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
values |
|---|---|---|---|---|---|---|
a shirt goes on sale for 50 % of its original price . one week later , the sale price is marked down 10 % . the final price is what percent of the original price ? | "just assume original price is 100 . sale price = 50 then it is marked down by 10 % = 50 - 5 = 45 . hence it is 45 % od the original price . hence answer is e ." | a ) 72 % , b ) 70 % , c ) 52 % , d ) 50 % , e ) 45 % | e | subtract(50, multiply(divide(50, const_100), 10)) | divide(n0,const_100)|multiply(n1,#0)|subtract(n0,#1)| | gain |
each child has 2 pencils and 13 skittles . if there are 8 children , how many pencils are there in total ? | 2 * 8 = 16 . answer is a . | a ) 16 , b ) 12 , c ) 18 , d ) 22 , e ) 08 | a | multiply(2, 8) | multiply(n0,n2)| | general |
in what time will a railway train 56 m long moving at the rate of 39 kmph pass a telegraph post on its way ? | "t = 56 / 39 * 18 / 5 = 5 sec answer : c" | a ) 3 sec , b ) 4 sec , c ) 5 sec , d ) 6 sec , e ) 7 sec | c | divide(56, multiply(39, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics |
sachin is younger than rahul by 9 years . if the ratio of their ages is 7 : 9 , find the age of sachin | "if rahul age is x , then sachin age is x - 9 , so ( x - 9 ) / x = 7 / 9 = > 9 x - 81 = 7 x = > 2 x = 81 = > x = 40.5 so sachin age is 40.5 - 9 = 31.5 answer : a" | a ) 31.5 , b ) 24.5 , c ) 24.3 , d ) 24.9 , e ) 24.1 | a | multiply(divide(9, subtract(9, 7)), 7) | subtract(n2,n1)|divide(n0,#0)|multiply(n1,#1)| | other |
a , b and c rent a pasture . if a puts 10 oxen for 7 months , b puts 12 oxen for 5 months and c puts 15 oxen for 3 months for grazing and the rent of the pasture is rs . 245 , then how much amount should c pay as his share of rent ? | "a : b : c = 10 × 7 : 12 × 5 : 15 × 3 = 2 × 7 : 12 × 1 : 3 × 3 = 14 : 12 : 9 amount that c should pay = 245 × 9 / 35 = 7 × 9 = 63 answer is d" | a ) 35 , b ) 45 , c ) 25 , d ) 63 , e ) 55 | d | multiply(245, divide(multiply(15, 3), add(add(multiply(10, 7), multiply(12, 5)), multiply(15, 3)))) | multiply(n4,n5)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|add(#3,#0)|divide(#0,#4)|multiply(n6,#5)| | general |
find the probability that a number selected from numbers 1 , 2 , 3 , . . . , 35 is a prime number , when each of the given numbers is equally likely to be selected ? | "let x be the event of selecting a prime number . x = { 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 } n ( x ) = 11 , n ( s ) = 35 hence , the required probability is 11 / 35 . answer : b" | a ) 10 / 35 , b ) 11 / 35 , c ) 8 / 35 , d ) 3 / 35 , e ) 7 / 35 | b | divide(multiply(3, const_5), 35) | multiply(n2,const_5)|divide(#0,n3)| | probability |
if 5 machines can produce 20 units in 10 hours , how long would it take 10 to produce 60 units ? | "5 machines would produce 60 units in 30 hours . increasing the amount of machines by 2 would mean dividing 30 hours by 2 . 30 / 2 = 15 answer : c" | a ) 9 , b ) 10 , c ) 15 , d ) 25 , e ) 30 | c | divide(60, multiply(divide(divide(10, 10), 5), 20)) | divide(n3,n2)|divide(#0,n0)|multiply(n1,#1)|divide(n4,#2)| | physics |
a and b started a business jointly a ' s investment was thrice the investment of b and the period of his investment was two times the period of investment of b . if b received rs . 4000 as profit , then their total profit is ? | suppose b invested rs . x for y months . then , a invested rs . 3 x for 2 y months . so , a : b = ( 3 x * 2 y ) : ( x * y ) = 6 xy : xy = 6 : 1 . b ' s profit : total profit = 1 : 7 . let the total profit be rs . x then , 1 / 7 = 4000 / x or x = 28000 . answer : b | a ) 28029 , b ) 28000 , c ) 28001 , d ) 28012 , e ) 28129 | b | multiply(add(multiply(const_2, const_3), const_1), 4000) | multiply(const_2,const_3)|add(#0,const_1)|multiply(n0,#1) | general |
the average ( arithmetic mean ) of 27 , 32 , and 64 is 6 less than the average of 29 , 42 , and x . what is x ? | "the average of 27 , 32 , and 64 is 41 . the average of 29 , 42 and x is 47 . then 29 + 42 + x = 141 . x = 70 . the answer is e ." | a ) 62 , b ) 64 , c ) 66 , d ) 68 , e ) 70 | e | subtract(add(add(27, 32), 64), add(add(multiply(6, const_3), 29), 42)) | add(n0,n1)|multiply(n3,const_3)|add(n2,#0)|add(n4,#1)|add(n5,#3)|subtract(#2,#4)| | general |
in a class of 40 students , 10 did not opt for math . 15 did not opt for science and 2 did not opt for either . how many students of the class opted for both math and science ? | 40 students total 10 did not opt for math 15 did not opt for science 2 did not opt for either total of 30 students in math and 13 did not opt for sci but did for math 30 - 13 = 7 7 students of the class opted for both math and science answer : c . 7 | a ) 13 , b ) 15 , c ) 7 , d ) 17 , e ) 18 | c | subtract(subtract(40, 10), subtract(subtract(40, 15), 2)) | subtract(n0,n1)|subtract(n0,n2)|subtract(#1,n3)|subtract(#0,#2) | other |
a car traveled 462 miles per tankful of gasoline on the highway and 336 miles per tankful of gasoline in the city . if the car traveled 15 fewer miles per gallon in the city than on the highway , how many miles per gallon did the car travel in the city ? | "i treat such problems as work ones . work = rate * time mileage ( m ) = rate ( mpg ) * gallons ( g ) x gallons is a full tank { 462 = rx { 336 = ( r - 15 ) x solve for r , r = 55 55 - 15 = 40 mpg e" | a ) 14 , b ) 16 , c ) 21 , d ) 22 , e ) 40 | e | divide(336, divide(subtract(462, 336), 15)) | subtract(n0,n1)|divide(#0,n2)|divide(n1,#1)| | physics |
sahil purchased a machine at rs 14000 , then got it repaired at rs 5000 , then gave its transportation charges rs 1000 . then he sold it with 50 % of profit . at what price he actually sold it . | "explanation : question seems a bit tricky , but it is very simple . just calculate all cost price , then get 150 % of cp . c . p . = 14000 + 5000 + 1000 = 20000 150 % of 20000 = 150 / 100 * 20000 = 30000 option d" | a ) rs . 22000 , b ) rs . 24000 , c ) rs . 26000 , d ) rs . 30000 , e ) none of these | d | add(add(add(14000, 5000), 1000), multiply(divide(add(add(14000, 5000), 1000), const_100), 50)) | add(n0,n1)|add(n2,#0)|divide(#1,const_100)|multiply(n3,#2)|add(#1,#3)| | gain |
the salary of a person was reduced by 50 % . by what percent should his reduced salary be raised so as to bring it at par with his original salary ? | "let the original salary be $ 100 new salary = $ 50 increase on 50 = 50 increase on 100 = 50 / 50 * 100 = 100 % ( approximately ) answer is e" | a ) 20 % , b ) 25 % , c ) 50 % , d ) 75 % , e ) 100 % | e | multiply(divide(multiply(const_100, divide(50, const_100)), subtract(const_100, multiply(const_100, divide(50, const_100)))), const_100) | divide(n0,const_100)|multiply(#0,const_100)|subtract(const_100,#1)|divide(#1,#2)|multiply(#3,const_100)| | gain |
the average of first seven multiples of 3 is : | "explanation : ( 3 ( 1 + 2 + 3 + 4 + 5 + 6 + 7 ) / 7 = 3 x 28 / 7 = 12 answer : c" | a ) 9 , b ) 16 , c ) 12 , d ) 8 , e ) 10 | c | add(3, const_1) | add(n0,const_1)| | general |
