Question 1

The smallest perfect square that is divisible by $7!$ is

Accepted Answer

### Concept & Formula A perfect square must have even powers for all of its prime factors. To find the smallest perfect square divisible by a number, break that number into its prime factorization and multiply by the exact missing prime factors needed to make all exponents even. $$ N = p_1^{e_1} 	imes p_2^{e_2} \dots $$ ### Step-by-Step Solution First, expand $7!$ (7 factorial): $7! = 7 	imes 6 	imes 5 	imes 4 	imes 3 	imes 2 	imes 1$ Break it down into its fundamental prime factors: $7! = 7 	imes (2 	imes 3) 	imes 5 	imes (2^2) 	imes 3 	imes 2$ Combine the prime factors by adding their exponents: $2^{(1 + 2 + 1)} 	imes 3^{(1 + 1)} 	imes 5^1 	imes 7^1$ $= 2^4 	imes 3^2 	imes 5^1 	imes 7^1$ Check the exponents for parity (even/odd). The powers of $2$ ($4$) and $3$ ($2$) are already even. The powers of $5$ ($1$) and $7$ ($1$) are odd. To make the entire expression a perfect square, we must multiply by $5^1$ and $7^1$ so their powers become $2$ (an even number). Multiplier needed $= 5 	imes 7 = 35$. Calculate the final perfect square: We know $7! = 5040$. Smallest Perfect Square $= 5040 	imes 35$. $5040 	imes 35 = 176400$. ### Exam Strategy & Shortcut $7!$ contains the prime numbers $2, 3, 5, 7$. Count the highest powers conceptually: $2$ appears in $2, 4, 6$ (total power 4, even). $3$ appears in $3, 6$ (total power 2, even). $5$ appears once (odd). $7$ appears once (odd). You immediately know you must multiply $7!$ by $35$. Multiplying $5040 	imes 35$ quickly gives $176400$. ### Common Pitfall Forgetting to break composite numbers like $4$ and $6$ down into their prime components when finding the factorization. This leads to incorrect power counts and a wrong final multiplier. ### Final Answer Therefore, the correct answer is **176400**.

Question 2

The least perfect square number divisible by $3$, $4$, $5$, $6$ and $8$ is

Accepted Answer

### Concept & Strategy To find a perfect square divisible by multiple numbers, first find the Least Common Multiple (LCM) of those numbers. Then, prime factorize the LCM and multiply it by any prime factors needed to make all exponents even. ### Step-by-Step Solution Find the LCM of $3$, $4$, $5$, $6$, and $8$. List their prime factorizations: $3 = 3^1$ $4 = 2^2$ $5 = 5^1$ $6 = 2^1 	imes 3^1$ $8 = 2^3$ The LCM takes the highest power of each prime present: LCM $= 2^3 	imes 3^1 	imes 5^1$ LCM $= 8 	imes 3 	imes 5 = 120$. Now, analyze the prime factorization of the LCM ($2^3 	imes 3^1 	imes 5^1$) for odd powers. The power of $2$ is $3$ (odd). The power of $3$ is $1$ (odd). The power of $5$ is $1$ (odd). To make it a perfect square, we need to multiply by one more of each prime factor to make their powers even ($2^4, 3^2, 5^2$). Multiplier needed $= 2 	imes 3 	imes 5 = 30$. Multiply the LCM by this required multiplier: Least Perfect Square $= 120 	imes 30 = 3600$. ### Exam Strategy & Shortcut Instead of calculating the LCM fully, look at the options. The number must be a perfect square. Options (b) $1200$ and (c) $2500$ are not divisible by $3$, so eliminate them. Option (a) $900$ is a perfect square and divisible by $3, 4, 5, 6$, but it is NOT divisible by $8$ ($900 / 8 = 112.5$). This leaves only $3600$ by process of elimination. ### Common Pitfall A common error is stopping once the LCM ($120$) is found and selecting the nearest multiple from the options without ensuring the final result is a perfect square. ### Final Answer Therefore, the correct answer is **3600**.

Question 3

$\sqrt{2 + \sqrt{3}} \cdot \sqrt{2 + \sqrt{2 + \sqrt{3}}} \cdot \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{3}}}} \cdot \sqrt{2 - \sqrt{2 + \sqrt{2 + \sqrt{3}}}}$ is equal to

