Tüm alıştırma soruları

1526 soru

Soru 101Soru

Tayo Traders had a trade debtors balance of 450,000\text{₦}450,000 before year-end adjustments for the financial year ended 31st December 2024. At the end of the year, additional unrecorded bad debts of 15,000\text{₦}15,000 were identified to be written off. During the same year, cash of 10,000\text{₦}10,000 was received from a customer whose debt of 10,000\text{₦}10,000 had been written off in 2022. What is the net trade debtors balance to be shown in the Statement of Financial Position as at 31st December 2024?

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Cevap: 435000

Cevap

The net trade debtors balance to be shown in the Statement of Financial Position at 31st December 2024 is 435,000\text{₦}435,000.
To calculate the net trade debtors figure for the Statement of Financial Position, only additional bad debts written off at year-end (15,000\text{₦}15,000) are deducted from the unadjusted trade debtors balance of 450,000\text{₦}450,000, resulting in 435,000\text{₦}435,000. Bad debts recovered (10,000\text{₦}10,000) are recognized as revenue in the Income Statement and received in cash, having no effect on closing trade debtors.

Adım Adım Çözüm

1
Determine the unadjusted trade debtors balance
Unadjusted Trade Debtors = 450,000\text{₦}450,000
This represents the recorded receivables prior to writing off year-end unrecorded bad debts.
2
Subtract additional bad debts written off at year-end
Adjusted Trade Debtors = 450,00015,000=435,000\text{₦}450,000 - \text{₦}15,000 = \text{₦}435,000
Bad debts written off represent irrecoverable amounts and must be deducted directly from trade debtors.
3
Evaluate the treatment of bad debts recovered
Effect on Trade Debtors = 0\text{₦}0
Bad debts recovered during the year are recorded by debiting Cash/Bank and crediting Bad Debts Recovered (an income account). Because the original personal account of the debtor was already cleared when written off in a previous period, recovery does not alter the closing trade debtors balance.

Anahtar Kavram

Treatment of Bad Debts Written Off and Bad Debts Recovered on the Statement of Financial Position
Tahmini Süre:1m 30s
Soru 102Soru

Given that PP varies directly as the square of rr, and P=48P = 48 when r=4r = 4, calculate the value of PP when r=6r = 6.

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Cevap: 108

Cevap

The value of PP when r=6r = 6 is 108108.
Since PP varies directly as r2r^2, the relationship is expressed as P=kr2P = k r^2. Substituting the given values P=48P = 48 and r=4r = 4 yields 48=16k48 = 16k, so k=3k = 3. Substituting k=3k = 3 and r=6r = 6 into the equation gives P=3×62=3×36=108P = 3 \times 6^2 = 3 \times 36 = 108.

Adım Adım Çözüm

1
Set up the variation equation using constant of variation kk
P=kr2P = k r^2
Direct variation with the square of a variable means PP is directly proportional to r2r^2.
2
Substitute the initial values P=48P = 48 and r=4r = 4 to determine kk
48=k×42    48=16k    k=348 = k \times 4^2 \implies 48 = 16k \implies k = 3
Finding the variation constant kk allows us to establish a specific relationship between PP and rr.
3
Calculate PP for r=6r = 6 using the specific equation P=3r2P = 3 r^2
P=3×62=3×36=108P = 3 \times 6^2 = 3 \times 36 = 108
Evaluating the formula with the new input r=6r = 6 yields the required value of PP.

Anahtar Kavram

Direct variation involving square powers
Soru 103Soru

Given that sinθ=513\sin \theta = \frac{5}{13}, where θ\theta is an acute angle, evaluate the value of 13cosθ12tanθ13 \cos \theta - 12 \tan \theta. What is the numerical value?

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Cevap: 7

Cevap

The numerical value of the expression 13cosθ12tanθ13 \cos \theta - 12 \tan \theta is 7.
For an acute angle θ\theta with sinθ=513\sin \theta = \frac{5}{13}, the corresponding right triangle has an opposite side of 5, a hypotenuse of 13, and an adjacent side of 13252=12\sqrt{13^2 - 5^2} = 12. Therefore, cosθ=1213\cos \theta = \frac{12}{13} and tanθ=512\tan \theta = \frac{5}{12}. Evaluating 13cosθ12tanθ13 \cos \theta - 12 \tan \theta yields 13(1213)12(512)=125=713\left(\frac{12}{13}\right) - 12\left(\frac{5}{12}\right) = 12 - 5 = 7.

Adım Adım Çözüm

1
Determine cosθ\cos \theta using the right triangle ratio or Pythagorean identity.
cosθ=1213\cos \theta = \frac{12}{13}
Since sinθ=oppositehypotenuse=513\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{5}{13}, the adjacent side is 13252=12\sqrt{13^2 - 5^2} = 12. Because θ\theta is acute, cosθ\cos \theta is positive.
2
Determine tanθ\tan \theta using the ratio of opposite to adjacent sides.
tantanθ=512\tan \tan \theta = \frac{5}{12}
\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{5}{12}$.
3
Substitute the evaluated ratios into 13cosθ12tanθ13 \cos \theta - 12 \tan \theta and simplify.
13\left(\frac{12}{13}\right) - 12\left(\frac{5}{12}\right) = 12 - 5 = 7
Multiplying clears the denominators, leaving 125=712 - 5 = 7.

