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
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how long does a train 110 m long traveling at 60 kmph takes to cross a bridge of 170 m in length ? | "d = 110 + 170 = 280 m s = 60 * 5 / 18 = 50 / 3 t = 280 * 3 / 50 = 16.8 sec answer a" | a ) 16.8 , b ) 17.8 , c ) 15.9 , d ) 13.2 , e ) 12.4 | a | divide(add(110, 170), multiply(60, const_0_2778)) | add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)| | physics |
for any positive integer n , the sum of the first n positive integers equals n ( n + 1 ) / 2 . what is the sum x of all the even integers between 99 and 301 ? | "i would solve it in a different way . first of all , total number of even integers between 99 and 301 are , ( 301 - 99 ) / 2 = 202 / 2 = 101 average = ( 301 + 99 ) / 2 = 400 / 2 = 200 sum x = average * total count = 200 * 101 = 20,200 answer is b" | a ) 10,100 , b ) 20,200 , c ) 22,650 , d ) 40,200 , e ) 45,150 | b | add(divide(subtract(subtract(301, 1), add(99, 1)), 2), 1) | add(n2,n0)|subtract(n3,n0)|subtract(#1,#0)|divide(#2,n1)|add(n0,#3)| | general |
in track last week , the boys ran 27 laps . the girls ran 9 more laps . each lap is a 3 - fourth of a mile . how many miles did the girls run ? | the girls ran 27 + 9 = 36 laps . 36 x 3 / 4 = 108 / 4 , which reduces to 27 . the girls ran 27 miles correct answer d | a ) 24 miles , b ) 25 miles , c ) 26 miles , d ) 27 miles , e ) 28 miles | d | multiply(divide(3, const_4), add(27, 9)) | add(n0,n1)|divide(n2,const_4)|multiply(#0,#1) | general |
if tim had lunch at $ 50.20 and he gave 20 % tip , how much did he spend ? | "the tip is 20 % of what he paid for lunch . hence tip = 20 % of 50.20 = ( 20 / 100 ) * 50.20 = $ 10.04 total spent 50.20 + 10.04 = $ 60.24 correct answer e" | a ) $ 30.60 , b ) $ 60.60 , c ) $ 70.60 , d ) $ 40.60 , e ) $ 60.24 | e | add(50.20, divide(multiply(50.20, 20), const_100)) | multiply(n0,n1)|divide(#0,const_100)|add(n0,#1)| | gain |
if the operation β¬ is defined for all x and y by the equation x β¬ y = 2 * x * y , then 8 β¬ ( 4 β¬ 5 ) = | "working inside out , ( 4 β¬ 5 ) = 2 * 4 * 5 = 40 8 β¬ 40 = 2 * 8 * 40 = 640 hence , answer is d" | a ) 80 , b ) 120 , c ) 160 , d ) 640 , e ) 760 | d | multiply(multiply(2, 8), multiply(multiply(2, 4), 5)) | multiply(n0,n1)|multiply(n0,n2)|multiply(n3,#1)|multiply(#0,#2)| | general |
a and b started a business in partnership investing rs . 20,000 and rs . 15,000 respectively . after 6 months , c joined them with rs . 20,000 . whatwill be b ' s share in total profit of rs . 24,000 earned at the end of 2 years from the startingof the business ? | "a : b : c = ( 20,000 x 24 ) : ( 15,000 x 24 ) : ( 20,000 x 18 ) = 4 : 3 : 3 . b ' s share = rs . 24000 x 3 / 10 = rs . 7,200 . e" | a ) s . 5,000 , b ) s . 5,500 , c ) s . 5,700 , d ) s . 6,500 , e ) s . 7,200 | e | multiply(multiply(add(const_4, const_1), const_4), multiply(2, multiply(const_3, const_4))) | add(const_1,const_4)|multiply(const_3,const_4)|multiply(#0,const_4)|multiply(n5,#1)|multiply(#2,#3)| | gain |
every year an amount increases by 1 / 6 th of itself . how much will it be after two years if its present value is rs . 64000 ? | "64000 * 7 / 6 * 7 / 6 = 87111 answer : a" | a ) 87111 , b ) 81007 , c ) 81008 , d ) 81066 , e ) 81022 | a | add(add(64000, multiply(divide(1, 6), 64000)), multiply(divide(1, 6), add(64000, multiply(divide(1, 6), 64000)))) | divide(n0,n1)|multiply(n2,#0)|add(n2,#1)|multiply(#2,#0)|add(#2,#3)| | general |
a factory has three types of machines , each of which works at its own constant rate . if 7 machine as and 11 machine bs can produce 305 widgets per hour , and if 8 machine as and 22 machine cs can produce 600 widgets per hour , how many widgets could one machine a , one machine b , and one machine c produce in one 8 -... | let machine a produce a widgets per hour . b produce b widgets per hour and c produce c widgets per hour . 7 a + 11 b = 305 - - - ( 1 ) 8 a + 22 c = 600 - - - ( 2 ) dividing ( 2 ) by 2 4 a + 11 c = 300 . . . . . ( 3 ) adding ( 1 ) ( 3 ) 11 a + 11 b + 11 c = 605 a + b + c = 55 per hour so for eight hrs = 55 * 8 = 440 = ... | a ) 400 , b ) 440 , c ) 550 , d ) 625 , e ) 700 | b | multiply(divide(600, 11), 8) | divide(n5,n1)|multiply(n3,#0) | physics |
how many odd numbers between 10 and 1200 are the squares of integers ? | "the numbers are the squares of 5 , 7 , 9 , . . . , 33 which includes 15 numbers . the answer is d ." | a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | d | add(10, const_4) | add(n0,const_4)| | geometry |
alok ordered 16 chapatis , 5 plates of rice , 7 plates of mixed vegetable and 5 ice - cream cups . the cost of each chapati is rs . 6 , that of each plate of rice is rs . 45 and that of mixed vegetable is rs . 70 . the amount that alok paid the cashier was rs . 961 . find the cost of each ice - cream cup ? | "explanation : let the cost of each ice - cream cup be rs . x 16 ( 6 ) + 5 ( 45 ) + 7 ( 70 ) + 6 ( x ) = 961 96 + 225 + 490 + 5 x = 961 5 x = 150 = > x = 30 . answer : b" | a ) 25 , b ) 30 , c ) 29 , d ) 12 , e ) 20 | b | divide(subtract(subtract(subtract(961, multiply(16, 5)), multiply(5, 45)), multiply(7, 70)), 5) | multiply(n0,n3)|multiply(n1,n5)|multiply(n2,n6)|subtract(n7,#0)|subtract(#3,#1)|subtract(#4,#2)|divide(#5,n3)| | general |
300 liters of a mixture contains milk and water in the ratio 3 : 2 if 15 liters of this mixture be replaced by 15 liters of milk , the ratio of milk to water in the new mixture would be ? | "quantity of milk in 300 liters if mix = 300 * 3 / 5 = 180 liters quantity of milk in 315 liters of new mix = 180 + 15 = 195 liters quantity of water in it = 315 - 195 = 120 liters ratio of milk and water in new mix = 90 : 20 = 13 : 8 answer isc" | a ) 12 : 8 , b ) 12 : 4 , c ) 13 : 8 , d ) 14 : 5 , e ) 13 : 5 | c | divide(multiply(subtract(15, const_1), 2), multiply(3, 2)) | multiply(n2,n1)|subtract(n3,const_1)|multiply(#1,n2)|divide(#2,#0)| | other |
a car started running at a speed of 30 km / hr and the speed of the car was increased by 2 km / hr at the end of every hour . find the total distance covered by the car in the first 12 hours of the journey . | "the total distance covered by the car in the first 12 hours = 30 + 32 + 34 + 36 + 38 + 40 + 42 + 44 + 46 + 48 + 50 + 52 = sum of 12 terms in ap whose first term is 30 and last term is 52 = 12 / 2 [ 30 + 52 ] = 492 km . answer : e" | a ) 342 km , b ) 352 km , c ) 642 km , d ) 742 km , e ) 492 km | e | divide(add(add(30, multiply(2, 12)), 30), 2) | multiply(n1,n2)|add(n0,#0)|add(n0,#1)|divide(#2,n1)| | physics |
a chemical supply company has 60 liters of a 25 % hno 3 solution . how many liters of pure undiluted hno 3 must the chemists add so that the resultant solution is a 50 % solution ? | "60 liters of a 25 % hno 3 solution means hno 3 = 15 liters in 60 liters of the solution . now , let x be the pure hno 3 added . as per question , 15 + x = 50 % of ( 60 + x ) or x = 30 . hence , e" | a ) 12 , b ) 15 , c ) 20 , d ) 24 , e ) 30 | e | multiply(subtract(divide(60, const_2), divide(multiply(60, 25), const_100)), const_2) | divide(n0,const_2)|multiply(n0,n1)|divide(#1,const_100)|subtract(#0,#2)|multiply(#3,const_2)| | gain |
