mathewhe/aqua_rat huggingface dataset:
Question: There are 100 students in a college class of which 36 are boys studying Tamil & 13 girls not studying Tamil. If there are 55 girls in all, then the probability that a boy picked up at random is not studying Tamil, is
(A) 1/5
(B) 2/3
(C) 3/5
(D) 3/7
(E) 3/8
Rationale: There are 55 girls and 45 boys in the college. Out of 45 boys, 36 are studying Tamil and 9 are not studying Tamil. The probability that a boy picked up at random is not studying Tamil= 9/45 = 1/5
A
Answer: (A) 1/5
Question: If the average cost of a house was $50,000 in 1960 and was x in 2005, what was the average increase in home prices per year over this period?
(A) A)(x+50,000)/40
(B) E)x/40
(C) C)(50,000/40)+x
(D) D)(50,000-x)/40
(E) B)(x-50,000)/45
Rationale: Average house price in 2005: x
Average house price in 1965: 50,000
2005-1960=45
Average increase per year: (x-50,000)/45
Answer is E
Answer: (E) B)(x-50,000)/45
Question: The total of the ages of Amar, Akbar and Anthony is 50 years. What was the total of their ages three years ago ?
(A) 48 years
(B) 42 years
(C) 41 years
(D) 47 years
(E) 45 years
Rationale: Explanation:
Required sum = (50 - 3 x 3) years = (50 - 9) years = 41 years.
Answer: Option C
Answer: (C) 41 years
Question: Vijay bought 160 shirts at the rate of Rs. 200 per shirt. The transport expenditure was Rs. 1400. He paid an octroi at the rate of Rs. 1.75 per shirt and labour charges were Rs. 320. What should be the selling price of one shirt, if he wants a profit of 20%?
(A) s.225
(B) s.288
(C) s.485
(D) s.255
(E) s.282
Rationale: Total CP per shirt = 200 + 1400/160 + 1.75 + 320/160 = Rs. 212.5
SP = CP[(100 + profit%)/100]
= 212.5 * [(100 + 20)/100] = Rs.255.
Answer:D
Answer: (D) s.255
Question: 864.942 - ? + 45.76 = 547.905
(A) A)899.015
(B) B)752.804
(C) C)714.642
(D) D)362.797
(E) of these
Rationale: Explanation:
362.797
Answer: Option D
Answer: (D) D)362.797
Question: In an examination, 300 students appeared. Out of these students; 30 % got first division, 54 % got second division and the remaining just passed. Assuming that no student failed; find the number of students who just passed.
(A) 33
(B) 37
(C) 54
(D) 99
(E) 48
Rationale: The number of students with first division = 30 % of 300 = 30/100 × 300 = 9000/100 = 90
And, the number of students with second division = 54 % of 300
= 54/100 × 300
=16200/100
= 162
Therefore, the number of students who just passed = 300 – (90 + 162) = 48
Answer:E
Answer: (E) 48
Question: A rock is dropped into a well and the distance traveled is 16 t 2 feet, where t is the time. If the water splash is heard 3 seconds after the rock was dropped, and that the speed of sound is 1100 ft / sec, approximate the height of the well
(A) 134.8 feet
(B) 136.2 feet
(C) 132.7 feet
(D) 132.2 feet
(E) 115.3 feet
Rationale: Let T1 be the time it takes the rock to reach the bottom of the well. If H is the height of the well, we can write
H = 16 T1 2
Let T2 be the time it takes sound wave to reach the top of the well. we can write
H = 1100 T2
The relationship between T1 and T2 is
T1 + T2 = 3
Eliminate H and combine the equations H = 16 T1 2 and H = 1100 T2 to obtain
16 T1 2 = 1100 T2
We now substitute T2 by 3 - T1 in the above equation
16 T1 2 = 1100 (3 - T1)
The above is a quadratic equation that may be written as
16 T1 2 + 1100 T1 - 3300 = 0
The above equations has two solutions and only one of them is positive and is given by
T1 = 2.88 seconds (2 decimal places)
We now calculate the height H of the well
H = 16 T1 2 = 16 (2.88) 2 = 132.7 feet
Answer C
Answer: (C) 132.7 feet
Question: If (10^4 * 3.456789)^10 is written as a single term, how many digits would be to the right of the decimal place?
