divelab/Sys2Bench huggingface dataset:
Question: In a Question paper there are 3 multiple choice questions. Each question has 4 choices with only one choice as the correct answer. What is the total number of ways in which a candidate will not get all the 3 answers correct?
Options:
A)19
B)85
C)63
D)54
E)64
Rationale: A candidate can answer the test in 4^3 number of ways (each question has 4 choices and we have total of 3 questions). Now, out of these cases there will be only one case when the candidate answered all the four questions correct. Therefore the total number of ways in which a candidate will NOT get all the four answers correct is 4^3-1=63.
Answer: D.
Answer: C
Question: An inspector rejects 0.08% of the meters as defective, How many meters he examine to reject 2 meteres
Options:
A)1200
B)2400
C)1400
D)2500
E)None of these
Rationale: Explanation:
It means that 0.08% of x = 2
=>(8/ 100×100×x)=2=>x=2×100×100/ 8=>x=2500
Option D
Answer: D
Question: 7^6n -1^ 6n When n is an integer > 0 is divisible by
Options:
A)119
B)127
C)134
D)145
E)154
Rationale: 127
B
Answer: B
Question: Henry professes to sell his goods at the cost price but he made use of 750 grams instead of a kg, what is the gain percent?
Options:
A)12 1/3%
B)13 1/3%
C)14 1/3%
D)15 1/3%
E)16 1/3%
Rationale: 750 --- 100
100 --- ? => 13 1/3%
Answer:B
Answer: B
Question: After doing 3/5 of the Biology homework on Monday night, Sanjay did 1/3 of the remaining homework on Tuesday night. What fraction of the original homework would Sanjay have to do on Wednesday night to complete the Biology assignment ?
Options:
A)1/15
B)2/15
C)4/15
D)2/5
E)None of these
Rationale: Sol.(c) Remaining homework on Monday night
= 1 - (3/5) = 2/5
Work done on Tuesday night
= 1/3 of 2/5 = 2/15
Remaining homework to complete the biology assignment
= 2/5 - 2/15 = (6 - 2)/15 = 4/15
Answer C
Answer: C
Question: In one hour,a boat goes 16 km along the stream and 6 km against the stream.The speed of the boat in still water (in km/hr) is :
Options:
A)3
B)11
C)8
D)9
E)10
Rationale: Solution
Speed in still water =1/2(16+6)kmph.= 11 kmph. Answer B
Answer: B
Question: The average score in an examination of 10 students of a class is 55. If the scores of the top five students are not considered, the average score of the remaining students falls by 5. The pass mark was 40 and the maximum mark was 100. It is also known that none of the students failed. If each of the top five scorers had distinct integral scores, the maximum possible score of the topper is
Options:
A)97
B)99
C)98
D)96
E)94
Rationale: The average score in an examination of 10 students of a class is 55 --> the total score = 10*55 = 550;
The 5 smallest scores have an average of 50 --> the total score of the 5 smallest scores = 250.
From above, the total score of the 5 largest scores = 550 - 250 = 300.
Say the 5 largest scores are a, b, c, d, and e (where a<b<c<d<e, since each of the top five scorers had distinct scores). We want to maximize e. To maximize e, we need to minimize a, b, c, and d. The least value of a, is 50 (The least score of the top 5 , a, should be equal to the highest of the bottom 5 and to minimize the highest of the bottom 5, all scores from the bottom 5 should be equal). In this case the least values of b, c, and d are 56, 57, and 58 respectively:
a + b + c + d + e = 50 + 51 +52 + 53 + e = 300;
e = 94.
Answer: E.
Answer: E
Question: A paint crew gets a rush order to paint 80 houses in a new development. They paint the first y houses at a rate of x houses per week. Realizing that they'll be late at this rate, they bring in some more painters and paint the rest of the houses at the rate of 1.25x houses per week. The total time C it takes them to paint all the houses under this scenario is what fraction of the time it would have taken if they had painted all the houses at their original rate of x houses per week?
Options:
A) 0.8(80 – y)
B) 0.8 + 0.0025y
C) 80/y – 1.25
D) 80/1.25y
E) 80 – 0.25y
Rationale: This may be a case of bad numberpicking but it worked for me.
80 houses. Let's say that y=40 houses are painted at the rate x=10 houses per week <=> 1/10 week per house. 40*1/10 = 4 houses per week will be painted at this rate.
