divelab/Sys2Bench huggingface dataset:
Question: The distance from City A to City B is 220 miles. While driving from City A to City B, Bob drives at a constant speed of 40 miles per hour. Alice leaves City A 30 minutes after Bob. What is the minimum constant speed in miles per hour that Alice must exceed in order to arrive in City B before Bob?
Options:
A)44
B)48
C)50
D)52
E)54
Rationale: The time it takes Bob to drive to City B is 220 / 40 = 5.5 hours.
Alice needs to take less than 5 hours for the trip.
Alice needs to exceed a constant speed of 220 / 5 = 44 miles per hour.
The answer is A.
Answer: A
Question: Of the goose eggs laid at a certain pond, 1/3 hatched and 3/4 of the geese that hatched from those eggs survived the first month. Of the geese that survived the first month, 3/5 did not survive the first year. If 120 geese survived the first year and if no more than one goose hatched from each egg, how many goose eggs were laid at the pond?
Options:
A)1000
B)1200
C)1400
D)1600
E)1800
Rationale: Let x be the number of eggs that were laid.
(2/5)(3/4)(1/3)x = 120
(6/60)x = 120
x = 1200
The answer is B.
Answer: B
Question: If log 64 = 1.8061, then the value of log 16 will be (approx)?
Options:
A)1.204
B)1.9874
C)1.3564
D)1.2547
E)1.6547
Rationale: Log 64 = 1.8061
log4^3 = 1.8061
log4 = 0.6020
2log4=1.2040
log4^2=1.2040
log16=1.2040(approx)
ANSWER A
Answer: A
Question: If x=kc and y=kt,then y+2x =?
Options:
A)k(t-c)
B)k(c-t)
C)k(t+2c)
D)t(k-c)
E)k(1-t)
Rationale: Just put in the given values of y and x
y +2x = kt +2 kc
Now take k common
y +2 x = k(t +2c)
You get (C) as your answer.
Answer: C
Question: A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw?
Options:
A)64
B)23
C)32
D)45
E)80
Rationale: We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).
Required number of ways	= (3C1 x 6C2) + (3C2 x 6C1) + (3C3)=>64
ANSWER A
Answer: A
Question: A and B began business with Rs.3000 and Rs.4000 after 8 months, A withdraws Rs.1000 and B advances Rs.1000 more. At the end of the year, their profits amounted to Rs.798 find the share of A.
Options:
A)240
B)388
C)379
D)304
E)122
Rationale: Explanation:
(3*8 + 2*4):(4*8 + 5*4)
8:13
8/21 * 798 = 304
Answer: D
Answer: D
Question: A shopkeeper sells two articles at Rs.1000 each, making a profit of 20% on the first article and a loss of 20% on the second article. Find the net profit or loss that he makes?
Options:
A)7%
B)3%
C)9%
D)4%
E)2%
Rationale: SP of first article = 1000
Profit = 20%
CP = (SP)*[100/(100+P)] = 5000/6 = 2500/3
SP of second article = 1000
Loss = 20%
CP = (SP)*[100/(100-L)] = 5000/4 = 1250
Total SP = 2000
Total CP = 2500/3 + 1250 = 6250/3
CP is more than SP, he makes a loss.
Loss = CP-SP = (6250/3)- 2000 = 250/3
Loss Percent = [(250/3)/(6250/3)]*100
=0.04 * 100 = 4%
Answer: D
Answer: D
Question: The wages of labourers in a factory increases in the ratio 22:25 and there was a reduction in the number of labourers in the ratio 15:11. Find the original wage bill if the present bill is Rs 5000?
Options:
A)6000
B)6500
C)7000
D)6300
E)None of these
Rationale: Explanation:
ratio of increase of wages=22:25
ratio of decrease of labourers=15:11
compound ratio of wages of labourers= 22*15:25*11=330:275
final bill = Rs 5000
for 275 ratio wages= Rs 5000
for 1 " " = 5000/275
for 330 " " = 5000/275*330 = Rs 6000
Answer : A
Answer: A
Question: Two consecutive traffic signals each show either red or green. How many different arrangements of the two signals are possible? (Note: “Red-green” is different from “green-red.”)
