- Without using table, evaluate (243)⅕ * (0.09)-1 * 125-⅔

(A) 4

(B) ¾

(C) 3

(D) (ANS) 4/3

- Three consecutive positive integers a, b and c are such that b2 = 4(a + c). Find the value of c

(A) 6

(B) 3

(C) (ANS) 9

(D) 5

- Find the principal which amount to =N= 4,400 at simple interest in 5years at 2% per annum. (A) =N=3,800

(B) =N=5.200

(C) =N=5,000

(D) (ANS) =N=4,000 - The sum of the first 20 terms of the progression 3, 6, 12, … is

(A) 3(221 – 1)

(B) (ANS) 3(220 – 1)

(C) 3(220 + 1)

(D) 3(221 + 1)

- Mr. Robson bought bags of orange for =N=2,500 each. He sold them for =N=90,000 at a loss of 20%. How many bags of orange did he buy?

(A) 60

(B) 80

(C) 50

(D) (ANS) 45 - The second and fifth terms of a geometric progression are 21 and 567 respectively find the first term and the common ratio of the progression.

(A) 3, 7

(B) (ANS) 7, 3

(C) -7, 3

(D) -3, 7

- Tunse bought a house for =N=1,250,000 and spent =N=350,000.00 to renovate it.He then sold the

house for =N=2,000,000.00. What is the percentage gain?

(A) 40%

(B) 65%

(C) 35%

(D) (ANS) 32% - List the integral values of which satisfy the inequality -2 < 7 – 3x ≤ 10

(A) (ANS) -1, 0, 1, 2

(B) -2, 0, 1

(C) 1, 2, 3

(D) 0, 1,2 - In a class, 120 students speak English or French or both. 70 speak English and 55 speak French. How many speak English but not French.

(A) 45

(B) 50

(C) (ANS) 55

(D) 60

- The lengths of the sides of a right angled triangle are ym, (3y – 1)m and (3y + 1)m, find y.

(A) (ANS) 12

(B) 9

(C) 8

(D) 4

- Uche, Adamu and Ope share profit on a business deal. Uche received ⅓ of the profit and Adamu received ⅔ of the remainder.If Ope received the remaining =N=12,000, how much profit did they share?

(A) (ANS) =N=54, 000

(B) =N=58,000

(C) =N=48, 000

(D) =N=42, 000 - A ladder resting on a vertical wall makes an angle whose tangent is 2.5 with the ground. If the distance between the foot of the ladder and the wall is 60cm, what is the length of the ladder?

(A) 2m

(B) (ANS) 1.6m

(C) 0.8m

(D) 3m - Three students share a bag of garri in such a way that the first student took¼ of the garri and the second¾ of the remainder. What fraction of the bag of garri did the third student take?

(A) (ANS) 3/16

(B) 2/9

(C) ⅗

(D) 4/15 - Solve the simultaneous linear equations 2x + 5y = 11, 7x + 4y = 2 (A) 27/34, 34/27

(B) (ANS) -34/27, 73/27

(C) ⅔, ⅘

(D) ⅓, ⅖ - If x + 1 is a factor of x3 + 3×2 + Kx + 4, find the value of K.

(A) (ANS) 6

(B) 4

(C) -4

(D) 3

- Paul and his friend, Peter went to buy an article costing =N=600. Peter had 10% of the cost and Paul 40% of the remainder. How much did they have altogether?

(A) =N=320

(B) =N=440

(C) (ANS) N=276

(D) =N=196 - Find all values of x satisfying the inequality -14 ≤ 4 – 3x ≤ 31 (A) -6 ≤ x ≤ 9

(B) (ANS) -9 ≤ x ≤ 6

(C) 5 ≤ x ≤ 8

(D) 6 ≤ x ≤ 12 - What factor is common to all the expressions x2 – x, 2×2 – x – 1 and x2 – 1? (A) (2x – 1)

(B) (x + 1)

(C) (ANS) (x – 1)

(D) (2x + 3) - Find the sum to infinity of the following series 3 + 2 + 4/3 + 8/9 + 16/27 + …

(A) 270

(B) (ANS) 9

(C) 27

(D) 90

- Find the number of sides of a regular polygon whose interior angle is twice the exterior angle.

