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Two sat math questions. Percentage and a graph

Two sat math questions. Percentage and a graph-example-1
User Hung Cao
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4 votes

Answer:

15) 300%

Explanation:

15)

Let the item originally cost n dollars.

The new incorrect price is (n-.6n).

So we want to find k such that (n-.6n)+k(n-.6n)=n+.6n

since we actually wanted it to be (n+.6n).

So we have (n-.6n)+k(n-.6n)=n+.6n

Distribute:

n-.6n+kn-.6nk=n+.6n

Subtract n on both sides

-.6n+kn-.6nk=.6n

We are trying to solve for k. So add .6n on both sides:

kn-.6nk=1.2n

Divide both sides by n:

k-.6k=1.2

.4k=1.2

Divide both sides by .4

k=1.2/.4

k=3

So 3=300%.

The incorrect price must be increased by 300% to get to the proper new price.

Here is an example:

Something cost $600.

It was reduce by 60% which means it cost 600-.6(600)=600-360=240

This was the wrong price.

We needed it to be increased by 60% which would have been 600+360=960.

So we need to figure out what to increase I wrong price 240 to to get to our right price of 960.

240+k(240)=960

1+k=960/240

1+k=4

k=3

So 240*300%+240 would give me my 960.

16) Speed=distance/time

In the first half hour, she traveled 5 miles (8:30 to 9).

In 1/3 hour she traveled (5-2)=3 miles (9 to 9:20).

We are told not to do anything where she stayed still.

In the last half hour, she traveled (2-0)=2 miles (9:30 to 10).

The average speed=
(5+3+2)/((1)/(2)+(1)/(3)+(1)/(2))=(10)/((4)/(3))=(10(3))/(4)=(30)/(4)=(15)/(2)=7(1)/(2).

User Morgoth
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