Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 164 | correct stringclasses 5 values | annotated_formula stringlengths 7 1.65k | linear_formula stringlengths 8 925 | category stringclasses 6 values | answer stringclasses 5 values |
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if 9 a - b = 10 b + 65 = - 12 b - 2 a , what is the value of 11 a + 11 b ? | "( i ) 9 a - 11 b = 65 ( ii ) 2 a + 22 b = - 65 adding ( i ) and ( ii ) : 11 a + 11 b = 0 the answer is c ." | a ) - 11 , b ) - 5 , c ) 0 , d ) 5 , e ) 11 | c | multiply(negate(multiply(divide(65, 2), 2)), 11) | divide(n2,n4)|multiply(#0,n4)|negate(#1)|multiply(n5,#2)| | general | C |
elena purchased brand x pens for $ 4.00 apiece and brand y for $ 2.20 apiece . if elena purchased a total of 12 of these pens for $ 42.00 , how many brand x pens did she purchase ? | 4 x + 2.8 y = 42 - - > multiply by 2.5 ( to get the integers ) - - > 10 x + 7 y = 105 - - > only one positive integers solutions x = 6 and y = 5 ( how to solve : 7 y must have the last digit of 5 in order the last digit of the sum to be 5 ) . answer : c . | a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | c | subtract(multiply(4, 12), 42) | multiply(n0,n2)|subtract(#0,n3) | general | C |
a retailer bought a machine at a wholesale price of $ 117 and later on sold it after a 10 % discount of the retail price . if the retailer made a profit equivalent to 20 % of the whole price , what is the retail price of the machine ? | "my solution : wholesale price = 117 retail price , be = x he provides 10 % discount on retail price = x - 10 x / 100 this retail price = 20 % profit on wholesale price x - 10 x / 100 = 117 + 1 / 5 ( 117 ) x = 156 ; answer : e" | a ) 81 , b ) 100 , c ) 120 , d ) 135 , e ) 159 | e | divide(multiply(add(117, divide(multiply(117, 20), const_100)), const_100), multiply(multiply(const_3, const_3), 10)) | multiply(n0,n2)|multiply(const_3,const_3)|divide(#0,const_100)|multiply(n1,#1)|add(n0,#2)|multiply(#4,const_100)|divide(#5,#3)| | gain | E |
express a speed of 56 kmph in meters per second ? | "56 * 5 / 18 = 15.56 mps answer : d" | a ) 10.56 mps , b ) 17.56 mps , c ) 97.56 mps , d ) 15.56 mps , e ) 18.56 mps | d | multiply(const_0_2778, 56) | multiply(n0,const_0_2778)| | physics | D |
a and b start walking towards each other at 1 pm at speed of 5 kmph and 7 kmph . they were initially 24 km apart . at what time do they meet ? | "time of meeting = distance / relative speed = 24 / 7 + 5 = 24 / 12 = 2 hrs after 1 pm = 3 pm answer is c" | a ) 4 pm , b ) 6 pm , c ) 3 pm , d ) 10 pm , e ) 5 pm | c | add(1, divide(24, add(5, 7))) | add(n1,n2)|divide(n3,#0)|add(n0,#1)| | physics | C |
the sum of squares of 3 numbers is 764 and sum of their products taken two at a time is 500 . what is their sum ? | description : let the numbers be x , y , z . = > x 2 + y 2 + z 2 = 764 = > 2 ( xy + yz + xz ) = 2 * 500 = 1000 = > ( x + y + z ) 2 = ( x 2 + y 2 + z 2 ) + 2 ( xy + yz + xz ) = 1764 = > ( x + y + z ) = sqrt ( 1764 ) = 42 answer b | a ) 48 , b ) 42 , c ) 52 , d ) 32 , e ) 33 | b | subtract(subtract(divide(divide(500, 3), 3), const_10), const_3) | divide(n2,n0)|divide(#0,n0)|subtract(#1,const_10)|subtract(#2,const_3) | general | B |
in an examination , 25 % of total students failed in hindi , 48 % failed in english and 27 % in both . the percentage of these who passed in both the subjects is : | "explanation : formula n ( a Γ’ Λ Βͺ b ) = n ( a ) + n ( b ) Γ’ Λ β n ( a Γ’ Λ Β© b ) fail in hindi or english = 25 + 48 Γ’ β¬ β 27 = 46 therefore students who passed = 100 Γ’ β¬ β 46 = 54 . answer : b" | a ) 23 , b ) 54 , c ) 28 , d ) 40 , e ) 81 | b | subtract(const_100, subtract(add(25, 48), 27)) | add(n0,n1)|subtract(#0,n2)|subtract(const_100,#1)| | general | B |
find the c . i . on a sum of rs . 10000 for 6 months at 25 % per annum , interest being compounded quarterly ? | c . i . = 2000 ( 21 / 20 ) ^ 2 - 1800 = 1289 answer : c | a ) 10290 , b ) 5290 , c ) 1289 , d ) 1290 , e ) 2290 | c | subtract(multiply(10000, multiply(add(const_1, divide(const_0_25, const_4)), add(const_1, divide(const_0_25, const_4)))), 10000) | divide(const_0_25,const_4)|add(#0,const_1)|multiply(#1,#1)|multiply(n0,#2)|subtract(#3,n0) | gain | C |
a man , a woman and a boy can together complete a piece of work in 3 days . if a man alone can do it in 6 days and a boy alone in 12 days , how long will a woman take to complete the work ? | "explanation : ( 1 man + 1 woman + 1 boy ) β s 1 day β s work = 1 / 3 1 man β s 1 day work = 1 / 6 1 boy β s 1 day β s work = 1 / 1 ( 1 man + 1 boy ) β s 1 day β s work = 1 / 6 + 1 / 12 = 1 / 4 therefore , 1 woman β s 1 day β s work = 1 / 3 β 1 / 4 = 1 / 12 therefore , the woman alone can finish the work in 12 days . answer : option a" | a ) 12 days , b ) 10 days , c ) 9 days , d ) 8 days , e ) 11 days | a | inverse(subtract(inverse(3), add(inverse(6), inverse(12)))) | inverse(n0)|inverse(n1)|inverse(n2)|add(#1,#2)|subtract(#0,#3)|inverse(#4)| | physics | A |
a car gets 40 kilometers per gallon of gasoline . how many gallons of gasoline would the car need to travel 140 kilometers ? | "each 40 kilometers , 1 gallon is needed . we need to know how many 40 kilometers are there in 140 kilometers ? 140 Γ· 40 = 3.5 Γ 1 gallon = 3.5 gallons correct answer e" | a ) 4.5 , b ) 5.5 , c ) 6.5 , d ) 7.5 , e ) 3.5 | e | divide(140, 40) | divide(n1,n0)| | physics | E |
the sum of the non - prime numbers between 60 and 70 , non - inclusive , is | "sum of consecutive integers from 61 to 69 , inclusive = = = = > ( a 1 + an ) / 2 * # of terms = ( 61 + 69 ) / 2 * 9 = 65 * 9 = 585 sum of non - prime numbers b / w 60 and 70 , non inclusive = = = > 585 - 128 ( i . e . , 61 + 67 , being the prime # s in the range ) = 457 answer : c" | a ) 387 , b ) 429 , c ) 457 , d ) 499 , e ) 536 | c | add(add(add(add(add(add(add(60, const_1), add(add(60, const_1), const_1)), add(add(add(60, const_1), const_1), const_2)), add(add(add(add(60, const_1), const_1), const_2), const_1)), add(add(add(add(add(60, const_1), const_1), const_2), const_1), const_1)), add(add(add(add(add(add(60, const_1), const_1), const_2), const_1), const_1), const_1)), add(add(add(add(add(add(add(60, const_1), const_1), const_2), const_1), const_1), const_1), const_1)) | add(n0,const_1)|add(#0,const_1)|add(#0,#1)|add(#1,const_2)|add(#2,#3)|add(#3,const_1)|add(#4,#5)|add(#5,const_1)|add(#6,#7)|add(#7,const_1)|add(#8,#9)|add(#9,const_1)|add(#10,#11)| | general | C |
john makes $ 60 a week from his job . he earns a raise andnow makes $ 68 a week . what is the % increase ? | "increase = ( 8 / 60 ) * 100 = ( 2 / 15 ) * 100 = 13.33 % . e" | a ) 16.12 % , b ) 16.66 % , c ) 16.56 % , d ) 17.66 % , e ) 13.33 % | e | multiply(divide(subtract(68, 60), 60), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain | E |
a sports retailer ordered white and yellow tennis balls in equal number but the dispatch clerk of the tennis ball company erred and dispatched 70 extra yellow balls and making ratio of white balls to yellow balls 8 / 13 . how many tennis balls did the retailer order originally . | "white : yellow = x : ( x + 70 ) = 8 : 13 - - > 13 x = 8 x + 560 - - > x = 112 . the total # of balls originally x + x = 112 + 112 = 224 . answer : e ." | a ) 180 , b ) 130 , c ) 140 , d ) 160 , e ) 224 | e | multiply(divide(multiply(8, 70), subtract(13, 8)), const_2) | multiply(n0,n1)|subtract(n2,n1)|divide(#0,#1)|multiply(#2,const_2)| | general | E |
