We need to solve:

   1/(x + 5) ≤ 1/(2x − 3)

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Step 1: Identify restrictions
Denominators ≠ 0
x ≠ −5,  x ≠ 3/2

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Step 2: Bring to common denominator
1/(x + 5) − 1/(2x − 3) ≤ 0

⇒ [ (2x − 3) − (x + 5) ] / [ (x + 5)(2x − 3) ] ≤ 0

⇒ (x − 8) / [ (x + 5)(2x − 3) ] ≤ 0

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Step 3: Critical points
x − 8 = 0   ⇒  x = 8  
x + 5 = 0   ⇒  x = −5  
2x − 3 = 0  ⇒  x = 3/2  

So, critical points: −5, 3/2, 8

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Step 4: Check intervals
Intervals are: (−∞, −5), (−5, 3/2), (3/2, 8), (8, ∞)

• For (−∞, −5): numerator (x−8) < 0, denominator (x+5)(2x−3) > 0 → fraction < 0 (valid)  
• For (−5, 3/2): numerator (x−8) < 0, denominator (x+5)(2x−3) < 0 → fraction > 0 (not valid)  
• For (3/2, 8): numerator (x−8) < 0, denominator (x+5)(2x−3) > 0 → fraction < 0 (valid)  
• For (8, ∞): numerator (x−8) > 0, denominator (x+5)(2x−3) > 0 → fraction > 0 (not valid)

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Step 5: Include equality
At x = 8 → numerator = 0 ⇒ inequality holds (included)

At x = −5, 3/2 → denominators = 0 ⇒ excluded

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Final Solution:
x < −5   or   3/2 < x ≤ 8

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Answer: Option A

1/A + 1/V= 1/ 150 ------------(1)
1/V + 1/S= 1/ 100 ------------(2)
75/A + 135/V+ 45/S= 1 ----(3)
Subtracting eqn (1) from eqn(2),
1/S – 1/A= 1/300 -----(4)
Multiplying eqn (1) with 135, we get 135/A + 135/V= 135/150= 9/10 --------(5)
Subracting eqn (5) from eqn (3), -60/A + 45/S= 1/10-----(6)
Solving, eqn (4) and eqn (6), we get 1/S= 1/150, 1/A= 1/300 . Substituting the value of 1/S in eqn (2) we get 1/V = 1/300.
Anil takes 300 days to finish the task.
Vimal takes 300 days to finish the task.
Sunil takes 150 days to finish the task.
Let total work= 300 units
Anil does 1 unit/day.
Vimal does 1 unit/day.
Sunil does 2 units/day.
If Anil works for 6k days, Vimal works for 3k days and Sunil works for 2k days.
If 6k= 138 days.
Anil would have finished 138 units, Vimal would have completed 69 units and Sunil would have finished 46*2= 92 units.
138+69+92= 299 units would be completed.
In 139th day, only Anil works and therefore (299+1= 300 units) work is finished.
Total work got completed in 139 days.

Let 'a' be sum of bonuses of 1 st 10 employees, ' b ' be the sum of bonuses of 2nd 10 employees, ' c ' be the sum of bonuses of 3rd 10 employees and ' d ' be the sum of bonuses of last 10 employees.

Adding the three equations we get,

In 2023, the average bonus of the first 10 employees increased by 100%.

In 2023, the average bonus of the last 10 employees increased by 200%.

Average in the year 2023 =(400k+1000k+2400k/40)=3800k/40=380k/4=95000

We can take AD to be x and CB to be 3-x; since the perimeter of ABCD is 6, and AB and CD are given as 2 and 1
cm, respectively, that leaves AD+BC as three only.
We can see that triangles AEB and DEC are similar, with the lengths of AB being double of CD, essentially
making D the mid-point of AE and C the mid-point of EB.
Through this, we get the length of DE and CE to be x and 3-x.
Give us the AE and BE lengths as 2x and 6-2x, respectively.
Giving the perimeter of AEB as 2x+6-2x+2 = 8 cm
Therefore, Option A is the correct answer.

Given: AB = BC = AC = a = 2r  
OB = 2r / √3  

Let x and y be the radii of the bigger and smaller circles.

