Cheukting/mathqa huggingface dataset:
Question (general): if money is invested at r percent interest , compounded annually , the amount of the investment will double in approximately 50 / r years . if luke ' s parents invested $ 12,500 in a long term bond that pays 12 percent interest compounded annually , what will be the approximate total amount of the investment 12 years later , when luke is ready for college ?
Options: a ) 62,000 , b ) 85,500 , c ) 95,500 , d ) 100,500 , e ) 100,000
Rationale: "since investment doubles in 70 / r years then for r = 8 it ' ll double in 70 / 8 = ~ 9 years ( we are not asked about the exact amount so such an approximation will do ) . thus in 18 years investment will double twice and become ( $ 5,000 * 2 ) * 2 = $ 20,000 ( after 9 years investment will become $ 5,000 * 2 = $ 10,000 and in another 9 years it ' ll become $ 10,000 * 2 = $ 20,000 ) . answer : a ."
Answer: a
Question (general): if 12 ! / 3 ^ x is an integer , what is the greatest possible value of x ?
Options: a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7
Rationale: "this question is asking how many power of 3 exists in 12 ! easiest way is to identify factor of three 12 - > 4 * 3 = > 1 power of 3 9 - > 3 * 3 = > 2 power of 3 6 - > 2 * 3 = > 1 power of 3 3 - > 1 * 3 = > 1 power of 3 total equal 5 answer : c"
Answer: c
Question (general): arnold and danny are two twin brothers that are celebrating their birthday . the product of their ages today is smaller by 13 from the product of their ages a year from today . what is their age today ?
Options: a ) 2 . , b ) 6 . , c ) 5 . , d ) 7 . , e ) 9 .
Rationale: ad = ( a + 1 ) ( d + 1 ) - 13 0 = a + d - 12 a + d = 12 a = d ( as they are twin brothers ) a = d = 6 b is the answer
Answer: b
Question (physics): a metal company ' s old machine makes bolts at a constant rate of 100 bolts per hour . the company ' s new machine makes bolts at a constant rate of 150 bolts per hour . if both machines start at the same time and continue making bolts simultaneously , how many minutes will it take the two machines to make a total of 350 bolts ?
Options: a ) 84 , b ) 72 , c ) 120 , d ) 144 , e ) 180
Rationale: old machine 100 bolts in 60 mins so , 5 / 3 bolts in 1 min new machine 150 bolts in 60 mins so , 5 / 2 bolts in 1 min together , 5 / 3 + 5 / 2 = 25 / 6 bolts in 1 min so , for 350 bolts 350 * 6 / 25 = 84 mins ans a
Answer: a
Question (general): what is the minimum possible range in scores of the 3 test - takers ? 3 people each took 5 tests . if the ranges of their scores in the 5 practice tests were 18 , 26 and 32 .
Options: a ) 26 , b ) 32 , c ) 18 , d ) 20 , e ) 30
Rationale: i simply looked at the 3 different possible scores for each individual test : 18 , 32,26 we have to find the minimum range : 32 - 18 = 14 32 - 26 = 6 26 - 18 = 8 the find the minimum range , you have to make the set of the 5 scores as small as possible . which means that 4 of the 5 scores of each individual person is zero . 6 * 5 = 30 answer : e
Answer: e
Question (physics): a leak in the bottom of a tank can empty the full tank in 8 hours . an inlet pipe fills water at the rate of 6 litres a minute . when the tank is full , the inlet is opened and due to the leak , the tank is empty in 12 hours . how many litres does the cistern hold ?
Options: a ) 7580 , b ) 7960 , c ) 8290 , d ) 8640 , e ) none
Rationale: "solution work done by the inlet in 1 hour = ( 1 / 8 - 1 / 12 ) = 1 / 24 . work done by the inlet in 1 min . = ( 1 / 24 × 1 / 60 ) = 1 / 1440 . volume of 1 / 1440 part = 6 litres . therefore , volume of whole = ( 1440 × 6 ) ‹ = › 8640 litres . answer d"
Answer: d
Question (physics): how many hours in one year ?
