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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if an object travels at ten feet per second , how many feet does it travel in two hours ? | "if an object travels at 10 feet per second it covers 10 x 60 feet in one minute , and 10 x 60 x 2 x 60 feet in two hour . answer = 72000 answer : d" | a ) 60000 , b ) 30000 , c ) 18000 , d ) 72000 , e ) 22200 | d | multiply(multiply(const_3, const_60), const_60) | multiply(const_3,const_60)|multiply(#0,const_60)| | physics |
3 medical experts , working together at the same constant rate , can write an anatomy textbook in 24 days . how many additional experts , working together at this same constant rate , are needed to write the textbook in 18 days ? | each expert can write 1 / 72 of the book per day . to complete the book in 18 days , we need 72 / 18 = 4 experts , thus 1 more expert is needed . the answer is a . | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | a | subtract(divide(multiply(3, 24), 18), 3) | multiply(n0,n1)|divide(#0,n2)|subtract(#1,n0) | physics |
the value of ( 34.31 * 0.473 * 1.567 ) / ( 0.0673 * 23.25 * 7.57 ) is close to | "( 34.31 * 0.473 * 1.567 ) / ( 0.0673 * 23.25 * 7.57 ) = 25.4303 / 11.845 = 2.15 answer : d" | a ) 2 , b ) 1.15 , c ) 2.05 , d ) 2.15 , e ) 2.35 | d | divide(divide(multiply(multiply(34.31, 0.473), 1.567), multiply(multiply(7.57, 23.25), 0.0673)), const_10) | multiply(n0,n1)|multiply(n4,n5)|multiply(n2,#0)|multiply(n3,#1)|divide(#2,#3)|divide(#4,const_10)| | general |
the cost of carpeting a room 15 meters long with a carpet 75 cm wide at 30 paise per meter is rs . 36 . the breadth of the room is ? | "length of carpet = total cost / rate = 3600 / 30 = 120 m area of carpet = ( 120 x 75 ) / 100 m 2 = 90 m 2 ∴ area of the room = 90 m 2 breadth of the room = area / length = 90 / 15 m = 6 m answer : a" | a ) 6 meters , b ) 7 meters , c ) 8 meters , d ) 9 meters , e ) 10 meters | a | divide(multiply(divide(multiply(36, const_100), 30), divide(75, const_100)), 15) | divide(n1,const_100)|multiply(n3,const_100)|divide(#1,n2)|multiply(#2,#0)|divide(#3,n0)| | physics |
an aeroplane covers a certain distance of 590 kmph in 8 hours . to cover the same distance in 2 3 / 4 hours , it must travel at a speed of | speed of aeroplane = 590 kmph distance travelled in 8 hours = 590 * 8 = 4720 km speed of aeroplane to acver 4720 km in 11 / 4 = 4720 * 4 / 11 = 1716 km answer b . | a ) 1780 , b ) 1716 , c ) 1890 , d ) 1980 , e ) 1450 | b | divide(multiply(590, 8), divide(add(multiply(4, 2), 3), 4)) | multiply(n0,n1)|multiply(n2,n4)|add(n3,#1)|divide(#2,n4)|divide(#0,#3) | physics |
a work can be completed by 12 boys in 24 days and 12 girls in 12 days . in how many days would the 6 boys and 6 girls working together complete the work ? | in 1 day , 12 boys does 1 / 24 of the total work . = > in 1 day , 1 boy does 1 / ( 24 * 12 ) of the total work = > in 1 days , 6 boys do 6 / ( 24 * 12 ) = 1 / 48 of the total work in 1 day , 12 girls do 1 / 12 of the total work . = > in 1 day , 1 girl does 1 / ( 12 * 12 ) of the total work = > in 1 day , 6 girls do 6 /... | a ) 12 , b ) 18 , c ) 16 , d ) 20 , e ) 24 | c | inverse(add(divide(6, multiply(12, 24)), divide(6, multiply(12, 12)))) | multiply(n0,n1)|multiply(n0,n0)|divide(n4,#0)|divide(n4,#1)|add(#2,#3)|inverse(#4) | physics |
walking at 75 % of his usual speed a man takes 24 minutes more to cover a distance . what is his usual time to cover this distance ? | "speed is inversly proprtional to time walking at 75 % of speed meand 3 / 4 s takes 4 / 3 t . it takes 24 minutes extra to cover the distance . then 4 / 3 t = t + 24 4 t = 3 t + 72 t = 72 option e is correct" | a ) 30 , b ) 36 , c ) 42 , d ) 48 , e ) 72 | e | divide(24, subtract(divide(const_1, divide(75, const_100)), const_1)) | divide(n0,const_100)|divide(const_1,#0)|subtract(#1,const_1)|divide(n1,#2)| | physics |
anusha , banu and esha run a running race of 100 meters . anusha is the fastest followed by banu and then esha . anusha , banu and esha maintain constant speeds during the entire race . when anusha reached the goal post , banu was 10 m behind . when banu reached the goal post esha was 10 m behind . how far was behind a... | by that time anusha covered 100 m , bhanu covered 90 m . so ratio of their speeds = 10 : 9 by that time bhanu reached 100 m , esha covered 90 m . so ratio of their speeds = 10 : 9 ratio of the speed of all the three = 100 : 90 : 81 by that time anusha covered 100 m , esha covers only 81 . answer : c | a ) 22 , b ) 88 , c ) 81 , d ) 66 , e ) 22 | c | multiply(subtract(100, 10), inverse(divide(100, subtract(100, 10)))) | subtract(n0,n1)|divide(n0,#0)|inverse(#1)|multiply(#2,#0) | physics |
the output of a factory was increased by 10 % to keep up with rising demand . to handle the holiday rush , this new output was increased by 30 % . by approximately what percent would the output now have to be decreased in order to restore the original output ? | "the original output increases by 10 % and then 30 % . total % change = a + b + ab / 100 total % change = 10 + 30 + 10 * 30 / 100 = 43 % now , you want to change it to 0 , so , 0 = 43 + x + 43 x / 100 x = - 43 ( 100 ) / 143 = 30 % approximately answer is c" | a ) 20 % , b ) 24 % , c ) 30 % , d ) 32 % , e ) 79 % | c | divide(multiply(subtract(add(add(const_100, 10), multiply(add(const_100, 10), divide(30, const_100))), const_100), const_100), add(add(const_100, 10), multiply(add(const_100, 10), divide(30, const_100)))) | add(n0,const_100)|divide(n1,const_100)|multiply(#0,#1)|add(#0,#2)|subtract(#3,const_100)|multiply(#4,const_100)|divide(#5,#3)| | general |
if the cost price is 89 % of selling price then what is the profit percentage . | "selling price = rs 100 : then cost price = rs 89 : profit = rs 11 . profit = { ( 11 / 89 ) * 100 } % = 12.35 % answer is e ." | a ) 8.35 , b ) 9.35 , c ) 10.35 , d ) 11.35 , e ) 12.35 | e | multiply(divide(subtract(const_100, 89), 89), const_100) | subtract(const_100,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
the rate of interest on a sum of money is 2 % p . a . for the first 3 years , 4 % p . a . for the next 4 years , and 5 % for the period beyond 7 years . if the s . i , occured on the sum for the total period of 8 years is rs . 540 / - , the sum is | "explanation : i 1 = ( p x 3 x 2 ) / 100 = 3 p / 50 i 2 = ( p x 4 x 4 ) / 100 = 4 p / 25 i 3 = ( p x 1 x 5 ) / 100 = p / 20 3 p / 50 + 4 p / 25 + p / 20 = 540 the l . c . m of 50 , 25 , 20 = 100 ( 6 p + 16 p + 5 p ) / 100 = 540 27 p / 100 = 540 27 p = 54000 p = 54000 / 27 p = 2000 answer : option b" | a ) 2,200 , b ) 2,000 , c ) 2,100 , d ) 2,250 , e ) 2,560 | b | divide(540, 540) | divide(n7,n7)| | general |
at what rate percent per annum will the simple interest on a sum of money be 7 / 5 of the amount in 10 years ? | "let sum = x . then , s . i . = 7 x / 5 , time = 10 years . rate = ( 100 * 7 x ) / ( x * 5 * 10 ) = 14 % answer : e" | a ) 4 % , b ) 7 % , c ) 9 % , d ) 3 % , e ) 14 % | e | divide(multiply(divide(7, 5), const_100), 10) | divide(n0,n1)|multiply(#0,const_100)|divide(#1,n2)| | gain |
two trains are moving in opposite directions at 60 km / hr and 90 km / hr . their lengths are 2.9 km and 1.6 km respectively . the time taken by the slower train to cross the faster train in seconds is ? | "relative speed = 60 + 90 = 150 km / hr . = 150 * 5 / 18 = 125 / 3 m / sec . distance covered = 2.9 + 1.6 = 4.5 km = 4500 m . required time = 4500 * 3 / 125 = 108 sec . answer : e" | a ) 99 , b ) 277 , c ) 48 , d ) 96 , e ) 108 | e | subtract(divide(multiply(2.9, const_1000), divide(multiply(60, const_1000), const_3600)), divide(multiply(1.6, const_1000), divide(multiply(90, const_1000), const_3600))) | multiply(n2,const_1000)|multiply(n0,const_1000)|multiply(n3,const_1000)|multiply(n1,const_1000)|divide(#1,const_3600)|divide(#3,const_3600)|divide(#0,#4)|divide(#2,#5)|subtract(#6,#7)| | physics |
