UNN Post UTME Past Questions for Mathematics (Authentic, PDF Available)

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The questions in the test will be from both JAMB and UNN Past questions.

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[ays_quiz id=’12’]

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UNN Post UTME Past Questions And Answers for Mathematics

NB: Make sure you have taken the test/quiz above. If you have, then go ahead and download the UNN Post UTME Past Questions in PDF or just view it directly on this website.

UNN POST UTME QUESTIONS FOR MATHEMATICS 2005/2006

1. Find n if 31410 – 2567 = 340n
   A. 7
   B. 8
   C. 9
   D. 10
2. What is the difference between 1.867551 correct to four significant figures and 1.867551 correct to four decimal places?
   A. 5×10^-3
   B. 4×10^-4
   C. 5×10^-4
   D. 10×10^-3
3. In an examination, all the candidates offered at least one of English and French, if 52% offered French and 65% offered English, what percentage offered French only?
   A. 17
   B. 35%
   C. 48%
   D. 45%
4. Simplify 6𝑥² +5𝑥² / 2𝑥²+𝑥−3
   A. 3x-1
   B. 1-3x
   C. 3x+1
   D. –(3×1)
5. Find the range of value of x satisfying the inequalities 2x – 5< 7 and 25 + 2x > 15
   A. 5<x<6
   B. -5<×<6
   C. -6<×<5
   D. -6<×<-5
6. If the 8^th term of an A.P is three times the second term and the sum of the first three terms is 18, find the first term of the A.P.
   A.4
   B. 2
   C. 8
   D. 3
7. Find the sum of infinity f the series 4+3 + 9/4+27/19+……
   A.16
   B. 16/3
   C. 1
   D. 8
8. A chord of a circle or radius 10cm is drawn 8cm from the center of circle. Find the length of the chord.
   A. 6cm
   B. 2√14𝑐m³
   C. 12cm
   D. √41𝑐m³
9. Find the equation of the line which passes through (-2,1) and is perpendicular to the line 4x-2y+1 = 0.
   A. 2y-x-4 = 0
   B. 2y + x = 0
   C. 2y-x = 0
   D. y-2x-5 = 0
10. If a is parallel to the line 2y-rx+4 = 0 and perpendicular to the line 4y+x-28 = 0, then the value of r is
    A.4
    B. 8
    C. -8
    D. -4
11. The distribution below shows the scores of sixty students in a class test. What percentage of the students scored at least 3?
    
    A.60%
    B. 36%
    C. 66%
    D. 40%
12. The first derivative of y = (2+3x)⁴ at x = -1 is
    A.12
    B. -12
    C. 4
    D. -4
13. The minimum value of (x) = x²-4x+5 in the interval [1, -1] is
    A.-2
    B. 10
    C. 4
    D. 5
14. The table below shows the marks scored by a group of students in a class test. If the mean score is 3.4, find x.
    
    A.3
    B. 4
    C. 5
    D. 2
15. A company is to select three different handset phones from five different types of Nokia brand and two different types of Samsung brand. In how many ways can the company choose the handsets, so as to include at least one Samsung brand?A.15
    B. 25
    C. 35
    D. 45

UNN POST UTME ANSWERS FOR MATHEMATICS 2005/2006

1. 314₁₀-256₇ = 340_n
314¹⁰ (2×7²+5×7¹ + 6×7⁰) = 3x
n² + 4×n¹+0×n
314-139 = 3n² + 4n
3n² + 4n-175 = 0
(n – 7) (n + 25/2) = 0
n = 7 or n = -25/2 (impossible)
∴ n = 7 (Ans. A)

2. 1.867551 to 4 sign. Fig. = 10868
1.867551 to 4 d.p. = 1.8676
Difference: 1.868-1.8676 = 0.0004 = 4×10^-4
Ans. B

5. 2x-5<7 and 25+2x>15
2x<12 and 2x>-10
X<6 and 2x> – 10(or x >-5)
-5< × <6
Ans. B

6. U₈ = 3U₂, S3 = 18, U₁ =?
a +7d = 3(a+d)→a + 7d = 3a + 3d
→ a +7d-3a – 3d = 0
→ 2a + 4d = 0→ 2a-4d = 0 ….. (1)
Sn = 𝑛/2 [2𝑎 + (𝑛 − 1)𝑑]
𝑆₃ = 18 = 3/2 [2𝑎 + (3 − 1)𝑑]
→2a + 2d = (18×2)/3
→2a + 2d = 12
∴ 𝑎 + 𝑑 = 6 … … … … (2)
From equation (2), a = 6-d
Substitute 6 – d for a into equation (1), 2(6-d) – 4d = 0
→ 12 – 2d – 4d = 0 → -6d = -12 ∴ 𝑑 = 2
Hence, a + 2 = 6 → a = 6-2 = 4
U₁ = a = 4
Ans. A

