Question Bank - Nitin Gupta, Alphanumeric

• The sum of Rs. 21 can be arrived at in two ways using 2 and 5 rupee coins. They are 5 + 16 (one 5 rupee coin and eight 2 rupee coins) and 15 + 6 (three 5 rupee coins and three 2 rupee coins).The sum of Rs. 65 can be arrived at using 2 and 5 rupee coins as 5 + 60 (one 5 rupee coin and thirty 2 rupee coins), 15 + 50 (three 5 rupee coins and twenty five 2 rupee coins), 25 + 40 (five 5 rupee coins and twenty 2 rupee coins), 35 + 30 (seven 5 rupee coins and fifteen 2 rupee coins), 45 + 20 (nine 5 rupee coins and ten 2 rupee coins), 55 + 10 (eleven 5 rupee coins and five 2 rupee coins) and 65 + 0 (thirteen 5 rupee coins and no 2 rupee coin).
Consider option 1.
If Ashok has only one five rupee coin, he would have at least thirty 2 rupee coins so that he can arrive at the value of Rs. 65. This would mean that he can also arrive at the value of Rs. 21 as he would have sufficient number of 5 and 2 rupee coins to do so. Hence this option cannot be true.
Consider option 2.
The same logic as that in option 1 would apply here, as he would have sufficient number of 5 and 2 rupee coins to arrive at the value of Rs. 21. Hence this option also cannot be true.
Consider option 3.
If Ashok has only five 2 rupee coins he would have at least eleven 5 rupee coins to arrive at the value of Rs. 65. This would mean that he can also arrive at the value of Rs. 21 as he would have sufficient number of 5 and 2 rupee coins to do so. Hence this option cannot be true.
Consider option 4.
If Ashok has only one 2 rupee coin, he would have at least thirteen 5 rupee coins to arrive at the value of Rs. 65. Now in order to arrive at the value of Rs. 21, he would need at least two more 2 rupee coins. Thus this option can be a possibility.
Consider option 5.
The same logic as that in option 1 would apply here, as he would have sufficient number of 5 and 2 rupee coins to arrive at the value of Rs. 21. Hence this option also cannot be true.
Hence option 4.

• Q95) Moreshwar and Ganesh started travelling towards each other from their hometowns, Hyderabad and Bangalore respectively. They met at point P in between for the first time. As soon as they met,they exchanged their cars (which could travel with their predefined speeds only) and turned back to travel towards their respective hometown cities. As soon as they reached their hometowns, they again started travelling back towards the other city and met at point Q for the second time. Note that after meeting at point P they did not meet each other before they reached their respective hometown cities. What was the ratio of their speeds such that the distance PQ was the highest?
a) 2 : 5
b) 1 : 2
c) 2 : 3
d) 5 : 6
e) 1 : 3

• Q96) In a shooting competition, four shooters participated in two rounds. In each round, the shooters took one shot each at the target. For hitting the bull’s eye, the shooters were awarded 10 points. If they did not hit the bull's eye, depending upon the distance of the shot from the bull's eye, they were awarded either 5 or 0 points. The following facts were known about the competition.

• At least 2 shooters in the first round and at least 3 shooters in the second round hit the bull's eye
• No shooter scored a zero in the competition
• The total score in both the rounds put together was 65

Which of the following cannot be a true statement.

a) There was only one shooter who hit the bull's eye in both the rounds
b) There were only two shooters who hit the bull's eye in both the rounds
c) Exactly one shooter had the same score in both the rounds
d) Exactly two shooters had the same score in both the rounds
e) Exactly three shooters had the same score in both the rounds

• Q97) If Z is a set of three digit numbers of base 10 which are divisible by 3 and end with 1 when written in base 5 and base 7 notations, then how many elements does Z has?
a) 18
b) 9
c) 12
d) 15
e) 6

• A number in base 10 has to leave remainder 1 when divided by 5 or 7 so that it ends in 1 when written in base 5 and base 7 notations.
∴ The numbers are of the form (5 × 7 × m) + 1 = 35m + 1 [Here, m is an integer]
Now all these numbers must be divisible by 3.
35m + 1 = 33m + 2m + 1
2m + 1 has to be divisible by 3. [i.e. odd multiples of 3]
Different values for 2m + 1 = 3, 9, 15, 21, 27, 33, 39, 45, 51, 57, 63 …
Corresponding values for m = 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31 …
Corresponding values for 35m + 1 = 36, 141, …, 981, 1089 …
∴ m can take values from 4, 7, 10, …, 28 which gives a three digit number for 35m + 1.
These are 9 values of m.
∴ There will be 9 elements of set Z.
Hence, option 2.