Accepted Answer

### Concept & Formula This problem uses the algebraic identity for the difference of two squares applied continuously from right to left under nested square roots. $$(a+b)(a-b) = a^2 - b^2$$ When numbers are multiplied under square roots of the same degree, we can combine them: $$\sqrt{x} \cdot \sqrt{y} = \sqrt{x \cdot y}$$ ### Step-by-Step Solution * **Calculation / Deduction:** * Focus on the last two terms on the right: $$\sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{3}}}} \cdot \sqrt{2 - \sqrt{2 + \sqrt{2 + \sqrt{3}}}}$$ * Combine them under a single square root. Let $a = 2$ and $b = \sqrt{2 + \sqrt{2 + \sqrt{3}}}$. * This forms $\sqrt{(a+b)(a-b)} = \sqrt{a^2 - b^2}$. $$= \sqrt{2^2 - (\sqrt{2 + \sqrt{2 + \sqrt{3}}})^2}$$ $$= \sqrt{4 - (2 + \sqrt{2 + \sqrt{3}})}$$ $$= \sqrt{2 - \sqrt{2 + \sqrt{3}}}$$ * Now, multiply this result by the second term in the original expression: $$\sqrt{2 + \sqrt{2 + \sqrt{3}}} \cdot \sqrt{2 - \sqrt{2 + \sqrt{3}}}$$ * Apply the same logic. Let $a = 2$ and $b = \sqrt{2 + \sqrt{3}}$. $$= \sqrt{2^2 - (\sqrt{2 + \sqrt{3}})^2}$$ $$= \sqrt{4 - (2 + \sqrt{3})}$$ $$= \sqrt{2 - \sqrt{3}}$$ * Finally, multiply this result by the first term in the original expression: $$\sqrt{2 + \sqrt{3}} \cdot \sqrt{2 - \sqrt{3}}$$ * Let $a = 2$ and $b = \sqrt{3}$. $$= \sqrt{2^2 - (\sqrt{3})^2}$$ $$= \sqrt{4 - 3}$$ $$= \sqrt{1} = 1$$ ### Exam Strategy & Shortcut For infinitely or finitely nested square roots of the form $(2 + \sqrt{\dots}) \cdot (2 - \sqrt{\dots})$, the pattern collapses rapidly inward. Each step eliminates one layer of the root. Recognize the $(x+y)(x-y)$ pattern immediately and collapse it backwards mentally to save writing time. ### Common Pitfall Getting intimidated by the length of the expression and attempting to calculate inner roots first. Always look for structural algebraic patterns (like difference of squares) before attempting brute-force calculation. ### Final Answer **Therefore, the correct answer is 1.**

Question 4

Given that $\sqrt{3} = 1.732$, the value of   $$ \frac{3 + \sqrt{6}}{5\sqrt{3} - 2\sqrt{12} - \sqrt{32} + \sqrt{50}} $$   is

Accepted Answer

### Concept & Formula The core strategy is simplifying compound surds by extracting perfect square factors from inside the square roots. Use the property: $$ \sqrt{a^2 	imes b} = a\sqrt{b} $$ After breaking down all terms into their simplest radical forms, combine like terms to simplify the fraction. ### Step-by-Step Solution Let's simplify the numerator and denominator separately. **Numerator:** $3 + \sqrt{6}$ We can extract a common factor of $\sqrt{3}$ because $3 = (\sqrt{3})^2$ and $\sqrt{6} = \sqrt{3}\sqrt{2}$: Numerator $= \sqrt{3}(\sqrt{3} + \sqrt{2})$ **Denominator:** $5\sqrt{3} - 2\sqrt{12} - \sqrt{32} + \sqrt{50}$ Simplify each term by factoring out perfect squares: $\sqrt{12} = \sqrt{4 	imes 3} = 2\sqrt{3}$ $\sqrt{32} = \sqrt{16 	imes 2} = 4\sqrt{2}$ $\sqrt{50} = \sqrt{25 	imes 2} = 5\sqrt{2}$ Substitute these back into the denominator: $= 5\sqrt{3} - 2(2\sqrt{3}) - 4\sqrt{2} + 5\sqrt{2}$ $= 5\sqrt{3} - 4\sqrt{3} - 4\sqrt{2} + 5\sqrt{2}$ Group the like terms (the $\sqrt{3}$ terms together and the $\sqrt{2}$ terms together): $= (5 - 4)\sqrt{3} + (5 - 4)\sqrt{2}$ $= \sqrt{3} + \sqrt{2}$ **Recombine Fraction:** Expression $= \frac{\sqrt{3}(\sqrt{3} + \sqrt{2})}{\sqrt{3} + \sqrt{2}}$ The term $(\sqrt{3} + \sqrt{2})$ cancels out from the top and bottom: Expression $= \sqrt{3}$ Substitute the given value $\sqrt{3} = 1.732$. ### Exam Strategy & Shortcut In competitive exams, complicated-looking surd fractions almost always simplify down to a single radical term. Once you spot that the denominator simplifies to $\sqrt{3} + \sqrt{2}$, look at the numerator $3 + \sqrt{6}$. Recognize instantly that $\sqrt{3}$ is a multiplier that maps the denominator to the numerator. Cancel them mentally to arrive at $\sqrt{3}$ directly. ### Common Pitfall A major trap is attempting to plug in $1.732$ for $\sqrt{3}$ and approximating other roots like $\sqrt{2}$ at the very beginning. This creates a nightmare of decimal arithmetic that wastes time and guarantees calculation errors. Always simplify variables and radicals fully before substituting numeric values. ### Final Answer **Therefore, the correct answer is 1.732.**

