# Cgi Group Aptitude Placement Papers - Cgi Group Aptitude Interview Questions and Answers updated on 04.Dec.2023

Number of ways of choosing 2 black pens from 5 black pens in 5C<sub>2</sub>ways.

Number of ways of choosing 2 white pens from 3 white pens in 3C<sub>2</sub>ways.

Number of ways of choosing 2 red pens from 4 red pens in 4C<sub>2</sub>ways.

By the Counting Principle, 2 black pens, 2 white pens, and 2 red pens can be chosen in 10 x 3 x 6 =180 ways.

Total CP = Rs. (500 X 10 + 2000) = Rs. 7000

SP = Rs. (5 X 750 + 5 X 550) = Rs. 6500

Loss = CP - SP = 7000 - 6500 = 500

Loss Percent = 500/7000 X 100 = 50/7.

4 novels can be selected out of 5 in 5C4 ways.

2 biographies can be selected out of 4 in 4C2 ways.

Number of ways of arranging novels and biographies = 5C4*4C2= 30

After selecting any 6 books (4 novels and 2 biographies) in one of the 30 ways, they can be arranged on the shelf in 6! = 720 ways.

By the Counting Principle, the total number of arrangements = 30 x 720 = 21600.

Let 4 girls be one unit and now there are 6 units in all.

They can be arranged in 6! ways.

In each of these arrangements 4 girls can be arranged in 4! ways.

Total number of arrangements in which girls are always together = 6! x 4!= 720 x 24 = 17280.

Given, Cost price (C.P) of 2 lemons = Rs. 1

=>C.P of 1 lemon = Rs. 1/2 = Rs. 0.50

Given, Selling price (S.P) of 5 lemons = Rs. 3

=>S.P of 1 lemon = Rs. 3/5 = 0.60

Gain = S.P of1 lemon -C.P of 1 lemon

= 0.60 - 0.50

= 0.10

=>Gain = Rs. 0.10

Gain % = (Gain / C.P of 1 lemon) * 100%

= (0.10 / 0.50)* 100%

= 20%

Thus, Gain % = 20%.

Cost of sugar = Rs 5.58 / kg

His lost percent =7 %

= 100 - 7

= 93.

Gain percent

= 100+ 7

= 107.

So, Gain = CP * gain / 100 - loss

= 5.58 * 107 / 100 - 7

= 5.58 * 107 / 93

= 597.06 / 93

= 6.42 .per kg

6.42 .per kg is to be sold to gain 7 % .

Given are the two AP'S:

15,12,9.... in which a=15, d=-3.............(1)

-15,-13,-11..... in which a'=-15 ,d'=2.....(2)

now using the nth term's formula,we get

a+(n-1)d = a'+(n-1)d'

substituting the value obtained in eq. 1 and 2,

15+(n-1) x (-3) = -15+(n-1) x 2

=> 15 - 3n + 3 = -15 + 2n - 2

=> 12 - 3n = -17 + 2n

=> 12+17 = 2n+3n

=> 29=5n

=> n= 29/5.

The first letter from the right can be chosen in 26 ways because there are 26 alphabets.

Having chosen this, the second letter can be chosen in 26 ways

The first two letters can chosen in 26 x 26 = 676 ways

Having chosen the first two letters, the third letter can be chosen in 26 ways.

All the three letters can be chosen in 676 x 26 =17576 ways.

It implies that the maximum possible number of five letter palindromes is 17576 because the fourth letter is the same as the second letter and the fifth letter is the same as the first letter.

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4) = (7C3*4C2

= 210.

Number of groups, each having 3 consonants and 2 vowels = 210.

Each group contains 5 letters.

Number of ways of arranging 5 letters among themselves = 5! = 120

Required number of ways = (210 x 120) = 25200.

A team of 6 members has to be selected from the 10 players.

This can be done in 10C6 or 210 ways.

Now, the captain can be selected from these 6 players in 6 ways.

Therefore, total ways the selection can be made is 210×6= 1260.

Given, Cost price = Rs. 1200

Profit = 30%

Selling price = {[100 + gain%] / 100} * Cost price

= [130 / 100] * 1200

= 130 * 12

= 1560

Therefore, Selling price = 1560.

There are 7 digits 1, 2, 0, 2, 4, 2, 4 in which 2 occurs 3 times, 4 occurs 2 times.

