Weekly Dose 2 - Maths Olympiad Practice

1. There are 20 red marbles, 30 white marbles and some blue marbles in a
box. If you draw one marble from the box, the probability of chance of
drawing one blue marble is
9
11
. How many blue marbles are there in the
box?
Solution 1:
With the probability of drawing one blue marbles is
9
11
, we can say that
the box has 11𝒙 marbles, 9𝒙 blue marbles.
And non-blue marbles = 𝟏𝟏𝒙 – 𝟗𝒙 = 𝟐𝒙 = 𝟐𝟎 + 𝟑𝟎
𝒙 = _____
Blue marbles = 𝟗𝒙 = ______
Solution 2:
Assuming there are 𝒙 blue marbles in the box.
𝒙
𝒙 + 𝟐𝟎 + 𝟑𝟎
=
𝟗
𝟏𝟏
𝟏𝟏𝒙 = 𝟗(𝒙 + 𝟓𝟎)
= 𝟗𝒙 + 𝟒𝟓𝟎
= …
Blue marbles = 𝒙 = ______

2. Fill in all the numbers below into circles A, B, C, such that all numbers in
circle A are divisible by 5, all numbers in circle B are divisible by 2, all
numbers in circle C are divisible by 3.
48, 102, 112, 207, 750, 930, 1348, 1605, 1749, 2001, 2025, 2030, 3250,
7893
Solution:
48, 102: divided by 2 and 3  B and C
B
C
A
48
112: divided by 2 only  B
207: divided by 3 only  C
750, 930: divided by 2,3 and 5  A, B and C
102
112
207
1348: divided by 2 only  B
1605: divided by 3 and 5  A and C
1749, 2001: divided by 3 only  C
2025: divided by 3 and 5  A and C
2030, 3250: divided by 2 and 5  A and B
7893: divided by 3 only  C
750
930
1348
1605
1749
2001
2025
2030
3250
7893

3. A boat is 50 kilometers away from the port. The boat is leaky, so water
flows into the boat at the rate of 2 tons per 5 minutes. If there were 90
tons of water in the boat, the boat would sink. If there is a pump in the
boat, pumping out 12 tons of water per hour , what should be the
minimum speed of the boat in km/h to avoid the boat from sinking?
Solution:
The pump pumps in water at 12 tons per hour ---- ①
Water flows into the boat at 2 tons per 5 minutes = ____ tons per hour --- ②
Net water comes into the boat at the rate of ② - ① = ____ tons per hour --- ③
The boat will sink in 90 ÷ ③ = ____ hours ---- ④
The boat needs to at least cover 50km in ④ hours to avoid sinking
∴ minimum speed of the boat = 50 ÷ ④ = ____ km/h

4. Solution:
Original C x (100% + 22%) = 488
Original C = 488 ÷ (122%) = 488 ×
100
122
= ____ ---- ①
Number of beads transferred to C = 488 - ① = ____ ---- ②
Bag A has twice the number of beads in bag B. 12% of beads in bag A are
removed and transferred to bag C. 20% of beads in bag B are removed and
transferred to bag C. After removing and transferring beads, there are now
488 beads in bag C which is 22% more than the original number of beads in
bag C. How many beads were there in the bag A at the beginning?
Because Original A = 2 times Original B
∴ 20% of Original B = 10% of Original A ---- ③
Number of beads transferred to C = 12% of Original A + 20% of Original B
(From ③ ) = 12% of Original A + 10% of Original A
= 22% of Original A
From ②, 22% of Original A = ____
Original A = ____ ÷ 22%
= ____ ×
100
22
= ______ beads