# Question Bank - Nitin Gupta, Alphanumeric

• Q89) My complete date of birth is represented as d/m/1976. Here, d, m and 1976 are the date, month and year of birth respectively. Now d is multiplied by the highest possible value of date in a month and m is multiplied by the highest possible value of month in a year. The sum of these two products is 487. What is the sum of my date and month of birth?
a) 23
b) 13
c) 7
d) 15
e) None of these

• The highest possible value of date and month in a calendar are 31 and 12 respectively.
∴ 31d + 12m = 487 …(i)
Since there is only one equation and number of variables is two, we cannot find the solution of this equation by normal method.
But, there is other information given in the question implicitly. The date can be any of the integers 1,2, …, 31 and the month can be any of the integers 1, 2, …, 12. By combining this information with equation (i) we can find a unique solution.
Consider equation (i), when left hand side and right hand side of this equation is divided by 12, the remainders should be same. [Note: we are taking 12 here, because 12 is the smaller coefficient]
Second term 12m will not give any remainder when divided by 12, and the remainder when 31d is divided by 12 will be same as the remainder when 7d is divided by 12. Also, the remainder when 487 is divided by 12 will be 7.
∴ From L.H.S. (the remainder when 7d is divided by 12) = from R.H.S. (the remainder when 487 is divided by 12)
∴ The remainder when 7d is divided by 12 = 7
To satisfy the above condition the minimum value of d could be 1. [Because when 7 is divided by 12 the remainder is 7 itself]
The corresponding value of m can be obtained by substituting d = 1 in equation (i).
∴ When d = 1, m = 38
We have to discard this pair of (d, m), because the month of birth cannot be 38.
To find the other pair of values for d and m, we increase the value of d by 12 (the coefficient of m) and decrease the value of m by 31 (the coefficient of d).
∴ The other pair for (d, m) will be (13, 7), (25, −24) and so on. Here also, we have to reject the second pair because month cannot be negative. Therefore no other pair is possible, and the only valid pair of date and month possible is d = 13 and m = 7.
∴ My complete date of birth is 13/7/1976.
∴ Sum of my date and month of birth = 13 + 7 = 20
Hence, option 5.

• Q90) Anand, Jaspal and Gaurav buy some fruits from a vendor. Anand buys 5 apples, 3 oranges and 6 pineapples at Rs. 118. Jaspal buys 8 pineapples, 6 apples and 4 oranges at Rs. 152. Gaurav buys 4 apples, definitely some oranges and more pineapples than oranges at Rs. 58. Unhappy with the quality of pineapples, Gaurav exchanges them for oranges and gets back Rs. 2 so that there is no loss of value. Gaurav now has 4 times the oranges he originally had. What is the cost of an apple, a pineapple and an orange?
a) 20
b) 24
c) 16
d) 26
e) Cannot be determined

• Let the cost of an apple, an orange and a pineapple be Rs. x, y and z respectively.
Therefore, for the purchase made by Anand, Jaspal and Gaurav, we can write the following equations:
5x + 3y + 6z = 118 ...(i)
6x + 4y + 8z = 152 ...(ii)
4x + my + nz = 58 ...(iii)
where m and n are the number of oranges and pineapples respectively bought by Gaurav.
Solving equations (i) and (ii), we get, x = 8
Substituting this value of x in equation (i), we get,
y + 2z = 26 ...(iv)
Now for equation (iii), if we substitute the value of x as 8 we get,32 + my + nz = 58
∴ my + nz = 26
Now the minimum value of m is one and since n is more than m the minimum value of n is 2.
Now one orange and two pineapples cost Rs. 26 [from equation (iv)]. Thus he buys one orange and two pineapples.
So when he exchanges 2 pineapples for oranges, he gets 3 oranges [because, at the end he is having 4 oranges] and Rs. 2.
∴ 2z = 3y + 2
We also know y + 2z = 26
Solving these two equations we get,
y = 6 and z = 10
Cost of an apple is Rs. 8.
Therefore the cost of an apple, a pineapple and an orange is Rs. 24.
Hence, option 2.