in a 500 m race , the ratio of the speeds of two contestants a and b is 3 : 4 . a has a start of 140 m . then , a wins by | "solution to reach the winning post a will have to cover a distance of ( 500 - 140 ) m , i . e , 360 m . while a covers 3 m , b covers 4 m . while a covers 360 m , b covers ( 4 / 3 x 360 ) m = 480 m thus , when a reaches the winning post , b covers 480 m and therefore remains 20 m behind . β΄ a wins by 20 m . answer c" | a ) 60 m , b ) 40 m , c ) 20 m , d ) 10 m , e ) 50 m | c | subtract(500, divide(multiply(subtract(500, 140), 4), 3)) | subtract(n0,n3)|multiply(n2,#0)|divide(#1,n1)|subtract(n0,#2)| | physics |
pipe a can fill a tank in 2 hours . due to a leak at the bottom , it takes 3 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 / 2 - 1 / x = 1 / 3 = > 1 / x = 1 / 2 - 1 / 3 = ( 3 - 2 ) / 6 = 1 / 6 = > x = 6 . answer : d" | a ) 10 , b ) 9 , c ) 7 , d ) 6 , e ) 5 | d | divide(multiply(3, 2), subtract(3, 2)) | multiply(n0,n1)|subtract(n1,n0)|divide(#0,#1)| | physics |
a certain bus driver is paid a regular rate of $ 16 per hour for any number of hours that does not exceed 40 hours per week . for any overtime hours worked in excess of 40 hours per week , the bus driver is paid a rate that is 75 % higher than his regular rate . if last week the bus driver earned $ 752 in total compens... | "for 40 hrs = 40 * 16 = 640 excess = 752 - 640 = 112 for extra hours = . 75 ( 16 ) = 12 + 16 = 28 number of extra hrs = 112 / 28 = 28 / 7 = 4 total hrs = 40 + 4 = 44 answer c 44" | a ) 36 , b ) 40 , c ) 44 , d ) 48 , e ) 52 | c | add(40, divide(subtract(752, multiply(16, 40)), divide(multiply(16, add(const_100, 75)), const_100))) | add(n3,const_100)|multiply(n0,n1)|multiply(n0,#0)|subtract(n4,#1)|divide(#2,const_100)|divide(#3,#4)|add(n1,#5)| | general |
if the weight of 12 meters long rod is 12.2 kg . what is the weight of 6 meters long rod ? | "answer β΅ weight of 12 m long rod = 12.2 kg β΄ weight of 1 m long rod = 12.2 / 12 kg β΄ weight of 6 m long rod = 12.2 x 6 / 12 = 6.1 kg option : d" | a ) 7.2 kg . , b ) 10.8 kg . , c ) 12.4 kg . , d ) 6.1 kg , e ) none | d | divide(multiply(6, 12.2), 12) | multiply(n1,n2)|divide(#0,n0)| | physics |
x starts a business with rs . 45000 . y joins in the business after 3 months with rs . 45000 . what will be the ratio in which they should share the profit at the end of the year ? | "ratio in which they should share the profit = ratio of the investments multiplied by the time period = 45000 Γ 12 : 45000 Γ 9 = 45 Γ 12 : 45 Γ 9 = 3 Γ 12 : 2 Γ 9 = 4 : 3 answer is e ." | a ) 1 : 2 , b ) 2 : 1 , c ) 3 : 2 , d ) 2 : 3 , e ) 4 : 3 | e | divide(multiply(45000, const_12), multiply(45000, add(const_4, const_3))) | add(const_3,const_4)|multiply(n0,const_12)|multiply(n2,#0)|divide(#1,#2)| | other |
find the fraction which has the same ratio to 2 / 5 that 3 / 7 has to 6 / 5 | "p : 2 / 5 = 3 / 7 : 6 / 5 as the product of the means is equal to the product of the extremes . p * 6 / 5 = 2 / 5 * 3 / 7 p * 6 / 5 = 6 / 35 p = 1 / 7 = > p = 1 / 7 answer : e" | a ) 2 / 4 , b ) 6 / 78 , c ) 4 / 7 , d ) 3 / 5 , e ) 1 / 7 | e | multiply(divide(divide(3, 7), divide(6, 5)), divide(2, 5)) | divide(n2,n3)|divide(n4,n5)|divide(n0,n1)|divide(#0,#1)|multiply(#3,#2)| | other |
a student traveled 25 percent of the distance of the trip alone , continued another 20 miles with a friend , and then finished the last half of the trip alone . how many miles long was the trip ? | "let x be the total length of the trip . 0.25 x + 20 miles + 0.5 x = x 20 miles = 0.25 x x = 80 miles the answer is b ." | a ) 60 , b ) 80 , c ) 100 , d ) 120 , e ) 150 | b | divide(20, subtract(subtract(const_1, inverse(25)), divide(const_1, const_2))) | divide(const_1,const_2)|inverse(n0)|subtract(const_1,#1)|subtract(#2,#0)|divide(n1,#3)| | physics |
david obtained 72 , 60 , 35 , 62 and 84 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology what are his average marks ? | explanation : average = ( 72 + 60 + 35 + 62 + 84 ) / 5 = 313 / 5 = 62.6 . answer : b | a ) 75 , b ) 62.6 , c ) 87 , d ) 165 , e ) 11 | b | divide(add(add(add(add(72, 60), 35), 62), 84), add(const_2, const_3)) | add(n0,n1)|add(const_2,const_3)|add(n2,#0)|add(n3,#2)|add(n4,#3)|divide(#4,#1) | general |
mohit sold an article for $ 12000 . had he offered a discount of 10 % on the selling price , he would have earned a profit of 8 % . what is the cost price of the article ? | "let the cp be $ x . had he offered 10 % discount , profit = 8 % profit = 8 / 100 x and hence his sp = x + 8 / 100 x = $ 1.08 x = 12000 - 10 / 100 ( 12000 ) = 12000 - 1200 = $ 10800 = > 1.08 x = 10800 = > x = 10000 c" | a ) 16000 , b ) 25000 , c ) 10000 , d ) 18000 , e ) 17000 | c | multiply(divide(subtract(12000, divide(multiply(12000, 10), const_100)), add(const_100, 8)), const_100) | add(n2,const_100)|multiply(n0,n1)|divide(#1,const_100)|subtract(n0,#2)|divide(#3,#0)|multiply(#4,const_100)| | gain |
a certain store sold pens for $ 0.35 each and pencils for $ 0.25 each . if a customer purchased both pens and pencils from the store for a total of $ 1.80 , what total number of pens and pencils did the customer purchase ? | "answer : algebraically , the question looks like this : 2.5 = 0.35 x + 0.25 y there are three possible numbers of pens that fit that requirement : 1 pen : $ 0.35 2 pens : $ 0.70 3 pens : $ 1.05 and 3 pencils for $ 0.25 each . that ' s 3 pens and 3 pencils for a total of 6 pens and pencils . choice ( a ) is correct ." | a ) 6 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | a | multiply(divide(1.80, add(0.35, 0.25)), const_2) | add(n0,n1)|divide(n2,#0)|multiply(#1,const_2)| | other |
what profit percent is made by selling an article at a certain price , if by selling at 1 / 2 rd of that price , there would be a loss of 20 % ? | "sp 2 = 1 / 2 sp 1 cp = 100 sp 2 = 80 1 / 2 sp 1 = 80 sp 1 = 160 100 - - - 160 = > 60 % answer : c" | a ) 20 % , b ) 29 % , c ) 60 % , d ) 27 % , e ) 28 % | c | subtract(divide(subtract(const_100, 20), divide(1, 2)), const_100) | divide(n0,n1)|subtract(const_100,n2)|divide(#1,#0)|subtract(#2,const_100)| | gain |
williams works at a resort from the beginning of march to the end of september . during the month of august this past year , he made 10 times the average ( arithmetic mean ) of his monthly totals in tips for the other months . his total tips for august were what fraction of his total tips for all of the months he worke... | "first notice the number of months for which he worked - march to sept i . e . 7 months avg of monthly totals in tips for months other than august = x tips in august = 10 x total tips for all months = 6 * x + 10 x = 16 x tips for august / total tips for all months = 10 x / 16 x = 5 / 8 answer : a" | a ) 5 / 8 , b ) 7 / 8 , c ) 6 / 8 , d ) 2 / 9 , e ) 1 / 7 | a | divide(10, add(10, subtract(add(const_3, const_4), const_1))) | add(const_3,const_4)|subtract(#0,const_1)|add(n0,#1)|divide(n0,#2)| | general |
