NSEJS-2017 (Maths Section)

Problem 1 of 20

How many four digit numbers are there such that when they are divided by $101$, they have $99$ as remainder?

Solution:

Any number of the form $101n + 99$, where $n$ is an integer, will leave a remainder of $99$ when divided by $101$

The smallest $4$ digit number is $1000$, and the largest $4$ digit number is $9999$.

Therefore, we can say:

$101n + 99 \ge 1000$

$\Rightarrow 101n \ge 1000 - 99$

$\Rightarrow 101n \ge 901$

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$\Rightarrow n \ge \dfrac{901}{101}$

$\Rightarrow n \ge 8\dfrac{93}{101}$

Since $n$ is an integer, the smallest value of $n$ is $9$

To find the upper limit of $n$, we use the condition:

$101n + 99 \le 9999$

$101n \le 9900$

$n \le \dfrac{9900}{101}$

$n \le 98\dfrac{2}{101}$

Therefore, the largest integer value of $n$ is $98$

Since, $n$ ranges from $9$ to $98$, both values inclusive, and is an integer, the number of possible values of $n$ is:

$98 - 9 + 1 = 90$

Each of these $90$ values of $n$ will give an integer which fulfils the given conditions.

Problem 2 of 20

$1\dfrac{1}{2} + 1\dfrac{1}{6} + 1\dfrac{1}{12} + 1\dfrac{1}{20} + 1\dfrac{1}{30} + ... + 1\dfrac{1}{380} = $________.

Solution:

Let us observe that each of the terms in the given series has $1$ as the integer part and for the fraction part we have $1$ as the numerator, and the product of two consecutive integers as the denominator, like

$\dfrac{1}{1.2}$, $\dfrac{1}{2.3}$ and so on.

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Let us break up the series as follows:

$1\dfrac{1}{2} + 1\dfrac{1}{6} + 1\dfrac{1}{12} + 1\dfrac{1}{20} + 1\dfrac{1}{30} + ... + 1\dfrac{1}{380}$

$= \left(1 + \dfrac{1}{1 \times 2}\right) + \left(1 + \dfrac{1}{2 \times 3}\right) + \left(1 + \dfrac{1}{3 \times 4}\right)$ $ + \left(1 + \dfrac{1}{4 \times 5}\right) + \left(1 + \dfrac{1}{5 \times 6}\right) + ... + \left(1 + \dfrac{1}{19 \times 20}\right)$

Since there are $19$ terms in the series, we can take out the $1$s and add $19$ to the final sum. For the fraction parts we get:

$= \left(\dfrac{1}{1 \times 2}\right) + \left(\dfrac{1}{2 \times 3}\right) + \left(\dfrac{1}{3 \times 4}\right) + \left(\dfrac{1}{4 \times 5}\right) $ $+ \left(\dfrac{1}{5 \times 6}\right) + ... + \left(\dfrac{1}{19 \times 20}\right)$

Using the technique of telescoping , we can split each of the terms as follows:

$= \left(\dfrac{2 - 1}{1 \times 2}\right) + \left(\dfrac{3 - 2}{2 \times 3}\right) + \left(\dfrac{4 - 3}{3 \times 4}\right) + \left(\dfrac{5 - 4}{4 \times 5}\right) $ $+ \left(\dfrac{6 - 5}{5 \times 6}\right) + ... + \left(\dfrac{20 - 19}{19 \times 20}\right)$

$= \left(\dfrac{1}{1} - \dfrac{1}{2}\right) + \left(\dfrac{1}{2} - \dfrac{1}{3}\right) + \left(\dfrac{1}{3} - \dfrac{1}{4}\right) + \left(\dfrac{1}{4} - \dfrac{1}{5}\right) $ $+ \left(\dfrac{1}{5} - \dfrac{1}{6}\right) + ... + \left(\dfrac{1}{19} - \dfrac{1}{20}\right)$

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Now, we can see that the second fraction of each term will cancel out with the first fraction of the next term, leaving us with:

$\dfrac{1}{1} - \dfrac{1}{20} = \dfrac{19}{20}$

Now, we can add back the $19$ which we had removed earlier, and get the result as:

$19\dfrac{19}{20} = 19.95$

Problem 3 of 20

Diagonals of a quadrilateral bisect each other. Therefore, the quadrilateral must be:

Solution:

We know that the diagonals of a parallelogram bisect, each other.

