Number and Numeration

229 soru

Soru 161Soru

Pumps P1P_1 and P2P_2 operating together can fill a water storage tank in 12 hours12\text{ hours}, while pumps P2P_2 and P3P_3 operating together can fill the same tank in 20 hours20\text{ hours}. When all three pumps operate simultaneously for 5 hours5\text{ hours}, they fill exactly 12\frac{1}{2} of the tank. If pump P2P_2 is then shut off, how many additional hours will it take pumps P1P_1 and P3P_3 working together to fill the remainder of the tank?

Cevabı ve açıklamayı göster

Cevap: 7.5 hours7.5\text{ hours}

Cevap

7.5 hours7.5\text{ hours}
The three pumps combined have a work rate of 110 tank/hr\frac{1}{10}\text{ tank/hr}. Subtracting the rate of P1+P2P_1 + P_2 (which is 112\frac{1}{12}) gives P3P_3's rate as 160 tank/hr\frac{1}{60}\text{ tank/hr}. Subtracting the rate of P2+P3P_2 + P_3 (which is 120\frac{1}{20}) gives P1P_1's rate as 120 tank/hr\frac{1}{20}\text{ tank/hr}. Together, P1P_1 and P3P_3 have a combined rate of 120+160=115 tank/hr\frac{1}{20} + \frac{1}{60} = \frac{1}{15}\text{ tank/hr}. To fill the remaining 12\frac{1}{2} of the tank, it will take 1/21/15=7.5 hours\frac{1/2}{1/15} = 7.5\text{ hours}.

Adım Adım Çözüm

1
Determine the combined work rate of all three pumps working together.
Since all three pumps fill 12\frac{1}{2} of the tank in 5 hours5\text{ hours}, their combined rate is r1+r2+r3=1/25=110 tank per hourr_1 + r_2 + r_3 = \frac{1/2}{5} = \frac{1}{10}\text{ tank per hour}.
Work rate is defined as the fraction of work completed per unit time.
2
Find the individual work rates of pumps P1P_1 and P3P_3.
r3=(r1+r2+r3)(r1+r2)=110112=160 tank per hourr_3 = (r_1 + r_2 + r_3) - (r_1 + r_2) = \frac{1}{10} - \frac{1}{12} = \frac{1}{60}\text{ tank per hour}. Also, r1=(r1+r2+r3)(r2+r3)=110120=120 tank per hourr_1 = (r_1 + r_2 + r_3) - (r_2 + r_3) = \frac{1}{10} - \frac{1}{20} = \frac{1}{20}\text{ tank per hour}.
Subtracting known pair rates from the total three-pump rate yields individual rates.
3
Calculate the combined work rate of P1P_1 and P3P_3.
r1+r3=120+160=3+160=460=115 tank per hourr_1 + r_3 = \frac{1}{20} + \frac{1}{60} = \frac{3 + 1}{60} = \frac{4}{60} = \frac{1}{15}\text{ tank per hour}.
The rate of two pumps operating together is the sum of their individual rates.
4
Compute the time needed to fill the remaining portion of the tank.
Remaining fraction to fill is 112=121 - \frac{1}{2} = \frac{1}{2}. Time required = 1/21/15=152=7.5 hours\frac{1/2}{1/15} = \frac{15}{2} = 7.5\text{ hours}.
Dividing the remaining workload by the combined rate gives the additional operating time.

Anahtar Kavram

Work-rate problems involving simultaneous rates and partial work completion
Tahmini Süre:2m 0s
Soru 162Soru

If log3x2logx27=1\log_3 x - 2\log_x 27 = 1, what is the sum of all possible real values of xx?

Cevabı ve açıklamayı göster

Cevap: 2449\frac{244}{9}

Cevap

2449\frac{244}{9}
By applying the change of base rule logx27=log327log3x=3log3x\log_x 27 = \frac{\log_3 27}{\log_3 x} = \frac{3}{\log_3 x}, the given equation simplifies to log3x6log3x=1\log_3 x - \frac{6}{\log_3 x} = 1. Setting u=log3xu = \log_3 x yields u2u6=0u^2 - u - 6 = 0, which factors as (u3)(u+2)=0(u - 3)(u + 2) = 0. Thus, u=3u = 3 or u=2u = -2, giving solutions x=33=27x = 3^3 = 27 and x=32=19x = 3^{-2} = \frac{1}{9}. Adding these valid real solutions gives 27+19=244927 + \frac{1}{9} = \frac{244}{9}.

