Several Christmas Presents

Let $A$ , $B$ and $C$ denote Alan, Brian and Charlie respectively.

Then,

$A + B = 18$

$B + C = 14$

$A + C = 22$

Summing the above 3 equations, we obtain

$2(A + B + C) = 54$

$A + B + C = \boxed{27}$

Let the presents received by Alan, Brian and Charlie be A,B and C respectively.

So, $A+B=18 , B+C=14 , C+A=22$

Adding all three, we get, $A+B+C+A+B+C=18+14+22 \implies 2(A+B+C)=54 \implies A+B+C=\boxed{27}$

So, the total presents received $=A+B+C=\boxed{27}$

Assumptions :

1) Alan - "A"

2) Brian - "B"

3) Charlie - "C"

Equation 1 : B + C = 14 -----> Sub-Equation 1------> B = 14 - C

Equation 2 : C + A = 22 -----> Sub-Equation 2 -----> C = 22 - A

Equation 3 : A + B = 18 -----> Sub-Equation 3 -----> A = 18 - B

Comparing Sub-Equations 1 & 2,

Sub-Equation 1 -----> B = 14 - ( C )

Sub-Equation 2 -----> ( C = 22 - A )

Substitute Sub-Equation 2 in Sub-Equation 1,

Hence We have got New Equation from Sub-Equation 1 and 2 as "B = 14 - (22 - A)" ----> i.e., "B = - 8 + A"

Comparing the New Equation and Equation 3,

New Equation -----> ( B = - 8 + A )

Equation 3 -----> A + ( B ) = 18

Substitute New Equation in Equation 3,

Hence we get the value of "A" as below,

A + ( B ) = 18

A + ( - 8 + A ) = 18

A - 8 + A = 18

2A - 8 = 18

2A = 18 + 8

2A = 26

A = 26 / 2

" A = 13 "

Substitute " A = 13 " in Equation 3 to get the value of "B" as below

Equation 3 -----> ( A ) + B = 18

                          ( 13 ) + B = 18

                                        B = 18 - 13

                                      " B = 5 "

Substitute " B = 5 " in Equation 1 to get the value of "C" as below

Equation 1 -----> ( B ) + C = 14

                            ( 5 ) + C = 14

                                       C = 14 - 5

                                      " C = 9 "

So , A = 13

B = 5

C = 9

So, A + B + C = 13 + 5 + 9

   A + B + C = 27

So, Your 3 Nephews Alan, Brian & Charlie would have received 27 presents altogether, assuming that they all telling the truth.

Since Brian and Charlie got 14 presents, Charlie and Alan got 18 presents, and Alan and Brian got 18 presents, this means that $2(A+B+C)=54$ . By dividing 54 by 2, you get your answer of 27.

Let x,y,z be the 3 nephews brian,charlie and alan respectively from 1st dialogue x+y=14 from 2nd y+z=22 from 3rd z+x=18 If we combine these 3 cases we will get x+y+y+z+z+x=14+22+18 implise 2(x+y+z)=54 implise X+Y+Z=54/2=17 therefor the answer is 17.

a + b = 14, b + c = 22, c + a = 18.

Combine all the three equations, we will find that 2 (a + b + c) = 54 => a + b + c = 27.

2(B+C+A)=54 So they all receive 27 presents together

Alan (a) +Brian (b) =18; Charlie (c) + a = 22; b + c = 14; Solve this three equation three variable and get a=13;b=5;c=9

B + C = 14 C + A = 22 A + B = 18

2C + (A + B) = 36 2C = 18 C = 9

Solve by substituting back in first 3 equations again.

Let Alan = A Brain = B and Cahrlie = C B + C =14 C + A = 22 A + B = 18 First of all we add these three groups (B+C) +(C+A) +( A+B) =14+22+18 2(A+B+C) =54 Dividing by 2 both sides A +B +C =27

If we add all of the sums you will get 54 which is two times Alan's, Brian's and Charlie's presents. So you have to divide 54 by 2 and you'll get 27 which is the answer.

let brian be b,

charlie be c,

and alan be a

b+c=14

c+a=22

a+b=18

2a+2b+2c=54

2(a+b+c)=54

a+b+c=27

Charlie and Alan have more presents than Alan and Brian, which means that Charie will have more presents than Brian, because Alan is the constant. Therefore Brian will have presents less than 7. Input values below 7 in the equations:

B + C = 14

C + A = 22

A + B = 18

Until you find a value which satisfies all three. Example,

Take B= 5

C = 14 - 5 = 9

A = 22 - 9 = 13

13 + 5 = 18

hence proved.

