Given that c is the third proportional of 15 and d. If d is the sum of first three odd natural numbers, then the value of c is:
\(\frac{29}{5}\)
\(\frac{27}{5}\)
\(\frac{26}{5}\)
\(\frac{28}{5}\)
The problem involves two main steps: first, finding the value of 'd' based on the sum of the first three odd natural numbers, and second, using 'd' to calculate 'c' which is the third proportional of 15 and 'd'.
The first three odd natural numbers are 1, 3, and 5.
The sum of these numbers is:
\[ d = 1 + 3 + 5 = 9 \]If 'c' is the third proportional of 'a' and 'b', it means that 'a', 'b', and 'c' are in continued proportion. This can be written as:
\[ \frac{a}{b} = \frac{b}{c} \]Given that 'c' is the third proportional of 15 and 'd'. Here, \(a = 15\) and \(b = d\). We found \(d = 9\).
So, the proportion is:
\[ \frac{15}{9} = \frac{9}{c} \]Cross-multiply the equation:
\[ 15 \times c = 9 \times 9 \]\[ 15c = 81 \]\[ c = \frac{81}{15} \]Both 81 and 15 are divisible by 3:
\[ c = \frac{81 \div 3}{15 \div 3} \]\[ c = \frac{27}{5} \]Correct Option: B) \(\frac{27}{5}\)
We need to deduce the relative rankings of individuals based on the given statements and then check which of the conclusions logically follow. We will represent "ranked higher than" as '>' and "ranked equal to" as '='.
A) Conclusion III follows
This is a circular seating arrangement puzzle. We need to deduce the positions of the eight girls (A, B, C, D, E, F, G, H) sitting around a circular table facing the center by following the given clues. We will start with a definite clue and build the arrangement step-by-step.
D) H and G are the immediate neighbours of D. Based on our deduction, positions 6 and 8 are occupied by G and H (in any order), and these are the immediate neighbours of D (at position 7).
A factory built in the shape of a cuboid except for its roof had its dimensions given as 140 m × 10 m × 10 m. The structure is then closed on the top with a semi-cylindrical structure to exactly cover the open top of the cuboidal structure. The volume (in m³) of the semi-cylindrical structure forming the roof is:
Take (π=\( \frac{22}{7} \))
The problem asks us to find the volume of a semi-cylindrical roof placed on top of a cuboidal structure. We need to identify the dimensions of the semi-cylinder from the given dimensions of the cuboid and then apply the formula for the volume of a semi-cylinder.
The dimensions of the cuboid are given as 140 m \(\times\) 10 m \(\times\) 10 m. Let's assume these are length (L), width (W), and height (H) respectively.
The semi-cylindrical structure covers the open top of the cuboidal structure. This means the length of the semi-cylinder will be equal to the length of the cuboid, and the diameter of the semi-cylinder will be equal to the width of the cuboid.
The radius (\(r\)) is half of the diameter.
\[ r = \frac{D}{2} = \frac{10 \text{ m}}{2} = 5 \text{ m} \]The formula for the volume of a cylinder is \(V_{\text{cylinder}} = \pi r^2 h_{\text{cylinder}}\).
\[ V_{\text{cylinder}} = \frac{22}{7} \times (5 \text{ m})^2 \times 140 \text{ m} \] \[ V_{\text{cylinder}} = \frac{22}{7} \times 25 \text{ m}^2 \times 140 \text{ m} \] \[ V_{\text{cylinder}} = 22 \times 25 \times \frac{140}{7} \text{ m}^3 \] \[ V_{\text{cylinder}} = 22 \times 25 \times 20 \text{ m}^3 \] \[ V_{\text{cylinder}} = 550 \times 20 \text{ m}^3 \] \[ V_{\text{cylinder}} = 11000 \text{ m}^3 \]Since the roof is a semi-cylinder, its volume will be half the volume of a full cylinder.
\[ V_{\text{semi-cylinder}} = \frac{1}{2} \times V_{\text{cylinder}} \] \[ V_{\text{semi-cylinder}} = \frac{1}{2} \times 11000 \text{ m}^3 \] \[ V_{\text{semi-cylinder}} = 5500 \text{ m}^3 \]Correct Option: B) 5500
This question asks about the trends in atomic properties as we move down a group in the modern periodic table. We need to consider how the addition of new electron shells affects atomic size and nuclear charge.
