Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
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a textile manufacturing firm employees 50 looms . it makes fabrics for a branded company . the aggregate sales value of the output of the 50 looms is rs 5 , 00,000 and the monthly manufacturing expenses is rs 1 , 50,000 . assume that each loom contributes equally to the sales and manufacturing expenses are evenly sprea... | "explanation : profit = 5 , 00,000 β ( 1 , 50,000 + 75,000 ) = rs . 2 , 75,000 . since , such loom contributes equally to sales and manufacturing expenses . but the monthly charges are fixed at rs 75,000 . if one loan breaks down sales and expenses will decrease . new profit : - = > 500000 Γ ( 49 / 50 ) β 150000 Γ ( 49... | a ) 13000 , b ) 7000 , c ) 10000 , d ) 5000 , e ) none of these | b | subtract(subtract(multiply(multiply(5, const_1000), const_100), add(multiply(75000, const_2), 75000)), subtract(subtract(multiply(divide(subtract(50, 1), 50), multiply(multiply(5, const_1000), const_100)), multiply(multiply(75000, const_2), divide(subtract(50, 1), 50))), 75000)) | multiply(n2,const_1000)|multiply(n6,const_2)|subtract(n0,n4)|add(n6,#1)|divide(#2,n0)|multiply(#0,const_100)|multiply(#4,#5)|multiply(#4,#1)|subtract(#5,#3)|subtract(#6,#7)|subtract(#9,n6)|subtract(#8,#10)| | general |
find large number from below question the difference of two numbers is 1365 . on dividing the larger number by the smaller , we get 4 as quotient and the 15 as remainder | "let the smaller number be x . then larger number = ( x + 1365 ) . x + 1365 = 4 x + 15 3 x = 1350 x = 450 large number = 450 + 1365 = 1815 e" | a ) 1235 , b ) 1346 , c ) 1378 , d ) 1635 , e ) 1815 | e | multiply(divide(subtract(1365, 15), subtract(4, const_1)), 4) | subtract(n0,n2)|subtract(n1,const_1)|divide(#0,#1)|multiply(n1,#2)| | general |
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 $ 216 per week . how much does she earn in dollars per hour ? | "let sheila earn x dollars per hour so , on monday , wednesday and friday , she earns 8 x each and , on tuesday and thursday , she earns 6 x each in total , over the week she should earn , 3 ( 8 x ) + 2 ( 6 x ) = 36 x she earns $ 216 per week 36 x = 216 x = 6 correct option : e" | a ) 11 , b ) 10 , c ) 9 , d ) 8 , e ) 6 | e | divide(216, add(multiply(8, const_3), multiply(6, const_2))) | multiply(n0,const_3)|multiply(n1,const_2)|add(#0,#1)|divide(n2,#2)| | physics |
by selling 15 pencils for a rupee a man loses 60 % . how many for a rupee should he sell in order to gain 60 % ? | "40 % - - - 15 160 % - - - ? 40 / 160 * 15 = 3.75 answer : d" | a ) 2.15 , b ) 4.2 , c ) 5.18 , d ) 3.75 , e ) 4.2 | d | multiply(divide(const_1, multiply(add(const_100, 60), divide(const_1, subtract(const_100, 60)))), 15) | add(n2,const_100)|subtract(const_100,n1)|divide(const_1,#1)|multiply(#0,#2)|divide(const_1,#3)|multiply(n0,#4)| | gain |
find the area of a cuboid of length 10 cm , breadth 8 cm . and height 6 cm . | "area of a cuboid = lxbxh = 10 cm x 8 cm x 6 cm = 480 cm cube answer : e" | a ) 400 cm cube , b ) 410 cm cube , c ) 420 cm cube , d ) 430 cm cube , e ) 480 cm cube | e | multiply(multiply(10, 8), 6) | multiply(n0,n1)|multiply(n2,#0)| | physics |
a shopkeeper bought 600 oranges and 400 bananas . he found 15 % of oranges and 8 % of bananas were rotten . find the percentage of fruits in good condition ? | "total number of fruits shopkeeper bought = 600 + 400 = 1000 number of rotten oranges = 15 % of 600 = 15 / 100 Γ 600 = 9000 / 100 = 90 number of rotten bananas = 8 % of 400 = 8 / 100 Γ 400 = 3200 / 100 = 32 therefore , total number of rotten fruits = 90 + 32 = 122 therefore number of fruits in good condition = 1000 - 1... | a ) 92.5 % , b ) 87.8 % , c ) 85.2 % , d ) 96.8 % , e ) 78.9 % | b | multiply(divide(subtract(add(600, 400), add(multiply(600, divide(15, const_100)), multiply(400, divide(8, const_100)))), add(600, 400)), const_100) | add(n0,n1)|divide(n2,const_100)|divide(n3,const_100)|multiply(n0,#1)|multiply(n1,#2)|add(#3,#4)|subtract(#0,#5)|divide(#6,#0)|multiply(#7,const_100)| | gain |
in an election , candidate a got 75 % of the total valid votes . if 15 % of the total votes were declared invalid and the total numbers of votes is 560000 , find the number of valid vote polled in favour of candidate | "total number of invalid votes = 15 % of 560000 = 15 / 100 Γ 560000 = 8400000 / 100 = 84000 total number of valid votes 560000 β 84000 = 476000 percentage of votes polled in favour of candidate a = 75 % therefore , the number of valid votes polled in favour of candidate a = 75 % of 476000 = 75 / 100 Γ 476000 = 35700000... | a ) 356000 , b ) 356500 , c ) 357000 , d ) 400000 , e ) none of these | c | multiply(multiply(560000, subtract(const_1, divide(15, const_100))), divide(75, const_100)) | divide(n0,const_100)|divide(n1,const_100)|subtract(const_1,#1)|multiply(n2,#2)|multiply(#0,#3)| | gain |
of 30 applicants for a job , 12 had at least 4 years ' experience , 19 had degrees , and 6 had less than 4 years ' experience and did not have a degree . how many of the applicants had at least 4 years ' experience and a degree ? | "set a : people with more than 4 years exp set b : people with degree aub = total - ( less than 4 exp and no degree ) aub = 30 - 6 = 24 aub = a + b - aib aib = 19 + 12 - 24 = 7 answer d" | a ) 14 , b ) 13 , c ) 9 , d ) 7 , e ) 5 | d | add(subtract(add(12, 19), subtract(30, 6)), subtract(19, 12)) | add(n1,n3)|subtract(n0,n4)|subtract(n3,n1)|subtract(#0,#1)|add(#3,#2)| | general |
the circumferences of two circles are 264 meters and 704 meters . find the difference between the areas of the larger and the smaller circles ? | "let the radii of the smaller and the larger circles be s m and l m respectively . 2 β s = 264 and 2 β l = 704 s = 264 / 2 β and l = 704 / 2 β difference between the areas = β l ^ 2 - β s ^ 2 = β { 132 ^ 2 / β ^ 2 - 352 ^ 2 / β ^ 2 } = 132 ^ 2 / β - 352 ^ 2 / β = ( 132 - 352 ) ( 132 + 352 ) / β = ( 220 ) ( 484 ) / ( 22... | a ) 29960 sq m , b ) 33880 sq m , c ) 43120 sq m , d ) 27680 sq m , e ) 12786 sq m | b | subtract(circle_area(divide(704, multiply(const_2, const_pi))), circle_area(divide(264, multiply(const_2, const_pi)))) | multiply(const_2,const_pi)|divide(n1,#0)|divide(n0,#0)|circle_area(#1)|circle_area(#2)|subtract(#3,#4)| | geometry |
two trains are moving in opposite directions at 60 km / hr and 90 km / hr . their lengths are 1.25 km and 1.00 km respectively . the time taken by the slower train to cross the faster train in seconds is ? | ": relative speed = 60 + 90 = 150 km / hr . = 150 * 5 / 18 = 125 / 3 m / sec . distance covered = 1.25 + 1.0 = 2.25 km = 2250 m . required time = 2250 * 3 / 125 = 54 sec . answer : e" | a ) 48 , b ) 9 , c ) 7 , d ) 67 , e ) 54 | e | subtract(divide(multiply(1.25, const_1000), divide(multiply(60, const_1000), const_3600)), divide(multiply(1.00, const_1000), divide(multiply(90, const_1000), const_3600))) | multiply(n2,const_1000)|multiply(n0,const_1000)|multiply(n3,const_1000)|multiply(n1,const_1000)|divide(#1,const_3600)|divide(#3,const_3600)|divide(#0,#4)|divide(#2,#5)|subtract(#6,#7)| | physics |
a reduction of 25 % in the price of salt enables a lady to obtain 10 kgs more for rs . 100 , find the original price per kg ? | "100 * ( 25 / 100 ) = 25 - - - 10 ? - - - 1 = > rs . 2.5 100 - - - 75 ? - - - 2.5 = > rs . 3.3 answer : d" | a ) 2.8 , b ) 2.3 , c ) 2.5 , d ) 3.3 , e ) 2.9 | d | multiply(divide(divide(multiply(divide(25, const_100), 100), 10), multiply(divide(25, const_100), 100)), const_100) | divide(n0,const_100)|multiply(n2,#0)|divide(#1,n1)|divide(#2,#1)|multiply(#3,const_100)| | gain |
