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a can do a piece of work 30 days . b can do work in 45 days . in how many days they will complete the work together ? | "lcm = 90 , ratio = 30 : 45 = 2 : 3 no of days = 90 / ( 2 + 3 ) = 90 / 5 = 18 days answer : e" | a ) 2.5 , b ) 6 , c ) 80 , d ) 50 % , e ) 18 days | e | divide(const_1, add(divide(const_1, 30), divide(const_1, 45))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)| | physics |
the ratio of 2 numbers is 3 : 7 and their h . c . f . is 12 . their l . c . m . is ? | "let the numbers be 3 x and 7 x their h . c . f . = 3 so the numbers are 3 * 12 , 7 * 12 = 36,84 l . c . m . = 252 answer is c" | a ) 20 , b ) 252 , c ) 9 , d ) x + 2 x + 2 , e ) 150 | b | sqrt(divide(12, add(power(7, 3), add(power(2, 3), power(3, 3))))) | power(n0,n1)|power(n1,n1)|power(n2,n1)|add(#0,#1)|add(#3,#2)|divide(n3,#4)|sqrt(#5)| | other |
average between two sets of numbers is closer to the set with morenumbers ? | "if on a test three people answered 90 % of the questions correctly and two people answered 80 % correctly , then the average for the group is not 85 % but rather 3 Γ 90 + 2 Γ 805 = 4305 = 86.3 Γ 90 + 2 Γ 805 = 4305 = 86 . here , 90 has a weight of 3 = > it occurs 3 times . whereas 80 has a weight of 2 = > it occurs 2 times . so the average is closer to 90 than to 80 as we have just calculated . b" | a ) 10.8 km / hr , b ) 23 , c ) $ 20 , d ) 80 , e ) 40.46 | d | multiply(multiply(const_2, const_4), const_10) | multiply(const_2,const_4)|multiply(#0,const_10)| | general |
a can do a piece of work in 20 days and b alone can do it in 10 days . b works at it for 5 days and then leaves . a alone can finish the remaining work in | "explanation : b ' s 5 days work = 1 / 10 β 5 = 1 / 2 remaining work = 1 β 1 / 2 = 1 / 2 a can finish work = 20 β 1 / 2 = 10 days option a" | a ) 71 % , b ) rs 10123.20 , c ) 10 days , d ) 54.2 % , e ) 80 % | c | divide(const_1, add(divide(const_1, 20), divide(const_1, 10))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)| | physics |
what is the probability that the sum of two dice will yield a 6 , and then when both are thrown again , their sum will again yield a 6 ? assume that each die has 5 sides with faces numbered 1 to 5 . | solution - rolling dices is an independent event . the combinations to get 6 are ( 1,5 ) , ( 5,1 ) , ( 2,4 ) , ( 4,2 ) , ( 3,3 ) and total combinations of both dices is 25 . the probability of getting 6 in first attempt is 5 / 25 = 1 / 5 . probability of getting 6 again in second attempt = ( 1 / 5 ) * ( 1 / 5 ) = 1 / 25 . ans b | a ) 10.4 , b ) 3 , c ) 1 / 25 , d ) 11.25 Β° , e ) 1575 | c | multiply(divide(5, power(5, const_2)), divide(5, power(5, const_2))) | power(n2,const_2)|divide(n2,#0)|multiply(#1,#1) | probability |
the average of runs of a cricket player of 10 innings was 34 . how many runs must he make in his next innings so as to increase his average of runs by 4 ? | "explanation : average = total runs / no . of innings = 34 so , total = average x no . of innings = 34 x 10 = 340 . now increase in avg = 4 runs . so , new avg = 34 + 4 = 38 runs total runs = new avg x new no . of innings = 38 x 11 = 418 runs made in the 11 th inning = 418 - 340 = 78 answer : a" | a ) 66 , b ) 3108 , c ) 78 , d ) 180 m , e ) 350 m | c | subtract(multiply(add(10, const_1), add(4, 34)), multiply(10, 34)) | add(n0,const_1)|add(n1,n2)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2)| | general |
a truck covers a distance of 288 km at a certain speed in 8 hours . how much time would a car take at an average speed which is 18 kmph more than that of the speed of the truck to cover a distance which is 90 km more than that travelled by the truck ? | "explanation : speed of the truck = distance / time = 288 / 8 = 36 kmph now , speed of car = ( speed of truck + 18 ) kmph = ( 36 + 18 ) = 54 kmph distance travelled by car = 288 + 90 = 378 km time taken by car = distance / speed = 378 / 54 = 7 hours . answer β c" | a ) 7 hours , b ) 1 : 2 , c ) $ 14.00 , d ) 2599980 , e ) 0.16 | a | divide(add(288, 90), add(divide(288, 8), 18)) | add(n0,n3)|divide(n0,n1)|add(n2,#1)|divide(#0,#2)| | physics |
a , b and c play a cricket match . the ratio of the runs scored by them in the match is a : b = 2 : 3 and b : c = 2 : 5 . if the total runs scored by all of them are 75 , the runs scored by b are ? a . 15 b . 18 | a : b = 2 : 3 b : c = 2 : 5 a : b : c = 4 : 6 : 15 6 / 25 * 75 = 18 answer : b | a ) 18 , b ) 9.87 % , c ) 4691530800 , d ) 100 kg , e ) 28000 | a | multiply(divide(75, add(add(multiply(divide(2, 3), divide(2, 5)), divide(2, 5)), const_1)), divide(2, 5)) | divide(n0,n3)|divide(n0,n1)|multiply(#1,#0)|add(#0,#2)|add(#3,const_1)|divide(n4,#4)|multiply(#5,#0) | general |
if 65 percent of a class answered the first question on a certain test correctly , 50 percent answered the second question on the test correctly , and 20 percent answered neither of the questions correctly , what percent answered both correctly ? | "65 % answered the first question correctly and 20 % answered neither correctly . then 15 % missed the first question but answered the second question correctly . then the percent who answered both correctly is 50 % - 15 % = 35 % . the answer is c ." | a ) 150 m , b ) 35 % , c ) $ 1050 , d ) $ 900 , e ) 550 sq . units', ' | b | subtract(add(add(65, 50), 20), const_100) | add(n0,n1)|add(n2,#0)|subtract(#1,const_100)| | other |
john and ingrid pay 30 % and 40 % tax annually , respectively . if john makes $ 56000 and ingrid makes $ 73000 , what is their combined tax rate ? | "( 1 ) when 30 and 40 has equal weight or weight = 1 / 2 , the answer would be 35 . ( 2 ) when 40 has larger weight than 30 , the answer would be in between 35 and 40 . unfortunately , we have 2 answer choices d and e that fit that condition so we need to narrow down our range . ( 3 ) get 73000 / 129000 = 73 / 129 is a little above 1 / 2 . thus , our answer is just a little above 35 . answer : d" | a ) 35.6 % , b ) 445 , c ) d , d ) 40 , e ) 75 | a | multiply(divide(add(multiply(divide(30, const_100), 56000), multiply(divide(40, const_100), 73000)), add(73000, 56000)), const_100) | add(n2,n3)|divide(n0,const_100)|divide(n1,const_100)|multiply(n2,#1)|multiply(n3,#2)|add(#3,#4)|divide(#5,#0)|multiply(#6,const_100)| | gain |
