GATE Overflow Test Series | Quantitative Aptitude | Test 1 | Question: 24

Question

It has been found that the number of active COVID-$19$ positive cases in New Delhi doubles every $14$ days. If the current number of hospital beds in New Delhi is $100,000$ and there are $20,000$ active cases, in how many days will New Delhi run out of hospital beds?

• A. $30$
• B. $31$
• C. $32$
• D. $33$

Comments

nice question!

Answer 1

Growth rate in number of days  of number of active cases is given as $2 = x^{14}$
$\implies x = 2^{1/14}$

Current number of active cases is $20,000$ and we need $100,000$ for the hospital beds to run out.

So, number of days $d$ within which hospital beds will run out is given by

$100,000 \leq 20,000 \times x^d$

$\implies 5 \leq \left(2^{1/14}\right)^d$

$\implies 5 \leq 2^{d/14}$

$\implies d \geq 14 \times \log_2 5 = 14 \times 2.32 = 32.48$
 
So, on the $33^{rd}$ day the hospital beds will run out.

Correct answer : D

Comments

Ignoring “000” for easy calculation.

100 → Beds

20 → Cases

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Doubles every 14 days.

Initially: 20

Day1 : 40 → +14 days

Day2 : 80 → +14 days

 Now at this time, 28 days are gone.

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14 days --→ +100%

1 day --→ +(100/14)% or +(1/14)

If we see answers we only need. 

Option A: +2 days

Option B: +3 days

Option C: +4 days

Option D: +5 days

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Currently, we have 80 cases & we need to do +[(100/14)]% increment in 80 and see if it crosses 100 or not.

80 + 80*(1/14) ≈  85.7 : Day 29th

85.7 + 85.7/14 ≈  91.8 not able to cross 100, Option A is not the answer. : Day 30th

91.8 + 91.8/14 ≈ 98.35 not able to cross 100, Option B is not the answer. : Day 31st

98.35 + 98.35/14 ≈ 105.37 crossed 100 hence Option C is the answer. : Day 32nd

@Deepak Poonia sir, is this correct.

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@Udhay_Brahmi Lets take the initial value as $3$ and assume $5$ days to double. So, as per your procedure in each day the value increases by $(100/5)\%$ or $0.2$ of the previous value.

So, at beginning we have $3$.

End of day $1 \to 3 \times 1.2 = 3.6$

End of day $2 \to 3.6 \times 1.2 = 4.32$

End of day $3 \to 4.32 \times 1.2 = 5.184$

End of day $4 \to 5.184 \times 1.2 = 6.228$

End of day $5 \to 6.228 \times 1.2 = 7.46496$ 

Instead of $2\times 3 = 6$ we got $7.46496$ 😼

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@gatecse bro, I got the point that my method is not giving proper results. 

but why which part of the method is causing the difference in the 2 results?

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Well, the method given in the answer is the standard geometric progression formula. In short you cannot assume uniform increase for every interval in a geometric series. If you draw a graph you'll understand it. Otherwise please comment.

Your solution is interesting and those who try methods like those are the ones needed for research.