NMTC KAPREKAR Sub-Junior 2026 Solved Question Paper

Since $$D$$ is the mid-point of $$BC$$ we get $$2a + 2 = 3a - 3$$, so $$a = 5$$ and $$BD = DC = 12$$, giving $$BC = 24$$. Also $$AB = BE + EA = 11 + 21 = 32$$, so $$AC = \sqrt{24^2 + 32^2} = \sqrt{1600} = 40$$. The semicircle on $$AC$$ has radius 20, so its area is $$\frac{1}{2}\pi (20)^2 = 200\pi$$.

Dividing the numerator and the denominator by $$x$$ turns the equation into $$\frac{x + \frac{1}{x} - 1}{x + \frac{1}{x} + 1} = \frac{11}{13}$$. Writing $$t = x + \frac{1}{x}$$ gives $$13(t - 1) = 11(t + 1)$$, so $$2t = 24$$ and $$t = 12$$.

From $$\frac{1}{2} \times AB \times 12 = 30$$ we get $$AB = 5$$, so $$AC = \sqrt{5^2 + 12^2} = 13$$. The square has side 3, so $$AD = 3$$ along $$AC$$, leaving $$DC = 13 - 3 = 10$$, and $$E$$ lies at a distance 3 from the line $$AC$$ because $$DE = 3$$ and $$DE$$ is perpendicular to $$AD$$. As $$A$$, $$D$$, $$C$$ are collinear, the quadrilateral $$ADEC$$ has the same area as $$\triangle DEC$$, which is $$\frac{1}{2} \times 10 \times 3 = 15$$.

In $$\triangle ABD$$, $$\angle BAD = 20^\circ + 60^\circ = 80^\circ$$ and $$\angle ADB = 50^\circ$$, so $$\angle ABD = 50^\circ$$ and hence $$AB = AD$$. In $$\triangle ACD$$, $$\angle CAD = 60^\circ$$ and $$\angle ADC = 50^\circ + 10^\circ = 60^\circ$$, so the triangle is equilateral and $$AC = AD$$. Therefore $$AB = AC$$, and in the isosceles triangle $$ABC$$ with $$\angle BAC = 20^\circ$$ we get $$\angle ACB = \frac{180^\circ - 20^\circ}{2} = 80^\circ$$.

From $$x^2 + x + 1 = 0$$ we get $$x^3 - 1 = (x - 1)(x^2 + x + 1) = 0$$, so $$x^3 = 1$$. Since $$48$$, $$51$$ are multiples of 3, the powers reduce to $$x + x^2 + 1 + x + x^2$$. Using $$x^2 + x + 1 = 0$$ this equals $$0 + (x + x^2) = -1$$.

TO BE FILLED - the printed condition must be checked against the paper, since as printed it does not lead to any of the four options. If the condition is $$a + \frac{1}{a} = \sqrt{2}$$, then $$a^3 + \frac{1}{a^3} = \left(a + \frac{1}{a}\right)^3 - 3\left(a + \frac{1}{a}\right) = 2\sqrt{2} - 3\sqrt{2} = -\sqrt{2}$$.

Put $$a = 3k$$, $$c = 4k$$ and $$b = 3m$$, $$d = 4m$$. Then $$\sqrt{a^2 + c^2} = 5k$$ and $$\sqrt{b^2 + d^2} = 5m$$, so $$5k - 5m = 25$$ and $$k - m = 5$$. Hence $$c - d = 4k - 4m = 4 \times 5 = 20$$.

Taking the successive segments along perpendicular directions, going from $$A$$ to $$E$$ the total horizontal displacement is $$BC + DE = 1 + 4 = 5$$ and the total vertical displacement is $$AB + CD = 5 + 4 = 9$$. By the Pythagoras theorem $$AE = \sqrt{5^2 + 9^2} = \sqrt{25 + 81} = \sqrt{106}$$.

The diagonals bisect each other at right angles, so if the other diagonal is $$d$$ and the side is $$s$$, then $$s^2 = 12^2 + \left(\frac{d}{2}\right)^2$$, that is $$(2s - d)(2s + d) = 576$$. Writing $$2s - d = 2p$$ and $$2s + d = 2q$$ gives $$pq = 144$$ with $$p + q$$ even, and the pairs $$(2, 72)$$, $$(4, 36)$$, $$(6, 24)$$, $$(8, 18)$$ work. These give $$d = 70, 32, 18, 10$$ with $$s = 37, 20, 15, 13$$, so 4 rhombuses are possible.