the average ( arithmetic mean ) of 20 , 40 , and 60 is 6 more than the average of 10 , 70 , and what number ? | "a 1 = 120 / 3 = 40 a 2 = a 1 - 6 = 34 sum of second list = 34 * 3 = 102 therefore the number = 102 - 80 = 22 answer : d" | a ) 22 , b ) rs . 27500 , c ) 6 , d ) 31.67 % , e ) 8 | a | subtract(add(add(20, 40), 60), add(add(multiply(6, const_3), 10), 70)) | add(n0,n1)|multiply(n3,const_3)|add(n2,#0)|add(n4,#1)|add(n5,#3)|subtract(#2,#4)| | general |
if 150 ! / 10 ^ n is an integer , what is the largest possible value of n ? | the question actually asks the highest power of 10 which divides 150 ! ( for a number to be an integer - without any remainder all the trailing zeroe ' s must be divided by the denominator ) 10 = 2 x 5 150 factorial will have 37 as - 150 / 5 = 30 30 / 5 = 6 6 / 5 = 1 so answer will be ( e ) 37 | a ) 48 days , b ) 3584 , c ) 20 , d ) 37 , e ) 660 | d | add(divide(divide(150, add(const_4, const_1)), add(const_4, const_1)), divide(150, add(const_4, const_1))) | add(const_1,const_4)|divide(n0,#0)|divide(#1,#0)|add(#2,#1) | general |
the average height of 20 students in a class was calculated as 175 cm . it has later found that the height of one of the students in the class was incorrectly written as 151 cm whereas the actual height was 131 cm . what was the actual average height of the students in the class ? | "the total height was 20 cm too much . the average height should be reduced by 20 cm / 20 = 1 cm the answer is a ." | a ) 683 , b ) 144 min , c ) 115 , d ) 40 , e ) 174 cm | e | divide(subtract(multiply(20, 175), subtract(151, 131)), 20) | multiply(n0,n1)|subtract(n2,n3)|subtract(#0,#1)|divide(#2,n0)| | general |
a certain number of workers can do a work in 85 days . if there were 10 workers more it could be finished in 10 days less . how many workers are there ? | "number of workers = 10 * ( 85 - 10 ) / 10 = 75 answer is a" | a ) 770 , b ) 75 , c ) 996004 , d ) 1070 , e ) 50', ' | b | divide(multiply(subtract(85, 10), 10), subtract(85, subtract(85, 10))) | subtract(n0,n1)|multiply(n1,#0)|subtract(n0,#0)|divide(#1,#2)| | physics |
in cliff β s impressive rock collection , there are half as many igneous rocks as sedimentary rocks . of the igneous rocks , 4 / 5 are shiny and the rest are matte , while 1 / 5 of the sedimentary rocks are shiny . if there are 40 shiny igneous rocks , how many total rocks does cliff have ? | "we can start with the known quantity and then go on to find the others . shiny igneous ricks are 40 . these are ( 4 / 5 ) of total igneous rocks . ( 4 / 5 ) * total igneous rocks = 40 total igneous rocks = 40 * ( 5 / 4 ) = 50 total sedimentary rocks = 2 * total igneous rocks = 2 * 50 = 100 total number of rocks = 50 + 100 = 150 answer ( b )" | a ) 150 , b ) 1 / 5 , c ) 4327 , d ) 1 : 2', ' , e ) 49 | a | divide(multiply(divide(multiply(multiply(4, 5), 40), 4), 5), 4) | multiply(n0,n1)|multiply(n4,#0)|divide(#1,n0)|multiply(n1,#2)|divide(#3,n0)| | general |
car z travels 48 miles per gallon of gasoline when driven at a constant rate of 45 miles per hour , but travels 20 percent fewer miles per gallon of gasoline when driven at a constant rate of 60 miles per hour . how many miles does car z travel on 10 gallons of gasoline when driven at a constant rate of 60 miles per hour ? | "the question stem asks us for the distance possible with 10 gallons of fuel at a constant speed of 60 miles per hour . we therefore first calculate the fuel efficiency at that speed . the stem tells us that at 45 miles / hour , the car will run 48 miles / gallon and at 60 miles / hour , that distance decreases by 20 % . we can therefore conclude that the car will travel 38.4 miles / gallon at a constant speed of 60 miles / gallon . with 10 gallons of fuel , the car can therefore travel 38.4 miles / gallon * 10 gallons = 384 miles . answer b ." | a ) 384 , b ) s . 200 , c ) 90 , d ) 35 km , e ) 2240 | a | multiply(multiply(subtract(const_1, divide(20, const_100)), 48), 10) | divide(n2,const_100)|subtract(const_1,#0)|multiply(n0,#1)|multiply(n4,#2)| | gain |
mr yadav spends 60 % of his monthly salary on consumable items and 50 % of the remaining on clothes and transport . he saves the remaining amount . if his savings at the end of the year were 24000 , how much amount per month would he have spent on clothes and transport ? | "β΅ amount , he have spent in 1 month on clothes transport = amount spent on saving per month β΅ amount , spent on clothes and transport = 24000 β 12 = 2000 answer a" | a ) 18 , b ) 3280 , c ) - 39 , d ) 2000 , e ) 272 | d | multiply(divide(divide(24000, divide(divide(multiply(subtract(const_100, 60), 50), const_100), const_100)), multiply(const_3, const_4)), divide(divide(multiply(subtract(const_100, 60), 50), const_100), const_100)) | multiply(const_3,const_4)|subtract(const_100,n0)|multiply(n1,#1)|divide(#2,const_100)|divide(#3,const_100)|divide(n2,#4)|divide(#5,#0)|multiply(#6,#4)| | general |
in the class of 50 students , 30 speak tamil and 40 speak telugu . what is the lowest possible number of students who speak both the languages ? | let the student who speaks tamil - x let the student who speaks telugu - y as ( xuy ) - ( xny ) = total 30 + 40 - ( xny ) = 50 = 20 c ) | a ) 20 , b ) rs . 1058 , c ) $ 65.00 , d ) 11 / 48,000 , e ) 1 / 6 | a | subtract(add(40, 30), 50) | add(n1,n2)|subtract(#0,n0) | other |
two trains start simultaneously from opposite ends of a 175 - km route and travel toward each other on parallel tracks . train x , traveling at a constant rate , completes the 175 - km trip in 4 hours . train y , travelling at a constant rate , completes the 175 - km trip in 3 hours . how many kilometers had train x traveled when it met train y ? | if the two trains cover a total distance d , then train x travels ( 3 / 7 ) * d while train y travels ( 4 / 7 ) * d . if the trains travel 175 km to the meeting point , then train x travels ( 3 / 7 ) * 175 = 75 km . the answer is d . | a ) 20 % , b ) rs . 5845 , c ) 75 , d ) 0.0375 days , e ) 136.8 | c | multiply(divide(175, 4), divide(175, add(divide(175, 4), divide(175, 3)))) | divide(n0,n2)|divide(n0,n4)|add(#0,#1)|divide(n0,#2)|multiply(#0,#3)| | physics |
