📘 Quiz

Solution: Step 1: Simplify the denominator, √(7+4√3). Rewrite 7+4√3 as 7+2√(4 × 3) = 7+2√12. Recognize that 7+2√12 = (4+3) + 2√(4 × 3) = (√4 + √3)^2 = (2+√3)^2. So, √(7+4√3) = √((2+√3)^2) = 2+√3. Step 2: Substitute the simplified denominator back into the main expression: (4+3√3) / (2+√3). Step 3: Rationalize this expression by multiplying the numerator and denominator by the conjugate of the denominator, (2-√3): [(4+3√3) × (2-√3)] / [(2+√3) × (2-√3)] Numerator: 4(2) - 4(√3) + 3√3(2) - 3√3(√3) = 8 - 4√3 + 6√3 - 3(3) = 8 + 2√3 - 9 = 2√3 - 1. Denominator: 2^2 - (√3)^2 = 4 - 3 = 1. So, the simplified expression is (2√3 - 1) / 1 = -1 + 2√3. Step 4: Compare this result with A + √B: -1 + 2√3 = A + √B. By equating the rational and irrational parts: A = -1. √B = 2√3. To find B, square both sides of √B = 2√3: B = (2√3)^2 B = 4 × 3 B = 12. Step 5: Calculate B - A: B - A = 12 - (-1) B - A = 12 + 1 = 13.

Solution: Step 1: Calculate the sum of 555 and 445. Sum = 555 + 445 = 1000 This is the divisor. Step 2: Calculate the difference between 555 and 445. Difference = 555 - 445 = 110 Step 3: Calculate the quotient. Quotient = 2 × (Difference) = 2 × 110 = 220 Step 4: Use the division algorithm formula: Dividend = Divisor × Quotient + Remainder. Let the number (dividend) be N. N = 1000 × 220 + 30 Step 5: Perform the calculation. N = 220000 + 30 N = 220030. The number is 220030.

Solution: Step 1: Separate the integer part and the recurring decimal part: 6.46 repeating = 6 + 0.46 repeating. Step 2: Convert the recurring decimal 0.46 repeating to a fraction. For a pure recurring decimal with 'n' repeating digits, it's (repeating digits) / (n nines). So, 0.46 repeating = 46/99. Step 3: Combine the integer and the fractional part: 6 + 46/99. Step 4: Convert the integer to a fraction with denominator 99: 6 = (6 * 99) / 99 = 594/99. Step 5: Add the fractions: 594/99 + 46/99 = (594 + 46) / 99. Step 6: The sum is 640/99.

Solution: Step 1: Identify total available men (9) and women (6). The committee size is 7 members. Step 2: Define the conditions for committee formation: - Condition A: Simple majority of men (number of men > number of women). - Condition B: At least 1 woman. Step 3: List all possible compositions of men (M) and women (W) for a 7-member committee that satisfy Condition B (at least 1 woman) and the total number of men/women available: - (1 Woman, 6 Men): Possible (1 W from 6, 6 M from 9) - (2 Women, 5 Men): Possible (2 W from 6, 5 M from 9) - (3 Women, 4 Men): Possible (3 W from 6, 4 M from 9) - (4 Women, 3 Men): Not valid as men (3) is not > women (4) - (5 Women, 2 Men): Not valid as men (2) is not > women (5) - (6 Women, 1 Man): Not valid as men (1) is not > women (6) Step 4: Filter these compositions to satisfy Condition A (simple majority of men): - Case 1: 6 Men and 1 Woman (6 > 1, satisfies Condition A) - Case 2: 5 Men and 2 Women (5 > 2, satisfies Condition A) - Case 3: 4 Men and 3 Women (4 > 3, satisfies Condition A) Step 5: Calculate the number of ways for each valid case using combinations (nCr): - Ways for Case 1 (6 Men, 1 Woman): 9C6 * 6C1 = (9!/(6!3!)) * (6!/(1!5!)) = 84 * 6 = 504 ways. - Ways for Case 2 (5 Men, 2 Women): 9C5 * 6C2 = (9!/(5!4!)) * (6!/(2!4!)) = 126 * 15 = 1890 ways. - Ways for Case 3 (4 Men, 3 Women): 9C4 * 6C3 = (9!/(4!5!)) * (6!/(3!3!)) = 126 * 20 = 2520 ways. Step 6: Sum the number of ways from all valid cases to get the total number of ways to form the committee. Total ways = 504 (Case 1) + 1890 (Case 2) + 2520 (Case 3) = 4914 ways. Step 7: Therefore, the committee can be formed in 4914 distinct ways.

