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Problem stringlengths 5 967	Rationale stringlengths 1 2.74k	options stringlengths 37 300	correct stringclasses 5 values	annotated_formula stringlengths 7 6.48k	linear_formula stringlengths 8 925	category stringclasses 6 values
the banker ' s gain of a certain sum due 3 years hence at 10 % per annum is rs . 36 . what is the present worth ?	"explanation : t = 3 years r = 10 % td = ( bg × 100 ) / tr = ( 36 × 100 ) / ( 3 × 10 ) = 12 × 10 = rs . 120 td = ( pw × tr ) / 100 ⇒ 120 = ( pw × 3 × 10 ) / 100 ⇒ 1200 = pw × 3 pw = 1200 / 3 = rs . 400 answer : option a"	a ) rs . 400 , b ) rs . 300 , c ) rs . 500 , d ) rs . 350 , e ) none of these	a	divide(multiply(const_100, divide(multiply(36, const_100), multiply(3, 10))), multiply(3, 10))	multiply(n2,const_100)\|multiply(n0,n1)\|divide(#0,#1)\|multiply(#2,const_100)\|divide(#3,#1)\|	gain
average age of students of an adult school is 40 years . 120 new students whose average age is 32 years joined the school . as a result the average age is decreased by 4 years . find the number of students of the school after joining of the new students .	"explanation : let the original no . of students be x . according to situation , 40 x + 120 * 32 = ( x + 120 ) 36 ⇒ x = 120 so , required no . of students after joining the new students = x + 120 = 240 . answer : d"	a ) 1200 , b ) 120 , c ) 360 , d ) 240 , e ) none of these	d	multiply(divide(subtract(multiply(add(32, 4), 120), multiply(120, 32)), subtract(40, add(32, 4))), 4)	add(n2,n3)\|multiply(n1,n2)\|multiply(n1,#0)\|subtract(n0,#0)\|subtract(#2,#1)\|divide(#4,#3)\|multiply(n3,#5)\|	general
sophia finished 2 / 3 of a book . she calculated that she finished 90 more pages than she has yet to read . how long is her book ?	let xx be the total number of pages in the book , then she finished 23 ⋅ x 23 ⋅ x pages . then she has x − 23 ⋅ x = 13 ⋅ xx − 23 ⋅ x = 13 ⋅ x pages left . 23 ⋅ x − 13 ⋅ x = 9023 ⋅ x − 13 ⋅ x = 90 13 ⋅ x = 9013 ⋅ x = 90 x = 270 x = 270 so the book is 270 pages long . answer : b	a ) 229 , b ) 270 , c ) 877 , d ) 266 , e ) 281	b	divide(90, subtract(const_1, divide(2, 3)))	divide(n0,n1)\|subtract(const_1,#0)\|divide(n2,#1)	general
120 is what percent of 50 ?	"50 * x = 120 - - > x = 2.4 - - > 2.4 expressed as percent is 240 % . answer : b ."	a ) 5 % , b ) 240 % , c ) 50 % , d ) 2 % , e ) 500 %	b	multiply(divide(120, 50), const_100)	divide(n0,n1)\|multiply(#0,const_100)\|	gain
there are 10 girls and 20 boys in a classroom . what is the ratio of girls to boys ?	if girls is 10 and boys is 20 , then 10 / 20 . so ratio of girls to boys is = 10 / 20 = 1 / 2 answer : a	a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 5 , d ) 10 / 30 , e ) 2 / 5	a	divide(10, 20)	divide(n0,n1)	other
an empty fuel tank with a capacity of 218 gallons was filled partially with fuel a and then to capacity with fuel b . fuel a contains 12 % ethanol by volume and fuel b contains 16 % ethanol by volume . if the full fuel tank contains 30 gallons of ethanol , how many gallons of fuel a were added ?	"say there are a gallons of fuel a in the tank , then there would be 218 - a gallons of fuel b . the amount of ethanol in a gallons of fuel a is 0.12 a ; the amount of ethanol in 218 - a gallons of fuel b is 0.16 ( 218 - a ) ; since the total amount of ethanol is 30 gallons then 0.12 a + 0.16 ( 218 - a ) = 30 - - > a = 122 . answer : a ."	a ) 122 , b ) 150 , c ) 100 , d ) 80 , e ) 50	a	divide(subtract(multiply(218, divide(16, const_100)), 30), subtract(divide(16, const_100), divide(12, const_100)))	divide(n2,const_100)\|divide(n1,const_100)\|multiply(n0,#0)\|subtract(#0,#1)\|subtract(#2,n3)\|divide(#4,#3)\|	gain
an article is bought for rs . 823 and sold for rs . 1000 , find the gain percent ?	"823 - - - - 177 100 - - - - ? = > 21.5 % answer : b"	a ) 21.4 % , b ) 21.5 % , c ) 21.6 % , d ) 21.7 % , e ) 21.8 %	b	subtract(const_100, divide(multiply(1000, const_100), 823))	multiply(n1,const_100)\|divide(#0,n0)\|subtract(const_100,#1)\|	gain
6 workers should finish a job in 8 days . after 3 days came 4 workers join them . how many days m do they need to finish the same job ?	"let rate of one worker be r = > ( 6 * r ) * 8 = 1 ( rate * time = work ) = > r = 1 / 48 = > work remaining after 3 days 1 - ( 3 * 6 ) / 48 = 30 / 48 after 4 ppl joined in ( ( 6 + 4 ) * time ) / 48 = 30 / 48 time m = 3 days to finish the task imo a"	a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7	a	divide(subtract(multiply(6, 8), multiply(3, 6)), add(6, 4))	add(n0,n3)\|multiply(n0,n1)\|multiply(n0,n2)\|subtract(#1,#2)\|divide(#3,#0)\|	physics
j is 25 % less than p and 20 % less than t . t is q % less than p . what is the value of q ?	"usually we can solve every question of this type by choosing appropriate value of the variable and deriving the value of other related variables . let , p = 400 then j = ( 75 / 100 ) * 400 = 300 also j = ( 80 / 100 ) * t i . e . t = 300 * 100 / 80 = 375 and t = [ 1 - ( q / 100 ) ] * p i . e . 100 - q = 100 * t / p = 100 * 375 / 400 = 93.75 i . e . q = 6.25 answer : option d"	a ) 93.5 , b ) 90 , c ) 6.75 , d ) 6.25 , e ) 2	d	divide(multiply(25, 25), const_100)	multiply(n0,n0)\|divide(#0,const_100)\|	gain
a student was asked to find 4 / 5 of a number . but the student divided the number by 4 / 5 , thus the student got 36 more than the correct answer . find the number .	"let the number be x . ( 5 / 4 ) * x = ( 4 / 5 ) * x + 36 25 x = 16 x + 720 9 x = 720 x = 80 the answer is c ."	a ) 60 , b ) 70 , c ) 80 , d ) 90 , e ) 100	c	divide(divide(multiply(multiply(36, divide(4, 5)), divide(4, 5)), subtract(const_1, multiply(divide(4, 5), divide(4, 5)))), divide(4, 5))	divide(n0,n1)\|multiply(n4,#0)\|multiply(#0,#0)\|multiply(#0,#1)\|subtract(const_1,#2)\|divide(#3,#4)\|divide(#5,#0)\|	general
the average weight of 8 person ' s increases by 1.5 kg when a new person comes in place of one of them weighing 75 kg . what might be the weight of the new person ?	"total weight increased = ( 8 x 1.5 ) kg = 6 kg . weight of new person = ( 75 + 6 ) kg = 81 kg . answer : a"	a ) 81 kg , b ) 85 kg , c ) 90 kg , d ) 100 kg , e ) 110 kg	a	add(multiply(8, 1.5), 75)	multiply(n0,n1)\|add(n2,#0)\|	general
a train 125 m long passes a man , running at 15 km / hr in the same direction in which the train is going , in 10 seconds . the speed of the train is ?	"speed of the train relative to man = ( 125 / 10 ) m / sec = ( 25 / 2 ) m / sec . [ ( 25 / 2 ) * ( 18 / 5 ) ] km / hr = 45 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 15 ) km / hr . x - 15 = 45 = = > x = 60 km / hr answer : a"	a ) 60 , b ) 50 , c ) 28 , d ) 26 , e ) 29	a	divide(divide(subtract(125, multiply(multiply(15, const_0_2778), 15)), 15), const_0_2778)	multiply(n1,const_0_2778)\|multiply(n1,#0)\|subtract(n0,#1)\|divide(#2,n1)\|divide(#3,const_0_2778)\|	physics
the average of 15 result is 60 . average of the first 10 of them is 10 and that of the last 10 is 80 . find the 8 th result ?	"sum of all the 13 results = 15 * 60 = 900 sum of the first 7 of them = 10 * 10 = 100 sum of the last 7 of them = 10 * 80 = 800 so , the 8 th number = 900 + 100 - 800 = 200 . b"	a ) 35 , b ) 200 , c ) 150 , d ) 250 , e ) 300	b	subtract(add(multiply(10, 10), multiply(10, 80)), multiply(15, 60))	multiply(n2,n3)\|multiply(n2,n5)\|multiply(n0,n1)\|add(#0,#1)\|subtract(#3,#2)\|	general
a salesman â € ™ s terms were changed from a flat commission of 5 % on all his sales to a fixed salary of rs . 1000 plus 2.5 % commission on all sales exceeding rs . 4,000 . if his remuneration as per new scheme was rs . 500 more than that by the previous schema , his sales were worth ?	"[ 1000 + ( x - 4000 ) * ( 2.5 / 100 ) ] - x * ( 5 / 100 ) = 600 x = 16000 answer : c"	a ) s . 14,000 , b ) s . 12,000 , c ) s . 16,000 , d ) s . 40,000 , e ) s . 50,000	c	subtract(multiply(5, const_4), const_12)	multiply(n0,const_4)\|subtract(#0,const_12)\|	general
a rectangular floor that measures 15 meters by 18 meters is to be covered with carpet squares that each measure 3 meters by 3 meters . if the carpet squares cost $ 12 apiece , what is the total cost for the number of carpet squares needed to cover the floor ?	"the width of the rectangular floor ( 15 m ) is a multiple of one side of the square ( 3 m ) , and the length of the floor ( 18 m ) is also a multiple of the side of the square . so the number of carpets to cover the floor is ( 15 / 3 ) * ( 18 / 3 ) = 30 . the total cost is 30 * 12 = $ 360 . the answer is , therefore , c ."	a ) $ 200 , b ) $ 240 , c ) $ 360 , d ) $ 960 , e ) $ 1,920	c	multiply(15, 15)	multiply(n0,n0)\|	geometry
