To find the least number that should be added to 1456 so that it is divisible by 6, 5, and 4, we first need to find the Least Common Multiple (LCM) of 6, 5, and 4. The number we are looking for must be a multiple of this LCM. We then find the smallest multiple of the LCM that is greater than or equal to 1456 and calculate the difference.
We want to find out how far 1456 is from the next multiple of 60.
\[ 1456 \div 60 \]So, \(1456 = 60 \times 24 + 16\). This means that 1456 is 16 more than a multiple of 60 (specifically, \(60 \times 24 = 1440\)).
To make 1456 divisible by 60, we need to add a number such that the sum is the next multiple of 60. The remainder is 16. The next multiple of 60 after 1440 is \(1440 + 60 = 1500\).
The number to be added is the difference between the next multiple of 60 and 1456, or simply \(60 - \text{remainder}\).
\[ \text{Number to be added} = 60 - 16 = 44 \]\(1456 + 44 = 1500\).
Is 1500 divisible by 6, 5, and 4?
Thus, 44 is the least number to be added.
Correct Option: C) 44
Choose the word which is different from the others.
The question asks to identify the word that is different from the others. This is a classification-based question where we need to find the odd one out based on a common category or characteristic shared by the other options.
Correct Option: A) Reptile is the correct answer because it is a class of animals, whereas Snake, Lizard, and Crocodile are specific animals belonging to that class.
To find the percentage of students who passed the examination, we first need to determine the number of boys and girls in the class. Then, we calculate the number of boys and girls who failed. Subtracting the failed students from the total in each group gives us the number of passed students. Finally, we sum the passed students and express it as a percentage of the total class strength.
Calculate the number of boys and girls:
Let the number of girls be \(G\).
The number of boys is twice that of girls, so the number of boys is \(B = 2G\).
The total number of students is 135.
\[ G + B = 135 \] \[ G + 2G = 135 \] \[ 3G = 135 \] \[ G = \frac{135}{3} = 45 \]So, the number of girls is 45.
The number of boys is \(B = 2 \times 45 = 90\).
Calculate the number of failed boys and girls:
One-sixth of the boys failed:
\[ \text{Failed Boys} = \frac{1}{6} \times 90 = 15 \]One-third of the girls failed:
\[ \text{Failed Girls} = \frac{1}{3} \times 45 = 15 \]Calculate the number of passed boys and girls:
Passed Boys = Total Boys - Failed Boys
\[ \text{Passed Boys} = 90 - 15 = 75 \]Passed Girls = Total Girls - Failed Girls
\[ \text{Passed Girls} = 45 - 15 = 30 \]Calculate the total number of students who passed:
\[ \text{Total Passed Students} = \text{Passed Boys} + \text{Passed Girls} = 75 + 30 = 105 \]Calculate the percentage of students who passed:
\[ \text{Percentage Passed} = \left( \frac{\text{Total Passed Students}}{\text{Total Students}} \right) \times 100 \] \[ \text{Percentage Passed} = \left( \frac{105}{135} \right) \times 100 \] \[ \text{Percentage Passed} = \frac{7}{9} \times 100 \] \[ \text{Percentage Passed} = 0.7777... \times 100 \approx 77.78\% \]C) 77.78% is the calculated percentage of students who passed the examination.
We are given the ratio of two numbers, p and q, and their Least Common Multiple (LCM). We need to find the sum of these numbers. We can represent the numbers using a common factor based on their ratio, then use the property that for two numbers, the product of the numbers is equal to the product of their HCF and LCM.
Represent the numbers using the given ratio:
Given that p and q are in the ratio 4 : 3, we can write them as:
where \(x\) is the common factor of p and q. This common factor \(x\) is also the Highest Common Factor (HCF) of p and q.
Use the relationship between numbers, HCF, and LCM:
For any two positive integers, the product of the numbers is equal to the product of their HCF and LCM. That is:
Substitute the values into the formula:
We know \(p = 4x\), \(q = 3x\), \(\text{HCF}(p, q) = x\), and \(\text{LCM}(p, q) = 36\). Substituting these into the formula:
Solve for \(x\):
Simplify the equation:
Since \(x\) is a common factor and numbers are positive, \(x \neq 0\). We can divide both sides by \(x\):
\[ 12x = 36 \]\[ x = \frac{36}{12} \]\[ x = 3 \]Find the values of p and q:
Now that we have \(x = 3\), we can find p and q:
Calculate p + q:
Finally, calculate the sum p + q:
B) 21 is the correct sum of p and q.
This is a letter series analogy question. We need to identify the pattern or relationship between the first pair of letter groups (EIMQUY : CGKOSW) and then apply the same pattern to the second letter group (BFJNRV) to find the missing group.
D) ZDHLPT is the correct option as it follows the established pattern of each letter being shifted backward by 2 positions.
