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 person travels from p to q a speed of 40 km / hr and returns by increasing his speed by 50 % . what is his average speed for both the trips ? | "speed on return trip = 150 % of 40 = 60 km / hr . average speed = ( 2 * 40 * 60 ) / ( 40 + 60 ) = 48 km / hr . answer : c" | a ) 27 , b ) 87 , c ) 48 , d ) 77 , e ) 51 | c | divide(add(multiply(40, add(const_1, divide(50, const_100))), 40), const_2) | divide(n1,const_100)|add(#0,const_1)|multiply(n0,#1)|add(n0,#2)|divide(#3,const_2)| | general |
the average of 5 numbers is 6.8 . if one of the numbers is multiplied by a factor of 2 , the average of the numbers increases to 9.2 . what number is multiplied by 2 ? | the average of 5 numbers is 6.8 the sum of 5 numbers will be 6.8 x 5 = 34 the average of 5 number after one of the number is multiplied by 2 is 9.2 the sum of the numbers will now be 9.2 x 5 = 46 so the sum has increased by 46 - 34 = 12 let the number multiplied by 2 be n then , 2 n = n + 12 or n = 12 answer : - a | a ) 12.0 , b ) 30.0 , c ) 30.9 , d ) 40.0 , e ) 60.0 | a | subtract(multiply(9.2, 5), multiply(6.8, 5)) | multiply(n0,n3)|multiply(n0,n1)|subtract(#0,#1) | general |
the sale price sarees listed for rs . 345 after successive discount is 12 % and 25 % is ? | "explanation : 345 * ( 88 / 100 ) * ( 75 / 100 ) = 228 answer : d" | a ) 321 , b ) 276 , c ) 342 , d ) 228 , e ) 245 | d | subtract(subtract(345, divide(multiply(345, 12), const_100)), divide(multiply(subtract(345, divide(multiply(345, 12), const_100)), 25), const_100)) | multiply(n0,n1)|divide(#0,const_100)|subtract(n0,#1)|multiply(n2,#2)|divide(#3,const_100)|subtract(#2,#4)| | gain |
the length and breadth of the floor of the room are 20 feet and 10 feet respectively . square tiles of 2 feet length of different colours are to be laid on the floor . black tiles are laid in the first row on all sides . if white tiles are laid in the one - third of the remaining and blue tiles in the rest , how many b... | area left after laying black tiles = [ ( 20 β 4 ) Γ ( 10 β 4 ) ] sq . ft . = 96 sq . ft . area under white tiles = ( 1 β 3 Γ 96 ) sq . ft = 32 sq . ft . area under blue tiles = ( 96 β 32 ) sq . ft = 64 sq . ft . number of blue tiles = 64 / 2 Γ 2 = 16 . answer a | ['a ) 16', 'b ) 24', 'c ) 32', 'd ) 48', 'e ) none of these'] | a | divide(subtract(subtract(multiply(20, 10), multiply(add(add(add(20, 20), add(const_3, const_3)), add(const_3, const_3)), const_2)), divide(subtract(multiply(20, 10), multiply(add(add(add(20, 20), add(const_3, const_3)), add(const_3, const_3)), const_2)), const_3)), const_4) | add(n0,n0)|add(const_3,const_3)|multiply(n0,n1)|add(#0,#1)|add(#3,#1)|multiply(#4,const_2)|subtract(#2,#5)|divide(#6,const_3)|subtract(#6,#7)|divide(#8,const_4) | geometry |
in a group of 89 students , 36 are taking history , and 32 are taking statistics . if 59 students are taking history or statistics or both , then how many students are taking history but not statistics ? | "number of students taking history = h = 36 number of students taking statistics = s = 32 total number of students = t = 89 number of students taking history or statistics or both = b = 59 number of students taking neither history nor statistics = n = 89 - 59 = 30 letxbe the number of students taking both history and s... | a ) 9 , b ) 19 , c ) 27 , d ) 26 , e ) 31 | c | subtract(36, subtract(add(36, 32), 59)) | add(n1,n2)|subtract(#0,n3)|subtract(n1,#1)| | other |
the workforce of company x is 60 % female . the company hired 22 additional male workers , and as a result , the percent of female workers dropped to 55 % . how many employees did the company have after hiring the additional male workers ? | "let ' s xx be total quantity of employees 0.6 x = females before adding men 0.55 ( x + 22 ) = females after adding men as quantity of women does n ' t change we can make an equation : 0.6 x = 0.55 ( x + 22 ) 0.05 x = 12.1 x = 242 - this is quantity of employees before adding 22 men so after adding it will be 264 answe... | a ) 160 , b ) 220 , c ) 264 , d ) 360 , e ) 420 | c | add(divide(multiply(divide(55, const_100), 22), subtract(divide(60, const_100), divide(55, const_100))), 22) | divide(n2,const_100)|divide(n0,const_100)|multiply(n1,#0)|subtract(#1,#0)|divide(#2,#3)|add(n1,#4)| | gain |
a fruit seller sells mangoes at the rate of rs . 8 per kg and thereby loses 20 % . at what price per kg , he should have sold them to make a profit of 5 % | explanation : 80 : 8 = 105 : x x = ( 8 Γ 105 / 80 ) = rs 10.5 option c | a ) rs 8.5 , b ) rs 9.5 , c ) rs 10.5 , d ) rs 11.5 , e ) none of these | c | divide(multiply(8, add(const_100, 5)), subtract(const_100, 20)) | add(n2,const_100)|subtract(const_100,n1)|multiply(n0,#0)|divide(#2,#1) | gain |
two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 21 seconds . the ratio of their speeds is ? | "let the speeds of the two trains be x m / sec and y m / sec respectively . then , length of the first train = 27 x meters , and length of the second train = 17 y meters . ( 27 x + 17 y ) / ( x + y ) = 21 = = > 27 x + 17 y = 21 x + 21 y = = > 6 x = 4 y = = > x / y = 2 / 3 . answer : c" | a ) 3 / 6 , b ) 3 / 2 , c ) 2 / 3 , d ) 3 / 5 , e ) 5 / 2 | c | divide(subtract(27, 21), subtract(21, 17)) | subtract(n0,n2)|subtract(n2,n1)|divide(#0,#1)| | physics |
if n = 9 ^ 11 β 9 , what is the units digit of n ? | always divide the power ( incase 11 ) by 4 and use the remainder as the new power . the question now becomes 9 ^ 3 - 9 . now 9 ^ 3 has last digit 9 . , we subtract 9 from 9 , thus 9 - 9 = 0 is the answer . option a | a ) 0 , b ) 1 , c ) 4 , d ) 6 , e ) 8 | a | divide(log(9), log(power(9, 11))) | log(n2)|power(n0,n1)|log(#1)|divide(#0,#2)| | general |
two trains each 210 m in length each , are running on two parallel lines in opposite directions . if one goes at the speed of 90 km / h while the other travels at 85 km / h . how long will it take for them to pass each other completely . | "explanation : d = 210 m + 210 m = 420 m rs = 90 + 85 = 175 * 5 / 18 = 243 / 5 t = 420 * 5 / 243 = 8.6 sec answer : option d" | a ) 3 sec , b ) 2.6 sec , c ) 3.6 sec , d ) 8.6 sec , e ) 7.6 sec | d | multiply(const_3600, divide(divide(add(210, 210), const_1000), add(90, 85))) | add(n0,n0)|add(n1,n2)|divide(#0,const_1000)|divide(#2,#1)|multiply(#3,const_3600)| | physics |
b completes a work in 8 days . a alone can do it in 10 days . if both work together , the work can be completed in how many days ? | "1 / 8 + 1 / 10 = 9 / 40 40 / 9 = 4.44 days answer : b" | a ) 3.75 days , b ) 4.44 days , c ) 5 days , d ) 6.44 days , e ) 7 days | b | divide(multiply(10, 8), add(8, 10)) | add(n0,n1)|multiply(n0,n1)|divide(#1,#0)| | physics |
a sum fetched a total simple interest of rs . 1000 at the rate of 10 p . c . p . a . in 4 years . what is the sum ? | sol . principal = rs . [ 100 * 1000 / 10 * 4 ] = rs . [ 100000 / 40 ] = rs . 2500 . answer a | a ) 2500 , b ) 2800 , c ) 3500 , d ) 1900 , e ) 1500 | a | divide(divide(multiply(1000, const_100), 10), const_4) | multiply(n0,const_100)|divide(#0,n1)|divide(#1,const_4) | gain |
