Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
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the population of a city increases by 7 % per year but due to migration it decrease by 1 % per years . what will be the percentage increase in population in 3 years ? | "actual increase in population = 6 % let , earlier population = 100 then the population after 3 years = 100 ( 1 + 6 / 100 ) ^ 3 = 119.1016 β΄ required percentage = 9.1016 % answer : b" | a ) 9 % , b ) 9.10 % , c ) 27 % , d ) 12 % , e ) none of these | b | add(divide(multiply(add(const_100, add(divide(multiply(add(const_100, subtract(7, 1)), subtract(7, 1)), const_100), subtract(7, 1))), subtract(7, 1)), const_100), add(divide(multiply(add(const_100, subtract(7, 1)), subtract(7, 1)), const_100), subtract(7, 1))) | subtract(n0,n1)|add(#0,const_100)|multiply(#1,#0)|divide(#2,const_100)|add(#3,#0)|add(#4,const_100)|multiply(#5,#0)|divide(#6,const_100)|add(#4,#7)| | general |
at a restaurant , glasses are stored in two different - sized boxes . one box contains 12 glasses , and the other contains 16 glasses . if the average number of glasses per box is 15 , and there are 16 more of the larger boxes , what is the total number of glasses r at the restaurant ? ( assume that all boxes are fille... | most test takers would recognize thesystemof equations in this prompt and just do algebra to get to the solution ( and that ' s fine ) . the wording of the prompt and the ' spread ' of the answer choices actually provide an interesting ' brute force ' shortcut that you can take advantage of to eliminate the 4 wrong ans... | a ) 96 , b ) 240 , c ) r = 256 , d ) r = 384 , e ) r = 480 | e | multiply(multiply(16, const_2), 15) | multiply(n1,const_2)|multiply(n2,#0) | general |
a factory has three types of machines , each of which works at its own constant rate . if 7 machine as and 11 machine bs can produce 360 widgets per hour , and if 8 machine as and 22 machine cs can produce 600 widgets per hour , how many widgets could one machine a , one machine b , and one machine c produce in one 8 -... | "let machine a produce a widgets per hour . b produce b widgets per hour and c produce c widgets per hour . 7 a + 11 b = 360 - - - ( 1 ) 8 a + 22 c = 600 - - - ( 2 ) dividing ( 2 ) by 2 4 a + 11 c = 300 . . . . . ( 3 ) adding ( 1 ) ( 3 ) 11 a + 11 b + 11 c = 660 a + b + c = 60 per hour so for eight hrs = 60 * 8 = 480 =... | a ) 400 , b ) 475 , c ) 480 , d ) 625 , e ) 700 | c | multiply(divide(600, 11), 8) | divide(n5,n1)|multiply(n3,#0)| | physics |
on increasing the number of lines in a page by 60 , they become 240 . what is the % of increase in the no . of lines in the page ? | "explanation : number of pages increased = 60 now , the number of pages of book = 240 number of pages of the books before increase = 240 β 60 = 180 % increase in the number of pages in the book = 60 / 180 x 100 % = 33.3 % a" | a ) 33.3 % , b ) 305 , c ) 50 % , d ) 55 % , e ) 60 % | a | subtract(multiply(divide(240, subtract(240, 60)), const_100), const_100) | subtract(n1,n0)|divide(n1,#0)|multiply(#1,const_100)|subtract(#2,const_100)| | general |
if ( 2 ^ 14 ) ( 25 ^ s ) = 5 ( 10 ^ m ) what is the value of m ? | "given 2 ^ 14 * 25 ^ s = 5 * 10 ^ 2 = > 2 ^ 14 * 5 ^ ( 2 s ) = 2 ^ m * 5 ^ ( m + 1 ) ans e on comparing the power of 2 = > m = 14" | a ) 7 , b ) 8 , c ) 15 , d ) 16 , e ) 14 | e | power(divide(10, 5), const_4) | divide(n4,n3)|power(#0,const_4)| | general |
in a box carrying one dozen of oranges , one - third have become bad . if 3 oranges are taken out from the box at random , what is the probability that at least one orange out of the 3 oranges picked up is good ? | n ( s ) = 12 c 3 = 12 Γ 11 Γ 10 / 3 Γ 2 = 2 Γ 11 Γ 10 = 220 no . of selection of 3 oranges out of the total 12 oranges = 12 c 3 = 2 Γ 11 Γ 10 = 220 no . of selection of 3 bad oranges out of the total 4 bad oranges = 4 c 3 = 4 \ n ( e ) = no . of desired selection of oranges = 220 β 4 = 216 \ p ( e ) = n ( e ) / n ( s )... | a ) 1 β 55 , b ) 54 β 55 , c ) 45 β 55 , d ) 3 β 55 , e ) none of these | b | divide(subtract(divide(multiply(multiply(add(const_10, const_1), const_12), const_10), multiply(3, const_2)), const_4), divide(multiply(multiply(add(const_10, const_1), const_12), const_10), multiply(3, const_2))) | add(const_1,const_10)|multiply(n0,const_2)|multiply(#0,const_12)|multiply(#2,const_10)|divide(#3,#1)|subtract(#4,const_4)|divide(#5,#4) | general |
a goods train runs at the speed of 72 km / hr and crosses a 220 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 + 220 ) / 26 = 20 x = 300 m . answer : c" | a ) 382 , b ) 782 , c ) 300 , d ) 270 , e ) 881 | c | subtract(multiply(multiply(divide(72, const_3600), const_1000), 26), 220) | divide(n0,const_3600)|multiply(#0,const_1000)|multiply(n2,#1)|subtract(#2,n1)| | physics |
a and b are two partially filled buckets of water . if 5 liters are transferred from a to b , then a would contain one - third of the amount of water in b . alternatively , if 3 liters are transferred from b to a , b would contain one - half of the amount of water in a . bucket a contains how many liters of water ? | "let a contains a , b contains b liters so , ( a - 5 ) / ( b + 5 ) = 1 / 3 . . . . . . . ( 1 ) again , ( b - 3 ) / ( a + 3 ) = 1 / 2 . . . . . . . . . . ( 2 ) from ( 1 ) ( 2 ) we find a = 49 / 5 ans : ( d )" | a ) 11 , b ) 13 , c ) 17 , d ) 49 / 5 , e ) 23 | d | divide(multiply(add(add(5, const_3), const_2), divide(5, const_2)), add(const_2, divide(const_1, const_2))) | add(n0,const_3)|divide(n0,const_2)|divide(const_1,const_2)|add(#0,const_2)|add(#2,const_2)|multiply(#3,#1)|divide(#5,#4)| | general |
two kinds of vodka are mixed in the ratio 1 : 2 and 2 : 1 and they are sold fetching the profit 10 % and 35 % respectively . if the vodkas are mixed in equal ratio and the individual profit percent on them are increased by 4 / 3 and 5 / 3 times respectively , then the mixture will fetch the profit of | answer : d . | a ) 18 % , b ) 20 % , c ) 21 % , d ) 35 % , e ) can not be determined | d | add(divide(multiply(10, 4), 2), add(10, 5)) | add(n4,n8)|multiply(n4,n6)|divide(#1,n1)|add(#0,#2) | general |
of 60 children , 30 are happy , 10 are sad , and 20 are neither happy nor sad . there are 18 boys and 42 girls . if there are 6 happy boys and 4 sad girls , how many boys are neither happy nor sad ? | "venn diagrams are useful for multiple values of a single variable e . g . state of mind - happy / sad / neither . when you have two or more variables such as here where you have gender - boy / girl too , it becomes unwieldy . in this case , either use the table or logic . table method is shown above ; here is how you ... | a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | c | subtract(subtract(18, subtract(10, 4)), 6) | subtract(n2,n7)|subtract(n4,#0)|subtract(#1,n6)| | other |
