mathewhe/aqua_rat huggingface dataset:
Question: Two trains are traveling on parallel tracks in the same direction. The faster train travels at 130 miles per hour, while the slower train travels at 100 miles per hour. At 2 o’clock the faster train is 30 miles behind the slower one. How far apart are the two trains at 5 o'clock?
(A) 60 miles
(B) 80 miles
(C) 90 miles
(D) 120 miles
(E) 400 miles
Rationale: Answer = A. 60 miles
Relational speed = 130 - 100 = 30 miles per hour
In 3 hours, difference = 30 * 3 = 90 miles
Fast train trailing 30 miles, so effective difference = 90 - 30 = 60 miles
Answer: (A) 60 miles
Question: How long does a train 100 m long travelling at 60 kmph takes to cross a bridge of 120 m in length?
(A) 15.8 sec
(B) 13.2 sec
(C) 12.4 sec
(D) 16.8 sec
(E) 11.8 sec
Rationale: B
13.2 sec
D = 100 + 120 = 220 m
S = 60 * 5/18 = 50/3
T = 220 * 3/50 = 13.2 sec
Answer is B
Answer: (B) 13.2 sec
Question: A coin is weighted so that the probability of heads on any flip is 0.1, while the probability of tails is 0.9. If the coin is flipped 5 times independently, which of the following represents the probability that tails will appear no more than twice?
(A) (0.1)^5 + 5(0.1)^4(0.9) + 10(0.1)^3(0.9)^2
(B) (0.1)^5 + 4(0.1)^4(0.9) + 6(0.1)^3(0.9)^2
(C) (0.6)^5 + 3(0.6)^4(0.4) + 2(0.6)^3(0.4)^2
(D) (0.6)^5 + 2(0.6)^4(0.4) + (0.6)^3(0.4)^2
(E) (0.6)^5 + (0.6)^4(0.4) + (0.6)^3(0.4)^2
Rationale: Probability of Head, P(H) = 0.1
Probability of Tail, P(T) = 0.9
Tail will appear NO more than twice
i.e. favourable cases
2 Tails and 3 Heads, Probability = 5C2*(0.1)^3*(0.9)^2
1 Tail and 4 Heads, Probability = 5C1*(0.1)^4*(0.9)^2
0 Tail and 5 Heads, Probability = (0.1)^5
Required Probability = Sum of all Favourable cases = (0.1)^5 + 5(0.1)^4(0.9) + 10(0.1)^3(0.9)^2
Answer: option A
Answer: (A) (0.1)^5 + 5(0.1)^4(0.9) + 10(0.1)^3(0.9)^2
Question: What is the positive difference between the sum of the squares of the first 8 positive integers and the sum of the prime numbers between the first square and fourth square?
(A) 161
(B) 162
(C) 163
(D) 164
(E) 165
Rationale: Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.
the sum of the squares of the first 4 positive integers = 1^2 + 2^2 + 3^2 +...+8^2= 204
the sum of the prime numbers between the first square(=1) and fourth square(=16) = 2+3+5+7+11+13=41.
So the difference between 41 and 204 is 163.
So the answer is (C).
Answer: (C) 163
Question: If (c - a)/(c - b) = 1, then (5b + 5a)/(c - a) =
(A) 10
(B) 1
(C) 1.5
(D) 2
(E) 2.5
Rationale: Let's say c=2, b=1, a=1 so that our 1st expression holds true. Now, ibsert those numbers in the second expression and we'll get10
Answer A (hopefully )))
Answer: (A) 10
Question: A man's speed with the current is 15 km/hr and the speed of the current is 2.8 km/hr. The man's speed against the current is?
(A) 9.4
(B) 20
(C) 50
(D) 30
(E) 40
Rationale: Man's speed with the current = 15 km/hr
=> speed of the man + speed of the current = 15 km/hr
speed of the current is 2.8 km/hr
Hence, speed of the man = 15 - 2.8 = 12.2 km/hr
man's speed against the current = speed of the man - speed of the current
= 12.2 - 2.8 = 9.4 km/hr
Answer is A.
Answer: (A) 9.4
Question: Akash's club with a total membership of 30 has formed 3 committees, M, S and R which have 8,12, and 5 members respectively. If no member of committee M is on either of the other 2 committees, what is the greatest possible number of members in the club who are on none of the committees.
