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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in a certain company , a third of the workers do not have a retirement plan . 20 % of the workers who do not have a retirement plan are women , and 40 % of the workers who do have a retirement plan are men . if 144 of the workers of that company are men , how many of the workers are women ? | "set up equation : x = total number of workers 144 = 0,4 * 2 / 3 * x + 0,8 * 1 / 3 * x 144 = 16 / 30 x x = 270 270 - 144 = 126 answer e" | a ) 80 , b ) 95 , c ) 105 , d ) 120 , e ) 126 | e | multiply(divide(144, add(subtract(divide(const_1, const_3), multiply(divide(const_1, const_3), divide(20, const_100))), multiply(subtract(const_1, divide(const_1, const_3)), divide(40, const_100)))), add(multiply(divide(const_1, const_3), divide(20, const_100)), subtract(subtract(const_1, divide(const_1, const_3)), mul... | divide(const_1,const_3)|divide(n0,const_100)|divide(n1,const_100)|multiply(#0,#1)|subtract(const_1,#0)|multiply(#2,#4)|subtract(#0,#3)|add(#5,#6)|subtract(#4,#5)|add(#3,#8)|divide(n2,#7)|multiply(#9,#10)| | gain |
95 white and black tiles will be used to form a 10 x 10 square pattern . if there must be at least one black tile in every row and at least one white tile in every column , what is the maximum difference between the number of black and white tiles that can be used ? | answer = c please refer diagram below 95 - 10 = 85 | a ) 75 , b ) 80 , c ) 85 , d ) 90 , e ) 95 | c | subtract(95, 10) | subtract(n0,n1) | general |
in what time will a railway train 20 m long moving at the rate of 36 kmph pass a telegraph post on its way ? | "t = 60 / 36 * 18 / 5 = 2 sec answer : e" | a ) 6 sec , b ) 7 sec , c ) 8 sec , d ) 9 sec , e ) 2 sec | e | divide(20, multiply(36, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics |
what is the rate percent when the simple interest on rs . 800 amount to rs . 400 in 2 years ? | interest for 2 yrs = 400 interest for 1 yr = 200 interest rate = 200 / 800 x 100 = 25 % answer : d | a ) 5 % , b ) 6 % , c ) 2 % , d ) 25 % , e ) 1 % | d | divide(multiply(const_100, 400), multiply(800, 2)) | multiply(n1,const_100)|multiply(n0,n2)|divide(#0,#1) | gain |
if the area of a square with sides of length 9 centimeters is equal to the area of a rectangle with a width of 3 centimeters , what is the length of the rectangle , in centimeters ? | let length of rectangle = l 9 ^ 2 = l * 3 = > l = 81 / 3 = 27 answer b | ['a ) 4', 'b ) 27', 'c ) 12', 'd ) 16', 'e ) 18'] | b | divide(power(9, const_2), const_3) | power(n0,const_2)|divide(#0,const_3) | geometry |
taequan is playing a dice game where he rolls two , fair , 6 - sided dice twice . if he gets the same roll on his second try as his first , he wins . if he rolled an 8 on his first try , what is the probability that he will win with his second roll ? | there are 5 ways to roll an 8 : 2 and 6 , 6 and 2 , 3 and 5 , 5 and 3 , 4 and 4 ( 4 and 4 is counted just once ) . there are 6 * 6 = 36 ways to roll two six - sided dice . the dice are fair , so the probability of rolling an 8 is 5 / 36 . e | a ) 1 / 18 , b ) 1 / 12 , c ) 1 / 9 , d ) 1 / 6 , e ) 5 / 36 | e | divide(subtract(6, const_1), multiply(6, 6)) | multiply(n0,n0)|subtract(n0,const_1)|divide(#1,#0) | general |
car a runs at the speed of 100 km / hr & reaches its destination in 5 hr . car b runs at the speed of 80 km / h & reaches its destination in 6 h . what is the respective ratio of distances covered by car a & car b ? | "sol . distance travelled by car a = 100 × 5 = 500 km distance travelled by car b = 80 × 6 = 480 km ratio = 500 / 480 = 25 : 24 answer : d" | a ) 11 : 5 , b ) 11 : 8 , c ) 13 : 7 , d ) 25 : 24 , e ) 16 : 9 | d | divide(multiply(100, 5), multiply(80, 6)) | multiply(n0,n1)|multiply(n2,n3)|divide(#0,#1)| | physics |
george baked a total of 100 pizzas for 7 straight days , beginning on saturday . he baked 3 / 5 of the pizzas the first day , and 3 / 5 of the remaining pizzas the second day . if each successive day he baked fewer pizzas than the previous day , what is the maximum number of pizzas he could have baked on wednesday ? | "3 / 5 of the 100 pizzas cooked on saturday = 60 pizzas 3 / 5 of the remaining pizzas on sunday = 24 pizzas we ' re left with ( 100 - 60 - 24 ) = 16 pizzas for the remaining 5 days . the prompt tells us that each day has fewer pizzas than the day before it , so we ca n ' t have duplicate numbers . m t w th f 6 4 3 2 1 ... | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | divide(subtract(subtract(subtract(100, multiply(100, divide(3, 7))), multiply(subtract(100, multiply(100, divide(3, 7))), divide(3, 7))), 3), subtract(subtract(subtract(100, multiply(100, divide(3, 7))), multiply(subtract(100, multiply(100, divide(3, 7))), divide(3, 7))), 3)) | divide(n2,n1)|multiply(n0,#0)|subtract(n0,#1)|multiply(#0,#2)|subtract(#2,#3)|subtract(#4,n2)|divide(#5,#5)| | general |
pipe a usually fills a tank in 2 hours . on account of a leak at the bottom of the tank , it takes pipe a 30 more minutes to fill the tank . how long will the leak take to empty a full tank if pipe a is shut ? | pipe a fills the tank normally in 2 hours . therefore , it will fill ½ of the tank in an hour . let the leak take x hours to empty a full tank when pipe a is shut . therefore , the leak will empty of the tank in an hour . the net amount of water that gets filled in the tank in an hour when pipe a is open and when there... | a ) 7 hours , b ) 8 hours , c ) 9 hours , d ) 10 hours , e ) 11 hours | d | multiply(subtract(divide(const_1, 2), divide(const_1, add(2, divide(const_1, 2)))), const_100) | divide(const_1,n0)|add(n0,#0)|divide(const_1,#1)|subtract(#0,#2)|multiply(#3,const_100) | physics |
walking at 5 / 6 th of its usual speed a cab is 5 mnts late . find its usual time to cover the journey ? | "new speed = 5 / 6 th of usual speed new time = 6 / 5 th of usual time 6 / 5 ut - ut = 5 m ut / 5 = 5 m ut = 25 m answer is a" | a ) 25 m , b ) 45 m , c ) 32 m , d ) 50 m , e ) 62 m | a | multiply(5, 5) | multiply(n0,n2)| | physics |
what is the 8 th digit to the right of the decimal point in the decimal expansion of 1 / 37 ? | "1 / 37 = 0.027027 . . . so , we have a repeating cycle of 027 . every third digit ( 2 nd , 4 th , 6 th , . . . ) to the right of the decimal point is alternating by 2 , 0,7 , thus 8 th digit to the right of the decimal point is 2 . answer : b ." | a ) 0 , b ) 2 , c ) 4 , d ) 7 , e ) 9 | b | divide(1, 37) | divide(n1,n2)| | general |
find the value of x : 3 x ^ 2 - 6 x + 3 = 0 | a = 3 , b = - 6 , c = 3 x 1,2 = ( 6 ± √ ( ( - 6 ) 2 - 4 × 3 × 3 ) ) / ( 2 × 3 ) = ( 6 ± √ ( 36 - 36 ) ) / 6 = ( 6 ± 0 ) / 6 x 1 = x 2 = 1 b | a ) - 1 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | b | divide(add(6, 3), power(3, 2)) | add(n0,n2)|power(n0,n1)|divide(#0,#1) | general |
in a kilometer race , a beats b by 50 meters or 5 seconds . what time does a take to complete the race ? | "time taken by b run 1000 meters = ( 1000 * 5 ) / 50 = 100 sec . time taken by a = 100 - 10 = 90 sec . answer : d" | a ) 199 sec , b ) 190 sec , c ) 726 sec , d ) 90 sec , e ) 128 sec | d | subtract(divide(multiply(const_1, const_1000), divide(50, 5)), 5) | divide(n0,n1)|multiply(const_1,const_1000)|divide(#1,#0)|subtract(#2,n1)| | physics |
