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Q:

How many number of times will the digit ‘7' be written when listing the integers from 1 to 1000?

A) 243 B) 300
C) 301 D) 290

Answer:   B) 300



Explanation:

7 does not occur in 1000. So we have to count the number of times it appears between 1 and 999. Any number between 1 and 999 can be expressed in the form of xyz where 0 < x, y, z < 9.

 

1. The numbers in which 7 occurs only once. e.g 7, 17, 78, 217, 743 etc

 

This means that 7 is one of the digits and the remaining two digits will be any of the other 9 digits (i.e 0 to 9 with the exception of 7)

 

You have 1*9*9 = 81 such numbers. However, 7 could appear as the first or the second or the third digit. Therefore, there will be 3*81 = 243 numbers (1-digit, 2-digits and 3- digits) in which 7 will appear only once.

 

In each of these numbers, 7 is written once. Therefore, 243 times.

 

 

2. The numbers in which 7 will appear twice. e.g 772 or 377 or 747 or 77

 

In these numbers, one of the digits is not 7 and it can be any of the 9 digits ( 0 to 9 with the exception of 7).

 

There will be 9 such numbers. However, this digit which is not 7 can appear in the first or second or the third place. So there are 3 * 9 = 27 such numbers.

 

In each of these 27 numbers, the digit 7 is written twice. Therefore, 7 is written 54 times.

 

 

3. The number in which 7 appears thrice - 777 - 1 number. 7 is written thrice in it.

 

Therefore, the total number of times the digit 7 is written between 1 and 999 is

 

243 + 54 + 3 = 300

Q:

In how many ways the letters of the word 'CIRCUMSTANCES' can be arranged such that all vowels came at odd places and N always comes at end?

A) 1,51,200 ways. B) 5,04,020 ways
C) 72,000 ways D) None of the above
 
Answer & Explanation Answer: A) 1,51,200 ways.

Explanation:

In circumcstances word there are 3C's, 2S's, I, U,R, T, A, N, E

Total = 13 letters

But last letter must be N

Hence, available places = 12

In that odd places = 1, 3, 5, 7, 9, 11

Owvels = 4

This can be done in 6P4 ways 

Remaining 7 letters can be arranged in 7!/3! x 2! ways

 

Hence, total number of ways = 6P4 x 7!/3! x 2! = 360 x 5040/12 = 1,51,200 ways.

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0 46
Q:

In how many different ways can the letters of the word 'RITUAL' be arranged?

A) 720 B) 5040
C) 360 D) 180
 
Answer & Explanation Answer: A) 720

Explanation:

The number of letters in the given word RITUAL = 6

Then, 

Required number of different ways can the letters of the word 'RITUAL' be arranged = 6!

=> 6 x 5 x 4 x 3 x 2 x 1 = 720

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2 158
Q:

How many four digits numbers greater than 6000 can be made using the digits 0, 4, 2, 6 together with repetition.

A) 64 B) 63
C) 62 D) 60
 
Answer & Explanation Answer: B) 63

Explanation:

Given digits are 0, 4, 2, 6

Required 4 digit number should be greater than 6000.

So, first digit must be 6 only and the remaining three places can be filled by one of all the four digits.

This can be done by

1x4x4x4 = 64

Greater than 6000 means 6000 should not be there.

Hence, 64 - 1 = 63.

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4 244
Q:

A card is drawn from a pack of 52 cards. What is the probability that either card is black or a king? 

A) 15/52 B) 17/26
C) 13/17 D) 15/26
 
Answer & Explanation Answer: D) 15/26

Explanation:

Number of cards in a pack of cards = 52

Number of black cards = 26

Number of king cards = 4 (2 Red, 2 Black)

 

Required, the probability that if a card is drawn either card is black or a king = 

2652 + 452 = 3052 = 1526

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4 352
Q:

A box contains 2 blue balls, 3 green balls and 4 yellow balls. In how many ways can 3 balls be drawn from the box, if at least one green ball is to be included in the draw?

A) 48 B) 24
C) 64 D) 32
 
Answer & Explanation Answer: C) 64

Explanation:

Total number of balls = 2 + 3 + 4 = 9

Total number of ways 3 balls can be drawn from 9 = 9C3

No green ball is drawn = 9 - 3 = 6 = 6C3

Required number of ways if atleast one green ball is to be included = Total number of ways - No green ball is drawn

= 9C3 - 6C3

= 9x8x7/3x2  -  6x5x4/3x2

= 84 - 20

= 64 ways.

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4 454
Q:

How many 4-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?

A) 60 B) 48
C) 36 D) 20
 
Answer & Explanation Answer: A) 60

Explanation:

Here given the required digit number is 4 digit.

It must be divisible by 5. Hence, the unit's digit in the required 4 digit number must be 0 or 5. But here only 5 is available.

x x x 5

The remaining places can be filled by remaining digits as 5 x 4 x 3 ways.

 

Hence, number 4-digit numbers can be formed are 5 x 4 x 3 = 20 x 3 = 60.

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1 443
Q:

In how many ways the word 'SCOOTY' can be arranged such that 'S' and 'Y' are always at two ends?

A) 720 B) 360
C) 120 D) 24
 
Answer & Explanation Answer: D) 24

Explanation:

Given word is SCOOTY

ATQ,

Except S & Y number of letters are 4(C 2O's T)

Hence, required number of arrangements = 4!/2! x 2! = 4!

= 4 x 3 x 2

= 24 ways.

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3 742
Q:

In how many ways word of 'GLACIOUS' can be arranged such that 'C' always comes at end?

A) 3360 B) 5040
C) 720 D) 1080
 
Answer & Explanation Answer: B) 5040

Explanation:

Given word is GLACIOUS has 8 letters.

=> C is fixed in one of the 8 places

Then, the remaining 7 letters can be arranged in 7! ways = 5040.

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2 898