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mounika tallapu reddy

· started a discussion

· 1 Months ago

reasoning q;6 (Question:103636.0)

this type of question asked in the exam of lic ado

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.

What is the difference between the number of \(\unicode{x20B9} \)2 coins with Ram and the number of \(\unicode{x20B9} \)5 coins with Mohan?
Options:
A) 5
B) 4
C) 3
D) 2
E) None of these
Solution:

Given that, the number of \(\unicode{x20B9} \)2 cons with Ram is six less than that of \(\unicode{x20B9} \)1 coins with Mohan. Let us look at the possibilities,


 

\(\unicode{x20B9} \)1

\(\unicode{x20B9} \)2

\(\unicode{x20B9} \)5

Total

Ram

 

1

 

32

Shyam

 

 

 

43

Mohan

7

 

 

46


In this case, the number of \(\unicode{x20B9} \)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

View detailed solution please by clicking report...
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