MATHEMATICS (POST JAMB)


MATHEMATICS (POST JAMB)

Question 1 of 50.

Make q the subject of the formula in the equation mna2−pqb2=1mna2−pqb2=1

1. q=b2(mn−a2)a2pq=b2(mn−a2)a2p
2. q=m2n−a2p2q=m2n−a2p2
3. q=mn−2b2a2q=mn−2b2a2
4. q=b2(n2−ma2)nq=b2(n2−ma2)n

Explanation to the correct answer

mna2−pqb2=1mna2−pqb2=1

mna2−1=pqb2mna2−1=pqb2

mn−a2a2=pqb2mn−a2a2=pqb2

pq=b2(mn−a2)a2pq=b2(mn−a2)a2

q=b2(mn−a2)a2pq=b2(mn−a2)a2p

Question 2 of 50.

The angle of elevation of the top of a tree from a point on the ground 60m away from the foot of the tree is 78°. Find the height of the tree correct to the nearest whole number.

1. 148m
2. 382m
3. 282m
4. 248m

Explanation to the correct answer

tan78=h60tan⁡78=h60

h=60tan78h=60tan⁡78

h=60×4.705=282.27mh=60×4.705=282.27m

≊≊ 282m to the nearest whole number.

Question 3 of 50.

A binary operation ⊗⊗ is defined by m⊗n=mn+m−nm⊗n=mn+m−n on the set of real numbers, for all m, n ∈∈ R. Find the value of 3 ⊗⊗ (2 ⊗⊗ 4).

1. 6
2. 25
3. 15
4. 18

Explanation to the correct answer

m⊗n=mn+m−nm⊗n=mn+m−n

3 ⊗⊗ (2 ⊗⊗ 4)

2 ⊗⊗ 4 = 2(4) + 2 - 4 = 6

3 otimesotimes 6 = 3(6) + 3 - 6  = 15

Question 4 of 50.

The table below shows the number of pupils in a class with respect to their ages. If a pie chart is constructed to represent the age, the angle corresponding to 8 years old is

Age in years 7 8 9 10 11

No of pupils 4 13 30 44 9

1. 48.6°
2. 56.3°
3. 46.8°
4. 13°

Explanation to the correct answer

Total number of pupils : 4 + 13 + 30 + 44 + 9 = 100
The number of 8 - year olds = 13
The angle represented by the 8-year olds on the pie chart = 13100×360°13100×360°
= 46.8°

Question 5 of 50.

In a class of 50 students, 40 students offered Physics and 30 offered Biology. How many offered both Physics and Biology?

1. 42
2. 20
3. 70
4. 54

Explanation to the correct answer

n(Total) = 50

n(Physics) = 40

n(Biology) = 30

Let n(Physics and Biology) = x

n(Physics only) = 40 -x

n(Biology only) = 30 - x

40 - x + 30 - x + x = 50

70 - x = 50

x = 20

 

Question 6 of 50.

Rationalize √2+√3√2−√32+32−3

1. −5−2√6−5−26
2. −5+3√2−5+32
3. 5−2√35−23
4. 5+2√65+26
Question 7 of 50.

Find the length of the chord |AB| in the diagram shown above.

1. 4.2 cm
2. 4.3 cm
3. 3.2 cm
4. 3.4 cm

Explanation to the correct answer

Length of chord = 2rsin(θ2)2rsin⁡(θ2)

= 2(3)sin(682)2(3)sin⁡(682)

= 6sin346sin⁡34

= 6×0.5596×0.559

= 3.354 cm ≊≊  3.4 cm

Question 8 of 50.

Given sin58°=cosp°sin⁡58°=cos⁡p°, find p.

1. 48°
2. 58°
3. 32°
4. 52°

Explanation to the correct answer

sinθ=cos(90−θ)sin⁡θ=cos⁡(90−θ)

sinθ=cos(90−58)sin⁡θ=cos⁡(90−58)

= cos32cos⁡32

Question 9 of 50.

23÷4514+35−1323÷4514+35−13

1. 31503150
2. 20312031
3. 31203120
4. 50315031

Explanation to the correct answer

23÷4514+35−1323÷4514+35−13

23÷45=23×5423÷45=23×54

= 5656

14+35−13=15+36−206014+35−13=15+36−2060

= 31603160

∴23÷4514+35−13=56÷3160∴23÷4514+35−13=56÷3160

= 56×603156×6031

= 50315031

 

Question 10 of 50.

