Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 164 | correct stringclasses 5 values | annotated_formula stringlengths 7 1.65k | linear_formula stringlengths 8 925 | category stringclasses 6 values | answer stringclasses 5 values |
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an employee β s annual salary was increased 50 % . if her old annual salary equals $ 80,000 , what was the new salary ? | "old annual salary = $ 80,000 salary increase = 50 % . original salary = $ 80,000 * 50 / 100 = $ 40,000 new salary = $ 80,000 + $ 40,000 = $ 120,000 hence b ." | a ) $ 128,000 , b ) $ 120,000 , c ) $ 110,000 , d ) $ 139,000 , e ) $ 125,000 | b | multiply(subtract(divide(multiply(subtract(const_100, const_10), const_1000), subtract(multiply(subtract(const_100, const_10), const_1000), multiply(multiply(const_0_25, const_100), const_1000))), const_1), const_100) | multiply(const_0_25,const_100)|subtract(const_100,const_10)|multiply(#1,const_1000)|multiply(#0,const_1000)|subtract(#2,#3)|divide(#2,#4)|subtract(#5,const_1)|multiply(#6,const_100)| | general | B |
what is the maximum number of pieces of birthday cake of size 10 β by 10 β that can be cut from a cake 20 β by 20 β ? | "the prompt is essentially asking for the maximum number of 10 x 10 squares that can be cut from a larger 20 by 20 square . since each ' row ' and each ' column ' of the larger square can be sub - divided into 2 ' pieces ' each , we have ( 2 ) ( 2 ) = 4 total smaller squares ( at maximum ) . a" | a ) 4 , b ) 10 , c ) 16 , d ) 20 , e ) 25 | a | divide(multiply(20, 20), multiply(10, 10)) | multiply(n2,n2)|multiply(n0,n0)|divide(#0,#1)| | physics | A |
a fill pipe can fill 1 / 2 of cistern in 20 minutes . in how many minutes , it can fill 1 / 2 of the cistern ? | required time = 20 * 2 * 1 / 2 = 20 minutes answer is d | a ) 5 min , b ) 10 min , c ) 15 min , d ) 20 min , e ) 25 min | d | divide(20, 1) | divide(n2,n0) | physics | D |
two digits in brother ' s age are the same as the digit in the sister ' s age , but in reverse order . in 20 9 years brother will be twice as old as sister will be then . what is the difference in their current age ? | brother ' s age = 10 x + y so sister ' s age = 10 y + x . . after 29 years , 10 x + y + 29 = 2 * ( 10 y + x + 29 ) . . . . so 29 + 19 y = 8 x . . . check for odd values of y , y = 1 satisfies the eqn with x = 6 . . . so ages are 61 and 16 and ans as found correctly by u is e . . 45 | a ) 34 , b ) 50 , c ) 32 , d ) 28 , e ) 45 | e | add(add(20, 20), subtract(9, const_4)) | add(n0,n0)|subtract(n1,const_4)|add(#0,#1) | general | E |
i chose a number and divide it by 6 . then i subtracted 189 from the result and got 3 . what was the number i chose ? | "let x be the number i chose , then x / 6 β 189 = 3 x / 6 = 192 x = 1152 answer is b ." | a ) 1200 , b ) 1152 , c ) 1189 , d ) 1190 , e ) 100 | b | multiply(add(189, 3), 6) | add(n1,n2)|multiply(n0,#0)| | general | B |
the current of a stream runs at the rate of 4 kmph . a boat goes 10 km and back to the starting point in 2 hours , then find the speed of the boat in still water ? | "s = 4 m = x ds = x + 4 us = x - 4 10 / ( x + 4 ) + 10 / ( x - 4 ) = 2 x = 11.4 answer : d" | a ) a ) 7 , b ) b ) 2 , c ) c ) 8.9 , d ) d ) 11.4 , e ) e ) 3 | d | divide(power(4, 2), 2) | power(n0,n2)|divide(#0,n2)| | physics | D |
the rate of spin of a certain gyroscope doubled every 10 seconds from the moment a particular stopwatch started . if after a minute and a half the gyroscope reached a speed of 3200 meters per second , what was the speed , in meters per second , when the stopwatch was started ? | "let x be the original speed when the stopwatch was started . in 90 seconds , the speed doubled 9 times . 2 ^ 9 * x = 3200 x = ( 2 ^ 7 * 25 ) / 2 ^ 9 = 25 / 4 the answer is b ." | a ) 25 / 3 , b ) 25 / 4 , c ) 25 / 8 , d ) 25 / 16 , e ) 25 / 32 | b | divide(divide(3200, power(const_2, subtract(divide(add(divide(const_60, const_2), const_60), 10), const_1))), const_2) | divide(const_60,const_2)|add(#0,const_60)|divide(#1,n0)|subtract(#2,const_1)|power(const_2,#3)|divide(n1,#4)|divide(#5,const_2)| | physics | B |
the main line train starts at 5.00 am and the harbor line train starts at 5.02 am . each train has the frequency of 10 minutes . if a guy goes in the morning at a random time what is the probability of he getting main line train ? | out of 10 minutes there is only 2 minutes chance of getting harbour line train and there is 8 minutes chance for getting main line train . so , probability = 8 / 10 = 0.8 answer : e | a ) . 5 , b ) 0.6 , c ) 0.4 , d ) 0.7 , e ) 0.8 | e | divide(subtract(10, const_2), 10) | subtract(n2,const_2)|divide(#0,n2) | probability | E |
geetha ' s home has one tap , working alone at its maximum constant rate , can fill a certain sink in 210 seconds . a second tap , working alone at its maximum constant rate , can fill the same sink in 214 seconds . if both taps work together at their respective maximum constant rates , the time it will take to fill the sink is closest to | tap 1 : 210 secs tap 2 : 214 secs considering the average of these 2 taps : 212 secs . so 1 tap can fill the tank in 212 secs , so 2 taps can fill the tank in 212 / 2 = 106 secs . closest answer is a . | a ) 106 seconds , b ) 130 seconds , c ) 177 seconds , d ) 200 seconds , e ) 270 seconds | a | inverse(add(divide(const_1, 210), divide(const_1, 214))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|inverse(#2) | physics | A |
a certain elevator has a safe weight limit of 2,500 pounds . what is the greatest possible number of people who can safely ride on the elevator at one time with the average ( arithmetic mean ) weight of half the riders being 150 pounds and the average weight of the others being 160 pounds ? | "lets assume there are 2 x people . half of them have average weight of 150 and other half has 160 . maximum weight is = 2500 so 150 * x + 160 * x = 2500 = > 310 x = 2500 = > x is approximately equal to 8 . so total people is 2 * 8 = 16 answer e ." | a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 16 | e | multiply(divide(multiply(const_10, 150), add(150, 160)), const_2) | add(n1,n2)|multiply(n1,const_10)|divide(#1,#0)|multiply(#2,const_2)| | general | E |
a large box contains 20 small boxes and each small box contains 25 chocolate bars . how many chocolate bars are in the large box ? | "the number of chocolate bars is equal to 20 * 25 = 500 correct answer b" | a ) 250 , b ) 500 , c ) 450 , d ) 550 , e ) 650 | b | multiply(20, 25) | multiply(n0,n1)| | general | B |
selling an kite for rs . 30 , a shop keeper gains 20 % . during a clearance sale , the shopkeeper allows a discount of 10 % on the marked price . his gain percent during the sale is ? | explanation : marked price = rs . 30 c . p . = 100 / 120 * 30 = rs . 25 sale price = 90 % of rs . 30 = rs . 27 required gain % = 0.2 / 25 * 100 = 8 % . answer : a | a ) 8 % , b ) 10 % , c ) 11 % , d ) 15 % , e ) 20 % | a | multiply(divide(subtract(multiply(divide(30, const_100), subtract(const_100, 10)), divide(multiply(30, const_100), add(20, const_100))), divide(multiply(30, const_100), add(20, const_100))), const_100) | add(n1,const_100)|divide(n0,const_100)|multiply(n0,const_100)|subtract(const_100,n2)|divide(#2,#0)|multiply(#1,#3)|subtract(#5,#4)|divide(#6,#4)|multiply(#7,const_100) | gain | A |
