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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what number is obtained by adding the units digits of 734 ^ 98 and 347 ^ 82 ? | "the units digit of 734 ^ 98 is 6 because 4 raised to the power of an even integer ends in 6 . the units digit of 347 ^ 82 is 9 because powers of 7 end in 7 , 9 , 3 , or 1 cyclically . since 82 is in the form 4 n + 2 , the units digit is 9 . then 6 + 9 = 15 . the answer is e ." | a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | e | subtract(subtract(98, 82), divide(98, add(const_1, const_10))) | add(const_1,const_10)|subtract(n1,n3)|divide(n1,#0)|subtract(#1,#2)| | general | E |
how much more would rs . 30000 fetch , after two years , if it is put at 20 % p . a . compound interest payable half yearly than if is put at 20 % p . a . compound interest payable yearly ? | 30000 ( 11 / 10 ) 4 - 30000 ( 6 / 5 ) 2 = 723 answer : e | a ) 482 , b ) 725 , c ) 992 , d ) 837 , e ) 723 | e | subtract(multiply(power(add(const_1, divide(divide(20, const_100), const_2)), const_4), 30000), multiply(power(add(divide(20, const_100), const_1), const_2), 30000)) | divide(n1,const_100)|add(#0,const_1)|divide(#0,const_2)|add(#2,const_1)|power(#1,const_2)|multiply(n0,#4)|power(#3,const_4)|multiply(n0,#6)|subtract(#7,#5) | gain | E |
find the simple interest on rs . 68,800 at 16 2 / 3 % per annum for 9 months . | "p = rs . 68800 , r = 50 / 3 % p . a and t = 9 / 12 years = 3 / 4 years . s . i . = ( p * r * t ) / 100 = rs . ( 68,800 * ( 50 / 3 ) * ( 3 / 4 ) * ( 1 / 100 ) ) = rs . 8600 answer is a ." | a ) s . 8600 , b ) s . 8000 , c ) s . 7500 , d ) s . 7000 , e ) s . 6500 | a | multiply(multiply(multiply(add(multiply(multiply(multiply(2, 3), const_100), const_100), multiply(multiply(multiply(3, 3), const_100), multiply(add(3, 2), 2))), divide(add(multiply(16, 3), 2), 3)), divide(multiply(3, 3), multiply(2, multiply(2, 3)))), divide(const_1, const_100)) | add(n2,n3)|divide(const_1,const_100)|multiply(n3,n3)|multiply(n2,n3)|multiply(n1,n3)|add(n2,#4)|multiply(n2,#3)|multiply(#3,const_100)|multiply(#2,const_100)|multiply(#0,n2)|divide(#2,#6)|divide(#5,n3)|multiply(#7,const_100)|multiply(#8,#9)|add(#12,#13)|multiply(#14,#11)|multiply(#10,#15)|multiply(#1,#16)| | gain | A |
9873 + x = 13200 , then x is ? | "answer x = 13200 - 9873 = 3327 option : a" | a ) 3327 , b ) 3237 , c ) 3337 , d ) 2337 , e ) none of these | a | subtract(13200, 9873) | subtract(n1,n0)| | general | A |
canister c is 1 / 2 full of water and canister d , which has twice the capacity of canister c , is 1 / 3 full of water . if the water in canister d is poured in canister c until canister c is completely full of water , canister d will still contain what fraction of its capacity of water ? | say canister c has a capacity of 6 liters . it ' s half full , thus there can be poured 3 liters of water . canister b is 12 liters and there are 4 liters of water . we can pour 3 liters from d to c and 1 liter will still be left in d , which is 1 / 12 of its total capacity . answer : c . | a ) 0 , b ) 1 / 36 , c ) 1 / 12 , d ) 1 / 6 , e ) 1 / 4 | c | subtract(divide(1, 3), multiply(divide(1, 2), divide(1, 2))) | divide(n0,n3)|divide(n0,n1)|multiply(#1,#1)|subtract(#0,#2) | general | C |
machine p and machine q are each used to manufacture 330 sprockets . it takes machine p 10 hours longer to produce 330 sprockets than machine q . machine q produces 10 % more sprockets per hour than machine a . how many sprockets per hour does machine a produce ? | "p makes x sprockets per hour . then q makes 1.1 x sprockets per hour . 330 / x = 330 / 1.1 x + 10 1.1 ( 330 ) = 330 + 11 x 11 x = 33 x = 3 the answer is c ." | a ) 5 , b ) 15 , c ) 3 , d ) 95 , e ) 125 | c | divide(subtract(330, divide(330, add(divide(10, const_100), const_1))), 10) | divide(n1,const_100)|add(#0,const_1)|divide(n0,#1)|subtract(n0,#2)|divide(#3,n1)| | gain | C |
a tree of height 36 m is on one edge of a road broke at a certain height . it fell in such a way that the top of the tree touches the other edge of the road . if the breadth of the road is 12 m , then what is the height at which the tree broke ? | let the tree was broken at x meters height from the ground and 36 - x be the length of other part of the tree . from the diagram , ( 36 β x ) 2 = x 2 + 122 ( 36 β x ) 2 = x 2 + 122 β 1296 β 72 x + x 2 = x 2 + 144 β 1296 β 72 x + x 2 = x 2 + 144 β 72 x = 1296 β 144 β 72 x = 1296 β 144 β x = 16 answer : a | a ) 16 , b ) 17 , c ) 18 , d ) 19 , e ) 20 | a | divide(subtract(power(36, const_2), power(12, const_2)), multiply(36, const_2)) | multiply(n0,const_2)|power(n0,const_2)|power(n1,const_2)|subtract(#1,#2)|divide(#3,#0) | physics | A |
if taxi fares were $ 1.00 for the first 1 / 5 mile and $ 0.30 for each 1 / 5 mile there after , then the taxi fare for a 3 - mile ride was | "in 3 miles , initial 1 / 5 mile charge is $ 1 rest of the distance = 3 - ( 1 / 5 ) = 14 / 5 rest of the distance charge = 14 ( 0.3 ) = $ 4.2 ( as the charge is 0.3 for every 1 / 5 mile ) = > total charge for 3 miles = 1 + 4.2 = 5.2 answer is d ." | a ) $ 1.56 , b ) $ 2.40 , c ) $ 3.80 , d ) $ 5.20 , e ) $ 2.80 | d | add(1.00, multiply(subtract(divide(1.00, divide(1, 5)), 1), 0.30)) | divide(n1,n2)|divide(n0,#0)|subtract(#1,n1)|multiply(n3,#2)|add(n0,#3)| | general | D |
what number has a 15 : 1 ratio to the number 10 ? | "15 : 1 = x : 10 x = 10 * 15 x = 150 answer : d" | a ) 130 , b ) 100 , c ) 200 , d ) 150 , e ) 120 | d | multiply(10, 15) | multiply(n0,n2)| | other | D |
a 270 m long train running at the speed of 120 km / hr crosses another train running in opposite direction at the speed of 80 km / hr in 9 sec . what is the length of the other train ? | "answer : option a explanation : relative speed = 120 + 80 = 200 km / hr . = 200 * 5 / 18 = 500 / 9 m / sec . let the length of the other train be x m . then , ( x + 270 ) / 9 = 500 / 9 = > x = 230 . answer a" | a ) 230 , b ) 200 , c ) 250 , d ) 300 , e ) 500 | a | subtract(multiply(multiply(add(120, 80), const_0_2778), 9), 270) | add(n1,n2)|multiply(#0,const_0_2778)|multiply(n3,#1)|subtract(#2,n0)| | physics | A |
x and y started a business by investing rs . 36000 and rs . 42000 respectively after 4 months z joined in the business with an investment of rs . 48000 , then find share of z in the profit of rs . 14080 ? | "ratio of investment , as investments is for different time . investment x number of units of time . ratio of investments x : y : z = 36000 : 42000 : 48000 = > 6 : 7 : 8 . x = 6 x 12 months = 72 , y = 7 x 12 = 84 , z = 8 x 8 = 64 = > 18 : 21 : 16 . ratio of investments = > x : y : z = 18 : 21 : 16 . investment ratio = profit sharing ratio . z = 14080 Γ£ β 16 / 55 = rs . 4096 . share of z in the profit is rs . 4096 . option e" | a ) 3200 , b ) 4000 , c ) 3250 , d ) 3825 , e ) 4096 | e | multiply(multiply(48000, subtract(multiply(const_3, 4), 4)), divide(14080, add(add(multiply(36000, multiply(const_3, 4)), multiply(42000, multiply(const_3, 4))), multiply(48000, subtract(multiply(const_3, 4), 4))))) | multiply(const_3,n2)|multiply(n0,#0)|multiply(n1,#0)|subtract(#0,n2)|add(#1,#2)|multiply(n3,#3)|add(#4,#5)|divide(n4,#6)|multiply(#7,#5)| | gain | E |
there are 280 female managers in a certain company . find the total number of female employees in the company , if 2 / 5 of all the employees are managers and 2 / 5 of all male employees are managers . | as per question stem 2 / 5 m ( portion of men employees who are managers ) + 280 ( portion of female employees who are managers ) = 2 / 5 t ( portion of total number of employees who are managers ) , thus we get that 2 / 5 m + 280 = 2 / 5 t , or 2 / 5 ( t - m ) = 280 , from here we get that t - m = 700 , that would be total number of female employees and the answer ( c ) | a ) 600 , b ) 650 , c ) 700 , d ) 750 , e ) none of these | c | divide(280, divide(2, 5)) | divide(n1,n2)|divide(n0,#0)| | general | C |
