## NUMBERS SET II MCQs

Total Questions : 90 | Page 1 of 9 pages
Question 1. Monish scored 50 marks, when each correct answer is awarded 4 marks and 1 mark is deducted for each wrong answer. Had 6 marks been awarded for each correct answer and 2 marks deducted for each incorrect answer, then Monish would have scored 60 marks. If all the questions are to be attempted compulsorily, then how many questions were there in the test?
1.    60
2.    20
3.    30
4.    50
:
D
Solution: - Let the number of correct answers marked by Monish be X and wrong answer be Y.
4X – Y = 50 ___________(1)
6X – 2Y= 60___________(2)
From Equation (1) and (2).
X = 20 and Y = 30
Total number of questions = 20 + 30 = 50.
Question 2.  A, B, C, D …………..X, Y, Z are the players who participated in a tournament. Everyone played with every other player exactly once.Team wins  2 points, a draw one point and a loss zero point. None of the matches ended in a draw. No two players scored the same score. At the end of the tournament, a ranking list is published which is in accordance with the alphabetical order, i.e. A is the top of the list. Then:
1.    M wins over N
2.    N wins over M
3.    M does not play with N
4.    M does not play with A
Answer: Option A. -> M wins over N
:
A
Option A.
It is given in the question that ranking is in accordance with the alphabetical order. It means, A occupies first, B second, C third, D fourth position and so on. In other words A wins all the matches, B wins all the matches except with A, C wins all the matches except with A and B and so on.
In view of the above order N wins all the matches except with A to M. Hence M wins over N.
Question 3. If we multiply first 50 prime numbers, then how many zeros will such a product have at the end?
1.    1
2.    0
3.    12
4.    10
:
A
We need 2's and 5's to get zeros. 2 is the only even prime number. So, the number of zeros is 1.
Question 4. N=23×32×53×10. If P is a natural number, then how many factors of N are of the form 2P+1?
1.    14
2.    15
3.    13
4.    12
:
A
N=24×32×54. Ultimately, we need to find out the number of odd factors of N except 1. Number of odd factors (including 1) =(2+1)×(4+1)=3×5=15.
Question 5. Given that p is a prime number. p2[p271p] is always divisible by?
1.    25
2.    23
3.    27
4.    none of these
Answer: Option D. -> none of these
:
D
According to fermat's theorem " If P is a prime number and N is prime to P, then NpN is divisible by P."
Open the brackets, p2[p271p]=p29p will always be divisible by 29, as it is prime.
Question 6. Multiplying prime numbers >1 to <100 consecutively, how many zeroes will result at the end?
1.    0
2.    2
3.    3
4.    None of these
Answer: Option D. -> None of these
:
D
option (d)
First prime number =2 .To need a zero, we need a 2 and a 5 as 2×5=10. this occurs only once,
hence there will be only one zero. Answer = option d
Question 7. Find the remainder when 9400+9401+9402.......9410 is divided by 6?
1.    0
2.    1
3.    3
4.    2
:
C
Any power of 9 when divided by 6, gives a remainder 3. hence answer =11×36((As there are 11 terms)= 336 = Remainder =3
Question 8. N=(323232.......50 digits)9. ie in base 9. Find the remainder when N is divided by 8?
1.    1
2.    -1
3.    0
4.    3
5.    None of these
:
D
In base 9, divisibility check for 8 is sum of the digits(digit sum) =Digit sum of N=(253+252)9=
First convert the numbers to base 10
25 in base 10=23, Thus the question changes to (233+232)10=11510
Converting it back to base 9 = 137, Now sum of digits in base 9=1+3+7=(4+7)9=121+28R=3
Question 9. (11100)2=(1001)x. What is the value of x?
1.    4
2.    5
3.    3
4.    10
5.    None of these
:
C
(11100)2=0+0+22+23+24=28
(1001)x=x0+x3=1+x3
1+x3=28=27+1=33+1,x=3.
Question 10. Find the sum of the series N=13+115+135.........100 terms
1.    99199
2.    1199
3.    100201
4.    1201
5.    1199