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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a man can row at 5 kmph in still water . if the velocity of current is 1 kmph and it takes him 1 hour to row to a plce and come back , how far is the place ? | "sol . speed downstream = ( 5 + 1 ) kmph = 6 kmph ; speed upstream = ( 5 - 1 ) kmph = 4 kmph . let the required distance be x km . then , x / 6 + x / 4 = 1 ⇔ 2 x + 3 x = 12 ⇔ 5 x = 12 ⇔ x = 2.4 km . answer b" | a ) 2 km , b ) 2.4 km , c ) 2.5 km , d ) 3.4 km , e ) none | b | divide(multiply(subtract(5, 1), const_3), 5) | subtract(n0,n1)|multiply(#0,const_3)|divide(#1,n0)| | physics |
a box contains 24 electric bulbs , out of which 4 are defective . two bulbs are chosen at random from this box . the probability that at least one of these is defective is | "solution p ( none is defective ) = 20 c 2 / 24 c 2 = 53 / 77 . p ( at least one is defective ) = ( 1 - 53 / 77 ) = 24 / 77 . answer d" | a ) 4 / 19 , b ) 7 / 19 , c ) 12 / 19 , d ) 24 / 77 , e ) none | d | subtract(const_1, divide(choose(subtract(24, 4), const_2), choose(24, const_2))) | choose(n0,const_2)|subtract(n0,n1)|choose(#1,const_2)|divide(#2,#0)|subtract(const_1,#3)| | probability |
one side of a rectangular field is 15 m and one of its diagonals is 17 m . find the area of the field in meter square | "other side = ( ( 17 ) 2 - ( 15 ) 2 ) ( 1 / 2 ) = ( 289 - 225 ) ( 1 / 2 ) = ( 64 ) ( 1 / 2 ) = 8 m . area = ( 15 x 8 ) m 2 = 120 m 2 . option c" | a ) 100 , b ) 110 , c ) 120 , d ) 130 , e ) 140 | c | rectangle_area(15, sqrt(subtract(power(17, const_2), power(15, const_2)))) | power(n1,const_2)|power(n0,const_2)|subtract(#0,#1)|sqrt(#2)|rectangle_area(n0,#3)| | geometry |
in a sample of christ college students , 30 percent are third year students and 90 percent are not second - year students . what fraction of those students who are not third - year students are second - year students ? | = 10 / 70 . = 1 / 7 my answer is e too | a ) 3 / 4 , b ) 2 / 3 , c ) 4 / 7 , d ) 1 / 2 , e ) 1 / 7 | e | divide(subtract(subtract(const_100, 30), subtract(90, 30)), subtract(const_100, 30)) | subtract(const_100,n0)|subtract(n1,n0)|subtract(#0,#1)|divide(#2,#0) | gain |
an electric pump can fill a tank in 3 hours . because of a leak in the tank , it took 12 hours to fill the tank . if the tank is full , how much time will the leak take to empty it ? | "work done by the leak in 1 hour = 1 / 3 - 1 / 12 = 1 / 4 the leak will empty the tank in 4 hours answer is a" | a ) 4 hours , b ) 12 hours , c ) 8 hours , d ) 5 hours , e ) 15 hours | a | divide(12, const_1) | divide(n1,const_1)| | physics |
in a school of 800 students , 45 % wear blue shirts , 23 % wear red shirts , 15 % wear green shirts , and the remaining students wear other colors . how many students wear other colors ( not blue , not red , not green ) ? | "45 + 23 + 15 = 83 % 100 – 83 = 17 % 800 * 17 / 100 = 136 the answer is c ." | a ) 120 , b ) 128 , c ) 136 , d ) 144 , e ) 152 | c | subtract(800, add(add(multiply(divide(45, const_100), 800), multiply(divide(23, const_100), 800)), multiply(divide(15, const_100), 800))) | divide(n1,const_100)|divide(n2,const_100)|divide(n3,const_100)|multiply(n0,#0)|multiply(n0,#1)|multiply(n0,#2)|add(#3,#4)|add(#6,#5)|subtract(n0,#7)| | gain |
what is the hcf of 2 / 3 , 4 / 12 and 6 / 5 | "explanation : hcf of fractions = hcf of numerators / lcm of denominators = ( hcf of 2 , 4 , 6 ) / ( lcm of 3 , 12 , 5 ) = 2 / 60 = 1 / 30 answer : option d" | a ) 7 / 45 , b ) 2 / 45 , c ) 4 / 15 , d ) 1 / 30 , e ) 9 / 45 | d | divide(2, 5) | divide(n0,n5)| | general |
michelle deposited a certain sum of money in a savings account on july 1 st , 2007 . she earns an 10 % interest compounded semiannually . the sum of money in the account on december 31 st , 2009 is approximately what percent of the initial deposit ? | "since michelle earns 10 % interest compounded semiannually , then she earns 5 % interest every 6 months . now , the simple interest earned in 5 periods ( 30 months = 5 * 6 months ) would be 5 % * 5 = 25 % . but , since the interest is compounded every 6 months , then there would be interest earned on interest ( very s... | a ) 127 % , b ) 120 % , c ) 121 % , d ) 135 % , e ) 140 % | a | multiply(power(add(1, divide(divide(10, const_100), const_2)), add(const_2, const_3)), const_100) | add(const_2,const_3)|divide(n2,const_100)|divide(#1,const_2)|add(#2,n0)|power(#3,#0)|multiply(#4,const_100)| | gain |
john want to buy a $ 100 trouser at the store , but he think it â € ™ s too expensive . finally , it goes on sale for $ 70 . what is the percent decrease ? | "the is always the difference between our starting and ending points . in this case , it â € ™ s 100 â € “ 70 = 30 . the â € œ original â € is our starting point ; in this case , it â € ™ s 100 . ( 30 / 100 ) * 100 = ( 0.3 ) * 100 = 30 % . b" | a ) 20 % , b ) 30 % , c ) 40 % , d ) 50 % , e ) 60 % | b | subtract(100, 70) | subtract(n0,n1)| | general |
the average salary of all the workers in a workshop is rs . 8000 . the average salary of 7 technicians is rs . 10000 and the average salary of the rest is rs . 6000 . the total number of workers in the workshop is : | "let the total number of workers be x . then , 8000 x = ( 10000 * 7 ) + 6000 ( x - 7 ) = 2000 x = 28000 = x = 14 . answer : e" | a ) 23 , b ) 19 , c ) 22 , d ) 20 , e ) 14 | e | add(7, divide(multiply(7, subtract(10000, 8000)), subtract(8000, 6000))) | subtract(n2,n0)|subtract(n0,n3)|multiply(n1,#0)|divide(#2,#1)|add(n1,#3)| | general |
a picture is copied onto a sheet of paper 8.5 inches by 10 inches . a 1.5 inch margin is left all around . what area in square inches does the picture cover ? | area covered by picture = ( 8.5 - 3 ) * ( 10 - 3 ) = 5.5 * 7 = 38.5 answer e | ['a ) 76', 'b ) 65', 'c ) 59.5', 'd ) 49', 'e ) 38.5'] | e | multiply(subtract(8.5, multiply(1.5, const_2)), subtract(10, multiply(1.5, const_2))) | multiply(n2,const_2)|subtract(n0,#0)|subtract(n1,#0)|multiply(#1,#2) | geometry |
a cistern 9 m long and 6 m wide contains water up to a breadth of 2 m 25 cm . find the total area of the wet surface . | "explanation : area of the wet surface = 2 [ lb + bh + hl ] - lb = 2 [ bh + hl ] + lb = 2 [ ( 6 * 2.25 + 9 * 2.25 ) ] + 9 * 6 = 122 m square option b" | a ) 142 m sqaure , b ) 122 m sqaure , c ) 135 m sqaure , d ) 188 m sqaure , e ) none of these | b | add(multiply(const_2, add(multiply(add(divide(25, const_100), 2), 6), multiply(add(divide(25, const_100), 2), 9))), multiply(6, 9)) | divide(n3,const_100)|multiply(n0,n1)|add(n2,#0)|multiply(n1,#2)|multiply(n0,#2)|add(#3,#4)|multiply(#5,const_2)|add(#6,#1)| | physics |
