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Numbers Questions

Home > Quantitative Aptitude > Numbers > General Questions
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If the Arithmetic mean is 34 and geometric mean is 16 then what is greates number in that series of numbers?


A55

B46

C65

D64

Answer: Option D

Explanation:

Lets assume two numbers be x, y;
Arthmetic mean=34=>( x+y)/2=34
=> x+y=68
geometric mean=16=>(xy)^(1/2) = 16
=> xy=16*16=256

By trail and error 16*16=64*4
And 64+4/2=34
So the greatest number int that series is 64.

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


A200

B450

C300

D230

Answer: Option C

Explanation:

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.

So, total number of times the digit 7 is written between 1 and 999 is 243 + 54 + 3 = 300

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On sports day,if 30 children were made to stand in a column,16 columns could be formed. If 24 children were made to stand in a column, how many columns could be formed?


A30

B25

C20

DNone of these

Answer: Option C

Explanation:

Total number of children=30*16=480
Number of columns of 24 children each =480/24=20.

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A zookeeper counted the heads of the animals in a zoo and found it to be 80. When he counted the legs of the animals he found it to be 260. If the zoo had either pigeons or horses, how many horses were there in the zoo?


A20

B50

C60

D100

Answer: Option B

Explanation:

Let the number of horses = x
Then the number of pigeons = (80 - x).
Each pigeon has 2 legs and each horse has 4 legs.
Therefore, total number of legs = 4x + 2(80-x) = 260
=>4x + 160 - 2x = 260
=>2x = 100
=>x = 50.

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In a college ,1/5 th of the girls and 1/8 th of the boys took part in a social camp.What of the total number of students in the college took part in the camp?


A2/13

B4/13

C2/15

D5/16

Answer: Option A

Explanation:

Out of 5 girls 1 took part in the camp
out of 8 boys 1 took part in the camp
so, out of 13 students 2 took part in the camp.
So, 2/13of the total strength took part in the camp.

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Finding the wrong term in the given series 7, 28, 63, 124, 215, 342, 511


A63

B28

C7

D511

Answer: Option B

Explanation:

Clearly, the correct sequence is
2^3 – 1, 3^3 – 1, 4^3 – 1, 5^3 – 1, ……….
Therefore, 28 is wrong and should be replaced by (3^3 – 1) i.e, 26.

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A farmer built a fence around his 17 cows,in a square shaped region.He used 27 fence poles on each side of the square. How many poles did he need altogether?


A108 poles

B114 poles

C104 poles

D100 poles

Answer: Option C

Explanation:

Here 25 poles Must be there on each side .And around four corners 4 poles will be present.
=> 4*25+4=100+4=104 poles.

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Divide 45 into four parts such that when 2 is added to the first part, 2 is subtracted from the second part, 2 is multiplied by the third part and the fourth part is divided by two, all result in the same number.


A8, 12, 5, 20

B18, 12, 5, 20

C8, 12, 15, 20

DNone of these

Answer: Option A

Explanation:

a + b + c + d =45;
a+2 = b-2 = 2c = d/2;
a=b-4; c = (b-2)/2;
d = 2(b-2);
b-4 + b + (b-2)/2 + 2(b-2) = 45;

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The least number which when divided by 48, 60, 72, 108 and 140 leaves 38, 50, 62, 98 and 130 as remainders respectively, is:


A4562

B15110

C2135

D7589

Answer: Option B

Explanation:

(48 - 38 ) = ( 60 - 50 ) = ( 72 - 62 ) = ( 108 - 98 ) = ( 140 - 130 ) = 10 in every case

Leat number will be LCM of ( 48, 60, 72, 108, 140 ) - 10

LCM = 15120

So, required number = 15120 - 10 = 15110

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Find the no of zeros in the product of 1^1*2^2*3^3*.....*49^49?


A250

B340

C130

D230

Answer: Option A

Explanation:

2*5=10
5^5 5 times 5
10^10 10 times 5
15^15 15
......
25^25 50
.....
5+10+15+20+2*25+30+35+40+45=250

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