Number and Numeration

229 soru

Soru 221Soru

A factory employs 1212 workers to produce 180180 identical wooden chairs in 55 days, working 88 hours per day. If 44 workers are reassigned to another department, how many days will the remaining workers take to produce 210210 such chairs if they work 77 hours per day?

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Cevap: 10 days10\text{ days}

Cevap

The remaining workers will take 10 days10\text{ days} to complete the task.
The correct answer is 10 days10\text{ days}. Using compound proportion, the relationship is given by W1×D1×H1C1=W2×D2×H2C2\frac{W_1 \times D_1 \times H_1}{C_1} = \frac{W_2 \times D_2 \times H_2}{C_2}. Substituting the given values 12×5×8180=8×D2×7210\frac{12 \times 5 \times 8}{180} = \frac{8 \times D_2 \times 7}{210} yields 480180=56D2210\frac{480}{180} = \frac{56 D_2}{210}, which simplifies to 83=56D2210\frac{8}{3} = \frac{56 D_2}{210}. Solving for D2D_2 gives D2=10 daysD_2 = 10\text{ days}.

Adım Adım Çözüm

1
Calculate the total man-hours required for the initial production batch.
Total man-hours = 12 workers×5 days×8 hours/day=480 man-hours12\text{ workers} \times 5\text{ days} \times 8\text{ hours/day} = 480\text{ man-hours}.
Determining total labor input needed for 180180 chairs.
2
Find the man-hours required per chair.
Man-hours per chair = 480180=83 hours/chair\frac{480}{180} = \frac{8}{3}\text{ hours/chair}.
Establishing the unit rate of work.
3
Calculate the total man-hours required for the new target of 210210 chairs.
Total man-hours needed = 210×83=560 man-hours210 \times \frac{8}{3} = 560\text{ man-hours}.
Scaling the unit work rate to the new quantity.
4
Determine the new workforce size and daily labor capacity.
Remaining workers = 124=8 workers12 - 4 = 8\text{ workers}. Daily man-hours = 8×7=56 man-hours/day8 \times 7 = 56\text{ man-hours/day}.
Accounting for the reduced workforce and new daily hours.
5
Calculate the number of days required.
Days = 560 man-hours56 man-hours/day=10 days\frac{560\text{ man-hours}}{56\text{ man-hours/day}} = 10\text{ days}.
Dividing total required work by daily capacity.

Anahtar Kavram

Compound Proportion and Work-Rate Relationships
Tahmini Süre:1m 30s
Soru 222Soru

Given the matrix A=(3142)A = \begin{pmatrix} 3 & 1 \\ 4 & 2 \end{pmatrix}, what is the determinant of the matrix 3A3A?

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Cevap: 1818

Cevap

The determinant of the matrix 3A3A is 1818.
For any 2×22 \times 2 matrix, scaling the matrix by a factor kk scales its determinant by k2k^2. Since det(A)=(3)(2)(1)(4)=2\det(A) = (3)(2) - (1)(4) = 2, multiplying the matrix by 33 yields det(3A)=32×2=9×2=18\det(3A) = 3^2 \times 2 = 9 \times 2 = 18. Alternatively, computing 3A=(93126)3A = \begin{pmatrix} 9 & 3 \\ 12 & 6 \end{pmatrix} directly gives det(3A)=(9)(6)(3)(12)=5436=18\det(3A) = (9)(6) - (3)(12) = 54 - 36 = 18.

Adım Adım Çözüm

1
Calculate the determinant of matrix AA.
\det(A) = (3)(2) - (1)(4) = 6 - 4 = 2.
The determinant of a 2×22 \times 2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix} is given by adbcad - bc.
2
Apply the determinant scalar multiplication property for an n×nn \times n matrix.
\det(3A) = 3^2 \det(A) = 9 \times 2 = 18.
For any n×nn \times n square matrix AA and scalar kk, the identity det(kA)=kndet(A)\det(kA) = k^n \det(A) holds. Here, k=3k=3 and n=2n=2.

Anahtar Kavram

Scalar Multiplication Property of Determinants
Tahmini Süre:1m 15s
Soru 223Soru

Given the matrix A=(y53y+2)A = \begin{pmatrix} y & 5 \\ 3 & y+2 \end{pmatrix}, if the determinant of AA is 99 and y>0y > 0, calculate the value of yy.

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Cevap: 4

Cevap

The value of yy is 4.
Expanding the determinant of AA yields det(A)=y(y+2)15=y2+2y15\det(A) = y(y+2) - 15 = y^2 + 2y - 15. Equating this to 9 gives y2+2y24=0y^2 + 2y - 24 = 0, which factors as (y+6)(y4)=0(y+6)(y-4) = 0. The roots are y=6y = -6 and y=4y = 4. Given that y>0y > 0, the required value is 4.

