Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 164 | correct stringclasses 5 values | annotated_formula stringlengths 7 1.65k | linear_formula stringlengths 8 925 | category stringclasses 6 values | answer stringclasses 5 values |
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the sum of number of boys and girls in a school is 900 . if the number of boys is x , then the number of girls becomes x % of the total number of students . the number of boys is ? | we have x + x % of 900 = 900 x + x / 100 * 900 = 900 10 * x = 900 x = 90 answer is d | a ) 50 , b ) 40 , c ) 60 , d ) 90 , e ) 70 | d | divide(900, add(divide(900, const_100), const_1)) | divide(n0,const_100)|add(#0,const_1)|divide(n0,#1) | general | D |
if 20 men can build a wall 66 metres long in 15 days , what length of a similar can be built by 86 men in 8 days ? | "if 20 men can build a wall 66 metres long in 15 days , length of a similar wall that can be built by 86 men in 8 days = ( 66 * 86 * 8 ) / ( 15 * 20 ) = 151.36 mtrs answer : a" | a ) 151.36 mtrs , b ) 378.4 mtrs , c ) 478.4 mtrs , d ) 488.4 mtrs , e ) 578.4 mtrs | a | multiply(66, divide(multiply(86, 8), multiply(20, 15))) | multiply(n3,n4)|multiply(n0,n2)|divide(#0,#1)|multiply(n1,#2)| | physics | A |
if k is the greatest positive integer such that 3 ^ k is a divisor of 18 ! then k = | "18 / 3 = 6 18 / 9 = 2 6 + 2 = 8 k = 8 answer : e" | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 8 | e | add(const_1, divide(18, 3)) | divide(n1,n0)|add(#0,const_1)| | general | E |
company z has 50 employees . if the number of employees having birthdays on wednesday is more than the number of employees having birthdays on any other day of the week , each of which have same number of birth - days , what is the minimum number of employees having birthdays on wednesday . | "say the number of people having birthdays on wednesday is x and the number of people having birthdays on each of the other 6 days is y . then x + 6 y = 50 . now , plug options for x . only a and e give an integer value for y . but only for e x > y as needed . answer : e ." | a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 12 | e | add(const_4, add(floor(divide(50, add(const_4, const_3))), const_1)) | add(const_3,const_4)|divide(n0,#0)|floor(#1)|add(#2,const_1)|add(#3,const_4)| | general | E |
gary ’ s gas station serves an average of 16 cars per hour on saturdays , 10 cars per hour on sundays , and 9 cars per hour on all other days of the week . if the station is open from 6 a . m . to 10 p . m . every day , how many cars does gary ’ s station serve over the course of a typical week ? | "6 a . m . to 10 p . m . = 16 hours number of cars serviced on weekdays = ( 16 * 9 * 5 ) number of cars serviced on saturday = ( 16 * 16 ) number of cars serviced on sunday = ( 16 * 10 ) number of cars served in a week = 16 ( 45 + 16 + 10 ) = 16 * 71 = 1136 answer : a" | a ) 1,136 , b ) 1,200 , c ) 1,240 , d ) 1,280 , e ) 1,320 | a | floor(divide(multiply(add(6, 10), add(add(16, 10), multiply(9, add(const_4, const_1)))), const_1000)) | add(n1,n3)|add(n0,n1)|add(const_1,const_4)|multiply(n2,#2)|add(#1,#3)|multiply(#0,#4)|divide(#5,const_1000)|floor(#6)| | physics | A |
two trains are moving in opposite directions at 60 km / hr and 90 km / hr . their lengths are 1.10 km and 0.4 km respectively . the time taken by the slower train to cross the faster train in seconds is ? | "relative speed = 60 + 90 = 150 km / hr . = 150 * 5 / 18 = 125 / 3 m / sec . distance covered = 1.10 + 0.4 = 1.5 km = 1500 m . required time = 1500 * 3 / 125 = 36 sec . answer : b" | a ) 65 sec , b ) 36 sec , c ) 48 sec , d ) 33 sec , e ) 12 sec | b | subtract(divide(multiply(1.10, const_1000), divide(multiply(60, const_1000), const_3600)), divide(multiply(0.4, const_1000), divide(multiply(90, const_1000), const_3600))) | multiply(n2,const_1000)|multiply(n0,const_1000)|multiply(n3,const_1000)|multiply(n1,const_1000)|divide(#1,const_3600)|divide(#3,const_3600)|divide(#0,#4)|divide(#2,#5)|subtract(#6,#7)| | physics | B |
a and b began business with rs . 7000 and rs . 15000 after 6 months , a advances rs . 3000 and b withdraws rs . 5000 . at the end of the year , their profits amounted to rs . 2940 find the share of a and b respectively . | "7 * 6 + 10 * 6 ) : ( 15 * 6 + 10 * 6 ) = 102 : 150 = 51 : 75 = 17 : 25 17 : 25 a ' s share = 17 / 42 * 2940 = 1190 . b ' s share = 25 / 42 * 2940 = 1750 . a and b shares respectively 1190 and 1750 . answer : d" | a ) 1100 , 1750 , b ) 1140 , 1800 , c ) 940 , 2000 , d ) 1190 , 1750 , e ) 1000 : 1690 | d | multiply(divide(2940, add(add(multiply(7000, 6), multiply(subtract(7000, 3000), subtract(const_12, 6))), add(multiply(15000, 6), multiply(add(15000, 3000), subtract(const_12, 6))))), add(multiply(7000, 6), multiply(subtract(7000, 3000), subtract(const_12, 6)))) | add(n1,n3)|multiply(n0,n2)|multiply(n1,n2)|subtract(n0,n3)|subtract(const_12,n2)|multiply(#3,#4)|multiply(#0,#4)|add(#1,#5)|add(#2,#6)|add(#7,#8)|divide(n5,#9)|multiply(#7,#10)| | gain | D |
a river 4 m deep and 65 m wide is flowing at the rate of 6 kmph the amount of water that runs into the sea per minute is ? | "rate of water flow - 6 kmph - - 6000 / 60 - - 100 m / min depth of river - - 4 m width of river - - 65 m vol of water per min - - 100 * 4 * 65 - - - 26000 answer b" | a ) 25000 , b ) 26000 , c ) 27000 , d ) 28000 , e ) 29000 | b | divide(multiply(multiply(4, 65), multiply(6, const_1000)), multiply(const_1, const_60)) | multiply(n0,n1)|multiply(n2,const_1000)|multiply(const_1,const_60)|multiply(#0,#1)|divide(#3,#2)| | physics | B |
bullock likes to keep a spare tyre in his car every time . on a certain day , he travels 1 , 60,000 km and just to make the most of all the tyres , he changes the tyres between his journey such that each tyre runs the same distance . what is the distance traveled by each tyre ? | the distance traveled by each tyre : 4 / 5 * 1 , 60 , 000 km = 128,000 km . c | a ) 70,000 , b ) 60,000 , c ) 128,000 , d ) 90,000 , e ) 10,000 | c | multiply(1, const_60) | multiply(n0,const_60) | physics | C |
a shoe merchant has declared a 10 % rebate in prices . how much rebate would one get if he purchases 5 pairs of shoes at rs . 28 each ? | explanation : 5 shirts * rs . 28 = rs . 140 rebate = 10 % * 140 = 14 correct option : d | a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | d | multiply(divide(10, const_100), multiply(5, 28)) | divide(n0,const_100)|multiply(n1,n2)|multiply(#0,#1) | gain | D |
150 ml of 30 % sulphuric acid was added to approximate 400 ml of 12 % sulphuric acid solution . find the approximate concentration c of the acid in the mixture ? | "do not need any computation 30 % - - - - - - - - - - - 21 % - - - - - - - - - 12 % if volume of both sol . were equal the concentration c would be 21 % = 1 / 5 , but 12 % is more than 3 times only possibility is 1 / 6 d" | a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 6 , e ) 1 / 5 | d | divide(add(divide(multiply(150, 30), const_100), divide(multiply(400, 12), const_100)), add(150, 400)) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|divide(#1,const_100)|divide(#2,const_100)|add(#3,#4)|divide(#5,#0)| | general | D |
the present worth of rs . 1183 due in 2 years at 4 % per annum compound interest is | "solution present worth = rs . [ 1183 / ( 1 + 4 / 100 ) ² ] = rs . ( 1183 x 25 / 26 x 25 / 26 ) = rs . 1093.75 answer d" | a ) rs . 150.50 , b ) rs . 154.75 , c ) rs . 156.25 , d ) rs . 1093.75 , e ) none | d | divide(1183, power(add(divide(4, const_100), const_1), 2)) | divide(n2,const_100)|add(#0,const_1)|power(#1,n1)|divide(n0,#2)| | gain | D |
if a car had traveled 35 kmh faster than it actually did , the trip would have lasted 30 minutes less . if the car went exactly 105 km , at what speed did it travel ? | "time = distance / speed difference in time = 1 / 2 hrs 105 / x - 105 / ( x + 35 ) = 1 / 2 substitute the value of x from the options . - - > x = 70 - - > 105 / 70 - 105 / 105 = 3 / 2 - 1 = 1 / 2 answer : e" | a ) 50 kmh , b ) 55 kmh , c ) 60 kmh , d ) 65 kmh , e ) 70 kmh | e | divide(subtract(sqrt(add(multiply(multiply(const_2, multiply(105, 35)), const_4), power(35, const_2))), 35), const_2) | multiply(n0,n2)|power(n0,const_2)|multiply(#0,const_2)|multiply(#2,const_4)|add(#3,#1)|sqrt(#4)|subtract(#5,n0)|divide(#6,const_2)| | physics | E |
