JAMB UTME - Mathematics - 2015
Question 1 Report
Answer Details
Question 2 Report
Answer Details
Question 3 Report
Answer Details
His allowances = ₦600
So his taxable income = Annual salary − allowance
= ₦2000 − ₦600
= ₦1400
He pays at 5%
Then, his allowance income tax 5/100 × 600 = ₦30
Question 4 Report
Answer Details
Question 5 Report
Simplify (0.09)2 and give your answer correct to 4 significant figures
Answer Details
(0.09)2 = 0.09 × 0.09
= 0.0081
= 0.008100 to 4 significant figures
Please, start counting first from the non-zero digits i.e. 8
Question 6 Report
Answer Details
Question 7 Report
Answer Details
The formula for finding the number of elements in the union of two sets X and Y is:
n(X U Y) = nX + nY - n(X ∩ Y)
where nX is the number of elements in set X, nY is the number of elements in set Y, and n(X ∩ Y) is the number of elements in the intersection of X and Y. In this case, nX = 15, nY = 12, and n(X ∩ Y) = 7, so we can plug in the values into the formula:
n(X U Y) = 15 + 12 - 7
n(X U Y) = 20
So, the number of elements in the union of X and Y is 20.
Question 8 Report
Answer Details
Question 9 Report
Answer Details
P(A ∪ B) = 1/12, P(A ∩ B)
P(A ∩ B) = P(A) + P(B) − P(A ∪ B)
1/6 + 1/4 − 1/12
= (2 + 3 − 1) ÷ 12
= 4/12
= 1/3
P(A ∩ B) = 1/3
Question 10 Report
Answer Details
6159626 R 364 R 20 R 4↑
15910 = 4236
Then Convert 0.75 to base 6
0.75 x 6 =4.5 i.e 4.5 - 3.00 = 1.5
0.5 x 6 = 3.00 423 + 1.5 =424.5
159.75 = (424.5)6
(159)10 = (424.5)6
Question 11 Report
Answer Details
Question 12 Report
Factorize \(a^2 - b^2 - 4a + 4\)
Answer Details
a2−b2−4a+4=(a2−4a+4)−b2
(a2−2a−2a+4)−b2
a(a−2)−2(a−2)−b2
(a−2)2−b2
(a−2+b)(a−2−b)
Question 13 Report
Evaluate \( \log_{5}\left(y^{2}x^{5} \div 125b\frac{1}{2}\right) \)
Answer Details
Question 14 Report
Answer Details
Radius = 4cm
π = 3.142
A = 3.142 × (4)2
= 3.142 × 16
= 50.272
A = 50.3cm2 (1dp)
Question 15 Report
One bag contain 3 blue and 5 red balls, another bag contain 2 blue and 4 red balls respectively. One ball is drawn for each bag. What is the probability both balls are blue
Answer Details
Pr (H) = 3/8Pr (r) 5/8
Another bag with 2 blue and 4 red (note one ball is drawn from each bag)
i.e. prob.=8 − 1 = 7
prob. (2 ball both are blue=pr (1st blue and 2nd blue)
prob. 2 balls both blue = 3/8 × 2/7
= 328
Question 16 Report
Given that A = {1, 5, 7}
B = {3, 9, 12, 15}
C = {2, 4, 6, 8}
Find (A ∪ B) ∪ C
Answer Details
Question 17 Report
In a town of 6250 inhabitants, there were 62 births during 1984. Find the percentage birth rate.
Answer Details
Question 18 Report
Answer Details
Question 19 Report
Factorize \(x^2 - 2x - 15\)
Answer Details
The expression x^2 - 2x - 15 can be factored into the product of two binomials. To find these binomials, we can use the "ac method," where we try to find two numbers that multiply to -15 and add up to -2. These numbers are -5 and 3. We can then write the expression as:
(x - 5)(x + 3)
So, the expression x^2 - 2x - 15 can be factored as (x - 5)(x + 3).
