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Re: On 1st January of a certain year, Jones invested a total of $10000 u [#permalink]
With the second statement, you are able to determine that one investment is for x value, and the second one for y value, but then you do not know which is which... am i correct? Therefore statement 2 would be insufficient.[/quote]

I think you're right! My apologies.
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Re: On 1st January of a certain year, Jones invested a total of $10000 u [#permalink]
commodork71 wrote:
Using statement #1 and the prompt, we can write 2 equations for the 2 unknowns: X(1.05)-300= Y(1.03) and X+Y=10,000. As a rule, you need x amount of distinct equations to solve for x amount of variables. Thus, without solving the equations, you know there is sufficient information.

Using statement #2 and the prompt you can do the same thing: X+Y=10,000 and X-Y=7,000. Again, you would only need to recognize that you have 2 equations for 2 unknowns to see that it's sufficient.

Note that this rule only applies for distinct equations. What I mean by that is that both offer information that is not contained in the other. For example, if the equations you were given were X+Y=10,000 and 2(X+Y)=20,000, you really only have one useful equation.

So here the answer is that either are sufficient on their own.

­From statement 1, we only know the difference which means, either X(1.05)-300= Y(1.03) OR X(1.05)+ 300= Y(1.03) which will give you two different answers. Isn't it ?
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Re: On 1st January of a certain year, Jones invested a total of $10000 u [#permalink]
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Nancykgp wrote:
Quote:
 On 1st January of a certain year, Jones invested a total of $10000 under simple interest in two different schemes. He invested one part at 5% p.a and the other part at 3% p.a. How much did he invest at 5% p.a?

(1) The difference in the total amounts of interests earned by the two investments at the end of 1 year from the date of the investments is $300.

(2) The difference of the amounts invested in the two schemes is $5000.

commodork71sing statement #1 and the prompt, we can write 2 equations for the 2 unknowns: X(1.05)-300= Y(1.03) and X+Y=10,000. As a rule, you need x amount of distinct equations to solve for x amount of variables. Thus, without solving the equations, you know there is sufficient information.

Using statement #2 and the prompt you can do the same thing: X+Y=10,000 and X-Y=7,000. Again, you would only need to recognize that you have 2 equations for 2 unknowns to see that it's sufficient.

Note that this rule only applies for distinct equations. What I mean by that is that both offer information that is not contained in the other. For example, if the equations you were given were X+Y=10,000 and 2(X+Y)=20,000, you really only have one useful equation.

So here the answer is that either are sufficient on their own.

­From statement 1, we only know the difference which means, either X(1.05)-300= Y(1.03) OR X(1.05)+ 300= Y(1.03) which will give you two different answers. Isn't it ?

From statement 1, we only know the difference in interest, which implies either 0.05x - 300 = 0.03y or 0.05x + 300 = 0.03y, alongside x + y = 10,000. This is because we are told the difference in the interest earned was $300, not the difference in the balance. So, we'd have:

1. 0.05x - 300 = 0.03y and x + y = 10,000
Or
2. 0.05x + 300 = 0.03y and x + y = 10,000

Next, solving the second set gives x = 0 and y = 10,000, which would be unrealistic since we can infer that at least some amount was invested at 5%.­
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Re: On 1st January of a certain year, Jones invested a total of $10000 u [#permalink]
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