there are 25 stations between hyderabad and bangalore . how many second class tickets have to be printed , so that a passenger can travel from any station to any other station ? | "the total number of stations = 27 from 27 stations we have to choose any two stations and the direction of travel ( i . e . , hyderabad to bangalore is different from bangalore to hyderabad ) in ² ⁰ p ₂ ways . ² ⁰ p ₂ = 27 * 26 = 702 . answer : e" | a ) 156 , b ) 167 , c ) 157 , d ) 352 , e ) 702 | e | multiply(add(25, const_1), add(add(25, const_1), const_1)) | add(n0,const_1)|add(#0,const_1)|multiply(#0,#1)| | physics |
look carefully for the pattern , and then choose which pair of numbers comes next . 28 25 5 21 18 5 14 | explanation : this is an alternating subtraction series with the interpolation of a random number , 5 , as every third number . in the subtraction series , 3 is subtracted , then 4 , then 3 , and so on . answer : option a | a ) 11 5 , b ) 10 7 , c ) 11 8 , d ) 5 10 , e ) 10 5 | a | add(multiply(subtract(14, const_3), const_10), 5) | subtract(n6,const_3)|multiply(#0,const_10)|add(n2,#1) | general |
if in a cricket one day match , there were no wide balls , no wides , no extras and no overthrows . what is the maximum number of runs that a batsman can score in 35 overs in an ideal case scenario ? | solution : 1158 explanation : for an ideal case , the batsman will hit a six on each ball . but if he hits six on the last ball of the over , the strike will change in the next over . thus , the best he can do in the last ball is run 3 runs so that he retains the strike even in the next over . thus the total runs that ... | a ) 1158 , b ) 1651 , c ) 1653 , d ) 1654 , e ) none | a | multiply(35, subtract(35, const_2)) | subtract(n0,const_2)|multiply(n0,#0) | physics |
there is a total of 90 marbles in a box , each of which is red , green , blue , or white . if one marble is drawn from the box at random , the probability that it will be white is 1 / 3 and the probability that it will be green is 1 / 5 . what is the probability that the marble will be either red or blue ? | "p ( red or blue ) = 1 - p ( white ) - p ( green ) = 15 / 15 - 5 / 15 - 3 / 15 = 7 / 15 the answer is c ." | a ) 1 / 3 , b ) 3 / 5 , c ) 7 / 15 , d ) 11 / 30 , e ) 17 / 30 | c | subtract(1, add(divide(1, 3), divide(1, 5))) | divide(n1,n2)|divide(n3,n4)|add(#0,#1)|subtract(n1,#2)| | general |
10 , 15 , 22.5 , 33.75 , 50.62 , ( . . . ) | "10 ( 10 ã — 3 ) ã · 2 = 15 ( 15 ã — 3 ) ã · 2 = 22.5 ( 22.5 ã — 3 ) ã · 2 = 33.75 ( 33.75 ã — 3 ) ã · 2 = 50.62 ( 50.62 ã — 3 ) ã · 2 = 75.93 answer is c" | a ) 60 , b ) 60.75 , c ) 75.93 , d ) 76.33 , e ) 70.1 | c | subtract(negate(33.75), multiply(subtract(15, 22.5), divide(subtract(15, 22.5), subtract(10, 15)))) | negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)| | general |
a coin is tossed twice if the coin shows head it is tossed again but if it shows a tail then a die is tossed . if 8 possible outcomes are equally likely . find the probability that the die shows a number greater than 4 , if it is known that the first throw of the coin results in a tail | sample space s = { hh , ht , t 1 , t 2 , t 3 , t 4 , t 5 , t 6 } let a be the event that the die shows a number greater than 4 and b be the event that the first throw of the coin results in a tail then , a = { t 5 , t 6 } b = { t 1 , t 2 , t 3 , t 4 , t 5 , t 6 } required probability = ( 2 / 8 ) / ( 6 / 8 ) = 1 / 3 ans... | a ) 2 / 4 , b ) 1 / 3 , c ) 2 / 3 , d ) 3 / 6 , e ) 1 / 4 | b | divide(divide(const_2, 8), divide(const_6, 8)) | divide(const_2,n0)|divide(const_6,n0)|divide(#0,#1) | probability |
john purchased a grinder and a mobile for rs . 15000 & rs . 10000 respectively . he sold the grinder at a loss of 4 % and the mobile phone at a profit of 10 % . overall how much he make a profit . | "let the sp of the refrigerator and the mobile phone be rs . r and rs . m respectively . r = 15000 ( 1 - 4 / 100 ) = 15000 - 600 m = 10000 ( 1 + 10 / 100 ) = 10000 + 1000 total sp - total cp = r + m - ( 15000 + 10000 ) = - 600 + 1000 = rs . 400 as this is positive , an overall profit of rs . 400 was made . a" | a ) s . 400 , b ) s . 120 , c ) s . 420 , d ) s . 450 , e ) s . 290 | a | subtract(multiply(10, divide(10000, const_100)), multiply(4, divide(15000, const_100))) | divide(n1,const_100)|divide(n0,const_100)|multiply(n3,#0)|multiply(n2,#1)|subtract(#2,#3)| | gain |
a boat crossed a lake from north to east at the speed of 10 km / h , entered a river and covered twice as much distance going upstream at 9 km / h . it then turned around and stopped at the south shore of the lake . if it averaged 3.8 km / h that day , what was its approximate downstream speed ? | one way of solving this is : speed of boat on still water ( lake ) = 10 kmph speed upstream = 9 kmph = speed in still water - speed of river = > speed of river = 1 kmph = > speed downstream = speed in still water + speed of river = 10 + 1 = 11 kmph ans is e | a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 11 | e | add(10, subtract(10, 9)) | subtract(n0,n1)|add(n0,#0) | general |
a train covers a distance of 20 km in 20 min . if it takes 9 sec to pass a telegraph post , then the length of the train is ? | "speed = ( 20 / 20 * 60 ) km / hr = ( 60 * 5 / 18 ) m / sec = 50 / 3 m / sec . length of the train = 50 / 3 * 9 = 150 m . answer : a" | a ) 150 m , b ) 200 m , c ) 120 m , d ) 225 m , e ) 160 m | a | divide(20, subtract(divide(20, 20), 9)) | divide(n0,n1)|subtract(#0,n2)|divide(n0,#1)| | physics |
a 180 m long train is running at 72 kmph . if it crossed the platform in 20 sec . then find the platform length ? | length = speed * time length = 72 km / hr * time length = [ 72 * ( 5 / 18 ) ] * 20 { ( ie ) ( x ) km / hr ) } = x * ( 5 / 18 ) = 20 * 20 length = 400 m length of platform = length - length of the train length of platform = 400 - 180 length of platform = 220 m option a | a ) 220 m , b ) 110 m , c ) 230 m , d ) 100 m , e ) 90 m | a | subtract(multiply(20, multiply(72, const_0_2778)), 180) | multiply(n1,const_0_2778)|multiply(n2,#0)|subtract(#1,n0) | physics |
two cars start at the same time from opposite ends of a highway that is 333 miles long . one car is riding at 54 mph and the second car is riding at 57 mph . how long after they begin will they meet ? | "as cars are moving in opposite directions their speeds will be added . so their relative speeds : 57 + 54 = 111 mph total distance to be covered = 333 miles . time taken would be : 333 miles / 111 mph = 3 hours c is the answer ." | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | divide(333, add(54, 57)) | add(n1,n2)|divide(n0,#0)| | physics |
it takes joey the postman 1 hours to run a 1 mile long route every day . he delivers packages and then returns to the post office along the same path . if the average speed of the round trip is 5 mile / hour , what is the speed with which joey returns ? | "let his speed for one half of the journey be 3 miles an hour let the other half be x miles an hour now , avg speed = 5 mile an hour 2 * 1 * x / 1 + x = 5 2 x = 5 x + 5 = > 3 x = 5 = 5 / 3 = 1.67 a" | a ) 1.67 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | a | divide(1, subtract(divide(multiply(const_2, 1), 5), 1)) | multiply(n1,const_2)|divide(#0,n2)|subtract(#1,n0)|divide(n1,#2)| | physics |