if 15 % of 30 % of 50 % of a number is 90 , then what is the number ? | "let the number be a given , 15 / 100 * 30 / 100 * 50 / 100 * a = 90 = > 3 / 20 * 3 / 10 * 1 / 2 * a = 90 = > a = 10 * 20 * 10 * 2 = 4000 . answer : a" | a ) 4000 , b ) 3050 , c ) 4400 , d ) 4500 , e ) none of these | a | divide(90, multiply(multiply(divide(50, const_100), divide(30, const_100)), divide(15, const_100))) | divide(n0,const_100)|divide(n2,const_100)|divide(n1,const_100)|multiply(#1,#2)|multiply(#0,#3)|divide(n3,#4)| | gain |
a certain box has 9 cards and each card has one of the integers from 1 to 9 inclusive . each card has a different number . if 2 different cards are selected at random , what is the probability that the sum of the numbers written on the 2 cards is less than the average ( arithmetic mean ) of all the numbers written on t... | "the average of the numbers is 5 . the total number of ways to choose 2 cards from 9 cards is 9 c 2 = 36 . the ways to choose 2 cards with a sum less than the average are : { 1,2 } , { 1,3 } the probability is 2 / 36 = 1 / 18 the answer is c ." | a ) 1 / 36 , b ) 1 / 15 , c ) 1 / 18 , d ) 1 / 9 , e ) 2 / 9 | c | divide(const_4, divide(factorial(9), multiply(factorial(2), factorial(subtract(9, 2))))) | factorial(n0)|factorial(n3)|subtract(n0,n3)|factorial(#2)|multiply(#1,#3)|divide(#0,#4)|divide(const_4,#5)| | general |
what is the 6 th digit to the right of the decimal point in the decimal equivalent of 16 / 45 ? | "16 / 45 = 0.3555 . . . . 35 is non - terminating repeating decimal . the 22 nd digit to the right of decimal point will be 5 . answer a" | a ) 5 , b ) 12 , c ) 2 , d ) 35 , e ) 4 | a | divide(16, 45) | divide(n1,n2)| | general |
a , b and c invested rs . 6300 , rs . 4200 and rs . 10500 respectively , in a partnership business . find the share of a in profit of rs . 13600 after a year ? | "6300 : 4200 : 10500 3 : 2 : 5 3 / 10 * 13600 = 4080 answer : e" | a ) 3630 , b ) 9232 , c ) 8828 , d ) 2387 , e ) 4080 | e | multiply(divide(6300, add(add(6300, 4200), 10500)), 13600) | add(n0,n1)|add(n2,#0)|divide(n0,#1)|multiply(n3,#2)| | gain |
two trains 120 meters and 165 meters in length respectively are running in opposite directions , one at the rate of 80 km and the other at the rate of 65 kmph . in what time will they be completely clear of each other from the moment they meet ? | "t = ( 120 + 165 ) / ( 80 + 65 ) * 18 / 5 t = 7.07 answer : c" | a ) 7.16 , b ) 7.16 , c ) 7.07 , d ) 7.15 , e ) 7.11 | c | divide(add(120, 165), multiply(add(80, 65), const_0_2778)) | add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)| | physics |
the average weight of 4 pigs is 15 kg . weight less than 16 kg is considered as under weight . what maximum no . of pig can be under weight . | sum of all four pigs weigh = 60 kg it is possible that all four pigs are underweight . because the average is 15 kg which itself underweight . answer e | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | e | multiply(subtract(16, 15), 4) | subtract(n2,n1)|multiply(n0,#0) | general |
a can do a job in 30 days and b in 40 days . if they work on it together for 10 days , then the fraction of the work that is left is ? | "a ' s 1 day work = 1 / 30 b ' s 1 day work = 1 / 40 a + b 1 day work = 1 / 30 + 1 / 40 = 7 / 120 a + b 4 days work = 7 / 120 * 10 = 7 / 12 remaining work = 1 - 7 / 12 = 5 / 12 answer is c" | a ) 2 / 11 , b ) 3 / 13 , c ) 5 / 12 , d ) 7 / 10 , e ) 6 / 15 | c | subtract(const_1, multiply(10, add(divide(const_1, 30), divide(const_1, 40)))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|multiply(n2,#2)|subtract(const_1,#3)| | physics |
a retail appliance store priced a video recorder at 20 percent above the wholesale cost of $ 200 . if a store employee applied the 15 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 = 15 emlpoyee paid = . 85 * 240 = 204 $ answer a" | a ) $ 204 , b ) $ 216 , c ) $ 220 , d ) $ 230 , e ) $ 240 | a | subtract(add(200, multiply(divide(200, const_100), 20)), multiply(divide(add(200, multiply(divide(200, const_100), 20)), const_100), 15)) | divide(n1,const_100)|multiply(n0,#0)|add(n1,#1)|divide(#2,const_100)|multiply(n2,#3)|subtract(#2,#4)| | gain |
the compound ratio of 1 : 4 , 2 : 5 and 1 : 6 ? | 1 / 4 * 2 / 5 * 1 / 6 = 1 / 60 1 : 60 answer : e | a ) 1 : 3 , b ) 1 : 4 , c ) 1 : 10 , d ) 1 : 30 , e ) 1 : 60 | e | divide(divide(multiply(1, 2), multiply(4, 5)), divide(multiply(2, 1), multiply(5, 6))) | multiply(n0,n2)|multiply(n1,n3)|multiply(n2,n4)|multiply(n3,n5)|divide(#0,#1)|divide(#2,#3)|divide(#4,#5)| | other |
the true discount on a bill due 9 months hence at 16 % per annum is rs . 189 . the amount of the bill is | "solution 32.5 let p . w . be rs . x . then , s . i . on rs . x at 16 % for 9 months = rs . 189 . β΄ x 16 x 9 / 12 x 1 / 100 } = 189 or x = 1575 . β΄ p . w . = rs . 1575 . answer b" | a ) rs . 1386 , b ) rs . 1764 , c ) rs . 1575 , d ) rs . 2268 , e ) none of these | b | add(divide(189, divide(multiply(divide(9, multiply(const_4, const_3)), 16), const_100)), 189) | multiply(const_3,const_4)|divide(n0,#0)|multiply(n1,#1)|divide(#2,const_100)|divide(n2,#3)|add(n2,#4)| | gain |
in a division sum , the remainder is 0 . as student mistook the divisor by 12 instead of 21 and obtained 35 as quotient . what is the correct quotient ? | 12 * 35 = 420 420 % 21 = 20 answer : d | a ) 0 , b ) 12 , c ) 13 , d ) 20 , e ) 25 | d | divide(multiply(12, 35), 21) | multiply(n1,n3)|divide(#0,n2) | general |
tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random . what is the probability that the ticket drawn bears a number which is a multiple of 3 ? | "here , s = { 1 , 2,3 , . . . . . . , 19,20 } e = event getting a multiple of 3 = { 3,6 , 9,12 , 15,18 } probability = 6 / 20 = 3 / 10 correct option is a" | a ) 3 / 10 , b ) 1 / 2 , c ) 2 / 5 , d ) 3 / 7 , e ) 1 / 6 | a | divide(3, const_10) | divide(n2,const_10)| | general |
the sum of the digits of [ ( 10 ^ x ) ^ y ] - 64 = 279 . what is the value of xy ? | 1000 β 64 = 936 . whatever xy is you finish with 36 = = > 3 + 6 = 9 therefore , 279 β 9 = 270 and 270 / 9 = 30 now you add the last two digits ( 3 and 6 ) answer is 30 + 2 = 32 answer : e | a ) 28 , b ) 29 , c ) 30 , d ) 31 , e ) 32 | e | divide(64, const_2) | divide(n1,const_2) | general |
a corporation paid $ 7 million in federal taxes on its first $ 50 million of gross profits and then $ 30 million in federal taxes on the next $ 150 million in gross profits . by approximately what percent did the ratio of federal taxes to gross profits increase from the first $ 50 million in profits to the next $ 150 m... | "initial ratio of federal taxes to gross profits : 7 / 50 = 0.14 final ratio : 30 / 150 = 0.2 thus , the percentage change : ( 0.2 - 0.14 ) / 0.14 * 100 = 6 / 14 * 100 = 3 / 7 * 100 = just less than 50 % . answer : e" | a ) 6 % , b ) 14 % , c ) 20 % , d ) 23 % , e ) 43 % | e | multiply(divide(subtract(divide(add(30, 7), 150), divide(7, 50)), divide(7, 50)), const_100) | add(n0,n2)|divide(n0,n1)|divide(#0,n3)|subtract(#2,#1)|divide(#3,#1)|multiply(#4,const_100)| | gain |
