RikoteMaster/aqua-rat-mcqa huggingface dataset:
Question: A question paper consists of two sections A and B having respectively 3 and 4 questions.Four questions are to be solved to qualify in that paper.It is compulsory to solve at least one question from section A and 2 questions from section B .In how many ways can a candidate select the questions to qualify in that paper?
A.30
B.40
C.42
D.44
Answer: A.30
Explanation: here two cases are possible.
case 1) 1 question from section A and 3 question from section B are solved = 3C1.4C3
case 2) 2 question from section A and 2 question from section B are solved = 3C2.4C2
thus total number of ways = 3C1.4C3 + 3C2.4C2 = 30
ANS:A
Question: Q-1) Of the 84 parents who attended a meeting at a
school, 35 volunteered to supervise children during
the school picnic and 11 volunteered both to supervise
children during the picnic and to bring refreshments to
the picnic. If the number of parents who volunteered
to bring refreshments was 2 times the number of
parents who neither volunteered to supervise children
during the picnic nor volunteered to bring
refreshments, how many of the parents volunteered to
bring refreshments?
A.25
B.40
C.38
D.42
Answer: B.40
Explanation: i used the following formula: total = group1 + group2 + neither - both
using the information from the question stem, we have:
84 total people
35 who supervise (group1)
x who neither supervise nor bring refreshments
2 x people who bring refreshments (group2)
11 who supervise AND bring refreshments (both)
therefore...
84 = 35 + 2x + x - 11
solving, we get x = 20
since we want the value of those who bring refreshments AND supervise:
2(20) = 40
answer is B
Question: A fruit seller had some mangoes. He sells 30% mangoes and still has 280 mangoes. Find the number of mangoes he had.
A.588 mangoes
B.400 mangoes
C.672 mangoes
D.700 mangoes
Answer: B.400 mangoes
Explanation: Explanation:
Suppose originally he had X mangoes.
Then, (100 – 30)% of X = 280
70/100 * X = 280
X = (280 * 100) / 70 = 400
ANSWER: B
Question: Rahul can finish one-fifth of hhis homework in one hour.Neha can finish three-seventh of her homework in one hout thirty minutes and riya can finish three fourth of her homework in three hours thirty minutes.If all of them start their homework at 12:00pm and can go to play as soon as they all finish their homework,when can they start to play,if they take a break at 3:30pm for thirty minutes?
A.5.00pm
B.5:30pm
C.4.40pm
D.6.30pm
Answer: B.5:30pm
Explanation: Let the work be 1
Then Rahul do 1/5 work in 60min
So Rahul do 1 work in 300 min
Then Neha do 3/7 work in 90min
So Neha do 1 work in 210min
Then Riya do 3/4 work in 210min
So Riya do 1 work in 280 min
By 300min i.e 5hr all will complete their work so
12-1,1-2,2-3,3-3:30, 4-4:30,4:30-5:30
By 5:30 they all go to play
ANSWER:B
Question: Five kilograms of oranges contained 98% of water. If the next day the concentration of water decreased by 2%, what was the new weight of the oranges, in kilograms?
A.4.9
B.4.8
C.2.5
D.2
Answer: C.2.5
Explanation: If C is the answer, then the question means that the concentration of water the newt day became 96%.
Out of 5 kilograms 98%, or 4.9 kilograms was water and 0.1 kilograms was non-water.
The next day, after some water evaporated, oranges became 96% water and 4% of non-water, so the next day 0.1 kilograms of non-water composed 4% of oranges, which means that the new weight of the oranges was x*0.04=0.1 --> x=2.5 kilograms.
Answer: C.
Question: Aravind had $ 2100 left after spending 30 % of the money he took for shopping. How much money did he
take along with him?
A.$ 3500
B.$ 3600
C.$ 3800
D.$ 3000
Answer: D.$ 3000
Explanation: Let the money he took for shopping be m.