the average weight of a class is x pounds . when a new student weighing 70 pounds joins the class , the average decreases by 1 pound . in a few months the student ’ s weight increases to 110 pounds and the average weight of the class becomes x + 4 pounds . none of the other students ’ weights changed . what is the valu... | "when the student weighs 80 pounds the average weight is x - 1 pounds ; when the student weighs 110 pounds the average weight is x + 4 pounds . so , the increase in total weight of 110 - 70 = 40 pounds corresponds to the increase in average weight of ( x + 4 ) - ( x - 1 ) = 5 pounds , which means that there are 40 / 5 ... | a ) 26 , b ) 86 , c ) 88 , d ) 90 , e ) 92 | a | add(add(70, 4), add(4, 1)) | add(n0,n3)|add(n1,n3)|add(#0,#1)| | general |
the market value of a 10.5 % stock , in which an income of rs . 756 is derived by investing rs . 6500 , brokerage being 1 / 4 % , is : | "face value = rs . 6500 . dividend = 10.5 % . annual income = rs . 756 . brokerage per rs . 100 = rs . 0.25 . dividend is always paid on the face value of a share . face value * dividend / ( market value + brokerage per rs . 100 ) = annual income . = 6500 * 10.5 / 756 = market value of rs . 100 stock + brokerage per rs... | a ) 90.03 , b ) 114 , c ) 114.75 , d ) 124 , e ) 124.75 | a | subtract(multiply(divide(6500, 756), 10.5), divide(1, 4)) | divide(n2,n1)|divide(n3,n4)|multiply(n0,#0)|subtract(#2,#1)| | gain |
a factory produces 5500 toys per week . if the workers at this factory work 5 days a week and if these workers make the same number of toys everyday , how many toys are produced each day ? | "to find the number of toys produced every day , we divide the total number of toys produced in one week ( of 5 days ) by 5 . 5500 / 5 = 1100 toys correct answer a" | a ) 1100 toys , b ) 5487 toys , c ) 6113 toys , d ) 2354 toys , e ) 1375 toys | a | divide(5500, 5) | divide(n0,n1)| | physics |
for any positive integer n , the sum of the first n positive integers equals n ( n + 1 ) / 2 . what is the sum of all the even integers between 1 and 100 ? | "average of the set : ( largest + smallest ) / 2 = ( 100 + 2 ) / 2 = 51 ; # of terms : ( largest - smallest ) / 2 + 1 = ( 100 - 2 ) / 2 + 1 = 50 the sum = 51 * 50 = 2550 answer : e" | a ) 1250 , b ) 1530 , c ) 2120 , d ) 2540 , e ) 2550 | e | add(divide(subtract(subtract(100, 1), add(1, 1)), 2), 1) | add(n2,n0)|subtract(n3,n0)|subtract(#1,#0)|divide(#2,n1)|add(n0,#3)| | general |
the average weight of a , b , c is 45 kg . the avg wgt of a & b be 40 kg & that of b , c be 43 kg . find the wgt of b . | "sol . let a , b , c represent their individual wgts . then , a + b + c = ( 45 * 3 ) kg = 135 kg a + b = ( 40 * 2 ) kg = 80 kg & b + c = ( 43 * 2 ) kg = 86 kg b = ( a + b ) + ( b + c ) - ( a + b + c ) = ( 80 + 86 - 135 ) kg = 31 kg . ans : b" | a ) 36 kg , b ) 84 kg , c ) 31 kg , d ) 45 kg , e ) 12 kg | b | subtract(add(multiply(40, const_2), multiply(43, const_2)), multiply(45, const_3)) | multiply(n1,const_2)|multiply(n2,const_2)|multiply(n0,const_3)|add(#0,#1)|subtract(#3,#2)| | general |
what is the area of a circle with the circumference of 88 cms ? | "circum = 2 * pi * r 88 = 2 * pi * r r = 44 / pi area of circle = pi * r 2 a = pi * ( 44 / pi ) 2 a = 616 answer : e" | a ) 618 sq cms , b ) 516 sq cms , c ) 600 sq cms , d ) 615 sq cms , e ) none of these | e | multiply(multiply(multiply(divide(divide(multiply(sqrt(88), const_4), const_pi), const_2), divide(divide(multiply(sqrt(88), const_4), const_pi), const_2)), const_pi), const_2) | sqrt(n0)|multiply(#0,const_4)|divide(#1,const_pi)|divide(#2,const_2)|multiply(#3,#3)|multiply(#4,const_pi)|multiply(#5,const_2)| | geometry |
how many different positive integers exist between 10 ^ 2 and 10 ^ 3 , the sum of whose digits is equal to 2 ? | "101 110 200 total no . is 3 a" | a ) 3 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | a | add(add(const_4, const_3), const_1) | add(const_3,const_4)|add(#0,const_1)| | general |
how many litres of pure acid are there in 8 litres of a 20 % solution | "explanation : question of this type looks a bit typical , but it is too simple , as below . . . it will be 8 * 20 / 100 = 1.6 answer : option b" | a ) 1.5 , b ) 1.6 , c ) 1.7 , d ) 1.8 , e ) 1.9 | b | multiply(divide(20, const_100), 8) | divide(n1,const_100)|multiply(n0,#0)| | gain |
evaluate : 1024 x 237 + 976 x 237 | "1024 x 237 + 976 x 237 = 237 x ( 1024 + 976 ) = 237 x 2000 = 474000 . answer is d ." | a ) 486000 , b ) 568000 , c ) 378000 , d ) 474000 , e ) none of them | d | subtract(1024, multiply(multiply(237, 976), 237)) | multiply(n1,n2)|multiply(n3,#0)|subtract(n0,#1)| | general |
gary ’ s gas station serves an average of 16 cars per hour on saturdays , 10 cars per hour on sundays , and 9 cars per hour on all other days of the week . if the station is open from 6 a . m . to 10 p . m . every day , how many cars does gary ’ s station serve over the course of a typical week ? | "6 a . m . to 10 p . m . = 16 hours number of cars serviced on weekdays = ( 16 * 9 * 5 ) number of cars serviced on saturday = ( 16 * 16 ) number of cars serviced on sunday = ( 16 * 10 ) number of cars served in a week = 16 ( 45 + 16 + 10 ) = 16 * 71 = 1136 answer : a" | a ) 1,136 , b ) 1,200 , c ) 1,240 , d ) 1,280 , e ) 1,320 | a | floor(divide(multiply(add(6, 10), add(add(16, 10), multiply(9, add(const_4, const_1)))), const_1000)) | add(n1,n3)|add(n0,n1)|add(const_1,const_4)|multiply(n2,#2)|add(#1,#3)|multiply(#0,#4)|divide(#5,const_1000)|floor(#6)| | physics |
what is the remainder when 81 + 82 + 83 … … + 815 is divided by 6 | "thus odd powers will have remainder 2 ; and even powers will have remainder 4 now , 81 + 82 + 83 … … + 81581 + 82 + 83 … … + 815 will have the following powers - odd = 1 , 3 , 5 , 7 , 9 , 11 , 13 , 15 ( 8 odd powers ) ; sum of remainder = 16 even powers = 2 , 4 , 6 , 8 , 10 , 12 , 14 ( 7 even powers ) ; sum of remaind... | a ) 0 , b ) 1 , c ) 2 , d ) 4 , e ) 5 | c | reminder(add(power(81, reminder(82, 83)), 83), 815) | reminder(n1,n2)|power(n0,#0)|add(n2,#1)|reminder(#2,n3)| | general |
the avg weight of a , b & c is 50 kg . if d joins the group , the avg weight of the group becomes 53 kg . if another man e who weights is 3 kg more than d replaces a , then the avgof b , c , d & e becomes 51 kg . what is the weight of a ? | "a + b + c = 3 * 50 = 150 a + b + c + d = 4 * 53 = 212 - - - - ( i ) so , d = 62 & e = 62 + 3 = 65 b + c + d + e = 51 * 4 = 204 - - - ( ii ) from eq . ( i ) & ( ii ) a - e = 212 – 204 = 8 a = e + 8 = 65 + 8 = 73 answer : c" | a ) 56 , b ) 65 , c ) 73 , d ) 89 , e ) 90 | c | subtract(multiply(53, const_4), subtract(multiply(51, const_4), add(3, subtract(multiply(53, const_4), multiply(50, 3))))) | multiply(n1,const_4)|multiply(n3,const_4)|multiply(n0,n2)|subtract(#0,#2)|add(n2,#3)|subtract(#1,#4)|subtract(#0,#5)| | general |
