JAMB UTME - Mathematics - 2006

Convert 22324 to base six

Bayanin Amsa

1st convert to base 10
22324 = 2x43 + 2x42 + 3x41 + 2x40
= 2x64 + 2x16 + 3x4 + 2x1
= 128 + 32 + 12 + 2
= 174 convert to base 6
6/174
6/29 R 0
6/4 R 5
6/0 R 5
= 4506

In triangle XYZ, ∠XYZ = 15${}^{\circ }$, ∠XZY = 45${}^{\circ }$ and lXYl = 7 cm. Find lYZl.

Bayanin Amsa

∠yxz = 180 - (45 + 15)

In the diagram, P, Q and R are three points in a plane such that the bearing of R from Q is 110o and the bearing of Q from P is 050o. Find angle PQR.

Bayanin Amsa

In the diagram above, KLMN is a cyclic quadrilateral. /KL/ = /KN/, ∠NKM = 55o and ∠KML = 40o. Find ∠LKM.

Bayanin Amsa

Calculate the logarithm to base 9 of 3-4 * 92 * (81)-1

Bayanin Amsa

3−4×92×81−1=log9(3−4×92×81−1)=log9(134×92×181)=log9(181×811×181)=log9181=log9192=log99−2=−2log99−2×1=−2

If the locus of the points which are equidistant from point P and Q meets line PQ at point N, then PN equals

Bayanin Amsa

The answer is NQ. By definition, the locus of points that are equidistant from two points P and Q is the perpendicular bisector of the line segment PQ. Therefore, if this locus intersects line PQ at point N, then N must be the midpoint of the segment PQ. So, PN and NQ are equal, and since N is the midpoint of PQ, PN is half of PQ. Thus, PN is equal to NQ.

A binary operation θ defined on the set of real number is such that xθy = xy/6 for all x, y ∈ R. Find the inverse of 20 under this operation when the identity element is 6

Bayanin Amsa

X ⊕ y = xy/6 = xe/6 = x
Xe = 6x
e = 6
X ⊕ y = xx1/6 where x1 = 20
Xx20/6 = 6
X = 36/20
X = 9/5

If tan θ = 5/4, find sin2θ - cos2θ.

Bayanin Amsa

Tan θ = 5/4
x2 = 52 + 42
= 25 + 16
= 41
x = √41
sin2θ - cos2θ = (5/√41)2 - (4/√41)2
= 25/41 - 16/41
= 9/41

In the diagram given PQ = 10cm, PS = 8cm and < PSR is 60 while < SRQ is a right angle. Find SR

Bayanin Amsa

Draw a line perpendicular to |SR| to form |PT| in $△$ PST, cos 60 = \(\frac{|ST|}{8}\(

|PT| = 8 cos 6 = 4cm

Since |TR| = |PQ|

SR = ST + TR

= (4 + 10)cm

= 14cm

If p varies inversely as the cube of q and q varies directly as the square of r, what is the relationship between p and r?

Bayanin Amsa

P ∝ 1/q3
P = K/q3
q3 = K/P
q = K/p1/3
But q ∝ r2
q = Kr2K/p1/3 = Kr2
r2 = K/p1/3 x 1/K
r2 = 1/p1/3
r = (1/p1/3)2
r = 1/p1/6
r ∝ 1/p1/6
∴ r varies inversely as 6√P

Find the variance of 2x, 2x-1 and 2x+1

Bayanin Amsa

To find the variance of a set of numbers, we first need to find the mean of the set. The mean of the set {2x, 2x-1, 2x+1} is: (2x + 2x-1 + 2x+1)/3 = 2x Next, we need to find the difference between each number and the mean, square these differences, and take the average of the squares. This gives us the variance. The differences between each number and the mean are: 2x - 2x = 0 2x-1 - 2x = -1 2x+1 - 2x = 1 Squaring these differences, we get: 0^2 = 0 (-1)^2 = 1 1^2 = 1 Taking the average of these squares, we get: (0 + 1 + 1)/3 = 2/3 Therefore, the variance is 2/3. Thus, the answer is option A: 2/3.

A bag contains 5 black, 4 white and x red marbles. If the probability of picking a red marble is 2/3, find the value of x

Bayanin Amsa

Let's start by using the total number of marbles in the bag which is the sum of the number of black, white, and red marbles. Total number of marbles = 5 (black) + 4 (white) + x (red) = 9 + x Now, we are given that the probability of picking a red marble is 2/3. This means that out of all the marbles in the bag, 2/3 of them are red. So, we can write an equation using this information: Number of red marbles / Total number of marbles = 2/3 Simplifying this equation, we get: x / (9 + x) = 2/3 Cross-multiplying, we get: 3x = 2(9 + x) Simplifying this equation, we get: 3x = 18 + 2x Subtracting 2x from both sides, we get: x = 18 Therefore, the value of x is 6. So, there are 6 red marbles in the bag.

