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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two good train each 500 m long , are running in opposite directions on parallel tracks . their speeds are 45 km / hr and 30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one . | sol . relative speed = ( 45 + 30 ) km / hr = ( 75 x 5 / 18 ) m / sec = ( 125 / 6 ) m / sec . distance covered = ( 500 + 500 ) m = 1000 m . required time = ( 1000 x 6 / 125 ) sec = 48 sec . answer c | a ) 12 sec , b ) 24 sec , c ) 48 sec , d ) 60 sec , e ) none | c | multiply(multiply(500, inverse(multiply(add(45, 30), const_0_2778))), const_2) | add(n1,n2)|multiply(#0,const_0_2778)|inverse(#1)|multiply(n0,#2)|multiply(#3,const_2) | physics |
in a tree , 2 / 5 of the birds are robins while the rest are bluejays . if 1 / 3 of the robins are female and 2 / 3 of the bluejays are female , what fraction of the birds in the tree are male ? | the fraction of birds that are male robins is ( 2 / 3 ) ( 2 / 5 ) = 4 / 15 . the fraction of birds that are male bluejays is ( 1 / 3 ) ( 3 / 5 ) = 1 / 5 . the total fraction of male birds is 4 / 15 + 1 / 5 = 7 / 15 . the answer is c . | a ) 3 / 5 , b ) 4 / 15 , c ) 7 / 15 , d ) 8 / 15 , e ) 17 / 30 | c | add(multiply(divide(2, 5), divide(2, 3)), multiply(divide(const_3, 5), divide(1, 3))) | divide(n0,n1)|divide(n0,n3)|divide(const_3,n1)|divide(n2,n3)|multiply(#0,#1)|multiply(#2,#3)|add(#4,#5) | general |
what distance will be covered by a bus moving at 90 kmph in 20 seconds ? | "90 kmph = 90 * 5 / 18 = 25 mps d = speed * time = 25 * 20 = 500 m . answer : a" | a ) 500 , b ) 600 , c ) 289 , d ) 276 , e ) 207 | a | multiply(multiply(90, const_0_2778), 20) | multiply(n0,const_0_2778)|multiply(n1,#0)| | physics |
in traveling from a dormitory to a certain city , a student went 1 / 2 of the way by foot , 1 / 4 of the way by bus , and the remaining 6 kilometers by car . what is the distance , in kilometers , from the dormitory to the city ? | "whole trip = distance by foot + distance by bus + distance by car x = 1 / 2 x + 1 / 4 x + 6 x - 1 / 2 x - 1 / 4 x = 6 x = 24 km option : d" | a ) 35 , b ) 24 , c ) 14 , d ) 24 , e ) 12 | d | multiply(6, inverse(subtract(1, add(divide(1, 2), divide(1, 4))))) | divide(n0,n1)|divide(n2,n3)|add(#0,#1)|subtract(n0,#2)|inverse(#3)|multiply(n4,#4)| | physics |
the speed of a car is 80 km in the first hour and 60 km in the second hour . what is the average speed of the car ? | "s = ( 80 + 60 ) / 2 = 70 kmph answer : b" | a ) 76 kmph , b ) 70 kmph , c ) 87 kmph , d ) 56 kmph , e ) 86 kmph | b | divide(add(80, 60), const_2) | add(n0,n1)|divide(#0,const_2)| | physics |
if a town of 15,000 people is growing at a rate of approx . 1 % per year , the population of the town in 5 years will be closest to ? | 1 % is quite small and hence the answer is a ) | a ) a . 16000 , b ) b . 17000 , c ) c . 18000 , d ) d . 19000 , e ) e . 20000 | a | multiply(multiply(multiply(5, add(const_3, const_4)), const_100), add(add(const_3, const_4), const_3)) | add(const_3,const_4)|add(#0,const_3)|multiply(n2,#0)|multiply(#2,const_100)|multiply(#1,#3)| | gain |
the ratio , by volume , of bleach ( b ) to detergent ( d ) to water in a certain solution is 2 : 40 : 100 . the solution will be altered so that the ratio of bleach to detergent is tripled while the ratio of detergent to water is halved . if the altered solution will contain 300 liters of water , how many liters of det... | b : d : w = 2 : 40 : 100 bnew / dnew = ( 1 / 3 ) * ( 2 / 40 ) = ( 1 / 60 ) dnew / wnew = ( 1 / 2 ) * ( 40 / 100 ) = ( 1 / 5 ) wnew = 300 dnew = wnew / 5 = 300 / 5 = 60 so , answer will be a | ['a ) 60', 'b ) 50', 'c ) 40', 'd ) 80', 'e ) 70'] | a | multiply(300, divide(divide(40, 100), const_2)) | divide(n1,n2)|divide(#0,const_2)|multiply(n3,#1) | other |
when 0.232323 . . . . . is converted into a fraction , then the result is | 0.232323 . . . = 0.23 = 23 / 99 answer is a . | a ) 23 / 99 , b ) 2 / 9 , c ) 1 / 5 , d ) 23 / 100 , e ) 23 / 98 | a | tangent(0.232323) | tangent(n0) | physics |
in a garden , 26 trees are planted at equal distances along a yard 400 metres long , one tree being at each end of the yard . what is the distance between two consecutive trees ? | 26 trees have 25 gaps between them . length of each gap = 400 / 25 = 16 i . e . , distance between two consecutive trees = 16 answer is e . | a ) 10 , b ) 8 , c ) 12 , d ) 14 , e ) 16 | e | divide(400, subtract(26, const_1)) | subtract(n0,const_1)|divide(n1,#0) | physics |
36 men can complete a piece of work in 18 days . in how many days will 108 men complete the same work ? | "explanation : less men , means more days { indirect proportion } let the number of days be x then , 108 : 36 : : 18 : x x = 6 answer : c ) 6 days" | a ) 24 , b ) 77 , c ) 6 , d ) 29 , e ) 21 | c | divide(multiply(18, 36), 108) | multiply(n0,n1)|divide(#0,n2)| | physics |
the length and breadth of a rectangle is increased by 11 % and 22 % respectively . what is the increase in the area ? | "100 * 100 = 10000 111 * 122 = 13542 - - - - - - - - - - - - 3542 10000 - - - - - - - 3542 100 - - - - - - - - - - 35.42 answer c" | a ) 35.22 , b ) 33.52 , c ) 35.42 , d ) 36.22 , e ) 38.42 | c | divide(multiply(subtract(rectangle_area(add(11, const_100), add(22, const_100)), rectangle_area(const_100, const_100)), const_100), rectangle_area(const_100, const_100)) | add(n0,const_100)|add(n1,const_100)|rectangle_area(const_100,const_100)|rectangle_area(#0,#1)|subtract(#3,#2)|multiply(#4,const_100)|divide(#5,#2)| | geometry |
the simple interest on a sum of money will be rs . 700 after 10 years . if the principal is trebled after 5 years what will be the total interest at the end of the tenth year ? | "p - - - 10 - - - - 700 p - - - 5 - - - - - 350 3 p - - - 5 - - - - - 1050 - - - - - - = > 1400 answer : c" | a ) 2261 , b ) 2888 , c ) 1400 , d ) 2699 , e ) 2771 | c | add(multiply(multiply(divide(700, 10), 5), const_3), multiply(divide(700, 10), 5)) | divide(n0,n1)|multiply(n2,#0)|multiply(#1,const_3)|add(#2,#1)| | general |
barbata invests $ 2000 in the national bank at 5 % . how much additional money must she invest at 8 % so that the total annual income will be equal to 6 % of her entire investment ? | "let the additional invested amount for 8 % interest be x ; equation will be ; 2000 + 0.05 * 2000 + x + 0.08 x = 2000 + x + 0.06 ( 2000 + x ) 0.05 * 2000 + 0.08 x = 0.06 x + 0.06 * 2000 0.02 x = 2000 ( 0.06 - 0.05 ) x = 2000 * 0.01 / 0.02 = 1000 ans : ` ` c ' '" | a ) 1200 , b ) 3000 , c ) 1000 , d ) 3600 , e ) 2400 | c | divide(subtract(multiply(divide(6, const_100), 2000), multiply(2000, divide(5, const_100))), subtract(divide(8, const_100), divide(6, const_100))) | divide(n3,const_100)|divide(n1,const_100)|divide(n2,const_100)|multiply(n0,#0)|multiply(n0,#1)|subtract(#2,#0)|subtract(#3,#4)|divide(#6,#5)| | general |
what is the greatest positive integer x such that 5 ^ x is a factor of 125 ^ 10 ? | "125 ^ 10 = ( 5 ^ 3 ) ^ 10 = 5 ^ 30 answer : e" | a ) 5 , b ) 9 , c ) 10 , d ) 20 , e ) 30 | e | multiply(subtract(125, 10), 10) | subtract(n1,n2)|multiply(n2,#0)| | general |
