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If n+10^3/ m+33=4 and m=10, then n/m=?
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29 Apr 2018, 18:02
29 Apr 2018, 18:02
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If \(\frac{(n+10^3)}{(m+10)}=4\) and m=10, then \(\frac{n}{m}\)= ?
(A) -92
(B) 4-10^-2
(C) 4x10^2
(D) 4
(E) 99
(A) -92
(B) 4-10^-2
(C) 4x10^2
(D) 4
(E) 99
If n+10^3/ m+33=4 and m=10, then n/m=?
[#permalink]
29 Apr 2018, 19:27
29 Apr 2018, 19:27
1
Kudos
Expert Reply
uzouka123 wrote:
If \(\frac{(n+10^3)}{(m+10)}=4\) and m=10, then \(\frac{n}{m}\)= ?
(A) -92
(B) 4-10^-2
(C) 4x10^2
(D) 4
(E) 99
(A) -92
(B) 4-10^-2
(C) 4x10^2
(D) 4
(E) 99
\(m = 10\)
\(\frac{(n+10^3)}{(m+10)}=4\)
\(\frac{(n+1,000)}{(10+10)}=4\)
\(n+1,000=80\)
\(n = -920\)
\(\frac{n}{m}=\frac{-920}{10}= -92\)
Answer A
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Re: If n+10^3/ m+33=4 and m=10, then n/m=?
[#permalink]
01 May 2018, 08:56
01 May 2018, 08:56
Expert Reply
uzouka123 wrote:
If \(\frac{(n+10^3)}{(m+10)}=4\) and m=10, then \(\frac{n}{m}\)= ?
(A) -92
(B) 4-10^-2
(C) 4x10^2
(D) 4
(E) 99
(A) -92
(B) 4-10^-2
(C) 4x10^2
(D) 4
(E) 99
Substituting 10 for m, we have:
(n + 10^3)/(10 + 10) = 4
(n + 1000)/20 = 4
n + 1000 = 80
n = -920
Thus,
n/m = -920/10 = -92
Answer: A
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Re: If n+10^3/ m+33=4 and m=10, then n/m=?
[#permalink]
16 May 2018, 20:21
16 May 2018, 20:21
Substituting m=10 in equation
We get
n-1000=80
N=-920
n/m=-920/10= -92
Posted from my mobile device
We get
n-1000=80
N=-920
n/m=-920/10= -92
Posted from my mobile device
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Re: If n+10^3/ m+33=4 and m=10, then n/m=?
[#permalink]
17 May 2018, 10:06
17 May 2018, 10:06
uzouka123 wrote:
If \(\frac{(n+10^3)}{(m+10)}=4\) and m=10, then \(\frac{n}{m}\)= ?
(A) -92
(B) 4-10^-2
(C) 4x10^2
(D) 4
(E) 99
(A) -92
(B) 4-10^-2
(C) 4x10^2
(D) 4
(E) 99
\(\frac{(n+10^3)}{(m+10)}=4\)
Or, \(n + 1000 = 4(10 + 10)\)
Or, \(n + 1000 = 80\)
So, \(n = -920\)
Now, \(\frac{n}{m}\)= \(\frac{-920}{10}\) = \(-92\) , Answer must be (A)
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Abhishek....
PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS
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Re: If n+10^3/ m+33=4 and m=10, then n/m=?
[#permalink]
13 Feb 2023, 16:30
13 Feb 2023, 16:30
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Re: If n+10^3/ m+33=4 and m=10, then n/m=? [#permalink]
13 Feb 2023, 16:30
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