Welcome to Test-paper.info
Sunday, January 21 2018 @ 04:11 PM CST
 Select a Forum » General Chat » Primary 1 Matters » Primary 2 Matters » Primary 3 Matters » Primary 4 Matters » Primary 5 Matters » Primary 6 Matters » Question, Feedback and Comments » Education Classified
 Forum Index >  Test Paper Related >  Primary 6 Matters 3 math questions(no answer key)
 | Printable Version
 By: vivilau1112 (offline)  Wednesday, June 05 2013 @ 07:16 AM CDT (Read 677 times)
vivilau1112

Hey people! Anyone mind helping me on these questions?

1)
The total number of stamps in Albums A,B,C was 444 at first.Dennis gave away 3/5 of the stamps from A,put 24 more stamps in B and put some stamps in C until the number of stamps became 3x it's original number. The ratio of A to B to C became 2:5:9 in the end.How many more stamps were there in C than A in the end?

2)
J bought some chocolates and gave 1/2 of them to K.K bought some sweets and gave 1/2 of them to J.J ate 12 sweets and K ate 18 chocolates. After that, the sweets and chocolates J had were in the ratio 1:7 and the number of sweets and chocolates K had were in the ratio 1:4. How many sweets did K buy?

3)
A and B shared the total cost of a present. A paid \$15 more than 3/8 of the cost of the present. B paid \$25. How much did the present cost?

Help in all the questions is appreciated. Thank you in advance

Junior

Registered: 04/13/11
Posts: 16

 By: echeewh (offline)  Wednesday, June 05 2013 @ 09:28 AM CDT
echeewh

Hey there,

Following pls find my worked solutions:

Q1.

<after>

A : B : C
2 : 5 : 9

<process>
A: -(3/5)A
B: +24
C: +(3)C

Using <work backwards> method, we have ...

<before>
C: (1/3)C = 9u ÷ 3 = 3u
B: 5u - 24
A: (2/5)A = 2u; (5/5)A = 2 × (5/2) = 5u

Given that total number of stamps in Albums A,B,C was 444 at first, we have ...

<before>
A + B + C = 444
5u + 5u - 24 + 3u = 444
13u --> 444 + 24 = 468
1u --> 36

C - A (end) = 9u - 2u = 7u
7u --> 7 × 36 = 252

=======

Q2.

<after>
Js : Jc
1 : 7

Ks : Kc
1 : 4

<process>
J: -12s
K: -18c

<before>
Let the <after> ratio for J be parts(p) and K be units(u).

Js: 1p + 12
Jc: 7p

Ks: 1u
Kc: 4u + 18

Given J bought some chocolates and gave 1/2 of them to K, and K bought some sweets and gave 1/2 of them to J, we have ...

Jc = Kc , and Js = Ks

7p = 4u + 18
1p + 12 = 1u

Rearranging these 2 equations, we have ...

7p - 4u = 18 -- (1)
1u - 1p = 12 -- (2)

As question is asking for number of sweets(s) K bought, we eliminate parts(p) qty as follows:

(2)x7:
7u - 7p = 84 -- (3)

(1)+(3):
7u - 4u = 18 + 84
3u --> 102
1u --> 34

Sweets bought by K (at first) = 2u
2u --> 2 × 34 = 68

=========

Q3.

A: (3/8)total + \$15
Remainder(R): (5/8)total - 15

Given that B paid \$25,

(5/8)total - 15 = 25
(5/8)total = 25 + 15 = 40
Total Cost of present = (8/8)total = 40 × (8/5)
= \$64

=========

Trust this helps.

Do let me know again if this is different from your <Answerkey> or if there's further clarification.

Cheers,
Edward

Active Member

Registered: 04/21/11
Posts: 627

 All times are CST. The time is now 04:11 pm.
 Normal Topic Locked Topic Sticky Topic
 New Post Sticky Topic w/ New Post Locked Topic w/ New Post
 View Anonymous Posts Able to Post HTML Allowed Censored Content