Numbers Set Ii(Exams > Cat > Quantitaitve Aptitude ) Questions and answers for exam Preparation

Exams > Cat > Quantitaitve Aptitude

NUMBERS SET II MCQs

Total Questions : 90 | Page 6 of 9 pages

Question 51.

What is the largest power of 2 that divides $2^{500} + 10^{501}$ ?

$2^{501}$
$2^{500}$
$2^{499}$
$2^{502}$

Discuss Question

Answer: Option B. ->

2^{500}

:
B

$2^{500} (1 + 2 \times 5^{501}      O d d)$ . Hence, $2^{500}$ is the answer.

Question 52.

What is the unit digit of $(6^{28} - 3^{22})$ ?

7
6
3
Cannot be determined

Discuss Question

Answer: Option A. -> 7
:
A

Unit digit of $6^{28}$ is 6 and unit digit of $3^{22}$ is 9 (since $22 = 4 k + 2$ ). So, unit digit of given expression is $(16 - 9) = 7$ .

Question 53.

If we multiply first 50 prime numbers, then how many zeros will such a product have at the end?

Discuss Question

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.

Question 54.

Parliament of India has 250 MPs. Each of them drink 25 cans of cold drinks per day. The cupboard in the canteen inside the parliament has “a” rows and “a” columns for storing cans of cold drinks. If in each row, we can put only 1000 cans of cold drinks. The cans will last for (Maximum) :-

Discuss Question

Answer: Option A. -> 160
:
A

Solution:
Total number of cans $= 1000 \times 1000 = 1, 00, 0000$

Daily consumption of cans $= 250 \times 25 = 6250$

The number of days, that the cans will last $= \frac{1, 00, 0000}{6250} = 160.$

Question 55.

$N = 2^{3} \times 3^{2} \times 5^{3} \times 10.$ If P is a natural number, then how many factors of N are of the form $2 P + 1$ ?

Discuss Question

Answer: Option A. -> 14
:
A

$N = 2^{4} \times 3^{2} \times 5^{4}$ . Ultimately, we need to find out the number of odd factors of N except 1. Number of odd factors (including 1) $= (2 + 1) \times (4 + 1) = 3 \times 5 = 15.$

Answer- $(15 - 1) = 14$ .

Question 56.

$S = 1 \times 6 + 2 \times 7 + 3 \times 8 + . . . . . . . . . . . . . . . + 15 \times 20$ , what is the value of S?

1841
1840
2871
2870

Discuss Question

Answer: Option B. -> 1840
:
B

$T_{n} = n (n + 5) = n^{2} + 5 n$

$S_{n} = \sum T_{n} = \sum n^{2} + 5 \sum n$

$= \frac{n (n + 1) (2 n + 1)}{6} + 5 \times \frac{n (n + 1)}{2}$ ; Put n = 15

$S_{n}$ = 1840

Question 57.

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?

Discuss Question

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.

Question 58.

A number when divided by 247 leaves a remainder 91. What will be the remainder if the number is divided by 19?

Discuss Question

Answer: Option C. -> 15
:
C

Conventional Solution: -Let the number be N

N = 247K + 91

247 is divisible by 19.

If N is divide by 19. Remainder will be 91 – 76 = 15.

Alternate solution :

247 is divisible by 19. The first number , N would be 247+91 = 338. 338 when divided by 19 gives remainder 15.

Question 59.

Five friends tried to guess the number of cows in a herd. Aamil guessed 32, Ankit guessed 38, Ashank guessed 34, Swapnil guessed 30, and Ajay guessed 36. Two were wrong by 2, and two were wrong by 4. The other one was correct.How many cows were there in the herd?

Discuss Question

Answer: Option C. -> 34
:
C

Go from answer options. The numbers are 30, 32, 34, 36 and 38. Only 34 satisfies these conditions.

Question 60.

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:

M wins over N
N wins over M
M does not play with N
M does not play with A

Discuss Question

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.