Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
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how many different pairs of positive integers ( a , b ) satisfy the equation 1 / a + 1 / b = 34 / 51 ? | a + b / ab = 2 / 3 3 = 1 * 3 or 3 * 1 a + b in both case = 4 correct 0 answer : e | a ) 6 , b ) 3 , c ) 2 , d ) 1 , e ) 0 | e | divide(34, 51) | divide(n2,n3) | general |
a telephone company needs to create a set of 3 - digit area codes . the company is entitled to use only digits 2 , 4 and 8 , which can be repeated . if the product of the digits in the area code must be even , how many different codes can be created ? | "total # of codes possible is 3 * 3 * 3 = 27 . oit of those 27 codes answer : d ." | a ) 20 , b ) 22 , c ) 24 , d ) 27 , e ) 30 | d | subtract(power(3, 3), const_1) | power(n0,n0)|subtract(#0,const_1)| | general |
the average age of one class strength 32 students is 16 yrs . if the staff ' s age is also included , the average increases by one year . find the age of the staff | total age of students = > 32 × 16 = 512 yrs total age inclusive staff = 33 × ( 16 + 1 ) = 561 = > staff ' s age 561 - 512 = 49 yrs answer c | a ) 50 , b ) 52 , c ) 49 , d ) 32 , e ) 35 | c | subtract(multiply(add(32, const_1), add(16, const_1)), multiply(32, 16)) | add(n0,const_1)|add(n1,const_1)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2) | general |
the area of a rectangular plot is 460 square metres . if the length is 15 % more than the breadth , what is the breadth of the plot ? | "lb = 460 m 2 let breath = b l = b * ( 100 + 15 ) / 100 = 115 b / 100 from these 115 b / 100 * b = 460 b 2 = 46000 / 115 = 400 b = root of 400 = 20 m answer b" | a ) 14 m , b ) 20 m , c ) 18 m , d ) 12 m , e ) 15 m | b | sqrt(divide(460, subtract(const_1, divide(15, const_100)))) | divide(n1,const_100)|subtract(const_1,#0)|divide(n0,#1)|sqrt(#2)| | geometry |
a man purchased 5 blankets @ rs . 100 each , 5 blankets @ rs . 150 each and two blankets at a certain rate which is now slipped off from his memory . but he remembers that the average price of the blankets was rs . 150 . find the unknown rate of two blankets ? | "explanation : 10 * 150 = 1500 5 * 100 + 5 * 150 = 1250 1500 – 1250 = 250 a" | a ) a ) 250 , b ) b ) 350 , c ) c ) 450 , d ) d ) 470 , e ) e ) 500 | a | subtract(multiply(const_10, 150), add(multiply(5, 100), multiply(5, 150))) | multiply(n3,const_10)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|subtract(#0,#3)| | general |
working simultaneously and independently at an identical constant rate , 5 machines of a certain type can produce a total of x units of product p in 8 days . how many of these machines , working simultaneously and independently at this constant rate , can produce a total of 3 x units of product p in 4 days ? | 5 machines do x units in 8 days we have x / 8 = > rate of the 5 machines we know that we need to have 3 x units in 4 days therefore , we need to get to 3 x / 4 rate of the machines . rate of one machine is x / 8 * 1 / 5 = x / 40 . now , we need to know how many machines need to work simultaneously , to get 3 x done in ... | a ) 30 , b ) 18 , c ) 16 , d ) 12 , e ) 8 | a | multiply(multiply(5, 3), divide(8, 4)) | divide(n1,n3)|multiply(n0,n2)|multiply(#0,#1) | general |
in a rectangular axis system , what is the area of a parallelogram with the coordinates : ( 2,2 ) , ( 8,2 ) , ( 3,5 ) , ( 9,5 ) ? | "delta x will give us the dimension of one side of the parallelogram = 8 - 2 = 6 unit delta y will give us the dimension of the other side of parallelogram = 5 - 2 = 3 unit area of parallelogram = 6 * 3 = 18 answer is a" | a ) 18 . , b ) 28 . , c ) 35 . , d ) 49 . , e ) 52 . | a | add(const_3, const_2) | add(const_2,const_3)| | geometry |
on a saturday night , each of the rooms at a certain motel was rented for either $ 40 or $ 60 . if 10 of the rooms that were rented for $ 60 had instead been rented for $ 40 , then the total rent the motel charged for that night would have been reduced by 10 percent . what was the total rent the motel actually charged ... | "let total rent the motel charge for all rooms = x if 10 rooms that were rented for 60 $ had instead been rented for 40 $ , then total difference in prices = 20 $ * 10 = 200 $ total rent the motel charged would have been reduced by 10 % . 10 x = 200 = > x = 2000 answer e" | a ) $ 600 , b ) $ 800 , c ) $ 1,000 , d ) $ 1,600 , e ) $ 2,000 | e | divide(multiply(10, subtract(60, 40)), divide(10, const_100)) | divide(n5,const_100)|subtract(n1,n0)|multiply(n2,#1)|divide(#2,#0)| | gain |
there is 60 lit of milk and water in which milk forms 84 % . howmuch water must be added to this solution to make it solution in which milk forms 50 % | "60 * 84 / 100 = 50.40 lit milk that is 9.60 lit water let x lit water will be added then ( 60 + x ) * 50 / 100 = 50.40 so x = 40.8 answer : a" | a ) 40.8 , b ) 19.75 , c ) 20.75 , d ) 21.75 , e ) 22.75 | a | subtract(multiply(divide(const_100, 50), divide(multiply(60, 84), const_100)), 60) | divide(const_100,n2)|multiply(n0,n1)|divide(#1,const_100)|multiply(#0,#2)|subtract(#3,n0)| | gain |
jill invests $ 10000 in an account that pays an annual rate of 3.96 % , compounding semi - annually . approximately how much q does she have in her account after two years ? | "ps . i guess one can use simple interest to solve cause the answer choices are quite spread between you can easily arrive at something near 8 % hence b the answer" | a ) $ 10079.44 , b ) q = $ 10815.83 , c ) $ 12652.61 , d ) $ 14232.14 , e ) $ 20598.11 | b | multiply(10000, power(add(const_1, divide(divide(3.96, const_100), const_2)), const_4)) | divide(n1,const_100)|divide(#0,const_2)|add(#1,const_1)|power(#2,const_4)|multiply(n0,#3)| | gain |
machine p and machine q are each used to manufacture 440 sprockets . it takes machine p 10 hours longer to produce 440 sprockets than machine q . machine q produces 10 % more sprockets per hour than machine a . how many sprockets per hour does machine a produce ? | "p makes x sprockets per hour . then q makes 1.1 x sprockets per hour . 440 / x = 440 / 1.1 x + 10 1.1 ( 440 ) = 440 + 11 x 11 x = 44 x = 4 the answer is b ." | a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | b | divide(subtract(440, divide(440, add(divide(10, const_100), const_1))), 10) | divide(n1,const_100)|add(#0,const_1)|divide(n0,#1)|subtract(n0,#2)|divide(#3,n1)| | gain |
a retailer buys 100 pens at the market price of 36 pens from a wholesaler , if he sells these pens giving a discount of 1 % , what is the profit % ? | "let the market price of each pen be $ 1 then , cost price of 100 pens = $ 36 selling price of 100 pens = 99 % of $ 100 = $ 99 profit % = ( ( 63 * 100 ) / 36 ) % = 175 % answer b" | a ) 110 , b ) 175 , c ) 212 , d ) 218 , e ) 220 | b | multiply(const_100, divide(subtract(subtract(1, divide(1, const_100)), divide(36, 100)), divide(36, 100))) | divide(n2,const_100)|divide(n1,n0)|subtract(n2,#0)|subtract(#2,#1)|divide(#3,#1)|multiply(#4,const_100)| | gain |