p and q are the only two applicants qualified for a short - term research project that pays 360 dollars in total . candidate p has more experience and , if hired , would be paid 50 percent more per hour than candidate q would be paid . candidate q , if hired , would require 10 hours more than candidate p to do the job ... | "let q ' s hourly wage be x , then p ' s hourly wage is 1.5 x let t be the number of hours that q needs , then p needs t - 10 hours to do the job . since they both are paid an equal total amount of $ 360 : x * t = 1.5 x * ( t - 10 ) t = 30 hours and q ' s hourly wage is 360 / 30 = $ 12 p ' s hourly wage is 360 / ( t - ... | a ) $ 3 , b ) $ 6 , c ) $ 9 , d ) $ 12 , e ) $ 15 | b | subtract(divide(360, divide(10, subtract(divide(const_3, const_2), const_1))), divide(360, add(divide(10, subtract(divide(const_3, const_2), const_1)), 10))) | divide(const_3,const_2)|subtract(#0,const_1)|divide(n2,#1)|add(n2,#2)|divide(n0,#2)|divide(n0,#3)|subtract(#4,#5)| | general |
find the area of trapezium whose parallel sides are 20 cm and 18 cm long , and the distance between them is 17 cm | "area of a trapezium = 1 / 2 ( sum of parallel sides ) * ( perpendicular distance between them ) = 1 / 2 ( 20 + 18 ) * ( 17 ) = 323 cm 2 answer : b" | a ) 178 cm 2 , b ) 323 cm 2 , c ) 285 cm 2 , d ) 167 cm 2 , e ) 197 cm 2 | b | quadrilateral_area(17, 18, 20) | quadrilateral_area(n2,n1,n0)| | physics |
a cylinder of height h is 7 / 8 of water . when all of the water is poured into an empty cylinder whose radius is 25 percent larger than that of the original cylinder , the new cylinder is 3 / 5 full . the height of the new cylinder is what percent of h ? | "basically we can disregard the radius is 25 % information , as we are only asked about the height of the original and the new cylinder . this is becausethe new cylinder is 3 / 5 fullmeans the same as that it ' s height is 3 / 5 . original cylinder 7 / 8 new cylinder 3 / 5 so 3 / 5 / 7 / 8 = 3 / 5 * 8 / 7 = 0.69 or 69 ... | a ) 25 % , b ) 50 % , c ) 69 % , d ) 80 % , e ) 100 % | c | multiply(divide(divide(3, 5), divide(7, 8)), const_100) | divide(n3,n4)|divide(n0,n1)|divide(#0,#1)|multiply(#2,const_100)| | gain |
linda spent 3 / 4 of her savings on furniture and the rest on a tv . if the tv cost her $ 250 , what were her original savings ? | "if linda spent 3 / 4 of her savings on furnitute , the rest 4 / 4 - 3 / 4 = 1 / 4 on a tv but the tv cost her $ 250 . so 1 / 4 of her savings is $ 250 . so her original savings are 4 times $ 250 = $ 1000 correct answer e" | a ) $ 500 , b ) $ 600 , c ) $ 700 , d ) $ 800 , e ) $ 1000 | e | divide(250, subtract(const_1, divide(3, 4))) | divide(n0,n1)|subtract(const_1,#0)|divide(n2,#1)| | general |
a particular store purchased a stock of turtleneck sweaters and marked up its cost by 20 % . during the new year season , it further marked up its prices by 25 % of the original retail price . in february , the store then offered a discount of 7 % . what was its profit on the items sold in february ? | "assume the total price = 100 x price after 20 % markup = 120 x price after 25 % further markup = 1.25 * 120 x = 150 x price after the discount = 0.93 * 150 x = 139.5 x hence total profit = 39.5 % option d" | a ) 27.5 % , b ) 30 % , c ) 35 % , d ) 39.5 % , e ) 40 % | d | subtract(multiply(divide(subtract(const_100, 7), const_100), multiply(add(const_100, 20), divide(add(const_100, 25), const_100))), const_100) | add(n0,const_100)|add(n1,const_100)|subtract(const_100,n2)|divide(#2,const_100)|divide(#1,const_100)|multiply(#0,#4)|multiply(#3,#5)|subtract(#6,const_100)| | gain |
bill made a profit of 10 % by selling a product . if he had purchased that product for 10 % less and sold it at a profit of 30 % , he would have received $ 56 more . what was his original selling price ? | "let p be the original purchase price of the product . bill originally sold the product for 1.1 * p . in the second scenario , the purchase price is 0.9 * p . a 30 % profit means the selling price would be 1.3 * 0.9 * p = 1.17 * p thus , according to the information in the question , 1.17 p - 1.1 p = 56 0.07 = 56 p = 8... | a ) $ 880 , b ) $ 950 , c ) $ 1100 , d ) $ 1240 , e ) $ 1360 | a | multiply(divide(56, subtract(multiply(subtract(const_1, divide(10, const_100)), add(const_1, divide(30, const_100))), add(const_1, divide(10, const_100)))), add(const_1, divide(10, const_100))) | divide(n0,const_100)|divide(n2,const_100)|add(#0,const_1)|add(#1,const_1)|subtract(const_1,#0)|multiply(#3,#4)|subtract(#5,#2)|divide(n3,#6)|multiply(#2,#7)| | general |
a train speeds past a pole in 10 seconds and a platform 100 m long in 30 seconds . its length is ? | "let the length of the train be x meters and its speed be y m / sec . they , x / y = 10 = > y = x / 10 x + 100 / 30 = x / 10 x = 50 m . answer : d" | a ) 188 m , b ) 876 m , c ) 251 m , d ) 50 m , e ) 45 m | d | multiply(100, subtract(const_2, const_1)) | subtract(const_2,const_1)|multiply(n1,#0)| | physics |
suraj has a certain average of runs for 16 innings . in the 17 th innings he scores 112 runs thereby increasing his average by 6 runs . what is his average after the 17 th innings ? | "to improve his average by 6 runs per innings he has to contribute 16 x 6 = 96 runs for the previous 16 innings . thus , the average after the 17 th innings = 112 - 96 = 16 . answer : c" | a ) 18 , b ) 25 , c ) 16 , d ) 72 , e ) 27 | c | divide(subtract(112, multiply(16, 6)), subtract(17, 16)) | multiply(n0,n3)|subtract(n1,n0)|subtract(n2,#0)|divide(#2,#1)| | general |
a man can row downstream at the rate of 16 km / hr and upstream at 8 km / hr . find man ' s rate in still water and the rate of current ? | "explanation : rate of still water = 1 / 2 ( 16 + 8 ) = 12 km / hr rate of current = 1 / 2 ( 16 - 8 ) = 4 km / hr answer : option a" | a ) 12,4 , b ) 6,3 , c ) 9,3 , d ) 6,6 , e ) none of these | a | divide(subtract(16, 8), const_2) | subtract(n0,n1)|divide(#0,const_2)| | gain |
the average of squares of first 11 consecutive even numbers is : | 2 ( n + 1 ) ( 2 n + 1 ) / 3 ( 2 * 12 * 23 ) / 3 = 2 * 4 * 23 = 184 answer : c | a ) 164 , b ) 174 , c ) 184 , d ) 194 , e ) 204 | c | divide(multiply(multiply(const_2, add(11, const_1)), add(multiply(11, const_2), const_1)), const_3) | add(n0,const_1)|multiply(n0,const_2)|add(#1,const_1)|multiply(#0,const_2)|multiply(#2,#3)|divide(#4,const_3) | general |
the length of a rectangular plot is 10 mtr more than its width . the cost of fencing the plot along its perimeter at the rate of rs . 6.5 mtr is rs . 1430 . the perimeter of the plot is ? | "sol . let width = x , length = ( 10 + x ) perimeter = 2 ( x + ( 10 + x ) ) = 2 ( 2 x = 10 ) & 2 ( 2 x + 10 ) * 6.5 = 1430 x = 50 required perimeter = 2 ( 50 + 60 ) = 220 e" | a ) 126 , b ) 156 , c ) 190 , d ) 321 , e ) 220 | e | multiply(add(divide(subtract(divide(divide(1430, 6.5), const_2), 10), const_2), add(divide(subtract(divide(divide(1430, 6.5), const_2), 10), const_2), 10)), const_2) | divide(n2,n1)|divide(#0,const_2)|subtract(#1,n0)|divide(#2,const_2)|add(#3,n0)|add(#4,#3)|multiply(#5,const_2)| | geometry |
rahul played weel in this season . his current batting averagge is 51 . if he score 69 runs in today match . his batting average will become 54 . how many matches had he played in this season . | 51 x + 69 = 54 ( x + 1 ) = > 3 x = 15 = > x = 5 answer : e | a ) 8 , b ) 10 , c ) 9 , d ) 6 , e ) 5 | e | divide(subtract(69, 54), subtract(54, 51)) | subtract(n1,n2)|subtract(n2,n0)|divide(#0,#1) | general |
1852 - 1230 Γ· 10.00 = ? | "answer given expression = 1852 - 1230 Γ· 10.00 = 1852 - 123 = 1729 correct option : d" | a ) 62.2 , b ) 2000 , c ) 1700 , d ) 1729 , e ) none | d | subtract(multiply(divide(1852, const_100), 1230), multiply(divide(const_1, const_3), multiply(divide(1852, const_100), 1230))) | divide(n0,const_100)|divide(const_1,const_3)|multiply(n1,#0)|multiply(#1,#2)|subtract(#2,#3)| | general |