10 play kabadi , 25 play kho kho only , 5 play both gmaes . then how many in total ? | "10 play kabadi = > n ( a ) = 10 , 5 play both gmaes . = > n ( anb ) = 5 25 play kho kho only , = > n ( b ) = n ( b only ) + n ( anb ) = 25 + 5 = 30 total = > n ( aub ) = n ( a ) + n ( b ) - n ( anb ) = 10 + 30 - 5 = 35 answer : b" | a ) 30 , b ) 35 , c ) 38 , d ) 40 , e ) 45 | b | subtract(add(10, add(25, 5)), 5) | add(n1,n2)|add(n0,#0)|subtract(#1,n2)| | general | B |
speed of a boat in standing water is 9 kmph and speed of the stream is 1.5 kmph . a man can rows to a place at a distance of 105 km and comes back to the starting point . the total time taken by him is ? | "speed upstream = 7.5 kmph speed downstream = 10.5 kmph total time taken = 105 / 7.5 + 105 / 10.5 = 24 hours answer is b" | a ) 12 hours , b ) 24 hours , c ) 36 hours , d ) 10 hours , e ) 15 hours | b | add(multiply(add(add(9, 1.5), subtract(9, 1.5)), 105), multiply(subtract(add(divide(105, add(9, 1.5)), divide(105, subtract(9, 1.5))), add(add(9, 1.5), subtract(9, 1.5))), 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 | B |
if a ^ 2 + b ^ 2 = 10 and ab = 10 , what is the value of the expression ( a - b ) ^ 2 + ( a + b ) ^ 2 ? | "( a - b ) ^ 2 = a ^ 2 + b ^ 2 - 2 ab = 10 - 20 = - 10 ( a + b ) ^ 2 = a ^ 2 + b ^ 2 + 2 ab = 10 + 20 = 30 so ( a + b ) ^ 2 + ( a - b ) ^ 2 = 30 - 10 = 20 b" | a ) 10 , b ) 20 , c ) 30 , d ) 60 , e ) 70 | b | add(multiply(sqrt(10), const_10), const_3) | sqrt(n2)|multiply(#0,const_10)|add(#1,const_3)| | general | B |
a scale 10 ft . 5 inches long is divided into 5 equal parts . find the length of each part . | "explanation : total length of scale in inches = ( 10 * 12 ) + 5 = 125 inches length of each of the 5 parts = 125 / 5 = 25 inches answer : a" | a ) 25 inches , b ) 77 inches , c ) 66 inches , d ) 97 inches , e ) 66 inches | a | divide(add(multiply(10, const_12), 5), 5) | multiply(n0,const_12)|add(n1,#0)|divide(#1,n2)| | general | A |
bookman purchased 60 copies of a new book released recently , 10 of which are hardback and sold for $ 20 each , and rest are paperback and sold for $ 10 each . if 14 copies were sold and the total value of the remaining books was 660 , how many paperback copies were sold ? | "the bookman had 10 hardback ad 60 - 10 = 50 paperback copies ; 14 copies were sold , hence 60 - 14 = 46 copies were left . let # of paperback copies left be p then 10 p + 20 ( 46 - p ) = 660 - - > 10 p = 260 - - > p = 26 # of paperback copies sold is 50 - 26 = 24 answer : b" | a ) 22 , b ) 24 , c ) 26 , d ) 28 , e ) 30 | b | divide(subtract(subtract(add(multiply(subtract(60, 10), 10), multiply(10, 20)), 660), multiply(gcd(60, 10), 20)), 10) | gcd(n0,n1)|multiply(n1,n2)|subtract(n0,n1)|multiply(n1,#2)|multiply(n2,#0)|add(#3,#1)|subtract(#5,n5)|subtract(#6,#4)|divide(#7,n1)| | general | B |
a reduction of 15 % in the price of oil enables a house wife to obtain 4 kgs more for rs . 1200 , what is the reduced price for kg ? | "1200 * ( 15 / 100 ) = 180 - - - - 4 ? - - - - 1 = > rs . 45 answer : d" | a ) 43 , b ) 55 , c ) 35 , d ) 45 , e ) 40 | d | divide(divide(multiply(1200, 15), const_100), 4) | multiply(n0,n2)|divide(#0,const_100)|divide(#1,n1)| | gain | D |
a class is 6 meters 24 centimeters in length and 4 meters 32 centimeters in width . find the least number of square tiles of equal size required to cover the entire floor of the class room ? | explanation : length = 6 m 24 cm = 624 cm width = 4 m 32 cm = 432 cm hcf of 624 and 432 = 48 number of square tiles required = ( 624 x 432 ) / ( 48 x 48 ) = 13 x 9 = 117 . answer is b | a ) 115 , b ) 117 , c ) 116 , d ) 114 , e ) 112 | b | divide(multiply(add(multiply(6, const_100), 24), add(multiply(4, const_100), 32)), multiply(add(24, 24), add(24, 24))) | add(n1,n1)|multiply(n0,const_100)|multiply(n2,const_100)|add(n1,#1)|add(n3,#2)|multiply(#0,#0)|multiply(#3,#4)|divide(#6,#5) | general | B |
a student scored an average of 85 marks in 3 subjects : physics , chemistry and mathematics . if the average marks in physics and mathematics is 90 and that in physics and chemistry is 70 , what are the marks in physics ? | given m + p + c = 85 * 3 = 255 - - - ( 1 ) m + p = 90 * 2 = 180 - - - ( 2 ) p + c = 70 * 2 = 140 - - - ( 3 ) where m , p and c are marks obtained by the student in mathematics , physics and chemistry . p = ( 2 ) + ( 3 ) - ( 1 ) = 180 + 140 - 255 = 65 answer : d | a ) 86 , b ) 16 , c ) 76 , d ) 65 , e ) 26 | d | subtract(add(multiply(90, const_2), multiply(70, const_2)), multiply(85, 3)) | multiply(n2,const_2)|multiply(n3,const_2)|multiply(n0,n1)|add(#0,#1)|subtract(#3,#2) | general | D |
when throwing 2 dices , and looking at the sum of numbers on the dices - what is the probability to have a sum which is smaller than 5 ? | a dice is composed of 6 numbers - 12 , 34 , 56 . when trowing 2 dices - there are 36 options ( 6 x 6 = 36 ) . if we want the sum of two dices to be < 5 there are those options : ( 11 ) , ( 12 ) , ( 21 ) , ( 13 ) , ( 31 ) , ( 22 ) - total of 6 options . therefore - the probability to have a sum which is smaller than 5 is 6 / 36 = 1 / 6 . answer : b | a ) 1 / 9 , b ) 1 / 6 , c ) 1 / 2 , d ) 1 / 3 , e ) 32 / 36 | b | divide(divide(2, const_2), add(5, const_1)) | add(n1,const_1)|divide(n0,const_2)|divide(#1,#0) | general | B |
a certain ball team has an equal number of right - and left - handed players . on a certain day , two - thirds of the players were absent from practice . of the players at practice that day , one - third were right handed . what is the ratio of the number of right - handed players who were not at practice that day to the number of left handed players who were not at practice ? | "say the total number of players is 18 , 9 right - handed and 9 left - handed . on a certain day , two - thirds of the players were absent from practice - - > 12 absent and 6 present . of the players at practice that day , one - third were right - handed - - > 6 * 1 / 3 = 2 were right - handed and 4 left - handed . the number of right - handed players who were not at practice that day is 9 - 2 = 7 . the number of left - handed players who were not at practice that days is 9 - 4 = 5 . the ratio = 7 / 5 . answer : c ." | a ) 1 / 3 , b ) 2 / 3 , c ) 7 / 5 , d ) 5 / 7 , e ) 3 / 2 | c | divide(subtract(divide(const_1, const_2), subtract(subtract(const_1, divide(const_2, const_3)), multiply(subtract(const_1, divide(const_2, const_3)), divide(const_2, const_3)))), subtract(divide(const_1, const_2), multiply(subtract(const_1, divide(const_2, const_3)), divide(const_2, const_3)))) | divide(const_1,const_2)|divide(const_2,const_3)|subtract(const_1,#1)|multiply(#1,#2)|subtract(#2,#3)|subtract(#0,#3)|subtract(#0,#4)|divide(#6,#5)| | general | C |
chlorination is used in industry for the purification of water ; 1 gm solution of chlorine will remove half of the germs in 1 liter of water . if water containing less than 0.09 % of germs is considered clean , what is the minimum amount of chlorine required to purify 150 liter of water containing 0.3 liter of harmful substances ? | assuming that it ' s supposed to state ` ` 0.3 ' ' liters of harmful substances , here ' s how you can answer this question . first , we have to figure out what . 09 % of 150 liters is . . . 1 % = . 01 . 1 % = . 001 . 09 % = . 0009 so , . 0009 ( 150 ) = . 135 liters we need enough doses of chlorine to get the 0.3 liters down to less than . 135 liters . each ` ` dose ' ' of chlorine will remove half of the harmful substances . . . start = 0.3 liters after 1 st dose = . 15 liters after 2 nd dose = . 075 liters we ' re now below the . 135 liters that we need to be less than . final answer : b | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | b | divide(multiply(multiply(150, 0.09), 0.3), const_2) | multiply(n2,n3)|multiply(n4,#0)|divide(#1,const_2) | gain | B |
