Problem stringlengths 5 628 | Rationale stringlengths 1 2.74k | options stringlengths 37 137 | correct stringclasses 5
values | annotated_formula stringlengths 6 848 | linear_formula stringlengths 7 357 | category stringclasses 6
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three cubes of iron whose edges are 6 cm , 8 cm and 10 cm respectively are melted and formed into a single cube . the edge of the new cube formed is | "relative speed = ( 126 - 54 ) * 5 / 18 = 4 * 5 = 20 mps . distance covered in 27 sec = 14 * 20 = 280 m . the length of the faster train = 280 m . answer : a" | a ) 280 , b ) 290 , c ) 300 , d ) 310 , e ) 320 | a | multiply(divide(subtract(126, 54), const_3_6), 14) | subtract(n0,n1)|divide(#0,const_3_6)|multiply(n2,#1)| | physics |
two trains are moving in the same direction at 126 kmph and 54 kmph . the faster train crosses a man in the slower train in 14 seconds . find the length of the faster train ? | "there are several algebraic ways to solve this question , but the easiest way is as follows : since we can not have two correct answers just pick a prime greater than 17 , square it and see what would be the remainder upon division of it by 12 . n = 19 - - > n ^ 2 = 361 - - > remainder upon division 361 by 12 is 1 . a... | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 5 | b | subtract(power(add(17, 2), 2), multiply(12, const_4)) | add(n0,n1)|multiply(n2,const_4)|power(#0,n1)|subtract(#2,#1)| | general |
if n is a prime number greater than 17 , what is the remainder when n ^ 2 is divided by 12 ? | "( 17 ^ 200 - 1 ^ 200 ) is completely divisible by ( 17 + 1 ) as 200 is even . = > ( 17 ^ 200 - 1 ) is completely divisible by 18 . hence , when 17 ^ 200 is divided by 18 , we will get 1 as remainder . answer is b" | a ) 3 , b ) 1 , c ) 2 , d ) 4 , e ) 7 | b | subtract(divide(18, const_2), multiply(17, 17)) | divide(n2,const_2)|multiply(n0,n0)|subtract(#0,#1)| | general |
what is the remainder when 17 ^ 200 is divided by 18 ? | "the area of each half is 100 + 4 ( 450 ) + 100 = 2000 the area that is not painted is 100 . the fraction that is not painted is 100 / 2000 = 1 / 20 = 5 % the answer is a ." | a ) 5 % , b ) 10 % , c ) 15 % , d ) 20 % , e ) 25 % | a | multiply(divide(const_100, add(add(multiply(multiply(const_4, const_100), const_4), const_100), const_100)), const_100) | multiply(const_100,const_4)|multiply(#0,const_4)|add(#1,const_100)|add(#2,const_100)|divide(const_100,#3)|multiply(#4,const_100)| | geometry |
a block of wood has dimensions 10 cm x 10 cm x 90 cm . the block is painted red and then cut evenly at the 45 cm mark , parallel to the sides , to form two rectangular solids of equal volume . what percentage of the surface area of each of the new solids is not painted red ? | "( 6 * 75 ) + 47 + x > 500 450 + 47 + x > 500 497 + x > 500 = > x > 3 option d" | a ) 5 , b ) 4 , c ) 7 , d ) 3 , e ) 8 | d | subtract(500, add(multiply(6, 75), 47)) | multiply(n0,n1)|add(n2,#0)|subtract(n4,#1)| | general |
after 6 games , team b had an average of 75 points per game . if it got only 47 points in game 7 , how many more points does it need to score to get its total above 500 ? | ( 2 ^ 5 ) x ( 6 ) x ( 7 ^ 3 ) has one appearance of 3 ( in the 6 ) and no appearances of 5 . thus n must include at least 3 ^ 2 * 5 ^ 2 = 9 * 25 = 225 the answer is e . | a ) 75 , b ) 125 , c ) 145 , d ) 175 , e ) 225 | e | add(add(add(add(add(multiply(multiply(5, 7), 2), multiply(multiply(5, 7), 2)), multiply(multiply(5, 7), 2)), 7), const_4), const_4) | multiply(n0,n7)|multiply(n1,#0)|add(#1,#1)|add(#2,#1)|add(n7,#3)|add(#4,const_4)|add(#5,const_4) | other |
if both 5 ^ 2 and 3 ^ 3 are factors of n x ( 2 ^ 5 ) x ( 6 ) x ( 7 ^ 3 ) , what is the smallest possible positive value of n ? | "b + c 1 day work = 1 / 9 + 1 / 12 = 7 / 36 work done by b and c in 3 days = 7 / 36 * 3 = 7 / 12 remaining work = 1 - 7 / 12 = 5 / 12 1 / 36 work is done by a in 1 day 5 / 12 work is done by a in 36 * 5 / 12 = 15 days answer is a" | a ) 15 days , b ) 12 days , c ) 6 days , d ) 7 days , e ) 8 days | a | multiply(divide(const_1, add(divide(const_1, 9), divide(const_1, 2))), 3) | divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|divide(const_1,#2)|multiply(n3,#3)| | physics |
a can finish a work in 36 days , b in 9 days and c in 2 days , b and c start the work but are forced to leave after 3 days . the remaining work was done by a in ? | "assuming the car starts at the first pole . to reach the 12 th pole , the car need to travel 11 poles ( the first pole does n ' t count , as the car is already there ) . 11 poles 26 seconds 1 pole ( 26 / 11 ) seconds to reach the last ( 20 th ) pole , the car needs to travel 19 poles . 19 pole 19 x ( 26 / 11 ) seconds... | a ) 44.4543 , b ) 44.9091 , c ) 44.95128 , d ) 44.91288 , e ) 44.91222 | b | multiply(divide(26, 12), 20) | divide(n1,n2)|multiply(n0,#0)| | physics |
there are 20 poles with a constant distance between each pole . a car takes 26 second to reach the 12 th pole . how much will it take to reach the last pole . | "p = ( 2 / 5 ) * p + 21 ( 3 / 5 ) * p = 21 p = 35 the answer is a ." | a ) $ 35 , b ) $ 36 , c ) $ 37 , d ) $ 38 , e ) $ 39 | a | divide(21, subtract(1, multiply(divide(1, 5), const_2))) | divide(n1,n2)|multiply(#0,const_2)|subtract(n1,#1)|divide(n0,#2)| | general |
p has $ 21 more than what q and r together would have had if both b and c had 1 / 5 of what p has . how much does p have ? | "the number of chocolate bars is equal to 18 ? 25 = 450 correct answer c" | a ) 350 , b ) 250 , c ) 450 , d ) 550 , e ) 650 | c | multiply(18, 25) | multiply(n0,n1)| | general |
a large box contains 18 small boxes and each small box contains 25 chocolate bars . how many chocolate bars are in the large box ? | "say morks income is - 100 so tax paid will be 40 say mindys income is 5 * 100 = 500 so tax paid is 30 % * 500 = 150 total tax paid = 40 + 150 = 190 . combined tax % will be 190 / 100 + 500 = 31.67 %" | a ) 32.5 % , b ) 34 % , c ) 35 % , d ) 31.67 % , e ) 37.5 % | d | multiply(const_100, divide(add(divide(40, const_100), multiply(5, divide(30, const_100))), add(const_1, 5))) | add(n2,const_1)|divide(n0,const_100)|divide(n1,const_100)|multiply(n2,#2)|add(#1,#3)|divide(#4,#0)|multiply(#5,const_100)| | gain |
because he ’ s taxed by his home planet , mork pays a tax rate of 40 % on his income , while mindy pays a rate of only 30 % on hers . if mindy earned 5 times as much as mork did , what was their combined tax rate ? | "solution : p ( a ∪ b ) = p ( a ) + p ( b ) - p ( a ∩ b ' ) = > 0.8 = 0.4 - p ( a ∩ b ) = > p ( a ∩ b ) = 0.2 p ( a ∩ b ' ) = p ( a ) - p ( a ∩ b ) = 0.4 - 0.2 = 0.2 answer b" | a ) 0.1 , b ) 0.2 , c ) 0.3 , d ) 0.4 , e ) none | b | multiply(multiply(0.4, 0.8), const_10) | multiply(n0,n2)|multiply(#0,const_10)| | general |
if p ( a ) = 0.4 , p ( b ) = 0.6 and p ( a ∪ b ) = 0.8 . what is the value of p ( a ∩ b ' ) = ? | "ratio of investments of a and b is ( 70000 * 12 ) : ( 120000 * 6 ) = 7 : 6 total profit = rs . 26000 share of b = 6 / 13 ( 26000 ) = rs . 12000 answer : b" | a ) s . 12028 , b ) s . 12000 , c ) s . 12003 , d ) s . 12029 , e ) s . 24029 | b | subtract(26000, multiply(const_60, const_100)) | multiply(const_100,const_60)|subtract(n3,#0)| | gain |
