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Nov 21st, 2021 - Daily Quiz -CSAT 2

1. In a hexadecimal system what is the value of “B” in equivalent of octal system? (a) 11                        (b) 10                           (c) 13 (d) None of these   2. Find the number of different 8-letter arrangements that can be made from the letters of the word DAUGHTER so that all vowels do not occur together. (a) 1300                       (b) 1700                       (c) 1800              (d) None of these   3. A truck driver driving at 90 km/h finds a red light 100 m ahead of him. He instantly applies brakes to stop the truck. The truck retards uniformly and stops just at the stop line of red light. How much time did the driver take to stop the truck? (a) 7 seconds             (b) 8 seconds               (c) 9 seconds     (d) 10 seconds   4. By selling 30m of a cloth, a shopkeeper makes a profit equal to the selling price of 10 m cloth, What is the profit%? (a) 15%             (b) 50%             (c) 25%     (d) Cannot be determined using the given data   5. Find the number of different 8-letter arrangements that can be made from the letters of the word DAUGHTER so that all vowels occur together. (a) 4310                       (b) 4320                           (c) 4300        (d) None of these   Answers
1. Solution:-
Decimal Binary Octal Hexa-decimal system
0 0 0 0
1 1 1 1
2 10 2 2
3 11 3 3
4 100 4 4
5 101 5 5
6 110 6 6
7 111 7 7
8 1000 10 8
9 1001 11 9
10 1010 12 A
11 10111 13 B
2. Solution:- If we have to count those permutations in which all vowels are never together, we first have to find all possible arrangements of 8 letters taken all at a time, which can be done in 8! ways. Then, we have to subtract from this number, the number of permutations in which the vowels are always together. Therefore, the required number 8 ! – 6 ! × 3 ! = 6 ! (7×8 – 6) = 2 × 6 ! (28 – 3) = 50 × 6 ! = 50 × 720 = 36000.
3. Solution:- Here, initial velocity = u = 90 km/h = 25 m/s. Let the retardation be a m/s2. Since the truck stops after travelling a distance of 100 metres, we have v2 = u2 – 2as or 02 = 252 – 2a × 100, or, a = 3.125m/ s2. Let the truck took t secs to stop at stop line. Then v = u – at, or, 0 = 25 – 3.125t; giving t = 8 s.
4. Solution: SP of 30m – CP of 30m = SP of 10m SP of 20m = CP of 30m \ profit % = {(30 – 20) / 20} x 100 = 50%
5. Solution:- There are 8 different letters in the word DAUGHTER, in which there are 3 vowels, namely, A, U and E. Since the vowels have to occur together, we can for the time being, assume them as a single object (AUE). This single object together with 5 remaining letters (objects) will be counted as 6 objects. Then we count permutations of these 6 objects taken all at a time. This number would be 6P6 = 6!. Corresponding to each of these permutations, we shall have 3! permutations of the three vowels A, U, E taken all at a time . Hence, by the multiplication principle the required number of permutations = 6 ! × 3 ! = 4320.






POSTED ON 21-11-2021 BY ADMIN
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