Instructions

For the following questions answer them individually

Question 11

Let $$n!=123* ...n$$ for integer $$n \geq 1$$.

If $$p = 1!+(22!)+(33!)+... +(1010!)$$, then $$p+2$$ when divided by 11! leaves a remainder of

Video Solution

Solution

According to given condiiton we have p = (1 × 1!) + (2 × 2!) + (3 × 3!) + (4 × 4!) + … + (10 × 10!) . So n × n! = [(n + 1) - 1] × n! = (n + 1)! - n!. So equation becomes p = 2! - 1! + 3! - 2! + 4! - 3! + 5! - 4! +… + 11! - 10!. So p = 11! - 1! = 11! - 1. p + 2 = 11! + 1 .So when it is divided by 11! gives a remainder of 1. Hence, option 4.

Question 12

Consider a triangle drawn on the X-Y plane with its three vertices at (41, 0), (0, 41) and (0, 0), each vertex being represented by its (X,Y) coordinates. The number of points with integer coordinates inside the triangle (excluding all the points on the boundary) is

Video Solution

Solution

The number of points on x = 1 is 39. The number of points on x = 2 is 38 and so on till x = 39, which has one point.

So, the total is 1+2+3+...+39 = $$\frac{39*40}{2}$$ = 780.

Question 13

The digits of a three-digit number A are written in the reverse order to form another three-digit number B. If B > A and B-A is perfectly divisible by 7, then which of the following is necessarily true?

Video Solution

Solution

Let A = 100x + 10y + z and B = 100z + 10y + x .According to given condition B - A = 99(z - x) As (B - A) is divisible by 7 . So clearly (z - x) should be divisible by 7. z and x can have values 8,1 or 9,2 , such that 8-2=9-2=7 and y can have value from 0 to 9.
So Lowest possible value of A lowest x,y and z which is is 108 and the highest possible value of A is 299.

Question 14

If $$a_1 = 1$$ and $$a_{n+1} - 3a_n + 2 = 4n$$ for every positive integer n, then $$a_{100}$$ equals

Video Solution

Solution

Using given condition we find $$a_2$$ = 5 and $$a_3$$ = 21 and so on.

We see that the numbers are of form $$3^n-(2*n)$$

So for 100 we have $$3^{100}-200$$

Question 15

Let S be the set of five-digit numbers formed by the digits 1, 2, 3, 4 and 5, using each digit exactly once such that exactly two odd positions are occupied by odd digits. What is the sum of the digits in the rightmost position of the numbers in S?

Video Solution

Solution

When the odd numbers occupy places 1 and 3, only 2 or 4 can be in the 5th place. Odd numbers can occupy places 1 and 3 in 3C2*2! = 6 ways. When 2 is at the 5th place, the other odd number and 4 can be arranged in the remaining places in 2 ways. So, 2 occurs at the end 6*2 = 12 times. Similarly, 4 occurs 12 times.

If odd numbers occupy places 1 and 5, then 2 or 4 should come in the 3rd place. The other two numbers can then be arranged in 2 ways in the remaining blanks. So, if 1 is in the first place and 5 is in the 5th place, the other numbers can be arranged in 2*2 = 4 ways. Similar for 1 and 3; 5 and 1; 3 and 1; 5 and 3; 3 and 5. So, 5 occurs 8 times, 1 8 times and 3 8 times. Similar is the case when odd numbers are placed in 3rd and 5th places.

On the whole, 4 occurs 12 times, 2 occurs 12 times, 5, 3 and 1 each occur 16 times. The total is, therefore, 48+24+80+48+16 = 216

Question 16

The rightmost non-zero digit of the number $$30^{2720}$$ is

Video Solution

Solution

Rightmost non-zero digit of $$30^{2720}$$ is same as rightmost non-zero digit of $$3^{272}$$.

272 is of the form 4k.

All $$3^{4k}$$ end in 1.

=> Right most non-zero digit is 1.

Question 17

Four points A, B, C, and D lie on a straight line in the X-Y plane, such that AB = BC = CD, and the length of AB is 1 metre. An ant at A wants to reach a sugar particle at D. But there are insect repellents kept at points B and C. The ant would not go within one metre of any insect repellent. The minimum distance in metres the ant must traverse to reach the sugar particle is

Video Solution

Solution

Since ant cant go more near than 1 m.

So it'll have to travel in circular path from A till it is south of pt.

B and hence travels a quarter circle with radius = 1 and center as B, so travel $$\frac{\pi }{2}$$ m distance.

Then from B to C it travels 1m in straight line till C.

Then again from C to D in circular path way with distance $$\frac{\pi }{2}$$ m.

Hence total distance traveled = $$\frac{\pi }{2}$$ + $$\frac{\pi }{2}$$ +1 = $$\pi $$ +1.

Alternate solution :

Let the given figure represent the situation. A,B,C,D are the points with distance between them 1 m.

The repellents are at B and C respectively. The circles drawn with centre at B and C and radius equal to AB = 1 m.

Let E and F be the points of intersection of the circles and G and H be the points on the circle perpendicular to B and C.

Therefore, the ant can only travel at the circumference of the circles.

The shortest path for the to take will take is : A-G-H-D.

This is because taking the curve G-E and E-H will be a longer path than travelling straight from G to H.

While travelling from G to H, the distance of the ant will be more than 1 m from the repellant.