a tailor trims 4 feet from opposite edges of a square piece of cloth , and 3 feet from the other two edges . if 120 square feet of cloth remain , what was the length of a side of the original piece of cloth ? | let the original side of the square be x . ( x - 8 ) * ( x - 6 ) = 120 = 10 * 12 x = 18 the answer is a . | ['a ) 18', 'b ) 20', 'c ) 24', 'd ) 26', 'e ) 40'] | a | divide(add(add(multiply(const_2, 4), multiply(const_2, 3)), sqrt(add(multiply(const_4, subtract(120, multiply(multiply(const_2, 3), multiply(const_2, 4)))), power(add(multiply(const_2, 4), multiply(const_2, 3)), const_2)))), const_2) | multiply(n0,const_2)|multiply(n1,const_2)|add(#0,#1)|multiply(#1,#0)|power(#2,const_2)|subtract(n2,#3)|multiply(#5,const_4)|add(#6,#4)|sqrt(#7)|add(#2,#8)|divide(#9,const_2) | geometry |
in a company , 54 percent of the employees are men . if 60 percent of the employees are unionized and 70 percent of these are men , what percent of the non - union employees are women ? | the percent of employees who are unionized and men is 0.7 * 0.6 = 42 % the percent of employees who are unionized and women is 60 - 42 = 18 % 46 % of all employees are women , so non - union women are 46 % - 18 % = 28 % 40 % of all employees are non - union . the percent of non - union employees who are women is 28 % /... | a ) 55 % , b ) 60 % , c ) 65 % , d ) 70 % , e ) 75 % | d | multiply(const_100, divide(subtract(subtract(const_100, 54), subtract(60, multiply(60, divide(70, const_100)))), subtract(const_100, 60))) | divide(n2,const_100)|subtract(const_100,n0)|subtract(const_100,n1)|multiply(n1,#0)|subtract(n1,#3)|subtract(#1,#4)|divide(#5,#2)|multiply(#6,const_100) | gain |
10 women can complete a work in 7 days & 10 children take 14 days to complete the work . how many days will 5 women and 10 children take to complete the work ? | 1 woman ' s 1 day ' s work = 1 / 70 1 child ' s 1 day ' s work = 1 / 140 5 women and 10 children 1 day work = ( 5 / 70 + 10 / 140 ) = 1 / 7 so 5 women and 10 children will finish the work in 7 days . b ) | a ) 5 days , b ) 7 days , c ) 8 days , d ) 9 days , e ) 11 days | b | inverse(add(multiply(divide(const_1, multiply(10, 7)), 5), multiply(divide(const_1, multiply(10, 14)), 10))) | multiply(n0,n1)|multiply(n0,n3)|divide(const_1,#0)|divide(const_1,#1)|multiply(n4,#2)|multiply(n0,#3)|add(#4,#5)|inverse(#6) | physics |
5 n + 7 > 12 and 7 n - 5 < 44 ; n must be between which numbers ? | "5 n > 5 - - > n > 1 7 n < 49 - - > n < 7 1 < n < 7 answer : d" | a ) 1 and 8 , b ) 2 and 6 , c ) 0 and 9 , d ) 1 and 7 , e ) 2 and 9 | d | add(multiply(7, const_10), divide(add(44, 5), 7)) | add(n4,n5)|multiply(const_10,n1)|divide(#0,n3)|add(#2,#1)| | general |
the total cost of a vacation was divided among 3 people . if the total cost of the vacation had been divided equally among 4 people , the cost per person would have been $ 30 less . what was the total cost cost of the vacation ? | "c for cost . p price per person . c = 3 * p c = 4 * p - 120 substituting the value of p from the first equation onto the second we get p = 120 . plugging in the value of p in the first equation , we get c = 360 . which leads us to answer choice b" | a ) $ 200 , b ) $ 360 , c ) $ 400 , d ) $ 500 , e ) $ 600 | b | multiply(multiply(4, 3), divide(30, subtract(4, 3))) | multiply(n0,n1)|subtract(n1,n0)|divide(n2,#1)|multiply(#2,#0)| | general |
a survey was sent to 75 customers , 7 of whom responded . then the survey was redesigned and sent to another 60 customers , 9 of whom responded . by approximately what percent did the response rate increase from the original survey to the redesigned survey ? | "rate of first survey = 7 / 75 rate of second survey = 9 / 60 % response rate increase ( 9 / 60 - 7 / 75 ) / ( 7 / 75 ) = 60 % answer is d = 60 %" | a ) 2 % , b ) 5 % , c ) 14 % , d ) 60 % , e ) 73 % | d | floor(multiply(subtract(divide(9, 60), divide(7, 75)), const_100)) | divide(n3,n2)|divide(n1,n0)|subtract(#0,#1)|multiply(#2,const_100)|floor(#3)| | gain |
mary ' s income is 60 % more than tim ' s income and tim ' s income is 20 % less than juan ' s income . what % of juan ' s income is mary ' s income . | even i got 96 % j = 100 t = 100 * 0.8 = 80 m = 80 * 1.6 = 128 if mary ' s income is x percent of j m = j * x / 100 x = m * 100 / j = 128 * 100 / 100 = 128 ans : a | a ) 128 % , b ) b . 120 % , c ) 96 % , d ) 80 % , e ) 64 % | a | add(subtract(const_100, 20), multiply(subtract(const_100, 20), divide(60, const_100))) | divide(n0,const_100)|subtract(const_100,n1)|multiply(#0,#1)|add(#2,#1) | general |
a train 125 m long passes a man , running at 9 km / hr in the same direction in which the train is going , in 10 seconds . the speed of the train is ? | "speed of the train relative to man = ( 125 / 10 ) m / sec = ( 25 / 2 ) m / sec . [ ( 25 / 2 ) * ( 18 / 5 ) ] km / hr = 45 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 9 ) km / hr . x - 9 = 45 = = > x = 54 km / hr . answer : e" | a ) 11 , b ) 50 , c ) 88 , d ) 65 , e ) 54 | e | divide(divide(subtract(125, multiply(multiply(9, const_0_2778), 9)), 9), const_0_2778) | multiply(n1,const_0_2778)|multiply(n1,#0)|subtract(n0,#1)|divide(#2,n1)|divide(#3,const_0_2778)| | physics |
steve traveled the first 2 hours of his journey at 40 mph and the last 3 hours of his journey at 80 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 ) 64 mph , e ) 66.67 mph | d | add(divide(add(multiply(80, 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 |
a brick measures 20 cm * 10 cm * 7.5 cm how many bricks will be required for a wall 25 m * 2 m * 0.75 m ? | "25 * 2 * 0.75 = 20 / 100 * 10 / 100 * 7.5 / 100 * x 25 = 1 / 100 * x = > x = 25000 answer : d" | a ) 24000 , b ) 23000 , c ) 22000 , d ) 25000 , e ) 26000 | d | divide(divide(divide(multiply(multiply(multiply(25, const_100), multiply(2, const_100)), multiply(0.75, const_100)), 20), 10), 7.5) | multiply(n3,const_100)|multiply(n4,const_100)|multiply(n5,const_100)|multiply(#0,#1)|multiply(#3,#2)|divide(#4,n0)|divide(#5,n1)|divide(#6,n2)| | physics |
the ratio of the incomes of a and b is 5 : 4 and the ratio of their expenditure is 3 : 2 . if at the end of the year , each saves $ 1600 then the income of a is ? | let the income of a and b be $ 5 x and $ 4 x let their expenditures be $ 3 y and $ 2 y 5 x - 3 y = 1600 - - - - - - - 1 ) 4 x - 2 y = 1600 - - - - - - - 2 ) from 1 ) and 2 ) x = 800 a ' s income = 5 x = 5 * 800 = $ 4000 answer is b | a ) $ 1500 , b ) $ 4000 , c ) $ 2000 , d ) $ 2500 , e ) $ 3200 | b | multiply(divide(5, 2), 1600) | divide(n0,n3)|multiply(n4,#0) | other |
a wooden box of dimensions 8 m x 7 m x 6 m is to carry rectangularboxes of dimensions 8 cm x 7 cm x 6 cm . the maximum number ofboxes that can be carried in the wooden box , is | explanation : number = ( 800 * 700 * 600 ) / 8 * 7 * 6 = 1000000 answer : b | a ) 9800000 , b ) 1000000 , c ) 7500000 , d ) 1200000 , e ) none of these | b | divide(multiply(multiply(multiply(const_4.0, const_100), multiply(7, const_100)), multiply(8, const_100)), multiply(multiply(8, 7), 7)) | multiply(n2,const_100)|multiply(n1,const_100)|multiply(n2,n5)|multiply(#0,#1)|multiply(n1,#2)|multiply(#3,#0)|divide(#5,#4)| | physics |
