A box contains 13 currency notes, which are either ₦50 or ₦20 notes.
The total value of the currency notes is ₦530. How many ₦50 notes are
in the box?
SOLUTION
Let the number of ₦50 notes be x and the number of ₦20 notes be y.
Number of notes = x + y = 13................(1).
Total value of the notes = 50x + 20y = 530................(2).
Multiply (1) by 20.
20x + 20y = 260................(3).
Subtract equation (3) from (2).
30x = 270.
X = 9.
Therefore,
9 + y = 13 ................(1).
Y= 13 - 9 = 4.
Hence, there are 9 ₦50 notes (and 4 ₦20 notes) in the box.
................................................
CHECK: 9 + 4 = 13, and (50 x 9) + (20 x 4) = 450 + 80 = 530.
Thursday, May 17, 2018
How to convert recurring decimals to common fractions
Converting recurring decimals to common fractions.
This requires a special process. Let the recurring decimal be equal to
a letter of the alphabet. Multiply both sides by 10 raised to power n,
where n is the number of recurring digits in the recurring decimal.
Subtract the first equation from the second and you will get the
fractional value of the recurring decimal.
EXAMPLES
(a) 0.111................
Solution.
R = 0.11... .............(1).
10R = 1.11... ................(2).
Subtract equation (1) from (2).
9R = 1.
R = 1/9.
(b) 0.1515................
Solution
X = 0.1515... .............(1).
100X = 15.1515... .............(2).
Subtract (1) from (2).
99X = 15.
X = 15/99 = 5/33.
(c)0.285714285714................
Solution
t = 0.285714... .............(1).
1 000 000 t = 285 714. 285714... .............(2).
Subtract (1) from (2).
999 999 t = 285 714.
t = 285 714 / 999 999 = 2/7.
This requires a special process. Let the recurring decimal be equal to
a letter of the alphabet. Multiply both sides by 10 raised to power n,
where n is the number of recurring digits in the recurring decimal.
Subtract the first equation from the second and you will get the
fractional value of the recurring decimal.
EXAMPLES
(a) 0.111................
Solution.
R = 0.11... .............(1).
10R = 1.11... ................(2).
Subtract equation (1) from (2).
9R = 1.
R = 1/9.
(b) 0.1515................
Solution
X = 0.1515... .............(1).
100X = 15.1515... .............(2).
Subtract (1) from (2).
99X = 15.
X = 15/99 = 5/33.
(c)0.285714285714................
Solution
t = 0.285714... .............(1).
1 000 000 t = 285 714. 285714... .............(2).
Subtract (1) from (2).
999 999 t = 285 714.
t = 285 714 / 999 999 = 2/7.
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