(A) 12
(B) 16
(C) 20
(D) 32
(E) 48
Rationale: 3.456789^10 has 6*10 = 60 decimal places.
10^40 moves the decimal place to the right 40 places.
(10^4 * 3.456789)^10 has 60-40 = 20 digits after the decimal point.
The answer is C.
Answer: (C) 20
Question: If 1 is added to the denominator of fraction, the fraction becomes
1/2. If 1 is added to the numerator, the fraction becomes 1. The fraction is
(A) 4/7
(B) 5/9
(C) 2/3
(D) 10/11
(E) None of these
Rationale: Explanation:
Let the required fraction be a/b. Then
⇒a/b+1=1/2⇒2a–b=1 ---- (1)
⇒a+1/b=1⇒a–b=–1⇒a–b=–1 ------- (2)
Solving (1) & (2) we get a = 2, b=3
Fraction = a/b=2/3
Correct Option: C
Answer: (C) 2/3
Question: 5 men and 12 boys finish a piece of work in 4 days, 7 men and 6 boys do it in 5 days. The ratio between the efficiencies of a man and boy is?
(A) 6:9
(B) 6:89
(C) 6:0
(D) 6:5
(E) 6:1
Rationale: 5M + 12B ----- 4 days
7M + 6B ------- 5 days
20M + 48B = 35M + 30B
18B = 15M => 5M = 6B
M: B = 6:5
Answer: D
Answer: (D) 6:5
Question: a lady's age is between 50 and 70.each of her sons has many sons as his
brothers. the sum of the sons and grandsons in the lady's age. what is her age?
(A) 61
(B) 62
(C) 63
(D) 64
(E) 65
Rationale: Let assume Lady has 'N' children.
Each child have (N-1) children. (Grandsons for that lady).
S0, total children = N+(N*(N-1))
By, putting N=6, we get 6+(6*(6-1)) = 36 childrens.
By, Putting N=7, we get 7+(7*(7-1)) = 49 childrens.
By, putting N=8, we get 8+(8*(8-1)) = 64 childrens.
By, putting N=9, we get 9+(9*(9-1)) = 81 childrens.
Here, Lady's age between 50 to 70.
So, answer is 64.
ANSWER:D
Answer: (D) 64
Question: If five persons sit in a row, then the probability that three particular persons are always together is?
(A) 3/5
(B) 3/4
(C) 1/9
(D) 1/6
(E) 3/10
Rationale: Five persons can be arranged in a row in 5! ways.
Treat the three persons to sit together as one unit then there three persons and they can be arranged in 3! ways.
Again three persons can be arranged among them selves in 3! ways.
Favourable outcomes = 3!3!
Required probability = 3!3!/5! = 3/10
Answer:E
Answer: (E) 3/10
Question: If a,b,c,d,e are five consecutive odd numbers, their average is :
(A) 22
(B) 77
(C) 288
(D) 22
(E) 20
Rationale: Correct Option: D
Explanation:
If the first number is a, then the remaining numbers a a +2, a+4, a + 6, a + 8
Average : = (a+(a+2)+(a+4)+(a+6)+(a+8)5)=(a+4)
Answer:A
Answer: (A) 22
Question: A pump can fill a tank with water in 2 hours. Because of a leak, it took 2 hours to fill the tank. The leak can drain all the water of the tank in?
(A) 7 hrs
(B) 8 hrs
(C) 10 hrs
(D) 12 hrs
(E) 14 hrs
Rationale: Work done by the leak in 1 hour = 1	-	3 =	1	.
2	7	14
Leak will empty the tank in 14 hrs.
E)
Answer: (E) 14 hrs
Question: A password of a computer used five digits where they are from 0 and 9. What is the probability that the first digit of the password is a 0 and the second digit is even ?
(A) 1/10
(B) 1/15
(C) 1/20
(D) 1/100
(E) 1/200
Rationale: There is one possibility out of ten that the first digit is zero, and five possibilities out of ten that the second digit is even. We don't care about the remaining digits, so the overall probability is 1/10 * 5/10 or 1/20 or answer C.