there has been successive increases of 25 % and then 15 % in the price of gas from the previous month . by what percentage should a driver reduce gas consumption so that the expenditure does not change ? | let p be the original price per unit of gas . let x be the original gas consumption . let y be the reduced gas consumption . y * 1.15 * 1.25 * p = x * p y = x / ( 1.15 * 1.25 ) which is about 0.7 x which is a decrease of about 30 % . the answer is a . | a ) 30 % , b ) 50 % , c ) 66.66 % , d ) 15 % , e ) 75 % | a | multiply(subtract(const_1, divide(const_100, add(add(const_100, 25), divide(multiply(add(const_100, 25), 15), const_100)))), const_100) | add(n0,const_100)|multiply(n1,#0)|divide(#1,const_100)|add(#0,#2)|divide(const_100,#3)|subtract(const_1,#4)|multiply(#5,const_100) | general | A |
a person was asked to state his age in years . his reply was , ` ` take my age 3 years hence , multiply it by 4 and subtract 4 times my age 3 years ago and you will know how old i am . ' ' what was the age of the person ? | explanation : let the present age of person be x years . then , 4 ( x + 3 ) - 4 ( x - 3 ) = x < = > ( 4 x + 12 ) - ( 4 x - 12 ) = x < = > x = 24 . . answer : d | a ) 18 , b ) 92 , c ) 27 , d ) 24 , e ) 19 | d | add(multiply(3, 4), multiply(3, 4)) | multiply(n0,n1)|add(#0,#0) | general | D |
in the storage room of a certain bakery , the ratio of sugar to flour is 5 to 4 , and the ratio of flour to baking soda is 10 to 1 . if there were 60 more pounds of baking soda in the room , the ratio of flour to baking soda would be 8 to 1 . how many pounds of sugar are stored in the room ? | sugar : flour = 5 : 4 = 25 : 20 ; flour : soda = 10 : 1 = 20 : 2 ; thus we have that sugar : flour : soda = 25 x : 20 x : 2 x . also given that 20 x / ( 2 x + 60 ) = 8 / 1 - - > x = 120 - - > sugar = 25 x = 3,000 answer : e . | a ) 600 , b ) 1200 , c ) 1500 , d ) 1600 , e ) 3000 | e | multiply(divide(5, 4), multiply(10, divide(multiply(8, 60), subtract(10, 8)))) | divide(n0,n1)|multiply(n4,n5)|subtract(n2,n5)|divide(#1,#2)|multiply(n2,#3)|multiply(#0,#4) | general | E |
in an examination , a student scores 4 marks for every correct answer and loses 1 mark for every wrong answer . if he attempts all 60 questions and secures 110 marks , the no of questions he attempts correctly is : | "explanation : let the number of correct answers be x . number of incorrect answers = ( 60 β x ) . 4 x β ( 60 β x ) = 110 = > 5 x = 170 = > x = 34 answer : a" | a ) 34 , b ) 38 , c ) 40 , d ) 42 , e ) 44 | a | divide(add(110, 60), add(4, 1)) | add(n2,n3)|add(n0,n1)|divide(#0,#1)| | physics | A |
in a certain animal shelter , the ratio of the number of dogs to the number of cats is 15 to 7 . if 16 additional cats were to be taken in by the shelter , the ratio of the number of dogs to the number of cats would be 15 to 11 . how many dogs are in the shelter ? | "this ratio question can be solved in a couple of different ways . here ' s an algebraic approach . . . we ' re told that the ratio of the number of dogs to the number of cats is 15 : 7 . we ' re then told that 16 more cats are added to this group and the ratio becomes 15 : 11 . we ' re asked for the number of dogs . algebraically , since the number of dogs is a multiple of 15 and the number of cats is a multiple of 7 , we can write this initial relationship as . . . 15 x / 7 x when we add the 12 cats and factor in the ' ending ratio ' , we have an equation . . . . 15 x / ( 7 x + 16 ) = 15 / 11 here we have 1 variable and 1 equation , so we can solve for x . . . . ( 15 x ) ( 11 ) = ( 7 x + 16 ) ( 15 ) ( x ) ( 11 ) = ( 7 x + 16 ) ( 1 ) 11 x = 7 x + 16 4 x = 16 x = 4 with this x , we can figure out the initial number of dogs and cats . . . initial dogs = 15 x = 15 ( 4 ) = 60 final answer : e" | a ) 15 , b ) 25 , c ) 30 , d ) 45 , e ) 60 | e | multiply(divide(divide(multiply(7, 16), subtract(11, 7)), 7), 15) | multiply(n1,n2)|subtract(n4,n1)|divide(#0,#1)|divide(#2,n1)|multiply(n0,#3)| | other | E |
the radius of a cylinder is 10 m , height 14 m . the volume of the cylinder is : | "cylinder volume = Ο r ( power 2 ) h = 22 / 7 Γ 10 Γ 10 Γ 14 = 4400 m ( power 3 ) answer is e ." | a ) 2200 , b ) 5500 , c ) 3300 , d ) 1100 , e ) 4400 | e | multiply(circumface(10), 14) | circumface(n0)|multiply(n1,#0)| | geometry | E |
some of 50 % - intensity red paint is replaced with 20 % solution of red paint such that the new paint intensity is 30 % . what fraction of the original paint was replaced ? | 30 % is 10 % - points above 20 % and 20 % - points below 50 % . thus the ratio of 25 % - solution to 50 % - solution is 2 : 1 . 2 / 3 of the original paint was replaced . the answer is c . | a ) 1 / 30 , b ) 1 / 5 , c ) 2 / 3 , d ) 3 / 4 , e ) 4 / 5 | c | divide(subtract(divide(30, const_100), divide(50, const_100)), subtract(divide(20, const_100), divide(50, const_100))) | divide(n2,const_100)|divide(n0,const_100)|divide(n1,const_100)|subtract(#0,#1)|subtract(#2,#1)|divide(#3,#4) | gain | C |
angelo and isabella are both salespersons . in any given week , angelo makes $ 550 in base salary plus 8 percent of the portion of his sales above $ 5,000 for that week . isabella makes 10 percent of her total sales for any given week . for what amount of weekly sales would angelo and isabella earn the same amount of money ? | "official solution : the problem asks for the amount of weekly sales it takes for angelo and isabella to earn the same amount of money . you can write an equation that sets angelo β s and isabella β s weekly earnings equal to each other , with x representing weekly sales . weekly earnings for each salesperson equal base salary plus commission . so angelo β s earnings are 550 + ( 0.08 ) ( x β 5,000 ) , and isabella β s are 0.10 x . set up the equation and solve : 550 + ( 0.08 ) ( x β 5,000 ) = 0.10 x distribute the 0.08 : 550 + 0.08 x β 400 = 0.10 x combine terms and subtract 0.08 x from both sides : 150 = 0.02 x divide both sides by 0.02 : 7,500 = x your answer is a ." | a ) 7,500 , b ) 24,500 , c ) 25,500 , d ) 26,500 , e ) 27,500 | a | floor(divide(divide(subtract(550, multiply(5,000, divide(8, const_100))), subtract(divide(10, const_100), divide(8, const_100))), 5,000)) | divide(n1,const_100)|divide(n3,const_100)|multiply(#0,n2)|subtract(#1,#0)|subtract(n0,#2)|divide(#4,#3)|divide(#5,n2)|floor(#6)| | general | A |
worker a takes 8 hours to do a job . worker b takes 6 hours to do the same job . how long it take both a & b , working together but independently , to do the same job ? | "one day work of a = 1 / 8 one day work of b = 1 / 6 so one day work of a and b together = 1 / 8 + 1 / 6 = 7 / 24 so total days required = 24 / 7 answer : d" | a ) 20 / 9 , b ) 40 / 9 , c ) 50 / 9 , d ) 24 / 7 , e ) 80 / 9 | d | divide(const_1, add(divide(const_1, 8), divide(const_1, 6))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)| | physics | D |
