1

Simplify 10\(\frac{2}{5} - 6 \frac{2}{3} + 3\)

A

6\(\frac{4}{15}\)

B

6\(\frac{11}{15}\)

C

7\(\frac{4}{15}\)

D

7\(\frac{11}{15}\)

CORRECT OPTION:
b

10\(\frac{2}{5} - 6 \frac{2}{3} + 3\)

\(\frac{52}{5} - \frac{20}{3} + \frac{3}{1}\)

= \(\frac{156 - 100 + 45}{15}\)

\(\frac{156 + 45 - 100}{15}\)

= \(\frac{201 - 100}{15}\)

= \(\frac{101}{15}\)

= 6\(\frac{11}{15}\)

\(\frac{52}{5} - \frac{20}{3} + \frac{3}{1}\)

= \(\frac{156 - 100 + 45}{15}\)

\(\frac{156 + 45 - 100}{15}\)

= \(\frac{201 - 100}{15}\)

= \(\frac{101}{15}\)

= 6\(\frac{11}{15}\)

2

If 23_{x} = 32_{5}, find the value of x

A

7

B

6

C

5

D

4

CORRECT OPTION:
a

23_{x} = 32_{5}

2 \(\times x^1 + 3 \times x^0 = 3 \times 5^1 + 2 \times 5^0\)

= 2x + 3 = 15 + 2

2x + 3 = 17

2x = 17 - 3

2x = 14

x = \(\frac{14}{2}\)

x = 7

2 \(\times x^1 + 3 \times x^0 = 3 \times 5^1 + 2 \times 5^0\)

= 2x + 3 = 15 + 2

2x + 3 = 17

2x = 17 - 3

2x = 14

x = \(\frac{14}{2}\)

x = 7

3

The volume of a cube is 512cm^{3}. Find the length of its side

A

6cm

B

7cm

C

8cm

D

9cm

CORRECT OPTION:
c

volume of cube = L x L x L

512cm^{3} = L^{3}

L^{3} = 512cm^{3}

L = 3\(\sqrt{512}\)

L (512)^{\(\frac{1}{3}\)}

= (2^{9})^{\(\frac{1}{3}\)}

2^{3} = 8cm

512cm

L

L = 3\(\sqrt{512}\)

L (512)

= (2

2

4

If one student is selected at random, find the probability that he/she scored at most 2 marks

A

\(\frac{11}{18}\)

B

\(\frac{11}{20}\)

C

\(\frac{7}{22}\)

D

\(\frac{5}{19}\)

CORRECT OPTION:
b

at most 2 marks = 5 + 2 + 4 students = 11 students

probability(at most 2 marks) = \(\frac{11}{20}\)

probability(at most 2 marks) = \(\frac{11}{20}\)

5

Simplify: \(\sqrt{12} ( \sqrt{48} - \sqrt{3}\))

A

18

B

16

C

14

D

12

CORRECT OPTION:
a

\(\sqrt{12} ( \sqrt{48} - \sqrt{3}\))

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

= 2\(\sqrt{3} \times \sqrt{3} (4 - 1) 2\sqrt{9}(3) = 2 \times 3 \times 3 = 18\)

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

= 2\(\sqrt{3} \times \sqrt{3} (4 - 1) 2\sqrt{9}(3) = 2 \times 3 \times 3 = 18\)

6

Given that x > y and 3 < y, which of the following is/are true? i. y > 3 ii. x < 3 iii. x > y > 3

A

i

B

i and ii

C

i and iii

D

i, ii and iii

CORRECT OPTION:
c

x > y and 3 < y; then 3 < y means that y > 3

x > 3 to give the possible

x > y > 3

x > 3 to give the possible

x > y > 3

7

Three quarters of a number added to two and a half of the number gives 13. Find the number

A

4

B

5

C

6

D

7

CORRECT OPTION:
a

let the number be x

2\(\frac{1}{2}x + \frac{3}{4}x = 13\)

