mathewhe/aqua_rat huggingface dataset:
Question: What was the day of the week on 4th April, 2006?
(A) Saturday
(B) Thursday
(C) Monday
(D) Friday
(E) Tuesday
Rationale: 28 May, 2006 = (2005 years + Period from 1.1.2006 to 28.5.2006)
Odd days in 1600 years = 0
Odd days in 400 years = 0
5 years = (4 ordinary years + 1 leap year) = (4 x 1 + 1 x 2) 6 odd days
Jan. Feb. March April
(31 + 28 + 31 + 4) = 94 days
94 days = (13 weeks + 3 day) 1 odd day.
Total number of odd days = (0 + 0 + 6 + 3) = 9 0 odd day.
Given day is Tuesday.
Answer: Option E
Answer: (E) Tuesday
Question: Each of the integers from 1 to 17 is written on the a seperate index card and placed in a box. If the cards are drawn from the box at random without replecement, how many cards must be drawn to ENSURE that the product of all the integers drawn is even?
(A) 19
(B) 12
(C) 11
(D) 10
(E) 3
Rationale: Out of the 17 integers: 9 are odd and 8 are even.
If we need to make sure that the product of all the integers withdrawn is even then we need to make sure that we have at least one even number. In the worst case:
1. we will end up picking odd numbers one by one, so we will pick all 9 odd numbers first
2. 10th number will be the first even number
So we need to withdraw at least 10 numbers to make sure that we get one even number and the product of all the integers picked is even.
So, answer will be 10. (D)
Answer: (D) 10
Question: In the last set of tests, a student scored marks in each of 5 subjects in the ratio 4:5:6:7:8 and got an overall average score of 54%. If a mark of 50% in a subject is a pass and the maximum marks of each subject is the same, how many subjects did the student pass?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
Rationale: The tests scores are 4k, 5k, 6k, 7k, and 8k.
The sum of the test scores is 30k.
The average test score is 30k/5 = 54. Then k = 9.
The test scores are 36, 45, 54, 63, and 72.
The student passed 3 subjects.
The answer is C.
Answer: (C) 3
Question: The average age of seven persons sitting in a row facing east is 28 years. If the average age of the first three persons is 21 years and the average age of the last three persons is 34 years, then find the age of the person sitting in the middle of the row?
(A) 15 years
(B) 31 years
(C) 27 years
(D) 82 years
(E) 12 years
Rationale: Total age seven persons = (28 * 7)years
Total age of the first three persons and the last three persons are (21 * 3) years and (34 * 3) years respectively.
Age of the person sitting in the middle of the row
= 28 * 7 - 21 * 3 - 34 * 3 = 196 - 63 - 102
= 31 years.
Answer: B
Answer: (B) 31 years
Question: Find the simple interest on the Rs. 2000 at 25/4% per annum for the period from 4th Feb 2005 to 18th
April 2005
(A) Rs 25
(B) Rs 30
(C) Rs 35
(D) Rs 40
(E) None of these
Rationale: Explanation:
One thing which is tricky in this question is to calculate the number of days.
Always remember that the day on which money is deposited is not counted while the day on which
money is withdrawn is counted.
So lets calculate the number of days now,
Time = (24+31+18) days = 73/365 years = 1/5 years
P = 2000
R = 25/4%
S.I. = =2000×254×5×100=25
Answer: A
Answer: (A) Rs 25
Question: If 'A' means '-', 'B' means '/', 'C' means '+', and 'D' means 'X' then
Find the value of the following:
21 B 7 C 9 D 10 A 13
(A) 80
(B) 40
(C) 60
(D) 70
(E) 50
Rationale: A
80
Explanation:
21 B 7 C 9 D 10 A 13
= 21 / 7 + 9 X 10 - 13
= 3 + 90 - 13
= 80
Answer: (A) 80
Question: At what rate of compound interest per annum will a sum of Rs. 1200 become Rs. 1348.32 in 2 years
(A) 3%
(B) 4%
(C) 5%
(D) 6%
(E) 7%
Rationale: Explanation:
Let Rate will be R%
1200(1+R/100)2=134832 /100
(1+R/100)2=134832/120000
(1+R/100)2=11236/10000
(1+R/100)=106/100
=>R=6%
Option D
Answer: (D) 6%
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 5w widgets?
(A) 4
(B) 30
(C) 8
(D) 10
(E) 12
Rationale: 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 5*6=30 days to produce 2w widgets.
