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# BStat Hons Objective Admission Test 2005

Here are the problems and their corresponding solutions from BStat Hons Objective Admission Test 2005. Try it yourself and then read the solutions.

Here are the problems and their corresponding solutions from BStat Hons Objective Admission Test 2005.

Problems: BStat Hons Objective Admission Test 2005

1. How many three-digit numbers of distinct digits can be formed by using the digits 1, 2, 3, 4, 5, 9 such that the sum of the digits is at least 12?

(A) 61    (B) 66   (C) 60   (D) 11

Discussion:

We will use the method of compliment of counting. First we count the number ways of selecting three distinct numbers from the six digits:

$${{6} \choose {3} }= \frac {6 \times 5 \times 4 }{3\times 2 \times 1} = 20$$

We want to count the cases that do not work (sum becomes less than 12).

Note that if 9 is one of the three digits then the sum of the digits will be 12 or more (least case is 9+1+2 =12).

Now if we restrict to the remaining five available digits, none of the cases work except {5, 4, 3} (as 5 + 4 + 3 = 12).

Therefore we select 3 distinct digits from the five digits and delete the one selection of {5, 4, 3}. This gives us the total number of bad cases.

$${{5}\choose {3}} – 1 = 9$$

Hence the good selections are: total – bad = 20 – 9 =11.

Next note that each selection of three distinct integers can be permuted in 3!= 6 ways. Hence the total number of three -digit numbers that can be formed with the desired property = $( 11 \times 6 = 66 )$

Ans: (B) 66

2. If $( \sqrt{3} + 1 )$ is a root of the equation $( 3x^3 + ax^2 + bx + 12 = 0 )$ where a and b are rational numbers, then b is equal to

(A) -6;   (B) 2;   (C) 6;   (D) 10;

Discussion:

Clearly if $( 1 + \sqrt {3} )$ is one of the roots, then it’s conjugate $(1 – \sqrt {3} )$ is another root (as all the coefficients are rational numbers).

Suppose the third root is $( \gamma )$. (By Fundamental theorem of algebra there are three roots of a cubic).

By Vieta’s Theorem, the product of the roots is $(\frac{-12}{3} = -4)$.

Hence:

$$(1 + \sqrt{3} ) \times (1 – \sqrt {3}) \times \gamma = – 4$$

This implies $( \gamma = 2 )$ .

Now Vieta’s Theorem says, $( \frac{b}{3} )$ is sum of the product of roots taken two at a time. Thus:

$$\frac{b}{3} = (1 + \sqrt{3} ) \times (1 – \sqrt {3}) + \gamma \times { (1 + \sqrt{3} ) + (1 – \sqrt {3}) } = 2$$.

This implies b = 6

Ans: (C) 6;

3. The sum of all integers from 1 to 1000 that are divisible by 2 or 5 but not divisible by 4 equals

(A) 175000;   (B) 225500;   (C) 149500;   (D) 124000;

Discussion:

This is a problem related to Inclusion and Exclusion Principle in Combinatorics.

First lets add numbers from 1 to 1000 divisible by 2 but not by 4. So starting from 2 we add every fourth number. It is an arithmetic progression with first term 2, last term 998 and number of terms 250 (there are 500 even numbers from 1 to 1000 and ( $\frac{1000}{4} = 250$ ) is divisible by 4; hence the remaining 250 are divisible by 2 and not 4).

Therefore sum of the terms is:

$$\frac {250}{2} (2 + 998) = 500 \times 250 = 125000$$

Next we add the odd numbers that are divisible by 5 (we have already added the even ones).

So starting from 5, 15, 25, … , upto 995.

There are 100 such numbers and this is an arithmetic progression with common difference 10, first term 5 and last term 995.

Hence the sum is:

$$\frac{100}{2}(5 + 995)= 500 \times 100 = 50000$$

Thus the total required sum is 125000+50000 = 175000.

Ans: (A) 175000;

4. The value of$( { \frac {1}{2} ( -1 + \sqrt{3} i ) }^{15} + { \frac {1}{2} ( -1 – \sqrt{3} i ) }^{15} )$is

(A) -1;   (B) 0;   (C) $( \frac{1}{2^{14}} )$;   (D) 2;

Discussion:

This is an application of De Moivre’s Theorem.

$\lbrace \frac {1}{2} ( -1 + \sqrt{3} i ) \rbrace ^{15} + \lbrace \frac {1}{2} ( -1 – \sqrt{3} i ) \rbrace^{15}$

$= \left (\cos \frac{2 \pi}{3} + i \sin \frac {2 \pi}{3} \right )^{15} + \left (\cos \frac{4\pi}{3} + i \sin \frac {4\pi}{3} \right )^{15}$

$= \cos \frac{2 \times 15 \times \pi}{3} + i \sin \frac {2 \times 15 \times \pi}{3} + \cos \frac{4\times 15\times \pi}{3} + i \sin \frac {4\times 15 \times \pi}{3}$

$= \cos 10 \pi + i \sin 10 \pi + \cos 20 \pi + i \sin 20 \pi = 2$

Ans: (D) 2;

5. 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;

Discussion:

Note that

$x(x+3) = y(y-1) – 2 \ \Rightarrow (x^2 + 3x) – (y^2 – y ) = -2$

$\Rightarrow (x^2 + 2 \left( \frac{3}{2} \right ) x + \left ( \frac{3}{2} \right )^2 ) – ( y^2 – 2 \left (\frac{1}{2} \right ) y + \left( \frac {1}{2} \right )^2 ) = 0$

$\Rightarrow \left (x + \frac{3}{2} \right )^2 = \left (y – \frac{1}{2} \right )^2$

$\Rightarrow x + \frac{3}{2} = y – \frac{1}{2} \text { or } x + \frac{3}{2} = -y + \frac{1}{2}$

Hence we get a pair of straight lines. Ans: (B) a pair of straight lines; ## By Dr. Ashani Dasgupta

Ph.D. in Mathematics, University of Wisconsin, Milwaukee, United States.

Research Interest: Geometric Group Theory, Relatively Hyperbolic Groups.

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