Answer: (C) 1/20
Question: Tickets to a certain concert sell for $20 each. The first 10 people to show up at the ticket booth received a 40% discount, and the next 20 received a 15% discount. If 56 people bought tickets to the concert, what was the total revenue from ticket sales?
(A) $600
(B) $740
(C) $850
(D) $980
(E) $1,140
Rationale: Price of 1 ticket = 20$
Revenue generated from sales of first 10 tickets = 10 * (60/100 * 20 )
= 10 * 12 = 120
Revenue generated from sales of next 20 tickets = 20 *(85/100 * 20 )
=20*17 = 340
Revenue generated from sales of last 26 tickets = 20 * 26
= 520
Revenue generated from sales of 56 tickets = 120 + 340 + 520
= 980 $
Answer D
Answer: (D) $980
Question: There is food for 760 men for 22 days. How many more men should join after two days so that the same food may last for 5 days more?
(A) 2280
(B) 2240
(C) 2299
(D) 2288
(E) 2266
Rationale: 760 ---- 22
760 ---- 20
x ----- 5
x*5 = 760*20
x = 3040
760
-------
2280
Answer: A
Answer: (A) 2280
Question: If xyz represents a positive three-digit number, where x, y and z are one-digit integers, which of the following CANNOT be true?
(A) y – x - z = 9
(B) x – y - z = 9
(C) x + y + z = 9
(D) xyz = 9
(E) x/y/z = 9
Rationale: The (hundreds place digit, x) cannot be zero, but (tens and ones place digits; y and z) can be zero.
A is NOT possible
B is possible
C is possible
D is possible
E is possible
Answer: A
Answer: (A) y – x - z = 9
Question: The denominator of a fraction is 7 greater than the numerator. If the numerator and the denominator are increased by 4, the resulting fraction is equal to 5â„6. What is the value of the original fraction?
(A) 23/30
(B) 25/32
(C) 31/38
(D) 37/44
(E) 47/54
Rationale: Let the numerator be x.
Then the denominator is x+7.
x+4 / x+11 = 5/6.
6x+24=5x+55.
x=31.
The original fraction is 31/38.
The answer is C.
Answer: (C) 31/38
Question: A single discount equivalent to the discount series of 20%, 10% and 5% is?
(A) 31.0
(B) 31.2
(C) 31.6
(D) 31.3
(E) 31.1
Rationale: 100*(80/100)*(90/100)*(95/100) = 68.4
100 - 68.4 = 31.6
Answer: C
Answer: (C) 31.6
Question: The original price of a car was $22,000. Because the car owner thought he could get more money for the car, he set a new price by increasing the original price of the car by 10%. After a week, the car had not sold, so the owner discounted the new price by 10%, and the car was finally sold. What price was the car sold for?
(A) $21,660
(B) $21,700
(C) $21,740
(D) $21,780
(E) $21,820
Rationale: The car was sold for (0.9)(1.1)($22,000)=$21,780.
The answer is D.
Answer: (D) $21,780
Question: Krishan and Nandan jointly started a business. Krishan invested three times as Nandan did and invested his money for double time as compared to Nandan. Nandan earned Rs. 4000. If the gain is proportional to the money invested and the time for which the money is invested then the total gain was?
(A) s.16000
(B) s.20000
(C) s.24000
(D) s.Rs 28000
(E) s.29000
Rationale: 3:1
2:1
------
6:1
1 ----- 4000
7 ----- ? => Rs.28,000
ANSWER: D
Answer: (D) s.Rs 28000
Question: A train 640 m in length crosses a telegraph post in 16 seconds. The speed of the train is?
(A) 22 kmph
(B) 77 kmph
(C) 144 kmph
(D) 71 kmph
(E) 88 kmph
Rationale: S = 640/16 * 18/5 = 144 kmph
Answer:C
Answer: (C) 144 kmph
Question: A train speeds past a pole in 15 sec and a platform 100 m long in 25 sec, its length is?
(A) 291
(B) 178
(C) 150
(D) 287
(E) 182
Rationale: Let the length of the train be x m and its speed be y m/sec.
