1)

You begin with 8 liters of a 15 percent solution. That means you have

15 percent acid and 85 percent water, right? Let’s figure out how

many liters of acid and water you have to begin with. 15 percent of 8

is 1.2 (.15*8) liters. So you’ve got 1.2 liters of acid and 6.8

liters of water (8-1.2 = 6.8). Now you want to add water until the

concentration is down to 10 percent. Note that the amount of acid will

remain constant..you’re only changing the amount of water. You want

1.2 liters of acid to be 10 percent of the entire quantity. Set up a

proportion to find out how much is 100 percent of the quantity.

10 percent 100 percent

———- = ———–

1.2 liters x liters

now cross multiply:

1.2 = .1*x

divide both sides by .1 and you get:

12 = x

So 12 liters is the entire amount of solution that makes 1.2 liters

10% of the concentration. Now read the problem. It asks how much

diltilled water must be added to make the concentration 10 percent.

You go from 8 liters to 12 liters so that’s 4 liters of water.

Number 2:

The family wants to save $120 over a period of two years. That means

they want to save $60 a year, right? The problem also states they want

to cut their yearly bill by 8 percent. Well, the two paragraphs above

tell you what you need to know…that $60 is 8 percent of the total

bill. Let’s set up a proportation and solve for the total bill:

60 x

——— = ———–

8 percent 100 percent

cross multiply:

60 = .08*x

divide both sides by .08

750 = x

The total yearly bill is $750.

2)

15×21,

greatest common divisor is 3, 5+7-1 = 11, 11×3 = 33, 33x 2 diagonals

= 66 tiles.(but 3 (GCD)common so answer = 63)

3) Solution: Let M stand for the man’s speed in mph. When the man

runs toward point A, the relative speed of the train with respect

to the man is the train’s speed plus the man’s speed (60 + M).

When he runs toward point B, the relative speed of the train is the

train’s speed minus the man’s speed (60 – M).

When he runs toward the train the distance he covers is 3 units.

When he runs in the direction of the train the distance he covers

is 5 units. We can now write that the ratio of the relative speed

of the train when he is running toward point A to the relative speed

of the train when he is running toward point B, is equal to the

inverse ratio of the two distance units or

(60 + M) 5

———– = —

(60 – M) 3

3(60 + M) = 5(60 – M)

3*60 + 3M = 5*60 – 5M

8M = 120

M = 15 mph

4)

I have seen problems like this before, but I have never taken the time

to try to finish working one of them out to the end. Having had some

free time since I first saw this question of yours, I have solved this

problem – not just your specific case, but the general case. Thanks

for sending the question; it gave me a lot of good (and enjoyable)

mental exercise.

I’ll present my solution roughly as I was able to figure it out. The

actual path I took to the solution was more convoluted than the

presentation below; I have cleaned things up a bit.

We start with

(1) a+b+c = 3

(2) a^2+b^2+c^2 = 5

(3) a^3+b^3+c^3 = 7

And we want to find the numerical value of

a^4+b^4+c^4 = ?

I first noticed that I could get an expression including the required

terms a^4, b^4, and c^4 by multiplying together either equations (1)

and (3) above, or by multiplying equation (2) above by itself. I

actually started down both paths more or less in parallel and chose

the latter path when it appeared to hold more promise than the former.

So we have

(a^2+b^2+c^2)^2 = (a^4+b^4+c^4)+2(a^2b^2+a^2c^2+b^2c^2)

and so

(a^4+b^4+c^4) = (a^2+b^2+c^2)^2 – 2(a^2b^2+a^2c^2+b^2c^2)

Then, substituting from equation (2), we have

(4) (a^4+b^4+c^4) = 25 – 2(a^2b^2+a^2c^2+b^2c^2)

Now, to get a numerical value for (a^4+b^4+c^4), we need to evaluate

the expression

(a^2b^2+a^2c^2+b^2c^2)

After some pondering, I determined that I could obtain an expression

including these terms by squaring the expression

(ab+ac+bc)

and that, in turn, I could obtain an expression including these terms

by squaring the given equation (1).

Note that I had no idea at this point whether this approach would lead

to expressions that I could evaluate using equations (1), (2), and (3)

– but, as you will see, it works out very nicely.

(a+b+c)^2 = (a^2+b^2+c^2)+2(ab+ac+bc)

and so

(ab+ac+bc) = [(a+b+c)^2 – (a^2+b^2+c^2)]/2

Then, substituting from equations (1) and (2), we have

(5) (ab+ac+bc) = (9-5)/2 = 2

Next

(ab+ac+bc)^2 = (a^2b^2+a^2c^2+b^2c^2)+2(a^2bc+ab^2c+abc^2)

and so

(a^2b^2+a^2c^2+b^2c^2) = (ab+ac+bc)^2 – 2(a^2bc+ab^2c+abc^2)

= (ab+ac+bc)^2 – 2abc(a+b+c)

Then, substituting from equations (1) and (5), we have

(6) (a^2b^2+a^2c^2+b^2c^2) = 2^2 – 2abc(3) = 4 – 6abc

And substituting (6) in (4), we now have

(a^4+b^4+c^4) = 25 – 2(4 – 6abc)

or

(7) (a^4+b^4+c^4) = 17 + 12abc

So now we can evaluate the desired expression (a^4+b^4+c^4) if we can

evaluate the expression abc.

