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Unconventional methods to orally solve problems from this year’s SimCATs and previous CAT papers under one minute

Wednesday 3 October 2012

1. A new flag is to be designed with six vertical stripes using some or all of the colours yellow, green, blue and red. Then, the number of ways this can be done such that no two adjacent stripes have the same colour is

(1)12 x 81(2)16 x 19(3)20 x 125(4)24 x 216

(CAT 2004)


For the first stripe there are 4 options and for each subsequent stripe you have 3 options excluding the colour to the left of it. So total number of ways = 4*3*3*3*3*3 = 12*81. Hence , (1).


2. Total expenses of a boarding house are partly fixed and partly varying linearly with the number of boarders. The average expense per boarder is Rs. 700 when there are 25 boarders and Rs. 600 when there are 50 boarders. What is the average expense per boarder when there are 100 boarders?

(1)540(2)550(3)560(4)570

(CAT 1999)


In calculating the price per boarder, the only thing that changes with the number of boarders is the fixed cost per boarder, since variable cost is always the same per boarder.

So if the fixed cost is F, the difference in the average cost per boarder at 25 boarders and 50 boarders, 700-600=100, is nothing but F/25-F/50.

The difference in the average cost per boarder at 50 boarders and 100 boarders = F/50-F/100 = (F/25-F/50)/2=100/2=50. Hence, (2).


3. There is a tunnel connecting city A & B. There is acat which is standing at 3/8th of the length of the tunnel from A. It hears the whistle of the train and starts running towards the entrance where, the train and the cat meet. In another case, the cat started running towards the exit and the train again met the cat at the exit. What is the ratio of their speeds?

(1)4:1(2)1:2(3)8:1(4)None of these

(CAT 2002)


The first case, running towards the entrance, the cat covers 3 parts. In the second case, running towards the exit, it covers 5 parts.

So in the case of the CAT the difference in the distances covered in the two cases is 2 parts.

In the case of the train, in the first case it reaches the entrance and in the second it reaches the exit of the tunnel.

So the difference in the two cases is the length of the tunnel, 8 parts.

So the required ratio is 4:1.Hence, (1).


4. X can do a piece of work in 20 days working 7 hours a day. The work is started by X and on the second day, one man whose capacity to do the work is twice that of X, joined in. On the third day, another man whose capacity is thrice that of X, joined in and this process continued till the work was completed. In how many days will the work be completed, if everyone works for four hours a day?

(1)5(2)6(3)7(4)8


7 hours a day takes 20 days, therefore 4 hours a day will take 35 man-days (20*7/4). Day 1 = 1 man-dayDay 2 = 1 + 2 = 3 man-daysDay 3 = 1 + 2 + 3 = 6 man-daysDay 4 = 10 Day 5 = 15 man-days At the end of 5 days the total number of man-days is 35. Hence, (1).


5. Working together, ‘A’ and ‘B’ can complete a piece of work in ‘t’ days. When ‘A’ works alone he takes 12 days more than ‘t’. When ‘B’ works alone he takes 3 days more than ‘t’. ‘A’ and ‘B’ work individually on alternate days and complete the work. If they are paid an amount of Rs. 1000 for the entire work and are to be paid in proportion to the amount of work done by each of them, then what amount does ‘A’ receive?

(1)Rs.300(2)Rs.300.33(3)Rs.400(4)Rs.400.33

(SimCAT 2012)


If A and B take a and b days more than the time taken, t, when they work together, then t is given by √ab. In this case they take 12 and 3 days more, so together they take 6 days. A takes 18 and B takes 9 days individually, so they do work in the ratio 1 : 2. So A gets 1/3 of the amount or 333.33. Hence, (2).


6. Aakash, Sameer and Sidhu can independently finish a piece of work in 12 days, 9 days and 18 days respectively. They start doing their work together and one of them quits after two days. Find who quits.

I. After one quits, the other two finish the work in 3 more days.

II. Had the two who had not quit, worked without the one who quit from the beginning, they would have completed the job in 6 days.

