mathewhe/aqua_rat huggingface dataset:
Question: Frances can complete a job in 16 hours, and Joan can complete the same job in 8 hours. Frances starts the job at 9 a.m., and stops working at 3 p.m. If Joan starts working at 4 p.m. to complete the job, at what time is the job finished?
(A) 9 p.m.
(B) 7 p.m.
(C) 8 p.m.
(D) 10 p.m.
(E) 12 p.m.
Rationale: Let the total work be 24 units.
Frances can complete 1.5 units/hr
Joan can complete 3 units/hr
Work done by Frances in 6 hours is 9 units ; work left for Joan is 15 units ( 24 - 9)
Time required by Jones to complete the remaining work will be 15/3 => 5 hours
Given Jones starts at 4PM , so she would complete the work at 9 PM
Answer will be (A)
Answer: (A) 9 p.m.
Question: A owes B Rs.50. He agrees to pay B over a number of consecutive days on a Monday, paying single note or Rs.10 or Rs.20 on each day. In how many different ways can A repay B.
(A) 2
(B) 8
(C) 9
(D) 6
(E) 5
Rationale: He can pay by all 10 rupee notes = 1 way
3 Ten rupee + 1 twenty rupee = 4!3!×1!4!3!×1! = 4 ways
1 Ten rupee + 2 twenty rupee notes = 3!2!×1!3!2!×1! = 3 ways
Total ways = 1 + 4 + 3 = 8
Answer:B
Answer: (B) 8
Question: Private Benjamin is a member of a squad of 10 soldiers, which must volunteer 4 of its members for latrine duty. If the members of the latrine patrol are chosen randomly, what is the probability Q that private Benjamin will be chosen for latrine duty?
(A) 1/10
(B) 1/5
(C) 2/5
(D) 3/5
(E) 4/5
Rationale: Easier way : Desired probability = 1- 'excluded' probability
In this case, Excluded probability = probability of Benjamin not being a part of the 4 volunteers. We can choose 4 out of 9 remaining soldiers in 9C4 ways. total ways possible = 10C4.
Thus excluded probability = 9C4/10C4 = 3/5
Thus, the desired probability Q= 1- 3/5 = 2/5. Thus C is the correct answer.
Answer: (C) 2/5
Question: A shopkeeper sells 20% of his stock at 10% profit ans sells the remaining at a loss of 5%. He incurred an overall loss of Rs. 400. Find the total worth of the stock?
(A) 30000
(B) 20000
(C) 23000
(D) 21020
(E) 28211
Rationale: Explanation:
Let the total worth of the stock be Rs. x.
The SP of 20% of the stock = 1/5 * x * 1.1 = 11x/50
The SP of 80% of the stock = 4/5 * x * 0.95 = 19x/25 = 38x/50
Total SP = 11x/50 + 38x/50 = 49x/50
Overall loss = x - 49x/50 = x/50
x/50 = 400 => x = 20000
Answer: B
Answer: (B) 20000
Question: In a certain brick wall, each row of bricks above the bottom row contains one less brick than the row just below it. If there are 5 rows in all and a total of 200 bricks in the wall, how many bricks does the bottom row contain ?
(A) 42
(B) 35
(C) 40
(D) 33
(E) 38
Rationale: The bottom row has x bricks
x+x-1+x-2+x-3+x-4 = 200
5x-10=200
5x=190
x=38
Answer:E
Answer: (E) 38
Question: For a certain set, the value range of its members is 108.4. A new set is created from the members of the old set as follows: 12 is subtracted from a member of the old set, and the result is divided by 4. The resulting value is a member of the new set. If this operation is done for each member of the old set, what is the range of values of the members of the new set?
(A) 24.1
(B) 27.1
(C) 54.4
(D) 98.4
(E) 108.4
Rationale: Let x and z be the smallest and largest of the original set, respectively.
z - x = 108.4
The smallest and largest members of the new set will be (x-12) / 4 and (z-12) / 4.
Then the range is (z-12) / 4 - (x-12) / 4 = (z-x) / 4 = 108.4 / 4 = 27.1
The answer is B.