two pipes p and q can fill a cistern in 11 and 15 minutes respectively . both are opened together , but at the end of 3 minutes the first is turned off . how much longer will the cistern take to fill ? | "3 / 11 + x / 15 = 1 x = 10 10 / 11 answer : d" | a ) 1 / 8 , b ) 1 / 4 , c ) 2 / 4 , d ) 10 / 11 , e ) 1 / 4 | d | multiply(subtract(const_1, multiply(add(divide(const_1, 11), divide(const_1, 15)), 3)), 15) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|multiply(#2,n2)|subtract(const_1,#3)|multiply(n1,#4)| | physics | D |
what is the speed of the stream if a canoe rows upstream at 6 km / hr and downstream at 10 km / hr | sol . speed of stream = 1 / 2 ( 10 - 6 ) kmph = 2 kmph . answer d | a ) 1 kmph , b ) 4 kmph , c ) 3 kmph , d ) 2 kmph , e ) 1.9 kmph | d | divide(subtract(10, 6), const_2) | subtract(n1,n0)|divide(#0,const_2) | physics | D |
a man rows his boat 90 km downstream and 5 ` km upstream , taking 3 hours each time . find the speed of the stream ? | "speed downstream = d / t = 90 / ( 3 ) = 30 kmph speed upstream = d / t = 51 / ( 3 ) = 17 kmph the speed of the stream = ( 30 - 17 ) / 2 = 7 kmph answer : a" | a ) 7 kmph , b ) 5 kmph , c ) 2 kmph , d ) 8 kmph , e ) 1 kmph | a | divide(subtract(divide(90, 3), divide(5, 3)), const_2) | divide(n0,n2)|divide(n1,n2)|subtract(#0,#1)|divide(#2,const_2)| | physics | A |
it takes joey the postman 1 hours to run a 3 mile long route every day . he delivers packages and then returns to the post office along the same path . if the average speed of the round trip is 6 mile / hour , what is the speed with which joey returns ? | "let his speed for one half of the journey be 3 miles an hour let the other half be x miles an hour now , avg speed = 6 mile an hour 2 * 3 * x / 3 + x = 6 6 x = 6 x + 18 = > x = 18 e" | a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 18 | e | divide(3, subtract(divide(multiply(const_2, 3), 6), 1)) | multiply(n1,const_2)|divide(#0,n2)|subtract(#1,n0)|divide(n1,#2)| | physics | E |
a man purchased 5 blankets @ rs . 100 each , 5 blankets @ rs . 150 each and two blankets at a certain rate which is now slipped off from his memory . but he remembers that the average price of the blankets was rs . 150 . find the unknown rate of two blankets ? | "explanation : 10 * 150 = 1500 5 * 100 + 5 * 150 = 1250 1500 β 1250 = 250 a" | a ) a ) 250 , b ) b ) 350 , c ) c ) 450 , d ) d ) 470 , e ) e ) 500 | a | subtract(multiply(const_10, 150), add(multiply(5, 100), multiply(5, 150))) | multiply(n3,const_10)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|subtract(#0,#3)| | general | A |
set a consists of the integers from 4 to 17 , inclusive , while set b consists of the integers from 6 to 20 , inclusive . how many distinct integers do belong to the both sets at the same time ? | "a = { 4 , 5,6 , 7,8 , 9,10 , 11,12 , . . . 17 } b = { 6 , 7,8 , 9,10 , 11,12 . . . , 20 } thus we see that there are 12 distinct integers that are common to both . e is the correct answer ." | a ) 5 , b ) 7 , c ) 8 , d ) 9 , e ) 12 | e | add(6, 4) | add(n0,n2)| | other | E |
a rectangular tank needs to be coated with insulation . the tank has dimensions of 4 feet , 5 feet , and 3 feet . each square foot of insulation costs $ 20 . how much will it cost to cover the surface of the tank with insulation ? | "the total surface area is 2 ( 3 * 4 + 4 * 5 + 3 * 5 ) = 94 square feet the total cost is 94 * $ 20 = $ 1880 the answer is e ." | a ) $ 1220 , b ) $ 1330 , c ) $ 1550 , d ) $ 1770 , e ) $ 1880 | e | multiply(surface_rectangular_prism(4, 5, 3), 20) | surface_rectangular_prism(n0,n1,n2)|multiply(n3,#0)| | geometry | E |
little john had $ 20.10 . he spent $ 1.05 on sweets and gave to his two friends $ 1.00 each . how much money was left ? | "john spent and gave to his two friends a total of 1.05 + 1.00 + 1.00 = $ 3.05 money left 20.10 - 3.05 = $ 17.05 answer : a" | a ) $ 17.05 , b ) $ 17.55 , c ) $ 17.15 , d ) $ 17.35 , e ) $ 17.25 | a | subtract(20.10, add(1.05, add(1.00, 1.00))) | add(n2,n2)|add(n1,#0)|subtract(n0,#1)| | general | A |
a boat can travel with a speed of 15 km / hr in still water . if the speed of the stream is 5 km / hr . find the time taken by the boat to go 68 km downstream ? | "solution speed downstream = ( 15 + 5 ) km / hr = 20 km / hr . time taken to travel 68 km downstream = ( 68 / 20 ) hrs = 3 hrs 24 minutes . answer d" | a ) 2 hours , b ) 3 hours , c ) 4 hours , d ) 3 hours 24 minutes , e ) none | d | divide(68, add(15, 5)) | add(n0,n1)|divide(n2,#0)| | physics | D |
what is the dividend ? the divisor is 17 , the quotient is 4 and the remainder is 8 . | "divided = divisor * quotient + remainder ? = 17 * 4 + 8 68 + 8 76 ( e . g . answer : a )" | a ) 76 , b ) 67 , c ) 176 , d ) 671 , e ) 0 | a | subtract(multiply(17, 4), 8) | multiply(n0,n1)|subtract(#0,n2)| | general | A |
a certain manufacturer of cake , muffin , and bread mixes has 100 buyers , of whom 50 purchases cake mix , 40 purchase muffin mix , and 17 purchase both cake mix and muffin mix . if a buyer is to be selected at random from the 100 buyers , what is the probability that the buyer selected will be one who purchases neither cake mix nor muffin mix ? | c + m + b - cm - mb - cb - 2 cmb = 100 c - cake buyers , m - muffin and b - bread buyers . cm , mb , cb and cmb are intersecting regions . the question asks for people who have bought only bread mixes = b - cb - mb - 2 cmb has to be found out . 50 + 40 + b - cb - mb - 17 - 2 cmb = 100 b - cb - mb - 2 cmb = 27 hence the probability = 27 / 100 d | a ) 1 / 10 , b ) 3 / 10 , c ) 1 / 2 , d ) 27 / 100 , e ) 9 / 100 | d | divide(subtract(100, subtract(add(50, 40), 17)), 100) | add(n1,n2)|subtract(#0,n3)|subtract(n0,#1)|divide(#2,n0) | other | D |
during a certain two - week period , 50 percent of the movies rented from a video store were comedies , and of the remaining movies rented , there were 9 times as many dramas as action movies . if no other movies were rented during that two - week period and there were a action movies rented , then how many comedies , in terms of a , were rented during that two - week period ? | "movies : 50 % comedies . 50 % remaining genre . now in this 50 % , there are only 2 categories . action movies and drama movies . if action = x ; drama movies = 9 x . total 10 x . 10 x = 50 ; x = 5 action movies : 5 % drama movies : 45 % we can say that out of 100 z , : comedies : 50 z action : 5 z drama : 45 z now action movies were a this means : a = 5 z . z = ( a / 5 ) comedies : 50 z = 50 * ( a / 5 ) 10 a b is the answer ." | a ) 5 a , b ) 10 a , c ) 20 a , d ) 25 a , e ) 28 a | b | floor(divide(50, add(9, const_1))) | add(n1,const_1)|divide(n0,#0)|floor(#1)| | general | B |
a , b and c are partners . a receives 2 / 3 of profits , b and c dividing the remainder equally . a ' s income is increased by rs . 400 when the rate to profit rises from 5 to 7 percent . find the capital of b ? | "a : b : c = 2 / 3 : 1 / 6 : 1 / 6 = 4 : 1 : 1 x * 2 / 100 * 2 / 3 = 400 b capital = 30000 * 1 / 6 = 5000 . answer : d" | a ) 3999 , b ) 7799 , c ) 2500 , d ) 5000 , e ) 2912 | d | divide(multiply(400, const_100), 2) | multiply(n2,const_100)|divide(#0,n0)| | general | D |
naomi drives to the beauty parlor in 60 minutes . on the way back , her average speed is half the average speed as it was to the way to the parlor . how much time will it take naomi to travel two round trips to the beauty parlor ? | s 1 = 2 s 2 since , speed is inversely proportional to time we have , s 1 / s 2 = 2 / 1 = t 2 / t 1 therefore , the time taken for one trip = 60 + 120 = 180 minutes total = 180 * 2 = 360 minutes = 6 hours answer : e | a ) 3 hours . , b ) 4 hours . , c ) 4 hours and 20 minutes . , d ) 5 hours and 50 minutes , e ) 6 hours . | e | divide(60, const_10) | divide(n0,const_10) | general | E |
a train 110 m long passes a man , running at 6 kmph in the direction opposite to that of the train , in 6 seconds . the speed of the train is | speed of train relative to man : 110 / 6 * 18 / 5 km / hr = 66 km / hr let speed of train = x therefore x + 6 = 66 x = 66 - 6 x = 60 km / hr answer : b | a ) 54 kmph , b ) 60 kmph , c ) 66 kmph , d ) 72 kmph , e ) 82 kmph | b | divide(divide(subtract(110, multiply(multiply(6, const_0_2778), 6)), 6), const_0_2778) | multiply(n1,const_0_2778)|multiply(n1,#0)|subtract(n0,#1)|divide(#2,n1)|divide(#3,const_0_2778) | physics | B |
