TCS NINJA APTITUDE QUESTIONS
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| TCS Ninja Aptitude test |
Q 1. 2/3rd of the balls in a bag is blue, and the rest are pink. if 5/9th of the blue balls and 7/8th of the pink balls are defective, find the total number of balls in the bag given that the number of non-defective balls is 146?
a) 216 b) 649 c) 432✅ d) 578
Solution:- let the total no of balls=x
blue=2x/3
pink=x/3
total no of defective balls = [5/9][2x/3]+[7/8][x/3]
=10x/27 +7x/24
=143x/216
Non - defective balls=x-143x/216=146 (Given)
73x/216 = 146
x= 2(216)
x=432
Q 2. Find no.of ways in which 4 particular person a,b,c,d, and 6 more persons can stand in a queue so that A always stands before B. B always stands before C And C always stands before D.
a)6! b)7!✅ c)1006*6! d)10046!
Solution:- a,b,c,d are grouped ie consider them as one and the remaining as 6. Total 6+1 = 7! Ways
4. There are 10 points on a straight line AB and 8 on another straight line AC none of them being point A. how many triangles can be formed with these points as vertices?
a. 680 b. 720 c. 816 d. 640✅
Solution:- To form a triangle we need 3 points
select 2 points from the 10 points of line AB & 1 from the 8 on AC = (10C2)*(8C1)
select 2 points from the 8 points of line AC & 1 from the 10 on AB=(8C2)*(10C1)
total no. of triangles = (10C2)*(8C1)+ (8C2)*(10C1) = 640
Q 5. From a bag containing 8 green and 5 red balls, three are drawn one after the other. the probability of all three balls beings green if the balls drawn are replaced before the next ball pick and the balls drawn are not replaced, respectively.
a)512/2197,336/2197
b) 512/2197,336/1716✅
c) 336/2197,512/2197
d) 336/1716,512/1716
Solution:- THE PROBABILITIES OF GETTING WITH REPLACEMENT=8/13*8/13*8/13=512/2197
THE PROBABILITIES OF GETTING WITHOUT REPLACEMENT=8/13*7/12*6/11=336/1716
Q 12. A bag contains 8 white balls and 3 blue balls. Another bag contains 7 white and 4 blue balls. What is the probability of getting a blue ball?
a)3/7 b)7/22✅ c)7/25 d)7/15
Solution:- First we have to select a bag and then we will draw a ball.
The probability of selection of both bags is equal =1/2
Now probability of blue ball taken from first bag = ( 1/2) x (3/11) and probability of blue ball taken from second bag = (1/2) x (4/11)
So probability of blue ball = ( 1/2) x (3/11) + (1/2) x (4/11) = 7/22
Q 19. How many 6-digit even numbers can be formed from digits 1 2 3 4 5 6 7 so that the digit should not repeat and the second last digit is even?
a)6480 b)320 c)2160 d)720✅
solution:- Given 6th digit even number, so last digit 2 or 4 or 6-> 3 ways
"5th digit should be even...so there will be 2 ways(rep. not allowed)
so, therefore we get 5*4*3*2*2*3=720 ways
Q 20. There are 5 letters and 5 addressed envelopes. If the letters are put at random in the envelopes, the probability that all the letters may be placed in wrongly addressed envelopes is.
a)119 b)44✅ c)53 d)40
Solution:- If there is one letter and one envelope then no way you can put it wrong(S1).
If there are 2 letters and 2 envelopes then you can put them wrong in 1 way(S2).
If there are 3 letters and 3 envelopes then you can put them wrong in 2 ways(S3).
If there are 4 then you can put them wrong in 9 ways(S4).
If there are 5 then you can put them wrong in 44 ways(S5).
If you observe you can find a pattern.