how many seconds will it take for a car that is traveling at a constant rate of 77 miles per hour to travel a distance of 21 yards ? ( 1 mile = 1,160 yards ) | "speed = 77 miles / hr = 37.64 yard / s distance = 21 yards time = distance / speed = 21 / 37.64 = 0.6 sec ans - c" | a ) . 1 , b ) 0.3 , c ) 0.6 , d ) 2 , e ) 1.6 | c | divide(21, multiply(divide(77, const_3600), multiply(subtract(add(77, 77), const_4), const_10))) | add(n0,n0)|divide(n0,const_3600)|subtract(#0,const_4)|multiply(#2,const_10)|multiply(#1,#3)|divide(n1,#4)| | physics |
by selling an article at rs . 400 , a profit of 25 % is made . find its cost price ? | "sp = 400 cp = ( sp ) * [ 100 / ( 100 + p ) ] = 400 * [ 100 / ( 100 + 25 ) ] = 400 * [ 100 / 125 ] = rs . 320 answer : b" | a ) 228 , b ) 320 , c ) 287 , d ) 480 , e ) 811 | b | divide(multiply(400, const_100), add(const_100, 25)) | add(n1,const_100)|multiply(n0,const_100)|divide(#1,#0)| | gain |
a certain galaxy is known to comprise approximately 4 x 10 ^ 12 stars . of every 50 million of these stars , one is larger in mass than our sun . approximately how many stars in this galaxy are larger than the sun ? | "total no . of stars on galaxy = 4 * 10 ^ 12 of every 50 million stars , 1 is larger than sun . 1 million = 10 ^ 6 therofore , 50 million = 50 * 10 ^ 6 total no . of stars larger than sun = 4 * 10 ^ 12 / 50 * 10 ^ 6 = 40 * 10 ^ 4 / 5 = 80000 therefore answer is c" | a ) 800 , b ) 1,250 , c ) 80,000 , d ) 12,000 , e ) 80,000 | c | multiply(divide(multiply(divide(multiply(4, 10), 50), power(10, const_4)), const_1000), 4) | multiply(n0,n1)|power(n1,const_4)|divide(#0,n3)|multiply(#2,#1)|divide(#3,const_1000)|multiply(n0,#4)| | general |
12.5 % of 192 = 50 % of ? | "answer let 12.5 % of 192 = 50 % of a , β ( 12.5 x 192 ) / 100 = ( 50 x a ) / 100 β΄ a = ( 12.5 x 192 ) / 50 = 48 correct option : a" | a ) 48 , b ) 96 , c ) 24 , d ) none of these , e ) can not be determined | a | divide(multiply(divide(multiply(50, 192), const_100), 12.5), const_100) | multiply(n1,n2)|divide(#0,const_100)|multiply(n0,#1)|divide(#2,const_100)| | gain |
a 9 % stock yields 25 % . the market value of the stock is : | "explanation : for an income of rs . 25 , investment = rs . 100 . for an income of rs 9 , investment = rs . 100 / 25 x 9 = rs 36 market value of rs . 100 stock = rs 36 answer is e" | a ) rs 45 , b ) rs 25 , c ) rs 40 , d ) rs 39 , e ) rs 36 | e | multiply(divide(const_100, 25), 9) | divide(const_100,n1)|multiply(n0,#0)| | gain |
find large number from below question the difference of two numbers is 1365 . on dividing the larger number by the smaller , we get 6 asquotient and the 15 as remainder ? | "let the smaller number be x . then larger number = ( x + 1365 ) . x + 1365 = 6 x + 15 5 x = 1350 x = 270 large number = 270 + 1365 = 1635 e" | a ) 1245 , b ) 1567 , c ) 1678 , d ) 1335 , e ) 1635 | e | multiply(divide(subtract(1365, 15), subtract(6, const_1)), 6) | subtract(n0,n2)|subtract(n1,const_1)|divide(#0,#1)|multiply(n1,#2)| | general |
β a β completes a work in 12 days . β b β completes the same work in 15 days . β a β started working alone and after 3 days b joined him . how many days will they now take together to complete the remaining work ? | work done by β a β in 3 days = 1 β 12 Γ 3 = 1 β 4 β΄ remaining work = 1 - 1 β 4 = 3 β 4 work done by a and b together = 12 Γ 15 / 27 = 20 / 3 β΄ remaining work done by a and b together in = 3 β 4 Γ 20 β 3 = 5 days answer a | a ) 5 , b ) 8 , c ) 6 , d ) 4 , e ) none of these | a | multiply(subtract(const_1, multiply(divide(const_1, 12), 3)), divide(multiply(12, 15), power(3, const_3))) | divide(const_1,n0)|multiply(n0,n1)|power(n2,const_3)|divide(#1,#2)|multiply(n2,#0)|subtract(const_1,#4)|multiply(#3,#5) | physics |
the sum of ages of 5 children born 3 years different each is 50 years . what is the age of the elder child ? | "let the ages of children be x , ( x + 3 ) , ( x + 6 ) , ( x + 9 ) and ( x + 12 ) years . then , x + ( x + 3 ) + ( x + 6 ) + ( x + 9 ) + ( x + 12 ) = 50 5 x = 20 x = 4 . x + 12 = 4 + 12 = 16 d" | a ) 8 , b ) 9 , c ) 10 , d ) 16 , e ) 17 | d | divide(add(add(add(add(3, const_4), add(3, const_4)), add(const_4, const_4)), 50), 5) | add(n1,const_4)|add(const_4,const_4)|add(#0,#0)|add(#2,#1)|add(n2,#3)|divide(#4,n0)| | general |
in a coconut grove , ( x + 2 ) trees yield 40 nuts per year , x trees yield 120 nuts per year and ( x β 2 ) trees yield 180 nuts per year . if the average yield per year per tree be 100 , find x . | "( x + 2 ) Γ 40 + x Γ 120 + ( x β 2 ) Γ 180 / ( x + 2 ) + x + ( x β 2 ) = 100 β 340 x β 280 / 3 x = 100 β 40 x = 280 β x = 7 answer e" | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | e | divide(subtract(multiply(180, 2), multiply(40, 2)), subtract(add(add(40, 120), 180), multiply(100, const_3))) | add(n1,n2)|multiply(n0,n4)|multiply(n0,n1)|multiply(n5,const_3)|add(n4,#0)|subtract(#1,#2)|subtract(#4,#3)|divide(#5,#6)| | general |
find value for x from given equation ? x + 1 = - 3 | "1 . subtract 1 from both sides : x + 1 - 1 = - 3 - 1 2 . simplify both sides : x = - 4 b" | a ) 1 , b ) x = - 4 , c ) - 3 , d ) - 5 , e ) 5 | b | subtract(3, 1) | subtract(n1,n0)| | general |
rectangular tile each of size 25 cm by 65 cm must be laid horizontally on a rectangular floor of size 150 cm by 390 cm , such that the tiles do not overlap and they are placed with edges jutting against each other on all edges . a tile can be placed in any orientation so long as its edges are parallel to the edges of f... | "area of tile = 25 * 65 = 1625 area of floor = 150 * 390 = 58500 no of tiles = 58500 / 1625 = 36 so , the no of tile = 36 answer : d" | a ) 50 , b ) 40 , c ) 25 , d ) 36 , e ) 45 | d | divide(multiply(150, 390), multiply(25, 65)) | multiply(n2,n3)|multiply(n0,n1)|divide(#0,#1)| | geometry |
of the 3,600 employees of company x , 1 / 6 are clerical . if the clerical staff were to be reduced by 1 / 4 , what percent of the total number of the remaining employees would then be clerical ? | "welcome , just post the question and the choices let ' s see , the way i did it was 1 / 6 are clerical out of 3600 so 600 are clerical 600 reduced by 1 / 4 is 1200 * 1 / 4 so it reduced 150 people , so there is 450 clerical people left but since 150 people left , it also reduced from the total of 3600 so there are 345... | a ) 13 % , b ) 22.2 % , c ) 20 % , d ) 12.5 % , e ) 11.1 % | a | multiply(divide(multiply(divide(1, 6), subtract(1, divide(1, 4))), add(multiply(divide(1, 6), subtract(1, divide(1, 4))), subtract(1, divide(1, 6)))), const_100) | divide(n1,n2)|divide(n3,n4)|subtract(n3,#1)|subtract(n3,#0)|multiply(#0,#2)|add(#4,#3)|divide(#4,#5)|multiply(#6,const_100)| | general |
the area of a triangle is with base 3 m and height 6 m ? | "1 / 2 * 3 * 6 = 9 m 2 answer : b" | a ) 88 m 2 , b ) 9 m 2 , c ) 66 m 2 , d ) 77 m 2 , e ) 31 m 2 | b | triangle_area(3, 6) | triangle_area(n0,n1)| | geometry |
9886 + x = 13200 , then x is ? | "answer x = 13200 - 9886 = 3314 option : e" | a ) 3327 , b ) 3237 , c ) 3337 , d ) 2337 , e ) none of these | e | subtract(13200, 9886) | subtract(n1,n0)| | general |