80-y = 80-40 = 40 houses are to be painted at the faster rate. X*1,25=12,5 houses per week <=> 1/12,5 weeks per house * 40 houses = 40/12,5 = 80/25 = 320/100 = 3,2 weeks.
Which means finishing all houses at normal rate C=> 2*4 = 8 weeks. Faster rate = 4+3,2 = 7,2 weeks.
7,2/8 = 9/10 = 0,9. Insert y=40 in equations and it is clear that only (B) gives us 0,9.
Answer: B
Question: X can finish a work in 21 days. Y can finish the same work in 15 days. Yworked for 10 days and left the job. how many days does X alone need to finish the remaining work?
Options:
A)3
B)5
C)7
D)8
E)9
Rationale: Work done by X in 1 day = 1/21
Work done by Y in 1 day = 1/15
Work done by Y in 10 days = 10/15 = 2/3
Remaining work = 1 – 2/3 = 1/3
Number of days in which X can finish the remaining work = (1/3) / (1/21) = 7
C
Answer: C
Question: There were 36,000 hardback copies of a certain novel sold before the paperback version was issued. From the time the first paperback copy was sold until the last copy of the novel was sold 9 times as many paperback copies as hardback copies were sold. If a total of 436,000 copies of the novel were sold in all, how many paperback copies were sold?
Options:
A)45,000
B)360,000
C)364,500
D)392,000
E)396,900
Rationale: Say x was the # of hardback copies sold from the time the first paperback copy was sold.
Then the total # of paperback copies sold was 9x;
Hence the total # of copies sold was (hardback)+(paperback)=(36+x)+(9x)=436 --> x=40.
So, the total # of paperback copies sold was 9x=9*40=360 or 360,000.
Answer: B.
Answer: B
Question: If Rs. 561 be divided among A, B, C in such a way that A gets 2/3 of what B gets and B gets 1/4 of what C gets, then their shares are respectively?
Options:
A)s.300
B)s.360
C)s.396
D)s.368
E)s.323
Rationale: (A = 2/3 B and B = 1/4 C) = A/B = 2/3 and B/C = 1/4
A:B = 2:3 and B:C = 1:4 = 3:12
A:B:C = 2:3:12
A;s share = 561 * 2/17 = Rs. 66
B's share = 561 * 3/17 = Rs. 99
C's share = 561 * 12/17 =Rs.396.
Answer:C
Answer: C
Question: What is the value of n if the sum of the consecutive odd intergers Y from 1 to n equals 169?
Options:
A)47
B)25
C)37
D)33
E)29
Rationale: Before you tackle this question you must first understand that the question is comprised of two key parts, 1st is finding out how manytermsis in that sequence and 2nd whatactual number valuethat term is. In an arithmetic progression, in this case consecutive odd integers 1, 3, 5, ...., there are two set of rules.
Rule #1 (Arithmetic Sequence): xn = a + d(n-1) Identifies what the actual # in the sequence would be. Each number in the sequence has a term such as 1(is the first term), 3(is the second term) and so on. So if I were to ask you to find out what the 10th term is of that sequence you would use that formula to find that value.
a=1 (first term)
d=2 (the common difference) remember in the sequence 1, 3, 5, 7 the common difference is always 2
*On a side note we use n-1 because we don't have d in the first term, therefore if we were solving for the first term we would get 0 as n-1 and 0 times d would give us 0, leaving only the first term. This works regardless what your first term is in any sequence.
But remember the question asksWhat is thevalueof n if the sum of the consecutive odd integers from 1 to n equals 169?which means we first need a consecutive sequence that sums up to 169 and than find what the value of the n is, in this case it would be the last number in that sequence. In order to find that we first need to knowhow many terms(how many of the n there is) in order to be able to plug n in this formula given we know what the sum is. For that to happen we need to use Rule #2.
Rule #2 (Summing an arithmetic series): 169 = n/2(2a+(n-1)d). Given the question gives us what the sum is (169 in this case) we would simply use this formula to solve for n. Once we solve for n (13 in this case) we can simply plug n into the first formula (rule 1) and find the value.
It feels very confusing and difficult at first, but once you identify the steps all you need to do is plug and play. We have the sum (169) of a sequence, the number of terms in that sequence is (unknown). Rule #2 tells us how many numbers there are in that sequence and Rule #1 gives us what that last term is.
Answer: B
Question: Find the largest 2 digit number which is exactly divisible by 88?