last year a certain bond price with a face value of 5000 yielded 8 % of its face value in interest . if that interest was approx 6.5 of the bond ' s selling price approx what was the bond ' s selling price ? | "interest = 0.08 * 5000 = 0.065 * selling price - - > selling price = 0.08 * 5000 / 0.065 - - > selling price = ~ 6,154 answer : e ." | a ) 4063 , b ) 5325 , c ) 5351 , d ) 6000 , e ) 6154 | e | divide(multiply(5000, divide(8, const_100)), divide(6.5, const_100)) | divide(n1,const_100)|divide(n2,const_100)|multiply(n0,#0)|divide(#2,#1)| | gain |
mixture contains alcohol and water in the ratio 4 : 3 . if 8 liters of water is added to the mixture , the ratio becomes 4 : 5 . find the quantity of alcohol in the given mixture . | "let the quantity of alcohol and water be 4 x litres and 3 x litres respectively 32 x = 4 ( 3 x + 5 ) 20 x = 20 x = 1 quantity of alcohol = ( 4 x 1 ) litres = 4 litres . answer : d" | a ) 10 , b ) 99 , c ) 27 , d ) 4 , e ) 29 | d | multiply(divide(multiply(4, 5), subtract(multiply(8, 4), multiply(3, 4))), 4) | multiply(n0,n4)|multiply(n0,n2)|multiply(n0,n1)|subtract(#1,#2)|divide(#0,#3)|multiply(n0,#4)| | general |
mary ' s income is 60 % more than tim ' s income and tim ' s income is 50 % less than juan ' s income . what % of juan ' s income is mary ' s income . | "even i got 96 % j = 100 t = 100 * 0.5 = 50 m = 50 * 1.6 = 80 if mary ' s income is x percent of j m = j * x / 100 x = m * 100 / j = 80 * 100 / 100 = 80 ans : d" | a ) 124 % , b ) b . 120 % , c ) 96 % , d ) 80 % , e ) 64 % | d | add(subtract(const_100, 50), multiply(subtract(const_100, 50), divide(60, const_100))) | divide(n0,const_100)|subtract(const_100,n1)|multiply(#0,#1)|add(#2,#1)| | general |
laxmi and prasanna set on a journey . laxmi moves northwards at a speed of 40 kmph and prasanna moves southward at a speed of 38 kmph . how far will be prasanna from laxmi after 60 minutes ? | "explanation : we know 60 min = 1 hr total northward laxmi ' s distance = 40 kmph x 1 hr = 40 km total southward prasanna ' s distance = 38 kmph x 1 hr = 38 km total distance between prasanna and laxmi is = 40 + 38 = 78 km . answer : c" | a ) 11 , b ) 50 , c ) 78 , d ) 27 , e ) 18 | c | add(40, 38) | add(n0,n1)| | physics |
a grocer has 400 pounds of coffee in stock , 20 percent of which is decaffeinated . if the grocer buys another 100 pounds of coffee of which 70 percent is decaffeinated , what percent , by weight , of the grocer β s stock of coffee is decaffeinated ? | "1 . 20 % of 400 = 80 pounds of decaffeinated coffee 2 . 70 % of 100 = 70 pounds of decaffeinated coffee 3 . wt have 150 pounds of decaffeinated out of 500 pounds , that means 150 / 500 * 100 % = 30 % . the correct answer is b ." | a ) 28 % , b ) 30 % , c ) 32 % , d ) 34 % , e ) 36 % | b | multiply(divide(add(multiply(divide(20, 100), 400), multiply(100, divide(70, 100))), add(400, 100)), 100) | add(n0,n2)|divide(n1,n2)|divide(n3,n2)|multiply(n0,#1)|multiply(n2,#2)|add(#3,#4)|divide(#5,#0)|multiply(#6,n2)| | gain |
the value of ( 34.31 * 0.5 * 1.567 ) / ( 0.0673 * 23.25 * 7.57 ) is close to | "( 34.31 * 0.5 * 1.567 ) / ( 0.0673 * 23.25 * 7.57 ) = 26.882 / 11.845 = 2.27 answer : d" | a ) 2 , b ) 1.15 , c ) 2.05 , d ) 2.27 , e ) 2.35 | d | divide(divide(multiply(multiply(34.31, 0.5), 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 average age of 15 students of a class is 15 years . out of these , the average age of 5 students is 14 years and that of the other 9 students is 16 years . the age of the 15 th student is | "solution age of the 15 th student = [ 15 x 15 - ( 14 x 5 + 16 x 9 ) ] = ( 225 - 214 ) = 11 years . answer b" | a ) 9 years , b ) 11 years , c ) 14 years , d ) 21 years , e ) 25 years | b | subtract(multiply(15, 15), add(multiply(5, 14), multiply(9, 16))) | multiply(n0,n0)|multiply(n2,n3)|multiply(n4,n5)|add(#1,#2)|subtract(#0,#3)| | general |
the population of a town is 10000 . it increases annually at the rate of 20 % p . a . what will be its population after 2 years ? | "formula : ( after = 100 denominator ago = 100 numerator ) 10000 Γ 120 / 100 Γ 120 / 100 = 14400 b )" | a ) 14300 , b ) 14400 , c ) 14500 , d ) 14600 , e ) 15400 | b | add(10000, multiply(divide(multiply(10000, 20), const_100), 2)) | multiply(n0,n1)|divide(#0,const_100)|multiply(#1,n2)|add(n0,#2)| | gain |
what is the decimal equivalent of ( 1 / 4 ) ^ 3 ? | ( 1 / 4 ) ^ 3 = 1 / 64 = 0.0156 answer : e | a ) 0.0016 , b ) 0.0625 , c ) 0.16 , d ) 0.25 , e ) 0.0156 | e | power(divide(1, 4), const_3) | divide(n0,n1)|power(#0,const_3) | general |
a student took 6 courses last year and received an average ( arithmetic mean ) grade of 100 points . the year before , the student took 5 courses and received an average grade of 40 points . to the nearest tenth of a point , what was the student β s average grade for the entire two - year period ? | "let the 6 courses that were taken last year be a 1 , a 2 , a 3 , a 4 , a 5 , a 6 a 1 + a 2 + a 3 + a 4 + a 5 + a 6 = 100 * 6 = 600 the year before , the 5 courses be b 1 , b 2 , b 3 , b 4 , b 5 b 1 + b 2 + b 3 + b 4 + b 5 = 40 * 5 = 200 student ' s average = ( 600 + 200 ) / 11 = 72.72 answer d" | a ) 79 , b ) 89 , c ) 95 , d ) 72.72 , e ) 97.2 | d | floor(divide(add(multiply(6, 100), multiply(5, 40)), add(6, 5))) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0)|floor(#4)| | general |
at what price must an article costing rs . 66.5 be marked in order that after deducting 5 % from the list price . it may be sold at a profit of 25 % on the cost price ? | "cp = 66.5 sp = 66.5 * ( 125 / 100 ) = 83.125 mp * ( 95 / 100 ) = 59.375 mp = 87.5 answer : a" | a ) 87.5 , b ) 89.5 , c ) 82.3 , d ) 82.1 , e ) 92.4 | a | divide(multiply(add(66.5, divide(multiply(66.5, 25), const_100)), const_100), subtract(const_100, 5)) | multiply(n0,n2)|subtract(const_100,n1)|divide(#0,const_100)|add(n0,#2)|multiply(#3,const_100)|divide(#4,#1)| | gain |
three 6 faced dice are thrown together . the probability that all the three show the same number on them is - . | "it all 3 numbers have to be same basically we want triplets . 111 , 222 , 333 , 444 , 555 and 666 . those are six in number . further the three dice can fall in 6 * 6 * 6 = 216 ways . hence the probability is 6 / 216 = 1 / 36 answer a" | a ) 1 / 36 , b ) 1 / 216 , c ) 1 / 48 , d ) 1 / 55 , e ) 1 / 90 | a | multiply(multiply(multiply(divide(const_1, 6), divide(const_1, 6)), divide(const_1, 6)), divide(const_1, 6)) | divide(const_1,n0)|multiply(#0,#0)|multiply(#0,#1)|multiply(#0,#2)| | probability |
the smallest number which when diminished by 7 , is divisible 12 , 16 , 18 , 21 and 28 is : | "explanation required number = ( l . c . m . of 12,16 , 18 , 21 , 28 ) + 7 = 1008 + 7 = 1015 answer b" | a ) 1008 , b ) 1015 , c ) 1022 , d ) 1032 , e ) none of these | b | add(multiply(multiply(power(const_3, const_2.0), power(const_2.0, const_4)), add(const_3, const_4)), 7) | add(const_3,const_4)|power(const_3,const_2.0)|power(const_2.0,const_4)|multiply(#1,#2)|multiply(#0,#3)|add(#4,n0)| | general |
the average of first 15 even numbers is ? | "sum of 15 even numbers = 15 * 16 = 240 average = 240 / 15 = 16 answer : d" | a ) 10 , b ) 11 , c ) 12 , d ) 16 , e ) 14 | d | add(15, const_1) | add(n0,const_1)| | general |