two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of 16 kmph and 19 kmph respectively . when they meet , it is found that one train has traveled 60 km more than the other one . the distance between the two stations is ? | "1 h - - - - - 3 ? - - - - - - 60 12 h rs = 16 + 19 = 35 t = 12 d = 35 * 12 = 420 answer : d" | a ) 565 , b ) 444 , c ) 676 , d ) 420 , e ) 663 | d | add(multiply(divide(60, subtract(19, 16)), 16), multiply(divide(60, subtract(19, 16)), 19)) | subtract(n1,n0)|divide(n2,#0)|multiply(n0,#1)|multiply(n1,#1)|add(#2,#3)| | physics |
the speed of the boat in still water in 12 kmph . it can travel downstream through 60 kms in 3 hrs . in what time would it cover the same distance upstream ? | "still water = 12 km / hr downstream = 60 / 3 = 20 km / hr upstream = > > still water = ( u + v / 2 ) = > > 12 = u + 20 / 2 = 4 km / hr so time taken in upstream = 60 / 4 = 15 hrs answer : d" | a ) 8 hours , b ) 6 hours , c ) 4 hours , d ) 15 hours , e ) 6 hours | d | divide(60, subtract(12, subtract(divide(60, 3), 12))) | divide(n1,n2)|subtract(#0,n0)|subtract(n0,#1)|divide(n1,#2)| | physics |
if there are 200 questions in a 3 hr examination . among these questions are 25 type a problems , which requires twice as much as time be spent than the rest of the type b problems . how many minutes should be spent on type a problems ? | "x = time for type b prolems 2 x = time for type a problem total time = 3 hrs = 180 min 175 x + 25 * 2 x = 180 x = 180 / 225 x = 0.8 time taken for type a problem = 25 * 2 * 0.8 = 40 min answer : a" | a ) 40 min , b ) 62 min , c ) 70 min , d ) 74 min , e ) 76 min | a | multiply(multiply(const_2, divide(multiply(3, const_60), add(subtract(200, 25), multiply(const_2, 25)))), 25) | multiply(n1,const_60)|multiply(n2,const_2)|subtract(n0,n2)|add(#1,#2)|divide(#0,#3)|multiply(#4,const_2)|multiply(n2,#5)| | general |
the size of a television screen is given as the length of the screen ' s diagonal . if the screens were flat , then the area of a square 20 - inch screen would be how many square inches greater than the area of a square 18 - inch screen ? | "pythogoras will help here ! let the sides be x and diagonal be d then d ^ 2 = 2 x ^ 2 and area = x ^ 2 now plug in the given diagonal values to find x values and then subtract the areas ans will be 20 ^ 2 / 2 - 18 ^ 2 / 2 = 76 / 2 = 38 ans d ." | a ) 2 , b ) 4 , c ) 16 , d ) 38 , e ) 40 | d | divide(subtract(power(20, const_2), power(18, const_2)), const_2) | power(n0,const_2)|power(n1,const_2)|subtract(#0,#1)|divide(#2,const_2)| | geometry |
in a group of 8 engineers , 3 engineers have a phd degree ; the others have only an ms or bs degree . a team of 3 engineers is to be chosen which must have at least 1 engineer with a phd , how many different teams can be chosen ? | the problem asks for a combination , since order does n ' t matter . now , selecting r items from a set of n gives the combination formula : ncr = n ! / r ! ( n - r ) ! n = 8 r = 3 so , total teams is 8 c 3 = 8 ! / ( 3 ! ( 8 - 3 ) ! ) = 56 , and n = 8 - 3 = 5 r = 3 for teams without a phd is 5 c 3 = 5 ! / ( 3 ! ( 5 - 3... | a ) 56 , b ) 46 , c ) 64 , d ) 65 , e ) 10 | b | add(1, multiply(multiply(subtract(8, 3), 3), 3)) | subtract(n0,n1)|multiply(n1,#0)|multiply(n1,#1)|add(n3,#2) | other |
in the standard formulation of a flavored drink the ratio by volume of flavoring to corn syrup to water is 1 : 12 : 30 . in the sport formulation , the ratio of flavoring to corn syrup is three times as great as in the standard formulation , and the ratio of flavoring to water is half that of the standard formulation .... | "f : c : w 1 : 12 : 30 sport version : f : c 3 : 12 f : w 1 : 60 or 3 : 180 so c : f : w = 12 : 3 : 180 c / w = 12 / 180 = 3 ounces / x ounces x = 4 * 180 / 12 = 60 ounces of water d" | a ) 45 , b ) 50 , c ) 55 , d ) 60 , e ) 63 | d | multiply(divide(multiply(divide(1, 12), const_3), divide(divide(1, 30), const_2)), 4) | divide(n0,n1)|divide(n0,n2)|divide(#1,const_2)|multiply(#0,const_3)|divide(#3,#2)|multiply(n3,#4)| | other |
in an election between two candidates , the winner has a margin of 20 % of the votes polled . if 1000 people change their mind and vote for the loser , the loser would have won by a margin of 20 % of the votes polled . find the total number of votes polled in the election ? | winner - looser 60 % - 40 % if 1000 people change their mind and vote for the loser : winner - looser 40 % - 60 % thus 1,000 people compose 20 % of all voters , which means that the total number of votes is 5,000 . answer : b | a ) 10000 , b ) 5000 , c ) 15000 , d ) 20000 , e ) 6000 | b | divide(1000, divide(20, const_100)) | divide(n0,const_100)|divide(n1,#0) | gain |
there are 600 students in a school . the ratio of boys and girls in this school is 3 : 5 . find the total of girls & boys are there in this school ? | in order to obtain a ratio of boys to girls equal to 3 : 5 , the number of boys has to be written as 3 x and the number of girls as 5 x where x is a common factor to the number of girls and the number of boys . the total number of boys and girls is 600 . hence 3 x + 5 x = 600 solve for x 8 x = 600 x = 75 number of boys... | a ) 243 , b ) 265 , c ) 375 , d ) 390 , e ) 400 | c | multiply(divide(600, 5), 3) | divide(n0,n2)|multiply(n1,#0) | other |
8 is 2 % of a , and 2 is 8 % of b . c equals b / a . what is the value of c ? | "2 a / 100 = 8 a = 400 8 b / 100 = 2 b = 25 c = b / a = 25 / 400 = 1 / 16 the answer is a ." | a ) 1 / 16 , b ) 1 / 4 , c ) 1 , d ) 4 , e ) 8 | a | divide(multiply(divide(2, 8), const_100), multiply(divide(8, 2), const_100)) | divide(n1,n0)|divide(n0,n1)|multiply(#0,const_100)|multiply(#1,const_100)|divide(#2,#3)| | general |
a certain psychologist charges $ 30 more for the first hour of therapy than for each additional hour . if the total charge to a patient who receives 5 hours of therapy is $ 400 , what is the total charge to a patient who receives only 3 hours of therapy ? | "let the charge for first hour = x + 30 then charge for each additional hour = x x + 30 + 4 x = 400 = > 5 x = 370 = > x = 74 total charge for patient for 3 hours of therapy = x + 30 + 2 x = 3 x + 30 = 252 $ answer e" | a ) $ 120 , b ) $ 135 , c ) $ 150 , d ) $ 165 , e ) $ 252 | e | add(multiply(divide(subtract(400, 30), 5), 3), 30) | subtract(n2,n0)|divide(#0,n1)|multiply(n3,#1)|add(n0,#2)| | general |
in a race , the speeds of a and b are in the ratio 3 : 4 . a takes 30 minutes more than b to reach the destination . the time taken by a to reach the destination is ? | explanation : ratio of speeds = 3 : 4 distance remaining constant , the ratio of time taken = 4 : 3 a takes 0.5 hours more than bhence time taken by a = 4 × 0.5 = 2 hour answer : b | a ) 1 hr , b ) 2 hr , c ) 1.5 hr , d ) 2.5 hr , e ) 3 hr | b | divide(add(30, multiply(3, 30)), const_60) | multiply(n0,n2)|add(n2,#0)|divide(#1,const_60) | physics |