7. 𝑆_∞ 𝑎/( 1−𝑟) ; 𝑟 = 𝑟[ (9/4) / 3] ; 𝑎 = 4
𝑆_∞ = 4 / (1 − 3/ 4) = 16 → 𝑆∞ = 16
Ans. A

9. Equation of the given line is
4x – y + 1 = 0
i.e. y = 2x + ½
Gradient of the line, m₁ = 2
Let the gradient of the perpendicular line be m₂. For perpendicular lines, m₁m₂ = -1→m₂ = -1/m
M₂ = 1/2 through (-2,1)
Hence, the required equation is y – 1 = -1/2 [x – (-2)]
y – 1 = -1/2(x + 2) or 2y + x = 0
Ans. B

10. Line 1 is 2y-rx+4i.e. y= r/2x – 2
Line 2 is 4y + x – 28 = 0 i.e. y =1/4x+7
Gradient of line 1 =m₁r/4 gradient of line 2 = M₂ = ¼
If line 1 and line 2 are perpendicular, then m₁m₂ = -1
→ 𝑟/2 x (−1/4) = −1 → 𝑟 = 8
Ans. B

11. No of student that scored at least 3 = 16 + 12 + 8 = 36 students
Total no of students = 60
Percentage that scored at least 3 = 36/60 × 100% = 60%
Ans. A

12. Y = (2 + 3x)⁴
= 4(2 + 3x)³ .3 =12 (2 + 3x)³
At x =-1, = 12(2+3(-1)3=-12)
Ans. B

13. F(x) = x2-4x + 5
f’(x) = 2x – 4
In the interval [1,1], f’(x) = 2(1) – 4 = -2
Ans. A

14.

Score (x)	Frequency (f)	Fx
1	3	3
2	6	12
3	7	21
5	X	5x
6	4	24
	∑𝑓 = 20 + 𝑥	∑𝑓𝑥 = 60 + 5𝑥

Mean = ∑ 𝑓𝑥 / ∑ 𝑓 = 3.4 → (60+5𝑥)/ (20+𝑥) = 3.4
60+5x = 3.4 (20 + x) →60 + 5x = 68 + 3.4x
5x – 3.4 = 68-60 → 1.6x = 8
Therefore, x = 5
Ans.  C

15. No of Nokia bran = 5, No of Samsung brand = 2 if 3 different handset phones are to be selected and at least 1 Samsung brand must be selected and at least 1 Samsung brand must be included, then it is either they select 2 Nokia brand from 5 and 2 Samsung brand from 2.

That is, ⁵C₂ x ²C₁ or ⁵C₁ x ²C₂ = 10×2+5×1=20+5=25 ways. Ans. B

SUMMARY OF ANSWERS (MATHEMATICS 2005/2006)

1. A 2.B. 3.B 4.C 5.B 6.A 7.A 8.C 9.B 10.B 11.A 12.B 13.1 14.C 15.B

UNN POST UTME QUESTIONS FOR MATHEMATICS 2007/2008

1. Express 8×10^-6 +2×10^-5 as a fraction
   A.1/4               B. 3/2
   C. 2/5               D. 1/5
2. Find the values of x for which 2^2x+3+ 33 × 2x + 4 = 0
   A. x=2, x= -3   
   B. x=-2, x= 3
   C. x=4, x=1/8  
   D. x=2, x=3
3. If 260₉ ÷ 100₂ = 66_n, find n
   A.7                   B. 9
   C. 10                D. 8
4. Find the values of x such that
   