what is the smallest number which when increased by 3 is divisible by 27 , 35 , 25 , & 21 ? | "here smallest number means lcm hence lcm of 27,35 , 25,21 = 3 ^ 3 x 5 ^ 2 x 7 = 4725 therefore required number is 4725 - 3 = 4722 ans - a" | a ) 4722 , b ) 4725 , c ) 4728 , d ) 4731 , e ) 4733 | a | add(lcm(lcm(27, 35), lcm(25, 21)), 3) | lcm(n1,n2)|lcm(n3,n4)|lcm(#0,#1)|add(n0,#2)| | general |
a train 150 m long crosses a platform 130 m long in 20 sec ; find the speed of the train ? | "d = 150 + 130 = 280 t = 20 s = 280 / 20 * 18 / 5 = 50 kmph . answer : a" | a ) 50 , b ) 60 , c ) 55 , d ) 45 , e ) 53 | a | subtract(multiply(20, multiply(130, const_0_2778)), 150) | multiply(n1,const_0_2778)|multiply(n2,#0)|subtract(#1,n0)| | physics |
the ages of patrick and michael are in the ratio of 3 : 5 and that of michael and monica are in the ratio of 3 : 5 . if the sum of their ages is 245 , what is the difference between the ages of patrick and monica ? | "ages of p and mi = 3 x : 5 x ages of mi and mo = 3 x : 5 x rationalizing their ages . ratio of their ages will be 9 x : 15 x : 25 x sum = 49 x = 245 x = 5 difference if ages of pa and mo = 25 x - 9 x = 16 x = 16 * 5 = 80 answer c" | a ) 27 , b ) 48 , c ) 80 , d ) 72 , e ) 18 | c | subtract(multiply(multiply(5, 5), divide(245, add(add(multiply(3, 3), multiply(3, 5)), multiply(5, 5)))), multiply(multiply(3, 3), divide(245, add(add(multiply(3, 3), multiply(3, 5)), multiply(5, 5))))) | multiply(n0,n0)|multiply(n1,n0)|multiply(n1,n3)|add(#0,#1)|add(#3,#2)|divide(n4,#4)|multiply(#5,#2)|multiply(#5,#0)|subtract(#6,#7)| | general |
when 6 + 2 = 85 , 8 + 3 = 115 , 7 + 5 = 125 , then 5 + 8 = ? | "6 + 2 = > 6 + 2 = 8 = > 8 ã — 10 + 5 = 85 8 + 3 = > 8 + 3 = 11 = > 11 ã — 10 + 5 = 115 7 + 5 = > 7 + 5 = 12 = > 12 ã — 10 + 5 = 125 then 5 + 8 = > 5 + 8 = 13 = > 13 ã — 10 + 5 = 135 answer : d" | a ) 145 , b ) 185 , c ) 245 , d ) 135 , e ) 140 | d | add(multiply(multiply(5, 8), const_10), 6) | multiply(n7,n10)|multiply(#0,const_10)|add(n0,#1)| | general |
pipe a can fill a tank in 4 hours . due to a leak at the bottom , it takes 8 hours for the pipe a to fill the tank . in what time can the leak alone empty the full tank ? | "let the leak can empty the full tank in x hours 1 / 4 - 1 / x = 1 / 8 = > 1 / x = 1 / 4 - 1 / 8 = ( 2 - 1 ) / 8 = 1 / 8 = > x = 8 . answer : d" | a ) 13 , b ) 17 , c ) 18 , d ) 8 , e ) 12 | d | divide(multiply(8, 4), subtract(8, 4)) | multiply(n0,n1)|subtract(n1,n0)|divide(#0,#1)| | physics |
in how many ways 7 boys and 7 girls can be seated in a row so that boys and girls are alternate ? | "number of ways = 2 * 7 ! * 7 ! = 50803200 answer is a" | a ) 50803200 , b ) 50328000 , c ) 50003280 , d ) 50703200 , e ) 50603200 | a | multiply(factorial(7), factorial(7)) | factorial(n0)|factorial(n1)|multiply(#0,#1)| | probability |
how many two - element subsets of < 1 , 23 , 45 , 67 > are there that do not contain the pair of elements 2 and 4 ? | 7 c 2 - 1 = 20 answer is e | a ) 11 , b ) 12 , c ) 14 , d ) 15 , e ) 20 | e | subtract(23, const_3) | subtract(n1,const_3) | general |
a certain junior class has 3000 students and a certain senior class has 800 students . among these students , there are 60 siblings pairs each consisting of 1 junior and 1 senior . if 1 student is to be selected at random from each class , what is the probability that the 2 students selected will be a sibling pair ? | "let ' s see pick 60 / 3000 first then we can only pick 1 other pair from the 800 so total will be 60 / 800 * 3000 simplify and you get 1 / 40000 answer is a" | a ) 1 / 40000 , b ) 1 / 3600 , c ) 9 / 2000 , d ) 1 / 60 , e ) 1 / 15 | a | divide(1, const_3) | divide(n3,const_3)| | probability |
there is a sufficient food for 200 men for 20 days after 15 days 100 men leave the place . for how many days will the rest of the food last for the rest of the men ? | explanation : after 13 days , 100 men left , therefore 5 days * 200 men = x days * 100 men x = 10 days answer : option a | a ) 10 days , b ) 5 days , c ) 15 days , d ) 12 days , e ) 10 days | a | multiply(subtract(20, 15), const_2) | subtract(n1,n2)|multiply(#0,const_2) | physics |
a motorcyclist started riding at highway marker a , drove 120 miles to highway marker b , and then , without pausing , continued to highway marker c , where she stopped . the average speed of the motorcyclist , over the course of the entire trip , was 30 miles per hour . if the ride from marker a to marker b lasted 3 t... | "a - b = 120 miles b - c = 60 miles avg speed = 30 miles time taken for a - b 3 t and b - c be t avg speed = ( 120 + 60 ) / total time 30 = 180 / 4 t t = 90 b - c = 90 mph answer a" | a ) 90 , b ) 45 , c ) 50 , d ) 55 , e ) 60 | a | multiply(divide(divide(add(divide(120, const_2), 120), 30), const_4), divide(120, const_2)) | divide(n0,const_2)|add(n0,#0)|divide(#1,n1)|divide(#2,const_4)|multiply(#3,#0)| | physics |
how much 60 % of 50 is greater than 30 % of 30 ? | "( 60 / 100 ) * 50 – ( 30 / 100 ) * 30 30 - 9 = 21 answer : a" | a ) 21 , b ) 91 , c ) 11 , d ) 17 , e ) 12 | a | subtract(divide(multiply(60, 50), const_100), divide(multiply(30, 30), const_100)) | multiply(n0,n1)|multiply(n2,n3)|divide(#0,const_100)|divide(#1,const_100)|subtract(#2,#3)| | gain |
the average of 50 numbers id 50 . if two numbers , namely 45 and 55 are discarded , the average of the remaining numbers is : | "explanation : total of 50 numbers = ( 50 × 50 ) = 2500 total of 48 numbers = ( 2500 - ( 45 + 55 ) ] = 2400 required average = 2400 / 48 = 50 answer : e" | a ) 22 , b ) 33 , c ) 22 , d ) 41 , e ) 50 | e | divide(subtract(multiply(50, 50), add(45, 55)), subtract(50, const_2)) | add(n2,n3)|multiply(n0,n1)|subtract(n0,const_2)|subtract(#1,#0)|divide(#3,#2)| | general |
a can do a work in 9 days . b can do the same work in 18 days . if both a & b are working together in how many days they will finish the work ? | "a rate = 1 / 9 b rate = 1 / 18 ( a + b ) rate = ( 1 / 9 ) + ( 1 / 18 ) = 1 / 6 a & b finish the work in 6 days correct option is e" | a ) 3 , b ) 5 , c ) 4 , d ) 2 , e ) 6 | e | divide(multiply(9, 18), add(9, 18)) | add(n0,n1)|multiply(n0,n1)|divide(#1,#0)| | physics |
how many numbers between 100 and 672 are divisible by 2 , 3 , and 7 together ? | "explanation : as the division is by 2 , 3 , 7 together , the numbers are to be divisible by : 2 * 3 * 7 = 42 the limits are 100 and 672 the first number divisible is 42 * 3 = 126 to find out the last number divisible by 42 within 672 : 672 / 42 = 14 hence , 42 * 16 = 672 is the last number divisible by 42 within 672 h... | a ) 112 , b ) 77 , c ) 267 , d ) 14 , e ) 99 | d | subtract(divide(672, multiply(multiply(2, 3), 7)), divide(100, multiply(multiply(2, 3), 7))) | multiply(n2,n3)|multiply(n4,#0)|divide(n1,#1)|divide(n0,#1)|subtract(#2,#3)| | general |