If (K2)6 * 36 = 35(K4)5, what is the value of k?

Bayanin Amsa

(K2)6 * 36 = 35(K4)5
k*61 + 2*60 + 3*50 = 3*50 + K*51 + 4*50
6K + 2 = 5K + 4
6K - 5k = 4 - 2
K = 2

For what of n is n+1C3 = 4(nC3)?

Bayanin Amsa

To solve the given equation, we can use the formula for combinations, which is: nCr = n! / r!(n-r)! Substituting n+1 for n and 3 for r, we get: (n+1)C3 = (n+1)! / 3!(n+1-3)! (n+1)C3 = (n+1)! / 6n! Simplifying the right side of the equation using the given information, we get: 4(nC3) = 4n! / 6n! (nC3) = n! / 6n! Now we have: (n+1)C3 = 4(nC3) Substituting the formulas we derived earlier, we get: (n+1)! / 3!(n+1-3)! = 4n! / 3!(n-3)! Simplifying and canceling the terms, we get: (n+1)(n)(n-1) = 4(n)(n-1)(n-2) Expanding and simplifying the equation, we get: n^3 - 3n^2 - 10n + 12 = 0 Factoring the equation, we get: (n-4)(n^2 - n - 3) = 0 The roots of the equation are n = 4, (1+sqrt(13))/2, and (1-sqrt(13))/2. Since n represents a counting number, the only possible solution is n = 4. Therefore, the answer is 4.

he pie chart above shows the expenditure of a family whose income is ₦30,000. If the expenditure on food is twice that on housing and that on school fees is twice that on transport, how much does the family spend on food?

Bayanin Amsa

Expenditure on housing = x
Expenditure on food = 2x
Expenditure on school fees = 90${}^{\circ }$
Expenditure on transport = 45${}^{\circ }$
x + 2x + 90 + 45 = 360${}^{\circ }$
3x + 135 = 360${}^{\circ }$
3x = 360 - 135
3x = 225
x = 225/3 = 75${}^{\circ }$
∴∠ for food = 2x = 2 * 75
= 150
360${}^{\circ }$ = N30,000
1${}^{\circ }$ = ?
1${}^{\circ }$ = 30,000/360
150${}^{\circ }$ = (30000/360) * (150/1)
= N12500

simplify $\frac{\frac{7}{9}?\frac{2}{3}}{\frac{1}{3}+\frac{\frac{2}{5}}{\frac{4}{5}}}$

Bayanin Amsa

In the parallelogram PQRS given, find angle < SQR

Bayanin Amsa

< QRS = 30${}^{\circ }$(corresponding < ), < QSR = 50${}^{\circ }$

< SQR = 180${}^{\circ }$ - (30 + 50)${}^{\circ }$ (Sum of < S in a $△$)

180${}^{\circ }$ - 80${}^{\circ }$ = 100${}^{\circ }$

The response of 160 pupils in a school asked to indicate their favourite subjects is given in the bar chart above. What percentage of the pupils have English English and Health education as the their favourite subjects?

Bayanin Amsa

($\frac{52}{160}=\frac{36}{160}$)% = $\frac{88}{160}$%

= 55%

If the mean of five consecutive integers is 30, find the largest of the numbers

Bayanin Amsa

Let's call the smallest of the five consecutive integers "x". Since the numbers are consecutive, we know that the next four integers will be x + 1, x + 2, x + 3, and x + 4. To find the mean of these five numbers, we add them up and divide by 5: (x + (x+1) + (x+2) + (x+3) + (x+4)) / 5 = 30 Simplifying the left side: (5x + 10) / 5 = 30 Canceling the 5's on the left side: x + 2 = 30 Subtracting 2 from both sides: x = 28 Therefore, the five consecutive integers are 28, 29, 30, 31, and 32, and the largest of these numbers is 32. So the correct option is: 32.

If y = x2 - x - 12, find the range of values of x for which y $\ge$ 0

Bayanin Amsa

y = x2 - x - 12
= (x - 4)(x - 3)
∴ x = 4 or x = -3
-3 < x < 4

But y $\ge$ 0
∴ -3 < x $\le$ 4

Two chords PQ and RS of a circle intersected at right angles at a point inside the circle. If ∠QPR = 35o,find ∠PQS

Bayanin Amsa

If m:n = 13:111, find m2 - n2 : (m + n)2

Bayanin Amsa

In the diagram above, QR is in the diameter of the semicircle QR. Find the areas of the figure to the nearest whole number.

