BHASKARA Contest - NMTC - Screening Test – 2025

Problem 1

The greatest 4 -digit number such that when divided by 16,24 and 36 leaves 4 as remainder in each case is
А) 9994
B) 9940
C) 9094
D) 9904

Problem 2

$A B C D$ is a rectangle whose length $A B$ is 20 units and breadth is 10 units. Also, given $A P=8$ units. The area of the shaded region is $\frac{p}{q}$ sq unit, where $p, q$ are natural numbers with no common factors other than 1 . The value of $p+q$ is
A) 167
В) 147
C) 157
D) 137

Problem 3

The solution of $\frac{\sqrt[7]{12+x}}{x}+\frac{\sqrt[7]{12+x}}{12}=\frac{64}{3}(\sqrt[7]{x})$ is of the form $\frac{a}{b}$ where $a, b$ are natural numbers with $\operatorname{GCD}(a, b)=1$; then $(b-a)$ is equal to
A) 115
B) 114
C) 113
D) 125

Problem 4

The value of $(52+6 \sqrt{43})^{3 / 2}-(52-6 \sqrt{43})^{3 / 2}$ is
A) 858
В) 918
C) 758
D) 828

Problem 5

In the adjoining figure $\angle D C E=10^{\circ}$, $\angle C E D=98^{\circ}, \angle B D F=28^{\circ}$
Then the measure of angle $x$ is
A) $72^{\circ}$
B) $76^{\circ}$
C) $44^{\circ}$
D) $82^{\circ}$

Problem 6

$A B C$ is a right triangle in which $\angle \mathrm{B}=90^{\circ}$. The inradius of the triangle is $r$ and the circumradius of the triangle is R . If $\mathrm{R}: r=5: 2$, then the value of $\cot ^2 \frac{A}{2}+\cot ^2 \frac{C}{2}$ is
A) $\frac{25}{4}$
B) 17
C) 13
D) 14

Problem 7

If $(\alpha, \beta)$ and $(\gamma, \beta)$ are the roots of the simultaneous equations:

\[
|x-1|+|y-5|=1 ; \quad y=5+|x-1|
\]

then the value of $\alpha+\beta+\gamma$ is
A) $\frac{15}{2}$
B) $\frac{17}{2}$
C) $\frac{14}{3}$
D) $\frac{19}{2}$

Problem 8

Three persons Ram, Ali and Peter were to be hired to paint a house. Ram and Ali can paint the whole house in 30 days, Ali and Peter in 40 days while Peter and Ram can do it in 60 days. If all of them were hired together, in how many days can they all three complete $50 \%$ of the work?
A) $24 \frac{1}{3}$
B) $25 \frac{1}{2}$
C) $26 \frac{1}{3}$
D) $26 \frac{2}{3}$

Problem 9

$\frac{\sqrt{a+3 b}+\sqrt{a-3 b}}{\sqrt{a+3 b}-\sqrt{a-3 b}}=x$, then the value of $\frac{3 b x^2+3 b}{a x}$ is
A) 1
B) 2
C) 3
D) 4

Problem 10

The number of integral solutions of the inequation $\left|\frac{2}{x-13}\right|>\frac{8}{9}$ is
A) 1
B) 2
C) 3
D) 4

Problem 11

In the adjoining figure, $P$ is the centre of the first circle, which touches the other circle in C . PCD is along the diameter of the second circle. $\angle \mathrm{PBA}=20^{\circ}$ and $\angle \mathrm{PCA}=30^{\circ}$.

