Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
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how many litres of pure acid are there in 10 litres of a 25 % solution | explanation : question of this type looks a bit typical , but it is too simple , as below . . . it will be 10 * 25 / 100 = 2.5 answer : option c | a ) 1.5 , b ) 1.6 , c ) 2.5 , d ) 1.8 , e ) 1.9 | c | multiply(divide(25, const_100), 10) | divide(n1,const_100)|multiply(n0,#0) | gain |
there are 15 teams in a soccer league and each team plays each of the other teams once . how many games will be played in total ? | "15 c 2 = 105 the answer is a ." | a ) 105 , b ) 110 , c ) 115 , d ) 120 , e ) 125 | a | divide(factorial(15), multiply(factorial(subtract(15, const_2)), factorial(const_2))) | factorial(n0)|factorial(const_2)|subtract(n0,const_2)|factorial(#2)|multiply(#3,#1)|divide(#0,#4)| | general |
rahim bought 65 books for rs . 6500 from one shop and 35 books for rs . 2000 from another . what is the average price he paid per book ? | "average price per book = ( 6500 + 2000 ) / ( 65 + 35 ) = 8500 / 100 = rs . 85 answer : e" | a ) 80 , b ) 138 , c ) 100 , d ) 129 , e ) 85 | e | divide(add(6500, 2000), add(65, 35)) | add(n1,n3)|add(n0,n2)|divide(#0,#1)| | general |
at a certain restaurant , the ratio of the number of cooks to the number of waiters is 3 to 10 . when 12 more waiters are hired , the ratio of the number of cooks to the number of waiters changes to 3 to 13 . how many cooks does the restaurant have ? | "originally there were 3 k cooks and 10 k waiters . 13 k = 10 k + 12 k = 4 there are 12 cooks . the answer is d ." | a ) 4 , b ) 6 , c ) 9 , d ) 12 , e ) 15 | d | multiply(divide(12, 13), divide(12, 13)) | divide(n2,n4)|multiply(#0,#0)| | other |
a man engaged a servant on the condition that he would pay him rs . 800 and a uniform after one year service . he served only for 9 months and received uniform and rs . 400 , find the price of the uniform ? | "9 / 12 = 3 / 4 * 800 = 600 400 - - - - - - - - - - - - - 200 1 / 4 - - - - - - - - 200 1 - - - - - - - - - ? = > rs . 800 answer : a" | a ) s . 800 , b ) s . 85 , c ) s . 90 , d ) s . 100 , e ) s . 120 | a | multiply(divide(subtract(multiply(9, 800), multiply(multiply(const_3, const_4), 400)), multiply(multiply(const_3, const_4), const_1)), const_4) | multiply(n0,n1)|multiply(const_3,const_4)|multiply(n2,#1)|multiply(#1,const_1)|subtract(#0,#2)|divide(#4,#3)|multiply(#5,const_4)| | general |
a customer purchased a package of ground beef at a cost of $ 1.96 per pound . for the same amount of money , the customer could have purchased a piece of steak that weighed 30 percent less than the package of ground beef . what was the cost per pound of the steak ? | "for simplicity , let ' s assume the customer bought 1 pound of ground beef for $ 1.96 . let x be the price per pound for the steak . then 0.7 x = 196 x = 196 / 0.7 = $ 2.80 the answer is e ." | a ) $ 4.65 , b ) $ 4.10 , c ) $ 3.60 , d ) $ 3.20 , e ) $ 2.80 | e | divide(1.96, add(multiply(const_0_25, const_2), multiply(const_0_33, const_1))) | multiply(const_0_25,const_2)|multiply(const_0_33,const_1)|add(#0,#1)|divide(n0,#2)| | general |
a computer manufacturer produces a certain electronic component at a cost of $ 80 per component . shipping costs for delivering the components are $ 7 per unit . further , the manufacturer has costs of $ 16,500 a month related to the electronic component regardless of how many it produces . if the manufacturer produces... | "by the question , the equation would be 150 p - 87 * 150 - 16500 = 0 p being the price we want to find and equation resulting zero means revenue and costs are equal so we can get the minimum price of the component . solving the equation , we get p = $ 197 . answer e for me ." | a ) $ 28 , b ) $ 82 , c ) $ 110 , d ) $ 138 , e ) $ 197 | e | divide(add(add(multiply(multiply(const_4, const_4), const_1000), multiply(7, const_100)), multiply(add(80, 7), 150)), 150) | add(n0,n1)|multiply(const_4,const_4)|multiply(n1,const_100)|multiply(#1,const_1000)|multiply(n3,#0)|add(#3,#2)|add(#5,#4)|divide(#6,n3)| | general |
what is the sum of the greatest common factor and the lowest common multiple of 24 and 36 ? | "prime factorization of 24 = 2 x 2 x 2 x 3 prime factorization of 36 = 2 x 2 x 3 x 3 gcf = 12 lcm = 72 sum = 84 answer b ." | a ) 92 , b ) 84 , c ) 24 , d ) 60 , e ) 184 | b | divide(multiply(24, 36), const_4) | multiply(n0,n1)|divide(#0,const_4)| | general |
a car travels at a speed of 65 miles per hour . how far will it travel in 5 hours ? | "during each hour , the car travels 65 miles . for 5 hours it will travel 65 + 65 + 65 + 65 + 65 = 5 Γ 65 = 325 miles correct answer is c ) 325 miles" | a ) 125 miles , b ) 225 miles , c ) 325 miles , d ) 425 miles , e ) 525 miles | c | multiply(65, 5) | multiply(n0,n1)| | physics |
in a race where 12 cars are running , the chance that car x will win is 1 / 6 , that y will win is 1 / 10 and that z will win is 1 / 8 . assuming that a dead heat is impossible , find the chance that one of them will win . | "required probability = p ( x ) + p ( y ) + p ( z ) ( all the events are mutually exclusive ) . = 1 / 6 + 1 / 10 + 1 / 8 = 47 / 120 answer : d" | a ) 1 / 140 , b ) 1 / 180 , c ) 27 / 410 , d ) 47 / 120 , e ) 57 / 120 | d | add(add(divide(1, 6), divide(1, 10)), divide(1, 8)) | divide(n1,n2)|divide(n1,n4)|divide(n1,n6)|add(#0,#1)|add(#3,#2)| | general |
in a box of 12 pens , a total of 3 are defective . if a customer buys 2 pens selected at random from the box , what is the probability that both pen will be defective ? | "method - 1 there are 9 fine pieces of pen and 3 defective in a lot of 12 pens i . e . probability of first pen being defective = ( 1 / 3 ) i . e . probability of second pen being defective = ( 1 / 2 ) probability of both pen being defective = ( 1 / 2 ) * ( 1 / 3 ) = 1 / 6 answer : option a" | a ) 1 / 6 , b ) 2 / 9 , c ) 6 / 11 , d ) 9 / 16 , e ) 3 / 4 | a | multiply(divide(subtract(12, 3), 12), divide(subtract(subtract(12, 3), const_1), subtract(12, const_1))) | subtract(n0,n1)|subtract(n0,const_1)|divide(#0,n0)|subtract(#0,const_1)|divide(#3,#1)|multiply(#2,#4)| | general |
the timing of a college is from 12 p . m to 4.20 p . m . five lectures are held in the given duration and a break of 5 minutes after each lecture is given to the students . find the duration of each lecture . | explanation : total time a student spends in college = 4 hours 20 minutes = 260 minutes as there are 5 lectures , the number of breaks between lectures is 4 . total time of the break = 20 minutes hence , the duration of each lecture is = ( 260 β 20 ) / 5 = 48 minutes answer d | a ) 52 minutes , b ) 45 minutes , c ) 30 minutes , d ) 48 minutes , e ) 44 minutes | d | divide(multiply(4.2, const_60), 5) | multiply(n1,const_60)|divide(#0,n2) | physics |