jack and jill are marathon runners . jack can finish a marathon ( 42 km ) in 5.5 hours and jill can run a marathon in 4.2 hours . what is the ratio of their average running speed ? ( jack : jill ) | "average speed of jack = distance / time = 42 / ( 11 / 2 ) = 84 / 11 average speed of jill = 42 / ( 4.2 ) = 10 ratio of average speed of jack to jill = ( 84 / 11 ) / 10 = 84 / 110 = 42 / 55 answer d" | a ) 14 / 15 , b ) 15 / 14 , c ) 4 / 5 , d ) 42 / 55 , e ) can not be determined | d | divide(divide(42, 5.5), divide(42, 4.2)) | divide(n0,n1)|divide(n0,n2)|divide(#0,#1)| | physics |
two trains a and b are 100 m and 150 m long and are moving at one another at 54 km / hr and 36 km / hr respectively . arun is sitting on coach b 1 of train a . calculate the time taken by arun to completely cross train b . | "detailed solution speed of a = 54 β 1000 / 60 β 60 = 15 m / s speed of b = 36 β 1000 / 60 β 60 = 10 m / s relative speed = s 1 + s 2 = 15 + 10 m / s = 25 m / s the length that needs to be crossed = length of train b = 150 m . therefore time taken = 150 / 25 = 6 s . what is the time taken for trains to completely cross... | a ) 10 s , b ) 6 s , c ) 4 s , d ) 8 s , e ) 12 s | b | divide(add(100, 150), add(divide(multiply(54, const_1000), const_3600), divide(multiply(36, const_1000), const_3600))) | add(n0,n1)|multiply(n2,const_1000)|multiply(n3,const_1000)|divide(#1,const_3600)|divide(#2,const_3600)|add(#3,#4)|divide(#0,#5)| | physics |
angelo and isabella are both salespersons . in any given week , angelo makes $ 550 in base salary plus 8 percent of the portion of his sales above $ 5,000 for that week . isabella makes 10 percent of her total sales for any given week . for what amount of weekly sales would angelo and isabella earn the same amount of m... | "official solution : the problem asks for the amount of weekly sales it takes for angelo and isabella to earn the same amount of money . you can write an equation that sets angelo β s and isabella β s weekly earnings equal to each other , with x representing weekly sales . weekly earnings for each salesperson equal bas... | a ) 7,500 , b ) 24,500 , c ) 25,500 , d ) 26,500 , e ) 27,500 | a | floor(divide(divide(subtract(550, multiply(5,000, divide(8, const_100))), subtract(divide(10, const_100), divide(8, const_100))), 5,000)) | divide(n1,const_100)|divide(n3,const_100)|multiply(#0,n2)|subtract(#1,#0)|subtract(n0,#2)|divide(#4,#3)|divide(#5,n2)|floor(#6)| | general |
15 % people of a village in sri lanka died by bombardment , 25 % of the remainder left the village on account of fear . if now the population is reduced to 4555 , how much was it in the beginning ? | "x * ( 85 / 100 ) * ( 75 / 100 ) = 4555 x = 7145 answer : b" | a ) 7400 , b ) 7145 , c ) 7200 , d ) 7323 , e ) 6030 | b | floor(divide(4555, multiply(divide(subtract(const_100, 15), const_100), divide(subtract(const_100, 25), const_100)))) | subtract(const_100,n0)|subtract(const_100,n1)|divide(#0,const_100)|divide(#1,const_100)|multiply(#2,#3)|divide(n2,#4)|floor(#5)| | general |
in a group of cows and hens , the number of legs are 18 more than twice the number of heads . the number of cows is | "explanation : let the number of cows be x and the number of hens be y . then , 4 x + 2 y = 2 ( x + y ) + 18 4 x + 2 y = 2 x + 2 y + 18 2 x = 18 x = 9 . answer : c" | a ) 5 , b ) 7 , c ) 9 , d ) 12 , e ) 14 | c | subtract(18, const_4) | subtract(n0,const_4)| | general |
the area of rectangular field is 460 square metres . if the length is 15 per cent more than the breadth , what is the breadth of the rectangular field ? | let the breadth of the rectangular field be β x β m . then , length of the field will be x + x Γ 15 / 100 = 23 x / 20 now , x Γ 23 x / 20 = 460 or , 23 x 2 = 460 Γ 20 or , x 2 = 20 Γ 20 or , x = 20 m answer e | ['a ) 15 metres', 'b ) 26 metres', 'c ) 34.5 metres', 'd ) can not be determined', 'e ) none of these'] | e | divide(multiply(460, multiply(const_2, const_10)), divide(multiply(460, multiply(const_2, const_10)), const_100)) | multiply(const_10,const_2)|multiply(n0,#0)|divide(#1,const_100)|divide(#1,#2) | geometry |
sum of the squares of 3 no . is 252 and the sum of their products taken two at a time is 116 . find the sum ? | "( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 ( ab + bc + ca ) = 252 + 2 * 116 a + b + c = Γ’ Λ Ε‘ 484 = 22 answer c" | a ) 24 , b ) 33 , c ) 22 , d ) 20 , e ) 21 | c | sqrt(add(252, multiply(116, const_2))) | multiply(n2,const_2)|add(n1,#0)|sqrt(#1)| | general |
a train 120 m long running at 72 kmph crosses a platform in 25 sec . what is the length of the platform ? | "d 350 d = 72 * 5 / 18 = 25 = 500 Γ’ β¬ β 120 = 380" | a ) 443 m , b ) 354 m , c ) 450 m , d ) 380 m , e ) 250 m | d | subtract(multiply(25, multiply(72, const_0_2778)), 120) | multiply(n1,const_0_2778)|multiply(n2,#0)|subtract(#1,n0)| | physics |
an exam consists of 8 true / false questions . brian forgets to study , so he must guess blindly on each question . if any score above 60 % is a passing grade , what is the probability that brian passes ? | "if you have 8 t or f and brian is going to guess then each question he has a 50 % chance of getting correct . if a passing score is 70 % it means brian needs to get 6 / 8 = 75 % , 7 / 8 = 87.5 % , or 8 / 8 = 100 % to pass . each is a possibility . if brian gets a 5 / 8 ( = 62.5 % ) or below he fails . so first figure ... | a ) 1 / 16 , b ) 37 / 256 , c ) 1 / 2 , d ) 35 / 256 , e ) 15 / 16 | d | add(divide(subtract(const_1, add(add(power(divide(const_1, const_2), 8), multiply(8, power(divide(const_1, const_2), 8))), multiply(multiply(8, const_3), power(divide(const_1, const_2), 8)))), const_10), subtract(const_1, add(add(power(divide(const_1, const_2), 8), multiply(8, power(divide(const_1, const_2), 8))), mult... | divide(const_1,const_2)|multiply(n0,const_3)|power(#0,n0)|multiply(n0,#2)|multiply(#1,#2)|add(#3,#2)|add(#5,#4)|subtract(const_1,#6)|divide(#7,const_10)|add(#8,#7)| | general |
the positive numbers w , x , y , and z are such that x is 35 percent greater than y , y is 20 percent greater than z , and w is 20 percent less than x . what percent greater than z is w ? | "my strategy is same as thedobermanbut instead take z = 100 , which makes life a bit easy . as : z = 100 y = 120 ( 20 % greater than z ) z = 144 ( 20 % greater than y ) now calculate w 20 % less than z = 144 * 80 / 100 = 115.2 now by just looking , relation between w and z : w - z / z * 100 = 20 - answer c" | a ) 15.2 % , b ) 16.0 % , c ) 20.0 % , d ) 23.2 % , e ) 24.8 % | c | multiply(const_100, subtract(multiply(multiply(divide(add(35, const_100), const_100), divide(add(35, const_100), const_100)), divide(subtract(const_100, 35), const_100)), const_1)) | add(n0,const_100)|subtract(const_100,n0)|divide(#1,const_100)|divide(#0,const_100)|multiply(#3,#3)|multiply(#2,#4)|subtract(#5,const_1)|multiply(#6,const_100)| | general |
evaluate : 22 + sqrt ( - 4 + 6 * 4 * 3 ) = ? | "according to order of operations , inner brackets first where 6 ? 4 ? 3 is first calculated since it has a multiplication and a division . 6 * 4 * 3 = 24 * 3 = 8 hence 22 + sqrt ( - 4 + 6 * 4 * 3 ) = 22 + sqrt ( - 4 + 8 ) = 22 + sqrt ( 4 ) = 22 + 2 = 24 correct answer c" | a ) 4 , b ) 14 , c ) 24 , d ) 34 , e ) 44 | c | add(22, sqrt(subtract(divide(multiply(6, 4), 3), 4))) | multiply(n1,n2)|divide(#0,n4)|subtract(#1,n1)|sqrt(#2)|add(n0,#3)| | general |
