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An army’s recruitment process included n rounds of selection tasks. Fo

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An army’s recruitment process included n rounds of selection tasks. Fo  [#permalink]

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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 100000 people who applied for the army and 1400 were finally selected, what was the value of n?

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

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An army’s recruitment process included n rounds of selection tasks. Fo  [#permalink]

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New post 05 Jun 2020, 02:55
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The selection % for first a rounds = 40% per round; The selection % for next b rounds = 50% per round; the selection percentage for remaining 'x' rounds = 70 percent per round

Total number of selections = \(100000*(\frac{2}{5})^a * (\frac{1}{2})^b * (\frac{7}{10})^x = 1400\)

\((\frac{2}{5})^a * (\frac{1}{2})^b * (\frac{7}{10})^x = \frac{14}{1000}\)

x must be equal to 1; since highest power of 7 that can divide numerator of RHS is 1.[ there is no way 7 gets cancelled out]

\((\frac{2}{5})^a * (\frac{1}{2})^b = \frac{1}{50}\)

\(\frac{2^{a-b} }{5^a} = \frac{2^{-1}}{5^2}\)

Comparing both sides, we get

a=2; b=3

n= a+b+x = 2+3+1=6
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Re: An army’s recruitment process included n rounds of selection tasks. Fo  [#permalink]

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New post 05 Jun 2020, 04:15
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\(100000(0.4)^a(0.5)^b(0.7)^{n-a-b}=1400\)

\((0.4)^a(0.5)^b(0.7)^{n-a-b}=0.014 = 0.4*0.7*0.5*0.5*0.2 = 0.4*0.7*0.5*0.5*0.4*0.5\)

So, \((0.4)^a(0.5)^b(0.7)^{n-a-b}=(0.4)^2(0.5)^3(0.7)^1\)

Therefore, \(a=2, b=3, n-a-b=1\), then \(n=6\)

Answer is (C)
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Re: An army’s recruitment process included n rounds of selection tasks. Fo  [#permalink]

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New post 05 Jun 2020, 07:08
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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 100000 people who applied for the army and 1400 were finally selected, what was the value of n?

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

n = a + b + c
Selection percentage in rounds a = 40% = \(\frac{2}{5}\)
Selection percentage in rounds b = 50% = \(\frac{1}{2}\)
Selection percentage in rounds c = 70% = \(\frac{7}{10}\)

Now,
\(100000(\frac{2}{5})^a(\frac{1}{2})^b(\frac{7}{10})^c = 1400\)
\((\frac{2}{5})^a(\frac{1}{2})^b(\frac{7}{10})^c = \frac{1400}{100000}\)
\((\frac{2}{5})^a(\frac{1}{2})^b(\frac{7}{10})^c = \frac{2*7}{(2*5)^3}\)
\((\frac{2}{5})^a(\frac{1}{2})^b(\frac{7}{10})^c = (\frac{2}{5})*(\frac{7}{10})*(\frac{1}{2})*\frac{2}{2*2*5}\)
\((\frac{2}{5})^a(\frac{1}{2})^b(\frac{7}{10})^c = (\frac{2}{5})^2*(\frac{7}{10})^1*(\frac{1}{2})^3\)

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

Answer C.
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Re: An army’s recruitment process included n rounds of selection tasks. Fo  [#permalink]

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New post 05 Jun 2020, 07:37
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IMO C

A/Q, 100000 x (40/100)^ a x (50/100)^ b x (70/100)^ n-a-b = 1400

(2/5)^a x (1/2)^b x (7/10)^ n-a-b = 7/500

7^(n-a-b) / 5^(n-a) x 2^(n-2a-2b) = 7/ 5^3 x 2^2

Equating powers,
I) n-a-b= 1
II) n-a = 3
III) n-2a-2b = 2

From I & II, b=2
From II & III, a=3
From II, n=6

Ans. C
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Re: An army’s recruitment process included n rounds of selection tasks. Fo  [#permalink]

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New post 05 Jun 2020, 08:42
1
Quote:
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 100000 people who applied for the army and 1400 were finally selected, what was the value of n?