3 litre of water is added to 11 litre of a solution containing 42 % of alcohol in the water . the percentage of alcohol in the new mixture is ? | we have a 11 litre solution containing 42 % of alcohol in the water . = > quantity of alcohol in the solution = 11 × 42 100 = 11 × 42100 now 3 litre of water is added to the solution . = > total quantity of the new solution = 11 + 3 = 14 percentage of alcohol in the new solution = 11 × 42 100 14 × 100 = 11 × 4210014 × ... | a ) 30 % , b ) 33 % , c ) 35 % , d ) 40 % , e ) 45 % | b | multiply(divide(divide(multiply(11, 42), const_100), add(3, 11)), const_100) | add(n0,n1)|multiply(n1,n2)|divide(#1,const_100)|divide(#2,#0)|multiply(#3,const_100) | general |
from a pack of 52 cards , 1 card is drawn at random . what is the probability that a red card is drawn ? | "the total number of cards is 52 . the number of red cards is 26 . p ( red card ) = 26 / 52 = 1 / 2 the answer is a ." | a ) 1 / 2 , b ) 1 / 4 , c ) 1 / 26 , d ) 1 / 13 , e ) 5 / 13 | a | divide(multiply(const_4, const_3), 52) | multiply(const_3,const_4)|divide(#0,n0)| | probability |
a trader sells 85 meters of cloth for rs . 8925 at the profit of rs . 10 per metre of cloth . what is the cost price of one metre of cloth ? | "sp of 1 m of cloth = 8925 / 85 = rs . 105 cp of 1 m of cloth = sp of 1 m of cloth - profit on 1 m of cloth = rs . 105 - rs . 10 = rs . 95 answer : b" | a ) 26 , b ) 95 , c ) 90 , d ) 42 , e ) 22 | b | subtract(divide(8925, 85), 10) | divide(n1,n0)|subtract(#0,n2)| | physics |
a trailer carries 3 , 4 and 5 crates on a trip . each crate weighs no less than 120 kg . what is the maximum weight of the crates on a single trip ? | max no . of crates = 5 . max weight = 120 kg max . weight carried = 5 * 120 = 600 kg = c | a ) 1250 , b ) 625 , c ) 600 , d ) 7500 , e ) 375 | c | multiply(5, 120) | multiply(n2,n3) | general |
a student completes a 3 part exam as follows . the student completes sections a in 24 minutes and takes an equal time to complete the two remaining sections . what is the ratio of time spent on section c to section a if the total duration is 1 hour ? | the time spent on section b and section c is 18 minutes each . the ratio of c to a is 18 : 24 = 3 : 4 the answer is e . | a ) 7 : 8 , b ) 6 : 7 , c ) 5 : 6 , d ) 4 : 5 , e ) 3 : 4 | e | divide(const_3, const_4) | divide(const_3,const_4) | physics |
the average of first 3 prime numbers greater than 20 is ? | 23 + 29 + 31 = 83 / 3 = 27.7 answer : d | a ) 10 , b ) 20 , c ) 30 , d ) 27.7 , e ) 50 | d | divide(add(add(add(20, const_3), add(add(add(20, const_3), const_4), const_2)), add(add(add(add(20, const_3), const_4), const_2), const_2)), 3) | add(n1,const_3)|add(#0,const_4)|add(#1,const_2)|add(#0,#2)|add(#2,const_2)|add(#3,#4)|divide(#5,n0) | general |
a reduction of 44 % in the price of bananas would enable a man to obtain 64 more for rs . 40 , what is reduced price per dozen ? | "explanation : 40 * ( 44 / 100 ) = 17.6 - - - 64 ? - - - 12 = > rs . 3.30 answer : d" | a ) 2.3 , b ) 8.3 , c ) 7.3 , d ) 3.3 , e ) 1.3 | d | multiply(const_12, divide(multiply(44, divide(44, const_100)), 64)) | divide(n0,const_100)|multiply(n0,#0)|divide(#1,n1)|multiply(#2,const_12)| | gain |
if x and y are integers , what is the least positive number of 24 x + 16 y ? | "24 x + 16 y = 8 ( 3 x + 2 y ) which will be a minimum positive number when 3 x + 2 y = 1 . 3 ( 1 ) + 2 ( - 1 ) = 1 then 8 ( 3 x + 2 y ) can have a minimum positive value of 8 . the answer is e ." | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 8 | e | subtract(24, 16) | subtract(n0,n1)| | general |
a certain company retirement plan has arule of 70 provision that allows an employee to retire when the employee ' s age plus years of employment with the company total at least 70 . in what year could a female employee hired in 1987 on her 32 nd birthday first be eligible to retire under this provision ? | "she must gain at least 70 points , now she has 32 and every year gives her two more points : one for age and one for additional year of employment , so 32 + 2 * ( # of years ) = 70 - - > ( # of years ) = 19 - - > 1987 + 19 = 2006 . answer : d ." | a ) 2003 , b ) 2004 , c ) 2005 , d ) 2006 , e ) 2007 | d | add(1987, divide(subtract(70, 32), const_2)) | subtract(n0,n3)|divide(#0,const_2)|add(n2,#1)| | general |
a cistern is filled by a tap in 3 1 / 2 hours . due to leak in the bottom of the cistern , it takes half an hour longer to fill the cistern . if the cistern is full how long will it take the leak to empty it ? | 2 / 7 - 1 / x = 1 / 4 x = 28 answer : d | a ) 11 , b ) 12 , c ) 29 , d ) 28 , e ) 29 | d | divide(const_1, subtract(divide(const_1, add(3, divide(1, 2))), divide(1, const_4))) | divide(n1,n2)|divide(n1,const_4)|add(n0,#0)|divide(const_1,#2)|subtract(#3,#1)|divide(const_1,#4) | physics |
in the coordinate plane , points ( x , 1 ) and ( 10 , y ) are on line k . if line k passes through the origin and has slope 1 / 3 , then x + y = | "line k passes through the origin and has slope 1 / 3 means that its equation is y = 1 / 3 * x . thus : ( x , 1 ) = ( 3 , 1 ) and ( 10 , y ) = ( 10,10 / 3 ) - - > x + y = 3 + 10 / 3 = 19 / 3 . answer : d ." | a ) 4.5 , b ) 7 , c ) 8 , d ) 19 / 3 , e ) 12 | d | multiply(multiply(10, 3), divide(1, 3)) | divide(n0,n3)|multiply(n1,n3)|multiply(#0,#1)| | general |
a 40 gallon solution of salt and water is 10 % salt . how many gallons of water must be added to the solution in order to decrease the salt to 8 % of the volume ? | "amount of salt = 4.0 assume x gallons of water are added . 4.0 / 40 + x = 8 / 100 400 = 8 x + 320 8 x = 80 x = 10 correct option : a" | a ) 810 , b ) 12 , c ) 13 , d ) 14 , e ) 16 | a | divide(multiply(40, subtract(10, 8)), 8) | subtract(n1,n2)|multiply(n0,#0)|divide(#1,n2)| | general |
if 10 boys meet at a reunion and each boy shakes hands exactly once with each of the others , then what is the total number of handshakes | n ( n - 1 ) / 2 = 10 * 9 / 2 = 45 answer : e | a ) 41 , b ) 42 , c ) 43 , d ) 44 , e ) 45 | e | divide(factorial(10), multiply(factorial(subtract(10, const_2)), factorial(const_2))) | factorial(n0)|factorial(const_2)|subtract(n0,const_2)|factorial(#2)|multiply(#3,#1)|divide(#0,#4) | general |
6 x – 5 y + 3 z = 22 4 x + 8 y – 11 z = 7 5 x – 6 y + 2 z = 12 given the equations above , x + y + z = ? | ( 6 x – 5 y + 3 z ) - ( 5 x – 6 y + 2 z ) = 22 - 12 or , x + y + z = 10 option a is the ans | a ) 10 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | a | multiply(5, const_2) | multiply(n1,const_2) | general |
