Entries Tagged 'Prep Material' ↓

5. Telephone Interview
Usually such interviews are used for selecting candidates from a large pool of people and also conducted on people who live far away from job location. Telephonic interview can be sued as initial screening of candidates. Telephonic interview is similar to face-to-face interview. Set up a place and time where there is no one to disturb or distract you. Focus on conversation as voice is the only key to make an impact. Keep your resume at hand. Disable the call waiting feature on your phone as your wouldn’t want to be disturbed during an interview.

6. Group Interview
Here, selected candidates are formed into an informal group. The interviewer will start the conversation and the candidates should take in over from there. The aim of this style of interview is to understand how can one can converse with other group members and influence or convince them. So, in short, it tries to gauge the leadership potential of each candidate.

7. Lunch/Dinner Interview
This is an interview planned over lunch or dinner. Though the setting at lunch or dinner may be casual, you should never get lax or complacent about yourself or the people around you. Since it’s a business meal, you need to maintain your composure and table etiquettes. For this, try to take the lead from the interviewer. Avoid consuming alcohol or eating messy food during the interview. That may leave to bad taste!

8. Stress Interview
This type of interview was very popular among sales position but rare today. Stress Interview tries to see how you handle pressure and under what circumstances will you give up. The interviewer may try tricks like being argumentative or may try to put you down or make a nasty comment – but you should realize it’s just a deliberate attempt to break your morale. So keep your cool and take time to answer each question without being in any hurry. The interviewer may also take into abrupt silence when asking a question; don’t get nervous and ask for clarification if needed.

•Whenever there appears any term of the type a3 + b3 + c3, do check for a + b + c being equal to zero. If a + b + c is indeed zero, then a3 + b3 + c3 = 3abc.

•The series 1, 3, 6, 10, 15 should immediately be recognized as series of sum of first n natural numbers.

•To form all natural numbers from 1 to N by adding any natural numbers, one would just need 1, 2, 4, 8, 16, 32, 64…..2n, where 2n is the largest power of 2 smaller than or equal to N. E.g. What is the minimum number of weights needed to be able to measure all natural numbers weights till 80, if weights can be kept only on one pan of the balance. One would need weights 1, 2, 4, 8, …64

•To form all natural numbers from 1 to N by adding or subtracting any natural numbers, one would just need 1, 3, 9, 27, 81, …..3n. Be careful of the largest number needed in this case. E.g. What is the minimum number of weights needed to be able to measure all natural numbers weights till 80, if weights can be kept on both pans of the balance. One would need weights 1, 3, 9, 27, 81.

•In questions of the type where certain flowers/sweets/points etc gets diminished and then increases and again diminishes and again increases……rather than forming an equation, see if one can work backwards if final quantity is given or else work with options. Also in such cases, if 1/3rd the objects are given away, work on the objects that are remaining i.e. 2/3rd to save time.

If working with options, select options intelligently e.g. I pick 1/3rd of the chocolates in a bowl and then return 3, next I pick 1/5th of the chocolates and then return five, next ……What is the number of chocolates in the bowl initially? The initial number of chocolates has to be a multiple of 3. Also 2/3rd of the initial number of chocolates plus 3 should be divisible by 5. This should be enough to reduce the possible options to just about 2.

•Remember that a2, b2, c2 or |a|, |b|, |c| are either zero or positive quantity. Thus solution to a2 + b2 + c2 = 0 or |a| + |b| + |c| = 0 is a = b = c =0

Integer solutions to equations of the type Ax ± By = C:

• If A, B and C have a common factor divide the entire equation by this common factor. The following matter considers that you have done this step and A and B are considered after the division.

• Now if A and B are not co-prime, the equation will not have any integer solution. Think why!!!

• Consecutive integer solutions to the equation will have x values differing by B and y values differing by A. For equation of type Ax + By = C, a increase (decrease) in x will cause a decrease (increase) in y and vice-versa. For equation of type Ax – By = C, a increase (decrease) in x will cause a increase (decrease) in y and vice-versa.