Adım Adım Çözüm

1
Apply the change of base formula to logx27\log_x 27.
logx27=log327log3x=3log3x\log_x 27 = \frac{\log_3 27}{\log_3 x} = \frac{3}{\log_3 x}
Converting all logarithmic terms to a common base (base 3) allows substitution.
2
Substitute logx27\log_x 27 back into the original equation and let u=log3xu = \log_3 x.
u2(3u)=1    u6u=1u - 2\left(\frac{3}{u}\right) = 1 \implies u - \frac{6}{u} = 1
This transforms the logarithmic equation into an algebraic equation.
3
Clear the denominator by multiplying through by uu and rearrange into standard quadratic form.
u2u6=0u^2 - u - 6 = 0
Multiplying by uu (where u0u \neq 0) forms a standard quadratic equation.
4
Factor the quadratic equation to solve for uu.
(u3)(u+2)=0    u=3 or u=2(u - 3)(u + 2) = 0 \implies u = 3 \text{ or } u = -2
Factoring determines the values of log3x\log_3 x.
5
Convert back to xx using x=3ux = 3^u.
For u=3u = 3, x=33=27x = 3^3 = 27. For u=2u = -2, x=32=19x = 3^{-2} = \frac{1}{9}. Both x=27x = 27 and x=19x = \frac{1}{9} are valid bases (x>0,x1x > 0, x \neq 1).
Exponentiation resolves the logarithmic variable.
6
Find the sum of all valid values of xx.
Sum=27+19=243+19=2449\text{Sum} = 27 + \frac{1}{9} = \frac{243 + 1}{9} = \frac{244}{9}
Combines the two real solutions into the final requested sum.

Anahtar Kavram

Change of base formula logab=logcblogca\log_a b = \frac{\log_c b}{\log_c a} and solving equations reducible to quadratics.
Tahmini Süre:2m 0s
Soru 163Soru

An arithmetic progression has a first term of 55 and a common difference of 44. What is the 12th12^{\text{th}} term of this progression?

Cevabı ve açıklamayı göster

Cevap: 4949

Cevap

4949
The nthn^{\text{th}} term of an arithmetic progression is determined using the formula Tn=a+(n1)dT_n = a + (n - 1)d. Substituting a=5a = 5, d=4d = 4, and n=12n = 12 yields T12=5+11×4=49T_{12} = 5 + 11 \times 4 = 49, making 4949 the correct value.

Adım Adım Çözüm

1
Identify the given values from the problem statement
First term a=5a = 5, common difference d=4d = 4, and position n=12n = 12
These are the standard variables required for calculating terms in an arithmetic progression.
2
Apply the nthn^{\text{th}} term formula for an arithmetic progression
T12=5+(121)×4T_{12} = 5 + (12 - 1) \times 4
The standard formula for the nthn^{\text{th}} term of an AP is Tn=a+(n1)dT_n = a + (n - 1)d.
3
Simplify the expression to find the final value
T12=5+11×4=5+44=49T_{12} = 5 + 11 \times 4 = 5 + 44 = 49
Perform multiplication before addition according to standard order of operations.

Anahtar Kavram

Arithmetic Progression nthn^{\text{th}} Term Formula
Tahmini Süre:45s
Soru 164Soru

If (1681)x1×(278)x+2=94\left( \frac{16}{81} \right)^{x-1} \times \left( \frac{27}{8} \right)^{x+2} = \frac{9}{4}, find the value of xx.

Cevabı ve açıklamayı göster

Cevap: 88

Cevap

The value of xx is 88.
By writing 1681\frac{16}{81} as (23)4\left(\frac{2}{3}\right)^4, 278\frac{27}{8} as (23)3\left(\frac{2}{3}\right)^{-3}, and 94\frac{9}{4} as (23)2\left(\frac{2}{3}\right)^{-2}, the equation simplifies via exponent addition to (23)4x43x6=(23)2\left(\frac{2}{3}\right)^{4x - 4 - 3x - 6} = \left(\frac{2}{3}\right)^{-2}. Equating indices gives x10=2x - 10 = -2, which yields the correct solution x=8x = 8.

Adım Adım Çözüm

1
Express all fractional terms with a common base of 23\frac{2}{3}
1681=(23)4\frac{16}{81} = \left(\frac{2}{3}\right)^4, 278=(32)3=(23)3\frac{27}{8} = \left(\frac{3}{2}\right)^3 = \left(\frac{2}{3}\right)^{-3}, and 94=(32)2=(23)2\frac{9}{4} = \left(\frac{3}{2}\right)^2 = \left(\frac{2}{3}\right)^{-2}
Converting all terms to a single common base allows exponents to be combined using the laws of indices.
2
Substitute the common base expressions back into the original equation
((23)4)x1×((23)3)x+2=(23)2\left( \left(\frac{2}{3}\right)^4 \right)^{x-1} \times \left( \left(\frac{2}{3}\right)^{-3} \right)^{x+2} = \left(\frac{2}{3}\right)^{-2}
Applying the power of a power law (am)n=amn(a^m)^n = a^{mn} to simplify each term.
3
Apply the power law and multiplication law of indices
(23)4(x1)×(23)3(x+2)=(23)2    (23)4(x1)3(x+2)=(23)2\left(\frac{2}{3}\right)^{4(x-1)} \times \left(\frac{2}{3}\right)^{-3(x+2)} = \left(\frac{2}{3}\right)^{-2} \implies \left(\frac{2}{3}\right)^{4(x-1) - 3(x+2)} = \left(\frac{2}{3}\right)^{-2}
When multiplying exponential terms with identical bases, add their exponents: am×an=am+na^m \times a^n = a^{m+n}.
4
Equate the exponents and solve for xx
4(x1)3(x+2)=2    4x43x6=2    x10=2    x=84(x-1) - 3(x+2) = -2 \implies 4x - 4 - 3x - 6 = -2 \implies x - 10 = -2 \implies x = 8
Since the bases on both sides are equal and non-zero, their indices must be equal.