after decreasing 50 % in the price of an article costs rs . 1200 . find the actual cost of an article ? | "cp * ( 50 / 100 ) = 1200 cp = 24 * 100 = > cp = 2400 answer : c" | a ) 1200 , b ) 50 , c ) 2400 , d ) 100 , e ) 600 | c | divide(1200, subtract(const_1, divide(50, const_100))) | divide(n0,const_100)|subtract(const_1,#0)|divide(n1,#1)| | gain |
mary ' s income is 60 % more than tim ' s income and tim ' s income is 30 % less than juan ' s income . what % of juan ' s income is mary ' s income . | "even i got 96 % j = 100 t = 100 * 0.7 = 70 m = 70 * 1.6 = 112 if mary ' s income is x percent of j m = j * x / 100 x = m * 100 / j = 112 * 100 / 100 = 112 ans : a" | a ) 112 % , b ) b . 100 % , c ) 96 % , d ) 80 % , e ) 64 % | a | add(subtract(const_100, 30), multiply(subtract(const_100, 30), divide(60, const_100))) | divide(n0,const_100)|subtract(const_100,n1)|multiply(#0,#1)|add(#2,#1)| | general |
a telephone company needs to create a set of 3 - digit area codes . the company is entitled to use only digits 2 , 4 and 5 , which can be repeated . if the product of the digits in the area code must be even , how many different codes can be created ? | we have 3 places ( - - - ) to fill & we can use only digits 2 , 4 , 5 first place can be filled in 3 ways 2 nd place can be filled in 3 ways because we can repeat the digits 3 rd place can be filled in 3 ways because we can repeat the digits total no of codes = 3 x 3 x 3 = 27 but one combination ( 555 ) will not be eve... | a ) 20 , b ) 22 , c ) 24 , d ) 26 , e ) 30 | d | subtract(power(3, const_3), const_1) | power(n0,const_3)|subtract(#0,const_1) | general |
which no . need to add to 859622 to get a number exactly divisible by 456 ? | "dividend = quotient * divisor + reminder 859622 / 456 gives quotient = 1885 and reminder = 62 . so , the next number divisible by 456 is 456 places infront of 456 * 1885 which means 456 β 62 = 394 should be added to 859622 . e" | a ) 345652 , b ) 547678 , c ) 356789 , d ) 778990 , e ) 859622 | e | subtract(859622, reminder(859622, 456)) | reminder(n0,n1)|subtract(n0,#0)| | general |
a 250 meter long train crosses a man standing on the platform in 12 sec . what is the speed of the train ? | "s = 250 / 12 * 18 / 5 = 75 kmph answer : d" | a ) 228 , b ) 108 , c ) 1266 , d ) 75 , e ) 211 | d | multiply(divide(250, 12), const_3_6) | divide(n0,n1)|multiply(#0,const_3_6)| | physics |
in what time will a train 120 m long cross an electric pole , it its speed be 185 km / hr ? | "speed = 185 * 5 / 18 = 51 m / sec time taken = 120 / 51 = 2.3 sec . answer : a" | a ) 2.3 sec , b ) 2.9 sec , c ) 2.7 sec , d ) 8.7 sec , e ) 8.5 sec | a | divide(120, multiply(185, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics |
a dishonest dealer professes to sell goods at the cost price but uses a weight of 750 grams per kg , what is his percent ? | "750 - - - 250 100 - - - ? = > 33.33 % answer : b" | a ) 25 % , b ) 33 % , c ) 29 % , d ) 55 % , e ) 45 % | b | subtract(multiply(divide(const_100, 750), multiply(const_100, multiply(add(const_3, const_2), const_2))), const_100) | add(const_2,const_3)|divide(const_100,n0)|multiply(#0,const_2)|multiply(#2,const_100)|multiply(#1,#3)|subtract(#4,const_100)| | gain |
what ratio must a shopkeeper mix peas and soybean of rs . 16 and rs . 25 / kg , as to obtain a mixture of rs . 21 ? | "correct option : ( c ) use rule of alligation , to determine the ratio the required ratio of soybean and peas = 4 : 5" | a ) 10 : 7 , b ) 9 : 8 , c ) 4 : 5 , d ) 13 : 11 , e ) 14 : 8 | c | divide(subtract(25, 21), subtract(21, 16)) | subtract(n1,n2)|subtract(n2,n0)|divide(#0,#1)| | general |
a part - time employee whose hourly wage was decreased by 20 percent decided to increase the number of hours worked per week so that the employee ' s total income did not change . by what percent e should the number of hours worked be increased ? | "correct answer : c solution : c . we can set up equations for income before and after the wage reduction . initially , the employee earns w wage and works h hours per week . after the reduction , the employee earns . 8 w wage and works x hours . by setting these equations equal to each other , we can determine the inc... | a ) 12.5 % , b ) 20 % , c ) 25 % , d ) 50 % , e ) 100 % | c | multiply(divide(subtract(divide(multiply(const_10, const_4), multiply(divide(subtract(const_100, 20), const_100), const_10)), const_4), const_4), const_100) | multiply(const_10,const_4)|subtract(const_100,n0)|divide(#1,const_100)|multiply(#2,const_10)|divide(#0,#3)|subtract(#4,const_4)|divide(#5,const_4)|multiply(#6,const_100)| | general |
the side of a cube is 8 m , find the voiume ? | "cube volume = a ( power ) 3 a ( power ) 3 = 8 ( power ) 3 = 512 m ( power ) 3 . answer is e ." | a ) 515 , b ) 509 , c ) 502 , d ) 508 , e ) 512 | e | rectangle_area(add(multiply(const_4, const_10), const_3), 8) | multiply(const_10,const_4)|add(#0,const_3)|rectangle_area(n0,#1)| | physics |
a rectangle has 20 meter length and 15 meter breadth . find ratio between area of rectangle and square whose side is same with rectangle ' s length . | area of rectangle = 20 * 15 = 300 area of square = 20 ^ 2 = 400 ratio of rectangle ' area and square ' s area = 300 : 400 = 3 : 4 answer : b | ['a ) 3 : 2', 'b ) 3 : 4', 'c ) 4 : 5', 'd ) 2 : 3', 'e ) 2 : 5'] | b | divide(rectangle_area(15, 20), square_area(20)) | rectangle_area(n0,n1)|square_area(n0)|divide(#0,#1) | geometry |
a number when divided by 221 gives a remainder 43 , what remainder will be obtained by dividing the same number 17 ? | 221 + 43 = 264 / 17 = 9 answer a | a ) 9 , b ) 11 , c ) 15 , d ) 5 , e ) 20 | a | subtract(43, multiply(17, const_2)) | multiply(n2,const_2)|subtract(n1,#0) | general |
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 8 % of the money for taxi fare and buys 20 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 8 % of 200 = 16 for taxi fare , so he is left with 184 he buys 20 mangoes ( @ 5 ) so he spends 100 money left is 84 ( 184 - 100 ) no of oranges he can buy = 84 / 4 = > 21 so , george can buy 21 oranges . a" | a ) 21 , b ) 30 , c ) 20 , d ) 16 , e ) 12 | a | multiply(subtract(subtract(const_1, divide(8, const_100)), divide(20, 50)), 50) | divide(n2,const_100)|divide(n3,n0)|subtract(const_1,#0)|subtract(#2,#1)|multiply(n0,#3)| | general |
a sum of 14,400 amounts to 16,600 in 3 years at the rate of simple interest . what is the rate of interest ? | "d 5 % s . i . = ( 16600 - 14400 ) = 2200 . rate = ( 100 x 2200 ) / ( 14400 x 3 ) % = 5 %" | a ) 2 % , b ) 1 % , c ) 6 % , d ) 5 % , e ) 8 % | d | multiply(divide(divide(const_3, 3), add(multiply(const_3, 3), add(const_0_25, const_0_25))), const_100) | add(const_0_25,const_0_25)|divide(const_3,n2)|multiply(const_3,n2)|add(#0,#2)|divide(#1,#3)|multiply(#4,const_100)| | gain |
