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1
A group consists of five women and six men. A committee of 4 members is to be selected from this group. In how many distinct ways can a committee be formed that includes exactly three women and one man?
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Solution: Step 1: Identify the total number of women (5) and men (6). Step 2: The committee needs to be formed with exactly 3 women and 1 man. Step 3: Calculate the number of ways to choose 3 women from 5 women: C(5, 3). C(5, 3) = (5 Ɨ 4 Ɨ 3) / (3 Ɨ 2 Ɨ 1) = 10 ways. Step 4: Calculate the number of ways to choose 1 man from 6 men: C(6, 1). C(6, 1) = 6 ways. Step 5: Since the selection of women and men are independent events, multiply the number of ways for each selection to find the total number of ways to form the committee. Step 6: Total ways = C(5, 3) Ɨ C(6, 1) = 10 Ɨ 6 = 60 ways. Step 7: Therefore, the committee can be formed in 60 different ways.
2
In how many ways can the letters of the word 'CREATE' be arranged so that all vowels come together?
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Solution: Step 1: Identify vowels in 'CREATE': E, A (2 vowels) Step 2: Treat the group of vowels (EA) as a single unit, so we have 5 units: (EA), C, R, T Step 3: Calculate permutations of 5 units: 5! = 120 Step 4: Within the vowel group (EA), calculate permutations: 2! = 2 Step 5: Total arrangements = 5! Ɨ 2! = 120 Ɨ 2 = 240
3
Given that the average of 'n' numbers is 'a'. If the first number is increased by 2, the second by 4, the third by 8, and so on (each increase being twice the previous one), what is the average of these modified 'n' numbers?
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Solution: Step 1: The initial sum of 'n' numbers is na (since average = a). Step 2: The increases applied to the numbers are 2, 4, 8, ..., which form a geometric progression (GP) with first term A=2 and common ratio R=2. Step 3: Calculate the sum of this GP for 'n' terms using the formula S_n = A * (R^n - 1) / (R - 1). Step 4: Substitute the values: S_n = 2 * (2^n - 1) / (2 - 1) = 2(2^n - 1). Step 5: The new total sum of the numbers is the original sum plus the sum of increases: na + 2(2^n - 1). Step 6: The average of the new numbers is the new total sum divided by 'n': [na + 2(2^n - 1)] / n. Step 7: Simplify the expression: a + [2(2^n - 1)] / n.
4
A rubber ball, dropped from a height of 120 meters, rebounds to 4/5th of its previous height after each bounce. Determine the total distance it travels before coming to rest.
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Solution: Step 1: The initial drop distance is 120 meters. Step 2: After the initial drop, the ball rebounds and falls in a continuous sequence. The first rebound height is (4/5) * 120 = 96 meters. Step 3: The subsequent distances form two infinite geometric progressions: one for the upward travel and one for the downward travel (after the initial drop). First term for these infinite series (a) = 96 meters. Common ratio (r) = 4/5. Step 4: The sum of an infinite geometric series is S = a / (1 - r), where |r| < 1. Step 5: Calculate the sum of all upward and downward distances *after* the initial drop (since it goes up 96m and then falls 96m, etc.): Sum of subsequent distances = 2 * (a / (1 - r)) = 2 * (96 / (1 - 4/5)) = 2 * (96 / (1/5)) = 2 * (96 * 5) = 2 * 480 = 960 meters. Step 6: Total distance traveled = Initial drop distance + Sum of subsequent distances. Total distance = 120 + 960 = 1080 meters. Step 7: The total distance the ball travels is 1080 meters.
5
For three numbers in an arithmetic progression (AP), their sum is 30 and their product is 910. Identify the largest number among them.
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Solution: Step 1: Represent the three numbers in AP. Let the numbers be a - d, a, and a + d, where 'a' is the middle term and 'd' is the common difference. Step 2: Use the sum condition. (a - d) + a + (a + d) = 30 3a = 30 a = 10. Step 3: Use the product condition. (a - d) Ɨ a Ɨ (a + d) = 910 Substitute a = 10: (10 - d) Ɨ 10 Ɨ (10 + d) = 910 Step 4: Simplify the product equation. 10 Ɨ (100 - d^2) = 910 100 - d^2 = 91 d^2 = 100 - 91 d^2 = 9 d = Ā±3. Step 5: Find the numbers and identify the greatest. If d = 3, the numbers are (10 - 3), 10, (10 + 3) ā†’ 7, 10, 13. If d = -3, the numbers are (10 - (-3)), 10, (10 + (-3)) ā†’ 13, 10, 7. In both cases, the greatest number is 13.
