mathewhe/aqua_rat huggingface dataset:
Question: In a competitive examination in State A, 6% candidates got selected from the total appeared candidates. State B had an equal number of candidates appeared and 7% candidates got selected with 82 more candidates got selected than A. What was the number of candidates appeared from each State?
(A) 8400
(B) 8200
(C) 6000
(D) 5000
(E) 4000
Rationale: State A and State B had an equal number of candidates appeared.
In state A, 6% candidates got selected from the total appeared candidates
In state B, 7% candidates got selected from the total appeared candidates
But in State B, 82 more candidates got selected than State A
From these, it is clear that 1% of the total appeared candidates in State B = 82
=> total appeared candidates in State B = 82 x 100 = 8200
=> total appeared candidates in State A = total appeared candidates in State B = 8200
Answer: (B) 8200
Question: The list price of an article is Rs.65. A customer pays Rs.56.16 for it. He was given two successive discounts, one of them being 10%. The other discount is?
(A) 3%
(B) 4%
(C) 5%
(D) 6%
(E) 7%
Rationale: 65*(90/100)*((100-x)/100) = 56.16
x = 4%
ANSWER:B
Answer: (B) 4%
Question: A container holding 12 ounces of a solution that is 1 part alcohol to 2 parts water is added to a container holding 4 ounces of a solution that is 1 part alcohol to 3 parts water. What is the ratio of alcohol to water in the resulting solution?
(A) 2:5
(B) 3:7
(C) 3: 5
(D) 5:11
(E) 7:3
Rationale: Container 1 has 12 ounces in the ratio 1:2
or,
x+2x=12 gives x(alcohol)=4 and remaining water =8
container 2 has 4 ounces in the ratio 1:3
or,
x+3x=4 gives x(alcohol)=1 and remaining water =3
mixing both we have alcohol=4+1 and water =8+3
ratio thus alcohol/water =5/11
Answer D
Answer: (D) 5:11
Question: By using the numbers 1, 2, 3, 5 and 9 only once, how many five digit numbers can be made that are divisible by 25?
(A) 0
(B) 1
(C) 4
(D) 8
(E) 6
Rationale: A number to be divisible by 25 must end with 00, 25, 50, or 75.
So, there are the following cases possible:
139-25 --> the first three digits can be arranged in 3=6 ways.
Total = 6.
Answer: E.
Answer: (E) 6
Question: A man can do a piece of work in 5 days, but with the help of his son he can do it in 2 days. In what time can the son do it alone ?
(A) 7 1/2 days
(B) 6 1/2 days
(C) 5 1/3 days
(D) 3 1/3 days
(E) None of these
Rationale: Explanation:
In this type of question, where we have one person work and together work done. Then we can easily
get the other person work just by subtracting them. As
Son's one day work =
(1/2−1/5)=(5−2)/10=3/10
So son will do whole work in 10/3 = 3 1/3 days
Answer: D
Answer: (D) 3 1/3 days
Question: If p and q are positive integers and q ≠ 1, then pq(q −1) is
(A) p+q
(B) p-q
(C) pq^2-pq
(D) pq^2
(E) pq
Rationale: pq(q-1) = pq^2-pq
Answer : C
Answer: (C) pq^2-pq
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 sold T was 25,200, how many of them were sold at full price?
(A) 21,000
(B) 19,800
(C) 16,500
(D) 15,800
(E) 8,700
Rationale: 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
Answer: (C) 16,500
Question: If 55 percent of congress are Democrats and the rest are Republicans. In a vote on a new law , if 40 percent of the members who are Democrats and 20 percent of the members who are Republicans are expected to vote for the new law , what percent of the members are expected to vote for the new law ?
(A)  30%
(B)  31%
(C)  33%
(D)  35%
(E)  39%
Rationale: Say there are total of 100 members in the congress . Thus 55 are Democrats and 45 are Republicans.
55*0.4=22 Democrats are expected to vote for the new law ;
45*0.20=9 Republicans are expected to vote for the new law.
Thus total of 22+9=31 members are expected to vote for the new law, which is 31% of the total number of congress members .
Answer: B.
Answer: (B)  31%
Question: In a competitive examination in a state A, 6% candidates got selected from the total appeared candidates. State B had an equal number of candidates appeared and 7% candidates got selected with 80 more candidates got selected than A. What was the number of candidates appeared from each state?
