Breakup G Strategy | Average + Percentage + Equations | Moderate
The average salary of 5 managers and 25 engineers in a company is Rs 60000. If each of the manager receive 20% hike in the salary while the salary of the engineer remains unchanged , the average salary of all 30 employees increased by 5%.The average salary, of engineer , in rs , is :
1.45000
2.54000
3.50000
4.40000
Solution & Explanation
Correct Answer: 2 (5400)
Total salary of 30 employees (5 managers and 25 engineers):
30 × 60,000 = 1,800,000
After 20% hike for managers, average salary increases by 5%, so new average salary is:
60,000 × 1.05 = 63,000
Total salary after the hike:
30 × 63,000 = 1,890,000
Before hike: 5M + 25E = 1,800,000
After hike: 6M + 25E = 1,890,000
Solve for M (manager’s salary):
(6M + 25E) – (5M + 25E) = 1,890,000 – 1,800,000
M = 90,000
Solve for E (engineer’s salary):
5 × 90,000 + 25E = 1,800,000
450,000 + 25E = 1,800,000
25E = 1,350,000 Thus E = 54,000
X maro G Strategy | Inequalities | Easy
For real values of x, the range of the function f(x) = (2x – 3)/(2x² + 4x – 6)
1.(- oo , 1/8] U [1 , oo)
2.(-oo , 1/4] U [1/2 , oo)
3.(-oo , 1/8] U [1/2 , oo)
4.(-oo , 1/4] U [1 , oo)
Solution & Explanation
Correct Answer: 3
We can factor the denominator:
2x² + 4x – 6 = 2(x + 3)(x – 1)
So, the function becomes:
f(x) = (2x – 3) / (2(x + 3)(x – 1)) , thus x ≠ -3 , 1
Range of f(x)
f(0) = 1/2
f(1) = not possible
f(2) = 1/10
f(3) = 1/8
Only option 3 includes f(0) and f(3)
Thus option 3 is the answer
Concepts of Algebra | Moderate
f(x) = (x²+ 3x)(x² + 3x + 2), then the sum of all real roots of the equation √f(x) + 1 = 9701, is
1.3
2.6
3.-3
4.-6
Solution & Explanation
Correct Answer: 3
Assume x² + 3x = a , so x² + 3x + 2 = a + 2
Fx = a.(a + 2) = a² + 2a
√fx + 1 = √a² + 2a + 1 = 9701
√(a+1)² = 9701 , a = 9700
Thus x² + 3x = 9700 , x² + 3x – 9700 = 0
Sum of roots = -b/a = -3/1 = -3
Thus option 3 is the answer
Square ±2 G Strategy | Easy
If (x²+ 1/x²) = 25 and x > 0, then the value of (x⁷ + 1/x⁷) is:
1.44856√3
2.44853√3
3.44859√3
4.44850√3
Solution & Explanation
Correct Answer: 44853√3
Given x² + 1/x² = 25 , x > 0
Using identity (a + b)² = a² + b² + 2ab
I. x + 1/x:
(x + 1/x)² = x² + 2 + 1/x² = 25 + 2 = 27
So, x + 1/x = 3√3
II. x³ + 1/x³:
x³ + 1/x³ = (x + 1/x) * (x² + 1/x²) – (x + 1/x)
x³ + 1/x³ = (3√3) * 25 – 3√3 = 72√3
III.x⁴ + 1/x⁴:
x⁴ + 1/x⁴ = (x² + 1/x²)² – 2
x⁴ + 1/x⁴ = 25² – 2 = 625 – 2 = 623
IV.x⁷ + 1/x⁷:
x⁷ + 1/x⁷ = (x³ + 1/x³) * (x⁴ + 1/x⁴) – (x + 1/x)
x⁷ + 1/x⁷ = (72√3) * 623 – 3√3 = 44853√3
Visual lens G Strategy | Set theory + Maximum/Minimum | Moderate
In a class of 150 students, 75 students chose physics, 111 students chose mathematics and 40 students chose chemistry. All students chose at least one of the three subjects and at least one student chose all three subjects. The number of students who chose both physics and chemistry is equal to the number of students who chose both chemistry and mathematics, and this is equal to half the number of students who chose both mathematics and physics. What is the maximum possible number of students who chose physics but not mathematics?
1.40
2.35
3.55
4.30
Solution & Explanation
Given Information: Physics students = 75, Mathematics students = 111 , Chemistry students = 40
Assume: x – number of students who chose both Physics and Mathematics , y – students who chose both Mathematics and Chemistry , z – student who chose both Physics and Chemistry , t – number of students who chose all three subjects
And z = y , y = x/2 (Given)
Using Set formula (the total number of students who chose at least one subject)
|P ∪ M ∪ C| = |P| + |M| + |C| − |P ∩ M| − |M ∩ C| − |P ∩ C| + |P ∩ M ∩ C|
Substitute the values:
150 = 75 + 111 + 40 – x – y – z + t
Now, substitute z = y and y = x/2:
150 = 75 + 111 + 40 – x – (x/2) – (x/2) + t
150 = 226 – x – x + t
150 = 226 – 2x + t
2x – 76 = t (Equation 1)
We want to maximize 75 – x , (students who chose only Physics but not mathematics). Using options :
1.40 , 75 – x = 40 , x = 35 , t can not be negative
2.35 , 75 – x = 35 , x = 40 , t = 4
3.55 , 75 – x = 55 , x= 20 , t can not be negative
4.30 , 75 – x = 30 , x = 45 , t = 14
Since 75 – x is maximum at option 2 thus option 2 is the answer
Breakup G Strategy | Linear Equation + Maximum/Minimum | Moderate
In a school with 1500 students , each students chooses one of the streams out of science , arts and commerce by paying a fee of rs 1100 , 1000 and 800 respectively. The total fees paid by all the students is RS. 15,50000.If the number of science students is not more than the number of arts students , then the maximum possible number of science students in school is : [TITA] _______
Answer & Explanation
Given S – science students , A – arts students , C – commerce students
S + A + C = 1500 —- I
1100S + 1000A + 800C = 1550000 becomes
11S + 10A + 8C = 15500 ——- II
Where S ≤ A , since we have to maximise S this S = A
Multiplying I by 8 :
8S + 8A + 8C = 12000 —— III
Subtracting III from I
3S + 2A = 3500 , Since S = A
5S = 3500 , S = 700
Answer : 700
