JAMB UTME - Mathematics - 2021

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

Bayanin Amsa

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.

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

Bayanin Amsa

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

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

Bayanin Amsa

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°

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

Bayanin Amsa

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)

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

Bayanin Amsa

The required locus is angle bisector of the two lines

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

Bayanin Amsa

Using trial expansion of each option

(m2 ${}^2$ + 2) (m - 1)

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?

Bayanin Amsa

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

Bayanin Amsa

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

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

Bayanin Amsa

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 (x + 2) and (x - 1) are factors of the expression Lx+2kx2+24 $Lx+2k{x}^2+24$, find the values of L and k.

Bayanin Amsa

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

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

Bayanin Amsa

If log9 ${}_9$x = 1.5,

91.5 ${}^{1.5}$ = x

9^32 $\frac{3}{2}$ = x

(√9)3 ${}^3$ = 3

∴ x = 27

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

Bayanin Amsa

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

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

Bayanin Amsa

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° 

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?

Bayanin Amsa

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

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.

Bayanin Amsa

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

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

Bayanin Amsa

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

Simplify and express in standard form 0.00275×0.006400.025×0.08

Bayanin Amsa

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.

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

Bayanin Amsa

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.

Find the derivative of the function y = 2x2 ${}^2$(2x - 1) at the point x = -1?

Bayanin Amsa

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

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

Bayanin Amsa

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

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

Bayanin Amsa

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.

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

Bayanin Amsa

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)

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

Bayanin Amsa

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?

Bayanin Amsa

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.

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

Bayanin Amsa

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".

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

Bayanin Amsa

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)

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

Bayanin Amsa

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.

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

Bayanin Amsa

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.

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

Bayanin Amsa

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

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

Bayanin Amsa

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

Simplify 27−−√ $\sqrt{27}$ + 33√

Bayanin Amsa

√ $27$ + 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–√

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

Bayanin Amsa

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%

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.

Bayanin Amsa

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

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?

Bayanin Amsa

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.

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

Bayanin Amsa

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.

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

Bayanin Amsa

-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

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?

Bayanin Amsa

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

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?

Bayanin Amsa

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