a dishonest milkman wants to make a profit on the selling of milk . he would like to mix water ( costing nothing ) with milk costing rs . 35 per litre so as to make a profit of 20 % on cost when he sells the resulting milk and water mixture for rs . 40 in what ratio should he mix the water and milk ? | "water = w ( liter ) milk = m ( liter ) = = > cost = price x quantity = 0.35 m = = > revenue = price x quantity = 0.40 ( m + w ) = = > profit = 0.40 ( m + w ) - 0.35 m = 0.2 * ( 0.35 m ) [ 20 % of cost ] = = > 0.40 m + 0.40 w - 0.35 m = 0.07 m = = > 0.02 m = 0.40 w = = > m / w = 0.40 / 0.02 = 20 - - or - - w / m = 1 / ... | a ) 1 : 20 , b ) 1 : 10 , c ) 1 : 8 , d ) 1 : 4 , e ) 6 : 11 | a | divide(const_1, divide(20, const_2)) | divide(n1,const_2)|divide(const_1,#0)| | gain |
two trains start simultaneously from opposite ends of a 150 - km route and travel toward each other on parallel tracks . train x , traveling at a constant rate , completes the 150 - km trip in 4 hours . train y , travelling at a constant rate , completes the 150 - km trip in 3.5 hours . how many kilometers had train x ... | if the two trains cover a total distance d , then train x travels ( 7 / 15 ) * d while train y travels ( 8 / 15 ) * d . if the trains travel 150 km to the meeting point , then train x travels ( 7 / 15 ) * 150 = 70 km . the answer is b . | a ) 68 , b ) 70 , c ) 72 , d ) 74 , e ) 76 | b | multiply(divide(150, 4), divide(150, add(divide(150, 4), divide(150, 3.5)))) | divide(n0,n2)|divide(n0,n4)|add(#0,#1)|divide(n0,#2)|multiply(#0,#3) | physics |
( ( 1 ^ ( m + 1 ) ) / ( 5 ^ ( m + 1 ) ) ) ( ( 1 ^ 18 ) / ( 4 ^ 18 ) ) = 1 / ( 2 ( 10 ) ^ 35 ) what is m ? | ( ( 1 ^ ( m + 1 ) ) / ( 5 ^ ( m + 1 ) ) ) ( ( 1 ^ 18 ) / ( 4 ^ 18 ) ) = 1 / ( 2 ( 10 ) ^ 35 ) ( ( 1 / 5 ) ^ ( m + 1 ) ) * ( ( 1 / 2 ) ^ 36 ) = 1 / ( 2 * ( 2 * 5 ) ^ 35 ) ) 2 ^ 36 will cancel out , since 1 can be written as 1 ^ 35 , so ( 1 / 5 ) ^ ( m + 1 ) = ( 1 / 5 ) ^ 35 ( ( 1 / 5 ) ^ ( m + 1 ) ) * ( ( 1 / 2 ) ^ 36 )... | a ) 17 , b ) 18 , c ) 34 , d ) 35 , e ) 36 | c | subtract(subtract(multiply(18, 2), const_1), 1) | multiply(n5,n9)|subtract(#0,const_1)|subtract(#1,n0) | general |
a student chose a number , multiplied it by 2 , then subtracted 138 from the result and got 106 . what was the number he chose ? | "let xx be the number he chose , then 2 â ‹ … x â ˆ ’ 138 = 106 x = 122 answer : d" | a ) 123 , b ) 267 , c ) 277 , d ) 122 , e ) 120 | d | divide(add(106, 138), 2) | add(n1,n2)|divide(#0,n0)| | general |
a honey bee flies for 10 seconds , from a daisy to a rose . it immediately continues to a poppy , flying for 6 additional seconds . the distance the bee passed , flying from the daisy to the rose is 18 meters longer than the distance it passed flying from the rose to the poppy . the bee flies to the poppy at 3 meters p... | "let the speed be ' s ' and let the distance between rose and poppy be ' x ' the problem boils down to : rose to poppy : s + 3 = x / 6 - - - - - - - 1 daisy to rose : s = ( x + 18 ) / 10 - - - - - - 2 so from 1 we can re write x as x = 6 s + 18 substitute the value of x in 2 gives us s = 1.8 m / s d" | a ) 1.4 , b ) 1.5 , c ) 1.6 , d ) 1.8 , e ) 1.0 | d | divide(add(multiply(6, 3), 18), 10) | multiply(n1,n3)|add(n2,#0)|divide(#1,n0)| | physics |
if the cost price is 98 % of sp then what is the profit % | "sol . sp = rs 100 : then cp = rs 98 : profit = rs 2 . profit = { ( 2 / 98 ) * 100 } % = 2.04 % answer is d ." | a ) 4.07 % , b ) 4 % , c ) 2.7 % , d ) 2.04 % , e ) 2.08 % | d | multiply(divide(subtract(const_100, 98), 98), const_100) | subtract(const_100,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
mary passed a certain gas station on a highway while traveling west at a constant speed of 50 miles per hour . then , 15 minutes later , paul passed the same gas station while traveling west at a constant speed of 60 miles per hour . if both drivers maintained their speeds and both remained on the highway for at least ... | "d = rt m : r = 50 mph , t = t + 1 / 4 hr d = 50 ( t + 1 / 4 ) p : r = 60 , t = t d = 60 t since they went the same distance : 50 t + 50 / 4 = 60 t 10 t = 50 / 4 t = 1.25 or 1 hr , 15 min c" | a ) 1.5 , b ) 1.3 , c ) 1.25 , d ) 1.6 , e ) 2 | c | multiply(const_60, divide(multiply(50, divide(15, const_60)), subtract(60, 50))) | divide(n1,const_60)|subtract(n2,n0)|multiply(n0,#0)|divide(#2,#1)|multiply(#3,const_60)| | physics |
the banker â € ™ s discount of a certain sum of money is rs . 78 and the true discount on the same sum for the same time is rs . 66 . the sum due is | sol . sum = b . d . * t . d . / b . d . - t . d . = rs . [ 78 * 66 / 78 - 66 ] = rs . [ 78 * 66 / 12 ] = rs . 429 answer a | a ) 429 , b ) 280 , c ) 360 , d ) 450 , e ) none | a | divide(multiply(78, 66), subtract(78, 66)) | multiply(n0,n1)|subtract(n0,n1)|divide(#0,#1) | gain |
right now , al and eliot have bank accounts , and al has more money than eliot . the difference between their two accounts is 1 / 10 of the sum of their two accounts . if al ’ s account were to increase by 10 % and eliot ’ s account were to increase by 20 % , then al would have exactly $ 22 more than eliot in his accou... | "lets assume al have amount a in his bank account and eliot ' s bank account got e amount . we can form an equation from the first condition . a - e = 1 / 10 * ( a + e ) = = > 9 a = 11 e - - - - - - - - - - - - ( 1 ) second condition gives two different amounts , al ' s amount = 1.1 a and eliot ' s amount = 1.2 e 1.1 a... | a ) $ 110 , b ) $ 152.3 , c ) $ 180 , d ) $ 220 , e ) $ 260 | b | divide(multiply(22, const_100), subtract(multiply(add(10, 1), 10), add(const_100, 20))) | add(n0,n1)|add(n3,const_100)|multiply(n4,const_100)|multiply(#0,n2)|subtract(#3,#1)|divide(#2,#4)| | general |
a charitable association sold an average of 66 raffle tickets per member . among the female members , the average was 70 raffle tickets . the male to female ratio of the association is 1 : 2 . what was the average number q of tickets sold by the male members of the association | "given that , total average q sold is 66 , male / female = 1 / 2 and female average is 70 . average of male members isx . ( 70 * f + x * m ) / ( m + f ) = 66 - > solving this equation after substituting 2 m = f , x = 58 . ans c ." | a ) 50 , b ) 56 , c ) 58 , d ) 62 , e ) 66 | c | subtract(multiply(66, add(1, 2)), multiply(70, 2)) | add(n2,n3)|multiply(n1,n3)|multiply(n0,#0)|subtract(#2,#1)| | general |
subtract the absolute value from the local value of 4 in 564823 | explanation : place value = local value face value = absolute value the place value of 4 in 564823 is 4 x 1000 = 4000 the face value of 4 in 564823 is nothing but 4 . = > 4000 - 4 = 3996 answer : option d | a ) 196 , b ) 1996 , c ) 2996 , d ) 3996 , e ) 6000 | d | multiply(multiply(4, const_100), const_10) | multiply(n0,const_100)|multiply(#0,const_10) | general |
