Nov 23rd, 2021 - Daily Quiz -CSAT 2
1. Twelve people, including two who do not wish to sit next to each other, are to be seated at a round table. How many circular seating plans can be made?.
(a) 9 x 10!
(b) 8 x 9!
(c) 9 x 11!
(d) 8 x 11!
2. Two friends, Alu and Blue, go to a bookshop, together with their sons Paul and Kaul. All four of them buy some books; each book costs a whole amount in rupees. When they leave the bookshop, they notice that both fathers have spent 21 rupees more than their respective sons. Moreover, each of them paid per book the same amount of rupees as books that he bought. The difference between the number of books of Alu and Paul is five. Who is the father of Kaul?.
(a) Alu
(b) Blue
(c) Can be either Alu or Blue
(d) Data insufficient
3. What is the number of ways to arrange the 26 alphabets so that no two of the vowels a, e, i, o, and u occur next to each other?.
(a) 21! 12! /17!
(b) 21! 22! /17!
(c) 21! 22! /18!
(d) 21! 12! /18!
4. In how many ways to make a basket of fruit from 6 oranges, 7 apples, and 8 bananas so that the basket contains at least two apples and one banana?.
(a) 224
(b) 234
(c) 334
(d) 336
5. A shopkeeper wants to expose his oranges neatly for sale. Doing this he discovers that one orange is left over when he places them in groups of three. The same happens if he tries to place them in groups of 5, 7, or 9 oranges. Only when he makes groups of 11 oranges, it fits exactly. How many oranges does the shopkeeper have at least?.
(a) 941
(b) 942
(c) 943
(d) None of these
Answers
| 1. We may have 11 people (including one of the two unhappy persons but not both) to sit first; there are 10! such seating plans. Next the second unhappy person can sit anywhere except the left side and right side of the first unhappy person; there are 9 choices for the second unhappy person.
Thus in 9 x 10! ways circular seating plans can be made. |
| 2. For each father-son couple holds: the father bought x books of x rupees, the son bought y books of y rupees. The difference between their expenses is 21 rupees, thus x2 - y2 = 21.
Since x and y are whole numbers (each book costs a whole amount of rupees), there are two possible solutions: (x=5, y=2) or (x=11, y=10). Because the difference between Alu and Paul is 5 books, this means that father Alu bought 5 books and son Peter 10.
i.e. the other son, Kaul, bought 2 books, and that his father is Alu. |
| 3. We first have the 21 consonants arranged arbitrarily and there are 21! ways to do so. For each such 21- permutation, we arrange the 5 vowels a,e,i,o,u in 22 positions between consonants; there are P (22, 5) ways of such arrangement.
Thus in 21! x P(22,5) =21! 22! /17! Ways we can arrange the 26 alphabets so that no two of the vowels a, e, i, o, and u occur next to each other. |
| 4. Number of ways = 7 x 6 x 8 = 336 |
| 5. Assume the number of oranges is A. Then A-1 is divisible by 3, 5, 7 and 9.
So, A-1 is a multiple of 5×7×9 = 315
(Note: 9 is also a multiple of 3, so 3 must not be included!).
We are looking for a value of N for which holds that 315×N + 1 is divisible by 11.
After some trying it turns out that N = 3. This means that the greengrocer has 946 oranges. |
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