Options: a ) 365 , b ) 52 , c ) 1024 , d ) 8760 , e ) no one knows
Rationale: the day has 24 hours , and the year contains 365 days . so the hours exist in one year is the multiplication of 24 by 365 . so the answer = 365 * 24 = 8760 hours so the correct answer is d
Answer: d
Question (general): if k is a non - negative integer and 15 ^ k is a divisor of 759,325 then 3 ^ k - k ^ 3 =
Options: a ) 0 , b ) 1 , c ) 37 , d ) 118 , e ) 513
Rationale: "at first we should understand whether 759325 divisible by 15 . 15 is equal to 3 * 5 so 759325 should be divisible on these integers 759325 is divisible by 5 because of last digit 5 but not divisible by 3 because sum of its digits does n ' t divisible by 3 : 7 + 5 + 9 + 3 + 2 + 5 = 31 and 31 does n ' t divisible by 3 so 759325 can be divisible by 15 ^ k only if k = 0 and 15 ^ k = 1 so 3 ^ k - k ^ 3 = 1 - 0 = 1 answer is b"
Answer: b
Question (general): when tossed , a certain coin has equal probability of landing on either side . if the coin is tossed 3 times , what is the probability that it will land once on heads and twice tails ?
Options: a ) 1 / 3 , b ) 1 / 6 , c ) 1 / 4 , d ) 1 / 8 , e ) 1 / 5
Rationale: "must be once on heads and twice on tails 1 / 2 * 1 / 2 * 1 / 2 = 1 / 8 answer : d"
Answer: d
Question (general): what is the smallest integer that is multiple of 7 , 9 and 20
Options: a ) 141 , b ) 180 , c ) 130 , d ) 1260 , e ) 1420
Rationale: "correct answer : d it is the lcm of 7 , 9 and 20 which is 1260"
Answer: d
Question (gain): if a man lost 4 % by selling oranges at the rate of 36 a rupee at how many a rupee must he sell them to gain 44 % ?
Options: a ) 1 , b ) 8 , c ) 9 , d ) 24 , e ) 3
Rationale: "96 % - - - - 36 144 % - - - - ? 96 / 144 * 36 = 24 answer : d"
Answer: d
Question (other): the fisherman sale , all of the prices of the fishes sold were different . if the price of a radio sold at the fisherman sale was both the 4 th highest price and the 13 th lowest price among the prices of the fishes sold , how many fishes were sold at the fisherman sale ?
Options: a ) 13 , b ) 14 , c ) 15 , d ) 41 , e ) 16
Rationale: 3 + 12 + 1 = 16 answer : e
Answer: e
Question (gain): in a school of 600 students , 45 % wear blue shirts , 23 % wear red shirts , 15 % wear green shirts , and the remaining students wear other colors . how many students wear other colors ( not blue , not red , not green ) ?
Options: a ) 102 , b ) 112 , c ) 122 , d ) 132 , e ) 142
Rationale: "45 + 23 + 15 = 83 % 100 – 83 = 17 % 600 * 17 / 100 = 102 the answer is a ."
Answer: a
Question (general): the average of the marks of 10 students in a class is 80 . if the marks of each student are doubled , find the new average ?
Options: a ) 80 , b ) 120 , c ) 160 , d ) 270 , e ) 110
Rationale: "sum of the marks for the 10 students = 10 * 80 = 800 . the marks of each student are doubled , the sum also will be doubled . the new sum = 800 * 2 = 1600 . so , the new average = 1600 / 10 = 160 . answer : c"
Answer: c
Question (general): in a forest 250 deer were caught , tagged with electronic markers , then released . a week later , 50 deer were captured in the same forest . of these 50 deer , it was found that 5 had been tagged with the electronic markers . if the percentage of tagged deer in the second sample approximates the percentage of tagged deer in the forest , and if no deer had either left or entered the forest over the preceding week , what is the approximate number of deer in the forest ?