the sum of all the integers k such that β 26 < k < 24 is | "since k defines a range between β 26 < k < 24 we can set 0 as the reference point for the negative values and positive values . the negative values will range from - 25 to 0 whereas the positive values will range from 0 - 23 . we can conclude that for all but - 25 and - 24 the number pairs will add to 0 . so we have l... | a ) 0 , b ) β 2 , c ) β 25 , d ) β 49 , e ) β 51 | d | add(add(negate(26), const_1), add(add(negate(26), const_1), const_1)) | negate(n0)|add(#0,const_1)|add(#1,const_1)|add(#1,#2)| | general |
evaluate 75 / . 05 | "explanation : 75 / . 05 = 7500 / 5 = 1500 option b" | a ) 1400 , b ) 1500 , c ) 1505 , d ) none of these , e ) 1506 | b | divide(const_100.0, divide(05, 75)) | divide(n1,const_100)|divide(n0,#0)| | general |
the length of a rectangle is three times of its width . if the length of the diagonal is 8 β 10 then find the perimeter of the rectangle . | let breadth = x cm , then , length = 3 x cm x 2 + ( 3 x ) 2 = ( 810 β β β ) 2 = > 10 x 2 = 640 = > x = 8 x 2 + ( 3 x ) 2 = ( 810 ) 2 = > 10 x 2 = 640 = > x = 8 so , length = 24 cm and breadth = 8 cm perimeter = 2 ( l + b ) = 2 ( 24 + 8 ) = 64 cm answer : c | ['a ) 77 cm', 'b ) 88 cm', 'c ) 64 cm', 'd ) 11 cm', 'e ) 18 cm'] | c | multiply(add(sqrt(power(8, const_2)), multiply(sqrt(power(8, const_2)), const_3)), const_2) | power(n0,const_2)|sqrt(#0)|multiply(#1,const_3)|add(#2,#1)|multiply(#3,const_2) | geometry |
225 metres long yard , 26 trees are palnted at equal distances , one tree being at each end of the yard . what is the distance between 2 consecutive trees | "26 trees have 25 gaps between them , required distance ( 225 / 25 ) = 10 a" | a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 17 | a | divide(225, add(subtract(26, 2), const_1)) | subtract(n1,n2)|add(#0,const_1)|divide(n0,#1)| | physics |
car a runs at the speed of 60 km / hr and reaches its destination in 8 hours . car b runs at the speed of 70 km / h and reaches its destination in 4 hours . what is the ratio of distances covered by car a and car b ? | car a travels 60 Γ 8 = 480 km car b travels 70 Γ 4 = 280 km the ratio is 480 : 280 = 48 : 28 = 12 : 7 the answer is c . | a ) 3 : 7 , b ) 4 : 9 , c ) 12 : 7 , d ) 5 : 7 , e ) 6 : 11 | c | divide(multiply(60, 8), multiply(70, 4)) | multiply(n0,n1)|multiply(n2,n3)|divide(#0,#1) | physics |
if an object travels at five feet per second , how many feet does it travel in one hour ? | "explanation : if an object travels at 5 feet per second it covers 5 x 60 feet in one minute , and 5 x 60 x 60 feet in one hour . answer = 1800 answer : d ) 1800" | a ) 3488 , b ) 3778 , c ) 12788 , d ) 1800 , e ) 2881 | d | multiply(multiply(const_3, const_60), const_60) | multiply(const_3,const_60)|multiply(#0,const_60)| | physics |
a house wife saved $ 4 in buying an item on sale . if she spent $ 29 for the item , approximately how much percent she saved in the transaction ? | "actual price = 29 + 4 = $ 33 saving = 4 / 33 * 100 = 12 % approximately answer is e" | a ) 8 % , b ) 9 % , c ) 10 % , d ) 11 % , e ) 12 % | e | multiply(divide(4, add(29, 4)), const_100) | add(n0,n1)|divide(n0,#0)|multiply(#1,const_100)| | general |
tough and tricky questions : exponents . if 2 ^ ( 3 x + 2 ) * 3 ^ ( 2 y - 3 ) = 32 ^ x * 9 ^ y , then x + y = | here is my solution . 2 ^ ( 3 x + 2 ) * 3 ^ ( y - 3 ) = 32 ^ x * 9 ^ y here rhs 32 ^ x * 9 ^ y = 2 ^ ( 5 x ) * 3 ^ ( 2 y ) equating powers on both sides - - > 3 x + 2 = 5 x , thus x = 1 and y - 3 = 2 y giving y = - 3 so , x + y = - 2 option : e | a ) - 3 , b ) - 1 , c ) 0 , d ) - 5 , e ) - 2 | e | add(add(multiply(subtract(const_2, 3), multiply(divide(const_3, 2), const_3)), divide(const_3, 2)), const_1) | divide(const_3,n0)|subtract(const_2,n1)|multiply(#0,const_3)|multiply(#2,#1)|add(#0,#3)|add(#4,const_1) | general |
what is the units digit of ( 147 ^ 25 ) ^ 49 ? | the units digit of the exponents of 7 repeat in a cycle of four , which is { 7,9 , 3,1 } . the number 25 has the form 4 n + 1 so the units digit is 7 inside the bracket . the exponent 49 has the form 4 n + 1 , so the units digit is 7 . the answer is d . | a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | d | add(add(const_4, const_3), const_2) | add(const_3,const_4)|add(#0,const_2)| | general |
simple interest on a certain sum of money for 3 years at 8 % per annum is half the compound interest on $ 4000 for 2 years at 10 % per annum . the sum placed on simple interest is : | "compound interest = ( 4000 x ( 1 + 10 / 100 ) ^ 2 - 4000 ) = 4000 x 11 / 10 x 11 / 10 - 4000 = 840 therefore the sum is 420 x 100 / ( 3 x 8 ) = 1750 . answer a ) $ 1750" | a ) $ 1750 , b ) $ 2250 , c ) $ 1250 , d ) $ 1500 , e ) $ 2000 | a | divide(multiply(divide(divide(add(divide(multiply(4000, 10), const_100), divide(multiply(add(4000, divide(multiply(4000, 10), const_100)), 10), const_100)), 2), 3), const_100), 8) | multiply(n2,n4)|divide(#0,const_100)|add(n2,#1)|multiply(n4,#2)|divide(#3,const_100)|add(#1,#4)|divide(#5,n3)|divide(#6,n0)|multiply(#7,const_100)|divide(#8,n1)| | gain |
in one hour , a boat goes 11 km along the stream and 3 km against the stream . the sped of the boat in still water ( in km / hr ) is : | "solution speed in still water = 1 / 2 ( 11 + 3 ) km / hr = 7 kmph . answer c" | a ) 3 , b ) 5 , c ) 7 , d ) 9 , e ) 10 | c | divide(add(11, 3), const_2) | add(n0,n1)|divide(#0,const_2)| | gain |
if a + b = β 6 , and a = 10 / b , what is the value of a ^ 2 + b ^ 2 ? | a ^ 2 + b ^ 2 should make you think of these formulas : ( a + b ) ( a + b ) = a ^ 2 + b ^ 2 + 2 ab we already know ( a + b ) = - 6 and a * b = 10 ( a + b ) ( a + b ) = ( - 6 ) ( - 6 ) = a ^ 2 + b ^ 2 + 2 * ( 10 ) a ^ 2 + b ^ 2 = 36 - 20 = 16 answer : c | a ) 22 , b ) 24 , c ) 16 , d ) 18 , e ) 20 | c | subtract(power(6, const_2), multiply(const_2, 10)) | multiply(n1,const_2)|power(n0,const_2)|subtract(#1,#0) | general |
a professional basketball player just signed a contract that has a 10 % raise in base salary from last year to this year . the contract also has provisions for a performance bonus of up to $ 10000 . if he earns the maximum performance bonus , his new effective salary will be a 10.5 % raise over last year ' s total sala... | let x = old salary then ( 1 + . 10 ) * x + 10000 = ( 1 + . 105 ) * x i . e . new base salary + 10000 = effective new salary thus 1.1 * x + 10000 = 1.105 * x 10000 = . 005 * x 2000000 = x thus the new effective salary is 1.105 * 2000000 = 2210000 e | a ) 2010000 , b ) 2110000 , c ) 2150000 , d ) 2200000 , e ) 2210000 | e | divide(const_4, const_2) | divide(const_4,const_2) | general |
how long does a train 110 m long running at the speed of 72 km / hr takes to cross a bridge 175 m length ? | "speed = 72 * 5 / 18 = 20 m / sec total distance covered = 110 + 175 = 285 m . required time = 285 / 20 = 14.25 sec . answer : e" | a ) 12.9 sec , b ) 12.1 sec , c ) 17.9 sec , d ) 16.8 sec , e ) 14.25 sec | e | divide(add(110, 175), multiply(72, const_0_2778)) | add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)| | physics |
find the length of the wire required to go 15 times round a square field containing 69696 m 2 . | "a 2 = 69696 = > a = 264 4 a = 1056 1056 * 15 = 15840 answer : a" | a ) 15840 , b ) 15848 , c ) 15828 , d ) 27892 , e ) 22892 | a | multiply(square_perimeter(square_edge_by_area(69696)), 15) | square_edge_by_area(n1)|square_perimeter(#0)|multiply(n0,#1)| | physics |
