Can You Answer In One Minute Or Less?

Relevant wiki: System of Equations - Substitution

Let $R$ = weight of the rabbit, $D$ = weight of the dog, and $C$ = weight of the cat. The pictures then give us the following equations: $R + C = 10 \\ R + D = 20 \\ C + D = 24$ If we add all three equations, we obtain $2R + 2C + 2D = 54 \longrightarrow R + C + D = \boxed{27}.$

$x+y=10$
$y+z=20$
$z+x=24$

Adding all three equations gives $2(x+y+z) = 54,$ so $x+y+z=27.$

Form some simultaneous equations: r+c=10 r+d=20 and c+d=24. Adding these together gives 2c+2r+2d=54, halving gives the weight of all three, namely 27kg.

• $A$ = Rabbit
• $B$ = Cat
• $C$ = Dog

Here we turn the question into maths

$A + B = 10$

$A + C = 20$

$B + C = 24$

We then rearrange the first two equations to give us A on one side

$A = 10 - B$

$A = 20 - C$

Make them equal each other

$10 - B = 20 - C$

We then arrange the equation so that variables and numbers are on different sides - then work through

$C - B = 10$

$B = C - 10$

$B = 24 - C$

Here we use the the first equation to compare against

$C - 10 = 24 - C$

$2C = 34$

$C = \boxed{17}$

Now we just substitute for A and B

$A + C = 20$

$A + 17 = 20$

$A = \boxed{3}$

$A + B = 10$

$3 + B = 10$

$B = \boxed{7}$

$A = 3$

$B = 7$

$C = 17$

Add them together...

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

At first, I was thinking you'd solve this mathematically where you create three equations with 3 variables and solve fore each with substitution. Then I realized it'd be easier to add all the total weights and divide by 2 since that weight is the weight of 2 of each animal.

I used more logic than actual math workout for this one:

By comparing the weight between D + C (24) and D + B (20), I knew that C = B + 4. Then I rewrote C + B = 10 into 2B + 4 =10, solving for B (3). Finally, I added B to C + D in order to get C + D + B (24 + 3 = 27).

a+b=10, b+c=20, a+c=24; b=10-a, c=24-a; 10-a+24-a=20, a=7; 7+b=10, b=3; 3+c=20, c=17; a+b+c=27.

let we name these 3 animal x=cat, y=rabbit,z=dog

     X+Y=10.....................................(1)
     Y+Z=20.....................................(2) 
     Z+X=24......................................(3)

then X+Y+Z=? let, X+Y=10 then X=10-Y substitute this in eqn3 Z+X=24 Z+(10-Y)=24 Z-Y=24+10=14 eqn2, Y+Z=20 -Y+Z=14 cancel y and we get 2Z=34 then Z=17 substitute Z=17 in eqn2 Y+17=20 then, Y=3 we found that Y=3, Z=17 substitute Y=3 in eqn1 X+3=10 then X=7 then X=7 Y=3 Z=17 X+Y+Z=27

Add the all the given weights to get 54. This means that two of each on a scale weighs 54 kg, and divide by 2 to get that one of each on the scale weighs 27 kg

C + R = 10 D + R = 20 C +D = 24

double the middle equation: 2D + 2R = 40 Add the first equation: C + R = 10 Subtract 3rd equation also: -C -D = -24

These get you D + 3R = 26 subtract 2nd equation -D - R = 20

Therefore 2R = 6 and R=3

From there c = 7 and d = 17 and the grand total is 27! QED

by implementing each animal as variables....we observe: figure 1:x+y=10 figure 2:y+z=20 figure 3:x+z=24 solving the above equations, we get: x=7 y=3 z=17 in the last figure,we observe,x+y+z=? and its 27

2 dog+2 cat+2*rabit=54
2(dog+cat+rabit)=54
dog+cat+rabit=27

keep it simple just see all the animals come twice; so divide the addition of all by 2 therefore [20+10+24] /2 =27!!!

Assume the cat weighs 7kg and therefore the rabbit is 3kg in scale number 1, hence in scale number 2 the dog is 17kg because the rabbit is 3kg. In scale number 3, the dog which is 17kg plus the cat which was assumed at 7kg totals 24kg putting their sum at scale number 4 to 27kg over all because the cat is 7kg, the rabbit is 3kg and the dog is 17kg!

There 3 animals in the picture - cat, rabbit, and dog Let c = cat r = rabbit d = dog

Now let's write a system of equations according to the scales c + r = 10 d + r = 20 c + d = 24

When we subtract c + r = 10 and d + r = 20, we get d - c =24 When we subtract d - c =24 from the 3rd equation we get 2d = 34 So d = 17 So you can plug d = 17 into the other equations, and you done!

c=cat

r=rabbit

d=dog

Now for the maths:

c+r=10

d+r=20

c+d=24

To work it out:

24-20=4

Which means:

Therefore c=r+4

So:

c+r=2r+4=10

10-4=6

6/2=3

We now know that r=3

And from that we can figure out:

10-3 =7 so c=7

20-3=17 so d=17

so the answer is:

c+d+r=3+7+17=27

Since the three animals has different weights and appears at least 3 times (including the 4th scale!), I will start with the first scale.

Labels:

c-cat

r-rabbit

d-dog

7+3=10 (c+r)

17+3=20 (d+r)

7+17=24 (c+d)

And we know the cat is 7 kg, the dog is 17 kg and the rabbit is 3 kg. Now add it up.

17+7+3=27

So the total weight of the animals is 27 kg.

Even simpler, each animal is on two scales. This means that the weight of each animal is counted twice. If you add up all the known weights, you get 54 kg. This is two bunnies, two cats, and two dogs. To find the mass of just one, divide by two and you get 27 kg.

10+20+24 = 54

54/2 = 27 --> Answer

Here's what I did:

1.) Add the Dog and the Bunnys weight (20kg) to the Dog and the cat's weight (24kg) to get 44. So far you've counted the dog twice, the bunny once and the cat once.

2.) I'm trying to get the dog weight by itself so next I subtract the cat and the bunny weight (10kg) from 44. You get 34 but this number represents having counted the dog twice. So we divide by 2 to get 17. Now we have the dogs weight alone.

3.) 17 + 10 = 27

add first two scales 10+20=30=1cat+1dog+2rabbits given 1dog+1cat=24; subtract we get 6=weight of two rabbits...so add 3 to 24(weight of cat nd dog) WE CAN CALCULATE IN OUR MIND BY THIS METHOD.

x = cat y = rabbit z = dog

x+y = 10 x+z = 24 z+y = 20

required : x+y+z = ?

substitute the following equations

x + y + 0 = 10 x + 0 + z = 24 0 + y + z = 20

solve for x+z first

(x + y + 0 = 10) - 1 0 + y + z = 20

-x - y + 0 = -10 0 + y + z = 20 cancel out y

new equation is

-x +z = 10 x + z = 24

2z = 34

z = 17

y = 20 - 17

y = 3

x = 24 - 17

x = 7 x + y +z = 17 + 3 + 7

answer = 27

If B+D =20 and C+D =24. Then C = B+4. Therefore if B = X, then C = X+4.
X+X+4 = 10.
2X+4 =10.
2X =6.
X= 3.
C = 3+4 =7.
D+3 = 20. D = 17.
B+C+D = 3+7+17 = 27