a started a business with an investment of rs . 70000 and after 6 months b joined him investing rs . 120000 . if the profit at the end of a year is rs . 26000 , then the share of b is ? | "sol . on dividing 3105 by 21 , we get 18 as remainder . number to be added to 3105 = ( 21 - 18 ) - 3 . hence , required number = 3105 + 3 = 3108 . option b" | a ) 1208 , b ) 3108 , c ) 241 , d ) 217 , e ) 3147 | b | add(3105, subtract(21, reminder(3105, 21))) | reminder(n0,n1)|subtract(n1,#0)|add(n0,#1)| | general |
find the number which is nearest to 3105 and is exactly divisible by 21 . | "980 x 436 + 980 x 764 = 986 x ( 436 + 664 ) = 986 x 1200 = 117600 . answer is a ." | a ) 1176000 , b ) 968000 , c ) 978000 , d ) 117000 , e ) none of them | a | subtract(980, multiply(multiply(436, 980), 764)) | multiply(n1,n2)|multiply(n3,#0)|subtract(n0,#1)| | general |
evaluate : 980 x 436 + 980 x 764 | you have 40 multiples of 5 , 8 of 25 and 1 of 125 . this will give 49 zeros . c | a ) 40 , b ) 48 , c ) 49 , d ) 55 , e ) 64 | c | add(divide(200, add(const_4, const_1)), divide(200, multiply(add(const_4, const_1), add(const_4, const_1)))) | add(const_1,const_4)|divide(n0,#0)|multiply(#0,#0)|divide(n0,#2)|add(#1,#3)| | other |
how many terminating zeroes r does 200 ! have ? | distance covered in ( 9 × 1000 ) / ( 3600 ) × 12 = 30 m diagonal of squarre field = 30 m . area of square field = 30 ( power ) 2 / 2 = 900 / 2 = 450 sq . m answer is c . | ['a ) 430 sq . m', 'b ) 425 sq . m', 'c ) 450 sq . m', 'd ) 475 sq . m', 'e ) 350 sq . m'] | c | divide(multiply(multiply(12, divide(multiply(9, const_1000), multiply(const_360, const_10))), multiply(12, divide(multiply(9, const_1000), multiply(const_360, const_10)))), const_2) | multiply(n0,const_1000)|multiply(const_10,const_360)|divide(#0,#1)|multiply(n1,#2)|multiply(#3,#3)|divide(#4,const_2) | geometry |
a girl walking at the rate of 9 km per hour crosses a square field diagonally in 12 seconds . the area of the field is : | "p ( x is prime ) = 1 / 3 p ( y is prime ) = 1 / 4 if y is prime , then z is not prime since y and z are unique . then the probability is 1 / 3 * 1 / 4 = 1 / 12 the answer is d ." | a ) 1 / 5 , b ) 3 / 20 , c ) 2 / 15 , d ) 1 / 12 , e ) 1 / 10 | d | multiply(divide(const_1, const_2), divide(const_1, const_4)) | divide(const_1,const_2)|divide(const_1,const_4)|multiply(#0,#1)| | probability |
x , y , and z are all unique numbers . if x is chosen randomly from the set { 6 , 7 , 8 , 9 , 10 , 11 } and y and z are chosen randomly from the set { 20 , 21 , 22 , 23 } , what is the probability that x and y are prime and z is not ? | "27 , 36 , 45 , . . . , 90,99 this is an equally spaced list ; you can use the formula : n = ( largest - smallest ) / ( ' space ' ) + 1 = ( 99 - 27 ) / ( 9 ) + 1 = 8 + 1 = 9 answer is e" | a ) 5 , b ) 15 , c ) 12 , d ) 7 , e ) 9 | e | add(divide(subtract(100, 20), 9), const_1) | subtract(n1,n0)|divide(#0,n2)|add(#1,const_1)| | general |
what is the total number of integers between 20 and 100 that are divisible by 9 ? | "explanation : area of a triangle = r * s where r is the inradius and s is the semi perimeter of the triangle . area of triangle = 2.5 * 40 / 2 = 50 cm 2 answer : option c" | a ) a ) 72 , b ) b ) 828 , c ) c ) 50 , d ) d ) 34 , e ) e ) 35 | c | triangle_area(2.5, 40) | triangle_area(n0,n1)| | geometry |
the perimeter of a triangle is 40 cm and the inradius of the triangle is 2.5 cm . what is the area of the triangle | answer let original price = rs . 100 . then , new price = rs . 90 . ∴ increased on rs . 90 = rs . 10 required increase % = ( 10 x 100 ) / 90 % = 111 / 9 % correct option : c | a ) 10 % , b ) 9 1 / 11 , c ) 11 1 / 9 , d ) 11 % , e ) none of these | c | add(subtract(const_100, subtract(const_100, 10)), const_2) | subtract(const_100,n0)|subtract(const_100,#0)|add(#1,const_2) | gain |
the price of an article is cut by 10 % . to restore it to the former value . the new price must be increased by ? | p ( a or b ) = p ( a ) + p ( b ) - p ( a n b ) 0.4 = 0.6 + p ( a ) - 0.25 p ( a ) = 0.05 ans : a | a ) 0.05 , b ) 0.15 , c ) 0.45 , d ) 0.5 , e ) 0.55 | a | subtract(add(0.25, 0.4), 0.6) | add(n1,n2)|subtract(#0,n0) | other |
the probability that event b occurs is 0.6 , and the probability that events a and b both occur is 0.25 . if the probability that either event a or event b occurs is 0.4 , what is the probability that event a will occur ? | since 2 xy prime e factors are x ^ 1 * y ^ 1 * 2 ^ 1 , its total number or factors must be ( 1 + 1 ) ( 1 + 1 ) ( 1 + 1 ) = 2 ^ 3 = 8 . thus , i think d would be the correct answer . | a ) 3 , b ) 4 , c ) 6 , d ) 8 , e ) 12 | d | multiply(multiply(2, add(const_1, const_1)), add(const_1, const_1)) | add(const_1,const_1)|multiply(n0,#0)|multiply(#0,#1)| | other |
if x and y are both odd prime numbers and x < y , how many distinct positive integer e factors does 2 xy have ? | two ways . . . 1 ) total ways = 10 c 3 = 10 ! / 7 ! 3 ! = 120 . . ways without python developer = 6 c 3 = 6 ! / 3 ! 3 ! = 20 . . ways of at least one python developer = 120 - 20 = 100 . . 2 ) ways of selecting only one = 4 * 6 c 2 = 4 * 15 = 60 . . ways of selecting only two = 4 c 2 * 6 c 1 = 6 * 6 = 36 . . ways of sel... | a ) 100 , b ) 40 , c ) 66 , d ) 80 , e ) 75 | a | subtract(divide(factorial(10), multiply(factorial(subtract(10, 3)), factorial(3))), divide(factorial(subtract(10, 4)), multiply(factorial(3), factorial(3)))) | factorial(n0)|factorial(n2)|subtract(n0,n2)|subtract(n0,n1)|factorial(#2)|factorial(#3)|multiply(#1,#1)|divide(#5,#6)|multiply(#4,#1)|divide(#0,#8)|subtract(#9,#7) | other |
in a certain group of 10 developers , 4 developers code only in python and the rest program in either ruby on rails or php - but not both . if a developer organization is to choose a 3 - member team , which must have at least 1 developer who codes in python , how many different programming teams can be chosen ? | "his percentage gain is 100 * 80 / 920 as he is gaining 80 units for his purchase of 920 units . so 8.69 % . answer : e" | a ) 5.26 % , b ) 5.36 % , c ) 4.26 % , d ) 6.26 % , e ) 8.69 % | e | multiply(subtract(inverse(divide(920, multiply(multiply(add(const_4, const_1), const_2), const_100))), const_1), const_100) | add(const_1,const_4)|multiply(#0,const_2)|multiply(#1,const_100)|divide(n0,#2)|inverse(#3)|subtract(#4,const_1)|multiply(#5,const_100)| | gain |
a dishonest shopkeeper professes to sell pulses at the cost price , but he uses a false weight of 920 gm . for a kg . his gain is … % . | "explanation : while calculation in terms of percentage we need to multiply by 100 , so 5.40 * 100 = 540 answer : option d" | a ) 5.04 % , b ) 50.4 % , c ) 209 % , d ) 540 % , e ) none of these | d | multiply(5.40, const_100) | multiply(n0,const_100)| | general |