the sum of two numbers is 16 . the difference is 4 . what are the two numbers ? let x be the first number . ley y be the second number x + y = 16 x - y = 4 | solution let x be the first number . ley y be the second number x + y = 16 x - y = 4 solve the system of equations by elimination adding the left sides and the right sides gives : x + x + y + - y = 16 + 4 2 x = 20 x = 10 since x + y = 16 , 10 + y = 16 10 + y = 16 10 - 10 + y = 16 - 10 y = 6 the numbers are 10 and 6 answer a | a ) 61 , b ) 10 and 6 , c ) 3 , 400,000 , d ) 18 , e ) 1 / 6 | b | add(multiply(divide(add(16, 4), const_2), const_10), subtract(16, divide(add(16, 4), const_2))) | add(n0,n1)|divide(#0,const_2)|multiply(#1,const_10)|subtract(n0,#1)|add(#2,#3) | general |
if 6 x = 8 y = 10 z , then what is a possible sum of positive integers x , y , and z ? | "6 x = 8 y = 10 z 3 x = 4 y = 5 z 3 ( 4 * 5 ) = 4 ( 3 * 5 ) = 5 ( 3 * 4 ) addition = 20 + 15 + 12 = 47 answer would be multiple of 47 which is 94 answer : d" | a ) 48 , b ) 44 , c ) 22.5 % , d ) 8 , e ) 94 | e | divide(multiply(multiply(8, 10), 6), const_4) | multiply(n1,n2)|multiply(n0,#0)|divide(#1,const_4)| | general |
what is the rate percent when the simple interest on rs . 2000 amount to rs . 320 in 4 years ? | "interest for 1 year = 320 / 4 = 80 interest on rs 2000 p / a = 80 interest rate = 80 / 2000 * 100 = 4 % answer : c" | a ) 29.8 % , b ) 13 , c ) 720 , d ) 45 , e ) 4 % | e | divide(multiply(const_100, 320), multiply(2000, 4)) | multiply(n1,const_100)|multiply(n0,n2)|divide(#0,#1)| | gain |
a ratio between two numbers is 4 : 5 and their l . c . m . is 200 . the first number is | "sol . let the required numbers be 4 x and 5 x . then , their l . c . m . is 20 x . β΄ 20 x = 200 β x = 10 . hence , the first number is 40 . answer b" | a ) 23 , b ) 75 , c ) 310 , d ) 66 , e ) 40 | e | multiply(divide(200, multiply(4, 5)), 4) | multiply(n0,n1)|divide(n2,#0)|multiply(n0,#1)| | other |
10 camels cost as much as 24 horses , 16 horses cost as much as 4 oxen and 6 oxen as much as 4 elephants . if the cost of 10 elephants is rs . 170000 , find the cost of a camel ? | "cost of the camel = p 10 camels = 24 horses 16 horses = 4 oxen 6 oxen = 4 elephants 10 elephants = rs . 170000 p = rs . [ ( 24 * 4 * 4 * 170000 ) / ( 10 * 16 * 6 * 10 ) ] p = rs . ( 65280000 / 9600 ) = > p = rs . 6800 answer : d" | a ) 384 , b ) 40 litres , c ) rs . 6800 , d ) 120 , e ) 400 % | c | divide(multiply(multiply(multiply(24, 4), 4), 170000), multiply(multiply(multiply(10, 16), 6), 10)) | multiply(n1,n3)|multiply(n0,n2)|multiply(n3,#0)|multiply(n4,#1)|multiply(n7,#2)|multiply(n0,#3)|divide(#4,#5)| | general |
a sum of money at simple interest amounts to rs . 835 in 3 years and to rs . 854 in 4 years . the sum is : | "s . i . for 1 year = rs . ( 854 - 835 ) = rs . 19 . s . i . for 3 years = rs . ( 19 x 3 ) = rs . 57 . principal = rs . ( 835 - 57 ) = rs . 778 . answer : c" | a ) $ 32500 , b ) 48 , c ) 1575 , d ) 4,000 , e ) 778 | e | subtract(835, divide(multiply(subtract(854, 835), 3), 4)) | subtract(n2,n0)|multiply(n1,#0)|divide(#1,n3)|subtract(n0,#2)| | gain |
when the positive integer k is divided by the positive integer n , the remainder is 11 . if k / n = 81.1 , what is the value of n ? | "here ' s an approach that ' s based on number properties and a bit ofbrute forcemath : we ' re told that k and n are both integers . since k / n = 81.2 , we can say that k = 81.2 ( n ) n has tomultiply outthe . 2 so that k becomes an integer . with the answers that we have to work with , n has to be a multiple of 5 . eliminate a and e . with the remaining answers , we can test the answers and find the one that fits the rest of the info ( k / n = 81.2 and k / n has a remainder of 11 ) answer b : if n = 20 , then k = 1624 ; 1624 / 20 has a remainder of 4 not a match answer c : if n = 55 , then k = 4466 ; 4466 / 55 has a remainder of 11 match . final answer : e" | a ) 8 , b ) rs . 937.5 , c ) 60 , d ) 272 , e ) 81 | e | divide(add(multiply(divide(subtract(multiply(81.1, const_10), const_2), const_10), divide(11, divide(const_2, const_10))), 11), divide(11, divide(const_2, const_10))) | divide(const_2,const_10)|multiply(n1,const_10)|divide(n0,#0)|subtract(#1,const_2)|divide(#3,const_10)|multiply(#4,#2)|add(n0,#5)|divide(#6,#2)| | general |
if a person walks at 10 km / hr instead of 8 km / hr , he would have walked 14 km more . the actual distance traveled by him is ? | "let the actual distance traveled be x km . then , x / 8 = ( x + 14 ) / 10 2 x - 112 = > x = 56 km . answer : c" | a ) 14', ' , b ) $ 1200 , c ) 24 , d ) 56 , e ) 4 years | d | multiply(8, divide(14, subtract(10, 8))) | subtract(n0,n1)|divide(n2,#0)|multiply(n1,#1)| | general |
if twice of a number divided by 3 d gives 20 as the remainder , and 5 times of the same number gives 32 as the remainder . what will be the value of d ? | remainder in second case is , 32 . so , 3 d > = 33 . so , minimum value of d should 11 . if number = 28 . double of number = 56 and take d = 12 so 56 / 36 remainder = 20 . now 5 times of number = 140 . so 140 / 36 remainder = 32 . that is , 3 d = 36 satisfy the conditions . so d = 12 . answer : b | a ) 240 , b ) 19 , c ) 563.2 m , d ) 1 / 6 , e ) 12 | e | subtract(multiply(divide(32, 5), const_2), divide(const_1, const_4)) | divide(n3,n2)|divide(const_1,const_4)|multiply(#0,const_2)|subtract(#2,#1) | general |
17 : 71 : : 18 : ? | "ans 81 reverse of 18 answer : d" | a ) $ 555.55 , b ) 84 , c ) 81 , d ) 0 , e ) 6 | c | multiply(18, divide(17, 71)) | divide(n0,n1)|multiply(n2,#0)| | general |
3 people candidates contested an election and they received 1136 , 7636 and 11628 votes respectively . what is the percentage of the total votes did the winning candidate get ? | tot no of votes = ( 1136 + 7636 + 11628 ) = 20400 req = > ( 11628 / 20400 * 100 ) = > 57 % answer c | a ) 273309 , b ) 200 , c ) 2.6 units , d ) 56 , e ) 57 % | e | multiply(divide(11628, add(add(1136, 7636), 11628)), const_100) | add(n1,n2)|add(n3,#0)|divide(n3,#1)|multiply(#2,const_100) | general |
in the faculty of reverse - engineering , 500 second year students study numeric methods , 300 second year students study automatic control of airborne vehicles and 300 second year students study them both . how many students are there in the faculty if the second year students are approximately 50 % of the total ? | "total number of students studying both are 500 + 300 - 300 = 500 ( subtracting the 300 since they were included in the both the other numbers already ) . so 50 % of total is 500 , so 100 % is 1000 answer is d" | a ) 5.2 , b ) 343 cc , c ) 8.4 sec , d ) 1000 , e ) 2160 | d | add(500, 300) | add(n0,n1)| | general |