Options:
A)10
B)9
C)8
D)7
E)4
Rationale: Since the answer choices to the question are so small (relatively speaking), you can get to the correct answer without having to do any fancy calculations. With a drawing and a bit of 'brute force', you can answer this question relatively quickly.
We're told that 2 consecutive traffic signals each show either red or green. We're asked how many different arrangements of the 2 signals are possible.
Total Options: 4
Final Answer:
E
Answer: E
Question: Two clocks are set correctly at 1 0'clock in the afternoon. One keeps correct time. The other gains 5 seconds per minute. That evening, when
the fast clock shows 6:20, what does the correct clock show?
Options:
A)4.0
B)5.0
C)6.0
D)7.2
E)3.5
Rationale: Faster clock gains 5 seconds per minute or 5*60=300 seconds or 5minutes per hour.
Both clocks are set at 1:00
At 2:00
correct clock is at 2:00
faster clock is at 2:05:00 (1 hour + gained 5 minutes)
At 3:00 -
correct clock is at 3:00
faster clock is at 3:10:00 (2 hours + gained 10 minutes)
.
In 5 hours, faster clock would gain 20 minutes. It will show 6:20. At that point, the correct clock would show 6:00.
Answer is C
Answer: C
Question: There are two divisions A and B of a class, consisting of 36 and 44 students respectively. If the average weight of divisions A is 40 kg and that of division b is 35 kg. What is the average weight of the whole class?
Options:
A)37.0
B)37.75
C)37.5
D)37.25
E)38.0
Rationale: Total weight of students in division A = 36 × 40
Total weight of students in division B = 44 × 35
Total students = 36 + 44 = 80
Average weight of the whole class
=((36×40)+(44×35))/80
=((9×40)+(11×35))/20
=((9×8)+(11×7))/4
=(72+77)/4
=149/4
=37.25
Answer is D.
Answer: D
Question: A school’s annual budget for the purchase of student computers increased by 44% this year over last year. If the price of student computers increased by 20% this year, then the number of computers it can purchase this year is what percent greater than the number of computers it purchased last year?
Options:
A)20%
B)30%
C)22%
D)28%
E)10%
Rationale: Lats year: budget = $100, price = $1 --> the number of computers purchased = 100/1 = 100;
This year: budget = $144, price = $1.2 --> the number of computers purchased = 144/1.2 = 120.
Increase in the number of computers purchased = 120%.
Answer: A.
Answer: A
Question: A began a business with Rs. 85000. He was joined afterwards by B with Rs. 42,500. For how much perod does B join, if the profits at the end of the year are divided in the ratio of 3:1 ?
Options:
A)8
B)9
C)1
D)6
E)5
Rationale: Suppose B joined for x months.
Given profit is divided in the ratio 3:1. Then,
= 8 months.
Answer: A
Answer: A
Question: What should be the least number to be added to the 4499 number to make it divisible by 9?
Options:
A)1
B)17
C)18
D)77
E)26
Rationale: Answer : 1
Option : A
Answer: A
Question: If 20 men can build a wall 66 metres long in 12 days, what length of a similar can be built by 86 men in 8 days?
Options:
A)189.2 mtrs
B)378.4 mtrs
C)478.4 mtrs
D)488.4 mtrs
E)578.4 mtrs
Rationale: If 20 men can build a wall 66 metres long in 12 days,
length of a similar wall that can be built by 86 men in 8 days = (66*86*8)/(12*20) = 189.2 mtrs
ANSWER:A
Answer: A
Question: During a certain week, a seal ate 60% of the first 80 smelt it came across, and 30% of the remaining smelt it came across. If the seal ate 40% of the smelt it came across during the entire week, how many smelt did it eat?
Options:
A)32
B)40
C)96
D)64
E)80
Rationale: Total smelt = x.