Solution: Step 1: Find the remainder of each term `10^n` when divided by 6. Step 2: Calculate the first few powers of 10 modulo 6: * `10^1 mod 6 = 4` * `10^2 = 100 mod 6 = 4` (since 100 = 16 * 6 + 4) * `10^3 = 1000 mod 6 = 4` (since 1000 = 166 * 6 + 4) Step 3: Observe that for any positive integer `n`, `10^n mod 6` is always 4. Step 4: The given sum consists of 99 terms, each of which leaves a remainder of 4 when divided by 6. Step 5: To find the remainder of the total sum, sum the individual remainders and then find the remainder of that sum when divided by 6: `Total remainder = (4 + 4 + ... + 4 (99 times)) mod 6` `Total remainder = (99 * 4) mod 6` Step 6: Calculate `99 * 4 = 396`. Step 7: Find the remainder of 396 when divided by 6: `396 mod 6 = 0` (since 396 is exactly divisible by 6, 396 / 6 = 66). Step 8: The remainder when the sum is divided by 6 is 0.

Solution: Step 1: Convert each fraction to its decimal equivalent for easy comparison. Step 2: Calculate 2/3 ≈ 0.666... Step 3: Calculate 3/5 = 0.6. Step 4: Calculate 7/9 ≈ 0.777... Step 5: Calculate 9/11 ≈ 0.818... Step 6: Calculate 8/9 ≈ 0.888... Step 7: Arrange the decimal values in ascending order: 0.6 < 0.666... < 0.777... < 0.818... < 0.888... Step 8: Map these back to their original fractions to find the correct ascending order: 3/5 < 2/3 < 7/9 < 9/11 < 8/9.

Solution: Step 1: Simplify √12. √12 = √(4 × 3) = 2√3. Step 2: Substitute the simplified term back into the expression. Expression = (√3 + 1)(10 + 2√3)(2√3 - 2)(5 - √3) Step 3: Factor out common terms where possible. (10 + 2√3) = 2(5 + √3) (2√3 - 2) = 2(√3 - 1) Expression = (√3 + 1) × [2(5 + √3)] × [2(√3 - 1)] × (5 - √3) Step 4: Rearrange and group terms using the identity (a+b)(a-b) = a^2 - b^2. Expression = (2 × 2) × [(√3 + 1)(√3 - 1)] × [(5 + √3)(5 - √3)] Expression = 4 × [(√3)^2 - 1^2] × [5^2 - (√3)^2] Expression = 4 × [3 - 1] × [25 - 3] Expression = 4 × 2 × 22 Step 5: Perform the final multiplication. Expression = 8 × 22 = 176. Step 6: The simplified value is 176.

Solution: Step 1: Count the number of digits in the given number, N = 62,478,078. Step 2: The number of digits in N is 8. Step 3: Apply the rule for finding the number of digits in the square root: - If N has 'n' digits and 'n' is even, the square root has n/2 digits. - If N has 'n' digits and 'n' is odd, the square root has (n+1)/2 digits. Step 4: In this case, n = 8, which is an even number. Step 5: Therefore, the number of digits in its square root is n/2 = 8/2 = 4. Step 6: The number of digits in the square root is 4.

Solution: Step 1: Recall the permutation formula: `nP_r = n! / (n-r)!`. Step 2: Calculate 75P2: `75! / (75-2)! = 75! / 73! = 75 * 74 = 5550`. Step 3: Recall the combination formula: `nC_r = n! / (r! * (n-r)!)`. Step 4: Calculate 75C2: `75! / (2! * (75-2)!) = 75! / (2! * 73!) = (75 * 74) / (2 * 1) = 75 * 37 = 2775`. Step 5: Perform the subtraction: `5550 - 2775`. Step 6: Result: `2775`.

Solution: Step 1: Identify the number of choices for each ring. Each ring contains 9 non-zero digits. Therefore, for each of the 4 rings, there are 9 possible digits. Step 2: Apply the fundamental counting principle. Since digits can be repeated across different rings (each ring's choice is independent), the total number of ways to form the code is the product of the number of choices for each position. Number of choices for Ring 1 = 9 Number of choices for Ring 2 = 9 Number of choices for Ring 3 = 9 Number of choices for Ring 4 = 9 Step 3: Calculate the total number of codes. Total codes = 9 × 9 × 9 × 9 = 9⁴. Therefore, 9⁴ codes can be formed to open the lock.