a sum of money is to be distributed among a , b , c , d in the proportion of 5 : 2 : 4 : 3 . if c gets $ 500 more than d , what is a ' s share ? | "let the shares of a , b , c and d be 5 x , 2 x , 4 x and 3 x respectively . then , 4 x - 3 x = 500 x = $ 500 a ' s share = 5 x = 5 * $ 500 = $ 2500 the answer is d ." | a ) $ 1200 , b ) $ 1600 , c ) $ 2000 , d ) $ 2500 , e ) $ 3000 | d | multiply(multiply(subtract(4, 3), 500), 3) | subtract(n2,n3)|multiply(n4,#0)|multiply(n3,#1)| | general | D |
a man goes from a to b at a speed of 21 kmph and comes back to a at a speed of 21 kmph . find his average speed for the entire journey ? | "distance from a and b be ' d ' average speed = total distance / total time average speed = ( 2 d ) / [ ( d / 21 ) + ( d / 24 ] = ( 2 d ) / [ 15 d / 168 ) = > 22.3 kmph . answer : d" | a ) 23.3 kmph , b ) 25.3 kmph , c ) 22.5 kmph , d ) 22.3 kmph , e ) 22.9 kmph | d | divide(add(21, 21), const_2) | add(n0,n1)|divide(#0,const_2)| | physics | D |
area of a square is 1 / 2 hectare . the diagonal of the square is ? | area = 1 / 2 hectare = 10000 / 2 m 2 = 5000 m 2 again area = 1 / 2 x ( diagonal ) 2 so 1 / 2 x ( diagonal ) 2 = 5000 m 2 diagonal 2 = 10000 diagonal = 100 answer : b | ['a ) 250 meter', 'b ) 100 meter', 'c ) 50 β 2 meter', 'd ) 50 meter', 'e ) 35 meter'] | b | sqrt(multiply(const_2, multiply(divide(1, 2), multiply(const_100, const_100)))) | divide(n0,n1)|multiply(const_100,const_100)|multiply(#0,#1)|multiply(#2,const_2)|sqrt(#3) | geometry | B |
a farmer used 1,034 acres of land for beans , wheat , and corn in the ratio of 5 : 2 : 4 , respectively . how many y acres were used for corn ? | "consider 5 x acres of land used for bean consider 2 x acres of land used for wheat consider 4 x acres of land used for corn total given is 1034 acres 11 x = 1034 x = 94 land used for corn y = 4 * 94 = 376 correct option - c" | a ) 188 , b ) 258 , c ) 376 , d ) 470 , e ) 517 | c | multiply(divide(add(multiply(const_1000, const_1), add(multiply(const_10, const_3), 4)), add(add(5, 2), 4)), 4) | add(n1,n2)|multiply(const_10,const_3)|multiply(const_1,const_1000)|add(n3,#1)|add(n3,#0)|add(#3,#2)|divide(#5,#4)|multiply(n3,#6)| | other | C |
a bag of potatoes weighs 28 lbs divided by half of its weight . how much does the bag of potatoes weight ? | "sol . 28 Γ· 2 = 14 . answer : b" | a ) 20 lb , b ) 14 lb , c ) 10 lb , d ) 15 lb , e ) 5 lb | b | divide(28, const_1) | divide(n0,const_1)| | general | B |
what is the range of all the roots of | x ^ 2 - 1 | = x ? | we get 2 quadratic equations here . . 1 ) x ^ 2 - x - 1 = 0 . . . . . . . roots 2 , - 1 2 ) x ^ 2 + x - 1 = 0 . . . . . . . . roots - 2 , 1 inserting each root in given equation , it can be seen that - 1 and - 2 do not satisfy the equations . so value of x for given equation . . . . x = 2 or x = 1 i guess range is 2 - 1 = 1 d | a ) 4 , b ) 3 , c ) 2 , d ) 1 , e ) 0 | d | sqrt(1) | sqrt(n1) | general | D |
in a group of 100 cars , 37 cars do not have air conditioning . if at least 41 cars have racing stripes , what is the greatest number of cars that could have air conditioning but not racing stripes ? | "lets assume ac = 63 ( includesonly ac carsandcars with ac and racing stripes ) lets assume rs ( racing stripes ) > = 41 ( includescars with ac and racing stripesandonly racing stripes ) . now since we want to maximize ( only ac ) we have to see to it thatcars with ac and racing stripesis minimal ( assume 0 ) but since rs > = 41 . . we have to assign atleast 4 tocars with ac and racing stripes . hence ac = 63 - 4 = 59 . the answer is" | a ) 45 , b ) 47 , c ) 59 , d ) 51 , e ) 53 | c | subtract(100, 41) | subtract(n0,n2)| | other | C |
gold is 10 times as heavy as water and copper is 5 times as heavy as water . in what ratio should these be mixed to get an alloy 6 times as heavy as water ? | g = 10 w c = 5 w let 1 gm of gold mixed with x gm of copper to get 1 + x gm of the alloy 1 gm gold + x gm copper = x + 1 gm of alloy 10 w + 5 wx = x + 1 * 6 w 10 + 5 x = 6 ( x + 1 ) x = 4 ratio of gold with copper = 1 : 4 = 1 : 4 answer is b | a ) 3 : 2 , b ) 1 : 4 , c ) 3 : 1 , d ) 5 : 2 , e ) 4 : 3 | b | divide(subtract(6, 5), subtract(10, 6)) | subtract(n2,n1)|subtract(n0,n2)|divide(#0,#1) | general | B |
an article is bought for rs . 775 and sold for rs . 900 , find the gain percent ? | "775 - - - - 125 100 - - - - ? = > 16.1 % answer : e" | a ) 11.1 % , b ) 12.1 % , c ) 13.1 % , d ) 15.1 % , e ) 16.1 % | e | subtract(const_100, divide(multiply(900, const_100), 775)) | multiply(n1,const_100)|divide(#0,n0)|subtract(const_100,#1)| | gain | E |
if the radius of a circle is decreased by 50 % , find the percentage decrease | "let original radius = r . new radius = = original area = and new area = decrease in area = = 75 % answer : c" | a ) 22 , b ) 099 , c ) 75 , d ) 26 , e ) 18 | c | multiply(subtract(divide(const_100, const_100), power(subtract(divide(const_100, const_100), divide(50, const_100)), const_2)), const_100) | divide(const_100,const_100)|divide(n0,const_100)|subtract(#0,#1)|power(#2,const_2)|subtract(#0,#3)|multiply(#4,const_100)| | gain | C |
what is the smallest positive integer x , such that 1152 x is a perfect cube ? | we need to make 1152 x a perfect cube , hence we need to have the powers a multiple of 3 1152 = 2 ^ 7 * 3 ^ 2 the minimum value of x for which 1152 x is a perfect cube = 2 ^ 2 * 3 = 12 correct option : d | a ) 4 , b ) 6 , c ) 8 , d ) 12 , e ) 18 | d | floor(add(power(1152, divide(const_1, const_3)), const_2)) | divide(const_1,const_3)|power(n0,#0)|add(#1,const_2)|floor(#2) | geometry | D |
when positive integer n is divided by 3 , the remainder is 2 . when n is divided by 6 , the remainder is 5 . how many values less than 100 can n take ? | "a quick approac to this q is . . the equation we can form is . . 3 x + 2 = 7 y + 5 . . 3 x - 3 = 7 y . . . 3 ( x - 1 ) = 7 y . . . so ( x - 1 ) has to be a multiple of 7 as y then will take values of multiple of 3 . . here we can see x can be 1 , 8,15 , 22,29 so 5 values till 100 is reached as ( 29 - 1 ) * 3 = 84 and next multiple of 7 will be 84 + 21 > 100 . . ans 4 . . d" | a ) 0 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | d | subtract(100, reminder(const_4.0, 6)) | reminder(const_4.0,n2)|subtract(n4,#0)| | general | D |
in 10 years , a will be twice as old 5 as b was 10 years ago . if a is now 11 years older than b , the present age of b is | "explanation : let b ' s age = x years . then , as age = ( x + 11 ) years . ( x + 11 + 10 ) = 2 ( x β 10 ) hence x = 41 . present age of b = 41 years answer : option d" | a ) 35 , b ) 37 , c ) 39 , d ) 41 , e ) 42 | d | add(multiply(const_2, 10), add(11, 10)) | add(n0,n3)|multiply(n0,const_2)|add(#0,#1)| | general | D |
a man saves 25 % of his monthly salary . if an account of dearness of things he is to increase his monthly expenses by 10 % , he is only able to save rs . 175 per month . what is his monthly salary ? | "income = rs . 100 expenditure = rs . 75 savings = rs . 25 present expenditure 75 + 75 * ( 10 / 100 ) = rs . 82.5 present savings = 100 β 82.50 = rs . 17.50 if savings is rs . 17.50 , salary = rs . 100 if savings is rs . 175 , salary = 100 / 17.5 * 175 = 1000 answer : a" | a ) rs . 1000 , b ) rs . 2000 , c ) rs . 1500 , d ) rs . 3000 , e ) rs . 3100 | a | divide(multiply(175, const_100), subtract(const_100, add(subtract(const_100, 25), multiply(subtract(const_100, 25), divide(10, const_100))))) | divide(n1,const_100)|multiply(n2,const_100)|subtract(const_100,n0)|multiply(#0,#2)|add(#3,#2)|subtract(const_100,#4)|divide(#1,#5)| | general | A |