If 6x3+2x2−5x+16x3+2x2−5x+1 divides x2−x−1x2−x−1, find the remainder.

1. 9x + 9
2. 6x + 8
3. 5x - 3
Question 11 of 50.

If a fair coin is tossed 3 times, what is the probability of getting at least two heads?

1. 2323
2. 4545
3. 2525
4. 1212

Explanation to the correct answer

The outcomes are {HHH, HHT, HTT, HTH, THH, THT, TTH, TTT}

P(at least two heads) = 4848

= 1212

 

Question 12 of 50.

In how many ways can the word MATHEMATICIAN be arranged?

1. 6794800 ways
2. 2664910 ways
3. 6227020800 ways
4. 129729600 ways

Explanation to the correct answer

MATHEMATICIAN = 13 letters with 2M, 3A, 2T, 2I.

Hence, the word MATHEMATICIAN can be arranged in 13!2!3!2!2!13!2!3!2!2!

= 129729600 ways

 

Question 13 of 50.

find MT+2MMT+2M

Given matrix M = ∣∣

∣∣−2040−16563∣∣

∣∣|−2040−16563|

 

1. ∣∣ ∣∣−421605062∣∣ ∣∣|−421605062|
2. ∣∣ ∣∣−60130−31814189∣∣ ∣∣|−60130−31814189|
3. ∣∣ ∣∣52601134−7∣∣ ∣∣|52601134−7|
4. ∣∣ ∣∣−4080−2−1610126∣∣ ∣∣|−4080−2−1610126|

Explanation to the correct answer

M = ∣∣

∣∣−2040−16563∣∣

∣∣|−2040−16563|

MTT = ∣∣

∣∣−2050−16463∣∣

∣∣|−2050−16463|

2M = ∣∣

∣∣−4080−21210126∣∣

∣∣|−4080−21210126|

MTT + 2M = ∣∣

∣∣−60130−31814189∣∣

∣∣|−60130−31814189|

Question 14 of 50.

Find the mean of the data.

Score (x) 0 1 2 3 4 5 6

Freq (f) 5 7 3 7 11 6 7

 

1. 3.26
2. 4.91
3. 6.57
4. 3.0

Explanation to the correct answer

Mean = ∑fx∑f∑fx∑f

= 1504615046

= 3.26

 

Question 15 of 50.

Find the variance

Score (x) 0 1 2 3 4 5 6

Freq (f) 5 7 3 7 11 6 7

 

 

1. 3.42
2. 4.69
3. 4.85
4. 3.72

Explanation to the correct answer

Variance = ∑f(x−¯x)∑f∑f(x−x¯)∑f

= 170.88846170.88846

= 3.72

 

Question 16 of 50.

The locus of a point which moves so that it is equidistant from two intersecting straight lines is the

1. bisector of the two lines
2. line parallel to the two lines
3. angle bisector of the two lines
4. perpendicular bisector of the two lines

Explanation to the correct answer

The locus of a points equidistant from two intersecting straight lines is a pair of bisectors that bisect the angles formed by the two intersecting lines.

Question 17 of 50.

From the cyclic quadrilateral MNOP above, find the value of x

1. 16°
2. 25°
3. 42°
4. 39°

Explanation to the correct answer

The sum of two opposite angles of a cyclic quadrilateral = 180°

∴∴ (2x + 18)° + 84° = 180°

2x + 102° = 180° ⟹⟹ 2x = 78°

x = 39°

Question 18 of 50.

If 4sin2x−3=04sin2⁡x−3=0, find the value of x, when 0° ≤≤ x ≤≤ 90°

1. 90°
2. 45°
3. 60°
4. 30°

Explanation to the correct answer

4sin2x−3=04sin2⁡x−3=0

4sin2x=3⟹sin2x=344sin2⁡x=3⟹sin2⁡x=34

sinx=√32sin⁡x=32

∴x=sin−1(√32)∴x=sin−1⁡(32)

x = 60°

Question 19 of 50.

