0
1360

Question 1:Â If a, b and c are 3 consecutive integers between -10 to +10 (both inclusive), how many integer values are possible for the expression?
$\frac{a^3+b^3+c^3+3abc}{(a+b+c)^2}$=?

a)Â 0

b)Â 1

c)Â 2

d)Â 3

e)Â 4

Question 2:Â Ln a bank the account numbers are all 8 digit numbers, and they all start with the digit 2. So, an account number can be represented as $2x_1x_2x_3x_4x_5x_6x_7$. An account number is considered to be a â€˜magicâ€™ number if $x_{1}x_{2}x_{3}$ is exactly the same as $x_{4}x_{5}x_{6}$ or $x_{5}x_{6}x_{7}$ or both. $X_{i}$ can take values from 0 to 9, but 2 followed by seven $0_{s}$ is not a valid account number. What is the maximum possible number of customers having a â€˜magicâ€™ account number?

a)Â 9989

b)Â 19980

c)Â 19989

d)Â 19999

e)Â 19990

Question 3:Â Little Pika who is five and half years old has just learnt addition. However, he does not know how to carry. For example, he can add 14 and 5, but he does not know how to add 14 and 7. How many pairs of consecutive integers between 1000 and 2000 (both 1000 and 2000 included) can Little Pika add?

a)Â 150

b)Â 155

c)Â 156

d)Â 258

e)Â None of the above

Question 4:Â If the last 6 digits of [(M)! – (N)!] are 999000, which of the following option is not possible for (M) Ã— (M – N)? Both (M) and (N) are positive integers and M > N. (M)! is factorial M.

a)Â 150

b)Â 180

c)Â 200

d)Â 225

e)Â 234

Question 5:Â There are 8436 steel balls, each with a radius of 1 centimeter, stacked in a pile, with 1 ball on top, 3 balls in the second layer, 6 in the third layer, 10 in the fourth, and so on. The number of horizontal layers in the pile is

a)Â 34

b)Â 38

c)Â 36

d)Â 32

Question 6:Â In a certain examination paper, there are n questions. For j = 1,2 â€¦n, there are $2^{n-j}$ students who answered j or more questions wrongly. If the total number of wrong answers is 4095, then the value of n is

a)Â 12

b)Â 11

c)Â 10

d)Â 9

Question 7:Â The number of positive integers n in the range $12 \leq n \leq 40$ such that the product (n -1)*(n – 2)*â€¦*3*2*1 is not divisible by n is

a)Â 5

b)Â 7

c)Â 13

d)Â 14

Question 8:Â Let T be the set of integers {3,11,19,27,â€¦451,459,467} and S be a subset of T such that the sum of no two elements of S is 470. The maximum possible number of elements in S is

a)Â 32

b)Â 28

c)Â 29

d)Â 30

Question 9:Â If a/b = 1/3, b/c = 2, c/d = 1/2 , d/e = 3 and e/f = 1/4, then what is the value of abc/def ?

a)Â 3/8

b)Â 27/8

c)Â 3/4

d)Â 27/4

e)Â 1/4

Question 10:Â On January 1, 2004 two new societies S1 and S2 are formed, each n numbers. On the first day of each subsequent month, S1 adds b members while S2 multiples its current numbers by a constant factor r. Both the societies have the same number of members on July 2, 2004. If b = 10.5n, what is the value of r?

a)Â 2.0

b)Â 1.9

c)Â 1.8

d)Â 1.7

Question 11:Â Three consecutive positive integers are raised to the first, second and third powers respectively and then added. The sum so obtained is perfect square whose square root equals the total of the three original integers. Which of the following best describes the minimum, say m, of these three integers?

a)Â $1 \leq m \leq 3$

b)Â $4 \leq m \leq 6$

c)Â $7 \leq m \leq 9$

d)Â $10 \leq m \leq 12$

e)Â $13 \leq m \leq 15$

Question 12:Â Twenty-seven persons attend a party. Which one of the following statements can never be true?

a)Â There is a person in the party who is acquainted with all the twenty-six others.

b)Â Each person in the party has a different number of acquaintances.

c)Â There is a person in the party who has an odd number of acquaintances.

d)Â In the party, there is no set of three mutual acquaintances.