6 % of customers that enter a store on any given day must not pay tax . if 1000 people shop every day , how many people pay taxes in the store every week | 6 % of customers do not pay taxes . 6 / 100 * 1000 customers do not pay taxes . 6 / 100 * 1000 = 60 therefore 1000 - 60 customers do pay taxes . 1000 - 60 = 940 940 * 7 days per week = 6580 b ) | a ) 6500 , b ) 6580 , c ) 7200 , d ) 7120 , e ) 6800 | b | multiply(add(6, const_1), subtract(1000, multiply(1000, divide(6, const_100)))) | add(n0,const_1)|divide(n0,const_100)|multiply(n1,#1)|subtract(n1,#2)|multiply(#0,#3) | gain | B |
if all the 6 are replaced by 9 , then the algebraic sum of all the numbers from 1 to 100 ( both inclusive ) varies by | 1 11 21 . . . . . . . 91 2 12 22 . . . . . . . . 92 3 13 23 . . . . . . . . 93 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . there are total 20 occurrence of 6 . once place at 6 th row in every column and tens place in 6 th column and diff of 6 and 9 is 3 so 3 * 1 * 10 + 3 * 10 * 10 = 330 answer : a | a ) 330 , b ) 340 , c ) 350 , d ) 360 , e ) 370 | a | add(multiply(multiply(subtract(9, 6), const_10), const_10), multiply(subtract(9, 6), const_10)) | subtract(n1,n0)|multiply(#0,const_10)|multiply(#1,const_10)|add(#2,#1) | other | A |
a number with an interesting property : when i divide it by 2 , the remainder is 1 . when i divide it by 3 , the remainder is 2 . when i divide it by 4 , the remainder is 3 . when i divide it by 5 , the remainder is 4 . when i divide it by 6 , the remainder is 5 . when i divide it by 7 , the remainder is 6 . when i divide it by 8 , the remainder is 7 . when i divide it by 9 , the remainder is 8 . when i divide it by 10 , the remainder is 9 . it ' s not a small number , but it ' s not really big , either . when i looked for a smaller number with this property i could n ' t find one . can you find it ? | d 2519 the number has to end in 9 . looked brute force for small numbers . 59 and 119 were promising , but no cigar . then looked for agreement among 39 + multiples of 40 , 69 + multiples of 70 and 89 + multiples of 90 smallest one was 2519 . | a ) 2588 , b ) 2474 , c ) 2254 , d ) 2519 , e ) 8725 | d | subtract(multiply(multiply(lcm(lcm(9, 10), 7), 2), 2), const_1) | lcm(n14,n16)|lcm(n10,#0)|multiply(n0,#1)|multiply(n0,#2)|subtract(#3,const_1) | gain | D |
one day a car rental agency rented 3 / 4 of its cars , including 4 / 5 of its cars with cd players . if 3 / 5 of its cars have cd players , what percent of the cars that were not rented had cd players ? | "the cars with cd players which were not rented is ( 1 / 5 ) ( 3 / 5 ) = 3 / 25 of all the cars . the cars which were not rented is 1 / 4 of all the cars . the percent of non - rented cars which had cd players is ( 3 / 25 ) / ( 1 / 4 ) = 12 / 25 the answer is c ." | a ) 23 / 40 , b ) 17 / 30 , c ) 12 / 25 , d ) 7 / 15 , e ) 3 / 10 | c | multiply(divide(subtract(divide(multiply(4, multiply(multiply(4, 4), 5)), 5), divide(multiply(divide(multiply(4, multiply(multiply(4, 4), 5)), 5), 4), 4)), subtract(multiply(multiply(4, 4), 5), divide(multiply(3, multiply(multiply(4, 4), 5)), 4))), const_100) | multiply(n1,n2)|multiply(n3,#0)|multiply(n1,#1)|multiply(#1,const_2)|divide(#2,n3)|divide(#3,n1)|multiply(n1,#4)|subtract(#1,#5)|divide(#6,n2)|subtract(#4,#8)|divide(#9,#7)|multiply(#10,const_100)| | general | C |
the s . i . on a certain sum of money for 6 years at 14 % per annum is half the c . i . on rs . 7000 for 2 years at 7 % per annum . the sum placed on s . i . is ? | c . i . = [ 7000 * ( 1 + 7 / 100 ) 2 - 7000 ] = ( 7000 * 11 / 10 * 11 / 10 - 7000 ) = rs . 1014.3 . sum = ( 507.15 * 100 ) / ( 6 * 14 ) = rs . 603.75 answer : a | a ) 603.75 , b ) 555.75 , c ) 569.55 , d ) 256.25 , e ) 563.23 | a | divide(divide(subtract(multiply(7000, power(add(const_1, divide(7, const_100)), 2)), 7000), 2), multiply(6, divide(14, const_100))) | divide(n4,const_100)|divide(n1,const_100)|add(#0,const_1)|multiply(n0,#1)|power(#2,n3)|multiply(n2,#4)|subtract(#5,n2)|divide(#6,n3)|divide(#7,#3)| | gain | A |
the present population of a town is 4032 . population increase rate is 20 % p . a . find the population of town before 2 years ? | "p = 4032 r = 20 % required population of town = p / ( 1 + r / 100 ) ^ t = 4032 / ( 1 + 20 / 100 ) ^ 2 = 4032 / ( 6 / 5 ) ^ 2 = 2800 ( approximately ) answer is b" | a ) 2500 , b ) 2800 , c ) 3500 , d ) 3600 , e ) 2050 | b | add(4032, divide(multiply(4032, 20), const_100)) | multiply(n0,n1)|divide(#0,const_100)|add(n0,#1)| | gain | B |
a shop owner sells 50 mtr of cloth and gains sp of 10 mtrs . find the gain % ? | here , selling price of 10 m cloth is obtained as profit . profit of 10 m cloth = ( s . p . of 50 m cloth ) β ( c . p . of 50 m cloth ) selling price of 40 m cloth = selling price of 50 m of cloth let cost of each metre be rs . 100 . therefore , cost price of 40 m cloth = rs . 4000 and s . p . of 40 m cloth = rs . rs . 5000 profit % = 10 / 40 Γ 100 = 25 % profit of 25 % was made by the merchant . a | a ) 25 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 70 % | a | multiply(divide(10, subtract(50, 10)), const_100) | subtract(n0,n1)|divide(n1,#0)|multiply(#1,const_100) | gain | A |
if the population of a certain country increases at the rate of two person every 60 seconds , by how many persons does the population increase in 100 minutes ? | "answer = 2 * 100 = 200 answer = e" | a ) 120 , b ) 150 , c ) 240 , d ) 220 , e ) 200 | e | multiply(divide(const_60, 60), 100) | divide(const_60,n0)|multiply(n1,#0)| | physics | E |
two pipes can fill a cistern in 14 and 16 hours respectively . the pipes are opened simultaneously and it is found that due to leakage in the bottom , 32 minutes extra are taken for the cistern to be filled up . if the cistern is full , in what time would the leak empty it ? | cistern filled by both pipes in one hour = 1 β 14 + 1 β 16 = 15 β 112 th β΄ both pipes filled the cistern in 112 β 15 hrs . now , due to leakage both pipes filled the cistern in 112 β 15 + 32 β 60 = 8 hrs . β΄ due to leakage , filled part in one hour 1 β 8 β΄ part of cistern emptied , due to leakage in one hour = 15 β 112 - 1 β 8 = 1 β 112 th β΄ in 112 hr , the leakage would empty the cistern . answer b | a ) 110 hours , b ) 112 hours , c ) 115 hours , d ) 100 hours , e ) none of these | b | divide(multiply(divide(multiply(14, 16), const_2), subtract(multiply(divide(add(14, 16), const_2), divide(16, const_2)), divide(multiply(14, 16), const_2))), subtract(multiply(divide(add(14, 16), const_2), divide(16, const_2)), divide(multiply(14, 16), const_2))) | add(n0,n1)|divide(n1,const_2)|multiply(n0,n1)|divide(#2,const_2)|divide(#0,const_2)|multiply(#4,#1)|subtract(#5,#3)|multiply(#3,#6)|divide(#7,#6) | physics | B |
a 30 % alcohol mixture is added to a 50 % alcohol mixture to form a 10 litre mixture of 45 % alcohol . how much of the 30 % mixture was used ? | solution a = 50 % solution solution b = 30 % solution 30 % - 45 % = 15 % solution b 50 % - 45 % = 5 % solution a so the ratio is 3 : 1 for 30 % : 50 % solutions 3 / 4 * 10 liter = 7.5 for 30 % solution and 2.5 for 50 % solution . . answer : b | a ) 1.5 liters , b ) 2.5 liters , c ) 3.5 liters , d ) 4.5 liters , e ) 5.5 liters | b | divide(multiply(divide(50, const_100), 10), const_2) | divide(n1,const_100)|multiply(n2,#0)|divide(#1,const_2) | general | B |
two employees m and n are paid a total of $ 605 per week by their employer . if m is paid 120 percent of the salary paid to n , how much is n paid per week ? | "1.2 n + n = 605 2.2 n = 605 n = 275 the answer is d ." | a ) $ 245 , b ) $ 255 , c ) $ 265 , d ) $ 275 , e ) $ 285 | d | divide(605, add(divide(120, const_100), const_1)) | divide(n1,const_100)|add(#0,const_1)|divide(n0,#1)| | general | D |