a train travels from new york to chicago , a distance of approximately 840 miles , at an average rate of 60 miles per hour and arrives in chicago at 7 : 00 in evening , chicago time . at what hour in the morning , new york time , did the train depart for chicago ? ( note : chicago time is one hour earlier than new york time ) | "7 : 00 in evening in chicago = 8 : 00 in evening in new york . so , the train was in chicago 8 : 00 in the evening , new york time . the trip took t = d / r = 840 / 60 = 14 hours . therefore , the train depart from new york at 8 : 00 - 14 hours = 6 : 00 in the morning , new york time . answer : d ." | a ) 3 : 00 , b ) 4 : 00 , c ) 5 : 00 , d ) 6 : 00 , e ) 7 : 00 | d | subtract(multiply(add(7, const_1), const_3), divide(840, 60)) | add(n2,const_1)|divide(n0,n1)|multiply(#0,const_3)|subtract(#2,#1)| | physics | D |
the average of runs of a cricket player of 10 innings was 32 . how many runs must he make in his next innings so as to increase his average of runs by 4 ? | "average after 11 innings = 36 required number of runs = ( 36 * 11 ) - ( 32 * 10 ) = 396 - 320 = 76 . answer : d" | a ) 87 , b ) 16 , c ) 10 , d ) 76 , e ) 17 | d | subtract(multiply(add(10, const_1), add(4, 32)), multiply(10, 32)) | add(n0,const_1)|add(n1,n2)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2)| | general | D |
if 85 percent of the test takers taking an old paper and pencil gmat exam answered the first question on a given math section correctly , and 80 percent of the test takers answered the second question correctly , and 5 percent of the test takers answered neither question correctly , what percent answered both correctly ? | "{ total } = { first correctly } + { second correctly } - { both correctly } + { neither correctly } 100 = 85 + 80 - { both correctly } + 5 { both correctly } = 70 . answer : c ." | a ) 60 % , b ) 65 % , c ) 70 % , d ) 75 % , e ) 80 % | c | subtract(add(add(85, 80), 5), const_100) | add(n0,n1)|add(n2,#0)|subtract(#1,const_100)| | other | C |
it is the new year and mandy has made a resolution to lose weight this year . she plans to exercise and do yoga . for exercise she plans to workout at the gym and ride her bicycle in the ratio of 2 : 3 everyday . she will also do yoga in the ratio , yoga : exercise = 2 : 3 . if she does yoga for 25 minutes , how much time will she spend in the gym ? | "the ratio is 2 : 3 = yoga : exer , so ( 25 ) ( 3 / 2 ) = 37.5 minutes for exercise , and 2 : 3 = gym : ride , and ( 2 / 3 ) ( 37.5 ) = 25 minutes in the gym . answer : d" | a ) 35 min , b ) 5 min , c ) 15 min , d ) 25 min , e ) 20 min | d | divide(multiply(25, divide(3, add(2, 3))), multiply(divide(3, add(2, 3)), divide(3, add(2, 3)))) | add(n0,n1)|divide(n1,#0)|multiply(n4,#1)|multiply(#1,#1)|divide(#2,#3)| | physics | D |
a man walks at a rate of 10 mph . after every ten miles , he rests for 6 minutes . how much time does he take to walk 40 miles ? | "to cover 40 miles the man needs ( time ) = ( distance ) / ( rate ) = 40 / 10 = 4 hours = 240 minutes . he will also rest 3 times ( after 10 , 20 , and 30 miles ) , so total resting time = 3 * 6 = 18 minutes . total time = 240 + 18 = 258 minutes . answer : a ." | a ) 258 , b ) 318 , c ) 322 , d ) 324 , e ) 330 | a | add(multiply(6, const_4), multiply(divide(40, 10), const_60)) | divide(n2,n0)|multiply(n1,const_4)|multiply(#0,const_60)|add(#1,#2)| | physics | A |
a train traveling at 100 kmph overtakes a motorbike traveling at 64 kmph in 18 seconds . what is the length of the train in meters ? | "train overtakes a bike means that we are talking about total length of the train . ( train ' s head is close to bike when it started and its tail crosses the bike when it overtakes the bike ) relative speed = 100 - 64 = 36 km / h = 36000 m / h time = 18 seconds distance = speed * time 36000 * 18 / 3600 = 180 meters . a is the answer ." | a ) 180 meters , b ) 1111 meters , c ) 1777 meters , d ) 1822 meters , e ) none of these | a | multiply(multiply(subtract(100, 64), const_0_2778), 18) | subtract(n0,n1)|multiply(#0,const_0_2778)|multiply(n2,#1)| | physics | A |
carl is facing very difficult financial times and can only pay the interest on a $ 30,000 loan he has taken . the bank charges him a quarterly compound rate of 10 % . what is the approximate interest he pays annually ? | "usually , you are given the annual rate of interest and it is mentioned that it is annual rate . the bank charges him a quarterly compounded annual rate of 40 % . here you find per quarter rate as ( 40 / 4 ) % = 10 % i have actually never seen a question with quarter rate given but since this question did not mentionannual rate of interestand since the options did not make sense with 5 % annual rate of interest , it is apparent that the intent was a 10 % quarterly rate . so the bank charges 10 % every quarter and compounds it in the next quarter . had it been a simple quarterly rate , we would have just found 4 * 10 % of 30,000 = $ 12000 as our answer . but since , the interest is compounded , it will be a bit more than $ 12000 . option ( e ) looks correct ." | a ) $ 1200 , b ) $ 2000 , c ) $ 2150 , d ) $ 2500 , e ) $ 12000 | e | subtract(multiply(multiply(const_100, const_100), power(add(const_1, divide(10, const_100)), const_4)), multiply(const_100, const_100)) | divide(n1,const_100)|multiply(const_100,const_100)|add(#0,const_1)|power(#2,const_4)|multiply(#1,#3)|subtract(#4,#1)| | gain | E |
5358 x 51 = ? | "5358 x 51 = 5358 x ( 50 + 1 ) = 5358 x 50 + 5358 x 1 = 267900 + 5358 = 273258 . c )" | a ) 272258 , b ) 272358 , c ) 273258 , d ) 274258 , e ) 274358 | c | multiply(divide(5358, 51), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general | C |
a person ' s present age is two - fifth of the age of his mother . after 8 years , he will be one - half of the age of his mother . what is the present age of the mother ? | "let present age of the mother = 5 x then , present age of the person = 2 x 5 x + 8 = 2 ( 2 x + 8 ) 5 x + 8 = 4 x + 16 x = 8 present age of the mother = 5 x = 40 answer is d ." | a ) 25 , b ) 30 , c ) 35 , d ) 40 , e ) 45 | d | divide(subtract(8, add(const_2, const_3)), subtract(divide(const_1, const_2), divide(const_2, add(const_2, const_3)))) | add(const_2,const_3)|divide(const_1,const_2)|divide(const_2,#0)|subtract(n0,#0)|subtract(#1,#2)|divide(#3,#4)| | general | D |
two men started from the same place walk at the rate of 10 kmph and 12 kmph respectively . what time will they take to be 30 km apart , if they walk in the same direction ? | "to be 2 km apart they take 1 hour to be 10 km apart they take = 1 / 2 * 30 = 15 hours answer is a" | a ) 15 hours , b ) 6 hours , c ) 8 hours , d ) 10 hours , e ) 12 hours | a | divide(subtract(12, 10), 30) | subtract(n1,n0)|divide(#0,n2)| | gain | A |
if x is a positive integer with fewer than 3 digits , what is the probability q that x * ( x + 1 ) is a multiple of either 4 or 5 ? | interesting question ! also one that we should be able to answer very quickly be keeping an eye on our best friends , the answer choices . we know that x belongs to the set { 1 , 2 , 3 , . . . , 99 } . we want to know the probability q that x ( x + 1 ) is a multiple of either 4 or 5 . when will this happen ? if either x or ( x + 1 ) is a multiple of 4 or 5 . since 4 * 5 is 20 , let ' s look at the first 20 numbers to get a rough idea of how often this happens . out of the numbers from 1 to 20 : 4 , 5 , 6 , 8 , 9 , 10 , 11 , 12 , 13 , 15 , 16 , 17 , 20 so , 14 out of the first 20 numbers match our criteria . since : probability = ( # of desired outcomes ) / ( total # of possibilities ) , we guesstimate the answer to be 14 / 20 . since ( e ) is the only answer greater than 1 / 2 , we go with ( e ) . | a ) 4 / 99 , b ) 2 / 25 , c ) 8 / 99 , d ) 49 / 100 , e ) 86 / 99 | e | divide(subtract(add(multiply(divide(const_100, 4), const_2), multiply(divide(const_100, 5), const_2)), 4), subtract(const_100, const_1)) | divide(const_100,n2)|divide(const_100,n3)|subtract(const_100,const_1)|multiply(#0,const_2)|multiply(#1,const_2)|add(#3,#4)|subtract(#5,n2)|divide(#6,#2) | general | E |