if x is a sum of all even integers on the interval 13 . . . 45 and y is their number , what is the gcd ( x , y ) ? | "x = 14 + 16 + . . . + 44 = ( largest + smallest ) / 2 * ( # of terms ) = ( 14 + 44 ) / 2 * 16 = 29 * 16 . gcd of 16 and 29 * 16 is 16 . answer : b ." | a ) 1 , b ) 16 , c ) 26 , d ) 52 , e ) 1014 | b | multiply(add(subtract(13, const_10), const_2), add(subtract(13, const_10), const_2)) | subtract(n0,const_10)|add(#0,const_2)|multiply(#1,#1)| | general |
in a stream running at 2 kmph , a motor boat goes 6 km upstream and back again to the starting point in 33 minutes . find the speed of the motorboat in still water . | let the speed of the motarboat in still water be x kmph . then , 6 / x + 2 + 6 / x - 2 = 33 / 60 11 x 2 - 240 x - 44 = 0 11 x 2 - 242 x + 2 x - 44 = 0 ( x - 22 ) ( 11 x + 2 ) = 0 x = 22 . answer is a | a ) 22 , b ) 24 , c ) 26 , d ) 28 , e ) none of them | a | divide(add(6, 6), divide(33, const_60)) | add(n1,n1)|divide(n2,const_60)|divide(#0,#1) | physics |
a cistern is filled by a tap in 5 1 / 2 hours . due to leak in the bottom of the cistern , it takes half an hour longer to fill the cistern . if the cistern is full how many hours will it take the leak to empty it ? | "filling rate - leak rate = net rate 1 / 5.5 - leak rate = 1 / 6 leak rate = 2 / 11 - 1 / 6 = 1 / 66 the answer is d ." | a ) 60 , b ) 62 , c ) 64 , d ) 66 , e ) 68 | d | divide(1, subtract(divide(1, add(5, divide(1, 2))), divide(1, const_4))) | divide(n1,n2)|divide(n1,const_4)|add(n0,#0)|divide(n1,#2)|subtract(#3,#1)|divide(n1,#4)| | physics |
ajay bought 15 kg of dal at the rate of rs 14.50 per kg and 10 kg at the rate of rs 13 per kg . he mixed the two and sold the mixture at the rate of rs 15 per kg . what was his total gain in this transaction ? | explanation : cost price of 25 kg = rs . ( 15 x 14.50 + 10 x 13 ) = rs . 347.50 . sell price of 25 kg = rs . ( 25 x 15 ) = rs . 375 . profit = rs . ( 375 β 347.50 ) = rs . 27.50 . answer : d | a ) rs 1.10 , b ) rs 11 , c ) rs 16.50 , d ) rs 27.50 , e ) none of these | d | subtract(multiply(15, add(10, 15)), add(multiply(15, 14.5), multiply(10, 13))) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|multiply(n0,#0)|subtract(#4,#3) | gain |
speed of a boat in standing water is 10 kmph and speed of the stream is 2 kmph . a man can rows to a place at a distance of 100 km and comes back to the starting point . the total time taken by him is ? | speed upstream = 8 kmph speed downstream = 12 kmph total time taken = 100 / 8 + 100 / 12 = 20 hours 50 min answer is c | a ) 24 hours , b ) 21 hr 30 min , c ) 20 hr 50 min , d ) 18 hr 15 min , e ) 22 hrs | c | add(multiply(add(add(10, 2), subtract(10, 2)), 100), multiply(subtract(add(divide(100, add(10, 2)), divide(100, subtract(10, 2))), add(add(10, 2), subtract(10, 2))), const_60)) | add(n0,n1)|subtract(n0,n1)|add(#0,#1)|divide(n2,#0)|divide(n2,#1)|add(#3,#4)|multiply(n2,#2)|subtract(#5,#2)|multiply(#7,const_60)|add(#6,#8) | physics |
shekar scored 76 , 65 , 82 , 67 and 95 marks in mathematics , science , social studies , english and biology respectively . what are his average marks ? | explanation : average = ( 76 + 65 + 82 + 67 + 95 ) / 5 = 385 / 5 = 77 hence average = 77 answer : b | a ) 65 , b ) 77 , c ) 75 , d ) 85 , e ) 90 | b | divide(add(add(add(add(76, 65), 82), 67), 95), add(const_1, const_4)) | add(n0,n1)|add(const_1,const_4)|add(n2,#0)|add(n3,#2)|add(n4,#3)|divide(#4,#1) | general |
if in a race of 110 m , a covers the distance in 20 seconds and b in 25 seconds , then a beats b by : | "explanation : the difference in the timing of a and b is 5 seconds . hence , a beats b by 5 seconds . the distance covered by b in 5 seconds = ( 110 * 5 ) / 25 = 22 m hence , a beats b by 22 m . answer d" | a ) 20 m , b ) 16 m , c ) 11 m , d ) 22 m , e ) 15 m | d | multiply(divide(subtract(25, 20), 25), 110) | subtract(n2,n1)|divide(#0,n2)|multiply(n0,#1)| | physics |
a small company reduced its faculty by approximately 20 percent to 195 employees . what was the original number of employees ? | "if x is the original number of employees , then after 20 % reduction in employees number is . 80 x but we are given . 80 x = 195 x = 244 so the original number of employees is 244 correct answer - e" | a ) a ) 182 , b ) b ) 208 , c ) c ) 220 , d ) d ) 224 , e ) e ) 244 | e | divide(195, divide(subtract(const_100, 20), const_100)) | subtract(const_100,n0)|divide(#0,const_100)|divide(n1,#1)| | gain |
a part of certain sum of money is invested at 9 % per annum and the rest at 15 % per annum , if the interest earned in each case for the same period is equal , then ratio of the sums invested is ? | "15 : 9 = 5 : 3 answer : c" | a ) 4 : 6 , b ) 4 : 9 , c ) 5 : 3 , d ) 4 : 1 , e ) 4 : 2 | c | multiply(divide(15, const_100), 9) | divide(n1,const_100)|multiply(n0,#0)| | gain |
if 3 men or 4 women can reap a field in 43 days how long will 7 men and 5 women take to reap it ? | "explanation : 3 men reap 1 / 43 field in 1 day 1 man reap 1 / ( 3 x 43 ) 4 women reap 1 / 43 field in 1 day 1 woman reap 1 / ( 43 x 4 ) 7 men and 5 women reap ( 7 / ( 3 x 43 ) + 5 / ( 4 x 43 ) ) = 1 / 12 in 1 day 7 men and 5 women will reap the field in 12 days answer : option b" | a ) 9 , b ) 12 , c ) 15 , d ) 18 , e ) 19 | b | divide(43, add(multiply(7, divide(43, 3)), multiply(5, divide(43, 4)))) | divide(n2,n0)|divide(n2,n1)|multiply(n3,#0)|multiply(n4,#1)|add(#2,#3)|divide(n2,#4)| | physics |
the smallest number when increased by ` ` 3 ` ` is exactly divisible by 510 , 4590 , and 105 is : | lcm = 630 630 - 3 = 627 answer : e | a ) 630 , b ) 633 , c ) 150 , d ) 140 , e ) 627 | e | subtract(add(multiply(divide(4590, 510), divide(105, 3)), multiply(divide(4590, 510), divide(105, 3))), const_3) | divide(n2,n1)|divide(n3,n0)|multiply(#0,#1)|add(#2,#2)|subtract(#3,const_3) | general |
if the perimeter of square region d and the perimeter of rectangular region r are equal and the sides of r are in the ratio 2 : 3 then the ratio of the area of r to the area of d | we know perimeter of a square ( pd ) = 4 * side perimeter of a rectangle ( pr ) = 2 ( length + breath ) let us assume 40 to be the perimeter of the square ( since we know each side of a square is equal and the perimeter is divisible by 4 , also take in to account the length and breadth of the rectangle is in the ration... | ['a ) 25 : 16', 'b ) 24 : 25', 'c ) 5 : 6', 'd ) 4 : 5', 'e ) 4 : 9'] | b | divide(multiply(3, 2), power(divide(multiply(add(3, 2), const_2), const_4), const_2)) | add(n0,n1)|multiply(n0,n1)|multiply(#0,const_2)|divide(#2,const_4)|power(#3,const_2)|divide(#1,#4) | geometry |
a waiter ' s salary consists of his salary and tips . during one week his tips were 5 / 3 of his salary . what fraction of his income came from tips ? | "income = salary ( s ) + tips = s + s * 5 / 3 = s * 8 / 3 tips = s * 8 / 3 fraction of his income came from tips = ( s * 5 / 3 ) / ( s * 8 / 3 ) = 5 / 8 answer : c" | a ) 4 / 9 , b ) 5 / 4 , c ) 5 / 8 , d ) 5 / 9 , e ) 6 / 9 | c | divide(divide(5, 3), add(divide(5, 3), const_1)) | divide(n0,n1)|add(#0,const_1)|divide(#0,#1)| | general |