5 n + 2 > 17 and 7 n - 5 < 44 ; n must be between which numbers ? | "5 n > 15 - - > n > 3 7 n < 49 - - > n < 7 3 < n < 7 answer : d" | a ) 1 and 8 , b ) 2 and 6 , c ) 0 and 9 , d ) 3 and 7 , e ) 2 and 9 | d | add(multiply(2, const_10), divide(add(44, 5), 7)) | add(n4,n5)|multiply(const_10,n1)|divide(#0,n3)|add(#2,#1)| | general |
find the average of first 6 multiples of 5 ? | "average = ( 5 + 10 + 15 + 20 + 25 + 30 ) / 6 = 17.5 answer is e" | a ) 10 , b ) 15 , c ) 12.5 , d ) 13 , e ) 17.5 | e | divide(add(add(add(6, const_1), add(add(6, const_1), const_2)), add(subtract(5, 6), subtract(5, const_2))), 6) | add(n0,const_1)|subtract(n1,n0)|subtract(n1,const_2)|add(#0,const_2)|add(#1,#2)|add(#0,#3)|add(#5,#4)|divide(#6,n0)| | general |
a large box contains 15 small boxes and each small box contains 20 chocolate bars . how many chocolate bars are in the large box ? | "the only thing you have to do is a simple multiply we have : 15 * 20 = 300 correct answer is : d ) 300" | a ) a ) 220 , b ) b ) 490 , c ) c ) 380 , d ) d ) 300 , e ) e ) 690 | d | multiply(15, 20) | multiply(n0,n1)| | general |
the current of a stream runs at the rate of 4 kmph . a boat goes 10 km and back to the starting point in 2 hours , then find the speed of the boat in still water ? | "s = 4 m = x ds = x + 4 us = x - 4 10 / ( x + 4 ) + 10 / ( x - 4 ) = 2 x = 11.4 answer : d" | a ) a ) 7 , b ) b ) 2 , c ) c ) 8.9 , d ) d ) 11.4 , e ) e ) 3 | d | divide(power(4, 2), 2) | power(n0,n2)|divide(#0,n2)| | physics |
the length of the bridge , which a train 170 metres long and travelling at 45 km / hr can cross in 30 seconds , is ? | "speed = [ 45 x 5 / 18 ] m / sec = [ 25 / 2 ] m / sec time = 30 sec let the length of bridge be x metres . then , ( 170 + x ) / 30 = 25 / 2 = > 2 ( 170 + x ) = 750 = > x = 205 m . answer : e" | a ) 245 , b ) 777 , c ) 282 , d ) 266 , e ) 205 | e | subtract(multiply(divide(multiply(45, speed(const_1000, const_1)), speed(const_3600, const_1)), 30), 170) | speed(const_1000,const_1)|speed(const_3600,const_1)|multiply(n1,#0)|divide(#2,#1)|multiply(n2,#3)|subtract(#4,n0)| | physics |
a retail appliance store priced a video recorder at 20 percent above the wholesale cost of $ 200 . if a store employee applied the 5 percent employee discount to the retail price to buy the recorder , how much did the employee pay for the recorder ? | "wholesale cost of video recorder = 200 $ video recorder was priced at 20 percent above 200 = 240 $ % discount given by store employee = 5 emlpoyee paid = . 95 * 240 = 228 $ answer b" | a ) $ 198 , b ) $ 228 , c ) $ 220 , d ) $ 230 , e ) $ 240 | b | subtract(add(200, multiply(divide(200, const_100), 20)), multiply(divide(add(200, multiply(divide(200, const_100), 20)), const_100), 5)) | divide(n1,const_100)|multiply(n0,#0)|add(n1,#1)|divide(#2,const_100)|multiply(n2,#3)|subtract(#2,#4)| | gain |
if a # b = ab β b + b ^ 2 , then 3 # 6 = | "solution - simply substitute 3 and 6 in equation in the place of a and b respectively . 3 # 6 = 3 * 6 - 6 + 6 ^ 2 = 18 - 6 + 36 = 48 . ans e" | a ) 2 , b ) 8 , c ) 15 , d ) 21 , e ) 48 | e | add(subtract(multiply(3, 6), 6), power(6, 2)) | multiply(n1,n2)|power(n2,n0)|subtract(#0,n2)|add(#1,#2)| | general |
a man engaged a servant on a condition that he ' ll pay rs 90 and also give him a bag at the end of the yr . he served for 9 months and was given a turban and rs 65 . so the price of turban is | let the value of turban = rs x for 1 year = 90 + x for 9 months ( i , e 3 / 4 year ) = 65 + x i , e 1 : ( 90 + x ) : : 3 / 4 : ( 65 + x ) i , e ( 90 + x ) / ( 65 + x ) = 1 / ( 3 / 4 ) 270 + 3 x = 260 + 4 x thus x = 10 answer : a | a ) rs 10 , b ) rs 9 , c ) rs 0 , d ) rs 55 , e ) rs 8 | a | subtract(multiply(divide(90, const_12), const_10), 65) | divide(n0,const_12)|multiply(#0,const_10)|subtract(#1,n2) | general |
if w is the set of all the integers between 29 and 99 , inclusive , that are either multiples of 3 or multiples of 2 or multiples of both , then w contains how many numbers ? | "multiples of 2 from 29 to 99 = multiples of 2 from 1 to 99 - multiples of 2 from 1 to 28 = [ 99 / 2 ] - [ 28 / 2 ] = 49 - 14 = 35 multiples of 3 from 29 to 99 = multiples of 3 from 1 to 99 - multiples of 3 from 1 to 28 = [ 99 / 3 ] - [ 28 / 3 ] = 33 - 9 = 24 multiples of 2 and 3 bothi . e . 6 from 29 to 99 = multiples... | a ) 26 , b ) 32 , c ) 33 , d ) 34 , e ) 47 | e | subtract(add(floor(divide(subtract(99, 29), 3)), divide(subtract(99, 29), 2)), floor(divide(subtract(99, 29), multiply(2, 3)))) | multiply(n3,n2)|subtract(n1,n0)|divide(#1,n3)|divide(#1,n2)|divide(#1,#0)|floor(#3)|floor(#4)|add(#2,#5)|subtract(#7,#6)| | other |
for a certain exam , a score of 86 was 7 standard deviations below mean and a score of 90 was 3 standard deviations above mean . what was the mean score for the exam ? | "mean - 7 sd = 86 mean + 3 sd = 90 by solving above equations we get , sd ( absolute value ) = 0.4 mean = 88.8 ans . d" | a ) 86.25 , b ) 84 , c ) 85 , d ) 88.8 , e ) 80 | d | divide(add(multiply(86, 3), multiply(90, 7)), add(7, 3)) | add(n1,n3)|multiply(n0,n3)|multiply(n1,n2)|add(#1,#2)|divide(#3,#0)| | general |
the area of a square is equal to three times the area of a rectangle of dimensions 49 cm * 27 cm . what is the perimeter of the square ? | "area of the square = s * s = 3 ( 49 * 27 ) = > s = 7 * 3 * 3 = 63 cm perimeter of the square = 4 * 63 = 252 cm . answer : c" | a ) 380 cm , b ) 264 cm , c ) 252 cm , d ) 324 cm , e ) 296 cm | c | multiply(sqrt(multiply(rectangle_area(49, 27), divide(const_10, const_2))), const_4) | divide(const_10,const_2)|rectangle_area(n0,n1)|multiply(#0,#1)|sqrt(#2)|multiply(#3,const_4)| | geometry |
in a division , a student took 72 as divisor instead of 36 . his answer was 24 . the correct answer is - | "x / 72 = 24 . x = 24 * 72 . so correct answer would be , ( 24 * 72 ) / 36 = 48 . answer : c" | a ) 42 , b ) 32 , c ) 48 , d ) 28 , e ) 38 | c | divide(multiply(72, 24), 36) | multiply(n0,n2)|divide(#0,n1)| | general |
a scientist used a unique two - color code to identify each of the test subjects involved in a certain study . if the scientist found that choosing from among 5 colors produced enough color codes to identify all but 6 of the test subjects , how many test subjects were in the study ? ( assume that the order of the color... | 5 c 2 = 10 the number of test subjects is 10 + 6 = 16 the answer is c . | a ) 14 , b ) 15 , c ) 16 , d ) 17 , e ) 18 | c | add(divide(factorial(5), multiply(factorial(subtract(5, const_2)), factorial(const_2))), 6) | factorial(n0)|factorial(const_2)|subtract(n0,const_2)|factorial(#2)|multiply(#3,#1)|divide(#0,#4)|add(n1,#5) | general |
