# Complex number- ISI entrance B. Stat. (Hons.) 2003- problem 5

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## Complex number

In complex number we have real and imaginary part mixed and $$\sqrt{-1}$$ is the basic unit and denoted by $$i$$. In the given question we have to find value of k for which the equation will be valid.

## Try the problem

How many integers $$k$$ are there for which $$(1-i)^k=2^k$$ ?

(A) One

(B) None

(C) Two

(D) More than one.

ISI entrance B. Stat. (Hons.) 2003 problem 5

Complex numbers

6 out of 10

Challenges and thrills of pre-college mathematics

## Use some hints

The complex number $$(1-i)$$ can be rationalized by multiplying numerator and denominator by $$1+i$$.

And we will get

$$(1-i)=(1-i)\frac{(1+i)}{1+i}=\frac{2}{1+i}$$

Now we will have $$(\frac{2}{1+i})^k=2^k$$

so, $$(1+i)^k=1$$, this is only possible when k=0,

So $$k$$ can have only one value, The option (A) is correct.

# Area of polygon - AMC 10B, 2019 Problem 8

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## Area of polygon

Polygon means shape composed of multiple sides , for example square , triangle, trapezium pentagon etc. A regular polygon means all of its sides have same length. Square ,equilateral triangle is a regular polygon. Let us learn to find the area of polygon.

## Try the problem

The figure below shows a square and four equilateral triangles, with each triangle having a side lying on a side of the square, such that each triangle has side length 2 and the third vertices of the triangles meet at the center of the square. The region inside the square but outside the triangles is shaded. What is the area of the shaded region?

AMC 10B, 2019 Problem 8

Area of polygon -square and triangle

6 out of 10

challenges and thrills of pre college mathematics

## Use some hints

Split the square in 4 identical parts by drawing two perpendicular (horizontal and vertical )lines from the center. And then we will get smaller square with two similar right angle triangles in it and one fourth part of the shaded region.

When we split an equilateral triangle in half, we get two triangles with angles 30,60 and 90 degrees. Therefore, the altitude, which is also the side length of one of the smaller squares, is $$\sqrt{3}$$.

the area of the two congruent triangles will be $2 \cdot \frac{1 \cdot \sqrt{3}}{2} = \sqrt{3}$.

.

The area of the each small squares is the square of the side length, i.e. $\left(\sqrt{3}\right)^2 = 3$. Therefore, the area of the shaded region in each of the four squares is $3 - \sqrt{3}$. Since there are $4$ of these squares, we multiply this by $4$ to get $4\left(3 - \sqrt{3}\right) = {\textbf{(B) } 12 - 4\sqrt{3}}$.

# Quadratic Equation ISI entrance B. Stat. (Hons.) 2003 problem 4

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The roots of the quadratic equation lie between some range and it depends upon the coefficients of the equation. or may be predicted using the relation between coefficients.

## Try the problem

Suppose $$a+b+c$$ and $$a-b+c$$ are positive and $$c<0$$. Then the equation $$ax^2+bx+c=0$$

(A) has exactly one root lying between -1 and +1

(B) has both the roots lying between -1 and +1

(C) has no root lying between -1 and +1

(D) nothing definite can be said about the roots without knowing the value of a,b and c.

ISI entrance B. Stat. (Hons.) 2003 problem 4

6 out of 10

challenges and thrills of pre college mathematics

## Use some hints

The question is based upon mean value theorem and the lower and upper limits are 1 and -1

So let $$f(x)=ax^2+bx+c$$ now find f(1) and f(-1), and try to use mean value theorem.

So $$f(1)=a+b+c$$ and $$f(-1)=a-b+c$$ and we can see that both of them are positive. so now using mean value theorem we can find c in $$f'(c)$$.

An now we can apply Sridharacharya's formula to find the valise(s) of x.

We know c in f'(c) represent the roots and the mean value theorem says c must lie between limits (-1 and +1 in this case)

So option B is correct one.

# Recursion - AMC 10B, 2019 Problem 25

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## What is Recursion

Recursion is basically an idea of connecting any term with the next or previous term of an series. The most famous example of recursion is Fibonacci series $$0,1,1,2,3,5,8,........$$ its recursion formula is $$t_{n+2}=t_{n+1}+t_n$$ for natural number n.

## Try the problem

How many sequences of $0$s and $1$s of length $19$ are there that begin with a $0$, end with a $0$, contain no two consecutive $0$s, and contain no three consecutive $1$s?

$\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75$

AMC 10B, 2019 Problem 25

Recursion

6 out of 10

challenges and thrills of pre college mathematics

## Use some hints

We can deduce, from the given restrictions, that any valid sequence of length $n$ will start with a $0$ followed by either $10$ or $110$. Thus we can define a recursive function $f(n) = f(n-3) + f(n-2)$, where $f(n)$ is the number of valid sequences of length $n$.

This is because for any valid sequence of length $n$, you can append either $10$ or $110$ and the resulting sequence will still satisfy the given conditions.

