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# Numbers Questions

Home > Quantitative Aptitude > Numbers > General Questions
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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

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

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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Find the total no of divisors of 1728 (including 1 and 1728)

A25

B22

C28

DNone of these

Explanation:

1728= 2^6 * 3^3
Hence the Number of factors = (6+1) x (3+1) = 7 x 4 = 28.
Imp: if a number represented in standard form (a^m *b^n) , then the number of factors Is given by (m+1)(n+1).
=> (6+1)(3+1) = 28

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How many prime factors are there in the expression (6)23 x (4)10 x (5)10 ?

A7

B76

C3

D9

Explanation:

(6)23 x (4)10 x (5)10
2*3*23*2*2*2*5*5*2*5
=> 2^5 * 3^1 * 5^3
=> 5+1+3 = 9

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0.0002 / 0.004 + 0.0002 = ?

A0.2002

B0.5002

C0.0052

D0.0502

ENone of these

Explanation:

0.0002/0.004 + 0.0002 = 0.0502

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(0.645 * 0.645 - 0.355 * 0.645 + 0.355*0.355)/(0.645 * 0.645*0.645 + 0.355 * 0.355*0.355) = ?

A10

B100

C0.01

D0.1

E1

Explanation:

a^3+b^3 = (a+b) (a^2 - ab +b^2)

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A number is 101102103104...150. As 101 102 103 103.... 150. What is reminder when divided by 3?

A0

B1

C4

D2

Explanation:

Divisibility rule for 3 are like if sum of the digits divisible by 3.
Aum of the digits = 380
Remainder = 380/3 = 2.

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what is the unit digit of(2^3)^123456

A2

B4

C8

D6

Explanation:

(2^3)^123456=2^370368
Take unit position of power
2^8=256
so unit position is 6

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Find the number of zeros in the expression 15*32*25*22*40*75*98*112*125.

A9

B7

C14

D12

Explanation:

Expen 15*32*25*22*40*75*98*112*125 as below
(3*5)*(2^5)*(5^2)*(11*2)*(5*2^3)*(3*5^2)*(49*2)*(7*2^4)*(5^3)
Total no. of 5 = 9 & n.o of 2=14
So, Total number of zero = 9

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How many prime numbers are there which are less than 100 and greater than 3 such that they are of the following forms 4x + 15y - 1

A11

B12

C2

D7

Explanation:

Let the number be N.
So N = 4x + 1 = 5y - 1
=> x=5y?24
y = 2 satisfies the equation. So minimum number satisfies both the equations is 9 and general format of the numbers which satisfies the equation = k. LCM (4, 5) + 9 = 20k + 9.
Now by putting values 1, 2, 3 . . . . for k, we get 29, 49, 69, 89. Of which only 29, 89 are primes.

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