(1)1 statement is sufficient but not the other(2)Either statement alone is sufficient(3)Both statements are required(4)The question cannot be answered

(SimCAT 2012)


Since all 3 take different number of days, 12, 9 and 18, the number of days any two of them will take together will be a unique value. Statements I and II give these unique values. Hence, a combination of any two can be taken and solved to determine if they result in the unique value given in each of the statements. Hence, (2).


7. A can complete a piece of work in 4 days. B takes double the time taken by A, C takes double that of B and D takes double that of C to complete the same task. They are paired in groups of two each. One pair takes two thirds the time needed by the second pair to complete the work. Which is the first pair?

(1)A,B(2)A,C(3)B,C(4)A,D


A, B, C and D take x, 2x, 4x, and 8x days, therefore their speeds will be in the ratio 8:4:2:1. If two of them working take 2/3 of the time, they will have 3/2 of the speed. A + D together (9) will be 3/2 times B and C (6). Hence, (4).


8. It takes 6 technicians a total of 10 hours to build a new server from Direct Computer, with each working at the same rate. If six technicians start to build the server at 11 am and one technician per hour is added beginning at 5 pm, at what time will the server be complete?

(1)6:40 PM(2)7:00 PM (3)7:20 PM (4)8:00 PM


Same method as question 4. Total man-hours required is 60. By 5 p.m, 36 man-hours are completed. Starting 5 p.m , every hour one additional man-hour is added. So the remaining 24 man-hours will be completed in the next 3 hours since 7 + 8 + 9 = 24. Hence, (4).


9. B moves by taking 3 steps forward and 1 step backward (each step in 1 sec). He walks up a stationary escalator and reaches the top of the escalator in 118 seconds. However on a moving escalator he takes 40 secs to reach the top of the escalator. Find the speed of the escalator (in steps/second.)

(1)1 step/sec(2)2 steps/sec(3)3 steps/sec (4)4 steps/sec


The time taken by the escalator decreases by 1/3, from 118 to 40 secs, therefore the speed has to increase by 3 times. When stationary, the man’s speed is 2 steps in 4 secs or .5 step/sec. So 3 times the speed will be 1.5 steps/sec of which the speed of the escalator will be 1 step/sec. Hence, (1).


10.Given a bus leaves from city A to B on June 4th at 10.00 PM on a 12 hour journey. On the same day, another bus leaves from city B to A at 11.00 AM. At what time will both the buses meet, given that they both travel (with no breaks in their journeys) with the same constant speed and along the same route?

(1)10.30 pm on June 4th(2)9.30 am on June 5th(3)11.00 pm on June 4th(4)They do not meet anywhere


At 10 pm the bus at A will be about to start and bus from B be will be 1 hour away from A. Since both of them have the same speed they will meet at 10:30. Hence, (1).


11. Two towns A and B are 180 kms apart. Car 1 and car 2 start travelling from A and B, respectively towards each other. Their speeds are in the ratio 1 : 2 and they start at 7:00 a.m. and 8:00 a.m., respectively. After meeting at C, they return to their starting positions and again start travelling towards each other. In order to meet car 1 for the second time at C, car 2 halts at C. For how much time does it halt?

(1)1 hour(2)1.5 hours(3)2 hours(4)2.5 hours


Both cars travel certain distances from A and B to meet at a point C.

A C B

In going back to their starting points and returning to meet at C they are traveling twice the distance they did the first time they met at C. The first time A needed a head-start of 1 hour, so to travel double the distance, A will take 2 hours more. Hence, (3).


12.Solution A contains 10% Alcohol and Solution B contains 9 litres of alcohol. When the two solutions are mixed, the resultant mixture contains 12 % Alcohol. If the volume of Solution B is one third the volume of Solution A, then what is the combined volume of the resultant mixture?

(1)200 litres(2)400/3 litres(3)250 litres(4)250/3 litres


Ratio of the volumes of A and B will be in the inverse ratio of the distance of their concentrations from the resultant concentration. A is 3 times in volume and 10% alcohol, B has 9 liters of alcohol, the resultant mixture has 12% alcohol. A is 2 less than 12%, let B be x more than 12%. Therefore, 3/1= x/2, x=6 and B is 18% alcohol. If 18% is 9 liters then 100% will be 50 liters, A will be 150 liters and total will be 200 liters. Hence, (1).