a particular library has 75 books in a special collection , all of which were in the library at the beginning of the month . these book are occasionally loaned out through an inter - library program . if , by the end of the month , 70 percent of books that were loaned out are returned and there are 66 books in the spec... | total = 75 books . 65 % of books that were loaned out are returned - - > 100 % - 70 % = 30 % of books that were loaned out are not returned . now , there are 66 books , thus 76 - 68 = 7 books are not returned . { loaned out } * 0.30 = 7 - - > { loaned out } = 30 . answer : b . | a ) 20 , b ) 30 , c ) 35 , d ) 40 , e ) 55 | b | divide(subtract(75, 66), subtract(const_1, divide(70, const_100))) | divide(n1,const_100)|subtract(n0,n2)|subtract(const_1,#0)|divide(#1,#2) | gain |
if a tire rotates at 400 revolutions per minute when the car is traveling 168 km / h , what is the circumference of the tire ? | "400 rev / minute = 400 * 60 rev / 60 minutes = 24,000 rev / hour 24,000 * c = 168,000 m : c is the circumference c = 3 meters correct answer a" | a ) 7 meters , b ) 9 meters , c ) 8 meters , d ) 5 meters , e ) 3 meters | a | multiply(divide(168, multiply(multiply(multiply(const_2, const_3), const_10), 400)), const_1000) | multiply(const_2,const_3)|multiply(#0,const_10)|multiply(n0,#1)|divide(n1,#2)|multiply(#3,const_1000)| | physics |
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 7 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 = 7 total people can seat = 81 + 7 = 88 answer : d" | a ) 52 , b ) 49 , c ) 95 , d ) 88 , e ) 66 | d | add(multiply(add(15, subtract(15, 3)), 3), 7) | subtract(n0,n1)|add(n0,#0)|multiply(n1,#1)|add(n3,#2)| | general |
a worker ' s take - home pay last year was the same each month , and she saved the same fraction of her take - home pay each month . the total amount of money that she had saved at the end of the year was 2 times the amount of that portion of her monthly take - home pay that she did not save . if all the money that she... | "let x be the fraction of her take - home pay that the worker saved . let p be the monthly pay . 12 xp = 2 ( 1 - x ) p 12 xp = 2 p - 2 xp 14 xp = 2 p x = 1 / 7 the answer is a ." | a ) 1 / 7 , b ) 1 / 6 , c ) 1 / 5 , d ) 1 / 4 , e ) 1 / 3 | a | inverse(add(divide(const_12, 2), const_1)) | divide(const_12,n0)|add(#0,const_1)|inverse(#1)| | general |
what is the sum of all remainders obtained when the first 120 natural numbers are divided by 9 ? | "a positive integer can give only the following 9 remainders when divided by 9 : 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , and 0 . 1 divided by 9 gives the remainder of 1 ; 2 divided by 9 gives the remainder of 2 ; . . . 8 divided by 9 gives the remainder of 8 ; 9 divided by 9 gives the remainder of 0 . we ' ll have 11 such bloc... | a ) 397 , b ) 401 , c ) 403 , d ) 405 , e ) 399 | d | add(const_3, multiply(divide(subtract(120, reminder(120, 9)), 9), add(add(add(const_1, const_2), const_3), add(add(const_1, const_4), add(const_4, add(add(const_1, const_2), const_3)))))) | add(const_1,const_2)|add(const_1,const_4)|reminder(n0,n1)|add(#0,const_3)|subtract(n0,#2)|add(#3,const_4)|divide(#4,n1)|add(#1,#5)|add(#3,#7)|multiply(#8,#6)|add(#9,const_3)| | general |
if a certain number of labours can dig earth 100 m deep , 25 m long , and 30 m broad in 12 days , then find the number of days the same number of labours will require to dig earth 75 m deep , 20 m long and 50 m broad . | explanation : more number of days means more length , more depth and more width . hence , its a direct proportion . ( 100 * 25 * 30 ) : ( 75 * 20 * 50 ) : : 12 : x 75000 : 75000 : : 12 : x x = 12 answer : a | a ) 12 , b ) 18 , c ) 6 , d ) 1 , e ) 2 | a | multiply(divide(multiply(multiply(75, 20), 50), multiply(multiply(100, 25), 30)), 12) | multiply(n4,n5)|multiply(n0,n1)|multiply(n6,#0)|multiply(n2,#1)|divide(#2,#3)|multiply(n3,#4) | physics |
a shopkeeper sold an article offering a discount of 5 % and earned a profit of 27.3 % . what would have been the percentage of profit earned if no discount had been offered ? | "giving no discount to customer implies selling the product on printed price . suppose the cost price of the article is 100 . then printed price = 100 Γ£ β ( 100 + 27.3 ) / ( 100 Γ’ Λ β 5 ) = 100 Γ£ β 127.3 / 95 = 134 hence , required % profit = 134 Γ’ β¬ β 100 = 34 % answer d" | a ) 28.5 , b ) 27.675 , c ) 30 , d ) 34 % , e ) none of these | d | subtract(divide(multiply(add(const_100, 27.3), const_100), subtract(const_100, 5)), const_100) | add(n1,const_100)|subtract(const_100,n0)|multiply(#0,const_100)|divide(#2,#1)|subtract(#3,const_100)| | gain |
in town x , 64 percent of the population are employed , and 45 percent of the population are employed males . what percent of the employed people in town x are females ? | "answer b male - employed - 64 % female - employed - 19 % total employed 64 % means total unemployed = 36 % therefore d and e are not correct because female unemployed has to be less than 36 % female - unemployed = 32 % male unemployed = 4 % 48 % + 4 % = 52 % 16 % + 32 % = 48 % plug in a and b in place of 32 % and the ... | a ) 16 % , b ) 45 % , c ) 32 % , d ) 40 % , e ) 28 % | e | multiply(divide(subtract(64, 45), 64), const_100) | subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)| | gain |
x and y invested in a business . they earned some profit which they divided in the ratio of 2 : 3 . if x invested rs . 40,000 , the amount invested by y is | "explanation : suppose y invested rs . y . then 40000 / y = 2 / 3 or y = 60000 . answer : c ) 60000" | a ) 33488 , b ) 63809 , c ) 60000 , d ) 37887 , e ) 77824 | c | divide(multiply(multiply(add(const_1, const_4), const_1000), 2), 3) | add(const_1,const_4)|multiply(#0,const_1000)|multiply(n0,#1)|divide(#2,n1)| | gain |
135 kg of an alloy a is mixed with 145 kg of alloy b . if alloy a has lead and tin in the ratio 3 : 5 and alloy b has tin and copper in the ratio 2 : 3 , then the amount of tin in the new alloy is ? | "quantity of tin in 135 kg of a = 135 * 5 / 8 = 84 kg quantity of tin in 145 kg of b = 145 * 2 / 5 = 58 kg quantity of tin in the new alloy = 84 + 58 = 142 kg answer is b" | a ) 100.6 kg , b ) 142 kg , c ) 135 kg , d ) 110.8 kg , e ) 114 kg | b | add(multiply(divide(135, add(3, 5)), 5), multiply(divide(145, add(2, 3)), 2)) | add(n2,n3)|add(n4,n5)|divide(n0,#0)|divide(n1,#1)|multiply(n3,#2)|multiply(n4,#3)|add(#4,#5)| | other |