• Thus in any set of A consecutive integers, y can assume one and only one value for which it will be a integer solution to the equation. And in any set of B consecutive integers, x can assume one and one one value for which it will be a integer solution to the eqn. (To understand this point, i.e. if you have’nt, refer to the concept explained in this article under the heading of LCM and HCF)

• E.g. For the equation 17x + 5y = 513:
There has to be a value of x from 1 to 5 (or from 0 to 4) that will satisfy this equation. For an integral solution 513 – 17x has to be divisible by 5 i.e. 513 – 17x has to end with a 0 or 5. Thus from 0 to 4, x can only take a value 4. Thus one can easily find that (4, 97) is a solution to this equation.

Further if the total number of natural number solution is asked, as x increases in steps of 5 starting with 4 (there is no question of x decreasing as it will not be a natural number then), y will decrease in steps of 17, starting from 97. Since 17 × 5 = 85, there can be 5 such steps of decrease and the number of natural number solutions is 6
If the question was asked as to how many integer solutions exists for the equation such that -43 = x = 67, from -43 to 67 there exists a total of 110 integers. Since x can take one and only one value in any given 5 consecutive integers, the number of solutions will be 22.

If the question was asked as to how many integer solutions exists for the equation such that 0 < x < 168, from 1 to 165 there exists a total of 165 integers. Since x can take one and only one value in any given 5 consecutive integers, the number of solutions will be for 0 < x = 165 will be 33. For x = 166 and 167 would not be a solution as x has to be of the type 5n – 1 to be a solution.

• Consider this question, how many non-negative integral solution exists for the equation 17x – 4y = 1 such that y = 1000.
y can take one and only one value out of 17 consecutive integers to be a solution to the equation. 17 × 60 = 1020. Thus 17 × 58 = 986. Thus from 1 to 986, y will take 58 values. Now all that remains to be checked is that from 987 to 1000 will there be a value that y can take to be a solution to the equation. One solution x = 1 and y = 4 is evident. Thus if we consider sets of 17 consecutive integers as 1 to 17, 18 to 34, 35 to 51, the fourth number in the set will be a solution. Thus from 987 to 1003, y = 991 will be the 59th solution.

• While solving simultaneous equations of the form x2 + y2 = 65 and x + y = 3, there is no real need to solve it theoretically, which turns out to be a laborious process. Consider perfect squares and by hit and trial see if two perfect squares add up to 65. x2 cannot be 4, 9, 25, 36 as then y2 has to be 61, 56, 40, 29 which are not perfect squares. Thus the possible values for x and y are (±1, ±8) and (±4, ±7). The values that satisfy x + y = 3 are 7 and -4.

The only assumption in the above is that x and y are rational. And this can be ascertained by looking at option choices or certain other data in the question or else one can simply take chances. If one does not get perfect squares adding up to 65, only then work theoretically.

Consider this question:
If x2 + y2 = 0.1 and |x – y| = 0.2, what is the value of |x| + |y|?
1. 0.3 2. 0.4 3. 0.6 4. 0.2
Consider the square of 0.1, 0.2, 0.3, 0.4 i.e. 0.01, 0.04, 0.09, 0.16 and it is obvious that 0.01 + 0.09 = 0.1. Thus x and y are 0.1 and 0.3. And these values also satisfy the second equation. Also the answer choices make it evident that one should not be thinking of multiple values or solutions. Thus answer has to be 0.4

• In simultaneous equations please remember that we need as many independent equations as the number of variables used ONLY when we have to find the unique value of all the variables. If we do not need the values of all the variables but only the value of an equation using the variables, we can possibly find the answer even when the number of equations is less than the number of variables. So do not be in a hurry to mark “Cannot be determined” as the answer in such case.

Even after having as many equations as the number of variables, it is not necessary that we will be able to fid the values of all the variables. To be able t do so, all the equations should be independent. Check to see if by adding or subtracting two of the equations (including use of multiples if needed) we can derive one of the given equation. If we can do this, all the given equations are not independent.

• Whenever the question deals with roots of any polynomial, axn + bxn-1 + cxn-2 + …… + px + q = 0, remember to check options such that sum of roots = -b/a and product of roots = ±q/a. Thus if a = 1 (as is usually the case or else you could make a = 1 by dividing the entire equation by a) and q is an integer and it is also known that the roots are also integers, then the roots have to be factors of q.