in an election between two candidates , the winner has a margin of 20 % of the votes polled . if 3000 people change their mind and vote for the loser , the loser would have won by a margin of 20 % of the votes polled . find the total number of votes polled in the election ? | "winner - looser 60 % - 40 % if 3000 people change their mind and vote for the loser : winner - looser 40 % - 60 % thus 3,000 people compose 20 % of all voters , which means that the total number of votes is 15,000 . answer : a" | a ) 15000 , b ) 20000 , c ) 30000 , d ) 60000 , e ) 45000 | a | divide(3000, divide(20, const_100)) | divide(n0,const_100)|divide(n1,#0)| | gain |
p , q and r have $ 5000 among themselves . r has two - thirds of the total amount with p and q . find the amount with r ? | "e 2000 let the amount with r be $ r r = 2 / 3 ( total amount with p and q ) r = 2 / 3 ( 5000 - r ) = > 3 r = 10000 - 2 r = > 5 r = 10000 = > r = 2000 ." | a ) 2400 , b ) 2403 , c ) 3998 , d ) 2539 , e ) 2000 | e | divide(multiply(5000, multiply(const_2, const_2)), add(add(multiply(divide(multiply(const_2, const_2), const_3), const_3), multiply(const_1, const_3)), multiply(const_1, const_3))) | multiply(const_2,const_2)|multiply(const_1,const_3)|divide(#0,const_3)|multiply(n0,#0)|multiply(#2,const_3)|add(#4,#1)|add(#5,#1)|divide(#3,#6)| | general |
a , b , c can do a piece of work in 20 days , 30 days and 40 days respectively , working alone . how soon can the work be done if a is assisted by b and c on alternate days ? | "a + b 1 day work = 1 / 20 + 1 / 30 = 1 / 12 a + c 1 day work = 1 / 20 + 1 / 40 = 3 / 40 work done in 2 days = 1 / 12 + 3 / 40 = 19 / 120 19 / 120 work is done by a in 2 days whole work will be done in 2 * 19 / 120 = 3 days approximately answer is a" | a ) 3 days , b ) 1 day , c ) 5 days , d ) 10 days , e ) 7 days | a | divide(const_2, add(add(divide(const_1, 20), divide(const_1, 30)), add(divide(const_1, 20), divide(const_1, 40)))) | divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|add(#0,#2)|add(#3,#4)|divide(const_2,#5)| | physics |
a $ 500 investment and a $ 1,500 investment have a combined yearly return of 16 percent of the total of the two investments . if the $ 500 investment has a yearly return of 7 percent , what percent yearly return does the $ 1,500 investment have ? | "the equation we can form the question : return on total investment = sum of individual investments ( 500 + 1500 ) ( 16 ) = ( 500 â ˆ — 7 ) + ( 1500 x ) , where x is the return on investment of 1500 . solving the equation , we get x = 19 % ( option e ) answer : e" | a ) 9 % , b ) 10 % , c ) 105 / 8 % , d ) 11 % , e ) 19 % | e | multiply(divide(subtract(multiply(add(500, multiply(const_3, 500)), divide(16, const_100)), multiply(500, divide(7, const_100))), multiply(const_3, 500)), const_100) | divide(n2,const_100)|divide(n4,const_100)|multiply(n0,const_3)|add(n0,#2)|multiply(n0,#1)|multiply(#3,#0)|subtract(#5,#4)|divide(#6,#2)|multiply(#7,const_100)| | gain |
an investment yields an interest payment of $ 216 each month . if the simple annual interest rate is 9 % , what is the amount of the investment ? | "let the principal amount = p simple annual interest = 9 % simple monthly interest = ( 9 / 12 ) = ( 3 / 4 ) % ( 3 / 4 ) * ( p / 100 ) = 216 = > p = ( 216 * 4 * 10 ^ 2 ) / 3 = 72 * 4 * 10 ^ 2 = 288 * 10 ^ 2 = 28800 answer c" | a ) $ 28,300 , b ) $ 30,400 , c ) $ 28,800 , d ) $ 32,500 , e ) $ 35,100 | c | multiply(divide(216, divide(9, multiply(const_3, const_4))), const_100) | multiply(const_3,const_4)|divide(n1,#0)|divide(n0,#1)|multiply(#2,const_100)| | gain |
you have a 6 - sided cube and 6 cans of paint , each a different color . you may not mix colors of paint . how many distinct ways can you paint the cube using a different color for each side ? ( if you can reorient a cube to look like another cube , then the two cubes are not distinct . ) | paint one of the faces red and make it the top face . 5 options for the bottom face . now , four side faces can be painted in ( 4 - 1 ) ! = 3 ! = 6 ways ( circular arrangements of 4 colors ) . total = 5 * 6 = 30 . answer : b . | a ) 24 , b ) 30 , c ) 48 , d ) 60 , e ) 120 | b | multiply(factorial(divide(6, const_2)), subtract(6, const_1)) | divide(n0,const_2)|subtract(n0,const_1)|factorial(#0)|multiply(#2,#1) | general |
sides of a rectangular park are in the ratio 3 : 2 and its area is 1536 sq m , the cost of fencing it at 50 ps per meter is ? | "3 x * 2 x = 1536 = > x = 16 2 ( 48 + 32 ) = 102 m 102 * 1 / 2 = rs . 51 answer : a" | a ) 51 , b ) 1287 , c ) 125 , d ) 988 , e ) 271 | a | divide(multiply(50, rectangle_perimeter(sqrt(divide(multiply(1536, 2), 3)), divide(1536, sqrt(divide(multiply(1536, 2), 3))))), const_100) | multiply(n1,n2)|divide(#0,n0)|sqrt(#1)|divide(n2,#2)|rectangle_perimeter(#3,#2)|multiply(n3,#4)|divide(#5,const_100)| | physics |
one half of a two digit number exceeds its one third by 8 . what is the sum of the digits of the number ? | "x / 2 – x / 3 = 8 = > x = 48 4 + 8 = 12 answer : c" | a ) a ) 3 , b ) b ) 5 , c ) c ) 12 , d ) d ) 9 , e ) e ) 11 | c | add(reminder(multiply(8, const_4), const_10), const_1) | multiply(n0,const_4)|reminder(#0,const_10)|add(#1,const_1)| | general |
one night 18 percent of the female officers on a police force were on duty . if 144 police officers were on duty that night and half of these were female officers , how many female officers were on the police force ? | "let x be the number of female police officers on the police force . the number of female police officers on duty was 72 . 0.18 x = 72 x = 400 the answer is b ." | a ) 300 , b ) 400 , c ) 500 , d ) 600 , e ) 700 | b | divide(divide(144, const_2), divide(18, const_100)) | divide(n1,const_2)|divide(n0,const_100)|divide(#0,#1)| | gain |
a shirt goes on sale for 70 % of its original price . one week later , the sale price is marked down 10 % . the final price is what percent of the original price ? | just assume original price is 100 . sale price = 70 then it is marked down by 10 % = 70 - 7 = 63 . hence it is 63 % od the original price . hence answer is a . | a ) 63 % , b ) 70 % , c ) 62 % , d ) 50 % , e ) 28 % | a | subtract(70, multiply(divide(70, const_100), 10)) | divide(n0,const_100)|multiply(n1,#0)|subtract(n0,#1) | gain |
a 250 m long train running at the speed of 120 km / hr crosses another train running in opposite direction at the speed of 80 km / hr in 9 sec . what is the length of the other train ? | "relative speed = 120 + 80 = 200 km / hr . = 200 * 5 / 18 = 500 / 9 m / sec . let the length of the other train be x m . then , ( x + 250 ) / 9 = 500 / 9 = > x = 250 . answer : option a" | a ) 250 , b ) 245 , c ) 240 , d ) 235 , e ) 230 | a | subtract(multiply(multiply(add(120, 80), const_0_2778), 9), 250) | add(n1,n2)|multiply(#0,const_0_2778)|multiply(n3,#1)|subtract(#2,n0)| | physics |
3 candidates in an election and received 2500 , 5000 and 15000 votes respectively . what % of the total votes did the winningcandidate got in that election ? | "total number of votes polled = ( 2500 + 5000 + 15000 ) = 22500 so , required percentage = 11628 / 20400 * 100 = 66.6 % d" | a ) 45 % , b ) 50 % , c ) 57 % , d ) 66.6 % , e ) 65.6 % | d | multiply(divide(15000, add(add(2500, 5000), 15000)), const_100) | add(n1,n2)|add(n3,#0)|divide(n3,#1)|multiply(#2,const_100)| | gain |