Money he spent = 30 % of m
= 30/100 × m
= 3/10 m
Money left with him = m – 3/10 m = (10m – 3m)/10 = 7m/10
But money left with him = $ 2100
Therefore 7m/10 = $ 2100
m = $ 2100× 10/7
m = $ 21000/7
m = $ 3000
Therefore, the money he took for shopping is $ 3000.
D)
Question: A train 140 m long is running with a speed of 50 km/hr. In what time will it pass a man who is running at 4 km/hr in the direction opposite to that in which the train is going?
A.8.1
B.7.0
C.9.33
D.8
Answer: C.9.33
Explanation: Speed of train relative to man = 50 +4 = 54 km/hr.
= 54 * 5/18 = 15 m/sec.
Time taken to pass the men = 140 * 1/15 = 9.33 sec.
Answer: Option C
Question: A contractor agreeing to finish a work in 150 days, employed 75 men each working 8 hours daily. After 90 days, only 2/7 of the work was completed. Increasing the number of men by­­ ________ each working now for 10 hours daily, the work can be completed in time.
A.140 workers
B.150 workers
C.160 workers
D.155 workers
Answer: B.150 workers
Explanation: total man work in 90 days is = 90*75*8==>54000 man work
remaining work =1-2/7=5/7
now, 2/7 work have 54000 man work
5/7 work will have (54000*7/2)*5/7 ==> 135000 man work
days left =60 and 10 hrs per day
so, no. of workers = 135000/(60*10)==>225
this is total workers.
so, workers needed ==>225-75= 150 workers
ANSWER:B
Question: Steve Purchased a house for $15000 and he like to resell it with 20% profit. And what price he want to sell that?
A.16000
B.17500
C.18000
D.19500
Answer: C.18000
Explanation: Explanation:
C.P= 15000
Profit% = 20% = (Profit*C.P) / 100
Profit= (Profit% * C.P) / 100 = (20*15000) / 100 = 3000
S.P = C.P + Profit = 15000+3000 = 18000 . Answer: C
Question: An article was sold after a discount of 20% and therewas a gain of 20%. If the profit madeon it was Rs.6 less thanthe discount offered on it, find its SP?
A.Rs. 60
B.Rs. 69
C.Rs. 72
D.Rs. 89
Answer: C.Rs. 72
Explanation: Let CP = Rs. 100x
SP = Rs. 120x
MP = 120x/80 * 100 = Rs. 150x
D = Rs. 150x - Rs. 120x = Rs. 30x
D - P = 30x - 20x = Rs. 6, 10x = Rs. 6
120x = 120/10 * 6 = Rs. 72
C
Question: A bag contains 6 red, 5 blue and 2 green balls. If 2 ballsare picked at random, what is the probability that both are red?
A.1/15
B.2/13
C.5/26
D.6/27
Answer: C.5/26
Explanation: P(Both are red),
=6C213C2=6C213C2
=5/26
C
Question: In how many ways can the letters of the word JUPITER be arranged in a row so that the vowels appear in alphabetic order?
A.736
B.768
C.792
D.840
Answer: D.840
Explanation: three letters can be arranged in 3! ways.
only one combination EIU is required.
7 letters can be arranged in 7! ways.
thus 7!/ 3! * 1 = 840.
D
Question: Eight identical machines can produce 360 aluminum cans per hour. If all of the machines work at the same constant rate, how many cans could 5 such machines produce in 2 hours?
A.450
B.750
C.1,800
D.5,900
Answer: A.450
Explanation: 8 machines/ 360 cans = 5 machines/ x cans
8x= 1800
x=225
(225)(2 hours)= 450 cans. The answer is A.
Question: If s and t are positive integers such that s/t = 64.28, which of the following could be the remainder when s is divided by t?
A.12
B.15
C.18
D.21
Answer: D.21
Explanation: 0.28 = 28/100 = 7/25
The remainder must be divisible by 7.
The answer is D.