Then, x/y = 15 => y = x/15
(x + 100)/25 = x/15 => x = 150 m.
Answer: C
Answer: (C) 150
Question: Find the next number in the series ?
2 3/4 5/7 7/10 11/13 13/16 ?
(A) 14/19
(B) 15/19
(C) 17/19
(D) 16/19
(E) 18/19
Rationale: Solution:
17/19
series is 2/1 3/4 5/7 7/10 11/13 13/16
Numerator is prime number series 2-3-5-7-11-13-17
and Denominator is +3 series 1-4-7-10-13-16-19
Answer C
Answer: (C) 17/19
Question: A and B are two partially filled buckets of water. If 4 liters are transferred from A to B, then A would contain one-third of the amount of water in B. Alternatively, if 4 liters are transferred from B to A, B would contain one-half of the amount of water in A. Bucket A contains how many liters of water?
(A) 44/5
(B) 13
(C) 17
(D) 21
(E) 23
Rationale: Let Bucket A be A and Bucket B be B
Scenario 1
A-4 = 1/3 (B+4) ----> 3A-12 =B+4
Scenario 2
B-4 = 1/2 (A+4) -----> 2B-8 = A+4
From Scenario 1, B = 3A-16
Substitute B with this information in Stmt 2
2 (3A-16) -8 = A+4 ------> 6A -32-8 = A+4 ------> 6A- A = 40+4 ---> 5A= 44
A= 44/5, Answer choice A
Answer: (A) 44/5
Question: 70, 75, 80, 85, 90, 105, 105, 130, 130, 130
The list shown consists of the times, in seconds, that it took each of 10 schoolchildren to run a distance of 400 on of meters. If the standard devastation of the 10 running times is 24.4 seconds, rounded to the nearest tenth of a second, how many of the 10 running times are more than 1 standard deviation below the mean of the 10 running times?
(A) one
(B) two
(C) three
(D) four
(E) five
Rationale: The most time consuming part in this question is to define the mean. Under exam pressure and time pressure it is very easy to make mistake.
it is easier to group numbers: 130*3=390; 105*2=210; 75+85=160; 70+80=150; 90;
Next stage combine results, again using more convenient ways to calculate: 390+210=600; 160+150=310; 90. 600+310+90=1000. Since there are 10 numbers the mean is 100. Questions asks to find the quantity of numbers one SD BELOW the mean, which is 100-24,4=75,6. There are only two numbers below 75,6. The answer is B
Answer: (B) two
Question: If p^2 – 13p + 40 = l, and p is a positive integer between 1 and 10, inclusive, what is the probability that l < 0?
(A) 1/10
(B) 1/5
(C) 2/5
(D) 3/5
(E) 3/10
Rationale: p2 – 13p + 40 = l
so (p – 8)(p – 5) = l
For q to be negative, the expressions (p – 8) and (p – 5) must have opposite signs. Which integers on the number line will yield opposite signs for the expressions (p – 8) and (p – 5)? Those integers in the range 5 < p < 8 (notice 5 and 8 are not included because they would both yield a value of zero and zero is a nonnegative integer). That means that there are only two integer values for p, 6 and 7, that would yield a negative q. With a total of 10 possible p values, only 2 yield a negative q, so the probability is 2/10 or 1/5.
The correct answer is B.
Answer: (B) 1/5
Question: The number that does not have a reciprocal is ____________.
(A) 0
(B) 1
(C) 2
(D) 3
(E) -1
Rationale: O does not have any reciprocal
as 1/0 is not defined ANSWER:A
Answer: (A) 0
Question: One packet of flower seeds was enough to just fill the area of the smaller garden in #3. How
many packets of seed are needed for the larger garden? Round each answer to the nearest square
foot before continuing.
