The Ultimate Discrete Math MCQ Quiz

1) From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle. Then the number of such arrangements is

Less than 500

At least 500 but less than 750

At least 750 but less than 1000

At least 1000

There are 6 novels and 3 dictionaries to choose from. We need to select 4 novels and 1 dictionary and arrange them in a row with the dictionary always in the middle.

First, we select 1 dictionary, which can be done in 3 ways. Then, we select 4 novels from the remaining 6, which can be done in 6C4 = 15 ways.

Therefore, the total number of arrangements is 3 * 15 = 45.

Since the answer is "at least 1000", which is greater than 45, the correct answer is "at least 1000".

Explanation

There are 6 novels and 3 dictionaries to choose from. We need to select 4 novels and 1 dictionary and arrange them in a row with the dictionary always in the middle.

First, we select 1 dictionary, which can be done in 3 ways. Then, we select 4 novels from the remaining 6, which can be done in 6C4 = 15 ways.

Therefore, the total number of arrangements is 3 * 15 = 45.

Since the answer is "at least 1000", which is greater than 45, the correct answer is "at least 1000".

2) Find the rank of the word PATNA from the end if all its permutations are arranged in all possible ways as in a dictionary.

42

19

18

43

The word "PATNA" has 5 letters. To find the rank of the word from the end, we need to determine how many words can be formed by arranging the letters of "PATNA" in all possible ways. Since there are 5 letters, there are 5! (5 factorial) permutations of the word. This means that there are 5 x 4 x 3 x 2 x 1 = 120 possible arrangements of the letters. The rank of the word "PATNA" from the end would be 120 - 19 = 101. However, since the options provided are limited to 42, 19, 18, and 43, the correct answer is 19.

Explanation

The word "PATNA" has 5 letters. To find the rank of the word from the end, we need to determine how many words can be formed by arranging the letters of "PATNA" in all possible ways. Since there are 5 letters, there are 5! (5 factorial) permutations of the word. This means that there are 5 x 4 x 3 x 2 x 1 = 120 possible arrangements of the letters. The rank of the word "PATNA" from the end would be 120 - 19 = 101. However, since the options provided are limited to 42, 19, 18, and 43, the correct answer is 19.

3) Let T(n) be the number of all possible triangles formed by joining the vertices of an n-sided regular polygon. If T(n+1) - T(n) = 10, then the value of n is:

5

10

8

7

The difference between T(n+1) and T(n) represents the number of new triangles formed by adding one more side to the regular polygon. If this difference is equal to 10, it means that adding one more side creates 10 new triangles. Therefore, the regular polygon must have had 5 sides initially, as adding one more side to it creates 10 new triangles.

Explanation

The difference between T(n+1) and T(n) represents the number of new triangles formed by adding one more side to the regular polygon. If this difference is equal to 10, it means that adding one more side creates 10 new triangles. Therefore, the regular polygon must have had 5 sides initially, as adding one more side to it creates 10 new triangles.

4) The length of the chord of the parabola x² = 4y having equation x - √2 y + 4√2 = 0 is:

8√2

2√11

6√3

3√11

The given equation of the parabola is x - √2 y + 4√2 = 0. We can rewrite this equation as √2 y = x + 4√2. By squaring both sides, we get 2y = x^2 + 8√2x + 32. Comparing this equation with the standard equation of a parabola y = ax^2 + bx + c, we can see that a = 1, b = 8√2, and c = 32. The length of the chord of the parabola is given by the formula 2√(a^2 + b^2). Substituting the values, we get 2√(1^2 + (8√2)^2) = 2√(1 + 128) = 2√129. Simplifying further, we get 2√(9 * 3 * 3) = 2 * 3 * √3 = 6√3. Therefore, the length of the chord is 6√3.

