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Numbers Questions

Home > Quantitative Aptitude > Numbers > General Questions
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Give the greatest pair of twin primes which are below 100?


A71,73

B97,99

C93,95

D87,89

Answer: Option A

Explanation:

A twin prime is a prime number that is either 2 less or 2 more than another prime number for example, either member of the twin prime pair (41, 43). In other words, a twin prime is a prime that has a prime gap of two.



The first few twin prime pairs are:



(3, 5), (5, 7), (11, 13), (17, 19), (29, 31), (41, 43), (59, 61), (71, 73), (101, 103), (107, 109) and so on



So the answer uis 71,73

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Find the remainder when 83*85*87*88*91 is divided by 41.


A32

B31

C29

D34

Answer: Option B

Explanation:

Note: % (Remainder)
83*85*87*88*91
83 % 41 = 1
85 % 41 = 3
87 % 41 = 5
88 % 41 = 6
91 % 41 = 9

So (1 * 3 * 5 * 6 * 9) % 41 = 31

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When a square of an odd number is divided by 8


AThe remainder could be any odd number.

Bnothing definite can be said about the remainder.

Cthe remainder is always 1.

Dthe remainder could be any even number.

Answer: Option C

Explanation:

Let's call the odd number 2a+1, where a is an integer.

Then (2a+1)^2 = 4a^2 + 4a + 1 = 4a(a+1) + 1.


Since a is odd, a+1 is even.


So, 4a(a+1) has to be a multiple of 8


Therefore 4a(a+1) + 1 must leave a remainder of 1 when divided by 8.

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How many ways can 360 be written as product of two numbers.


A10

B11

C12

D18

Answer: Option C

Explanation:

1*360, 2*180, 3*120, 4*90, 5*72, 6*60, 8*45, 9*40, 10*36, 12*30, 15*24, 18*20
SOL 2) 360 => 2^3 * 3^2 * 5^1 [i.e. a^x * b^y * c^z)
No. of ways can be written as a prod of 2 no's is
(x+1)*(y+1)(z+1)/2
=> (3+1)*(2+1)*(1+1)/2
=> 24/2 = 12

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How many prime numbers between 1 and 100 are factors of 7150?


A5

B3

C4

D7

Answer: Option C

Explanation:

Solution: 7,150 = 2 * 5 * 5 * 11 * 13



so there are 4 distinct prime numbers that are below 100, so the answer is 4.

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2345 23455 234555 234555........... what was last 2 numbers at 200th digit?


A44

B55

C54

D66

Answer: Option B

Explanation:

the given series is 2345 23455 234555......



So the number of digits in each term are 4, 5, 6, ... or (3  1), (3  2), (3  3),......upto nth terms are=3n [n(n 1)/2]



so 3n [n(n 1)/2]>=200



For n = 16, We get 184 in the left hand side. So after 16 terms the number of digits equal to 184. And 16 them contains 16  3 = 19 digits.



Now 17 term contains 20 digits and 123444......4(17times)...  last two digits are 55.

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What is the remainder of (32^31^301) when it is divided by 9?


A3

B5

C2

D1

Answer: Option B

Explanation:

32^31^301
when 31 divided by 9 gives remainder 5
5 5^2 5^3 all gives the same unit digit 5
so 31^32 gives unit digit 5
same rule applicable to 31^301
when 31 divided by 9 gives remainder 4
4 4^2 4^3 4^4 =4 6 4 6 unit place repeats for every 2 times i,e for even power its uint place is 6 and for odd its 4
as 301 is odd its unit place is 4
so 31^32^301=31^4=5^4=5

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How many positive integers less than 500 can be formed using the numbers 1,2,3,and 5 for digits, each digit being used only once.


A52

B68

C66

D34

Answer: Option D

Explanation:

Single digit number = 4
Double digit number = 4*3 = 12
Three digit numbers = 3*3*2= 18
Total = 34.

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The sum of two numbers is 45. Sum of their quotient and reciprocal is 2.05, Find the product of the numbers.


A340

B140

C500

D450

Answer: Option C

Explanation:

a + b = 45
a/b+b/a = 2.05
=>(a^2+b^2)/ab = 2.05
=>((a+b)^2?2ab)/ab=2.05
=>(a+b)^2 = 2.05ab + 2ab = 4.05ab
=> ab = 45^2 / 4.05 = 500

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7^1+7^2+7^3+.......+7^205. Find out how many numbers present which unit place contain 3?


A51

B49

C65

D45

Answer: Option A

Explanation:

Units digits of first 4 terms are 7, 9, 3, 1. and this pattern repeats. So for every 4 terms we get one term with 3 in its unit digit. So there are total of 205/4 = 51 sets and each set contains one terms with 3 in its unit digit.

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