1. The lowest common multiple of 12, 15 and x is 180. Find the
greatest possible value of x which is an odd number.
How do I present the working and do this question?
2. Below is the stem and leaf diagram of the marks of a sec 1
class of 25 pupils.
Stem |
Leaf
0
| 2 5 9
1
| 0 1 1 4 8 9
2
| 1 3 6 6 8 8 9
3
| 2 8 9
4 |0
1 2 3 3 7
Find the interquartile range. I got 26 for IQR, and also 26 for
median. But answer for IQR is 27...
The second part of question says
if a pupil is taken out from the group, what is the minimum
score that he could have such that the median remains the same?
What is the question asking for actually? If the pupil is taken
out of the group, why does he still need to have a "score"that
makes the median the same? Anyway, answer is 28.
Thanks everybody.
x is the largest odd factor of 180, thus you can use 2 methods
to solve this question.
Method 1 (
Laying out of factors )
Factors of 180 are : 1, 2, 3, 4, 5, 6, 10, 12, 15, 18, 30, 36,
45, 60,
90,180
Method 2 (
Trial and Error )
180/1 = 180
180/2 = 90
180/3 = 60
180/4 = 45
IQR = upper quartile - lower quartile
Upper quartile = ( 39 + 40 ) / 2 = 39.5
Lower quartile = ( 11+14 ) / 2 = 12.5
Hence, IQF = 39.5 - 12.5 = 27
The last part is very simple. Study the diagram carefully and
you will realise that if you want the median to remain the same,
you can only take out the score of a pupil which is higher than the
current median.
Hence the lowest number that you can take out will be 28. The
trick part of this question is that some pupils will opt to take
out one of the two 26 instead, and end up getting it wrong.