enrollment in city college in 1980 was 85 / 3 percent of enrollment in 1990 . what was the percent increase in the college β s enrollment from 1980 to 1990 ? | assume 100 enrollments present in 1990 85 / 3 % of 1990 = enrollments on 1980 enrollments on 1980 = 28.33 % change = ( enrollment in 1990 - enrollment in 1980 ) * 100 / ( enrollment in 1980 ) = ( 100 - 28.33 ) * 100 / 28.333 = 7166.67 / 28.333 = 253 % ans - d | a ) 100 % , b ) 16 2 / 3 % , c ) 200 % , d ) 253 % , e ) 183 1 / 3 % | d | multiply(divide(subtract(const_100, divide(85, 3)), divide(85, 3)), const_100) | divide(n1,n2)|subtract(const_100,#0)|divide(#1,#0)|multiply(#2,const_100) | general |
there were two candidates in an election . winner candidate received 54 % of votes and won the election by 288 votes . find the number of votes casted to the winning candidate ? | "w = 54 % l = 46 % 54 % - 46 % = 8 % 8 % - - - - - - - - 288 54 % - - - - - - - - ? = > 1944 answer : b" | a ) 776 , b ) 1944 , c ) 2199 , d ) 2157 , e ) 2125 | b | divide(multiply(divide(288, divide(subtract(54, subtract(const_100, 54)), const_100)), 54), const_100) | subtract(const_100,n0)|subtract(n0,#0)|divide(#1,const_100)|divide(n1,#2)|multiply(n0,#3)|divide(#4,const_100)| | gain |
a person bought 115 glass bowls at a rate of rs . 18 per bowl . he sold 104 of them at rs . 20 and the remaining broke . what is the percentage gain for a ? | "cp = 115 * 18 = 2070 and sp = 104 * 20 = 2080 gain % = 100 * ( 2080 - 2070 ) / 2070 = 100 / 207 answer : e" | a ) 40 , b ) 300 / 11 , c ) 243 / 7 , d ) 279 / 11 , e ) 100 / 207 | e | multiply(divide(subtract(multiply(104, 20), multiply(115, 18)), multiply(115, 18)), const_100) | multiply(n2,n3)|multiply(n0,n1)|subtract(#0,#1)|divide(#2,#1)|multiply(#3,const_100)| | gain |
lisa and robert have taken the same number of photos on their school trip . lisa has taken 3 times as many photos as claire and robert has taken 10 more photos than claire . how many photos has claire taken ? | "l = r l = 3 c r = c + 10 3 c = c + 10 c = 5 the answer is a ." | a ) 5 , b ) 8 , c ) 11 , d ) 12 , e ) 14 | a | divide(10, subtract(3, const_1)) | subtract(n0,const_1)|divide(n1,#0)| | general |
how many even multiples of 45 are there between 449 and 901 ? | 450 = 10 * 45 900 = 20 * 45 the even multiples are 45 multiplied by 10 , 12 , 14 , 16 , 18 , and 20 for a total of 6 . the answer is b . | a ) 5 , b ) 6 , c ) 9 , d ) 10 , e ) 11 | b | add(divide(subtract(901, 449), multiply(45, const_2)), const_1) | multiply(n0,const_2)|subtract(n2,n1)|divide(#1,#0)|add(#2,const_1) | general |
19 chess players take part in a tournament . every player plays twice with each of his opponents . how many games are to be played ? | "2 * 19 c 2 = 2 * 171 = 342 the answer is b ." | a ) 286 , b ) 342 , c ) 422 , d ) 540 , e ) 618 | b | multiply(19, subtract(19, const_1)) | subtract(n0,const_1)|multiply(n0,#0)| | general |
rs . 585 is divided amongst a , b , c so that 4 times a ' s share , 6 times b ' s share and 3 times c ' s share are all equal . find c ' s share ? | a + b + c = 585 4 a = 6 b = 3 c = x a : b : c = 1 / 4 : 1 / 6 : 1 / 3 = 3 : 2 : 4 4 / 9 * 585 = rs . 260 answer : a | a ) 260 , b ) 150 , c ) 817 , d ) 716 , e ) 616 | a | divide(585, add(add(divide(3, 4), divide(3, 6)), const_1)) | divide(n3,n1)|divide(n3,n2)|add(#0,#1)|add(#2,const_1)|divide(n0,#3)| | general |
in 10 years , a will be twice as old as b was 10 years ago . if a is now 8 years older than b the present age of b is | "let present age of a be a and b be b a + 10 = 2 * ( b - 10 ) = > 2 b - a = 30 . . . . . . ( i ) a = b + 8 = > 2 b - b - 8 = 30 b = 38 so the present age of b is 38 years answer : b" | a ) 37 , b ) 38 , c ) 39 , d ) 40 , e ) 41 | b | add(add(multiply(const_2, 10), 10), 8) | multiply(n0,const_2)|add(n0,#0)|add(n2,#1)| | general |
a die is rolled twice . what is the probability of getting a sum equal to 12 ? | "total number of outcomes possible when a die is rolled = 6 ( β΅ any one face out of the 6 faces ) hence , total number of outcomes possible when a die is rolled twice , n ( s ) = 6 Γ 6 = 36 e = getting a sum of 12 when the two dice fall = { ( 6 , 6 ) } hence , n ( e ) = 1 p ( e ) = n ( e ) / n ( s ) = 1 / 36 = 1 / 36 a... | a ) 1 / 36 , b ) 1 / 9 , c ) 1 / 7 , d ) 1 / 24 , e ) 2 / 36 | a | divide(const_1, 12) | divide(const_1,n0)| | general |
a man can row his boat with the stream at 14 km / h and against the stream in 4 km / h . the man ' s rate is ? | "ds = 14 us = 4 s = ? s = ( 14 - 4 ) / 2 = 5 kmph answer : e" | a ) 1 kmph , b ) 3 kmph , c ) 8 kmph , d ) 7 kmph , e ) 5 kmph | e | divide(subtract(14, 4), const_2) | subtract(n0,n1)|divide(#0,const_2)| | gain |
cost is expressed by the formula tb ^ 4 . if b is doubled , the new cost is what percent t of the original cost ? | "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 ) t = 200 , b ) t = 600 , c ) t = 800 , d ) t = 1600 , e ) t = 50 | d | multiply(power(const_2, 4), const_100) | power(const_2,n0)|multiply(#0,const_100)| | general |
in a certain egg - processing plant , every egg must be inspected , and is either accepted for processing or rejected . for every 96 eggs accepted for processing , 4 eggs are rejected . if , on a particular day , 12 additional eggs were accepted , but the overall number of eggs inspected remained the same , the ratio o... | straight pluggin in for me . as usual , i started with c and got the answer . lets ' back calculate and see what we get let us consider eggs processed each day to be 400 so initial ratio of eggs processed and rejected is 96 : 4 or 24 : 1 so out of 400 eggs , there will be 384 eggs processed and 16 rejected . now if the... | a ) 100 , b ) 300 , c ) 400 , d ) 3000 , e ) 4000 | c | multiply(divide(12, subtract(99, 96)), const_100) | subtract(n3,n0)|divide(n2,#0)|multiply(#1,const_100) | general |
on her annual road trip to visit her family in seal beach , california , traci stopped to rest after she traveled 1 β 2 of the total distance and again after she traveled 1 β 4 of the distance remaining between her first stop and her destination . she then drove the remaining 400 miles and arrived safely at her destina... | "let d = total distance traci traveled 1 / 2 = d / 2 i . e . remaining distance = d / 2 she traveled 1 / 4 th of d / 2 = d / 8 thus : d = ( d / 2 ) + ( d / 8 ) + 400 d = 3200 / 3 answer : b" | a ) 250 , b ) 3200 / 3 , c ) 3150 / 3 , d ) 4002 / 3 , e ) 550 | b | multiply(400, const_2) | multiply(n4,const_2)| | physics |
a certain country had a total annual expenditure of $ 1.2 x 10 ^ 11 last year . if the population of the country was 240 million last year , what was the per capita expenditure ? | "total expenditure / population = per capita expenditure hence , ( 1,2 x 10 ^ 11 ) / 240 000 000 = ( 12 x 10 ^ 10 ) / ( 2,4 x 10 ^ 8 ) = 5 x 10 ^ ( 10 - 8 ) = 5 x 10 ^ 2 = 500 . answer is a ." | a ) $ 500 , b ) $ 1,000 , c ) $ 2,000 , d ) $ 3,000 , e ) $ 5,000 | a | divide(multiply(1.2, power(10, 11)), multiply(240, power(10, add(const_4, const_2)))) | add(const_2,const_4)|power(n1,n2)|multiply(n0,#1)|power(n1,#0)|multiply(n3,#3)|divide(#2,#4)| | general |