Question 13:Â The number of non-negative real roots of $2^x – x – 1 = 0$ equals

a)Â 0

b)Â 1

c)Â 2

d)Â 3

Question 14:Â Twenty-seven persons attend a party. Which one of the following statements can never be true?

a)Â There is a person in the party who is acquainted with all the twenty-six others.

b)Â Each person in the party has a different number of acquaintances.

c)Â There is a person in the party who has an odd number of acquaintances.

d)Â In the party, there is no set of three mutual acquaintances.

Question 15:Â How many three digit positive integers, with digits x, y and z in the hundred’s, ten’s and unit’s place respectively, exist such that x < y, z < y and x $\neq$ 0 ?

a)Â 245

b)Â 285

c)Â 240

d)Â 320

Question 16:Â In a tournament, there are n teams $T_1 , T_2 ….., T_n$ with $n > 5$. Each team consists of k players, $k > 3$. The following pairs of teams have one player in common: $T_1$ & $T_2$ , $T_2$ & $T_3$ ,……, $T_{n-1}$ & $T_n$ , and $T_n$ & $T_1$ . No other pair of teams has any player in common. How many players are participating in the tournament, considering all the n teams together?

a)Â n (k – 1)

b)Â k (n – 1)

c)Â n (k – 2)

d)Â k (k – 2)

e)Â (n – 1)(k – 1)

Question 17:Â Consider four digit numbers for which the first two digits are equal and the last two digits are also equal. How many such numbers are perfect squares?

a)Â 3

b)Â 2

c)Â 4

d)Â 0

e)Â 1

Question 18:Â The integers 1, 2, â€¦, 40 are written on a blackboard. The following operation is then repeated 39 times: In each repetition, any two numbers, say a and b, currently on the blackboard are erased and a new number a + b – 1 is written. What will be the number left on the board at the end?

a)Â 820

b)Â 821

c)Â 781

d)Â 819

e)Â 780

Question 19:Â A shop stores x kg of rice. The first customer buys half this amount plus half a kg of rice. The second customer buys half the remaining amount plus half a kg of rice. Then the third customer also buys half the remaining amount plus half a kg of rice. Thereafter, no rice is left in the shop. Which of the following best describes the value of x?

a)Â $2 \leq x \leq 6$

b)Â $5 \leq x \leq 8$

c)Â $9 \leq x \leq 12$

d)Â $11 \leq x \leq 14$

e)Â $13 \leq x \leq 18$

Question 20:Â How many pairs of positive integers m, n satisfy 1/m + 4/n = 1/12 , where n is an odd integer less than 60?

a)Â 6

b)Â 4

c)Â 7

d)Â 5

e)Â 3

Since a,b,c are consecutive integers

=> $a = b-1$ and $c = b+1$

ExpressionÂ :Â $\frac{a^3+b^3+c^3+3abc}{(a+b+c)^2}$

= $\frac{(b – 1)^3 + b^3 + (b + 1)^3 + 3 (b – 1) b (b + 1)}{(b – 1 + b + b + 1)^2}$

= $\frac{b^3 + 3b + b^3 + b^3 + 3b + 3b^3 – 3b}{9 b^2}$

= $\frac{6 b^3 + 3 b}{9 b^2} = \frac{2 b^2 + 1}{3 b}$

Putting different values of b from – 10 to 10, we can verify that only – 1 and 1 satisfies to get integer values for the expression.

Ans – (C)

Account number = $2 x_1x_2x_3x_4x_5x_6x_7$

Case 1Â : $x_1x_2x_3$ is exactly same as $x_4x_5x_6$

=>Â $x_1x_2x_3 = x_4x_5x_6 = 000$ and $x_7$ =Â 1 to 9 Â  (as ‘20000000’ is not valid)

=> 9 possibilities

Now,Â $x_1x_2x_3 =Â x_4x_5x_6$ = 001 to 999 andÂ $x_7$ =Â 0 to 9

=> $999 \times 10 = 9990$ possibilities.