Thus, the distance travelled by the ant on the circumference of the circle ie

arc AG = $$\dfrac{90}{360} \times 2 \times \pi 1 = \dfrac{\pi}{2} $$

Length of arc HD = length of arc AG = $$\dfrac{\pi}{2} $$

Total length = $$(2 \times \dfrac{\pi}{2})+ BC = \pi + 1$$

Question 18

If x >= y and y > 1, then the value of the expression $$log_x (x/y) + log_y (y/x)$$ can never be

Video Solution

Solution

$$log_x (x/y) + log_y (y/x)$$ = $$1 - log_x (y) + 1 - log_y (x)$$
= $$2 - (log_x y + 1/log_x y)$$ <= 0 (Since $$log_x y + 1/log_x y$$ >= 2)
So, the value of the expression cannot be 1.

Question 19

For a positive integer n, let $$P_n$$ denote the product of the digits of n, and $$S_n$$ denote the sum of the digits of n. The number of integers between 10 and 1000 for which $$P_n$$ + $$S_n$$ = n is

Video Solution

Solution

Let n can be a 2 digit or a 3 digit number.

First let n be a 2 digit number.
So n = 10x + y and Pn = xy and Sn = x + y
Now, Pn + Sn = n
Therefore, xy + x + y = 10x + y , we have y = 9 .
Hence there are 9 numbers 19, 29,.. ,99, so 9 cases .

Now if n is a 3 digit number.
Let n = 100x + 10y + z
So Pn = xyz and Sn = x + y + z
Now, for Pn + Sn = n ; xyz + x + y + z = 100x + 10y + z ; so. xyz = 99x + 9y .
For above equation there is no value for which the above equation have an integer (single digit) value.

Hence option D.

Question 20

Rectangular tiles each of size 70 cm by 30 cm must be laid horizontally on a rectangular floor of size 110 cm by 130 cm, such that the tiles do not overlap. A tile can be placed in any orientation so long as its edges are parallel to the edges of the floor. No tile should overshoot any edge of the floor. The maximum number of tiles that can be accommodated on the floor is

Video Solution

Solution

As seen from the fig. If following configuration is used max 6 number of tiles that can be accommodated on the floor.

CAT Content

CAT 2005 Question Paper

Let $$n!=123* ...n$$ for integer $$n \geq 1$$.

If $$p = 1!+(22!)+(33!)+... +(1010!)$$, then $$p+2$$ when divided by 11! leaves a remainder of

Consider a triangle drawn on the X-Y plane with its three vertices at (41, 0), (0, 41) and (0, 0), each vertex being represented by its (X,Y) coordinates. The number of points with integer coordinates inside the triangle (excluding all the points on the boundary) is

The digits of a three-digit number A are written in the reverse order to form another three-digit number B. If B > A and B-A is perfectly divisible by 7, then which of the following is necessarily true?

If $$a_1 = 1$$ and $$a_{n+1} - 3a_n + 2 = 4n$$ for every positive integer n, then $$a_{100}$$ equals

Let S be the set of five-digit numbers formed by the digits 1, 2, 3, 4 and 5, using each digit exactly once such that exactly two odd positions are occupied by odd digits. What is the sum of the digits in the rightmost position of the numbers in S?

The rightmost non-zero digit of the number $$30^{2720}$$ is

If x >= y and y > 1, then the value of the expression $$log_x (x/y) + log_y (y/x)$$ can never be

For a positive integer n, let $$P_n$$ denote the product of the digits of n, and $$S_n$$ denote the sum of the digits of n. The number of integers between 10 and 1000 for which $$P_n$$ + $$S_n$$ = n is

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CAT Content

CAT 2005 Question Paper

Let $$n!=1*2*3* ...*n$$ for integer $$n \geq 1$$. If $$p = 1!+(2*2!)+(3*3!)+... +(10*10!)$$, then $$p+2$$ when divided by 11! leaves a remainder of

Consider a triangle drawn on the X-Y plane with its three vertices at (41, 0), (0, 41) and (0, 0), each vertex being represented by its (X,Y) coordinates. The number of points with integer coordinates inside the triangle (excluding all the points on the boundary) is

The digits of a three-digit number A are written in the reverse order to form another three-digit number B. If B > A and B-A is perfectly divisible by 7, then which of the following is necessarily true?

If $$a_1 = 1$$ and $$a_{n+1} - 3a_n + 2 = 4n$$ for every positive integer n, then $$a_{100}$$ equals

Let S be the set of five-digit numbers formed by the digits 1, 2, 3, 4 and 5, using each digit exactly once such that exactly two odd positions are occupied by odd digits. What is the sum of the digits in the rightmost position of the numbers in S?

The rightmost non-zero digit of the number $$30^{2720}$$ is

If x >= y and y > 1, then the value of the expression $$log_x (x/y) + log_y (y/x)$$ can never be

For a positive integer n, let $$P_n$$ denote the product of the digits of n, and $$S_n$$ denote the sum of the digits of n. The number of integers between 10 and 1000 for which $$P_n$$ + $$S_n$$ = n is

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Let $$n!=123* ...n$$ for integer $$n \geq 1$$.

If $$p = 1!+(22!)+(33!)+... +(1010!)$$, then $$p+2$$ when divided by 11! leaves a remainder of