Answer: (B) 27.1
Question: A present value of a machine is $1000. Its value depletiation rate is 10% per annum then find the machine value after 2 years?
(A) $900
(B) $810
(C) $915
(D) $715
(E) $795
Rationale: P= $1000
R = 10%
T = 2 years
Machine value after 2 years = P[(1-R/100)^T]
=1000*9/10 * 9/10
=$810
Answer is B
Answer: (B) $810
Question: Month Noah Landing Bellaville
1 3 2
2 2 0
3 0 4
4 2 X
5 1 1
Monthly rainfall was recorded for two towns over a five-month period using the chart above. If the median rainfall in Noah’s Landing was equal to the average (arithmetic mean) rainfall in Bellaville, what is the value of x?
(A) 1
(B) 8/5
(C) 2
(D) 3
(E) 10
Rationale: Arrange Noah's landing rainfall in order which is 0,1,2,2,3 and hence median is 2
Given median of Noah's landing = AM of Bellaville
So 5 * 2 is the sum of all values,
to find x , 10 = 1 + 2+ 4 + x
gives x = 3 ,
Answer is D
Answer: (D) 3
Question: If a natural number p has 8 factors, then which of the following cannot be the difference between the number of factors of p^3 and p?
(A) 14
(B) 30
(C) 32
(D) 56
(E) None of these
Rationale: p has 8 factors. That gives following possibilities:
a) p has one prime factor with power 7
=> p^3 will have 22 factors. Difference with number of factors of p = 22-8 = 14: A is ok
b) p has two prime factors with powers 3 and 1
=> p^3 will have 40 factors. Difference with number of factors of p =40-8 = 32 : C is ok
c) p has three prime factors with powers 1 each
=> p^3 will have 64 factors. Difference with number of factors of p=64-8 = 56 : D is ok
There are no other possibilities. Hence remaining answer choice B is not possible.
Answer B
Answer: (B) 30
Question: If t is an even integer and q is an odd integer, which of the following must be an odd integer?
(A) t/q
(B) tq
(C) 2t+q
(D) 2(t+q)
(E) 3t/q
Rationale: this is a must question:
A. t/q
can be a non-integer
B. tq
always even
C. 2t+q
2t = even -> even+odd = odd
D. 2(t+q)
always even - t+q = odd *2 -> even
E. 3t/q
might be a non integer.
C
Answer: (C) 2t+q
Question: On a game show, a contestant spins three different wheels; each wheel is numbered 1 to 12. What is the probability that the contestant will spin all tens?
(A) 52
(B) 1/1728
(C) 67
(D) 2/58
(E) 9/7
Rationale: All 12s:
1st wheel = 12= 1/12
2nd wheel = 12 = 1/12
3rd wheel = 12 = 1/12
(1/12)(1/12)(1/12) = 1/1728
Answer: B
Answer: (B) 1/1728
Question: Find the value of * in the following
(1 5/3) ÷ 3/11 × */11 = (2 2/3 × 7/5 × 6/7)
(A) 1
(B) 1.5
(C) 3.6
(D) 4
(E) 4.1
Rationale: Option 'C'
Answer: (C) 3.6
Question: The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one of the numbers is 275, then the other is:
(A) 306
(B) 308
(C) 310
(D) 311
(E) 315
Rationale: LCM X HCF = First No. X Second No.
LCM=7700
HCF=11
First No.=275
7700 X 11 = 275 X Second No.
Second No.= 308
answer :B
Answer: (B) 308
Question: Two trains 200m and 250m long run at the speed of 72kmph and 18kmph in opposite directions in parallel tracks. The time which they take to cross each other is?
(A) 5.6sec
(B) 8.9sec
(C) 10.8sec
(D) 12.6sec
(E) 18 sec
Rationale: Relative speed = 72+18 = 90 kmph * 5/18 = 25 m/s
Distance covered in crossing each other = 200+250 = 450m
Required time = 450*1/25 = 18 sec
Answer is E
Answer: (E) 18 sec
Question: An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other
without replacement. What is the probability that both drawn balls are black?
