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### Exams > Cat > Quantitaitve Aptitude

## NUMBERS SET II MCQs

Total Questions : 90
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**1**of 9 pages**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?**

__Question 1.__**Answer: Option D. ->**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.

**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:**

__Question 2.__**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.

**Answer: Option A. ->**1

:

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.

**Answer: Option A. ->**14

:

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.

Answer- (15−1)=14.

**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 Np−N is divisible by P."

Open the brackets, p2[p27−1p]=p29−p will always be divisible by 29, as it is prime.

**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

**Answer: Option C. ->**3

:

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

**Answer: Option D. ->**3

:

D

In base 9, divisibility check for 8 is sum of the digits(digit sum) =Digit sum of N=(25∗3+25∗2)9=

First convert the numbers to base 10

25 in base 10=23, Thus the question changes to (23∗3+23∗2)10=11510

Converting it back to base 9 = 137, Now sum of digits in base 9=1+3+7=(4+7)9=12⇒1+28∣R=3

**Answer: Option C. ->**3

:

C

(11100)2=0+0+22+23+24=28

(1001)x=x0+x3=1+x3

1+x3=28=27+1=33+1,x=3.

**Answer: Option C. ->**100201

:

C

First term of the series (13)=1(3×1)=12[1−13]

Second term of the series (115)=1(3×5)=12[13−15]

Third term of the series (135)=1(7×5)=12[15−17]

..............................................................................

100th term of the series (1199×201)=12[1199−1201]

12[1−13+13−15+15−17+17...................+1199−1201]

12[1−1201]=200(2×201)=100201