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 $ 3,000 for that week . isabella makes 10 percent of her total sales for any given week . for what amount of weekly sales would angelo and isabella earn the same amount of m... | "official solution : the problem asks for the amount of weekly sales it takes for angelo and isabella to earn the same amount of money . you can write an equation that sets angelo ’ s and isabella ’ s weekly earnings equal to each other , with x representing weekly sales . weekly earnings for each salesperson equal bas... | a ) 23,500 , b ) 15,500 , c ) 25,500 , d ) 26,500 , e ) 27,500 | b | floor(divide(divide(subtract(550, multiply(3,000, divide(8, const_100))), subtract(divide(10, const_100), divide(8, const_100))), 3,000)) | 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 girl scout was selling boxes of cookies . in a month , she sold both boxes of chocolate chip cookies ( $ 1.25 each ) and boxes of plain cookies ( $ 0.75 each ) . altogether , she sold 1,585 boxes for a combined value of $ 1 , 586.75 . how many boxes of plain cookies did she sell ? | "let # plain cookies sold be x then # chocolate cookies = ( total cookies - x ) equating for x ( 0.75 ) * x + ( 1.25 ) * ( 1585 - x ) = 1586.75 = > x = 789" | a ) 0 , b ) 233 , c ) 500 , d ) 695 , e ) 789 | e | divide(add(const_1000, 586.75), const_2) | add(n4,const_1000)|divide(#0,const_2)| | other |
a man purchased 3 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 . 157 . find the unknown rate of two blankets ? | "10 * 157 = 1570 3 * 100 + 5 * 150 = 1050 1570 – 1050 = 520 answer : e" | a ) 420 , b ) 550 , c ) 490 , d ) 450 , e ) 520 | e | subtract(multiply(const_10, 150), add(multiply(3, 100), multiply(5, 150))) | multiply(n3,const_10)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|subtract(#0,#3)| | general |
0.0004 ? = 0.02 | "explanation : required answer = 0.0004 / 0.02 = 0.04 / 2 = 0.02 . answer : option a" | a ) 0.02 , b ) 0.2 , c ) 2 , d ) 20 , e ) none of these | a | multiply(divide(0.0004, 0.02), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
a bag contains 10 red jellybeans and 10 blue jellybeans . if 3 jellybeans are removed one at a time , at random and are not replaced , what is the probability t that all 3 jellybeans removed from the bag are blue ? | "method - 1 10 red jellybeans and 10 blue jellybeans total outcomes = no . of ways to choose 3 jelly bean at random out of a total 20 jellybeans = 20 c 3 = 1140 favourable outcomes = no . of ways to choose 3 jelly bean such that they are all blue out of 10 blue = 10 c 3 = 120 probability = favourable outcomes / total o... | a ) 9 / 100 , b ) 2 / 19 , c ) 1 / 8 , d ) 3 / 20 , e ) 3 / 10 | b | divide(choose(10, 3), choose(add(10, 10), 3)) | add(n0,n0)|choose(n0,n2)|choose(#0,n2)|divide(#1,#2)| | probability |
the length of a rectangle is reduced by 25 % . by what % would the width have to be increased to maintainthe original area ? | "sol . required change = ( 25 * 100 ) / ( 100 - 25 ) = 25 % d" | a ) 15 % , b ) 20 % , c ) 25 % , d ) 33 % , e ) 40 % | d | multiply(divide(subtract(const_1, divide(subtract(const_100, 25), const_100)), divide(subtract(const_100, 25), const_100)), const_100) | subtract(const_100,n0)|divide(#0,const_100)|subtract(const_1,#1)|divide(#2,#1)|multiply(#3,const_100)| | geometry |
a survey was sent to 95 customers , 7 of whom responded . then the survey was redesigned and sent to another 63 customers , 9 of whom responded . by approximately what percent did the response rate increase from the original survey to the redesigned survey ? | "case 1 : ( 7 / 95 ) = x / 100 x = 7 % case 2 : ( 9 / 63 ) = y / 100 y = 14 % so percent increase is = ( y - x ) = ( 14 - 7 ) % = 7 % answer is c" | a ) 2 % , b ) 5 % , c ) 7 % , d ) 28 % , e ) 63 % | c | floor(multiply(subtract(divide(9, 63), divide(7, 95)), const_100)) | divide(n3,n2)|divide(n1,n0)|subtract(#0,#1)|multiply(#2,const_100)|floor(#3)| | gain |
tough and tricky questions : word problems . ak car company wants to manufacture a new car known as model x , and it is trying to determine how many cars it needs to sell in order to make an annual profit of $ 30 , 500000 . the annual fixed costs for model x total $ 50 , 200000 . in addition , each model x car has an a... | the ak company plans to make a profit of $ 30 , 500000 and it needs to earn the fixed costs of $ 50 , 200000 , so it needs to earn $ 80 , 700000 on top of the costs for the production of the 20000 cars . therefore , price of each car needs to be the unit costs plus $ 80 , 700000 / 20000 . after removing the 0 ' s , we ... | a ) $ 4035 , b ) $ 4036 , c ) $ 9035 , d ) $ 16140 , e ) $ 36140 | c | divide(add(add(add(multiply(50, multiply(const_1000, const_1000)), 200000), multiply(20000, 5000)), add(multiply(30, multiply(const_1000, const_1000)), 500000)), 20000) | multiply(const_1000,const_1000)|multiply(n4,n5)|multiply(n2,#0)|multiply(n0,#0)|add(n3,#2)|add(n1,#3)|add(#4,#1)|add(#6,#5)|divide(#7,n5) | general |
according to the directions on a packet of smoothie mix , 1 3 - ounce packet of smoothie mix is to be combined with 15 ounces of water to make a smoothie . how many 3 - ounce packets of smoothie mix are required to prepare 150 12 - ounce smoothies ? | "this question was n ' t particularly grueling , but i think it ' s the first where i had the opportunity to solve it via theory andinspectionthat many on this board suggest as strategy on the gmat . it actually came to me by accident . basically , if we thought that the 3 packets of powder were included in the 12 ounc... | a ) 120 , b ) 150 , c ) 180 , d ) 240 , e ) 600 | c | add(150, multiply(3, const_10)) | multiply(n1,const_10)|add(n4,#0)| | general |
how many 3 - digit numbers are divisible by 6 in all ? | solution required numbers are 102 , 018114 , . . . . 996 . this is an a . p with a = 102 , d = 6 . let the number of its terms be n . then a + ( n - 1 ) d ‹ = › 102 + ( n - 1 ) × 6 = 996 ‹ = › n = 150 . answer b | a ) 149 , b ) 150 , c ) 151 , d ) 166 , e ) none | b | subtract(divide(const_1000, 6), divide(const_100, 6)) | divide(const_1000,n1)|divide(const_100,n1)|subtract(#0,#1) | general |
two pipes can fill a tank in 10 minutes and 25 minutes . an outlet pipe can empty the tank in 45 minutes . if all the pipes are opened when the tank is empty , then how many minutes will it take to fill the tank ? | "part of the filled by all the three pipes in one minute = 1 / 10 + 1 / 25 - 1 / 45 = ( 45 + 18 - 10 ) / 450 = 53 / 450 so , the tank becomes full in 7 minutes . answer : c" | a ) 30 minutes , b ) 17 minutes , c ) 7 minutes , d ) 10 minutes , e ) 12 minutes | c | subtract(add(divide(const_1, 10), divide(const_1, 25)), divide(const_1, 45)) | divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|subtract(#3,#2)| | physics |
if 7 ^ k = 2 , then 7 ^ ( 4 k + 2 ) = | "7 ^ k = 2 7 ^ 4 k = 2 ^ 4 7 ^ 4 k = 16 7 ^ ( 4 k + 2 ) = 7 ^ 4 k * 7 ^ 2 = 16 * 49 = 784 answer : c" | a ) 729 , b ) 754 , c ) 784 , d ) 783 , e ) 108 | c | multiply(power(2, 4), power(7, 2)) | power(n1,n3)|power(n0,n4)|multiply(#0,#1)| | general |