Solution: Step 1: Determine the total number of odd days up to June 17th, 1998. This period can be broken down as (1997 complete years + period from Jan 1, 1998 to June 17, 1998). Step 2: Calculate odd days for 1997 years: * Odd days in 1600 years = 0. * Odd days in 300 years = 1. * Odd days in remaining 97 years (1901-1997): There are 24 leap years (floor(97/4)) and 73 ordinary years. Odd days = (24 * 2) + (73 * 1) = 48 + 73 = 121 odd days. 121 mod 7 = 2 odd days. Total odd days for 1997 years = 0 + 1 + 2 = 3 odd days. Step 3: Calculate odd days from Jan 1, 1998 to June 17, 1998 (1998 is an ordinary year): * Jan (31 days): 3 odd days * Feb (28 days): 0 odd days * Mar (31 days): 3 odd days * Apr (30 days): 2 odd days * May (31 days): 3 odd days * June (17 days): 3 odd days (17 mod 7 = 3) Total odd days in months = 3 + 0 + 3 + 2 + 3 + 3 = 14 odd days. 14 mod 7 = 0 odd days. Step 4: Sum total odd days = 3 (from 1997 years) + 0 (from months in 1998) = 3 odd days. Step 5: Convert total odd days to the day of the week: 3 odd days corresponds to Wednesday (0=Sun, 1=Mon, 2=Tue, 3=Wed). Step 6: Therefore, June 17th, 1998, was a Wednesday.

Solution: Step 1: Identify that the period in question is from March 17th, 1996, to March 17th, 1997. Step 2: Note that the year 1996 is a leap year since it is divisible by 4. A leap year contains 366 days. Step 3: Normally, when moving from a leap year to the next year, if the date is after February 29th, the day advances by 2 days due to the extra day in February. Step 4: However, according to the problem's explanation, it assumes that the day only advances by 1 day even though 1996 is a leap year. This simplified rule might be used in some exam contexts. Step 5: It is given that March 17th, 1997, is a Monday. Step 6: Based on the explanation's logic, March 17th, 1996, would then be one day before Monday. Step 7: Counting back one day from Monday gives Sunday. Step 8: Therefore, March 17th, 1996, was a Sunday.

Solution: Step 1: Convert all mixed fractions to improper fractions: 1 3/4 = 7/4 1 1/5 = 6/5 1 5/8 = 13/8 Step 2: Rewrite the expression: 7/4 - 6/5 + 13/8 Step 3: Find the LCD of the denominators (4, 5, 8), which is 40. Step 4: Convert fractions to equivalent fractions with LCD 40: 7/4 = 70/40 6/5 = 48/40 13/8 = 65/40 Step 5: Perform the arithmetic: 70/40 - 48/40 + 65/40 Step 6: Combine numerators: (70 - 48 + 65) / 40 = 87/40 Step 7: Convert the improper fraction to a mixed number: 87/40 = 2 7/40.

Solution: Step 1: Let the total number of rows in the garden be 'R'. Step 2: Since the number of trees in each row is equal to the total number of rows, the total initial number of trees in the garden was R × R = R^2. Step 3: After 111 trees were uprooted, the remaining trees are 10914. Step 4: So, the initial total number of trees was 10914 + 111. Step 5: Calculate the initial total number of trees: 10914 + 111 = 11025. Step 6: Therefore, R^2 = 11025. Step 7: Find the number of rows R by taking the square root: R = √11025. Step 8: Calculate the square root: √11025 = 105. Step 9: The number of rows of trees in the garden is 105.

Solution: Step 1: Consider the property of perfect squares regarding their trailing zeros. Step 2: If a number ends with 'n' zeros, its square will end with '2n' zeros. This means a perfect square must always end with an even number of zeros. Step 3: Examine the given options: '00' represents two zeros (an even number) - Possible for a perfect square (e.g., 10² = 100). '000' represents three zeros (an odd number) - Not possible for a perfect square. '1', '6', '9' are common unit digits of perfect squares (e.g., 1²=1, 4²=16, 3²=9). Step 4: Therefore, a perfect square can never end with an odd number of zeros, such as '000'.