We also know that all the remaining figures given are special cases of parallelograms.

Therefore, the minimum requirement to fulfil the given diagonal bisect condition is a parallelogram.

Problem 4 of 20

By which smallest number should we divide $198396198$ to get a perfect square?

Solution:

We can solve this problem easily by elimination.

The last two digits of the given number, that is $98$ is not divisible by $4$.

Hence the number itself is not divisible by $4$ but by $2$ being a even number.

Hence we need to divide by $2$ to eliminate $2$ as a factor.

Since the number is not divisible by $4$ dividing by $28$ is not possible.

Also, we need not divide by a perfect square, like $9$, since it won't help making the number a perfect square.

Therefore, we are left with options $14$ and $22$.

We can check the divisibility by $11$ by adding the alternate digits. We get:

$1 + 8 + 9 + 1 + 8 = 27$ and

$9 + 3 + 6 + 9 = 27$

The given number is divisible by $11$

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We also need to check the divisibility by $7$ by actual division, and we get a remainder of $0$

The most efficient way now is to keep dividing the number by $11$ and check if $11$ occurs odd number of times or even number of times.

$198396198 \div 11 = 18036018$

This number is also divisible by $11$, and dividing by $11$ we get:

$18036018 \div 11 = 1639638$

This result again is divisible by $11$, and dividing by $11$ again we get:

$1639638 \div 11 = 149058$

This number is no longer divisible by $11$. Therefore we know that the given number is divisible by $11^3$ and we need to take out one $11$ to make it a perfect square.

Therefore, we need to divide by $2 \times 11 = 22$ to get a perfect square.

Problem 5 of 20

If $x^2 + xy + xz = 135$, $y^2 + yz + xy = 351$ and $z^2 + xz + yz = 243$, then $x^2 + y^2 + z^2 = $__________.

Solution:

We can write the three given equations as:

$x(x + y + z) = 135$

$y(x + y + z) = 351$

$z(x + y + z) = 243$

Adding these three equations we get:

$(x + y + z)(x + y + z) = 729$

$\therefore (x + y + z) = 27$

Therefore,

$x = \dfrac{135}{x + y + z} = \dfrac{135}{27} = 5$

$y = \dfrac{351}{x + y + z} = \dfrac{351}{27} = 13$

$z = \dfrac{243}{x + y + z} = \dfrac{351}{27} = 9$

Therefore,

$x^2 + y^2 + z^2 = 5^2 + 13^2 + 9^2 = 25 + 169 + 81 = 275$

Problem 6 of 20

If $p + q + r = 2$, $p^2 + q^2 + r^2 = 30$ and $pqr = 10$, the value of $(1-p)(1-q)(1-r) = $______.

Solution:

For this problem we start with the unknown expression first and simplify it.

$(1-p)(1-q)(1-r)$

$= (1 - p - q + pq)(1 - r)$

$= 1 - p - q + pq - r + pr + qr - pqr$

$= 1 - (p + q + r) + (pq + pr + qr) - pqr$

We can see that we already know the values of all the terms in the above expression except:

$pq + pr + qr$ and we can easily find its value using:

$2(pq + pr + qr) = (p + q + r)^2 - (p^2 + q^2 + r^2)$

$\Rightarrow 2(pq + pr + qr) = 2^2 - 30 = -26$

$\therefore (pq + pr + qr) = -13$

Therefore, our simplified expression becomes:

$1 - (p + q + r) + (pq + pr + qr) - pqr$

$= 1 - 2 + (-13) - 10$

$= -24$

Problem 7 of 20

$\left(x + \dfrac{1}{x}\right) = 5$, then

$\left(x^3 + \dfrac{1}{x^3}\right) - 5\left(x^2 + \dfrac{1}{x^2}\right) + \left(x + \dfrac{1}{x}\right) = $_________.

