11221.How many divisors (including 1, but excluding 1000) are there for the number 1000?
15
16
31
10
11222.A call center agent has a list of 305 phone numbers of people in alphabetic order of names (but she does not have any of the names). She needs to quickly contact Deepak Sharma to convey a message to him. If each call takes 2 minutes to complete, and every call is answered, what is the minimum amount of time in which she can guarantee to deliver the message to Mr Sharma.
180 minutes
18 minutes
206 minutes
34 minutes
11223.The times taken by a phone operator to complete a call are 2,9,3,1,5 minutes respectively. What is the average time per call?
1 minutes
7 minutes
4 minutes
5 minutes
11224.The times taken by a phone operator to complete a call are 2,9,3,1,5 minutes respectively. What is the median time per call?
5 minutes
None of these
1 minutes
4 minutes
Explanation:
NO option is correct. Median is 3
11225.Eric throws two dice, and his score is the sum of the values shown. Sandra throws one die, and her score is the square of the value shown. What is the probability that Sandra?s score will be strictly higher than Eric?s score?
5/6
17/36
173/216
137/216
11226.What is the largest integer that divides all three numbers 23400,272304,205248 without leaving a remainder?
48
96
24
72
11227.Of the 38 people in my office, 10 like to drink chocolate, 15 are cricket fans, and 20 neither like chocolate nor like cricket. How many people like both cricket and chocolate?
10
7
15
18
11229.If f(x) = 7 x +12, what is f-1(x) (the inverse function)?
No inverse exists
7x+12
1/(7x+12)
(x-12)/7
11230.A permutation is often represented by the cycles it has. For example, if we permute the numbers in the natural order to 2 3 1 5 4, this is represented as (1 3 2) (5 4). In this the (132) says that the first number has gone to the position 3, the third number has gone to the position 2, and the second number has gone to position 1, and (5 4) means that the fifth number has gone to position 4 and the fourth number has gone to position 5. The numbers with brackets are to be read cyclically. If a number has not changed position, it is kept as a single cycle. Thus 5 2 1 3 4 is represented as (1345)(2). We may apply permutations on itself If we apply the permutation (132)(54) once, we get 2 3 1 5 4. If we apply it again, we get 3 1 2 4 5 , or (1 2 3)(4) (5) If we consider the permutation of 7 numbers (1457)(263), what is its order (how many times must it be applied before the numbers appear in their original order)?
12
None of these
7! (factorial of 7)
14
11231.What is the maximum value of $x^3y^3 + 3 x \times y$ when x+y = 8?
4144
256
8192
102
Explanation:
The question probably be $x^3y^3 + 3 x \times y$Sustitute x = 4 and y = 4
11232.A white cube[with six faces] is painted red on two different faces. How many different ways can this be done [two paintings are considered same if on a suitable rotation of the cube one painting can be carried to the other]?
2
15
4
30
11233.In Goa beach, there are three small picnic tables. Tables 1 and 2 each seat three people. Table 3 seats only one person, since two of its seats are broken. Akash, Babu, Chitra, David, Eesha, Farooq, and Govind all sit at seats at these picnic tables. Who sits with whom and at which table are determined by the following constraints:1. Chitra does not sit at the same table as Govind.2. Eesha does not sit at the same table as David.3. Farooq does not sit at the same table as Chitra.4. Akash does not sit at the same table as Babu.5. Govind does not sit at the same table as Farooq.Which of the following is a list of people who could sit together at table 2?
Govind, Eesha, Akash
Babu, Farooq, Chitra
Chitra, Govind, David.
Farooq, David, Eesha.
11234.There are a number of chocolates in a bag. If they were to be equally divided among 14 children, there are 10 chocolates left. If they were to be equally divided among 15 children, there are 8 chocolates left. Obviously, this can be satisfied if any multiple of 210 chocolates are added to the bag. What is the remainder when the minimum feasible number of chocolates in the bag is divided by 9?
2
5
4
6
11235.Let f(m,n) =$45 \times m$ + $36 \times n$, where m and n are integers [positive or negative] What is the minimum positive value for f(m,n) for all values of m,n [this may be achieved for various values of m and n]?
9
6
5
18
11236.What is the largest number that will divide 90207, 232585 and 127986 without leaving a remainder?