if ( c - a ) / ( c - b ) = 3 , then ( 5 b - 5 a ) / ( c - a ) = | "let ' s say c = 7 , b = 5 , a = 1 so that our 1 st expression holds true . now , ibsert those numbers in the second expression and we ' ll get 3.33 answer d ( hopefully ) ) )" | a ) 0.5 , b ) 1 , c ) 1.5 , d ) 3.33 , e ) 2.5 | d | multiply(subtract(5, const_2.0), 5) | subtract(n1,const_2.0)|multiply(n2,#0)| | general |
rachel and sarah spent $ 8 for gasoline , $ 15.65 for their lunch , and $ 5 apiece for gifts for grandma . grandma gave each of them $ 10 . if the girls left home with a total of $ 50 , how much do they have for the return trip ? | $ 50 + $ 20 = $ 70 what they left with plus $ 10 to each $ 8 + $ 15.65 + $ 10 = $ 33.65 what they spent ( gifts are $ 5 + $ 5 ) $ 70 - $ 33.65 = $ 36.35 left correct answer a | a ) $ 36.35 , b ) $ 26.35 , c ) $ 16.35 , d ) $ 46.35 , e ) $ 56.35 | a | add(10, subtract(50, subtract(add(5, add(8, 15.65)), 5))) | add(n0,n1)|add(n2,#0)|subtract(#1,n2)|subtract(n4,#2)|add(n3,#3) | general |
in the fifth grade at parkway elementary school there are 420 students . 320 students are boys and 250 students are playing soccer . 86 % of the students that play soccer are boys . how many girl student are in parkway that is not playing soccer ? | "total students = 420 boys = 320 , girls = 100 total playing soccer = 250 86 % of 250 = 215 are boys who play soccer . girls who play soccer = 35 . total girls who do not play soccer = 100 - 35 = 65 . correct option : d" | a ) 69 . , b ) 73 . , c ) 81 , d ) 65 , e ) 108 | d | subtract(subtract(420, 320), subtract(250, divide(multiply(250, 86), const_100))) | multiply(n2,n3)|subtract(n0,n1)|divide(#0,const_100)|subtract(n2,#2)|subtract(#1,#3)| | gain |
p and q invested in a shop . the profits were divided in the ratio of 2 : 4 respectively . if p invested rs . 500000 , the amount invested by q is : | "suppose q invested rs . y . then , 500000 / y = 2 / 4 or y = [ 500000 x 4 / 2 ] = 1000000 . answer c" | a ) 5000000 , b ) 4000000 , c ) 1000000 , d ) 6000000 , e ) none | c | multiply(divide(500000, 2), 4) | divide(n2,n0)|multiply(n1,#0)| | gain |
1939392 * 625 | "explanation : trick : when multiplying with 5 n 5 n then put n zeros to the right of multiplicand and divide the number with 2 n 2 n so using this we can solve this question in much less time . 1939392 Γ 5 ( 4 ) = 1939392000016 = 1212120000 option b" | a ) 1212120010 , b ) 1212120000 , c ) 1212120011 , d ) 1212121010 , e ) 1212121000 | b | multiply(divide(1939392, 625), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
if ( m - 8 ) is a factor of m ^ 2 - qm - 24 , then q = | "( m - 8 ) ( m - a ) = m ^ 2 - qm - 24 a = - 3 q = 8 + a = 5 = b" | a ) 3 , b ) 5 , c ) 6 , d ) 11 , e ) 16 | b | subtract(8, divide(24, 8)) | divide(n2,n0)|subtract(n0,#0)| | general |
peter invests a sum of money and gets back an amount of $ 815 in 3 years . david invests an equal amount of money and gets an amount of $ 870 in 4 years . if both amounts were invested at the same rate ( simple interest ) what was the sum of money invested ? | "since both peter and david invested the same amount of money at the same rate , they would earn same interest per year . david invested for one year more than peter and hence he got interest amount for one more year . interest earned per year = amount received by david - amount received by peter = 870 - 815 = 55 inter... | a ) 670 , b ) 650 , c ) 698 , d ) 744 , e ) 700 | b | subtract(815, multiply(divide(subtract(870, 815), subtract(divide(4, const_100), divide(3, const_100))), divide(3, const_100))) | divide(n3,const_100)|divide(n1,const_100)|subtract(n2,n0)|subtract(#0,#1)|divide(#2,#3)|multiply(#4,#1)|subtract(n0,#5)| | gain |
the ratio of the adjacent angles of a parallelogram is 5 : 13 . also , the ratio of the angles of quadrilateral is 5 : 6 : 7 : 12 . what is the sum of the smaller angle of the parallelogram and the second largest angle of the quadrilateral ? | "the measures of the adjacent angles of a parallelogram add up to be 180 Β° given so , 5 x + 13 x = 180 Β° or , 18 x = 180 Β° or , x = 10 Β° hence the angles of the parallelogram are 50 Β° and 130 Β° further it is given we know sum of all the four angles of a quadrilateral is 360 Β° so , 5 y + 6 y + 7 y + 12 y = 360 Β° or , 5 ... | a ) 168 Β° , b ) 134 Β° , c ) 156 Β° , d ) 224 Β° , e ) none of these | b | multiply(divide(multiply(add(multiply(multiply(5, const_2), const_10), const_100), const_2), add(add(add(5, 6), 7), 12)), 13) | add(n2,n3)|multiply(const_2,n0)|add(n4,#0)|multiply(#1,const_10)|add(#3,const_100)|add(n5,#2)|multiply(#4,const_2)|divide(#6,#5)|multiply(n1,#7)| | other |
in a bag containing 3 balls , a white ball was placed and then 1 ball was taken out at random . what is the probability that the extracted ball would turn on to be white , if all possible hypothesis concerning the color of theballs that initiallyin the bag were equally possible ? | "since , all possible hypothesis regarding the colour of the balls are equally likely , therefore these could be 3 white balls , initially in the bag . β΄ required probability = 1 / 4 [ 1 + 3 / 4 + 1 / 2 + 1 / 4 ] = 1 / 4 [ ( 4 + 3 + 2 + 1 ) / 4 ] = 5 / 8 b" | a ) 2 / 3 , b ) 5 / 8 , c ) 5 / 9 , d ) 3 / 7 , e ) 4 / 7 | b | divide(add(1, add(divide(1, const_2), add(divide(1, add(1, 3)), divide(3, add(1, 3))))), add(1, 3)) | add(n0,n1)|divide(n1,const_2)|divide(n1,#0)|divide(n0,#0)|add(#2,#3)|add(#4,#1)|add(n1,#5)|divide(#6,#0)| | probability |
half a number plus 5 is 13 . what is the number ? | "let x be the number . always replace ` ` is ' ' with an equal sign ( 1 / 2 ) x + 5 = 13 ( 1 / 2 ) x = 13 - 5 ( 1 / 2 ) x = 8 x = 16 correct answer is b" | a ) 8 , b ) 16 , c ) 10 , d ) 11 , e ) 12 | b | multiply(subtract(13, 5), const_2) | subtract(n1,n0)|multiply(#0,const_2)| | general |
a hare and a jackal are running a race . 3 leaps of the hare are equal to 4 leaps of the jackal . for every 6 leaps of the hare , the jackal takes 7 leaps . find the ratio of the speed of the hare to the speed of the jackal . | the hare takes 6 leaps and the jackal takes 7 leaps . 1 hare leap = 4 / 3 jackal leaps thus the hare ' s 6 leaps = 6 * ( 4 / 3 ) = 8 jackal leaps . the ratio of their speeds is 8 : 7 . the answer is b . | a ) 64 : 25 , b ) 8 : 7 , c ) 5 : 8 , d ) 25 : 64 , e ) 6 : 7 | b | divide(add(const_4, const_4), 7) | add(const_4,const_4)|divide(#0,n3) | other |
on dividing 109 by a number , the quotient is 9 and the remainder is 1 . find the divisor . | "d = ( d - r ) / q = ( 109 - 1 ) / 9 = 108 / 9 = 12 c" | a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 15 | c | floor(divide(109, 9)) | divide(n0,n1)|floor(#0)| | general |
a , b and c enter into a partnership by investing $ 15000 , $ 21000 and $ 27000 respectively . at the end of 8 months , b receives $ 1540 as his share . find the share of a . | "the ratio of capital of a , b and c = 15000 : 21000 : 27000 = 5 : 7 : 9 a ' s share = ( 5 / 7 ) * 1540 = $ 1100 the answer is d ." | a ) $ 800 , b ) $ 900 , c ) $ 1000 , d ) $ 1100 , e ) $ 1200 | d | divide(multiply(multiply(15000, divide(1540, divide(multiply(21000, 8), const_100))), 8), const_100) | multiply(n1,n3)|divide(#0,const_100)|divide(n4,#1)|multiply(n0,#2)|multiply(n3,#3)|divide(#4,const_100)| | gain |
