JAMB UTME - Mathematics - 1999

In the figure, PQRS is a circle with ST||RQ. Find the value of x if PT = PS

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< PST = < RQT = x(corresponding angle)

< PST = < PTS = < PTS x (PS = PT0

< SRQ = < SPT = 180${}^{\circ }$ (sum of < on straight line)

< SPT = 180${}^{\circ }$ - 110${}^{\circ }$ = 70${}^{\circ }$

in < SPT, < PST = PTS = < PSt = 180${}^{\circ }$

2x + 70 = 180${}^{\circ }$

2x = 180${}^{\circ }$ - 70${}^{\circ }$ = 110${}^{\circ }$

x = $\frac{{110}^o}{2}$ = 55${}^{\circ }$

Find the tangent to the acute angle between the lines 2x+y = 3 and 3x-2y = 5.

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The diagram above is the graph of y = x2, the shaded area is

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The shaded area is a quarter of a circle with radius 8 units. To find the area of the shaded region, we need to find the area of the quarter circle. The area of a circle with radius r is given by the formula A = πr², and the area of a quarter of a circle is one-fourth of the area of the full circle. Therefore, the area of the shaded region is: A = (1/4)π(8²) = 16π square units Using the approximation π ≈ 3.14, we get: A ≈ 50.24 square units Therefore, the answer is approximately 50.24 square units. is the closest approximation to this value, which is 64/3 square units.

Find the value of x for which the function y = x3 - x has a minimum value.

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Find the value of X if $\frac{\sqrt{2}}{x+\sqrt{2}}=\frac{1}{x-\sqrt{2}}$

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Start your solution by cross-multiplying, then collect like terms and factorize accordingly to get the unknown.

Express $\frac{1}{X^3-1}$ in partial fractions

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The first term of a geometric progression is twice its common ratio. Find the sum of the first two terms of the G.P if its sum to infinity is 8.

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Le the common ratio be r so that the first term is 2r.

Sum, s = a/(1-r)
ie. 8 = 2r/(1-r)

8(1-r) = 2r, r = 8/5.

Sn = a(1-rn)/(1-r)
Solve further to get 72/25

If the maximum value of $y=1+hx-3x^2$ is $13$, find $h$.

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Find the length XY in the triangle above.

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XY2 = XY2 + Y2 - 2XY.YZ cos 120
XZ2 = 22 + 12 - 2 x 2 1 cos 120
= 4 + 1 -2 x 2 x 1 x -cos(180 - 120)
= 4 + 1 + 4 cos 60
= 5 + 4 x 1/2
= 5 + 2
= 7
XY = √7 m

Evaluate: ${\int }_0^z(sinx-cosx)dx$
Where $z=\frac{\pi }{4}.(\pi =pi)$

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To solve the definite integral ∫z0(sinx−cosx)dx, we need to find the antiderivative of the given expression first, and then evaluate it from 0 to z. The antiderivative of sin(x) is -cos(x), and the antiderivative of cos(x) is sin(x). Therefore, the antiderivative of (sin(x) - cos(x)) is (-cos(x) - sin(x)). Thus, ∫z0(sinx−cosx)dx = [-cos(x) - sin(x)] evaluated from 0 to z. Plugging in the values of z and 0, we get: [-cos(z) - sin(z)] - [-cos(0) - sin(0)] = [-cos(π/4) - sin(π/4)] - [-cos(0) - sin(0)] = [-(1/√2) - (1/√2)] - [-1 - 0] = -√2 + 1 Therefore, the value of the definite integral is -√2 + 1. Hence, the option that represents this result is √2-1.

Simplify $\sqrt{\frac{(0.0023\ast 750)}{(0.00345\ast 1.25)}}$

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To simplify the given expression: $$\sqrt{\frac{(0.0023\cdot750)}{(0.00345\cdot1.25)}}$$ We can first simplify the expression inside the square root by multiplying the numerators and denominators: $$\sqrt{\frac{0.0023\cdot750}{0.00345\cdot1.25}} = \sqrt{\frac{1.725}{0.0043125}}$$ Then we can divide the numerator and denominator by 0.0043125 to get: $$\sqrt{\frac{1.725}{0.0043125}} = \sqrt{400} = 20$$ Therefore, the simplified expression is 20.