the output of a factory is increased by 10 % to keep up with rising demand . to handle the holiday rush , this new output is increased by 60 % . by approximately what percent would the output of the factory now have to be decreased in order to restore the original output ? | "take it as original output = 100 . to meet demand increase by 10 % , then output = 110 . to meet holiday demand , new output increase by 60 % then output equals 176 to restore new holidy demand output to original 100 . final - initial / final * 100 = 76 / 176 * 100 = 8 / 33 * 100 = 43 % approxiamately . option d is co... | a ) 20 % , b ) 24 % , c ) 30 % , d ) 43 % , e ) 79 % | d | multiply(divide(subtract(multiply(divide(add(const_100, 10), const_100), divide(add(const_100, 60), const_100)), const_1), multiply(divide(add(const_100, 10), const_100), divide(add(const_100, 60), const_100))), const_100) | add(n0,const_100)|add(n1,const_100)|divide(#0,const_100)|divide(#1,const_100)|multiply(#2,#3)|subtract(#4,const_1)|divide(#5,#4)|multiply(#6,const_100)| | general |
car x began traveling at an average speed of 35 miles per hour . after 72 minutes , car y began traveling at an average speed of 38 miles per hour . when both cars had traveled the same distance , both cars stopped . how many miles did car x travel from the time car y began traveling until both cars stopped ? | "car y began travelling after 72 minutes or 1.2 hours . let t be the time for which car y travelled before it stopped . both cars stop when they have travelled the same distance . so , 35 ( t + 1.2 ) = 38 t t = 14 distance traveled by car x from the time car y began traveling until both cars stopped is 35 x 14 = 490 mi... | a ) 405 , b ) 420 , c ) 440 , d ) 447 , e ) 490 | e | multiply(35, divide(multiply(divide(72, const_60), 35), subtract(38, 35))) | divide(n1,const_60)|subtract(n2,n0)|multiply(n0,#0)|divide(#2,#1)|multiply(n0,#3)| | physics |
a certain car ' s price decreased by 2.5 % ( from the original price ) each year from 1996 to 2002 , during that time the owner of the car invested in a new carburetor and a new audio system for the car , which increased car ' s price by $ 2,500 . if the price of the car in 1996 was $ 22,000 , what is the car ' s price... | "price in 96 = 22000 price decrease each year = 2.5 / 100 * 22000 = 550 price in 97 = 22000 - 550 price in 98 = 22000 - 2 * 550 price in 99 = 22000 - 3 * 550 price in 00 = 22000 - 4 * 550 price in 01 = 22000 - 5 * 550 price in 02 = 22000 - 6 * 550 = 18700 investment in the car = 2500 net price of the car in 02 = 18700 ... | a ) $ 18,400 , b ) $ 19,500 , c ) $ 20,200 , d ) $ 20,400 , e ) $ 21,200 | e | multiply(const_2, const_10) | multiply(const_10,const_2)| | gain |
jerry travels 8 miles at an average speed of 40 miles per hour , stops for 15 minutes , and then travels another 20 miles at an average speed of 60 miles per hour . what is jerry ’ s average speed , in miles per hour , for this trip ? | total time taken by jerry = ( 8 / 40 ) * 60 minutes + 15 minutes + ( 20 / 60 ) * 60 minutes = 47 minutes average speed = total distance / total time = ( 8 + 20 ) miles / ( 47 / 60 ) hours = 28 * 60 / 47 = 35 miles per hour answer : option a | a ) 35 , b ) 42.5 , c ) 44 , d ) 50 , e ) 52.5 | a | divide(add(8, 20), add(add(divide(8, 40), divide(15, const_60)), divide(20, 60))) | add(n0,n3)|divide(n0,n1)|divide(n2,const_60)|divide(n3,n4)|add(#1,#2)|add(#4,#3)|divide(#0,#5) | physics |
john makes $ 40 a week from his job . he earns a raise and now makes $ 60 a week . what is the % increase ? | "increase = ( 20 / 40 ) * 100 = 50 % . e" | a ) 16 % , b ) 16.66 % , c ) 18 % , d ) 21 % , e ) 50 % | e | multiply(divide(subtract(60, 40), 40), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
- 69 * 39 + 450 = ? | "= > - 69 * ( 40 - 1 ) + 450 ; = > - ( 69 * 40 ) + 69 + 450 ; = > - 2760 + 519 = - 2241 . correct option : d" | a ) 2736 , b ) 2309 , c ) - 2801 , d ) - 2241 , e ) none of these | d | add(multiply(negate(69), 39), 450) | negate(n0)|multiply(n1,#0)|add(n2,#1)| | general |
when n is divided by 24 , the remainder is 4 . what is the remainder when 3 n is divided by 8 ? | "let n = 4 ( leaves a remainder of 4 when divided by 24 ) 3 n = 3 ( 4 ) = 12 , which leaves a remainder of 4 when divided by 8 . answer b" | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | b | subtract(4, reminder(3, 8)) | reminder(n2,n3)|subtract(n1,#0)| | general |
two pipes a and b can separately fill a cistern in 45 minutes and 60 minutes respectively . there is a third pipe in the bottom of the cistern to empty it . if all the three pipes are simultaneously opened , then the cistern is full in 40 minutes . in how much time , the third pipe alone can empty the cistern ? | "1 / 40 - ( 1 / 45 + 1 / 60 ) = - 1 / 72 third pipe can empty in 72 minutes answer : c" | a ) 90 min , b ) 100 min , c ) 72 min , d ) 75 min , e ) 130 min | c | inverse(subtract(add(inverse(45), inverse(60)), inverse(40))) | inverse(n0)|inverse(n1)|inverse(n2)|add(#0,#1)|subtract(#3,#2)|inverse(#4)| | physics |
a car takes 6 hours to cover a distance of 450 km . how much should the speed in kmph be maintained to cover the same direction in 3 / 2 th of the previous time ? | "time = 6 distance = 450 3 / 2 of 6 hours = 6 * 3 / 2 = 9 hours required speed = 450 / 9 = 50 kmph answer b ." | a ) 60 , b ) 50 , c ) 40 , d ) 70 , e ) 65 | b | divide(450, divide(multiply(6, 3), 2)) | multiply(n0,n2)|divide(#0,n3)|divide(n1,#1)| | physics |
a letter lock consists of 3 rings each marked with 6 different letters . the number of distinct unsuccessful attempts to open the lock is at the most - . | since each ring consists of six different letters , the total number of attempts possible with the three rings is = 6 * 6 * 6 = 216 . of these attempts , one of them is a successful attempt . maximum number of unsuccessful attempts = 216 - 1 = 215 . answer : c | a ) 216 , b ) 243 , c ) 215 , d ) 729 , e ) 728 | c | subtract(multiply(multiply(6, 6), 6), const_1) | multiply(n1,n1)|multiply(n1,#0)|subtract(#1,const_1) | probability |
the ratio between the length and the breadth of a rectangular park is 3 : 2 . if a man cycling along the boundary of the park at the speed of 12 km / hr completes one round in 10 minutes , then the area of the park ( in sq . m ) is | "perimeter = distance covered in 10 min . = 12000 x 10 m = 2000 m . 60 let length = 3 x metres and breadth = 2 x metres . then , 2 ( 3 x + 2 x ) = 2000 or x = 200 . length = 600 m and breadth = 400 m . area = ( 600 x 400 ) m 2 = 240000 m 2 . b" | a ) 153200 , b ) 240000 , c ) 153600 , d ) 154000 , e ) 154200 | b | rectangle_area(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 10), const_1000), add(3, 2)), const_2), multiply(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 10), const_1000), add(3, 2)), const_2), 2)) | 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 |
average of money that group of 4 friends pay for rent each month is $ 800 . after one persons rent is increased by 16 % the new mean is $ 850 . what was original rent of friend whose rent is increased ? | "0.16 x = 4 ( 850 - 800 ) 0.16 x = 200 x = 1250 answer e" | a ) 800 , b ) 900 , c ) 1000 , d ) 1100 , e ) 1250 | e | divide(multiply(subtract(850, 800), 4), divide(16, const_100)) | divide(n2,const_100)|subtract(n3,n1)|multiply(n0,#1)|divide(#2,#0)| | general |