the width of a rectangle is 10 mtr and its area is 150 square meter . if length of the rectangle is increased , then its new area is 1 ( 1 / 3 ) times of the original area . what is the new perimeter of the rectangle ? | ans . ( d ) sol . length of original rectangle = 150 / 10 = 15 ( 10 * ( 15 + x ) ) / ( 10 * 15 ) = 4 / 3 x = 5 perimeter of new rectangle = 2 ( 10 + 20 ) = 60 | ['a ) 45', 'b ) 54', 'c ) 67', 'd ) 60', 'e ) 69'] | d | add(multiply(const_2, divide(divide(multiply(150, const_4), 3), 10)), multiply(const_2, 10)) | multiply(n1,const_4)|multiply(n0,const_2)|divide(#0,n4)|divide(#2,n0)|multiply(#3,const_2)|add(#4,#1) | geometry |
how long will a boy take to run round a square field of side 40 meters , if he runs at the rate of 9 km / hr ? | "speed = 9 km / hr = 9 * 5 / 18 = 5 / 2 m / sec distance = 40 * 4 = 160 m time taken = 160 * 2 / 5 = 64 sec answer is a" | a ) 64 sec , b ) 45 sec , c ) 1 min , d ) 32 sec , e ) 25 sec | a | divide(multiply(40, const_4), multiply(9, divide(const_1000, const_3600))) | divide(const_1000,const_3600)|multiply(n0,const_4)|multiply(n1,#0)|divide(#1,#2)| | gain |
the side of a square is increased by 40 % then how much % does its area increases ? | "a = 100 a 2 = 10000 a = 140 a 2 = 19600 - - - - - - - - - - - - - - - - 10000 - - - - - - - - - 9600 100 - - - - - - - ? = > 96 % answer : a" | a ) 96 , b ) 56.25 , c ) 50.75 , d ) 42.75 , e ) 52.75 | a | divide(multiply(subtract(square_area(add(const_100, 40)), square_area(const_100)), const_100), square_area(const_100)) | add(n0,const_100)|square_area(const_100)|square_area(#0)|subtract(#2,#1)|multiply(#3,const_100)|divide(#4,#1)| | geometry |
a runs twice as fast as b and gives b a start of 83 m . how long should the racecourse be so that a and b might reach in the same time ? | "ratio of speeds of a and b is 2 : 1 b is 83 m away from a but we know that a covers 1 meter ( 2 - 1 ) more in every second than b the time taken for a to cover 83 m is 83 / 1 = 83 m so the total time taken by a and b to reach = 2 * 83 = 166 m answer : e" | a ) 75 m . , b ) 80 m . , c ) 150 m . , d ) 100 m . , e ) none of the above | e | multiply(83, const_2) | multiply(n0,const_2)| | physics |
two trains 190 m and 160 m long run at the speed of 60 km / hr and 40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? | "relative speed = 60 + 40 = 100 km / hr . = 100 * 5 / 18 = 250 / 9 m / sec . distance covered in crossing each other = 190 + 160 = 350 m . required time = 350 * 9 / 250 = 12.6 sec . answer : d" | a ) 10.7 , b ) 10.9 , c ) 10.6 , d ) 12.6 , e ) 18.8 | d | divide(add(190, 160), multiply(add(60, 40), const_0_2778)) | add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)| | physics |
if a square mirror has a 20 - inch diagonal , what is the approximate perimeter w of the mirror , in inches ? | "if you draw the square and diagonal inside the square . u can see square becomes part of two triangles opposite to each other . and we know the property of the triangle , addition of two sides of triangle must be greater than its diagonal in order to complete the triangle . and each side must be less than 20 and perim... | a ) 40 , b ) 60 , c ) 80 , d ) 100 , e ) 120 | b | square_perimeter(divide(20, power(add(const_1, const_1), inverse(const_2)))) | add(const_1,const_1)|inverse(const_2)|power(#0,#1)|divide(n0,#2)|square_perimeter(#3)| | geometry |
10 litres of water are poured into an aquarium of dimensions 50 cm length , 20 cm breadth , and 40 cm height . how high ( in cm ) will the water rise ? ( 1 litre = 1000 cm Β³ ) | lxbxh = 10000 h = 10000 / 50 * 20 = 10 cm ' c ' is the answer . | a ) 6 , b ) 8 , c ) 10 , d ) 20 , e ) 40 | c | divide(40, divide(divide(multiply(multiply(50, 20), 40), 1000), 10)) | multiply(n1,n2)|multiply(n3,#0)|divide(#1,n5)|divide(#2,n0)|divide(n3,#3) | physics |
a retailer sells 10 shirts . the first 3 he sells for $ 82 , $ 100 and $ 90 . if the retailer wishes to sell the 10 shirts for an overall average price of over $ 100 , what must be the minimum average price of the remaining 7 shirts ? | first 3 shirts are sold for $ 82 , $ 100 and $ 90 = $ 272 . to get average price of $ 100 , total sale should be 10 * $ 100 = $ 1000 so remaining 7 shirts to be sold for $ 1000 - $ 272 = $ 728 answer should be 728 / 7 = $ 104.00 that is b | a ) $ 100.00 , b ) $ 104.00 , c ) $ 110.00 , d ) $ 115.00 , e ) $ 119.00 | b | divide(subtract(multiply(10, 100), add(add(82, 100), 90)), 7) | add(n2,n3)|multiply(n0,n3)|add(n4,#0)|subtract(#1,#2)|divide(#3,n7) | general |
a train running at the speed of 90 km / hr crosses a pole in 9 seconds . find the length of the train . | "speed = 90 * ( 5 / 18 ) m / sec = 25 m / sec length of train ( distance ) = speed * time ( 25 ) * 9 = 225 meter answer : a" | a ) 225 meter , b ) 200 meter , c ) 250 meter , d ) 210 meter , e ) 230 meter | a | multiply(divide(multiply(90, const_1000), const_3600), 9) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics |
what is the total number of integers between 20 and 80 ( exclusive ) that are divisible by 10 ? | "30 , 40,50 , 60,70 total 5 answer is a" | a ) 5 , b ) 7 , c ) 4 , d ) 9 , e ) 3 | a | add(divide(subtract(80, 20), 10), const_1) | subtract(n1,n0)|divide(#0,n2)|add(#1,const_1)| | general |
a mixture of 85 kg contains milk and water in the ratio 27 : 7 . how much more water is to be added to get a new mixture containing milk and water in the ratio 3 : 1 ? | explanation : expl : milk in 85 kg of mix = 85 * ( 27 / 34 ) = 135 / 2 kg . water in it = 85 - 135 / 2 = 35 / 2 kg . let x kg of water be added to it . then , ( 135 / 2 ) / ( 35 / 2 + x ) = 3 / 1 ; 135 / ( 35 + 2 x ) = 3 / 1 ; 105 + 6 x = 135 ; 6 x = 30 ; x = 5 the quantity of water to be added = 5 kg answer : a | a ) 5 kg , b ) 6.5 kg , c ) 7.5 kg , d ) 8 kg , e ) 9 kg | a | divide(subtract(multiply(85, divide(27, add(27, 7))), multiply(3, multiply(85, divide(7, add(27, 7))))), 3) | add(n1,n2)|divide(n1,#0)|divide(n2,#0)|multiply(n0,#1)|multiply(n0,#2)|multiply(n3,#4)|subtract(#3,#5)|divide(#6,n3) | general |
what is the sum of all the multiples of 25 between 30 and 100 ? | "you first have to know all the multiples of 25 between 30 and 100 . they are 25 , 50,75 and 100 . if you add all these numbers together , you get 250 . final answer : d" | a ) 800 , b ) 625 , c ) 600 , d ) 250 , e ) 200 | d | add(add(add(add(add(add(const_12, const_2), const_1), add(add(const_12, const_2), add(add(add(add(add(const_2, const_4), const_4), subtract(const_10, const_1)), add(add(const_2, const_4), const_4)), add(const_10, const_2)))), add(add(add(const_12, const_2), const_1), const_1)), 25), add(const_2, const_4)) | add(const_12,const_2)|add(const_2,const_4)|add(const_10,const_2)|subtract(const_10,const_1)|add(#0,const_1)|add(#1,const_4)|add(#5,#3)|add(#4,const_1)|add(#6,#5)|add(#8,#2)|add(#0,#9)|add(#4,#10)|add(#11,#7)|add(n0,#12)|add(#13,#1)| | general |
the product of a and b is equal to 12 more than twice the sum of a and b . if b = 10 , what is the value of b - a ? | ab = 12 + 2 ( a + b ) 10 a = 12 + 2 a + 20 8 a = 32 a = 4 b - a = 10 - 4 = 6 d is the answer | a ) 2 , b ) 5 , c ) 7 , d ) 6 , e ) 35 | d | subtract(10, divide(add(multiply(10, const_2), 12), subtract(10, const_2))) | multiply(n1,const_2)|subtract(n1,const_2)|add(n0,#0)|divide(#2,#1)|subtract(n1,#3) | general |
a 600 meter long train crosses a signal post in 40 seconds . how long will it take to cross a 5.4 kilometer long bridge , at the same speed ? | "s = 600 / 40 = 15 mps s = 5400 / 15 = 360 sec = 6 min . answer : d" | a ) 4 min , b ) 2 min , c ) 8 min , d ) 6 min , e ) 3 min | d | divide(divide(multiply(5.4, const_1000), speed(600, 40)), const_60) | multiply(n2,const_1000)|speed(n0,n1)|divide(#0,#1)|divide(#2,const_60)| | physics |