the edge of a cube is 3 a cm . find its surface ? | "6 a 2 = 6 * 3 a * 3 a = 54 a 2 answer : d" | a ) 24 a 8 , b ) 24 a 4 , c ) 24 a 1 , d ) 54 a 2 , e ) 24 a 7 | d | surface_cube(3) | surface_cube(n0)| | geometry |
how many integers between 324,700 and 458,600 have a 2 in the tens digit and a 1 in the units digit ? | "there is one number in hundred with 2 in th tens digit and 1 in the units digit : 21 , 121 , 221 , 321 , . . . the difference between 324,700 and 458,600 is 458,600 - 324,700 = 133,900 - one number per each hundred gives 133,900 / 100 = 1,339 numbers . answer : 1,339 a" | a ) 1339 , b ) 2300 , c ) 4200 , d ) 1340 , e ) 2414 | a | subtract(458,600, add(add(multiply(const_2, const_100), multiply(add(const_3, const_4), const_10)), const_2)) | add(const_3,const_4)|multiply(const_100,const_2)|multiply(#0,const_10)|add(#1,#2)|add(#3,const_2)|subtract(n1,#4)| | general |
in a graduating class of 242 students , 144 took geometry and 119 took biology . what is the difference between the greatest possible number and the smallest possible number of students that could have taken both geometry and biology ? | "greatest possible number taken both should be 144 ( as it is maximum for one ) smallest possible number taken both should be given by total - neither = a + b - both both = a + b + neither - total ( neither must be 0 to minimize the both ) so 144 + 119 + 0 - 242 = 21 greatest - smallest is 144 - 21 = 123 so answer must... | a ) 144 , b ) 119 , c ) 113 , d ) 88 , e ) 123 | e | subtract(119, subtract(add(144, 119), 242)) | add(n1,n2)|subtract(#0,n0)|subtract(n2,#1)| | other |
the average mark of the students of a class in a particular exam is 80 . if 5 students whose average mark in that exam is 20 are excluded , the average mark of the remaining will be 95 . find the number of students who wrote the exam . | let the number of students who wrote the exam be x . total marks of students = 80 x . total marks of ( x - 5 ) students = 95 ( x - 5 ) 80 x - ( 5 * 20 ) = 95 ( x - 5 ) 375 = 15 x = > x = 25 answer : d | a ) 15 , b ) 45 , c ) 35 , d ) 25 , e ) 55 | d | divide(subtract(multiply(95, 5), multiply(5, 20)), subtract(95, 80)) | multiply(n1,n3)|multiply(n1,n2)|subtract(n3,n0)|subtract(#0,#1)|divide(#3,#2) | general |
real - estate salesman z is selling a house at a 30 percent discount from its retail price . real - estate salesman x vows to match this price , and then offers an additional 30 percent discount . real - estate salesman y decides to average the prices of salesmen z and x , then offer an additional 30 percent discount .... | "let the retail price be = x selling price of z = 0.70 x selling price of x = 0.70 * 0.80 x = 0.56 x selling price of y = ( ( 0.70 x + 0.56 x ) / 2 ) * 0.70 = 0.63 x * 0.70 = 0.44 x 0.44 x = k * 0.56 x k = 0.44 / 0.56 = 44 / 56 = 11 / 14 answer : d" | a ) 12 / 13 , b ) 11 / 15 , c ) 11 / 17 , d ) 11 / 14 , e ) 11 / 7 | d | multiply(divide(divide(multiply(divide(add(subtract(const_100, 30), multiply(subtract(const_100, 30), divide(subtract(const_100, 30), const_100))), const_2), subtract(const_100, 30)), const_100), multiply(subtract(const_100, 30), divide(subtract(const_100, 30), const_100))), const_10) | subtract(const_100,n1)|subtract(const_100,n0)|subtract(const_100,n2)|divide(#0,const_100)|multiply(#3,#1)|add(#4,#1)|divide(#5,const_2)|multiply(#6,#2)|divide(#7,const_100)|divide(#8,#4)|multiply(#9,const_10)| | general |
a student has to obtain 33 % of the total marks to pass . he got 92 marks and failed by 40 marks . the maximum marks are ? | "let the maximum marks be x then , 33 % of x = 92 + 40 33 x / 100 = 132 x = 400 answer is a" | a ) 400 , b ) 300 , c ) 500 , d ) 610 , e ) 175 | a | divide(add(92, 40), divide(33, const_100)) | add(n1,n2)|divide(n0,const_100)|divide(#0,#1)| | general |
p works 25 % more efficiently than q and q works 50 % more efficiently than r . to complete a certain project , p alone takes 50 days less than q alone . if , in this project p alone works for 60 days and then q alone works for 150 days , in how many days can r alone complete the remaining work ? | "p works 25 % more efficiently than q : something that takes q 5 days , takes p 4 days q works 50 % more efficiently than r : something that takes r 7.5 days , takes q 5 days p alone takes 50 days less than q : for every 4 days p works , q has to work an extra day . hence p alone can do it in 200 days and q alone in 25... | a ) 50 days , b ) 75 days , c ) 120 days , d ) 150 days , e ) 80 days | c | subtract(multiply(const_4, 50), multiply(divide(150, const_100), 60)) | divide(n4,const_100)|multiply(n1,const_4)|multiply(n3,#0)|subtract(#1,#2)| | physics |
what is the sum of the multiples of 6 from 48 to 72 , inclusive ? | "the formula we want to use in this type of problem is this : average * total numbers = sum first , find the average by taking the sum of the f + l number and divide it by 2 : a = ( f + l ) / 2 second , find the total numbers in our range by dividing our f and l numbers by 7 and add 1 . ( 72 / 6 ) - ( 48 / 6 ) + 1 mult... | a ) 500 , b ) 600 , c ) 700 , d ) 300 , e ) 400 | d | multiply(divide(add(subtract(72, const_3), add(48, const_2)), const_2), add(divide(subtract(subtract(72, const_3), add(48, const_2)), 6), const_1)) | add(n1,const_2)|subtract(n2,const_3)|add(#0,#1)|subtract(#1,#0)|divide(#3,n0)|divide(#2,const_2)|add(#4,const_1)|multiply(#6,#5)| | general |
if d = 1 / ( 2 ^ 3 * 5 ^ 10 ) is expressed as a terminating decimal , how many nonzero digits will d have ? | another way to do it is : we know x ^ a * y ^ a = ( x * y ) ^ a given = 1 / ( 2 ^ 3 * 5 ^ 10 ) = multiply and divide by 2 ^ 7 = 2 ^ 7 / ( 2 ^ 3 * 2 ^ 7 * 5 ^ 10 ) = 2 ^ 7 / 10 ^ 10 = > non zero digits are 128 = > ans c | a ) one , b ) two , c ) three , d ) seven , e ) ten | c | add(const_1, const_2) | add(const_1,const_2) | general |
a , band c enter into partnership . a invests 3 times as much as b and b invests two - third of what c invests . at the end of the year , the profit earned is rs . 3300 . what is the share of b ? | "let c ' s capital = rs . x . then , b ' s capital = rs . ( 2 / 3 ) x a ’ s capital = rs . ( 3 x ( 2 / 3 ) . x ) = rs . 2 x . ratio of their capitals = 2 x : ( 2 / 3 ) x : x = 6 : 2 : 3 . hence , b ' s share = rs . ( 3300 x ( 2 / 11 ) ) = rs . 600 answer is d ." | a ) 1100 , b ) 800 , c ) 1400 , d ) 600 , e ) none of them | d | multiply(3300, divide(const_2, add(add(multiply(const_2, 3), multiply(divide(const_2, 3), 3)), 3))) | divide(const_2,n0)|multiply(const_2,n0)|multiply(#0,n0)|add(#1,#2)|add(#3,n0)|divide(const_2,#4)|multiply(n1,#5)| | gain |
tanks a and b are each in the shape of a right circular cylinder . the interior of tank a has a height of 10 meters and a circumference of 11 meters , and the interior of tank b has a height of 11 meters and a circumference of 10 meters . the capacity of tank a is what percent of the capacity of tank b ? | "the radius of tank a is 11 / ( 2 * pi ) . the capacity of tank a is 10 * pi * 121 / ( 4 * pi ^ 2 ) = 605 / ( 2 * pi ) the radius of tank b is 10 / ( 2 * pi ) . the capacity of tank b is 11 * pi * 100 / ( 4 * pi ^ 2 ) = 550 / ( 2 * pi ) tank a / tank b = 605 / 550 = 11 / 10 = 110 % the answer is d ." | a ) 75 % , b ) 80 % , c ) 100 % , d ) 110 % , e ) 125 % | d | multiply(multiply(power(divide(11, 10), const_2), divide(10, 11)), const_100) | divide(n0,n2)|divide(n1,n3)|power(#1,const_2)|multiply(#0,#2)|multiply(#3,const_100)| | physics |