two pipes a and b can separately fill a tank in 8 and 10 minutes respectively . a third pipe c can drain off 15 liters of water per minute . if all the pipes are opened , the tank can be filled in 10 minutes . what is the capacity of the tank ? | "1 / 8 + 1 / 10 - 1 / x = 1 / 10 x = 8 8 * 15 = 120 answer : e" | a ) 5 % , b ) 25 % , c ) 120 liters , d ) $ 12.48 , e ) 786 | c | multiply(multiply(inverse(subtract(add(add(divide(const_1, 8), divide(const_1, 10)), divide(const_1, 15)), divide(const_1, 10))), const_3), 10) | divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|divide(const_1,n3)|add(#0,#1)|add(#4,#2)|subtract(#5,#3)|inverse(#6)|multiply(#7,const_3)|multiply(n1,#8)| | physics |
what annual installment will discharge a debt of $ 1092 due in 3 years at 12 % simple interest ? | "let each installment be x [ x + ( x * 12 * 1 / 100 ) ] + [ x + ( x * 12 * 2 / 100 ) ] + x = 1092 28 x / 25 + 31 x / 25 + x = 1092 x = 325 answer is d" | a ) 1095 , b ) 4.25 hr , c ) $ 325 , d ) 227.04 mtrs , e ) 80 % | c | multiply(multiply(const_100.0, divide(12, 1092)), 3) | divide(n2,const_100)|multiply(n0,#0)|multiply(n1,#1)| | gain |
a spirit and water solution is sold in a market . the cost per liter of the solution is directly proportional to the part ( fraction ) of spirit ( by volume ) the solution has . a solution of 1 liter of spirit and 1 liter of water costs 40 cents . how many cents does a solution of 1 liter of spirit and 2 liters of water cost ? | "yes , ensure that you understand the relation thoroughly ! cost per liter = k * fraction of spirit 40 cents is the cost of 2 liters of solution ( 1 part water , 1 part spirit ) . so cost per liter is 20 cents . fraction of spirit is 1 / 2 . 20 = k * ( 1 / 2 ) k = 40 cost per liter = 40 * ( 1 / 3 ) ( 1 part spirit , 2 parts water ) cost for 3 liters = 40 * ( 1 / 3 ) * 3 = 40 cents a . 40 cents" | a ) 50', ' , b ) 40 , c ) 43 , d ) 40 % , e ) 12.27 | b | multiply(multiply(40, divide(1, add(1, 2))), add(1, 2)) | add(n0,n4)|divide(n0,#0)|multiply(n2,#1)|multiply(#0,#2)| | geometry |
if 12 men or 20 women can do a piece of work in 54 days , then in how many days can 9 men and 12 women together do the work ? | "e 40 days given that 12 m = 20 w = > 3 m = 5 w 9 men + 12 women = 15 women + 12 women = 27 women 20 women can do the work in 54 days . so , 27 women can do it in ( 20 * 54 ) / 27 = 40 days ." | a ) 4096 , b ) 838 , c ) $ 1.75 , d ) 40 days , e ) 40 | d | inverse(add(divide(9, multiply(12, 54)), divide(12, multiply(20, 54)))) | multiply(n0,n2)|multiply(n1,n2)|divide(n3,#0)|divide(n4,#1)|add(#2,#3)|inverse(#4)| | physics |
set q consists of integers { 3 , - 8 , y , 19 , - 6 } and set b consists of integers { k , - 3 , 0 , 16 , - 5 , 9 } . number l represents the median of set q , number m represents mode of set b and number z = l ^ m . if y is an integer greater than 21 , for what value of k will z be a divisor of 26 ? | a = { 3 , - 8 , y , 19 , - 6 } b = { k , - 3 , 0,16 , - 5,9 } y > 21 so l = median of q = 3 m = mode of set b z = ( 3 ) ^ m if z is a divisor of 26 , ( 3 ) ^ m = 1 because 26 does not have 3 as a factor = > m = 0 hence k = 0 , as m is mode and 0 will be the most frequently occuring number in set b . answer - c | a ) 2700 , b ) 0 , c ) 15 , d ) - 2407 , e ) 25 | b | log(3) | log(n0) | general |
gold is 19 times as heavy as water and copper is 9 times as heavy as water . in what ratio should these be mixed to get an alloy 13 times as heavy as water ? | "g = 19 w c = 9 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 19 w + 9 wx = x + 1 * 13 w 19 + 9 x = 13 ( x + 1 ) x = 3 / 2 ratio of gold with copper = 1 : 3 / 2 = 2 : 3 answer is b" | a ) 43 , b ) 567 , c ) 2 : 3 , d ) 1 / 3 , e ) 9.5 gallons | c | divide(subtract(13, 9), subtract(19, 13)) | subtract(n2,n1)|subtract(n0,n2)|divide(#0,#1)| | general |
x and y are both integers . if x / y = 59.60 , then what is the sum of all the possible two digit remainders of x / y ? | "remainder = 0.60 - - > 60 / 100 - - > can be written as ( 60 / 4 ) / ( 100 / 4 ) = 15 / 25 so remainders can be 15 , 30 , 45 , 60 , . . . . . 90 . we need the sum of only 2 digit remainders - - > 15 + 30 + 45 + 60 + 75 + 90 = 315 answer : b" | a ) 315 , b ) 70 m , c ) 5 , d ) 170 m , e ) 33 % | a | add(multiply(divide(const_3, const_2), const_100), add(multiply(add(const_2, const_3), 59.60), const_3)) | add(const_2,const_3)|divide(const_3,const_2)|multiply(n0,#0)|multiply(#1,const_100)|add(#2,const_3)|add(#4,#3)| | general |
a and b started a business jointly a ' s investment was thrice the investment of b and the period of his investment was two times the period of investment of b . if b received rs . 4000 as profit , then their total profit is | "explanation : suppose b invested rs . x for y months . then , a invested rs . 3 x for 2 y months . so , a : b = ( 3 x * 2 y ) : ( x * y ) = 6 xy : xy = 6 : 1 . b ' s profit : total profit = 1 : 7 . let the total profit be rs . x then , 1 / 7 = 4000 / x or x = 28000 . answer : b ) 28000" | a ) 270 m , b ) 28000 , c ) 400 , d ) 14 , e ) 80 % | b | multiply(add(multiply(const_2, const_3), const_1), 4000) | multiply(const_2,const_3)|add(#0,const_1)|multiply(n0,#1)| | general |
excluding stoppages , the average speed of a bus is 100 km / hr and including stoppages , the average speed of the bus is 40 km / hr . for how many minutes does the bus stop per hour ? | in 1 hr , the bus covers 100 km without stoppages and 40 km with stoppages . stoppage time = time take to travel ( 100 - 40 ) km i . e 60 km at 100 km / hr . stoppage time = 60 / 100 hrs = 36 min answer : e | a ) 104 meters', ' , b ) 67 % , c ) 9.9 % , d ) 26 , e ) 36 min | e | subtract(multiply(const_1, const_60), multiply(divide(40, 100), const_60)) | divide(n1,n0)|multiply(const_1,const_60)|multiply(#0,const_60)|subtract(#1,#2) | general |
