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a bullock cart has to cover a distance of 80 km in 10 hrs . if it covers half of the journey in 3 / 5 th time . what should be its speed to cover the remaining distance in the time left . | "let rate = r % and time = r years . then , 1200 x r x r / 100 = 300 12 r 2 = 300 r 2 = 25 r = 5 . answer : a" | a ) 5 , b ) 6 , c ) 18 , d ) can not be determined , e ) none of these | a | sqrt(divide(multiply(300, const_100), 1200)) | multiply(n1,const_100)|divide(#0,n0)|sqrt(#1)| | gain |
reena took a loan of $ . 1200 with simple interest for as many years as the rate of interest . if she paid $ 300 as interest at the end of the loan period , what was the rate of interest ? | "multiplication rule of 11 : ( sum of digits at odd places - sum of digits at even places ) should be divisible by 11 given number : 872 , 152,24 x sum of digits at odd places = 8 + 2 + 5 + 2 + x = 17 + x ( i ) sum of digits at even places = 7 + 1 + 2 + 4 = 14 ( ii ) ( i ) - ( ii ) = 17 + x - 14 = x - 3 hence x should be = 3 to make this a multiple of 11 ( 0 ) option c" | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | multiply(multiply(multiply(add(multiply(const_3, const_10), const_1), const_2), const_4), 11) | multiply(const_10,const_3)|add(#0,const_1)|multiply(#1,const_2)|multiply(#2,const_4)|multiply(n2,#3)| | general |
if the number 872 , 152,24 x is divisible by 11 , what must be the value of x ? | "the time it takes darcy to walk to work is ( 1.5 / 3 ) * 60 = 30 minutes the time it takes darcy to take the train is ( 1.5 / 20 ) * 60 + x = 4.5 + x minutes it takes 15 minutes longer to walk , so 30 = 4.5 + x + 20 x = 5.5 minutes answer : a" | a ) 5.5 , b ) 15 , c ) 25.5 , d ) 30 , e ) 60 | a | subtract(subtract(divide(const_60, const_2), 20), divide(const_60, divide(20, 1.5))) | divide(const_60,const_2)|divide(n2,n0)|divide(const_60,#1)|subtract(#0,n3)|subtract(#3,#2)| | physics |
darcy lives 1.5 miles from work . she can walk to work at a constant rate of 3 miles per hour , or she can ride the train to work at a constant rate of 20 miles per hour . if she rides the train , there is an additional x minutes spent walking to the nearest train station , waiting for the train , and walking from the final train station to her work . if it takes darcy a total of 20 more minutes to commute to work by walking than it takes her to commute to work by riding the train , what is the value of x ? | "when ball comes down , then i have indicated the distance covered in green when ball goes up , then i have indicated the distance covered in red distance travelled uptil the ball touches the floor 3 rd time : h + 0.5 h + 0.5 h + 0.5 * 0.5 h + 0.5 * 0.5 h h + 2 * 0.5 * h + 2 * 0.25 * h = h ( 1 + 2 * 0.5 + 2 * 0.25 ) = h ( 1 + 1 + 0.5 ) = 150 2.5 h = 150 h = 60 . a is the answer ." | a ) 60 cm , b ) 90 cm , c ) 100 cm , d ) 120 cm , e ) 130 cm | a | divide(150, add(const_2, divide(50, const_100))) | divide(n0,const_100)|add(#0,const_2)|divide(n1,#1)| | gain |
one ball will drop from a certain height . the height it will reach after rebounding from the floor is 50 percent of the previous height . the total travel is 150 cm when it touches the floor on third time . what is the value of the original height ? | ' approximately ' could actually make such a question ambiguous . not this one though but a similar question with the answer as 9.2 days . you round off 8.89 days as 9 days and everything is fine in this question . what do you do when you get 9.2 days ? do you need 9 days or 10 days ? can you round off 9.2 as 9 even though that is what you do with numbers ? no , because in 9 days your work is not over . you do need 10 days . to finish a work machine w say you need to work full 9 days and a part of the 10 th day . if i ask you how many days do you need to complete the work , will you say 9 or 10 ? you will say 10 even if you do n ' t use the 10 th day fully = d | a ) 14 , b ) 13 , c ) 11 , d ) 9 , e ) 7 | d | inverse(add(inverse(divide(multiply(12, const_2), add(const_1, divide(const_1, const_2)))), inverse(divide(multiply(15, const_2), add(const_1, divide(const_1, const_2)))))) | divide(const_1,const_2)|multiply(n0,const_2)|multiply(n1,const_2)|add(#0,const_1)|divide(#1,#3)|divide(#2,#3)|inverse(#4)|inverse(#5)|add(#6,#7)|inverse(#8) | physics |
a furniture manufacturer has two machines , but only one can be used at a time . machine w is utilized during the first shift and machine b during the second shift , while both work half of the third shift . if machine w can do the job in 12 days working two shifts and machine b can do the job in 15 days working two shifts , how many days will it take to do the job with the current work schedule ? | "when n is even , ( x ^ n - a ^ n ) is completely divisible by ( x - a ) . ( 17 ^ 200 - 1 ^ 200 ) is completely divisible by ( 17 + 1 ) , ( 17 ^ 200 - 1 ) is completely divisible by 18 . on dividing 17 ^ 200 by 18 , we get 1 as remainder . answer is d" | a ) 3 , b ) 8 , c ) 5 , d ) 1 , e ) 7 | d | power(17, 17) | power(n0,n0)| | general |
what will be the remainder when 17 ^ 200 is divided by 18 ? | "answer 469200 x 9999 = 469200 x ( 10000 - 1 ) = 4692000000 - 469200 = 4691100843 . option : e" | a ) 4586970843 , b ) 4686970743 , c ) 4691100843 , d ) 4586870843 , e ) 4691530800 | e | multiply(469200, 9999) | multiply(n0,n1)| | general |
calculate 469200 x 9999 = ? | "explanation : d = 100 + 130 = 230 s = 36 * 5 / 18 = 10 mps t = 230 / 10 = 23 sec answer : option d" | a ) 26 , b ) 72 , c ) 25 , d ) 23 , e ) 27 | d | divide(add(130, 100), multiply(36, const_0_2778)) | add(n0,n1)|multiply(n2,const_0_2778)|divide(#0,#1)| | physics |
how many seconds will a train 100 meters long take to cross a bridge 130 meters long if the speed of the train is 36 kmph ? | "explanation : first six prime numbers which are between 60 and 90 = 61 , 67 , 71 , 73 , 79 , 83 average = ( 61 + 67 + 71 + 73 + 79 + 83 ) / 6 = 72.33 answer : e" | a ) 35.4 , b ) 42 , c ) 45.7 , d ) 57 , e ) 72.33 | e | add(60, const_1) | add(n0,const_1)| | general |
the average of first six prime numbers which are between 60 and 90 is | "c age x , then b age is 2 x so a age is 2 x + 2 . ( 2 x + 2 ) + 2 x + x = 27 5 x = 25 x = 5 so b is 2 x = 2 ( 5 ) 2 x 5 = 10 answer : b" | a ) 12 years , b ) 10 years , c ) 8 years , d ) 14 years , e ) 16 years | b | divide(multiply(subtract(27, const_2), const_2), add(const_4, const_1)) | add(const_1,const_4)|subtract(n0,const_2)|multiply(#1,const_2)|divide(#2,#0)| | general |
a is two years older than b who is twice as old as c . if the total ages of a , b and c be 27 . what is the age of b ? | total number of installments = 52 payment per installment for the first 22 installments = 410 payment per installment for the remaining 30 installments = 410 + 65 = 475 average = ( 22 * 410 + 30 * 475 ) / 52 = 447.50 answer a | a ) 447.5 , b ) 450 , c ) 465 , d ) 468 , e ) 475 | a | divide(add(multiply(22, 410), multiply(add(410, 65), subtract(52, 22))), 52) | add(n4,n5)|multiply(n3,n4)|subtract(n0,n3)|multiply(#0,#2)|add(#1,#3)|divide(#4,n0) | general |
a certain debt will be paid in 52 installments from january 1 to december 31 of a certain year . each of the first 22 payments is to be $ 410 ; each of the remaining payments is to be $ 65 more than each of the first 22 payments . what is the average ( arithmetic mean ) payment that will be made on the debt for the year ? | "let the numbers be 5 x and 6 x . then their h . c . f = x . so , x = 4 . so , the numbers are 20 and 24 . l . c . m of 20 and 24 = 120 . answer : e" | a ) 12 , b ) 16 , c ) 24 , d ) 48 , e ) 120 | e | lcm(multiply(5, 4), multiply(6, 4)) | multiply(n0,n2)|multiply(n1,n2)|lcm(#0,#1)| | other |
the ratio of numbers is 5 : 6 and their h . c . f is 4 . their l . c . m is : | "easy solution : n = dq 1 + 241 2 n = 2 dq 1 + 482 - ( 1 ) 2 n = dq 2 + 102 - ( 2 ) as ( 1 ) = ( 2 ) = 2 n d * ( q 2 - 2 q 1 ) = 380 d * some integer = 380 checking all options only ( c ) syncs with it . answer c" | a ) 370 , b ) 365 , c ) 380 , d ) 456 , e ) 460 | c | subtract(multiply(241, const_2), 102) | multiply(n0,const_2)|subtract(#0,n1)| | general |