a little girl is numbering her collection of butterflies by placing single - digit stickers under each specimen . if the girl has 330 butterflies , how many stickers does she need ? ( the numbers are consecutive and the number of the first butterfly is 1 ) . | for the first 9 butterflies we need 9 stickers . for the next 90 we need 2 stickers each or 180 stickers . for the next 900 butterflies we need 3 stickers each . the first 99 butterflies correspond to 189 stickers . subtract 99 from the total number of butterflies ( 330 ) . this leaves 231 butterflies that use 3 stickers each . so , multiply 231 by 3 to get 693 . add 693 to 189 stickers to get 882 stickers . answer : a | a ) 882 , b ) 901 , c ) 1001 , d ) 1003 , e ) 1346 | a | add(multiply(add(subtract(330, const_100), const_1), const_3), add(multiply(const_3, const_3), multiply(multiply(multiply(const_3, const_3), const_10), const_2))) | multiply(const_3,const_3)|subtract(n0,const_100)|add(#1,const_1)|multiply(#0,const_10)|multiply(#3,const_2)|multiply(#2,const_3)|add(#0,#4)|add(#6,#5) | physics | A |
what is the next number : 3 , 11 , 83 , __ | "3 ^ 0 + 2 = 3 3 ^ 2 + 2 = 11 3 ^ 4 + 2 = 83 3 ^ 6 + 2 = 731 the answer is b ." | a ) 631 , b ) 731 , c ) 831 , d ) 849 , e ) 901 | b | subtract(subtract(subtract(multiply(83, const_10), const_100), 3), 3) | multiply(n2,const_10)|subtract(#0,const_100)|subtract(#1,n0)|subtract(#2,n0)| | general | B |
on rainy mornings , mo drinks exactly n cups of hot chocolate ( assume that n is an integer ) . on mornings that are not rainy , mo drinks exactly 4 cups of tea . last week mo drank a total of 42 cups of tea and hot chocolate together . if during that week mo drank 14 more tea cups than hot chocolate cups , then how many rainy days were there last week ? | "t = the number of cups of tea c = the number of cups of hot chocolate t + c = 42 t - c = 14 - > t = 28 . c = 14 . mo drinks 4 cups of tea a day then number of days that are not rainy = 28 / 4 = 7 so number of rainy days = 7 - 7 = 0 a is the answer ." | a ) 0 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | subtract(add(const_4, 4), divide(divide(add(42, 14), const_2), 4)) | add(n0,const_4)|add(n1,n2)|divide(#1,const_2)|divide(#2,n0)|subtract(#0,#3)| | general | A |
the cost price of a radio is rs . 1500 and it was sold for rs . 1245 , find the loss % ? | "1500 - - - - 255 100 - - - - ? = > 17 % answer : a" | a ) 17 % , b ) 16 % , c ) 19 % , d ) 78 % , e ) 28 % | a | multiply(divide(subtract(1500, 1245), 1500), const_100) | subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)| | gain | A |
the average monthly income of p and q is rs . 2050 . the average monthly income of q and r is rs . 5250 and the average monthly income of p and r is rs . 6200 . the monthly income of p is : | "explanation : let p , q and r represent their respective monthly incomes . then , we have : p + q = ( 2050 x 2 ) = 4100 . . . . ( i ) q + r = ( 5250 x 2 ) = 10500 . . . . ( ii ) p + r = ( 6200 x 2 ) = 12400 . . . . ( iii ) adding ( i ) , ( ii ) and ( iii ) , we get : 2 ( p + q + r ) = 27000 or p + q + r = 13500 . . . . ( iv ) subtracting ( ii ) from ( iv ) , we get p = 3000 . p ' s monthly income = rs . 3000 . answer : b" | a ) 3050 , b ) 3000 , c ) 4000 , d ) 4050 , e ) 3100 | b | subtract(add(2050, 6200), 5250) | add(n0,n2)|subtract(#0,n1)| | general | B |
each day a man meets his wife at the train station after work , and then she drives him home . she always arrives exactly on time to pick him up . one day he catches an earlier train and arrives at the station an hour early . he immediately begins walking home along the same route the wife drives . eventually his wife sees him on her way to the station and drives him the rest of the way home . when they arrive home the man notices that they arrived 12 minutes earlier than usual . how much time did the man spend walking ? | "as they arrived 12 minutes earlier than usual , they saved 12 minutes on round trip from home to station ( home - station - home ) - - > 6 minutes in each direction ( home - station ) - - > wife meets husband 6 minutes earlier the usual meeting time - - > husband arrived an hour earlier the usual meeting time , so he must have spent waking the rest of the time before their meeting , which is hour - 6 minutes = 54 minutes . answer : d" | a ) 45 minutes , b ) 50 minutes , c ) 40 minutes , d ) 54 minutes , e ) 35 minutes | d | subtract(const_60, divide(12, const_2)) | divide(n0,const_2)|subtract(const_60,#0)| | physics | D |
at a restaurant , glasses are stored in two different - sized boxes . one box contains 12 glasses , and the other contains 16 glasses . if the average number of glasses per box is 15 , and there are 16 more of the larger boxes , what is the total number of glasses t at the restaurant ? ( assume that all boxes are filled to capacity . ) | "most test takers would recognize the system of equations in this prompt and just do algebra to get to the solution ( and that ' s fine ) . the wording of the prompt and the ' spread ' of the answer choices actually provide an interesting ' brute force ' shortcut that you can take advantage of to eliminate the 4 wrong answers . . . . we ' re told that there are 2 types of boxes : those that hold 12 glasses and those that hold 16 glasses . since the average number of boxes is 15 , we know that there must be at least some of each . we ' re also told that that there are 16 more of the larger boxes . this means , at the minimum , we have . . . 1 small box and 17 large boxes = 1 ( 12 ) + 17 ( 16 ) = 12 + 272 = 284 glasses at the minimum since the question asks for the total number of glasses , we can now eliminate answers a , b and c . . . . the difference in the number of boxes must be 16 though , so we could have . . . . 2 small boxes and 18 large boxes 3 small boxes and 19 large boxes etc . with every additional small box + large box that we add , we add 12 + 16 = 28 more glasses . thus , we can justadd 28 suntil we hit the correct answer . . . . 284 + 28 = 312 312 + 28 = 340 340 + 28 = 368 368 + 28 = 396 at this point , we ' ve ' gone past ' answer d , so the correct answer must be answer e . . . . . but here ' s the proof . . . . 396 + 28 = 424 424 + 28 = 452 452 + 28 = 480 final answer : e" | a ) 96 , b ) 240 , c ) t = 256 , d ) t = 384 , e ) t = 480 | e | multiply(multiply(16, const_2), 15) | multiply(n1,const_2)|multiply(n2,#0)| | general | E |
the average weight of 4 persons increases by 1.5 kg . if a person weighing 65 kg is replaced by a new person , what could be the weight of the new person ? | "total weight increases = 4 × 1.5 = 6 kg so the weight of new person = 65 + 6 = 71 kg answer a" | a ) 71 kg , b ) 77 kg , c ) 76.5 kg , d ) data inadequate , e ) none of these | a | add(65, multiply(4, 1.5)) | multiply(n0,n1)|add(n2,#0)| | general | A |
how much greater is the combined area in square inches of the front and back of a rectangular sheet of paper measuring 11 inches by 19 inches than that of a rectangular sheet of paper measuring 9.5 inches by 11 inches ? | "let ' s just look at the dimensions ( no calculation needed ) . with dimension 11 the same , the other dimension 19 is twice 9.5 then the area will be double which means 100 % greater . the answer is c ." | a ) 50 % , b ) 87 % , c ) 100 % , d ) 187 % , e ) 200 % | c | multiply(divide(subtract(multiply(rectangle_area(11, 19), const_2), multiply(rectangle_area(9.5, 11), const_2)), rectangle_area(11, 19)), const_100) | rectangle_area(n0,n1)|rectangle_area(n0,n2)|multiply(#0,const_2)|multiply(#1,const_2)|subtract(#2,#3)|divide(#4,#0)|multiply(#5,const_100)| | geometry | C |
how long will a boy take to run round a square field of side 40 meters , if he runs at the rate of 12 km / hr ? | speed = 12 km / hr = 12 * 5 / 18 = 10 / 3 m / sec distance = 40 * 4 = 160 m time taken = 160 * 3 / 10 = 48 sec answer is b | a ) 52 sec , b ) 48 sec , c ) 60 sec , d ) 25 sec , e ) 39 sec | b | divide(multiply(40, const_4), multiply(12, divide(const_1000, const_3600))) | divide(const_1000,const_3600)|multiply(n0,const_4)|multiply(n1,#0)|divide(#1,#2) | gain | B |
