NMTC BHASKARA Junior Level 2026 Solved Question Paper PDF

For the following questions answer them individually

A right circular cone is cut into three parts of volume $$v_1, v_2, v_3$$ from the vertex respectively, by giving two cuts parallel to the base, which trisect the altitude of the cone. Then $$v_1 \colon v_2 \colon v_3 =$$

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In the adjoining figure, AOB is a diameter of the circle with center O. Point M is the mid-point of chord CD. $$\angle CDO = 32^\circ$$, $$\angle DBE = 35^\circ$$ and $$\angle BEO = x^\circ$$, then $$x =$$

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During a wartime, a group of $$n$$ soldiers had enough food to last for 30 days. After 10 days, 50 more soldiers joined. If the food now can only last for 16 days, the value of $$n$$ is

Let $$\triangle ABC$$ be a right-angled triangle, $$\angle ABC = 90^\circ$$, $$AB = 5$$ cm, $$BC = 12$$ cm. Squares ABPD and BCEQ are drawn in the exterior of the $$\triangle ABC$$. DR is drawn perpendicular to line AC at R and ES is perpendicular to the line AC at S. Let $$DR = h_1$$ and $$ES = h_2$$, then $$\frac{h_1 + h_2}{AC} =$$

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In the adjoining figure, AOB is a diameter of the circle with center O. QR is a chord and the tangents to the circle at points Q and R include an angle of measure $$72^\circ$$. Chords BQ and AR intersect at P, then the measure of $$\angle RPB$$ is

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If the arithmetic mean of two numbers is 25 and their geometric mean is 24, then an equation with the given numbers as its roots, is

The average marks obtained by Sunil in 6 subjects is 88. On subsequent verification, it was found that the marks obtained by him in a subject was wrongly copied as 86 instead of 68. The correct average marks obtained by him is $$(ab)_{10}$$, that is a two-digit number in base 10. Then $$a^2 + b^2 =$$

If $$\frac{\sqrt{60} + \sqrt{95} + \sqrt{228} + 12}{\sqrt{5} + 2\sqrt{12} + \sqrt{19}} = \frac{\sqrt{a} + \sqrt{b}}{c}$$, where $$a, b$$ are square-free natural numbers and $$c$$ is a natural number, then the value of $$a + b + 7c =$$

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In the adjoining figure, ABC is a right-angled triangle. $$\angle ABC = 90^\circ$$, $$AB = 18$$ cm, $$BC = 24$$ cm. There are three squares drawn externally on the hypotenuse AC as shown, such that AC would be the sum of the side lengths of these squares. Area of two of these squares is $$225$$ $$\text{cm}^2$$ and $$64$$ $$\text{cm}^2$$ respectively, as indicated in the figure. Then the area of the shaded region = _______ $$cm^{2}$$

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We say that a positive integer is arithmetically sequenced if its digits, in order, form an arithmetic sequence. Then the number of 4-digit positive integers which are arithmetically sequenced, is

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The smallest positive integer $$(abc)_{10}$$, in base 10, $$a, b, c$$ are digits, with $$500 > (abc)_{10} > 150$$ such that the number formed by all possible arrangements (for example $$(bac)_{10}$$, $$(cba)_{10}$$) of all the digits of $$(abc)_{10}$$, is prime, is

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Let circle-1 and circle-2 are orthogonal circles. The length of an arc measuring $$60^\circ$$ of circle-1 is equal to the length of an arc measuring $$45^\circ$$ of circle-2. Then the ratio of the area of square on the line-segment joining their centres to the area of square on the radius of circle-1, in its simplest form, is

[Note: Let two distinct circles are intersecting in different points. If their tangents at the point of intersection are perpendicular to each other, then the circles are said to be Orthogonal circles]

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P is a point in the interior of an equilateral $$\triangle ABC$$. Line-segments PD, PE, PF are the perpendiculars drawn from P to the sides BC, CA, AB respectively with D lies on side BC, E lies on side CA and F lies on side AB. If $$PD = 6$$ cm, $$PE = 8$$ cm, $$PF = 10$$ cm, then area of the triangle ABC is _____ $$cm^2$$

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Let $$a_1, a_2, a_3, \ldots, a_n$$ be all natural numbers and also $$n \in \mathbb{N}$$ such that $$a_1 + a_2 + a_3 + \cdots + a_n = 1000$$. When $$a_1 \times a_2 \times a_3 \times \cdots \times a_n$$ would be maximum, then $$n =$$

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If $$x = 2026$$ and $$y = \frac{1}{1013}$$, then the value of $$\left( \frac{x^2 + 1}{y}\right)\left(\frac{y^2 + 1}{x}\right) + \left( \frac{x^2 - 1}{y}\right)\left( \frac{y^2 - 1}{x}\right)$$ is

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$$n$$ is a natural number such that $$(n + 17)$$ and $$(n - 40)$$ both are perfect square numbers, then the square-root of greatest possible value of $$(n - 40)$$ is

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