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12 Oct 2021, 06:00
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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 100,000 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
(A) 4
(B) 5
(C) 6
(D) 8
(E) 10
sanjitscorps18
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Schools: Fuqua '24 HKUST Johnson '24 Kellogg '24 LBS '24 Sloan '24 Stanford '24 Wharton '24 Yale '24 Harvard '24
Posts: 498
12 Oct 2021, 07:29
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As per the question the selection percentages in each round reduces the number of candidates. Hence the equation
100000 * (2/5)^a * (1/2)^b * (7/10)^c = 1400
=> 10^3 * (2/5)^a * (1/2)^b * (7/10)^c = 14
=> 2^3 * 5^3 * (2/5)^a * (1/2)^b * (7/10)^c = 2 * 7
=> 2^(3+a-b-c) * 5^(3-a-c) * 7^c = 2 * 7
Equating both sides we get c = 1
Since 5 does not have any representation on the RHS we have 3 - a - c = 0 => a = 2
Now 3+a-b-c = 1 => b = 3
So total rounds = a + b+ c = 6
Option C
100000 * (2/5)^a * (1/2)^b * (7/10)^c = 1400
=> 10^3 * (2/5)^a * (1/2)^b * (7/10)^c = 14
=> 2^3 * 5^3 * (2/5)^a * (1/2)^b * (7/10)^c = 2 * 7
=> 2^(3+a-b-c) * 5^(3-a-c) * 7^c = 2 * 7
Equating both sides we get c = 1
Since 5 does not have any representation on the RHS we have 3 - a - c = 0 => a = 2
Now 3+a-b-c = 1 => b = 3
So total rounds = a + b+ c = 6
Option C
General Discussion
27 Aug 2024, 07:42
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Let the remaining rounds be c
100,000 × (0.4)^a × (0.5)^b × (0.7)^c = 1400
where a+b+c=n
(2^2×10^-1)^a ×(5×10^-1)^b ×(7×10^-1)^c= 2 × 7 × 10^-3
Therefore C= 1, and
(2^2×10^-1)^a ×(5×10^-1)^b= 2 ×10^-2
2^2a×5^b×10^-a-b = 2×10^-2
2^2a×5^b = 2×10^a+b-2
here a=1 won't satisfy, so let's try with a= 2
2^4×5^b=2×10^b
2^3×5^b=10^b
so b has to be 3
and a+b+c=n, 2+3+1=6
100,000 × (0.4)^a × (0.5)^b × (0.7)^c = 1400
where a+b+c=n
(2^2×10^-1)^a ×(5×10^-1)^b ×(7×10^-1)^c= 2 × 7 × 10^-3
Therefore C= 1, and
(2^2×10^-1)^a ×(5×10^-1)^b= 2 ×10^-2
2^2a×5^b×10^-a-b = 2×10^-2
2^2a×5^b = 2×10^a+b-2
here a=1 won't satisfy, so let's try with a= 2
2^4×5^b=2×10^b
2^3×5^b=10^b
so b has to be 3
and a+b+c=n, 2+3+1=6
