The average salary of 5 managers and 25 engineers in a company is 60000 rupees. If each of the managers received 20% salary increase while the salary of the engineers remained unchanged, the average salary of all 30 employees would have increased by 5%. The average salary, in rupees, of the engineers is
Show Hint
Set up equations using total sums (average $\times$ number of people).
Apply the percentage increase only to the relevant group.
Use the new average to form a second equation and solve the system.
- \(40000\)
- \(54000\)
- \(50000\)
- \(45000\)
The Correct Option is B
Approach Solution - 1
Let's denote the average salary of the 25 engineers as \( x \) rupees. Therefore, the total salary of the engineers is \( 25x \).
We are given that the average salary of all 30 employees (both managers and engineers) is 60000 rupees. Thus, the total salary of all employees is:
\[ 30 \times 60000 = 1800000 \text{ rupees} \]
The total salary of the managers is a part of this total. Let's denote the original average salary of the 5 managers as \( y \). Thus their total original salary is:
\[ 5y \]
Therefore, the equation representing the sum of the salaries for the managers and engineers is:
\[ 5y + 25x = 1800000 \]
After a 20% increase in the manager's salary, each manager gets \( 1.2y \). Thus, their total new salary becomes:
\[ 5 \times 1.2y = 6y \]
The new average salary of all employees increases by 5%, making the new total salary:
\[ 1800000 \times 1.05 = 1890000 \text{ rupees} \]
The new equation, after the salary increment, is:
\[ 6y + 25x = 1890000 \]
We now have two equations:
\[ \begin{align*} 5y + 25x &= 1800000 \\ 6y + 25x &= 1890000 \end{align*} \]
Subtract the first equation from the second:
\[ (6y + 25x) - (5y + 25x) = 1890000 - 1800000 \]
Leading to:
\[ y = 90000 \]
Substitute \( y = 90000 \) back into the first equation:
\[ 5 \times 90000 + 25x = 1800000 \]
Solving that gives:
\[ 450000 + 25x = 1800000 \]
\[ 25x = 1800000 - 450000 = 1350000 \]
\[ x = \frac{1350000}{25} = 54000 \]
Therefore, the average salary of the engineers is \(\boxed{54000}\) rupees.
Approach Solution -2
Step 1: Use the initial average. There are 5 managers and 25 engineers, total 30 employees. \[ \frac{5M + 25E}{30} = 60000 \quad \Rightarrow \quad 5M + 25E = 60000 \times 30 = 1800000. \tag{1} \]
Step 2: Use the new average after managers get 20% hike. Each manager’s new salary: \[ 1.2M. \] Engineers’ salary remains \(E\). New overall average increases by 5%: \[ 60000 \times 1.05 = 63000. \] So, \[ \frac{5(1.2M) + 25E}{30} = 63000 \quad \Rightarrow \quad 6M + 25E = 63000 \times 30 = 1890000. \tag{2} \]
Step 3: Solve the system for \(M\) and \(E\). Subtract (1) from (2): \[ (6M + 25E) - (5M + 25E) = 1890000 - 1800000 \] \[ M = 90000. \] Substitute \(M = 90000\) into (1): \[ 5(90000) + 25E = 1800000 \Rightarrow 450000 + 25E = 1800000 \Rightarrow 25E = 1350000 \Rightarrow E = 54000. \] Thus, the average salary of the engineers is: \[ \boxed{54000}. \]
Top Questions on Averages
- The ages of a group of people are: 34, 27, 36, 18, 15, 72, 67, 12, 24, 45. What percentage of people in the group are above the average age for the group?
- A class of 25 students took an English test. 15 students had an average score of 80. The other students had an average score of 60. What is the average score of the whole class?
- The average of the first 7 numbers in a series is 60. When the 8th number is added, the average of the first 8 numbers becomes 63. The 9th number is 11 more than the 8th number. It is also given that the average of the 2nd to the 9th numbers is 66. Find the value of the 1st number in the series.
- There are three groups of students M1, M2, and M3, having 20, 30, and 40 students respectively. If one student each from M1 and M3 leave their respective groups and join M2, the average weight of M1 and M3 decreases by 2 kg each and the average weight of M2 increases by 3 kg. By how much (in kg) does twice the average weight of M2 differ from the sum of the average weights of M1 and M3?
- A desert outpost has a water supply that is sufficient to last 21 days for 15 people. At the same average rate of water consumption per person, how many days would the water supply last for 9 people?
Questions Asked in CAT exam
- The passage given below is followed by four summaries. Choose the option that best captures the essence of the passage.
In the dynamic realm of creativity, artists often find themselves at the crossroads between drawing inspiration from diverse cultures and inadvertently crossing into the territory of cultural appropriation. Inspiration is the lifeblood of creativity, driving artists to create works that resonate across borders. In a globalized era of the modern world, artists draw from a vast array of cultural influences. When approached respectfully, inspiration becomes a bridge, fostering understanding and appreciation of cultural diversity. However, the line between inspiration and cultural appropriation can be thin and easily blurred.
Cultural appropriation occurs when elements from a particular culture are borrowed without proper understanding, respect, or acknowledgment. This leads to the commodification of sacred symbols, the reinforcement of stereotypes, and the erasure of the cultural context from which these elements originated. It is essential to recognize that the impact of cultural appropriation extends beyond the realm of artistic expression, influencing societal perceptions and perpetuating power imbalances.- CAT - 2025
- Para Summary
- The number of distinct integers $n$ for which $\log_{\left(\frac14\right)}(n^2 - 7n + 11)>0$ is:
- CAT - 2025
- Linear Inequalities
- In the set of consecutive odd numbers $\{1, 3, 5, \ldots, 57\}$, there is a number $k$ such that the sum of all the elements less than $k$ is equal to the sum of all the elements greater than $k$. Then, $k$ equals?
- CAT - 2025
- Number Systems
- The number of distinct pairs of integers $(x, y)$ satisfying the inequalities $x>y \ge 3$ and $x + y<14$ is:
- CAT - 2025
- Number Systems
For any natural number $k$, let $a_k = 3^k$. The smallest natural number $m$ for which \[ (a_1)^1 \times (a_2)^2 \times \dots \times (a_{20})^{20} \;<\; a_{21} \times a_{22} \times \dots \times a_{20+m} \] is:
- CAT - 2025
- Linear Inequalities