one day a car rental agency rented 3 / 4 of its cars , including 3 / 5 of its cars with cd players . if 3 / 5 of its cars have cd players , what percent of the cars that were not rented had cd players ? | "the cars with cd players which were not rented is ( 2 / 5 ) ( 3 / 5 ) = 6 / 25 of all the cars . the cars which were not rented is 1 / 4 of all the cars . the percent of non - rented cars which had cd players is ( 6 / 25 ) / ( 1 / 4 ) = 24 / 25 the answer is b ." | a ) 29 / 30 , b ) 24 / 25 , c ) 19 / 20 , d ) 14 / 15 , e ) 9 / 10 | b | multiply(divide(subtract(divide(multiply(4, multiply(multiply(4, 3), 5)), 5), divide(multiply(divide(multiply(4, multiply(multiply(4, 3), 5)), 5), 4), 3)), subtract(multiply(multiply(4, 3), 5), divide(multiply(3, multiply(multiply(4, 3), 5)), 4))), const_100) | multiply(n1,n2)|multiply(n3,#0)|multiply(n1,#1)|multiply(#1,const_2)|divide(#2,n3)|divide(#3,n1)|multiply(n1,#4)|subtract(#1,#5)|divide(#6,n2)|subtract(#4,#8)|divide(#9,#7)|multiply(#10,const_100)| | general |
little john had $ 5.10 . he spent $ 1.05 on sweets and gave to his two friends $ 1.00 each . how much money was left ? | "john spent and gave to his two friends a total of 1.05 + 1.00 + 1.00 = $ 3.05 money left 5.10 - 3.05 = $ 2.05 answer : b" | a ) $ 2.15 , b ) $ 2.05 , c ) $ 2.45 , d ) $ 2.25 , e ) $ 2.85 | b | subtract(5.10, add(1.05, add(1.00, 1.00))) | add(n2,n2)|add(n1,#0)|subtract(n0,#1)| | general |
a fair coin is tossed 3 times . what is the probability of getting at least 2 heads ? | "let ' s find the probability of 2 heads , 3 heads p ( hhh ) = ( ( 1 / 2 ) ^ 3 = 1 / 8 . p ( hht ) = ( 3 ! / 2 ! ) * ( 1 / 2 ) ^ 3 = 3 / 8 total probablity = 1 / 8 + 3 / 8 = 1 / 2 answer ( d )" | a ) 3 / 4 , b ) 3 / 2 , c ) 1 / 4 , d ) 1 / 2 , e ) 1 / 5 | d | divide(add(add(add(choose(3, const_2), choose(3, const_3)), choose(3, const_4)), choose(3, 3)), power(const_2, 3)) | choose(n0,const_2)|choose(n0,const_3)|choose(n0,const_4)|choose(n0,n0)|power(const_2,n0)|add(#0,#1)|add(#5,#2)|add(#6,#3)|divide(#7,#4)| | probability |
a , b and c enter into partnership . a invests some money at the beginning , b invests double the amount after 6 months , and c invests thrice the amount after 8 months . if the annual gain be rs . 18900 . a ' s share is ? | "x * 12 : 2 x * 6 : 3 x * 4 1 : 1 : 1 1 / 3 * 18900 = 6300 answer : a" | a ) 6300 , b ) 2765 , c ) 6000 , d ) 1298 , e ) 1261 | a | multiply(multiply(const_1, const_12), divide(18900, add(add(multiply(const_1, const_12), multiply(subtract(const_12, 6), const_2)), multiply(subtract(const_12, 8), const_3)))) | multiply(const_1,const_12)|subtract(const_12,n0)|subtract(const_12,n1)|multiply(#1,const_2)|multiply(#2,const_3)|add(#0,#3)|add(#5,#4)|divide(n2,#6)|multiply(#7,#0)| | gain |
bottle r contains 250 capsules and costs $ 8.25 . bottle t contains 130 capsules and costs $ 2.99 . what is the difference between the cost per capsule for bottle r and the cost per capsule for bottle t ? | "cost per capsule in r is 8.25 / 250 = 0.825 / 25 = 0.033 cost per capsule in t is 2.99 / 130 = 0.023 the difference is 0.01 the answer is b" | a ) $ 0.25 , b ) $ 0.01 , c ) $ 0.05 , d ) $ 0.03 , e ) $ 0.002 | b | subtract(divide(2.99, 130), divide(8.25, 250)) | divide(n3,n2)|divide(n1,n0)|subtract(#0,#1)| | general |
if albert ’ s monthly earnings rise by 30 % , he would earn $ 598 . if , instead , his earnings rise by only 35 % , how much ( in $ ) would he earn this month ? | "= 598 / 1.3 ∗ 1.35 = 621 = 621 answer is a" | a ) 621 , b ) 652 , c ) 660 , d ) 690 , e ) 693 | a | multiply(divide(598, add(const_1, divide(30, const_100))), add(const_1, divide(35, const_100))) | divide(n2,const_100)|divide(n0,const_100)|add(#0,const_1)|add(#1,const_1)|divide(n1,#3)|multiply(#2,#4)| | gain |
the unit digit in the product 3421 * 6721 * 4584 * 6784 is ? | "unit digit in the given product = unit digit in 1 * 1 * 4 * 4 = 6 answer is c" | a ) 0 , b ) 1 , c ) 6 , d ) 3 , e ) 4 | c | subtract(multiply(multiply(multiply(3421, 6721), 4584), 6784), subtract(multiply(multiply(multiply(3421, 6721), 4584), 6784), add(const_4, const_4))) | add(const_4,const_4)|multiply(n0,n1)|multiply(n2,#1)|multiply(n3,#2)|subtract(#3,#0)|subtract(#3,#4)| | general |
the sum of the present ages of two persons a and b is 60 . if the age of a is twice that of b , find the sum of their ages 4 years hence ? | "a + b = 60 , a = 2 b 2 b + b = 60 = > b = 20 then a = 40 . 4 years , their ages will be 44 and 24 . sum of their ages = 44 + 24 = 68 . answer : d" | a ) 22 , b ) 77 , c ) 70 , d ) 68 , e ) 71 | d | add(add(multiply(divide(60, 4), const_2), 4), add(divide(60, 4), 4)) | divide(n0,n1)|add(#0,n1)|multiply(#0,const_2)|add(#2,n1)|add(#3,#1)| | general |
robert is travelling on his cycle andhas calculated to reach point a at 2 pm . if he travels at 10 kmph , he will reach there at 12 pm if he travels at 15 kmph . at what speed must he travel to reach a at 1 pm ? | let the distance travelled by x km . then , x - x = 2 10 15 3 x - 2 x = 60 x = 60 km . time taken to travel 60 km at 10 km / hr = 60 hrs = 6 hrs . 10 so , robert started 6 hours before 2 p . m . i . e . , at 8 a . m . required speed = 60 kmph . = 12 kmph . 5 c | a ) 8 kmph , b ) 9 kmph , c ) 12 kmph , d ) 14 kmph , e ) 16 kmph | c | divide(divide(2, subtract(divide(const_1, 10), divide(const_1, 15))), subtract(divide(divide(2, subtract(divide(const_1, 10), divide(const_1, 15))), 10), 1)) | divide(const_1,n1)|divide(const_1,n3)|subtract(#0,#1)|divide(n0,#2)|divide(#3,n1)|subtract(#4,n4)|divide(#3,#5) | physics |
let f ( x , y ) be defined as the remainder when ( x – y ) ! is divided by x . if x = 36 , what is the maximum value of y for which f ( x , y ) = 0 ? | "the question is finding y such that ( 36 - y ) ! is a multiple of 36 . that means we need to have 2 ^ 2 * 3 ^ 2 in ( 36 - y ) ! 6 ! is the smallest factorial number with 2 ^ 2 * 3 ^ 2 as a factor . 36 - y = 6 y = 30 the answer is e ." | a ) 9 , b ) 12 , c ) 18 , d ) 20 , e ) 30 | e | subtract(36, multiply(const_2, divide(divide(36, const_2), add(const_1, const_4)))) | add(const_1,const_4)|divide(n0,const_2)|divide(#1,#0)|multiply(#2,const_2)|subtract(n0,#3)| | general |
a computer factory produces 4032 computers per month at a constant rate , how many computers are built every 30 minutes assuming that there are 28 days in one month ? | number of hours in 28 days = 28 * 24 number of 30 mins in 28 days = 28 * 24 * 2 number of computers built every 30 mins = 4032 / ( 28 * 24 * 2 ) = 3 answer b | a ) 2.25 . , b ) 3.00 . , c ) 4.5 . , d ) 5.225 . , e ) 6.25 . | b | divide(divide(4032, 28), multiply(subtract(28, const_4), const_2)) | divide(n0,n2)|subtract(n2,const_4)|multiply(#1,const_2)|divide(#0,#2) | physics |
the ratio of the radius of two circles is 1 : 4 , and then the ratio of their areas is ? | "r 1 : r 2 = 1 : 4 π r 12 : π r 22 r 12 : r 22 = 1 : 16 answer : a" | a ) 1 : 16 , b ) 1 : 6 , c ) 1 : 9 , d ) 1 : 3 , e ) 1 : 2 | a | divide(circle_area(1), circle_area(4)) | circle_area(n0)|circle_area(n1)|divide(#0,#1)| | geometry |
the average age of a class of 39 students is 15 years . if the age of the teacher be included , then the average increased by 3 months . find the age of the teacher ? | "total age of 39 persons = 39 * 15 = 585 years average age of 40 persons = 15 yrs 3 months = 61 / 4 years total age of 40 persons = 61 / 4 * 40 = 610 years age of the teacher = 610 - 585 = 25 years answer is a" | a ) 25 , b ) 28 , c ) 30 , d ) 35 , e ) 39 | a | subtract(multiply(add(39, 3), add(15, 3)), multiply(39, 15)) | add(n0,n2)|add(n1,n2)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2)| | general |