a person started from point a to point b and returned back to a with a speed of 60 kmh . if the distance between a and b is 120 km , calculate the time he travelled | time = distance / speed total distance = ( from a - > b and from b - > a ) = 120 + 120 = 240 km speed = 60 kmh 240 / 60 = 4 answer : a | a ) 4 hours , b ) 2 hours , c ) 3 hours , d ) 6 hours , e ) 5 hours | a | divide(multiply(120, const_2), 60) | multiply(n1,const_2)|divide(#0,n0) | physics |
a meeting has to be conducted with 5 managers . find the number of ways in which the managers may be selected from among 7 managers , if there are 2 managers who refuse to attend the meeting together . | "the total number of ways to choose 5 managers is 7 c 5 = 21 we need to subtract the number of groups which include the two managers , which is 5 c 3 = 10 . 21 - 10 = 11 the answer is d ." | a ) 6 , b ) 8 , c ) 10 , d ) 11 , e ) 15 | d | subtract(choose(7, 5), choose(subtract(7, 2), 2)) | choose(n1,n0)|subtract(n1,n2)|choose(#1,n2)|subtract(#0,#2)| | probability |
ravi and sunil are partners in a business . ravi invests rs . 15,000 for 8 months and sunil invested rs . 8000 for 10 months then after one year ratio of their profits will be | "= ( 15000 * 8 ) : ( 8000 * 10 ) = 120000 : 80000 = 3 : 2 answer : c" | a ) 1 : 3 , b ) 2 : 3 , c ) 3 : 2 , d ) 3 : 1 , e ) 3 : 4 | c | divide(multiply(multiply(multiply(multiply(const_4, const_3), multiply(10, const_2)), const_100), const_4.0), multiply(8000, 10)) | multiply(const_3,const_4)|multiply(n3,const_2)|multiply(n2,n3)|multiply(#0,#1)|multiply(#3,const_100)|multiply(const_4.0,#4)|divide(#5,#2)| | gain |
on a saturday night , each of the rooms at a certain motel was rented for either $ 50 or $ 60 . if 10 of the rooms that were rented for $ 60 had instead been rented for $ 50 , then the total rent the motel charged for that night would have been reduced by 25 percent . what was the total rent the motel actually charged ... | let total rent the motel charge for all rooms = x if 10 rooms that were rented for 60 $ had instead been rented for 50 $ , then total difference in prices = 10 $ * 10 = 100 $ total rent the motel charged would have been reduced by 25 % . 25 x = 100 = > x = 400 answer a | a ) $ 400 , b ) $ 800 , c ) $ 1,000 , d ) $ 1,600 , e ) $ 2,400 | a | divide(multiply(10, subtract(60, 50)), divide(25, const_100)) | divide(n5,const_100)|subtract(n1,n0)|multiply(n2,#1)|divide(#2,#0) | gain |
if price of t . v set is reduced by 10 % , then its sale increases by 85 % , find net effect on sale value | - a + b + ( ( - a ) ( b ) / 100 ) = - 10 + 85 + ( - 10 * 85 ) / 100 = - 10 + 85 - 8.5 = 67 answer : d | a ) 44 , b ) 45 , c ) 46 , d ) 67 , e ) 48 | d | multiply(subtract(multiply(divide(subtract(const_100, 10), const_100), divide(add(const_100, 85), const_100)), const_1), const_100) | add(n1,const_100)|subtract(const_100,n0)|divide(#1,const_100)|divide(#0,const_100)|multiply(#2,#3)|subtract(#4,const_1)|multiply(#5,const_100) | gain |
a retailer buys a radio for rs 225 . his overhead expenses are rs 20 . he sellis the radio for rs 300 . the profit percent of the retailer is | "explanation : cost price = ( 225 + 20 ) = 245 sell price = 300 gain = ( 55 / 245 ) * 100 = 22.4 % . answer : a" | a ) 22.4 % , b ) 50 % , c ) 25 % , d ) 52 % , e ) none of these | a | subtract(multiply(divide(300, add(225, 20)), const_100), const_100) | add(n0,n1)|divide(n2,#0)|multiply(#1,const_100)|subtract(#2,const_100)| | gain |
which number should replace both the asterisks in ( * / 18 ) x ( * / 162 ) = 1 ? | let ( y / 18 ) x ( y / 162 ) = 1 y ^ 2 = 18 x 162 = 18 x 18 x 9 y = ( 18 x 3 ) = 54 the answer is c . | a ) 27 , b ) 36 , c ) 54 , d ) 72 , e ) 90 | c | sqrt(multiply(162, 18)) | multiply(n0,n1)|sqrt(#0) | general |
anne bought doughnuts for a class breakfast party . she bought 5 chocolate doughnuts , 6 coconut doughnuts , and 11 jam - filled doughnuts . how many doughnuts did anne buy in all ? | "add the numbers of doughnuts . 5 + 6 + 11 = 22 . answer is a ." | a ) 22 , b ) 36 , c ) 39 , d ) 21 , e ) 11 | a | add(add(5, 6), 11) | add(n0,n1)|add(n2,#0)| | general |
a certain music store stocks 800 cellos and 600 violas . of these instruments , there are 120 cello - viola pairs , such that a cello and a viola were both made with wood from the same tree ( each tree can make at most one viola and one cello , so there are no pairs other than these 90 ) . if one viola and one cello ar... | "solution provided by stanford 2012 is correct : 120 / 800 choosing one of the cellos which has a pair viola , 1 / 600 choosing the viola which is the pair of chosen cello - - > p = 120 / 800 * 1 / 600 = 1 / 8,000 . answer : b ." | a ) 3 / 16,000 , b ) 1 / 8,000 , c ) 3 / 1,600 , d ) 1 / 90 , e ) 2 / 45 | b | multiply(divide(120, 800), divide(const_1, 600)) | divide(n2,n0)|divide(const_1,n1)|multiply(#0,#1)| | other |
a man is walking at the rate of 10 km / hr crosses a bridge in 15 minutes . the length of the bridge is | "explanation : we need to get the answer in meters . so we will first of change distance from km / hour to meter / sec by multiplying it with 5 / 18 and also change 15 minutes to seconds by multiplying it with 60 . speed = 10 â ˆ — 5 / 18 = 25 / 9 m / sec time = 15 â ˆ — 60 seconds = 900 seconds distance = time â ˆ — s... | a ) 1000 meters , b ) 1050 meters , c ) 1200 meters , d ) 1250 meters , e ) 2500 meters | e | multiply(multiply(10, divide(15, const_60)), const_1000) | divide(n1,const_60)|multiply(n0,#0)|multiply(#1,const_1000)| | gain |
826924 × 9999 = ? | "b 8268413076 826924 × 9999 = 826924 × ( 10000 - 1 ) = 826924 × 10000 - 826924 × 1 = 8269240000 - 826924 = 8268413076" | a ) 8248402936 , b ) 8268413076 , c ) 8268413071 , d ) 8064329671 , e ) 8244563371 | b | multiply(divide(826924, 9999), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
the megatek corporation is displaying its distribution of employees by department in a circle graph . the size of each sector of the graph representing a department is proportional to the percentage of total employees in that department . if the section of the circle graph representing the manufacturing department take... | answer : d 144 ° divided by 360 ° equals 0.4 , therefore the sector is equal to 40 % of the total | a ) 20 % , b ) 25 % , c ) 30 % , d ) 40 % , e ) 72 % | d | multiply(divide(144, divide(const_3600, const_10)), const_100) | divide(const_3600,const_10)|divide(n0,#0)|multiply(#1,const_100)| | physics |
a train speeds past a pole in 15 seconds and a platfrom 100 m long in 25 seconds . its length is : | "sol . let the length of the train be x metres and its speed be y m / sec . they , x / y = 15 ⇒ y = x / 15 ∴ x + 100 / 25 = x / 15 ⇔ x = 150 m . answer d" | a ) 100 m , b ) 125 m , c ) 130 m , d ) 150 m , e ) none | d | multiply(100, subtract(const_2, const_1)) | subtract(const_2,const_1)|multiply(n1,#0)| | physics |