by selling an article at rs . 800 , a profit of 25 % is made . find its cost price ? | sp = 800 cp = ( sp ) * [ 100 / ( 100 + p ) ] = 800 * [ 100 / ( 100 + 25 ) ] = 800 * [ 100 / 125 ] = rs . 640 answer : c | a ) s . 486 , b ) s . 455 , c ) s . 640 , d ) s . 480 , e ) s . 489 | c | divide(multiply(800, const_100), add(const_100, 25)) | add(n1,const_100)|multiply(n0,const_100)|divide(#1,#0) | gain |
kanul spent $ 500 in buying raw materials , $ 400 in buying machinery and 10 % of the total amount he had as cash with him . what was the total amount ? | "let the total amount be x then , ( 100 - 10 ) % of x = 500 + 400 90 % of x = 900 90 x / 100 = 9000 / 9 x = $ 1000 answer is e" | a ) $ 1010 , b ) $ 1025 , c ) $ 1125 , d ) $ 1100 , e ) $ 1000 | e | divide(add(500, 400), subtract(const_1, divide(10, const_100))) | add(n0,n1)|divide(n2,const_100)|subtract(const_1,#1)|divide(#0,#2)| | gain |
the area of one square is x ^ 2 + 8 x + 16 and the area of another square is 4 x ^ 2 − 20 x + 25 . if the sum of the perimeters of both squares is 44 , what is the value of x ? | "the areas are ( x + 4 ) ^ 2 and ( 2 x - 5 ) ^ 2 . the lengths of the sides are x + 4 and 2 x - 5 . if we add the two perimeters : 4 ( x + 4 ) + 4 ( 2 x - 5 ) = 44 12 x = 48 x = 4 the answer is b ." | a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | b | divide(subtract(44, subtract(multiply(4, divide(8, 2)), 8)), 8) | divide(n1,n0)|multiply(#0,n3)|subtract(#1,n1)|subtract(n7,#2)|divide(#3,n1)| | general |
what is the probability of getting exactly three heads on five flips of a fair coin ? | "5 flips of a fair coin to get = hhhtt = no . of ways this can be achieved = 5 ! / 3 ! x 2 ! = 10 probability to get any of the above 10 arrangements ( hhhtt ) = ( 1 / 2 ) ^ 5 = 1 / 32 total probability = 1 / 32 x 10 = 5 / 16 answer : d" | a ) 1 / 32 , b ) 3 / 32 , c ) 1 / 4 , d ) 5 / 16 , e ) 1 / 2 | d | divide(const_2, choose(add(const_3, const_3), const_3)) | add(const_3,const_3)|choose(#0,const_3)|divide(const_2,#1)| | probability |
simple interest on a certain sum of money for 4 years at 10 % per annum is half the compound interest on rs . 3000 for 3 years at 12 % per annum . the sum placed on simple interest is | "solution c . i . = rs [ 3000 x ( 1 + 12 / 100 ) 3 - 3000 ] rs . ( 3000 x 112 / 100 x 112 / 100 x 112 / 100 - 3000 ) = rs . 1214.78 . sum = rs . [ 607.39 x 100 / 4 x 10 ] = rs . 1518.48 . answer d" | a ) rs . 3000 , b ) rs . 607.39 , c ) rs . 1214.78 , d ) rs . 1518.48 , e ) none | d | divide(multiply(divide(divide(add(divide(multiply(3000, 12), const_100), divide(multiply(add(3000, divide(multiply(3000, 12), const_100)), 12), const_100)), 3), 4), const_100), 10) | multiply(n2,n4)|divide(#0,const_100)|add(n2,#1)|multiply(n4,#2)|divide(#3,const_100)|add(#1,#4)|divide(#5,n3)|divide(#6,n0)|multiply(#7,const_100)|divide(#8,n1)| | gain |
in a certain lottery , the probability that a number between 14 and 20 , inclusive , is drawn is 1 / 6 . if the probability that a number 14 or larger is drawn is 2 / 3 , what is the probability that a number less than or equal to 20 is drawn ? | "you can simply use sets concept in this question . the formula total = n ( a ) + n ( b ) - n ( a and b ) is applicable here too . set 1 : number 14 or larger set 2 : number 20 or smaller 1 = p ( set 1 ) + p ( set 2 ) - p ( set 1 and set 2 ) ( combined probability is 1 because every number will be either 14 or moreor 2... | a ) 1 / 18 , b ) 1 / 6 , c ) 1 / 2 , d ) 1 / 3 , e ) 5 / 6 | c | subtract(add(14, divide(14, 3)), divide(2, 6)) | divide(n0,n3)|divide(n5,n6)|add(n0,#0)|subtract(#2,#1)| | general |
x is the product of each integer from 1 to 50 , inclusive and y = 100 ^ k , where k is an integer . what is the greatest value of k for which y is a factor of x ? | x = 1 * 2 * 3 * 4 * 5 * . . . . . . * 49 * 50 = 50 ! y = 100 ^ k 100 = 10 ^ 2 = 2 ^ 2 * 5 ^ 2 on prime factorization of x , power of 5 will always be greater than power of 2 concept : in every factorial value ( x ! ) , on prime factorization , the power of bigger prime number > the power of smaller prime number also , ... | a ) 0 , b ) 5 , c ) 6 , d ) 10 , e ) 12 | c | add(divide(50, const_10), const_1) | divide(n1,const_10)|add(#0,const_1) | general |
in a can , there is a mixture of milk and water in the ratio 3 : 7 . if it is filled with an additional 8 litres of milk the can would be full and ratio of milk and water would become 4 : 7 . find the capacity of the can ? | "let the capacity of the can be t litres . quantity of milk in the mixture before adding milk = 3 / 10 ( t - 8 ) after adding milk , quantity of milk in the mixture = 4 / 11 t . 4 t / 11 - 8 = 3 / 10 ( t - 8 ) 37 t = 880 - 264 = > t = 16.6 . answer : a" | a ) 16.6 , b ) 44 , c ) 27 , d ) 50 , e ) 56 | a | add(add(multiply(7, divide(8, subtract(multiply(divide(4, 7), 7), 3))), divide(8, subtract(multiply(divide(4, 7), 7), 3))), 8) | divide(n3,n4)|multiply(n1,#0)|subtract(#1,n0)|divide(n2,#2)|multiply(n1,#3)|add(#3,#4)|add(n2,#5)| | general |
dawson and henry are in a relay race . dawson runs the first leg of the course in 38 seconds . henry runs the second leg of the course in 7 seconds . what was the average time they took to run a leg of the course ? | ( 38 seconds + 7 seconds ) / 2 = 22.5 seconds correct option is : c | a ) 20.5 , b ) 21.5 , c ) 22.5 , d ) 23.5 , e ) 24.5 | c | divide(add(38, 7), const_2) | add(n0,n1)|divide(#0,const_2) | general |
if 1500 is increased by 20 % , and then reduced by y % , yielding 1080 , what is y ? | soln : - 1,500 * 1.2 = 1,800 . 1,800 ( 1 - y / 100 ) = 1,080 - - > ( 1 - y / 100 ) = 0.6 - - > y = 40 % . answer : b | a ) 20 % , b ) 40 % , c ) 60 % , d ) 80 % , e ) 100 % | b | multiply(subtract(const_1, divide(1080, multiply(1500, divide(add(const_100, 20), const_100)))), const_100) | add(n1,const_100)|divide(#0,const_100)|multiply(n0,#1)|divide(n2,#2)|subtract(const_1,#3)|multiply(#4,const_100) | gain |
if a is thrice as fast as b and together can do a work in 24 days . in how many days a alone can do the work ? | "a ’ s one day ’ s work = 1 / x b ’ s one day ’ s work = 1 / 3 x a + b ’ s one day ’ s work = 1 / x + 1 / 3 x = 1 / 24 = 3 + 1 / 3 x = 4 / 3 x = 1 / 24 x = 24 * 4 / 3 = 32 answer : d" | a ) 36 , b ) 42 , c ) 28 , d ) 32 , e ) 45 | d | inverse(divide(const_3, multiply(24, const_4))) | multiply(n0,const_4)|divide(const_3,#0)|inverse(#1)| | physics |
the figure above shows the dimensions of a semicircular cross section of a one - way tunnel . the single traffic lane is 12 feet wide and is equidistant from the sides of the tunnel . if vehicles must clear the top of the tunnel by at least ½ foot when they are inside the traffic lane , what should be the limit u on th... | "let ' s label the midpoint of the circle o . since the base of the semi - circle is 20 , we know that the diameter is 20 and , accordingly , the radius is 10 . we also know that the traffic lane is 12 feet long and there ' s an equal amount of space on either side , so the traffic lane extends 6 feet on either side of... | a ) 5 ½ ft , b ) 7 ½ ft , c ) 8 ½ ft , d ) 9 ½ ft , e ) 10 ft | b | subtract(divide(12, const_2), divide(const_1, const_2)) | divide(n0,const_2)|divide(const_1,const_2)|subtract(#0,#1)| | physics |
