Breakup G Strategy | Simple Equations | EASY | CAT 2025 Slot 1
Stocks A, B and C are priced at rupees 120, 90 and 150 per share, respectively. A trader holds a portfolio consisting of 10 shares of stock A, and 20 shares of stocks B and C put together. If the total value of her portfolio is rupees 3300, then the number of shares of stock B that she holds, is _______
Answer
Let the trader hold x shares of stock B.
Then stock C shares = 20 − x.
Prices:
A = 120
B = 90
C = 150
Portfolio value equation:
10(120) + x(90) + (20 − x)(150) = 3300
1200 + 90x + 3000 − 150x = 3300
1200 + 3000 + (90x − 150x) = 3300
4200 − 60x = 3300
60x = 900
x = 15
Answer: 15 shares of stock B.
Breakup G Strategy | Simple Equations | MEDIUM | CAT 2025 Slot 1
If a -6b + 6c = 4 and 6a + 3b – 3c = 50, where a, b, and c are real numbers, the value of 2a + 3b – 3c is
1. 14
2. 15
3. 16
4. 18
Answer
Step 1 From eq(2): 6a + 3b − 3c = 50
→ take one-third → 2a + b − c = 50/3
Step 2 Compare target with this:
Target = (2a + b − c) + 2(b − c)
Step 3 Find (b − c) from eq(1):
a − 6b + 6c = 4 → rewrite → a = 4 + 6(b − c)
Step 4 Substitute into step 1:
2(4 + 6x) + x = 50/3
8 + 12x + x = 50/3
13x = 26/3 → x = 2/3
So (b − c) = 2/3
Step 5 Plug in:
Target = 50/3 + 2*(2/3) = 18
X maro G Strategy | Concept of Quadratic Equations | CAT 2025 Slot 1
The number of non-negative integer values of k for which the quadratic equation x² – 5x + k = 0 has only integer roots, is
Answer
Quadratic: x² – 5x + k = 0.
For integer roots, discriminant must be a square:
D = 25 – 4k = n².
Try n = 1, 3, 5:
n = 1 → k = 6
n = 3 → k = 4
n = 5 → k = 0
All are non-negative.
Total values = 3.
Answer: 3
X maro | Visual | Inequality | Sets | EASY | CAT 2025 Slot 1
Let 3 ≤ x ≤ 6 and [x²] = ([x])², where [x] is the greatest integer not exceeding x.
If S represents all feasible values of x, then a possible subset of S is
1. [3, √10] ∪ [4, √17] ∪ {6}
2. [3, √10) ∪ [5, √26)
3. (4, √18] ∪ [5, √27) ∪ {6}
4. (3, √10] ∪ [5, √26) ∪ {6}
Answer
Using X maro
For n = 3 → x must be before 4 → [3, √10)
For n = 4 → x must be before 5 → [4, √17)
For n = 5 → x must be before 6 → [5, √26)
For n = 6: only x = 6 (since x ≤ 6)
Only Option 4 matches: (3, √10] ∪ [5, √26) ∪ {6}
Note: ideally Set should include 4 as well.
X maro Functions + Maxima Minima | EASY | CAT 2025 Slot 1
A value of c for which the minimum value of f(x) = x² – 4cx + 8c is greater than the maximum value of
g(x) = –x² + 3cx – 2c, is
A. 1/2
B. –1/7
C. 2
D. –2
Answer
Let B = number of boys, G = number of girls (both integers), and B > 10.
After departures: remaining girls = 60% of G = 3G/5, remaining boys = 40% of B = 2B/5.
Given 3G/5 = 2B/5 + 8. Multiply by 5: 3G = 2B + 40. (Equation ★)
For 3G to be integer, G must be integer. For the remaining counts 3G/5 and 2B/5 to be integers (no fractional people), G and B should both be multiples of 5.
Let B = 5m (m integer) with B > 10 ⇒ m ≥ 3. Try smallest multiples of 5 for B:
• B = 15 ⇒ 2B + 40 = 70 ⇒ G = 70/3 (not integer)
• B = 20 ⇒ 2B + 40 = 80 ⇒ G = 80/3 (not integer)
• B = 25 ⇒ 2B + 40 = 90 ⇒ G = 90/3 = 30 (integer, works)
So the smallest B (multiple of 5) that satisfies (★) is B = 25, with G = 30.
Total initial students = B + G = 25 + 30 = 55.
Therefore the minimum possible number of students initially in the class (with realistic integer people and integer remaining counts) is 55.
