Direct, Inverse, Joint and Partial Variation

24 soru

Soru 1Soru

A quantity PP varies partially as xx and partially as the square of yy. When x=2x = 2 and y=3y = 3, P=24P = 24, and when x=5x = 5 and y=1y = 1, P=17P = 17. What is the value of PP when x=4x = 4 and y=3y = 3?

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Cevap: 30

Cevap

The value of PP is 30.
The relationship follows the partial variation formula P=k1x+k2y2P = k_1 x + k_2 y^2. Substituting the given conditions gives 2k1+9k2=242k_1 + 9k_2 = 24 and 5k1+k2=175k_1 + k_2 = 17. Solving these simultaneous equations yields k1=3k_1 = 3 and k2=2k_2 = 2. Evaluating P=3(4)+2(32)P = 3(4) + 2(3^2) produces 12+18=3012 + 18 = 30.

Adım Adım Çözüm

1
Set up the general formula for partial variation.
P=k1x+k2y2P = k_1 x + k_2 y^2, where k1k_1 and k2k_2 are constants.
Partial variation combines terms linearly with separate variation constants.
2
Substitute the given pairs of values to form simultaneous linear equations.
Equation (1): 2k1+9k2=242k_1 + 9k_2 = 24; Equation (2): 5k1+k2=175k_1 + k_2 = 17.
Plugging in (x=2,y=3,P=24)(x=2, y=3, P=24) and (x=5,y=1,P=17)(x=5, y=1, P=17) creates a system of equations in terms of k1k_1 and k2k_2.
3
Solve the simultaneous linear equations for k1k_1 and k2k_2.
From Equation (2), k2=175k1k_2 = 17 - 5k_1. Substitute into Equation (1): 2k1+9(175k1)=24    43k1=129    k1=32k_1 + 9(17 - 5k_1) = 24 \implies -43k_1 = -129 \implies k_1 = 3. Then k2=175(3)=2k_2 = 17 - 5(3) = 2.
Finding the specific values of the variation constants is required to complete the formula.
4
Calculate PP for x=4x = 4 and y=3y = 3 using the complete formula P=3x+2y2P = 3x + 2y^2.
P=3(4)+2(32)=12+2(9)=12+18=30P = 3(4) + 2(3^2) = 12 + 2(9) = 12 + 18 = 30.
Evaluating the relationship with the target parameters produces the final answer.

Anahtar Kavram

Partial Variation and Simultaneous Linear Equations
Tahmini Süre:2m 0s
Soru 2Soru

The variable yy varies inversely as xx. If y=6y = 6 when x=4x = 4, what is the value of yy when x=8x = 8?

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Cevap: 3

Cevap

The value of yy when x=8x = 8 is 3.
Since yy varies inversely as xx, the formula connecting them is y=kxy = \frac{k}{x}. Substituting y=6y = 6 when x=4x = 4 gives 6=k46 = \frac{k}{4}, which yields k=24k = 24. Using this constant, when x=8x = 8, y=248=3y = \frac{24}{8} = 3.

Adım Adım Çözüm

1
Write the relationship for inverse variation.
y=kxy = \frac{k}{x}
Inverse variation means yy is inversely proportional to xx, where kk is the constant of variation.
2
Substitute given values y=6y = 6 and x=4x = 4 to solve for kk.
6=k4    k=6×4=246 = \frac{k}{4} \implies k = 6 \times 4 = 24
To complete the equation of variation, the constant kk must be determined.
3
Substitute k=24k = 24 and x=8x = 8 into the variation equation to find yy.
y=248=3y = \frac{24}{8} = 3
Evaluating the relationship at x=8x = 8 gives the requested value.

Anahtar Kavram

Inverse Variation
Soru 3Soru

A quantity yy is partly constant and partly varies directly as xx. When x=2x = 2, y=10y = 10, and when x=5x = 5, y=19y = 19. What is the value of yy when x=8x = 8?

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Cevap: 28

Cevap

The value of yy when x=8x = 8 is 28.
By representing partial variation as y=c+kxy = c + kx, substituting the given conditions gives two simultaneous linear equations: 10=c+2k10 = c + 2k and 19=c+5k19 = c + 5k. Subtracting the first equation from the second yields 3k=93k = 9, so k=3k = 3. Substituting k=3k = 3 back into the first equation yields c=4c = 4. The general equation is y=4+3xy = 4 + 3x. Evaluating at x=8x = 8 gives y=4+3(8)=28y = 4 + 3(8) = 28.