(A) 8000
(B) 7540
(C) 6500
(D) 9100
(E) 6000
Rationale: Let the number of candidates appeared be x
then, 7% of x - 6% of x = 80
1% of x = 80
x = 80*100 = 8000
Answer is A
Answer: (A) 8000
Question: If one person completes a journey in 10 hrs. He travels 1st half of the journey at the rate of 21 km/hr and 2nd half at therate of 24 km/hr. Find the total journey in km.
(A) 200 km
(B) 220 km
(C) 224 km
(D) 230 km
(E) 254 km
Rationale: distance = speed x time
Let time taken to travel the first half = x hr
then time taken to travel the second half = (10 - x) hr
Distance covered in the the first half = 21x
Distance covered in the the second half = 24(10 - x)
But distance covered in the the first half = Distance covered in the the second half
=> 21x = 24(10 - x)
=> 21x = 240 - 24x
=> 45x = 240
=> 9x = 48
=> 3x = 16
⇒x=163Hence Distance covered in the the first half = 21x=21×163=7×16=112 kmTotal distance = 2×112=224 km
C
Answer: (C) 224 km
Question: A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required?
(A) 40 ft
(B) 34 ft
(C) 68 ft
(D) 78 ft
(E) 88 ft
Rationale: We have: l = 20 ft and lb = 680 sq. ft.
So, b = 34 ft.
Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.
ANSWER:E
Answer: (E) 88 ft
Question: Train X crosses a stationary train Y in 60 seconds and a pole in 25 seconds with the same speed. The length of the train X is 300 m. What is the length of the stationary train Y?
(A) 228
(B) 420
(C) 882
(D) 661
(E) 222
Rationale: Let the length of the stationary train Y be LY
Given that length of train X, LX = 300 m
Let the speed of Train X be V.
Since the train X crosses train Y and a pole in 60 seconds and 25 seconds respectively.
=> 300/V = 25 ---> ( 1 )
(300 + LY) / V = 60 ---> ( 2 )
From (1) V = 300/25 = 12 m/sec.
From (2) (300 + LY)/12 = 60
=> 300 + LY = 60 (12) = 720
=> LY = 720 - 300 = 420 m
Length of the stationary train = 420 m
Answer:B
Answer: (B) 420
Question: Two persons start at the same point, walk in opposite directions with 6 km/hr and 4 km/hr respectively.
What is the distance separated after 4 and half hrs?
(A) 45 km
(B) 48 km
(C) 50 km
(D) 29.250 km
(E) 25.250 km
Rationale: As the two persons are moving in the opposite direction, so they will be separateed in 1 hour = 6 + 4 = 10 km.
They will be separated in 4.5 hours = 10 * 4.5 = 45 km
ANSWER:A
Answer: (A) 45 km
Question: Simple interest on a sum at 4% per annum for 2 years is Rs.80. The C.I. on the same sum for the same period is?
(A) 81.6
(B) 88.6
(C) 81.1
(D) 81.5
(E) 81.2
Rationale: SI = 40 + 40
CI = 40 + 40 + 1.6 = 81.6
Answer: A
Answer: (A) 81.6
Question: In a BCCI meeting there are 11 people. All shake hands with each other once. How many handshakes will be there altogether?
(A) 28
(B) 10
(C) 11
(D) 19
(E) 18
Rationale: There will be n-1 handshakes as A person will shake hands with all the other members.
=11-1=10
Answer:B
Answer: (B) 10
Question: Simplify:(4x2 - 2x) - (-5x2 - 8x)
(A)  2x(3x + 2)
(B)  4x(3x + 2)
(C)  2-(3x + 2)
(D)  3x(3x - 2)
(E)  3x(3x + 2)
Rationale: (4x2 - 2x) - (-5x2 - 8x)
= 4x2 - 2x + 5x2 + 8x.
= 4x2 + 5x2 - 2x + 8x.
= 9x2 + 6x.
= 3x(3x + 2).
Answer: (E) 3x(3x + 2)
Answer: (E)  3x(3x + 2)
Question: A train moves fast a telegraph post and a bridge 260 m long in 8 sec and 20 sec respectively. What is the speed of the train?
(A) 79.5
(B) 39.5
(C) 79.9
(D) 78.0
(E) 79.0
Rationale: Let the length of the train be x m and its speed be y m/sec.