In the figure above, |CD| is the base of the triangle CDE. Find the area of the figure to the nearest whole number.

Area of rectangle ABCD = length x breadth

= 7 x 4 

= 28 cm22

Area of triangle CDE = 1212 base x height

= 12×3×412×3×4

= 6 cm22

Area of the figure = 28 cm22 + 6 cm22

= 34 cm22

 

1. 56 cm22
2. 24 cm22
3. 42 cm22
4. 34 cm22

Explanation to the correct answer

Question 20 of 50.

The marks scored by 30 students in a Mathematics test are recorded in the table below:

Scores (Mark) 0 1 2 3 4 5

No of students 4 3 7 8 6 2

What is the total number of marks scored by the children?

 

1. 82
2. 15
3. 63
4. 75

Explanation to the correct answer

Scores (Mark)012345

No of students437862

fx 0 3 14 24 24 10 75

 

Question 21 of 50.

If given two points A(3, 12) and B(5, 22) on a x-y plane. Find the equation of the straight line with intercept at 2.

1. y = 5x + 2
2. y = 5x + 3
3. y = 12x + 2
4. y = 22x + 3

Explanation to the correct answer

The equation of a straight line is given as y=mx+by=mx+b 

where m = the slope of the line

b = intercept

Given points A(3, 12) and B(5, 22), the slope = 22−125−322−125−3

= 102102 = 5

Hence, the equation of the line is y=5x+2y=5x+2.

Question 22 of 50.

If P(2, m) is the midpoint of the line joining Q(m, n) and R(n, -4), find the values of m and n.

1. m = 0, n = 4
2. m = 4, n = 0
3. m = 2, n = 2
4. m = -2, n = 4

Explanation to the correct answer

Q(m, n) and R(n, -4)

Midpoint : P(2, m)

⟹(m+n2,n−42)=(2,m)⟹(m+n2,n−42)=(2,m)

m+n=2×2⟹m+n=4...(i)m+n=2×2⟹m+n=4...(i)

n−4=2×m⟹n−4=2m...(ii)n−4=2×m⟹n−4=2m...(ii)

Solving (i) and (ii) simultaneously,

m = 0 and n = 4.

 

Question 23 of 50.

If ∣∣∣2−4x9∣∣∣=58|2−4x9|=58, find the value of x.

1. 10
2. 30
3. 14
4. 28

Explanation to the correct answer

∣∣∣2−4x9∣∣∣=58|2−4x9|=58

⟹(2×9)−(−4×x)=58⟹(2×9)−(−4×x)=58

18+4x=58⟹4x=58−18=4018+4x=58⟹4x=58−18=40

x=10x=10

 

Question 24 of 50.

If y=6x3+2x−2−x−3y=6x3+2x−2−x−3, find dydxdydx.

1. dydx=15x2−4x−2−3x−2dydx=15x2−4x−2−3x−2
2. dydx=6x+4x−1−3x−4dydx=6x+4x−1−3x−4
3. dydx=18x2−4x−3+3x−4dydx=18x2−4x−3+3x−4
4. dydx=12x2+4x−1−3x−2dydx=12x2+4x−1−3x−2

Explanation to the correct answer

y=6x3+2x−2−x−3y=6x3+2x−2−x−3

dydx=18x2−4x−3+3x−4dydx=18x2−4x−3+3x−4

 

Question 25 of 50.

ddx[log(4x3−2x)]ddx[log⁡(4x3−2x)] is equal to

1. 12x−24x212x−24x2
2. 43x2−2x7x43x2−2x7x
3. 4x2−27x+64x2−27x+6
4. 12x2−24x3−2x12x2−24x3−2x

Explanation to the correct answer

ddx[log(4x3−2x)]ddx[log⁡(4x3−2x)] ... (1)

Let u = 4x33 - 2x.

ddx(log(4x3−2x))=(ddu)(dudx)ddx(log⁡(4x3−2x))=(ddu)(dudx)

ddu(logu)ddu(log⁡u) = 1u1u

dudx=12x2−2dudx=12x2−2

∴ddx[log(4x3−2x)]=12x2−2u∴ddx[log⁡(4x3−2x)]=12x2−2u

= 12x2−24x3−2x12x2−24x3−2x

 

Question 26 of 50.