in the fifth grade at parkway elementary school there are 470 students . 300 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 = 470 boys = 300 , girls = 170 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 = 170 - 35 = 135 . correct option : b" | a ) 69 . , b ) 135 . , c ) 81 . , d ) 91 . , e ) 108 . | b | subtract(subtract(470, 300), 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 | B |
water boils at 212 Β° f or 100 Β° c and melts at 32 Β° f or 0 Β° c . if the temperature of the particular day is 35 Β° c , it is equal to | let f and c denotes the temparature in fahrenheit anid celcsius respectively . then , ( f - 32 ) / ( 212 - 32 ) = ( c - 0 ) / ( 100 - 0 ) , if c = 35 , then f = 95 . c | a ) 50 Β° f , b ) 76 Β° f , c ) 95 Β° f , d ) 110 Β° f , e ) 120 Β° f | c | add(multiply(divide(35, 100), subtract(212, 32)), 32) | divide(n4,n1)|subtract(n0,n2)|multiply(#0,#1)|add(n2,#2) | physics | C |
if the selling price of 50 articles is equal to the cost price of 40 articles , then the loss or gain percent is : | "c c . p . of each article be re . 1 . then , c . p . of 50 articles = rs . 50 ; s . p . of 50 articles = rs . 40 . loss % = 10 / 50 * 100 = 20 %" | a ) 45 % , b ) 23 % , c ) 20 % , d ) 60 % , e ) 56 % | c | subtract(50, 40) | subtract(n0,n1)| | gain | C |
the average of first 50 non - zero positive integers is | "explanation : sum of first n non - zero positive integers n ( n + 1 ) / 2 so , average of first n non - zero positive integers n ( n + 1 ) / 2 n = ( n + 1 ) / 2 = > ( 50 + 1 ) / 2 = 25.5 answer : b" | a ) 11.5 , b ) 25.5 , c ) 22 , d ) 25 , e ) 27 | b | add(50, const_1) | add(n0,const_1)| | general | B |
if 10 gallons of grape juice are added to 40 gallons of a mixture , which contains 30 percent grape juice then what percent of the resulting mixture is grape juice ? | "official solution : if we start with 40 gallons of a mixture that is 10 % grape juice , then we have : 40 Γ 0.30 = 12 gallons of grape juice . 40 Γ 0.70 = 28 gallons of other components . if we add 10 gallons of grape juice , we will end up with 24 gallons of grape juice and 36 gallons of other components , and we will have a total of 50 gallons of the mixture . so 24 / 50 of the new mixture is grape juice . now we convert this to a percent : percent grape juice = 24 / 50 = 48 / 100 = 48 % . the correct answer is choice ( c )" | a ) 14 % , b ) 25 % , c ) 48 % , d ) 34 % , e ) 50 % | c | multiply(divide(add(multiply(divide(30, const_100), 40), 10), add(40, 10)), const_100) | add(n0,n1)|divide(n2,const_100)|multiply(n1,#1)|add(n0,#2)|divide(#3,#0)|multiply(#4,const_100)| | general | C |
mahesh can do a piece of work in 40 days . he works at it for 20 days and then rajesh finished it in 30 days . how long will y take to complete the work ? | "work done by mahesh in 60 days = 20 * 1 / 40 = 1 / 2 remaining work = 1 - 1 / 2 = 1 / 2 1 / 2 work is done by rajesh in 30 days whole work will be done by rajesh is 30 * 2 = 60 days answer is e" | a ) 45 , b ) 25 , c ) 37 , d ) 41 , e ) 60 | e | divide(const_1, divide(subtract(const_1, multiply(20, divide(const_1, 40))), 30)) | divide(const_1,n0)|multiply(n1,#0)|subtract(const_1,#1)|divide(#2,n2)|divide(const_1,#3)| | physics | E |
the cost of one photocopy is $ 0.02 . however , a 25 % discount is offered on orders of more than 100 photocopies . if steve and danny have to make 80 copies each , how much will each of them save if they submit a single order of 160 copies ? | "if steve and danny submit separate orders , each would be smaller than 100 photocopies , so no discount . each would pay ( 80 ) * ( $ 0.02 ) = $ 1.60 , or together , a cost of $ 3.20 - - - that ' s the combined no discount cost . if they submit things together as one big order , they get a discount off of that $ 3.20 price - - - - 25 % or 1 / 4 of that is $ 0.80 , the discount on the combined sale . they each effective save half that amount , or $ 0.40 . answer = ( b ) ." | a ) $ 0.32 , b ) $ 0.40 , c ) $ 0.45 , d ) $ 0.48 , e ) $ 0.54 | b | divide(subtract(multiply(const_2, multiply(80, 0.02)), multiply(multiply(160, divide(subtract(100, 25), 100)), 0.02)), const_2) | multiply(n0,n3)|subtract(n2,n1)|divide(#1,n2)|multiply(#0,const_2)|multiply(n4,#2)|multiply(n0,#4)|subtract(#3,#5)|divide(#6,const_2)| | gain | B |
a shopkeeper sold an article for rs 2524.36 . approximately what was his profit percent if the cost price of the article was rs 2400 | "explanation : gain % = ( 125.36 * 100 / 2400 ) = 5.2 % = 5 % approx option b" | a ) 4 % , b ) 5 % , c ) 6 % , d ) 7 % , e ) 8 % | b | floor(multiply(const_100, divide(subtract(2524.36, 2400), 2400))) | subtract(n0,n1)|divide(#0,n1)|multiply(#1,const_100)|floor(#2)| | gain | B |
when n is divided by 24 , the remainder is 3 . find thee difference between previous remainder and the remainder when 9 n is divided by 7 ? | "let n = 3 ( leaves a remainder of 3 when divided by 24 ) 9 n = 9 ( 3 ) = 27 , which leaves a remainder of 6 when divided by 7 . difference = 6 - 3 = 3 . answer d" | a ) 2 , b ) 7 , c ) 5 , d ) 3 , e ) 6 | d | subtract(3, reminder(9, 7)) | reminder(n2,n3)|subtract(n1,#0)| | general | D |
divide $ 1000 among a , b in the ratio 1 : 4 . how many $ that b get ? | "sum of ratio terms = 1 + 4 = 5 a = 1000 * 4 / 5 = $ 800 answer is b" | a ) $ 500 , b ) $ 800 , c ) $ 1000 , d ) $ 200 , e ) $ 750 | b | divide(1000, 1) | divide(n0,n1)| | other | B |
find the average of all the numbers between 11 and 34 which are divisible by 5 ? | "average = ( 15 + 20 + 25 + 30 ) 4 = 90 / 4 = 22 answer is c" | a ) 10 , b ) 20 , c ) 22 , d ) 30 , e ) 15 | c | divide(add(add(11, const_4), subtract(34, const_4)), const_2) | add(n0,const_4)|subtract(n1,const_4)|add(#0,#1)|divide(#2,const_2)| | general | C |
a frog can climb up a well at 3 ft per min but due to slipperiness of the well , frog slips down 2 ft before it starts climbing the next minute . if the depth of the well is 57 ft , how much time will the frog take to reach the top ? | explanation : as per given , in 1 min , frog climbs up 3 ft and slips down by 2 ft . so the frog climbs only 1 ft in 1 min so after 54 mins , it would have climbed 54 ft . at the end of 55 mins it climbs up 3 ft to make it 57 ft and come out of the well . once it had reached the destination , it will not slip . so the frog will take only 55 minutes to climb up the well . answer : c | a ) 29 , b ) 27 , c ) 55 , d ) 17 , e ) 10 | c | divide(57, subtract(3, 2)) | subtract(n0,n1)|divide(n2,#0) | general | C |
in a family 11 people eat only vegetarian , 6 people eat only non veg . , 9 people eat both veg and non veg . . how many people eat veg in the family ? | "total people eat veg = only veg + both veg and non veg total = 11 + 9 = 20 answer = a" | a ) 20 , b ) 11 , c ) 9 , d ) 31 , e ) 21 | a | add(11, 9) | add(n0,n2)| | other | A |
ramesh , xyz and rajeev put a partnership . profit is 36000 , if ramesh and xyz ratio is 5 : 4 and xyz and rajeev 8 : 9 . find rajeev ' s share . | ramesh : xyz = 5 : 4 xyz : rajeev = 8 : 9 then ramesh : xyz / rajeev : xyz = 5 : 4 / 9 : 8 ramesh / rajeev = 10 / 9 therefore ramesh : xyz : rajeev = 10 : 8 : 9 then rajeev share = ( 9 / 27 ) * 36000 = 12000 answer : d | a ) 14000 , b ) 15000 , c ) 16000 , d ) 12000 , e ) 17000 | d | multiply(36000, divide(9, add(add(divide(multiply(5, 8), 4), 8), 9))) | multiply(n1,n3)|divide(#0,n2)|add(n3,#1)|add(n4,#2)|divide(n4,#3)|multiply(n0,#4) | other | D |
how many 4 digit numbers are there , if it is known that the first digit is odd , the second is even , the third is prime , the fourth ( units digit ) is divisible by 3 , and the digit 2 can be used only once ? | "4 options for the first digit : 1 , 3 , 5 , 7 , 9 ; 5 options for the second digit : 0 , 2 , 4 , 6 , 8 ; 4 options for the third digit : 2 , 3 , 5 , 7 ; 4 options for the fourth digit : 0 , 3 , 6 , 9 . four digit # possible without the restriction ( about the digit 2 ) : 5 * 5 * 4 * 4 = 400 numbers with two 2 - s , 2 x 2 x 5 * 1 * 1 * 4 = 20 . thus there are 400 - 20 = 380 such numbers . answer : e ." | a ) 20 , b ) 150 , c ) 225 , d ) 300 , e ) 380 | e | subtract(multiply(multiply(add(4, const_1), add(4, const_1)), multiply(4, 4)), multiply(multiply(add(4, const_1), add(4, const_1)), 4)) | add(n0,const_1)|multiply(n0,n0)|multiply(#0,#0)|multiply(#2,#1)|multiply(#2,n0)|subtract(#3,#4)| | physics | E |