Options: a ) 150 , b ) 750 , c ) 1,250 , d ) 1,500 , e ) 2,500
Rationale: "the percentage of tagged deer in the second sample = 5 / 50 * 100 = 10 % . so , 250 tagged deers comprise 10 % of total # of deers - - > total # of deers = 250 * 10 = 2,500 . answer : e"
Answer: e
Question (physics): the area of a sector of a circle of radius 5 cm formed by an arc of length 5.5 cm is ?
Options: a ) 13.78 , b ) 13.67 , c ) 13.75 , d ) 13.98 , e ) 13.28
Rationale: "( 5 * 5.5 ) / 2 = 13.75 answer : c"
Answer: c
Question (physics): jane makes toy bears . when she works with an assistant , she makes 60 percent more bears per week and works 10 percent fewer hours each week . having an assistant increases jane ’ s output of toy bears per hour by what percent ?
Options: a ) 70 % , b ) 80 % , c ) 100 % , d ) 180 % , e ) 200 %
Rationale: "we can use fractional equivalents here to solve the problem 80 % = 4 / 5 ; this means that in 1 st case if she prepares 5 bears , in 2 nd case she prepares 9 bears 10 % = 1 / 10 ; this means that in 1 st case if she needs 10 hours , in 2 nd case she needs 9 hours now we come to productivity based on above fractional values the productivity in 1 st case is 0.5 bears / hour and in the 2 nd case it is 1 bear / hour hence the productivity is double with the assistant i . e . the increase in productivity is 70 % a"
Answer: a
Question (general): there were totally 100 men . 83 are married . 75 have t . v , 85 have radio , 70 have a . c . how many men have t . v , radio , a . c and also married ?
Options: a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15
Rationale: 100 - ( 100 - 83 ) - ( 100 - 75 ) - ( 100 - 85 ) - ( 100 - 70 ) = 100 - 17 - 25 - 15 - 30 = 100 - 87 = 13 answer : c
Answer: c
Question (physics): a 450 m long train crosses a platform in 39 sec while it crosses a signal pole in 18 sec . what is the length of the platform ?
Options: a ) 600 , b ) 525 , c ) 360 , d ) 370 , e ) 380
Rationale: "speed = 450 / 18 = 25 m / sec . let the length of the platform be x meters . then , ( x + 450 ) / 39 = 25 = > x = 975 m . l = 975 - 450 = 525 answer : option b"
Answer: b
Question (physics): the distance between west - town to east - town is 15 kilometers . two birds start flying simultaneously towards one another , the first leaving from west - town at a speed of 4 kilometers per minute and the second bird , leaving from east - town , at a speed of 1 kilometers per minute . what will be the distance , in kilometers , between the meeting point and west - town ?
Options: a ) 3 . , b ) 7 . , c ) 10 . , d ) 12 , e ) 15 .
Rationale: "by end of 1 st min : distance covered by 1 st bird is 4 km and 2 nd is 1 km . total distance = 5 km . by end of 2 nd min : distance covered by 1 st bird is 8 km and 2 nd is 2 km . total distance = 10 km . by end of 3 rd min : distance covered by 1 st bird is 12 km and 2 nd is 3 km . total distance = 15 km . so , meeting point is 12 kms from west - town ( from where 1 st bird started ) hence , answer will be d ."
Answer: d
Question (general): if an integer n is to be chosen at random from the integers 1 to 94 , inclusive , what is the probability that n ( n + 1 ) ( n + 2 ) will be divisible by 8 ?
Options: a ) 1 / 4 , b ) 3 / 8 , c ) 1 / 2 , d ) 5 / 8 , e ) 59 / 94
Rationale: "n ( n + 1 ) ( n + 2 ) will be divisible by 8 when n is a multiple of 2 or when ( n + 1 ) is a multiple of 8 . thus when n is even , this whole expression will be divisible by 8 . from 1 to 96 , there are 47 even integers . now when ( n + 1 ) is multiple by 8 , we have 12 such values for ( n + 1 ) probability that n ( n + 1 ) ( n + 2 ) will be divisible by 8 = ( 47 + 12 ) / 94 = 59 / 94 = 5 / 8 ans is e"
Answer: e
Question (general): what is x if x + 2 y = 10 and y = 1 ?