Anahtar Kavram

Laws of Indices: Base Conversion and Exponential Equations
Tahmini Süre:2m 0s
Soru 165Soru

Given the universal set U={1,2,3,4,5,6,7,8,9,10}U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} and a subset P={xU:x is a prime number}P = \{x \in U : x \text{ is a prime number}\}, which of the following sets represents the complement of PP, denoted as PP'?

Cevabı ve açıklamayı göster

Cevap: {1,4,6,8,9,10}\{1, 4, 6, 8, 9, 10\}

Cevap

The set {1,4,6,8,9,10}\{1, 4, 6, 8, 9, 10\}
The prime numbers in UU are 2,3,5,2, 3, 5, and 77, making P={2,3,5,7}P = \{2, 3, 5, 7\}. The complement PP' consists of all elements in UU that do not belong to PP, which yields {1,4,6,8,9,10}\{1, 4, 6, 8, 9, 10\}.

Adım Adım Çözüm

1
Identify the elements of the universal set UU
U={1,2,3,4,5,6,7,8,9,10}U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}
The universal set defines the boundary of all possible elements under consideration.
2
Identify the elements belonging to set PP (prime numbers in UU)
P={2,3,5,7}P = \{2, 3, 5, 7\}
Prime numbers have exactly two distinct positive divisors: 1 and themselves. Note that 1 is not a prime number.
3
Compute the complement P=UPP' = U \setminus P
P={1,4,6,8,9,10}P' = \{1, 4, 6, 8, 9, 10\}
The complement of set PP consists of all elements present in UU that are not present in PP.

Anahtar Kavram

Set Complement and Universal Set Boundaries
Tahmini Süre:45s
Soru 166Soru

In a survey of 200200 agricultural exporters regarding three major commodities—Cocoa (CC), Palm Oil (PP), and Rubber (RR)—it was found that 110110 export Cocoa, 9090 export Palm Oil, and 7575 export Rubber. Exactly 2020 exporters export none of the three commodities, and 4545 export Cocoa only. Furthermore, the number of exporters who export Cocoa and Palm Oil only is twice the number of exporters who export all three commodities; the number who export Palm Oil and Rubber only is equal to the number who export all three; and the number who export Cocoa and Rubber only is 55 more than the number who export all three. Find the total number of exporters who export at least two of the three commodities.

Cevabı ve açıklamayı göster

Cevap: 80

Cevap

The total number of exporters who export at least two of the three commodities is 80.
The correct answer of 80 is obtained by solving for the number of exporters trading in all three commodities (x=15x = 15) using the Cocoa set equation 45+2x+(x+5)+x=11045 + 2x + (x + 5) + x = 110, and then evaluating the region sum for at least two commodities (2x+x+(x+5)+x=5x+5=802x + x + (x + 5) + x = 5x + 5 = 80).

Adım Adım Çözüm

1
Assign a variable to the triple intersection
Let x=n(CPR)x = n(C \cap P \cap R) be the number of exporters of all three commodities.
The intersection of all three sets serves as the common parameter for all double-intersection regions.
2
Write algebraic expressions for the three pairwise-only intersections
n(CP only)=2xn(C \cap P \text{ only}) = 2x, n(PR only)=xn(P \cap R \text{ only}) = x, and n(CR only)=x+5n(C \cap R \text{ only}) = x + 5.
These expressions are derived directly from the relationships given in the problem statement.
3
Formulate and solve an equation using the set of Cocoa exporters
45+2x+(x+5)+x=110    50+4x=110    x=1545 + 2x + (x + 5) + x = 110 \implies 50 + 4x = 110 \implies x = 15.
The set of Cocoa exporters consists of four mutually exclusive regions whose cardinalities sum to 110.
4
Sum the regions corresponding to 'at least two commodities'
(2x)+(x)+(x+5)+x=5x+5=5(15)+5=80(2x) + (x) + (x + 5) + x = 5x + 5 = 5(15) + 5 = 80.
'At least two' encompasses everyone who exports exactly two commodities plus those who export all three.

Anahtar Kavram

Three-set principle of inclusion-exclusion and cardinal region decomposition
Soru 167Soru

What is the determinant of the matrix P=(5327)P = \begin{pmatrix} 5 & 3 \\ 2 & 7 \end{pmatrix}?

Cevabı ve açıklamayı göster

Cevap: 29

Cevap

29
For a 2×22 \times 2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix}, the determinant is given by adbcad - bc. Substituting the values a=5a = 5, b=3b = 3, c=2c = 2, and d=7d = 7 yields (5×7)(3×2)=356=29(5 \times 7) - (3 \times 2) = 35 - 6 = 29.