• For a quadratic expression used in an inequality and that cannot be factorized, find the determinant and see if the quadratic expression takes only positive or negative values or takes all values:

For a quadratic ax2 + bx + c with imaginary roots i.e. b2 – 4ac < 0,
(ax2 + bx + c) is always positive if a is positive OR
(ax2 + bx + c) is always negative if a is negative

When you take CAT, what makes a crucial difference in determining whether you crack CAT and get into the IIMs or not, is how quick you can attempt questions.

One of the key observations that might throw light on CAT cracking methodologies is that - in most cases students spend about 60% of the time in quant and DI section on calculations. If you can make a difference to your calculation speed, even if it is marginal, it will make a substantial difference to your getting into an IIM. After all people make it to the cut off by a margin of 0.25 marks.

Practice speed calculation whenever you have to calculate anything. Engineers particularly, should try and throw their fx 100 calculators away, at least, for the time being. And commerce graduates do not go anywhere near your calculators for the next 6 months.

Number System Tips

If you have to find the square of numbers ending with ‘5′.

e.g 25 * 25. Find the square of the units digit (which is 5) = 25. Write this down.
Then take the tenths digit (2 in this case) and increment it by 1 (therefore, 2 becomes 3). Now multiply 2 with 3 = 6.
Write ‘6′ before 25 and you get the answer = 625.

For instance, if you have to find the square of 45.
45 * 45. The square of the units digit = 25
Increment 4 by 1. It will give you ‘5′. Now multiply 4 * 5 = 20.
Write 20 before 25. The answer is 2025.

In the case of numbers like 125. The rule applies without an iota of difference.
The square of the units digit = 25.
Increment 12 by 1. It will give you 13. Now multiply 12*13 = 156.
Write 156 before 25. The answer is 15625.

Try out the above technique with the following numbers.

Questions:

1. 225 * 225
2. 75 * 75
3. 145 * 145
4. 35 * 35
5. 375 * 375.

Answers and Solution:

1. 225 * 225. Square of units digits is 25. Increment 22 by 1 to get 23. The product of 22 and 23 is 506. Therefore, the answer is 50625.

2. 75 * 75. Square of the units digits is 25. Increment 7 by 1 to get 8. The product of 7*8 = 56. Therefore, the answer is 5625.

3. 145 * 145. Square of the units digits is 25. Increment 14 by 1 to get 15. The product of 14*15 = 210. Therefore, the answer is 21025.

4. 35 * 35. Square of the units digits is 25. Increment 3 by 1 to get 4. The product of 3*4=12. Therefore, the answer is 1225.

5. 375 * 375. Square of the units digits is 25. Increment 37 by 1 to get 38. The product of 37*38=1406. Therefore, the answer is 140625.

Exercise1 + Answers + Explanations

Question:
The sum of the first 100 numbers, 1 to 100 is divisible by

(1) 2, 4 and 8
(2) 2 and 4
(3) 2 only
(4) None of these

Correct Answer - (3)

Explanation:
The sum of the first 100 natural numbers is given by (n(n + 1))/2 = (100(101))/2 = 50(101).
101 is an odd number and 50 is divisible by 2. Hence, 50 (101) will be divisible by 2.

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Question:
What is the minimum number of square marbles required to tile a floor of length 5 metres 78 cm and width 3 metres 74 cm?

(1) 176
(2) 187
(3) 54043
(4) 748

Correct Answer - (2)

Explanation:
The marbles used to tile the floor are square marbles. Therefore, the length of the marble = width of the marble.As we have to use whole number of marbles, the side of the square should a factor of both 5 m 78 cm and 3m 74. And it should be the highest factor of 5 m 78 cm and 3m 74. 5 m 78 cm = 578 cm and 3 m 74 cm = 374 cm.
The HCF of 578 and 374 = 34.

Hence, the side of the square is 34.

The number of such square marbles required = = 187 marbles.

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What is the remainder when 9^1 + 9^2 + 9^3 + …. + 9^8 is divided by 6?

