Maths 003

StudyPin past Jamb questions and answers - Mathematics 

Question 1 

Find the value of 1101112 + 101002 

A. 11010112 

B. 1001012 

C. 10010112 

D. 10011112 

Answer C 

Solution: 

1101112 

+101002 

10010112 

Question 2 

A woman bought a grinder of N60,000. She sold it as loss 15%. How much 

did she sell it? 

A. N53,000 

B. N52,000 

C. N51,000 

D. N50,000 

Answer C 

Solution: 

Selling price, SP = Ny 

Cost price CP = N60,000 

Percentage loss = 15% 

Percentage = 

CP− SP 

CP 

x 100% 

15% = 60000−y 

x 100% 

60000 

15 (60000) = (60000 – y) 100 

60000-y = 

15 x 60000 

100 

60000-y = 9000 

y=60000 – 9000 

y= N51, 000 

Study Pin Jamb solution | Mathematics | www.studypin.com | 1

Question 3 

Express the product of 0.00043 and 2000 in standard form. 

A. 8.6 x 10 -3 

B. 8.3 x 10 -2 

C. 8.6 x 10 -1 

D. 8.6 x 10 

Answer C 

Solution: 

0.00043 = 43 x 10 -5 

2000 = 2x10 3 

Product = 43 x 10 -5 x 2 x 10 3 

= 86 x 10 -5+3 

= 86 x 10 -2 = 0.86 

= 8.6 x 10 -1 

See NCM 1 page 47 – 48 

Question 5 

A man donates 10% for his monthly net earnings to church. If it amounts to 

N4,500, what is his net monthly income? 

A. N40500 

B. N45000 

C. N52500 

D. N62000 

Answer B 

Solution: 

Amount given to church = 

N 4500 = 10 

100 

x monthly income 

Percentage of amount 

100% 

x monthly income 

Monthly income = 

N4500 x 100 

100 

= N45000 

Question 5 

If log 7.5 = 0.8751, evaluate 2 log 75 + log 750 

A. 6.6252 

B. 6.6253 

C. 66.252 

D. 66.253 

Answer B 

Solution: 


Log 7.5 = 0.8751 

But 2Log 75 + log 750 

2log 7.5x10 + log 7.5 x 100 

2(log7.5 + log10) + Log 7.5 + Log100 

2(0.8751 + 1) + 0.8751 + 2 

2 x 1.8751 + 2.8751 

2.7502 + 2.8751 

=> 6.6253 

See comprehensive maths page 215 

Question 6 

Solve for x in n 8x-2 = 2/25 

A. 4 

B. 6 

C. 8 

D. 10 

Answer D 

Solution: 

8x -2 = 2/25 

Divide by 81 

x -2 = 

2 

= 1 

8 x 25 4 x 25 

x -2 = 1 

100 = 1 x 2 = 1 

100 

Taking reciprocal 

x 2 = 100 

x = √100 = 10 

Question 7 

Simplify 2√2−√3 

√2+ √3 

A. 3√6 − 7 

B. 3√6 + 7 

C. 3√6 − 1 

D. 3√6 + 1 

Answer A 


Solution: 

Simplify 2√2−√3 

√2+ √3 

Taking the conjugate of 

√2 + √3 = √2 - √3 

Then, multiply the denominator and numerator by he conjugate 

2√2−√3 

√2+ √3 x 2√2−√3 

√2− √3 

⇒ 

2√2 x √2 −2√2 x √3 x √2+ √32 

√2 2 + √2 x √3 + √2 x √3− √2 2 

4 − 2√6 − √6 + 3 

2 − √6 + √6 − 3 

7 − 3√6 

−1 

? 

