From a total of six men and four ladies a committee of three is to be formed. If Mrs. X is not willing to join the committee in which Mr. Y is a member, whereas Mr.Y is willing to join the committee only if Mrs Z is included, how many such committee are possible?
Answer: D We first count the number of committee in which i). Mr. Y is a member (ii). the ones in which he is not case (i) :- As Mr. Y agrees to be in committee only where Mrs. Z is a member. Now we are left with (6-1) men and (4-2) ladies (Mrs. X is not willing to join). We can choose 1 more in (5+2)C1=7 ways. case (ii):- If Mr. Y is not a member then we left with (6+4-1) people. we can select 3 from 9 in 9C3 = 84 ways. Thus, total number of ways is 7+84 =91 ways.
Q. No. 8:
There are five cards lying on the table in one row. Five numbers from among 1 to 100 have to be written on them, one number per card, such that the difference between the numbers on any two adjacent cards is not divisible by 4. The remainder when each of the 5 numbers is divided by 4 is written down on another card (the 6th card) in order. How many sequences can be written down on the 6th card?
Answer: C The remainder on the first card can be 0,1,2 or 3 i.e 4 possibilities. The remainder of the number on the next card when divided by 4 can have 3 possible vales (except the one occurred earlier). For each value on the card the remainder can have 3 possible values. The total number of possible sequences is 4(3)4
Q. No. 9:
How many natural numbers less than a lakh can be formed with the digits 0,6 and 9?
Answer: A The digits to be used are 0,6 and 9. The required numbers are from 1 to 99999. The numbers are five digit numbers. Therefore, every place can be filled by 0,6 and 9 in 3 ways. Total number of ways = 3*3*3*3*3 = 35 But 00000 is also a number formed and has to be excluded. Total number of numbers = 35-1 = 243-1 =242
Q. No. 10:
In how many ways can seven friends be seated in a row having 35 seats, such that no two friends occupy adjacent seats?
Answer: A first let us consider the 28 unoccupied seats. They create 29 slots- one on the left of each seat and one on the right of the last one. We can place the 7 friends in any of these 29 slots i.e 29P7 ways.
Q. No. 11:
Jay wants to buy a total of 100 plants using exactly a sum of Rs 1000. He can buy Rose plants at Rs 20 per plant or marigold or Sun flower plants at Rs 5 and Re 1 per plant respectively. If he has to buy at least one of each plant and cannot buy any other type of plants, then in how many distinct ways can Jay make his purchase?
Answer: B Let the number of Rose plants be a let number of marigold plants be b Let the number of Sunflower plants be c. 20a+5b+1c=1000 ; a+b+c=100 Solving the above two equations by eliminating c, 19a+4b=900 b= (900-19a)/4 = 225 - 19a/4...(1) b being the number of plants, is a positive integer, and is less than 99, as each of the other two types have at least one plant in the combination i.e , 0<b<99.....(2) Substituting (1) in (2), 0 < 225 - 19a/4 < 99 => 225 < -19a/4 < (99-225) => 4*225 >19a > 126*4 =>900/19 > a >504 a is the integer between 47 and 27.....(3) From (1), it is clear , a should be multiple of 4. hence possible values of a are (28,32,36,40,44) For a=28 and 32, a+b>100 For all other values of a, we get the desired solution a=36, b=54, c=10 a=40, b=35, c=25 a=44, b=16, c=40 Three solutions are possible.
Q. No. 12:
How many arrangements of four 0's (zeroes), two 1's and two 2's are there in which the first 1 occur before the first 2?
Answer: D Total number of arrangements = 8!/(4!*2!*2!) = 420. Since, there are two 1's and two 0's, the number of arrangements in which the first 1 is before the first 2 is same as the number of arrangement in which the first 2 is before the first 1 and they are each equal to half the total number of arrangements = 210.