(A) 5
(B) 7
(C) 8
(D) 10
(E) 12
Rationale: Akash's club with a total membership of 30 has formed 3 committees Given no member of committee M is on either of the other 2 committees -- hence (M n R) and (M n S) and (M n R n S) is zero.
Hence only M is 8. Now we need to consider only S, (S n R) and R.
(M U R U S) = M + R + S - (M n R) - (M n S) - (R n S) + (M n R n S) + Neither.
30 = 8 + (12 - x) + (5 - x) + 0 + 0 + x + 0 + Neither.
22 = (12-x) + (5-x) + x + Neither.
Now the max value of x could be 5 and the min value of x could be 0.
When x is 5 (max), Neither is 10.
When x is 0 (min), Neither is 5. We need maximum no of people who do not belong to any group. Hence max value of neither is 10.
Answer 10 (D).
Answer: (D) 10
Question: Calculate the ratio between x and y if 15% of x equal to 50% of y?
(A) 3 : 5
(B) 3 : 7
(C) 3 : 8
(D) 3 : 10
(E) 3 : 9
Rationale: Explanation:
15 x = 50 y
x: y = 15: 50 = 3 : 10
Answer:D
Answer: (D) 3 : 10
Question: The wages earned by Robin is 30% more than that earned by Erica. The wages earned by Charles is 70% more than that earned by Erica. How much percent is the wages earned by Charles more than that earned by Robin?
(A) 18.75%
(B) 23%
(C) 30%
(D) 31%
(E) 100%
Rationale: Let wage of Erica = 10
Wage of Robin = 1.3*10 = 13
Wage of Charles = 1.7 *10 = 17
Percentage by which wage earned by Charles is more than that earned by Robin = (17-13)/13 * 100%
= 4/13 * 100%
=31%
Answer D
Answer: (D) 31%
Question: Two trains of length 140 m and 200 m are 100 m apart. They start moving towards each other on parallel tracks, at speeds 54 kmph and 72 kmph. In how much time will the trains cross each other?
(A) 100/7 sec
(B) 108/7 sec
(C) 57/7 sec
(D) 120/7 sec
(E) 50/7 sec
Rationale: Relative speed = (54 + 72)* 5/18 = 7 * 5 = 35 mps.
The time required = d/s = (140 + 100 + 300)/35
= 540/35 = 108/7 sec.
ANSWER:B
Answer: (B) 108/7 sec
Question: Out of 8 persons working on a project, 4 are graduates. If 3 are selected, what is the probability that there is at least one graduate among them?
(A) A)5/7
(B) B)5/4
(C) C)5/14
(D) D)5/6
(E) E)9/14
Rationale: P(at least one graduate) = 1 - P(no graduates)
= 1 - ⁶C₃/¹⁰C₃
= 1 - (6 * 5 * 4)/(6 * 7 * 8)
= 9/14
Answer:E
Answer: (E) E)9/14
Question: A has to pay Rs. 220 to B after 1 year.B asks A to pay Rs. 110 in cash and defer the payment of Rs. 110 for 2 years. A agrees to it. If the rate of interest be 10% per annum, in this mode of payment :
(A) There is no gain or loss to any one
(B) A gains Rs. 7.34
(C) A loses Rs. 7.34
(D) A gains Rs. 11
(E) None of these
Rationale: A has to pay = P.W. of Rs. 220 due 1 year hence = Rs.[ 220 x100/100+( 10x1)]= Rs. 200.
A actually pays = Rs. 110 + P.W. of Rs. 110 due 2 years hence
[110 + 110x 100/100+( 10x2 ) ] = Rs. 192.66.
∴ A gains = Rs. (200- 192.66)= Rs. 7.34.
Answer B
Answer: (B) A gains Rs. 7.34
Question: A team has to design a flag. The team has four yellow strips of cloth and five green strips of cloth that they must use as is to make the flag. How many different flags can the team design with the materials at hand?
(A)  24
(B)  120
(C)  126
(D)  140
(E)  156
Rationale: The # of permutations of total 4+5=9 strips where 4 are identical yellow and 5 are identical green is 9!/(4!*5!)=126 (note that they must use all the materials at hand, also I guess that the strips must be either only vertical or only horizontal).