a is twice as good as workman as b and together they finish a piece of work in 18 days . in how many days will a alone finish the work ? | "a : b = 2 : 1 ab ' s together wrk in 18 days a ; s alone is 1 / 18 * 2 / 3 = 27 days answer : a" | a ) 27 , b ) 26 , c ) 29 , d ) 25 , e ) 24 | a | multiply(18, const_3) | multiply(n0,const_3)| | physics |
when a number is divided by 6 & then multiply by 12 the answer is 15 what is the no . ? | "if $ x $ is the number , x / 6 * 12 = 15 = > 2 x = 15 = > x = 7.5 d" | a ) 4.5 , b ) 5 , c ) 5.6 , d ) 7.5 , e ) 6.5 | d | multiply(divide(15, 12), 6) | divide(n2,n1)|multiply(n0,#0)| | general |
by selling 15 pens for a rupee a woman loses 25 % . how many for a rupee should he sell in order to gain 25 % ? | "d 85 % - - - 15 125 % - - - ? 85 / 125 * 15 = 10" | a ) 12 , b ) 14 , c ) 45 , d ) 10 , e ) 65 | d | inverse(add(divide(divide(const_100, subtract(const_100, 25)), 15), divide(multiply(divide(divide(const_100, subtract(const_100, 25)), 15), 25), const_100))) | subtract(const_100,n1)|divide(const_100,#0)|divide(#1,n0)|multiply(n2,#2)|divide(#3,const_100)|add(#2,#4)|inverse(#5)| | gain |
if the perimeter of square region z and the perimeter of rectangular region r are equal and the sides of r are in the ratio 2 : 3 then the ratio of the area of r to the area of z | we know perimeter of a square ( pz ) = 4 * side perimeter of a rectangle ( pr ) = 2 ( length + breath ) let us assume 40 to be the perimeter of the square ( since we know each side of a square is equal and the perimeter is divisible by 4 , also take in to account the length and breadth of the rectangle is in the ration... | ['a ) 25 : 16', 'b ) 24 : 25', 'c ) 5 : 6', 'd ) 4 : 5', 'e ) 4 : 9'] | b | divide(multiply(2, 3), power(divide(add(2, 3), const_2), const_2)) | add(n0,n1)|multiply(n0,n1)|divide(#0,const_2)|power(#2,const_2)|divide(#1,#3) | geometry |
the sum of the present ages of two persons a and b is 60 . if the age of a is twice that of b , find the sum of their ages 9 years hence ? | "a + b = 60 , a = 2 b 2 b + b = 60 = > b = 20 then a = 40 . 9 years , their ages will be 49 and 29 . sum of their ages = 49 + 29 = 78 . answer : c" | a ) 22 , b ) 77 , c ) 78 , d ) 98 , e ) 71 | c | add(add(multiply(divide(60, 9), const_2), 9), add(divide(60, 9), 9)) | divide(n0,n1)|add(#0,n1)|multiply(#0,const_2)|add(#2,n1)|add(#3,#1)| | general |
a , band c can do a piece of work in 35 days , 20 days and 55 days respectively , working alone . how soon can the work be done if a is assisted by band c on alternate days ? | "( a + b ) ' s 1 day ' s work = 1 / 35 + 1 / 20 = 11 / 140 ( a + c ) ' s 1 day ' s work = 1 / 35 + 1 / 55 = 18 / 385 work done in 2 day ' s = 11 / 140 + 18 / 385 = 48 / 383 48 / 383 th work done in 2 days work done = 383 / 48 * 2 = 16 days ( approx ) answer : e" | a ) 7 days , b ) 8 days , c ) 9 days , d ) 10 days , e ) 16 days | e | divide(55, divide(add(add(divide(55, 35), divide(55, 20)), add(divide(55, 35), divide(55, 55))), const_2)) | divide(n2,n0)|divide(n2,n1)|divide(n2,n2)|add(#0,#1)|add(#0,#2)|add(#3,#4)|divide(#5,const_2)|divide(n2,#6)| | physics |
aram β s car was fined when he gave joe and peter a ride , so they decided to help aram pay the fine . joe paid $ 3 more than 1 / 4 of the fine and peter paid $ 3 less than 1 / 3 of the fine , leaving pay $ 4 less than 1 / 2 the fine to complete the payment . what fraction of the fine did aram pay ? | as given aram ' payment is [ 1 ] [ / 2 ] * fine - 4 i . e total fine = 2 * ( sam ' s payment + 4 ) now , as everything looks integer , fine should be able to divide by 4 and 3 . because joe ' s contribution has 1 / 4 of the fine and peter ' s contribution has 1 / 3 of the fine so , that leads me to following : options ... | a ) $ 13 , b ) $ 15 , c ) $ 20 , d ) $ 28 , e ) $ 48 | c | subtract(divide(add(add(3, 4), 3), subtract(add(add(divide(1, 4), divide(1, 3)), divide(1, 2)), 1)), const_100) | add(n0,n2)|divide(n1,n2)|divide(n1,n0)|divide(n1,n8)|add(n0,#0)|add(#1,#2)|add(#5,#3)|subtract(#6,n1)|divide(#4,#7)|subtract(#8,const_100) | general |
what is 992 * 992 ? | "if you take a base of 1000 then 992 is 8 less than 1000 to get the product of 992 x 992 write like this 992 - 8 ( as 8 less than base 1000 ) 992 - 8 now 8 x 8 = 64 and 992 - 8 = 984 so 992 x 992 = 984064 . . . ( bingo the answer is c . you can even have a shortcut . . . . . . 8 x 8 = 64 . . . only answer choice has la... | a ) 974,169 , b ) 974,219 , c ) 984,064 , d ) 985,219 , e ) 985,369 | c | circle_area(divide(992, multiply(const_2, const_pi))) | multiply(const_2,const_pi)|divide(n0,#0)|circle_area(#1)| | general |
the mean of 50 observations was 36 . it was found later that an observation 43 was wrongly taken as 23 . the corrected new mean is | "solution correct sum = ( 36 x 50 + 43 - 23 ) = 1820 . Γ’ Λ Β΄ correct mean = 1820 / 50 = 36.4 . answer d" | a ) 35.2 , b ) 36.1 , c ) 36.2 , d ) 36.4 , e ) none | d | divide(add(multiply(36, 50), subtract(subtract(50, const_2), 23)), 50) | multiply(n0,n1)|subtract(n0,const_2)|subtract(#1,n3)|add(#0,#2)|divide(#3,n0)| | general |
due to construction , the speed limit along an 5 - mile section of highway is reduced from 52 miles per hour to 36 miles per hour . approximately how many minutes more will it take to travel along this section of highway at the new speed limit than it would have taken at the old speed limit ? | "old time in minutes to cross 5 miles stretch = 5 * 60 / 52 = 5 * 15 / 13 = 5.76 new time in minutes to cross 5 miles stretch = 5 * 60 / 36 = 5 * 5 / 3 = 8.33 time difference = 2.57 ans : d" | a ) a ) 3.12 , b ) b ) 8 , c ) c ) 10 , d ) d ) 2.57 , e ) e ) 24 | d | max(multiply(subtract(add(52, 5), const_1), subtract(divide(5, 36), divide(5, 52))), const_4) | add(n0,n1)|divide(n0,n2)|divide(n0,n1)|subtract(#0,const_1)|subtract(#1,#2)|multiply(#3,#4)|max(#5,const_4)| | physics |
it is currently 7 : 16 pm . at what time in the morning was it exactly 19,443 minutes earlier ? | converting 19,443 minutes to hours , we get 19,443 / 60 = 324 r 3 ; that is 324 hours and 3 minutes . all of the answers are during the same hour of the morning , thus the hours can be assumed to bring us into the 7 am hour evenly from 7 : 16 pm . thus 324 hours ago was 7 : 16 am . take an extra 3 minutes off , and it ... | a ) 7 : 11 , b ) 7 : 13 , c ) 7 : 17 , d ) 7 : 19 , e ) 7 : 21 | b | divide(7, subtract(16, const_3)) | subtract(n1,const_3)|divide(n0,#0) | physics |
52 is to be divided into two parts such that the sum of 10 times the first and 22 times the second is 780 . the bigger part is : | "explanation : let the two parts be ( 52 - x ) and x . then , 10 ( 52 - x ) + 22 x = 780 = > 12 x = 260 = > x = 21.66 . bigger part = ( 52 - x ) = 30.33 . answer : a ) 30.33" | a ) 30.33 , b ) 34.44 , c ) 26.1 , d ) 28.0 , e ) 21.0 | a | subtract(52, divide(subtract(780, multiply(52, 10)), subtract(22, 10))) | multiply(n0,n1)|subtract(n2,n1)|subtract(n3,#0)|divide(#2,#1)|subtract(n0,#3)| | general |