if 3 ^ 8 x 3 ^ 7 = 3 ^ n what is the value of n ? | "3 ^ 8 * 3 ^ 7 = 3 ^ n or 3 ^ 8 + 7 = 3 ^ n n = 15 e" | a ) 20 , b ) 18 , c ) 17 , d ) 16 , e ) 15 | e | divide(log(multiply(power(3, 7), power(3, 8))), log(3)) | log(n4)|power(n2,n3)|power(n0,n1)|multiply(#1,#2)|log(#3)|divide(#4,#0)| | general | E |
a certain sum of money is divided among a , b and c such that a gets one - third of what b and c together get and b gets two - seventh of what a and c together get . if the amount received by a is $ 20 more than that received by b , find the total amount shared by a , b and c . | "a = 1 / 3 ( b + c ) = > c = 3 a - b - - - ( 1 ) b = 2 / 7 ( a + c ) = > c = 3.5 b - a - - ( b ) a - b = $ 20 a = 20 + b ( 1 ) = = = > c = 60 + 3 b - b = 2 b + 60 = = > 2 b - c = - 60 - - - ( 3 ) ( 2 ) = = = > c = 3.5 b - b - 20 = 2.5 b - 20 = = > 2.5 b - c = 20 - - - ( 4 ) from ( 4 ) and ( 3 ) 0.5 b = 80 b = $ 160 a = $ 180 c = 540 - 160 = $ 380 total amount = 180 + 160 + 380 = $ 720 answer : c" | a ) $ 320 , b ) $ 420 , c ) $ 720 , d ) $ 220 , e ) $ 200 | c | add(add(add(multiply(multiply(const_3, 20), const_2), multiply(const_2, 20)), add(add(multiply(multiply(const_3, 20), const_2), multiply(const_2, 20)), 20)), subtract(multiply(add(add(multiply(multiply(const_3, 20), const_2), multiply(const_2, 20)), 20), const_3), add(multiply(multiply(const_3, 20), const_2), multiply(const_2, 20)))) | multiply(n0,const_3)|multiply(n0,const_2)|multiply(#0,const_2)|add(#2,#1)|add(n0,#3)|add(#3,#4)|multiply(#4,const_3)|subtract(#6,#3)|add(#5,#7)| | general | C |
a train covers a distance of 18 km in 10 min . if it takes 6 sec to pass a telegraph post , then the length of the train is ? | "speed = ( 18 / 10 * 60 ) km / hr = ( 108 * 5 / 18 ) m / sec = 30 m / sec . length of the train = 30 * 6 = 180 m . answer : e" | a ) 177 , b ) 168 , c ) 120 , d ) 882 , e ) 180 | e | divide(18, subtract(divide(18, 10), 6)) | divide(n0,n1)|subtract(#0,n2)|divide(n0,#1)| | physics | E |
a bus covered a distance of 250 km , partly at an average speed of 40 kmph and partly at 60 kmph . if the total time taken is 5.2 hours , then the distance covered at 40 kmph is | "let the partial distance covered at 40 kmph be x let the another partial distance covered at 60 kmph be ( 250 - x ) thus , x / 40 - ( 250 - x ) / 60 = 5.2 or , x / 40 + ( 250 - x ) / 60 = 5.2 or , ( 3 x + 500 - 2 x ) / / 120 = 5.2 or 500 + x = 624 x = 124 answer : d" | a ) 130 km , b ) 120 km , c ) 100 km , d ) 124 km , e ) none of these | d | multiply(divide(subtract(multiply(60, 5.2), 250), negate(subtract(40, 60))), 40) | multiply(n2,n3)|subtract(n1,n2)|negate(#1)|subtract(#0,n0)|divide(#3,#2)|multiply(n1,#4)| | physics | D |
a card game called “ high - low ” divides a deck of 52 playing cards into 2 types , “ high ” cards and “ low ” cards . there are an equal number of “ high ” cards and “ low ” cards in the deck and “ high ” cards are worth 2 points , while “ low ” cards are worth 1 point . if you draw cards one at a time , how many ways can you draw “ high ” and “ low ” cards to earn 6 points if you must draw exactly 4 “ low ” cards ? | "great question ravih . this is a permutations problem ( order matters ) with repeating elements . given thatlowcards are worth 1 pt andhigh cards 2 pts , and you must draw 3 low cards , we know that you must also draw 1 high card . the formula for permutations problems with repeating elements isn ! / a ! b ! . . . where n represents the number of elements in the group and a , b , etc . represent the number of times that repeating elements are repeated . here there are 4 elements and thelowcard is repeated 3 times . as a result , the formula is : 5 ! / 4 ! which represents ( 5 * 4 * 3 * 2 * 1 ) / ( 4 * 3 * 2 * 1 ) which simplifies to just 5 , giving you answer e ." | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | e | divide(multiply(multiply(multiply(const_4, 4), 2), 1), multiply(multiply(4, 2), 1)) | multiply(n5,const_4)|multiply(n1,n5)|multiply(n1,#0)|multiply(n3,#1)|multiply(n3,#2)|divide(#4,#3)| | general | E |
evaluate 28 % of 550 + 45 % of 280 | "explanation : = ( 28 / 100 ) * 550 + ( 45 / 100 ) * 280 = 154 + 126 = 280 answer : option e" | a ) 232 , b ) 242 , c ) 252 , d ) 262 , e ) 280 | e | divide(28, divide(550, 28)) | divide(n1,n0)|divide(n0,#0)| | gain | E |
a man rows his boat 75 km downstream and 45 km upstream , taking 2 1 / 2 hours each time . find the speed of the stream ? | "speed downstream = d / t = 75 / ( 2 1 / 2 ) = 30 kmph speed upstream = d / t = 45 / ( 2 1 / 2 ) = 18 kmph the speed of the stream = ( 30 - 18 ) / 2 = 6 kmph answer : a" | a ) 6 , b ) 7 , c ) 5 , d ) 8 , e ) 9 | a | divide(subtract(divide(75, 2), divide(45, 2)), const_2) | divide(n0,n2)|divide(n1,n2)|subtract(#0,#1)|divide(#2,const_2)| | physics | A |
if the radius of a circle that centers at the origin is 5 , how many w points on the circle have integer coordinates ? | "i understand this might not be required but i used the equation of a circle . since the origin is at 0 , x ^ 2 + y ^ 2 = 5 ^ 2 . x , y could be + / - ( 0,5 or 5,0 ) - 4 possibilities . x , y could be + / - ( 3,4 or 4,3 ) - 8 possibilities . ans : w = c" | a ) 4 , b ) 8 , c ) 12 , d ) 15 , e ) 20 | c | subtract(subtract(multiply(5, 5), add(5, const_4)), power(const_2, const_2)) | add(n0,const_4)|multiply(n0,n0)|power(const_2,const_2)|subtract(#1,#0)|subtract(#3,#2)| | geometry | C |
a student committee on academic integrity has 72 ways to select a president and vice president from a group of candidates . the same person can not be both president and vice president . how many candidates are there ? | "xc 1 * ( x - 1 ) c 1 = 72 x ^ 2 - x - 72 = 0 ( x - 9 ) ( x + 8 ) = 0 x = 9 , - 8 - 8 ca n ' t possible . c" | a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | c | divide(add(const_1, sqrt(add(multiply(const_4, 72), power(negate(const_1), const_2)))), const_2) | multiply(n0,const_4)|negate(const_1)|power(#1,const_2)|add(#0,#2)|sqrt(#3)|add(#4,const_1)|divide(#5,const_2)| | other | C |
if ( 18 ^ a ) * 9 ^ ( 3 a – 1 ) = ( 2 ^ 5 ) ( 3 ^ b ) and a and b are positive integers , what is the value of a ? | "( 18 ^ a ) * 9 ^ ( 3 a – 1 ) = ( 2 ^ 5 ) ( 3 ^ b ) = 2 ^ a . 9 ^ a . 9 ^ ( 3 a – 1 ) = ( 2 ^ 5 ) ( 3 ^ b ) just compare powers of 2 from both sides answer = 5 = e" | a ) 22 , b ) 11 , c ) 9 , d ) 6 , e ) 5 | e | multiply(3, 1) | multiply(n2,n3)| | general | E |
working alone , printers x , y , and z can do a certain printing job , consisting of a large number of pages , in 12 , 40 , and 50 hours , respectively . what is the ratio of the time it takes printer x to do the job , working alone at its rate , to the time it takes printers y and z to do the job , working together at their individual rates ? | "p 1 takes 12 hrs rate for p 2 p 3 together = 1 / 40 + 1 / 50 = 9 / 200 therefore they take 200 / 9 ratio = 200 / 9 = d" | a ) 4 / 11 , b ) 1 / 2 , c ) 15 / 22 , d ) 200 / 9 , e ) 11 / 4 | d | divide(12, divide(const_1, add(divide(const_1, 40), divide(const_1, 50)))) | divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|divide(const_1,#2)|divide(n0,#3)| | general | D |
the length of a rectangular field is 7 / 5 its width . if the perimeter of the field is 360 meters , what is the width of the field ? | "let l be the length and w be the width . l = ( 7 / 5 ) w perimeter : 2 l + 2 w = 360 , 2 ( 7 / 5 ) w + 2 w = 360 solve the above equation to find : w = 75 m and l = 105 m . correct answer e ) 75" | a ) 50 , b ) 60 , c ) 70 , d ) 80 , e ) 75 | e | divide(360, add(add(divide(7, 5), divide(7, 5)), const_2)) | divide(n0,n1)|add(#0,#0)|add(#1,const_2)|divide(n2,#2)| | geometry | E |