Anahtar Kavram

Basic Trigonometric Ratios and Pythagorean Triples
Tahmini Süre:1m 15s
Soru 104Soru

If sinθ=35\sin \theta = \frac{3}{5} for an acute angle θ\theta, what is the exact value of 5cosθ+4tanθ5 \cos \theta + 4 \tan \theta?

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Cevap: 7

Cevap

The exact value of 5cosθ+4tanθ5 \cos \theta + 4 \tan \theta is 7.
For an acute angle θ\theta with sinθ=35\sin \theta = \frac{3}{5}, the corresponding right triangle has opposite side = 3, hypotenuse = 5, and adjacent side = 4. Using trig definitions, cosθ=45\cos \theta = \frac{4}{5} and tanθ=34\tan \theta = \frac{3}{4}. Evaluating 5cosθ+4tanθ5 \cos \theta + 4 \tan \theta gives 5(45)+4(34)=4+3=75\left(\frac{4}{5}\right) + 4\left(\frac{3}{4}\right) = 4 + 3 = 7.

Adım Adım Çözüm

1
Find the adjacent side of the right-angled triangle.
Adjacent side = 5232=4\sqrt{5^2 - 3^2} = 4.
By the Pythagorean theorem (a2+b2=c2a^2 + b^2 = c^2), where opposite = 3 and hypotenuse = 5.
2
Determine the values of cosθ\cos \theta and tanθ\tan \theta.
cosθ=45\cos \theta = \frac{4}{5} and tanθ=34\tan \theta = \frac{3}{4}.
Using fundamental trigonometric definitions: cosine is adjacent/hypotenuse and tangent is opposite/adjacent.
3
Substitute these values into 5cosθ+4tanθ5 \cos \theta + 4 \tan \theta and simplify.
5(45)+4(34)=4+3=75\left(\frac{4}{5}\right) + 4\left(\frac{3}{4}\right) = 4 + 3 = 7.
Performing simple multiplication and addition gives 7.

Anahtar Kavram

Basic Trigonometric Ratios in Right Triangles
Soru 105Soru

Convert the fractional binary number 0.110120.1101_2 to its equivalent base 10 (decimal) value. What is the decimal value?

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Cevap: 0.8125

Cevap

The base 10 value of 0.110120.1101_2 is 0.81250.8125.
To convert a fractional binary number to decimal, expand each digit after the radix point using decreasing negative powers of 2 (21,22,23,242^{-1}, 2^{-2}, 2^{-3}, 2^{-4}). Evaluating 1(0.5)+1(0.25)+0(0.125)+1(0.0625)1(0.5) + 1(0.25) + 0(0.125) + 1(0.0625) yields 0.81250.8125.

Adım Adım Çözüm

1
Write the given binary fraction in place-value expansion using powers of 2
0.11012=121+122+023+1240.1101_2 = 1 \cdot 2^{-1} + 1 \cdot 2^{-2} + 0 \cdot 2^{-3} + 1 \cdot 2^{-4}
Positions after the binary point represent negative powers of 2 starting from 212^{-1}.
2
Evaluate each fractional component
21=0.52^{-1} = 0.5, 22=0.252^{-2} = 0.25, 23=0.1252^{-3} = 0.125, 24=0.06252^{-4} = 0.0625
Calculating standard decimal values for binary fractional places.
3
Add the non-zero fractional terms together
0.5+0.25+0.0625=0.81250.5 + 0.25 + 0.0625 = 0.8125
Summing the decimal values gives the complete converted decimal representation.

Anahtar Kavram

Conversion of fractional numbers from base 2 to base 10
Soru 106Soru

The electrical resistance RR of a wire varies directly as its length LL and inversely as the square of its diameter dd. If a wire of length 36 m36\text{ m} and diameter 3 mm3\text{ mm} has a resistance of 16 Ω16\ \Omega, what is the resistance, in ohms, of a wire of the same material with a length of 45 m45\text{ m} and a diameter of 5 mm5\text{ mm}?

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Cevap: 7.2

Cevap

The resistance of the wire is 7.2 ohms.
The equation governing the relation is R=kLd2R = \frac{kL}{d^2}. Substituting the initial parameters R=16 ΩR=16\ \Omega, L=36 mL=36\text{ m}, and d=3 mmd=3\text{ mm} gives 16=36k9=4k16 = \frac{36k}{9} = 4k, which yields k=4k = 4. Using k=4k = 4 with the new dimensions L=45 mL=45\text{ m} and d=5 mmd=5\text{ mm} gives R=4×4552=18025=7.2 ΩR = \frac{4 \times 45}{5^2} = \frac{180}{25} = 7.2\ \Omega.