the cost price of a radio is rs . 1500 and it was sold for rs . 1335 , find the loss % ? | "explanation : 1500 - - - - 165 100 - - - - ? = > 11 % answer : e" | a ) 18 , b ) 16 , c ) 26 , d ) 17 , e ) 11 | e | multiply(divide(subtract(1500, 1335), 1500), const_100) | subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)| | gain |
each of the products produced yesterday was checked by worker x or worker y . 0.5 % of the products checked by worker x are defective and 0.8 % of the products checked by worker y are defective . if the total defective rate of all the products checked by worker x and worker y is 0.6 % , what fraction of the products wa... | "x : 0.5 % is 0.1 % - points from 0.6 % . y : 0.8 % is 0.2 % - points from 0.6 % . therefore the ratio of products checked by y : x is 1 : 2 . thus , worker y checked 1 / 3 of the products . the answer is c ." | a ) 1 / 5 , b ) 1 / 4 , c ) 1 / 3 , d ) 2 / 5 , e ) 3 / 7 | c | divide(subtract(0.6, 0.5), subtract(0.8, 0.5)) | subtract(n2,n0)|subtract(n1,n0)|divide(#0,#1)| | general |
in 1998 the profits of company n were 10 percent of revenues . in 1999 , the revenues of company n fell by 30 percent , but profits were 15 percent of revenues . the profits in 1999 were what percent of the profits in 1998 ? | "0,105 r = x / 100 * 0.1 r answer b" | a ) 80 % , b ) 105 % , c ) 120 % , d ) 124.2 % , e ) 138 % | b | multiply(divide(multiply(subtract(const_1, divide(30, const_100)), divide(15, const_100)), divide(10, const_100)), const_100) | divide(n4,const_100)|divide(n3,const_100)|divide(n1,const_100)|subtract(const_1,#1)|multiply(#0,#3)|divide(#4,#2)|multiply(#5,const_100)| | gain |
a shopkeeper sells 300 metres of cloth for rs . 18000 at a loss of rs . 5 per metre . find his cost price for one metre of cloth ? | sp per metre = 18000 / 300 = rs . 60 loss per metre = rs . 5 cp per metre = 60 + 5 = rs . 65 . answer : e | a ) 12 , b ) 27 , c ) 29 , d ) 50 , e ) 65 | e | add(divide(18000, 300), 5) | divide(n1,n0)|add(n2,#0)| | gain |
? % of 360 = 144 | "? % of 360 = 144 or , ? = 144 × 100 / 360 = 40 answer d" | a ) 277 , b ) 36 , c ) 64 , d ) 40 , e ) none of these | d | divide(multiply(144, const_100), 360) | multiply(n1,const_100)|divide(#0,n0)| | gain |
linda spent 5 / 6 of her savings on furniture and the rest on a tv . if the tv cost her $ 500 , what were her original savings ? | "if linda spent 5 / 6 of her savings on furniture , the rest 6 / 6 - 5 / 6 = 1 / 6 on a tv but the tv cost her $ 500 . so 1 / 6 of her savings is $ 500 . so her original savings are 6 times $ 500 = $ 3000 correct answer b" | a ) $ 9000 , b ) $ 3000 , c ) $ 6000 , d ) $ 7000 , e ) $ 8000 | b | divide(500, subtract(const_1, divide(5, 6))) | divide(n0,n1)|subtract(const_1,#0)|divide(n2,#1)| | general |
how many bricks , each measuring 40 cm x 11.25 cm x 6 cm , will be needed to build a wall of 8 m x 6 m x 22.5 cm ? | "number of bricks = volume of the wall / volume of 1 brick = ( 800 x 600 x 22.5 ) / ( 40 x 11.25 x 6 ) = 4000 answer : c" | a ) 1400 , b ) 2400 , c ) 4000 , d ) 7000 , e ) 3400 | c | divide(multiply(multiply(multiply(8, const_100), multiply(6, const_100)), 22.5), multiply(multiply(40, 11.25), 6)) | multiply(n3,const_100)|multiply(n4,const_100)|multiply(n0,n1)|multiply(#0,#1)|multiply(n2,#2)|multiply(n5,#3)|divide(#5,#4)| | physics |
find the simple interest on rs . 64,000 at 16 2 / 3 % per annum for 9 months . | "p = rs . 64000 , r = 50 / 3 % p . a and t = 9 / 12 years = 3 / 4 years . s . i . = ( p * r * t ) / 100 = rs . ( 64,000 * ( 50 / 3 ) * ( 3 / 4 ) * ( 1 / 100 ) ) = rs . 8000 answer is b ." | a ) s . 8500 , b ) s . 8000 , c ) s . 7500 , d ) s . 7000 , e ) s . 6500 | b | multiply(multiply(multiply(add(multiply(multiply(multiply(2, 3), const_100), const_100), multiply(multiply(multiply(3, 3), const_100), multiply(add(3, 2), 2))), divide(add(multiply(16, 3), 2), 3)), divide(multiply(3, 3), multiply(2, multiply(2, 3)))), divide(const_1, const_100)) | add(n2,n3)|divide(const_1,const_100)|multiply(n3,n3)|multiply(n2,n3)|multiply(n1,n3)|add(n2,#4)|multiply(n2,#3)|multiply(#3,const_100)|multiply(#2,const_100)|multiply(#0,n2)|divide(#2,#6)|divide(#5,n3)|multiply(#7,const_100)|multiply(#8,#9)|add(#12,#13)|multiply(#14,#11)|multiply(#10,#15)|multiply(#1,#16)| | gain |
three unbiased coins are tossed . what is the probability of getting at most 2 heads ? | "here s = { ttt , tth , tht , htt , thh , hth , hht , hhh } let e = event of getting at least two heads = { ttt , tth , tht , htt , thh , hth , hht } p ( e ) = n ( e ) / n ( s ) = 7 / 8 . answer a ." | a ) 7 / 8 , b ) 5 / 8 , c ) 3 / 8 , d ) 7 / 5 , e ) 3 / 5 | a | negate_prob(divide(const_1, power(const_2, const_3))) | power(const_2,const_3)|divide(const_1,#0)|negate_prob(#1)| | probability |
the average height of 15 girls out of a class of 60 is 138 cm . and that of the remaining girls is 142 cm . the average height of the whole class is : | "explanation : average height of the whole class = ( 15 × 138 + 45 × 142 / 60 ) = 141 cms answer b" | a ) 132 cms , b ) 141 cms , c ) 142 cms , d ) 152 cms , e ) 161 cms | b | divide(add(multiply(138, 15), multiply(142, const_10)), 60) | multiply(n0,n2)|multiply(n3,const_10)|add(#0,#1)|divide(#2,n1)| | general |
a cistern can be filled by a tap in 2 hours while it can be emptied by another tap in 9 hours . if both the taps are opened simultaneously , then after how much time will the cistern get filled ? | "net part filled in 1 hour = 1 / 2 - 1 / 9 = 7 / 18 therefore the cistern will be filled in 18 / 7 hours or 2.57 hours . answer : a" | a ) 2.57 hrs , b ) 5 hrs , c ) 6.57 hrs , d ) 7.2 hrs , e ) 9.27 hrs | a | divide(const_1, subtract(divide(const_1, 2), divide(const_1, 9))) | divide(const_1,n0)|divide(const_1,n1)|subtract(#0,#1)|divide(const_1,#2)| | physics |
the edges of a cuboid are 6 cm , 5 cm and 6 cm . find the volume of the cuboid ? | "6 * 5 * 6 = 180 answer : e" | a ) 120 , b ) 160 , c ) 210 , d ) 368 , e ) 180 | e | volume_rectangular_prism(6, 5, 6) | volume_rectangular_prism(n0,n1,n2)| | physics |
a began business with rs . 45000 and was joined afterwards by b with rs . 27000 . when did b join if the profits at the end of the year were divided in the ratio of 2 : 1 ? | "45 * 12 : 27 * x = 2 : 1 x = 10 12 - 10 = 2 answer : e" | a ) 1 , b ) 6 , c ) 7 , d ) 8 , e ) 2 | e | subtract(multiply(const_4, const_3), divide(divide(multiply(45000, multiply(const_4, const_3)), 27000), 2)) | multiply(const_3,const_4)|multiply(n0,#0)|divide(#1,n1)|divide(#2,n2)|subtract(#0,#3)| | other |
1 / 3 + 1 / 2 - 5 / 6 + 1 / 5 + 1 / 4 - 9 / 20 - 9 / 20 = | "we need to determine the result of 1 / 3 + 1 / 2 - 5 / 6 + 1 / 5 + 1 / 4 - 9 / 20 let ’ s add the given fractions in two groups . in the group of the first three fractions , notice that 1 / 3 and 1 / 2 share a common denominator of 6 with 5 / 6 . 1 / 2 + 1 / 3 = 3 / 6 + 2 / 6 = 5 / 6 thus , 5 / 6 – 5 / 6 = 0 looking a... | a ) 0 , b ) 2 / 15 , c ) 2 / 5 , d ) 9 / 20 , e ) 5 / 6 | d | divide(9, 20) | divide(n12,n13)| | general |
