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mikemcgarry
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Could someone post the algebraic way please
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chibimoon
Could someone post the algebraic way please
A+B=100k
0.17B=0.23A+200=> -23A+17B=20000

from first equation 23A+23B=2300K
add above and eq2
40B=2320K
B=58K

A=100k-58k=42k
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chibimoon
Could someone post the algebraic way please

A+B=100,000

0.17B=0.23A+200

You now have two equations, so you can either substitute or eliminate. In the explanation above, elimination is used, here I use substitution (elimination is easier in this case)

Take away decimals first: 17B=23A+20,000

Isolate first equation to solve for A (your goal): B=100,000-A

Plug in for B:
17(100,000-A)=23A+20,000
1,700,000-17A=23A+20,000
1,680,000=40A
1,680,000/40=A
A=42,000=Answer Choice D
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Here's another problem from that same article, one that's even less amenable to algebraic treatment.

If the sequence a(n) is defined as a(n) = \(n^2 + n + \sqrt{n+3}\), then which of the following values of n represents the first terms such that a(n) > 500?
(A) 13
(B) 22
(C) 33
(D) 46
(E) 78


OA = (B)

A full solution is shown at that that article:
https://magoosh.com/gmat/2012/gmat-plugg ... -choice-c/

Mike :-)
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took some time as the numbers were high...
0.17B-0.23A = 200 | * 100
17B-23A=20,000
A+B=100,000
B=100,000 - A

17(100,000 - A) - 23A = 20,000
1,700,000 - 17A - 23A = 20,000
1,680,000 = 40A
divide by 40
first two digits: 42...so answer is D.
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\(A + B = 100,000\)

\(0.17B = 0.23A + 200 ==> -23A + 17B = 20,000\)

For making our calculation easy, we can divide constant by factor of 1000 and get the result.


\(A + B = 100\)
==> \(17A + 17B = 1700\) --{i)
==> \(-23A + 17B = 20\) --(ii)

Subtract (ii) from (i)

17A - (-23A) = 1700-20

40A = 1680

A =\(\frac{168}{4}\)= 42.

A = 42,000
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