JAMB UTME - Mathematics - 2015
Frage 1 Bericht
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
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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
Frage 2 Bericht
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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
Frage 3 Bericht
Factorize \(a^2 - b^2 - 4a + 4\)
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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)
Frage 4 Bericht
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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
Frage 8 Bericht
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Frage 9 Bericht
Make x the subject of the equation
\(s = 2 + \frac{t}{5}(x + \frac{3}{5}y)\)
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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
Frage 10 Bericht
Given that \( \tan x = \frac{2}{3} \), where \(0^\circ < x < 90^\circ\), Find the value of \(2\sin x\).
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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
Frage 11 Bericht
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= √(30 × 40)
= √(3 × 10 × 4 × 10)
= √(400 × 3)
= √400 × √3
= 20 × √3
= 20√3
Frage 12 Bericht
Find the value of x if \( [1 \div 64^{(x+2)}] = [4^{(x-3)} \div 16^x] \)
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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
Frage 13 Bericht
Simplify \(6\frac{1}{12} - 2\frac{3}{4} + 1\frac{1}{2}\)
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73/12 − 11/4 + 3/2
[(73 − 33 + 18) ÷ (12)]
58/12
29/6
456
Frage 14 Bericht
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Frage 15 Bericht
Simplify \( [1 \div (x^2 + 3x + 2)] + [1 \div (x^2 + 5x + 6)] \)
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= 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)
Frage 16 Bericht
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Frage 17 Bericht
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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
Frage 18 Bericht
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The internal radius = 10cm
The area of cross section = π(π2 − 102)
= π (11 + 10)(11 − 10)
= π(21)(1)
= 21πcm2
Frage 20 Bericht
Simplify \( \frac{1}{(x+1)} + \frac{1}{(x-1)} \)
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= ((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)
Frage 21 Bericht
Simplify (0.09)2 and give your answer correct to 4 significant figures
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(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
Frage 22 Bericht
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Frage 23 Bericht
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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
Frage 24 Bericht
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Frage 25 Bericht
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[(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
Frage 26 Bericht
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Radius = 4cm
π = 3.142
A = 3.142 × (4)2
= 3.142 × 16
= 50.272
A = 50.3cm2 (1dp)
Frage 27 Bericht
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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
Frage 28 Bericht
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Frage 29 Bericht
Solve the equation
\[ 5^{(x-2)}=\left(1\div125\right)^{(x+3)} \]Antwortdetails
Frage 30 Bericht
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Frage 31 Bericht
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Frage 32 Bericht
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Frage 33 Bericht
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Frage 34 Bericht
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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
Frage 35 Bericht
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Frage 36 Bericht
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Frage 37 Bericht
Solve the equation \(3x^2 - 4x - 5 = 0\)
Frage 38 Bericht
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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.
Frage 39 Bericht
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Frage 40 Bericht
Factorize \(x^2 - 2x - 15\)
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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).
Frage 41 Bericht
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Frage 42 Bericht
Given that A = {1, 5, 7}
B = {3, 9, 12, 15}
C = {2, 4, 6, 8}
Find (A ∪ B) ∪ C
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Frage 43 Bericht
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Frage 44 Bericht
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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.
Frage 45 Bericht
If \(y = (x^2 + 3x - 1) \div (3x + 4)\). Find dy/dx
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= 12x2+6x+2
⇒ 12x2+6+2
Divide both sides by 2
12x22−6x2−22
= 6x2+3x+1
dy/dx
= 6x2+3x+1
Frage 46 Bericht
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Frage 47 Bericht
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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
Frage 48 Bericht
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Frage 49 Bericht
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Frage 50 Bericht
In a town of 6250 inhabitants, there were 62 births during 1984. Find the percentage birth rate.