if one length of a square is tripled , and the original parameter is 4 y . what is the new length of one of the sides ? | if the original parameter is 4 y , each length of the square is 4 y / 4 . if the parameter is tripled , then each length becomes 3 ( 4 y ) / 4 . this is 12 y / 4 . answer option d . | ['a ) 12 y / 3', 'b ) 2 y / 2', 'c ) 6 y / 3', 'd ) 12 y / 4', 'e ) 4 y / 3'] | d | divide(square_area(multiply(4, const_3)), square_area(4)) | multiply(n0,const_3)|square_area(n0)|square_area(#0)|divide(#2,#1) | physics | D |
when a merchant imported a certain item , he paid a 9 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 $ 76.50 , what was the total value of the item ? | "let x be the value in excess of $ 1,000 . 0.09 x = 76.5 x = $ 850 the total value was $ 850 + $ 1,000 = $ 1,850 . the answer is d ." | a ) $ 1,250 , b ) $ 1,450 , c ) $ 1,650 , d ) $ 1,850 , e ) $ 2,050 | d | floor(divide(add(divide(76.50, divide(9, const_100)), 1,000), 1,000)) | divide(n0,const_100)|divide(n2,#0)|add(#1,n1)|divide(#2,n1)|floor(#3)| | general | D |
the sum of two numbers is 62 , and one of them is 12 more than the other . what are the two numbers ? | "in this problem , we are asked to find two numbers . therefore , we must let x be one of them . let x , then , be the first number . we are told that the other number is 12 more , x + 12 . the problem states that their sum is 62 : word problem = 62 the line over x + 12 is a grouping symbol called a vinculum . it saves us writing parentheses . we have : 2 x = 62 â ˆ ’ 12 = 50 . x = 50 / 2 = 25 . this is the first number . therefore the other number is x + 12 = 25 + 12 = 37 . the sum of 25 + 37 is 62 . e" | a ) 36 - 48 , b ) 50 - 34 , c ) 60 - 24 , d ) 42 - 42 , e ) 25 - 37 | e | divide(subtract(62, 12), const_2) | subtract(n0,n1)|divide(#0,const_2)| | general | E |
in the rectangular coordinate system , points ( 5 , 0 ) and ( – 5 , 0 ) both lie on circle c . what is the maximum possible value of the radius of c ? | "the answer is b it takes 3 distinct points to define a circle . only 2 are given here . the two points essentially identify a single chord of the circle c . since no other information is provided , however , the radius of the circle can essentially be anything . all this information tell us is that the radius isgreater 5 b" | a ) 2 , b ) 4 , c ) 8 , d ) 16 , e ) none of the above | b | sqrt(power(5, const_2)) | power(n0,const_2)|sqrt(#0)| | geometry | B |
the units digit of ( 35 ) ^ ( 87 ) + ( 93 ) ^ ( 53 ) is : | the units digit of powers of 3 , cycles in a group of 4 : { 3 , 9 , 7 , 1 } 53 has the form 4 k + 1 , so the units digit of 93 ^ 53 is 3 . the units digit of powers of 5 is always 5 . 3 + 5 = 8 , so the units digit is 8 . the answer is d . | a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 0 | d | add(reminder(multiply(reminder(53, const_4), 93), const_10), reminder(35, const_10)) | reminder(n3,const_4)|reminder(n0,const_10)|multiply(n2,#0)|reminder(#2,const_10)|add(#3,#1) | general | D |
what are the last two digits of ( 301 * 402 * 503 * 604 * 646 * 547 * 448 * 349 ) ^ 2 | "( ( 301 * 402 * 503 * 604 * 646 ) * ( 547 * 448 * 349 ) ) ^ 2 if you observe above digits , last digit are : 1,2 , 3,4 , 6,7 , 8,9 ; 5 is missing ; so i have rearranged them so that multiplication will be easy for me as initial 4 digits have last two digits as 01 , 02,03 , 04,46 and final three as 47 * 48 * 49 . solving for only last two digits and multiplying them we get : ( ( 06 * 04 * 46 ) ( 56 * 49 ) ) ^ 2 = ( 44 * 44 ) ^ 2 = 36 ^ 2 = 76 hence answer is b" | a ) 96 , b ) 76 , c ) 56 , d ) 36 , e ) 16 | b | multiply(add(divide(const_100, const_4), multiply(multiply(multiply(multiply(2, const_3), subtract(const_1, 2)), subtract(const_1, 2)), subtract(const_1, 2))), const_4) | divide(const_100,const_4)|multiply(n8,const_3)|subtract(const_1,n8)|multiply(#1,#2)|multiply(#3,#2)|multiply(#4,#2)|add(#0,#5)|multiply(#6,const_4)| | general | B |
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 + 21 ) both lie on the line defined by the equation x = ( y / 7 ) - ( 2 / 5 ) , what is the value of p ? | x = ( y / 7 ) - ( 2 / 5 ) , and so y = 7 x + 14 / 5 . the slope is 7 . ( n + 21 - n ) / ( m + p - m ) = 7 p = 3 the answer is c . | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | divide(21, 7) | divide(n0,n1) | general | C |
in 2008 , the profits of company n were 10 percent of revenues . in 2009 , the revenues of company n fell by 20 percent , but profits were 10 percent of revenues . the profits in 2009 were what percent of the profits in 2008 ? | "x = profits r = revenue x / r = 0,1 x = 10 r = 100 2009 : r = 80 x / 80 = 0,10 = 10 / 100 x = 80 * 10 / 100 x = 8 8 / 10 = 0.8 = 800 % , answer a" | a ) 80 % , b ) 105 % , c ) 120 % , d ) 124.2 % , e ) 138 % | a | multiply(divide(multiply(10, subtract(const_1, divide(20, const_100))), 10), const_100) | divide(n3,const_100)|subtract(const_1,#0)|multiply(n4,#1)|divide(#2,n1)|multiply(#3,const_100)| | gain | A |
what is the least number should be added to 1100 , so the sum of the number is completely divisible by 23 ? | "( 1100 / 23 ) gives remainder 19 19 + 4 = 23 , so we need to add 4 answer : d" | a ) 1 , b ) 2 , c ) 5 , d ) 4 , e ) 8 | d | subtract(multiply(add(floor(divide(1100, 23)), const_1), 23), 1100) | divide(n0,n1)|floor(#0)|add(#1,const_1)|multiply(n1,#2)|subtract(#3,n0)| | general | D |
the length of the longest tape in cm which can be used to measure exactly , the length 6 m ; 5 m ; and 12 m is : | the three lengths in cm are 600,500 & 1200 . hcf of 600 , 500 & 1200 is 100 hence , the answer is 100 cm . answer : b | a ) 200 , b ) 100 , c ) 50 , d ) 1200 , e ) 500 | b | multiply(multiply(multiply(const_2, const_2), 5), 5) | multiply(const_2,const_2)|multiply(n1,#0)|multiply(n1,#1) | physics | B |
the amount of an investment will double in approximately 70 / p years , where p is the percent interest , compounded annually . if thelma invests $ 50,000 in a long - term cd that pays 5 percent interest , compounded annually , what will be the approximate total value of the investment when thelma is ready to retire 42 years later ? | "the amount of an investment will double in approximately 70 / p years , where p is the percent interest , compounded annually . if thelma invests $ 50,000 in a long - term cd that pays 5 percent interest , compounded annually , what will be the approximate total value of the investment when thelma is ready to retire 42 years later ? the investment gets doubled in 70 / p years . therefore , the investment gets doubled in 70 / 5 = every 14 years . after 42 years , the investment will get doubled 42 / 14 = 3 times . so the amount invested will get doubled thrice . so , 50000 * 2 ^ 3 = 400000 hence , the answer is d ." | a ) $ 280,000 , b ) $ 320,000 , c ) $ 360,000 , d ) $ 400,000 , e ) $ 540,000 | d | divide(const_3600, const_10) | divide(const_3600,const_10)| | general | D |
the radius of a cone is 14 m , slant height is 20 m . find the curved surface area ? | "cone curved surface area = π rl 22 / 7 × 14 × 120 = 44 × 20 = 880 m ( power 2 ) answer is d ." | a ) 770 , b ) 550 , c ) 110 , d ) 880 , e ) 330 | d | volume_cone(14, 20) | volume_cone(n0,n1)| | geometry | D |
11 + 12 + 13 + . . . 51 + 52 + 53 = ? | "sum = 11 + 12 + 13 + . . . 51 + 52 + 53 sum of n consecutive positive integers starting from 1 is given as n ( n + 1 ) / 2 sum of first 53 positive integers = 53 * 54 / 2 sum of first 10 positive integers = 11 * 10 / 2 sum = 11 + 12 + 13 + . . . 51 + 52 + 53 = 53 * 54 / 2 - 11 * 10 / 2 = 1376 answer : b" | a ) 1361 , b ) 1376 , c ) 1363 , d ) 1364 , e ) 1365 | b | divide(multiply(51, add(51, 11)), 12) | add(n0,n3)|multiply(n3,#0)|divide(#1,n1)| | general | B |
alok ordered 16 chapatis , 5 plates of rice , 7 plates of mixed vegetable and 6 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 . 1111 . find the cost of each ice - cream cup ? | "let the cost of each ice - cream cup be rs . x 16 ( 6 ) + 5 ( 45 ) + 7 ( 70 ) + 6 ( x ) = 1111 96 + 225 + 490 + 6 x = 1111 6 x = 300 = > x = 50 . answer : d" | a ) 25 , b ) 66 , c ) 77 , d ) 50 , e ) 91 | d | divide(subtract(subtract(subtract(1111, multiply(16, 6)), multiply(5, 45)), multiply(7, 70)), 6) | multiply(n0,n3)|multiply(n1,n5)|multiply(n2,n6)|subtract(n7,#0)|subtract(#3,#1)|subtract(#4,#2)|divide(#5,n3)| | general | D |