−1 − 3√6 

−1 

⇒ 3√6 – 7 

Reference: Comprehensive Maths page 30 

Question 8 

Evaluate Log 28 + log216 - log24 

Answer C 

Solution: 

Log 28 + log216 - log24 

Log22 3 + log 22 4 – log2 2 2 

3log2 2 + 4log2 2 – 2log2 2 

But log2 2 = 1 

3 + 4 – 2 = 5 

Question 9 

P = {1, 2, 3, 4, 5} and P ∩ Q = {1, 2, 3, 4, 5, 6, 7}, list the element in Q 

A. {6} 

B. {7} 


C. {6,7} 

D. {5,6} 

Answer C 

Solution: 

P = {1, 2, 3, 4, 5} 

PuQ = {1, 2, 3, 4, 5, 6, 7} 

∴ P+Q = PuQ 

Q = PuQ – P 

= {6, 7} 

Comprehensive Maths page 49 

Question 10 

From the vein diagram below, the shaded parts represents 

A. (P ∩ Q) ∪ (P ∩ R) 

B. (P ∪ Q) ∩ (P ∪ R) 

C. (P ∪ Q) ∪ (PER) 

D. (P ∩ Q) ∩ (P ∩ R) 

Answer D 

Solution: 

P 

Q 

R 

Consider the shaded portion: 

First PnQ then, PnR 

Again PnQnR 

∴ (PnQ) n (PnR) 

Question 11 

gt 2 – K – w = 0, make g the subject of the formular. 

A. K+w 

t 2 


B. K−w 

t 2 

C. K+w 

t 

D. K−w 

t 

Answer A 

Solution: 

gt 2 – K – w = 0 

gt 2 = k+ w 

g = 

k+ w 

t 2 

See NCM page 50 – 51 

Question 12 

Factorize 2y 2 – 15xy + 18x2 

A. (2y – 3x)(y - 6x) 

B. (3y – 2x) (y – 6x) 

C. (2y – 3x) (y - 6x) 

D. (2y - 3x) (y - 6x) 

Answer B 

Solution: 

2y2 – 15xy + 18x2 

Product = 2 x 18 = 36 

Sum = - 15 

Factors of 36 and -15 are – 12 and -3 

∴ Fractioning 

2y 2 – 12xy – 3xy + 18x 2 

2y (y – 6x) -3x (y – 6x) 

(2y – 3x)(y - 6x) 

See comprehensive maths page 61 

Question 13 

Find the value of k if y-1 is factor of y 3 + 4y 2 + ky – 6 

A. -6 

B. -4 

C. 0 

D. 1 


Answer D 

Solution: 

f(y) = y 3 + 4y 2 + ky – 6 

For y - 1 to be a factor then, 

y – 1 = 0, y=1 

Substitute y=1 into f(y) =0 

f(1) = 1 3 + 4(1) 2 ⦤ (1) – 6 = 0 

1 + 4 x 1 + k – 6 = 0 

1 + 4 + k – 6 = 0 

K =1 

Question 13 

y varies directly as w2. When y=8, w=2, find y when w = 3 

A. 18 

B. 12 

C. 9 

D. 6 

Answer A 

Solution: 

y⋉ w 2 

y = kw 2 (k= constant of variation) 

when y = 8, w = 2 

8 = K x 2 2 = k= 4 

K = 8/4 = 2 

∴ y = 2w 2 

When w = 3 

y = 2 x 3 2 = 2 x 9 = 18 

Question 15 

P varies directly as Q and inversely as R. When Q = 36 and R = 16, P27. Find 

the relationship between P,Q AND R. 

Q 

A. 

12R 

B. 12Q 

R 

C. 12QR 

D. 12/QR 

Answer B 

Solution: 

P⋉Q, P⋉ 1 R 

∴ P⋉ Q/R (K = constant) 


When Q = 36, R = 16, P = 27 

K x 36 

27 = 

16 

K x 36 = 27 x 16 

K = 27x16 

36 

K = 12 

∴ P = 12Q 

R 

Question 16 

Evaluate the inequality 

x 

2 + 3 4 ≤ 5x 6 − 7 12 

A. x ≥ 4 

B. x ≤ 3 

C. x ≥ −3 

D. x ≤ −4 

Answer A 

Solution: 

x 

2 + 3 4 ≤ 5x 6 − 7 12 

x 

2 − 5x 6 ≤ − 7 12 − 3 4 

3x − 5x 

6 

≤ − −7 − 9 

12 

−2x 

6 

≤ −16 

12 

−x 

3 ≤ −4 

3 

Cancel the denominator 

−x ≤ −4 

x ≥ −4 

−1 

= x ≥ 4 


Comprehensive maths Page 56 

Question 17 

The nth term of anA.P. is 13 while the 10 th term is 31. Find the 24 th term. 

