Approximation and Errors Questions and Answers - Form 3 Topical Mathematics

Questions 

1. A rectangular room has a length of 12.0 meters and width 8.0 meters. Find the maximum percentage error in estimating the perimeter of the room.
2. In this question, mathematical tables or calculators should not be used. The base and perpendicular height of a triangle measured to the nearest centimeters are 12cm and 8cm respectively;
   Find;
   1. The absolute error in calculating the are of the triangle
   2. The percentage error in the area, giving the answer to 1 decimal place
3. A rectangular plate has a perimeter of 28cm. determine the dimensions of the plate that give the maximum area
4. A wire of length 5.2m is cut into two pieces without wastage. One of the pieces is 3.08m long. What is the shortest possible length of the second piece?
5. The dimensions of a rectangle are 10cm and 15cm. If there is an error of 5% in each of the Measurements. Find the percentage error in the area of the rectangle.
6. Find the products of 17.3 and 13.8. Find also the percentage error in getting the product.
7. The mass of a metal is given as 14kg to the nearest l0g. Find the percentage error in this measurement.
8. Complete the table below for the functions y = cos x and y = 2 cos (x +30^°) for 0^° ≤ X ≤ 360^°
   
   
   1. On the same axis, draw the graphs of y = cosx and y = 2 cos (x + 30^°) for 0^°≤ X≤ 360^°
   2. 1. State the amplitude of the graph y = cos x^°
      2. State the period of the graph y = 2cos (x + 30^°)
   3. Use your graph to solve
      cos x = 2 cos (x + 30^°)
9. Given that 8≤ y ≤ 12 and 1 ≤ x ≤ 6, find the maximum possible value of:
   ^{(y + x)}/_{(y – x)} 

Answers

1. Maximum perimeter = 2(12.05 + 8.05) = 40.2cm
   Actual perimeter = 2(12.0 + 18.0)= 40.0cm
   Error = 40.2cm – 40.0cm = 0.2cm
   %error = ^{(0.2 x100)}/₄₀
   = 0.5%
2. A = ½ x 12 x 8 = 48
   1. Absolute error
      = ^{[ ½ x12.5 x 8.5 – ½ x 11.5 x 7.5 ]}/₂
      = 5
   2. % error = ⁵/₂₄ X 100%
      = 10.4%
3. A = L x W
   A = x (14−x) = 14x − x^{²
   dA}/_dx = 14 – 2x = 0
    14 = 2x,  »x = 7
   Maximum area = 7(14 -7)
   = 7x 7 = 49cm^²
4. 
   
   Shortest possible length of 2^nd piece
   = 5.15 – 3.085= 2.065m
5. Absolute error 10 ± 0.05 and 15 ± 0.05
   Max area = 10..5 x 15.05
   Min area = 9.95 x 14.95 = 148.7525
   a.e = ^{(150.2525 – 15 + 150 – 148.7525)}/₂
   = 1.25
   % error = ^1.25/₁₅₀ x 100
   = 0.8333%
6. 17.35 X 13.85 = 240.3
   17.35 X 13.75 = 237.2
   »17.3 X 13.8 = 238.7
   Max err            240.3 – 238.7 = 1.5
   Min err             238.7 – 237.2 = 1.6
   Max err            = ^{(1.6 + 1.5)}/₂ = ^3.1/₂ = 1.55
   Product            238.7 ± 1.55
   Last product     240
   Max err =        1.55
   Relative err =   1.55
                        »28.1%
   error = 1.55 x 100 = 0.6% 28.1
   Relative err = ^1.55/_238.7
7. 14 Kg to the nearest ¹⁰/₁₀₀₀ Kg
   A.E = 0.01
   % E = ^0.01/₁₄ x 100
   =0.07
8. 1. 
      
   2. 1. Amplitude of y = cos x is 1 unit
         And Y = 2cos (x +30) 2 units
      2. period of y = 2 cos (x + 30^°)
         =330^°
   3. Cos x = 2 cos (x + 30^°)
      x = 40^° ± 1
      x = 219^° ± 1
9. ^{(y + x)}/_{(y – x)} = ^{(12 + 6)}/_{(8 – 6)}
   = ¹⁸/₂
   = 9