there are 7 stores in town that had a total of 21 visitors on a particular day . however , only 11 people went shopping that day ; some people visited more than one store . if 7 people visited exactly two stores each , and everyone visited at least one store , what is the largest number of stores anyone could have visi... | "7 people visited 2 stores each for 14 visits . to maximize the number of stores that one person visited , let ' s assume that 3 people visited 1 store each . the number of remaining visits is 21 - 14 - 3 = 4 , which is the maximum that one person could have visited . the answer is c ." | a ) 6 , b ) 8 , c ) 4 , d ) 9 , e ) 2 | c | subtract(subtract(21, multiply(7, const_2)), subtract(subtract(11, 7), const_1)) | multiply(n0,const_2)|subtract(n2,n0)|subtract(n1,#0)|subtract(#1,const_1)|subtract(#2,#3)| | general |
what is the largest 4 digit number exactly divisible by 12 ? | "largest 4 digit number = 9999 9999 Γ· 12 = 833 , remainder = 3 hence largest 4 digit number exactly divisible by 12 = 9999 - 3 = 9996 answer : e" | a ) 4676 , b ) 4678 , c ) 8888 , d ) 9504 , e ) 9996 | e | multiply(floor(divide(power(const_10, 4), 12)), 12) | power(const_10,n0)|divide(#0,n1)|floor(#1)|multiply(n1,#2)| | general |
the length of a rectangulat plot is 20 metres more than its breadth . if the cost of fencing the plot @ rs . 26.50 per metre is rs . 5300 , what is the length of the plot in metres ? | solution let breadth = x metres then , length = ( x + 20 ) metres . perimeter = ( 5300 / 26.50 ) m = 200 m answer c | a ) 40 , b ) 50 , c ) 200 , d ) 120 , e ) none | c | divide(5300, 26.5) | divide(n2,n1) | physics |
if two positive numbers are in the ratio 1 / 8 : 1 / 7 , then by what percent is the second number more than the first ? | given ratio = 1 / 8 : 1 / 7 = 7 : 8 let first number be 7 x and the second number be 8 x . the second number is more than first number by 1 x . required percentage = 1 x / 7 x * 100 = 14.3 % . answer : a | a ) 14.3 % . , b ) 70 % . , c ) 60 % . , d ) 68 % . , e ) 80 % . | a | multiply(divide(1, 7), const_100) | divide(n0,n3)|multiply(#0,const_100) | general |
ashok secured average of 79 marks in 6 subjects . if the average of marks in 5 subjects is 74 , how many marks did he secure in the 6 th subject ? | "explanation : number of subjects = 6 average of marks in 6 subjects = 79 therefore total marks in 6 subjects = 79 * 6 = 474 now , no . of subjects = 5 total marks in 5 subjects = 74 * 5 = 370 therefore marks in 6 th subject = 474 β 370 = 104 answer : a" | a ) 104 , b ) 27 , c ) 99 , d ) 17 , e ) 80 | a | subtract(multiply(79, 6), multiply(74, 5)) | multiply(n0,n1)|multiply(n2,n3)|subtract(#0,#1)| | general |
johnny makes $ 4.75 per hour at his work . if he works 6 hours , how much money will he earn ? | 4.75 * 6 = 28.50 . answer is c . | a ) $ 30 , b ) $ 54 , c ) $ 28.50 , d ) $ 12 , e ) $ 9.60 | c | multiply(4.75, 6) | multiply(n0,n1)| | physics |
the sale price sarees listed for rs . 150 after successive discount is 20 % and 10 % is ? | "150 * ( 80 / 100 ) * ( 90 / 100 ) = 108 answer : c" | a ) 187 , b ) 120 , c ) 108 , d ) 178 , e ) 175 | c | subtract(subtract(150, divide(multiply(150, 20), const_100)), divide(multiply(subtract(150, divide(multiply(150, 20), const_100)), 10), const_100)) | multiply(n0,n1)|divide(#0,const_100)|subtract(n0,#1)|multiply(n2,#2)|divide(#3,const_100)|subtract(#2,#4)| | gain |
the difference between a number and its three fifth is 50 . what is the number ? | "solution let the number be x . then , x - 3 / 5 x = 50 βΉ = βΊ 2 / 5 x = 50 βΉ = βΊ x = ( 50 x 5 / 2 ) βΉ = βΊ 125 . answer c" | a ) 75 , b ) 100 , c ) 125 , d ) 180 , e ) none | c | power(add(50, const_4), const_4) | add(n0,const_4)|power(#0,const_4)| | general |
a rectangular plot measuring 90 metres by 50 metres is to be enclosed by wire fencing . if the poles of the fence are kept 20 metres apart , how many poles will be needed ? | "solution perimeter of the plot = 2 ( 90 + 50 ) = 280 m . β΄ number of poles = [ 280 / 20 ] = 14 m answer c" | a ) 55 , b ) 56 , c ) 14 , d ) 58 , e ) none of these | c | divide(rectangle_perimeter(90, 50), 20) | rectangle_perimeter(n0,n1)|divide(#0,n2)| | physics |
how many liters of water must be added to 14 liters of milk and water containing 10 % water to make it 30 % water ? | by rule of alligation : 30 % - 10 % = 20 % 100 % - 30 % = 70 % quantity of pure water : quantity of the mixture = 2 : 7 there are 14 liters of mixture , so we need to add 4 liters of pure water . the answer is d . | a ) 1 liter , b ) 2 liters , c ) 3 liters , d ) 4 liters , e ) 5 liters | d | multiply(divide(14, 10), divide(30, 10)) | divide(n0,n1)|divide(n2,n1)|multiply(#0,#1) | gain |
in the x - y plane , there are 4 points ( 0,0 ) , ( 0,4 ) , ( 4,4 ) , and ( 4,0 ) . if these 4 points makes a rectangle , what is the probability that x + y < 4 ? | "if x + y < 4 , then y < - x + 4 . the line y = - x + 4 intersects the rectangle and these three points of intersection ( 0,0 ) , ( 0,4 ) and ( 4,0 ) form a triangle . the points below the line y = - x + 4 satisfy x + y < 4 . the area of this triangle is ( 1 / 2 ) ( 4 ) ( 4 ) = 8 the area of the rectangle is 16 . p ( x... | a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 5 , e ) 1 / 6 | a | divide(divide(multiply(4, 4), const_2), multiply(4, add(const_1, 4))) | add(const_1,n0)|multiply(n0,n0)|divide(#1,const_2)|multiply(n0,#0)|divide(#2,#3)| | general |
the population of a town is 10000 . it increases annually at the rate of 20 % p . a . what will be its population after 6 years ? | "formula : ( after = 100 denominator ago = 100 numerator ) 10000 Γ 120 / 100 ^ 6 = 29859 b )" | a ) 14300 , b ) 29859 , c ) 14500 , d ) 14600 , e ) 15400 | b | add(10000, multiply(divide(multiply(10000, 20), const_100), 6)) | multiply(n0,n1)|divide(#0,const_100)|multiply(#1,n2)|add(n0,#2)| | gain |
john and ingrid pay 30 % and 40 % tax annually , respectively . if john makes $ 58000 and ingrid makes $ 72000 , what is their combined tax rate ? | ( 1 ) when 30 and 40 has equal weight or weight = 1 / 2 , the answer would be 35 . ( 2 ) when 40 has larger weight than 30 , the answer would be in between 35 and 40 . unfortunately , we have 2 answer choices d and e that fit that condition so we need to narrow down our range . ( 3 ) get 72000 / 130000 = 36 / 65 . 36 /... | a ) 32 % , b ) 34.4 % , c ) 35 % , d ) 35.6 % , e ) 37.4 % | d | multiply(divide(add(multiply(divide(30, const_100), 58000), multiply(divide(40, const_100), 72000)), add(72000, 58000)), const_100) | add(n2,n3)|divide(n0,const_100)|divide(n1,const_100)|multiply(n2,#1)|multiply(n3,#2)|add(#3,#4)|divide(#5,#0)|multiply(#6,const_100) | gain |
a basketball player is shooting hoops . he shoots a free throw , a high school 3 - pointer , and then a professional 3 - pointer . the probability that he makes the free throw is 4 / 5 , the probability that he makes the high school 3 - pointer is 1 / 2 , and the probability that he makes the professional 3 - pointer i... | the probability that he makes at least one of the three shots = 1 - ( the probability that he misses all of the shots ) the probability that he misses all shots = ( prob that he misses free throw ) * ( prob that he misses 3 - pointer ) * ( prob that he misses the professional 3 - pointer ) prob that he misses the free ... | a ) 14 / 15 , b ) 15 / 16 , c ) 17 / 18 , d ) 9 / 10 , e ) 5 / 6 | a | divide(add(const_10, 4), multiply(5, 3)) | add(n2,const_10)|multiply(n0,n3)|divide(#0,#1) | general |