Options: a ) a ) 10 , b ) b ) 8 , c ) c ) 6 , d ) d ) 4 , e ) e ) 2
Rationale: x = 10 - 2 y x = 10 - 2 . x = 8 answer : b
Answer: b
Question (general): what is the median of a set of consecutive integers if the sum of nth number from the beginning and nth number from the end is 50 ?
Options: a ) 10 , b ) 25 , c ) 50 , d ) 75 , e ) 100
Rationale: "surprisingly no one answered this easy one . property of a set of consecutive integerz . mean = median = ( first element + last element ) / 2 = ( second element + last but one element ) / 2 = ( third element + third last element ) / 2 etc . etc . so mean = median = 50 / 2 = 25 answer is b"
Answer: b
Question (general): set j consists of 15 consecutive even numbers . if the smallest term in the set is - 10 , what is the range of the positive integers in set j ?
Options: a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 20
Rationale: "since there are only 15 integers , another approach is the just list all 15 . we get : - 10 , - 8 , - 6 , - 4 , - 2 , 0,2 , 4 , 6 , 8 , 10 , 12 , 14 , 16 , 18 range of positive integers = 18 - 2 = 16 answer : d"
Answer: d
Question (probability): a number is selected at random from the first 40 natural numbers . what is the probability that the number is a multiple of either 2 or 21 ?
Options: a ) 21 / 40 , b ) 2 / 5 , c ) 7 / 15 , d ) 4 / 15 , e ) 11 / 30
Rationale: "number of multiples of 2 from 1 through 40 = 40 / 2 = 20 number of multiples of 21 from 1 through 40 = 40 / 21 = 1 number of multiples of 2 and 21 both from 1 through 40 = number of multiples of 21 * 2 ( = 42 ) = 0 total favourable cases = 20 + 1 - 0 = 21 probability = 21 / 40 answer : option a"
Answer: a
Question (general): if a , b , c and d are positive integers less than 4 , and 4 ^ a + 3 ^ b + 2 ^ c + 1 ^ d = 78 then what is the value of b / c ?
Options: a ) 3 , b ) 2 , c ) 1 , d ) 1 / 2 , e ) 1 / 3
Rationale: since a , b and c are positive integers less than 4 we have the following possibilities : 4 ^ a = 416 , or 64 3 ^ b = 39 , or 27 2 ^ c = 24 , or 8 trial and error gives us quite quick the solution of 4 ^ a = 64 3 ^ b = 9 2 ^ c = 4 64 + 9 + 4 = 77 i . e . c = 2 and b = 2 - - - - > b / c = 1 / 1 = 1 the correct answer is c
Answer: c
Question (general): on a school â € ™ s annual day sweets were to be equally distributed amongst 112 children . but on that particular day , 32 children were absent . thus the remaining children got 6 extra sweets . how many sweets was each child originally supposed to get ?
Options: a ) 28 , b ) 27 , c ) 11 , d ) 15 , e ) 19
Rationale: explanation : let ' k ' be the total number of sweets . given total number of students = 112 if sweets are distributed among 112 children , let number of sweets each student gets = ' l ' = > k / 112 = l . . . . ( 1 ) but on that day students absent = 32 = > remaining = 112 - 32 = 80 then , each student gets ' 6 ' sweets extra . = > k / 80 = l + 6 . . . . ( 2 ) from ( 1 ) k = 112 l substitute in ( 2 ) , we get 112 l = 80 l + 480 32 l = 480 l = 15 therefore , 15 sweets were each student originally supposed to get . answer : d
Answer: d
Question (gain): the difference between compound interest and simple interest on a certain amount of money at 5 % per annum for 2 years is 16 . find the sum :
Options: a ) 4500 , b ) 7500 , c ) 5000 , d ) 6400 , e ) none of these
Rationale: "sol . ( d ) let the sum be 100 . therefore , si = 100 × 5 × 2 / 100 = 10 and ci = 100 ( 1 + 5 / 100 ) 2 − 100 ∴ = 100 × 21 × 21 / 20 × 20 − 100 = 41 / 4 difference of ci and si = 41 ⁄ 4 - 10 = 1 ⁄ 4 if the difference is 1 ⁄ 4 , the sum = 100 = > if the difference is 16 , the sum = 400 × 16 = 6400 answer d"
Answer: d
Question (general): there are two numbers . if 50 % of the first number is added to the second number , then the second number increases to its 5 - fourth . find the ratio of the first number to the second number ?