(A) 5

(B) (ANS) 6

(C) 8

(D) 9

- Find the value of p, if the line which passes through (-1, -p) and (-2p, 2) is parallel to the line 2y + 8x

– 17 = 0

(A) (ANS) 6/7

(B) 4/7

(C) ⅖

(D) –6/7

- Obtain a maximum value of the function f(x) = x3 – 12x + 11

(A) -15

(B) (ANS) 27

(C) 15

(D) 20

- If y = 5cos (-6x), dx/dy is

(A) (ANS) 30sin (-6x)

(B) 5sin (-6x)

(C) -30sin (-6x)

(D) -30cos (-6x)

- Find the variance of the numbers k, k + 1, k + 2

(A) ⅓

(B) 3

(C) (ANS) ⅔

(D) 1

- At what value of x is the function y = x2 – 6x – 7 minimum

(A) (ANS) 3

(B) 5

(C) 6

(D) 2

- The chances of three independent events P, Q, R occurring are ½, ⅔, ¼ respectively. What are the chances of P and Q only occurring?

(A) (ANS) ¼

(B) ⅔

(C) 2/9

(D) ½ - If x and y represent the mean and the median respectively of the following marks of students in a Mathematics test 11, 12, 13, 14, 15, 16, 17, 18, 19, 21. Find x/y correct to 1 decimal places.

(A) (ANS) 1

(B) 3

(C) 5

(D) 2 - Find the point on the Euclidean plane where the curve y = 2×2 – 2x + 9 has 2 as gradient. (A) (2, 4)

(B) (3, 5)

(C) (ANS) (1, 3)

(D) (1, 4) - Find the equation of the line through (5, 7) parallel to the line 7x + 5y – 5 = 0

(A) (ANS) 5y + 7x= 70

(B) 7x + 5y= 7

(C) 5x + 7y= 110

(D) y + 5x= 70

- AAUA allocations to various Faculties in a school budget are as follows:

Science =N=35,000,000.00

Education =N=25,000,000.00

Law =N=20,000,000.00

Social and Management- =N=20,000,000.00

In a pie chart to represent this information the corresponding angle to Science is

(A) 900

(B) (ANS) 1260

(C) 450

(D) 1000

- “implify 3 ⅓ – 1¼ x ⅔ + 1 ⅖

(A) 3 ⅟10

(B) (ANS) 3 9/10

(C) 3 ⅖

(D) 33 ⅕ - Find the distance between points A(-4, 5) and B(-3, 2)

(A) 10

(B) 2√5

(C) 3

(D) (ANS) √10

No of children 0 1 2 3 4 5 6

No of families 7 11 6 7 7 5 3

Find the mode and median respectively of the distribution above.

(A) 3, 1

(B) (ANS) 1, 2

(C) 1, 6

(D) 1, 1

- Rationalize (4√7 + √2)/(√2 – √7)

(A) (ANS) -√14 – 6

(B) 2√14 – 6

(C) -3√14 – 28

(D) √14 – 8

- A housewife bought five yams at Y34 per yam and three oranges at Y5 each where Y is the currency reckoned in base six. The total amount spent by the housewife is

(A) Y2256

(B) Y2526

(C) (ANS)

(D) Y3526

- Justice Starts a 5km walk from P on a bearing 0230 . He then walks 4km on a bearing of 1130 to Q. What is the bearing of Q from P?

(A) 67.40

(B) 230

(C) (ANS) 90.40

(D) 1130 - Divide the L.C.M. of 36, 54 and 90 by their HCF

(A) 15

(B) 20

(C) (ANS) 30

(C) 45

- If 125x = 2010 find x

(A) 4

(B) 5

(C) (ANS) 3

(D) 2

- Simplify 4⅓ – (2½ – 1 ⅗) (A) 7/30

(B) 2 ½

(C) 3 ⅟ 10

(D) (ANS) 3 13/30 - If x : y = ⅔ : ⅚ and y : z = ¾ : ½, find x : y : z (A) 21 : 15 : 10

(B) 15 : 12 : 10

(C) (ANS) 12 : 15 : 10

(D) 10 : 15 : 12 - The number 327,036 was corrected to 327,000. Which of the following can exactly describe the degree of approximation used?