S3=(S1+S2)*2
S4=(S2+S3)*3
S5=(S3+S4)*4
S6=(S4+S5)*5
In general, Sn=(Sn-2 + Sn-1)*(n-1)
So, if there are 5 letters then S5=(S3+S4)*4=(2+9)*4 = 44
Q 29. In a horse racing competition, there were 18 numbered 1 to 18. The organizers assigned a probability of winning the race to each horse based on the horse's health and
training the probability that horse one would win is 1/7, that 2 would win is 1/8, and that 3 would win is 1/7.Assuming that tie is impossible Find the chance that one of these three will win the race?
a)22/392 b)1/392 c)23/56✅ d)391/392
Solution:- HORSE 1: 1/7 WINNING PROBABILITY
HORSE 2: 1/8 WINNING PROBABILITY
HORSE 3: 1/7 WINNING PROBABILITY
ONE OF THESE WINS THE RACE:
=> 1/7 + 1/8 + 1/7
=> 8/56 +7/56 + 8/56 (TAKING LCM)
=> 23/56
Q 31. How many vehicle registration plate numbers can be formed with digits 1,2,3,4,5(no digits being repeated)if it is given that the registration number can have 1 to 5 digits?
a) 205 b) 100 c) 325✅ d) 105
Solution:- you can have registration plates of 5,4,3,2 or 1 digits
So, it's 5*4*3*2*1 + 5*4*3*2 + 5*4*3 +5*4 + 5
=120 + 120 + 60 + 20 +5 => 325
Q 40. There are 20 people sitting in a circle. In that, there are 18 men and 2 sisters. How many arrangements are possible, in which the two sisters are always separated by a man?
a. 18!*2✅ b. 17! c. 17!*2 d. 12
Solution:- 18!*2 Consider 1 man along with two sisters as one group..so they can be arranged in 17! ways as circular. The one man in between the two sisters can be out of any 18 men..so,17!*18.. and the two sisters can be arranged in 2ways..so 18!*2
Q 41. a number plate can be formed with two alphabets followed by two digits with no repetition. then how many possible combinations can we get?
a) 58500✅ b) 67600 c) 57850 d) 58761
Solution:- no.of alphabets=26 (a-z), no. of digits=10(0-9).
ways of arranging two alphabets without repetition=26*25;
ways of forming two digits without repetition=10*9
no.of combinations of forming the number on the number plate= 26*25*10*9 = 58500
Q 42. The letters in the word "PLACES" are permuted in all possible ways and arranged in alphabetical order. Find the word at the 48 positions.
a)AESPCL b)ALCEPS c)ALSCEP d)AESPLC✅
Solution:- answer = (d) for words AC**** possible ways for arranging * will be 4!=24
now next seq in alphabetical order will be AE**** so...
same way for AE**** possible ways for arranging * will be 4!=24
Thus, the 48th element will be the last element in AE**** which means AE followed by reverse alphabetical order! which is AESPLC
Q 45. In how many ways a team of 11 must be selected a team of 5 men and 11 women such that the team must comprise not more than 3 men.
a) 1565 b) 2256✅ c) 2456 d) 1243
Solution:- A maximum of 3 men can be played which means there can be 0, 1, 2, 3 men in the team.
(5C0×11C11)+(5C1×11C10)+(5C2×11C9)+(5C3×11C8) = 2256
Q 48. Three dice are rolled. What is the probability of getting a sum of 13?
a) 19/216 b) 21/216✅ c) 17/216 d) 23/216
Solution:- Just count the number of ways to get 13. We just need to count possibilities for two dice because the third dice value is fixed. For two dice, the sum can be anywhere from 7 to 12 and that would be 6 + 5 + 4 + 3 + 2 + 1 = 21. So, the probability is 21/216
Q 58. In a staircase, there are 10 steps. A child is attempting to climb the staircase. Each time, she can either make 1 step or 2 steps. In how many different ways can she climb the staircase?
a). 10 b). 21 c). 36 d). None of these.✅
Solution:- she can go by 1 steps-1 way that is choosing 1 two-step in 9 moves:9C1: 9 ways//2 two-steps: choosing 2 two-steps in 8 moves:8C2 = 28 ways// 3 two-steps 7C3 = 35 ways// 4 two-steps// 6C4 = 15 ways// 5 two-steps// 5C5 = 1 way which covers all the 10 stairs.. that means only one way 2 2 2 2 2 = 1 way// Adding all the ways: 1 + 9 + 28 + 35 + 15 + 1 = 89 ways
Q 60. Find the 32nd word in the list where the word MONOS IS permuted in all possible ways and arranged in alphabetical order.