Correct Option: D) increase in atomic size and nuclear charge
This is a logical reasoning problem involving sequencing and deduction. We need to determine the graduation year for each of the six individuals based on the given clues and then identify who graduated in 2006.
List the years and individuals:
Place the known information:
Let's create a table:
| Year | Graduate |
|---|---|
| 2004 | Midhun (M) |
| 2005 | |
| 2006 | |
| 2007 | Komal (K) |
| 2008 | |
| 2009 |
Apply the clue: "Jayanth graduated in the year immediately before the year Navan graduated."
This means J and N are consecutive. Possible pairs of years for (J, N):
Apply the clue: "Only one engineering student graduated between Komal and Ojaswi, where Komal was Ojaswi's senior."
Update the table:
| Year | Graduate |
|---|---|
| 2004 | Midhun (M) |
| 2005 | |
| 2006 | |
| 2007 | Komal (K) |
| 2008 | |
| 2009 | Ojaswi (O) |
Re-evaluate the Jayanth-Navan clue:
Update the table:
| Year | Graduate |
|---|---|
| 2004 | Midhun (M) |
| 2005 | Jayanth (J) |
| 2006 | Navan (N) |
| 2007 | Komal (K) |
| 2008 | |
| 2009 | Ojaswi (O) |
Identify the remaining graduate and year:
Final table:
| Year | Graduate |
|---|---|
| 2004 | Midhun (M) |
| 2005 | Jayanth (J) |
| 2006 | Navan (N) |
| 2007 | Komal (K) |
| 2008 | Lasya (L) |
| 2009 | Ojaswi (O) |
Answer the question: "who graduated in 2006?"
From the final table, Navan graduated in 2006.
Correct Option: C) Navan
The problem requires us to transform the letters of the word "GRADIENT" based on specific rules: vowels are changed to the next letter in the alphabet, and consonants are changed to the preceding letter. After applying these rules, we need to identify which letter appears twice in the newly formed group of letters.
A) The letter 'F' appears twice in the transformed group of letters (F, Q, B, C, J, F, M, S). This matches our finding.
The question asks to identify the type of chemical reaction represented by the given equation. We need to analyze the reactants and products to determine how the atoms rearrange and classify the reaction based on standard chemical reaction types.
A) displacement reaction — As explained above, a more reactive element (Aluminium) displaces a less reactive element (Iron) from its compound (Iron(III) oxide). This is the definition of a displacement reaction.
Which of the following fractions is the largest?
\(\frac{5}{7}\), \(\frac{9}{14}\), \(\frac{16}{21}\), \(\frac{29}{42}\)
\(\frac{16}{21}\)
\(\frac{29}{42}\)
\(\frac{5}{7}\)
\(\frac{9}{14}\)
To compare fractions and find the largest one, we need to convert them to a common denominator. The least common multiple (LCM) of the denominators will be the most efficient common denominator.
Correct Option: A) \(\frac{16}{21}\)
This problem involves setting up equations based on the given age relationships and then solving for the difference between the ages of A and C. We will represent the present ages of A, B, and C with variables and form an equation from the given statement.
Let the present age of A be \(A\).
Let the present age of B be \(B\).
Let the present age of C be \(C\).
According to the problem statement, "The present total age of A and B is 12 years more than the present total age of B and C."
We can write this as an equation:
\[ A + B = (B + C) + 12 \]Now, we need to simplify this equation to find the relationship between A and C. Subtract B from both sides of the equation:
\[ A + B - B = B + C - B + 12 \]\[ A = C + 12 \]To find out by how many years C is younger than A, we need to calculate the difference \(A - C\). Rearrange the equation from step 6:
\[ A - C = 12 \]This means that A is 12 years older than C, or equivalently, C is 12 years younger than A.
D) 12. The calculation shows that \(A - C = 12\), which means C is 12 years younger than A.