# DILR – Logical Reasoning – Set The following set contains four questions related to Logical Reasoning. Choose the best answer to each question.

An ATM dispenses exactly Rs. 5000 per withdrawal using 100, 200 and 500 rupee notes. The ATM requires every customer to give her preference for one of the three denominations of notes. It then dispenses notes such that the number of notes of the customer’s preferred denomination exceeds the total number of notes of other denominations dispensed to her.

1) In how many diﬀerent ways can the ATM serve a customer who gives 500 rupee notes as her preference?

Solution: As it is given that the number of notes of the customer’s preferred denomination exceeds the total number of notes of other denominations dispensed to her So the following cases are possible,
 Cases No. of 500 rupee notes No. of 200 rupee notes No. of 100 rupee notes 1 10 0 0 2 9 2 1 3 9 1 3 4 9 0 5 5 8 5 0 6 8 4 2 7 8 3 4

So total number of required ways = 7

2) If the ATM could serve only 10 customers with a stock of fifty 500 rupee notes and a sufficient number of notes of other denominations, what is the maximum number of customers among these 10 who could have given 500 rupee notes as their preferences?

Solution: As total number of 500 notes = 50

From the solution of the previous question we can see if the customers could have given 500 rupee notes as their preferences minimum number of 500 rupee notes they will need is 8 . So maximum number of who could have given 500 rupee notes as their preferences = [50/8 ] = 6

3) What is the maximum number of customers that the ATM can serve with a stock of fifty 500 rupee notes and a sufficient number of notes of other denominations, if all the customers are to be served with at most 20 notes per withdrawal?

a) 13
b) 10
c) 16
d) 12
Solution: To maximize the number of customers we need to minimize the number of 500 rupee notes per customers.
Such that x+y+z ≤ 20——1)
and 500x+200y+100z=5000—–2)
Where x = number of 500 rupee notes
y = number of 200 rupee notes
z = number of 100 rupee notes
So we need to minimize x. and from above two equations minimum value of x =  4.
Thus maximum number of customers that the ATM can serve [50/4] = 12

4.) What is the number of 500 rupee notes required to serve 50 customers with 500 rupee notes as their preferences and another 50 customers with 100 rupee notes as their preferences, if the total number of notes to be dispensed is the smallest possible?

a) 800
b) 900
c) 1400
d) 750

Solution:

Case 1) When with 500 rupee notes is their preference , to minimize the number of notes we need to serve  maximum possible 500 notes, minimum number of notes required = [5000/500] = 10
So total number of 500 notes required to to serve 50 customers = 50*10 = 500
Case 2) When with 100 rupee notes is their preference, to minimize the number of notes we need to serve  maximum possible 500 notes. So
100z+500x=5000———1) and
z>x such that (z+x)isminimum .——-2)
From eq 1) & eq 2) only possible solution is x = 8 and z = 10
Thus total number of 500 notes required to to serve 50 customers = 50*8 = 400
So required number of 500 notes = case 1) + case 2) = 500 + 400 = 900

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