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Bunuel
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M01-31 [#permalink] New post  15 Sep 2014, 23:16
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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 annual savings is 3:2, and combined they spend $20,000, annually.
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Bunuel
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Re M01-31 [#permalink] New post  15 Sep 2014, 23:16
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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
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Re: M01-31 [#permalink] New post  24 Sep 2014, 15:29
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
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Re: M01-31 [#permalink] New post  22 Nov 2014, 14:06
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?
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Re: M01-31 [#permalink] New post  23 Nov 2014, 06:10
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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 ?
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Re: M01-31 [#permalink] New post  26 Nov 2014, 08:51
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Re: M01-31 [#permalink] New post  31 Jul 2015, 10:49
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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 ?
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Re: M01-31 [#permalink] New post  11 Nov 2015, 02:17
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
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Re: M01-31 [#permalink] New post  10 Mar 2016, 15:28
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
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Re: M01-31 [#permalink] New post  20 Jul 2016, 09:16
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
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Re: M01-31 [#permalink] New post  20 Jul 2016, 10:26
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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.
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Re: M01-31 [#permalink] New post  20 Jul 2016, 19:48
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!
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Re: M01-31 [#permalink] New post  23 Oct 2016, 18:46
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
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Re: M01-31 [#permalink] New post  23 Oct 2016, 22:01
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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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Re: M01-31 [#permalink] New post  29 Mar 2019, 12:03
Yes, I agree with explanation, as their total income and the expenditures of Sarah and Mary respectively are unknown.
INCOME - EXPENDITURE =SAVINGS
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Re: M01-31 [#permalink] New post  01 Apr 2019, 04:13
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
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Re: M01-31 [#permalink] New post  07 Nov 2019, 02:44
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)
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Re: M01-31 [#permalink] New post  01 Dec 2019, 03:59
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)
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Re: M01-31 [#permalink] New post  01 Dec 2019, 22:17
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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
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Re: M01-31 [#permalink] New post  15 Jan 2020, 08:14
Bunuel wrote:
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.


Each individual is clearly insufficient. I don't think there is much to talk about with that respect.

Combined:

Income : S/M = 4x/3x

Savings: S/M = 3y/2y.

Let's say Spending is defined as (Income - Savings)

We will have something like (4x-3y)+(3x-2y)=20000.

We cannot find individual values of x and y. Hence (E)

MentorTutoring Hey Andrew, I am not too confident on my approach here. Though I would have marked the answer as E regardless.

Can you take a look and help?
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