How good are you with figures and formulas? Find out with these Mathematics past questions and answers. This Test is useful for both job aptitude test candidates and students preparing for JAMB, WAEC, NECO or Post UTME.
The ages of students in a small primary school were recorded in the table below.
Age | 5 - 6 | 7 - 8 | 9 -10 |
Frequency | 29 | 40 | 38 |
Estimate the mean
7.7
7.5
7.8
7.6
Correct answer is A
Class Interval | Class Mark | Frequency (f) | fx |
5 - 6 | 5.5 | 29 | 5.5 x 29 = 159.5 |
7 - 8 | 7.5 | 40 | 7.5 x 40 = 300 |
9 - 10 | 9.5 | 38 | 9.5 x 38 = 361 |
∑f=107 | ∑fx=820.5 |
Mean = ∑fx∑f=820.5107 = 7.7 (1 d.p)
Evaluate the following limit: limx→2x2+4x−12x2−2x
4
8
0
2
Correct answer is A
limx→2x2+4x−12x2−2x = limx→2(x−2)(x+6)x(x−2)
limx→2x+6x
2+62=82 = 4
15 years
25 years
20 years
30 years
Correct answer is C
Amount (A) = Principal (P) + Interest (I) i.e. A = P + I
1 = PTR100
A = 3P; T = 10 years (Given)
Since A = P + I
⟹3P=P+P×10×R100
⟹3P=P+10PR100
⟹3P=P+PR10
⟹3P−P=PR10
⟹2P=PR10
⟹2P1=PR10
⟹ PR = 20P
∴ R = 20%
Since the rate stays the same
A = 5P; R = 20%;T =?; A =P + I
\implies 5P = P + \frac {p \times T \times 20}{100}
\implies 5P - P = \frac {2PT}{10}
\implies 4P = \frac {2PT}{10}
\implies 2P = \frac {PT}{10}
\implies PT = 20P
\therefore T = 20 years
Find the value of t, if the distance between the points P(–3, –14) and Q(t, –5) is 9 units.
3
2
-3
-2
Correct answer is C
Let the given points be:
P(-3, -14) = (x_1, y_1)
Q(t, -5) = (x_2, y_2)
PQ = 9 units (given)
Using the distance formula,
d = √ [ (x_2 - x_1)^2 + (y_2 - y_1)^2]
PQ = √ [ (t - (-3))^2 + (-5 + 14)^2]
\implies √ [ (t + 3)^2 + 81] = 9
Squaring on both sides,
⇒ (t + 3)^2 + 81 = 81
⇒ (t + 3)^2 = 0
⇒ t + 3 = 0
∴ t = -3