If f(x)=3x3+4x2+x−8f(x)=3x3+4x2+x−8, what is the value of f(-2)?

1. -24
2. 30
3. -18
4. -50

Explanation to the correct answer

f(x)=3x3+4x2+x−8f(x)=3x3+4x2+x−8

f(−2)=3(−2)3+4(−2)2+(−2)−8f(−2)=3(−2)3+4(−2)2+(−2)−8

= −24+16−2−8−24+16−2−8

= -18

 

Question 27 of 50.

Solve for x in 4x−63≤3+2x24x−63≤3+2x2

1. x≤112x≤112
2. x≤212x≤212
3. x≥212x≥212
4. x≥112x≥112

Explanation to the correct answer

4x−63≤3+2x24x−63≤3+2x2

2(4x - 6) ≤≤ 3(3 + 2x)

8x - 12 ≤≤ 9 + 6x

8x - 6x ≤≤ 9 + 12

2x ≤≤ 21

x≤212x≤212

 

Question 28 of 50.

Solve the inequality: -7 ≤≤ 9 - 8x < 16 - x

1. -1 ≤≤ x ≤≤ 2
2. -1 ≤≤ x < 2
3. -1 < x < 2
4. -1 < x ≤≤ 2

Explanation to the correct answer

 -7 ≤≤ 9 - 8x < 16 - x

-7 ≤≤ 9 - 8x and 9 - 8x < 16 - x

-7 - 9 ≤≤ -8x and -8x + x < 16 - 9

-16 ≤≤ -8x and -7x < 7

∴∴ x ≤≤ 2 and -1 < x

-1 < x ≤≤ 2.

 

Question 29 of 50.

The nth term of a sequence is given by 22n−12n−1. Find the sum of the first four terms.

1. 74
2. 32
3. 42
4. 170

Explanation to the correct answer

Tn=22n−1Tn=22n−1

T1=22(1)−1T1=22(1)−1

= 2

T2=22(2)−1T2=22(2)−1

= 8

T3=22(3)−1T3=22(3)−1

= 32

T4=22(4)−1T4=22(4)−1

= 128

T1+T2+T3+T4=2+8+32+128T1+T2+T3+T4=2+8+32+128

= 170

 

Question 30 of 50.

Integrate ∫2−1(2x2+x)dx∫−12(2x2+x)dx

1. 412412
2. 312312
3. 712712
4. 514514

Explanation to the correct answer

∫2−1(2x2+x)dx∫−12(2x2+x)dx

= [2x2+13+x1+12]2−1[2x2+13+x1+12]−12

= [2x33+x22]2−1[2x33+x22]−12

= (2(2)33+222)−(2(−1)33+(−1)22)(2(2)33+222)−(2(−1)33+(−1)22)

= (163+2)−(−23+12)(163+2)−(−23+12)

= 223−(−16)223−(−16)

= 223+16223+16

= 152152

= 712

 

Question 31 of 50.

If P varies inversely as the square root of q, where p = 3 and q = 16, find the value of q when p = 4.

1. 12
2. 8
3. 9
4. 16

Explanation to the correct answer

p∝1√qp∝1q

⟹p=k√q⟹p=kq

when p = 3, q = 16.

3=k√163=k16

k=3×4=12k=3×4=12

∴p=12√q∴p=12q

when p = 4,

4=12√q⟹√q=1244=12q⟹q=124

√q=3⟹q=32q=3⟹q=32

q=9q=9

 

Question 32 of 50.

Tade bought 200 mangoes at 4 for ₦2.50. 30 out of the mangoes got spoilt and the remaining were sold at 2 for ₦2.40. Find the percentage profit or loss.

1. 43.6% loss
2. 35% profit
3. 63.2% profit
4. 28% loss

Explanation to the correct answer

 200 mangoes at 4 for N2.50

⟹⟹ Total cost price = 2004×N2.502004×N2.50

= N 125.00

Since 30 mangoes got spoilt ⟹⟹ Left over = 200 - 30

= 170 mangoes 

170 mangoes at 2 for N 2.40

⟹⟹ Total selling point = 1702×N2.401702×N2.40

= N 204.00

Profit : N (204.00 - 125.00) = N 79.00

% profit = 79125×10079125×100

= 63.2% profit.

