Ibps po Answer key for oct 3,4|po preparation tips & shortcuts

IBPS PO Preparation Tips and Shortcuts

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IBPS PO Exam Short Cuts

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IBPS PO CWE V Answer Key

Number System & Arithmetic

Natural Numbers: 1, 2, 3, 4…..

Whole Numbers: 0, 1, 2, 3, 4…..

Integers: ….-2, -1, 0, 1, 2 …..

Rational Numbers: Any number which can be expressed as a ratio of two integers for example a p/q format where ‘p’ and ‘q’ are integers. Proper fraction will have (p<q) and=”” improper=”” fraction=”” will=”” have=”” (p=””>q).

Factors: A positive integer ‘f’ is said to be a factor of a given positive integer ‘n’ if f divides n without leaving a remainder. e.g. 1, 2, 3, 4, 6 and 12 are the factors of 12.

Prime Numbers: A prime number is a positive number which has no factors besides itself and unity.

Composite Numbers: A composite number is a number which has other factors besides itself and unity.

Factorial: For a natural number ‘n’, its factorial is defined as: n! = 1 x 2 x 3 x 4 x …. x n (Note: 0! = 1)

Absolute value: Absolute value of x (written as |x|) is the distance of ‘x’ from 0 on the number line. |x| is always positive. |x| = x for x > 0 OR -x for x < 0

Law of Indices:

Tip: The fifth power of any number has the same units place digit as the number itself.

How to Clear IBPS PO Exam

HCF and LCM:

For two numbers, HCF x LCM = product of the two.

Tip: If a, b and c give remainders p, q and r respectively, when divided by the same number H, then H is HCF of (a-p), (b-q), (c-r)

Tip: If the HCF of two numbers ‘a’ and ‘b’ is H, then, the numbers (a+b) and (a-b) are also divisible by H.

Tip: If a number N always leaves a remainder R when divided by the numbers a, b and c, then N = LCM (or a multiple of LCM) of a, b and c + R.

Relatively Prime or Co-Prime Numbers:

Two positive integers are said to be relatively prime to each other if their highest common factor is 1.

If N = x a y b z c where x, y, z are prime factors. Then, Number of factors of N = P = (a + 1)(b + 1)(c + 1)

Sum of Factors of N is:

IBPS PO Preparation Tips: Number of ways N can be written as a product of two factors = P/2 or (P+1)/2 if P is even or odd respectively.

The number of ways in which a composite number can be resolved into two co-prime factors is 2m-1, where m is the number of different prime factors of the number.

Number of numbers which are less than N and co-prime to

Tip: If N = (2)a (y)b (z)c where x, y, z are prime factors Number of even factors of N = (a)(b+1)(c+1) Number of odd factors of N = (b+1)(c+1).

Divisibility Rules

A number is divisible by:

2, 4 & 8 when the number formed by the last, last two, last three digits are divisible by 2,4 & 8 respectively.

3 & 9 when the sum of the digits of the number is divisible by 3 & 9 respectively.

11 when the difference between the sum of the digits in the odd places and of those in even places is 0 or a multiple of 11.

6, 12 & 15 when it is divisible by 2 and 3, 3 and 4 & 3 and 5 respectively.

7, if the number of tens added to five times the number of units is divisible by 7.

13, if the number of tens added to four times the number of units is divisible by 13.

19, if the number of tens added to twice the number of units is divisible by 19.

For ex., check divisibility of 312 by 7, 13 & 19

For 7: 31 + 2 x 5 = 31 + 10 = 41 Not divisible
For 13: 31 + 2 x 4 = 31 + 8 = 39 Divisible.
For 19: 31 + 2 x 2 = 31 + 4 = 35 Not divisible.

IBPS PO Preparation Tips: If f(a) = 0, (x-a) is a factor of f(x)

Remainder Related Theorems

IBPS PO Preparation Tips: Any single digit number written (P-1) times is divisible by P, where P is a prime number >5. Examples: 222222 is divisible by 7 444444….. 18 times is divisible by 19.

Converting from base ‘n’ to decimal ->(pqrst)n = pn4 + qn3 + rn2 + sn + t.

Converting from decimal to base ‘n’ # The example given below is converting from 156 to binary.

For this we need to keep dividing by 2 till we get the quotient as 0.

2)156 0

2)78 0

2)39 1

2)19 1

2)9 1

2)4 0

2)2 0

2)1

Starting with the bottom remainder, we read the sequence of remainders upwards to the top. By that, we get 15610 = 100111002

Tip: (pqrst)n x n2 = (pqrst00)n

(pqrst)n x n3 = (pqrst000)n.

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Median of a finite list of numbers can be found by arranging all the observations from lowest value to highest value and picking the middle one.
Mode is the value that occurs most often.