A. 89 

B. 75 

C. 73 

D. 69 

Answer C 

Solution: 

4 th term, T4 = 13 

10 th term, T10 = 31 

Using Tn = a + (n-1) d 

When a= first term 

d = common difference 

T4 = a + (4 – 1)d = 13 _______(1) 

T10 = a + (10-1) = 31 _______(2) 

a + 3d = 13 ________(3) 

a + 9d = 31 _________(4) 

(4) – (3) 

a – a + 9d – 3d = 31 – 13 

6d = 18 

d = 18/6 = 3 

Substitute d = 3 into (3) 

a + 3d = 13 

a + 3 x 3 = 13 

a = 13 – 9 = 4 

The T2 = a + (n-1)d 

T24 = 4 + 23 x 2 

= 4 + 23 x 2 

= 4 69 = 73 

Question 18 

What is the common ratio of the G.P. (√10 + √5) + (√10 + 2√5)+ ...? 

A. √2 

B. √5 


C. 3 

D. 5 

Answer A 

Solution: 

Common ratio 

r = 

√10 +2√5 

√10+ √5 

= 

= 

Second term 

first term 

Third term 

Fourth term 

or 

Taking conjugate of √10 + √5 

= √10 - √5 

∴ r = 

√10 +2√5 

√10+ √5 

x 

√10 −2√5 

√10− √5 

r = √102 − √50 + 2√50−2 √5 x √5 

√10 2 − √50 + √50 − √5 2 

r = 

r = 

10+ √50 −10 

10−5 

√25 x √2 

5 

= 5√2 

5 

= r = √50 

5 

r = √2 

See comprehensive maths page 100, 30 

Question 19 

A binary operation * is defined by x*y= x y . if x*2=12-x, find the possible 

value of x. 

A. 3,4 

B. 3,-4 

C. -3,4 

D. -3,-4 

Answer B 

Solution: 

X × Y = x y 

X × 2 = 12 – x 

Compare x × y to x × 2 

∴ x × 2 = x 2 


x 2 = 12 = x 

x 2 + x − 12 = 0 

x 2 + 4x − 3x − 12 = 0 

x(x + 4) -3 (x + 4) = 0 

(x – 3)(x +4) 

= x- 3 = 0 or x + 4 = 0 

x = 3 or x = -4 

Question 20 

Find y, if ( 5 

−6 

2 7 ) (x y ) =( 7 

A. 8 

B. 5 

C. 3 

D. 2 

Answer C 

Solution: 

( 5 −6 

2 7 ) (x y ) =( 7 

−11 ) 

It becomes 

5x – 6y = 7_______(1) x2 

2x – 7y = -11_____(2) x5 

−11 ) 

10x – 12y = 14 ____(3) 

10x – 35y = -55____(4) 

(4) – (3) 

10x – 10x -35y – (-12y) = -55-14 

-35y + 12y = -69 

-23y = -69 

y = 69/23 = 3 

Question 21 

−x 12 

If | | = -12 find x. 

−1 4 

A. -6 

B. -2 

C. 3 

D. 6 


Answer A 

Solution: 

−x 12 

| 

−1 4 | = -12 

−x × 4 − (−1 × 12) = −12 

−4x + 12 = −12 

−4x = −12 − 12 

−4x = −24 

x = −24/4 

x = −6 

Question 22 

Fin the value of 

A. 12 

B. 10 

C. -1 

D. -2 

Answer D 

Solution: 

0 3 

1 7 

0 5 

2 

8 

4 

0 3 

1 7 

0 5 

2 

8 

4 

Bringing the determinant method 

0 | 7 8 8 7 

| -3 |1 | +2 |1 

5 4 0 4 0 5 | 

0(7 × 4 − 5 × 8) − 3 (1 × 4 − 0 × 8) + 2 (1 × 5 − 0 × 7) 

0 × −12 − 3 × 4 + 2 × 5 

0 − 12 + 10 = −2 

Question 23 

How many side has a regular polygon whose interior angle is 135° each? 

