RikoteMaster/aqua-rat-mcqa huggingface dataset:
Question: A student took five papers in an examination, where the full marks were the same for each paper. His marks in these papers were in the proportion of 5 : 6 : 7 : 8 : 9. In all papers together, the candidate obtained 60% of the total marks. Then the number of papers in which he got more than 50% marks is
A.2
B.3
C.4
D.5
Answer: D.5
Explanation: Here we can assume that one subject is of 100 marks so total there are 5 subjects => 100*5 = 500.
Now according to the question he secured 60 % of those which is 60%of 500 = 300 marks in total.
The ratio between the marks is given as 5 : 6 : 7 : 8 : 9 , so now we can distribute 300 marks according to the ratio.
Total ratio = 35
For 6 : (300/35)*6 = 51.4
Similarly , we will get for others as 60 , 68.57 , 77.14
Hence , there are 5 subject where he secured more that 50 %.
The answer is 5.
D
Question: 20, 19, 17, ?, 10, 5
A.15
B.14
C.13
D.12
Answer: B.14
Explanation: The Pattern is - 1, - 2, -3, ...
Answer : B.
Question: .05% of seeds in a packet do not germinate. How many seeds must be in the packet for 6 seeds not to germinate?
A.10000
B.11000
C.12000
D.14000
Answer: C.12000
Explanation: Let the number of total seeds be x
Then, .05% of x=6
(5/100)*(1/100)*x=6
x=12000
Answer is C
Question: Two pipes A and B can separately fill a tank in 2 minutes and 15 minutes respectively. Both the pipes are opened together but 4 minutes after the start the pipe A is turned off. How much time will it take to fill the tank?
A.15
B.10
C.81
D.71
Answer: B.10
Explanation: 4/12 + x/15 = 1
x = 10
Answer: B
Question: Lucy invested $10,000 in a new mutual fund account exactly three years ago. The value of the account increased by 10 percent during the first year, increased by 9 percent during the second year, and decreased by 10 percent during the third year. What is the value of the account today?
A.$10,350
B.$10,395
C.$10,791
D.$11,500
Answer: C.$10,791
Explanation: the amount at the end of first year will become the principle for 2nd year, applying this trick and calculating we get the amount at the end of third year to be 10791. So the answer is (C).
Question: The average age of 25 students in a class is 12years. If teacher's age is also included then average increases 1year then find the teacher's age?
A.40
B.35
C.29
D.38
Answer: D.38
Explanation: Total age of 25 students = 25*12 = 300
Total age of 26 persons = 26*13 = 338
Age of teacher = 338-300 = 38years
Answer is D
Question: Uncle Bruce is baking chocolate chip cookies. He has 36 ounces of dough (with no chocolate) and 13 ounces of chocolate. How many ounces of chocolate are left over if he uses all the dough but only wants the cookies to consist of 20% chocolate?
A. 4
B. 6
C. 7.2
D. 7.8
Answer: A. 4
Explanation: Answer is A.
x/x+36=1/5
x=9
13-9=4
Question: In a certain production lot, 40 percent of the toys are red and the remaining toys are green. Half of toys are small and half are large. If 10 percent of the toys are red and small, and 40 toys are green and large. How many of the toys are red and large?
A.40
B.60
C.70
D.80
Answer: B.60
Explanation: 40% toys are red.
50% toys are small - 50% are large
Red and small - 10%
Red and large must be 30% (since there are total 40% red toys)
But 50% toys are large. If 30% are red and large, the rest 20% large toys must be green. So large and green toys are 20%.
These 20% large and green toys are actually 40 so total number of toys must be 200.
Since red and large toys are 30%, they must be 60.
ANSWER:B
Question: Find the area of right angled triangle whose hypotenuse is 15 cm and one of the sides is 12 cm.