(1) 3
(2) 2
(3) 0
(4) 5

Correct Answer - (3)

Explanation:
6 is an even multiple of 3. When any even multiple of 3 is divided by 6, it will leave a remainder of 0. Or in other words it is perfectly divisible by 6.
On the contrary, when any odd multiple of 3 is divided by 6, it will leave a remainder of 3. For e.g when 9 an odd multiple of 3 is divided by 6, you will get a remainder of 3.

9 is an odd multiple of 3. And all powers of 9 are odd multiples of 3. Therefore, when each of the 8 powers of 9 listed above are divided by 6, each of them will leave a remainder of 3.

The total value of the remainder = 3 + 3 + …. + 3 (8 remainders) = 24. 24 is divisible by 6. Hence, it will leave no remainder.

Hence the final remainder when the expression 9^1 + 9^2 + 9^3 + ….. + 9^8 is divided by 6 will be equal to ‘0′.

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Question:
What is the reminder when 91 + 92 + 93 + …… + 99 is divided by 6?

(1) 0
(2) 3
(3) 4
(4) None of these

Correct Answer - (2)

Explanation:
Any number that is divisible by ‘6′ will be a number that is divisible by both ‘2′ and ‘3′. i.e. the number should an even number and should be divisible by three (sum of its digits should add to a multiple of ‘3′). Any number that is divisible by ‘9′ is also divisible by ‘3′ but unless it is an even number it will not be divisibly by ‘6′. In the above case, ‘9′ is an odd number. Any power of ‘9′ which is an odd number will be an odd number which is divisible by ‘9′.

Therefore, each of the terms 91, 92 etc are all divisibly by ‘9′ and hence by ‘3′ but are odd numbers.

Any multiple of ‘3′ which is odd when divided by ‘6′ will leave a reminder of ‘3′. For example 27 is a multiple of ‘3′ which is odd. 27/6 will leave a reminder of ‘3′. Or take 45 which again is a multiple of ‘3′ which is odd. 45/6 will also leave a reminder of ‘3′.

Each of the individual terms of the given expression 91 + 92 + 93 + …… + 99 when divided by 6 will leave a reminder of ‘3′. There are 9 such terms. The sum of all the reminders will therefore be equal to 9*3 = 27.

27/6 will leave a reminder of ‘3′.

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Question:
Find the value of 1.1! + 2.2! + 3.3! + ……+n.n!

(1) n! +1
(2) (n+1)!
(3) (n+1)!-1
(4) (n+1)!+1
Correct Answer - (3)

Explanation:
1.1! can be written as (2!-1!), likewise 2.2! = (3!-2!) , 3.3! = (4!-3!) and n.n! = (n+1)!-n!.
Thus by summing it up we get the result as given below.
1.1! + 2.2! + 3.3! + ……+n.n! = (2!-1!) + (3!-2!) + (4!-3!) +……+ (n+1)!-n! = (n+1)!-1.

Exercise2 + Answers + Explanations

Question:
‘a’ and ‘b’ are the lengths of the base and height of a right angled triangle whose hypotenuse is ‘h’. If the values of ‘a’ and ‘b’ are positive integers, which of the following cannot be a value of the square of the hypotenuse?

(1) 13
(2) 23
(3) 37
(4) 41

Correct Answer - (2)

Explanation:
The value of the square of the hypotenuse = h2 = a2 + b2

As the problem states that ‘a’ and ‘b’ are positive integers, the values of a2 and b2 will have to be perfect squares. Hence we need to find out that value amongst the four answer choices which cannot be expressed as the sum of two perfect squares.

Choice 1 is 13. 13 = 9 + 4 = 32 + 22. Therefore, Choice 1 is not the answer as it is a possible value of h2

Choice 2 is 23. 23 cannot be expressed as the sum two numbers, each of which in turn happen to be perfect squares. Therefore, Choice 2 is the answer.

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Question:
Two numbers when divided by a certain divisor leave remainders of 431 and 379 respectively. When the sum of these two numbers is divided by the same divisor, the remainder is 211. What is the divisor?