a number when divided by a certain divisor left remainder 241 , when twice the number was divided by the same divisor , the remainder was 102 . find the divisor ? | "sol . sum = b . d . * t . d . / b . d . - t . d . = rs . [ 90 * 60 / 90 - 60 ] = rs . [ 90 * 60 / 30 ] = rs . 180 answer b" | a ) 210 , b ) 180 , c ) 360 , d ) 450 , e ) none | b | divide(multiply(90, 60), subtract(90, 60)) | multiply(n0,n1)|subtract(n0,n1)|divide(#0,#1)| | gain |
the banker Γ’ β¬ β’ s discount of a certain sum of money is rs . 90 and the true discount on the same sum for the same time is rs . 60 . the sum due is | it is essential to recognize that the remainder when an integer is divided by 10 is simply the units digit of that integer . to help see this , consider the following examples : 4 / 10 is 0 with a remainder of 4 14 / 10 is 1 with a remainder of 4 5 / 10 is 0 with a remainder of 5 105 / 10 is 10 with a remainder of 5 it is also essential to remember that the s is a positive integer and multiple of 2 . any integer that is a multiple of 2 is an even number . so , s must be a positive even integer . with these two observations , the question can be simplified to : what is the units digit of 4 raised to an even positive integer ? the units digit of 4 raised to an integer follows a specific repeating pattern : 4 ^ 1 = 4 4 ^ 2 = 16 4 ^ 3 = 64 4 ^ 4 = 256 4 ^ ( odd number ) - - > units digit of 4 4 ^ ( even number ) - - > units digit of 6 there is a clear pattern regarding the units digit . 4 raised to any odd integer has a units digit of 4 while 4 raised to any even integer has a units digit of 6 . since s must be an even integer , the units digit of p = 4 ^ s will always be 6 . consequently , the remainder when p = 4 ^ s is divided by 10 will always be 6 . in case this is too theoretical , consider the following examples : s = 2 - - > p = 4 ^ z = 16 - - > s / 10 = 1 with a remainder of 6 s = 4 - - > p = 4 ^ z = 256 - - > s / 10 = 25 with a remainder of 6 s = 6 - - > p = 4 ^ z = 4096 - - > s / 10 = 409 with a remainder of 6 s = 8 - - > p = 4 ^ z = 65536 - - > s / 10 = 6553 with a remainder of 6 answer : b . | a ) 10 , b ) 6 , c ) 4 , d ) 0 , e ) it can not be determined | b | reminder(power(4, 2), 10) | power(n1,n0)|reminder(#0,n2) | general |
s is a positive integer and multiple of 2 ; p = 4 ^ s , what is the remainder when p is divided by 10 ? | "sol . let the numbers be x and y . then , ( x + y ) = 22 and x 2 + y 2 = 386 . now , 2 xy = ( x + y ) 2 - ( x 2 + y 2 ) = ( 22 ) 2 - 386 = 484 - 386 = 98 xy = 49 . answer b" | a ) 40 , b ) 49 , c ) 80 , d ) 88 , e ) 90 | b | divide(subtract(power(22, const_2), 386), const_2) | power(n0,const_2)|subtract(#0,n1)|divide(#1,const_2)| | general |
if the sum of two numbers is 22 and the sum of their squares is 386 , then the product of the numbers is | "largest 4 - digit number = 9999 88 ) 9999 ( 113 88 - - - - 119 88 - - - - 319 264 - - - 55 - - - required number = ( 9999 - 55 ) = 9944 . e )" | a ) 9934 , b ) 9938 , c ) 9940 , d ) 9942 , e ) 9944 | e | square_area(const_pi) | square_area(const_pi)| | general |
the largest 4 digit number exactly divisible by 88 is ? | "round trip means 2 trips i . e . to and fro . he has completed one i . e 50 % completed . then he traveled another 80 % of 50 % i . e 40 % . so he completed 50 + 40 = 90 % of total trip . e" | a ) 50 , b ) 60 , c ) 70 , d ) 80 , e ) 90 | e | add(divide(const_100, const_2), divide(multiply(80, divide(const_100, const_2)), const_100)) | divide(const_100,const_2)|multiply(n0,#0)|divide(#1,const_100)|add(#0,#2)| | gain |
a technician makes a round - trip to and from a certain service center by the same route . if the technician completes the drive to the center and then completes 80 percent of the drive from the center , what percent of the round - trip has the technician completed ? | "( - 2 ) ^ ( 2 m ) = 4 ^ m and 2 ^ ( 15 - m ) = 4 ^ ( ( 15 - m ) / 2 ) therefore , m = ( 15 - m ) / 2 2 m = 15 - m m = 5 answer d" | a ) 1 , b ) 2 , c ) 3 , d ) 5 , e ) 6 | d | divide(15, add(2, const_1)) | add(n0,const_1)|divide(n3,#0)| | general |
if m is an integer such that ( - 2 ) ^ 2 m = 2 ^ ( 15 - m ) then m = ? | "perimeter = distance covered in 10 min . = ( 15000 x 8 ) / 60 m = 2000 m . let length = 3 x metres and breadth = 4 x metres . then , 2 ( 3 x + 4 x ) = 2000 or x = 142.86 . length = 428.58 m and breadth = 571.44 m . area = ( 428.58 x 571.44 ) m 2 = 244907.76 m a" | a ) 244907.04 m , b ) 244000 m , c ) 244900 m , d ) 400000 m , e ) 154300 m | a | rectangle_area(divide(divide(multiply(multiply(divide(15, multiply(const_10, multiply(const_3, const_2))), 10), const_1000), add(3, 4)), const_2), multiply(divide(divide(multiply(multiply(divide(15, multiply(const_10, multiply(const_3, const_2))), 10), const_1000), add(3, 4)), const_2), 4)) | add(n0,n1)|multiply(const_2,const_3)|multiply(#1,const_10)|divide(n2,#2)|multiply(n3,#3)|multiply(#4,const_1000)|divide(#5,#0)|divide(#6,const_2)|multiply(n1,#7)|rectangle_area(#7,#8)| | physics |
the ratio between the length and the breadth of a rectangular park is 3 : 4 . if a man cycling along the boundary of the park at the speed of 15 km / hr completes one round in 10 min , then the area of the park ( in sq . m ) is ? | "let the shares for a , b , and c be x , 2 x , and 4 x respectively . 7 x = 378 x = 54 4 x = 216 the answer is c ." | a ) $ 200 , b ) $ 208 , c ) $ 216 , d ) $ 224 , e ) $ 232 | c | multiply(divide(378, add(add(divide(const_1, const_2), const_1), const_2)), const_2) | divide(const_1,const_2)|add(#0,const_1)|add(#1,const_2)|divide(n0,#2)|multiply(#3,const_2)| | general |
$ 378 is divided among a , b , and c so that a receives half as much as b , and b receives half as much as c . how much money is c ' s share ? | "cost of 8 kg grapes = 75 Γ 8 = 600 . cost of 9 kg of mangoes = 55 Γ 9 = 495 . total cost he has to pay = 600 + 495 = 1095 . e )" | a ) a ) 1000 , b ) b ) 1055 , c ) c ) 1065 , d ) d ) 1075 , e ) e ) 1095 | e | add(multiply(8, 75), multiply(9, 55)) | multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)| | gain |
harkamal purchased 8 kg of grapes at the rate of 75 per kg and 9 kg of mangoes at the rate of 55 per kg . how much amount did he pay to the shopkeeper ? | "let x be the side length of square a . then the area of square a is x ^ 2 . the area of square b is ( sqrt ( 5 ) x ) ^ 2 = 5 x ^ 2 . the answer is e ." | a ) 1.5 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | e | power(sqrt(5), const_2) | sqrt(n0)|power(#0,const_2)| | geometry |
if the side length of square b is sqrt ( 5 ) times that of square a , the area of square b is how many times the area of square a ? | "let ' s say that the total original mixture a is 100 ml the original mixture a thus has 50 ml of alcohol out of 100 ml of solution you want to replace some of that original mixture a with another mixture b that contains 25 ml of alcohol per 100 ml . thus , the difference between 50 ml and 25 ml is 25 ml per 100 ml of mixture . this means that every time you replace 100 ml of the original mixture a by 100 ml of mixture b , the original alcohol concentration will decrease by 25 % . the question says that the new mixture , let ' s call it c , must be 35 % alcohol , a decrease of only 15 % . therefore , 15 out of 25 is 3 / 5 and e is the answer ." | a ) 1 / 4 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 3 / 5 | e | inverse(add(divide(subtract(35, 25), subtract(50, 35)), const_1)) | subtract(n2,n1)|subtract(n0,n2)|divide(#0,#1)|add(#2,const_1)|inverse(#3)| | other |
