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The Enforcement Directorate (ED) of the Government of India had for a long time been keeping watch over the Education Minister of West Bengal in Kolkata regarding a scam of recruitment of Staff Selection Commission candidates. 

The Director of ED, Mr Sanjay Kr Mishra (IRS), with the help of his department sleuths, had observed a very interesting trend in the activity of the Education Minister :

Once every day of a week, except Sunday, the personal assistant of the Education Minister took delivery of sealed envelopes of money from the minister personally at his office to deliver to the minister’s lady-friend at her residence. It was also known from a secret source that for two fixed days of a week he delivered a certain number of envelopes less than the previous day, and for three fixed days a week, he delivered the same number of envelopes (as mentioned in the previous line) more than the previous day. However, it was not possible to get information as to what those days were. This practice continued for exactly twenty-seven consecutive days from the first Monday of delivery. From high-definition telescopic photographs of the envelopes, it was surely concluded that each envelope held a hundred crisp two thousand rupee notes. It was also known from the secret source that every week the greatest and least number of envelopes that the personal assistant delivered was eighty and twenty respectively.

Mr Mishra took the help of a professional data interpreter to analyze the collected data and to provide the ED Department with an idea of the details and enormity of the scam. What would be the answers found out by the data-interpreter to the following questions asked to him by Mr Mishra ?

Question :

What could be the lowest possible quantum of the total delivery of money from the Education Minister’s office to the residence of his lady-friend, by the personal assistant of the minister ?

Option: 1

Rs 5.40 crores

 


Option: 2

Rs 14.75 crores


Option: 3

Rs 21.60 crores


Option: 4

Rs 24.00 crores


Answers (1)

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The deliveries of sealed envelopes of money from the Education Minister’s office to the residence of his lady-friend, by the personal assistant of the minister, were done from Monday to Saturday.

The data interpreter assumed that the number of envelopes delivered on Monday to be E.

Since it was known by Mr Sanjay Kr Mishra (IRS) from a secret source that for two fixed days of a week the delivery was a certain number of envelopes less than the previous day, and for three fixed days a week the delivery was the same number of envelopes more than the previous day, he understood that the number of different permutations of those 2+3 = 5 days from Tuesday to Saturday would be 5C2, that is 10 ways.

He also assumed that the number of envelopes less or more than the previous day be x.

Hence he tabulated the 10 probable ways in which the permutation could take place and the vis-à-vis total delivery of the week as given below. The yellow shade indicated an increase and the blue a decrease in the number of envelopes :

Cases Count of Envelopes delivered
Mon Tues Wed Thurs Frid Sat Total in 6 days
1 E (E-x) (E-x-x)=(E-2x) (E-2x+x)=(E-x) (E-x+x)=E (E+x) (6E-3x)
2 E (E-x) (E-x+x)=E (E-x) (E-x+x)=E (E+x) (6E-x)
3 E (E-x) (E-x+x)=E (E+x) (E+x-x)=E (E+x) (6E+x)
4 E (E-x) (E-x+x)=E (E+x) (E+x+x)=(E+2x) (E+2x-x)=(E+x) (6E+3x)
5 E (E+x) (E+x-x)=E (E-x) (E-x+x)=E (E+x) (6E+x)
6 E (E+x) (E+x-x)=E (E+x) (E+x-x)=E (E+x) (6E+3x)
7 E (E+x) (E+x-x)=E (E+x) (E+x+x)=(E+2x) (E+2x-x)=(E+x) (6E+5x)
8 E (E+x) (E+x+x)=(E+2x) (E+2x-x)=(E+x) (E+x-x)=E (E+x) (6E+5x)
9 E (E+x) (E+x+x)=(E+2x) (E+2x-x)=(E+x) (E+x+x)=(E+2x) (E+2x-x)=(E+x) (6E+7x)
10 E (E+x) ???????(E+x+x)=(E+2x) (E+2x+x)=(E+3x) (E+3x-x)=(E+2x) (E+2x-x)=(E+x) (6E+9x)

From the secret source, it was also known that every week the greatest and least number of envelopes that were delivered was eighty and twenty respectively. Identifying them in each case (greatest number in green and least number in pink), he got :