5.40 can be expressed in terms of percentage as | "for retail price = $ 25 first maximum discounted price = 25 - 30 % of 25 = 25 - 7.5 = 17.5 price after additional discount of 20 % = 17.5 - 20 % of 17.5 = 17.5 - 3.5 = 14 answer : option a" | a ) $ 14.00 , b ) $ 11.20 , c ) $ 14.40 , d ) $ 16.00 , e ) $ 18.00 | a | multiply(divide(subtract(const_100, 20), const_100), multiply(divide(subtract(const_100, 30), const_100), 25.00)) | subtract(const_100,n2)|subtract(const_100,n1)|divide(#0,const_100)|divide(#1,const_100)|multiply(n3,#3)|multiply(#2,#4)| | gain |
a pet store regularly sells pet food at a discount of 10 percent to 30 percent from the manufacturer ’ 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 ’ s sugge... | lets take a number 20 20 / 5 = 4 20 * 5 = 100 diff = 100 - 4 = 96 % answer : a | a ) 96 % , b ) 95 % , c ) 2400 % , d ) 200 % , e ) 400 % | a | multiply(subtract(multiply(5, 5), const_1), divide(const_100, multiply(5, 5))) | multiply(n0,n0)|divide(const_100,#0)|subtract(#0,const_1)|multiply(#1,#2) | general |
a number is mistakenly divided by 5 instead of being multiplied by 5 . find the percentage change in the result due t this mistake . | explanation : let the fourth term be x . thus 56 , 16 , 49 , x are in proportion . product of extreme terms = 56 x product of mean terms = 16 x 49 since , the numbers make up a proportion therefore , 56 x = 16 49 or , x = ( 16 49 ) / 56 or , x = 14 therefore , the fourth term of the proportion is 14 . answer : b | a ) 10 , b ) 14 , c ) 40 , d ) 50 , e ) 60 | b | divide(multiply(49, 16), 56) | multiply(n1,n2)|divide(#0,n0) | physics |
the first , second and third terms of the proportion are 56 , 16 , 49 . find the fourth term . | "assume the price = 100 price during sale = 90 price after sale = 100 percent increase = 10 / 90 * 100 = 11 % approx . correct option : c" | a ) 9 % , b ) 10 % , c ) 11 % , d ) 15 % , e ) 90 % | c | divide(multiply(10, const_100), subtract(const_100, 10)) | multiply(n0,const_100)|subtract(const_100,n0)|divide(#0,#1)| | gain |
during a sale , the price of a pair of shoes is marked down 10 % from the regular price . after the sale ends , the price goes back to the original price . what is the percent of increase to the nearest percent from the sale price back to the regular price for the shoes ? | "explanation : total hours worked = 8 x 3 + 6 x 2 = 36 total earned = 324 . hourly wage = 324 / 36 = 9 answer : c ) 9" | a ) 2 , b ) 8 , c ) 9 , d ) 1 , e ) 2 | c | divide(324, add(multiply(8, const_3), multiply(6, const_2))) | multiply(n0,const_3)|multiply(n1,const_2)|add(#0,#1)|divide(n2,#2)| | physics |
sheila works 8 hours per day on monday , wednesday and friday , and 6 hours per day on tuesday and thursday . she does not work on saturday and sunday . she earns $ 324 per week . how much does she earn in dollars per hour ? | solution : both fractions should be reduced before performing arithmetic operations . we get 3 * 27 / 31 + 3.27 / 3.31 = 3 * 27 / 31 + 27 / 31 = 4 * 27 / 31 = 151 / 31 answer d | a ) 0 , b ) 156 / 31 , c ) 123 / 31 , d ) 151 / 31 , e ) none | d | divide(add(subtract(add(81, multiply(27, 3)), subtract(93, 81)), const_1), 31) | multiply(n0,n1)|subtract(n4,n3)|add(n3,#0)|subtract(#2,#1)|add(#3,const_1)|divide(#4,n2) | general |
determine the value of 3 * 27 / 31 + 81 / 93 | "1 . we are given the following percentages : 30 ( 70 ) , 40 ( 60 ) , 25 ( 75 ) . there are two threads from here . first starts at 30 % and finishes there . second one starts at 70 , then 40 , and then 25 . we need a value that is divisible by 7 , 2 , and 5 at least once . lets pick a number now , say 700 . so say if ... | a ) 25 % , b ) 35 % , c ) 45 % , d ) 70 % , e ) 80 % | b | add(const_10, divide(add(25, 25), const_2)) | add(n2,n2)|divide(#0,const_2)|add(#1,const_10)| | general |
in goshawk - eurasian nature reserve 30 percent of the birds are hawks , and 40 percent of the non - hawks are paddyfield - warblers . if there are 25 percent as many kingfishers as paddyfield - warblers in the reserve , then what percent of the birds e in the nature reserve are not hawks , paddyfield - warblers , or k... | or u can just use the answer choices here . since the answers are already arranged in ascending order , the first number which gives remainder t as 1 for all three is the correct answer . in the given question , the first number which gives a remainder of 1 for 6,8 and 10 is 121 . c | a ) 21 , b ) 41 , c ) t = 121 , d ) 241 , e ) 481 | c | add(lcm(lcm(6, 8), 10), 1) | lcm(n2,n3)|lcm(n4,#0)|add(n0,#1) | general |
what is the smallest integer t greater than 1 that leaves a remainder of 1 when divided by any of the integers 6 , 8 , and 10 ? | "e 250 e = 72 * 5 / 18 = 20 = 400 â € “ 150 = 250" | a ) 443 m , b ) 354 m , c ) 450 m , d ) 350 m , e ) 250 m | e | subtract(multiply(20, multiply(72, const_0_2778)), 150) | multiply(n1,const_0_2778)|multiply(n2,#0)|subtract(#1,n0)| | physics |
a train 150 m long running at 72 kmph crosses a platform in 20 sec . what is the length of the platform ? | e their ages were respectively 58 and 28 | a ) 60 and 23 , b ) 66 and 25 , c ) 29 and 56 , d ) 71 and 43 , e ) 58 and 28 | e | divide(divide(multiply(1624, 30), const_4), const_2) | multiply(n0,n1)|divide(#0,const_4)|divide(#1,const_2) | general |
in bangalore there is a well known science institute . during a visit i asked two of the men to tell me their ages . one replied , ' one of our ages subtracted from the other ' s equal 30 . ' then the other man spoke . ' our ages multiplied together equal 1624 . ' what were their ages ? | lcm = 72 72 - 1 = 71 answer : a | a ) 71 , b ) 70 , c ) 72 , d ) 73 , e ) 36 | a | subtract(lcm(24, 36), 1) | lcm(n3,n4)|subtract(#0,n0) | general |
the smallest number when increased by ` ` 1 ` ` is exactly divisible by 2 , 8 , 24 , 36 is : | "cp = 135 * 15 = 2025 and sp = 115 * 18 = 2070 gain % = 100 * ( 2070 - 2025 ) / 2025 = 20 / 9 answer : c" | a ) 40 , b ) 30 / 11 , c ) 20 / 9 , d ) 27 / 11 , e ) 29 / 8 | c | multiply(divide(subtract(multiply(115, 18), multiply(135, 15)), multiply(135, 15)), const_100) | multiply(n2,n3)|multiply(n0,n1)|subtract(#0,#1)|divide(#2,#1)|multiply(#3,const_100)| | gain |
a person bought 135 glass bowls at a rate of rs . 15 per bowl . he sold 115 of them at rs . 18 and the remaining broke . what is the percentage gain for a ? | "let initial price be 100 price in day 1 after 3 % discount = 97 price in day 2 after 3 % discount = 94.09 price in day 3 after 10 % discount = 84.68 so , price in day 3 as percentage of the sale price on day 1 will be = 84.68 / 97 * 100 = > 87.3 % answer will definitely be ( c )" | a ) 85.1 % , b ) 86.9 % , c ) 87.3 % , d ) 88.8 % , e ) 89.5 % | c | add(multiply(divide(divide(10, const_100), subtract(1, divide(1, 3))), const_100), 2) | divide(n5,const_100)|divide(n1,n0)|subtract(n1,#1)|divide(#0,#2)|multiply(#3,const_100)|add(n2,#4)| | gain |