Answer: C.
Answer: (C)  126
Question: Five years ago, the average age of A, B, C and D was 40 yr. with E joining them now, the average of all the five is 49 yr. How old is E?
(A) 25 yr
(B) 40 yr
(C) 45yr
(D) 65 yr
(E) NOne
Rationale: Solution:
Total present age of A, B, C and D,
= (40*4)+(4*5) = 180 yr;
Total age present age of A, B, C, D and E,
= 49*5 = 245 yr.
So, age of E = 65 yr.
Answer: Option D
Answer: (D) 65 yr
Question: How many times in a day, the hands of a clock are straight?
(A) 44
(B) 40
(C) 30
(D) 25
(E) 20
Rationale: 12 hrs ==> 22 times
24 hrs ==> 44 times
ANSWER A
Answer: (A) 44
Question: What is the tens digit of 6^22?
(A) 1
(B) 3
(C) 5
(D) 7
(E) 9
Rationale: The tens digit of 6 in integer power starting from 2 (6^1 has no tens digit) repeats in a pattern of 5: {3, 1, 9, 7, 5}:
The tens digit of 6^2=36 is 3.
The tens digit of 6^3=216 is 1.
The tens digit of 6^4=...96 is 9.
The tens digit of 6^5=...76 is 7.
The tens digit of 6^6=...56 is 5.
The tens digit of 6^7=...36 is 3 again.
etc...
22 has the form 5n+2, so the tens digit of 6^22 is 3.
The answer is B.
Answer: (B) 3
Question: If julie rows 32 km upstream and 72 km down steam taking 4 hours each, then the speed of the stream
(A) 1.5kmph
(B) 0.5kmph
(C) 0.4kmph
(D) 1.6kmph
(E) 2.5kmph
Rationale: Speed upstream = 32/4 = 8 kmph
Speed down stream = 72/8 = 9 kmph
Speed of stream = ½ (9-8) = 0.5kmph
ANSWER:B
Answer: (B) 0.5kmph
Question: A boy is standing in one corner of a square ground of side 90m.He can throw a stone upto maximum 70m. Find the area in which that boy can throw that stone in that ground?
(A) 2850
(B) 3150
(C) 3250
(D) 3450
(E) 3850
Rationale: Area covered by stone = Pi*r^2/4 ( here we divide by 4 because the boy is standing in the corner of the ground and only 1/4 part of the ground , the stone can reach)
where r= 70 m = length of throw
So area = (22/7)*70*70/4 = 3850 sq m
answer :E
Answer: (E) 3850
Question: A man spends 10% of his income in house rent, 20% of the rest on his children’s education, 25% of the rest miscellaneous causes. If he now posses Rs. 1944 then his income is?
(A) Rs.3600
(B) Rs.4000
(C) Rs.4500
(D) Rs.3000
(E) Rs.5000
Rationale: X * (90/100) * (80/100) * (75/100) = 1944
X * 0.9 * 0.8 * 0.75
X = 1944/0.54
X = 3600
ANSWER:A
Answer: (A) Rs.3600
Question: If r is the smallest positive integer that is not prime and not a factor of 50!, what is the sum of the factors of r?
(A) 51
(B) 162
(C) 72
(D) 54
(E) 50!+2
Rationale: In fact 51=3*17 IS a factor of 50!.
The smallest positive integer that is NOT prime and NOT a factor of 50! is 106=2*53. The sum of the factors of 106 is 162.
Answer: B.
Answer: (B) 162
Question: The area of a square is 625 sq cm. Find the ratio of the breadth and the length of a rectangle whose length is thrice the side of the square and breadth is 20 cm less than the side of the square.
(A) 5 : 12
(B) 5 : 13
(C) 2:9
(D) 1:15
(E) 1:5
Rationale: Let the length and the breadth of the rectangle be l cm and b cm respectively. Let the side of the square be a cm.
a2 = 625
a = 25
L = 3a and b = a - 20
b : l = a - 20 : 3a = 5 : 75 = 1:15
Answer: D
Answer: (D) 1:15
Question: If x*y = xy – 2(x + y) for all integers x and y, then 3* (–3) =
(A) –16
(B) –9
(C) –4
(D) 4
(E) 16
Rationale: 3*(-3) = 3*(-3) - 2 (3 + (-3) ) = -9 -0 = -9
Option (B)
Answer: (B) –9
Question: If 32 men and 24 days can reap a field in 9 days in how many days will 28 men and 12 boys reap it supposing that 4 men can do as much as 6 boys?