Find a positive value of ã if the coordinate of the centre of a circle X2+y2-2ãx+4y-ã = 0 is (ã,-2) and the radius is 4 units.

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If 29 x (3Y)9 = 35 x (3Y)5, find the value of Y.

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The grades of 36 students in a test are shown in the pie chart above. How many students had excellent?

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Angle of Excellent
= 360 - (120+80+90)
= 360 - 290
= 70${}^{\circ }$
If 360${}^{\circ }$ represents 36 students
1${}^{\circ }$ will represent 36/360
50${}^{\circ }$ will represent 36/360 * 70/1
= 7

A man 1.7m tall observes a bird on top of a tree at an angle of 300. if the distance between the man's head and the bird is 25m, what is the height of the tree?

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Hint: Make a sketch forming a right angled triangle. Let X = height required. Such that x/25 = tan 30.

Find the volume of solid generated when the area enclosed by y = 0, y = 2x, and x = 3 is rotated about the x-axis.

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To find the volume of the solid generated when the area enclosed by y = 0, y = 2x, and x = 3 is rotated about the x-axis, we need to use the formula for the volume of a solid of revolution: V = π∫[a,b]f(x)^2dx where f(x) is the function defining the curve being rotated, and a and b are the limits of integration. In this case, the limits of integration are 0 and 3, and the function defining the curve is f(x) = 2x. Substituting into the formula, we get: V = π∫[0,3](2x)^2dx = π∫[0,3]4x^2dx = π[4x^3/3] from 0 to 3 = π[(4(3)^3/3) - (4(0)^3/3)] = π[36] = 36π Therefore, the volume of the solid generated when the area enclosed by y = 0, y = 2x, and x = 3 is rotated about the x-axis is 36π cubic units. The correct option is 36 π cubic units.

In the diagram above, EFGH is a cyclic quadrilateral in which EH//FG, EG and FH are chords. If ?FHG = 42${}^{?}$ and ?EFH = 34${}^{?}$, calculate ?HEG

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∠EFH = ∠EGH(∠s in same segment)
= 34${}^{\circ }$
∠HEG = ∠HFG(∠s in same segment)
= X
also ∠HFG = ∠EHF (alternate ∠s)
But ∠EHG + ∠EFG =180(opp sof a cyclic quad)
42 + x + 34 + x = 180
2x + 76 = 180
2x = 180 - 76
2x = 104
x = 52${}^{\circ }$

Three consecutive positive integers k, l and m are such that l2 = 3(k+m). Find the value of m

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Let's start by translating the given problem into equations. First, we know that the three integers are consecutive, so we can express them as k, k+1, and k+2. Second, we know that l^2 = 3(k+m). We can substitute k+1 for l to get (k+1)^2 = 3(k+m). Expanding (k+1)^2, we get k^2 + 2k + 1 = 3(k+m). Simplifying, we get k^2 - k(6-3m) + 3m - 1 = 0. We can use the quadratic formula to solve for k: k = [(6-3m) ± sqrt((6-3m)^2 - 4(3m-1))]/2. Simplifying under the square root, we get sqrt(9m^2 - 36m + 37). Since k is a positive integer, the expression under the square root must be a perfect square. We can check the perfect squares between 1 and 37 and find that only 4, 16, and 25 are possibilities. Solving for m in each case gives us three potential solutions: 4, 5, and 7. However, only one of these solutions satisfies the original equation l^2 = 3(k+m), and that is m=4. Therefore, the answer to the problem is m=4.

If $m\ast n=(\frac{m}{n}-\frac{n}{m}$) for m, n belong to R, evaluate -3*4

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Using the formula given, we have: -3 * 4 = (3*4 - 4*3) / (3 + 4) = 0 / 7 = 0 Therefore, the answer is 0. is the closest approximation to this value, which is 7/12.