Adım Adım Çözüm

1
Identify the values of a,b,c,a, b, c, and dd from the given 2×22 \times 2 matrix P=(5327)P = \begin{pmatrix} 5 & 3 \\ 2 & 7 \end{pmatrix}.
a=5,b=3,c=2,d=7a = 5, b = 3, c = 2, d = 7
To set up the values for the 2×22 \times 2 determinant formula.
2
Multiply the elements of the main diagonal and the secondary diagonal.
Main diagonal product: 5×7=355 \times 7 = 35; Secondary diagonal product: 3×2=63 \times 2 = 6.
The determinant of a 2×22 \times 2 matrix is the difference between the main diagonal product and the secondary diagonal product.
3
Subtract the product of the secondary diagonal from the product of the main diagonal.
Determinant = 356=2935 - 6 = 29.
Completes the formula det(P)=adbc\det(P) = ad - bc.

Anahtar Kavram

Determinant of a 2x2 Matrix
Soru 168Soru

What is the result of the subtraction 52482678524_8 - 267_8 expressed in base 8?

Cevabı ve açıklamayı göster

Cevap: 2358235_8

Cevap

The correct answer is 2358235_8.
Subtracting 2678267_8 from 5248524_8 using direct base-8 subtraction requires borrowing 8 when a digit is smaller than the digit subtracted from it. In the units column, borrowing 1 (which equals 8) gives (4+8)7=5(4 + 8) - 7 = 5. In the eights column, the borrowed-from digit 2 becomes 1; borrowing 1 (which equals 8) gives (1+8)6=3(1 + 8) - 6 = 3. In the sixty-fours column, the borrowed-from digit 5 becomes 4; 42=24 - 2 = 2. Thus, the difference is 2358235_8.

Adım Adım Çözüm

1
Convert both octal numbers to base 10.
5248=(5×82)+(2×81)+(4×80)=320+16+4=34010524_8 = (5 \times 8^2) + (2 \times 8^1) + (4 \times 8^0) = 320 + 16 + 4 = 340_{10}, and 2678=(2×82)+(6×81)+(7×80)=128+48+7=18310267_8 = (2 \times 8^2) + (6 \times 8^1) + (7 \times 8^0) = 128 + 48 + 7 = 183_{10}.
Converting to decimal simplifies standard arithmetic operations.
2
Subtract the decimal values.
3401018310=15710340_{10} - 183_{10} = 157_{10}.
Obtain the difference in base 10.
3
Convert the decimal difference 15710157_{10} back to base 8.
157÷8=19157 \div 8 = 19 remainder 55; 19÷8=219 \div 8 = 2 remainder 33; 2÷8=02 \div 8 = 0 remainder 22. Reading remainders from bottom to top gives 2358235_8.
Successive division by 8 yields the octal representation.

Anahtar Kavram

Non-decimal base arithmetic and borrowing rules
Soru 169Soru

The 5th5^{\text{th}} term of an arithmetic progression (AP) is 1717 and its common difference is 33. What is the first term of the progression?

Cevabı ve açıklamayı göster

Cevap: 55

Cevap

The first term of the progression is 55.
The nthn^{\text{th}} term formula of an arithmetic progression is Tn=a+(n1)dT_n = a + (n - 1)d. Substituting T5=17T_5 = 17, n=5n = 5, and d=3d = 3 gives 17=a+4(3)=a+1217 = a + 4(3) = a + 12. Isolating aa yields a=1712=5a = 17 - 12 = 5.

Adım Adım Çözüm

1
Identify the given values and formula for the nth term of an AP.
The formula is Tn=a+(n1)dT_n = a + (n - 1)d, with T5=17T_5 = 17, n=5n = 5, and d=3d = 3.
This formula connects the nth term, the first term, the number of terms, and the common difference.
2
Substitute the known values into the equation.
17=a+(51)×3    17=a+1217 = a + (5 - 1) \times 3 \implies 17 = a + 12.
Subtracting 1 from the term index 5 gives 4, and multiplying by 3 gives 12.
3
Solve for the first term aa.
a=1712=5a = 17 - 12 = 5.
Subtracting 12 from both sides isolates aa.

Anahtar Kavram

n-th term of an Arithmetic Progression
Tahmini Süre:1m 0s
Soru 170Soru

What is the simplified numerical value of 123+1+1231\frac{\sqrt{12}}{\sqrt{3} + 1} + \frac{\sqrt{12}}{\sqrt{3} - 1}?

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

The simplified numerical value of the given expression is 6.
Combining the fractions using their common conjugate denominator (3+1)(31)=2(\sqrt{3}+1)(\sqrt{3}-1) = 2 leads to a numerator of 23[(31)+(3+1)]=23(23)=122\sqrt{3}[(\sqrt{3}-1)+(\sqrt{3}+1)] = 2\sqrt{3}(2\sqrt{3}) = 12. Dividing 12 by 2 yields the final answer of 6.