the price of 2 sarees and 4 shirts is $ 1600 . with the same money one can buy 1 saree and 6 shirts . if one wants to buy 12 shirts , how much shall he have to pay ? | c $ 2400 let the price of a saree and a shirt be $ x and $ y respectively . then , 2 x + 4 y = 1600 . . . . ( i ) and x + 6 y = 1600 . . . . ( ii ) divide equation ( i ) by 2 , we get the below equation . = > x + 2 y = 800 . - - - ( iii ) now subtract ( iii ) from ( ii ) x + 6 y = 1600 ( - ) x + 2 y = 800 - - - - - - -... | a ) 3800 , b ) 3300 , c ) 2400 , d ) 4200 , e ) 1400 | c | multiply(divide(subtract(multiply(2, 1600), 1600), subtract(multiply(2, 6), 4)), 12) | multiply(n0,n2)|multiply(n0,n4)|subtract(#0,n2)|subtract(#1,n1)|divide(#2,#3)|multiply(n5,#4) | general |
on dividing 271 by a number , the quotient is 9 and the remainder is 1 . find the divisor ? | "d = ( d - r ) / q = ( 271 - 1 ) / 9 = 270 / 9 = 30 e )" | a ) a ) 12 , b ) b ) 15 , c ) c ) 16 , d ) d ) 17 , e ) e ) 30 | e | floor(divide(271, 9)) | divide(n0,n1)|floor(#0)| | general |
working alone at their respective constant rates , a can complete a task in β a β days and b in β b β days . they take turns in doing the task with each working 2 days at a time . if a starts they finish the task in exactly 12 days . if b starts , they take half a day more . how long does it take to complete the task i... | "work done by ab in a day = xy respectively . when a starts : no . of days when a works = 6 no . of days when b works = 6 β 6 x + 6 y = 1 when b starts : no . of days when a works = 6.5 no . of days when b works = 5.5 β 6.5 x + 5.5 y = 1 solving the above two equations for xy x = 1 / 12 y = 1 / 12 β total work done by ... | a ) 4 , b ) 4.5 , c ) 6 , d ) 8 , e ) 6.5 | c | inverse(add(inverse(subtract(12, const_1)), inverse(add(12, 2)))) | add(n0,n1)|subtract(n1,const_1)|inverse(#1)|inverse(#0)|add(#2,#3)|inverse(#4)| | physics |
on dividing a number by 5 , we get 3 as remainder . what will the remainder when the square of the this number is divided by 5 ? | "explanation : let the number be x and on dividing x by 5 , we get k as quotient and 3 as remainder . x = 5 k + 3 x ^ 2 = ( 5 k + 3 ) ^ 2 = ( 25 k ^ 2 + 30 k + 9 ) = 5 ( 5 k ^ 2 + 6 k + 1 ) + 4 on dividing x 2 by 5 , we get 4 as remainder . e )" | a ) 0 , b ) 5 , c ) 8 , d ) 9 , e ) 4 | e | multiply(subtract(divide(power(3, const_2), 5), floor(divide(power(3, const_2), 5))), 5) | power(n1,const_2)|divide(#0,n0)|floor(#1)|subtract(#1,#2)|multiply(n0,#3)| | geometry |
find the area of a rhombus one side of which measures 20 cm and one diagonal is 24 cm . | "explanation : let other diagonal = 2 x cm . since diagonals of a rhombus bisect each other at right angles , we have : ( 20 ) 2 = ( 12 ) 2 + ( x ) 2 = > x = β ( 20 ) 2 β ( 12 ) 2 = β 256 = 16 cm . _ i so , other diagonal = 32 cm . area of rhombus = ( 1 / 2 ) x ( product of diagonals ) = ( 1 / 2 Γ 24 x 32 ) cm 2 = 384 ... | a ) 370 cm 2 , b ) 365 cm 2 , c ) 380 cm 2 , d ) 384 cm 2 , e ) 394 cm 2 | d | add(multiply(multiply(divide(const_1, const_2), 24), sqrt(subtract(multiply(multiply(20, 20), const_4), multiply(24, 24)))), 24) | divide(const_1,const_2)|multiply(n0,n0)|multiply(n1,n1)|multiply(n1,#0)|multiply(#1,const_4)|subtract(#4,#2)|sqrt(#5)|multiply(#3,#6)|add(n1,#7)| | geometry |
a taxi leaves point a 4 hours after a bus left the same spot . the bus is traveling 30 mph slower than the taxi . find the speed of the taxi , if it overtakes the bus in two hours . | "let the speed of bus be v - 30 , speed of taxi be v the bus travelled a total of 6 hrs and taxi a total of 2 hrs . hence 6 * ( v - 30 ) = 2 v 6 v - 180 = 2 v 4 v = 180 v = 45 mph c" | a ) 40 , b ) 42 , c ) 45 , d ) 48 , e ) 49 | c | divide(add(multiply(4, 30), multiply(4, 30)), subtract(add(const_3.0, 4), 4)) | add(n0,n0)|multiply(const_3.0,n1)|add(#1,#1)|subtract(#0,n0)|divide(#2,#3)| | physics |
m and n are the x and y coordinates , respectively , of a point in the coordinate plane . if the points ( m , n ) and ( m + p , n + 15 ) both lie on the line defined by the equation x = ( y / 5 ) - ( 2 / 5 ) , what is the value of p ? | "x = ( y / 5 ) - ( 2 / 5 ) , and so y = 5 x + 2 . the slope is 5 . ( n + 15 - n ) / ( m + p - m ) = 5 p = 3 the answer is c ." | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | divide(15, 5) | divide(n0,n1)| | general |
in a certain pet shop , the ratio of dogs to cats to bunnies in stock is 7 : 7 : 8 . if the shop carries 330 dogs and bunnies total in stock , how many dogs are there ? | "let us assume the number of dogs , cats and bunnies to be 7 x , 7 x and 8 x total dogs and bunnies = 15 x . and we are given that 15 x = 330 . hence x = 22 . dogs = 7 x = 7 * 22 = 154 ( option e )" | a ) 42 , b ) 66 , c ) 98 , d ) 112 , e ) 154 | e | multiply(divide(330, add(7, 8)), 7) | add(n0,n2)|divide(n3,#0)|multiply(n0,#1)| | other |
of the 150 employees at company x , 80 are full - time , and 100 have worked at company x for at least a year . there are 10 employees at company x who aren β t full - time and haven β t worked at company x for at least a year . how many full - time employees of company x have worked at the company for at least a year ... | "full time employee who have not worked for at least one year = a full time employee who have worked for at least one year = b non full time employee who have worked for at least one year = c non full time employee who have not worked for at least one year = d a + b + c + d = 150 a + b = 80 i . e . c + d = 70 b + c = 1... | a ) 10 , b ) 30 , c ) 50 , d ) 80 , e ) 100 | a | subtract(subtract(150, 80), 10) | subtract(n0,n1)|subtract(#0,n3)| | general |
if y is the smallest positive integer such that 7,700 multiplied by y is the square of an integer , then y must be | "7700 = 5 * 5 * 2 * 2 * 7 * 11 , so we need one 7 and one 11 to make it a square of a number . so 7 * 11 = 77 ans : e" | a ) 29 , b ) 58 , c ) 63 , d ) 74 , e ) 77 | e | multiply(add(const_2, const_3), const_2) | add(const_2,const_3)|multiply(#0,const_2)| | geometry |
a bookstore has a shelf that contains biographies which normally sell for $ 20 each and mysteries that normally sell for $ 12 each . during a sale , the biographies and mysteries are discounted at different rates so that a customer saves a total of $ 19 from the normal price by buying 5 discounted biographies and 3 dis... | "let b be the discount on biographies and m be the discount on mysteries so . , b + m = 0.38 - - - - - ( 1 ) and ( 20 * 5 + 12 * 3 ) - ( 20 * 5 * ( 1 - b ) + 12 * 3 * ( 1 - m ) ) = 19 - - > 100 ( 1 - ( 1 - b ) ) + 36 ( 1 - ( 1 - m ) = 19 100 b + 36 m = 19 - - - - - - ( 2 ) solving 12 . , we get m = 0.2968 = 29.68 % b" | a ) 18 % , b ) 29.68 % , c ) 19.68 % , d ) 29 % , e ) 68 % | b | multiply(subtract(const_1, divide(multiply(20, const_2), add(add(20, 38), const_1))), const_100) | add(n0,n5)|multiply(n0,const_2)|add(#0,const_1)|divide(#1,#2)|subtract(const_1,#3)|multiply(#4,const_100)| | gain |