A.81
B.91
C.99
D.108
Answer: A.81
Explanation: AB² = AC² - BC²
= 15² - 12²
= 225 - 144
= 81
ANSWER :OPTION A
Question: During a two-week period, the price of an ounce of silver increased by 8 percent by the end of the first week and then decreased by 12 percent of this new price by the end of the second week. If the price of silver was x dollars per ounce at the beginning of the two-week period, what was the price, in dollars per ounce, by the end of the period?
A.0.93x
B.0.95x
C.0.97x
D.0.99x
Answer: B.0.95x
Explanation: The price at the end is 0.88(1.08x)=0.9504x which is approximately 0.95x
The answer is B.
Question: Soja finished 2/3 of a book. She calculated that she finished 100 more pages than she has yet to read. How long is her book?
A.A)300
B.B)150
C.C)130
D.D)90
Answer: A.A)300
Explanation: Let x be the total number of pages in the book, then she finished 2/3*x pages.
Then she has x−2/3*x=1/3*x pages left.
2/3*x−1/3*x=100
1/3*x=100
x=300
So the book is 300 pages long.
Answer is A.
Question: A clock is started at noon. By 10 minutes past 2, the hour hand has turned through how many degrees?
A.55Â°
B.60Â°
C.65Â°
D.70Â°
Answer: C.65Â°
Explanation: The angle traced by the hour hand in 12 hrs is 360Â°
The angle traced by the hour hand in 1 hour is 30Â°
The angle traced by the hour hand in 10 minutes is 5Â°
The angle traced by the hour hand in 2 hours and 10 minutes is 2*30Â° + 5Â° = 65Â°
The answer is C.
Question: A lent Rs. 5000 to B for 2 years and Rs. 3000 to C for 4 years on simple interest at the same rate of interest and received Rs. 2200 in all from both of them as interest. The rate of interest per annum is?
A.11
B.66
C.88
D.10
Answer: D.10
Explanation: Let the rate be R% p.a. Then,
(5000 * R * 2)/100 + (3000 * R * 4)/100 = 2200
100 R + 120 R = 2200
R = 10%
Answer:D
Question: A shopkeeper sold 30 articles at the cost price of 35 articles. Then find the profit% or lost%
A.12%
B.15%
C.16.67%
D.18%
Answer: C.16.67%
Explanation: here 30 articles selling price = 35 articles cost price
so the difference = 35-30 = 5
% of profit = 5*100/30 = 16.67%
correct option is C
Question: A can do a piece of work in 15 days and B in 20 days. They began the work together but 5 days before the completion of the work, A leaves. The work was completed in?
A.8 days
B.10 days
C.15 days
D.11 3/7 days
Answer: D.11 3/7 days
Explanation: Explanation:
(x – 5)/15 + x/20 = 1
x = 11 3/7 days
Answer: D
Question: If n = (33)^43 + (53)^33 what is the units digit of n?
A.0
B.2
C.4
D.6
Answer: C.4
Explanation: First of all, the units digit of (33)^53 is the same as that of 3^53 and the units digit of (53)^33 is the same as that of 3^33. So, we need to find the units digit of 3^53 + 3^33.
Next, the units digit of 3 in positive integer power repeats in blocks of four {3, 9, 7, 1}:
3^1=3 (the units digit is 3)
3^2=9 (the units digit is 9)
3^3=27 (the units digit is 7)
3^4=81 (the units digit is 1)
3^5=243 (the units digit is 3 again!)
...
Thus:
The units digit of 3^53 is the same as the units digit of 3^3, so 7 (53 divided by the cyclicity of 4 gives the remainder of 3).
The units digit of 3^33 is the same as the units digit of 3^1, so 3 (33 divided by the cyclicity of 4 gives the remainder of 1).
Therefore the units digit of (33)^53 + (53)^33 is 7 + 7 = 4.
Answer: C.
Question: The annual interest rate earned by an investment increased by 10 percent from last year to this year. If the annual interest rate earned by the investment this year was 1 percent, what was the annual interest rate last year?
