
More Difficult Problems in mathematics From post utme (futa 2007)
Mathematics
Chima Sunday (Tutor)
08042016 10:06:00 +0000About 80% of students that tried solving these questions failed it, try and see if you can get it right for practice can make perfect.
(Question 1) If log10 base 8 = x, find in terms of x log25 base 8
Option A. x/2
Option B. (x1)/2
Option C. 2x2/3
Option D. 2x3/2Updated Solution
log_{8}^{25} = log_{8}^{52}
applying logarithm law = 2log_{8}^{5}
note that 10/2 = 5
2log_{8}^{10/2}
applying a reverse of logarithm law to have
2(log_{8}^{10}  log_{8}^{2})
recall that log_{8}^{10} = x and also applying log on log_{8}^{2} = log_{23}^{2} = 1/3log_{2}^{2} = 1/3
so log_{8}^{25} = 2(x  1/3) = (2x  2/3)
So answer is Option C(Question 2) The first term of a geometrical progression is twice its common ratio. Find the sum of the first two terms of the progression if its sum to infinity is 8.
Option A 8/5
Option B 8/3
Option C 56/9
Option D 72/25Updated Solution
sum to infinity is given as
S_{∞} = a/(1  r) = 8
where a = first term and r = common ratio. but from question a = 2r
so 2r/(1  r) = 8
cross multiplying to obtain 2r = 8  8r
2r + 8r = 8
factorizing, r + 4r = 4
then 5r = 4
r = 4/5
and a = 8/5
but S_{n} = a(1  r^{n})/1  r
at n = 2
S_{2} = 8/5[1  (4/5)S^{2}]/(1  4/5)
applying difference of two sqaure to have
S_{2} = 8/5(1  4/5)(1 + 4/5)/(1  4/5) = 8/5(1 + 4/5) = 72/25
Answer is Option D

Chima Sunday (Tutor)
An update of this post has been updated with its detailed solution so try and check it up. God bless you my dear for attempting the above questions. keep it up and also keep the fire burning
0 13062016 16:07:00 +0000

{[reply.name]}
{[reply.voteCount]} {[reply.voteCount]} {[reply.created]}
{[reply.voteCount]} {[reply.created]}