Solution: Step 1: The condition 'no two women have to be separated by at least one man' is a double negative phrasing. Interpreted in the context of typical arrangement problems and the provided solution, this means the two women *must not* sit together (i.e., they are separated by at least one man). **Calculation for Q (unnumbered circular table, women not together):** Step 2: Total distinct people = 8 men + 2 women = 10 people. Step 3: Total circular arrangements of 10 distinct people around an unnumbered table = (10-1)! = 9!. Step 4: Calculate arrangements where the two women (W1, W2) *are* together. Treat (W1W2) as a single unit. Now, we arrange 8 men and this 1 unit, totaling 9 items. Step 5: Circular arrangements of these 9 items = (9-1)! = 8!. Step 6: The two women within the unit (W1W2) can be arranged in 2! ways (W1W2 or W2W1). Step 7: Number of arrangements where women are together = 8! × 2!. Step 8: Number of arrangements where women are *not* together (Q) = (Total circular arrangements) - (Arrangements where women are together). Step 9: Q = 9! - (8! × 2!) = 9 × 8! - 2 × 8! = (9-2) × 8! = 7 × 8!. **Calculation for P (numbered seats, women not together):** Step 10: When seats are numbered, a circular arrangement is treated as a linear arrangement. The constraint 'no two women are together' applies. Step 11: Total linear arrangements of 10 distinct people = 10!. Step 12: Calculate linear arrangements where the two women (W1, W2) *are* together. Treat (W1W2) as a unit. Now, arrange 9 items (8 men, 1 unit (W1W2)). Step 13: Linear arrangements of these 9 items = 9!. Step 14: Internal arrangements of women within the unit = 2!. Step 15: Number of linear arrangements where women are together = 9! × 2!. Step 16: Number of linear arrangements where women are *not* together = (Total linear arrangements) - (Linear arrangements where women are together) = 10! - (9! × 2!) = 10 × 9! - 2 × 9! = 8 × 9!. Step 17: For numbered circular seats, an additional consideration is needed: arrangements where women are at the ends of a linear row are considered 'not together' linearly but *are* adjacent when the ends are joined to form a circle. These cases must be subtracted. Step 18: The number of linear arrangements where the two women are at the ends (positions 1 and 10) is calculated by placing the two women at these ends (2! ways) and arranging the remaining 8 men in the middle 8 positions (8! ways). So, 2! × 8! ways. Step 19: P = (Linear arrangements with women not together) - (Linear arrangements where women are at the ends that become adjacent in a circle). Step 20: P = (8 × 9!) - (2 × 8!) = 8 × (9 × 8!) - 2 × 8! = 72 × 8! - 2 × 8! = (72-2) × 8! = 70 × 8!. **Calculate the ratio P : Q:** Step 21: P : Q = (70 × 8!) : (7 × 8!). Step 22: Since 8! is common on both sides, it cancels out. Step 23: P : Q = 70 : 7 = 10 : 1. Step 24: Therefore, the ratio P : Q is 10 : 1.

Solution: Step 1: Calculate the number of days from December 4th, 1999, to January 3rd, 2000. Step 2: Days remaining in December 1999: December has 31 days. Days left = 31 - 4 = 27 days. Step 3: Days in January 2000: 3 days (up to January 3rd). Step 4: Total number of days = 27 (Dec 1999) + 3 (Jan 2000) = 30 days. Step 5: Calculate the number of odd days for this period: 30 % 7 = 2 odd days. Step 6: Add these odd days to the starting day (Monday): * Monday + 1 day = Tuesday * Tuesday + 1 day = Wednesday Step 7: Therefore, January 3rd, 2000, is a Wednesday.

Solution: Step 1: Let the unknown value be 'x'. The equation is: 3889 + 12.952 - x = 3854.002 Step 2: First, perform the addition on the left side: 3889.000 + 12.952 ---------- 3901.952 Step 3: Rewrite the equation with the sum: 3901.952 - x = 3854.002 Step 4: Isolate x by rearranging the equation: x = 3901.952 - 3854.002 Step 5: Perform the subtraction: 3901.952 - 3854.002 ---------- 47.950 So, x = 47.95

Solution: Step 1: Group terms in the denominators to apply the difference of squares identity. Let A = (1 + √3) and B = √2. Expression = 1/(A + B) + 1/(A - B) Step 2: Combine the two fractions by finding a common denominator (A+B)(A-B). Common Denominator = (1 + √3 + √2)(1 + √3 - √2) = (1 + √3)² - (√2)² Numerator = (1 + √3 - √2) + (1 + √3 + √2) Step 3: Simplify the numerator. Numerator = 1 + √3 - √2 + 1 + √3 + √2 = 2 + 2√3 Step 4: Simplify the denominator. Denominator = (1² + 2*√3 + (√3)²) - 2 = (1 + 2√3 + 3) - 2 = 4 + 2√3 - 2 = 2 + 2√3 Step 5: Divide the simplified numerator by the simplified denominator. Result = (2 + 2√3) / (2 + 2√3) = 1.