CTS Placement Paper General - Other -01 Jun 2008
COGNIZANT PAPER ON 2nd JUNE 2008
If all the 6 are replaced by 9, then the algebraic sum of all the numbers from 1 to 100(both inclusive) varies by The solution is as follows: 1} Sum of numbers from 1 to 100 is 5050 2} Numbers having 6 are 6,16,..56,60,61,...69,76,86,96 whose sum is 1089. 3) Remove the above sum from 5050 we get 3961. 4) Replace 6 in numbers in step 2 with 9,which we get 9,19,..59,90,91,..99,79,89,99 whose sum is 1419. 5) Add above sum to 3961 we get 5380. 6) So increase in sum is by 5380-5050=330.
B Is 50% faster than A. If A starts at 9 A.M. and B starts at 10 A.M. A travels at a speed of 50 km/hr. If A and Bare 300 kms apart, The time when they meet when-they travel In opposite direction is Ans: 12 noon speed of A=50 so speed of B=75. Let at T A.M. they meet. So distance travelled by A=50(T-9} ABD distance by B=75(T-10). Total = 50(T-9) + 75(T-10) = 300 so T=12 noon.
A, B, C, can do a work in 8,14,16 days respectively. A does the work for 2 days. B continues from it and finishes till 25% of the remaining work. C finishes the remaining work. How many days would have taken to complete the work I' A completes 1/4th in 2 days and remaining is 3/4th.
B completes 3/16th work in 14*3/16 21/8 days ~.
C compietes 9/161h work in 9 days. So total no of days =2+21/8+9=13 5/8 days.
A starts from a place at 11.00 A.M. and travels at a speed of 4 kmph, B starts at 1.00 P.M. and travels with speeds of 1 kmph for 1 hour, 2 kmph for the next '/ hour, 3 kmph for the next 1 hour and so on. At what time will B catch up with A a) 9.24 b) 9.32 c) 9.48 d) none ' .
Answer is 9.48. A starts at 11.00 am with 4kmph speed that means he covers 8 km in 2 hrs i.e., at 1 pm. Now at 1 pm 8 starts, his speed increases by 1 kmph for every hour. Therefore they both can catch up at speeds for 7 hrs 4+4+4+4+4+4+4=28 for A; 1+2+3+4+5+6+7=28 for 8 But A has a head start of 2 hrs this can be covered in the next hour but again A covers 4hrs in this 1 hr. In this way it can be worked out and the answer is 9.48hrs.
At a certain moment a watch shows 2 min lag although it Is running fast. If It showed a 3 min lag at that moment, but also gains by 1/2 min more a day than Its current speed It would show the true time one day sooner than it usually does. How many minutes does the watch gain per day. Let fast per day be x, no of days required to show correct time For condition 1 is 21 x. For condition 2 is 3 I (x + 1/2) Condition 2 is one day sooner hence 2 I x=3 I (x + 1/2) +1 Solving this we will get x"2 +1.5 x -1 = 0 x=-2 or +1/2. It is running fast hence x = 112 Hence the answer is 112 min per day. To verify this for condition 1 it requlre 4 days ( 2 % 1/2 ) Condition 2 it require 3 days [ 3 % (112 + 1/2)] i.e 1 day less than condition 1.
Swetha and Chaitanya went to a bookshop. Swetha purchased 5 pens, 3 note books and 9 pencils and used up all her money. Chaitanya purchased 6 pens, 6 note books and 18 pencils and paid 50% more than what Swetha paid. What % of Swetha1s money was spent on pens? Ans: 62.5% There r two jars. In the first jar there is 25%Milk and in the second jar there is 50% milk. What ratio of milk should be added to get 12 lit of milk from the two jars such that we get 62.5% Milk In the whole mixture? x=2 + 21\(113) + 21\(2/3). Find the value of xl\3 -6x1\2 -8x. 62.5% of 121ts=12/1 00*62.5=7.5Its.Ratioofmilkinjars=25%:50%=1:2ratio of milk from 2jars to be taken to get a total of 7.51ts of milk = (1/3)*7.5 and (2/3)*7.5 = 2.5 Its and 51ts respectively
A man bought at the cost of 5 plums a rupee and 2 oranges a rupee. HI:! sells 10 plums and 6 oranges at the selling price of 4 plums a rupee and 3 oranges a rupee. What is his gain or loss? ' Cost Price of 1 plum = Re. 1 15 = 20 paise Cost Price of 1 orange = Re. 112 = 50 paise Cost Price of 10 plums &6 oranges = 10*0.20 + 6*0.50 = Rs. 5/Selling Price of 1 plum = Re. 1/4 = 25 paise Selling Price of 1 orange = Re. 1/3 = 33 paise Selling Price of 10 plums &6 oranges = 10*0.25 + 6*0.33 = Rs. 4.501 So Selling Price -Cost Price = Rs. 4.50 -Rs. 5 =50 paise Loss = 50 paise
In an examination a student must answer 65% of the questions correctly. But it was found t hat after answer 7 questions correctly out of the first 14, the student qualified. What is the min no. of questions in the paper. (a) 22 (b) 18 (c) 20 (d) Can't say The Answer is 20. Explanation: What is asked is the min no. of questions.... so what we need to do is assuming that he answers every question from now on correctly, we need to find the no. of questions. First let us take 18. We find that 65% of 18 are 11.7. Out of the 18 questions, he has finished answering 14 questions already out of which 7 is correct. So even if he answers the remaining 4 questions correctly, he gets to only 61.11%.... So answer is not 18. Now let us take 20. 65% of 20 is 13. 14 questions are already answered and 7 correct out of them. So remaining is 6Jquestions. And if he answers all the 6 questions correctly, he gets to 7+6 = 13 which is the required % to pass in the exam. So the min. no. of questions is 20.