Adım Adım Çözüm

1
Formulate the joint variation equation
R=kLd2R = \frac{kL}{d^2}
Direct variation places length LL in the numerator and inverse variation of the square of diameter dd places d2d^2 in the denominator.
2
Determine the variation constant kk
k=4k = 4
Substituting R=16R = 16, L=36L = 36, and d=3d = 3 gives 16=36k9    16=4k    k=416 = \frac{36k}{9} \implies 16 = 4k \implies k = 4.
3
Calculate the new resistance
R=7.2 ΩR = 7.2\ \Omega
Substituting k=4k = 4, L=45L = 45, and d=5d = 5 into R=kLd2R = \frac{kL}{d^2} yields R=4×4525=7.2R = \frac{4 \times 45}{25} = 7.2.

Anahtar Kavram

Joint Variation involving direct proportionality and inverse square law
Soru 107Soru

Given that θ\theta is an acute angle satisfying the relationship secθ+tanθ=3\sec \theta + \tan \theta = 3, what is the exact value of 5sinθ5\sin \theta?

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Cevap: 4

Cevap

The exact value of 5sinθ5\sin \theta is 4.
Using the identity sec2θtan2θ=1\sec^2 \theta - \tan^2 \theta = 1, we deduce (secθtanθ)(secθ+tanθ)=1(\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1. Given secθ+tanθ=3\sec \theta + \tan \theta = 3, it follows that secθtanθ=13\sec \theta - \tan \theta = \frac{1}{3}. Solving the system of equations yields secθ=53\sec \theta = \frac{5}{3} and tanθ=43\tan \theta = \frac{4}{3}, which gives sinθ=45\sin \theta = \frac{4}{5}. Multiplying by 5 gives the final answer of 4.

Adım Adım Çözüm

1
Apply the trigonometric Pythagorean identity
\sec^2 \theta - \tan^2 \theta = 1
This relates secant and tangent functions directly.
2
Factorize the identity and solve for secθtanθ\sec \theta - \tan \theta
(\sec \theta - \tan \theta)(3) = 1 \implies \sec \theta - \tan \tan \theta = \frac{1}{3}
Using the algebraic identity a2b2=(ab)(a+b)a^2 - b^2 = (a - b)(a + b).
3
Set up a linear system to solve for secθ\sec \theta and tanθ\tan \theta
\sec \theta = \frac{5}{3}, \quad \tan \theta = \frac{4}{3}
Adding and subtracting the equations secθ+tanθ=3\sec \theta + \tan \theta = 3 and \sec \theta - \tan \theta = \frac{1}{3} gives the individual function values.
4
Calculate sinθ\sin \theta and evaluate 5sinθ5\sin \theta
\sin \theta = \frac{\tan \theta}{\sec \theta} = \frac{4/3}{5/3} = \frac{4}{5} \implies 5\sin \theta = 4
The quotient of tangent and secant gives sine.

Anahtar Kavram

Pythagorean Trigonometric Identities

Alternatif Yöntem

Draw a right-angled triangle where hypotenuse over adjacent plus opposite over adjacent equals 3: c+ab=3\frac{c + a}{b} = 3. By Pythagorean theorem c2a2=b2c^2 - a^2 = b^2, so cab=13\frac{c - a}{b} = \frac{1}{3}. Solving yields a/c=4/5a/c = 4/5, hence sinθ=4/5\sin \theta = 4/5 and 5sinθ=45\sin \theta = 4.
Tahmini Süre:2m 0s
Soru 108Soru

The hourly operational cost, CC Naira, of an industrial water pump is partly constant and partly varies jointly as the flow rate, rr in litres per second, and the square of the pressure head, hh in metres. When r=10 L/sr = 10\text{ L/s} and h=4 mh = 4\text{ m}, the operational cost is N620\text{N}620. When r=15 L/sr = 15\text{ L/s} and h=2 mh = 2\text{ m}, the operational cost is N380\text{N}380. What is the operational cost in Naira when r=20 L/sr = 20\text{ L/s} and h=3 mh = 3\text{ m}?

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Cevap: 668

Cevap

The operational cost when r=20r = 20 and h=3h = 3 is 668 Naira.
The partial and joint variation relationship is defined by C=k1+k2rh2C = k_1 + k_2 r h^2. Substituting the two given states gives the simultaneous equations 620=k1+160k2620 = k_1 + 160k_2 and 380=k1+60k2380 = k_1 + 60k_2. Subtracting these equations gives 100k2=240100k_2 = 240, so k2=2.4k_2 = 2.4. Substituting k2=2.4k_2 = 2.4 into the second equation yields k1=236k_1 = 236. Finally, evaluating CC for r=20r = 20 and h=3h = 3 gives C=236+2.4(20)(32)=236+432=668C = 236 + 2.4(20)(3^2) = 236 + 432 = 668.