Options: a ) a ) 1 / 2 , b ) b ) 5 / 8 , c ) c ) 6 , d ) d ) 6 / 7 , e ) e ) 7
Rationale: let the two numbers be x and y . 50 / 100 * x + y = 5 / 4 y = > 1 / 2 x = 1 / 4 y = > x / y = 1 / 2 a )
Answer: a
Question (general): if you wish to live to be 100 years old ( hypothetically ) , you must consume 500 calories less than your average daily allowance for your age . if you are in your 60 ' s , and your average daily allowance is 2000 calories per day , how many calories are you allowed in a week ?
Options: a ) 7000 , b ) 10500 , c ) 14000 , d ) 12000 , e ) 3500
Rationale: to determine the amount of calories allowed per day , you need to subtract 500 from 2000 , for a total of 1500 to determine the amount of calories allowed per week , you need to multiply the daily amount allowed ( 1500 ) by the number of days in a week ( 7 ) for a total of 10500 the correct answer is b
Answer: b
Question (physics): 250 metres long yard , 51 trees are palnted at equal distances , one tree being at each end of the yard . what is the distance between 2 consecutive trees
Options: a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 5
Rationale: "51 trees have 50 gaps between them , required distance ( 250 / 50 ) = 15 e"
Answer: e
Question (gain): a sum of money deposited at c . i . amounts to rs . 2420 in 2 years and to rs . 2783 in 3 years . find the rate percent ?
Options: a ) 15 , b ) 10 , c ) 28 , d ) 24 , e ) 82
Rationale: "explanation : 2420 - - - 363 100 - - - ? = > 15 % answer : option a"
Answer: a
Question (general): if an integer n is to be selected at random from 1 to 100 , inclusive , what is probability n ( n + 1 ) will be divisible by 16 ?
Options: a ) 1 / 4 , b ) 1 / 8 , c ) 1 / 2 , d ) 2 / 3 , e ) 3 / 4
Rationale: "because n ( n + 1 ) is always an even product of even * odd or odd * even factors , there is a probability of 1 that that it will be divisible by 2 , and , thus , a probability of 1 / 2 that it will be divisible by 4 and , thus , a probability of 1 / 4 that it will be divisible by 8 and , thus , a probability of 1 / 8 that it will be divisible by 16 1 * 1 / 8 = 1 / 8 answer : b"
Answer: b
Question (other): how many zeros are there in 50 !
Options: a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14
Rationale: "to find the number of trailing zeroes in 50 ! factorial we must find the number of integral values in : 50 / 5 + 50 / 5 ^ 2 + 50 / 5 ^ 3 + . . . . . . . . . = 10 + 2 + 0 + 0 . . . . = 12 answer : c"
Answer: c
Question (physics): eddy and freddy start simultaneously from city a and they travel to city b and city c respectively . eddy takes 3 hours and freddy takes 4 hours to complete the journey . if the distance between city a and city b is 450 kms and city a and city c is 300 kms . what is the ratio of their average speed of travel ? ( eddy : freddy )
Options: a ) 8 / 3 , b ) 2 / 1 , c ) 8 / 5 , d ) 5 / 8 , e ) 5 / 3
Rationale: "distance traveled by eddy = 600 km time taken by eddy = 3 hours average speed of eddy = 450 / 3 = 150 km / hour distance traveled by freddy = 300 km time taken by freddy = 4 hours average speed of freddy = 300 / 4 = 75 km / hour ratio of average speed of eddy to freddy = 150 / 75 = 2 / 1 answer b"
Answer: b
Question (physics): pipe x that can fill a tank in an hour and pipe y that can fill the tank in half an hour are opened simultaneously when the tank is empty . pipe y is shut 15 minutes before the tank overflows . when will the tank overflow ?