a sum of money lent out at s . i . amounts to a total of $ 520 after 2 years and to $ 820 after a further period of 5 years . what was the initial sum of money that was invested ? | "s . i for 5 years = $ 820 - $ 520 = $ 300 the s . i . is $ 60 / year s . i . for 2 years = $ 120 principal = $ 520 - $ 120 = $ 400 the answer is d ." | a ) $ 340 , b ) $ 360 , c ) $ 380 , d ) $ 400 , e ) $ 420 | d | subtract(520, multiply(subtract(820, 520), divide(2, 5))) | divide(n1,n3)|subtract(n2,n0)|multiply(#0,#1)|subtract(n0,#2)| | general |
a person borrows rs . 5000 for 2 years at 4 % p . a . simple interest . he immediately lends it to another person at 61 β 4 % p . a for 2 years . find his gain in the transaction per year . | "explanation : the person borrows rs . 5000 for 2 years at 4 % p . a . simple interest simple interest that he needs to pay = prt / 100 = 5000 Γ 4 Γ 2 / 100 = 400 he also lends it at 6 1 β 4 % p . a for 2 years simple interest that he gets = prt / 100 = 5000 Γ 254 Γ 21 / 00 = 625 his overall gain in 2 years = rs . 625 ... | a ) 167.5 , b ) 150 , c ) 225 , d ) 112.5 , e ) 212.5 | d | divide(subtract(divide(multiply(multiply(5000, 61), 2), const_100), divide(multiply(multiply(5000, 4), 2), const_100)), 2) | multiply(n0,n3)|multiply(n0,n2)|multiply(n1,#0)|multiply(n1,#1)|divide(#2,const_100)|divide(#3,const_100)|subtract(#4,#5)|divide(#6,n1)| | gain |
a small water pump would take 4 hours to fill an empty tank . a larger pump would take 1 / 2 hour to fill the same tank . how many hours would it take both pumps , working at their respective constant rates , to fill the empty tank if they began pumping at the same time ? | "rate of the small pump is 1 / 4 tank / hour rate of the larger pump is 1 / ( 1 / 2 ) or 2 tank / hour ; combined rate of the two pumps is 1 / 4 + 2 = 9 / 4 tank / hour , together they will fill the empty tank in 1 / ( 9 / 4 ) ) = 4 / 9 hours ( time = job / rate ) . answer : b" | a ) 1 / 6 , b ) 4 / 9 , c ) 1 / 4 , d ) 5 / 9 , e ) 1 / 3 | b | inverse(add(inverse(4), 2)) | inverse(n0)|add(n2,#0)|inverse(#1)| | physics |
a man cycling along the road noticed that every 3 minutes a bus overtakes him and every 9 minutes he meets an oncoming bus . if all buses and the cyclist move at a constant speed , what is the time interval between consecutive buses ? | "let ' s say the distance between the buses is d . we want to determine interval = \ frac { d } { b } , where b is the speed of bus . let the speed of cyclist be c . every 9 minutes a bus overtakes cyclist : \ frac { d } { b - c } = 9 , d = 9 b - 9 c ; every 3 minutes cyclist meets an oncoming bus : \ frac { d } { b + ... | a ) 5 minutes , b ) 6 minutes , c ) 8 minutes , d ) 9 / 2 minutes , e ) 10 minutes | d | divide(subtract(3, divide(3, divide(add(9, 3), subtract(3, 9)))), const_1) | add(n0,n1)|subtract(n0,n1)|divide(#0,#1)|divide(n0,#2)|subtract(n0,#3)|divide(#4,const_1)| | physics |
a pump can fill a tank with water in 2 hours . because of a leak , it took 2 2 / 3 hours to fill the tank . the leak can drain all the water of the tank in | "explanation : solution 1 the pump can fill the tank in 2 hour . but , because of the leak , it took 2 2 / 3 hours to fill the tank . i . e . , the additional water filled by the pump in 2 / 3 hour is emptied by the leak in 2 2 / 3 = 8 / 3 hours . therefore , the water filled by pump in 2 bour can be emptied by the lea... | a ) 6 hours , b ) 8 hours , c ) 9 hours , d ) 10 hours , e ) 12 hours | b | inverse(subtract(divide(2, 2), inverse(divide(add(multiply(2, 3), 2), 3)))) | divide(n2,n0)|multiply(n0,n3)|add(n2,#1)|divide(#2,n3)|inverse(#3)|subtract(#0,#4)|inverse(#5)| | physics |
a goods train runs at the speed of 72 kmph and crosses a 250 m long platform in 26 seconds . what is the length of the goods train ? | "speed = ( 72 x ( 5 / 18 ) ) m / sec = 20 m / sec . time = 26 sec . let the length of the train be x metres . then , ( ( x + 250 ) / 26 ) = 20 x + 250 = 520 x = 270 . d" | a ) 230 m , b ) 240 m , c ) 260 m , d ) 270 m , e ) none of the above | d | multiply(subtract(26, divide(250, multiply(const_0_2778, 72))), multiply(const_0_2778, 72)) | multiply(n0,const_0_2778)|divide(n1,#0)|subtract(n2,#1)|multiply(#0,#2)| | physics |
if teena is driving at 55 miles per hour and is currently 7.5 miles behind joe , who is driving at 40 miles per hour in the same direction then in how many minutes will teena be 22.5 miles ahead of joe ? | "this type of questions should be solved without any complex calculations as these questions become imperative in gaining that extra 30 - 40 seconds for a difficult one . teena covers 55 miles in 60 mins . joe covers 40 miles in 60 mins so teena gains 15 miles every 60 mins teena need to cover 7.5 + 22.5 miles . teena ... | a ) 15 , b ) 60 , c ) 75 , d ) 90 , e ) 120 | e | multiply(divide(add(subtract(55, 40), 7.5), subtract(55, 40)), const_60) | subtract(n0,n2)|add(n1,#0)|divide(#1,#0)|multiply(#2,const_60)| | physics |
what is the average of first 21 multiples of 5 ? | required average = 7 ( 1 + 2 + . . . . + 21 ) / 21 ( 5 / 21 ) x ( ( 21 x 22 ) / 2 ) ( because sum of first 21 natural numbers ) = 55 c | a ) a ) 70 , b ) b ) 77 , c ) c ) 55 , d ) d ) 81 , e ) e ) 82 | c | multiply(divide(divide(multiply(21, add(21, const_1)), const_2), 21), 5) | add(n0,const_1)|multiply(n0,#0)|divide(#1,const_2)|divide(#2,n0)|multiply(n1,#3) | general |
the sale price sarees listed for rs . 560 after successive discount is 20 % and 30 % is ? | "explanation : 560 * ( 80 / 100 ) * ( 70 / 100 ) = 314 answer : b" | a ) 321 , b ) 314 , c ) 316 , d ) 265 , e ) 162 | b | subtract(subtract(560, divide(multiply(560, 20), const_100)), divide(multiply(subtract(560, divide(multiply(560, 20), const_100)), 30), const_100)) | multiply(n0,n1)|divide(#0,const_100)|subtract(n0,#1)|multiply(n2,#2)|divide(#3,const_100)|subtract(#2,#4)| | gain |
45 girls and 60 boys are present at a party . there are 17 girl - boy are in relationship among them . if a girl and a boy is selected at random , what is the probability that they will be a couple ? | "in how many ways we can select a girl and a boy from 45 lady and 60 gentlemen ? in 45 * 60 = 2,700 ways . we have a total of 17 couples so , the probability of selecting a couple is 17 / 2,700 = 17 / 2700 . ans - d" | a ) 1 / 200 , b ) 1 / 100 , c ) 90 / 2900 , d ) 17 / 2700 , e ) 1 / 20 | d | divide(subtract(multiply(choose(45, const_1), choose(60, const_1)), choose(17, const_1)), multiply(choose(45, const_1), choose(60, const_1))) | choose(n0,const_1)|choose(n1,const_1)|choose(n2,const_1)|multiply(#0,#1)|subtract(#3,#2)|divide(#4,#3)| | probability |
a can complete a work in 12 days and b can do the same work in 8 days . if a after doing 2 days , leaves the work , find in how many days b will do the remaining work ? | the required answer = ( 12 - 2 ) * 8 / 12 = 80 / 12 = 6 1 / 2 days answer is c | a ) 2 days , b ) 4 1 / 2 days , c ) 6 1 / 2 days , d ) 7 3 / 4 days , e ) 10 days | c | add(multiply(12, 2), divide(12, 2)) | divide(n0,n2)|multiply(n0,n2)|add(#0,#1) | physics |
