Question 1: Recurrence Relation
If x₀ = 1, x₁ = 2, and
xₙ₊₂ = (1 + xₙ₊₁) / xₙ for n = 0, 1, 2, …,
then what is the value of x₂₀₂₁?
Options:
- 1 2) 2 3) 3 4) 4
Answer:
We calculate the first few values using the recurrence:
- x₀ = 1
- x₁ = 2
- x₂ = (1 + 2) / 1 = 3
- x₃ = (1 + 3) / 2 = 2
- x₄ = (1 + 2) / 3 = 1
- x₅ = (1 + 1) / 2 = 1
- x₆ = (1 + 1) / 1 = 2
- x₇ = (1 + 2) / 1 = 3
- x₈ = (1 + 3) / 2 = 2
- x₉ = (1 + 2) / 3 = 1
From x₄ onward, the values repeat in a cycle: [1, 1, 2, 3, 2]
This cycle has a length of 5.
Now, to find x₂₀₂₁:
- Offset from x₄ = 2021 – 4 = 2017
- 2017 mod 5 = 2 → 3rd term in the cycle
The cycle is [1, 1, 2, 3, 2], so the 3rd term is 2.
Correct Option: 2) 2
Question 2: Natural Numbers Grouping
The natural numbers are grouped as:
(1), (2,3,4), (5,6,7,8,9), ………
Find the sum of all terms in the 15th group.
Answer:
Each group nn has 2n−12n – 1 terms.
We calculate the starting number (first term) of each group recursively:
- Group 1: starts at 1 (1 term)
- Group 2: starts at 2 (3 terms)
- Group 3: starts at 5 (5 terms)
- Group 4: starts at 10 (7 terms)
- Group 5: starts at 17 (9 terms)
- Group 6: starts at 26 (11 terms)
- Group 7: starts at 37 (13 terms)
- Group 8: starts at 50 (15 terms)
- Group 9: starts at 65 (17 terms)
- Group 10: starts at 82 (19 terms)
- Group 11: starts at 101 (21 terms)
- Group 12: starts at 122 (23 terms)
- Group 13: starts at 145 (25 terms)
- Group 14: starts at 170 (27 terms)
- Group 15: starts at 197 (29 terms)
So, 15th group is an AP from 197 to 197+28=225197 + 28 = 225
Sum = (Number of terms)/2 × (First term + Last term)
= 29/2 × (197 + 225)
= 29 × 211 = 6119
Correct Answer: 6119
Question 3: Onion Price – Weighted Average
Onion is sold for 5 consecutive months at the rates:
₹10, ₹20, ₹25, ₹25, ₹50 per kg.
A family spends a fixed amount on onion for each of the first 3 months,
and then spends half that amount on onion for each of the last 2 months.
What is the average price per kg of onion for the family?
Options:
- 16 2) 26 3) 20 4) 18
Answer:
Assume spend in first 3 months is ₹100 each,
then spend in month 4 and 5 is ₹50 each.
| Month | Price (₹/kg) | Spend (₹) | Quantity Bought (kg) |
|---|---|---|---|
| 1 | 10 | 100 | 10 |
| 2 | 20 | 100 | 5 |
| 3 | 25 | 100 | 4 |
| 4 | 25 | 50 | 2 |
| 5 | 50 | 50 | 1 |
Total Spend = ₹400
Total Quantity = 10 + 5 + 4 + 2 + 1 = 22 kg
Average price per kg = 400 ÷ 22 ≈ 18.18
Correct Option: 4) 18
Question 4: Conditional Group Formation
The number of groups of three or more distinct numbers that can be chosen from the set:
{1, 2, 3, 4, 5, 6, 7, 8}
such that:
- Each group always includes 3 and 5
- 7 and 8 are never included together
What is the total number of such valid groups?
Options:
- 32 2) 40 3) 48 4) 56
Answer:
We must form groups:
- Of at least 3 elements
- Must include both 3 and 5
- Must not include both 7 and 8 together
Step 1: Fix 3 and 5
We will always include 3 and 5, so the group size becomes ≥ 3 → we need to choose at least 1 more element from the remaining set:
Remaining elements to choose from:
{1, 2, 4, 6, 7, 8} → 6 elements
Let’s call this set S = {1, 2, 4, 6, 7, 8}
We need to choose at least 1 element from S, but 7 and 8 cannot be together.
Step 2: Count total subsets from S (excluding 7 and 8 together)
We’ll count valid subsets of S of size ≥ 1, with the condition that 7 and 8 cannot appear together.
Total subsets of S (size ≥ 1):
= Total subsets of 6 elements – empty set = 2⁶ – 1 = 64 – 1 = 63
Subsets that include both 7 and 8:
Fix 7 and 8 → remaining choices from {1, 2, 4, 6} → 2⁴ = 16 subsets
So, 16 subsets include both 7 and 8
Valid subsets = 63 – 16 = 47
Step 3: Ensure total group size is ≥ 3
We already have 3 and 5 → so subset size from S must be ≥ 1
(That is already handled, as we only considered non-empty subsets of S)
Final Answer = 47 valid groups
✅ Correct Option: 3) 48
⚠ Wait: 47 is the number of valid extra elements, each combined with fixed (3,5)
So total groups = 47
But if any group formed is of size ≥ 3 (i.e., (3,5) + at least 1 more), that satisfies the condition.
So, the correct answer is 47.
But that does not match any option directly. Possibly there’s a mismatch.
Let’s double-check the logic using direct counting:
Alternate: Count valid subsets of size ≥ 3 that include 3 and 5 and not both 7 and 8
We need to count all subsets of size k ≥ 3 that include 3 and 5, and never include both 7 and 8.
So, let’s fix (3,5). Remaining from S = {1,2,4,6,7,8} → 6 elements.
We need to choose r elements from S where r = 1 to 6 (since 3+5 are fixed → total group size ≥ 3)
Total combinations = sum of all valid combinations from r = 1 to 6 excluding both 7 and 8
Let’s calculate that properly:
| r (size from S) | Total ways from 6 | Remove (with both 7 & 8) |
|---|---|---|
| 1 | C(6,1) = 6 | 0 |
| 2 | C(6,2) = 15 | C(4,0) = 1 if choosing both 7 & 8 |
| 3 | C(6,3) = 20 | C(4,1) = 4 |
| 4 | C(6,4) = 15 | C(4,2) = 6 |
| 5 | C(6,5) = 6 | C(4,3) = 4 |
| 6 | C(6,6) = 1 | C(4,4) = 1 |
| Total | 63 | 16 |
So, valid = 63 – 16 = 47
✅ Final Answer: 47