Adım Adım Çözüm

1
Set up the partial variation equation
y=c+kxy = c + kx, where cc is the constant part and kk is the constant of variation.
Partial variation consists of a fixed term plus a variable term.
2
Substitute given values to form simultaneous equations
Equation 1: 10=c+2k10 = c + 2k
Equation 2: 19=c+5k19 = c + 5k
Two pairs of (x,y)(x, y) values are provided to solve for the two unknown constants cc and kk.
3
Solve for kk and cc
Subtract Equation 1 from Equation 2: 9=3k    k=39 = 3k \implies k = 3.
Substitute k=3k = 3 into Equation 1: 10=c+2(3)    c=410 = c + 2(3) \implies c = 4.
Thus, y=4+3xy = 4 + 3x.
Eliminating cc yields the value of kk, which is then used to find cc.
4
Calculate yy for x=8x = 8
y=4+3(8)=4+24=28y = 4 + 3(8) = 4 + 24 = 28.
Substitute the required value of xx into the established formula.

Anahtar Kavram

Partial Variation and Simultaneous Equations
Soru 4Soru

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 5Soru

Given that zz varies directly as x2x^2 and inversely as y\sqrt{y}, and z=12z = 12 when x=2x = 2 and y=9y = 9, what is the value of zz when x=3x = 3 and y=16y = 16?

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Cevap: 814\frac{81}{4}

Cevap

814\frac{81}{4}
The joint variation formula is z=kx2yz = \frac{k x^2}{\sqrt{y}}. Substituting the given values x=2,y=9,z=12x = 2, y = 9, z = 12 gives 12=4k312 = \frac{4k}{3}, which yields k=9k = 9. Evaluating zz for x=3x = 3 and y=16y = 16 gives z=9×3216=814z = \frac{9 \times 3^2}{\sqrt{16}} = \frac{81}{4}.

Adım Adım Çözüm

1
Set up the joint variation equation
z=kx2yz = \frac{k x^2}{\sqrt{y}}
Direct variation means x2x^2 is in the numerator, and inverse variation means y\sqrt{y} is in the denominator.
2
Substitute the initial values to solve for the constant of variation kk
12=k(2)29    12=4k3    4k=36    k=912 = \frac{k (2)^2}{\sqrt{9}} \implies 12 = \frac{4k}{3} \implies 4k = 36 \implies k = 9
Using x=2x = 2, y=9y = 9, and z=12z = 12 allows us to find the constant kk.
3
Calculate the new value of zz using x=3x = 3 and y=16y = 16
z=9(3)216=9×94=814z = \frac{9 (3)^2}{\sqrt{16}} = \frac{9 \times 9}{4} = \frac{81}{4}
Substitute k=9k = 9, x=3x = 3, and y=16y = 16 into the variation formula.

Anahtar Kavram

Joint Variation involving powers and roots
Tahmini Süre:1m 30s
Soru 6Soru

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 7Soru

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 8Soru

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 9Soru

The time, tt hours, required to complete a road maintenance project varies inversely as the number of workers, ww, assigned to the project. If 8 workers can finish the project in 15 hours, calculate the time, in hours, required for 12 workers to finish the same project.

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Cevap: 10

Cevap

10 hours
Because the time tt varies inversely as the number of workers ww, the total worker-hours required for the project is constant: k=8×15=120k = 8 \times 15 = 120 worker-hours. Dividing this total work by 12 workers gives 12012=10\frac{120}{12} = 10 hours.

Adım Adım Çözüm

1
Set up the inverse variation equation
t=kwt = \frac{k}{w}, where kk is the constant of variation.
Inverse variation implies that as the number of workers increases, the time required decreases proportionally.
2
Determine the value of the constant of variation kk
k=t×w=15×8=120k = t \times w = 15 \times 8 = 120.
Substitute the known pair of values (w=8,t=15w = 8, t = 15) into the equation.
3
Compute the new value of tt for 12 workers
t=12012=10t = \frac{120}{12} = 10 hours.
Substitute w=12w = 12 and k=120k = 120 into t=kwt = \frac{k}{w}.