a certain quantity of 50 % solution is replaced with 25 % solution such that the new concentration is 35 % . what is the fraction of the solution that was replaced ? | "let x be the amount of money paid for the first purchase . the second time , the customer paid 0.8 ( 1.2 x ) = 0.96 x . the difference is 4 % . the answer is d ." | a ) 10 % , b ) 8 % , c ) 6 % , d ) 4 % , e ) 2 % | d | multiply(subtract(const_1, multiply(add(divide(20, const_100), const_1), divide(80, const_100))), const_100) | divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|multiply(#2,#1)|subtract(const_1,#3)|multiply(#4,const_100)| | general |
a customer bought a product at the shop . however , the shopkeeper increased the price of the product by 20 % so that the customer could not buy the required amount of the product . the customer managed to buy only 80 % of the required amount . what is the difference in the amount of money that the customer paid for the second purchase compared to the first purchase ? | amount left = 0.5 x β 4 for fifth store this is zero . so x = 8 . that means he entered fifth store with 8 . now for fourth store , amount left = 8 so 0.5 x β 4 = 8 β x = 24 for third store , amount left = 24 so 12 x β 4 = 24 β x = 56 for second store , amount left = 56 so 0.5 x β 4 = 56 β x = 120 for first store , amount left = 120 so 0.5 x β 4 = 120 β x = 248 so he entered first store with 248 . answer : a | a ) 248 , b ) 120 , c ) 252 , d ) 250 , e ) 350 | a | multiply(add(multiply(add(multiply(add(multiply(add(multiply(4, const_2), 4), const_2), 4), const_2), 4), const_2), 4), const_2) | multiply(n1,const_2)|add(n1,#0)|multiply(#1,const_2)|add(n1,#2)|multiply(#3,const_2)|add(n1,#4)|multiply(#5,const_2)|add(n1,#6)|multiply(#7,const_2) | general |
there is a lot of speculation that the economy of a country depends on how fast people spend their money in addition to how much they save . auggie was very curious to test this theory . auggie spent all of his money in 5 stores . in each store , he spent rs . 4 more than one - half of what he had when he went in . how many rupees did auggie have when he entered the first store ? | "area of the poster is 2.4 x 2.4 = 5.76 1 / 2 the area = 2.88 pi * r ^ 2 = 2.88 r ^ 2 = 2.88 / pi r = sqrt ( 2.88 / pi ) answer ( b )" | a ) 1 / pi , b ) sqrt ( 2.88 / pi ) , c ) 1 , d ) 2 / sqrt ( pi ) , e ) pi / 2 | b | sqrt(divide(divide(square_area(2.4), 2), const_pi)) | square_area(n0)|divide(#0,n2)|divide(#1,const_pi)|sqrt(#2)| | geometry |
an artist wishes to paint a circular region on a square poster that is 2.4 feet on a side . if the area of the circular region is to be 1 / 2 the area of the poster , what must be the radius of the circular region in feet ? | "5359 x 51 = 5359 x ( 50 + 1 ) = 5359 x 50 + 5359 x 1 = 267950 + 5359 = 273309 e )" | a ) 273232 , b ) 273243 , c ) 273247 , d ) 273250 , e ) 273309 | e | multiply(divide(5359, 51), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
5359 x 51 = ? | original share of 1 person = 1 / 3 new share of 1 person = 1 / 2 increase = ( 1 / 2 - 1 / 3 = 1 / 6 ) therefore , required fraction = ( 1 / 6 ) / ( 1 / 3 ) = ( 1 / 6 ) x ( 3 / 1 ) = 1 / 2 answer is a . | a ) 1 / 2 , b ) 2 / 7 , c ) 3 / 2 , d ) 4 / 7 , e ) none of them | a | divide(subtract(divide(const_1, const_2), divide(const_1, 3)), divide(const_1, 3)) | divide(const_1,const_2)|divide(const_1,n0)|subtract(#0,#1)|divide(#2,#1) | general |
3 people are planning to share equally the cost of a rental car . if one person withdraws from the arrangement and the others share equally the entire cost of the car , then the share of each of the remaining persons increased by : | solution relative speed = ( 45 - 40 ) kmph = 5 kmph = ( 5 x 5 / 18 ) m / sec = ( 25 / 18 ) m / sec time taken = ( 350 x 18 / 25 ) sec = 252 sec . answer d | a ) 72 sec , b ) 132 sec , c ) 192 sec , d ) 252 sec , e ) none | d | multiply(const_3600, divide(divide(add(200, 150), const_1000), subtract(45, 40))) | add(n0,n1)|subtract(n3,n2)|divide(#0,const_1000)|divide(#2,#1)|multiply(#3,const_3600) | physics |
two trains 200 m and 150 m long are running on parallel rails at the rate of 40 kmph and 45 kmph respectively . in how much time will they cross each other , if they are running in the same direction ? | "l . c . m of 15 and 12 = 60 cp of 60 articles = rs . 100 ( 25 * 4 ) sp of 60 articles = rs . 175 ( 35 * 5 ) profit percentage = ( 175 - 100 ) / 100 * 100 = 75 % answer : c" | a ) 80 % , b ) 50 % , c ) 75 % , d ) 40 % , e ) 53 % | c | subtract(multiply(35, add(const_4, const_1)), multiply(25, const_4)) | add(const_1,const_4)|multiply(n1,const_4)|multiply(n3,#0)|subtract(#2,#1)| | gain |
a dealer purchases 15 articles for rs . 25 and sells 12 articles for rs . 35 . find the profit percentage ? | "required number = ( 75 / 2 ) / ( 1 / 8 ) = ( 75 / 2 x 8 / 1 ) = 300 . answer : a" | a ) 300 , b ) 400 , c ) 500 , d ) 600 , e ) 700 | a | divide(add(37, divide(1, 2)), divide(1, 8)) | divide(n0,n4)|divide(n0,n1)|add(n2,#0)|divide(#2,#1)| | general |
how many 1 / 8 s are there in 37 1 / 2 ? | "suppose q can drain in 1 hr . so , rq = 1 / 1 = 1 so , rp = 1 / [ ( 2 / 4 ) rq ] = 4 / 2 also , rp = rr / ( 2 / 3 ) = > 2 = rr / ( 2 / 3 ) = > rr = 4 / 3 let h is the time it takes to drain by running all 3 pipes simultaneously so combined rate = rc = 1 / h = 1 + 2 + 4 / 3 = 13 / 3 = 1 / ( 3 / 13 ) thus running simultaneously , pipe q will drain 3 / 13 of the liquid . thus answer = a ." | a ) 3 / 13 , b ) 8 / 23 , c ) 3 / 8 , d ) 17 / 29 , e ) 3 / 4 | a | divide(multiply(2, 2), add(multiply(multiply(4, 2), const_2), multiply(2, 2))) | multiply(n0,n0)|multiply(n0,n1)|multiply(#1,const_2)|add(#2,#0)|divide(#0,#3)| | physics |
pipe p can drain the liquid from a tank in 2 / 4 the time that it takes pipe q to drain it and in 2 / 3 the time that it takes pipe r to do it . if all 3 pipes operating simultaneously but independently are used to drain liquid from the tank , then pipe q drains what portion of the liquid from the tank ? | "look at the diagram below : now , in case when qy is perpendicular to pr , two right triangles pqr and pqy are similar : qy : qp = qr : pr - - > qy : 5 = 8 : 10 - - > qy = 4.0 . answer : c ." | a ) 3.6 cm , b ) 2.4 cm , c ) 4.0 cm , d ) 2.16 cm , e ) 3.2 cm | c | divide(multiply(5, 8), const_10) | multiply(n1,n2)|divide(#0,const_10)| | geometry |
in triangle pqr , the angle q = 90 degree , pq = 5 cm , qr = 8 cm . x is a variable point on pq . the line through x parallel to qr , intersects pr at y and the line through y , parallel to pq , intersects qr at z . find the least possible length of xz | "explanation : average after 11 innings = 24 required number of runs = ( 24 * 11 ) β ( 20 * 10 ) = 264 β 200 = 64 answer : c" | a ) 22 , b ) 77 , c ) 64 , d ) 19 , e ) 17 | c | subtract(multiply(add(10, const_1), add(20, 4)), multiply(20, 10)) | add(n0,const_1)|add(n1,n2)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2)| | general |
the average runs of a cricket player of 10 innings was 20 . how many runs must he make in his next innings so as to increase his average of runs by 4 ? | "ratio of the capitals of a , b and c = 20000 Γ£ β 10 + 15000 Γ£ β 2 : 20000 Γ£ β 10 + 16000 Γ£ β 2 : 20000 Γ£ β 10 + 26000 Γ£ β 2 = 230000 : 232000 : 252000 = 230 : 232 : 252 . b Γ’ β¬ β’ s share = ( 71400 Γ£ β 232 Γ’ Β β 714 ) = 23200 ; answer c" | a ) 20,000 , b ) 21,200 , c ) 23,200 , d ) 20,500 , e ) none of these | c | divide(add(multiply(subtract(const_12, 10), 4000), multiply(multiply(const_2, multiply(const_100, const_100)), 10)), multiply(const_100, const_10)) | multiply(const_100,const_100)|multiply(const_10,const_100)|subtract(const_12,n1)|multiply(n3,#2)|multiply(#0,const_2)|multiply(n1,#4)|add(#3,#5)|divide(#6,#1)| | gain |
a , b and c start a business each investing 20,000 . after 10 months a withdrew 5000 , b withdrew 4000 and c invests 6000 more . at the end of the year , a total profit of 71400 was recorded . find the share of b . | "the number of terms in this set would be : n = ( k - 1 ) / 2 ( as k is odd ) last term : k - 1 average would be first term + last term / 2 = ( 2 + k - 1 ) / 2 = ( k + 1 ) / 2 also average : sum / number of terms = 79 * 80 / ( ( k - 1 ) / 2 ) = 158 * 80 / ( k - 1 ) ( k + 1 ) / 2 = 158 * 80 / ( k - 1 ) - - > ( k - 1 ) ( k + 1 ) = 158 * 160 - - > k = 159 answer e ." | a ) 79 , b ) 80 , c ) 81 , d ) 157 , e ) 159 | e | add(multiply(79, const_2), 1) | multiply(n1,const_2)|add(#0,n0)| | general |