how many 3 x 3 x 3 cubes could fit in a 10 x 24 x 16 box ? | the answer is d ) 142 . the 10 x 24 x 16 box has an area of 3840 . if you divide it by 27 ( the total of the 3 x 3 x 3 smaller cube ) you get 142.2 . this means you could fit 142 whole cubes in the larger box . | a ) 150 , b ) 145 , c ) 138 , d ) 142 , e ) 175 | d | volume_rectangular_prism(10, 24, 16) | volume_rectangular_prism(n3,n4,n5)| | geometry | D |
the sum of three consecutive even numbers is 246 . what are the numbers ? | "first x make the first x second x + 2 even numbers , sowe add 2 to get the next third x + 4 add 2 more ( 4 total ) to get the third f + s + t = 246 summeansaddfirst ( f ) plussecond ( s ) plusthird ( t ) ( x ) + ( x + 2 ) + ( x + 4 ) = 246 replace each f , s , and t withwhatwe labeled them x + x + 2 + x + 4 = 246 here the parenthesis are not needed 3 x + 6 = 246 combine like terms x + x + x and 2 + 4 β 6 β 6 subtract 6 fromboth sides 3 x = 240 the variable ismultiplied by 3 3 3 divide both sides by 3 x = 80 our solution for x first 80 replace x in the origional listwith 80 . second ( 80 ) + 2 = 82 the numbers are 80 , 82 , and 84 . third ( 80 ) + 4 = 84 correct answer a" | a ) 80 , 82,84 , b ) 65 , 68,71 , c ) 45 , 47,42 , d ) 100 , 102,49 , e ) 87 , 88,89 | a | add(add(power(add(add(divide(subtract(subtract(246, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(246, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(246, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(246, const_10), const_2), const_4), const_2), const_2))) | subtract(n0,const_10)|subtract(#0,const_2)|divide(#1,const_4)|add(#2,const_2)|power(#2,const_2)|add(#3,const_2)|power(#3,const_2)|add(#5,const_2)|add(#4,#6)|power(#5,const_2)|power(#7,const_2)|add(#9,#10)|add(#11,#8)| | physics | A |
log 3 n + log 6 n what is 3 digit number n that will be whole number | "no of values n can take is 1 6 ^ 3 = 216 answer : e" | a ) 629 , b ) 729 , c ) 829 , d ) 929 , e ) 216 | e | power(6, 3) | power(n1,n0)| | other | E |
how many digits will be there to the right of the decimal point in the product of 98 and . 08216 ? | "product of 98 and . 08216 is 8.05168 . therefore number of digits to right of decimal point is 5 answer is a ." | a ) 5 , b ) 6 , c ) 9 , d ) 7 , e ) 8 | a | subtract(subtract(const_100, 98), const_1) | subtract(const_100,n0)|subtract(#0,const_1)| | general | A |
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 $ 109.90 , what was the total value of the item ? | let x be the value of the item . 0.07 * ( x - 1000 ) = 109.90 x = 2570 the answer is d . | a ) $ 1940 , b ) $ 2150 , c ) $ 2360 , d ) $ 2570 , e ) $ 2780 | d | add(const_1000, divide(109.9, divide(7, const_100))) | divide(n0,const_100)|divide(n2,#0)|add(#1,const_1000) | general | D |
if x > 3000 , then the value of ( 3 x ) / ( 2 x ^ 1.21 - 1111 ^ 1.3 ) is closest to ? | "assume x = 3002 ( 3 x ) / ( 2 x ^ 1.21 - 1111 ^ 1.3 ) = 9006 / ( 2 * ( 3002 ) ^ 1.21 - 1111 ^ 1.3 ) = 9006 / 23154 = = 4 / 10 d" | a ) 1 / 6 , b ) 1 / 3 , c ) 10 / 21 , d ) 4 / 10 , e ) 3 / 2 | d | multiply(subtract(const_2, 2), const_2) | subtract(const_2,n2)|multiply(#0,const_2)| | general | D |
if 2994 Γ· 14.5 = 172 , then 29.94 Γ· 1.45 = ? | "29.94 / 1.45 = 299.4 / 14.5 = ( 2994 / 14.5 ) x 1 / 10 ) [ here , substitute 172 in the place of 2994 / 14.5 ] = 172 / 10 = 17.2 answer is a ." | a ) 17.2 , b ) 18.2 , c ) 19.2 , d ) 15.2 , e ) 16.2 | a | divide(29.94, 1.45) | divide(n3,n4)| | general | A |
the ratio of male to female in a class is 2 : 7 . the career preference of the students in the class are to be represented in a circle graph . if the area of the graph allocated to each career preference is to be proportional to the number of students who have that career preference , how many degrees of the circle should be used to represent a career that is preferred by one third of the males and two - third of the females in the class ? | here is my approach = > males = > 2 x and females = 7 x = > total = 9 x now 9 x = > 360 therefore 16 x / 3 = > 213 degree . p . s = > 16 x / 3 is nothing but total number of students with the given preference answer e | ['a ) a ) 160 degree', 'b ) b ) 168 degree', 'c ) c ) 191 degree', 'd ) d ) 192 degree', 'e ) e ) 213 degree'] | e | add(multiply(const_360, divide(divide(const_2, add(2, 7)), const_3)), multiply(const_360, multiply(divide(7, add(2, 7)), divide(const_2, const_3)))) | add(n0,n1)|divide(const_2,const_3)|divide(const_2,#0)|divide(n1,#0)|divide(#2,const_3)|multiply(#3,#1)|multiply(#4,const_360)|multiply(#5,const_360)|add(#6,#7) | geometry | E |
two trains are moving in opposite directions @ 60 kmph and 90 kmph . their length are 1.10 km and 0.9 km . the time taken by the slower train to cross faster train in second is ? | "relative speed = 60 + 90 = 150 * 5 / 18 = 125 / 3 m / s distance covered = 1.10 + 0.9 = 2 km = 2000 m required time = 2000 * 3 / 125 = 48 sec answer is c" | a ) 25 sec , b ) 30 sec , c ) 48 sec , d ) 36 sec , e ) 40 sec | c | divide(add(1.10, 0.9), multiply(add(const_100.0, 90), const_0_2778)) | add(n2,n3)|add(const_100.0,n1)|multiply(#1,const_0_2778)|divide(#0,#2)| | physics | C |
sachin is younger than rahul by 6 years . if the ratio of their ages is 7 : 9 , find the age of sachin | "if rahul age is x , then sachin age is x - 6 , so ( x - 6 ) / x = 7 / 9 = > 9 x - 42 = 7 x = > 2 x = 42 = > x = 21 so sachin age is 21 - 6 = 15 answer : c" | a ) 24.58 , b ) 14 , c ) 15 , d ) 24.9 , e ) 24.1 | c | multiply(divide(6, subtract(9, 7)), 7) | subtract(n2,n1)|divide(n0,#0)|multiply(n1,#1)| | other | C |
the average student age of a certain class which has 45 students is 14 . if one student aged 15 go to another class and the age of the class teacher is included the average changes to 14.66 . what is the age of class teacher ? | sum of ages of class before replacement = 45 x 14 = 630 sum of ages of class without teacher = 630 - 15 = 615 sum of ages of class after replacement = 45 x 14.66 = 660 age of teacher = 660 - 615 = 45 answer : e | a ) 30 , b ) 35 , c ) 38 , d ) 40 , e ) 45 | e | subtract(multiply(45, 14.66), subtract(multiply(45, 14), 15)) | multiply(n0,n3)|multiply(n0,n1)|subtract(#1,n2)|subtract(#0,#2) | general | E |
a jogger is running at 9 kmph alongside a railway track in 240 meters ahead of the engine of a 120 meters long train . the train is running at 45 kmph in the same direction . how much time does it take for the train to pass the jogger ? | explanation : distance to be covered = 240 + 120 = 360 m relative speed = 36 km / hr = 36 Γ 10 / 36 = 10 m / s time = distance / speed = 360 / 10 = 36 seconds answer : option b | a ) 46 , b ) 36 , c ) 18 , d ) 22 , e ) 23 | b | divide(add(240, 120), divide(multiply(subtract(45, 9), const_1000), const_3600)) | add(n1,n2)|subtract(n3,n0)|multiply(#1,const_1000)|divide(#2,const_3600)|divide(#0,#3) | physics | B |
working together , jose and jane can complete an assigned task in 15 days . however , if jose worked alone and complete half the work and then jane takes over the task and completes the second half of the task , the task will be completed in 45 days . how long will jose take to complete the task if he worked alone ? assume that jane is more efficient than jose | "assume : jose does 1 job in x days , so jose does 1 / x job in a day jane does 1 job in y days , so jane does 1 / y job in a day together , they does ( x + y ) / xy job in a day . this is equals to 1 / 20 . so ( x + y ) / xy = 1 / 15 15 ( x + y ) = xy next , we ' re told 1 job takes 45 days to complete if jose and jane each does half the work . so since jose does 1 job in x days , he wil need x / 2 days to do half the job . jane similarly will need y / 2 days to do the other half . x / 2 + y / 2 = 45 x + y = 90 so xy = 1350 the answer choices are : 25 days 30 days 60 days 65 days 36 days from the answer choices , so i ' ll go for 50 days for jose and 27 days for jane . answer : c" | a ) 26 days , b ) 31 days , c ) 50 days , d ) 65 days , e ) 36 days | c | multiply(const_3, 15) | multiply(n0,const_3)| | physics | C |