The tangents at B and D meet at E . The measure of the angle $x$ is
A) $75^{\circ}$
B) $80^{\circ}$
C) $70^{\circ}$
D) $85^{\circ}$

Problem 12

If $\alpha, \beta$ are the values of $x$ satisfying the equation $3 \sqrt{\log _2 x}-\log _2 8 x+1=0$, where $\alpha<\beta$, then the value of $\left(\frac{\beta}{\alpha}\right)$ is
A) 2
B) 4
C) 6
D) 8

Problem 13

When a natural number is divided by 11 , the remainder is 4 . When the square of this number is divided by 11 , the remainder is
A) 4
B) 5
C) 7
D) 9

Problem 14

The unit's digit of a 2-digit number is twice the ten's digit. When the number is multiplied by the sum of the digits the result is 144 . For another 2-digit number, the ten's digit is twice the unit's digit and the product of the number with the sum of its digits is 567 . Then the sum of the two 2 -digit numbers is
A) 68
В) 86
C) 98
D) 87

Problem 15

$A B C D E$ is a pentagon. $\angle A E D=126^{\circ}, \angle B A E=\angle C D E$ and $\angle A B C$ is $4^{\circ}$ less than $\angle B A E$ and $\angle B C D$ is $6^{\circ}$ less than $\angle C D E . P R, Q R$ the bisectors of $\angle B P C, \angle E Q D$ respectively, meet at $R$. Points $\mathrm{P}, \mathrm{C}, \mathrm{D}, \mathrm{Q}$ are collinear. Then measure of $\angle P R Q$ is
A) $151^{\circ}$
B) $137^{\circ}$
C) $141^{\circ}$
D) $143^{\circ}$

Problem 16

$a, b, c$ are real numbers such that $b-c=8$ and $b c+a^2+16=0$.
The numerical value of $a^{2025}+b^{2025}+c^{2025}$ is $\rule{2cm}{0.2mm}$.

Problem 17

Given $f(x)=\frac{2025 x}{x+1}$ where $x \neq-1$. Then the value of $x$ for which $f(f(x))=(2025)^2$ is $\rule{2cm}{0.2mm}$.

Problem 18

The sum of all the roots of the equation $\sqrt[3]{16-x^3}=4-x$ is $\rule{2cm}{0.2mm}$.

Problem 19

In the adjoining figure, two
Quadrants are touching at $B$.
$C E$ is joined by a straight line, whose mid-point is $F$.

The measure of $\angle C E D$ is $\rule{2cm}{0.2mm}$.

Problem 20

The value of $k$ for which the equation $x^3-6 x^2+11 x+(6-k)=0$ has exactly three positive integer solutions is $\rule{2cm}{0.2mm}$.

Problem 21

The number of 3-digit numbers of the form $a b 5$ (where $a, b$ are digits) which are divisible by 9 is $\rule{2cm}{0.2mm}$.

Problem 22

If $a=\sqrt{(2025)^3-(2023)^3}$, the value of $\sqrt{\frac{a^2-2}{6}}$ is $\rule{2cm}{0.2mm}$.

Problem 23

In a math Olympiad examination, $12 \%$ of the students who appeared from a class did not solve any problem; $32 \%$ solved with some mistakes. The remaining 14 students solved the paper fully and correctly. The number of students in the class is $\rule{2cm}{0.2mm}$.

Problem 24

When $a=2025$, the numerical value of
$\left|2 a^3-3 a^2-2 a+1\right|-\left|2 a^3-3 a^2-3 a-2025\right|$ is $\rule{2cm}{0.2mm}$.

Problem 25

A circular hoop and a rectangular frame are standing on the level ground as shown. The diagonal $A B$ is extended to meet the circular hoop at the highest point $C$. If $A B=18 \mathrm{~cm}, B C=32 \mathrm{~cm}$, the radius of the hoop (in cm ) is $\rule{2cm}{0.2mm}$.

Problem 26

' $n$ ' is a natural number. The number of ' $n$ ' for which $\frac{16\left(n^2-n-1\right)^2}{2 n-1}$ is a natural number is $\rule{2cm}{0.2mm}$.

Problem 27

The number of solutions $(x, y)$ of the simultaneous equations $\log _4 x-\log _2 y=0, \quad x^2=8+2 y^2$ is $\rule{2cm}{0.2mm}$.