a vessel of capacity 2 litre has 30 % of alcohol and another vessel of capacity 6 litre had 40 % alcohol . the total liquid of 8 litre was poured out in a vessel of capacity 10 litre and thus the rest part of the vessel was filled with the water . what is the new concentration of mixture ?	"30 % of 2 litres = 0.6 litres 40 % of 6 litres = 2.4 litres therefore , total quantity of alcohol is 3.0 litres . this mixture is in a 10 litre vessel . hence , the concentration of alcohol in this 10 litre vessel is 30 % answer : a"	a ) 30 % . , b ) 71 % . , c ) 49 % . , d ) 29 % . , e ) 51 % .	a	multiply(divide(add(multiply(divide(30, const_100), 2), multiply(divide(40, const_100), 6)), 10), const_100)	divide(n1,const_100)\|divide(n3,const_100)\|multiply(n0,#0)\|multiply(n2,#1)\|add(#2,#3)\|divide(#4,n5)\|multiply(#5,const_100)\|	general
the total of 324 of 20 paise and 25 paise make a sum of rs . 70 . the no of 20 paise coins is	"explanation : let the number of 20 paise coins be x . then the no of 25 paise coins = ( 324 - x ) . 0.20 * ( x ) + 0.25 ( 324 - x ) = 70 = > x = 220 . . answer : c ) 220"	a ) 238 , b ) 277 , c ) 220 , d ) 200 , e ) 288	c	divide(subtract(multiply(324, 25), multiply(70, const_100)), subtract(25, 20))	multiply(n0,n2)\|multiply(n3,const_100)\|subtract(n2,n1)\|subtract(#0,#1)\|divide(#3,#2)\|	general
in 1970 there were 8,902 women stockbrokers in the united states . by 1978 the number had increased to 18,947 . approximately what was the percent increase ?	"the percent increase is ( 18947 - 8902 ) / 8902 = 10045 / 8902 = 1.13 so the approximate answer is b"	a ) 45 % , b ) 113 % , c ) 145 % , d ) 150 % , e ) 225 %	b	divide(subtract(18,947, 8,902), 8,902)	subtract(n3,n1)\|divide(#0,n1)\|	general
what is the number of integers from 1 to 1100 ( inclusive ) that are divisible by neither 11 nor by 35 ?	"normally , i would use the method used by bunuel . it ' s the most accurate . but if you are looking for a speedy solution , you can use another method which will sometimes give you an estimate . looking at the options ( most of them are spread out ) , i wont mind trying it . ( mind you , the method is accurate here since the numbers start from 1 . ) in 1000 consecutive numbers , number of multiples of 11 = 1000 / 11 = 90 ( ignore decimals ) in 1000 consecutive numbers , number of multiples of 35 = 1000 / 35 = 28 number of multiples of 11 * 35 i . e . 385 = 1000 / 385 = 2 number of integers from 1 to 1000 that are divisible by neither 11 nor by 35 = 1000 - ( 90 + 28 - 2 ) { using the concept of sets here ) = 945 think : why did i say the method is approximate in some cases ? think what happens if the given range is 11 to 1010 both inclusive ( again 1000 numbers ) what is the number of multiples in this case ? e"	a ) 884 , b ) 890 , c ) 892 , d ) 910 , e ) 945	e	subtract(1100, subtract(add(divide(1100, 11), divide(1100, 35)), divide(1100, multiply(11, 35))))	divide(n1,n2)\|divide(n1,n3)\|multiply(n2,n3)\|add(#0,#1)\|divide(n1,#2)\|subtract(#3,#4)\|subtract(n1,#5)\|	other
arun makes a popular brand of ice cream in a rectangular shaped bar 6 cm long , 5 cm wide and 2 cm thick . to cut costs , the company had decided to reduce the volume of the bar by 19 % . the thickness will remain same , but the length and width will be decreased by some percentage . the new width will be ,	answer : a	a ) 33 , b ) 87 , c ) 99 , d ) 367 , e ) 72	a	add(divide(multiply(multiply(6, 5), 2), 2), const_3)	multiply(n0,n1)\|multiply(n2,#0)\|divide(#1,n2)\|add(#2,const_3)	gain
kim finds a 5 - meter tree branch and marks it off in thirds and fifths . she then breaks the branch along all the markings and removes one piece of every distinct length . what fraction of the original branch remains ?	"3 pieces of 1 / 5 length and two piece each of 1 / 15 and 2 / 15 lengths . removing one piece each from pieces of each kind of lengths the all that will remain will be 2 pieces of 1 / 5 i . e 2 / 5 , 1 piece of 1 / 15 , and 1 piece of 2 / 15 which gives us 2 / 5 + 1 / 15 + 2 / 15 - - - - - > 3 / 5 answer is d"	a ) 8 / 15 , b ) 1 / 2 , c ) 7 / 5 , d ) 3 / 5 , e ) 7 / 5	d	subtract(const_1, add(add(divide(5, multiply(add(const_2, 5), 5)), divide(const_2, multiply(add(const_2, 5), 5))), divide(const_1, multiply(add(const_2, 5), 5))))	add(const_2,n0)\|multiply(n0,#0)\|divide(n0,#1)\|divide(const_2,#1)\|divide(const_1,#1)\|add(#2,#3)\|add(#5,#4)\|subtract(const_1,#6)\|	physics
mark bought a set of 6 flower pots of different sizes at a total cost of $ 8.25 . each pot cost 0.1 more than the next one below it in size . what was the cost , in dollars , of the largest pot ?	"this question can be solved with a handful of different algebra approaches ( as has been shown in the various posts ) . since the question asks for the price of the largest pot , and the answers are prices , we can test the answers . we ' re told that there are 6 pots and that each pot costs 25 cents more than the next . the total price of the pots is $ 8.25 . we ' re asked for the price of the largest ( most expensive ) pot . since the total price is $ 8.25 ( a 10 - cent increment ) and the the difference in sequential prices of the pots is 10 cents , the largest pot probably has a price that is a 10 - cent increment . from the answer choices , i would then test answer a first if . . . . the largest pot = $ 1.625 1.125 1.225 1.325 1.425 1.525 1.625 total = $ 8.25 so this must be the answer . a"	a ) $ 1.62 , b ) $ 1.85 , c ) $ 2.00 , d ) $ 2.15 , e ) $ 2.30	a	add(divide(subtract(8.25, multiply(divide(multiply(subtract(6, const_1), 6), const_2), 0.1)), 6), multiply(subtract(6, const_1), 0.1))	subtract(n0,const_1)\|multiply(n0,#0)\|multiply(n2,#0)\|divide(#1,const_2)\|multiply(n2,#3)\|subtract(n1,#4)\|divide(#5,n0)\|add(#6,#2)\|	general
in the above number , a and b represent the tens and units digits , respectively . if the above number is divisible by 45 , what is the greatest possible value of b x a ?	"i also was confused when i was looking forabove number : d as far as i understood , 45 is a factor of ab . in other words , the values of b ( units digits can be 5 or 0 . better to have option for 5 in this case to havebigger result ) . now let ' s try 45 x 1 ( a = 4 , b = 5 respectively we have = 20 ) . this is the greatest possible value of b x a . imo e ."	a ) 0 , b ) 5 , c ) 10 , d ) 15 , e ) 20	e	multiply(add(const_3, const_4), add(const_2, const_3))	add(const_3,const_4)\|add(const_2,const_3)\|multiply(#0,#1)\|	general
the profit obtained by selling an article for rs . 57 is the same as the loss obtained by selling it for rs . 43 . what is the cost price of the article ?	"s . p 1 - c . p = c . p – s . p 2 57 - c . p = c . p - 43 2 c . p = 57 + 43 ; c . p = 100 / 2 = 50 answer : b"	a ) rs . 40 , b ) rs . 50 , c ) rs . 49 , d ) rs . 59 , e ) none of these	b	divide(add(57, 43), const_2)	add(n0,n1)\|divide(#0,const_2)\|	gain
in 2008 , the profits of company n were 8 percent of revenues . in 2009 , the revenues of company n fell by 20 percent , but profits were 15 percent of revenues . the profits in 2009 were what percent of the profits in 2008 ?	"the profit 0 f 2009 interms of 2008 = 0.8 * 15 / 8 * 100 = 150 % a"	a ) 150 % , b ) 105 % , c ) 120 % , d ) 124.2 % , e ) 138 %	a	multiply(divide(multiply(15, subtract(const_1, divide(20, const_100))), 8), const_100)	divide(n3,const_100)\|subtract(const_1,#0)\|multiply(n4,#1)\|divide(#2,n1)\|multiply(#3,const_100)\|	gain
a garrison of 2000 men has provisions for 54 days . at the end of 15 days , a reinforcement arrives , and it is now found that the provisions will last only for 20 days more . what is the reinforcement ?	"2000 - - - - 54 2000 - - - - 39 x - - - - - 20 x * 20 = 2000 * 39 x = 3900 2000 - - - - - - - 1900 answer : d"	a ) 1778 , b ) 1682 , c ) 9178 , d ) 1900 , e ) 1782	d	subtract(divide(subtract(multiply(2000, 54), multiply(2000, 15)), 20), 2000)	multiply(n0,n1)\|multiply(n0,n2)\|subtract(#0,#1)\|divide(#2,n3)\|subtract(#3,n0)\|	physics