6
A businessman's income on his first day is Re. 1. Each subsequent day, his earnings are double the amount from the previous day. What is his income on the 10th day of business?
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Solution: Step 1: Identify the income pattern for each day: Day 1 income = Re. 1 Day 2 income = Re. 1 * 2 = Rs. 2 Day 3 income = Rs. 2 * 2 = Rs. 4 Step 2: Observe that the income on the 'n'th day follows a geometric progression with a first term (a) of 1 and a common ratio (r) of 2. The formula for the nth term of a geometric progression is a * r^(n-1). Step 3: Apply this formula to find the income on the 10th day (n=10): Income on 10th day = 1 * 2^(10-1) = 2^9. Step 4: Compute the value of 2^9: 2^1 = 2 2^2 = 4 2^3 = 8 2^4 = 16 2^5 = 32 2^6 = 64 2^7 = 128 2^8 = 256 2^9 = 512. Step 5: The businessman's income on the 10th day is Rs. 512.
7
Calculate the sum of squares from 11 squared to 20 squared: (11^2 + 12^2 + 13^2 + ... + 20^2).
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Solution: Step 1: The sum of squares of the first 'n' natural numbers is given by the formula: S_n = n(n+1)(2n+1) / 6. Step 2: The required sum (11^2 + 12^2 + ... + 20^2) can be calculated by finding the sum of squares up to 20 and subtracting the sum of squares up to 10. Required Sum = (1^2 + 2^2 + ... + 20^2) - (1^2 + 2^2 + ... + 10^2). Step 3: Calculate the sum for n = 20: S_20 = 20 * (20+1) * (2*20+1) / 6 = 20 * 21 * 41 / 6. S_20 = 17220 / 6 = 2870. Step 4: Calculate the sum for n = 10: S_10 = 10 * (10+1) * (2*10+1) / 6 = 10 * 11 * 21 / 6. S_10 = 2310 / 6 = 385. Step 5: Subtract S_10 from S_20 to get the desired sum: Required Sum = 2870 - 385 = 2485.
8
Calculate the sum of the series: (1 - 1/n) + (1 - 2/n) + (1 - 3/n) + ... up to n terms.
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Solution: Step 1: Identify the pattern of the series. Each term is of the form (1 - k/n), where k ranges from 1 to n. Step 2: Rewrite the sum by separating the integer '1' part and the fractional part: Sum = (1 + 1 + ... + 1 (n times)) - (1/n + 2/n + 3/n + ... + n/n). Step 3: The sum of 'n' ones is simply n. Step 4: Factor out 1/n from the fractional sum: Fractional sum = (1/n) * (1 + 2 + 3 + ... + n). Step 5: Apply the formula for the sum of the first 'n' natural numbers, which is Sum_k = k(k+1)/2. For 'n' terms, the sum is n(n+1)/2. Fractional sum = (1/n) * [n(n+1)/2]. Step 6: Simplify the fractional sum: n cancels out, leaving (n+1)/2. Step 7: Substitute these results back into the overall sum equation: Sum = n - (n+1)/2. Step 8: Combine the terms by finding a common denominator: Sum = (2n)/2 - (n+1)/2 = (2n - (n+1))/2. Step 9: Simplify the numerator: (2n - n - 1)/2 = (n - 1)/2. Step 10: Therefore, the sum of the series is (n - 1)/2.
9
Given A = 2^65 and B = (2^64 + 2^63 + 2^62 + ... + 2^0), determine the correct relationship between A and B.
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Solution: Step 1: Identify the series for B. B = 2^64 + 2^63 + 2^62 + ... + 2^0 This is a Geometric Progression (G.P.) in reverse order, or summing from 2^0 to 2^64. Step 2: Identify the parameters of the G.P. First term (a) = 2^0 = 1 Common ratio (r) = 2 (since each term is multiplied by 2 to get the next term if read from 2^0 upwards) Number of terms (n) = 64 - 0 + 1 = 65 Step 3: Use the formula for the sum of a finite G.P. S_n = a * (r^n - 1) / (r - 1) Step 4: Calculate the sum B. B = 1 * (2^65 - 1) / (2 - 1) B = 2^65 - 1 Step 5: Compare B with A. Given A = 2^65. So, B = A - 1. This means A is larger than B by 1.
10
If the product (1/4) Ɨ (2/6) Ɨ (3/8) Ɨ (4/10) Ɨ ... Ɨ (31/64) equals 1/2^x, determine the value of x.