Bayanin Amsa

Area of rectangle PQRS
= 10 x 7 = 70 cm2
Area of semi circle
QR = Dπ = 7 x 22/7 = 22cm2
∴Area of the figure = 70 + 22
= 92cm2
= 90cm2

The table above shows the distribution of recharge cards of four major GSM operators. What is the probability that a recharge card selected at random will be GTN or Qtel?

Bayanin Amsa

P(GTN) = 5/20
P(Qtel) = 3/20
∴P(GTN or Qtel) = (5/20) + (3/20)
= 8/20
= 2/5

Differentiate (x2 - 1/x)2 with respect to x

Bayanin Amsa

The given expression is (x^2 - 1)/x^2. To differentiate this expression with respect to x, we can use the quotient rule of differentiation. The quotient rule states that the derivative of (f(x)/g(x)) = [g(x)*f'(x) - f(x)*g'(x)]/[g(x)]^2. Applying the quotient rule, we get: [(x^2)*2x - (x^2 - 1)*2x]/(x^2)^2 Simplifying the expression further, we get: [2x^3 - 2x^3 + 2x]/x^4 = 2/x^3 Therefore, the answer is option C: 4x^2 - 2 - 2/x^3.

THe cost of renovating a 6 m square room is N540. What is the cost of renovating a 9 m square room?

Bayanin Amsa

The cost of renovating a room is directly proportional to the area of the room. If the cost of renovating a 6 m square room is N540, then we can find the cost of renovating a 9 m square room using the proportionality constant. Let C be the cost of renovating a room and A be the area of the room. Then, we have: C = kA where k is the proportionality constant. We can find k by using the given information that the cost of renovating a 6 m square room is N540: 540 = k(6) k = 90 Now, we can use this value of k to find the cost of renovating a 9 m square room: C = kA = 90(9) = N810 Therefore, the cost of renovating a 9 m square room is N810.

Find the value of x for which the function 3x3 - 9x is minimum

Bayanin Amsa

The gradient of a curve is 2x + 7 and the curve passes through point (2, 0). find the equation of the curve.

Bayanin Amsa

To find the equation of the curve, we need to integrate the gradient. Since the gradient is 2x + 7, the equation of the curve will be of the form y = x^2 + 7x + C, where C is a constant of integration. We can find the value of C by using the fact that the curve passes through the point (2, 0). Substituting x = 2 and y = 0 into the equation, we get: 0 = 2^2 + 7(2) + C 0 = 4 + 14 + C C = -18 Therefore, the equation of the curve is y = x^2 + 7x - 18. The answer is.

A final examination requires that a student answer any 4 out of 6 questions. In how many ways can this be done?

Bayanin Amsa

To calculate the number of ways a student can answer any 4 out of 6 questions, we can use the combination formula, which is: nCk = n! / (k! * (n-k)!) where n is the total number of items, k is the number of items to be selected, and ! represents the factorial function. In this case, we have n = 6 (the total number of questions on the exam) and k = 4 (the number of questions the student must answer). So we can calculate the number of ways to select 4 questions out of 6 using the combination formula as follows: 6C4 = 6! / (4! * (6-4)!) = (6 * 5 * 4 * 3) / (4 * 3 * 2 * 1) = 15 Therefore, the answer is 15, which means that a student can answer any 4 out of 6 questions in 15 different ways.

How many terms of the series 3, -6, +12, - 24, + ..... are needed to make a total of 1-28?

Bayanin Amsa

3, -6, +12, -24
a = 3, r = -2
$8n=\frac{a(1-r^n)}{1-r}\therefore 1-2^8=\frac{3(1-(-2^{n-1}\left)\right)}{1-(-2)}1-2^8=\frac{3(1-(-2^{n-1}\left)\right)}{3}$
1-28 = 1-(-2)n-1
-28 = -2n-1
8 = n-1
n = 9

A binary operation * on the set of rational numbers is defined as $x\times y=\frac{x^2-y^2}{2xy}find-5\times 3$

Bayanin Amsa

To evaluate -5 * 3 using the binary operation * on the set of rational numbers, we substitute x = -5 and y = 3 in the given expression: -5 * 3 = (-5)^2 - 3^2 / 2(-5)(3) Simplifying the numerator, we get: (-5)^2 - 3^2 = 25 - 9 = 16 Simplifying the denominator, we get: 2(-5)(3) = -30 Substituting these values back into the expression, we get: -5 * 3 = 16 / -30 Simplifying the fraction, we get: -5 * 3 = -8/15 Therefore, the value of -5 * 3 using the binary operation * on the set of rational numbers is -8/15.