Then, 0.6*80+0.3(x-80)=0.4*x --> x=240 --> 0.4*x=96.
Answer: C.
Answer: C
Question: How many words can be formed by using all letters of the word ‘DAUGHTER’ so that the vowels always come together?
Options:
A)3561
B)7845
C)5410
D)4320
E)1245
Rationale: Given word contains 8 different letters. When the vowels AUE are always together, we may suppose them to form an entity, treated as one letter.
Then, the letters to be arranged are DGNTR (AUE).
Then 6 letters to be arranged in 6p6 = 6! = 720 ways.
The vowels in the group (AUE) may be arranged in 3! = 6 ways.
Required number of words = (720x6) = 4320.
Ans: D
Answer: D
Question: The length and width of a rectangle are integer values. What is the area of the smallest such rectangle that can be inscribed in a circle whose radius is also an integer
Options:
A) 42
B) 24
C) 36
D) 48
E) 60
Rationale: If we consider all pythagorean triplets with integral values, we have the following results:-
(3,4,5),(6,8,10)........
The diameter of the circle is 5 in the 1st case and 10 in the 2nd.
But the question says that radius of the circle is also an integer hence the 1st triplet will be ruled out.
We are left with the 2nd triplet where radius of the circle=5,
Hence the sides of the rectangle are 6 and 8 making the area =42.
A
Answer: A
Question: During a trip, Francine traveled x percent of the total distance at an average speed of 20 miles per hour and the rest of the distance at an average speed of 80 miles per hour. In terms of x, what was Francineâ€™s average speed for the entire trip?
Options:
A)(180-x)/2
B)(x+60)/4
C)(300-x)/5
D)8,000/(100+3x)
E)12,000/(x+200)
Rationale: D....
X% traveeled at 20 mph
100-X% traveeled at 80 mph
total timt taken = x/20 +(100-x)/80 = (100+3x)/80
Total distance = x+(100-x) = 100
Av speed = Tot Dis/Time Taken which is D
Answer: D
Question: Rs.1500 is divided into two parts such that if one part is invested at 6% and the other at 5% the whole annual interest from both the sum is Rs.75. How much was lent at 5%?
Options:
A)388
B)1500
C)277
D)500
E)271
Rationale: (x*5*1)/100 + [(1500 - x)*6*1]/100 = 75
5x/100 + 90 – 6x/100 = 75
x/100 = 15
=> x = 1500.Answer: B
Answer: B
Question: The average age of the mother and her six children is 12 year which is reduced by 5 years if the age of the mother is excluded. how old is the mother ?
Options:
A)40 year
B)42 year
C)48 year
D)50 year
E)None of these
Rationale: Solution
age of the mother= ( 12 × 7 - 7 × 6 ) year = 42 year. Answer B
Answer: B
Question: The edge of a cube is 3a cm. Find its surface?
Options:
A)64a2 cm2
B)54a2 cm2
C)84a2 cm2
D)24a2 cm2
E)34a2 cm2
Rationale: 6a2 = 6 * 3a * 3a = 54a2
Answer:B
Answer: B
Question: Two pipes can fill the cistern in 10hr and 12 hr respectively, while the third empty it in 20hr.
If all pipes are opened simultaneously, then the cistern will be filled in
Options:
A)10.5
B)8
C)6
D)7.5
E)8.5
Rationale: First pipe fills cistern in 10 hours.
So, in 1 hour, pipe fills 1/10 of the cistern capacity
2nd pipe fills cistern in 12 hours.
So, in 1 hour, 2nd pipe fills 1/12 of the cistern capacity
3rd pipe empties cistern in 20 hours
So, in 1 hour, 3rd pipe empties 1/20 of the cistern capacity
Net volume added to cistern per hour = Volume filled by 1st pipe per hour + Volume filled by 2nd pipe per hour - Volume emptied by 3rd pipe per hour
= 1/10 + 1/12 - 1/20
=(6 + 5 - 3) / 60 = 8/60
=> Net volume added to cistern per hour = 2/15
So, Cistern will be fill in 15/2 hours = 7.5 hours
So, answer is D
Answer: D
Question: Rahul went to a shop and bought things worth Rs. 25, out of which 30 paise went on sales tax on taxable purchases. If the tax rate was 6%, then what was the cost of the tax free items?