Then, x/y = 8 => x = 8y
(x + 260)/20 = y
y = 21.67
Speed = 21.67m/sec = 21.67 * 18/5 = 78 km/hr.
Answer:D
Answer: (D) 78.0
Question: 20 is subtracted from 60% of a number, the result is 88. Find the number?
(A) 120
(B) 300
(C) 180
(D) 170
(E) 148
Rationale: (60/100) * X â€“ 20 = 88
6X = 1080
X = 180
Answer:C
Answer: (C) 180
Question: The sum of the ages of a father and son is 45 years. Five years ago, the product of their ages was four times the father’s age at that time. The present age of father and son, repectively are
(A) 25 years , 10 years
(B) 36 years, 9 years
(C) 39 years, 6 years
(D) 40 years, 6 years
(E) None of these
Rationale: Let son’s age = x years. The father’s age = (45 - x) years
we have (x - 5)(45 - x - 5) = 4(45 - x - 5)
=>x - 5 = 4
=> x = 9
Thus, ages are 36 years and 9 years.
ANSWER:B
Answer: (B) 36 years, 9 years
Question: A tank can supply water to a village for 80 days. If a leak at the bottom of the tank drains out 10 liters per day, the supply lasts for 60 days only. For how many days will the supply last if the leak drains out 20 liters per day?
(A) 42
(B) 45
(C) 48
(D) 50
(E) 55
Rationale: Losing 10 liters per day results in a loss of 600 liters in 60 days. So, those 600 liters were for 20 days, making daily consumption of the village 30 liters per day. Thus the capacity of the tank is 30*80 = 2400 liters.
Losing 20 liters plus 30 liters gives 50 liters per day. At this rate the supply will last 2400/50 = 48 days.
The answer is C.
Answer: (C) 48
Question: The ratio by weight, measured in pounds, of books to clothes to electronics in a suitcase initially stands at 7:4:3. Someone removes 6 pounds of clothing from the suitcase, thereby doubling the ratio of books to clothes. How many pounds do the electronics in the suitcase weigh?
(A) 1
(B) 3
(C) 5
(D) 7
(E) 9
Rationale: The weights of the items in the suitcase are 7k, 4k, and 3k.
If removing 6 pounds of clothes doubles the ratio of books to clothes, then 6 pounds represents half the weight of the clothes.
2k = 6 pounds and then k = 3 pounds.
The electronics weigh 3(3) = 9 pounds.
The answer is E.
Answer: (E) 9
Question: The rate of spin of a certain gyroscope doubled every 10 seconds from the moment a particular stopwatch started. If after a minute and a half the gyroscope reached a speed of 1600 meters per second, what was the speed, in meters per second, when the stopwatch was started?
(A) 25/3
(B) 25/4
(C) 25/8
(D) 25/16
(E) 25/32
Rationale: Let x be the original speed when the stopwatch was started.
In 90 seconds, the speed doubled 9 times.
2^9 * x = 1600
x = (2^6 * 25) / 2^9 = 25/8
The answer is C.
Answer: (C) 25/8
Question: In a market, a dozen eggs cost as much as a pound of rice, and a half-liter of kerosene costs as much as 8 eggs. If the cost of each pound of rice is $0.33, then how many E cents does a liter of kerosene cost? [One dollar has 100 cents.]
(A)  0.33
(B)  0.44
(C)  0.55
(D)  44
(E)  55
Rationale: Main thing to remember is answer is asked in cents, however when we calculate, it comes up as 0.44 $
Just multiply by 100, Answer E= 44.D
Answer: (D)  44
Question: If two dice are thrown together, the probability of getting an even number on one die and an odd number on the other is ?
(A) 1/9
(B) 1/2
(C) 1/6
(D) 3/2
(E) 1/4
Rationale: The number of exhaustive outcomes is 36.
Let E be the event of getting an even number on one die and an odd number on the other. Let the event of getting either both even or both odd then = 18/36 = 1/2
P(E) = 1 - 1/2 = 1/2.
Answer: B
Answer: (B) 1/2
Question: Students of three different classes appeared in common examination. Pass average of 10 students of first class was 55%, pass average of 15 students of second class was 60% and pass average of 25 students of third class was 80% then what will be the pass average of all students of three classes?
