1)The distance S metres of an object varies partly with time t seconds and partly with the square root of time. Given that s=16 when t=4 and s=48 when t=16,write an equation connecting s and t hence find s when t=9
2)A quantity P varies jointly as the square root of Q and as the inverse of the square of R. Determine the percentage change in P when Q is increased by 44% and R decreased by 28%
3)The mass of wire m grams (g) is partly constant and partly varies as the square of its thickness t mm when t=2mm,m=40g and when t=4mm,m=100g.Determine the value of m when t=7mm
4)The quantities P, Q and R are such that P varies directly as the square of Q and inversely as the square root of R.P=8 when Q=2 and R=9.Determine the equation connecting P, Q and R
5)Two variables A and B are such that A varies partly as the square root of B.Given that A=8.58 when B=1.69 and A=9.52 when B=1.96,find the law connecting A and B hence determine A when B=2.89
6)A quantity Q is partly constant and partly varies as the square of E.When E=2,Q=560 and when E =3,Q=510.Find an equation connecting Q and E hence determine E given Q=537.5
7)A quantity P is partly constant and partly varies as the inverse of Q^2.When Q=2,P=49 and 1/4 ,and when Q=3P=49 and 2/3.Determine the value of P when Q=5
8)A quantity y varies inversely as the square of x.The difference between the value of y when x=6 and when x=10 is 16.Find the law connecting x and y
9)Three quantities P, Q and R are such that P varies as the square of Q and inversely as the square root of R.Given that P=20 when Q=5 and R=9,find P when Q=7 and R=25
10)H varies as V and inversely as the square of r.Find the percentage change in H if V is increased by 20% and at the same time,r is increased by 50%
11)Two quantities M and N are such that M varies partly as N and partly as the square of N.Determine the relationship between M and N given that M=1050,N=10 and when M=1272,N=12
12)A cone is made by cutting off a sector from a circle and gluing the straight edges of the sector. The cone formed has a slant height of 14 cm and circular base of perimeter 11cm;
a)Determine the value of the angle
b)The radius of the cones circular base
c)The height of the cone
d)The cone is cut uniformly on a horizontal plane 1cm below the apex.Calculate the slant height of the frustum formed correct to 2d.p
13)A right conical frustum of base radius 7cm and top radius 3.5cm,and height of 6cm is stuck onto a cylinder of base radius 7cm and height of 5cm which is further attached to a hemisphere to form a closed solid;
a)Find the volume of the solid
b)Find the surface area of the solid
14)The ends of the roof of a workshop are segment of a circle of radius 10m.The roof is 20m long.The angle of the centre is 120°.Calculate;
a)The area of one end of the roof
b)The area of the curved surface of the roof
c)What would be the cost to the nearest shilling of covering the two ends and the curved surface with galvanized iron sheet costing 80shs per square metre
15)A wooden drum closed at the top and bottom has a top radius of 28cm and bottom radius 21cm,its height is 90cm.The drum is covered with leather excluding the top and bottom.Calculate;
a)The volume of the drum in litres
b)The area of the leather covering the drum excluding the top and bottom.
1) The equation connecting s and t is: s = kt + a√t
Using the given values of s and t:
16 = k(4) + a√4 -> 16 = 4k + 2a
48 = k(16) + a√16 -> 48 = 16k + 4a
Solving these two equations simultaneously, we get:
k = 4 and a = 4
Thus, the equation connecting s and t is: s = 4t + 4√t
Substituting t = 9 into the equation, we can find s:
s = 4(9) + 4√9 -> s = 36 + 12 -> s = 48
2) The equation connecting P, Q, and R is: P = k√Q/R^2
When Q is increased by 44%: Q' = Q + 0.44Q = 1.44Q
When R is decreased by 28%: R' = R - 0.28R = 0.72R
Substituting these new values into the equation, we get:
P' = k√(1.44Q)/(0.72R)^2
Calculating the percentage change in P:
((P' - P)/P) * 100 = ((k√(1.44Q)/(0.72R)^2 - k√Q/R^2)/k√Q/R^2) * 100
3) Based on the given data, we can find the values of k and a in the equation: m = kt + a√t
Substituting t = 2 and t = 4 into the equation, we get:
40 = 2k + 2a (equation 1)
100 = 4k + 4a (equation 2)
Solving these two equations simultaneously, we get:
k = 10 and a = 10
Substituting t = 7 into the equation, we can find m:
m = 10(7) + 10√7 -> m = 70 + 10√7
4) The equation connecting P, Q, and R is: P = kQ^2/√R
Using the given values of P, Q, and R:
8 = k(2)^2/√9 -> 8 = 4k/3 -> k = 6
Thus, the equation connecting P, Q, and R is: P = 6Q^2/√R
Now we can find P when Q = 7 and R = 25:
P = 6(7)^2/√25 -> P = 6(49)/5 -> P = 294/5
5) The equation connecting A and B is: A = kB^0.5
Using the given values of A and B:
8.58 = k(1.69)^0.5 -> 8.58 = 1.3k
9.52 = k(1.96)^0.5 -> 9.52 = 1.4k
Dividing the second equation by the first equation, we get:
9.52/8.58 = 1.4k/1.3k -> 1.109 = 1.077k/k -> 1.109 = 1.077
Thus, the equation connecting A and B is: A = 1.077B^0.5
Substituting B = 2.89 into the equation, we can find A:
A = 1.077(2.89)^0.5 -> A = 1.077(1.7) -> A ≈ 1.83
6) The equation connecting Q and E is: Q = kE^2
Using the given values of Q and E:
560 = k(2)^2 -> 560 = 4k -> k = 140
510 = k(3)^2 -> 510 = 9k -> k = 56.6667
Since the values of k don't match, the equation is not consistent. Please check the given values again.
