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1984-15: P varies directly as the square of q and inversely as r. If p=36 when q=3 and r=4, find p when q=5 and r=2
Solutions
p\propto \frac { { q }^{ 2 } }{ r } \\ p=\frac { k{ q }^{ 2 } }{ r } \\ when\quad p=36,\quad q=3\quad and\quad r=4\\ 36=\frac { 9k }{ 4 } \\ k=16\\ \therefore \quad p=\frac { 16{ q }^{ 2 } }{ r } \\ when\quad q=5,\quad r=2\quad and\quad k=16\\ p=\frac { 16\times 25 }{ 2 } \\ p=200
1985-16: In a restaurant, the cost of providing a particular type of food is partly constant and partly inversely proportional to the number of people. If the cost per head for 100 people is 30k and the cost for 40 people is 60k. Find the cost for 50 people.
Solution
Let the cost be C, and the number of people be N
C=A+\frac { B }{ N } \quad where\quad A\quad and\quad B\quad are\quad constants\\ 30=A+\frac { B }{ 100 } \quad (if\quad 100\quad people\quad cost\quad 30k)\\ 300=100A+B\quad ---------eq(1)\\ 60=A+\frac { B }{ 40 } \quad (if\quad 40\quad people\quad cost\quad 60k)\\ 2400=40A+B\quad ---------eq(2)\\ by\quad subtracting\quad eq(2)\quad from\quad eq(1)\\ 600=60A\\ A=10\\ put\quad A=10\quad into\quad eq(2)\\ 2400=40(10)+B\\ 2400=400+B\\ B=2000\\ \therefore \quad C=10+\frac { 2000 }{ N } \\ where\quad N=50,\quad B=2000,\quad A=10\\ C=10+\frac { 2000 }{ 50 } \\ C=10+40\\ C=50\\ \therefore \quad the\quad cost\quad for\quad 50\quad people\quad is\quad 50k
1983-14: Y varies partly as the square of x and partly as the inverse of the square root of x. Write down the expression for y if y=2 when x=1 and y=6 when x=4
Solution
y=A{ x }^{ 2 }+\frac { B }{ \sqrt { x } } \quad where\quad A\quad and\quad B\quad are\quad constant\\ if\quad y=2\quad and\quad x=1\\ 2=A+B---------eq(1)\\ if\quad y=6\quad and\quad x=4\\ 6=16A+\frac { B }{ 2 } \\ 12=32A+B------eq(2)\\ subtracting\quad eq(1)\quad from\quad equation(2)\\ 10=31A\\ A=\frac { 10 }{ 31 } \\ from\quad eq(2)\\ 2=A+B\\ 2=B+\frac { 10 }{ 31 } \\ B=\frac { 52 }{ 31 } \\ \therefore \quad Y=\frac { 10 }{ 31 } { x }^{ 2 }+\frac { 52 }{ 31\sqrt { x } }
1983-23: If w varies directly as v and u varies directly as { w }^{ 3 }, find the relationship between u and v given that u=1 when v=2
Solution
w\propto v,\quad u\propto { w }^{ 3 }\\ w={ k }_{ w }v\quad ------eq(1)\\ u={ k }_{ u }{ w }^{ 3 }\quad ------eq(2)\\ w=\sqrt [ 3 ]{ \frac { u }{ { k }_{ u } } } \\ equating\quad eq(1)\quad and\quad eq(2)\\ { k }_{ w }v=\sqrt [ 3 ]{ \frac { u }{ { k }_{ u } } } \\ { { k }_{ w } }^{ 3 }{ v }^{ 3 }=\frac { u }{ { k }_{ u } } \\ { k }_{ u }{ { k }_{ w } }^{ 3 }{ v }^{ 3 }=u\\ { k }_{ u }{ { k }_{ w } }^{ 3 }=k
operations between constants will also give constant, let’s call it k
u=k{ v }^{ 3 }\quad \\ where\quad u=1\quad and\quad v=2\\ 1=k{ (2) }^{ 3 }\\ 1=8k\\ k=\frac { 1 }{ 8 } \\ \therefore \quad u=\frac { { v }^{ 3 } }{ 8 }
1978-8: The number of telephone calls N between two cities A and B varies directly as the population { P }_{ A }, { P }_{ B } in A and B respectively and inversely as the square of the distance D between A and B. Write the equation of the relationship.