Options:
A)88
B)89
C)99
D)54
E)90
Rationale: Largest 2 digit number is 99
After doing 99 ÷ 88 we get remainder 11
Hence largest 2 digit number exactly divisible by 88 = 99 - 11 = 88
A
Answer: A
Question: A certain sum of money is divided among A, B and C so that for each Rs. A has, B has 65 paisa and C 40 paisa. If C's share is Rs.56, find the sum of money?
Options:
A)288
B)287
C)72
D)205
E)267
Rationale: A:B:C = 100:65:40
= 20:13:8
8 ---- 56
41 ---- ? => Rs.287
Answer: B
Answer: B
Question: Three partners started a business with Rs. 80000. At the end of the year they receive Rs. 1800. Rs. 3000 and Rs. 4800 as profit. Find the investment of the second person.
Options:
A)Rs. 2700
B)Rs. 40000
C)Rs. 25000
D)Rs. 32000
E)None of these
Rationale: Explanation:
The ratio of profit of the three persons
= 1800 : 3000 : 4800 = 3 : 5 : 8
.: investment of the second person = Rs.[(5/16) × 80000]
= Rs. 250000
Answer: Option C
Answer: C
Question: There are 35 students in a hostel. If the no. of students increases by 6, the expenses of the mess increase by Rs.42/day while the average expenditure per head diminishes by Rs1. Find the original expenditure of the mess?
Options:
A)s. 34.83
B)s. 39.83
C)s. 42.83
D)s. 48.83
E)s. 54.83
Rationale: Suppose the average join the mess , total expenditure = 35x + 42
Now, the average expenditure = (35x + 42)/ (35 + 6) = x – 1
or, 35x + 42 = 41x – 41
or, x = 13.83
Thus the original expenditure of the mess = 35 x 13.83=484.167
Therefore x-1 = 12.83
hence 42 x 12.83 =539
539 - 484.17 = Rs. 54.83
E
Answer: E
Question: if Juan takes 13 seconds to run y yards, how many seconds will it take him to run x yards at the same rate?
Options:
A)11x/y
B)11y/x
C)x/11y
D)11/xy
E)13x/y
Rationale: Answer is E
The rate of running hards: 13 minutes per y hards.
Thus the time for running x hards = x hards * 13 minutes/y hards = 13x/y
Answer: E
Question: In a survey of political preferences, 78% of those asked were in favour of at least one of the proposals: I, II and III. 50% of those asked favoured proposal I, 30% favoured proposal II, and 20% favoured proposal III. If 5% of those asked favoured all three of the proposals, what Z percentage of those asked favoured more than one of the 3 proposals.
Options:
A) 10
B) 12
C) 17
D) 22
E) 30
Rationale: Bunuel, my answer for exactly 2 people was 17 and this was my approach:
100%=(A+B+C)-(AnB+AnC+BnC)-5%+22% which leads me to
Z=100%=(50+30+20)-(at least 2 people)-5%+22%.C
Answer: C
Question: If taxi fares were $8.00 for the first 1/5 mile and $0.8 for each 1/5 mile there after, then the taxi fare for a 8-mile ride was
Options:
A)$39.00
B)$39.10
C)$39.25
D)$39.30
E)$39.20
Rationale: In 8 miles , initial 1/5 mile charge is $8
rest of the distance = 8 - (1/5) = 39/5
rest of the distance charge = 39(0.8) = $31.2 (as the charge is 0.8 for every 1/5 mile)
=> total charge for 4 miles = 8+31.2 = 39.2
Answer is E
Answer: E
Question: A crate of mangoes contains one bruised mango for every thirty mango in the crate. If three out of every four bruised mango are considerably unsaleable and there are 12 unsaleable mangoes in the crate then how many mango are there in the crate?
Options:
A)480
B)680
C)980
D)280
E)380
Rationale: Let the total no of mangoes in the crate be x
Then the no of bruised mango = 1/30 x
Let the no of unsalable mangoes =3/4 (1/30 x)
1/40 x =12
x=480
Answer is A.
Answer: A
Question: A train is moving at 6/7 of its usual speed. The train is 15 minutes too late. What is the usual time (in hours) for the train to complete the journey?
Options:
A)1
B)1.5
C)2
D)2.5
E)3
Rationale: new time = d / (6v/7) = 7/6 * usual time
15 minutes represents 1/6 of the usual time.
The usual time is 1.5 hours.
The answer is B.