Adım Adım Çözüm

1
Set up the variation equation
C=k1+k2rh2C = k_1 + k_2 r h^2, where k1k_1 is the constant part and k2k_2 is the constant of joint variation.
The problem states that CC is partly constant (k1k_1) and partly varies jointly as rr and h2h^2 (k2rh2k_2 r h^2).
2
Form simultaneous linear equations using the given data points
(1) 620=k1+160k2620 = k_1 + 160k_2 and (2) 380=k1+60k2380 = k_1 + 60k_2
Substituting r=10,h=4,C=620r = 10, h = 4, C = 620 gives 10×42=16010 \times 4^2 = 160. Substituting r=15,h=2,C=380r = 15, h = 2, C = 380 gives 15×22=6015 \times 2^2 = 60.
3
Solve for the constants k1k_1 and k2k_2
k2=2.4k_2 = 2.4 and k1=236k_1 = 236
Subtracting equation (2) from (1) eliminates k1k_1, giving 100k2=240    k2=2.4100k_2 = 240 \implies k_2 = 2.4. Substituting back into equation (2) gives k1=38060(2.4)=236k_1 = 380 - 60(2.4) = 236.
4
Calculate the operational cost for the target parameters
C=236+2.4×20×32=668C = 236 + 2.4 \times 20 \times 3^2 = 668
Substitute k1=236k_1 = 236, k2=2.4k_2 = 2.4, r=20r = 20, and h=3h = 3 into the variation formula.

Anahtar Kavram

Partial and Joint Variation
Soru 109Soru

A regular hexagon has a side length of 6 cm6\text{ cm}. At each vertex of the hexagon, a circular sector of radius 3 cm3\text{ cm} is formed inside the figure. Taking π=227\pi = \frac{22}{7} and 3=1.732\sqrt{3} = 1.732, what is the area of the remaining region inside the hexagon not covered by the sectors, in cm2\text{cm}^2, correct to two decimal places?

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Cevap: 36.96

Cevap

The area of the remaining region inside the hexagon is 36.96 cm236.96\text{ cm}^2.
The total area of the regular hexagon is computed by multiplying the area of one equilateral triangle of side 6 cm6\text{ cm} by 6, yielding 543=54×1.732=93.528 cm254\sqrt{3} = 54 \times 1.732 = 93.528\text{ cm}^2. Each interior angle of a regular hexagon is 120120^\circ, so each vertex sector has a central angle of 120120^\circ and radius 3 cm3\text{ cm}. The area of one sector is 120360×227×32=667 cm2\frac{120^\circ}{360^\circ} \times \frac{22}{7} \times 3^2 = \frac{66}{7}\text{ cm}^2. The total area for all six sectors is 6×667=396756.5714 cm26 \times \frac{66}{7} = \frac{396}{7} \approx 56.5714\text{ cm}^2. Subtracting this from the total area gives 93.52856.5714=36.9566 cm293.528 - 56.5714 = 36.9566\text{ cm}^2, which rounds to 36.96 cm236.96\text{ cm}^2.

Adım Adım Çözüm

1
Determine the interior angle of the regular hexagon.
Each interior angle is 120120^\circ.
The formula for the interior angle of a regular polygon with nn sides is (n2)×180n\frac{(n-2) \times 180^\circ}{n}.
2
Calculate the total area of the 6 circular sectors at the vertices.
Total sector area is 396756.5714 cm2\frac{396}{7} \approx 56.5714\text{ cm}^2.
Each sector has a central angle of 120120^\circ and radius 3 cm3\text{ cm}. With 6 sectors, the total area is 6×120360×227×32=18×227=3967 cm26 \times \frac{120^\circ}{360^\circ} \times \frac{22}{7} \times 3^2 = 18 \times \frac{22}{7} = \frac{396}{7}\text{ cm}^2.
3
Calculate the total area of the regular hexagon.
Hexagon area is 93.528 cm293.528\text{ cm}^2.
A regular hexagon consists of 6 equilateral triangles of side length 6 cm6\text{ cm}. Area = 6×(34×62)=543=54×1.732=93.528 cm26 \times \left(\frac{\sqrt{3}}{4} \times 6^2\right) = 54\sqrt{3} = 54 \times 1.732 = 93.528\text{ cm}^2.
4
Subtract the sector area from the total hexagon area.
93.52856.5714=36.9566 cm236.96 cm293.528 - 56.5714 = 36.9566\text{ cm}^2 \approx 36.96\text{ cm}^2.
The remaining area is the total area minus the area occupied by the six corner sectors.

Anahtar Kavram

Area of regular polygons and circular sectors
Tahmini Süre:2m 30s
Soru 110Soru

Find the number of integers that satisfy the compound inequality 3<2x+19-3 < 2x + 1 \le 9.

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Cevap: 6

Cevap

The number of integer solutions satisfying the inequality is 6.
Subtracting 1 across 3<2x+19-3 < 2x + 1 \le 9 gives 4<2x8-4 < 2x \le 8. Dividing by 2 yields 2<x4-2 < x \le 4. The integers in this interval are 1,0,1,2,3,-1, 0, 1, 2, 3, and 44, giving a total of 6 integer solutions.