(A) 1 seed packet
(B) 2 seed packets
(C) 3 seed packets
(D) 4 seed packets
(E) 5 seed packets
Rationale: 3.14 x 32 = 28.26 sq. ft. area of small garden
28 sq. ft. rounded
3.14 x 62 = 113.04 sq. ft. area of large garden
113 sq. ft. rounded
113 ÷ 28 = 4.04 or 4 seed packets (rounded)
correct answer D
Answer: (D) 4 seed packets
Question: Due to construction, the speed limit along an 10-mile section of highway is reduced from 55 miles per hour to 35 miles per hour. Approximately how many minutes more will it take to travel along this section of highway at the new speed limit than it would have taken at the old speed limit ?
(A)  6.24
(B)  8
(C)  10
(D)  15
(E)  24
Rationale: Old time in minutes to cross 10 miles stretch = 10*60/55 = 10*12/11 = 10.9
New time in minutes to cross 10 miles stretch = 10*60/35 = 10*12/7 = 17.14
Time difference = 6.24
Ans:A
Answer: (A)  6.24
Question: How many seconds does a train 125 meters long, traveling at 72 km/h, take to completely cross a bridge of 205 meters in length?
(A) 12.5
(B) 14.5
(C) 16.5
(D) 18.5
(E) 20.5
Rationale: 72 km/h = 72000/3600 = 20 m/s
Time = 330 / 20 = 16.5 seconds
The answer is C.
Answer: (C) 16.5
Question: If f(x) = k(x - k) and k is a constant, what is the value of f(5) - f(2), in terms of k?
(A)  3k
(B)  k
(C)  7k - 1
(D)  k^2 + k
(E)  k^2 - k
Rationale: First, distribute f(x). k(x - k) = kx - k^2. Now evaluate f(5) and f(2):
f(5) = k(5) - k^2 = 5k - k^2
f(2) = 2k - k^2
f(5)-f(2)= 3k... Answer A
Answer: (A)  3k
Question: Two goods trains each 500 m long are running in opposite directions on parallel tracks. Their speeds are 70 km/hr and 30 km/hr respectively. Find the time taken by the slower train to pass the driver of the faster one?
(A) 12 sec
(B) 24 sec
(C) 36 sec
(D) 60 sec
(E) 62 sec
Rationale: Relative speed = 70 + 30 = 100 km/hr.
100 * 5/18 = 250/9 m/sec.
Distance covered = 500 + 500 = 1000 m.
Required time = 1000 * 9/250 = 36 sec.
ANSWER:C
Answer: (C) 36 sec
Question: I. a2 - 9a + 20 = 0,
II. 2b2 - 5b - 12 = 0 to solve both the equations to find the values of a and b?
(A) If a < b
(B) If a ≤ b
(C) If the relationship between a and b cannot be established
(D) If a > b
(E) If a ≥ b
Rationale: I. (a - 5)(a - 4) = 0
=> a = 5, 4
II. (2b + 3)(b - 4) = 0
=> b = 4, -3/2 => a ≥ b
ANSWER:E
Answer: (E) If a ≥ b
Question: If a number p is prime, and 2p+3 = q, where q is also prime, then the decimal expansion of 1/q will produce a decimal with q-1 digits. If this method produces a decimal with 166 digits, what is the units digit of the product of p and q?
(A) 1
(B) 3
(C) 5
(D) 7
(E) 9
Rationale: 1/5 = 0.2.. (a repeating pattern one digit long)
A
Answer: (A) 1
Question: Find the odd man out. 15, 25, 30, 51, 85, 90, 115
(A) 15
(B) 25
(C) 51
(D) 90
(E) 115
Rationale: Explanation :
All except 51 are multiples of 5
Answer : Option C
Answer: (C) 51
Question: Find large no. from below question The difference of two numbers is 1365. On dividing the larger number by the smaller, we get 6 as quotient and the 15 as remainder
(A) 1235
(B) 1456
(C) 1567
(D) 1678
(E) 1635
Rationale: Let the smaller number be x. Then larger number = (x + 1365).
x + 1365 = 6x + 15
5x = 1350
x = 270
Large number = 270+1365 = 1635
E
Answer: (E) 1635
Question: Plane leaves from Chelsea every 40 minutes from West Ham airport. Air hostess told Frank Lampard that the plane had left 25 minutes ago and the next plane for Chelsea will leave at 7:00 p.m.
At what time air hostess convey this information to Frank Lampard?