gopi gives rs . 90 plus one turban as salary to his servant for one year . the servant leaves after 9 months and receives rs . 55 and the turban . find the price of the turban . | let the price of turban be x . thus , for one year the salary = ( 90 + x ) for 9 months he should earn 3434 ( 90 + x ) . now he gets one turban and rs . 55 . thus , 3434 ( 90 + x ) = 55 + x or 270 + 3 x = 220 + 4 x or x = 50 answer : d | a ) 27 , b ) 36 , c ) 29 , d ) 50 , e ) 11 | d | subtract(multiply(divide(subtract(90, 55), subtract(const_12, 9)), const_12), 90) | subtract(n0,n2)|subtract(const_12,n1)|divide(#0,#1)|multiply(#2,const_12)|subtract(#3,n0) | general |
in a ratio which is equal to 2 : 4 , if the antecedent is 50 , then the consequent is ? | "we have 2 / 4 = 50 / x 2 x = 50 * 4 x = 100 consequent = 100 answer is c" | a ) 72 , b ) 85 , c ) 100 , d ) 51 , e ) 80 | c | add(50, divide(50, const_2)) | divide(n2,const_2)|add(n2,#0)| | other |
before leaving home for the town of madison , pete checks a map which shows that madison is 5 inches from his current location , gardensquare . pete arrives in madison 1.5 hours later and drove at an average speed of 60 miles per hour . at what scale , in inches per mile , is the map drawn ? | "pete covered 1.5 * 60 = 90 miles which correspond to 5 inches on the map - - > scale in inches per mile is 5 / 90 = 1 / 18 . answer : b ." | a ) 1 / 3 , b ) 1 / 18 , c ) 1 / 10 , d ) 2 , e ) 30 | b | divide(const_1, multiply(divide(1.5, 5), 60)) | divide(n1,n0)|multiply(#0,n2)|divide(const_1,#1)| | physics |
the average of 50 numbers is 38 . if two numbers 45 and 55 are discarded , the average of the remaining set of numbers is | average of the remaining set of numbers [ 50 x 38 - ( 45 + 55 ) ] / ( 50 - 2 ) = ( 1900 - 100 ) / 48 = 37.5 answer : b | a ) 38.5 , b ) 37.5 , c ) 37.0 , d ) 36.5 , e ) none of these | b | divide(subtract(subtract(multiply(50, 38), 45), 55), subtract(50, const_2)) | multiply(n0,n1)|subtract(n0,const_2)|subtract(#0,n2)|subtract(#2,n3)|divide(#3,#1) | general |
i bought two books ; for rs . 420 . i sold one at a loss of 15 % and other at a gain of 19 % and then i found each book was sold at the same price . find the cost of the book sold at a loss ? | "x * ( 85 / 100 ) = ( 420 - x ) 119 / 100 x = 245 answer : b" | a ) 197 , b ) 245 , c ) 189 , d ) 278 , e ) 268 | b | divide(multiply(420, add(const_100, 19)), add(subtract(const_100, 15), add(const_100, 19))) | add(n2,const_100)|subtract(const_100,n1)|add(#0,#1)|multiply(n0,#0)|divide(#3,#2)| | gain |
the ratio of the ages of mini and minakshi is 4 : 3 . the sum of their ages is 28 years . the ratio of their ages after 6 years will be | "let mini ’ s age = 4 x and minakshi ’ s age = 3 x then 4 x + 3 x = 28 x = 4 mini ’ s age = 16 years and minakshi ’ s age = 12 years ratio of their ages after 8 years = ( 16 + 6 ) : ( 12 + 6 ) = 22 : 18 = 11 : 9 answer : e" | a ) 2 : 3 , b ) 7 : 11 , c ) 5 : 4 , d ) 4 : 5 , e ) 11 : 9 | e | divide(add(multiply(4, divide(28, add(4, 3))), 6), add(multiply(3, divide(28, add(4, 3))), 6)) | add(n0,n1)|divide(n2,#0)|multiply(n0,#1)|multiply(n1,#1)|add(n3,#2)|add(n3,#3)|divide(#4,#5)| | general |
50 men took a dip in a water tank 40 m long and 20 m broad on a religious day . if the average displacement of water by a man is 5 m 3 , then the rise in the water level in the tank will be : | "explanation : total volume of water displaced = ( 5 x 50 ) m 3 = 250 m 3 rise in water level = 250 / 40 ã — 20 = 0.3125 m = 31.25 cm answer : a" | a ) 31.25 cm , b ) 25 cm , c ) 35 cm , d ) 50 cm , e ) none of these | a | divide(multiply(5, 50), multiply(40, 20)) | multiply(n0,n3)|multiply(n1,n2)|divide(#0,#1)| | physics |
if the price of petrol increases by 20 , by how much must a user cut down his consumption so that his expenditure on petrol remains constant ? | "explanation : let us assume before increase the petrol will be rs . 100 . after increase it will be rs ( 100 + 20 ) i . e 120 . now , his consumption should be reduced to : - = ( 120 − 100 ) / 120 ∗ 100 . hence , the consumption should be reduced to 16.7 % . answer : a" | a ) 16.7 % , b ) 20 % , c ) 24.67 % , d ) 33.33 % , e ) none of these | a | multiply(subtract(const_1, divide(const_100, add(const_100, 20))), const_100) | add(n0,const_100)|divide(const_100,#0)|subtract(const_1,#1)|multiply(#2,const_100)| | general |
thirty percent of the women in a college class are science majors , and the non - science majors make up 60 % of the class . what percentage of the men are science majors if 40 % of the class are men ? | science majors make up 0.4 of the class . 60 % of the class are women and 0.3 * 0.6 = 0.18 of the class are female science majors . then 0.22 of the class are male science majors . 0.4 x = 0.22 x = 0.55 = 55 % the answer is c . | a ) 35 % , b ) 45 % , c ) 55 % , d ) 65 % , e ) 75 % | c | multiply(divide(subtract(subtract(const_100, 60), multiply(divide(multiply(const_10, const_3), const_100), subtract(const_100, 40))), 40), const_100) | multiply(const_10,const_3)|subtract(const_100,n0)|subtract(const_100,n1)|divide(#0,const_100)|multiply(#3,#2)|subtract(#1,#4)|divide(#5,n1)|multiply(#6,const_100) | gain |
a cargo ship carrying 4 kinds of items , doohickies , geegaws , widgets , and yamyams , arrives at the port . each item weighs 2 , 11 , 5 , and 7 pounds , respectively , and each item is weighed as it is unloaded . if , in the middle of the unloading process , the product of the individual weights of the unloaded items... | we need to know the number of widgets ( which weigh 5 pounds each ) . the number of times that 5 divides the number is related to the number of times that 10 divides the number . when we divide 104 , 350,400 by 100 , we get 104 , 350,400 = 1 , 043,504 * 5 ^ 2 * 2 ^ 2 . 1 , 043,504 is not divisible by 5 , thus there are... | a ) 2 , b ) 3 , c ) 4 , d ) 25 , e ) 20 , 870,080 | a | add(const_1, const_1) | add(const_1,const_1) | general |
what is the sum of all the composite numbers up to 18 ? | explanation : 4 + 6 + 8 + 9 + 10 + 12 + 14 + 15 + 16 + 18 = 112 answer : b | a ) 228 , b ) 112 , c ) 80 , d ) 96 , e ) 100 | b | 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)), 18), 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 |
if 10,000 microns = 1 decimeter , and 1 , 000,000 angstroms = 1 decimeter , how many angstroms equal 1 micron ? | "10,000 microns = 1 decimeter , and 1 , 000,000 angstroms = 1 decimeter 10,000 microns = 1 , 000,000 angstroms 1 micron = 100 answer : a" | a ) 100 , b ) 10,000 , c ) 100,000 , d ) 0.01 , e ) 0.0001 | a | multiply(divide(1, multiply(const_100, const_100)), multiply(const_100, const_100)) | multiply(const_100,const_100)|divide(n1,#0)|multiply(#1,#0)| | general |