country x imported approximately $ 1.44 billion of goods in 1996 . if country x imported $ 488 million of goods in the first two months of 1997 and continued to import goods at the same rate for the rest of the year , by how much would country xs 1997 imports exceed those of 1996 ? | convert units to millions as answer is in millions 1996 imports = $ 1.44 bill = $ 1440 mill i . e . 1440 / 12 = $ 120 mill / month 1997 imports = $ 488 mill / 2 month i . e . $ 244 mill / month difference / month = 244 - 120 = 124 difference / year = $ 124 mill * 12 = $ 1488 mill answer : e | a ) $ 1124 million , b ) $ 1120 million , c ) $ 1144 million , d ) $ 1240 million , e ) $ 1488 million | e | subtract(multiply(488, divide(const_12, const_2)), multiply(1.44, const_1000)) | divide(const_12,const_2)|multiply(n0,const_1000)|multiply(n2,#0)|subtract(#2,#1) | general |
the ratio of 2 numbers is 2 : 7 and their h . c . f . is 18 . their l . c . m . is ? | "let the numbers be 2 x and 7 x their h . c . f . = 18 so the numbers are 2 * 18 , 7 * 18 = 36,126 l . c . m . = 252 answer is a" | a ) 252 , b ) 243 , c ) 521 , d ) 361 , e ) 600 | a | sqrt(divide(18, add(power(7, 2), add(power(2, 2), power(2, 2))))) | power(n0,n1)|power(n1,n1)|power(n2,n1)|add(#0,#1)|add(#3,#2)|divide(n3,#4)|sqrt(#5)| | other |
110 people are attending a newspaper conference . 45 of them are writers and more than 38 are editors . of the people at the conference , x are both writers and editors and 2 x are neither . what is the largest possible number of people who are both writers and editors ? | "{ total } = { writers } + { editors } - { both } + { neither } . { total } = 110 ; { writers } = 45 ; { editors } > 38 ; { both } = x ; { neither } = 2 x ; 110 = 45 + { editors } - x + 2 x - - > x = 65 - { editors } . we want to maximize x , thus we should minimize { editors } , minimum possible value of { editors } i... | a ) 34 , b ) 32 , c ) 30 , d ) 26 , e ) 28 | d | subtract(110, add(add(38, const_1), 45)) | add(n2,const_1)|add(n1,#0)|subtract(n0,#1)| | other |
in may mrs lee ' s earnings were 70 percent of the lee family ' s total income . in june mrs lee earned 20 percent more than in may . if the rest of the family ' s income was the same both months , then , in june , mrs lee ' s earnings were approximately what percent of the lee family ' s total income ? | "let in may lee family ' s total income = 100 in may mrs lee ' s income = 70 in may rest of the family ' s income = 30 in june mrs lees income = 70 * 120 / 100 = 84 in june total income = 84 + 30 = 114 % of mrs lee ' s income = 84 / 114 = 73.68 ( d )" | a ) 64 % , b ) 68 % , c ) 72 % , d ) 74 % , e ) 83 % | d | multiply(divide(add(const_100, 20), add(add(const_100, 20), const_100)), const_100) | add(n1,const_100)|add(#0,const_100)|divide(#0,#1)|multiply(#2,const_100)| | general |
one year ago , the ratio of roonie ’ s and ronaldo ’ s age was 6 : 7 respectively . 4 years hence , this ratio would become 7 : 8 . how old is ronaldo ? | 7 ( x - 1 ) = 6 ( y - 1 ) = > 7 x - 6 y - 1 = 0 - - - - eq 1 8 ( x + 4 ) = 7 ( y + 4 ) = > 8 x - 7 y + 4 = 0 - - - - eq 2 on solving eq 1 n eq 2 y = 36 answer : e | a ) 32 , b ) 33 , c ) 34 , d ) 35 , e ) 36 | e | add(divide(multiply(7, subtract(multiply(8, add(4, const_1)), multiply(7, add(4, const_1)))), subtract(multiply(7, 7), multiply(6, 8))), const_1) | add(n2,const_1)|multiply(n1,n1)|multiply(n0,n4)|multiply(n4,#0)|multiply(n1,#0)|subtract(#1,#2)|subtract(#3,#4)|multiply(n1,#6)|divide(#7,#5)|add(#8,const_1) | other |
a sum of money is to be distributed among a , b , c , d in the proportion of 5 : 2 : 4 : 3 . if c gets rs . 900 more than d , what is b ' s share ? | "let the shares of a , b , c and d be rs . 5 x , rs . 2 x , rs . 4 x and rs . 3 x respectively . then , 4 x - 3 x = 900 x = 900 . b ' s share = rs . 2 x = rs . ( 2 x 900 ) = rs . 1800 . answer = b" | a ) rs . 500 , b ) rs . 1800 , c ) rs . 2000 , d ) rs . 2500 , e ) none of the above | b | multiply(multiply(subtract(4, 3), 900), 3) | subtract(n2,n3)|multiply(n4,#0)|multiply(n3,#1)| | general |
the speed at which a man can row a boat in still water is 14 kmph . if he rows downstream , where the speed of current is 2 kmph , what time will he take to cover 40 metres ? | "speed of the boat downstream = 14 + 2 = 16 kmph = 16 * 5 / 18 = 40 / 9 m / s hence time taken to cover 40 m = 40 * 9 / 40 = 9 seconds . answer : c" | a ) 16 seconds , b ) 34 seconds , c ) 9 seconds , d ) 12 seconds , e ) 15 seconds | c | divide(40, multiply(add(14, 2), const_0_2778)) | add(n0,n1)|multiply(#0,const_0_2778)|divide(n2,#1)| | physics |
a , b and c started a business with a total investment of rs . 90000 . a invests rs . 6000 more than b and b invests rs . 3000 less than c . if the total profit at the end of a year is rs . 8640 , find c ' s share . | "explanation : let c ' s investment = rs . x b ' s investment = rs . ( x - 3000 ) a ' s investment = rs . ( x - 3000 + 6000 ) = rs . ( x + 3000 ) now , ( a + b + c ) ' s investment = rs . 90000 = > x + ( x - 3000 ) + ( x + 3000 ) = 90000 = > 3 x = 90000 = > x = 30000 hence , a ' s investment = rs . 33000 b ' s investme... | a ) rs . 3240 , b ) rs . 2520 , c ) rs . 2880 , d ) rs . 3360 , e ) none of these | c | multiply(8640, divide(add(divide(subtract(90000, add(6000, 3000)), const_3), 6000), 90000)) | add(n1,n2)|subtract(n0,#0)|divide(#1,const_3)|add(n1,#2)|divide(#3,n0)|multiply(n3,#4)| | general |
the ratio of the present age of sandy to that of molly is 9 : 2 . sixteen years from now , the ratio of the ages of sandy to molly will be 5 : 2 . what was sandy ' s age 6 years ago ? | "let the present age of sandy be 9 x years and that of molly be 2 x years . ( 9 x + 16 ) / ( 2 x + 16 ) = 5 / 2 8 x = 48 x = 6 six years ago , sandy ' s age was 9 ( 6 ) - 6 = 48 the answer is d ." | a ) 30 , b ) 36 , c ) 42 , d ) 48 , e ) 54 | d | subtract(divide(multiply(subtract(multiply(5, add(const_10, 6)), multiply(2, add(const_10, 6))), 9), add(2, 2)), 6) | add(n4,const_10)|add(n1,n1)|multiply(n2,#0)|multiply(n1,#0)|subtract(#2,#3)|multiply(n0,#4)|divide(#5,#1)|subtract(#6,n4)| | general |
how many 3 - digits number are palindromic numbers ? a palindromic number reads the same forward and backward , example 121 . | "take the task of building palindromes and break it intostages . stage 1 : select the hundreds digit we can choose 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , or 9 so , we can complete stage 1 in 9 ways stage 2 : select the tens digit we can choose 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , or 9 so , we can complete stage 2 in 10 ways im... | a ) 90 , b ) 610 , c ) 729 , d ) 900 , e ) 1000 | a | multiply(multiply(multiply(multiply(3, const_10), const_10), const_10), const_10) | multiply(n0,const_10)|multiply(#0,const_10)|multiply(#1,const_10)|multiply(#2,const_10)| | general |
