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

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.

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

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)

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

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.

Posted from my mobile device

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 .

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.

Posted from my mobile device

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

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

Nvm. Mi process and answer are correct, thanks!

Posted from my mobile device

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

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

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


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

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

Answer (C).

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

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.

Posted from my mobile device

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