the price of an item is discounted 3 percent on day 1 of a sale . on day 2 , the item is discounted another 3 percent , and on day 3 , it is discounted an additional 10 percent . the price of the item on day 3 is what percentage of the sale price on day 1 ? | "number of pens = 1200 number of pencils = 820 required number of students = h . c . f . of 1200 and 820 = 20 answer is b" | a ) 40 , b ) 20 , c ) 60 , d ) 80 , e ) 65 | b | gcd(1200, 820) | gcd(n0,n1)| | general |
the maximum number of students among them 1200 pens and 820 pencils can be distributed in such a way that each student get the same number of pens and same number of pencils ? | this is a rather straight forward ratio problem . 1 . 80 fish tagged 2 . 2 out of the 50 fish caught were tagged thus 2 / 50 2 / 50 = 80 / x thus , x = 2000 think of the analogy : 2 fish is to 50 fish as 50 fish is to . . . ? you ' ve tagged 50 fish and you need to find what that comprises as a percentage of the total ... | a ) 400 , b ) 625 , c ) 1,250 , d ) 2,000 , e ) 10,000 | d | divide(80, divide(2, 50)) | divide(n2,n1)|divide(n0,#0) | gain |
in a certain pond , 80 fish were caught , tagged , and returned to the pond . a few days later , 50 fish were caught again , of which 2 were found to have been tagged . if the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond , what is the approximate number of fish in the p... | "r is the set of positive odd integers less than 100 , and s is the set of the squares of the integers in r . how many elements does the intersection of r and s contain ? r = 1,3 , 5,7 , 9,11 , 13,15 . . . s = 1 , 9,25 , 49,81 . . . numbers : 1 , 9 , 25 , 49 , and 81 are odd integers ( less than 100 ) that are in both ... | a ) none , b ) two , c ) four , d ) five , e ) seven | d | subtract(subtract(100, const_4), const_4) | subtract(n0,const_4)|subtract(#0,const_4)| | physics |
r is the set of positive odd integers less than 100 , and s is the set of the squares of the integers in r . how many elements does the intersection of r and s contain ? | "let the list price be 2 x for min sale price , the first discount given should be 50 % , 2 x becomes x here now , during summer sale additional 20 % off is given ie sale price becomes 0.8 x it is given lise price is $ 80 = > 2 x = 80 = > x = 40 and 0.8 x = 32 so lowest sale price is 32 , which w is 40 % of 80 hence , ... | a ) 20 , b ) 25 , c ) 30 , d ) 40 , e ) 50 | d | divide(80, const_2) | divide(n3,const_2)| | general |
a soccer store typically sells replica jerseys at a discount of 30 percent to 50 percent off list price . during the annual summer sale , everything in the store is an additional 20 percent off the original list price . if a replica jersey ' s list price is $ 80 , approximately what w percent of the list price is the l... | "28 * 2 * 0.75 = 20 / 100 * 10 / 100 * 7.5 / 100 * x 28 = 1 / 100 * x = > x = 28000 answer : a" | a ) 28000 , b ) 27908 , c ) 78902 , d ) 25000 , e ) 27991 | a | divide(divide(divide(multiply(multiply(multiply(28, const_100), multiply(2, const_100)), multiply(0.75, const_100)), 20), 10), 7.5) | multiply(n3,const_100)|multiply(n4,const_100)|multiply(n5,const_100)|multiply(#0,#1)|multiply(#3,#2)|divide(#4,n0)|divide(#5,n1)|divide(#6,n2)| | physics |
a brick measures 20 cm * 10 cm * 7.5 cm how many bricks will be required for a wall 28 m * 2 m * 0.75 m ? | "explanation : the difference of two successive amounts must be the simple interest in 1 year on the lower amount of money . s . i = 20286 / - - 17640 / - = rs . 2646 / - rate of interest = ( 2646 / 17640 ) × ( 100 / 1 ) = > 15 % answer : option d" | a ) 5 % , b ) 7 % , c ) 9 % , d ) 15 % , e ) 12 % | d | multiply(divide(subtract(20286, 17640), 17640), const_100) | subtract(n2,n0)|divide(#0,n0)|multiply(#1,const_100)| | general |
an amount at compound interest sums to rs . 17640 / - in 2 years and to rs . 20286 / - in 3 years at the same rate of interest . find the rate percentage ? | let the rate be x , then population of the bacteria after each hour can be given as 600,600 x , 600 ( x ^ 2 ) , 600 ( x ^ 3 ) now population at 4 pm = 4800 thus we have 600 ( x ^ 3 ) = 4800 = 8 thus x = 2 therefore population at 3 pm = 600 ( 4 ) = 2400 answer : a | a ) 2400 , b ) 3600 , c ) 3000 , d ) 2800 , e ) 2500 | a | multiply(multiply(power(divide(multiply(multiply(2, 4), 600), 600), const_0_33), 600), power(divide(multiply(multiply(2, 4), 600), 600), const_0_33)) | multiply(n2,n4)|multiply(n1,#0)|divide(#1,n1)|power(#2,const_0_33)|multiply(n1,#3)|multiply(#4,#3) | physics |
david works at a science lab that conducts experiments on bacteria . the population of the bacteria multiplies at a constant rate , and his job is to notate the population of a certain group of bacteria each hour . at 1 p . m . on a certain day , he noted that the population was 600 and then he left the lab . he return... | solution 5 years ago average age of a , b , c , d = 45 years = > 5 years ago total age of a , b , c , d = 45 x 4 = 180 years = > total present age of a , b , c , d = 180 + 5 x 4 = 200 years if e ' s present age is x years = 200 + x / 5 = 50 x = 50 years . answer a | a ) 50 , b ) 47 , c ) 48 , d ) 49 , e ) 46 | a | subtract(multiply(50, 5), add(multiply(45, multiply(const_2, const_2)), multiply(5, const_4))) | multiply(n0,n3)|multiply(const_2,const_2)|multiply(n0,const_4)|multiply(n1,#1)|add(#3,#2)|subtract(#0,#4) | general |
5 years ago , the average age of a , b , c and d was 45 years . with e joining them now , the average of all the 5 is 50 years . the age of e is ? | "average price per book = ( 1180 + 860 ) / ( 65 + 55 ) = 2040 / 120 = $ 17 the answer is c ." | a ) $ 13 , b ) $ 15 , c ) $ 17 , d ) $ 19 , e ) $ 21 | c | divide(add(1180, 860), add(65, 55)) | add(n1,n3)|add(n0,n2)|divide(#0,#1)| | general |
sandy bought 65 books for $ 1180 from one shop and 55 books for $ 860 from another shop . what is the average price that sandy paid per book ? | "total no of rocks = 45 probability of choosing 1 st slate rock = 15 / 45 probability of choosing 2 nd slate rock = 14 / 44 ( without replacement ) so combined probability = 15 / 45 * 14 / 44 = 7 / 66 so , answer d ." | a ) 1 / 3 , b ) 7 / 22 , c ) 1 / 9 , d ) 7 / 66 , e ) 2 / 45 | d | multiply(divide(15, add(add(15, 20), 10)), divide(subtract(15, const_1), subtract(add(add(15, 20), 10), const_1))) | add(n0,n1)|subtract(n0,const_1)|add(n2,#0)|divide(n0,#2)|subtract(#2,const_1)|divide(#1,#4)|multiply(#3,#5)| | other |
there are 15 slate rocks , 20 pumice rocks , and 10 granite rocks randomly distributed in a certain field . if 2 rocks are to be chosen at random and without replacement , what is the probability that both rocks will be slate rocks ? | "net part filled in 1 hour 1 / 4 - 1 / 9 = 5 / 36 the cistern will be filled in 36 / 5 hr = 7.2 hr answer is d" | a ) 6 hr , b ) 5.6 hr , c ) 9.5 hr , d ) 7.2 hr , e ) 4 hr | d | divide(const_1, subtract(divide(const_1, 4), divide(const_1, 9))) | divide(const_1,n0)|divide(const_1,n1)|subtract(#0,#1)|divide(const_1,#2)| | physics |
a cistern can be filled by a tap in 4 hours while it can be emptied by another tap in 9 hours . if both the taps are opened simultaneously then after how much time will the cistern get filled ? | "answer should be e . v = \ pir ^ 2 h = \ pi * 20 ^ 2 * 7 = approximately 9000" | a ) 700 , b ) 1500 , c ) 3000 , d ) 5000 , e ) 9000 | e | volume_cylinder(divide(40, const_2), 7) | divide(n0,const_2)|volume_cylinder(#0,n1)| | geometry |