find the ones digit of 73 ^ 320 | "cyclicity of 3 is 3,9 , 7,1 after 4 multiplication again the cycle repeats . so divide 320 by 4 and we get 87 as quotient and 2 as remainder . so cycle will run for 87 times and then 2 times more . so pick up the 2 nd item from the cycle . hence answer d ." | a ) 3 , b ) 5 , c ) 6 , d ) 7 , e ) 9 | d | subtract(power(subtract(73, multiply(add(const_3, const_4), const_10)), subtract(320, multiply(floor(divide(320, const_4)), const_4))), multiply(const_2, const_10)) | add(const_3,const_4)|divide(n1,const_4)|multiply(const_10,const_2)|floor(#1)|multiply(#0,const_10)|multiply(#3,const_4)|subtract(n0,#4)|subtract(n1,#5)|power(#6,#7)|subtract(#8,#2)| | general |
the ratio of ducks and frogs in a pond is 37 : 39 respectively . the average number of ducks and frogs in the pond is 152 . what is the number of frogs in the pond ? | "solution : ratio of ducks and frogs in pond , = 37 : 39 . average of ducks and frogs in pond , = 152 . so , total number of ducks and frogs in the pond , = 2 * 152 = 304 . therefore , number of frogs , = ( 304 * 39 ) / 76 = 156 . answer : option c" | a ) 148 , b ) 152 , c ) 156 , d ) 144 , e ) none | c | multiply(39, divide(152, divide(add(add(37, 39), const_2), const_2))) | add(n0,n1)|add(#0,const_2)|divide(#1,const_2)|divide(n2,#2)|multiply(n1,#3)| | general |
the side of a square is increased by 30 % then how much % does its area increases ? | "a = 100 a 2 = 10000 a = 130 a 2 = 16900 - - - - - - - - - - - - - - - - 10000 - - - - - - - - - 6900 100 - - - - - - - ? = > 69 % answer : b" | a ) 52.65 , b ) 69 , c ) 50.75 , d ) 42.75 , e ) 52.75 | b | divide(multiply(subtract(square_area(add(const_100, 30)), square_area(const_100)), const_100), square_area(const_100)) | add(n0,const_100)|square_area(const_100)|square_area(#0)|subtract(#2,#1)|multiply(#3,const_100)|divide(#4,#1)| | geometry |
the β connection β between any two positive integers a and b is the ratio of the smallest common multiple of a and b to the product of a and b . for instance , the smallest common multiple of 8 and 12 is 24 , and the product of 8 and 12 is 96 , so the connection between 8 and 12 is 24 / 96 = 1 / 4 the positive integer ... | since β connection β between y and 6 is 1 / 1 then lcm ( 6 , y ) = 6 y , which means that 6 and y are co - prime ( they do not share any common factor but 1 ) , because if the had any common factor but 1 then lcm ( 6 , y ) would be less than 6 y . so , we should check how many integers less than 20 are co - prime with ... | a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | a | subtract(subtract(20, const_1), 12) | subtract(n12,const_1)|subtract(#0,n1) | general |
the probability of two events a and b are 0.15 and 0.40 respectively . the probability that both a and b occur is 0.15 . the probability that neither a nor b occur is _________ | "we are apply that formula . . . . . . . . . . . . . . p ( aorb ) = p ( a ) + p ( b ) - p ( a and b ) = . 15 + . 40 - . 15 = . 40 but the probability of neither a nor b = 1 - . 40 = 0.60 answer : e" | a ) 0.45 , b ) 0.4 , c ) 0.5 , d ) 0.05 , e ) 0.6 | e | subtract(const_1, subtract(add(0.15, 0.40), 0.15)) | add(n0,n1)|subtract(#0,n2)|subtract(const_1,#1)| | other |
( 0.8 ) ( power 3 ) - ( 0.5 ) ( power 3 ) / ( 0.8 ) ( power 2 ) + 0.40 + ( 0.5 ) ( power 2 ) is : | "given expression = ( 0.8 ) ( power 3 ) - ( 0.5 ) ( power 3 ) / ( 0.8 ) ( power 2 ) + ( 0.8 x 0.5 ) + ( 0.5 ) ( power 2 ) = a ( power 3 ) - b ( power 3 ) / a ( power 2 ) + ab + b ( power 2 ) = ( a - b ) = ( 0.8 - 0.5 ) = 0.30 answer is d ." | a ) 0.6 , b ) 0.5 , c ) 0.35 , d ) 0.3 , e ) none of them | d | divide(subtract(power(0.8, 3), power(0.5, 3)), add(add(power(0.8, 2), 0.40), power(0.5, 2))) | power(n0,n1)|power(n2,n1)|power(n0,n5)|power(n2,n5)|add(n6,#2)|subtract(#0,#1)|add(#4,#3)|divide(#5,#6)| | general |
if the length of the sides of two cubes are in the ratio 7 : 1 , what is the ratio of their total surface area ? | "let x be the length of the small cube ' s side . the total surface area of the small cube is 6 x ^ 2 . the total surface area of the large cube is 6 ( 7 x ) ^ 2 = 294 x ^ 2 . the ratio of surface areas is 49 : 1 . the answer is e ." | a ) 7 : 1 , b ) 14 : 1 , c ) 21 : 1 , d ) 42 : 1 , e ) 49 : 1 | e | multiply(7, 7) | multiply(n0,n0)| | geometry |
when a merchant imported a certain item , she paid a 7 percent import tax on the portion of the total value of the item in excess of $ 1,000 . if the amount of the import tax that the merchant paid was $ 110.60 , what was the total value of the item ? | "let x be the value of the item . 0.07 * ( x - 1000 ) = 110.60 x = 2580 the answer is c ." | a ) $ 2160 , b ) $ 2370 , c ) $ 2580 , d ) $ 2790 , e ) $ 2950 | c | add(1,000, divide(110.60, divide(7, const_100))) | divide(n0,const_100)|divide(n2,#0)|add(#1,n1)| | general |
on a trip covering 400 km , a bicyclist travels the first 100 km at 20 km per hour and the remainder of the distance at 15 km per hour . what is the average speed , in km per hour , for the entire trip ? | time 1 = 100 / 20 = 5 hours time 2 = 300 / 15 = 20 hours total time = 25 hours average speed = 400 / 25 = 16 km / hr the answer is b . | a ) 15.5 , b ) 16.0 , c ) 16.5 , d ) 17.0 , e ) 17.5 | b | speed(400, add(divide(100, 20), divide(subtract(400, 100), 15))) | divide(n1,n2)|subtract(n0,n1)|divide(#1,n3)|add(#0,#2)|speed(n0,#3) | physics |
eighteen years ago , a father was 3 times as old as his son . now the father is only twice as old his son . then the sum of the present ages of the son and the father is : | explanation : let the present ages of the father and son be 2 x and x years respectively . then , ( 2 x - 18 ) = 3 ( x - 18 ) = > x = 36 required sum = ( 2 x + x ) = 108 years . answer : option d | a ) 53 , b ) 108 , c ) 77 , d ) 107 , e ) 109 | d | add(subtract(multiply(multiply(add(const_3, const_3), 3), 3), multiply(add(const_3, const_3), 3)), multiply(subtract(multiply(multiply(add(const_3, const_3), 3), 3), multiply(add(const_3, const_3), 3)), const_2)) | add(const_3,const_3)|multiply(n0,#0)|multiply(n0,#1)|subtract(#2,#1)|multiply(#3,const_2)|add(#4,#3) | general |
how many pieces can a square pizza be cut into by making 5 linear cuts ? | "1 cut can make 2 pieces a second cut can make 4 pieces by cutting through 2 pieces a third cut can make 7 pieces by cutting through 3 of the pieces a fourth cut can make 11 pieces by cutting through 4 of the pieces a fifth cut can make 16 pieces by cutting through 5 of the pieces a" | a ) 16 , b ) 17 , c ) 13 , d ) 14 , e ) 15 | a | subtract(multiply(5, const_3), const_1) | multiply(n0,const_3)|subtract(#0,const_1)| | geometry |
the area of a parallelogram is 128 sq m . then the area of a triangle formed by its diagonal is - - - - - - - - - - ? | b * h / 2 = 128 / 2 = 64 m . ans : ( b ) | ['a ) 128 m', 'b ) 64 m', 'c ) 64 m ^ 2', 'd ) 128 m ^ 2', 'e ) none'] | b | divide(128, const_2) | divide(n0,const_2) | geometry |
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