## SOLVING WORD PROBLEMS IN EVALUATING FUNCTIONS

Example 1  :

The formula for finding the circumference C of a circle with radius r is C  =  2∏r.

Find :

a) The circumference of a circle of radius 4.2 cm.

b) The radius of a circle with circumference 112 cm.

c) The diameter of a circle with circumference 400 meters.

Solution :

(a)  Radius (r)  =  4.2 cm

Circumference of a circle C  =  2∏r

=  2 × 22/7 × 4.2

=  26.4 cm

So, circumference of a circle C is 26.4 cm.

(b)  Circumference C  =  112 cm

112  =  2 × 22/7 × r

r  =  17.8 cm

So, radius of a circle r is 17.8 cm.

(c)  Circumference C  =  400 m

400  =  2 × 22/7 × r

r  =  63.63 m

diameter (d)  =  2r

d  =  2(63.63)m

d  =  127.3 m

So, diameter of a circle r is 127.3 m.

Example 2  :

When a stone is dropped from the top of a cliff, the total distance fallen is given by the formula

D  =  1/2gt2

where D is the distance in meters and t is the time taken in seconds. Given that g  =  9.8 ms-2,

Find :

a) The total distance fallen in the first 2 seconds of fall

b) The height of the cliff, to the nearest metre, if the stone takes 4.8 seconds to hit the ground.

Solution :

(a)  Time t  =  2 seconds and g  =  9.8 ms-2

D  =  1/2gt2

D  =  1/2  (9.8)  22

=  1/2 (39.2)

D  =  19.6 m

So, the total distance D is 19.6 m.

Time t  =  4.8 seconds

D  =  1/2gt2

D  =  1/2  (9.8)  (4.8)2

=  (1/2) (225.79)

D  =  112.9 m

So, height of the cliff is 112.9 m

Example 3  :

When a car travels a distance d kilometers in time t hours, the average speed for the journey is given by the formula

s  =  d/t kmh-1

Find :

(a) The distance travelled by a car in 2 3/4 hours if its average speed is 80 kmh-1.

(b) The time taken, to the nearest minute, for a car travel 790 km at an average speed of 95 kmh-1

Solution :

(a)  t  =  11/4 hrs and s  =  80 kmh-1

s  =  d/t kmh-1

80  =  d/(11/4)

d  =  (880/4)

=  220 km

So, distance travelled by a car is 220 km.

(b)  d  =  790 km and s  =  95 kmh-1

s  =  (d/t) kmh-1

95  =  790/t

t  =  8.316 hrs

t  =  (8 + 0.316) hrs

t  =  8 hr (0.316 × 60) min

t  =  8 hr 19 min

So, time taken 8 hr 19 min

Example 4 :

A circle’ s area A is given by

A  =  ∏r2

where r is the length of its radius.

Find :

(a) the area of a circle radius 6.4 cm

(b) the radius of a circular swimming pool which has an area of 160 m2

Solution :

(a)  radius (r)  =  6.4 cm

Area of a circle A  =  ∏r2

A  =  (22/7) × 6.4 × 6.4

A  =  128.7 cm2

So, area of a circle is 128.7 cm2

(b)  Area  =  160 m2

Area of a circle A  =  ∏r2

160  =  (22/7) × r2

r2  =  50.90

r  =  7.14 m

So, the radius of a circular swimming pool is 7.14 m

Example 5 :

A cylinder of radius r and height h has volume given by

V  =  ∏r2h

Find :

(a) the volume of a cylindrical tin can of radius 8 cm and height 21.2 cm.

(b) the height of a cylinder of radius 6 cm and volume 120 cm3

(c) the radius, in mm, of a copper pipe of volume 470 cm3 and length 6 m

Solution :

radius (r)  =  8 cm and height (h)  =  21.2 cm

Volume of cylinder  =  ∏r2h

V  =  22/7 × 82 × 21.2

V  =  4264.2 cm3

So, the volume of a cylindrical tin is 4264.2 cm3

(b)  radius  =  6 cm and volume  =  120 cm3

Volume of cylinder  =  ∏r2h

120  =  22/7 × 62 × h

(840/792)  =  h

h  =  1.06 cm

So, the height of a cylinder is 1.06 cm

volume  =  470 cm3

V  =  470 cm3

height  =  6 m, h  =  600 cm

Volume of cylinder  =  ∏r2h

470  =  22/7 × r2 × 600

(470 × 7)/(22 × 600)  =  r2

r2  =  0.2492

r  =  0.499 cm

r  =  4.99 mm

So, the radius is 4.99 mm.

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