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 |
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evaluate : 983 x 207 - 983 x 107 | "983 x 207 - 983 x 107 = 983 x ( 207 - 107 ) = 983 x 100 = 98300 . answer is d ." | a ) 93700 , b ) 97300 , c ) 93800 , d ) 98300 , e ) none of them | d | subtract(983, multiply(multiply(207, 983), 107)) | multiply(n1,n2)|multiply(n3,#0)|subtract(n0,#1)| | general |
find value of x : 3 x ^ 2 + 5 x + 2 = 0 | "a = 3 , b = 5 , c = 2 x 1,2 = ( - 5 Β± β ( 52 - 4 Γ 3 Γ 2 ) ) / ( 2 Γ 3 ) = ( - 5 Β± β ( 25 - 24 ) ) / 6 = ( - 5 Β± 1 ) / 6 x 1 = ( - 5 + 1 ) / 6 = - 4 / 6 = - 2 / 3 x 2 = ( - 5 - 1 ) / 6 = - 6 / 6 = - 1 a" | a ) - 1 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | a | multiply(add(add(3, 3), divide(multiply(3, 3), const_2)), const_10) | add(n0,n0)|multiply(n0,n0)|divide(#1,const_2)|add(#0,#2)|multiply(#3,const_10)| | general |
the probability that a computer company will get a computer hardware contract is 3 / 4 and the probability that it will not get a software contract is 3 / 5 . if the probability of getting at least one contract is 4 / 5 , what is the probability that it will get both the contracts ? | "let , a β‘ event of getting hardware contract b β‘ event of getting software contract ab β‘ event of getting both hardware and software contract . p ( a ) = 3 / 4 , p ( ~ b ) = 5 / 9 = > p ( b ) = 1 - ( 3 / 5 ) = 2 / 5 . a and b are not mutually exclusive events but independent events . so , p ( at least one of a and b )... | a ) 4 / 11 , b ) 5 / 12 , c ) 6 / 17 , d ) 7 / 20 , e ) 11 / 30 | d | subtract(add(divide(3, 4), subtract(const_1, divide(3, 5))), divide(5, 5)) | divide(n0,n1)|divide(n0,n3)|divide(n3,n5)|subtract(const_1,#1)|add(#0,#3)|subtract(#4,#2)| | other |
if a coin has an equal probability of landing heads up or tails up each time it is flipped , what is the probability that the coin will land tails up exectly twice in 3 consecutive flips ? | "total number of ways in which h or t can appear in 3 tosses of coin is = 2 * 2 * 2 = 8 ways for 2 t and 1 th thus probability is = p ( htt ) + p ( tth ) + p ( tht ) = 1 / 8 + 1 / 8 + 1 / 8 = 3 / 8 = . 375 answer : a" | a ) 0.375 , b ) 0.25 , c ) 0.325 , d ) 0.5 , e ) 0.666 | a | multiply(power(divide(const_1, const_2), 3), 3) | divide(const_1,const_2)|power(#0,n0)|multiply(n0,#1)| | general |
if the average ( arithmetic mean ) of 10 consecutive odd integers is 154 , then the least of these integers is | "a very helpful rule to know in arithmetic is the rule that in evenly spaced sets , average = median . because the average will equal the median in these sets , then we quickly know that the median of this set of consecutive odd integer numbers is 154 . there are 10 numbers in the set , and in a set with an even number... | a ) a ) 51 , b ) b ) 58 , c ) c ) 145 , d ) d ) 190 , e ) e ) 210 | c | add(subtract(154, 10), const_1) | subtract(n1,n0)|add(#0,const_1)| | general |
the average weight of a class is x pounds . when a new student weighing 100 pounds joins the class , the average decreases by 1 pound . in a few months the student β s weight increases to 110 pounds and the average weight of the class becomes x + 4 pounds . none of the other students β weights changed . what is the val... | "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 - 80 = 30 pounds corresponds to the increase in average weight of ( x + 4 ) - ( x - 1 ) = 5 pounds , which means that there are 30 / 5 ... | a ) 85 , b ) 86 , c ) 88 , d ) 90 , e ) 92 | e | add(add(100, 4), add(4, 1)) | add(n0,n3)|add(n1,n3)|add(#0,#1)| | general |
each of 3 investments has a 20 % of becoming worthless within a year of purchase , independently of what happens to the other two investments . if simone invests an equal sum v in each of these 3 investments on january 1 , the approximate chance that by the end of the year , she loses no more than 1 / 3 of her original... | the problem asks for the approximate chance that no more than 1 / 3 of the original investment is lost . we can apply the β 1 β x β technique : what β s the chance that more than 1 / 3 of the original investment is lost ? there are two outcomes we have to separately measure : ( a ) all 3 investments become worthless . ... | a ) 90 % , b ) 80 % , c ) 70 % , d ) 60 % , e ) 40 % | a | multiply(add(1, negate(add(power(divide(20, const_100), 3), multiply(multiply(multiply(subtract(const_1, divide(20, const_100)), divide(20, const_100)), divide(20, const_100)), 3)))), const_100) | divide(n1,const_100)|power(#0,n0)|subtract(const_1,#0)|multiply(#0,#2)|multiply(#0,#3)|multiply(n0,#4)|add(#5,#1)|negate(#6)|add(n3,#7)|multiply(#8,const_100) | general |
a man owns 2 / 3 of market reserch beauro buzness , and sells 3 / 4 of his shares for 75000 rs , what is the value of buzness ? | "if value of business = x total sell ( 2 x / 3 ) ( 3 / 4 ) = 75000 - > x = 150000 answer : a" | a ) 150000 , b ) 160000 , c ) 170000 , d ) 190000 , e ) 250000 | a | divide(75000, multiply(divide(2, 3), divide(3, 4))) | divide(n0,n1)|divide(n1,n3)|multiply(#0,#1)|divide(n4,#2)| | general |
on a certain day , joey , the ice - cream seller sold his ice creams to 4 different kids in a manner that each of the kids purchased half of the remaining ice creams and half ice - cream more . if we tell you that the fourth kid bought just a single ice cream , can you find out how many ice creams were sold by joey tha... | a joey sold 15 ice creams that day . the fourth kid bought a single ice cream . therefore , we have the following equation : total - ( total / 2 + 1 / 2 ) = 1 solving it , we get the total as 3 . let ' s work this method till we reach the first kid . the first kid bought 15 / 2 + 1 / 2 = 8 ( leaving 7 ) the second kid ... | a ) 15 , b ) 20 , c ) 22 , d ) 12 , e ) 8 | a | subtract(add(multiply(4, const_2), multiply(4, const_2)), const_1) | multiply(n0,const_2)|add(#0,#0)|subtract(#1,const_1) | general |
the population of a town is 7900 . it decreases annually at the rate of 10 % p . a . what was its population 2 years ago ? | "formula : ( after = 100 denominator ago = 100 numerator ) 7900 Γ£ β 100 / 90 Γ£ β 100 / 90 = 9753 a )" | a ) 9750 , b ) 8000 , c ) 8500 , d ) 9500 , e ) 10000 | a | subtract(subtract(7900, multiply(7900, divide(10, const_100))), multiply(subtract(7900, multiply(7900, divide(10, const_100))), divide(10, const_100))) | divide(n1,const_100)|multiply(n0,#0)|subtract(n0,#1)|multiply(#0,#2)|subtract(#2,#3)| | gain |
a 160 meter long train crosses a man standing on the platform in 12 sec . what is the speed of the train ? | "s = 160 / 12 * 18 / 5 = 48 kmph answer : a" | a ) 48 kmph , b ) 54 kmph , c ) 92 kmph , d ) 86 kmph , e ) 76 kmph | a | multiply(divide(160, 12), const_3_6) | divide(n0,n1)|multiply(#0,const_3_6)| | physics |
9 log 9 ( 5 ) = ? | "exponential and log functions are inverse of each other . hence aloga ( x ) = x , for all x real and positive . and therefore 9 log 9 ( 5 ) = 5 correct answer d" | a ) 1 , b ) 2 , c ) 3 , d ) 5 , e ) 8 | d | divide(log(multiply(9, 9)), log(const_10)) | log(const_10)|multiply(n0,n0)|log(#1)|divide(#2,#0)| | other |
the unit digit in the product ( 124 * 812 * 816 * 467 ) is : | "explanation : unit digit in the given product = unit digit in ( 4 * 2 * 6 * 7 ) = 6 answer : c" | a ) 2 , b ) 7 , c ) 6 , d ) 8 , e ) 1 | c | subtract(multiply(multiply(multiply(124, 812), 816), 467), subtract(multiply(multiply(multiply(124, 812), 816), 467), add(const_4, const_4))) | add(const_4,const_4)|multiply(n0,n1)|multiply(n2,#1)|multiply(n3,#2)|subtract(#3,#0)|subtract(#3,#4)| | general |
