Numbers- Aptitude Questions and Answers

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The number of prime numbers which are less then 100 is?

A)

24


B)

25


C)

26


D)

27



Correct Answer :

25


Explanation :

Prime num bers less than 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

Thus, there are 25 prime numbers less than 100

In an arithmetic progression, if 17 is the 3rd term, - 25 is the 17th term, then -1 is which term?

A)

10


B)

11


C)

9


D)

12



Correct Answer :

9


Explanation :

a3 = a + 2d = 17 …(i)
a17 = a + 16d = - 25 …(ii)
From Eqs(i) and (ii)

- 14d = 42
d = - 3
So by Eq(i), a = 23

Let, nth term of the given AP is - 1 then
an = a + (n - 1)d = - 1
23 + (n - 1)(- 3) = - 1
23 - 3x + 3 = - 1
- 3n = - 1 - 26
- 3n = - 27
n = 9

Twice the first of three consecutive even integer is 6 more than the second integer. What is the third integers?

A)

6


B)

8


C)

10


D)

12



Correct Answer :

12


Explanation :

Let the three consecutive even integers are X, X+2, X+4
As is given, 2X = (X+2)+6
2X = X+8
X = 8

Hence, third integer = X + 4 = 8 + 4 = 12

If a, b and c are digits, then abc + bca + cab is divisible by?

A)

22


B)

37


C)

99


D)

101



Correct Answer :

37


Explanation :

If a ,b ,c are digits then the numbers formed can be:
abc = 100a + 10b + c
bca = 100b + 10c + a
cab = 100c + 10a + b

The sum is abc + bca + cab= (100a + 10b + c)+(100b + 10c + a)+(100c + 10a + b)
=111*a + 111*b + 111*c
=111*( a + b + c)
So it's divisible by 111.
Factors of 111 are 1, 3, 37, 111
Therfore, number formed can be divisible by 37 (37*4 = 111)

The common difference of the AP √12,√27,√48,√75 is ?

A)

√2


B)

3


C)

√3


D)

√12



Correct Answer :

√3


Explanation :

The given AP is √12,√27,√48,√75
After simplification, it can be written as 2√3,3√3,4√3,5√3
d1= 3√3-2√3=√3
d2= 4√3-3√3=√3
So common difference is √3

There are 22 terms in an arithmetic sequence. The sum of the first and last terms is 94. If the 11th term is 45, then the 12th term is ?

A)

48


B)

49


C)

50


D)

51



Correct Answer :

49


Explanation :

tn = a + (n – 1)d where tn = nth term, a= the first term , d= common difference
As given, a + (a + 21 d) = 94
2a + 21 d = 94 ------eq1

11th term, a + 10 d = 45
2a + 20 d = 90 ------eq2

from eq1 - eq2,
d = 4
after substitute value of d in eq1,
a = 5

Hence 12th term is, (a + 11 d)= 5 + 11*4 = 49

The difference between the place value of 4 and 2 in the number 833749502 is ?

A)
39098

B)
39998

C)
49998

D)
30098


Correct Answer :

39998


Explanation :

Place value of digit 4 in the number = 40000
Face value of digit 2 in the number = 2

Their difference = 40000 - 2 = 39998

The number of natural numbers less than 100 which have exactly 3 factors is ?

A)
10

B)
4

C)
12

D)
33


Correct Answer :

4


Explanation :

Numbers less than 100 that have exactly three factors are 4, 9, 25 and 49. All of these numbers are squares of prime numbers, which means that their only factors are one, themselves and their square roots.


2² = 4 = factors(1,2,4)
3² = 9 = factors(1,3,9)
5² = 25 = factors(1,5,25)
7² = 49 = factors(1,7,49)

Find the sum of the first 20 terms of the series : 8, -3, 11, -6, 14, -12, 17, -24,....... ?

A)
-2854

B)
2854

C)
-3069

D)
3069


Correct Answer :

-2854


Explanation :

8, -3, 11, -6, 14, -12, 17, -24,.......

odd term: next value is +3 of previous one
8, (+3) 11,(+3)14,(+3) 17...-3, -6, -12, -24, -48, -96, -192, -384, -768, -1536
8, 11, 14, 17, 20, 23, 26, 29, 32, 35
total => 215

even term: next value is double of previous one
-3, (-3*2) -6, (-6*2) -12, (-12*2) -24....
-3, -6, -12, -24, -48, -96, -192, -384, -768, -1536
total => -3069

total => 215 + (-3069) = -2854

Find the total number of terms in the series : -3, 1, 5, 9,............., 101 ?