\(\frac{5}{2}x + \frac{3}{4}x = 13\)

multiply through by 4

4(\(\frac{5}{2}\))x + 4(\(\frac{3}{4}\))x = 13 x 4

2(5x) + 3x = 52

10x + 3x = 52

13x = 52

x = \(\frac{52}{13}\)

x = 4

2\(\frac{1}{2}x + \frac{3}{4}x = 13\)

\(\frac{5}{2}x + \frac{3}{4}x = 13\)

multiply through by 4

4(\(\frac{5}{2}\))x + 4(\(\frac{3}{4}\))x = 13 x 4

2(5x) + 3x = 52

10x + 3x = 52

13x = 52

x = \(\frac{52}{13}\)

x = 4

8

If x = {0, 2, 4, 6}, y = {1, 2, 3, 4} and z = {1, 3} are subsets of u = {x:0 \(\geq\) x \(\geq\) 6}, find x \(\cap\) (Y' \(\cup\) Z)

A

{0, 2, 6}

B

{1, 3}

C

{0, 6)

D

{9}

CORRECT OPTION:
c

x = {0, 2, 4, 6}; y = {1, 2, 3, 4}; z = {1, 3}

u = {0, 1, 2, 3, 4, 5, 6}

y' = {0, 5, 6}

to find x \(\cap\) (Y' \(\cup\) Z)

first find y' \(\cup\) z = {0, 1, 3, 5, 6}

then x \(\cap\) (Y' \(\cup\) Z) = {0, 6}

u = {0, 1, 2, 3, 4, 5, 6}

y' = {0, 5, 6}

to find x \(\cap\) (Y' \(\cup\) Z)

first find y' \(\cup\) z = {0, 1, 3, 5, 6}

then x \(\cap\) (Y' \(\cup\) Z) = {0, 6}

9

Find the truth set of the equation x^{2} = 3(2x + 9)

A

{x : x = 3, x = 9}

B

{x : x = -3, x = -9}

C

{x : x = 3, x = -9}

D

{x : x = -3, x = 9}

CORRECT OPTION:
d

x^{2} = 3(2x + 9)

x^{2} = 6x + 27

x^{2} - 6x - 27 = 0

x^{2} - 9x + 3x - 27 = 0

x(x - 9) + 3(x - 9) = (x + 3)(x - 9) = 0

x + 3 = 0 or x - 9 = 0

x = -3 or x = 9

x = -3, x = 9

x

x

x

x(x - 9) + 3(x - 9) = (x + 3)(x - 9) = 0

x + 3 = 0 or x - 9 = 0

x = -3 or x = 9

x = -3, x = 9

10

The coordinates of points P and Q are (4, 3) and (2, -1) respectively. Find the shortest distance between P and Q.

A

10\(sqrt{2}\)

B

4\(sqrt{5}\)

C

5\(sqrt{2}\)

D

2\(sqrt{5}\)

CORRECT OPTION:
d

p(4, 3) Q(2 - 1)

distance = \(\sqrt{(x_2 - x_1)^2 + (Y_2 - y_1)^2}\)

= \(\sqrt{(2 - 4)^2 + (-1 - 3)^2}\)

= \(\sqrt{(-2)^2 = (-4)^2}\)

= \(\sqrt{4 + 16}\)

= \(\sqrt{20}\)

= \(\sqrt{4 \times 5}\)

= 2\(\sqrt{5}\)

distance = \(\sqrt{(x_2 - x_1)^2 + (Y_2 - y_1)^2}\)

= \(\sqrt{(2 - 4)^2 + (-1 - 3)^2}\)

= \(\sqrt{(-2)^2 = (-4)^2}\)

= \(\sqrt{4 + 16}\)

= \(\sqrt{20}\)

= \(\sqrt{4 \times 5}\)

= 2\(\sqrt{5}\)

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