Question 21 Report
Answer Details
Question 22 Report
Simplify \(6\frac{1}{12} - 2\frac{3}{4} + 1\frac{1}{2}\)
Answer Details
73/12 − 11/4 + 3/2
[(73 − 33 + 18) ÷ (12)]
58/12
29/6
456
Question 23 Report
Simplify \( [1 \div (x^2 + 3x + 2)] + [1 \div (x^2 + 5x + 6)] \)
Answer Details
= 1÷(x2+3x+2)+[1÷(x2+5x+6)]
= [1÷((x2+x)+(2x+2))]+[1÷((x2+3x)+(2x+6))]
= [1 ÷ (x(x + 2) + 2(x +1))] + [1 ÷ (x(x + 3) +2(x + 3) )]
= [1 ÷ (x + 1)(x + 2)] + [1 ÷ ((x + 3) + (x + 2))]
=((x + 3) + (x + 1)) ÷ (x + 1)(x + 2)(x + 3)
Using the L.C.M
=((x + x + 3 + 1)) ÷ (x + 1)(x + 2)(x + 3)
=(2x+4)/(x+1)(x+2)(x+3) =2(x+2)/(x+1)(x+2)(x+3)
= 2(x+1)(x+3)
Question 24 Report
Answer Details
1×x2+3×x1+2×x0
7×81+0×80
x2+3x+2−56=0
x2+3x−54=0
x(x + 9)−6(x + 9) = 0
(x + 9)(x − 6) = 0
Either (x + 9) = 0 or (x − 6) = 0
x = − 9 or +6
The positive values for x = 6
The base number for 132x = 1326
Question 25 Report
If \(y = (x^2 + 3x - 1) \div (3x + 4)\). Find dy/dx
Answer Details
= 12x2+6x+2
⇒ 12x2+6+2
Divide both sides by 2
12x22−6x2−22
= 6x2+3x+1
dy/dx
= 6x2+3x+1
Question 27 Report
Find the value of x if \( [1 \div 64^{(x+2)}] = [4^{(x-3)} \div 16^x] \)
Answer Details
64−(x+2)=[4(x−3)]÷[16x]
Break down 4, 16 and 64 into smaller index numbers:
2−6(x+2)=22(x−3)÷24(x)
2−6x−12=22x−4x−6
2−6x−12=2−2x−6
− 6x − 12 = − 2x − 6
Collect the like terms:
−6x + 2x = −6 + 12
−4x =6
x = 64
x = −32
Question 28 Report
Answer Details
Question 29 Report
Answer Details
Question 30 Report
Answer Details
Question 31 Report
Answer Details
Question 32 Report
Answer Details
Question 33 Report
Answer Details
The internal radius = 10cm
The area of cross section = π(π2 − 102)
= π (11 + 10)(11 − 10)
= π(21)(1)
= 21πcm2
Question 34 Report
Answer Details
log101.5 + log1023 − log100.3
[log10(1.5 × 23) ÷ log100.3]
[log10(15/10 × 8 ) ÷ log10( 3/10)]
log10(15/10 × 8 × 10/3 )
log1040
Question 35 Report
Answer Details
Question 36 Report
Simplify \( \frac{1}{(x+1)} + \frac{1}{(x-1)} \)
Answer Details
= ((x − 1) + [(x + 1)) ÷ (x+1)(x − 1)]
Using the L.C.M.
= (x − 1 + x + 1) ÷ (x + 1)(x − 1)
= (x + 2 − 1 + 1) ÷ (x + 1)(x − 1)
= 2x ÷ (x + 1)(x − 1) =2x ÷ (x + 1)(x − 1)
Question 37 Report
Answer Details
The formula for finding the simple interest on a principal amount P for a period of t years at an interest rate of r per annum is:
simple interest = P * r * t / 100
In this case, P = 325, t = 5, and r = 3, so we can plug in the values into the formula:
simple interest = 325 * 3 * 5 / 100
simple interest = 48.75
So, the simple interest on ₦325 in 5 years at 3% per annum is ₦48.75.
Question 38 Report
Answer Details
Question 39 Report
Answer Details
= √(30 × 40)
= √(3 × 10 × 4 × 10)
= √(400 × 3)
= √400 × √3
= 20 × √3
= 20√3
Question 40 Report
Answer Details
y−y1=m(x−x1)
Where m is the slope or gradient
Don't forget, the line passes through the origin. That means it has a coordinates of {0,0} and {−3,−4} . So to calculate m
m=ΔyΔx=0−(−4)0−(−3)=43
Equation of straight line is therefore given as:
y−(−4)=43[x−(−3)]
3(y+4)=4(x+3)
3y+12=4x+12
3y=4x
∴y=4x3
Question 41 Report
Solve the equation
\[ 5^{(x-2)}=\left(1\div125\right)^{(x+3)} \]Answer Details
Question 42 Report
Answer Details
Question 43 Report
Answer Details
Question 44 Report
Make x the subject of the equation
\(s = 2 + \frac{t}{5}(x + \frac{3}{5}y)\)
Answer Details
t/5(x + 3/5y ) = s − 2
[(t(5x + 3y) ÷ 25)] = s − 2
t(5x + 3y) = 25 (s − 2)
5x + 3y = [(25(s − 2))÷ t]
5x = [(25(s − 2))÷ t] − 3y
Divide both sides by 5
x = [25(s − 2) − 3ty)]÷ 5t
Question 45 Report
Answer Details
[(390625T − 8)½]¼
[((390625)½ × T − 8)½]¼
(√ 390625 × T − 4 )¼
(625 × T −¼)¼
(54 × T −4)¼
= (54)¼ × (T − 4 )¼
= 5 × T − 1
= 5 × 1/T
= 5/T
Question 46 Report
Given that \( \tan x = \frac{2}{3} \), where \(0^\circ < x < 90^\circ\), Find the value of \(2\sin x\).
Answer Details
thus m2 = 22 + 32
= 4 + 9 = 13
m = √13
Hence, 2sin x = 2 x 2m
2 x2√13
= 4√13
= 4√13=√13√13
= 4√1313
Question 47 Report
Answer Details
Question 49 Report
Answer Details
Question 50 Report
Answer Details
2nd G.P. = ar − 1 = 54
2(r) = 54
r = 54/2 = 27
r = 27
3rd term = ar2 = (2) (27)2
2 × 27 × 27
= 1458