approximately how many cubic feet of water are needed to fill a circular swimming pool that is 40 feet across and 7 feet deep ? | "let assume there are 100 students of which 70 are male and 30 are females if 40 are married then 60 will be single . now its given that two - sevenths of the male students are married that means 2 / 7 of 70 = 20 males are married if 40 is the total number of students who are married and out of that 20 are males then t... | a ) 2 / 7 , b ) 5 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 1 / 3 | e | divide(const_10, 40) | divide(const_10,n1)| | gain |
in a graduate physics course , 70 percent of the students are male and 40 percent of the students are married . if two - sevenths of the male students are married , what fraction of the female students is single ? | "expanding we have 6 x ^ 2 - 3 x + 4 x - 2 6 x ^ 2 + x - 2 taking coefficients , a = 6 , k = 1 , n = - 2 therefore a - n + k = 6 - ( - 2 ) + 1 = 8 + 1 = 9 the answer is c ." | a ) 5 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | c | add(add(multiply(3, 2), multiply(1, 2)), subtract(multiply(2, 2), multiply(1, 3))) | multiply(n0,n1)|multiply(n1,n3)|multiply(n1,n1)|multiply(n0,n3)|add(#0,#1)|subtract(#2,#3)|add(#4,#5)| | general |
( 3 x + 2 ) ( 2 x - 1 ) = ax ^ 2 + kx + n . what is the value of a - n + k ? | "60 % of oranges = 600 100 % of oranges = ( 600 × 100 ) / 6 = 1000 total oranges = 1000 answer : c" | a ) 700 , b ) 710 , c ) 1000 , d ) 730 , e ) 740 | c | add(600, multiply(600, divide(40, const_100))) | divide(n0,const_100)|multiply(n1,#0)|add(n1,#1)| | gain |
a fruit seller had some oranges . he sells 40 % oranges and still has 600 oranges . how many oranges he had originally ? | "the triangle with sides 196 cm , 81 cm and 277 cm is right angled , where the hypotenuse is 277 cm . area of the triangle = 1 / 2 * 81 * 196 = 7938 cm 2 answer : option e" | a ) 5000 , b ) 5656 , c ) 7878 , d ) 7900 , e ) 7938 | e | divide(multiply(81, 277), const_2) | multiply(n1,n2)|divide(#0,const_2)| | geometry |
if the sides of a triangle are 196 cm , 81 cm and 277 cm , what is its area ? | so addition of all term - 10 , 20 , 30 , . . . . . . . 90 so average = ( 10 + 20 + 30 + 40 + 50 + 60 + 70 + 80 + 90 ) / 9 = ( 450 ) / 9 = 50 hence , the correct answer is e . | a ) 90 , b ) 95 , c ) 70 , d ) 85 , e ) 50 | e | subtract(divide(add(add(add(add(10, 2030), 4050), 6070), 8090), add(const_4, const_1)), multiply(multiply(const_100, const_10), const_4)) | add(n0,n1)|add(const_1,const_4)|multiply(const_10,const_100)|add(n2,#0)|multiply(#2,const_4)|add(n3,#3)|add(n4,#5)|divide(#6,#1)|subtract(#7,#4) | general |
what is the average ( arithmetic mean ) of 10 , 2030 , 4050 , 6070 , 8090 ? | "100 - - - 10 ds = 10 ? - - - - 1 30 - - - - 10 us = 3 ? - - - - 1 s = ? s = ( 10 - 3 ) / 2 = 3.5 answer : b" | a ) 2.5 , b ) 3.5 , c ) 4.5 , d ) 5.3 , e ) 3.4 | b | divide(add(divide(30, 10), divide(100, 10)), const_2) | divide(n1,n2)|divide(n0,n2)|add(#0,#1)|divide(#2,const_2)| | physics |
a man swims downstream 100 km and upstream 30 km taking 10 hours each time ; what is the speed of the current ? | "let the breadth of the plot be b m . length of the plot = 3 b m ( 3 b ) ( b ) = 1323 3 b 2 = 1323 b 2 = 441 = 21 ( b > 0 ) b = 21 m . answer : d" | a ) 11 , b ) 17 , c ) 18 , d ) 21 , e ) 1322 | d | sqrt(divide(1323, const_3)) | divide(n0,const_3)|sqrt(#0)| | geometry |
the length of a rectangular plot is thrice its breadth . if the area of the rectangular plot is 1323 sq m , then what is the breadth of the rectangular plot ? | "let initial volume of ice be = x ice remaining after 1 hour = x - 0.75 x = 0.25 x ice remaining after 2 hour = ( 1 / 4 ) x - ( 3 / 4 * 1 / 4 * x ) = ( 1 / 16 ) x ( 1 / 16 ) x = 0.3 x = 4.8 alternate solution : try to backsolve . initial volume = 4.8 after one hour - - > ( 1 / 4 ) 4.8 = 1.2 after two hours - - > ( 1 / ... | a ) 2.5 , b ) 3.0 , c ) 4.8 , d ) 6.5 , e ) 8.0 | c | divide(divide(0.3, const_0_25), const_0_25) | divide(n4,const_0_25)|divide(#0,const_0_25)| | physics |
after an ice began to melt out from the freezer , in the first hour lost 3 / 4 , in the second hour lost 3 / 4 of its remaining . if after two hours , the volume is 0.3 cubic inches , what is the original volume of the cubic ice , in cubic inches ? | "area of a triangle = r * s where r is the inradius and s is the semi perimeter of the triangle . area of triangle = 3.5 * 22 / 2 = 38.5 cm 2 answer : e" | a ) 22 , b ) 35 , c ) 77 , d ) 54 , e ) 38 | e | triangle_area(3.5, 22) | triangle_area(n0,n1)| | geometry |
the perimeter of a triangle is 22 cm and the inradius of the triangle is 3.5 cm . what is the area of the triangle ? | "index for females = ( 20 - 6 ) / 20 = 7 / 10 = 0.7 index for males = ( 20 - 14 / 20 = 3 / 10 = 0.3 index for females exceeds males by 0.7 - 0.3 = 0.4 answer : a" | a ) 0.4 , b ) 0.0625 , c ) 0.2 , d ) 0.25 , e ) 0.6 | a | subtract(divide(subtract(20, 6), 20), divide(6, 20)) | divide(n1,n0)|subtract(n0,n1)|divide(#1,n0)|subtract(#2,#0)| | general |
for a group of n people , k of whom are of the same sex , the ( n - k ) / n expression yields an index for a certain phenomenon in group dynamics for members of that sex . for a group that consists of 20 people , 6 of whom are females , by how much does the index for the females exceed the index for the males in the gr... | "x ' s 1 day ' s work = 1 / 40 y ' s 1 day ' s work = 1 / 60 ( x + y ) ' s 1 day ' s work = ( 1 / 40 + 1 / 60 ) = 1 / 24 both together will finish the work in 24 days . correct option is c" | a ) 10 , b ) 12 , c ) 24 , d ) 30 , e ) 15 | c | inverse(add(divide(const_1, 40), divide(const_1, 60))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|inverse(#2)| | physics |
x does a work in 40 days . y does the same work in 60 days . in how many days they together will do the same work ? | "let the number be xy . given xy – yx = 36 . this means the number is greater is than the number got on reversing the digits . this shows that the ten ’ s digit x > unit digit y . also given ratio between digits is 1 : 2 = > x = 2 y ( 10 x + y ) – ( 10 y + x ) = 36 = > x – y = 4 = > 2 y – y = 4 . hence , ( x + y ) – ( ... | a ) 6 , b ) 8 , c ) 10 , d ) 14 , e ) 15 | b | multiply(divide(36, subtract(multiply(subtract(const_10, 1), multiply(2, 1)), subtract(const_10, 1))), 2) | multiply(n0,n2)|subtract(const_10,n2)|multiply(#0,#1)|subtract(#2,#1)|divide(n1,#3)|multiply(#4,n0)| | general |
the difference of 2 digit number & the number obtained by interchanging the digits is 36 . what is the sum and the number if the ratio between the digits of the number is 1 : 2 ? | explanation : solution : let the required number of carpets be x . more weavers , more carpets ( direct proportion ) more days , more carpets ( direct proportion ) weavers 7 : 14 } : : 7 : x days 7 : 14 . ' . 7 * 7 * x = 14 * 14 * 7 < = > x = 14 * 14 * 7 / 7 * 7 = 28 . answer : b | a ) 14 , b ) 28 , c ) 21 , d ) 35 , e ) none of these | b | add(14, add(7, 7)) | add(n0,n0)|add(n3,#0) | gain |