Answer : B
Answer: (B) 30
Question: In the quadratic equation ax2+ 1x - 6 = 0, if the sum of two roots is -1, what is the product of the two roots?
(A) 4
(B) 5
(C) 6
(D) -6
(E) 8
Rationale: Explanation:
The sum of the roots of the quadratic equation ax2 + bx + c = 0 are (-b/a) and the product of the roots are (c/a).
Thus, in the equation ax2+ x - 6 = 0, where a = a, b = 1 and c = -6.
we get, sum of the roots = -1 / a = -1
a = 1
Product of the roots = -6 / 1 = -6
ANSWER: D
Answer: (D) -6
Question: If r>0.9, which of the following options can be the value of r?
(A) 0.9^2
(B) √(0.9)
(C) 0.09
(D) 0.9
(E) 0.9^3
Rationale: Lets take a few examples
When .2 is squared it gives .04
when .5 is squared it gives .25
When .9 is squared it gives .81
So raising a fraction to its power of 2 or 3 only reduces the value of the fraction
This is because
.2 = 2/10
.2^2= 4/100
.2^3=8/1000
Since the denominator gets an extra 0 each time u increase the power , it is going to drag the value down
But if you just think about the opposite case :
Then we can see , if squaring a fraction diminish its value, then taking the square root of it will again increase its value
that is √(0.04) will give you back 0.2
so √(0.04) > 0.04
so obviously diminishing the power will appreciate the value
So r √(0.9) > 0.9
Hence Correct Answer is (B)
Answer: (B) √(0.9)
Question: Five friends go to see a movie at the local movie theater and sit in a row together. If there are two friends who refuse to sit next to each other, how many 5-person seating arrangements are possible?
(A) 72
(B) 90
(C) 120
(D) 140
(E) 180
Rationale: The number of ways to arrange 5 people in a row is 5! = 120
From this we want to subtract all the arrangements where the two friends are next to each other. If we consider them as one unit, then the number of arrangements of this 2-person unit and the 3 other friends is 4! And the number of arrangements of the two friends within this unit are 2 because the two friends can switch places. So the total number of arrangements of the 5 friends with the 2-person unit together is 4!*2
The total number of permutations where these two friends are not seated next to each other is 5! - 4!*2 = 120 - 48 = 72
The answer is A.
Answer: (A) 72
Question: 5020−(502÷100.4)=?
(A) 15
(B) 20
(C) 5015
(D) 25
(E) 35
Rationale: Explanation:
=5020−(502/1004×10)
=5020−5=5015
Option C
Answer: (C) 5015
Question: P software has coding line 5% more than N, N software has coding line 5/2 more than M. M software has 108 lines of coding. find P lines.
(A) 106
(B) 107
(C) 397
(D) 109
(E) 110
Rationale: M s/w has 108 line of code
N s/w has = 108+108*5/2= 378 line of code
P s/w 5% more N' code 378+18.9=396.9
or
397 line of code
ANSWER:C
Answer: (C) 397
Question: When 5 fair coins are tossed simultaneously, in how many outcomes will the first and the last coin turn up heads?
(A) 1
(B) 3
(C) 8
(D) 31
(E) 30
Rationale: With 5 coins being tossed, you might expect this question to be more difficult than it actually is. The prompt asks for the number of outcomes in which the first and fifth coins are both heads. To start, we can 'lock in' those options.
H _ _ _ H
Now we just have to focus on the possible outcomes for the remaining 3 coins. Mathematically, since each coin has 2 possible outcomes, there are (2)(2)(2) = 8 options.
You could also 'map out' the possibilities:
HHH
HHT
HTH
THH
TTT
TTH
THT
HTT
8 total options.
Final Answer:C
Answer: (C) 8
Question: If z is a multiple of 8902, what is the remainder when z^2 is divided by 4?
(A) 0
(B) 1
(C) 2
(D) 3
(E) It depends on z.
Rationale: 2 is a factor of 8902, so 2 is a factor of z.
Then 2^2 = 4 is a factor of z^2.
Then the remainder when z^2 is divided by 4 is 0.
The answer is A.
Answer: (A) 0
Question: Sunita goes 30 km towards North from a fixed point, then after turning to her right she goes 15 km. After this she goes 30 km after turning to her right. How far and in what direction is she from her starting point?
(A) 45 km, East
(B) 15 km, East
(C) 45 km, West
(D) 45 Km, North
(E) None of these
Rationale: Sunita moves 15 km right after she was moving towards north. That is she walked 15 kms towards east. Then she moves 30 kms right i.e. se comes in the same direction from which she started. Hence the displacement is only 15 km and that is towards east.
