1983 MATHEMATICS 1 to 10

## 1. If M represents the median and D the mode of the measurements 5, 9, 3, 5, 8 then (M,D) is

A. (6,5) B. (5,8) C. (5,7) D. (5,5) E. (7,5)

Solution:

Arrange the measurements in either ascending order or descending order of magnitude

3, , 5, 5, 8, 9

Median = 5

Mode = 5

So, (M,D) = (5,5)

## 2. A construction company is owned by two partners X and Y and it is agreed that their profit will be divided in the ratio 4:5. at the end of the year. Y received #5,000 more than x. what is the total profit of the company for the year?

A. #20,000.00 B. #25,000.00 C. #30,000.00 D. #15,000.003 E. #45,000.00

Solution:

Let the total profit be “P”

4 + 5 = 9

X received 4/9 x P = (4P)/9

Y received 5/9 x P = (5P)/9

(5P)/9 = (4P)/9 + #5000

(5P)/9 = [4P + 9(5000)]/9

5P = 4P + 45000

5P – 4P = 45000

P = 45000

## 3. Given a regular hexagon, calculate each interior angle of the hexagon.

A. 60° B. 30° C. 120° D. 45° E. 135°

Solution:

A hexagon has 6 straight sides and 6 angles

The total internal angles of a simple hexagon is 720°

Let each of the internal angles be a.

a + a + a + a + a + a = 720

6a = 720

a = 720/6 = 120°

## 4. Solve the following equations

4x – 3 = 3x + y = 2y + 5x – 12

A. x=5, y= 2 B. x=2, y=5 C. x=-2, y=-5 D. x=5, y=-2 E. x=-5, y=-2

Solution:

4x – 3 = 3x + y………………..(1)

3x + y = 2y + 5x – 12……….(2)

4x – 3x – y = 3

3x – 5x + y – 2y = -12

x – y = 3………….(1)

-2x – y = -12…….(2)

x = 3 + y…………(3)

Substitute equa. 3 in equa. 2

-2(3 + y) – y = -12

-6 – 2y – y = -12

-3y = -12 + 6 = -6

y = -6/-3 = 2

Substitute y = 2 in equa. 1

x – (2) = 3

x = 3 + 2 = 5

x = 5, y = 2

## 5. If x = 1 is root of the equation x³ – 2x² – 5x + 6, find the other roots

A. -3 and 2 B. –2 and 2 C. 3 and –2 D. 1 and 3 E. –3 and 1

Solution: Use division method

So, (x – 1) and (x² – x – 6) are factors of x³ – 2x² – 5x + 6

x² – x – 6 = x² – 3x + 2x – 6 = x(x – 3) + 2(x – 3)

x² – x – 6 = (x – 3)(x + 2)

So, the other roots are x = 3 and -2

## If x is jointly proportional to the cube of y and the fourth power of z. In what ratio is x increased or decreased when y is halved and z is doubled?

A. 4:1 increase B. 2:1 increase C. 1:4 decrease D. 1:1 no change E. 3:4 decrease

Solution:

Let ~ be the proportionality sign.

x ~ y³ and x ~ z⁴

x ~ y³z⁴

x = y³z⁴k, where k is the constant

When y is halved and z is doubled

x ~ (y/2)³ and x ~ (2z)⁴

x ~ (y³/8)(16z⁴)

x = (y³/8)(16z⁴)k = (16ky³z⁴)/8

The ratio now (y³z⁴k) ÷ (16ky³z⁴)/8

= ky³z⁴ ÷ 2ky³z⁴ = 1:2

Therefore x increases by the Ratio 2:1

## 7. In the above figure PQR= 60°, QPR= 90°, PRS = 90°, RPS = 45°, QR= 8cm. Determine PS

A. 2√3cm B. 4√6cm C. 2√6cm D. 8√6cm E. 8cm

Solution:

Sin 60 = PR/8

PR = 8 x sin60

= 8 x √3/2 = 4√3

Cos 45 = (4√3)/PS

PS = (4√3)/Cos45

= 4√3 ÷ 1/√2

= 4√3 x √2 = 4√6

## 8. Given that cos z = L, where z is an acute angle find an expression for

(Cot z – cosec z) ÷ (sec z + tan z)

Solution:

Since it’s an acute angle, the hypotenuse is 1, adjacent is L, now to get the opposite of the triangle we have to use Pythagoras theorem

Opp = √(1² – L²) = √(1 – L²)

Sin z = √(1 – L²)

Cos z = L

Tan z = sin z/cos z

Tan z = √(1 – L²)/L

Cot z = cos z/Sin z

Cosec z = 1/sin z

Sec z = 1/cos z

(Cot z – cosec z) ÷ (sec z + tan z)

= [(cos z/sin z) – (1/sin z)] ÷ [(1/cos z) + (sin z/cos z)]

= [(cos z – 1)/sin z] ÷ [(1 + sin z)/cos z]

= [(L – 1)/√(1 – L²)] ÷ [1 + √(1 – L²)]/L

= [(L – 1)/√(1 – L²)] x [L/ 1 + √(1 – L²)

= [L(L – 1)]/[1 – L² + √(1 – L²)] = E

## 9. If 0.0000152 x 0.00042 = A x 10^{B}, where 1 ≤ A < 10, find A and B.

A. A= 9, B= 6.38 B. A= 6.38, B = -9 C. A= 6.38, B = 9 D. A= 6.38, B = -1 E. A= 6.38, B= 1

^{B}

Solution:

0.0000152 x 0.00042 = 1.52 x 10^{-5} X 4.2 x 10^{-4}

= 6.384 x 10^{-9}

A = 6.38, B = -9

Click here to learn more on standard form

## 10. If x + 2 and x – 1 are factors of the expressions lx³ + 2kx² + 24, find the values of l and k

A. l = -6, k = -9 B. l=-2,k= 1 C. l=-2,k=-1 D. l=0,k= 1 E. l=6,k= 0

Solution:

When x + 2 = 0, x = -2

I(-2)³ + 2k(-2)² + 24 = 0

-8I + 2k(4) + 24 = 0

-8I + 8k = -24

-I + k = -3………………(1)

When x -1 = 0, x = 1

I(1)³ + 2k(1)² + 24 = 0

I + 2k = -24………………..(2)

I = -24 – 2k………………….(3)

Substitute I = -24 – 2k in equa. 1

-I + k = -3

-(-24 – 2k) + k = -3

24 + 2k + k = -3

3k = -3 – 24 = -27

k = -27/3 = -9

Substitute k = -9 in equa. 2

I + 2k = -24

I + 2(-9) = -24

I – 18 = -24

I = -24 + 18 = -6

Click here to see 1983 Mathematics 11 to 20

Learn Mathematics on intellectsolver.com

## Leave a Reply