in 100 litres mixture milk and water are in the ratio 3 : 2 . after adding how many liters of water its ratio will become 3 : 4 | milk quantity = 3 / 5 * 60 = 36 water quantity = 100 - 36 = 64 new ratio of m : w = 36 : 64 + x = 3 : 4 192 + 3 x = 144 x = 48 answer is a | a ) 48 , b ) 50 , c ) 62 , d ) 78 , e ) 100 | a | multiply(subtract(divide(multiply(divide(3, add(3, 2)), 100), divide(3, add(3, 4))), 100), divide(add(const_10, const_2), const_10)) | add(const_10,const_2)|add(n1,n2)|add(n1,n4)|divide(#0,const_10)|divide(n1,#1)|divide(n1,#2)|multiply(n0,#4)|divide(#6,#5)|subtract(#7,n0)|multiply(#3,#8) | general |
a sum fetched a total simple interest of 4034.25 at the rate of 9 % . p . a . in 5 years . what is the sum ? | "principal = ( 100 x 4034.25 ) / ( 9 x 5 ) = 403425 / 45 = 8965 . answer c" | a ) 5768 , b ) 8925 , c ) 8965 , d ) 6474 , e ) 8723 | c | divide(divide(multiply(4034.25, const_100), 9), 5) | multiply(n0,const_100)|divide(#0,n1)|divide(#1,n2)| | gain |
rani bought more apples than oranges . she sells apples at βΉ 23 apiece and makes 15 % profit . she sells oranges at βΉ 10 apiece and makes 25 % profit . if she gets βΉ 653 after selling all the apples and oranges , find her profit percentage w . | "given : selling price of an apple = 23 - - > cost price = 23 / 1.15 = 20 selling price of an orange = 10 - - > cost price = 10 / 1.25 = 8 a > o 23 * ( a ) + 10 * ( o ) = 653 653 - 23 * ( a ) has to be divisible by 10 - - > units digit has to be 0 values of a can be 1 , 11 , 21 , 31 , . . . . - - > 1 can not be the val... | a ) 16.8 % , b ) 17.4 % , c ) 17.9 % , d ) 18.5 % , e ) 19.1 % | b | multiply(divide(subtract(653, add(multiply(multiply(const_2, 10), add(multiply(const_2, 10), const_1)), multiply(divide(10, add(divide(25, const_100), const_1)), add(15, const_2)))), add(multiply(multiply(const_2, 10), add(multiply(const_2, 10), const_1)), multiply(divide(10, add(divide(25, const_100), const_1)), add(1... | add(n1,const_2)|divide(n3,const_100)|multiply(n2,const_2)|add(#2,const_1)|add(#1,const_1)|divide(n2,#4)|multiply(#3,#2)|multiply(#0,#5)|add(#6,#7)|subtract(n4,#8)|divide(#9,#8)|multiply(#10,const_100)| | gain |
cereal a is 9 % sugar by weight , whereas healthier but less delicious cereal b is 2 % sugar by weight . to make a delicious and healthy mixture that is 4 % sugar , what should be the ratio of cereal a to cereal b , by weight ? | "( 9 / 100 ) a + ( 2 / 100 ) b = ( 4 / 100 ) ( a + b ) 5 a = 2 b = > a / b = 2 / 5 answer is b ." | a ) 2 : 9 , b ) 2 : 5 , c ) 1 : 6 , d ) 1 : 4 , e ) 1 : 3 | b | divide(subtract(4, 2), subtract(9, 4)) | subtract(n2,n1)|subtract(n0,n2)|divide(#0,#1)| | general |
how long does a train 180 meters long running at the rate of 54 kmph take to cross a bridge 660 meters in length ? | "explanation : t = ( 660 + 180 ) / 54 * 18 / 5 t = 56 answer : option b" | a ) 33 , b ) 56 , c ) 55 , d ) 82 , e ) 62 | b | divide(add(180, 660), multiply(54, const_0_2778)) | add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)| | physics |
a profit of rs . 500 is divided between x and y in the ratio of 1 / 2 : 1 / 3 . what is the difference between their profit shares ? | "a profit of rs . 500 is divided between x and y in the ratio of 1 / 2 : 1 / 3 or 3 : 2 . so profits are 300 and 200 . difference in profit share = 300 - 200 = 100 answer : b" | a ) rs . 200 , b ) rs . 100 , c ) rs . 300 , d ) rs . 50 , e ) rs . 90 | b | subtract(divide(divide(500, add(divide(1, 2), divide(1, 3))), 2), divide(divide(500, add(divide(1, 2), divide(1, 3))), 3)) | divide(n1,n2)|divide(n1,n4)|add(#0,#1)|divide(n0,#2)|divide(#3,n2)|divide(#3,n4)|subtract(#4,#5)| | general |
a man buys rs . 56 shares paying 9 % dividend . the man wants to have an interest of 12 % on his money . the market value of each share is : | "dividend on rs . 56 = rs . 9 / 100 x 56 = rs . 5.04 . rs . 12 is an income on rs . 100 . rs . 5.04 is an income on rs . 100 / 12 x 5.04 = rs . 42 . answer : option e" | a ) s . 12 , b ) s . 15 , c ) s . 18 , d ) s . 21 , e ) s . 42 | e | multiply(divide(const_100, 12), multiply(divide(9, const_100), 56)) | divide(const_100,n2)|divide(n1,const_100)|multiply(n0,#1)|multiply(#0,#2)| | gain |
of 440 surveyed students , 20 % of those who read book a also read book b and 25 % of those who read book b also read book a . if each student read at least one of the books , what is the difference between the number of students who read only book a and the number of students who read only book b ? | "say the number of students who read book a is a and the number of students who read book b is b . given that 20 % of those who read book a also read book b and 25 % of those who read book b also read book a , so the number of students who read both books is 0.2 a = 0.25 b - - > a = 1.25 b . since each student read at ... | a ) 20 , b ) 25 , c ) 30 , d ) 35 , e ) 55 | e | subtract(multiply(multiply(divide(440, subtract(add(divide(divide(25, const_100), divide(20, const_100)), const_1), divide(25, const_100))), divide(divide(25, const_100), divide(20, const_100))), subtract(const_1, divide(20, const_100))), multiply(divide(440, subtract(add(divide(divide(25, const_100), divide(20, const_... | divide(n2,const_100)|divide(n1,const_100)|divide(#0,#1)|subtract(const_1,#1)|subtract(const_1,#0)|add(#2,const_1)|subtract(#5,#0)|divide(n0,#6)|multiply(#7,#2)|multiply(#7,#4)|multiply(#8,#3)|subtract(#10,#9)| | gain |
on a certain road , 10 % of the motorists exceed the posted speed limit and receive speeding tickets , but 30 % of the motorists who exceed the posted speed limit do not receive speeding tickets . what percent of the motorists on that road exceed the posted speed limit ? | "suppose there are x motorists . 10 % of them exceeded the speed limit and received the ticket , i . e . x / 10 . again , suppose total no . of motorists who exceeded the speed limit are y . 30 % of y exceeded the speed limit but did n ' t received the ticket , i . e . 3 y / 10 . it means 7 y / 10 received the ticket .... | a ) 10.5 % , b ) 12.5 % , c ) 14.3 % , d ) 22 % , e ) 30 % | c | multiply(divide(10, subtract(const_100, 30)), const_100) | subtract(const_100,n1)|divide(n0,#0)|multiply(#1,const_100)| | gain |
ajay can ride 50 km in 1 hour . in how many hours he can ride 750 km ? | "1 hour he ride 50 km he ride 750 km in = 750 / 50 * 1 = 15 hours answer is b" | a ) 10 hrs , b ) 15 hrs , c ) 20 hrs , d ) 25 hrs , e ) 18 hrs | b | divide(750, 50) | divide(n2,n0)| | physics |
a number of 47 marbles is to be divided and contain with boxes . if each box is to contain 3 , 4 , or 5 marbles , what is the largest possible number of boxes ? | "to maximize # of boxes we should minimize marbles per box : 14 * 3 + 1 * 5 = 47 - - > 14 + 1 = 15 . answer : c ." | a ) 10 , b ) 12 , c ) 15 , d ) 16 , e ) 17 | c | divide(47, 3) | divide(n0,n1)| | general |
12 is subtracted from 95 % of a number , the result is 178 . find the number ? | "( 95 / 100 ) * x Γ’ β¬ β 12 = 178 9.5 x = 1900 x = 200 answer : a" | a ) 200 , b ) 240 , c ) 198 , d ) 190 , e ) 322 | a | divide(add(12, 178), divide(95, const_100)) | add(n0,n2)|divide(n1,const_100)|divide(#0,#1)| | gain |
a man invests rs . 8000 at the rate of 5 % per annum . how much more should he invest at the rate of 8 % , so that he can earn a total of 6 % per annum ? | explanation : interest on rs . 8000 at 5 % per annum = ( 8000 Γ 5 Γ 1 ) / 100 = rs . 400 let his additional investment at 8 % = x interest on rs . x at 8 % per annum = ( x Γ 8 Γ 1 ) / 100 = 2 x / 25 . to earn 6 % per annum for the total , interest = ( 8000 + x ) Γ 6 Γ 1 / 100 . = > 400 + 2 x / 25 = ( 8000 + x ) Γ 6 Γ 1... | a ) rs . 1200 , b ) rs . 1300 , c ) rs . 1500 , d ) rs . 4000 , e ) none of these | d | divide(subtract(multiply(8000, 6), multiply(8000, 5)), subtract(8, 6)) | multiply(n0,n3)|multiply(n0,n1)|subtract(n2,n3)|subtract(#0,#1)|divide(#3,#2) | gain |