the ratio of incomes of two person p 1 and p 2 is 5 : 4 and the ratio of their expenditures is 3 : 2 . if at the end of the year , each saves rs . 1600 , then what is the income of p 1 ? | let the income of p 1 and p 2 be rs . 5 x and rs . 4 x respectively and let their expenditures be rs . 3 y and 2 y respectively . then , 5 x β 3 y = 1600 β¦ ( i ) and 4 x β 2 y = 1600 β¦ β¦ . . ( ii ) on multiplying ( i ) by 2 , ( ii ) by 3 and subtracting , we get : 2 x = 1600 - > x = 800 p 1 β s income = rs 5 * 800 = rs . 4000 answer : c | a ) rs . 800 , b ) rs . 2400 , c ) rs . 4000 , d ) rs . 3200 , e ) rs . 4200 | c | multiply(subtract(multiply(divide(1600, 2), 3), 1600), 5) | divide(n6,n1)|multiply(n4,#0)|subtract(#1,n6)|multiply(n2,#2) | other | C |
a can do a piece of work in 8 days . b can do it in 16 days . with the assistance of c they completed the work in 4 days . find in how many days can c alone do it ? | "c = 1 / 4 - 1 / 8 - 1 / 16 = 1 / 16 = > 16 days answer : e" | a ) 87 days , b ) 20 days , c ) 26 days , d ) 19 days , e ) 16 days | e | divide(multiply(8, 16), divide(subtract(multiply(8, 16), multiply(add(divide(multiply(8, 16), 8), divide(multiply(8, 16), 16)), 4)), 4)) | multiply(n0,n1)|divide(#0,n0)|divide(#0,n1)|add(#1,#2)|multiply(n2,#3)|subtract(#0,#4)|divide(#5,n2)|divide(#0,#6)| | physics | E |
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 | C |
a case of 12 rolls of paper towels sells for $ 9 . the cost of one roll sold individually is $ 1 . what is the percent t of savings per roll for the 12 - roll package over the cost of 12 rolls purchased individually ? | "cost of 12 paper towels individually = 1 * 12 = 12 cost of a set of 12 paper towels = 9 cost of one roll = 9 / 12 = 3 / 4 = 0.75 savings per roll = 1 - . 75 = 0.25 % of savings is t = . 25 / 1 * 100 = 25 % d is the answer ." | a ) 9 % , b ) 11 % , c ) 15 % , d ) 25 % , e ) 90 % | d | subtract(const_100, multiply(divide(9, 12), const_100)) | divide(n1,n0)|multiply(#0,const_100)|subtract(const_100,#1)| | general | D |
a firm is comprised of partners and associates in a ratio of 2 : 63 . if 50 more associates were hired , the ratio of partners to associates would be 1 : 34 . how many partners are currently in the firm ? | the ratio 1 : 34 = 2 : 68 so the ratio changed from 2 : 63 to 2 : 68 . 68 - 63 = 5 which is 1 / 10 of the increase in 50 associates . the ratio changed from 20 : 630 to 20 : 680 . thus the number of partners is 20 . the answer is d . | a ) 5 , b ) 10 , c ) 12 , d ) 20 , e ) 25 | d | multiply(divide(50, subtract(multiply(34, 2), 63)), 2) | multiply(n0,n4)|subtract(#0,n1)|divide(n2,#1)|multiply(n0,#2) | other | D |
if the population of a certain country increases at the rate of one person every 30 seconds , by how many persons does the population increase in 10 minutes ? | "answer = 2 * 10 = 20 answer = a" | a ) 20 , b ) 30 , c ) 15 , d ) 10 , e ) 80 | a | multiply(divide(const_60, 30), 10) | divide(const_60,n0)|multiply(n1,#0)| | physics | A |
in a survey of political preferences , 78 % of those asked were in favour of at least one of the proposals : i , ii and iii . 50 % of those asked favoured proposal i , 30 % favoured proposal ii , and 20 % favoured proposal iii . if 5 % of those asked favoured all 3 of the proposals , what x percentage of those asked favoured more than one of the 3 proposals . | bunuel , my answer for exactly 2 people was 17 and this was my approach : 100 % = ( a + b + c ) - ( anb + anc + bnc ) - 5 % + 22 % which leads me to x = 100 % = ( 50 + 30 + 20 ) - ( at least 2 people ) - 5 % + 22 % . c | a ) 10 , b ) 12 , c ) 17 , d ) 22 , e ) 30 | c | add(subtract(subtract(add(subtract(add(50, 30), 5), 20), 5), 78), 5) | add(n1,n2)|subtract(#0,n4)|add(n3,#1)|subtract(#2,n4)|subtract(#3,n0)|add(n4,#4) | general | C |
in a public show 62 % of the seats were filled . if there were 600 seats in the hall , how many seats were vacant ? | "75 % of 600 = 62 / 100 Γ 600 = 372 therefore , the number of vacant seats = 600 - 372 = 228 . answer : e" | a ) 100 , b ) 110 , c ) 120 , d ) 140 , e ) 228 | e | divide(multiply(600, subtract(const_100, 62)), const_100) | subtract(const_100,n0)|multiply(n1,#0)|divide(#1,const_100)| | gain | E |
a boat takes 90 minutes less to travel 36 miles downstream than to travel the same distance upstream . if the speed of the boat in still water is 10 mph , the speed of the stream is : | explanation : speed of the boat in still water = 10 mph let speed of the stream be x mph then , speed downstream = ( 10 + x ) mph speed upstream = ( 10 - x ) mph time taken to travel 36 miles upstream - time taken to travel 36 miles downstream = 90 / 60 hours = > 36 / ( 10 β x ) β 36 / ( 10 + x ) = 3 / 2 = > 12 / ( 10 β x ) β 12 / ( 10 + x ) = 1 / 2 = > 24 ( 10 + x ) β 24 ( 10 β x ) = ( 10 + x ) ( 10 β x ) = > 240 + 24 x β 240 + 24 x = ( 100 β x 2 ) = > 48 x = 100 β x 2 = > x 2 + 48 x β 100 = 0 = > ( x + 50 ) ( x β 2 ) = 0 = > x = - 50 or 2 . answer : option d | a ) 4 mph , b ) 2.5 mph , c ) 3 mph , d ) 2 mph , e ) none of these | d | divide(subtract(sqrt(add(multiply(power(10, const_2), const_4), power(multiply(divide(36, divide(90, const_60)), const_2), const_2))), multiply(divide(36, divide(90, const_60)), const_2)), const_2) | divide(n0,const_60)|power(n2,const_2)|divide(n1,#0)|multiply(#1,const_4)|multiply(#2,const_2)|power(#4,const_2)|add(#3,#5)|sqrt(#6)|subtract(#7,#4)|divide(#8,const_2) | physics | D |
the timing of a college is from 12 p . m to 4.20 p . m . five lectures are held in the given duration and a break of 5 minutes after each lecture is given to the students . find the duration of each lecture . | explanation : total time a student spends in college = 4 hours 20 minutes = 260 minutes as there are 5 lectures , the number of breaks between lectures is 4 . total time of the break = 20 minutes hence , the duration of each lecture is = ( 260 β 20 ) / 5 = 48 minutes answer d | a ) 52 minutes , b ) 45 minutes , c ) 30 minutes , d ) 48 minutes , e ) 44 minutes | d | divide(multiply(4.2, const_60), 5) | multiply(n1,const_60)|divide(#0,n2) | physics | D |
the decimal 0.1 is how many times greater than the decimal ( 0.001 ) ^ 4 ? | "0.1 = 10 ^ - 1 ( 0.001 ) ^ 4 = ( 10 ^ - 3 ) ^ 4 = 10 ^ - 12 10 ^ 11 * 10 ^ - 12 = 10 ^ - 1 the answer is d ." | a ) 10 ^ 8 , b ) 10 ^ 9 , c ) 10 ^ 10 , d ) 10 ^ 11 , e ) 10 ^ 12 | d | divide(0.1, power(0.001, 4)) | power(n1,n2)|divide(n0,#0)| | general | D |
total 52 matches are conducted in knockout match type . how many players will be participated in that tournament ? | "51 players answer : b" | a ) 60 , b ) 51 , c ) 53 , d ) 56 , e ) 50 | b | add(52, const_1) | add(n0,const_1)| | general | B |