machine a and machine g are each used to manufacture 660 sprockets . it takes machine a 10 hours longer to produce 660 sprockets than machine g . machine g produces 10 percent more sprockets per hour than machine a . how many sprockets per hour does machine a produces ? | [ reveal ] spoiler : timeg : 660 / x timea : [ 660 / x + 10 ] 660 / x = [ 660 / x + 10 ] * 110 / 100 660 / x = 66 * 11 / x + 10 660 x + 10 = 66 * 11 * x 660 x + 6600 = 66 * 11 * x x = 100 plug in back to timea 660 / 100 + 10 = > 660 / 110 = 6 | a ) 6 , b ) 6.6 , c ) 60 , d ) 100 , e ) 110 | a | divide(660, divide(multiply(multiply(add(divide(10, const_100), const_1), 660), 10), subtract(multiply(add(divide(10, const_100), const_1), 660), 660))) | divide(n1,const_100)|add(#0,const_1)|multiply(n0,#1)|multiply(n1,#2)|subtract(#2,n0)|divide(#3,#4)|divide(n0,#5) | gain |
for all positive integers m , [ m ] = 3 m when m is odd and [ m ] = ( 1 / 2 ) * m when m is even . what is [ 7 ] * [ 4 ] equivalent to ? | "[ 7 ] * [ 4 ] = 21 * 2 = 42 = ( 1 / 2 ) ( 84 ) = [ 84 ] the answer is b ." | a ) [ 28 ] , b ) [ 84 ] , c ) [ 42 ] , d ) [ 48 ] , e ) [ 14 ] | b | divide(multiply(divide(4, 2), multiply(3, 7)), 3) | divide(n4,n2)|multiply(n0,n3)|multiply(#0,#1)|divide(#2,n0)| | general |
what is the tens digit of 6 ^ 21 ? | "the tens digit of 6 in integer power starting from 2 ( 6 ^ 1 has no tens digit ) repeats in a pattern of 5 : { 3 , 1 , 9 , 7 , 5 } : the tens digit of 6 ^ 2 = 36 is 3 . the tens digit of 6 ^ 3 = 216 is 1 . the tens digit of 6 ^ 4 = . . . 96 is 9 . the tens digit of 6 ^ 5 = . . . 76 is 7 . the tens digit of 6 ^ 6 = . .... | a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | c | floor(divide(reminder(power(6, reminder(21, add(const_4, const_1))), const_100), const_10)) | add(const_1,const_4)|reminder(n1,#0)|power(n0,#1)|reminder(#2,const_100)|divide(#3,const_10)|floor(#4)| | general |
machine x and machine b are each used to manufacture 660 sprockets . it takes machine x 10 hours longer to produce 660 sprockets than machine b . machine b produces 10 percent more sprockets per hour than machine x . how many sprockets per hour does machine x produces ? | i think the correct answer is a . machine x produces at a speed of x sp / hour and b at a speed of b sp / hour . so , 660 / x = ( 660 / b ) + 10 and x = 1,1 x - - - > 1,1 * 660 = 660 + 11 x - - - > a = 6 , so answer a is correct | a ) 6 , b ) 6.6 , c ) 60 , d ) 100 , e ) 110 | a | divide(660, multiply(add(const_1, divide(10, const_100)), const_100)) | divide(n1,const_100)|add(#0,const_1)|multiply(#1,const_100)|divide(n0,#2) | general |
a can do a piece of work in 2 hours ; b and c together can do it in 3 hours , while a and c together can do it in 2 hours . how long will b alone take to do it ? | "a ' s 1 hour ' s work = 1 / 2 ; ( b + c ) ' s 1 hour ' s work = 1 / 3 ; ( a + c ) ' s 1 hour ' s work = 1 / 2 . ( a + b + c ) ' s 1 hour ' s work = ( 1 / 2 + 1 / 3 ) = 5 / 6 . b ' s 1 hour ' s work = ( 5 / 6 - 1 / 2 ) = 1 / 3 . therefore a alone will take 3 hours to do the work . e" | a ) 12 , b ) 9 , c ) 8 , d ) 6 , e ) 3 | e | divide(const_1, subtract(divide(const_1, 3), subtract(divide(const_1, 2), divide(const_1, 2)))) | divide(const_1,n1)|divide(const_1,n2)|divide(const_1,n0)|subtract(#1,#2)|subtract(#0,#3)|divide(const_1,#4)| | physics |
the ages of two persons differ by 16 years . 6 years ago , the elder one was 3 times as old as the younger one . what are their present ages of the elder person | "let ' s take the present age of the elder person = x and the present age of the younger person = x – 16 ( x – 6 ) = 3 ( x - 16 - 6 ) = > x – 6 = 3 x – 66 = > 2 x = 60 = > x = 60 / 2 = 30 answer : d" | a ) 11 , b ) 66 , c ) 28 , d ) 30 , e ) 99 | d | add(divide(add(multiply(3, 6), subtract(16, 6)), const_2), 16) | multiply(n1,n2)|subtract(n0,n1)|add(#0,#1)|divide(#2,const_2)|add(n0,#3)| | general |
in one alloy there is 12 % chromium while in another alloy it is 8 % . 10 kg of the first alloy was melted together with 30 kg of the second one to form a third alloy . find the percentage of chromium in the new alloy . | "the amount of chromium in the new 10 + 30 = 40 kg alloy is 0.12 * 10 + 0.08 * 30 = 3.6 kg , so the percentage is 3.6 / 40 * 100 = 9 % . answer : c" | a ) 9.4 % , b ) 9.6 % , c ) 9 % , d ) 9.8 % , e ) 10 % | c | multiply(divide(add(divide(multiply(12, 10), const_100), divide(multiply(8, 30), const_100)), add(10, 30)), const_100) | add(n2,n3)|multiply(n0,n2)|multiply(n1,n3)|divide(#1,const_100)|divide(#2,const_100)|add(#3,#4)|divide(#5,#0)|multiply(#6,const_100)| | gain |
an automotive test consisted of driving a car the same distance 3 separate times , first at an average rate of 4 miles per hour , then 5 miles per hour , then 6 miles per hour . if the test took 37 hours to complete , how many miles was the car driven during the entire test ? | x is the segment to be traveled x / 4 + x / 5 + x / 6 = 37 or 74 x / 120 = 37 x = 60 miles . total distance = 3 * 60 = 180 miles c | a ) 100 , b ) 120 , c ) 180 , d ) 200 , e ) 240 | c | multiply(divide(37, add(divide(const_1, 6), add(divide(const_1, 4), divide(const_1, 5)))), const_3) | divide(const_1,n1)|divide(const_1,n2)|divide(const_1,n3)|add(#0,#1)|add(#3,#2)|divide(n4,#4)|multiply(#5,const_3) | physics |
a can do a work in 12 days . b can do the same work in 24 days . if both a & b are working together in how many days they will finish the work ? | "a rate = 1 / 12 b rate = 1 / 24 ( a + b ) rate = ( 1 / 12 ) + ( 1 / 24 ) = 1 / 8 a & b finish the work in 8 days correct option is c" | a ) 3 , b ) 5 , c ) 8 , d ) 2 , e ) 6 | c | divide(multiply(12, 24), add(12, 24)) | add(n0,n1)|multiply(n0,n1)|divide(#1,#0)| | physics |
a , b and c invests rs . 2000 , rs . 3000 and rs . 4000 in a business . after 2 year a removed his money ; b and c continued the business for two more year . if the net profit after 3 years be rs . 4000 , then a ' s share in the profit is ? | "2 * 24 : 3 * 36 : 4 * 36 4 : 9 : 12 4 / 25 * 4000 = 640 answer : e" | a ) 460 , b ) 289 , c ) 220 , d ) 400 , e ) 640 | e | multiply(divide(2000, add(2000, add(multiply(3000, 2), multiply(4000, 2)))), 2000) | multiply(n1,n3)|multiply(n2,n3)|add(#0,#1)|add(n0,#2)|divide(n0,#3)|multiply(n0,#4)| | gain |
balls of equal size are arranged in rows to form an equilateral triangle . the top most row consists of one ball , the 2 nd row of two balls and so on . if 19 balls are added , then all the balls can be arranged in the shape of square and each of the sides of the square contain 8 balls less than the each side of the tr... | "as expected , this question boils down to 2 equation , consider total number of balls in triangle = t and number of balls in last row = x . 1 + 2 + 3 + . . . + x = t x ( x + 1 ) / 2 = t - - - - ( a ) as mentioned in the question , side of a square will be ( x - 8 ) and total number of balls in square will be ( t + 19 ... | a ) 465 , b ) 2209 , c ) 2878 , d ) 1210 , e ) 1560 | a | multiply(subtract(19, multiply(const_4, const_100)), add(multiply(subtract(19, multiply(const_4, const_100)), 2), const_1)) | multiply(const_100,const_4)|subtract(n1,#0)|multiply(n0,#1)|add(#2,const_1)|multiply(#3,#1)| | general |