(A) 1/5
(B) 2/7
(C) 3/7
(D) 4/5
(E) 5/4
Rationale: Let E and F denote respectively the events that first and second ball drawn
are black. We have to find P(E n F) or P (EF).
Now P(E) = P (black ball in first draw) = 10/15
Also given that the first ball drawn is black, i.e., event E has occurred, now there are 9 black balls and five white balls left in the urn. Therefore, the probability that the second ball drawn is black, given that the ball in the first draw is black, is nothing but the conditional probability of F given that E has occurred.
That is P(F|E) = 9/14
By multiplication rule of probability, we have
P (E n F) = P(E) P(F|E)
= 10/15 × 9/14 = 3/7
C
Answer: (C) 3/7
Question: The simple interest on Rs.12000 at a certain rate of interest in five years is Rs.7200. Find the compound interest on the same amount for five years at the same rate of interest.
(A) Rs.3502.80
(B) Rs.3052.40
(C) Rs.3054.8
(D) Rs.3052.80
(E) None of these
Rationale: R = 100 I / PT
=> R = (100 * 7200)/ (12000 * 5) = 12%
CI = P{ [1 + R /100]n - 1}
= 12000 { [ 1 + 12 / 100]2 - 1} = Rs.3052.80
ANSWER:D
Answer: (D) Rs.3052.80
Question: a, b, c, and d are positive consecutive even integers and a < b < c < d.
If the product of the highest and lowest numbers is six more than the sum of the middle numbers, what is the product of the smallest two numbers?
(A) 2
(B) 4
(C) 8
(D) 12
(E) 20
Rationale: We know the numbers must be consecutive even numbers, from the list 2, 4, 6, 8, 10, 12, 14, 16, and so on.
The product ad is 6 more than the sum bc. By testing some possible ad products (even numbers separated by two other even numbers), we can quickly see that the numbers have to be quite small: 6x12 is 72, much higher than any of the possible sums, and the products will go up much faster than the sums will.
However, the lowest possible sequence gives an ad product of 2x8, or 16. And summing the middle numbers gives 10, exactly six less than the product. We have found the sequence: 2-4-6-8.
The answer, then, is the product of the smallest two numbers, or 2x4, which is 8 (option C).
Answer: (C) 8
Question: A radio station has to choose a week of the four in a month to broadcast a certain program, and that set will repeat each month. The program can be broadcast equally on any of the four week —- 1st week vs. last week don’t matter at all —- nor does it matter whether the week the program airs are adjacent or not. Absolutely any week of the 4 weeks can be chosen. How many different week combinations of the 4 weeks can be constructed?
(A) 4
(B) 15
(C) 21
(D) 35
(E) 56
Rationale: Pick 1 week out of 4.. 4C1 =4
Answer : A
Answer: (A) 4
Question: A TV is bought for Rs.6000/- and sold at a loss of 25% find its selling price?
(A) s.4000/-
(B) s.5560/-
(C) s.3580/-
(D) s.4500/-
(E) s.5600/-
Rationale: 100 % ------> 6000 (100 * 60 = 600)
75 % ------> 4500 (75 * 60 = 4500)
Selling price = Rs.4500/-
Option 'D'
Answer: (D) s.4500/-
Question: Of the 720 students at a certain university, 25 percent are seniors. If x seniors were to graduate early and leave the university and no additional students entered or left the university, what value of x would reduce the number of seniors at the university to 20 percent?
(A) 30
(B) 36
(C) 42
(D) 45
(E) 48
Rationale: Sol:
180-x= 20% of (720-x)
4x=180
x=45
ANSWER:D
Answer: (D) 45
Question: Find the value of x. (47% of 1442 - 36% of 1412) + 66 = x?
(A) 1
(B) 2
(C) 3
(D) 6
(E) 5
Rationale: D) 6
Answer: (D) 6
Question: You forgot the three digit code of your safe. But you have written some hints to calculate the code if you forgot the code.
Below are the hints that you have written
Hint-1 : One of the numbers is correct and is placed in its correct position
5 4 7
Hint-2: Nothing is correct
5 3 0
Hint-3: Two numbers are correct but not placed at its correct position.
