Title: SNAP 2024 Slot 1 Paper: Quantitative Reasoning Solutions
Content: English Solutions for Indian Competitive Exam Problems
Problem 1: Arrangement of Tokens
Question:
"In a SNAP 2024 game, players must arrange 5 distinct tokens in a row. Each token can be either red or blue. How many different arrangements are possible if exactly 3 tokens are red?"
Solution:
Understanding the Problem:
Total tokens = 5 (distinct shapes, but colors are binary: red/blue).
Constraint: Exactly 3 tokens must be red.
Need to find the number of valid arrangements.
Step 1: Choose Positions for Red Tokens
We need to select 3 positions out of 5 for red tokens.
Number of ways to choose positions: ( \binom{5}{3} = 10 ).
Step 2: Assign Colors to Remaining Tokens
The remaining 2 positions must be blue.
Since tokens are distinct, each arrangement of colors corresponds to a unique permutation of shapes.
Total permutations for tokens: ( 5! = 120 ).
Step 3: Combine Choices and Permutations
Total arrangements = (Number of color choices) × (Permutations of tokens).
However, since colors are fixed (3 red, 2 blue), the correct approach is:
First choose positions for red tokens (( \binom{5}{3} )), then permute the 5 distinct tokens in the selected positions.
Total arrangements = ( \binom{5}{3} \times 5! = 10 \times 120 = 1200 ).
Answer: 1200 different arrangements.
Problem 2: Ratio and Proportion
Question:
"A mixture contains milk and water in the ratio 7:3. If 10 liters of water are added, the ratio becomes 7:5. Find the original quantity of the mixture."
Solution:
Let the Original Quantities:
Let milk = ( 7x ) liters, water = ( 3x ) liters.
Total mixture = ( 10x ) liters.
After Adding 10 Liters of Water:
New water quantity = ( 3x + 10 ).
New ratio = 7:5 (milk:water).
Equation: ( \frac{7x}{3x + 10} = \frac{7}{5} ).
Solve for ( x ):
Cross-multiply: ( 35x = 21x + 70 ).
Simplify: ( 14x = 70 ) → ( x = 5 ).
Original Mixture:
Total = ( 10x = 10 \times 5 = 50 ) liters.
Answer: 50 liters.
Problem 3: Algebraic Equations
Question:
"Solve for ( x ): ( \sqrt{2x + 3} + \sqrt{x - 1} = 4 )."
Solution:
Isolate One Radical:
( \sqrt{2x + 3} = 4 - \sqrt{x - 1} ).
Square Both Sides:
( 2x + 3 = 16 - 8\sqrt{x - 1} + (x - 1) ).
Simplify: ( 2x + 3 = x + 15 - 8\sqrt{x - 1} ).
Rearrange and Square Again:
( x - 12 = -8\sqrt{x - 1} ).
Square: ( x^2 - 24x + 144 = 64(x - 1) ).
Simplify: ( x^2 - 88x + 208 = 0 ).
Solve Quadratic Equation:
( x = \frac{88 \pm \sqrt{7744 - 832}}{2} = \frac{88 \pm 84}{2} ).
Solutions: ( x = 86 ) or ( x = 2 ).
Check for Extraneous Solutions:
( x = 86 ): ( \sqrt{175} + \sqrt{85} \neq 4 ) (invalid).
( x = 2 ): ( \sqrt{7} + \sqrt{1} \approx 2.645 + 1 = 3.645 \neq 4 ) (invalid).
Conclusion: No real solution exists.
Answer: No real solution.
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