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nidhi pandey

· started a discussion

· 1 Months ago

Puzzle 2nd no. (Question:103636.3)

What will approach to solve that

Question:

DIRECTIONS: These questions are based on the information given below.


Three persons Ram, Shyam and Mohan have \(\unicode{x20B9} \)32, \(\unicode{x20B9} \)43 and \(\unicode{x20B9} \)46 with them respectively. Each of them has only coins of denominations \(\unicode{x20B9} \)1, \(\unicode{x20B9} \)2, and \(\unicode{x20B9} \)5. The number of coins with any of them of any denomination is a distinct single digit natural number i.e. these will he nine distinct single digit natural numbers (or nine possible combinations of person and denomination). Further, the number of \(\unicode{x20B9} \)2 coins with Ram is six less than the number of \(\unicode{x20B9} \)1 coins Mohan.

How many ₹5 coins are there with Ram, Shyam and Mohan altogether.

Options:
A) 12
B) 16
C) 10
D) 21
E) Can't be determined
Solution:
Given that, the number of ₹2 cons with Ram is six less than that of ₹1 coins with Mohan. Let us look at the possibilities.


-

₹ 1

₹2

₹5

Total

Ram

-

1

-

32

Shyam

-

-

-

43

Mohan

7

-

-

46


In this case, the number of ₹5 coins with Mohan has to be an odd number. But for any of the three possible values of 5, I or 9, there will be repetitions. Similarly, no arrangement is possible for the values of 3 and 9 also. The only possible arrangement is.


-

₹1

₹2

₹ 5

Ram

3

2

5

Shyam

6

1

7

Mohan

8

9

4

Knowledge Expert

· commented

· 1 Months ago

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