(A) 5:35 p.m
(B) 6:45 p.m
(C) 7:35 p.m
(D) 8:35 p.m
(E) 9:35 p.m
Rationale: Solution:
Explanation:
Time of Frank Lampard last missed plane = 7:00 p.m - 40 min = 6:20 p.m
As this information has been given to Frank Lampard 25 minutes after the plane has already left, the air hostess gave the information at 6:20 p.m + 25 min = 6:45 p.m
Answer B
Answer: (B) 6:45 p.m
Question: 100 identical coins each with probability 'pp' showing up heads & tossed. If 0<p<10<p<1 and the probabilityof heads showing on 50coins is equal tothat of heads on 51 coins, then the value of pp is?
(A) 50/107
(B) 53/103
(C) 51/101
(D) 65/101
(E) 89/101
Rationale: Let aa be the number of coins showing heads
Then, P(A=50)=P(A=51)P(A=50)=P(A=51)
⇒100C50×P50×(1−P)50⇒100C51×P51×(1−P)49⇒100C50×P51×(1−P)50=100C51×P⇒51(1âˆ′P)=50P⇒P=51/101
C
Answer: (C) 51/101
Question: All of the stocks on the over-the-counter market are designated by either a 7-letter or a 6-letter code that is created by using the 26 letters of the alphabet. Which of the following gives the maximum number of different stocks that can be designated with these codes?
(A) a. 2(26^5)
(B) b. 26(26^4)
(C) c. 27(26^4)
(D) d. 27(26^6)
(E) e. 27(26^5)
Rationale: no of different stocks for 7-letter code = 26^7
no of different stocks for 6-letter code = 26^6
total = 26^7 + 26^6 = 26^6(26+1) = 27(26^6)
[Reveal]Spoiler:
Option D
Answer: (D) d. 27(26^6)
Question: A certain junior class has 1,000 students and a certain senior class has 400 students. Among these students, there are 60 siblings pairs, each consisting of 1 junior and 1 senior. If 1 student is to be selected at random from each class, what is the probability that the 2 students selected at will be a sibling pair?
(A) 3/20,000
(B) 1/3,600
(C) 9/2,000
(D) 1/60
(E) 1/15
Rationale: Total number of ways of choosing one student from each group is = 400 * 1000
Number of cases in which a sibling pair will be got is = 60
Thus the probability that the 2 students selected will be a sibling pair is
= 60 /(400 * 1000)
= 3/20,000
A
Answer: (A) 3/20,000
Question: A started a business with an investment of Rs. 70000 and after 6 months B joined him investing Rs. 120000. If the profit at the end of a year is Rs. 52000, then the share of B is?
(A) Rs.24098
(B) Rs.24090
(C) Rs.24000
(D) Rs.24006
(E) Rs.24202
Rationale: Ratio of investments of A and B is (70000 * 12) : (120000 * 6) = 7 : 6
Total profit = Rs. 52000
Share of B = 6/13 (52000)
= Rs.24000
Answer: C
Answer: (C) Rs.24000
Question: If a and b are positive integers such that a – b and a/b are both even integers, which of the following must be an odd integer?
(A) a/2
(B) b/2
(C) b/2
(D) (a+b)/2
(E) (b+2)/2
Rationale: A and B both have to be even to fulfill the condition.
a could be = 4, 6, 8 ...., b = 2,4, ....
so a - b = 2 (Even)
a/b= 2k (Even)
a/2=2k
a = 4k
so a must be multiple of 4
now try the options
Ans. D
Answer: (D) (a+b)/2
Question: A man purchased 3 blankets @ Rs.100 each, 5 blankets @ Rs.150 each and two blankets at a certain rate which is now slipped off from his memory. But he remembers that the average price of the blankets was Rs.150. Find the unknown rate of two blankets?
(A) 288
(B) 299
(C) 266
(D) 450
(E) 291
Rationale: 10 * 150 = 1500
3 * 100 + 5 * 150 = 1050
1500 – 1050 = 450
Answer: D
Answer: (D) 450
Question: If x is an integer, and 196 < x^2 <= 361, which of the following is the sum of all possible values of x?