ANSWER:B
Answer: (B) 15 km, East
Question: If 4 men or 6 boys can finish a work in 20 days. How long will 6 men and 11 boys take to finish the same work?
(A) 10 days
(B) 6 days
(C) 4 days
(D) 3 days
(E) 8 days
Rationale: 4 men = 6 boys
Then, 1 boy = 4/6 men = 2/3 men
11 boys = 2/3 × 11 men = 22/3 men
Then, 6 men + 11 boys = 6 + 22/3 men = 40/3 men
1 work done = 4 men × 20days
That is, 4 × 20 = 40/3 × ? days
? days = 4 × 20 × 3/40 = 6 days
Answer : B
Answer: (B) 6 days
Question: Mr. James sold two pipes at $1.20 each. Based on the cost, his profit one was 20% and his loss on the other was 20%. On the sale of the pipes, he:
(A) broke even
(B) lost 4 cents
(C) gained 4 cents
(D) lost 10 cents
(E) gained 10 cents
Rationale: A
broke even
20 % profit on $ 1.20
= $ 20/100 × 1.20
= $ 0.20 × 1.20
= $ 0.24
Similarly, 20 % loss on $ 1.20
= $ 20/100 × 1.20
= $ 0.20 × 1.20
= $ 0.24
Therefore, in one pipe his profit is $ 0.24 and in the other pipe his loss is $ 0.24.
Since both profit and loss amount is same so, it’s broke even.
Answer: (A) broke even
Question: A man sitting in a train which is traveling at 40 kmph observes that a goods train, traveling in opposite direction, takes 9 seconds to pass him. If the goods train is 280 m long, find its speed.?
(A) 50 kmph
(B) 58 kmph
(C) 62 kmph
(D) 65 kmph
(E) 72 kmph
Rationale: Relative speed = 280/9 m / sec = ((280/9)*(18/5)) kmph = 112 kmph.
Speed of goods train = (112 - 40) kmph = 72 kmph.
Answer : E
Answer: (E) 72 kmph
Question: A train travels m miles in n hours. How many miles will it travel in x hours at the same rate?
(A) mnx
(B) m⁄n
(C) mx⁄n
(D) mx
(E) mn⁄x
Rationale: Given Speed = m/n
To Find distance in x hours with same speed
Distance = speed * time
Distance = m/n*x = mx/n
Answer C
Answer: (C) mx⁄n
Question: A tradesman by means of his false balance defrauds to the extent of 20%? in buying goods as well as by selling the goods. What percent does he gain on his outlay?
(A) 44%
(B) 48%
(C) 94%
(D) 45%
(E) 34%
Rationale: g% = 20 + 20 + (20*20)/100
= 44%
Answer: A
Answer: (A) 44%
Question: If a and b are positive integers and x = 4^a and y = 3^b, which of the following is a possible units digit of xy?
(A) 0
(B) 1
(C) 5
(D) 7
(E) 8
Rationale: The units digit of 4^a is either 4 or 6.
The units digit of 3^b is 3, 9, 7, or 1.
We can multiply the units digits together to see that the units digits of xy could be 2, 4, 6, or 8.
The answer is E.
Answer: (E) 8
Question: A, B and C are partners. A receives 2/3 of profits, B and C dividing the remainder equally. A's income is increased by Rs.300 when the rate to profit rises from 5 to 7 percent. Find the Capital of A?
(A) 5000
(B) 10000
(C) 15000
(D) 2778
(E) 1991
Rationale: A:B:C = 2/3:1/6:1/6 = 4:1:1
x * 2/100 * 2/3 = 300
A's capital = 22500*2/3 = 15000
Answer: C
Answer: (C) 15000
Question: At what rate percent per annum will the simple interest on a sum of money be 2/5 of the amount in 10 years?
(A) 4 %
(B) 5 %
(C) 6 %
(D) 7 %
(E) 8 %
Rationale: Sol.
Let sum = x. Then, S.I. = 2x/5, Time = 10 years.
∴ Rate = [100 * 2x / x*5*10]% = 4%.
Answer A
Answer: (A) 4 %
Question: The calendar of the year 2028 can be used again in the year?
(A) 2041
(B) 1951
(C) 2056
(D) 1971
(E) 1973
Rationale: Explanation:
Given year 2028 when divided by 4, leaves a remainder 0.
NOTE: When remainder is 0, 28 is added to the given year to get the result.
