Screening Test – Bhaskara Contest(NMTC JUNIOR LEVEL—IX and X Grades)2024-2025
Join Trial or Access Free ResourcesJoin 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 $\rule{2cm}{0.2mm}$ .
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