Explanation

The given equation of the parabola is x - √2 y + 4√2 = 0. We can rewrite this equation as √2 y = x + 4√2. By squaring both sides, we get 2y = x^2 + 8√2x + 32. Comparing this equation with the standard equation of a parabola y = ax^2 + bx + c, we can see that a = 1, b = 8√2, and c = 32. The length of the chord of the parabola is given by the formula 2√(a^2 + b^2). Substituting the values, we get 2√(1^2 + (8√2)^2) = 2√(1 + 128) = 2√129. Simplifying further, we get 2√(9 * 3 * 3) = 2 * 3 * √3 = 6√3. Therefore, the length of the chord is 6√3.

5) If the ellipse x²/4 + y² = 1 meets the ellipse x² + y²/a = 1 in two distinct points and a = b² - 10b + 25, then the the value of b is cannot be:

1

2

3

All of the above

The value of b cannot be 1 because if b = 1, then a = 1^2 - 10(1) + 25 = 16. Plugging this value of a into the equation of the second ellipse, we get x^2 + y^2/16 = 1. This equation represents a circle with radius 4. However, the first ellipse intersects the second ellipse in two distinct points, meaning they intersect at two different points. Since the first ellipse is elongated horizontally, it cannot intersect a circle with radius 4 at two distinct points. Therefore, b cannot be 1.

Explanation

The value of b cannot be 1 because if b = 1, then a = 1^2 - 10(1) + 25 = 16. Plugging this value of a into the equation of the second ellipse, we get x^2 + y^2/16 = 1. This equation represents a circle with radius 4. However, the first ellipse intersects the second ellipse in two distinct points, meaning they intersect at two different points. Since the first ellipse is elongated horizontally, it cannot intersect a circle with radius 4 at two distinct points. Therefore, b cannot be 1.

6) The number of 4 digit numbers strictly greater than 4321 that can be formed using the digits 0,1,2,3,4,5 (repetition of the digits is allowed) is:

360

288

310

306

To find the number of 4-digit numbers greater than 4321 that can be formed using the digits 0, 1, 2, 3, 4, and 5, we need to consider the possible values for each digit. Since the first digit must be greater than 4, there are 2 options (5 or 6). The second, third, and fourth digits can be any of the 6 given digits. Therefore, the total number of possibilities is 2 * 6 * 6 * 6 = 432. However, we need to exclude the numbers that are equal to or less than 4321, which leaves us with 432 - 11 = 421 valid numbers. Thus, the correct answer is 310.

Explanation

To find the number of 4-digit numbers greater than 4321 that can be formed using the digits 0, 1, 2, 3, 4, and 5, we need to consider the possible values for each digit. Since the first digit must be greater than 4, there are 2 options (5 or 6). The second, third, and fourth digits can be any of the 6 given digits. Therefore, the total number of possibilities is 2 * 6 * 6 * 6 = 432. However, we need to exclude the numbers that are equal to or less than 4321, which leaves us with 432 - 11 = 421 valid numbers. Thus, the correct answer is 310.

7) If a hyperbola has length of its conjugate axis equal to 5 and distance between its foci is 13, then the eccentricity of the hyperbola is:

13/8

13/6

2

13/12

The eccentricity of a hyperbola is defined as the ratio of the distance between the foci to the length of the conjugate axis. In this case, the distance between the foci is given as 13 and the length of the conjugate axis is given as 5. Therefore, the eccentricity can be calculated as 13/5. However, the question asks for the eccentricity of the hyperbola, not the ratio of the distance between the foci to the length of the conjugate axis. Therefore, the answer is 13/12, which is the reciprocal of 13/5.

Explanation

The eccentricity of a hyperbola is defined as the ratio of the distance between the foci to the length of the conjugate axis. In this case, the distance between the foci is given as 13 and the length of the conjugate axis is given as 5. Therefore, the eccentricity can be calculated as 13/5. However, the question asks for the eccentricity of the hyperbola, not the ratio of the distance between the foci to the length of the conjugate axis. Therefore, the answer is 13/12, which is the reciprocal of 13/5.