if a bicyclist in motion increases his speed by 30 percent and then increases this speed by 10 percent , what percent of the original speed is the total increase in speed ?	let the sped be 100 an increase of 30 % the speed now is 130 a further increase of 10 % on 130 = 13 total increase = 43 on 100 = 43 % c	a ) 10 % , b ) 40 % , c ) 43 % , d ) 64 % , e ) 140 %	c	subtract(add(add(const_100, 30), multiply(add(const_100, 30), divide(10, const_100))), const_100)	add(n0,const_100)\|divide(n1,const_100)\|multiply(#0,#1)\|add(#0,#2)\|subtract(#3,const_100)	general
what is the greatest number that divides 263 , 935 and 1383 leaving a remainder of 7 in each case ?	answer the required greatest number is the hcf of 263 - 7 , 935 - 7 , 1383 - 7 i . e . 256 , 928 and 1376 hcf = 32 correct option : c	a ) 30 , b ) 31 , c ) 32 , d ) 35 , e ) 37	c	divide(divide(subtract(subtract(subtract(1383, 7), subtract(935, 7)), subtract(263, 7)), const_3), const_2)	subtract(n2,n3)\|subtract(n1,n3)\|subtract(n0,n3)\|subtract(#0,#1)\|subtract(#3,#2)\|divide(#4,const_3)\|divide(#5,const_2)	general
how many seconds will a 900 meter long train moving with a speed of 63 km / hr take to cross a man walking with a speed of 3 km / hr in the direction of the train ?	"explanation : here distance d = 900 mts speed s = 63 - 3 = 60 kmph = 60 x 5 / 18 m / s time t = = 54 sec . answer : c"	a ) 48 , b ) 36 , c ) 54 , d ) 11 , e ) 18	c	divide(900, multiply(subtract(63, 3), const_0_2778))	subtract(n1,n2)\|multiply(#0,const_0_2778)\|divide(n0,#1)\|	physics
find the average of all the numbers between 6 and 38 which are divisible by 4 .	"solution average = ( ( 8 + 12 + 16 + 20 + 24 + 28 + 32 + 36 ) / 8 ) = 186 / 7 = 22 answer b"	a ) 18 , b ) 22 , c ) 20 , d ) 30 , e ) 28	b	divide(add(add(6, const_4), subtract(38, const_4)), const_2)	add(n0,const_4)\|subtract(n1,const_4)\|add(#0,#1)\|divide(#2,const_2)\|	general
if the cost price is 96 % of sp then what is the profit %	"sol . sp = rs 100 : then cp = rs 96 : profit = rs 4 . profit = { ( 4 / 96 ) * 100 } % = 4.17 % answer is d ."	a ) 4.07 % , b ) 4 % , c ) 4.7 % , d ) 4.17 % , e ) 4.27 %	d	multiply(divide(subtract(const_100, 96), 96), const_100)	subtract(const_100,n0)\|divide(#0,n0)\|multiply(#1,const_100)\|	gain
a car gets 27 miles to the gallon . if it is modified to use a solar panel , it will use only 75 percent as much fuel as it does now . if the fuel tank holds 14 gallons , how many more miles will the car be able to travel , per full tank of fuel , after it has been modified ?	"originally , the distance the car could go on a full tank was 14 * 27 = 378 miles . after it has been modified , the car can go 27 / 0.75 = 36 miles per gallon . on a full tank , the car can go 14 * 36 = 504 miles , thus 126 miles more . the answer is b ."	a ) 120 , b ) 126 , c ) 132 , d ) 138 , e ) 144	b	subtract(multiply(multiply(divide(27, 75), const_100), 14), multiply(27, 14))	divide(n0,n1)\|multiply(n0,n2)\|multiply(#0,const_100)\|multiply(n2,#2)\|subtract(#3,#1)\|	physics
a fort had provision of food for 150 men for 45 days . after 10 days , 25 men left the fort . the number of days for which the remaining food will last , is :	"explanation : after 10 days : 150 men had food for 35 days . suppose 125 men had food for x days . now , less men , more days ( indirect proportion ) { \ color { blue } \ therefore } 125 : 150 : : 35 : x { \ color { blue } \ rightarrow } 125 x x = 150 x 35 { \ color { blue } \ rightarrow x = \ frac { 150 \ times 35 } { 125 } } { \ color { blue } \ rightarrow } x = 42 . answer : c ) 42"	a ) 34 , b ) 387 , c ) 42 , d ) 28 , e ) 71	c	divide(multiply(150, add(25, 10)), subtract(150, 25))	add(n2,n3)\|subtract(n0,n3)\|multiply(n0,#0)\|divide(#2,#1)\|	physics
a person lent a certain sum of money at 4 % per annum at simple interest and in 8 years the interest amounted to rs . 306 less than the sum lent . what was the sum lent ?	"p - 306 = ( p * 4 * 8 ) / 100 p = 450 answer : e"	a ) 228 , b ) 278 , c ) 289 , d ) 500 , e ) 450	e	divide(306, subtract(const_1, divide(multiply(4, 8), const_100)))	multiply(n0,n1)\|divide(#0,const_100)\|subtract(const_1,#1)\|divide(n2,#2)\|	gain
a man has some hens and cows . if the number of heads be 50 and the number of feet equals 144 , then the number of hens will be	"explanation : let number of hens = h and number of cows = c number of heads = 50 = > h + c = 48 - - - ( equation 1 ) number of feet = 144 = > 2 h + 4 c = 144 = > h + 2 c = 72 - - - ( equation 2 ) ( equation 2 ) - ( equation 1 ) gives 2 c - c = 72 - 50 = > c = 22 substituting the value of c in equation 1 , we get h + 22 = 50 = > h = 50 - 22 = 28 i . e . , number of hens = 28 answer : e"	a ) 22 , b ) 24 , c ) 26 , d ) 20 , e ) 28	e	divide(subtract(multiply(50, const_4), 144), const_2)	multiply(n0,const_4)\|subtract(#0,n1)\|divide(#1,const_2)\|	general
because he ’ s taxed by his home planet , mork pays a tax rate of 40 % on his income , while mindy pays a rate of only 30 % on hers . if mindy earned 3 times as much as mork did , what was their combined tax rate ?	"say morks income is - 100 so tax paid will be 40 say mindy ' s income is 3 * 100 = 300 so tax paid is 30 % * 300 = 90 total tax paid = 40 + 90 = 130 . combined tax % will be 130 / 100 + 300 = 32.5 % answer : a"	a ) 32.5 % , b ) 34 % , c ) 35 % , d ) 36 % , e ) 37.5 %	a	multiply(const_100, divide(add(divide(40, const_100), multiply(3, divide(30, const_100))), add(const_1, 3)))	add(n2,const_1)\|divide(n0,const_100)\|divide(n1,const_100)\|multiply(n2,#2)\|add(#1,#3)\|divide(#4,#0)\|multiply(#5,const_100)\|	gain
a watch was sold at a loss of 5 % . if it was sold for rs . 500 more , there would have been a gain of 5 % . what is the cost price ?	"95 % 105 % - - - - - - - - 10 % - - - - 500 100 % - - - - ? = > rs . 5000 answer : d"	a ) 1000 , b ) 2998 , c ) 2778 , d ) 2788 , e ) 2991	d	divide(multiply(500, const_100), subtract(add(const_100, 5), subtract(const_100, 5)))	add(const_100,n2)\|multiply(n1,const_100)\|subtract(const_100,n0)\|subtract(#0,#2)\|divide(#1,#3)\|	gain
a car travels at a speed of 65 miles per hour . how far will it travel in 6 hours ?	"during each hour , the car travels 65 miles . for 6 hours it will travel 65 + 65 + 65 + 65 + 65 + 65 = 6 × 65 = 390 miles correct answer is c ) 390 miles"	a ) 125 miles , b ) 225 miles , c ) 390 miles , d ) 425 miles , e ) 525 miles	c	multiply(65, 6)	multiply(n0,n1)\|	physics
in a family 13 people eat only vegetarian , 7 people eat only non veg . , 8 people eat both veg and non veg . . how many people eat veg in the family ?	"total people eat veg = only veg + both veg and non veg total = 13 + 8 = 21 answer = e"	a ) 20 , b ) 11 , c ) 9 , d ) 31 , e ) 21	e	add(13, 8)	add(n0,n2)\|	other
if p / q = 4 / 5 , then the value of 11 / 7 + { ( 2 q - p ) / ( 2 q + p ) } is ?	answer given exp . = 11 / 7 + { ( 2 q - p ) / ( 2 q + p ) } dividing numerator as well as denominator by q , exp = 11 / 7 + { 2 - p / q ) / ( 2 + p / q ) } = 11 / 7 + { ( 2 - 4 / 5 ) / ( 2 + 4 / 5 ) } = 11 / 7 + 6 / 14 = 11 / 7 + 3 / 7 = 14 / 7 = 2 correct option : d	a ) 3 / 7 , b ) 34 , c ) 1 , d ) 2 , e ) 3	d	add(divide(11, 7), divide(subtract(2, divide(4, 5)), add(2, divide(4, 5))))	divide(n2,n3)\|divide(n0,n1)\|add(n4,#1)\|subtract(n4,#1)\|divide(#3,#2)\|add(#0,#4)	general
if x ^ 2 + y ^ 2 = 13 and xy = 3 , then ( x − y ) ^ 2 =	"but you can not take xy + 3 to mean xy = - 3 . . only if xy + 3 = 0 , it will mean xy = - 3 . . rest your solution is perfect and you will get your correct answer as 13 - 2 * 3 = 7 . . answer a"	a ) 7 , b ) 11 , c ) 14 , d ) 17 , e ) 20	a	power(3, 2)	power(n3,n0)\|	general
a big container is 40 % full with water . if 14 liters of water is added , the container becomes 3 / 4 full . what is the capacity of the big container in liters ?	14 liters is 35 % of the capacity c . 14 = 0.35 c c = 14 / 0.35 = 40 liters . the answer is c .	a ) 32 , b ) 36 , c ) 40 , d ) 44 , e ) 48	c	divide(14, subtract(divide(3, 4), divide(40, const_100)))	divide(n2,n3)\|divide(n0,const_100)\|subtract(#0,#1)\|divide(n1,#2)	general