7 carpet - weavers can weave 7 carpets in 7 days . at the same rate , how many carpets would be woven by 14 carpet - weavers in 14 days ? | "sol . speed of stream = 1 / 2 ( 12 - 6 ) kmph = 3 kmph . answer c" | a ) 1 kmph , b ) 4 kmph , c ) 3 kmph , d ) 2 kmph , e ) 1.9 kmph | c | divide(subtract(12, 6), const_2) | subtract(n1,n0)|divide(#0,const_2)| | physics |
what is the speed of the stream if a canoe rows upstream at 6 km / hr and downstream at 12 km / hr | i year best is to give a number to his income , say 100 . . and let saving be x . . so expenditure = 100 - x next year - income = 140 savings = 2 x expenditure = 140 - 2 x . . now 140 - 2 x + 100 - x = 2 ( 100 - x ) . . . 240 - 3 x = 200 - 2 x . . . . . . . . . . . . . . . . x = 40 . . . saving % = 40 / 100 * 100 = 40 ... | a ) 45 % , b ) 40 % , c ) 25 % , d ) 28 % , e ) 33.33 % | b | multiply(divide(subtract(add(add(100, 40), 100), multiply(2, 100)), const_100), const_100) | add(n0,n1)|multiply(n1,n2)|add(n1,#0)|subtract(#2,#1)|divide(#3,const_100)|multiply(#4,const_100) | general |
a man saves a certain portion of his income during a year and spends the remaining portion on his personal expenses . next year his income increases by 40 % but his savings increase by 100 % . if his total expenditure in 2 years is double his expenditure in 1 st year , what % age of his income in the first year did he ... | "speed = 54 * 5 / 18 = 15 m / sec . length of the train = 15 * 20 = 300 m . let the length of the platform be x m . then , ( x + 300 ) / 40 = 15 = > x = 180 m answer : c" | a ) 615 m , b ) 240 m , c ) 180 m , d ) 197 m , e ) 691 m | c | multiply(20, multiply(54, const_0_2778)) | multiply(n2,const_0_2778)|multiply(n1,#0)| | physics |
a train passes a station platform in 40 sec and a man standing on the platform in 20 sec . if the speed of the train is 54 km / hr . what is the length of the platform ? | "speed upstream = 8.5 kmph speed downstream = 11.5 kmph total time taken = 105 / 8.5 + 105 / 11.5 = 21.48 hours answer is b" | a ) 20.48 hours , b ) 21.48 hours , c ) 22.48 hours , d ) 23.48 hours , e ) 24.48 hours | b | add(multiply(add(add(10, 1.5), subtract(10, 1.5)), 105), multiply(subtract(add(divide(105, add(10, 1.5)), divide(105, subtract(10, 1.5))), add(add(10, 1.5), subtract(10, 1.5))), const_60)) | add(n0,n1)|subtract(n0,n1)|add(#0,#1)|divide(n2,#0)|divide(n2,#1)|add(#3,#4)|multiply(n2,#2)|subtract(#5,#2)|multiply(#7,const_60)|add(#6,#8)| | physics |
speed of a boat in standing water is 10 kmph and speed of the stream is 1.5 kmph . a man can rows to a place at a distance of 105 km and comes back to the starting point . the total time taken by him is ? | "let # plain cookies sold be x then # chocolate cookies = ( total cookies - x ) equating for x ( 0.75 ) * x + ( 1.25 ) * ( 1585 - x ) = 1587.75 = > x = 787 e" | a ) 0 , b ) 233 , c ) 500 , d ) 695 , e ) 787 | e | divide(add(const_1000, 587.75), const_2) | add(n4,const_1000)|divide(#0,const_2)| | other |
a girl scout was selling boxes of cookies . in a month , she sold both boxes of chocolate chip cookies ( $ 1.25 each ) and boxes of plain cookies ( $ 0.75 each ) . altogether , she sold 1,585 boxes for a combined value of $ 1 , 587.75 . how many boxes of plain cookies did she sell ? | "let the smaller number be x . then larger number = ( x + 1365 ) . x + 1365 = 6 x + 10 5 x = 1355 x = 271 large number = 271 + 1365 = 1636 a" | a ) 1636 , b ) 1346 , c ) 1378 , d ) 1635 , e ) 1489 | a | multiply(divide(subtract(1365, 10), subtract(6, const_1)), 6) | subtract(n0,n2)|subtract(n1,const_1)|divide(#0,#1)|multiply(n1,#2)| | general |
find large number from below question the difference of two numbers is 1365 . on dividing the larger number by the smaller , we get 6 as quotient and the 10 as remainder | widgets inspected by lauren : ( ( 95 - 5 ) / 5 ) + 1 = 18 + 1 = 19 widgets inspected by steven : ( ( 96 - 4 ) / 4 ) + 1 = 23 + 1 = 24 widgets inspected by both : ( ( 96 / 12 ) + 1 = 9 total : 19 + 24 - 9 = 34 hence , widgets not inspected : 98 - 34 = 64 option d | a ) 66 , b ) 68 , c ) 70 , d ) 64 , e ) 72 | d | subtract(98, subtract(add(floor(divide(98, add(const_4, const_1))), floor(divide(98, const_4))), floor(divide(98, add(const_10, add(const_4, const_1)))))) | add(const_1,const_4)|divide(n0,const_4)|add(#0,const_10)|divide(n0,#0)|floor(#1)|divide(n0,#2)|floor(#3)|add(#6,#4)|floor(#5)|subtract(#7,#8)|subtract(n0,#9) | other |
two assembly line inspectors , lauren and steven , inspect widgets as they come off the assembly line . if lauren inspects every fifth widget , starting with the fifth , and steven inspects every fourth , starting with the fourth , how many of the 98 widgets produced in the first hour of operation are not inspected by ... | "on dividing 3000 by 19 , we get 17 as remainder . number to be added = ( 19 - 17 ) = 2 . answer a 2" | a ) 2 , b ) 1 , c ) 4 , d ) 18 , e ) 17 | a | subtract(multiply(add(multiply(const_4, const_10), const_2), 19), 3000) | multiply(const_10,const_4)|add(#0,const_2)|multiply(n1,#1)|subtract(#2,n0)| | general |
what least number must be added to 3000 to obtain a number exactly divisible by 19 ? | "p ( a or b ) = p ( a ) + p ( b ) - p ( a n b ) 0.6 = 0.4 + p ( b ) - 0.45 p ( b ) = 0.55 ans : e" | a ) 0.05 , b ) 0.15 , c ) 0.45 , d ) 0.5 , e ) 0.55 | e | subtract(add(0.6, 0.45), 0.4) | add(n1,n2)|subtract(#0,n0)| | other |
the probability that event a occurs is 0.4 , and the probability that events a and b both occur is 0.45 . if the probability that either event a or event b occurs is 0.6 , what is the probability that event b will occur ? | "speed = [ 36 x 5 / 18 ] m / sec = 10 m / sec time = 45 sec let the length of bridge be x metres . then , ( 130 + x ) / 45 = 10 = > 130 + x = 450 = > x = 320 m . answer : a" | a ) 320 m , b ) 225 m , c ) 245 m , d ) 250 m , e ) 240 m | a | subtract(multiply(divide(multiply(36, speed(const_1000, const_1)), speed(const_3600, const_1)), 45), 130) | speed(const_1000,const_1)|speed(const_3600,const_1)|multiply(n1,#0)|divide(#2,#1)|multiply(n2,#3)|subtract(#4,n0)| | physics |
the length of the bridge , which a train 130 metres long and travelling at 36 km / hr can cross in 45 seconds , is : | "total distance travelled in 12 hours = ( 35 + 37 + 39 + . . . upto 12 terms ) . this is an a . p . with first term , a = 35 , number of terms , n = 12 , common difference d = 2 required distance = 12 / 2 ( 2 * 35 + ( 12 - 1 ) * 2 ) = 6 ( 70 + 22 ) = 552 km . correct option : c" | a ) 456 kms , b ) 482 kms , c ) 552 kms , d ) 556 kms , e ) none of these | c | multiply(add(multiply(2, 35), multiply(subtract(12, const_1), 2)), divide(12, 2)) | divide(n2,n0)|multiply(n0,n1)|subtract(n2,const_1)|multiply(n0,#2)|add(#1,#3)|multiply(#4,#0)| | physics |