2 5 9
Hint-4: One number is correct but not placed at its correct position.
7 0 6
Hint-5: One number is correct but not placed at its correct position.
6 4 9
What's is the code.
(A) 927
(B) 937
(C) 947
(D) 957
(E) 967
Rationale: Solution:
927
1) 5 4 7 : 7 is placed correctly
2) 5 3 0 : None of the numbers is in the code
3) 2 5 9 : 2 & 9 are correct number of code but placed at wrong position
4) 7 0 6: Number 7 is there but placed at wrong position
5) 6 4 9: Number 9 is there but placed at wrong position
Answer A
Answer: (A) 927
Question: A box contains 30 mangoes out of which 7 are spoilt. If seven mangoes are chosen at random, find the probability that A number is selected at random from first ninty natural numbers. What is the chance that it is a multiple of either 9 or 11?
(A) 1/5
(B) 3/5
(C) 2/5
(D) 4/15
(E) 7/5
Rationale: The probability that the number is a multiple of 9 is 10/90. (Since 9*10 = 90).
Similarly the probability that the number is a multiple of 11 is 8/90. {Since 11*8 =
88).
Neither 9 nor 11 has common multiple from 1 to 90. Hence these events are mutually exclusive events. Therefore chance that the selected number is a multiple of 9 or 11 is (10+8)/90= 1/5.
ANSWER:A
Answer: (A) 1/5
Question: A motorist travels for 6 hours, the first half at 60 Kmph and the rest at 48 Kmph. Find the distance Traveled by him.
(A) 234
(B) 124
(C) 324
(D) 452
(E) 352
Rationale: Distance = 3 * 60 + 3 * 48
= 180 + 144
= 324 Km
Answer C.
Answer: (C) 324
Question: A person want to give his money of $1000 to his 3 children A,B,C in the ratio 6:4:5 What is the A+B share?
(A) $660.67
(B) $646.67
(C) $666.67
(D) $500.56
(E) $600.24
Rationale: A's share = 1000*6/15= $400
B's share = 1000*4/15= $266.67
A+B = $666.67
Answer is C
Answer: (C) $666.67
Question: Tough and Tricky questions: Combinations.
Seven students - 4 boys and 3 girls - are to sit side by side for a makeup exam. How many ways could they arrange themselves given that no two boys and no two girls can sit next to one another?
(A) 12
(B) 36
(C) 72
(D) 288
(E) 720
Rationale: 4!*3!*2 = 24*6*2 = 288 ways.
Ans. D) 288
Answer: (D) 288
Question: Which of the following is the ninth to the right of the twentieth element from the right end of the above arrangement?
(A) I only
(B) II only
(C) I and II
(D) WEWW
(E) II and III
Rationale: Answer: Option D
Answer: (D) WEWW
Question: he L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is:
(A) 30
(B) 40
(C) 38
(D) 36
(E) 39
Rationale: Let the numbers be 2x and 3x.
Then, their L.C.M. = 6x.
So, 6x = 48 or x = 8
The numbers are 16 and 24.
Hence, required sum = (16 + 24) = 40. Answer: Option B
Answer: (B) 40
Question: A University library checks out books for a 5-week period, but allows them to be renewed up to 6 times. What is the Maximum no.of Weeks the library will allow a book to be checked out?
(A) 33
(B) 35
(C) 37
(D) 39
(E) 41
Rationale: Library checkout is 5 week period and can be renewed 6 times. So he can keep for a maximum of 8 weeks.
when he is allotted the book
0-5 weeks (1st renew)
5-10 weeks (2nd renew)
10-15 weeks (3rd renew)
15-20 weeks (4th renew)
20-25 weeks (5th renew)
25-30 weeks (6th renew)- maximum renew
after that,
30-35 (submission)
So 35 weeks is maximum.
ANSWER:B
Answer: (B) 35
Question: X is a positive number which when increased by 17 is equal to 60 times the reciprocal of the number. Find value of X?
(A) 3
(B) 4
(C) 6
(D) 7
(E) 8
Rationale: Let the number be x.
