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concerned1970:

Pls help me solve this ques without guess n chk method. Ty in adv.


A bill of $110 was paid with $10, $5 and $2 notes. If 28 notes were used, how many of each kind were used?

Stack the notes to every $10 value per stack for each type:
[$10] or [$5][$5] or [$2][$2][$2][$2][$2]

We'll need 11 stacks to reach $110.

If we try to use all fairly, then we use each of the 3 types of stacks which will be a total of 3x$10=$30 that will contain 1+2+5=8 notes. Let's call this a bundle.

For $110, we can only use 3 bundles which will have a total value of 3x$30 = $90 and containing 3x8 = 24 notes.

So for remainder of $110 - $90 = $20, we can only use 28-24=4 notes.
Taking $20 / 4 = $5, the remaining 4 notes are $5 notes.

3 bundles + 4 notes of $5
= 3 x ( 1x[$10] + 2x[$5] + 5x[$2]) + 4x[$5]
= 3x[$10] + 10x[$5] + 15x[$2]
ANSWER: 3 ten dollar notes, 10 five dollar notes, 15 two dollar notes

Mary Joy:

Pls help me with this question....Thank you..


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*represent the group at the beginning,
#represent the group after alighting

At the beginning,
Adults* = Children* + 250

After alighting,
Children# = Children* - 20%xChildren*

Ratio of (Adult* - Adult#):Adult# is 2:3, so
Adult# = Adult* - (2/5)xAdult*

Modeling method:
Children* = [u][u][u][u][u] = 5[u]
Adults* = 5[u] + 250 = [u+50][u+50][u+50][u+50][u+50] = 5[u+50]

Children# = 5[u] - [u] = 4[u]
Adults# = 5[u+50] - 2[u+50] = 3[u+50]

Children#-Adults# = 30
4[u] - 3[u+50] = 30
[u] - 150 = 30
[u] = 180
Adults in the end, Adults# = 3[180 + 50] = 690

Mary Joy:

Hi pls help me with this question...Thanks in advance..


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If FangLing used 12, Shanti would have half as many as FangLing:
FangLing = [V][V][12] or FL=2V+12 ......equation (1)
Shanti = [V] or S=V ......equation (2)

If Shanti used 18, FangLing would have 5 times as many as Shanti:
FangLing=[u][u][u][u][u] or FL=5u ......equation (3)
Shanti=[u][18] or S=u+18 ......equation (4)

From equation (1) and (3), we get: 5u=2V+12
From equation (2) and (4), we get: V=u+18

Therefore, 5u = 2(u+18) +12
5u = 2u +36 +12
3u = 48
u = 16

How many stickers did they have altogether?
Answer: FL + S = 5u + u + 18 = 6*16 + 18 = 114 stickers

Mary Joy:

Hello Everyone,


Pls help me with this question taken from Singapore Hokkien Huay Kuan Prelims....Thanks in advance....

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Here's my answer. I hope I'm using only accepted Maths theories for PSLE to solve this question.

QS = PR;
Since PR is also the radius of the semicircle, QS = 5cm

XZ = RY;
Since RY is also the radius of the semicircle, XZ = 5cm

Using pythagoras theorem:
QR*QR + RS*RS = QS*QS = 5*5 .... equation (1)
RZ*RZ + XR*XR = XZ*XZ = 5*5 .... equation (2)

Equating equation (1) with (2):
QR*QR + RS*RS = RZ*RZ + XR*XR .... equation (3)

Given that Perimeter of PQRS = Perimeter of XYZR = 14cm
2*(QR+RS) = 2*(RZ+XR) = 14cm
Therefore, QR +RS = RZ + XR .... equation (4)

Also given that XS = 1cm:
XR = RS + XS
XR = RS +1 .... equation (5)

Substituting equation (5) into equation (4),
QR + RS = RZ + (RS + 1)
QR = RZ + 1 .... equation (6)

Substitution equations (5) and (6) into equation (3),
(RZ+1)*(RZ+1) + RS*RS = RZ*RZ + (RS+1)*(RS+1)
RZ*RZ + 2RZ + 1 + RS*RS = RZ*RZ + RS*RS + 2RS + 1
2RZ = 2RS
RZ = RS .... equation (7)

From the given perimeter of 14cm ,
2*(RZ + XR) = 14
RZ + XR = 7 .... equation (8)

Substituting equation (5) into equation (8),
RZ + (RS+1) = 7
RZ + RS = 6 .... equation (9)

Substitution equation (7) into equation (9),
RZ + (RZ) = 6
RZ = 3cm
The rest can be obtained from this RZ=3cm substituted into the above equations:
RS = 3cm; XR = 4cm; QR =4cm

Double checking again:
Triangle sides: 3*3 + 4*4 = 9 + 16 = 25 = 5*5 .... so this is correct
For perimeter, 2*(3+4) = 2*7 = 14 .... also correct

Perimeter of shaded
= Half perimeter of circle + QS + XS + XZ + (diameter - QR - RZ)
= 3.142*5 + 5 + 1 + 5 + (10 - 4 - 3)
= 15.71 + 11 + (3)
= 29.71 .... ANSWER

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