the average of 5 quantities is 10 . the average of 3 of them is 4 . what is the average of remaining 2 numbers ? | answer : a ( 5 x 10 - 3 x 4 ) / 2 = 19 | a ) 19 , b ) 14 , c ) 8 , d ) 9.5 , e ) none of these | a | divide(subtract(multiply(5, 10), multiply(3, 4)), 2) | multiply(n0,n1)|multiply(n2,n3)|subtract(#0,#1)|divide(#2,n4) | general |
a tank is 25 m long 12 m wide and 6 m deep . the cost of plastering its walls and bottom at 70 paise per sq m is | "explanation : area to be plastered = [ 2 ( l + b ) ã — h ] + ( l ã — b ) = [ 2 ( 25 + 12 ) ã — 6 ] + ( 25 ã — 12 ) = 744 sq m cost of plastering = 744 ã — ( 70 / 100 ) = rs . 520.80 answer : d" | a ) rs . 258.80 , b ) rs . 358.80 , c ) rs . 458.80 , d ) rs . 520.80 , e ) none of these | d | multiply(divide(70, const_100), add(multiply(25, 12), add(multiply(const_2, multiply(25, 6)), multiply(multiply(12, 6), const_2)))) | divide(n3,const_100)|multiply(n0,n2)|multiply(n1,n2)|multiply(n0,n1)|multiply(#1,const_2)|multiply(#2,const_2)|add(#4,#5)|add(#6,#3)|multiply(#7,#0)| | physics |
how many odd factors does 520 have ? | "start with the prime factorization : 520 = 2 * 5 * 13 for odd factors , we put aside the factor of two , and look at the other prime factors . set of exponents = { 1 , 1 } plus 1 to each = { 2 , 2 } product = 2 * 2 = 4 therefore , there are 4 odd factors of 520 . answer : b ." | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 8 | b | add(add(add(const_4, const_2), const_1), const_1) | add(const_2,const_4)|add(#0,const_1)|add(#1,const_1)| | other |
if the sum of two numbers is 22 and the sum of their squares is 460 , then the product of the numbers is | "according to the given conditions x + y = 22 and x ^ 2 + y ^ 2 = 460 now ( x + y ) ^ 2 = x ^ 2 + y ^ 2 + 2 xy so 22 ^ 2 = 460 + 2 xy so xy = 24 / 2 = 12 answer : c" | a ) 40 , b ) 44 , c ) 12 , d ) 88 , e ) 48 | c | divide(subtract(power(22, const_2), 460), const_2) | power(n0,const_2)|subtract(#0,n1)|divide(#1,const_2)| | general |
what is the value of n if the sum of the consecutive odd integers from 1 to n equals 169 ? | # of terms = ( n - 1 / 2 ) + 1 { ( last term - first term ) / 2 + 1 | sum = ( 1 + n ) / 2 * # of terms = ( n + 1 ) ^ 2 / 4 = 169 n + 1 = 13 * 2 n + 1 = 26 n = 25 . answer : b | a ) 47 , b ) 25 , c ) 37 , d ) 33 , e ) 29 | b | add(subtract(multiply(sqrt(169), const_2), multiply(const_2, 1)), 1) | multiply(n0,const_2)|sqrt(n1)|multiply(#1,const_2)|subtract(#2,#0)|add(n0,#3) | general |
the total number of plums that grew during each year on a certain plum tree was equal to the number of plums that grew during the previous year , less the age of the tree in years ( rounded to the nearest lower integer . ) during its 3 rd year , this plum tree grew 50 plums . how many plums did it grow during its 6 th ... | the answer shud be 38 . ( 50 - 3 - 4 - 5 = 38 ) yes , if you go the 50 - 3 way , then it will be 38 . if instead you go the 50 - 4 way ( which they have suggested ) , the answer will be 35 . as i said , there is ambiguity here . answer is c | a ) a ) 41 , b ) b ) 38 , c ) c ) 35 , d ) d ) 29 , e ) e ) 28 | c | subtract(subtract(subtract(50, const_4), add(const_4, const_1)), 6) | add(const_1,const_4)|subtract(n1,const_4)|subtract(#1,#0)|subtract(#2,n2) | general |
a can finish a work in 8 days and b can do same work in half the time taken by a . then working together , what part of same work they can finish in a day ? | "explanation : please note in this question , we need to answer part of work for a day rather than complete work . it was worth mentioning here because many do mistake at this point in hurry to solve the question so lets solve now , a ' s 1 day work = 1 / 8 b ' s 1 day work = 1 / 4 [ because b take half the time than a... | a ) 3 / 5 , b ) 3 / 6 , c ) 3 / 7 , d ) 3 / 8 , e ) none of these | d | add(divide(const_1, 8), multiply(divide(const_1, 8), const_2)) | divide(const_1,n0)|multiply(#0,const_2)|add(#0,#1)| | physics |
donovan and michael are racing around a circular 500 - meter track . if donovan runs each lap in 45 seconds and michael runs each lap in 40 seconds , how many laps will michael have to complete in order to pass donovan , assuming they start at the same time ? | "one way of approaching this question is by relative speed method 1 . speed / rate of donovan = distance / time = > 500 / 45 = > 100 / 9 2 . speed / rate of michael = distance / time = > 500 / 40 = > 50 / 4 relative speed between them = 50 / 4 - 100 / 9 = > 25 / 36 ( we subtract the rates if moving in the same directio... | a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 13 | a | divide(divide(500, subtract(divide(500, 40), divide(500, 45))), 40) | divide(n0,n2)|divide(n0,n1)|subtract(#0,#1)|divide(n0,#2)|divide(#3,n2)| | physics |
if m = | | n – 3 | – 2 | , for how many values of n is m = 6 ? | "m = | | n – 3 | – 2 | can be 4 only and only when n - 3 = + / - 8 . so there are 2 values of n answer : e" | a ) 1 , b ) 3 , c ) 4 , d ) 5 , e ) 2 | e | subtract(add(6, 2), 6) | add(n1,n2)|subtract(#0,n2)| | physics |
in order to obtain an income of rs . 650 from 10 % stock at rs . 96 , one must make an investment of : | "explanation : to obtain rs . 10 , investment = rs . 96 . to obtain rs . 650 , investment = = rs . 6240 answer : b ) 6240" | a ) 3488 , b ) 6240 , c ) 2776 , d ) 2889 , e ) 7721 | b | multiply(divide(96, 10), 650) | divide(n2,n1)|multiply(n0,#0)| | gain |
in a group of 68 students , each student is registered for at least one of 3 classes – history , math and english . 20 - one students are registered for history , 20 - 6 students are registered for math , and thirty - 4 students are registered for english . if only 3 students are registered for all 3 classes , how many... | a u b u c = a + b + c - ab - bc - ac + abc 68 = 21 + 26 + 34 - ab - bc - ac + 3 = > ab + bc + ac = 16 exactly two classes = ab + bc + ac - 3 abc = 16 - 3 * 3 = 7 hence e | a ) 13 , b ) 10 , c ) 9 , d ) 8 , e ) 7 | e | subtract(subtract(add(add(add(20, const_1), add(20, 6)), add(add(20, const_10), 4)), 68), multiply(const_2, 3)) | add(n2,const_1)|add(n2,n4)|add(n2,const_10)|multiply(n1,const_2)|add(#0,#1)|add(n5,#2)|add(#4,#5)|subtract(#6,n0)|subtract(#7,#3) | general |
210 reputed college students were asked in a survey if they preferred windows or mac brand computers . 60 students claimed that they preferred mac to windows brand computers . one third as many of the students who preferred mac to windows , equally preferred both brands . 90 of the students had no preference . how many... | we are told that 60 students claimed that they preferred mac to windows , which means that 60 preferred mac but not windows , so # of students who preferred mac ( p ( a ) as you wrote ) , does not equal to 60 , it equals to 60 + 20 ( 20 is # of students who equally preferred both brands ) . also we are asked to find # ... | a ) 25 , b ) 40 , c ) 50 , d ) 60 , e ) 75 | b | subtract(210, add(add(60, 90), divide(60, const_3))) | add(n1,n2)|divide(n1,const_3)|add(#0,#1)|subtract(n0,#2) | other |