Adım Adım Çözüm

1
Subtract 1 from all parts of the compound inequality.
4<2x8-4 < 2x \le 8
Isolate the variable term 2x2x in the middle.
2
Divide all parts by 2.
2<x4-2 < x \le 4
Solve for xx by undoing the coefficient of 2.
3
Identify the set of integer solutions within the interval (2,4](-2, 4].
x{1,0,1,2,3,4}x \in \{-1, 0, 1, 2, 3, 4\}
The endpoint 2-2 is excluded due to the strict inequality (<<), while the endpoint 44 is included due to the inclusive inequality (le\\le).
4
Count the elements in the solution set.
6
There are 6 distinct integer values in the set.

Anahtar Kavram

Solving compound linear inequalities and identifying integer solution sets.
Soru 111Soru

What is the exact numerical value of the trigonometric expression 4cos230+2sin245tan260csc30\frac{4\cos^2 30^\circ + 2\sin^2 45^\circ}{\tan^2 60^\circ - \csc 30^\circ}?

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Cevap: 4

Cevap

4
By evaluating the special angles directly: 4cos230=34\cos^2 30^\circ = 3, 2sin245=12\sin^2 45^\circ = 1, tan260=3\tan^2 60^\circ = 3, and csc30=2\csc 30^\circ = 2. Substituting these into the expression yields 3+132=41=4\frac{3 + 1}{3 - 2} = \frac{4}{1} = 4.

Adım Adım Çözüm

1
Substitute the exact value of cos30\cos 30^\circ
4cos230=4(32)2=4(34)=34\cos^2 30^\circ = 4 \left(\frac{\sqrt{3}}{2}\right)^2 = 4 \left(\frac{3}{4}\right) = 3
The exact value of cos30\cos 30^\circ is 32\frac{\sqrt{3}}{2}.
2
Substitute the exact value of sin45\sin 45^\circ
2sin245=2(12)2=2(12)=12\sin^2 45^\circ = 2 \left(\frac{1}{\sqrt{2}}\right)^2 = 2 \left(\frac{1}{2}\right) = 1
The exact value of sin45\sin 45^\circ is 12\frac{1}{\sqrt{2}}.
3
Substitute the exact values of tan60\tan 60^\circ and csc30\csc 30^\circ
tan260=(3)2=3\tan^2 60^\circ = (\sqrt{3})^2 = 3 and csc30=1sin30=2\csc 30^\circ = \frac{1}{\sin 30^\circ} = 2
The exact value of tan60\tan 60^\circ is 3\sqrt{3} and csc30\csc 30^\circ is the reciprocal of sin30=12\sin 30^\circ = \frac{1}{2}.
4
Simplify the entire fraction
3+132=41=4\frac{3 + 1}{3 - 2} = \frac{4}{1} = 4
Dividing the simplified numerator (4) by the simplified denominator (1) gives 4.

Anahtar Kavram

Evaluation of Special Angle Trigonometric Ratios and Reciprocal Functions
Soru 112Soru

A variable yy is partly constant and partly varies directly as x\sqrt{x}. Given that y=26y = 26 when x=16x = 16, and y=38y = 38 when x=49x = 49, what is the value of yy when x=64x = 64?

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Cevap: 42

Cevap

The value of yy when x=64x = 64 is 4242.
The relationship is given by the partial variation equation y=a+bxy = a + b\sqrt{x}. Substituting the pairs (16,26)(16, 26) and (49,38)(49, 38) yields the linear system a+4b=26a + 4b = 26 and a+7b=38a + 7b = 38. Solving this system gives the constants a=10a = 10 and b=4b = 4. Substituting x=64x = 64 into y=10+464y = 10 + 4\sqrt{64} results in y=10+4(8)=42y = 10 + 4(8) = 42.

Adım Adım Çözüm

1
Formulate the partial variation equation.
y=a+bxy = a + b\sqrt{x}, where aa and bb are constants of variation.
Partial variation implies yy is the sum of a constant term aa and a term directly proportional to x\sqrt{x}.
2
Set up simultaneous equations using the given pairs of (x,y)(x, y).
Equation 1: a+4b=26a + 4b = 26
Equation 2: a+7b=38a + 7b = 38
Evaluating 16=4\sqrt{16} = 4 and 49=7\sqrt{49} = 7 simplifies the relationship into two linear equations in two unknowns.
3
Solve for constants aa and bb.
b=4b = 4 and a=10a = 10
Subtracting Equation 1 from Equation 2 yields 3b=12    b=43b = 12 \implies b = 4, and substituting b=4b = 4 back into Equation 1 gives a=10a = 10.
4
Calculate yy when x=64x = 64.
y=10+4(8)=42y = 10 + 4(8) = 42
Using the specific formula y=10+4xy = 10 + 4\sqrt{x} for x=64x = 64 gives y=10+32=42y = 10 + 32 = 42.

