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##### Mathematics (2017) QUIZ STUDY

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Given T = { even numbers from 1 to 12 }

N = {common factors of 6, 8 and 12}

Find T ∩ N

T = {evenn numbers from 1 to 12}

N = {common factors of 6,8 and 12}

Find T ∩ N

T = {2, 4, 6, 8, 10, 12}

N = {2}

T ∩ N = {2} i.e value common to T & N

What is the next number in the series 2, 1, \(\frac{1}{2}\), \(\frac{1}{4}\)...

\(\frac{1}{3}\)

\(\frac{2}{8}\)

\(\frac{3}{7}\)

\(\frac{1}{8}\)

There are 4 terms in the series

Therefore the next number will be the 5th term

T

_{n}= ar\(^{n - 1}\) (formular for geometric series)

a = first term = 2

r = common rate = \(\frac{\text{next term}}{\text{previous term}}\) = \(\frac{1}{2}\)

n = number of terms

T

_{5}= 5th term = ?

T

_{5}= ar\(^{5 - 1}\)

= ar\(^4\)

= 2 \(\times\) (ar\(^{n - 1}\))

^{4}

= 2 \(\times\) \(\frac{1}{16}\)

= \(\frac{1}{8}\)

E

^{1}= {x: x is a multiple of 3}

E

^{2}= {x: x is a multiple of 4} and an integer is picked at random from U, find the probability that it is not in E

^{2}

E

_{1}= {3, 6, 9, 12, 15, 18}

E

_{2}= {4, 8, 12, 16, 20}

Probability of E

_{2}= \(\frac{5}{20}\) i.e \(\frac{\text{Total number in}E_2}{\text{Entire number in set}}\)

Probability of set E

_{2}= 1 ? \(\frac{5}{20}\)

= \(\frac{15}{20}\)

= \(\frac{3}{4}\)

The curved surface area of a cylinder 5cm high is 110cm^{2}. Find the radius of its base

π = \(\frac{22}{7}\)

2.6cm

3.5cm

3.6cm

7.0cm

110 = 2 \(\times\) \(\frac{22}{7}\) \(\times\) r \(\times\) 5

r = \(\frac{110 \times 7}{44 \times 5}\)

= 3.5cm

If two graphs Y = px^{2} + q and y = 2x^{2} - 1 intersect at x =2, find the value of p in terms of q

q - \(\frac{8}{7}\)

7 - \(\frac{q}{4}\)

8 - \(\frac{q}{2}\)

7 + \(\frac{q}{8}\)

^{2}+ q

Y = 2x

^{2}- 1

Px

^{2}+ q = 2x

^{2}- 1

Px

^{2}= 2x

^{2}- 1 - q

p = \(\frac{2x^2 - 1 - q}{x^2}\)

at x = 2

P = \(\frac{2(2)^2 - 1 - q}{2^2}\)

= \(\frac{2(4) - 1 -q}{4}\)

= \(\frac{8 - 1 - q}{4}\)

P = \(\frac{7 - q}{4}\)

Evaluate (\(\sin\)45^{o} + \(\sin\)30^{o} ) in surd form

\(\frac{\sqrt{3}}{2\sqrt{2}}\)

-3 - \(\frac{1}{2}\)

\(\frac{1}{2}\)-2

1 + \(\frac{\sqrt{2}}{2}\)

sin = \(\frac{1}{2}\)

\(\sin45 = \frac{1}{\sqrt{2}}\)

= \(\frac{2}{2}\)

= (sin45 + sin30)

= \(\frac{1}{\sqrt{2}} + \frac{1}{2}\)

= \(\frac{\sqrt{2}}{2}\) + \(\frac{1}{2}\)

= \(\frac{\sqrt{2} + 1}{2}\)

= \(\frac{1 + \sqrt{2}}{2}\)

\(\frac{dy}{dx}\) = \(1 \sin x + x \cos x\)

= \(\sin x + x \cos x\)

At x = \(\frac{\pi}{2}\)

= sin\(\frac{\pi}{r}\) + \(\frac{\pi}{2} \cos {\frac{\pi}{2}}\)

= 1 + \(\frac{\pi}{2}\) \(\times\) 10

= 1

If temperature t is directly proportional to heat h, and when t = 20^{o}C, h = 50 J, find t when h = 60J

24^{o}C

20^{o}C

34^{o}C

30^{o}C

t = ? h = 60

t = kh where k is constant

20 = 50k

k = \(\frac{20}{50}\)

k = \(\frac{2}{5}\)

when h = 60, t = ?

