FORMULA AND VALUE OF EQUATION

QUESTION

1. (a) Transpose the following formula to make v the subject
(b) Solve the following equation to find the value of x:
(c) In the formula the value of
θ = 58, V = 255, R = 0.1
and L = 0.5. Find the corresponding value of t.
(d)
⎟
1
h
L
L
ω =
⎛
ln –
⎝
⎜
0
34 85
θ = Ve
1
Find L if
ω = –2.6, L
23
..
()
=
⎞
⎠
+x
0
= 16 and h = 1.5.
2. (a) Use polynomial long division to determine the quotient when
3 5 10 4 3 1
xx x x– ++ + is divided by
32
(b) Show, by polynomial long division that
2
––
3125
2
xx x
x
f
uv
Rt
L
–
32
2
Teesside University Open Learning
(Engineering)
uv
=
,
3
–
–
+
10
15
–
+
=+
()
+
xx
x
© Teesside University 2011
3. A ball is thrown down at 72 km h
speed from the top of a building. The
building is 125 metres tall. The distance travelled before it reach the
ground is as follows,
where: initial velocity m s
accele
u
g
0
–1
sut gt=+
=
()
–1
= rration due to gravity 10 m s
time
=
()
ts..
4
1
2
0
2
(a) Find the time for the ball to drop one fifth of the height of the
building.
(b) Find the time for the ball to reach the ground.
()
–2
4. Water fills a tank at a rate of 150 litres during the first hour, 350 litres
during the second hour, 550 litres during the third hour and so on. Find the
number of hours necessary to fill a rectangular tank 16 m × 9 m × 9 m.
5. (a) A firm starts work with 110 employees for the 1st week. The number
of the employees rises by 6% per week. How many persons will be
employed in the 20th week if the present rate of expansion continues.
(b) A contractor hires out machinery. In the first year of hiring out one
piece of equipment the profit is £6000, but this diminishes by 5% in
successive years. Show that the annual profits form a geometric
progression and find the total of all the profit for the first 5 years.

SOLUTION

1. a)

f = uv/(u+v)

Transposing to make v the subject, we get,

ð f = uv/ (u+v)

ð f (u+v) = uv

ð fu+ fv = uv

ð v(u-f) = fu

ð v = fu/ (u-f)

ANS: v= fu/ (u-f)

b)

(3.4)^2x+3 = 8.5

Taking log on both sides, we get,

ð (2x+3) ln3.4 = ln8.5

ð (2x+3) = ln8.5/ln3.4

ð 2x+3 = 1.7487

ð x = (1.7487-3)/2

ð x = -0.62565

ANS: x= -0.62565

c)

θ = Ve^{-Rt/ L}

Where, θ = 58, V = 255, R = 0.1, and L = 0.5.

ð  58 = 255 e^{-.01 t/0.5}

ð  e^-t/5 = 4.3966 (approx)

Taking log on both sides, we get,

ð  -t/5 = ln(4.3966)

ð  -t = 7.40415

ð  t = -7.40415

ANS: t= -7.40415

d)

ω = ln(L/Lo – 1)

h

where, ω = -2.6, Lo = 16 and h = 1.5

ð  -2.6 = ln(L/16 -1)

1.5

ð  -3.9 = ln(L/16 -1)

Taking e on both sides, we get,

ð  e^-3.9 = L/16 -1

ð  0.02024 = L/16 -1

ð  1.02024 = L/16

ð  L = 16.32384

ANS: L= 16.32384

2. a)

3x³ – 5x²+10x+4 is divided by 3x+1

3x+1    3x³ – 5x²    +10x +4    x² -2x +4

3x³ + x²

– 6x² +10x +4

– 6x² – 2x

12x +4

12x +4

0

QUOTIENT: x² -2x +4

b)

x-2       x³ -3x² +12x -5      x² -x +10

x³ -2x²

–  x² +12x -5

–  x² +2x

10x -5

10x -20

15

Now, since Dividend = Quotient * Divisor + Remainder

ð  x³ -3x² +12x -5 = (x-2) * (x² -x +10) + 15

ð  x³ -3x² +12x -5 = (x² -x +10) +   15

                                    x-2                                    x-2

Hence shown.

3. A ball is thrown down at 72km/hr =20m/s

Height of the building =125m

Distance travelled

S= Uot+ (1/2)gt²

a)      S= 1/5 th of 125 = 25 m

Uo= 20m/s

ð   25= 20*t + 10*t²/2

ð  t² +4t -5 =0

ð  t² -t +5t -5 =0

ð  t (t-1) +5(t-1) =0

ð  t= 1, -5

Since, t cannot be negative, t= 1sec

Thus, time = 1sec

b)      S= 125m

125 = 20t +10*^t2/2

t² +4t -25 =0

solving quadratic equation,

t = -4 (+-) (16+100)^1/2 /2

t= -4 – (116)^1/2/2, -4+(116)^1/2/2

Since t cannont be negative,

t= 3.385sec (approx)

Thus, time to reach the ground is 3.385 sec

4. Amount of water filled in 1^st hour (a) = 150 ltr.

Amount of water filled in 1^st hour (a) = 350 ltr.

Amount of water filled in 1^st hour (a) = 550 ltr.

So on..

Volume of the rectangular tank is = length * breadth * height

= 16 * 9 * 9 = 1296m3

Amount of water to completely fill the rectangular tank (S) = 1296000 ltr.        [1m3=1000ltr]

Since the amount of water fills with a common difference (d) of 200ltr/hr, it follows arithmetic progression.

Thus using the formula,

S= (n/2)*[2a+ (n-1)d]

ð  1296000 = [300+ (n-1)*200] * (n/2)

ð  200n² +100n -2592000 = 0

ð  2n²+n-25920=0

ð  n = (-1 (+-) (1+4*2*25920)^1/2) /4

ð  n= 113.59, -114.09

Since n cannot be negative, time is 113.59hrs

• a) Number of employees for the 1^st week = 110

Rate of increase = 6 % = 0.06 (r)

Number of employees in 2^nd week

= 110+ 6%of 110 = 110 (1+ r)

Similarly, number of employees in 3^rd week

= 110 (1+r) + 110 (1+r) * r

= 110 (1+r)²

Thus, in the 20^th week, number of employees will be

= 110 (1+r)¹⁹

= 110 (1.06)¹⁹

= 332.82

= 333 (approx)

          ANS: Number of employees in the 20^th week are 333

b) Profit in 1^st year (p) = 60,000

Rate of decrease = 5% = 0.05 (r)

Total profit at the end of 5 years

= p + p(1-r) + p(1-r)² + p(1-r)³ + p(1-r)⁴

= p [1 + (1-r) + (1-r)² + (1-r)³ + (1-r)⁴ ]

Thus, it forms a geometric series with common ratio = 1-r

= p [1-(1-r)⁵]/[1-(1-r)]

= 60,000 * (1-0.95⁵) / 0.95

= 14287.5 (approx)

ANS: The total of all the profit for the first 5 years is £14287.5

GF64

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