So, 2028 + 28 = 2056
Answer: C
Answer: (C) 2056
Question: If the cost price of 60 articles is equal to the selling price of 50 articles, then what is the percent profit?
(A) 10%
(B) 15%
(C) 20%
(D) 25%
(E) 30%
Rationale: Let x be the cost price of one article.
Let y be the selling price of one article.
50y = 60x
y = 1.2x
The answer is C.
Answer: (C) 20%
Question: Tabby is training for a triathlon. She swims at a speed of 1 mile per hour. She runs at a speed of 11 miles per hour. She wants to figure out her average speed for these two events. What is the correct answer for her?
(A) 6 mph
(B) 5.25 mph
(C) 3.5 mph
(D) 4 mph
(E) 0.5 mph
Rationale: (1 mph +11 mph)/2 = 6 mph
Correct option is: A
Answer: (A) 6 mph
Question: The sides of a triangle are in the ratio 5: 12: 13 and its perimeter is 300 m, its area is?
(A) 3000 sq m
(B) 1878 sq m
(C) 2789 sq m
(D) 26811 sq m
(E) 29762 sq m
Rationale: 5x + 12x + 13x = 300 => x = 10
a = 50, b = 120, c = 130
S = (50 + 120 + 130)/2 = 150
3000 sq m
Answer:A
Answer: (A) 3000 sq m
Question: A dealer sold two of his cattle for Rs. 500 each. On one of them he lost 10% on the other, he gained 10%. His gain or loss percent in the entire transaction was:
(A) 2
(B) 1
(C) 7
(D) 9
(E) 0
Rationale: Explanation:
Loss%= (common gain or loss % / 10)2 = (10/10)² % = 1%.
Answer: B) 1% loss
Answer: (B) 1
Question: 47% of the students in a class of 34 students has glasses or contacts. How many students in the class have either glasses or contacts?
(A) 15
(B) 16
(C) 17
(D) 18
(E) 19
Rationale: a=r⋅b
a=r⋅b
47%=0.47a
47%=0.47a
=0.47⋅34
=0.47⋅34
a=15.98≈16
Option B is correct.
Answer: (B) 16
Question: A certain clock marks every hour by striking a number of times equal to the hour, and the time require for a stroke is exactly equal to the time interval between strokes. At 6:00 the time lapse between the beginning of the first stoke and the end of the last stroke is 22 seconds. At 7:00, how many seconds elapse between the beginning of the first stroke and the end of the last stroke?
(A) 26
(B) 29
(C) 19
(D) 30
(E) 40
Rationale: at 6'o clock, there would be 6 strikes. First strike, then a short interval, the second strike, then a short interval and so on till the 6th strike. So there would be in all 5 intervals between 6 strikes. Similarly, between 7 strikes, there would be 6 intervals.
According to the question, the time spent in the strike and the interval is same.
At 6'o clock, the 6 strikes and the 5 intervals together take 22 sec so each strike and each interval takes 2 secs.
At 7'o clock, the 7 strikes and 6 intervals will take 2*(7+6) = 26 secs
A
Answer: (A) 26
Question: A does a work in 15 days and B does the same work in 30 days. In how many days they together will do the same work ?
(A) 5 days
(B) 6 days
(C) 8 days
(D) 10 days
(E) None of these
Rationale: Explanation:
Firstly we will find 1 day work of both A and B, then by adding we can get collective days for them,
So,
A's 1 day work = 1/15
B's 1 day work = 1/30
(A+B)'s 1 day work =
(1/15+1/30)=1/10
So together they can complete work in 10 days.
Answer: D
Answer: (D) 10 days
Question: What is theleast number should be added to 1056, so the sum of the number is completely divisible by 23?
(A) 20
(B) 23
(C) 25
(D) 27
(E) 28
Rationale: (1056 / 23) gives remainder 21
21 + 2 = 23, So we need to add 2
B
Answer: (B) 23
Question: We give the following information about a race hedge.
There are 10 hurdles. The distance between consecutive two lines (that is to say which follow) is 9.14 m. There are 13.72 m between the starting line and the first line and the last hurdle between 14.02 m and line arrival. Each hurdle measures 106 cm in height.
What is the length(in centimeters) of the track?
(A) 14060cm.
(B) 12060cm.
(C) 15060cm.
(D) 11060cm.
(E) None
Rationale: Solution:
[13.72+(9.14*9)+14.02+[(106÷100)*10]] =>L=[13.72+82.26+14.02+10.6] =>L=120.6m = 12060cm.
Answer B
Answer: (B) 12060cm.