t = \(\frac{2}{5}\) \(\times\) 60

t = 24

^{o}C

make T subject of formula square both sides

M

^{2}= \(\frac{N^2SL}{T}\)

TM

^{2}= N

^{2}SL

T = \(\frac{N^2SL}{M^2}\)

Simplify 3 \(^{n + 1}\) \(\times\) 27 \(\frac{n + 1}{81^n}\)

3\(^{2n}\)

9

3^{n}

3 \(^{n + 1}\)

= 3\(^{n + 1}\) \(\times\) 3 \(\frac{3^{(n + 1)}}{3^{4n}}\)

= 3\(^{n - 1 + 3n + 3 - 4n}\)

= 3\(^{4n - 4n - 1 + 3}\)

= 3

^{2}

= 9

The locus of points at a fixed distance from the point Q is a circle with the given point Q at its centre

The locus of points equidistant from two points P and Q i.e line PQ is the perpendicular bisector of the segment determined by the points

Hence, The locus of a point which is equidistant from the line PQ forms a perpendicular line to PQ

N = {common factors of 6, 8 and 12} Find T, N

= { 2, 4, 6, 8,10, 12}

N = {common factors of 6, 8 and 12}

= {2} Find T n N = {2}

Given the quadrilateral RSTO inscribed in the circle with O as centre. Find the size angle x and given RST = 60^{o}

100^{o}

140^{o}

120^{o}

10^{o}

^{o}

RXT = 2 \(\times\) RST

(angle at the centre twice angle at the circumference)

RXT = 2 \(\times\) 60

= 120

^{o}

Find the sum of the range and the mode of the set of numbers 10, 9, 10, 9, 8, 7, 7, 10, 8, 10, 8, 4, 6, 9, 10, 9, 7, 10, 6, 5

16

14

12

10

Range = Highest Number - Lowest Number

Mode is the number with highest occurrence

10, 9, 10, 9, 8, 7, 7, 10, 8, 4, 6,, 9, 10, 9, 7, 10, 6, 5

Range = 10 - 4 = 6

Mode = 10

Sum of range and mode = range + mode = 6 + 10

= 16

\(\frac{1}{4}\), \(\frac{1}{8}\), \(\frac{1}{16}\),..........

? = arn ? 1

= \(\frac{a}{1}\) ? r

a = \(\frac{1}{4}\)

r = \(\frac{1}{8}\) · \(\frac{1}{4}\)

r = \(\frac{1}{s}\) \(\times\) \(\frac{4}{1}\)

= \(\frac{1}{2}\)

S = \(\frac{1 \div 4}{1}\) ? \(\frac{1}{2}\)

= \(\frac{1}{4}\) · \(\frac{1}{2}\)

= \(\frac{1}{4}\) \(\times\) \(\frac{2}{1}\)

= \(\frac{1}{2}\)

The base in which the operation was performed was

6

2

4

5

^{x}) when x = 2 is

^{o}

3x = 180

^{o}

x = 60

^{o}(exterior angle of the polygon)

angle = \(\frac{\text{total angle}}{\text{number of sides}}\)

60 = \(\frac{360}{n}\)

n = \(\frac{360}{60}\)

n = 6 sides

A cylindrical tank has a capacity of 3080m^{3}. What is the depth of the tank if the diameter of its base is 14m? Take pi = 22/7.

23m

25m

20m

22m

^{3}

base diameter = 14m

radius = \(\frac{\text{diameter}}{2}\)

= 7m

Volume of Cylidner = Capacity of cylinder

πr

^{2}h = 3080

\(\frac{22}{7}\) \(\times\) 7 \(\times\) 7 \(\times\) h = 3080

h = \(\frac{3080}{22 \times 7}\)

h = 20m

Simplify 4\(\sqrt{27}\) + 5\(\sqrt{12}\) - 3\(\sqrt{75}\)

7

-7

- 7\(\sqrt{3}\)

7\(\sqrt{3}\)

= 4\(\sqrt{3}\) \(\times\) 9 + 5\(\sqrt{3}\) \(\times\) 4 - 3\(\sqrt{3}\) \(\times\) 25

= 4 \(\times\) 3\(\sqrt{3}\) + 5 \(\times\) 2\(\sqrt{3}\) - 3 \(\times\) 5\(\sqrt{3}\)

= 12\(\sqrt{3}\) + 10\(\sqrt{3}\) - 15\(\sqrt{3}\)

= (12 + 10 - 15)\(\sqrt{3}\)

= 7\(\sqrt{3}\)

and the speed of the 2nd trip be 3x miles/hr

Speed = distance/time

? Time taken to cover a distance of 50 miles on the 1st trip

= \(\frac{50}{xhr}\)

time taken to cover a distance of 300 miles on the next trip

= \(\frac{300}{3xhr}\)

