# Free Number Set I - 01 Practice Test - CAT

The number of positive integers not greater than 100, which are not divisible by 2, 3 or 5 is : (CAT 1993)

A. 26
B. 18
C. 31
D. None of these

#### SOLUTION

Solution : A

Ans:(a)

There are 50 odd numbers less than 100 which are not divisible by 2. Out of these 50 there are 17 numbers 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

Alternate Approach:

Taking the Euler's number approach:

If we take the prime factors of a number then we get all the numbers that are co-prime to that number, which means all the multiples of the prime factors are removed, using the same formula we can find out the numbers that are not divisible by a particular set of numbers.

100=29

As we don't want the number 2, 3, and 5 also to in the list remove these 3 numbers, hence option will be 26

Which is the least number that must be subtracted from 1856 so that the remainder when divided by 7, 12, and 16 is 4?

(CAT 1994)

A. 137
B. 1361
C. 140
D. 172

#### SOLUTION

Solution : D

Ans: (d)

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 least number to be subtracted = (176 - 4) = 172.

Alternate Approach:

In case you don't remember this method: go from answer options by subtracting the answer option from 1856 and divide by each of 7,12, and 16 to get a reminder of 4.

Two positive integers differ by 4 and sum of their reciprocals is 1021. Then one of the numbers is: (CAT 1995)

A. 3
B. 1
C. 5
D. 21

#### SOLUTION

Solution : A

Ans: (a)

Let one number be x, then second number will be (x+4);
Thus :1x+(1(x+4))=1021;
(x+x+4)(x(x+4))=1021;
(2x+4)x(x+4)=1021;
x=3

Elimination strategy:

By looking at the answer option as the sum of reciprocal of the numbers will have to give 1021The number has to be a factor or multiple of 21. So directly option (c) can be eliminated.

Now the numbers that we will have to check will be 3 & 7 , 1 & 5, 17 & 21, 21 &25

The only possibility from here is 3 & 7. Hence option (a).

The remainder obtained when a prime number greater than 6 is divided by 6 is :

(CAT 1995)

A. 1 or 3
B. 1 or 5
C. 3 or5
D. 4 or 5

#### SOLUTION

Solution : B

Ans: (b)

This is a generalized question: the important word "ANY" can be used here. Let us solve the question for some prime numbers greater than 6 i.e., 7, 11, 13 and 17. If these numbers are divided by 6, the remainder is always either 1 or 5.

If a number 774958A96B is to be divisible by 8 and 9, the respective values of A and B will be :

(CAT 1996)

A. 7 and 8
B. 8 and 0
C. 5 and 8
D. None of these

#### SOLUTION

Solution : B

Ans: (b)

The number 774958A96B is divisible by 8 if 96B is divisible by 8. And 96B is divisible by 8 if B is either 0 or 8. Now to make the same number divisible by 9 sum of all the digits should be divisible by 9. Hence (55 + A + B) is divisible by 9 if (A + B) is either 0 or 8 either A = 0 or B = 8, or A = 8 or B = 0. Since the number is divisible by both A and B. Hence A and B may take either values, i.e., 8 or 0.

A student instead of finding the value of 78 of the number, found the value of 718 of the number. If his answer differed from the actual one by 770, find the number. (CAT 1997)

A. 1584
B. 2520
C. 1728
D. 1656

#### SOLUTION

Solution : A

Ans: (a)
7Dx71Dx=770=(12656144)x=770=x=1584

Elimination strategy:

As 718 is slightly less than half of 78,we have to find the answer which is slightly greater than twice of 770, hence check the given conditions for 1584 and 1656, the other options can be easily eliminated. Check manually for 1584. we find that the difference is 770, hence that is the option.

If n is an integer, how many values of n will give an integral value of(16nz+7n+6)n? (CAT 1997)

A. 2
B. 3
C. 4
D. None of these

#### SOLUTION

Solution : C

Ans: (c)

(16nz+7n+6)n=16n+7+6n;
Since n is an integer, hence for the entire expression to become an integer, (6n) should be an integer. And (6n) can be integer for n=1,2,3,6. Hence n has 4 values.

A is set of positive integers such that when divided by 2, 3, 4, 5, 6 leaves the remainders 1, 2, 3, 4, 5 respectively. How many integers between 0 and 100 belong to set A?

(CAT 1998)

A. 0
B. 1
C. 2
D. None of these

#### SOLUTION

Solution : B

Ans: (b)

This can be solved by Chinese remainder theorem, but as the common difference is constant, it is a special case of Chinese remainder concept.

Required number of the set is calculated by the LCM of (2, 3, 4, 5, 6) - (common difference)

In this case, common difference = (2-1) = (3-2) = (4-3) = (5- 4) = (6 - 5) = 1.

