(1) Two numbers when divided by a certain divisor leave remainders of 431 and 379 respectively. When the sum of these two numbers is divided by the same divisor, the remainder is 211. What is the divisor? (1) 599 (2) 1021 (3) 263 (4) Cannot be determined
Correct Answer - (1)
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Solution:
Two numbers when divided by a common divisor, if they leave remainders of x and y and when their sum is divided by the same divisor leaves a remainder of z, the divisor is given by x + y - z.
In this case, x and y are 431 and 379 and z = 211.
Hence the divisor is 431 + 379 - 211 = 599
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(2) What is the least number that should be multiplied to 100! to make it perfectly divisible by 350? (1) 144 (2) 72 (3) 108 (4) 216
Correct Answer - (2)
Solution:
100! has 348 as the greatest power of 3 that can divide it. Similarly, the greatest power of 2 that can divide 100! is 297. 297 = 448 * 21.
Therefore, the largest power of 12 that can divide 100! is 48.
Therefore, for 350 to be included in 100!, 100! needs to be multiplied by 32 * 23 = 9 * 8 = 72.
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(3) Two squares are chosen at random on a chessboard. What is the probability that they have a side in common? (1) 1 / 18 (2) 64 / 4032 (3) 63 / 64 (4) 1 / 9
Correct Answer - (1)
Solution:
The number of ways of choosing the first square is 64. The number of ways of choosing the second square is 63. There are a total of 64 * 63 = 4032 ways of choosing two squares.
If the first square happens to be any of the four corner ones, the second square can be chosen in 2 ways. If the first square happens to be any of the 24 squares on the side of the chess board, the second square can be chosen in 3 ways. If the first square happens to be any of the 36 remaining squares, the second square can be chosen in 4 ways. Hence the desired number of combinations = (4 * 2) + (24 * 3) + (36 * 4) = 224. Therefore, the required probability = 224/4032 = 1/18
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(4) Mark 1
If the question can be answered by using one of the statements alone, but cannot be answered by using the other statement alone.
Mark 2
If the question can be answered by using either statement alone.
Mark 3
If the question can be answered by using both statements together, but cannot be answered by using either statement alone.
Mark 4
If the question cannot be answered even by using both the statements together.
Question
A number is divisible by 33 if
the number is simultaneously divisible 3 and 11.
a number formed by reversing the positions of the digits of the same number is divisible by 33.
Correct Answer - (2)
Solution:
Statement A. If a number is divisible by 33, it should be simultaneously divisible by 3 and 11.
Statement B. The property of any number divisible by 11 is that the palindrome of the number is also divisible by 11. The test of divisibility of 11 is that the difference between the sums of the alternate digits should be 0 or a multiple of 11. Therefore, Statement B also holds true.
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(5) Each interior angle of a regular polygon is 120 degrees greater than each exterior angle. How many sides are there in the polygon? (1) 6 (2) 8 (3) 12 (4) 3
Correct Answer - (3)
Solution:
Let an exterior angle Ao. Then each interior angle will be 120 + Ao.
We know that in any regular polygon, the sum of an exterior and interior angle is always = 180o.
Therefore, A + 120 + A = 180 => A = 30o.
No. of sides of a polygon = = 12 sides.
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(6 If the diagonal and the area of a rectangle are 25 m and 168 m2, what is the length of the rectangle? (1) 17 m (2) 31 m (3) 12 m (4) 24 m
Correct Answer - (4)
Solution:
The diagonal d = 25m. and area A = 168 m2.
Let 'l' be the length and 'b' be the width of the rectangle.
Therefore, l2 + b2 = d2. and lb = A
We can therefore write (l + b)2 = d2 + 2A and (l - b)2 = d2 - 2A.
Substituting and solving we get, l + b = 31 and l - b = 17. Hence l = 24 and b = 7
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(7) A phenyl bottle is full of phenyl. One-third of it is taken out and then an equal amount of water is poured into the bottle. The operation is repeated four times. What is the ratio of dettol to water after the last operation? (1) 27 : 91 (2) 16 : 65 (3) 14 : 29 (4) None of these
Correct Answer - (2)
Solution:
Amount of dettol remaining / Amount of dettol original = (1-1/3)*(1-1/3)*(1-1/3)*(1-1/3)=16/81
Therefore, the ratio of dettol to water is 16 : 65.
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( A regular hexagon is inscribed in a circle of radius r cms. What is the perimeter of the regular hexagon? (1) 3r (2) 6r (3) r (4) 9r
Correct Answer - (2)
Solution:
A regular hexagon comprises 6 equilateral triangles, each of them having one of their vertices at the center of the hexagon. The sides of the equilateral triangle are equal to the radius of the smallest circle inscribing the hexagon. Hence, each of the side of the hexagon is equal to the radius of the hexagon and the perimeter of the hexagon is 6r.
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(9) If the product of n positive numbers is unity, then their sum is
(1) a positive integer
(2) divisible by n
(3) equal to n + (1/n)
(4) never less than n
Correct Answer - (4)
Solution:
Let the n positive numbers be a1, a2, ... an. The product is unity.