find the product of the local value and absolute value of 4 in 20468 ? | "local value of 4 = 4 x 100 = 400 place value of 4 = 4 there fore = 4 x 400 = 1600 e" | a ) 1000 , b ) 1200 , c ) 1340 , d ) 1450 , e ) 1600 | e | multiply(multiply(4, const_1000), 4) | multiply(n0,const_1000)|multiply(n0,#0)| | general |
a train 500 meters long completely crosses a 300 meters long bridge in 45 seconds . what is the speed of the train is ? | "s = ( 500 + 300 ) / 45 = 800 / 45 * 18 / 5 = 64 answer : d" | a ) 32 , b ) 28 , c ) 49 , d ) 64 , e ) 21 | d | divide(divide(add(500, 300), const_1000), divide(45, const_3600)) | add(n0,n1)|divide(n2,const_3600)|divide(#0,const_1000)|divide(#2,#1)| | physics |
a and b invests rs . 3000 and rs . 4000 respectively in a business . if a doubles his capital after 6 months . in what ratio should a and b divide that year ' s profit ? | "( 3 * 6 + 6 * 6 ) : ( 4 * 12 ) 54 : 48 = > 9 : 8 answer : b" | a ) 9 : 6 , b ) 9 : 8 , c ) 9 : 0 , d ) 9 : 2 , e ) 9 : 1 | b | divide(add(multiply(3000, 6), multiply(multiply(3000, const_2), 6)), multiply(4000, add(6, 6))) | add(n2,n2)|multiply(n0,n2)|multiply(n0,const_2)|multiply(n2,#2)|multiply(n1,#0)|add(#1,#3)|divide(#5,#4)| | gain |
we have boxes colored either red or blue . in each blue box there is a fixed number of blueberries . in each red box there is a fixed number of strawberries . if we dispose of one blue box for one additional red box , the total number of berries would increase by 30 , and the difference between the total number of stra... | "let x be the number of blueberries in each blue box . then there are x + 30 strawberries in each red box . x + ( x + 30 ) = 100 x = 35 the answer is a ." | a ) 35 , b ) 40 , c ) 45 , d ) 50 , e ) 55 | a | divide(subtract(100, 30), const_2) | subtract(n1,n0)|divide(#0,const_2)| | general |
there are an average of 117 jellybeans in each of 34 bags of various sizes . after adding another bag , the average number of jellybeans per bag increases by 7 . how many jellybeans are in the new bag ? | total jellybeans in 34 bags = 34 * 117 = 3978 total jellybeans in 35 bags = 35 * 124 = 4340 number of jellybeans in new bag = 3978 - 4340 = 362 jellybeans answer is e | a ) 124 , b ) 97 , c ) 117 , d ) 229 , e ) 362 | e | add(117, multiply(7, add(34, const_1))) | add(n1,const_1)|multiply(n2,#0)|add(n0,#1) | general |
a student has to obtain 33 % of the total marks to pass . he got 175 marks and failed by 89 marks . the maximum marks are ? | "let the maximum marks be x then , 33 % of x = 175 + 89 33 x / 100 = 264 x = 800 answer is c" | a ) 450 , b ) 300 , c ) 800 , d ) 610 , e ) 175 | c | divide(add(175, 89), divide(33, const_100)) | add(n1,n2)|divide(n0,const_100)|divide(#0,#1)| | general |
a solution of 90 litres contains milk and water in the ratio 7 : x . if 10 litres of water is added to the solution , the ratio becomes 3 : 2 , find the value of x ? | exp . total new quantity = original sol + water = 90 + 10 = 100 new ratio = 3 : 2 , new quantity of milk = 3 / 5 * 100 = 60 lit , new quantity of water = 2 / 5 * 100 = 40 lit water present initially = ( 40 - 10 ) = 30 lit ratio = 60 / 30 = 7 / x there for x = 3.5 answer : c | a ) 2 , b ) 1.5 , c ) 3.5 , d ) 2.5 , e ) 3 | c | divide(subtract(divide(divide(multiply(90, 7), const_10), divide(3, 2)), 10), divide(90, const_10)) | divide(n3,n4)|divide(n0,const_10)|multiply(n0,n1)|divide(#2,const_10)|divide(#3,#0)|subtract(#4,n2)|divide(#5,#1) | general |
a rectangle has a perimeter of 176 inches . the length of the rectangle is 8 inches more than its width . what is the area of the rectangle ? | explanation : we have : ( l - b ) = 8 and 2 ( l + b ) = 176 or ( l + b ) = 88 . solving the two equations , we get : l = 48 and b = 40 . area = ( l x b ) = ( 48 x 40 ) in 2 = 1920 in 2 . correct answer : b | ['a ) 1,408 square inches', 'b ) 1,920 square inches', 'c ) 1,936 square inches', 'd ) 2,304 square inches', 'e ) none of these'] | b | rectangle_area(add(divide(subtract(divide(176, const_2), 8), const_2), 8), divide(subtract(divide(176, const_2), 8), const_2)) | divide(n0,const_2)|subtract(#0,n1)|divide(#1,const_2)|add(n1,#2)|rectangle_area(#3,#2) | geometry |
in an election between the two candidates , the candidates who gets 70 % of votes polled is winned by 240 vote ’ s majority . what is the total number of votes polled ? | "explanation : note : majority ( 40 % ) = difference in votes polled to win ( 70 % ) & defeated candidates ( 30 % ) 40 % = 70 % - 30 % 40 % - - - - - > 280 ( 40 * 6 = 240 ) 100 % - - - - - > 600 ( 100 * 6 = 600 ) answer : option a" | a ) 600 , b ) 700 , c ) 800 , d ) 850 , e ) none of these | a | divide(multiply(const_100, 240), subtract(70, subtract(const_100, 70))) | multiply(n1,const_100)|subtract(const_100,n0)|subtract(n0,#1)|divide(#0,#2)| | gain |
in covering a distance of 42 km , abhay takes 2 hours more than sameer . if abhay doubles his speed , then he would take 1 hour less than sameer . abhay ' s speed is : | "let abhay ' s speed be x km / hr . then , 42 / x - 42 / 2 x = 3 6 x = 42 x = 7 km / hr . answer : option d" | a ) 5 kmph , b ) 6 kmph , c ) 6.25 kmph , d ) 7 kmph , e ) 7.8 kmph | d | divide(subtract(42, divide(42, 2)), add(1, 2)) | add(n1,n2)|divide(n0,n1)|subtract(n0,#1)|divide(#2,#0)| | physics |
a store sells chairs and tables . if the price of 2 chairs and 1 table is 60 % of the price of 1 chair and 2 tables , and the price of 1 table and 1 chair is $ 96 , what is the price , in dollars , of 1 table ? ( assume that every chair has the same price and every table has the same price . ) | let c = chair ; t = table 2 c + 1 t = 0.6 ( 1 c + 2 t ) or c ( 2 - 0.6 ) = 1.2 t - 1 t or 1.4 c = 0.2 t therefore c = 0.2 / 1.4 t = 1 / 7 t ic + 1 t = 96 or 1 / 7 t + 1 t = 96 therefore t = 96 * 7 / 8 = 84 b | a ) 81 , b ) 84 , c ) 85 , d ) 88 , e ) 90 | b | divide(subtract(multiply(96, const_2), multiply(96, divide(60, const_100))), add(divide(60, const_100), const_1)) | divide(n2,const_100)|multiply(n7,const_2)|add(#0,const_1)|multiply(n7,#0)|subtract(#1,#3)|divide(#4,#2) | gain |
the length of the bridge , which a train 140 meters long and travelling at 45 km / hr can cross in 30 seconds , is ? | speed = ( 45 * 5 / 18 ) m / sec = ( 25 / 2 ) m / sec . time = 30 sec . let the length of bridge be x meters . then , ( 140 + x ) / 30 = 25 / 2 = = > 2 ( 140 + x ) = 750 = = > x = 235 m . answer : c | a ) 328 , b ) 279 , c ) 235 , d ) 288 , e ) 211 | c | subtract(multiply(divide(multiply(45, speed(const_1000, const_1)), speed(const_3600, const_1)), 30), 140) | speed(const_1000,const_1)|speed(const_3600,const_1)|multiply(n1,#0)|divide(#2,#1)|multiply(n2,#3)|subtract(#4,n0)| | physics |
