To solve this problem, we need to understand the concept of average. The average is calculated by summing all the values and dividing by the number of values. We are given the marks in four subjects and the desired average across five subjects. We can set up an equation to find the unknown mark in the fifth subject.
First, calculate the sum of the marks obtained in the four subjects:
\[ \text{Sum of 4 subjects} = 78 + 85 + 69 + 90 \] \[ \text{Sum of 4 subjects} = 322 \]Next, determine the total sum of marks required to achieve an average of 80 across five subjects. The formula for average is:
\[ \text{Average} = \frac{\text{Total Sum of Marks}}{\text{Number of Subjects}} \]We want the average to be 80 and the number of subjects to be 5. So,
\[ 80 = \frac{\text{Total Sum of Marks}}{5} \]Multiply both sides by 5 to find the required total sum:
\[ \text{Required Total Sum of Marks} = 80 \times 5 \] \[ \text{Required Total Sum of Marks} = 400 \]Finally, to find the score needed in the fifth subject, subtract the sum of the first four subjects from the required total sum:
\[ \text{Score in 5th subject} = \text{Required Total Sum of Marks} - \text{Sum of 4 subjects} \] \[ \text{Score in 5th subject} = 400 - 322 \] \[ \text{Score in 5th subject} = 78 \]Therefore, the student needs to score 78 marks in the fifth subject to achieve an average of 80 across all five subjects.
C) 78 — This is the score required in the fifth subject to bring the overall average to 80.
A) 88 — If the student scored 88 in the fifth subject, the total sum would be \(322 + 88 = 410\), and the average would be \(410 / 5 = 82\), which is higher than the desired average of 80.
B) 94 — If the student scored 94 in the fifth subject, the total sum would be \(322 + 94 = 416\), and the average would be \(416 / 5 = 83.2\), which is also higher than the desired average of 80.
D) 82 — If the student scored 82 in the fifth subject, the total sum would be \(322 + 82 = 404\), and the average would be \(404 / 5 = 80.8\), which is higher than the desired average of 80.
Find the median from the following data.
| Age in years | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | 80-90 |
| No. of patients | 12 | 16 | 30 | 20 | 17 | 38 | 15 |
To find the median for grouped data, we first need to calculate the cumulative frequencies. Then, we determine the median class using the formula \(N/2\), where \(N\) is the total number of observations. Finally, we apply the median formula for grouped data.
Construct the cumulative frequency table:
| Age in years (Class Interval) | No. of patients (Frequency, \(f\)) | Cumulative Frequency (\(cf\)) |
|---|---|---|
| 20-30 | 12 | 12 |
| 30-40 | 16 | 12 + 16 = 28 |
| 40-50 | 30 | 28 + 30 = 58 |
| 50-60 | 20 | 58 + 20 = 78 |
| 60-70 | 17 | 78 + 17 = 95 |
| 70-80 | 38 | 95 + 38 = 133 |
| 80-90 | 15 | 133 + 15 = 148 |
Calculate \(N/2\):
The total number of patients, \(N = \sum f = 148\).
Identify the median class:
The median class is the class interval whose cumulative frequency is just greater than or equal to \(N/2\). Here, \(N/2 = 74\). The cumulative frequency just greater than 74 is 78, which corresponds to the class interval 50-60. Therefore, the median class is 50-60.
Identify the values for the median formula:
Apply the median formula for grouped data:
\[ \text{Median} = L + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h \] \[ \text{Median} = 50 + \left( \frac{74 - 58}{20} \right) \times 10 \] \[ \text{Median} = 50 + \left( \frac{16}{20} \right) \times 10 \] \[ \text{Median} = 50 + \left( \frac{4}{5} \right) \times 10 \] \[ \text{Median} = 50 + 4 \times 2 \] \[ \text{Median} = 50 + 8 \] \[ \text{Median} = 58 \text{ years} \]Correct Option: C) 58 years
We need to find the length of the longer side of a rectangle given two conditions: the relationship between its sides and the relationship between its area and perimeter. We will set up equations based on these conditions and solve them simultaneously.
Let the length of the longer side of the rectangle be \(L\) cm and the length of the shorter side be \(W\) cm.
