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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.
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
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?
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?
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
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
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
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.
_________________
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.
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
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 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
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)
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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72% (01:17) correct 