We have:  
r + y = (2r / √3)  
2r + y = x  

So,  
2r + (2r / √3) − r = x  
r + (2r / √3) = x  

Also,  
(2r / √3) − r = y  

Thus,  
x / y = [ r + (2r / √3) ] / [ (2r / √3) − r ]  

= { (2/√3) + 1 } / { (2/√3) − 1 }  

= (2 + √3) / (2 − √3)  

= (2 + √3)(2 + √3) / (2 − √3)(2 + √3)  

= (4 + 3 + 4√3) / (4 − 3)  

= (7 + 4√3) : 1  

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The question is:  
"Three circles of equal radii touch (but not cross) each other externally.  
Two other circles, X and Y, are drawn such that both touch (but not cross) each of the three previous circles.  
If the radius of X is more than that of Y, the ratio of the radii of X and Y is ?"

Final Answer:  
(7 + 4√3) : 1

Let's take Rajesh and Garima's ages to be R and G, respectively.

From the given ratio, we can see that Rajesh is older than Garima, 
so let's take R = G + x.

When Rajesh was of age G, which was x years ago, 
Garima was of (G − x) years old.

So the ratio is:  
(G + x − x) : (G − x) = G : (G − x)

This gives G = 3x, which in turn gives R = 4x.

We are asked the ratio when Garima becomes 4x years old.  
At that time, Rajesh will be 5x years old.

Therefore, their ratio is:  

5x : 4x = 5 : 4

The roots α, β of the equation 3x² + λx − 1 = 0.

Sum of the roots: α + β = −λ/3  
Product of the roots: αβ = −1/3  

Given: 1/α² + 1/β² = 15  

=> (α² + β²) / (αβ)² = 15  

Now,  
α² + β² = (α + β)² − 2αβ  
          = (λ² / 9) − 2(−1/3)  
          = (λ² / 9) + (2/3)  
          = (λ² + 6) / 9  

Substitute into the condition:  
(λ² + 6) / (9 × 1/9) = 15  
=> λ² + 6 = 15  
=> λ² = 9  
=> λ = +3 or −3  

Now,  
α³ + β³ = (α + β)³ − 3(αβ)(α + β)  
         = (−λ³ / 27) − 3(−1/3)(−λ/3)  
         = (−λ³ / 27) − (λ / 3)  
         = −(λ³ / 27 + λ / 3)  

Case 1: λ = +3  
α³ + β³ = −(27/27 + 3/3) = −(1 + 1) = −2  
(α³ + β³)² = (−2)² = 4  

Case 2: λ = −3  
α³ + β³ = −(−27/27 + (−3)/3) = −(−1 − 1) = 2  
(α³ + β³)² = (2)² = 4  

Final Answer:  
(α³ + β³)² = 4

Looking at the additional information about the prime numbers should make one realise 
that they are the key to solving the question.

f(16000) can be written as

Now, we can try to find these individual values:

For any prime p: f(p) = 1

This way, we can find the function output for any prime number raised to a power.

We can see that each new exponent is twice the previous output + 1, solving this way 
till prime raised to power 8.

f(p^2) = f(p) f(p) + f(p) + f(p) = 1 + 1 + 1 = 3  
f(p^3) = f(p^2) f(p) + f(p^2) + f(p) = 3 + 3 + 1 = 7  
f(p^4) = 7 + 7 + 1 = 15  
f(p^5) = 15 + 15 + 1 = 31  
f(p^6) = 31 + 31 + 1 = 63  
f(p^7) = 63 + 63 + 1 = 127  
f(p^8) = 127 + 127 + 1 = 255  

So, f(2^8 × 5^4) = (255 × 15) + 255 + 15 = 4095

Let the total number of fruits= 5x.
Therefore, the number of mangoes in the stock= (40/100)*5x= 2x.
Let the number of apples= 5y.
He sold x mangoes, 96 bananas and 2y apples which is equal to 50% of the total fruits.
x+ 96+ 2y= 2.5x
2x+192+4y= 5x
192+ 4y= 3x
The smallest value of y which will satisfy the equation is y= 3.
192+ 12= 204= 3x
x= 68
The smallest possible total number of fruits in the stock at the beginning of the day is 5x= 5*68= 340.

The question is "A fruit seller has a stock of mangoes, bananas and apples with at least one fruit of each type. At the beginning of a day, the number of mangoes make up 40% of his stock. That day, he sells half of the mangoes, 96 bananas and 40% of the apples. At the end of the day, he ends up selling 50% of the fruits. The smallest possible total number of fruits in the stock at the beginning of the day is"

Hence, the answer is '340'