Question: At what rate percent on simple interest will Rs.750 amount to Rs.900 in 5 years?
A.7%
B.9%
C.4%
D.2%
Answer: C.4%
Explanation: 150 = (750*5*R)/100
R = 4%
Answer: C
Question: If Bill can buy 3 pairs of jeans and 2 shirts for $69 or 2 pairs of jeans and 3 shirts for $81, how much does one shirt cost?
A.$10
B.$12
C.$13.20
D.$21
Answer: D.$21
Explanation: 3J + 2S = 69
2J + 3S = 81
----------------
5J + 5S = 150 ----(divide by 5)---> J + S = 30
3J + 2S = J + 2(J + S) = J + 60 = 69 ---> J = 9
3*9 + 2S = 69
27 + 2S = 69
2S = 42
S = 21
Answer: D
Question: In May, the grounds keeper at Spring Lake Golf Club built a circular green with an area of 49π square feet. In August, he doubled the distance from the center of the green to the edge of the green. What is the total area of the renovated green?
A.100PI
B.144PI
C.196 PI
D.80
Answer: C.196 PI
Explanation: AREA OF CIRCLE 49PI SQ FEET =PI R^2 THEREFORE R =7
NOW GREEN RADI DOUBLED I.E R =14 AREA =196 PI
C
Question: A wheel makes 1200 revolutions in covering a distance of 60 km. Find the radius of the wheel.
A.10.5m
B.11m
C.11.25m
D.11.45m
Answer: D.11.45m
Explanation: Explanation: Distance covered in one revolution =((60X 1200)/1000)= 72m.
2πR = 72 => 2 x (22/7) x R = 72 => R = 72 x (7/44) = 11.45 m.
Answer: Option D
Question: (2^2-1)(2^2+1)(2^4+1)(2^11+1)=
A.2^19 - 1
B.2^16 + 1
C.2^32 - 1
D.2^128 - 1
Answer: A.2^19 - 1
Explanation: A: is correct
(2^2-1)* (2^2+1)=(2^4-1)
(2^4-1)*(2^4+1)=2^8-1
(2^8-1)*(2^11 +1)=2^19-1
You can apply to this formula: a^2- b^2=(a-b)*(a+b)
Question: Pipe A can fill a tank in 4 hours, pipe B in 8 hours, and pipe C in 24 hours. If all the pipes are open, in how many hours will the tank be filled?
A.2.4 hours
B.2.8 hours
C.3.2 hours
D.4.8 hours
Answer: A.2.4 hours
Explanation: The part filled by A + B + C in 1 hour
= 1/4 + 1/8 + 1/24
= 10/24
All the three pipes together will fill the tank in 24/10 = 2.4 hours.
The answer is A.
Question: The average age of husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 24 years. The present age of the husband is
A.40
B.32
C.45
D.55
Answer: B.32
Explanation: Explanation:
Sum of the present ages of husband, wife and child = (27 * 3 + 3 * 3) years = 90 years.
Sum of the present ages of wife and child = (24 * 2 + 5 * 2) years = 58 years.
Husband's present age = (90 - 58) years = 32 years
Option B
Question: In a hostel there were 100 students. To accommodate 20 more students the average is decreased by rupees 5. But total expenditure increased by Rs.400. Find the total expenditure of the hostel now?
A.Rs.5800
B.Rs.5480
C.Rs.5400
D.Rs.5900
Answer: C.Rs.5400
Explanation: 100x + 400
= 12(x – 5)
x = 50
100 * 50 + 400
= 5400
Answer:C
Question: From a group of 6boys&4girls a Committee of 4 persons is to be formed. In how many different ways can it be done, so that the committee has at least 2 boys?
