The inflation index for the year 1989 relative to the year

Hi,
I didn't understand this part, is there any theory behind two distinct linear equation with two unknowns?
Please feel free to share any resources

Let´s call
a = 1970
b = 1989

Given information
b/a = 3.56/1

Statement 1
b - a = 102.4
We have 2 different, linear equations. Therefore, we can solve. Sufficient.

Statement 2
b = 142.4
We have 2 different, linear equations. Therefore, we can solve. Sufficient.

OA = D

Solution:

We are given that the inflation for the year 1989 relative to the year 1970 was 3.56, which means the price of an item in 1989 was 3.56 times the price of the same item in 1970. We are also given that the price of a Model K mixer increased according to the given inflation index, and from that we must determine the price of the mixer in 1970.

If we let x = the price of the Model K mixer in 1970, then the price of the Model K mixer in 1989 is 3.56x. Thus, we must determine the value of x.

Statement One Alone:

The price of the Model K mixer was $102.40 more in 1989 than in 1970.

Using the information in statement one, we can create the following equation:

3.56x = 102.40 + x

We see that we have enough information to determine the value of x. Statement one alone is sufficient to answer the question. We can eliminate answer choices B, C, and E.

Statement Two Alone:

The price of the Model K mixer was $142.40 in 1989.

Using the information in statement two we can create the following equation:

3.56x = 142.40

We see that we have enough information to determine the value of x.

Answer: D

\({{{\rm{pric}}{{\rm{e}}_{\,70}}} \over {{\rm{price}}{\,_{89}}}} = {1 \over {3.56}}\,\,\,\,\,\,\left( * \right)\)


\(?\,\, = {\rm{pric}}{{\rm{e}}_{\,70}}\)


\(\left( 1 \right)\,\,\,{\rm{pric}}{{\rm{e}}_{\,89}} - {\rm{pric}}{{\rm{e}}_{\,70}} = 102.40\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,3.56 \cdot {\rm{pric}}{{\rm{e}}_{\,70}} - {\rm{pric}}{{\rm{e}}_{\,70}} = 102.40\,\,\,\,\, \Rightarrow \,\,\,\,{\rm{SUFF}}.\,\)


\(\left( 2 \right)\,\,\,{\rm{pric}}{{\rm{e}}_{\,89}} = 142.40\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,3.56 \cdot {\rm{pric}}{{\rm{e}}_{\,70}}\, = 142.40\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{SUFF}}.\,\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

One query:
Could you explain the highlighted part, please? What do these 2 same figure (one is dollar figure, the other has no $ sign) mean, actually?
Thanks__

Hi....

First understand the meaning of "The inflation index for the year 1989 relative to the year 1970 was 3.56"


This is telling us that \(\frac{(Inflation Index 1989 Let suppose in dollars)}{(Inflation Index 1970 Let suppose in dollars)}=\frac{3.56}{1}\) (Dollars to dollars cancel out)

means $1 in 1970 = $ 3.56 in 1989

Now the second sentence, "indicating that, on the average, for each dollar spent in 1970 for goods, $3.56 had to be spent for the same goods in 1989"

Now if u read closely, it start with 'indicating that'---means supporting the sentence before it. This above telling us the same thing but in dollars

each dollar spent in 1989 for goods = each dollar spent in 1970 for goods * $ 3.56


Hope this helps....

Thanks

1. The price of the Model K mixer was $102.40 more in 1989 than in 1970

let the price of Model K be x$ in the 70's then price in 1980 = x+ 102.40 $

therefore, we get the following relation:

\(1/3.56=x/x+102.40\)----- clearly, we can find x from this. suff.

(2) The price of the Model K mixer was $142.40 in 1989

this one is straight forward, we got the rate in the '89, we also know every dollar in '70= 3.56$ in '89

thus, \(142.40/3.56= $cost\) in '70. Also, sufficient.

D

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