How do you solve these linear math problems?

Question

Problem number 1: A number consists of two digits followed by the digit 4. A second number starts with the digit 4 followed by the same two digits as used in the first number.

The equation for this problem would be 10x + 4 (first number) and (400 + x) - 400 (second number).

So the question is, if the second number is as much greater than 400 as the the first number is smaller than 400, what is the first number?

Problem number 2: Sea water contains 5% of salt (by weight). How many kg of fresh water should be added to 40 kg of sea water for the latter to contain 2% salt?

Problem 3: The length of a rectangle is increased by 20% while its with by 10%. What percentage increase is there in its area?

Problem 4: Emily wishes to mix two types of tea. Type A costs $0.36 per kg while type B costs $0.50 per kg. She then sells the mixture at $0.46 per kg and makes a profit of 20% of the total cost. What percentage of the total mixture is made up of type A?

Answer 1

Problem 1:

First number as you said is 10x+4 = A
Second number is however 400 + x = B

The word problem says:
B - 400 = 400 - A

so ...

(400 + x) - 400 = 400 - (10x + 4)

x = 396 - 10x
11x = 396

x = 36

CHECK:
First number is 364
Second number is 436

364 is 36 LESS than 400
436 is 36 MORE than 400


Problem 2:

40 kg of sea water contains both salt and Fresh Water

We know that S + W = 40 (ie total of water and salt is 40 kg)
and that S / 40 = 0.05 (ie ratio of salt to 40 kg is 5% or .05)

So S = 40 * .05 = 2
and W = 38

(or we have 2 kg of salt, and 38 of water)

Now we want to have 2% salt in the final mix. We have 2 kg of salt and will be adding NO MORE.

So we know the total weight of the mixture will be 100 kg (since 2 kg is 2% of 100 kg)

We currently have a total weight of 40 kg, so will be adding
60 kg of fresh water.

Problem 3:

Assume the initial dimensions of the rectangle are L by W

This means the area would have been LW

The new length is 1.20 L and the new width 1.10 W

So the new area would be 1.2 L * 1.1 W or
1.32 LW

This new area is 32% larger than the original


Problem 4:

Assume that that we use a kg of type A tea, and b kg of type B tea.

We know that
a + b = 1 (since we sell one kg of tea)

The cost of type A tea in this blend is .36 * a
and of type B tea is .50 * b

for a total cost of making this tea of (.36a + .50b)

Emily though makes 20% profit

0.46 = 1.20 * (.36a + .50b)
0.46 = .432 a + .60 b

We now have two equations:

a + b = 1
and
0.46 = .432 a + .60 b

Rearranging the first gives
a = 1 - b

Substituting for a in the second gives
0.46 = .432 (1 -b) + .60 b
or
0.46 = .432 - .432 b + .60 b
or
0.46 = .432 + .168 b

.168 b = .46 - .432
b = 1/6
so a = 5/6

Answer 2

100h +10t +4 is the first number
400 +10h+t is the second number
So 400+ 10h+t -400 = 100h +10t+4 -400
10h +t = 100h+10t-396
-90h -9t = -396
10h +t = 44
h = (44-t)/10t = 4 and h= 4 are the only solutions
so the number is 444.

40 +60 = 100 2/100 = .02 = 2%
Thus 60 kg pure water must be added

1.1*1.2 = 1.32 = 32%

x = amount of type A
Then 1-x = amount of type B
So .36x + .5(1-x) = .46
.36x+.5 -.5x = .46
-.14x = -.04
x = .286 Kg

Answer 3

y = -11 skill y coordinate will proceed to be -11 for all values of x that enables you to acquire an horizontal line this is parallel to x axis and perpendicular to y axis its slope would be 0 and y intercept would be (0,-11) further x=3 mean x coordinate will proceed to be 3 for all values of y so will acquire vertical line parallel to y axis and perpendicular to x axis its slope infinity x intercept is(3,0)