two students appeared at an examination . one of them secured 9 marks more than the other and his marks was 56 % of the sum of their marks . the marks obtained by them are : | let their marks be ( x + 9 ) and x . then x + 9 = 56 / 100 ( x + 9 + x ) 3 x = 99 x = 33 their marks 42 and 33 answer c | a ) 39,30 , b ) 41,32 , c ) 42,33 , d ) 43,34 , e ) 35,36 | c | add(divide(subtract(multiply(9, const_100), multiply(56, 9)), subtract(multiply(56, const_2), const_100)), 9) | multiply(n0,const_100)|multiply(n0,n1)|multiply(n1,const_2)|subtract(#0,#1)|subtract(#2,const_100)|divide(#3,#4)|add(n0,#5) | general |
aaron will jog from home at 4 miles per hour and then walk back home by the same route at 8 miles per hour . how many miles from home can aaron jog so that he spends a total of 3 hours jogging and walking ? | "xyt / ( x + y ) x = 4 , y = 8 t = 3 4 * 8 * 3 / 4 + 8 = 96 / 12 = 8 answer : d" | a ) 2 , b ) 10 , c ) 12 , d ) 8 , e ) 15 | d | divide(multiply(multiply(const_2.0, const_4.0), 3), multiply(4, 8)) | multiply(const_2.0,const_4.0)|multiply(n2,#0)|divide(#1,#0)| | physics |
if x is the product of the positive integers from 1 to 8 , inclusive , and if i , k , m , and p are positive integers such that x = 2 ^ i β 3 ^ k β 5 ^ m β 7 ^ p , then i + k + m + p = | "here the idea is to go expressing each factor from one to eight , decomposed into its primes : 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 = 1 * 2 * 3 * ( 2 * 2 ) * 5 + ( 2 * 3 ) * 7 * ( 2 * 2 * 2 ) = 2 ^ 7 * 3 ^ 2 * 5 ^ 1 * 7 ^ 1 then i = 7 k = 2 m = 1 p = 1 i + k + m + p = 11 correct answer d" | a ) 4 , b ) 7 , c ) 8 , d ) 11 , e ) 12 | d | add(add(add(7, 2), divide(5, 5)), divide(3, 3)) | add(n2,n5)|divide(n4,n4)|divide(n3,n3)|add(#0,#1)|add(#3,#2)| | general |
in right triangle abc , ac is the hypotenuse . if ac is 10 and ab + bc = 20 , what is the area of the triangle abc ? | "square ab + bc = 20 : ( ab ) ^ 2 + 2 * ab * bc + ( bc ) ^ 2 = 400 . since ( ac ) ^ 2 = ( ab ) ^ 2 + ( bc ) ^ 2 = 10 ^ 2 = 100 , then ( ab ) ^ 2 + 2 * ab * bc + ( bc ) ^ 2 = 100 + 2 * ab * bc = 400 . 100 + 2 * ab * bc = 400 . ab * bc = 150 . the area = 1 / 2 * ab * bc = 75 . answer : c ." | a ) 225 , b ) 450 , c ) 75 , d ) 200 , e ) 200 β 2 | c | triangle_area_three_edges(10, multiply(const_3, const_10), multiply(const_4, const_10)) | multiply(const_10,const_3)|multiply(const_10,const_4)|triangle_area_three_edges(n0,#0,#1)| | geometry |
each side of a rectangular field diminished by 40 % . by how much per cent is the area of the field diminished ? | solution : let the original length of the rectangle be 20 unit and breadth be 10 unit . then original area = length * breadth = 20 * 10 = 200 square unit . 40 % decrease in each side , then length = ( 20 - 40 % of 20 ) = 12 unit . breadth = ( 10 - 40 % of 10 ) = 6 unit . now , area = 12 * 6 = 72 square unit . decrease ... | ['a ) 32', 'b ) 64', 'c ) 25', 'd ) 16', 'e ) none'] | b | multiply(divide(subtract(multiply(const_100, const_100), multiply(subtract(const_100, 40), subtract(const_100, 40))), multiply(const_100, const_100)), const_100) | multiply(const_100,const_100)|subtract(const_100,n0)|multiply(#1,#1)|subtract(#0,#2)|divide(#3,#0)|multiply(#4,const_100) | geometry |
the average of 10 numbers is calculated as 14 . it is discovered later on that while calculating the average , one number namely 36 was wrongly read as 26 . the correct average is ? | "explanation : 10 * 14 + 36 β 26 = 150 = > 150 / 10 = 15 b )" | a ) a ) 16 , b ) b ) 15 , c ) c ) 19 , d ) d ) 22 , e ) e ) 24 | b | add(14, divide(subtract(36, 26), 10)) | subtract(n2,n3)|divide(#0,n0)|add(n1,#1)| | general |
there are two groups of students in the sixth grade . there are 20 students in group a , and 80 students in group b . if , on a particular day , 20 % of the students in group a forget their homework , and 15 % of the students in group b forget their homework , then what percentage of the sixth graders forgot their home... | "number of students in group a = 20 students who forget homework in group a = 20 % of 20 = 4 students number of students in group b = 80 students who forget homework in group b = 15 % of 80 = 12 students total number of students = 20 + 80 = 100 students who forgot homework = 4 + 12 = 16 students percentage of students ... | a ) 13 % , b ) 14 % , c ) 15 % , d ) 16 % , e ) 17 % | d | multiply(divide(add(divide(multiply(20, 20), const_100), divide(multiply(80, 15), const_100)), add(20, 80)), const_100) | add(n0,n1)|multiply(n0,n2)|multiply(n1,n3)|divide(#1,const_100)|divide(#2,const_100)|add(#3,#4)|divide(#5,#0)|multiply(#6,const_100)| | gain |
the average of 6 quantities is 8 . the average of 4 of them is 5 . what is the avg . of the remaining 2 numbers ? | "sum of 8 quantities = total no of quantities x average of them = 6 x 8 = 48 sum of 4 quantities = 4 x 5 = 20 sum of 4 quantities + sum of 2 quantities = sum of 6 quantities 20 + sum of 2 quantities = 48 sum of 2 quantities = 28 average of 2 quantities = sum of 2 quantities / 2 = 28 / 2 = 14 answer : d" | a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | d | divide(subtract(multiply(6, 8), multiply(4, 5)), 2) | multiply(n0,n1)|multiply(n2,n3)|subtract(#0,#1)|divide(#2,n4)| | general |
suresh started a business , investing rs . 18000 . after 3 months and 4 months respectively , rohan and sudhir joined him with capitals of 12000 and 9000 . at the end of the year the total profit was rs . 4048 . what is the difference between rohan β s and sudhir β s share in the profit ? | "suresh : rohan : sudhir ratio of their investments = 18000 Γ 12 : 12000 Γ 9 : 9000 Γ 8 = 6 : 3 : 2 the difference between rohan β s and sudhir β s share = 1 share : . i . e . = rs . 4048 Γ 1 / 11 = rs . 368 . e" | a ) s . 345 , b ) s . 350 , c ) s . 352 , d ) s . 362 , e ) s . 368 | e | multiply(subtract(multiply(12000, divide(subtract(multiply(3, 4), 3), multiply(3, 4))), multiply(9000, divide(subtract(multiply(3, 4), 4), multiply(3, 4)))), divide(4048, add(add(18000, multiply(12000, divide(subtract(multiply(3, 4), 3), multiply(3, 4)))), multiply(9000, divide(subtract(multiply(3, 4), 4), multiply(3, ... | multiply(n1,n2)|subtract(#0,n1)|subtract(#0,n2)|divide(#1,#0)|divide(#2,#0)|multiply(n3,#3)|multiply(n4,#4)|add(n0,#5)|subtract(#5,#6)|add(#7,#6)|divide(n5,#9)|multiply(#10,#8)| | gain |
a man has rs . 10350 in the form of rs . 50 notes and rs . 500 notes . the total number of notes are 90 . find the number of notes of rs . 50 denomination . | total money = rs . 10350 . let 50 rupees note was x . then 500 rupees note = 90 - x now , 50 * x + 500 * ( 90 - x ) = 10350 50 x + 45000 - 500 x = 10350 - 450 x = - 34650 x = 77 . no . of 50 rupees note = 77 . answer : option e | a ) 15 , b ) 21 , c ) 17 , d ) 57 , e ) 77 | e | divide(subtract(multiply(500, 90), 10350), subtract(500, 50)) | multiply(n2,n3)|subtract(n2,n1)|subtract(#0,n0)|divide(#2,#1) | general |
during 2003 , a company produced an average of 4,000 products per month . how many products will the company need to produce from 2004 through 2007 in order to increase its monthly average for the period from 2003 through 2007 by 150 % over its 2003 average ? | "company produced 12 * 4000 = 48,000 products in 2003 . if company produces x products from 2004 to 2007 , then total amount of product produced in 4 years ( 2003 through 2007 ) is x + 48,000 . the gives the average of ( x + 48,000 ) / 4 . this average needs to be 300 % higher than that in 2003 . in math terms , 48,000... | a ) 435,000 , b ) 440,000 , c ) 450,000 , d ) 480,000 , e ) 432,000 | e | multiply(150, divide(subtract(2007, 2003), const_2)) | subtract(n3,n0)|divide(#0,const_2)|multiply(n6,#1)| | general |
