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We can consider the 10 percent increase as of following: \(\frac{11}{10} *\frac{11}{10}*\frac{11}{0} or (\frac{11}{10})^3\)


\((\frac{11}{10}*\frac{11}{10}*\frac{11}{10})*1000=1331\)

The correct option: C
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Bunuel - How can I apply this consecutive interest/rate changes formula below to problems like this?

a+b+\(\frac{ab}{100}\). Or is it that this formula applies only when there are two consecutive rate/interest changes? TIA.
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John received a 10% raise each month for three consecutive months. What was his salary after the three raises if his starting salary was $1,000 per month?

A. $1,248
B. $1,300
C. $1,331
D. $1,410
E. $1,463


My answer is C- 1331.

We can apply the concept of CI as the 10% raise each month is on the salary of that month. And each month's salary includes last month's salary plus the 10% hike.

Thus, applying CI formula,
Amount = P[1+(10/100)]^3
= 1000(11/10)^ 3 = 1331.




Sent from my iPhone using GMAT Club Forum mobile app
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Bunuel - How can I apply this consecutive interest/rate changes formula below to problems like this?

a+b+\(\frac{ab}{100}\). Or is it that this formula applies only when there are two consecutive rate/interest changes? TIA.

Yes, \(a+b+\frac{ab}{100}\) gives the effective percentage change over TWO periods, not three. So, for this problem if it were "...for TWO consecutive months...", then the effective percentage change would be \(a+b+\frac{ab}{100}=10+10+\frac{100}{100}=21\%\). But you can find it directly with 1.1^2 = 1.21. So, you should use this formula if there are two periods and a and b are different from each other.

For more check "Successive Percentage Changes" chapter in Percentages, Interest and More.

Hope it helps.
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Bunuel
John received a 10% raise each month for three consecutive months. What was his salary after the three raises if his starting salary was $1,000 per month?

A. $1,248
B. $1,300
C. $1,331
D. $1,410
E. $1,463
1,000*1.1^3=1,331
Answer C
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Bunuel
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Bunuel - How can I apply this consecutive interest/rate changes formula below to problems like this?

a+b+\(\frac{ab}{100}\). Or is it that this formula applies only when there are two consecutive rate/interest changes? TIA.

Yes, \(a+b+\frac{ab}{100}\) gives the effective percentage change over TWO periods, not three. So, for this problem if it were "...for TWO consecutive months...", then the effective percentage change would be \(a+b+\frac{ab}{100}=10+10+\frac{100}{100}=21\%\). But you can find it directly with 1.1^2 = 1.21. So, you should use this formula if there are two periods and a and b are different from each other.

For more check "Successive Percentage Changes" chapter in Percentages, Interest and More.

Hope it helps.

Thank you for the explanation.
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Bunuel
John received a 10% raise each month for three consecutive months. What was his salary after the three raises if his starting salary was $1,000 per month?

A. $1,248
B. $1,300
C. $1,331
D. $1,410
E. $1,463

John’s salary after 3 months is 1000(11/10)(11/10)(11/10) = 11 x 11 x 11 = $1331.

Answer: C
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Bunuel
John received a 10% raise each month for three consecutive months. What was his salary after the three raises if his starting salary was $1,000 per month?

A. $1,248
B. $1,300
C. $1,331
D. $1,410
E. $1,463

1.1*1.1*1.1*1=1.331

Since the base amount is 1000, the increased value will be 1331
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