(A) 20 days
(B) 30 days
(C) 12 days
(D) 24 days
(E) 25 days
Rationale: Explanation:
(32m+24b)9=(28m+12b)x
[Since 24b(6 * 4)b
=> 6 boys = 4 men
=> 24 boys = 16 men]
(32m+16m)9=(28m+8m)x
[since 4 men = 6 boys]
48m*9/36m = 12 days
Answer: Option C
Answer: (C) 12 days
Question: If a(a - 9) = 10 and b(b -9) = 10, where a ≠ b, then a + b =
(A)  −48
(B)  −2
(C)  2
(D)  9
(E)  48
Rationale: i.e. if a = -1 then b = 10
or if a = 10 then b = -1
But in each case a+b = -1+10 = 9
Answer: option D
Answer: (D)  9
Question: In how many different ways can 5 boys and 3 girls sit on a bench such that the girls always sit together?
(A) 360
(B) 720
(C) 1440
(D) 4320
(E) 5040
Rationale: We can think of the girls as one unit.
The number of ways to arrange 6 units in order is 6!
The girls can have different arrangements, so we multiply this by 3!
The number of ways to sit is 6!*2!=4320
The answer is D.
Answer: (D) 4320
Question: If a and b are positive integers such that a/b = 4.88, which of the following must be a divisor of a
(A) 3
(B) 4
(C) 11
(D) 12
(E) 13
Rationale: a/b = 4.88
a/b = 488/100
a/b = 244/50
a = 244/50*b
We require to find from the available 5 options that must be a divisor of a"
This also means which of the following 5 options can divide 143 evenly
(Ignore b as its value unknown; Ignore 50 as in denominator)
Only option B = 4 best fits in
Answer = 4
ANSWER:B
Answer: (B) 4
Question: There were 35 students in a hostel. If the number of students increases by 7, the expenses of the mess increase by Rs. 42 per day while the average expenditure per head diminishes by Re 1. Find the original expenditure of the mess.
(A) Rs. 480
(B) Rs. 520
(C) Rs. 420
(D) Rs. 460
(E) None of these
Rationale: Let d be the average daily expenditure
Original expenditure = 35 × d
New expenditure = 35 × d + 42
New average expenditure will be :
(35 × d + 42)/42 = d - 1
On solving, we get d = 12
Therefore original expenditure = 35 × 12 = 420
Answer: C
Answer: (C) Rs. 420
Question: What is the remainder when 1274 * 1275 * 1277 *1285 is divided by 12?
(A)  0
(B)  6
(C)  11
(D)  9
(E)  7
Rationale: 1271*1275*1277*1285 = (1272 - 2)(1272 + 3)(1284 - 7)(1284 + 1) = (multiple of 12 - 1)(multiple of 12 + 3)(multiple of 12 - 7)(multiple of 12 + 1)
Now, if we expand this expression all terms but the last one will have a multiple of 12 in them, thus will leave no remainder upon division by 12, the last term will be (-2)(3)(-7)(1) = 42. Thus the question becomes: what is the remainder upon division of 42 by 12? The answer to this question is 6.
Answer: B
Answer: (B)  6
Question: A box contains seven bulbs out of which 3 are defective. If three bulbs are chosen at random, find the probability that all the four bulbs are defective?
(A) 1/35
(B) 1/12
(C) 1/18
(D) 1/198
(E) 1/126
Rationale: Out of seven, four are good and three are defective. Required probability
= 3C3/7C3
= 1/35
Answer: A
Answer: (A) 1/35
Question: If each day a tree loses half its remaining leaves, what fraction of its leaves will it have lost after 4 days ?
(A) X-1/8
(B) X-1/16
(C) X/8
(D) X/16
(E) X/32
Rationale: LET NO OF LEAVES IX X ..IST DAY X/2 ..2ND DAY X/4 ..3RD DAY X/8 , FOURTH DAY X/16 , 5TH DAY X/32 , 6TH DAY X/64 SO ON SO FOURTH.