Number of 7 digit numbers = 7!3!×2! = 420

But out of these 420 numbers, there are some numbers which begin with '0' and they are not 7-digit numbers.

The number of such numbers beginning with '0'.

=6!3!×2! = 60

Hence the required number of 7 digits numbers = 420 - 60 = 360.

Let the Cost Price(C.P) be Rs. 100

Given,tradesman's prices are 20% above C.P

=> Marked Price (M.P) = 20% more than C.P

=> M.P = Rs. 120

Given,profit = 8%

=> Selling price (S.P) = 8% more than C.P

=> S.P = Rs. 108

Rate of Discount = {(M.P - S.P) / M.P} * 100%

= {(120 - 108) / 120} * 100%

= (12 / 120) * 100%

= 10 %

Thus,Rate of Discount = 10%.

There are total 9 places out of which 4 are even and rest 5 places are odd.

4 women can be arranged at 4 even places in 4! ways.

and 5 men can be placed in remaining 5 places in 5! ways.

Hence, the required number of permutations  = 4! x 5! = 24 x 120 = 2880.

1 million distinct 3 digit initials are needed.

Let the number of required alphabets in the language be ‘n’.

Therefore, using ‘n’ alphabets we can form n * n * n = n3 distinct 3 digit initials.

Note distinct initials is different from initials where the digits are different.

For instance, AAA and BBB are acceptable combinations in the case of distinct initials while they are not permitted when the digits of the initials need to be different.

This n3 different initials = 1 million

i.e. n3=106  (1 million = 106)

=> n = 102 = 100

Hence, the language needs to have a minimum of 100 alphabets to achieve the objective.

'LOGARITHMS' contains 10 different letters.

Required number of words = Number of arrangements of 10 letters, taking 4 at a time.

= 10P4

= 5040.

Here the order of choosing the elements doesn’t matter and this is a problem in combinations.

We have to find the number of ways of choosing 4 elements of this set which has 11 elements.

This can be done in 11C4ways = 330 ways.

S.P = 75 % of CP

=> 75 x CP /100= 1500

=> CP = 2000

20 % of CP = (20/100) x 2000 = 400

SP = 2000 + 400 = 2400.

Let the cost price be Rs. k

Now, as per the question,

600 - k = k - 400

=> 2k = 1000

=> k = 500

Again, selling price of the article for making 25 % profit

= (500 x 125) /200

= 125 * 5

= Rs. 625.

Given, Cost price (C.P) of 8 toffees = Re. 1

Gain = 60%

So, Selling price, (S.P) = {[100 + Gain%] / 100} * C.P

= Rs. (160 / 100) x 1

= Rs. 8 / 5

For Rs. 8 / 5, toffees sold = 8

For Re. 1, toffees sold = (8 x 5) / 8 = 5

So, to gain 60%, toffees must be sold at 5 for Re. 1.

There are 8 students and the maximum capacity of the cars together is 9.

We may divide the 8 students as follows

Case I: 5 students in the first car and 3 in the second

Case II: 4 students in the first car and 4 in the second

Hence, in Case I: 8 students are divided into groups of 5 and 3 in8C3 ways.

Similarly, in Case II: 8 students are divided into two groups of 4 and 4 in 8C4ways.

Therefore,

the total number of ways in which 8 students can travel is:

8C3+8C4=56+70= 126.

Let x be the total profit

A’s share in profit = Rs. 3x/5

Remaining Profit = x - (3x/5) = 2x/5

So, B’s share in profit = Rs. x/5

C’s share in profit = Rs. x/5

Given,(3x/5 – x/5) = 400

=> 2x/5 = 400

=> x = (400×5)/ 2

=> x = Rs. 1000

Therefore, Total Profit = x =Rs. 1000.

The 5 letter word can be rearranged in 5!=120 Ways without any of the letters repeating.

Then the 25th word will start will CA _ _ _.

The remaining 3 letters can be rearranged in 3!=6 Ways. i.e. 6 words exist that start with CA.

The next word starts with CH and then A, i.e., CHA _ _.

The first of the words will be CHAMS. The next word will be CHASM.

Therefore, the rank of CHASM will be 24+6+2= 32.

fix one person and the brothers B1 P B2 = 2 ways to do so.

other 17 people= 17!