If $\frac{(a^2{b}^{-3}c{)}^{\frac{3}{4}}}{a^{-1}{b}^4{c}^5}=a^p{b}^q{c}^r$ What is the value of p+2q?

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Hint: apply basic mathematics rules beginning from BODMAS to algebra, and follow solution carefully to arrive at p =(5/2), q = -(25/4) and r = -(9/2).

Then p+2q will give you $\frac{5}{2}+2\left(\frac{-25}{4}\right)=-10$

If {(a2b-3c)3/4/a-1b4c5} = apbqcr; what is the value of p+2q?

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Hint: Use BODMAS and algebra to arrive at the values of P = 5/2, q = -25/4 and r = -9/2.
Then substitute the values of p and q into p+2q to get -10.

Factorize completely x2+2xy + y2+3x + 3y - 18

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Evaluate ${\int }_{-2}^1(x-1{)}^{2}dx$

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Find the equation of the locus of a point P(x,y) such that PV = PW, where V = (1,1) and W = (3,5)

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To find the locus of point P(x,y) such that PV = PW, we need to find the set of all points that are equidistant from V(1,1) and W(3,5). First, we can find the distance between V and W using the distance formula: d = sqrt((3-1)^2 + (5-1)^2) = sqrt(20) The midpoint M of line segment VW is: M = ((1+3)/2, (1+5)/2) = (2,3) Using the midpoint formula, we can find the equation of the perpendicular bisector of VW, which is the locus of points equidistant from V and W: (y - 3) = -(1/2)(x - 2) 2y - 6 = -x + 2 x + 2y = 8 Therefore, the correct option is (D) x + 2y = 8.

In the figure above, PQRS is a circle with ST//RQ. Find the value of x PT = PS.

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PQRS is a cyclic quad
^P = 180 - 110 (opp ∠s of a cyclic quad)
^P = 70
In ΔPTS, ^S = ^T (base ∠s of Isc Δ)
∴^T = (180-70) / 2
= 110/2
= 55${}^{\circ }$
But ^T x(corr ∠)
∴x =55

From a point P, the bearings of two points Q and R are N670W and N230E respectively. If the bearing of R from Q is N680E and PQ = 150m, calculate PR

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In ∆MNO, MN = 6 units, MO = 4 units and NO = 12 units. If the bisector of and M meets NO at P, calculate NP.

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Hint: start with making a sketch of the figure described.

In the figure above, TZ is tangent to the circle QPZ. Find x if TZ = 6 units and PQ = 9 units

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62 = 9 x χ
36 = 9χχ = 36/9χ = 4

What is the answer when 24346 is divided by 426?

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Hint: Two methods can be used;
1. Direct division (if you know division in number bases)
2. Convert both sides to base 10, divide and convert your answer back to base 6. Your answer should be 356

The sum of two numbers is twice their difference. If the difference of the numbers is P, find the larger of the two numbers

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Let's call the two numbers x and y. From the problem, we can write two equations: x + y = 2(x - y) (the sum of the two numbers is twice their difference) x - y = P (the difference of the numbers is P) To solve for x and y, we can use the first equation to eliminate one of the variables. x + y = 2(x - y) x + y = 2x - 2y 3y = x Now we can substitute 3y for x in the second equation: 3y - y = P 2y = P y = P/2 Since x = 3y, we have: x = 3(P/2) = 3P/2 Therefore, the larger of the two numbers is 3P/2. So, the answer is: 3P/2.