• Q91) A town is protected by a circular wall which has three doors A, B and C. The door A is at a distance of 3 km from both the doors B and C. The perpendicular distance between the line joining doors A and B and the centre of the town is 2 km. What is the distance between the doors B and C?
a) 4.8 km
b) 4 km
c) 3.9 km
d) 4.2 km
e) 5.1 km

• Q92) Following inequalities represent length of sides of ΔABC,

1. |a − 7| ≤ 2
2. ||b − 4| − 7| ≤ 3
3. c^2 − 2c − 24 < 0

If a, b and c are integers, what is the difference between the minimum possible and maximum possible perimeter of the ΔABC?
a) 14
b) 11
c) 10
d) 12
e) 13

• Consider the first inequality |a − 7| ≤ 2
∴ −2 ≤ a − 7 ≤ 2
∴ 5 ≤ a ≤ 9
∴ Possible values of a are 5, 6, 7, 8 and 9
||b − 4| − 7| ≤ 3
−3 ≤ |b − 4| − 7 ≤ 3
4 ≤ |b − 4| ≤ 10
Consider |b − 4| ≥ 4
b − 4 ≥ 4 or b − 4 ≤ −4
∴ b ≥ 8 or b ≤ 0
∵ Side of triangle cannot be 0 or negative
∴ b ≥ 8 ......(i)
Consider |b − 4| ≤ 10
− 10 ≤ b − 4 ≤ 10
∴ −6 ≤ b ≤14
∵ Side of triangle cannot be 0 or negative
∴ b ≤ 14 ......(ii)
From (i) and (ii),
8 ≤ b ≤ 14
∴ Possible values of b are 8, 9, 10, 11, 12, 13 and 14
c2 − 2c − 24 < 0
(c − 6)(c + 4) < 0
∴ −4 < c < 6
∵ Side of triangle cannot be 0 or negative
∴ c < 6
∴ Possible values of c are 1, 2, 3, 4 and 5.
Now we have all the possible values of the sides a, b, c. We need to consider an additional condition that in a triangle the sum of any two sides is greater than the third side.
∴ The possible triangle which satisfies the given inequalities and has the minimum perimeter will be having side lengths 5, 4, 8. (Other combinations are also possible like 8, 8, 1)
∴ Minimum possible perimeter = 5 + 4 + 8 = 17
Similarly,Maximum possible perimeter = 9 + 13 + 5 = 27
Required Difference = 27 − 17 = 10
∴ The difference between the minimum possible and maximum possible perimeter of the ΔABC is 10 units.
Hence, option 3.

• Q93) 1000 students appeared for the talent search examination NCERT. The examination had 1000 questions. During checking of the papers, a very peculiar trend was noticed. The first student attempted all the questions, the second student attempted questions with question numbers which were multiples of 2, the third student attempted the questions with question numbers which were multiples of 3 and so on.
Let f(n) = the number of students who attempted the question number n
For example, f(10) represents the number of students who attempted the question number 10 and f(10) = 4 because four students (1st, 2nd, 5th and 10th) attempted the question.For how many questions was f(n) odd?
a) 234
b) 31
c) 78
d) 101
e) None of these

• The first student attempted all the questions i.e. question number 1, 2, 3, ..., 999, 1000.
The second student attempted the questions with numbers which were multiple of 2 i.e. question
number 2, 4, 6, ..., 998, 1000.
The third student attempted the questions with numbers which were multiple of 3 i.e. question
number 3, 6, 9, 12, ..., 999.
And so on.
Here we can see that the number of students who have attempted a particular question is the
number of factors (including the number itself and 1) it has. For example, f(24) = 8 as the question
number 24 was attempted by 8 students (1st, 2nd, 3rd, 4th, 6th, 8th, 12th and 24th student).
∴ We have to find the question numbers for which the number of factors are odd.
We know that any number 'n' can be expressed in the form,
n = 2(b)3(c)5(d)... where 2, 3, 5,.... are prime numbers
The number of factors for n = (b + 1)(c + 1)(d + 1)...
To have odd number of factors each of (b + 1), (c + 1), (d + 1), ... should be odd.
∴ b, c, d, ... all should be even.
∴ n is a perfect square.
∴ The questions with numbers which were perfect squares were attempted by odd number of students.
∴ The problem reduces to finding out the number of perfect squares in the range 1 - 1000.
There are 31 perfect squares between 1 to 1000 including 1 and 1000.
Hence, option 2.

• Q94) Ashok has some 5 rupee and some 2 rupee coins. He finds that he cannot choose a set of coins out of his collection whose total value is Rs. 21. However, he can choose a set of coins whose total value is Rs. 65. Which of these statements could be true?
a) Ashok has only one 5 rupee coin
b) Ashok has only two 5 rupee coins
c) Ashok has only five 2 rupee coins
d) Ashok has only one 2 rupee coin
e) Ashok has only seven 5 rupee coins

• 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

22

123

200

127

145

207

151

155