41.John and Peter are among the nine players a basketball coach can choose from to field a five-player team. If all five players are chosen at random, what is the probability of choosing a team that includes John and Peter? A. 1/8 B. 1/6 C. 2/9 D. 5/18 E. 1/3
Solution:
Total Players to choose from: 9 No of players to choose: 5 Restriction: Team with John and Peter - - - - - The order in which the team members are selected doesnt matter. No of ways of selecting John: 1 No of ways of selecting Peter: 1 No of ways of selecting the remaining 3: 7c3 Hence P (team with Jon and Peter) = 1*1*7c3/9c5 = 5/18
42.John and Mary were each paid x dollars in advance to do a certain job together. John worked on the job for 10 hours and Mary worked 2 hours less than John. If Mary gave john y dollars of her payment so that they would have received the same hourly wage, what was the dollar amount, in terms of y, that john was paid in advance? 4y 5y 6y 8y 9y
Solution:
The key point to solve is 'they have same hourly wage after Mary shares Y dollars. Hence (x+y)/10 = (x-y)/8 Solving we get 18y = 2x, which is E
43.What is the sum of all 4-digit numbers that can be formed using the digits 1,2,3,4 where repetition of digits is allowed?
Solution:
Approach #1
We know the smallest number we can make is 1111 and the largest number we can make is 4444. We also know that our numbers will be evenly distributed in the middle (i.e. 1112 is balanced by 4443; 1113 is balanced by 4442). So, we can solve using the average formula. Finally, we know that there are 4*4*4*4 = 256 numbers in our set. Average = sum of terms/# of terms sum of terms = average * # of terms sum of terms = (1111+4444)/2 * 256 = 5555/2 * 256 = 5555*128 = 711040
Approach #2
We have 64 numbers with each of unit digit 1,2,3,4 Sum of values of all these unit digits = 64(4+3+2+1)=640 We have 64 numbers with each 1,2,3,4 at tens digit Sum of values of all these at tens digits = 64(40+30+20+10)=6400 Sum of all digits for hundred place=64000 Sum of all digits at thousand place =640000
Sum all the numbers which equals 711040
Solution:
Total Players to choose from: 9 No of players to choose: 5 Restriction: Team with John and Peter - - - - - The order in which the team members are selected doesnt matter. No of ways of selecting John: 1 No of ways of selecting Peter: 1 No of ways of selecting the remaining 3: 7c3 Hence P (team with Jon and Peter) = 1*1*7c3/9c5 = 5/18
42.John and Mary were each paid x dollars in advance to do a certain job together. John worked on the job for 10 hours and Mary worked 2 hours less than John. If Mary gave john y dollars of her payment so that they would have received the same hourly wage, what was the dollar amount, in terms of y, that john was paid in advance? 4y 5y 6y 8y 9y
Solution:
The key point to solve is 'they have same hourly wage after Mary shares Y dollars. Hence (x+y)/10 = (x-y)/8 Solving we get 18y = 2x, which is E
43.What is the sum of all 4-digit numbers that can be formed using the digits 1,2,3,4 where repetition of digits is allowed?
Solution:
Approach #1
We know the smallest number we can make is 1111 and the largest number we can make is 4444. We also know that our numbers will be evenly distributed in the middle (i.e. 1112 is balanced by 4443; 1113 is balanced by 4442). So, we can solve using the average formula. Finally, we know that there are 4*4*4*4 = 256 numbers in our set. Average = sum of terms/# of terms sum of terms = average * # of terms sum of terms = (1111+4444)/2 * 256 = 5555/2 * 256 = 5555*128 = 711040
Approach #2
We have 64 numbers with each of unit digit 1,2,3,4 Sum of values of all these unit digits = 64(4+3+2+1)=640 We have 64 numbers with each 1,2,3,4 at tens digit Sum of values of all these at tens digits = 64(40+30+20+10)=6400 Sum of all digits for hundred place=64000 Sum of all digits at thousand place =640000
Sum all the numbers which equals 711040
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