the cost per unit of electricity during a 4 year period is as follows : year 1 , $ 3 per unit ; year 2 , $ 4 per unit ; year 3 , $ 5 per unit ; year 4 , $ 6 per unit . if a factory spent equal dollars on electricity each year , what was the average cost of a unit of electricity during the 4 year period ? | let money spent each year = $ 60 units bought first year = 20 units bought second year = 15 units bought third year = 12 units bought fourth year = 10 total number of units = 57 total money spent = 60 * 4 = 240 average = 240 / 57 = 4.21 answer : c | a ) $ 4.31 , b ) $ 4.51 , c ) $ 4.21 , d ) $ 5.21 , e ) $ 5.61 | c | divide(4, add(add(add(divide(const_1, 3), divide(const_1, 4)), divide(const_1, 5)), divide(const_1, 6))) | divide(const_1,n2)|divide(const_1,n0)|divide(const_1,n6)|divide(const_1,n8)|add(#0,#1)|add(#4,#2)|add(#5,#3)|divide(n0,#6) | general |
in what time will a train 45 m long cross an electric pole , it its speed be 108 km / hr ? | "speed = 108 * 5 / 18 = 30 m / sec time taken = 45 / 30 = 1.5 sec . answer : c" | a ) 0.5 , b ) 1.25 , c ) 1.5 , d ) 2.5 , e ) 3.5 | c | divide(45, multiply(108, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics |
in township k , 1 / 5 of the housing units are equiped with cable tv . if 1 / 10 of the housing units , including 1 / 4 of those that are equiped with cable tv , are equipped with videocassette recorders , what fraction of the housing units have neither cable tv nor videocassette recorders ? | 1 / 5 - - cable tv ( this includes some data from video cassette recorder ) 1 / 10 - - video cassette recorder including 1 / 4 ( equiped with cable tv ) i . e . 1 / 4 ( 1 / 5 ) = 1 / 20 therefore only video cassette recorder = 1 / 10 - 1 / 20 = 1 / 20 total = 1 / 5 + 1 / 20 + neither cable tv nor videocassette recorder... | a ) 3 / 4 , b ) 11 / 15 , c ) 7 / 10 , d ) 1 / 6 , e ) 2 / 15 | a | divide(add(10, 5), multiply(10, const_2)) | add(n1,n3)|multiply(n3,const_2)|divide(#0,#1) | other |
two trains of equal are running on parallel lines in the same direction at 46 km / hr and 36 km / hr . the faster train passes the slower train in 45 sec . the length of each train is ? | "let the length of each train be x m . then , distance covered = 2 x m . relative speed = 46 - 36 = 10 km / hr . = 10 * 5 / 18 = 25 / 9 m / sec . 2 x / 45 = 25 / 9 = > x = 65.5 . answer : d" | a ) 50 , b ) 26 , c ) 27 , d ) 62.5 , e ) 21 | d | divide(multiply(45, divide(multiply(subtract(46, 36), const_1000), const_3600)), const_2) | subtract(n0,n1)|multiply(#0,const_1000)|divide(#1,const_3600)|multiply(n2,#2)|divide(#3,const_2)| | general |
a batsman makes a score of 87 runs in the 17 th inning and thus increases his averages by 3 . find his average after 17 th inning ? | "solution let the average after 17 th inning = x . then , average after 16 th inning = ( x - 3 ) average = 16 ( x - 3 ) + 87 = 17 x or x = ( 87 - 48 ) = 39 . answer c" | a ) 19 , b ) 29 , c ) 39 , d ) 49 , e ) 59 | c | add(subtract(87, multiply(17, 3)), 3) | multiply(n1,n2)|subtract(n0,#0)|add(n2,#1)| | general |
in what time will a train 100 meters long cross an electric pole , if its speed is 360 km / hr | "explanation : first convert speed into m / sec speed = 360 * ( 5 / 18 ) = 100 m / sec time = distance / speed = 100 / 100 = 1 second answer : a" | a ) 1 second , b ) 4.5 seconds , c ) 3 seconds , d ) 2.5 seconds , e ) none of these | a | divide(100, multiply(360, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics |
a 340 - liter solution of kola is made from 64 % water , 9 % concentrated kola and the rest is made from sugar . if 3.2 liters of sugar , 8 liter of water and 6.8 liters of concentrated kola were added to the solution , what percent of the solution is made from sugar ? | "denominator : 340 + 8 + 3.2 + 6.8 = 358 numerator : 340 ( 1 - . 64 - . 09 ) + 3.2 340 ( 0.27 ) + 3.2 91.8 + 3.2 95 ratio : 95 / 358 = 0.265 answer : e" | a ) 6 % . , b ) 7.5 % . , c ) 9.2 % . , d ) 10.5 % . , e ) 26.5 % . | e | multiply(divide(add(subtract(subtract(340, multiply(340, divide(64, const_100))), multiply(340, divide(9, const_100))), 3.2), add(add(add(340, 3.2), 8), 6.8)), const_100) | add(n0,n3)|divide(n1,const_100)|divide(n2,const_100)|add(n4,#0)|multiply(n0,#1)|multiply(n0,#2)|add(n5,#3)|subtract(n0,#4)|subtract(#7,#5)|add(n3,#8)|divide(#9,#6)|multiply(#10,const_100)| | gain |
if ( 5 ^ 11 ) ( 9 ^ 6 ) = 3 ( 15 ^ x ) , what is the value of x ? | "( 5 ^ 11 ) ( 9 ^ 6 ) = 3 ( 15 ^ x ) = > 5 ^ 11 * 3 ^ 12 = 3 * 3 ^ x * 5 ^ x = > 5 ^ 11 * 3 ^ 12 = 3 ^ ( x + 1 ) * 5 ^ x value of x = 11 answer c" | a ) 7 , b ) 9 , c ) 11 , d ) 13 , e ) 15 | c | divide(log(divide(multiply(power(5, 11), power(9, 6)), 3)), log(15)) | log(n5)|power(n0,n1)|power(n2,n3)|multiply(#1,#2)|divide(#3,n4)|log(#4)|divide(#5,#0)| | general |
find the total number of prime factors in the expression ( 4 ) ^ 11 x ( 7 ) ^ 5 x ( 11 ) ^ 5 | "( 4 ) ^ 11 x ( 7 ) ^ 5 x ( 11 ) ^ 5 = ( 2 x 2 ) ^ 11 x ( 7 ) ^ 5 x ( 11 ) ^ 5 = 2 ^ 11 x 2 ^ 11 x 7 ^ 5 x 11 ^ 5 = 2 ^ 22 x 7 ^ 5 x 11 ^ 5 total number of prime factors = ( 22 + 5 + 5 ) = 32 . answer is b ." | a ) 26 , b ) 32 , c ) 25 , d ) 27 , e ) 29 | b | add(add(multiply(5, 11), 5), 5) | multiply(n1,n5)|add(n3,#0)|add(n5,#1)| | general |
a rectangular photograph is surrounded by a border that is 1 inch wide on each side . the total area of the photograph and the border is m square inches . if the border had been 5 inches wide on each side , the total area would have been ( m + 180 ) square inches . what is the perimeter of the photograph , in inches ? | "let length and breadth of photograph be l and b respectively . perimeter is given by 2 * ( l + b ) - - - - - ( 1 ) according to the question : ( l + 2 ) ( b + 2 ) = m - - - - ( 2 ) and ( l + 10 ) ( b + 10 ) = m + 180 - - - - - - - - - > ( l + 10 ) ( b + 10 ) - 180 = m - - - - - - ( 3 ) equating ( 2 ) and ( 3 ) ( l + 2... | a ) 21 , b ) 36 , c ) 38 , d ) 40 , e ) 42 | a | divide(subtract(180, subtract(power(multiply(5, const_2), const_2), power(multiply(1, const_2), const_2))), const_2) | multiply(n1,const_2)|multiply(n0,const_2)|power(#0,const_2)|power(#1,const_2)|subtract(#2,#3)|subtract(n2,#4)|divide(#5,const_2)| | geometry |
a student took 6 courses last year and received an average ( arithmetic mean ) grade of 100 points . the year before , the student took 5 courses and received an average grade of 50 points . to the nearest tenth of a point , what was the student β s average grade for the entire two - year period ? | "let the 6 courses that were taken last year be a 1 , a 2 , a 3 , a 4 , a 5 , a 6 a 1 + a 2 + a 3 + a 4 + a 5 + a 6 = 100 * 6 = 600 the year before , the 5 courses be b 1 , b 2 , b 3 , b 4 , b 5 b 1 + b 2 + b 3 + b 4 + b 5 = 50 * 5 = 250 student ' s average = ( 600 + 250 ) / 11 = 77.27 answer d" | a ) 79 , b ) 89 , c ) 95 , d ) 77.27 , e ) 97.2 | d | floor(divide(add(multiply(6, 100), multiply(5, 50)), add(6, 5))) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0)|floor(#4)| | general |