a certain car dealership sells economy cars , luxury cars , and sport utility vehicles . the ratio of economy to luxury cars is 4 : 3 . the ratio of economy cars to sport utility vehicles is 6 : 5 . what is the ratio of luxury cars to sport utility vehicles ? | "the ratio of economy to luxury cars is 4 : 3 - - > e : l = 4 : 3 = 24 : 18 . the ratio of economy cars to sport utility vehicles is 6 : 5 - - > e : s = 6 : 5 = 24 : 20 . thus , l : s = 18 : 20 = 9 : 10 . answer : a ." | a ) 9 : 10 , b ) 8 : 9 , c ) 3 : 2 , d ) 2 : 3 , e ) 1 : 2 | a | divide(divide(multiply(const_4, const_3.0), multiply(5, 5)), divide(multiply(5, const_4), multiply(3, const_4))) | multiply(const_3.0,const_4)|multiply(n3,n3)|multiply(n1,const_4)|divide(#0,#1)|divide(#0,#2)|divide(#3,#4)| | other |
which digits should come in place of @ and # if the number 62684 @ # is divisible by both 8 and 5 ? | explanation : since the given number is divisible by 5 , so 0 or 5 must come in place of # . but , a number ending with 5 is never divisible by 8 . so , 0 will replace # . now , the number formed by the last three digits is 4 @ 0 , which becomes divisible by 8 , if @ is replaced by 4 . hence , digits in place of @ and ... | a ) 4,0 , b ) 0,4 , c ) 4,4 , d ) 0,0 , e ) none | a | divide(divide(multiply(lcm(8, 5), const_2), const_10), const_2) | lcm(n1,n2)|multiply(#0,const_2)|divide(#1,const_10)|divide(#2,const_2) | other |
a invested $ 300 in a business after 6 months b invested $ 200 in the business . end of the year if they got $ 100 as profit . find b ' s shares ? | "a : b = 300 * 12 : 200 * 6 a : b = 1 : 1 b ' s share = 100 * 1 / 4 = $ 25 answer is c" | a ) $ 100 , b ) $ 75 , c ) $ 25 , d ) $ 120 , e ) $ 50 | c | multiply(100, subtract(const_1, divide(divide(200, const_2), add(300, divide(200, const_2))))) | divide(n2,const_2)|add(n0,#0)|divide(#0,#1)|subtract(const_1,#2)|multiply(n3,#3)| | gain |
ashok secured average of 76 marks in 6 subjects . if the average of marks in 5 subjects is 74 , how many marks did he secure in the 6 th subject ? | "explanation : number of subjects = 6 average of marks in 6 subjects = 76 therefore total marks in 6 subjects = 76 * 6 = 456 now , no . of subjects = 5 total marks in 5 subjects = 74 * 5 = 370 therefore marks in 6 th subject = 456 – 370 = 86 answer : e" | a ) 38 , b ) 27 , c ) 99 , d ) 17 , e ) 86 | e | subtract(multiply(76, 6), multiply(74, 5)) | multiply(n0,n1)|multiply(n2,n3)|subtract(#0,#1)| | general |
a man has a certain number of small boxes to pack into parcles . if he packs 3 , 4 , 5 or 6 in a parcel , he is left with one over ; if he packs 7 in a parcle , none is left over . what is the number of boxes , he may have to pack ? | explanation : clearly , the required number would be such that it leaves a remainder of 1 when divided by 3 , 4 , 5 , or 6 and no remainder when divided by 7 . thus , the number must be of the form ( l . c . m of 3 , 4 , 5 , 6 ) x + 1 i . e . , ( 60 x + 1 ) and a multiple of 7 . clearly , for x = 5 , the number is a mu... | a ) 106 , b ) 301 , c ) 309 , d ) 400 , e ) 450 | b | add(multiply(lcm(lcm(lcm(3, 4), 5), 6), 5), const_1) | lcm(n0,n1)|lcm(n2,#0)|lcm(n3,#1)|multiply(n2,#2)|add(#3,const_1) | general |
find b and c so that the parabola with equation y = 4 x 2 - bx - c has a vertex at ( 2 , 4 ) ? | h = b / 8 = 2 : formula for x coordinate of vertex b = 16 : solve for b y = 4 for x = 2 : the vertex point is a solution to the equation of the parabola 4 ( 2 ) 2 - 16 ( 2 ) - c = 4 c = - 20 : solve for c correct answer b | a ) - 10 , b ) - 20 , c ) - 30 , d ) - 40 , e ) - 50 | b | multiply(2, multiply(2, 4)) | multiply(n0,n1)|multiply(n1,#0) | general |
pipe a can fill a tank in 8 minutes and pipe b cam empty it in 24 minutes . if both the pipes are opened together after how many minutes should pipe b be closed , so that the tank is filled in 30 minutes ? | let the pipe b be closed after x minutes . 30 / 8 - x / 24 = 1 = > x / 24 = 30 / 8 - 1 = 11 / 4 = > x = 11 / 4 * 24 = 66 . answer : e | a ) 18 , b ) 27 , c ) 98 , d ) 27 , e ) 66 | e | multiply(subtract(divide(30, 8), const_1), 24) | divide(n2,n0)|subtract(#0,const_1)|multiply(n1,#1) | physics |
a call center has two teams . each member of team a was able to process 2 / 5 calls as compared to each member of team b . if team a has 5 / 8 as many number of call center agents as team b , what fraction of the total calls was processed by team b ? | let team b has 8 agents , so team a has 5 agents let each agent of team b picked up 5 calls , so total calls by team b = 40 so , each agent in team a picked up 2 calls , so total calls for team a = 10 fraction for team b = 40 / ( 40 + 10 ) = 4 / 5 = answer = c | a ) 3 / 2 , b ) 3 / 4 , c ) 4 / 5 , d ) 1 / 2 , e ) 1 / 5 | c | divide(multiply(8, 5), add(multiply(8, 5), multiply(5, 2))) | multiply(n1,n3)|multiply(n0,n1)|add(#0,#1)|divide(#0,#2) | general |
if the wheel is 15 cm then the number of revolutions to cover a distance of 1056 cm is ? | "2 * 22 / 7 * 15 * x = 1056 = > x = 11.2 answer : b" | a ) 18 , b ) 11.2 , c ) 14 , d ) 12 , e ) 91 | b | divide(1056, multiply(multiply(const_2, divide(add(add(multiply(const_3, const_100), multiply(const_1, const_10)), const_4), const_100)), 15)) | multiply(const_100,const_3)|multiply(const_1,const_10)|add(#0,#1)|add(#2,const_4)|divide(#3,const_100)|multiply(#4,const_2)|multiply(n0,#5)|divide(n1,#6)| | physics |
a sum of rs . 1190 has been divided among a , b and c such that a gets of what b gets and b gets of what c gets . b ’ s share is : | explanation let c ’ s share = rs . x then , b ’ s share = rs . x / 4 , a ’ s share = rs . ( 2 / 3 x x / 4 ) = rs . x / 6 = x / 6 + x / 4 + x = 1190 = > 17 x / 12 = 1190 = > 1190 x 12 / 17 = rs . 840 hence , b ’ s share = rs . ( 840 / 4 ) = rs . 210 . answer b | a ) rs . 120 , b ) rs . 210 , c ) rs . 240 , d ) rs . 300 , e ) none | b | subtract(subtract(multiply(divide(1190, const_10), const_2), const_12), const_12) | divide(n0,const_10)|multiply(#0,const_2)|subtract(#1,const_12)|subtract(#2,const_12) | general |
a number increased by 40 % gives 1680 . the number is ? | "formula = total = 100 % , increase = ` ` + ' ' decrease = ` ` - ' ' a number means = 100 % that same number increased by 40 % = 140 % 140 % - - - - - - - > 1680 ( 140 ã — 12 = 1680 ) 100 % - - - - - - - > 1200 ( 100 ã — 12 = 1200 ) option ' c '" | a ) 1680 , b ) 1600 , c ) 1200 , d ) 1500 , e ) 600 | c | divide(1680, add(const_1, divide(40, const_100))) | divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)| | gain |
if each digit in the set a = { 1 , 2 , 3 , 4 , 5 } is used exactly once , in how many ways can the 5 digits be arranged ? | "use the slot method for all the possible arrangements : we have 5 options for the 1 st slot , 4 for the 2 nd , 3 for 3 rd , 2 for 4 th and 1 for the 5 th , giving the total number of arrangements = 5 * 4 * 3 * 2 * 1 = 120 , e is the correct answer ." | a ) 6 , b ) 24 , c ) 72 , d ) 96 , e ) 120 | e | factorial(5) | factorial(n5)| | general |
two pipes a and b can fill a tank in 12 hours and 15 hours respectively . if both the pipes are opened simultaneously , how much time will be taken to fill the tank ? | "part filled by a in 1 hour = 1 / 12 part filled by b in 1 hour = 1 / 15 part filled by ( a + b ) in 1 hour = 1 / 12 + 1 / 15 = 9 / 60 both the pipes together fill the tank in 60 / 9 = 7 4 / 9 hours answer is e" | a ) 20 hours , b ) 15 hours , c ) 10 hours , d ) 12 hours , e ) 7 4 / 9 hours | e | divide(const_1, add(divide(const_1, 12), divide(const_1, 15))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)| | physics |
the sides of the triangle are in the ratio 5 : 6 : 7 and its perimeter is 720 cm . the length of the longest side is ? | ratio of sides = 5 : 6 : 7 largest side = 720 * 7 / 18 = 280 cm answer is e | ['a ) 150 cm', 'b ) 200 cm', 'c ) 162 cm', 'd ) 220 cm', 'e ) 280 cm'] | e | divide(multiply(7, 720), add(7, add(5, 6))) | add(n0,n1)|multiply(n2,n3)|add(n2,#0)|divide(#1,#2) | geometry |