Adım Adım Çözüm

1
Find the expression for the determinant of matrix AA
det(A)=y(y+2)(5)(3)=y2+2y15\det(A) = y(y+2) - (5)(3) = y^2 + 2y - 15
The determinant of a 2×22 \times 2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix} is given by adbcad - bc.
2
Form and solve the quadratic equation
y2+2y15=9    y2+2y24=0    (y+6)(y4)=0y^2 + 2y - 15 = 9 \implies y^2 + 2y - 24 = 0 \implies (y + 6)(y - 4) = 0
Set the determinant expression equal to the given determinant value of 9 and rearrange into standard quadratic form.
3
Apply the given domain restriction y>0y > 0
y=4y = 4
The root y=6y = -6 is discarded because yy must be strictly positive.

Anahtar Kavram

Determinant of a 2x2 matrix and solving non-linear determinant equations
Soru 224Soru

At a regional agricultural exhibition, 150150 farmers registered their crop cultivation. 8585 farmers grow cassava (CC), 7070 grow maize (MM), and 6060 grow yam (YY). 3535 farmers grow both cassava and maize, 2525 grow both maize and yam, and 3030 grow both cassava and yam. If 1515 farmers grow all three crops, how many farmers grow none of these three crops?

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Cevap: 10

Cevap

The number of farmers who grow none of the three crops is 10.
Using the Principle of Inclusion-Exclusion for three sets, the number of farmers growing at least one of cassava, maize, or yam is given by 85+70+60352530+15=14085 + 70 + 60 - 35 - 25 - 30 + 15 = 140. Since there are 150150 farmers in total, the number of farmers who grow none of these three crops is 150140=10150 - 140 = 10.

Adım Adım Çözüm

1
Calculate the cardinality of the union of the three sets using the Principle of Inclusion-Exclusion.
CMY=85+70+60352530+15=140|C \cup M \cup Y| = 85 + 70 + 60 - 35 - 25 - 30 + 15 = 140
Summing the three individual set counts overcounts elements in pairwise intersections, which must be subtracted. The central triple intersection is then added back because it was subtracted once too often.
2
Subtract the size of the union from the size of the universal set.
(CMY)=150140=10|(C \cup M \cup Y)'| = 150 - 140 = 10
The complement of the union represents the set of farmers outside all three crop categories.

Anahtar Kavram

Principle of Inclusion-Exclusion for Three Sets
Soru 225Soru

Given the matrices M=(2134)M = \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix} and N=(1023)N = \begin{pmatrix} 1 & 0 \\ 2 & 3 \end{pmatrix}, what is the determinant of the matrix product MNMN?

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Cevap: 33

Cevap

33
Using the property that the determinant of a product of matrices equals the product of their individual determinants, det(M)=(2)(4)(1)(3)=11\det(M) = (2)(4) - (-1)(3) = 11 and det(N)=(1)(3)(0)(2)=3\det(N) = (1)(3) - (0)(2) = 3. Multiplying these gives det(MN)=11×3=33\det(MN) = 11 \times 3 = 33.

Adım Adım Çözüm

1
Calculate the determinant of matrix MM
\det(M) = (2)(4) - (-1)(3) = 8 + 3 = 11
The determinant of a 2×22 \times 2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix} is given by adbcad - bc.
2
Calculate the determinant of matrix NN
\det(N) = (1)(3) - (0)(2) = 3 - 0 = 3
Apply the determinant formula adbcad - bc to matrix NN.
3
Apply the determinant product property to find det(MN)\det(MN)
\det(MN) = \det(M) \times \det(N) = 11 \times 3 = 33
For any two square matrices of the same order, det(MN)=det(M)det(N)\det(MN) = \det(M) \cdot \det(N).

Anahtar Kavram

Determinant Property of Matrix Products
Tahmini Süre:1m 30s
Soru 226Soru

A poultry farmer in Ogun State took a loan of 120,000\text{₦}120,000 from a cooperative society at an annual interest rate of 5%5\%, compounded annually. What is the total interest paid by the farmer at the end of 22 years?

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Cevap: 12,300\text{₦}12,300

Cevap

The total compound interest paid at the end of 2 years is 12,300\text{₦}12,300.
The total compound interest is determined by calculating the accumulated amount A=120,000×(1.05)2=132,300A = 120,000 \times (1.05)^2 = \text{₦}132,300 and subtracting the original principal of 120,000\text{₦}120,000, yielding 12,300\text{₦}12,300.

Adım Adım Çözüm

1
Identify the given values for principal, rate, and time
Principal P=120,000P = \text{₦}120,000, Rate r=5%=0.05r = 5\% = 0.05, Time t=2 yearst = 2\text{ years}
These parameters are required to substitute into the compound amount formula.
2
Calculate the total accumulated amount AA after 2 years
A=P(1+r)t=120,000×(1+0.05)2=120,000×1.1025=132,300A = P(1 + r)^t = 120,000 \times (1 + 0.05)^2 = 120,000 \times 1.1025 = \text{₦}132,300
The compound interest formula yields the total balance including the initial principal.
3
Subtract the initial principal from the total accumulated amount to determine interest
Compound Interest =AP=132,300120,000=12,300= A - P = 132,300 - 120,000 = \text{₦}12,300
Interest is the extra amount generated beyond the original loan amount.