the average age of an adult class is 40 years . 12 new students with an avg age of 32 years join the class . thereforedecreasing the average by 4 year . find what was theoriginal strength of class ? | "let original strength = y then , 40 y + 12 x 32 = ( y + 12 ) x 36 β 40 y + 384 = 36 y + 432 β 4 y = 48 β΄ y = 12 b" | a ) 10 , b ) 12 , c ) 16 , d ) 20 , e ) 22 | b | divide(subtract(multiply(12, subtract(40, 4)), multiply(12, 32)), 4) | multiply(n1,n2)|subtract(n0,n3)|multiply(n1,#1)|subtract(#2,#0)|divide(#3,n3)| | general |
the distance from the x - axis to point p is half the distance from the y - axis to point p . if the coordinates of p are ( x , - 5 ) , how many units is p from the y - axis ? | "the x - axis is 5 units from the point p . thus the y - axis is 10 units from the point p . the answer is d ." | a ) 2.5 , b ) 5 , c ) 7.5 , d ) 10 , e ) 12 | d | multiply(5, const_2) | multiply(n0,const_2)| | general |
two numbers are respectively 20 % and 60 % more than a third number . the percentage that is first of the second is ? | "i ii iii 120 160 100 160 - - - - - - - - - - 120 100 - - - - - - - - - - - ? = > 75 % answer : d" | a ) 34 % , b ) 68 % , c ) 51 % , d ) 75 % , e ) 83 % | d | subtract(const_100, multiply(divide(add(20, const_100), add(60, const_100)), const_100)) | add(n0,const_100)|add(n1,const_100)|divide(#0,#1)|multiply(#2,const_100)|subtract(const_100,#3)| | gain |
we bought a total of 90 books at the store . math books cost $ 4 and history books cost $ 5 . the total price was $ 397 . how many math books did we buy ? | "let m be the number of math books and let h be the number of history books . m + h = 90 . h = 90 - m . 4 m + 5 h = 397 . 4 m + 5 ( 90 - m ) = 397 . - m + 450 = 397 . m = 450 - 397 = 53 . the answer is b ." | a ) 47 , b ) 53 , c ) 56 , d ) 61 , e ) 64 | b | subtract(90, subtract(397, multiply(90, 4))) | multiply(n0,n1)|subtract(n3,#0)|subtract(n0,#1)| | general |
two employees a and b are paid a total of rs . 580 per week by their employer . if a is paid 150 percent of the sum paid to b , how much is b paid per week ? | "let the amount paid to a per week = x and the amount paid to b per week = y then x + y = 580 but x = 150 % of y = 150 y / 100 = 15 y / 10 β΄ 15 y / 10 + y = 580 β y [ 15 / 10 + 1 ] = 580 β 25 y / 10 = 580 β 25 y = 5800 β y = 5800 / 25 = rs . 232 d )" | a ) s . 130 , b ) s . 140 , c ) s . 150 , d ) s . 232 , e ) s . 280 | d | divide(580, add(divide(150, const_100), const_1)) | divide(n1,const_100)|add(#0,const_1)|divide(n0,#1)| | general |
at a certain committee meeting only associate professors and assistant professors are present . each associate professor has brought 2 pencils and 1 chart to the meeting , while each assistant professor has brought 1 pencil and 2 charts . if a total of 7 pencils and 11 charts have been brought to the meeting , how many... | "say there are ' a ' associate professors . so we have 2 a pencils and a charts . say there are ' b ' assistant professors . so we have b pencils and 2 b charts . total pencils are 10 so 2 a + b = 7 total charts are 11 so a + 2 b = 11 add both : 3 a + 3 b = 18 so a + b = 6 total number of people = 6 a" | a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | a | divide(add(11, 7), add(2, 1)) | add(n4,n5)|add(n0,n1)|divide(#0,#1)| | general |
the difference between the value of a number increased by 12.5 % and the value of the original number decreased by 25 % is 30 . what is the original number r ? | "( 1 + 1 / 8 ) x - ( 1 - 1 / 4 ) x = 30 ( 9 / 8 ) x - ( 3 / 4 ) x = 30 x = 80 = r answer : b" | a ) 60 , b ) 80 , c ) 40 , d ) 120 , e ) 160 | b | divide(30, subtract(add(const_1, divide(12.5, const_100)), subtract(const_1, divide(25, const_100)))) | divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|subtract(const_1,#1)|subtract(#2,#3)|divide(n2,#4)| | general |
the average height of 35 boys in a class was calculated as 182 cm . it has later found that the height of one of the boys in the class was wrongly written as 166 cm whereas his actual height was 106 cm . find the actual average height of the boys in the class ( round off your answer to two decimal places ) . ? | "calculated average height of 35 boys = 182 cm . wrong total height of 35 boys = 182 * 35 cm . this was as a result of an actual height of 106 cm being wrongly written as 166 cm . correct total height of 35 boys = 182 cm - ( 166 cm - 106 cm ) / 35 = 182 cm - 60 / 35 cm = 182 cm - 1.71 cm = 180.29 cm . answer : b" | a ) 180.27 cm , b ) 180.29 cm , c ) 978.29 cm , d ) 178.89 cm , e ) 176.29 cm | b | floor(divide(add(subtract(multiply(35, 182), 166), 106), 35)) | multiply(n0,n1)|subtract(#0,n2)|add(n3,#1)|divide(#2,n0)|floor(#3)| | general |
the salaries of a and b together amount to $ 3000 . a spends 95 % of his salary and b , 85 % of his . if now , their savings are the same , what is a ' s salary ? | "let a ' s salary is x b ' s salary = 3000 - x ( 100 - 95 ) % of x = ( 100 - 85 ) % of ( 3000 - x ) x = $ 2250 answer is d" | a ) $ 1000 , b ) $ 1250 , c ) $ 2500 , d ) $ 2250 , e ) $ 1200 | d | subtract(3000, divide(3000, add(divide(subtract(const_100, 85), subtract(const_100, 95)), const_1))) | subtract(const_100,n2)|subtract(const_100,n1)|divide(#0,#1)|add(#2,const_1)|divide(n0,#3)|subtract(n0,#4)| | gain |
if a number is subtracted from its square we get 272 . find the number ? | x ^ 2 Γ’ β¬ β x = 272 = > x = 17 answer e | ['a ) 7', 'b ) 21', 'c ) 19', 'd ) 15', 'e ) 17'] | e | divide(add(const_1, sqrt(add(const_1, multiply(const_4, 272)))), const_2) | multiply(n0,const_4)|add(#0,const_1)|sqrt(#1)|add(#2,const_1)|divide(#3,const_2) | geometry |
a train 120 m long crosses a platform 220 m long in 20 sec ; find the speed of the train ? | "d = 120 + 220 = 340 t = 20 s = 340 / 20 * 18 / 5 = 61 kmph answer : d" | a ) 87 kmph , b ) 65 kmph , c ) 54 kmph , d ) 61 kmph , e ) 18 kmph | d | subtract(multiply(20, multiply(220, const_0_2778)), 120) | multiply(n1,const_0_2778)|multiply(n2,#0)|subtract(#1,n0)| | physics |
for all positive integers m and v , the expression m ΞΈ v represents the remainder when m is divided by v . what is the value of ( ( 88 ΞΈ 33 ) ΞΈ 17 ) - ( 90 ΞΈ ( 33 ΞΈ 17 ) ) ? | "( ( 88 ΞΈ 33 ) ΞΈ 17 ) the remainder of 88 divided by 33 is 22 ; the remainder of 22 divided by 17 is 5 ; ( 98 ΞΈ ( 33 ΞΈ 17 ) ) the remainder of 33 divided by 17 is 16 ; the remainder of 98 divided by 16 is 2 . 5 - 2 = 3 . answer : d ." | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | d | subtract(reminder(reminder(88, 33), 17), reminder(90, reminder(33, 17))) | reminder(n0,n1)|reminder(n1,n2)|reminder(#0,n2)|reminder(n3,#1)|subtract(#2,#3)| | general |