an auction house charges a commission of 15 % on the first $ 50000 of the sale price of an item , plus 10 % on the amount of of the sale price in excess of $ 50000 . what was the price of a painting for which the house charged a total commission of $ 24000 ? | say the price of the house was $ x , then 0.15 * 50,000 + 0.1 * ( x - 50,000 ) = 24,000 - - > x = $ 215,000 ( 15 % of $ 50,000 plus 10 % of the amount in excess of $ 50,000 , which is x - 50,000 , should equal to total commission of $ 24,000 ) . answer : c | a ) $ 115,000 , b ) $ 160,000 , c ) $ 215,000 , d ) $ 240,000 , e ) $ 365,000 | c | divide(subtract(add(divide(subtract(24000, multiply(50000, divide(15, const_100))), divide(const_10, const_100)), 50000), 50000), const_1000) | divide(n0,const_100)|divide(const_10,const_100)|multiply(n1,#0)|subtract(n4,#2)|divide(#3,#1)|add(n1,#4)|subtract(#5,n1)|divide(#6,const_1000) | general |
laura took out a charge account at the general store and agreed to pay 4 % simple annual interest . if she charges $ 35 on her account in january , how much will she owe a year later , assuming she does not make any additional charges or payments ? | "principal that is amount taken by laura at year beginning = 35 $ rate of interest = 4 % interest = ( 4 / 100 ) * 35 = 1.4 $ total amount that laura owes a year later = 35 + 1.4 = 36.4 $ answer a" | a ) $ 36.4 , b ) $ 37.10 , c ) $ 37.16 , d ) $ 38.10 , e ) $ 38.80 | a | add(multiply(divide(4, const_100), 35), 35) | divide(n0,const_100)|multiply(n1,#0)|add(n1,#1)| | general |
a boat running up stram takes 6 hours to cover a certain distance , while it takes 10 hours to cover the same distance running down stream . what is the ratio between the speed of the boat and the speed of water current respectively ? | explanation : let speed of boat is x km / h and speed stream is y km / hr 6 ( x + y ) = 10 ( x - y ) 6 x + 6 y = 10 x - 10 y 16 y = 4 x 4 y = x x / y = 4 / 1 4 : 1 answer : option c | a ) 2 : 3 , b ) 5 : 6 , c ) 4 : 1 , d ) 7 : 1 , e ) 8 : 1 | c | subtract(10, 6) | subtract(n1,n0) | physics |
how many of the integers between 10 and 40 are even ? | "number start between 10 to 40 is 30 numbers half of them is even . . which is 15 answer : c" | a ) 21 , b ) 20 , c ) 15 , d ) 10 , e ) 9 | c | divide(subtract(40, 10), const_2) | subtract(n1,n0)|divide(#0,const_2)| | general |
for a race a distance of 224 meters can be covered by p in 28 seconds and q in 32 seconds . by what distance does p defeat q eventually ? | "explanation : this is a simple speed time problem . given conditions : = > speed of p = 224 / 28 = 8 m / s = > speed of q = 224 / 32 = 7 m / s = > difference in time taken = 4 seconds therefore , distance covered by p in that time = 8 m / s x 4 seconds = 32 metres answer : b" | a ) 26 m , b ) 32 m , c ) 24 m , d ) 28 m , e ) none of these | b | subtract(224, multiply(28, speed(224, 32))) | speed(n0,n2)|multiply(n1,#0)|subtract(n0,#1)| | physics |
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 12 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 . b = 60 - 12 = 48 . answer : c" | a ) 18 , b ) 20 , c ) 48 , d ) 36 , e ) 42 | c | multiply(const_3, divide(60, const_10)) | divide(n2,const_10)|multiply(#0,const_3)| | gain |
in a class of 35 students , 2 students did not borrow any books from the library , 12 students each borrowed 1 book , 10 students each borrowed 2 books , and the rest borrowed at least 3 books . if the average number of books per student was 2 , what is the maximum number of books any single student could have borrowed... | "total # of students = 35 avg # of books per student = 2 total # of books = 35 * 2 = 70 # of student borrowed at least 3 books = 35 - 2 - 12 - 10 = 11 # of books borrowed by above 11 students = 70 - ( 12 * 1 ) + ( 10 * 2 ) = 38 considering that 10 out of above 11 students borrowed only 3 books , # of books borrowed = 1... | a ) 3 , b ) 5 , c ) 8 , d ) 13 , e ) 15 | c | subtract(multiply(35, 2), add(multiply(subtract(subtract(35, add(add(multiply(12, 1), 10), 2)), 1), 3), add(multiply(12, 1), multiply(10, 2)))) | multiply(n0,n1)|multiply(n2,n3)|multiply(n1,n4)|add(#1,#2)|add(n4,#1)|add(n1,#4)|subtract(n0,#5)|subtract(#6,n3)|multiply(n6,#7)|add(#3,#8)|subtract(#0,#9)| | general |
find avrg speed if a man travels at speed of 24 km / hr up and 36 km / hr dawn at an altitude of 200 m . | avg speed = 2 * x * y / ( x + y ) = 2 * 24 * 36 / ( 24 + 36 ) = 28.8 answer : d | a ) 25.8 , b ) 26.8 , c ) 27.8 , d ) 28.8 , e ) 29.8 | d | divide(multiply(multiply(const_2, 24), 36), add(24, 36)) | add(n0,n1)|multiply(n0,const_2)|multiply(n1,#1)|divide(#2,#0) | physics |
a watch was sold at a loss of 36 % . if it was sold for rs . 140 more , there would have been a gain of 4 % . what is the cost price ? | "64 % 104 % - - - - - - - - 40 % - - - - 140 100 % - - - - ? = > rs . 350 answer : a" | a ) 350 , b ) 288 , c ) 799 , d ) 778 , e ) 901 | a | divide(multiply(140, const_100), subtract(add(const_100, 4), subtract(const_100, 36))) | add(const_100,n2)|multiply(n1,const_100)|subtract(const_100,n0)|subtract(#0,#2)|divide(#1,#3)| | gain |
what is the probability that jamie and tom will get selected for a job they both applied for in a company , given that their chances of getting selected is 2 / 3 and 5 / 7 respectively ? | explanation : p ( jamie ) = 2 / 3 p ( tom ) = 5 / 7 e = { jamie and tom both get selected } p ( e ) = p ( jamie ) * p ( tom ) = 2 / 3 * 5 / 7 = 10 / 21 answer : e | a ) 13 / 25 , b ) 2 / 5 , c ) 9 / 20 , d ) 11 / 20 , e ) 10 / 21 | e | multiply(divide(2, 3), divide(5, 7)) | divide(n0,n1)|divide(n2,n3)|multiply(#0,#1) | probability |
weights of two friends ram and shyam are in the ratio 4 : 5 . if ram ' s weight is increased by 10 % and total weight of ram and shyam become 82.8 kg , with an increases of 15 % . by what percent did the weight of shyam has to be increased ? | "solution : given ratio of ram and shayam ' s weight = 4 : 5 hence , ( x - 15 ) / ( 15 - 10 ) = 4 / 5 or , x = 19 % . answer : option a" | a ) 19 % , b ) 10 % , c ) 21 % , d ) 16 % , e ) none | a | add(15, multiply(subtract(15, 10), divide(4, 5))) | divide(n0,n1)|subtract(n4,n2)|multiply(#0,#1)|add(n4,#2)| | gain |
a man invests in a 16 % stock at 128 . the interest obtained by him is : | "explanation : by investing rs 128 , income derived = rs . 16 by investing rs . 100 , income derived = = rs . 12.5 interest obtained = 12.5 % answer : c ) rs . 12.5 %" | a ) 12.9 % , b ) 92.5 % , c ) 12.5 % , d ) 12.6 % , e ) 12.7 % | c | multiply(divide(16, 128), const_100) | divide(n0,n1)|multiply(#0,const_100)| | gain |
if a certain computer is capable of printing 4900 monthly credit card bills per hour , while a new model is capable of printing at a rate of 6600 per hour , the old model will take approximately how much longer than the new model to print 10000 bills ? | "explanation : old model is capable of printing at a rate of 4900 per hour new model is capable of printing at a rate of 6600 per hour old model time taken to print 10000 cards = 10000 / 4900 = 100 / 49 new model time taken to print 10000 cards = 10000 / 6600 = 100 / 66 old model – new model : 100 / 49 - 100 / 66 = 170... | a ) 31 , b ) 76 , c ) 88 , d ) 26 , e ) 81 | a | multiply(subtract(divide(10000, 4900), divide(10000, 6600)), const_60) | divide(n2,n0)|divide(n2,n1)|subtract(#0,#1)|multiply(#2,const_60)| | general |