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


100,000*0.4^a*0.5^b*0.7^c=1400
0.4^a*0.5^b*0.7^c=14/1000=0.014
0.014 is a multiple of 4 and 7
0.4^{a: 1,2,3}=0.4,0.16,0.064
0.5^{b: 1,2,3}=0.5,0.25,0.125
0.7^{c: 1,2,3}=0.7,0.49,0.0343

all^{1}=0.5*0.4*0.7=0.14
we need to divide this by 10 or 2*5 to get 0.014

playing around we find that
a=2, b=3, c=1, n=6

ans (C)
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Re: An army’s recruitment process included n rounds of selection tasks. Fo  [#permalink]

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New post 05 Jun 2020, 12:27
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IMO C

For the first a rounds, the rejection percentage was 60 percent per round
For the first a rounds, the selection percentage was 40 percent per round

For the next b rounds, the rejection percentage was 50 percent per round
For the next b rounds, the selection percentage was 50 percent per round

For the remaining rounds, the selection percentage was 70 percent per round

Problem can be solved using successive %ge change

Let the count of a rounds be a
Count of b rounds = b
Count of remaining rounds = c


Total 100000 people

==> 100000 * \((0.4)^a\) * \((0.5)^b\) * \((0.7)^c\) = 1400

Converting to prime factors

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

==> factorizing 10 = 2*5 and combining the factors
==> \(2^(5+x-y-z)\) * \(5^(5-x-z)\) * \(7^z\) = \(2^3\) * \(5^2\) * \(7^1\)

Comparing the powers of each prime factors:

z =1
& 5-x-z = 2
==> x = 2

& 5+x-y-z = 3
y =3

Toatl rounds = x+y+z = 6
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Re: An army’s recruitment process included n rounds of selection tasks. Fo  [#permalink]

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New post 05 Jun 2020, 13:00
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I took a lot of time doing this.

IMO C

If out of 100 , 60 are rejected then selected are 100(0.4). And if i were to follow this logic for each round I keep multiplyinf with 0.4 until round A is over.

I break it down to :

100000(0.4)^a

Similarly for other rounds

Finally equation looks like -->

100000(0.4^a)(0.5^b) (0.7^c) = 1400

Note a+b+c = n. (total rounds)

(0.4^a)(0.5^b) (0.7^c) = 14/1000

Notice that 4x5x7 = 140

So if we put a,b,c as 1 , we get 140/1000. But we need 14/1000. (Consider this as point 1)

So we need to acquire 1/10 more from the (0.4^a)(0.5^b) (0.7^c).

I did hit and trial. Took me a lot of time. Therefore, I welcome better approach from someone.

4/10 x 5/10 give 2/10. I need 1/10. I multiply one more 5/10 so i get 10/100.

Therefore 0.4 x 0.5 * 0.5 gives 1/10.

So final powers are as follows.

0.4 --> 2 one from 1/10 calc. And one from (point 1 above)

0.5 ---> 3

0.7 ---> 1

Total n = 6

Please tell me a shorter approach. I spent almost 10 minutes at this problem.

Also, do we really get such problems of hit and trial?? I suppose that could take a lot of time.

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Re: An army’s recruitment process included n rounds of selection tasks. Fo  [#permalink]

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New post 05 Jun 2020, 19:49
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Quote:
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 100000 people who applied for the army and 1400 were finally selected, what was the value of n?
A. 4
B. 5
C. 6
D. 8
E. 10


Tough one for me ..

c imo

------Prob (a) = (40%)^a --------|---------Prob (b)=(50%)^b--------|---------Prob (c)=(70%)^c-------

Selection percentage in first a rounds= 100-60=40
Selection percentage in next b rounds= 100-50=50
Selection percentage in remaining c rounds =70

a+b+c =n
Now probability of selection = 1400 / 100k = 14 / 1000

one will be selected if he is selected for all the tasks .

so total prob. of selection = (40%)^a * (50%)^b * (70%)^b = 14/1000
(2/5)^a∗(1/2)^b∗(7/10)^c=(2/5)^2∗(1/2)^3∗(7/10)^1
so
a=2, b=3 and c=1
n=2+3+1=6
Hence C is my ans .
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Re: An army’s recruitment process included n rounds of selection tasks. Fo  [#permalink]

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New post 06 Jun 2020, 15:10
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0.014 portion was selected
There were first 2 rounds, so 0.4x0.4 were selected
Second 3 rounds, so 0.5x0.5x0.5 were selected
Finally there was only 1 round i.e., 0.7 were selected
Overall 0.4x0.4x0.5x0.5x0.5x0.7=0.014

Total of 6 rounds.