Options: a ) 35 mins , b ) 30 mins , c ) 40 mins , d ) 32 mins , e ) 36 mins
Rationale: the last 15 minutes only pipe x was open . since it needs 1 hour to fill the tank , then in 15 minutes it fills 1 / 4 th of the tank , thus 3 / 4 of the tank is filled with both pipes open . the combined rate of two pipes is 1 + 2 = 3 tanks / hour , therefore to fill 3 / 4 th of the tank they need ( time ) = ( work ) / ( rate ) = ( 3 / 4 ) / 3 = 1 / 4 hours = 15 minutes . total time = 15 + 15 = 30 minutes . answer : b
Answer: b
Question (physics): a and b can complete a work in 15 days and 10 day . they started doing the work together but after 2 days b had to leave and a alone completed the remaining work . the whole work was completed in ?
Options: a ) 10 days , b ) 12 days , c ) 15 days , d ) 8 days , e ) 16 days
Rationale: "a + b 1 day work = 1 / 15 + 1 / 10 = 1 / 6 work done by a and b in 2 days = 1 / 6 * 2 = 1 / 3 remaining work = 1 - 1 / 3 = 2 / 3 now 1 / 15 work is done by a in 1 day 2 / 3 work will be done by a in 15 * 2 / 3 = 10 days total time taken = 10 + 2 = 12 days answer is b"
Answer: b
Question (physics): the speeds of three asteroids were compared . asteroids x - 13 and y - 14 were observed for identical durations , while asteroid z - 15 was observed for 2 seconds longer . during its period of observation , asteroid y - 14 traveled three times the distance x - 13 traveled , and therefore y - 14 was found to be faster than x - 13 by 6000 kilometers per second . asteroid z - 15 had an identical speed as that of x - 13 , but because z - 15 was observed for a longer period , it traveled five times the distance x - 13 traveled during x - 13 ' s inspection . asteroid x - 13 traveled how many kilometers during its observation ?
Options: a ) 500 , b ) 1,600 / 3 , c ) 1,000 , d ) 1,500 , e ) 2,500
Rationale: "x 13 : ( t , d , s ) y 14 : ( t , 3 d , s + 6000 mi / hour ) z 15 : ( t + 2 seconds , s , 5 d ) d = ? distance = speed * time x 13 : d = s * t x 14 : 3 d = ( s + 6000 ) * t = = = > 3 d = ts + 6000 t z 15 : 5 d = s * ( t + 2 t ) = = = > 5 d = st + 2 st = = = > 5 d - 2 st = st 3 d = 5 d - 2 st + 6000 t - 2 d = - 2 st + 6000 t 2 d = 2 st - 6000 t d = st - 3000 t x 13 : d = s * t st - 3000 t = s * t s - 3000 = s - 1500 = s i got to this point and could n ' t go any further . this seems like a problem where i can set up individual d = r * t formulas and solve but it appears that ' s not the case . for future reference how would i know not to waste my time setting up this problem in the aforementioned way ? thanks ! ! ! the distance of z 15 is equal to five times the distance of x 13 ( we established that x 13 is the baseline and thus , it ' s measurements are d , s , t ) s ( t + 2 ) = 5 ( s * t ) what clues would i have to know to set up the equation in this fashion ? is it because i am better off setting two identical distances together ? st + 2 s = 5 st t + 2 = 5 t 2 = 4 t t = 1 / 2 we are looking for distance ( d = s * t ) so we need to solve for speed now that we have time . speed y 14 - speed x 13 speed = d / t 3 d / t - d / t = 6000 ( remember , t is the same because both asteroids were observed for the same amount of time ) 2 d = 6000 2 = 3000 d = s * t d = 3000 * ( 1 / 2 ) d = 1500 answer : d"
Answer: d
Question (general): working alone , printers x , y , and z can do a certain printing job , consisting of a large number of pages , in 12 , 30 , and 40 hours , respectively . what is the ratio of the time it takes printer x to do the job , working alone at its rate , to the time it takes printers y and z to do the job , working together at their individual rates ?