The area is $$\frac{1}{2}(p + q)h$$, and each of $$p$$ and $$q$$ becomes $$1.1$$ times its value while $$h$$ is unchanged. So the new area is $$\frac{1}{2}(1.1p + 1.1q)h = 1.1 \times \frac{1}{2}(p + q)h$$, an increase of 10%.

From $$ab + \frac{1}{ab} = \frac{37}{6} = 6 + \frac{1}{6}$$ and $$ab$$ an integer we get $$ab = 6$$. The integer pairs with product 6 that also satisfy $$a + \frac{1}{b} = \frac{7}{2}$$ force $$a = 3$$ and $$b = 2$$. Hence $$\frac{a^2 + b^2}{a^2 - b^2} = \frac{9 + 4}{9 - 4} = \frac{13}{5}$$.

Since $$GEF$$ is perpendicular to the line $$ACE$$ at $$E$$, $$\angle GEA = 90^\circ$$, so $$5x - y = 90$$. As $$BCD$$ and $$ACE$$ are straight lines, $$\angle DCE$$ is vertically opposite $$\angle ACB = 60^\circ$$, and the right angle of $$\triangle CDE$$ at $$D$$ gives $$\angle DEC = 30^\circ$$. Because $$EF$$ is perpendicular to $$EC$$, $$\angle DEF = 90^\circ - 30^\circ = 60^\circ$$, so $$x + y = 60$$. Solving gives $$x = 25$$ and $$y = 35$$, hence $$4x + 3y = 100 + 105 = 205$$.

Cancelling $$y$$ from $$x + y < y$$ gives $$x < 0$$, and cancelling $$x$$ from $$x - y > x$$ gives $$-y > 0$$, that is $$y < 0$$. Nothing more can be said about which of $$x$$ and $$y$$ is larger, so the correct relation is $$x < 0$$ and $$y < 0$$.

The first part is $$\left(\frac{1207}{30} - \frac{461}{12}\right) \div \frac{109}{10} = \frac{109}{60} \times \frac{10}{109} = \frac{1}{6}$$. The second part is $$\frac{77}{120} \times \frac{20}{11} = \frac{7}{6}$$, so the bracket equals $$\frac{1}{6} + \frac{7}{6} = \frac{4}{3}$$. Since $$\frac{4.2}{0.008} = 525$$, the value is $$\frac{4}{3} \times 525 = 700$$.

Using $$a^3 + b^3 = (a + b)^3 - 3ab(a + b)$$ we get $$1^3 - 3 \times 2 \times 1 = 1 - 6 = -5$$.

In a mirror the four digits appear in reverse order and each digit is itself reflected, so only $$0, 1, 2, 5, 8$$ remain readable, with 2 and 5 interchanging. Trying the afternoon hours whose digits come from this set, the display 12.21 is reflected as 15.51, and 15.51 is exactly three and a half hours after 12.21. So the actual time was 12.21 in the afternoon.

The area gives $$\frac{1}{2}(10 + 20)h = 180$$, so the height is $$h = 12$$. Dropping perpendiculars from $$A$$ and $$B$$ to $$CD$$ leaves $$\frac{20 - 10}{2} = 5$$ on each side. Hence $$BC = \sqrt{12^2 + 5^2} = \sqrt{169} = 13$$.

Each interior angle of a regular 90-gon is $$180^\circ - \frac{360^\circ}{90} = 176^\circ$$. At each end of the chosen edge the extension of the adjacent side makes an angle of $$180^\circ - 176^\circ = 4^\circ$$ with that edge, so both base angles are $$4^\circ$$. The apex angle is therefore $$180^\circ - 8^\circ = 172^\circ$$, which is the largest angle.