tom traveled the entire 60 miles trip . if he did the first 15 miles of at a constant rate 24 miles per hour and the remaining trip of at a constant rate 45 miles per hour , what is the his average speed , in miles per hour ? | avg speed = total distance / total time = ( d 1 + d 2 ) / ( t 1 + t 2 ) = ( 15 + 45 ) / ( ( 15 / 24 ) + ( 45 / 45 ) ) = 60 * 2 / 3 = 36.92 mph d | a ) 40 , b ) 36.92 mph , c ) c : 37.5 , d ) 250 , e ) 3108 | b | divide(60, add(divide(45, subtract(60, 15)), divide(15, 24))) | divide(n1,n2)|subtract(n0,n1)|divide(n3,#1)|add(#2,#0)|divide(n0,#3) | physics |
a new home buyer pays 4 % annual interest on her first mortgage and 9 % annual interest on her second mortgage . if she borrowed a total of $ 325,000 , 80 % of which was in the first mortgage , what is her approximate monthly interest payment ? | "0.04 x + 0.09 y = 325000 [ 1 ] 0.04 x = 0.80 * 325000 = 260000 [ 2 ] 260000 + 0.09 y = 325000 - - > 0.09 y = 65000 [ 3 ] 260000 / 12 = 21666.67 [ 4 ] 65000 / 12 = 5416.667 [ 5 ] adding [ 4,5 ] we get : 27083.33 [ 6 ] dividing [ 6 ] / 2 to get an average we get 1.354 , ans a" | a ) 7 , b ) 11 , c ) 1 , d ) 2 , e ) $ 1,354 | e | subtract(9, multiply(4, const_2)) | multiply(n0,const_2)|subtract(n1,#0)| | general |
80 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 = e please refer diagram below 70 - 10 = 60" | a ) 3.696 kg , b ) 2 , c ) 12 , d ) 60 , e ) 45 | d | subtract(80, 10) | subtract(n0,n1)| | general |
if w is the set of all the integers between 49 and 114 , inclusive , that are either multiples of 3 or multiples of 2 or multiples of both , then w contains how many numbers ? | "official solution : number of multiples of 3 step 1 . subtract the extreme multiples of 3 within the range ( the greatest is 114 , the smallest is 51 ) : 114 - 51 = 63 step 2 . divide by 3 : 63 / 3 = 21 step 3 . add 1 : 21 + 1 = 22 . so there are 22 multiples of 3 within the range : examples are 51 , 54 , 57 , 60 , etc . number of multiples of 2 step 1 . subtract the extreme multiples of 2 within the range ( the greatest is 114 , the smallest is 50 ) : 114 - 50 = 64 step 2 . divide by 2 : 64 / 2 = 32 step 3 . add 1 : 32 + 1 = 33 . so there are 33 multiples of 2 within the range : examples are 50 , 52 , 54 , 56 , 58 , 60 etc . add the 22 multiples of 3 and the 33 multiples of 2 : 22 + 33 = 55 . however , by adding the multiples of 2 and the multiples of 3 , we are effectively counting several numbers twice : for example , 54 and 60 are parts of both the lists above . so we ca n ' t just take 22 + 33 = 55 . find the number of multiples of 6 ( which are the double counted , as 6 is divisible by both 2 and 3 ) , and subtract it from 55 : step 1 . subtract the extreme multiples of 6 within the range ( the greatest is 72 , the smallest is 54 ) : 114 - 54 = 60 step 2 . divide by 6 : 60 / 6 = 10 step 3 . add 1 : 10 + 1 = 11 . so there are 11 multiples of 6 within the range : we counted 11 numbers twice . subtract the 11 multiples of 6 from the sum of the multiples of 2 and 3 : = 22 + 33 - 11 = 55 - 11 = 44 therefore , the final number of multiples of 2 , 3 or 6 is 44 . hence , this is the correct answer . ( b )" | a ) 12 , b ) 2 , c ) 44 , d ) 4 , e ) 8 min | c | subtract(add(floor(divide(subtract(114, 49), 3)), divide(subtract(114, 49), 2)), floor(divide(subtract(114, 49), multiply(2, 3)))) | multiply(n3,n2)|subtract(n1,n0)|divide(#1,n3)|divide(#1,n2)|divide(#1,#0)|floor(#3)|floor(#4)|add(#2,#5)|subtract(#7,#6)| | other |
four machines , each working at the same constant rate , together can complete a certain job in 12 days . how many additional machines , each working at the same constant rate , will be needed to complete the job in 8 days ? | "another solution which is faster is since each machine works at a constant rate . the time needs to bought down from 12 to 8 . so the new time is 2 / 3 of the original time . thus to achieve this we need the rate to be 3 / 2 of original . so 3 / 2 * 4 = 6 so we need 6 - 4 = 2 more machines . answer : a" | a ) 1 β 10 , b ) 40.2 degrees c , c ) 20 % , d ) 40 % , e ) 2 | e | subtract(divide(multiply(12, add(const_4, const_1)), 8), add(const_4, const_1)) | add(const_1,const_4)|multiply(n0,#0)|divide(#1,n1)|subtract(#2,#0)| | physics |
a shopkeeper sold 8 articles at the cost price of 10 articles . then find the profit % or lost % | "here 8 articles selling price = 10 articles cost price so the difference = 10 - 8 = 2 % of profit = 2 * 100 / 8 = 25 % correct option is b" | a ) 15 , b ) 605.03 , c ) 25 % , d ) $ 23400 , e ) 200 | c | multiply(divide(subtract(10, 8), 8), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
what is the maximum number of pieces that a circular pizza can be cut into with 5 linear cuts ? | "1 cut makes 2 pieces a second cut can make 4 pieces a third cut can make 7 pieces by cutting through 3 of the pieces a fourth cut can make 11 pieces by cutting through 4 of the pieces a fifth cut can make 16 pieces by cutting through 5 of the pieces d" | a ) 63 , b ) 120 km , c ) 16 , d ) 32 , e ) 600 cm 2 | c | multiply(5, const_3) | multiply(n0,const_3)| | geometry |
the general hospital is comprised of , 3 / 5 pediatricians , 1 / 4 surgeons , and the rest are gp doctors . if 1 / 4 of the surgeons are heart surgeons , and the hospital doubles the number of gp doctors , what proportion of the hospital are now heart surgeons ? | ped = 3 / 5 = 12 / 20 sur = 1 / 4 = 4 / 20 gp = 1 - ( 12 / 20 + 4 / 20 ) = 1 - 16 / 20 = 4 / 20 hsur = ( 1 / 4 ) ( 4 / 20 ) = 1 / 20 if gp doubled = > ( 2 ) ( 4 ) = 8 ; then , total = 12 + 4 + 8 = 24 , and 1 is hsur = > proportion = 1 / 24 . answer : d | a ) 1 / 24 , b ) 20 , c ) $ 216 , d ) 44 % , e ) 22.85 % | a | divide(subtract(3, add(1, 1)), add(multiply(4, 5), 4)) | add(n2,n2)|multiply(n1,n3)|add(n3,#1)|subtract(n0,#0)|divide(#3,#2) | general |
45 pupil , out of them 12 in debate only and 22 in singing only . then how many in both ? | the intersection for two = 45 - 12 - 22 = 11 play both games . answer : c | a ) 1 , b ) 30 , c ) 30 % , d ) 11 , e ) $ 14.00 | d | subtract(45, add(12, 22)) | add(n1,n2)|subtract(n0,#0) | other |
of the 75 cars on a car lot , 45 have air - conditioning , 35 have power steering , and 12 have both air - conditioning and power steering . how many of the cars on the lot have neither air - conditioning nor power steering ? | "total - neither = all air conditioning + all power steering - both or 75 - neither = 45 + 35 - 12 = 68 . = > neither = 7 , hence a . answer : a" | a ) s : 1300 , b ) 4 cm , c ) 244907.04 m , d ) 77 , e ) 7 | e | subtract(75, subtract(add(45, 35), 12)) | add(n1,n2)|subtract(#0,n3)|subtract(n0,#1)| | other |