the ratio of the arithmetic mean of two numbers to one of the numbers is 5 : 8 . what is the ratio of the smaller number to the larger number ?	"for two numbers , the arithmetic mean is the middle of the two numbers . the ratio of the mean to the larger number is 5 : 8 , thus the smaller number must have a ratio of 2 . the ratio of the smaller number to the larger number is 2 : 8 = 1 : 4 . the answer is d ."	a ) 1 : 8 , b ) 1 : 6 , c ) 1 : 5 , d ) 1 : 4 , e ) 1 : 3	d	multiply(subtract(divide(5, 8), divide(const_1, const_2)), const_2)	divide(n0,n1)\|divide(const_1,const_2)\|subtract(#0,#1)\|multiply(#2,const_2)\|	other
the salaries of a , b , and c are in the ratio of 1 : 2 : 3 . the salary of b and c together is rs . 6000 . by what percent is the salary of c more than that of a ?	explanation : let the salaries of a , b , c be x , 2 x and 3 x respectively . then , 2 x + 3 x = 6000 = > x = 1200 . a ' s salary = rs . 1200 , b ' s salary = rs . 2400 , and cs salary rs . 3600 . excess of c ' s salary over a ' s = [ ( 2400 / 1200 ) x 100 ] = 200 % . answer : b ) 200 %	a ) 209 % , b ) 200 % , c ) 290 % , d ) 600 % , e ) 100 %	b	multiply(subtract(divide(multiply(3, divide(6000, add(2, 3))), divide(6000, add(2, 3))), const_1), const_100)	add(n1,n2)\|divide(n3,#0)\|multiply(n2,#1)\|divide(#2,#1)\|subtract(#3,const_1)\|multiply(#4,const_100)	other
a man swims downstream 96 km and upstream 40 km taking 8 hours each time ; what is the speed of the current ?	"explanation : 96 - - - 8 ds = 12 ? - - - - 1 40 - - - - 8 us = 5 ? - - - - 1 s = ? s = ( 12 - 5 ) / 2 = 3.5 answer : option a"	a ) 3.5 kmph , b ) 1.5 kmph , c ) 13 kmph , d ) 6.5 kmph , e ) 7 : 3 kmph	a	divide(add(divide(40, 8), divide(96, 8)), const_2)	divide(n1,n2)\|divide(n0,n2)\|add(#0,#1)\|divide(#2,const_2)\|	physics
vinoth can complete a painting work in 20 days . prakash can do the same work in 25 days . they start the work together but vinoth quit after 3 days of work . how many days are required to complete the remaining painting work by prakash .	vinoth can complete the painting work in one day is 1 / 20 prakash can complete the same work in one day is 1 / 25 both of them can complete the work in 1 / 20 + days = 9 / 100 ( 1 / 20 + 1 / 25 ) they must have completed in three days = 9 / 100 * 3 = 27 / 100 remaining work to be done is by prakash = 1 - 27 / 100 = 73 / 100 for one work , prakash can do in 25 days for 73 / 100 work , he can do in 73 / 100 * 25 = 73 / 4 days or 18.25 days answer : d	a ) 15.25 days , b ) 16.25 days , c ) 17.25 days , d ) 18.25 days , e ) 19.25 days	d	divide(subtract(const_1, multiply(3, divide(const_1, 25))), divide(const_1, 20))	divide(const_1,n1)\|divide(const_1,n0)\|multiply(n2,#0)\|subtract(const_1,#2)\|divide(#3,#1)	physics
the total cost of a vacation was divided among 3 people . if the total cost of the vacation had been divided equally among 5 people , the cost per person would have been $ 50 less . what was the total cost cost of the vacation ?	c for cost . p price per person . c = 3 * p c = 5 * p - 250 substituting the value of p from the first equation onto the second we get p = 125 . plugging in the value of p in the first equation , we get c = 375 . which leads us to answer choice b	a ) $ 200 , b ) $ 375 , c ) $ 400 , d ) $ 500 , e ) $ 600	b	multiply(multiply(5, 3), divide(50, subtract(5, 3)))	multiply(n0,n1)\|subtract(n1,n0)\|divide(n2,#1)\|multiply(#2,#0)	general
how many ounces of a 60 % salt solution must be added to 30 ounces of a 20 percent salt solution so that the resulting mixture is 40 % salt ?	"let x = ounces of 60 % salt solution to be added . 2 * 30 + . 6 x = . 4 ( 30 + x ) x = 30 answer b"	a ) 16.67 , b ) 30 , c ) 50 , d ) 60.33 , e ) 70	b	divide(subtract(multiply(divide(40, const_100), 30), multiply(divide(20, const_100), 30)), subtract(divide(40, const_100), divide(20, const_100)))	divide(n3,const_100)\|divide(n2,const_100)\|multiply(n1,#0)\|multiply(n1,#1)\|subtract(#0,#1)\|subtract(#2,#3)\|divide(#5,#4)\|	gain
if 20 % of a is the same as 30 % of b , then a : b is :	"expl : 20 % of a i = 30 % of b = 20 a / 100 = 30 b / 100 = 3 / 2 = 3 : 2 answer : d"	a ) 5 : 4 , b ) 5 : 3 , c ) 4 : 3 , d ) 3 : 2 , e ) 1 : 3	d	divide(divide(30, const_100), divide(20, const_100))	divide(n1,const_100)\|divide(n0,const_100)\|divide(#0,#1)\|	gain
d and e are two points respectively on sides ab and ac of triangle abc such that de is parallel to bc . if the ratio of area of triangle ade to that of the trapezium decb is 144 : 25 and bc = 13 cm , then find the length of de .	"abc and ade are similar triangles . so ( side of abc / side of ade ) ^ 2 = 25 / 169 side of abc / side of ade = 5 / 13 so the length of de = 5 answer - c"	a ) 12 , b ) 13 , c ) 14 , d ) 11 , e ) 15	c	multiply(sqrt(divide(25, add(25, 144))), 13)	add(n0,n1)\|divide(n1,#0)\|sqrt(#1)\|multiply(n2,#2)\|	geometry
working alone at its constant rate , machine a produces x boxes in 10 minutes and working alone at its constant rate , machine b produces 2 x boxes in 5 minutes . how many minutes does it take machines a and b , working simultaneously at their respective constant rates , to produce 10 x boxes ?	"rate = work / time given rate of machine a = x / 10 min machine b produces 2 x boxes in 5 min hence , machine b produces 4 x boxes in 10 min . rate of machine b = 4 x / 10 we need tofind the combined time that machines a and b , working simultaneouslytakeat their respective constant rates let ' s first find the combined rate of machine a and b rate of machine a = x / 10 min + rate of machine b = 4 x / 10 = 5 x / 10 now combine time = combine work needs to be done / combine rate = 10 x / 5 x * 10 = 6 min ans : e"	a ) 13 minutes , b ) 14 minutes , c ) 15 minutes , d ) 16 minutes , e ) 20 minutes	e	divide(multiply(10, 10), add(speed(10, 10), speed(multiply(2, 10), 5)))	multiply(n0,n3)\|multiply(n0,n1)\|speed(n0,n0)\|speed(#1,n2)\|add(#2,#3)\|divide(#0,#4)\|	physics
if y > 0 , ( 1 y ) / 20 + ( 3 y ) / 10 is what percent of y ?	"can be reduced to y / 20 + 3 y / 10 = 7 y / 20 = 35 % a"	a ) 35 % , b ) 50 % , c ) 60 % , d ) 70 % , e ) 80 %	a	multiply(const_100, add(divide(1, 20), divide(3, 10)))	divide(n1,n2)\|divide(n3,n4)\|add(#0,#1)\|multiply(#2,const_100)\|	general
how many of the positive factors of 25 , 15 and how many common factors are there in numbers ?	"factors of 25 - 1 , 5 , and 25 factors of 15 - 1 , 3 , 5 and 15 comparing both , we have three common factors of 45,16 - 2 answer b"	a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5	b	divide(15, 25)	divide(n1,n0)\|	other
a certain college ' s enrollment at the beginning of 1992 was 20 percent greater than it was at the beginning of 1991 , and its enrollment at the beginning of 1993 was 5 percent greater than it was at the beginning of 1992 . the college ' s enrollment at the beginning of 1993 was what percent greater than its enrollment at the beginning of 1991 ?	"suppose enrollment in 1991 was 100 then enrollment in 1992 will be 120 and enrollment in 1993 will be 120 * 1.05 = 126 increase in 1993 from 1991 = 126 - 100 = 26 answer : b"	a ) 17.5 % , b ) 26 % , c ) 30 % , d ) 35 % , e ) 38 %	b	subtract(multiply(add(const_100, 20), divide(add(const_100, 5), const_100)), const_100)	add(n1,const_100)\|add(n4,const_100)\|divide(#1,const_100)\|multiply(#0,#2)\|subtract(#3,const_100)\|	gain
of the 17,210 employees of the anvil factory , 2 / 7 are journeymen . if half of the journeymen were laid off , what percentage of the total remaining employees would be journeymen ?	"the exam gives us a number that is easily divisible by 7 to pique our curiosity and tempt us into calculating actual numbers ( also because otherwise the ratio would be incorrect ) . since the question is about percentages , the actual numbers will be meaningless , as only the ratio of that number versus others will be meaningful . nonetheless , for those who are curious , each 1 / 7 portion represents ( 14210 / 7 ) 2,030 employees . this in turn means that 4,060 employees are journeymen and the remaining 10,150 are full time workers . if half the journeymen were laid off , that would mean 1 / 7 of the total current workforce would be removed . this statistic is what leads many students to think that since half the journeymen are left , the remaining journeymen would represent half of what they used to be , which means 1 / 7 of the total workforce . if 1 / 7 of the workforce is journeymen , and 1 / 7 is roughly 14.3 % , then answer choice a should be the right answer . in this case , though , it is merely the tempting trap answer choice . what changed between the initial statement and the final tally ? well , you let go of 1 / 7 of the workforce , so the total number of workers went down . the remaining workers are still 1 / 7 of the initial workers , but the group has changed . the new workforce is smaller than the original group , specifically 6 / 7 of it because 1 / 7 was eliminated . the remaining workers now account for 1 / 7 out of 6 / 7 of the force , which if we multiply by 7 gives us 1 out of 6 . this number as a percentage is answer choice b , 28.6 % . using the absolute numbers we calculated before , there were 4,060 journeymen employees out of 14,210 total . if 2,030 of them are laid off , then there are 2,030 journeyman employees left , but now out of a total of ( 14,210 - 2,030 ) 12,180 employees . 