the time taken by a man to row his boat upstream is twice the time taken by him to row the same distance downstream . if the speed of the boat in still water is 18 kmph , find the speed of the stream ? | "the ratio of the times taken is 2 : 1 . the ratio of the speed of the boat in still water to the speed of the stream = ( 2 + 1 ) / ( 2 - 1 ) = 3 / 1 = 3 : 1 speed of the stream = 18 / 3 = 6 kmph . answer : d" | a ) 8 kmph , b ) 9 kmph , c ) 7 kmph , d ) 6 kmph , e ) 5 kmph | d | subtract(18, divide(multiply(18, const_2), const_3)) | multiply(n0,const_2)|divide(#0,const_3)|subtract(n0,#1)| | physics |
a mathematics teacher tabulated the marks secured by 35 students of 8 th class . the average of their marks was 62 . if the marks secured by reema was written as 50 instead of 85 then find the correct average marks up to two decimal places . | "total marks = 35 x 62 = 2170 corrected total marks = 2170 - 50 + 85 = 2205 correct average = 2205 / 35 = 63 answer : a" | a ) 63 , b ) 74.31 , c ) 72.43 , d ) 73.43 , e ) can not be determined | a | divide(subtract(multiply(35, 62), subtract(85, 50)), 35) | multiply(n0,n2)|subtract(n4,n3)|subtract(#0,#1)|divide(#2,n0)| | general |
on the xy - coordinate plane , point a lies on the y - axis and point b lies on the x - axis . points a , b , and c form a right triangle with a 90 - degree angle at point c and the triangle has an area of 20 . if ac is parallel to the x - axis , and bc is parallel to the y - axis , which of the following could be the ... | the area of the triangle formed will be : 1 / 2 * base * height = 1 / 2 * | ( x - coordinate of point c ) | * | ( y - coordinate of point c ) | = 20 thus | product of coordinates of point c | = 40 the answer is d . | ['a ) ( 8 , - 3 )', 'b ) ( 4 , 5 )', 'c ) ( - 5 , 7 )', 'd ) ( - 2 , - 20 )', 'e ) ( - 15 , - 3 )'] | d | divide(20, const_2) | divide(n1,const_2) | geometry |
63 men working 8 hours per day dig 30 m deep . how many extra men should be put to dig to a depth of 50 m working 6 hours per day ? | "( 63 * 8 ) / 30 = ( x * 6 ) / 50 = > x = 140 140 – 63 = 77 answer : b" | a ) a ) 87 , b ) b ) 77 , c ) c ) 66 , d ) d ) 55 , e ) e ) 97 | b | subtract(divide(multiply(divide(multiply(63, 8), 30), 50), 6), 63) | multiply(n0,n1)|divide(#0,n2)|multiply(n3,#1)|divide(#2,n4)|subtract(#3,n0)| | physics |
steve traveled the first 2 hours of his journey at 40 mph and the last 3 hours of his journey at 90 mph . what is his average speed of travel for the entire journey ? | "answer average speed of travel = total distance travelled / total time taken total distance traveled by steve = distance covered in the first 2 hours + distance covered in the next 3 hours . distance covered in the first 2 hours = speed * time = 40 * 2 = 80 miles . distance covered in the next 3 hours = speed * time =... | a ) 60 mph , b ) 56.67 mph , c ) 53.33 mph , d ) 70 mph , e ) 66.67 mph | d | add(divide(add(multiply(90, 3), multiply(40, 2)), add(3, 2)), subtract(divide(const_100, 3), const_0_33)) | add(n0,n2)|divide(const_100,n2)|multiply(n2,n3)|multiply(n0,n1)|add(#2,#3)|subtract(#1,const_0_33)|divide(#4,#0)|add(#6,#5)| | physics |
if x is to be chosen at random from the set { 1 , 2 , 3 , 4 } and y is to be chosen at random from the set { 5 , 6 , 7 } , what is the probability that xy will be odd ? | "probably the best way to solve would be to use 1 - p ( opposite event ) = 1 - p ( odd ) = 1 - p ( odd ) * p ( odd ) = 1 - 2 / 4 * 2 / 3 = 8 / 12 = 1 / 3 . answer : b ." | a ) 1 / 6 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 5 / 6 | b | divide(power(2, 2), multiply(choose(4, 2), choose(2, 1))) | choose(n3,n1)|choose(n1,n0)|power(n1,n1)|multiply(#0,#1)|divide(#2,#3)| | probability |
in march mrs bil ' s earnings were 80 percent of the bil family ' s total income . in april mrs bil earned 10 percent more than in may . if the rest of the family ' s income was the same both months , then , in june , mrs bil ' s earnings were approximately what percent of the bil family ' s total income ? | lets say the family income is 100 in may , lee earned 80 family income is 20 in june , lee earned 20 % more than may , so it is ( 80 + 20 * 80 / 100 = 96 ) family income is same 20 in june lee ' s income percent is 96 * 100 / 116 ~ 83 ans is a | a ) 83 , b ) 56 , c ) 34 , d ) 12 , e ) 26 | a | add(multiply(divide(multiply(divide(subtract(const_100, 10), const_100), 80), add(subtract(const_100, 80), multiply(divide(subtract(const_100, 10), const_100), 80))), const_100), const_4) | subtract(const_100,n1)|subtract(const_100,n0)|divide(#0,const_100)|multiply(n0,#2)|add(#3,#1)|divide(#3,#4)|multiply(#5,const_100)|add(#6,const_4) | general |
how many prime numbers are between 29 / 4 and 87 / 5 ? | 29 / 4 = 7 . xxx 87 / 5 = 17 . xxx so we need to find prime numbers between 7 ( exclusive ) - 16 ( inclusive ) there are 2 prime numbers 1113 hence answer will be ( b ) 2 b | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | b | floor(const_2) | floor(const_2) | general |
in digging a pond 25 m * 10 m * 5 m the volumes of the soil extracted will be ? | "25 * 10 * 5 = 1750 answer : a" | a ) 1750 , b ) 1000 , c ) 2866 , d ) 2787 , e ) 1121 | a | multiply(multiply(25, 10), 5) | multiply(n0,n1)|multiply(n2,#0)| | general |
the time taken by a man to row his boat upstream is twice the time taken by him to row the same distance downstream . if the speed of the boat in still water is 48 kmph , find the speed of the stream ? | "the ratio of the times taken is 2 : 1 . the ratio of the speed of the boat in still water to the speed of the stream = ( 2 + 1 ) / ( 2 - 1 ) = 3 / 1 = 3 : 1 speed of the stream = 48 / 3 = 16 kmph . answer : a" | a ) 16 kmph , b ) 19 kmph , c ) 14 kmph , d ) 11 kmph , e ) 13 kmph | a | subtract(48, divide(multiply(48, const_2), const_3)) | multiply(n0,const_2)|divide(#0,const_3)|subtract(n0,#1)| | physics |
what is the total number of integers between 100 and 500 ( exclusive ) that are divisible by 5 ? | 105 , 110 , 115 , . . . , 490,495 this is an equally spaced list ; you can use the formula : n = ( largest - smallest ) / ( ' space ' ) + 1 = ( 495 - 105 ) / ( 5 ) + 1 = 78 + 1 = 79 answer is c | a ) 51 , b ) 63 , c ) 79 , d ) 66 , e ) 42 | c | add(divide(subtract(500, 100), 5), const_1) | subtract(n1,n0)|divide(#0,n2)|add(#1,const_1) | general |
a shopkeeper sold 10 articles at the cost price of 12 articles . then find the profit % or lost % | here 10 articles selling price = 12 articles cost price so the difference = 12 - 10 = 2 % of profit = 2 * 100 / 10 = 20 % correct option is d | a ) - 10 % , b ) 10 % , c ) 15 % , d ) 20 % , e ) 25 % | d | multiply(divide(subtract(12, 10), 10), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100) | gain |
venus travels around the sun at a speed of approximately 21.9 miles per second . the approximate speed is how many miles per hour ? | 21.9 * 3600 = 78,840 the answer is b . | a ) 64,420 , b ) 78,840 , c ) 83,160 , d ) 88,970 , e ) 94,320 | b | multiply(21.9, const_3600) | multiply(n0,const_3600) | physics |