Problem 28

In the adjoining figure,
$P A, P B$ are tangents.
$A R$ is parallel to $P B$

$P Q=6 ; Q R=18 .$

Length $S B= \rule{2cm}{0.2mm}$.

Problem 29

A large watermelon weighs 20 kg with $98 \%$ of its weight being water. It is left outside in the sunshine for some time. Some water evaporated and the water content in the watermelon is now $95 \%$ of its weight in water. The reduced weight in kg is $\rule{2cm}{0.2mm}$.

Problem 30

In a geometric progression, the fourth term exceeds the third term by 24 and the sum of the second and third term is 6 . Then, the sum of the second, third and fourth terms is $\rule{2cm}{0.2mm}$.

The Organic Math of Origami

Did you know that there exists a whole set of seven axioms of Origami Geometry just like that of the Euclidean Geometry?

Related image — The Origami in building the solar panel maximizing the input of Solar Power and minimizing the Volume of the satellite

Instead of being very mathematically strict, today we will go through a very elegant result that arises organically from Origami.

Before that, let us travel through some basic terminologies. Be patient for a few more minutes and wait for the gem to arrive.

In case you have forgotten what Origami is, the following pictures will remove the dust from your memories.

Origami (from the Japanese oru, “to fold,” and kami, “paper”) is a traditional Japanese art of folding a sheet of paper, usually square, into a representation of an object such as a bird or flower.

Flat origami refers to configurations that can be pressed flat, say between the pages of a book, without adding any new folds or creases.

Image result for flat origami — Flat Origami

When an origami object is unfolded, the resulting diagram of folds or creases on the paper square is called a crease pattern.

We denote mountain folds by unbroken lines and valley folds by dashed
lines. A vertex of a crease pattern is a point where two or more folds intersect, and a flat vertex fold is a crease pattern with just one vertex.

In a crease pattern, we see two types of folds, called mountain folds and valley folds.

Now, if you get to play with your hands, you will get to discover a beautiful pattern.

The positive difference between the dotted lines and the full lines is always 2 at a given vertex. The dotted line denotes the mountain fold and the full line denotes the valley fold.

This is encoded in the following theorem.

Maekawa's Theorem: The difference between the number of mountain
folds and the number of valley folds in a flat vertex fold is two.

Isn't it strange ?

Those who are familiar graph theory may think it is related to the Euler Number.

We will do the proof step by step but you will weave together the steps to understand it yourself. The proof is very easy.

Step 1:

Let n denote the number of folds that meet at the vertex, m of which
are mountain folds and v that are valley folds, so that n = m + v. (m for mountain folds and v for valley folds.)

Step 2:

Consider the cross section of a flat vertex.

Step 3:

Consider the creases as shown and fold it accordingly depending on the type of fold - mountain or valley.

Folding the Cross Section of a Flat Vertex along the creases

Step 4:

Now observe that we get the following cross section. Observe that the number of sides of the formed polygon is n = m + v.

Step 5:

We will also count the angle sum of the n sided polygon in the following way. Consider the polygon formed.

We get a four sided polygon here with internal angles 0 degree and 360 degrees.

An upper view of the four sided triangle formed.

Observe that the vertex 2 is the Valley Fold and other vertices are Mountain Fold. Also the angle subtended the vertex due to valley fold is 360 degrees and that of due to the mountain fold is 0 degrees.

Therefore, the sum of the internal angles is v.360 degrees.

Step 6:

Now we also know that the sum of internal angles formed by n vertices is 180.(n-2), which is = v.360.

Hence we get by replacing n by m+v, that m - v = 2.

QED

So simple and yet so beautiful and magical right?

But it is just the beginning!

There are lot more to discover ...

Do you observe any pattern or any different symmetry about the creases or even in other geometry while playing with just paper and folding them?

We will love to hear it from you in the comments.

Do you know Cheenta is bringing out their third issue of the magazine "Reason, Debate and Story" this summer?

Do you want to write an article for us?

Email us at babinmukherjee08@gmail.com.