I. to 3 significant figures

II. to 4 significant figures III to the nearest hundred IV to the nearest thousand

(A) I and III only

(B) II and IV only

(C) I, II and III only

(D) (ANS) I, II, III and IV

- If 1010102 = x10 + 11112 the value of x is

(A) 32

(B) (ANS) 27

(C) 19

(D) 13

- The population of a school is 1,376. Express this to three significant figures (A) 1,370

(B) 1,376

(C) (ANS) 1,380

(D) 138 - A labourer’s daily wage is =N=80 for the first 10 days and =N=100 for the next 9 days. Find the daily wage for the remaining 6 working days of the month, if his average daily wage for the month is

=N=90.80

(A) =N=110

(B) =N=105

(C) (ANS) =N=95

(D) =N=92.50

- Find the value of x if 4logx + 5logx – 7logx = log16

(A) 2

(B) (ANS) 4

(C) 8

(D) 16

- A motorist drives from P to Q at an average speed of 80 km/hr and immediately returns from Q to P through the same route at an average speed of 50km/hr. The average speed for the round-trip journey is

(A) (ANS) 53.3 km/hr

(B) 60.0 km/hr

(C) 62.5 km/hr

(D) 65.0 km/hr - If X = {1, 2, 3, 4, 5, 6}, Y = {2, 4, 5, 7} and Z = {1, 4, 5} which of the following is (are) correct?

I. n(X U Y U Z) = 13

II. (X ∩ Y) U Z = X ∩ (Y U Z)

III. n(X ∩ Y ∩ Z) = 2

(A) I only

(B) II only

(C) I and III only

(D) (ANS) II and III only

- X sold a radio set to Y at profit of 10% and Y sold it for =N=2, 612.50 at a loss of 5%. The cost of the radio to X was

(A) (ANS) =N=2,500

(B) =N=2,488

(C) =N=2,375

(D) =N=2,272 - Factorize: 62m + 1 + 7(6m) – 5

(A) (ANS) [3(6m) + 5[2(6m) – 1

(B) [3(6m) – 5[2(6m) + 1

(C) [3(6m) – 5[3(6m) + 1

(D) [2(6m) + 5[3(6m) – 1

- The minimum point on the curve y = x2 – 4x + 3 is

(A) (ANS) (2, -1)

(B) (0, 3)

(C) (1, 0)

(D) (2, -4)

## PRACTICE 2

**Question 1**

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

Options

A) q=b2(mn−a2)a2pq=b2(mn−a2)a2p

B) q=m2n−a2p2q=m2n−a2p2

C) q=mn−2b2a2q=mn−2b2a2

D) q=b2(n2−ma2)nq=b2(n2−ma2)n

The correct answer is A.

Explanation:

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**

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.

Options

A) 148m

B) 382m

C) 282m

D) 248m

The correct answer is C.

Explanation:

tan78=h60tan78=h60

h=60tan78h=60tan78

h=60×4.705=282.27mh=60×4.705=282.27m

≊≊ 282m to the nearest whole number.**Question 3**

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).

Options

A) 6

B) 25

C) 15

D) 18

The correct answer is C.

Explanation:

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**

Age in years 7 8 9 10 11

No of pupils 4 13 30 44 9

The table above 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

Options

A) 48.6°

B) 56.3°

C) 46.8°

D) 13°

The correct answer is C.

Explanation:

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**

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

Options

A) 42

B) 20

C) 70

D) 54

The correct answer is B.

Explanation:

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**

Rationalize √2+√3√2−√32+32−3

Options

A) −5−2√6−5−26

B) −5+3√2−5+32

C) 5−2√35−23

D) 5+2√65+26

The correct answer is A.