All integers of the set will be given by (60n - 1) ; If n = 1, (60 - 1) = 59; If n = 2,((60*2) - 1) = 119; Since range of the set A is between 0 and 100, hence there will exist only one number i.e., 59.

Number of students who have opted for the subjects A, B and C are 60, 84 and 108 respectively. The examination is to be conducted for these students such that only the students of the same subject are allowed in one room. Also the number of students in each room must be same. What is the minimum number of rooms that should be arranged to meet all these conditions? (CAT 1998)

A. 28
B. 60
C. 12
D. 21

#### SOLUTION

Solution : D

Ans: (d)

Number of students which should be seated in each room is the HCF of 60, 84 and 108 which is 12.

Number of rooms required for subject A, subject B and subject C=6012=5 rooms, 8412=7rooms and 10812=9 rooms respectively. Hence minimum number of rooms required to satisfy our condition =(5+7+9)=21.

All values in S1 are changed in sign, while those in S2 remain unchanged. Which of the following statements is true?

A. Every member of S1 is greater than or equal to every member of S2.
B. G is in S1.
C. If all numbers originally in S1 and S2 had the same sign, then after the change of sign, the largest number of S1 and S2 is in S1.
D. None of the above

#### SOLUTION

Solution : D

Ans: (d)

None of the options (a), (b) and (c) is necessarily true. Hence option (d) is an answer.

As this is a variable based question: the word "ANY" can be used

Let the series of integers a1,a2,.......,a50 be 1,2,3,4,5,.......,50.

S1=1,2,3,4,.......24, S2=25,26,27............50

Elements of S1 are in ascending order, and those of S2 are in descending order. a24 and a35 are interchanged, then which of the following statements is true?

A. S1 continues to be in ascending order.
B. S2 continues to be in descending order.
C. S1 continues to be in ascending order and s2 in descending order.
D. None of the above

#### SOLUTION

Solution : A

Ans: (a)

Let S1 = 1, 2, 3, 4,......., 24,   S2 = 50, 49,........., 25;

New series after interchange S1 = 1, 2, 3, 4........, 25,  S2 = 50, 49,.......24

It is therefore clear that S1 continues to be in ascending order.

As this is a variable based question: the word "ANY" can be used

Let the series of integers a1,a2,.......,a50 be 1,2,3,4,5,......,50.

S1=1,2,3,4,......24, S2=25,26,27,.........50

Every element of S1 is made greater than or equal to every element of S2 by adding to each element of S1 an integer x. Then, x cannot be less than: (CAT 1999)

A. 2 10
B. the smallest value of S2
C. the largest value of S2
D. (G- L)

#### SOLUTION

Solution : D

Ans: (d)

The smallest integer of the series is 1 and greatest integer is 50. If each element of S1 is made greater of equal to every element of S2, then the smallest element 1 should be added to (50 - 1) = 49. Hence option (G-L) is the correct answer.

As this is a variable based question: the word "ANY" can be used

Let the series of integers a1,a2,.......,a50 be 1,2,3,4,5,.......,50.

S1=1,2,3,4,.........24, S2=25,26,27,..........50

If Roopa leaves home with 30 flowers, the number of flowers she offers to each deity is : (CAT 1999)

A. 16
B. 24
C. 32
D. 30

#### SOLUTION

Solution : C

Ans: (c)

If x=30,

y=16×2=32

Let Roopa have x flowers with her; then

Balance before offeringFlowers offeredBalance after offeringIst Place2xy2xyIInd Place4x2yy4x3yIIIrd Place8x6yy8x7yIVth Place16x14yy16x15y

16x15y=016x=15yy=16x15

The minimum number of flowers that could be offered to each deity is : (CAT 1999)

A. 0
B. 15
C. 16
D. cannot be determined

#### SOLUTION

Solution : C

Ans: (c)

Minimum value for y is available for x=15; y=x15y=16;

Let Roopa have x flowers with her; then
Balance before offeringFlowers offeredBalance after offeringIst Place2xy2xyIInd Place4x2yy4x3yIIIrd Place8x6yy8x7yIVth Place16x14yy16x15y

16x15y=016x=15yy=16x15y

Therefore minimum value of x for which y is an integer is 15 hence y = 16.

The minimum number of flowers with which Roopa leaves home is : (CAT 1999)

A. 16
B. 15
C. 0
D. Cannot be determined

#### SOLUTION

Solution : B

Ans: (b)

Minimum value of x is available for y = 16. x=1516xy=1516×16=15

Let Roopa have x flowers with her; then

Balance before offeringFlowers offeredBalance after offeringIst Place2xy2xyIInd Place4x2yy4x3yIIIrd Place8x6yy8x7yIVth Place16x14yy16x15y

16x15y=016x=15yy=16x15