Help ASAP!! Thanks!

1. We bought 2 pens and 6 pencils, the ratio of the amount we spent on pencils to the amount we spent on pens was 7:4. Later we went back and bought 4 pencils and 3 pens and spent 1 dollar less. How much money did we spend that time?

Let the cost of a pen  = x  and the cost of  a pencil  = y

So we have that  

6y / 2x  = 7 / 4

3y / x  = 7/4

3y  = (7/4)x

y = (7/12)x

And we know that

2x + 6y   = A        where A  is in dollars

2x + 6(7/12)x  = A

2x + (42/12)x  = A

2x + 3.5x  = A

5.5x   = A  = (11/2)x    (1)

And we also know that

3x + 4y  =  A  - 1

3x + 4(7/12)x  = A  - 1

3x + (28/12)x  = A  - 1

3x + (7/3)x   = A  - 1

(16/3)x  = A  - 1      sub( 1)  in for  A

(16/3)x = (11/2)x- 1    subtract (11/2)x from both sides

-1/6x  = -1

x  = 6 (dollars)  = cost of a pen

And y = (7/12)(6)  = 42/12 =  3.5  dollars = cost of a pencil

So...the first time we spent

2(6) + 6(3.5) = $33

And the second time we spent

3(6) + 4(3.5)  = $32

2. An ammonia and water mixture fills a 5 gallon container. 80% of the mixture is ammonia but some of the mixture will be drained and replaced with pure water. If a 5 gallon mixture of fifty percent ammonia is desired, how many quarts of the mixture need to be drained before the water is added? (A quart is one-fourth of a gallon.) Express your answer as a decimal to the nearest tenth.

5 gallons  = 20 qts  ....and 80% ammonia   = 20% water

Let x be the number of quarts of mixture that we need to drain

So...when this is done we have    [20 -x] qts of 20% water

[20 (.20) - x(.20)]  =

[20 - x] .20

To this, we obviously need to add back  "x" quarts of  pure water  to get 20 qts of  50% water [ = 50% ammonia ]

So we have

[20 - x ] .20  +  x (1)  = 20(.50)  simplify

4  - .2x + x  = 10

.8x + 4  = 10    subtract  4 from each side

.8x + 6     dibide both sides by .8

6 / [8/10]  = x

60/ 8  = x

x  = 7.5

So...we need to drain 7.5 qts  of the 20%water soluiton...and then add back this many qts of pure water to get a 20 qt solution of 50% water

I think 98 kg for the 3rd one 

Here is my naive approach to this:
Weight of apples =99% moisture + 1%"F", for "fibre". Originally
After 2 days =98% moisture + 1%F =99 kg of their original weight.
The loss of 1 kg was solely due to moisture loss in the sun!!.

^{There was a flat containing boxes of apples having a total weight of 100 kg. An analysis showed that the apples were 99% moisture, by weight. After two days in the sun, a second analysis showed that the moisture content of the apples was only 98%, by weight. What was the total weight of the apples after 2 days, in kg?}

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My statistics professor often posed questions like this to reinforce the importance of paying attention to how a question presents data and correctly interpreting data that results from analysis.

Solution:

\(\text {The first analysis shows the apples are 99% water. The weight of the water is then}\\ \left(0.99\cdot 100\right) = 99 ~ kg\\ \text {Let x be the weight of the water lost after exposure to the sun. }\\ \left(0.99 \cdot 100-0.98(100-x)\right)=x\\ \begin{aligned}99-(98-0.98x)&=x\\99-98+0.98x&=x\\1+0.98x&=x\end{aligned}\\ \begin{aligned}1+0.98x-0.98x&=x-0.98x\\1&=0.02x\\x&=50\\\end{aligned}\\ \text { }\\ 100-x=100-50=50 ~ kg \leftarrow \text { The total weight of the apples after 2 days}\\\)

The correct solution appears paradoxical. It’s not, but it is counter-intuitive.

This question is known as the Potato Paradox. We genetically enhanced Chimps refer to it as the Banana Paradox.

GA