(A) 74%
(B) 75%
(C) 69%
(D) 72%
(E) None
Rationale: Solution:
Sum of pass students of first, second and third class,
= (55% of 10) + (60% of 15)+ (80% of 25)
= 5.5+9+20 = 34.5
Total students appeared,
= 10+15+25 = 50
Pass average,
= 34.5*100/50 = 69%.
Answer: Option C
Answer: (C) 69%
Question: If a farmer sells 5 of his chickens, his stock of feed will last for 4 more days than planned, but if he buys 20 more chickens, he will run out of feed 1 day earlier than planned. If no chickens are sold or bought, the farmer will be exactly on schedule. How many chickens does the farmer have?
(A) 15
(B) 20
(C) 20/3
(D) 25
(E) 10
Rationale: Let x = total feed required for the planned period
n= number of chicken
t = total time of the planned feed
x = nt
1) x = (n-5) * (t+4)
2) x = (n+20) * (t-1)
equating 1 &2
(n-5) * (t+4) = (n+20) * (t-1)
5n = 25t
n =5t
x= n * n/5
substituting this value in 1
n * n/5 = (n-5) * (n/5+4)
15n = 100
n =20/3
C
Answer: (C) 20/3
Question: A can do a piece of work in 20 days which B can do in 12 days. B worked at it for 9 days. A can finish the remaining work in ?
(A) 3 days
(B) 5 days
(C) 6 days
(D) 7 days
(E) 11 days
Rationale: B's 9 day's work = 9 x (1/12) = 3/4
Remaining work = (1 - 3/4) = 1/4
1/4 work is done by A in = 20 x (1/4) = 5 days.
answer : B
Answer: (B) 5 days
Question: Solve 8z – 5z + 5 = 2z + 11
(A) 5
(B) 2
(C) 10
(D) 8
(E) 6
Rationale: Sol.
8z – 5z + 5 = 2z + 11
=> 8z - 5z - 2z = 11 - 5
=> z = 6
Answer E
Answer: (E) 6
Question: If xy+3z = x(y+z), which of the following must be true?
(A) 1. x=3 and z=0
(B) 2. x=1 and y=1
(C) 3. y=1 and z=0
(D) 4. x=1 or y=0
(E) 5. x=1 or z=0
Rationale: xy+3z=x(y+z)
xy+3z=xy+xz
3z=xz
it means x=3 or z=0
Answer A
Answer: (A) 1. x=3 and z=0
Question: The difference between the local value and the face value of 4 in the numeral 98625147 is
(A) 36
(B) 63
(C) 34
(D) 30
(E) 33
Rationale: Explanation:
(Local value of 4) - (Face value of 4) = (40 - 4) = 36
A)
Answer: (A) 36
Question: 3 buffaloes eat as much as 4 cows or 2 oxen. At a farm there are 15 buffaloes, 8 oxen and 24 cows. The fodder for these cattle is available for 24days. If 8 more cows & 9 more buffaloes are brought in, how long will the fodder last?
(A) 10
(B) 13
(C) 15
(D) 18
(E) 21
Rationale: Sol. 2 oxen = 3 buffaloes = 4 cows 8 oxen = 16 cows and 15 buffaloes = 20 cows 16 cows + 20 cows + 24 cows = 60 cows Since fodder is available for 60 cows for 24 days
So fodder will be available for (60 + 8 + 12) = 80 cows c 24x60 .0 .
for-^—=18 days.
D
Answer: (D) 18
Question: Consider below info to be a table:
AMOUNT OF BACTERIA
TIME - AMOUNT
1PM - 10gms
4PM - x gms
7PM - 12.1 gms
Data for a certain biology experiment are given in the table above. If the amount of bacteria present increased by the same factor during each of the two 3-hour periods shown, how many grams of bacteria were present at 4PM ?
(A) 12.0
(B) 12.1
(C) 12.2
(D) 12.3
(E) 11.0
Rationale: let say the factor is x, the at 7PM quantity = 10 (i.e. quantity @1PM)* x^2 =12.1 => x^2 =1.21 or x = 1.1 => @4PM quantity = 10x =10*1.1 =11.
Answer E
Answer: (E) 11.0
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 8 times as many paperback copies as hardback copies were sold. If a total of 441,000 copies of the novel were sold in all, how many paperback copies were sold?
(A) 45,000
(B) 360,000
(C) 364,500
(D) 392,000
(E) 405,000
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)+(8x)=441 --> x=45.
So, the total # of paperback copies sold was 9x=9*45=405 or 405,000.