the sum of the even numbers between 1 and k is 79 * 80 , where k is an odd number , then k = | "speed = 78 * 5 / 18 = 21.7 m / sec total distance covered = 125 + 125 = 250 m . required time = 250 / 21.7 ' = 11.5 sec . answer : c" | a ) 12.7 sec , b ) 10.1 sec , c ) 11.5 sec , d ) 12.1 sec , e ) 11.7 sec | c | divide(add(125, 125), multiply(78, const_0_2778)) | add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)| | physics |
how long does a train 125 m long running at the speed of 78 km / hr takes to cross a bridge 125 m length ? | ele , the circus elephant , is currently three times older than lyn , the circus lion . ele = 3 * lyn usually , ages are integers so there is a good possibility that the age of ele is 45 ( the only option that is a multiple of 3 ) . then age of lyn would be 15 . in 15 yrs , ele would be 60 and lyn would be 30 - so lyn would be half as old as ele . answer ( d ) | a ) 40 , b ) 48 , c ) 43 , d ) 45 , e ) 41 | d | multiply(subtract(multiply(const_2, 15), 15), 3) | multiply(n1,const_2)|subtract(#0,n1)|multiply(n0,#1) | general |
ele , the circus elephant , is currently 3 times older than lyn , the circus lion . in 15 years from now , lyn the circus lion will be exactly half as old as ele , the circus elephant . how old is ele today ? | "explanation : let ' s son age is x , then father age is x + 30 . = > 2 ( x + 2 ) = ( x + 30 + 2 ) = > 2 x + 4 = x + 32 = > x = 28 years option a" | a ) 28 years , b ) 22 years , c ) 23 years , d ) 24 years , e ) 26 years | a | divide(subtract(30, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | multiply(const_2,const_2)|subtract(const_2,const_1)|subtract(#0,const_2)|subtract(n0,#2)|divide(#3,#1)| | general |
a man is 30 years older than his son . in two years , his age will be twice the age of his son . the present age of this son is | "total land of sukhiya = \ inline \ frac { 480 x } { 0.6 } = 800 x \ therefore cultivated land of village = 394000 x \ therefore required percentage = \ inline \ frac { 800 x } { 394000 } \ times 100 = 0.20304 e" | a ) 0.20833 , b ) 0.14544 , c ) 0.25632 , d ) 0.35466 , e ) 0.20304 | e | multiply(divide(divide(480, divide(60, const_100)), add(multiply(multiply(3, const_100), const_1000), multiply(add(multiply(const_4, const_10), const_4), const_1000))), const_100) | divide(n0,const_100)|multiply(n1,const_100)|multiply(const_10,const_4)|add(#2,const_4)|divide(n3,#0)|multiply(#1,const_1000)|multiply(#3,const_1000)|add(#5,#6)|divide(#4,#7)|multiply(#8,const_100)| | general |
lagaan is levied on the 60 percent of the cultivated land . the revenue department collected total rs . 3 , 94,000 through the lagaan from the village of mutter . mutter , a very rich farmer , paid only rs . 480 as lagaan . the percentage of total land of mutter over the total taxable land of the village is : | "speed = ( 45 * 5 / 18 ) m / sec = ( 25 / 2 ) m / sec . time = 30 sec . let the length of bridge be x meters . then , ( 110 + x ) / 30 = 25 / 2 = = > 2 ( 110 + x ) = 750 = = > x = 265 m . answer : option a" | a ) 265 , b ) 244 , c ) 245 , d ) 238 , e ) 236 | a | subtract(multiply(divide(multiply(45, speed(const_1000, const_1)), speed(const_3600, const_1)), 30), 110) | speed(const_1000,const_1)|speed(const_3600,const_1)|multiply(n1,#0)|divide(#2,#1)|multiply(n2,#3)|subtract(#4,n0)| | physics |
the length of the bridge , which a train 110 meters long and travelling at 45 km / hr can cross in 30 seconds , is : | "let the distance one way be x time from home to work = x / 60 time from work to home = x / 100 total time = 2 hrs ( x / 60 ) + ( x / 100 ) = 2 solving for x , we get x = 75 time from home to work in minutes = ( 75 ) * 60 / 60 = 75 minutes ans = d" | a ) 66 , b ) 70 , c ) 72 , d ) 75 , e ) 78 | d | multiply(divide(multiply(100, 2), add(60, 100)), const_60) | add(n0,n1)|multiply(n1,n2)|divide(#1,#0)|multiply(#2,const_60)| | physics |
cole drove from home to work at an average speed of 60 kmh . he then returned home at an average speed of 100 kmh . if the round trip took a total of 2 hours , how many minutes did it take cole to drive to work ? | "area of square = pi * radius ^ 2 new radius = 0.5 * old radius so new area = ( 0.5 ) ^ 2 old area = > 0.25 of old area = > 25 % old area ans : c" | a ) 10 % decrease , b ) 20 % decrease , c ) 75 % decrease , d ) 40 % decrease , e ) 50 % decrease | c | subtract(const_100, multiply(power(divide(50, const_100), const_2), const_100)) | divide(n0,const_100)|power(#0,const_2)|multiply(#1,const_100)|subtract(const_100,#2)| | geometry |
if the radius of a circle is decreased 50 % , what happens to the area ? | "solution ratio of times taken by sakshi and tanya = 125 : 100 = 5 : 4 . suppose tanya taken x days to do the work . 5 : 4 : : 30 : x β x = ( 30 x 4 / 5 ) β x = 24 days . hence , tanya takes 16 days is complete the work . answer c" | a ) 15 , b ) 16 , c ) 14 , d ) 25 , e ) 10 | c | divide(30, add(const_1, divide(25, const_100))) | divide(n1,const_100)|add(#0,const_1)|divide(n0,#1)| | physics |
sakshi can do a piece of work in 30 days . tanya is 25 % more efficient than sakshi . the number of days taken by tanya to do the same piece of work : | "3 ^ 8 - 1 = ( 3 ^ 4 - 1 ) ( 3 ^ 4 + 1 ) = 80 * 82 = 8 * 10 * 2 * 41 the answer is d ." | a ) 29 , b ) 31 , c ) 37 , d ) 41 , e ) 43 | d | floor(divide(3, divide(8, const_2))) | divide(n1,const_2)|divide(n0,#0)|floor(#1)| | general |
what is the greatest prime factor of 3 ^ 8 - 1 ? | "explanation : 90 x = 60 y x : y = 90 : 60 = 3 : 2 answer : c" | a ) 1 : 2 , b ) 3 : 5 , c ) 3 : 2 , d ) 3 : 4 , e ) 5 : 2 | c | divide(90, 60) | divide(n0,n1)| | general |
calculate the ratio between x and y if 90 % of x equal to 60 % of y ? | "1 / 2 * 50 ( 15 + 5 ) = 500 cm 2 answer : e" | a ) 100 cm 2 , b ) 150 cm 2 , c ) 200 cm 2 , d ) 250 cm 2 , e ) 500 cm 2 | e | multiply(multiply(divide(const_1, const_2), add(5, 15)), 50) | add(n1,n2)|divide(const_1,const_2)|multiply(#0,#1)|multiply(n0,#2)| | geometry |
find the area of the quadrilateral of one of its diagonals is 50 cm and its off sets 15 cm and 5 cm ? | "speed = 78 * 5 / 18 = 65 / 3 m / sec . time = 1 min = 60 sec . let the length of the train be x meters . then , ( 1200 + x ) / 60 = 65 / 3 x = 100 m . answer : d" | a ) 2898 , b ) 277 , c ) 500 , d ) 100 , e ) 435 | d | divide(1200, multiply(subtract(78, 1), const_0_2778)) | subtract(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
a train 1200 m long is running at a speed of 78 km / hr . if it crosses a tunnel in 1 min , then the length of the tunnel is ? | "7 + 11 + 13 + 17 + 19 = 67 / 5 = 13.40 answer : a" | a ) 13.4 , b ) 12.26 , c ) 12.3 , d ) 32.8 , e ) 32.4 | a | add(5, const_1) | add(n0,const_1)| | general |
the average of first five prime numbers greater than 5 is ? | edge of the cube = 3 β 334 = 7 cm β΄ radius of cone = 3.5 cm height = 7 cm volume of cone = 1 β 3 Ο r 2 h 1 β 3 Ο r 2 h = 1 β 3 Γ 22 β 7 Γ ( 3.5 ) 2 Γ 7 = 1 β 3 Γ 22 Γ 12.25 β 90 sec answer b | ['a ) 80 cc', 'b ) 90 cc', 'c ) 110 cc', 'd ) 105 cc', 'e ) 100 cc'] | b | volume_cone(divide(cube_edge_by_volume(343), const_2), cube_edge_by_volume(343)) | cube_edge_by_volume(n0)|divide(#0,const_2)|volume_cone(#1,#0) | geometry |
a right circular cone is exactly fitted inside a cube in such away that the edges of the base of the cone are touching the edges of one of the faces of the cube and the vertex is on the opposite face of the cube . if the volume of the cube is 343 cc , what approximately is the volume of the cone ? | "here , l = 7 and m = 28 therefore , first number = l / m x 100 % of second number = 7 / 28 x 100 % of second number = 25 % of second number answer : b" | a ) 20 % , b ) 25 % , c ) 18 % , d ) 30 % , e ) none of these | b | multiply(divide(divide(7, const_100), divide(28, const_100)), const_100) | divide(n0,const_100)|divide(n1,const_100)|divide(#0,#1)|multiply(#2,const_100)| | gain |