A.1%
B.1.1%
C.9.1%
D.10%
Answer: C.9.1%
Explanation: let i = interest rate
i(this year) = i(last year) + 0.10i(last year) = 1.1i(last year)
1 = 1.1 x i(last year)
i(last year) = 1/1.1 = 10/11 = 0.90909...
or i(last year) = 0.91%
ANSWER:C
Question: The average employee age of a certain company which has 80 employees is 38. One employee aged 60 got retired and replaced by a fresher of age 22. What is the new average age of employees?
A.35.5
B.36.2
C.36.8
D.37.5
Answer: D.37.5
Explanation: Sum of ages of employees before replacement= 80x38=3040
Sum of ages of employees after replacement=(80x38)+22-60=3002
The new average age=3002/80=37.5
Answer: D
Question: A watch was sold at a loss of 10%. If it was sold for Rs.140 more, there would have been a gain of 4%. What is the cost price?
A.Rs:1000
B.Rs:1067
C.Rs:1278
D.Rs:1028
Answer: A.Rs:1000
Explanation: 90%
104%
--------
14% ---- 140
100% ---- ? =>
Rs:1000
Answer: A
Question: A committee of 4 people is to be selected from 8 people. How many different committees are possible?
A.28
B.40
C.56
D.420
Answer: D.420
Explanation: Because the order does not matter in selecting 4 people for the committee, this is a combination question.
The number of ways to select 4 people from 8 = 8C4 = 8!/4!*4! = 420. Hence D.
Question: A vendor bought toffees at 6 for a rupee. How many for a rupee must he sell to gain 20%?
A.6
B.9
C.5
D.1
Answer: C.5
Explanation: C.P. of 6 toffees = Re. 1
S.P. of 6 toffees = 120% of Re. 1 = Rs. 6/5
For Rs. 6/5, toffees sold = 6. For re. 1.
Toffees sold = 6 * 5/6 = 5
AnswerC
Question: A question paper consists of five problems, each problem having three internal choices. In how many ways can a candidate attempt one or more problems?
A.676
B.833
C.1023
D.839
Answer: C.1023
Explanation: Given that, the question paper consists of five problems. For each problem, one or two or three or none of the choices can be attempted.
Hence, the required number of ways = 45 - 1.
= 210 - 1 = 1024 - 1 = 1023
Answer:C
Question: 7122,6521,8332,4743,9911,387?
find the value of the question tag.
A.1
B.2
C.3
D.4
Answer: D.4
Explanation: 7+1+2+2=12
6+5+2+1=14
so
9+9+1+1=20
3+8+7+x=22
so x=4
ANSWER:D
Question: The average of 25 result is 18.The average of 1st 12 of them is 14 & that of last 12 is 17.Find the 13th result.
A.68
B.80
C.78
D.60
Answer: C.78
Explanation: Clearly 13th result=(sum of 25 results)-(sum of 24 results)
=(18*25)-(14*12)+(17*12)
=450-(168+204)
=450-372
=78.
Answer is C.
Question: Evaluate: 13 + sqrt(- 4 + 5Ã—3Ã·3) =?
A.12
B.13
C.14
D.15
Answer: C.14
Explanation: According to order of operations, inner brackets first where 5x3Ã·3 is first calculated since it has a multiplication and a division.
5Ã—3Ã·3 = 15Ã·3 = 5
Hence
13 + sqrt(- 4 + 5Ã—3Ã·3)= 13 + sqrt(- 4 + 5)
= 13 + sqrt(1) = 13 + 1 = 14
correct answer C)14
Question: How many 5-digit positive integers exist the sum of whose digits are odd?
A.36000
B.38000
C.45000
D.90000
Answer: C.45000
Explanation: Solution:
There are 9�104 = 90000, 5-digit positive integers.
Out of these 90000 positive integers, the sum of the digits of half of the numbers will add up to an odd number and the remaining half will add up to an even number.
Hence, there are 90000 /2= 45000, 5-digit positive integers whose sum add up to an odd number.
Answer: Option C
Question: In a division sum, the remainder is 6 and the divisor is 5 times the quotient and is obtained by adding 2 to the thrice of the remainder. The dividend is:
A.86
B.90
C.92
D.94
Answer: A.86
Explanation: Divisor = (6 * 3) + 2 = 20
5 * Quotient = 20
Quotient = 4.