FINALLY ON LEAVE WILL BE LEFT ..SO ON 3RD DAY LEAVE LOST IS X-1/16 ANSWER B
Answer: (B) X-1/16
Question: The sum of four consecutive odd numbers is equal to the sum of 3 consecutive even numbers. Given that the middle term of the even numbers is greater than 101 and lesser than 204, how many such sequences can be formed?
(A) 25
(B) 17
(C) 26
(D) 33
(E) 50
Rationale: Four consecutive odd numbers: k-2, k, k+2, k+4
Three consecutive even numbers: n-2, n, n+2
k-2+k+k+2+k+4=n-2+n+n+2
4k+4=3n
4(k+1)=3n
k+1=(3/4)n
k=(3/4)n-1
All n's that's divisible by 4 will have an integral k. So, we need to find out how many such n's are available within given range:
We know,
101<n<204
104<=n<=200
Count=(200-104)/4+1=96/4+1=24+1=25
Ans: 25.
Answer: (A) 25
Question: A sum of Rs.1480 is divided into 13 students so that each boy gets Rs.120 while each girl gets Rs.100. Find the number of boys and number of girls
(A) 8B, 5G
(B) 5B, 8G
(C) 4B, 9G
(D) 9B, 4G
(E) 9B, 5G
Rationale: 120B+100G=1480-----i
B+G=13-------ii
from i
100B+20B+100G=1480
100(B+G)+20B=1480
100(13)+20B=1480
By simplifing we get
B=9;G=4
ANSWER:D
Answer: (D) 9B, 4G
Question: An investor can sell her MicroTron stock for 36$ per share and her Dynaco stock for 56$ per share, If she sells 300 shares altogether, some of each stock, at an average price per share of 40$, how many shares of Dynaco stock has she sold?
(A) 52
(B) 75
(C) 60
(D) 136
(E) 184
Rationale: w1/w2 = (A2 - Aavg)/(Aavg - A1) = (56 - 40)/(40 - 36) = 16/4 = 4/1 = Number of MicroTron stocks/Number of Dynaco stocks
So for every 4 MicroTron stock, she sold 1 Dynaco stock.
So out of 300 total stocks, (1/5)th i.e. 300/5 = 60 must be Dynaco stock.
Answer (C)
Answer: (C) 60
Question: The ratio of men to women in the Snyder community choir is 6 to 5. The ratio of men to women in the Leigh community choir is 5 to 6. If the two choirs merged, the ratio of men to women in the combined choir would be 22 to 27. If Snyder has 4 more men and 6 more women than Leigh, how many men are in the Snyder choir?
(A) 20
(B) 22
(C) 32
(D) 36
(E) 34
Rationale: This question has a number of patterns in it that we can take advantage of. We can also TEST THE ANSWERS (but we can actually eliminate most of the answer choices rather quickly.
We're told that the ratio of the Men to Women in the Snyder choir is 6 to 5, so the number of men MUST be a multiple of 4 and the number of women MUST be a multiple of 6 The question asks how many MEN are in the SNYDER choir, so the answer MUST be a multiple of 6.. We can eliminate Answers B,A, C and E (since they're NOT multiples of 6.
Answer D
Answer: (D) 36
Question: The L.C.M of 22, 54, 108, 135 and 198 is
(A) 5942
(B) 2887
(C) 5940
(D) 2888
(E) 28881
Rationale: Answer: C) 5940
Answer: (C) 5940
Question: When 1/10 percent of 5,000 is subtracted from 1/10 of 5,000, the difference is
(A) 0
(B) 50
(C) 450
(D) 495
(E) 500
Rationale: 1/10 of 5,000 is 500;
1/10 percent of 5,000 is 100 times smaller than the above, so it's 5;
Thus the difference is 500-5=495.
Answer: D.
Answer: (D) 495
Question: What is the sum of all digits for the number 10^26 - 51?
(A) 199
(B) 209
(C) 219
(D) 229
(E) 239
Rationale: 10^26 is a 27-digit number: 1 followed by 26 zeros.
10^26 - 51 is a 26-digit number: 24 9's and 49 at the end.
The sum of the digits is 24*9+4+9=229.
