Exams > Cat > Quantitaitve Aptitude
NUMBER SET I MCQs
Total Questions : 90
| Page 4 of 9 pages
Answer: Option A. -> 81
:
A
Sol:
(a)10<n<1000
Let n is two digit number
n=10a+b^a†′Pn=ab,Sn=a+b
then, ab+a+b=10a+b
^a†′ab=9a^a†′b=9
There are 9 such numbers 19,29,33,....,99.
Then let n is three digit number
^a†′n=100a+10b+c^a†′Pn=abc,Sn=a+b+c
Then abc+a+b+c=100a+10b+c
^a†′abc=99a+9b
^a†′bc=99+9
But the maximum value for bc=81
And RHS is more than 99. Hence, no such number is possible.
:
A
Sol:
(a)10<n<1000
Let n is two digit number
n=10a+b^a†′Pn=ab,Sn=a+b
then, ab+a+b=10a+b
^a†′ab=9a^a†′b=9
There are 9 such numbers 19,29,33,....,99.
Then let n is three digit number
^a†′n=100a+10b+c^a†′Pn=abc,Sn=a+b+c
Then abc+a+b+c=100a+10b+c
^a†′abc=99a+9b
^a†′bc=99+9
But the maximum value for bc=81
And RHS is more than 99. Hence, no such number is possible.
Answer: Option C. -> a solution for 250
:
C
Soln:
(c) 5x + 19y = 64
We see that if y = 1, we get an integer solution for x = 9, now if y changes (increases or decreases) by 5, x will change (decrease or increase) by 19.
Looking at options, if x = 256 we get y = 64.
Using these values we see option 1, 2 and 4 are eliminated and also that these exists a solution for 250<x<300.
:
C
Soln:
(c) 5x + 19y = 64
We see that if y = 1, we get an integer solution for x = 9, now if y changes (increases or decreases) by 5, x will change (decrease or increase) by 19.
Looking at options, if x = 256 we get y = 64.
Using these values we see option 1, 2 and 4 are eliminated and also that these exists a solution for 250<x<300.
Answer: Option D. -> 1
:
D
Soln:
Using the technique of finding squares of numbers upto 100 by keeping the base of 50 and 100( refer to the e-booklet for further details). We see that only squares of 12 will end with 144 (or the last 2 digits being the same). Hence we will have to find out if the first 2 digits will be the same for squares of 38, 62 (absolute difference of 12 from 50) and also for 88 (absolute difference of 12 from 100), we see that only 88 satisfies the condition, hence option 4.
:
D
Soln:
Using the technique of finding squares of numbers upto 100 by keeping the base of 50 and 100( refer to the e-booklet for further details). We see that only squares of 12 will end with 144 (or the last 2 digits being the same). Hence we will have to find out if the first 2 digits will be the same for squares of 38, 62 (absolute difference of 12 from 50) and also for 88 (absolute difference of 12 from 100), we see that only 88 satisfies the condition, hence option 4.
Answer: Option B. -> 1
:
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.
:
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.
Answer: Option B. -> 8 and 0
:
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.
:
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.
Answer: Option A. -> 3
:
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).
:
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).
Answer: Option A. -> 1584
:
A
Ans: (a)
7Dx−71Dx=770=(126−56144)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.
:
A
Ans: (a)
7Dx−71Dx=770=(126−56144)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.
Answer: Option D. -> None of the above
:
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
:
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
Answer: Option D. -> (G- L)
:
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
:
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
Question 40. 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)
Answer: Option D. -> 21
:
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.
:
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.
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