Options:
A)15
B)12.1
C)19.7
D)16.0
E)20.0
Rationale: Total cost of item he pirchased = 25
tot tax = 30 paise = 30/100
tax free = x
given tax rate = 6%
25-30/100 - x6/100 = 30/100
6(25-0.3-x) = 30
x=19.7
ANSWER C
Answer: C
Question: Buffalo gives 4 liter milk, cow gives (1/2) liter milk and goat gives 1/4 liter milk. you have to get 20 liter milk by 20 animals.what is the number of buffalos?
Options:
A)1
B)2
C)3
D)4
E)5
Rationale: Assume number of respective animals are x,y,z.
x+y+z=20---(1) as the total number of animal has to be 20
Amt of milk will be 4x+(1/2)y+(1/4)z=20---(2)
Solving equation 1 and 2 we get
15x+y=60 ----(3)
Since buffalo gives 4litre and total milk is 20, x < 5
But from eq 3, x cannot be more than 4;
Further if x=1 or 2; y>20... Not possible, since total animal is 20
Thus, x=3, y=15, z=2
Ans: 3 buffalos
ANSWER:C
Answer: C
Question: Jane Odd is the owner of Oddly Odd Book Store. Jane Odd loves odd things, but her book seller only sells books in even numbers. If there are an even number of books in the book store (b = an even number), and Jane divides the books into groups of 18 books, which results in an odd number of groupings (b/18 = an odd number). Now, if Jane doubles the number of books in the store, how can she divide the books up to keep an odd number of groupings (b^2/X = an odd number)?
Options:
A)b^2/12
B)b^2/18
C)b^2/9
D)b^2/6
E)b^2/2
Rationale: Divisible by 2 (Parity Mathematics) states: an even number is an integer of the form n = 2k, where k is an integer; an odd number is an integer of the form n = 2k + 1.
Since a is an even number, a = 2k, a^2 = 4k^2 = (2)(2)(k)^2, and a/18 is an odd number 2k/18.
18 = (2)(9). Now (2k)/((2)(9)) = k/9. So, we are looking for a (2)(2) factor in the denominator.
Factors are:
12 = (2)(2)(3)
18 = (2)(9)
9 = (3)(3)
6 = (2)(3)
2 = (2)
So, 12 has more than 1 (2).
Answer: A
Answer: A
Question: A car traveled 462 miles per tankful of gasoline on the highway and 336 miles per tankful of gasoline in the city. If the car traveled 3 fewer miles per gallon in the city than on the highway, how many miles per gallon did the car travel in the city?
Options:
A)8
B)16
C)21
D)22
E)27
Rationale: I treat such problems as work ones.
Work=Rate*Time
Mileage(m)=Rate(mpg)*Gallons(g)
X gallons is a full tank
{462=RX
{336=(R-3)X solve for R, R=11
11-3=8 mpg
A
Answer: A
Question: What is the remainder when 2^20 is divided by 10 ?
Options:
A)0
B)2
C)4
D)6
E)8
Rationale: 2^20 = 2^10 x 2^10
And 2^10 = 1024
4*4= 16, so the unit digit of this multiplication is 6
The remainder is 6
ANSWER:D
Answer: D
Question: If x/y = 2/7, then (x + y)/(x - y) = ?
Options:
A)5
B)1/5
C)-1/6
D)-9/5
E)-5
Rationale: Any x and y satisfying x/y = 2/7 should give the same value for (x + y)/(x - y).
Say x=2 and y=7, then (x + y)/(x - y) = (2 + 7)/(2-7) = -9/5.
Answer: D.