the difference of two numbers is 1385 . on dividing the larger number by the smaller , we get 6 as quotient and the 15 as remainder . what is the smaller number ? | "let the smaller number be x . then larger number = ( x + 1385 ) . x + 1385 = 6 x + 15 5 x = 1370 x = 274 smaller number = 274 . answer a" | a ) 274 , b ) 270 , c ) 295 , d ) 360 , e ) 252 | a | divide(add(1385, 15), subtract(6, const_1)) | add(n0,n2)|subtract(n1,const_1)|divide(#0,#1)| | general |
if a book is sold at 8 % profit instead of 8 % loss , it would have brought rs 12 more . find out the cost price of the book | "let c . p . of the book be rs . ’ x ’ given , 1.08 x - 0.92 x = 12 = > 0.16 x = 12 = 12 / 0.16 = rs 75 answer : a" | a ) rs 75 , b ) rs 72 , c ) rs 60 , d ) rs 70 , e ) rs 80 | a | divide(multiply(const_100, divide(8, const_2)), 8) | divide(n0,const_2)|multiply(#0,const_100)|divide(#1,n0)| | gain |
if a is the smallest positive integer such that 3150 multiplied by a is the square of an integer , then a must be | solution : this problem is testing us on the rule that when we express a perfect square by its unique prime factors , every prime factor ' s exponent is an even number . let ’ s start by prime factorizing 3150 . 3150 = 315 x 10 = 5 x 63 x 10 = 5 x 7 x 3 x 3 x 5 x 2 3150 = 2 ^ 1 x 3 ^ 2 x 5 ^ 2 x 7 ^ 1 ( notice that the... | a ) 2 , b ) 5 , c ) 6 , d ) 7 , e ) 14 | e | multiply(const_2, divide(divide(divide(divide(3150, multiply(add(const_1, const_4), add(const_1, const_4))), const_2), const_3), const_3)) | add(const_1,const_4)|multiply(#0,#0)|divide(n0,#1)|divide(#2,const_2)|divide(#3,const_3)|divide(#4,const_3)|multiply(#5,const_2) | geometry |
calculate how long it will take a swimmer to swim a distance of 3 km against the current of a river which flows at 1.7 km / hr , given that he can swim in still water at 3 km / h | "swim in still water at = 3 speed of river = 1.7 us = 3 - 1.7 = 1.3 distance = 3 t = 3 / 1.3 = 2.31 answer : a" | a ) 2.31 , b ) 9.31 , c ) 1.31 , d ) 4.31 , e ) 3.31 | a | divide(3, subtract(3, 1.7)) | subtract(n2,n1)|divide(n0,#0)| | gain |
crazy eddie has a key chain factory . eddie managed to decrease the cost of manufacturing his key chains while keeping the same selling price , and thus increased the profit from the sale of each key chain from 35 % of the selling price to 50 % of the selling price . if the manufacturing cost is now $ 50 , what was it ... | "deargoodyear 2013 , i ' m happy to help . this is a relatively straightforward problem , not very challenging . btw , crazy eddiewas the actually name of an electronics chain on the east coast of the usa back in the 1970 s . manufacturing now is $ 50 . they now are making a 50 % profit , so the selling price must be $... | a ) $ 20 , b ) $ 65 , c ) $ 50 , d ) $ 80 , e ) $ 100 | b | subtract(divide(50, divide(50, const_100)), multiply(divide(50, divide(50, const_100)), divide(35, const_100))) | divide(n1,const_100)|divide(n0,const_100)|divide(n1,#0)|multiply(#2,#1)|subtract(#2,#3)| | general |
find a two digit number , given that the sum of the digits is 13 and the difference of the digits is 5 . ? | "using elimination method find which of the options fit the description of the number . . . from the option only 94 meets this description sum of digits - - - 9 + 4 = 13 difference of digits - - - 9 - 4 = 5 answer e ." | a ) 74 , b ) 82 , c ) 95 , d ) 76 , e ) 94 | e | multiply(multiply(5, 5), add(5, const_4)) | add(n1,const_4)|multiply(n1,n1)|multiply(#0,#1)| | general |
a can do a work in 24 days and b can do it in 16 days . in how many days a and b can do the work ? | "explanation : a ' s 1 day ' s work = 1 / 24 b ' s 1 day ' s work = 1 / 16 they work together = 1 / 24 + 1 / 16 = 5 / 48 = 48 / 5 = 9 3 / 5 days answer : option d" | a ) 20 days , b ) 10 days , c ) 6 days , d ) 9 3 / 5 days , e ) 7 days | d | divide(const_1, subtract(divide(const_1, 16), divide(const_1, 24))) | divide(const_1,n1)|divide(const_1,n0)|subtract(#0,#1)|divide(const_1,#2)| | physics |
mike earns $ 14 per hour and phil earns $ 7 per hour . approximately how much less , as a percentage , does phil earn than mike per hour ? | what % less of 14 is 7 let it be x % less , then = 14 ( 1 - x / 100 ) = 7 1 - x / 100 = 7 / 14 x = 100 / 2 x = 50 % ans d | a ) 25 % , b ) 32.5 % , c ) 37 % , d ) 50 % , e ) 40 % | d | multiply(divide(7, 14), const_100) | divide(n1,n0)|multiply(#0,const_100) | general |
albert is 2 times mary ’ s age and 4 times as old as betty . mary is 8 years younger than albert . how old is betty ? | a = 2 m = m + 8 m = 8 a = 16 a = 4 b , and so b = 4 the answer is a . | a ) 4 , b ) 6 , c ) 10 , d ) 16 , e ) 18 | a | divide(multiply(2, 8), 4) | multiply(n0,n2)|divide(#0,n1) | general |
a manufacturer is using glass as the surface for the multi - touch screen of its smartphone . the glass on the manufactured phone has a 4 % probability of not passing quality control tests . the quality control manager bundles the smartphone in groups of 10 . if that bundle has any smartphone that does not pass the qua... | find the probability of the opposite event and subtract from 1 . the opposite event is that bundle will not be rejected by quality control , which will happen if all 10 phones pass the test , so p ( all 10 phones pass test ) = 0.96 ^ 10 . p ( at least one phone do not pass the test ) = 1 - p ( all 10 phones pass test )... | a ) 0.25 , b ) . 05 ^ 10 , c ) 1 - 0.95 ^ 10 , d ) 1 - 0.05 ^ 10 , e ) 1 - 0.96 ^ 10 | e | subtract(const_1, power(subtract(const_1, divide(4, const_100)), 10)) | divide(n0,const_100)|subtract(const_1,#0)|power(#1,n1)|subtract(const_1,#2) | general |
the ratio between the perimeter and the width of a rectangle is 5 : 1 . if the area of the rectangle is 150 sq . cm , what is the length of the rectangle ? | 2 l + 2 w = 5 w l = 3 w / 2 w * l = 150 3 w ^ 2 / 2 = 150 w ^ 2 = 100 w = 10 l = 3 ( 10 ) / 2 = 15 the answer is b . | ['a ) 12 cm', 'b ) 15 cm', 'c ) 18 cm', 'd ) 21 cm', 'e ) 24 cm'] | b | divide(150, const_10) | divide(n2,const_10) | geometry |
in a village there are 150 men and 90 women in present time . if in next year population will be p = ( a ^ 2 + b ^ 2 ) ^ 1 / 2 , and in every year men are reduces 6 % . what is population of after 2 year . | "next year total population = [ 150 ^ 2 + 90 ^ 2 ] ^ . 5 = 174.92 = 175 man decreased by 6 % so total man = 150 * . 94 = 141 women will be = 175 - 141 = 34 so population after two years = [ 135 ^ 2 + 34 ^ 2 ] ^ . 5 = 145.04 = 145 so population after two year = 145 answer : e" | a ) 140 , b ) 141 , c ) 142 , d ) 143 , e ) 145 | e | power(add(power(multiply(subtract(1, divide(1, 6)), 150), 2), power(subtract(power(add(power(150, 2), power(90, 2)), divide(1, 2)), multiply(subtract(1, divide(1, 6)), 150)), 2)), divide(1, 2)) | divide(n4,n6)|divide(n4,n2)|power(n0,n2)|power(n1,n2)|add(#2,#3)|subtract(n4,#0)|multiply(n0,#5)|power(#4,#1)|power(#6,n2)|subtract(#7,#6)|power(#9,n2)|add(#8,#10)|power(#11,#1)| | general |