 

Question 33 of 50.

The simple interest on ₦8550 for 3 years at x% per annum is ₦4890. Calculate the value of x to the nearest whole number.

1. 19%
2. 20%
3. 25%
4. 16.3%

Explanation to the correct answer

S.I = PRT100PRT100

⟹⟹ N 4890 = 8550×3×x1008550×3×x100

x=4890×1008550×3x=4890×1008550×3

x=19.06x=19.06

x≊19x≊19

 

Question 34 of 50.

Simplify 81−34−34 x 251212 x 2432525

1. 2525
2. 3535
3. 5252
4. 5353

Explanation to the correct answer

81−34−34 x 251212 x 2432525

= (4√81)−3×√25×(5√243)2(814)−3×25×(2435)2

= 5×323−35×323−3

= 5353

 

Question 35 of 50.

Find the value of (0.5436)30.017×0.219(0.5436)30.017×0.219 to 3 significant figures.

1. 46.2
2. 43.1
3. 534
4. 431

Explanation to the correct answer

(0.5436)30.017×0.219(0.5436)30.017×0.219

= 0.160630.017×0.2190.160630.017×0.219

= 43.1 (to 3 s.f)

 

Question 36 of 50.

If S = (4t + 3)(t - 2), find ds/dt when t = 5 secs.

1. 50 units per sec
2. 35 units per sec
3. 22 units per sec
4. 13 units per sec

Explanation to the correct answer

s=(4t+3)(t−2)s=(4t+3)(t−2)

dsdt=(4t+3)(1)+(t−2)(4)dsdt=(4t+3)(1)+(t−2)(4)

= 4t+3+4t−84t+3+4t−8

= 8t - 5

dsdt(t=5secs)=8(5)−5dsdt(t=5secs)=8(5)−5

= 40 - 5 

= 35 units per sec

 

Question 37 of 50.

The angles of a polygon are given by 2x, 5x, x and 4x respectively. The value of x is

1. 31°
2. 30°
3. 26°
4. 48°

Explanation to the correct answer

Since there are 4 angles given, the polygon is a quadrilateral.

Sum of angle in a quadrilateral = 360°

∴∴ 2x + 5x + x + 4x = 360°

12x = 360°

x = 30°

 

Question 38 of 50.

The weight of a day-old chick was measured to be 0.21g. If the actual weight of the chick is 0.18g, what was the percentage error in the measurement?

1. 15.5%
2. 18.2%
3. 14.8%
4. 16.7%

Explanation to the correct answer

Actual weight = 0.18g

Error = 0.21g - 0.18g

= 0.03g

% error = 0.030.18×1000.030.18×100

= 16.7%

 

Question 39 of 50.

Evaluate (60.32÷20.084)−1(60.32÷20.084)−1 correct to 1 decimal place.

1. 1.3
2. 2.5
3. 4.6
4. 3.2

Explanation to the correct answer

(60.32÷20.084)−1(60.32÷20.084)−1

= (60032÷200084)−1(60032÷200084)−1

= (60032×842000)−1(60032×842000)−1

= (6380)−1(6380)−1

= 80638063

= 1.3 (to 1 decimal place)

 

Question 40 of 50.

If 2x+yx+y = 16 and 4x−y=132x−y=132, find the values of x and y.

1. x = 3434, y = 114114
2. x = 3434, y = 134134
3. x = 2323, y = 4545
4. x = 2323, y = 134134

Explanation to the correct answer

2x+yx+y = 16 ; 4x−yx−y = 132132.

⟹2x+y=24⟹2x+y=24

x+y=4...(1)x+y=4...(1)

22(x−y)=2−522(x−y)=2−5

22x−2y=2−522x−2y=2−5

⟹2x−2y=−5...(2)⟹2x−2y=−5...(2)

Solving the equations (1) and (2) simultaneously, we have

x = 3434 and y = 134

 

Question 41 of 50.

Simplify 0.0839×6.3815.44 to 2 significant figures.

1. 0.2809
2. 2.51
3. 3.5
4. 0.098
Question 42 of 50.

