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# A man has Rs.480 in the denominations of one-rupee notes, five-rupee notes and ten-rupee notes. The number of notes of each denomination is equal. What is the total number of notes that he has ?

A. 45
B. 90
C. 34
D. 65

Solution:

Let number of notes of each denomination be x.

$\left. \begin{array} { l } { \text { Then } x + 5 x + 10 x = 480 } \\ { \Rightarrow 16 x = 480 } \\ { \therefore \quad x = 30 } \\ { \text { Hence, total number of notes } = 3 x}\\ {= 90 \text { Answer } } \end{array} \right.$

# There are two examinations rooms A and B. If 10 students are sent from A to B, then the number of students in each room is the same. If 20 candidates are sent from B to A, then the number of students in A is double the number of students in B. The number of students in room A is:

A. 125
B. 164
C. 143
D. 100

Solution:

Let the number of students in rooms A and B be x and y respectively.

$\left. \begin{array} { l } { \text { Then, } x – 10 = y + 10 \Rightarrow x – y = 20 \ldots \text { (i) } } \\ { \text { and } x + 20 = 2 ( y – 20 ) \Rightarrow x – 2 y = – 60 \ldots \text { (ii) } } \\ { \text { Solving (i) and (ii) we get: } x = 100 , y = 80 \text { } } \\ { \therefore \text { The required answer } A = 100 \text { Answer} } \end{array} \right.$

# The price of 10 chairs is equal to that of 4 tables. The price of 15 chairs and 2 tables together is Rs. 4000. The total price of 12 chairs and 3 tables is:

A. RS. 3900
B. RS. 3000
C. RS. 3281
D. RS. 3891

Solution:

Let the cost of a chair and that of a table be Rs. x and Rs. y respectively.

$\left. \begin{array} { l } { \text { Then, } 10 x = 4 y \text { or } y = \frac { 5 } { 2 } x . } \\ { \therefore 15 x + 2 y = 4000 } \\ { \Rightarrow 15 x + 2 \times \frac { 5 } { 2 } x = 4000 } \\ { \Rightarrow 20 x = 4000 } \\ { \therefore x = 200 } \\ { \text { So, } y = ( \frac { 5 } { 2 } \times 200 ) = 500 } \\ { \text { Hence, the cost of } 12 \text { chairs and } 3 \text { tables :} }\\ {= 12 x + 3 y } \\ { \quad = \text { Rs. } ( 2400 + 1500 ) } \\ { \quad = \text { Rs. } 3900 } \end{array} \right.$

# If a – b = 3 and a2 + b2 = 29, find the value of ab.

A. 30
B. 20
C. 15
D. 10

Solution:

$\left. \begin{array} { l } { 2 a b = ( a ^ { 2 } + b ^ { 2 } ) – ( a – b ) ^ { 2 } } \\ { \quad = 29 – 9 = 20 } \\ { \Rightarrow a b = 10 } \end{array} \right.$

# A sum of Rs. 1360 has been divided among A, B and C such that A gets $$\frac{2}{3}$$ of what B gets and B gets $$\frac{1}{4}$$ of what C gets. B’s share is:

A. RS. 240
B. RS. 200
C. RS. 278
D. RS. 125

Solution:

$\left. \begin{array} { l } { \text { Let } C ^ { \prime } s \text { share } = Rs . x } \\ { \text { Then, } B \text { ‘s share } = Rs \cdot \frac { x } { 4 } , A \text { ‘s share :}}\\ {= Rs \cdot ( \frac { 2 } { 3 } \times \frac { x } { 4 } ) = Rs \cdot \frac { x } { 6 } } \\ { \therefore \frac { x } { 6 } + \frac { x } { 4 } + x = 1360 } \\ { \Rightarrow \frac { 17 x } { 12 } = 1360 } \\ { \Rightarrow x = \frac { 1360 \times 12 } { 17 } = Rs .960 } \\ { \text { Hence, } B ^ { \prime } s \text { share } = Rs . ( \frac { 960 } { 4 } ) = Rs .240 } \end{array} \right.$

# One-third of Rahul’s savings in National Savings Certificate is equal to one-half of his savings in Public Provident Fund. If he has Rs. 1,50,000 as total savings, how much has he saved in Public Provident Fund ?

A. RS. 40000
B. RS. 50000
C. RS. 60000
D. RS. 70000

Solution:

Let savings in N.S.C and P.P.F. be Rs. x and Rs. (150000 – x) respectively. Then,

$\left. \begin{array} { l } { \frac { 1 } { 3 } x = \frac { 1 } { 2 } ( 150000 – x ) } \\ { \Rightarrow \frac { x } { 3 } + \frac { x } { 2 } = 75000 } \\ { \Rightarrow \frac { 5 x } { 6 } = 75000 } \\ { \Rightarrow x = \frac { 75000 \times 6 } { 5 } = 90000 } \\ { \therefore \text { Savings in Public Provident Fund :}}\\ {= Rs . ( 150000 – 90000 ) = Rs .60000 } \end{array} \right.$

# A fires 5 shots to B’s 3 but A kills only once in 3 shots while B kills once in 2 shots. When B has missed 27 times, A has killed:

A. 40 birds
B. 30 birds
C. 32 birds
D. 39 birds

Solution:

$\left. \begin{array} { l } { \text { Let the total number of shots be } x \text { . Then, } } \\ { \text { Shots fired by } A = \frac { 5 } { 8 } x } \\ { \text { Shots fired by } B = \frac { 3 } { 8 } x } \\ { \text { Killing shots by } A = \frac { 1 } { 3 } \text { of } \frac { 5 } { 8 } x = \frac { 5 } { 24 } x } \\ { \text { Shots missed by } B = \frac { 1 } { 2 } \text { of } \frac { 3 } { 8 } x = \frac { 3 } { 16 } x } \\ { \therefore \frac { 3 x } { 16 } = 27 \text { or } x = ( \frac { 27 \times 16 } { 3 } ) = 144 } \\ { \text { Birds killed by } A = \frac { 5 x } { 24 } = ( \frac { 5 } { 24 } \times 144 ) = 30 } \end{array} \right.$

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