Dividend = (Divisor * Quotient) + Remainder
Dividend = (20 * 4) + 6 = 86.
A
Question: Machines A and B produce bottles at their respective constant rates. Machine A produces k bottles in 3 hours and machine B produces k bottles in 5 hours. How many hours does it take machines A and B, working simultaneously, to produce 16k bottles?
A.24
B.26
C.28
D.30
Answer: D.30
Explanation: A's rate = k/3
B's rate = k/5
k/3 + k/5 = 8k/15
16k / (8k/15) = 30 hours
The answer is D.
Question: The credit card and a global payment processing companies have been suffering losses for some time now. A well known company recently announced its quarterly results. According to the results, the revenue fell to $48.0 billion from $69.0 billion, a year ago. By what percent did the revenue fall?
A.29.43
B.30.43
C.31.43
D.32.43
Answer: B.30.43
Explanation: $69-$48=21$
(21/69)*100=30.434%
ANSWER:B
Question: The closest approximation of A(69.28 × 0.004)/0.03 is
A.0.092
B.0.92
C.9.2
D.92
Answer: C.9.2
Explanation: A(69.28 × 0.004)/0.03
1. 0.004 = 4 × 10^(-3)
2. 0.03 = 3 × 10^(-2)
3. (a × b)/c = a × (b/c)
4. 0.004/0.03 = 4 × 10^(-3)/ (3 × 10^(-2)) = 4 × 10^( -3 - (-2))/3 = 4 × 10^(-1) / 3 = (4/3) × 10^(-1) = 1.333 × 10^(-1)
Therefore,
(69.28 × 0.004)/0.03 = 69.28 × (0.004/0.03) = 69.28 × 1.33 × 10^(-1) = 69.28 × 1.33/10 = 6.928*1.33
Now,
7 × 2 = 14
7 × 1 = 7
or better:
6.9 × 1 = 6.9
6.9 × 2 = 13.8
hence,
6.9 <6.928 × 1.33 < 13.8
9.2 is the only answer that satisfies this condition.C
Question: 97 students take part in a test and the median score is 85. Which of the following descriptions must be correct?
I.At least 49 students’ score are less than 85.
II. At least 49 students’ score are equal to or more than 85
III. At most 49 students’ score are equal to or more than 85.
A.I
B.II
C.III
D.III
Answer: B.II
Explanation: If we order the scores increasingly, then on the 49th place we have the score of 85. There must be 48 scores (those on the right of the 85 score) greater or equal to 85. So, I is necessarily correct.
The scores on the left of the 49th place, all have to be equal or less than 85. Since we can have the extreme case of all the scores 85, neither II, nor III is correct. The question asks for descriptions thatmustbe correct.
Answer B.
Question: A certain hall contains two cuckoo clocks. If the first clock chirps 20 times per hour and the second clock chirps 9 times per hour, and both clocks chirp for the first time at 2:03 pm, at what time will the first clock have chirped three times as many times as the second clock can chirp per hour?
A.2:42 pm
B.2:50 pm
C.3:07 pm
D.3:21 pm
Answer: D.3:21 pm
Explanation: The question is asking when the first clock will have chirped 27 times (3*9 chirps).
The first clock chirps once every 3 minutes.
If the first chirp is at 2:03, then the 27th chirp is at 81 minutes after 2:00.
The answer is D.
Question: Each factor of 200 is inscribed on its own plastic ball, and all of the balls are placed in a jar. If a ball is randomly selected from the jar, what is the probability that the ball is inscribed with a multiple of 40?
A.1/6
B.5/42
C.1/8
D.3/16
Answer: A.1/6
Explanation: 200=the # of factors 200 has is=12 (see below);
so out of 12 factors only two are multiples of 40: 40 and 200, itself;
So, the probability is 2/12=1/6.
Answer: A.