a leak in the bottom of a tank can empty the full tank in 6 hours . an inlet pipe fills water at the rate of 4 litres a minute . when the tank is full , the inlet is opened and due to the leak the tank is empty in 8 hours . the capacity of the tank ( in litres ) is | "explanation : work done by the inlet in 1 hour = 1 / 6 β 1 / 8 = 1 / 24 work done by inlet in 1 min = 1 / 24 β 1 / 60 = 1 / 1440 = > volume of 1 / 1440 part = 4 liters volume of whole = ( 1440 * 4 ) litres = 5760 litres . option c" | a ) 5780 litres , b ) 5770 litres , c ) 5760 litres , d ) 5750 litres , e ) 5740 litres | c | divide(multiply(4, multiply(8, const_60)), subtract(divide(multiply(8, const_60), multiply(6, const_60)), const_1)) | multiply(n2,const_60)|multiply(n0,const_60)|divide(#0,#1)|multiply(n1,#0)|subtract(#2,const_1)|divide(#3,#4)| | physics | C |
an equilateral triangle and three squares are combined as shown above , forming a shape of area 48 + 4 β 3 . what is the perimeter of the shape formed by the triangle and squares ? | triangle area = root ( 3 ) s ^ 2 / 4 area of 3 squares together = 3 s ^ 2 root ( 3 ) s ^ 2 / 4 + 3 s ^ 2 = 48 + 4 root ( 3 ) root ( 3 ) / 4 s ^ 2 = 4 root ( 3 ) s ^ 2 = 16 = > s = 4 there are 3 sides of each square = 3 ( 4 ) ( 3 ) = 36 option c | ['a ) 18', 'b ) 27', 'c ) 36', 'd ) 48', 'e ) 64'] | c | multiply(power(3, const_2), 4) | power(n2,const_2)|multiply(n1,#0) | geometry | C |
two diesel trains of length 120 m and 280 m are running towards each other on parallel lines at 42 kmph and 30 kmph respectively . in what time will they be clear of each other from the moment they meet ? | d relative speed = ( 42 + 30 ) * 5 / 18 = 4 * 5 = 20 mps . distance covered in passing each other = 120 + 280 = 400 m . the time required = d / s = 400 / 20 = 20 sec . | a ) 10 sec , b ) 30 sec , c ) 40 sec , d ) 20 s , e ) 50 sec | d | divide(add(120, 280), divide(multiply(add(42, 30), const_1000), const_3600)) | add(n0,n1)|add(n2,n3)|multiply(#1,const_1000)|divide(#2,const_3600)|divide(#0,#3) | physics | D |
the least number which when divided by 5 , 67 and 8 leaves a remainder 3 , but when divided by 9 leaves no remainder , is : | solution l . c . m . of 5 , 6 , 7 , 8 = 840 . so , required number is of the form 840 k + 3 . least value of k for which ( 840 k + 3 ) is divisible by 9 is k = 2 . so , required number = ( 840 Γ 2 + 3 ) = 1683 . answer b | a ) 1677 , b ) 1683 , c ) 2523 , d ) 3363 , e ) none of these | b | add(multiply(multiply(multiply(5, add(5, const_1)), add(5, const_2)), 8), 3) | add(n0,const_2)|add(n0,const_1)|multiply(n0,#1)|multiply(#0,#2)|multiply(n2,#3)|add(n3,#4) | general | B |
how many multiples of 4 are there between 16 and 112 , inclusive ? | "the multiples of 4 are from 4 * 4 up to 4 * 28 . 28 - 4 + 1 = 25 . the answer is e ." | a ) 21 , b ) 22 , c ) 23 , d ) 24 , e ) 25 | e | add(divide(subtract(112, 16), 4), const_1) | subtract(n2,n1)|divide(#0,n0)|add(#1,const_1)| | general | E |
if the function q is defined by the formula q = 5 w / ( 4 vf ( z ^ 2 ) ) , by what factor will q be multiplied if w is quadrupled , f is doubled , and z is tripled ? | "we just need to find the factor thats all , w - > quadrupled - > 4 w f - > doubled - > 2 f z - > tripled - > 3 z hence , z ^ 2 = 9 z ^ 2 w is in numerator , and f * z in denominator . hence , additional factor being introduced = 4 / 2 * 9 = 4 / 18 = 2 / 9 = b" | a ) 1 / 9 , b ) 2 / 9 , c ) 4 / 9 , d ) 3 / 9 , e ) 2 / 27 | b | divide(4, multiply(2, power(const_3, 2))) | power(const_3,n2)|multiply(n2,#0)|divide(n1,#1)| | general | B |
john purchased a grinder and a mobile for rs . 15000 & rs . 8000 respectively . he sold the grinder at a loss of 4 % and the mobile phone at a profit of 15 % . overall how much he make a profit . | "let the sp of the refrigerator and the mobile phone be rs . r and rs . m respectively . r = 15000 ( 1 - 4 / 100 ) = 15000 - 600 m = 8000 ( 1 + 15 / 100 ) = 8000 + 120 total sp - total cp = r + m - ( 15000 + 8000 ) = - 600 + 1200 = rs . 600 as this is positive , an overall profit of rs . 600 was made . c" | a ) s . 90 , b ) s . 620 , c ) s . 600 , d ) s . 650 , e ) s . 690 | c | subtract(multiply(15, divide(8000, const_100)), multiply(4, divide(15000, const_100))) | divide(n1,const_100)|divide(n0,const_100)|multiply(n3,#0)|multiply(n2,#1)|subtract(#2,#3)| | gain | C |
in a bus left side are 15 seats available , 3 few seats in right side because in rear exit door . each seat hold 3 people . in addition , there is a seat back can sit 10 people all together . how many people can sit in a bus ? | right side = 15 seat left side = 15 - 3 ( 3 few seat in right side ) = 12 seat total = 15 + 12 = 27 people can seat in 27 seat = 27 * 3 = 81 people can seat in last seat = 10 total people can seat = 81 + 10 = 91 answer : c | a ) 52 , b ) 49 , c ) 91 , d ) 88 , e ) 66 | c | add(multiply(add(15, subtract(15, 3)), 3), 10) | subtract(n0,n1)|add(n0,#0)|multiply(n1,#1)|add(n3,#2) | general | C |
a and b go around a circular track of length 900 m on a cycle at speeds of 36 kmph and 54 kmph . after how much time will they meet for the first time at the starting point ? | "time taken to meet for the first time at the starting point = lcm { length of the track / speed of a , length of the track / speed of b } = lcm { 900 / ( 36 * 5 / 18 ) , 900 / ( 54 * 5 / 18 ) } = lcm ( 90 , 60 ) = 180 sec . answer : d" | a ) 120 sec , b ) 165 sec , c ) 186 sec , d ) 180 sec , e ) 168 sec | d | divide(900, subtract(multiply(54, const_0_2778), multiply(36, const_0_2778))) | multiply(n2,const_0_2778)|multiply(n1,const_0_2778)|subtract(#0,#1)|divide(n0,#2)| | physics | D |
the difference in compound interest earned on a deposit ( compounded annually ) in year 1 and year 2 is $ 40 . had the interest rate been 3 times its present value , the difference c would have been how much ? | case 1 : deposit = $ x ; rate of increase = r . interest yearned in 1 year = xr . deposit in 1 year = x + xr . interest yearned in 2 year = ( x + xr ) r . the difference c = ( x + xr ) r - xr = xr ^ 2 = 40 . case 2 : deposit = $ x ; rate of increase = 3 r . interest yearned in 1 year = x ( 3 r ) . deposit in 1 year = x + 3 xr . interest yearned in 2 year = ( x + 3 xr ) 3 r . the difference = ( x + 3 xr ) 3 r - 3 xr = 9 xr ^ 2 . since from case 1 we know that xr ^ 2 = 40 , then 9 xr ^ 2 = 9 * 40 = 360 . answer : d . | a ) 40 / 3 , b ) 40 , c ) 120 , d ) 360 , e ) 420 | d | multiply(multiply(3, 3), 40) | multiply(n3,n3)|multiply(n2,#0) | general | D |
if neha is 10 both sonali and priyanka is 15 both sadaf and tanu is 10 . how much is prinka by the same system ? | c prinka is 10 , in a system that awards five for each vowel . | a ) 8 , b ) 11 , c ) 10 , d ) 15 , e ) 6 | c | multiply(subtract(15, 10), const_2) | subtract(n1,n0)|multiply(#0,const_2) | other | C |
the difference between simple and compound interest on rs . 1200 for one year at 10 % per annum reckoned half - yearly is ? | "s . i . = ( 1200 * 10 * 1 ) / 100 = rs . 120 c . i . = [ 1200 * ( 1 + 5 / 100 ) 2 - 1200 ] = rs . 123 difference = ( 123 - 120 ) = rs . 3 . answer : b" | a ) rs . 7 , b ) rs . 3 , c ) rs . 8 , d ) rs . 1 , e ) rs . 4 | b | multiply(subtract(power(add(divide(divide(10, const_2), const_100), const_1), const_2), add(divide(10, const_100), const_1)), 1200) | divide(n1,const_2)|divide(n1,const_100)|add(#1,const_1)|divide(#0,const_100)|add(#3,const_1)|power(#4,const_2)|subtract(#5,#2)|multiply(n0,#6)| | gain | B |
tourist purchased a total of 30 travelers checks in $ 50 and $ 100 denominations . the total worth of the travelers checks is $ 1800 . how many checks of $ 50 denominations can he spend so that average amount ( arithmetic mean ) of the remaining travelers checks is $ 75 ? | you could set - up a quick table and brute force the answer . a 4 * 50 200 1800 - 200 1600 26 61.54 b 18 * 50 600 1800 - 900 0900 12 75.00 c 15 * 50 750 1800 - 750 1050 15 70.00 d 20 * 50 1000 1800 - 1000 800 10 80.00 e 24 * 50 1200 1800 - 1200 600 6 100.00 answer is b | a ) 4 , b ) 18 , c ) 15 , d ) 20 , e ) 24 | b | divide(subtract(multiply(75, 30), 1800), subtract(75, 50)) | multiply(n0,n5)|subtract(n5,n1)|subtract(#0,n3)|divide(#2,#1) | general | B |