this year , mbb consulting fired 5 % of its employees and left remaining employee salaries unchanged . sally , a first - year post - mba consultant , noticed that that the average ( arithmetic mean ) of employee salaries at mbb was 10 % more after the employee headcount reduction than before . the total salary pool all... | "100 employees , each earning 1 $ . after the cut , 95 are left , each earning 1.1 $ . 1.1 * 95 = 104.5 100 * 1 = 100 this gives you the answer . answer : c" | a ) 98.5 % , b ) 100.0 % , c ) 102.8 % , d ) 104.5 % , e ) 105.0 % | c | divide(multiply(add(const_100, multiply(const_100, 10)), add(subtract(const_100, 5), const_4)), multiply(const_100, 10)) | multiply(n1,const_100)|subtract(const_100,n0)|add(#0,const_100)|add(#1,const_4)|multiply(#2,#3)|divide(#4,#0)| | general |
during a certain two - week period , 75 percent of the movies rented from a video store were comedies , and of the remaining movies rented , there were 2 / 3 times as many dramas as action movies . if no other movies were rented during that two - week period and there were a action movies rented , then how many comedie... | "movies : 75 % comedies . 25 % remaining genre . now in this 25 % , there are only 2 categories . action movies and drama movies . if action = x ; drama movies = 2 x / 3 . total 5 x / 3 . 5 x / 3 = 25 ; x = 15 action movies : 15 % drama movies : 10 % we can say that out of 100 z , : comedies : 75 z action : 15 z drama ... | a ) 5 a , b ) 10 a , c ) 15 a , d ) 18 a , e ) 25 a | a | floor(divide(75, add(2, const_1))) | add(n1,const_1)|divide(n0,#0)|floor(#1)| | general |
rs 60000 is divided into two parts one part is given to a person with 10 % interest and another part is given to a person with 20 % interest . at the end of first year he gets profit 7000 find money given by 10 % ? | "let first parrt is x and second part is y then x + y = 60000 - - - - - - - - - - eq 1 total profit = profit on x + profit on y 7000 = ( x * 10 * 1 ) / 100 + ( y * 20 * 1 ) / 100 70000 = x + 2 y - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - eq 2 70000 = 60000 + y so y = 10000 then x = 50000 - 10... | a ) 30000 , b ) 40000 , c ) 50000 , d ) 60000 , e ) 70000 | b | divide(subtract(divide(multiply(60000, 20), const_100), 7000), divide(10, const_100)) | divide(n1,const_100)|multiply(n0,n2)|divide(#1,const_100)|subtract(#2,n3)|divide(#3,#0)| | gain |
in measuring the sides of a rectangle , one side is taken 5 % in excess , and the other 4 % in deficit . find the error percent in the area calculated from these measurements . | "let x and y be the sides of the rectangle . then , correct area = xy . calculated area = ( 105 / 100 ) * x * ( 96 / 100 ) * y = ( 504 / 500 ) ( xy ) error in measurement = ( 504 / 500 ) xy - xy = ( 4 / 500 ) xy error % = [ ( 4 / 500 ) xy * ( 1 / xy ) * 100 ] % = ( 4 / 5 ) % = 0.8 % . answer d 0.8 %" | a ) 0.2 % , b ) 0.4 % , c ) 0.6 % , d ) 0.8 % , e ) 1 % | d | subtract(subtract(5, 4), divide(multiply(5, 4), const_100)) | multiply(n0,n1)|subtract(n0,n1)|divide(#0,const_100)|subtract(#1,#2)| | geometry |
a box contains 100 balls , numbered from 1 to 100 . if 3 balls are selected at random and with replacement from the box . if the 3 numbers on the balls selected contain two odd and one even . what is the probability g that the first ball picked up is odd numbered ? | "answer - d selecting the balls either even or odd is having probability 50 / 100 = 1 / 2 we have already selected 3 balls with 2 odd numbers and 1 even number . so we have 3 combinations ooe , oeo , eoo . we have 3 outcomes and 2 are favourable as in 2 cases 1 st number is odd . so probability g is 2 / 3 . d" | a ) 0 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 1 | d | divide(const_2, 3) | divide(const_2,n3)| | probability |
a gardener wants to plant trees in his garden in such a way that the number of trees in each row should be the same . if there are 2 rows or 4 rows or 4 rows , then no tree will be left . find the least number of trees required | "explanation : the least number of trees that are required = lcm ( 2 , 4,4 ) = 4 . answer : c" | a ) 22 , b ) 60 , c ) 4 , d ) 8 , e ) 10 | c | divide(multiply(multiply(4, 4), 2), const_2) | multiply(n1,n2)|multiply(n0,#0)|divide(#1,const_2)| | general |
bert left the house with n dollars . he spent 1 / 4 of this at the hardware store , then $ 9 at the dry cleaners , and then half of what was left at the grocery store . when he got home , he had $ 15 left in his pocket . what was the value of n ? | started to test answer c if he had 52 , then he spent 13 at hardware store now he was left with 39 $ he spent 9 dollars on cleaning , thus he remained with 30 $ he then spent 1 / 2 of 30 , or 15 , and was left with 15 . hence , the only option that can be right is c . | a ) $ 36 , b ) $ 44 , c ) $ 52 , d ) $ 60 , e ) $ 68 | c | multiply(multiply(inverse(subtract(1, divide(1, 4))), add(divide(9, const_2), 15)), const_2) | divide(n2,const_2)|divide(n0,n1)|add(n3,#0)|subtract(n0,#1)|inverse(#3)|multiply(#2,#4)|multiply(#5,const_2) | general |
a company has $ 329864 in its account . what is the least amount of money ( in whole number of dollars ) that it must add to the account if the money is paid evenly among 9 of its vendors ? | to find the least amount that must be added to the account to split the money evenly among 9 of its vendors , the total divisible by 9 simply add the individual digits of the total = 3 + 2 + 9 + 8 + 6 + 4 = 32 if you add 4 , the number is divisible by 9 ( 32 + 4 ) correct option : d | a ) $ 1 , b ) $ 2 , c ) $ 3 , d ) $ 4 , e ) $ 6 | d | subtract(9, reminder(329864, 9)) | reminder(n0,n1)|subtract(n1,#0) | general |
an escalator moves towards the top level at the rate of 10 ft . sec and its length is 112 feet . if a person walks on the moving escalator at the rate of 4 feet per second towards the top level , how much time does he take to cover the entire length . | "time taken to cover the entire length = tot . dist / resultant speed = 112 / ( 10 + 4 ) = 8 sec answer : d" | a ) 14 sec , b ) 10 sec , c ) 12 sec , d ) 8 sec , e ) 9 sec | d | divide(112, add(10, 4)) | add(n0,n2)|divide(n1,#0)| | gain |
if - 2 / ( a - 6 ) = 4 / ( a + 4 ) , then a = ? | "multiply all terms of the given equation by ( a - 6 ) ( a + 4 ) , simplify and solve ( a - 6 ) ( a + 4 ) [ - 2 / ( a - 6 ) ] = ( a - 6 ) ( a + 4 ) [ 4 / ( a + 4 ) ] - 2 ( a + 4 ) = 4 ( a - 6 ) a = 8 / 3 correct answer d" | a ) 2 / 3 , b ) 1 / 2 , c ) 5 / 3 , d ) 8 / 3 , e ) 7 / 3 | d | divide(2, multiply(2, 4)) | multiply(n0,n3)|divide(n0,#0)| | general |
what is the total surface area in square meters of a rectangular solid whose length is 5 meters , width is 4 meters , and depth is 1 meters ? | "surface area of a cuboid = 2 ( lb + bh + lh ) = 2 ( 5 * 4 + 4 * 1 + 5 * 1 ) = 2 ( 20 + 4 + 5 ) = 2 * 29 = 58 m 2 answer : d" | a ) 48 m 2 , b ) 40 m 2 , c ) 50 m 2 , d ) 58 m 2 , e ) 62 m 2 | d | surface_rectangular_prism(4, 5, 1) | surface_rectangular_prism(n0,n1,n2)| | geometry |