jackie has two solutions that are 2 percent sulfuric acid and 12 percent sulfuric acid by volume , respectively . if these solutions are mixed in appropriate quantities to produce 60 liters of a solution that is 10 percent sulfuric acid , approximately how many liters of the 2 percent solution will be required ? | "let a = amount of 2 % acid and b = amount of 12 % acid . now , the equation translates to , 0.02 a + . 12 b = . 1 ( a + b ) but a + b = 60 therefore . 02 a + . 12 b = . 1 ( 60 ) = > 2 a + 12 b = 600 but b = 60 - a therefore 2 a + 12 ( 60 - a ) = 600 = > 10 a = 120 hence a = 12 . answer : b" | a ) 18 , b ) 12 , c ) 24 , d ) 36 , e ) 42 | b | multiply(const_3, divide(60, const_10)) | divide(n2,const_10)|multiply(#0,const_3)| | gain |
a man saves 20 % of his monthly salary . if an account of dearness of things he is to increase his monthly expenses by 20 % , he is only able to save rs . 220 per month . what is his monthly salary ? | "income = rs . 100 expenditure = rs . 80 savings = rs . 20 present expenditure 80 * ( 20 / 100 ) = rs . 96 present savings = 100 â € “ 96 = rs . 4 100 - - - - - - 4 ? - - - - - - - - - 220 = > 5500 answer : a" | a ) 5500 , b ) 2999 , c ) 2878 , d ) 2990 , e ) 2771 | a | divide(multiply(220, const_100), subtract(const_100, add(subtract(const_100, 20), multiply(subtract(const_100, 20), divide(20, const_100))))) | divide(n1,const_100)|multiply(n2,const_100)|subtract(const_100,n0)|multiply(#0,#2)|add(#3,#2)|subtract(const_100,#4)|divide(#1,#5)| | general |
triangle atriangle b are similar triangles with areas 1536 units square and 2166 units square respectively . the ratio of there corresponding height would be | let x be the height of triangle a and y be the height of triangle of b . since triangles are similar , ratio of area of a and b is in the ratio of x ^ 2 / y ^ 2 therefore , ( x ^ 2 / y ^ 2 ) = 1536 / 2166 ( x ^ 2 / y ^ 2 ) = ( 16 * 16 * 6 ) / ( 19 * 19 * 6 ) ( x ^ 2 / y ^ 2 ) = 17 ^ 2 / 19 ^ 2 x / y = 16 / 19 ans = e | ['a ) 9 : 10', 'b ) 17 : 19', 'c ) 23 : 27', 'd ) 13 : 17', 'e ) 16 : 19'] | e | sqrt(divide(1536, 2166)) | divide(n0,n1)|sqrt(#0) | geometry |
the megatek corporation is displaying its distribution of employees by department in a circle graph . the size of each sector of the graph representing a department is proportional to the percentage of total employees in that department . if the section of the circle graph representing the manufacturing department take... | "answer : c 108 ° divided by 360 ° equals 0.3 , therefore the sector is equal to 30 % of the total" | a ) 20 % , b ) 25 % , c ) 30 % , d ) 35 % , e ) 72 % | c | multiply(divide(108, divide(const_3600, const_10)), const_100) | divide(const_3600,const_10)|divide(n0,#0)|multiply(#1,const_100)| | physics |
the toll t , in dollars , for a truck using a certain bridge is given by the formula t = 2.50 + 0.50 ( x − 2 ) , where x is the number of axles on the truck . what is the toll for an 18 - wheel truck that has 2 wheels on its front axle and 4 wheels on each of its other axles ? | "number of wheels in truck = 18 number of wheels on its front axle = 2 number of wheels remaining = 16 number of axles remaining axles = 16 / 4 = 4 total number of axles = 5 t = 2.50 + 0.50 ( x − 2 ) = 2.50 + . 5 * 3 = 2.5 + 1.5 = 4 $ answer d" | a ) $ 2.50 , b ) $ 3.00 , c ) $ 3.50 , d ) $ 4.00 , e ) $ 5.00 | d | add(2.50, multiply(0.50, subtract(add(divide(subtract(18, 2), 4), const_1), 2))) | subtract(n3,n2)|divide(#0,n5)|add(#1,const_1)|subtract(#2,n2)|multiply(n1,#3)|add(n0,#4)| | general |
in 2008 , the profits of company n were 10 percent of revenues . in 2009 , the revenues of company n fell by 20 percent , but profits were 20 percent of revenues . the profits in 2009 were what percent of the profits in 2008 ? | "x = profits r = revenue x / r = 0,1 x = 10 r = 100 2009 : r = 80 x / 80 = 0,2 = 15 / 100 x = 80 * 20 / 100 x = 16 16 / 10 = 1,6 = 160 % , answer e" | a ) 80 % , b ) 105 % , c ) 120 % , d ) 124.2 % , e ) 160 % | e | multiply(divide(multiply(20, subtract(const_1, divide(20, const_100))), 10), const_100) | divide(n3,const_100)|subtract(const_1,#0)|multiply(n4,#1)|divide(#2,n1)|multiply(#3,const_100)| | gain |
a man walking at a constant rate of 6 miles per hour is passed by a woman traveling in the same direction along the same path at a constant rate of 12 miles per hour . the woman stops to wait for the man 10 minutes after passing him , while the man continues to walk at his constant rate . how many minutes must the woma... | "when the woman passes the man , they are aligned ( m and w ) . they are moving in the same direction . after 5 minutes , the woman ( w ) will be ahead the man ( m ) : m - - - - - - m - - - - - - - - - - - - - - - w w in the 5 minutes , after passing the man , the woman walks the distance mw = ww , which is 10 * 12 / 6... | a ) 5 , b ) 15 , c ) 10 , d ) 20 , e ) 25 | c | multiply(const_60, divide(multiply(divide(10, const_60), subtract(12, 6)), 6)) | divide(n2,const_60)|subtract(n1,n0)|multiply(#0,#1)|divide(#2,n0)|multiply(#3,const_60)| | physics |
a number when divided by 219 gives a remainder 32 , what remainder will be obtained by dividing the same number 12 ? | "explanation : 219 + 32 = 251 / 12 = 9 ( remainder ) answer : b" | a ) 7 , b ) 11 , c ) 9 , d ) 2 , e ) 3 | b | subtract(32, multiply(12, const_2)) | multiply(n2,const_2)|subtract(n1,#0)| | general |
what is the difference between the c . i . on rs . 7000 for 1 1 / 2 years at 4 % per annum compounded yearly and half - yearly ? | "c . i . when interest is compounded yearly = [ 7000 * ( 1 + 4 / 100 ) * ( 1 + ( 1 / 2 * 4 ) / 100 ] = 7000 * 26 / 25 * 51 / 50 = rs . 7425.6 c . i . when interest is compounded half - yearly = [ 7000 * ( 1 + 2 / 100 ) 2 ] = ( 7000 * 51 / 50 * 51 / 50 * 51 / 50 ) = rs . 7428.46 difference = ( 7428.46 - 7425.6 ) = rs . ... | a ) s . 2.04 , b ) s . 2.08 , c ) s . 2.02 , d ) s . 2.86 , e ) s . 2.42 | d | subtract(multiply(7000, multiply(multiply(add(1, divide(2, const_100)), add(1, divide(2, const_100))), add(1, divide(2, const_100)))), multiply(7000, multiply(add(1, divide(2, const_100)), add(1, divide(4, const_100))))) | divide(n3,const_100)|divide(n4,const_100)|add(#0,n1)|add(#1,n1)|multiply(#2,#2)|multiply(#2,#3)|multiply(#2,#4)|multiply(n0,#5)|multiply(n0,#6)|subtract(#8,#7)| | general |
mary ' s income is 70 percent more than tim ' s income , and tim ' s income is 40 percent less than juan ' s income . what percent of juan ' s income is mary ' s income ? | juan ' s income = 100 ( assume ) ; tim ' s income = 60 ( 40 percent less than juan ' s income ) ; mary ' s income = 102 ( 70 percent more than tim ' s income ) . thus , mary ' s income ( 102 ) is 102 % of juan ' s income ( 100 ) . answer : b . | a ) 124 % , b ) 102 % , c ) 96 % , d ) 80 % , e ) 64 % | b | multiply(multiply(subtract(const_1, divide(40, const_100)), add(const_1, divide(70, const_100))), const_100) | divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|multiply(#4,const_100) | general |
a sum of money lent out at s . i . amounts to a total of $ 480 after 2 years and to $ 680 after a further period of 5 years . what was the initial sum of money that was invested ? | "s . i for 5 years = $ 680 - $ 480 = $ 200 the s . i . is $ 40 / year s . i . for 2 years = $ 80 principal = $ 480 - $ 80 = $ 400 the answer is c ." | a ) $ 360 , b ) $ 380 , c ) $ 400 , d ) $ 420 , e ) $ 440 | c | subtract(480, multiply(subtract(680, 480), divide(2, 5))) | divide(n1,n3)|subtract(n2,n0)|multiply(#0,#1)|subtract(n0,#2)| | general |