the average of the 5 numbers is 6.8 . if one of the numbers is multiplied by 3 , the average of the numbers increases to 9.2 . which of the 5 numbers is multiplied by 3 ? | initial avg is 6.8 so the summation of the nos is 6.8 * 5 = 34 now the new avg is 9.2 so summation will be 9.2 * 5 = 46 . so change in summation is 12 . this comes when a number is multiplied by 3 . in the original sum we already have that number added once . to bring additional 12 increment by multiplying by 3 , we ge... | a ) 1.5 , b ) 3.0 , c ) 3.9 , d ) 4.0 , e ) 6.0 | e | divide(subtract(multiply(9.2, 5), multiply(6.8, 5)), const_2) | multiply(n0,n3)|multiply(n0,n1)|subtract(#0,#1)|divide(#2,const_2) | general |
( 3192 + 6160 ) / 28 | "explanation : as per bodmas rule , first we will solve the equation in bracket then we will go for division = ( 9352 ) / 28 = 334 option c" | a ) 380 , b ) 350 , c ) 334 , d ) 310 , e ) none of these | c | multiply(3192, power(add(const_4, const_1), const_4)) | add(const_1,const_4)|power(#0,const_4)|multiply(n0,#1)| | general |
a ferry can transport 100 tons of vehicles . automobiles range in weight from 1,800 to 3,200 pounds . what is the greatest number of automobiles that can be loaded onto the ferry ? | "to get maximum vehicles we must take into consideration the minimum weight i . e 1800 pounds here since , 1 ton = 2000 pounds 78 tons will be 200,000 pounds from the answer choices : let max number of vehicles be 110 total weight will be = 110 * 1800 = 198000 pounds , which is lesser than the maximum weight allowed . ... | a ) 115 , b ) 120 , c ) 110 , d ) 100 , e ) 90 | c | divide(multiply(multiply(100, const_2), const_1000), add(add(add(add(add(add(const_1000, const_100), const_100), const_100), const_100), const_100), const_100)) | add(const_100,const_1000)|multiply(n0,const_2)|add(#0,const_100)|multiply(#1,const_1000)|add(#2,const_100)|add(#4,const_100)|add(#5,const_100)|add(#6,const_100)|divide(#3,#7)| | general |
when positive integer n is divided by positive integer j , the remainder is 25 . if n / j = 204.05 , what is value of j ? | "when a number is divided by another number , we can represent it as : dividend = quotient * divisor + remainder so , dividend / divisor = quotient + remainder / divisor given that n / j = 204.05 here 204 is the quotient . given that remainder = 25 so , 204.05 = 204 + 25 / j so , j = 500 answer : b" | a ) 600 , b ) 500 , c ) 800 , d ) 200 , e ) 900 | b | divide(25, subtract(204.05, add(const_100, add(multiply(const_4, const_10), const_2)))) | multiply(const_10,const_4)|add(#0,const_2)|add(#1,const_100)|subtract(n1,#2)|divide(n0,#3)| | general |
a , b , c subscribe rs . 50000 for a business . a subscribes rs . 4000 more than b and b rs . 5000 more than c . out of a total profit of rs . 30000 , b receives : | let c = x . then , b = x + 5000 and a = x + 5000 + 4000 = x + 9000 . so , x + x + 5000 + x + 9000 = 50000 3 x = 36000 x = 12000 a : b : c = 21000 : 17000 : 12000 = 21 : 17 : 12 . b ' s share = rs . ( 30000 x 17 / 50 ) = rs . 10,200 . b | a ) s . 10,000 , b ) s . 10,200 , c ) s . 10,400 , d ) s . 10,700 , e ) s . 10,800 | b | multiply(30000, divide(divide(add(50000, subtract(5000, 4000)), const_3), 50000)) | subtract(n2,n1)|add(n0,#0)|divide(#1,const_3)|divide(#2,n0)|multiply(n3,#3) | general |
a man swims downstream 48 km and upstream 18 km taking 3 hours each time , what is the speed of the man in still water ? | "48 - - - 3 ds = 16 ? - - - - 1 18 - - - - 3 us = 6 ? - - - - 1 m = ? m = ( 16 + 6 ) / 2 = 11 answer : d" | a ) 2 , b ) 8 , c ) 9 , d ) 11 , e ) 14 | d | divide(add(divide(18, 3), divide(48, 3)), const_2) | divide(n1,n2)|divide(n0,n2)|add(#0,#1)|divide(#2,const_2)| | physics |
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 can complete a project in 20 days and b can complete the same project in 30 days . if a and b start working on the project together and b quits 15 days before the project is completed , in how many days total will the project be completed ? | "a ' s rate is 1 / 20 of the project per day . b ' s rate is 1 / 30 of the project per day . the combined rate is 1 / 12 of the project per day . in the last 15 days , a can do 3 / 4 of the project . thus a and b must complete 1 / 4 of the project , which takes 3 days . the total number of days is 3 + 15 = 18 . the ans... | a ) 12 , b ) 14 , c ) 16 , d ) 18 , e ) 20 | d | add(divide(subtract(const_1, multiply(divide(const_1, 30), 15)), add(divide(const_1, 20), divide(const_1, 30))), 15) | divide(const_1,n1)|divide(const_1,n0)|add(#1,#0)|multiply(n2,#0)|subtract(const_1,#3)|divide(#4,#2)|add(n2,#5)| | physics |
a meeting has to be conducted with 4 managers . find the number of ways in which the managers be selected from among 7 managers , if 2 managers will not attend the meeting together ? | "we can either choose all 4 people from 5 manager who have no problems or choose 3 from the 5 and 1 from the 2 managers who have a problem sitting together so 5 c 4 + ( 5 c 3 * 2 c 1 ) this is 5 + 20 = 25 answer : a" | a ) 25 , b ) 20 , c ) 35 , d ) 36 , e ) 38 | a | subtract(choose(7, 4), choose(subtract(7, 2), 2)) | choose(n1,n0)|subtract(n1,n2)|choose(#1,n2)|subtract(#0,#2)| | probability |
two trains each 150 m in length each , are running on two parallel lines in opposite directions . if one goes at the speed of 95 km / h while the other travels at 85 km / h . how long will it take for them to pass each other completely . | "explanation : d = 150 m + 150 m = 300 m rs = 95 + 85 = 180 * 5 / 18 = 50 t = 300 * 1 / 50 = 6 sec answer : option e" | a ) 25 sec , b ) 23 sec , c ) 34 sec , d ) 22 sec , e ) 6 sec | e | multiply(const_3600, divide(divide(add(150, 150), const_1000), add(95, 85))) | add(n0,n0)|add(n1,n2)|divide(#0,const_1000)|divide(#2,#1)|multiply(#3,const_3600)| | physics |
p is 10 % more efficient than q . p can complete a work in 21 days . if p and q work together , how many days will it take to complete the same work ? | "the work done by p in 1 day = 1 / 21 let work done by q in 1 day = q q × ( 110 / 100 ) = 1 / 21 q = 100 / ( 21 × 110 ) = 10 / ( 21 × 11 ) the work done by p and q in 1 day = 1 / 21 + 10 / ( 21 × 11 ) = 21 / ( 21 × 11 ) = 1 / 11 p and q together can do the work in 11 days . the answer is c ." | a ) 7 , b ) 9 , c ) 11 , d ) 13 , e ) 15 | c | divide(multiply(add(divide(10, const_100), const_1), 21), add(add(divide(10, const_100), const_1), const_1)) | divide(n0,const_100)|add(#0,const_1)|add(#1,const_1)|multiply(n1,#1)|divide(#3,#2)| | physics |
the average weight of a , b and c is 44 kg . if the average weight of a and b be 40 kg and that of b and c be 43 kg , then the weight of b is : | "let d sum of a , b , c is 3 * 44 = 132 and sum of a and b s 2 * 40 = 80 sum of b and c is 2 * 43 = 86 hence 80 + 86 - 132 = 28 ans = 28 answer : b" | a ) 17 kg , b ) 28 kg , c ) 26 kg , d ) 31 kg , e ) 35 kg | b | subtract(add(multiply(40, const_2), multiply(43, const_2)), multiply(44, const_3)) | multiply(n1,const_2)|multiply(n2,const_2)|multiply(n0,const_3)|add(#0,#1)|subtract(#3,#2)| | general |