Anahtar Kavram

Partial Variation with Simultaneous Equations
Soru 113Soru

In ΔPQR\Delta PQR, side p=3 cmp = 3\text{ cm}, side q=8 cmq = 8\text{ cm}, and the included angle R=60\angle R = 60^\circ. What is the length of side rr in cm?

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Cevap: 7

Cevap

The length of side rr is 7 cm7\text{ cm}.
Using the Cosine Rule r2=p2+q22pqcosRr^2 = p^2 + q^2 - 2pq \cos R with p=3p=3, q=8q=8, and R=60\angle R=60^\circ yields r2=9+6448(0.5)=49r^2 = 9 + 64 - 48(0.5) = 49, which gives r=7 cmr = 7\text{ cm}.

Adım Adım Çözüm

1
State the Cosine Rule formula for side rr
r2=p2+q22pqcosRr^2 = p^2 + q^2 - 2pq \cos R
The Cosine Rule allows calculating the third side of a triangle when two sides and the included angle are given.
2
Substitute the given values into the formula
r2=32+822(3)(8)cos60r^2 = 3^2 + 8^2 - 2(3)(8) \cos 60^\circ
We are given p=3p = 3, q=8q = 8, and R=60\angle R = 60^\circ.
3
Evaluate the trigonometric expression and simplify
r2=9+6448(0.5)=7324=49r^2 = 9 + 64 - 48(0.5) = 73 - 24 = 49
Since cos60=0.5\cos 60^\circ = 0.5, multiplying 2×3×8×0.52 \times 3 \times 8 \times 0.5 yields 2424.
4
Solve for rr by taking the positive square root
r=7 cmr = 7\text{ cm}
Length must be a positive value, and 49=7\sqrt{49} = 7.

Anahtar Kavram

Cosine Rule for finding an unknown side given two sides and the included angle (SAS).
Soru 114Soru

A trader estimated the mass of a bag of rice to be 25 kg25\text{ kg}, but the actual mass of the bag was 20 kg20\text{ kg}. Calculate the percentage error in the trader's estimate.

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Cevap: 25

Cevap

The percentage error in the trader's estimate is 25%25\%.
The absolute error is 25 kg20 kg=5 kg25\text{ kg} - 20\text{ kg} = 5\text{ kg}. Evaluating the error relative to the actual value gives 5 kg20 kg=0.25\frac{5\text{ kg}}{20\text{ kg}} = 0.25, which equals 25%25\%.

Adım Adım Çözüm

1
Calculate the absolute error in measurement
Error = 2520=5 kg|25 - 20| = 5\text{ kg}
Absolute error is the absolute difference between the estimated value and the true value.
2
Calculate the percentage error relative to the actual value
\text{Percentage Error} = \frac{5}{20} \times 100\% = 25\%
Percentage error must always be calculated using the actual (true) value as the denominator.

Anahtar Kavram

Percentage Error Calculation
Tahmini Süre:45s
Soru 115Soru

Two ships, PP and QQ, leave a port OO at the same time. Ship PP sails on a bearing of 040040^\circ at a constant speed of 25 km/h25\text{ km/h}, while Ship QQ sails on a bearing of 100100^\circ at a constant speed of 40 km/h40\text{ km/h}. What is the distance in kilometers between the two ships after 22 hours?

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Cevap: 70

Cevap

The distance between the two ships after 2 hours is 70 km.
The distance traveled by Ship P in 2 hours is 50 km50\text{ km} and by Ship Q is 80 km80\text{ km}. The angle between their paths is 100040=60100^\circ - 040^\circ = 60^\circ. Applying the Cosine Rule yields PQ2=502+8022(50)(80)cos(60)=2500+64004000=4900PQ^2 = 50^2 + 80^2 - 2(50)(80)\cos(60^\circ) = 2500 + 6400 - 4000 = 4900, giving a distance of 4900=70 km\sqrt{4900} = 70\text{ km}.

Adım Adım Çözüm

1
Calculate the distances traveled by Ship P and Ship Q after 2 hours
OP=50 kmOP = 50\text{ km} and OQ=80 kmOQ = 80\text{ km}
Distance equals speed multiplied by time.
2
Find the angle between the lines of travel from port O
POQ=10040=60\angle POQ = 100^\circ - 40^\circ = 60^\circ
The angle between two bearings from a common origin is the difference between their bearing angles.
3
Use the Cosine Rule to calculate the side length PQ
PQ2=502+8022(50)(80)cos(60)=4900PQ^2 = 50^2 + 80^2 - 2(50)(80)\cos(60^\circ) = 4900
The Cosine Rule c2=a2+b22abcosCc^2 = a^2 + b^2 - 2ab\cos C calculates the unknown opposite side given two side lengths and their included angle.
4
Take the square root to find PQ
PQ=70 kmPQ = 70\text{ km}
Taking the principal square root gives the final linear distance.

