Latest Elitmus Aptitude Question SOLUTION: Sherlock homes and Dr Watson have to travel from Rajiv Gandhi Chowk to Indira Gandhi International airport via metro.they have enough coins of 1,5,10,25 paise. Sherlock agrees to

Let the number of 1, 5, 10 and 50 paise coins required are x, y, z and w respectively.

So when paid total amount is: x + 5y + 10z + 25w, which is equal to the fare paid.

So, x + 5y + 10z + 25w = 50

The number of non negative integral solution of this equation will be equal to the number of ways the fare can be paid.



x + 5y + 10z + 25w = 50

In this equation w can be either 0, 1 or 2



Case 1: When w is 2

No other coins are required to pay

So there is only one way to pay: (0, 0, 0, 2)

Total no of ways : 1


Case 2: When w is 1

x + 5y + 10z = 25

In this case z can have only three values possible: 0, 1 or 2

Case I: When z is 0

x + 5y = 25

possible values are: (25, 0, 0, 1), (20, 1, 0, 1), (15, 2, 0, 1), (10, 3, 0, 1), (5, 4, 0, 1), (0, 5, 0, 1)

Case II: When z is 1

x + 5y = 15

possible values are: (15, 0, 1, 1), (10, 1, 1, 1), (5, 2, 1, 1), (0, 3, 1, 1)

Case III: When z is 2

x + 5y = 5

possible values are: (5, 0, 2, 1), (0, 1, 2, 1)

Total no of ways : 12


Case 3: When w is 0

x + 5y + 10z = 50

In this case z can have only three values possible: 0, 1, 2, 3, 4 or 5

Case I: When z is 0

x + 5y = 50

possible values are: (50, 0, 0, 0), (45, 1, 0, 0), (40, 2, 0, 0), (35, 3, 0, 0), (30, 4, 0, 0), (25, 5, 0, 0), (20, 6, 0, 0), (15, 7, 0, 0), (10, 8, 0, 0), (5, 9, 0, 0), (0, 10, 0, 0)

Case II: When z is 1

x + 5y = 40

possible values are: (40, 0, 1, 0), (35, 1, 1, 0), (30, 2, 1, 0), (25, 3, 1, 0), (20, 4, 1, 0), (15, 5, 1, 0), (10, 6, 1, 0), (5, 7, 1, 0), (0, 8, 1, 0)

Case III: When z is 2

x + 5y = 30

possible values are: (30, 0, 2, 0), (25, 1, 2, 0), (20, 2, 2, 0), (15, 3, 2, 0), (10, 4, 2, 0), (5, 5, 2, 0), (0, 6, 2, 0)

Case IV: When z is 3

x + 5y = 20

possible values are: (20, 0, 3, 0), (15, 1, 3, 0), (10, 2, 3, 0), (5, 3, 3, 0), (0, 4, 3, 0)

Case V: When z is 4

x + 5y = 10

possible values are: (10, 0, 4, 0), (5, 1, 4, 0), (0, 2, 4, 0)

Case VI: When z is 5

x + 5y = 0

possible values are: (0, 0, 5, 0)

Total no of ways : 36


So total no of ways = 1 + 12 + 36 = 49

49 is answr

49 as a+5b+10c+25d=50 ;; there are 49 combinations

set on
p+5p+10p+25p=50
take combination of
take 1p and rest will zero
simlrly take 5p
simlry take 10p
simlrly take 125p
again make comniation of 2 diffrnt
agin make combination of 3 diffrnt
last combination of 4
then u will get total combination =52

@aparna bajpai
how is it 49? can u explain?