13.A dealer bought rice worth Rs.1700 and sold 50% of it at the marked price gaining 8% in a particular month. There was an inflation of 4% in the price of rice the following month. What percentage discount (on the inflated M.P.) has to be given to make an overall profit of 9% on the rice bought?

(1)2.1%(2)2.6% (3)3% (4)3.4%


50 percent is sold at 8% profit and total profit is 9%, so the rest of the 50% is sold at 10% profit. If CP is 100, then the two SPs are 108(same as MP) and 110. In the second case there is an inflation of 4% on the MP, so 108 increases by 4% to 108+4 +.32=112.32(inflated MP). Discount is 2.32, which will be 2.32 percent of 100, so on the inflated MP of 112.32 it will be less than that and there is only one answer less than 2.32. Hence, (1).


14.I have to arrange six books on different subjects in a row such that the Math book is always to the left of the History book (not necessarily adjacent). In how many ways can I do this?

(1)5!(2)5! × 2 (3)60 (4)3 × 5!


6 books can be arranged in 6! ways. The history book is always either to the right or to the left of the Math book. In half of the cases the History book will be left of the Math book and in the other half it will be to the right. So required number of arrangements, 6!/2=5!*3. Hence, (4).


15.How many different words (not necessarily meaningful) can be formed if all the letters of the word ‘RAVINA’ are to be arranged such that the ‘I’ is always somewhere between the 2 ‘A’s?

(1)24(2)64(3)120(4)60


Similar to the previous problem. The word RAVINA can be arranged in 6!/2! or 360 ways. In all of these cases the letters A, I and A can be arranged in 3 ways relative to each other AIA, IAA and AAI (irrespective of how all the other alphabets are arranged). So in 1/3 of the total number of cases I will be between the two As. Hence, (3).


16. In a building with 8 floors (excluding the ground floor), Mr. ‘X’ along with 4 other people enters the lift on the ground floor. Given that each person gets off the lift on a different floor, what is the probability that Mr. X gets off the lift before exactly 3 people?

(1)1/5(2)2/45(3)32/35(4)3/7

(SimCAT 2012)


There are five people and each person gets off at a different floor. Since any person can get off at any floor, the probability that X gets off second out of a possible 5 positions(1, 2, 3, 4 or 5)is 1/5. Hence, (1).


17.The values of ‘x’ in the equation log10(x2 - 7x) = 1/log12010 are:

(1)7 + √10(2)–7, 4 (3)(15 + √5) / 2(4)15, -8


logba = 1/ logab. So the RHS will become log10120. Therefore, x(x-7) = 120. Hence, (4).


18. Find the value of ‘x’, if x = 1 .

2 + 1 .

4 + 1 .

2 + 1 .

4 + …

(1)√6 + 2 (2)2 - √2 (3)2 + √2 (4)√6 + 2


RHS is 1/(2 + ...), so it is less than .5. Options (3) and (4) are straightaway eliminated and (2) is 2 - 1.414 = .586 Hence , (1).


19.The sum of the first ‘n’ terms of the series :1/2 + 3/4 + 7/8 + 15/16 + ……. is equal to:

(1)2n – n + 1 (2)n – 2-n + 1 (3)n + 2-n – 1(4)2n – n – 1


Substitute n = 1 in check in which option the value is 1/2. Hence, (3).


20.Let n = 999 ... 99 be an integer consisting of a string of 2009 nines and no other digits Find the sum of digits of n2.

(1)18072(2)18081(3)18080(4)18073


If n = 9(1 nine), sum of digits of n2 (81) = 9. If n = 99 (2 nine’s), sum of digits of n2 (9801) = 18. Based on this pattern, in the given case the required sum will be 2009 *9 = 18081. Hence, (2).


21.If the product of n positive real numbers is unity, then their sum is necessarily

(1)a multiple of n(2)equal to n + 1/n(3)never less than n (4)a positive integer

(CAT 2008)


Take two cases where n= 2, 2 & 1/2 and 3 & 1/3, and you will see that only (3) holds true in both cases. Hence, (3).