Anahtar Kavram

Applying the Cosine Rule to solve bearing and distance non-right triangle problems
Tahmini Süre:2m 0s
Soru 116Soru

Find the total number of distinct solutions to the trigonometric equation 2cos2(2x)+sin(2x)1=02\cos^2(2x) + \sin(2x) - 1 = 0 in the interval 0x3600^\circ \le x \le 360^\circ.

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Cevap: 6

Cevap

The total number of distinct solutions is 6.
Substituting cos2(2x)=1sin2(2x)\cos^2(2x) = 1 - \sin^2(2x) gives the quadratic 2sin2(2x)sin(2x)1=02\sin^2(2x) - \sin(2x) - 1 = 0, which factors into (2sin(2x)+1)(sin(2x)1)=0(2\sin(2x) + 1)(\sin(2x) - 1) = 0. For 0x3600^\circ \le x \le 360^\circ, the angle argument 2x2x covers 02x7200^\circ \le 2x \le 720^\circ. The equation sin(2x)=1\sin(2x) = 1 provides 2 values for xx (45,22545^\circ, 225^\circ), while sin(2x)=12\sin(2x) = -\frac{1}{2} provides 4 values for xx (105,165,285,345105^\circ, 165^\circ, 285^\circ, 345^\circ). Summing these gives 6 distinct solutions in total.

Adım Adım Çözüm

1
Use the Pythagorean trigonometric identity cos2(2x)=1sin2(2x)\cos^2(2x) = 1 - \sin^2(2x) to express the entire equation in terms of sin(2x)\sin(2x).
2(1sin2(2x))+sin(2x)1=0    22sin2(2x)+sin(2x)1=02(1 - \sin^2(2x)) + \sin(2x) - 1 = 0 \implies 2 - 2\sin^2(2x) + \sin(2x) - 1 = 0
Converting all trigonometric terms to a single function allows the equation to be solved as a polynomial.
2
Rearrange and factorize the resulting quadratic equation in terms of sin(2x)\sin(2x).
2sin2(2x)sin(2x)1=0    (2sin(2x)+1)(sin(2x)1)=02\sin^2(2x) - \sin(2x) - 1 = 0 \implies (2\sin(2x) + 1)(\sin(2x) - 1) = 0
Factorization splits the quadratic trigonometric equation into two simple linear trigonometric equations.
3
Determine the expanded domain for 2x2x given 0x3600^\circ \le x \le 360^\circ.
02x7200^\circ \le 2x \le 720^\circ
Multiplying the bounds of xx by 2 accounts for two full rotations in the unit circle.
4
Solve the first linear equation sin(2x)=1\sin(2x) = 1 within 02x7200^\circ \le 2x \le 720^\circ.
2x=90,450    x=45,2252x = 90^\circ, 450^\circ \implies x = 45^\circ, 225^\circ (2 distinct solutions)
The sine function equals 1 at 9090^\circ in the first revolution and at 90+360=45090^\circ + 360^\circ = 450^\circ in the second revolution.
5
Solve the second linear equation sin(2x)=12\sin(2x) = -\frac{1}{2} within 02x7200^\circ \le 2x \le 720^\circ.
2x=210,330,570,690    x=105,165,285,3452x = 210^\circ, 330^\circ, 570^\circ, 690^\circ \implies x = 105^\circ, 165^\circ, 285^\circ, 345^\circ (4 distinct solutions)
The sine function is negative in the 3rd and 4th quadrants of both revolutions.
6
Combine the solution counts from both cases.
Total number of solutions = 2+4=62 + 4 = 6.
Adding the valid solutions from both factor equations gives the complete set of roots.

Anahtar Kavram

Solving quadratic trigonometric equations across multiple revolutions
Tahmini Süre:3m 0s
Soru 117Soru

The table below shows the distribution of masses (in kg) of cocoa bags harvested on a farm:

Mass (kg)Frequency (ff)
101910 - 1966
202920 - 29kk
303930 - 391515
404940 - 491212
505950 - 5977

If the estimated mean mass of the distribution is 35.3 kg35.3\text{ kg}, calculate the value of the missing frequency kk.

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Cevap: 10

Cevap

The value of the missing frequency is 10.
To find the missing frequency, compute the midpoints (xx) of each mass class interval: 14.5, 24.5, 34.5, 44.5, and 54.5. Next, express the sum of frequencies as f=40+k\sum f = 40 + k and the sum of products of frequency and midpoint as fx=6(14.5)+k(24.5)+15(34.5)+12(44.5)+7(54.5)=1520+24.5k\sum fx = 6(14.5) + k(24.5) + 15(34.5) + 12(44.5) + 7(54.5) = 1520 + 24.5k. Equating the mean expression fxf\frac{\sum fx}{\sum f} to 35.335.3 gives 1520+24.5k40+k=35.3\frac{1520 + 24.5k}{40 + k} = 35.3. Cross-multiplying and solving yields 1520+24.5k=1412+35.3k1520 + 24.5k = 1412 + 35.3k, which simplifies to 10.8k=10810.8k = 108, giving k=10k = 10.

