Problem 1 of 20
Let $AB$ be a diameter of a circle $C_1$ of radius $30\ cm$ and with center $O$. Two circles $C_2$ and $C_3$ of radii $15\ cm$
and $10\ cm$ touch $C_1$ internally at $A$ and $B$ respectively. $A$ fourth circle $C_4$ touches $C_1$, $C_2$ and $C_3$. What is
the largest possible radius of $C_4$
Solution:
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We should observe that the radius of $C_1$ is $30\ cm$ and the radius of $C_2$ is $15\ cm$, which makes the radius of $C_1$ equal to the diameter of $C_2$. Therefore $C_2$ will pass through the centre of $C_1$, that is $O$.
There are two ways to draw the circle $C_4$. We can draw it touching $C_1$ internally and $C_2$ and $C_3$ externally as shown in the figure below.
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The second way to draw $C_4$ is to make $C_2$ touch it internally and $C_3$ touch it externally, as shown below.
The diameter of $C_2$ is equal to the radius of $C_1$. In the first case the diameter of $C_4$ is less than that of $C_2$ while in the second case it is greater than that of $C_2$.
Clearly the size of $C_4$ is greater in the second case, therefore we will ignore the first case and find the diameter of $C_4$ in the second case.
$PB = 2 \times 10 = 20$
$AO = 2 \times 15 = 30$
$\therefore OP = 2 \times 30 - 20 - 30 = 10$
$\therefore$ Diameter of $C_4 = AP = AO + OP = 30 + 10 = 40$
$\therefore$ Radius of $C_4 = \dfrac{40}{2} = 20$
Problem 2 of 20
A $5 \times 5 \times 5$ cube is built using unit cubes. How many different cuboids (that differ in at least one unit cube)
can be formed using the same number of unit cubes ?
Solution:
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Let's draw the cube as shown below:
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The way of selecting any cuboid which is part of this cube is to select any two planes parallel to the $X-Y$ plane, two planes parallel to the $Y-Z$ plane and two planes parallel to the $Z-X$ plane.
Suppose you were to select the cuboid formed by the two gray cubes shown in the above figure, you will need to select the two planes shown in red, which are parallel to the $X-Y$ plane, two planes shown in blue parallel to the $Y-Z$ plane, one of them being the right outer surface of the large cube, and two planes parallel to the $Z-X$ plane shown in green, one of them being the front surface facing us.
Observe that including the outer surfaces, there are $6$ planes to choose from in each direction, and we have to choose two from each set of $6$.
We can choose two in a particular direction in $\xacomb{6}{2}$ ways, and for each of these selections, we can choose a set two in the other two directions.
Therefore, the total number of ways a cuboid can be chosen, is:
$\xacomb{6}{2} \times \xacomb{6}{2} \times \xacomb{6}{2}$
$= \dfrac{6 \times 5}{2} \times \dfrac{6 \times 5}{2} \times \dfrac{6 \times 5}{2}$
$= 15 \times 15 \times 15$
$= 3375$
Problem 3 of 20
What is the largest value of the positive integer $k$ such that $k$ divides $n^2 (n^2 – 1) (n^2 – n – 2)$ for every natural number $n$?
Solution:
We will factorise the given expression into possible smallest factor.$n^2(n^2 - 1)(n^2 - n - 2)$
$= n.n.(n-1)(n+1)(n-2)(n+1)$
Arranging the factors in ascending order, we get:
$(n-2)(n-1)n.n(n+1)(n+1)$
First let us take the duplicate factors $n$ and $n+1$ out and analyse the remaining part $(n-2)(n-1)n(n+1)$
These are $4$ consecutive integers.
If you take any $3$ consecutive integers, it is easy to show that one of them is divisible by $3$.
Similarly, since these are $4$ consecutive integers, $2$ of them will be even number, and any one of them has to be divisible by $4$.
Hence, we can guarantee that the given number $(n-2)(n-1)n(n+1)$ is always divisible by $2 \times 4 \times 3 = 24$
The factors which we earlier removed, that is, $n(n+1)$ are again two consecutive integers and one of them has to an even number, therefore, divisible by $2$.
We should also observe the special case for natural numbers $n$, where $1 \le n \le 2$, the given expression will become $0$, but $0$ is considered to be divisible by every other integer.
Therefore, the given number $n^2(n^2 - 1)(n^2 - n - 2)$ has to be divisible by $24 \times 2 = 48$.
Problem 4 of 20
A person kept rolling a regular (six faced) die until one of the numbers appeared third time on the top. This happened in $12\xasuper{th}$ throw and the sum of all the numbers in $12$ throws was $46$. Which number appeared least number of times?
(Select all possible values)
Solution:
Let the values of the rolls be:$a_1,\ a_2,\ a_3...\ a_{12}$
Let us list down the conditions given in the problem:
- No value is repeated more than thrice.