in a group of ducks and cows , the total number of legs are 34 more than twice the no . of heads . find the total no . of buffaloes . | "let the number of buffaloes be x and the number of ducks be y = > 4 x + 2 y = 2 ( x + y ) + 34 = > 2 x = 34 = > x = 17 d" | a ) 11 , b ) 12 , c ) 14 , d ) 17 , e ) 18 | d | divide(34, const_2) | divide(n0,const_2)| | general |
what is the units digit of ( 7 ! * 6 ! + 6 ! * 4 ! ) / 2 ? | "( 7 ! * 6 ! + 6 ! * 4 ! ) / 2 = 6 ! ( 7 ! + 4 ! ) / 2 = 240 ( 1680 + 24 ) / 2 = 204480 units digit of the above product will be equal to 0 answer a" | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | a | divide(add(multiply(factorial(7), factorial(6)), multiply(factorial(7), factorial(4))), 7) | factorial(n0)|factorial(n1)|factorial(n3)|multiply(#0,#1)|multiply(#0,#2)|add(#3,#4)|divide(#5,n0)| | general |
in a division sum , the remainder is 0 . as student mistook the divisor by 12 instead of 21 and obtained 63 as quotient . what is the correct quotient ? | "12 * 63 = 756 756 % 21 = 36 answer : b" | a ) 0 , b ) 36 , c ) 13 , d ) 20 , e ) 25 | b | divide(multiply(12, 63), 21) | multiply(n1,n3)|divide(#0,n2)| | general |
a cube is painted red on all faces . it is then cut into 27 equal smaller cubes . how many v cubes are painted on only 2 faces ? | "1 ) draw a simple cube 2 ) draw 9 squares on each face of the cube ( so that it looks like a rubik ' s cube ) - this is what the cube will look like when it ' s cut into 27 equal smaller cubes . 3 ) remember that the outside of the cube is the part that ' s painted . . . . the mini - cubes with 2 painted sides are all... | a ) 12 , b ) 8 , c ) 6 , d ) 10 , e ) 16 | a | multiply(const_4, power(27, divide(const_1, const_3))) | divide(const_1,const_3)|power(n0,#0)|multiply(#1,const_4)| | geometry |
a tank can supply water to a village for 30 days . if a leak at the bottom of the tank drains out 15 liters per day , the supply lasts for 20 days only . for how many days will the supply last if the leak drains out 20 liters per day ? | "losing 15 liters per day results in a loss of 300 liters in 20 days . so , those 300 liters were for 10 days , making daily consumption of the village 30 liters per day . thus the capacity of the tank is 30 * 30 = 900 liters . losing 20 liters plus 30 liters gives 50 liters per day . at this rate the supply will last ... | a ) 12 , b ) 14 , c ) 15 , d ) 16 , e ) 18 | e | divide(multiply(divide(multiply(15, 20), subtract(30, 20)), 30), add(20, divide(multiply(15, 20), subtract(30, 20)))) | multiply(n1,n2)|subtract(n0,n2)|divide(#0,#1)|add(n3,#2)|multiply(n0,#2)|divide(#4,#3)| | physics |
the volume of a sphere with radius r is ( 4 / 3 ) * pi * r ^ 3 and the surface area is 4 * pi * r ^ 3 . if a sperical balloon has a volume of 12348 pi cubic centimeters , what is hte surface area of the balloon in square centimeters ? | "the surface area is 4 . pi . r ^ 2 ( its area remember not volume ) as 4 / 3 . pi . r ^ 3 = 12348 pi r = 21 so area = 4 . pi . r ^ 2 = 1764 . pi = 1764 x 3.14 = 5000 ( approx ) e" | a ) a . 40 , b ) b . 100 , c ) c . 400 , d ) d . 1,000 , e ) e . 5,000 | e | divide(const_180, const_1000) | divide(const_180,const_1000)| | geometry |
a room of 2 m 40 cm long and 4 m 20 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 = 240 * 420 sq cm size of largest square tile = h . c . f of 240 cm and 420 cm = 60 cm area of 1 tile = 60 * 60 sq cm no . of tiles required = ( 240 * 420 ) / ( 60 * 60 ) = 28 answer : c ) 28" | a ) 23 , b ) 25 , c ) 28 , d ) 60 , e ) 240 | c | divide(multiply(add(multiply(2, const_100), 40), add(multiply(4, const_100), 20)), power(divide(add(multiply(2, const_100), 40), power(const_2, const_4)), const_2)) | multiply(n0,const_100)|multiply(n2,const_100)|power(const_2,const_4)|add(n1,#0)|add(n3,#1)|divide(#3,#2)|multiply(#3,#4)|power(#5,const_2)|divide(#6,#7)| | physics |
how many times in a day , the hands of a clock are shows opposite directions ? | "sol . in 12 hours , the hands coincide or are in opposite direction 11 times . â ˆ ´ in 24 hours , the hands coincide or are in opposite direction 22 times a day . ( between 5 and 7 they point in opposite directions at 6 o ' clock only ) answer e" | a ) 12 , b ) 20 , c ) 21 , d ) 23 , e ) 22 | e | divide(subtract(multiply(const_2, multiply(const_2, const_12)), const_4), const_2) | multiply(const_12,const_2)|multiply(#0,const_2)|subtract(#1,const_4)|divide(#2,const_2)| | physics |
the difference between a positive proper fraction and its reciprocal is 9 / 20 . then the fraction is : | "explanation : let the required fraction be x . then , ( 1 / x ) - x = 9 / 20 1 - x ^ ( 2 ) / x = 9 / 20 = > 20 - 20 * x ^ ( 2 ) = 9 * x . 20 * x ^ ( 2 ) + 9 * x - 20 = 0 . = > ( 4 * x + 5 ) ( 5 * x - 4 ) = 0 . = > x = 4 / 5 . answer : c ) 4 / 5" | a ) 4 / 0 , b ) 4 / 1 , c ) 4 / 5 , d ) 4 / 2 , e ) 4 / 3 | c | divide(const_4, subtract(9, const_4)) | subtract(n0,const_4)|divide(const_4,#0)| | general |
a sum of money is to be distributed among a , b , c , d in the proportion of 6 : 3 : 5 : 4 . if c gets rs . 1000 more than d , what is b ' s share ? | "let the shares of a , b , c and d be rs . 5 x , rs . 3 x , rs . 5 x and rs . 4 x respectively . then , 5 x - 4 x = 1000 x = 1000 . b ' s share = rs . 3 x = rs . ( 3 x 1000 ) = rs . 3000 . answer : e" | a ) 1500 , b ) 2000 , c ) 2500 , d ) 2800 , e ) 3000 | e | multiply(multiply(subtract(5, 4), 1000), const_3.0) | subtract(n2,n3)|multiply(n4,#0)|multiply(const_3.0,#1)| | general |
a sports equipment store sold ping pong rackets for a total of $ 686 . if the average ( arithmetic mean ) price of a pair of rackets is $ 9.8 , how many pairs were sold ? | average price for a pair of rackets = $ 9.8 total cost = $ 9.8 * x = $ 686 x = 70 pairs were sold . answer : c | a ) 50 , b ) 100 , c ) 70 , d ) 500 , e ) 980 | c | divide(686, 9.8) | divide(n0,n1) | general |
one woman and one man can build a wall together in four hours , but the woman would need the help of two girls in order to complete the same job in the same amount of time . if one man and one girl worked together , it would take them eight hours to build the wall . assuming that rates for men , women and girls remain ... | "solution : let work done by man , women and girl per hour be m , w , g respectively . then , m + w = 1 / 4 - - > ( 1 ) , w + 2 g = 1 / 4 - - > ( 2 ) and m + g = 1 / 8 - - > ( 3 ) . no . of hours it would take forone woman , one man , and one girl , working together , to build the wall , n = 1 / m + w + g from ( 1 ) an... | a ) 5 / 7 , b ) 1 , c ) 10 / 7 , d ) 12 / 7 , e ) 24 / 7 | e | inverse(add(divide(subtract(divide(add(inverse(add(add(const_4, const_1), const_4)), multiply(inverse(add(const_4, const_1)), const_2)), const_2), inverse(add(const_4, const_1))), divide(const_3, const_2)), inverse(add(const_4, const_1)))) | add(const_1,const_4)|divide(const_3,const_2)|add(#0,const_4)|inverse(#0)|inverse(#2)|multiply(#3,const_2)|add(#4,#5)|divide(#6,const_2)|subtract(#7,#3)|divide(#8,#1)|add(#9,#3)|inverse(#10)| | physics |