Case 2 : $x_1x_2x_3$ is exactly same as $x_5x_6x_7$

=>Â $x_1x_2x_3 = x_5x_6x_7 = 000$ and $x_4$ =Â 1 to 9 Â  (as ‘20000000’ is not valid)

=> 9 possibilities

Now,Â $x_1x_2x_3 = x_5x_6x_7$ = 001 to 999 andÂ $x_4$ =Â 0 to 9

=> $999 \times 10 = 9990$ possibilities.

Subtracting common possibilities in above cases.

=> $x_4x_5x_6 = x_5x_6x_7$

=> $x_4 = x_5 = x_6 = x_7$

Except ‘0’, the possibilities are 1111,2222,….,9999 => 9 possibilities.

$\therefore$ Maximum possible number of customers having a â€˜magicâ€™ account number

= $(9 + 9990) + (9 + 9990) – (9)$

= $19989$

Little Pika can addÂ  Â (1000, 1001), (1001, 1002), (1002, 1003), (1003, 1004), (1004, 1005) and (1009, 1010).

Similarly, he can add (1010, 1011),Â (1011, 1012),Â (1012, 1013),Â (1013, 1014), (1014, 1015) and (1019, 1020).

Similarly, he can add (1020, 1021),Â (1021, 1022),Â (1022, 1023),Â (1023, 1024), (1024, 1025) and (1029, 1030).

Similarly, he can add (1030, 1031),Â (1031, 1032),Â (1032, 1033),Â (1033, 1034), (1034, 1035) and (1039, 1040).

Similarly, he can add (1040, 1041),Â (1041, 1042),Â (1042, 1043),Â (1043, 1044) , (1044, 1045) and (1049, 1050).

We can see that there are 30 cases when we have changed unit and tens digit. Now the hundreds digit can be anything from {0, 1, 2, 3, 4}.

Hence, total number of such pairs which Pika can add = 5*30 = 150.

He can also number of form, (1099, 1100), (1199, 1200), (1299, 1300), (1399, 1400) (1499, 1500) and (1999, 2000)

Therefore, we can say that Pika can add 150+6 = 156 numbers. Hence, option C is the correct answer.

None of the answers given are correct. The reasoning is as given below.

999000 is a multiple of 8 but not of 16. If N! is a multiple of 16, M! would also be a multiple of 16 and hence M!-N! would be a multiple of 16.

Hence, as M!-N! = 999000, it would imply that N! is a multiple of 8 and not of 16. Therefore, N is either 4 or 5. So, N! is either 24 or 120. So, it would imply that M! is either 999024 or 999120. Both of which are not factorials for any natural number.

Hence, the given question is wrong.

For the given problem ,

$\sum {n(n+1)/2} = 8436$ which is

$\sum {n^2/2} + \sum{n/2} = 8436$ which is equal to

n*(n+1)(2n+1)/12 + n*(n+1)/4 = 8436 , solving we get n=36.

Solving the equation might be lengthy. you can substitute the values in the optionsÂ to arrive at the answer.

Let there only be 2 questions.

Thus there are $2^{2-1}$ = 2 students who have done 1 or more questions wrongly, 2$^{2-2}$ = 1 students who have done all 2 questions wrongly .

Thus total number of wrong answers = 2 + 1 = 3= $2^n – 1$.

Now let there be 3 questions. Then j = 1,2,3

Number of students answering 1 or more questions incorrectly = 4

Number of students answering 2 or more questions incorrectly = 2

Number of students answering 3 or more questions incorrectly = 1

Total number of incorrect answers = 1(3)+(2-1)*2+(4-2)*1 = 7 = $2^3-1$

According to the question , the total number of wrong answers = 4095 = $2^{12} – 1$.

Hence Option A.

positive integers n in the range $12 \leq n \leq 40$ such that the product (n -1)*(n – 2)*â€¦*3*2*1 is not divisible by n, implies that n should be a prime no. So there are 7 prime nos. in given range. Hence option B.

No. of terms in series T , 3+(n-1)*8 = 467 i.e. n=59.

Now S will have atleast have of 59 terms i.e 29 .

Also the sum of 29th term and 30th term is less than 470.

