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by chintudave » Sun Jan 25, 2009 10:07 pm
(1/5)^m * (1/4)^18 = 1/2(10)^35

(1/5)^m = (1/2(10)^35) / (1/4)^18

(1/5)^m = (4^18) / 2(10)^35

(1/5)^m = (2^2^18) / 2(10)^35

(1/5)^m = 2^36 / 2(10)^35

(1/5)^m = 2^35 / 10 ^ 35

(1/5)^m = 2^35 / 2^35 * 5^35

(1/5)^m = 1/5^35

Hence m = 35
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by chintudave » Sun Jan 25, 2009 10:16 pm
Gear Revolutions

P - 10 revolutions per minute i.e. 6 seconds per revolution

Q - 40 revolutions per minute i.e. 1.5 seconds per revolution

So in 6 seconds Q has 3 additional revolutions than P.

So for 6 additional revolutions the time would be 12 seconds.
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by chintudave » Sun Jan 25, 2009 10:26 pm
Q: Sum a1 + a2 + ... an is either ....

I would start with dividing 350 / 7 = 50. Since 50 is not an option there has to be at least one term that is 77.

So 350 - 77 = 273

273/7 = 39 times

so the total is 39 times 7 and 1 time 77 so n is 39 + 1 = 40

However I would also negate the option of having two terms as 77

350 - 154 = 196 ... 196/7 = 28 times + 2 (77s)

So n=30 which is not an option.

Hence C.
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Two type of machines... Type R & S

by chintudave » Sun Jan 25, 2009 10:44 pm
Type R does the work in 36 hours. So in 1 hour R does 1/36th portion of the work.

Type S does the work in 18 hours. So in 1 hour S does 1/18th portion of the work.

So together R & S in an hour will do

1/36 + 1/18 = 3/36 = 1/12th portion of the work. i.e They take 12 hours to complete the work.

Now to complete the work in 2 hours you will need 6 pairs of R & S working together.

Hence C
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Grades of Milk

by chintudave » Sun Jan 25, 2009 10:50 pm
This is very straight forward.

0.1x + 0.2y + 0.3z = 0.15(x + y + z)

0.1x + 0.2y + 0.3z = 0.15x + 0.15y + 0.15z

Solving the equation...
0.05x = 0.05y + 0.15z

x = y + 3z
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by gaggleofgirls » Mon Jan 26, 2009 9:04 am
For question 2:

Gear P turns at 1/6 rev per sec
Gear Q tuns at 4/6 rev per sec

4/6x - 1/6x = 6

4x-1x / 6 = 6
3x / 6 = 6
3x = 36
x = 36 / 3 = 12

Answer is D

-Carrie
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by gaggleofgirls » Mon Jan 26, 2009 9:13 am
For Question 4:

Machine R puts out 1/36 job per hour
Machine S puts out 1/18 of a job per hour (or 2/36)

Combined rate is 1/12 job per hour (for each pair of R+S)

To get the job done in 2 hours, you only need to get 1/2 the job done in 1 hour.

x/12 = 1/2
2x = 12
x = 6

You will need 6 pairs of machines R and S, so there are 6 Rs and 6 Ss

Answer is C

-Carrie
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by gaggleofgirls » Mon Jan 26, 2009 9:25 am
For Question 6:

If the car uses 1 gal for every 30 miles at 50 mph, then it uses 1 2/3 gal for every 50 miles at 50 mph.

5 hours at 1 2/3 gal = 25/3 gal used

25/3 of the 12 gal tank = 25/3 / 12 = 25/3 * 1/12 = 25/36

Answer is E.

-Carrie
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by tj123 » Mon Jan 26, 2009 11:08 pm
for question 6.
i did the following:

50mph at 5 hours = 250 miles driven
30 miles per gallon at 50mph means:

250/30 = approx 8 gallons

8 out of 12 gallons comes out to 2/3.

why is 2/3 not right
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by Zipper » Tue Jan 27, 2009 12:34 am
tj123 wrote:for question 6.
i did the following:

50mph at 5 hours = 250 miles driven
30 miles per gallon at 50mph means:

250/30 = approx 8 gallons

8 out of 12 gallons comes out to 2/3.

why is 2/3 not right
coz it's an approximation. it's a little over 2/3 not exactly 2/3

25/36 is a little over 2/3 hence closer to the truth
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by krisraam » Tue Jan 27, 2009 4:35 am
For Gears problem

P rotates 10 revs per min ie 1/6 rotaion per second

Q rotates 40 revs per min ie 4/6 rotation per second.

Ecevry second Q is ahead of P by 4/6-1/6 = 1/2 rotation

So after 12 rotations Q will be 6 revolutions ahead of P.

For Series problem

Lets say 7m + 77n = 350

-> m + 11n = 50

If you observe when 11 is multiplied by n the last digit is n. example 11 *1 = 11, 11* 2 = 22.....
For m + 11n = 50. m + n should be multiple of 10. The only choice is C
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Grades of milk

by tj123 » Tue Feb 03, 2009 10:24 pm
why do you use .1 +.2 + .3

1% = .01
2%= .02
3% = .03

and 1.5% = .015
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