if the given two numbers are respectively 7 % and 28 % of a third number , then what percentage is the first of the second ? | "let there be x pupils in the class . total increase in marks = ( x * 1 / 2 ) = x / 2 x / 2 = ( 83 - 53 ) = > x / 2 = 30 = > x = 60 . answer : c" | a ) 18 , b ) 82 , c ) 60 , d ) 27 , e ) 29 | c | multiply(subtract(83, 53), const_2) | subtract(n0,n1)|multiply(#0,const_2)| | general |
a pupil ' s marks were wrongly entered as 83 instead of 53 . due to the average marks for the class got increased by half . the number of pupils in the class is ? | area of rectangle = 20 x 10 = 200 cm Γ’ Β² let ' l ' the length of other diagonal = 0.5 x 32 xl = 200 which gives x = 12.5 cm answer : b | ['a ) 10', 'b ) 12.5', 'c ) 15', 'd ) 16', 'e ) 17.5'] | b | divide(multiply(rectangle_area(20, 10), const_2), 32) | rectangle_area(n0,n1)|multiply(#0,const_2)|divide(#1,n2) | geometry |
the area of a rhombus is equal to the area of a rectangle whose length is 20 cm and the breadth is 10 cm . if one of the diagonals is 32 cm what is the length of other diagonal ? | p - 2160 = ( p * 5 * 4 ) / 100 p = 2700 answer : c | a ) 1500 , b ) 2500 , c ) 2700 , d ) 3200 , e ) 11500 | c | divide(2160, subtract(const_1, divide(multiply(4, 5), const_100))) | multiply(n0,n1)|divide(#0,const_100)|subtract(const_1,#1)|divide(n2,#2) | gain |
if the simple interest on a certain amount in at 4 % rate 5 years amounted to rs . 2160 less than the principal . what was the principal ? | "for 40 hrs = 40 * 16 = 640 excess = 948 - 640 = 308 for extra hours = . 75 ( 16 ) = 12 + 16 = 28 number of extra hrs = 308 / 28 = 77 / 7 = 11 total hrs = 40 + 11 = 51 answer e 51" | a ) 36 , b ) 40 , c ) 44 , d ) 48 , e ) 51 | e | add(40, divide(subtract(948, multiply(16, 40)), divide(multiply(16, add(const_100, 75)), const_100))) | add(n3,const_100)|multiply(n0,n1)|multiply(n0,#0)|subtract(n4,#1)|divide(#2,const_100)|divide(#3,#4)|add(n1,#5)| | general |
a certain bus driver is paid a regular rate of $ 16 per hour for any number of hours that does not exceed 40 hours per week . for any overtime hours worked in excess of 40 hours per week , the bus driver is paid a rate that is 75 % higher than his regular rate . if last week the bus driver earned $ 948 in total compensation , how many total hours did he work that week ? | "assume that the equation of the line is y = mx + c , where m and c are the slope and y - intercept . you are also given that the line crosses the point ( - 12 , - 39 ) , this means that this point will also lie on the line above . thus you get - 39 = m * ( - 12 ) + c , with m = 3 / 4 as the slope is given to be 3 / 4 . after substituting the above values , you get c = - 30 . thus the equation of the line is y = 0.75 * x - 30 and the point where it will intersect the x - axis will be with y coordinate = 0 . put y = 0 in the above equation of the line and you will get , x = 40 . thus , the point w of intersection is ( 40,0 ) . a is the correct answer ." | a ) ( 40,0 ) , b ) ( 30,0 ) , c ) ( 0,40 ) , d ) ( 40,30 ) , e ) ( 0,30 ) | a | multiply(negate(divide(subtract(negate(39), multiply(negate(12), divide(3, 4))), divide(3, 4))), const_10) | divide(n0,n1)|negate(n3)|negate(n2)|multiply(#0,#2)|subtract(#1,#3)|divide(#4,#0)|negate(#5)|multiply(#6,const_10)| | general |
a line has a slope of 3 / 4 and intersects the point w ( - 12 , - 39 ) . at which point does this line intersect the x - axis ? | let his total sales be x total sales - commission = $ 31100 x - [ ( 5 % of 10000 ) + 4 % of ( x - 10000 ) ] = 31100 x - 500 - ( x - 10000 ) / 25 = 31100 x = 32500 answer is b | a ) $ 30000 , b ) $ 32500 , c ) $ 35120 , d ) $ 41520 , e ) $ 25460 | b | divide(add(31100, subtract(divide(multiply(5, 10000), const_100), divide(multiply(4, 10000), const_100))), subtract(const_1, divide(4, const_100))) | divide(n2,const_100)|multiply(n0,n1)|multiply(n1,n2)|divide(#1,const_100)|divide(#2,const_100)|subtract(const_1,#0)|subtract(#3,#4)|add(n3,#6)|divide(#7,#5) | general |
a salesman commission is 5 % on all sales upto $ 10000 and 4 % on all sales exceeding this . he remits $ 31100 to his parent company after deducting his commission . find the total sales ? | ( 60 / 100 ) * 50 β ( 34 / 100 ) * 30 30 - 10.2 = 19.8 answer : b | a ) 18 , b ) 19.8 , c ) 11 , d ) 17 , e ) 12 | b | subtract(divide(multiply(60, 50), const_100), divide(multiply(34, 30), const_100)) | multiply(n0,n1)|multiply(n2,n3)|divide(#0,const_100)|divide(#1,const_100)|subtract(#2,#3) | gain |
how much 60 % of 50 is greater than 34 % of 30 ? | "m = 5 s = 2 ds = 7 us = 3 x / 7 + x / 3 = 1 x = 2.1 d = 2.1 * 2 = 4.2 answer : d" | a ) 5.75 , b ) 5.7 , c ) 5.76 , d ) 4.2 , e ) 5.71 | d | multiply(divide(multiply(add(5, 2), subtract(5, 2)), add(add(5, 2), subtract(5, 2))), const_2) | add(n0,n1)|subtract(n0,n1)|add(#0,#1)|multiply(#0,#1)|divide(#3,#2)|multiply(#4,const_2)| | physics |
a man can row 5 kmph in still water . when the river is running at 2 kmph , it takes him 1 hour to row to a place and black . what is the total distance traveled by the man ? | d = 200 + 180 = 380 m s = 60 * 5 / 18 = 50 / 3 t = 380 * 3 / 50 = 22.8 sec answer : c | a ) 36.7 , b ) 26.8 , c ) 22.8 , d ) 21.1 , e ) 16.2 | c | divide(add(200, 180), multiply(60, const_0_2778)) | add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1) | physics |
how long does a truck of 200 m long traveling at 60 kmph takes to cross a bridge of 180 m in length ? | solving the two equations , we get : l = 30 and b = 40 . area = ( l x b ) = ( 30 x 40 ) m 2 = 1200 m ^ 2 d | ['a ) 2400 m ^ 2', 'b ) 1500 m ^ 2', 'c ) 2520 m ^ 2', 'd ) 1200 m ^ 2', 'e ) 2580 m ^ 2'] | d | multiply(multiply(10, const_4), multiply(10, const_3)) | multiply(n0,const_4)|multiply(n0,const_3)|multiply(#0,#1) | geometry |
difference between the length & breadth of a rectangle is 10 m . if its perimeter is 206 m , then its area is ? | "1370 - 50 * ( 5 ^ 0 ) = 1320 1320 - 50 * ( 5 ^ 1 ) = 1070 1070 - 50 * ( 5 ^ 2 ) = - 180 - 180 - 50 * ( 5 ^ 3 ) = - 6430 answer : a ." | a ) 1070 , b ) 6530 , c ) 6630 , d ) 6730 , e ) 6830 | a | subtract(negate(6430), multiply(subtract(1320, 180), divide(subtract(1320, 180), subtract(1370, 1320)))) | negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)| | general |
1370 , 1320 , x , - 180 , - 6430 | "machine y produces 20 percent more widgets in an hour than machine x does in an hour . so if machine x produces 100 widgets , then machine y produces 120 widgets . ratio of 120 / 100 = 6 / 5 . this is their speed of work ( y : x ) . i . e . speed of their work ( x : y ) = 5 / 6 now , time is inversely proportional to speed . hence the ratio of the time spent ( x : y ) = 6 / 5 let us assume that they spend 6 x and 5 x hours . given that 6 x - 5 x = 50 so , x = 50 . hence 6 x = 6 * 50 = 300 hours . hence x takes 120 hours to produce 1080 widgets . so , in 1 hour , it can produce ( 1 * 1080 ) / 300 = 3.6 hence option ( b ) ." | a ) 100 , b ) 3.6 , c ) 25 , d ) 11 , e ) 9 | b | divide(1080, multiply(divide(50, const_10), 50)) | divide(n0,const_10)|multiply(n0,#0)|divide(n1,#1)| | general |
machine x takes 50 hours longer than machine y to produce 1080 widgets . machine y produces 20 percent more widgets in an hour than machine x does in an hour . how many widgets per hour does machine x produce | "factorization of 45 = 3 * 3 * 5 factorization of 80 = 5 * 2 ^ 4 means the smallest value of k is \ sqrt { 3 ^ 2 * 5 * 2 ^ 4 } = 3 * 5 * 2 ^ 2 k ^ 3 = 3 ^ 3 * 5 ^ 3 * 2 ^ 6 * x = 3 ^ 3 * 2 * 4000 * x where x can be any integer k ^ 3 / 4000 = 3 ^ 3 * 2 * x = 54 x which obviously has different units digit depending on x answer : d" | a ) 0 , b ) 1 , c ) 27 , d ) 54 , e ) can not be determined | d | divide(multiply(multiply(multiply(3, 2), multiply(3, 2)), multiply(3, 2)), const_4) | multiply(n6,n9)|multiply(#0,#0)|multiply(#1,#0)|divide(#2,const_4)| | general |