jim β s taxi service charges an initial fee of $ 2.35 at the beginning of a trip and an additional charge of $ 0.35 for each 2 / 5 of a mile traveled . what is the total charge for a trip of 3.6 miles ? | "let the fixed charge of jim β s taxi service = 2.35 $ and charge per 2 / 5 mile ( . 4 mile ) = . 35 $ total charge for a trip of 3.6 miles = 2.35 + ( 3.6 / . 4 ) * . 35 = 2.35 + 9 * . 35 = 5.5 $ answer d" | a ) $ 3.15 , b ) $ 4.45 , c ) $ 4.80 , d ) $ 5.5 , e ) $ 5.40 | d | add(2.35, multiply(0.35, divide(3.6, divide(2, 5)))) | divide(n2,n3)|divide(n4,#0)|multiply(n1,#1)|add(n0,#2)| | general |
a miniature roulette wheel is divided into 9 equal sectors , each bearing a distinct integer from 1 to 9 , inclusive . each time the wheel is spun , a ball randomly determines the winning sector by settling in that sector . if the wheel is spun three times , approximately what is the probability that the product of the... | "the only way to have an odd product is if all 3 integers are odd . p ( odd product ) = 5 / 9 * 5 / 9 * 5 / 9 = 125 / 729 p ( even product ) = 1 - 125 / 729 = 604 / 729 which is about 83 % the answer is c ." | a ) 50 % , b ) 67 % , c ) 83 % , d ) 90 % , e ) 96 % | c | multiply(subtract(1, power(divide(divide(9, const_2), 9), const_2)), const_100) | divide(n0,const_2)|divide(#0,n0)|power(#1,const_2)|subtract(n1,#2)|multiply(#3,const_100)| | general |
local kennel has cats and dogs in the ratio of 2 : 3 . if there are 6 fewer cats than dogs , how many dogs are in the kennel ? | lets work with the data given to us . we know that there ratio of cats to dogs is 2 : 3 or cats 2 dogs 3 we can write number of cats as 2 x and number of dogs as 3 x and we know that 3 x - 2 x = 6 ( therefore x = 6 ) then # of dogs = 3 x 6 = 18 answer is b | a ) 24 , b ) 18 , c ) 30 , d ) 15 , e ) 12 | b | multiply(6, 3) | multiply(n1,n2) | other |
in the coordinate plane a slope of the line k is 3 / 2 times the y - intercept of the line k . what is the x - intercept of the line k ? | as y = 3 / 2 mx + m , from 0 = 3 / 2 mx + m we get x = 2 / 3 . hence , the correct answer choice is b . | a ) - 2 , b ) 2 / 3 , c ) - 2 / 3 , d ) 1 / 4 , e ) 3 / 2 | b | divide(const_1, 3) | divide(const_1,n0)| | general |
10 camels cost as much as 24 horses , 16 horses cost as much as 4 oxen and 6 oxen as much as 4 elephants . if the cost of 10 elephants is rs . 140000 , find the cost of a camel ? | cost of the camel = p 10 camels = 24 horses 16 horses = 4 oxen 6 oxen = 4 elephants 10 elephants = rs . 140000 p = rs . [ ( 24 * 4 * 4 * 140000 ) / ( 10 * 16 * 6 * 10 ) ] p = rs . ( 53760000 / 9600 ) = > p = rs . 5600 answer : e | a ) s . 9800 , b ) s . 3800 , c ) s . 9800 , d ) s . 6800 , e ) s . 5600 | e | divide(multiply(multiply(multiply(24, 4), 4), 140000), multiply(multiply(multiply(10, 16), 6), 10)) | multiply(n1,n3)|multiply(n0,n2)|multiply(n3,#0)|multiply(n4,#1)|multiply(n7,#2)|multiply(n0,#3)|divide(#4,#5) | general |
a watch was sold at a loss of 10 % . if it was sold for rs . 140 more , there would have been a gain of 5 % . what is the cost price ? | "explanation : 90 % 105 % - - - - - - - - 15 % - - - - 140 100 % - - - - ? = > rs . 933.33 answer : a" | a ) s . 933.33 , b ) s . 1009 , c ) s . 1007 , d ) s . 1006 , e ) s . 1002 | a | divide(multiply(140, const_100), subtract(add(const_100, 5), subtract(const_100, 10))) | add(const_100,n2)|multiply(n1,const_100)|subtract(const_100,n0)|subtract(#0,#2)|divide(#1,#3)| | gain |
how many positive integers less than 15 can be expressed as the sum of a positive multiple of 2 and a positive multiple of 3 ? | the number = 2 a + 3 b < 20 when a = 1 , b = 1 , 2 , 3 , 4 , 5 - > 2 a = 2 ; 3 b = 3 , 6 , 9 - > the number = 5 , 8 , 11 - - > 3 numbers when a = 2 , b = 1 , 2,3 - > . . . . - - > 3 numbers when a = 3 , b = 1,2 , 3,4 - - > . . . . - - > 2 numbers total number is already 8 . look at the answer there is no number greater... | a ) 14 , b ) 13 , c ) 12 , d ) 8 , e ) 10 | d | subtract(subtract(subtract(15, const_2), const_4), const_1) | subtract(n0,const_2)|subtract(#0,const_4)|subtract(#1,const_1) | general |
the sum of all consecutive odd integers from β 31 to 41 , inclusive , is | "the sum of the odd numbers from - 31 to + 31 is 0 . let ' s add the remaining numbers . 33 + 35 + 37 + 39 + 41 = 5 ( 37 ) = 185 the answer is e ." | a ) 110 , b ) 135 , c ) 150 , d ) 165 , e ) 185 | e | add(add(add(add(31, const_2), add(add(31, const_2), const_2)), add(add(add(31, const_2), const_2), const_2)), 41) | add(n0,const_2)|add(#0,const_2)|add(#0,#1)|add(#1,const_2)|add(#2,#3)|add(n1,#4)| | physics |
find the compound ratio of ( 2 : 5 ) , ( 6 : 11 ) and ( 11 : 2 ) is | "required ratio = 2 / 5 * 6 / 11 * 11 / 2 = 2 / 1 = 6 : 5 answer is d" | a ) 3 : 2 , b ) 2 : 1 , c ) 1 : 2 , d ) 6 : 5 , e ) 2 : 3 | d | multiply(divide(2, 5), multiply(divide(2, 5), divide(6, 5))) | divide(n0,n1)|divide(n2,n1)|multiply(#0,#1)|multiply(#0,#2)| | other |
if a man lost 4 % by selling oranges at the rate of 24 a rupee at how many a rupee must he sell them to gain 44 % ? | "96 % - - - - 24 144 % - - - - ? 96 / 144 * 24 = 16 answer : a" | a ) 16 , b ) 8 , c ) 9 , d ) 4 , e ) 3 | a | divide(multiply(subtract(const_100, 4), 24), add(const_100, 44)) | add(n2,const_100)|subtract(const_100,n0)|multiply(n1,#1)|divide(#2,#0)| | gain |
the ratio of the number of females to males at a party was 1 : 2 but when 6 females and 6 males left , the ratio became 1 : 3 . how many people were at the party originally ? | "the total number of people are x females + 2 x males . 3 * ( x - 6 ) = 2 x - 6 x = 12 there were 3 x = 36 people at the party originally . the answer is d ." | a ) 24 , b ) 28 , c ) 32 , d ) 36 , e ) 40 | d | add(divide(subtract(multiply(multiply(6, 2), 3), multiply(6, 2)), 2), subtract(multiply(multiply(6, 2), 3), multiply(6, 2))) | multiply(n1,n2)|multiply(n5,#0)|subtract(#1,#0)|divide(#2,n1)|add(#3,#2)| | other |
one robot builds a robot in 1 hour , another one builds a robot in 2 hours . the 2 robots work together and when a new robot is complete , it joins the others , working at a constant rate of one robot in 2 hours . how much time will it take until there are 6 robots altogether , if the robots build only one robot at a t... | "first the 2 robots work at the rate of 1 + 1 / 2 = 3 / 2 so they complete one robot in 2 / 3 rd of an hour = 40 minutes - ( 1 ) now the 3 robots work together at the rate of 1 + 1 / 2 + 1 / 2 = 4 / 2 = 2 / 1 so they complete one robot in 1 / 2 an hour , i . e 30 minutes - ( 2 ) now the 4 robots work together at the ra... | a ) 70 min , b ) 94 min , c ) 109 min , d ) 114 min , e ) 120 min | d | add(inverse(add(add(inverse(1), inverse(2)), inverse(2))), inverse(add(inverse(1), inverse(2)))) | inverse(n0)|inverse(n1)|add(#0,#1)|add(#2,#1)|inverse(#2)|inverse(#3)|add(#5,#4)| | physics |
john had a stock of 1200 books in his bookshop . he sold 75 on monday , 50 on tuesday , 64 on wednesday , 78 on thursday and 135 on friday . what percentage of the books were not sold ? | let n be the total number of books sold . hence n = 75 + 50 + 64 + 78 + 135 = 402 let m be the books not sold m = 1200 - n = 1200 - 402 = 798 percentage books not sold / total number of books = 798 / 1200 = 0.665 = 66.5 % correct answer c | a ) 50.4 % , b ) 23.8 % , c ) 66.5 % , d ) 34.9 % , e ) 43.5 % | c | multiply(divide(subtract(1200, add(add(add(75, 50), add(64, 78)), 135)), 1200), const_100) | add(n1,n2)|add(n3,n4)|add(#0,#1)|add(n5,#2)|subtract(n0,#3)|divide(#4,n0)|multiply(#5,const_100) | gain |