Options: a ) 4 / 11 , b ) 1 / 2 , c ) 15 / 22 , d ) 120 / 7 , e ) 11 / 4
Rationale: "p 1 takes 12 hrs rate for p 2 p 3 together = 1 / 30 + 1 / 40 = 7 / 120 therefore they take 120 / 7 ratio = 120 / 7 = d"
Answer: d
Question (geometry): the roof of an apartment building is rectangular and its length is 3 times longer than its width . if the area of the roof is 768 feet squared , what is the difference between the length and the width of the roof ?
Options: a ) 32 . , b ) 40 . , c ) 42 . , d ) 44 . , e ) 46 .
Rationale: "answer is a : 42 let w be the width , so length is 3 w . therefore : w * 4 w = 768 , solving for , w = 16 , so 3 w - w = 2 w = 2 * 16 = 32"
Answer: a
Question (general): the sum of two numbers is 528 and their h . c . f . is 33 . the number of pairs of numbers satisfying the above conditions is :
Options: a ) 4 , b ) 6 , c ) 8 , d ) 12 , e ) 15
Rationale: "explanation : let the required numbers be 33 a and 33 b . then , 33 a + 33 b = 528 . = > a + b = 16 . now , co - primes with sum 16 are ( 1 , 15 ) , ( 3 , 13 ) , ( 5 , 11 ) and ( 7 , 9 ) . â ˆ ´ required numbers are ( 33 * 1 , 33 * 15 ) , ( 33 * 3 , 33 * 13 ) , ( 33 * 5 , 33 * 11 ) , ( 33 x 7 , 33 x 9 ) . the number of such pairs is 4 . answer is a"
Answer: a
Question (general): a certain fraction is equivalent to 2 / 5 . if the numerator of the fraction is increased by 3 and the denominator is doubled , the new fraction is equivalent to 1 / 3 . what is the sum of the numerator and denominator of the original fraction ?
Options: a ) 21 / 4 , b ) 35 , c ) 28 , d ) 26 , e ) 21
Rationale: x / y = 2 / 5 - > 1 ( x + 3 ) / 2 y = 1 / 3 - > 2 divide 1 by 2 : = > 2 x / ( x + 3 ) = 6 / 5 = > 10 x = 6 x + 18 = > x = 3 / 2 = > y = 5 / 2 * 3 / 2 = 15 / 4 so x + y = 21 / 4 the answer is a
Answer: a
Question (general): it takes 30 days to fill a laboratory dish with bacteria . if the size of the bacteria doubles each day , how long did it take for the bacteria to fill 1 / 64 of the dish ?
Options: a ) 20 , b ) 22 , c ) 24 , d ) 26 , e ) 28
Rationale: the bacteria doubles each day , so after 29 days , the dish was half full . after 28 days , the dish was one quarter full . after 27 days , the dish was one eighth full . after 26 days , the dish was one sixteenth full . after 25 days , the dish was 1 / 32 full . after 24 days , the dish was 1 / 64 full . the answer is c .
Answer: c
Question (physics): in what time will a train 95 m long cross an electric pole , it its speed be 214 km / hr ?
Options: a ) 1.6 sec , b ) 2.9 sec , c ) 2.7 sec , d ) 8.7 sec , e ) 8.5 sec
Rationale: "speed = 214 * 5 / 18 = 59 m / sec time taken = 95 / 59 = 1.6 sec . answer : a"
Answer: a