A.126
B.138
C.185
D.198
Answer: C.185
Explanation: The committee of 4 persons is to be so formed that it has at least 2 men. The different ways that we can choose to form such a committee are:
(i) 2m. 2w in 6C2 x 4 C1=6X5 - 2X1 X 3X3 2X1 =n an (ii) 3m. 1w in 6C3 x 4C1 =6X5X4 3X2X1 X4 _n - a ''''
(iii) 4m in 6C4 -2x1 = 15 Total no. of different ways in which a committee of 4 persons can be formed so that it has at least 2 men. = 90 + 18 + 15 = 185
C
Question: I bought two books; for Rs.480. I sold one at a loss of 15% and other at a gain of 19% and then I found each book was sold at the same price. Find the cost of the book sold at a loss?
A.167
B.280
C.187
D.178
Answer: B.280
Explanation: x*(85/100)
= (480 - x)119/100
x =280
Answer:B
Question: Jar Y has 10% more marbles than Jar X . What percent of marbles from Jar Y need to be moved into X so that both jars have equal marbles
A.4.54%
B.9.09%
C.10%
D.11%
Answer: A.4.54%
Explanation: No; the choices will be approximated only when GMAT question explicitly mentions that;
Here the answer is very close 4.54, not 5.
Jar Y has 10% more marbles than X
Y = 1.1X
If k marbles are taken out of Y. The remaining marbles in Jar Y= 1.1X-k
If k marbles are added to X. The new marbles' count in Jar X= X+k
And we know they are equal;
X+k = 1.1X-k
2k = 0.1X
k= 0.05X
The marbles taken out of Y = 0.05X
Question is: 0.05X is what percent of marbles that was originally there in Y i.e. 1.1X
0.05X1.1X∗1000.05X1.1X∗100
0.05X1.1X∗1000.05X1.1X∗100
5011=4.54¯≈4.545011=4.54¯≈4.54
ANSWER:A
Question: During the first week of performances of a certain play, 5,400 tickets were sold, all at reduced price. During the remaining weeks of performances, 5 times as many tickets were sold at full price as were sold at reduced price. If the total number of tickets E sold was 25,200, how many of them were sold at full price?
A.21,000
B.19,800
C.16,500
D.15,800
Answer: C.16,500
Explanation: Given:
The question tells us about the number of tickets sold in the first week at reduced price = 5,400. It also tells us that during the remaining weeks, the number of tickets sold at full price was 5 times the no. of tickets sold at reduced price. The total tickets sold was 25,200, and we are asked to find the number of tickets that where sold at full price.
Approach:
The question asks us to find the number of ticket sold at full price. We know that the number of tickets sold at full price was 5 times the no. of tickets sold at reduced price in remaining weeks. Hence, assuming tickets sold at reduced price in remaining weeks to be x would give us the no. of tickets sold at full price to be 5x.
Tickets sold in first week + Tickets sold in remaining weeks = Total no. of tickets soldi.e.
Tickets sold in first week + Tickets sold at full price in remaining weeks + Tickets sold at reduced price in remaining weeks = Total no. of tickets sold
We know the no. of tickets sold during first week and the total no. of tickets sold. We will use this relation to get to our answer.
Working Out:
Tickets sold in first week + Tickets sold at full price in remaining weeks + Tickets sold at reduced price in remaining weeks = Total no. of tickets sold
Substituting values in the equation, we get
5,400 + 5x + x = 25,200
x= 3,300
Tickets sold at full price = 5x = 16,500
Answer: Option C
Question: What is the sum which earned interest?
I. The total simple interest was Rs. 7000 after 7 years.
II. The total of sum and simple interest was double of the sum after 5 years.
A.I alone sufficient while II alone not sufficient to answer
B.II alone sufficient while I alone not sufficient to answer
C.Either I or II alone sufficient to answer
D.Both I and II are necessary to answer
Answer: D.Both I and II are necessary to answer
Explanation: Explanation:
Let the sum be Rs. x.
I gives, S.I. = Rs. 7000 and T = 7 years.
II gives, Sum + S.I. for 5 years = 2 x Sum Sum = S.I. for 5 years.