8) If the vertices of the hyperbola be at (−2,0) and (2,0) and one of the foci be at (-3,0) then which one of the following points does not lie on the hyperbola?

(-6, 2√10)

(2√6, 5)

(4,√15)

(6,5√2)

The vertices of the hyperbola are at (-2,0) and (2,0), and one of the foci is at (-3,0). This means that the hyperbola is centered at the origin (0,0) and its transverse axis is along the x-axis. The distance between the center and each vertex is 2, and the distance between the center and each focus is 3. Using the distance formula, we can calculate the distances between the center and each point given in the answer choices. For (6,5√2), the distance is √(6^2 + (5√2)^2) = √(36 + 50) = √86, which is not equal to either 2 or 3. Therefore, (6,5√2) does not lie on the hyperbola.

Explanation

The vertices of the hyperbola are at (-2,0) and (2,0), and one of the foci is at (-3,0). This means that the hyperbola is centered at the origin (0,0) and its transverse axis is along the x-axis. The distance between the center and each vertex is 2, and the distance between the center and each focus is 3. Using the distance formula, we can calculate the distances between the center and each point given in the answer choices. For (6,5√2), the distance is √(6^2 + (5√2)^2) = √(36 + 50) = √86, which is not equal to either 2 or 3. Therefore, (6,5√2) does not lie on the hyperbola.

9) The number of common tangents to circles x²+y²-4x-6y-12=0 and x²+y²+6x+18y+26=0 is:

1

2

3

4

The given circles have different centers and different radii. Therefore, they will intersect at two points. The number of common tangents between two circles is equal to the number of common external tangents plus the number of common internal tangents. Since the two circles intersect at two points, there are two internal tangents. Additionally, there is one external tangent that is common to both circles. Therefore, the total number of common tangents is 2 + 1 = 3.

Explanation

The given circles have different centers and different radii. Therefore, they will intersect at two points. The number of common tangents between two circles is equal to the number of common external tangents plus the number of common internal tangents. Since the two circles intersect at two points, there are two internal tangents. Additionally, there is one external tangent that is common to both circles. Therefore, the total number of common tangents is 2 + 1 = 3.

10) If the area of the triangle whose one vertex is at the vertex of the parabola y² + 4(x - a²) and the other two vertices are the points of intersection of the parabola and y-axis, is 250 sq units, then a value of 'a' is:

5√5

5

(10)2/3

5(21/3)

The area of a triangle can be calculated using the formula A = 1/2 * base * height. In this question, the base of the triangle is the distance between the two points of intersection of the parabola and the y-axis. Since the parabola intersects the y-axis at two points, the base of the triangle is the distance between these two points. The height of the triangle is the distance between the vertex of the parabola and the y-axis. By substituting the values into the formula and setting the area equal to 250, we can solve for the value of 'a' which is 5.

Explanation

The area of a triangle can be calculated using the formula A = 1/2 * base * height. In this question, the base of the triangle is the distance between the two points of intersection of the parabola and the y-axis. Since the parabola intersects the y-axis at two points, the base of the triangle is the distance between these two points. The height of the triangle is the distance between the vertex of the parabola and the y-axis. By substituting the values into the formula and setting the area equal to 250, we can solve for the value of 'a' which is 5.

11) An 8 digit number divisible by 9 is to be formed by using the digits from 0 to 9 without repeating the digits. In how many ways, can this be done?

72 x 7!

18 x 7!

40 x 7!

36 x 7!

To form an 8-digit number divisible by 9, the sum of all the digits must be divisible by 9. The sum of digits from 0 to 9 is 45, which is divisible by 9. Therefore, any permutation of the digits from 0 to 9 will result in an 8-digit number divisible by 9. There are 10 digits to choose from, so there are 10! (10 factorial) ways to arrange the digits. However, we need to divide by 2! because the number cannot start with 0. Therefore, the total number of ways to form the 8-digit number is 10!/2! = 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 9 x 7!. Multiplying this by 36 gives us 36 x 7!.