A. 12 

B. 10 

C. 9 

D. 8 


Answer D 

Solution: 

Interior angle = 135 

Exterior angle 180-interior angle 

Exterior angle = 180 – 135 

= 45° 

∴ Number of sides (n) 

n = 360 

exterior = 360 

45 = 8 

Comprehensive maths page 140 

Question 24 

In the figure below, KL//NM, LN bisects

Question 25 

q 30 

P+2q 

From the figure above, what is the value of p? 

A. 135° 

B. 90° 

C. 60° 

D. 45° 

Answer B 

Solution: 

q=30° (Vertically opposite angels) 

∴ P + 2q = P+ 2 x 30 = P + 60 

P + 2q + 30 = 180 (sum of angles on a straight line) 

P + 60 + 30 = 180 

P = 180 – 60 – 30 = 90 

Comprehensive page 125 

Question 26 

Find the value of x in the figure above. 

A. 20√3cm 

B. 10√3cm 

C. 5√3cm 

D. 5√3cm 

Answer B 

Solution: 

30° 

60° 

Since the angles and one side is given, x can be found using sine rule. 

x 

sin60 = 10 

sin30 

Cross multiply 

xsin30 = 10sin60 


x = 10sin60 

sin30 

x = 10 × √3 

2 = 5√3 

1 

2 

x = 5√3 

1 × 2 1 

x = 10√3cm 

Question 27 

If the angle of a sector of a circle with radius 10.5cm is 120°, find the 

perimeter of the sector. 

A. 40cm 

B. 43cm 

C. 45cm 

D. 48cm 

Answer B 

Solution: 

r = 10.5cm, angle ϴ=120° 

Perimeter of a sector 

= 2r + ϴ/360 × 2πr 

2 × 10.5 + 120 

360 × 22 7 × 10.5 

21 + 1 3 × 2 × 22 7 × 10.5 

= 21 + 22 = 43cm 

Question 28 

A cylindrical tank has a capacity of 6160m3. What is the depth of the tank if 

the radius of its base is 28cm? 

A. 8.0m 

B. 7.5m 

C. 5.0m 

D. 2.5m 

Answer D 

Solution: 

Capacity = Volume of tank (v) 

v = 6160cm 3 

Radius, r =28cm 


V = πr 2 h 

6160 x 7 = 22 x 28h 

h = 6160 × 7 

22 × 28 

h = 2.5m 

Question 29 

The locus of a dog tethered to a pole with a rope of 4m is 

A. Circle with diameter 4m 

B. Circle with radius 4m 

C. Semi-circle with diameter 4m 

D. Semi-circle with radius 4m 

Answer B 

Solution: 

The locus of a point to a pole with a rope of 4m is a circle of radius 4m 

Question 30 

Find the mid-point of S(-5,4) and T(-3-2) 

A. -4,2 

B. 4,-2 

C. -4, 1 

D. 41 -1 

Answer C 

Solution: 

(x1, y1) = (-5, 4), (x2, y2) = (-3, -2) 

∴ mid − point (x, y) 

x = x 1+x 2 

2 

= 

x = −5 − 3 

2 

y = y 1+y 2 

2 

−5+ −3 

2 

= −8 

2 = −4 

= 4+−2 

2 

∴ (x, y) = (−4,1) 

= 4−2 

2 = 2 2 = 1 


Question 31 

The gradient of the line joining (x,3) and (1,2) is ½. Find the value of x. 

A. 5 

B. 3 

C. -3 

D. -5 

Answer A 

Solution: 

Gradient (m) = ½ 

(x 1 y 2 ) = (x 1 4), (x 2 y 2 ) = (1,2) 

M = (y 2y 1 ) 

= −2 

(x 2 x 1 ) 1−x 

1 

2 = 2 − 4 

1 − x = −2 

1 − x 

1(1 − x) = 2x − 2 

1 − x = −4 

x = 1 + 4 = +5 

Question 32 

In the figure below, what is the equation of the line that passes the y-axis at 

(0,5) and passes the x-axis at (5,0)? 

