Q&A - P5 Math

[Jamesbond]

MathIzzzFun:

Jamesbond:

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Pl help.... :nailbite:

wow...still working on maths on CNY eve ? ... have a kit-kat...

4-3 =1
since there are 8 divisions from 3 to 4,
each division = 1/8 = 0.125
3 divisions = 3 x 0.125 = 0.375
3 + 0.375 = 3.375

Have a break !

Happy Chinese New Year to ALL KSPians !

cheers.

Thx for ur fast reply....Poor me... CA1 maths on 19th....

[Jamesbond]

Jar Y has 22 more sweets than Jar Z and Jar Z has 16 more sweets than Jar X. The total number of sweets in Jars Y and Z is 5 times the number of sweets in Jar X.

a) How many sweets are there altogether?
b) How many sweets should be transferred from Jar Y to the other two jars so that there is an equal number of sweets in all three jars?

Pl help..... :rotflmao:

[MathIzzzFun]

Jamesbond:

Jar Y has 22 more sweets than Jar Z and Jar Z has 16 more sweets than Jar X. The total number of sweets in Jars Y and Z is 5 times the number of sweets in Jar X.

a) How many sweets are there altogether?
b) How many sweets should be transferred from Jar Y to the other two jars so that there is an equal number of sweets in all three jars?

Pl help..... :rotflmao:

Jar X --> 1 unit
Jar Z --> 1 unit + 16
Jar Y --> 1 unit + 16 + 22 = 1 unit + 38

Jar Y + Jar Z --> 2 units + 54 = 5 units
3 units --> 54, 1 unit --> 18

Jar X --> 18
Jar Z --> 34
Jar Y --> 56
Total --> 108

108 / 3 = 36
Number transferred from Jar Y to
Jar X --> 36 - 18 = 18
Jar Z --> 36 - 34 = 2

cheers.

[Jamesbond]

MathIzzzFun:

Jamesbond:

Jar Y has 22 more sweets than Jar Z and Jar Z has 16 more sweets than Jar X. The total number of sweets in Jars Y and Z is 5 times the number of sweets in Jar X.

a) How many sweets are there altogether?
b) How many sweets should be transferred from Jar Y to the other two jars so that there is an equal number of sweets in all three jars?

Pl help..... :rotflmao:

Jar X --> 1 unit
Jar Z --> 1 unit + 16
Jar Y --> 1 unit + 16 + 22 = 1 unit + 38

Jar Y + Jar Z --> 2 units + 54 = 5 units
3 units --> 54, 1 unit --> 18

Jar X --> 18
Jar Z --> 34
Jar Y --> 56
Total --> 108

108 / 3 = 36
Number transferred from Jar Y to
Jar X --> 36 - 18 = 18
Jar Z --> 36 - 34 = 2

cheers.

:salute: :thankyou:

[Jamesbond]

Cali's spent $1485 on 9 identical handbags and 3 identical pairs of shoes. Sherlyn spent $ 867 less than Calise on 2 such handbags and 6 such pairs of shoes.what was the total cost of 3 such handbags and 2 such pairs of shoes?

Pl help :?

[MathIzzzFun]

Jamesbond:

Cali's spent $1485 on 9 identical handbags and 3 identical pairs of shoes. Sherlyn spent $ 867 less than Calise on 2 such handbags and 6 such pairs of shoes.what was the total cost of 3 such handbags and 2 such pairs of shoes?

Pl help :?

Elimination method :

9 handbags + 3 pairs of shoes --> $1485 -- (1)
2 handbags + 6 pairs of shoes --> $ 1485 - $ 867 = $618 -- (2)

(1)x2 --- 18 handbags + 6 pairs of shoes --> $ 2970 -- (3)
(3)-(2) -- 16 handbags --> $ 2970 - $ 618 = $ 2352
1 handbag --> $ 147
from (1) -- 1 pair of shoes --> $(1485-9 x 147)/3 = $ 54

3 handbags + 2 pairs of shoes --> 3 x $147 + 2 x $ 54 = $ 549

grouping method:

9 handbags + 3 pairs of shoes --> $1485
divide by 3 -- 3 handbags + 1 pair of shoes --> $495 --- (1)

2 handbags + 6 pairs of shoes --> $ 1485 - $ 867 = $618
divide by 2 -- 1 handbag + 3 pairs of shoes --> $309 -- (2)

(1)+(2) -- 4 handbags + 4 pairs of shoes --> $ 495 + $ 309= $ 804
divide by 4 -- 1 handbag + 1 pair of shoes = $ 201

from (2) -- 2 pairs of shoes --> $ 309 - $ 201 = $ 108
1 pair of shoes --> $ 54

3 handbags + 2 pairs of shoes
= $ 495 (from (1)) + $ 54 (1 pair of shoes)
= $ 549


cheers.

[Oracle]

How about this:


There are some 20-cent and 50-cent coins in a piggy bank. The coins add up to $17.30. There are 6 more 20-cent coins than 50-cent coins. How many 50-cent coins are there in the piggy bank?

[MathIzzzFun]

Oracle:

How about this:


There are some 20-cent and 50-cent coins in a piggy bank. The coins add up to $17.30. There are 6 more 20-cent coins than 50-cent coins. How many 50-cent coins are there in the piggy bank?

make the number of coins same ie same number of 20-cent and 50-cent coins -- either add 6 more 50-cent coins or remove 6 20-cent coins

add 6 more 50-cent coins --> 6 x $0.50 = $3.00
Total amount of money = $ 17.30 + $3.00 = $20.30
1 set of 1x20-cent and 1x50-cent coins --> $0.70
$20.30 / $0.70 --> 29 sets of 1x20-cent and 1x50-cent coins

Number of 50-cents --> 29 - 6 = 23


remove 6 20-cent coins --> 6 x $0.20 = $ 1.20
Total amount of money --> $ 17.30 - $ 1.20 = $ 16.10
1 set of 1x20-cent and 1x50-cent coins --> $0.70
$16.10 / $0.70 --> 23 sets of 1x20-cent and 1x50-cent coins

Number of 50-cent coins = 23

cheers.

[Oracle]

Thanks MathIzzzFun, inspired by your workings, I managed to conceptualize the problem sum. Basically,


Total amount of money → $17.30
6 more 20¢ coins than 50¢ coins → (20c x 7) + 1x50¢ → $1.90
One set of 1x20¢ and 1x50¢ coins → $0.70
$17.30 - $1.90 = $15.40
$15.40 ÷ $0.70 = 22 sets of 20¢ and 50¢

Therefore, number of 50¢ coins
= twenty-two 50¢ + one 50¢
= 23

Thank you very much indeed.

[chloecube]

Borsche manufactures 3 cars daily while Percedes manufatures 2 more cars everyday. Even though Percedes started production 12 days later than Borsche, Percedes has now produced 14 more cars than Borsche.

a) how many cars did Borsche produce by Day 12?
b) how many days did Percedes take to produce 14 more cars than Borsche?

(no listing method)