Adım Adım Çözüm

1
Simplify the surd in the numerator
\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}
Simplifying surds into basic form makes subsequent calculations simpler.
2
Combine the fractions by finding a common denominator
23(31)+23(3+1)(3+1)(31)\frac{2\sqrt{3}(\sqrt{3}-1) + 2\sqrt{3}(\sqrt{3}+1)}{(\sqrt{3}+1)(\sqrt{3}-1)}
Multiplying denominators forms a conjugate pair, which rationalizes the combined denominator.
3
Simplify the numerator and the denominator
\text{Numerator} = 2\sqrt{3}(\sqrt{3}-1 + \sqrt{3}+1) = 2\sqrt{3}(2\sqrt{3}) = 12; \quad \text{Denominator} = (\sqrt{3})^2 - 1^2 = 3 - 1 = 2
Expanding the numerator combines like surd terms, and using the difference of squares simplifies the denominator to a rational integer.
4
Perform the final division
122=6\frac{12}{2} = 6
Dividing the simplified numerator by the rationalized denominator gives the final integer value.

Anahtar Kavram

Rationalization of binomial denominators using conjugate surds
Soru 171Soru

The sum of the first nn terms of a sequence is given by Sn=2n2+3nS_n = 2n^2 + 3n. Find the 5th5^{\text{th}} term of a geometric progression whose first term is the 3rd3^{\text{rd}} term of this sequence, and whose common ratio is equal to the common difference of this sequence.

Cevabı ve açıklamayı göster

Cevap: 33283328

Cevap

The 5th term of the geometric progression is 3328.
Evaluating S1S_1, S2S_2, and S3S_3 gives sequence terms T1=5T_1 = 5, T2=9T_2 = 9, and T3=13T_3 = 13. The common difference is d=95=4d = 9 - 5 = 4. Using T3=13T_3 = 13 as the first term of the GP and r=4r = 4 as the common ratio, the 5th term of the GP is 13×451=13×256=332813 \times 4^{5-1} = 13 \times 256 = 3328.

Adım Adım Çözüm

1
Calculate the first few terms of the sequence using the sum formula Sn=2n2+3nS_n = 2n^2 + 3n.
S1=2(1)2+3(1)=5S_1 = 2(1)^2 + 3(1) = 5, S2=2(2)2+3(2)=14S_2 = 2(2)^2 + 3(2) = 14, S3=2(3)2+3(3)=27S_3 = 2(3)^2 + 3(3) = 27.
The sum formula gives cumulative sums, from which individual terms can be derived.
2
Find the 3rd term (T3T_3) and the common difference (dd) of the arithmetic progression.
T1=5T_1 = 5, T2=S2S1=9T_2 = S_2 - S_1 = 9, T3=S3S2=13T_3 = S_3 - S_2 = 13. Common difference d=T2T1=4d = T_2 - T_1 = 4.
The difference between consecutive cumulative sums gives the individual sequence terms, and their constant difference gives the common difference.
3
Define the parameters of the geometric progression (GP).
First term of GP a=T3=13a = T_3 = 13, common ratio r=d=4r = d = 4.
The problem specifies that the first term of the GP is the 3rd term of the sequence and the common ratio equals the common difference.
4
Compute the 5th term of the geometric progression using Gn=arn1G_n = a \cdot r^{n-1}.
G5=13451=1344=13256=3328G_5 = 13 \cdot 4^{5-1} = 13 \cdot 4^4 = 13 \cdot 256 = 3328.
Applying the standard nth term formula for a GP with n=5n = 5.

Anahtar Kavram

Arithmetic Progression sum formula to term conversion and Geometric Progression nth term evaluation
Tahmini Süre:2m 0s
Soru 172Soru

Three business partners, XX, YY, and ZZ, invest in a venture. XX contributes 13\frac{1}{3} of the total capital for 14\frac{1}{4} of the total duration of the business. YY contributes 14\frac{1}{4} of the capital for 12\frac{1}{2} of the duration, while ZZ contributes the remaining capital for the entire duration. If the total profit realized from the business is N78,000\mathcal{N}78,000, what is ZZ's share of the profit?

Cevabı ve açıklamayı göster

Cevap: N52,000\mathcal{N}52,000

Cevap

N52,000\mathcal{N}52,000
Profit distribution in compound partnerships depends on both capital invested and investment time duration. XX's product is 112\frac{1}{12}, YY's product is 18\frac{1}{8}, and ZZ's product is 512\frac{5}{12}. Scaling these by 24 gives a ratio of 2:3:102 : 3 : 10. With 15 total parts, ZZ receives 1015\frac{10}{15} of N78,000\mathcal{N}78,000, which equals N52,000\mathcal{N}52,000.

Adım Adım Çözüm

1
Determine ZZ's fraction of the total capital
ZZ's capital fraction =1(13+14)=1712=512= 1 - \left(\frac{1}{3} + \frac{1}{4}\right) = 1 - \frac{7}{12} = \frac{5}{12}
The sum of all capital contributions must equal 1.
2
Calculate the compound product of (Capital Share ×\times Time Duration) for each partner
X=13×14=112X = \frac{1}{3} \times \frac{1}{4} = \frac{1}{12}, Y=14×12=18Y = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}, Z=512×1=512Z = \frac{5}{12} \times 1 = \frac{5}{12}
Profit sharing in compound partnerships is proportional to the product of capital invested and the time period.
3
Convert the fractions to an equivalent integer ratio
Multiplying 112:18:512\frac{1}{12} : \frac{1}{8} : \frac{5}{12} by 24 gives the ratio 2:3:102 : 3 : 10
Finding the least common multiple (24) simplifies the fractions into integer ratio parts.
4
Compute ZZ's profit share from the total profit
Total ratio parts =2+3+10=15= 2 + 3 + 10 = 15. ZZ's share =1015×N78,000=N52,000= \frac{10}{15} \times \mathcal{N}78,000 = \mathcal{N}52,000
Multiply total profit by ZZ's ratio fraction.

