To find the LCM (Least Common Multiple) of prime numbers between 1 and 12, we first need to identify all prime numbers in that range. Then, since all these numbers are prime, their LCM will simply be their product.
C) 2310 is the correct answer because it is the product of all prime numbers between 1 and 12.
This problem involves work and time, where the efficiency of different types of workers is given. We need to standardize the work units based on a single type of worker and then calculate the total work and the time required by the combined team.
Correct Option: A) 12
Simplify∶ \( \frac{5}{8} \)+\( \frac{3}{4} \) of \(\frac{4}{11}\div\frac{3}{16}-\frac{1}{2}\times\frac{3}{4}\) =?
3 \( \frac{2}{13} \)
3\( \frac{4}{11} \)
1\( \frac{31}{44} \)
1
To simplify the given expression, we must follow the order of operations (BODMAS/PEMDAS): Brackets, Orders (powers/roots), Division, Multiplication, Addition, Subtraction. The expression involves fractions and the "of" operator, which implies multiplication.
C) \( 1 \frac{31}{44} \)
To find the largest size of square tiles that can cover the floor of a room without cutting any tiles, we need to find the Highest Common Factor (HCF) of the room's dimensions. First, convert the dimensions from meters to centimeters to work with whole numbers.
The dimensions of the room are given as 5.25 m \(\times\) 4.55 m.
We need to find the HCF of 525 cm and 455 cm. We can use the prime factorization method or the Euclidean algorithm.
Using prime factorization:
The common prime factors are 5 and 7. The HCF is the product of the lowest powers of the common prime factors.
\[ \text{HCF}(525, 455) = 5^1 \times 7^1 = 35 \]The HCF, 35 cm, represents the largest possible side length of a square tile that can perfectly fit both the length and the width of the room without any cutting.
B) 35 cm — The HCF of 525 cm and 455 cm is 35 cm, which is the largest side length for square tiles that can cover the floor without cutting.
An athlete runs 400 m in 20 s. Calculate the average speed of the athlete.
To calculate the average speed of the athlete, we need to use the fundamental formula for speed, which relates distance covered and the time taken. The units must be consistent for the final answer.
Average speed is defined as the total distance traveled divided by the total time taken.
\[ \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} \]The unit for speed is meters per second (\(\text{m/s}\) or \(\text{ms}^{-1}\)).
Correct Option: C) 20 ms-1
To find the minimum number of sections, we need to maximize the number of students in each section. Since every section in every grade must have the same number of students, this number must be a common divisor of the number of students in each grade. To maximize this common divisor, we need to find the Highest Common Factor (HCF) or Greatest Common Divisor (GCD) of the number of students in Grades 4, 5, and 6.
Identify the number of students in each grade:
Find the HCF of 215, 258, and 301:
We can use the prime factorization method or the division method.
Let's use the prime factorization method:
The common prime factor among 215, 258, and 301 is 43.
So, the HCF is 43. This means each section will have 43 students.
Calculate the number of sections for each grade:
Calculate the total minimum number of sections:
Total sections = (Sections in Grade 4) + (Sections in Grade 5) + (Sections in Grade 6)
Total sections = \(5 + 6 + 7 = 18\)
C) 18 is the correct answer. This is the minimum total number of sections possible when each section has the maximum possible number of students (which is the HCF of the students in each grade).
Dilip invested Rs. 100 for three years at a compound interest rate of 10%. Haru also invested Rs. 100 for three years at a simple interest rate of x%. At the end of three years, both received the same amount of money. What is the value of x?
The problem involves comparing compound interest and simple interest. We need to calculate the amount received by Dilip (compound interest) and Haru (simple interest) after three years. Since both amounts are equal, we can set up an equation to solve for the unknown simple interest rate, x.
Rounding to two decimal places, \(x \approx 11.03\).
A) 11.03 is the correct value for x, calculated by equating the compound amount for Dilip and the simple amount for Haru.
If sinA = \( \frac{12}{13}\), then the value of (secA + cosecA) is∶
\(\frac{67}{20}\)
\(\frac{221}{60}\)
\(\frac{11}{3}\)
\(\frac{211}{60}\)
To find the value of \((\sec A + \operatorname{cosec} A)\) given \(\sin A = \frac{12}{13}\), we first need to determine the values of \(\cos A\), \(\sec A\), and \(\operatorname{cosec} A\). We can use the Pythagorean identity \(\sin^2 A + \cos^2 A = 1\) to find \(\cos A\), and then calculate \(\sec A\) and \(\operatorname{cosec} A\) using their definitions.
Given \(\sin A = \frac{12}{13}\).
Recall the Pythagorean identity: \(\sin^2 A + \cos^2 A = 1\).
