JAMB UTME - Mathematics - 2021

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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

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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

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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.

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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° 

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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.

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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

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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

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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

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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.

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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

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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

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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)

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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.

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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

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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.

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Using trial expansion of each option

(m2 ${}^2$ + 2) (m - 1)

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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

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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".

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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

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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)

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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

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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

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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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To simplify the expression 2log 25 - log 72125 + log 9, we need to use the logarithmic properties of logarithms. One of the properties of logarithms is log a^b = b log a, which allows us to rewrite log 25 as 5 log 5, log 72125 as 5 log 72, and log 9 as 2 log 3. Using this property, the expression becomes 2 log 5 - 5 log 72 + 2 log 3. Another property of logarithms is log a / log b = log a / b, which allows us to simplify the logarithm of the ratio of two numbers as the logarithm of the first number divided by the logarithm of the second number. Using this property, the expression becomes 2 log 5 - (5 / 5) log 72 + 2 log 3, which simplifies to 2 log 5 - log 72 + 2 log 3. Finally, we can simplify the expression further by using the property log a^m = m log a, which allows us to rewrite log 72 as log 2^6, so that the expression becomes 2 log 5 - 6 log 2 + 2 log 3. So, the simplified expression is 2 log 5 - 6 log 2 + 2 log 3 = 2 log (5 * 3^2 / 2^6) = 2 log (3^2 / 2^4) = 2 log 3 - 4 log 2 = 2 * 1.0986 - 4 * 0.3010 = 2.1976 - 1.2040 = 1 - 2 log 2. So, the answer is (D) 1 - 2 log 2.

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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%

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y = 2x2 ${}^2$(2x - 1)
y = 4x3 ${}^3$ - 2x2 ${}^2$
dy/dx = 12x2 ${}^2$ - 4x
at x = -1
dy/dx = 12(-1)2 ${}^2$ - 4(-1)
= 12 + 4
= 16

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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

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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

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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

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Given (x + 2) and (x - 1), i.e. x = -2 or +1

when x = -2

L(-2) + 2k(-2)2 ${}^2$ + 24 = 0

f(-2) = -2L + 8k = -24...(i)

And x = 1
L(1) + 2k(1) + 24 = 0

f(1):L + 2k = -24...(ii)

Subst, L = -24 - 2k in eqn (i)

-2(-24 - 2k) + 8k = -24

+48 + 4k + 8k = -24

12k = -24 - 48 = -72

k = frac−7212 $frac-7212$

k = -6

where L = -24 - 2k

L = -24 - 2(-6)

L = -24 + 12

L = -12

That is; K = -6 and L = -12

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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°

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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.

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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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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.

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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)

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√ $27$ + 33√ $\frac{3}{\sqrt{3}}$

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