if @ is a binary operation defined as the difference between an integer n and the product of n and 5 , then what is the largest positive integer n such that the outcome of the binary operation of n is less than 16 ? | "@ ( n ) = 5 n - n we need to find the largest positive integer such that 5 n - n < 16 . then 4 n < 16 and n < 4 . the largest possible integer is n = 3 . the answer is c ." | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | floor(divide(16, subtract(5, const_1))) | subtract(n0,const_1)|divide(n1,#0)|floor(#1)| | general |
find the average of all numbers between 2 and 63 which are divisible by 5 | "explanation : average = ( 5 + 10 + 15 + 20 + 25 + 30 + 35 + 40 + 45 + 50 + 55 + 60 ) / 12 = 390 / 12 = 32.5 answer : option d" | a ) 15 , b ) 20 , c ) 25 , d ) 32.5 , e ) 35 | d | divide(add(add(add(multiply(5, const_3), add(5, multiply(5, const_2))), multiply(5, const_4)), multiply(add(const_4, const_1), 5)), 5) | add(const_1,const_4)|multiply(n2,const_2)|multiply(n2,const_3)|multiply(n2,const_4)|add(n2,#1)|multiply(n2,#0)|add(#4,#2)|add(#6,#3)|add(#7,#5)|divide(#8,n2)| | general |
a mixture contains alcohol and water in the ratio 4 : 3 . if 7 litres of water is added to the mixture , the ratio becomes 4 : 5 . find the quantity of alcohol in the given mixture | "let the quantity of alcohol and water be 4 x litres and 3 x litres respectively 4 x / ( 3 x + 7 ) = 4 / 5 20 x = 4 ( 3 x + 7 ) 8 x = 28 x = 3.5 quantity of alcohol = ( 4 x 3.5 ) litres = 14 litres . answer is b ." | a ) 15 litres , b ) 14 litres , c ) 30 litres , d ) 22 litres , e ) 8 litres | b | multiply(7, const_1) | multiply(n2,const_1)| | general |
a sum fetched total simple interest of 4016.25 at the rate of 12 p . c . p . a . in 5 years . what is the sum ? | "let the sums be p . now , 60 % of p = 4016.25 or , p = 6693.75 answer a" | a ) 6693.75 , b ) 8032.5 , c ) 4462.5 , d ) 8900 , e ) none of these | a | divide(multiply(const_100, 4016.25), multiply(12, 5)) | multiply(n0,const_100)|multiply(n1,n2)|divide(#0,#1)| | gain |
a man buys 50 pens at marked price of 46 pens from a whole seller . if he sells these pens giving a discount of 1 % , what is the profit percent ? | "explanation : let marked price be re . 1 each c . p . of 50 pens = rs . 46 s . p . of 50 pens = 99 % of rs . 50 = rs . 49.50 profit % = ( profit / c . p . ) x 100 profit % = ( 3.50 / 46 ) x 100 = 7.6 % answer b" | a ) 5.3 % , b ) 7.6 % , c ) 10.6 % , d ) 12 % , e ) none of these | b | multiply(const_100, divide(subtract(multiply(50, subtract(const_100, 1)), multiply(46, const_100)), multiply(46, const_100))) | multiply(n1,const_100)|subtract(const_100,n2)|multiply(n0,#1)|subtract(#2,#0)|divide(#3,#0)|multiply(#4,const_100)| | gain |
a dog is tied to a tree by a long nylon cord . if the dog runs from the due north side of the tree to the due south side of the tree with the cord extended to its full length at all items , and the dog ran approximately 30 feet , what was the approximate length of the nylon cord w , in feet ? | "because the cord was extended to its full length at all items , the dog ran along a semi - circular path , from north to south . the circumference of a full circle is 2 * pi * r , but since we only care about the length of half the circle , the semi - circle path is pi * r . w = pi * r = 30 . round pi = 3 , then r = 1... | a ) 30 , b ) 25 , c ) 15 , d ) 10 , e ) 5 | d | divide(30, const_3) | divide(n0,const_3)| | general |
what is the least value of x , so that 2 x 5475 is divisible by 9 | "explanation : the sum of the digits of the number is divisible by 9 . then the number is divisible by 9 . 2 + x + 5 + 4 + 7 + 5 = 23 + x least value of x may be ' 4 ' , so that the total 23 + 4 = 27 is divisible by 9 . answer : option c" | a ) 7 , b ) 8 , c ) 4 , d ) 3 , e ) 2 | c | divide(divide(divide(lcm(2, 5475), 5475), const_4), const_4) | lcm(n0,n1)|divide(#0,n1)|divide(#1,const_4)|divide(#2,const_4)| | general |
in the rectangular coordinate system , if the line x = 5 y + 5 passes through points ( m , n ) and ( m + 2 , n + p ) , what is the value of p ? | "x = 5 y + 5 , and thus y = x / 5 - 1 the slope is 1 / 5 . the slope of a line through points ( m , n ) and ( m + 2 , n + p ) is ( n + p - n ) / ( m + 2 - m ) = p / 2 p / 2 = 1 / 5 and thus p = 2 / 5 the answer is d ." | a ) 1 / 2 , b ) 2 / 3 , c ) 1 / 4 , d ) 2 / 5 , e ) 3 / 7 | d | divide(2, 5) | divide(n2,n0)| | general |
in digging a pond 20 m * 15 m * 5 m the volumes of the soil extracted will be ? | "20 * 15 * 5 = 1500 answer : d" | a ) 3387 , b ) 1000 , c ) 2866 , d ) 1500 , e ) 1121 | d | multiply(multiply(20, 15), 5) | multiply(n0,n1)|multiply(n2,#0)| | general |
a train overtakes two persons who are walking in the same direction in which the train is going , at the rate of 2 kmph and 4 kmph and passes them completely in 9 and 10 seconds respectively . the length of the train is : | 2 kmph = ( 2 x 5 / 18 ) m / sec = 5 / 9 m / sec . 4 kmph = ( 4 x 5 / 18 ) m / sec = 10 / 9 m / sec . let the length of the train be x metres and its speed by y m / sec . then , ( x / y - 5 / 9 ) = 9 and ( x / y - 10 / 9 ) = 10 . 9 y - 5 = x and 10 ( 9 y - 10 ) = 9 x = > 9 y - x = 5 and 90 y - 9 x = 100 . on solving , w... | a ) 10 , b ) 100 , c ) 50 , d ) 200 , e ) 300 | c | multiply(multiply(subtract(divide(subtract(multiply(2, divide(9, const_3600)), multiply(4, divide(10, const_3600))), subtract(divide(9, const_3600), divide(10, const_3600))), 2), divide(9, const_3600)), const_1000) | divide(n2,const_3600)|divide(n3,const_3600)|multiply(n0,#0)|multiply(n1,#1)|subtract(#0,#1)|subtract(#2,#3)|divide(#5,#4)|subtract(#6,n0)|multiply(#0,#7)|multiply(#8,const_1000) | physics |
the sum of the present ages of two persons a and b is 45 . if the age of a is twice that of b , find the sum of their ages 5 years hence ? | "a + b = 45 , a = 2 b 2 b + b = 45 = > b = 15 then a = 30 . 5 years , their ages will be 35 and 20 . sum of their ages = 35 + 20 = 55 . answer : a" | a ) 55 , b ) 60 , c ) 70 , d ) 80 , e ) 90 | a | add(add(multiply(divide(45, 5), const_2), 5), add(divide(45, 5), 5)) | divide(n0,n1)|add(#0,n1)|multiply(#0,const_2)|add(#2,n1)|add(#3,#1)| | general |
the length of a rectangle is two third of the radius of a circle . the radius of the circle is equal to the side of the square , whose area is 4761 sq . units . what is the area ( in sq . units ) of the rectangle if the rectangle if the breadth is 13 units ? | given that the area of the square = 4761 sq . units = > side of square = Γ’ Λ Ε‘ 4761 = 69 units the radius of the circle = side of the square = 69 units length of the rectangle = 2 / 3 * 69 = 46 units given that breadth = 13 units area of the rectangle = lb = 46 * 13 = 598 sq . units answer : a | ['a ) 598 sq . units', 'b ) 556 sq . units', 'c ) 589 sq . units', 'd ) 958 sq . units', 'e ) 985 sq . units'] | a | rectangle_area(divide(multiply(sqrt(4761), const_2), const_3), 13) | sqrt(n0)|multiply(#0,const_2)|divide(#1,const_3)|rectangle_area(n1,#2) | geometry |
a car traveling at a certain constant speed takes 2 seconds longer to travel 1 kilometer than it would take to travel 1 kilometer at 90 kilometers per hour . at what speed , in kilometers per hour , is the car traveling ? | "many approaches are possible , one of them : let the distance be 1 kilometer . time to cover this distance at 90 kilometers per hour is 1 / 90 hours = 3,600 / 90 seconds = 40 seconds ; time to cover this distance at regular speed is 40 + 2 = 42 seconds = 42 / 3,600 hours = 7 / 600 hours ; so , we get that to cover 1 k... | a ) 71.5 , b ) 72 , c ) 72.5 , d ) 73 , e ) 85.7 | e | divide(1, divide(add(multiply(const_3600, divide(1, 90)), 2), const_3600)) | divide(n1,n3)|multiply(#0,const_3600)|add(n0,#1)|divide(#2,const_3600)|divide(n1,#3)| | physics |
