Latest Elitmus Aptitude Question SOLUTION: Q. Sherlock and Waston have to travel from Rajiv Gandhi Chowk to airport in metro. they have enough coins 1,5,10,25 paise. Sherlock agree to pay for Waston only if he tells all the

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(#M40031036) ELITMUS Numerical Ability question problem solving Keep an EYE

Q. Sherlock and Waston have to travel from Rajiv Gandhi Chowk to airport in metro. they have enough coins 1,5,10,25 paise. Sherlock agree to pay for Waston only if he tells all the possible combination of coins that can used to pay for the ticket.

1. how many combination are possible if the fare is 50 paise.
2. how many combination are possible if the discount on fair is 10% ie fare is 45paise

Asked In Elitmus Lilly (10 years ago)

Unsolved Read Solution (8)

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Read Solution (8) :

Bhaskar pal  (10 years ago)

For 1st case: Answer is 49.
25|10|5|1|Total
----------------
0|0|0|50--1 way
0|0|1-10|45-0--10 ways
0|1|0-8|40-0--9 ways
0|2|0-6|30-0--7 ways
0|3|0-4|20-0--5 ways
0|4|0-2|10-0--3 ways
0|5|0|0--1 way
1|0|0-5|25-0--6 ways
1|1|0-3|15-0--4 ways
1|2|0-1|5-0--2 ways
2|0|0|0--1 way

so, the total no of ways is- 49 ways.

Is this solution Helpfull? Yes (3) | No   

kushal jain  (10 years ago)

I think for 1 its 30 possible cases as follow

cases: 25 including (13)

25+25 (25+1*25)
25+10+10+5(25+10+10+1*5)(25+10+5+1*10)(1*25+10+10+5)
25+10+5+5+5(again 3 case)
25+5+5+5+5+5(2 case as it has two coins two replace by 1)

cases: excluding 25 only (14)

10+10+10+10+10 (10+10+10+10+1*10)
10+10+10+10+5+5 (10+10+10+10+5+1*5)(10+10+10+1*10+5+5)
10+10+10+5+5+5+5 (again 2 case)
10+10+5+5+5+5+5+5 (again 2 case)
10+5+5+5+5+5+5+5+5 (again 2 case)

cases: excluding 25,10 (2)

5+5+5+5+5+5+5+5+5+5 (only 1 case)

cases : only 1rs coin (1)

1*50

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Avigyan  (10 years ago)

in no 1 question ... total possible cases are 49.
in no.2 question... total possible cases are 26.

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pavithra  (10 years ago)

@ bhaskar pal couldnt understand

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jasna  (10 years ago)

pls tell clearly

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Kuldeep singh  (10 years ago)

25+25
25+10+10+5
25+10+5+5+5
25+5+5+5+5+5
25+5+5+5+5+1*5
25+5+5+5+1*5+1*5
.
.
.
.
10+10+10+10+10
10+10+10+10+5+5
10+10+10+5+5+5+5
.
.
.
.
10+1*5+1*5........
.
.
.
5*10
5*9+1*5
5*8+1*5+1*5
;.
.
.
.
.
1*50
answer is 33.

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Shubham Saxena  (10 years ago)

case 25(fixed)::
+25 1
cases number of cases
+10+10+5(/or)1*5 2
+10+5+5+5(/)1*5 2
+10+5+1*5+5/1*5 2
+10+1*5+(5+5(/)1*5)(/or)(1*5+5/1*5) 4

and so on will give 50 possibilities and then if we vary 25(first half) again then it will give 50*50=2500.

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will  (9 years ago)

for case 1
answr should b 53c3
becoz we tak it as 1x+5y+10z+25w=50
hence total numbr of solutions are 53c3

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