a) OSMON b) OSNOM c) OSMNO d) ONMSO✅
Solution:- Arranging in alphabetical order MNOOS
M_ _ _ _ CAN BE ARRANGED IN 4!/2!=12 WAYS
N_ _ _ _ CAN BE ARRANGED IN 4!/2!=12 WAYS (12+12=24)
O_ _ _ _ CAN BE ARRANGED IN 4! =24 WAYS (12+12+24=48) OUT OF BOUND
OM_ _ _CAN BE ARRANGED IN 3!=6 WAYS(12+12+6=30)(25-30)
ON_ _ _CAN BE ARRANGED IN 3!=6 WAYS(31-36)
ONM _ _CAN BE ARRANGED IN 2!=2 WAYS(31,32)
31ST WORD IS ONMOS
32ND WORD IS ONMSO
Q 61. One card is lost out of 52 cards. two cards are drawn randomly. They are spades. What is the probability that the lost card is also a spade?
a)1/52 b)1/13 c) 1/4 d) 11/50✅
Solution:- (13 - 2) / (52 - 2) = 11 / 50
Q 65. The sum of 5 numbers in AP is 30 and the sum of their squares is 190. Which of the following is the third term?
a)5 b) 6✅ c) 8 d) 9
Solution:- consider the 5 numbers in AP as a-2d,a-d,a,a+d,a+2d;
given,a-2d+a-d+a+a+d+a+2d=30;
5a=30==>a=6 here a is the 3 rd term so.. the third term is 6.
Q 69. how many parallelograms are formed by a set of 4 parallel lines intersecting another set of 7 parallel lines?
a) 125 b)126✅ c) 127 d) 128
Solution:- Let there be 4 horizontal sets of parallel lines and 7 vertical sets of parallel lines. (U can also consider vice-versa) Now, for a parallelogram, u need 2 horizontally parallel and 2 vertically parallel lines ie. we need to choose 2 lines from each set. So the solution will be 7C2*4C2=126
Q 71. a person starts writing all 4 digits numbers. how many times had he written the digit 2?
a) 4200 b) 4700 c) 3700✅ d) 3200
Solution:- 1)when 2 is at unit place =9*10*10*1
2)when 2 is at tenth place=9*10*1*10
3)when 2 is at hundred place=9*1*10*10
4)when 2 is at thousand place=1*10*10*10
so total no. of 2s=900+900+900+1000=3700
Q 75. there are 16 teams divided into 4 groups. Every team from each group will play with each other once. The top 2 teams will go to the next round and so on the top two teams will play the final match. Minimum how many matches will be played in that tournament?
a) 43✅ b) 40 c) 14 d) 50
Solution:- For the first round 4*4C2=24
second round 2*4C2=12
Third round 1*4C2=6
FINAL round =1
TOTAL=43
Or Total no.of ways = (6+6+6+6)+(6+6)+6+1 =36+7 =43
Q 76. According to the question, there is 16 team divided into 4 groups, which means each group has 4 teams every team from each group will play with each other once.
let the first group contains P, Q, R, and S team they can play with themself =4c2 ways =4x3/2x1=6 ways.
similarly, the second group team can play with themself=6ways
parallaly 3rd group, a team can play with themself=6ways
parallaly 4th group 6ways
now, from every group, the top 2 teams are selected. so, a total of 8 teams are selected again they
form two groups each group has 4 teams. they can again arrange.
first group team can play with themself=6ways
2nd group team can play by themself=6ways
From these two groups again selected the top two teams, a total of 4 teams again arrange in 6 different
ways. from this team selected and there 2 teams play the final match eg. 1 way. now, total no.of ways = (6+6+6+6)+(6+6)+6+1=24+12+7=36+7=43
Q 77. Tickets are numbered from 1,2....1100 and one card is drawn randomly what is the probability of having 2 as a digit?
a) 29/110✅ b) 32/110 c) 30/110 d) 22/110
Solution:- For every 100 probability of having 2 as a digit is 19 and 2 is used 20 times for every 100. ( Here 22,122 These digits will be counted as 1 not 2 ) According to the question, there are 10 100's having 19 digits as 2, and 200 to 299 have 100 digits having 2. Hence,((10 *19)+100)/1100=29/110
Q 85. a number plate can be formed with two alphabets followed by two digits with no repetition. then how many possible combinations can we get?