what is the least number to be subtracted from 1300 to make it a perfect square ? | "the numbers less than 1300 and are squares of certain number is 1225 . the least number that should be subtracted from 1300 to make it perfect square = 1300 - 1225 = 75 . answer : c" | a ) 80 , b ) 25 , c ) 75 , d ) 63 , e ) 90 | c | subtract(multiply(const_100, add(const_10, multiply(const_3, const_2))), 1300) | multiply(const_2,const_3)|add(#0,const_10)|multiply(#1,const_100)|subtract(#2,n0)| | general |
the average weight of 8 person ' s increases by 3.5 kg when a new person comes in place of one of them weighing 65 kg . what might be the weight of the new person ? | "total weight increased = ( 8 x 3.5 ) kg = 28 kg . weight of new person = ( 65 + 28 ) kg = 93 kg . e )" | a ) kg , b ) 70 kg , c ) 80 kg , d ) 90 kg , e ) 93 kg | e | add(multiply(8, 3.5), 65) | multiply(n0,n1)|add(n2,#0)| | general |
the area of a square is 6400 sq cm . find the ratio of the breadth and the length of a rectangle whose length is thrice the side of the square and breadth is 30 cm less than the side of the square ? | "let the length and the breadth of the rectangle be l cm and b cm respectively . let the side of the square be a cm . a 2 = 6400 a = ( 6400 ) ^ 1 / 2 = 80 l = 3 a and b = a - 30 b : l = a - 30 : 2 a = 50 : 240 = 5 : 24 answer : a" | a ) 5 : 24 , b ) 5 : 19 , c ) 5 : 13 , d ) 5 : 22 , e ) 5 : 18 | a | divide(subtract(sqrt(6400), 30), multiply(sqrt(6400), const_2)) | sqrt(n0)|multiply(#0,const_2)|subtract(#0,n1)|divide(#2,#1)| | geometry |
if xy > 0 , 1 / x + 1 / y = 15 , and 1 / xy = 5 , then ( x + y ) / 5 = ? | ( 1 / x + 1 / y ) = 15 canbe solved as { ( x + y ) / xy } = 5 . substituting for 1 / xy = 5 , we get x + y = 15 / 5 = = > ( x + y ) / 5 = 15 / ( 5 * 5 ) = 3 / 5 . c | a ) 1 / 25 , b ) 1 / 6 , c ) 3 / 5 , d ) 5 , e ) 6 | c | divide(divide(15, 5), 5) | divide(n3,n5)|divide(#0,n5) | general |
a man complete a journey in 10 hours . he travels first half of the journey at the rate of 21 km / hr and second half at the rate of 26 km / hr . find the total journey in km . | "0.5 x / 10 + 0.5 x / 10 = 20 - - > x / 10 + x / 10 = 40 - - > 2 x = 10 x 40 - - > x = ( 10 x 40 ) / 2 = 200 km . answer : d ." | a ) 220 km , b ) 224 km , c ) 230 km , d ) 200 km , e ) 234 km | d | multiply(const_2, divide(multiply(multiply(21, 26), 10), add(21, 26))) | add(n1,n2)|multiply(n1,n2)|multiply(n0,#1)|divide(#2,#0)|multiply(#3,const_2)| | physics |
a bookseller sells his books at a 10 % markup in price . if he sells a book for $ 11.00 , how much did he pay for it ? | "let the cost price of book = x selling price of book = 11 $ markup % = 10 ( 110 / 100 ) x = 11 = > x = 10 answer c" | a ) $ 14.40 , b ) $ 14.00 , c ) $ 10.00 , d ) $ 9.60 , e ) $ 5.00 | c | subtract(11.00, multiply(divide(10, const_100), 11.00)) | divide(n0,const_100)|multiply(n1,#0)|subtract(n1,#1)| | gain |
there are 60 doors marked with numbers 1 to 60 . there are 60 individuals marked 1 to 60 . an operation on a door is defined as changing the status of the door from open to closed or vice versa . all the doors are closed to start with . one at a time , one randomly chosen individual goes and operates the doors . the in... | if a door is closed at the start , it requires an odd number of people to operate to be open at the end . only the perfect squares have an odd number of factors . the doors which are open at the end are : 1 , 4 , 9 , 16 , 25 , 36 , 49 for a total of 7 doors . the answer is d . | a ) 0 , b ) 2 , c ) 5 , d ) 7 , e ) 9 | d | add(divide(60, const_10), 1) | divide(n0,const_10)|add(n1,#0) | general |
a train 220 m long is running with a speed of 59 kmph . in what time will it pass a man who is running at 7 kmph in the direction opposite to that in which the train is going ? | "speed of the train relative to man = ( 59 + 7 ) kmph = 66 * 5 / 18 m / sec = 55 / 3 m / sec . time taken by the train to cross the man = time taken by it to cover 220 m at ( 55 / 3 ) m / sec = ( 220 * 3 / 55 ) sec = 12 sec answer : d ." | a ) 7 sec , b ) 8 sec , c ) 10 sec , d ) 12 sec , e ) 14 sec | d | multiply(divide(divide(220, const_1000), subtract(59, 7)), const_3600) | divide(n0,const_1000)|subtract(n1,n2)|divide(#0,#1)|multiply(#2,const_3600)| | physics |
the perimeter of a triangle is 28 cm and the inradius of the triangle is 2.0 cm . what is the area of the triangle ? | "area of a triangle = r * s where r is the inradius and s is the semi perimeter of the triangle . area of triangle = 2.0 * 28 / 2 = 28 cm 2 answer : a" | a ) 28 cm 2 , b ) 27 cm 2 , c ) 29 cm 2 , d ) 25 cm 2 , e ) 35 cm 2 | a | triangle_area(2.0, 28) | triangle_area(n0,n1)| | geometry |
due to construction , the speed limit along an 9 - mile section of highway is reduced from 55 miles per hour to 20 miles per hour . approximately how many minutes more will it take to travel along this section of highway at the new speed limit than it would have taken at the old speed limit ? | "old time in minutes to cross 9 miles stretch = 9 * 60 / 55 = 9 * 12 / 11 = 9.81 new time in minutes to cross 9 miles stretch = 9 * 60 / 20 = 9 * 3 / 1 = 27 time difference = 17.19 ans : d" | a ) a ) 5.61 , b ) b ) 8 , c ) c ) 10 , d ) d ) 17.19 , e ) e ) 24 | d | max(multiply(subtract(add(55, 9), const_1), subtract(divide(9, 20), divide(9, 55))), const_4) | add(n0,n1)|divide(n0,n2)|divide(n0,n1)|subtract(#0,const_1)|subtract(#1,#2)|multiply(#3,#4)|max(#5,const_4)| | physics |
the area of a triangle will be when a = 1 m , b = 2 m , c = 7 m , a , b , c being lengths of respective sides ? | s = ( 1 + 2 + 7 ) / 2 = 5 answer : c | a ) 3 , b ) 6 , c ) 5 , d ) 9 , e ) 1 | c | divide(add(add(1, 2), 7), 2) | add(n0,n1)|add(n2,#0)|divide(#1,n1)| | geometry |
a train 125 m long takes 6 sec to cross a man walking at 5 kmph in a direction opposite to that of the train . find the speed of the train ? | "let the speed of the train be x kmph speed of the train relative to man = x + 5 = ( x + 5 ) * 5 / 18 m / sec 125 / [ ( x + 5 ) * 5 / 18 ] = 6 30 ( x + 5 ) = 2250 x = 70 kmph answer is b" | a ) 65 kmph , b ) 70 kmph , c ) 75 kmph , d ) 80 kmph , e ) 85 kmph | b | subtract(divide(divide(125, 6), const_0_2778), 5) | divide(n0,n1)|divide(#0,const_0_2778)|subtract(#1,n2)| | physics |
what is the units digit of the expression 14 ^ 7 − 15 ^ 4 ? | i think answer on this one should be b too . since we know that 14 ^ 7 > 15 ^ 4 , as will said one should always check if the number is positive . | a ) 0 , b ) 8 , c ) 4 , d ) 6 , e ) 3 | b | add(add(const_4, const_3), const_2) | add(const_3,const_4)|add(#0,const_2)| | general |
at what rate percent on simple interest will rs . 750 amount to rs . 1050 in 5 years ? | "300 = ( 750 * 5 * r ) / 100 r = 8 % . answer : a" | a ) 8 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | a | multiply(divide(divide(subtract(1050, 750), 750), 5), const_100) | subtract(n1,n0)|divide(#0,n0)|divide(#1,n2)|multiply(#2,const_100)| | gain |
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