This is a blog created to the Etwinning Project Fun and Math. A project directed to mathematics, focused on funny and stimulating activities for children, enabling them to gain some interest and pleasure for this subject, but also explore their logical-mathematical thinking abilities and skills. With this project we intend to share challenges, problems, online games and tasks that allow children to improve their performance in Math.

Monday, 2 April 2012

Kieron's cats

Kieron has three cats. Each cat is a different weight. The first and second weigh 7kg altogether. The second and third weigh 8kg altogether. The first and 3rd weigh 11kg altogether. What is the weight of each cat?

Cat 1=5kg; Cat 2=2kg and Cat 3=6kg Step 1: In a single column assign weights 1-6kg for cat 1 (it cannot weigh 0). Step 2: In a column next to Step 1 calculate the possible weights of cat 2 by subtracting each weight of cat 1 from Step 1 from 7kg. Ex. if cat 1 is 1kg then cat 2 is 6kg; if cat 1 is 2kg then cat 2 is 5kg; if cat 1 is 3kg then cat 2 is 4kg; etc. Step 3: In a column next to Step 2 calculate the possible weights of cat 3 by subtracting each weight of cat 1 from Step 1 from 11kg. Ex. If cat 1 is 1kg then cat 3 is 10kg; if cat 1 is 2kg then cat 3 is 9kg; etc. Step 4: From the table just created, there is only one line where cat 2 plus cat 3 equals 8kg: If cat 1=5kg; cat 2=2kg and cat 3=6kg.

While it was not necessary in this particular puzzle, it is worth noting that cat 3 must weigh 4kg more than cat 2. If cats 1&2 weigh 7kg and cats 1&3 weigh 11kg since cat 1 is a constant, the difference between cats 2&3 must be the difference between the combined weights (11kg-7kg=4kg) Sometimes these kinds of observations are necessary to solve puzzles like this.

omg this is so easy..... ok so let the first, second and third be x,y,z respectively. if x and y is 9 and y and z is 10, that means z is one higher than x and can be written as (x+1) also if x+z=13, then:

x+(x+1)=13 2x+1=13 2x=12 x=6

and we know x+1=z so z is 7. using this we can solve for y with either two of the remaining equations.

very hard man

ReplyDeletecan someone help me on this

ReplyDeleteWe're going to post some solutions from our 3rd graders!

ReplyDeleteCat 1=5kg; Cat 2=2kg and Cat 3=6kg Step 1: In a single column assign weights 1-6kg for cat 1 (it cannot weigh 0). Step 2: In a column next to Step 1 calculate the possible weights of cat 2 by subtracting each weight of cat 1 from Step 1 from 7kg. Ex. if cat 1 is 1kg then cat 2 is 6kg; if cat 1 is 2kg then cat 2 is 5kg; if cat 1 is 3kg then cat 2 is 4kg; etc. Step 3: In a column next to Step 2 calculate the possible weights of cat 3 by subtracting each weight of cat 1 from Step 1 from 11kg. Ex. If cat 1 is 1kg then cat 3 is 10kg; if cat 1 is 2kg then cat 3 is 9kg; etc. Step 4: From the table just created, there is only one line where cat 2 plus cat 3 equals 8kg: If cat 1=5kg; cat 2=2kg and cat 3=6kg.

ReplyDeleteThis comment has been removed by the author.

ReplyDeleteWhile it was not necessary in this particular puzzle, it is worth noting that cat 3 must weigh 4kg more than cat 2. If cats 1&2 weigh 7kg and cats 1&3 weigh 11kg since cat 1 is a constant, the difference between cats 2&3 must be the difference between the combined weights (11kg-7kg=4kg) Sometimes these kinds of observations are necessary to solve puzzles like this.

ReplyDeletex + y = 7

ReplyDeletey + z = 8

x + z = 11

so x + y + 2z = 19

substitute (x + y ) with 7 (from first equation) so we got

7 + 2z = 19

z = 6

thus

y = 2

x = 5

easy

A + B = 7

ReplyDeleteB + C = 8

A + C = 11

2A + 2B + 2C = 26

A + B + C = 13

Therefore

If A + B = 7 then 7 + C = 13 so C = 6

If B + C = 8 then B + 6 = 8 so B = 2

And, A + C = 11 then A + 6 = 11 so A = 5

This is sooo easy.

ReplyDeleteAssume:

x + y = 9

y + z = 10

x + z = 13

Therefore, when we add x + y + y + z + x + z, we get:

2x + 2y + 2x = 32

Therefore, x + y + z = 16

Therefore, if x + y = 9, then z = 16 - 9 = 7

Therefore, if z = 7, then x + 7 = 13, x = 6

Therefore, if x = 6, then 6 + y = 9, y = 3

Therefore, x = 6

y = 3

z = 7

omg this is so easy.....

ReplyDeleteok so let the first, second and third be x,y,z respectively.

if x and y is 9 and y and z is 10, that means z is one higher than x and can be written as (x+1)

also if x+z=13, then:

x+(x+1)=13

2x+1=13

2x=12

x=6

and we know x+1=z so z is 7.

using this we can solve for y with either two of the remaining equations.

x+y=9

6+y=9

y=9-6

y=3

Respectively:

x=6

y=3

z=7