(1) 599
(2) 1021
(3) 263
(4) Cannot be determined

Correct Answer - (1)

Explanation:
Two numbers when divided by a common divisor, if they leave remainders of x and y and when their sum is divided by the same divisor leaves a remainder of z, the divisor is given by x + y - z.

In this case, x and y are 431 and 379 and z = 211. Hence the divisor is 431 + 379 - 211 = 599.

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Question:
What is the least number that should be multiplied to 100! to make it perfectly divisible by 350?

(1) 144
(2) 72
(3) 108
(4) 216

Correct Answer - (2)

Explanation:
100! has 348 as the greatest power of 3 that can divide it. Similarly, the greatest power of 2 that can divide 100! is 297. 297 = 448 * 21.
Therefore, the largest power of 12 that can divide 100! is 48.
Therefore, for 350 to be included in 100!, 100! needs to be multiplied by 32 * 23 = 9 * 8 = 72.

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Question:
A certain number when successfully divided by 8 and 11 leaves remainders of 3 and 7 respectively. What will be remainder when the number is divided by the product of 8 and 11, viz 88?

(1) 3
(2) 21
(3) 59
(4) 68

Correct Answer - (3)

Explanation:
When a number is successfully divided by two divisors d1 and d2 and two remainders r1 and r2 are obtained, the remainder that will be obtained by the product of d1 and d2 is given by the relation d1r2 + r1.

Where d1 and d2 are in ascending order respectively and r1 and r2 are their respective remainders when they divide the number.

In this case, the d1 = 8 and d2 = 11. And r1 = 3 and r2 = 7.
Therefore, d1r2 + r1 = 8*7 + 3 = 59.

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Question:
What is the total number of different divisors including 1 and the number that can divide the number 6400?

(1) 24
(2) 27
(3) 27
(4) 68

Correct Answer - (2)

Explanation:
To find the number of divisors for a number, express then number as the product of the prime numbers like ax * by * cz .

In this case, 6400 can be expressed as 64*100 = 26 * 4 * 25 = 28 * 52.

Having done that, the way to find the number of divisors is by multiplying the indices of each of the prime numbers after incrementing the indices by 1.
i.e. the number of divisors = (x+1)(y+1)(z+1).
In this case, (8 + 1)(2 + 1) = 9*3 = 27.

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Exercise3 + Answers + Explanations

Question:
When 26854 and 27584 are divided by a certain two digit prime number, the remainder obtained is 47. Which of the following choices is a possible value of the divisor?

(1) 61
(2) 71
(3) 73
(4) 89

Correct Answer - (3)

Explanation:
When both 26854 and 27584 are divided by a divisor, the remainder obtained happens to be the same value 47.

Which means that the difference between 26854 and 27584 is a multiple of the divisor. Else the remainders would have been different.

=> 27584 - 26854 = 730

Now we need to find out a two digit prime number, whose multiple will be 730. 730 can be written as 73 * 10, where 73 is a prime number and greater than 47.
Hence, the two digit prime divisor is 73.

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Question:
How many times will the digit ‘0′ appear between 1 and 10,000?

(1) 4000
(2) 4003
(3) 2893
(4) 3892
Correct Answer - (3)

Explanation:

In two digit numbers, ‘0′ appears 9 times.
In 3-digit numbers, either 1 ‘0′ or 2 ‘0s’ can appear.
When one ‘0′ appears: The first digit of a 3-digit number can be any of the 9 digits other than ‘0′. The second digit can be any of the ‘9′ digits and the third digit be ‘0′.Hence there will 9*9 = 81 such combinations. The ‘0′ can either be the 2nd digit or the third digit, hence there are 2 arrangements possible.Therefore, there are a total of 81 * 2 = 162 three digit numbers with ‘0′ as one of the digits.

When two ‘0s’ appear: The first digit of the number can be any of the 9 digits. The second and third digit are 0s.
Hence there are 9 such numbers and 0 appears 9*2 = 18 times. Hence, ‘0′ appears 18 + 162 = 180 times in three digit numbers. In 4-digit numbers, 0 might be either 1 or 2 or 3 of the digits.