- Exactly one value repeats thrice
- The sum of all the values is $46$
Let us see what happened after the $11\xasuper{th}$ roll of the dice.
Since no number repeated thrice, then $5$ numbers repeated exactly twice, the $6\xasuper{th}$ appeared once.
On the twelfth roll, one of the $5$ repeated numbers showed up, hence the game stopped.
Let the number which did not repeat be $x$ and the number which appeared thrice be $y$.
If all the number repeated exactly twice in $12$ throws, the total would be $21 \times 2 = 42$
Therefore, $42 - x + y = 46$
Therefore, $y - x = 4$
Therefore the possible values of $x$ and $y$ are $1,\ 5$ and $2,\ 6$
Therefore, the number that appeared least number of times can be either $1$ or $2$.
N.B: For the original question as appeared in NSEJS, any one of the two correct options was treated as correct, we altered this question to require both correct options.
Problem 5 of 20
In a square $ABCD$, a point $P$ is inside the square such that $ABP$ is an equilateral triangle. The segment $AP$
cuts the diagonal $BD$ in $E$. Suppose $AE = 2$. The area of $ABCD$ is:
Solution:
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We can draw the diagram for the given problem as shown below:
Considering $\triangle AEB$,
$\angle EAB = 60^\circ$
$\angle EBA = 45^\circ$
$\therefore \angle AEB = 180 - (60 + 45) = 75^\circ$
And it is given that $AE = 2$
Using the $sine$ rule from , we get:
$\dfrac{sin(75)}{AB} = \dfrac{sin(45)}{AE}$
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$\Rightarrow \dfrac{sin(75)}{AB} = \dfrac{sin(45)}{2}$ $...eqn(i)$
We can derive the value of $sin(75)$ as:
$sin(75) = sin(45 + 30)$
$= sin(45).cos(30) + sin(30)cos(45)$
$= \dfrac{1}{\sqrt{2}}\dfrac{\sqrt{3}}{2} + \dfrac{1}{2}\dfrac{1}{\sqrt{2}}$
$= \dfrac{\sqrt{3} + 1}{2\sqrt{2}}$
From $eqn(i)$
$AB = \dfrac{2sin(75)}{sin(45)}$
$= 2 \left(\dfrac{\sqrt{3} + 1}{2\sqrt{2}}\right) \div \dfrac{1}{\sqrt{2}}$
$= 2 \left(\dfrac{\sqrt{3} + 1}{2\sqrt{2}}\right) \times \dfrac{\sqrt{2}}{1}$
$= \sqrt{3} + 1$
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Therefore, the area of $ABCD$
$= (\sqrt{3} + 1)^2$
$ = 3 + 1 + 2\sqrt{3}$
$ = 4 + 2\sqrt{3}$
Problem 6 of 20
Let $n$ be a positive integer not divisible by $6$. Suppose $n$ has $6$ positive divisors. The number of positive
divisors of $9n$ is:
Solution:
As we learnt , the number of factors of any number is given by $(c_1 + 1)(c_2 + 1)...(c_n + 1)$, where$c_1... c_n$ are the count of prime factors of the given number.
Since the given number, $n$ has $6$ factors, and $6$ can be written as either $(5+1)$ or or $(1 + 1) \times (2 + 1)$
Therefore, $n$ is composed of a single prime number, $p_1$ repeated $5$ times, that is:
$n = p_1 \times p_1 \times p_1 \times p_1 \times p_1$
or two distinct prime numbers, one of them repeated twice, that is:
$n = p_1 \times p_2 \times p_2$
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Since $9$ is composed of only one prime factor, that is $3$, we are going to consider three different cases here
$I.\ n$ is composed of only $3$
$II.\ n$ is composed of one $3$ and some prime factor, other than $2$, repeated twice (since $n$ is not divisible by $6$)
$II.\ n$ is composed of two $3$s and some prime factor, other than $2$, occurring once (since $n$ is not divisible by $6$)
$IV.\ n$ is composed of any prime number other than $3$.
$\underline{Case:\ I}$
The only number that is composed of $3$ as prime factor, and has $6$ divisors is $3^5$
In this case $9n$ can be written as $3^7$, which will have $8$ prime factors.
$\underline{Case:\ II}$
Let us say $n = 3^1 \times m^2$ where $m$ is a prime number other than $2$.
In this case $9n = 3^3 \times m^2$ which will have $(3 + 1)(2+1) = 12$ divisors.
$\underline{Case:\ III}$
Let us say $n = 3^2 \times m^1$ where $m$ is a prime number other than $2$.
In this case $9n = 3^4 \times m^1$ which will have $(4 + 1)(1+1) = 10$ divisors.