Solution:

$x^2 + \dfrac{1}{x^2}$

$= x^2 + \dfrac{1}{x^2} + 2x\dfrac{1}{x} - 2x\dfrac{1}{x}$

$= \left(x + \dfrac{1}{x}\right)^2 - 2$

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$= (5)^2 - 2$

$= 23$

$x^3 + \dfrac{1}{x^3}$

$= x^3 + \dfrac{1}{x^3} + 3x + \dfrac{3}{x} - 3x - 3\dfrac{1}{x}$

$= x^3 + \dfrac{1}{x^3} + 3x + \dfrac{3}{x} - 3\left(x - 3\dfrac{1}{x}\right)$

$= x^3 + 3x + \dfrac{3}{x}+ \dfrac{1}{x^3} - 3\left(x - 3\dfrac{1}{x}\right)$

$= x^3 + 3x^2.\dfrac{1}{x} + 3x\dfrac{1}{x^2} + \dfrac{1}{x^3} - 3\left(x - 3\dfrac{1}{x}\right)$

$= \left(x + \dfrac{1}{x}\right)^3 - 3\left(x - 3\dfrac{1}{x}\right)$

$= (5)^3 - 3 \times 5 = 110$

For the given expression we get:

$\left(x^3 + \dfrac{1}{x^3}\right) - 5\left(x^2 + \dfrac{1}{x^2}\right) + \left(x + \dfrac{1}{x}\right)$

$= 110 - 5 \times 23 + 5$

$= 110 - 115 + 5$

$= 0$

Problem 8 of 20

If $x = (\sqrt{21} - \sqrt{20})$ and $y = (\sqrt{18} - \sqrt{17})$, then:

Solution:

Rationalising the numerator of $x$ we get:

$x = \sqrt{21} - \sqrt{20}$

$= \dfrac{21 - 20}{\sqrt{21} + \sqrt{20}}$

$= \dfrac{1}{\sqrt{21} + \sqrt{20}}$

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Similarly for $y$ we get:

$y = \sqrt{18} - \sqrt{17}$

$= \dfrac{18 - 17}{\sqrt{21} + \sqrt{20}}$

$= \dfrac{1}{\sqrt{18} + \sqrt{17}}$

Now, for both numbers we have the numerators as $1$ and the denominators as the sum of two irrational numbers.

Comparing the denominators, we get:

$\sqrt{21} \gt \sqrt{18}$, and

$\sqrt{20} \gt \sqrt{17}$

Adding the two, we get:

$\sqrt{21} + \sqrt{20} \gt \sqrt{18} + \sqrt{17}$

Therefore,

$\dfrac{1}{\sqrt{21} + \sqrt{20}} \lt \sqrt{18} + \sqrt{17}$

$\Rightarrow x \lt y$

Problem 9 of 20

A train is travelling at a speed of $54\ km/hr$. It is not stopping at a certain station. It crosses the person showing green flag in $20$ seconds and crosses the platform in $36$ seconds. What is the length of the train?

Solution:

The train is travelling at $54\ km/h$.

It took $20\ secs$ to cross the man showing the flag.

This mean it took $20\ secs$ to travel a distance equal to its own length.

In $20\ secs$ the train travelled $54 \times \dfrac{1000}{3600} \times 20 = 300\ m$

Therefore, the length of the train is $300\ m$

Problem 10 of 20

If $(a + b + c + d) = 4$, then

$\dfrac{1}{(1-a)(1-b)(1-c)} + \dfrac{1}{(1-b)(1-c)(1-d)}$ $ + \dfrac{1}{(1-c)(1-d)(1-a)} + \dfrac{1}{(1-d)(1-a)(1-b)} = $_________.