=> a1.a2....an = 1
We know that AM > GM
=> a1 + a2 + ... an > n
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(10) A 4 cm cube is cut into 1 cm cubes. What is the percentage increase in the surface area after such cutting? (1) 4% (2) 300% (3) 75% (4) 400%
Correct Answer - (2)
Solution:
Volume of 4 cm cube = 64 cc. When it is cut into 1 cm cube, the volume of each of the cubes = 1cc
Hence, there will be 64 such cubes. Surface area of small cubes = 6 (12) = 6 sqcm.
Therefore, the surface area of 64 such cubes = 64 * 6 = 384 sqcm.
The surface area of the large cube = 6(42) = 6*16 = 96.
% increase = = 300%
(11) There are 12 pipes attached to a tank. Some of them are fill pipes and some are drain pipes. Each of the fill pipes can fill the tank in 12 hours, while each of the drain pipes will take 24 hours to drain a full tank completely. If all the pipes are kept open when the tank was empty, it takes 2 hours for the tank to overflow. How many of these pipes are drain pipes? (1) 6 (2) 11 (3) 4 (4) 7
Correct Answer - (3)
Solution:
There are 12 pipes attached to the tank. Let ‘n’ of them be fill pipes. Therefore, there will be 12-n drain pipes.
Each fill pipe, fills the tank in 12 hours. Therefore, th of the tank gets filled every hour by one fill pipe.
‘n’ fill pipes will, therefore, fill th of the tank in an hour.
Each drain pipe drains the tank in 24 hours. That is, th of the tank gets drained by one drain pipe every hour.
12-n drain pipes, will therefore, drain th of the tank in an hour.
When all the pipes are open when the tank is empty, it takes 2 hours for the tank to overflow. i.e. ½ the tank gets filled every hour.
Equating the information, we get
=> => 3n - 12 = 12 or 3n = 24 or n = 8.
Therefore, there are 8 fill pipes and (12 - = 4 drain pipes.
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(12) What is the longest gap that any person will have to wait between two birthdays? (1) 365 days (2) 366 days (3) 4 years (4) 8 years
Correct Answer - (4)
Solution:
Any person who is born on February 29th will normally celebrate his birthday once in four years whenever a leap year happens.
However, years like 1700, 1800, 1900 which are divisible by ‘4’ but not by ‘400’ are not leap years. 2000 was a leap year, 2400 will be a leap year.
So, for a person born on Feb 29, 1892, for instance, will have celebrated a birthday on Feb 29, 1896 and then the next birthday will be only on Feb 29, 1904. That is a 8-year gap.
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(13) What is the area of the largest triangle that can be fitted into a rectangle of length 'l' units and width 'w' units? (1) lw/3 (2) (2lw)/3 (3) (3lw)/4 (4) (lw)/2
Correct Answer - (4)
Solution:
The triangle which has its base as the length of the rectangle and its height as the width of the rectangle or the triangle which has its base as the width of the rectangle and its height as the length of the rectangle will be the largest triangle that can be fitted in the rectangle.
If the base of the triangle is 'l' and its height 'w', then its area is units.
Similarly, if the base of the triangle is 'w' units and its height is 'l' units, then its area is units.
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(14) A predator is chasing its prey. The predator takes 4 leaps for every 6 leaps of the prey and the predator covers as much distance in 2 leaps as 3 leaps of the prey. Will the predator succeed in getting its food? (1) Yes
(2) In the 6th leap
(3) Never
(4) Cannot determine
Correct Answer - (4)
Solution:
Distance covered in 2 leaps by predator = 3 leaps of the prey.
Distance covered in 1 leap of predator = 3/2 leaps of prey. ----(1)
4 leaps of predator : 6 leaps of prey ----(2)
Using (1) and (2), we get
4*3/2 leaps of predator : 6 leaps of prey.
=> 1:1
If the predator and prey start simultaneously at the same point, the predator will catch the prey immediately. If not so, then the predator will never catch the prey as it was running at the same speed.
As it was not mentioned in the question that they start simultaneously from the same point or not, we can't determine the answer. Therefore, the answer choice is (4).
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(15) In what ratio must a person mix three kinds of tea costing Rs.60/kg, Rs.75/kg and Rs.100 /kg so that the resultant mixture when sold at Rs.96/kg yields a profit of 20%? (1) 1 : 2 : 4 (2) 3 : 7 : 6 (3) 1 : 4 : 2 (4) None of these
Correct Answer - (3)
Solution
The resultant mixture is sold at a profit of 20% at Rs.96/kg
i.e. 1.2 (cost) = Rs.96 => Cost = = Rs.80 / kg.
Let the three varities be A, B, and C costing Rs.60, Rs.75 and Rs.100 respectively.
The mean price falls between B and C.
Hence the following method should be used to find the ratio in which they should be mixed.
Step 1. Find out the ratio of QA : QC using alligation rule
Step 2. Find out the ratio of QB : QC using alligation rule
Step 3. QC, the resultant ratio of variety c can be found by adding the value of QC in step 1 and step 2 = 1 + 1 = 2.
However, in CAT if you try and solve the problem using the above method, you will end up spending more than 2, and may be 3 minutes on this problem, which is a criminal mismanagement of time.