$\underline{Case:\ IV}$
Let $n = p^1 \times q^2$ where $p$ and $q$ are any prime numbers other than $3$
Therefore, $9n = 3^2 \times p^1 \times q^2$, and this number will have $(2 + 1)(1+1)(2 + 1) = 18$ divisors.
N.B: In the NSEJS paper this question was published as a single choice question, with $12$ and $18$ amongst the option values. Selecting any one of them would give full credit. We have modified this question to add the options $8$ and $10$ and made it a multiple correct option question, so that the student is able to analyse all the possible scenarios.
Problem 7 of 20
The value of $\dfrac{\sqrt{a + x} - \sqrt{a - x}}{\sqrt{a + x} + \sqrt{a - x}}$
when $x = \dfrac{2ab}{b^2 + 1}$
Solution:
Given that $x = \dfrac{2ab}{b^2 + 1}$
$\Rightarrow \dfrac{x}{a} = \dfrac{2b}{b^2 + 1}$
$\Rightarrow \dfrac{a}{x} = \dfrac{b^2 + 1}{2b}$ $...eqn(i)$
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Adding $1$ to both sides of $eqn(i)$, we get:
$\Rightarrow \dfrac{a}{x} + 1= \dfrac{b^2 + 1}{2b} + 1$
$\Rightarrow \dfrac{a + x}{x} = \dfrac{b^2 + 1 + 2b}{2b}$ $...eqn(ii)$
Again subtracting $1$ from both sides of $eqn(i)$,
$\Rightarrow \dfrac{a}{x} - 1= \dfrac{b^2 + 1}{2b} - 1$
$\Rightarrow \dfrac{a - x}{x} = \dfrac{b^2 + 1 - 2b}{2b}$ $...eqn(iii)$
Dividing $eqn(ii)$ by $eqn(iii)$, we get:
$\dfrac{a + x}{a - x} = \dfrac{b^2 + 2b + 1}{b^2 - 2b + 1}$
$\Rightarrow \dfrac{a + x}{a - x} = \dfrac{(1 + b)^2}{(1 - b)^2}$
$\Rightarrow \dfrac{\sqrt{a + x}}{\sqrt{a - x}} = \dfrac{1 + b}{1 - b}$
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Again performing the same steps as we did in $eqn(i)$, $(ii)$ and $(iii)$, that is adding $1$, subtracting $1$ and then dividing the two equattions, we get:
$\Rightarrow \dfrac{\sqrt{a + x} - \sqrt{a - x}}{\sqrt{a + x} + \sqrt{a - x}} = \dfrac{(1 + b) - (1 - b)}{(1 + b) + (1 - b)}$
$\Rightarrow \dfrac{\sqrt{a + x} - \sqrt{a - x}}{\sqrt{a + x} + \sqrt{a - x}} = \dfrac{2b}{2} = b$
The correct answer is $b$.
Problem 8 of 20
Two regular polygons of different number of sides are taken. In one of them, its sides are coloured red and diagonals are coloured green; in the other, sides are coloured green and diagonals are coloured red. Suppose there are $103$ red lines and $80$ green lines. The total number of sides the two polygons together have is:
Solution:
Let the polygons have $m$ and $n$ sides respectively.We know that an $p$ sided polygon has $\dfrac{p^2 - 3p}{2}$ diagonals.
Therefore, the first polygon has:
$m$ red sides and $\dfrac{m^2 - 3m}{2}$ green diagonals, while the second polygon has
$n$ green sides and $\dfrac{n^2 - 3n}{2}$ red diagonals.
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Therefore, the total number of the red lines is:
$m + \dfrac{n^2 - 3n}{2}$ and the total number of green lines is $n + \dfrac{m^2 - 3m}{2}$
Therefore,
$m + \dfrac{n^2 - 3n}{2} = 103$
$\Rightarrow \dfrac{2m + n^2 - 3n}{2} = 103$
$\Rightarrow$ $2m + n^2 - 3n = 206$ $... eqn(i)$
Similarly, using the red lines we get:
$2n + m^2 - 3m = 160$ $... eqn(ii)$
Subtracting $eqn(ii)$ from $eqn(i)$ we get:
$2m - 2n + n^2 - m^2 - 3n + 3m = 46$
$\Rightarrow 5(m - n) - (m^2 - n^2) = 46$
$5(m - n) - (m-n)(m + n) = 46$
$(m - n)\{5 - (m + n)\} = 46$
Now we can solve this equation the conventional way, which will make it quite complex. But we can use common observation to solve this equation much more easily.
$m$ and $n$ are the number of sides of two polygons, hence each of them has to be a positive integer, greater than or equal to $3$.
