Q. For which value of k does the equation x^2 - 4x + k = 0 have roots that differ by 2?
Solution
Let the roots be r and r+2. Then, r + (r+2) = 4 and r(r+2) = k leads to k = 4.
Correct Answer: C — 4
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Q. For which value of k does the equation x^2 - kx + 9 = 0 have roots that are both positive?
-
A.
k < 6
-
B.
k > 6
-
C.
k = 6
-
D.
k = 0
Solution
For both roots to be positive, k must be greater than 6, as the sum of the roots must be positive.
Correct Answer: B — k > 6
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Q. For which value of k does the quadratic equation x^2 - kx + 4 = 0 have no real roots?
-
A.
k < 4
-
B.
k = 4
-
C.
k > 4
-
D.
k ≤ 4
Solution
The discriminant must be negative: k^2 - 16 < 0, hence k > 4.
Correct Answer: C — k > 4
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Q. From a group of 10 people, how many ways can a committee of 4 be formed?
-
A.
210
-
B.
120
-
C.
150
-
D.
180
Solution
The number of ways to choose 4 people from 10 is given by 10C4 = 210.
Correct Answer: A — 210
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Q. From a set of 8 different books, how many ways can you choose 3 books?
Solution
The number of ways to choose 3 books from 8 is given by 8C3 = 56.
Correct Answer: B — 84
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Q. How many different ways can 3 red, 2 blue, and 1 green balls be arranged in a line?
Solution
The total arrangements = 6! / (3! * 2! * 1!) = 60.
Correct Answer: B — 120
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Q. How many different ways can 4 people be seated at a round table?
Solution
The number of arrangements of n people at a round table is (n-1)!. For 4 people, it is 3! = 6.
Correct Answer: B — 12
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Q. How many different ways can 4 students be selected from a group of 10?
-
A.
210
-
B.
120
-
C.
100
-
D.
90
Solution
The number of ways to choose 4 from 10 is given by 10C4 = 210.
Correct Answer: A — 210
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Q. How many different ways can 6 people be seated at a round table?
-
A.
720
-
B.
120
-
C.
600
-
D.
480
Solution
The number of arrangements of 6 people at a round table is (6-1)! = 5! = 120.
Correct Answer: A — 720
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Q. How many subsets can be formed from the set S = {a, b, c, d}?
Solution
The number of subsets of a set with n elements is 2^n. Here, n = 4, so the number of subsets is 2^4 = 16.
Correct Answer: B — 8
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Q. How many subsets does the set A = {a, b, c} have?
Solution
The number of subsets of a set with n elements is 2^n. Here, n = 3, so the number of subsets is 2^3 = 8.
Correct Answer: D — 8
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Q. How many subsets does the set {a, b, c} have?
Solution
The number of subsets of a set with n elements is 2^n. Here, n = 3, so the number of subsets is 2^3 = 8.
Correct Answer: D — 8
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Q. How many ways can 2 boys and 2 girls be selected from 5 boys and 4 girls?
Solution
The number of ways = 5C2 * 4C2 = 10 * 6 = 60.
Correct Answer: A — 60
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Q. How many ways can 2 boys and 2 girls be selected from 6 boys and 4 girls?
Solution
The number of ways is C(6,2) * C(4,2) = 15 * 6 = 90.
Correct Answer: A — 60
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Q. How many ways can 3 different books be chosen from a set of 7 books?
Solution
The number of ways to choose 3 books from 7 is 7C3 = 7! / (3! * 4!) = 35.
Correct Answer: A — 35
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Q. How many ways can 3 different fruits be chosen from 8 fruits?
Solution
The number of ways is C(8,3) = 56.
Correct Answer: B — 84
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Q. How many ways can 3 different fruits be selected from 5 available fruits?
Solution
The number of ways to choose 3 from 5 is given by 5C3 = 10.
Correct Answer: B — 15
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Q. How many ways can 3 different letters be chosen from the word 'COMBINATION'?
-
A.
120
-
B.
220
-
C.
60
-
D.
80
Solution
The number of ways is C(11, 3) = 165.
Correct Answer: C — 60
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Q. How many ways can 3 letters be chosen from the word 'COMBINATION'?
Solution
The number of ways to choose 3 letters from 11 distinct letters is 11C3 = 165.
Correct Answer: B — 60
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Q. How many ways can 3 men and 2 women be arranged in a line if the men must be together?
Solution
Treat the 3 men as one unit. So, we have 3 units (MMM, W, W). Arrangements = 4! * 3! = 24 * 6 = 144.
Correct Answer: B — 120
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Q. How many ways can 3 red balls and 2 blue balls be arranged in a row?
Solution
The arrangements = 5! / (3! * 2!) = 10.
Correct Answer: A — 10
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Q. How many ways can 3 red, 2 blue, and 1 green balls be arranged in a line?
Solution
The arrangements = 6! / (3! * 2! * 1!) = 60.
Correct Answer: A — 60
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Q. How many ways can 3 red, 2 blue, and 1 green balls be arranged in a row?
Solution
The total arrangements = 6! / (3! * 2! * 1!) = 60.
Correct Answer: A — 60
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Q. How many ways can 4 different books be chosen from a shelf of 10 books?
-
A.
210
-
B.
120
-
C.
240
-
D.
300
Solution
The number of ways to choose 4 books from 10 is C(10, 4) = 210.
Correct Answer: A — 210
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Q. How many ways can 4 different colored balls be arranged in a line?
Solution
The number of arrangements is 4! = 24.
Correct Answer: B — 24
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Q. How many ways can 4 different colored balls be placed in 3 different boxes?
Solution
Each ball can go into any of the 3 boxes, so the total ways = 3^4 = 81.
Correct Answer: A — 81
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Q. How many ways can 4 different fruits be selected from a basket of 10 fruits?
-
A.
210
-
B.
120
-
C.
300
-
D.
150
Solution
The number of ways to choose 4 fruits from 10 is given by 10C4 = 210.
Correct Answer: A — 210
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Q. How many ways can 4 different letters be chosen from the word 'COMBINATION'?
-
A.
210
-
B.
126
-
C.
70
-
D.
84
Solution
The number of ways to choose 4 letters from 11 distinct letters is 11C4 = 330.
Correct Answer: A — 210
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Q. How many ways can 4 different letters be selected from the word 'COMBINATION'?
-
A.
210
-
B.
120
-
C.
60
-
D.
30
Solution
The number of ways to choose 4 letters from 11 distinct letters is 11C4 = 330.
Correct Answer: A — 210
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Q. How many ways can 4 different prizes be awarded to 3 students?
Solution
The number of ways to award 4 different prizes to 3 students is 3^4 = 81.
Correct Answer: C — 36
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