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Two men X and Y started working for a certain company at

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CEO
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Two men X and Y started working for a certain company at [#permalink] New post 13 Sep 2003, 03:14
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43. Two men X and Y started working for a certain company at similar
jobs on January 1, 1950. X asked for an initial salary of $ 300 with an
annual increment of $ 30. Y asked for an initial salary of $ 200 with a raise
of $ 15 every six months. Assume that the arrangements remained unaltered till December 31, 1959. Salary is paid on the last day of the month. What is the total amount paid to them as salary during the period?

a) $ 93,300
b) $ 93,200
c) $ 93,100
d) $ 93,150
e) $ 93,175
CEO
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Kudos [?]: 670 [0], given: 781

 [#permalink] New post 13 Sep 2003, 04:40
stolyar wrote:
we have two ariphmetical progressions

(X) the period is 10 years; x1=300 for the first year, +30 for each of the next 9 years. x10=300+9*30=300+270=570. The sum=[300+570]*10/2=4350

(Y) the period is 20 half-years; y1=200 for the first half-year; y20=200+19*15=200+285=485; the sum=[200+485]*20/2=6850

the total is 11200. no such options.
I have assumed that 200 and 300 are yearly salaries. Let's make them monthly ones.

(X) x1=3600; x10=3600+270=3870; the sum=[3600+3870]*10/2=37350
(Y) y1=2400; y20=2400+285=2685; the sum=[2400+2685]*20/2=50850

the total is 88200. no such option again.

I do not know what to do...



Actually i wanted to know if we can do this without a "formula".

Anyways...

The problem does mention that the salary is paid at the end of the month

The formula for SUM of an AP is => N/2 *[ 2*a + (N-1) *d]

X : a = 300, d = 30, N = 10;

S1 = 5(600 + 9 ├Ч (30 ├Ч 12)) = 870 ├Ч 5 ├Ч 12 = 52,200.

Y : a = 200, d = 15, N = 20;

S2 = 10(400 + 19 ├Ч (15 ├Ч 6)) = 41,100.

Total amount = 52,200 + 41,100 = 93,300

Thanks
Praetorian
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 [#permalink] New post 15 Sep 2003, 06:47
What is an AP? And where does this formula come from?

Thanks

Martin
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 [#permalink] New post 15 Sep 2003, 07:21
MartinMag wrote:

What is an AP? And where does this formula come from?
Thanks
Martin



Martin

A.P is arithmetic progression

6,12,18,24,30 are all in A.P...

The next # is obtained by addition of a fixed #..that is 6.

Here $30 is a fixed increment for X and $ 20 is a fixed increment for Y..

Now, this was a question..where i wanted to see if we can use traditional methods to solve this one.. but i guess no...

Sum of an AP = n/2 * [ 2*a + ( n-1) d ]

First term a1 = 6
Number of terms n = 5
Common increment d = 6

Sum = 90

To find the 100th term in the series...

n=100

a100 = a1 + (n-1)*d
= 6 + 99*6
a100 = 600

Hope that helps

Thanks
Praetorian
  [#permalink] 15 Sep 2003, 07:21
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