22.A, B and C play a card game once every day for thirty days. The sole loser of every game gives the other two one rupee each. They keep a record of their wins and losses and at the end of thirty days, A and C together owe B Rs.6. In how many games was B not the loser?

(1)18(2)20(3)22(4)24


If B did not lose a single game, he would have made Rs.30. If one of these games changes to a loss he will lose the Rs.1 he made and will have to give Rs.2. So the net loss if a game changes from a win to a loss is Rs.3. B makes only Rs.6. This means a net change of 24 from 30 or a loss in 24/3 = 8 games. Hence, (3).


23.If a1 = 1 and an+1 – 3an + 2 = 4n for every positive integer n, then a100 equals

(1)399 – 200 (2)399 + 200(3)3100 – 200 (4)3100 + 200

(CAT 2005)


In option 1, if a100 = 399 - 200 then a1 = 30 - 2 = -2. The same pattern applies in all the other options, find out in which option a1 = 1. Hence, (3).


24.Let x = √(4 + √(4 - √(4 + √(4 - …)))) to infinity, then x equals

(1)3 (2)(√13 – 1) / 2(3)(√13 + 1) / 2(4)√13


If all the 4s apart from the first two are not considered (√4 + √4 - ...), it can be seen that the maximum value of sequence can be √6, whose value is between 2 and 3 (√4 and √9). So option (1) is ruled out; option (4), √13, is between 3 and 4, so it is ruled out; option (2) will be between 1 and 1.5. Hence (3).


25.Find the sum √ (1+1/12+1/22) + √(1+1/22+1/32) + ...+ √(1+1/20072+1/20082)

(1)2008 – 1/2008(2)2007 – 1/2007(3)2007 – 1/2008(4)2008 – 1/2009

(CAT 2008)


The given equation is in the form of a series of n terms with n being equal to 2008. If we take only the first term, n = 2 , the value is 3/2. The options is given in the format n-1/n, (n-1) - 1/(n-1), (n-1) - 1/n, n - 1/(n+1). Substitute 2 in the place is 2008 and 1 in the place of 2007 in each of the options and see in which case the value is 3/2. Hence, (1).


26.A two digit number is 18 less than the square of the sum of its digits. How many such numbers are there?

(1)1(2)2(3)3(4)4


Start with taking squares and subtracting 18 to shortlist 2-digit numbers. Below 36 all squares will give a 1-digit number when 18 is subtracted from them. So the numbers are 36-18 = 18, 49-18=31, 64-18=46, 81-18=63, 100-18=82. Of these only in the case of 63 and 82 is the square of sum of the digits 18 more than than number. Hence, (2).


27. The integers 1, 2, …, 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?

(1)820(2)819(3)781(4)780

(CAT 2008)


Let us simplify the problem by saying that in each operation two numbers are replaced by the sum of the numbers. So if we take any two numbers, say 39 and 2, we will be replacing it with 41. Now if we take 41 and 5, we will be replacing it with 46. In these two operations we just added 39 , 2 & 5. At the end of 39 operations we would have added all the numbers from 1 to 40, which is (40*41)/2=820. In the given problem apart from adding two numbers we are also subtracting 1, so in 39 operations we will have subtracted 39 one’s. So the answer will be 820-39=781. Hence, (3).


28. A rich merchant had collected many gold coins. He did not want anybody to know about them. One day, his wife asked. "How many gold coins do we have?" After pausing a moment, he replied, "Well! If I divide the coins into two unequal numbers, then 48 times the difference of the numbers is equal to the difference of their squares. The wife looked puzzled. Can you help the merchant's wife by finding out how many gold coins the merchant has?

(1)48(2)96(3)32(4)36

(CAT 2002)


If the merchant divides his coins into two groups with a and b coins then according to him 48(a-b)=a2-b2, so a + b = 48. Hence, (1).


29.Let D be a recurring decimal of the form D = 0.a1a2a1a2a1a2… where digits a1 and a2 lie between 0 and 9. Furthermore, at most one of them is zero. Then which of the following numbers, when multiplied by D, necessarily produces an integer?