how many trailing zeroes does 53 ! + 54 ! have ? | 53 ! + 54 ! = 53 ! + 54 * 53 ! = 53 ! ( 1 + 54 ) = 53 ! * 55 number of trailing 0 s in 53 ! = number of 5 s in the expansion of 53 ! = 10 + 2 = 12 there is 1 more 5 in 55 . hence , total number of trailing 0 s = 12 + 1 = 13 answer ( b ) . in most cases , when we are adding multiple terms , all of which have trailing 0 s , the sum will have as many trailing 0 s as that in the lowest term . example : 20 + 2300 + 34210000 = 34212320 - > one 0 because lowest term 20 has one 0 . this case has an exception because of an additional 5 in 55 . answer is b | a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | b | add(divide(add(54, const_1), add(const_1, const_4)), const_2) | add(n1,const_1)|add(const_1,const_4)|divide(#0,#1)|add(#2,const_2) | other | B |
how many books each of volume 200 meter cube can be packed into a crate of volume 4000 meter cube ? | "gud question with a simple concept . in geo if we want to insert one shape into another we need to know the dimensions of the two shapes . in above with volume given , we can come up with different shapes , so we cant know the answer for ex : 4000 m 3 can be 400 * 10 or 40 * 100 or just 4000 * 1 we do n ' t know , so we cant calculate answer : e" | a ) 20 , b ) 40 , c ) 60 , d ) 80 , e ) can not be determined | e | multiply(4000, const_1) | multiply(n1,const_1)| | geometry | E |
a shopkeeper buys two articles for rs . 1000 each and then sells them , making 75 % profit on the first article and 75 % loss on second article . find the net profit or loss percent ? | "profit on first article = 75 % of 1000 = 750 . 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 ) 768 , c ) 276 , d ) 750 , e ) 279 | d | multiply(divide(multiply(subtract(add(multiply(divide(const_100, subtract(const_100, 75)), 1000), multiply(divide(const_100, add(const_100, 75)), 1000)), add(1000, 1000)), const_100), add(multiply(divide(const_100, subtract(const_100, 75)), 1000), multiply(divide(const_100, add(const_100, 75)), 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 |
what sum of money put at c . i amounts in 2 years to rs . 8820 and in 3 years to rs . 9261 ? | "explanation : 8820 - - - - 441 100 - - - - ? = > 5 % x * 105 / 100 * 105 / 100 = 8820 x * 1.1025 = 8820 x = 8820 / 1.1025 = > 8000 answer is a" | a ) rs . 8000 , b ) rs . 8400 , c ) rs . 7500 , d ) rs . 7800 , e ) none of these | a | divide(8820, power(add(subtract(divide(9261, 8820), const_1), const_1), 2)) | divide(n3,n1)|subtract(#0,const_1)|add(#1,const_1)|power(#2,n0)|divide(n1,#3)| | general | A |
a textile manufacturing firm employees 70 looms . it makes fabrics for a branded company . the aggregate sales value of the output of the 70 looms is rs 5 , 00,000 and the monthly manufacturing expenses is rs 1 , 50,000 . assume that each loom contributes equally to the sales and manufacturing expenses are evenly spread over the number of looms . monthly establishment charges are rs 75000 . if one loom breaks down and remains idle for one month , the decrease in profit is : | explanation : profit = 5 , 00,000 Γ’ Λ β ( 1 , 50,000 + 75,000 ) = rs . 2 , 75,000 . since , such loom contributes equally to sales and manufacturing expenses . but the monthly charges are fixed at rs 75,000 . if one loan breaks down sales and expenses will decrease . new profit : - = > 500000 Γ£ β ( 69 / 70 ) Γ’ Λ β 150000 Γ£ β ( 69 / 70 ) Γ’ Λ β 75000 . = > rs 2 , 70,000 . decrease in profit = > 2 , 75,000 Γ’ Λ β 2 , 70,000 = > rs . 5,000 . answer : d | a ) 13000 , b ) 7000 , c ) 10000 , d ) 5000 , e ) none of these | d | subtract(subtract(multiply(multiply(5, const_1000), const_100), add(multiply(75000, const_2), 75000)), subtract(subtract(multiply(divide(subtract(70, 1), 70), multiply(multiply(5, const_1000), const_100)), multiply(multiply(75000, const_2), divide(subtract(70, 1), 70))), 75000)) | multiply(n2,const_1000)|multiply(n6,const_2)|subtract(n0,n4)|add(n6,#1)|divide(#2,n0)|multiply(#0,const_100)|multiply(#4,#5)|multiply(#4,#1)|subtract(#5,#3)|subtract(#6,#7)|subtract(#9,n6)|subtract(#8,#10) | general | D |
every student of a certain school must play an instrument . in last year , 1 / 2 of the students picked a woodwind instrument , 2 / 5 of the students picked a brass instrument , and all of the other students took percussion . in this year , 1 / 2 of the students who play woodwind and 1 / 4 of the students who play brass instruments left school , other students did not leave , and no fresh student come in . what fraction of all students play either a woodwind or brass instrument ? | lets pick smart numbers . total number of students : 20 woodwind ( 1 / 2 ) : 10 brass ( 2 / 5 ) : 8 percussion ( 1 / 10 ) : 2 after leaving school woodwind : 5 brass : 6 percussion : 2 new total number of students : 13 woodwind and brass : 11 answer 11 / 13 or d | a ) 7 / 13 , b ) 4 / 5 , c ) 9 / 13 , d ) 11 / 13 , e ) 7 / 10 | d | divide(add(subtract(subtract(1, divide(const_1, const_2)), divide(divide(const_1, const_2), 2)), subtract(divide(const_2, 5), divide(divide(const_2, 5), 4))), subtract(subtract(1, subtract(subtract(1, divide(const_1, const_2)), divide(divide(const_1, const_2), 2))), subtract(subtract(1, divide(const_1, const_2)), divide(const_2, 5)))) | divide(const_1,const_2)|divide(const_2,n3)|divide(#0,n1)|divide(#1,n7)|subtract(n0,#0)|subtract(#4,#2)|subtract(#1,#3)|subtract(#4,#1)|add(#5,#6)|subtract(n0,#5)|subtract(#9,#7)|divide(#8,#10) | general | D |
a sum fetched a total simple interest of rs . 4016.25 at the rate of 1 % p . a . in 3 years . what is the sum ? | "principal = ( 100 * 4016.25 ) / ( 1 * 3 ) = rs . 133875 . answer : d" | a ) 122762 , b ) 132877 , c ) 122882 , d ) 133875 , e ) 132887 | d | divide(divide(multiply(4016.25, const_100), 1), 3) | multiply(n0,const_100)|divide(#0,n1)|divide(#1,n2)| | gain | D |
the divisor is 21 , the quotient is 14 and the remainder is 7 . what is the dividend ? | d = d * q + r d = 21 * 14 + 7 d = 294 + 7 d = 301 | a ) 201 , b ) 394 , c ) 302 , d ) 301 , e ) 294 | d | add(multiply(21, 14), 7) | multiply(n0,n1)|add(n2,#0) | general | D |
excluding stoppages , the average speed of a bus is 50 km / hr and including stoppages , the average speed of the bus is 40 km / hr . for how many minutes does the bus stop per hour ? | "in 1 hr , the bus covers 50 km without stoppages and 40 km with stoppages . stoppage time = time take to travel ( 50 - 40 ) km i . e 10 km at 50 km / hr . stoppage time = 10 / 50 hrs = 12 min answer : a" | a ) 12 min , b ) 18 min , c ) 16 min , d ) 20 min , e ) 26 min | a | subtract(multiply(const_1, const_60), multiply(divide(40, 50), const_60)) | divide(n1,n0)|multiply(const_1,const_60)|multiply(#0,const_60)|subtract(#1,#2)| | general | A |
if a is an integer greater than 6 but less than 17 and b is an integer greater than 3 but less than 29 , what is the range of a / b ? | "the way to approach this problem is 6 < a < 17 and 3 < b < 29 minimum possible value of a is 7 and maximum is 16 minimum possible value of b is 4 and maximum is 28 range = max a / min b - min a / max b ( highest - lowest ) 16 / 4 - 7 / 28 = 15 / 4 hence a" | a ) 15 / 4 , b ) 13 / 2 , c ) 9 / 7 , d ) 1 / 5 , e ) 7 / 6 | a | subtract(divide(subtract(17, const_1), add(3, const_1)), divide(add(6, const_1), subtract(29, const_1))) | add(n2,const_1)|add(n0,const_1)|subtract(n1,const_1)|subtract(n3,const_1)|divide(#2,#0)|divide(#1,#3)|subtract(#4,#5)| | general | A |
the point a and point b is 30 miles apart , the point b and point c is 25 miles apart . a car travels from point a to point b in 30 min and point b to point c in 40 min . what is the average speed of the car ? | average speed = total distance covered / total time taken = 55 miles / 70 min = 0.79 miles / min = 47.14 miles / hour answer : c | a ) 34.15 , b ) 40.35 , c ) 47.14 , d ) 50.12 , e ) 56.66 | c | multiply(divide(add(divide(30, 30), divide(25, 40)), const_2), const_60) | divide(n0,n0)|divide(n1,n3)|add(#0,#1)|divide(#2,const_2)|multiply(#3,const_60) | physics | C |