= \(\frac{100}{xhr}\)

?the new time compared with the old time is twice as much

In the figure, find x

40^{o}

55^{o}

50^{o}

60^{o}

^{o}

2x + 3x + 4x = 360

9x = 360

x = \(\frac{360}{9}\)

x = 40

^{o}

^{3}- 3x + 1 by 2x - 1

^{2}-x + 1

^{2}- x -1

^{2}+ x + 1

^{2}+ x -1

= N 32,000

Selling Price = N 400,000(Given)

Gain = Selling Price - Cost Price

= 400,000 - 300,000

= 80,000

% gain = \(\frac{\text{Gain}}{\text{Cost Price}}\) \(\times\) 100

= \(\frac{80,000}{320,000}\) \(\times\) 100

Gain % = 25%

It is a ten letter word = 10!

Since we have repeating letters, we have to divide to remove the duplicates accordingly. There are 4 Es, 2 Cs, 2 Ls

? there are

\(\frac{10!}{4!2!2!}\) ways to arrange

Evaluate \(\frac{0.00000231}{0.007}\) and leave the answer in standard form

3.3 x 10^{4}

3.3 x 10^{-3}

3.3 x 10^{-4}

3.3 x 10^{-8}

= \(\frac{231 \times 10^{-8}}{7 \times 10^{-3}}\)

= 33 \(\times\) 10\(^{-8 - (-3)}\)

= 33 \(\times\) 10\(^{- 8 + 3}\)

= 33 \(\times\) 10

^{-5}

If a rod 10cm in length was measured as 10.5cm, calculate the percentage error

5%

10%

8%

7%

approximated value of measurement = 10.5cm

% error = \(\frac{\text{Actual measurement - Approximated}}{\text{Actual measure}}\) \(\times\) 100

= \(\frac{10 - 10.5}{10}\) \(\times\) 100

= \(\frac{-0.5}{10}\) \(\times\) 100

ignore -sign i.e take absolute value

= \(\frac{0.5}{10}\) \(\times\) 100

= 5 %

Find the principal which amounts to ~~N~~ 5,500 at a simple interest in 5 years at 2% per annum

~~N~~ 4,900

~~N~~ 500

~~N~~ 4,700

~~N~~ 400

Amount = Principal + Simple Interest

I = \(\frac{PRT}{100}\)

R = rate, T = time

I = \(\frac{P \times 5 \times 2}{100}\)

I = \(\frac{10P}{100}\)

I = \(\frac{P}{10}\)

Amount A = P + I

5500 = P + \(\frac{P}{10}\)

Multiply through by 100

5500 = 10P + P

5500 = 11P

p = \(\frac{5500}{11}\)

p =

The pie chart shows the allocation of money to each sector in a farm. The total amount allocated to the farm is ~~N~~ 80 000. Find the amount allocated to fertilizer

~~N~~ 35, 000

~~N~~ 40,000

~~N~~ 25,000

~~N~~ 20,000

Total angle at a point = 3600

∴ To get the angle occupied by fertilizer we have,

40 + 50 + 80 + 70 + 30 + fertilizer(x) = 360

270 + x = 360

x = 360 -270

x = 90o

Total amount allocated to the farm

= ~~N~~ 80,000

∴Amount allocated to the fertilizer

= \(\frac{\text{fertilizer (angle) × Total amount}}{\text{total angle}}\)

= \(\frac{90}{360}\) × 80,000

= ~~N~~20,000

There are 2Ms and 2As and 2Es

Divide the number of repeating letters

= \(\frac{11!}{2!2!2!}\)

Since we have repeating letters, we have to divide to remove duplicates accordingly. There are 2A, 2C, 2I

? \(\frac{8!}{2! 2! 2!}\)

y is inversely proportional to x and y and 6 when x = 7. Find the constant of the variation

47

42

54

46

Y = 6, X = 7

Y = \(\frac{k}{x}\) where k is constant

6 = \(\frac{k}{7}\)

k = 42

In the diagram MN, PQ and RS are parallel lines. What is the value of the angle marked X?