if solomon distributes his toys in the ratio of 1 / 4 : 1 / 6 : 1 / 10 : 1 / 9 between his 4 friends a , b , c and d , then find the total number of toys solomon should have ? | here , a : b : c : d = 1 / 4 : 1 / 6 : 1 / 10 : 1 / 9 1 ) l . c . m of 4 : 6 : 10 : 9 is 180 2 ) find the number of pens each friend received - - - - - - - - - ( to find no . of pens each friend has , multiply the ratio with the l . c . m . calculated ) a = ( 1 / 4 ) x 180 = 45 b = ( 1 / 6 ) x 180 = 30 c = ( 1 / 10 ) x... | a ) 111.7 , b ) 118.7 , c ) 113.7 , d ) 218.7 , e ) 418.7 | b | add(add(multiply(10, 6), multiply(10, 4)), multiply(6, const_3)) | multiply(n3,n5)|multiply(n1,n5)|multiply(n3,const_3)|add(#0,#1)|add(#3,#2) | general |
the area of a circular field is 13.86 hectares . find the cost of fencing it at the rate of rs . 4.40 per metre . | "explanation : area = ( 13.86 x 10000 ) sq . m = 138600 sq . m circumference = cost of fencing = rs . ( 1320 x 4.40 ) = rs . 5808 . answer : d ) 5808" | a ) 2399 , b ) 3888 , c ) 2999 , d ) 5808 , e ) 2888 | d | multiply(circumface(multiply(sqrt(divide(13.86, const_pi)), const_100)), 4.40) | divide(n0,const_pi)|sqrt(#0)|multiply(#1,const_100)|circumface(#2)|multiply(#3,n1)| | geometry |
an article is bought for rs . 675 and sold for rs . 1000 , find the gain percent ? | "675 - - - - 325 100 - - - - ? = > = 48 % answer : c" | a ) 34 % , b ) 44 % , c ) 48 % , d ) 32 % , e ) 46 % | c | subtract(const_100, divide(multiply(1000, const_100), 675)) | multiply(n1,const_100)|divide(#0,n0)|subtract(const_100,#1)| | gain |
a 13 % stock yields 8 % . the market value of the stock is : | "solution to obtain rs . 8 , investment = rs . 100 . to obtain rs . 13 , investment = rs . ( 100 / 8 x 13 ) = rs . 162.50 β΄ market value of rs . 100 stock = rs . 162.50 answer c" | a ) rs . 72 , b ) rs . 92 , c ) rs . 162.50 , d ) rs . 116.50 , e ) none of these | c | multiply(divide(const_100, 8), 13) | divide(const_100,n1)|multiply(n0,#0)| | gain |
from a container , 6 liters milk was drawn out and was replaced by water . again 6 liters of mixture was drawn out and was replaced by the water . thus the quantity of milk and water in the container after these two operations is 9 : 16 . the quantity of mixture is : | explanation : let quantity of mixture be x liters . suppose a container contains x units of liquid from which y units are taken out and replaced by water . after operations , the quantity of pure liquid = units , where n = no of operations . so , quantity of milk = given that , milk : water = 9 : 16 milk : ( milk + wat... | a ) 15 , b ) 26 , c ) 22 , d ) 27 , e ) 11 | a | divide(multiply(6, sqrt(add(9, 16))), const_2) | add(n2,n3)|sqrt(#0)|multiply(n0,#1)|divide(#2,const_2) | other |
a train travels from new york to chicago , a distance of approximately 480 miles , at an average rate of 60 miles per hour and arrives in chicago at 5 : 00 in evening , chicago time . at what hour in the morning , new york time , did the train depart for chicago ? ( note : chicago time is one hour earlier than new york... | 5 : 00 in evening in chicago = 6 : 00 in evening in new york . so , the train was in chicago 6 : 00 in the evening , new york time . the trip took t = d / r = 480 / 60 = 8 hours . therefore , the train depart from new york at 6 : 00 - 8 hours = 10 : 00 in the morning , new york time . answer : a . | a ) 10 : 00 , b ) 4 : 00 , c ) 5 : 00 , d ) 6 : 00 , e ) 7 : 00 | a | subtract(multiply(add(5, const_1), const_3), divide(480, 60)) | add(n2,const_1)|divide(n0,n1)|multiply(#0,const_3)|subtract(#2,#1) | physics |
the speed of a subway train is represented by the equation z = s ^ 2 + 2 s for all situations where 0 β€ s β€ 7 , where z is the rate of speed in kilometers per hour and s is the time in seconds from the moment the train starts moving . in kilometers per hour , how much faster is the subway train moving after 7 seconds t... | "given : z = s ^ 2 + 2 s for 0 β€ s β€ 7 z ( 2 ) = 2 ^ 2 + 2 * 2 = 8 z ( 7 ) = 7 ^ 2 + 2 * 7 = 63 therefore z ( 7 ) - z ( 3 ) = 63 - 8 = 55 km / hr option b" | a ) 4 , b ) 55 , c ) 15 , d ) 48 , e ) 63 | b | subtract(add(power(7, 2), multiply(7, 2)), add(power(2, 2), multiply(2, 2))) | multiply(n0,n4)|multiply(n0,n5)|power(n4,n0)|power(n5,n0)|add(#0,#2)|add(#1,#3)|subtract(#4,#5)| | physics |
the diagonals of a rhombus are 11 cm and 16 cm . find its area ? | "1 / 2 * 11 * 16 = 88 answer : d" | a ) 277 , b ) 266 , c ) 150 , d ) 88 , e ) 212 | d | rhombus_area(11, 16) | rhombus_area(n0,n1)| | geometry |
in 1950 , richard was 4 times as old as robert . in 1955 , richard was 3 times as old as robert . in which year was richard 1.6 times as old as robert ? | "in 1950 : ri = 4 ro - - - - - - - - - - - - - - eq 1 in 1955 : ri + 5 = 3 ( ro + 5 ) - - - - - - - - - eq 2 thus in 1950 , solving eq 1 and eq 2 1950 ro = 10 , ri = 40 1990 ro = 50 , ri = 80 = 1.6 times thus ans : e" | a ) 1960 , b ) 1965 , c ) 1970 , d ) 1975 , e ) 1990 | e | add(1950, add(1.6, add(4, 3))) | add(n3,n1)|add(n4,#0)|add(n0,#1)| | general |
a room is 8 meters 16 centimeters in length and 4 meters 32 centimeters in width . find the least number of square tiles of equal size required to cover the entire floor of the room . | "let us calculate both the length and width of the room in centimeters . length = 8 meters and 16 centimeters = 816 cm width = 4 meters and 32 centimeters = 432 cm as we want the least number of square tiles required , it means the length of each square tile should be as large as possible . further , the length of each... | a ) 107 , b ) 153 , c ) 178 , d ) 198 , e ) 165 | b | divide(multiply(add(multiply(8, const_100), 16), add(multiply(4, const_100), 32)), multiply(gcd(add(multiply(8, const_100), 16), add(multiply(4, const_100), 32)), gcd(add(multiply(8, const_100), 16), add(multiply(4, const_100), 32)))) | multiply(n0,const_100)|multiply(n2,const_100)|add(n1,#0)|add(n3,#1)|gcd(#2,#3)|multiply(#2,#3)|multiply(#4,#4)|divide(#5,#6)| | general |
there are 8 executives , including the ceo and cfo , that are asked to form a small team of 5 members . however , the ceo and cfo may not both be assigned to the team . given this constraint , how many ways are there to form the team ? | "the total number of ways to form a team of 5 is 8 c 5 = 56 . we need to subtract the number of teams that have both the ceo and the cfo . the number of teams with both the ceo and cfo is 6 c 3 = 20 . the number of ways to form an acceptable team is 56 - 20 = 36 . the answer is c ." | a ) 34 , b ) 35 , c ) 36 , d ) 37 , e ) 38 | c | subtract(factorial(subtract(8, 5)), add(multiply(5, const_3), const_3)) | multiply(n1,const_3)|subtract(n0,n1)|add(#0,const_3)|factorial(#1)|subtract(#3,#2)| | other |
victor gets 92 % marks in examinations . if these are 460 marks , find the maximum marks . | "let the maximum marks be m then 92 % of m = 460 β 92 / 100 Γ m = 460 β m = ( 460 Γ 100 ) / 92 β m = 46000 / 92 β m = 500 therefore , maximum marks in the examinations are 500 . answer : b" | a ) 334 , b ) 500 , c ) 376 , d ) 288 , e ) 271 | b | divide(460, divide(92, const_100)) | divide(n0,const_100)|divide(n1,#0)| | gain |
