Are you wondering where you can find NECO GCE General Mathematics Questions to guide you before participating in the NECO exams? Don't fret! We've curated a comprehensive set of past questions along with detailed answers to help you prepare effectively.
NECO GCE General Mathematics Questions:
(1a)
Using the sine rule, Solve the equation b/sin20 = 6/sin30 for the side length ‘b’.
(b) (i) Given P = N15000, R = 10%, and T = 3 years, calculate the simple interest (I) using the formula.
(2a) (Solve: 7x + 4 < (4x + 3).
(b) Salem, Sunday, and Shaka shared a sum of N 1,100.00. For every N2.00 Salem gets, Sunday gets fifty kobo and for every N 4.00, Sunday gets, Shaka gets N 2.00. Find Shaka’s share.
(3a) The present ages of a father and his son are in the ratio of 10: 3. If the son is 15 years old now, in how many years will the ratio of their ages be 2: 1?
(b) The arithmetic mean of x, y, and z is 6, while that of x, y, z, t, u, v and w is 9. Calculate the arithmetic mean of t, u, v, and w.
(4a) A boy, 1.2 m tall, stands 6 m away from the foot of a vertical lamp pole 4.2 m long. If the lamp is at the tip of the pole,
Calculate the:
(i) length of the shadow of the boy cast by the lamp;
(ii) angle of elevation of the lamp from the boy, correct to the nearest degree.
Answers for the NECO GCE General Mathematics Questions:
Here are the answers:
Answer One
(1a) Using sine rule
b/sin20 = 6/sin30
bsin30 = 6sin120
b 6sin120/sin30
= 6×0.2511/0.4540
= 5.7063/0.4540
b = 12.57 ≠ 12.6cm
(b) I = PRT/100, p=N15000 R=10% and I=3years
A = P+ I
where I = 15000*10*3/100=N4500
A=4500+15000 =N19500
Answer Two
(2a) To solve the inequality 7x+4<4x+3, let’s isolate x on one side of the inequality.
First, subtract 4x from both sides:
7x −4x +4<4x−4x 3
This simplifies to:
3x+4<3
Now, subtract 4 from both sides:
3x+4−4<3−4
Which simplifies to:
3x<−1
Finally, divide both sides by 3 to solve for x
x< −1/3
So, the solution to the inequality 7x+4<4x+3 is x < −1/3
(b) Salem and Sunday’s Share: For every ?2 Salem gets, Sunday gets fifty kobo (?0.50).
Sunday and Shaka’s Share: For every ?4 Sunday gets, Shaka gets ?2.
From the first part, the ratio of Salem’s share to Sunday’s share is ?2 to ?0.50, or simplified, 4:1 (since ?2 / ?0.50 = 4).
So, for every 4 parts Salem gets, Sunday gets 1 part.
From the second part, the ratio of Sunday’s share to Shaka’s share is ?4 to ?2, or simplifying, 2:1. So, for every 2 parts Sunday gets, Shaka gets 1 part.
Let’s combine these ratios to find the overall distribution among Salem, Sunday, and Shaka.
Since Sunday’s share is a common element between the two ratios, we can multiply the first ratio by 2 and keep the second ratio as is.
This gives us:
Salem: Sunday = 4:1 becomes 8:2 after multiplying by 2 (to make Sunday’s share compatible with the second ratio).
Sunday: Shaka = 2:1 as given.
So now, we have Salem:Sunday: Shaka = 8:2:1.
The total parts shared among them equals 8 (for Salem) + 2 (for Sunday) + 1 (for Shaka) = 11 parts.
They share a total of ?1,100.00, and we’ve established 11 parts.
Each part is worth ?1,100.00 / 11 = ?100.00.
Since Shaka’s share is 1 part, Shaka gets ?100.00.
Answer Three
(3a) Father and Son’s Age Problem
Given:
The present ratio of the ages of a father and his son is 10:3. The son is 15 years old.
Let’s first find the father’s current age.
Since the ratio of their ages is 10:3, and the son is 15 years old, which corresponds to the ‘3’ part of the ratio:
Son’s age per part = 15/3 = 5 years
Therefore, for the ’10’ part of the ratio, the father’s age is:
Father’s age = 5 × 10 = 50 years
Now, let x be the number of years in the future when their ages will have a 2:1 ratio.
The son’s age will be 15+x and the father’s age will be 50+x.
Setting up the ratio for their future ages to be 2:1:
15+x
50+x = 2/1
?Solving for
50+x=2(15+x)
50+x=30+2x
50−30=2x−x
20=x
So, in 20 years, the ratio of their ages will be 2:1.
(b) Arithmetic Mean Problem
Given:
The arithmetic mean of x, y, and z is 6.
The arithmetic mean of x, y, z, t, u, v, and w is 9.
The formula for the arithmetic mean is the sum of the elements divided by the number of elements.
For x, y, and z:
Mean = 6 = x+y+z/3
?Multiplying both sides by 3:
x+y+z=18
For x, y, z, t, u, v, and w:
Mean = 9 = x+y+z+t+u+v+w/7
?Multiplying both sides by 7:
x+y+z+t+u+v+w=63
Subtracting the sum of x+y+z from this:
63−18=t+u+v+w
45=t+u+v+w
To find the arithmetic mean of t, u, v, and w:
The mean of t,u,v, and w= 45/4 =11.25
Answer Four
(4) The situation forms a right triangle with the boy, the lamp pole, and the shadow. The height of the boy is the opposite side, the lamp pole is the hypotenuse, and the shadow is the adjacent side.
Given:
Height of the boy (opposite side) = 1.2 m
Distance from the boy to the lamp pole (adjacent side) = 6 m
Length of the lamp pole (hypotenuse) = 4.2 m
Length of the Shadow:
(i) Using the tangent function:
tan(θ)= Adjacent/Opposite
?For the boy and the shadow:
tan(θ) = 1.2/6
?Now, solve for θ:
θ=arctan( 1.2/6 )
θ≈11.54
(ii) Length of the Shadow:
To find the length of the shadow, we can use the tangent of tan(θ) = Adjacent/Opposite
?Let x be the length of the shadow: tan (θ) = 1.2/x
?Now, solve for x = 1.2/tan(θ)
Substitute the value of θ we found earlier:
x= 1.2/ tan(11.54)
?x≈ 1.2/0.2126
?x≈5.64
These past questions and answers are invaluable resources to aid your preparation for the NECO GCE General Mathematics exam. Familiarize yourself with the formats and concepts to boost your confidence on exam day.
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