Therefore,
$m + n \ge 6$, hence $5 - (m + n)$ is a negative number. Since the right hand side of $eqn(iii)$ is positive, we can rewrite it as:
$(n - m)(m + n - 5) = 46$
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Since $n$ and $m$ are positive integers both the factors on the left hand side must be integers, and we know that the only ways to factor $46$ are as follows:
$I)\ 1 \times 46$
$II)\ 2 \times 23$
$III)\ 23 \times 2$
$IV)\ 46 \times 1$
$\underline{Case:\ 1}$
$n - m = 1$
$m + n - 5 = 46$
Adding the two we get:
$2n - 5 = 47$
$n = 26$
$m = 25$
Substituting the values of $m$ and $n$ on the left hand side of $eqn(i)$ we get:
$2m + n^2 - 3n$
$= 2(25) + 26^2 - 3(26)$
$= 50 + 676 - 52$
$= 674$
Since this values do not satisfy our equations, we will not consider these values.
$\underline{Case:\ 2}$
$n - m = 2$
$m + n - 5 = 23$
Therefore,
$2n = 30 \Rightarrow n = 15$
$m = 13$
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Substituting these values of $m$ and $n$ on the left hand side of $eqn(i)$ we get:
$2m + n^2 - 3n$
$= 2(13) + 15^2 - 3(15)$
$= 26 + 225 - 45$
$= 206$
These value satisfy the equation.
Again substituting these values in $eqn(ii)$ we get:
$2n + m^2 - 3m$
$= 2(15) + 13^2 - 3(13)$
$= 30 + 169 - 39$
$= 30 + 169 - 39$
$= 160$
These values satisfy this equation as well.
Since these values satisfy both equations, these are valid values.
$\underline{Case:\ 3}$
$n - m = 23$
$m + n - 5 = 2$
Therefore,
$n = 15$
$m = -8$
In this case $m$ is negative, which is not a valid value for sides of a polygon. Hence we will not consider this case.
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$\underline{Case:\ 4}$
$n - m = 46$
$m + n - 5 = 1$
Therefore,
$n = 26$
$m = -8$
In this case $m$ is negative, which is not a valid value for sides of a polygon. Hence we will not consider this case.
Since out of the four cases, only $Case:\ 2$ satisfies both the equations the valid values for $m$ and $n$ are $13$ and $15$
Therefore the correct answer is $13 + 15 = 28$
Problem 9 of 20
A box contains some red and some yellow balls. If one red ball is removed, one seventh of the remaining
balls would be red ; if one yellow ball is removed, one-sixth of the remaining balls would be red. If $n$
denotes the total number of balls in the box, then the sum of the digits of $n$ is:
Solution:
Let $r$ and $y$ be the number of red and yellow balls respectively.$\therefore r + y = n$
Based on the given condition:
$\dfrac{r-1}{r + y - 1} = \dfrac{1}{7}$
$\Rightarrow 7r - 7 = r + y - 1$
Similarly,
$\dfrac{r}{r + y - 1} = \dfrac{1}{6}$
$\Rightarrow 6r = r + y - 1$
Therefore,
$7r - 7 = 6r$
$\Rightarrow r = 7$
Therefore,
$6\times 7 = 7 + y - 1$
$42 = 6 + y$
$y = 36$
Therefore total number of balls $= n = 36 + 7 = 43$
The sum of the digits of $n = 4 + 3 = 7$
Problem 10 of 20
Let $ABCD$ be a rectangle. Let $X$ and $Y$ be points respectively on $AB$ and $CD$ such that
$AX : XB = 1 : 2 = CY : YD$. Join $AY$ and $CX$; let $BY$ intersect $CX$ in $K$; let $DX$ intersect $AY$ in $L$.
If $\dfrac{m}{n}$ denotes the ratio of the area $XKYL$ to that of $ABCD$, then $m + n$ equals:
Solution:
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Let us draw the diagram for the given problem as shown below.Additionally let us construct a straight line parallel to $BC$, passing to $K$ and meeting $AB$ and $CD$ at points $P$ and $Q$ respectively.
Considering $\triangle KBX$ and $\triangle KYC$
$\because DC \parallel AB$, $\angle KCY = \angle KXB$, $\angle KYC = \angle KBX$.
$\angle XKB = \angle CKY$ (vertically opposite angles).
Therefore, $\triangle KBX \sim \triangle KYC$
Given that $AX = \dfrac{1}{3} AB$ and $XB = \dfrac{2}{3} AB$, and,
$CY = \dfrac{1}{3} CD$ and $YD = \dfrac{2}{3} CD$
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$\because CD = AB$, $AX = CY$ and $BX = DY$
$\therefore BX:CY = 2:1$
Since, $\triangle KBX \sim \triangle KYC$, their corresponding altitudes $KP$ and $KQ$ are also in proportion.