a train 800 m long is running at a speed of 78 km / hr . if it crosses a tunnel in 2 min , then the length of the tunnel is ? | "speed = 78 * 5 / 18 = 65 / 3 m / sec . time = 2 min = 120 sec . let the length of the train be x meters . then , ( 800 + x ) / 120 = 65 / 3 x = 1800 m . answer : b" | a ) 2898 , b ) 1800 , c ) 500 , d ) 297 , e ) 435 | b | divide(800, multiply(subtract(78, 2), const_0_2778)) | subtract(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
given that e and f are events such that p ( e ) = 25 , p ( f ) = 75 and p ( e n f ) = 75 , find p ( e | f ) and p ( f | e ) | here , e and f are events p ( e | f ) = p ( enf ) / p ( f ) = 75 / 25 = 3 p ( f | e ) = p ( enf ) / p ( e ) = 75 / 75 = 1 . e ) | a ) 5 , b ) 46 , c ) 52 , d ) 12 , e ) 3 | e | divide(75, 25) | divide(n1,n0) | general |
a train 2500 m long can cross an electric pole in 100 sec and then find the speed of the train ? | "length = speed * time speed = l / t s = 2500 / 100 s = 25 m / sec speed = 25 * 18 / 5 ( to convert m / sec in to kmph multiply by 18 / 5 ) speed = 90 kmph answer : a" | a ) 90 , b ) 60 , c ) 120 , d ) 180 , e ) 360 | a | divide(divide(2500, const_1000), divide(100, const_3600)) | divide(n0,const_1000)|divide(n1,const_3600)|divide(#0,#1)| | physics |
if integer k is equal to the sum of all even multiples of 35 between 280 and 630 , what is the greatest prime factor of k ? | "if we break down what the stem is asking what is the sum of all mult of 30 between 300 and 600 . using arithmetic progression to find n : 630 = 280 + ( n - 1 ) 35 350 + 35 = 35 n 385 = 35 n = > n = 11 the sum would be : 11 * mean mean = [ 630 + 280 ] / 2 = 455 11 * 455 = e" | a ) 5 , b ) 7 , c ) 11 , d ) 13 , e ) 11 * 455 | e | divide(add(divide(subtract(multiply(floor(divide(630, 35)), 35), multiply(floor(divide(280, 35)), 35)), 35), const_1), const_2) | divide(n2,n0)|divide(n1,n0)|floor(#0)|floor(#1)|multiply(n0,#2)|multiply(n0,#3)|subtract(#4,#5)|divide(#6,n0)|add(#7,const_1)|divide(#8,const_2)| | general |
if 360 is divided into 4 parts proportional to 5 , 7 , 4 , 8 then the smallest part is : | given ratio = 5 : 7 : 4 : 8 , sum of ratio = 24 : . the smallest part = ( 360 * 4 / 24 ) = 60 answer : b | a ) 50 , b ) 60 , c ) 40 , d ) 18 , e ) 22 | b | multiply(divide(4, add(add(add(5, 7), 4), 8)), 360) | add(n2,n3)|add(n1,#0)|add(n5,#1)|divide(n1,#2)|multiply(n0,#3) | general |
1370 , 1320 , 1070 , - 180 , - 6430 , x | "1370 - 50 * ( 5 ^ 0 ) = 1320 1320 - 50 * ( 5 ^ 1 ) = 1070 1070 - 50 * ( 5 ^ 2 ) = - 180 - 180 - 50 * ( 5 ^ 3 ) = - 6430 - 6430 - 50 * ( 5 ^ 4 ) = - 37680 answer : a ." | a ) - 37680 , b ) 6530 , c ) 6630 , d ) 6730 , e ) 6830 | a | subtract(negate(180), multiply(subtract(1320, 1070), divide(subtract(1320, 1070), subtract(1370, 1320)))) | negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)| | general |
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 train 150 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 : 150 / 6 * 18 / 5 km / hr = 90 km / hr let speed of train = x therefore x + 6 = 90 x = 90 - 6 x = 84 km / hr answer : e" | a ) 54 kmph , b ) 60 kmph , c ) 66 kmph , d ) 72 kmph , e ) 84 kmph | e | divide(divide(subtract(150, 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 |
a thief runs away from a with a speed of 20 km / hr . a police from a station 60 km away started chasing him after 1 hr . with a speed of 40 km / hr . time taken by the police to catch the thief is : | explanation : solution : distance traveled by the thief in 1 hr . = 20 km . total distance of them after 1 hr = ( 60 + 20 ) km = 80 km . their relative speed = ( 40 - 20 ) = 20 km / hr . time taken to catch the thief = 80 / 20 = 4 hrs . answer : a | a ) 4 hrs , b ) 4.5 hrs , c ) 5.4 hrs , d ) 6 hrs . , e ) none of these | a | divide(add(60, multiply(20, 1)), subtract(60, 40)) | multiply(n0,n2)|subtract(n1,n3)|add(n1,#0)|divide(#2,#1) | physics |
what is the square root of 16 , divided by 2 ? | "square root is a number times itself square root of 16 = 4 , 4 / 2 = 2 ( b ) 2" | a ) 9 , b ) 2 , c ) 122 , d ) 6 , e ) 1 | b | divide(sqrt(16), 2) | sqrt(n0)|divide(#0,n1)| | other |
the value of ( 4.7 Γ 13.26 + 4.7 Γ 9.43 + 4.7 Γ 77.31 ) is : | solution given expression = 4.7 Γ ( 13.26 + 9.43 + 77.31 ) = 4.7 Γ 100 = 470 . answer c | a ) 0.47 , b ) 47 , c ) 470 , d ) 4700 , e ) none of these | c | multiply(4.7, add(add(13.26, 9.43), 77.31)) | add(n1,n3)|add(n5,#0)|multiply(n0,#1) | general |
an ant walks an average of 500 meters in 60 minutes . a beetle walks 15 % less distance at the same time on the average . assuming the beetle walks at her regular rate , what is its speed in km / h ? | "the ant walks an average of 500 meters in 60 minutes 500 meters in 1 hours the beetle walks 15 % less distance = 500 - 75 = 425 meters in 60 minutes 0.425 km in 1 hours speed = 0.425 * 1 = 0.425 km / h i guess option d should be 0.425" | a ) 1.0 , b ) 0.5 , c ) 0.75 , d ) 0.425 , e ) 0.3 | d | multiply(divide(divide(500, const_1000), divide(60, const_60)), subtract(const_1, divide(15, const_100))) | divide(n0,const_1000)|divide(n1,const_60)|divide(n2,const_100)|divide(#0,#1)|subtract(const_1,#2)|multiply(#3,#4)| | general |
a tank holds x gallons of a saltwater solution that is 20 % salt by volume . one fourth of the water is evaporated , leaving all of the salt . when 10 gallons of water and 20 gallons of salt are added , the resulting mixture is 33 1 / 3 % salt by volume . what is the value of x ? | "original mixture has 20 % salt and 80 % water . total = x out of which salt = 0.2 x and water = 0.8 x now , 1 / 4 water evaporates and all salt remains . so what remains is 0.2 x salt and 0.6 x water . now 20 gallons salt is added and 10 gallons of water is added . so salt now becomes - > ( 0.2 x + 20 ) and water - - ... | a ) 37.5 , b ) 75 , c ) 100 , d ) 150 , e ) 175 | d | divide(subtract(multiply(20, const_2), 10), subtract(subtract(subtract(1, divide(20, const_100)), multiply(subtract(1, divide(20, const_100)), divide(1, const_4))), multiply(const_2, divide(20, const_100)))) | divide(n0,const_100)|divide(n4,const_4)|multiply(n2,const_2)|multiply(#0,const_2)|subtract(#2,n1)|subtract(n4,#0)|multiply(#1,#5)|subtract(#5,#6)|subtract(#7,#3)|divide(#4,#8)| | general |
