
By: AngelLim (offline) Tuesday, October 23 2012 @ 03:26 AM CDT (Read 1203 times)



AngelLim 
Hi everyone.
I am new to this forum and I would like to ask help for some Maths questions for my daughter who is having her Maths exam next Monday. Here are the questions. I would appreciate it if you can reply ASAP. Thanks in advance.
Q1. There are some pens and pencils in a box. 2/3 of the number of pens is equal to 3/5 of the number of pencils. If there were 15 fewer pens than pencils, what was the total number of pens and pencils in the box?
Q2. Isabella spent $897 on entertainment and food and 0.25 of the remainder on clothes. If she had 3/7 of her salary left, how much is her salary?
Q3. If Chris gives Ben 8 stickers, they will have the same amount of stickers each. If Ben gives Chris 8 stickers, Chris will have 5 times as many stickers as Ben.
a) How many stickers does Chris have?
b) Express the number of stickers Ben has as a fraction of the number of stickers Chris has. Leave your answer in simplest form.
Q4. There were some sweets in Boxes X, Y and Z. Box X contained 1/5 of the total number of sweets in X, Y and Z. The number of sweets in Y was twice the total number of sweets in X and Z. If there were 32 more sweets in Box Y than in Box Z, find the total number of sweets in the 3 boxes.

Newbie
Registered: 10/22/12 Posts: 1





By: echeewh (offline) Tuesday, October 23 2012 @ 04:20 AM CDT



echeewh 
Hey Angel,
Following pls find my worked solutions: ( Will complete remaining by tomorrow )
Q1.
P=Pens; C=Pencils
Apply <Equal Concept>  make numerator of the 2 fractions to be same; then denominator will be the starting/initial number of units of respective object/subject.
(2/3)P = (3/5)C
(2/3) × (3/3) P = (3/5) × (2/2) C
(6/9)P = (6/10)C
Hence, P = 9u; C = 10u
Given there were 15 fewer pens than pencils, we have ..
C  P = 15
10u  9u > 15
1u > 15
Total number of P, C = 9u + 10u = 19u > 19 × 15 = 285
=================
Q2
***  for alignment purpose
********< E & F >< C >< 3/7total >
Salary < 897 >
****************************< 1u>< 3u >
3u > (3/7) total
1u > (1/7) total
Hence, E & F = (3/7) total > 897
Salary = (7/7) total > 897 ÷ (3/7) = 897 × (7/3)
= $2093
==================
Trust these help.
Do let me know again if these are different from your Answerkey or if there's further clarification.
Cheers and all the best to ur gal,
Edward

Active Member
Registered: 04/21/11 Posts: 623





By: ystella (offline) Tuesday, October 23 2012 @ 09:34 AM CDT



ystella 
Hi Angel,
Q3,
first scenario
C [][][ ][ 8 ]
B [][][+8]
second scenario
C [ ][ ][ ][ ][+8 ]
B [ ][ 8]
a) Chris has 32 stickers
b) 16/32 = 1/2
Hope this help.

Newbie
Registered: 04/03/09 Posts: 5





By: echeewh (offline) Tuesday, October 23 2012 @ 08:11 PM CDT



echeewh 
Hey Angel,
As promised,
Following pls find my worked solutions to Q3 and Q4.
***  for alignment purpose
Q3.
<after>
B 
C 
<process>
Apply <work backwards> method,
B: 8
C: +8
<Before>
B <8>
C <8>
<process>
B: 8
C: +8
<after>
**<1u>
B <8>
C <16><8>
**< 5u >
Apply <Gap & Difference> method, we have ...
5u  1u > 8 + 16 + 8 = 32
4u > 32
1u > 32 ÷ 4 = 8
(a)
C (at first) = 5u  8 > (5 × 8)  8
= 40  8 = 32
(b)
B (at first) = 1u + 8 > 8 + 8 = 16
(B / C) = 16/32 = 1/2
==================
Q4.
X : (Y + Z) Total
1 : ** 4 ***** 5  (1)
Y : (X + Z) Total
2 : ** 1 ***** 3  (2)
Since total number of sweets in X,Y,Z remain unchanged in the two ratios above <Unchanged Total>,
multiply (1) x 3; (2) x 5 ( common multiple) as follows:
(1)x3:
X : (Y + Z) Total
3 : ** 12 *** 15
(2)x5:
Y : (X + Z) Total
10 : ** 5 **** 15
So, X = 3u; Y = 10u; Z = 5u  3u = 2u
If there were 32 more sweets in Box Y than in Box Z,
Y  Z = 32
10u  2u > 32
8u > 32
1u > 4
Total sweets in X,Y,Z = 3u + 10u + 2u = 15u
15u > 15 × 4 = 60
====================
Trust the above helps.
Do let me know again if these answers are different from your Answerkey or if there's further clarification.
Cheers,
Edward

Active Member
Registered: 04/21/11 Posts: 623