a corporation 5 times its annual bonus to 10 of its employees . what percent of the employees β new bonus is the increase ? | let the annual bonus be x . a corporation triples its annual bonus . so new bonus = 5 x . increase = 5 x - x = 4 x the increase is what percent of the employees β new bonus = ( 4 x / 5 x ) * 100 = 80 % hence c . | a ) 2 , b ) 82 , c ) 5 % , d ) 80 % , e ) 3 | d | multiply(divide(subtract(5, const_1), 5), const_100) | subtract(n0,const_1)|divide(#0,n0)|multiply(#1,const_100) | general |
the sum of ages of 5 children born at the intervals of 3 years each is 50 years . what is the age of the youngest child ? | "explanation let the ages of children be x , ( x + 3 ) , ( x + 6 ) , ( x + 9 ) and ( x + 12 ) years . then , x + ( x + 3 ) + ( x + 6 ) + ( x + 9 ) + ( x + 12 ) = 50 5 x = 20 x = 4 . age of the youngest child = x = 4 years . answer a" | a ) 100 , b ) 54.4 cm , c ) 54 , d ) 1405.25 , e ) 4 years | e | subtract(subtract(divide(50, 5), 3), 3) | divide(n2,n0)|subtract(#0,n1)|subtract(#1,n1)| | general |
yearly subscription to professional magazines cost a company $ 940.00 . to make a 25 % cut in the magazine budget , how much less must be spent ? | total cost 940 940 * 25 / 100 = 235 so the cut in amount is 235 the less amount to be spend is 940 - 235 = 705 answer : a | a ) rs . 98.56 , b ) 46 , c ) 705 , d ) $ 20.00 , e ) 14 | c | multiply(divide(subtract(const_100, 25), const_100), 940) | subtract(const_100,n1)|divide(#0,const_100)|multiply(n0,#1) | general |
mr . evans will states that each of his children will receive an equal share of his estate and that his grandchildren will split a portion of the estate that is equal to the share received by each of his children . if mr . evans has 4 children and 5 grandchildren , then approximately what percentage of mr . evans estate will each grandchild receive ? | "20 / 5 = 4 answer : c" | a ) 1 / 12 , b ) 180 ares . , c ) 4.0 % , d ) 4 , e ) 12 days | c | multiply(divide(divide(divide(const_100, const_3), 5), const_100), const_100) | divide(const_100,const_3)|divide(#0,n1)|divide(#1,const_100)|multiply(#2,const_100)| | general |
if the selling price of 50 articles is equal to the cost price of 20 articles , then the loss or gain percent is : | "c . p . of each article be re . 1 . then , c . p . of 50 articles = rs . 50 ; s . p . of 50 articles = rs . 20 . loss % = 30 / 50 * 100 = 60 % answer d" | a ) $ 23,760 , b ) 60 % , c ) 331 times , d ) 19 , e ) 1764 | b | subtract(50, 20) | subtract(n0,n1)| | gain |
the mean of 50 observations was 40 . it was found later that an observation 48 was wrongly taken as 23 . the corrected new mean is | "sol . therefore correct sum = ( 40 Γ 50 + 48 β 23 ) = 2023 . therefore correct mean = 2023 / 50 = 40.46 . answer a" | a ) 1 / 216 , b ) 40 , c ) 40.46 , d ) rs . 960 , e ) 318 $ | c | divide(add(multiply(40, 50), subtract(subtract(50, const_2), 23)), 50) | multiply(n0,n1)|subtract(n0,const_2)|subtract(#1,n3)|add(#0,#2)|divide(#3,n0)| | general |
a certain tests consists 8 sections with 25 questions , numbered from 1 to 25 , in each section . if a student answered all of the even - numbered questions correctly and 3 / 4 of the odd - numbered questions correctly , what was the total number of questions he answered correctly ? a . 150 b . 172 c . 174 d . 175 e . 176 | each set has 12 even and 13 odd numbered questions leading to total 96 even and 104 odd questions . 96 + 3 / 4 Γ’ Λ β 104 = 96 + 78 = 17496 + 3 / 4 Γ’ Λ β 104 = 96 + 78 = 174 answer : a | a ) 174 , b ) 200 sq feet , c ) 25 , d ) 149 , e ) 24 metre | a | add(divide(multiply(8, 25), const_2), multiply(divide(multiply(8, 25), const_2), divide(3, 4))) | divide(n4,n5)|multiply(n0,n1)|divide(#1,const_2)|multiply(#2,#0)|add(#2,#3) | general |
a train running at the speed of 60 km / hr crosses a pole in 30 seconds . what is the length of the train ? | "speed = ( 60 * 5 / 18 ) m / sec = ( 50 / 3 ) m / sec length of the train = ( speed x time ) = ( 50 / 3 * 30 ) m = 500 m . answer : a" | a ) 4.0 % , b ) 500 m , c ) $ 144 , d ) 146 1 / 7 min , e ) 17 | b | multiply(divide(multiply(60, const_1000), const_3600), 30) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics |
when greenville state university decided to move its fine arts collection to a new library , it had to package the collection in 20 - inch by 20 - inch by 15 - inch boxes . if the university pays $ 0.60 for every box , and if the university needs 3.06 million cubic inches to package the collection , what is the minimum amount the university must spend on boxes ? | "total no . of boxes = 3060000 / ( 20 Γ 20 Γ 15 ) = 510 total cost = 510 Γ $ 0.6 = $ 306 answer a" | a ) 501 , b ) $ 306 , c ) 13122 , d ) 4 , e ) 2056 | b | multiply(divide(multiply(3.06, multiply(const_1000, const_1000)), multiply(multiply(20, 20), 15)), 0.60) | multiply(const_1000,const_1000)|multiply(n0,n0)|multiply(n4,#0)|multiply(n2,#1)|divide(#2,#3)|multiply(n3,#4)| | general |
if 2 x + y = 7 and x + 2 y = 5 , then 10 xy / 3 = ? | "2 * ( x + 2 y = 5 ) equals 2 x + 4 y = 10 2 x + 4 y = 10 - 2 x + y = 7 = 3 y = 3 therefore y = 1 plug and solve . . . 2 x + 1 = 7 2 x = 6 x = 3 ( 10 * 3 * 1 ) / 3 = 30 / 3 = 10 a" | a ) 35 : 2 , b ) 25 % , c ) rs . 640 , d ) 10 , e ) 254 m 2 | d | divide(add(divide(subtract(multiply(7, 2), 5), subtract(multiply(2, 2), const_1)), subtract(7, multiply(2, divide(subtract(multiply(7, 2), 5), subtract(multiply(2, 2), const_1))))), 10) | multiply(n0,n1)|multiply(n0,n0)|subtract(#0,n3)|subtract(#1,const_1)|divide(#2,#3)|multiply(n0,#4)|subtract(n1,#5)|add(#4,#6)|divide(#7,n4)| | general |
the sum of ages of 5 children born at the intervals of 3 years each is 80 years . what is the age of the youngest child ? | "let the ages of children be x , ( x + 3 ) , ( x + 6 ) , ( x + 9 ) and ( x + 12 ) years . then , x + ( x + 3 ) + ( x + 6 ) + ( x + 9 ) + ( x + 12 ) = 80 5 x = 50 x = 10 . age of the youngest child = x = 10 years . e )" | a ) 22.37 , b ) 40 , c ) 26 % , d ) 10 years , e ) 1804 | d | subtract(subtract(divide(80, 5), 3), 3) | divide(n2,n0)|subtract(#0,n1)|subtract(#1,n1)| | general |
