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 Forum Index >  Test Paper Related >  Primary 5 Matters
 CHIJ P5 maths- need help please
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By: Phoenix (offline)  Wednesday, October 24 2012 @ 03:47 AM CDT (Read 817 times)  
Phoenix

A coach can carry a total of total of 35 adults or 49 children. If there were already 135 adults and 87 children seated in 6 such coaches, how many more children could be seated in these 6 coaches?

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By: echeewh (offline)  Wednesday, October 24 2012 @ 06:18 AM CDT  
echeewh

hey Phoenix,

Following pls find my worked solution:

Number of coaches for A:

135 ÷ 35 = 4 (round up)

Remaining seats from 4 coaches:

(4 × 35) - 135 = 5 (adults)

Number of Coaches for C:

87 ÷ 49 = 2 (round up)

Remaining seats from 2 coaches:

(2 × 49) - 87 = 11 (children)

Coach can take either 35A or 49C

equivalent of 1A --> (49/35) = (7/5) C
************** 5A --> 7C

Hence, number of C that could still be seated = 11 + 7 = 18

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

Trust this helps.

Do let me know again if there's further clarification.

Cheers,
Edward

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By: Phoenix (offline)  Wednesday, October 24 2012 @ 08:12 AM CDT  
Phoenix

Thanks Edward.

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