Then, x + 17 =	60
x
x2 + 17x - 60 = 0
(x + 20)(x - 3) = 0
x = 3
A
Answer: (A) 3
Question: 25, 24, 22, ?, 15, 10
(A) 19
(B) 15
(C) 13
(D) 18
(E) 17
Rationale: The Pattern is - 1, - 2, -3, ...
22-3=19
Answer :A
Answer: (A) 19
Question: A train 120 m long is running with a speed of 66 km/hr. In what time will it pass a man who is roller skating at 12 km/hr in the direction opposite to that in which the train is going?
(A) 4.92
(B) 6.92
(C) 7.92
(D) 4.92
(E) 2.92
Rationale: Speed of train relative to man = 66 + 12 = 78 km/hr.
= 78 * 5/18 = 65/3 m/sec.
Time taken to pass the man = 150 * 3/65
= 6.92 sec.
Answer: B
Answer: (B) 6.92
Question: A train 300 m long can cross an electric pole in 10 sec and then find the speed of the train?
(A) 88 Kmph
(B) 89 Kmph
(C) 72 Kmph
(D) 16 Kmph
(E) 108 Kmph
Rationale: Length = Speed * time
Speed = L/T
S = 300/10
S = 30 M/Sec
Speed= 30*18/5 (To convert M/Sec in to Kmph multiply by 18/5)
Speed = 108 Kmph
Answer: E
Answer: (E) 108 Kmph
Question: In a class of 20 students in an examination in Mathematics 2 students scored 100 Marks each, 3 get
zero each and the average of the rest was 40. What is the average Of the whole class?
(A) 46
(B) 49
(C) 40
(D) 38
(E) 35
Rationale: Total marks obtained by a class of 20 students
= 2 * 100 + 3 * 0 + 15 * 40
= 200 + 600 = 800
: Average marks of whole class = 800/20 = 40
Answer is C.
Answer: (C) 40
Question: In how many ways can A, B, C, D and E be seated in a row such that A and B are not seated next to each other?
(A) 24
(B) 72
(C) 56
(D) 76
(E) 96
Rationale: When the constraint on an arrangement says,Two people should not be seated together,we do just the opposite. We make them sit together! Kind of tie them with a thread and assume they are one unit!
Let's see why....
These 5 people can be arranged in 5! ways. These are the total number of ways you get.
Now, when we tie 2 people together, we have only 4 entities to arrange. We can do this in 4! ways. But in each of these entities, the two people can sit in two different ways (AB and BA). So number of ways in which these two people sit together is 4!*2!.
Now, the ways in which these two people will not be together will be 5!- 4!*2! = 4!(5 - 2) = 72
Answer :B
Answer: (B) 72
Question: A bowl was filled with 10 ounces of water, and 0.006 ounce of the water evaporated each day during a 50-day period. What percent of the original amount of water evaporated during this period?
(A) 0.004%
(B) 0.04%
(C) 0.40%
(D) 3%
(E) 40%
Rationale: Total amount of water evaporated each day during a 50-day period = .006 * 50
=.006 * 100/2
= .6/2
= .3
percent of the original amount of water evaporated during this period = (.3/10) * 100%
= 3 %
Answer D
Answer: (D) 3%
Question: Two trains of length 100 m and 200 m are 100 m apart. They start moving towards each other on parallel tracks, at speeds 90 kmph and 72 kmph. In how much time will the trains cross each other?
(A) 80/6
(B) 80/7
(C) 80/9
(D) 80/2
(E) 80/1
Rationale: Relative speed = (90 + 72)* 5/18 = 45 mps.
The time required = d/s = (100 + 100 + 200)/45
= 400/45 = 80/9 sec.
Answer:C
Answer: (C) 80/9
Question: If x > 3000, then the value of (1.2x)/(0.5x-406) is closest to?
(A) 1/6
(B) 1/3
(C) 10/21
(D) 1/2
(E) 3/1
Rationale: assume x = 3002
(1.2 x)/(0.5 x-406) = 3602.4 / (3002*0.5-406)
=3602.4 / 1095
= = 3 /1
E
Answer: (E) 3/1
Question: A, B and C can do a piece of work in 9 days, 14 days and 28 days respectively. How long will they taken, if all the three work together?