max planned to buy bike , sales man advised him to buy within this month , if he purchase next month means state tax will be increasing 10 % than its cost and it would be 82500 . what is the bike ' s selling price in this month | sp * ( 110 / 100 ) = 82500 sp = 750 * 100 = > sp = 75000 answer : c | a ) 70000 , b ) 73000 , c ) 75000 , d ) 77000 , e ) 79000 | c | divide(82500, add(divide(10, const_100), const_1)) | divide(n0,const_100)|add(#0,const_1)|divide(n1,#1) | general |
a boy goes to his school from his house at a speed of 3 km / hr and returns at a speed of 2 km / hr . if he takes 5 hours in going and coming . the distance between his house and school is ? | "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 : c" | a ) 3 , b ) 4 , c ) 6 , d ) 5 , e ) 7 | c | multiply(divide(5, add(divide(3, 2), const_1)), 3) | divide(n0,n1)|add(#0,const_1)|divide(n2,#1)|multiply(n0,#2)| | physics |
find the perimeter of a triangle with sides measuring 7 centimeters , 10 centimeters and 15 centimeters . | perimeter of triangle = a + b + c perimeter = 7 cm + 10 cm + 15 cm = 32 cm answer : c . | ['a ) 10 cm', 'b ) 15 cm', 'c ) 32 cm', 'd ) 35 cm', 'e ) 20 cm'] | c | triangle_perimeter(7, 10, 15) | triangle_perimeter(n0,n1,n2) | geometry |
in a 500 m race , ravi beats ramesh by 5 seconds or 100 m . they decide to run another race and this time ravi gives ramesh a head tart of 200 m . if ravi β s speed is twice his previous speed and ramesh β s speed is one and half times his previous speed , how far from the starting point should the winning post be so t... | ravi beats ramesh by 5 sec or 100 mtr . . . . means ramesh cover 100 mtr in 5 sec . . . . by this we can get ramesh speed in 20 m / sec 20 . . . . . . . . . . . . . in 1 sec 500 . . . . . . . . in 500 / 20 = 25 sec ( ramesh ' s time ) so ravi take 20 sec to cover 500 mtr . . . . . . by this we can get that ravis speed ... | a ) 500 m , b ) 1000 m , c ) 1500 m , d ) 2000 m , e ) 2500 m | a | multiply(500, const_1) | multiply(n0,const_1) | physics |
a store owner estimates that the average price of type a products will increase by 25 % next year and that the price of type b products will increase by 10 % next year . this year , the total amount paid for type a products was $ 4500 and the total price paid for type b products was $ 8300 . according to the store owne... | "cost of type a products next year = 1.25 * 4500 = 5625 cost of type b products next year = 1.1 * 8300 = 9130 total 5625 + 9130 = 14755 answer : a" | a ) $ 14,755 , b ) $ 15,325 , c ) $ 16,000 , d ) $ 16,225 , e ) $ 17,155 | a | multiply(divide(const_3, const_4), const_1000) | divide(const_3,const_4)|multiply(#0,const_1000)| | general |
two isosceles triangles have equal vertical angles and their areas are in the ratio 25 : 49 . find the ratio of their corresponding heights . | "we are basically given that the triangles are similar . in two similar triangles , the ratio of their areas is the square of the ratio of their sides and also , the square of the ratio of their corresponding heights . therefore , area / area = height ^ 2 / height ^ 2 = 25 / 49 - - > height / height = 5 / 7 . answer : ... | a ) 4 / 5 , b ) 5 / 4 , c ) 3 / 2 , d ) 5 / 7 , e ) 2 / 3 | d | divide(sqrt(25), sqrt(49)) | sqrt(n0)|sqrt(n1)|divide(#0,#1)| | geometry |
a , b and c rent a pasture . if a puts 10 oxen for 7 months , b puts 12 oxen for 5 months and c puts 15 oxen for 3 months for grazing and the rent of the pasture is rs . 280 , then how much amount should c pay as his share of rent ? | a : b : c = 10 Γ 7 : 12 Γ 5 : 15 Γ 3 = 2 Γ 7 : 12 Γ 1 : 3 Γ 3 = 14 : 12 : 9 amount that c should pay = 280 Γ 9 / 35 = 8 Γ 9 = 72 answer is d | a ) 35 , b ) 45 , c ) 25 , d ) 72 , e ) 55 | d | multiply(280, divide(multiply(15, 3), add(add(multiply(10, 7), multiply(12, 5)), multiply(15, 3)))) | multiply(n4,n5)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|add(#3,#0)|divide(#0,#4)|multiply(n6,#5) | general |
exactly 30 % of the reporters for a certain wire service cover local politics in country x . if 25 % of the reporters who cover politics for the wire service do not cover local politics in country x , what percent of the reporters for the wire service do not cover politics ? | let ' s assume there are 100 reporters - - > 30 reporters cover local politics . now , as 25 % of the reporters who cover all politics do not cover local politics then the rest 75 % of the reporters who cover politics do cover local politics , so if there are x reporters who cover politics then 75 % of them equal to 30... | a ) 20 % , b ) 42 % , c ) 60 % , d ) 80 % , e ) 84 % | c | multiply(subtract(const_1, divide(30, subtract(const_100, 25))), const_100) | subtract(const_100,n1)|divide(n0,#0)|subtract(const_1,#1)|multiply(#2,const_100) | gain |
positive integer y is 50 percent of 50 percent of positive integer x , and y percent of x equals 16 . what is the value of x ? | "y = 0.5 * 0.5 * x = x / 4 y % * x = 16 ( y / 100 ) * x = 16 ( x / 400 ) * x = 16 x ^ 2 = 16 * 400 x = 80 the answer is c ." | a ) 20 , b ) 40 , c ) 80 , d ) 160 , e ) 320 | c | multiply(multiply(divide(50, 16), divide(50, 16)), const_1000) | divide(n0,n2)|divide(n1,n2)|multiply(#0,#1)|multiply(#2,const_1000)| | general |
excluding stoppages , the speed of a train is 48 kmph and including stoppages it is 36 kmph . of how many minutes does the train stop per hour ? | "t = 12 / 48 * 60 = 15 answer : c" | a ) 16 , b ) 17 , c ) 15 , d ) 18 , e ) 12 | c | subtract(const_60, multiply(const_60, divide(36, 48))) | divide(n1,n0)|multiply(#0,const_60)|subtract(const_60,#1)| | physics |
difference between the length & breadth of a rectangle is 26 m . if its perimeter is 208 m , then the ratio of area and perimeter is ? ? we have : ( l - b ) = 26 and 2 ( l + b ) = 208 or ( l + b ) = 104 ? | "solving the two equations , we get : l = 63 and b = 40 . area = ( l x b ) = ( 63 x 40 ) m 2 = 2520 m ^ 2 the required ratio is 2520 : 208 = 315 : 26 a" | a ) 315 : 26 , b ) 315 : 25 , c ) 315 : 30 , d ) 415 : 35 , e ) 425 : 35 | a | divide(const_3, 2) | divide(const_3,n3)| | geometry |
everyone shakes hands with everyone else in a room . total number of handshakes is 105 . number of persons = ? | "in a room of n people , the number of possible handshakes is c ( n , 2 ) or n ( n - 1 ) / 2 so n ( n - 1 ) / 2 = 105 or n ( n - 1 ) = 210 or n = 15 answer is ( d )" | a ) 14 , b ) 12 , c ) 11 , d ) 15 , e ) 16 | d | divide(divide(multiply(105, const_2), const_3), const_4) | multiply(n0,const_2)|divide(#0,const_3)|divide(#1,const_4)| | general |
the price of a t . v . set worth rs . 12000 is to be paid in 20 installments of rs . 1200 each . if the rate of interest be 6 % per annum , and the first installment be paid at the time of purchase , then the value of the last installment covering the interest as well will be ? | money paid in cash = rs . 1200 balance payment = ( 12000 - 1200 ) = rs . 10800 . answer : b | a ) 29997 , b ) 10800 , c ) 27098 , d ) 19000 , e ) 2799 | b | subtract(12000, 1200) | subtract(n0,n2) | gain |
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