Question 5

$\frac{12}{3 + \sqrt{5} + 2\sqrt{2}}$ is equal to

Accepted Answer

### Concept & Formula When a denominator contains three surd terms, you must rationalize it by strategically grouping the terms into two parts to create a binomial structure, and then multiplying by its conjugate. $$ (a+b)(a-b) = a^2 - b^2 $$ ### Step-by-Step Solution **Given:** The algebraic fraction $\frac{12}{3 + \sqrt{5} + 2\sqrt{2}}$ **Calculation / Deduction:** * Group the denominator into a single binomial structure to prepare for rationalization: $((3 + \sqrt{5}) + 2\sqrt{2})$ * Rationalize by multiplying the numerator and the denominator by the conjugate $((3 + \sqrt{5}) - 2\sqrt{2})$: Denominator $= (3 + \sqrt{5})^2 - (2\sqrt{2})^2$ * Expand the first term of the denominator using $(a+b)^2 = a^2 + 2ab + b^2$: Denominator $= (9 + 5 + 6\sqrt{5}) - 8 = 14 + 6\sqrt{5} - 8 = 6 + 6\sqrt{5} = 6(1 + \sqrt{5})$ * Update the entire fraction with the simplified denominator: Fraction $= \frac{12(3 + \sqrt{5} - 2\sqrt{2})}{6(1 + \sqrt{5})} = \frac{2(3 + \sqrt{5} - 2\sqrt{2})}{1 + \sqrt{5}}$ * The denominator still contains a surd, so rationalize again by multiplying the top and bottom by $(\sqrt{5} - 1)$: New Denominator $= (\sqrt{5})^2 - (1)^2 = 5 - 1 = 4$ * Multiply the numerator expressions out carefully: Numerator $= 2(3 + \sqrt{5} - 2\sqrt{2})(\sqrt{5} - 1)$ $= 2(3\sqrt{5} - 3 + 5 - \sqrt{5} - 2\sqrt{10} + 2\sqrt{2})$ $= 2(2 + 2\sqrt{5} + 2\sqrt{2} - 2\sqrt{10})$ Factor out the common $2$: Numerator $= 4(1 + \sqrt{5} + \sqrt{2} - \sqrt{10})$ * Divide the final numerator by the new denominator ($4$): Result $= \frac{4(1 + \sqrt{5} + \sqrt{2} - \sqrt{10})}{4} = 1 + \sqrt{5} + \sqrt{2} - \sqrt{10}$ ### Exam Strategy & Shortcut For three-term rationalization, always group the two terms whose squares sum closest to the square of the third term. Here, $3^2 + (\sqrt{5})^2 = 9 + 5 = 14$, and $(2\sqrt{2})^2 = 8$. This keeps the resulting coefficients manageable. Keep constants factored out (like keeping the 12 outside the brackets) until the very end to maximize cancellation opportunities. ### Common Pitfall Students often distribute the numerator's constant immediately instead of looking for structural cancellations. Distributing the $12$ into the numerator right away makes the subsequent multiplication by $(\sqrt{5} - 1)$ much harder and severely increases the likelihood of simple arithmetic errors. ### Final Answer **Therefore, the correct answer is $1 + \sqrt{5} + \sqrt{2} - \sqrt{10}$.**

Question 6

One-fourth of a herd of camels was seen in the forest. Twice the square root of the herd had gone to mountains and the remaining 15 camels were seen on the bank of a river. Find the total number of camels.