2 is what percent of 40 ? | "2 = x * 40 / 100 x = 5 % ans ; c" | a ) 52 , b ) 48 , c ) 5 % , d ) 5 , e ) 0.0012 | c | multiply(divide(2, 40), const_100) | divide(n0,n1)|multiply(#0,const_100)| | gain |
a no . when divided by 221 gives a remainder 43 , what remainder will beobtained by dividing the same number 19 ? | "221 + 43 = 264 / 17 = 9 ( remainder ) c" | a ) 4 and 12 , b ) 10 , c ) 27 / 64 , d ) 1528 , e ) 9 | e | divide(add(221, 43), 19) | add(n0,n1)|divide(#0,n2)| | general |
tea worth rs . 126 per kg are mixed with a third variety in the ratio 1 : 1 : 2 . if the mixture is worth rs . 153 per kg , the price of the third variety per kg | "explanation : since first second varieties are mixed in equal proportions , so their average price = rs . ( 126 + 135 / 2 ) = rs . 130.50 so , the mixture is formed by mixing two varieties , one at rs . 130.50 per kg and the other at say , rs . x per kg in the ratio 2 : 2 , i . e . , 1 : 1 . we have to find x . cost of 1 kg tea of 1 st kind cost of 1 kg tea of 2 nd kind x - 153 / 22.50 = 1 = > x - 153 = 22.50 = > x = 175.50 . hence , price of the third variety = rs . 175.50 per kg . answer : c ) rs . 175.50" | a ) 9 , b ) 175.57 , c ) 6.6 kmph , d ) 5 , e ) 123 | b | divide(subtract(multiply(153, add(add(1, 1), 2)), add(126, 126)), 2) | add(n1,n1)|add(n0,n0)|add(n3,#0)|multiply(n4,#2)|subtract(#3,#1)|divide(#4,n3)| | other |
in an examination , a student scores 4 marks for every correct answer and loses 1 mark for every wrong answer . if he attempts all 80 questions and secures 120 marks , the number of questions he attempts correctly , is : | "let the number of correct answers be x . number of incorrect answers = ( 80 Γ’ β¬ β x ) . 4 x Γ’ β¬ β 1 ( 80 Γ’ β¬ β x ) = 120 or 5 x = 200 or x = 40 . answer : e" | a ) 855220 , b ) 32 , c ) 55 cm 2 , d ) 787 , e ) 40 | e | divide(add(120, 80), add(4, 1)) | add(n2,n3)|add(n0,n1)|divide(#0,#1)| | physics |
find the compound ratio of ( 2 : 3 ) , ( 6 : 11 ) and ( 11 : 4 ) is | "required ratio = 2 / 3 * 6 / 11 * 11 / 4 = 2 / 1 = 1 : 1 answer is c" | a ) 11 am , b ) 1 : 1 , c ) 12 , d ) 450 , e ) 14 : 00 | b | multiply(divide(2, 11), multiply(divide(2, 3), divide(6, 3))) | divide(n0,n5)|divide(n2,n1)|multiply(#0,#1)|multiply(#0,#2)| | other |
a no . when divided by the sum of 555 and 445 gives 2 times their difference as quotient & 10 as remainder . find the no . is ? | "( 555 + 445 ) * 2 * 110 + 10 = 220000 + 10 = 220010 a" | a ) 2 , b ) 18 : 13 , c ) 8 , d ) 220010 , e ) s . 295.45 | d | add(multiply(multiply(add(555, 445), 2), subtract(555, 445)), 10) | add(n0,n1)|subtract(n0,n1)|multiply(n2,#0)|multiply(#2,#1)|add(n3,#3)| | general |
in a sports club with 55 members , 23 play badminton and 29 play tennis and 7 do not play either . how many members play both badminton and tennis ? | "23 + 29 = 52 but where as total number is 55 - 7 = 48 therefore answer is 52 - 48 = 4 hence answer is c" | a ) 81 , b ) 40 % , c ) 12 , d ) 9 , e ) 4 | e | subtract(add(add(23, 29), 7), 55) | add(n1,n2)|add(n3,#0)|subtract(#1,n0)| | other |
a garrison of 500 men has provisions for 20 days . at the end of 14 days , a reinforcement arrives , and it is now found that the provisions will last only for 3 days more . what is the reinforcement ? | "500 - - - - 20 500 - - - - 6 x - - - - - 3 x * 3 = 500 * 6 x = 1000 500 - - - - - - - 500 answer : a" | a ) 500 , b ) 70 , c ) 15 litres , d ) 0.2 , e ) β 14 | a | subtract(divide(subtract(multiply(500, 20), multiply(500, 14)), 3), 500) | multiply(n0,n1)|multiply(n0,n2)|subtract(#0,#1)|divide(#2,n3)|subtract(#3,n0)| | physics |
how long does a train 165 meters long running at the rate of 54 kmph take to cross a bridge 660 meters in length ? | "t = ( 660 + 165 ) / 54 * 18 / 5 t = 55 answer a" | a ) 463 , b ) 10,800 , c ) 82 , d ) 14 m , e ) 55 sec | e | divide(add(165, 660), multiply(54, const_0_2778)) | add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)| | physics |
in covering a distance of 54 km , abhay takes 2 hours more than sameer . if abhay doubles his speed , then he would take 1 hour less than sameer . abhay ' s speed is : | "let abhay ' s speed be x km / hr . then , 54 / x - 54 / 2 x = 3 6 x = 54 x = 9 km / hr . answer : option e" | a ) 310 , b ) 27 , c ) 245 , d ) 9 kmph , e ) 4 | d | divide(subtract(54, divide(54, 2)), add(1, 2)) | add(n1,n2)|divide(n0,n1)|subtract(n0,#1)|divide(#2,#0)| | physics |
the length of minute hand of a clock is 5.6 cm . what is the area covered by this in 10 minutes | area of circle is pi * r ^ 2 but in 10 minutes area covered is ( 10 / 60 ) * 360 = 60 degree so formula is pi * r ^ 2 * ( angle / 360 ) = 3.14 * ( 5.6 ^ 2 ) * ( 60 / 360 ) = 16.41 cm ^ 2 answer : d | a ) 67 % , b ) 30 kmph , c ) 14 days , d ) 16.41 , e ) 11 | d | multiply(divide(add(multiply(const_2, const_10), const_2), add(const_3, const_4)), multiply(multiply(5.6, 5.6), divide(multiply(const_1, const_60), multiply(const_100, const_3_6)))) | add(const_3,const_4)|multiply(const_10,const_2)|multiply(const_1,const_60)|multiply(const_100,const_3_6)|multiply(n0,n0)|add(#1,const_2)|divide(#2,#3)|divide(#5,#0)|multiply(#6,#4)|multiply(#7,#8) | physics |
john has $ 1600 at the beginning of his trip , after spending money , he still has exactly $ 600 less than he spent on the trip . how much money does john still have ? | suppose total money spent = x not spend ( money he still has ) = x - 600 x + x - 600 = 1600 x = 1100 money not spend = 1100 - 600 = 500 answer : e | a ) s . 5100 , b ) $ 500 , c ) 644 . , d ) 13 , e ) 16 / 625 | b | divide(subtract(1600, 600), const_2) | subtract(n0,n1)|divide(#0,const_2) | general |
a train 1000 m long is running at a speed of 78 km / hr . if it crosses a tunnel in 1 min , then the length of the tunnel is ? | "speed = 78 * 5 / 18 = 65 / 3 m / sec . time = 1 min = 60 sec . let the length of the train be x meters . then , ( 1000 + x ) / 60 = 65 / 3 x = 300 m . answer : c" | a ) 5 , b ) 75 kg , c ) 1000 m , d ) 40 , e ) 13 | c | divide(1000, multiply(subtract(78, 1), const_0_2778)) | subtract(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