if a truck is traveling at a constant rate of 72 kilometers per hour , how many seconds will it take the truck to travel a distance of 600 meters ? ( 1 kilometer = 1000 meters ) | "speed = 72 km / hr = > 72,000 m / hr in one minute = > 72000 / 60 = 1200 meters in one sec = > 1200 / 60 = 20 meters time = total distance need to be covered / avg . speed = > 600 / 20 = 30 and hence the answer : c" | a ) 18 , b ) 24 , c ) 30 , d ) 36 , e ) 48 | c | multiply(divide(divide(600, 1000), 72), const_3600) | divide(n1,n3)|divide(#0,n0)|multiply(#1,const_3600)| | physics |
the total cost of a vacation was divided among 3 people . if the total cost of the vacation had been divided equally among 4 people , the cost per person would have been $ 40 less . what was the total cost cost of the vacation ? | "c for cost . p price per person . c = 3 * p c = 4 * p - 160 substituting the value of p from the first equation onto the second we get p = 160 . plugging in the value of p in the first equation , we get c = 480 . which leads us to answer choice c" | a ) $ 200 , b ) $ 300 , c ) $ 480 , d ) $ 500 , e ) $ 600 | c | multiply(multiply(4, 3), divide(40, subtract(4, 3))) | multiply(n0,n1)|subtract(n1,n0)|divide(n2,#1)|multiply(#2,#0)| | general |
a certain mixture of nuts consists of 5 parts almonds to 2 parts walnuts , by weight . what is the number of pounds of almonds in 150 pounds of the mixture ? | "almonds : walnuts = 5 : 2 total mixture has 7 parts in a 150 pound mixture , almonds are 5 / 7 ( total mixture ) = 5 / 7 * 150 = 107 pounds answer ( a )" | a ) 107 , b ) 84 , c ) 40 , d ) 28 , e ) 20 | a | divide(multiply(5, 150), add(5, 2)) | add(n0,n1)|multiply(n0,n2)|divide(#1,#0)| | general |
a , b and c , each working alone can complete a job in 6 , 8 and 12 days respectively . if all three of them work together to complete a job and earn $ 2340 , what will be c ' s share of the earnings ? | "explanatory answer a , b and c will share the amount of $ 2340 in the ratio of the amounts of work done by them . as a takes 6 days to complete the job , if a works alone , a will be able to complete 1 / 6 th of the work in a day . similarly , b will complete 1 / 8 th and c will complete 1 / 12 th of the work . so , t... | a ) $ 1100 , b ) $ 520 , c ) $ 1080 , d ) $ 1170 , e ) $ 630 | b | multiply(2340, divide(inverse(8), add(inverse(12), add(inverse(6), inverse(8))))) | inverse(n1)|inverse(n0)|inverse(n2)|add(#1,#0)|add(#3,#2)|divide(#0,#4)|multiply(n3,#5)| | physics |
in the coordinate plane , points ( x , 1 ) and ( 5 , y ) are on line k . if line k passes through the origin and has slope 1 / 5 , then what are the values of x and y respectively ? | "line k passes through the origin and has slope 1 / 5 means that its equation is y = 1 / 5 * x . thus : ( x , 1 ) = ( 5 , 1 ) and ( 5 , y ) = ( 5,1 ) - - > x = 5 and y = 1 answer : c" | a ) 4 and 1 , b ) 1 and 5 , c ) 5 and 1 , d ) 3 and 5 , e ) 5 and 3 | c | multiply(multiply(5, 5), divide(1, 5)) | divide(n0,n3)|multiply(n1,n3)|multiply(#0,#1)| | general |
one fourth of a solution that was 13 % salt 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 salt by weight ? | "consider total solution to be 100 liters and in this case you ' ll have : 75 * 0.13 + 25 * x = 100 * 0.16 - - > x = 0.25 . answer : a ." | a ) 25 % , b ) 34 % , c ) 22 % , d ) 18 % , e ) 8.5 % | a | multiply(subtract(multiply(divide(16, const_100), const_4), subtract(multiply(divide(13, const_100), const_4), divide(13, const_100))), const_100) | divide(n1,const_100)|divide(n0,const_100)|multiply(#0,const_4)|multiply(#1,const_4)|subtract(#3,#1)|subtract(#2,#4)|multiply(#5,const_100)| | gain |
if daily wages of a man is double to that of a woman , how many men should work for 25 days to earn rs . 14400 ? given that wages for 40 women for 30 days are rs . 21600 . | explanation : wages of 1 woman for 1 day = 21600 / 40 Γ 30 wages of 1 man for 1 day = 21600 Γ 2 / 40 Γ 30 wages of 1 man for 25 days = 21600 Γ 2 Γ 25 / 40 Γ 30 number of men = 14400 / ( 21600 Γ 2 Γ 25 / 40 Γ 30 ) = 144 / ( 216 Γ 50 / 40 Γ 30 ) = 144 / 9 = 16 answer : option c | a ) 12 , b ) 14 , c ) 16 , d ) 18 , e ) 20 | c | divide(14400, multiply(25, multiply(const_2, divide(21600, multiply(40, 30))))) | multiply(n2,n3)|divide(n4,#0)|multiply(#1,const_2)|multiply(n0,#2)|divide(n1,#3) | physics |
what is the probability of randomly selecting one of the shortest diagonals from all the diagonals of a regular hexagon ? | "it has two vertices on sides , which do not make a diagonal but a side . . so remaining 3 vertices make diagonals . . . only the opposite vertex will make largest diagonal and other two smaller ones . . so prob = 2 / ( 2 + 1 ) = 2 / 3 answer : d" | a ) 1 / 4 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 7 / 9 | d | divide(multiply(const_3, const_2), multiply(divide(const_26, const_2), multiply(multiply(const_3, const_2), const_2))) | divide(const_26,const_2)|multiply(const_2,const_3)|multiply(#1,const_2)|multiply(#0,#2)|divide(#1,#3)| | probability |
what is the compound interest on rs : 30,000 for 4 months at the rate of 5 % per annum | it is monthly compound rate = 5 / 12 % per month 30000 * ( 1 + 5 / 1200 ) ^ 4 - 30000 = 503.13 answer : c | a ) 501.13 , b ) 502.13 , c ) 503.13 , d ) 504.13 , e ) 505.13 | c | divide(multiply(multiply(multiply(const_3, const_100), const_100), multiply(5, divide(4, multiply(const_4, const_3)))), const_100) | multiply(const_100,const_3)|multiply(const_3,const_4)|divide(n1,#1)|multiply(#0,const_100)|multiply(n2,#2)|multiply(#3,#4)|divide(#5,const_100) | gain |
the speed of a car is 90 km in the first hour and 40 km in the second hour . what is the average speed of the car ? | "s = ( 90 + 40 ) / 2 = 65 kmph answer : d" | a ) 12 , b ) 75 , c ) 88 , d ) 65 , e ) 15 | d | divide(add(90, 40), const_2) | add(n0,n1)|divide(#0,const_2)| | physics |
how many trailing zeroes does 53 ! + 54 ! have ? | 53 ! + 54 ! = 53 ! + 54 * 53 ! = 53 ! ( 1 + 54 ) = 53 ! * 55 number of trailing 0 s in 53 ! = number of 5 s in the expansion of 53 ! = 10 + 2 = 12 there is 1 more 5 in 55 . hence , total number of trailing 0 s = 12 + 1 = 13 answer ( b ) . in most cases , when we are adding multiple terms , all of which have trailing 0 ... | a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | b | add(divide(add(54, const_1), add(const_1, const_4)), const_2) | add(n1,const_1)|add(const_1,const_4)|divide(#0,#1)|add(#2,const_2) | other |
an empty fuel tank with a capacity of 208 gallons was filled partially with fuel a and then to capacity with fuel b . fuel a contains 12 % ethanol by volume and fuel b contains 16 % ethanol by volume . if the full fuel tank contains 30 gallons of ethanol , how many gallons of fuel a were added ? | "say there are a gallons of fuel a in the tank , then there would be 208 - a gallons of fuel b . the amount of ethanol in a gallons of fuel a is 0.12 a ; the amount of ethanol in 208 - a gallons of fuel b is 0.16 ( 208 - a ) ; since the total amount of ethanol is 30 gallons then 0.12 a + 0.16 ( 208 - a ) = 30 - - > a =... | a ) 160 , b ) 150 , c ) 82 , d ) 80 , e ) 50 | c | divide(subtract(multiply(208, divide(16, const_100)), 30), subtract(divide(16, const_100), divide(12, const_100))) | divide(n2,const_100)|divide(n1,const_100)|multiply(n0,#0)|subtract(#0,#1)|subtract(#2,n3)|divide(#4,#3)| | gain |