According to the first condition, "The shorter side of a rectangle is 21 cm less than the longer side."
\[ W = L - 21 \quad \text{(Equation 1)} \]The area of the rectangle is given by:
\[ \text{Area} = L \times W \]The perimeter of the rectangle is given by:
\[ \text{Perimeter} = 2(L + W) \]According to the second condition, "The numerical value of its area is equal to 7 times the numerical value of its perimeter."
\[ \text{Area} = 7 \times \text{Perimeter} \]\[ L \times W = 7 \times 2(L + W) \]\[ LW = 14(L + W) \quad \text{(Equation 2)} \]Now, substitute Equation 1 into Equation 2:
\[ L(L - 21) = 14(L + (L - 21)) \]\[ L^2 - 21L = 14(2L - 21) \]\[ L^2 - 21L = 28L - 294 \]Rearrange the terms to form a quadratic equation:
\[ L^2 - 21L - 28L + 294 = 0 \]\[ L^2 - 49L + 294 = 0 \]Solve the quadratic equation for \(L\). We can use factorization. We need two numbers that multiply to 294 and add up to -49. These numbers are -42 and -7.
\[ L^2 - 42L - 7L + 294 = 0 \]\[ L(L - 42) - 7(L - 42) = 0 \]\[ (L - 42)(L - 7) = 0 \]This gives two possible values for \(L\):
\[ L = 42 \quad \text{or} \quad L = 7 \]If \(L = 7\) cm, then from Equation 1, \(W = L - 21 = 7 - 21 = -14\) cm. A side length cannot be negative, so \(L = 7\) is not a valid solution.
If \(L = 42\) cm, then from Equation 1, \(W = L - 21 = 42 - 21 = 21\) cm. Both length and width are positive, and \(L > W\), which is consistent with \(L\) being the longer side. This is a valid solution.
Therefore, the length of the longer side is 42 cm.
D) 42
The volume and height of a cylinder are 8470 cm3 and 2.2 cm, respectively. Find the radius (in cm) of the cylinder. (Use π = \( \frac{22}{7} \))
To find the radius of the cylinder, we will use the formula for the volume of a cylinder, which is \(V = \pi r^2 h\). We are given the volume (\(V\)) and the height (\(h\)), and the value of \(\pi\). We can rearrange the formula to solve for the radius (\(r\)).
Correct Option: A) 35
The radius of the cylinder is 35 cm.
The problem requires us to arrange seven people (P, Q, R, S, T, U, V) for exams on seven different days of a week, from Monday to Sunday, based on given clues. We will use a table to represent the days of the week and fill in the names as we deduce their positions.
| Day | Person |
|---|---|
| Monday | |
| Tuesday | |
| Wednesday | |
| Thursday | |
| Friday | |
| Saturday | |
| Sunday | Q |
| Day | Person |
|---|---|
| Monday | U |
| Tuesday | |
| Wednesday | |
| Thursday | |
| Friday | |
| Saturday | P |
| Sunday | Q |
| Day | Person |
|---|---|
| Monday | P |
| Tuesday | |
| Wednesday | |
| Thursday | |
| Friday | |
| Saturday | U |
| Sunday | Q |
| Day | Person |
|---|---|
| Monday | P |
| Tuesday | T |
| Wednesday | |
| Thursday | |
| Friday | |
| Saturday | U |
| Sunday | Q |
| Day | Person |
|---|---|
| Monday | P |
| Tuesday | T |
| Wednesday | V |
| Thursday | R |
| Friday | S |
| Saturday | U |
| Sunday | Q |
A) One. As determined in the step-by-step solution, T is the only person between P (Monday) and V (Wednesday).
The problem involves calculating the individual investment of Anjali in a partnership. We are given the profit earned, which is a percentage of the total investment, and the ratio of investments among the partners. We will first find the total investment, then use the investment ratio to determine Anjali's share.
Calculate the Total Investment: We are given that the profit of Rs. 28,800 is 25% of the total investment.
\[ \text{Profit} = 25\% \text{ of Total Investment} \] \[ \text{Rs. } 28,800 = 0.25 \times \text{Total Investment} \] \[ \text{Total Investment} = \frac{28,800}{0.25} \] \[ \text{Total Investment} = 28,800 \times 4 \] \[ \text{Total Investment} = \text{Rs. } 1,15,200 \]Determine Anjali's Investment: The investment ratio of Vikram, Anjali, and Raj is 7 : 3 : 20.