Solution:

The given expression can be written as:

$\dfrac{1}{(1-a)(1-b)(1-c)} + \dfrac{1}{(1-b)(1-c)(1-d)}$ $ + \dfrac{1}{(1-c)(1-d)(1-a)} + \dfrac{1}{(1-d)(1-a)(1-b)}$

$= \dfrac{(1-d) + (1-a) + (1-b) + (1-c)}{(1-a)(1-b)(1-c)(1-d)}$

$= \dfrac{4 - (a + b + c + d)}{(1-a)(1-b)(1-c)(1-d)}$

$= \dfrac{4 - 4}{(1-a)(1-b)(1-c)(1-d)}$

$= 0$

Problem 11 of 20

What will be the remainder if the number $(7)^{2017}$ is divided by $25$?

Solution:

From the concept of number congruence explained , we know that:

$7^2 = 49 \equiv 24\ (mod\ 25) \equiv -1\ (mod\ 25)$

$7^{2017} = 7^{(2016 + 1)} = 7^{2016}.7 = (7^2)^{1008}.7$

$\therefore 7^{2017} \equiv (-1)^{1008}.7\ (mod\ 25) \equiv 1 \times 7\ (mod\ 25) \equiv 7\ (mod\ 25)$

Therefore, $7^{2017}$ will leave a remainder of $7$ when divided by $25$

Problem 12 of 20

What is the radius of the circumcircle of a triangle whose sides are $30\ cm$, $36\ cm$ and $30\ cm$.

Solution:

We know from , that the area of a triangle, its sides and its circum-radius are related by the equation:

$4 \times Area \times R = product\ of\ the\ 3\ sides$ where $R$ is the circum-radius.

The given triangle is an isosceles triangle. Let the altitude corresponding to the unequal base be $h$.

$\therefore h^2 = 30^2 - 18^2$

$\Rightarrow h = \sqrt{(30 - 18)(30 + 18)}$

$\Rightarrow h = \sqrt{12 \times 48} = \sqrt{12 \times 12 \times 4} = 12 \times 2 = 24$

$\therefore [\triangle ABC] = \dfrac{1}{2}\times 24 \times 36 = 12 \times 36$

$4 \times 12 \times 36 \times R = 30 \times 36 \times 30$

$R = \dfrac{30 \times 30}{4 \times 12} = \dfrac{15 \times 5}{4} = 18\dfrac{3}{4} = 18.75$

Note: The original question published for NSEJS did not have the correct option $18.75$. We have modified the question to include the correct answer and added also added $16.5$ as an incorrect decimal option.

Problem 13 of 20

The mean of the following frequency distribution is ____________.

$Class\ Interval$	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$
$Frequency$	$4$	$6$	$8$	$10$	$12$

Solution:

In this problem, the class intervals are specified, and we need to find the class marks to be able to find the mean for the distribution.

For each given interval we find the class mark using:

$class\ mark = \dfrac{upper\ limit + lower\ limit}{2}$

The computed values of $x_i$ are shown in the following table:

$x_i$	$f_i$	$x_if_i$
$5$	$4$	$20$
$15$	$6$	$90$
$25$	$8$	$200$
$35$	$10$	$350$
$45$	$12$	$540$
Sum	$40$	$1200$

Therefore, $\overline{x} = \dfrac{\Sigma x_if_1}{\Sigma f_i} = \dfrac{1200}{40} = 30$

Problem 14 of 20

If $x^2 – 3x + 2$ is a factor of $x^4 – px^2 + q$, then $p,\ q$ are:

Solution:

Using the method of quadratic factorisation explained , we can factorise $x^2 - 3x + 2$ as $(x - 2)(x-1)$

Since $x^4 - px^2 + q$ is divisible by $x^2 - 3x + 2$, we can say that it divisible by $(x-2)$ and $(x-1)$

Therefore substituting $x = 2$ or $x = 1$ in the given expression, we should get $0$.