a confectioner decides to sell all of his pastry due to the coming holiday . his pastry goods are equally divided among a group of 28 regular customers . if only 49 customers come to the bakery , each one will receive 6 less pastry goods . how much pastry does the confectioner needs to sell ? | "solution 1 : given that number of pastries are equally divided in to 28 groups , so the answer should be a multiple of 28 . and in the given choices , only 392 is divisible by 28 with remainder 0 ( make sure others are no multiples of 28 ) . so this is good enough to stop here , but to verify , 392 / 28 = 14 and 392 /... | a ) 392 , b ) 412 . , c ) 432 . , d ) 502 , e ) 522 | a | multiply(divide(multiply(49, 6), subtract(49, 28)), 28) | multiply(n1,n2)|subtract(n1,n0)|divide(#0,#1)|multiply(n0,#2)| | general |
what decimal of an hour is a second | explanation : 1 / ( 60 * 60 ) = 1 / 3600 = . 0027 option b | a ) . 0028 , b ) . 0027 , c ) . 0026 , d ) . 0025 , e ) none of these | b | divide(divide(const_1, const_60), const_60) | divide(const_1,const_60)|divide(#0,const_60) | physics |
a train of 24 carriages , each of 60 meters length , when an engine also of 60 meters length is running at a speed of 60 kmph . in what time will the train cross a bridge 1.5 km long ? | "d = 25 * 60 + 1500 = 3000 m t = 3000 / 60 * 18 / 5 = 180 sec = 3 mins answer : b" | a ) 6 , b ) 3 , c ) 4 , d ) 9 , e ) 2 | b | add(divide(multiply(add(24, const_1), 60), const_1000), 1.5) | add(n0,const_1)|multiply(n1,#0)|divide(#1,const_1000)|add(n4,#2)| | physics |
the total cost of 100 paper plates and 200 paper cups is $ 7.50 at the same rates what is the total cost of 20 of the plates and 40 of the cups ? | "u dont need to go through all this what u have with u is 100 p + 200 c = $ 7.50 just divide the equation by 5 and you will get what u are looking for 20 p + 40 c = $ 1.50 therefore oa is d" | a ) $ . 90 , b ) $ 1.00 , c ) $ 1.20 , d ) $ 1.50 , e ) $ 1.60 | d | multiply(divide(20, 100), 7.50) | divide(n3,n0)|multiply(n2,#0)| | gain |
the average age of 20 students in a class is 15 years . if the age of teacher is also included , the average becomes 16 years , find the age of the teacher . | "explanation : if teacher ' s age is 15 years , there is no change in the average . but teacher has contributed 1 year to all the students along with maintaining his age at 16 . age of teacher = average age of all + total increase in age = 16 + ( 1 x 20 ) = 36 years answer : d" | a ) 35 , b ) 38 , c ) 45 , d ) 36 , e ) 60 | d | subtract(add(add(multiply(20, 15), 16), 20), multiply(20, 15)) | multiply(n0,n1)|add(n2,#0)|add(n0,#1)|subtract(#2,#0)| | general |
in how many t ways can 4 white and 3 black chess pieces be arranged in a row such that they occupy alternate places ? assume that the pieces are distinct . | 4 white chess pieces can be arranged in 4 ! ways and 4 black pieces can be arranged in 3 ! ways . w _ w _ w _ w number of ways = 4 ! * 3 ! = 24 * 6 = 144 answer b | a ) t = 288 , b ) t = 144 , c ) t = 12 , d ) 48 , e ) 96 | b | multiply(factorial(3), factorial(4)) | factorial(n1)|factorial(n0)|multiply(#0,#1) | general |
a baseball card decreased in value 20 % in its first year and 10 % in its second year . what was the total percent decrease of the card ' s value over the two years ? | "let the initial value of baseball card = 100 after first year , value of baseball card = ( 1 - 20 / 100 ) * 100 = 80 after second year , value of baseball card = ( 1 - 10 / 100 ) * 80 = 72 total percent decrease of the card ' s value over the two years = ( 100 - 72 ) / 100 * 100 % = 28 % answer a" | a ) 28 % , b ) 30 % , c ) 32 % , d ) 36 % , e ) 72 % | a | subtract(const_100, multiply(multiply(subtract(const_1, divide(10, const_100)), subtract(const_1, divide(20, const_100))), const_100)) | divide(n1,const_100)|divide(n0,const_100)|subtract(const_1,#0)|subtract(const_1,#1)|multiply(#2,#3)|multiply(#4,const_100)|subtract(const_100,#5)| | gain |
1393 x 1393 = ? | "1393 x 1393 = ( 1393 ) 2 = ( 1400 - 7 ) 2 = ( 1400 ) 2 + ( 7 ) 2 - ( 2 x 1400 x 7 ) = 1960000 + 49 - 19600 = 1960049 - 19600 = 1940449 . answer : c" | a ) a ) 1951609 , b ) b ) 1951601 , c ) c ) 1940449 , d ) d ) 1951603 , e ) e ) 1951604 | c | multiply(divide(1393, 1393), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
two alloys a and b are composed of two basic elements . the ratios of the compositions of the two basic elements in the two alloys are 5 : 3 and 5 : 1 , respectively . a new alloy x is formed by mixing the two alloys a and b in the ratio 4 : 3 . what is the ratio of the composition of the two basic elements in alloy x ... | "mixture a has a total of 5 + 3 = 8 parts . if in the final mixture this represents 4 parts , then the total number of parts in mixture b should be ( 8 / 4 ) * 3 = 6 . so , we should take of mixture b a quantity with 5 and 1 parts , respectively . this will give us in the final mixture ( 5 + 5 ) : ( 3 + 1 ) , which mea... | a ) 1 : 1 , b ) 2 : 3 , c ) 5 : 2 , d ) 4 : 3 , e ) 7 : 9 | c | divide(add(multiply(4, 5), multiply(3, divide(add(5, 3), const_2))), add(multiply(4, 3), multiply(3, divide(add(5, 3), const_2)))) | add(n0,n1)|multiply(n0,n4)|multiply(n1,n4)|divide(#0,const_2)|multiply(n1,#3)|add(#1,#4)|add(#2,#4)|divide(#5,#6)| | other |
in the rectangular coordinate system , what is the x - intercept of a line passing through ( 10 , 3 ) and ( β 4 , β 4 ) ? | "slope = rise / run = 7 / 14 = 1 / 2 the equation of the line is y = ( 1 / 2 ) x + b 3 = ( 1 / 2 ) ( 10 ) + b b = - 2 the equation of the line is y = ( 1 / 2 ) x - 2 to find the x - intercept , let y = 0 : 0 = ( 1 / 2 ) x - 2 x = 4 the answer is a ." | a ) 4 , b ) 2 , c ) 0 , d ) β 2 , e ) β 4 | a | divide(subtract(multiply(divide(subtract(negate(4), 3), subtract(negate(4), 10)), 10), 3), divide(subtract(negate(4), 3), subtract(negate(4), 10))) | negate(n3)|negate(n2)|subtract(#0,n1)|subtract(#1,n0)|divide(#2,#3)|multiply(n0,#4)|subtract(#5,n1)|divide(#6,#4)| | general |
how many liters of pure alcohol must be added to a 100 - liter solution that is 20 percent alcohol in order to produce a solution that is 24 percent alcohol ? | "20 % alcohol solution means ; in the 100 liter solution , 20 liters of solution is alcohol and 80 liters other solvents . if we addxliters of alcohol to the solution , the solution becomes 100 + xliters and alcohol , which was 20 liters , becomes 20 + x liters . according to the statement ; 20 + x = 24 % of ( 100 + x ... | a ) 7 / 2 , b ) 5 , c ) 20 / 3 , d ) 100 / 19 , e ) 39 / 4 | d | divide(100, add(const_4, const_3)) | add(const_3,const_4)|divide(n0,#0)| | general |
a room 8 m 47 cm long and 7 m 77 cm broad is to be paved with square tiles . find the least number of square tiles required to cover the floor . | "explanation : area of the room = ( 847 x 777 ) cm 2 . size of largest square tile = h . c . f . of 847 cm and 777 cm = 7 cm . area of 1 tile = ( 7 x 7 ) cm 2 . number of tiles required = ( 847 Γ 777 ) / ( 7 Γ 7 ) = 13431 answer : option d" | a ) 13636 , b ) 13440 , c ) 13647 , d ) 13431 , e ) 13675 | d | divide(multiply(add(multiply(8, const_100), 47), add(multiply(7, const_100), 77)), multiply(subtract(47, add(multiply(const_2, const_4), const_2)), subtract(47, add(multiply(const_2, const_4), const_2)))) | multiply(n0,const_100)|multiply(n2,const_100)|multiply(const_2,const_4)|add(n1,#0)|add(n3,#1)|add(#2,const_2)|multiply(#3,#4)|subtract(n1,#5)|multiply(#7,#7)|divide(#6,#8)| | physics |
