A) 20 | B) 25 |

C) 30 | D) Data inadequate |

Explanation:

Let the number of boys and girls participating in sports be 3x and 2x respectively.

Then, 3x = 15 or x = 5.

So, number of girls participating in sports = 2x = 10.

Number of students not participating in sports = 60 - (15 + 10) = 35.

Let number of boys not participating in sports be y.

Then, number of girls not participating in sports = (35 -y).

Therefore (35 - y) = y + 5 2y 30 y = 15.

So, number of girls not participating in sports = (35 - 15) = 20.

Hence, total number of girls in the class = (10 + 20) = 30.

A) 125 | B) 96 |

C) 35 | D) 25 |

Explanation:

We know that, to be in the form that the number of rows and the number of columns to be equal the number should be a perfect square ().

Given number is 1200

The perfect number which is above and near to 1200 is 1225 which is 35 35.

The minimum number of coins he need is 25.

A) 60 | B) 30 |

C) 70 | D) 90 |

Explanation:

Let 'x' be the number of new Rs.2000 notes.

Given for one new 2000 note there are three old 500 notes.

3x

Given 10 more new Rs.2000 notes are added to the collection

and the ratio of new Rs.2000 notes to old Rs.500 notes

Therefore, number of new Rs.2000 notes x+10 = 30.

old Rs.500 notes 3x = 60.

Thus, the total number of notes in the collection 30 + 60 = 90.

A) 5/18 | B) 4/9 |

C) 11/18 | D) 17/36 |

Explanation:

Let the total number of workers be x. Then,

Number of women = and number of men = =

Number of women having children =

Number of men having children =

Number of workers having children =

workers having no children = of all workers

A) 6555 | B) 5685 |

C) 1705 | D) 870 |

Explanation:

Candidates passed in atleast four subjects

= (Candidates passed in 4 subjects) + (Candidates Passed in all 5 subjects)

= (Candidates failed in only 1 subject ) + ( Candidates passed in all subjects)

= (78 + 275 + 149 + 147 + 221) + 5685 = 870 + 5685 = 6555

A) 106 | B) 301 |

C) 309 | D) 400 |

Explanation:

Clearly, the required number would be such that it leaves a remainder of 1 when divided by 3, 4, 5, or 6 and no remainder when divided by 7. Thus, the number must be of the form (L.C.M of 3, 4, 5, 6) x + 1 i.e., (60x + 1 ) and a multiple of 7. Clearly, for x = 5, the number is a multiple of 7. So the number is 301.