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Re: An army’s recruitment process included n rounds of selection tasks. Fo  [#permalink]

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New post 06 Jun 2020, 18:52
I will focus on the selected ones:

After the first "a" rounds --> Blue
After the next "b" rounds --> Green
After the remaining rounds --> Red

\(0.7^{n-a-b}\) x \(0.5^b\) x \(0.4^a\) x 100000 = 1400

then we convert the decimals into fractions, simplify and we will express everything in terms of their prime factors

\(\frac{(7^n)}{(2^n x 5^n)}\) x \(\frac{(5^b)}{(7^b)}\) x \(\frac{(2^{2a})}{(7^a)}\) = \(\frac{7}{(2^2x5^3)}\)

we operate and group similar terms together

\(\frac{7^{n-a-b}}{2^{n-2a}x5^{n-b}}\) = \(\frac{7}{2^2x5^3}\)

Comparing the two expressions term by term and we obtain 3 equations from the exponents

\(n-a-b = 1\)
\(n-b=3\)
\(n-2a=2\)

We replace the second one on the first one to obtain: a = 2
Then we replace that value on the third one to obtain: n = 6
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Re: An army’s recruitment process included n rounds of selection tasks. Fo  [#permalink]

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New post 08 Jun 2020, 00:13
no. of rounds = n
"a" rounds - 60% rejection rate
"b" rounds = 50% rejection rate
"n-a-b" rounds = 70% rejection rate
total students= 100000
selected students = 1400

a*.60+ .50*b+(n-a-b)*.70 = 100000-1400
.60*a+.50*b + .70*n - .70*a - .70*b
-.10*a -.20*b +.70*n = 98600


.40* a + .50*b + (n-a-b)*.30 = 1400
40a+50b+30n-30a-30b = 1400*100
30n+10a+20b = 1400*100
3n + a+2b = 14000


reached to this point, Bunuel kindly suggest
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An army’s recruitment process included n rounds of selection tasks. Fo  [#permalink]

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New post Updated on: 10 Jun 2020, 11:09
Nvm. Mi process and answer are correct, thanks!

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Originally posted by CesarAMSTech on 08 Jun 2020, 03:04.
Last edited by CesarAMSTech on 10 Jun 2020, 11:09, edited 1 time in total.
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An army’s recruitment process included n rounds of selection tasks. Fo  [#permalink]

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New post Updated on: 09 Jun 2020, 10:47
\(100000 (\frac{2}{5})^{a} (\frac{1}{5})^{b} (\frac{7}{10})^{n-a-b} = 1400\)

\((\frac{2}{5})^{a} (\frac{1}{5})^{b} (\frac{7}{10})^{n-a-b}= \frac{1400}{100000}= \frac{14}{1000}= \frac{(2*7)}{(2^{3} 5^{3})} = \frac{7}{ (2^{2}*5^{3})}\)

\(\frac{7^{n -a -b} }{ 2^{n -2a }*5^{n -b}} = \frac{7}{ (2^{2}*5^{3})}\)


----> \(n -a -b = 1\)
----> \(n -2a = 2\)
----> \(n -b = 3\)

\(a= 2\) ---> \(n =6\)

Answer (C).

Originally posted by lacktutor on 08 Jun 2020, 21:54.
Last edited by lacktutor on 09 Jun 2020, 10:47, edited 1 time in total.
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Re: An army’s recruitment process included n rounds of selection tasks. Fo  [#permalink]

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New post 09 Jun 2020, 04:28
Bunuel wrote:

Competition Mode Question



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 100000 people who applied for the army and 1400 were finally selected, what was the value of n?

