Sign Up!
Login
Welcome to Test-paper.info
Sunday, November 19 2017 @ 11:34 AM CST
 Forum Index >  Test Paper Related >  Primary 6 Matters
 maths problem sum p6
   |  Printable Version
By: thayal (offline)  Sunday, February 19 2012 @ 07:37 AM CST (Read 1388 times)  
thayal

Hi pls could anyone help me solve this problem

Martin and Jack baked some cookies each. If Martin gives Jack 26 cookies, Jack will have 3/4 as many cookies as Martin. If Jack gives Martin 30 cookies, Jack will have 25% as many cookies as Martin. How many cookies did Martin bake?

Forum Junior
Junior

Registered: 04/18/09
Posts: 21

Profile Email    
   
By: awyw1201 (offline)  Sunday, February 26 2012 @ 07:40 AM CST  
awyw1201

If Martin gives Jack 26 cookies,
J : M---3u : 4u
This means, J should have (3u-26) at first, and
M should have (4u+26) at first.

If Jack gives Martin 30 cookies,
J:M--- 1p : 4p
This means, J should have (1p+30) at first, and
M should have (4p-30) at first.

Therefore, for J------- 3u-26 = 1p + 30
Simplifying, we have 3u = 1p + 56------Eqn (1)

As for M-----4u + 26 = 4p - 30
Simplifying, we have 4u + 56 = 4p----Eqn (2)

Divide Eqn (2) by 4====1u + 14 = 1p---Eqn (3)

Substitute Eqn (3) into Eqn (1), we have
3u = 1u + 14 + 56
2u = 70
1u=35

35 x 4 + 26 = 166

Ans: 166 cookies

Hope this is not too late

Confused

Forum Regular Member
Regular Member

Registered: 07/08/11
Posts: 87

Profile Email    
   
By: echeewh (offline)  Monday, February 27 2012 @ 06:28 PM CST  
echeewh

Another method is to use Before and After model concept.
(Note: * are used to fill up whitespace & realign labels in model drawing here)

Case1 (1st IF cond)

<After>
J=3u, M=4u

M |-----|-----|-----|-----|

J |-----|-----|-----|

<Before>

M |-----|-----|-----|-----|--|
********************* -->
******************** +26

**************1u
*********** <---->
J |-----|-----|---|--|
********** <-><-
***********1p*-26

From J's model, 1u --> 1p + 26; hence, J's model (at start) = 2u + 1p = [2 x (1p + 26)] + 1p
= 2p + 52 + 1p = 3p + 52
From M's model, there are 4 such identical units plus 26 = 4u + 26 = [4 x (1p + 26)] + 26
= 4p + 104 + 26 = 4p + 130

Case 2 ( 2nd IF cond)

*****************130
***************<------->
M |---|---|---|---|----------|--|
*********************** -->
*********************** +30
** <-------------------------->
************ 4u

***********52
***********<-->
J |---|---|---|--|--|
** <----------><-
*******1u*** -30

Hence, from J's model, 1u ---> 3p + 52 - 30 = 3p + 22
From M's model, 4u ---> 4p + 130 + 30 = 4p + 160
Substituting 1u into here , we have 4 x (3p + 22) ---> 4p + 160
12p + 88 ---> 4p + 160
8p ---> 160 - 88 = 72
1p ---> 9
M (at start) = 4p + 130 ---> (4 x 9) + 130 = 36 + 130
= 166

Martin baked 166 cookies

Forum Active Member
Active Member

Registered: 04/21/11
Posts: 627

Profile Email    
   


 All times are CST. The time is now 11:34 am.
Normal Topic Normal Topic
Locked Topic Locked Topic
Sticky Topic Sticky Topic
New Post New Post
Sticky Topic w/ New Post Sticky Topic w/ New Post
Locked Topic w/ New Post Locked Topic w/ New Post
View Anonymous Posts 
Able to Post 
HTML Allowed 
Censored Content