Anahtar Kavram

Compound Interest vs Total Accumulated Amount
Tahmini Süre:1m 30s
Soru 227Soru

A boutique owner in Enugu borrowed 80,000\text{₦}80,000 to expand her business at an annual interest rate of 15%15\%, compounded annually. If she repays the loan in full after 22 years, what is the total interest she paid on the loan?

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Cevap: 25,800\text{₦}25,800

Cevap

The total interest paid on the loan after 2 years is 25,800\text{₦}25,800.
The compound interest is calculated year by year. In the first year, the interest paid is 15%15\% of 80,000=12,000\text{₦}80,000 = \text{₦}12,000, raising the balance to 92,000\text{₦}92,000. In the second year, the interest paid is 15%15\% of 92,000=13,800\text{₦}92,000 = \text{₦}13,800. Adding the two yearly interest payments gives 12,000+13,800=25,800\text{₦}12,000 + \text{₦}13,800 = \text{₦}25,800.

Adım Adım Çözüm

1
Calculate the interest for the first year.
Interest for Year 1 = 15%15\% of 80,000=15100×80,000=12,000\text{₦}80,000 = \frac{15}{100} \times 80,000 = \text{₦}12,000.
In compound interest, interest for the first period is calculated on the initial principal.
2
Determine the amount at the end of the first year, which becomes the principal for the second year.
Principal for Year 2 = 80,000+12,000=92,000\text{₦}80,000 + \text{₦}12,000 = \text{₦}92,000.
Compound interest adds earned interest to the principal for subsequent period calculations.
3
Calculate the interest for the second year.
Interest for Year 2 = 15%15\% of 92,000=15100×92,000=13,800\text{₦}92,000 = \frac{15}{100} \times 92,000 = \text{₦}13,800.
Interest in year 2 is computed on the updated principal balance of 92,000\text{₦}92,000.
4
Sum the interest amounts from both years to find total interest paid.
Total Interest = 12,000+13,800=25,800\text{₦}12,000 + \text{₦}13,800 = \text{₦}25,800.
The total interest is the sum of interest accumulated across each compounding period.

Anahtar Kavram

Compound Interest Calculation
Tahmini Süre:1m 30s
Soru 228Soru

A cooperative society in Akure granted a loan of 160,000\text{₦}160,000 to a farmer at an interest rate of 12%12\% per annum, compounded annually. What is the total compound interest owed at the end of 22 years?

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Cevap: 40,704\text{₦}40,704

Cevap

The total compound interest owed at the end of 2 years is 40,704\text{₦}40,704.
The compound interest is obtained by calculating the accumulated amount A=160,000×(1.12)2=200,704A = 160,000 \times (1.12)^2 = \text{₦}200,704 and then subtracting the principal of 160,000\text{₦}160,000 to get 40,704\text{₦}40,704.

Adım Adım Çözüm

1
Identify the given financial variables
Principal P=160,000P = \text{₦}160,000, Rate R=12%R = 12\%, Time n=2n = 2 years.
Establishing the parameters needed for the compound interest formula.
2
Calculate the total accumulated amount AA using the formula A=P(1+R100)nA = P\left(1 + \frac{R}{100}\right)^n
A=160,000×(1+0.12)2=160,000×(1.12)2=160,000×1.2544=200,704A = 160,000 \times \left(1 + 0.12\right)^2 = 160,000 \times (1.12)^2 = 160,000 \times 1.2544 = \text{₦}200,704.
Determines the total value of the loan including principal and accrued compound interest after 2 years.
3
Subtract the principal PP from the total amount AA to find the compound interest CICI
CI=AP=200,704160,000=40,704CI = A - P = 200,704 - 160,000 = \text{₦}40,704.
Isolates the interest portion from the total accumulated balance.

Anahtar Kavram

Compound Interest Calculation
Soru 229Soru

If the matrix P=(k20141302)P = \begin{pmatrix} k & 2 & 0 \\ 1 & 4 & 1 \\ 3 & 0 & 2 \end{pmatrix} has a determinant equal to 1818, find the value of kk.

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Cevap: 2

Cevap

The value of kk is 22.
Expanding matrix PP along the first row gives det(P)=k(80)2(23)+0(012)=8k+2\det(P) = k(8 - 0) - 2(2 - 3) + 0(0 - 12) = 8k + 2. Setting 8k+2=188k + 2 = 18 yields 8k=168k = 16, which simplifies to k=2k = 2.

Adım Adım Çözüm

1
Expand the matrix determinant along the first row.
\det(P) = k(8 - 0) - 2(2 - 3) + 0 = 8k + 2
Applying the cofactor expansion formula for a 3×33 \times 3 matrix across row 1.
2
Set the resulting determinant expression equal to the given value of 18.
8k + 2 = 18
The question specifies that the determinant of PP is equal to 1818.
3
Solve the linear equation for kk.
k = 2
Subtracting 2 from both sides yields 8k=168k = 16, and dividing by 8 gives k=2k = 2.

Anahtar Kavram

Determinant of a 3×33 \times 3 Matrix
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