123^{o}

170^{o}

117^{o}

137^{o}

MN = PQ = RS (parallel lines)

Label the angle in the lines

a = i (corresponding angles are equal)

b = x (corresponding angles are equal)

If |MN| = |RS|

If a = i

and a = 63 = i

a + b = 180 (Adjacent interior angles are supplementary i.e add to 180)

? i + x = 180

63 + x = 180

x = 180 - 63

x = 1170

and w = (3, 5). This means that the point p moves so that its distance from v and w are equidistance

\(\sqrt{(x ? x_1)^2 + (y ? y_1)^2}\) = \(\sqrt{(x ? x_2)^2 + (y ? y_2)^2}\)

\(\sqrt{(x -1)^2 + (y - 1)^2}\) = \(\sqrt{(x - 3)^2 + (y - 5)^2}\)

square both sides

(x - 1)

^{2}+ (y - 1)

^{2}= (x - 3)

^{2}+ (y - 5)

^{2}

x

^{2}- 2x + 1 + y2 - 2y + 1 = x

^{2}- 6x + 9 + y2 - 10y + 25

x

^{2}+ y

^{2}-2x -2y + 2 = x

^{2}+ y

^{2}- 6x - 10y + 34

Collecting like terms

x

^{2}- x

^{2}+ y

^{2}- y

^{2}- 2x + 6x -2y + 10y = 34 - 2

4x + 8y = 32

Divide through by 4

x + 2y = 8

Find -(x^{2} + 3x - 5)dx

\(\frac{x_3}{3}\) - \(\frac{3x_2}{2}\) - 5x + k

\(\frac{x_3}{3}\) - \(\frac{3x_2}{2}\) + 5x + k

\(\frac{x_3}{3}\) + \(\frac{3x_2}{2}\) - 5x + k

\(\frac{x_3}{3}\) + \(\frac{3x_2}{2}\) + 5x + k

-x^{n}dx = \(\frac{x_{n + 1}}{n + 1}\)

-dx = x + k

where k is constant

-(x^{2} + 3x - 5)dx

-x^{2} dx + -3xdx - -5dx

\(\frac{2_{2 + 1}}{2 + 1}\) + \(\frac{3x^{1 + 1}}{1 + 1}\) - 5x + k

\(\frac{x_3}{3}\) + \(\frac{3x_2}{2}\) - 5x + k

Sin 70

^{o}

x = 10 Sin 70

^{o}

= 9.3969

Hence, length of chord MN = 2x

= 2 \(\times\) 9.3969

= 18.79

= 19cm (nearest cm)

If m * n = [m_{n} - n_{m}] for m, n belong to R, evaluate - 3 * 4

3

4

5

6

m * n = \(\frac{m}{n}\) - \(\frac{m}{n}\)

m = - 3

n = 4

- - 3 \(\times\) 4 = \(\frac{-3}{4}\) - \(\frac{-4}{-3}\)

= \(\frac{3(-3) - (- 4 \(\times\) 4)}{12}\)

= \(\frac{- 9 + 16}{12}\)

= \(\frac{7}{12}\)

^{2}+ 12xy + y

^{}+ 3x + 3y - 18

^{}+ 2xy + y

^{}+ 3x + 3y - 18

x

^{}+ 2xy + 3x + y

^{}+ 3y -18

x

^{}+ 2xy - 3x + 6x + y

^{}-3y + 6y -18

x

^{}+ 2xy -3x + y

^{}-3y + 6x + 6y -18

x

^{}+ xy -3x + xy + y

^{}- 3y + 6x + 6y -18

x(x + y - 3) + y(x + y - 3) + 6(x + y - 3)

= (x + y - 3)(x + y + 6)

= (x + y + 6)(x + y -3)

p = s + \(\frac{sm^2}{nr}\)

^{2}

^{2}

p = s + ( 1 + \(\frac{m^2}{nr}\))

p = s (1 + \(\frac{nr + m^2}{nr}\))

nr \(\times\) p = s (nr + m

^{2})

s = \(\frac{nrp}{nr + m^2}\)

The operation * on the set R of real number is defined by x * y = 3x + 2y - 1, find 3* - 1

9

-9

6

-6

x * y is an operation on 3x + 2y - 1

Find 3A - 1

x = 3, y = -1

3 * - 1 on 3x + 2y - 1

3(3) + 2(-1) -1

= 9 - 2 - 1

= 6

Find the gradient of the line joining the points (3, 2) and (1, 4)

3-Feb

2-Jan

-1

3-Feb

Gradient of line joining points (3, 2), (1, 4)

Gradient = \(\frac{\text{Change in Y}}{\text{Change in X}}\)

= \(\frac{y_2 - Y_1}{x_2 - x_1}\)

(X1, Y1) = (3, 2)

(X2, Y2) = (1, 4)

Gradient = \(\frac{4 - 2}{1 + 3}\)

= \(\frac{2}{-2}\)

= -1

Simplify (3-64a^{3})\(^{?1}\)

4a

\(\frac{1}{8a}\)

