Bhaskara Contest (NMTC Junior 2018 - IX and X Grades) - Stage I- Problems and Solution
Join Trial or Access Free ResourcesJoin Trial or Access Free ResourcesThe value of $\frac{3+\sqrt{6}}{8 \sqrt{3}-2 \sqrt{12}-\sqrt{32}+\sqrt{50}-\sqrt{27}}$ is
(A) $\sqrt{2}$
(B) $\sqrt{3}$
(C) $\sqrt{6}$
(D) $\sqrt{18}$
A train moving with a constant speed crosses a stationary pole in 4 seconds and a platform $75 \mathrm{~m}$ long in 9 seconds. The length of the train is (in meters)
(A) 56
(B) 58
(C) 60
(D) 62
One of the factors of $9 x^2-4 z^2-24 x y+16 y^2+20 y-15 x+10$ is
(A) $3 x-4 y-2 z$
(B) $3 x+4 y-2 z$
(C) $3 x+4 y+2 z$
(D) $3 x-4 y+2 z$
The natural number which is subtracted from each of the four numbers $17,31,25,47$ to give four numbers in proportion is
(A) 1
(B) 2
(C*) 3
(D) 4
The solution to the equation $5\left(3^x\right)+3\left(5^x\right)=510$ is
(A) 2
(B) 4
(C) 5
(D) No solution
If $(x+1)^2=x$, the value of $11 x^3+8 x^2+8 x-2$ is
(A) 1
(B) 2
(C) 3
(D) 4
There are two values of $m$ for which the equation $4 x^2+m x+8 x+9=0$ has only one solution for $x$. The sum of these two value of $m$ is
(A) 1
(B) 2
(C) 3
(D) 4
The number of zeros in the product of the first 100 natural numbers is
(A) 12
(B) 15
(C) 18
(D) 24
The length of each side of a triangle in increased by $20 \%$ then the percentage increase of area is
(A) $60 \%$
(B) $120 \%$
(C) $80 \%$
(D) $44 \%$
The number of pairs of relatively prime positive integers $(a, b)$ such that $\frac{a}{b}+\frac{15 b}{4 a}$ is an integer is
(A) 1
(B) 2
(C) 3
(D) 4
The four digit number $8 a b 9$ is a perfect square. The value of $a^2+b^2$ is
(A) 52
(B) 62
(C) 54
(D) 68
$a, b$ are positive real numbers such that $\frac{1}{a}+\frac{9}{b}=1$. The smallest value of $a+b$ is
(A) 15
(B) 16
(C) 17
(D) 18
$a, b$ real numbers. The least value of $a^2+a b+b^2-a-2 b$ is
(A) 1
(B) 0
(C) -1
(D) 2
I is the incenter of a triangle $\mathrm{ABC}$ in which $\angle \mathrm{A}=80^{\circ} . \angle \mathrm{BIC}=$
(A) $120^{\circ}$
(B) $110^{\circ}$
(C) $125^{\circ}$
(D) $130^{\circ}$
In the adjoining figure $A B C D$ is a square and DFEB is a rhombus $\angle C D F=$
(A) $15^{\circ}$
(B) $18^{\circ}$
(C) $20^{\circ}$
(D) $30^{\circ}$
$A B C D$ is a square $E, F$ are point on $B C, C D$ respectively and $E A F=45^{\circ}$. The value of $\frac{E F}{B E+D F}$ is $\rule{1cm}{0.15mm}$
The average of 5 consecutive natural numbers is 10 . The sum of the second and fourth of these numbers is $\rule{1cm}{0.15mm}$
The number of natural number $n$ for which $n^2+96$ is a perfect square is $\rule{1cm}{0.15mm}$
$n$ is an integer and $\sqrt{\frac{3 n-5}{n+1}}$ is also an integer. The sum of all such $n$ is $\rule{1cm}{0.15mm}$
$\frac{a}{b}$ is a fraction where $a, b$ have no common factors other 1 . b exceeds a by 3 . If the numerator is increased by 7 , the fraction is increased by unity. The value of $a+b$ $\rule{1cm}{0.15mm}$
If $x=\sqrt[3]{2}+\frac{1}{\sqrt[3]{2}}$ then the value of $2 x^3-6 x$ is $\rule{1cm}{0.15mm}$
The angle of a heptagon are $160^{\circ}, 135^{\circ}, 185^{\circ}, 140^{\circ}, 125^{\circ}, x^{\circ}, x^{\circ}$. The value of $x$ is $\rule{1cm}{0.15mm}$
$A B C$ is a triangle and $A D$ is its altitude. If $B D=5 D C$, then the value of $\frac{3\left(A B^2-A C^2\right)}{B C^2}$ is $\rule{1cm}{0.15mm}$
As sphere is inscribed in a cube that has surface area of $24 \mathrm{~cm}^2$. A second cube is then inscribed within the sphere. The surface area of the inner cube $\left(\right.$ in $\left.\mathrm{cm}^2\right)$ is $\rule{1cm}{0.15mm}$
A positive integer $n$ is multiple of 7 . If $\sqrt{n}$ lies between 15 and 16 , the number of possible values (s) of n is $\rule{1cm}{0.15mm}$
The value of $x$ which satisfies the equation $\frac{\sqrt{x+5}+\sqrt{x-16}}{\sqrt{x+5}-\sqrt{x-16}}=\frac{7}{3}$ is $\rule{1cm}{0.15mm}$
$\mathrm{M}$ man do a work in $\mathrm{m}$ days. If there had been $\mathrm{N}$ men more, the work would have been finished $\mathrm{n}$ days earlier, then the value of $\frac{m}{n}-\frac{M}{N}$ is $\rule{1cm}{0.15mm}$
The sum of the digit of a two number is 15 . If the digits of the given number are reversed, the number is increased by the square of 3 . The original number is $\rule{1cm}{0.15mm}$
When expanded the units place of $(3127)^{173}$ is $\rule{1cm}{0.15mm}$
If $a:(b+c)=1: 3$ and $c:(a+b)=5: 7$, then $b:(c+a)$ is $\rule{1cm}{0.15mm}$