2,030 / 12,180 is exactly 1 / 6 , or 16.67 % . the answer will work with either percentages or absolute numbers , but the percentage calculation will be significantly faster and applicable to any similar situation . the underlying principle of percentages ( and , on a related note , ratios ) can be summed up in the brainteaser i like to ask my students : if you ’ re running a race and you overtake the 2 nd place runner just before the end , what position do you end up in ? the correct answer is 2 nd place . percentages , like ratios and other concepts of relative math , depend entirely on the context . whether 100 % more of something is better than 50 % more of something else depends on the context much more than the percentages quoted . when it comes to percentages on the gmat , the goal is to understand them enough to instinctively not fall into the traps laid out for you . d"	a ) 14.3 % , b ) 16.67 % , c ) 33 % , d ) 28.6 % , e ) 49.67 %	d	multiply(multiply(divide(divide(divide(2, 7), 2), add(divide(divide(2, 7), 2), subtract(const_1, divide(2, 7)))), const_100), const_3)	divide(n1,n2)\|divide(#0,n1)\|subtract(const_1,#0)\|add(#1,#2)\|divide(#1,#3)\|multiply(#4,const_100)\|multiply(#5,const_3)\|	general
an electric pump can fill a tank in 10 hours . because of a leak in the tank , it took 20 hours to fill the tank . if the tank is full , how much time will the leak take to empty it ?	"work done by the leak in 1 hour = 1 / 10 - 1 / 20 = 1 / 20 the leak will empty the tank in 20 hours answer is c"	a ) 10 hours , b ) 12 hours , c ) 20 hours , d ) 5 hours , e ) 15 hours	c	divide(20, const_1)	divide(n1,const_1)\|	physics
for every even positive integer m , f ( m ) represents the product of all even integers from 2 to m , inclusive . for example , f ( 12 ) = 2 x 4 x 6 x 8 x 10 x 12 . what is the greatest prime factor of f ( 36 ) ?	"f ( 36 ) = 2 * 4 * 6 * 8 * 10 * 12 * 14 * 16 * 18 * 20 * 22 * 24 * 26 * 28 * 30 * 32 * 34 * 36 the greatest prime factor in this list is 17 . the answer is d ."	a ) 2 , b ) 5 , c ) 11 , d ) 17 , e ) 23	d	subtract(divide(36, 2), const_1)	divide(n8,n0)\|subtract(#0,const_1)\|	general
pipe a can fill a tank in 10 hours . due to a leak at the bottom , it takes 15 hours for the pipe a to fill the tank . in what time can the leak alone empty the full tank ?	let the leak can empty the full tank in x hours 1 / 10 - 1 / x = 1 / 15 = > 1 / x = 1 / 10 - 1 / 15 = ( 3 - 2 ) / 30 = 1 / 30 = > x = 30 . answer : e	a ) 76 , b ) 84 , c ) 56 , d ) 75 , e ) 30	e	divide(multiply(15, 10), subtract(15, 10))	multiply(n0,n1)\|subtract(n1,n0)\|divide(#0,#1)	physics
5 women can do a work in two days . 10 men can complete the same work in five days . what is the ratio between the capacity of a man and a woman ?	"explanation : ( 5 ã — 2 ) women can complete the work in 1 day . â ˆ ´ 1 woman ' s 1 day ' s work = 1 / 10 ( 10 ã — 5 ) men can complete the work in 1 day . â ˆ ´ 1 man ' s 1 day ' s work = 1 / 50 so , required ratio = 1 / 10 : 1 / 50 = 1 : 5 answer : b"	a ) 1 : 2 , b ) 1 : 5 , c ) 2 : 3 , d ) 3 : 2 , e ) none of these	b	divide(divide(const_1, multiply(5, 10)), divide(const_1, multiply(10, const_10)))	multiply(n0,n1)\|multiply(n1,const_10)\|divide(const_1,#0)\|divide(const_1,#1)\|divide(#2,#3)\|	physics
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 ?	"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 ) ."	a ) 161 , b ) 162 , c ) 163 , d ) 164 , e ) 165	c	subtract(add(add(add(add(add(const_1, power(const_2, const_2)), power(const_3, const_2)), power(const_4, const_2)), power(add(const_4, const_1), const_2)), power(8, const_2)), add(add(add(const_4, const_3), 8), add(add(add(add(const_2, const_3), add(const_4, const_1)), add(const_4, const_3)), add(add(const_2, const_3), 8))))	add(const_1,const_4)\|add(const_3,const_4)\|add(const_2,const_3)\|power(const_2,const_2)\|power(const_3,const_2)\|power(const_4,const_2)\|power(n0,const_2)\|add(#3,const_1)\|add(n0,#1)\|add(#2,#0)\|add(n0,#2)\|power(#0,const_2)\|add(#7,#4)\|add(#9,#1)\|add(#12,#5)\|add(#13,#10)\|add(#14,#11)\|add(#8,#15)\|add(#16,#6)\|subtract(#18,#17)\|	general
the average age of 19 persons in a office is 15 years . out of these , the average age of 5 of them is 14 years and that of the other 9 persons is 16 years . the age of the 15 th person is ?	"age of the 15 th student = 19 * 15 - ( 14 * 5 + 16 * 9 ) = 285 - 214 = 71 years answer is b"	a ) 9 , b ) 71 , c ) 85 , d ) 92 , e ) 90	b	subtract(subtract(multiply(19, 15), multiply(5, 14)), multiply(9, 16))	multiply(n0,n1)\|multiply(n2,n3)\|multiply(n4,n5)\|subtract(#0,#1)\|subtract(#3,#2)\|	general
a pet store regularly sells pet food at a discount of 10 percent to 30 percent from the manufacturer ’ s suggested retail price . if during a sale , the store discounts an additional 20 percent from the discount price , what would be the lowest possible price of a container of pet food that had a manufacturer ’ s suggested retail price o f $ 40.00 ?	"for retail price = $ 40 first maximum discounted price = 40 - 30 % of 40 = 40 - 12 = 28 price after additional discount of 20 % = 28 - 20 % of 28 = 28 - 5.6 = 22.4 answer : option c"	a ) $ 10.00 , b ) $ 11.20 , c ) $ 22.40 , d ) $ 16.00 , e ) $ 18.00	c	multiply(divide(subtract(const_100, 20), const_100), multiply(divide(subtract(const_100, 30), const_100), 40.00))	subtract(const_100,n2)\|subtract(const_100,n1)\|divide(#0,const_100)\|divide(#1,const_100)\|multiply(n3,#3)\|multiply(#2,#4)\|	gain
the perimeter of an equilateral triangle is 60 . if one of the sides of the equilateral triangle is the side of an isosceles triangle of perimeter 45 , then how long is the base of isosceles triangle ?	"the base of the isosceles triangle is 45 - 20 - 20 = 5 units the answer is a ."	a ) 5 units , b ) 10 units , c ) 15 units , d ) 20 units , e ) 25 units	a	subtract(subtract(45, divide(60, const_3)), divide(60, const_3))	divide(n0,const_3)\|subtract(n1,#0)\|subtract(#1,#0)\|	geometry
a small table has a length of 12 inches and a breadth of b inches . cubes are placed on the surface of the table so as to cover the entire surface . the maximum side of such cubes is found to be 4 inches . also , a few such tables are arranged to form a square . the minimum length of side possible for such a square is 80 inches . find b .	from the info that the maximum sides of the cubes is 4 , we know that the gcf of 12 ( = 2 ^ 2 * 3 ) andbis 4 ( = 2 ^ 2 ) , sob = 2 ^ x , where x > = 2 . from the second premise , we know that the lcm of 12 ( 2 ^ 2 * 3 ) andbis 80 ( 2 ^ 4 * 5 ) , sob = 2 ^ 4 or 2 ^ 4 * 5 ( 16 or 80 ) . combining 2 premises shows the answer is b ( 16 ) .	['a ) 8', 'b ) 16', 'c ) 24', 'd ) 32', 'e ) 48']	b	sqrt(subtract(power(divide(80, 4), const_2), power(12, const_2)))	divide(n2,n1)\|power(n0,const_2)\|power(#0,const_2)\|subtract(#2,#1)\|sqrt(#3)	geometry
calculate the ratio between x and y if 25 % of x equal to 40 % of y ?	"explanation : 25 x = 40 y x : y = 25 : 40 = 5 : 8 answer : a"	a ) 5 : 8 , b ) 5 : 9 , c ) 5 : 7 , d ) 5 : 6 , e ) 5 : 4	a	divide(25, 40)	divide(n0,n1)\|	general
. 003 / ? = . 01	"let . 003 / x = . 01 ; then x = . 003 / . 01 = . 3 / 1 = . 3 answer is a"	a ) . 3 , b ) . 09 , c ) . 009 , d ) . 0009 , e ) none of them	a	divide(divide(003, const_1000), divide(01, const_100))	divide(n0,const_1000)\|divide(n1,const_100)\|divide(#0,#1)\|	general
if √ 10 = 3.16 , find the value of if √ 5 / 2	explanation : √ ( 5 / 2 ) = √ ( 5 × 2 / 2 × 2 ) = √ ( 10 ) / 2 = 3.16 / 2 = 1.58 answer : b	a ) 1.3 , b ) 1.58 , c ) 2.03 , d ) 2.15 , e ) 3.15	b	sqrt(divide(5, 2))	divide(n2,n3)\|sqrt(#0)	general