Each person out of 18 can be fixed between the two=18, thus, 2 x 17! x 18=2 x 18!.

From 5 consonants, 3 consonants can be selected in 5C3 ways.

From 4 vowels, 2 vowels can be selected in 4C2ways.

Now with every selection, number of ways of arranging 5 letters is 5P5ways.

Total number of words = 5C3*4C2*5P5= 10x 6 x 5 x 4 x 3 x 2 x 1= 7200.

We are to choose 11 players including 1 wicket keeper and 4 bowlers  or, 1 wicket keeper and 5 bowlers.

Number of ways of selecting 1 wicket keeper, 4 bowlers and 6 other players in 2C1*5C4*9C6= 840

Number of ways of selecting 1 wicket keeper, 5 bowlers and 5 other players in 2C1*5C5*9C5=252

Total number of ways of selecting the team = 840 + 252 = 1092.

The bus fromA to B can be selected in 3 ways.

The bus from B to C can be selected in 4 ways.

The bus from C toD can be selected in 2 ways.

The bus fromD to E can be selected in 3 ways.

So, by the General Counting Principle, one can travel fromA to E in 3 x 4 x 2 x 3 ways = 72.

In the given problem, let C.P denote the cost price,

then (100 +10)% of CP -(100-6) % of CP = Rs. 96

=>(110)% of CP - (94) % of CP = Rs.96

=>16 % of CP = 96

=> 16 / 100 = 96

=> CP = 96 x 100 / 16

=> 9600 / 16

= 600 Rs

Rs 600is cost price.

We have to arrange 6 books.

The number of permutations of n objects is n! = n. (n – 1) . (n – 2) ... 2.1

Here n = 6 and therefore, number of permutations is 6.5.4.3.2.1 = 720.

In the given problem,

Let C.P denote the cost price,

Then (100+12)% of CP = (100-4) % of Cost

=> Rs. 128 (112)% of CP = (96) % of Cost = Rs.128

16 % of CP = 128

=> CP = 128 x 100 / 16

= 12800 / 16

= 800.

There are seven positions to be filled.

The first position can be filled using any of the 7 letters contained in PROBLEM.

The second position can be filled by the remaining 6 letters as the letters should not repeat.

The third position can be filled by the remaining 5 letters only and so on.

Therefore, the total number of ways of rearranging the 7 letter word = 7*6*5*4*3*2*1 = 7! ways.

Let last digit is 2

when second last digit is 4 remaining 4 digits can be filled in 120 ways,

similarly second last digit is 6 remained 4 digits can be filled in 120 ways.

so for last digit = 2, total numbers=240

Similarly for 4 and 6

When last digit = 4, total no. of ways =240

and last digit = 6, total no. of ways =240

so total of 720 even numbers are possible.

Since each ring consists of six different letters,

the total number of attempts possible with the three rings is = 6 x 6 x 6 = 2@Of these attempts,

one of them is a successful attempt.

Maximum number of unsuccessful attempts = 216 - 1 = 215.

We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).

Required number of ways=(3C1*6C2)+(3C2*6C1)+3C3 = (45 + 18 + 1) =64.

The first letter is E and the last one is R.

Therefore, one has to find two more letters from the remaining 11 letters.

Of the 11 letters, there are 2 Ns, 2Es and 2As and one each of the remaining 5 letters.

The second and third positions can either have two different letters or have both the letters to be the same.

Case 1: When the two letters are different. One has to choose two different letters from the 8 available different choices. This can be done in 8 * 7 = 56 ways.

Case 2: When the two letters are same. There are 3 options - the three can be either Ns or Es or As. Therefore, 3 ways.

Total number of possibilities = 56 + 3 = 59.

When 4 dice are rolled simultaneously, there will be a total of 6 x 6 x 6 x 6 = 1296 outcomes.

The number of outcomes in which none of the 4 dice show 3 will be 5 x 5 x 5 x 5 = 625 outcomes.

Therefore, the number of outcomes in which at least one die will show 3 = 1296 – 625 = 671.

Let the price be Rs 100

After discount of 20% we get = 100-20 = 80

shirt costs Rs. 64

Let x be the cost price of the shirt,

x * 80/100 = 64

x = (64 x 100) / 80 = 80

Original price of shirt in Rs. 80.

