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An army’s recruitment process included n rounds of selecti

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by duahsolo » Thu Jul 20, 2017 2:16 am
An army's recruitment process included n rounds of selection tasks. For the first a rounds, the rejection percentage was 60 percent per round. For the next b rounds, the rejection percentage was 50 percent per round and for the remaining rounds, the selection percentage was 70 percent per round. If there were 100,000 people who applied for the army and 1,400 were finally selected, what was the value of n? (Source: e-gmat)

A) 4
B) 5
C) 6
D) 8
E) 10

[spoiler]Correct Answer: C[/spoiler][/spoiler]
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by Jay@ManhattanReview » Thu Jul 20, 2017 3:25 am
duahsolo wrote:An army's recruitment process included n rounds of selection tasks. For the first a rounds, the rejection percentage was 60 percent per round. For the next b rounds, the rejection percentage was 50 percent per round and for the remaining rounds, the selection percentage was 70 percent per round. If there were 100,000 people who applied for the army and 1,400 were finally selected, what was the value of n? (Source: e-gmat)

A) 4
B) 5
C) 6
D) 8
E) 10

[spoiler]Correct Answer: C[/spoiler][/spoiler]
Given that the rejection rate in the first 'a' rounds = 60% per round, thus the selection rate per round = 40% = 2/5 per round for the first 'a' rounds

Thus, after the first 'a' rounds, the number of selections = 1,00,000*(2/5)^a ---(1)

Given that the rejection rate in the next 'b' rounds = 50% per round, thus the selection rate per round = 50% = 1/2 per round for the next 'b' rounds

Thus, after '(a + b)' rounds, the number of selections = 1,00,000*(2/5)^a*(1/2)^b ---(2)

It is given that the selection rate per round for the remaining rounds (n - a - b) = 70% per round = 7/10 per round for the next '(n - a - b)' rounds

Thus, after 'n' rounds, the number of selections = 1,00,000*(2/5)^a*(1/2)^b*(7/10)^(n - a - b) ---(3)

Thus, we have,

1,00,000*(2/5)^a*(1/2)^b*(7/10)^(n - a - b) = 1400

1000*2^a / 5^a * 1^b / 2^b * 7^(n - a - b) / 10^(n - a - b) = 14

500*2^(a-b) / 5^a * 1^b * 7^(n - a - b) / [2^(n - a - b)*5^(n - a - b)] = 7

500*2^(a-b) * 5^(-a) * 7^(n - a - b) * 2^(-n + a + b)*5^(-n + a + b) = 7^1

5^3 * 2^(2+a-b-n + a + b) * 5^(-a-n + a + b) * 7^(n - a - b) = 7^1

2^(2+a-n + a ) * 5^(3-n + b) * 7^(n - a - b) = 2^0 * 5^0 * 7^1

Equating the respective exponents of 2, 3, and 5, we get

2+a-n + a = 0 ---(1)
3-n + b = 0 --(2)
n - a - b = 1 ---(3)

From (1), we get, a = (n - 2)/2 and from (2), we get b = n - 3

By plugging in the values of a and b in (3), we get

n - (n-2)/2 - (n-3) = 1

2n - n + 2 - 2n + 6 = 2

-n + 8 = 2

[spoiler]n = 6.[/spoiler]

The correct answer: C

Hope this helps!

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by Ian Stewart » Thu Jul 20, 2017 5:57 am
duahsolo wrote:An army's recruitment process included n rounds of selection tasks. For the first a rounds, the rejection percentage was 60 percent per round. For the next b rounds, the rejection percentage was 50 percent per round and for the remaining rounds, the selection percentage was 70 percent per round. If there were 100,000 people who applied for the army and 1,400 were finally selected, what was the value of n? (Source: e-gmat)

A) 4
B) 5
C) 6
D) 8
E) 10

[spoiler]Correct Answer: C[/spoiler][/spoiler]
In the first a rounds, we keep 2/5 of applicants; in the next b rounds, we keep 1/2 of applicants, and in the final c rounds we keep 7/10 of applicants. So:

(2/5)^a * (1/2)^b * (7/10)^c * 10,000 = 1400

Let's break these numbers down into primes, get integers on both sides, and cancel what we can:

2^a * 7^c * 10^5 = 5^a * 2^b * 10^c * 14 * 10^2
2^a * 7^c * 10^3 = 5^a * 2^b * 10^c * 2 * 7
2^(a + 3) * 5^3 * 7^c = 2^(b + c + 1) * 5^(a + c) * 7^1

The left side and right side of the equation above are prime factorizations of the same number - the exponents on each prime must be identical. Looking first at the 7, we find c = 1. Then looking at the 5, we find a + c = 3, and since c=1, we find a = 2. Lastly, looking at the 2, we have a+3 = b + c + 1, so 5 = b + 2, and b = 3. We want the value of a+b+c = 2 + 3 + 1 = 6.