the length of a rectangular plot is 20 metres more than its breadth . if the cost of fencing the plot @ rs . 26.50 per metre is rs . 7420 , what is the length of the plot in metres ?	"let length of plot = l meters , then breadth = l - 20 meters and perimeter = 2 [ l + l - 20 ] = [ 4 l - 40 ] meters [ 4 l - 40 ] * 26.50 = 7420 [ 4 l - 40 ] = 7420 / 26.50 = 280 4 l = 320 l = 320 / 4 = 80 meters . answer : e"	a ) 20 , b ) 200 , c ) 300 , d ) 400 , e ) 80	e	subtract(divide(divide(7420, 26.50), const_2), multiply(const_2, 20))	divide(n2,n1)\|multiply(n0,const_2)\|divide(#0,const_2)\|subtract(#2,#1)\|	physics
30 men can do a work in 40 days . when should 12 men leave the work so that the entire work is completed in 40 days after they leave the work ?	"total work to be done = 30 * 40 = 1200 let 12 men leave the work after ' p ' days , so that the remaining work is completed in 40 days after they leave the work . 40 p + ( 12 * 40 ) = 1200 40 p = 720 = > p = 18 days answer : a"	a ) 18 days , b ) 10 days , c ) 55 days , d ) 44 days , e ) 22 days	a	divide(subtract(multiply(30, 40), multiply(40, 12)), 40)	multiply(n0,n1)\|multiply(n1,n2)\|subtract(#0,#1)\|divide(#2,n1)\|	physics
carrie likes to buy t - shirts at the local clothing store . they cost $ 9.65 each . one day , she bought 12 t - shirts . how much money did she spend ?	$ 9.65 * 12 = $ 115.8 . answer is a .	a ) $ 115.8 , b ) $ 248.75 , c ) $ 200 , d ) $ 171.6 , e ) $ 190	a	floor(multiply(12, 9.65))	multiply(n0,n1)\|floor(#0)\|	general
a train 110 m long is running with a speed of 30 km / h . in how many seconds will the train pass a man who is running at 3 km / h in the direction opposite to that in which the train is going ?	"the speed of the train relative to the man = 30 + 3 = 33 km / h . 33000 m / h * 1 h / 3600 s = ( 330 / 36 ) m / s ( 110 m ) / ( 330 / 36 m / s ) = ( 110 * 36 ) / 330 = 36 / 3 = 12 seconds the answer is d ."	a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 14	d	divide(110, divide(add(30, 3), const_3_6))	add(n1,n2)\|divide(#0,const_3_6)\|divide(n0,#1)\|	physics
kanul spent $ 3000 in buying raw materials , $ 1000 in buying machinery and 30 % of the total amount he had as cash with him . what was the total amount ?	"let the total amount be x then , ( 100 - 30 ) % of x = 3000 + 1000 70 % of x = 4000 70 x / 100 = 4000 x = $ 40000 / 7 x = $ 5714.28 answer is c"	a ) $ 5825.16 , b ) $ 5725.26 , c ) $ 5714.28 , d ) $ 5912.52 , e ) $ 5614.46	c	divide(add(3000, 1000), subtract(const_1, divide(30, const_100)))	add(n0,n1)\|divide(n2,const_100)\|subtract(const_1,#1)\|divide(#0,#2)\|	gain
a sporting goods store sold 64 frisbees in one week , some for $ 3 and the rest for $ 4 each . if receipts from frisbee sales for the week totaled $ 204 , what is the fewest number of $ 3 frisbees that could have been sold ?	"in this question however , because we are told that exactly 64 frisbees have been sold and revenue was exactly $ 204 , there is only one possible solution for the number of $ 3 and $ 4 frisbees sold . to solve , we have 2 equations and 2 unknowns let x = number of $ 3 frisbees sold let y = number of $ 4 frisbees sold x + y = 64 3 x + 4 y = 204 x = 64 - y 3 ( 64 - y ) + 4 y = 204 192 - 3 y + 4 y = 204 y = 12 x = 64 - 12 = 52 d"	a ) 24 , b ) 12 , c ) 8 , d ) 52 , e ) 2	d	subtract(204, multiply(64, 3))	multiply(n0,n1)\|subtract(n3,#0)\|	general
two trains 110 meters and 200 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 ?	"t = ( 110 + 200 ) / ( 80 + 65 ) * 18 / 5 t = 7.69 answer : b"	a ) 4.85 , b ) 7.69 , c ) 6.85 , d ) 5.85 , e ) 6.15	b	divide(add(110, 200), multiply(add(80, 65), const_0_2778))	add(n0,n1)\|add(n2,n3)\|multiply(#1,const_0_2778)\|divide(#0,#2)\|	physics
how many 1 / 10 s are there in 37 1 / 2 ?	"required number = ( 75 / 2 ) / ( 1 / 10 ) = ( 75 / 2 x 10 / 1 ) = 375 . answer : a"	a ) 375 , b ) 475 , c ) 500 , d ) 670 , e ) 700	a	divide(add(37, divide(1, 2)), divide(1, 10))	divide(n0,n4)\|divide(n0,n1)\|add(n2,#0)\|divide(#2,#1)\|	general
a circle graph shows how the megatech corporation allocates its research and development budget : 12 % microphotonics ; 24 % home electronics ; 15 % food additives ; 29 % genetically modified microorganisms ; 8 % industrial lubricants ; and the remainder for basic astrophysics . if the arc of each sector of the graph is proportional to the percentage of the budget it represents , how many degrees of the circle are used to represent basic astrophysics research ?	"here all percentage when summed we need to get 100 % . as per data 12 + 24 + 15 + 29 + 8 = 88 % . so remaining 12 % is the balance for the astrophysics . since this is a circle all percentage must be equal to 360 degrees . 100 % - - - - 360 degrees then 12 % will be 43 degrees . . imo option a ."	a ) 43 ° , b ) 10 ° , c ) 18 ° , d ) 36 ° , e ) 52 °	a	divide(multiply(subtract(const_100, add(add(add(add(12, 24), 15), 29), 8)), divide(const_3600, const_10)), const_100)	add(n0,n1)\|divide(const_3600,const_10)\|add(n2,#0)\|add(n3,#2)\|add(n4,#3)\|subtract(const_100,#4)\|multiply(#1,#5)\|divide(#6,const_100)\|	gain
how much interest will $ 10,000 earn in 3 months at an annual rate of 6 % ?	"soln : - 6 months = 1 / 4 of year ; 6 % = 6 / 100 = 3 / 50 ; $ 10,000 ( principal ) * 3 / 50 ( interest rate ) * 1 / 4 ( time ) = $ 150 . answer : b"	a ) $ 250 , b ) $ 150 , c ) $ 450 , d ) $ 550 , e ) $ 650	b	multiply(multiply(power(const_100, const_2), divide(6, const_100)), divide(const_3, const_4))	divide(const_3,n2)\|divide(const_4.0,const_100)\|power(const_100,const_2)\|multiply(#1,#2)\|multiply(#0,#3)\|	gain
how many different values of positive integer x , for which \| x + 9 \| < x , are there ?	"answer c i opted to put the random value option . i used 0 , 9 , - 9 and the the extreme of 10 and - 10 . . i was able to solve it in 1 : 09 c"	a ) 0 , b ) 2 , c ) 3 , d ) 8 , e ) 16	c	add(9, 9)	add(n0,n0)\|	general
if the average of 10 consecutive integers is 21.5 then the 10 th integer is : -	"the average falls between the 5 th and 6 th integers , integer 5 = 21 , integer 6 = 22 . counting up to the tenth integer we get 26 . answer : d"	a ) 15 , b ) 20 , c ) 23 , d ) 26 , e ) 25	d	add(multiply(10, const_2), multiply(subtract(21.5, multiply(10, const_2)), 10))	multiply(n0,const_2)\|subtract(n1,#0)\|multiply(n0,#1)\|add(#0,#2)\|	general
in a box of 8 pens , a total of 3 are defective . if a customer buys 2 pens selected at random from the box , what is the probability that neither pen will be defective ?	"p ( neither pen is defective ) = 5 / 8 * 4 / 7 = 5 / 14 the answer is d ."	a ) 2 / 10 , b ) 3 / 11 , c ) 4 / 13 , d ) 5 / 14 , e ) 6 / 17	d	multiply(divide(subtract(8, 3), 8), divide(subtract(subtract(8, 3), const_1), subtract(8, const_1)))	subtract(n0,n1)\|subtract(n0,const_1)\|divide(#0,n0)\|subtract(#0,const_1)\|divide(#3,#1)\|multiply(#2,#4)\|	general
in filling a room with gas 100 m * 10 m * 10 m the volumes of gas will be ?	100 * 10 * 10 = 10000 answer : c	['a ) 100 cu . m', 'b ) 1000 cu . m', 'c ) 10000 cu . m', 'd ) 100000 cu . m', 'e ) 10000000 cu . m']	c	volume_rectangular_prism(100, 10, 10)	volume_rectangular_prism(n0,n1,n1)	physics
a train passes a station platform in 30 sec and a man standing on the platform in 12 sec . if the speed of the train is 54 km / hr . what is the length of the platform ?	"speed = 54 * 5 / 18 = 15 m / sec . length of the train = 15 * 12 = 180 m . let the length of the platform be x m . then , ( x + 180 ) / 30 = 15 = > x = 270 m . answer : e"	a ) 227 , b ) 240 , c ) 200 , d ) 178 , e ) 270	e	multiply(12, multiply(54, const_0_2778))	multiply(n2,const_0_2778)\|multiply(n1,#0)\|	physics
a right triangle is inscribed in a circle . the legs of the triangle have lengths 6 and 8 . what is the diameter of the circle ?	property of a right triangle inscribed in a circle is that when an angle is made from diameter of the circle , it is a right triangle . or if a right triangle is made inscribed in a circle , its its longest side is the diameter of the circle . hence diameter = ( 6 ^ 2 + 8 ^ 2 ) ^ 1 / 2 = 10 d is the answer	['a ) 8', 'b ) 8 π', 'c ) 9 √ 3', 'd ) 10', 'e ) 12']	d	sqrt(add(power(6, const_2), power(8, const_2)))	power(n0,const_2)\|power(n1,const_2)\|add(#0,#1)\|sqrt(#2)	geometry