5020 β ( 502 Γ· 100.4 ) = ? | "explanation : = 5020 β ( 502 / 1004 Γ 10 ) = 5020 β 5 = 5015 option c" | a ) 15 , b ) 20 , c ) 5015 , d ) 25 , e ) 35 | c | subtract(5020, divide(502, 100.4)) | divide(n1,n2)|subtract(n0,#0)| | general |
the average of 10 numbers is calculated as 46 . it is discovered later on that while calculating the average , the number 65 was incorrectly read as 25 , and this incorrect number was used in the calculation . what is the correct average ? | "the total sum of the numbers should be increased by 40 . then the average will increase by 40 / 10 = 4 . the correct average is 50 . the answer is b ." | a ) 48 , b ) 50 , c ) 54 , d ) 66 , e ) 86 | b | divide(add(subtract(multiply(46, 10), 25), 65), 10) | multiply(n0,n1)|subtract(#0,n3)|add(n2,#1)|divide(#2,n0)| | general |
a part - time employee whose hourly wage was decreased by 20 percent decided to increase the number of hours worked per week so that the employee ' s total income did not change . by what percent w should the number of hours worked be increased ? | "correct answer : c solution : c . we can set up equations for income before and after the wage reduction . initially , the employee earns w wage and works h hours per week . after the reduction , the employee earns . 8 w wage and works x hours . by setting these equations equal to each other , we can determine the inc... | a ) 12.5 % , b ) 20 % , c ) 25 % , d ) 50 % , e ) 100 % | c | multiply(divide(subtract(divide(multiply(const_10, const_4), multiply(divide(subtract(const_100, 20), const_100), const_10)), const_4), const_4), const_100) | multiply(const_10,const_4)|subtract(const_100,n0)|divide(#1,const_100)|multiply(#2,const_10)|divide(#0,#3)|subtract(#4,const_4)|divide(#5,const_4)|multiply(#6,const_100)| | general |
john spends $ 320 buying his favorite dolls . if he buys only small monkey dolls , which are $ 4 cheaper than the large monkey dolls , he could buy 40 more dolls than if he were to buy only large monkey dolls . how much does a large monkey doll cost ? | a and b is not an integer . so we start with c if large doll costs $ 8 , then he can buy 320 / 8 = 40 large dolls and 320 / 4 = 80 small dolls . difference is 40 , which is we wanted . answer c . | a ) $ 7.2 , b ) $ 7.5 , c ) $ 8 , d ) $ 9 , e ) $ 10 | c | add(divide(320, add(divide(subtract(320, multiply(4, 40)), 4), 40)), 4) | multiply(n1,n2)|subtract(n0,#0)|divide(#1,n1)|add(n2,#2)|divide(n0,#3)|add(n1,#4) | general |
the number 110 can be written as sum of the squares of 3 different positive integers . what is the sum of these 3 different integers ? | "sum of the squares of 3 different positive integers = 110 5 ^ 2 + 6 ^ 2 + 7 ^ 2 = 110 now , sum of these 3 different integers = 5 + 6 + 7 = 18 ans - c" | a ) 10 , b ) 12 , c ) 18 , d ) 14 , e ) 17 | c | multiply(const_2, sqrt(divide(110, const_2))) | divide(n0,const_2)|sqrt(#0)|multiply(#1,const_2)| | geometry |
two trains , each 100 m long , moving in opposite directions , cross other in 12 sec . if one is moving twice as fast the other , then the speed of the faster train is ? | "let the speed of the slower train be x m / sec . then , speed of the train = 2 x m / sec . relative speed = ( x + 2 x ) = 3 x m / sec . ( 100 + 100 ) / 12 = 3 x = > x = 50 / 9 . so , speed of the faster train = 100 / 9 = 100 / 9 * 18 / 5 = 40 km / hr . answer : b" | a ) 26 km / hr , b ) 40 km / hr , c ) 60 km / hr , d ) 77 km / hr , e ) 46 km / hr | b | divide(multiply(multiply(divide(add(100, 100), multiply(12, add(const_1, const_2))), const_2), const_3600), const_1000) | add(n0,n0)|add(const_1,const_2)|multiply(n1,#1)|divide(#0,#2)|multiply(#3,const_2)|multiply(#4,const_3600)|divide(#5,const_1000)| | physics |
a profit of rs . 1000 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 . 1000 is divided between x and y in the ratio of 1 / 2 : 1 / 3 or 3 : 2 . so profits are 600 and 400 . difference in profit share = 600 - 400 = 200 answer : a" | a ) s . 200 , b ) s . 100 , c ) s . 300 , d ) s . 50 , e ) s . 90 | a | subtract(divide(divide(1000, add(divide(1, 2), divide(1, 3))), 2), divide(divide(1000, 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 |
there is a total of 120 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 / 4 and the probability that it will be green is 1 / 3 . what is the probability that the marble will be either red or blue ? | "total marbles in the box = 120 white marbles = 120 / 4 = 30 green marbles = 120 / 3 = 40 w + g = 70 red + blue = 50 p ( red or blue ) = 50 / 120 = 5 / 12 answer : e" | a ) 1 / 6 , b ) 1 / 4 , c ) 2 / 7 , d ) 1 / 3 , e ) 5 / 12 | e | subtract(1, add(divide(1, 4), divide(1, 3))) | divide(n1,n2)|divide(n3,n4)|add(#0,#1)|subtract(n1,#2)| | general |
three pipes of same capacity can fill a tank in 8 hours . if there are only two pipes of same capacity , the tank can be filled in ? | "the part of the tank filled by three pipes in one hour = 1 / 8 = > the part of the tank filled by two pipes in 1 hour = 2 / 3 * 1 / 8 = 1 / 12 . the tank can be filled in 12 hours . answer : b" | a ) 08 hours , b ) 12 hours , c ) 03 hours , d ) 04 hours , e ) 21 hours | b | inverse(multiply(divide(const_2, const_3), divide(const_1, 8))) | divide(const_2,const_3)|divide(const_1,n0)|multiply(#0,#1)|inverse(#2)| | physics |
if a 2 + b 2 + c 2 = 941 and ab + bc + ca = 10 , then a + b + c is | "by formula , ( a + b + c ) ^ 2 = a ^ 2 + b ^ 2 + c ^ 2 + 2 ( ab + bc + ca ) , since , a ^ 2 + b ^ 2 + c ^ 2 = 941 and ab + bc + ca = 10 , ( a + b + c ) ^ 2 = 941 + 2 ( 10 ) = 961 = 31 ^ 2 therefore : a + b + c = 31 answer : e" | a ) 44 , b ) 56 , c ) 16 , d ) 32 , e ) 31 | e | sqrt(add(941, multiply(10, 2))) | multiply(n4,n0)|add(n3,#0)|sqrt(#1)| | general |
what is the tens digit of 6 ^ 18 ? | "the tens digit of 6 in integer power starting from 2 ( 6 ^ 1 has no tens digit ) repeats in a pattern of 5 : { 3 , 1 , 9 , 7 , 5 } : the tens digit of 6 ^ 2 = 36 is 3 . the tens digit of 6 ^ 3 = 216 is 1 . the tens digit of 6 ^ 4 = . . . 96 is 9 . the tens digit of 6 ^ 5 = . . . 76 is 7 . the tens digit of 6 ^ 6 = . .... | a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | a | floor(divide(reminder(power(6, reminder(18, add(const_4, const_1))), const_100), const_10)) | add(const_1,const_4)|reminder(n1,#0)|power(n0,#1)|reminder(#2,const_100)|divide(#3,const_10)|floor(#4)| | general |
if dev works alone he will take 20 more hours to complete a task than if he worked with tina to complete the task . if tina works alone , she will take 5 more hours to complete the complete the task , then if she worked with dev to complete the task ? what is the ratio of the time taken by dev to that taken by tina if ... | let time taken by dev to complete the work alone be x days and that by tina be y days work done by dev in 1 day = 1 / x work done by tina in 1 day = 1 / y work done by devtina in 1 day = 1 / x + 1 / y = ( x + y ) / xy thus working together they will complete the work in xy / ( x + y ) days acc . to the ques : 1 ) if de... | a ) 4 : 1 , b ) 2 : 1 , c ) 10 : 1 , d ) 3 : 1 , e ) 1 : 2 | b | divide(add(20, sqrt(multiply(20, 5))), add(5, sqrt(multiply(20, 5)))) | multiply(n0,n1)|sqrt(#0)|add(n0,#1)|add(n1,#1)|divide(#2,#3) | physics |
a train 140 m long is running with a speed of 50 km / hr . in what time will it pass a man who is running at 4 km / hr in the direction opposite to that in which the train is going ? | "speed of train relative to man = 50 + 4 = 54 km / hr . = 54 * 5 / 18 = 15 m / sec . time taken to pass the men = 140 * 1 / 15 = 9.33 sec . answer : option c" | a ) 8.1 , b ) 7.0 , c ) 9.33 , d ) 8 , e ) 9 | c | divide(140, multiply(add(50, 4), const_0_2778)) | add(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