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Solution: Step 1: Write the given product as P = (1/4) Ɨ (2/6) Ɨ (3/8) Ɨ (4/10) Ɨ ... Ɨ (31/64). Step 2: Rewrite each term in the form n / (2(n+1)) for n from 1 to 31. P = (1/(2 Ɨ 2)) Ɨ (2/(2 Ɨ 3)) Ɨ (3/(2 Ɨ 4)) Ɨ ... Ɨ (31/(2 Ɨ 32)). Step 3: Group the numerators and denominators: P = (1 Ɨ 2 Ɨ 3 Ɨ ... Ɨ 31) / [(2 Ɨ 2) Ɨ (2 Ɨ 3) Ɨ (2 Ɨ 4) Ɨ ... Ɨ (2 Ɨ 32)]. Step 4: Separate the powers of 2 in the denominator: The denominator has 31 factors of '2' (one from each `2n` term) and then the product (2 Ɨ 3 Ɨ 4 Ɨ ... Ɨ 32). P = (1 Ɨ 2 Ɨ ... Ɨ 31) / [2^31 Ɨ (2 Ɨ 3 Ɨ ... Ɨ 32)]. Step 5: Cancel common terms (2 Ɨ 3 Ɨ ... Ɨ 31) in the numerator and denominator: P = 1 / (2^31 Ɨ 32). Step 6: Express 32 as a power of 2: 32 = 2^5. P = 1 / (2^31 Ɨ 2^5). Step 7: Apply the exponent rule a^m Ɨ a^n = a^(m+n): P = 1 / 2^(31+5) = 1 / 2^36. Step 8: Compare with the given 1/2^x: 1/2^36 = 1/2^x Step 9: Therefore, x = 36.
11
Calculate the sum of all positive integers that are multiples of 3 and are less than 50.
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Solution: Step 1: Identify the series of multiples of 3 less than 50. This is an arithmetic progression (AP): 3, 6, 9, ..., 48. Step 2: Determine the first term (a), common difference (d), and the last term (l). a = 3 d = 3 l = 48 Step 3: Calculate the number of terms (n) in the AP using the formula: l = a + (n-1)d. 48 = 3 + (n-1)3 45 = (n-1)3 15 = n-1 n = 16 Step 4: Calculate the sum of the AP using the formula: Sum = n/2 * (a + l). Sum = 16/2 * (3 + 48) Sum = 8 * 51 Sum = 408
12
If the 7th term of an A.P. is 34 and the 13th term is 64, what is its 18th term?
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Solution: Step 1: Write the formulas for the 7th and 13th terms of an A.P. Let 'a' be the first term and 'd' be the common difference. The nth term formula is a_n = a + (n-1)d. 7th term: a_7 = a + (7-1)d = a + 6d = 34 (Equation 1) 13th term: a_13 = a + (13-1)d = a + 12d = 64 (Equation 2) Step 2: Solve the system of linear equations to find 'a' and 'd'. Subtract Equation 1 from Equation 2: (a + 12d) - (a + 6d) = 64 - 34 6d = 30 d = 30 / 6 d = 5 Step 3: Substitute the value of 'd' into Equation 1 to find 'a'. a + 6(5) = 34 a + 30 = 34 a = 34 - 30 a = 4 Step 4: Calculate the 18th term. Using the nth term formula: a_18 = a + (18-1)d a_18 = 4 + 17(5) a_18 = 4 + 85 a_18 = 89
13
A railway compartment features two rows of seats facing each other, with each row accommodating 5 passengers, totaling 10 seats. Among 10 passengers, 4 prefer to sit facing forward, 3 prefer to sit facing the rear, and the remaining 3 are indifferent to their seating direction. In how many ways can these 10 passengers be seated?
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Solution: Step 1: The compartment has 10 seats: 5 seats facing forward and 5 seats facing the rear. Step 2: Categorize the 10 passengers: Group A: 4 passengers who prefer to sit facing forward. Group B: 3 passengers who prefer to sit facing the rear. Group C: 3 indifferent passengers (10 - 4 - 3 = 3). Step 3: Arrange Group A passengers: The 4 passengers from Group A must occupy 4 of the 5 forward-facing seats. The number of ways to do this is P(5, 4) = 5! / (5-4)! = 5! = 120 ways. Step 4: Arrange Group B passengers: The 3 passengers from Group B must occupy 3 of the 5 rear-facing seats. The number of ways to do this is P(5, 3) = 5! / (5-3)! = 5! / 2! = (5 Ɨ 4 Ɨ 3) = 60 ways. Step 5: Arrange Group C passengers: After placing Group A and Group B, there are 1 (5-4) forward-facing seat left and 2 (5-3) rear-facing seats left, totaling 3 remaining seats. The 3 indifferent passengers can sit in these 3 remaining seats in P(3, 3) = 3! = 6 ways. Step 6: Multiply the number of ways for each step to find the total number of distinct seating arrangements. Step 7: Total ways = P(5, 4) Ɨ P(5, 3) Ɨ P(3, 3) = 120 Ɨ 60 Ɨ 6. Step 8: Total ways = 43200.