if the average of 201 , 202 , 204 , 205 , 206 , 209 , 209 , 210 , 212 and x is 207 , what is the value of x ? | sum of the deviations of the numbers in the set from the mean is always zero 201 , 202 , 204 , 205 , 206 , 209 , 209 , 210 , 212 mean is 207 so the list is - 6 - 5 - 3 - 2 - 1 + 2 + 2 + 3 + 5 . . . this shud total to zero but this is - 5 , hence we need a number that is 5 more than the mean to get a + 5 and make it zero hence the answer is 207 + 5 = 212 d | a ) 207 , b ) 209 , c ) 211 , d ) 212 , e ) 213 | d | subtract(multiply(207, const_10), add(add(add(add(add(add(add(add(201, 202), 204), 205), 206), 209), 209), 210), 212)) | add(n0,n1)|multiply(n9,const_10)|add(n2,#0)|add(n3,#2)|add(n4,#3)|add(n5,#4)|add(n5,#5)|add(n7,#6)|add(n8,#7)|subtract(#1,#8) | general | D |
in an examination 35 % of the students passed and 351 failed . how many students appeared for the examination ? | let the number of students appeared be x then , 65 % of x = 351 65 x / 100 = 351 x = 351 * 100 / 65 = 540 answer is a | a ) a ) 540 , b ) b ) 400 , c ) c ) 700 , d ) d ) 650 , e ) e ) 840 | a | divide(multiply(351, const_100), subtract(const_100, 35)) | multiply(n1,const_100)|subtract(const_100,n0)|divide(#0,#1) | gain | A |
find the ratio in which wheat of inferior quality ( rs . 14 / kg ) be mixed with wheat of superior quality ( rs . 28 / kg ) so that the shopkeeper gains rs . 2 by selling the resulting mixture at rs . 20 / kg . | explanation : let the resulting mixture be 1 kg , and x kg be the amount of wheat of inferior quality . therefore , ( 1 - x ) kg is the amount of wheat of superior quality . as the shopkeeper gains rs . 2 , the cost of the mixture is rs . 18 14 * x + 28 * ( 1 - x ) = 18 14 x - 28 x + 28 = 18 14 x = 10 x = 5 / 7 ( 1 β x ) = 2 / 7 x : ( 1 - x ) = 5 / 7 : 2 / 7 = 5 : 2 answer : a | a ) 5 : 2 , b ) 5 : 9 , c ) 5 : 3 , d ) 5 : 1 , e ) 5 : 8 | a | divide(divide(divide(20, 2), const_2), 2) | divide(n3,n2)|divide(#0,const_2)|divide(#1,n2) | general | A |
a man sold 18 toys for rs . 25200 , gaining thereby the cost price of 3 toy find the cost price of a toy | let the cost of one toy = x . then , cost of 18 toys = 18 x . gain = 3 x . sp of 18 toys = rs . 25200 . gain = sp Γ’ β¬ β cp 3 x = 25200 Γ’ β¬ β 18 x 21 x = 25200 x = rs . 1200 . answer : option e | a ) s . 600 , b ) s . 800 , c ) s . 500 , d ) s . 900 , e ) s . 1200 | e | divide(25200, add(18, 3)) | add(n0,n2)|divide(n1,#0) | gain | E |
for any integer k > 1 , the term β length of an integer β refers to the number of positive prime factors , not necessarily distinct , whose product is equal to k . for example , if k = 24 , the length of k is equal to 4 , since 24 = 2 Γ 2 Γ 2 Γ 3 . if x and y are positive integers such that x > 1 , y > 1 , and x + 3 y < 960 , what is the maximum possible sum of the length of x and the length of y ? | "we know that : x > 1 , y > 1 , and x + 3 y < 960 , and it is given that length means no of factors . for any value of x and y , the max no of factors can be obtained only if factor is smallest no all factors are equal . hence , lets start with smallest no 2 . 2 ^ 1 = 2 2 ^ 2 = 4 2 ^ 3 = 8 2 ^ 4 = 16 2 ^ 5 = 32 2 ^ 6 = 64 2 ^ 7 = 128 2 ^ 8 = 256 2 ^ 9 = 512 2 ^ 10 = 1024 ( it exceeds 960 , so , x ca n ' t be 2 ^ 10 ) so , max value that x can take is 2 ^ 9 , for which has length of integer is 9 . now , since x = 512 , x + 3 y < 960 so , 3 y < 448 = = > y < 448 / 3 so , y can take any value which is less than 448 / 3 . and to get the maximum no of factors of smallest integer , we can say y = 2 ^ 7 for 2 ^ 7 has length of integer is 7 . so , combined together : 9 + 7 = 16 . b" | a ) 10 , b ) 16 , c ) 12 , d ) 14 , e ) 19 | b | add(add(4, 3), add(add(4, 4), 1)) | add(n2,n7)|add(n2,n2)|add(n0,#1)|add(#0,#2)| | general | B |
the instructions state that deepthi needs 4 / 17 square yards of one type of material and 3 / 10 square yards of another type of material for a project . she buys exactly that amount . after finishing the project , however , she has 9 / 30 square yards left that she did not use . what is the total amount of square yards of material deepthi used ? | total bought = 4 / 17 + 3 / 10 left part 9 / 30 - - - > 3 / 10 so used part 4 / 17 + 3 / 10 - 3 / 10 = 4 / 17 answer : b | a ) 1 / 12 , b ) 4 / 17 , c ) 2 / 3 , d ) 1 1 / 9 , e ) 2 1 / 9 | b | subtract(add(divide(4, 17), divide(const_3, const_10)), divide(9, 30)) | divide(n0,n1)|divide(const_3,const_10)|divide(n4,n5)|add(#0,#1)|subtract(#3,#2) | general | B |
how many times will the digit 7 be written when listing the integers from 1 to 1000 ? | another approach : in the range 0 - 100 : 7 as units digit - 10 times ( 7 , 17 , 27 , . . . , 97 ) ; 7 as tens digit - 10 time ( 71 , 72 , 73 , . . . , 79 ) ; so in first one hundred numbers 7 is written 10 + 10 = 20 times . in 10 hundreds 7 as units or tens digit will be written 10 * 20 = 200 times . plus 100 times when 7 is written as hundreds digit ( 700 , 701 , 702 , . . . , 799 ) . total 200 + 100 = 300 . answer : d . | a ) 110 , b ) 111 , c ) 271 , d ) 300 , e ) 304 | d | multiply(multiply(multiply(1, const_10), const_10), const_3) | multiply(n1,const_10)|multiply(#0,const_10)|multiply(#1,const_3) | general | D |
working at constant rate , pump x pumped out half of the water in a flooded basement in 3 hours . the pump y was started and the two pumps , working independently at their respective constant rates , pumped out rest of the water in 3 hours . how many hours would it have taken pump y , operating alone at its own constant rate , to pump out all of the water that was pumped out of the basement ? | "rate of x = 1 / 8 rate of x + y = 1 / 6 rate of y = 1 / 6 - 1 / 8 = 1 / 24 20 hours a" | a ) a . 20 , b ) b . 12 , c ) c . 14 , d ) d . 18 , e ) e . 24 | a | add(add(add(add(add(add(add(add(add(add(add(add(const_1, add(3, 3)), const_1), const_1), const_1), const_1), 3), const_1), const_1), const_1), const_1), const_1), const_1) | add(n0,n1)|add(#0,const_1)|add(#1,const_1)|add(#2,const_1)|add(#3,const_1)|add(#4,const_1)|add(#5,n0)|add(#6,const_1)|add(#7,const_1)|add(#8,const_1)|add(#9,const_1)|add(#10,const_1)|add(#11,const_1)| | physics | A |
a 6 by 8 rectangle is inscribed in circle . what is the circumference of the circle ? | "the diagonal of the rectangle will be the diameter of the circle . and perimeter = 2 * pi * r ans : b" | a ) 5 Ο , b ) 10 Ο , c ) 15 Ο , d ) 20 Ο , e ) 25 Ο | b | circumface(divide(sqrt(add(power(6, const_2), power(8, 8))), 8)) | power(n0,const_2)|power(n1,n1)|add(#0,#1)|sqrt(#2)|divide(#3,n1)|circumface(#4)| | geometry | B |
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 | C |
a merchant sells an item at a 20 % discount , but still makes a gross profit of 20 percent of the cost . what percent t of the cost would the gross profit on the item have been if it had been sold without the discount ? | "let the market price of the product is mp . let the original cost price of the product is cp . selling price ( discounted price ) = 100 % of mp - 20 % mp = 80 % of mp . - - - - - - - - - - - - - - - - ( 1 ) profit made by selling at discounted price = 20 % of cp - - - - - - - - - - - - - - ( 2 ) apply the formula : profit t = selling price - original cost price = > 20 % of cp = 80 % of mp - 100 % cp = > mp = 120 cp / 80 = 3 / 2 ( cp ) now if product is sold without any discount , then , profit = selling price ( without discount ) - original cost price = market price - original cost price = mp - cp = 3 / 2 cp - cp = 1 / 2 cp = 50 % of cp thus , answer should bec ." | a ) 20 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 75 % | c | subtract(const_100, subtract(subtract(const_100, 20), 20)) | subtract(const_100,n0)|subtract(#0,n1)|subtract(const_100,#1)| | gain | C |