7) The equation connecting P and Q is: P = k/Q^2
Using the given values of P and Q:
49 = k/4^2 -> 49 = k/16 -> k = 784
When Q = 5:
P = 784/5^2 -> P = 784/25 -> P = 31.36
8) The equation connecting x and y is: y = k/x^2
Using the given difference in y when x = 6 and x = 10:
y(6) - y(10) = k/6^2 - k/10^2 = k/36 - k/100 = 16
Simplifying and solving for k:
k(100 - 36)/(36*100) = 16 -> 64k/3600 = 16 -> k = 900
Thus, the law connecting x and y is: y = 900/x^2
9) The equation connecting P, Q, and R is: P = kQ^2/√R
Using the given values of P, Q, and R:
20 = k(5)^2/√9 -> 20 = 25k/3 -> k = 240/25 = 9.6
Now we can find P when Q = 7 and R = 25:
P = 9.6(7)^2/√25 -> P = 9.6(49)/5 -> P = 470.4/5 = 94.08
10) The equation connecting H, V, and r is: H = kV/r^2
When V is increased by 20%: V' = V + 0.2V = 1.2V
When r is increased by 50%: r' = r + 0.5r = 1.5r
Substituting these new values into the equation, we get:
H' = k(1.2V)/(1.5r)^2
Calculating the percentage change in H:
((H' - H)/H) * 100 = ((k(1.2V)/(1.5r)^2 - kV/r^2)/(kV/r^2)) * 100
11) The equation connecting M and N is: M = kN + aN^2
Using the given values of M and N:
1050 = k(10) + a(10)^2 -> 1050 = 10k + 100a
1272 = k(12) + a(12)^2 -> 1272 = 12k + 144a
Solving these two equations simultaneously, we get:
k = 42 and a = 6
Thus, the relationship between M and N is: M = 42N + 6N^2
12) a) To determine the value of the angle, we can use the formula for the perimeter of a circle: P = 2πr, where P = 11 cm.
11 = 2πr -> r = 11/(2π)
The length of the curved edge of the sector is equal to the circumference of the circle, which is 2πr.
Since the angle of the sector is 120°, the curved edge is equal to 1/3 of the circumference of the entire circle.
Therefore, the curved edge of the sector is: (1/3)(2πr) = (1/3)(2π)(11/(2π)) = (11/3) cm
b) The radius of the cone's circular base is equal to the radius of the large circle, which is 10 cm.
c) The height of the cone is equal to the slant height of the frustum, which is given as 14 cm.
d) To find the slant height of the frustum, we can use the Pythagorean theorem. Let x be the height of the frustum:
(x + 1)^2 + (14)^2 = (10)^2
x^2 + 2x + 1 + 196 = 100
x^2 + 2x + 97 = 0
Using the quadratic formula, we get:
x = (-2 ± √(2^2 - 4(1)(97)))/(2(1))
x ≈ -9.14 or x ≈ 7.14 (we ignore the negative value)
Therefore, the slant height of the frustum is approximately 7.14 cm.
13) a) The volume of the solid can be calculated by finding the sum of the individual volumes of the cone, cylinder, and hemisphere.
Volume of cone = (1/3)π(14^2)(14) = (1/3)(196π)(14) = 8232π/3 cm^3
Volume of cylinder = π(7^2)(5) = 245π cm^3
Volume of hemisphere = (2/3)π(7^3) = (2/3)(343π) = 686π/3 cm^3
Volume of solid = (8232π/3) + (245π) + (686π/3) = (8232π + 735π)/3 = 8967π/3 cm^3
b) The surface area of the solid can be calculated by finding the sum of the individual surface areas of the cone, cylinder, and hemisphere.
Surface area of cone = π(14)(14) + π(14)(14) = 4π(14^2) = 784π cm^2
Surface area of cylinder = (2π(7)(5)) + (π(7^2)) = (70π + 49π) = 119π cm^2
Surface area of hemisphere = 2π(7^2) = 98π cm^2
Surface area of solid = 784π + 119π + 98π = 1001π cm^2
14) a) The area of one end of the roof is equal to the area of a segment of a circle.
To find the area of the segment, we can use the formula: (θ/360)πr^2, where θ is the angle of the center, which is 120°, and r = 10 m.
Area of the end of the roof = (120/360)π(10^2) = (1/3)(100π) = 100π/3 m^2
b) The area of the curved surface of the roof is equal to the lateral surface area of a cone.
Using the formula: πrs, where r = 10 m and s = slant height = 20 m.
Area of the curved surface of the roof = π(10)(20) = 200π m^2
c) To find the cost of covering the two ends and the curved surface, we need to find the total area and multiply it by the cost per square meter.
Total area = 2(area of end of the roof) + (area of curved surface)
Total area = 2(100π/3) + 200π
Total area ≈ 660.8 m^2
Cost = 660.8 * 80 = 52864 shs. Answer: 52864 shillings