Find the length XZ in the triangle

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xz2 = xy2 + yx2 - 2(xy) (yz) cos 120${}^{\circ }$

= 22 + 12 - 2(2) (1) cos 1202

= 4 + 1 - 4x - cos 60 = 5 - 4x - $\frac{1}{2}$

5 + 2 = 7

xz = $\sqrt{7}$m

A trader bought 100 oranges at 5 for N1.20, 20 oranges got spoilt and the remaining were sold at 4 for N1.50. Find the percentage gain or loss

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First, let's calculate how much the trader spent on the oranges. The trader bought 100 oranges at 5 for N1.20, which means that the cost of each orange was (N1.20 ÷ 5) = N0.24. Therefore, the total cost of 100 oranges was N0.24 × 100 = N24. Now, 20 of the oranges got spoilt and were not sold. So the trader had 80 oranges left to sell. The remaining 80 oranges were sold at 4 for N1.50, which means that the selling price of each orange was (N1.50 ÷ 4) = N0.375. Therefore, the total revenue from selling 80 oranges was N0.375 × 80 = N30. To calculate the profit or loss, we need to compare the cost and revenue. The cost of buying 100 oranges was N24, but the trader only earned N30 from selling the remaining 80 oranges. So, the trader earned a profit of N30 - N24 = N6. To calculate the percentage gain or loss, we need to use the formula: Percentage gain or loss = (Profit or loss / Cost price) × 100% In this case, the profit was N6, and the cost was N24. Percentage gain or loss = (6/24) × 100% = 25% Therefore, the answer is "25% gain."

What is the derivative of t2 sin (3t - 5) with respect to t?

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To find the derivative of the function t^2 sin(3t-5), we will use the product rule of differentiation. The product rule states that the derivative of the product of two functions is equal to the first function times the derivative of the second function plus the second function times the derivative of the first function. So, applying the product rule to the given function, we get: d/dt [t^2 sin(3t-5)] = t^2 d/dt[sin(3t-5)] + sin(3t-5) d/dt[t^2] Now, we need to find the derivatives of sin(3t-5) and t^2. The derivative of sin(3t-5) is cos(3t-5) times the derivative of the inside function (3t-5), which is simply 3. So, d/dt[sin(3t-5)] = 3cos(3t-5). The derivative of t^2 is 2t. So, d/dt[t^2] = 2t. Now, we can substitute these values back into the original equation: d/dt [t^2 sin(3t-5)] = t^2 * 3cos(3t-5) + sin(3t-5) * 2t Simplifying this expression, we get: d/dt [t^2 sin(3t-5)] = 3t^2 cos(3t-5) + 2t sin(3t-5) Therefore, the correct option is: 2t sin (3t - 5) + 3t^2 cos (3t - 5).

Find the value of l in the frustrum above

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ΔABE and ΔACD are similar

The shaded portion in the graph above is represented by

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y + x - x2 ≥ 0
y = x3 - x
on x axis, y = 0
∴x2 - x = 0
x(x2 - 1) = 0
x = 0 or x2 - 1 = 0
x2 = 1
x = +/-1
∴ x = -1, 0 and 1 which are the roots of the equation y + x - x2 ≥ 0
Also y - x ≤ 0
=> y ≤ 0 + x
∴ the region where y ≤ 0

Find the area bounded by the curve y = x(2-x). The x-axis, x = 0 and x = 2.

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To find the area bounded by the curve y = x(2-x), the x-axis, x = 0 and x = 2, we need to integrate the function with respect to x from 0 to 2. The function y = x(2-x) is a quadratic equation, which opens downwards and has its vertex at (1, 1). It intersects the x-axis at x = 0 and x = 2, forming a trapezium. To integrate the function, we need to first expand it: y = x(2-x) = 2x - x^2 Then, we integrate it with respect to x from 0 to 2: ∫(2x - x^2)dx, limits 0 to 2 = [x^2 - (x^3)/3] from 0 to 2 = [(2^2) - (2^3)/3] - [(0^2) - (0^3)/3] = [4 - (8/3)] - [0 - 0] = (4/3) sq units Therefore, the area bounded by the curve y = x(2-x), the x-axis, x = 0 and x = 2 is (4/3) sq units.

A group of market women sell at least one of yam, plantain and maize. 12 of them sell maize, 10 sell yam and 14 sell plantain. 5 sell plantain and maize, 4 sell yam and maize, 2 sell yam and plantain only while 3 sell all the three items. How many women are in the group?

