Below are 50 useful tricks & shortcuts for Quantitative Aptitude for CAT and other MBA exams:
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 infinity
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]
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 Number
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
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 digits
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
16. Consider the equation x^n + y^n = z^n
As 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:
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 infinity
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]
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 Number
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
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 digits
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
16. Consider the equation x^n + y^n = z^n
As 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:
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