Answer: D
Question: Ravi purchased 20 dozens of toys at the rate of Rs. 320 per dozen. He sold each one of them at the rate of Rs. 33. What was his percentage profit?
Options:
A)23.0%
B)21.2%
C)55.6%
D)34.12%
E)23.75%
Rationale: E
C.P. of 1 toy = 320/12 = Rs. 26.66
S.P of 1 toy = Rs. 33
Profit = 6.33/31.25 * 100 = 23.75%
Answer is E
Answer: E
Question: A certain quantity is measured on two different scales, the R-scale and the S-scale, that are related linearly. Measurements on the R-scale of 6 and 24 correspond to measurements on the S-scale of 30 and 60, respectively. What measurement on the R-scale corresponds to a measurement of 100 on the S-scale?
Options:
A)20
B)36
C)48
D)60
E)84
Rationale: Question says they are both related linearly so they the relationships can be represented by standard linear definition y=mx+b.
Let S scale = y. and R scale = x.
30=6m+b EQN1
60=24m+b EQN2
EQN2-EQN1 => 30=18m => m=5/3
Solving for b using either equation gives us b=20
linear relationship is represented by y=(5/3)x+20
Solve for 100:
100=5/3x+20
80=5/3x
x=48
ANS = C
Answer: C
Question: 7 does not occur in 1000. So we have to count the number of times it appears between 1 and 999. Any number between 1 and 999 can be expressed in the form of xyz where 0 < x, y, z < 9.
Options:
A)500
B)300
C)250
D)415
E)236
Rationale: 1. The numbers in which 7 occurs only once. e.g 7, 17, 78, 217, 743 etc
This means that 7 is one of the digits and the remaining two digits will be any of the other 9 digits (i.e 0 to 9 with the exception of 7)
You have 1*9*9 = 81 such numbers. However, 7 could appear as the first or the second or the third digit. Therefore, there will be 3*81 = 243 numbers (1-digit, 2-digits and 3- digits) in which 7 will appear only once.
In each of these numbers, 7 is written once. Therefore, 243 times.
2. The numbers in which 7 will appear twice. e.g 772 or 377 or 747 or 77
In these numbers, one of the digits is not 7 and it can be any of the 9 digits ( 0 to 9 with the exception of 7).
There will be 9 such numbers. However, this digit which is not 7 can appear in the first or second or the third place. So there are 3 * 9 = 27 such numbers.
In each of these 27 numbers, the digit 7 is written twice. Therefore, 7 is written 54 times.
3. The number in which 7 appears thrice - 777 - 1 number. 7 is written thrice in it.
Therefore, the total number of times the digit 7 is written between 1 and 999 is 243 + 54 + 3 = 300
Ans: B
Answer: B
Question: A tradesman by means of his false balance defrauds to the extent of 40%? in buying goods as well as by selling the goods. What percent does he gain on his outlay?
Options:
A)17%
B)77%
C)44%
D)96%
E)47%
Rationale: g% = 40 + 40 + (40*40)/100
= 96%
Answer: D
Answer: D
Question: A bag contains ten marbles of the same size: 3 are identical green marbles, 2 are identical red marbles, and the other 5 are five distinct colors. If 5 marbles are selected at random, how many distinct combinations W of five marbles could be drawn?
Options:
A) 41
B) 51
C) 62
D) 72
E) 82
Rationale: there will be different cases, which we will take one by one..
1)case 1 when all five colours are different-7c5=21..
2)case 2 when three different colours and 2 of same type=6c3*2=30..
3)case 3 when two different colours and 3 of same type=6c2=15..
4)case 4 when one different colours and 2 of two different colors=5c1*4!/2!2!=15...
5)case 4 when 3 of one type and 2 of other colour=1...
total W-82 ans E
Answer: E
Question: Diana took out a charge account at the General Store and agreed to pay 7% simple annual interest. If she charges $75 on her account in January, how much will she owe a year later, assuming she does not make any additional charges or payments?