Anahtar Kavram

Inverse Variation
Soru 10Soru

The total energy loss EE in Joules per minute in a magnetic core circuit is partly constant and partly varies directly as the square of the frequency ff in Hz of the alternating current. Given that E=120 JE = 120\text{ J} when f=10 Hzf = 10\text{ Hz}, and E=360 JE = 360\text{ J} when f=20 Hzf = 20\text{ Hz}, what is the value of EE in Joules when f=15 Hzf = 15\text{ Hz}?

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Cevap: 220

Cevap

220
The relation describes a partial variation model E=c+kf2E = c + k f^2. Substituting the two given conditions (f=10,E=120f=10, E=120 and f=20,E=360f=20, E=360) yields the system of equations c+100k=120c + 100k = 120 and c+400k=360c + 400k = 360. Solving this system gives k=0.8k = 0.8 and c=40c = 40. Evaluating E=40+0.8(15)2E = 40 + 0.8(15)^2 results in 220 J220\text{ J}.

Adım Adım Çözüm

1
Write the general equation for partial variation involving a constant term and a term proportional to f2f^2
E=c+kf2E = c + k f^2
Partial variation consists of a sum of a constant component and a variable component.
2
Substitute the known conditions into the variation equation to create a system of linear equations
Equation 1: c+100k=120c + 100k = 120; Equation 2: c+400k=360c + 400k = 360
Two pairs of values are provided to determine the two unknown constants cc and kk.
3
Solve the system of simultaneous equations for kk and cc
k=0.8k = 0.8 and c=40c = 40
Subtracting Equation 1 from Equation 2 eliminates cc, allowing direct calculation of kk, after which cc is found by substitution.
4
Calculate the required value of EE when f=15f = 15
E=40+0.8(152)=40+0.8(225)=220E = 40 + 0.8(15^2) = 40 + 0.8(225) = 220
Applying the discovered constants to the target frequency yields the final energy loss value.

Anahtar Kavram

Partial Variation and Simultaneous Equations
Tahmini Süre:2m 0s
Soru 11Soru

If yy varies directly as xx, and y=20y = 20 when x=4x = 4, what is the value of yy when x=10x = 10?

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Cevap: 50

Cevap

The value of yy when x=10x = 10 is 50.
Direct variation means y=kxy = kx. Substituting y=20y = 20 and x=4x = 4 yields k=5k = 5. Substituting k=5k = 5 and x=10x = 10 gives y=50y = 50.

Adım Adım Çözüm

1
Set up the equation for direct variation
y=kxy = kx, where kk is the constant of variation.
Direct variation implies that yy is directly proportional to xx.
2
Substitute the given initial values (y=20y = 20, x=4x = 4) to find kk
20=k(4)    k=204=520 = k(4) \implies k = \frac{20}{4} = 5.
Determining the constant of variation allows us to write the specific relationship equation.
3
Calculate yy when x=10x = 10 using the constant k=5k = 5
y=5×10=50y = 5 \times 10 = 50.
Substituting x=10x = 10 into y=5xy = 5x yields the required value.

Anahtar Kavram

Direct Variation (y=kxy = kx)
Tahmini Süre:45s
Soru 12Soru

The power consumption PP (in watts) of a variable-speed motor is partly constant and partly varies directly as the square of its operational speed vv (in revolutions per second). If P=250 WP = 250\text{ W} when v=10 rev/sv = 10\text{ rev/s} and P=700 WP = 700\text{ W} when v=20 rev/sv = 20\text{ rev/s}, calculate the value of PP (in watts) when v=15 rev/sv = 15\text{ rev/s}.

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Cevap: 437.5

Cevap

437.5 W
The partial variation equation is P=k1+k2v2P = k_1 + k_2 v^2. Setting up simultaneous equations k1+100k2=250k_1 + 100 k_2 = 250 and k1+400k2=700k_1 + 400 k_2 = 700 yields k1=100k_1 = 100 and k2=1.5k_2 = 1.5. Substituting v=15v = 15 into P=100+1.5(152)P = 100 + 1.5(15^2) gives P=437.5 WP = 437.5\text{ W}.