for any positive number x , the function [ x ] denotes the greatest integer less than or equal to x . for example , [ 1 ] = 1 , [ 1.367 ] = 1 and [ 1.999 ] = 1 . if k is a positive integer such that k ^ 2 is divisible by 45 and 80 , what is the units digit of k ^ 3 / 4000 ? | make r = 10 p = 0.6 r = 6 q = 0.8 p = 4.8 s = 0.6 p = 3.6 for that we get s = 7.6 and q 10.8 so 10.8 / 7.6 = 2.7 / 1.9 ans : e | a ) 11 / 12 , b ) 13 / 17 , c ) 13 / 19 , d ) 12 / 19 , e ) 27 / 19 | e | divide(add(multiply(60, const_100), multiply(60, subtract(const_100, 20))), add(multiply(40, const_100), multiply(add(40, 20), 60))) | add(n0,n1)|multiply(n4,const_100)|multiply(n0,const_100)|subtract(const_100,n1)|multiply(n4,#3)|multiply(n4,#0)|add(#1,#4)|add(#2,#5)|divide(#6,#7) | general |
paul ' s income is 40 % less than rex ' s income , quentin ' s income is 20 % less than paul ' s income , and sam ' s income is 40 % less than paul ' s income . if rex gave 40 % of his income to sam and 60 % of his income to quentin , quentin ' s new income would be what fraction of sam ' s new income ? | "let p = 3 , q = 7 then 2 * 3 + 7 = 13 so 2 p + q = 13 . answer : c" | a ) 12 , b ) 14 , c ) 13 , d ) 15 , e ) 16 | c | add(multiply(3, 2), 7) | multiply(n0,n2)|add(n1,#0)| | general |
if p / q = 3 / 7 , then 2 p + q = ? | "# defective pens = 2 # good pens = 7 probability of the 1 st pen being good = 7 / 9 probability of the 2 nd pen being good = 6 / 8 total probability = 7 / 9 * 6 / 8 = 7 / 12 the answer is d ." | a ) 4 / 7 , b ) 5 / 9 , c ) 6 / 11 , d ) 7 / 12 , e ) 8 / 15 | d | multiply(divide(subtract(9, 2), 9), divide(subtract(subtract(9, 2), const_1), subtract(9, const_1))) | subtract(n0,n1)|subtract(n0,const_1)|divide(#0,n0)|subtract(#0,const_1)|divide(#3,#1)|multiply(#2,#4)| | general |
in a box of 9 pens , a total of 2 are defective . if a customer buys 2 pens selected at random from the box , what is the probability that neither pen will be defective ? | "explanation : let the numbers be x and y according to the problem x + y = 63 x - y = 1 / 8 ( x + y ) x - y = 1 / 8 * 63 , x - y = 9 2 x = 72 , x = 36 and y = 27 l . c . m of 36 and 27 is 351 answer : option c" | a ) 231 , b ) 153 , c ) 351 , d ) 345 , e ) 355 | c | subtract(63, divide(subtract(63, divide(1, const_2)), const_2)) | divide(n1,const_2)|subtract(n0,#0)|divide(#1,const_2)|subtract(n0,#2)| | general |
sum of two numbers is 63 . their difference is 1 / 8 of their sum . their l . c . m is | "explanation : let the required no of hours be x . then less men , more hours ( indirct proportion ) \ inline \ fn _ jvn \ therefore 15 : 36 : : 25 : x \ inline \ fn _ jvn \ leftrightarrow ( 15 x x ) = ( 36 x 25 ) \ inline \ fn _ jvn \ leftrightarrow \ inline \ fn _ jvn x = \ frac { 36 \ times 25 } { 15 } = 60 hence , 15 men can do it in 60 hours . answer : c ) 60" | a ) 22 , b ) 38 , c ) 60 , d ) 88 , e ) 72 | c | divide(multiply(36, 25), 15) | multiply(n0,n1)|divide(#0,n2)| | physics |
if 36 men can do a piece of work in 25 hours , in how mwny hours will 15 men do it ? | "d 5 % increase in 10 years = ( 262500 - 175000 ) = 87500 . increase % = ( 87500 / 175000 x 100 ) % = 50 % . required average = ( 50 / 10 ) % = 5 % ." | a ) 1 % , b ) 2 % , c ) 3 % , d ) 5 % , e ) 4 % | d | add(multiply(divide(subtract(divide(subtract(subtract(subtract(multiply(multiply(const_10, const_1000), const_10), const_1000), const_1000), multiply(add(2, const_3), const_100)), multiply(add(multiply(add(const_3, const_4), const_10), add(2, const_3)), const_1000)), 1), const_10), const_100), const_4) | add(n2,const_3)|add(const_3,const_4)|multiply(const_10,const_1000)|multiply(#2,const_10)|multiply(#0,const_100)|multiply(#1,const_10)|add(#0,#5)|subtract(#3,const_1000)|multiply(#6,const_1000)|subtract(#7,const_1000)|subtract(#9,#4)|divide(#10,#8)|subtract(#11,n0)|divide(#12,const_10)|multiply(#13,const_100)|add(#14,const_4)| | general |
the population of a town increased from 1 , 75,000 to 2 , 62,500 in a decade . the average percent increase of population per year is : | let the lowest price be x . therefore , highest price is x + 100 . now price of each variety is increased by 10 % . therefore the price will remain arranged in the same order as before . or lowest price = 1.1 x and highest = 1.1 * ( x + 100 ) or range = highest - lowest = 1.1 * ( x + 100 ) - 1.1 x = 110 , hence , c | a ) $ 50 , b ) $ 100 , c ) $ 110 , d ) $ 600 , e ) $ 300 | c | multiply(100, divide(add(10, const_100), const_100)) | add(n3,const_100)|divide(#0,const_100)|multiply(n0,#1) | general |
last year the price range ( per kg ) for 100 varieties of apples in wholesale market was $ 100 . if the prices of each of the 100 varieties increased by 10 percent this year over what it was last year , what is the range of the wholesale prices of the 1000 varieties of apples this year ? | "explanation : let the c . p . of the article be rs . x given that % profit earned by selling article at rs . 1920 = % loss incurred by selling article at rs . 1280 ( 1920 β x / x ) β 100 = ( x β 1280 / x ) β 100 = > 1920 - x = x - 1280 = > 2 x = 3200 = > x = 1600 s . p . for 25 % profit = rs . 1600 + 25 % of rs . 1600 = rs . 1600 * ( 125 / 100 ) = rs . 2000 answer : a" | a ) rs . 2000 , b ) rs . 2200 , c ) rs . 2400 , d ) data inadequate , e ) can not be determined | a | multiply(divide(add(const_100, 25), const_100), divide(add(1920, 1280), const_2)) | add(n2,const_100)|add(n0,n1)|divide(#0,const_100)|divide(#1,const_2)|multiply(#2,#3)| | gain |
the percentage profit earned by selling an article for rs . 1920 is equal to the percentage loss incurred by selling the same article for rs . 1280 . at what price should the article be sold to make 25 % profit ? | the tailors would need to cut the fabric 49 times thus the total amount spent would be 245 minutes . the answer is d | a ) 150 , b ) 200 , c ) 188 , d ) 245 , e ) 123 | d | multiply(subtract(divide(multiply(10, const_1000), 200), const_1), 5) | multiply(n0,const_1000)|divide(#0,n1)|subtract(#1,const_1)|multiply(n3,#2) | physics |
the tailor has a 10 meter long piece of fabric for which to sew a ball room dress . she has to cuts this fabric into strips of 200 centimeters each . how long will it take the tailor to complete this tasks if each 200 centimeter took 5 minutes to cut ? | "answer = 60 / 42 * 4 = 60 / 168 = 0.3571 option d is correct" | a ) 14.7653 , b ) 0.5897 , c ) 3.6974 , d ) 0.3571 , e ) 1.2587 | d | divide(60, 42) | divide(n0,n1)| | general |
find 60 / 42 * 4 | "there are five prime numbers between 30 and 50 . they are 31,37 , 41,43 and 47 . therefore the required average = ( 31 + 37 + 41 + 43 + 47 ) / 5 ο³ 199 / 5 ο³ 39.8 answer c 39.8" | a ) 18.9 , b ) 19.8 , c ) 39.8 , d ) 29.8 , e ) 20.8 | c | divide(add(add(add(30, const_1), add(add(30, const_1), const_2)), add(subtract(50, 30), subtract(50, const_2))), 30) | 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 |
find the average of all prime numbers between 30 and 50 ? | "let the total number of votes polled in the election be 100 k . number of valid votes = 100 k - 20 % ( 100 k ) = 80 k let the number of votes polled in favour of a and b be a and b respectively . a - b = 15 % ( 100 k ) = > a = b + 15 k = > a + b = b + 15 k + b now , 2 b + 15 k = 80 k and hence b = 32.5 k it is given that 100 k = 4720 32.5 k = 32.5 k / 100 k * 4720 = 1534 the number of valid votes polled in favour of b is 1534 . answer : a" | a ) 1534 , b ) 2999 , c ) 2834 , d ) 2777 , e ) 2991 | a | divide(subtract(multiply(subtract(const_1, divide(20, const_100)), 4720), divide(multiply(15, 4720), const_100)), const_2) | divide(n1,const_100)|multiply(n0,n2)|divide(#1,const_100)|subtract(const_1,#0)|multiply(n2,#3)|subtract(#4,#2)|divide(#5,const_2)| | general |
in an election between two candidates a and b , the number of valid votes received by a exceeds those received by b by 15 % of the total number of votes polled . if 20 % of the votes polled were invalid and a total of 4720 votes were polled , then how many valid votes did b get ? | "worker b will get 11 / 16 = 68.75 % the answer is d ." | a ) 62.45 % , b ) 64.55 % , c ) 66.65 % , d ) 68.75 % , e ) 70.85 % | d | divide(1088, add(5, 11)) | add(n1,n2)|divide(n0,#0)| | other |