Answer: B
Question: In a certain region, the number of children who have been vaccinated against rubella is twice the number who have been vaccinated against mumps. The number who have been vaccinated against both is twice the number who have been vaccinated only against mumps. If 7,000 have been vaccinated against both, how many have been vaccinated only against rubella?
Options:
A)2,5000
B)7,500
C)10,000
D)14,000
E)17,500
Rationale: Use a single variable as far as possible.
Number vaccinated only against mumps = x
Number vaccinated against both = 2x = 7000 (so x = 3500)
Then, number vaccinated against mumps (including both) = x + 2x = 3x
Number vaccinated against rubella = 2*3x = 6x
Then, number vaccinated against only rubella = 6x - 2x = 4x = 4*3500 = 14,000
Answer (D)
Answer: D
Question: A sum of Rs. 2717 is lent into two parts so that the interest on the first part for 8 years at 3% per annum may be equal to the interest on the second part for 3 years at 5% per annum. Find the second sum?
Options:
A)1672
B)1640
C)1677
D)1698
E)1679
Rationale: (x*8*3)/100 = ((2717 - x)*3*5)/100
24x/100 = 40755/100 - 15x/100
39x = 40755 => x = 1045
Second sum = 2717 â€“ 1045 = 1672
Answer:A
Answer: A
Question: An art gallery owner is hanging paintings for a new show. Of the six paintings she has to choose from, she can only hang two on the main wall of the gallery. Assuming that she hangs as many as possible on that wall, in how many ways can she arrange the paintings?
Options:
A)18
B)30
C)64
D)120
E)216
Rationale: Arrangment means permutations questions..
6P2 = 6!/2! = 30.
IMO option B is correct answer..
Answer: B
Question: If the average of 8x and 8y is greater than 120, and x is twice y, what is the least integer value of x ?
Options:
A)20
B)25
C)30
D)32
E)42
Rationale: Substitution can be used in the following way:
Always start with the equation: x = 2y. It is more straight forward to manage as compared to the inequality.
Substitute y = x/2, not the other way because you need to find the minimum value of x. So you can get rid of y.
Now go on to the inequality. So 8y = 8x/2 = 4x
Now average of 8x and 4x is greater than 120. Average of 8x and 4x is 6x. So,
6x > 120
x > 20
Answer : A
Answer: A
Question: Sum of the squares of 3 no.'s is 267 and the sum of their products taken two at a time is 131. Find the sum?
Options:
A)20
B)24
C)26
D)23
E)30
Rationale: (a + b + c)2 = a2 + b2 + c2 + 2(ab +bc + ca) = 267 + 2* 131
a + b + c = √529 = 23
D
Answer: D
Question: A train of 24 carriages, each of 60 meters length, when an engine also of 60 meters length is running at a speed of 60 kmph. In what time will the train cross a bridge 1.5 km long?
Options:
A)7 mins
B)3 mins
C)8 mins
D)9 mins
E)6 mins
Rationale: D = 25 * 60 + 1500
= 3000 m
T = 3000/60 * 18/5
= 180 sec = 3 mins
Answer:B
Answer: B
Question: It was Sunday on Jan. 1,2006.What was the day of the week on Jan 1,2010 ?
Options:
A)Sunday
B)Saturday
C)Friday
D)Wednesday
E)None of these
Rationale: Solution
On 31st December,2005 it was Saturday.
Number of odd days from the year 2006 to the year 2009
=(1 + 1 + 2 + 1)=5 days
On 31st December 2009,it was Thursday.
Thus,on 1st Jan,2010 it is Friday. Answer C
Answer: C
Question: A 300 meter long train running at the speed of 120 kmph crosses another train running in the opposite direction at the speed of 80 kmph in 9 seconds.What is the lenght of other train.
Options:
A)210m
B)220m
C)230m
D)240m
E)200m
Rationale: Relative speeds=(120+80)km/hr
=200km/hr=(200*5/18)m/s=(500/9)m/s
let length of train be xm
x+300/9=500/9
x=200
Ans is 200m
ANSWER:E
Answer: E
Question: Pipe A that can fill a tank in an hour and pipe B that can fill the tank in half an hour are opened simultaneously when the tank is empty. Pipe B is shut 10 minutes before the tank overflows. When will the tank overflow?
Options:
A)37 mins
B)35 mins
C)40 mins
D)32 mins
E)36 mins
Rationale: The last 10 minutes only pipe A was open. Since it needs 1 hour to fill the tank, then in 10 minutes it fills 1/5th of the tank, thus 5/6 of the tank is filled with both pipes open.