$\therefore KP:KQ = BX:CY = 2:1$
$\because PQ = BC$, $KP = \dfrac{2}{3} BC$ and $KQ = \dfrac{1}{3} BC$
$\therefore ar[\triangle KYC] = \dfrac{1}{2} KQ \times CY = \dfrac{1}{2} \left(\dfrac{1}{3}BC\right) \times \left(\dfrac{1}{3}CD\right)$
$= \dfrac{1}{18} BC \times CD = \dfrac{1}{18} ar[ABCD]$
$ar[XKYL]$
$= ar[ABCD] - \left(ar[\triangle ADY] + ar[\triangle BCX] + ar[\triangle LXA] + ar[\triangle KYC]\right)$
$ar[\triangle ADY] = ar[\triangle BCX] = \dfrac{1}{2}BC \times BX = \dfrac{1}{2}BC \times \left(\dfrac{2}{3} AB\right)$
$= \dfrac{1}{3} ar[ABCD]$
$ar[\triangle LXA] = ar[\triangle KYC] = \dfrac{1}{18} [ABCD]$ (shown earlier)
Therefore,
$ar[XKYL] = ar[ABCD] - \left(ar[\triangle ADY] + ar[\triangle BCX] + ar[\triangle LXA]\right.$
$\left.+ ar[\triangle KYC]\right)$
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$= ar[ABCD] - \left(\dfrac{1}{3} ar[ABCD] + \dfrac{1}{3} ar[ABCD] + \dfrac{1}{18} ar[ABCD]\right.$
$\left.+ \dfrac{1}{18} ar[ABCD]\right)$
$= ar[ABCD] - \dfrac{7}{9} ar[ABCD]$
$= \dfrac{2}{9} ar[ABCD]$
$\therefore \dfrac{ar[XKYL]}{ar[ABCD]} = \dfrac{2}{9} = \dfrac{m}{n}$
$\therefore m + n = 2 + 9 = 11$
Problem 11 of 20
Let $ABC$ be an equilateral triangle. The bisector of $\angle BAC$ meets the circumcircle of $ABC$ in $D$. Suppose
$DB + DC = 4$. The diameter of the circumcircle of $ABC$ is:
Solution:
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We will solve this problem using two different approaches, the first one is Euclidean geometry based approach and the second one is a trigonometric approach.
Let us draw the diagram for the problem as follows.
As an additional construction, let us draw join the vertex $C$ with the centre $O$ of the circumcircle.
(Note that this construction is not required for the trigonometric approach)
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$\underline{Approach\ I\ (Euclidean\ Geometry):}$
Since $\triangle ABC$ is equilateral, its angle bisectors coincide with the perpendicular bisector of the sides, therefore the circumcentre lies on the angle bisector $AD$, therefore $AD$ is the diameter of the circle, which makes $O$ the mid-point of $AD$.
$\because $ $AD$ bisects $\angle BAC$, $\angle OAC = 30^\circ$.
$\because$ $OA$, $OC$ are radii of the same circle, $\triangle OAC$ is isosceles, and $\angle OCA = 30^\circ$
$\therefore \angle DOC = 60^\circ$ (exterior angle of $\triangle OAC$)
$\angle ACD = 90^\circ$
$\therefore \angle OCD = 90 - \angle OCA = 60^\circ$
$\therefore \triangle OCD$ is equilateral.
Since, $\angle BAD = \angle DAC$, chord $BD$ = chord $DC$.
$\therefore BD = DC = 4 \div 2 = 2$
$\because \triangle OCD$ is equilateral, $OD = OC$
$\therefore OD = 2$
Since, $OD$ is the radius of the circumcircle, the diameter of the circle is $4$.
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$\underline{Approach\ II\ (Trigonometry):}$
$\triangle ACD$ is right angled, with $\angle ACD = 90^\circ$.
$\angle DAC = \dfrac{1}{2} \times \angle BAC = 30^\circ$
$\dfrac{DC}{AD} = sin30^\circ = \dfrac{1}{2}$
(You can read about the basic trigonometric ratios ).
$\therefore \dfrac{2}{AD} = \dfrac{1}{2}$
$\Rightarrow AD = 4$
Problem 12 of 20
Let $T_k$ denote the $k\xasuper{th}$ term of an arithmetic progression. Suppose there are positive integers $m \ne n$ such
that $T_m = \dfrac{1}{n}$
and $T_n = \dfrac{1}{m}$
Then $T_{mn}$ equals:
Solution:
We will solve this problem using the concepts of $AP$ series explained .
Let $d$ be the common difference.Therefore,
$\dfrac{T_m - T_n}{m - n} = d$
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$\Rightarrow \dfrac{\frac{1}{n} - \frac{1}{m}}{m - n} = d$
$\Rightarrow \dfrac{\frac{m - n}{mn}}{m - n} = d$
$\Rightarrow \dfrac{1}{mn} = d$
Let $a$ be the first term.