The sum of the ratios is:
\[ 7 + 3 + 20 = 30 \]Anjali's share in the investment is 3 parts out of a total of 30 parts. Therefore, Anjali's investment is:
\[ \text{Anjali's Investment} = \frac{\text{Anjali's Ratio Share}}{\text{Total Ratio Share}} \times \text{Total Investment} \] \[ \text{Anjali's Investment} = \frac{3}{30} \times 1,15,200 \] \[ \text{Anjali's Investment} = \frac{1}{10} \times 1,15,200 \] \[ \text{Anjali's Investment} = \text{Rs. } 11,520 \]Correct Option: C) Rs.11,520
The question asks to identify the hill station located in the Nilgiri Hills. This requires knowledge of the geographical locations of prominent Indian hill stations.
Correct Option: B) Ooty is a well-known hill station located in the Nilgiri Hills, a mountain range in the Western Ghats of Tamil Nadu, India.
The problem asks us to identify the number of symbols in the given series that are immediately preceded by a number and immediately followed by another number. We need to scan the series from left to right and look for the pattern: Number - Symbol - Number.
D) More than three
There are 5 such symbols in the series that are immediately preceded by a number and immediately followed by another number: @ (in 5 @ 6), # (in 6 # 8), ^ (in 9 ^ 7), < (in 7 < 2), and > (in 3 > 4). Since 5 is greater than 3, the correct option is "More than three".
The problem asks us to identify numbers in a given series that are immediately preceded by a symbol and immediately followed by a symbol. We need to scan the series from left to right and apply this condition to each number.
D) 1 is the correct answer because there is only one number (8) in the series that is immediately preceded by a symbol (£) and immediately followed by a symbol (%).
A group of numbers and symbols is coded using letter codes as per the codes given below and the conditions that follow. Study the given codes and conditions and answer the question that follows.
Note: If none of the conditions apply, then codes for the respective number/symbol are to be followed directly as given in the table.
| Number/symbol | 6 | ÷ | 7 | % | @ | + | 8 | 4 | - | 2 | # | 3 | 9 | & |
| Code | P | E | S | H | W | Q | G | T | I | M | Z | A | V | R |
Conditions:
If the first element is a prime number and the last element is an even number, then the codes for these two (first and last elements) are to be interchanged.
If the third element is a symbol, then it is to be coded as alpha.
If both the second and the fourth elements are perfect numbers, then the fourth number is to be coded as the code for the second element. What will be the code for the following group?
3 ÷ 7 @ 4
The problem requires us to code a given group of numbers and symbols based on a provided code table and a set of conditions. We need to apply the conditions sequentially and then use the direct codes if no conditions apply to a specific element.
D) TESWA — Upon re-checking the question and the provided options, there seems to be a mismatch between my derived answer and the options. Let's re-evaluate the direct codes carefully from the table.
Number/symbol: 6 ÷ 7 % @ + 8 4 - 2 # 3 9 &
Code: P E S H W Q G T I M Z A V R
Let's re-map the codes for \(3 \div 7 @ 4\):
The group is \(3 \div 7 @ 4\).
So, the final codes are:
\(3 \rightarrow R\)
\(\div \rightarrow E\)
\(7 \rightarrow S\)
\(@ \rightarrow W\)
\(4 \rightarrow A\)
The combined code is \(R E S W A\).
None of the options match \(RESWA\). Let's re-examine the question and options provided. It's possible there's a typo in the question's provided correct answer or the options themselves.
However, if we assume there might be a misunderstanding of the table or a typo in the question's intended answer, let's look at the options and try to work backwards or see if any other interpretation fits.
Let's assume the provided correct answer D) TESWA is correct and try to deduce why.
If the answer is TESWA:
This implies that the first and last elements' codes were interchanged, which is consistent with Condition 1. The original code for \(3\) is \(A\) and for \(4\) is \(R\). If they are interchanged, \(3\) gets \(R\) and \(4\) gets \(A\). So, the first element should be \(R\), not \(T\).
Let's re-read the code table very carefully:
Number/symbol: 6 ÷ 7 % @ + 8 4 - 2 # 3 9 &
Code: P E S H W Q G T I M Z A V R
Mapping:
Now, let's re-apply the steps with the correct mapping for \(4\).