if an average hard drive had a capacity of 0.7 tb in 2000 , and average hard drive capacities double every 5 years , what will be the average hard drive capacity in 2050 ? | 0.7 * 2 ^ 10 = 0.7 * 1024 = 716.8 the answer is e . | a ) 256 , b ) 512 , c ) 768 , d ) 1024 , e ) 7168 | e | multiply(add(multiply(multiply(multiply(multiply(multiply(multiply(multiply(multiply(multiply(multiply(0.7, const_2), const_2), const_2), const_2), const_2), const_2), const_2), const_2), const_2), const_2), 0.7), const_10) | multiply(n0,const_2)|multiply(#0,const_2)|multiply(#1,const_2)|multiply(#2,const_2)|multiply(#3,const_2)|multiply(#4,const_2)|multiply(#5,const_2)|multiply(#6,const_2)|multiply(#7,const_2)|multiply(#8,const_2)|add(n0,#9)|multiply(#10,const_10) | general | E |
a small island country of 5000 people has an emigration and immigration statistic per 500 persons of 50.4 and 15.4 respectively . how long will it take for the population to be half of its current number . | annual decrease is ( 500 + 15.4 - 50.4 ) * 10 = 4650 hence every year there is a decrease of 350 for the population to become half 2500 must emigrate therefore 2500 / 350 = 133.3 correct option is e ) 7.14 | a ) 6.8 , b ) 8.5 , c ) 6.1 , d ) 9.1 , e ) 7.14 | e | divide(divide(5000, const_2), multiply(subtract(50.4, 15.4), divide(5000, 500))) | divide(n0,const_2)|divide(n0,n1)|subtract(n2,n3)|multiply(#1,#2)|divide(#0,#3) | other | E |
if price of t . v set is reduced by 22 % , then its sale increases by 86 % , find net effect on sale value | "- a + b + ( ( - a ) ( b ) / 100 ) = - 22 + 86 + ( - 22 * 86 ) / 100 = - 22 + 86 - 19 = 45 answer : b" | a ) 44 , b ) 45 , c ) 46 , d ) 47 , e ) 48 | b | multiply(subtract(multiply(divide(subtract(const_100, 22), const_100), divide(add(const_100, 86), const_100)), const_1), const_100) | add(n1,const_100)|subtract(const_100,n0)|divide(#1,const_100)|divide(#0,const_100)|multiply(#2,#3)|subtract(#4,const_1)|multiply(#5,const_100)| | gain | B |
operation # is defined as adding a randomly selected two digit multiple of 10 to a randomly selected two digit prime number and reducing the result by half . if operation # is repeated 10 times , what is the probability that it will yield at least two integers ? | "any multiple of 10 is even . any two - digit prime number is odd . ( even + odd ) / 2 is not an integer . thus # does not yield an integer at all . therefore p = 0 . answer : a ." | a ) 0 % , b ) 10 % , c ) 20 % , d ) 30 % , e ) 40 % | a | divide(add(10, add(10, const_3)), const_2) | add(n1,const_3)|add(n0,#0)|divide(#1,const_2)| | general | A |
a car traveling at a certain constant speed takes 10 seconds longer to travel 1 km than it would take to travel 1 km at 80 km / hour . at what speed , in km / hr , is the car traveling ? | time to cover 1 kilometer at 80 kilometers per hour is 1 / 80 hours = 3,600 / 80 seconds = 45 seconds ; time to cover 1 kilometer at regular speed is 45 + 10 = 55 seconds = 55 / 3,600 hours = 1 / 65 hours ; so , we get that to cover 1 kilometer 1 / 65 hours is needed - - > regular speed 65 kilometers per hour ( rate is a reciprocal of time or rate = distance / time ) . answer : d | a ) 70 , b ) 72 , c ) 74 , d ) 65 , e ) 78 | d | divide(1, divide(add(multiply(const_3600, divide(1, 80)), 10), const_3600)) | divide(n1,n3)|multiply(#0,const_3600)|add(n0,#1)|divide(#2,const_3600)|divide(n1,#3) | physics | D |
in a school of 850 boys , 40 % of muslims , 28 % hindus , 10 % sikhs and the remaining of other communities . how many belonged to the other communities ? | "40 + 28 + 10 = 78 % 100 β 78 = 22 % 850 * 22 / 100 = 187 answer : c" | a ) a ) 125 , b ) b ) 627 , c ) c ) 187 , d ) d ) 721 , e ) e ) 159 | c | divide(multiply(850, subtract(const_100, add(add(40, 28), 10))), const_100) | add(n1,n2)|add(n3,#0)|subtract(const_100,#1)|multiply(n0,#2)|divide(#3,const_100)| | gain | C |
the lowest number which should be added to 5915 so that the sum is exactly divisible by 3 , 5 , 8 and 7 is : | "l . c . m . of 3 , 5 , 8 and 7 = 840 . on dividing 5915 by 840 , the remainder is 35 . number to be added = ( 840 - 35 ) = 805 . answer : option ' c '" | a ) 705 , b ) 825 , c ) 805 , d ) 905 , e ) 810 | c | divide(5915, 5) | divide(n0,n2)| | general | C |
the average of marks obtained by 120 boys was 37 . if the average of marks of passed boys was 39 and that of failed boys was 15 , the number of boys who passed the examination is ? | let the number of boys who passed = x . then , 39 x x + 15 x ( 120 - x ) = 120 x 37 24 x = 4440 - 1800 = > x = 2640 / 24 x = 110 . hence , the number of boys passed = 110 . answer : b | a ) 100 , b ) 110 , c ) 120 , d ) 130 , e ) 140 | b | divide(subtract(multiply(37, 120), multiply(120, 15)), subtract(39, 15)) | multiply(n0,n1)|multiply(n0,n3)|subtract(n2,n3)|subtract(#0,#1)|divide(#3,#2) | general | B |
a feed store sells two varieties of birdseed : brand a , which is 65 % millet and 35 % sunflower , and brand b , which is 40 % millet and 60 % safflower . if a customer purchases a mix of the two types of birdseed that is 50 % millet , what percent of the mix is brand a ? | "yes there is a simple method : consider the following method brand a : 65 % millet and 35 % sunflower brand b : 40 % millet and 60 % safflower mix : 50 % millet here the weighted average is 50 % , now brand a has 65 % millet , which is 15 % more than the weighted average of mix = + 0.15 a - - - - - - - - - - - - - - - i similarly , brand b has 40 % millet , which is 10 % less than the weighted average of mix = - 0.10 b - - - - - - - - - - - - ii now , both brand a and brand b are combined to give a 50 % mix containing millet , so equate i and ii implies , 0.15 a = 0.10 b therefore a / b = 0.10 / 0.15 = 2 / 3 a : b : ( a + b ) = 2 : 3 : ( 3 + 2 ) = 2 : 3 : 5 we have to find , percent of the mix is brand a i . e . a : ( a + b ) = 2 : 5 = ( 2 / 5 ) * 100 = 40 % here is a pictorial representation : brand a = 65 % - - - - - - - - - - - - - - - - - - - - - - - - 15 % or 0.15 above average , a times - - - - - - - - - - - - - - - - - total below = + 0.15 a - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - average = 50 % or 0.50 brand b = 40 % - - - - - - - - - - - - - - - - - - - - - - - - - - 10 % or 0.10 below average , b times - - - - - - - - - - - - - - - - - total above = - 0.10 b since the amount below the average has to equal the average above the average ; therefore , 0.15 a = 0.10 b a / b = 2 / 3 a : b : total = 2 : 3 : 5 therefore a / total = 2 : 5 = 40 %" | a ) 40 % , b ) 45 % , c ) 50 % , d ) 60 % , e ) 55 % | a | multiply(divide(const_3, add(const_3, const_2)), const_100) | add(const_2,const_3)|divide(const_3,#0)|multiply(#1,const_100)| | gain | A |
a can do a work in 15 days and b in 20 days . if they work on it together for 4 days then the fraction of the work that is left is ? | "explanation : a β s one day work = 1 / 15 b β s one day work = 1 / 20 ( a + b ) β s one day work is = 1 / 15 + 1 / 20 = 7 / 60 their 4 days work = 4 x 7 / 60 = 7 / 15 remaining work = 8 / 15 answer : option d" | a ) 1 / 4 , b ) 1 / 10 , c ) 7 / 15 , d ) 8 / 15 , e ) 9 / 13 | d | subtract(const_1, multiply(add(divide(const_1, 15), divide(const_1, 20)), 4)) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|multiply(n2,#2)|subtract(const_1,#3)| | physics | D |
sum of the squares of 3 no . ' s is 99 and the sum of their products taken two at a time is 131 . find the sum ? | "( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 ( ab + bc + ca ) = 99 + 2 * 131 a + b + c = β 361 = 19 c" | a ) 20 , b ) 24 , c ) 19 , d ) 28 , e ) 30 | c | sqrt(add(multiply(131, const_2), 99)) | multiply(n2,const_2)|add(n1,#0)|sqrt(#1)| | general | C |
