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CrackVerbalGMAT
Simple Interest = \(\frac{(P * r * t) }{ 100}\)

Where P is the Principal, r = rate and t = time in years

Here Time is assumed for a year, t = 1

Let Connies Investment in B = x
Then her Investment in A = (750 + x)



Income from A = Interest = \(\frac{(750+x) * 6 * 1}{100}\)

Income from B = \(\frac{(x * 4 * 1)}{100}\)

Difference in Incomes = 10000 = \(\frac{(750+x) * 6 * 1}{100}\) - \(\frac{(x * 4 * 1)}{100}\)

Solving for x:

10000 * 100 = 6(750 + x) - 4x

1000000 = 4500 + 6x - 4x

2x = 995500

x = 497750

Therefore Total Amount Invested = x + 750 + x = 2x + 750 = (2 * 497750) + 750 = 996250

Option C

Arun Kumar

In the question, it is mentioned that Connie has two investments, A and B. Not necessarily, these two investment amounts are equal. So, ideally we should be considering the investment from B as y. Am I missing something here?
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elavendan1
CrackVerbalGMAT
Simple Interest = \(\frac{(P * r * t) }{ 100}\)

Where P is the Principal, r = rate and t = time in years

Here Time is assumed for a year, t = 1

Let Connies Investment in B = x
Then her Investment in A = (750 + x)



Income from A = Interest = \(\frac{(750+x) * 6 * 1}{100}\)

Income from B = \(\frac{(x * 4 * 1)}{100}\)

Difference in Incomes = 10000 = \(\frac{(750+x) * 6 * 1}{100}\) - \(\frac{(x * 4 * 1)}{100}\)

Solving for x:

10000 * 100 = 6(750 + x) - 4x

1000000 = 4500 + 6x - 4x

2x = 995500

x = 497750

Therefore Total Amount Invested = x + 750 + x = 2x + 750 = (2 * 497750) + 750 = 996250

Option C

Arun Kumar

In the question, it is mentioned that Connie has two investments, A and B. Not necessarily, these two investment amounts are equal. So, ideally we should be considering the investment from B as y. Am I missing something here?

______________________________________________________________

Hi elavendan1. In the question it says Connie has 750 more invested in A than in B. This helps us have the investments related through a common variable x.

If B is x A will be 750 + x.

Arun Kumar
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