## NUMBER SET I MCQs

Total Questions : 90 | Page 1 of 9 pages
Question 1. The sum of four consecutive two digit odd numbers, when divided by 10, becomes a perfect square. Which of the following can possibly be one of these four members? (CAT 2006)
1.    21
2.    25
3.    41
4.    67
5.    73
:
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.
Question 2. A group of 630 children is arranged in rows for a group photograph session. Each row contains three fewer children than the row in front of it. What number of rows is not possible? (CAT 2006)
1.    3
2.    4
3.    5
4.    6
5.    7
:
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(n1)]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
Question 3. Let N=553+173723. N is divisible by : (CAT 2000)
1.    both 7 and 13
2.    both 3 and 13
3.    both 17 and 7
4.    both 3 and 17
Answer: Option D. -> both 3 and 17
:
D
Ans: (d).
N=553+173723=(54+1)3+(181)3723
=(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)
Question 4. How many pairs of positive integers m, n satisfy (1m)+(4n)=(112), where n is an odd integer less than 60? (CAT 2007)
1.    4
2.    7
3.    5
4.    3
5.    6
:
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(n48);
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
Question 5. 76n66n, where n is an integer >0, is divisible by (CAT 2002)
1.    13
2.    127
3.    559
4.    All of these
5.    81
Answer: Option D. -> All of these
:
D
Ans: (d)
For n=1,7666=(73)2(63)2;
=(7363)(73+63)=(343216)(343+216);=127559=1271343. Hence it is divisible by 127,13,559.
Question 6. The minimum number of flowers with which Roopa leaves home is : (CAT 1999)
1.    16
2.    15
3.    0
4.    Cannot be determined
:
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
BalancebeforeofferingFlowersofferedBalanceafterofferingIstPlace2xy2xyIIndPlace4x2yy4x3yIIIrdPlace8x6yy8x7yIVthPlace16x14yy16x15y
16x15y=016x=15yy=16x15
Question 7. The minimum number of flowers that could be offered to each deity is : (CAT 1999)
1.    0
2.    15
3.    16
4.    cannot be determined
:
C
Ans: (c)
Minimum value for y is available for x=15;y=x15y=16;
Let Roopa have x flowers with her; then
BalancebeforeofferingFlowersofferedBalanceafterofferingIstPlace2xy2xyIIndPlace4x2yy4x3yIIIrdPlace8x6yy8x7yIVthPlace16x14yy16x15y
16x15y=016x=15yy=16x15y
Therefore minimum value of x for which y is an integer is 15 hence y = 16.
Question 8. If n is an integer, how many values of n will give an integral value of(16nz+7n+6)n? (CAT 1997)
1.    2
2.    3
3.    4
4.    None of these
:
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.
Question 9. L Let D be a recurring decimal of the form D=0. a1a2a1a2a1a2............; where digits a1 and a2lie between 0 and 9. Further, at most one of them is zero. Which of the following numbers necessarily produces an integer, when multiplied by D? (CAT 2000)
1.    18
2.    108
3.    198
4.    288
:
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.
Question 10. Let S be a set of positive integers such that every element n of S satisfies the conditions
I. 1000n1200
II. Every digit in n is odd
Then how many elements of S are divisible by 3? (CAT 2005)
1.    9
2.    10
3.    11
4.    12