There are two bridges at a distance of 1km. A person starts rowing upstream from 1st bridge and when he reaches the second bridge he losses his cap. After 15mlns he realizes it and turns back and rows downstream and catches his hat at the bottom of 1st bridge. Find: The speed of boat in still water The equation is like this: let u kmlhr : speed of water, v kmlhr : speed of boat in still water 1/u=1/4+1/(v+u)+(v-u)/4(v+u) 15 mins=1/4 hr solving, u=2km/hr
There are 10 coins. 6 coins showing head and 4 showing tail. Each coin was randomly flipped (not tossed) seven times successively. After flipping the coins are 5 heads 4 tails one is hided the hided coin will have what. After flipping the coins, there are 6 heads and 4 tails. ***KCT***
A and B run in opposite directions from a point P on a circle with different but constant speeds. A runs in anticlockwise direction. They meet for the first time at a distance of 900 m in clockwise direction from P and for the second time at a distance of 800 m in anticlockwise direction from P. If B is yet to complete one round, the circumference of the circle is a)1700m b)1250m c)1300m d)1200m i. Ans: 1200
Some of the friends decided to go on a picnic they decided to go on a bike if the biker decided to take one pillion with him then there is no means of travel for the people as equal to the no of bike + 1. if they decided to go on two trip three blker left unused in the second trip find the no oJ person having bike? Let the no of bikes be 'x' Let the no of friends be fly' At a time... 2 can travel in a bike... so the total no of friends who can go in the bike at a time = 2x. Then the number of friends left over are (no of bikes" 1) ie (x+1 )... so let us equate y=2x+(x+1 )
y=3x+1----------EQ 1 Take the 2nd case.. 2 trips r made.. In the 1st trip... 2x travel and in the 2nd trip 3 bikes r left alone.. So only 2(x~3) travel [assume that the bikes return on their own.] let us equate now y=2x+2(x-3) ie y=4x-6------~---EQ 2 Solving the two equations, y=22 and x=7 ie.. There were 22 friends all together and there were 7 bikes ***KCT*** What Is the co-efficient of the term Independent of x in the series [(1/2)(1/x)"1/3 +(1/x)"(-1/5)]"8. Where" denotes "to the power of". (1/2x"-1/3 + x"1/5)"8 is a binomial expression n the general term is c(8,r)(1/2x"·r/3)(x"r/5) so if we want the term independent f x the x coefficient of x =0 so we have -5r+24-3r=0. So x=3 hence the coefficient of the independent r term is 7 I A red cube Is cut Into 1000 pieces. cubes from 2nd column are removed and remaining Is painted black. 1.How many cubes will have more than one surface red? 2.How many cube will have three surface black? 3.How many cubes will have one side as red and atleast two sides as black?
1.4 cubes (of 2 sides which is painted red). 2.8 cubes (of 3 sides which is painted black). 3.0 (red and black already separated).
A squarer side is 5cm. If a square of side 10cm Is hinged at the center of the previous square. When they r rotated common area to both squares Ans is 25
4 weights are weighed in pairs. Weights of pairs are determined as 103, 105, 106, 106, 107, and 109. What is the minimum weight? Minimum weight is 51
Diameter of circle Is d, Find length of string. (outer string that covers the circle) Answer is 1=2(22/7)*r l=d(22/7) 1=3.14*d
Diamond's value Is proportional to its weight2 .When the diamond broke weights of pieces in ratio 1:2:3; 4:5.Totalloss in value is 85,000. What is the value of the diamond twice the weight of the original diamond is 1, 80,000
40 shots taken. 50p for a hit. 10p for a miss. (he have to give). Finally he has Rs.5.How many hits.? i x*50~(40-x)1 0=500
Solid cube of 6· 6· 6. This cube is cut into to 216 small cubes. (1 ? 1 ? 1).the big cube is painted in all its faces. Then how many of cubes are painted at least 2 sides.