in a triangle , one side is 6 cm and another side is 8 cm . which of the following can be the perimeter of the triangle ? | given : one side is 6 cm and another side is 8 cm . so the 3 rd side will be > 3 and < 15 . thus the perimeter will be : 18 < perimeter < 30 . only option satisfying this condition is 22 . hence a . | ['a ) 22 .', 'b ) 25 .', 'c ) 30 .', 'd ) 32 .', 'e ) 34 .'] | a | subtract(triangle_perimeter(6, 8, sqrt(add(power(6, const_2), power(8, const_2)))), const_2) | power(n0,const_2)|power(n1,const_2)|add(#0,#1)|sqrt(#2)|triangle_perimeter(n0,n1,#3)|subtract(#4,const_2) | geometry | A |
there are 18 stations between ernakulam and chennai . how many second class tickets have to be printed , so that a passenger can travel from one station to any other station ? | "the total number of stations = 20 from 20 stations we have to choose any two stations and the direction of travel ( ernakulam to chennai is different from chennai to ernakulam ) in 20 p 2 ways . 20 p 2 = 20 * 19 = 380 answer : b" | a ) 800 , b ) 380 , c ) 360 , d ) 478 , e ) 566 | b | multiply(add(18, const_2), subtract(add(18, const_2), const_1)) | add(n0,const_2)|subtract(#0,const_1)|multiply(#0,#1)| | physics | B |
what is the sum of all possible solutions to | x - 3 | ^ 2 + | x - 3 | = 30 ? | "denote | x - 3 | as y : y ^ 2 + y = 30 - - > y = - 6 or y = 5 . discard the first solution since y = | x - 3 | , so it ' s an absolute value and thus can not be negative . y = | x - 3 | = 5 - - > x = 8 or x = - 3 . the sum = 5 . answer : b ." | a ) 4 , b ) 5 , c ) 3 , d ) 2 , e ) 1 | b | add(add(const_4, 3), subtract(3, const_4)) | add(n0,const_4)|subtract(n0,const_4)|add(#0,#1)| | general | B |
the speed of a car is 120 km in the first hour and 60 km in the second hour . what is the average speed of the car ? | "s = ( 120 + 60 ) / 2 = 90 kmph c" | a ) 89 kmph , b ) 92 kmph , c ) 90 kmph , d ) 65 kmph , e ) 77 kmph | c | divide(add(120, 60), const_2) | add(n0,n1)|divide(#0,const_2)| | physics | C |
the area of rectangular field is 460 square metres . if the length is 15 per cent more than the breadth , what is the breadth of the rectangular field ? | let the breadth of the rectangular field be β x β m . then , length of the field will be x + x Γ 15 / 100 = 23 x / 20 now , x Γ 23 x / 20 = 460 or , 23 x 2 = 460 Γ 20 or , x 2 = 20 Γ 20 or , x = 20 m answer e | ['a ) 15 metres', 'b ) 26 metres', 'c ) 34.5 metres', 'd ) can not be determined', 'e ) none of these'] | e | divide(multiply(460, multiply(const_2, const_10)), divide(multiply(460, multiply(const_2, const_10)), const_100)) | multiply(const_10,const_2)|multiply(n0,#0)|divide(#1,const_100)|divide(#1,#2) | geometry | E |
how many distinct integer values of n satisfy the inequality | | n - 3 | + 4 | β€ 17 ? | "so i can write this as in - 3 i + 4 < = 17 or in - 3 i < = 13 so n can have - 10 to 16 = 15 true values . . . . a" | a ) 15 , b ) 16 , c ) 17 , d ) 18 , e ) 19 | a | add(multiply(3, 4), const_1) | multiply(n0,n1)|add(#0,const_1)| | general | A |
if $ 120 invested at a certain rate of simple interest amounts to $ 160 at the end of 3 years , how much will $ 150 amount to at the same rate of interest in 6 years ? | "120 amounts to 160 in 3 years . i . e ( principal + interest ) on 120 in 3 years = 160 120 + 120 * ( r / 100 ) * ( 3 ) = 160 = > r = 100 / 9 150 in 6 years = principal + interest = 150 + 150 * ( r / 100 ) * ( 6 ) 250 answer is e ." | a ) $ 190 , b ) $ 180 , c ) $ 200 , d ) $ 240 , e ) $ 250 | e | add(150, divide(multiply(multiply(150, 6), divide(divide(multiply(subtract(160, 120), 120), 120), 3)), 120)) | multiply(n3,n4)|subtract(n1,n0)|multiply(#1,n0)|divide(#2,n0)|divide(#3,n2)|multiply(#4,#0)|divide(#5,n0)|add(n3,#6)| | gain | E |
marcella has 27 pairs of shoes . if she loses 9 individual shoes , what is the greatest number of matching pairs she could have left ? | "marcella has 27 pairs of shoes and loses 9 shoes . to minimize the loss of identical pairs of shoes we want marcella to lose as many identical pairs as possible . this would yield 4 identical pairs and 1 additional shoe ( destroying 5 pairs of shoes ) . the 27 pairs of shoes minus the 5 ' destroyed ' pairs yields 22 pairs that still fulfill the requirements . answer : a" | a ) 22 , b ) 20 , c ) 19 , d ) 16 , e ) 15 | a | subtract(27, add(floor(divide(9, const_2)), const_1)) | divide(n1,const_2)|floor(#0)|add(#1,const_1)|subtract(n0,#2)| | general | A |
80 % of 5 / 8 = | "should be simple . 0.8 * 5 / 8 = 4 / 8 = 1 / 2 = 0.5 correct option : b" | a ) 0.2 , b ) 0.5 , c ) 0.6 , d ) 0.75 , e ) 1.0 | b | divide(multiply(divide(multiply(8, 5), const_100), 80), const_100) | multiply(n1,n2)|divide(#0,const_100)|multiply(n0,#1)|divide(#2,const_100)| | general | B |
18 men can complete a piece of work in 30 days . in how many days can 15 men complete that piece of work ? | "b 36 days 18 * 30 = 15 * x = > x = 36 days" | a ) 23 days , b ) 36 days , c ) 22 days , d ) 29 days , e ) 20 days | b | divide(multiply(30, 18), 15) | multiply(n0,n1)|divide(#0,n2)| | physics | B |
a student gets 60 % in one subject , 80 % in the other . to get an overall of 75 % how much should get in third subject . | let the 3 rd subject % = x 60 + 80 + x = 3 * 75 140 + x = 225 x = 225 - 140 = 85 answer : c | a ) 75 % , b ) 25 % , c ) 85 % , d ) 55 % , e ) 65 % | c | subtract(multiply(75, const_3), add(60, 80)) | add(n0,n1)|multiply(n2,const_3)|subtract(#1,#0) | gain | C |
in a bag containing 3 balls , a white ball was placed and then 1 ball was taken out at random . what is theprobability that the extracted ball would turn onto be white , if all possible hypothesisconcerning the color of the balls that initially in the bag were equally possible ? | "since , all possible hypothesis regarding the colour of the balls are equally likely , therefore these could be 3 white balls , initially in the bag . β΄ required probability = 1 / 4 [ 1 + 3 / 4 + 1 / 2 + 1 / 4 ] = 1 / 4 [ ( 4 + 3 + 2 + 1 ) / 4 ] = 5 / 8 b" | a ) 3 / 5 , b ) 5 / 8 , c ) 6 / 7 , d ) 7 / 8 , e ) 6 / 7 | b | divide(add(1, add(divide(1, const_2), add(divide(1, add(1, 3)), divide(3, add(1, 3))))), add(1, 3)) | add(n0,n1)|divide(n1,const_2)|divide(n1,#0)|divide(n0,#0)|add(#2,#3)|add(#4,#1)|add(n1,#5)|divide(#6,#0)| | probability | B |
two - third of a positive number and 49 / 216 of its reciprocal are equal . the number is : | "let the number be x . then , 2 / 3 x = 49 / 216 * 1 / x x 2 = 49 / 216 * 3 / 2 = 49 / 144 x = 7 / 12 answer : b" | a ) 5 / 12 , b ) 7 / 12 , c ) 25 / 144 , d ) 144 / 25 , e ) 146 / 25 | b | sqrt(divide(multiply(49, const_3), multiply(216, const_2))) | multiply(n0,const_3)|multiply(n1,const_2)|divide(#0,#1)|sqrt(#2)| | general | B |
in a hostel there were 100 students . to accommodate 25 more students the average is decreased by some rupees . but total expenditure increased by rs . 500 . if the total expenditure of the hostel now 8500 , find decrease of average budget ? | "let average is x 100 x + 500 = 8500 x = 80 let decrease = y 125 ( 80 β y ) = 8500 y = 12 answer : c" | a ) 20 , b ) 15 , c ) 12 , d ) 18 , e ) 24 | c | multiply(subtract(divide(add(multiply(add(100, 25), 500), 8500), 25), 500), add(100, 25)) | add(n0,n1)|multiply(n2,#0)|add(n3,#1)|divide(#2,n1)|subtract(#3,n2)|multiply(#0,#4)| | general | C |