The best way to solve a problem of this kind in CAT is to go from the answer choices as shown below
The resultant ratio QA : QB : QC :: 1 : 4 : 2.
1 kg of variety A at Rs.60 is mixed with 4 kgs of variety B at Rs.75 and 2 kgs of variety C at Rs.100.
The total cost for the 7 kgs = 60 + (4 * 75) + (2 * 100) = 60 +300 + 200 = 560.
Cost per kg of the mixture = 560/7 = 80 kgs.
Part III
(16) What is the reminder when 9^1 + 9^2 + 9^3 + ...... + 9^9 is divided by 6? (1) 0 (2) 3 (3) 4 (4) None of these
Correct Answer - (2)
Solution:
Any number that is divisible by ‘6' will be a number that is divisible by both ‘2' and ‘3'. i.e. the number should an even number and should be divisible by three (sum of its digits should add to a multiple of ‘3').
Any number that is divisible by ‘9' is also divisible by ‘3' but unless it is an even number it will not be divisibly by ‘6'.
In the above case, ‘9' is an odd number. Any power of ‘9' which is an odd number will be an odd number which is divisible by ‘9'.
Therefore, each of the terms 91, 92 etc are all divisibly by ‘9' and hence by ‘3' but are odd numbers.
Any multiple of ‘3' which is odd when divided by ‘6' will leave a reminder of ‘3'.
For example 27 is a multiple of ‘3' which is odd. 27/6 will leave a reminder of ‘3'. Or take 45 which again is a multiple of ‘3' which is odd. 45/6 will also leave a reminder of ‘3'.
Each of the individual terms of the given expression 9^1 + 9^2 + 9^3 + ...... + 9^9 when divided by 6 will leave a reminder of ‘3'. There are 9 such terms. The sum of all the reminders will therefore be equal to 9*3 = 27.
27/6 will leave a reminder of ‘3'.
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(17) ‘a’ and ‘b’ are the lengths of the base and height of a right angled triangle whose hypotenuse is ‘h’. If the values of ‘a’ and ‘b’ are positive integers, which of the following cannot be a value of the square of the hypotenuse? (1) 13 (2) 23 (3) 37 (4) 41
Correct Answer - (2)
Solution
The value of the square of the hypotenuse = h^2 = a^2 + b^2
As the problem states that ‘a’ and ‘b’ are positive integers, the values of a2 and b2 will have to be perfect squares. Hence we need to find out that value amongst the four answer choices which cannot be expressed as the sum of two perfect squares.
Choice 1 is 13. 13 = 9 + 4 = 32 + 22. Therefore, Choice 1 is not the answer as it is a possible value of h2
Choice 2 is 23. 23 cannot be expressed as the sum two numbers, each of which in turn happen to be perfect squares. Therefore, Choice 2 is the answer.
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(1 Two trains A and B start simultaneously from stations X and Y towards each other respectively. After meeting at a point between X and Y, train A reaches station Y in 9 hours and train B reaches station X in 4 hours from the time they have met each other. If the speed of train A is 36 km/hr, what is the speed of train B? (1) 24 km/hr (2) 54 km/hr (3) 81 km/hr (4) 16 km/hr
Correct Answer - (2)
Solution
The ratio of the speed of the two trains S_a and S_b is given by
S_a/S_b = root(b/a)
, where b is the time taken by train B to reach its destination after meeting train A and a is the time taken by train A to reach its destination after meeting train B.
In this case,
=> Speed of train A/ Speed of train B = root( 4/9)
Speed of B = 3/2*36 =54 kmph
An interesting exercise that you can try is to find out how this formula was derived. You can check out for the derivation tomorrow on our site to verify if your approach was right.
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(19) What is the value of 1*1! + 2*2! + 3!*3! + ............ n*n!,
where n! means n factorial or n(n-1)(n-2)...1 (1) n(n-1)(n-1)! (2) (n+1)!/(n(n-1)) (3) (n+1)! - n! (4) (n + 1)! - 1!
Correct Answer - (4)
Solution
1*1! = (2 -1)*1! = 2*1! - 1*1! = 2! - 1!
2*2! = (3 - 1)*2! = 3*2! - 2! = 3! - 2!
3*3! = (4 - 1)*3! = 4*3! - 3! = 4! - 3!
..
..
..
n*n! = (n+1 - 1)*n! = (n+1)(n!) - n! = (n+1)! - n!
Summing up all these terms, we get (n+1)! - 1!
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(20) The largest number amongst the following that will perfectly divide 101100 - 1 is (1) 100 (2) 10,000 (3) 100100 (4) 100,000
Correct Answer - (2)
Solution
The easiest way to solve such problems for CAT puposes is trial and error.
101^2 = 10201. 101^2 - 1 = 10200. This is divisible by 100. Similarly try for 101^3 - 1 = 1030301 - 1 = 1030300.
So you can safely conclude that (101^1 - 1) to (101^9 - 1) will be divisible by 100 & then (101^10 - 1) to (101^99 - 1) will be divisible by 1000 & therefore (101^100 - 1) will be divisible by 10,000.
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