Let required fraction be then

............. (i)

and ............ (ii)

Solving (i) and (ii) we get x=3, y=5

Required fraction =

A) 7, 5 | B) 9, 3 |

C) 8, 4 | D) 6, 6 |

Explanation:

Let the number of girls = x

=> x + x + 4 = 12

=> 2x = 8

=> x = 4

=> Number of girls = 4

=> Number of boys = 4 + 4 = 8

A) 7 | B) 23 |

C) 9 | D) 21 |

Explanation:

Let the greater and smaller number be p and q respectively.

4q = p + 5 ------ (I)

p = 3q+2 ------- (II)

From equation (I) and (II)

q = 7

p = 23

A) 1 | B) 2 |

C) 3 | D) 0 |

Explanation:

A number is divisible by 9 only if the sum of the digits of the number is divisible by 9.

Here 86236 = 8 + 6 + 2 + 3 + 6 = 25

We must add 2 to 25 to become 27 which is divisible by 9.

A) 18 | B) 20 |

C) 19 | D) 17 |

A) 57 | B) 75 |

C) 48 | D) 39 |

Explanation:

Let the two-digit number be 10a + b

a + b = 12 --- (1)

If a>b, a - b = 6

If b>a, b - a = 6

If a - b = 6, adding it to equation (1), we get

2a = 18 => a =9

so b = 12 - a = 3

Number would be 93.

if b - a = 6, adding it to the equation (1), we get

2b = 18 => b = 9

a = 12 - b = 3.

Number would be 39.

There fore, Number would be 39 or 93.