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.

Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. Problems on Finding Derivative of a Function

    Mar 29, 24 12:11 AM

    Problems on Finding Derivative of a Function

    Read More

  2. How to Solve Age Problems with Ratio

    Mar 28, 24 02:01 AM

    How to Solve Age Problems with Ratio

    Read More

  3. AP Calculus BC Integration of Rational Functions by Partical Fractions

    Mar 26, 24 11:25 PM

    AP Calculus BC Integration of Rational Functions by Partical Fractions (Part - 1)

    Read More