Exams > Cat > Quantitaitve Aptitude
NUMBER SET I MCQs
Total Questions : 90
| Page 3 of 9 pages
Answer: Option C. -> 01
:
C
Ans: (c) 01
The last two digits of a number is nothing but the remainder obtained when the number is divided by 100.This number leaves a remainder 1 when divided by 4 as well as 25. Hence the remainder obtained when this number is divided by 100 is also 1. Hence the last two digits of this number are 01.
Alternatively
Using Chinese remainder theorem: as we have to divide by 100 and find the reminder by 4 and then by 25 ,
i.e., 4A+1=25B+1, finding integer solutions we see that A=25 and B=4 will hence give a remainder of 01.
Alternatively, :
Use the last 2 digit rule for 7. Even here u will get 01 as the last 2 digits.
:
C
Ans: (c) 01
The last two digits of a number is nothing but the remainder obtained when the number is divided by 100.This number leaves a remainder 1 when divided by 4 as well as 25. Hence the remainder obtained when this number is divided by 100 is also 1. Hence the last two digits of this number are 01.
Alternatively
Using Chinese remainder theorem: as we have to divide by 100 and find the reminder by 4 and then by 25 ,
i.e., 4A+1=25B+1, finding integer solutions we see that A=25 and B=4 will hence give a remainder of 01.
Alternatively, :
Use the last 2 digit rule for 7. Even here u will get 01 as the last 2 digits.
Answer: Option A. -> 10
:
A
We know, 992250 = 21 x 34 x 53 x 72.
The value of a + b + c + d = 1 + 3 + 4 + 2 = 10
:
A
We know, 992250 = 21 x 34 x 53 x 72.
The value of a + b + c + d = 1 + 3 + 4 + 2 = 10
Answer: Option E. -> 97 : 84
:
E
Ans:(e)
Using options, we find that sum of numerator and denominator or 97 : 84 is (97 + 84) = 181 which is a prime number. Hence, it is the appropriate answer.
:
E
Ans:(e)
Using options, we find that sum of numerator and denominator or 97 : 84 is (97 + 84) = 181 which is a prime number. Hence, it is the appropriate answer.
Answer: Option D. -> 4
:
D
Ans: (d)
42=4 (mod 6)
44=4 (mod 6)
46=4 (mod 6)
and so on. The answer will remain the same.
:
D
Ans: (d)
42=4 (mod 6)
44=4 (mod 6)
46=4 (mod 6)
and so on. The answer will remain the same.
Question 25. Three consecutive positive integers are raised to the first, second and third powers respectively and then added. The sum so obtained is a perfect square whose square root equals the total of the three original integers. Which of the following best describes the minimum value, say m, of these three integers?(CAT 2008)
Answer: Option A. -> 1
:
A
Ans: a
It's an easy question. Trial with some numbers will give you the solution
1<=m<=3
31+42+53=144=122=(3+4+5)2.
:
A
Ans: a
It's an easy question. Trial with some numbers will give you the solution
1<=m<=3
31+42+53=144=122=(3+4+5)2.
Question 26. For two positive integers a and b, define the function h (a,b) as the greatest common factor (GCF) of a, b. Let A be a set of n positive integers G(A), the GCF of the elements of set A is computed by repeatedly using the function h. The minimum number of times h is required to be used to compute G is : (CAT 1999)
Answer: Option B. -> (n - 1)
:
B
Ans:
It is clear that for n positive integers function h (a,b) has to be used one time less than the number of integers, i.e., (n-1) times.
:
B
Ans:
It is clear that for n positive integers function h (a,b) has to be used one time less than the number of integers, i.e., (n-1) times.
Answer: Option E. -> 55
:
E
Sol:
e. 55
From the definition of "seed"?, it is clear that we have to count number of integers between 1 and 500, which are divisible by 9. The smallest is 9 and the largest is 495. 9∼A__1=9 and 9∼A__55=495. Hence there are 55 such numbers.
:
E
Sol:
e. 55
From the definition of "seed"?, it is clear that we have to count number of integers between 1 and 500, which are divisible by 9. The smallest is 9 and the largest is 495. 9∼A__1=9 and 9∼A__55=495. Hence there are 55 such numbers.
Answer: Option A. -> 1
:
A
Ans: (a)
2256 can be written as (24)64=(17−1)64.
In the expansion of (17−1)64every term is divisible by 17 except (−1)64. Hence remainder is 1.
Or directly:
Euler's number of 17 is 16 and 256 is a multiple of 16, hence the remainder is 1.
:
A
Ans: (a)
2256 can be written as (24)64=(17−1)64.
In the expansion of (17−1)64every term is divisible by 17 except (−1)64. Hence remainder is 1.
Or directly:
Euler's number of 17 is 16 and 256 is a multiple of 16, hence the remainder is 1.
Answer: Option C. -> 0
:
C
Ans: (c)
an+bn is always divisible by a+b when n is odd.
523+2323 is always divisible by 15+23=38. As 38 is a multiple of 19,1523+2323 is divisible by 19.
Hence we get a remainder of 0.
:
C
Ans: (c)
an+bn is always divisible by a+b when n is odd.
523+2323 is always divisible by 15+23=38. As 38 is a multiple of 19,1523+2323 is divisible by 19.
Hence we get a remainder of 0.
Answer: Option B. -> 31
:
B
Ans: (b)
(56−1)=(53)2−(1)2=(125)2−(1)2=(125+1)(125−1)=126∗124=31∗4∗126. It is therefore clear that the expression is divisible by 31.
:
B
Ans: (b)
(56−1)=(53)2−(1)2=(125)2−(1)2=(125+1)(125−1)=126∗124=31∗4∗126. It is therefore clear that the expression is divisible by 31.
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