if $ 1088 are divided between worker a and worker b in the ratio 5 : 11 , what is the share that worker b will get ? | "20 + 24 + 6 * 2 = 56 / 2 = 28 answer : e" | a ) 24 , b ) 25 , c ) 26 , d ) 27 , e ) 28 | e | divide(add(add(20, 24), multiply(6, 2)), const_2) | add(n2,n3)|multiply(n0,n1)|add(#0,#1)|divide(#2,const_2)| | general |
the average age of 6 men increases by 2 years when two women are included in place of two men of ages 20 and 24 years . find the average age of the women ? | official solution : ( b ) we know that a # b = 100 and a # b = 4 a Β² + 4 b Β² + 8 ab . so 4 a Β² + 4 b Β² + 8 ab = 100 we can see that 4 a Β² + 4 b Β² + 8 ab is a well - known formula for ( 2 a + 2 b ) Β² . therefore ( 2 a + 2 b ) Β² = 100 . ( 2 a + 2 b ) is non - negative number , since both a and b are non - negative numbers . so we can conclude that 2 ( a + b ) = 10 . ( a + b ) + 4 = 10 / 2 + 4 = 9 . the correct answer is b . | a ) 5 , b ) 9 , c ) 10 , d ) 13 , e ) 17 | b | add(sqrt(divide(100, 4)), 4) | divide(n6,n0)|sqrt(#0)|add(n0,#1) | general |
operation # is defined as : a # b = 4 a ^ 2 + 4 b ^ 2 + 8 ab for all non - negative integers . what is the value of ( a + b ) + 4 , when a # b = 100 ? | "area = 5.5 Γ 3.75 sq . metre . cost for 1 sq . metre . = $ 500 hence , total cost = 5.5 Γ 3.75 Γ 500 = $ 10312.50 a" | a ) $ 10312.50 , b ) $ 13512.50 , c ) $ 16512.50 , d ) $ 14512.50 , e ) $ 18512.50 | a | multiply(500, multiply(5.5, 3.75)) | multiply(n0,n1)|multiply(n2,#0)| | physics |
the length of a room is 5.5 m and width is 3.75 m . what is the cost of paying the floor by slabs at the rate of $ 500 per sq . metre . | "we have : l = 20 ft and lb = 10 sq . ft . so , b = 0.5 ft . length of fencing = ( l + 2 b ) = ( 20 + 1 ) ft = 21 ft . answer : d" | a ) 34 , b ) 40 , c ) 68 , d ) 21 , e ) 78 | d | add(multiply(divide(10, 20), const_2), 20) | divide(n1,n0)|multiply(#0,const_2)|add(n0,#1)| | geometry |
a rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered . if the area of the field is 10 sq . feet , how many feet of fencing will be required ? | the problem asks for the approximate chance that no more than 1 / 3 of the original investment is lost . we can apply the β 1 β x β technique : what β s the chance that more than 1 / 3 of the original investment is lost ? there are two outcomes we have to separately measure : ( a ) all 3 investments become worthless . ( b ) 2 of the 3 investments become worthless , while 1 doesn β t . outcome ( a ) : the probability is ( 0.2 ) ( 0.2 ) ( 0.2 ) = 0.008 , or a little less than 1 % . outcome ( b ) : call the investments x , y , and z . the probability that x retains value , while y and z become worthless , is ( 0.8 ) ( 0.2 ) ( 0.2 ) = 0.032 . now , we have to do the same thing for the specific scenarios in which y retains value ( while x and z don β t ) and in which z retains value ( while x and y don β t ) . each of those scenarios results in the same math : 0.032 . thus , we can simply multiply 0.032 by 3 to get 0.096 , or a little less than 10 % . the sum of these two probabilities is 0.008 + 0.096 = 0.104 , or a little more than 10 % . finally , subtracting from 100 % and rounding , we find that the probability we were looking for is approximately 90 % . the correct answer is a . this problem illustrates the power of diversification in financial investments . all else being equal , it β s less risky to hold a third of your money in three uncorrelated ( independent ) but otherwise equivalent investments than to put all your eggs in one of the baskets . that said , be wary of historical correlations ! housing price changes in different us cities were not so correlated β and then they became highly correlated during the recent housing crisis ( they all fell together ) , fatally undermining spreadsheet models that assumed that these price changes were independent . | a ) 90 % , b ) 80 % , c ) 70 % , d ) 60 % , e ) 40 % | a | subtract(add(multiply(20, const_2), multiply(20, 3)), const_10) | multiply(n1,const_2)|multiply(n0,n1)|add(#0,#1)|subtract(#2,const_10) | general |
each of 3 investments has a 20 % of becoming worthless within a year of purchase , independently of what happens to the other two investments . if simone invests an equal sum k in each of these 3 investments on january 1 , the approximate chance that by the end of the year , she loses no more than 1 / 3 of her original investment is | "multiples of 4 less than 250 = { 4 , 8,12 , . . . . . . 248 } = 62 numbers multiples of 4 which are multiples of 3 too = { 12 , 24,36 . . . . . 240 } = 20 numbers so required number = 62 - 20 = 42 choice c" | a ) 20 , b ) 31 , c ) 42 , d ) 53 , e ) 64 | c | divide(factorial(subtract(add(const_4, 4), const_1)), multiply(factorial(4), factorial(subtract(const_4, const_1)))) | add(n1,const_4)|factorial(n1)|subtract(const_4,const_1)|factorial(#2)|subtract(#0,const_1)|factorial(#4)|multiply(#1,#3)|divide(#5,#6)| | general |
how many positive integers less than 250 are multiple of 4 but not multiples of 6 ? | to get maximum vehicles we must take into consideration the minimum weight i . e 1600 pounds here since , 1 ton = 2000 pounds 50 tons will be 100,000 pounds from the answer choices : let max number of vehicles be 62 total weight will be = 62 * 1600 = 99200 pounds , which is lesser than the maximum weight allowed . ans : d | a ) 23 , b ) 41 , c ) 48 , d ) 62 , e ) 86 | d | divide(multiply(multiply(50, const_2), const_1000), add(add(add(add(add(add(const_1000, const_100), const_100), const_100), const_100), const_100), const_100)) | add(const_100,const_1000)|multiply(n0,const_2)|add(#0,const_100)|multiply(#1,const_1000)|add(#2,const_100)|add(#4,const_100)|add(#5,const_100)|add(#6,const_100)|divide(#3,#7) | general |
a ferry can transport 50 tons of vehicles . automobiles range in weight from 1,600 to 3,200 pounds . what is the greatest number of automobiles that can be loaded onto the ferry ? | "d = 8 d d = d a β 2 = 8 d a β 2 = d a = 8 d / β 2 a = d / β 2 = > 8 : 1 answer : d" | a ) 3 : 6 , b ) 3 : 3 , c ) 3 : 8 , d ) 8 : 1 , e ) 3 : 2 | d | divide(8, divide(8, 8)) | divide(n0,n0)|divide(n0,#0)| | geometry |
what is the ratio between perimeters of two squares one having 8 times the diagonal then the other ? | bank balance is doubled with accumulation of interest tp 5080 . . this means interest is 5080 / 2 = 2540 for entire year . . although since interest is 10 % of avg monthly balance , it becomes 25400 . . d | a ) 2840 , b ) 5680 , c ) 6840 , d ) 25400 , e ) 28400 | d | multiply(5080, divide(10, const_2)) | divide(n0,const_2)|multiply(n1,#0) | general |
a bank pays interest to its customers on the last day of the year . the interest paid to a customer is calculated as 10 % of the average monthly balance maintained by the customer . john is a customer at the bank . on the last day , when the interest was accumulated into his account , his bank balance doubled to $ 5080 . what is the average monthly balance maintained by john in his account during the year ? | "speed of bus relative to woman = 21 + 3 = 24 km / hr . = 24 * 5 / 18 = 20 / 3 m / sec . time taken to pass the woman = 75 * 3 / 20 = 11.25 sec . answer : c" | a ) 5.75 , b ) 7.62 , c ) 11.25 , d ) 4.25 , e ) 3.25 | c | divide(divide(multiply(75, const_3600), add(21, 3)), const_1000) | add(n1,n2)|multiply(n0,const_3600)|divide(#1,#0)|divide(#2,const_1000)| | physics |