This is a lot more awkward than any real GMAT problem testing these concepts.
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by GMATGuruNY » Thu Jul 20, 2017 6:30 am
duahsolo wrote:An army's recruitment process included n rounds of selection tasks. For the first a rounds, the rejection percentage was 60 percent per round. For the next b rounds, the rejection percentage was 50 percent per round and for the remaining rounds, the selection percentage was 70 percent per round. If there were 100,000 people who applied for the army and 1,400 were finally selected, what was the value of n? (Source: e-gmat)

A) 4
B) 5
C) 6
D) 8
E) 10
Round a: Since 60% = 3/5 are rejected, 2/5 are selected.
Round b: Since 50% = 1/2 are rejected, 1/2 are selected.
Round c: 70% = 7/10 are selected.

Since there must be at least one of each round, calculate the remaining number of applicants after 2/5 are selected in round a, 1/2 are selected in round b, and 7/10 are selected in round c:
(100,000)(2/5)(1/2)(7/10) = (100,000)(14/100) = 14,000.

Since the desired number of applicants -- 1400 -- is equal to 1/10 of 14,000, 1/10 of the remaining 14,000 applicants must be selected.

Fraction selected after another round a and round b = (2/5)(1/2) = 1/5.
For the fraction in red to decrease to 1/10, it must be multiplied by 1/2, the fraction attributed to round b:
(1/5)(1/2) = 1/10.
Thus, 1/10 of the 14,000 remaining applicants will be selected after another round a and round b and one more round b:
(14,000)(2/5 * 1/2)(1/2) = 1400.

n = (one of each of the 3 rounds ) + (another round a and round b) + (one more round b) = 6 rounds.

The correct answer is C.
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by [email protected] » Thu Jul 20, 2017 11:14 am
Hi duahsolo,

While this prompt 'looks' complex, you can solve it with a bit of 'brute force' math and a little logic - but you do have to really pay attention to the numbers involved (and you need to know how to could "halves" and 'doubles').

From the prompt, we know that we start with 100,000 people and that we end with 1,400 people. With each 'round' we lose a certain percentage of people.

In the first "A" rounds, we lose 60% of the total people remaining each round.
In the next "B" rounds, we lose 50% of the total people remaining each round.
In the final "C" rounds, we lose 30% of the total people remaining each round.

From the answer choices, we know that there are at least 4 total rounds but no more than 10 total rounds. We also know what happens in the "first" and "last" rounds for sure (re: a 60% drop and a 30% drop), so let's start by 'mapping' those outcomes:

Start = 100,000
1st round = lose 60% of 100,000 = lose 60,000.... 40,000 remain
...
Last round = lose 30% of X .... 1,400 remain... X = 2,000

Thus, we have to find a way to get from 40,000 down to 2,000 using the 'rounds' described above. As an estimate, I'm going to assume that we lose 50% each round.....
40,000 to 20,000 to 10,000 to 5,000 to 2,500

Thus, it certainly looks like it will take about 4 additional rounds (along with the two rounds that I already listed) to get to 2,000 people. Let's try 'counting up' from 2,000 using 'doubles'....
2,000 to 4,000 to 8,000 to 16,000 to 32,000

So, can we get from 40,000 to either 16,000 or 32,000? YES we can - if we remove 60% of 40,000, then we end up with 16,000. Thus, the rounds would be...
Start = 100,000
1st round = lose 60% = lose 60,000... 40,000 remain
2nd round = lose 60% = lose 24,000... 16,000 remain
3rd round = lose 50% = lose 8,000... 8,000 remain
4th round = lose 50% = lose 4,000... 4,000 remain
5th round = lose 50% = lose 2,000... 2,000 remain
6th round = lose 30% = lose 600... 1,400 remain

Final Answer:
C

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by Matt@VeritasPrep » Sun Jul 23, 2017 4:43 pm
Let's start with everything in terms of selection percentage.

First a rounds: 40% selected
Next b rounds: 50% selected
Last c rounds: 70% selected

We know that in the end 1.4% were selected, so

.4ᵃ * .5ᵇ * .7ᶜ = .014

From here, we can cheat a bit. .4 * .5 * .7 = .14, so we're almost there, we just need to multiply by .1.

Luckily for us, .4 * .5 * .5 = .1!

So a = 2, b = 3, c = 1 is a solution, and we're done, a + b + c = 6.
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by Matt@VeritasPrep » Sun Jul 23, 2017 4:56 pm
It might also help to see this with fractions:

(2/5)ᵃ * (1/2)ᵇ * (7/10)ᶜ = 1400/100000 = 7/500

From here, notice that c = 1, since we have exactly one 7 in the numerator. That gives us

(2/5)ᵃ * (1/2)ᵇ * (7/10) = 7/500

or

(2/5)ᵃ * (1/2)ᵇ = 70/3500 = 1/50

From here, notice that (2/5) * (1/2) = 1/5: the numerator 2 and the denominator 2 cancel out. Since we're left with 1/50, we should have MORE 1/2s than 2/5s, since there is no 2 left in the numerator of 1/50.

From there, we should have (1/5) to some power, times (1/2) to some power. Luckily for us, 1/50 breaks down nicely: 1/5 * 1/5 * 1/2.

That leaves us with a = 2, b = 2 + 1, and we're set:

a = 2, b = 3, c = 1, a + b + c = 6
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