Substituting $x = 2$ we get:

$(2)^4 - p(2)^2 + q = 0$

$\Rightarrow 4p - q = 16$ $...eqn\ (i)$

Substituting $x = 1$ we get:

$(1)^4 - p(1)^2 + q = 0$

$\Rightarrow p - q = 1$ $...eqn\ (ii)$

Subtracting $eqn\ (ii)$ from $eqn\ (i)$, we get:

$3p = 15$

$\Rightarrow p = 5$

$\therefore q = p - 1 = 4$

Problem 15 of 20

What is the sum of all odd numbers between $500$ and $600$?

Solution:

The first odd number after $500$ is $501$, and the last odd number before $600$ is $599$.

What we have here is an $AP$ series with $501$ as the first term and $2$ as the common difference, therefore, we will use the concepts of $AP$ series explained to solve this problem.

Let $599$ be the $n\xasuper{th}$ term of the given series.

$\therefore 599 = 501 + (n - 1) 2$

$\Rightarrow n = 1 + \dfrac{599 - 501}{2}$

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$\Rightarrow n = 1 + \dfrac{98}{2}$

$\Rightarrow n = 1 + 49 = 50$

$\therefore S_{50} = \dfrac{n}{2} \left\{2a + (n-1)d \right\}$

$= \dfrac{50}{2}\left\{2\times 501 + (50-1)2 \right\}$

$= \dfrac{50}{2} \times 2 (501 + 49)$

$= 50 \times 550$

$= 27500$

Problem 16 of 20

In $\triangle ABC$, segment $AD$, segment $BE$ and segment $CF$ are altitudes. If $AB \times AC = 172.8\ cm^2$ and $BE × CF = 108.3\ cm^2$ then $AD \times BC =$

Solution:

From the given data it is possible to infer that:

$ar[\triangle ABC] = \dfrac{1}{2} \times AD \times BC = \dfrac{1}{2} \times BE \times AC = \dfrac{1}{2} \times CF \times AB$

$\therefore \left(\dfrac{1}{2} \times CF \times AB \right) \times \left(\dfrac{1}{2} \times BE \times AC \right) = (ar[\triangle ABC])^2$

$\therefore CF \times BE \times AB \times AC = 4 \times (ar[\triangle ABC])^2 = 172.8 \times 108.3$

$\Rightarrow (ar[\triangle ABC])^2 = \dfrac{1}{4} \times 172.8 \times 108.3$

$\Rightarrow ar[\triangle ABC] = \dfrac{1}{2} \times \sqrt{172.8 \times 108.3}$

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$\therefore \dfrac{1}{2} \times AD \times BC = ar[\triangle ABC] = \dfrac{1}{2} \times \sqrt{172.8 \times 108.3}$

$\Rightarrow AD \times BC = \sqrt{172.8 \times 108.3}$

$= \sqrt{\dfrac{1728 \times 1083}{100}}$

$= \sqrt{\dfrac{3 \times 576 \times 3 \times 361}{100}}$

$= \sqrt{\dfrac{24 \times 24 \times 3 \times 3 \times 19 \times 19}{100}}$

$= \dfrac{24 \times 3 \times 19}{10}$

$= 136.8$

Problem 17 of 20

The sum of two numbers is $13$ and the sum of their cubes is $1066$. Find the product of those two numbers.

Solution:

Let the two numbers by $a$ and $b$.

Therefore,

$a + b = 13$

$\Rightarrow (a + b)^2 = 169$

$\Rightarrow a^2 + 2ab + b^2 = 169$ $...eqn\ (i)$

And,

$a^3 + b^3 = 1066$

$\Rightarrow (a + b)(a^2 - ab + b^2) = 1066$

$\Rightarrow 13(a^2 - ab + b^2) = 1066$

$\Rightarrow a^2 - ab + b^2 = \dfrac{1066}{13} = 82$ $...eqn\ (ii)$

Subtracting $eqn\ (ii)$ from $eqn\ (i)$, we get:

$3ab = 169 - 82 = 87$

$\therefore ab = 29$

Problem 18 of 20

If $ABCD$ is a cyclic quadrilateral, $AB = 204$, $BC = 104$, $CD= 195$, $DA = 85$ and $BD = 221$, then $AC =$_________.

