2013 DSE Maths Paper 1 Q10
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10
(a) :
Required Median = 31
Required Mode = 23
(b)(i):
(6x10+b) - (2x10+a) = 47
40+a+b = 47
a+b = 7
∵ 0 ≦ a ≦ 5
7≦b ≦ 9
∴The solutions are a=0,b=7 / a=1,b=8 / a=2,b=9
(b)(ii):
Required probability =
3+3+2+2+2+2+2x3+4x3÷13x20 = 8/65
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