Lets take, Alan, Brian and Charlie = X+Y+Z

X+Y=14

Y+Z=22

X+Z=18

2(X+Y+Z) = 14 + 22 + 18 = 54

(X+Y+Z) = 54 / 2

(X+Y+Z) = 27

Let's take A for Alan , B for Brian and C for Charlie

Now from the information we can form equations like B+C=14 ------- (1) c+A=22 ------- (2) A+B=18 ------- (3)

from the (1) equation C = B-14

now put the C value in (2) there by you will get the A-B=8 -------(4)

now solve (4) & (3) you will get A=13 , B= 5 and put the b value in (1) then you will get C=9 Sum up all the value asnwer so that you will get 27

First, I figured out the possibilities of the number of presents that Brian and Charlie could receive that would equal 14 presents. Then, I figured out the possibilities of the number of presents that Charlie and Alan could receive that would equal 22 presents. Doing this, I could figure out how many presents Charlie got, by seeing which number appeared in both lists. When I figured out what Charlie had, I just subtracted it by 22 and figured out what Alan had. Then, I figured out how many Brian had by subtracting what Alan had by 18. When I was done, I added together the number of presents that each nephew got and my final answer was 27.

Let Alan be a, Brian be b and Charlie be C. From the Question, b+c=14, a+c=22, a+b=18. Add the 3 statements together, you get 2a+2b+2c=54. So a+b+c=27.

let gifts received by brian=x gifts recieved by charlie=14-x and gifts received by alan=y now we know 14-x+y=22 and y+x=18 solving for x and y gives u answer.

ans:27

Alan says Brain+Charlie = 14, Brain says Charlie+Alan =22, Charlie says Alan +brian =18, BY ADDING all three =54, so 2Alan+2Brain+2Charlie =54, 2(Alan+Brain+Charlie) =54, Alan+Brain+Charlie =54/2=27

Let x,y,z are the numbers of parents recieved by Brain, Charlie and Alan individually... so,x+y=14; y+z=22; z+x=18; eq(2) - eq(3) y-x=4; eq(1)+eq(4) 2y=18; y=9 then x+y+z=18+9 =27 [from eq(3)] so total is 27

Let Alan=A; Brian=B; Charlie=C. Using the trial and error... A+B=18. B+C=14. C+A=22. And I get A=13 B=5 C=9. And then add.

A+B=18 A+C=22 B+C=14 by solving, A=13 B=5 C=9 hence A+B+C=27

PRESENTS RECEIVED BY ALAN BE [A] , PRESENTS RECEIVED BY BRAIN BE [B] , PRESENTS RECEIVED BY CHARLIE BE [ C] , GIVEN A+B =18,B+C=14,C+A=22..............THEN 2 (A+B+C)=54..................A+B+C=27

Let Brian's presents are b , Charlie's presents c , Alan's presents a b+c = 14 c+a = 22 a+b =18 By adding up these we get (b+c) + (c+a ) + (a+b) = 14+22+18

2 (a+b+c) =54

a+b+c+27

suppose brian and charlie (x+y) charlie and alan (y+z) and alan and brian (z+x). so we have x+y=14 y+z=22 z+x=18 by adding all of them x+y+y+z+z+x=14+22+18 2x+2y+2z=54 x+y+z=54/2 x+y+z=27

In the first Part , B(brian) + C(charlie) = 14 ----> B = x, C = 14-x (eq.1)

In the second Part, C(charlie) + A(alan) = 22 ----> C = x, A = 22-x (eq.2)

In the third Part, A(alan) + B(brian) = 18 (eq.3) -----> A = x, B = 18-x

To find the total number of presents, we need to find how many presents each has received(induvidually).

Lets start with Brian. According to eq.1, B = x. According to eq.3, B = 18-x.----> x = 18-x. Therefore 2x=18, x =9. BRIAN GETS 9 GIFTS

Now, lets go to Charlie. According to eq.1, C = 14-x.According to eq.2, C =x. ---> 14-x=x. 2x = 14. x= 7.CHARLIE GETS 7 GIFTS

And finally, Alan. According to eq.2,A = 22-x and eq.3 says A =x.--->22-x=x. 2x = 22. x=11. ALAN GETS 11 GIFTS

To find the total, simply add the number of gifts each gets induvidually, ie, 9+11+7=27 TOTAL NUMBER OF PRESENTS = 27

b+c=14------1 c+a=22------2 a+b=18------3

from 1, c=14-b

from 2, c=22-a

14-b=22-a -b+a=8

a=8+b (8+b)+b=18 2b=10 b=5

a=8+5 a=13

c=14-5 c=9

a+b+C= 27

First we assume alan as x, then brian as y and charlie as z, the accorindly we sort out what ech of them says and we get 3 two variable expressions i.e, y+z=14; x+z=22; & x+y=18. Then we say that let z=14 - y and put that value in second eq. and thus we solve 2nd and 3rd eq. getting the value of x and finding out the rest

Assume A for Alan, B for Brian and C for Charlie, then A says B+C=14 B says C+A=22 C says A+B=18 2(A+B+C)=54 A+B+C = 27 so they received 27 presents altogether.

assume alan as a ,brian as b, and charlie as c. alan says ........................b+c=14 brian says.......................c+a=22 charlie says....................a+b=18 sum all of them comes 2a+2b+2c+54 take 2 common 2(a+b+c)=54 then a+b+c=54/2 a+b+c=27 which is the answer

A + B = 14

B + C = 22

A + B = 18

Then

A - B = 6

A + B = 18

2A = 26

A = 13

B = 5

C = 9

A + B + C = $\boxed{27}$

It's simple Add all the numbers = 14+22+18 = 54 Divide the result by 2 = 54/2
= 27