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# Elitmus Number System Questions And Answers

## Elitmus Number System Questions And Answers

1.If n is any odd number greater than 1, then n(n2 – 1) is always divisible by?
A) 96
B) 48
C) 24
D) None of these

Explaination :
We have use put different values of n ( odd numbers ) greater than 1.
i.e. n=3, 5, 7, 9
When n=3 n(n2 – 1)= 24
When n=5 n(n2 – 1)= 120
When n=7 n(n2 – 1)=336
using options we find that all the numbers are divisible by 24

2.Three consecutive positive even numbers are such that thrice the first number exceeds double the third by 2, the third number is?
A) 10
B) 14
C) 16
D) 12

Explaination :
Let the three even number are ( x – 2), x, (x + 2)
Then, 3(x – 2) – 2(x + 2) = 2
3x – 6 – 2x – 4 = 2 i.e. x=12
Hence, the third number is (12 + 2) = 14

3.Which is the least number that must be subtracted from 1856 so that the remainder when divided by 7, 12, 16, is 4?
A) 137
B) 1361
C) 140
D) 172

Explaination :
Firstly, we have to take LCM of 7, 12, 16 = 336
If we divide 1856 by 336, then remainder is 176.
Since it is given that remainder in this condition is 4. Hence, the lease number to be subtracted = (176 – 4)172.

4.For the product n(n+1)(2n+1) , which one of the following is not necessarily true?
A) It is even
B) Divisible by 3
C) divisible by (n(n+1)(2n+1))/2
D) Never divisible by 237

Explaination :
We have to check for each option separately by taking the different values of n.
Option (A) : Check for n= 3, 4, 5, 6…
Option (B) : Check for n= 3, 4, 5, 7…
Option (C) : Divisible
Option (D) : For n= 237 : n(n + 1)(2n + 1) is divisible.

5.The integers 1, 2, 3, … 40 are written on blackboard. The following operation is then repeated 39 times. In each repetition, any two numbers, say a and b, currently on the blackboard are erased and a new number a + b – 1 is written. What will be the number left on the board at the end ?
A) 820
B) 821
C) 781
D) 819

Explaination :

According to question, if two numbers say a and b are erased and replaced by a new number a + b – 1, then in every repetition, the number of integers gets reduced by 1 and consequently at the last repetition there will be only one number left.
Whatever may be our selection of two numbers a and b. In any and every repetition, the final number so arrived will not changes.
Now, the sum of integers from 1 to 40= n(n + 1)/2 = 820
As, discussed above the sum of integers of the first, second, third ……… repetitions will be 819, 818, 817, ………….. so on respectively.
Therefore, after 39 operations there will be only 1 number left and that will be 820 – 39=781.

6.The number of common terms in the two sequences 17, 21, 25, ….. , 417 and 16, 21, 26, …,466 is ?
A) 19
B) 20
C) 77
D) 22

Explaination :

Both the sequences ( 17, 21, 25 ……………… and (16, 21, 26….. are arithmetic progression with a common difference of 4 and 5 respectively.
In both the sequence first common term is 21.
Hence a new arithmetic sequence containing the common terms of both the series can be formed with a common difference of LCM of ( 4, 5) is 20
New sequence will be 21, 41, 61, ….401
nth term = a + (n-1)d
401= 21 + (n-1)20
n-1=19
Hence, n=20

7.Let U(n+1)= 2Un + 1, ( n=0, 1, 2, …………) U0=0 then U(10) would be nearest to ?
A) 1023
B) 2047
C) 4095
D) 8195

Explaination :

U(n+1)= 2Un + 1, ( n=0, 1, 2, …………) U0=0
Put
n=0, U1=1
n=1, U2=3
n=2, U3=7
n=4, U5=31
Seeing this pattern we can conclude i.e. U(n)= 2n -1
Hence U(10)= (2)10 – 1= 1023

8.A young girl counted in the following way on the fingers of her left hand. She started calling the thumb 1, the index finger 2, middle finger 3, ring finger 4, little finger 5, then reversed direction, calling the ring finger 6, middle finger 7, index finger 8 and thumb 9 and then back to the index finger for 10, middle finder for 11 and so on. She counted up to 1994. She ended on her ?
A) thumb
B) index finger
C) middle finger
D) ring finger

Explaination :

Thumb Finger :1, 9, 17, ….
Index Finger: 2, 8, 10, 16, 18, ….
Middle Finger: 3, 7, 11, 16, 19, ….
Ring Finger : 4, 6, 12, 14, 20, ……
Little Finger : 5, 13, 21, ….
Numbers on thumb forms an AP with common difference=8
Numbers on middle forms an AP with common difference=4
1993 will be on the thumb.
Hence, 1994 will be on the index finger.

9.The product of all integers from 1 to 100 will have the following numbers of zeros at the end ?
A) 20
B) 24
C) 19
D) 22

Explaination :

Every combination of 5 and 2 will give one zero, and number of zero in the product of any number is decided by the number of 2 and 5, whichever is less.
Hence, this problem can be solved by determining the number of 2 and 5 between 1 to 100.
Clearly, there are 20 numbers which are divisible by 5. Besides, there are four numbers 25, 50, 75, and 100 which will have one addition 5.
Hence, number of zeroes in the product of all the numbers from 1 to 100 is 21.

10.The number of positive integers not greater than 100, which are not divisible by 2, 3 or 5 is ?
A) 26
B) 18
C) 31
D) None of these

Explaination :
There are 50 odd numbers less than 100 which are not divisible by 2.
Out of these 50 there are 17 number which are divisible by 3.
Out of remaining there are 7 numbers which are divisible by 5.
Hence, numbers which are not divisible by 2, 3, 5 = 50 – 17 – 7 = 26

## Elitmus Quantitative Aptitude Question And Answers

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