150 ml of 30 % sulphuric acid was added to approximate 400 ml of 12 % sulphuric acid solution . find the approximate concentration of the acid in the mixture ? | 150 ml solution contains 45 ml sulphuric acid 400 ml solution contains 48 ml sulphuric acid adding 550 ml solution contains 93 ml sulphuric acid 93 / 550 ≈ 90 / 540 = 1 / 6 answer is d | a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 6 , e ) 1 / 5 | d | divide(add(divide(multiply(150, 30), const_100), divide(multiply(400, 12), const_100)), add(150, 400)) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|divide(#1,const_100)|divide(#2,const_100)|add(#3,#4)|divide(#5,#0) | general |
ratio between rahul and deepak is 4 : 3 , after 6 years rahul age will be 18 years . what is deepak present age ? | "present age is 4 x and 3 x , = > 4 x + 6 = 18 = > x = 3 so deepak age is = 3 ( 3 ) = 9 answer : d" | a ) 22 , b ) 15 , c ) 77 , d ) 9 , e ) 182 | d | divide(multiply(subtract(18, 6), 3), 4) | subtract(n3,n2)|multiply(n1,#0)|divide(#1,n0)| | other |
a and b can do a piece of work in 8 days . b and c can do it in 12 days and a and c in 16 days . working together they will complete the work in how many days ? | a + b = 1 / 8 b + c = 1 / 12 c + a = 1 / 16 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 2 ( a + b + c ) = 1 / 8 + 1 / 12 + 1 / 16 = 13 / 48 a + b + c = 13 / 96 96 / 13 = 7.38 days answer : d | a ) 3.33 days , b ) 8.33 days , c ) 3.99 days , d ) 7.38 days , e ) 2.88 days | d | inverse(divide(add(inverse(16), add(inverse(8), inverse(12))), const_2)) | inverse(n0)|inverse(n1)|inverse(n2)|add(#0,#1)|add(#3,#2)|divide(#4,const_2)|inverse(#5) | physics |
a circle graph shows how the megatech corporation allocates its research and development budget : 14 % microphotonics ; 19 % home electronics ; 10 % food additives ; 24 % genetically modified microorganisms ; 8 % industrial lubricants ; and the remainder for basic astrophysics . if the arc of each sector of the graph i... | "14 % microphotonics ; 19 % home electronics ; 10 % food additives ; 24 % genetically modified microorganisms ; 8 % industrial lubricants ; 100 - ( 14 + 19 + 10 + 24 + 8 ) = 25 % basic astrophysics . 25 % of 360 ° is 90 ° . answer : b ." | a ) 8 ° , b ) 90 ° , c ) 18 ° , d ) 36 ° , e ) 52 ° | b | divide(multiply(subtract(const_100, add(add(add(add(14, 19), 10), 24), 8)), divide(const_3600, const_10)), const_100) | add(n0,n1)|divide(const_3600,const_10)|add(n2,#0)|add(n3,#2)|add(n4,#3)|subtract(const_100,#4)|multiply(#1,#5)|divide(#6,const_100)| | gain |
a will was written such that an estate would be divided among 4 beneficiaries with the remainder donated to charity . one stipulation in the will is that no one beneficiary can receive an amount within 20 % of another ' s amount . if one of the beneficiaries received $ 40000 , what is the smallest possible range betwee... | 1 st - 40,000 . 2 nd - 0.8 * 40,000 = 32,000 3 rd - 0.8 * 32,000 = 25,600 4 th - 0.8 * 25,600 = ~ 20,480 range = 40,000 - 20,480 = 19520 answer : e . | a ) $ 12436 , b ) $ 13344 , c ) $ 71234 , d ) $ 15736 , e ) $ 19403 | e | subtract(40000, multiply(multiply(multiply(40000, subtract(const_1, divide(20, const_100))), subtract(const_1, divide(20, const_100))), subtract(const_1, divide(20, const_100)))) | divide(n1,const_100)|subtract(const_1,#0)|multiply(n2,#1)|multiply(#2,#1)|multiply(#3,#1)|subtract(n2,#4) | general |
a box contains 2 red balls and 5 black balls . one by one , every ball is selected at random without replacement . what is the probability that the fourth ball selected is black ? | "my complicated version of your simple approach let the 5 black balls be bbbbb and 2 red balls be rrr they can be arranged in 8 slots _ _ _ _ _ _ _ _ in ( 7 ! ) / ( 5 ! x 2 ! ) if the fourth slot is black ball then the arrangement will be to fill _ _ _ b _ _ _ _ we have 6 slots and 4 black ( bbbb ) and 2 red ( rrr ) th... | a ) 1 / 4 , b ) 7 / 5 , c ) 1 / 2 , d ) 5 / 8 , e ) 2 / 3 | b | divide(add(2, 5), 5) | add(n0,n1)|divide(#0,n1)| | probability |
sachin is younger than rahul by 4 years . if their ages are in the respective ratio of 7 : 9 , how old is sachin ? | "let rahul ' s age be x years . then , sachin ' s age = ( x - 7 ) years . ( x - 7 ) / x = 7 / 9 2 x = 63 = > x = 31.5 hence , sachin ' s age = ( x - 7 ) = 24.5 years . answer : e" | a ) 24.9 , b ) 24.6 , c ) 24.2 , d ) 24.8 , e ) 24.5 | e | multiply(divide(4, subtract(9, 7)), 7) | subtract(n2,n1)|divide(n0,#0)|multiply(n1,#1)| | other |
if a ( a + 5 ) = 6 and b ( b + 5 ) = 6 , where a ≠ b , then a + b = | "i . e . if a = 1 then b = - 6 or if a = - 6 then b = 1 but in each case a + b = - 6 + 1 = - 5 answer : option a" | a ) − 5 , b ) − 9 , c ) 2 , d ) 46 , e ) 48 | a | add(divide(6, const_10), divide(6, divide(6, const_10))) | divide(n1,const_10)|divide(n1,#0)|add(#0,#1)| | general |
the l . c . m . of two numbers is 36 . the numbers are in the ratio 2 : 3 . then sum of the number is : | "let the numbers be 2 x and 3 x . then , their l . c . m . = 6 x . so , 6 x = 36 or x = 6 . the numbers are 12 and 18 . hence , required sum = ( 12 + 18 ) = 30 . answer : option b" | a ) 28 , b ) 30 , c ) 40 , d ) 53 , e ) 54 | b | add(multiply(divide(36, multiply(2, 3)), 3), multiply(divide(36, multiply(2, 3)), 2)) | multiply(n1,n2)|divide(n0,#0)|multiply(n2,#1)|multiply(n1,#1)|add(#2,#3)| | other |
a man can do a piece of work in 6 days , but with the help of his son , he can do it in 3 days . in what time can the son do it alone ? | "son ' s 1 day ' s work = ( 1 / 3 ) - ( 1 / 6 ) = 1 / 6 the son alone can do the work in 6 days answer is c" | a ) 5 , b ) 7 , c ) 6 , d ) 8 , e ) 9 | c | divide(multiply(6, 3), subtract(6, 3)) | multiply(n0,n1)|subtract(n0,n1)|divide(#0,#1)| | physics |
a rectangular wall is covered entirely with two kinds of decorative tiles : regular and jumbo . 1 / 3 of the tiles are jumbo tiles , which have a length three times that of regular tiles and have the same ratio of length to width as the regular tiles . if regular tiles cover 70 square feet of the wall , and no tiles ov... | "the number of jumbo tiles = x . the number of regular tiles = 2 x . assume the ratio of the dimensions of a regular tile is a : a - - > area = a ^ 2 . the dimensions of a jumbo tile is 3 a : 3 a - - > area = 9 a ^ 2 . the area of regular tiles = 2 x * a ^ 2 = 70 . the area of jumbo tiles = x * 9 a ^ 2 = 4.5 ( 2 x * a ... | a ) 160 , b ) 240 , c ) 385 , d ) 440 , e ) 560 | c | add(70, multiply(divide(multiply(70, 3), const_2), 3)) | multiply(n2,n1)|divide(#0,const_2)|multiply(n1,#1)|add(n2,#2)| | geometry |
what is the remainder when 7 ^ 74 - 5 ^ 74 is divided by 8 ? | "easiest way for me : 7 ^ 74 - 5 ^ 74 = ( 49 ) ^ 37 - 25 ^ 37 = ( 48 + 1 ) ^ 37 - ( 24 + 1 ) ^ 37 - > remainder is 1 ^ 37 - 1 ^ 37 = 0 ans : c" | a ) 0 , b ) 2 , c ) 0 , d ) 3 , e ) none of these | c | reminder(multiply(74, 7), 5) | multiply(n0,n1)|reminder(#0,n2)| | general |
the sum of the present ages of a , b , c is 57 years . 3 years ago their ages were in the ratio 1 : 2 : 3 . what is the present age of a ? | three years ago : a : b : c = 1 : 2 : 3 let a = 1 x , b = 2 x , and c = 3 x . today : ( x + 3 ) + ( 2 x + 3 ) + ( 3 x + 3 ) = 57 x = 8 so the present age of a is x + 3 = 11 the answer is d . | a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | d | add(divide(subtract(57, multiply(3, const_3)), add(add(1, 2), 3)), 3) | add(n2,n3)|multiply(n1,const_3)|add(n1,#0)|subtract(n0,#1)|divide(#3,#2)|add(n1,#4) | general |