Explanation:

√2+√3√2−√32+32−3

= (√2+√3√2−√3)(√2+√3√2+√3)(2+32−3)(2+32+3)

= 2+√6+√6+32−√6+√6−32+6+6+32−6+6−3

= 5+2√6−15+26−1

= −5−2√6−5−26**Question 7**

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

Options

A) 4.2 cm

B) 4.3 cm

C) 3.2 cm

D) 3.4 cm

The correct answer is D.

Explanation:

Length of chord = 2rsin(θ2)2rsin(θ2)

= 2(3)sin(682)2(3)sin(682)

= 6sin346sin34

= 6×0.5596×0.559

= 3.354 cm ≊≊ 3.4 cm**Question 8**

Given sin58°=cosp°sin58°=cosp°, find p.

Options

A) 48°

B) 58°

C) 32°

D) 52°

The correct answer is C.

Explanation:

sinθ=cos(90−θ)sinθ=cos(90−θ)

sinθ=cos(90−58)sinθ=cos(90−58)

= cos32cos32**Question 9**

23÷4514+35−1323÷4514+35−13

Options

A) 31503150

B) 20312031

C) 31203120

D) 50315031

The correct answer is D.

Explanation:

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**

If 6×3+2×2−5x+16×3+2×2−5x+1 divides x2−x−1×2−x−1, find the remainder.

Options

A) 9x + 9

B) 6x + 8

C) 5x – 3

The correct answer is A.

**Question 11**

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

Options

A) 2323

B) 4545

C) 2525

D) 1212

The correct answer is D.

Explanation:

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

P(at least two heads) = 4848

= 1212**Question 12**

In how many ways can the word MATHEMATICIAN be arranged?

Options

A) 6794800 ways

B) 2664910 ways

C) 6227020800 ways

D) 129729600 ways

The correct answer is D.

Explanation:

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**

Given matrix M = ∣∣

∣∣−2040−16563∣∣

∣∣|−2040−16563|, find MT+2MMT+2M

Options

A) ∣∣

∣∣−421605062∣∣

∣∣|−421605062|

B) ∣∣

∣∣−60130−31814189∣∣

∣∣|−60130−31814189|

C) ∣∣

∣∣52601134−7∣∣

∣∣|52601134−7|

D) ∣∣

∣∣−4080−2−1610126∣∣

∣∣|−4080−2−1610126|

The correct answer is B.

Explanation:

M = ∣∣

∣∣−2040−16563∣∣

∣∣|−2040−16563|

MTT = ∣∣

∣∣−2050−16463∣∣

∣∣|−2050−16463|

2M = ∣∣

∣∣−4080−21210126∣∣

∣∣|−4080−21210126|

MTT + 2M = ∣∣

∣∣−60130−31814189∣∣

∣∣|−60130−31814189|**Question 14**

Score (x) 0 1 2 3 4 5 6

Freq (f) 5 7 3 7 11 6 7

Find the mean of the data.

Options

A) 3.26

B) 4.91

C) 6.57

D) 3.0

The correct answer is A.

Explanation:

Mean = ∑fx∑f∑fx∑f

= 1504615046

= 3.26

**Question 15**

Score (x) 0 1 2 3 4 5 6

Freq (f) 5 7 3 7 11 6 7

Find the variance

Options

A) 3.42

B) 4.69

C) 4.85

D) 3.72

The correct answer is D.

Explanation:

Variance = ∑f(x−¯x)∑f∑f(x−x¯)∑f

= 170.88846170.88846

= 3.72

**Question 16**

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

Options

A) bisector of the two lines

B) line parallel to the two lines

C) angle bisector of the two lines

D) perpendicular bisector of the two lines

The correct answer is A.**Question 17**

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

Options

A) 16°

B) 25°

C) 42°

D) 39°

The correct answer is D.

Explanation:

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

∴∴ (2x + 18)° + 84° = 180°

2x + 102° = 180° ⟹⟹ 2x = 78°

x = 39°

**Question 18**

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

Options

A) 90°

B) 45°

C) 60°

D) 30°

The correct answer is C.

