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 A* Maths Problems
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By: Rohana (offline)  Saturday, February 23 2013 @ 07:03 AM CST (Read 736 times)  
Rohana

Hi can anyone help me with these 3 questions!

Q1. When Kok Hwa's age was twice Bala's age, Mike's age was 30. When Mike's age was twice Kok Hwa's age, Bala's age was 21. Given that Mike was the oldest among the three persons, find Bala's age when Mike was 62 years old.

Q2. Four files cost as much as 3 storybooks. Five files cost $7.50 more than 2 storybooks. Mary spent $30 on an equal number of files and storybooks. How many files did she buy?

Q3. Mrs Hodge bought three times as many key chains as stuffed toys and spent $124 in total. She spent $64 more on the stuffed toys than the key chains. Given that a stuffed toy cost $8.40 more than a key chain, find the cost of a key chain.


Thankyou in advance.

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By: echeewh (offline)  Saturday, February 23 2013 @ 08:12 PM CST  
echeewh

Hey Rohana,

In view that there are 3Q here, I will only post the solutions to Q1, Q2 here first while the remaining one will be posted in due course.

<Updates>
25/2 - Corrected answer for Q2
25/2 - Q3 solution was posted



Following pls find my worked solutions:

Q1

Apply <Age Difference> concept - where age difference between 2 persons remains constant / same now and then.

<1st When statement>

K |---|---|
B |---|
M |<--- 30 --->|

Age Diff (between K,B ) = 2p - 1p = 1p
Age Diff (between M,K) = 30 - 2p

<2nd When statement>

K |------------|
B |<- 21 ->|
M |------------|------------|

Age Diff (between K,B ) = 1u - 21
Age Diff (between M,K) = 2u - 1u = 1u

We now have ...

Age Diff (between K,B ):
1p = 1u - 21
1u - 1p = 21 -- (1)

Age Diff (between M,K):
30 - 2p = 1u
1u + 2p = 30 -- (2)

Using <Simultaneous Eqn> method, eliminate <1u> from (1),(2).

(2)-(1):
2p + 1p --> 9
3p --> 9
1p --> 3

B = 1p --> 3, when M = 30

Age Diff (between M,B ) = M - B = 30 - 3 = 27

When M = 62, B = 62 - 27 = 35 yrs old

=========================

Q2.

4F = 3S
5F - 2S = 750 -- (1)
4F - 3S = 0 -- (2)

(1)-(2):
F + S = 750

Given that 1u of F and 1u of S cost $30, we have ...

3000 ÷ 750 = 4

i.e. 4F + 4S = 3000

Hence, 1u of F = 4

=========================

Q3

<Value>

T |----|<- 64 ->|
C |----|

Total spending: 124

124 - 64 = 60
2u --> 60
1u --> 30

Hence, T = 1u + 64 --> 30 + 64 = 94
C = 1u --> 30

<Qty>

C |---|---|---|
T |---|

1u of T = $94
1u of C = 30 ÷ 3 = $10

Difference in amount between same qty of T and C = 94 - 10 = $84

Given that T - C = $8.40, the amount diff above for the same qty/number of T and C purchased is a result of this.

Hence, 84 ÷ 8.4 = 10

This means there were 10 T and 10 C in that <1u> qty of T and C.

Qty of C (purchased) = 3u --> 3 × 10 = 30

Amount spent on C = $30

Hence, cost of 1C = 30 ÷ 30 = $1

==========================


Trust this helps.

Do let me know again if this is different from your Answerkey or if there are further clarifications.

Cheers,
Edward

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