(A) 4.12 days
(B) 4.52 days
(C) 3.52 days
(D) 2.52 days
(E) 4.92 days
Rationale: 1/9 + 1/14 + 1/28 = 55/252
=> 252/55 days
=>4.52 days
ANSWER:B
Answer: (B) 4.52 days
Question: Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and they are sold fetching the profit 10% and 40% respectively. If the vodkas are mixed in equal ratio and the individual profit percent on them are increased by 4/3 and 5/3 times respectively, then the mixture will fetch the profit of
(A) 18%
(B) 20%
(C) 21%
(D) 40%
(E) Cannot be determined
Rationale: Answer: D.
Answer: (D) 40%
Question: The table below shows the enrollment in various classes at a certain college.
Class | Number of Students
Biology | 55
Physics | 35
Calculus | 40
Although no student is enrolled in all three classes, 15 are enrolled in both Biology and Physics, 10 are enrolled in both Biology and Calculus, and 12 are enrolled in both Physics and Calculus. How many different students are in the three classes?
(A) 51
(B) 88
(C) 93
(D) 125
(E) 162
Rationale: No. of total students = no of students of biology + no of students of physics but not biology + no of students of calculus but not biology and physics
No. of total students = 55 + (35-15) + (40-10-12) = 55+20+18 = 93
Hence, Answer will be C.
Answer: (C) 93
Question: Set A contains three different positive odd integers and six different positive even integers. Set B contains two different positive odd integers and three different positive even integers. If one integer from set A and one integer from set B are chosen at random, what is the probability that the product of the chosen integers is even?
(A) 3/5
(B) 7/10
(C) 13/15
(D) 23/30
(E) 37/45
Rationale: There can be an odd product only if both integers are odd.
P(odd product) = 3/9*2/5 = 2/15
P(even product) = 1 - 2/15 = 13/15
The answer is C.
Answer: (C) 13/15
Question: Last year a certain bond yielded 15 percent of its face value in interest. If that interest was approximately 4 percent of the bonds selling price of $7,500, what is the bonds face value?
(A) $2,000
(B) $6,750
(C) $7,425
(D) $7,500
(E) $9,375
Rationale: Interest = 0.15*face value = 0.04*7,500 --> face value = 0.04*7,500/0.15 = 2,000.
Answer: A.
Answer: (A) $2,000
Question: The sum of five numbers is 655. The average of the first two numbers is 85 and the third number is 125. Find the average of the two numbers?
(A) 180
(B) 88
(C) 278
(D) 268
(E) 267
Rationale: Let the five numbers be P, Q, R, S and T.
=> P + Q + R + S + T = 655.
(P + Q)/2 = 85 and R = 125
P + Q = 170 and R = 125
P + Q + R = 295
S + T = 655 - (P + Q + R) = 360
Average of the last two numbers = (S + T)/2 = 180.
Answer: A
Answer: (A) 180
Question: When a is divided by b, the quotient is c with remainder 3. What is a in terms of 3, b, and c?
(A) b+c+3
(B) b+c
(C) bc
(D) bc+3
(E) bc-3
Rationale: a/b =c with remainder 3
i.e, a-3 = bc
a= bc+3
Answer : D
Answer: (D) bc+3
Question: A train passes a station platform in 36 seconds and a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, what is the length of the platform?
(A) 299
(B) 240
(C) 288
(D) 161
(E) 921
Rationale: Speed = (54 * 5/18) m/sec = 15 m/sec. Length of the train = (15 x 20)m = 300 m. Let the length of the platform be x meters. Then, (x + 300)/36 = 15
==> x + 300 = 540 ==> x
= 240 m.
Answer:B
Answer: (B) 240
Question: The Computer Store has reduced the price of a computer by 15%.
What is the original price of the computer if the sale price is $1275?