Explanation:

4sin2x−3=04sin2x−3=0

4sin2x=3⟹sin2x=344sin2x=3⟹sin2x=34

sinx=√32sinx=32

∴x=sin−1(√32)∴x=sin−1(32)

x = 60°

**Question 19**

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

Options

A) 56 cm22

B) 24 cm22

C) 42 cm22

D) 34 cm22

The correct answer is D.

Explanation:

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**Question 20**

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?

Options

A) 82

B) 15

C) 63

D) 75

The correct answer is D.

Explanation:

Scores (Mark) 0 1 2 3 4 5

No of students 4 3 7 8 6 2

fx 0 3 14 24 24 10 75

**Question 21**

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.

Options

A) y = 5x + 2

B) y = 5x + 3

C) y = 12x + 2

D) y = 22x + 3

The correct answer is A.

Explanation:

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**

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.

Options

A) m = 0, n = 4

B) m = 4, n = 0

C) m = 2, n = 2

D) m = -2, n = 4

The correct answer is A.

Explanation:

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**

If ∣∣∣2−4×9∣∣∣=58|2−4×9|=58, find the value of x.

Options

A) 10

B) 30

C) 14

D) 28

The correct answer is A.

Explanation:

∣∣∣2−4×9∣∣∣=58|2−4×9|=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**

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

Options

A) dydx=15×2−4x−2−3x−2dydx=15×2−4x−2−3x−2

B) dydx=6x+4x−1−3x−4dydx=6x+4x−1−3x−4

C) dydx=18×2−4x−3+3x−4dydx=18×2−4x−3+3x−4

D) dydx=12×2+4x−1−3x−2dydx=12×2+4x−1−3x−2

The correct answer is C.

Explanation:

y=6×3+2x−2−x−3y=6×3+2x−2−x−3

dydx=18×2−4x−3+3x−4dydx=18×2−4x−3+3x−4**Question 25**

ddx[log(4×3−2x)ddx[log(4×3−2x) is equal to

Options

A) 12x−24x212x−24×2

B) 43×2−2x7x43x2−2x7x

C) 4×2−27x+64×2−27x+6

D) 12×2−24×3−2x12x2−24×3−2x

The correct answer is D.

Explanation:

ddx[log(4×3−2x)ddx[log(4×3−2x) … (1)

Let u = 4×33 – 2x.

ddx(log(4×3−2x))=(ddu)(dudx)ddx(log(4×3−2x))=(ddu)(dudx)

ddu(logu)ddu(logu) = 1u1u

dudx=12×2−2dudx=12×2−2

∴ddx[log(4×3−2x)=12×2−2u∴ddx[log(4×3−2x)=12×2−2u

= 12×2−24×3−2x12x

**Question 26**

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

Options

A) -24

B) 30

C) -18

D) -50

The correct answer is C.

Explanation:

f(x)=3×3+4×2+x−8f(x)=3×3+4×2+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**

Solve for x in 4x−63≤3+2x24x−63≤3+2×2

Options

A) x≤112x≤112

B) x≤212x≤212

C) x≥212x≥212

D) x≥112x≥112

The correct answer is B.

Explanation:

4x−63≤3+2x24x−63≤3+2×2

2(4x – 6) ≤≤ 3(3 + 2x)

8x – 12 ≤≤ 9 + 6x

8x – 6x ≤≤ 9 + 12

2x ≤≤ 21

x≤212x≤212**Question 28**

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

Options

A) -1 ≤≤ x ≤≤ 2

B) -1 ≤≤ x < 2

C) -1 < x < 2

D) -1 < x ≤≤ 2

The correct answer is D.

Explanation:

-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**

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

Options

A) 74

B) 32

C) 42

D) 170

The correct answer is D.

Explanation:

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**

Integrate ∫2−1(2×2+x)dx∫−12(2×2+x)dx

Options

A) 412412

B) 312312

C) 712712

D) 514514

The correct answer is C.

Explanation:

∫2−1(2×2+x)dx∫−12(2×2+x)dx

= [2×2+13+x1+122−1[2×2+13+x1+12−12

= [2×33+x222−1[2×33+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**

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

Options

A) 12

B) 8

C) 9

D) 16

The correct answer is C.

