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. |
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