Simplify (25)-1/2 x (27)1/3 + (121)-1/2 x (625)-1/4

Bayanin Amsa

What is the locus of points equidistant from the lines ax + bc + c = 0?

Bayanin Amsa

The given equation of a line can be rewritten in the standard form of a line, which is y = mx + b, where m is the slope and b is the y-intercept. To rewrite the equation, we can solve for y: ax + bc + c = 0 ax + bc = -c y = -(a/b)x - (c/b) The locus of points equidistant from the lines y = -(a/b)x - (c/b) is a line perpendicular to the given lines, with the distance from the origin equal to the absolute value of c/b. This is because the perpendicular bisector of a line segment is equidistant from the endpoints of the segment, and in this case the "endpoints" are the given lines. Since the slope of the given lines is -(a/b), the slope of the perpendicular bisector is b/a. The perpendicular bisector passes through the origin (0,0), so we can find its equation by using the point-slope form of a line: y - 0 = (b/a)x y = (b/a)x This is equivalent to the equation y = mx, where m = b/a, which is the equation of a line passing through the origin with slope b/a. Therefore, the answer is: a line ax - ay + q = 0.

Compute 1100112 + 111112

Bayanin Amsa

1100112 + 111112 = 10100102

If T = 2 π √1/g, make g the subject of the formula

Bayanin Amsa

T = 2 π √1/gT/2π = √1/g
(T/2π)2 = (√1/g)2T2/4π2 = 1/g
T2 x g = 4π2 x l
G = (4π2 x l) / T2

Evaluate ${\int }_{-4}^0(1-2x)dx$

Bayanin Amsa

∫0−4(1−2x)dx=[x−x2]0−4=(0−0+C)−(−4(−4)2+C)=C−(−4−16+C)=C−(−20+C)=C+20−C=20

In a small village of 500 people, 350 speak the local language while 200 speak pidgin English. What percentage of the population speak both.

Bayanin Amsa

To find the percentage of the population that speaks both the local language and pidgin English, we need to use the formula: Percentage = (Number of people who speak both / Total population) x 100% We are given that there are 500 people in the village, 350 speak the local language, and 200 speak pidgin English. However, we do not know the number of people who speak both. To find the number of people who speak both, we can add the number of people who speak the local language and the number of people who speak pidgin English and subtract it from the total population. Number of people who speak both = Total population - (Number of people who speak local language + Number of people who speak pidgin English) Number of people who speak both = 500 - (350 + 200) Number of people who speak both = 50 So, 50 people speak both languages. Now we can substitute this value in the formula to find the percentage. Percentage = (Number of people who speak both / Total population) x 100% Percentage = (50 / 500) x 100% Percentage = 0.1 x 100% Percentage = 10% Therefore, the percentage of the population that speaks both the local language and pidgin English is 10%. Answer is correct.

If dy/dx = x + cos x, find y

Bayanin Amsa

We need to integrate the given expression to find the value of y. Integrating both sides with respect to x, we get: ∫dy = ∫(x + cos x)dx y = 1/2 * x^2 + sin x + c where c is the constant of integration. Therefore, the correct answer is: - x^2/2 + sin x + c

In the diagram, the tangent MN makes an angle of 55o with the chord PS. IF O is the centre of the circle, find

Bayanin Amsa

To solve this problem, we need to use the fact that the angle between a tangent and a chord of a circle is equal to the angle formed by the chord in the opposite segment. Let's call the point where the tangent MN touches the circle point T. Then, angle MOT is 90 degrees because OT is a radius of the circle and MT is a tangent, and these two lines are perpendicular at point T. Also, angle MTS is 55 degrees because MN is tangent to the circle and makes an angle of 55 degrees with PS. Therefore, angle MTN is 180 - 90 - 55 = 35 degrees because the angles in a triangle add up to 180 degrees. Now, let's consider the triangle OTN. We know that angle OTN is 90 degrees because OT is a radius of the circle and TN is a tangent, and these two lines are perpendicular at point T. We also know that angle MTN is 35 degrees. Therefore, angle OTM is 180 - 90 - 35 = 55 degrees because the angles in a triangle add up to 180 degrees. Finally, we can see that angle PTS is equal to angle OTM because they both intercept the same arc PS. Therefore, angle PTS is also 55 degrees. Since angle PTS is half of angle POS (which is equal to 110 degrees because it intercepts a semicircle), angle POS is 2 * 110 = 220 degrees. Finally, the angle between the tangent MN and the chord PS is equal to angle MTS, which is 55 degrees. Therefore, the answer is option (C) 35o.