Adım Adım Çözüm

1
Determine the class midpoints (xx) for all intervals.
Class midpoints are 14.5, 24.5, 34.5, 44.5, and 54.5.
Grouped mean calculations require representative midpoint values for each class interval.
2
Formulate expressions for total frequency f\sum f and total weighted sum fx\sum fx.
\sum f = 40 + k and \sum fx = 1520 + 24.5k.
These algebraic expressions are necessary to substitute into the mean formula.
3
Set up and solve the linear equation using the given mean of 35.3.
\frac{1520 + 24.5k}{40 + k} = 35.3 \implies 10.8k = 108 \implies k = 10.
Equating the algebraic mean expression to the numerical mean allows solving for the unknown frequency k.

Anahtar Kavram

Measures of Central Tendency for Grouped Data
Soru 118Soru

In triangle LMNLMN, the side lengths are given as l=7 cml = 7\text{ cm}, m=8 cmm = 8\text{ cm}, and n=13 cmn = 13\text{ cm}. What is the measure of angle NN in degrees?

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Cevap: 120

Cevap

The measure of angle NN is 120120^\circ.
Using the Cosine Rule for angle NN, cosN=l2+m2n22lm=72+821322(7)(8)=56112=0.5\cos N = \frac{l^2 + m^2 - n^2}{2lm} = \frac{7^2 + 8^2 - 13^2}{2(7)(8)} = \frac{-56}{112} = -0.5. The angle whose cosine is 0.5-0.5 within the interior angles of a triangle (0<N<1800^\circ < N < 180^\circ) is 120120^\circ.

Adım Adım Çözüm

1
Set up the Cosine Rule formula for angle NN
\cos N = \frac{l^2 + m^2 - n^2}{2lm}
The Cosine Rule relates the three sides of any triangle to the cosine of one of its interior angles.
2
Substitute the known side lengths into the formula
\cos N = \frac{7^2 + 8^2 - 13^2}{2 \times 7 \times 8} = \frac{49 + 64 - 169}{112}
Side n=13 cmn = 13\text{ cm} is opposite to angle NN and must be subtracted in the numerator.
3
Simplify the fraction
\cos N = \frac{-56}{112} = -0.5
Evaluating the numerical expression gives a negative value, indicating that angle NN is obtuse.
4
Calculate the principal inverse cosine angle for the triangle
N=120N = 120^\circ
Since cos60=0.5\cos 60^\circ = 0.5, cos(18060)=0.5\cos(180^\circ - 60^\circ) = -0.5, giving N=120N = 120^\circ.

Anahtar Kavram

Applying the Cosine Rule to find an obtuse angle given three side lengths (SSS)
Soru 119Soru

The fourth term of an arithmetic progression (A.P.) is 1515 and the ninth term is 3535. What is the common difference of the progression?

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Cevap: 4

Cevap

The common difference of the arithmetic progression is 44.
Using the A.P. term formula Tn=a+(n1)dT_n = a + (n-1)d, the fourth term gives a+3d=15a + 3d = 15 and the ninth term gives a+8d=35a + 8d = 35. Subtracting the two equations yields 5d=205d = 20, which simplifies directly to d=4d = 4.

Adım Adım Çözüm

1
Express the given terms using the n-th term formula Tn=a+(n1)dT_n = a + (n-1)d
a+3d=15a + 3d = 15 and a+8d=35a + 8d = 35
The nthn^{\text{th}} term formula relates any term to the first term (aa) and common difference (dd).
2
Subtract the equation for the fourth term from the ninth term
5d=205d = 20
Subtracting eliminates the first term aa and leaves a simple equation in terms of dd.
3
Divide by 55 to solve for dd
d=4d = 4
Dividing both sides of 5d=205d = 20 by 55 gives the common difference.

Anahtar Kavram

Finding the common difference of an Arithmetic Progression given two non-consecutive terms
Soru 120Soru

What is the mean deviation of the data set 4,7,8,11,154, 7, 8, 11, 15?

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Cevap: 3.2

Cevap

The mean deviation of the data set is 3.2.
To find the mean deviation, first calculate the mean of the dataset, which is 9. Then, compute the absolute difference of each number from 9, obtaining values of 5, 2, 1, 2, and 6. Finally, divide the sum of these absolute values (16) by the total number of items (5) to get 3.2.

Adım Adım Çözüm

1
Calculate the arithmetic mean of the given data set.
The mean xˉ=9\bar{x} = 9.
The mean is needed to evaluate how far each data point deviates from the central value.
2
Find the absolute difference between each value and the mean.
The absolute deviations are 5, 2, 1, 2, and 6.
Mean deviation measures dispersion using absolute distances, ignoring negative signs.
3
Calculate the mean of the absolute deviations.
Mean deviation = 3.2.
Dividing the total sum of absolute deviations (16) by the number of observations (5) gives the mean deviation.

Anahtar Kavram

Mean Deviation for Ungrouped Data
ÖncekiSayfa 6 / 77Sonraki
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