5.005 / 2.002 = | "5.005 / 2.002 = 5005 / 2002 = 5 ( 1001 ) / 2 ( 1001 ) = 5 / 2 = 2.5 the answer is e ." | a ) 2.05 , b ) 2.50025 , c ) 2.501 , d ) 2.5025 , e ) 2.5 | e | multiply(divide(5.005, 2.002), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
irin , ingrid and nell bake chocolate chip cookies in the ratio of 9.18 : 5.17 : 2.05 . if altogether they baked a batch of 148 cookies , what percent of the cookies did irin bake ? | "9.18 x + 5.17 x + 2.05 x = 16.4 x = 150 cookies x = 150 / 16.4 = 9.14 ( approx ) so , irin baked 9.14 * 9.18 cookies or 84 cookies ( approx ) % share = 84 / 150 = 56 approx hence , answer is c ." | a ) 0.125 % , b ) 1.25 % , c ) 56 % , d ) 125 % , e ) 0.152 % | c | multiply(divide(divide(multiply(148, 5.17), add(add(9.18, 5.17), 2.05)), 148), const_100) | add(n0,n1)|multiply(n1,n3)|add(n2,#0)|divide(#1,#2)|divide(#3,n3)|multiply(#4,const_100)| | other |
a man is 24 years older than his son . in two years , his age will be twice the age of his son . the present age of the son is ? | "let the son ' s present age be x years . then , man ' s present age = ( x + 24 ) years . ( x + 24 ) + 2 = 2 ( x + 2 ) x + 26 = 2 x + 4 = > x = 22 . answer : d" | a ) 11 , b ) 25 , c ) 27 , d ) 22 , e ) 91 | d | divide(subtract(24, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | multiply(const_2,const_2)|subtract(const_2,const_1)|subtract(#0,const_2)|subtract(n0,#2)|divide(#3,#1)| | general |
a dress on sale in a shop is marked at $ d . during the discount sale its price is reduced by 55 % . staff are allowed a further 50 % reduction on the discounted price . if a staff member buys the dress what will she have to pay in terms of d ? | "effective discount = a + b + ab / 100 = - 55 - 50 + ( - 55 ) ( - 50 ) / 100 = 77.5 sale price = d * ( 1 - 77.5 / 100 ) sale price = . 225 * d answer ( b )" | a ) 0.275 d , b ) 0.225 d , c ) 0.265 d , d ) 0.245 d , e ) 0.205 d | b | subtract(divide(subtract(const_100, 55), const_100), multiply(divide(subtract(const_100, 55), const_100), divide(50, const_100))) | divide(n1,const_100)|subtract(const_100,n0)|divide(#1,const_100)|multiply(#2,#0)|subtract(#2,#3)| | gain |
if 100 < x < 190 and 10 < y < 100 , then the product xy can not be equal to : | "correct answer : ( d ) determine the range of xy by multiplying the two extremes of each individual range together . the smallest value of xy must be greater than 100 * 10 . the largest value must be less than 190 * 100 . this means that 1000 < xy < 19,000 . ( d ) is outside of this range , so it is not a possible pro... | a ) 18,104 , b ) 18,303 , c ) 18 , 356.732 , d ) 19,502 , e ) 18,909 | d | add(100, const_1) | add(n3,const_1)| | general |
if the cost price of 20 articles is equal to the selling price of 40 articles , what is the % profit or loss made by the merchant ? | "let the cost price of 1 article be $ 1 . therefore , cost price of 20 articles = 20 * 1 = $ 20 the selling price of 40 articles = cost price of 40 articles = $ 40 . now , we know the selling price of 40 articles . let us find the cost price of 40 articles . cost price of 40 articles = 40 * 1 = $ 40 . therefore , profi... | a ) 25 % loss , b ) 25 % profit , c ) 20 % loss , d ) 20 % profit , e ) 50 % loss | e | multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 40), 20)), divide(multiply(const_100, 40), 20))) | multiply(n1,const_100)|divide(#0,n0)|subtract(const_100,#1)|divide(#2,#1)|multiply(#3,const_100)| | gain |
if x < y < z and y - x > 5 , where x is an even integer and y and z are odd integers , what is the least possible value of z - x ? | "we want to minimize z − xz − x , so we need to maximize xx . say z = 11 = oddz = 11 = odd , then max value of yy will be 9 ( as yy is also odd ) . now , since y − 5 > xy − 5 > x - - > 9 − 5 > x 9 − 5 > x - - > 4 > x 4 > x , then max value of xx is 2 ( as xx is even ) . hence , the least possible value of z − xz − x is... | a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | d | add(add(5, const_2), const_2) | add(n0,const_2)|add(#0,const_2)| | general |
reeya obtained 65 , 67 , 76 , 82 and 85 out of 100 in different subjects , what will be the average | "explanation : ( 65 + 67 + 76 + 82 + 855 ) = 75 answer : option b" | a ) 70 , b ) 75 , c ) 80 , d ) 85 , e ) 60 | b | divide(add(add(add(add(65, 67), 76), 82), 85), add(const_4, const_1)) | add(n0,n1)|add(const_1,const_4)|add(n2,#0)|add(n3,#2)|add(n4,#3)|divide(#4,#1)| | general |
if the sides of a square are multiplied by sqrt ( 2 ) , the area of the original square is how many times as large as the area of the resultant square ? | "let x be the original length of one side . then the original area is x ^ 2 . the new square has sides of length sqrt ( 2 ) * x , so the area is 2 x ^ 2 . the area of the original square is 1 / 2 = 50 % times the area of the new square . the answer is c ." | a ) 2 % , b ) 4 % , c ) 50 % , d ) 100 % , e ) 200 % | c | square_perimeter(2) | square_perimeter(n0)| | geometry |
a certain bus driver is paid a regular rate of $ 15 per hour for any number of hours that does not exceed 40 hours per week . for any overtime hours worked in excess of 40 hours per week , the bus driver is paid a rate that is 75 % higher than his regular rate . if last week the bus driver earned $ 982 in total compens... | "for 40 hrs = 40 * 15 = 600 excess = 982 - 600 = 382 for extra hours = . 75 ( 15 ) = 11.25 + 16 = 27.25 number of extra hrs = 382 / 27.25 = 14 total hrs = 40 + 14 = 56 answer a 56" | a ) 56 , b ) 40 , c ) 44 , d ) 48 , e ) 52 | a | add(40, divide(subtract(982, multiply(15, 40)), divide(multiply(15, add(const_100, 75)), const_100))) | add(n3,const_100)|multiply(n0,n1)|multiply(n0,#0)|subtract(n4,#1)|divide(#2,const_100)|divide(#3,#4)|add(n1,#5)| | general |
how long does a train 165 meters long running at the rate of 54 kmph take to cross a bridge 720 meters in length ? | "t = ( 720 + 165 ) / 54 * 18 / 5 t = 59 answer : d" | a ) 28 , b ) 27 , c ) 55 , d ) 59 , e ) 12 | d | divide(add(165, 720), multiply(54, const_0_2778)) | add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)| | physics |
90 students represent x percent of the boys at jones elementary school . if the boys at jones elementary make up 30 % of the total school population of x students , what is x ? | "90 = x / 100 * 30 / 100 * x = > x ^ 2 = 9 * 10000 / 3 = > x = 173 b" | a ) 125 , b ) 173 , c ) 225 , d ) 250 , e ) 500 | b | sqrt(divide(multiply(90, const_100), divide(30, const_100))) | divide(n1,const_100)|multiply(n0,const_100)|divide(#1,#0)|sqrt(#2)| | gain |
last year , company x paid out a total of $ 1 , 050,000 in salaries to its 21 employees . if no employee earned a salary that is more than 8 % greater than any other employee , what is the lowest possible salary that any one employee earned ? | employee 1 earned $ x ( say ) employee 2 will not earn more than $ 1.08 x therfore , to minimize the salary of any one employee , we need to maximize the salaries of the other 20 employees ( 1.08 x * 20 ) + x = 1 , 050,000 solving for x = $ 46 , 460.17 answer d | a ) $ 40,000 , b ) $ 41,667 , c ) $ 42,000 , d ) $ 46 , 460.17 , e ) $ 60,000 | d | add(divide(divide(divide(multiply(add(divide(8, const_100), const_1), multiply(subtract(add(const_1000, const_60), const_10), const_1000)), add(multiply(subtract(21, const_1), add(divide(8, const_100), const_1)), 1)), add(divide(8, const_100), const_1)), const_100), add(multiply(const_100, const_2), const_3)) | add(const_1000,const_60)|divide(n3,const_100)|multiply(const_100,const_2)|subtract(n2,const_1)|add(#2,const_3)|add(#1,const_1)|subtract(#0,const_10)|multiply(#6,const_1000)|multiply(#5,#3)|add(n0,#8)|multiply(#5,#7)|divide(#10,#9)|divide(#11,#5)|divide(#12,const_100)|add(#4,#13) | general |
