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 By: shirley wee (offline)  Monday, January 07 2013 @ 08:04 AM CST (Read 725 times)
shirley wee

Hi everyone! I am stuck with this question. May you help me? Please show your workings clearly.

1. Schools A, B, C and D have a total of 6600 pupils. School A has 3/7 as many pupils as the other 3 schools. School B has 5/6 as many pupils as the other 3 schools. School C has 2/3 as many pupils as School D. How many pupils are there in School D?

Chatty

Registered: 02/28/12
Posts: 68

 By: echeewh (offline)  Monday, January 07 2013 @ 10:30 PM CST
echeewh

Hi Shirley,

Following pls find my worked solution:

A : (B + C + D) -- (1)
3 : 7

B : (A + C + D) -- (2)
5 : 6

C : D -- (3)
2 : 3

Apply <Unchanged Total> concept to (1), (2) since total units in (1) must be the same as (2).

To do this, multiply (1) by 11 , (2) by 10 as follows:

(1) × 11
A : (B + C + D) -- (4)
33 : 77

(2) × 10
B : (A + C + D) -- (5)
50 : 60

From (4), we have ... {the same for (5)}
B + C + D = 77
C + D = 77 - 50 = 27

From (3), C + D = 5. Hence, multiply this ratio by 27, and multiply (4), (5) by 5. These gave the following:

(4) × 5
A : (B + C + D) -- (6)
165 : 385

(5) × 5
B : (A + C + D) -- (7)
250 : 300

(3) × 27
C : D -- (3)
54 : 81

Hence, A = 165; B = 250, C = 54, D = 81

Given A + B + C + D = 6600, we have ...

165u + 250u + 54u + 81u --> 6600
550u --> 6600
1u --> 12

D = 81u --> 81 × 12 = 972

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

Trust this helps.

Do let me know again if this is different from your Answerkey or if theres further clarification.

Cheers,
Edward

Active Member

Registered: 04/21/11
Posts: 627

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