Options:
A)$79.25
B)$79.50
C)$79.75
D)$80.00
E)$80.25
Rationale: 1.07*$75=$80.25
The answer is E.
Answer: E
Question: A grocer has a sales of Euro 520 , Euro 320 , Euro 480 , Euro 120 and Euro 720 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Euro 620?
Options:
A)uro 4691
B)uro 1591
C)uro 2591
D)uro 1560
E)uro 1991
Rationale: D
Euro 1560
Total sale for 5 months = Euro (520 + 320 + 480 + 120 + 720) = Euro 2160.
Required sale = Euro [ (620 x 6) - 2160]
= Euro (3720- 2160)
= Euro 1560.
Answer is D
Answer: D
Question: If S is the sum of the reciprocals of the consecutive integers from 51 to 60, inclusive, which of the following is less than S?
I. 1/5
II. 1/6
III. 1/7
IV. 1/5>S >1/6
Options:
A)I
B)II
C)III
D)IV
E)I & II
Rationale: Since we summarize the reciprocals from 60 to 51, we can say also that we add ten numbers who are all (with one exception 1/60) greater than 1/60, so that the sum must be greater than 1/6.
On the other side we can say that we add the reciprocals from 51 to 60, so that the sum has to be less than the sum of ten times 1/51.
sum has to be greater than 1/6 and less than 1/5
D
Answer: D
Question: The variable x is inversely proportional to the square of the variable y. If y is divided by 6a, then x is multiplied by which of the following?
Options:
A)1/36a
B)1/36a^2
C)1/6a
D)36a
E)36a^2
Rationale: initially - did x= 1/y^2
But this yielded the wrong answer,
I then re examined the question and tried y^2/x =1
Which I then got y = SQR(X)
subbing into y/6a I got answer E)
Answer: E
Question: A certain electric-company plan offers customers reduced rates for electricity used between 8 p.m. and 8 a.m. weekdays and 24 hours a day Saturdays and Sundays. Under this plan, the reduced rates X apply to what fraction of a week?
Options:
A) 1/2
B) 5/8
C) 9/14
D) 16/21
E) 9/10
Rationale: Number of hours between 8 PM to 8 AM = 12
Number of hours with reduced rates = (12*5)+(24*2)
Hours with reduced rates X/Total number of hours in a week = (12*5)+(24*2)/(24*7) = 108/(24*7) = 9/14
Answer: C
Answer: C
Question: For a certain art exhibit, a museum sold admission tickets to a group of 30 people every 5 minutes from 9:00 in the morning to 5:55 in the afternoon, inclusive. The price of a regular admission ticket was $8 and the price of a student ticket was $4. If on one day 3 times as many regular admission tickets were sold as student tickets, what was the total revenue from ticket sales that day?
Options:
A)$20960
B)$21920
C)$22680
D)$28500
E)$29160
Rationale: From 9:00 in the morning to 5:55 in the afternoon, inclusive there are 9*12=108 five-minute intervals, thus total of 108*30 tickets were sold.
Say x student and 3x regular tickets were sold, then x+3x=108*30 --> x=27*30 and 3x=3*(27*30)=27*90.
Therefore, the total revenue from ticket sales that day was 27*30*4+27*90*8=$22,680.
Answer: C.
Answer: C
Question: A dishonest dealer professes to sell goods at the cost price but uses a weight of 800 grams per kg, what is his percent?
Options:
A)22
B)25
C)77
D)99
E)12
Rationale: 800 --- 200
100 --- ? => 25%
Answer: B
Answer: B
Question: The five sides of a pentagon have lengths of 2, 3, 4, 5 and 6 inches. Two pentagons are considered different only when the positions of the side lengths are different relative to each others. What is the total number of different possible pentagons that could be drawn using these five side lengths ?
Options:
A)5
B)12
C)24
D)32
E)120
Rationale: The pentagon in the problem above can be flipped over (Clockwise Arrangement of Sides = Counter-clockwise arrangement), the number of distinct ways to arrange the 5 sides = (5-1)!/2 = 12.