a) 52500 b) 58500✅ c) 56500 d) 56800
Solution:- no.of alphabets=26 (a-z), no.of digits=10(0-9). ways of arranging two alphabets without repetition=26*25; ways of forming two digits without repetition=10*9
no.of combinations of forming the number on the number plate= 26*25*10*9=58500
Q 92. A certain function f satisfies the equation f(x)+2*f(6-x) = x for all real numbers x. The value of f(1) is
a) 2 b) can't determine c) 1 d) 3✅
Solution:- f(1)+2*f(6-1)=1......... (1)
f(5)+2*f(6-5)=5......... (2)
substituting we have (2) in (1) we have: -3f(1)=-9, hence answer f(1)=3
Q 115. How many 6-digit even numbers can be formed from digits 1 2 3 4 5 6 7 so that the digit should not repeat and the second last digit is even?
a)6480 b)320 c)2160 d)720✅
Solution:- Given the 6th digit even number, so the last digit 2 or 4 or 6-> 3 ways "5th digit should be even...so there will be 2 ways(rep. not allowed) so, therefore we get 5*4*3*2*2*3=720 ways
Q 129. A bag contains six sticks of the following lengths 1 cm, 3 cm, 5cm,7 cm, 11 cm, and 13 cm. three sticks are drawn at random from the bag. What is the probability that we can form a triangle with those sticks?
a) 11/20 b) 1 c) 1/4✅ d) 2/5
Solution:- A bag contains six sticks of the following lengths 1 cm, 3 cm, 7 cm, 5cm,11 cm, and 13 cm. three sticks are drawn at random from the bag. the probability that we can form a triangle with those sticks total possibilities=6c3=20
now for making a triangle,, the sum of any 2 sides must be greater than 3rd side
so 3 5 7
3 11 13
5 7 11
7 11 13
5 11 13
5/20=1/4
Q 131. how many number x (x being an integer) with 10<=x<=99 are 18 more than sum of their digits?
a)12 b)9 c)18 d)10✅
Solution:- Let the no. be 10y+z. then, 10y+z= y+z+18 => Y=2 So the no. are 20, 21, 22, 23, 24, 25, 26, 27, 28, 29. Total nos. 10
Q 133. In how many ways can the digit of the number 2233558888 be arranged so that the odd digits are placed in even positions?
a) 900 b) 450✅ c) 225 d) 360
Solution:- odd place combination = 5!/2!*2!=30 even place combination= 6!/2!*4!=15 to total=30*15=450
Q 134. Find the probability that a leap year chosen at random will contain 53 Sundays.
a) 2/7✅ b) 3/7 c) 1/49 d) 1/7
Solution:- In leap year, we have 366 days. 366/7=52Weeks+2days we have already 52 sundays,mondays,etc.
then we have 2 days may be {sunday,monday} , {monday,tuesday} , {tuesday,wednesday},...{saturday,sunday}
we have 2 possible chances out of 7. hence ans:2/7.
Q 140. The figure shows an equilateral triangle on the side of length 5 which is divided into several unit triangles. A valid path is a path from the triangle in the top row to the middle triangle in the bottom row such that the adjacent triangles in our path share a common edge and the path never travels up (from a lower row to a higher row) or revisits a triangle. An example is given below. How many such valid paths are there?
a) 120 b) 16 c) 23 d) 24✅
Solution:- Given side length=5 .....so (n-1)!=(5-1)!=4! ans is 24
Q 153. In how many possible ways can write 3240 as a product of 3 positive integers a,b, and c
a) 450 b) 420✅ c) 350 d) 320
Solution:- 3240=2*2*2*5*3*3*3*3 so, no. of ways=8!/(3!*4!)=420.
Q 157. A manufacturer of chocolates makes 6 different flavors of chocolates. The chocolates are sold in boxes of 10. How many different boxes of chocolates can be made?
(NOTE: A box is considered different from another only if, regardless of the order, the box contains a different number of chocolates of at least one type)
a) 3003✅ b) 10^6 c) 3000 d) 6^10
Solution:- If n similar articles are to be distributed to r persons, x1+x2+x3......+xr=n each person is eligible to take any number of articles then the total ways are n+r-1Cr-1
In this case x 1+x2+x3......x6=10
in such a case the formula for non-negative integral solutions is n+r-1Cr-1
Here n =10 and r=6. So total ways are 10+6-1C6-1 =15C5 =3003
Q 158. 4 men throw a die each simultaneously. Find the probability that at least 2 people get the same number.
a) 5/18 b)13/18✅ c) 1/36 d) ½
Solution:- 13/18
As 1-(probability of not getting same no.)