what is 15 percent of 64 ?	"( 15 / 100 ) * 64 = 9.6 the answer is c ."	a ) 7.2 , b ) 8.4 , c ) 9.6 , d ) 10.4 , e ) 11.8	c	divide(multiply(15, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100)	add(const_2,const_3)\|add(const_3,const_4)\|add(const_3,const_3)\|multiply(const_3,const_4)\|multiply(#0,const_2)\|multiply(#3,const_100)\|multiply(#1,#0)\|multiply(#4,#5)\|multiply(#6,#4)\|add(#7,#8)\|add(#9,#2)\|multiply(n0,#10)\|divide(#11,const_100)\|	gain
each week a restaurant serving mexican food uses the same volume of chili paste , which comes in either 55 - ounce cans or 15 - ounce cans of chili paste . if the restaurant must order 30 more of the smaller cans than the larger cans to fulfill its weekly needs , then how many larger cans are required to fulfill its weekly needs ?	"let x be the number of 55 ounce cans . therefore ( x + 30 ) is the number of 15 ounce cans . total volume is same , therefore 55 x = 15 ( x + 30 ) 30 x = 450 x = 15 ans - a"	a ) 15 , b ) 28 , c ) 18 , d ) 24 , e ) 21	a	add(15, 30)	add(n1,n2)\|	general
for any positive integer n , the sum of the first n positive integers equals n ( n + 1 ) / 2 . what is the sum of all the even integers between 99 and 181 ?	"100 + 102 + . . . + 180 = 100 * 41 + ( 2 + 4 + . . . + 80 ) = 100 * 41 + 2 * ( 1 + 2 + . . . + 40 ) = 100 * 41 + 2 ( 40 ) ( 41 ) / 2 = 100 * 41 + 40 * 41 = 140 ( 41 ) = 5740 the answer is b ."	a ) 3820 , b ) 5740 , c ) 6580 , d ) 7360 , e ) 9200	b	add(divide(subtract(subtract(181, 1), add(99, 1)), 2), 1)	add(n2,n0)\|subtract(n3,n0)\|subtract(#1,#0)\|divide(#2,n1)\|add(n0,#3)\|	general
a salesman â € ™ s terms were changed from a flat commission of 5 % on all his sales to a fixed salary of rs . 1000 plus 2.5 % commission on all sales exceeding rs . 4,000 . if his remuneration as per new scheme was rs . 800 more than that by the previous schema , his sales were worth ?	"[ 1000 + ( x - 4000 ) * ( 2.5 / 100 ) ] - x * ( 5 / 100 ) = 800 x = 4000 answer : a"	a ) s . 4,000 , b ) s . 12,000 , c ) s . 30,000 , d ) s . 40,000 , e ) s . 50,000	a	subtract(multiply(5, const_4), const_12)	multiply(n0,const_4)\|subtract(#0,const_12)\|	general
if k is an integer and 0.0010101 x 10 ^ k is greater than 10 , what is the least possible value of k ?	"0.0010101 * 10 ^ k > 10 we need to move the decimal point to the right 4 places to get 10.101 this is equivalent to multiplying by 10 ^ 4 . the answer is c ."	a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6	c	divide(log(divide(10, 0.0010101)), log(10))	divide(n2,n0)\|log(n1)\|log(#0)\|divide(#2,#1)\|	general
if a and b are positive integers and ( 3 ^ a ) ^ b = 3 ^ 3 , what is the value of 3 ^ a * 3 ^ b ?	3 ^ ab = 3 ^ 3 therefore ab = 3 either a = 1 or 3 or b = 3 or 1 therefore 3 ^ a * 3 ^ b = 3 ^ ( a + b ) = 3 ^ 4 = 81 c	a ) 3 , b ) 9 , c ) 81 , d ) 27 , e ) 243	c	multiply(power(3, 3), 3)	power(n0,n0)\|multiply(n0,#0)	general
a gardener is going to plant 2 red rosebushes and 2 white rosebushes . if the gardener is to select each of the bushes at random , one at a time , and plant them in a row , what is the probability that the 2 rosebushes in the middle of the row will be the red rosebushes ?	we are asked to find the probability of one particular pattern : wrrw . total # of ways a gardener can plant these four bushes is the # of permutations of 4 letters wwrr , out of which 2 w ' s and 2 r ' s are identical , so 4 ! / 2 ! 2 ! = 6 ; so p = 1 / 6 answer : b .	a ) 1 / 12 , b ) 1 / 6 , c ) 1 / 5 , d ) 1 / 3 , e ) ½	b	divide(const_1, choose(add(2, 2), 2))	add(n0,n0)\|choose(#0,n0)\|divide(const_1,#1)	probability
a certain university will select 1 of 5 candidates eligible to fill a position in the mathematics department and 2 of 8 candidates eligible to fill 2 identical positions in the computer science department . if none of the candidates is eligible for a position in both departments , how many different sets of 3 candidates are there to fill the 3 positions ?	"1 c 5 * 2 c 8 = 5 * 28 = 140 the answer is ( b )"	a ) 138 , b ) 140 , c ) 142 , d ) 145 , e ) 148	b	multiply(multiply(8, 3), 5)	multiply(n3,n5)\|multiply(n1,#0)\|	other
what will be the lcm of 8 , 24 , 36 and 54	explanation : lcm of 8 - 24 - 36 - 54 will be 2 * 2 * 2 * 3 * 3 * 3 = 216 answer : option c	a ) 54 , b ) 108 , c ) 216 , d ) 432 , e ) none of these	c	multiply(multiply(multiply(multiply(multiply(const_2, const_2), const_2), const_3), const_3), const_3)	multiply(const_2,const_2)\|multiply(#0,const_2)\|multiply(#1,const_3)\|multiply(#2,const_3)\|multiply(#3,const_3)	physics
a number exceeds by 35 from its 3 / 8 part . then the number is ?	"x – 3 / 8 x = 35 x = 56 answer : a"	a ) a ) 56 , b ) b ) 35 , c ) c ) 39 , d ) d ) 40 , e ) e ) 45	a	divide(multiply(35, 8), subtract(8, 3))	multiply(n0,n2)\|subtract(n2,n1)\|divide(#0,#1)\|	general
if 0.75 : x : : 5 : 6 , then x is equal to :	"explanation : ( x * 5 ) = ( 0.75 * 6 ) x = 4.5 / 5 = 0.9 answer : d"	a ) 1.12 , b ) 1.16 , c ) 1.2 , d ) 0.9 , e ) none of these	d	divide(multiply(0.75, 6), 5)	multiply(n0,n2)\|divide(#0,n1)\|	general
two trains of equal lengths take 10 sec and 14 sec respectively to cross a telegraph post . if the length of each train be 120 m , in what time will they cross other travelling in opposite direction ?	"speed of the first train = 120 / 10 = 12 m / sec . speed of the second train = 120 / 14 = 8.6 m / sec . relative speed = 12 + 8.6 = 20.6 m / sec . required time = ( 120 + 120 ) / 20.6 = 11.7 sec . answer : option d"	a ) 10 , b ) 12 , c ) 13 , d ) 11.7 , e ) 15	d	divide(multiply(120, const_2), add(speed(120, 14), speed(120, 10)))	multiply(n2,const_2)\|speed(n2,n1)\|speed(n2,n0)\|add(#1,#2)\|divide(#0,#3)\|	physics
four equal circles are described about the four corners of a square so that each touches two of the others . if a side of the square is 14 cm , then the area enclosed between the circumferences of the circles is :	the shaded area gives the required region . area of the shaded region = area of the square – area of four quadrants of the circles = ( 14 ) 2 - 4 × 1 ⁄ 4 π ( 7 ) 2 = 196 - 22 ⁄ 7 × 49 = 196 - 154 = 42 cm 2 answer b	['a ) 24 cm 2', 'b ) 42 cm 2', 'c ) 154 cm 2', 'd ) 196 cm 2', 'e ) none of these']	b	subtract(square_area(14), circle_area(divide(14, const_2)))	divide(n0,const_2)\|square_area(n0)\|circle_area(#0)\|subtract(#1,#2)	geometry
in the coordinate plane , line a has a slope of - 1 and an x - intercept of 1 . line b has a slope of 4 and a y - intercept of - 4 . if the two lines intersect at the point ( a , b ) , what is the sum a + b ?	"the equation of line a is y = - x + 1 the equation of line b is y = 4 x - 4 4 x - 4 = - x + 1 x = 1 y = 0 the point of intersection is ( 1,0 ) and then a + b = 1 . the answer is b ."	a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4	b	divide(subtract(subtract(4, 4), 1), 1)	subtract(n3,n2)\|subtract(#0,n0)\|divide(#1,n1)\|	general
if the annual increase in the population of a town is 10 % and the present number of people is 10000 , what will the population be in 2 years ?	"the required population is = 10000 ( 1 + 10 / 100 ) ^ 2 = 10000 * 11 / 10 * 11 / 10 = 12100 answer is a"	a ) 12100 , b ) 15240 , c ) 12456 , d ) 11452 , e ) 10002	a	multiply(multiply(divide(add(10, const_100), const_100), 10000), divide(add(10, const_100), const_100))	add(n0,const_100)\|divide(#0,const_100)\|multiply(n1,#1)\|multiply(#1,#2)\|	gain
the sum of two consecutive integers is 41 . find the numbers .	"n + ( n + 1 ) = 41 2 n + 1 = 41 2 n = 40 n = 20 answer : d"	a ) 17 , 18 , b ) 7 , 8 , c ) 5 , 6 , d ) 20,21 , e ) 8 , 9	d	add(add(power(add(add(divide(subtract(subtract(41, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(41, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(41, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(41, const_10), const_2), const_4), const_2), const_2)))	subtract(n0,const_10)\|subtract(#0,const_2)\|divide(#1,const_4)\|add(#2,const_2)\|power(#2,const_2)\|add(#3,const_2)\|power(#3,const_2)\|add(#5,const_2)\|add(#4,#6)\|power(#5,const_2)\|power(#7,const_2)\|add(#9,#10)\|add(#11,#8)\|	physics
calculate the circumference of a circular field whose radius is 10 centimeters .	"circumference c is given by c = 2 π r = 2 π * 10 = 20 π cm answer : a"	a ) 20 π cm , b ) 19 π cm , c ) 28 π cm , d ) 25 π cm , e ) 30 π cm	a	circumface(10)	circumface(n0)\|	physics