14
A square with a side length of 4 cm is given. A new square is drawn by connecting the midpoints of the sides of the previous square, and this process continues indefinitely. What is the sum of the areas of all such squares?
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Solution: Step 1: Calculate the area of the first square. Side of first square (S1) = 4 cm. Area of first square (A1) = S1^2 = 4^2 = 16 cm^2. Step 2: Determine the side length of the second square. When midpoints of a square are joined, the new square's vertices are at the midpoints. The side of this new square is the hypotenuse of a right-angled triangle formed by half the side of the previous square. Let 's' be the side of the previous square. The legs of the right triangle are s/2 and s/2. The new side (S_new) = sqrt((s/2)^2 + (s/2)^2) = sqrt(s^2/4 + s^2/4) = sqrt(s^2/2) = s/sqrt(2). For the second square, s = 4 cm. So, S2 = 4/sqrt(2) = 2*sqrt(2) cm. Step 3: Calculate the area of the second square. Area of second square (A2) = S2^2 = (2*sqrt(2))^2 = 8 cm^2. Step 4: Find the common ratio (r) of the areas. The areas form a geometric progression. r = A2 / A1 = 8 / 16 = 1/2. Step 5: Use the formula for the sum of an infinite geometric series: S = a / (1 - r), where 'a' is the first term and 'r' is the common ratio (for |r| < 1). Here, a = A1 = 16 cm^2, and r = 1/2. Step 6: Calculate the sum of all areas: S = 16 / (1 - 1/2) S = 16 / (1/2) S = 16 * 2 = 32 cm^2. Step 7: The sum of the areas of all the squares is 32 cm^2.
15
In how many distinct ways can 8 Indians, 4 Americans, and 4 Englishmen be arranged in a single row such that all individuals of the same nationality are seated adjacently?
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Solution: Step 1: Treat each group of the same nationality as a single block. We have 3 blocks: [8 Indians], [4 Americans], [4 Englishmen]. Step 2: Arrange these 3 blocks among themselves. These 3 blocks can be arranged in 3! ways. 3! = 3 * 2 * 1 = 6 ways. Step 3: Arrange the individuals within each block. - The 8 Indians within their block can be arranged in 8! ways. - The 4 Americans within their block can be arranged in 4! ways. - The 4 Englishmen within their block can be arranged in 4! ways. Step 4: Multiply the arrangements from all steps to get the total number of ways. Total ways = (Arrangement of blocks) * (Arrangement within Indians) * (Arrangement within Americans) * (Arrangement within Englishmen) Total ways = 3! * 8! * 4! * 4!.
16
Calculate the sum of integers from 51 to 100, inclusive: (51 + 52 + 53 + ... + 100).
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Solution: Step 1: The sum of the first 'n' natural numbers (1 + 2 + ... + n) is given by the formula S_n = n(n+1)/2. Step 2: To find the sum of numbers from 51 to 100, we can calculate the sum of numbers from 1 to 100 and subtract the sum of numbers from 1 to 50. Required Sum = (1 + 2 + ... + 100) - (1 + 2 + ... + 50). Step 3: Calculate the sum for n = 100: S_100 = 100 * (100 + 1) / 2 = 100 * 101 / 2 = 50 * 101 = 5050. Step 4: Calculate the sum for n = 50: S_50 = 50 * (50 + 1) / 2 = 50 * 51 / 2 = 25 * 51 = 1275. Step 5: Subtract the sum of the first 50 numbers from the sum of the first 100 numbers: Required Sum = S_100 - S_50 = 5050 - 1275 = 3775.
17
Calculate the sum of the series: 1/15 + 1/35 + 1/63 + 1/99 + 1/143.