Solution:

From Ptolemy's theorem explained , we know that for a cyclic quadrilateral the product of the diagonals is equal to the sum of the product of the opposite sides.

Therefore,

$BD \times AC = AB\times CD + BC \times DA$

$\Rightarrow 221 \times AC = 204\times 195 + 104 \times 85$

$\Rightarrow 221 \times AC = 2 \times 5 \times (102 \times 39 + 52 \times 17)$

$\Rightarrow 221 \times AC = 2 \times 5 \times 13(102 \times 3 + 4 \times 17)$

$\Rightarrow 221 \times AC = 2 \times 5 \times 13 \times 2 (51 \times 3 + 2 \times 17)$

$\Rightarrow 221 \times AC = 2 \times 5 \times 13 \times 2 \times 17(3 \times 3 + 2 \times 1)$

$\Rightarrow 221 \times AC = 2 \times 5 \times 13 \times 2 \times 17 \times (11)$

$\Rightarrow AC = 2 \times 5 \times 2 \times 11 = 220$

Problem 19 of 20

The seventy first $Independence\ day$ was on a $Tuesday$. After how many years $Independence\ day$ will again be on a $Tuesday$?

(Note for international students: India became Independent on $15\xasuper{th}\ August, 1947$)

Solution:

The seventy first independence day was on $August\ 15$, in the year $1947 + 71 - 1= 2017$.

We know that $365$ modulo $7$ is $1$ for a normal year, and $366$ modulo $7$ is $2$ for a leap year.

Each year the calendar day for the same date will advance by the elapsed number of days modulo $7$.

In $2018$ the independence day would be on a Wednesday.

In $2019$ the independence day would be on a Thursday.

In $2020$ being a leap year the independence day would be on a Saturday.

Likewise Starting with $2018$, if we keep adding the number of elapsed days modulo $7$ till we get a number divisible by $7$ we get:

$1 + 1 + 2 + 1 + 1 + 1 = 7$

Therefore the next Independence day that will be on a Tuesday, will come after $6$ years.

Problem 20 of 20

If the roots of the equation

$\dfrac{x^2 - bx}{ax - c} = \dfrac{m - 1}{m + 1}$ are equal and of opposite signs; then the value of $m$ is:

Solution:

Given that:

$\dfrac{x^2 - bx}{ax - c} = \dfrac{m - 1}{m + 1}$

$\Rightarrow (x^2 - bx)(m + 1) = (ax - c)(m-1)$

$\Rightarrow (m+1)x^2 - b(m+1)x = a(m-1)x - c(m-1)$

$(m+1)x^2 - \{b(m+1) + a(m-1)\}x + c(m-1) = 0$

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Since the two roots have the same magnitude and opposite signs, their sum must be $0$, using the concepts of quadratic equations from , we can conclude that the coefficient of the $x$ term has to be $0$.

$\therefore b(m+1) + a(m-1) = 0$

$\Rightarrow \dfrac{a}{b} = -\dfrac{m+1}{m-1} = \dfrac{1 + m}{1 - m}$

Applying componendo and dividendo we get:

$\dfrac{a - b}{a + b} = \dfrac{(1 + m) - (1-m)}{(1 + m) + (1-m)}$

$\Rightarrow \dfrac{a - b}{a + b} = \dfrac{1 + m - 1+m}{1 + m + 1-m}$

$\Rightarrow \dfrac{a - b}{a + b} = \dfrac{2m}{2} = m$

$\therefore m = \dfrac{a - b}{a + b}$