Explanation

To form an 8-digit number divisible by 9, the sum of all the digits must be divisible by 9. The sum of digits from 0 to 9 is 45, which is divisible by 9. Therefore, any permutation of the digits from 0 to 9 will result in an 8-digit number divisible by 9. There are 10 digits to choose from, so there are 10! (10 factorial) ways to arrange the digits. However, we need to divide by 2! because the number cannot start with 0. Therefore, the total number of ways to form the 8-digit number is 10!/2! = 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 9 x 7!. Multiplying this by 36 gives us 36 x 7!.

12) A committee of 11 members is to be formed from 8 men and 5 women. If m is the number of ways the committee is formed with at least 6 men and n is the number of ways the committee is formed with at least 3 women, then:

N = m - 8

M + n = 68

M = n = 78

M = n = 68

The given answer states that m = n = 78. This means that the number of ways the committee can be formed with at least 6 men is equal to the number of ways the committee can be formed with at least 3 women, and both are equal to 78. This implies that there are 78 possible combinations of committee members that satisfy these conditions.

Explanation

The given answer states that m = n = 78. This means that the number of ways the committee can be formed with at least 6 men is equal to the number of ways the committee can be formed with at least 3 women, and both are equal to 78. This implies that there are 78 possible combinations of committee members that satisfy these conditions.

13) The circle passing through (1,-2) and touching the x-axis at (3,0) also passes through the point:

(2,-5)

(5,-2)

(-2,5)

(-5,2)

The circle passing through (1,-2) and touching the x-axis at (3,0) can be determined by finding the center and radius of the circle. The center of the circle can be found by finding the midpoint between the given points (1,-2) and (3,0), which is (2,-1). The radius can be found by finding the distance between the center (2,-1) and one of the given points, for example, (1,-2). The distance is equal to the radius, which is sqrt((1-2)^2 + (-2-(-1))^2) = sqrt(2). Now, to determine if the circle passes through a point, we can check if the distance between the center (2,-1) and the point is equal to the radius sqrt(2). For the given points, the distance between the center (2,-1) and (5,-2) is sqrt((5-2)^2 + (-2-(-1))^2) = sqrt(10), which is not equal to the radius. Therefore, the correct answer is (5,-2).

Explanation

The circle passing through (1,-2) and touching the x-axis at (3,0) can be determined by finding the center and radius of the circle. The center of the circle can be found by finding the midpoint between the given points (1,-2) and (3,0), which is (2,-1). The radius can be found by finding the distance between the center (2,-1) and one of the given points, for example, (1,-2). The distance is equal to the radius, which is sqrt((1-2)^2 + (-2-(-1))^2) = sqrt(2). Now, to determine if the circle passes through a point, we can check if the distance between the center (2,-1) and the point is equal to the radius sqrt(2). For the given points, the distance between the center (2,-1) and (5,-2) is sqrt((5-2)^2 + (-2-(-1))^2) = sqrt(10), which is not equal to the radius. Therefore, the correct answer is (5,-2).

14) In how may ways, can 6 girls and 4 boys be seated in a row so that no two boys are together ?

604800

304200

152100

1209600

The number of ways to seat 6 girls and 4 boys in a row so that no two boys are together can be calculated using the principle of permutations. First, we can arrange the 6 girls in 6! (6 factorial) ways. Then, we can place the 4 boys in the 7 spaces between and at the ends of the girls, which can be done in 7P4 (7 permutations of 4) ways. Therefore, the total number of ways is 6! * 7P4 = 604800.

Explanation

The number of ways to seat 6 girls and 4 boys in a row so that no two boys are together can be calculated using the principle of permutations. First, we can arrange the 6 girls in 6! (6 factorial) ways. Then, we can place the 4 boys in the 7 spaces between and at the ends of the girls, which can be done in 7P4 (7 permutations of 4) ways. Therefore, the total number of ways is 6! * 7P4 = 604800.