in a certain corporation , there are 300 male employees and 150 female employees . it is known that 10 % of the male employees have advanced degrees and 40 % of the females have advanced degrees . if one of the 450 employees is chosen at random , what is the probability this employee has an advanced degree or is female ? | p ( female ) = 150 / 450 = 1 / 3 p ( male with advanced degree ) = 0.1 * 300 / 450 = 30 / 450 = 1 / 15 the sum of the probabilities is 6 / 15 = 2 / 5 the answer is c . | a ) 1 / 5 , b ) 5 / 15 , c ) 6 / 15 , d ) 3 / 10 , e ) 4 / 15 | c | add(divide(multiply(subtract(const_1, divide(40, multiply(10, 10))), 150), 450), divide(add(multiply(divide(10, multiply(10, 10)), 300), multiply(divide(40, multiply(10, 10)), 150)), 450)) | multiply(n2,n2)|divide(n3,#0)|divide(n2,#0)|multiply(n0,#2)|multiply(n1,#1)|subtract(const_1,#1)|add(#3,#4)|multiply(n1,#5)|divide(#7,n4)|divide(#6,n4)|add(#8,#9) | other | C |
total dinning bill of 8 people was $ 139.00 and 10 % tip divided the bill evenly ? what is the bill amount each person shared . | "dinner bill of 8 person = 139 + 10 % tip so , 10 % of 139 = ( 139 * 10 ) / 100 = 13.9 so , the actual total amount = 139 + 13.9 = $ 152.9 so per head bill = 152.9 / 8 = $ 19.11 answer : d" | a ) 11.84 , b ) 12.84 , c ) 23.84 , d ) 19.11 , e ) 10.12 | d | divide(multiply(139.00, add(divide(const_1, 10), const_1)), 8) | divide(const_1,n2)|add(#0,const_1)|multiply(n1,#1)|divide(#2,n0)| | general | D |
a person borrows 5000 for 2 years at 4 % p . a . simple interest . he immediately lends it to another person at 61 β 4 % p . a . for 2 years . find his gain in the transaction per year . | "gain in 2 years = [ ( 5000 Γ 25 / 4 Γ 2 / 100 ) β ( 5000 Γ 4 Γ 2 / 100 ) ] = ( 625 β 400 ) = 225 . β΄ gain in 1 year = ( 225 β 2 ) = 112.50 answer a" | a ) 112.50 , b ) 125 , c ) 150 , d ) 167.5 , e ) none of these | a | multiply(5000, divide(const_1, const_100)) | divide(const_1,const_100)|multiply(n0,#0)| | gain | A |
which number should replace both the asterisks in ( * / 21 ) x ( * / 189 ) = 1 ? | "answer let ( y / 21 ) x ( y / 189 ) = 1 then , y 2 = 21 x 189 = 21 x 21 x 9 β΄ y = ( 21 x 3 ) = 63 . option : b" | a ) 21 , b ) 63 , c ) 3969 , d ) 147 , e ) 167 | b | sqrt(multiply(189, 21)) | multiply(n0,n1)|sqrt(#0)| | general | B |
if the cost price of 12 pencils is equal to the selling price of 8 pencils , the gain percent is : | c 50 % let c . p . of each pencil be $ 1 . then , c . p . of 8 pencils = $ 8 ; s . p . of 8 pencils = $ 12 . gain % = 4 / 8 * 100 = 50 % | a ) 30 , b ) 40 , c ) 50 , d ) 60 , e ) 70 | c | divide(const_100, divide(8, subtract(12, 8))) | subtract(n0,n1)|divide(n1,#0)|divide(const_100,#1) | gain | C |
the population of a town increases 25 % and 30 % respectively in two consecutive years . after the growth the present population of the town is 1950 . then what is the population of the town 2 years ago ? | "explanation : formula : ( after = 100 denominator ago = 100 numerator ) 1950 * 100 / 125 * 100 / 130 = 1200 answer : option e" | a ) a ) 800 , b ) b ) 900 , c ) c ) 1000 , d ) d ) 1100 , e ) e ) 1200 | e | divide(1950, multiply(add(const_1, divide(25, const_100)), add(const_1, divide(30, const_100)))) | divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|add(#1,const_1)|multiply(#2,#3)|divide(n2,#4)| | gain | E |
a train covers a distance of 14 km in 10 min . if it takes 6 sec to pass a telegraph post , then the length of the train is ? | "speed = ( 14 / 10 * 60 ) km / hr = ( 84 * 5 / 18 ) m / sec = 70 / 3 m / sec . length of the train = 70 / 3 * 6 = 140 m . answer : c" | a ) m , b ) m , c ) m , d ) m , e ) m | c | divide(14, subtract(divide(14, 10), 6)) | divide(n0,n1)|subtract(#0,n2)|divide(n0,#1)| | physics | C |
if n = 5 ^ 11 β 5 , what is the units digit of n ? | always divide the power ( incase 11 ) by 4 and use the remainder as the new power . the question now becomes 5 ^ 3 - 5 . now 5 ^ 3 has last digit 5 . we subtract 5 from 5 = 0 is the answer . option a | a ) 0 , b ) 1 , c ) 4 , d ) 6 , e ) 8 | a | divide(log(5), log(power(5, 11))) | log(n2)|power(n0,n1)|log(#1)|divide(#0,#2)| | general | A |
a fruit seller had some apples . he sells 35 % and still has 6500 apples . originally , he had ? | "answer Γ’ Λ Β΅ 65 % of n = 800 Γ’ Λ Β΄ n = ( 650 x 100 ) / 65 = 1000 correct option : e" | a ) 650 apples , b ) 600 apples , c ) 772 apples , d ) 700 apples , e ) none | e | original_price_before_loss(35, 6500) | original_price_before_loss(n0,n1)| | gain | E |
f ( x ) is a function such that f ( x ) + 3 f ( 8 - x ) = x for all real numbers x . find the value of f ( 2 ) . | f ( x ) + 3 f ( 8 - x ) = x : given f ( 2 ) + 3 f ( 6 ) = 2 : x = 2 above f ( 6 ) + 3 f ( 2 ) = 6 : x = 6 above f ( 6 ) = 6 - 3 f ( 2 ) : solve equation c for f ( 6 ) f ( 2 ) + 3 ( 6 - 3 f ( 2 ) ) = 2 : substitute f ( 2 ) = 2 : solve above equation . correct answer is d ) 2 | a ) 5 , b ) 4 , c ) 3 , d ) 2 , e ) 1 | d | divide(subtract(2, multiply(multiply(3, 3), const_2)), subtract(const_1, multiply(3, 3))) | multiply(n0,n0)|multiply(#0,const_2)|subtract(const_1,#0)|subtract(n2,#1)|divide(#3,#2) | general | D |
what is 50 % of 40 % of 1200 grams ? | "50 / 100 Γ 40 / 100 Γ 1200 = 240 answer : d" | a ) 450 gms , b ) 100 gms , c ) 300 gms , d ) 240 gms , e ) none of these | d | divide(multiply(50, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100) | add(const_2,const_3)|add(const_3,const_4)|add(const_3,const_3)|multiply(const_3,const_4)|multiply(#0,const_2)|multiply(#3,const_100)|multiply(#1,#0)|multiply(#4,#5)|multiply(#6,#4)|add(#7,#8)|add(#9,#2)|multiply(n0,#10)|divide(#11,const_100)| | gain | D |
calculate 85184 Γ· ? = 352 | "answer let 85184 Γ· x = 242 then x = 85184 / 242 = 352 . option : b" | a ) 241 , b ) 242 , c ) 244 , d ) 247 , e ) 240 | b | multiply(85184, 352) | multiply(n0,n1)| | general | B |
a can do a work in 9 days and b can do the same work in 18 days . if they work together , in how many days will they complete the work ? | one day ' s work of a and b = 1 / 9 + 1 / 8 = ( 2 + 1 ) / 18 = 1 / 6 so , the time taken to complete the work is 6 days . answer : a | a ) 6 days , b ) 7 days , c ) 5 days , d ) 3 days , e ) 2 days | a | divide(const_1, add(divide(const_1, 9), divide(const_1, 18))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2) | physics | A |
a number when divided by 899 gives a remainder 63 . what remainder will be obtained by dividing the same number by 29 | 63 / 29 , thereforerequired number is : 5 , correct answer ( b ) | a ) 8 , b ) 5 , c ) 7 , d ) 6 , e ) 9 | b | subtract(63, multiply(29, const_2)) | multiply(n2,const_2)|subtract(n1,#0)| | general | B |
in a coconut grove , ( x + 4 ) trees yield 60 nuts per year , x trees yield 120 nuts per year and ( x β 4 ) trees yield 180 nuts per year . if the average yield per year per tree be 100 , find x . | ( x + 4 ) Γ 60 + x Γ 120 + ( x β 4 ) Γ 180 / ( x + 4 ) + x + ( x β 4 ) = 100 β 360 x β 480 / 3 x = 100 β 60 x = 480 β x = 8 answer c | a ) 3 , b ) 4 , c ) 8 , d ) 9 , e ) none of the above | c | divide(subtract(multiply(180, 4), multiply(60, 4)), subtract(add(add(60, 120), 180), multiply(100, const_3))) | add(n1,n2)|multiply(n0,n4)|multiply(n0,n1)|multiply(n5,const_3)|add(n4,#0)|subtract(#1,#2)|subtract(#4,#3)|divide(#5,#6) | general | C |
