In the sum 
for which digit does the symbol stand?
The given problem involves finding the digit represented by the symbol ⊗ in the equation. We need to replace ⊗ with a digit from the options to satisfy the equation.
Correct Option: B) 3
To find the minimum length that can be measured using two sticks of given lengths, we need to find the greatest common divisor (GCD) of the two lengths. The GCD represents the largest unit length that can precisely divide both given lengths. Any measurement made using these sticks will be an integer multiple of this GCD. Therefore, the smallest non-zero measurement we can make is this GCD itself.
B) 0.25 foot — As calculated, the greatest common divisor of 7.5 feet and 3.25 feet is 0.25 feet. This is the smallest non-zero length that can be measured using combinations of these two sticks.
The problem asks us to find the next number in a given sequence by applying a simple mathematical operation to each number, such that the results form a sequence related to prime numbers. We need to identify this operation and the resulting prime number sequence to predict the next term.
Thus, the next number in the sequence is 38.
Correct Option: C) 38
The problem involves finding the numbers on a torn page from a booklet. We are given the sum of the numbers on the remaining pages. We will use the formula for the sum of an arithmetic series to determine the total number of pages in the booklet and then identify the torn page numbers.
Correct Option: B) 7, 8
Consider the following arrangement that has some missing letters:
abab_b_bcb_dcdcded_d
The missing letters which complete the arrangement are
The problem presents a sequence of letters with missing elements. The goal is to identify the pattern in the sequence and fill in the missing letters from the given options. We will analyze the given partial sequence to find a repeating or logical pattern.
abab_b_bcb_dcdcded_dabab_b_bcb_dcdcded_dabab, it suggests a pattern of two alternating letters._b. If the pattern is abab, then the missing letter could be 'a' to form ababab. Let's try to extend this.ababa b_bcb_dcdcded_d.b_bcb. This looks like bcbcb. So the second missing letter could be 'c'. The sequence becomes ababa bcbcb _dcdcded_d._dcdcded. This looks like cdcdcded. So the third missing letter could be 'c'. The sequence becomes ababa bcbcb cdcdcded_d._d. This looks like ede. But the pattern is dcdcded. So the last missing letter should complete the pattern. The sequence dcdcded is followed by _d. If we consider the pattern to be ababa, bcbcb, cdcdc, deded, then the last missing letter should be 'e' to complete deded.abab_b_bcb_dcdcded_dababa b_bcb_dcdcded_dababa bcbcb _dcdcded_dababa bcbcb cdcdcded_dababa bcbcb cdcdcded edababa bcbcb cdcdc dede d. This does not seem to be a consistent pattern.abab_b_bcb_dcdcded_dabab_ -> ababa (if the first missing is 'a')b_bcb -> bcbcb (if the second missing is 'c')_dcdc -> cdcdc (if the third missing is 'c')ded_d -> deded (if the fourth missing is 'e')ababa bcbcb cdcdc dededCorrect Option: C) a,c,c,e
ababa bbcbc cdcdcdedd. This does not form a consistent pattern of repeating blocks like ababa, bcbcb, etc.ababa bbdcd deded. This also does not form a consistent pattern.ababb bccdc deded. This does not form a consistent pattern.The problem involves two four-digit numbers, A3BC and DE2F, where A, B, C, D, E, F are distinct digits greater than 3. We are given their sum is 15902. We need to find the difference between the values of A and D. We will use column addition to deduce the values of the unknown digits.
The given sum is:
\[ \begin{array}{c} \quad \text{A}3\text{BC} \ + \quad \text{DE}2\text{F} \ \hline 15902 \end{array} \]From the units column (C + F):
\(C + F\) must end in 2. Since C and F are digits greater than 3, the minimum sum is \(4+5=9\) and the maximum is \(8+9=17\). So, \(C+F\) could be 12. This means there is a carry-over of 1 to the tens column.
\[ C + F = 12 \]From the tens column (B + 2 + carry-over 1):
\(B + 2 + 1 = B + 3\) must end in 0. Since B is a digit greater than 3, the minimum value for B is 4. So, \(B+3\) could be 10. This means \(B = 7\). There is a carry-over of 1 to the hundreds column.
\[ B + 3 = 10 \implies B = 7 \]From the hundreds column (3 + E + carry-over 1):
\(3 + E + 1 = E + 4\) must end in 9. Since E is a digit greater than 3, the minimum value for E is 4. So, \(E+4\) could be 9. This means \(E = 5\). There is no carry-over to the thousands column (or a carry-over of 0).
\[ E + 4 = 9 \implies E = 5 \]From the thousands column (A + D + carry-over 0):
\(A + D\) must be 15. A and D are distinct digits greater than 3. Also, A, B, C, D, E, F must be distinct digits. We have \(B=7\) and \(E=5\).