Question: The mass of the sun is approximately 2 × 10^30 kg and the mass of the moon is approximately 9 × 10^12 kg. The mass of the sun is approximately how many times the mass of the moon?
(A) 4.0 × 10^(−18)
(B) 2.5 × 10^17
(C) 4.0 × 10^18
(D) 2.5 × 10^19
(E) 3.33 × 10^17
Rationale: Mass of sun = x * Mass of moon
x = Mass of sun / Mass of moon = (3 × 10^30) / (9 × 10^12) = 3*3^-2*10^18 = 10^18/3 = 3.33*10^17
Ans. E) 3.33 × 10^17
Answer: (E) 3.33 × 10^17
Question: Diana took out a charge account at the General Store and agreed to pay 6% simple annual interest. If she charges $60 on her account in January, how much will she owe a year later, assuming she does not make any additional charges or payments?
(A) $62.40
(B) $62.80
(C) $63.20
(D) $63.60
(E) $64.00
Rationale: 1.06*$60=$63.60
The answer is D.
Answer: (D) $63.60
Question: There are 7 girls and 3 chairs. In how many ways girls can occupy the chairs?
(A) 760
(B) 210
(C) 560
(D) 740
(E) 800
Rationale: total girls = 7
total chairs = 3
1st chair can be occupy in 7ways
2nd chair can be occupy in 6ways
3rd chair can be occupy in 5ways
total number of ways = 7*6*5 = 210 ways
correct option is B
Answer: (B) 210
Question: Find the simple interest on $500 for 3 years at 10% per annum?
(A) $250
(B) $300
(C) $500
(D) $150
(E) $100
Rationale: SI = PTR/100
= 500*3*10/100 = $150
Answer is D
Answer: (D) $150
Question: A bag contains 64 balls of eight different colours. There are eight of each colour (including red). What is the least number you would have to pick, without looking, to be sure of selecting 3 red balls?
(A) 54
(B) 60
(C) 45
(D) 38
(E) 59
Rationale: E
59
The first 56 balls could be of all colours except red. This would leave 8 balls, all of which are red, so any three chosen would be red.
Answer: (E) 59
Question: In a group of 24 students, each student is registered for at least one of three classes – History, Math and English. 15 students are registered for History, 15 students are registered for Math, and 20 students are registered for English. If only three students are registered for all three classes, how many students are registered for exactly two classes?
(A) 13
(B) 10
(C) 9
(D) 20
(E) 7
Rationale: Each student is registered forat least one of three classesmeans thatthere are no students who are registered for none of the classes.
Total = {people in group A} + {people in group B} + {people in group C} - {people in exactly 2 groups} - 2*{people in exactly 3 groups} + {people in none of the groups}:
24 = 15 + 15 + 20 - {people in exactly 2 groups} - 2*3 + 0-->{people in exactly 2 groups}=20
Answer: D
Answer: (D) 20
Question: The average weight of A, B and C is 45 kg. If the average weight of A and B be 40 kg and that of B and C be 46 kg, then the weight of B is:
(A) 17 kg
(B) 20 kg
(C) 26 kg
(D) 31 kg
(E) 37 kg
Rationale: let d sum of a, b, c is 3*45=135
and sum of a and b s 2*40=80
sum of b and c is 2 *46=92
hence 80+92-144=37
ans=37
ANSWER:E
Answer: (E) 37 kg
Question: From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?
(A) 756
(B) 800
(C) 956
(D) 850
(E) 650
Rationale: We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).
Required number of ways	= (7C3 x 6C2) + (7C4 x 6C1) + (7C5)
= (525 + 210 + 21)
= 756.
ANSWER A
Answer: (A) 756
Question: If you divide 7^115 by 5, which remainder do you get?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
Rationale: The units digit of 7^1 is 7.
The units digit of 7^2 is 9.
The units digit of 7^3 is 3.
The units digit of 7^4 is 1.
The units digit of 7^5 is 7.
etc...
The units digit of 7^n repeats in blocks of 4: {7, 9, 3, 1}
The remainder of 115/4 is 3.
So the units digit of 7^115 is 3.
The remainder of 7^115 / 5 is 3.
The answer is D.
Answer: (D) 3
Question: There is a train and car. The ratio between the speed of a train & a car is 24:19 respectively. Also, a bus covered a distance of 320 km in 5 hours. The speed of the bus is 2/3rd of the speed of the train. How many kilometers will the car cover in 5 hours?
