JAMB UTME - Mathematics - 2021

Factorize completely 81a4 ${}^4$ - 16b4

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81a4 ${}^4$ - 16b4 ${}^4$ = (9a2 ${}^2$)2 ${}^2$ - (4b2 ${}^2$)2 ${}^2$

= (9a2 ${}^2$ + 4b2 ${}^2$)(9a2 ${}^2$ - 4b2 ${}^2$)

N:B 9a2 ${}^2$ - 4b2 ${}^2$ = (3a - 2b)(3a + 2b)

Solve the following equation: 2(2r−1) $\frac{2}{(2r-1)}$ - 53 $\frac{5}{3}$ =  1(r+2)

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2(2r−1) $\frac{2}{(2r-1)}$ - 53 $\frac{5}{3}$ =  1(r+2) $\frac{1}{(r+2)}$

2(2r−1) $\frac{2}{(2r-1)}$ - 1(r+2) $\frac{1}{(r+2)}$  = 53 $\frac{5}{3}$

The L.C.M.: (2r - 1) (r + 2) 

2(r+2)−1(2r−1)(2r−1)(r+2) $\frac{2(r+2)-1(2r-1)}{(2r-1)(r+2)}$ = 53 $\frac{5}{3}$

2r+4−2r+1(2r−1)(r+2) $\frac{2r+4-2r+1}{(2r-1)(r+2)}$ = 53 $\frac{5}{3}$

cross multiply the solution

3 = (2r - 1) (r + 2) or 2r2 ${}^2$ + 3r - 2 (when expanded)

collect like terms

2r2 ${}^2$ + 3r - 2 - 3 = 0

2r2 ${}^2$ + 3r - 5 = 0

Factorize to get x = 1 or - 52

Each of the interior angles of a regular polygon is 140°. How many sides has the polygon?

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A regular polygon is a polygon with equal sides and equal interior angles. The formula to find the interior angle of a regular polygon is given by: (180 * (n-2)) / n, where n is the number of sides in the polygon. So, if each interior angle of a regular polygon is 140°, we can plug that value into the formula and solve for n: 140 = (180 * (n-2)) / n Multiplying both sides by n gives: 140n = 180 * (n-2) Expanding the right side gives: 140n = 180n - 360 Subtracting 140n from both sides gives: 0 = 40n - 360 Adding 360 to both sides gives: 360 = 40n Dividing both sides by 40 gives: n = 9 So, the polygon has 9 sides.

If w varies inversely as uvu+v $\frac{uv}{u+v}$ and w = 8 when

u = 2 and v = 6, find a relationship between u, v, w.

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W α $\alpha$ 1uvu+v $\frac{\frac{1}{uv}}{u+v}$

∴ w = kuvu+v $\frac{\frac{k}{uv}}{u+v}$

= k(u+v)uv $\frac{k(u+v)}{uv}$

w = k(u+v)uv $\frac{k(u+v)}{uv}$

w = 8, u = 2 and v = 6

8 = k(2+6)2(6) $\frac{k(2+6)}{2(6)}$

= k(8)12 $\frac{k(8)}{12}$

k = 12

i.e 12 ( u + v) = uwv

What is the n-th term of the sequence 2, 6, 12, 20...?

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Given that 2, 6, 12, 20...? the nth term = n2 ${}^2$ + n

check: n = 1, u1 = 2

n = 2, u2 = 4 + 2 = 6

n = 3, u3 = 9 + 3 = 12

∴ n = 4, u4 = 16 + 4 = 20

wo numbers are removed at random from the numbers 1, 2, 3 and 4. What is the probability that the sum of the numbers removed is even?

