Join Trial or Access Free ResourcesIf $x^2+x=1$, then the value of $\frac{x^7+34}{x+2}$ is equal to
a) 7
b) 1
c) 13
d) 17
The angle between the hour hand and the minute hand of a clock at the time $9: 38 \mathrm{pm}$ is
a) $60^{\circ}$
b) $61^{\circ}$
c) $59^{\circ}$
d) $62^{\circ}$

In the adjoining figure, $A O B$ is a diameter of the circle with centre O. PC and PD are two tangents. Then the measure of $\angle E P D$ is $\qquad$
a) $15^{\circ}$
b) $10^{\circ}$
c) $12^{\circ}$
d) $20^{\circ}$
The value of $x$ satisfying $4^x-3^{x-1 / 2}=3^{x+1 / 2}-2^{2 x-1}$ is of the form $\frac{a}{b}$ where $\operatorname{gcd}(a, b)=1$. Then the value of $\left(\frac{a+b}{a-b}\right)$ is equal to
a) 7
b) -5
c) 4
d) 5
The number of polynomials of the form $\left(x^3+a x^2+b x+c\right)$ which are divisible by $x^2+1$ where $a, b, c \in{1,2,3,4, \ldots, 12}$ is
a) $12^3$
b) $12^2$
c) 12
d) 1
The number of real solutions of the equation $\frac{(x+2)(x+3)(x+4)(x+5)}{(x-2)(x-3)(x-4)(x-5)}=1$ is
a) 1
b) 2
c) 3
d) 0
If $a=\sqrt{23 a+b}, b=\sqrt{23 b+a}, a \neq b$, then the value of $\sqrt{a^2+b^2+48}$ is
a) 30
b) 25
c) 24
d) 23

In the adjoining figure, PA and PB are tangents to the circle.
$A C$ is parallel to $P B$.
Then measure of $\angle C D A$ is
a) $118^{\circ}$
b) $108^{\circ}$
c) $98^{\circ}$
d) $88^{\circ}$
If $\sqrt{\frac{19^8+19^x}{19^x+1}}=361$, then $x$ satisfies the equation
a) $4 x^2-7 x-15=0$
b) $2 x^2-9 x-5=0$
c) $3 x^2+11 x-4=0$
d) $3 x^2-11 x-4=0$
If $S=4^2+2.5^2+3.6^2+\ldots \ldots \ldots+25.28^2$, then the value of $\frac{S}{325}$ is equal to
a) 436
b) 326
c) 346
d) 324
A sequence $\{a_n\}, n \geq 1$ with $a_1=\frac{1}{2}$ and $a_n=\frac{a_{n-1}}{2 n a_{n-1}+1}$ is given. Then the value of $a_1+a_2+a_3+\ldots \ldots \ldots+a_{2024}$ is equal to
a) $\frac{2025}{2024}$
b) $\frac{2024}{2025}$
c) 2025
d) $\frac{1}{2025}$
If $\alpha$ and $\beta(\alpha>\beta)$ satisfy the equation $x^{1+\log _{10} x}=10 x$ then the value of $\alpha+\frac{1}{\beta}$ is equal to
a) 100
b) 20
c) 10
d) $\frac{1}{100}$

In the adjoining figure, four successively touching circles are placed in the interior of $\angle A O B$. 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
a) 42
b) 48
c) 52
d) 56
The coefficient of $x$ in the equation $x^2+p x+q=0$ 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
a) $30 x^2+109 x+30=0$
b) $20 x^2-107 x+20=0$
c) $30 x^2-109 x+30=0$
d) $20 x^2+107 x+20=0$
If $(1+x y+x+y)^2-(1-x y+x-y)^2=k y(1+x)^2$, then $k$ equals to
a) 1
b) 2
c) 3
d) 4
When $x^{10}+1$ is divided by $x^2+1$, we get
$$
a x^8+b x^7+c x^6+d x^5+e x^4+f x^3+g x^2+h x+k
$$
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 $\rule{2cm}{0.2mm}$
The equation $x^4-4 x^3+a x^2+b x+1=0$ has 4 positive roots. Then $a+b=$ $\rule{2cm}{0.2mm}$