The answer is D.
Answer: (D) 229
Question: If the wheel is 15 cm then the number of revolutions to cover a distance of 1540 cm is?
(A) 11
(B) 14
(C) 18
(D) 12
(E) 16
Rationale: 2 * 22/7 * 15 * x = 1540 => x = 16
Answer: E
Answer: (E) 16
Question: John started driving on a highway at a constant speed of R miles per hour at 13:00. Then, 2 hours later, Tom started driving on the same highway at a constant speed of 3R/2 miles per hour at 15:00. If both drivers maintained their speed, how many miles had John driven on the highway when Tom caught up with John?
(A) 6R
(B) 7R
(C) 8R
(D) 9R
(E) 10R
Rationale: In 2 hours, John drove 2R miles.
Tom can catch John at a rate of 3R/2 - R = R/2 miles per hour.
The time it takes Tom to catch John is 2R / (R/2) = 4 hours.
In 4 hours, John can drive another 4R miles for a total of 6R miles.
The answer is A.
Answer: (A) 6R
Question: If the cost price is 96% of sp then what is the profit %
(A) 4.07%
(B) 4%
(C) 4.7%
(D) 4.17%
(E) 4.27%
Rationale: Sol. sp=Rs100 : then cp=Rs 96:profit =Rs 4.
Profit={(4/96)*100}%=4.17%
Answer is D.
Answer: (D) 4.17%
Question: Two trains 131 meters and 165 meters in length respectively are running in opposite directions, one at the rate of 80 km and the other at the rate of 65 kmph. In what time will they be completely clear of each other from the moment they meet?
(A) 7.16
(B) 7.16
(C) 7.12
(D) 7.35
(E) 7.11
Rationale: T = (131 + 165)/ (80 + 65) * 18/5
T = 7.35
Answer: D
Answer: (D) 7.35
Question: A factory has a fixed cost of $55,000 a month, and a cost of $2.5 for every item produced. If the selling price of a single item is $5, what is the number of items must be sold monthly for the factory to cover its cost exactly?
(A) 9,000
(B) 14,000
(C) 18,000
(D) 22,000
(E) 27,000
Rationale: selling price - Cost price = 2.5 so per unit profit =2.5
cost to recover =55000
no of items required = 55000/2.5 = 22000
Ans D
Answer: (D) 22,000
Question: What is the greatest value of positive integer x such that 2^x is a factor of 100^60?
(A) 100
(B) 110
(C) 120
(D) 105
(E) 115
Rationale: Put in simple words, we need to find the highest power of 2 in 100^60
100 = 2^2*5^2
Therefore 100^60 = (2^2*5^2)^60 = 2^120*5^120
Answer : C
Answer: (C) 120
Question: What is the difference between the place value and the face value of 8 in the numeral 296487?
(A) None of these
(B) 82
(C) 78
(D) 52
(E) 72
Rationale: Explanation:
Place value of 8 = 80
Face value of 8 = 8
Difference = 80 - 8 = 72
Answer: E
Answer: (E) 72
Question: A man can row upstream at 25 kmph and downstream at 35 kmph, and then find the speed of the man in still water?
(A) 27
(B) 29
(C) 30
(D) 20
(E) 24
Rationale: US = 25
DS = 35
M = (35 + 25)/2
= 30
Answer:C
Answer: (C) 30
Question: There are, in a certain league, 20 teams, and each team face another team for a total of 10 times. How many games are played in the season?
(A) 1900
(B) 3600
(C) 2400
(D) 180
(E) 5400
Rationale: 1st team:190
2nd team:180
3rd team :170
Similarly the last team will play 0
therefore(190+180+170....10)
10(19+18+17+16....1)
10 (n(n+1)/2)=10(19*20/2)=1900
ANSWER:A
Answer: (A) 1900
Question: Jim takes a seconds to swim u meters at a constant rate from point P to point Q in a pool. Roger, who is faster than Jim, can swim the same distance in v seconds at a constant rate. If Jim leaves point P the same time that Roger leaves point Q, how many fewer meters will Jim have swum than Roger when the two swimmers pass each other?