x and y are positive integers . when x is divided by 17 , the remainder is 3 , and when x is divided by 19 , the remainder is 1 . when y is divided by 14 , the remainder is 5 , and when y is divided by 21 , the remainder is 12 . what is the least possible value of y - x ? | when x is divided by 17 , the remainder is 3 : so , the possible values of x are : 3 , 20 , 37,54 , etc . when x is divided by 19 , the remainder is 1 : so , the possible values of x are : 1,20 . . . stop . since both lists include 20 , the smallest possible value of x is 20 . when y is divided by 14 , the remainder is 5 : so , the possible values of y are : 5 , 19 , 33,47 etc . when y is divided by 21 , the remainder is 12 : so , the possible values of y are : 12 , 33 , . . . stop . since both lists include 33 , the smallest possible value of y is 33 since the smallest possible values of x and y are 20 and 33 respectively , the smallest possible value of y - x is 13 . so , b is the correct answer to the original question . | a ) 15 , b ) 13 , c ) 11 , d ) 14 , e ) 18 | b | subtract(add(multiply(14, const_2), 5), add(3, 17)) | add(n0,n1)|multiply(n4,const_2)|add(n5,#1)|subtract(#2,#0) | general | B |
suraj has a certain average of runs for 14 innings . in the 15 th innings he scores 140 runs thereby increasing his average by 8 runs . what is his average after the 15 th innings ? | "to improve his average by 8 runs per innings he has to contribute 14 x 8 = 112 runs for the previous 14 innings . thus , the average after the 15 th innings = 140 - 112 = 28 . answer : b" | a ) 48 , b ) 28 , c ) 36 , d ) 72 , e ) 27 | b | divide(subtract(140, multiply(14, 8)), subtract(15, 14)) | multiply(n0,n3)|subtract(n1,n0)|subtract(n2,#0)|divide(#2,#1)| | general | B |
andrew purchased 11 kg of grapes at the rate of 98 per kg and 7 kg of mangoes at the rate of 50 per kg . how much amount did he pay to the shopkeeper ? | "cost of 11 kg grapes = 98 Γ 11 = 1078 . cost of 7 kg of mangoes = 50 Γ 7 = 350 . total cost he has to pay = 1078 + 350 = 1428 c" | a ) 1000 , b ) 1055 , c ) 1428 , d ) 1500 , e ) 1080 | c | add(multiply(11, 98), multiply(7, 50)) | multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)| | gain | C |
a certain car traveled twice as many miles from town a to town b as it did from town b to town c . from town a to town b , the car averaged 12 miles per gallon , and from town b to town c , the car averaged 16 miles per gallon . what is the average miles per gallon that the car achieved on its trip from town a through town b to town c ? | "ans is c given d _ ab = 2 * d _ bc let d _ ab = d and d _ bc = x so d = 2 x for average miles per gallon = ( d + x ) / ( ( d / 12 ) + ( x / 16 ) ) = 14.4 ( formula avg speed = total distance / total time )" | a ) 13 , b ) 13.5 , c ) 14.4 , d ) 14.5 , e ) 15 | c | divide(add(multiply(16, const_10), divide(multiply(16, const_10), const_2)), add(divide(multiply(16, const_10), 12), divide(divide(multiply(16, const_10), const_2), 16))) | multiply(n1,const_10)|divide(#0,const_2)|divide(#0,n0)|add(#1,#0)|divide(#1,n1)|add(#2,#4)|divide(#3,#5)| | general | C |
if a rectangular room measures 8 meters by 5 meters by 4 meters , what is the volume of the room in cubic centimeters ? ( 1 meter = 100 centimeters ) | b . 160 , 000,000 8 * 100 * 5 * 100 * 4 * 100 = 160 , 000,000 | ['a ) 24,000', 'b ) 16 , 000,000', 'c ) 2 , 400,000', 'd ) 24 , 000,000', 'e ) 240 , 000,000'] | b | multiply(multiply(multiply(multiply(4, 100), divide(1, const_10)), multiply(5, 100)), multiply(8, 100)) | divide(n3,const_10)|multiply(n2,n4)|multiply(n1,n4)|multiply(n0,n4)|multiply(#0,#1)|multiply(#4,#2)|multiply(#5,#3) | geometry | B |
a not - so - good clockmaker has four clocks on display in the window . clock # 1 loses 17 minutes every hour . clock # 2 gains 15 minutes every hour relative to clock # 1 ( i . e . , as clock # 1 moves from 12 : 00 to 1 : 00 , clock # 2 moves from 12 : 00 to 1 : 15 ) . clock # 3 loses 20 minutes every hour relative to clock # 2 . finally , clock # 4 gains 20 minutes every hour relative to clock # 3 . if the clockmaker resets all four clocks to the correct time at 12 noon , what time will clock # 4 display after 6 actual hours ( when it is actually 6 : 00 pm that same day ) ? | c 1 loses 15 minutes every hour . so after 60 minutes have passed , c 1 displays that 60 - 15 = 45 minutes have passed . c 2 gains 15 minutes for every 60 minutes displayed on c 1 . thus , the time displayed on c 2 is 75 / 60 = 5 / 4 the time displayed on c 1 . so after 60 minutes have passed , c 2 displays the passing of ( 5 / 4 * 45 ) minutes . c 3 loses 20 minutes for every 60 minutes displayed on c 2 . thus , the time displayed on c 3 is 40 / 60 = 2 / 3 the time displayed on c 2 . so after 60 minutes have passed , c 3 displays the passing of ( 2 / 3 * 5 / 4 * 45 ) minutes . c 4 gains 20 minutes for every 60 minutes displayed on c 3 . thus , the time displayed on c 4 is 80 / 60 = 4 / 3 the time displayed on clock 3 . so after 60 minutes have passed , c 4 displays the passing of 4 / 3 * 2 / 3 * 5 / 4 * 45 = 50 minutes . c 4 loses 10 minutes every hour . in 6 hours , c 4 will lose 6 * 10 = 60 minutes = 1 hour . since the correct time after 6 hours will be 6 pm , c 4 will show a time of 6 - 1 = 5 : 34 pm . the correct answer is b . | a ) 5 : 00 , b ) 5 : 34 , c ) 5 : 42 , d ) 6 : 00 , e ) 6 : 24 | b | subtract(multiply(6, const_10), multiply(multiply(multiply(divide(add(const_60, 15), const_60), divide(subtract(const_60, 20), const_60)), divide(add(const_60, 20), const_60)), subtract(const_60, 17))) | add(n16,const_60)|add(n3,const_60)|multiply(n23,const_10)|subtract(const_60,n16)|subtract(const_60,n1)|divide(#0,const_60)|divide(#1,const_60)|divide(#3,const_60)|multiply(#6,#7)|multiply(#5,#8)|multiply(#9,#4)|subtract(#2,#10) | physics | B |
in a 500 m race , the ratio of the speeds of two contestants a and b is 2 : 4 . a has a start of 300 m . then , a wins by : | "to reach the winning post a will have to cover a distance of ( 500 - 300 ) m , i . e . , 200 m . while a covers 2 m , b covers 4 m . while a covers 200 m , b covers 4 x 200 / 2 m = 400 m . thus , when a reaches the winning post , b covers 400 m and therefore remains 100 m behind . a wins by 100 m . answer : e" | a ) 60 m , b ) 20 m , c ) 43 m , d ) 20 m , e ) 100 m | e | subtract(500, divide(multiply(subtract(500, 300), 4), 2)) | subtract(n0,n3)|multiply(n2,#0)|divide(#1,n1)|subtract(n0,#2)| | physics | E |
a number of 47 marbles is to be divided and contain with boxes . if each box is to contain 3 , 4 , or 5 marbles , what is the largest possible number of boxes ? | "to maximize # of boxes we should minimize marbles per box : 14 * 3 + 1 * 5 = 47 - - > 14 + 1 = 15 . answer : c ." | a ) 10 , b ) 12 , c ) 15 , d ) 16 , e ) 17 | c | divide(47, 3) | divide(n0,n1)| | general | C |
on a race track a maximum of 5 horses can race together at a time . there are a total of 25 horses . there is no way of timing the races . what is the minimum number r of races we need to conduct to get the top 3 fastest horses ? | "r = 7 is the correct answer . good solution buneul . b" | a ) 5 , b ) 7 , c ) 8 , d ) 10 , e ) 11 | b | add(divide(25, 5), const_2) | divide(n1,n0)|add(#0,const_2)| | general | B |
a , band c enter into partnership . a invests 3 times as much as b and b invests two - third of what c invests . at the end of the year , the profit earned is rs . 7700 . what is the share of b ? | "let c ' s capital = rs . x . then , b ' s capital = rs . ( 2 / 3 ) x a β s capital = rs . ( 3 x ( 2 / 3 ) . x ) = rs . 2 x . ratio of their capitals = 2 x : ( 2 / 3 ) x : x = 6 : 2 : 3 . hence , b ' s share = rs . ( 7700 x ( 2 / 11 ) ) = rs . 1400 . answer is c" | a ) 1100 , b ) 800 , c ) 1400 , d ) 1200 , e ) none of them | c | multiply(7700, divide(const_2, add(add(multiply(const_2, 3), multiply(divide(const_2, 3), 3)), 3))) | divide(const_2,n0)|multiply(const_2,n0)|multiply(#0,n0)|add(#1,#2)|add(#3,n0)|divide(const_2,#4)|multiply(n1,#5)| | gain | C |
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