(A)  135
(B)  124
(C)  116
(D)  105
(E)  None of the above
Rationale: The range of x is from 15 .. to 19 and -15 to -19 so the sum is 0 hence ans E, I guess, unless I am missing something!
Answer: (E)  None of the above
Question: A = {2, 3, 5, 7, 17}
B = {2, 4, 6, 13}
Two integers will be randomly selected from sets A and B, one integer from set A and one from set B, and then multiplied together. How many different products can be obtained?
(A) 15
(B) 16
(C) 19
(D) 20
(E) 36
Rationale: Step 1: Find out the number of products you get.
5 distinct numbers in set A and 4 distinct in set B so number of products = 5*4 = 20
Step 2: Remove the products that appear more than once.
Notice that 5, 7, 11 and 13 are primes and none of their multiples are in either set. So ignore them. We just need to focus on 2 and 3 of set A and 2, 4 and 6 of set B.
2, 3
2, 4, 6
The only product repeated when you take a number from each set is 12. (3*4 and 2*6) Rest all are distinct.
Answer = 20 - 1 = 36
Note here that the second step will involve manual calculation since it will depend on the specific numbers you have in the two sets.
E
Answer: (E) 36
Question: The numbers 272738 and 232342, when divided by n, a two digit number, leave a remainder of 13 and 17 respectively. Find the sum of the digits of n?
(A) 7
(B) 8
(C) 9
(D) 4
(E) 5
Rationale: From the given information, (272738 - 13, 232342 - 17) are exactly divisible by that two digit number.
We have to find the HCF of the given numbers 272725, 232325.
HCF = 25.
So sum of the digits = 7.
Answer:A
Answer: (A) 7
Question: Two letters from the word KYANTKAR are selected at random. What is the probability that at least one of the selected letters is a consonant?
(A) 14/27
(B) 15/28
(C) 27/28
(D) 16/28
(E) 30/28
Rationale: There are 6 consonants and 2 vowels in KYANTKAR .
probability that at least one of the selected letters is a consonant = 1 - (Probability of both vowels)
Probability of both vowels = 2C2/8C2 = 1/28
So,Answer is 1 - 1/28 = 27/28
Answer : C
Answer: (C) 27/28
Question: In a stockpile of products produced by three machines M1, M2 and M3, 40% and 30% were manufactured by M1 and M2 respectively. 3% of the products of M1 are defective, 1% of products of M2 defective, while 95% of the products of M3 III are not defective. What is the percentage of defective in the stockpile?
(A) 3%
(B) 5%
(C) 2.5%
(D) 4%
(E) 6%
Rationale: Explanation :
Let there be 100 products in the stockpile. 40% and 30% were manufactured by M1 and M2 respectively. So, 30% are manufactured by M3.
Products from M1 = 40, from M2 = 30 and from M3 = 30.
Number of defective products from M1 = 0.03 x 40 = 1.2, from M2 = 0.01 x 30 = 0.3 and from M3 = 0.05 x 30 = 1.5
Therefore, total number of defective products = 3.
Answer : A
Answer: (A) 3%
Question: If 5% more is gained by selling an article for Rs. 350 than by selling it for Rs. 340, the cost of the article is:
(A) Rs. 50
(B) Rs. 160
(C) Rs. 200
(D) Rs. 225
(E) Rs. 325
Rationale: Let C.P. be Rs. x.
Then, 5% of x = 350 - 340 = 10
x/20 = 10 => x = 200
ANSWER:C
Answer: (C) Rs. 200
Question: A particular library has 75 books in a special collection, all of which were in the library at the beginning of the month. These book are occasionally loaned out through an inter-library program. If, by the end of the month, 65 percent of books that were loaned out are returned and there are 68 books in the special collection at that time, how many books of the special collection were loaned out during that month?
(A) 20
(B) 30
(C) 35
(D) 40
(E) 55
Rationale: Total = 75 books.
65% of books that were loaned out are returned --> 100%-65%=35% of books that were loaned out are NOT returned.
Now, there are 68 books, thus 76-68=7 books are NOT returned.
{loaned out}*0.35=7 --> {loaned out}=20.
Answer: A.
Answer: (A) 20