JAMB UTME - Mathematics - 1999

Akọwa Nkọwa

Akọwa Nkọwa

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 = -$\frac{2}{3}$

Akọwa Nkọwa

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.

Akọwa Nkọwa

62 = 9 x χ
36 = 9χχ = 36/9χ = 4

Akọwa Nkọwa

Akọwa Nkọwa

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

Akọwa Nkọwa

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).

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

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

Akọwa Nkọwa

Akọwa Nkọwa

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."

Akọwa Nkọwa

∠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${}^{\circ }$

Akọwa Nkọwa

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.

Akọwa Nkọwa

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.

Akọwa Nkọwa

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

Akọwa Nkọwa

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.

Akọwa Nkọwa

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.

Akọwa Nkọwa

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

Akọwa Nkọwa

Akọwa Nkọwa

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.

Akọwa Nkọwa

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.

Akọwa Nkọwa

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

Akọwa Nkọwa

Akọwa Nkọwa

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.

Akọwa Nkọwa

< 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${}^{\circ }$

Akọwa Nkọwa

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.

Akọwa Nkọwa

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

Akọwa Nkọwa

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.

Akọwa Nkọwa

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.

Akọwa Nkọwa

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 $\frac{5}{2}+2\left(\frac{-25}{4}\right)=-10$

Akọwa Nkọwa

< 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${}^{\circ }$

Akọwa Nkọwa

ΔABE and ΔACD are similar

Akọwa Nkọwa

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

Akọwa Nkọwa

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

Akọwa Nkọwa

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

Akọwa Nkọwa

Akọwa Nkọwa

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.

Akọwa Nkọwa

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

Akọwa Nkọwa

Akọwa Nkọwa

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

Akọwa Nkọwa

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.

Akọwa Nkọwa