To find the cost price, we can use the formula relating selling price, cost price, and profit percentage. If the selling price (SP) and profit percentage (P%) are known, the cost price (CP) can be calculated using the formula: \[\text{CP} = \frac{\text{SP}}{1 + \frac{\text{P}}{100}}\] or \[\text{CP} = \frac{\text{SP} \times 100}{100 + \text{P}}\].
The selling price is the cost price plus the profit. If the profit is 13% of the cost price, then:
\[ \text{SP} = \text{CP} + \left( \frac{13}{100} \times \text{CP} \right) \]\[ \text{SP} = \text{CP} \left( 1 + \frac{13}{100} \right) \]\[ \text{SP} = \text{CP} \left( \frac{100 + 13}{100} \right) \]\[ \text{SP} = \text{CP} \left( \frac{113}{100} \right) \]First, divide 27120 by 113:
\[ \frac{27120}{113} = 240 \]Then, multiply by 100:\[ \text{CP} = 240 \times 100 \]\[ \text{CP} = 24000 \]The cost price of the product is Rs. 24,000.
The problem asks us to find the difference in interest earned when calculated using compound interest versus simple interest for a given principal, rate, and time. First, we will use the simple interest information to find the principal amount. Then, we will calculate the compound interest for the same principal, rate, and time. Finally, we will find the difference between the compound interest and the simple interest.
Calculate the Principal (P) using Simple Interest:
Given:
The formula for Simple Interest is:
\[ SI = \frac{P \times R \times T}{100} \]\[ 4000 = \frac{P \times 10 \times 2}{100} \]\[ 4000 = \frac{20P}{100} \]\[ 4000 = \frac{P}{5} \]\[ P = 4000 \times 5 \]\[ P = \text{Rs. } 20,000 \]Calculate the Compound Interest (CI) for the same Principal:
Using the principal \(P = \text{Rs. } 20,000\), rate \(R = 10\%\) p.a., and time \(T = 2\) years, compounded annually.
The formula for Amount \((A)\) under compound interest is:
\[ A = P \left(1 + \frac{R}{100}\right)^T \]\[ A = 20000 \left(1 + \frac{10}{100}\right)^2 \]\[ A = 20000 \left(1 + \frac{1}{10}\right)^2 \]\[ A = 20000 \left(\frac{11}{10}\right)^2 \]\[ A = 20000 \times \frac{121}{100} \]\[ A = 200 \times 121 \]\[ A = \text{Rs. } 24,200 \]Calculate the Compound Interest (CI):
Compound Interest \((CI) = A - P\)
\[ CI = 24200 - 20000 \]\[ CI = \text{Rs. } 4,200 \]Find the difference between Compound Interest and Simple Interest:
Difference \(= CI - SI\)
\[ \text{Difference} = 4200 - 4000 \]\[ \text{Difference} = \text{Rs. } 200 \]A) Rs. 200 is the correct answer, as calculated above.
This problem involves calculating the cost price of an item given the difference in its selling prices at two different profit percentages. We can set up equations for the selling price at each profit margin and then use the given difference to find the cost price.
Let the Cost Price (CP) of the item be \( \text{Rs.} \, x \).
When the item is sold at a profit of 9%, the Selling Price (SP1) is given by:
\[ \text{SP1} = \text{CP} + 9\% \, \text{of} \, \text{CP} = x + \frac{9}{100}x = \frac{109}{100}x \]When the item is sold at a profit of 7%, the Selling Price (SP2) is given by:
\[ \text{SP2} = \text{CP} + 7\% \, \text{of} \, \text{CP} = x + \frac{7}{100}x = \frac{107}{100}x \]The difference between these two selling prices is given as Rs. 190. So, we can write the equation:
\[ \text{SP1} - \text{SP2} = 190 \]Substitute the expressions for SP1 and SP2 into the equation:
\[ \frac{109}{100}x - \frac{107}{100}x = 190 \]Simplify the equation:
\[ \frac{(109 - 107)}{100}x = 190 \]\[ \frac{2}{100}x = 190 \]Solve for \( x \):
\[ x = \frac{190 \times 100}{2} \]\[ x = 95 \times 100 \]\[ x = 9500 \]Therefore, the cost price of the item is Rs. 9,500.
This is a blood relation problem. We need to deduce the relationship between the person in the photograph and the lady by breaking down the statement given by the lady into smaller, understandable parts.
Correct Option: D) Brother
The question asks to identify the Brazilian international football star among the given options. This requires knowledge of the nationality of prominent football players.
Correct Option: A) Neymar
Neymar da Silva Santos Júnior, commonly known as Neymar, is a renowned professional footballer who plays as a forward for Al Hilal and the Brazil national team. He is widely regarded as one of the best players in the world.