a , b , k start from the same place and travel in the same direction at speeds of 30 km / hr , 40 km / hr , 100 km / hr respectively . b starts two hours after a . if b and k overtake a at the same instant , how many hours after a did k start ? | "in 2 hours , a travels 60 km . b can catch a at a rate of 10 km / hr , so b catches a 6 hours after b starts . so a and b both travel a distance of 240 km . c needs 2.4 hours to travel 240 km , so c leaves 5.6 hours after a . the answer is c ." | a ) 4.2 , b ) 4.8 , c ) 5.6 , d ) 6.4 , e ) 7.2 | c | subtract(40, 30) | subtract(n1,n0)| | physics |
a shopkeeper bought 600 oranges and 400 bananas . he found 15 % of oranges and 4 % of bananas were rotten . find the percentage of fruits in good condition ? | "total number of fruits shopkeeper bought = 600 + 400 = 1000 number of rotten oranges = 15 % of 600 = 15 / 100 × 600 = 9000 / 100 = 90 number of rotten bananas = 4 % of 400 = 16 therefore , total number of rotten fruits = 90 + 16 = 106 therefore number of fruits in good condition = 1000 - 106 = 894 therefore percentage... | a ) 92.5 % , b ) 89.4 % , c ) 85.2 % , d ) 96.8 % , e ) 78.9 % | b | multiply(divide(subtract(add(600, 400), add(multiply(600, divide(15, const_100)), multiply(400, divide(4, const_100)))), add(600, 400)), const_100) | add(n0,n1)|divide(n2,const_100)|divide(n3,const_100)|multiply(n0,#1)|multiply(n1,#2)|add(#3,#4)|subtract(#0,#5)|divide(#6,#0)|multiply(#7,const_100)| | gain |
james took a 3 - hour bike ride . in the second hour he travlled 18 miles , which was 20 percent farther than he traveled the first hour . if he traveled 25 percent farther in the third hour than he did in the second hour , how many miles did jose travel during the entire ride ? | let the distance travelled in the first hour be x . thus , 1.2 x = 18 , x = 15 . now , the distance travelled in the 3 rd hour = 18 + 1 / 4 ∗ 18 = the only option ending with a 0.5 in the decimal place is c . answer : c | a ) 54.0 , b ) 54.9 , c ) 55.5 , d ) 57.0 , e ) 63.0 | c | add(add(divide(multiply(multiply(const_1, const_100), 18), add(multiply(const_1, const_100), 20)), divide(multiply(25, 18), 20)), 18) | multiply(const_1,const_100)|multiply(n1,n3)|add(n2,#0)|divide(#1,n2)|multiply(n1,#0)|divide(#4,#2)|add(#5,#3)|add(n1,#6) | physics |
by weight , liquid x makes up 2 percent of solution p and 7.5 percent of solution q . if 400 grams of solution p are mixed with 600 grams of solution q , then liquid x accounts for what percent of the weight of the resulting solution ? | "the number of grams of liquid x is 2 ( 400 ) / 100 + 7.5 ( 600 ) / 100 = 8 + 45 = 53 grams . 53 / 1000 = 5.3 % the answer is e ." | a ) 4.5 % , b ) 4.7 % , c ) 4.9 % , d ) 5.1 % , e ) 5.3 % | e | multiply(divide(add(const_1, divide(multiply(7.5, 600), const_100)), const_1000), const_100) | multiply(n1,n3)|divide(#0,const_100)|add(#1,const_1)|divide(#2,const_1000)|multiply(#3,const_100)| | gain |
a train traveling at 216 kmph crosses a platform in 30 seconds and a man standing on the platform in 12 seconds . what is the length of the platform in meters ? | "answer distance covered by the train when crossing a man and when crossing a platform when a train crosses a man standing on a platform , the distance covered by the train is equal to the length of the train . however , when the same train crosses a platform , the distance covered by the train is equal to the length o... | a ) 240 meters , b ) 360 meters , c ) 420 meters , d ) 720 meters , e ) can not be determined | d | subtract(multiply(divide(multiply(216, const_1000), const_3600), 30), multiply(divide(multiply(216, const_1000), const_3600), 12)) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)|multiply(n2,#1)|subtract(#2,#3)| | physics |
for my anniversary , i decided to surprise my wife . since she is a voracious reader , i decided to collect a lot of books for her . on the first day of the month , i bought one book , on the second , i bought two and on the third , i bought three . this process went on till the anniversary and on the day , i had 190 b... | c 19 rd day 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 = 190 thus our anniversary falls on the 19 rd day of the month . you can apply other formulas to shorten the process but that is the simplest way to do it . | a ) 31 , b ) 28 , c ) 19 , d ) 33 , e ) 25 | c | divide(190, const_10) | divide(n0,const_10) | physics |
at a tanning salon , customers are charged $ 10 for their first visit in a calendar month and $ 3 for each visit after that in the same calendar month . in the last calendar month , 100 customers visited the salon , of which 30 made a second visit , and 10 made a third visit . all other customers made only one visit . ... | "i get b . this question seems too straightforward for 600 + . am i missing something ? 100 first - time visits - - > 100 ( 10 ) = $ 1000 30 + 10 = 40 subsequent visits - - > 40 ( 3 ) = $ 120 total revenue : 1000 + 120 = $ 1120 the answer is b ." | a ) $ 1220 , b ) $ 1120 , c ) $ 1300 , d ) $ 1340 , e ) $ 1880 | b | add(multiply(add(10, 3), 30), multiply(subtract(100, 30), 10)) | add(n0,n1)|subtract(n2,n3)|multiply(n3,#0)|multiply(n0,#1)|add(#2,#3)| | physics |
there are 7 players in a bowling team with an average weight of 94 kg . if two new players join the team , one weighs 110 kg and the second weighs 60 kg , what will be the new average weight ? | "the new average will be = ( 94 * 7 + 110 + 60 ) / 9 = 92 kgs e is the answer" | a ) 75 kg . , b ) 80 kg . , c ) 86 kg . , d ) 90 kg . , e ) 92 kg . | e | divide(add(multiply(7, 94), add(110, 60)), add(7, const_2)) | add(n2,n3)|add(n0,const_2)|multiply(n0,n1)|add(#0,#2)|divide(#3,#1)| | general |
the average weight of 12 person ' s increases by 4 kg when a new person comes in place of one of them weighing 58 kg . what is the weight of the new person ? | "total increase in weight = 12 × 4 = 48 if x is the weight of the new person , total increase in weight = x − 58 = > 48 = x - 58 = > x = 48 + 58 = 106 answer : e" | a ) 85 , b ) 80 , c ) 95 , d ) 90 , e ) 106 | e | add(multiply(12, 4), 58) | multiply(n0,n1)|add(n2,#0)| | general |
a cube 125 mt is colored pink on two opposite surfaces and then cut in to 125 smaller cubes then find how many number of cubes have pink color ? | answer : b | ['a ) 3', 'b ) 4', 'c ) 5', 'd ) 6', 'e ) 7'] | b | floor(cube_edge_by_volume(125)) | cube_edge_by_volume(n0)|floor(#0) | geometry |
6 ) a marketing firm determined that , of 180 households surveyed , 80 used neither brand a nor brand b soap . 60 used only brand a soap and for every household that used both brands of soap , 3 used only brand b soap . how many of the 200 household surveyed used both brands of soap ? | 100 = at least one of soap a or b both brands = x brand b = 3 x = > 60 + x + 3 x = 100 = > 4 x = 40 = > x = 10 answer - d | a ) a ) 15 , b ) b ) 20 , c ) c ) 30 , d ) d ) 10 , e ) e ) 45 | d | divide(subtract(subtract(180, 80), 60), const_4) | subtract(n1,n2)|subtract(#0,n3)|divide(#1,const_4) | other |
one machine takes 3 minutes to check fault in 100 km fiber optic cable . another machine does the same job in 3 seconds . what time will it take to do the same job when both machines are put to work ? | by guess it is clear that the time taken will be less than 3 seconds and very near to it . therefore , answer 2.95 seconds will be correct . answer - d | a ) 2.00 seconds , b ) 2.20 seconds , c ) 2.50 seconds , d ) 2.95 seconds , e ) 3.00 seconds | d | inverse(add(inverse(multiply(3, const_60)), inverse(3))) | inverse(n0)|multiply(n0,const_60)|inverse(#1)|add(#2,#0)|inverse(#3) | physics |
