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An epidemic is reported to have broken out in Florida.

Expert replies
by Uva@90 » Sat Jul 04, 2015 7:37 am
An epidemic is reported to have broken out in Florida. The number of detected
instances of a certain disease is reported to have increased by 85% in
the last year. What is the lowest number of newly detected instances possible?
(A) 1
(B) 5
(C) 11
(D) 15
(E) 17

OA E


Could you correct my mistakes,

Let X be the number of OLD Instances,
and N be the New Instances
and it has been increased by 85%

So N = X(185/100) ==> N = X(37/20)
to make N as low, then X should be 20 so N is 37 right ??

Regards,
Uva.
Known is a drop Unknown is an Ocean
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Source: — Problem Solving |

by theCEO » Sat Jul 04, 2015 7:58 am
Uva@90 wrote:An epidemic is reported to have broken out in Florida. The number of detected
instances of a certain disease is reported to have increased by 85% in
the last year. What is the lowest number of newly detected instances possible?
(A) 1
(B) 5
(C) 11
(D) 15
(E) 17

OA E


Could you correct my mistakes,

Let X be the number of OLD Instances,
and N be the New Instances
and it has been increased by 85%

So N = X(185/100) ==> N = X(37/20)
to make N as low, then X should be 20 so N is 37 right ??

Regards,
Uva.
X = no. of old instances = x
n = no. of this years instances = 1.85x

newly detected instances = 1.85x - x = 0.85x
0.85x = 85x/100 = 17x/20

since 17X/20 has to be an integer, the lowest number occurs when x =20
17x20/20 = 17
ans = e
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by theCEO » Sat Jul 04, 2015 8:15 am
Another way to look at this......

Old instances detected - x
New instances detected - 1.85x

1.85x = a + b
a = old instances that we detected previously
b = new instances that didnt exist previously

In order to find the lowest number of new instances (b), we have to maximize the no. of old instances (a).
a = x
therefore the lowest new instances occurs when 1.85x = x + b
b = 0.85x
b =(17/20)x
since b is an integer lowest occurs when x = 20
b =(17/20)*20 = 17

Hope this helps!



The question asks us to find the lowest no. of detected instances
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by [email protected] » Sat Jul 04, 2015 9:59 am
Hi Uva@90,

Your approach is perfect, but you have to remember to answer the question that is ASKED.

Here, the question is: what is the lowest possible number of NEWLY detected instances of the disease. Using your example, there were 20 cases LAST YEAR and 17 more THIS YEAR. Thus, the number of NEWLY detected cases would be 17.

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
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by Uva@90 » Sat Jul 04, 2015 7:17 pm
[email protected] wrote:Hi Uva@90,

Your approach is perfect, but you have to remember to answer the question that is ASKED.

Here, the question is: what is the lowest possible number of NEWLY detected instances of the disease. Using your example, there were 20 cases LAST YEAR and 17 more THIS YEAR. Thus, the number of NEWLY detected cases would be 17.

GMAT assassins aren't born, they're made,
Rich
Rich,

Wow! You got my mistake....

You are the BEST!

Thanks a lot.

Regards,
Uva.
Known is a drop Unknown is an Ocean
Join the discussion

by GMATGuruNY » Sun Jul 05, 2015 6:19 am
Uva@90 wrote:An epidemic is reported to have broken out in Florida. The number of detected
instances of a certain disease is reported to have increased by 85% in
the last year. What is the lowest number of newly detected instances possible?
(A) 1
(B) 5
(C) 11
(D) 15
(E) 17
Let x = the number of cases at the beginning of this year.
Let n = the number of new cases.
Since the total number of cases increases by 85%, the number of new cases is equal to 85% of the number of cases at the beginning of this year:
n = (85/100)x = (17/20)x.

We can PLUG IN THE ANSWERS, which represent the least possible value of n.
When the correct value for n is plugged into n = (17/20)x, the value of x will be an INTEGER.

A: 1 = (17/20)x, implying that x = 20/17.
B: 5 = (17/20)x, implying that x = 100/17.
C: 11 = (17/20)x, implying that x = 220/17.
D: 15 = (17/20)x, implying that x = 300/17.
E: 17 = (17/20)x, implying that x = 20.

The correct answer is E.
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by nikhilgmat31 » Thu Jul 09, 2015 3:32 am
great, you guys catch the statement correctly as number of newly cases.

it different new - old = 1.85x -x = .85

n=.85x

n must atleast 17
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by Jeff@TargetTestPrep » Fri May 04, 2018 9:31 am
Uva@90 wrote:An epidemic is reported to have broken out in Florida. The number of detected
instances of a certain disease is reported to have increased by 85% in
the last year. What is the lowest number of newly detected instances possible?
(A) 1
(B) 5
(C) 11
(D) 15
(E) 17
The key to this problem is realizing that we must have whole number values for both the original number of detected diseases and this year's new number of cases. If we let the original number of detected diseases = n, then the new number of cases is:

1.85n = 185/100 x n = 37/20 x n = 37n/20

Since the fraction has been reduced to lowest terms, we see that for 37n/20 to be a whole number, then n must be at least 20.

Since 20 x 1.85 = 37, the minimum number of newly detected cases is 37 - 20 = 17 cases.

Answer: E

Jeffrey Miller
Head of GMAT Instruction
[email protected]

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