A) 2414 | B) 204 |

C) 87 | D) 8 |

Explanation:

Suppose there are 9 balls

Let us give name to each ball B1 B2 B3 B4 B5 B6 B7 B8 B9

Now we will divide all the balls into 3 groups.

Group1 - B1 B2 B3

Group2 - B4 B5 B6

Group3 - B7 B8 B9

Step1 - Now weigh any two groups. Let's assume we choose Group1 on left side of the scale and Group2 on the right side.

So now when we weigh these two groups we can get 3 outcomes.

Weighing scale tilts on left - Group1 has a heavy ball.

Weighing scale tilts on right - Group2 has a heavy ball.

Weighing scale remains balanced - Group3 has a heavy ball.

Lets assume we got the outcome as 3. i.e Group 3 has a heavy ball.

Step2 - Now weigh any two balls from Group3. Lets assume we keep B7 on left side of the scale and B8 on right side.

So now when we weigh these two balls we can get 3 outcomes.

Weighing scale tilts on left - B7 is the heavy ball.

Weighing scale tilts on right - B8 is the heavy ball.

Weighing scale remains balanced - B9 is the heavy ball.

The conclusion we get from this Problem is that each time weigh. We element 2/3 of the balls.

As we came to conclusion that Group3 has the heavy ball from Step1, we remove 6 balls from the equation i.e (2/3) of 9.

Simillarly we do the ame thing for the Step2.

Now going with this conclusion. We have 6561 balls.

Step - 1

Divided into 3 groups

Group1 - 2187Balls

Group2 - 2187Balls

Group3 - 2187Balls

Taking the similar steps as we did in the above example, we come to the conclusion that Group1 has the heavy ball.

Step - 2

Divided into 3 groups

Group1 - 729Balls

Group2 - 729Balls

Group3 - 729Balls

Taking the similar steps as we did in the above example, we come to the conclusion that Group1 has the heavy ball.

Step - 3

Divided into 3 groups

Group1 - 243Balls

Group2 - 243Balls

Group3 - 243Balls

Taking the similar steps as we did in the above example, we come to the conclusion that Group1 has the heavy ball.

Step - 4

Divided into 3 groups

Group1 - 81Balls

Group2 - 81Balls

Group3 - 81Balls

Step - 5

Divided into 3 groups

Group1 - 27Balls

Group2 - 27Balls

Group3 - 27Balls

Step - 6

Divided into 3 groups

Group1 - 9Balls

Group2 - 9Balls

Group3 - 9Balls

Step - 7

Divided into 3 groups

Group1 - 3Balls

Group2 - 3Balls

Group3 - 3Balls

Step - 8

So now when we weigh 2 balls out of 3 we can get 3 outcomes.

Weighing scale tilts on left - left side placed is the heavy ball.

Weighing scale tilts on right - right side placed is the heavy ball.

Weighing scale remains balanced - remaining ball is the heavy ball.

So the general answer to this question is, it is always multiple of 3 steps.

For 9 balls ${3}^{2}$= 9. therefore 2 steps

For 6561 balls ${3}^{8}$ = 6561 therefore 8 steps

A) 13 | B) 26 |

C) 39 | D) 52 |

Explanation:

The total cards in the deck are** 52**. These 52 cards are divided into 4 suits of 13 cards in each suit. Two Red suits and Two black suits.

**Red suits ::** Heart suit and Diamond suit **= 26**

**Black suits ::** Spade suit and Club suit **= 26.**

A) 20 | B) 30 |

C) 18 | D) 24 |

Explanation:

6 choose 3 means number of possible unordered combinations when 3 items are selected from 6 available items i.e, nothing but **6C3.**

Now **6C3 = 6 x 5 x 4/3 x 2 x 1 = 120/6 = 20.**

A) 0 | B) 19 |

C) 29 | D) 91 |

Explanation:

Here in the given numbers 91 is a Composite number. Since it has factors of 7 and 13 other than 1 and itself.

**Composite Numbers :**

A composite number is a positive integer which is not prime (i.e., which has factors other than 1 and itself).

Examples :: 4, 6, 8, 12, 14, 15, 18, 20, ...

A) B and C together are sufficient | B) Any one pair of A and B, B and C or C and A is sufficient |

C) C and A together are sufficient | D) A and B together are sufficient |

Explanation:

From the given data,

Let the two gits of a number be x & y

A) x + y = 15

B) ${x}^{2}-{y}^{2}=45$

(x+y) (x - y) = 45

C) x - y = 3

From any 2 of the given 3 statements, we can find that 2 digit number as

2x = 18 => x = 9

=> y = 6

Hence, 2 digit number is 96.

Any one pair of A and B, B and C or C and A is sufficient to find.

A) -ve | B) +ve |

C) 0 | D) Can't be determined |

Explanation:

We know the Mathematical rules that

$\frac{\mathbf{+}}{\mathbf{+}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{+}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{\mathbf{+}}{\mathbf{-}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{-}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{\mathbf{-}}{\mathbf{+}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{-}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{\mathbf{-}}{\mathbf{-}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{+}$

A) TRUE | B) FALSE |

Explanation:

We know that,

**Alternate Exterior Angles Theorem::**

The Alternate Exterior Angles Theorem states that, when two parallel lines are cut by a transversal, the resulting alternate exterior angles are congruent .

**Converse of the Alternate Exterior Angles Theorem ::**

Converse of the Alternate Exterior Angles Theorem states that, If two lines are cut by a transversal and the alternate exterior angles are congruent, then the lines are parallel.

A) 4 | B) 3 |

C) 2 | D) 1 |