Adım Adım Çözüm

1
Express the partial variation mathematically.
P=k1+k2v2P = k_1 + k_2 v^2, where k1k_1 and k2k_2 are constants.
The total power consumption is the sum of a fixed baseline constant k1k_1 and a variable component proportional to v2v^2.
2
Form simultaneous linear equations using the provided data points.
k1+100k2=250k_1 + 100 k_2 = 250 and k1+400k2=700k_1 + 400 k_2 = 700.
Substituting v=10v = 10 gives 102=10010^2 = 100, and substituting v=20v = 20 gives 202=40020^2 = 400.
3
Solve for the constants k1k_1 and k2k_2.
k2=1.5k_2 = 1.5 and k1=100k_1 = 100.
Subtracting the two equations eliminates k1k_1, yielding 300k2=450    k2=1.5300 k_2 = 450 \implies k_2 = 1.5. Substituting k2=1.5k_2 = 1.5 back into k1+100k2=250k_1 + 100 k_2 = 250 gives k1=100k_1 = 100.
4
Calculate the value of PP at v=15 rev/sv = 15\text{ rev/s}.
P=437.5 WP = 437.5\text{ W}.
Substitute v=15v = 15, k1=100k_1 = 100, and k2=1.5k_2 = 1.5 into the governing formula P=100+1.5(152)=100+337.5=437.5P = 100 + 1.5(15^2) = 100 + 337.5 = 437.5.

Anahtar Kavram

Partial Variation and Simultaneous Linear Equations
Tahmini Süre:1m 30s
Soru 13Soru

The total cost CC per trip of operating a high-speed passenger ferry consists of a fixed administrative overhead cost and an operational cost that varies directly as the cube of its speed vv in knots. Given that the total cost per trip is $1,400\$1,400 when the ferry travels at 10 knots10\text{ knots} and $4,200\$4,200 when it travels at 20 knots20\text{ knots}, what is the total cost per trip when the ferry operates at a speed of 15 knots15\text{ knots}?

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Cevap: \\ 2,350$

Cevap

\\ 2,350$
The partial variation formula is C=k1+k2v3C = k_1 + k_2 v^3. Substituting the given points yields k1+1000k2=1400k_1 + 1000 k_2 = 1400 and k1+8000k2=4200k_1 + 8000 k_2 = 4200. Subtracting these equations gives 7000k2=28007000 k_2 = 2800, so k2=0.4k_2 = 0.4 and k1=1000k_1 = 1000. Evaluating at v=15v = 15 gives C=1000+0.4(3375)=1000+1350=C = 1000 + 0.4(3375) = 1000 + 1350 = \\ 2,350$.

Adım Adım Çözüm

1
Set up the partial variation equation
C=k1+k2v3C = k_1 + k_2 v^3
Partial variation combines a fixed constant k1k_1 with a variable term k2v3k_2 v^3.
2
Form simultaneous linear equations using given conditions
Equation (1): k1+1000k2=1400k_1 + 1000 k_2 = 1400; Equation (2): k1+8000k2=4200k_1 + 8000 k_2 = 4200
Substitute v=10v = 10, C=1400C = 1400 and v=20v = 20, C=4200C = 4200 into the variation model.
3
Solve for the constants k1k_1 and k2k_2
k2=0.4k_2 = 0.4 and k1=1000k_1 = 1000
Subtract Equation (1) from Equation (2): 7000k2=2800    k2=0.47000 k_2 = 2800 \implies k_2 = 0.4. Substitute k2=0.4k_2 = 0.4 into Equation (1) to get k1=1400400=1000k_1 = 1400 - 400 = 1000.
4
Calculate total cost CC for speed v=15 knotsv = 15\text{ knots}
C=1000+0.4(15)3=1000+0.4(3375)=1000+1350=2350C = 1000 + 0.4(15)^3 = 1000 + 0.4(3375) = 1000 + 1350 = 2350
Substitute k1=1000k_1 = 1000, k2=0.4k_2 = 0.4, and v=15v = 15 back into the formula.

Anahtar Kavram

Partial Variation with non-linear powers solved via simultaneous linear equations
Tahmini Süre:2m 0s
Soru 14Soru

The monthly operating cost CC (in naira) of a commercial power generator is partly constant and partly varies directly as the square of its daily operating time hh (in hours). When the generator operates for 33 hours per day, the monthly cost is ₦5,5005,500. When it operates for 55 hours per day, the monthly cost is ₦13,50013,500. What is the daily operating time, in hours, when the monthly operating cost is ₦33,00033,000?