Question: The flow of water through a drainage pipe was monitored for a 3-hour period. In the second hour, the rate of flow was 24 gallons per hour, which was 50 percent faster than the rate of flow for the first hour. If 25 percent more water flowed through the pipe in the third hour than it did in the second, how many gallons of water flowed through the pipe during the entire three hours?
A.61.25
B.62.5
C.63.75
D.70.0
Answer: D.70.0
Explanation: Rate of flow of water in second hour = 24 gallons per hour
Rate of flow of water in first hour = 24/(3/2) = 16 gallons per hour
Rate of flow of water in third hour = (125/100)* 24 = (5/4) * 24 = 30 gallons per hour
Number of gallons of water that flowed through the pipe during the entire 3 hours = 16+24+30 = 70 gallons
Answer D
Question: 5862 is to 714
and 3498 is to 1113
and 9516 is to 156
therefore 8257 is to ?
A.157
B.236
C.274
D.123
Answer: A.157
Explanation: A
157
7 + 8 = 15,
2 + 5 = 7;
Question: A grocer purchased a quantity of bananas at 3 pounds for $0.50 and sold the entire quantity at 4 pounds for $1.00. How many pounds did the grocer purchase if the profit from selling the bananas was $7.00?
A. 40
B. 84
C. 90
D. 120
Answer: B. 84
Explanation: Cost price of 1 pound of bananas = 0.5/3 = 1/6
Selling price of 1 pound of bananas = 1/4
Profit per pound = (1/4 - 1/6) = (1/12)
Total profit is given as 7
(1/12)*x = 7
x = 84
Answer: B
Question: The original price of a certain TV set is discounted by f percent, and the reduced price is then discounted by 2f percent. If P is the original price of the TV Set, which of the following represents the price of the television set after the two successive discounts?
A.P(1 - 0.03x + 0.02x^2)
B.P(1 - 0.03f + 0.0002f^2)
C.P(1 - 0.03x + 0.002x^2)
D.P(1 - 2x^2)
Answer: B.P(1 - 0.03f + 0.0002f^2)
Explanation: Youcansolve this problem with number plugging.
Say the original price was $10 and f=50. Then after the first reduction the price would become $5 and after the second reduction of 2*50=100% the rprice would become $0.
Now, since P is not zero, then the expression in the brackets must be zero for f=50. Only answer choice B works.
Answer: B.
On this problem - number pluggin is not giving me the answer.. I initially used f = 10, then 2f = 20 and P = 100. Answer should after both consecutive discounts = 72. I plug in the respective values and I keep getting 68. Can you double check my math.
100 (1-0.03(10) + 0.0002 (10)^2)
100 (1-0.3 + 0.0002 (100))
100 (0.7 + 0.02)
100 (0.68) = 68????
double check my math. Am I missing something? I also plugged in your numbers and still did not get zero as final answer with choice B..
Question: A jogger running at 9 km/hr along side a railway track is 200 m ahead of the engine of a 120 m long train running at 45 km/hr in the same direction. In how much time will the train pass the jogger?
A.32
B.20
C.36
D.88
Answer: A.32
Explanation: Speed of train relative to jogger = 45 - 9 = 36 km/hr.
= 36 * 5/18 = 10 m/sec.
Distance to be covered = 200 + 120 = 320 m.
Time taken = 320/10 = 32 sec.
Answer: Option A
Question: At a conference, the only events scheduled for the day are seminars and breaks. All seminars are the same length and are 10 times as long as each break. A break must be between two consecutive seminars. If the first seminar begins at 9:00 a.m. and the third one ends at 10:40 a.m., how long will the conference last if there are a total of 6 seminars that day?
A.hours and 15 minutes
B.hours and 20 minutes
C.hours and 25 minutes
D.hours and 30 minutes
Answer: D.hours and 30 minutes
Explanation: Well its like this S B S B S B S B S B S
the interval between 9 am to 10:40 am is 160 min having 3 seminars and 2 breaks .If 1 break is of 5 min than time for 3 seminars = 160 - 2*5=150
Time of 1 seminar =150 /3=50 min .