8a

\(\frac{1}{4a}\)

(3-64a^{3})\(^{-1}\)

\(\frac{1}{(3-64a^3)}\)

= \(\frac{1}{4a}\)

If \(\frac{2 \sqrt{3} - \sqrt{2}}{\sqrt{3} + 2 \sqrt{2}}\) = m + n - 6,

find the values of m and n respectively

1, - 2

- 2, n = 1

\(\frac{-2}{5}\), 1

\(\frac{2}{3}\)

\(\frac{2 \sqrt{3} - \sqrt{2}}{\sqrt{3} + 2 \sqrt{2}}\)= m + n-6

\(\frac{2 \sqrt{3} - \sqrt{2}}{\sqrt{3} + 2 \sqrt{2}}\) x \(\frac{\sqrt{3} - 2 \sqrt{2}}{\sqrt{3} - \sqrt{2}}\)

\(\frac{2 \sqrt{3} (\sqrt{3} - 2 \sqrt{2}) - \sqrt{2}(\sqrt{3} - 2 \sqrt{2})}{\sqrt{3}(\sqrt{3} - 2 \sqrt{2}) + 2 \sqrt{2}(\sqrt{3} - 2 \sqrt{2})}\)

\(\frac{2 \times 3 - 4\sqrt{6} - 6 + 2 \times 2}{3 - 2 \sqrt{6} + 2 \sqrt{6} - 4 \times 2}\)

= \(\frac{6 - 4 \sqrt{6} - \sqrt{6} + 4}{3 - 8}\)

= \(\frac{0 - 4 \sqrt{6} - 6}{5}\)

= \(\frac{10 - 5 \sqrt{6}}{5}\)

= - 2 + -6

- m + n\(\sqrt{6}\) = - 2 + -6

m = - 2, n = 1

If α and β are the roots of the equation 3x^{2} + bx - 2 = 0. Find the value of \(\frac{1}{\alpha}\) + \(\frac{1}{\beta}\)

\(\frac{-5}{3}\)

\(\frac{-2}{3}\)

\(\frac{1}{2}\)

\(\frac{5}{2}\)

\(\frac{1}{\alpha}\) + \(\frac{1}{\beta}\) = \(\frac{\beta -\alpha}{\alpha \beta}\)

3x^{2} + 5x + 5x - 2 = 0.

Sum of root = α + β

Product of root = αβ

x^{2} + \(\frac{5x}{3}\) - \(\frac{2}{3}\) = 0

αβ = - \(\frac{-2}{3}\)

α + β = \(\frac{5}{3}\)

- \(\frac{\alpha + \beta}{\alpha \beta}\) = - \(\frac{\frac{5}{3}}{\frac{2}{3}}\)

= - \(\frac{2}{3}\) \(\times\) \(\frac{3}{3}\)

= \(\frac{5}{2}\)

Find the range of the following set of numbers 0.4, -0.4, 0.3, 0.47, -0.53, 0.2 and -0.2

1.03

0.07

0.03

1

0.4, -0.4, 0.3, 0.47, -0.53, 0.2, -0.2

Range is the difference between the highest and lowest value

i.e Highest - Lowest

- 0.53, -0.4, -0.2, 0.2, 0.3, 0.4, 0.47

0.47 is the highest

- 0.53 is the lowest

- = 0.47 - (- 0.53)

-0.47 + 0.53

= 1.0

Evaluate 1 - (\(\frac{1}{5}\) + 1\(\frac{2}{3}\)) + (5 + 1\(\frac{2}{3}\))

4

3

2\(\frac{2}{3}\)

3 \(\frac{2}{3}\)

1 - (\(\frac{1}{5}\) + 1\(\frac{2}{3}\)) + (5 + 1\(\frac{2}{3}\))

1 - (\(\frac{1}{5}\) \(\times\) \(\frac{5}{3}\)) + (5 + \(\frac{5}{3}\))

1 - \(\frac{1}{3}\) + \(\frac{20}{3}\)

= \(\frac{22}{3}\)

What is the product of 2x^{2} - x + 1 and 3 - 2x

4x^{3} - 8x^{2} + 5x + 3

-4x^{3} + 8x^{2} - 5x + 3

-4x^{3} - 8x^{2} + 5x + 3

4x^{3} + 8x^{2} - 5x + 3

(2x^{2} - x + 1) \(\times\) (3 - 2x);

3(2x^{2} - x + 1) - 2x (2x^{2} - x + 1)

6x^{2} - 3x + 3 - 4x^{3} + 2x^{2} - 2x

-4x^{3} + 8x^{2} -5x + 3