A) 13 | B) 15 |

C) 17 | D) 19 |

Explanation:

The horizontal lines are IJ, AB, EF, MN, HG, DC and LK i.e. 7 in number.

The vertical lines are AD, EH, IL, FG, BC and JK i.e. 6 in number.

Thus, there are 7 + 6 = 13 straight lines in the figure.

A) 76 | B) 78 |

C) 80 | D) 82 |

A) I, L | B) J, L |

C) J, M | D) I, J |

Explanation:

This question concerns a committee’s decision about which five of eight areas of expenditure to reduce. The question requires you to suppose that K and N are among the areas that are to be reduced, and then to determine which pair of areas could not also be among the five areas that are reduced.

The fourth condition given in the passage on which this question is based requires that exactly two of K, N, and J are reduced. Since the question asks us to suppose that both K and N are reduced, we know that J must not be reduced:

**Reduced :: K, N****Not reduced :: J**

The second condition requires that if L is reduced, neither N nor O is reduced. So L and N cannot both be reduced. Here, since N is reduced, we know that L cannot be. Thus, adding this to what we’ve determined so far, we know that J and L are a pair of areas that cannot both be reduced if both K and N are reduced:

**Reduced :: K, N****Not reduced :: J, L**

Answer choice (**B**) is therefore the correct answer.

A) K & L will always be together | B) K is not there, then L will not be there |

C) k is there, then L will also be there | D) K & L will always be not together |

Explanation:

This would not mean that **K** and **L** will always be together. It just implies that, if **K** is there, then **L** will also be there.

At the same time, it can happen that **L** is there but **K** isn't.

Remember, the condition is on **K**, not on **L**.

A) 4 | B) 5 |

C) 6 | D) 0 |

Explanation:

A regular Pentagon have 5 sides and 5 lines of symmetry.

- The number of lines of symmetry in a regular polygon is equal to the number of sides.

A) 5 | B) 2 |

C) 3 | D) 6 |

Explanation:

The given figure can be labelled as shown :

The spaces P, Q and R have to be shaded by three different colours definitely (since each of these three spaces lies adjacent to the other two).

Now, in order that no two adjacent spaces be shaded by the same colour, the spaces T, U and S must be shaded with the colours of the spaces P, Q and R respectively.

Also the spaces X, V and W must be shaded with the colours of the spaces S, T and U respectively i.e. with the colours of the spaces R, P and Q respectively. Thus, minimum three colour pencils are required.

A) 20 | B) 19 |

C) 17 | D) 15 |

Explanation:

The given figure can be labelled as :

**Straight lines** :

The number of straight lines are 19

i.e. BC, CD, BD, AF, FE, AE, AB, GH, IJ, KL, DE, AG, BH, HI, GJ, IL, JK, KE and DL.

A) 18 | B) 17 |

C) 14 | D) 16 |

Explanation:

The given figure can be labelled as shown :

The horizontal lines are AK, BJ, CI, DH and EG i.e. 5 in number.

The vertical lines are AE, LF and KG i.e. 3 in number.

The slanting lines are LC, CF, FI, LI, EK and AG i.e. 6 in number.

Thus, there are 5 + 3 + 6 = 14 straight lines in the figure.