Exams > Cat > Quantitaitve Aptitude
ALGEBRA MCQs
Total Questions : 240
| Page 7 of 24 pages
Answer: Option A. -> 6666600
:
A
Keeping one digit in fixed position, another four can be arranged in 4 ! Ways = 24 ways. Thus each of 5 digit will occur in each of five places 4! Times. Hence, the sum of digits in each position is 24 (1+3+5+9) = 600. So, the sum of all numbers = 6000 (1+10+100+1000+10000) = 6666600.SHORTCUTLet n=number of digitsThen, The sum of all possible numbers is given by (n)!(sum of all the digits)(1111….n times)= 5!(18)(11111)= 6666600
:
A
Keeping one digit in fixed position, another four can be arranged in 4 ! Ways = 24 ways. Thus each of 5 digit will occur in each of five places 4! Times. Hence, the sum of digits in each position is 24 (1+3+5+9) = 600. So, the sum of all numbers = 6000 (1+10+100+1000+10000) = 6666600.SHORTCUTLet n=number of digitsThen, The sum of all possible numbers is given by (n)!(sum of all the digits)(1111….n times)= 5!(18)(11111)= 6666600
Question 62. Once I had been to the post office to buy five-rupee, two-rupee and one-rupee stamps. I paid the clerk Rs. 20, and since he had no change, he gave me three more one-rupee stamps. If the number of stamps of each type that I had ordered initially was more than one, what was the total number of stamps that I bought?
Answer: Option A. -> 10
:
A
Option (a)
The number of stamps that were initially bought were more than one of each type. Hence the total number of stamps
2(5 rupees) + 2(2 rupees) + 3(1 rupee) + 3(1 rupee) = 10 tickets
:
A
Option (a)
The number of stamps that were initially bought were more than one of each type. Hence the total number of stamps
2(5 rupees) + 2(2 rupees) + 3(1 rupee) + 3(1 rupee) = 10 tickets
Answer: Option D. -> None of these
:
D
Option (d) n=1 is a root of the equation(n - 1)(n2 – 6n + 5) = (n - 1)2(n - 5)Now, (n - 1)2 is always positive.Now, for n < 5, the expression gives a negative quantity.Therefore, the least value of n will be 6. Hence m = 7 .
:
D
Option (d) n=1 is a root of the equation(n - 1)(n2 – 6n + 5) = (n - 1)2(n - 5)Now, (n - 1)2 is always positive.Now, for n < 5, the expression gives a negative quantity.Therefore, the least value of n will be 6. Hence m = 7 .
Question 64. You can collect as many rubies and emeralds as you can. Each ruby is worth Rs. 4 crores and each emerald is worth of Rs. 5 crore. Each ruby weights 0.3 kg and each emerald weights 0.4 kg. Your bag can carry at the most 12 kg. What you should collect to get the maximum wealth ? ( CAT 1998)
Answer: Option B. -> 40 rubies
:
B
Option (b) Number of emeralds which can be carried in total = 30. Value = 30 x 5 = 150 crNumber of rubies which can be carried in total = 40. Value = 40 x 4 = 160 cr. Only rubies will give the maximum wealth
:
B
Option (b) Number of emeralds which can be carried in total = 30. Value = 30 x 5 = 150 crNumber of rubies which can be carried in total = 40. Value = 40 x 4 = 160 cr. Only rubies will give the maximum wealth
Answer: Option B. -> 160
:
B
Option (b) 3 girls can be selected out of 5 girls in 5C3 ways. Since the number of boys to be invited is not to be given, he can invite or not invite them in 24 ways.
Hence required numbers of ways =5C3×24=10×16=160.
:
B
Option (b) 3 girls can be selected out of 5 girls in 5C3 ways. Since the number of boys to be invited is not to be given, he can invite or not invite them in 24 ways.
Hence required numbers of ways =5C3×24=10×16=160.
Answer: Option D. -> 768
:
D
Option (d)
There are 32 white and 32 black squares on the chessboard
Number of ways of choosing the white square = 32
When a white square is selected, we cannot select the black square lying on the row or column of the white
Square. We have 8 such black squares for every white square selected.
Hence we have 32-8=24 black squares which can be selected for every white square selected
Total number of possibilities= 32 × 24= 768. Answer is option (d)
:
D
Option (d)
There are 32 white and 32 black squares on the chessboard
Number of ways of choosing the white square = 32
When a white square is selected, we cannot select the black square lying on the row or column of the white
Square. We have 8 such black squares for every white square selected.
Hence we have 32-8=24 black squares which can be selected for every white square selected
Total number of possibilities= 32 × 24= 768. Answer is option (d)
Answer: Option B. -> 60
:
B
Option (b)Case 1- units place= 1 (0 possibilities)Case 2- units place = 2 (tens place=1 possibility (2) and 3! Arrangements for the remaining places) = 1x3!Case 3- units place = 3 (tens place=2 possibilities (1,2) and 3! Arrangements for the remaining places) = 2x3!Case 4- units place = 4 (tens place=3 possibilities (1,2,3) and 3! Arrangements for the remaining places) = 3x3!Case 5- units place = 5 (tens place=4 possibilities (1,2,3,4) and 3! Arrangements for the remaining places) = 4x3!Total = 60
:
B
Option (b)Case 1- units place= 1 (0 possibilities)Case 2- units place = 2 (tens place=1 possibility (2) and 3! Arrangements for the remaining places) = 1x3!Case 3- units place = 3 (tens place=2 possibilities (1,2) and 3! Arrangements for the remaining places) = 2x3!Case 4- units place = 4 (tens place=3 possibilities (1,2,3) and 3! Arrangements for the remaining places) = 3x3!Case 5- units place = 5 (tens place=4 possibilities (1,2,3,4) and 3! Arrangements for the remaining places) = 4x3!Total = 60
Question 68. Three Englishmen and three Frenchmen work for the same company. Each of them knows a secret not known to others. They need to exchange these secrets over phone-to-phone calls so that eventually each person knows all six secrets. None of the Frenchmen knows English and only one Englishman knows French. What is the minimum number of calls needed for the above purpose?
___.
___.
:
(c) There have to be 2 calls from each person to the Englishman who knows French to get all the information. So, there should be 10 calls. But when the fifth guy call he would get all the information of the previous 4 guys alongwith Englishman’s information. Hence, 1 call can be saved. So, the total number of calls = 9.
Answer: Option D. -> 728
:
D
Option (d)
Since we are looking for numbers < a million, we can represent all the numbers in this range as “abcdef” . each
of the digits can be selected in 3 ways.
Total possible cases = 36 -1 = 728. We subtract one case where each of “abcdef” take 0.
:
D
Option (d)
Since we are looking for numbers < a million, we can represent all the numbers in this range as “abcdef” . each
of the digits can be selected in 3 ways.
Total possible cases = 36 -1 = 728. We subtract one case where each of “abcdef” take 0.
Question 70. A survey of 200 people in a community who watched at least one of the three channels – BBC, CNN and DD-showed that 80% of the people watched DD, 22% watched BBC, and 15% watched CNN.
If 5% of people watched DD and CNN. 10% watched DD and BBC, what is the maximum percentage of people who can watch all the three channels?
If 5% of people watched DD and CNN. 10% watched DD and BBC, what is the maximum percentage of people who can watch all the three channels?
Answer: Option B. -> 8.5%
:
B
Option (b)
Use the maxima of all shortcut
S=117% and X=100%
S−Xn−1=172=8.5
:
B
Option (b)
Use the maxima of all shortcut
S=117% and X=100%
S−Xn−1=172=8.5
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