So calculating for 6 seminars and 5 breaks ,total time comes out to be 5 hrs and 30 min .
Answer is D.
Question: If Sn is the sum of the first n terms of a certain sequence and if Sn = n(n+6) for all positive integers n, what is the third term of the sequence?
A.10
B.4
C.6
D.8
Answer: A.10
Explanation: Tn = nth term
T1 = 1st term
T2 = 2nd term
and so on...
Sn = Sum of first n terms of the sequence
Sn = 1st term + 2nd term + 3rd term + ... + nth term
Sn = T1 + T2 + T3 + ....Tn
You are given here thatSn is the SUM of first n terms...So you have
Sn = T1 + T2 + T3 + ....Tn = n(n+1)
So S1 = T1
S2 = T1 + T2
S3 = T1 + T2 + T3
and so on
S1 = T1 = 1*(1+1) = 2
S2 = T1 + T2 = 2 + T2 = 2*(2+1) = 6
So T2 = 4
S3 = T1 + T2 + T3 = 2 + 4 + T3 = 3*(3+1) = 12
So T3 = 10
(This is what we wanted)
The third term is 10.
A
Question: A train moves fast a telegraph post and a bridge 264 m long in 8 sec and 20 sec respectively. What is the speed of the train?
A.45
B.46
C.47
D.79.2
Answer: D.79.2
Explanation: Let the length of the train be x m and its speed be y m/sec.
Then, x/y = 8 => x = 8y
(x + 264)/20 = y
y = 22
Speed = 22 m/sec = 22 * 18/5 = 79.2 km/hr.
Answer: Option D
Question: A train is 360 meter long is running at a speed of 48 km/hour. In what time will it pass a bridge of 140
meter length.
A.27.5 seconds
B.37.5 seconds
C.47.5 seconds
D.57.5 seconds
Answer: B.37.5 seconds
Explanation: Explanation:
Speed = 48 Km/hr = 48*(5/18) m/sec
= 40/3 m/sec
Total distance = 360+140 = 500 meter
Time = Distance/speed
=500/40/3 = 37.5 seconds
Answer: B
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.200
B.205
C.209
D.250
Answer: C.209
Explanation: =>1 boy 3 girls or 2 boys 2 girls or 3 boys 1 girl or 4 boys
=> (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4)=> 24+90+80+15 = >209
ANSWER C
Question: In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together ?
A.58300
B.60800
C.55600
D.50400
Answer: D.50400
Explanation: In the word 'CORPORATION', we treat the two vowels OOAIO as one letter. Thus,we have CRPRTN (OOAIO)
This has 7 letters of which R occurs 2 times and the rest are different.
Number of ways of arranging these letters= 7!/(2!)=2520.
Now 5 vowels in which O occurs 3 times and the rest are different, can be arranged in 5!/3!=20 ways.
Required number of ways =(2520*20)=50400.
Answer is D
Question: 20 litres of water are poured into an aquarium of dimensions 50cm length, 20cm breadth, and 40cm height. How high (in cm) will the water rise?
(1 litre = 1000cm³)
A.6
B.8
C.10
D.20
Answer: D.20
Explanation: lXbXh= 12000
h= 20000/50*20= 20 cm
'D' is the answer.
Question: A cistern 8m long and 4 m wide contains water up to a depth of 1 m 25 cm. The total area of the wet surface is:
A.62 m2
B.50 m2
C.53.5 m2
D.55 m2
Answer: A.62 m2
Explanation: Area of the wet surface	= [2(lb + bh + lh) - lb]
= 2(bh + lh) + lb
= [2 (4 x 1.25 + 8 x 1.25) + 8 x 4] m2
= 62 m2.
Answer: Option A
Question: Arrange the words given below in a meaningful sequence.