roses can be purchased individually for $ 2.30 , one dozen for $ 36 , or two dozen for $ 50 . what is the greatest number of roses that can be purchased for $ 680 ? | "buy as many $ 50 deals as possible . we can by 650 / 50 = 13 two dozen roses , thus total of 13 * 24 = 312 roses . we are left with 680 - 650 = $ 30 . we can buy 30 / 2.3 = ~ 13 roses for that amount . total = 312 + 13 = 325 . answer : c ." | a ) 156 , b ) 162 , c ) 325 , d ) 324 , e ) 326 | c | add(multiply(floor(divide(680, 50)), multiply(const_12, const_2)), floor(divide(subtract(680, multiply(floor(divide(680, 50)), 50)), 2.30))) | divide(n3,n2)|multiply(const_12,const_2)|floor(#0)|multiply(n2,#2)|multiply(#2,#1)|subtract(n3,#3)|divide(#5,n0)|floor(#6)|add(#7,#4)| | general |
in the land of oz only one or two - letter words are used . the local language has 65 different letters . the parliament decided to forbid the use of the seventh letter . how many words have the people of oz lost because of the prohibition ? | the answer to the question is indeed e . the problem with above solutions is that they do not consider words like aa , bb , . . . the number of 1 letter words ( x ) that can be made from 65 letters is 65 ; the number of 2 letter words ( xx ) that can be made from 65 letters is 65 * 65 , since each x can take 65 values ... | a ) 65 , b ) 66 , c ) 67 , d ) 131 , e ) 130 | e | add(add(const_1, 65), 65) | add(n0,const_1)|add(n0,#0) | general |
baker ram is storing cake boxes in cartons that measure 25 inches by 42 inches by 60 inches . if the measurement of each cake box is 8 inches by 7 inches by 5 inches , then what is the maximum number of cake boxes that can be placed in each carton ? | baker ram is storing cake boxes in cartons the dimensions of the carton are 25 x 42 x 60 and those of each cake box are 8 x 7 x 5 . to maximize number of boxes try and fit maximum into the base of the box . if you take the base as 25 x 42 and fit 5 x 6 boxes of 5 x 7 , then the total number of boxes is 5 x 6 x integer ... | a ) 330 , b ) 300 , c ) 252 , d ) 225 , e ) 210 | e | multiply(multiply(add(const_4, const_2), 7), 5) | add(const_2,const_4)|multiply(n4,#0)|multiply(n5,#1) | general |
a 40 - liter solution of alcohol and water is 5 percent alcohol . if 4.5 liters of alcohol and 5.5 liters of water are added to this solution , what percent of the solution produced is alcohol ? | the percent of alcohol in the solution is ( 0.05 ( 40 ) + 4.5 ) / 50 = 6.5 / 50 = 13 % the answer is c . | a ) 7 % , b ) 10 % , c ) 13 % , d ) 16 % , e ) 19 % | c | multiply(divide(add(multiply(40, divide(5, const_100)), 4.5), add(40, add(4.5, 5.5))), const_100) | add(n2,n3)|divide(n1,const_100)|add(n0,#0)|multiply(n0,#1)|add(n2,#3)|divide(#4,#2)|multiply(#5,const_100) | general |
a snail , climbing a 20 feet high wall , climbs up 4 feet on the first day but slides down 2 feet on the second . it climbs 4 feet on the third day and slides down again 2 feet on the fourth day . if this pattern continues , how many days will it take the snail to reach the top of the wall ? | "total transaction in two days = 4 - 2 = 2 feet in 16 days it will climb 16 feet on the 17 th day , the snail will climb 4 feet , thus reaching the top therefore , total no of days required = 17 answer : c" | a ) 12 , b ) 16 , c ) 17 , d ) 20 , e ) 21 | c | subtract(20, 4) | subtract(n0,n1)| | physics |
if n is the smallest integer such that 432 times n is the square of an integer , what is the value of n ? | the prime factorization of a square has to have even powers of all its prime factors . if the original number has a factor , say of 7 , then when it β s squared , the square will have a factor of 7 ^ 2 . another way to say that is : any positive integer all of whose prime factors have even powers must be a perfect squa... | ['a ) 2', 'b ) 3', 'c ) 6', 'd ) 12', 'e ) 24'] | b | divide(divide(divide(divide(divide(divide(432, const_2), const_2), const_2), const_2), const_3), const_3) | divide(n0,const_2)|divide(#0,const_2)|divide(#1,const_2)|divide(#2,const_2)|divide(#3,const_3)|divide(#4,const_3) | geometry |
the average of the marks of 12 students in a class is 36 . if the marks of each student are doubled , find the new average ? | "sum of the marks for the 12 students = 12 * 36 = 432 . the marks of each student are doubled , the sum also will be doubled . the new sum = 432 * 2 = 864 . so , the new average = 864 / 12 = 72 . answer : a" | a ) 72 , b ) 23 , c ) 29 , d ) 26 , e ) 13 | a | multiply(36, const_2) | multiply(n1,const_2)| | general |
a trader sells 85 meters of cloth for rs . 8925 at the profit of rs . 20 per metre of cloth . what is the cost price of one metre of cloth ? | "explanation : sp of 1 m of cloth = 8925 / 85 = rs . 105 cp of 1 m of cloth = sp of 1 m of cloth - profit on 1 m of cloth = rs . 105 - rs . 20 = rs . 85 . answer : d" | a ) 21 , b ) 28 , c ) 90 , d ) 85 , e ) 11 | d | subtract(divide(8925, 85), 20) | divide(n1,n0)|subtract(#0,n2)| | physics |
if the simple interest on $ 1500 increase by $ 20 , when the time increase by 2 years . find the rate percent per annum . | applying the above formula , we have 20 = 1500 * r * 2 / 100 r = 20 * 100 / 1500 * 2 = 0.7 % . answer is e | a ) 0.6 % , b ) 0.5 % , c ) 0.1 % , d ) 0.2 % , e ) 0.7 % | e | multiply(divide(divide(20, 2), 1500), const_100) | divide(n1,n2)|divide(#0,n0)|multiply(#1,const_100) | gain |
a man sells an article at a profit of 25 % . if he had bought it at 20 % less and sold it for rs . 8.40 less , he would have gained 30 % . find the cost of the article . | "let c . p = 100 gain = 25 % s . p = 125 supposed c . p = 80 gain = 30 % s . p = ( 130 * 80 ) / 100 = 104 diff = ( 125 - 104 ) = 21 diff 21 when c . p = 100 then diff 8.40 when c . p = ( 100 * 8.40 ) / 21 = 40 answer : b" | a ) 30 , b ) 40 , c ) 50 , d ) 60 , e ) 70 | b | divide(multiply(8.40, const_100), subtract(add(25, const_100), divide(multiply(add(30, const_100), subtract(const_100, 20)), const_100))) | add(n0,const_100)|add(n3,const_100)|multiply(n2,const_100)|subtract(const_100,n1)|multiply(#1,#3)|divide(#4,const_100)|subtract(#0,#5)|divide(#2,#6)| | gain |
the sum of two numbers is 29 and the difference of their squares is 145 . the difference between the number is | "explanation : let the numbers be x and y . we know that , ( x β y ) = x 2 β y 2 / ( x + y ) = 145 / 29 = 5 correct option : b" | a ) 13 , b ) 5 , c ) 8 , d ) 11 , e ) 12 | b | divide(subtract(145, power(29, const_2)), const_2) | power(n0,const_2)|subtract(n1,#0)|divide(#1,const_2)| | general |
the cube root of what integer power of 2 is closest to 50 ? | integer powers of 2 = 2 , 2 ^ 2 , 2 ^ 3 , 2 ^ 4 , 2 ^ 5 , 2 ^ 6 , 2 ^ 7 etc . let , cube root of x is closest to 50 i . e . x is closest to cube of 50 = 50 * 50 * 50 = 125000 but x must be equivalent to an integer power of 2 2 ^ 10 = 1024 2 ^ 7 = 128 i . e . 2 ^ 17 = 128 * 1024 = 128000 ( approx . ) i . e . 2 ^ 18 = 25... | a ) 16 , b ) 17 , c ) 18 , d ) 19 , e ) 20 | b | subtract(divide(50, 2), add(const_4, const_4)) | add(const_4,const_4)|divide(n1,n0)|subtract(#1,#0) | geometry |