a part - time employee ’ s hourly wage was increased by 40 % . she decided to decrease the number of hours worked per week so that her total income did not change . by approximately what percent should the number of hours worked be decreased ? | "let ' s plug in somenicenumbers and see what ' s needed . let ' s say the employee used to make $ 1 / hour and worked 100 hours / week so , the total weekly income was $ 100 / week after the 40 % wage increase , the employee makes $ 1.40 / hour we want the employee ' s income to remain at $ 100 / week . so , we want (... | a ) 9 % , b ) 15 % , c ) 29 % , d ) 50 % , e ) 100 % | c | multiply(divide(divide(40, const_100), divide(add(40, const_100), const_100)), const_100) | add(n0,const_100)|divide(n0,const_100)|divide(#0,const_100)|divide(#1,#2)|multiply(#3,const_100)| | general |
a boat moves upstream at the rate of 1 km in 20 minutes and down stream 1 km in 9 minutes . then the speed of the current is : | "rate upstream = ( 1 / 20 * 60 ) = 3 kmph rate dowm stream = 1 / 9 * 60 = 6.7 kmph rate of the current = ½ ( 6.7 - 3 ) = 1.85 kmph answer : e" | a ) 1 kmph , b ) 2 kmph , c ) 3 kmph , d ) 2.5 kmph , e ) 1.85 kmph | e | divide(subtract(multiply(divide(1, 9), const_60), multiply(divide(1, 20), const_60)), const_2) | divide(n0,n3)|divide(n0,n1)|multiply(#0,const_60)|multiply(#1,const_60)|subtract(#2,#3)|divide(#4,const_2)| | physics |
if the selling price of 10 articles is same as the cost price of 12 articles . find the gain or loss percentage ? | "let the c . p of each article be re 1 . then , s . p of 10 articles = c . p of 12 articles = rs . 12 now , c . p of 10 articles = rs . 10 , s . p of 10 articles = rs 12 gain = rs ( 12 - 10 ) = rs 2 . gain % = ( 2 / 10 × 100 ) % = 20 % b )" | a ) 10 % , b ) 20 % , c ) 30 % , d ) 32 % , e ) none | b | subtract(10, 12) | subtract(n0,n1)| | gain |
a man swims downstream 51 km and upstream 18 km taking 3 hours each time , what is the speed of the man in still water ? | "51 - - - 3 ds = 17 ? - - - - 1 18 - - - - 3 us = 6 ? - - - - 1 m = ? m = ( 17 + 6 ) / 2 = 11.5 answer : e" | a ) 2 , b ) 8.5 , c ) 9.5 , d ) 6.5 , e ) 11.5 | e | divide(add(divide(18, 3), divide(51, 3)), const_2) | divide(n1,n2)|divide(n0,n2)|add(#0,#1)|divide(#2,const_2)| | physics |
a jar of 312 marbles is divided equally among a group of marble - players today . if 2 people joined the group in the future , each person would receive 1 marble less . how many people are there in the group today ? | "312 = 24 * 13 = 26 * 12 there are 24 people in the group today . the answer is d ." | a ) 18 , b ) 20 , c ) 22 , d ) 24 , e ) 26 | d | divide(subtract(sqrt(add(multiply(multiply(312, 2), const_4), power(2, 2))), 2), 2) | multiply(n0,n1)|power(n1,n1)|multiply(#0,const_4)|add(#2,#1)|sqrt(#3)|subtract(#4,n1)|divide(#5,n1)| | general |
find the simple interest on rs . 422 for 3 months at 2 paisa per month ? | "explanation : i = ( 422 * 3 * 2 ) / 100 = 25.32 answer : option e" | a ) s . 27.5 , b ) s . 34 , c ) s . 26 , d ) s . 25.28 , e ) s . 25.32 | e | multiply(422, divide(3, const_100)) | divide(n1,const_100)|multiply(n0,#0)| | gain |
amit and ian paint a wall in alternating shifts . first amit paints alone , then ian paints alone , then amit paints alone , etc . during each of his shifts , amit paints 1 / 2 of the remaining unpainted area of the wall , while ian paints 1 / 3 of the remaining unpainted area of the wall during each of his shifts . if... | fraction of area unpainted after first shift of amit = 1 - ( 1 / 2 ) = 1 / 2 fraction of area unpainted after first shift of ian = ( 1 / 2 ) - ( 1 / 3 ) ( 1 / 2 ) = ( 2 / 3 ) ( 1 / 2 ) fraction of area unpainted after second shift of amit = ( 1 / 2 ) ( 2 / 3 ) ( 1 / 2 ) fraction of area unpainted after second shift of ... | ['a ) 1 / 27', 'b ) 1 / 54', 'c ) 1 / 81', 'd ) 1 / 162', 'e ) 1 / 486'] | e | divide(subtract(divide(subtract(divide(subtract(divide(subtract(divide(subtract(subtract(1, divide(1, 2)), divide(subtract(1, divide(1, 2)), 3)), 2), divide(divide(subtract(subtract(1, divide(1, 2)), divide(subtract(1, divide(1, 2)), 3)), 2), 3)), 2), divide(divide(subtract(divide(subtract(subtract(1, divide(1, 2)), di... | divide(n0,n1)|subtract(n0,#0)|divide(#1,n3)|subtract(#1,#2)|divide(#3,n1)|divide(#4,n3)|subtract(#4,#5)|divide(#6,n1)|divide(#7,n3)|subtract(#7,#8)|divide(#9,n1)|divide(#10,n3)|subtract(#10,#11)|divide(#12,n1)|divide(#13,n3)|subtract(#13,#14)|divide(#15,n1) | geometry |
the radius of a semi circle is 6.7 cm then its perimeter is ? | "36 / 7 r = 6.7 = 34.45 answer : d" | a ) 32.75 , b ) 32.45 , c ) 22.45 , d ) 34.45 , e ) 32.15 | d | add(divide(circumface(6.7), const_2), multiply(6.7, const_2)) | circumface(n0)|multiply(n0,const_2)|divide(#0,const_2)|add(#2,#1)| | physics |
john was 19 years old when he married betty . they just celebrated their fifth wedding anniversary , and betty ' s age is now 7 / 8 of john ' s . how old is betty ? | "assume betty ' s age on marriage = x years . john ' s age on marriage = 19 john ' s age after 5 years = 24 years . betty ' s age after 5 years = x + 5 given : x + 5 = 7 / 8 ( 24 ) = 21 therefore betty ' s current age = 21 option a" | a ) 21 , b ) 26 , c ) 28 , d ) 30 , e ) 32 | a | multiply(divide(7, 8), add(19, const_4)) | add(n0,const_4)|divide(n1,n2)|multiply(#0,#1)| | general |
a train 70 m long , running with a speed of 63 km / hr will pass a tree in ? | "speed = 63 * 5 / 18 = 35 / 2 m / sec time taken = 70 * 2 / 35 = 140 sec answer : a" | a ) 140 sec , b ) 160 sec , c ) 176 sec , d ) 150 sec , e ) 170 sec | a | multiply(divide(70, multiply(63, const_1000)), const_3600) | multiply(n1,const_1000)|divide(n0,#0)|multiply(#1,const_3600)| | physics |
a box has exactly 100 balls , and each ball is either red , blue , or white . if the box has 10 more blue balls than white balls , and thrice as many red balls as blue balls , how many white balls does the box has ? | "x = the number of red balls y = the number of blue balls z = the number of white balls from the first sentence we have equation # 1 : x + y + z = 100 . . . the box has 10 more blue balls than white balls . . . equation # 2 : y = 10 + z . . . thrice as many red balls as blue balls . . . equation # 3 : x = 3 y solve equ... | a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 16 | c | divide(subtract(100, multiply(10, const_3)), const_4) | multiply(n1,const_3)|subtract(n0,#0)|divide(#1,const_4)| | general |
right now , al and eliot have bank accounts , and al has more money than eliot . the difference between their two accounts is 1 / 12 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 $ 21 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 / 12 * ( a + e ) = = > 11 a = 13 e - - - - - - - - - - - - ( 1 ) second condition gives two different amounts , al ' s amount = 1.1 a and eliot ' s amount = 1.2 e 1.1 ... | a ) $ 110 , b ) $ 120 , c ) $ 180 , d ) $ 220 , e ) $ 210 | e | divide(multiply(21, const_100), subtract(multiply(add(12, 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 |