Possible pairs for \((A, D)\) such that \(A+D=15\) and \(A, D > 3\):
However, B is 7, so A and D cannot be 7. This eliminates \((7, 8)\) and \((8, 7)\).
So, the possible pairs for \((A, D)\) are \((6, 9)\) or \((9, 6)\).
Now let's check the distinct digit condition for C and F. We know \(C+F=12\). The digits used so far are A, D (either 6,9 or 9,6), B=7, E=5. So the digits 5, 6, 7, 9 are already used by A, B, D, E. C and F must be distinct and greater than 3, and not any of 5, 6, 7, 9.
Possible pairs for \((C, F)\) such that \(C+F=12\), \(C, F > 3\), and \(C \neq F\):
These pairs are valid as 4 and 8 are not among 5, 6, 7, 9.
We need to find the difference between A and D. In both valid cases for \((A, D)\), i.e., \((6, 9)\) or \((9, 6)\), the difference \(|A - D|\) is:
\[ |9 - 6| = 3 \]C) 3 — The difference between A and D is 3, as derived from the column addition and distinct digit constraints.
Two Statements S1 and S2 are given below followed by a Question:
S1: There are not more than two figures on any page of a 51-page book.
S2: There is at least one figure on every page.
Question:
Are there more than 100 figures in that book?
Which one of the following is correct in respect of the above Statements and the Question?
Both S1 and S2 are sufficient to answer the Question, but neither S1 alone nor S2 alone is sufficient to answer the Question.
S1 alone is sufficient to answer the Question.
S1 and S2 together are not sufficient to answer the Question.
S2 alone is sufficient to answer the Question.
The question asks whether there are more than 100 figures in a 51-page book, based on two statements. We need to determine if the given statements (S1 and S2), either individually or together, provide enough information to definitively answer the question.
Correct Option: C) S1 and S2 together are not sufficient to answer the Question.
Consider the following data:
What is the value of x in the above table?
To find the value of x in the table, we need to understand how overall averages are calculated using the averages of different groups. The table provides average marks in English and Hindi for girls and boys, along with the overall average marks in English. We need to calculate the overall average marks in Hindi.
Let the number of girls and boys be equal. Then the overall average marks in Hindi can be calculated as:
\[ \text{Overall average marks in Hindi} = \frac{\text{Average marks for girls} + \text{Average marks for boys}}{2} \]
Substituting the values:
\[ \text{Overall average marks in Hindi} = \frac{8 + 7}{2} = \frac{15}{2} = 7.5 \]
Since the table shows the overall average marks in English as 8.8, the overall average marks in Hindi should be slightly lower than this value.
Therefore, the closest option to this calculation is 7.8.
Correct Option: A) 7.8
To solve this problem, we need to calculate the number of odd days between January 12th and July 12th/15th for both a leap year and a non-leap year. The day of the week advances by the number of odd days. If the number of odd days is a multiple of 7, the day of the week remains the same.
Days in January = 31
Days remaining after Jan 12th = \(31 - 12 = 19\) days.
January: 19 days
February: 28 days
March: 31 days
April: 30 days
May: 31 days
June: 30 days
July: 12 days
Total days = \(19 + 28 + 31 + 30 + 31 + 30 + 12 = 181\) days.
Odd days = \(181 \div 7\)
\[ 181 = 7 \times 25 + 6 \]So, there are 6 odd days.
January: 19 days
February: 29 days (since it's a leap year)
March: 31 days
April: 30 days
May: 31 days
June: 30 days
July: 12 days
Total days = \(19 + 29 + 31 + 30 + 31 + 30 + 12 = 182\) days.
Odd days = \(182 \div 7\)
\[ 182 = 7 \times 26 + 0 \]So, there are 0 odd days.
From Jan 12th to July 15th, the total days would be \(181 + 3 = 184\) days.
Odd days = \(184 \div 7\)
\[ 184 = 7 \times 26 + 2 \]So, there are 2 odd days. July 15th will be Sunday + 2 days = Tuesday.
From Jan 12th to July 15th, the total days would be \(182 + 3 = 185\) days.
Odd days = \(185 \div 7\)
\[ 185 = 7 \times 26 + 3 \]So, there are 3 odd days. July 15th will be Sunday + 3 days = Wednesday.
C) 12th July is a Sunday if the year is a leap year.
As calculated in step 3, if the year is a leap year, there are 0 odd days between January 12th and July 12th. Therefore, if January 12th is a Sunday, July 12th will also be a Sunday.