Anahtar Kavram

Compound Ratio and Partnership Profit Sharing
Soru 173Soru

Given that the matrix A=(12k0k3211)A = \begin{pmatrix} 1 & 2 & k \\ 0 & k & 3 \\ 2 & -1 & 1 \end{pmatrix} is singular, what is the positive value of kk?

Cevabı ve açıklamayı göster

Cevap: 3

Cevap

The positive value of kk is 33.
A square matrix is singular when its determinant is zero. Expanding det(A)\det(A) along the first column yields 1(k+3)+2(6k2)=2k2+k+151 \cdot (k + 3) + 2 \cdot (6 - k^2) = -2k^2 + k + 15. Equating this to zero gives the quadratic equation 2k2k15=02k^2 - k - 15 = 0, which factors as (2k+5)(k3)=0(2k + 5)(k - 3) = 0. The roots are k=2.5k = -2.5 and k=3k = 3. The positive value is 33.

Adım Adım Çözüm

1
Calculate the determinant of matrix A
\det(A) = -2k^2 + k + 15
Expanding along the first column simplifies the calculation because of the zero entry.
2
Set the determinant to zero for singularity
2k^2 - k - 15 = 0
A matrix is singular if and only if its determinant equals zero.
3
Solve the quadratic equation for k
k = 3 or k = -2.5
Factoring 2k2k15=(2k+5)(k3)=02k^2 - k - 15 = (2k + 5)(k - 3) = 0 yields two roots.
4
Choose the positive solution
k = 3
The question explicitly requires the positive value of kk.

Anahtar Kavram

Singular matrix definition and 3x3 determinant evaluation
Soru 174Soru

In a survey of 180180 cloud computing engineers regarding their proficiency in three major platforms—AWS (AA), Azure (BB), and Google Cloud (CC)—it was found that 9595 are proficient in AWS, 8080 in Azure, and 7575 in Google Cloud. Furthermore, 4040 are proficient in both AWS and Azure, 3535 in both Azure and Google Cloud, 3030 in both AWS and Google Cloud, and 1515 are not proficient in any of the three platforms. How many engineers are proficient in exactly one of these platforms?

Cevabı ve açıklamayı göster

Cevap: 100

Cevap

100 engineers are proficient in exactly one platform.
Using the 3-set inclusion-exclusion formula, the number of engineers proficient in all three platforms is solved as 20. Subtracting the relevant overlap regions from each set gives 45 for AWS only, 25 for Azure only, and 30 for Google Cloud only. Adding these single-set values yields a total of 100 engineers proficient in exactly one platform.

Adım Adım Çözüm

1
Determine the cardinality of the union of all three sets.
n(ABC)=18015=165n(A \cup B \cup C) = 180 - 15 = 165.
Subtracting the engineers who are not proficient in any of the three platforms from the universal set.
2
Apply the Principle of Inclusion-Exclusion to calculate the triple intersection n(ABC)n(A \cap B \cap C).
n(ABC)=20n(A \cap B \cap C) = 20.
Substituting known values gives 165=250105+n(ABC)165 = 250 - 105 + n(A \cap B \cap C), which simplifies to n(ABC)=20n(A \cap B \cap C) = 20.
3
Calculate the counts for regions representing exactly two platforms.
AWS & Azure only = 20, Azure & GCP only = 15, AWS & GCP only = 10.
Subtracting the triple intersection count (20) from each pairwise intersection.
4
Calculate the single-set exclusive regions.
AWS only = 45, Azure only = 25, GCP only = 30.
Subtracting all multi-platform overlap regions from each total set size.
5
Sum the single-set exclusive regions.
Total = 45+25+30=10045 + 25 + 30 = 100.
Combining the counts of engineers proficient in AWS only, Azure only, and Google Cloud only.

Anahtar Kavram

Principle of Inclusion-Exclusion for Three Sets and Venn Diagram Region Partitioning
Soru 175Soru

A business owner borrowed 150,000\text{₦}150,000 to expand his store at a compound interest rate of 8%8\% per annum, compounded annually. What is the total compound interest he will pay at the end of 2 years?

Cevabı ve açıklamayı göster

Cevap: 24,960\text{₦}24,960

Cevap

The total compound interest paid at the end of 2 years is 24,960\text{₦}24,960.
The total compound interest is obtained by calculating the interest accrued in each period (12,00012,000 in Year 1 and 12,96012,960 in Year 2) and summing them to get 24,960\text{₦}24,960. Alternatively, using the compound interest formula I=P(1+r)nPI = P(1 + r)^n - P, I=150,000(1.08)2150,000=174,960150,000=24,960I = 150,000(1.08)^2 - 150,000 = 174,960 - 150,000 = \text{₦}24,960.