Count the products $$pq \le 100$$ with $$p < q$$ both prime. For $$p = 2$$ the prime $$q$$ can be any of the 14 primes from 3 to 47, for $$p = 3$$ any of the 9 primes from 5 to 31, for $$p = 5$$ any of $$7, 11, 13, 17, 19$$, and for $$p = 7$$ either 11 or 13. No larger $$p$$ works, so the total is $$14 + 9 + 5 + 2 = 30$$.

If the number is $$x$$, the correct answer is $$5x$$ and the wrong answer is $$\frac{x}{5}$$, so the error is $$5x - \frac{x}{5} = \frac{24x}{5}$$. The percentage error is $$\frac{24x/5}{5x} \times 100 = \frac{24}{25} \times 100 = 96$$.

The first radicand is $$\frac{121}{49} + \frac{3600}{49} = \frac{3721}{49} = \left(\frac{61}{7}\right)^2$$ and the second is $$\frac{225}{49} + \frac{64}{49} = \frac{289}{49} = \left(\frac{17}{7}\right)^2$$. So the expression equals $$\frac{61}{7} - \frac{17}{7} = \frac{44}{7}$$, where 44 and 7 are coprime. Hence $$a - b = 44 - 7 = 37$$.

The rule simplifies to $$f(n + 1) = f(n) + \frac{2}{3}$$, so the values form an arithmetic progression with common difference $$\frac{2}{3}$$. Therefore $$f(2026) = f(1) + 2025 \times \frac{2}{3} = 2 + 1350 = 1352$$.

The new area is $$\left(1 + \frac{x}{100}\right)^2$$ times the old area, and this must equal 2, so $$1 + \frac{x}{100} = \sqrt{2}$$. Hence $$x = 100(\sqrt{2} - 1) \approx 41.42$$, independent of the given side lengths.

TO BE FILLED - the printed expression contains no $$x$$ and its computed value is not $$\frac{17}{27}$$, so the statement must be re-read from the paper before the answer is filled in. The denominator does simplify neatly, since $$(0.5 - 0.375) \div 0.125 = 1$$ and $$\frac{1}{4} \div (0.358 - 0.108) = 1$$, giving 2.

Take each vessel to hold 3 units. The second vessel has 1 unit of water and receives 2 units of milk from the first, leaving 1 unit of milk in the first vessel, while the second holds milk and water in the ratio 2 to 1. Pouring back 2 units to fill the first vessel adds $$\frac{4}{3}$$ of milk and $$\frac{2}{3}$$ of water, so the first vessel has $$\frac{7}{3}$$ milk and $$\frac{2}{3}$$ water, a ratio of 7 to 2. Hence $$a + b = 9$$.

TO BE FILLED - as printed the equation reduces to $$x^2(n^2 + 1) - 3x(n^2 - 1) + (2n^2 - 3) = 0$$, whose roots depend on $$n$$ and are not even real for $$n = 2$$, so the statement must be re-read from the paper before the answer is filled in.

Put $$m = n + 12$$, so that $$12n = 12m - 144$$ and $$m$$ must divide 144. Since $$n \ge 1$$ we need $$m > 12$$, and the divisors of 144 greater than 12 are $$16, 18, 24, 36, 48, 72, 144$$. These give $$n = 4, 6, 12, 24, 36, 60, 132$$, so there are 7 possible values.

The number $$2k$$ is written $$2k$$ times, so after $$k$$ blocks the number of terms written is $$2 + 4 + \cdots + 2k = k(k + 1)$$. For $$k = 9$$ this is 90 and for $$k = 10$$ it is 110, so the 100th term lies in the tenth block. That block consists of the number $$2 \times 10 = 20$$.

Write $$N = 10a + b$$, so the condition becomes $$10a + b = ab + 3(a + b)$$, which simplifies to $$7a = b(a + 2)$$, that is $$b = \frac{7a}{a + 2}$$. Testing $$a = 1$$ to $$9$$, only $$a = 5$$ gives an integer, namely $$b = 5$$. So $$N = 55$$, and indeed $$25 + 3 \times 10 = 55$$.

The semi-perimeter is $$s = \frac{17 + 25 + 26}{2} = 34$$, so $$s - a = 17$$, $$s - b = 9$$ and $$s - c = 8$$. By Heron's formula the area is $$\sqrt{34 \times 17 \times 9 \times 8} = \sqrt{41616} = 204$$.