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Cevap: 8 hours

Cevap

8 hours
The relationship is modeled by partial variation C=k1+k2h2C = k_1 + k_2 h^2. Solving the simultaneous equations derived from h=3h = 3 (C=5500C = 5500) and h=5h = 5 (C=13500C = 13500) yields k2=500k_2 = 500 and k1=1000k_1 = 1000. Substituting C=33000C = 33000 into 33000=1000+500h233000 = 1000 + 500 h^2 gives 500h2=32000500 h^2 = 32000, so h2=64h^2 = 64, which yields h=8h = 8 hours.

Adım Adım Çözüm

1
Set up the general formula for partial variation.
C=k1+k2h2C = k_1 + k_2 h^2, where k1k_1 and k2k_2 are constants.
The cost consists of a constant part (k1k_1) and a part that varies directly as the square of daily hours (k2h2k_2 h^2).
2
Form simultaneous linear equations using the given data points.
Equation (1): 5500=k1+9k25500 = k_1 + 9 k_2
Equation (2): 13500=k1+25k213500 = k_1 + 25 k_2
Substitute h=3,C=5500h = 3, C = 5500 and h=5,C=13500h = 5, C = 13500 into the variation equation.
3
Solve for the variation constants k1k_1 and k2k_2.
Subtract Equation (1) from Equation (2): 8000=16k2    k2=5008000 = 16 k_2 \implies k_2 = 500.
Substitute k2=500k_2 = 500 into Equation (1): 5500=k1+9(500)    k1=10005500 = k_1 + 9(500) \implies k_1 = 1000.
Eliminating k1k_1 gives k2k_2, which is then used to find the constant part k1k_1.
4
Calculate hh when C=33000C = 33000.
33000=1000+500h2    32000=500h2    h2=64    h=833000 = 1000 + 500 h^2 \implies 32000 = 500 h^2 \implies h^2 = 64 \implies h = 8 hours.
Substitute the values of k1,k2,k_1, k_2, and target CC into the relation equation to solve for hh.

Anahtar Kavram

Partial Variation with Simultaneous Equations
Tahmini Süre:2m 30s
Soru 15Soru

The total cost CC of printing a school magazine is partly constant and partly varies directly as the number of copies nn printed. If it costs N70,000\text{N}70,000 to print 500500 copies and N140,000\text{N}140,000 to print 1,2001,200 copies, what is the cost of printing 2,0002,000 copies?

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Cevap: N220,000\text{N}220,000

Cevap

N220,000\text{N}220,000
In partial variation, the total cost CC is represented by C=k1+k2nC = k_1 + k_2 n. Subtracting the two linear equations 70,000=k1+500k270,000 = k_1 + 500 k_2 and 140,000=k1+1,200k2140,000 = k_1 + 1,200 k_2 gives 700k2=70,000700 k_2 = 70,000, so k2=100k_2 = 100. Substituting k2=100k_2 = 100 yields the fixed constant k1=20,000k_1 = 20,000. Substituting n=2,000n = 2,000 into C=20,000+100nC = 20,000 + 100n yields 20,000+200,000=N220,00020,000 + 200,000 = \text{N}220,000.

Adım Adım Çözüm

1
Set up the partial variation equation
C=k1+k2nC = k_1 + k_2 n, where k1k_1 is the fixed cost and k2k_2 is the rate per copy
Partial variation consists of a constant term and a term that varies directly with the independent variable.
2
Substitute given values to form simultaneous equations
Equation 1: 70,000=k1+500k270,000 = k_1 + 500 k_2
Equation 2: 140,000=k1+1,200k2140,000 = k_1 + 1,200 k_2
Using the two given data points (n=500,C=70,000)(n=500, C=70,000) and (n=1,200,C=140,000)(n=1,200, C=140,000) creates a system of linear equations.
3
Solve for the variation constants k1k_1 and k2k_2
Subtracting Equation 1 from Equation 2 gives 70,000=700k2    k2=10070,000 = 700 k_2 \implies k_2 = 100.
Substituting k2=100k_2 = 100 into Equation 1 gives 70,000=k1+500(100)    k1=20,00070,000 = k_1 + 500(100) \implies k_1 = 20,000.
Determines the specific values for the fixed overhead and rate per copy.
4
Calculate the total cost for 2,0002,000 copies
C=20,000+100(2,000)=20,000+200,000=N220,000C = 20,000 + 100(2,000) = 20,000 + 200,000 = \text{N}220,000
Evaluates the completed formula C=20,000+100nC = 20,000 + 100n at n=2,000n = 2,000.

