1
Q:

The Number of times the digit 8 will be written when listing the integers from 1 to 1000 is :

A) 100 B) 200
C) 300 D) 400

Answer:   C) 300

Explanation:

 Since 8 does not occur in 1000, we have to count the number of times 8 occurs when we list the integers from 1 to 999. Any number between 1 and 999 is of the form xyz, where \inline 0\leq x, y,z\leq 9.

Let us first count the numbers in which 8 occurs exactly once.

Since 8 can occur atone place in \inline ^{3}\textrm{C}_{1} ways. There are \inline ^{3}\textrm{C}_{1}(9\times 9)=3\times 9^{2} such numbers.

Next, 8 can occur in exactly two places in \inline ^{3}\textrm{C}_{2}(9)=3\times 9 such numbers. Lastly, 8 can occur in all three digits in one number only.

Hence, the number of times 8 occur is 

\inline 1\times (3\times 9^{2})+2\times (3\times 9)+3\times 1=300

Q:

There are 3 sections with 5 questions each. If four questions are selected from each section, the chance of getting different questions is ?

A) 1000 B) 625
C) 525 D) 125
 
Answer & Explanation Answer: D) 125

Explanation:

Methods for selecting 4 questions out of 5 in the first section = 5x4x3x2x1/4x3x2x1 = 5
similarly for other 2 sections also i.e 5 and 5
so total methods = 5 x 5 x 5 = 125.

Report Error

View Answer Workspace Report Error Discuss

2 13
Q:

If two cards are taken one after another without replacing from a pack of 52 cards. What is the probability for the two cards be Ace ?

A) 51/1221 B) 42/221
C) 1/221 D) 52/1245
 
Answer & Explanation Answer: C) 1/221

Explanation:

Total Combination of getting a card from 52 cards = 52C1
Because there is no replacement, so number of cards after getting first card= 51
Now, Combination of getting an another card= 51C1
Total combination of getting 2 cards from 52 cards without replacement= (52C1)x(51C1)
There are total 4 Ace in stack. Combination of getting 1 Ace is = 4C1
Because there is no replacement, So number of cards after getting first Ace = 3
Combination of getting an another Ace = 3C1
Total Combination of getting 2 Ace without replacement=(4C1)x(3C1)
Now,Probability of getting 2 cards which are Ace = (4C1)x(3C1)/(52C1)x(51C1) = 1/221.

Report Error

View Answer Workspace Report Error Discuss

3 29
Q:

A Cricket team of 23 people all shake hands with each other exactly once. How many hand shakes occur ?

A) 142 B) 175
C) 212 D) 253
 
Answer & Explanation Answer: D) 253

Explanation:

The first person shakes hands with 22 different people, the second person also shakes hands with 22 different people, but one of those handshakes was counted in the 22 for the first person, so the second person actually shakes hands with 21 new people. The third person, 20 people, and so on...
So,
22 + 21 + 20 + 19 + 18 + 17 + 16 + 15 + 14 + 13 + 12 + 11 + 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1
= n(n+1)/2 = 22 x 23 /2 = 11 x 23 = 253.

Report Error

View Answer Workspace Report Error Discuss

3 34
Q:

There are three rooms in a Hotel: one single, one double and one for four persons. How many ways are there to house seven persons in these rooms ?

A) 105 B) 7! x 6!
C) 7!/5! D) 420
 
Answer & Explanation Answer: A) 105

Explanation:

Choose 1 person for the single room & from the remaining choose 2 for the double room & from the remaining choose 4 people for the four person room,

Then, 7C1 x 6C2 x 4C4
     = 7 x  x 1
     = 7 x 15 = 105.
Report Error

View Answer Workspace Report Error Discuss

4 92
Q:

How many parallelograms will be formed if 7 parallel horizontal lines intersect 6 parallel vertical lines?

A) 215 B) 315
C) 415 D) 115
 
Answer & Explanation Answer: B) 315

Explanation:
Parallelograms are formed when any two pairs of parallel lines (where each pair is not parallel to the other pair) intersect.
Hence, the given problem can be considered as selecting pairs of lines from the given 2 sets of parallel lines.
Therefore, the total number of parallelograms formed = 7C2 x 6C2 = 315
Report Error

View Answer Workspace Report Error Discuss

4 76