Explanation:

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**

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.

Options

A) 43.6% loss

B) 35% profit

C) 63.2% profit

D) 28% loss

The correct answer is C.

Explanation:

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**

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

Options

A) 19%

B) 20%

C) 25%

D) 16.3%

The correct answer is A.

Explanation:

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**

Simplify 81−34−34 x 251212 x 2432525

Options

A) 2525

B) 3535

C) 5252

D) 5353

The correct answer is D.

Explanation:

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**

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

Options

A) 46.2

B) 43.1

C) 534

D) 431

The correct answer is B.

Explanation:

(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**

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

Options

A) 50 units per sec

B) 35 units per sec

C) 22 units per sec

D) 13 units per sec

The correct answer is B.

Explanation:

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**

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

Options

A) 31°

B) 30°

C) 26°

D) 48°

The correct answer is B.

Explanation:

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**

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?

Options

A) 15.5%

B) 18.2%

C) 14.8%

D) 16.7%

The correct answer is D.

Explanation:

Actual weight = 0.18g

Error = 0.21g – 0.18g

= 0.03g

% error = 0.030.18×1000.030.18×100

= 16.7%**Question 39**

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

Options

A) 1.3

B) 2.5

C) 4.6

D) 3.2

The correct answer is A.

Explanation:

(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**

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

Options

A) x = 3434, y = 114114

B) x = 3434, y = 134134

C) x = 2323, y = 4545

D) x = 2323, y = 134134

The correct answer is B.

Explanation:

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**

Simplify 0.0839×6.3815.44 to 2 significant figures.

Options

A) 0.2809

B) 2.51

C) 3.5

D) 0.098

The correct answer is D.**Question 42**

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

Options

A) x = 3 or 7, y = 12 or 8

B) x = 6 or 1, y = 11 or 5

C) x = 2 or 5, y = 15 or 6

D) x = 1 or 5, y = 10 or 7

The correct answer is C.

Explanation:

3x + y = 21 … (i);

xy = 30 … (ii)

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

3x + 30x = 21

⟹ 3×2 + 30 = 21x

3×2 – 21x + 30 = 0

3×2 – 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**

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.

Options

A) 24°

B) 56°

C) 127°

D) 156°

The correct answer is D.

Explanation:

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**

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

Options

A) T=15R−QP3

B) T=R−15P3Q

C) T=R−15P3Q

D) T=R+P315Q

The correct answer is B.

Explanation:

P=(Q(R−T)15)13

P3=Q(R−T)15

Q(R−T)=15P3

R−T=15P3Q

T=R−15P3Q**Question 45**

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

Options

A) 26°

B) 13°

C) 80°

D) 102°

The correct answer is D.

Explanation:

< BAC = 1302 (angle subtended at the centre)

< BAC = 65°

Also, x = 26° (theorem)

y = 65° – 26° = 39°

< AOC = 180° – (39° + 39°)

= 102°

**Question 46**

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

Options

A) 1080°

B) 1260°

C) 2160°

D) 1800°

The correct answer is B.

Explanation:

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**

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?

Options

A) 43.7% loss

B) 13.2% gain

C) 13.2% loss

D) 43.7% gain

The correct answer is C.

Explanation:

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**

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

Options

A) πh3(2R+r)

B) 2R+r+πh3

C) πh3(2R2+r+2r)

D) 2R23πh

The correct answer is A.

Explanation:

V=πh3(R2+Rr+r2)

V=πR2h3+πRrh3+πr2h3

dVdR=2πRh3+πrh3

= π3(2R+r)**Question 49**

Express (0.0439÷3.62) as a fraction.

Options

A) 21100

B) 211000

C) 121000

D) 12100

The correct answer is C.

Explanation:

(0.0439÷3.62)

= 0.01213

≊ 0.012

= 121000**Question 50**

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

Options

A) x = -4

B) x = 2

C) x = -2

D) x = 4

The correct answer is A.

Explanation:

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