(A) 320
(B) 340
(C) 360
(D) 380
(E) 400
Rationale: The speed of the bus is 320/5 = 64 km/hr
The speed of the train is (64*3)/2 = 96 km/hr
The speed of the car is 96/24 * 19 = 76 km/hr
The distance covered by the car in 5 hours is 76 × 5 = 380 km
The answer is D.
Answer: (D) 380
Question: A reduction of 5% in the price of sugar enables a lady to obtain 5kgs more for Rs.500, find the original price per kg?
(A) 5.26
(B) 5.0
(C) 5.52
(D) 6.2
(E) 5.8
Rationale: Answer:	500*(5/100) = 25 --- 5
? --- 1 => Rs.5
500 --- 475
? --- 5 => Rs.5.26.Answer: A
Answer: (A) 5.26
Question: If 6 persons can do 6 times of a particular work in 6 days, then, 8 persons can do 8 times of that work in?
(A) 2 days
(B) 6 days
(C) 5 days
(D) 7 days
(E) 9 days
Rationale: That is, 1 person can do one time of the work in 3 days.
Therefore, 8 persons can do 8 times work in the same 6 days itself.
B)
Answer: (B) 6 days
Question: The roots of the equation ax2 + bx + c = 0 will be reciprocal if:
(A) c = b
(B) c = 0
(C) c = a
(D) c = U
(E) c = K
Rationale: Let k be one of the root of the given equation.
According to the problem,
1/k will be the other root of the given equation.
We know that, product of the roots of the equation = c/a.
Therefore, k × 1/k = c/a.
or, 1 = c/a.
or, a = c [multiplying a on both sides].
The roots of the equation ax2 + bx + c = 0 will be reciprocal if a = c.
Therefore, a = c or c = a.
Answer: C
Answer: (C) c = a
Question: A man is running at 9 kmph alongside a railway track in 240 metres ahead of the engine of a
120 metres long train. The train is running at 45 kmph in the same direction. How much time does it take for the train to pass the man?
(A) 46 sec
(B) 36 sec
(C) 18 sec
(D) 22 sec
(E) 30 sec
Rationale: Distance covered = (240+120)=360
relative speed = (45-9) = 36 km/hr
36*5/18 = 10 m/s
req time = 360/10 = 36 sec
ANSWER B
Answer: (B) 36 sec
Question: A farmer used 1,034 acres of land for beans, wheat, and corn in the ratio of 5 : 2 : 4, respectively. How many W acres were used for corn?
(A)  188
(B)  258
(C)  376
(D)  470
(E)  517
Rationale: consider 5x acres of land used for bean
consider 2x acres of land used for wheat
consider 4x acres of land used for corn
Total given is 1034 acres
11x =1034
x = 94
land used for corn W= 4 * 94 = 376
correct option - C
Answer: (C)  376
Question: Edward invested five-ninths of his money at an annual rate of 2r% compounded semi-annually, and the remaining money at an annual rate of r% compounded annually. If after one year, Edward’s money had grown by one-thirds, the value of r is equal to which of the following?
(A) 10%
(B) 15%
(C) 20%
(D) 25%
(E) 33%
Rationale: X=Amount invested
5/9*X*(1+2r/2)^2+4/9*X*(1+r)=4/3X
Set X = 9 because we can pick any value
5*(1+r)^2+4(1+r)=12
5r^2+14r-3=0
Here you can use the formula for roots of a quadratic, but it is faster to guess and check using the answers
5*(1/5)^2+14(1/5)-3=5*(1/25)+14/5-3=1/5+14/5-3=0
ANSWER:C
Answer: (C) 20%
Question: A sum of money is borrowed and paid back in two annual installments of Rs.882 each allowing 5% compound interest. The sum borrowed was :
(A) Rs.1620
(B) Rs. 1640
(C) Rs.1680
(D) Rs.1700
(E) Rs.1800
Rationale: Principal = 882/(1+(5/100)) + 882/(1+(5/100)
= (882*20)/21+ (882*400)/441
=840 +800
= Rs.1640
ANSWER:B
Answer: (B) Rs. 1640
Question: One ton has 2300 pounds, and one pound has 16 ounces. How many packets containing wheat weighing 16 pounds and 4 ounces each would totally fill a gunny bag of capacity 13 tons?
(A)  1840
(B)  1700
(C)  2350
(D)  2500
(E)  8000
Rationale: 16 pounds and 4 ounces= 16*16 + 4 = 260 ounces.
13 tons = 13*2300 pound = 13*2300*16 ounces.
Hence the answer is (13*2300*16)/260 = 1840.
Answer: A.
Answer: (A)  1840