A)
25

B)
26

C)
27

D)
28


Correct Answer :

27


Explanation :

the common diffrence is d= 4 and first term a = -3
1-(-3)=4 ,5-1 =4 , 9-5 =4

nth term of AP, tn = a+(n-1)d
101 = -3 +(n-1)4
26 = n-1
number of term, n = 27

How many numbers between 1200 to 1500 are multiple of 13?

A)
21

B)
22

C)
23

D)
24


Correct Answer :

23


Explanation :

The first term divisible by 13 after 1200 is 1209, the last term before 1500 divisble by 13 is 1495

Let n be the number of terms multiple by 13 between 1200 and 1500
as per Tn=A+(n-1)d
1495=1209+(n-1)*13
286= (n-1)*13
n-1 = 22
n = 23
number of terms multiple by 13, n= 23

How many 2 digit numbers are divisible by 6?

A)
12

B)
13

C)
14

D)
15


Correct Answer :

15


Explanation :

The first two digit number divisible by 6 is 12, and last two digit divisible by 6 is 96

Tn=a1+(n-1)d
Tn = the nᵗʰ term in the sequence
a1 = the first term in the sequence
d = the common difference between term

Let a=12 , d=6, Tn=96
Now we have
96=12+(n-1)6
84=(n-1)6
n-1=14
n=15

The geometric mean of three observations 40, 50 and X is 10, then the value of X is ?

A)
1/2

B)
1

C)
2

D)
3


Correct Answer :

1/2


Explanation :

Geometric mean observations=40,50,x
=> (40*50*X)1/3 =10
=> 40*50*X=(10)3
=> X= 1000/(40*50)
=> X= 1/2

The value of product 6 1/2 * 6 1/4 * 6 1/8 * 6 1/16 * -----∞ upto infinite terms?

A)
1

B)
6

C)
36

D)
216


Correct Answer :

6


Explanation :

6 1/2 * 6 1/4 * 6 1/8 * 6 1/16 * -----∞ => 6 (1/2 + 1/4 + 1/8 + 1/16 -----∞)

1/2 + 1/4 + 1/8 + 1/16 -----∞ is geometric series [a = 1/2, r = 1/2]
The sum of geometric series = a / 1-r
=> 1/2 / 1− 1/2
=> 1/2 / 1/2
=> 1

Hence , 6 (1/2 + 1/4 + 1/8 + 1/16 -----∞) => 6 (1) => 6

A person is to count 4500 currency notes. Let an denote the number of notes he counts in the nth minute. If a1 = a2 = .... = a10 = 150 and a10 , a11 ,... are in an AP with common difference – 2, then the time taken by him to count all notes is ?

A)
24 minutes

B)
34 minutes

C)
125 minutes

D)
135 minutes


Correct Answer :

34 minutes


Explanation :

Suppose he takes n minutes to count 4500 notes.
Number of notes he counts in the 10 minute
=> a1+a2+........+a10 = 150*10 =1500 (a1 = a2 = .... = a10 = 150)
As is given
a1+a2+........+an=4500
Hence, a11+a12+.....+an = 4500-1500 = 3000

a10 , a11 ,... are in an AP with common difference −2 then a10=150 and a11=148
Now, 3000 = Sum of n terms of an A.P.(first term=148 , difference= −2)
Sum of first n terms of an AP: S =(n/2)[2a + (n- 1)d]
⇒3000 = (n/2)[2*148+(n−1)*(−2)]
⇒3000 = n(148−(n−1))
⇒3000=n(149−n)
⇒n2−149n+3000=0
⇒(n−125)(n−24)=0
⇒n=125,24
n=125 is not possible because it is negative[ a125= 148+(125-1)*-2 =148 - 252 < 0 ]
Hence, time taken=(10+24)=34 min.

Find the sum of the given series: 3 + 9 + 27 + 81 + 243 + 729 + 2187 + 6561

A)
9840

B)
9855

C)
7960

D)
8892


Correct Answer :

9840


Explanation :

3 + 9 + 27 + 81 + 243 + 729 + 2187 + 6561
31 + 32 + 33 + 34 + 35 + 36 + 37 + 38

a{(rn - 1)/(r - 1)}, a is first term, r is common ratio, n is number of terms
a = 3, r = 3, n = 8.

3 * {(38 - 1)/(3 - 1)}
3 *(6561-1)/2
3*3280
9840

The sum of the series is 9840

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