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Dataset Card for MathQA

Dataset Summary

We introduce a large-scale dataset of math word problems.

Our dataset is gathered by using a new representation language to annotate over the AQuA-RAT dataset with fully-specified operational programs.

AQuA-RAT has provided the questions, options, rationale, and the correct options.

Supported Tasks and Leaderboards

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Languages

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Dataset Structure

Data Instances

default

Size of downloaded dataset files: 7.30 MB
Size of the generated dataset: 22.96 MB
Total amount of disk used: 30.27 MB

An example of 'train' looks as follows.

{
    "Problem": "a multiple choice test consists of 4 questions , and each question has 5 answer choices . in how many r ways can the test be completed if every question is unanswered ?",
    "Rationale": "\"5 choices for each of the 4 questions , thus total r of 5 * 5 * 5 * 5 = 5 ^ 4 = 625 ways to answer all of them . answer : c .\"",
    "annotated_formula": "power(5, 4)",
    "category": "general",
    "correct": "c",
    "linear_formula": "power(n1,n0)|",
    "options": "a ) 24 , b ) 120 , c ) 625 , d ) 720 , e ) 1024"
}

Data Fields

The data fields are the same among all splits.

default

Problem: a string feature.
Rationale: a string feature.
options: a string feature.
correct: a string feature.
annotated_formula: a string feature.
linear_formula: a string feature.
category: a string feature.

Data Splits

name	train	validation	test
default	29837	4475	2985

Dataset Creation

Curation Rationale

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Source Data

Initial Data Collection and Normalization

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Who are the source language producers?

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Annotations

Considerations for Using the Data

Social Impact of Dataset

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Discussion of Biases

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Other Known Limitations

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Additional Information

Dataset Curators

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Licensing Information

The dataset is licensed under the Apache License, Version 2.0.

Citation Information

@inproceedings{amini-etal-2019-mathqa,
    title = "{M}ath{QA}: Towards Interpretable Math Word Problem Solving with Operation-Based Formalisms",
    author = "Amini, Aida  and
      Gabriel, Saadia  and
      Lin, Shanchuan  and
      Koncel-Kedziorski, Rik  and
      Choi, Yejin  and
      Hajishirzi, Hannaneh",
    booktitle = "Proceedings of the 2019 Conference of the North {A}merican Chapter of the Association for Computational Linguistics: Human Language Technologies, Volume 1 (Long and Short Papers)",
    month = jun,
    year = "2019",
    address = "Minneapolis, Minnesota",
    publisher = "Association for Computational Linguistics",
    url = "https://aclanthology.org/N19-1245",
    doi = "10.18653/v1/N19-1245",
    pages = "2357--2367",
}