= 1 - ( (6*5*4*3)/(6*6*6*6))
= 13/18
Q 161. The letters in the word ROADIE are permuted in all possible ways and arranged in alphabetic order. Find the word in the 44th rank.
a) AERIOD b) AERDOI✅ c) AERODI d) AEODRI
Solution- A----- => 5!=120
AD---- => 4!=24
AED--- => 3!=6
AEI--- => 3!=6
AEO--- => 3!=6
24+6+6+6=42
AERDIO => 43th
AERDOI => 44th
Q 171. two dice are thrown. find the probability of getting a multiple of 3 or 4 as the sum.
a) 5/9✅ b) 4/9 c) 2/9 d) 1/9
Solution:- 20/36=5/9
(1,2),(1,3),(1,5),(2,1)(2,2)(2,4)(2,6)(3,1)(3,3)(3,5)(3,6)(4,2)(4,4)(4,5)(5,1)(5,3)(5,4)(6,2)(6,3)(6,6) as its some is multiple of 3 or 4
Q 176. 0>a>b>c>d. Which is largest
a) (b+d)/(a+c)✅ b) (a+d)/(b+c) c) (b+c)/(a+d) d) (c+d)/(a+d)
Solution- given statement is 0>a>b>c>d
that means all the values of a,b,c,d are less than ZERO
so lets consider a=-1 b=-2, c=-3 and d=-4 so that 0>a>b>c>d will satisfy
by solving the options we get the values as follows
a.(b+d)/(a+c)= 1.5
b. 1
c.1
d.1.4
among all of these options, a is the highest so the ans is option a.
Q 180. How many 5’s will be there in the number 121122123...till 356?
a) 51✅ b) 54 c) 50 d) 49
Solution:- 121122123... till 356 121,122,123,124... till 356
121 to 200 => 125,135,145, from 150 to 159 => 11 5's , 165,175,185,195
[from 150 to 159, each no. has 1 no. of 5 except 155, 155 has 2 no. of 5]
total=18 201 to 300 => 205,215,225,235,245, 250 to 259 => 11 5's , 265,275,285,295
=> total=20 301 to 356 => 305,315,325,335,345,350,351,352,353,354,355(2 5's),356
=> total=13 total 5's = 18+20+13 = 51
Q 213. 10 people are there, and they are shaking hands together, how many handshakes are possible, if they are in no pair of cyclic sequences.
a) 45✅ b) 9 c) 12 d) 10
Solution:- 10c2= n!/(n-r)!.r! =10!/8!.2! =10*9/2 =45
(Or)
1st person shakes hand with 9 others=9 handshakes then 2nd person shakes hand with 8 others(ie excluding 1st person)=8 handshakes .... finally 9th person=1 handshake and 10th person=0 handshake so total=9+8+7+6+5+4+3+2+1=45 handshakes
Q 219. When all possible six-letter arrangements of the letters of the word "MASTER" are sorted in alphabetical order, what will be the 49th word?
a) AREMST✅ b) ARMEST c) AMERST d) ARMSET
Solution:- ARRANGING THE LETTERS IN MASTER ALPHABETICALLY WE GET...
AEMRST
NOW A E _ _ _ _ can be filled in 4! ways
A M _ _ _ _ can be filled in 4! ways
so words upto 4!+4! =48 words are of AE series and am series
So the 49th word can be A R E M S T
Q 223. If ABERSU is sorted in alphabetical order, if 24 sortings are req for ABUSRE, 25 - AEBRSU, 49- ARBESU, den how many for AEUSRB.
a) 45 b) 48✅ c) 47 d) 46
Solution:- AEUSRB
ALPHA OREDER ABERSU
A B _ _ _ _ =4! =24
A E B _ _ _ =3!=6
A E R _ _ _ =3!=6
A E S _ _ _ =3! =6
A E U B _ _ =2! =2
A E U R _ _=2!=2
A E U S B _=1
A E U S R B=1
24+6+6+6+2+2+1+1=48
Q 259. 3 white chips, 7 blue chips, 16 green chips, and 2 chips drawn from the box in succession what is the probability that one is blue and the other is white?
a) 7/50 b) 8/30 c) 7/25 d) 21/25*13✅
Solution:- no.of ways that 2 chips selected from a total of 26 are 26C2=26*25/2=13*25
no.of the ways that 1 white and 1 blue is selected is 3C1*7C1=21
total probability=21/13*25
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