5 avg. rating (100% score) - 1 votes
Answer :
Given, Cost price = Rs. 1200
Profit = 30%
Selling price = {[100 + gain%] / 100} * Cost price
= [130 / 100] * 1200
= 130 * 12
= 1560
Therefore, Selling price = 1560.
Answer :
S.P = 75 % of CP
=> 75 x CP /100= 1500
=> CP = 2000
20 % of CP = (20/100) x 2000 = 400
SP = 2000 + 400 = 2400.
Answer :
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.
Answer :
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.
Answer :
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.
Answer :
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%.
Answer :
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.
Answer :
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%.
Answer :
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 % .
Question 10. Toffee Are Bought At A Rate Of 8 For One Rupee. To Gain 60% They Must Be Sold At?
Answer :
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.
Answer :
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%.
Answer :
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.
Answer :
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.
Answer :
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.
Answer :
Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4) = (7C_{3}*4C_{2})
= 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.
Answer :
When at least 2 women are included.
The committee may consist of 3 women, 2 men : It can be done in 4C_{3}*6C_{2}ways
or, 4 women, 1 man : It can be done in 4C_{4}*6C_{1}ways
or, 2 women, 3 men : It can be done in 4C_{2}*6C_{3}ways.
Total number of ways of forming the committees
= 4C_{2}*6C_{3}+4C_{3}*6C_{2}+4C_{4}*6C_{1}
= 6 x 20 + 4 x 15 + 1x 6
= 120 + 60 + 6 =186.
Answer :
If the word started with the letter A then the remaining 5 positions can be filled in 5! Ways.
If it started with c then the remaining 5 positions can be filled in 5! Ways.Similarly if it started with H,I,N the remaining 5 positions can be filled in 5! Ways.
If it started with S then the remaining position can be filled with A,C,H,I,N in alphabetical order as on dictionary.
The required word SACHIN can be obtained after the 5X5!=600 Ways i.e. SACHIN is the 601th letter.
Answer :
A team of 6 members has to be selected from the 10 players.
This can be done in 10C_{6} 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.
Answer :
'LOGARITHMS' contains 10 different letters.
Required number of words = Number of arrangements of 10 letters, taking 4 at a time.
= 10P_{4}
= 5040.
Answer :
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 2C_{1}*5C_{4}*9C_{6}= 840
Number of ways of selecting 1 wicket keeper, 5 bowlers and 5 other players in 2C_{1}*5C_{5}*9C_{5}=252
Total number of ways of selecting the team = 840 + 252 = 1092.
Answer :
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.
Answer :
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.
Answer :
The 5 letter word can be rearranged in 5!=120 Ways without any of the letters repeating.
The first 24 of these words will start with A.
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.
Answer :
There are 7 letters in the word Bengali of these 3 are vowels and 4 consonants.
There are 4 odd places and 3 even places.
3 vowels can occupy 4 odd places in 4P_{3} ways and 4 constants can be arranged in 4P_{4}ways.
Number of words =4P_{3}x4P_{4}= 24 x 24 = 576.
Question 25. How Many 7 Digit Numbers Can Be Formed Using The Digits 1, 2, 0, 2, 4, 2, 4?
Answer :
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.
Answer :
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.
Answer :
From 5 consonants, 3 consonants can be selected in 5C_{3} ways.
From 4 vowels, 2 vowels can be selected in 4C_{2}ways.
Now with every selection, number of ways of arranging 5 letters is 5P5ways.
Total number of words = 5C_{3}*4C_{2}*5P_{5}= 10x 6 x 5 x 4 x 3 x 2 x 1= 7200.
Answer :
Since each ring consists of six different letters,
the total number of attempts possible with the three rings is = 6 x 6 x 6 = 216. Of these attempts,
one of them is a successful attempt.
Maximum number of unsuccessful attempts = 216 - 1 = 215.
Answer :
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.
Answer :
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 in8C_{3} 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:
8C_{3}+8C_{4}=56+70= 126.
Answer :
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.
Answer :
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.
Question 33. Find The Number Of Subsets Of The Set {1,2,3,4,5,6,7,8,9,10,11} Having 4 Elements?
Answer :
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 11C_{4}ways = 330 ways.
Answer :
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.
Answer :
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, 7C_{1} x 6C_{2} x 4C_{4}
= 7 x 15 x 1 = 105.
Answer :
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.
Answer :
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.
Answer :
We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).
Required number of ways=(3C_{1}*6C_{2})+(3C_{2}*6C_{1})+3C_{3} = (45 + 18 + 1) =64.
Answer :
4 novels can be selected out of 5 in 5C_{4} ways.
2 biographies can be selected out of 4 in 4C_{2} ways.
Number of ways of arranging novels and biographies = 5C_{4}*4C_{2}= 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.
Answer :
The number of arrangements of 4 different digits taken 4 at a time is given by 4P_{4} = 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.
Answer :
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.
Answer :
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+2C_{1}=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 9C_{3}=84 ways.
Thus, total number of ways is 7+84= 91 ways.
Answer :
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.
Answer :
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.
Answer :
Out of 26 alphabets two distinct letters can be chosen in 26P_{2} 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 6P_{2} X 100 ways = 65000 ways to choose vehicle numbers.
Answer :
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.
Answer :
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.
Answer :
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!.
Question 49. In How Many Ways Can 5 Letters Be Posted In 4 Letter Boxes?
Answer :
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.
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