When one ‘0′ appears: The remaining three digits can be any of the 9 digits.i.e. 93 combinations = 729. The ‘0′ can either be the second or third or fourth digit. Hence 3*729 numbers = 2187.

When two 0s appear: The remaining two digits can be any of the 9 digits.Hence 92 = 81 possible combinations. The two 0s can be in two of the three positions = 3*81 = 243 such numbers and 2*243 = 486 number of times 0 will be appear.

When three 0s appear: There are 9 such numbers and 0 appears thrice in each. Hence 9*3 = 27 times.
Hence the number of 0s in 4-digit numbers = 2187 + 486 + 27 = 2700 times.
There is one 5-digit number - 10,000 and 0 appears four times in this number.
Hence the total number of times 0 appears between 1 and 10,000 = 9 + 180 + 2700 + 4 = 2893.

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Question:
What number should be subtracted from x^3 + 4x^2 – 7x + 12 if it is to be perfectly divisible by x + 3?

(1) 42
(2) 39
(3) 13
(4) None of these
Correct Answer - (1)

Explanation:
According to remainder theorem when, then the remainder is f(-a). In this case, as x + 3 divides x3 + 4×2 – 7x + 12 – k perfectly (k being the number to be subtracted), the remainder is 0 when the value of x is substituted by –3.

=> (-3)3 + 4(-3)2 – 7(-3) + 12 – k = 0
=> -27 + 36 + 21 + 12 = k
=> k = 42

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Question:
What is the value of M and N respectively? If M39048458N is divisible by 8 and 11; Where M and N are single digit integers?

(1) 7, 8
(2) 8, 6
(3) 6, 4
(4) 5, 4

Correct Answer - (3)

Explanation:
If the last three digits of a number is divisible by 8, then the number is divisible by 8 (test of divisibility by 8).
Here, last three digits 58N is divisible by 8 if N = 4. (Since 584 is divisible by 8.)

For divisibility by 11. If the digits at odd and even places of a given number are equal or differ by a number divisible by 11, then the given number is divisible by 11.

Therefore, (M+9+4+4+8)-(3+0+8+5+N)=(M+5) should be divisible by 11 => when M = 6.

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Question:
How many zeros contained in 100!?

(1) 100
(2) 24
(3) 97
(4) Cannot be determined

Correct Answer - (2)

Explanation:
If the number N = 2×1.3×2.5×3 ….. pxq. It is clear that each pair of prime factors 2 and 5 will generate one zero in the number N, because 10 = 2X5.

For finding out the number of zeros in the number it is sufficient to find out how many two’s and five’s appear in the expansion of each product N.

No. of five’s = No. of two’s = Hence it is evident that there are only 24 pairs of primes 2 and 5 and therefore 100! ends in 24 zeros.

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Exercise4 + Answers + Explanation

Question:
When 242 is divided by a certain divisor the remainder obtained is 8. When 698 is divided by the same divisor the remainder obtained is 9. However, when the sum of the two numbers 242 and 698 is divided by the divisor, the remainder obtained is 4. What is the value of the divisor?

(1) 11
(2) 17
(3) 13
(4) 23

Correct Answer - (3)

Explanation:
Let the divisor be d.
When 242 is divided by the divisor, let the quotient be ‘x’ and we know that the remainder is 8. Therefore, 242 = xd + 8

Similarly, let y be the quotient when 698 is divided by d.
Then, 698 = yd + 9.

242 + 698 = 940 = xd + yd + 8 + 9
940 = xd + yd + 17
As xd and yd are divisible by d, the remainder when 940 is divided by d should have been 17.
However, as the question states that the remainder is 4, it would be possible only when leaves a remainder of 4.
If the remainder obtained is 4 when 17 is divided by d, then d has to be 13.

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Question:
What is the total number of different divisors of the number 7200?

(1) 20
(2) 4
(3) 54
(4) 32

Correct Answer - (3)

Explanation:
Express 7200 as a product of prime numbers. We get 7200 = 25 * 32 * 52.
The number of different divisors of 7200 including 1 and 7200 are (5 + 1)(2 + 1)(2 + 1) = 6 * 3 * 3 = 54

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Question:
When a number is divided by 36, it leaves a remainder of 19. What will be the remainder when the number is divided by 12?