Answer: E.
Answer: (E) 405,000
Question: If two dice are thrown together, the probability of getting a doublet on the dice is
(A) 1/6
(B) 1/5
(C) 1/4
(D) 1/3
(E) 1/2
Rationale: The number of exhaustive outcomes is 36.
Let E be the event of getting doublet on the dies is 6/36 = 1/6
P(E) = 1/6.
A)
Answer: (A) 1/6
Question: The number obtained by interchanging the two digits of a two-digit number is less than the original number by 45. If the sum of the two digits of the number so obtained is 13, then what is the original number?
(A) 22
(B) 94
(C) 27
(D) 19
(E) 11
Rationale: Explanation:
Let the number be in the form of 10a + b
Number formed by interchanging a and b = 10b + a.
a + b = 13 --- (1)
10b + a = 10a + b - 45
45 = 9a - 9b => a - b = 5 --- (2)
Adding (1) and (2), we get
2a = 18 => a = 9 and b = 4
The number is: 94.
Answer:B
Answer: (B) 94
Question: An office manager must choose a five-digit lock code for the office door. The first and last digits of the code must be odd, and no repetition of the digits is allowed. How many different lock codes are possible.?
(A) 3120
(B) 3250
(C) 3275
(D) 3860
(E) 3980
Rationale: However, I think 6720 (the answer given in MGMAT) is wrong. Here's my reasoning. If you have (an) odd number(s) for your 2nd, 3rd and/or 4th digits, you cannot repeat that(those) numbers for the last digit. Therefore, this problem needs to be partitioned into 4 groups.
1) ODD (1st digit); 3 EVENs (2nd, 3rd, 4th digits); ODD (5th digit)
5 x 5x 4 x 3 x 4 = 1200
2) ODD (1st digit); 2 EVEN, 1 ODD (2nd, 3rd, 4th digits); ODD (5th digit)
5 x 5 x 4 x 4 x 3 = 1200
3) ODD (1st digit); 1 EVEN, 2 ODDs (2nd, 3rd, 4th digits); ODD (5th digit)
5 x 5x 4 x 3 x 2 = 600
4) ODD (1st digit); 3 ODDs (2nd, 3rd, 4th digits); ODD (5th digit)
5 x 4 x 3 x 2 x 1 = 120
Then the total # of combinations would be 1200 + 1200 + 600 + 120 = 3120.
ANS:A
Answer: (A) 3120
Question: In the number 11,0AB, A and B represent the tens and units digits, respectively. If 11,0AB is divisible by 52, what is the greatest possible value of B × A?
(A) 0
(B) 5
(C) 10
(D) 15
(E) 22
Rationale: You should notice that 52*2=110 so 11,000 is divisible by 55: 55*200=11,000 (or you can notice that 11,000 is obviously divisible by both 5 and 11 so by 55) --> B*A=0*0=0. Next number divisible by 55 is 11,000+55=11,055: B*A=5*5=22 (next number won't have 110 as the first 3 digits so we have only two options 0 and 25).
Answer: E.
!
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E
Answer: (E) 22
Question: For an employee to qualify for early retirement at a certain company, the sum of the employee’s age and years of service must be at least 78, If Sue was K years old when she was hired by the company, what is the minimum age at which she could possibly qualify for early retirement?
(A) K+39
(B) 2K+39
(C) (78+K)/2
(D) (78-K)/2
(E) 2(78-K)
Rationale: Let y be the years of service at the company.
After y years of service, Sue's age will be K + y.
Sue can retire when (K+y) + y = 78
y = (78 - K)/2
Then Sue's age will be K + y = K + (78 - K)/2 = (78+K) / 2
The answer is C.
Answer: (C) (78+K)/2
Question: A dealer offers a cash discount of 20% and still makes a profit of 20% when he further allows 10 articles to be sold at the cost price of 9 articles to a particular sticky bargainer. How much percent above the cost price were his articles listed?
(A) 50%
(B) 55%
(C) 60%
(D) 66+(2/3)%
(E) 80%
Rationale: Given
Cash Discount - 20%
Profit - 20%
Items Sold - 10
Price Sold at = List Price of 9
Assume List price = $10
Total Invoice = $90
-20% Cash Discount = $72
Let cost price of 10 items be x
So total cost = 10*x
Given the shopkeeper had a profit of 20%
10 * x * 120/100 = 72
or x = $6
Which means his products were listed at $10 which is a 66+(2/3)% markup over $6
Answer D
Answer: (D) 66+(2/3)%
Question: Juan is a gold medalist in athletics. In the month of May, if Juan takes 8 seconds to run y yards, how many seconds will it take him to run x yards at the same rate?