(A) 1600
(B) 2500
(C) 1500
(D) 1200
(E) 2100
Rationale: Selling price = 1275/-
Reduction rate : 15%
Orginal price is : 1500/-
Answer is Option C
Answer: (C) 1500
Question: A certain sum is invested at simple interest at 14% p.a. for two years instead of investing at 12% p.a. for the same time period. Therefore the interest received is more by Rs. 840. Find the sum?
(A) s. 7000
(B) s. 9000
(C) s. 14000
(D) s. 17000
(E) s. 42000
Rationale: Let the sum be Rs. x.
(x * 14 * 2)/100 - (x * 12 * 2)/100 = 840
=> 28x/100 - 24x/100 =840
=> 4x/100 = 840
=> x = 42000.
ANSWER:E
Answer: (E) s. 42000
Question: The length of the train which is moving at 66 km/hr, can crosses a bridge of length 345 m in 60 sec, is ?
(A) 575 m.
(B) 645 m.
(C) 555 m.
(D) 655 m.
(E) 755 m.
Rationale: Speed = [66 X 5/18] m/sec = [55/3] m/sec Time = 60 sec Let the length of train be x metres. Then, (345 + x)/60= 55/3
345+x=1100
755 = x
x= 755 m.
Answer:E
Answer: (E) 755 m.
Question: A boy has 4 trousers and 15 shirts. In how many different ways can he select a trouser and a shirt?
(A) 299
(B) 60
(C) 67
(D) 89
(E) 271
Rationale: The boy can select one trouser in 4 ways.
The boy can select one shirt in 15 ways.
The number of ways in which he can select one trouser and one shirt is 4 * 15
= 60 ways.
Answer: B
Answer: (B) 60
Question: Two trains 140 m and 160 m long run at the speed of 60 km/hr and 40 km/hr respectively in opposite directions on parallel tracks. The time which they take to cross each other is?
(A) 10.6
(B) 10.5
(C) 10.1
(D) 10.8
(E) 10.2
Rationale: Relative speed = 60 + 40 = 100 km/hr.
= 100 * 5/18 = 250/9 m/sec.
Distance covered in crossing each other = 140 + 160 = 300 m.
Required time = 300 * 9/250 = 54/5 = 10.8 sec.
Answer: D
Answer: (D) 10.8
Question: How many positive integers less than 5,000 are evenly divisible by neither 13 nor 21?
(A) 4,514
(B) 4,475
(C) 4,521
(D) 4,428
(E) 4,349
Rationale: integers less than 5000 divisible by 13 5000/13 =333.something , so 333
integers less than 5000 divisible by 21 5000/21 = 238.## , so 238
we have double counted some, so take LCM of 13 and 21 =105 and divide by 5000, we get 47. so all numbers divisible by 13 and 21 = 333 +238 -47 =524
now subtract that from 4999. 4999- 524 = 4349 answer E.
Answer: (E) 4,349
Question: The ratio of incomes of C and D is 3 : 4 and their expenditures is 4 : 5. If the savings of C is one fourth of his income then what is the ratio of savings of C and D?
(A) 11/19
(B) 12/19
(C) 13/19
(D) 14/19
(E) 15/19
Rationale: According to the given condition
3x - 4y = (1/4)3x
=> (3/4)3x = 4y
=> 9x = 16y
=> x/y = 16/9
so ratio of savings = (3x - 4y)/(4x - 5y) = (3*16 - 4*9)/(4*16 - 5*9) = 12/19
ANSWER:B
Answer: (B) 12/19
Question: A train 140 meters long takes 6 seconds to cross a man walking at 5 kmph in the direction opposite to that of the train. Find the speed of the train.
(A) 45 kmph
(B) 50 kmph
(C) 55 kmph
(D) 60 kmph
(E) 79 kmph
Rationale: Explanation:
Let the speed of the train be x kmph.
Speed of the train relative to man = (x + 5) kmph = (x + 5) ×5/18 m/sec.
Therefore 140/((x+5)×5/18)=6 <=> 30 (x + 5) = 2520 <=> x = 79
Speed of the train is 79 kmph.
Answer: Option E
Answer: (E) 79 kmph