a train 110 m long is running with a speed of 40 km / hr . in what time will it pass a man who is running at 5 km / hr in the direction opposite to that in which the train is going ? | "speed of train relative to man = 40 + 5 = 45 km / hr . = 45 * 5 / 18 = 25 / 2 m / sec . time taken to pass the men = 110 * 2 / 25 = 8.8 sec . answer : option d" | a ) 5 , b ) 6 , c ) 7.5 , d ) 8.8 , e ) 9 | d | divide(110, multiply(add(40, 5), const_0_2778)) | add(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
john and steve are speed walkers in a race . john is 15 meters behind steve when he begins his final push . john blazes to the finish at a pace of 4.2 m / s , while steve maintains a blistering 3.8 m / s speed . if john finishes the race 2 meters ahead of steve , how long was john β s final push ? | "let t be the time that john spent for his final push . thus , per the question , 4.2 t = 3.8 t + 15 + 2 - - - > 0.4 t = 17 - - - > t = 42.5 seconds . e is the correct answer ." | a ) 13 seconds , b ) 17 seconds , c ) 26 seconds , d ) 34 seconds , e ) 42.5 seconds | e | divide(add(divide(multiply(3.8, add(15, 2)), subtract(4.2, 3.8)), add(15, 2)), 4.2) | add(n0,n3)|subtract(n1,n2)|multiply(n2,#0)|divide(#2,#1)|add(#0,#3)|divide(#4,n1)| | physics |
an error 2.5 % in excess is made while measuring the side of a square . the percentage of error in the calculated area of the square is | "100 cm is read as 102 cm . a 1 = ( 100 x 100 ) cm 2 and a 2 ( 102.5 x 102.5 ) cm 2 . ( a 2 - a 1 ) = [ ( 102.5 ) 2 - ( 100 ) 2 ] = ( 102.5 + 100 ) x ( 102.5 - 100 ) = 506.25 cm 2 . percentage error = 5.0625 d" | a ) 6.04 % , b ) 5.14 % , c ) 5.23 % , d ) 5.0625 % , e ) 5.5 % | d | divide(multiply(subtract(square_area(add(const_100, 2.5)), square_area(const_100)), const_100), square_area(const_100)) | add(n0,const_100)|square_area(const_100)|square_area(#0)|subtract(#2,#1)|multiply(#3,const_100)|divide(#4,#1)| | gain |
the average of 10 consecutive integers is 20 . then , 9 is deducted from the first consecutive number , 8 is deducted from the second , 7 is deducted form the third , and so on until the last number which remains unchanged . what is the new average ? | "the total subtracted is ( 9 + 8 + . . . + 1 ) = ( 9 * 10 ) / 2 = 45 on average , each number will be reduced by 45 / 10 = 4.5 therefore , the overall average will be reduced by 4.5 the answer is d ." | a ) 14 , b ) 14.5 , c ) 15 , d ) 15.5 , e ) 16 | d | divide(subtract(multiply(10, 20), multiply(add(const_4, const_1), 9)), 10) | add(const_1,const_4)|multiply(n0,n1)|multiply(n2,#0)|subtract(#1,#2)|divide(#3,n0)| | general |
a person decided to build a house in 100 days . he employed 100 men in the beginning and 100 more after 10 days and completed the construction in stipulated time . if he had not employed the additional men , how many days behind schedule would it have been finished ? | "200 men do the rest of the work in 100 - 10 = 90 days 100 men can do the rest of the work in 90 * 200 / 100 = 180 days required number of days = 180 - 90 = 90 days answer is e" | a ) 40 , b ) 80 , c ) 70 , d ) 60 , e ) 90 | e | subtract(100, 10) | subtract(n0,n3)| | general |
suresh can complete a job in 15 hours . ashutosh alone can complete the same job in 30 hours . suresh works for 9 hours and then the remaining job is completed by ashutosh . how many hours will it take ashutosh to complete the remaining job alone ? | "the part of job that suresh completes in 9 hours = 9 Γ’ Β β 15 = 3 Γ’ Β β 5 remaining job = 1 - 3 Γ’ Β β 5 = 2 Γ’ Β β 5 remaining job can be done by ashutosh in 2 Γ’ Β β 5 Γ£ β 30 = 12 hours answer d" | a ) 4 , b ) 5 , c ) 6 , d ) 12 , e ) none of these | d | multiply(subtract(const_1, divide(9, 15)), 30) | divide(n2,n0)|subtract(const_1,#0)|multiply(n1,#1)| | physics |
if a rectangular room measures 10 meters by 5 meters by 4 meters , what is the volume of the room in cubic centimeters ? ( 1 meter = 100 centimeters ) | "d . 200 , 000,000 10 * 100 * 5 * 100 * 4 * 100 = 200 , 000,000" | a ) 24,000 , b ) 240,000 , c ) 2 , 400,000 , d ) 200 , 000,000 , e ) 240 , 000,000 | d | multiply(multiply(multiply(multiply(4, 100), divide(1, const_10)), multiply(5, 100)), multiply(10, 100)) | divide(n3,const_10)|multiply(n2,n4)|multiply(n1,n4)|multiply(n0,n4)|multiply(#0,#1)|multiply(#4,#2)|multiply(#5,#3)| | geometry |
only a single rail track exists between stations a and b on a railway line . one hour after the north bound super fast train n leaves station a for station b , a south - bound passenger train s reaches station a from station b . the speed of the super fast train is twice that of a normal express train e , while the spe... | explanation : if speed of n = 4 , speed of s = 1 . = > average speed = ( 2 x 4 x 1 ) / ( 4 + 1 ) = 1.6 . because time available is 2 / 3 , speed = 3 / 2 . now average speed = 2.4 now speed of n = 8 . speed of s = y . = > ( 2 x 8 x y ) / ( 8 + y ) = 2.4 = > y = 1.3 the required ratio is 1.3 : 8 i . e 1 : 6 . answer : d | a ) 1 : 3 , b ) 1 : 4 , c ) 1 : 5 , d ) 1 : 6 , e ) 1 : 7 | d | divide(const_2, subtract(subtract(20, divide(const_12, const_2)), const_2)) | divide(const_12,const_2)|subtract(n0,#0)|subtract(#1,const_2)|divide(const_2,#2) | general |
a can complete a project in 20 days and b can complete the same project in 30 days . if a and b start working on the project together and a quits 15 days before the project is completed , in how many days total will the project be completed ? | a ' s rate is 1 / 20 of the project per day . b ' s rate is 1 / 30 of the project per day . the combined rate is 1 / 12 of the project per day . in the last 15 days , b can do 1 / 2 of the project . thus a and b must complete 1 / 2 of the project , which takes 6 days . the total number of days is 6 + 15 = 21 . the answ... | a ) 18 , b ) 21 , c ) 24 , d ) 27 , e ) 30 | b | add(divide(subtract(const_1, multiply(divide(const_1, 30), 15)), add(divide(const_1, 20), divide(const_1, 30))), 15) | divide(const_1,n1)|divide(const_1,n0)|add(#1,#0)|multiply(n2,#0)|subtract(const_1,#3)|divide(#4,#2)|add(n2,#5) | physics |
2.2 cubic dm of lead is to be drawn into a cylindrical wire 0.50 cm diameter . find the length of the wire in meters . | let the length of the wire be h meters . then , β ( 0.50 / ( 2 x 100 ) ) ^ 2 x h = 2.2 / 1000 = h = ( ( 2.2 / 1000 ) x ( 100 x 100 ) / ( 0.25 x 0.25 ) x ( 7 / 22 ) ) = 112 m . answer is d . | ['a ) 110', 'b ) 111', 'c ) 113', 'd ) 112', 'e ) none of them'] | d | divide(divide(multiply(2.2, const_1000), circle_area(divide(0.5, const_2))), const_100) | divide(n1,const_2)|multiply(n0,const_1000)|circle_area(#0)|divide(#1,#2)|divide(#3,const_100) | geometry |
how many positive integers less than 200 are there such that they are multiples of 13 or multiples of 14 ? | "200 / 13 = 15 ( plus remainder ) so there are 15 multiples of 13 200 / 14 = 14 ( plus remainder ) so there are 14 multiples of 14 we need to subtract 1 because 13 * 14 is a multiple of both so it was counted twice . the total is 15 + 14 - 1 = 28 the answer is d ." | a ) 25 , b ) 26 , c ) 27 , d ) 28 , e ) 29 | d | divide(factorial(subtract(add(const_4, 13), const_1)), multiply(factorial(13), factorial(subtract(const_4, const_1)))) | add(n1,const_4)|factorial(n1)|subtract(const_4,const_1)|factorial(#2)|subtract(#0,const_1)|factorial(#4)|multiply(#1,#3)|divide(#5,#6)| | general |