yearly subscription to professional magazines cost a company $ 940.00 . to make a 30 % cut in the magazine budget , how much less must be spent ? | "total cost 940 940 * 30 / 100 = 282 so the cut in amount is 282 the less amount to be spend is 940 - 282 = 658 answer : e" | a ) 654 , b ) 655 , c ) 656 , d ) 657 , e ) 658 | e | multiply(divide(subtract(const_100, 30), const_100), 940.00) | subtract(const_100,n1)|divide(#0,const_100)|multiply(n0,#1)| | general | E |
two equally sized jugs full of water are each emptied into two separate unequally sized empty jugs , x and y . now , jug x is 1 / 7 full , while jug y is 2 / 3 full . if water is poured from jug x into jug y until jug y is filled , what fraction of jug x then contains water ? | suppose the water in each jug is l liters cx x ( 1 / 7 ) = l cx = 7 l liters cx is capacity of x cy x ( 2 / 3 ) = l cy = 3 l / 2 liters cy is capacity of y now , y is 3 l / 2 - l empty = l / 2 empty so , we can put only l / 2 water in jug y from jug x jug x ' s remaining water = l - l / 2 = l / 2 fraction of x which contains water = water / cx = ( l / 2 ) / 7 l = 1 / 14 answer will be b | a ) 0 , b ) 1 / 14 , c ) 2 / 15 , d ) 1 / 10 , e ) 2 / 10 | b | multiply(divide(1, 7), divide(subtract(const_1, divide(2, 3)), divide(2, 3))) | divide(n0,n1)|divide(n2,n3)|subtract(const_1,#1)|divide(#2,#1)|multiply(#0,#3) | general | B |
difference between the length & breadth of a rectangle is 26 m . if its perimeter is 208 m , then the ratio of area and perimeter is ? ? we have : ( l - b ) = 26 and 2 ( l + b ) = 208 or ( l + b ) = 104 ? | "solving the two equations , we get : l = 63 and b = 40 . area = ( l x b ) = ( 63 x 40 ) m 2 = 2520 m ^ 2 the required ratio is 2520 : 208 = 315 : 26 a" | a ) 315 : 26 , b ) 315 : 25 , c ) 315 : 30 , d ) 415 : 35 , e ) 425 : 35 | a | divide(const_3, 2) | divide(const_3,n3)| | geometry | A |
if 4 men can reap 40 acres of land in 15 days , how many acres of land can 16 men reap in 30 days ? | "4 men 40 acres 15 days 16 men ? 30 days 40 * 16 / 4 * 30 / 15 40 * 4 * 2 40 * 8 = 320 answer : a" | a ) 320 , b ) 240 , c ) 369 , d ) 489 , e ) 125 | a | multiply(40, multiply(divide(16, 4), divide(30, 15))) | divide(n3,n0)|divide(n4,n2)|multiply(#0,#1)|multiply(n1,#2)| | physics | A |
1 / 4 of all the juniors and 2 / 3 of all the seniors are going on a trip . if there are 2 / 3 as many juniors as seniors , what fraction of the students are not going on the trip ? | j be the number of juniors and s be the number of seniors . there are 2 / 3 as many juniors as seniors 23 β s = j j / s = 23 going on trip = j 4 + 23 β s fraction = ( j 4 + 23 s ) / ( j + s ) = ( 2 / 3 * 1 / 4 + 2 / 3 ) / ( 2 / 3 + 1 ) = 1 / 2 fraction not on trip = 1 / 2 answer : b | a ) 4 / 9 , b ) 1 / 2 , c ) 2 / 3 , d ) 1 / 3 , e ) 5 / 6 | b | divide(add(multiply(divide(2, 3), multiply(4, 3)), multiply(multiply(divide(2, 3), multiply(4, 3)), divide(1, 4))), add(multiply(4, 3), multiply(divide(2, 3), multiply(4, 3)))) | divide(n2,n3)|divide(n0,n1)|multiply(n1,n3)|multiply(#0,#2)|add(#2,#3)|multiply(#1,#3)|add(#3,#5)|divide(#6,#4) | general | B |
a man can row upstream at 25 kmph and downstream at 65 kmph , and then find the speed of the man in still water ? | "us = 25 ds = 65 m = ( 65 + 25 ) / 2 = 45 answer : a" | a ) 45 , b ) 86 , c ) 30 , d ) 78 , e ) 38 | a | divide(add(25, 65), const_2) | add(n0,n1)|divide(#0,const_2)| | physics | A |
an art gallery has only paintings and sculptures . currently , 1 / 3 of the pieces of art are displayed , and 1 / 6 of the pieces on display are sculptures . if 1 / 3 of the pieces not on display are paintings , and 400 sculptures are not on display , how many pieces of art does the gallery have ? | "too many words and redundant info there . ( i ) 1 / 3 of the pieces of art are displayed , hence 2 / 3 of the pieces of art are not displayed . ( ii ) 1 / 6 of the pieces on display are sculptures , hence 5 / 6 of the pieces on display are paintings . ( iii ) 1 / 3 of the pieces not on display are paintings , hence 2 / 3 of the pieces not on display are sculptures . 400 sculptures are not on display , so according to ( iii ) 2 / 3 * { not on display } = 400 - - > { not on display } = 600 . according to ( i ) 2 / 3 * { total } = 600 - - > { total } = 900 . answer : b ." | a ) 360 , b ) 900 , c ) 540 , d ) 640 , e ) 720 | b | divide(divide(400, subtract(1, divide(1, 3))), subtract(1, divide(1, 3))) | divide(n0,n1)|subtract(n0,#0)|divide(n6,#1)|divide(#2,#1)| | general | B |
the principal that amounts to rs . 3913 in 3 years at 6 1 / 4 % per annum c . i . compounded annually , is ? | "principal = [ 4913 / ( 1 + 25 / ( 4 * 100 ) ) 3 ] = 3913 * 16 / 17 * 16 / 17 * 16 / 17 = rs . 3096 . answer : a" | a ) s . 3096 , b ) s . 4076 , c ) s . 4085 , d ) s . 4096 , e ) s . 5096 | a | divide(3913, power(add(1, divide(add(6, divide(1, 4)), const_100)), 3)) | divide(n3,n4)|add(n2,#0)|divide(#1,const_100)|add(#2,n3)|power(#3,n1)|divide(n0,#4)| | gain | A |
how many integers from 0 to 44 , inclusive , have a remainder of 1 when divided by 3 ? | "explanation : 1 also gives 1 remainder when divided by 3 , another number is 4 , then 7 and so on . hence we have an arithmetic progression : 1 , 4 , 7 , 10 , . . . . . 43 , which are in the form 3 n + 1 . now we have to find out number of terms . tn = a + ( n - 1 ) d , where tn is the nth term of an ap , a is the first term and d is the common difference . so , 43 = 1 + ( n - 1 ) 3 or , ( n - 1 ) 3 = 42 or , n - 1 = 12 or , n = 13 a" | a ) 13 , b ) 16 , c ) 17 , d ) 18 , e ) 19 | a | divide(44, const_10) | divide(n1,const_10)| | general | A |
mahesh marks an article 15 % above the cost price of rs . 540 . what must be his discount percentage if he sells it at rs . 459 ? | "cp = rs . 540 , mp = 540 + 15 % of 540 = rs . 621 sp = rs . 459 , discount = 621 - 459 = 162 discount % = 162 / 621 * 100 = 26.1 % answer : d" | a ) 18 % , b ) 21 % , c ) 20 % , d ) 26.1 % , e ) none of these | d | divide(multiply(subtract(multiply(540, divide(add(const_100, 15), const_100)), 459), const_100), multiply(540, divide(add(const_100, 15), const_100))) | add(n0,const_100)|divide(#0,const_100)|multiply(n1,#1)|subtract(#2,n2)|multiply(#3,const_100)|divide(#4,#2)| | gain | D |
the difference between a two - digit number and the number obtained by interchanging the digits is 36 . what is the difference between the sum and the difference of the digits of the number if the ratio between the digits of the number is 1 : 2 ? | "since the number is greater than the number obtained on reversing the digits , so the ten ' s digit is greater than the unit ' s digit . let ten ' s and unit ' s digits be 2 x and x respectively . then , ( 10 x 2 x + x ) - ( 10 x + 2 x ) = 36 9 x = 36 x = 4 . required difference = ( 2 x + x ) - ( 2 x - x ) = 2 x = 8 . answer : b" | a ) 4 , b ) 8 , c ) 16 , d ) 12 , e ) 14 | b | subtract(add(divide(36, subtract(subtract(add(multiply(2, const_10), 1), const_10), 2)), multiply(divide(36, subtract(subtract(add(multiply(2, const_10), 1), const_10), 2)), 2)), subtract(multiply(divide(36, subtract(subtract(add(multiply(2, const_10), 1), const_10), 2)), 2), divide(36, subtract(subtract(add(multiply(2, const_10), 1), const_10), 2)))) | multiply(n2,const_10)|add(n1,#0)|subtract(#1,const_10)|subtract(#2,n2)|divide(n0,#3)|multiply(n2,#4)|add(#4,#5)|subtract(#5,#4)|subtract(#6,#7)| | general | B |
