Total Sweet Distribution

Let $a$ be number of students in class A and $x$ be the number of sweets given per student in class A. Then, $ax = (x+3)(a-4)+10=(a-8)(x+8)+24$ . Expanding and subtracting $ax$ from each equation, we get $0 = 3a-4x-2 = 8a-8x-40$ . Solving this, we obtain $a=18, x=13$ . Required answer is $13\times 18+16\times 14+21\times 10 = 668$ .

[Latex edits - Calvin]

Define variables $N_A$ , $N_B$ , $N_C$ as the number of students in class A, B, C, respectively. Define variables $S_A$ , $S_B$ , $S_C$ as the number of sweets given to EACH student in class A, B, C respectively.

Then the total number of sweets given out is equal to $N_A S_A + N_B S_B + N_C S_C$ , and is what we are looking for.

The information in the problem yields the following 6 equations:

$N_A = N_B + 4$ ;

$N_B = N_C + 4$ ;

$S_C = S_B + 5$ ;

$S_B = S_A + 3$ ;

$N_A S_A = N_B S_B + 10$ (E1);

$N_B S_B = N_C S_C + 14$ (E2).

There are 6 unknowns, and 6 (independent) linear equations, so we can solve for all of the unknowns. One procedure for solving is as follows:

Convert all of the variables in the equations marked (E1) and (E2) into the variables for class B. Then (E1) becomes:

$(N_B + 4)(S_B - 3) = 10 + N_B S_B$

$N_B S_B - 3N_B + 4S_B - 12 = 10 + N_B S_B$

$-3N_B + 4S_B = 22$ (E3)

And (E2) becomes:

$N_B S_B = (N_B - 4)(S_B + 5) + 14$

$N_B S_B = N_B S_B + 5N_B - 4S_B - 20 + 14$

$5N_B - 4S_B = 6$ (E4)

Now add equations (E3) and (E4):

$2N_B = 28$

$N_B = 14$

Plugging this back into either (E3) or (E4) yields $S_B = 16$ .

Then we can use the first 4 equations above to find the remaining variables:

$N_A = N_B + 4 = 18$ ;

$N_C = N_B - 4 = 10$ ;

$S_C = S_B + 5 = 21$ ;

$S_A = S_B - 3 = 13$ ;

So our final answer (the total number of sweets) is

$N_A S_A + N_B S_B + N_C S_C = (18)(13) + (14)(16) + (10)(21) = 668$ .

p.s. Does the Align environment not work in your LaTeX interpreter?

let 'a' be the number of students in class A. let 'b' be the number of students in class B. let 'c' be the number of students in class C.

let 'x' be the number of sweets each student got in class A. let 'y' be the number of sweets each student got in class B. let 'z' be the number of sweets each student got in class C.

then total sweets distributed in class A will be a x. total sweets distributed in class B will be b y. total sweets distributed in class C will be c*z.

Now as per the question , a=c+8; b=c+4; c=c;

x=x; y=x+3; z=x+8;

b y + 10 = a x --------(1). c z + 14 = b y --------(2).

by solving this equation separatly we get (1) x=(3 c + 22)/4; c=(4 x - 22)/3;

(2) x=(5 c + 2)/4; c=(4 x - 2)/5;

On solving those to results we get, x=13; c=10; then substitute for the above equation. then the total of a x + b y + c*z = 668.

x = base # of students y = base # of treats Class A = (x+8)(y+1) Class B = (x+4)(y+4) Class C = (x)(y+9) Class A = Class B + 10 Class B = Class C + 14 Class A = Class C + 24 Thus (x+8)(y+1) = (x+4)(y+4) + 10 (x+4)(y+4) = (x)(y+9) + 14 (x+8)(y+1) = (x)(y+9) + 24 Solve for y and y = x+2 y = .75x+4.5 y = 1.25x-.5 Set two equations equal to each other x+2 = .75x+4.5 .25x+2 = 4.5 .25x = 2.5 x = 10 Then solve for y y = (10)+2 y = 12 Then plug back into original equations Class A = (10+8)(12+1) Class B = (10+4)(12+4) Class C = (10)(12+9) Class A + Class B + Class C = (18)(13) + (14)(16) + (10)(21) Or 234 + 224 + 210 = 668

let students in C= x so in B= x+4, in A= x+8 Now let each students of A get no. of sweets= y, in B = y+3, in C= y+8 (x+8)y-(x+4)(y+3)=10 (x+4)(y+3)-x(y+8)=14 Now, solve the two equations and get x=10, y=13 now total no. of sweets=(x+8)y+(x+4)(y+3)+(y+8)x putting the value of x and y, we get the result=668.

Let x+4 be the number of students in A, x in B, and x-4 in C. Let y-3 be the number of candies each student received in A, y in B, y+5 in C.