the speed of a car increases by 2 kms after every one hour . if the distance travelled in the first one hour was 35 kms , what was the total distance travelled in 12 hours ? | "according to order of operations , 24 ? 3 ? 10 ( division and multiplication ) is done first from left to right 24 / 2 = 8 * 10 = 80 hence 11110 + 24 * 3 * 10 = 11110 + 80 = 11190 correct answer c" | a ) 90111 , b ) 52631 , c ) 11190 , d ) 65321 , e ) 11133 | c | subtract(11110, multiply(multiply(24, 3), 10)) | multiply(n1,n2)|multiply(n3,#0)|subtract(n0,#1)| | general |
evaluate : 11110 + 24 * 3 * 10 = ? | "360 . answer d" | a ) 720 , b ) 120 , c ) 300 , d ) 360 , e ) 333 | d | subtract(subtract(subtract(divide(divide(divide(factorial(15), factorial(subtract(15, 3))), factorial(3)), const_2), 15), 15), const_10) | factorial(n0)|factorial(n1)|subtract(n0,n1)|factorial(#2)|divide(#0,#3)|divide(#4,#1)|divide(#5,const_2)|subtract(#6,n0)|subtract(#7,n0)|subtract(#8,const_10)| | physics |
mariah has decided to hire three workers . to determine whom she will hire , she has selected a group of 15 candidates . she plans to have one working interview with 3 of the 15 candidates every day to see how well they work together . how many days will it take her to have working interviews with all the different com... | "explanation : 2356 6532 - - - - - - - - - - - - 4176 answer : b" | a ) 6084 , b ) 4176 , c ) 2077 , d ) 2721 , e ) 1812 | b | subtract(add(add(add(multiply(multiply(6, const_100), const_10), multiply(5, const_100)), multiply(3, const_10)), 2), add(add(add(const_1000, multiply(3, const_100)), multiply(5, const_10)), 6)) | multiply(n0,const_100)|multiply(n3,const_100)|multiply(n1,const_10)|multiply(n1,const_100)|multiply(n3,const_10)|add(#3,const_1000)|multiply(#0,const_10)|add(#6,#1)|add(#5,#4)|add(#7,#2)|add(n0,#8)|add(n2,#9)|subtract(#11,#10)| | general |
what is the difference between the largest number and the least number written with the digits 6 , 3 , 2 , 5 ? | ( 1000 xtx 4 / 100 ) + ( 1400 xtx 5 / 100 ) = 350 â † ’ t = 3.2 answer a | a ) 3.2 , b ) 3.75 , c ) 4 , d ) 4.25 , e ) 4.5 | a | divide(350, add(divide(multiply(4, 1000), const_100), divide(multiply(1400, 5), const_100))) | multiply(n0,n1)|multiply(n2,n3)|divide(#0,const_100)|divide(#1,const_100)|add(#2,#3)|divide(n4,#4)| | gain |
a money lender lent rs . 1000 at 4 % per year and rs . 1400 at 5 % per year . the amount should be returned to him when the total interest comes to rs . 350 . find the number of years . | "n = 4 k + 1 = 5 j + 4 let ' s start at 1 = 4 ( 0 ) + 1 and keep adding 4 until we find a number in the form 5 j + 4 . 1 , 5 , 9 = 5 ( 1 ) + 4 the next such number is 9 + 4 * 5 = 29 . 9 + 29 = 38 the answer is c ." | a ) 30 , b ) 34 , c ) 38 , d ) 42 , e ) 46 | c | add(add(multiply(5, const_2), 4), add(multiply(5, multiply(const_2, 4)), 4)) | multiply(n2,const_2)|multiply(const_2,n3)|add(n3,#0)|multiply(n2,#1)|add(n3,#3)|add(#2,#4)| | general |
a group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 5 with 4 students left over . what is the sum of the two smallest possible values of n ? | divisor = ( 6 * 3 ) + 7 = 25 5 * quotient = 25 quotient = 5 . dividend = ( divisor * quotient ) + remainder dividend = ( 20 * 5 ) + 6 = 106 . e ) | a ) 74 , b ) 78 , c ) 86 , d ) 92 , e ) 106 | e | add(multiply(add(multiply(6, const_3), 7), divide(add(multiply(6, const_3), 7), 5)), 6) | multiply(n0,const_3)|add(n2,#0)|divide(#1,n1)|multiply(#1,#2)|add(n0,#3) | general |
in a division sum , the remainder is 6 and the divisor is 5 times the quotient and is obtained by adding 7 to the thrice of the remainder . the dividend is | "b = 1 / 8 – 1 / 22 = 0.075 days answer : a" | a ) 0.075 days , b ) 0.45 days , c ) 0.55 days , d ) 0.25 days , e ) 0.15 days | a | inverse(subtract(inverse(8), inverse(20))) | inverse(n0)|inverse(n1)|subtract(#0,#1)|inverse(#2)| | physics |
a and b together can do a piece of work in 8 days . if a alone can do the same work in 20 days , then b alone can do the same work in ? | "gear l - - 20 rotations per 60 seconds - - 2 rotation per 6 seconds . gear r - - 60 rotations per 60 seconds - - 6 rotations per 6 seconds . first 6 seconds - - gear l makes 1 rotation . - - gear r makes 4 rotations - - net difference - - 4 rotations hence every 6 seconds the difference between the number of rotations... | a ) a ) 6 , b ) b ) 8 , c ) c ) 10 , d ) d ) 14 , e ) e ) 12 | e | divide(divide(8, subtract(divide(60, const_60), divide(20, const_60))), const_3) | divide(n1,const_60)|divide(n0,const_60)|subtract(#0,#1)|divide(n2,#2)|divide(#3,const_3)| | physics |
circular gears l and r start to rotate at the same time at the same rate . gear l makes 20 complete revolutions per minute and gear r makes 60 revolutions per minute . how many seconds after the gears start to rotate will gear r have made exactly 8 more revolutions than gear l ? | ( ( 90 θ 33 ) θ 17 ) the remainder of 90 divided by 33 is 24 ; the remainder of 24 divided by 17 is 7 ; ( 97 θ ( 33 θ 17 ) ) the remainder of 33 divided by 17 is 16 ; the remainder of 97 divided by 16 is 1 . 7 - 1 = 6 . answer : d . | a ) 0 , b ) 2 , c ) 4 , d ) 6 , e ) 8 | d | subtract(reminder(reminder(90, 33), 17), reminder(97, reminder(33, 17))) | reminder(n0,n1)|reminder(n1,n2)|reminder(#0,n2)|reminder(n3,#1)|subtract(#2,#3) | general |
for all positive integers m and v , the expression m θ v represents the remainder when m is divided by v . what is the value of ( ( 90 θ 33 ) θ 17 ) - ( 97 θ ( 33 θ 17 ) ) ? | "we get 2 quadratic equations here . . 1 ) x ^ 2 - x - 3 = 0 . . . . . . . roots 2 , - 1 2 ) x ^ 2 + x - 3 = 0 . . . . . . . . roots - 2 , 1 inserting each root in given equation , it can be seen that - 1 and - 2 do not satisfy the equations . so value of x for given equation . . . . x = 3 or x = 1 i guess range is 3 -... | a ) 4 , b ) 3 , c ) 2 , d ) 1 , e ) 0 | c | sqrt(3) | sqrt(n1)| | general |
what is the range of all the roots of | x ^ 2 - 3 | = x ? | "given that 1,000 microns = 1 decimeter = 1,000 , 000,000 angstroms so , 1 micron = 1,000 , 000,000 / 1,000 = 1 , 000,000 answer : c" | a ) 1,000 , b ) 100 , c ) 1 , 000,000 , d ) 10 , e ) 10,000 | c | multiply(divide(1, multiply(const_100, const_100)), multiply(const_100, const_100)) | multiply(const_100,const_100)|divide(n1,#0)|multiply(#1,#0)| | general |
if 1,000 microns = 1 decimeter , and 1,000 , 000,000 angstroms = 1 decimeter , how many angstroms equal 1 micron ? | "required difference = [ 3 1 / 2 % of $ 8400 ] – [ 3 1 / 3 % of $ 8400 ] = [ ( 7 / 20 ) - ( 10 / 3 ) ] % of $ 8400 = 1 / 6 % of $ 8400 = $ [ ( 1 / 6 ) * ( 1 / 100 ) * 8400 ] = $ 14 . answer a ." | a ) 14 , b ) 24 , c ) 34 , d ) 12 , e ) 13 | a | divide(multiply(subtract(add(divide(1, 2), 3), add(divide(1, 3), 3)), 8400), const_100) | divide(n1,n2)|divide(n1,n0)|add(n0,#0)|add(n0,#1)|subtract(#2,#3)|multiply(n6,#4)|divide(#5,const_100)| | general |
if the sales tax reduced from 3 1 / 2 % to 3 1 / 3 % , then what difference does it make to a person who purchases an article with market price of $ 8400 ? | "let the numbers be x , x + 1 and x + 2 then , x + ( x + 1 ) + ( x + 2 ) = 63 3 x = 60 x = 20 greatest number , ( x + 2 ) = 22 . answer : d" | a ) 26 , b ) 28 , c ) 29 , d ) 22 , e ) 31 | d | divide(add(63, const_1), const_2) | add(n0,const_1)|divide(#0,const_2)| | physics |