Solution
Number\quad of\quad telephone\quad calls=N\\ Population\quad of\quad city\quad A={ { P }_{ A } }\\ Population\quad of\quad city\quad B={ P }_{ B }\\ Distance\quad between\quad city\quad A\quad and\quad city\quad B=D\\ N\propto { P }_{ A }{ P }_{ B }\\ N\propto \frac { 1 }{ { D }^{ 2 } } \\ \therefore \quad N\propto \frac { { P }_{ A }{ P }_{ B } }{ { D }^{ 2 } } \\ N=\frac { K{ P }_{ A }{ P }_{ B } }{ { D }^{ 2 } } \quad where\quad k\quad is\quad proportionality\quad constant
1979-11: x is directly proportional to y and inversely proportional to z. If x=9 when y=24 and z=8, what is the value of x when y=5 and z=6?
Solution
x\propto \frac { y }{ z } \\ x=\frac { ky }{ z } \quad where\quad k\quad is\quad the\quad proportionality\quad constant\\ 9=\frac { 24k }{ 8 } \\ k=3\\ x=\frac { 3y }{ z } \quad (relationship\quad between\quad x,\quad y\quad and\quad z)\\ when\quad y=5\quad and\quad z=6\\ x=\frac { 3\times 5 }{ 6 } \\ x=2.5\\
1986-16: If x varies directly as { y }^{ 3 } and x=2 when y=1. Find x when y=5
Solution
x\propto { y }^{ 3 }\\ x=k{ y }^{ 3 }\\ when\quad x=2\quad and\quad y=1\\ 2=k\\ \therefore \quad k=2\\ x=2{ y }^{ 3 }\\ when\quad y=5\\ x=2\times { (5) }^{ 3 }\\ x=250
1987-19: If p varies inversely as v and v varies directly as { r }^{ 2 }, find the relationship between p and r given that r=7, when p=2
Solution
p\propto \frac { 1 }{ v } ,\quad v\propto { r }^{ 2 }\\ p=\frac { k }{ v } \\ \therefore \quad v=\frac { k }{ p } \quad -------eq(1)\\ v=k{ r }^{ 2 }\quad -----------eq(2)\\ \frac { k }{ p } =k{ r }^{ 2 }
note: when a constant operate on another constant, it just becomes another constant
\frac { 1 }{ p } =k{ r }^{ 2 }\\ when\quad p=2\quad and\quad r=7\\ 2=\frac { k }{ { 7 }^{ 2 } } \\ 2=\frac { k }{ 49 } \\ k=98\\ p=\frac { 98 }{ { r }^{ 2 } }
1988-14: If x varies inversely as the cube root of y and x=1 when y=8. Find y when x=3
Solution
x\propto \frac { 1 }{ \sqrt [ 3 ]{ y } } \\ x=\frac { k }{ \sqrt [ 3 ]{ y } } \\ when\quad x=1\quad and\quad y=8\\ 1=\frac { k }{ \sqrt [ 3 ]{ 8 } } \\ 1=\frac { k }{ 2 } \\ k=2\\ x=\frac { k }{ \sqrt [ 3 ]{ y } } \\ when\quad x=3\\ 3=\frac { 2 }{ \sqrt [ 3 ]{ y } } \\ \sqrt [ 3 ]{ y } =\frac { 2 }{ 3 } \\ y={ (\frac { 2 }{ 3 } ) }^{ 3 }\\ y=\frac { 8 }{ 27 }
1990-13: y varies inversely as { x }^{ 2 } and x varies directly as { z }^{ 2 }. Find the relationship between y and z if c is constant.
Solution
y\propto \frac { 1 }{ { x }^{ 2 } } ,\quad x\propto { z }^{ 2 }\\ y=\frac { { C }_{ xy } }{ { x }^{ 2 } } \quad (where\quad { C }_{ xy } is the constant of proportionality between x and y)
{ x }^{ 2 }=\frac { { C }_{ xy } }{ y } \\ x=\frac { \sqrt { { C }_{ xy } } }{ \sqrt { y } } \quad ----------eq(1)\\ x={ C }_{ xz }{ z }^{ 2 }\quad ----------eq(2)\quad (where\quad { C }_{ xz } is the constant of proportionality between x and z)
equating\quad both\quad equations\\ \frac { \sqrt { { C }_{ xy } } }{ \sqrt { y } } ={ C }_{ xz }{ z }^{ 2 }\\ \sqrt { y } =\frac { \sqrt { { C }_{ xy } } }{ { C }_{ xz }{ Z }^{ 2 } } \\ y=\frac { { C }_{ xy } }{ { { C }_{ xz } }^{ 2 }{ z }^{ 4 } } \\
like I mentioned earlier constants operate on each other to give a constant
y=\frac { c }{ { z }^{ 4 } } \\ \\ \\
This is as much as we can take on this topic. If you feel there’s any other question on this topic giving you problem feel free to drop it. Be rest assured that it will be solved. No one is above errors, notify us if you feel there was an error in any of the calculations
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