Animals on Scales

Suppose the weight of the cat be $x$ , weight of the rabbit be $y$ and weight of the dog be $z$ :

Hence according to the given figure,

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

$\Rightarrow 2x+2y+2z=54$

$\Rightarrow 2(x+y+z)=54$

$\Rightarrow x+y+z=27$ .

Hence the total weight of all the three animals is 27 kg. $\square$

x=(weight of cat); y=(weight of rabbit); z=(weight of dog)

From the figure,

x+y=10

z+y=20

x+z=24

Adding,

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

2x+2y+2z=54

2(x+y+z)=54

x+y+7=27

i.e (weight of cat)+(weight of rabbit)+(weight of dog)=27

Apparently, I took the less beaten path of finding each animal's weight... the other solution is way simpler, but still, here it is.

• b + c = 10
• b + d = 20
• c + d = 24

First thing to notice is that comparing the weigh-ins with the dog, the cat weighs 4 kg more than the bunny, so...

• c = b + 4
• plug it in to the first equation
• 2b + 4 = 10
• -4 on both sides
• 2b = 6
• /2
• b = 3
• plug it back in to the first equation
• 3 + c = 10
• -3
• c = 7
• 3 + d =20
• 7 + d = 24
• d = 17
• 3 + 7 + 17 = 27

Add all the three total is 54. U have counted all the animals twice so half is 27 & that is d answer.

If you plus the Dog and rabbit(20kg) + the dog and the cat (24kg) = adds to 44kg now u know that cat and rabbit weights 10kg together. Minus that 10 kg from 44kg gives u 34kg 2 dogs left. Now u want to find only one of the dog means u ÷ it by 2 = 17kg one dog

So u can work out that the: Rabbit weights 3kg Cat weights 7kg Dog 17kg

Add all of them total is 27kg

add the weights of the two two animals (cat+rabbit=10) (dog+rabbit=20) so that total weight of a cat, a dog n 2 rabbits become 30 ( 10+20=30) as the third picture says the weight of a cat+dog= 24 so we deduct 24 from 30 (30-24=6) thus 6 will be the weight of TWO RABBITS as these rabbits are identical they must have the same weight so when we divide 6 between 2 it is 3 of each. now for the fourth picture we know that the weight of the cat n dog is 24 n if we add 3 (the weight of rabbit ) it becomes 27. that's it.

suppose, Cat =C

Dog=D

and Mouse=M

now, C +M =10 .............(1)

and, D + M=20...................(2)

subtracting (1) from (2) we get, D-C =10................(3)

again, D+C =24..................(4)

from ,(3)+(4), we get, 2D =34, or, D =17

co, C= 24 -17 =7 and M =10-7=3

now, M+D+C =3+17+7=27kg

Cat add Bunny is 10. Cat is 7.
Bunny is 3.

7 add 3 is 10.

bunny add dog is 20. If bunny is 3,add on 17 for the dog.

3 add 17 is 20.

Dog add cat is 24 because 17 add 7 is 24.

All them together is 27. All you need to do is add the 3 for the bunny onto the cat and dog. 24 add 3 is 27.😊

My solution is to try to find the rabbit's weight, and then add it to the cat and dog's weight. If we add the first row of animals' weight, the sum is 30, and the third one is 24kg. As you can see, the two on top has two more rabbits than the third one, which has no rabbits. 30-24=6, so the two rabbits are 6kg. now we divide 6 by two because there are two rabbits and the answer is three. We add the rabbit's weight to the cat and dogs', and we get 27.

dog + cat= 24, dog+ rabbit= 20, so cat is 4 kg heavier than rabbit. cat+rabbit= 10 kg, so cat= 7, dog= 17, rabbit=3, so the solution is 27

• Cat + Rabbit = 10KG ⇥ ①
• Dog + Rabbit = 20KG ⇥ ②
• ②-① = Cat + Rabbit -Dog - Rabbit = Cat - Dog = 10 ⇥ ③
• Cat + Dog = 24 ⇥ ④
• ③-④ = Cat + Cat + Dog - Dog = 2Cat = 14
• Cat = 7
• Dog = 24 - 7 = 17
• Rabbit = 20 - 17 = 3

Suppose the weight of the cat is $x$ (kg), the weight of the rabbit is $y$ (kg) and the weight of the dog is $z$ (kg).