When I got to this point, I realized I could get an expression

involving the term abc by multiplying equation (1) by itself three

times….

(a+b+c)^3 = (a+b+c)(a+b+c)^2

= (a+b+c)(a^2+b^2+c^2+2ab+2ac+2bc)

= a^3+ ab^2+ ac^2+2a^2b+2a^2c+2abc

+ a^2b +2abc+b^3+ bc^2+2b^2c

+2ab^2+2ac^2 + a^2c+2abc +2bc^2+ b^2c+c^3

—————————————————–

= a^3+3ab^2+3ac^2+3a^2b+3a^2c+6abc+b^3+3bc^2+3b^2c+c^3

= (a^3+b^3+c^3)+3(ab^2+ac^2+a^2b+bc^2+a^2c+b^2c)+6abc

and so

6abc = (a+b+c)^3 – (a^3+b^3+c^3) – 3(ab^2+ac^2+a^2b+bc^2+a^2c+b^2c)

Looking at this, I first tried grouping some terms…

6abc = (a+b+c)^3 – (a^3+b^3+c^3) – 3[a(b^2+c^2)+b(a^2+c^2)+c(a^2+b^2)]

and then, after some examination of this expression, I saw that I

could get clever by adding and subtracting 3(a^3+b^3+c^3) to the

expression on the right:

6abc = (a+b+c)^3 – (a^3+b^3+c^3) + 3(a^3+b^3+c^3)

– 3[a(b^2+c^2)+b(a^2+c^2)+c(a^2+b^2)] – 3(a^3+b^3+c^3)

6abc = (a+b+c)^3 + 2(a^3+b^3+c^3)

– 3[a(a^2+b^2+c^2)+b(a^2+b^2+c^2)+c(a^2+b^2+c^2)]

(8) 6abc = (a+b+c)^3 + 2(a^3+b^3+c^3)

– 3(a+b+c)(a^2+b^2+c^2)

Substituting from equations (1), (2), and (3), we have

6abc = 3^3 + 2(7) – 3(3)(5) = 27 + 14 – 45 = -4

and so

(9) abc = -4/6 = -2/3

Then, finally, substituting this in equation (7), we have

(a^4+b^4+c^4) = 17 + 12abc = 17 + 12(-2/3) = 17-8

and we finally have our result:

a^4+b^4+c^4 = 9

********************************************

After going through the algebra for your particular case, I went back

and worked out the general case:

(1) a+b+c = x

(2) a^2+b^2+c^2 = y

(3) a^3+b^3+c^3 = z

I will spare you the details of the algebra for this general case (if

you really love algebra, you might want to try to work it through for

yourself). I came up with the following expression for a^4+b^4+c^4:

a^4+b^4+c^4 = y^2 – 2[((x^2-y)/2)^2 – (x^4-3x^2y+2xz)/3]

I checked this result using the values from your problem. With x=3,

y=5, and z=7, we get

a^4+b^4+c^4 = 25 – 2[((9-5)/2)^2 – (3^4-3(3^2)(5)+2(3)(7))/3]

= 25 – 2[4 – (81-135+42)/3]

= 25 – 2[4 – (-12/3)]

= 25 – 2(4+4)

= 25 – 2(8)

= 25 – 16

= 9

You can also check the general result by choosing numbers for

a, b, and c. For example, if a = 1, b = 2, c = 3, then x = a+b+c = 6,

y = a^2+b^2+c^2 = 14, and z = a^3+b^3+c^3 = 36; the formula should

give us a^4+b^4+c^4 = 81+16+1 = 98.

a^4+b^4+c^4 = y^2 – 2[((x^2-y)/2)^2 – (x^4-3x^2y+2xz)/3]

= 196 – 2[((36-14)/2)^2 – (1296-3(36)(14)+2(6)(36))/3]

= 196 – 2[121 – (1296-1512+432)/3]

= 196 – 2[121 – 216/3)

= 196 – 2(121-72)

= 196 – 2(49)

= 196 – 98

= 98

5)

Let x be the number of stairs visible on the escalator when it is

stopped. Let r be the rate (number of stairs per second) at which the

escalator moves when it is running.

The woman walks down the escalator at one rate and runs back up the

escalator at a rate 5 times as fast; as we showed before, the time she

takes walking down is twice as much as the time she takes running back up.

Let t be the time (seconds) she takes to run up; then 2t is the time she

takes to walk down.

In walking down the escalator, the number of stairs she walks down is

equal to the number of stairs on the stopped escalator, minus the

number of stairs that the escalator moves in the time 2t; we are told

she walks down 10 steps. So we have:

x – (r)(2t) = 10 …………………………[1]

In running up the escalator, the number of stairs she runs up is equal

to the number of stairs on the stopped escalator, plus the number of

stairs that the escalator moves in time t; we are told that she runs

up 25 stairs. So we have:

x + (r)(t) = 25 ………………………….[2]

We can solve equations [1] and [2] simultaneously by getting the “rt”

terms to drop out:

x – 2rt = 10

2x + 2rt = 50

————–

3x = 60

x = 20

6)

Let’s forget about the total number of pearls for a moment, and assume

that we have one pearl, whose value is x. Then let’s add pairs of pearls,

following the formula given by the problem:

x

/ \

(x-100) (x-150)

Hw did u got the answer for question no.2 …related to tiles…..

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