Numbers- Aptitude Questions and Answers

The sum is = 2+3+5+7+11+13+17= 58

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.

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

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 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

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.

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

(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

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

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

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

4th term of AP, a4 = a + 3d where a is st 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 value in (2) we get : a + 6*3 = 26 or, a = 8
Therefore 12th term a12 = a + 11d
=> 8 + 11*3
=> 8 + 33
=>41

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

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

Only 143616 is a multiple of both 11 and 8 so it is a multiple of 88.

These numbers are following
5, 15, 25, 35, 45, 50, 55, 65, 75, 85, 95
So, total 11 such type of numbers.

1+2+3+4....+100= n(n+1)/2
hence x= 100*(100+1)/2 =50*101 =5050

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