a bus 75 m long is running with a speed of 21 km / hr . in what time will it pass a woman who is walking at 3 km / hr in the direction opposite to that in which the bus is going ? | "in real basic terms , we ' re asked to find all of the ' 3 s ' in 200 ! we can figure out that 200 / 3 = 66 , so we know that there are at least 66 ' 3 s ' in 200 ! while that answer is among the 5 choices , it seems a bit too ' easy ' , so let ' s do a bit more work and list out the first few numbers that we know have a ' 3 ' in them : 3 = 3 x 1 6 = 3 x 2 9 = 3 x 3 notice how both 3 and 6 have just one 3 in them , but 9 has two 3 s ( there ' s an ' extra ' 3 that we have to account for ) . this implies that there are probably other numbers that include ' extra 3 s ' that we have to figure out : to find those extra 3 s , we have to look at numbers that contain ' powers of 3 ' . . . 3 ^ 2 = 9 3 ^ 3 = 27 3 ^ 4 = 81 3 ^ 5 = 243 , but that ' s too big ( we ' re only going up to 200 ) . keep in mind that a multiple of 81 is also a multiple of 9 and 27 , so we do n ' t want to count any of those values more than once . 200 / 9 = 22 , so we know that there are at least 22 extra 3 s ( and certainly more because of the 27 and 81 ) . with the 66 3 s that we already have , those 22 extra 3 s increase the total to 88 . with the other extra 3 s , we ' ll end up with more than 88 3 s . there ' s only one answer that fits that logic . . . answer : d" | a ) 88 , b ) 48 , c ) 66 , d ) 97 , e ) 39 | d | multiply(floor(divide(power(const_10, 3), 200)), 200) | power(const_10,n0)|divide(#0,n1)|floor(#1)|multiply(n1,#2)| | other |
what is the largest power of 3 contained in 200 ! | "explanation : ( 40 + 60 + 70 + 80 + 80 / 5 ) = 66 option a" | a ) 66 , b ) 75 , c ) 80 , d ) 85 , e ) 90 | a | divide(add(add(add(add(40, 60), 70), 80), 80), add(const_4, const_1)) | add(n0,n1)|add(const_1,const_4)|add(n2,#0)|add(n3,#2)|add(n4,#3)|divide(#4,#1)| | general |
reeya obtained 40 , 60 , 70 , 80 and 80 out of 100 in different subjects , what will be the average | "the time it takes bob to drive to city b is 100 / 40 = 2.5 hours . alice needs to take less than 2 hours for the trip . alice needs to exceed a constant speed of 100 / 2 = 50 miles per hour . the answer is c ." | a ) 45 , b ) 48 , c ) 50 , d ) 52 , e ) 54 | c | divide(100, subtract(divide(100, 40), divide(30, const_60))) | divide(n0,n1)|divide(n2,const_60)|subtract(#0,#1)|divide(n0,#2)| | physics |
the distance from city a to city b is 100 miles . while driving from city a to city b , bob drives at a constant speed of 40 miles per hour . alice leaves city a 30 minutes after bob . what is the minimum constant speed in miles per hour that alice must exceed in order to arrive in city b before bob ? | "on the first day 1 unit of orange juice and 1 unit of water was used to make 2 units of orangeade ; on the second day 1 unit of orange juice and 3 units of water was used to make 4 units of orangeade ; so , the ratio of the amount of orangeade made on the first day to the amount of orangeade made on the second day is 2 to 4 . naturally the ratio of the # of glasses of orangeade made on the first day to the # of glasses of orangeade made on the second day is 2 to 4 . we are told thatthe revenue from selling the orangeade was the same for both daysso the revenue from 2 glasses on the first day equals to the revenue from 4 glasses on the second day . say the price of the glass of the orangeade on the second day was $ x then 2 * 0.6 = 4 * x - - > x = $ 0.3 . answer : c ." | a ) $ 015 , b ) $ 0.20 , c ) $ 0.30 , d ) $ 0.40 , e ) $ 0.45 | c | divide(multiply(add(const_1, const_1), 0.60), add(const_1, const_2)) | add(const_1,const_1)|add(const_1,const_2)|multiply(n0,#0)|divide(#2,#1)| | general |
on a certain day , orangeade was made by mixing a certain amount of orange juice with an equal amount of water . on the next day , orangeade was made by mixing the same amount of orange juice with thrice the amount of water . on both days , all the orangeade that was made was sold . if the revenue from selling the orangeade was the same for both days and if the orangeade was sold at $ 0.60 per glass on the first day , what was the price per glass on the second day ? | "d 7429 - 4358 = 3071 let 8597 - x = 3071 then , x = 8597 - 3071 = 5526" | a ) 3567 , b ) 6424 , c ) 6835 , d ) 5526 , e ) none of these | d | subtract(multiply(divide(8597, const_100), 7429), multiply(divide(const_1, const_3), multiply(divide(8597, const_100), 7429))) | divide(n0,const_100)|divide(const_1,const_3)|multiply(n1,#0)|multiply(#1,#2)|subtract(#2,#3)| | general |
8597 - ? = 7429 - 4358 | "cant think of a straight approach but here is how i solved it : k is divided by 5 and remainder is 2 . this means k = 5 n + 2 ( n is an integer ) so the possible values of k = { 2 , 7 , 12 , 17 , 22 } ( less than 24 ) secondly , if k is divided by 6 , the remainder is 5 = > k = 6 m + 5 so the possible value set for k = { 5 , 11 , 17 , 23 } ( less than 24 ) 17 is the only common number in both the sets . hence k = 17 answer : d" | a ) 2 , b ) 5 , c ) 6 , d ) 3 , e ) 8 | d | reminder(add(const_12, 5), 7) | add(n0,const_12)|reminder(#0,n5)| | general |
when positive integer k is divided by 5 , the remainder is 2 . when k is divided by 6 , the remainder is 5 . if k is less than 24 , what is the remainder when k is divided by 7 ? | "let c ' s age be x years . then , b ' s age = 2 x years . a ' s age = ( 2 x + 2 ) years . ( 2 x + 2 ) + 2 x + x = 32 5 x = 30 = > x = 6 hence , b ' s age = 2 x = 12 years . answer : e" | a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 12 | e | multiply(divide(subtract(32, const_2), add(const_3, const_2)), const_2) | add(const_2,const_3)|subtract(n0,const_2)|divide(#1,#0)|multiply(#2,const_2)| | general |
is two years older than b who is twice as old as c . if the total of the ages of a , b and c be 32 , then how old is b ? | "speed = 72 * 5 / 18 = 20 m / sec total distance covered = 110 + 132 = 242 m . required time = 242 / 20 = 12.1 sec . answer : b" | a ) 12.6 , b ) 12.0 , c ) 12.1 , d ) 12.3 , e ) 12.2 | b | divide(add(110, 132), multiply(72, const_0_2778)) | add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)| | physics |
how long does a train 110 m long running at the speed of 72 km / hr takes to cross a bridge 132 m length ? | "speed = ( 12 / 10 * 60 ) km / hr = ( 72 * 5 / 18 ) m / sec = 20 m / sec . length of the train = 20 * 6 = 120 m . answer : c" | a ) 298 , b ) 288 , c ) 120 , d ) 776 , e ) 991 | c | divide(const_100.0, subtract(divide(const_100.0, 10), 6)) | divide(const_100.0,n1)|subtract(#0,n2)|divide(const_100.0,#1)| | physics |
a train covers a distance of 12 km in 10 min . if it takes 6 sec to pass a telegraph post , then the length of the train is ? | "even i got 96 % j = 100 t = 100 * 0.4 = 40 m = 40 * 1.6 = 64 if mary ' s income is x percent of j m = j * x / 100 x = m * 100 / j = 64 * 100 / 100 = 64 ans : c" | a ) 124 % , b ) b . 120 % , c ) 64 % , d ) 80 % , e ) 64 % | c | add(subtract(const_100, 60), multiply(subtract(const_100, 60), divide(60, const_100))) | divide(n0,const_100)|subtract(const_100,n1)|multiply(#0,#1)|add(#2,#1)| | general |
mary ' s income is 60 % more than tim ' s income and tim ' s income is 60 % less than juan ' s income . what % of juan ' s income is mary ' s income . | "no . of ways of picking 2 biology books ( from 11 books ) = 11 c 2 = ( 11 * 10 ) / 2 = 55 no . of ways of picking 2 chemistry books ( from 8 books ) = 8 c 2 = ( 8 * 7 ) / 2 = 28 total ways of picking 2 books of each type = 55 * 28 = 1540 ( option e )" | a ) 80 , b ) 160 , c ) 720 , d ) 1100 , e ) 1540 | e | multiply(divide(divide(factorial(11), factorial(subtract(11, 2))), 2), divide(divide(factorial(8), factorial(subtract(8, 2))), 2)) | factorial(n0)|factorial(n1)|subtract(n0,n2)|subtract(n1,n2)|factorial(#2)|factorial(#3)|divide(#0,#4)|divide(#1,#5)|divide(#6,n2)|divide(#7,n2)|multiply(#8,#9)| | other |
11 different biology books and 8 different chemistry books lie on a shelf . in how many ways can a student pick 2 books of each type ? | pipe a fills 1 / 30 th of the tank in a minute and pipe b empties 1 / 90 th of the tank ( 1 / 30 ) - ( 1 / 90 ) = ( 1 / x ) 2 / 90 = 1 / x = > x = 45 answer : d | a ) 30 , b ) 35 , c ) 40 , d ) 45 , e ) 50 | d | divide(const_1, subtract(divide(const_1, 30), divide(const_1, 90))) | divide(const_1,n0)|divide(const_1,n1)|subtract(#0,#1)|divide(const_1,#2) | physics |