a parking garage rents parking spaces for $ 12 per week or $ 30 per month . how much does a person save in a year by renting by the month rather than by the week ? | "12 $ per week ! an year has 52 weeks . annual charges per year = 52 * 12 = 624 $ 30 $ per month ! an year has 12 months . annual charges per year = 12 * 30 = 360 $ 624 - 360 = 264 ans e" | a ) $ 140 , b ) $ 160 , c ) $ 220 , d ) $ 240 , e ) $ 264 | e | subtract(multiply(add(multiply(12, add(const_3, const_2)), const_2), 12), multiply(30, 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 |
the profit earned by selling an article for rs . 753 is equal to the loss incurred when the same article is sold for rs . 455 . what should be the sale price for making 50 % profit ? | "let c . p = rs . x then as given , ( 753 - x ) = ( x - 455 ) 2 x = 1208 x = 604 there fore s . p = 150 % of 604 = > rs . = rs . 906 answer : c" | a ) 706 , b ) 786 , c ) 906 , d ) 896 , e ) 696 | c | multiply(subtract(753, divide(subtract(753, 455), const_2)), add(const_1, divide(50, const_100))) | divide(n2,const_100)|subtract(n0,n1)|add(#0,const_1)|divide(#1,const_2)|subtract(n0,#3)|multiply(#2,#4)| | gain |
the sum of a number and its square is 1406 . what is the number ? | explanation : by option verification 37 + 37 × 37 = 1406 answer : d | ['a ) 1409', 'b ) 1402', 'c ) 1221', 'd ) 1406', 'e ) 8828'] | d | divide(add(sqrt(subtract(const_1, multiply(multiply(add(add(const_1000, multiply(const_4, const_100)), multiply(const_2, const_3)), subtract(const_1, const_2)), const_4))), subtract(const_1, const_2)), const_2) | multiply(const_100,const_4)|multiply(const_2,const_3)|subtract(const_1,const_2)|add(#0,const_1000)|add(#3,#1)|multiply(#4,#2)|multiply(#5,const_4)|subtract(const_1,#6)|sqrt(#7)|add(#8,#2)|divide(#9,const_2) | geometry |
1 , 1 , 2 , 1 , 2 , 3 , 1 , 2 , 3 , 4 , 1 , 2 , __ ? | "1,1 , 2,1 , 2,3 , 1,2 , 3,4 , 1,2 , __ ? looking at the series 1 12 123 1234 12 ( 3 ) 45 so the answer is 3 . answer : c" | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | multiply(const_3.0, 1) | multiply(n5,n0)| | general |
the banker ' s gain on a sum due 6 years hence at 12 % per annum is rs . 756 . what is the banker ' s discount ? | "explanation : td = ( bg × 100 ) / tr = ( 756 × 100 ) / ( 6 × 12 ) = rs . 1050 bg = bd – td = > 756 = bd - 1050 = > bd = 1050 + 756 = 1806 answer : option d" | a ) 1240 , b ) 1120 , c ) 1190 , d ) 1806 , e ) none of these | d | add(756, divide(multiply(756, const_100), multiply(12, 6))) | multiply(n2,const_100)|multiply(n0,n1)|divide(#0,#1)|add(n2,#2)| | gain |
one fourth of one third of two fifth of a number is 35 . what will be 40 % of that number | "explanation : ( 1 / 4 ) * ( 1 / 3 ) * ( 2 / 5 ) * x = 35 then x = 35 * 30 = 1050 40 % of 1050 = 420 answer : option b" | a ) a ) 140 , b ) b ) 420 , c ) c ) 180 , d ) d ) 200 , e ) e ) 220 | b | divide(multiply(divide(35, multiply(multiply(divide(const_1, const_4), divide(const_1, const_3)), divide(const_2, add(const_2, const_3)))), 40), const_100) | add(const_2,const_3)|divide(const_1,const_4)|divide(const_1,const_3)|divide(const_2,#0)|multiply(#1,#2)|multiply(#3,#4)|divide(n0,#5)|multiply(n1,#6)|divide(#7,const_100)| | gain |
a flagstaff 17.5 m high casts a shadow of length 40.25 m . what will be the height of a building , which casts a shadow of length 28.75 m under similar conditions ? | explanation : let the required height of the building be x meter more shadow length , more height ( direct proportion ) hence we can write as ( shadow length ) 40.25 : 28.75 : : 17.5 : x ⇒ 40.25 × x = 28.75 × 17.5 ⇒ x = 28.75 × 17.5 / 40.25 = 2875 × 175 / 40250 = 2875 × 7 / 1610 = 2875 / 230 = 575 / 46 = 12.5 answer : ... | a ) 12.5 m , b ) 10.5 m , c ) 14 , d ) 12 , e ) 10 | a | multiply(28.75, divide(17.5, 40.25)) | divide(n0,n1)|multiply(n2,#0) | physics |
in a party every person shakes hands with every other person . if there were a total of 153 handshakes in the party then what is the number of persons present in the party ? | "explanation : let the number of persons be n â ˆ ´ total handshakes = nc 2 = 153 n ( n - 1 ) / 2 = 153 â ˆ ´ n = 18 answer : option d" | a ) 15 , b ) 16 , c ) 17 , d ) 18 , e ) 19 | d | divide(add(sqrt(add(multiply(multiply(153, const_2), const_4), const_1)), const_1), const_2) | multiply(n0,const_2)|multiply(#0,const_4)|add(#1,const_1)|sqrt(#2)|add(#3,const_1)|divide(#4,const_2)| | general |
in the xy - coordinate system , if ( m , n ) and ( m + 1 , n + k ) are two points on the line with the equation x = 2 y + 5 , then k = | "since ( m , n ) and ( m + 2 , n + k ) are two points on the line with the equation x = 2 y + 5 they should satisfy m = 2 n + 5 and m + 1 = 2 * ( n + k ) + 5 . by 1 st equation we have m - 2 n = 5 and by 2 nd equation m - 2 n = 2 k + 4 - - - > 5 = 2 k + 4 - - - > k = 1 / 2 . the answer is , therefore , ( a ) ." | a ) 1 / 2 , b ) 1 , c ) 2 , d ) 5 / 2 , e ) 4 | a | divide(subtract(add(5, 1), 5), 2) | add(n0,n2)|subtract(#0,n2)|divide(#1,n1)| | general |
if a lends rs . 25000 to b at 10 % per annum and b lends the same sum to c at 11.5 % per annum then the gain of b in a period of 3 years is ? | "( 25000 * 1.5 * 3 ) / 100 = > 1125 answer : a" | a ) 1125 , b ) 1225 , c ) 1325 , d ) 1145 , e ) 2125 | a | subtract(divide(multiply(multiply(25000, 11.5), 3), const_100), divide(multiply(multiply(25000, 10), 3), const_100)) | multiply(n0,n2)|multiply(n0,n1)|multiply(#0,n3)|multiply(n3,#1)|divide(#2,const_100)|divide(#3,const_100)|subtract(#4,#5)| | gain |
50 is what percent of 125 ? | 50 / 125 × 100 = 50 % answer : b | a ) 35 % , b ) 40 % , c ) 45 % , d ) 50 % , e ) 55 % | b | multiply(divide(50, 125), const_100) | divide(n0,n1)|multiply(#0,const_100)| | gain |
in a certain corporation , there are 300 male employees and 150 female employees . it is known that 10 % of the male employees have advanced degrees and 40 % of the females have advanced degrees . if one of the 450 employees is chosen at random , what is the probability this employee has an advanced degree or is female... | p ( female ) = 150 / 450 = 1 / 3 p ( male with advanced degree ) = 0.1 * 300 / 450 = 30 / 450 = 1 / 15 the sum of the probabilities is 6 / 15 = 2 / 5 the answer is c . | a ) 1 / 5 , b ) 5 / 15 , c ) 6 / 15 , d ) 3 / 10 , e ) 4 / 15 | c | add(divide(multiply(subtract(const_1, divide(40, multiply(10, 10))), 150), 450), divide(add(multiply(divide(10, multiply(10, 10)), 300), multiply(divide(40, multiply(10, 10)), 150)), 450)) | multiply(n2,n2)|divide(n3,#0)|divide(n2,#0)|multiply(n0,#2)|multiply(n1,#1)|subtract(const_1,#1)|add(#3,#4)|multiply(n1,#5)|divide(#7,n4)|divide(#6,n4)|add(#8,#9) | other |
