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# M01-31

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Math Expert
Joined: 02 Sep 2009
Posts: 59587

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16 Sep 2014, 00:16
1
9
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25% (medium)

Question Stats:

72% (01:17) correct 28% (01:15) wrong based on 230 sessions

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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. _________________ Math Expert Joined: 02 Sep 2009 Posts: 59587 M01-31 [#permalink] ### Show Tags 16 Sep 2014, 00:16 2 2 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: $$\frac{S}{M} = \frac{4}{3}$$ $$\frac{S-s}{M-m} = \frac{3}{2}$$ $$s+m = 20000$$ 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 GMAT 1: 460 Q35 V20 GPA: 3.6 WE: Consulting (Computer Software) Re: M01-31 [#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 GMAT 1: 650 Q39 V41 WE: Analyst (Commercial Banking) Re: M01-31 [#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: M01-31 [#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 GMAT 1: 650 Q39 V41 WE: Analyst (Commercial Banking) Re: M01-31 [#permalink] ### Show Tags 26 Nov 2014, 09:51 True Manager Joined: 18 Mar 2014 Posts: 226 Location: India Concentration: Operations, Strategy GMAT 1: 670 Q48 V35 GPA: 3.19 WE: Information Technology (Computer Software) Re: M01-31 [#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: M01-31 [#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: M01-31 [#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, 7x-5x=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)

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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, 7x-5x=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: M01-31 [#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, 7x-5x=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)

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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, 7x-5x=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
GMAT 1: 600 Q49 V23
GPA: 3.5

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23 Oct 2016, 19:46
I think this is a high-quality 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 = 4x-3y +3x-2y=7x-5y
20,000=7x-5y

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

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23 Oct 2016, 23:01
Ayush Mishra wrote:
I think this is a high-quality 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 = 4x-3y +3x-2y=7x-5y
20,000=7x-5y

therefore only Values of X and Y which satisfies these equation are X=5000 and Y=3000

20,000=7x-5y has infinitely many solutions. You cannot solve two variable linear equation to get only one solution (assuming you don't have any other constraints).
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Joined: 28 Apr 2018
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29 Mar 2019, 13:03
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)

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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

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07 Nov 2019, 03:44
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 7I-5S= 20,000

My query can we write expenditure by taking /considering both ratios as

(4I-3S) and (3I - 2S)
Manager
Joined: 16 Jul 2018
Posts: 69

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01 Dec 2019, 04:59
Bunuel wrote:
Ayush Mishra wrote:
I think this is a high-quality 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 = 4x-3y +3x-2y=7x-5y
20,000=7x-5y

therefore only Values of X and Y which satisfies these equation are X=5000 and Y=3000

20,000=7x-5y 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

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01 Dec 2019, 23:17
UNSTOPPABLE12 wrote:
Bunuel wrote:
Ayush Mishra wrote:
I think this is a high-quality 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 = 4x-3y +3x-2y=7x-5y
20,000=7x-5y

therefore only Values of X and Y which satisfies these equation are X=5000 and Y=3000

20,000=7x-5y 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
_________________
Re: M01-31   [#permalink] 01 Dec 2019, 23:17
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# M01-31

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