# A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is:

A.  $$\frac{2}{91}$$
B.  $$\frac{3}{21}$$
C.  $$\frac{12}{31}$$
D.  $$\frac{1}{13}$$

Solution:

Let S be the sample space.

Then, n(S) = number of ways of drawing 3 balls out of 15
\left. \begin{array} { l } { \quad = ^ { 15 } C _ { 3 } } \\ { = \frac { ( 15 \times 14 \times 13 ) } { ( 3 \times 2 \times 1 ) } } \\ { = 455 } \\ { \text { Let } E = \text { event of getting all the } 3 \text { red balls. } } \\ { \therefore \quad n ( E ) = ^ { 5 } C _ { 3 } = ^ { 5 } C _ { 2 } = \frac { ( 5 \times 4 ) } { ( 2 \times 1 ) } = 10 } \\ { \therefore P ( E ) = \frac { n ( E ) } { n ( S ) } = \frac { 10 } { 455 } = \frac { 2 } { 91 } } \end{array}

# A batsman scored 110 runs which included 3 boundaries and 8 sixes. What percent of his total score did he make by running between the wickets?

A.  40%
B.  $$44\frac{4}{6}$$%
C.  $$45\frac{9}{8}$$%
D.  $$45\frac{5}{11}$$%

Solution:

$\left. \begin{array} { l } { \text { Number of runs made by running :} }\\ {= 110 – ( 3 \times 4 + 8 \times 6 ) } \\ { = 110 – ( 60 ) } \\ { = 50 } \\ { \therefore \text { Required percentage } = ( \frac { 50 } { 110 } \times 100 ) \% }\\ {= 45 \frac { 5 } { 11 } \% } \end{array} \right.$

# Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. The marks obtained by them are:

A. 42:33
B. 49:41
C. 41:39
D. 44:11

Solution:

\begin{array} { l } { \text { Let their marks be } ( x + 9 ) \text { and } x . } \\ { \text { Then, } x + 9 = \frac { 56 } { 100 } ( x + 9 + x ) } \\ { \Rightarrow 25 ( x + 9 ) = 14 ( 2 x + 9 ) } \\ { \Rightarrow 3 x = 99 } \\ { \Rightarrow x = 33 } \\ { \text { So, their marks are 42 and 33. } } \end{array}

# A fruit seller had some apples. He sells 40% apples and still has 420 apples. Originally, he had:

A. 696 apples
B. 700  apples
C. 740 apples
D. 800 apples

Solution:

$\left. \begin{array} { l } { \text { Suppose originally he had } x \text { apples } } \\ { \text { Then, } ( 100 – 40 ) \% \text { of } x = 420 \text { . } } \\ { \Rightarrow \frac { 60 } { 100 } \times x = 420 } \\ { \Rightarrow x = ( \frac { 420 \times 100 } { 60 } ) = 700 } \end{array} \right.$

# What percentage of numbers from 1 to 70 have 1 or 9 in the unit’s digit?

A. 20%
B. 22%
C. 20%
D. 30%

Solution:

Clearly, the numbers which have 1 or 9 in the unit’s digit, have squares that end in the digit 1. Such numbers from 1 to 70 are 1, 9, 11, 19, 21, 29, 31, 39, 41, 49, 51, 59, 61, 69.
$\left. \begin{array} { l } { \text { Number of such number } = 14 } \\ { \therefore \text { Required percentage } = ( \frac { 14 } { 70 } \times 100 ) \% = 20 \% } \end{array} \right.$

# If 20% of a = b, then b% of 20 is the same as:

A. 9% of a
B. 2% of a
C. 4% of a
D. 7% of a

Solution:

$\left. \begin{array} { l } { 20 \% \text { of } a = b \Rightarrow \frac { 20 } { 100 } a = b } \\ { \therefore b \% \text { of } 20 = ( \frac { b } { 100 } \times 20 ) =}\\ { ( \frac { 20 } { 100 } a \times \frac { 1 } { 100 } \times 20 ) = \frac { 4 } { 100 } a = 4 \% \text { of } a } \end{array} \right.$

# In a certain school, 20% of students are below 8 years of age. The number of students above 8 years of age is $$\frac{2}{3}$$ of the number of students of 8 years of age which is 48. What is the total number of students in the school?

A. 100
B. 120
C. 130
D. 150

Solution:

$\left. \begin{array} { l } { \text { Let the number of students be } x \text { . Then, } } \\ { \text { Number of students above } 8 \text { years of age :} }\\ {= ( 100 – 20 ) \% \text { of } x = 80 \% \text { of } x \text { . } } \\ { \therefore 80 \% \text { of } x = 48 + \frac { 2 } { 3 } \text { of } 48 } \\ { \Rightarrow \frac { 80 } { 100 } x = 80 } \\ { \Rightarrow x = 100 } \end{array} \right.$

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