a watch was sold at a loss of 10 % . if it was sold for rs . 190 more , there would have been a gain of 4 % . what is the cost price ? | "explanation : 90 % 104 % - - - - - - - - 14 % - - - - 190 100 % - - - - ? = > rs . 1357.1 answer : d" | a ) s . 1000 , b ) s . 1009 , c ) s . 1007 , d ) s . 1357.1 , e ) s . 1002 | d | divide(multiply(190, const_100), subtract(add(const_100, 4), subtract(const_100, 10))) | add(const_100,n2)|multiply(n1,const_100)|subtract(const_100,n0)|subtract(#0,#2)|divide(#1,#3)| | gain |
if 3 ^ x * 4 ^ y = 19683 and x β y = 9 , then x = ? | since 19,683 is an odd number , it is not a multiple of 4 . this means that y must equal zero . since x - y = 9 and y = 0 , then x = 9 . the answer is b . | a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | b | subtract(9, divide(log(divide(power(3, 9), 19683)), log(multiply(3, 4)))) | multiply(n0,n1)|power(n0,n3)|divide(#1,n2)|log(#0)|log(#2)|divide(#4,#3)|subtract(n3,#5) | general |
in triangle pqr , the angle q = 90 degree , pq = 2 cm , qr = 8 cm . x is a variable point on pq . the line through x parallel to qr , intersects pr at y and the line through y , parallel to pq , intersects qr at z . find the least possible length of xz | "look at the diagram below : now , in case when qy is perpendicular to pr , two right triangles pqr and pqy are similar : qy : qp = qr : pr - - > qy : 2 = 8 : 10 - - > qy = 1.6 . answer : a ." | a ) 1.6 cm , b ) 2.4 cm , c ) 4.8 cm , d ) 2.16 cm , e ) 3.2 cm | a | divide(multiply(2, 8), const_10) | multiply(n1,n2)|divide(#0,const_10)| | geometry |
5 men are equal to as many women as are equal to 8 boys . all of them earn rs . 210 only . men Γ’ β¬ β’ s wages are ? | "5 m = xw = 8 b 5 m + xw + 8 b - - - - - 210 rs . 5 m + 5 m + 5 m - - - - - 210 rs . 15 m - - - - - - 210 rs . = > 1 m = 14 rs . answer : e" | a ) 6 rs , b ) 2 rs , c ) 4 rs , d ) 9 rs , e ) 14 rs | e | divide(210, multiply(const_3, 5)) | multiply(n0,const_3)|divide(n2,#0)| | general |
in may mrs lee ' s earnings were 60 percent of the lee family ' s total income . in june mrs lee earned 10 percent more than in may . if the rest of the family ' s income was the same both months , then , in june , mrs lee ' s earnings were approximately what percent of the lee family ' s total income ? | "let in may lee family ' s total income = 100 in may mrs lee ' s income = 60 in may rest of the family ' s income = 40 in june mrs lees income = 60 * 110 / 100 = 66 in june total income = 66 + 40 = 106 % of mrs lee ' s income = 66 / 106 = 62.26 ( a )" | a ) 62 % , b ) 68 % , c ) 72 % , d ) 76 % , e ) 80 % | a | multiply(divide(add(const_100, 10), add(add(const_100, 10), const_100)), const_100) | add(n1,const_100)|add(#0,const_100)|divide(#0,#1)|multiply(#2,const_100)| | general |
express a speed of 60 kmph in meters per second ? | "60 * 5 / 18 = 16.67 mps answer : b" | a ) 10 mps , b ) 16.67 mps , c ) 97.67 mps , d ) 17.67 mps , e ) 18 mps | b | multiply(const_0_2778, 60) | multiply(n0,const_0_2778)| | physics |
a person decided to build a house in 100 days . he employed 100 men in the beginning and 100 more after 50 days and completed the construction in stipulated time . if he had not employed the additional men , how many days behind schedule would it have been finished ? | 200 men do the rest of the work in 100 - 50 = 50 days 100 men can do the rest of the work in 50 * 200 / 100 = 100 days required number of days = 100 - 50 = 50 days answer is c | a ) 40 , b ) 80 , c ) 50 , d ) 60 , e ) 50 | c | subtract(100, 50) | subtract(n0,n3) | general |
how many multiples of 6 are there between 30 and 140 , 30 and 140 inclusive ? | "6 multiples are . . . 30 , 36,42 , 48,54 , 60,66 , 72,78 , 84,90 , 96,102 , 108,114 , 120,126 , 132,138 , . . . , . . . , the answer is = 19 answer is d" | a ) 18 , b ) 16 , c ) 17 , d ) 19 , e ) 15 | d | add(divide(subtract(140, 30), 6), const_1) | subtract(n2,n1)|divide(#0,n0)|add(#1,const_1)| | general |
how many seconds will a 400 metre long train take to cross a man running with a speed of 6 km / hr in the direction of the moving train if the speed of the train is 46 km / hr ? | "explanation : speed of train relatively to man = ( 46 - 6 ) km / hr = 40 km / hr = ( 40 x 5 / 18 ) m / sec = 11.11 m / sec time taken to pass the man = ( 400 / 11.11 ) sec = 36 sec . answer : e" | a ) 25 , b ) 30 , c ) 40 , d ) 45 , e ) 36 | e | divide(400, multiply(const_0_2778, subtract(46, 6))) | subtract(n2,n1)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
if x + y = 24 , x - y = 40 , for integers of x and y , x = ? | "x + y = 24 x - y = 40 2 x = 64 x = 32 answer is d" | a ) 20 , b ) 23 , c ) 25 , d ) 32 , e ) 42 | d | divide(add(24, 40), const_2) | add(n0,n1)|divide(#0,const_2)| | general |
a car traveling at a certain constant speed takes 12 seconds longer to travel 1 kilometer than it would take to travel 1 kilometer at 75 kilometers per hour . at what speed , in kilometers per hour , is the car traveling ? | "many approaches are possible , one of them : let the distance be 1 kilometer . time to cover this distance at 75 kilometers per hour is 1 / 75 hours = 3,600 / 75 seconds = 48 seconds ; time to cover this distance at regular speed is 48 + 12 = 60 seconds = 60 / 3,600 hours = 1 / 60 hours ; so , we get that to cover 1 k... | a ) 60 , b ) 72 , c ) 72.5 , d ) 73 , e ) 73.5 | a | divide(1, divide(add(multiply(const_3600, divide(1, 75)), 12), const_3600)) | divide(n1,n3)|multiply(#0,const_3600)|add(n0,#1)|divide(#2,const_3600)|divide(n1,#3)| | physics |
if 222 = 6 333 = 12 444 = 20 then , 555 = ? ? | 222 = 2 * 2 + 2 = 6 333 = 3 * 3 + 3 = 12 444 = 4 * 4 + 4 = 20 555 = 5 * 5 + 5 = 30 answer : a | a ) 30 , b ) 35 , c ) 40 , d ) 45 , e ) 50 | a | multiply(add(const_2, const_3), 6) | add(const_2,const_3)|multiply(n1,#0) | general |
if n = 6 ^ 11 β 6 , what is the units digit of n ? | always divide the power ( incase 11 ) by 4 and use the remainder as the new power . the question now becomes 6 ^ 3 - 6 . now 6 ^ 3 has last digit 6 . , we subtract 6 from 6 = 0 is the answer . option a | a ) 0 , b ) 1 , c ) 4 , d ) 6 , e ) 8 | a | divide(log(6), log(power(6, 11))) | log(n2)|power(n0,n1)|log(#1)|divide(#0,#2)| | general |
a certain industrial loom weaves 0.128 meters of cloth every second . approximately how many seconds will it take for the loom to weave 25 meters of cloth ? | "explanation : let the required number of seconds be x more cloth , more time , ( direct proportion ) hence we can write as ( cloth ) 0.128 : 25 : : 1 : x β 0.128 x = 25 β x = 25 / 0.128 = 25000 / 128 = 3125 / 16 β 195 answer : option d" | a ) 205 , b ) 200 , c ) 180 , d ) 195 , e ) 175 | d | divide(25, 0.128) | divide(n1,n0)| | physics |