Now, S.I. for 7 years = Rs. 7000.
therefore, S.I. for 1 year = Rs. 1000.
Thus, I and II both are needed to get the answer.
Answer: D
Question: A is thrice as good a workman as B and therefore is able to finish a job in 40 days less than B. working together, they can do it in
A.days
B.½ days
C.days
D.days
Answer: A.days
Explanation: Explanation:
B = 3A
3A – A =40 days
A = 20days
Then, B = 90 days
(A + B) = 20 × 90/ 120 = 20 days .
Answer: Option A
Question: The sum of 36 hours 45 minutes and 12 hours 15 minutes is approximately what percent of a week?
A. 24%
B. approx 29%
C. approx 24%
D. approx 18%
Answer: B. approx 29%
Explanation: Since the question is asking for an approximate percentage
36:45+12:15 ~ 49 hours
% of day = (49*100)/(24*7) ~ 29%
Answer is B
Question: Positive integer p has n factors; 3p has 3 factors; Which of the following values can n take?
I. 1
II. 2
III. 3
A.I or II
B.II or III
C.only I is true
D.only II is true
Answer: D.only II is true
Explanation: Squares have odd number of factors.The square which is a multiple of 3 is 9.
3 has factors 1,3 .
Therefore 2 factors
Answer D
Question: Sum of the numbers from 1 to 29 is
A.400
B.430
C.435
D.405
Answer: C.435
Explanation: Explanation: Sum of first n natural numbers = 1 + 2 + 3 + ..... n = n(n+1)/2
Substitute n = 29.
So S20=29Ã—30/2=435
Correct Option: C
Question: If p is a positive integer and 10p/96 is an integer, then the minimum number of unique prime factors p could have is T:
A.T=1
B.T=2
C.T=3
D.T=4
Answer: B.T=2
Explanation: 10p/96 should resolve to a integer => 5p/48 should resolve to an integer.
Hence p/48 should resolve to a integer.
48 -- > 2^4 * 3.
Hence p should have atleast 2^4*3 for p/48 to resolve to a int.
The unique prime factors in 2^4 * 3 is 2, 3 and hence the answer is B(2).
Question: When 81 is divided by the cube of positive integer z, the remainder is 17. Which of the following could be the value of z?
I.	2
II.	4
III.	8
A.I only
B.II only
C.III only
D.I and II only
Answer: B.II only
Explanation: 81 divided by z^3 leaves a remainder of 17.
81 = q * 2^3 + 17
64 = q+z^3
Then I tried answer possibilities of 2, 4 and 8.
Only 2^3 = 64 fulfills the equation and yields a remainder of 17.
ANSWER:B
Question: Magnabulk Corp sells boxes holding d magnets each. The boxes are shipped in crates, each holding c boxes. What is the price charged per magnet, if Magnabulk charges m dollars for each crate?
A.100cd/m
B.100m/(cd)
C.cd/(100m)
D.m/(cd)
Answer: D.m/(cd)
Explanation: crate has = cd magnates
each magnate cost for each crate = m / cd
Ans; D
Question: In a group of ducks and cows, the total number of legs are 32 more than twice the number of heads. Find the total number of cows.
A.A)14
B.B)12
C.C)16
D.D)8
Answer: C.C)16
Explanation: Explanation :
Let the number of ducks be d
and number of cows be c
Then, total number of legs = 2d + 4c = 2(d + 2c)
total number of heads = c + d
Given that total number of legs are 32 more than twice the number of heads
=> 2(d + 2c) = 32 + 2(c + d)
=> d + 2c = 16 + c + d
=> 2c = 16 + c
=> c = 14
i.e., total number of cows = 16
Answer : C
Question: If the sum of six numbers is between 135 and 164, then the average (arithmetic mean) of the six numbers could be which of the following?
A.15.4.
B.20.5
C.25.5
D.31.25
Answer: C.25.5
Explanation: The problem is actually testing the concept of average and Maxima/Minima....
