Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 59587

Question Stats:
72% (01:17) correct 28% (01:15) wrong based on 230 sessions
HideShow timer Statistics
What is Sarah's annual income? (1) The ratio of Sarah's and Mary's annual income is 4:3. (2) The ratio of Sarah's and Mary's savings is 3:2, and combined they spend $20,000, annually.
Official Answer and Stats are available only to registered users. Register/ Login.
_________________



Math Expert
Joined: 02 Sep 2009
Posts: 59587

Official Solution: Statement (1) by itself is not sufficient. Statement (2) by itself is not sufficient. Statements (1) and (2) combined are not sufficient. If \(s\) and \(m\) denote Sarah's and Mary's spending and \(S\) and \(M\) denote their incomes, then combining the two statements gives: \(\frac{S}{M} = \frac{4}{3}\)
\(\frac{Ss}{Mm} = \frac{3}{2}\)
\(s+m = 20000\) This linear system has 3 equations and 4 unknowns. Consequently, we cannot solve the system of equations. Answer: E
_________________



Manager
Status: Do till 740 :)
Joined: 13 Jun 2011
Posts: 79
Concentration: Strategy, General Management
GPA: 3.6
WE: Consulting (Computer Software)

Re: M0131
[#permalink]
Show Tags
24 Sep 2014, 16:29
Hi Bunuel, Thanks for the Solution. I framed teh equations but spent lot of time to figure whether it could be solved or not. can you please help me with the theory or logic behind it? How do we decide that there are 3 equations and 4 unknowwns hence not solvable.
Please provide any linke for this theory if you have any
Thanks



Manager
Joined: 08 Feb 2014
Posts: 200
Location: United States
Concentration: Finance
WE: Analyst (Commercial Banking)

Re: M0131
[#permalink]
Show Tags
22 Nov 2014, 15:06
shankar245 wrote: Hi Bunuel, Thanks for the Solution. I framed teh equations but spent lot of time to figure whether it could be solved or not. can you please help me with the theory or logic behind it? How do we decide that there are 3 equations and 4 unknowwns hence not solvable.
Please provide any linke for this theory if you have any
Thanks Typically, if you have x number of variables, you will only be able to solve for each variable if you have x number of equations. This is because you need to substitute. for instance if asked to solve for x, and given x+y=4 alone, we cannot solve. But if we are also told that y+3=4, we can use this eq. to find y=1, then replace y with 1 in the first eq. and we'll have 1+x=4. now we can solve for x, x=3. Just the rule. Question: should S/M = 3/4 be S/M = (4x)/(3x), since this is just a proportion and the numbers can take various values, so long as they remain proportionate?



Math Expert
Joined: 02 Sep 2009
Posts: 59587

Re: M0131
[#permalink]
Show Tags
23 Nov 2014, 07:10
JackSparr0w wrote: shankar245 wrote: Hi Bunuel, Thanks for the Solution. I framed teh equations but spent lot of time to figure whether it could be solved or not. can you please help me with the theory or logic behind it? How do we decide that there are 3 equations and 4 unknowwns hence not solvable.
Please provide any linke for this theory if you have any
Thanks Question: should S/M = 3/4 be S/M = (4x)/(3x), since this is just a proportion and the numbers can take various values, so long as they remain proportionate? Does not 3x/4x = 3/4 ?
_________________



Manager
Joined: 08 Feb 2014
Posts: 200
Location: United States
Concentration: Finance
WE: Analyst (Commercial Banking)

Re: M0131
[#permalink]
Show Tags
26 Nov 2014, 09:51
True



Manager
Joined: 18 Mar 2014
Posts: 226
Location: India
Concentration: Operations, Strategy
GPA: 3.19
WE: Information Technology (Computer Software)

Re: M0131
[#permalink]
Show Tags
31 Jul 2015, 11:49
i solve it like Income 4x and 3x Savings 4y and 3y Expenditure total for both = 20000
Income = saving + expenditure so 7x=7y + 20000
Hence 1 and 2 toghether insufficient Is this correct way of solving ?