there are 2 sections a and b in a class , consisting of 36 and 24 students respectively . if the average weight of section a is 30 kg and that of section b is 30 kg , find the average of the whole class ? | "total weight of 36 + 44 students = 36 * 30 + 24 * 30 = 1800 average weight of the class is = 1800 / 60 = 30 kg answer is a" | a ) 30 kg , b ) 37 kg , c ) 42 kg , d ) 55.12 kg , e ) 29.78 kg | a | divide(add(multiply(36, 30), multiply(24, 30)), add(36, 24)) | add(n1,n2)|multiply(n1,n3)|multiply(n2,n4)|add(#1,#2)|divide(#3,#0)| | general |
sarah is driving to the airport . after driving at 25 miles per hour for one hour , she realizes that if she continues at that same average rate she will be an hour late for her flight . she then travels 50 miles per hour for the rest of the trip , and arrives 30 minutes before her flight departs . how many miles did s... | "after driving at 25 miles per hourfor one hour , this distance left to cover is d - 25 . say this distance is x miles . now , we know that the difference in time between covering this distance at 25 miles per hour and 50 miles per hour is 1 + 1 / 2 = 3 / 2 hours . so , we have that x / 25 - x / 50 = 3 / 2 - - > 2 x / ... | a ) 100 , b ) 175 , c ) 210 , d ) 245 , e ) 280 | a | subtract(30, subtract(25, 30)) | subtract(n0,n2)|subtract(n2,#0)| | physics |
two trains start from p and q respectively and travel towards each other at a speed of 50 km / hr and 40 km / hr respectively . by the time they meet , the first train has travelled 100 km more than the second . the distance between p and q is : | "sol . at the time of meeting , let the distane travelled byb the second train be x km . then , distance covered by the first train is ( x + 100 ) km β΄ x / 40 = ( x + 100 ) / 50 β 50 x = 40 x 4000 β x = 400 . so , distance between p and q = ( x + x + 100 ) km = 900 km . answer c" | a ) 500 km , b ) 630 km , c ) 900 km , d ) 660 km , e ) none | c | add(add(divide(multiply(100, 40), subtract(50, 40)), 100), divide(multiply(100, 40), subtract(50, 40))) | multiply(n1,n2)|subtract(n0,n1)|divide(#0,#1)|add(n2,#2)|add(#3,#2)| | physics |
a is thrice as good a workman as b and takes 10 days less to do a piece of work than b takes . b alone can do the whole work in | explanation : ratio of times taken by a and b = 1 : 3 means b will take 3 times which a will do in 1 time if difference of time is 2 days , b takes 3 days if difference of time is 10 days , b takes ( 3 / 2 ) * 10 = 15 days option a | a ) 15 days , b ) 10 days , c ) 9 days , d ) 8 days , e ) 7 days | a | divide(multiply(const_3, 10), const_2) | multiply(n0,const_3)|divide(#0,const_2) | physics |
the difference between simple and compound interest on rs . 1500 for one year at 10 % per annum reckoned half - yearly is ? | "s . i . = ( 1500 * 10 * 1 ) / 100 = rs . 150 c . i . = [ 1500 * ( 1 + 5 / 100 ) 2 - 1500 ] = rs . 153.75 difference = ( 153.75 - 150 ) = rs . 3.75 answer : d" | a ) 8.0 , b ) 3.0 , c ) 9.5 , d ) 3.75 , e ) 2.15 | d | multiply(subtract(power(add(divide(divide(10, const_2), const_100), const_1), const_2), add(divide(10, const_100), const_1)), 1500) | divide(n1,const_2)|divide(n1,const_100)|add(#1,const_1)|divide(#0,const_100)|add(#3,const_1)|power(#4,const_2)|subtract(#5,#2)|multiply(n0,#6)| | gain |
in a class , 12 students like to play basketball and 8 like to play cricket . 3 students like to play on both basketball and cricket . how many students like to play basketball or cricket or both ? | "draw a venn diagram yourself ! b + c - bc = number of students that play either basketball or cricket 12 + 8 - 3 = 17 c )" | a ) 12 , b ) 15 , c ) 17 , d ) 18 , e ) 22 | c | subtract(add(12, 8), 3) | add(n0,n1)|subtract(#0,n2)| | other |
in a graduate physics course , 70 percent of the students are male and 30 percent of the students are married . if one - sevenths of the male students are married , what fraction of the female students is single ? | let assume there are 100 students of which 70 are male and 30 are females if 30 are married then 70 will be single . now its given that two - sevenths of the male students are married that means 1 / 7 of 70 = 10 males are married if 30 is the total number of students who are married and out of that 10 are males then th... | a ) 1 / 3 , b ) 1 / 7 , c ) 1 / 2 , d ) 2 / 3 , e ) 5 / 7 | a | divide(const_10, 30) | divide(const_10,n1) | gain |
a train ride from two p to town q costs $ 6.35 more than does a bus ride from town p to town q . together , the cost of one train ride and one bus ride is $ 9.85 . what is the cost of a bus ride from town p to town q ? | "let x be the cost of a bus ride . x + ( x + 635 ) = 985 2 x = 350 x = $ 1.75 the answer is a ." | a ) $ 1.75 , b ) $ 2.50 , c ) $ 4.10 , d ) $ 4.70 , e ) $ 8.20 | a | divide(subtract(9.85, 6.35), const_2) | subtract(n1,n0)|divide(#0,const_2)| | general |
a drink vendor has 10 liters of maaza , 144 liters of pepsi and 368 liters of sprite . he wants to pack them in cans , so that each can contains the same number of liters of a drink , and does n ' t want to mix any two drinks in a can . what is the least number of cans required ? | "the number of liters in each can = hcf of 10 , 144 and 368 = 2 liters . number of cans of maaza = 10 / 2 = 5 number of cans of pepsi = 144 / 2 = 72 number of cans of sprite = 368 / 2 = 184 the total number of cans required = 5 + 72 + 184 = 261 cans . answer : b" | a ) 135 , b ) 261 , c ) 422 , d ) 430 , e ) 438 | b | add(divide(368, gcd(gcd(10, 144), 368)), add(divide(10, gcd(gcd(10, 144), 368)), divide(144, gcd(gcd(10, 144), 368)))) | gcd(n0,n1)|gcd(n2,#0)|divide(n0,#1)|divide(n1,#1)|divide(n2,#1)|add(#2,#3)|add(#5,#4)| | general |
if 40 % of a certain number is 160 , then what is 30 % of that number ? | "explanation : 40 % = 40 * 4 = 160 30 % = 30 * 4 = 120 answer : option c" | a ) 270 , b ) 380 , c ) 120 , d ) 360 , e ) 180 | c | multiply(divide(160, divide(40, const_100)), divide(30, const_100)) | divide(n0,const_100)|divide(n2,const_100)|divide(n1,#0)|multiply(#2,#1)| | gain |
if the sum of two numbers is 45 and the l . c . m and sum of the reciprocal of the numbers are 120 and 11 / 120 then hcf of numbers is equal to : | let the numbers be a and b and hcf is x , then , a + b = 45 and ab = x * 120 = 120 x required sum = 1 / a + 1 / b = ( a + b ) / ab = 45 / 120 x = 11 / 120 . x = 5 . answer : b | a ) 55 , b ) 5 , c ) 10 , d ) 45 , e ) 4 | b | add(floor(divide(45, 11)), const_1) | divide(n0,n2)|floor(#0)|add(#1,const_1) | general |
find the missing figures : 0.1 % of ? = 0.24 | "let 0.1 % of x = 0.24 . then , 0.1 * x / 100 = 0.24 x = [ ( 0.24 * 100 ) / 0.1 ] = 240 . answer is d ." | a ) 12 , b ) 120 , c ) 24 , d ) 240 , e ) 36 | d | divide(0.24, divide(0.1, const_100)) | divide(n0,const_100)|divide(n1,#0)| | gain |
there has been successive increases of 25 % and then 10 % in the price of gas from the previous month . by what percentage should a driver reduce gas consumption so that the expenditure does not change ? | "let p be the original price per unit of gas . let x be the original gas consumption . let y be the reduced gas consumption . y * 1.1 * 1.25 * p = x * p y = x / ( 1.1 * 1.25 ) which is about 0.73 x which is a decrease of about 27 % . the answer is b ." | a ) 23 % , b ) 27 % , c ) 31 % , d ) 35 % , e ) 39 % | b | multiply(subtract(const_1, divide(const_100, add(add(const_100, 25), divide(multiply(add(const_100, 25), 10), const_100)))), const_100) | add(n0,const_100)|multiply(n1,#0)|divide(#1,const_100)|add(#0,#2)|divide(const_100,#3)|subtract(const_1,#4)|multiply(#5,const_100)| | general |
