Q&A - P5 Math

[pixiedust]

NPYS 2012 P5 SA1 Paper2


Q9) Any solution without using guess and check ?

http://s1197.photobucket.com/user/pixie_dust8/media/nypsq9.jpg.html

Q17) I think the answer key is wrong, hence would like to check for a 2nd opinion.
http://s1197.photobucket.com/user/pixie_dust8/media/nypsq17.jpg.html

[smartmummy]

pixiedust:

NPYS 2012 P5 SA1 Paper2



Q17) I think the answer key is wrong, hence would like to check for a 2nd opinion.
http://s1197.photobucket.com/user/pixie_dust8/media/nypsq17.jpg.html

Pls refer here,
http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=68&t=25129&start=2250

[MathIzzzFun]

pixiedust:

NPYS 2012 P5 SA1 Paper2


Q9) Any solution without using guess and check ?

http://s1197.photobucket.com/user/pixie_dust8/media/nypsq9.jpg.html

http://i46.tinypic.com/pb0gx.png\">

cheers.

[pixiedust]

Thank you smartmummy ! (helping my niece and find my maths rusty !)


Thanks MathIzzzFun ! Ingenious !!

[S-H]

Hi, can anyone help me with these 2 questions, thanks!


1) Jeremy had a container filled with a number of 20cents, 50cents and $1 coins. 1/5 of the 20cents conins is equal to 2/3 of the 50cents coins. There were 3/8 as many 50cents coins as $1 coins. If the total value of these coins was $6900, how many coins did Jeremy have altogether?

2) Moses had a total of 408 blue and green counters at first. He lost 3/5 of his blue counters and bought another 32 counters. Then the number of green counters he had was 1/4 of the number of the blue counters left. How many more blue counters than green counters did Moses have at first?

[MathIzzzFun]

S-H:

Hi, can anyone help me with these 2 questions, thanks!


1) Jeremy had a container filled with a number of 20cents, 50cents and $1 coins. 1/5 of the 20cents conins is equal to 2/3 of the 50cents coins. There were 3/8 as many 50cents coins as $1 coins. If the total value of these coins was $6900, how many coins did Jeremy have altogether?

1/5 of number of 20 cent coins = 2/3 of number of 50 cents coins

2/10 of number of 20 cent coins = 2/3 of number of 50 cents coins

Number of 20 cents coins --> 10 units, value of 20 cents coins --> $ 2 units
Number of 50 cents coins --> 3 units, value of 50 cents coins --> $ 1.5 units
Number of $1 coins --> 8 units, value of $1 coins --> $8 units

Total number of coins --> 21 units

Total value --> 11.5 units -- $6900
1 unit --> 600

Total number of coins --> 21 x 600 = 12600

cheers.

[S-H]

MathIzzzFun:

S-H:

Hi, can anyone help me with these 2 questions, thanks!


1) Jeremy had a container filled with a number of 20cents, 50cents and $1 coins. 1/5 of the 20cents conins is equal to 2/3 of the 50cents coins. There were 3/8 as many 50cents coins as $1 coins. If the total value of these coins was $6900, how many coins did Jeremy have altogether?

1/5 of number of 20 cent coins = 2/3 of number of 50 cents coins

2/10 of number of 20 cent coins = 2/3 of number of 50 cents coins

Number of 20 cents coins --> 10 units, value of 20 cents coins --> $ 2 units
Number of 50 cents coins --> 3 units, value of 50 cents coins --> $ 1.5 units
Number of $1 coins --> 8 units, value of $1 coins --> $8 units

Total number of coins --> 21 units

Total value --> 11.5 units -- $6900
1 unit --> 600

Total number of coins --> 21 x 600 = 12600

cheers.

Thank you MathIzzzFun, but I don't understand what is $2 units, $1.5 units and $8 units? Can you explain, thanks!

[MathIzzzFun]

S-H:

MathIzzzFun:

Hi, can anyone help me with these 2 questions, thanks!


1) Jeremy had a container filled with a number of 20cents, 50cents and $1 coins. 1/5 of the 20cents conins is equal to 2/3 of the 50cents coins. There were 3/8 as many 50cents coins as $1 coins. If the total value of these coins was $6900, how many coins did Jeremy have altogether?


1/5 of number of 20 cent coins = 2/3 of number of 50 cents coins

2/10 of number of 20 cent coins = 2/3 of number of 50 cents coins

Number of 20 cents coins --> 10 units, value of 20 cents coins --> $ 2 units
Number of 50 cents coins --> 3 units, value of 50 cents coins --> $ 1.5 units
Number of $1 coins --> 8 units, value of $1 coins --> $8 units

Total number of coins --> 21 units

Total value --> 11.5 units -- $6900
1 unit --> 600

Total number of coins --> 21 x 600 = 12600

cheers.

Thank you MathIzzzFun, but I don't understand what is $2 units, $1.5 units and $8 units? Can you explain, thanks!

Number of $1 coins --> 8 units, value of $1 coin --> $1
$1 x 8 units = $8 units

Alternatively,
ratio of 20c : 50c : $1 = 10:3:8 --- 10+3+8 = 21

For 1 set of 21 coins comprising:
10 x 20c --> $2
3 x 50c --> $1.50
8 x $1 --> $8
Total $11.50

Total number of sets --> $6900 / $11.50 = 600 sets
TOtal number of coins = 21 x 600 = 21 600

cheers.

[S-H]

Hi MathIzzzFun, I think the 2nd method is more clear to me, thank you so much!!

[S-H]

Sorry MathizzzFun, can you help me with the second question:-


2) Moses had a total of 408 blue and green counters at first. He lost 3/5 of his blue counters and bought another 32 counters. Then the number of green counters he had was 1/4 of the number of the blue counters left. How many more blue counters than green counters did Moses have at first?

Thank you.