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Solution: Step 1: Express each denominator as a product of two consecutive odd numbers or integers with a constant difference: 15 = 3 Ɨ 5 35 = 5 Ɨ 7 63 = 7 Ɨ 9 99 = 9 Ɨ 11 143 = 11 Ɨ 13 Step 2: Rewrite each fraction using partial fraction decomposition. The general form for 1/(n(n+k)) is (1/k) Ɨ (1/n - 1/(n+k)). Here k=2. 1/15 = 1/(3 Ɨ 5) = (1/2) Ɨ (1/3 - 1/5) 1/35 = 1/(5 Ɨ 7) = (1/2) Ɨ (1/5 - 1/7) 1/63 = 1/(7 Ɨ 9) = (1/2) Ɨ (1/7 - 1/9) 1/99 = 1/(9 Ɨ 11) = (1/2) Ɨ (1/9 - 1/11) 1/143 = 1/(11 Ɨ 13) = (1/2) Ɨ (1/11 - 1/13) Step 3: Sum all these decomposed terms. Factor out the (1/2): Sum = (1/2) Ɨ [(1/3 - 1/5) + (1/5 - 1/7) + (1/7 - 1/9) + (1/9 - 1/11) + (1/11 - 1/13)] Step 4: Notice that this is a telescoping series, where intermediate terms cancel out: Sum = (1/2) Ɨ [1/3 - 1/13] Step 5: Perform the subtraction inside the bracket: 1/3 - 1/13 = (13 - 3) / (3 Ɨ 13) = 10/39. Step 6: Multiply by 1/2: Sum = (1/2) Ɨ (10/39) = 10/78. Step 7: Simplify the fraction: 10/78 = 5/39. Step 8: The value of the expression is 5/39.
18
At a gathering, every person present shakes hands with every other person exactly once. If a total of 105 handshakes occurred, how many people were in attendance at the party?
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Solution: Step 1: Let 'x' be the total number of persons present at the party. Step 2: A handshake involves two people. Since the order of people in a handshake does not matter, this is a combination problem. Step 3: The number of handshakes is given by the combination formula C(x, 2) = x Ɨ (x - 1) / 2. Step 4: Set up the equation using the given number of handshakes: x Ɨ (x - 1) / 2 = 105. Step 5: Multiply both sides by 2: x Ɨ (x - 1) = 210. Step 6: Expand and rearrange into a quadratic equation: x^2 - x - 210 = 0. Step 7: Factorize the quadratic equation: (x - 15)(x + 14) = 0. Step 8: This yields two possible values for x: x = 15 or x = -14. Step 9: Since the number of persons cannot be negative, discard x = -14. Step 10: Therefore, 15 persons were present at the party.
19
Six fair coins are tossed concurrently. How many of the possible outcomes will result in at most three coins showing heads?
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Solution: Step 1: We are tossing 6 fair coins. Each toss can result in a Head (H) or a Tail (T). Step 2: The condition "at most three of the coins turn up as heads" means we need to count the outcomes where the number of heads is 0, 1, 2, or 3. Step 3: This is a combination problem for each case, as the order of individual coin outcomes for a given number of heads does not matter. Step 4: Calculate the number of ways for each case: Case 1: 0 Heads (all 6 are tails). Number of ways = C(6, 0) = 1. Case 2: 1 Head (and 5 tails). Number of ways = C(6, 1) = 6. Case 3: 2 Heads (and 4 tails). Number of ways = C(6, 2) = (6 Ɨ 5) / (2 Ɨ 1) = 15. Case 4: 3 Heads (and 3 tails). Number of ways = C(6, 3) = (6 Ɨ 5 Ɨ 4) / (3 Ɨ 2 Ɨ 1) = 20. Step 5: Since these cases are mutually exclusive, add the number of ways for each case to find the total number of outcomes. Step 6: Total outcomes = 1 (for 0 heads) + 6 (for 1 head) + 15 (for 2 heads) + 20 (for 3 heads) = 42. Step 7: Therefore, there are 42 outcomes where at most three of the coins turn up as heads.
20
Determine the sum of the series 2 - 2 + 2 - 2 + ... for 101 terms.
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Solution: Step 1: Observe the pattern in the series. Each pair of consecutive terms sums to zero (2 - 2 = 0). Step 2: The series has 101 terms, which is an odd number. Step 3: The first 100 terms can be grouped into 50 pairs: (2 - 2) + (2 - 2) + ... (50 times). Step 4: The sum of these 50 pairs is 50 Ɨ 0 = 0. Step 5: The 101st term is the last term, which is 2 (since the terms alternate starting with 2). Step 6: The total sum of the 101 terms is the sum of the first 100 terms plus the 101st term, which is 0 + 2 = 2.
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