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The probability that the sum of two numbers removed is even can be found by counting the number of even sums and dividing it by the total number of possible sums. Out of the 4 numbers, 2 are even (2 and 4) and 2 are odd (1 and 3). If we take one even and one odd number, the sum will be odd. If we take two even numbers, the sum will be even. And if we take two odd numbers, the sum will be even. So, there are 3 ways to get an even sum (taking 2 and 4, taking 2 and 2, or taking 1 and 3) and 6 total ways to choose two numbers (since order doesn't matter). So, the probability of getting an even sum is 3/6 or 1/2. In simple terms, there's a 50-50 chance that the sum of two numbers removed will be even.

factorize m3 ${}^3$ - m2 ${}^2$ + 2m - 2

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Using trial expansion of each option

(m2 ${}^2$ + 2) (m - 1)

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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To find the total number of market women, we need to make sure that we don't count the same person twice. So, let's break it down: - 12 women sell maize only. - 10 women sell yam only. - 14 women sell plantain only. - 5 women sell both plantain and maize. - 4 women sell both yam and maize. - 2 women sell both yam and plantain. - 3 women sell all three items. Since we counted the women who sell both plantain and maize twice, we need to subtract 5-2=3 from the total number of women. And since we counted the women who sell both yam and maize twice, we need to subtract 4-2=2 from the total number of women. Finally, we need to add back the 3 women who sell all three items because we subtracted them twice (once for each combination of two items). So the total number of women is 12 + 10 + 14 + 3 - 3 + 2 + 3 = 25. Therefore, there are 25 market women in the group.

A cylinder pipe, made of metal is 3cm thick.If the internal radius of the pipe is 10cm.Find the volume of metal used in making 3m of the pipe.

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Volume of a cylinder = πr2 ${}^2$h

First convert 3m to cm by multiplying by 100

Volume of External cylinder = π×132×300 $\pi \times {13}^2\times 300$

Volume of Internal cylinder = π×102×300 $\pi \times {10}^2\times 300$

Hence; Volume of External cylinder - Volume of Internal cylinder

Total volume (v) = π(169−100)×300 $\pi (169-100)\times 300$

V = π×69×300 $\pi \times 69\times 300$

V = 20700πcm3

The locus of a point which moves so that it is equidistant from two intersecting straight lines is the?

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The required locus is angle bisector of the two lines

If the binary operation ∗ $\ast$ is defined by m ∗ $\ast$ n = mn + m + n for any real number m and n, find the identity of the elements under this operation

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The identity element of a binary operation is an element "e" in a set such that for any element "x" in the same set, the result of the binary operation "e * x" or "x * e" is equal to "x". In other words, it's an element that doesn't change the result when it's multiplied with any other element. In this case, if we set "m = e" and "n = 0", the equation becomes: "e * 0 + e + 0 = e". This means that the identity element for this binary operation is "e = 0". So, the answer is: "e = 0".

X is due east point of y on a coast. Z is another point on the coast but 6.0km due south of Y. If the distance ZX is 12km, calculate the bearing of Z from X

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Sinθ = 612 $\frac{6}{12}$

Sinθ = 12 $\frac{1}{2}$

θ =  Sin0.5 ${}^0.5$
θ = 30°

Bearing of Z from X, (270 - 30)° = 240°

List all integers satisfying the inequality in -2 < 2x-6 < 4 

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-2 < 2x - 6   AND  2x - 6 < 4

-2 + 6 <2x  AND  2x < 4 + 6

4 <2x  AND  2x < 10

: 2 <x  AND x <5

2 < x < 5 

As 3 and 4

Simplify 27−−√ $\sqrt{27}$ + 33√

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√ $27$ + 33√ $\frac{3}{\sqrt{3}}$

= 9×3−−−−√ $\sqrt{9\times 3}$ + 3×3√3√×3√ $\frac{3\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}}$

= 33–√ $\sqrt{3}$ + 3–√ $\sqrt{3}$

= 43–√

In how many ways can 2 students be selected from a group of 5 students in a debating competition?

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The number of ways to select 2 students from a group of 5 students is 10. Imagine you're picking two students from a lineup of 5 students. You can pick the first student in 5 different ways (student 1, student 2, student 3, student 4, or student 5). Once you've picked the first student, there are only 4 students left in the lineup, so you can only pick the second student in 4 different ways. So, to find the total number of ways to select 2 students from a group of 5, you multiply the number of ways to pick the first student by the number of ways to pick the second student: 5 * 4 = 20. However, since order doesn't matter (it doesn't matter if you pick student 1 first or student 2 first), you divide the total number of ways by 2 to account for the overcounting. So, the final answer is 20 / 2 = 10.