Find the value of k if the expression kx3 + x2 - 5x - 2 leaves a remainder 2 when it is divided by 2x + 1

Bayanin Amsa

F(x) = Q x D + R
Kx3 + x2 - 5x – 2 = Q(2x+1)+R
If 2x+1 = 0 implies x = -1/2
∴k(-1/2)3 + (-1/2)2 -5(-1/2) -2
= Q(2(-1/2) + 1) +2
K(-1/8) + 1/4 + 5/2 - 2 = Q(-1+1)+2
-k/8 + 1/4 + 5/2 - 2 = 0+2
(-k+2+20-16) / 8 = 2
(-k+6) / 8 = 2
-k+6 = 2*8
-k+6 = 16
-k = 16-6
-k = 10
∴k = -10

The table above shows the scores of a group of students in a physics test. If the mode is m and the number of students who scored 4 or more is n, what is (n, m)?

Bayanin Amsa

To find the values of n and m, we need to first determine the mode and the number of students who scored 4 or more. The mode is the most frequently occurring score in the set. In this case, the score of 4 appears most frequently, with a total of 12 students achieving that score. Therefore, the mode is 4. To find the number of students who scored 4 or more, we need to add up the frequencies of all scores that are 4 or greater. Adding the frequencies of scores 4, 5, 6, and 7 gives a total of 33 students. Therefore, the values of (n, m) are (33, 4).

PQ and RS are two parallel lines. If the coordinates of P, Q, R, S are (1,q), (3,2), (3,4), (5,2q) respectively, find the value of q

Bayanin Amsa

Given that PQ and RS are two parallel lines, we can observe that their slopes are equal. The slope of PQ can be found using the coordinates of points P and Q: slope of PQ = (2 - q)/(3 - 1) = (2 - q)/2 Similarly, the slope of RS can be found using the coordinates of points R and S: slope of RS = (2q - 4)/(5 - 3) = (2q - 4)/2 Since PQ and RS are parallel, their slopes are equal: (2 - q)/2 = (2q - 4)/2 Simplifying the equation above, we get: 2 - q = q - 2 2q = 4 q = 2 Therefore, the value of q is 2, which corresponds to.

The sum of the first n positive integers is

Bayanin Amsa

Let the positive integers be 1, 2, ,3, 4, .....n
∴ a = 1, d = 1 and n = n
Sn = n/2(2a + (n-1)d)
= n/2 (2 + n – 1)
= 1/2n(n + 1)

The solution set of the shaded area above is

Bayanin Amsa

Y = x implies y ≤ x
Y + x = 4 implies y = 4 – x
∴y = 4 – x
∴y ≤ x, y + x ≤ 4 and y ≥ 0

Evaluate $\frac{2}{6-5\sqrt{3}}$

Bayanin Amsa

In the diagram above, PQ = 10 cm, PS = 8 cm and ∠PSR is 60${}^{\circ }$ while ∠SRQ is a right angle. Find SR

Bayanin Amsa

In ΔPST; cos 60 = ST/8
ST = 8cos 60
ST = 8 x 1/2 = 4
TR = 10 cm (opp. sides of a rectangle PQRT)
SR = ST + TR
SR = 4 + 10
SR = 14 cm

Find the tax on an income of N20,000 if no tax is paid on the first N10,000 and tax is paid at N50 in N1000 on the next N5000 and at N55 in N1000 on the remainder

Bayanin Amsa

To calculate the tax on an income of N20,000 using the given tax rules, we need to break down the income into three parts: 1. The first N10,000: No tax is paid on this amount. 2. The next N5,000: Tax is paid at N50 for every N1000, so the tax on N5,000 would be (N5,000/N1,000) x N50 = N250. 3. The remainder of N5,000: Tax is paid at N55 for every N1000, so the tax on N5,000 would be (N5,000/N1,000) x N55 = N275. Therefore, the total tax on an income of N20,000 would be N0 + N250 + N275 = N525. In summary, the tax on an income of N20,000 using the given tax rules is N525, which is option (D).

Differentiate (cos θ - sin θ)2

Bayanin Amsa

The solution set of the shaded area above is

Bayanin Amsa

In the diagram P, Q, R, S are points on the circle RQS = 30o. PRS = 50o and PSQ = 20o. What is the value of xo + yo?

Bayanin Amsa

Draw a line from P to Q

< PQS = < PRS (angle in the sam segment)

< PQS = 50o

Also, < QSR = < QPR(angles in the segment)

< QPR = xo

x + y + 5= = 180(angles in a triangle)

x + y = 180 - 50

x + y = 130o