Find the value of x and y in the simultaneous equation: 3x + y = 21; xy = 30

1. x = 3 or 7, y = 12 or 8
2. x = 6 or 1, y = 11 or 5
3. x = 2 or 5, y = 15 or 6
4. x = 1 or 5, y = 10 or 7

Explanation to the correct answer

3x + y = 21 ... (i);

xy = 30 ... (ii)

From (ii), y=30x. Putting the value of y in (i), we have

3x + 30x = 21

⟹ 3x2 + 30 = 21x

3x2 - 21x + 30 = 0

3x2 - 15x - 6x + 30 = 0

3x(x - 5) - 6(x - 5) = 0

(3x - 6)(x - 5) = 0

3x - 6 = 0 ⟹ x = 2.

x - 5 = 0 ⟹ x = 5.

If x = 2, y = 302 = 15;

If x = 5, y = 305 = 6.

 

Question 43 of 50.

Points X and Y are 20km North and 9km East of point O, respectively. What is the bearing of Y from X? Correct to the nearest degree.

1. 24°
2. 56°
3. 127°
4. 156°

Explanation to the correct answer

tanθ=920=0.45

θ=tan−1(0.45)

= 24.23°

∴ The bearing of Y from X = 180° - 24.23°

= 155.77°

= 156° (to the nearest degree)

 

Question 44 of 50.

If P=(Q(R−T)15)13, make T the subject of the formula.

1. T=15R−QP3
2. T=R−15P3Q
3. T=R−15P3Q
4. T=R+P315Q

Explanation to the correct answer

P=(Q(R−T)15)13

P3=Q(R−T)15

Q(R−T)=15P3

R−T=15P3Q

T=R−15P3Q

 

Question 45 of 50.

In the diagram above, O is the centre of the circle ABC, < ABO = 26° and < BOC = 130°. Calculate < AOC.

1. 26°
2. 13°
3. 80°
4. 102°

Explanation to the correct answer

< BAC = 1302 (angle subtended at the centre)

< BAC = 65°

Also, x = 26° (theorem)

y = 65° - 26° = 39°

< AOC = 180° - (39° + 39°)

= 102°

 

Question 46 of 50.

Each of the interior angles of a regular polygon is 140°. Calculate the sum of all the interior angles of the polygon.

1. 1080°
2. 1260°
3. 2160°
4. 1800°

Explanation to the correct answer

Since each interior angle = 140°;

Each exterior angle = 180° - 140° = 40°

Number of sides of the polygon = 360°40°

= 9 

Sum of angles in the polygon = 140° x 9

= 1260°

 

Question 47 of 50.

A man bought a car newly for ₦1,250,000. He had a crash with the car and later sold it at the rate of ₦1,085,000. What is the percentage gain or loss of the man?

1. 43.7% loss
2. 13.2% gain
3. 13.2% loss
4. 43.7% gain

Explanation to the correct answer

Cost price of the car = N 1,250.00

Selling price = N 1,085.00

Loss = N (1250 - 1085)

= N 165.00

% loss = 1651250×100

= 13.2% loss

 

Question 48 of 50.

If the volume of a frustrum is given as V=πh3(R2+Rr+r2), find dVdR.

1. πh3(2R+r)
2. 2R+r+πh3
3. πh3(2R2+r+2r)
4. 2R23πh

Explanation to the correct answer

V=πh3(R2+Rr+r2)

V=πR2h3+πRrh3+πr2h3

dVdR=2πRh3+πrh3

= π3(2R+r)

 

Question 49 of 50.

Express (0.0439÷3.62) as a fraction.

1. 21100
2. 211000
3. 121000
4. 12100

Explanation to the correct answer

(0.0439÷3.62)

= 0.01213

≊ 0.012

= 121000

 

Question 50 of 50.

If 251−x×5x+2÷(1125)x=625−1, find the value of x.

1. x = -4
2. x = 2
3. x = -2
4. x = 4

Explanation to the correct answer

251−x×5x+2÷(1125)x=625−1

(52)(1−x)×5(x+2)÷(5−3)x=(54)−1

52−2x×5x+2÷5−3x=5−4

5(2−2x)+(x+2)−(−3x)=5−4

Equating bases, we have

2−2x+x+2+3x=−4

4+2x=−4⟹2x=−4−4

2x=−8

x=−4

 

Next question 1 of 50

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