Let the Cost Price be Rs. 100,

Sincethere is a profit of 20% on the Cost Price,

then Selling Price = C.P + 20% of C.P

= 100 + 20

= Rs. 120

=>Selling Price =Rs. 120

Gain = SP - CP

= 120 - 100

= 20

Gain % on S.P = (Gain / S.P) * 100%

= (20/120) x 100

= 50/3%.

When at least 2 women are included.

The committee may consist of 3 women, 2 men : It can be done in  4C3*6C2ways

or, 4 women, 1 man : It can be done in  4C4*6C1ways

or, 2 women, 3 men : It can be done in 4C2*6C3ways.

Total number of ways of forming the committees

= 4C2*6C3+4C3*6C2+4C4*6C1

= 6 x 20 + 4 x 15 + 1x 6

= 120 + 60 + 6 =186.

We first count the number of committee in which

(i). Mr. Y is a member

(ii). the ones in which he is not

Case (i): As Mr. Y agrees to be in committee only where Mrs. Z is a member.

Now we are left with (6-1) men and (4-2) ladies (Mrs. X is not willing to join).

We can choose 1 more in5+2C1=7 ways.

Case (ii): If Mr. Y is not a member then we left with (6+4-1) people.

we can select 3 from 9 in 9C3=84 ways.

Thus, total number of ways is 7+84= 91 ways.

First letter can be posted in 4 letter boxes in 4 ways.

Similarly second letter can be posted in 4 letter boxes in 4 ways and so on.

Hence all the 5 letters can be posted in = 4 x 4 x 4 x 4 x 4 = 1024.

Out of 26 alphabets two distinct letters can be chosen in 26P2 ways.

Coming to numbers part, there are 10 ways.

(any number from 0 to 9 can be chosen) to choose the first digit and similarly another 10ways to choose the second digit.

Hence there are totally 10X10 = 100 ways.

Combined with letters there are 6P2 X 100 ways = 65000 ways to choose vehicle numbers.

Choose 1 person for the single room & from the remaining choose 2 for the double room & from the remaining choose 4 people for the four person room,

Then, 7C1 x 6C2 x 4C4

= 7 x 15 x 1 = 105.

The number of arrangements of 4 different digits taken 4 at a time is given by 4P4 = 4! = 24.

All the four digits will occur equal number of times at each of the position,namely ones,tens,hundreds,thousands.

Thus,each digit will occur 24/4 = 6 times in each of the position.

The sum of digits in one's position will be 6 x (1+3+5+7) = 96.

Similar is the case in ten's,hundred's and thousand's places.

Therefore,the sum will be 96 + 96 x 10 + 96 x 100 + 96 x 100 = 106656.

Let, cost price of the article = Rs.100x

Then selling price = 5% loss of Cost price

= C.P - loss

= 100x - (5/100)*100x

= 100x - 5x

= 95x

=> selling price = 95x

But if he sold the product for Rs.30 more, ==> his profit is 1.25%.

In this case his selling price = 100x + (1.25/100) * 100x

= 100x + 1.25x

= 101.25x

=> selling price = 101.25x

Difference in two selling prices = Rs.30

=> 101.25x - 95x = Rs.30

=> 6.25x = Rs.30

=> x = Rs.30 / 6.25

=> x = Rs.4.8 --->Substitutingin cost price, we get

Cost Price of the article = Rs. 100x = Rs. 100 * 4.8 = Rs. 480

Therefore, Cost Price of the article = Rs. 480.

There are total 9 letters in the word COMMITTEE in which there are 2M's, 2T's, 2E's.

The number of ways in which 9 letters can be arranged = 9!2!×2!×2! = 45360

There are 4 vowels O,I,E,E in the given word. If the four vowels always come together, taking them as one letter we have to arrange 5 + 1 = 6 letters which include 2Ms and 2Ts and this be done in 6!2!×2! = 180 ways.

In which of 180 ways, the 4 vowels O,I,E,E remaining together can be arranged in 4!2! = 12 ways.

The number of ways in which the four vowels always come together = 180 x 12 = 2160.

Hence, the required number of ways in which the four vowels do not come together = 45360 - 2160 = 43200.

In a 3 digit number one’s place can be filled in 5 different ways with (0,2,4,6,8)

10’s place can be filled in 10 different ways

100’s place can be filled in 9 different ways

There fore total number of ways = 5X10X9 = 450.