The combined rate of two pipes is 1 + 2 = 3 tanks/hour, therefore to fill 5/6th of the tank they need (time) = (work)/(rate) = (5/6)/3 = 5/18 hours = 27 minutes.
Total time = 10+ 27= 37 minutes.
Answer: A.
Answer: A
Question: If a car went the first third of the distance at 80 kmh, the second third at 24 kmh, and the last third at 60 kmh, what was the average speed of the car for the entire trip?
Options:
A)36 kmh
B)40 kmh
C)42 kmh
D)42 kmh
E)50 kmh
Rationale: Assume D/3 = 240 (this number is convenient because it is divisible by 80, 24 and 60)
So:
240 = 80*T1 = 3 hrs
240 = 24*T2 = 10 hrs
240 = 60*T3 = 4 hrs
T = T1 + T2 + T3 = 17 hrs
D = RT
(240*3) = R*17
R = 42
ANSWER: D
Answer: D
Question: 16 people can write 52 book in 12 days working 8 hour a day.then in how many day 206 can be written by 64 people?
Options:
A)309/26
B)309/28
C)309/22
D)319/26
E)339/26
Rationale: work per day epr hour per person= 52/(12*8*16) // eq-1
people= 64; let suppose day=p; per day work for 8 hours
acc. to condition
work per day epr hour per person= 206/(p*8*64) // eq-2
eq-1==eq-2;
p= 309/26
ANSWER:A
Answer: A
Question: The average age of 18 students of a class is 18 years. Out of these, the average age of 5 students is 14 years and that of the other 9 students is 16 years, The age of the 18th student is
Options:
A)101
B)66
C)110
D)160
E)12
Rationale: Explanation:
Age of the 18th student = [18 * 18 - (14 * 5 + 16 * 9)] = (324-214) = 110 years.
Answer: C
Answer: C
Question: A container contains 1000 liters of milk, From this container 10 liters of milk was taken out and replaced by water. This process was repeated further 2 times. How much milk is now contained by the container?
Options:
A)970.3liters
B)1000.45liters
C)879.65liters
D)1020.56liters
E)910.95liters
Rationale: Amount of milk left after 3 operations = 1000 (1 - 10/1000)^3 = 1000 * 99/100 * 99/100 * 99/100 = 970.3 liters
Answer is A
Answer: A
Question: What is the two-digit number?
I. Sum of the digits is 7.
II. Difference between the number and the number obtained by interchanging the digits is 9.
III. Digit in the ten's place is bigger than the digit in the unit's place by 1.
Options:
A)I and II only
B)II and III only
C)I and III only
D)All I, II and III
E)I and II as well as I and III
Rationale: Let the tens and units digit be x and y respectively.
I. x + y = 7.
II. (10x + y) - (10y + x) = 9 x - y = 1.
III. x - y = 1.
Thus, I and II as well as I and III give the answer.
Answer: Option E
Explanation:
Let the tens and units digit be x and y respectively.
I. x + y = 7.
II. (10x + y) - (10y + x) = 9 x - y = 1.
III. x - y = 1.
Thus, I and II as well as I and III give the answer.
Correct answer is (E).
ANSWER E I and II as well as I and III
Answer: E
Question: If x is an integer, then x(x – 1)(x – k) must be evenly divisible by two when k is any of the following values EXCEPT
Options:
A)-1
B)-2
C)-1
D)2
E)5
Rationale: We have the product of 3 integers: (x-1)x(x-k).
Note that the product of 3 integers is divisible by 2 if at least one multiple is divisible by 2. Now, to guarantee that at least one integer out of x, (x – 1), and (x – k) is divisible by 2 these numbers must have different remainders upon division by 2, meaning that one of them should have remainder of 1, another reminder of 2 and the last one remainder of 0, so be divisible by 2.
Next, if k=-1 then we'll have (x-1)x(x+1)=(x-1)x(x+1) --> which means that (x-1) and (x+1) will have the same remainder upon division by 2. Thus for k=-1 we won't be sure whether (x-1)x(x-k) is divisible by 2.
Answer: A.
Answer: A
Question: How many three digits number can be formed by using the digits 4,5,6,7,8,9 if the repetition of digits not allowed?