crazy eddie has a key chain factory . eddie managed to decrease the cost of manufacturing his key chains while keeping the same selling price , and thus increased the profit from the sale of each key chain from 30 % of the selling price to 50 % of the selling price . if the manufacturing cost is now $ 50 , what was it ... | "deargoodyear 2013 , i ' m happy to help . this is a relatively straightforward problem , not very challenging . btw , crazy eddiewas the actually name of an electronics chain on the east coast of the usa back in the 1970 s . manufacturing now is $ 50 . they now are making a 50 % profit , so the selling price must be $... | a ) $ 20 , b ) $ 40 , c ) $ 50 , d ) $ 70 , e ) $ 100 | d | subtract(divide(50, divide(50, const_100)), multiply(divide(50, divide(50, const_100)), divide(30, const_100))) | divide(n1,const_100)|divide(n0,const_100)|divide(n1,#0)|multiply(#2,#1)|subtract(#2,#3)| | general |
if ( 1 / 2 ) ^ 16 ( 1 / 81 ) ^ k = 1 / 18 ^ 16 , then k = | "i ' m going to focus on denominator only . . ( 2 ^ 16 ) . ( ( 3 ^ 4 ) ^ k = 18 ^ 16 ( 2 ^ 16 ) . ( ( 3 ^ 4 k ) = ( 2 . 3 ^ 2 ) ^ 16 ( 2 ^ 16 ) . ( ( 3 ^ 4 k ) = ( 2 ^ 24 ) . ( 3 ^ 2 ) ^ 16 hence 4 k = 32 k = 8 answer a i hope it ' s quite clear" | a ) 8 , b ) 12 , c ) 16 , d ) 24 , e ) 36 | a | divide(multiply(16, 2), const_4) | multiply(n1,n2)|divide(#0,const_4)| | general |
excluding stoppages , the speed of a bus is 50 kmph and including stoppages , it is 35 kmph . for how many minutes does the bus stop per hour ? | "explanation : in one hour due to stoppages , it covers 15 km less . time taken to cover 15 km = [ 15 / 50 * 60 ] min = 18 min answer : b" | a ) 17 minutes , b ) 18 minutes , c ) 19 minutes , d ) 16 minutes , e ) none of these | b | multiply(const_60, divide(subtract(50, 35), 50)) | subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_60)| | physics |
in triangle pqr , the angle q = 90 degree , pq = 3 cm , qr = 8 cm . x is a variable point on pq . the line through x parallel to qr , intersects pr at y and the line through y , parallel to pq , intersects qr at z . find the least possible length of xz | "look at the diagram below : now , in case when qy is perpendicular to pr , two right triangles pqr and pqy are similar : qy : qp = qr : pr - - > qy : 3 = 8 : 10 - - > qy = 2.4 . answer : b ." | a ) 3.6 cm , b ) 2.4 cm , c ) 4.8 cm , d ) 2.16 cm , e ) 3.2 cm | b | divide(multiply(3, 8), const_10) | multiply(n1,n2)|divide(#0,const_10)| | geometry |
an engineer undertakes a project to build a road 15 km long in 300 days and employs 40 men for the purpose . after 100 days , he finds only 2.5 km of the road has been completed . find the ( approximate ) number of extra men he must employ to finish the work in time . | "40 workers working already let x be the total men required to finish the task in next 200 days 2.5 km done hence remaining is 12.5 km also , work has to be completed in next 200 days ( 300 - 100 = 200 ) we know that , proportion of men to distance is direct proportion and , proportion of men to days is inverse proport... | a ) a . 43 , b ) b . 45 , c ) c . 55 , d ) d . 68 , e ) e . 60 | e | subtract(divide(multiply(multiply(40, subtract(15, 2.5)), 100), multiply(2.5, subtract(300, 100))), 40) | subtract(n0,n4)|subtract(n1,n3)|multiply(n2,#0)|multiply(n4,#1)|multiply(n3,#2)|divide(#4,#3)|subtract(#5,n2)| | physics |
the average age of 25 students of a class is 25 years . out of these , the average age of 10 students is 22 years and that of the other 14 students is 28 years . the age of the 25 th student is ? | "age of the 25 th student = 25 * 25 - ( 10 * 22 + 14 * 28 ) = 625 - 612 = 13 years answer is c" | a ) 10 , b ) 15 , c ) 13 , d ) 11 , e ) 9 | c | subtract(multiply(25, 25), add(multiply(10, 22), multiply(14, 28))) | multiply(n0,n0)|multiply(n2,n3)|multiply(n4,n5)|add(#1,#2)|subtract(#0,#3)| | general |
how many numbers between 1000 and 5000 are exactly divisible by 225 ? | explanation : first multiple of 225 after 1000 is 1125 ( 225 ã — 5 ) and last multiple of 225 before 5000 is 4950 ( 225 ã — 22 ) total number are l + a / d = 4950 - 1125 / 225 + 1 = 18 answer is b | a ) 16 , b ) 18 , c ) 19 , d ) 12 , e ) 21 | b | divide(subtract(5000, 1000), 225) | subtract(n1,n0)|divide(#0,n2) | general |
a small table has a length of 12 inches and a breadth of b inches . cubes are placed on the surface of the table so as to cover the entire surface . the maximum side of such cubes is found to be 4 inches . also , a few such tables are arranged to form a square . the minimum length of side possible for such a square is ... | "from the info that the maximum sides of the cubes is 4 , we know that the gcf of 12 ( = 2 ^ 2 * 3 ) andbis 4 ( = 2 ^ 2 ) , sob = 2 ^ x , where x > = 2 . from the second premise , we know that the lcm of 12 ( 2 ^ 2 * 3 ) andbis 20 ( 2 ^ 2 * 5 ) , sob = 2 ^ 2 or 2 ^ 2 * 5 ( 4 or 20 ) . combining 2 premises shows the ans... | a ) 8 , b ) 16 , c ) 4 , d ) 32 , e ) 48 | c | sqrt(subtract(power(divide(20, 4), const_2), power(12, const_2))) | divide(n2,n1)|power(n0,const_2)|power(#0,const_2)|subtract(#2,#1)|sqrt(#3)| | geometry |
anita has 360 apples . there are 60 students are studying in her class . she would like to give each student the same amount of apples , how much should she give to each student ? | 360 / 60 = 6 the answer is b . | a ) 12 , b ) 6 , c ) 9 , d ) 15 , e ) 7 | b | divide(360, 60) | divide(n0,n1) | general |
a number is increased by 60 % and then decreased by 60 % . find the net increase or decrease per cent . | "let the number be 100 . increase in the number = 60 % = 60 % of 100 = ( 60 / 100 ã — 100 ) = 60 therefore , increased number = 100 + 60 = 160 this number is decreased by 60 % therefore , decrease in number = 60 % of 160 = ( 60 / 100 ã — 160 ) = 9600 / 100 = 96 therefore , new number = 160 - 96 = 64 thus , net decrease... | a ) 19 % , b ) 18 % , c ) 27 % , d ) 33 % , e ) 36 % | e | divide(multiply(add(60, const_100), subtract(const_100, 60)), const_100) | add(n0,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)| | general |
the area of sector of a circle whose radius is 12 metro and whose angle at the center is 36 â ° is ? | "36 / 360 * 22 / 7 * 12 * 12 = 45.3 m 2 answer : b" | a ) 52.6 , b ) 45.3 , c ) 52.8 , d ) 52.1 , e ) 52.2 | b | multiply(multiply(power(12, const_2), divide(add(multiply(const_2, const_10), const_2), add(const_4, const_3))), divide(36, divide(const_3600, const_10))) | add(const_3,const_4)|divide(const_3600,const_10)|multiply(const_10,const_2)|power(n0,const_2)|add(#2,const_2)|divide(n1,#1)|divide(#4,#0)|multiply(#6,#3)|multiply(#5,#7)| | geometry |