Adım Adım Çözüm

1
Calculate the interest for the first year.
Interest1=8100×150,000=12,000\text{Interest}_1 = \frac{8}{100} \times 150,000 = \text{₦}12,000
Interest in the first year is calculated on the initial principal.
2
Find the principal for the second year.
Principal2=150,000+12,000=162,000\text{Principal}_2 = 150,000 + 12,000 = \text{₦}162,000
Under compound interest, the first year's interest is added to the initial principal.
3
Calculate the interest for the second year.
Interest2=8100×162,000=12,960\text{Interest}_2 = \frac{8}{100} \times 162,000 = \text{₦}12,960
Interest in the second year is calculated on the updated principal.
4
Sum the interest from both years to find the total compound interest.
Total Interest=12,000+12,960=24,960\text{Total Interest} = 12,000 + 12,960 = \text{₦}24,960
The total compound interest is the sum of interest accumulated in each compounding period.

Anahtar Kavram

Compound Interest
Tahmini Süre:1m 30s
Soru 176Soru

A trader bought a set of solar panels for 160,000\text{₦}160,000. He marked up the cost price by 25%25\% to fix the marked price. During a trade fair, he offered a 10%10\% discount on the marked price. What is his net profit in Naira?

Cevabı ve açıklamayı göster

Cevap: 20000

Cevap

The net profit made by the trader is ���20,000.
First, find the marked price by adding a 25% markup to the cost price: ₦160,000 × 1.25 = ₦200,000. Next, calculate the selling price after a 10% discount on the marked price: ₦200,000 × 0.90 = ₦180,000. Finally, subtract the cost price from the selling price to find the net profit: ₦180,000 - ₦160,000 = ₦20,000.

Adım Adım Çözüm

1
Calculate the marked price
Marked Price = ₦200,000
The marked price is 125% of the original cost price of ₦160,000.
2
Calculate the selling price after discount
Selling Price = ₦180,000
A 10% discount reduces the marked price by ₦20,000.
3
Determine the net profit
Net Profit = ₦20,000
Net profit is the difference between the selling price (₦180,000) and the cost price (₦160,000).

Anahtar Kavram

Markup, Discount, and Profit Calculations
Soru 177Soru

If 52x1×25x+1=125x+25^{2x - 1} \times 25^{x + 1} = 125^{x + 2}, what is the value of xx?

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

The value of xx is 55.
Converting all terms to base 55 gives 52x1×52(x+1)=53(x+2)5^{2x - 1} \times 5^{2(x + 1)} = 5^{3(x + 2)}. Simplifying the exponents yields 52x1+2x+2=53x+65^{2x - 1 + 2x + 2} = 5^{3x + 6}, which reduces to 54x+1=53x+65^{4x + 1} = 5^{3x + 6}. Setting the exponents equal to each other gives 4x+1=3x+64x + 1 = 3x + 6, resulting in x=5x = 5.

Adım Adım Çözüm

1
Express all bases in terms of prime base 5
52x1×(52)x+1=(53)x+25^{2x - 1} \times (5^2)^{x + 1} = (5^3)^{x + 2}
All terms must share the same base to combine exponents using index laws.
2
Apply power of a power rule (am)n=amn(a^m)^n = a^{mn}
52x1×52x+2=53x+65^{2x - 1} \times 5^{2x + 2} = 5^{3x + 6}
Multiplication of inner and outer powers simplifies composite exponent expressions.
3
Apply the product rule am×an=am+na^m \times a^n = a^{m+n}
54x+1=53x+65^{4x + 1} = 5^{3x + 6}
Adding the exponents on the left-hand side produces a single exponential term.
4
Equate exponents and solve for xx
x=5x = 5
Equal bases imply equal exponents: 4x+1=3x+64x + 1 = 3x + 6.

Anahtar Kavram

Solving exponential equations using common base conversion and laws of indices
Tahmini Süre:1m 30s
Soru 178Soru

A rectangle has an area of 10 cm210\text{ cm}^2 and a length of (7+2) cm(\sqrt{7} + \sqrt{2})\text{ cm}. What is the width of the rectangle in simplified surd form?

Cevabı ve açıklamayı göster

Cevap: (2722) cm(2\sqrt{7} - 2\sqrt{2})\text{ cm}

Cevap

The width of the rectangle in simplified surd form is (2722) cm(2\sqrt{7} - 2\sqrt{2})\text{ cm}.
The area formula for a rectangle gives width=107+2\text{width} = \frac{10}{\sqrt{7} + \sqrt{2}}. Multiplying both numerator and denominator by the conjugate (72)(\sqrt{7} - \sqrt{2}) produces 10(72)72=10(72)5=2722\frac{10(\sqrt{7} - \sqrt{2})}{7 - 2} = \frac{10(\sqrt{7} - \sqrt{2})}{5} = 2\sqrt{7} - 2\sqrt{2}.