let p be a prime number greater than 2 and let n = 14 p . how many even numbers divide n ? | answer : a . there is exactly one . the prime factorization of 14 is 14 = 2 * 7 , so n = 2 * 7 * p = 7 * 2 * p = 7 * 2 p , so 2 p is the only even number that divides n . | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | a | subtract(2, const_1) | subtract(n0,const_1) | general | A |
in 130 m race , a covers the distance in 36 seconds and b in 45 seconds . in this race a beats b by : | "distance covered by b in 9 sec . = 130 / 45 x 9 m = 26 m . a beats b by 26 metres . answer : option d" | a ) 20 m , b ) 25 m , c ) 22.5 m , d ) 26 m , e ) 12 m | d | multiply(divide(130, 45), subtract(45, 36)) | divide(n0,n2)|subtract(n2,n1)|multiply(#0,#1)| | physics | D |
dividing by 3 ⁄ 8 and then multiplying by 5 ⁄ 9 is the same as dividing by what number ? | say x / 3 / 8 * 5 / 9 = x * 8 / 3 * 5 / 9 = x * 40 / 27 d | a ) 31 ⁄ 5 , b ) 16 ⁄ 5 , c ) 20 ⁄ 9 , d ) 40 / 27 , e ) 5 ⁄ 16 | d | multiply(divide(8, 3), divide(5, 9)) | divide(n1,n0)|divide(n2,n3)|multiply(#0,#1) | general | D |
at a florist shop on a certain day , all corsages sold for either $ 20 or $ 30 . if 8 of the corsages that sold for $ 30 had instead sold for $ 20 , then the store ' s revenue from corsages that day would have been reduced by 35 percent . what was the store ' s actual revenue from corsages that day ? | i am doing it elaborately , hope it will help you . let , no . of corsages @ $ 20 = x , no . of corsages @ $ 30 = y and revenue = r so , 20 x + 30 y = r . . . . . . . . . ( 1 ) now , given the situation , 20 ( x + 8 ) + 30 ( y - 8 ) = r - . 25 r = > 20 x + 160 + 30 y - 240 = . 75 r = > 20 x + 30 y = . 75 r + 80 . . . . . . . . . . . . ( 2 ) so , r = . 75 r + 80 = > r = 320 the answer is a . | a ) $ 320 , b ) $ 400 , c ) $ 600 , d ) $ 800 , e ) $ 1000 | a | subtract(divide(negate(subtract(multiply(20, 8), multiply(30, 8))), divide(35, const_100)), subtract(multiply(20, 8), multiply(30, 8))) | divide(n5,const_100)|multiply(n0,n2)|multiply(n1,n2)|subtract(#1,#2)|negate(#3)|divide(#4,#0)|subtract(#5,#3) | gain | A |
the least perfect square , which is divisible by each of 21 , 36 and 66 is : | "explanation : l . c . m . of 21 , 36 , 66 = 2772 . now , 2772 = 2 x 2 x 3 x 3 x 7 x 11 to make it a perfect square , it must be multiplied by 7 x 11 . so , required number = 22 x 32 x 72 x 112 = 213444 answer is a" | a ) 213444 , b ) 214344 , c ) 214434 , d ) 231444 , e ) 233444 | a | multiply(multiply(multiply(power(add(const_3, const_4), const_2), power(divide(divide(36, const_3), const_2), const_2)), power(const_2, const_2)), power(const_3, const_2)) | add(const_3,const_4)|divide(n1,const_3)|power(const_2,const_2)|power(const_3,const_2)|divide(#1,const_2)|power(#0,const_2)|power(#4,const_2)|multiply(#5,#6)|multiply(#7,#2)|multiply(#8,#3)| | geometry | A |
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 = √ ( 20 ) 2 – ( 12 ) 2 = √ 256 = 16 cm . _ i so , other diagonal = 32 cm . area of rhombus = ( 1 / 2 ) x ( product of diagonals ) = ( 1 / 2 × 24 x 32 ) cm 2 = 384 cm 2 answer : option d" | 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 | D |
in the first 10 overs of a cricket game , the run rate was only 4.6 . what should be the run rate in the remaining 40 overs to reach the target of 282 runs ? | "required run rate = 282 - ( 4.6 x 10 ) = 236 236 / 40 = 5.9 b )" | a ) 4.25 , b ) 5.9 , c ) 6.25 , d ) 7 , e ) 7.5 | b | divide(subtract(282, multiply(10, 4.6)), 40) | multiply(n0,n1)|subtract(n3,#0)|divide(#1,n2)| | gain | B |
a group of 55 adults and 70 children go for trekking . if there is meal for either 70 adults or 90 children and if 42 adults have their meal , find the total number of children that can be catered with the remaining food . | explanation : as there is meal for 70 adults and 42 have their meal , the meal left can be catered to 28 adults . now , 70 adults = 90 children 7 adults = 9 children therefore , 28 adults = 36 children hence , the meal can be catered to 36 children . answer : c | a ) 33 , b ) 54 , c ) 36 , d ) 17 , e ) 01 | c | multiply(subtract(70, 42), divide(90, 70)) | divide(n3,n1)|subtract(n1,n4)|multiply(#0,#1) | general | C |
in the rectangular coordinate system , what is the x - intercept of a line passing through ( 10 , 3 ) and ( − 6 , − 5 ) ? | first , solve for the slope , m : m = ( y 2 - y 1 ) / ( x 2 - x 1 ) = ( - 5 - 3 ) / ( - 6 - 10 ) = - 8 / - 16 or 1 / 2 then you can re - write the equation as y = ( 1 / 2 ) x + b and plug in one of the points to find b , the y intercept . using point ( 10 , 3 ) : 3 = ( 1 / 2 ) ( 10 ) + b , which can be simplified to 3 = 5 + b . solve for b and you get b = - 2 . the equation now reads : y = ( 1 / 2 ) x - 2 . to find the x intercept , plug in 0 for y : 0 = ( 1 / 2 ) x - 2 , which simplifies to 2 = ( 1 / 2 ) x . from this , we can see that the x intercept = 4 . answer is a . | a ) 4 , b ) 2 , c ) 0 , d ) − 2 , e ) − 4 | a | divide(subtract(multiply(divide(subtract(negate(5), 3), subtract(negate(6), 10)), 10), 3), divide(subtract(negate(5), 3), subtract(negate(6), 10))) | negate(n3)|negate(n2)|subtract(#0,n1)|subtract(#1,n0)|divide(#2,#3)|multiply(n0,#4)|subtract(#5,n1)|divide(#6,#4) | general | A |
a man rows his boat 105 km downstream and 45 km upstream , taking 2 1 / 2 hours each time . find the speed of the stream ? | "speed downstream = d / t = 105 / ( 2 1 / 2 ) = 42 kmph speed upstream = d / t = 45 / ( 2 1 / 2 ) = 18 kmph the speed of the stream = ( 42 - 18 ) / 2 = 12 kmph answer : c" | a ) 5 kmph , b ) 7 kmph , c ) 12 kmph , d ) 8 kmph , e ) 1 kmph | c | divide(subtract(divide(105, 2), divide(45, 2)), const_2) | divide(n0,n2)|divide(n1,n2)|subtract(#0,#1)|divide(#2,const_2)| | physics | C |
the cost price of 12 articles is equal to the selling price of 8 articles . what is the profit percent ? | "8 * sp = 12 * cp sp = 1.5 * cp the profit percent is 50 % . the answer is c ." | a ) 40 % , b ) 45 % , c ) 50 % , d ) 55 % , e ) 60 % | c | divide(multiply(12, const_4), add(const_4, const_1)) | add(const_1,const_4)|multiply(n0,const_4)|divide(#1,#0)| | gain | C |
a number x is 8 times another number y . the percentage that y is less than x is | "say y = 1 and x = 8 . then y = 1 is less than x = 8 by ( 8 - 1 ) / 8 * 100 = 7 / 8 * 100 = 87.5 % . answer : b ." | a ) 12.5 % , b ) 87.5 % , c ) 80 % , d ) 11 % , e ) 1 % | b | multiply(divide(subtract(8, const_1), 8), const_100) | subtract(n0,const_1)|divide(#0,n0)|multiply(#1,const_100)| | general | B |
when n is divided by 24 , the remainder is 5 . what is the remainder when 4 n is divided by 8 ? | "let n = 5 ( leaves a remainder of 5 when divided by 24 ) 4 n = 4 ( 5 ) = 20 , which leaves a remainder of 4 when divided by 8 . answer c" | a ) 3 , b ) 5 , c ) 4 , d ) 6 , e ) 7 | c | subtract(5, reminder(4, 8)) | reminder(n2,n3)|subtract(n1,#0)| | general | C |
if a card is drawn from a well shuffled deck of cards , what is the probability of drawing a black card or a face card ? | "p ( b á ´ œ f ) = p ( b ) + p ( f ) - p ( b â ˆ © f ) , where b denotes black cards and f denotes face cards . p ( b á ´ œ f ) = 26 / 52 + 12 / 52 - 6 / 52 = 8 / 13 answer : d" | a ) 7 / 13 , b ) 15 / 26 , c ) 9 / 12 , d ) 8 / 13 , e ) 29 / 52 | d | add(divide(const_3, const_52), divide(divide(const_52, const_4), const_52)) | divide(const_3,const_52)|divide(const_52,const_4)|divide(#1,const_52)|add(#0,#2)| | probability | D |
there are 20 poles with a constant distance between each pole . a car takes 20 second to reach the 12 th pole . how much will it take to reach the last pole . | "assuming the car starts at the first pole . to reach the 12 th pole , the car need to travel 11 poles ( the first pole does n ' t count , as the car is already there ) . 11 poles 20 seconds 1 pole ( 20 / 11 ) seconds to reach the last ( 20 th ) pole , the car needs to travel 19 poles . 19 pole 19 x ( 20 / 11 ) seconds = 34.5455 seconds answer : b" | a ) 34.4543 , b ) 34.5455 , c ) 34.45128 , d ) 34.51288 , e ) 34.41222 | b | multiply(divide(20, 12), 20) | divide(n1,n2)|multiply(n0,#0)| | physics | B |