If g(x) = x2 ${}^2$ + 3x  find g(x + 1) - g(x)

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g(x) = x2 + 3x

When g(x + 1) = (x + 1)^2 + 3(x + 1) 

= x2 ${}^2$ + 1 + 2x + 3x + 3 

= x2 ${}^2$ + 5x + 4

g(x + 1) - g(x) = x2 + 5x + 8 - (x2 ${}^2$ + 3x)

= x2 ${}^2$ + 5x + 4 - x2 -3x 

= 2x + 4 or 2(x + 4)

= 2(x + 2)

Three brothers in a business deal share the profit at the end of a contract. The first received 13 of the profit and the second 23 of the remainder. If the third received the remaining N12000.00 how much profit did they share?

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use "T" to represent the total profit. The first receives 13 $\frac{1}{3}$ T

remaining, 1 - 13 $\frac{1}{3}$

= 23 $\frac{2}{3}$T

The seconds receives the remaining, which is 23 $\frac{2}{3}$ also

23 $\frac{2}{3}$ x 23 $\frac{2}{3}$ x 49 $\frac{4}{9}$

The third receives the left over, which is 23 $\frac{2}{3}$T - 49 $\frac{4}{9}$T = (6−49 $\frac{6-4}{9}$)T

= 29 $\frac{2}{9}$T

The third receives 29 $\frac{2}{9}$T which is equivalent to N12000

If 29 $\frac{2}{9}$T = N12, 000

T = 1200029 $\frac{12000}{\frac{2}{9}}$

= N54, 000

find the value of p if the line which passes through (-1, -p) and (-2,2) is parallel to the line 2y+8x-17=0?

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Line:  2y+8x-17=0 

recall y = mx + c

2y = -8x + 17

y = -4x  + 172 $\frac{17}{2}$

Slope m1 ${}_1$ = 4

parallel lines: m1 ${}_1$. m2 ${}_2$ = -4

where Slope ( -4) = y2−y1x2−x1 $\frac{y_2-y_1}{x_2-x_1}$ at points (-1, -p) and (-2,2)

-4( x2−x1 $x_2-x_1$ ) = y2−y1 $y_2-y_1$ 

-4 ( -2 - -1) = 2 - -p

p = 4 - 2 = 2

The mean of ten positive numbers is 16. When another number is added, the mean becomes 18. Find the eleventh number

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The mean of ten positive numbers is the sum of all ten numbers divided by ten. So if the mean of the ten numbers is 16, then the sum of all ten numbers must be 160 (16 x 10). When an eleventh number is added, the mean becomes 18, so the sum of all eleven numbers must be 198 (18 x 11). We can find the eleventh number by subtracting the sum of the first ten numbers from the sum of all eleven numbers: 198 (sum of all eleven numbers) - 160 (sum of first ten numbers) = 38 So the eleventh number is 38.

Correct 241.34(3 x 10−3 ${}^{-3}$)2 ${}^2$ to 4 significant figures

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first work out the expression and then correct the answer to 4 s.f = 241.34..............(A)

(3 x 10-3 ${}^3$)2 ${}^2$............(B)

= 32 ${}^2$x2 ${}^2$

= 1103 $\frac{1}{{10}^3}$ x 1103 $\frac{1}{{10}^3}$

(Note that x2 ${}^2$ = 1x3 $\frac{1}{x^3}$)

= 24.34 x 32 ${}^2$ x 1106 $\frac{1}{{10}^6}$

= 2172.06106 $\frac{2172.06}{{10}^6}$

= 0.00217206

= 0.002172(4 s.f)

Three children shared a basket of mangoes in such a way that the first child took 14 $\frac{1}{4}$ of the mangoes and the second 34 $\frac{3}{4}$ of the remainder. What fraction of the mangoes did the third child take?

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You can use any whole numbers (eg. 1. 2. 3) to represent all the mangoes in the basket.

