A) Rs. 4, Rs. 23 | B) Rs. 13, Rs. 17 |

C) Rs. 15, Rs. 14 | D) Rs. 17, Rs. 13 |

Explanation:

Let Rs. x be the fare of city B from city A and Rs. y be the fare of city C from city A.

Then, 2x + 3y = 77 ...(i) and

3x + 2y = 73 ...(ii)

Multiplying (i) by 3 and (ii) by 2 and subtracting, we get: 5y = 85 or y = 17.

Putting y = 17 in (i), we get: x = 13.

A) 10 | B) 6 |

C) 35/4 | D) 45/4 |

Explanation:

As clear from the figure itself, triangle OQX and triangle O'PX are similar.

So,

OQ/O'P = OX/O'X

=> OX = (7/9)xO'X

And since, OX + O'X = 20

=> (7/9)xO'X + O'X = 20

=> O'X = 45/4

A) only 1 is needed | B) only 1 is needed |

C) Both and 2 are needed | D) None |

Explanation:

Given that 50 paise and 25 paise coins are in the ratio of 7:3, they can be either 7 & 3 or 14 & 6.

As total 20 coins also includes Rs. 1 coins, so 50 paise & 25 paise coins are 7 & 3 respectively & Rs. 1 coins are 10.

Total amount = (100x10) + (50x7) + (25x3)=1425 paise or Rs.14.25.

A) 22 mts | B) 28 mts |

C) 32 mts | D) 34 mts |

Explanation:

let 'b' be the Length of bridge from cow to the near end of the bridge and 'a' be the distance of the train from the bridge.

'x' be speed of cow => '4x' speed of train

Then the total length of the bridge 2b + 10.

(a-2)/4x = b/x

=> a-2 = 4b........(1)

Now if it had run in opposite direction

(a+2b+10-2)/4x = (b+10-2)/x

=> a - 2b = 24......(2)

Solving (1) and (2)

b = 11 ,

Therefore length of the bridge is 2 x 11 + 10 = 32mts.

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 = 9. therefore 2 steps

For 6561 balls = 6561 therefore 8 steps

A) 17 & 9 | B) 9 & 11 |

C) 9 & 9 | D) 11 & 9 |

Explanation:

Let the number of apples in each tree of the 5 farmers be a, b, c, d,e respectively. Therefore total no of apples are 7a, 9b, 11c, 13d and 14e respectively.

Given,

7a+1 = 9b+3 = 11c-1 = 13d+3 = 14e-6 = x

9b+3 = 13d+3

===> 9b = 13d

so take b = 13 and d = 9

9b+3 = 9*13+3 = 120

ie, x=120

substituting

7a+1 = 120

7a = 119

a = 17

11c-1 = 120

11c = 121

c=11

14e-6 = 120

14e = 126

e = 9

Yields per tree in the orchards of d 3rd & 4th farmers= 11,9