if x and y are numbers such that ( x + 8 ) ( y - 8 ) = 0 , what is the smallest possible value of x ^ 2 + y ^ 2 | "from ( x + 8 ) ( y - 8 ) = 0 it follows that either , x = - 8 or y = 8 . thus either x ^ 2 = 64 or y ^ 2 = 64 . now , if x ^ 2 = 64 , then the least value of y ^ 2 is 0 , so the least value of x ^ 2 + y ^ 2 = 64 + 0 = 64 . similarly if y ^ 2 = 64 , then the least value of x ^ 2 is 0 , so the least value of x ^ 2 + y ^... | a ) 4 , b ) 24 , c ) 54 , d ) 64 , e ) 94 | d | power(8, 2) | power(n0,n3)| | general |
what will be the compound interest on rs . 50000 after 3 years at the rate of 12 % per annum | "explanation : ( 50000 Γ ( 1 + 12 / 100 ) 3 ) = > 50000 Γ 28 / 25 Γ 28 / 25 Γ 28 / 25 = > 70246.4 so compound interest will be 70246.4 - 50000 = rs 20246.4 option b" | a ) rs 20123.20 , b ) rs 20246.4 , c ) rs 20123.40 , d ) rs 20123.50 , e ) none of these | b | subtract(multiply(multiply(multiply(const_4, const_100), const_100), power(add(const_1, divide(12, const_100)), 3)), multiply(multiply(const_4, const_100), const_100)) | divide(n2,const_100)|multiply(const_100,const_4)|add(#0,const_1)|multiply(#1,const_100)|power(#2,n1)|multiply(#3,#4)|subtract(#5,#3)| | gain |
dick and jane each saved $ 1000 in 1989 . in 1990 dick saved 15 percent more than in 1989 , and together he and jane saved a total of $ 2000 . approximately what percent less did jane save in 1990 than in 1989 ? | 1990 dick saved = $ 1150 jane saved = $ 850 ( jane saved $ 150 less than she did the prior year ) jane saved approximately $ 150 / 1000 $ ( 15 % ) less in 1990 answer : d | a ) 3 % , b ) 5 % , c ) 10 % , d ) 15 % , e ) 18 % | d | divide(subtract(1000, subtract(2000, add(multiply(divide(15, const_100), 1000), 1000))), divide(1000, const_100)) | divide(n3,const_100)|divide(n0,const_100)|multiply(n0,#0)|add(n0,#2)|subtract(n5,#3)|subtract(n0,#4)|divide(#5,#1) | general |
calculate the share of y , if rs . 2690 is divided among x , y and z in the ratio 5 : 7 : 9 ? | "5 + 7 + 9 = 21 2690 / 21 = 128.1 y ' s share = 7 * 128.1 = 896.7 answer : d" | a ) 890.7 , b ) 826.7 , c ) 895.7 , d ) 896.7 , e ) 816.7 | d | multiply(divide(2690, add(add(5, 7), 9)), 5) | add(n1,n2)|add(n3,#0)|divide(n0,#1)|multiply(n1,#2)| | general |
the perimeter of an equilateral triangle is 45 . if one of the side is the side of an isoceles triangle of perimeter 40 then what is the lenght of base of isoceles triangle . | "base of isoceles triangle is 40 - 15 - 15 = 10 units . answer : a" | a ) 10 units , b ) 20 units , c ) 30 units , d ) 40 units , e ) 15 units | a | subtract(subtract(40, divide(45, const_3)), divide(45, const_3)) | divide(n0,const_3)|subtract(n1,#0)|subtract(#1,#0)| | geometry |
a lemonade stand sold only small and large cups of lemonade on tuesday . 3 / 5 of the cups sold were small and the rest were large . if the large cups were sold for 5 / 6 as much as the small cups , what fraction of tuesday ' s total revenue was from the sale of large cups ? | "let the total # of cups be 10 . # of small cups sold 3 / 5 * 10 = 6 ; # of large cups sold 10 - 6 = 4 ; let the price of small cup be $ 6 , then the price of larges cup would be 5 / 6 * 6 = $ 5 ; revenue from small cups : 6 * $ 6 = $ 36 ; revenue from large cups cups : 4 * $ 5 = $ 20 ; fraction of total revenue from l... | a ) 7 / 16 , b ) 7 / 15 , c ) 10 / 21 , d ) 17 / 35 , e ) 5 / 14 | e | divide(multiply(subtract(const_10, multiply(divide(3, 5), const_10)), multiply(divide(5, 6), multiply(divide(3, 5), const_10))), add(multiply(multiply(divide(3, 5), const_10), multiply(divide(3, 5), const_10)), multiply(subtract(const_10, multiply(divide(3, 5), const_10)), multiply(divide(5, 6), multiply(divide(3, 5), ... | divide(n2,n3)|divide(n0,n1)|multiply(#1,const_10)|multiply(#0,#2)|multiply(#2,#2)|subtract(const_10,#2)|multiply(#3,#5)|add(#4,#6)|divide(#6,#7)| | general |
after 10 % of the inhabitants of a village disappeared , a panic set in during which 25 % of the remaining inhabitants left the village . at that time , the population was reduced to 4860 . what was the number of original inhabitants ? | let the total number of original inhabitants be x . ( 75 / 100 ) * ( 90 / 100 ) * x = 4860 ( 27 / 40 ) * x = 4860 x = 4860 * 40 / 27 = 7200 the answer is b . | a ) 7000 , b ) 7200 , c ) 7400 , d ) 7600 , e ) 7800 | b | divide(4860, subtract(subtract(const_1, divide(10, const_100)), multiply(subtract(const_1, divide(10, const_100)), divide(25, const_100)))) | divide(n0,const_100)|divide(n1,const_100)|subtract(const_1,#0)|multiply(#1,#2)|subtract(#2,#3)|divide(n2,#4) | gain |
worker a takes 7 hours to do a job . worker b takes 10 hours to do the same job . how long it take both a & b , working together but independently , to do the same job ? | "one day work of a = 1 / 7 one day work of b = 1 / 10 so one day work of a and b together = 1 / 7 + 1 / 10 = 17 / 70 so total days required = 70 / 17 answer : a" | a ) 70 / 17 , b ) 40 / 9 , c ) 50 / 9 , d ) 60 / 9 , e ) 80 / 9 | a | divide(const_1, add(divide(const_1, 7), divide(const_1, 10))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)| | physics |
in what time will a train 200 meters long cross an electric pole , if its speed is 128 km / hr | "explanation : first convert speed into m / sec speed = 128 * ( 5 / 18 ) = 35 m / sec time = distance / speed = 200 / 35 = 5.7 seconds option b" | a ) 5 seconds , b ) 5.7 seconds , c ) 3 seconds , d ) 2.5 seconds , e ) none of these | b | divide(200, multiply(128, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics |
4 , 12 , 48 , 240 , 1440 , 10080,80640 , 725760 ( . . . ) | "explanation : go on multiplying the given numbers by 3 , 4 , 5 , 6 , 7,8 , 9,10 answer : option a" | a ) 7257600 , b ) 10080 , c ) 6200 , d ) 10020 , e ) 12500 | a | subtract(negate(240), multiply(subtract(12, 48), divide(subtract(12, 48), subtract(4, 12)))) | negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)| | general |
if rs . 10 be allowed as true discount on a bill of rs . 110 at the end of a certain time , then the discount allowed on the same sum due at the end of double the time is ? | explanation : present worth = amount - truediscount = 110 - 10 = rs . 100 si on rs . 100 for a certain time = rs . 10 si on rs . 100 for doube the time = rs . 20 truediscount on rs . 120 = 120 - 100 = rs . 20 truediscount on rs . 110 = = rs . 18.33 answer : b | a ) 68.33 , b ) 18.33 , c ) 28.33 , d ) 48.33 , e ) 98.33 | b | multiply(110, divide(subtract(add(10, 110), const_100), add(10, 110))) | add(n0,n1)|subtract(#0,const_100)|divide(#1,#0)|multiply(n1,#2) | general |
a train 120 m long running at 60 kmph crosses a platform in 20 sec . what is the length of the platform ? | "d = 60 * 5 / 18 = 20 = 333 Γ’ β¬ β 120 = 213 answer : a" | a ) 213 , b ) 298 , c ) 350 , d ) 726 , e ) 267 | a | subtract(multiply(20, multiply(60, const_0_2778)), 120) | multiply(n1,const_0_2778)|multiply(n2,#0)|subtract(#1,n0)| | physics |
