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# Problems on Numbers Questions

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A boy was asked to find the value of 7/12 of a sum of money. Instead of multiplying The sum by 7/12 he divided it by 7/12 and thus his answer exceeded the correct Answer By Rs.95. Find the correct answer.

A38

B48

C39

D49

Answer: Option D

Explanation:

Lets assume the sum = x
So as per the ques.
x*12/7 - x*7/12 = 95
x*(144-49) = 84*95
x = 84*95 / 95 = 84.

Therefore, correct answer = 84*7/12 = 49

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Find the difference of minimum number of number required to form every number from 1 to 1234 and 1 to 4321.

A9

B7

C4

D3

Answer: Option D

Explanation:

In 1234,
Unit place 1 to 1234 can be formed using 0 to 9 digits. i.e., 10 digits.
10's can be formed using 0 to 9 digits. i.e., 10 digits.
100's can be formed using 0 to 9 digits. i.e., 10 digits.
1000's can be formed using only one digit, 1. i.e., 1 digit.
Total no. of numbers required is 10 + 10 + 10 + 1 = 31.
In 4321, it is 10 + 10 + 10 + 4 = 34.
Therefore, difference = 34-31=3.

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The minimum number of numbers required to form a number from 9 to 9000 which are multiples of 5 is:

A31

B41

C42

D32

Answer: Option A

Explanation:

The multiples of 5 must end with 0 or 5.
Therefore, units place of a number from 9 to 9000 can be filled by only two digits 0 and 5. i.e., 2 digits.
10's place of a number from 9 to 9000 can be formed by 0 to 9 digits. i.e., 10 digits.
100's place of a number from 9 to 9000 can be formed by 0 to 9 digits. i.e., 10 digits.
1000's place of a number from 9 to 9000 can be formed by by 1 to 9 digits 1. i.e., 9 digits.
Therefore, total number of numbers required is 2 + 10 + 10 + 9 = 31.

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If abc4d is divisible by 4, then d cannot be:

A0

B4

C8

D2

Answer: Option D

Explanation:

40,44,48 is divisible by 4.
not 42. hence d cannot be 2

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The mean of 100 abservations was calculated was 40. It was found later on that one of the observations was misread as 83 instead of 53. The correct mean is :

A39

B39.7

C40.3

D42.7

Answer: Option B

Explanation:

correct sum =(40*100 +53 - 83) = 3970
correct mean = 3970/100 = 39.7

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The minimum number of digits required to form every number which is greater than 900 and less than 9,000 is:-

A39

B72

C49

D81

Answer: Option A

Explanation:

We have to find the minimum number of numbers required to form every number from 900 to 9,000.
Unit place of a number from 900 to 9,000 can be formed using 0 to 9 digits. i.e., 10 digits.
10's place of a number from 900 to 9,000 can be formed using 0 to 9 digits. i.e., 10 digits.
100's place of a number from 900 to 9,000 can be formed using 0 to 9 digits. i.e., 10 digits.
1000's place of a number from 900 to 9,000 can be formed using 1 to 9 digits. i.e., 9 digits.
Therefore, total number of numbers required is 10 + 10 + 10 + 9 = 39.

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In a group of cows and hens, the number of legs are 14 more than twice the number of heads. The number of cows is :

A5

B10

C7

D12

Answer: Option C

Explanation:

Lets assume no. of cows = x, and no. of hens = y.
So heads = (x+y)
Legs = (4x+2y)
As per condition:
4x+2y = 2(x+y)+ 14
2x = 14 => x = 7.

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Some people are standing square. From them 32 are removed and then form a small square and then 8 more removed. now we cannot form the square. So how many people are there at the beginning?

A64

B81

C110

DNone of these

Answer: Option B

Explanation:

Initial 81 form square and removing 32 gives 49(7*7) which also give square. removing 8 results in 41 which is not perfect square

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There are 20,6 and 9 rupees notes. you have to form 99 rupees with minimum number of notes. At what minimum number of notes you can achieve this?

A9 notes

B7 notes

C8 notes

DNone of these

Answer: Option A

Explanation:

20*3 = 60
6*2 = 12
9*2 = 18
9*1 = 9
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8 notes (3+2+2+1)

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A Bacteria is doubling at every 4 min. After 40 min 1024 bacteria. At what time there will be 256 bacteria?

A32 min.

B33 min.

C36 min.

D23 min.

Answer: Option A

Explanation:

Calculating backwards...
40 min - 1024
36 min - 512
32 min - 256

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