a meal cost $ 35.50 adn there was no tax . if the tip was more than 2 pc but less than 12 pc of the price , then the total amount paid should be : | "2 % ( 35.5 ) = 0.71 12 % ( 35.5 ) = 4.26 total amount could have been 35.5 + 0.71 and 35.5 + 4.26 = > could have been between 36.21 and 39.76 = > approximately between 36 and 40 answer is e ." | a ) 40 - 43 , b ) 39 - 42 , c ) 39 - 42 , d ) 38 - 41 , e ) 36 - 40 | e | add(divide(multiply(35.50, 2), const_100), 35.50) | multiply(n0,n1)|divide(#0,const_100)|add(n0,#1)| | general | E |
o ( x ) represents the least odd integer greater than x , whereas o ( x ) represents the greatest odd integer less than x . likewise , e ( x ) represents the least even integer greater than x , whereas e ( x ) represents the greatest even integer less than x . according to these definitions , the value of o ( 5.3 ) + e ( β 6.7 ) + o ( β 3.3 ) + e ( 4.7 ) is : | "o ( 5.3 ) + e ( β 6.7 ) + o ( β 3.3 ) + e ( 4.7 ) = 7 + ( - 6 ) + ( - 5 ) + 4 = 0 the answer is c ." | a ) - 3 , b ) - 1 , c ) 0 , d ) 1 , e ) 3 | c | add(add(add(multiply(5.3, const_1), multiply(negate(6.7), const_1)), negate(3.3)), multiply(4.7, const_1)) | multiply(n0,const_1)|multiply(n3,const_1)|negate(n1)|negate(n2)|multiply(#2,const_1)|add(#0,#4)|add(#5,#3)|add(#6,#1)| | general | C |
george went to a fruit market with certain amount of money . with this money he can buy either 50 oranges or 40 mangoes . he retains 8 % of the money for taxi fare and buys 20 mangoes . how many oranges can he buy ? | "let the amount of money be 200 let cost of 1 orange be 4 let cost of 1 mango be 5 he decides to retain 8 % of 200 = 16 for taxi fare , so he is left with 184 he buys 20 mangoes ( @ 5 ) so he spends 100 money left is 84 ( 184 - 100 ) no of oranges he can buy = 84 / 4 = > 21 so , george can buy 21 oranges . a" | a ) 21 , b ) 30 , c ) 20 , d ) 16 , e ) 12 | a | multiply(subtract(subtract(const_1, divide(8, const_100)), divide(20, 50)), 50) | divide(n2,const_100)|divide(n3,n0)|subtract(const_1,#0)|subtract(#2,#1)|multiply(n0,#3)| | general | A |
preethi has 6 flavors of ice cream in his parlor . how many options are there for dorathy to pick a one - flavor , two - flavor , 3 - flavor , 4 - flavor , 5 flavor or 6 flavor order ? | 6 c 1 + 6 c 2 + 6 c 3 + 6 c 4 + 6 c 5 + 6 c 6 = 63 . answer : e | a ) 26 , b ) 36 , c ) 64 , d ) 67 , e ) 63 | e | add(add(add(add(divide(factorial(6), factorial(5)), divide(factorial(6), multiply(const_2, factorial(4)))), divide(factorial(6), multiply(factorial(3), factorial(3)))), divide(factorial(6), multiply(const_2, factorial(4)))), divide(factorial(6), factorial(5))) | factorial(n0)|factorial(n3)|factorial(n2)|factorial(n1)|divide(#0,#1)|multiply(#2,const_2)|multiply(#3,#3)|divide(#0,#5)|divide(#0,#6)|add(#4,#7)|add(#9,#8)|add(#10,#7)|add(#11,#4) | probability | E |
to fill a tank , 20 buckets of water is required . how many buckets of water will be required to fill the same tank if the capacity of the bucket is reduced to one - fifth of its present ? | "let the capacity of 1 bucket = x . then , the capacity of tank = 20 x . new capacity of bucket = 1 / 5 x therefore , required number of buckets = ( 20 x ) / ( 1 x / 5 ) = ( 20 x ) x 5 / 1 x = 100 / 1 = 100 answer is a ." | a ) 100 , b ) 200 , c ) 300 , d ) 400 , e ) 500 | a | divide(multiply(20, add(const_4, const_1)), const_2) | add(const_1,const_4)|multiply(n0,#0)|divide(#1,const_2)| | physics | A |
a goods train runs at the speed of 72 km / hr and crosses a 80 m long platform in 26 sec . what is the length of the goods train ? | "speed = 72 * 5 / 18 = 20 m / sec . time = 26 sec . let the length of the train be x meters . then , ( x + 80 ) / 26 = 20 x = 440 m . answer : b" | a ) 382 , b ) 440 , c ) 278 , d ) 270 , e ) 881 | b | subtract(multiply(multiply(divide(72, const_3600), const_1000), 26), 80) | divide(n0,const_3600)|multiply(#0,const_1000)|multiply(n2,#1)|subtract(#2,n1)| | physics | B |
a certain machine produces 550 units of product p per hour . working continuously at this constant rate , this machine will produce how many units of product p in 5 days ? | "since 5 days consist of 24 * 5 hours the total is 120 hours . since every hour the machine produces 550 units of product p the total product during 120 hours is 120 * 550 = 66,000 . correct option : c" | a ) 7,000 , b ) 24,000 , c ) 66,000 , d ) 100,000 , e ) 168,000 | c | multiply(const_4, const_10) | multiply(const_10,const_4)| | physics | C |
if a boat is rowed downstream for 500 km in 5 hours and upstream for 240 km in 6 hours , what is the speed of the boat and the river ? | "explanation : if x : speed of boats man in still water y : speed of the river downstream speed ( ds ) = x + y upstream speed ( us ) = x Γ’ β¬ β y x = ( ds + us ) / 2 y = ( ds Γ’ β¬ β us ) / 2 in the above problem ds = 100 ; us = 40 x = ( 100 + 40 ) / 2 = 140 / 2 = 70 km / hr y = ( 100 - 40 ) / 2 = 60 / 2 = 30 km / hr answer : a" | a ) 70 , 30 , b ) 30 , 70 , c ) 70 , 70 , d ) 30 , 30 , e ) 80 , 50 | a | divide(add(divide(240, 6), divide(500, 5)), const_2) | divide(n2,n3)|divide(n0,n1)|add(#0,#1)|divide(#2,const_2)| | physics | A |
the product of the squares of two positive integers is 900 . how many pairs of positive integers satisfy this condition ? | "ans : e - 4 pairs ( x Λ 2 ) ( y Λ 2 ) = 900 [ square root both sides ] xy = 30 20 = 1 x 30 , 3 x 10 , 6 x 5 , 5 x 6 , 10 x 3 , 30 x 1 , 15 x 2 , 2 x 15 cancel the repeats this leaves us with exactly 4 options . hence , e" | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | e | subtract(add(const_2, const_3), const_2) | add(const_2,const_3)|subtract(#0,const_2)| | geometry | E |
in what ratio should a 20 % methyl alcohol solution be mixed with a 50 % methyl alcohol solution so that the resultant solution has 40 % methyl alcohol in it ? | two things to remember : - 1 ) the ratio of quantity mixed to get an average is related to the ratio of each qty ' s distance from average . 2 ) more the qty of x from that of y , more closer will be average to x . . qty of 20 % / qty of 50 % = ( 50 β 40 ) / ( 40 β 20 ) = 1 / 2 = 1 : 2 answer : b | a ) 1 : 3 , b ) 1 : 2 , c ) 2 : 3 , d ) 2 : 1 , e ) 3 : 1 | b | divide(subtract(50, 40), subtract(40, 20)) | subtract(n1,n2)|subtract(n2,n0)|divide(#0,#1) | other | B |
in a certain quiz that consists of 10 questions , each question after the first is worth 4 points more than the preceding question . if the 10 questions on the quiz are worth a total of 320 points , how many points is the third question worth ? | "x x + 4 x + 8 x + 12 x + 16 x + 20 x + 24 x + 28 x + 32 x + 36 10 x + 180 = 320 10 x = 140 x = 14 3 rd question = x + 8 = 14 + 8 = 22 answer c" | a ) 39 , b ) 24 , c ) 22 , d ) 20 , e ) 18 | c | add(divide(320, 10), subtract(subtract(10, 4), const_1)) | divide(n3,n2)|subtract(n0,n1)|subtract(#1,const_1)|add(#0,#2)| | general | C |
total dinning bill for 5 people was $ 211.00 . if they add 15 % tip and divided the bill evenly , approximate . what was each persons find share | "211 * 15 = 3165 / 100 = 31.65 211 + 31.65 = 242.65 242.65 / 5 = 48.53 answer : b" | a ) $ 30.14 , b ) 48.53 , c ) 34.66 , d ) 32.29 , e ) 33.16 | b | divide(add(211.00, divide(multiply(15, 211.00), const_100)), 5) | multiply(n1,n2)|divide(#0,const_100)|add(n1,#1)|divide(#2,n0)| | general | B |