a doctor prescribed 18 cubic centimeters of a certain drug to a patient whose body weight was 120 pounds . if the typical dosage is 2 cubic centimeters per 10 pounds of the body weight , by what percent was the prescribed dosage lesser than the typical dosage ? | typical dosage per 10 pound of the body weight = 2 c . c typical dosage per 120 pound of the body weight = 2 * ( 120 / 10 ) = 2 * 12 = 24 c . c dosage prescribed by doctor for 120 pound patient = 18 c . c % prescribed dosage greater than the typical dosage = ( 18 - 24 / 16 ) * 100 % = ( - 6 / 16 ) * 100 % = - 37.5 % answer b | a ) 8 % , b ) - 37.5 % , c ) 11 % , d ) 12.5 % , e ) 14.8 % | b | multiply(divide(subtract(multiply(divide(2, 10), 120), 18), multiply(divide(2, 10), 120)), const_100) | divide(n2,n3)|multiply(n1,#0)|subtract(#1,n0)|divide(#2,#1)|multiply(#3,const_100) | general | B |
given that 100.48 = x , 100.70 = y and xz = y Β² , then the value of z is close to | sol . xz = y Β² β ( 10 0.48 ) z ( 100.70 ) 2 ( 10 0.48 z ) = 10 ( 2 Γ 0.70 ) = 101.40 β 0.48 z = 1.40 β z = 140 / 48 = 35 / 12 = 2.9 ( approx ) . answer d | a ) 1.2 , b ) 1.45 , c ) 2.2 , d ) 2.9 , e ) none | d | divide(multiply(subtract(100.7, const_100), const_2), subtract(100.48, const_100)) | subtract(n1,const_100)|subtract(n0,const_100)|multiply(#0,const_2)|divide(#2,#1) | general | D |
a tempo is insured to an extent of 5 / 7 of its original value . if the premium on it at the rate of 3 % amounts to $ 300 , the original value of the tempo is ? | "let the original value of the tempo is $ x 3 % of 5 / 7 of x = 300 ( 3 / 100 ) * ( 5 / 7 ) x = 300 x = $ 14000 answer is d" | a ) 12500 , b ) 13280 , c ) 17520 , d ) 14000 , e ) 12560 | d | divide(300, divide(multiply(divide(5, 7), 3), const_100)) | divide(n0,n1)|multiply(n2,#0)|divide(#1,const_100)|divide(n3,#2)| | gain | D |
at the end of year x , automobile installment credit accounted for 43 % of all outstanding consumer installment credit . at that time automobile finance companies extended $ 50 billion of credit , or 1 / 4 of the automobile installment credit . how many billion dollars of consumer installment credit was outstanding at that time ? | system of equations a = ( 43 / 100 ) c ( 1 / 4 ) a = 50 - - > a = 200 substitution 200 = ( 43 / 100 ) c c = ( 100 / 43 ) 200 calculate 200 / 43 * 100 - the the correct answer is 465.12 . the correct answer is b . | a ) 450 , b ) 465.12 , c ) 475 , d ) 500.15 , e ) 200 | b | divide(multiply(50, 4), divide(43, const_100)) | divide(n0,const_100)|multiply(n1,n3)|divide(#1,#0) | general | B |
the arithmetic mean and standard deviation of a certain normal distribution are 10.5 and 1 , respectively . what value is exactly 2 standard deviations less than the mean ? | "mean = 10.5 two standard deviations is 1 + 1 = 2 there could be two calues for this . mean + two standard deviations = 12.5 mean - two standard deviations = 8.5 answer choice has 8.5 and so b is the answer ." | a ) 6 , b ) 8.5 , c ) 11.5 , d ) 12 , e ) 12.5 | b | subtract(10.5, multiply(2, 1)) | multiply(n1,n2)|subtract(n0,#0)| | general | B |
a dishonest dealer professes to sell goods at the cost price but uses a weight of 850 grams per kg , what is his percent ? | "850 - - - 150 100 - - - ? = > 17.64 % answer : b" | a ) 25 % , b ) 17 % , c ) 29 % , d ) 55 % , e ) 45 % | b | subtract(multiply(divide(const_100, 850), multiply(const_100, multiply(add(const_3, const_2), const_2))), const_100) | add(const_2,const_3)|divide(const_100,n0)|multiply(#0,const_2)|multiply(#2,const_100)|multiply(#1,#3)|subtract(#4,const_100)| | gain | B |
a bowl of fruit contains 14 apples and 23 oranges . how many oranges must be removed so that 70 % of the pieces of fruit in the bowl will be apples ? | number of apples = 14 number of oranges = 23 let number of oranges that must be removed so that 70 % of pieces of fruit in bowl will be apples = x total number of fruits after x oranges are removed = 14 + ( 23 - x ) = 37 - x 14 / ( 37 - x ) = 7 / 10 = > 20 = 37 - x = > x = 17 answer d | a ) 3 , b ) 6 , c ) 14 , d ) 17 , e ) 20 | d | divide(subtract(multiply(add(14, 23), divide(70, const_100)), 14), divide(70, const_100)) | add(n0,n1)|divide(n2,const_100)|multiply(#0,#1)|subtract(#2,n0)|divide(#3,#1) | gain | D |
what is the sum of 60 consecutive integers from - 30 inclusive , in a increasing order ? | from - 39 to - 1 - - > 30 nos . zero - - > 1 number from + 1 to + 29 - - > 29 nos . when we add up nos . from - 30 to + 29 sum will be - 30 for total 60 numbers . c is the answer . | a ) - 29 , b ) 29 , c ) - 30 , d ) 30 , e ) 60 | c | add(30, const_1) | add(n1,const_1) | general | C |
p and q together can do a work in 6 days . if p alone can do it in 15 days . in how many days can q alone do it ? | 1 / 6 - 1 / 15 = = > 1 / 10 = = > 10 answer c | a ) 5 , b ) 4 , c ) 10 , d ) 11 , e ) 12 | c | multiply(subtract(inverse(6), inverse(15)), const_100) | inverse(n0)|inverse(n1)|subtract(#0,#1)|multiply(#2,const_100) | physics | C |
according to the direction on a can of frozen orange juice concentrate is to be mixed with 3 cans of water to make orange juice . how many 20 - ounce cans of the concentrate are required to prepare 200 6 - ounce servings of orange juice ? | "orange juice concentrate : water : : 1 : 3 total quantity of orange juice = 200 * 6 = 1200 oz so orange juice concentrate : water : : 300 oz : 900 oz no . of 20 oz can = 300 oz / 20 oz = 15 answer a , 15 cans" | a ) 15 , b ) 34 , c ) 50 , d ) 67 , e ) 100 | a | divide(multiply(6, 200), divide(200, const_10)) | divide(n2,const_10)|multiply(n2,n3)|divide(#1,#0)| | general | A |
cost is expressed by the formula tb ^ 4 . if b is doubled , the new cost r is what percent of the original cost ? | original cost c 1 = t 1 * b 1 ^ 4 new cost c 2 = t 2 * b 2 ^ 4 . . . . only b is doubled so t 2 = t 1 and b 2 = 2 b 1 c 2 = t 2 * ( 2 b 1 ) ^ 4 = 16 ( t 1 * b 1 ^ 4 ) = 16 c 1 16 times c 1 = > 1600 % of c 1 ans d = 1600 | a ) r = 200 , b ) r = 600 , c ) r = 800 , d ) r = 1600 , e ) r = 50 | d | multiply(power(const_2, 4), const_100) | power(const_2,n0)|multiply(#0,const_100) | general | D |
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 [ 1.899 ] = 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 ? | "k = [ lcm of 80 and 45 ] * ( any integer ) however minimum value of k is sq . rt of 3 ^ 2 * 4 ^ 2 * 5 ^ 2 = 60 * any integer for value of k ( 60 ) * any integer unit value will be always zero . e" | a ) 0 , b ) 1 , c ) 27 , d ) 54 , e ) 0 | e | 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 | E |
if a and b are integers and ( a * b ) ^ 5 = 128 y , y could be : | distribute the exponent . a ^ 5 * b ^ 5 = 128 y find the prime factorization of 96 . this is 2 ^ 5 * 2 ^ 2 . we need 2 ^ 3 ( or some other power of 2 that will give us a multiple of 2 ^ 5 as our second term ) . 2 ^ 3 = 8 the answer is d . | a ) 1 , b ) 2 , c ) 4 , d ) 8 , e ) 16 | d | multiply(divide(128, power(const_2, 5)), const_2) | power(const_2,n0)|divide(n1,#0)|multiply(#1,const_2) | general | D |
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