a man goes from a to b at a speed of 14 kmph and comes back to a at a speed of 10 kmph . find his average speed for the entire journey ? | "distance from a and b be ' d ' average speed = total distance / total time average speed = ( 2 d ) / [ ( d / 14 ) + ( d / 10 ] = ( 2 d ) / [ 12 d / 70 ) = > 11.6 kmph . answer : a" | a ) 11.6 kmph , b ) 12.9 kmph , c ) 12.8 kmph , d ) 11.9 kmph , e ) 12.6 kmph | a | divide(add(14, 10), const_2) | add(n0,n1)|divide(#0,const_2)| | physics |
p software has coding line 5 % more than n , n software has coding line 4 / 20 more than m . m software has 80 lines of coding . find p lines . | "m s / w has 80 line of code n s / w has = 80 + 80 * 4 / 20 = 96 line of code p s / w 5 % more n ' code 96 + 4.8 = 100.8 or 101 line of code answer : a" | a ) 101 , b ) 107 , c ) 108 , d ) 109 , e ) 110 | a | add(80, divide(80, 20)) | divide(n3,n2)|add(n3,#0)| | general |
if f ( x ) = 5 - 2 x , then f - 1 ( - 3 ) = ? | find f - 1 ( x ) and then find f - 1 ( - 3 ) y = 5 - 2 x , given x = 5 - 2 y , interchange x and y 2 y = 5 - x , y = log 2 ( 5 - x ) , solve for y f - 1 ( x ) = log 2 ( 5 - x ) , inverse function f - 1 ( - 3 ) = log 2 ( 5 + 3 ) = log 2 ( 23 ) = 3 correct answer e | a ) 1 , b ) 8 , c ) 6 , d ) 4 , e ) 3 | e | subtract(5, multiply(divide(subtract(3, 1), 2), 2)) | subtract(n3,n2)|divide(#0,n1)|multiply(n1,#1)|subtract(n0,#2) | general |
the weight of every type a widget is the same , the weight of every type b widget is the same , and the weight of every type c widget is the same . if the weight of 7 type a widgets is equal to the weight of 2 type b widgets , and the weight of 4 type b widgets is equal to the weight of 7 type c widgets . what is the r... | "4 b = 7 c and so b = 7 c / 4 7 a = 2 b and so a = 2 b / 7 = c / 2 a + b = c / 2 + 7 c / 4 = 9 c / 4 b + c = 7 c / 4 + c = 11 c / 4 the ratio of a + b : b + c = 9 : 11 the answer is d ." | a ) 3 : 5 , b ) 5 : 7 , c ) 7 : 9 , d ) 9 : 11 , e ) 11 : 13 | d | divide(add(1, divide(7, 2)), add(divide(7, 2), divide(4, 2))) | divide(n0,n1)|divide(n2,n1)|add(n4,#0)|add(#0,#1)|divide(#2,#3)| | general |
( 500 + 200 ) Γ 4 Γ ( 3 + 2 ) = ? | "( 500 + 200 ) Γ 4 Γ ( 3 + 2 ) = ? or , ? = 700 Γ 4 Γ 5 = 14000 answer b" | a ) 2902 , b ) 14000 , c ) 8402 , d ) 16800 , e ) none of these | b | divide(add(500, 200), 4) | add(n0,n1)|divide(#0,n2)| | general |
how many seconds will a 700 meter long train moving with a speed of 63 km / hr take to cross a man walking with a speed of 3 km / hr in the direction of the train ? | "explanation : here distance d = 700 mts speed s = 63 - 3 = 60 kmph = 60 x 5 / 18 m / s time t = = 42 sec . answer : c" | a ) 48 , b ) 36 , c ) 42 , d ) 11 , e ) 18 | c | divide(700, multiply(subtract(63, 3), const_0_2778)) | subtract(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
joan and karl each bought a telescopes and the sum of their purchases was $ 400 . if twice of what joan paid was $ 74 more than what karl paid , what did joan pay for her telescope ? | j + k = 400 so k = 400 - j 2 j = k + 74 2 j = ( 400 - j ) + 74 3 j = 474 j = 158 the answer is e . | a ) $ 150 , b ) $ 152 , c ) $ 154 , d ) $ 156 , e ) $ 158 | e | subtract(400, divide(subtract(multiply(400, const_2), 74), const_3)) | multiply(n0,const_2)|subtract(#0,n1)|divide(#1,const_3)|subtract(n0,#2) | general |
how many odd prime numbers are there less than 150 ? | "odd prime number less than 150 : 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 , 41 , 43 , 47 , 53 , 59 , 61 , 67 , 71 , 73 , 79 , 83 , 89 , 97 , 101 , 103 , 107 , 109 , 113 , 127 , 131 , 137 , 139 , 149 there is 34 the odd prime number answer is b" | a ) 24 , b ) 34 , c ) 44 , d ) 54 , e ) 64 | b | add(subtract(150, const_100), const_4) | subtract(n0,const_100)|add(#0,const_4)| | general |
a can do a piece of work in 8 days . he works at it for 4 days and then b finishes it in 6 days . in how many days can a and b together finish the work ? | 4 / 8 + 6 / x = 1 = > x = 12 1 / 8 + 1 / 12 = 5 / 24 = > 4.8 days answer : b | a ) 4.1 days , b ) 4.8 days , c ) 3.8 days , d ) 2.8 days , e ) 4.5 days | b | inverse(add(divide(const_1, 8), divide(const_1, multiply(6, const_2)))) | divide(const_1,n0)|multiply(n2,const_2)|divide(const_1,#1)|add(#0,#2)|inverse(#3) | physics |
a boat moves upstream at the rate of 1 km in 20 minutes and down stream 1 km in 12 minutes . then the speed of the current is : | rate upstream = ( 1 / 20 * 60 ) = 3 kmph rate dowm stream = 1 / 12 * 60 = 5 kmph rate of the current = Β½ ( 5 - 3 ) = 1 kmph answer : a | a ) 1 kmph , b ) 2 kmph , c ) 3 kmph , d ) 2.5 kmph , e ) 3.5 kmph | a | divide(subtract(multiply(divide(1, 12), const_60), multiply(divide(1, 20), const_60)), const_2) | divide(n0,n3)|divide(n0,n1)|multiply(#0,const_60)|multiply(#1,const_60)|subtract(#2,#3)|divide(#4,const_2) | physics |
a parking garage rents parking spaces for $ 10 per week or $ 40 per month . how much does a person save in a year by renting by the month rather than by the week ? | "10 $ per week ! an year has 52 weeks . annual charges per year = 52 * 10 = 520 $ 30 $ per month ! an year has 12 months . annual charges per year = 12 * 40 = 480 $ 520 - 480 = 40 ans a" | a ) $ 40 , b ) $ 160 , c ) $ 220 , d ) $ 240 , e ) $ 260 | a | subtract(multiply(add(multiply(10, add(const_3, const_2)), const_2), 10), multiply(40, const_12)) | add(const_2,const_3)|multiply(n1,const_12)|multiply(#0,n0)|add(#2,const_2)|multiply(n0,#3)|subtract(#4,#1)| | general |
what sum of money will produce rs . 70 as simple interest in 4 years at 4 1 / 2 percent ? | "70 = ( p * 4 * 9 / 2 ) / 100 p = 388.9 answer : c" | a ) 263 , b ) 500 , c ) 388.9 , d ) 368 , e ) 339 | c | divide(70, divide(multiply(4, add(4, divide(1, 2))), const_100)) | divide(n3,n4)|add(n2,#0)|multiply(n1,#1)|divide(#2,const_100)|divide(n0,#3)| | gain |
in a dairy farm , 34 cows eat 34 bags of husk in 34 days . in how many days one cow will eat one bag of husk ? | "explanation : less cows , more days ( indirect proportion ) less bags , less days ( direct proportion ) [ cows 1 34 bags 34 1 ] : : 34 : x = > x β 34 β 1 = 34 β 1 β 34 = > x = 34 option a" | a ) 34 days , b ) 40 days , c ) 46 days , d ) 50 days , e ) none of these | a | multiply(divide(34, 34), 34) | divide(n0,n0)|multiply(n0,#0)| | physics |
if the sides of a triangle are 28 cm , 24 cm and 15 cm , what is its area ? | "the triangle with sides 28 cm , 24 cm and 15 cm is right angled , where the hypotenuse is 28 cm . area of the triangle = 1 / 2 * 24 * 15 = 180 cm 2 answer : a" | a ) 180 cm 2 , b ) 112 cm 2 , c ) 776 cm 2 , d ) 666 cm 2 , e ) 886 cm 2 | a | divide(multiply(24, 15), const_2) | multiply(n1,n2)|divide(#0,const_2)| | geometry |
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