The formula for the cube problem is n"3=((n-2)"3)+6((n-2)"2)+12(n·2)+88-~->3 sides painted12(n2)--->2 sides painted 6((n-2)"2)----> 1 side painted (n-2)"3-----> no sides painted here, at least 2 sides should be painted. So, 2 side +3 side = 48+8 = 56 ***KCT*** A Bacteria is doubling at every 4 min. After 40 min 1024 bacteria's formed. Then 256 when Ans. 32 min ' '. 4 minutes less then 40 will have half the no. of bacteria, l.e., 1024 Bacteria in 40 min 512 Bacteria in 36 min 256 Bacteria in 32 min
A work in 12 days B in 15 days. Find the no of days if they work on alternate days Answer will be 71/4, it is as A+B in one day can work as 1/12+1/15=9/60 So in 6 days they work = 6* 9/60= 54/60 7th day has to be done by A, then = 54/60 + 1/12= 59/60 Remaining is =1-59/60= 1/60 On 8th day B has to do 1/60th of work, but he does 1/15 in one day, so in order to complete 1/60, he takes 1/4th of a day So answer will be 7 + 1/4= 7 1/4 days
A,B,C are positive integer. Out of them 2 are odd. Then 52a + (b-5) 3(c-3~ 2 =Ans is 52.
The ratio of white balls and black balls is 1:2. If 9 gray balls is added it becomes 2:4:3. Then what is number of black balls? The number of balls should be divisible by both 3 and 4. The numbers which are divisible by 3 and 4. 12 -> here the no of urns should be 3 at first case, but it is not suitable for the next case. 24 -> here the no of urns should be 7 at the first case, and it also satisfies the second case 24/4=6, so 1 urn will be remaining. Hence Answer is: There are 24 Balls and 7 Urns.
X and Y live in a North-South parallel street. X travels 10 km towards North to reach the east-West Street. Y travels 6 km towards south to reach the east -West Street. X travels now 4km towards east and y travels 8km towards west and they met each other. What Is the distance between x and y? Answer is 20.7, as X moves to north then east it makes a angle of 90deg so we can get the length of hypotaneous, similarly for y we can get the length of hypotaneous.. Adding up these two length gives us the actual distance between X and Yi.e 10+10.7=20.7 Ans:20 ***KCT*** Two cubes cut into 216 pieces and merged, later painted find how many cubes are un coloured, only two face colored There is an easy formula One side painted =[(x-2)"2]*6 Two side painted =[x-2]*12 Three side painted=8(always) (A cube has 8 corners) Unpainted=(x-2)"3 6"3=216 1 side painted = 96 2" = 48 3" = 8 Unpainted = 64 Total = 216
A family went for a picnic. 20% of them ate apples, 30% for them ate oranges, 40% of the age grapes. 10% of ate all the three. 12 % ate apples and oranges. 14% ate apples and grapes. 10% of they ate grapes and oranges. 1.How many ate only apples 2.How many ate only grapes 3.How many ate none Draw the three circle name as apple grapes orange They have to be intersect each other Only apple given =20% Only oranges given=30% ,only grapes given=40% All three=1O%,apple & orange=12% Apple & grape=14%;orange&grapes=10%
Cognizant Answers: Only apple=20%-10% (b'cos all)-(12%-10%)(apple & orange)(2%)-(14%-10%) (apple & grapes)(4%) =20%-( 10+2+4 )=4% only apple: like this all For none add all only apply, orange, grapes= 48%+16%(from apple & orange(2)apple &grape(4)grape &orange(O)all(10)) Total=64 100-64=36% not.ate the apple
A bag contains 3 balls of 11 different colors each. Find the minimum n'ulnber of chances to find at least 3 balls of same color? Ans: 23 We have to pick 3 balls of same colour from a bag containing 11 different coloured balls. So 1st pick11 balls of 11 different colours. Now pick again the 11 different coloured balls. Now pick 1 ball of any colour. So the answer is 23. (Now you get 3 balls of same); **"KCT*** Weights are weighed in pairs. Weights of pairs are determined as 103,105,106,106,107,109 What is the minimum weight?