if g is the smallest positive integer such that 3150 multiplied by g is the square of an integer , then g 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 β 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 exponents of both 2 and 7 are not even numbers . this tells us that 3150 itself is not a perfect square . ) we also are given that 3150 multiplied by g is the square of an integer . we can write this as : 2 ^ 1 x 3 ^ 2 x 5 ^ 2 x 7 ^ 1 x g = square of an integer according to our rule , we need all unique prime factors ' exponents to be even numbers . thus , we need one more 2 and one more 7 . therefore , g = 7 x 2 = 14 answer is e . | a ) 2 , b ) 5 , c ) 6 , d ) 7 , e ) 14 | e | multiply(const_2, divide(divide(divide(divide(divide(3150, const_2), add(const_2, const_3)), add(const_2, const_3)), const_3), const_3)) | add(const_2,const_3)|divide(n0,const_2)|divide(#1,#0)|divide(#2,#0)|divide(#3,const_3)|divide(#4,const_3)|multiply(#5,const_2) | geometry | E |
a pair of articles was bought for $ 810 at a discount of 10 % . what must be the marked price of each of the article ? | "s . p . of each of the article = 810 / 2 = $ 405 let m . p = $ x 90 % of x = 405 x = 405 * 100 / 90 = $ 450 answer is d" | a ) $ 300 , b ) $ 500 , c ) $ 350 , d ) $ 450 , e ) $ 600 | d | divide(multiply(subtract(const_100, 10), divide(810, const_2)), const_100) | divide(n0,const_2)|subtract(const_100,n1)|multiply(#0,#1)|divide(#2,const_100)| | gain | D |
if x is an integer and 2.134 Γ 10 ^ x is less than 220,000 , what is the greatest possible value for x ? | "x is an integer and 2.134 Γ 10 x is less than 220,000 , what is the greatest possible value for x ? for 2.134 Γ 10 x is less than 220,000 to remain true , the greatest number is 213,400 , which makes x = 5 c . 5" | a ) 7 , b ) 6 , c ) 5 , d ) 4 , e ) 3 | c | floor(divide(log(divide(220,000, 2.134)), log(10))) | divide(n2,n0)|log(n1)|log(#0)|divide(#2,#1)|floor(#3)| | general | C |
find the slope of the line perpendicular to the line y = ( 1 / 4 ) x - 7 | "two lines are perpendicular if the product of their slopes is equal to - 1 . the slope of the given line is equal to 1 / 4 . if m is the slope of the line perpendicular to the given line , then m Γ ( 1 / 4 ) = - 1 solve for m m = - 4 correct answer c ) - 4" | a ) 1 , b ) 2 , c ) - 4 , d ) 4 , e ) 5 | c | divide(1, 4) | divide(n0,n1)| | general | C |
the present population of a town is 1000 . population increase rate is 25 % p . a . find the population of town before 1 years ? | "p = 1000 r = 25 % required population of town = p / ( 1 + r / 100 ) ^ t = 1000 / ( 1 + 25 / 100 ) = 1000 / ( 5 / 4 ) = 800 answer is a" | a ) 800 , b ) 1500 , c ) 1000 , d ) 750 , e ) 500 | a | add(1000, divide(multiply(1000, 25), const_100)) | multiply(n0,n1)|divide(#0,const_100)|add(n0,#1)| | gain | A |
on a game show , a contestant is given 3 keys , each of which opens exactly one of the 3 identical boxes . the contestant assigns each key to one of the boxes and wins the amount of money contained in any box that is opened by the key assigned to it . the first box contains $ 4 , the second $ 400 , and the third $ 4000 . what is the probability that a contestant will win more than $ 4000 ? | let ' s call the boxes that contain $ 4 , $ 400 , and $ 4000 , respectively , box a , box b , box c . these are opened , respectively , by key a , key b , and key c . we want to know the probability of winning more than $ 4000 . notice that if the distribution of keys is : box a = key b box b = key a box c = key c then the contestant wins exactly $ 4000 , not more than $ 4000 . the only configuration that leads to winning more than $ 1000 is : box a = key a box b = key b box c = key c i . e . , getting all three keys correct . that ' s the only way to be more than $ 4000 . so , really , the question can be rephrased : what is the probability of guessing the order of keys so that each key matches the correct box ? well , for a set of three items , the number of possible permutations is 3 ! = 3 * 2 * 1 = 6 . of those 6 possible permutations , only one of them leads to all three keys being paired with the right box . so , the answer is probability = 1 / 6 answer : b | a ) 2 / 6 , b ) 1 / 6 , c ) 1 / 8 , d ) 1 / 3 , e ) 1 / 2 | b | divide(const_1, multiply(3, const_2)) | multiply(n0,const_2)|divide(const_1,#0) | general | B |
what number is obtained by adding the units digits of 734 ^ 99 and 347 ^ 83 ? | the units digit of 734 ^ 99 is 4 because 4 raised to the power of an odd integer ends in 4 . the units digit of 347 ^ 83 is 3 because powers of 7 end in 7 , 9 , 3 , or 1 cyclically . since 83 is in the form 4 n + 3 , the units digit is 3 . then 4 + 3 = 7 . the answer is a . | a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | a | subtract(subtract(99, 83), divide(99, add(const_1, const_10))) | add(const_1,const_10)|subtract(n1,n3)|divide(n1,#0)|subtract(#1,#2) | general | A |
seed mixture x is 40 % ryegrass and 60 % bluegrass by weight ; seed mixture y is 25 % ryegrass and 75 % fescue . if a mixture of x and y contains 34 % ryegrass , what percent of the weight of the mixture is from mixture x ? | "34 % is 9 % - points above 25 % and 6 % - points below 40 % . thus the ratio of mixture y to mixture x is 2 : 3 . the percent of mixture x is 3 / 5 = 60 % . the answer is d ." | a ) 25 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 75 % | d | divide(subtract(34, add(25, const_1)), subtract(divide(40, const_100), divide(add(25, const_1), const_100))) | add(n2,const_1)|divide(n0,const_100)|divide(#0,const_100)|subtract(n4,#0)|subtract(#1,#2)|divide(#3,#4)| | gain | D |
a semicircular cubicle has a radius of 14 . what is the approximate perimeter of the cubicle ? | perimeter of a circle = 2 pi * r perimeter of a semicircle = pi * r + 2 r aprox perimiter = 3.14 * 14 + 2 * 14 = 71.96 approximately 72 answer d | ['a ) 55', 'b ) 86', 'c ) 25', 'd ) 72', 'e ) 35'] | d | add(multiply(const_pi, 14), multiply(const_2, 14)) | multiply(n0,const_pi)|multiply(n0,const_2)|add(#0,#1) | geometry | D |
a classroom has equal number of boys and girls . 8 girls left to play kho - kho , leaving twice as many boys as girls in the classroom . what was the total number of girls and boys present initially ? | after 8 girls left remaining 8 girls now boys 16 are twice as many as remaining girls . initially boys = 16 and girls = 16 . answer : c | a ) 16 , b ) 24 , c ) 32 , d ) 48 , e ) 54 | c | multiply(multiply(const_2, 8), const_2) | multiply(n0,const_2)|multiply(#0,const_2) | general | C |
a batsman makes a score of 56 runs in the 17 th inning and thus increases his averages by 3 . what is his average after 17 th inning ? | "let the average after 17 innings = x total runs scored in 17 innings = 17 x average after 16 innings = ( x - 3 ) total runs scored in 16 innings = 16 ( x - 3 ) total runs scored in 16 innings + 56 = total runs scored in 17 innings = > 16 ( x - 3 ) + 56 = 17 x = > 16 x - 48 + 56 = 17 x = > x = 8 answer is b ." | a ) 25 , b ) 8 , c ) 27 , d ) 29 , e ) 39 | b | add(subtract(56, multiply(17, 3)), 3) | multiply(n1,n2)|subtract(n0,#0)|add(n2,#1)| | general | B |
how long does a train 110 m long running at the speed of 90 km / hr takes to cross a bridge 132 m length ? | "speed = 90 * 5 / 18 = 25 m / sec total distance covered = 110 + 132 = 242 m . required time = 242 / 25 = 9.68 sec . answer : b" | a ) 12.9 sec , b ) 9.68 sec , c ) 17.9 sec , d ) 16.8 sec , e ) 14.9 sec | b | divide(add(110, 132), multiply(90, const_0_2778)) | add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)| | physics | B |
a cyclist rides a bicycle 7 km at an average speed of 10 km / hr and again travels 10 km at an average speed of 7 km / hr . what is the average speed for the entire trip ? | distance = 17 km time = 7 / 10 + 10 / 7 = ( 49 + 100 ) / 70 = 149 / 70 hours average speed = ( 17 * 70 ) / 149 = 7.99 km / h the answer is c . | a ) 7.59 , b ) 7.79 , c ) 7.99 , d ) 8.19 , e ) 8.39 | c | divide(add(7, 10), add(divide(7, 10), divide(10, 7))) | add(n0,n1)|divide(n0,n1)|divide(n1,n0)|add(#1,#2)|divide(#0,#3) | general | C |