the sum of three consecutive numbers is 63 . the greatest among these three number is : | "p makes x sprockets per hour . then q makes 1.1 x sprockets per hour . 770 / x = 770 / 1.1 x + 10 1.1 ( 770 ) = 770 + 11 x 11 x = 77 x = 7 the answer is c ." | a ) 3 , b ) 5 , c ) 7 , d ) 9 , e ) 11 | c | divide(subtract(770, divide(770, add(divide(10, const_100), const_1))), 10) | divide(n1,const_100)|add(#0,const_1)|divide(n0,#1)|subtract(n0,#2)|divide(#3,n1)| | gain |
machine p and machine q are each used to manufacture 770 sprockets . it takes machine p 10 hours longer to produce 770 sprockets than machine q . machine q produces 10 % more sprockets per hour than machine a . how many sprockets per hour does machine a produce ? | explanation : 20 cp = 30 sp 30 - - - 10 cp loss 100 - - - ? = > 33.33 % loss answer : c | a ) 63.33 % , b ) 34.33 % , c ) 33.33 % , d ) 31.33 % , e ) 36.33 % | c | multiply(subtract(const_1, divide(20, 30)), const_100) | divide(n0,n1)|subtract(const_1,#0)|multiply(#1,const_100) | gain |
the c . p of 20 books is equal to the s . p of 30 books . find his gain % or loss % ? | "solution : given ratio of ram and shayam ' s weight = 3 : 5 hence , ( x - 15 ) / ( 15 - 10 ) = 3 / 5 or , x = 18 % . answer : option a" | a ) 18 % , b ) 10 % , c ) 21 % , d ) 16 % , e ) none | a | add(15, multiply(subtract(15, 10), divide(3, 5))) | divide(n0,n1)|subtract(n4,n2)|multiply(#0,#1)|add(n4,#2)| | gain |
weights of two friends ram and shyam are in the ratio 3 : 5 . if ram ' s weight is increased by 10 % and total weight of ram and shyam become 82.8 kg , with an increases of 15 % . by what percent did the weight of shyam has to be increased ? | "percent of defective produced = 7 % percent of the defective units that are shipped for sale = 4 % percent of units produced are defective units that are shipped for sale = ( 4 / 100 ) * ( 7 / 100 ) * 100 % = ( 28 / 10000 ) * 100 % = ( 28 / 100 ) % = . 28 % answer b" | a ) 0.125 % , b ) 0.28 % , c ) 0.8 % , d ) 1.25 % , e ) 2.0 % | b | multiply(7, divide(4, const_100)) | divide(n1,const_100)|multiply(n0,#0)| | gain |
in the manufacture of a certain product , 7 percent of the units produced are defective and 4 percent of the defective units are shipped for sale . what percent of the units produced are defective units that are shipped for sale ? | "solution sum of the ages of 14 students = ( 16 x 35 ) - ( 14 x 21 ) = 560 - 294 . = 266 . ∴ required average = 266 / 7 = 38 years . answer d" | a ) 14 years , b ) 17 years , c ) 19 years , d ) 38 years , e ) none | d | subtract(add(add(multiply(35, 16), 21), 35), multiply(35, 16)) | multiply(n0,n1)|add(n2,#0)|add(n0,#1)|subtract(#2,#0)| | general |
the average age of 35 students in a class is 16 years . the average age of 21 students is 14 . what is the average age of remaining 7 students ? | "original price = 100 cp = 70 s = 70 * ( 180 / 100 ) = 126 100 - 126 = 26 % answer : b" | a ) 18 % , b ) 26 % , c ) 12 % , d ) 32 % , e ) 15 % | b | multiply(subtract(divide(divide(multiply(subtract(const_100, 30), add(const_100, 80)), const_100), const_100), const_1), const_100) | add(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|divide(#3,const_100)|subtract(#4,const_1)|multiply(#5,const_100)| | gain |
a trader bought a car at 30 % discount on its original price . he sold it at a 80 % increase on the price he bought it . what percent of profit did he make on the original price ? | average speed = 2 * 3 * 2 / 3 + 2 = 12 / 5 km / hr distance traveled = 12 / 5 * 5 = 12 km distance between house and school = 12 / 2 = 6 km answer is b | a ) 5 km , b ) 6 km , c ) 10 km , d ) 12 km , e ) 8 km | b | multiply(divide(5, add(divide(3, 2), const_1)), 3) | divide(n0,n1)|add(#0,const_1)|divide(n2,#1)|multiply(n0,#2) | physics |
a boy goes to his school from his house at a speed of 3 km / hr and return at a speed of 2 km / hr . if he takes 5 hours in going and coming , the distance between his house and school is ? | let length of tunnel is x meter distance = 600 + x meter time = 1 minute = 60 seconds speed = 78 km / hr = 78 * 5 / 18 m / s = 65 / 3 m / s distance = speed * time 600 + x = ( 65 / 3 ) * 60 600 + x = 20 * 65 = 1300 x = 1300 - 600 = 700 meters answer : a | a ) 700 , b ) 288 , c ) 500 , d ) 277 , e ) 121 | a | multiply(multiply(subtract(divide(600, multiply(subtract(63, 3), const_0_2778)), const_1), const_10), const_2) | subtract(n2,n1)|multiply(#0,const_0_2778)|divide(n0,#1)|subtract(#2,const_1)|multiply(#3,const_10)|multiply(#4,const_2) | physics |
how many seconds will a 600 meter long train take to cross a man walking with a speed of 3 km / hr in the direction of the moving train if the speed of the train is 63 km / hr ? | a ' s 1 hour work = 1 / 4 ; ( b + c ) ' s 1 hour work = 1 / 3 ; ( a + b ) ' s 1 hour work = 1 / 2 ( a + b + c ) ' s 1 hour work = ( 1 / 4 + 1 / 3 ) = 7 / 12 c ' s 1 hour work = ( 7 / 12 - 1 / 2 ) = 1 / 12 c alone will take 12 hours to do the work . answer : a | a ) 12 hours , b ) 10 hours , c ) 6 hours , d ) 8 hours , e ) 4 hours | a | divide(const_1, subtract(divide(const_1, 3), subtract(divide(const_1, 2), divide(const_1, 4)))) | divide(const_1,n1)|divide(const_1,n2)|divide(const_1,n0)|subtract(#1,#2)|subtract(#0,#3)|divide(const_1,#4) | physics |
a can do a piece of work in 4 hours ; b and c together can do it in 3 hours , which a and b together can do it in 2 hours . how long will c alone take to do it ? | "total = 75 books . 60 % of books that were loaned out are returned - - > 100 % - 60 % = 40 % of books that were loaned out are not returned . now , there are 68 books , thus 75 - 65 = 10 books are not returned . { loaned out } * 0.4 = 10 - - > { loaned out } = 25 . answer : b ." | a ) 20 , b ) 25 , c ) 35 , d ) 40 , e ) 55 | b | divide(subtract(75, 65), subtract(const_1, divide(60, const_100))) | divide(n1,const_100)|subtract(n0,n2)|subtract(const_1,#0)|divide(#1,#2)| | gain |
a particular library has 75 books in a special collection , all of which were in the library at the beginning of the month . these book are occasionally loaned out through an inter - library program . if , by the end of the month , 60 percent of books that were loaned out are returned and there are 65 books in the spec... | "original cost c 1 = t 1 * b 1 ^ 4 new cost c 2 = t 2 * b 2 ^ 4 . . . . only b is doubled so t 2 = t 1 and b 2 = 2 b 1 c 2 = t 2 * ( 2 b 1 ) ^ 4 = 16 ( t 1 * b 1 ^ 4 ) = 16 c 1 16 times c 1 = > 1600 % of c 1 ans d = 1600" | a ) q = 200 , b ) q = 600 , c ) q = 800 , d ) q = 1600 , e ) q = 50 | d | multiply(power(const_2, 4), const_100) | power(const_2,n0)|multiply(#0,const_100)| | general |
cost is expressed by the formula tb ^ 4 . if b is doubled , the new cost q is what percent of the original cost ? | "a = 2 m = m + 12 m = 12 a = 24 a = 4 b , and so b = 6 the answer is a ." | a ) 6 , b ) 12 , c ) 10 , d ) 15 , e ) 18 | a | divide(multiply(2, 12), 4) | multiply(n0,n2)|divide(#0,n1)| | general |
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