Substitute the value of \(\sin A\) into the identity:
\[ \left(\frac{12}{13}\right)^2 + \cos^2 A = 1 \]\[ \frac{144}{169} + \cos^2 A = 1 \]Solve for \(\cos^2 A\):
\[ \cos^2 A = 1 - \frac{144}{169} \]\[ \cos^2 A = \frac{169 - 144}{169} \]\[ \cos^2 A = \frac{25}{169} \]Take the square root to find \(\cos A\). Assuming A is an acute angle (or in a quadrant where cosine is positive):
\[ \cos A = \sqrt{\frac{25}{169}} = \frac{5}{13} \]Now, find \(\sec A\) and \(\operatorname{cosec} A\):
\[\sec A = \frac{1}{\cos A} = \frac{1}{\frac{5}{13}} = \frac{13}{5}\]
\[\operatorname{cosec} A = \frac{1}{\sin A} = \frac{1}{\frac{12}{13}} = \frac{13}{12}\]
Finally, calculate \((\sec A + \operatorname{cosec} A)\):
\[ \sec A + \operatorname{cosec} A = \frac{13}{5} + \frac{13}{12} \]To add these fractions, find a common denominator, which is \(5 \times 12 = 60\):
\[ \sec A + \operatorname{cosec} A = \frac{13 \times 12}{5 \times 12} + \frac{13 \times 5}{12 \times 5} \]\[ \sec A + \operatorname{cosec} A = \frac{156}{60} + \frac{65}{60} \]\[ \sec A + \operatorname{cosec} A = \frac{156 + 65}{60} \]\[ \sec A + \operatorname{cosec} A = \frac{221}{60} \]Correct Option: B) \(\frac{221}{60}\)
This problem involves work and time. We are given the combined work rate of Ratul and Praful, and information about a scenario where one person leaves early, leading to an extended completion time. We need to find the time Praful would take alone. We will use the concept of work done being equal to the product of efficiency and time.
Let \(R\) be the number of days Ratul takes to complete the work alone, and \(P\) be the number of days Praful takes to complete the work alone.
Their one-day work rates are \(\frac{1}{R}\) and \(\frac{1}{P}\) respectively.
Given that Ratul and Praful together can complete a task in 18 days:
\[ \frac{1}{R} + \frac{1}{P} = \frac{1}{18} \quad \text{(Equation 1)} \]The total time taken to complete the task is 23 days. Ratul left 7 days before the work got over. This means Ratul worked for \(23 - 7 = 16\) days.
Praful worked for all 23 days.
The total work done can be expressed as the sum of the work done by Ratul and Praful:
\[ \text{Work done by Ratul} + \text{Work done by Praful} = 1 \]\[ 16 \times \frac{1}{R} + 23 \times \frac{1}{P} = 1 \]\[ \frac{16}{R} + \frac{23}{P} = 1 \quad \text{(Equation 2)} \]From Equation 1, we can write \[\frac{1}{R} = \frac{1}{18} - \frac{1}{P}\].
Substitute this into Equation 2:
\[ 16 \left( \frac{1}{18} - \frac{1}{P} \right) + \frac{23}{P} = 1 \]\[ \frac{16}{18} - \frac{16}{P} + \frac{23}{P} = 1 \]\[ \frac{8}{9} + \frac{7}{P} = 1 \]Now, solve for \(P\):
\[ \frac{7}{P} = 1 - \frac{8}{9} \]\[ \frac{7}{P} = \frac{9 - 8}{9} \]\[ \frac{7}{P} = \frac{1}{9} \]\[ P = 7 \times 9 \]\[ P = 63 \]So, Praful would take 63 days to complete the work by himself.
A) 63
You are given question and three statements. Identify which of the statements is/are sufficient to answer the question.
Question∶
There are 3 shifts in a company. How many employees come in the 3rd shift?
Statements∶
1. Total employees are 500.
2. One-fourth of the total employees come in shift 1.
3. The number of employees coming in shift 2 is twice that of those coming in shift 1.
Statements 1, 2 and 3 together are sufficient.
Only statements 1 and 2 together are sufficient.
Only statements 2 and 3 together are sufficient.
All the statements are insufficient.
The problem asks to determine the number of employees in the 3rd shift. We need to evaluate if the given statements, individually or in combination, provide enough information to find this value. We will use variables to represent the number of employees in each shift and form equations based on the statements.
Let \(E_1\), \(E_2\), and \(E_3\) be the number of employees in Shift 1, Shift 2, and Shift 3, respectively.
The total number of employees is \(E_1 + E_2 + E_3\).
Statement 1: Total employees are 500.
This means \(E_1 + E_2 + E_3 = 500\).
Statement 2: One-fourth of the total employees come in shift 1.
From Statement 1, total employees are 500. So, \(E_1 = \frac{1}{4} \times 500 = 125\).
Statement 3: The number of employees coming in shift 2 is twice that of those coming in shift 1.
From Statement 2, \(E_1 = 125\). So, \(E_2 = 2 \times E_1 = 2 \times 125 = 250\).
Now, let's combine all three statements:
Substitute the values of \(E_1\) and \(E_2\) into the equation from Statement 1:
\[ 125 + 250 + E_3 = 500 \]\[ 375 + E_3 = 500 \]\[ E_3 = 500 - 375 \]\[ E_3 = 125 \]Since we were able to find the number of employees in the 3rd shift (\(E_3 = 125\)) using all three statements, all three statements together are sufficient.
A) Statements 1, 2 and 3 together are sufficient. This is correct because by combining all three statements, we can uniquely determine the number of employees in the 3rd shift.
B) Only statements 1 and 2 together are sufficient. If we only use statements 1 and 2, we know \(E_1 = 125\) and \(E_1 + E_2 + E_3 = 500\). This gives \(125 + E_2 + E_3 = 500\), or \(E_2 + E_3 = 375\). We cannot find \(E_3\) without knowing \(E_2\).
C) Only statements 2 and 3 together are sufficient. If we only use statements 2 and 3, we know \[E_1 = \frac{1}{4} \times \text{Total Employees}\] and \(E_2 = 2 \times E_1\). While we can find \(E_1\) and \(E_2\) in terms of the total employees, we don't know the total number of employees, so we cannot find the absolute number of employees in Shift 3, nor the total number of employees to calculate \(E_1\) and \(E_2\) numerically.
D) All the statements are insufficient. This is incorrect because, as shown above, all three statements together are sufficient to find the number of employees in the 3rd shift.