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I would like some help with these two questions.

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I would like some help with these two questions.

by aalamer6 » Sat Jan 01, 2011 4:30 pm
Q10, P153 of the OG12: When 1/10 percent of 5.000 is subtracted from 1/10 of 5.000, the difference is : A) 0 b)50 c) 450 D)490 E)500

I could not even understand the question rather than ansewring it.

My second question is Q66,P161 of the OG12: At a certain school, the ratio of the number of the second graders to the number of the forth grader is 8 to 5, and the ratio of the number of the first garder to the numebr of the second grader is 3 to 4. If the ratio of the number of the forth garder is 3 to 2, what is the ratio of the numebr of the first grader to the number of the third grader: A)16 to 15 B)9 to 5 C)5 to 16 D) 5 to 4 E) 4 to 5

Would you please help me to understand these questions and explain the answers because I didn't find the OG12's explainations helpful.

Thank you so much.
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by gmat7202011 » Sat Jan 01, 2011 4:45 pm
For 1)

1/10 percent of 5,000 = 0.1% of 5000 = 0.1/100 * 5000 = 5

1/10 of 5000 = 500

500 - 5 = 495
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by anshumishra » Sat Jan 01, 2011 6:04 pm
aalamer6 wrote: Q10, P153 of the OG12: When 1/10 percent of 5.000 is subtracted from 1/10 of 5.000, the difference is : A) 0 b)50 c) 450 D)490 E)500

I could not even understand the question rather than ansewring it.

My second question is Q66,P161 of the OG12: At a certain school, the ratio of the number of the second graders to the number of the forth grader is 8 to 5, and the ratio of the number of the first garder to the numebr of the second grader is 3 to 4. If the ratio of the number of the forth garder is 3 to 2, what is the ratio of the numebr of the first grader to the number of the third grader: A)16 to 15 B)9 to 5 C)5 to 16 D) 5 to 4 E) 4 to 5

Would you please help me to understand these questions and explain the answers because I didn't find the OG12's explainations helpful.

Thank you so much.
For the 1st question, the solution above is right.
Please check the OA, It should be 495.

Solution to 2nd question :

F-> No. of First grade
S-> No. of second grade
T-> No. of third grade
Z -> No. of fourth grade

Given : S/Z = 8/5; F/S = 3/4; T/Z = 3/2
Required : F/T = ?

F/T = F/S * S/Z * Z/T = 3/4 * 8/5 * 2/3 = 4/5

Hence answer is E.
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Anshu

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by aalamer6 » Sat Jan 01, 2011 6:34 pm
Thank you so much, but can you tell me why you mulptiplied the ratios?
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by anshumishra » Sat Jan 01, 2011 6:49 pm
aalamer6 wrote:Thank you so much, but can you tell me why you mulptiplied the ratios?
You can multiply the ratios to get the fraction which you are looking for.

Let me give you a simpler example to show this : You can actually solve it without pen-paper. The idea is to see how both the methods can work.

If the ratio of A's earning to B's earning is 2:1 and the ratio of B's earning to C's earning is 2:1. What is the ratio of C's earning to A's earning ?

Solution without ratio :
Lets assume A earns x
So, B earns -> x/2 (so that the ratio of A's earning to B's earning is 2:1)
C earns -> x/4 (so that the ratio of B's earning to A's earning is 2:1)

Hence, the ratio of C's earning to A's earning = (x/4)/x = 1/4

Solution with ratio :
A/B = 2/1
B/C = 2/1
C/A = ?

C/A = C/B * B/A = 1/2 * 1/2 = 1/4

Now you can see, that both the methods give the same answer. In fact in a complex question, the method where you cancel the ratio would be faster.
Hope that helps !
Thanks
Anshu

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