Numbers- Aptitude Questions and Answers

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Shantanu was given a task of adding a certain number of consecutive natural numbers starting from 1. By mistake, he missed a number during addition. He obtained the sum as 800. Find the number that he missed?

A)
15

B)
20

C)
25

D)
10


Correct Answer :

20


Explanation :

Sum of n natural numbers is n(n+1)/2 ,and let missing number is x (x<=n) ,then
=> n(n+1)/2 = 800+x
n(n+1) = 1600+2x

It seems that n2 is approximate to 1600, then
n2=402
n=40 and , n+1=41

Therefore n(n+1) = 1600+2x
=> 40*41 =1600+2x
1640 =1600+2x
x = 20
Thus, 20 is the number that he missed.

A two digit number is such that the product of its digits is 12. When 9 is added to the number, the digits interchanges their places, find the number?

A)
62

B)
34

C)
2

D)
43


Correct Answer :

34


Explanation :

lets the digits of number are x and y, where x is tenth digit.then number would be 10x+y
as given
x * y = 12 ........................... (eq 1)
When 9 is added to the number ,the digits interchanges their places. i.e.
10x+y+9 = 10y+x
x + 1 = y ............................ (eq 2)

After putting eq2 in eq1 , the eq1 would be
x(x+1) = 12
x² + x - 12 = 0
(x + 4)(x - 3) = 0
then,x = -4 or 3

Suppose x= 3, then y=4 .Then number would be 34.when add 9 in 34 it would be 43, which is reversed of 34.
So correct answer is 34

What is the sum of first 10 prime numbers?

A)
Even

B)
Odd

C)
Can’t say

D)
None of these


Correct Answer :

Odd


Explanation :

first ten prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, and 29.
Their sum is (2+3+5+7+11+13+17+19+23+29)=129, that is odd.

If 43571A98B is a ‘9’ digit number divisible by 72. Which of the following is true for A and B ?

A)
A>B

B)
A=B

C)
A<B

D)
Can’t be determined


Correct Answer :

A=B


Explanation :

if 43571A98B is divisible by 72 then it should be divisible by co-prime of 72.
There, 9x8=72 So 8 and 9 are co-prime numbers of 72.

43571A98B divisible by 8.
that means last 3 digit, 98B should be divisible by 8. it possible when B=4.

now, 43571A984 to be divisible by 9.
that means sum of digits (4+3+5+7+1+A+9+8+4) <=> (41+A) should be divisible by 9 and that would be possible if A is 4.
that is (41+4)=45

Therefore, 43571A98B is divisible by 72 when A and B shuld be 4 , i.e. A=B<=>4

A 4 digit number is formed by repeating a 2 digit number such as 2424, 1212 etc. Any number of this form is always exactly divisible by?

A)
7

B)
11

C)
13

D)
101


Correct Answer :

101


Explanation :

Let the unit digit x and tenth digit y ,
then 4 digit number formed by repeating a 2 digit number as = 1000y + 100x + 10y + x
=> 1010y + 101x
=> 101 (10y + x)

So, that number must be divisible by 101, also 101 is the smallest three-digit prime number.

The product of two numbers is 192 and the difference is 4. These two numbers are?

A)
16, 12

B)
18, 14

C)
17, 13

D)
15, 11


Correct Answer :

16, 12


Explanation :

let the numbers are x and x-4
product of number is x*(x-4) =192
x2-4x-192=0
x2-4x-192=0
x2-16x+12x-192=0
(x-16)(x+12)=0
So x either 16 or -12 and other number either 12 or -16

that both sets of answers will work. (12 * 16) = 192 , (-12 * -16) = 192

(422+404)2−(4×422×404) = ?

A)
344

B)
342

C)
324

D)
312


Correct Answer :

324


Explanation :

(a+b)2 - (a-b)2 = 4ab
(a+b)2 - 4ab = (a-b)2

Hence , (422+404)2−(4×422×404) => (422 - 404)2 => (18)2 => 324

What is the largest 4 digit number exactly divisible by 88?

A)
9944

B)
9999

C)
9988

D)
9900


Correct Answer :

9944


Explanation :

Largest 4 digit number = 9999
9999 / 88 = 113, remainder = 55

Hence largest 4 digit number exactly divisible by 88 = 9999 - 55 = 9944

The sum of, 1+2+3+⋯+12=?

A)
66

B)
68

C)
76

D)
78


Correct Answer :

78


Explanation :

Sum=1+2+3+⋯+n= n(n+1)/2
Hence, 1+2+3+⋯+12= 12(12+1)/2 = 6*13 = 78

If the 4th term of an arithmetic progression is 17 and the 7th term is 26, what is the 12th term?

A)
33

B)
37

C)
41

D)
53


Correct Answer :

41


Explanation :

4th term of AP , a4 = a + 3d where a is 1st term & d is common difference,then-
a+3d = 17-------(1) 4th term
a+6d = 26 --------(2) 7th term

Subtracting (2) from (1) we get: 3d = 9 or, d = 3

Putting in (2) we get : a + 6*3 = 26 or, a = 8

So 12th term a12 = a + 11d
= 8 + 11*.3 = 8+33 =41

The 17th term of the arithmetic progression whose first two term are –3 and 2 is?

A)
136

B)
82

C)
77

D)
120


Correct Answer :

77


Explanation :

as given first two term of AP are –3 and 2 , So a is 1st term that is -3 and the 2nd term a+d = 2 (d is common difference), then-
-3 + d= 2
d=5

So 17th term of AP is = a + 16d
= -3 + 16*5
= -3 +80
= 77

The 7th term of an AP is 5 times the first term and its 9th term exceeds twice the 4th term by 1. The first term of the AP is ?

A)
151

B)
-39

C)
3

D)
-124


Correct Answer :

3


Explanation :

7th term , A + (7–1)d = 5A
A + 6d = 5A
4A – 6d = 0 ...(i)

as given, A + (9–1)d = 2[A + (4–1)d] + 1
A + 8d = 2A + 6d + 1
A –2d +1 = 0 ....(ii)

after solving (i) and (ii) we get d = 2, A = 3
hence first term = 3

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