• Q98) A frog is sitting at one vertex (A) of a hexagonal pond ABCDEF. It wants to reach the opposite vertex (D). However, it can only jump from one vertex to the adjacent vertices (i.e. from A to B or A to F,from B to C or B to A, etc.). Let J(n) represent the number of ways in which the frog can reach D from A in n jumps. Therefore, J(1) = 0, J(2) = 0, J(3) = 2 ... and so on. Then what is the value of J(2n)? (n is a natural number)
a) 0
b) 1
c) 2n – 1
d) (2n – 1)!
e) Cannot be determined

• The minimum number of jumps the frog needs to take is 3, as mentioned in the question.
Now, if the frog jumps in the 'wrong' direction before reaching its destination, it will have to jump once more in the 'right' direction to compensate for that jump.
∴ The number of jumps required to reach D will increase in steps of 2 (For example: suppose the frog's route is A-B-C-B-C-D, then it has taken 2 'extra' jumps, C to B and back to C).
∴ The frog can only reach D from A in an odd number of jumps.
∴ J(2n) = 0
Hence, option 1.

• Q99) A palindromic number is a positive integer which is unchanged on reversal of its digits. For example, 393 is the same even when its digits are reversed and is hence a palindromic number.
Let X = (number of 6 digit palindromic numbers) – (number of 5 digit palindromic numbers).
What is the value of X?
a) 0
b) 1
c) 50
d) 99
e) 100

• First, we need to find the number of 6 digit palindromic numbers. We can do this by considering the first three digits. Consider the list of three digit numbers 100 to 999. There are 900 numbers in this list. Now, the first 3 digits of all 6 digit numbers must be one of these numbers. Also, only one 6-digit
palindrome corresponds to any given starting three digit number. This can be easily seen.
Assume that our starting three digit number is 100. The six-digit palindrome starting with 100 has its 4th, 5th and 6th digits determined by the 3rd, 2nd and 1st digits of 100 respectively. Thus, there is one and only one 6 digit palindrome starting with each of 900, 3 digit numbers. The total number of 6-digit palindromic numbers is therefore 900.
Now, we need to find the number of 5 digit palindromic numbers. Again, we consider the first three
digits. We can prove, just as in the 6 digit case, that there is one and only one, 5 digit palindrome starting with each of 900 three digit numbers. For example, corresponding to the 3 digit number 654 is the five digit palindrome 65456. Thus, there are 900, 5-digit palindromic numbers.
∴ X = (number of 6 digit palindromic numbers) – (number of 5 digit palindromic numbers)
= 900 – 900 = 0
Hence, option 1.

• Q100) How many 2 digit numbers are there such that absolute difference of the square of the 2 digits is a prime number?
a) 21
b) 12
c) 10
d) 14
e) 16

• Let the two digit number be represented as ab.
a^2 – b^2 = (a – b)(a + b)
Now, |a^2 – b^2| will be a prime number only when |a – b| = 1 and (a + b) comes out to be a prime number
Two digit numbers in which |a – b| = 1 are 10, 12, 21, 23, 32, 34, 43, 45, 54, 56, 65, 67, 76, 78, 87,89, 98
From these numbers, the numbers in the form of 'ab' in which (a + b) comes out to be a prime number are 12, 21, 23, 32, 34, 43, 56, 65, 67, 76, 89 and 98.
There are 12 two digit numbers such that absolute difference of the square of the two digits results in a prime number.

207

136

106

149

160

199

92

119