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;

Solutions for Test of Mathematics at the 10 +2 Level