in what time will a railway train 60 m long moving at the rate of 54 kmph pass a telegraph post on its way ? | "t = 60 / 54 * 18 / 5 = 4 sec answer : a" | a ) 4 sec , b ) 7 sec , c ) 2 sec , d ) 6 sec , e ) 9 sec | a | divide(60, multiply(54, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics | A |
what is the ratio between perimeters of two squares one having 2 times the diagonal then the other ? | "d = 2 d d = d a β 2 = 2 d a β 2 = d a = 2 d / β 2 a = d / β 2 = > 2 : 1 answer : d" | a ) 3 : 8 , b ) 3 : 6 , c ) 3 : 7 , d ) 2 : 1 , e ) 3 : 3 | d | divide(2, divide(2, 2)) | divide(n0,n0)|divide(n0,#0)| | geometry | D |
let the number which when multiplied by 13 is increased by 180 . | "solution let the number be x . then , 13 x - x = 180 βΉ = βΊ 12 x = 180 x βΉ = βΊ 15 . answer c" | a ) 14 , b ) 20 , c ) 15 , d ) 28 , e ) 30 | c | divide(180, subtract(13, const_1)) | subtract(n0,const_1)|divide(n1,#0)| | general | C |
if the a radio is sold for rs 490 and sold for rs 465.50 . find loss percentage . | cost price = rs 490 , selling price = 465.50 . loss = rs ( 490 - 465.50 ) = rs 24.50 . loss % = [ ( 24.50 / 490 ) * 100 ] % = 5 % answer is b | a ) 3 , b ) 5 , c ) 7 , d ) 9 , e ) none of them | b | multiply(divide(subtract(490, 465.5), 490), const_100) | subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100) | gain | B |
a number exceeds by 30 from its 3 / 8 part . then the number is ? | "x β 3 / 8 x = 30 x = 48 answer : e" | a ) a ) 32 , b ) b ) 35 , c ) c ) 39 , d ) d ) 40 , e ) e ) 48 | e | divide(multiply(30, 8), subtract(8, 3)) | multiply(n0,n2)|subtract(n2,n1)|divide(#0,#1)| | general | E |
the price of a jacket is reduced by 20 % . during a special sale the price of the jacket is reduced another 25 % . by approximately what percent must the price of the jacket now be increased in order to restore it to its original amount ? | "1 ) let the price of jacket initially be $ 100 . 2 ) then it is decreased by 20 % , therefore bringing down the price to $ 80 . 3 ) again it is further discounted by 25 % , therefore bringing down the price to $ 60 . 4 ) now 60 has to be added byx % in order to equal the original price . 60 + ( x % ) 60 = 100 . solving this eq for x , we get x = 66.67 ans is e ." | a ) 32.5 , b ) 35 , c ) 48.1 , d ) 65 , e ) 66.67 | e | multiply(const_100, divide(subtract(const_100, subtract(subtract(const_100, 20), multiply(subtract(const_100, 20), divide(25, const_100)))), subtract(subtract(const_100, 20), multiply(subtract(const_100, 20), divide(25, const_100))))) | divide(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|subtract(#1,#2)|subtract(const_100,#3)|divide(#4,#3)|multiply(#5,const_100)| | gain | E |
pipe a can fill a tank in 5 hr , pipe b in 10 hr and pipe c in 30 hr . if all the pipes are open in how many hours will tank be filled ? | "part filled by ( a + b + c ) in 1 hour = 1 / 5 + 1 / 10 + 1 / 30 = 1 / 3 all the 3 pipes together will fill the tank in 3 hours answer is c" | a ) 1 hr , b ) 2 hr , c ) 3 hr , d ) 4 hr , e ) 5 hr | c | multiply(add(add(divide(5, 5), divide(5, 10)), divide(5, 30)), 5) | divide(n0,n0)|divide(n0,n1)|divide(n0,n2)|add(#0,#1)|add(#3,#2)|multiply(n0,#4)| | physics | C |
two trains running in opposite directions cross a man standing on the platform in 67 seconds and 17 seconds respectively and they cross each other in 73 seconds . the ratio of their speeds is : | "let the speeds of the two trains be x m / sec and y m / sec respectively . then , length of the first train = 27 x meters , and length of the second train = 17 y meters . ( 27 x + 17 y ) / ( x + y ) = 23 = = > 27 x + 17 y = 23 x + 23 y = = > 4 x = 6 y = = > x / y = 3 / 2 . answer : a" | a ) 3 / 2 , b ) 3 / 9 , c ) 3 / 6 , d ) 3 / 1 , e ) 3 / 5 | a | divide(subtract(67, 73), subtract(73, 17)) | subtract(n0,n2)|subtract(n2,n1)|divide(#0,#1)| | physics | A |
jim is able to sell a hand - carved statue for $ 540 which was a 35 % profit over his cost . how much did the statue originally cost him ? | 540 = 1.35 * x x = 540 / 1.35 = 400 $ 400 , which is ( a ) . | a ) $ 400.00 , b ) $ 412.40 , c ) $ 455.40 , d ) $ 474.90 , e ) $ 488.20 | a | divide(540, add(divide(35, const_100), const_1)) | divide(n1,const_100)|add(#0,const_1)|divide(n0,#1) | gain | A |
in an election between two candidates , one got 45 % of the total valid votes , 20 % of the votes were invalid . if the total number of votes was 2000 , the number of valid votes that the other candidate got , was : | "number of valid votes = 80 % of 4000 = 3200 . valid votes polled by other candidate = 55 % of 4000 = ( 55 / 100 ) x 8000 = 2200 answer = a" | a ) 2200 , b ) 3600 , c ) 3270 , d ) 3700 , e ) 4000 | a | multiply(multiply(subtract(const_1, divide(20, const_100)), subtract(const_1, divide(45, const_100))), 2000) | divide(n1,const_100)|divide(n0,const_100)|subtract(const_1,#0)|subtract(const_1,#1)|multiply(#2,#3)|multiply(n2,#4)| | gain | A |
after decreasing 56 % in the price of an article costs rs . 4400 . find the actual cost of an article ? | "cp * ( 44 / 100 ) = 4400 cp = 100 * 100 = > cp = 10000 answer : e" | a ) 12000 , b ) 15000 , c ) 1200 , d ) 1000 , e ) 10000 | e | divide(4400, subtract(const_1, divide(56, const_100))) | divide(n0,const_100)|subtract(const_1,#0)|divide(n1,#1)| | gain | E |
a , b and c are partners . a receives 2 / 3 of profits , b and c dividing the remainder equally . a ' s income is increased by rs . 500 when the rate to profit rises from 5 to 7 percent . find the capital of b ? | "a : b : c = 2 / 3 : 1 / 6 : 1 / 6 = 4 : 1 : 1 x * 2 / 100 * 2 / 3 = 500 b capital = 37500 * 1 / 6 = 6250 . answer : e" | a ) 3999 , b ) 7799 , c ) 2500 , d ) 2772 , e ) 6250 | e | divide(multiply(500, const_100), 2) | multiply(n2,const_100)|divide(#0,n0)| | general | E |
two trains of length 100 m and 200 m are 100 m apart . they start moving towards each other on parallel tracks , at speeds 54 kmph and 36 kmph . in how much time will the trains cross each other ? | "d relative speed = ( 54 + 36 ) * 5 / 18 = 5 * 5 = 25 mps . the time required = d / s = ( 100 + 100 + 200 ) / 25 = 400 / 25 = 16 sec ." | a ) 12 sec , b ) 13 sec , c ) 15 sec , d ) 16 sec , e ) 17 sec | d | divide(100, multiply(add(54, 36), const_0_2778)) | add(n3,n4)|multiply(#0,const_0_2778)|divide(n2,#1)| | physics | D |
the grade point average of one third of the classroom is 30 ; the grade point average of the rest is 33 . what is the grade point average of the whole class ? | "let n = total students in class total points for 1 / 3 class = 30 n / 3 = 10 n total points for 2 / 3 class = 33 * 2 n / 3 = 22 n total points for whole class = 10 n + 22 n = 32 n 32 n total class points / n total students = 32 grade point average for total class answer : e" | a ) 44 , b ) 41 , c ) 38 , d ) 35 , e ) 32 | e | add(multiply(divide(33, const_3), const_2), divide(30, const_3)) | divide(n0,const_3)|divide(n1,const_3)|multiply(#1,const_2)|add(#0,#2)| | general | E |
at what rate percent on simple interest will rs . 800 amount to rs . 950 in 5 years ? | "150 = ( 800 * 5 * r ) / 100 r = 3.75 % answer : a" | a ) 3.75 % , b ) 5.93 % , c ) 4.75 % , d ) 5.33 % , e ) 6.33 % | a | multiply(divide(divide(subtract(950, 800), 800), 5), const_100) | subtract(n1,n0)|divide(#0,n0)|divide(#1,n2)|multiply(#2,const_100)| | gain | A |
