Q&A - PSLE Math

[Nebbermind]

Need help with this question. Thank you.

Since MR = RO, Angle NOR = 21
Therefore Angle ORQ = 42

Angle RMN + Angle RQO + Angle OPQ×2 = 180
Angle OPQ = (180-21-42)/2

[EC72]

Thank you Beautyful Minds! Can you please explain to me how did you get the ratio in the distance travelled 2:3?

Speed is proportional to distance travelled

1.2 : 1.8 = 2 : 3

Beautyful Minds

Beautyful Minds

[Ilye]

Need help with this question. Thank you.

Angle QCB = 70
Angle ASQ = 20
Therefore Angle DSQ = 160

Hi, may I know how to see angle ASQ=20? Is there any quick way to see this? Thanks.

[Nebbermind]

If you extend the line RC to tbe right, the angle below QCB is 20.

[EC72]

The distance between Town A and Town C is 825 km. Bob set off from Town A at 1.30 p.m., at an average speed of 60 km/h. At 2.45 p.m., Carl also set off from Town A to Town B at an average speed of 75 km/h. Carl arrived at Town B at 10.45 p.m.


(a) At what time did Carl overtake Bob?


(b) How long did Bob take to travel from Town B to Town C if he increased his speed by 50% starting from Town B?

[DearSonR]

The distance between Town A and Town C is 825 km. Bob set off from Town A at 1.30 p.m., at an average speed of 60 km/h. At 2.45 p.m., Carl also set off from Town A to Town B at an average speed of 75 km/h. Carl arrived at Town B at 10.45 p.m.


(a) At what time did Carl overtake Bob?


(b) How long did Bob take to travel from Town B to Town C if he increased his speed by 50% starting from Town B?

a)
1.30p.m. --> 2.45p.m. 1h15min
1h15min*60km/h=75km
75km/h-60km/h=15km/h
75km/15km/h=5h
2.45p.m-->5h-->7.45p.m.
b)
2.45p.m. --> 10.45p.m. 8h
8h*75km/h=600km
825km-600km=225km
75km/h*150%=112.5kmh
225km/112.5kmh=2h

a)7.45p.m.
b)2h

Sorry if it is a little messy.

[DearSonR]

Need help with this question. Thank you.

Angle QCB = 70
Angle ASQ = 20
Therefore Angle DSQ = 160

Hi, may I know how to see angle ASQ=20? Is there any quick way to see this? Thanks.

A few things you should note before my number statements:
1. QCB+ABC=180
2. The angle at the extended line from QC in the parallelogram and QCB when added together is 180 too.
3. Angles on a straight line add up to 180.
4. Angles in a triangle add up to 180.

To get ASQ:
QCB-->180-110=70
Angle at the extended line from QC in the parallelogram-->180-70=110
Angle at the extended line from QC in the triangle (not the right angle)-->180-110=70
ASQ-->180-70-90=20


If you knew that angles which are opposite of one another in a parallelogram and rhombus are equal, you could save the first 2 steps and just go on to the 3rd.

Hope this helps you!!!

[DearSonR]

Wondering if anyone knows some 'cheat methods' in solving angle questions.

[Daddy]

Thank Nebbermind.

Hi all, need help on these questions.

The shaded triangles form a square of side 14.
Radius of circle is 14
Perimeter = (3/4 * 2 * pi * 14) + 14 + 14

Why u square root 196,even if 196 is a triangle?And how is 14 the radius of the circle?

[DearSonR]

Thank Nebbermind.

Hi all, need help on these questions.

The shaded triangles form a square of side 14.
Radius of circle is 14
Perimeter = (3/4 * 2 * pi * 14) + 14 + 14

Why u square root 196,even if 196 is a triangle?And how is 14 the radius of the circle?

I believe that the radius of the circle should be square root of 98.
Because the length of 1 shaded triangle is twice the height of 1 shaded triangle. So, if you let the height of 1 shaded triangle be x,
x*2x*0.5*2=196=2x^2
x^2=98
x=square root of 98
(square root of 98*2*3/4*22/7)+square root of 98+square root of 98=66.4680374315354673

Sorry if this did not solve your question... Did the question tell the student to round off to nearest whole number?