Anahtar Kavram

Partial Variation and Simultaneous Equations
Tahmini Süre:1m 30s
Soru 16Soru

The electrical resistance RR of a wire varies directly as its length LL and inversely as the square of its diameter dd. A wire of length 50 m50\text{ m} and diameter 2 mm2\text{ mm} has a resistance of 5 ohms5\text{ ohms}. Calculate the resistance (in ohms) of a wire made of the same material with a length of 80 m80\text{ m} and a diameter of 4 mm4\text{ mm}.

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Cevap: 2

Cevap

The resistance of the wire is 2 ohms2\text{ ohms}.
The variation model is R=kLd2R = \frac{k L}{d^2}. Substituting R=5 ohmsR = 5\text{ ohms}, L=50 mL = 50\text{ m}, and d=2 mmd = 2\text{ mm} gives 5=50k45 = \frac{50k}{4}, so k=0.4k = 0.4. Substituting k=0.4k = 0.4, L=80 mL = 80\text{ m}, and d=4 mmd = 4\text{ mm} yields R=0.4×8042=3216=2 ohmsR = \frac{0.4 \times 80}{4^2} = \frac{32}{16} = 2\text{ ohms}.

Adım Adım Çözüm

1
Set up the variation formula
R=kLd2R = \frac{k L}{d^2}
Resistance varies directly as length LL and inversely as the square of diameter dd.
2
Calculate the constant of variation kk
k=0.4k = 0.4
Substitute the initial values R=5R = 5, L=50L = 50, and d=2d = 2 into the variation equation.
3
Compute the new resistance RR
R=2 ohmsR = 2\text{ ohms}
Substitute k=0.4k = 0.4, L=80L = 80, and d=4d = 4 into the formula.

Anahtar Kavram

Direct and Inverse Joint Variation
Tahmini Süre:1m 30s
Soru 17Soru

The rate of heat transfer QQ across a building wall varies directly as the surface area AA of the wall and the temperature difference ΔT\Delta T between the interior and exterior, and inversely as the wall thickness dd. When the surface area is 4 m24\text{ m}^2, the temperature difference is 15C15^\circ\text{C}, and the thickness is 0.05 m0.05\text{ m}, the heat transfer rate is 1200 W1200\text{ W}. What is the heat transfer rate in watts when the surface area is 6 m26\text{ m}^2, the temperature difference is 20C20^\circ\text{C}, and the thickness is 0.08 m0.08\text{ m}?

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Cevap: 1500

Cevap

The heat transfer rate is 1500 W1500\text{ W}.
Establishing the variation constant k=1k = 1 using the initial given values and substituting the new parameters yields Q=1×6×200.08=1500 WQ = \frac{1 \times 6 \times 20}{0.08} = 1500\text{ W}.

Adım Adım Çözüm

1
Formulate the variation equation
Q=kAΔTdQ = \frac{k A \Delta T}{d}
Direct variation means multiplying by AA and ΔT\Delta T, while inverse variation means dividing by dd.
2
Calculate the constant of variation kk
k=1k = 1
Substituting Q=1200Q=1200, A=4A=4, ΔT=15\Delta T=15, and d=0.05d=0.05 gives 1200=60k0.05=1200k1200 = \frac{60k}{0.05} = 1200k, so k=1k = 1.
3
Compute the target heat transfer rate QQ
1500 W1500\text{ W}
Substituting k=1k=1, A=6A=6, ΔT=20\Delta T=20, and d=0.08d=0.08 gives Q=1×6×200.08=1200.08=1500 WQ = \frac{1 \times 6 \times 20}{0.08} = \frac{120}{0.08} = 1500\text{ W}.