From the given figure, we have:

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

$\Rightarrow$ $2x + 2y + 2z = 54$

$\Rightarrow$ $2(x + y + z) = 54$

$\Rightarrow$ $x + y + z = \frac {54}{2}$

$\Rightarrow$ $x + y + z = 27$

So the total weight of all the three animals is $\color{#69047E}\boxed{27 kg}$

(10+20+24)/2

Picture

Let $a$ is a dog, $b$ is a rabbit & $c$ a cat

$a+b=10,b+c=20,a+c=24 \implies a+b+b+c+c+a=10+20+24=2(a+b+c)=54 \implies a+b+c=\boxed{\large{27}}$ (kg)s

We can add the weights of the first three scales . so 2 dogs + 2 cats + 2 bunnies will have have a weight of 54 . So 1 dog + 1 cat + 1 bunny will be 27

A simple way to solve these problems is to simply add together the three total weights and then divide by 2.

An important thing to note: the kilogram is a unit of mass, not weight.

You already know that the mass of the cat and the dog is 24kg, so all you really need to figure out is the mass of the rabbit.

Let $C$ , $D$ , and $R$ be the masses of the cat, the dog, and the rabbit, respectively.

According to the diagram,

$C + R = 10$

$D + R = 20$

Therefore, $C + D + 2R = 30$

Given $C + D = 24$ from the third part of the figure,

$C + D + 2R = 30$

$\Rightarrow 24 + 2R = 30$

$\Rightarrow 2R = 6$

$\Rightarrow R = 3$

Adding $R$ to $C + D$ will yield 27kg, the sum of the animals' masses.

let r be the weight of rabbit, d be the weight of dog, and c be the weight of cat.

the given equations show:

c+r=10

d+r=20

d+c= 24

Note that:

d=20-r

d=24-c

Meaning,

20-r=24-c

Also note that c+r=10 which gives us c=10-r

So, by substitution,

20-r=24-c

20-r=24-(10-r)

20-r=24-10+r

20-14=2r

6=2r

$r=3$

d=20-3= $17$

c=10-3= $7$

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

C+r=10.  C= r+4 R+4+r=10 2r+4=10 2r=6 R=3 C=3+4=7 D=17

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

So if we : C+R+C+D = 10+24 2C+R+D = 34

We add minus both parts -R + (-D) = 20 So: 2C+R+D-R-D= 34-20 2C=14 2C/2 = 14/2 C =7 So:R+7 =10 R= 3 D=17 C+R+D= 7+3+17 = 27

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

y=10-x, z=20-x,

10-x + 20-x = 24, -2x + 30 - 24 =0, -2x + 6=0, -2x=-6, x=3, y=7, z=17, so x + y + z = 27

Solving simultaneously by letting Rabbit = r, Dog = d and Cat = c

c + r = 10 (i)

d + r = 20 (ii)

d + c = 24 (iii)

Now from equ. (i) and (ii) solving for r gives : 10 - c = 20 - d

☞ d - c = 10 solving this with equ.(iii) simultaneously results Cat = 7 kg, Rabbit = 3 kg and Dog = 17 kg

total weight = r + d + c = 27 kg

dog+rabbit= 20 dog+cat=24 (dog+cat)-(dog+rabbit)=4 So, Cat-Rabbit= 4.. Cat+Rabbit= 10, so 7+3 then dog is 17, rabbit 3, cat 7

x+y=10 y+z=20 x+z=24 Solving x,y,z we get x=7 y=3 z=17 x+y+z= 27

Let,

Weight of Rabbit = $R$

Weight of Cat = $C$

Weight of Dog = $D$

Given,

$C+R$ $= 10$ $......(i)$

$D+R$ $= 20$ $......(ii)$

$D+C$ $= 24$ $......(iii)$

Adding all equations we get,

$2(C+D+R) = 54$

$C+D+R = \frac{54}{2}$

$C+D+R = 27$ = $Answer$