A. y=x+5 

B. y=-x+5 

C. y=x-5 

D. y=-x-5 

Answer B 

Solution: 

x 1 y 1 = (0, 5), (x 2 , y 2 ) = (5, 0) 

From equation of a line through two points 

y − y 1 

x − x 1 

= y 2 − y 1 

x 2 − x 1 

y − 5 

x − 0 = 0 − 5 

5 − 0 

5 

4 

3 

2 

1 

6 5 4 3 2 1 1 2 3 4 5 

1 

2 

3 

4 

5 

y − 5 

x 

= −5 

5 

5y – 25 = −5x 


Divided through h by 5 

y − 5 = −x 

y = −x + 5 

NCM 2 page 69 

Question 33 

Calculate the mid point of the line segment y-4x+3=0 which lies between 

the x-axis and y-axis. 

A. ( 3 −3 

8 2 ) 

B. ( 3 3 

8 2 ) 

C. ( −2 2 

2 2 ) 

D. ( −2 3 

3 2 ) 

Answer A 

Solution: 

Mid-point of a line segment y − 4x + 3 = 0 

When y = 0 

0 – 4x = -3 

-4x = -3 

x= −3 ⁄ 

−4 

= 3 4 

∴ (x 1 y 1 ) = ( 3 ⁄ 4, 

0) 

When x=0 

y-4 (0) + 3= 0 

y=-3 

(x 2 y 2 ) = (0, -3) 

∴mid-point coordinates (x, y) 

x = x 1+x 2 

2 

= 3⁄ 4 +0 

2 

= 3/8 

y = y 1 + y 2 

2 

= 0 ± 3 

2 

= −3 

2 


(x, y) = (3/4, −3/2) 

Question 34 

Find the equation of straight line through (-2, 3) and perpendicular to 

4x+3y-5=0 

A. 3x - 4y + 18 = 0 

B. 3x + 2y - 8=0 

C. 4x + 5y + 3=0 

D. 5x - 2y - 11=0 

Answer C 

Solution: 

(x, y) = (−2, 3) 

Equation of line perpendicular to 4x + 3y – 5 = 0 

Gradient of line, m 

4x + 3y -5 = 0 

y + 4.3 x − 5/3 

Perpendicularity, m1 m2 = -1 

m2 = -1/m1 = −1 

4 = ⁄ (3) 

3 4 

Using equation 

y = mx + c 

y = 3 4 x 

4y = 3x 

3x − 4y = C 

For (x, y) = (-2, 3) 

3(-2) – 4(3) = c 

-6 – 12 = c 

C = -18 

Substituting c = -18 

3x – 4y = c = −18 

Question 35 

Find the value of 1+cosθ 

A. 25/13 

B. 18/13 

C. 8/13 

D. 5/13 


Answer B 

Solution: 

sinϴ = 12/13 

12 

13 

Using Pythagoras rule 

13 2 =12 2 +x 2 

x 2 -13 2 – 12 2 = 169 – 144 

x 2 = 25 

x = √25 =5 

x 

ϴ 

∴ + cosϴ = 1 + x 13 = 1 1 + 5 13 

= 13 + 5/13 

= 18 

13 

Question 37 

Find the minimum value of y = x 2 − 2x – 2 

A. 4 

B. 1 

C. -1 

D. -4 

Answer D 

Solution: 

y = x 2 − 2x – 2 

dy 

dx = 2x − 2 

At minimum point 

dy 

dx = 0 

2x= 2 


x = 2 2 = 1 

Substitution into (1) 

y=1 2 -2 (1) – 3 

y=1-2-3 = -4 

Question 38 

Evaluate ∫ sin2 dx 

A. Cos2x+k 

B. 1/2cos 2x + k 

C. -1/2cos 2x + k 

D. –cos 2x + k 

Answer C 

Solution: 

∫ sin2xdx = 1 (−cos2x) + k 

2 

= 1 2 

cos2x + k 

Question 39 

Evaluate ∫(2x + 3) 1⁄ 2 dx 

A. 1/12(2x+3) 6 + k 

B. 1/3(2x+3) 1/2 +k 

C. 1/3 (2x +3) 3/2 + k 

D. 1/12(2x + 3) 1/4 +k 

Answer C 

Solution: 

∫(2x + 3) 1⁄ 2 dx 

Let u = 2x+3 

dy 

dx = 2, dx = du 2 

∴ ∫ u du = 1 2 u 

∫ u 1⁄ 2 × 1 2 du = 1 2 ∫ u1 ⁄ 2du 


1 

2 u1 ⁄ 2 × 2 3 + k 

1 

3 u1 ⁄ 2 + k 

1 

3 (2x + 3)3 ⁄ 2 + k 

Question 40 

The pie chart above shows the monthly distribution of a man’s salary on 

food items. If he spent N8,000 on rice, how much did he spend on yam? 