Therefore, $T_m = a + (m-1)d$
$\Rightarrow \dfrac{1}{n} = a + \dfrac{m-1}{mn}$
$\Rightarrow a = \dfrac{1}{n} - \dfrac{m-1}{mn}$
Therefore,
$T_{mn} = a + (mn - 1)d$
$\Rightarrow T_{mn} = \dfrac{1}{n} - \dfrac{m-1}{mn} + (mn - 1)\times\dfrac{1}{mn}$
$\Rightarrow T_{mn} = \dfrac{m}{mn} - \dfrac{m-1}{mn} + \dfrac{mn - 1}{mn}$
$T_{mn} = \dfrac{m - m + 1 + mn - 1}{mn} = 1$
Problem 13 of 20
In a triangle $ABC$, let $AD$ be the median from $A$; Let $E$ be a point on $AD$ such that $AE : ED = 1 : 2$; and let
$BE$ extended meets $AC$ in $F$. The ratio of $\dfrac{AF}{FC}$ is:
Solution:
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We draw the diagram for the problem as shown below.As a construction, we also draw a line parallel to $BF$ and passing through $D$. Let this line meet the side $AC$ at $G$.
Considering $\triangle CBF$, we have $D$ as the mid point of $CB$ and $DG$ parallel to $BF$.
$\therefore$ $G$ is the midpoint of $FC$, and $FG = GC$
Similarly, considering $\triangle ADG$, we have $EF \parallel DG$ and $AE:ED = 1:2$
$\therefore AF:FG = 1:2$
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$\Rightarrow FG = 2AF$
and since $FG = GC$, $GC = 2AF$
$\therefore AF:FC = AF:(FG + GC) = AF:(2AF + 2AF)$
$= AF:4AF = 1:4$
Problem 14 of 20
If $sin \theta$ and $cos \theta$ are roots of the equation $px^2 + qx +r = 0$, then:Solution:
We know from , that for the given quadratic equation the sum of the roots will be:
$-\dfrac{q}{p}$ and the product of the roots will be $\dfrac{r}{p}$
Using this rule for sums and products of quadratic equations, we get
$sin\theta + cos\theta = -\dfrac{q}{p}$
and $sin\theta . cos\theta = \dfrac{r}{p}$
From the basic trigonometric identities explained , we know that:
$sin^2\theta + cos^2\theta = 1$
$\Rightarrow (sin\theta + cos\theta)^2 - 2sin\theta . cos\theta = 1$
$\Rightarrow (-\dfrac{q}{p}) - 2\dfrac{r}{p} = 1$
$\Rightarrow \dfrac{q^2}{p^2} - 2\dfrac{r}{p} = 1$
$\Rightarrow \dfrac{q^2 - 2pr}{p^2} = 1$
$\Rightarrow q^2 - 2pr = p^2$
$\Rightarrow p^2 - q^2 + 2pr = 0$
Problem 15 of 20
For a regular $k$-sided polygon, let $\alpha (k)$ denotes its interior angle. Suppose $n \gt 4$ is such that $\alpha (n – 2)$,
$\alpha (n)$, $\alpha (n + 3)$ forms an arithmetic progression. The sum of digits of $n$ is
Solution:
We know that the sum of the angles of a polygon with $n$ sides is $(n-2)180^\circ$,therefore, if the polygon is a regular polygon, each of its angle is $\dfrac{n-2}{n}\times 180^\circ$
Therefore, based on the problem:
$\alpha (k) = \dfrac{k-2}{k}\times 180$
and $\dfrac{(n-2)-2}{n-2}\times 180$, $\dfrac{n-2}{n}\times 180$ and $\dfrac{n+3-2}{n+3}\times 180$ are in arithmetic progression.
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Therefore,
$\dfrac{n-2}{n}\times 180 - \dfrac{(n-2)-2}{n-2}\times 180 = \dfrac{n+3-2}{n+3}\times 180 - \dfrac{n-2}{n}\times 180$
$\Rightarrow \dfrac{n-2}{n} - \dfrac{n-4}{n-2} = \dfrac{n+1}{n+3} - \dfrac{n-2}{n}$
$\Rightarrow \dfrac{(n-2)^2 - n(n-4)}{n(n-2)} = \dfrac{n(n+1) - (n+3)(n-2)}{n(n+3)}$
$\Rightarrow \dfrac{n^2 - 4n + 4 - n^2 + 4n}{n - 2} = \dfrac{n^2 + n -n^2 -n + 6}{n + 3}$
$\Rightarrow \dfrac{4}{n - 2} = \dfrac{6}{n + 3}$
$\Rightarrow 4n + 12 = 6n - 12$
$\Rightarrow 2n = 24$
$\Rightarrow n = 12$
Therefore, the sum of the digits in $n$ is $3$.