pipe a can fill a tank in 60 min . there is a second pipe in the bottom of the cistern to empty it . if all the two pipes are simultaneously opened , then the cistern is full in 180 min . in how much time , the second pipe alone can empty the cistern ? | work done by the third pipe in 1 min = 1 / 180 - ( 1 / 60 ) = - 1 / 45 . [ - ve sign means emptying ] the third pipe alone can empty the cistern in 45 min . answer : b | a ) 60 min , b ) 45 min , c ) 90 min , d ) 70 min , e ) 30 min | b | inverse(add(divide(const_1, 180), divide(const_1, 60))) | divide(const_1,n1)|divide(const_1,n0)|add(#0,#1)|inverse(#2) | physics |
a dishonest dealer claims to sell a product at its cost price . he uses a counterfeit weight which is 20 % less than the real weight . further greed overtook him and he added 35 % impurities to the product . find the net profit percentage of the dealer ? | "the dealer uses weight which is 20 % less than the real weight . or ( 1 - 1 / 5 ) or 4 / 5 of real weight . it means that he is selling $ 4 worth of product for $ 5 . the dealer then further added 35 % impurities to the product . it means that he is selling $ 5 worth of product for $ 6.75 . so his profit is $ 6.75 - $... | a ) 44 % , b ) 40 % , c ) 68.75 % , d ) 56.25 % , e ) 36 % | c | multiply(subtract(divide(add(const_100, 35), subtract(const_100, 20)), const_1), const_100) | add(n1,const_100)|subtract(const_100,n0)|divide(#0,#1)|subtract(#2,const_1)|multiply(#3,const_100)| | general |
nina has exactly enough money to purchase 6 widgets . if the cost of each widget were reduced by $ 2 , then nina would have exactly enough money to purchase 8 widgets . how much money does nina have ? | e its is . let price = x ( x - 2 ) 8 = 6 x x = 8 hence total money = 6 * 8 = 48 | a ) $ 22 , b ) $ 24 , c ) $ 30 , d ) $ 36 , e ) $ 48 | e | multiply(divide(multiply(2, 8), subtract(8, 6)), 6) | multiply(n1,n2)|subtract(n2,n0)|divide(#0,#1)|multiply(n0,#2) | general |
according to a recent student poll , 3 / 4 out of 20 members of the finance club are interested in a career in investment banking . if two students are chosen at random , what is the probability that at least one of them is interested in investment banking ? | "15 students are interested , 5 are not interested prob = 1 - 5 c 2 / 20 c 2 = 1 - ( 5 * 4 / ( 20 * 19 ) ) = 1 - 1 / 19 = 18 / 19 answer : c" | a ) 1 / 14 , b ) 4 / 49 , c ) 18 / 19 , d ) 45 / 49 , e ) 13 / 14 | c | divide(subtract(choose(20, const_2), choose(subtract(20, multiply(20, divide(3, 4))), const_2)), choose(20, const_2)) | choose(n2,const_2)|divide(n0,n1)|multiply(n2,#1)|subtract(n2,#2)|choose(#3,const_2)|subtract(#0,#4)|divide(#5,#0)| | gain |
there are 4 red shoes & 4 green shoes . if two of red shoes are drawn what is the probability of getting red shoes | "taking 2 red shoe the probability is 4 c 2 from 8 shoes probability of taking 2 red shoes is 4 c 2 / 8 c 2 = 3 / 14 answer : c" | a ) 1 / 3 , b ) 1 / 14 , c ) 3 / 14 , d ) 1 / 15 , e ) 1 / 16 | c | divide(choose(4, const_2), choose(add(4, 4), const_2)) | add(n0,n1)|choose(n0,const_2)|choose(#0,const_2)|divide(#1,#2)| | probability |
ram , who is half as efficient as krish , will take 30 days to complete a task if he worked alone . if ram and krish worked together , how long will they take to complete the task ? | "number of days taken by ram to complete task = 30 since ram is half as efficient as krish , amount of work done by krish in 1 day = amount of work done by ram in 2 days if total work done by ram in 30 days is 30 w amount of work done by ram in 1 day = w amount of work done by krish in 1 day = 2 w total amount of work ... | a ) 16 days , b ) 10 days , c ) 8 days , d ) 6 days , e ) 18 days | b | inverse(add(divide(const_1, 30), divide(const_1, divide(30, const_2)))) | divide(const_1,n0)|divide(n0,const_2)|divide(const_1,#1)|add(#0,#2)|inverse(#3)| | physics |
of the final grades received by the students in a certain math course , 1 / 5 are a ' s , 1 / 4 are b ' s , 1 / 2 are c ' s , and the remaining 5 grades are d ' s . what is the number of students in the course ? | "we start by creating a variable for the total number of students in the math course . we can say : t = total number of students in the math course next , we can use variable t in an equation that we translate from the given information . we are given that , of the final grades received by the students in a certain mat... | a ) 80 , b ) 110 , c ) 160 , d ) 100 , e ) 400 | d | divide(5, subtract(1, add(add(divide(1, 5), divide(1, 4)), divide(1, 2)))) | divide(n0,n1)|divide(n0,n3)|divide(n0,n5)|add(#0,#1)|add(#3,#2)|subtract(n0,#4)|divide(n6,#5)| | general |
the captain of a cricket team of 11 members is 26 years old and the wicket keeper is 11 years older . if the ages of these two are excluded , the average age of the remaining players is one year less than the average age of the whole team . what is the average age of the team ? | "explanation let the average age of the whole team by x years . 11 x Γ’ β¬ β ( 26 + 40 ) = 9 ( x - 1 ) 11 x Γ’ β¬ β 9 x = 54 2 x = 54 x = 27 . so , average age of the team is 27 years . answer e" | a ) 23 years , b ) 24 years , c ) 25 years , d ) 26 years , e ) 27 years | e | divide(subtract(add(26, add(26, 11)), multiply(11, 11)), const_2) | add(n1,n2)|multiply(n2,n2)|add(n1,#0)|subtract(#2,#1)|divide(#3,const_2)| | general |
the area of one square is x ^ 2 + 12 x + 36 and the area of another square is 4 x ^ 2 β 12 x + 9 . if the sum of the perimeters of both squares is 64 , what is the value of x ? | spotting the pattern of equations both are in form of ( x + c ) ^ 2 so a 1 = ( x + 6 ) ^ 2 a 2 = ( 2 x - 3 ) ^ 2 l 1 = x + 6 l 2 = 2 x - 3 p 1 = 4 ( x + 6 ) p 2 = 4 ( 2 x - 3 ) p 1 + p 2 = 64 4 ( x + 6 ) + 4 ( 2 x - 3 ) = 64 . . . . . . . . . . . . . . > x = 4.3 answer : b | a ) 0 , b ) 4.3 , c ) 2.5 , d ) 4.67 , e ) 10 | b | divide(subtract(64, subtract(multiply(const_4, divide(12, const_2)), 12)), 12) | divide(n1,const_2)|multiply(#0,const_4)|subtract(#1,n1)|subtract(n7,#2)|divide(#3,n1) | general |
in one hour , a boat goes 6 km along the stream and 2 km against the stream . the sped of the boat in still water ( in km / hr ) is : | "solution speed in still water = 1 / 2 ( 6 + 2 ) km / hr = 4 kmph . answer a" | a ) 4 , b ) 5 , c ) 8 , d ) 9 , e ) 10 | a | divide(add(6, 2), const_2) | add(n0,n1)|divide(#0,const_2)| | gain |
on the first day of her vacation , louisa traveled 200 miles . on the second day , traveling at the same average speed , she traveled 350 miles . if the 200 - mile trip took 3 hours less than the 350 - mile trip , what was the average speed , in miles per hour ? | "( time ) * ( rate ) = ( distance ) - - > ( rate ) = ( distance ) / ( time ) - - > given : ( rate ) = 200 / t = 350 / ( t + 3 ) - - > 4 / t = 7 / ( t + 3 ) - - > 4 t + 12 = 7 t - - - - > 3 t = 12 . t = 4 - - - - > ( rate ) = 200 / 4 = 50 answer : e" | a ) 45 , b ) 55 , c ) 60 , d ) 65 , e ) 50 | e | divide(subtract(350, 200), 3) | subtract(n1,n0)|divide(#0,n3)| | physics |