a rectangle having length 100 cm and width 40 cm . if the length of the rectangle is increased by fifty percent then how much percent the breadth should be decreased so as to maintain the same area . | "explanation : solution : ( 50 / ( 100 + 50 ) * 100 ) % = 33.33 % answer : b" | a ) 25 % , b ) 33.33 % , c ) 40 % , d ) 75 % , e ) none of these | b | multiply(add(const_1, divide(divide(multiply(100, 40), add(100, divide(multiply(multiply(const_3, const_10), 100), const_100))), 40)), const_10) | multiply(n0,n1)|multiply(const_10,const_3)|multiply(n0,#1)|divide(#2,const_100)|add(n0,#3)|divide(#0,#4)|divide(#5,n1)|add(#6,const_1)|multiply(#7,const_10)| | geometry | B |
in a certain diving competition , 5 judges score each dive on a scale from 1 to 10 . the point value of the dive is obtained by dropping the highest score and the lowest score and multiplying the sum of the remaining scores by the degree of difficulty . if a dive with a degree of difficulty of 3.2 received scores of 7.5 , 7.8 , 9.0 , 6.0 , and 8.5 , what was the point value of the dive ? | degree of difficulty of dive = 3.2 scores are 6.0 , 7.5 , 8.0 , 8.5 and 9.0 we can drop 6.0 and 9.0 sum of the remaining scores = ( 7.5 + 7.8 + 8.5 ) = 23.8 point of value of the dive = 23.8 * 3.2 = 76.16 answer d | a ) 68.8 , b ) 73.6 , c ) 75.2 , d ) 76.16 , e ) 81.6 | d | multiply(add(add(7.5, 7.8), 8.5), 3.2) | add(n4,n5)|add(n8,#0)|multiply(n3,#1) | general | D |
a certain shade of gray paint is obtained by mixing 3 parts of white paint with 5 parts of black paint . if 2 gallons of the mixture is needed and the individual colors can be purchased only in one gallon or half gallon cans , what is the least amount of paint , in gallons , that must be purchased in order to measure out the portions needed for the mixture ? | "ratio ; white : black : gray = 3 : 5 : ( 3 + 5 ) = 3 : 5 : 8 if 2 gallons of the mixture is needed < = = > need 2 gallons of gray paint . then , 3 : 5 : 8 = x : y : 2 3 / 8 = x / 2 ; x = 3 / 4 5 / 8 = y / 2 ; y = 5 / 4 you need 3 / 4 gallon of white . you have to buy 1 gallons . you need 5 / 4 gallons of black . you have to buy 1 gallon and another 1 / 2 gallon of black . the least amount of paint = 1 + 1 + 1 / 2 = 2.5 . answer : b" | a ) 2 , b ) 2 1 / 2 , c ) 3 , d ) 3 1 / 2 , e ) 4 | b | add(multiply(const_0_25, 2), multiply(2, 5)) | multiply(const_0_25,n2)|multiply(n1,n2)|add(#0,#1)| | general | B |
the simple form of the ratio 5 / 3 : 2 / 5 is ? | "5 / 3 : 2 / 5 = 25 : 6 answer : c" | a ) 10 : 6 , b ) 10 : 3 , c ) 25 : 6 , d ) 25 : 3 , e ) 30 : 3 | c | divide(2, 5) | divide(n2,n3)| | other | C |
find 20 % of 240 | "we know that r % of m is equal to r / 100 Γ m . so , we have 20 % of 240 20 / 100 Γ 240 = 48 answer : e" | a ) 96 , b ) 94 , c ) 86 , d ) 74 , e ) 48 | e | divide(20, 240) | divide(n0,n1)| | gain | E |
a candidate appearing for an examination has to secure 42 % marks to pass paper i . but he secured only 42 marks and failed by 22 marks . what is the maximum mark for paper i ? | "he secured 42 marks nd fail by 22 marks so total marks for pass the examinatn = 64 let toal marks x x * 42 / 100 = 64 x = 152 answer : d" | a ) 110 , b ) 120 , c ) 130 , d ) 152 , e ) 150 | d | divide(add(42, 22), divide(42, const_100)) | add(n1,n2)|divide(n0,const_100)|divide(#0,#1)| | gain | D |
a certain number of men can finish a piece of work in 10 days . however , if there were 10 less men it will take 10 days more for the work to be finished . how many men were there originally ? | "let x be the number of men originally thus , 10 x = ( 10 + 10 ) ( x - 10 ) 10 x = 20 ( x - 10 ) x = 2 ( x - 10 ) x = 20 answer : d" | a ) 10 men , b ) 30 men , c ) 40 men , d ) 20 men , e ) none of these | d | divide(10, subtract(const_1, divide(10, add(10, 10)))) | add(n0,n1)|divide(n0,#0)|subtract(const_1,#1)|divide(n1,#2)| | physics | D |
if x and y are integers , what is the least positive number of 24 x + 16 y ? | "24 x + 16 y = 8 ( 3 x + 2 y ) which will be a minimum positive number when 3 x + 2 y = 1 . 3 ( 1 ) + 2 ( - 1 ) = 1 then 8 ( 3 x + 2 y ) can have a minimum positive value of 8 . the answer is e ." | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 8 | e | subtract(24, 16) | subtract(n0,n1)| | general | E |
in the graduating class of a certain college , 48 percent of the students are male and 52 percent are female . in this class 50 percent of the male and 20 percent of the female students are 25 years old or older . if one student in the class is randomly selected , approximately what is the probability that he or she will be less than 25 years old ? | percent of students who are 25 years old or older is 0.5 * 48 + 0.2 * 52 = ~ 34 , so percent of people who are less than 25 years old is 100 - 3 = 66 . answer : b . | a ) a ) 0.9 , b ) b ) 0.6 , c ) c ) 0.45 , d ) d ) 0.3 , e ) e ) 0.25 | b | subtract(const_1, multiply(divide(50, const_100), divide(subtract(const_100, 20), const_100))) | divide(n2,const_100)|subtract(const_100,n3)|divide(#1,const_100)|multiply(#0,#2)|subtract(const_1,#3) | general | B |
a company has $ 329864 in its account . what is the least amount of money ( in whole number of dollars ) that it must add to the account if the money is paid evenly among 9 of its vendors ? | to find the least amount that must be added to the account to split the money evenly among 9 of its vendors , the total divisible by 9 simply add the individual digits of the total = 3 + 2 + 9 + 8 + 6 + 4 = 32 if you add 4 , the number is divisible by 9 ( 32 + 4 ) correct option : d | a ) $ 1 , b ) $ 2 , c ) $ 3 , d ) $ 4 , e ) $ 6 | d | subtract(9, reminder(329864, 9)) | reminder(n0,n1)|subtract(n1,#0) | general | D |
what is the greatest positive integer w such that 3 ^ w is a factor of 9 ^ 10 ? | "what is the greatest positive integer w such that 3 ^ w is a factor of 9 ^ 10 ? 9 ^ 10 = ( 3 ^ 2 ) ^ 10 = 3 ^ 20 d . 20" | a ) 5 , b ) w = 9 , c ) w = 10 , d ) w = 20 , e ) w = 30 | d | multiply(subtract(9, 10), 10) | subtract(n1,n2)|multiply(n2,#0)| | other | D |
of the 3600 employees of company x , 12 / 25 are clerical . if the clerical staff were to be reduced by 1 / 4 , what percent of the total number of the remaining employees would then be clerical ? | let ' s see , the way i did it was 12 / 25 are clerical out of 3600 so 1728 are clerical 1728 reduced by 1 / 4 is 1728 * 1 / 4 so it reduced 432 people , so there is 1296 clerical people left but since 432 people left , it also reduced from the total of 3600 so there are 3168 people total since 1296 clerical left / 3168 people total you get ( a ) 40 % | a ) 40 % , b ) 22.2 % , c ) 20 % , d ) 12.5 % , e ) 11.1 % | a | multiply(divide(subtract(divide(multiply(3600, 12), 25), divide(divide(multiply(3600, 12), 25), 4)), subtract(3600, divide(divide(multiply(3600, 12), 25), 4))), const_100) | multiply(n0,n1)|divide(#0,n2)|divide(#1,n4)|subtract(#1,#2)|subtract(n0,#2)|divide(#3,#4)|multiply(#5,const_100) | general | A |
243 has been divided into 3 parts such that half of the first part , one - third of the second part and one - fourth of the third part are equal . the largest part is ? | explanation : ratio of their shares = ( 35000 * 8 ) : ( 42000 * 10 ) = 2 : 3 . reena ' s share rs . 31570 * ( 2 / 5 ) = rs . 12628 . answer : a ) 12628 | a ) 12628 , b ) 23488 , c ) 26688 , d ) 26667 , e ) 12672 | a | multiply(multiply(multiply(multiply(243, 3), 3), 3), const_2) | multiply(n0,n1)|multiply(n1,#0)|multiply(n1,#1)|multiply(#2,const_2) | general | A |
a train 50 m long passes a platform 100 m long in 10 seconds . the speed of the train in m / sec is ? | speed of train = distance covered / time . = ( 50 + 100 ) / 10 = 15 m / sec . answer : d | a ) 150 , b ) 50 , c ) 10 , d ) 15 , e ) 12 | d | divide(add(50, 100), 10) | add(n0,n1)|divide(#0,n2) | physics | D |
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