(A) c(v-u)/ (v+U)
(B) c(a-b)/a+b
(C) c(a+b)/a-b
(D) ab(a-b)/a+b
(E) ab(b-a)/a+b
Rationale: Both JimRoger are travelling at constant speedin opposite direction: and c is the distance
So, speed of Jim = c/u speed of Roger = c/v
Let say Jim travelled distance x from P where it met Roger, it means that Roger travelled (c-x) from point Q
[x would be less than (c-x) as Jim is travelling slow]
From above, time taken by Jim to travel x = xu/c....................... (1)
Also, time taken by Roger to travel (c-x) = (c-x)v/c.....................(2)
Time taken by both JimRoger is same, so (1) = (2)
xu/c = (c-x)v/c,
Solving further, x = vc/(u+v).................... (3)
We require to find how many fewer meters will Jim have swum i.e
additional distance travelled by Roger = (c - x) - x
= c-2x
Substituting value of x from (3)solving the equation further, we get Answer = c(v-u)/v+u
Answer = (A)
Answer: (A) c(v-u)/ (v+U)
Question: There are three prizes to be distributed among five students. If no students gets more than one prize, then this can be done in:
(A) 10 ways
(B) 30 ways
(C) 60 ways
(D) 80 ways
(E) None
Rationale: Solution:
3 prize among 5 students can be distributed in,
5C3 ways = 10 ways.
Answer: Option A
Answer: (A) 10 ways
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) 24008
(B) 24000
(C) 24005
(D) 24009
(E) 24002
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: B
Answer: (B) 24000
Question: On July 1 of last year, total employees at company E was decreased by 10 percent. Without any change in the salaries of the remaining employees, the average (arithmetic mean) employee salary was 10 percent more after the decrease in the number of employees than before the decrease. The total of the combined salaries of all the employees at Company E after July 1 last year was what percent of that before July 1 last year?
(A) 90%
(B) 99%
(C) 100%
(D) 101%
(E) 110%
Rationale: Let the number of employees before Jul 1 be 100
Then Av sal1 = Total1/100
After Jul 1 employees = 90
Av Sal2 = Total 2/90
Now from the question
Av2 = Av1 + 10%
Total 2/90 = Total1/100 + 1/10 * Total1/100
Total 2/90 = 11 Total1/1000
Total 2 = 99 total1/100
Now Percent = Total2/Total1 * 100
= 99 Total1/Total1
= 99
ANSWER:B
Answer: (B) 99%
Question: The probability that a visitor at the mall buys a pack of candy is 60%. If three visitors come to the mall today, what is the probability that exactly two will buy a pack of candy?
(A) 0.284
(B) 0.326
(C) 0.432
(D) 0.548
(E) 0.652
Rationale: One case is: candy - candy - no candy
The probability is 6/10*6/10*4/10 = 144/1000
There are 3 such cases, so we should multiply this probability by 3.
P(exactly 2 buy candy) = 144/1000 * 3 = 432/1000 = 0.432
The answer is C.
Answer: (C) 0.432
Question: Two numbers are in the ratio 3 : 4. If their L.C.M. is 84. what is sum of the numbers?
(A) 49
(B) 56
(C) 76
(D) 84
(E) 85
Rationale: Explanation:
Let the numbers be 3x and 4x
LCM of 3x and 4x = 12x (Since LCM of 3 and 4 is 12. Hence LCM of 3x and 4x is 12x)
Given that LCM of 3x and 4x is 84
=> 12x =84
=> x = 84/12 = 7
Sum of the numbers
= 3x + 4x
= 7x
= 7 x 7 = 49
Answer: Option A
Answer: (A) 49
Question: Annie has 100 cards numbered 1 through 100. If she deals six cards to Alex, without replacing any of them, what is the probability that Alex will get five consecutive numbers?
(A) 95!/100!
(B) 96!/100!
(C) (95! X 6!)/100!
(D) (96! X 5!)/100!
(E) (97! X 4!)/100!
Rationale: Alex to get 6 consecutive cards in ascending order, to have 6 consecutive at the end is good enough.
There are 95 consecutive numbers in 100: {1, 2, 3, 4, 5,6}, {2, 3, 4, 5, 6,7}, ..., {95,96, 97, 98, 99, 100};
Total ways to pick 6 cards out of 100 is {6!*95!}{100!}.
Answer: C
Answer: (C) (95! X 6!)/100!