Intern
Joined: 07 Dec 2014
Posts: 9

Re: M0131
[#permalink]
Show Tags
11 Nov 2015, 03:17
Hi Could you please tell me.. where did I go wrong (1) S & M Income in the ratio of 4:3; I equated it to a constant K; 7K ..(Insufficient) (2) s & m saving in the ratio of 3:2; I equated it to a constant K; total saving 5K Now total income = 5K + Total expenses Total Income = 5K + 20000 Combining (1) & (2) 7K=5K + 20000 K = 10000; S = 40000, M=30000 So where did I go wrong. I presume equating both the equations to same constant is where I went wrong. Is it so, could you please clarify



Intern
Joined: 15 Nov 2015
Posts: 2

Re: M0131
[#permalink]
Show Tags
10 Mar 2016, 16:28
Why is the answer not C?
ratio of incomes is 4:3, using an unknown multiplier it is 4x:3x, therefore total income is 7x ratio of savings is 3:2, using an unknown multiplier it is 3x:2x, therefore total savings is 5x combining the statements we have total income  total savings = total spending, 7x5x=20,000 x=10,000 Sarah's annual income is 4*10,000 = $40,000



Senior Manager
Status: Always try to face your worst fear because nothing GOOD comes easy. You must be UNCOMFORTABLE to get to your COMFORT ZONE
Joined: 15 Aug 2014
Posts: 268
Concentration: Marketing, Technology
GMAT 1: 570 Q44 V25 GMAT 2: 600 Q48 V25
WE: Information Technology (Consulting)

Re: M0131
[#permalink]
Show Tags
20 Jul 2016, 10:16
Can someone explain what's wrong with below approach. ratio of incomes is 4:3, 4x:3x, therefore total income is 7x ratio of savings is 3:2, 3x:2x, therefore total savings is 5x combining the statements we have total income  total savings = total spending, 7x5x=20,000 x=10,000 Sarah's annual income is 4*10,000 = $40,000
_________________
"When you want to succeed as bad as you want to breathe, then you’ll be successful.”  Eric Thomas
I need to work on timing badly!!



Math Expert
Joined: 02 Sep 2009
Posts: 59587

Re: M0131
[#permalink]
Show Tags
20 Jul 2016, 11:26
smartguy595 wrote: Can someone explain what's wrong with below approach.
ratio of incomes is 4:3, 4x:3x, therefore total income is 7x ratio of savings is 3:2, 3x:2x, therefore total savings is 5x combining the statements we have total income  total savings = total spending, 7x5x=20,000 x=10,000 Sarah's annual income is 4*10,000 = $40,000 You cannot use same multiplier x for both income and savings.
_________________



Senior Manager
Status: Always try to face your worst fear because nothing GOOD comes easy. You must be UNCOMFORTABLE to get to your COMFORT ZONE
Joined: 15 Aug 2014
Posts: 268
Concentration: Marketing, Technology
GMAT 1: 570 Q44 V25 GMAT 2: 600 Q48 V25
WE: Information Technology (Consulting)

Re: M0131
[#permalink]
Show Tags
20 Jul 2016, 20:48
Bunuel wrote: smartguy595 wrote: Can someone explain what's wrong with below approach.
ratio of incomes is 4:3, 4x:3x, therefore total income is 7x ratio of savings is 3:2, 3x:2x, therefore total savings is 5x combining the statements we have total income  total savings = total spending, 7x5x=20,000 x=10,000 Sarah's annual income is 4*10,000 = $40,000 You cannot use same multiplier x for both income and savings. Thanks Bunuel. I realized my mistake!
_________________
"When you want to succeed as bad as you want to breathe, then you’ll be successful.”  Eric Thomas
I need to work on timing badly!!