It is easy to find $f(5) = 1$ with the only possible sequence being $01010$ and $f(6) = 2$ with the only two possible sequences being $011010$ and $010110$ by hand, and then by the recursive formula, we have $$f(19)=65$$ so option C is correct option.

# Coordinate Geometry - B.Stat. (Hons.) Admission Test 2005 – Objective Problem 5

## Competency in Focus: Coordinate Geometry

This problem from B.Stat. (Hons.) based on coordinate geometry Admission Test 2005 – Objective Problem 5  is based nature of curve.

## Next understand the problem

The equation $x(x+3)=y(y-1)-2$ represents
(A) a hyperbola

(B) a pair of straight lines
(C) a point

(D) none of the foregoing curves

[/et_pb_text][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="4.3.1" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" _builder_version="4.3.1" open="on"]B.Stat. (Hons.) Admission Test 2005 – Objective problem 5[/et_pb_accordion_item][et_pb_accordion_item title="Key Competency" open="off" _builder_version="4.3.1" inline_fonts="Abhaya Libre"]

### Coordinate Geometry

[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.1" open="off"]4/10[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.3.1" open="off"]Challenges and Thrills in Pre College Mathematics

[/et_pb_text][et_pb_tabs _builder_version="4.2.2"][et_pb_tab title="HINT 0" _builder_version="4.0.9"]Do you really need a hint? Try it first![/et_pb_tab][et_pb_tab title="HINT 1" _builder_version="4.2.2"]

It cannot be a straight line because Straight line are linear equation of the form. $$ax+by+c=0$$.

[/et_pb_tab][et_pb_tab title="HINT 2" _builder_version="4.2.2"]

So it may be a circle or a point if radius is zero, But when we generalized it to the standard form of circle we get negative radius. so it cant be either of this one.

Hint: Stanrd form of circle is $(x-h)^{2}+(y-k)^{2}=r^{2}$.

[/et_pb_tab][et_pb_tab title="HINT 3" _builder_version="4.2.2"]

Now lets try to factorize to find the product of two linear equations so as we can verify the pair of straight line.

$a x^{2}+2 h x y+b y^{2}+2 g x+2 f y+c=0$
This equation represents two straight lines, if $\Delta=a b c+2 f g h-a t^{2}-b g^{2}-c h^{2}=0$
or $\left|\begin{array}{lll}{a} & {h} & {g} \\ {h} & {b} & {f} \\ {g} & {f} & {c}\end{array}\right|=0$.

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The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

# Mean-median - Statistics - AMC 10B, 2019 Problem 13

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## Mean-median of some numbers

Mean, median, and mode are three kinds of "averages". … The "mean" is the "average" you're used to, where you add up all the numbers and then divide by the number of numbers. The "median" is the "middle" value in the list of numbers, And the mode is the number repeated most number of times in the given list. Let's see how to find the mean-median of some numbers.

## Try the problem

What is the sum of all real numbers $x$ for which the median of the numbers $4,6,8,17,$ and $x$ is equal to the mean of those five numbers?

$\textbf{(A) } -5 \qquad\textbf{(B) } 0 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } \frac{15}{4} \qquad\textbf{(E) } \frac{35}{4}$

AMC 10B, 2019 Problem 13

Statistics (Mean-median)

6 out of 10

challenges and thrills of pre college mathematics

## Use some hints

The mean is $\frac{4+6+8+17+x}{5}=\frac{35+x}{5}$.

Now there are only three possibilities for the median, It can be either 6,8 or x. It is because 4 is the smallest number and 17 cannot fit in the middle for any possible value of x.

Now if we consider 6 to be median then we must have to get 6 as the mean also. And we will verify this condition for each of the 6,8, and x.

See the final step for more hints.

$\frac{35+x}{5}=6$ has solution $x=-5$, and the sequence is $-5, 4, 6, 8, 17$, which does have median $6$, so this is a valid solution.

Now let the median be $8$.

$\frac{35+x}{5}=8$ gives $x=5$, so the sequence is $4, 5, 6, 8, 17$, which has median $6$, so this is not valid.

Finally we let the median be $x$.

Finally we let the median be $x$.

and the sequence is $4, 6, 8, 8.75, 17$, which has median $8$. This case is therefore again not valid.

Hence the only possible value of $x$ is $$(A) -5$$.

# Geometric Progression- ISI Entrance B. Stat (Hons) 2003- Problem 3

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## Geometric Progression

A geometric progression is a sequence in which each term is derived by multiplying or dividing the preceding term by a fixed number called the common ratio. E.g., the height to which a ball rises in each successive bounce follows a geometric progression. The sequence 4, -2, 1,... is a Geometric Progression (GP) for which (-1/2) is the common ratio. We can use the concept to find an arbitrary term, a finite or infinite sum of the series, and apply them in various contexts, including some difficult problems.