Let all the values be 135 , so the average will be 135/6 => 22.5 ( Minima )
Let all the values be 164, so the average will be 164/6 => 27.3 ( Minima )
Hence answer will be (C) { any number between the maximum and minimum value.... }
Question: 70% of 300 + 25% of 400 - ? = 256
A.48
B.68
C.54
D.60
Answer: C.54
Explanation: Explanation :
Solution: Let 70% of 300 + 25% of 400 - x = 256.
then, x = (70/100 * 300) + (25/100 * 400) - 256 = 210+100 -256 = 54.
Answer : C
Question: Indu gave Bindu Rs.6250 on compound interest for 2 years at 4% per annum. How much loss would Indu has suffered had she given it to Bindu for 2 years at 4% per annum simple interest?
A.5
B.2
C.10
D.5
Answer: C.10
Explanation: 6250
= D(100/4)2
D = 10
Answer: C
Question: For what value of â€œkâ€ will the equation (2kx2 + 5kx +1)=0 have equal roots?
A.2/7
B.9/4
C.8/25
D.7/1
Answer: C.8/25
Explanation: for a 2nd degree equation ax2+bx_c=0 has equal roots the condition is b2-4ac =0
in the given equation
(5k)^2-4*2k*1=0
by solving this equation
we get k=0, k=8/25
ANSWER:C
Question: In a college, 95 percent of students Passed in math, 97 percent of students Passed in English, 99 percent Passed in French and 92 percent Passed in German.
Can you find out the percentage of students who Passed in all 4 subjects ?
A.92%
B.90%
C.96%
D.83%
Answer: D.83%
Explanation: D
83% of student Passed in all four subjects.
Question: Three candidates contested an election and received 1130, 7530 and 11620 votes respectively. What percentage of the total votes did the winning candidate get?
A.57%
B.57.3%
C.65%
D.90%
Answer: B.57.3%
Explanation: total votes are 20280 ( sum of votes obtained by 3 candidates)
percentage of the total votes recd by the winning candidate = 100*11620/20280 = 57.3%
ANSWER:B
Question: A and B can do a piece of work in 10 days, B and C in 15 days, C and A in 12 days. How long will C take to do it?
A.70 days
B.55 days
C.24 days
D.40 days
Answer: D.40 days
Explanation: 2c = 1/15+ 1/12– 1/10 = 1/20
c = 1/40 => 40 days
Answer: D
Question: C, D and E invested Rs.4500, Rs.9500 and Rs.11000 respectively, in a partnership business. Find the share of E in profit of Rs.25000 after a year?
A.10000
B.11000
C.12000
D.13000
Answer: B.11000
Explanation: Explanation:
4500:9500:11000
9:19:22
22/50 * 25000 = 11000
B)
Question: If 0 < a < b and k = (2a + 9b)/b , which of the following must be true?
A.k < 2
B.k < 7
C.k < 9
D.k > 9
Answer: D.k > 9
Explanation: Please follow posting guidelines (link in my signature) especially writing the correct topic title. Also do mention the source of the question if you select the tag:Source-Other Please Specify.
As for your question, it is very simple and straightforward with number plug in.
Assume a=1 and b=2 as a<b, you get k = (2+14)/2 = 10. This eliminates all but option D , making it the correct answer.
Question: There are A girls in a class, and their average height is M inches. In the same class, there are B boys with an average height of N inches. What is the average height of all the students in the class?
A.(X + Y)/(M + N)
B.(M + N)/(X + Y)
C.(AM + BN)/(A + B)
D.(XM + YN)/(X + Y)
Answer: C.(AM + BN)/(A + B)
Explanation: Total height = AM+BN
No of students = A+B
Avg = (AM+BN)/(A+B)
Answer is C
Question: In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?
A.80
B.90
C.15
D.209
Answer: D.209
Explanation: We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).
Required number
of ways	= (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4)
= (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2)
= (24 + 90 + 80 + 15)
= 209.