If the first child takes 14 $\frac{1}{4}$ it will remain 1 - 14 $\frac{1}{4}$ = 34 $\frac{3}{4}$

Next, the second child takes 34 $\frac{3}{4}$ of the remainder

which is 34 $\frac{3}{4}$ i.e. find 34 $\frac{3}{4}$ of 34 $\frac{3}{4}$

= 34 $\frac{3}{4}$ x 34 $\frac{3}{4}$

= 916 $\frac{9}{16}$

the fraction remaining now = 34 $\frac{3}{4}$ - 916 $\frac{9}{16}$

= 12−916 $\frac{12-9}{16}$

= 316

The angles of a quadrilateral are 5x-30, 4x+60, 60-x and 3x+61.find the smallest of these angles​

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Sum of all 4 angles of a quadrilateral = 360°

(5x-30) + (3x + 61) + (60-x) + (4x+ 60) = 360°

11x + 151 = 360°

11x = 360 - 151 = 209

x = 209/11 = 19°

Each angles is :

5x - 30 =  65°

4x+ 60 = 136°

60 - x =41°

3x + 61 = 118°

Smallest of these angles is 41° 

Find x if log9 ${}_9$x = 1.5

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If log9 ${}_9$x = 1.5,

91.5 ${}^{1.5}$ = x

9^32 $\frac{3}{2}$ = x

(√9)3 ${}^3$ = 3

∴ x = 27

Find the probability that a number selected at random from 41 to 56 is a multiple of 9

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The numbers between 41 and 56 that are multiples of 9 are 45 and 54. So, there are 2 numbers out of 16 total numbers that are multiples of 9. To find the probability, we divide the number of favorable outcomes by the number of total outcomes. In this case, the probability is 2/16, which can be simplified to 1/8. So, the probability that a number selected at random from 41 to 56 is a multiple of 9 is 1/8.

A car travels from calabar to Enugu, a distance of P km with an average speed of U km per hour and continues to benin, a distance of Q km, with an average speed of Wkm per hour. Find its average speed from Calabar to Benin

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Average speed = totalDistanceTotalTime $\frac{totalDistance}{TotalTime}$

from Calabar to Enugu in time t1, hence

t1 = PU $\frac{P}{U}$

Also from Enugu to Benin

t2 qw $\frac{q}{w}$

Av. speed = p+qt1+t2 $\frac{p+q}{t_1+t_2}$

= p+qp/u+q/w $\frac{p+q}{p/u+q/w}$

= p + q x uwpw+qu $\frac{uw}{pw+qu}$

= uw(p+q)pw+qu

What is the rate of change of the volume V of a hemisphere with respect to its radius r when r = 2?

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The formula for the volume of a hemisphere is given by (2/3)πr^3, where r is the radius. To find the rate of change of the volume with respect to the radius, we need to take the derivative of this formula with respect to r. So, dV/dr = (2/3) * π * 3r^2 * dr/dr = (2/3) * π * 3r^2 = 2πr^2. Now, when r = 2, the rate of change of the volume with respect to the radius is given by 2π * 2^2 = 8π. So, the answer is 8π.

determine the maximum value of y=3x2 ${}^2$ + 5x - 3

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y=3x2 ${}^2$ + 5x - 3

dy/dx = 6x + 5

as dy/dx = 0

6x + 5 = 0

x = −56 $\frac{-5}{6}$

Maximum value: 3 2−56 ${}^2\frac{-5}{6}$  + 5 −56 $\frac{-5}{6}$ - 3

3 7536 $\frac{75}{36}$ - 256 $\frac{25}{6}$ - 3

Using the L.C.M. 36

= 25−50−3636 $\frac{25-50-36}{36}$

= −6136 $\frac{-61}{36}$

No correct option

P(-6, 1) and Q(6, 6) are the two ends of the diameter of a given circle. Calculate the radius.

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The radius of a circle is the distance from the center of the circle to any point on the circumference of the circle. In this question, we are given the two ends of the diameter of a circle, so the radius is half of the diameter. The distance between two points (x1, y1) and (x2, y2) can be calculated using the Pythagorean theorem: √((x2 - x1)^2 + (y2 - y1)^2). Applying this formula to points P and Q, we have: √((6 - (-6))^2 + (6 - 1)^2) = √(12^2 + 5^2) = √(144 + 25) = √169 = 13.0 So, the diameter is 13.0 units, and the radius is half of that, which is 6.5 units. Therefore, the answer is 6.5 units.