(A) 8x/y
(B) 11x/y
(C) 9x/y
(D) 12x/y
(E) 13x/y
Rationale: velocity=distance/time;
since he ran at same rate both d tyms,
y/8=x/t
so t=8x/y
ANSWER:A
Answer: (A) 8x/y
Question: Two pipes P and Q can fill a cistern in 11 and 15 minutes respectively. Both are opened together, but at the end of 3 minutes the first is turned off. How much longer will the cistern take to fill?
(A) 1/8
(B) 1/4
(C) 2/4
(D) 10/11
(E) 1/4
Rationale: 3/11 + x/15
= 1
x= 10 10/11
Answer: D
Answer: (D) 10/11
Question: Find the average of all even numbers upto 75.
(A) 35
(B) 36
(C) 38
(D) 34
(E) 37
Rationale: Average of all even numbers upto 75 = [ 35/2* ( first even number +
greatest even number before 75) ]/35
= ½ *(2 +74 )
= 76/2
= 38.
ANSWER:C
Answer: (C) 38
Question: The weights of three boys are in the ratio 4 : 5 : 6. If the sum of the weights of the heaviest and the lightest boy is 45 kg more than the weight of the third boy, what is the weight of the lightest boy?
(A) 39 kg
(B) 36 kg
(C) 32 kg
(D) 31 kg
(E) 96 kg
Rationale: Let the weights of the three boys be 4k, 5k and 6k respectively.
4k + 6k = 5k + 45
=> 5k = 45 => k = 9
Therefore the weight of the lightest boy
= 4k = 4(9) = 36 kg.
Answer:B
Answer: (B) 36 kg
Question: The simple interest and the true discount on a certain sum for a given time and at a given rate are Rs. 88 and Rs. 80 respectively. The sum is:
(A) 880
(B) 1450
(C) 1600
(D) 1800
(E) None
Rationale: Sol.
Sum = S.I. * T.D. / (S.I)-(T.D.) = 88 * 80 / (88 - 80) = Rs.880.
Answer A
Answer: (A) 880
Question: Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 21 seconds. The ratio of their speeds is?
(A) 3/6
(B) 3/2
(C) 2/3
(D) 3/5
(E) 5/2
Rationale: Let the speeds of the two trains be x m/sec and y m/sec respectively. Then, length of the first train
= 27 x meters, and length of the second train
= 17 y meters. (27 x + 17 y) / (x + y) = 21 ==> 27 x + 17 y = 21 x + 21 y
==> 6 x = 4 y ==> x/y
= 2/3.
Answer:C
Answer: (C) 2/3
Question: 6 letters have to be put into their 6 respective envelopes. If the letters are inserted randomly into the envelopes, what is the probability that all the letters go into their respective envelopes?
(A) 1/720
(B) 1/60
(C) 1/120
(D) 1/240
(E) 3/5
Rationale: A-B-C-D-E (envelopes)
a-b-c-d-e (letters)
6 letters (a, b, c, d, and e) can be assigned to 6 envelopes (arranged) in 6! ways, out of which only one will be correct, thus the probability is 1/6! = 1/720.
Answer: A
Answer: (A) 1/720
Question: In a bag there are coins of 50 paisa, 25 paisa and one rupee in the proportion 5:6:2. If there are in all Rs.34, the number of 25 paisa coins is?
(A) 30
(B) 31
(C) 32
(D) 34
(E) 38
Rationale: 5x 6x 2x
50 25 100
250x + 150x + 200x = 3400
600x = 3400
x = 5.66 => 6x = 34
ANSWER:D
Answer: (D) 34
Question: The expression (11.98 × 11.98 + 11.98 × x + 0.02 × 0.02) will be a perfect square for x equal to
(A) 0.02
(B) 0.2
(C) 0.04
(D) 0.4
(E) 4
Rationale: Solution
Given expression = (11.98)2 + (0.02)2 + 11.98 × x.
For the given express to be a perfect square, we must have
11.98 × x = 2 × 11.98 × 0.02 or x = 0.04. Answer C
Answer: (C) 0.04