Accepted Answer

### Concept & Logic This word problem is solved by translating the given conditions into an algebraic equation. Since the problem involves the square root of the total number of camels, it will naturally form a quadratic equation when formalized. ### Step-by-Step Solution * **Given:** Let the total number of camels in the herd be $x$. * Number of camels in the forest = $\frac{x}{4}$ * Number of camels gone to the mountains = $2\sqrt{x}$ * Number of remaining camels on the bank = $15$ * The sum of these three groups must equal the total number of camels: * $\frac{x}{4} + 2\sqrt{x} + 15 = x$ * Rearrange the terms to group the $x$ variables on one side: * $x - \frac{x}{4} - 2\sqrt{x} = 15$ * $\frac{3x}{4} - 2\sqrt{x} = 15$ * Multiply the entire equation by 4 to clear the fraction: * $3x - 8\sqrt{x} = 60$ * Let $y = \sqrt{x}$. This implies $y^2 = x$. Substitute this into the equation to form a standard quadratic: * $3y^2 - 8y - 60 = 0$ * Factor the quadratic equation. We need two numbers that multiply to $(3 	imes -60) = -180$ and add to $-8$. Those numbers are $-18$ and $10$. * $3y^2 - 18y + 10y - 60 = 0$ * $3y(y - 6) + 10(y - 6) = 0$ * $(3y + 10)(y - 6) = 0$ * This gives $y = -\frac{10}{3}$ or $y = 6$. * Since $y$ represents a square root of a physical count of camels, it must be positive. Thus, $y = 6$. * Substitute back to find $x$: $\sqrt{x} = 6$, so $x = 36$. ### Exam Strategy & Shortcut **Option Elimination is the fastest method here.** The problem states "Twice the square root of the herd...". This strongly implies that the total number of camels must be a perfect square integer (since you can't have a fractional number of live camels). Look at the options: (a) 32 - Not a perfect square. (b) 34 - Not a perfect square. (c) 35 - Not a perfect square. (d) 36 - Perfect square! ($\sqrt{36} = 6$) You can solve this question in exactly 2 seconds without writing a single equation. ### Common Pitfall The most common mistake is setting up the equation correctly but struggling to factor the resulting quadratic with surds. Students often panic when they see $3x - 8\sqrt{x} = 60$. Using substitution ($y = \sqrt{x}$) simplifies it, but relying on option elimination bypasses the risk of arithmetic errors entirely. ### Final Answer **Therefore, the correct answer is 36.**

Question 7

A mobile company offered to pay the Indian Cricket Team as much money per run scored by the side as the total number it gets in a one-dayer against Australia.   Which one of the following cannot be the total amount to be spent by the company in this…

Accepted Answer

Concept & Logic The company pays an amount equal to the number of runs multiplied by the number of runs. Thus, the total amount spent must be a perfect square. The question is asking us to identify the single option that is **not** a perfect square. Step-by-Step Solution * **Given**: Total amount = (Runs Scored) $	imes$ (Runs Scored) = $R^2$. * We need to inspect the unit digits of the given options to rule out non-perfect squares. * A fundamental rule of mathematics: A perfect square can never end in $2$, $3$, $7$, or $8$. * Let us check the last digit of each given option: - Option (a) $21,904$ ends in $4$ (possible perfect square). - Option (b) $56,169$ ends in $9$ (possible perfect square). - Option (c) $1,01,761$ ends in $1$ (possible perfect square). - Option (d) $1,21,108$ ends in $8$ (impossible). * Since no integer squared ends in $8$, $1,21,108$ mathematically cannot be the total amount spent. Exam Strategy & Shortcut Always check the unit digits first when asked if a number is a perfect square! Perfect squares only end in $0, 1, 4, 5, 6,$ or $9$. Option (d) ends in $8$, making it instantly incorrect as a perfect square. Absolutely no calculation is required. Common Pitfall Wasting valuable exam time trying to calculate the exact square root of each large 5- or 6-digit number. Recognizing unit digit rules saves massive amounts of time on these types of questions. Final Answer **Therefore, the correct answer is 1,21,108.**

Question 8

$\sqrt{24 \div 0.5 + 1} + \sqrt{18 \div 0.6 + 6} = $ $x$

Accepted Answer

Concept & Logic This problem tests your ability to apply the BODMAS rule (specifically handling division with decimals) inside radical expressions. You must completely simplify the expressions under each square root before adding them together. $$ \sqrt{a} + \sqrt{b} 
eq \sqrt{a+b} $$ Step-by-Step Solution * **Given:** $\sqrt{24 \div 0.5 + 1} + \sqrt{18 \div 0.6 + 6} = x$ * **Calculation:** First, evaluate the expression inside the first square root: $24 \div 0.5 = 24 \div \frac{1}{2} = 24 	imes 2 = 48$ $48 + 1 = 49$ So, the first term is $\sqrt{49} = 7$. * Next, evaluate the expression inside the second square root: $18 \div 0.6 = \frac{180}{6} = 30$ $30 + 6 = 36$ So, the second term is $\sqrt{36} = 6$. * Finally, add the simplified terms: $7 + 6 = 13$ Exam Strategy & Shortcut Division by a decimal like 0.5 is exactly the same as multiplying by 2. Division by 0.6 is the same as multiplying by $\frac{10}{6}$. Recognizing these fraction equivalents allows you to do the math mentally: $24 	imes 2 = 48$, and $18 	imes \frac{10}{6} = 30$. This skips the need for long division. Common Pitfall A frequent mistake is adding the numbers inside the roots before performing the division, or incorrectly evaluating $18 \div 0.6$ as $3$ instead of $30$ due to decimal misplacement. Always align your decimal points carefully! Final Answer **Therefore, the correct answer is 13.**

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