Cases Count of Envelopes delivered
Mon Tues Wed Thurs Frid Sat Total in 6 days
1 E (E-x) (E-x-x)=(E-2x) (E-2x+x)=(E-x) ???????(E-x+x)=E (E-x) (6E - 3x)
    20     ???????80  
2 E (E-x) (E-x+x)=E (E-x) ???????(E-x+x)=E (E-x) (6E - x)
  20   20   ???????80  
3 E (E-x) ???????(E-x+x)=E (E-x) (E+x-x)=E (E-x) (6E+x)
  20   80   ???????80  
4 E (E-x) ???????(E-x+x)=E (E-x) (E+x+x)=(E+2x) (E+2x-x)=(E+x) (6E+3x)
  20     ???????80    
5 E (E-x) (E+x-x)=E (E-x) (E-x+x)=E (E-x) (6E+x)
  ?80   20   ???????80  
6 E (E+x) (E+x-x)=E (E+x) (E+x-x)=E (E-x) (6E+3x)
20 80 20 ?80 20 ???????80  
7 E (E+x) (E+x-x)=E (E+x) (E+x+x)=(E+2x) (E+2x-x)=(E+x) (6E+5x)
20   20   ???????80    
8 E (E+x) (E+x+x)=(E+2x) (E+2x-x)=(E+x) (E+x-x)=E (E-x) (6E+5x)
20   80   20    
9 E (E+x) (E+x+x)=(E+2x) (E+2x-x)=(E+x) (E+x+x)=(E+2x) (E+2x-x)=(E+x) (6E+7x)
20   80   ???????80    
10 E (E+x) (E+x+x)=(E+2x) (E+x+x)=(E+3x) (E+3x-x)=(E+2x) (E+2x-x)=(E+x) (6E+9x)
20     80      

Now, as a result of the above the data interpreter could deduce the following :

Case 1 : E + x = 80 and E – 2x = 20. Solving, he got x = 20 and E = 60

Case 2 : E + x = 80 and E – x = 20. Solving, he got x = 30 and E = 50

Case 3 : E + x = 80 and E – x = 20. Solving, he got x = 30 and E = 50

Case 4 : E + 2x = 80 and E – x = 20. Solving, he got x = 20 and E = 40

Case 5 : E + x = 80 and E – x = 20. Solving, he got x = 30 and E = 50

Case 6 : E + x = 80 and E = 20. Solving, he got x = 60 and E was already 20

Case 7 : E + 2x = 80 and E = 20. Solving, he got x = 30 and E was already 20

Case 8 : E + 2x = 80 and E = 20. Solving, he got x = 30 and E was already 20

Case 9 : E + 2x = 80 and E = 20. Solving, he got x = 30 and E was already 20

Case 10 : E + 3x = 80 and E = 20. Solving, he got x = 20 and E was already 20

Hence, he could exactly find out the individual number of envelopes that were delivered from the Education Minister’s office to the residence of his lady-friend by the personal assistant of the minister, for each day of each of the 10 probable cases. Also the total number of envelopes delivered in a week of each of the 10 probable cases. They were as below :

Cases Count of Envelopes delivered
Mon Tues Wed Thurs Frid Sat Total in 6 days 
1 E (E-x) (E-x-x)=(E-2x) (E-2x+x)=(E-x) (E-x+x)=E (E+x) (6E-3x)
60 40 20 40 60 80 300
2 E (E-x) (E-x+x)=E (E-x) (E-x+x)=E (E+x) (6E-x)
50 20 50 20 50 80 270
3 E (E-x) (E-x+x)=E (E+x) (E+x-x)=E (E+x) (6E+x)
50 20 50 80 50 80 330
4 E (E-x) (E-x+x)=E (E+x) (E+x+x)=(E+2x) (E+2x-x)=(E+x) (6E+3x)
40 20 40 60 80 60 300
5 E (E+x) (E+x-x)=E (E-x) (E-x+x)=E (E+x) (6E+x)
50 80 50 20 50 80 330
6 E (E+x) ???????(E+x-x)=E (E+x) (E+x-x)=E (E+x) (6E+3x)
20 80 20 80 20 80 300
7 E (E+x) ???????(E+x-x)=E (E+x) (E+x+x)=(E+2x) (E+2x-x)=(E+x) (6E+3x)
20 50 20 50 80 50 270
8 E (E+x) (E+x+x)=(E+2x) (E+2x-x)=(E+x) (E+x+x)=(E+2x) (E+2x-x)=(E+x) (6E+5x)
20 50 80 50 20 50 270
9 E (E+x) ???????(E+x+x)=(E+2x) (E+2x-x)=(E+x) (E+x+x)=(E+2x) (E+2x-x)=(E+x) (6E+7x)
20 50 80 50 80 50 330
10 E (E+x) ???????(E+x+x)=(E+2x) (E+x+x)=(E+3x) (E+3x-x)=(E+2x) (E+2x-x)=(E+x) (6E+9x)
20 40 60 80 60 40 300

From the deductions made by the data interpreter, it could be seen that in Case 2, Case 7 and Case 8, the total number of envelopes delivered from the Education Minister’s office to the residence of his lady-friend, by the personal assistant of the minister was 270 each, which was the lowest number among all the ten cases. 

From photographs of the envelopes, it was known that each envelope held hundred two thousand rupee notes, that is each envelope contained (100*2000) = Rs 2,00,000.

Hence, 270 envelopes would contain 200000*270 = Rs 5,40,00,000, that is, Rs 5.40 crores

This practice had continued for exactly twenty-seven consecutive days from the first Monday of delivery, that is for four Monday to Saturday time spans.

Hence, the lowest possible quantum of the total delivery of money from the Education Minister’s office to the residence of his lady-friend, by the personal assistant of the minister, could be 4*5.40 = Rs 21.60 crores

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