(1) 10
(2) 7
(3) 192
(4) None of these

Correct Answer - (2)

Explanation:
Let the number be ‘a’.
When ‘a’ is divided by 36, let the quotient be ‘q’ and we know the remainder is 19
=> and remainder is 19
=> a = 36q + 19

now, a is divided by 12

=> 36q is perfectly divided by 12
Therefore, remainder = 7

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Question:
A person starts multiplying consecutive positive integers from 20. How many numbers should he multiply before the will have result that will end with 3 zeroes?

(1) 11
(2) 10
(3) 6
(4) 5

Correct Answer - (3)

Explanation:
A number will end in 3 zeroes when it is multiplied by 3 10s. To get a 10, one needs a 5 and a 2.
Therefore, this person should multiply till he encounters three 5s and three 2s.
20 has one 5 (5 * 4) and 25 has two 5s (5 * 5).
20 has two 2s (5 * 2 * 2) and 22 has one 2 (11 * 2).

Therefore, he has to multiply till 25 to get three 5s and three 2s, that will make three 10s. So, he has to multiply from 20 to 25 i.e. 6 numbers.

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Question:
How many four digit numbers exist which can be formed by using the digits 2, 3, 5 and 7 once only such that they are divisible by 25?

(1) 4! - 3!
(2) 4
(3) 8
(4) 6

Correct Answer - (2)

Explanation:
The numbers that are divisible by 25 are the numbers whose last two digits are ‘00′, ‘25′, ‘50′ and ‘75′.
However as ‘0′ is not one of the four digits given in the problem, only numbers ending in ‘25′ and ‘75′ need to be considered.

Four digit number whose last two digits are ‘25′ using 3 and 7 as the other two digits are 3725 and 7325.
Similarly, four digit numbers whose last two digits are ‘75′ using 2 and 3 as the other two digits are 2375 and 3275.

Hence there are four 4-digit numbers that exist.

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Tips For Number System:

This rule applies to numbers whose units digits add up to ten (2 and 8 in this case) and the tenth digits are same.
If you have a pair of numbers like 22 and 28 and you need to multiply the two of them.

Step 1:
Multiply the unit digits
2 * 8 = 16

Step 2:
Increment the tenth digit by 1.
2 + 1 = 3

Step 3:
Multiply the incremented number with the original tenth digit number
3 * 2 = 6

Step 4:
Concatenate the results got from step 1 and step 3
6 and 16 => 616

Note: Remember to put the products of unit digits of the given numbers in the unit and tenth digit of your result.

Example 2: 36 * 34 = ?

Step 1: Multiply the unit digits
6 * 4 = 24

Step 2: Increment the tenth digit by 1.
3 + 1 = 4

Step 3: Multiply the incremented number with the original tenth digit number
4 * 3 = 12

Step 4: Concatenate the results got from step 1 and step 3
12 and 24 => 1224

Therefore, 36 * 34 = 1224

Example 3: 61 * 69 = ?

Step 1: Multiply the unit digits
1 * 9 = 9. In this case remember to write 9 as 09

Step 2: Increment the tenth digit by 1.
6 + 1 = 7

Step 3: Multiply the incremented number with the original tenth digit number
6 * 7 = 42

Step 4: Concatenate the results got from step 1 and step 3
42 and 09 => 4209 (Since, we have to put the product of the unit digits only in the unit and tenth digit of the result)

Therefore, 61 * 69 = 4209

Example 4: 127 * 123 = ?

Step 1: Multiply the unit digits
7 * 3 = 21

Step 2: Increment the tenth digit by 1.
12 + 1 = 13

Step 3: Multiply the incremented number with the original tenth digit number
12 * 13 = 156

Step 4: Concatenate the results got from step 1 and step 3
156 and 21 => 15621

Therefore, 127 * 123 = 15621

Try out the above technique with the following problems.

Questions:

1. 54 * 56
2. 79 * 71
3. 152 * 158
4. 33 * 37
5. 111 * 119

Correct Answers:

1. 3024
2. 5609
3. 24016
4. 1221
5. 13209