In the adjoining figure, $B O C$ is the diameter of the semicircle with centre O.
DE is the tangent at D .
If $\mathrm{AB}=k(\mathrm{AE})$, then the numerical value of $k$ is $\rule{2cm}{0.2mm}$
In triangle $A B C$,
$\tan A: \tan B: \tan C=1: 2: 3$.
If $\frac{A C}{A B}=\frac{p \sqrt{q}}{r}$, where $q$ is Square free and $\operatorname{gcd}(p, r)=1$ then the value of $p+q+r$ is $\rule{2cm}{0.2mm}$
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 $\rule{2cm}{0.2mm}$ .
The greatest number that leaves the same remainder when it divides 30,53 and 99 is$\rule{2cm}{0.2mm}$ ـ.
If $f(x+1)=x^2-3 x+2$ and if the roots of the equation $f(x)=0$ are $\alpha$ and $\beta$, then the value of $\alpha^2+\beta^2$ is $\rule{2cm}{0.2mm}$ .
The maximum volume of a cylinder is cut from a cube of edge $a$. The volume of the remaining solid is $k a^3$, where $k=\frac{p}{q}, \operatorname{gcd}(p, q)=1$. Taking $\pi=\frac{22}{7}$, the value of $p+q$ is $\rule{2cm}{0.2mm}$ .
If the irreducible quadratic factor of $5 x^4+9 x^3-2 x^2-4 x-8$ is $a x^2+b x+c$, then the value of $a^2+b^2-c^2$ is $\rule{2cm}{0.2mm}$ .

In the adjoining figure, POQ is the diameter of the semicircle with centre O.
OABC is a square whose area is $36 \mathrm{~cm}^2$. If $\mathrm{QD}=x \mathrm{~cm}$, the value of $x \sqrt{3}$ is $\rule{2cm}{0.2mm}$ .
If $a=\sqrt{2024}, b=\sqrt{2025}$, the value of $2(a b)^{1 / 2}(a+b)^{-1}\left\{1+\frac{1}{4}\left(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\right)^2\right\}^{1 / 2}$ is $\rule{2cm}{0.2mm}$
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 $\rule{2cm}{0.2mm}$
The value of $\left(\frac{\sqrt{10}}{10}\right)^{\left(\log _{10} 9\right)-2}$ is of the form $\frac{a}{b}$, where $a, b$ are relatively prime to each other. Then $a-b$ is equal to ـ.$\rule{2cm}{0.2mm}$

ABCD is a square. BE is the tangent to the semicircle on AD as diameter. The area of the triangle BCE is $216 \mathrm{~cm}^2$. The radius of the semicircle (in cm ) is $\rule{2cm}{0.2mm}$
$a, b, c, d$ are real constants in a $f(x)=a x^{2025}+b x^{2023}+c x^{2021}+d x^{2019}$ and $f(-4)=18$. Then the maximum value of $|f(4)|+|2 \cos x|$ is $\rule{2cm}{0.2mm}$ .

In 2025, 8 students from Cheenta Academy cracked the prestigious Regional Math Olympiad. In this post, we will share some of their success stories and learning strategies. The Regional Mathematics Olympiad (RMO) and the Indian National Mathematics Olympiad (INMO) are two most important mathematics contests in India.These two contests are for the students who are […]

Cheenta Academy proudly celebrates the success of 27 current and former students who qualified for the Indian Olympiad Qualifier in Mathematics (IOQM) 2025, advancing to the next stage — RMO. This accomplishment highlights their perseverance and Cheenta’s ongoing mission to nurture mathematical excellence and research-oriented learning.

Cheenta students shine at the Purple Comet Math Meet 2025 organized by Titu Andreescu and Jonathan Kanewith top national and global ranks.

Celebrate the success of Cheenta students in the Stanford Math Tournament. The Unified Vectors team achieved Top 20 in the Team Round.
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