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Let angle C =  x

Then angle B = 2x + 22

The angles sum  to 180....so....

A +  B + C  =180

86   + 2x + 22  + x = 180

3x + 108 = 180

3x = 180 - 108

3x =72

x = 72 / 3

x = 24 = angle C

Hi GA, have you considered getting a life instead of obsessing over forum trolls?

This answer is wildly incorrect. A constant function cannot return two different values.

Yep !!!!

Cphil!! You're back!!

Let the amount of money that you have at first =  X

Let the amount of money that your brother has = Y

So.....you give him 5......you have 5 less ad he has  5 more....so

X - 5  = Y + 5    →    X  = Y + 10      (1)

And after he gives you $25 you have 2  times as much as he does.... so.we have

X + 25  = 2 ( Y - 25)          (2)

Sub (1) into (2)

Y + 10 +25  = 2 (Y - 25)

Y+35  = 2Y - 50       rearrange as

50 + 35  =  2Y - Y

85  = Y        (he started with $85)

And you stared with Y + 10 = $95

Proof

You started with  95  he starts with 85

When you give him 5 you both have the same

If he gives you 25 you have 95 + 25 =   120   and he has   85 - 25 =  60   (you have twice as much !!! )

Assumig that the berries themselves lost no weight.....and I'm assuming that the "90%"  is 90% of the original water weight 

If the box of berries was 99%  water.....the weight of the water was  .99 * 200  =  198 kg

And the berries must weigh 200 - 198 = 2kg

If after two days, the water content was 90%, then the weight of the water = .90 * 200 = 180 kg

So.....the weight of the berries after two days =  180 + 2 =  182 kg

Using some trig

Note that cos (angle BDA)  =  -cos ( angle BDA)  = cos (angle CDA) 

 also  let  BD = CD  = x

and  5sqrt (2)  = sqrt (50)

Using the Law of Cosines (twice)

AB^2  =  x^2 + AD^2  - 2 (AD)(x) cos (BDA)

AC^2 = x^2 + AD^2 - 2(AD)(cos CDA)                simplify these

50 = x^2 + 53  - 2(AD)(X) cos (BDA)

70 = x^2 + 53 - 2(AD)(x) (-cos (BDA) 

50 = x^2 + 53 - 2(AD)(cos BDA)

70 = x^2 + 53 + 2(AD)( cos BDA)     add these equations

120 =  2x^2 + 106

120 - 106 = 2x^2

14 = 2x^2

7 = x^2

sqrt (7)  = x

So  BC  = 2x  =  2sqrt (7) = sqrt (  28)

Using Heron's formula to find the area

s = [ AB + BC + CD ] /  2  =  [ sqrt (50) + sqrt (28) + sqrt ( 70)  ] / 2

Area of ABC  =  sqrt [ s ( s - sqrt (50)) ( s - sqrt (28)) ( s -sqrt (70) ) ]

Using WolframAlpha to evaluate :

Area of ABC ≈   sqrt (346) ≈  18.6 units^2

Call the total amount of money that he has at first =  M

The backpack costs (3/7)M

The amount that remains = (4/7)M

(2/3)  of this =  (2/3)(4/7)M  = (8/21)M = money spent on purification bottles

Money spent of first aid kits = (M - 3/7 M - 8/21 M ) =   (4/21)M

Note that  the total difference between the cost of the purification bottles and the first aid  kits = 3 (14.80)

So

(8/21)M - (4/21) M = 3 (14.80)

(4/21)M = 44.40

M = 44.4 *(21/4)  ≈ $233.10

Amt spent on backpack ≈  (3/7)(233.10) ≈ $99.90

Amt spent on purification bottles = (8/21)(233.10) ≈ $88.80

I believe there is an error with the question.

If 6 more juniors join and 1 senior leaves, there is no way the ratio would go from j:s 3:2 to 2:3, because no juniors left and no seniors joined.

The total number of yards lost on the two plays is 5+2=7 yards lost.

I've had some  health problems.....thanks  for  thinking about me.....GOOD TO BE BACK!!!

Hi Chris.

I'm so glad to see that you're back.  We've all missed you, everybody has been asking about you.  

_.

Let Jim's salary be J.

Jim spends 3/5 of his salary on transport. J-3/5J=2/5J is left.

1/8 of his remaining salary is (1/8)*(2/5J)=2/40J=1/20J, which is how much he spends on food.

He gives 1/10J to his mother.

So J-3/5J-1/20J-1/10J=5/20J=1/4J left, and this amount is equal to $1325.

(a) Jim spends 1/20 of his salary on food ($265)

(b) Jim's salary is 4*1325=$5300.

Heck, Bosco.....you beat me to it  !!!

Here's an approach that only uses a little trial and error

Note that  999/9 = 111

So....let's assume that our number must have the form   1 0  __  ___  where the last two digits are unique

Note that successive multiples of 9 (after 999) are

1008

1017

1026 =  our number

What is the smallest four digit multiple of 9 with four distinct digits?    

I did a little bit of logic and a little bit of brute force.    

We want the lowest four digit number.  Well, it has to start with a 1.  So we've got that.

What second digit would be the lowest.  Obviously, 0.  So we've got 10 _ _.   

For the third digit, it can't be 0 or 1 so let's go with 2.  Now we have 102 _.

The fourth digit can't be 0 or 1 or 2.  It has to be 3 or larger. 

That's the end of the logic.  Now for the brute force. 

Tried 1023, not evenly divisible by 9.       

Tried 1024, not evenly divisible by 9.   

Tried 1025, not evenly divisible by 9.  

Tried 1026, Bingo.  It's 114.  So that's your number right there:  1026.  

_.

Let the number of appes transferred for A to B =  A

Let the number of pears transferred from A to B = P

Now  A contains (70 - A) apples and (100 - P) pears

So

Number of apples now in A / ( total fruits in A)  =  .40.....in equation form we have

(70 - A) / ( 170 - A - P)  = .40    simplify

(70 - A) =  .40 (170 - A - P)

70 -A = 68 - .40A -.40P

70 - 68 = A - .40A - .40P

2 = .60A - .40P     (1)

And in B....number of pears in B / (total fruits in B)  =  .30.....in equation form...

(20 + P)  / (110 + A + P)  = .30   simplify

20 + P =  .30 ( 110 + A + P)

20+ P  = 33 + .30A + .30P

33-20 = P -.30P - .30A

13 = .70P - .30A    (2)

Combining (1) and (2)  in a system  we have

.60A - .40P  =  2

.70P -.30A = 13       simplify

.6A - .4P =  2

-.3A + .7P =  13         multiply this equation through by 2

.6A - .4P =  2

-.6A + 1.4P  = 26       add these

P =  28 =  the number of pears transferred

Using either equation

.6A - .4(28)  = 2

.6A - 11.2 = 2

.6A = 2 + 11.2

,6A = 13.2

A = 13.2 / .6 =  22  =  the number of apples transferred

Proof    after   22 appples and 28 pears are transferred from A to B

( 70 - 22)  / ( 170 - 28 - 22)

48 / 120 =  40%  of the fruit in A  are apples

And in B

(20 + 28)  /( 110 + 22 + 28)

48 / 160  = 30% are pears