NMTC BHASKARA Junior Level 2024 Solved Question Paper PDF

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The value of $$x$$ satisfying $$4^x - 3^{x - 1/2} = 3^{x + 1/2} - 2^{2x - 1}$$ is of the form $$\frac{a}{b}$$ where $$\gcd(a, b) = 1$$. Then the value of $$\left(\frac{a+b}{a-b}\right)$$ is equal to

A sequence $$\{a_n\}$$, $$n \geq 1$$ with $$a_1 = \frac{1}{2}$$ and $$a_n = \frac{a_{n-1}}{2na_{n-1} + 1}$$ is given. Then the value of $$a_1 + a_2 + a_3 + \ldots + a_{2024}$$ is equal to

In the adjoining figure, four successively touching circles are placed in the interior of $$\angle AOB$$. The first (smallest) has a radius 7 cm . The third circle has a radius 28 cm . Then the radius of the largest circle (in cm ) is

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The coefficient of $$x$$ in the equation $$x^2 + px + q = 0$$ was taken as 17 , in place of 13 and its roots were found to be -2 and -15 . If $$\alpha, \beta$$ are the roots of the original equation, then the equation whose roots are $$\frac{\alpha}{\beta}$$ and $$\frac{\beta}{\alpha}$$ is

When $$x^{10} + 1$$ is divided by $$x^2 + 1$$, we get 

$$ax^8 + bx^7 + cx^6 + dx^5 + ex^4 + fx^3 + gx^2 + hx + k$$ 

as quotient. Then the value of 

$$a^{2024} + b^{2024} + c^{2024} + d^{2024} + e^{2024} + f^{2024} + g^{2024} + h^{2024} + k^{2024}$$ is

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In the adjoining figure, $$BOC$$ is the diameter of the semicircle with centre O. DE is the tangent at D . If $$AB = k(AE)$$, then the numerical value of $$k$$ is

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In triangle $$ABC$$, 

$$\tan A \colon \tan B \colon \tan C = 1 \colon 2 \colon 3$$. 
If $$\frac{AC}{AB} = \frac{p\sqrt{q}}{r}$$, where $$q$$ is Square free and $$\gcd(p, r) = 1$$ then the value of $$p + q + r$$ is

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Simon was given a number and asked to divide it by 120. He divided the number by 5,6 and 7 and got 3,2 and 2 as remainders respectively. The remainder when the number is divided by 120 is

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The maximum volume of a cylinder is cut from a cube of edge $$a$$. The volume of the remaining solid is $$ka^3$$, where $$k = \frac{p}{q}$$, $$\gcd(p, q) = 1$$. Taking $$\pi = \frac{22}{7}$$, the value of $$p + q$$ is

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In the adjoining figure, POQ is the diameter of the semicircle with centre O. OABC is a square whose area is $$36 \text{ cm}^2$$. If $$QD = x \text{ cm}$$, the value of $$x\sqrt{3}$$ is

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If $$a = \sqrt{2024}$$, $$b = \sqrt{2025}$$, the value of $$2(ab)^{1/2}(a + b)^{-1}\left\{1 + \frac{1}{4}\left(\sqrt{\frac{a}{b}} - \sqrt{\frac{b}{a}}\right)^2\right\}^{1/2}$$ is

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In a decreasing geometric progression, the $$2^{\text{nd}}$$ term is 6. The sum of all infinite terms of the progression is one-eighth of the sum to infinity of the squares of the terms. The sum of the $$1^{\text{st}}$$ and the $$4^{\text{th}}$$ terms is $$\frac{p}{q}$$ where $$p, q$$ are relatively prime to each other. Then the value of $$\left[\frac{p}{q}\right]$$, where $$[x]$$ represents the greatest integer not exceeding $$x$$ is

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The value of $$\left(\frac{\sqrt{10}}{10}\right)^{(\log_{10} 9) - 2}$$ is of the form $$\frac{a}{b}$$, where $$a, b$$ are relatively prime to each other. Then $$a - b$$ is equal to

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ABCD is a square. BE is the tangent to the semicircle on AD as diameter. The area of the triangle BCE is $$216 \text{ cm}^2$$. The radius of the semicircle (in cm ) is

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$$a, b, c, d$$ are real constants in a $$f(x) = ax^{2025} + bx^{2023} + cx^{2021} + dx^{2019}$$ and $$f(-4) = 18$$. Then the maximum value of $$|f(4)| + |2\cos x|$$ is

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