A. N24,000 

B. N18,000 

C. N16,000 

D. N12,000 

Answer C 

Solution: 

Sectorial angle for yam 

=360 – (70 + 800+ 50)° 

=360 – 200 = 160° 

Rice = 80/360 x total amount(T) 

= 80 

360 × T 

800 = 80 

360 × T 

T = 

360 ×8000 

80 

T = N36,000 

Amount on yam = 160 

360° × 16000 

N16000 

Question 41 

The mean of 2-t, 4+t 3-2t and t-1 is 

A. t 

B. –t 


C. 2 

D. -2 

Answer C 

Solution: 

Mean x̅ = 

10 

5 = 2 

(2−7) + (4+t)+ (3−2t) + (2+t) + (t−1) 

5 

Question 42 

Values 0 1 2 3 4 

Frequency 1 2 2 1 9 

Find the mode of the distribution above. 

A. 1 

B. 2 

C. 3 

D. 4 

Answer D 

Solution: 

From the table, mode is the value with the highest frequency 

∴ i.e. =4 

Comprehensive mathematics page 244 

Question 43 

Find the median of 5,9,1,10,3,8,9,2,4,5,5,5,7,3 and 6. 

A. 6 

B. 5 

C. 4 

D. 3 

Answer B 

Solution: 

Rearranging in order of increasing magnitude 

1, 2, 3, 3, 4, 5, 5, 5, 5, 6, 7, 8, 9, 9, 10 

100k out for the middle number by calculating left and right 

Hence median = 5 


Question 44 

Calculate the standard deviation of 5,4,3,4 and 1 

A. √2 

B. √3 

C. √6 

D. √10 

Answer A 

Solution: 

Mean x̅ = 5+4+3+4+1 

5 

15 

15 = 3 

x d=x-x̅ d 2 

5 2 4 

4 1 1 

3 0 0 

2 -1 1 

1 -2 4 

∑ d 2 = 10 

Standard deviation = √ ∑ d2 

= √ 10 5 = √2 

n 

See comprehensive mathematics page 252 

Question 45 

In how many ways can a team of 3 girls be selected from 7 girls? 

A. 7!/3! 

B. 7!/4! 

C. 7!/4!3! 

D. 7!/2!5! 

Answer C 

Solution: 

Selection is a combination 

n Cr = 

n! 

(n−r)!r! 

n = 7 r= 3 

7 C3= 7! 

(7−3)!3! = 7! 

4!3! 


7 C3= 7! 

4!3! ways 

Question 46 

Number 1 2 3 4 5 6 

Frequency 18 22 20 16 10 14 

The table above represents the outcome of throwing a die 100time. What is 

the probability of obtaining at least 4? 

A. 1/5 

B. ½ 

C. 2/5 

D. 3/4 

Answer C 

Solution: 

At least 4 = 4, 5, 6 

Total frequency for, 4, 5, 6 

= 16 + 10 + 14 = 40 

∴ P (at least 4) = 

At leat 4 

Total frequency 

= 40 

100 = 2 5 

Comprehensive mathematics page 218 

Question 47 

A number is chosen at random from 10 to 30 both inclusive, what is the 

probability that the number is divisible by 3? 

A. 2/15 

B. 1/10 

C. 1/3 

D. 2/5 

Answer 

Solution: 

10 – 30 inclusive 

= {10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 

30} 

Divisible by 3= {12, 15, 18, 21, 21, 24, 27, 30} 

Total number = 21 

∴ P (divisible by 3) = 7 21 = 1/3 

See comprehensive page 218 - 220

Maths 003

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