there are 7 players in a bowling team with an average weight of 94 kg . if two new players join the team , one weighs 110 kg and the second weighs 60 kg , what will be the new average weight ? | "the new average will be = ( 94 * 7 + 110 + 60 ) / 9 = 92 kgs e is the answer" | a ) 75 kg . , b ) 80 kg . , c ) 86 kg . , d ) 90 kg . , e ) 92 kg . | e | divide(add(multiply(7, 94), add(110, 60)), add(7, const_2)) | add(n2,n3)|add(n0,const_2)|multiply(n0,n1)|add(#0,#2)|divide(#3,#1)| | general | E |
jane and thomas are among the 7 people from which a committee of 3 people is to be selected . how many different possible committees of 3 people can be selected from these 7 people if at least one of either jane or thomas is to be selected ? | "the total number of ways to choose 3 people from 7 is 7 c 3 = 35 . the number of committees without jane or thomas is 5 c 3 = 10 . there are 35 - 10 = 25 possible committees which include jane and / or thomas . the answer is b ." | a ) 20 , b ) 25 , c ) 30 , d ) 35 , e ) 40 | b | add(add(choose(7, const_1), choose(7, const_1)), choose(const_4, const_1)) | choose(n0,const_1)|choose(const_4,const_1)|add(#0,#0)|add(#2,#1)| | probability | B |
there are 456 doctors and nurses in a hospital . if the ratio of the doctors to the nurses is 8 : 11 , then how many nurses are there in the hospital ? | given , the ratio of the doctors to the nurses is 8 : 11 number of nurses = 11 / 19 x 456 = 264 answer : a | a ) 264 , b ) 209 , c ) 57 , d ) 171 , e ) 181 | a | multiply(multiply(8, subtract(11, 8)), 11) | subtract(n2,n1)|multiply(n1,#0)|multiply(n2,#1) | other | A |
a , b , c can complete a piece of work in 21 , 6,12 days . working together , they complete the same work in how many days ? | "a + b + c 1 day work = 1 / 21 + 1 / 6 + 1 / 12 = 25 / 84 a , b , c together will complete the job in 84 / 25 days answer is e" | a ) 2 , b ) 4 / 5 , c ) 7 / 9 , d ) 10 , e ) 84 / 25 | e | divide(const_1, add(add(divide(const_1, 21), divide(const_1, add(const_4, const_2))), divide(const_1, multiply(const_2, add(const_4, const_2))))) | add(const_2,const_4)|divide(const_1,n0)|divide(const_1,#0)|multiply(#0,const_2)|add(#1,#2)|divide(const_1,#3)|add(#4,#5)|divide(const_1,#6)| | physics | E |
jerry ’ s average ( arithmetic mean ) score on the first 3 of 4 tests is 85 . if jerry wants to raise his average by 3 points , what score must he earn on the fourth test ? | "total score on 3 tests = 85 * 3 = 255 jerry wants the average to be = 88 hence total score on 4 tests should be = 88 * 4 = 352 score required on the fourth test = 352 - 255 = 97 option a" | a ) 97 , b ) 89 , c ) 90 , d ) 93 , e ) 95 | a | subtract(multiply(4, add(85, 3)), multiply(85, 3)) | add(n2,n3)|multiply(n0,n2)|multiply(n1,#0)|subtract(#2,#1)| | general | A |
a metallic sphere of radius 12 cm is melted and drawn into a wire , whose radius of cross section is 12 cm . what is the length of the wire ? | "volume of the wire ( in cylindrical shape ) is equal to the volume of the sphere . π ( 12 ) ^ 2 * h = ( 4 / 3 ) π ( 12 ) ^ 3 = > h = 16 cm answer : c" | a ) 6 cm , b ) 14 cm , c ) 16 cm , d ) 23 cm , e ) 29 cm | c | divide(multiply(const_4, divide(power(12, const_3), power(12, const_2))), const_3) | power(n0,const_3)|power(n1,const_2)|divide(#0,#1)|multiply(#2,const_4)|divide(#3,const_3)| | physics | C |
in a sample of 800 high school students in which all students are either freshmen , sophomores , juniors , or seniors , 23 percent are juniors and 75 percent are not sophomores . if there are 160 seniors , how many more freshmen than sophomores are there among the sample of students ? | 200 are sophomores . the number of freshmen is 600 - 160 - 0.23 ( 800 ) = 256 the answer is c . | a ) 42 , b ) 48 , c ) 56 , d ) 64 , e ) 72 | c | subtract(divide(multiply(subtract(subtract(subtract(const_100, multiply(divide(160, 800), const_100)), 23), subtract(const_100, 75)), 800), const_100), divide(multiply(subtract(const_100, 75), 800), const_100)) | divide(n3,n0)|subtract(const_100,n2)|multiply(#0,const_100)|multiply(n0,#1)|divide(#3,const_100)|subtract(const_100,#2)|subtract(#5,n1)|subtract(#6,#1)|multiply(n0,#7)|divide(#8,const_100)|subtract(#9,#4) | gain | C |
the sale price of a trolley bag including the sale tax is rs . 1232 . the rate of sale tax is 10 % . if the shopkeeper has made a profit of 12 % , the cost price of the trolley bag is : | "explanation : 110 % of s . p . = 1232 s . p . = rs . ( 1232 x 100 / 110 ) = rs . 1120 . c . p . = rs ( 100 / 112 x 1120 ) = rs 1000 answer : a" | a ) rs 1000 , b ) rs 1515 , c ) rs 1525 , d ) rs 1250 , e ) none of these | a | divide(subtract(1232, multiply(1232, divide(10, const_100))), add(divide(12, const_100), const_1)) | divide(n1,const_100)|divide(n2,const_100)|add(#1,const_1)|multiply(n0,#0)|subtract(n0,#3)|divide(#4,#2)| | gain | A |
a number x is 18 times another number y . the percentage that y is less than x is | "say y = 1 and x = 18 . then y = 1 is less than x = 18 by ( 18 - 1 ) / 18 * 100 = 17 / 18 * 100 = 94.4 % . answer : d ." | a ) 12.5 % , b ) 87.5 % , c ) 80 % , d ) 94.4 % , e ) 1 % | d | multiply(divide(subtract(18, const_1), 18), const_100) | subtract(n0,const_1)|divide(#0,n0)|multiply(#1,const_100)| | general | D |
the length of a room is 5.5 m and width is 3.75 m . what is the cost of paying the floor by slabs at the rate of rs . 800 per sq . meter . | "explanation : area = 5.5 × 3.75 sq . metre . cost for 11 sq . metre . = rs . 800 hence , total cost = 5.5 × 3.75 × 800 = 5.5 × 3000 = rs . 16500 answer : option d" | a ) rs . 12000 , b ) rs . 19500 , c ) rs . 18000 , d ) rs . 16500 , e ) rs . 17500 | d | multiply(800, multiply(5.5, 3.75)) | multiply(n0,n1)|multiply(n2,#0)| | physics | D |
a starts business with rs . 3500 and after 5 months , b joins with a as his partner . after a year , the profit is divided in the ratio 2 : 3 . what is b ’ s contribution in the capital ? | "sol . let b ’ s capital be rs . x . then , 3500 * 12 / 7 x = 2 / 3 ⇔ 14 x = 126000 ⇔ x = 9000 . answer b" | a ) 6000 , b ) 9000 , c ) 9500 , d ) 12000 , e ) none | b | divide(multiply(multiply(3500, const_12), 3), multiply(subtract(const_12, 5), 2)) | multiply(n0,const_12)|subtract(const_12,n1)|multiply(n3,#0)|multiply(n2,#1)|divide(#2,#3)| | other | B |
x is a positive integer less than 100 . when x is divided by 7 , the remainder is 1 ; when x is divided by 3 , the remainder is 2 . how many x are there ? | "the nubmer which when divided by 7 leaves remainder 1 should be of the form 7 k + 1 this number when divided by 3 leaves remainder 2 . so , ( 7 k + 1 ) - 2 should be divisible by 3 or 7 k - 1 should be divisible by 3 . we now put the values of k starting from 0 to find first number divisible by 3 we find 1 st number at k = 1 thus smallest number will be 7 ( 1 ) + 1 = 8 now , next number will be = 8 + lcm of 37 i . e 29 now we will find number of all such values less than 500 by using the formula for last term of an a . p 8 + ( n - 1 ) 21 = 100 n = 14.42 or n = 14 answer : - a" | a ) 14 , b ) 22 , c ) 23 , d ) 24 , e ) 25 | a | subtract(add(multiply(reminder(7, 100), 3), reminder(3, 100)), reminder(1, 100)) | reminder(n1,n0)|reminder(n3,n0)|reminder(n2,n0)|multiply(n3,#0)|add(#3,#1)|subtract(#4,#2)| | general | A |
a question paper has 2 parts , a & b , each containing 7 questions . if a student has to choose 5 from part a & 4 from part b , in how many ways can he choose the questions ? | "there 7 questions in part a out of which 5 question can be chosen as = 7 c 5 . similarly , 4 questions can be chosen from 7 questions of part b as = 7 c 4 . hence , total number of ways , = 7 c 5 * 7 c 4 = [ 7 ! / ( 2 ! 5 ! ) ] * [ 7 ! / ( 3 ! * 4 ! ) ] = { 21 } * { 7 * 6 * 5 * 4 / ( 4 * 3 * 2 * 1 ) } = 735 . e" | a ) 1100 , b ) 1200 , c ) 1235 , d ) 1354 , e ) 735 | e | divide(multiply(choose(7, 5), choose(7, 4)), 7) | choose(n1,n2)|choose(n1,n3)|multiply(#0,#1)|divide(#2,n1)| | probability | E |