Problem 16 of 20
The sum of $5$ numbers in geometric progression is $24$. The sum of their reciprocals is $6$. The product of the
terms of the geometric progression is:
Solution:
We will solve this problem using two approaches, the conventional way and the smarter approach based on some observations. Both approaches involve concepts of geometric progression explained .
$\underline{Approach\ I:}$
We can solve this problem taking the first term as $a$ and the common ratio as $r$ and taking the terms as $a$, $ar$.... $ar^4$.
Therefore,
$a + ar + ar^2 + ar^3 + ar^4 = 24$
$\Rightarrow a(1 + r + r^2 + r^3 + r^4) = 24$
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$\dfrac{1}{a} + \dfrac{1}{ar} + \dfrac{1}{ar^2} + \dfrac{1}{ar^3} + \dfrac{1}{ar^4} = 6$
$\Rightarrow \dfrac{1}{a}\left(1 + \dfrac{1}{r} + \dfrac{1}{r^2} + \dfrac{1}{r^3} + \dfrac{1}{r^4}\right) = 6$
$\Rightarrow \dfrac{1}{a}\left(\dfrac{r^4 + r^3 + r^2 + r + 1}{r^4}\right) = 6$
$\Rightarrow \dfrac{r^4 + r^3 + r^2 + r + 1}{a}\left(\dfrac{1}{r^4}\right) = 6$
$\Rightarrow \dfrac{\frac{24}{a}}{a}\left(\dfrac{1}{r^4}\right) = 6$
$\Rightarrow \dfrac{1}{a^2} \dfrac{1}{r^4} = \dfrac{6}{24} = \dfrac{1}{4}$
$\Rightarrow ar^2 = 2$
We are supposed to find the product of the five terms which is $a \times ar \times ar^2 \times ar^3 \times ar^4 = a^5r^{10}$, that is, $(ar^2)^5$, which is $2^5 = 32$
$\underline{Approach\ II:}$
We can adopt a smarter and simpler approach by observing that that number of terms is odd, hence the numbers are in symmetry about the middle number, therefore, in the product all the common ratio terms will cancel out.
So, we can take the middle term as $a$ and the common ratio as $r$.
Therefore our terms become $\dfrac{a}{r^2}$, $\dfrac{a}{r}$, $a$, $ar$ and $ar^2$,
and we have to find:
$\dfrac{a}{r^2} \times \dfrac{a}{r} \times a \times ar \times ar^2 = a^5$
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Therefore,
$\dfrac{a}{r^2} + \dfrac{a}{r} + a + ar + ar^2 = 24$
$\Rightarrow a\left( \dfrac{1 + r + r^2 + r^3 + r^4}{r^2}\right) = 24$ $...eqn\ (i)$
Adding the reciprocals, we get:
$\dfrac{1}{a} \left(\dfrac{r^4 + r^3 + r^2 + r + 1}{r^2} \right) = 6$ $...eqn\ (ii)$
Dividing $eqn(i)$ by $eqn(ii)$ we get:
$a^2 = 4$
$\Rightarrow a = 2$
$\Rightarrow a^5 = 32$
Problem 17 of 20
Digits $a$ and $b$ are such that the product $4a1 \times 25b$ is divisible by $36$ (in base $10$). The number of ordered
pairs $(a, b)$ is:
Solution:
When a number is divisible by $36$, we can say that the number is divisible by $4$ and $9$ separately.Let us look at the given numbers.
The number $2a1$ is an odd number, no matter what the value of $a$ is.
Hence the number $25b$ has to be divisible by $4$.
Therefore, based on the divisibility rules described , the last two digits of $25b$ has to be divisible by $4$, that is $5b$ has to be divisible by $4$.
The numbers $52$ and $56$ are divisible by $4$, therefore, $b$ can be either $2$ or $6$.
If $b = 2$, then $252$ is divisible by $9$, which makes $2a1 \times 252$ divisible by $36$ irrespective of the value of $a$, which means $a$ can be any one of $0,\ 1,\ 2,\ ...9$. Hence there are $10$ possible values of $a$
If $b = 6$, that means $256$ is not divisible by $9$.
Therefore, the number $2a1$ has to be divisible by $9$.
Note: In the original paper published for NSEJS this question did not have the correct option $11$, therefore, this question was marked as a bonus question later on. We have replaced one of the incorrect options with the correct option.
Problem 18 of 20
The integer closest to $\sqrt{111...1– 222...2}$ , where there are $2018$ ones and $1009$ twos, is:Solution:
We know that if we have any number of the form:$99....9$ comprising of the digit $9$ repeated $n$ times, we can write the number is $10^n - 1$
For example,
$9 = 10^1 - 1$
$99 = 10^2 - 1$
$999 = 10^3 - 1$
$...$
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Let us try to represent the numbers given under the square root sign, in the shorter form using exponent.