6 persons standing in queue with different age group , after two years their average age will be 43 and seventh person joined with them . hence the current average age has become 45 . find the age of seventh person ? | explanation : let the sum of current ages of 6 persons = x given average age of 6 person after 2 years = 43 = > x + 6 ( 2 ) / 6 = 43 = > x + 12 = 258 = x = 246 let the seventh ' s person age will be y given current average age of 7 persons = 45 [ sum of current 6 person ' s age ( x ) + seventh person ' s age ( y ) ] / ... | a ) 65 , b ) 67 , c ) 68 , d ) 69 , e ) 50 | d | subtract(multiply(45, add(6, const_1)), multiply(subtract(43, const_2), 6)) | add(n0,const_1)|subtract(n1,const_2)|multiply(n2,#0)|multiply(n0,#1)|subtract(#2,#3) | general |
one fourth of a solution that was 15 % sugar by weight was replaced by a second solution resulting in a solution that was 16 percent sugar by weight . the second solution was what percent sugar by weight ? | "instead of using complex calculations and remembering formulae , why dont u directly get to weighted average . 3 parts of 15 % + 1 part of x ( unknown ) % = 4 parts of 16 % = > x % = 64 % - 45 % = 16 % ans d it is ." | a ) 34 % , b ) 24 % , c ) 22 % , d ) 16 % , e ) 8.5 % | d | multiply(divide(subtract(multiply(const_100, divide(16, const_100)), multiply(subtract(const_100, multiply(divide(const_1, const_4), const_100)), divide(15, const_100))), multiply(divide(const_1, const_4), const_100)), const_100) | divide(n1,const_100)|divide(n0,const_100)|divide(const_1,const_4)|multiply(#0,const_100)|multiply(#2,const_100)|subtract(const_100,#4)|multiply(#1,#5)|subtract(#3,#6)|divide(#7,#4)|multiply(#8,const_100)| | gain |
every year an amount increases by 1 / 8 th of itself . how much will it be after two years if its present value is rs . 1600 ? | "1800 * 9 / 8 * 9 / 8 = 2025 answer : c" | a ) 3000 , b ) 4200 , c ) 2025 , d ) 1205 , e ) 3251 | c | add(add(1600, multiply(divide(1, 8), 1600)), multiply(divide(1, 8), add(1600, multiply(divide(1, 8), 1600)))) | divide(n0,n1)|multiply(n2,#0)|add(n2,#1)|multiply(#2,#0)|add(#2,#3)| | general |
how many unique positive odd integers less than 110 are equal to the product of a positive multiple of 5 and an odd number ? | "the question basically asks how many positive odd integers less than 110 are odd multiples of 5 so we have 5 , 15,25 , 35,45 , . . and 105 = 11 ans c" | a ) 4 , b ) 6 , c ) 11 , d ) 12 , e ) 15 | c | divide(divide(110, 5), const_2) | divide(n0,n1)|divide(#0,const_2)| | general |
if the difference between the length and breadth of a rectangle is 23 m and its perimeter is 186 m , what is its area ? | "length = breadth + 23 . therefore , 4 × breadth + 2 × 23 = 186 m ⇒ breadth = 35 m length = 35 + 23 = 58 m area = 58 × 35 = 2030 m 2 answer is d ." | a ) 2510 , b ) 2535 , c ) 2530 , d ) 2030 , e ) 2520 | d | rectangle_area(add(divide(subtract(186, multiply(const_2, 23)), const_4), 23), divide(subtract(186, multiply(const_2, 23)), const_4)) | multiply(n0,const_2)|subtract(n1,#0)|divide(#1,const_4)|add(n0,#2)|rectangle_area(#3,#2)| | geometry |
every disk in a bag is either blue , yellow or green . the ratio of blue disks to yellow disks to green disks in this bag is 3 : 7 : 8 . if the total number of disks in the bag is 72 , how many more green disks than blue disks are in the bag ? | "let b : y : g = 3 x : 7 x : 8 x . 3 x + 7 x + 8 x = 18 x = 72 - - > x = 4 . g - b = 8 x - 3 x = 5 x = 20 . the answer is a ." | a ) 20 , b ) 28 , c ) 30 , d ) 35 , e ) 40 | a | multiply(divide(72, add(add(3, 7), 8)), subtract(8, 3)) | add(n0,n1)|subtract(n2,n0)|add(n2,#0)|divide(n3,#2)|multiply(#3,#1)| | general |
an engineer designed a ball so that when it was dropped , it rose with each bounce exactly one - half as high as it had fallen . the engineer dropped the ball from a 16 - meter platform and caught it after it had traveled 40.5 meters . how many times did the ball bounce ? | "division of total distance travelled will be 16 + 16 + 8 + 0.5 ans : a" | a ) 3 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | a | divide(divide(16, const_2), const_2) | divide(n0,const_2)|divide(#0,const_2)| | general |
what is the probability of getting atmost one head when two unbiased coin are tossed ? | "here s = { hh , ht , th , tt } let ee = event of getting one head e = { tt , ht , th } p ( e ) = n ( e ) / n ( s ) = 3 / 4 option a" | a ) 3 / 4 , b ) 3 / 2 , c ) 4 / 3 , d ) 1 / 2 , e ) 2 / 6 | a | divide(const_2, choose(add(const_3, const_3), const_3)) | add(const_3,const_3)|choose(#0,const_3)|divide(const_2,#1)| | probability |
if 5 x + 3 = 10 x – 22 , what is the value of x ? | solve for x : 5 x + 3 = 10 x - 22 25 = 5 x 5 = x correct answer e ) 5 | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | e | divide(add(3, 22), subtract(10, 5)) | add(n1,n3)|subtract(n2,n0)|divide(#0,#1) | general |
a jogger running at 9 km / hr along side a railway track is 270 m ahead of the engine of a 120 m long train running at 45 km / hr in the same direction . in how much time will the train pass the jogger ? | "speed of train relative to jogger = 45 - 9 = 36 km / hr . = 36 * 5 / 18 = 10 m / sec . distance to be covered = 270 + 120 = 390 m . time taken = 390 / 10 = 39 sec . answer : c" | a ) 67 sec , b ) 89 sec , c ) 39 sec , d ) 87 sec , e ) 45 sec | c | divide(add(270, 120), multiply(subtract(45, 9), divide(divide(const_10, const_2), divide(subtract(45, 9), const_2)))) | add(n1,n2)|divide(const_10,const_2)|subtract(n3,n0)|divide(#2,const_2)|divide(#1,#3)|multiply(#4,#2)|divide(#0,#5)| | general |
the speed of a boat in still water in 26 km / hr and the rate of current is 6 km / hr . the distance travelled downstream in 20 minutes is : | "explanation : speed downstream = ( 26 + 6 ) = 32 kmph time = 20 minutes = 20 / 60 hour = 1 / 3 hour distance travelled = time × speed = ( 1 / 3 ) × 32 = 10.66 km answer : option b" | a ) 10.4 km , b ) 10.66 km , c ) 11.4 km , d ) 11.22 km , e ) 12 km | b | multiply(add(26, 6), divide(20, const_60)) | add(n0,n1)|divide(n2,const_60)|multiply(#0,#1)| | physics |
find the average of all the numbers between 6 and 39 which are divisible by 5 ? | "average = ( 10 + 15 + 20 + 25 + 30 + 35 ) 6 = 135 / 6 = 23 answer is e" | a ) 10 , b ) 20 , c ) 25 , d ) 30 , e ) 23 | e | divide(add(add(6, const_4), subtract(39, const_4)), const_2) | add(n0,const_4)|subtract(n1,const_4)|add(#0,#1)|divide(#2,const_2)| | general |
add : + 47 and - 27 | the absolute values of + 47 and - 27 are 47 and 27 respectively ; and their difference = 47 – 27 = 30 . ( + 47 ) + ( - 27 ) = + 30 answer : b | a ) - 30 , b ) + 30 , c ) 0 , d ) 27 , e ) 47 | b | add(27, const_3) | add(n1,const_3) | general |
a small , rectangular park has a perimeter of 560 feet and a diagonal measurement of 400 feet . what is its area , in square feet ? | "you can avoid a lot of work in this problem by recognizing that , with the info provided , the diagonal forms a triangle inside the rectangle with sides that have a 3 : 4 : 5 ratio . diagonal = 200 2 x + 2 y = 560 , or x + y = 280 a ^ 2 + b ^ 2 = c ^ 2 for each the sides of the triangle using the ratio 3 : 4 : 5 for s... | a ) 76,800 , b ) 19,600 , c ) 20,000 , d ) 20,400 , e ) 20,800 | a | multiply(multiply(divide(400, add(const_2, const_3)), const_3), multiply(divide(400, add(const_2, const_3)), const_4)) | add(const_2,const_3)|divide(n1,#0)|multiply(#1,const_3)|multiply(#1,const_4)|multiply(#2,#3)| | geometry |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.