a searchlight on top of the watchtower makes 2 revolutions per minute . what is the probability that a man appearing near the tower will stay in the dark for at least 5 seconds ? | "the searchlight completes one revolution every 30 seconds . the probability that the man ' s area will be lit up is 5 / 30 = 1 / 6 . the probability that he will stay in the dark is 1 - 1 / 6 = 5 / 6 the answer is a ." | a ) 5 / 6 , b ) 5 / 12 , c ) 7 / 12 , d ) 11 / 12 , e ) 11 / 15 | a | subtract(const_1, divide(5, multiply(5, const_3))) | multiply(n1,const_3)|divide(n1,#0)|subtract(const_1,#1)| | physics |
on a map , 1 inch represents 28 miles . how many v inches would be necessary to represent a distance of 383.6 miles ? | "v inches necessary to represent a distance of 383.6 miles = 383.6 / 28 = 13.7 answer c" | a ) 5.2 , b ) 7.4 , c ) 13.7 , d ) 21.2 , e ) 28.7 | c | divide(383.6, 28) | divide(n2,n1)| | physics |
36 welders work at a constant rate they complete an order in 5 days . if after the first day , 12 welders start to work on the other project , how many more days the remaining welders will need to complete the rest of the order ? | "1 . we need to find out the time taken by 24 workers after day 1 . 2 . total no . of wokers * total time taken = time taken by 1 worker 3 . time taken by 1 worker = 36 * 5 = 180 days 4 . but on day 1 thirty - six workers had already worked finishing 1 / 5 of the job . so 24 workers have to finish only 4 / 5 of the job... | a ) 5 , b ) 2 , c ) 8 , d ) 4 , e ) 6 | e | divide(divide(subtract(const_1, multiply(divide(const_1, multiply(36, 5)), 36)), subtract(36, 12)), divide(const_1, multiply(36, 5))) | multiply(n0,n1)|subtract(n0,n2)|divide(const_1,#0)|multiply(n0,#2)|subtract(const_1,#3)|divide(#4,#1)|divide(#5,#2)| | physics |
there are 23 students in a class . in how many different ways can a committee of 3 students be formed ? | "23 c 3 = 23 * 22 * 21 / 6 = 1771 the answer is c ." | a ) 1254 , b ) 1482 , c ) 1771 , d ) 1875 , e ) 1923 | c | multiply(subtract(const_1, divide(3, 23)), 23) | divide(n1,n0)|subtract(const_1,#0)|multiply(#1,n0)| | probability |
the volume of the sphere qq is ( dfrac { 37 } { 64 } % ) less than thevolume of sphere pp and the volume of sphere rr is ( dfrac { 19 } { 27 } % ) less than that of sphere qq . by what is the surface areaof sphere rr less than the surfacearea of sphere pp ? | let the volume of sphere pp be 64 parts . therefore volume of sphere qq = 64 β 3764 % = 64 β 3764 % of 6464 = 64 β 37 = 27 = 64 β 37 = 27 parts . the volume of rr = 27 β 1927 Γ 27 = 27 β 1927 Γ 27 = 27 β 19 = 8 = 27 β 19 = 8 parts . volume ratio : = p : q : r = 64 : 27 : 8 = p : q : r = 64 : 27 : 8 radius ratio : = p :... | ['a ) 40 %', 'b ) 50 %', 'c ) 75 %', 'd ) 80 %', 'e ) 90 %'] | c | multiply(divide(subtract(add(subtract(add(multiply(power(64, divide(const_1, const_3)), power(27, divide(const_1, const_3))), power(64, divide(const_1, const_3))), power(64, divide(const_1, const_3))), power(64, divide(const_1, const_3))), power(64, divide(const_1, const_3))), add(subtract(add(multiply(power(64, divide... | divide(const_1,const_3)|power(n1,#0)|power(n3,#0)|multiply(#1,#2)|add(#3,#1)|subtract(#4,#1)|add(#1,#5)|subtract(#6,#1)|divide(#7,#6)|multiply(#8,const_100) | geometry |
the length of the bridge , which a train 150 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 , ( 150 + x ) / 30 = 25 / 2 = = > 2 ( 150 + x ) = 750 = = > x = 225 m . answer : option b" | a ) 230 , b ) 225 , c ) 245 , d ) 238 , e ) 236 | b | subtract(multiply(divide(multiply(45, speed(const_1000, const_1)), speed(const_3600, const_1)), 30), 150) | speed(const_1000,const_1)|speed(const_3600,const_1)|multiply(n1,#0)|divide(#2,#1)|multiply(n2,#3)|subtract(#4,n0)| | physics |
speed of a boat in standing water is 7 kmph and speed of the stream is 1.5 kmph . a man can rows to a place at a distance of 105 km and comes back to the starting point . the total time taken by him is ? | "speed upstream = 5.5 kmph speed downstream = 8.5 kmph total time taken = 105 / 5.5 + 105 / 8.5 = 31.44 hours answer is b" | a ) 32.44 hours , b ) 31.44 hours , c ) 30.44 hours , d ) 29.44 hours , e ) 28.44 hours | b | add(multiply(add(add(7, 1.5), subtract(7, 1.5)), 105), multiply(subtract(add(divide(105, add(7, 1.5)), divide(105, subtract(7, 1.5))), add(add(7, 1.5), subtract(7, 1.5))), const_60)) | add(n0,n1)|subtract(n0,n1)|add(#0,#1)|divide(n2,#0)|divide(n2,#1)|add(#3,#4)|multiply(n2,#2)|subtract(#5,#2)|multiply(#7,const_60)|add(#6,#8)| | physics |
excluding stoppages , the speed of a train is 45 kmph and including stoppages it is 37 kmph . of how many minutes does the train stop per hour ? | "explanation : t = 8 / 45 * 60 = 11 answer : option e" | a ) 73 , b ) 83 , c ) 12 , d ) 83 , e ) 11 | e | subtract(const_60, multiply(const_60, divide(37, 45))) | divide(n1,n0)|multiply(#0,const_60)|subtract(const_60,#1)| | physics |
a man traveled a total distance of 1800 km . he traveled one - third of the whole trip by plane and the distance traveled by train is one - half of the distance traveled by bus . if he traveled by train , plane and bus , how many kilometers did he travel by bus ? | "total distance traveled = 1800 km . distance traveled by plane = 600 km . distance traveled by bus = x distance traveled by train = x / 2 x + x / 2 + 600 = 1800 3 x / 2 = 1200 x = 800 km the answer is b ." | a ) 700 , b ) 800 , c ) 900 , d ) 1000 , e ) 1100 | b | divide(multiply(divide(multiply(1800, const_2), const_3), const_3), add(const_2, const_3)) | add(const_2,const_3)|multiply(n0,const_2)|divide(#1,const_3)|multiply(#2,const_3)|divide(#3,#0)| | physics |
the jogging track in a sports complex is 726 m in circumference . suresh and his wife start from the same point and walk in opposite direction at 4.5 km / hr and 3.75 km / hr respectively . they will meet for the first time in : | "let both of them meet after t min 4500 m are covered by suresh in 60 m . in t min he will cover 4500 t / 60 likewise , in t min suresh ' s wife will cover 3750 t / 60 given , ( 4500 t / 60 ) + ( 3750 t / 60 ) = 726 t = 5.28 minutes answer : d" | a ) 5.5 min . , b ) 6 min . , c ) 4.9 min . , d ) 5.28 min . , e ) 5 min . | d | divide(726, divide(add(multiply(4.5, const_1000), multiply(3.75, const_1000)), const_60)) | multiply(n1,const_1000)|multiply(n2,const_1000)|add(#0,#1)|divide(#2,const_60)|divide(n0,#3)| | general |
a and b have monthly incomes in the ratio 5 : 6 and monthly expenditures in the ratio 3 : 4 . if they save rs . 1800 and rs . 1600 respectively , find the monthly income of b | explanation : incomes of a and b = 5 x and 6 x and expenses of a and b = 3 y and 4 y then , savings of a = 5 x - 3 y = 1800 β ? ( 1 ) savings of b = 6 x - 4 y = 1600 β ? ( 2 ) by solving equations ( 1 ) and ( 2 ) y = 1400 monthly income of b = expenses of b + savings of b = 4 y + 1600 = 4 ( 1400 ) + 1600 = rs . 7200 an... | a ) rs . 3400 , b ) rs . 2700 , c ) rs . 1720 , d ) rs . 7200 , e ) rs . 8200 | d | multiply(divide(subtract(divide(multiply(4, 1800), 3), 1600), subtract(divide(multiply(4, 5), 3), 6)), 6) | multiply(n3,n4)|multiply(n0,n3)|divide(#0,n2)|divide(#1,n2)|subtract(#2,n5)|subtract(#3,n1)|divide(#4,#5)|multiply(n1,#6) | other |
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