ANSWER D 209
Question: 85% of a number is added to 24, the result is the same number. Find the number?
A.228
B.267
C.277
D.160
Answer: D.160
Explanation: (85/100) * X + 24 = X
3X = 480
X = 160
Answer: D
Question: Which of the following cannot be true if the sum of k consecutive integers is 0?
I. The product of the k integers is positive
II. The smallest of the k integers is zero
III. The largest of the k integers is negative
A.I only
B.II only
C.III only
D.I and III
Answer: D.I and III
Explanation: For the sum of k consecutive integers to be 0 we must have one of the following cases: {0}, {-1, 0, 1}, {-2, -1, 0, 1, 2}, ..., {..., -3, -2, -1, 0, 1, 2, 3, ...}.
I. The product of the k integers is positive. As you can see the product of the integers in either of these sets is always 0, thus I cannot be true.
II. The smallest of the k integers is zero. If the set is {0}, then it's possible.
III. The largest of the k integers is negative. As you can see the largest of the integers is always more than or equal to zero, thus III cannot be true.
Answer: D.
Question: The population of a town increases by 10% per annum for two years then decreases by 10% per annum for the next two years, if the population of the town four year ago was 10000, what would it be today?
A.9500
B.9801
C.10100
D.9000
Answer: B.9801
Explanation: Explanation :
Increase of population = (100+10)% = 110% = 1.1
Reduction of population = (100 -10)% = 90% = 0.9
Total effect = 1.1 X 1.1 X 0.9 X 0.9 = 0.9801,
Population = 10000 X 0.9801 = 9801
Answer : B
Question: Which of the following options is the correct value of z if the greatest common divisor of (z-1)!,(z+2)!, and (z+4)! is 120.
A.9
B.5
C.6
D.7
Answer: C.6
Explanation: GCD of (z-1)!,(z+2)!, and (z+4)! is 120. Find z.
Best way to find n is substitute each of the answer and find out the correct one.
Only option C gave the correct answer.
Option C: 6 and find GCD or 5!, 8! and 10!. 5! itself is 120 and 5! which is contained in the other two factorials. This satisfies the condition, hence answer is 6.
Answer is C.
Question: The speed of a bus including stoppages is 35 kmph and excluding stoppages is 70 kmph. Of how many minutes does the train stop per hour?
A.7
B.15
C.25
D.30
Answer: D.30
Explanation: Explanation:
T = 35/70 * 60 = 30
Answer: Option D
Question: A number when divided by 45, gives 432 as quotient and 0 as remainder. What will be the remainder when dividing the same number by 21
A.A)15
B.B)3
C.C)4
D.D)6
Answer: A.A)15
Explanation: Explanation:
P ÷ 45 = 432
=> P = 432 * 45 = 19440
P / 31 = 19440 / 21 = 925, remainder = 15
Answer: Option A
Question: A certain no. whendivided by 35leaves a remainder 25, what is the remainder if the same no.be dividedby 15?
A.2
B.4
C.7
D.8
Answer: B.4
Explanation: Explanation:
35 + 25 = 60/15 = 4 (Remainder)
B
Question: How many S ways are there of placing 6 marbles in 4 bowls, if any number of them can be placed in each bowl?
A.6C4
B.6P4
C.4^6
D.6^4
Answer: C.4^6
Explanation: Each marble has 4 options, so there are total of S= 4*4*4*4*4*4=4^6 ways.
Answer: C.
The total number of ways of dividing n identical items among r persons, each one of whom, can receive 0,1,2 or more items is (n+r -1)C(r-1).
Question: A and B can do a piece of work in 8 days. With the help of C they finish the work in 5 days. C alone can do that piece of work in?
A.30 DAYS
B.35 DAYS
C.40 DAYS
D.8 DAYS
Answer: C.40 DAYS
Explanation: C = 1/5 – 1/8= 1/40 => 40 days
Answer:C