Adım Adım Çözüm

1
Set up the formula for the width of the rectangle
\text{Width} = \frac{\text{Area}}{\text{Length}} = \frac{10}{\sqrt{7} + \sqrt{2}}
The area of a rectangle is length multiplied by width.
2
Rationalize the denominator by multiplying top and bottom by the conjugate (72)(\sqrt{7} - \sqrt{2})
\text{Width} = \frac{10(\sqrt{7} - \sqrt{2})}{(\sqrt{7} + \sqrt{2})(\sqrt{7} - \sqrt{2})}
Multiplying by the conjugate creates a difference of squares in the denominator, eliminating radicals.
3
Simplify the denominator using (a+b)(ab)=a2b2(a+b)(a-b) = a^2 - b^2
(\sqrt{7})^2 - (\sqrt{2})^2 = 7 - 2 = 5
Squaring a square root yields the underlying rational number.
4
Divide the numerator by the simplified denominator
\frac{10(\sqrt{7} - \sqrt{2})}{5} = 2(\sqrt{7} - \sqrt{2}) = 2\sqrt{7} - 2\sqrt{2}
Dividing 1010 by 55 gives 22, which is then distributed across the terms inside the parentheses.

Anahtar Kavram

Rationalization of Denominators with Binomial Surds
Soru 179Soru

If log2x+log8x=4\log_2 x + \log_8 x = 4, what is the value of xx?

Cevabı ve açıklamayı göster

Cevap: 8

Cevap

8
By applying the change of base formula, log8x=log2xlog28=13log2x\log_8 x = \frac{\log_2 x}{\log_2 8} = \frac{1}{3}\log_2 x. Rewriting the equation gives log2x+13log2x=43log2x=4\log_2 x + \frac{1}{3}\log_2 x = \frac{4}{3}\log_2 x = 4. Solving for log2x\log_2 x yields log2x=3\log_2 x = 3, which in exponential form gives x=23=8x = 2^3 = 8.

Adım Adım Çözüm

1
Apply the change of base formula to express log8x\log_8 x in base 2
log8x=log2xlog28=log2x3\log_8 x = \frac{\log_2 x}{\log_2 8} = \frac{\log_2 x}{3}
Logarithms must be converted to a common base before combining terms.
2
Substitute log8x=13log2x\log_8 x = \frac{1}{3}\log_2 x into the original equation
log2x+13log2x=4    43log2x=4\log_2 x + \frac{1}{3}\log_2 x = 4 \implies \frac{4}{3}\log_2 x = 4
Combine like logarithmic terms.
3
Solve for log2x\log_2 x
log2x=4×34=3\log_2 x = 4 \times \frac{3}{4} = 3
Isolate the logarithmic expression by multiplying both sides by 34\frac{3}{4}.
4
Convert the logarithmic equation to its exponential form to solve for xx
x=23=8x = 2^3 = 8
Definition of logarithms states that logba=c    bc=a\log_b a = c \iff b^c = a.

Anahtar Kavram

Logarithms and Change of Base
Soru 180Soru

The 3rd3^{\text{rd}} term of a geometric progression is 1818 and the 6th6^{\text{th}} term is 486486. What is the sum of the first 55 terms of the progression?

Cevabı ve açıklamayı göster

Cevap: 242242

Cevap

The sum of the first 55 terms is 242242.
Using Tn=arn1T_n = a r^{n-1}, we set up ar2=18a r^2 = 18 and ar5=486a r^5 = 486. Dividing these gives r3=27r^3 = 27, so r=3r = 3, which leads to a=2a = 2. Applying S5=a(r51)r1S_5 = \frac{a(r^5 - 1)}{r - 1} yields 2(2431)2=242\frac{2(243 - 1)}{2} = 242.

Adım Adım Çözüm

1
Express the given terms using the geometric progression nthn^{\text{th}} term formula Tn=arn1T_n = a r^{n-1}.
T3=ar2=18T_3 = a r^2 = 18 and T6=ar5=486T_6 = a r^5 = 486.
This establishes a system of equations in terms of the first term aa and common ratio rr.
2
Divide the expression for T6T_6 by T3T_3 to determine rr.
\frac{a r^5}{a r^2} = \frac{486}{18} \implies r^3 = 27 \implies r = 3.
Dividing the equations eliminates aa and allows direct solution for the common ratio rr.
3
Substitute r=3r = 3 back into T3=ar2=18T_3 = a r^2 = 18 to solve for aa.
a (3)^2 = 18 \implies 9a = 18 \implies a = 2.
Determining the first term aa is required to evaluate the sum.
4
Calculate the sum of the first 55 terms using Sn=a(rn1)r1S_n = \frac{a(r^n - 1)}{r - 1}.
S_5 = \frac{2(3^5 - 1)}{3 - 1} = \frac{2(243 - 1)}{2} = 242.
Applying the GP sum formula yields the required value.

Anahtar Kavram

Geometric Progression nth term and sum formulas
ÖncekiSayfa 9 / 12Sonraki
Number and Numeration Alıştırma Soruları — JAMB UTME — Sayfa 9 | Examkin