Multiplying the given to get the total number of candies, two equations are derived:

xy-3x+4x-12 = xy xy = xy+5x-4y-6

Solving the system, x = 14 y= 16

Substituting the answers to the given: There are 234 candies in A, 224 in B, and 210 in C. Therefore, a total of 668candies were distributed

I set that A (students of class A) =B+4 and B=C+4, therefore A=C+8; Then i set that As( sweets given to each class member of class A)=B-3 and Bs=Bs-5, therefore As=Cs-8. As As A, Bs B and Cs C are the numbers of sweets given rispectively to the whole class A, B and C, I know that As A=Bs B+10 and Bs B=Cs C+14 and I put Both equation in a system. Now I can substitute the values of A and B in funtion of C and the values of As and Bs in function of Cs optaining (C+8) (Cs-8)=(C+4) (Cs-5)+10 and (C+4) (Cs-5)=C Cs+14. Now i do a little bit of algebra and solve the equation of the sistem (as i'm not able to write them with the right sistem simbols i'm going to solve the first one first, than the second sobstituting C in function of Cs and at the end finding the value of C (after having found the value of Cs in the second equation). (C+8) (Cs-8)=(C+4) (Cs-5)+10 =(C+4+4) (Cs-8)-(C+4) (Cs-5)=10 =(C+4) (Cs-8-Cs+5)+4 (Cs-8)=10= (C+4) (-3)+(4) (Cs-8)=10= -3C-12+4Cs-32=10=-3C=54-4Cs=C=4/3Cs-18. Now the second equatinon: (C+4) (Cs-5)=C Cs+14= (4/3Cs-18+4) (Cs-5)=(4/3Cs-18) Cs+14=(4/3Cs-18) (Cs-5-Cs)+4(Cs-5)=14= -20/3Cs+90+4Cs-20=14=-8/3Cs=-56=Cs=21.If Cs=21, C=10, Bs=16 and As=13. If C=10, B=14 and A=18. The total number of sweets given is equal to As A+Bs B+Cs C, therefore 13 18+16 14+21 10=210+224+234=668.

The challenge here is to find a suitable (algebraic) representation for the given information. After that step, the rest is simply computational.

Let $A, B,$ and $C$ denote the number of students in classes A, B, and C, respectively, and let $a, b,$ and $c$ denote the number of candies each student in classes A, B, and C got, respectively.

We know the following according to the problem: $E_1: A = B + 4 \\ E_2: B = C +4 \\ E_3: c = b + 5 \\ E_4: b = a + 3 \\ E_5: Aa = Bb + 10 \ \text{and} \\ E_6: Bb = Cc + 14$

Our ultimate goal is to solve for $Aa + Bb + Cc.$ Using $E_5$ and $E_6,$ $Aa + Bb + Cc \\ = (Bb + 10) + (Cc + 14) + Cc \\ = [(Cc + 14) + 10] + (Cc + 14) + Cc \\ = 3Cc +38.$

So if we can solve for $C$ and $c$ , then we are done.

Now, we write the equations in terms of $C$ and $c. \\$ $E_1: A = C + 8 \\ E_2: B = C + 4 \\ E_3: b = c - 5 \\ E_4: a = c - 8 \\ E_5: Aa = Cc + 24 \ \text{and} \\ E_6: Bb = Cc + 14$

Using $E_1$ to $E_4$ , we can plug-in the values of $A, B, a,$ and $b$ to $E_5$ and $E_6$ to get: $E_5 : (C+8)(c-8)=Cc +24 \ \text{and} \\ E_6: (C+4)(c-5)=Cc +14$

Simplifying, we get $E_5: Cc - 8C + 8c -64 = Cc +24 \\ -8C + 8c = 88 \\ -C + c = 11 \\ \\ E_6: Cc - 5C + 4c -20 = Cc + 14 \\ -5C + 4c = 34$

We get this system of equations: \{ \begin{array}{2 1} - C + c =11 \\ -5C + 4c = 34 \\ \end{array}

Solving the system, we get $C = 10$ and $c=21.$ Finally, $Aa + Bb + Cc. \\ = 3Cc +42 \\ = 3*10*21+38 \\ = 668$

                           A       B    C

no.of students x+4 x x-4 choc to individual y-3 y y+5 total chocs(A) =(x+4)(y-3) total chocs(B)=xy total chocs(C)= (x-4)(y+5) (x+4)(y-3)=xy+10;
xy= (x-4)(y+5)+14; solving both equations x=14,y=16 total chocs= total chocs(a)+total chocs(b)+total chocs(c); total chocs=668

Let Class B have $p$ students and let each student have $q$ sweets. Thus Class A has $p+4$ students and each student has $q-3$ sweets and Class C has $p-4$ students and each student has $q+5$ sweets each.

The last two statements in the question are equivalent to
$(p+4)(q-3) - pq = 10$ $\Rightarrow 4q-3p = 22$ and $pq - (p-4)(q+5) = 14 \Rightarrow 4q-5p=-6$ .
Subtracting the two equations we have $2p = 28 \Rightarrow p=14$ .
Substituting $p=14$ into the first equation gives $q=\frac{22+3(14)}{4} = 16$ , so Class B got $pq = 14(16) = 224$ sweets.

Hence the total number of sweets given out is $(pq + 10) + pq + (pq-14) = (224+10)+224+(224-14)= 668.$