**INDUSTRY AND COMPANY AWARENESS (ICA)**

### Exams > Cat > Quantitaitve Aptitude

## NUMBER SET I MCQs

Total Questions : 90
| Page

**1**of 9 pages**Answer: Option C. ->**41

:

C

Option (c)

Sum of 4 consecutive odd numbers should result in a zero in the end, so as to be divisible by 10.

This happens in 7+9+1+3. Using options, we find that four consecutive odd numbers are 37, 39, 41 and 43, as the sum of these 4 numbers is 160, When divided by 10, we get 16 which is a perfect square.

Thus, 41 is one of the odd numbers.

**Answer: Option D. ->**6

:

D

Let there be n rows and a students in the first row.

Number of students in the second row = + 3a Number of students in the third row = a+ 6 and so on. aThe number of students in each row forms an arithmetic progression with common difference = 3. The total number of students = The sum of all terms in the arithmetic progression.

=n[2a+3(n−1)]2=630

Now consider options.

1.n=3,a=207

2.n=4,a=153

3.n=5,a=120

4.n=6,a=1952

5.n=7,a=8

Hence the only option not possible is when n=6

**Answer: Option D. ->**both 3 and 17

:

D

Ans: (d).

N=553+173−723=(54+1)3+(18−1)3−723

=(51+4)3+173−(68+4)3.

These two different forms of the expression are divisible by 3 and 17 both.

Elimination strategy:

Check the divisibility by 13, it is seen that it is not divisible, now check for 3, it is

divisible, hence the only option will be (d)

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

:

D

Soln:d.

Given conditions :

(1m)+(4n)=(112)

M and n are positive integers

N is an odd integer less than 60.

On solving the equation, we get m=12n(n−48);

Since m and n are positive integers, n>48, and from the question, 48<n<60.

Hence possible values on n=49,51,53,55,57,59.

For n=49,51 and 57 we get integral values of m. Hence the required no of integral pairs = 3

**Answer: Option D. ->**All of these

:

D

Ans: (d)

For n=1,76−66=(73)2−(63)2;

=(73−63)(73+63)=(343−216)(343+216);=127∗559=127∗13∗43. Hence it is divisible by 127,13,559.

**Answer: Option B. ->**15

:

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

BalancebeforeofferingFlowersofferedBalanceafterofferingIstPlace2xy2x−yIIndPlace4x−2yy4x−3yIIIrdPlace8x−6yy8x−7yIVthPlace16x−14yy16x−15y

16x−15y=0⇒16x=15y⇒y=16x15

**Answer: Option C. ->**16

:

C

Ans: (c)

Minimum value for y is available for x=15;y=x∗15⇒y=16;

Let Roopa have x flowers with her; then

BalancebeforeofferingFlowersofferedBalanceafterofferingIstPlace2xy2x−yIIndPlace4x−2yy4x−3yIIIrdPlace8x−6yy8x−7yIVthPlace16x−14yy16x−15y

16x−15y=0⇒16x=15y⇒y=16x15y

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

**Answer: Option C. ->**4

:

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.

**Answer: Option C. ->**198

:

C

Ans: (c)

D=0.a1a2;100D=a1a2.a1a2 (recurring); Thus, 99D=a1a2;⇒D=(a1a299)

Required number should thus be a multiple of 99. Hence 198 is the required number.

**Answer: Option A. ->**9

:

A

Ans: (a)

The 100th and 1000th position value will be only 1.Divisibility rule for 3 is sum of digits. We already have a sum of 2, we need to make it to 6, 9 or another multiple of 3.

Units digit= a

Tens digit= b

Possible combinations of a+b= 4, 7, 10, 13, 16; where both "a and b" are odd

Thus, the possibility of unit and tens digits are (1, 3), (1, 9), (3, 1), (3, 7), (5, 5), (7, 3),(7, 9), (9, 1), (9, 7).