   A. x=y=2         
   B. x=2, y=-2
   C. x=12, y=2    
   D. x=y=-2
5. A chord of a circle of radius 13cm is drawn 5cm from the centre of the circle. Find the length of the chord.
   A. 12cm           B.24cm
   C. 18cm           D. √194cm
6. If x-2 is a factor of px³ + 2x²-2p+12, find the value of p.
   A. 8/5              B. -10/3
   C. 2                  D. -2
7. In a regular pentagon ABCDE, AC intersects BD at P. calculate < CPD.
   A.108^°             B. 36^°
   C. 72^°               D. 48^°
8. 
   The table above shows the marks obtained by a student in an examination. If the total mark obtained is 300, what is the angle corresponding to the mark obtained in Chemistry if the information is represented in a pie chart?
   A.120°            B.144°
   C. 48°            D. 108°
9. A ladder 17m rests against a vertical wall so that its foot is 8.5m from the wall. Find the angle of inclination of the ladder to the horizontal floor.
   A.30°               B.45°
   C. 60°               D. 55
10. 
    A.0                   B. 5
    C.                 D. 1
11. If 𝑑𝑦/𝑑𝑥 = 6x-3 and y(-1) = 8, find y (x)
    A.3x²-3x-8       B. 3x²-3x+8
    C. 3x²-3x-2       D. 3x²-3x+2
12. The minimum of the function f(x) = 2x²-12x + 5 is
    A.59                 B. -59
    C.3                   D. -3
13. A basket contains 5 MTN cards, 6 GLO cards, 3 MTEL cards and 6 Vmobiles cards. What is the probability that a card selected from the basket at random will be MTN or MTEL card?
    A. 3/20            B. ¾
    C. ¼                  D. 2/5
14. Find the range of the numbers 1/3, ½, 3/5, 4/5, 2/3, 6/7, 8/9
    A.7/27             B.13/45
    C. 9/5               D. 5/9
15. If the mean of the numbers 4, 3, 5, x, 7, is 5, find the variance:
    A.2                   B.10
    C.√2                D. 5

UNN POST UTME ANSWERS FOR MATHEMATICS 2007/2008

1. 8 × 10⁶ ÷ 2 × 10⁻⁵
= 8/10⁶ ÷ 2/10⁵
= 8/10⁶ × 10⁵/2
= 8/2 × 10⁵/10⁶
= 4/10
= 2/5
Ans. C

2. 2^2x+3 – 33 × 2^x + 4 = 0
2^2x × 2³ – 33 × 2^x + 4 = 0
8 × (2^x)² – 33(2^x) + 4 = 0
Let 2^x = p
8p²-33p + 4 = 0
(8p-1) (p-4) = 0
P = 1/8 or p = 4
But = p = 1/8
= 8^-1 = 2^-3 × = -3 OR
= p = 4 = 2², x = 2
∴ 𝑥 = 2 𝑜𝑟 − 3
𝑨𝒏𝒔. A

4. 2x-7y = 10…………(1)
3x + ½x = -7……….(2)
Solving equations 1 and 2
Simultaneously gives x = -2 and y = -2
Ans. D

6. Let f(x) = px³ + 2x² – 2p + 12
If x-2 is a factor of f(x), f(2) = 0
F(2) = p(2)³ + 2(2)² – 2p + 12 = 0
8p + 8 – 2p + 12 = 0
8p – 2p + 12 = 0
6p = -20
P = -20/6 = -10/3
Ans. B

7. A regular pentagon has five equal sides and all the angles are equal. Sum of interior angles = (2n – 4) × 90^o
Where n = 5, sum of interior angles = [(2 × 5)4] × 90^o = 540
Therefore, each angle = 540^o/5 = 108^o
Required angle = 180^o – 108^o = 72^o
Ans. C

8. 95 + 2x + 10 + x + 75 = 300
3x + 180 = 300
3x = 300 – 180 = 120
X = 120/3 = 40
Mark obtained in chemistry = 2x + 40 = 2 × 40 + 10 = 90
Angle of mark = =90/300 x 360^o = 108^o
Ans. D

Ans. B

12. 𝑓(𝑥) = 2𝑥 2 − 12𝑥 −5 ?
𝑓(𝑥) = 4𝑥 − 12 = 0?
4𝑥 − 12 = 0
4𝑥=12
𝑥=3
Ans. C

13. 𝑀𝑇𝑁 Cards = 5,
Glo Cards = 6,
MTEL Cards =3,
Vmobile Cards = 6
Total Cards = 20
Prob (MTN or MTEL) = 5/20
= 8/20
= 2/5
Ans D

14. 𝑅𝑎𝑛𝑔𝑒 = 8/9 − 1/3
= 5/9
Ans D

SUMMARY OF ANSWERS (MATHEMATICS 2007/2008)

1.C 2.A 3.D 4.B 5.B 6.B 7.C 8.D 9.B. 10.B 11.D 12.C 13.D 14.D 15.A

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