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Numbers Questions

Home > Quantitative Aptitude > Numbers > General Questions
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Choose the correct option.

The value of 3/4 + 5 / 36 + 7 / 144 + ...................+17 / 5184 + 19 / 8100 is ??


A0.99

B0.98

C0.95

DNone of these

Answer: Option A

Explanation:

Convert it into this series

[1-1/4]+ [1/4-1/9]+[1/9-1/16]+.........[1/81-1/100]

=1-1/100

=0.99

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Choose the correct option.

What is the value of

(44444445*88888885*44444442+44444438)/444444442


A88888883

B88888884

C88888888

D44444443

Answer: Option A

Explanation:

On taking x=44444444
The equation can be written as- (x+1) (2x-3) (x-2)+(x-6) / x^2

Solving equation we get 2x-5

Substitute x=44444444 in above equation we get- 88888883

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The addition 457 + 982 + 896 = 2345 is incorrect. What is the least digit that can be changed to make the addition correct?


A5

B7

C6

D3

Answer: Option A

Explanation:

457+982+896= 2335 But in question its given 2345
If we change 5 of 457 and make it 467 then the equation will become correct. So the least possible number to be changed is 5.

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Find the value of "n" where 3^48 + 3^1996 + 3^3943+33^n is a perfect cube?


A1963

B1964

C1960

D1991

Answer: Option A

Explanation:

 



see subtracting powers of each coefficient 1996-48 = 1948

3943-1996= 1947

similary 3n-3943=1946



solving for n , n comes out to be 1963.


 

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If 28a + 30b+ 31c=365, find the value of a+b+c, if a, b, and c are natural numbers.


A4

B12

C10

DCannot be determined

Answer: Option B

Explanation:

Because in a year there are 7 months of 31 days and 4 months of 30 days and 1 month of 29 or 28 days so

28*1+30*4+31*7=365 (There is no other trick to solve this question)
1+4+7=12 Answer

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Let X be a four-digit number with exactly three consecutive digits being same and is a multiple of 9. How many such X?s are possible?


A22

B19

C20

D10

Answer: Option C

Explanation:

Let the four digit number be ?aaab? or baaa?.

Since, the number has to be a multiple of 9, therefore 3a+b should be either 9, 18 or 27.

Case I: 3a+b=9
Possible cases are: (1116,6111,2223,3222,3330,9000)

Case II: 3a+b=18
Possible cases are: (3339,9333,4446,6444,5553,3555,6660)

Case III: 3a+b=27
Possible cases are: (6669,9666,8883,3888,7776,6777,9990)

Hence, total number of cases is 20.

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Find the smallest 8-digit number which gives 15 as a remainder when divided by 38, 22 and 16.


A10002004

B10002015

C10001919

D10000015

Answer: Option C

Explanation:

Solution: Go with the options. The smaller numbers are option c and d.



Now divide them and check for the remainders.



10001919, when divided by 16, gives remainder 15

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The sum of odd number between 20 and 30 is


A125

B140

C120

D105

Answer: Option A

Explanation:

a = 21,l = 29 Clearly n = 5;
sum = n(a+l)/2 = 5*(21+29)/2 = 125

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The sum of odd numbers between 0 and 10 is:


A15

B20

C25

D30

E35

Answer: Option C

Explanation:

Formula for sum of numbers = n/2(a+l)
Here series is 1 3 5 7 9
a=1,n=5,l=9
So, 5/2(1+9)=25

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Which is the middle even number between 9 and 23?


A14

B16

C18

D20

Answer: Option B

Explanation:

9+x = 23-x 2x = 14
X = 7
Now 9+x = 16

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