## Try the problem

Suppose that the three distinct real numbers $$a,b \text{ and } c$$ are in G.P. and $$a+b+c=xb$$. Then

(A) $$-3<x<1 ;$$

(B) $$x>1$$ or $$x<-3 ;$$

(C) $$x>3$$ or $$x<-1 ;$$

(D) $$-1<x<3 ;$$

ISI entrance B. Stat. (Hons.) 2003 problem 3

Geometric Progression

6 out of 10

challenges and thrills of pre college mathematics

## Use some hints

If any three quantity are in GP then we have a relation between them, in this case $$a,b,c$$ are in G.P. so we have

$$b^2=ac$$ or $$b= \sqrt{ac}$$.

We also have $$a+b+c=xb$$ so

which is equal to $$a+c=b(x-1)$$
=$$\frac{a}{b}+\frac{c}{b}=(x-1)$$
since $$b=\sqrt{ac}$$ we will get
=$$\sqrt{\frac{a}{c}}+\sqrt{\frac{c}{a}}=(x-1)$$
=$$\sqrt{\frac{a}{c}}+\frac{1}{\sqrt{\frac{a}{c}}}=(x-1)$$
Let $$\sqrt{\frac{a}{c}}=k$$, then we have the form of $$k+\frac{1}{k}$$ which we know has a value either greater than 2 or less than -2.
so we can write
either, $$x-1 > 2$$ or $$x-1 < -2$$
and now you can easily get the answer.

So the answer is x>3 or x<-1.

# Decimal system conversion AMC 10B, 2019 Problem 12

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## Octal to Decimal system conversion

This problem is based on decimal system conversion. In the given problem we have to convert the given number to the another number system from decimal and finding the sum of the digits later on.

## Try the problem

What is the greatest possible sum of the digits in the base-seven representation of a positive integer less than $2019$?

$\textbf{(A) } 11 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 22 \qquad\textbf{(D) } 23 \qquad\textbf{(E) } 27$

AMC 10B, 2019 Problem 12

Decimal system conversion

6 out of 10

challenges and thrills of pre college mathematics

## Use some hints

First of all 2019 is in base 10, see below

$$2019=2*10^3+0*10^2+1*10^1+9+10^0$$.

so we have to convert it in the system having base 7.

After converting we will get

But we have to maximize the sum of digits. So we need to increase the number of 6 in the converted number(in base 7) and it is because 6 is the largest number in the number system having base 7.

The conversion of maximize 6 in the number will occur with either of the numbers $4666_7$ or $5566_7$.

and now we can simply find the sum of the digits.

# Logarithm ISI entrance B. Stat. (Hons.) 2003 problem 2

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## Logarithm

The problem is based upon logarithm in which we find the value of an unknown quantity in an equation. And understanding whether the root or (value of x) is real or not and if yes, then how many real roots exist.

## Try the problem

The equation $$\log_3 x-\log_x 3 =2$$ has

(A) no real solution

(B) exactly one real solution

(C) exactly two real solution

(D) infinitely many real solution.

ISI entrance B. Stat. (Hons.) 2003 problem 2

Logarithm

6 out of 10

challenges and thrills of pre college mathematics

## Use some hints

we know when logarithm base and value are interchanged then the whole quantity is equal to the reciprocal of the previous logarithm.

i.e. $$\log_3 x$$ = $$\frac{1}{log_x 3}$$

Now we can assume the value of $$\log_3 x$$ is $$a$$ and the equation will reduce to $$a^2 -1/a=2$$ , or $$a^2-2a-1=0$$.

An now we can apply Sridharacharya's formula to find the valise(s) of x.

After solving we will get two values of a and they are

$$1\pm \sqrt{2}$$

And now these values will be equal to a or $$\log_3 x$$ , And from here we will get two values of x which are real.

So option (C) is the correct option.

# Ratio and Proportion , 2019 AMC 10B Problem 11

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## Ratio and Proportion

The given problem is based upon calculating the number of marbles in jars of specific color, to do so we have to use ratios of the marbles of different colors and use the ratio to calculate the actual number of marbles of required color.

## Try the problem

Two jars each contain the same number of marbles, and every marble is either blue or green. In Jar $1$ the ratio of blue to green marbles is $9:1$, and the ratio of blue to green marbles in Jar $2$ is $8:1$. There are $95$ green marbles in all. How many more blue marbles are in Jar $1$ than in Jar $2$?

$\textbf{(A) } 5\qquad\textbf{(B) } 10 \qquad\textbf{(C) }25 \qquad\textbf{(D) } 45 \qquad \textbf{(E) } 50$

2019 AMC 10B Problem 11

Ratio and proportion

6 out of 10

Secrets in Inequalities.

## Use some hints

Let $$2x$$ is the total no of marbles in both the jars. so each of the jar have $$x$$ marbles.

Thus, $\frac{x}{10}$ is the number of green marbles in Jar $1$, and $\frac{x}{9}$ is the number of green marbles in Jar $2$.

Since $\frac{x}{9}+\frac{x}{10}=\frac{19x}{90}$, we have $\frac{19x}{90}=95$, so there are $x=450$ marbles in each jar.

Since $$\frac{9}{10}th$$ of the jar 1 marbles are of blue color and $$\frac{8}{9}th$$ of the jar 2 marbles are of blue color.

Now we can easily find the no of blue marbles in both the jars and then we can subtract them to get the amount by which one exceed the other.