the least number by which 72 must be multiplied in order to produce a multiple of 112 , is | "solution required numbers is divisible by 72 as well as by 112 , if it is divisible by their lcm , which is 1008 . now , 1008 when divided by 72 , gives quotient = 14 . required number = 14 . answer c" | a ) 6 , b ) 12 , c ) 14 , d ) 18 , e ) 20 | c | divide(112, gcd(72, 112)) | gcd(n0,n1)|divide(n1,#0)| | general |
find the number , difference between number and its 3 / 5 is 58 . | "explanation : let the number = x , then , x - ( 3 / 5 ) x = 58 , = > ( 2 / 5 ) x = 58 = > 2 x = 58 * 5 , = > x = 145 answer : option d" | a ) 140 , b ) 143 , c ) 144 , d ) 145 , e ) 146 | d | multiply(divide(58, const_2), 5) | divide(n2,const_2)|multiply(n1,#0)| | general |
the unit digit in the product ( 891 * 781 * 912 * 463 ) is : | "explanation : unit digit in the given product = unit digit in ( 1 * 1 * 2 * 3 ) = 6 answer : c" | a ) 2 , b ) 5 , c ) 6 , d ) 8 , e ) 10 | c | subtract(multiply(multiply(multiply(891, 781), 912), 463), subtract(multiply(multiply(multiply(891, 781), 912), 463), add(const_4, const_4))) | add(const_4,const_4)|multiply(n0,n1)|multiply(n2,#1)|multiply(n3,#2)|subtract(#3,#0)|subtract(#3,#4)| | general |
find the value of 3 + 2 . ( 8 - 3 ) | "3 + 2 . ( 8 - 3 ) = 3 + 2 ( 5 ) , = 3 + 2 * 5 , = 3 + 10 , = 13 , correct answer : ( d )" | a ) 25 , b ) 13 , c ) 17 , d ) 24 , e ) 19 | d | multiply(add(divide(2, 3), 3), 8) | divide(n1,const_100.0)|add(n0,#0)|multiply(#1,n2)| | general |
two numbers a and b are such that the sum of 5 % of a and 4 % of b is two - third of the sum of 6 % of a and 8 % of b . find the ratio of a : b . | "5 / 100 a + 4 / 100 b = 2 / 3 ( 6 / 100 a + 8 / 100 b ) = > 1 / 20 a + 1 / 25 b = 1 / 25 a + 4 / 75 b = > ( 1 / 20 - 1 / 25 ) b = 1 / 25 a + 4 / 75 b ( 1 / 20 - 1 / 25 ) a = ( 4 / 75 - 1 / 25 ) b 1 / 100 a = 1 / 75 b a / b = 100 / 75 = 4 / 3 req ratio = 4 : 3 answer d" | a ) 2 : 3 , b ) 1 : 1 , c ) 3 : 4 , d ) 4 : 3 , e ) 5 : 5 | d | divide(subtract(divide(multiply(divide(8, const_100), 4), const_3), divide(4, const_100)), subtract(divide(5, const_100), divide(multiply(divide(6, const_100), const_2), const_3))) | divide(n3,const_100)|divide(n1,const_100)|divide(n0,const_100)|divide(n2,const_100)|multiply(#0,n1)|multiply(#3,const_2)|divide(#4,const_3)|divide(#5,const_3)|subtract(#6,#1)|subtract(#2,#7)|divide(#8,#9)| | general |
when jessica withdrew $ 200 from her bank account , her account balance decreased by 2 / 5 . if she deposits an amount equal to 1 / 5 of the remaining balance , what will be the final balance in her bank account ? | "as per the question 200 = 2 a / 5 thus - a which is the total amount = 500 the amount thus left = 300 she then deposited 1 / 5 of 300 = 60 total amount in her account = 360 answer b" | a ) 300 , b ) 360 , c ) 400 , d ) 500 , e ) 575 | b | multiply(subtract(divide(200, subtract(1, divide(const_3, 5))), 200), add(1, divide(1, 5))) | divide(n3,n4)|divide(const_3,n2)|add(n3,#0)|subtract(n3,#1)|divide(n0,#3)|subtract(#4,n0)|multiply(#2,#5)| | general |
in a certain company , the ratio of the number of managers to the number of non - managers in any department must always be greater than 7 : 32 . in the company , what is the maximum number of non - managers in a department that has 9 managers ? | "9 / 7 * 32 = 41.1 the answer is c ." | a ) 39 , b ) 40 , c ) 41 , d ) 42 , e ) 43 | c | floor(divide(multiply(32, 9), 7)) | multiply(n1,n2)|divide(#0,n0)|floor(#1)| | general |
a man ' s speed with the current is 22 km / hr and the speed of the current is 5 km / hr . the man ' s speed against the current is | "man ' s rate in still water = ( 22 - 5 ) km / hr = 17 km / hr . man ' s rate against the current = ( 17 - 5 ) km / hr = 12 km / hr . answer : d" | a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 13 | d | subtract(subtract(22, 5), 5) | subtract(n0,n1)|subtract(#0,n1)| | gain |
machine z takes 2 more hours than machine b to make 20 widgets . if working together , the machines can make 25 widgets in 3 hours , how long will it take machine z to make 40 widgets ? | i approached this one by plugging in numbers . . . started with c . if 40 are made in 8 hours , then 20 are made in 4 hours . so time of z is 4 , and time of b is 2 . rate together : 20 / 4 + 20 / 2 = 5 + 10 = 15 . so in 1 hour , together make 15 widgets . in 3 hours = 45 . way too much . we can eliminate right away c ... | a ) 5 , b ) 6 , c ) 8 , d ) 10 , e ) 12 | e | multiply(divide(40, 20), multiply(2, 3)) | divide(n4,n1)|multiply(n0,n3)|multiply(#0,#1) | physics |
a certain pair of used shoes can be repaired for $ 13.50 and will last for 1 year . a pair of the same kind of shoes can be purchased new for $ 32.00 and will last for 2 years . the average cost per year of the new shoes is what percent greater than the cost of repairing the used shoes ? | "1 ) cost of repairing = 13.5 ( for one year ) , therefore for 2 years it would be $ 27 . 2 ) cost of new pair which will last for 2 years is $ 32 . percentage change formula = ( final value - initial value ) / ( initial value ) * 100 . in this case the final value would be the price of new shoe initial value would be ... | a ) 17.69 , b ) 14.85 , c ) 21.3 , d ) 16.5 , e ) 18.51 | e | multiply(const_100, divide(subtract(32.00, multiply(13.50, 2)), multiply(13.50, 2))) | multiply(n0,n3)|subtract(n2,#0)|divide(#1,#0)|multiply(#2,const_100)| | general |
the average weight of a group of boys is 20 kg . after a boy of weight 31 kg joins the group , the average weight of the group goes up by 1 kg . find the number of boys in the group originally ? | "let the number off boys in the group originally be x . total weight of the boys = 20 x after the boy weighing 31 kg joins the group , total weight of boys = 20 x + 31 so 20 x + 31 = 21 ( x + 1 ) = > x = 10 . answer : e" | a ) 12 , b ) 14 , c ) 18 , d ) 24 , e ) 10 | e | add(subtract(31, add(20, 1)), 1) | add(n0,n2)|subtract(n1,#0)|add(#1,n2)| | general |
if x = 55 + 2 q and y = 4 q + 41 , then for what value of q , x is equal to y ? | explanation : x = y < = > 55 + 2 q = 4 q + 41 < = > 2 q = 14 < = > q = 7 . answer : d | a ) - 14 , b ) 14 , c ) - 7 , d ) 7 , e ) 0 | d | divide(multiply(2, add(const_4, const_3)), 2) | add(const_3,const_4)|multiply(n1,#0)|divide(#1,n1) | general |
a man can swim in still water at 4 km / h , but takes twice as long to swim upstream than downstream . the speed of the stream is ? | "m = 4 s = x ds = 4 + x us = 4 + x 4 + x = ( 4 - x ) 2 4 + x = 8 - 2 x 3 x = 4 x = 0.75 answer : e" | a ) 1.7 , b ) 1.9 , c ) 1.1 , d ) 1.5 , e ) 0.75 | e | divide(4, const_3) | divide(n0,const_3)| | general |
there are 10 teams in a certain league and each team plays each of the other teams exactly twice . if each game is played by 2 teams , what is the total number of games played ? | "every team plays with 9 teams . . . so total no of matches = 10 x 9 = 90 . now , each match is played twice = > 90 x 2 but 2 teams play a match = > 90 x 2 / 2 = 90 . answer : e" | a ) 15 , b ) 16 , c ) 28 , d ) 56 , e ) 90 | e | divide(multiply(10, subtract(10, const_1)), const_2) | subtract(n0,const_1)|multiply(n0,#0)|divide(#1,const_2)| | general |
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