Anahtar Kavram

Joint and Inverse Variation
Soru 18Soru

The maximum safe load LL supported by a horizontal wooden beam varies directly as its width ww and the square of its depth dd, and inversely as its length ll. A beam of width 6 cm6\text{ cm}, depth 10 cm10\text{ cm}, and length 4 m4\text{ m} can support a maximum safe load of 900 kg900\text{ kg}. What is the maximum safe load that can be supported by a beam of the same material having a width of 4 cm4\text{ cm}, depth 12 cm12\text{ cm}, and length 6 m6\text{ m}?

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Cevap: 576 kg576\text{ kg}

Cevap

576 kg576\text{ kg}
The relationship is modeled by L=kwd2lL = \frac{k w d^2}{l}. Using the initial parameters (w=6w=6, d=10d=10, l=4l=4, L=900L=900), we find k=6k = 6. Substituting w=4w=4, d=12d=12, and l=6l=6 into the equation gives L=6×4×1446=576 kgL = \frac{6 \times 4 \times 144}{6} = 576\text{ kg}.

Adım Adım Çözüm

1
Formulate the variation equation
L=kwd2lL = \frac{k \cdot w \cdot d^2}{l}
Direct variation means multiplying factors in the numerator, while inverse variation places the variable in the denominator.
2
Calculate the constant of variation kk using initial conditions
900=k61024    900=600k4=150k    k=6900 = \frac{k \cdot 6 \cdot 10^2}{4} \implies 900 = \frac{600 k}{4} = 150 k \implies k = 6
Substitute L=900L = 900, w=6w = 6, d=10d = 10, and l=4l = 4 to solve for kk.
3
Calculate the new load LL for the new dimensions
L=641226=641446=576 kgL = \frac{6 \cdot 4 \cdot 12^2}{6} = \frac{6 \cdot 4 \cdot 144}{6} = 576\text{ kg}
Substitute k=6k = 6, w=4w = 4, d=12d = 12, and l=6l = 6 into the variation formula.

Anahtar Kavram

Joint and Inverse Variation
Tahmini Süre:1m 30s
Soru 19Soru

If yy varies inversely as xx, and y=8y = 8 when x=3x = 3, what is the value of yy when x=6x = 6?

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Cevap: 4

Cevap

4
For inverse variation, y=kxy = \frac{k}{x}. Given y=8y = 8 when x=3x = 3, the constant of variation k=8×3=24k = 8 \times 3 = 24. Substituting x=6x = 6 gives y=246=4y = \frac{24}{6} = 4.

Adım Adım Çözüm

1
Set up the equation for inverse variation
y=kxy = \frac{k}{x}
Inverse variation means yy is inversely proportional to xx with constant kk.
2
Solve for the variation constant kk using y=8y = 8 and x=3x = 3
k=8×3=24k = 8 \times 3 = 24
Multiply both sides of the equation by xx.
3
Calculate yy when x=6x = 6
y=246=4y = \frac{24}{6} = 4
Substitute the constant k=24k = 24 and x=6x = 6 into the inverse variation equation.

Anahtar Kavram

Inverse Variation
Soru 20Soru

The electric power PP dissipated in a resistor varies directly as the square of the current II flowing through it. If a current of 3 A3\text{ A} produces a power of 45 W45\text{ W}, what is the power dissipated, in watts, when the current is 5 A5\text{ A}?

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Cevap: 125

Cevap

The power dissipated when the current is 5 A is 125 W.
Because electric power varies directly as the square of the current, the formula is P=kI2P = k I^2. Substituting the initial conditions gives 45=k(32)=9k45 = k(3^2) = 9k, so k=5k = 5. Evaluating at I=5 AI = 5\text{ A} gives P=5(52)=125 WP = 5(5^2) = 125\text{ W}.

Adım Adım Çözüm

1
Set up the variation equation
P=kI2P = k I^2
Power varies directly as the square of current.
2
Find the constant of variation kk
k=5k = 5
Substitute P=45P = 45 and I=3I = 3 into the variation equation: 45=k(32)    45=9k    k=545 = k(3^2) \implies 45 = 9k \implies k = 5.
3
Calculate the required power for I=5 AI = 5\text{ A}
P=125 WP = 125\text{ W}
Substitute k=5k = 5 and I=5I = 5 into P=kI2P = k I^2: P=5(52)=125P = 5(5^2) = 125.

Anahtar Kavram

Direct variation involving a squared quantity
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Direct, Inverse, Joint and Partial Variation Alıştırma Soruları — JAMB UTME | Examkin