Quantitative Aptitude -Develop-Enhance-Enjoy

  Permutation, Combination and Inclusion and Exclusion M. Velrajan 1. How many five digit numbers can be formed with the digits 0, 1, 2, 4, 6 and 8? (a) 800                            (b) 600 (c) 500 (d) 400 In a five digit number the first digit should not be 0, so it can be selected in 5 ways.  If the digits are not repeating, in the second digit choice 0 has to be included with the remaining 4 given digits, so that the second one can also be selected in 5 ways and the third, fourth and fifth digits can be selected in 4, 3 and 2 ways. The number of 5 digit numbers is 5 × 5 × 4 × 3 × 2 = 600.  In this we assumed that the digits are not repeating.  If the digits can be repeated then each of the second, third, fourth and fifth digit can be selected in 6 ways, so the number of 5 digits numbers is 5 × 6 × 6 × 6 × 6 = 6480 2. How many 3 digit numbers can be generated from 1, 2, 3, 4, ...

  Linear Equations and Polynomials M. Velrajan 1. Two railway tickets from city A to B and three tickets from city A to C cost Rs.177. Three tickets from city A to B and two tickets from city A to C cost Rs.173. The fare for city B from city A will be Rs.       (1)25  (2)27  (3)30  (4)33 Let x, y be the fare for A to B and A to C respectively. Then 2x + 3y = 177 and 3x + 2y = 173. To get x from the above equations we have to eliminate y by multiplying the first equation by 2 and the second by 3.  4x + 6y = 177 × 2 9x + 6y = 173 × 3 Subtracting the first from the second equation 5x = 173 × 3 - 177 × 2 = 165, hence x = 33. Aliter :   Instead of writing the two equations and multiplying the first equation by 2 and the second by 3 and then subtracting the first from second, by working out it in mind we have to write directly 5AB = 173 × 3 - 177 × 2 = 165, AB = 33 2. In a class if 5 students are seated in a bench then 2 benches will be vacant. I...