# Free Numbers Set II - 03 Practice Test - CAT

### Question 1

What is the numerical value of M+F?

#### SOLUTION

Solution :A

Option A.Given that the combined average is M+F, hence

(M+F)=(2MF+42){(M+F)+1}

Or (M+F)2+(M+F)=2MF+42

Or M2+F2+M+F=42

Or M(M+1)+F(F+1)=42, since M and F are integers, there is only one possibility that M(M+1) =12 or 30 and F(F+1) =30 or 12. Hence M can be 3 or 5 and F can be 5 or 3. So the sum of M and F is 8. Thus

option (a).

### Question 2

What is the numerical value of F?

#### SOLUTION

Solution :D

Option D :

The value of F = 5 or 3; so, we can’t determine exact value of F. Answer is can’t be determined

### Question 3

If ab×cd=3337 and ba×cd=799, find the value of (ab+cd) given that ab, ba and cd are all two digit positive integers.

#### SOLUTION

Solution :B

1st Method: - 71×47=3337 and 17×47=799, so ab+cd=118 (Option b).

2nd Method: - Odd × Odd = Odd and Odd + Odd = Even.

From the question we can conclude that all the four numbers are odd numbers. So the required answer has to be even, only option (b) is even.

### Question 4

If 5y–5(y−1)=100, then what is the value of (3y)y?

#### SOLUTION

Solution :D

5y−5y5=100; 5y(1−15)=52×4; 5y(45)=52×4

⟹5(y−1)×4=52×4;y−1=2; y=3. Answer is (3y)y=(9)3=729.

### Question 5

When (11111+511111) is divided by 4, what is the remainder?

#### SOLUTION

Solution :A

(11111)4+5111114=34+(4+1)111114=34+14=44. Thus the required remainder is 0.

### Question 6

What is the largest power of 2 that divides 2500+10501?

#### SOLUTION

Solution :B

2500(1+2×5501Odd). Hence,

2500 is the answer.

### Question 7

What is the unit digit of (628−322)?

#### SOLUTION

Solution :A

Unit digit of 628 is 6 and unit digit of 322 is 9 (since 22=4k+2). So, unit digit of given expression is (16−9)=7.

### Question 8

N=23×32×53×10. If P is a natural number, then how many factors of N are of the form 2P+1?

#### SOLUTION

Solution :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.

### Question 9

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

#### SOLUTION

Solution :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 10

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) :-

#### SOLUTION

Solution :A

Solution:

Total number of cans =1000×1000=1,00,0000Daily consumption of cans =250×25=6250

The number of days, that the cans will last =1,00,00006250=160.

### Question 11

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?

#### SOLUTION

Solution :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 12

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

#### SOLUTION

Solution :B

Tn=n(n+5)=n2+5n

Sn=∑Tn=∑n2+5∑n

=n(n+1)(2n+1)6+5×n(n+1)2; Put n = 15

Sn= 1840

### Question 13

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

#### SOLUTION

Solution :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 14

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?

#### SOLUTION

Solution :C

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

### Question 15

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:

#### SOLUTION

Solution :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.