Below are 50 useful tricks & shortcuts for Quantitative Aptitude for CAT and other MBA exams:

Note: 2 <>23. log(1+x) = x - (x^2)/2 + (x^3)/3 - (x^4)/4 .........to infinity [Note the alternating sign . .Also note that the logarithm is with respect to base e]

Let the number be XYZ

Steps

a.Last digit = Last digit of Sq(Z)

b. Second last digit = 2*Y*Z + any carryover from STEP 1

c. Third last digit 2*X*Z+ Sq(Y) + any carryover from STEP 2

d. Fourth last digit is 2*X*Y + any carryover from STEP 3

e. Beginning of result will be Sq(X) + any carryover from Step 4

Eg) Let us find the square of 431

Step

a. Last digit = Last digit of Sq(1) = 1

b. Second last digit = 2*3*1 + any carryover from STEP 1=6+0=6

c. Third last digit 2*4*1+ Sq(3) + any carryover from STEP 2 = 8+9+0 = 17 i.e. 7 with carry over of 1

d. Fourth last digit is 2*4*3 + any carryover from STEP 3 = 24+1 = 25 i.e. 5 with carry over of 2

e. Beginning of result will be Sq(4) + any carryover from Step 4 = 16+2 = 18

THUS SQ(431) = 185761

-> The sum of first n natural numbers = n(n+1)/2

-> The sum of squares of first n natural numbers is n(n+1)(2n+1)/6

-> The sum of cubes of first n natural numbers is (n(n+1)/2)2/4

-> The sum of first n even numbers= n (n+1)

-> The sum of first n odd numbers= n2

-> the total number of factors is (x+1)(y+1)(z+1) ....

-> the total number of relatively prime numbers less than the number is N * (1-1/a) * (1-1/b) * (1-1/c)....

-> the sum of relatively prime numbers less than the number is N/2 * N * (1-1/a) * (1-1/b) * (1-1/c)....

-> the sum of factors of the number is {a^(x+1)} * {b^(y+1)} * ...../(x * y *...)

-> Total no. of prime numbers between 1 and 50 is 15

-> Total no. of prime numbers between 51 and 100 is 10

-> Total no. of prime numbers between 101 and 200 is 21

-> The number of squares in n*m board is given by m*(m+1)*(3n-m+1)/6

-> The number of rectangles in n*m board is given by n+1C2 * m+1C2

-> The number of rectangles in n*m board is given by n+1C2 * m+1C2

-> 2^10 = 4^5 = 32^2 = 1024

-> 3^8 = 9^4 = 81^2 = 6561

-> 7 * 11 * 13 = 1001

-> 11 * 13 * 17 = 2431

-> 13 * 17 * 19 = 4199

-> 19 * 21 * 23 = 9177

-> 19 * 23 * 29 = 12673

This is given by (sum of digits) * (no. of digits-1)! * 1111…1 (i.e. based on the no. of digits)

Eg) Find the sum of all 3-digit nos. formed using the digits 2, 3, 5, 7 & 8.

Sum = (2+3+5+7+8) * (5-1)! * 11111 (since 5 digits are there)

= 25 * 24 * 11111

=6666600

As per Fermat’s Last Theorem, the above equation will not have any solution whenever n>=3.

145 = 1! + 4! + 5!

The no. is always divisible by a - b

Further, the no. is divisible by a + b when n is even and not divisible by a + b when n is odd

The no. is usually not divisible by a - b

However, the no. is divisible by a + b when n is odd and not divisible by a + b when n is even

**1.**x^n -a^n = (x-a)(x^(n-1) + x^(n-2) + .......+ a^(n-1) ) ......Very useful for finding multiples. For example (17-14=3 will be a multiple of 17^3 - 14^3)**2.**e^x = 1 + (x)/1! + (x^2)/2! + (x^3)/3! + ........to infinityNote: 2 <>23. log(1+x) = x - (x^2)/2 + (x^3)/3 - (x^4)/4 .........to infinity [Note the alternating sign . .Also note that the logarithm is with respect to base e]

**3.**(m + n)! is divisible by m! * n!**4.**When a three digit number is reversed and the difference of these two numbers is taken, the middle number is always 9 and the sum of the other two numbers is always 9.**5.**Any function of the type y=f(x)=(ax-b)/(bx-a) is always of the form x=f(y)**6.**To Find Square of a 3-Digit NumberLet the number be XYZ

Steps

a.Last digit = Last digit of Sq(Z)

b. Second last digit = 2*Y*Z + any carryover from STEP 1

c. Third last digit 2*X*Z+ Sq(Y) + any carryover from STEP 2

d. Fourth last digit is 2*X*Y + any carryover from STEP 3

e. Beginning of result will be Sq(X) + any carryover from Step 4

Eg) Let us find the square of 431

Step

a. Last digit = Last digit of Sq(1) = 1

b. Second last digit = 2*3*1 + any carryover from STEP 1=6+0=6

c. Third last digit 2*4*1+ Sq(3) + any carryover from STEP 2 = 8+9+0 = 17 i.e. 7 with carry over of 1

d. Fourth last digit is 2*4*3 + any carryover from STEP 3 = 24+1 = 25 i.e. 5 with carry over of 2

e. Beginning of result will be Sq(4) + any carryover from Step 4 = 16+2 = 18

THUS SQ(431) = 185761

**7.**If the answer choices provided are such that the last two digits are different, then, we need to carry out only the first two steps only.-> The sum of first n natural numbers = n(n+1)/2