Answer: option B
Answer: B
Question: Vijay bought 160 shirts at the rate of Rs. 225 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%?
Options:
A)285
B)665
C)999
D)255
E)121
Rationale: Total CP per shirt = 225 + 1400/160 + 1.75 + 320/160 = Rs. 237.5
SP = CP[(100 + profit%)/100]
= 237.5 * [(100 + 20)/100] = Rs. 285.
Answer: D
Answer: D
Question: Sara bought both German chocolate and Swiss chocolate for some cakes she was baking. The Swiss chocolate cost $3.5 per pound, and German chocolate cost $2.5 per pound. if the total the she spent on chocolate was $70 and both types of chocolate were purchased in whole number of pounds, how many total pounds of chocolate she purchased?
Options:
A)7
B)8
C)10
D)24
E)15
Rationale: If there were all the expensive ones,3.5....
There would be 70/3.5 or 20 of them but since 2.5$ ones are also there, answer has to be >20....
If all were 2.5$ ones, there will be 70/2.5 or 28..
So ONLY 24 is left
Ans D..
Answer: D
Question: A company selected 250 zip codes in Northeastern, Florida, Arizona and California in order to mail a product catalog. The number of Northeastern zip codes was twice Florida's, and four times Arizona's, and 4/3 California's. 225,000 catalogues were mailed to California and 25% resulted in product orders. What is the ratio of the number of orders to the number of California's zip codes?
Options:
A)a) 500
B)b) 750
C)c) 1,000
D)d) 1,250
E)e) 1,500
Rationale: Since N = 2F, no of Florida zip codes (F) will be N/2.
Since N = 4A, no of Arizona zip codes (A) will be N/4
Since N = (4/3)C, no of California zip codes (C) will be 3N/4
N + N/2 + N/4 + 3N/4 = 5N/2 = 250
N = 100
C = 3N/4 = 75
No of orders is 25% of 225,000 = 56,250
So required ratio = 56,250/75 = 750
Answer B
Answer: B
Question: What was the day of the week on, 17th July, 1776?
Options:
A)Tuesday
B)Wednesday
C)Monday
D)Saturday
E)SUNDAY
Rationale: 16th July, 1776 = (1775 years + Period from 1st Jan, 1776 to 16th July, 1776)
Counting of odd days :
1600 years have 0 odd day.
100 years have 5 odd days.
75 years = (18 leap years + 57 ordinary years) = [(18 x 2) + (57 x 1)] = 93 (13 weeks + 2 days) = 2 odd days.
1775 years have (0 + 5 + 2) odd days = 7 odd days = 0 odd day.
Jan Feb Mar Apr May Jun Jul
31 + 29 + 31 + 30 + 31 + 30 + 16 = 198 days= (28 weeks + 3 days)
Total number of odd days = (0 + 3) = 3.
Required day was 'Wednesday'.
Answer:B
Answer: B
Question: Two trains 110 meters and 140 meters long are running in the same direction with speeds of 102 km/hr, 66 km/hr. In how much time will the first train cross the second
Options:
A)22 sec
B)25 sec
C)27 sec
D)23 sec
E)30 sec
Rationale: Explanation:
Relative speed of the trains = (102 - 66) km/hr = 36 km/hr
= (36Ã— 5/18) m/sec = 10 m/sec.
Time taken by the trains to cross each other
= Time taken to cover (110 + 140) m at 10 m /sec = (250/10) sec = 25 sec.
Answer: Option B
Answer: B
Question: f (x + y)/(2x + y) = 4/5, then find (2x + y)/(3x + y) ?
Options:
A)5/9
B)5/6
C)5/7
D)5/4
E)5/1
Rationale: (x + y)/(2x + y) = 4/5
=> 5x + 5y = 8x + 4y => y = 3x
Now, (2x + y)/(3x + y) = 5x/6x = 5/6.
Answer: B
Answer: B