Question

What is the minimum number of students who appeared in the Mock test.

• A. 105
• B. 110
• C. 115
• D. None of these
• A. 5
• B. 15
• C. 20
• D. None of these
• A. 3:1
• B. 3:2
• C. 1:3
• D. None of these
• A. 5
• B. 15
• C. 20
• D. None of these

Answer 1

The Correct Option is A

Given (from the passage):
DI = 80, VA = 70, QA = 60; All three (DI ∩ VA ∩ QA) = 40; Outside all three = 10;
VA ∩ QA (total) = 50; Only QA = 5.

Notation for regions:
Only DI = x; Only VA = y; Only QA = 5;
(DI ∩ VA) only = p; (DI ∩ QA) only = q; (VA ∩ QA) only = r; All three = 40.

Step 1 — Use VA ∩ QA = 50:
VA ∩ QA includes the triple intersection, so r = (VA ∩ QA) − (All three) = 50 − 40 = 10.

Step 2 — Use subject totals to find unknown pairwise-only parts:
QA total: 60 = (Only QA) + (DI ∩ QA only) + (VA ∩ QA only) + (All three)
60 = 5 + q + 10 + 40 ⇒ q = 60 − 55 = 5.

Step 3 — Express constraints for DI and VA:
DI total: 80 = (Only DI) + (DI ∩ VA only) + (DI ∩ QA only) + (All three)
80 = x + p + 5 + 40 ⇒ x + p = 35. …(1)

VA total: 70 = (Only VA) + (DI ∩ VA only) + (VA ∩ QA only) + (All three)
70 = y + p + 10 + 40 ⇒ y + p = 20. …(2)

Step 4 — Write the union size in terms of p:
|DI ∪ VA ∪ QA| = (Only DI) + (Only VA) + (Only QA) + (DI ∩ VA only) + (DI ∩ QA only) + (VA ∩ QA only) + (All three)
= x + y + 5 + p + 5 + 10 + 40.
From (1): x = 35 − p; from (2): y = 20 − p.
So |Union| = (35 − p) + (20 − p) + 5 + p + 5 + 10 + 40 = 115 − p.

Step 5 — Minimize the total appeared students:
Total appeared = |Union| + (Outside all three). To minimize total appeared, minimize |Union|.
From x = 35 − p ≥ 0 and y = 20 − p ≥ 0, we must have p ≤ 35 and p ≤ 20 ⇒ p ≤ 20.
Thus the maximum feasible p is 20, which minimizes |Union|.
Minimum |Union| = 115 − p = 115 − 20 = 95.

Step 6 — Add the 10 who failed all:
Minimum total appeared = 95 (in at least one subject) + 10 (in none) = 105.

Answer: (A) 105.

The correct option is (A): 105

Answer 2

We’re asked for: the minimum number of students who did not clear cutoff in exactly two subjects (i.e., they cleared exactly one subject).

Set up the Venn regions (DI, VA, QA):
Let Only DI = x, Only VA = y, Only QA = 5 (given).
Let (DI ∩ VA only) = p, (DI ∩ QA only) = q, (VA ∩ QA only) = r, and (All three) = 40 (given).
Also given: VA ∩ QA (including the triple) = 50, total QA = 60, total DI = 80, total VA = 70, and 10 students are outside all three.

Step 1 — Resolve pairwise regions using the givens:
VA ∩ QA (including triple) = 50 ⇒ (VA ∩ QA only) + (All three) = r + 40 = 50 ⇒ r = 10.
QA total: 60 = (Only QA) + (DI ∩ QA only) + (VA ∩ QA only) + (All three)
60 = 5 + q + 10 + 40 ⇒ q = 5.

Step 2 — Express DI and VA totals to relate x, y, p:
DI total: 80 = (Only DI) + (DI ∩ VA only) + (DI ∩ QA only) + (All three)
80 = x + p + 5 + 40 ⇒ x + p = 35 ⇒ x = 35 − p.

VA total: 70 = (Only VA) + (DI ∩ VA only) + (VA ∩ QA only) + (All three)
70 = y + p + 10 + 40 ⇒ y + p = 20 ⇒ y = 20 − p.

Step 3 — Link “failed in exactly two subjects” to “cleared exactly one”:
Failing in exactly two subjects means clearing exactly one subject. Hence the required count = (Only DI) + (Only VA) + (Only QA).
Therefore, Required = x + y + 5 = (35 − p) + (20 − p) + 5 = 60 − 2p.

Step 4 — Minimize the required count by maximizing p (feasibility constraints):
We must keep each region non-negative: x ≥ 0 ⇒ 35 − p ≥ 0 ⇒ p ≤ 35; y ≥ 0 ⇒ 20 − p ≥ 0 ⇒ p ≤ 20; also p ≥ 0.
Thus the maximum feasible p is p = 20 (the tighter bound).

Step 5 — Compute the minimum:
Minimum (x + y + 5) = 60 − 2p = 60 − 2·20 = 20.

Conclusion: The minimum number of students who did not clear cutoff in exactly two subjects (i.e., cleared exactly one subject) is 20.

The correct option is (C): 20

Answer 3

We need: Ratio = (Students who didn’t clear DI but cleared QA) : (Students who didn’t clear only VA).

Step 1 — Recall Venn regions and known values:
Only DI = x = 35 − p;
Only VA = y = 20 − p;
Only QA = 5;
(DI ∩ VA only) = p;
(DI ∩ QA only) = q = 5;
(VA ∩ QA only) = r = 10;
All three = 40.

Step 2 — Students who didn’t clear DI but cleared QA:
This group = those inside QA but outside DI.
= (Only QA) + (VA ∩ QA only) + (All three?) No, careful: “didn’t clear DI” means they must NOT be in DI, so exclude anything involving DI.

Therefore include: (Only QA) + (VA ∩ QA only).
= 5 + 10 = 15.

Step 3 — Students who didn’t clear only VA:
“Didn’t clear only VA” means they failed VA but cleared both DI and QA.
That corresponds exactly to (DI ∩ QA only).
= q = 5.

Step 4 — Form the ratio:
Required ratio = (15 : 5) = 3 : 1.

Answer: The correct option is (A): 3:1.

Answer 4

We need: Number of students who failed in exactly one subject, given that those who didn’t clear at least two subjects is maximized.

Step 1 — Recall what “failed in at least two subjects” means:
It includes students who failed in exactly two subjects (i.e., cleared exactly one) + those who failed in all three (10 students).
So, to maximize this category, we want to maximize the number who cleared exactly one subject.

Step 2 — Expression for students clearing exactly one subject:
Cleared exactly one = (Only DI) + (Only VA) + (Only QA).
= (35 − p) + (20 − p) + 5 = 60 − 2p.

Step 3 — Maximize “failed in at least two subjects”:
Since “at least two fails” = (Cleared exactly one) + 10, we maximize this by maximizing (Cleared exactly one).
The expression (60 − 2p) is largest when p = 0 (its minimum possible).
Then Cleared exactly one = 60 − 0 = 60.
So “failed in at least two” = 60 + 10 = 70 (maximum).

Step 4 — Now find students who failed in exactly one subject:
“Failed in exactly one subject” = those who cleared exactly two subjects.
Cleared exactly two = (DI ∩ VA only) + (DI ∩ QA only) + (VA ∩ QA only).
= p + q + r = p + 5 + 10 = p + 15.

Step 5 — With p = 0 (from maximum case):
Cleared exactly two = 0 + 5 + 10 = 15.
Thus, students who failed in exactly one subject = 15.

Answer: The correct option is (B): 15.