1983 mathematics 31 to 40

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## 31. If a rod of length 250cm is measured as 255cm longer in error, what is the percentage error in measurement?

### A. 55 B. 10 C. 5 D. 4 E. 2

Solution:

Actual Value = 250cm

Measured Value = 255cm

percentage Error= (measured Value – Actual Value)/Actual Value x 100

= (255 – 250)/250 x 100 = 5/250 x 100

Percentage error = 2%

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## 32. If (2/3)^{m } ^{ }(3/4)^{n} = 256/729, find the values of m and n

### A. m = 4, n = 2 B. m = -4, n = -2 C. m = -4, n = 2 D. m = 4, n = -2 E. m = -2, n = 4

Solution:

(2/3)^{m} (3/4)^{n} = 256/729

(2^{m} /3^{m}) x (3^{n} /4^{n}) = 256/729

(2^{m}/4^{n}) x (3^{n}/3^{m}) = 256/729

(2^{m} ÷ 2^{2n}) x (3^{n} ÷ 3^{m}) = 256/729

2^{m-2n} x 3^{n-m} = 256/729

2^{m-2n} = 256 and 3^{n-m} = 729

2^{m-2n} = 2⁸ and 3^{n-m} = 3^{-6}

m – 2n = 8……..i

n – m = – 6……….ii

substitute n = – 6 + m in equation i

m – 2(-6 + m) = 8

m + 12 – 2m = 8

m – 2m = 8 – 12

– m = – 4

m = 4

Substitute m = 4 in equation ii

n – m = – 6

n – (4) = – 6

n = – 6 + 4

n = -2

Therefore, m = 4, n = -2

## 33. Without using tables find the numerical value of log_{ 7} 49 + log_{ 7} (1/7)

### A. 1 B. 2 C. 3 D. 7 E. 0

Solution:

log_{ 7} 49 + log_{ 7} (1/7) = log_{ 7} 7² + log_{ 7} 7^{-1}

= 2log_{ 7} 7 + (-1)log_{ 7} 7

= 2(1) – (1) = 2 – 1

= 1

## 34. Factorize completely 81a⁴ –16b⁴

### A. (3a + 2b)(2a – 3b)(9a² + 4b²)

B. (3a – 2b)(2a – 3b)(4a² – 9b²)

C. (3a – 2b)(3a + 2b)(9a² + 4b²)

D. (3a – 2b)(2a – 3b)(9a² + 4b²)

E. (3a – 2b)(2a – 3b)(9a² – 4b²)

Solution:

81a⁴ –16b⁴ = 3⁴a⁴ – 2⁴b⁴

= (3²a²)² – (2²b²)² = (9a²)² – (4b²)²

Using formula for difference of two squares

(a – b)(a + b)

= (9a² – 4b²)(9a² + 4b²)

= (3²a² – 2²b²)(9a² + 4b²)

= [(3a)² – (2b)²](9a² + 4b²)

(3a – 2b)(3a + 2b)(9a² + 4b²)

## 35. One interior angle of a convex hexagon is 170° and each of the remaining interior angles is equal to x. find x

### A. 120° B. 110° C.105° D. 102° E. 100°

Solution:

Hexagon is known to have 6 angles and 6 sides

The sum of interior angles of an n sided polygon = (2n – 4) x 90°

One of the angles has been given which is 170°, each of the remaining five angles is x°

(2n – 4) x 90 = 170 + 5x

(2 x 6 – 4) x 90 = 170 + 5x

(12 – 4) x 90 = 170 + 5x

8 x 90 = 170 + 5x

5x = 720 – 170 = 550

x = 550/5

x = 110°

## 36. PQRS is a cyclic quadrilateral in which PQ = PS. PT is a tangent to the circle and PQ makes an angle 50° with the tangent as shown in the figure below. What is the size of QRS?

### A. 50° B. 40° C. 110° D. 80° E. 100°

Solution:

Since PQ = PS and PQ makes an angle of 50° with the Tangent, PS also makes an angle of 50° with the Tangent.

SPQ = 180 – (50 + 50) = 80° (angle on a straight line)

QRS + SPQ = 180 (sum of opposite angles in a cyclic quadrilateral)

QRS = 180 – 80

= 100°

## 37. A ship H leaves a port P and sails 30km due South. Then it sails 60km due west. What is the bearing of H from P?

### A. 26° 34’ B.243° 26’ C.116° 34’ D.63° 26’ E.240°

Solution:

Solving for Tita

Tan @ = 60/30 = 2

@ = tan–¹ 2 = 63.43°

Therefore the bearing of H from P = 90 + 90 + 63.43 = 243.43°

= 243° 26′

## 38. In a sample survey of a university community, the following table shows the percentage distribution of the number of members per household. What is the median?

### A. 4 B. 3 C. 5 D. 4.5 E. None

Solution:

Form a cumulative frequency

No of members Per household | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |

Number of households | 3 | 12 | 15 | 28 | 21 | 10 | 7 | 4 |

Cumulative freq. | 3 | 15 | 30 | 58 | 79 | 89 | 96 | 100 |

From the table, there are 100 households as seen by the last cumulative frequency

Since the 100 is even,

Median= [(N/2)^{th} + (N/2 + 1)^{th}]/2

N = 100

Median = [(100/2)^{th} + (100/2 + 1)^{th}]/2

= [(50)^{th} + (50 + 1)^{th}]/2

= (50^{th} + 51^{st})/2

The 50th member is 4

The 51st member is still 4

So, Median = (4 + 4)/2 = 8/2 = 4

## 39. On a square paper of length 2.524375cm is inscribed a square diagram of length 0.524375cm. Find the area of the paper no covered by the diagram correct to 3 significant figures.

### A. 6.00cm² B. 6.10cm² C. 6.cm² D. 6.09cm² E. 4.00cm²

Solution:

Paper length = 2.524375cm

Paper area = L x B = 2.524375 x 2.524375 = 6.372469cm²

Diagram length = 0.524375cm

Diagram area = L x B = 0.524375 x 0.524375 = 0.274969cm²

Area of the paper that is not covered by the Diagram = 6.372469cm² – 0.274969cm²

= 6.0975cm² = 6.10cm² (3 s.f)

## 40. If f(X) = 1/(x-1) + (x-1)/(x²-1). Find f(1-x)

### A. 1/x + 1/(x+2) B. x + 1/(2x-1) C.-1/x – 1/(x-2) D. -1/x + 1/(x²-1)

Solution:

f(X) = 1/(x-1) + (x-1)/(x²-1) = 1/(x-1) + (x-1)/(x-1)(x+1)

= 1/(x-1) + 1/(x+1)

f(1-x) = 1/[(1-x)-1] + 1/[(1-x) +1]

= 1/-x + 1/(2-x) = -1/x – 1/(x-2)

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