Diamond's value Is proportional to its weight .When the diamond broke weights of pieces in ratio 1:2:3; 4:5. Total loss in value Is 85,000. What Is the value of thEt diamond twice the wt of the original diamond? We can assume the weights as a.b.c and d and as he mentioned that they are taken in pairs we can relate it as (a+b)+(b+c)+(c+d)+(b+d)+(a+d)+(c+a)=103+105+106+106+107+109 3(a+b+c+d) = 636 a+b+c+d = 2'12 Every sum of two numbers is crossing 100 so every no. is greater than or near to 50 so weight can be 51,52,54,55 so minimum is 51
Four Members A, B, C, D are playing a game .A person losing a game should double the amount of others .B,C,O are losing in order after three games .The amount after 3 games are A&B having 40,0 is having 16&C 80. Each questions carry one mark: a) Who started with small amount of money? b) Who started with greatest amount of money? : I c) What amount did B have? Initial values of each player A -5; B -93; C -54; 0 -24.reason 1) 'A' did not lose any match so his money gets double in each game 5 --> 10 --> 20 --> 402) 'B' lost the first game so he gave 5 to A; 46 to C; 32 to 0; A -->10; COO> 108; 0--> 48 and B--> 10 (to arrive at B's value always remember total doesn't change. so subtract A,C,D amounts with total (176))3) similar procedure for the remaining two rounds. A--> 20 --> 40B--> 20 --> 40C--> 40 --> 800--> 96 -->16. So answers are a) A b) B c) 93
For algebraic expressions: Replace all 6 with 9 from 1 to 100, What is the algebraic sum of It. Answer is 330 ,\
Sum of first 100 natural numbers = 5050
Now consider the number series 0·10, 11-20,21-30,31-40,41-50,71-80,81190,91-100.
When 6 is replaced by 9 in all the above series the sum increases by 3 in each one.
So the sum increases by 3*8 = 24
Now consider the series 51-60
Here 56 will become 59 and 60 will become 90.
So the increase in sum is 3 + 30 = 33
Now the last series 61-70 Except for 66 when 6 is replaced by 9 the sum increases by 30 * 8 = 24066ib~comes 99.
So the increase is 33
Total =5050+24+33+240+33 =5380
Why Result is Different for 1) lnt x=5; Int y; y=++x + ++x + ++x 2) Int x=5; Int y=++x + ++x + ++x
In Type 1: Line 3 is treated as an expression, and based on the "Operator Precedence" the increment [ ++ ] operator takes precedence over summation [ + ] operator. Hence in the expression, the increment operation is done first, and then the summation operation is done, thus the output of the expression will be "24 ".In Type 2: Line 2 is treated as an assignment statement, and is hence evaluated from left to right. Thus the value of 1 y , in this case will be 21.
In a certain code, the symbol for 0 (zero) is. * and that for 1 Is $. The numbers greater than 1 are to be written only by using the two symbols given above. The value of the symbol for 1 doubles Itself every time it shifts one place to the left. (For example, 4 is written as $**; and; 3 is writte~ as $$) 1.(6) 260 can be represented as: A) $****$** B) $$*$$$$$ C) $$*$$$$** D) $*****$** 2.(7) 60 / 17 can also be represented as: A) $$$*$*** / $$**$$ B) $$$***** / $$**$$ C) s-ss-s-/$$**$$ D) $$*$*$** / $$**$$ 3.(8) $***$ can be represented as: A) $$$ / $* B) $*$**-$$ C) $*$*$-$$ D) $$$***$ -$$ 4.(9) 30"2 can be represented as: A) ($$*$$ ) $*+ $*$*$$*$ B) ($$*$$ ) $* ~ $$****$ C) ( $$*$$ ) $$ + $*$**** D) ( $$*$$ ) $$ + $*$**$ 5.(10) 11x 17/10 + 2 x 5 + 3 /10 can also be represented as: A) $*$$* B) $*$$$ C) $$$*$ D) s-ss ***KCT*** For a nine digit number, the values $ will take at each position will be 256 1286432168421 This is because, the value doubles itself with every left shift. Now in the problem, if there is a '$', take its corresponding value from the above code and find the sum of all such '$' values. For eg, to represent 260, $*****$** The first $ from left has value 256 and the second $ has value 4. 256+4=260.
If a+b+c=c+d+e=e+f+g=g+h+i=13 all are unique and between 0-9. Find e1 Ans: 4 This problem is similar to solving sudoku... The problem is like this... _+_ +_C_= _C_+_ +_E_= _E_+_ +_G_ =_G_+_ +_ = 13. G + _ + _ = 13 starting from G=1 it can be assumed as 1+3+9, 1+4+8 and 1+5+7 i've put the first possibility and now we get
+ + C =C++ E = E+ +1= 1+3+9= 13. then I've input thenext possibility 1:a,fforE+F+G now we have
_+_ +_C_= _C_+_+ 4= 4+8+1 =1+3+9= 13.
***KCT*" The numbers we have currently used now are 1,3,4,8 and 9. The remaining numbers are 2,5,6,7 and the possibilities for getting 13 are only two. They are 2+7+4 and 5+6+2 Therefore the resultis5+6+2=2+7+ 4-=4+8+1= 1+3+9= 13. So the value of e is 4.
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