maths , physics and chemistry books are stored on a library shelf that can accommodate 25 books . currently , 20 % of the shelf spots remain empty . there are twice as many maths books as physics books and the number of physics books is 4 greater than that of chemistry books . among all the books , 12 books are soft co... | "first phase of this problem requires you to determine how many mathematics and chemistry books are even on the shelf . to do so , you have the equations : m + p + c = 20 ( since 4 / 5 of the 25 spots are full of books ) m = 2 p p = 4 + c from that , you can use substitution to get everything down to one variable . c =... | a ) 1 / 10 , b ) 3 / 20 , c ) 1 / 5 , d ) 1 / 4 , e ) 9 / 20 | e | subtract(add(divide(const_2, add(add(const_2, add(const_2, 4)), 12)), divide(add(const_2, add(const_2, 4)), add(add(const_2, add(const_2, 4)), 12))), multiply(divide(const_2, add(add(const_2, add(const_2, 4)), 12)), divide(add(const_2, add(const_2, 4)), add(add(const_2, add(const_2, 4)), 12)))) | add(n2,const_2)|add(#0,const_2)|add(n3,#1)|divide(const_2,#2)|divide(#1,#2)|add(#3,#4)|multiply(#3,#4)|subtract(#5,#6)| | general |
evaluate : 86 x 863 - 86 x 237 | "86 x 863 + 986 x 237 = 986 x ( 863 - 237 ) = 86 x 626 = 53836 . answer is b" | a ) 45823 , b ) 53836 , c ) 44824 , d ) 66524 , e ) none of them | b | subtract(86, multiply(multiply(863, 86), 237)) | multiply(n1,n2)|multiply(n3,#0)|subtract(n0,#1)| | general |
a man complete a journey in 40 hours . he travels first half of the journey at the rate of 20 km / hr and second half at the rate of 30 km / hr . find the total journey in km . | "0.5 x / 20 + 0.5 x / 30 = 40 - - > x / 20 + x / 30 = 80 - - > 5 x = 60 x 80 - - > x = ( 60 x 80 ) / 5 = 960 km . answer : a" | a ) 960 km , b ) 224 km , c ) 230 km , d ) 232 km , e ) 234 km | a | multiply(const_2, divide(multiply(multiply(20, 30), 40), add(20, 30))) | add(n1,n2)|multiply(n1,n2)|multiply(n0,#1)|divide(#2,#0)|multiply(#3,const_2)| | physics |
a large box contains 19 small boxes and each small box contains 25 chocolate bars . how many chocolate bars are in the large box ? | "the number of chocolate bars is equal to 19 * 25 = 475 correct answer a" | a ) 475 , b ) 350 , c ) 450 , d ) 550 , e ) 650 | a | multiply(19, 25) | multiply(n0,n1)| | general |
a store has 10 bottles of juice , including 6 bottles of apple juice . in the evening , 6 bottles of juice are sold one by one . what is the probability of selling 4 bottles of apple juice among the 6 bottles ? assume that every bottle has an equal chance of being bought . | "the total number of ways to sell 6 bottles from 10 is 10 c 6 = 210 . the number of ways to sell 4 bottles of apple juice is 6 c 4 * 4 c 2 = 15 * 6 = 90 p ( selling 4 bottles of apple juice ) = 90 / 210 = 9 / 21 = 3 / 7 the answer is e ." | a ) 2 / 5 , b ) 3 / 5 , c ) 2 / 3 , d ) 1 / 2 , e ) 3 / 7 | e | divide(choose(6, 4), choose(10, 6)) | choose(n1,n3)|choose(n0,n1)|divide(#0,#1)| | probability |
in a class of 50 students , 18 take chorus , 26 take band , and 2 take both chorus and band . how many students in the class are not enrolled in either chorus or band ? | 16 + 2 + 24 + x = 50 42 + x = 50 x = 8 students a ) | a ) 8 , b ) 12 , c ) 15 , d ) 16 , e ) 18 | a | subtract(50, add(add(subtract(18, 2), subtract(26, 2)), 2)) | subtract(n1,n3)|subtract(n2,n3)|add(#0,#1)|add(n3,#2)|subtract(n0,#3) | other |
a shopkeeper sold an article offering a discount of 5 % and earned a profit of 34.9 % . 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 ã — ( 100 + 34.9 ) / ( 100 â ˆ ’ 5 ) = 142 hence , required % profit = 142 â € “ 100 = 42 % answer e" | a ) 28.5 , b ) 27.675 , c ) 30 , d ) data inadequate , e ) 42 | e | subtract(divide(multiply(add(const_100, 34.9), 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 |
how long will a boy take to run round a square field of side 30 meters , if he runs at the rate of 12 km / hr ? | "speed = 12 km / hr = 12 * 5 / 18 = 10 / 3 m / sec distance = 30 * 4 = 120 m time taken = 120 * 3 / 10 = 36 sec answer is e" | a ) 52 sec , b ) 45 sec , c ) 60 sec , d ) 25 sec , e ) 36 sec | e | divide(multiply(30, const_4), multiply(12, divide(const_1000, const_3600))) | divide(const_1000,const_3600)|multiply(n0,const_4)|multiply(n1,#0)|divide(#1,#2)| | gain |
the number of timeshare condos available at sunset beach is 2 / 5 the number of timeshare condos available at playa del mar . if the total number of timeshare condos available at the two beaches combined is 280 , what is the difference between the number of condos available at sunset beach and the number of condos avai... | "let x be the number of timeshare condos available at playa del mar . then number of timeshare condos available at sunset beach = 3 / 5 x we know , x + 2 / 5 x = 280 hence , x = 200 . so , number of timeshare condos available at playa del mar = 200 the difference between the number of condos available at sunset beach a... | a ) 60 , b ) 90 , c ) 120 , d ) 150 , e ) 240 | c | add(divide(multiply(280, 2), 5), multiply(2, 5)) | multiply(n0,n2)|multiply(n0,n1)|divide(#0,n1)|add(#2,#1)| | general |
the price of a coat in a certain store is $ 500 . if the price of the coat is to be reduced by $ 350 , by what percent is the price to be reduced ? | "price of a coat in a certain store = $ 500 the price of the coat is to be reduced by $ 350 % change = ( final value - initial value ) * 100 / initial value % reduction = ( reduction in price ) * 100 / initial value i . e . % reduction = ( 350 ) * 100 / 500 = 70 % answer : option a" | a ) 70 % , b ) 15 % , c ) 20 % , d ) 25 % , e ) 30 % | a | multiply(divide(350, 500), const_100) | divide(n1,n0)|multiply(#0,const_100)| | gain |
ratio of two numbers x and y is 7 : 9 . if x is increased by 20 % and y is increased by 33 then the new ratio becomes 1 : 5 . what is the ratio 2 y : ( x + 4 ) | let : x = 7 n y = 9 n 8.4 n / ( 9 n + 33 ) = 1 / 5 42 n = 9 n + 33 n = 1 so , x = 7 ; y = 9 2 y / ( x + 4 ) = 18 / 11 answer = a | a ) 18 : 11 , b ) 8 : 1 , c ) 11 : 18 , d ) 1 : 8 , e ) 1 : 1 | a | divide(multiply(const_2, multiply(divide(33, subtract(multiply(divide(add(const_100, 20), const_100), 5), divide(9, 7))), divide(9, 7))), add(divide(33, subtract(multiply(divide(add(const_100, 20), const_100), 5), divide(9, 7))), 4)) | add(n2,const_100)|divide(n1,n0)|divide(#0,const_100)|multiply(n5,#2)|subtract(#3,#1)|divide(n3,#4)|add(n7,#5)|multiply(#5,#1)|multiply(#7,const_2)|divide(#8,#6) | general |
of 70 players on a football team , 52 are throwers . the rest of the team is divided so one third are left - handed and the rest are right handed . assuming that all throwers are right handed , how many right - handed players are there total ? | "total = 70 thrower = 52 rest = 70 - 52 = 18 left handed = 18 / 3 = 6 right handed = 12 if all thrower are right handed then total right handed is 52 + 12 = 64 so d . 64 is the right answer" | a ) 54 , b ) 59 , c ) 63 , d ) 64 , e ) 92 | d | add(multiply(subtract(const_1, divide(const_1, const_3)), subtract(70, 52)), 52) | divide(const_1,const_3)|subtract(n0,n1)|subtract(const_1,#0)|multiply(#2,#1)|add(n1,#3)| | general |