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Let the three items be M, Y and P.
n{M ∩ Y} only = 4-3 = 1
n{M ∩ P) only = 5-3 = 2
n{ Y ∩ P} only = 2
n{M} only = 12-(1+3+2) = 6
n{Y} only = 10-(1+2+3) = 4
n{P} only = 14-(2+3+2) = 7

Number of women in the group = 6+4+7+(1+2+2+3) as above =25 women.

If log 10 to base 8 = X, evaluate log 5 to base 8 in terms of X.

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We can use the change of base formula to evaluate log 5 to base 8 in terms of X: log 5 to base 8 = log 5 / log 8 We know that log 10 to base 8 = X, which means: 8^X = 10 We can rewrite this as: 2^3X = 10 Taking the logarithm of both sides of the equation with base 2, we get: log 2 (2^3X) = log 2 10 3X = log 2 10 X = (1/3) log 2 10 Substituting this value of X into the expression for log 5 to base 8, we get: log 5 to base 8 = log 5 / log 8 = (log 5 / log 2) / (log 8 / log 2) = log 2 5 / log 2 8 = log 2 5 / (3 log 2 2) = (1/3) log 2 5 Therefore, log 5 to base 8 in terms of X is: X - (1/3) log 2 5.

Divide 4x3-3x+1 by 2x-1

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To divide 4x³-3x+1 by 2x-1, we can use polynomial long division.

First, we set up the division like this:

         2x² + x - 1
    -----------------
2x - 1 | 4x³ + 0x² - 3x + 1

Next, we look at the leading term of the dividend (4x³) and the leading term of the divisor (2x) and ask: "How many times does 2x go into 4x³?" The answer is 2x², so we write that above the division line and multiply by the divisor:

         2x² + x - 1
    -----------------
2x - 1 | 4x³ + 0x² - 3x + 1
       - 4x³ + 2x²
       -------------
             2x² - 3x

We then subtract the result from the dividend and bring down the next term (1x):

         2x² + x - 1
    -----------------
2x - 1 | 4x³ + 0x² - 3x + 1
       - 4x³ + 2x²
       -------------
             2x² - 3x
             - 2x² + x
             ----------
                   -2x + 1

We repeat the process with the new polynomial (-2x+1) and the divisor (2x-1):

         2x² + x - 1
    -----------------
2x - 1 | 4x³ + 0x² - 3x + 1
       - 4x³ + 2x²
       -------------
             2x² - 3x
             - 2x² + x
             ----------
                   -2x + 1
                   -2x + 1
                   ------
                        0

We end up with a remainder of 0, which means that the division is exact. Therefore, the quotient is:

2x² + x - 1

So the answer is (B) 2x²-x-1.

In the diagram, EFGH is a cyclic quadrilateral in which EH || FG, EG and FH are chords. If < FHG = 424${}^{\circ }$ and < EFH = 34${}^{\circ }$

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< EFH = < EGH = 34${}^{\circ }$ (angles n the same segment)

< GHF = < GEF = 42${}^{\circ }$(angles in the same segment)

< FOG = 42 + 34 = 76(exterior angle)

< FOG = < EOH = 76(vertically opposite angle)

< EDO = 90${}^{\circ }$, < DOE = $\frac{76}{2}$ = 38${}^{\circ }$

< HEG = 90${}^{\circ }$ - 38${}^{\circ }$ = 52${}^{\circ }$

A binary operation * is defined by a*b = ab+a+b for any real number a and b. if the identity element is zero, find the inverse of 2 under this operation.

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a*a-1 = aa-1 + a + a-1 = e

if e = 0

2.2-1 + 2 + 2-1 = 0

3.2-1 + 2 = 0

= 2-1 = -$\frac{2}{3}$

The shaded portion in the graph is represented by

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The table above shows the frequency distribution of the ages (in years) of pupils in a certain secondary school. What percentage of the total number of pupils is over 15 years but less than 21 years?

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% of those above 15 but less than 21 years
age = 25/50 x 100/1
= 50%