a certain telescope increases the visual range at a particular location from 60 kilometers to 150 kilometers . by what percent is the visual range increased by using the telescope ? | "original visual range = 60 km new visual range = 150 km percent increase in the visual range by using the telescope = ( 150 - 60 ) / 60 * 100 % = 3 / 2 * 100 % = 150 % answer e" | a ) 30 % , b ) 33 1 / 2 % , c ) 40 % , d ) 60 % , e ) 150 % | e | multiply(divide(subtract(150, 60), 60), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)| | general |
a small company reduced its faculty by approximately 13 percent to 181 employees . what was the original number of employees ? | "if x is the original number of employees , then after 13 % reduction in employees number is . 87 x but we are given . 87 x = 181 x = 208 so the original number of employees is 208 correct answer - b" | a ) a ) 182 , b ) b ) 208 , c ) c ) 220 , d ) d ) 224 , e ) e ) 302 | b | divide(181, divide(subtract(const_100, 13), const_100)) | subtract(const_100,n0)|divide(#0,const_100)|divide(n1,#1)| | gain |
66 small identical cubes are used to form a large cube . how many more cubes are needed to add one top layer of small cube all over the surface of the large cube ? | "66 small cube will make a large cube with 4 cubes in each line i . e . adding one layer will require one cube at each end and hence new cube will have 6 cubes in each line . total number of small cubes in new cube = 6 ^ 3 = 216 extra cube required = 216 - 66 = 150 hence , b is the answer ." | a ) 64 , b ) 150 , c ) 152 , d ) 216 , e ) 256 | b | subtract(volume_cube(add(cube_edge_by_volume(66), const_2)), 66) | cube_edge_by_volume(n0)|add(#0,const_2)|volume_cube(#1)|subtract(#2,n0)| | geometry |
0.9 * 0.007 = _________ | 9 * 7 = 63 sum of decimal places = 4 so , 0.9 * 0.007 = 0.0063 answer : b | a ) 0.063 , b ) 0.0063 , c ) 0.63 , d ) 0.00063 , e ) 63 | b | multiply(0.9, 0.007) | multiply(n0,n1) | general |
a train passes a station platform in 31 sec and a man standing on the platform in 20 sec . if the speed of the train is 54 km / hr . what is the length of the platform ? | "speed = 54 * 5 / 18 = 15 m / sec . length of the train = 15 * 20 = 300 m . let the length of the platform be x m . then , ( x + 300 ) / 31 = 15 = > x = 165 m . answer : c" | a ) 228 , b ) 240 , c ) 165 , d ) 166 , e ) 1811 | c | multiply(20, multiply(54, const_0_2778)) | multiply(n2,const_0_2778)|multiply(n1,#0)| | physics |
the length of a rectangular garden is three times its width . if the area of the rectangular garden is 768 square meters , then what is the width of the rectangular garden ? | "let x be the width of the garden . 3 x ^ 2 = 768 x ^ 2 = 256 x = 16 the answer is a ." | a ) 16 , b ) 17 , c ) 18 , d ) 19 , e ) 20 | a | sqrt(divide(768, const_3)) | divide(n0,const_3)|sqrt(#0)| | geometry |
mr . hernandez , who was a resident of state x for only 9 months last year , had a taxable income of $ 42,500 for the year . if the state tax rate were 4 percent of the year ’ s taxable income prorated for the proportion of the year during which the taxpayer was a resident , what would be the amount of mr . hernandez ’... | total tax for the year = 42,500 x 4 % = 1700 as stated annual tax is prorated as per the duration of stay . prorated tax = 1700 ( 9 / 12 ) = 1275 answer b | a ) $ 900 , b ) $ 1275 , c ) $ 1200 , d ) $ 1300 , e ) $ 1060 | b | divide(multiply(multiply(divide(add(multiply(multiply(4, const_100), const_100), multiply(multiply(const_100, const_0_25), const_100)), const_100), const_4), 9), multiply(const_3, const_4)) | multiply(n2,const_100)|multiply(const_0_25,const_100)|multiply(const_3,const_4)|multiply(#0,const_100)|multiply(#1,const_100)|add(#3,#4)|divide(#5,const_100)|multiply(#6,const_4)|multiply(n0,#7)|divide(#8,#2) | gain |
if a man lost 10 % by selling oranges at the rate of 10 a rupee at how many a rupee must he sell them to gain 50 % ? | "90 % - - - - 10 150 % - - - - ? 90 / 150 * 10 = 6 answer : a" | a ) 6 , b ) 8 , c ) 7 , d ) 4 , e ) 2 | a | divide(multiply(subtract(const_100, 10), 10), add(const_100, 50)) | add(n2,const_100)|subtract(const_100,n0)|multiply(n1,#1)|divide(#2,#0)| | gain |
a candidate appearing for an examination has to secure 42 % marks to pass paper i . but he secured only 42 marks and failed by 22 marks . what is the maximum mark for paper i ? | "he secured 42 marks nd fail by 22 marks so total marks for pass the examinatn = 64 let toal marks x x * 42 / 100 = 64 x = 152 answer : d" | a ) 110 , b ) 120 , c ) 130 , d ) 152 , e ) 150 | d | divide(add(42, 22), divide(42, const_100)) | add(n1,n2)|divide(n0,const_100)|divide(#0,#1)| | gain |
if ( - 3 ) ^ ( 2 x ) = 3 ^ ( 12 - x ) and x is an integer , what is the value of x ? | since x is an integer , ( - 3 ) ^ ( 2 x ) is always positive . so , 3 ^ 2 x = 3 ^ ( 12 - x ) 2 x = 12 - x 3 x = 12 x = 4 answer : b | a ) 5 , b ) 4 , c ) 3 , d ) 2 , e ) 1 | b | divide(12, 3) | divide(n3,n0) | general |
in a division sum , the divisor is 10 times the quotient and 5 times the remainder . if the remainder is 46 , what is the dividend ? | "divisor = ( 5 * 46 ) = 230 10 * quotient = 230 quotient = 230 / 10 = 23 . dividend = ( divisor * quotient ) + remainder = ( 230 * 23 ) + 46 = ( 5290 + 46 ) = 5336 . answer is a" | a ) 5336 , b ) 5242 , c ) 6524 , d ) 3526 , e ) 4526 | a | multiply(multiply(5, 46), divide(multiply(5, 46), 10)) | multiply(n1,n2)|divide(#0,n0)|multiply(#1,#0)| | general |
find the perimeter and area of a square of side 11 cm . | "we know that the perimeter of square = 4 × side side = 11 cm therefore , perimeter = 4 × 11 cm = 44 cm now , area of the square = ( side × side ) sq . units = 11 × 11 cm ² = 121 cm ² answer : option e" | a ) 11 cm ² , b ) 21 cm ² , c ) 22 cm ² , d ) 113 cm ² , e ) 121 cm ² | e | square_area(11) | square_area(n0)| | geometry |
a company plans to gradually replace its fleet of 20 cars with newer models . at the end of every year starting with 2000 it retires 5 of its old cars and buys 6 new models . how many years did it take before the number of older cars left in the company ' s fleet was less than 50 percent of the fleet ? | total fleet - - 20 end of year 2000 - - ( retires 5 and buys 6 ) total fleet - - 21 ( 15 old 6 new ) . end of year 2001 - - ( retires 5 and buys 6 ) total fleet - - 22 ( 10 old 12 new ) . answer is c - - after 2 years the company has a total of 22 cars ( 10 old 12 new ) , the old cars are already less than 50 % of its ... | a ) 6 , b ) 3 , c ) 2 , d ) 4 , e ) 5 | c | floor(add(divide(20, add(const_10, const_1)), const_1)) | add(const_1,const_10)|divide(n0,#0)|add(#1,const_1)|floor(#2) | gain |
the ratio of radius of a circle and the side of a square is 2 : 3 . find the ratio of their areas : | "radius / side = 2 / 3 â ‡ ’ area of circle / area of square = 4 / 9 answer : b" | a ) 2 : 1 , b ) 4 : 9 , c ) 8 : 77 , d ) 4 : 121 , e ) none | b | power(divide(2, 3), 2) | divide(n0,n1)|power(#0,n0)| | geometry |
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