a man can row upstream at 15 kmph and downstream at 35 kmph , and then find the speed of the man in still water ? | "us = 15 ds = 35 m = ( 35 + 15 ) / 2 = 25 answer : e" | a ) 65 , b ) 86 , c ) 30 , d ) 78 , e ) 25 | e | divide(add(15, 35), const_2) | add(n0,n1)|divide(#0,const_2)| | physics |
39 ! is divided by 41 . then what is the remainder ? | "substituting p = 41 in the wilson ' s theorem , we get 40 ! + 141 = 0 40 × 39 ! + 141 = 0 − 1 × 39 ! 41 = − 1 cancelling - 1 on both sides , 39 ! 41 = 1 a" | a ) 1 , b ) 3 , c ) 4 , d ) 6 , e ) 9 | a | reminder(factorial(39), 41) | factorial(n0)|reminder(#0,n1)| | other |
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 | "solution suppose y invested rs . y then , 40000 / y = 2 / 3 ‹ = › y = ( 40000 × 3 / 2 ) . ‹ = › y = 60000 . answer c" | a ) rs . 45,000 , b ) rs . 50,000 , c ) rs . 60,000 , d ) rs . 80,000 , e ) none | 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 |
if the product of two numbers is 82500 and their h . c . f . is 55 , find their l . c . m . | "explanation : hcf * lcm = 82500 , because we know product of two numbers = product of hcf and lcm lcm = 82500 / 55 = 1500 option c" | a ) 2574 , b ) 2500 , c ) 1500 , d ) 1574 , e ) none of these | c | divide(82500, 55) | divide(n0,n1)| | physics |
what is the area of an obtuse angled triangle whose two sides are 8 and 12 and the angle included between the two sides is 150 o ? | explanatory answer if two sides of a triangle and the included angle ' y ' is known , then the area of the triangle = 1 / 2 * ( product of sides ) * sin y substituting the values in the formula , we get 1 / 2 * 8 * 12 * sin 150 o = 1 / 2 * 8 * 12 * 1 / 2 . note : sin 150 o = 1 / 2 area = 24 sq units . choice a | ['a ) 24 sq units', 'b ) 48 sq units', 'c ) 24 under root 3', 'd ) 48 under root 3', 'e ) such a triangle does not exist'] | a | subtract(triangle_area(add(multiply(8, divide(sqrt(const_3), const_2)), 12), divide(8, const_2)), triangle_area(divide(8, const_2), multiply(8, divide(sqrt(const_3), const_2)))) | divide(n0,const_2)|sqrt(const_3)|divide(#1,const_2)|multiply(n0,#2)|add(n1,#3)|triangle_area(#0,#3)|triangle_area(#4,#0)|subtract(#6,#5) | geometry |
tabby is training for a triathlon . she swims at a speed of 1 mile per hour . she runs at a speed of 8 miles per hour . she wants to figure out her average speed for these two events . what is the correct answer for her ? | "( 1 mph + 8 mph ) / 2 = 4.5 mph correct option is : d" | a ) 8 mph , b ) 5.25 mph , c ) 3.5 mph , d ) 4.5 mph , e ) 0.5 mph | d | divide(add(1, 8), const_2) | add(n0,n1)|divide(#0,const_2)| | physics |
the average age of boys in a class is 16 years and that of the girls is 15 years . what is the average age for the whole class ? | explanation : we do not have the number of boys and girls . hence we can not find out the answer . answer : option d | a ) 15 , b ) 16 , c ) 15.5 , d ) insufficient data , e ) 17 | d | subtract(16, 15) | subtract(n0,n1) | general |
there were totally 100 men . 84 are married . 75 have t . v , 85 have radio , 70 have a . c . how many men have t . v , radio , a . c and also married ? | "100 - ( 100 - 84 ) - ( 100 - 75 ) - ( 100 - 85 ) - ( 100 - 70 ) = 100 - 16 - 25 - 15 - 30 = 100 - 86 = 14 answer : d" | a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | d | subtract(100, add(add(add(subtract(100, 84), subtract(100, 75)), subtract(100, 85)), subtract(100, 70))) | subtract(n0,n1)|subtract(n0,n2)|subtract(n0,n3)|subtract(n0,n4)|add(#0,#1)|add(#4,#2)|add(#5,#3)|subtract(n0,#6)| | general |
excluding stoppages , the average speed of a bus is 60 km / hr and including stoppages , the average speed of the bus is 35 km / hr . for how many minutes does the bus stop per hour ? | "in 1 hr , the bus covers 60 km without stoppages and 35 km with stoppages . stoppage time = time take to travel ( 60 - 35 ) km i . e 25 km at 60 km / hr . stoppage time = 25 / 60 hrs = 24 min . answer : c" | a ) 22 , b ) 88 , c ) 24 , d ) 20 , e ) 99 | c | subtract(multiply(const_1, const_60), multiply(divide(35, 60), const_60)) | divide(n1,n0)|multiply(const_1,const_60)|multiply(#0,const_60)|subtract(#1,#2)| | general |
one paper sheet can be used to create 8 paper boats or 6 paper planes . if total exact 80 paper toys are to be made , minimum how many sheets need to be used to make paper boats ? | let us say x sheets are used to create paper boats and y sheets are used to create paper planes . since total 80 paper toys are to be created , 8 x + 6 y = 80 . = > 8 x = 80 - 6 y = > x = 10 - ( 6 / 8 ) y = > x = 10 - ( 3 / 4 ) y to minimize x , we need to maximize y . and y needs to be multiple of 4 . y = 4 , x = 10 -... | a ) 5 , b ) 4 , c ) 3 , d ) 2 , e ) 1 | e | divide(divide(80, 8), const_10) | divide(n2,n0)|divide(#0,const_10) | general |
the h . c . f . of two numbers is 12 and their l . c . m . is 600 . if one of the number is 60 , find the other ? | "other number = 12 * 600 / 60 = 120 answer is c" | a ) 100 , b ) 240 , c ) 120 , d ) 200 , e ) 150 | c | multiply(12, 60) | multiply(n0,n2)| | physics |
1604 x 1604 = ? | "1604 x 1604 = ( 1604 ) 2 = ( 1600 + 4 ) 2 = ( 1600 ) 2 + ( 4 ) 2 + ( 2 x 1600 x 4 ) = 2560000 + 16 + 12800 . = 2572816 . a )" | a ) a ) 2572816 , b ) b ) 3624216 , c ) c ) 3624316 , d ) d ) 3625116 , e ) e ) 3625216 | a | multiply(divide(1604, 1604), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
if abc is a quarter circle and a smaller circle is inscribed in it ; if radius of quarter circle is 1.414 units . find the radius of smaller circle | if r be radius of smaller circle then distance b / n centres = √ 2 * r so , r + r √ 2 = r here , r = 1.414 = √ 2 = > r = ( √ 2 - 1 ) r r = 0 . 586 unit answer : a | ['a ) 0.586', 'b ) 0.687', 'c ) 0.784', 'd ) 0.874', 'e ) 0.974'] | a | subtract(const_2, 1.414) | subtract(const_2,n0) | geometry |
a bag of potatoes weighs 16 kg divided by a quarter of its weight . how much does the bag weigh ? | b 8 kg 8 kg : 16 / 2 | a ) 4 kg , b ) 8 kg , c ) 7 kg , d ) 3 kg , e ) 9 kg | b | divide(16, const_2) | divide(n0,const_2) | general |
the cost price of a radio is rs . 1590 and it was sold for rs . 1450 , find the loss % ? | "1590 - - - - 140 100 - - - - ? = > 8 % answer : a" | a ) 8 % , b ) 12 % , c ) 10 % , d ) 14 % , e ) 21 | a | multiply(divide(subtract(1590, 1450), 1590), const_100) | subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)| | gain |
