## Di And Lr Clubbed

Find out the sum of all the boxes of the set E.

**Options:**

A. | 61 |

B. | 60 |

C. | 55 |

D. | 49 |

E. | 31 |

**Answer: Option E**

: E

Sum of the top most boxes (119) is 10 more than the sum of bottom most boxes (109); it means that the total number of the boxes between the bottom most box and top most box in all the five sets is 119 – 109 = 10. As each set has different number of boxes so this sum 10 should be a sum of five different natural numbers i.e. 0+1+2+3+4 = 10. So, the number of boxes between the bottom most boxes and the top most boxes for the 5 sets should be 0, 1, 2, 3 & 4. Hence there are 1, 2, 3, 4 & 5 boxes in the sets A, B, C, D, and E. Sum of the boxes of set B is 81 which is a sum of 1 or 2 or 3 or 4 or 5 consequent natural numbers. If B is having 1 or 2 boxes than it would be impossible to make a sum of 109 from all bottom most boxes. So they can be 26, 27 & 28 only i.e. B has only 3 boxes. Now as no two consecutive sets have consecutive number of boxes, so A and C cannot have 2 or 4 boxes. Let us assume that A has only one box so its number would be 21. Also one of the sets will have a bottom most box numbered 29 (as B’s top most box is numbered 28 and counting cannot stop there as in that case sum of the bottom most boxes will not be 109) and one will have 22 as the bottom most box number. So, the bottom most box number of the 5th set should be 109 - (21+ 22 + 26 + 29) = 11 which is impossible because in that case the sum of top most boxes will not be 119, so set A has 5 boxes and hence C, D, E have 1, 4 and 2 boxes respectively. As the top most number of A is 21 and bottom most number of B is 26 that can be reached only if the counting from A goes to the set with 4 boxes. So, the 3 bottom most box numbers obtained are 17, 22 and 26. Hence the other two bottom most box numbers are 29 and 15. So, the table is like:- Answer option E

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