We are asked to find the number of factors of 1800 that are multiples of 6.

Step 1: Prime factorisation of 1800

1800 = 18 × 100

18 = 2 × 3²
100 = 2² × 5²

So,

1800 = 2³ × 3² × 5²

Step 2: Condition for being a multiple of 6

6 = 2 × 3

So any factor of 1800 that is a multiple of 6 must contain:

• at least one 2

• at least one 3

Step 3: Count valid exponents

For a general factor of 1800:

2ᵃ × 3ᵇ × 5ᶜ

Where:

• a = 0 to 3

• b = 0 to 2

• c = 0 to 2

But for multiples of 6:

• a ≥ 1 → a = 1, 2, 3 → 3 choices

• b ≥ 1 → b = 1, 2 → 2 choices

• c can be anything → c = 0, 1, 2 → 3 choices

Step 4: Total number of such factors

Total = 3 × 2 × 3 = 18

Final Answer

18

We focus only on Company A and Company B, because the ages of A are strictly less than the ages of B, and the question asks for the highest possible age in A.

Step 1: Understand the median of Company A

Company A has 32 employees (even number).

Median age = 24

For an even number of observations, the median is the average of the 16th and 17th values.

So:

• 16th age = 24

• 17th age = 24

This means:

• First 16 employees have age ≤ 24

• Last 16 employees have age ≥ 24

So, the maximum age in A can be greater than 24, but it will be restricted by Company B.

Step 2: Understand the median of Company B

Company B has 28 employees.
Median age = 30

So:

• 14th and 15th observations determine the median

• To keep the median exactly 30 (with integer ages), the safest configuration is:

  • 14th age = 30

  • 15th age = 30

This ensures the median is 30.

Hence, the minimum possible age in Company B is 30.

Step 3: Use the strict inequality condition

It is given that:

• Every employee in A is strictly younger than every employee in B

So:

Maximum age in A<Minimum age in B\text{Maximum age in A} < \text{Minimum age in B}Maximum age in A<Minimum age in B

Minimum age in B = 30
Therefore:

Maximum age in A=29\text{Maximum age in A} = 29Maximum age in A=29

FINAL ANSWER

29

Given:

• Total students = 150

• Like tea = 52

• Like juice = 48

• Like coffee = 62

• Every student likes at least one drink

Step 1: Add total likes

Total likes = 52 + 48 + 62 = 162

If everyone liked exactly one drink, total likes would be 150.

But actual likes are 162.

Extra likes = 162 − 150 = 12

These extra likes come only because some students like more than one drink.

Step 2: Maximise number of students who like more than one drink

To maximize the number of students who like more than one drink:

• Each such student should contribute only one extra like

• So, each of them should like exactly two drinks

• No one should like all three (because that would use up extra likes without increasing the count of students)

Thus:

• Each student liking more than one drink accounts for 1 extra like

So,
Maximum number of students liking more than one drink = 12

Final Answer
12

Given:

• Oranges : Apples : Bananas = 3 : 6 : 7

• Number of oranges is a multiple of both 5 and 6

• We need the minimum total number of fruits

Step 1: Let the common multiplier be kkk

Then:

• Oranges = 3k3k3k

• Apples = 6k6k6k

• Bananas = 7k7k7k

Step 2: Apply the condition on oranges

Oranges = 3k3k3k must be a multiple of both 5 and 6.

LCM of 5 and 6 = 30

So,

3k=30⇒k=103k = 30 \Rightarrow k = 103k=30⇒k=10

(This is the smallest possible value of kkk)

Step 3: Find the number of each fruit

• Oranges = 3×10=303 \times 10 = 303×10=30

• Apples = 6×10=606 \times 10 = 606×10=60

• Bananas = 7×10=707 \times 10 = 707×10=70

Step 4: Total number of fruits

30+60+70=16030 + 60 + 70 = \boxed{160}30+60+70=160​

Final Answer
160

We are given the inequality:

|x − 5| + |y − 5| ≤ 6

This represents all integer points inside and on a diamond (rhombus) centered at (5, 5) in the coordinate plane.

Step 1: Shift the origin

Let:

• X = x − 5

• Y = y − 5

Then the inequality becomes:

|X| + |Y| ≤ 6

Now we just need to count integer solutions (X, Y) satisfying this.

Step 2: Count integer solutions

For |X| + |Y| ≤ n, the number of integer solutions is given by:

Number = 1 + 4(1 + 2 + … + n)
= 1 + 4 × (n(n + 1)/2)

Here, n = 6.

So:
1 + 4 × (6 × 7 / 2)
= 1 + 4 × 21
= 1 + 84
= 85

Step 3: Interpretation

Each solution (X, Y) corresponds to exactly one solution (x, y), since the shift is one-to-one.

Final Answer

85

Let the original price of the chocolate be 100.

Step 1: Increase by x%

After an increase of x%, the price becomes:

100 × (1 + x/100)

Step 2: Reduce by x%

Now this new price is reduced by x%, so the final price becomes:

100 × (1 + x/100) × (1 − x/100)

Step 3: Use the given condition

The final price is given as 96.76% of the original price, i.e.,

100 × (1 + x/100)(1 − x/100) = 96.76

Divide both sides by 100:

(1 + x/100)(1 − x/100) = 0.9676

Step 4: Simplify

(1 − (x/100)²) = 0.9676

So,

(x/100)² = 1 − 0.9676
(x/100)² = 0.0324

Step 5: Solve for x

x/100 = 0.18
x = 18

Final Answer
18

We are asked to find the median age across all employees of the eight companies, given:

• Ages are integers

• Employees in A < B < C < D < E < F < G < H (strictly increasing company-wise)

• Number of employees and median age of each company are given

Step 1: Total number of employees

A = 32
B = 28
C = 43
D = 39
E = 35
F = 29
G = 23
H = 16

Total =
32 + 28 + 43 + 39 + 35 + 29 + 23 + 16
= 245 employees

Step 2: Position of overall median

Since total employees = 245 (odd),

Overall median position =

245+12=123rd employee\frac{245 + 1}{2} = 123^\text{rd} \text{ employee}2245+1​=123rd employee

Step 3: Cumulative employee count (in age order)

Because all employees in one company are younger than the next company, we can cumulate company-wise.

• Up to A: 32

• Up to B: 32 + 28 = 60

• Up to C: 60 + 43 = 103

• Up to D: 103 + 39 = 142

So:

• 103rd employee → last of Company C

• 104th to 142nd employees → Company D

The 123rd employee lies in Company D.

Step 4: Median age of Company D

Company D:

• Number of employees = 39 (odd)

• Median age = 45

Since all employees in Company D are clustered around this median and Company D fully contains the overall median position, the overall median age must be 45.

Final Answer

45