a retailer bought a machine at a wholesale price of $ 90 and later on sold it after a 10 % discount of the retail price . if the retailer made a profit equivalent to 20 % of the whole price , what is the retail price of the machine ? | "my solution : wholesale price = 90 retail price , be = x he provides 10 % discount on retail price = x - 10 x / 100 this retail price = 20 % profit on wholesale price x - 10 x / 100 = 90 + 1 / 5 ( 90 ) x = 120 ; answer : c" | a ) 81 , b ) 100 , c ) 120 , d ) 135 , e ) 160 | c | divide(multiply(add(90, divide(multiply(90, 20), const_100)), const_100), multiply(multiply(const_3, const_3), 10)) | multiply(n0,n2)|multiply(const_3,const_3)|divide(#0,const_100)|multiply(n1,#1)|add(n0,#2)|multiply(#4,const_100)|divide(#5,#3)| | gain |
what is the least number should be added to 1056 , so the sum of the number is completely divisible by 27 ? | "( 1056 / 27 ) gives remainder 3 24 + 3 = 27 , so we need to add 24 d" | a ) 21 , b ) 22 , c ) 23 , d ) 24 , e ) 25 | d | subtract(multiply(add(floor(divide(1056, 27)), const_1), 27), 1056) | divide(n0,n1)|floor(#0)|add(#1,const_1)|multiply(n1,#2)|subtract(#3,n0)| | general |
if 30 % of the 880 students at a certain college are enrolled in biology classes , how many students at the college are not enrolled in a biology class ? | "we know 37.5 % people study biology , therefore the no of people not studying = 100 - 30 = 70 % > therefore the people not studying biology out of a total 880 people are = 70 % of 880 > ( 70 / 100 ) * 880 = 616 people c" | a ) 620 , b ) 630 , c ) 616 , d ) 600 , e ) 650 | c | multiply(divide(880, const_100), subtract(const_100, 30)) | divide(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)| | other |
the average of 9 observations was 9 , that of the 1 st of 5 being 10 and that of the last 5 being 8 . what was the 5 th observation ? | "1 to 9 = 9 * 9 = 81 1 to 5 = 5 * 10 = 50 5 to 9 = 5 * 8 = 40 5 th = 50 + 40 = 90 – 81 = 9 answer : b" | a ) 6 , b ) 9 , c ) 8 , d ) 2 , e ) 3 | b | subtract(add(multiply(10, 5), multiply(9, 5)), multiply(9, 9)) | multiply(n3,n4)|multiply(n1,n3)|multiply(n0,n1)|add(#0,#1)|subtract(#3,#2)| | general |
in a certain company 20 % of the men and 40 % of the women attended the annual company picnic . if 30 % of all the employees are men . what % of all the employee went to the picnic ? | "total men in company 30 % means total women in company 70 % ( assume total people in company 100 % ) no of men employees attended picnic = 30 x ( 20 / 100 ) = 6 no of women employees attended picnic = 70 x ( 40 / 100 ) = 28 total percentage of employees attended the picnic = 6 + 28 = 34 % answer : b" | a ) 33 % , b ) 34 % , c ) 35 % , d ) 36 % , e ) 37 % | b | multiply(add(multiply(divide(30, const_100), divide(20, const_100)), multiply(divide(subtract(const_100, 30), const_100), divide(40, const_100))), const_100) | divide(n2,const_100)|divide(n0,const_100)|divide(n1,const_100)|subtract(const_100,n2)|divide(#3,const_100)|multiply(#0,#1)|multiply(#4,#2)|add(#5,#6)|multiply(#7,const_100)| | gain |
30 is what percent of 100 ? | 30 / 100 × 100 = 30 % answer : b | a ) 35 % , b ) 30 % , c ) 45 % , d ) 50 % , e ) 55 % | b | multiply(divide(30, 100), const_100) | divide(n0,n1)|multiply(#0,const_100)| | gain |
how many liters of a 40 % iodine solution need to be mixed with 35 liters of a 20 % iodine solution to create a 31 % iodine solution ? | "solution 1 : assume the iodine solution to be mixed = x lts . iodine = 0.4 x lts , water = 0.6 x lts . solution 2 : 35 liters of a 20 % iodine solution iodine = 7 lts , water = 28 lts . total iodine = 0.4 x + 7 total water = 0.6 x + 28 the resultant is a 35 % idoine solution . hence ( 0.4 x + 7 ) / ( x + 35 ) = 31 / 1... | a ) 35 , b ) 42 , c ) 100 , d ) 105 , e ) 140 | b | add(divide(subtract(multiply(divide(multiply(35, 20), const_100), 31), multiply(divide(multiply(35, 20), const_100), 20)), subtract(multiply(20, divide(40, const_100)), divide(multiply(35, 20), const_100))), 35) | divide(n0,const_100)|multiply(n1,n2)|divide(#1,const_100)|multiply(n2,#0)|multiply(n3,#2)|multiply(n2,#2)|subtract(#3,#2)|subtract(#4,#5)|divide(#7,#6)|add(n1,#8)| | gain |
if k ^ 3 is divisible by 180 , what is the least possible value of integer k ? | "180 = 2 ^ 2 * 3 ^ 2 * 5 therefore k must include at least 2 * 3 * 5 = 30 . the answer is b ." | a ) 12 , b ) 30 , c ) 60 , d ) 90 , e ) 120 | b | divide(divide(180, const_2), const_2) | divide(n1,const_2)|divide(#0,const_2)| | general |
a clock shows the time as 8 a . m . if the minute hand gains 5 minutes every hour , how many minutes will the clock gain by 6 p . m . ? | "there are 10 hours in between 8 a . m . to 6 p . m . 10 * 5 = 50 minutes . answer : e" | a ) 45 minutes , b ) 55 minutes , c ) 35 minutes , d ) 25 minutes , e ) 50 minutes | e | multiply(add(const_3, 6), 5) | add(const_3,n2)|multiply(n1,#0)| | physics |
a watch was sold at a loss of 12 % . if it was sold for rs . 140 more , there would have been a gain of 4 % . what is the cost price ? | "explanation : 88 % 104 % - - - - - - - - 16 % - - - - 140 100 % - - - - ? = > rs . 875 answer : b" | a ) s . 1000 , b ) s . 875 , c ) s . 1007 , d ) s . 1006 , e ) s . 1002 | b | divide(multiply(140, const_100), subtract(add(const_100, 4), subtract(const_100, 12))) | add(const_100,n2)|multiply(n1,const_100)|subtract(const_100,n0)|subtract(#0,#2)|divide(#1,#3)| | gain |
a vendor bought toffees at 6 for a rupee . how many for a rupee must he sell to gain 20 % ? | "explanation : c . p . of 6 toffees = re . 1 s . p . of 6 toffees = 120 % of re . 1 = rs . 6 / 5 for rs . 6 / 5 , toffees sold = 6 . for re . 1 . toffees sold = 6 * 5 / 6 = 5 answer : c" | a ) 7 , b ) 6 , c ) 5 , d ) 9 , e ) 1 | c | multiply(6, add(const_1, divide(20, const_100))) | divide(n1,const_100)|add(#0,const_1)|multiply(n0,#1)| | gain |
ratio between rahul and deepak is 4 : 3 , after 10 years rahul age will be 26 years . what is deepak present age | "explanation : present age is 4 x and 3 x , = > 4 x + 10 = 26 = > x = 4 so deepak age is = 3 ( 4 ) = 12 answer : option a" | a ) 12 , b ) 15 , c ) 20 , d ) 22 , e ) 23 | a | divide(multiply(subtract(26, 10), 3), 4) | subtract(n3,n2)|multiply(n1,#0)|divide(#1,n0)| | other |
a man can row downstream at the rate of 10 km / hr and upstream at 4 km / hr . find man ' s rate in still water and the rate of current ? | "explanation : rate of still water = 1 / 2 ( 10 + 4 ) = 7 km / hr rate of current = 1 / 2 ( 10 - 4 ) = 3 km / hr answer : option b" | a ) 9,6 , b ) 7,3 , c ) 9,3 , d ) 6,6 , e ) none of these | b | divide(subtract(10, 4), const_2) | subtract(n0,n1)|divide(#0,const_2)| | gain |
worker a takes 8 hours to do a job . worker b takes 12 hours to do the same job . how long it take both a & b , working together but independently , to do the same job ? | "a ' s one hour work = 1 / 8 . b ' s one hour work = 1 / 12 . ( a + b ) ' s one hour work = 1 / 8 + 1 / 12 = 5 / 24 . both a & b can finish the work in 24 / 5 days b" | a ) 40 days , b ) 24 / 5 days , c ) 39 / 9 days , d ) 30 / 9 days , e ) 60 / 9 days | b | divide(const_1, add(divide(const_1, 8), divide(const_1, 12))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)| | physics |
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