Intern
Joined: 08 Jan 2016
Posts: 8
GPA: 3.5

Re M0131
[#permalink]
Show Tags
23 Oct 2016, 19:46
I think this is a highquality question and I don't agree with the explanation. i Think these eqation can be solved Let salary be 4x and 3x Let Saving be 3y and 2y Expenditure = 4x3y +3x2y=7x5y 20,000=7x5y
therefore only Values of X and Y which satisfies these equation are X=5000 and Y=3000



Math Expert
Joined: 02 Sep 2009
Posts: 59587

Re: M0131
[#permalink]
Show Tags
23 Oct 2016, 23:01
Ayush Mishra wrote: I think this is a highquality question and I don't agree with the explanation. i Think these eqation can be solved Let salary be 4x and 3x Let Saving be 3y and 2y Expenditure = 4x3y +3x2y=7x5y 20,000=7x5y
therefore only Values of X and Y which satisfies these equation are X=5000 and Y=3000 20,000=7x5y has infinitely many solutions. You cannot solve two variable linear equation to get only one solution (assuming you don't have any other constraints).
_________________



Intern
Joined: 28 Apr 2018
Posts: 23

Yes, I agree with explanation, as their total income and the expenditures of Sarah and Mary respectively are unknown. INCOME  EXPENDITURE =SAVINGS



Manager
Joined: 24 Dec 2011
Posts: 55
Location: India
GPA: 4
WE: General Management (Health Care)

Re: M0131
[#permalink]
Show Tags
01 Apr 2019, 05:13
arunmandapaka wrote: Hi Could you please tell me.. where did I go wrong (1) S & M Income in the ratio of 4:3; I equated it to a constant K; 7K ..(Insufficient) (2) s & m saving in the ratio of 3:2; I equated it to a constant K; total saving 5K Now total income = 5K + Total expenses Total Income = 5K + 20000 Combining (1) & (2) 7K=5K + 20000 K = 10000; S = 40000, M=30000 So where did I go wrong. I presume equating both the equations to same constant is where I went wrong. Is it so, could you please clarify u cannot use constant 'k' for both income and savings. both are two different entities



Intern
Joined: 04 Jun 2019
Posts: 10

Hi Bunnel,
Can you please explain whats wrong in my thought process.
Income :4I:3I SAVING RATIO : 3S: 2S
Therefore expenditure ratio :
combined expenditure is 7I5S= 20,000
My query can we write expenditure by taking /considering both ratios as
(4I3S) and (3I  2S)



Manager
Joined: 16 Jul 2018
Posts: 69

Bunuel wrote: Ayush Mishra wrote: I think this is a highquality question and I don't agree with the explanation. i Think these eqation can be solved Let salary be 4x and 3x Let Saving be 3y and 2y Expenditure = 4x3y +3x2y=7x5y 20,000=7x5y
therefore only Values of X and Y which satisfies these equation are X=5000 and Y=3000 20,000=7x5y has infinitely many solutions. You cannot solve two variable linear equation to get only one solution (assuming you don't have any other constraints). Hi Bunuel , A prompt question, if the statement told us that both incomes should be integers would then the answer be C? (because only x=5000 and y=3000 would satisfy the equation)



Math Expert
Joined: 02 Sep 2009
Posts: 59587

Re: M0131
[#permalink]
Show Tags
01 Dec 2019, 23:17
UNSTOPPABLE12 wrote: Bunuel wrote: Ayush Mishra wrote: I think this is a highquality question and I don't agree with the explanation. i Think these eqation can be solved Let salary be 4x and 3x Let Saving be 3y and 2y Expenditure = 4x3y +3x2y=7x5y 20,000=7x5y
therefore only Values of X and Y which satisfies these equation are X=5000 and Y=3000 20,000=7x5y has infinitely many solutions. You cannot solve two variable linear equation to get only one solution (assuming you don't have any other constraints). Hi Bunuel , A prompt question, if the statement told us that both incomes should be integers would then the answer be C? (because only x=5000 and y=3000 would satisfy the equation) There are many other solutions. For example: y = 11 and x = 2865 y = 18 and x = 2870 y = 25 and x = 2875
_________________