while working alone at their respective constant rates , server g uploads 480 files in 4 hours and server y uploads 480 files in 8 hours . if all files uploaded by these servers are the same size , how long would it take the two servers , working at the same time and at their respective constant rates , to process a to... | server g processes 480 / 4 files per hour = 120 per hour server y processes 180 / 8 files per hour = 60 per hour total files processed per hour when g and y work together = 120 + 60 per hour = 180 files per hour 480 / 180 = 2 2 / 3 hours = c | a ) 2 hr , b ) 2 hr 20 min , c ) 2 hr 40 min , d ) 5 hr 40 min , e ) 6 hr | c | add(multiply(const_2, const_100), multiply(const_60, subtract(divide(480, add(divide(480, 4), divide(480, 8))), const_2))) | divide(n0,n1)|divide(n0,n3)|multiply(const_100,const_2)|add(#0,#1)|divide(n0,#3)|subtract(#4,const_2)|multiply(#5,const_60)|add(#2,#6) | physics |
two numbers are respectively 40 % and 60 % more than a third number . what percentage is the first of the second ? | "required answer = ( 100 + x ) / ( 100 + y ) * 100 = 140 * 100 / 160 = 87.5 % answer is b" | a ) 72 % , b ) 87.5 % , c ) 63.7 % , d ) 56.9 % , e ) 90 % | b | subtract(const_100, multiply(divide(add(40, const_100), add(60, const_100)), const_100)) | add(n0,const_100)|add(n1,const_100)|divide(#0,#1)|multiply(#2,const_100)|subtract(const_100,#3)| | gain |
if rs . 578 be divided among a , b , c in such a way that a gets 2 / 3 of what b gets and b gets 1 / 4 of what c gets , then their shares are respectively ? | "( a = 2 / 3 b and b = 1 / 4 c ) = a / b = 2 / 3 and b / c = 1 / 4 a : b = 2 : 3 and b : c = 1 : 4 = 3 : 12 a : b : c = 2 : 3 : 12 a ; s share = 578 * 2 / 17 = rs . 68 b ' s share = 578 * 3 / 17 = rs . 102 c ' s share = 578 * 12 / 17 = rs . 408 . answer : b" | a ) s . 300 , b ) s . 408 , c ) s . 389 , d ) s . 368 , e ) s . 323 | b | divide(578, add(add(multiply(divide(2, 3), divide(1, 4)), divide(1, 4)), 1)) | divide(n3,n4)|divide(n1,n2)|multiply(#1,#0)|add(#0,#2)|add(#3,n3)|divide(n0,#4)| | general |
the ratio between the number of sheep and the number of horses at the stewar farm is 6 to 7 . if each of horse is fed 230 ounces of horse food per day and the farm needs a total 12880 ounces of horse food per day . what is number sheep in the form ? ? | "et no of sheep and horses are 6 k and 7 k no of horses = 12880 / 230 = 56 now 7 k = 56 and k = 8 no of sheep = ( 6 * 8 ) = 48 answer : d" | a ) 18 , b ) 28 , c ) 32 , d ) 48 , e ) 58 | d | multiply(divide(divide(12880, 230), 7), 6) | divide(n3,n2)|divide(#0,n1)|multiply(n0,#1)| | other |
a man swims downstream 40 km and upstream 56 km taking 8 hours each time , what is the speed of the man in still water ? | "40 - - - 8 ds = 5 ? - - - - 1 56 - - - - 8 us = 7 ? - - - - 1 m = ? m = ( 5 + 7 ) / 2 = 6 answer : a" | a ) 6 , b ) 24 , c ) 8 , d ) 12 , e ) 44 | a | divide(add(divide(56, 8), divide(40, 8)), const_2) | divide(n1,n2)|divide(n0,n2)|add(#0,#1)|divide(#2,const_2)| | physics |
4 / 6 of the population of the country of venezia lives in montague province , while the rest lives in capulet province . in the upcoming election , 80 % of montague residents support romeo , while 70 % of capulet residents support juliet ; each resident of venezia supports exactly one of these two candidates . rounded... | "total population = 60 ( assume ) . 4 / 6 * 60 = 40 people from montague . 2 / 6 * 60 = 20 people from capulet . 0.2 * 40 = 8 people from montague support juliet . 0.7 * 20 = 14 people from capulet support juliet . the probability that a juliet supporter chosen at random resides in capulet is 14 / ( 8 + 14 ) = ~ 64 . a... | a ) 28 % , b ) 41 % , c ) 45 % , d ) 64 % , e ) 78 % | d | multiply(divide(multiply(divide(70, const_100), divide(subtract(6, 4), 6)), add(multiply(divide(subtract(const_100, 80), const_100), divide(4, 6)), multiply(divide(70, const_100), divide(subtract(6, 4), 6)))), const_100) | divide(n3,const_100)|divide(n0,n1)|subtract(n1,n0)|subtract(const_100,n2)|divide(#2,n1)|divide(#3,const_100)|multiply(#0,#4)|multiply(#5,#1)|add(#7,#6)|divide(#6,#8)|multiply(#9,const_100)| | gain |
if a certain number x is divided by 82 , the reminder is 5 . what is the reminder when x + 3 is divided by 41 ? | "x can be written as 82 k + 5 or x = 5 , 87,169 , etc . x + 3 = 82 k + 5 + 3 = 82 k + 8 or x + 3 = 8 , 90,172 etc . when divided by 41 , we will get the remainder 8 . e" | a ) 3 , b ) 5 , c ) 6 , d ) 16 , e ) 18 | e | add(5, 3) | add(n1,n2)| | general |
alfred buys an old scooter for $ 4400 and spends $ 800 on its repairs . if he sells the scooter for $ 5800 , his gain percent is ? | "c . p . = 4400 + 800 = $ 5200 s . p . = $ 5800 gain = 5800 - 5200 = $ 600 gain % = 600 / 5200 * 100 = 11.5 % answer is e" | a ) 5.45 % , b ) 6.23 % , c ) 7 % , d ) 8.12 % , e ) 11.5 % | e | multiply(divide(subtract(5800, add(4400, 800)), add(4400, 800)), const_100) | add(n0,n1)|subtract(n2,#0)|divide(#1,#0)|multiply(#2,const_100)| | gain |
a 1100 m long train crosses a tree in 110 sec , how much time will it take to pass a platform 700 m long ? | l = s * t s = 1100 / 110 s = 10 m / sec . total length ( d ) = 1800 m t = d / s t = 1800 / 10 t = 180 sec answer : d | a ) 288 , b ) 190 , c ) 188 , d ) 180 , e ) 12 | d | divide(add(1100, 700), divide(1100, 110)) | add(n0,n2)|divide(n0,n1)|divide(#0,#1) | physics |
if rs . 440 amount to rs . 540 in 4 years , what will it amount to in 6 years at the same rate % per annum ? | "80 = ( 440 * 4 * r ) / 100 r = 5.68 % i = ( 440 * 6 * 5.68 ) / 100 = 150 440 + 150 = 590 answer : c" | a ) s . 575 , b ) s . 595 , c ) s . 590 , d ) s . 580 , e ) s . 585 | c | subtract(multiply(subtract(540, 440), 6), subtract(540, 440)) | subtract(n1,n0)|multiply(n3,#0)|subtract(#1,#0)| | gain |
what is the remainder when 43 ^ 92 is divided by 5 ? | "the units digit of the exponents of 3 cycle in a group of 4 : { 3 , 9 , 7 , 1 } 92 has the form 4 k so the units digit of 43 ^ 92 is 1 . the remainder when dividing by 5 is 1 . the answer is b ." | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | b | subtract(divide(5, const_2), multiply(43, 43)) | divide(n2,const_2)|multiply(n0,n0)|subtract(#0,#1)| | general |
the average weight of 29 students is 28 kg . by the admission of a new student , the average weight is reduced to 27.8 kg . the weight of the new student is | "the total weight of 29 students = 29 * 28 the total weight of 30 students = 30 * 27.8 weight of the new student = ( 30 * 27.8 β 29 * 28 ) = 834 - 812 = 22 answer : a" | a ) 22 kg , b ) 21.6 kg , c ) 22.4 kg , d ) 21 kg , e ) 23 kg | a | subtract(multiply(add(29, const_1), 27.8), multiply(29, 28)) | add(n0,const_1)|multiply(n0,n1)|multiply(n2,#0)|subtract(#2,#1)| | general |
in the quadratic equation ax 2 - x - 12 = 0 , if the sum of two roots is 1 , what is the product of the two roots ? | explanation : the sum of the roots of the quadratic equation ax 2 + bx + c = 0 are ( - b / a ) and the product of the roots are ( c / a ) . thus , in the equation ax 2 - 11 x + 40 = 0 , where a = a , b = - 1 and c = - 12 . we get , sum of the roots = 1 / a = 1 a = 1 product of the roots = - 12 / 1 = - 12 answer : a | a ) - 12 , b ) - 6 , c ) 12 , d ) 6 , e ) 24 | a | multiply(divide(12, const_3), negate(divide(12, const_4))) | divide(n1,const_3)|divide(n1,const_4)|negate(#1)|multiply(#0,#2) | general |
a book is bought for $ 60 and sold for $ 63 . what is the profit in percentage ? | "63 / 60 = 1.05 the answer is b ." | a ) 3 , b ) 5 , c ) 7 , d ) 9 , e ) 11 | b | multiply(divide(subtract(63, 60), 60), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