Hence, maximum possible elements in S is 30.

a/d = a/b * b/c * c/d = 1/3 * 2 * 1/2 = 1/3

Similarly, b/e and c/f are 3 and 3/8 respectively.

b/e = b/c*c/d*d/e = 3

c/f = c/d*d/e*e/f = 3/8

=> Value of abc/def = 1/3 * 3 * 3/8 = 3/8

According to given condition we have ,

n+6b =$nr^6$ and b=10.5n ,

63n+n = $nr^6$

$r^6 = 64$

r = 2

Let us assume that three positive consecutive integers are x, x+1, x+2. They are raised to first, second and third powers respectively.

$x^{1} + (x+1)^{2} + (x+2)^{3} = (x + (x+1) +(x+2))^{2}$

$x^{1} + (x+1)^{2} + (x+2)^{3}$ = $(3x + 3)^{2}$

$x^{3} + 7x^{2} + 15x + 9$ = $9x^{2} + 9 + 18x$

After simplifying you get,

$x^{3} – 2x^{2} – 3x = 0$

=> x=0,3,-1

Since x is a positive integer, it can only be 3.

So, the minimum of the three integers is 3. Option a) is the correct answer.

From the options a, c and d all can possibly occur. Hence option b. Besides, if all people have different number of acquaintances, then first person will have 26 acquaintance, second person will have 25 acquaintance, third person will have 24 and so on till 27 th person will have 0 acquaintance. 0 acquaintance is practically not possible.

$2^x – x – 1 = 0$ for this equation only 0 and 1 i.e 2 non-negative solutions are possible. Or we can plot the graph of $2^x$ and x+1 and determine the number of points of intersection and hence the solutin.

From the options a, c and d all can possibly occur. Hence option b. Besides, if all people have different number of acquaintances, then first person will have 26 acquaintance, second person will have 25 acquaintance, third person will have 24 and so on till 27 th person will have 0 acquaintance. 0 acquaintance is practically not possible.

x, y and z in the hundred’s, ten’s and unit’s place. So y should start from 2

If y=2 , possible values of x=1 and z = 0,1 .So 2 cases 120,121.

Also if y=3 , possible values of x=1,2 and z=0,1,2.

Here 6 three digit nos. possible .

Similarly for next cases would be 3*4=12,4*5=20,5*6=30,…..,8*9=72 . Adding all we get 240 cases.

The number of players in all the teams put together = k * n

The number of players that are common is 1*n = n

So, the number of players in the tournament = kn – n = n(k-1)

Let the number be xxyy
xxyy = 1000x + 100x + 10y +y = 1100x+11y = 11(100x+y)
Since xxyy is a perfect square, and 11 is one of the factors, it should be a multiple of 121
So, xxyy = 121k, where k is also a perfect square.
For k = 4, xxyy is a 3 digit number and for k > 82, xxyy is a five digit number
Between 4 and 82, only for k = 64, the number is of the form xxyy
121*64 = 7744
So, there is only 1 number 7744 which is of the form xxyy and a perfect square.

Alternatively:

The number should be definitely more than 32 and less than 100 as the square is a two digit number.

A number of such form can be written as $(50 \pm a)$ and $100 – a$ where $0 \leq a \leq 100$
So, the square would be of form $(50 \pm a)^2 = 2500 + a^2 \pm 100a$ or $(100 – a)^2$ i.e. $10000 + a^2 + 200 a$

In both cases, only $a^2$ contributes to the tens and ones digit. Among squares from 0 to 25, only 12 square i.e. 144 has repeating tens and ones digit. So, the number can be 38, 62, or 88. Checking these squares only 88 square is in the form of xxyy i.e. 7744.

Let the first operation be (1+40-1) = 40, the second operation be (2+39-1) = 40 and so on

So, after 20 operations, all the numbers are 40. After 10 more operations, all the numbers are 79

Proceeding this way, the last remaining number will be 781

After the first sale, the remaining quantity would be (x/2)-0.5 and after the second sale, the remaining quantity is 0.25x-0.75

After the last sale, the remaining quantity is 0.125x-(7/8) which will be equal to 0

SoÂ 0.125x-(7/8) = 0 => x = 7