-> The sum of squares of first n natural numbers is n(n+1)(2n+1)/6

-> The sum of cubes of first n natural numbers is (n(n+1)/2)2/4

-> The sum of first n even numbers= n (n+1)

-> The sum of first n odd numbers= n2

**8.**If a number ‘N’ is represented as a^x * b^y * c^z… where {a, b, c, …} are prime numbers, then-> the total number of factors is (x+1)(y+1)(z+1) ....

-> the total number of relatively prime numbers less than the number is N * (1-1/a) * (1-1/b) * (1-1/c)....

-> the sum of relatively prime numbers less than the number is N/2 * N * (1-1/a) * (1-1/b) * (1-1/c)....

-> the sum of factors of the number is {a^(x+1)} * {b^(y+1)} * ...../(x * y *...)

-> Total no. of prime numbers between 1 and 50 is 15

-> Total no. of prime numbers between 51 and 100 is 10

-> Total no. of prime numbers between 101 and 200 is 21

-> The number of squares in n*m board is given by m*(m+1)*(3n-m+1)/6

-> The number of rectangles in n*m board is given by n+1C2 * m+1C2

**9.**If ‘r’ is a rational no. lying between 0 and 1, then, r^r can never be rational.**10.**The number of squares in n*m board is given by m*(m+1)*(3n-m+1)/6-> The number of rectangles in n*m board is given by n+1C2 * m+1C2

**11.**If ‘r’ is a rational no. lying between 0 and 1, then, r^r can never be rational.**12.**Certain nos. to be remembered-> 2^10 = 4^5 = 32^2 = 1024

-> 3^8 = 9^4 = 81^2 = 6561

-> 7 * 11 * 13 = 1001

-> 11 * 13 * 17 = 2431

-> 13 * 17 * 19 = 4199

-> 19 * 21 * 23 = 9177

-> 19 * 23 * 29 = 12673

**13.**Where the digits of a no. are added and the resultant figure is 1 or 4 or 7 or 9, then, the no. could be a perfect square.**14.**If a no. ‘N’ has got k factors and a^l is one of the factors such that l>=k/2, then, a is the only prime factor for that no.**15.**To find out the sum of 3-digit nos. formed with a set of given digitsThis is given by (sum of digits) * (no. of digits-1)! * 1111…1 (i.e. based on the no. of digits)

Eg) Find the sum of all 3-digit nos. formed using the digits 2, 3, 5, 7 & 8.

Sum = (2+3+5+7+8) * (5-1)! * 11111 (since 5 digits are there)

= 25 * 24 * 11111

=6666600

**16.**Consider the equation x^n + y^n = z^nAs per Fermat’s Last Theorem, the above equation will not have any solution whenever n>=3.

**17.**Further as per Fermat, where ‘p’ is a prime no. and ‘N’ is co-prime to p, then, N^(p-1) – 1 is always divisible by p.**18.**145 is the 3-digit no. expressed as sum of factorials of the individual digits i.e.145 = 1! + 4! + 5!

**19.**Where a no. is of the form a^n – b^n, then,The no. is always divisible by a - b

Further, the no. is divisible by a + b when n is even and not divisible by a + b when n is odd

**20.**Where a no. is of the form a^n + b^n, then,The no. is usually not divisible by a - b

However, the no. is divisible by a + b when n is odd and not divisible by a + b when n is even

**Also Read:**
fuddu ekdum

ReplyDeleteFor learning quantitative aptitude shortcuts and tricks i found the best course. Affordable and helps in solving the above mentioned questions easily improving calculation speed :)

ReplyDeletewww.wiziq.com/course/5553-speed-calculations-shortcuts-tricks-for-cat-competitive-exams

Preparing for CAT- MBA 2014? Cracking CAT Exam is not a difficulty when one gets the guidance of highly qualified and experienced scholars. if you are wondering how to prepare for CAT Exam 2014, prepare Online in the best way for Cat Exam 2014 . Get the guidance of CAT gurus, and crack the mba entrance exam . To learn more, check : www.wiziq.com/course/9277-lr-vr-di-ds-speed-calculations-quant-general-awareness

ReplyDeleteThankyou for the above information.

ReplyDeleteAll the CAT 2014 aspirants who are heading to start their CAT 2014 preparation, one of the best source of CAT preparation i.e CAT coaching is now at your doorstep with the Online CAT preparation coaching under the guidance of great scholars. For more information, check Comprehensive CAT-MBA 2014 Online Coaching Course- Ravi Handa

Quality articles is the secret to invite the users to pay a quick visit

ReplyDeletethe web site, that's what this site is providing.

Here is my website; Assitir Breaking bad no Netflix breaking bad wiki