the average of 6 numbers is 30 . if the average of first 4 is 25 and that of last 3 is 35 , the fourth number is : | let the six numbers be , a , b , c , d , e , f . a + b + c + d + e + f = 30 × 6 = 180 - - - - ( 1 ) a + b + c + d = 25 × 4 = 100 - - - - ( 2 ) d + e + f = 35 × 3 = 105 - - - - ( 3 ) add 2 nd and 3 rd equations and subtract 1 st equation from this . d = 25 answer : a | a ) 25 , b ) 26 , c ) 18 , d ) 19 , e ) 10 | a | subtract(multiply(3, 35), subtract(multiply(6, 30), multiply(4, 25))) | multiply(n4,n5)|multiply(n0,n1)|multiply(n2,n3)|subtract(#1,#2)|subtract(#0,#3) | general | A |
one fourth of one third of two fifth of a number is 15 . what will be 40 % of that number | "explanation : ( 1 / 4 ) * ( 1 / 3 ) * ( 2 / 5 ) * x = 15 then x = 15 * 30 = 450 40 % of 450 = 180 answer : option c" | a ) 140 , b ) 150 , c ) 180 , d ) 200 , e ) 220 | c | divide(multiply(divide(15, multiply(multiply(divide(const_1, const_4), divide(const_1, const_3)), divide(const_2, add(const_2, const_3)))), 40), const_100) | add(const_2,const_3)|divide(const_1,const_4)|divide(const_1,const_3)|divide(const_2,#0)|multiply(#1,#2)|multiply(#3,#4)|divide(n0,#5)|multiply(n1,#6)|divide(#7,const_100)| | gain | C |
a shopkeeper buys two articles for rs . 1000 each and then sells them , making 40 % profit on the first article and 40 % loss on second article . find the net profit or loss percent ? | "profit on first article = 40 % of 1000 = 400 . this is equal to the loss he makes on the second article . that , is he makes neither profit nor loss . answer : d" | a ) 200 , b ) 278 , c ) 282 , d ) 400 , e ) 270 | d | multiply(divide(multiply(subtract(add(multiply(divide(const_100, subtract(const_100, 40)), 1000), multiply(divide(const_100, add(const_100, 40)), 1000)), add(1000, 1000)), const_100), add(multiply(divide(const_100, subtract(const_100, 40)), 1000), multiply(divide(const_100, add(const_100, 40)), 1000))), const_100) | add(n1,const_100)|add(n0,n0)|subtract(const_100,n1)|divide(const_100,#2)|divide(const_100,#0)|multiply(n0,#3)|multiply(n0,#4)|add(#5,#6)|subtract(#7,#1)|multiply(#8,const_100)|divide(#9,#7)|multiply(#10,const_100)| | gain | D |
when a random experiment is conducted , the probability that event a occurs is 1 / 5 . if the random experiment is conducted 4 independent times , what is the probability that event a occurs exactly twice ? | "one case is : 1 / 5 * 1 / 5 * 4 / 5 * 4 / 5 = 16 / 625 the total number of possible cases is 4 c 2 = 6 p ( event a occurs exactly twice ) = 6 * ( 16 / 625 ) = 96 / 625 the answer is c ." | a ) 16 / 625 , b ) 56 / 625 , c ) 96 / 625 , d ) 126 / 625 , e ) 156 / 625 | c | subtract(1, divide(const_2, 5)) | divide(const_2,n1)|subtract(n0,#0)| | general | C |
evaluate ( 51 + 52 + 53 + . . . + 100 ) | sn = ( 1 + 2 + 3 + . . . + 50 + 51 + 52 + . . . + 100 ) - ( 1 + 2 + 3 + . . . + 50 ) = ( 50 x 101 ) - ( 25 x 51 ) = ( 5050 - 1275 ) = 3775 . option a | a ) 3775 , b ) 3665 , c ) 3456 , d ) 3459 , e ) 3569 | a | multiply(divide(add(100, 51), const_2), subtract(51, const_1)) | add(n0,n3)|subtract(n0,const_1)|divide(#0,const_2)|multiply(#2,#1) | general | A |
one fourth of a solution that was 10 % sugar by weight was replaced by a second solution resulting in a solution that was 14 percent sugar by weight . the second solution was what percent sugar by weight ? | say the second solution ( which was 1 / 4 th of total ) was x % sugar , then 3 / 4 * 0.1 + 1 / 4 * x = 1 * 0.14 - - > x = 0.26 . alternately you can consider total solution to be 100 liters and in this case you ' ll have : 75 * 0.1 + 25 * x = 100 * 0.14 - - > x = 0.26 . answer : e . | a ) 74 % , b ) 64 % , c ) 42 % , d ) 28 % , e ) 26 % | e | multiply(divide(subtract(multiply(const_100, divide(14, const_100)), multiply(subtract(const_100, multiply(divide(const_1, const_4), const_100)), divide(10, const_100))), multiply(divide(const_1, const_4), const_100)), const_100) | divide(n1,const_100)|divide(n0,const_100)|divide(const_1,const_4)|multiply(#0,const_100)|multiply(#2,const_100)|subtract(const_100,#4)|multiply(#1,#5)|subtract(#3,#6)|divide(#7,#4)|multiply(#8,const_100) | gain | E |
if f ( x ) = 12 - x ^ 2 / 2 and f ( 2 k ) = 6 k , what is one possible value for k ? | "first of all , see thisgmat blog postand check the related lesson linked below for some background on function notation . we can plug anything in for x and get a result . you can find f ( 1 ) , for example , by plugging in 1 where x is , and you would get 12 - 1 / 2 = 11.5 . or we could find f ( 2 ) , which would be 12 - 4 / 2 = 10 . so the notation f ( 2 k ) means that we are going to plug a 2 k in for x everywhere in the formula for f ( x ) . that would be : f ( 2 k ) = 12 - ( 2 k ) ^ 2 / 2 = 12 - 2 k ^ 2 . remember that we have to square both the 2 and the k , to get 4 k 2 . now , this expression , the output , we will set equal to 2 k . 12 - 2 k ^ 2 = 2 k - - > k = - 3 or k = 6 . all the answers are positive , so we choose k = 2 . answer = d" | a ) 2 , b ) 3 , c ) 4 , d ) 6 , e ) 8 | d | add(divide(power(2, 2), 2), const_1) | power(n1,n1)|divide(#0,n1)|add(#1,const_1)| | general | D |
two trains , one from howrah to patna and the other from patna to howrah , start simultaneously . after they meet , the trains reach their destinations after 16 hours and 4 hours respectively . the ratio of their speeds is : | "let us name the trains as a and b . then , ( a ' s speed ) : ( b ' s speed ) = â ˆ š b : â ˆ š a = â ˆ š 4 : â ˆ š 16 = 2 : 4 . answer a" | a ) 2 : 4 , b ) 4 : 3 , c ) 6 : 7 , d ) 9 : 16 , e ) none of these | a | divide(sqrt(4), sqrt(16)) | sqrt(n1)|sqrt(n0)|divide(#0,#1)| | physics | A |
how many digits 2 ^ 400 has ? | "2 ^ 10 = 1.024 * 10 ^ 3 = > 2 ^ 100 = ( 1.024 ) ^ 10 * 10 ^ 120 therefore 121 digits would be my best guess e" | a ) 31 , b ) 35 , c ) 50 , d ) 99 , e ) 121 | e | floor(add(const_1, multiply(divide(log(2), log(const_10)), 400))) | log(n0)|log(const_10)|divide(#0,#1)|multiply(n1,#2)|add(#3,const_1)|floor(#4)| | general | E |
george went to a fruit market with certain amount of money . with this money he can buy either 50 oranges or 40 mangoes . he retains 15 % of the money for taxi fare and buys 15 mangoes . how many oranges can he buy ? | "let the amount of money be 200 let cost of 1 orange be 4 let cost of 1 mango be 5 he decides to retain 15 % of 200 = 30 for taxi fare , so he is left with 170 he buys 20 mangoes ( @ 5 ) so he spends 100 money left is 70 ( 170 - 100 ) no of oranges he can buy = 70 / 4 = > 17,5 so , george can buy 15 oranges . b" | a ) 25 , b ) 17.5 , c ) 20 , d ) 16 , e ) 12 | b | multiply(subtract(subtract(const_1, divide(15, const_100)), divide(15, 50)), 50) | divide(n2,const_100)|divide(n3,n0)|subtract(const_1,#0)|subtract(#2,#1)|multiply(n0,#3)| | general | B |
when a mobile is sold for rs . 24000 , the owner loses 40 % . at what price must that mobile be sold in order to gain 40 % ? | "60 : 24000 = 140 : x x = ( 24000 x 140 ) / 60 = 56000 . hence , s . p . = rs . 56,000 . answer : option b" | a ) 54,000 , b ) 56,000 , c ) 58,000 , d ) 60,000 , e ) 62,000 | b | floor(multiply(divide(divide(divide(multiply(divide(multiply(24000, const_100), subtract(const_100, 40)), add(const_100, 40)), const_100), const_100), 40), const_2)) | add(n1,const_100)|multiply(n0,const_100)|subtract(const_100,n1)|divide(#1,#2)|multiply(#0,#3)|divide(#4,const_100)|divide(#5,const_100)|divide(#6,n1)|multiply(#7,const_2)|floor(#8)| | gain | B |
if 3 women can color 180 m long cloth in 2 days , then 5 women can color 300 m long cloth in ? | "the length of cloth painted by one woman in one day = 180 / 3 × 2 = 30 m no . of days required to paint 300 m cloth by 5 women = 300 / 5 × 30 = 2 days answer : a" | a ) 2 days , b ) 3 days , c ) 5 days , d ) 1 day , e ) 4 days | a | divide(300, multiply(5, divide(180, multiply(3, 2)))) | multiply(n0,n2)|divide(n1,#0)|multiply(n3,#1)|divide(n4,#2)| | physics | A |
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