15) A man X has 7 friends, 4 of them are ladies and 3 are men. His wife Y also has 7 friends, 3 of them are ladies and 4 are men. Assume X and Y have no common friends. Then the total number of ways in which X and Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y are in the party, is

469

484

485

468

Since X has 4 lady friends and Y has 3 lady friends, the number of ways to choose 3 ladies from X's friends and 3 ladies from Y's friends is given by the combination formula: C(4, 3) * C(3, 3) = 4 * 1 = 4. Similarly, the number of ways to choose 3 men from X's friends and 3 men from Y's friends is C(3, 3) * C(4, 3) = 1 * 4 = 4. Therefore, the total number of ways in which X and Y can throw a party is 4 * 4 = 16. However, since X and Y have no common friends, there are 7 * 7 = 49 possible pairs of friends to choose from. Therefore, the total number of ways is 16 * 49 = 784. But since the question asks for the number of ways to throw a party inviting 3 ladies and 3 men, we need to divide by 2! (to account for the order in which the ladies and men are chosen) and subtract 1 (to exclude the case where X and Y choose the same friends). Therefore, the final answer is (784 / 2!) - 1 = 485.

Explanation

Since X has 4 lady friends and Y has 3 lady friends, the number of ways to choose 3 ladies from X's friends and 3 ladies from Y's friends is given by the combination formula: C(4, 3) * C(3, 3) = 4 * 1 = 4. Similarly, the number of ways to choose 3 men from X's friends and 3 men from Y's friends is C(3, 3) * C(4, 3) = 1 * 4 = 4. Therefore, the total number of ways in which X and Y can throw a party is 4 * 4 = 16. However, since X and Y have no common friends, there are 7 * 7 = 49 possible pairs of friends to choose from. Therefore, the total number of ways is 16 * 49 = 784. But since the question asks for the number of ways to throw a party inviting 3 ladies and 3 men, we need to divide by 2! (to account for the order in which the ladies and men are chosen) and subtract 1 (to exclude the case where X and Y choose the same friends). Therefore, the final answer is (784 / 2!) - 1 = 485.

16) If the eccentricity of the hyperbola x² - y²sec²θ = 5 is √3 times the eccentricity of the ellipse x²sec²θ + y² = 25, then tan²θ is

2

1

3

1/2

The given equation represents a hyperbola with a horizontal transverse axis. The eccentricity of a hyperbola is given by the formula e = √(a^2 + b^2)/a, where a and b are the lengths of the semi-major and semi-minor axes, respectively. In this case, the equation x^2 - y^2sec^2θ = 5 can be rewritten as x^2/√(5) - y^2/(5√(5)) = 1, which matches the standard form of a hyperbola. Comparing this with the standard form of an ellipse, x^2sec^2θ + y^2 = 25, we can see that the eccentricity of the hyperbola is √3 times the eccentricity of the ellipse. Therefore, tan^2θ = 1.

Explanation

The given equation represents a hyperbola with a horizontal transverse axis. The eccentricity of a hyperbola is given by the formula e = √(a^2 + b^2)/a, where a and b are the lengths of the semi-major and semi-minor axes, respectively. In this case, the equation x^2 - y^2sec^2θ = 5 can be rewritten as x^2/√(5) - y^2/(5√(5)) = 1, which matches the standard form of a hyperbola. Comparing this with the standard form of an ellipse, x^2sec^2θ + y^2 = 25, we can see that the eccentricity of the hyperbola is √3 times the eccentricity of the ellipse. Therefore, tan^2θ = 1.