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



Solution:

We can create the equation:

100,000 x 0.4^a x 0.5^b x 0.7^(n - a - b) = 1,400

0.4^a x 0.5^b x 0.7^(n - a - b) = 0.014

We see that n - a - b must be 1 since then we have one and only one factor of 0.7 (notice 7 divides into 14 and any higher power of 7 does not divide into 14). So we have:

0.4^a x 0.5^b x 0.7 = 0.014

0.4^a x 0.5^b = 0.02

Now let’s rewrite the decimals as fractions:

(2/5)^a x (1/2)^b = 1/50

Since 50 = 2 x 5 x 5, we see that a ≥ 2 and b ≥ 1. However, if a = 2 and b = 1, we have:

(2/5)^2 x (1/2)^1 = 4/25 x 1/2 = 4/50, which is not 1/50.

Since 4/50 is 4 times 1/50, we need to increase the value of b to 3 so that we can have:

(2/5)^2 x (1/2)^3 = 4/25 x 1/8 = 1/25 x 1/2 = 1/50

So a = 2 and b = 3, since n - a - b = 1, we have:

n - 2 - 3 = 1

n = 6

Answer: C
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Re: An army’s recruitment process included n rounds of selection tasks. Fo  [#permalink]

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New post 09 Jun 2020, 08:29
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lacktutor wrote:
\(100000 (\frac{2}{5})^{a} (\frac{1}{5})^{b} (\frac{7}{10})^{n-a-b} = 1400\)

\((\frac{2}{5})^{a} (\frac{1}{5})^{b} (\frac{7}{10})^{n-a-b}= \frac{1400}{100000}= \frac{14}{1000}= \frac{(2*7)}{(2^{3} 5^{3})} = \frac{7}{ (2^{4}*5^{3})}\)

\(\frac{7^{n -a -b} }{ 2^{n -2a }*5^{n -b}} = \frac{7}{ (2^{4}*5^{3})}\)


----> \(n -a -b = 1\)
----> \(n -2a = 4\)
----> \(n -b = 3\)

\(a= 2\) ---> \(n =6\)

Answer (C).


Hello dear lacktutor, there's a small mistake on this expression:

\(\frac{7}{ (2^{4}*5^{3})}\)

the power of the 2 must be two instead of four, i mean:

\(\frac{7}{ (2^{2}*5^{3})}\)

that would change your second equation \(n-2a = 4\) ----> \(n-2a = 2\) which leads to the answer that you gave, n = 6
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Re: An army’s recruitment process included n rounds of selection tasks. Fo  [#permalink]

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New post 09 Jun 2020, 10:49
Maver1ck3 wrote:
lacktutor wrote:
\(100000 (\frac{2}{5})^{a} (\frac{1}{5})^{b} (\frac{7}{10})^{n-a-b} = 1400\)

\((\frac{2}{5})^{a} (\frac{1}{5})^{b} (\frac{7}{10})^{n-a-b}= \frac{1400}{100000}= \frac{14}{1000}= \frac{(2*7)}{(2^{3} 5^{3})} = \frac{7}{ (2^{4}*5^{3})}\)

\(\frac{7^{n -a -b} }{ 2^{n -2a }*5^{n -b}} = \frac{7}{ (2^{4}*5^{3})}\)


----> \(n -a -b = 1\)
----> \(n -2a = 4\)
----> \(n -b = 3\)

\(a= 2\) ---> \(n =6\)

Answer (C).


Hello dear lacktutor, there's a small mistake on this expression:

\(\frac{7}{ (2^{4}*5^{3})}\)

the power of the 2 must be two instead of four, i mean:

\(\frac{7}{ (2^{2}*5^{3})}\)

that would change your second equation \(n-2a = 4\) ----> \(n-2a = 2\) which leads to the answer that you gave, n = 6


hi, Maver1ck3
Good catch!!!
Kudos for you
Corrected it.

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Re: An army’s recruitment process included n rounds of selection tasks. Fo   [#permalink] 09 Jun 2020, 10:49

An army’s recruitment process included n rounds of selection tasks. Fo

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