at a florist shop on a certain day , all corsages sold for either $ 20 or $ 30 . if 8 of the corsages that sold for $ 30 had instead sold for $ 20 , then the store ' s revenue from corsages that day would have been reduced by 35 percent . what was the store ' s actual revenue from corsages that day ? | i am doing it elaborately , hope it will help you . let , no . of corsages @ $ 20 = x , no . of corsages @ $ 30 = y and revenue = r so , 20 x + 30 y = r . . . . . . . . . ( 1 ) now , given the situation , 20 ( x + 8 ) + 30 ( y - 8 ) = r - . 25 r = > 20 x + 160 + 30 y - 240 = . 75 r = > 20 x + 30 y = . 75 r + 80 . . . .... | a ) $ 320 , b ) $ 400 , c ) $ 600 , d ) $ 800 , e ) $ 1000 | a | subtract(divide(negate(subtract(multiply(20, 8), multiply(30, 8))), divide(35, const_100)), subtract(multiply(20, 8), multiply(30, 8))) | divide(n5,const_100)|multiply(n0,n2)|multiply(n1,n2)|subtract(#1,#2)|negate(#3)|divide(#4,#0)|subtract(#5,#3) | gain |
in a group of houses , 40 had dogs , 30 had cats and 10 houses having both dogs and cats . what is the number of houses ? | make a venn diagram , and enter your data . let the number of houses be x 30 + 10 + 20 = x x = 60 so number of houses were = 60 answer b | a ) 30 , b ) 60 , c ) 40 , d ) 45 , e ) 50 | b | subtract(add(40, 30), 10) | add(n0,n1)|subtract(#0,n2) | other |
cricket match is conducted in us . the run rate of a cricket game was only 3.2 in first 10 over . what should be the run rate in the remaining 40 overs to reach the target of 282 runs ? | "required run rate = 282 - ( 3.2 x 10 ) = 250 = 6.25 40 40 b" | a ) 6 , b ) 6.25 , c ) 7.25 , d ) 7.5 , e ) 8 | b | divide(subtract(282, multiply(3.2, 10)), 40) | multiply(n0,n1)|subtract(n3,#0)|divide(#1,n2)| | gain |
the product of x and y is a constant . if the value of x is increased by 30 % , by what percentage must the value of y be decreased ? | "x * y = constt . let x = y = 100 in beginning i . e . x * y = 100 * 100 = 10000 x ( 100 ) - - - becomes - - - > 1.3 x ( 130 ) i . e . 130 * new ' y ' = 10000 i . e . new ' y ' = 10000 / 130 = 76.92 i . e . y decreases from 100 to 76.92 i . e . decrease of 23.07 % b" | a ) 16 % , b ) 23.07 % , c ) 25 % , d ) 30 % , e ) 35 % | b | multiply(subtract(const_1, divide(const_100, add(const_100, 30))), const_100) | add(n0,const_100)|divide(const_100,#0)|subtract(const_1,#1)|multiply(#2,const_100)| | general |
two employees a and b are paid a total of rs . 550 per week by their employer . if a is paid 120 percent of the sum paid to b , how much is b paid per week ? | "let the amount paid to a per week = x and the amount paid to b per week = y then x + y = 550 but x = 120 % of y = 120 y / 100 = 12 y / 10 ∴ 12 y / 10 + y = 550 ⇒ y [ 12 / 10 + 1 ] = 550 ⇒ 22 y / 10 = 550 ⇒ 22 y = 5500 ⇒ y = 5500 / 22 = 500 / 2 = rs . 250 d )" | a ) rs . 150 , b ) rs . 190 , c ) rs . 200 , d ) rs . 250 , e ) rs . 300 | d | divide(550, add(divide(120, const_100), const_1)) | divide(n1,const_100)|add(#0,const_1)|divide(n0,#1)| | general |
the average age of students of a class is 15.8 years . the average age of boys in the class is 16.4 years and that of the girls is 15.6 years , the ratio of the number of boys to the number of girls in the class is | "explanation : let the ratio be k : 1 . then , k * 16.4 + 1 * 15.6 = ( k + 1 ) * 15.8 < = > ( 16.4 - 15.8 ) k = ( 15.8 - 15.6 ) < = > k = 0.2 / 0.6 = 1 / 3 . required ratio = 1 / 3 : 1 = 1 : 3 . answer : b" | a ) 7 : 3 , b ) 1 : 3 , c ) 9 : 3 , d ) 6 : 3 , e ) 2 : 5 | b | divide(subtract(15.8, 15.6), subtract(16.4, 15.8)) | subtract(n0,n2)|subtract(n1,n0)|divide(#0,#1)| | general |
jack , jill , and sandy each have one try to make a basket from half court . if their individual probabilities of making the basket are 1 / 6 , 1 / 7 , and 1 / 8 respectively , what is the probability that all three will miss ? | "the probability that all three will miss is 5 / 6 * 6 / 7 * 7 / 8 = 5 / 8 . the answer is b ." | a ) 3 / 8 , b ) 5 / 8 , c ) 7 / 16 , d ) 9 / 16 , e ) 23 / 32 | b | multiply(multiply(divide(1, 7), divide(1, 8)), subtract(1, divide(1, 6))) | divide(n0,n3)|divide(n0,n5)|divide(n0,n1)|multiply(#0,#1)|subtract(n0,#2)|multiply(#3,#4)| | general |
in an election between two candidates a and b , the number of valid votes received by a exceeds those received by b by 15 % of the total number of votes polled . if 20 % of the votes polled were invalid and a total of 9720 votes were polled , then how many valid votes did b get ? | "let the total number of votes polled in the election be 100 k . number of valid votes = 100 k - 20 % ( 100 k ) = 80 k let the number of votes polled in favour of a and b be a and b respectively . a - b = 15 % ( 100 k ) = > a = b + 15 k = > a + b = b + 15 k + b now , 2 b + 15 k = 80 k and hence b = 32.5 k it is given t... | a ) 1888 , b ) 2999 , c ) 3159 , d ) 2777 , e ) 2991 | c | divide(subtract(multiply(subtract(const_1, divide(20, const_100)), 9720), divide(multiply(15, 9720), const_100)), const_2) | divide(n1,const_100)|multiply(n0,n2)|divide(#1,const_100)|subtract(const_1,#0)|multiply(n2,#3)|subtract(#4,#2)|divide(#5,const_2)| | general |
if 2 ^ ( 2 w ) = 8 ^ ( w − 3 ) , what is the value of w ? | "2 ^ ( 2 w ) = 8 ^ ( w − 3 ) 2 ^ ( 2 w ) = 2 ^ ( 3 * ( w − 3 ) ) 2 ^ ( 2 w ) = 2 ^ ( 3 w - 9 ) let ' s equate the exponents as the bases are equal . 2 w = 3 w - 9 w = 9 the answer is c ." | a ) 3 , b ) 6 , c ) 9 , d ) 12 , e ) 15 | c | divide(multiply(const_12, log(2)), log(2)) | log(n0)|multiply(#0,const_12)|divide(#1,#0)| | general |
the average runs scored by a batsman in 20 matches is 30 . in the next 10 matches the batsman scored an average of 15 runs . find his average in all the 30 matches ? | "total score of the batsman in 20 matches = 600 . total score of the batsman in the next 10 matches = 150 . total score of the batsman in the 30 matches = 750 . average score of the batsman = 750 / 30 = 25 . answer : a" | a ) 25 , b ) 46 , c ) 88 , d ) 13 , e ) 12 | a | divide(add(multiply(30, 20), multiply(15, 10)), add(20, 10)) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0)| | general |
a luxury liner , queen marry ii , is transporting several cats as well as the crew ( sailors , a cook , and one - legged captain ) to a nearby port . altogether , these passengers have 14 heads and 41 legs . how many cats does the ship host ? | "sa ' s + co + ca + cats = 14 . sa ' s + 1 + 1 + cats = 14 or sa ' s + cats = 12 . sa ' s ( 2 ) + 2 + 1 + cats * 4 = 41 sa ' s * 2 + cats * 4 = 38 or sa ' s + cats * 2 = 19 or 12 - cats + cat * 2 = 19 then cats = 7 c" | a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | c | multiply(divide(subtract(subtract(41, const_1), multiply(subtract(14, const_1), const_2)), subtract(multiply(subtract(14, const_1), const_4), multiply(subtract(14, const_1), const_2))), subtract(14, const_1)) | subtract(n1,const_1)|subtract(n0,const_1)|multiply(#1,const_2)|multiply(#1,const_4)|subtract(#0,#2)|subtract(#3,#2)|divide(#4,#5)|multiply(#6,#1)| | general |
a ladder learning against a wall makes an angle of 60 ° with the ground . if the length of the ladder is 19 m , find the distance of the foot of the ladder from the wall . | let ab be the wall and bc be the ladder . then , < abc = 60 ° and , bc = 19 m . ; ac = x metres ac / bc = cos 60 ° = x / 19 = 1 / 2 x = 19 / 2 = 9.5 m . answer b | a ) 9 m , b ) 9.5 m , c ) 10.5 m , d ) 12 m , e ) none | b | multiply(cosine(divide(multiply(60, divide(add(19, const_3), add(const_4, const_3))), multiply(const_60, const_3))), 19) | add(n1,const_3)|add(const_3,const_4)|multiply(const_3,const_60)|divide(#0,#1)|multiply(n0,#3)|divide(#4,#2)|cosine(#5)|multiply(n1,#6) | physics |
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