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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