(1)18(2)108(3)198(4)288


Any recurring decimal of the form 0.a1a2a1a2a1a2

has to be of the form n/99 (Take any decimal of this form, say D = 0.646464..., 100D = 64.6464..., 99D= 64, D = 64/99) . So to get an integer it has to be multiplied by a multiple of 99. Hence, (3).


30.Let A and B be two solid spheres such that the surface area of B is 300% higher than the surface area of A. The volume of A is found to be k% lower than the volume of B. The value of k must be____?

(1)85.5 (2)92.5 (3)90.5 (4)87.5

(CAT 2000)


According to the problem, if surface area of A = 1, then surface area of B = 4, so if volume of A = 1 then volume of B = 8, therefore k = 7/8% or 87.5. Hence, (4).


31. ABCDEF is a regular hexagon, of side 5 units. Two diagonals are drawn inside the hexagon dividing it into parts such that no part is a quadrilateral. Let P and Q be two such parts formed inside the hexagon such that the area of part P is maximum and the area of part Q is minimum amongst the areas of all the parts. What is the ratio of the area of P to that of Q?

(1)2:1 (2)11:3(3)13:1 (4)13:3


For any trapezium, if the parallel sides are in the ratio p : q, then the triangles formed by the diagonals will be in the ratio as shown in the figure below.

A hexagon is nothing but two such trapeziums with p:q being 1:2. So the required ratio will be 13:1

Hence, (3).


32.A stick of length 7P is cut into three parts and they form an isosceles triangle. What is the range of ‘x’ if it is the length of the equal side?

(1)7P/3 < x < 7P/2(2)7P/4 < x < 7P/2(3)7P/4 < x < 7P/3(4)None of these


The problem is solved better through visualization.

The largest possible length of the equal sides, they can together be almost 7P with the third side as negligible as possible. So, the maximum possible length of the equal side will be just less than 7P/2. Sum of two sides has to be greater than the third side, so the maximum length of the third and unequal side can be just less than 7P/2, so the us of equal sides will be just greater than 7P/4 and each side just greater than 7P/4. Hence, (2).


33.Given ABCD and A1B1C1D1 are both squares where the length of each side of both the squares is 4 units. Find (XY).

(1)√2 units (2)4 / (√2 + 1) units (3)2√2 / (√2 + 1)units (4)√2 + 1 units


Since the circle is inscribed in both squares, the figure that circumscribes the circle is a regular octagon.

Let the side of the regular octagon be a, it is also the hypotenuse if a isosceles right triangle. From the figure it can be seen that a/√2+ a + a/√2 = a + √2 a = a( 1+ √2 ) = 4. Hence, (2).


34.Find the approximate area of the shaded region in the figure, if the side of the largest square is 4 cm and the other squares are the largest squares that can be drawn inside the respective quarter circles.

(1)7.5 cm2(2)8 cm2 (3)4.4 cm2 (4)6.4 cm2


The area outermost shaded region is 16 - 4pi = 3.44.

The diagonal of the inner, smaller square is equal to the side of the outer, larger square since both of them are the radii of the circle that lies between them.

So the side of the smaller square is 1/√2 of the side of the larger square and hence its area will be 1/2 of the larger square.

So each successive shaded area will be 1/2 preceding shaded area.

So total shaded area = 3.44(1 + 1/2+ 1/4+1/8) which slightly less than 2*3.44. Only one option, (4), satisfies. Hence, (4).


35.In the adjoining figure, the ratio of the sides of the two squares ABCD and PQRC is 3 : 1. Also, l(AR) = 24 cm.Then l(PS) : l(SC) = ?

(1)1 : 3(2)1 : 4(3)2 : 3(4)1 : 2


The triangles RSC and RAD are similar. So, RD/AD= RC/SC = 4/3 = PC/SC = (PS + SC)/SC.

Therefore, PC/SC = 1/3. Hence, (1).


The preceding methods have been cracked by professionals, so please try them at home first before you do so in the CAT.

ALL THE BEST!


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