1. Poverty 2. Population 3. Death 4. Unemployment 5. Disease
A.2, 4, 1, 5, 8
B.2, 4, 1, 5, 7
C.2, 4, 1, 5, 3
D.2, 4, 1, 5, 9
Answer: C.2, 4, 1, 5, 3
Explanation: Answer: C) 2, 4, 1, 5, 3
Explanation:
The correct order is :
Population Unemployment Poverty Disease Death
2 4 1 5 3
Answer:C
Question: What is the least number which when doubled will be exactly divisible by 12, 14, 18 and 22 ?
A.1286
B.1436
C.1216
D.1386
Answer: D.1386
Explanation: LCM of 12, 14, 18 and 22 = 2772
Hence the least number which will be exactly divisible by 12, 14, 18 and 22 = 2772
2772 ÷ 2 = 1386
1386 is the number which when doubled, we get 2772 Answer: Option D
Hence, 1386 is the least number which when doubled will be exactly divisible by 12, 14, 18 and 22.
Question: The average monthly salary of laborers and supervisors in a factory is Rs.1250 per month; where as the average monthly salary of 6 supervisors is Rs.2450. If the average monthly salary of the laborers is Rs.950 find the number of laborers?
A.22
B.42
C.44
D.55
Answer: B.42
Explanation: 5x 6x 2x
50 25 100
250x + 150x + 200x = 4200
600x = 4200
x = 7 => 6x = 42
Answer: B
Question: Running at their respective constant rates, machine X takes 2 days longer to produce w widgets than machine Y. At these rates, if the two machines together produce 5/4 w widgets in 3 days, how many days would it take machine X alone to produce 3w widgets?
A.4
B.6
C.18
D.10
Answer: C.18
Explanation: Let Y produce w widgets in y Days
Hence, in 1 day Y will produce w/y widgets.
Also, X will produce w widgets in y+2 days (given, X takes two more days)
Hence, in 1 day X will produce w/y+2 widgets.
Hence together X and Y in 1 day will produce {w/y + w/y+2} widgets.
Together X and Y in 3 days will produce = 3 * [{w/y + w/y+2}] widgets.
It is given that in 3 days together they produce (5/4)w widgets.
Equating, 3 * [{w/y + w/y+2}] = (5/4)w
Take out w common and move 3 to denominator of RHS
w{1/y + 1/(y+2)} = (5/12)w
Canceling w from both sides
{1/y + 1/(y+2)} = 5/12
2y+2/y(y+2) = 5/12
24y+24=5y^2 + 10y
5y^2-14y-24=0
5y^2-20y+6y-24=0
5y(y-4)+6(y-4)=0
(5y+6)+(y-4)=0
y=-6/5 or y=4
Discarding y=-6/5 as no of days cannot be negative
y=4
hence it takes Y, 4 days to produce w widgets.
Therefore, it will take X (4+2)=6 days to produce w widgets.
Hence it will take X 3*6=18 days to produce 3w widgets.
Answer : C
Question: How many different possible arrangements X can be obtained from the letters G, M, A, T, I, I, and T, such that there is at least one character between both I's?
A.360
B.720
C.900
D.1800
Answer: C.900
Explanation: Total ways for arranging without restriction = 7!/ (2!2!) { 2! is becoz of two T's and other 2! for two I's)
Restriction : atleast one character between I's = Possible ways - both I's together i.e.o character between I's
_ _ _ _ _ (I I)
Both I's Together = 6! (Assuming 2 I's as one unit) /2!(for 2 T's) * 2! (No of arrangements of 2 I's)/2! (for 2 I's)
=6!/2!
Therefore ans X= 7!/ (2!2!) -6!/2! = 900
HENCE C.
Question: Average age of 18 men is decreased by 1 year when one man whose age is 49 years is replaced by a new man. Find age of the new man
A.31
B.77
C.88
D.29
Answer: A.31
Explanation: Explanation:
If the replaced persons age is same with the existing average, there is no change in the average. But by replacement overall decrease in the age is 18 × 1 = 18 years. This is the change bought by the new man.
Age of new man = Age of man replaced - Total decrease in age
= 49 - (1 × 18) = 31 years
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Answer:A