Solution:

Let \(x\) represent Felicia’s present age.
Let \(y\) represent Asare’s present age.

\(\Rightarrow 3x=y+4\)

\(\Rightarrow 3x-y=4 ……… (1)\)

In three years time, Felicia’s age will be \((x+3)\)
Asare’s age will be \((y+3)\)

\(\Rightarrow (x+3)+(y+3)=30\\x+y+6=30\\x+y=30-6\\x+y=24 …….. (2)\)

eqn(1) + eqn(2)

\((3x-y=4)\)

\(+(x+y=24)\)

\(\Rightarrow 4x=28\\\frac{4x}{4}=\frac{28}{4}\\x=7\)

put \(x=7\) into eqn(2)

\(\Rightarrow 7+y=24\\y=24-7\\y=17\)

Hence Felicia’s age is 7ys and Asare is 17yrs old.

By: Abdul Karim

Solution:

Let \(x\) represent the number for bags of rice bought.
Let \(y\) represent the number of bags of beans bought.

\(x+y=17 …….. (1)\)

\(2250x+2400y=39600 ……. (2)\)

from eqn(1) \(x=17-y ……. (3)\)

put \(x=17-y\) into eqn(2)

\(\Rightarrow 2250(17-y)+2400y=39600\\38250-2250y+2400y=39600\\150y+38250=39600\\150y=39600-38250\\150y=1350\\\frac{150y}{150}=\frac{1350}{150}\)

\(y=9\)

from \(x=17-y\)

\(\Rightarrow x=17-9\)

\(x=8\)

Therefore he bought 9 bags of beans.

By: Abdul Karim

Solution:

Total number of students = 52.
Number of science students = 16.

Let \(x\) represent number of boys in the class.
Let \(y\) represent number of girls in the class.

\(\Rightarrow \frac{1}{3}x+\frac{1}{4}y=16\)

\(4x+3y=92 ……. (1)\)

\(x+y=52 ……. (2)\)

eqn(2)\(\times\)4

\(\Rightarrow 4x+4y=208 ……. (3)\)

eqn(1)\(-\)eqn(3)

\((4x+3y=192)\\-(4x+4y=208)\)

\(\Rightarrow -y=-16\\y=16\)

put \(y=16\) into eqn(2)

\(\Rightarrow x+16=52\\x=52-16\\\\\\x=36\)

Hence there are 36 boys in the class.

By: Abdul Karim

Solution:

\(\frac{1}{x}+\frac{1}{y}=5 ……. (1)\)

\(\frac{1}{y}+\frac{1}{x}=1 ……. (2)\)

from eqn(2) \(\frac{1}{y}=1+\frac{1}{x} ……. (3)\)

put eqn(3) into eqn(1)

\(\Rightarrow \frac{1}{x}+1+\frac{1}{x}=5\)

\(\frac{1}{x}+\frac{1}{x}=5-1\)

\(\frac{2}{x}=4\)

\(2=4x\) HintAfter multiplying through by \(x\)

\(\frac{2}{4}=\frac{4x}{4}\)

\(x=\frac{1}{2}\)

put \(x=\frac{1}{2}\) into eqn(2)

\(\Rightarrow \frac{1}{y}=1+\frac{1}{\frac{1}{2}}\)

\(\frac{1}{y}=1+2\)

\(\frac{1}{y}=3\)

\(3y=1\)

\(\frac{3y}{3}=\frac{1}{3}\)

\(y=\frac{1}{3}.\)

By: Abdul Karim

Solution:

(i) Let \(x\) represent the fare for a child and \(y\) represent the fare for a teacher.

\(x=\frac{1}{2}y\)

\(2x=y\) HintAfter multiplying through by \(2\)

\(2x-y=0 ……. (1)\)

\(120x+15y=180 ……. (2)\)

devide eqn(2) by 15

\(\Rightarrow 8x+y=12 ……. (3)\)

from eqn(1), \(y=2x ……. (4)\)

put \(y=2x\) into eqn(3)

\(Rightarrow 8x+2x=12\)

\(10x=12\)

\(\frac{10x}{10}=\frac{12}{10}\)

\(x=1.20\)

\(y=2x\\\Rightarrow y=2(1.2)\\y=2.4\)

Hence the fare for a child is GH₵1.20.

(ii) \(x(1.2)+20(2.4)=240\)

\(1.2x+48=240\)

\(1.2x=240-48\)

\(1.2x=192\)

\(\frac{1.2x}{1.2}=\frac{192}{1.2}=160\)

Hence 160 children.

By: Abdul Karim