$11...11$ with $2018$ $1$s can be written as:
$\dfrac{1}{9} \times (99... 99)$ with $2018$ $9$s, which is equal to:
$\dfrac{1}{9} \times (10^{2018} - 1)$
Similarly $22... 22$ with $1009$ $2$s can be written as:
$2 \times (11... 11)$
$= \dfrac{2}{9} \times (99... 99)$
$= \dfrac{2}{9} \times (10^{1009}-1)$
Therefore, the given expression becomes:
$\sqrt{\dfrac{1}{9} \times (10^{2018} - 1) - \dfrac{2}{9} \times (10^{1009}-1)}$
$= \sqrt{\dfrac{1 \times (10^{2018} - 1) - {2} \times (10^{1009}-1)}{9}}$
$= \sqrt{\dfrac{10^{2018} - 1 - {2} \times 10^{1009}+2}{9}}$
$= \sqrt{\dfrac{(10^{1009})^2 - {2} \times 10^{1009}+1}{9}}$
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$= \sqrt{\dfrac{(10^{1009} - 1)^2}{3^2}}$
$= \dfrac{10^{1009} - 1}{3}$
Problem 19 of 20
In a triangle $ABC$, a point $D$ on $AB$ is such that $AD : AB = 1 : 4$ and $DE$ is parallel to $BC$ with $E$ on $AC$. Let $M$ and $N$ be the mid points of $DE$ and $BC$ respectively. What is the ratio of the area of the quadrilateral $BNMD$ to that of triangle $ABC$?
Solution:
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Let us draw the diagram for the given problem as below:
Given that $DE \parallel BC$,
$\therefore \triangle ADE \sim \triangle ABC$
Given that $\dfrac{AD}{AB} = \dfrac{1}{4}$, using the rules of triangle similarity from , we get:
$[\triangle ADM] = \left(\dfrac{1}{4}\right)^2 [\triangle ABN] = \dfrac{1}{16} [\triangle ABN]$
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$\therefore [BNMD] = [\triangle ABN] - [\triangle ADM] = [\triangle ABN] - \dfrac{1}{16} [\triangle ABN]$
$= \dfrac{15}{16}[\triangle ABN]$
Since $M$ is the midpoint of $DE$ and $N$ is the midpoint of $DE$ we can prove that the point $M$ lies on the median $AN$
$[\triangle ABN] = \dfrac{1}{2}[\triangle ABC]$
$\therefore [BNMD] = \dfrac{15}{16}[\triangle ABN] = \dfrac{15}{16} \times \dfrac{1}{2} [\triangle ABC] $
$= \dfrac{15}{32} [\triangle ABC]$
$\therefore \dfrac{[BNMD]}{[\triangle ABC]} = \dfrac{15}{32}$
Problem 20 of 20
The number of distinct integers in the collection$\left\lfloor\dfrac{10^2}{1} \right\rfloor$, $\left\lfloor\dfrac{10^2}{2} \right\rfloor$, $\left\lfloor\dfrac{10^2}{3} \right\rfloor$, $...$, $\left\lfloor\dfrac{10^2}{20} \right\rfloor$
where $\left\lfloor x \right\rfloor$ denotes the largest integer not exceeding $x$, is:
Solution:
This problem can be solved by dividing $100$ by each of the given divisors and then taking the integer part of the result.However that will be a brute force approach, and will not help us understand the behaviour of the floor function.
Let us take a smarter approach to this problem.
The highest divisor is $20$, and $100 \div 20 = 5$
It is evident that the other quotients will be greater than $5$ since the divisors are less than $20$.
And the highest quotient will be $100$, where the divisor is $1$
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$100 \div 5 = 20$
$100 \div 6 = 16\dfrac{2}{3}$
Therefore, for divisors greater than $16$ and less than or equal to $20$ the quotient will be $5.xxx$ and the floor function will be $5$
Similarly,
$100 \div 7 = 14\dfrac{2}{7}$
Therefore, all divisors greater than $14$ and less than or equal to $16$, will be $6.xxx$ and the floor function will be $6$
$100 \div 8 = 12\dfrac{1}{2}$
Therefore, all divisors greater than $12$ and less than or equal to $14$ will be $7.xxx$ and the floor function will be $7$
$100 \div 9 = 11\dfrac{1}{11}$
Therefore, all divisors greater than $11$ and less than or equal to $12$ will be $8.xxx$ and the floor function will be $8$
For all divisors $d \le \sqrt{100}$, each of $\left\lfloor\dfrac{10^2}{d} \right\rfloor$ will map to a unique integer.
Hence, the total number of distinct integers $= 10 + 1 + 1 + 1 + 1 + 1 = 15$