number systems level 2 special equations 4

1. NUMBER SYSTEMS- SPECIAL EQUATIONS C.S.VEERARAGAVAN By Veeraragavanaa 2. Number of Positive Integer Solutions The number of positive integer solutions of the equation 2x + 3y =15 is a) 0 b) 1 c) 2 d) 3 The number of positive integer solutions = constant term lcm of coefficients of x & y = 15 6 =2. They are x = 3, y = 2 and x = 6, y = 1. 3. The number of nonnegative integral solutions of the equation 4x + 5y = 60 is 1) 1 2) 2 3) 3 4)4 LCM OF 4 & 5 IS 20. The no. of positive integral solutions = 60 20 = 3. Allowing zero (non negative integer), we get two more solutions. Hence Answer is 5. Number of Positive Integer Solutions 4. The number of nonnegative integral solutions of the equation 12x + 7y = 35 is 1) 0 2) 1 3) 2 4)3 LCM OF 12 & 7 IS 84. The no. of positive integral solutions = 35 84 = 0. Allowing zero, we get one solution. Hence Answer is 1. Number of Positive Integer Solutions 5. Number of Positive Integer Solutions A student purchases gel pens, ballpoint pens and pencils by spending a total of `.28. Each gel pen, ball point pen and pencil cost `.15, `.5 and `.3 respectively. In how many combinations can he purchase them, If he buys at least one of each item? 1) 1 2) 2 3) 3 4) 0 LCM OF 15, 5 & 3 IS 15. No of possible integral solution = 28 15 = 1. He purchases one gel pen, two ball point pens and one pencil. 6. Number of Positive Integer Solutions A student purchases gel pens, ballpoint pens and pencils by spending a total of `.28. Each gel pen, ball point pen and pencil cost `.15, `.5 and `.3 respectively. The number of ball point pens the student purchases, If he buys at least one of each item? 1) 1 2) 2 3) 3 4) 0 LCM OF 15, 5 & 3 IS 15. No of possible integral solution = 28 15 = 1. He purchases one gel pen, two ball point pens and one pencil. 7. Number of Positive Integer Solutions On a particular day, a salesman sold three types of toys. Each toy of the 3 varieties costs `.100, `.50 and `.25 respectively. If the total sale on that day was `300 and the salesman sold at least one toy of each variety, find the maximum number of toys he could have sold. He sell atleast one toy of each variety. He obtain 100 + 50 + 25 = 175. In the balance amount ( 300 175 = 125), maximum number could be sold third variety Hence 5 toys of third variety could be sold. Totally EIGHT TOYS HE COULD HAVE SOLD 8. Number of Positive Integer Solutions On a particular day, a salesman sold three types of toys. Each toy of the 3 varieties costs `.100, `.50 and `.25 respectively. If the total sale on that day was `300 and the salesman sold at least one toy of each variety, find the minimum number of toys he could have sold. He sell atleast one toy of each variety. He obtain 100 + 50 + 25 = 175. In the balance amount ( 300 175 = 125), to be minimum number could be sold include first variety Hence 2 toys of first & third and one toy of second variety should be sold. Totally FIVE TOYS HE COULD HAVE SOLD 9. Value of x should be ? If 7x + 4y = 64 and x,y are positive integers, then the values of x are 1) Multiples of 7 2) Multiples of 4 3) Multiples of 4 & 7 4) Any integer 7x = 64 4y. Both RHS terms are multiples of 4. Hence 7x also should be multiple of 4. 6/12/2015 C.S.VEERARAGAVAN 98948 34264 [email protected] 9 10. Value of b should be? Given 3a + 7b = 72, where a, b are positive integers. Which of the following is a possible value of b? 1) 2 2) 4 3)5 4) 6 7b = 72 3a. Both the RHS terms are multiples of 3. Hence the possible value of b should be 6 out of given choices. 6/12/2015 C.S.VEERARAGAVAN 98948 34264 [email protected] 10 11. Value of one number should be? Seven times a number plus eleven times another is equal to 61. Then a possible value of one of the numbers is 1) 4 2) 2 3) 1 4) 6 Let the numbers be x and y. Given 7x + 11y = 61. The only possible solution is x = 4 and y = 3. Hence answer is 1) 4. 6/12/2015 C.S.VEERARAGAVAN 98948 34264 [email protected] 11 12. Remainder Theorem If the remainder obtained when 4p is divided by 9 is 5, then a possible value of p is 1) 3 2) 4 3) 6 4) 8 4p = 9k + 5. P = 9k+5 4 = 9k+1 4 + 1. Hence k = 3,7,11,. And P = 8,17,26, 6/12/2015 C.S.VEERARAGAVAN 98948 34264 [email protected] 12 13. Remainder Theorem If the remainder of 4x 5 is 4, one value of x that satisfies the given equation is 1, and k is any integer then the other values of x are 1) 5k + 1 2) 4k 1 3) 5k 1 4) 4k + 1 4x = 5n + 4. X = 5n+4 4 = 5n 4 + 1 = 5 n 4 + 1 = 5k + 1. 6/12/2015 C.S.VEERARAGAVAN 98948 34264 [email protected] 13 14. Remainder Theorem The number 3x divided by 7 leaves 6 as its remainder. The values of x form an arithmetic progression whose common difference is 1) 7 2) 3 3) 2 4) 6 3x = 7k + 6. X = 7k+6 3 = 7(3p)+6 3 = 7(3p)21+27 3 = 21(p1)+27 3 = 7(p 1) + 9 Hence the common difference is 5. 6/12/2015 C.S.VEERARAGAVAN 98948 34264 [email protected] 14 15. Numbers The sum of a twodigit number formed by reversing its digit is equal to 88. The number of such numbers, in which the tens digit is greater than the units digit is 1) 1 2) 2 3) 3 4) 4 Let the number be 10x + y. Given 10x + y + 10y + x = 88. Hence 11x + 11y = 88. Hence x + y = 8. Given x >y (8,0), (7,1), (6,2) , (5,3). Hence 4 numbers only available. 6/12/2015 C.S.VEERARAGAVAN 98948 34264 [email protected] 15