Find the length of a side of a rhombus whose diagonals are 6cm and 8cm

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Let's call the length of one side of the rhombus "s". We know that the diagonals of a rhombus bisect each other at 90 degrees, which means that each diagonal cuts the rhombus into two congruent right triangles. If we take one of these triangles and drop a perpendicular from the corner to the midpoint of the opposite side, we can use the Pythagorean theorem to find the length of "s". The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides. In this case, the hypotenuse is one of the diagonals, and the other two sides are half of "s" and half of the other diagonal. So, we have: (1/2)^2 * 6^2 + (1/2)^2 * 8^2 = s^2 Simplifying this expression, we find: 9 + 16 = s^2 25 = s^2 And finally, taking the square root of both sides: s = 5 So, the length of one side of the rhombus is 5cm.

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Area of Trapezium = 1/2(sum of parallel sides) * h
91 = 12 $\frac{1}{2}$ (5 + 9)h

cross multiply
91 = 7h
h = 917 $\frac{91}{7}$
h = 13cm

At what rate would a sum of N100.00 deposited for 5 years raise an interest of N7.50?

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nterest I = PRT100 $\frac{PRT}{100}$

∴ R = 100×1100×5 $\frac{100\times 1}{100\times 5}$

= 100×7.50500×5 $\frac{100\times 7.50}{500\times 5}$

= 750500 $\frac{750}{500}$

= 32 $\frac{3}{2}$

= 1.5%

Find the mean deviation of 1, 2, 3 and 4

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Mean deviation = Σ|x - x|
                                  n
_
x = 2.5
= |1 - 2.5| + |2 - 2.5| + |3 - 2.5| + |4 - 2.5|
                               4
= 4/4 = 1

The ratio of the length of two similar rectangular blocks is 2 : 3. If the volume of the larger block is 351cm3 ${}^3$, then the volume of the other block is?

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Let x represent total vol. 2 : 3 = 2 + 3 = 5

35 $\frac{3}{5}$x = 351

x = 351×53 $\frac{351\times 5}{3}$

= 585

Volume of smaller block = 25 $\frac{2}{5}$ x 585

= 234.00cm3

Musa borrows N10.00 at 2% per month simple interest and repays N8.00 after 4 months. How much does he still owe?

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To calculate the simple interest that Musa owes, we can use the following formula: Simple Interest = Principal * Rate * Time Where Principal is the amount of the loan, Rate is the interest rate per month and Time is the number of months the loan is taken for. In this case, the Principal is N10.00, the Rate is 2% per month, and the Time is 4 months. So, the simple interest would be: N10.00 * 2% * 4 = N0.80 This means that after 4 months, Musa would owe N10.00 + N0.80 = N10.80 Now, if we subtract the amount that he repaid (N8.00) from the total amount he owes (N10.80), we get N10.80 - N8.00 = N2.80 So, Musa still owes N2.80. Answer: N2.80

Simplify and express in standard form 0.00275×0.006400.025×0.08

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The expression 0.00275 × 0.00640 ÷ 0.025 × 0.08 can be simplified to 8.8 × 10^-3, which is the same as 8.8 x 10^-3. This is the standard form for writing very small numbers, and it means 8.8 times 0.001 or 0.0088.

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Given that 4, 16, 30, 20, 10, 14 and 26

Adding up = 120

nos ≥ $\ge$ 16 are 16 + 30 + 20 + 26 = 92

The requires sum of angles = 92120 $\frac{92}{120}$ x 360o1 $\frac{{360}^o}{1}$

= 276o

The angle of a sector of a circle, radius 10.5cm, is 48°, Calculate the perimeter of the sector

Antwortdetails

Explanation

Length of Arc AB = θ360 $\frac{\theta }{360}$ 2π $\pi$r

= 48360 $\frac{48}{360}$ x 2227 $\frac{22}{7}$ x 212 $\frac{21}{2}$

= 4×22××330 $\frac{4\times 22\times \times 3}{30}$ 8810 $\frac{88}{10}$ = 8.8cm

Perimeter = 8.8 + 2r

= 8.8 + 2(10.5)

= 8.8 + 21

= 29.8cm

Simplify 2log 25 $\frac{2}{5}$ - log72125 $\frac{72}{125}$ + log 9