two pipes a and b can fill a tank in 12 hours and 15 hours respectively . if both pipe open together in tank , then how much time it take to fill the tank ? | in 1 hour pipe a fill = 1 / 12 in 1 hour pipe b fill = 1 / 15 pipe ( a + b ) in 1 hour = [ 1 / 12 + 1 / 15 ] = 3 / 20 total time taken to fill the tank = 20 / 3 hour answer c | a ) 15 , b ) 18 , c ) 20 / 3 , d ) 19 / 4 , e ) 25 / 4 | c | divide(const_1, add(divide(const_1, 12), divide(const_1, 15))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2) | physics |
when a number is divided by 13 , the remainder is 6 . when the same number is divided by 7 , then remainder is 1 . what is the number ? | "explanation : take 243 243 Γ· 7 = 34 , remainder = 5 hence this is not the answer take 312 312 Γ· 7 = 44 , remainder = 4 hence this is not the answer take 253 253 Γ· 7 = 36 , remainder = 1 . 253 Γ· 13 = 19 , remainder = 6 this satisfies both the conditions given in the question . hence it is the answer . answer : b" | a ) 243 , b ) 253 , c ) 312 , d ) 432 , e ) none of these | b | add(multiply(13, add(add(const_10, const_12), const_4)), 6) | add(const_10,const_12)|add(#0,const_4)|multiply(n0,#1)|add(n1,#2)| | general |
for the past n days , the average ( arithmetic mean ) daily production at a company was 60 units . if today ' s production of 90 units raises the average to 62 units per day , what is the value of n ? | "( average production for n days ) * n = ( total production for n days ) - - > 60 n = ( total production for n days ) ; ( total production for n days ) + 90 = ( average production for n + 1 days ) * ( n + 1 ) - - > 60 n + 90 = 62 * ( n + 1 ) - - > n = 14 . or as 30 extra units increased the average for n + 1 days by 2 ... | a ) 30 , b ) 14 , c ) 10 , d ) 9 , e ) 7 | b | subtract(divide(subtract(90, 60), subtract(62, 60)), const_1) | subtract(n1,n0)|subtract(n2,n0)|divide(#0,#1)|subtract(#2,const_1)| | general |
james took a 3 - hour bike ride . in the second hour he travlled 24 miles , which was 20 percent farther than he traveled the first hour . if he traveled 25 percent farther in the third hour than he did in the second hour , how many miles did jose travel during the entire ride ? | "let the distance travelled in the first hour be x . thus , 1.2 x = 24 , x = 20 . now , the distance travelled in the 3 rd hour = 24 + 1 / 4 Γ’ Λ β 24 = 30 . = 24 + 20 + 30 = 74 answer : a" | a ) 74.0 , b ) 54.9 , c ) 55.5 , d ) 57.0 , e ) 63.0 | a | add(add(divide(multiply(multiply(const_1, const_100), 24), add(multiply(const_1, const_100), 20)), divide(multiply(25, 24), 20)), 24) | multiply(const_1,const_100)|multiply(n1,n3)|add(n2,#0)|divide(#1,n2)|multiply(n1,#0)|divide(#4,#2)|add(#5,#3)|add(n1,#6)| | physics |
a rectangular block 6 cm by 12 cm by 15 cm is cut up into an exact number of equal cubes . find the least possible number of cubes . | "explanation : volume of the block = ( 6 x 12 x 15 ) cm 3 = 1080 cm 3 side of the largest cube = h . c . f . of 6 cm , 12 cm , 15 cm = 3 cm . volume of this cube = ( 3 x 3 x 3 ) cm 3 = 27 cm 3 number of cubes = ( 1080 / 27 ) = 40 . answer : b" | a ) 30 , b ) 40 , c ) 10 , d ) 20 , e ) 50 | b | divide(add(subtract(divide(rectangle_area(const_360, const_1000), const_10), multiply(const_1000, multiply(const_3, const_2))), add(multiply(const_3, const_1000), multiply(15, const_10))), divide(add(subtract(divide(rectangle_area(const_360, const_1000), const_10), multiply(const_1000, multiply(const_3, const_2))), add... | multiply(const_1000,const_3)|multiply(n2,const_10)|multiply(const_2,const_3)|rectangle_area(const_1000,const_360)|add(#0,#1)|divide(#3,const_10)|multiply(#2,const_1000)|subtract(#5,#6)|add(#4,#7)|divide(#8,const_10)|divide(#8,#9)| | geometry |
a train 360 m long is running at a speed of 45 km / hr . in what time will it pass a bridge 390 m long ? | "speed = 45 * 5 / 18 = 25 / 2 m / sec total distance covered = 360 + 140 = 750 m required time = 750 * 2 / 25 = 40 sec answer : b" | a ) 40 sec , b ) 60 sec , c ) 45 sec , d ) 48 sec , e ) 50 sec | b | divide(360, multiply(subtract(45, 390), const_0_2778)) | subtract(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
an empty fuel tank with a capacity of 200 gallons was filled partially with fuel a and then to capacity with fuel b . fuel a contains 12 % ethanol by volume and fuel b contains 16 % ethanol by volume . if the full fuel tank contains 20 gallons of ethanol , how many gallons of fuel a were added ? | "say there are a gallons of fuel a in the tank , then there would be 200 - a gallons of fuel b . the amount of ethanol in a gallons of fuel a is 0.12 a ; the amount of ethanol in 200 - a gallons of fuel b is 0.16 ( 200 - a ) ; since the total amount of ethanol is 20 gallons then 0.12 a + 0.16 ( 200 - a ) = 20 - - > a =... | a ) 300 , b ) 150 , c ) 100 , d ) 80 , e ) 50 | a | divide(subtract(multiply(200, divide(16, const_100)), 20), subtract(divide(16, const_100), divide(12, const_100))) | divide(n2,const_100)|divide(n1,const_100)|multiply(n0,#0)|subtract(#0,#1)|subtract(#2,n3)|divide(#4,#3)| | gain |
the circus sells two kinds of tickets : lower seats for $ 30 and upper seats for $ 20 . on a certain night , the circus sells 80 tickets and gets $ 2100 in revenue from the sales . how many tickets for lower seats did they sell ? | let l be the number of lower seat tickets . let u be the number of upper seat tickets . l + u = 80 and u = 80 - l . 30 l + 20 u = 2100 . 30 l + 20 ( 80 - l ) = 2100 10 l + 1600 = 2100 . 10 l = 500 . l = 50 . the answer is c . | ['a ) 40', 'b ) 45', 'c ) 50', 'd ) 55', 'e ) 60'] | c | subtract(80, divide(subtract(multiply(30, 80), 2100), subtract(30, 20))) | multiply(n0,n2)|subtract(n0,n1)|subtract(#0,n3)|divide(#2,#1)|subtract(n2,#3) | geometry |
the average weight of 8 persons increases by 1.5 kg . if a person weighing 65 kg is replaced by a new person , what could be the weight of the new person ? | "total weight increases = 8 Γ 1.5 = 12 kg so the weight of new person = 65 + 12 = 77 kg answer b" | a ) 76 kg , b ) 77 kg , c ) 76.5 kg , d ) data inadequate , e ) none of these | b | add(65, multiply(8, 1.5)) | multiply(n0,n1)|add(n2,#0)| | general |
the speed at which a man can row a boat in still water is 26 kmph . if he rows downstream , where the speed of current is 4 kmph , what time will he take to cover 150 metres ? | speed of the boat downstream = 26 + 4 = 30 kmph = 30 * 5 / 18 = 8.33 m / s hence time taken to cover 150 m = 150 / 8.33 = 18 seconds . answer : a | a ) 18 , b ) 27 , c ) 28 , d ) 12 , e ) 25 | a | divide(150, multiply(add(26, 4), const_0_2778)) | add(n0,n1)|multiply(#0,const_0_2778)|divide(n2,#1)| | physics |
a mathematics teacher tabulated the marks secured by 35 students of 8 th class . the average of their marks was 72 . if the marks secured by reema was written as 35 instead of 99 then find the correct average marks up to two decimal places . | "total marks = 35 x 72 = 2520 corrected total marks = 2520 - 35 + 99 = 2584 correct average = 2584 / 35 = 73.82 answer : e" | a ) 73.41 , b ) 74.31 , c ) 72.43 , d ) 73.43 , e ) 73.82 | e | divide(subtract(multiply(35, 72), subtract(99, 35)), 35) | multiply(n0,n2)|subtract(n4,n3)|subtract(#0,#1)|divide(#2,n0)| | general |
two employees m and n are paid a total of $ 572 per week by their employer . if m is paid 120 percent of the salary paid to n , how much is n paid per week ? | "1.2 n + n = 572 2.2 n = 572 n = 260 the answer is c ." | a ) $ 220 , b ) $ 240 , c ) $ 260 , d ) $ 300 , e ) $ 320 | c | divide(572, add(divide(120, const_100), const_1)) | divide(n1,const_100)|add(#0,const_1)|divide(n0,#1)| | general |
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