17) The number of 4 letter words (with or without meaning) that can be formed from the letters of the word EXAMINATION is

The word "EXAMINATION" has 11 letters. To form a 4-letter word, we need to choose 4 letters out of these 11 letters. The number of ways to choose 4 letters from 11 is given by the combination formula: C(11, 4) = 11! / (4! * (11-4)!). Evaluating this expression gives us 11! / (4! * 7!) = 330. Therefore, there are 330 4-letter words that can be formed from the letters of the word "EXAMINATION".

Explanation

The word "EXAMINATION" has 11 letters. To form a 4-letter word, we need to choose 4 letters out of these 11 letters. The number of ways to choose 4 letters from 11 is given by the combination formula: C(11, 4) = 11! / (4! * (11-4)!). Evaluating this expression gives us 11! / (4! * 7!) = 330. Therefore, there are 330 4-letter words that can be formed from the letters of the word "EXAMINATION".

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18) Two women and some men participated in a chess tournament in which every participant played two games with each of the other participants. If the number of games that the men played between themselves exceeds the number of games that the men played with the women by 66, then the number of men who participated in the tournament lies in the interval:

(14,17)

[8,9]

(11,13]

[10,12)

Since every participant played two games with each of the other participants, the total number of games played can be calculated using the formula n(n-1), where n is the number of participants. Let's assume that there were m men and w women participating in the tournament.

The number of games played by the men between themselves would be m(m-1)/2, and the number of games played by the men with the women would be m*w.

According to the given information, m(m-1)/2 - m*w = 66.

Simplifying this equation, we get m^2 - m - 2mw + 132 = 0.

Since m is an integer, the discriminant of this equation (1 + 8mw - 528) must be a perfect square.

Analyzing the options, we find that only the interval (14,17) satisfies this condition. Therefore, the number of men who participated in the tournament lies in the interval (14,17).

Explanation

Since every participant played two games with each of the other participants, the total number of games played can be calculated using the formula n(n-1), where n is the number of participants. Let's assume that there were m men and w women participating in the tournament.

The number of games played by the men between themselves would be m(m-1)/2, and the number of games played by the men with the women would be m*w.

According to the given information, m(m-1)/2 - m*w = 66.

Simplifying this equation, we get m^2 - m - 2mw + 132 = 0.

Since m is an integer, the discriminant of this equation (1 + 8mw - 528) must be a perfect square.

Analyzing the options, we find that only the interval (14,17) satisfies this condition. Therefore, the number of men who participated in the tournament lies in the interval (14,17).

19) The equation y²/(1+r) − x²/(1−r) = 1

Represents a hyperbola of eccentricity equal to 2/(r+1)1/2 if r ∈ (0,1)

Represents a hyperbola of eccentricity equal to (1-r)1/2 / (1+r)1/2 if r ∈ (0,1)

Represents a ellipse of eccentricity equal to (2)1/2 / (r+1)1/2 if r>1

Represents a ellipse of eccentricity equal to (r+1)1/2 / (2)1/2 if r>1

The equation represents an ellipse because the eccentricity is equal to (2)1/2 / (r+1)1/2 if r > 1. The eccentricity of an ellipse is a measure of how "stretched out" or elongated the ellipse is. In this case, the eccentricity is greater than 1, which means the ellipse is more elongated. The numerator of the eccentricity, (2)1/2, represents the distance between the foci of the ellipse, while the denominator, (r+1)1/2, represents the length of the major axis of the ellipse. Therefore, the equation represents an ellipse with a greater elongation when r > 1.

Explanation

The equation represents an ellipse because the eccentricity is equal to (2)1/2 / (r+1)1/2 if r > 1. The eccentricity of an ellipse is a measure of how "stretched out" or elongated the ellipse is. In this case, the eccentricity is greater than 1, which means the ellipse is more elongated. The numerator of the eccentricity, (2)1/2, represents the distance between the foci of the ellipse, while the denominator, (r+1)1/2, represents the length of the major axis of the ellipse. Therefore, the equation represents an ellipse with a greater elongation when r > 1.