257
905
351
498
11237.We have an equal arms two pan balance and need to weigh objects with integral weights in the range 1 to 40 kilo grams. We have a set of standard weights and can place the weights in any pan. . [i.e] some weights can be in a pan with objects and some weights can be in the other pan. The minimum number of standard weights required is:
4
10
5
6
42407.In particular language if A=0, B=1, C=2,........ , Y=24, Z=25 then what is the value of ONE+ONE (in the form of alphabets only)
BDAI
ABDI
DABI
CIDA
Explanation:
This problem is based on Base 26 rather than regular base 10 (decimal system) that we normally use. In base 10 there are 10 digits 0 to 9 exist. In base 26 there are 26 digits 0 to 25 exist. To convert any number into base 26, we have to divide the number with 26 and find the remainder.
Here, ONE + ONE =
E has value of 4. So E + E = 8 which is equal to I.
Now N + N = 13 + 13 = 26. But in base 26, there is no 26. So
$(26)_{10}=(10)_{26}$
So we put 0 and 1 carry over. But 0 in this system is A.
Now O + O + 1 = 14 + 14 + 1 = 29
Therefore,
$(29)_{10}=$(13)_{26}
But 1 = B and 3 = D in that system. So ONE + ONE = BDAI
This problem is based on Base 26 rather than regular base 10 (decimal system) that we normally use. In base 10 there are 10 digits 0 to 9 exist. In base 26 there are 26 digits 0 to 25 exist. To convert any number into base 26, we have to divide the number with 26 and find the remainder.
Here, ONE + ONE =
E has value of 4. So E + E = 8 which is equal to I.
Now N + N = 13 + 13 = 26. But in base 26, there is no 26. So
$(26)_{10}=(10)_{26}$
So we put 0 and 1 carry over. But 0 in this system is A.
Now O + O + 1 = 14 + 14 + 1 = 29
Therefore,
$(29)_{10}=$(13)_{26}
But 1 = B and 3 = D in that system. So ONE + ONE = BDAI
42408.Find the number of perfect squares in the given series 2013, 2020, 2027,................, 2300 (Hint 44^2=1936)
1
2
3
Cant be determined
Explanation:
The given series is an AP with common difference of 7. So the terms in the above series are in the form of 2013 + 7k. We have to find the perfect squares in this format in the given series.
Given that 44^2 = 1936.
Shortcut: To find the next perfect square, add 45th odd number to 44^2.
So 45^2 = 1936 + (2 x 45 -1) = 2025
46^2 = 2025 + (2 x 46 - 1) = 2116
47^2 = 2116 + (2 x 47 - 1) = 2209
Now subtract 2013 from the above numbers and divide by 7. Only 2209 is in the format of 2013 + 7k. One number satisfies.
The given series is an AP with common difference of 7. So the terms in the above series are in the form of 2013 + 7k. We have to find the perfect squares in this format in the given series.
Given that 44^2 = 1936.
Shortcut: To find the next perfect square, add 45th odd number to 44^2.
So 45^2 = 1936 + (2 x 45 -1) = 2025
46^2 = 2025 + (2 x 46 - 1) = 2116
47^2 = 2116 + (2 x 47 - 1) = 2209
Now subtract 2013 from the above numbers and divide by 7. Only 2209 is in the format of 2013 + 7k. One number satisfies.
42409.What is in the 200th position of 1234 12344 123444 1234444....?
3
4
5
7
Explanation:
The given series is 1234, 12344, 123444, 1234444, .....
So the number of digits in each term are 4, 5, 6, ... or (3 + 1), (3 + 2), (3 + 3), .....upto n terms =$3n+\frac{n(n+1)}{2}$
So $3n+\frac{n(n+1)}{2}$ <=200
For n = 16, We get 184 in the left hand side. So after 16 terms the number of digits equal to 184. And 16 them contains 16 + 3 = 19 digits.
Now 17 term contains 20 digits and 123444......4(17times) So last digit is 4 and last two digits are 44.
The given series is 1234, 12344, 123444, 1234444, .....
So the number of digits in each term are 4, 5, 6, ... or (3 + 1), (3 + 2), (3 + 3), .....upto n terms =$3n+\frac{n(n+1)}{2}$
So $3n+\frac{n(n+1)}{2}$ <=200
For n = 16, We get 184 in the left hand side. So after 16 terms the number of digits equal to 184. And 16 them contains 16 + 3 = 19 digits.
Now 17 term contains 20 digits and 123444......4(17times) So last digit is 4 and last two digits are 44.