Options:
A)80
B)70
C)60
D)40
E)30
Rationale: For a number to be odd we must have 5,7 or 9 at the unit place. So there 3 ways of filling the unit place. Since the repetion of digits is not allowed, the tenths place can be filled with any of the remaining 5 digits in 5 ways. Now hundreds place can be filled in 4 ways by the remaining 4 numbers. So the required number of numbers is 3x5x4 =60 Answer : C
Answer: C
Question: Grace has 20 jellybeans in her pocket. She has 12 red ones, 4 green ones, and 4 blue ones. What is the minimum number of jellybeans she must take out of her pocket to ensure that she has one of each color?
Options:
A)4
B)8
C)12
D)13
E)17
Rationale: The worst scenario is when she takes 12 red and 4 green, total of 16 jellybeans, and still doesn't have 3 distinct colors. But the next draw (17th) will surely be the third color jellybean (blue) as there is no other color jellybeans left in pocket.
Answer: E.
Answer: E
Question: What will come in place of the x in the following Number series? 46080, ?, 384, 48, 8, 2, 1
Options:
A)1
B)3840
C)5
D)7
E)9
Rationale: 46080 /12 = 3840
3840 /10 = 384
384 /8 = 48
48 /6 = 8
8/4 = 2
2 /2 = 1
B
Answer: B
Question: A can do a piece of work in 80 days. He works at it for 10 days & then B alone finishes the remaining work in 42 days. In how much time will A and B, working together, finish the work?
Options:
A)10 days
B)230 days
C)50 days
D)40 days
E)20 days
Rationale: B
30 days
Work done by A in 10 days=10/80=1/8
Remaining work=(1-(1/8))=7/8
Now, work will be done by B in 42 days.
Whole work will be done by B in (42*8/7)=48 days
Therefore, A's one day's work=1/80
B’s one day's work=1/48
(A+B)'s one day's work=1/80+1/48=8/240=1/30
Hence, both will finish the work in 30 days.
Answer: B
Question: A man took loan from a bank at the rate of 12% p.a. S.I. After 3 years he had to pay Rs. 5400 interest only for the period. The principal amount borrowed by him was?
Options:
A)15002
B)15001
C)15000
D)15007
E)15009
Rationale: Principal = (100 * 5400) / (12 * 3) = Rs. 15000.Answer:C
Answer: C
Question: An auto assembly plant performs six functions with each frame that arrives: add axles, add wheels to the axles, install the windshield to the frame, install the instrument panel, install the steering wheel, and install the interior seating. Once those six tasks are performed, each car goes to a separate building for finishing touches. If these tasks can be arranged along a linear assembly line in any order, except that the axles must be installed before the wheels can be added, how many T ways can the assembly line be arranged?
Options:
A)120
B)240
C)360
D)480
E)720
Rationale: C) 360
short way: there are 6C! ways to do the six tasks. Half will have wheels before axles and half will have axles before wheels. So we want T=6C!/2 -> 720/2 = 360
Answer: C
Question: 6x − 12 = 6y
5y + 5x = 5
Which of the following is the number of solutions to the system of equations shown above?
Options:
A)More than three
B)Exactly three
C)Exactly two
D)Exactly one
E)None
Rationale: 6x − 12 = 6y
=> 6x - 6y = 12
=> x- y =2 -- 1
5y + 5x = 5
=>x + y = 1 -- 2
From equation 1 and 2 , we get
2x = 3
=> x=1.5
y=-0.5
Therefore , the given system will have Exactly one solution
D
Answer: D
Question: A train starts full of passengers. At the first station, it drops one-third of the passengers and takes 280 more. At the second station, it drops one-half of the new total and takes 12 more. On arriving at the third station, it is found to have 248 passengers. Find the number of passengers in the beginning.
Options:
A)240
B)248
C)280
D)288
E)290
Rationale: Let x be no. of passengers at beginning
At, 1st station is drops 1/3rd so remaining 2/3rd are in train only plus 280 i.e., 2x/3+280
at 2nd stop 1/2 of new total and 12 more i.e., (2x/3+280)/2 + 12 = 248
on solving above equation we get x as 288
ANSWER:D
Answer: D
Question: A and B’s salaries together amount to Rs. 3,000. A spends 95% of his salary and B spends 85% of his. If now their savings are the same, what is A’s salary?
Options:
A)2777
B)2250
C)2789
D)2776
E)2881
Rationale: (5/100) A = (15/100) B
A = 3B
A + B = 3000
4B = 3000 => B = 750
A = 2250
Answer B
Answer: B