A) Rs.2500 | B) Rs.2000 |

C) Rs.4000 | D) None of these |

Explanation:

Maximum earning will be only when he will won on the maximum yielding table.

A ----> 10:1

B ----> 20:1

C ----> 30:1

i.e, he won on B and C but lost on A

20 x 200 + 30 x 200 -1 x 200 = 9800

minimum earning will be when he won on table A and B and lose on that table 3.

10 x 200 + 20 x 200 - 1 x 200

6000-200 = 5800

Difference= 9800 - 5800 = 4000

A) 6548 | B) 5667 |

C) 7556 | D) 8457 |

Explanation:

Let ratio of the incomes of Pavan and Amar be 4x and 3x

and Ratio of their expenditures be 3y and 2y

4x - 3y = 1889 ......... I

and

3x - 2y = 1889 ...........II

I and II

y = 1889

and x = 1889

**Pavan's income = 7556**

A) 7:1 | B) 13:5 |

C) 15:7 | D) 2:7 |

A) 4,12,480 | B) 3,67,500 |

C) 5,44,700 | D) 2,98,948 |

Explanation:

Let the total population be 'p'

Given ratio of male and female in a city is 7 : 8

In that percentage of children among male and female is **25%** and **20%**

=> Adults male and female % = **75% & 80%**

But given adult females is = 156800

=> **80%(8p/15) = 156800**

=> 80 x 8p/15 x 100 = 156800

=> p = 156800 x 15 x 100/80 x 8

=> p = 367500

Therefore, the total population of the city = **p = 367500**

A) 11 & 17 | B) 7 & 17 |

C) 9 & 21 | D) 13 & 23 |

Explanation:

Let the two numbers be x and y

Given x : y = 3 : 7 .....(1)

Now, x+6 : y+6 = 5 : 9 .....(2)

From (1), x = 3y/7

From (2), 5y - 9x = 24

=> 5y - 9(3y/7) = 24

=> y = 21

=> From(1), x = 9

Hence, the two numbers be **9** and **21**

A) 175 | B) 165 |

C) 155 | D) 145 |

Explanation:

Let the number of failed students be x

=> Number of passed students = 4x

So total number of students was 5x

From the given data,

If total number of students be 5x – 35

=> 4x-35-9/x+9 = 2/1

=> 4x – 44 = 2(x + 9)

=> 4x – 2x = 18 + 44

=> x = 31

Total number = 31×5

= 155