for any integer n greater than 1 , n * denotes the product of all the integers from 1 to n , inclusive . how many prime numbers e are there between 6 * + 2 and 6 * + 6 , inclusive ? | "given that n * denotes the product of all the integers from 1 to n , inclusive so , 6 * + 2 = 6 ! + 2 and 6 * + 6 = 6 ! + 6 . now , notice that we can factor out 2 our of 6 ! + 2 so it can not be a prime number , we can factor out 3 our of 6 ! + 3 so it can not be a prime number , we can factor out 4 our of 6 ! + 4 so... | a ) none , b ) one , c ) two , d ) three , e ) four | a | divide(add(factorial(6), 6), add(factorial(6), 6)) | factorial(n2)|add(n2,#0)|divide(#1,#1)| | general |
in the science city , kolkata the rate of the ticket is increased by 50 % to increased the revenue but simultaneously 20 % of the visitor decreased . what is percentage change in the revenue . if it is known that the science city collects one revenue only from the visitors and it has no other financial supports : | solution : let the initial revenue be 100 . 100 - - - - - 50 % β ( ticket up ) - - - - - > 150 - - - - - 20 % β ( visitors down ) - - - - - > 120 . there is 20 % increase in the revenue . answer : option a | a ) + 20 % , b ) - 25 % , c ) + 305 , d ) - 30 % , e ) can not determined | a | divide(subtract(multiply(add(const_100, 50), subtract(const_100, 20)), multiply(const_100, const_100)), const_100) | add(n0,const_100)|multiply(const_100,const_100)|subtract(const_100,n1)|multiply(#0,#2)|subtract(#3,#1)|divide(#4,const_100) | gain |
81 * 82 * 83 * 84 * 85 * 86 * 87 * 89 . what should be in the unit place in this product ? | it should be zero in the unit place in this product . 85 * 82 will give zero in unit place . answer : a | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | a | multiply(add(const_3, const_4), const_2) | add(const_3,const_4)|multiply(#0,const_2) | general |
how many zeros does 500 ! end with ? | "according to above 500 ! has 500 / 5 + 500 / 25 + 500 / 125 = 100 + 20 + 4 = 124 trailing zeros . answer : c ." | a ) 201 , b ) 240 , c ) 124 , d ) 125 , e ) 180 | c | add(add(divide(500, add(const_4, const_1)), divide(subtract(500, add(const_4, const_1)), power(add(const_4, const_1), const_2))), divide(subtract(500, add(const_4, const_1)), power(add(const_4, const_1), const_3))) | add(const_1,const_4)|divide(n0,#0)|power(#0,const_2)|power(#0,const_3)|subtract(n0,#0)|divide(#4,#2)|divide(#4,#3)|add(#1,#5)|add(#7,#6)| | other |
if the sum of two positive integers is 12 and the difference of their squares is 48 , what is the product of the two integers ? | "let the 2 positive numbers x and y x + y = 12 - - 1 x ^ 2 - y ^ 2 = 48 = > ( x + y ) ( x - y ) = 48 - - 2 using equation 1 in 2 , we get = > x - y = 4 - - 3 solving equation 1 and 3 , we get x = 8 y = 4 product = 8 * 4 = 32 answer a" | a ) 32 , b ) 119 , c ) 128 , d ) 135 , e ) 143 | a | multiply(divide(subtract(12, divide(48, 12)), divide(48, 12)), add(divide(subtract(12, divide(48, 12)), divide(48, 12)), divide(48, 12))) | divide(n1,n0)|subtract(n0,#0)|divide(#1,#0)|add(#2,#0)|multiply(#3,#2)| | general |
john makes $ 50 a week from his job . he earns a raise and now makes $ 80 a week . what is the % increase ? | "increase = ( 30 / 50 ) * 100 = ( 3 / 5 ) * 100 = 60 % . e" | a ) 16 % , b ) 16.66 % , c ) 17.9 % , d ) 18.12 % , e ) 60 % | e | multiply(divide(subtract(80, 50), 50), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
a man can row downstream at 16 kmph and upstream at 10 kmph . find the speed of the man in still water and the speed of stream respectively ? | "let the speed of the man in still water and speed of stream be x kmph and y kmph respectively . given x + y = 16 - - - ( 1 ) and x - y = 10 - - - ( 2 ) from ( 1 ) & ( 2 ) 2 x = 26 = > x = 13 , y = 3 . answer : a" | a ) 3 , b ) 5 , c ) 6 , d ) 4 , e ) 9 | a | divide(divide(add(16, 10), const_2), const_2) | add(n0,n1)|divide(#0,const_2)|divide(#1,const_2)| | physics |
a pet store regularly sells pet food at a discount of 10 percent to 30 percent from the manufacturer β s suggested retail price . if during a sale , the store discounts an additional 20 percent from the discount price , what would be the lowest possible price of a container of pet food that had a manufacturer β s sugge... | "for retail price = $ 45 first maximum discounted price = 45 - 30 % of 45 = 45 - 13.5 = 31.5 price after additional discount of 20 % = 31.5 - 20 % of 31.5 = 31.5 - 6.3 = 25.2 answer : option d" | a ) $ 10.00 , b ) $ 11.20 , c ) $ 14.40 , d ) $ 25.20 , e ) $ 18.00 | d | multiply(divide(subtract(const_100, 20), const_100), multiply(divide(subtract(const_100, 30), const_100), 45.00)) | subtract(const_100,n2)|subtract(const_100,n1)|divide(#0,const_100)|divide(#1,const_100)|multiply(n3,#3)|multiply(#2,#4)| | gain |
in a group of 150 people , 90 have an age of more 30 years , and the others have an age of less than 20 years . if a person is selected at random from this group , what is the probability the person ' s age is less than 20 ? | "number of people whose age is less than 20 is given by 150 - 90 = 60 probability p that a person selected at random from the group is less than 20 is given by 60 / 150 = 0.4 correct answer a" | a ) 0.4 , b ) 0.5 , c ) 0.65 , d ) 0.75 , e ) 0.85 | a | divide(subtract(150, 90), 150) | subtract(n0,n1)|divide(#0,n0)| | general |
the sum of the first 50 positive even integers is 2550 . what is the sum of the even integers from 202 to 300 inclusive ? | "2 + 4 + 6 + 8 + . . . + 100 = 2550 202 + 204 + . . . + 300 = 50 ( 200 ) + ( 2 + 4 + . . . + 100 ) = 10,000 + 2550 = 12,550 the answer is d ." | a ) 5,150 , b ) 7,550 , c ) 10,150 , d ) 12,550 , e ) 20,150 | d | multiply(divide(add(300, 202), const_2), add(divide(subtract(300, 202), const_2), const_1)) | add(n2,n3)|subtract(n3,n2)|divide(#1,const_2)|divide(#0,const_2)|add(#2,const_1)|multiply(#4,#3)| | general |
the average weight of 8 person ' s increases by 3 kg when a new person comes in place of one of them weighing 65 kg . what might be the weight of the new person ? | "total weight increased = ( 8 x 3 ) kg = 24 kg . weight of new person = ( 65 + 24 ) kg = 89 kg . d )" | a ) 60 kg , b ) 70 kg , c ) 80 kg , d ) 89 kg , e ) 95 kg | d | add(multiply(8, 3), 65) | multiply(n0,n1)|add(n2,#0)| | general |
a dishonest dealer claims to sell a product at its cost price . he uses a counterfeit weight which is 20 % less than the real weight . further greed overtook him and he added 15 % impurities to the product . find the net profit percentage of the dealer ? | the dealer uses weight which is 20 % less than the real weight . or ( 1 - 1 / 5 ) or 4 / 5 of real weight . it means that he is selling $ 4 worth of product for $ 5 . the dealer then further added 20 % impurities to the product . it means that he is selling $ 5 worth of product for $ 5.75 . so his profit is $ 5.75 - $ ... | a ) 46.85 % , b ) 43.75 % , c ) 50 % , d ) 56.25 % , e ) 36 % | b | multiply(subtract(divide(add(const_100, 15), subtract(const_100, 20)), const_1), const_100) | add(n1,const_100)|subtract(const_100,n0)|divide(#0,#1)|subtract(#2,const_1)|multiply(#3,const_100) | general |
if it is assumed that 65 percent of those who receive a questionnaire by mail will respond and 300 responses are needed , what is the minimum number of questionnaires that should be mailed ? | "minimum no of mail to be sent for getting 300 responses at 65 % = 300 / 0.65 = 461.5 option c" | a ) 400 , b ) 420 , c ) 461.5 , d ) 500 , e ) 600 | c | divide(300, divide(65, const_100)) | divide(n0,const_100)|divide(n1,#0)| | gain |
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