students at a school were on average 180 cm tall . the average female height was 170 cm , and the average male height was 181 cms . what was the ratio of men to women ? | we ' re given a few facts to work with : 1 ) the average height of the females is 170 cm 2 ) the average height of the males is 181 cm 3 ) the average of the group is 180 cm we ' re asked for the ratio of men to women . w = number of women m = number of men ( 170 w + 181 m ) / ( w + m ) = 180 170 w + 181 m = 180 w + 180 m 1 m = 10 w m / w = 10 / 1 the ratio of men to women is 10 to 1 . e | a ) 192 , b ) 8 , c ) 0.16 , d ) 0.1388 , e ) 10 : 1 | e | divide(subtract(180, 170), subtract(181, 180)) | subtract(n0,n1)|subtract(n2,n0)|divide(#0,#1) | general |
if the length of the longest chord of a certain circle is 18 , what is the radius of that certain circle ? | "longest chord of a circle is the diameter of the circle diameter = 2 * radius if diameter of the circle is given as 18 = 2 * 9 so radius of the circle = 9 correct answer - b" | a ) 3 , b ) 40 min , c ) 6.7 kg . , d ) 9 , e ) 55 / 71 | d | divide(18, const_2) | divide(n0,const_2)| | geometry |
a cistern 10 meters long and 4 meters wide contains water up to a depth of 1 meter 25 cm . what is the total area of the wet surface ? | area of the wet surface = [ 2 ( lb + bh + lh ) - lb ] = 2 ( bh + lh ) + lb = [ 2 ( 4 x 1.25 + 10 x 1.25 ) + 10 x 4 ] = 75 the answer is d . | a ) 2000 , b ) 75 , c ) 5 % , d ) $ 35 , e ) 2 / 9 | b | add(add(multiply(10, 4), multiply(const_2, multiply(4, add(1, divide(25, const_100))))), multiply(const_2, multiply(10, add(1, divide(25, const_100))))) | divide(n3,const_100)|multiply(n0,n1)|add(n2,#0)|multiply(n1,#2)|multiply(n0,#2)|multiply(#3,const_2)|multiply(#4,const_2)|add(#1,#5)|add(#7,#6) | physics |
a certain junior class has 1,000 students and a certain senior class has 800 students . among these students , there are 20 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 at will be a sibling pair ? | "total number of ways of choosing one student from each group is = 800 * 1000 number of cases in which a sibling pair will be got is = 60 thus the probability that the 2 students selected will be a sibling pair is = 20 / ( 800 * 1000 ) = 1 / 40,000 d" | a ) 82 % , b ) 12800 , c ) 0.18 , d ) 1 / 40,000 , e ) 1954404 | d | divide(1, const_3) | divide(n3,const_3)| | probability |
if 50 % of ( x - y ) = 30 % of ( x + y ) then what percent of x is y ? | "50 % of ( x - y ) = 30 % of ( x + y ) ( 50 / 100 ) ( x - y ) = ( 30 / 100 ) ( x + y ) 5 ( x - y ) = 3 ( x + y ) 2 x = 8 y x = 4 y therefore required percentage = ( ( y / x ) x 100 ) % = ( ( y / 4 y ) x 100 ) = 25 % answer is e ." | a ) 62 mph , b ) 3 , c ) 5.6 , d ) 25 % , e ) 11 | d | multiply(divide(subtract(50, 30), add(50, 30)), const_100) | add(n0,n1)|subtract(n0,n1)|divide(#1,#0)|multiply(#2,const_100)| | general |
how many seconds does sandy take to cover a distance of 600 meters , if sandy runs at a speed of 18 km / hr ? | 18 km / hr = 18000 m / 3600 s = 5 m / s time = 600 / 5 = 120 seconds the answer is b . | a ) $ 2570 , b ) 150 , c ) 55 , d ) 120 , e ) 1200 | d | divide(600, multiply(18, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0) | physics |
calculate how many days it will take for 10 boys to paint a 80 m long wall if 6 boys can paint a 70 m long wall in 8 days , | "the length of wall painted by one boy in one day = 70 / 6 * 1 / 8 = 1.46 m no . of days required to paint 50 m cloth by 8 boys = 80 / 10 * 1 / 1.46 = 5.48 days . d" | a ) 1.52 % , b ) 11 , c ) 1080 kmph , d ) 629 , e ) 5.48 days | e | divide(multiply(multiply(6, 8), 80), multiply(70, 10)) | multiply(n2,n4)|multiply(n0,n3)|multiply(n1,#0)|divide(#2,#1)| | physics |
a no . when divided by the sum of 555 and 445 gives 2 times their difference as quotient & 60 as remainder . find the no . is ? | "( 555 + 445 ) * 2 * 110 + 60 = 220000 + 60 = 220060 e" | a ) 270 , b ) 4 , c ) 20 , d ) 220060 , e ) 3 | d | add(multiply(multiply(add(555, 445), 2), subtract(555, 445)), 60) | add(n0,n1)|subtract(n0,n1)|multiply(n2,#0)|multiply(#2,#1)|add(n3,#3)| | general |
a monkey start climbing up a tree 18 ft tall . each hour it hops 3 ft and slips back 2 ft . how much time would it take the monkey to reach the top . | "if monkey hops 3 ft and slips back 2 ft in a hour , it means the monkey hops ( 3 ft - 2 ft ) = 1 ft / hr . similarly in 15 hrs it wil be 15 ft . bt since the height of the tree is 18 ft , so if the monkey hops up the tree in the next hr i . e 16 th hr then it reaches at the top of the tree . hence it takes 16 hrs for monkey to reach at the top answer : e" | a ) 8 , b ) 4 dm , c ) 4 , d ) 16 hrs , e ) s . 19,800 | d | subtract(divide(18, subtract(3, 2)), 2) | subtract(n1,n2)|divide(n0,#0)|subtract(#1,n2)| | physics |
jane and ashley take 8 days and 40 days respectively to complete a project when they work on it alone . they thought if they worked on the project together , they would take fewer days to complete it . during the period that they were working together , jane took an eight day leave from work . this led to jane ' s working for four extra days on her own to complete the project . how long did it take to finish the project ? | "let us assume that the work is laying 40 bricks . jane = 5 bricks per day ashley = 1 brick per day together = 6 bricks per day let ' s say first 8 days ashley works alone , no of bricks = 8 last 4 days jane works alone , no . of bricks = 20 remaining bricks = 40 - 28 = 12 so together , they would take 12 / 6 = 2 total no . of days = 8 + 4 + 2 = 14 answer is a" | a ) none of these , b ) 5 kmph , c ) 6 cm', ' , d ) 14 days , e ) 75 | d | add(add(divide(subtract(subtract(const_1, multiply(const_4, divide(const_1, 8))), multiply(add(const_4, const_4), divide(const_1, 40))), add(divide(const_1, 8), divide(const_1, 40))), add(const_4, const_4)), const_4) | add(const_4,const_4)|divide(const_1,n0)|divide(const_1,n1)|add(#1,#2)|multiply(#1,const_4)|multiply(#0,#2)|subtract(const_1,#4)|subtract(#6,#5)|divide(#7,#3)|add(#0,#8)|add(#9,const_4)| | physics |
if an integer n is to be selected at random from 1 to 100 , inclusive , what is probability n ( n + 1 ) will be divisible by 32 ? | "because n ( n + 1 ) is always an even product of even * odd or odd * even factors , there is a probability of 1 that that it will be divisible by 2 , and , thus , a probability of 1 / 2 that it will be divisible by 4 and , thus , a probability of 1 / 4 that it will be divisible by 8 and , thus , a probability of 1 / 8 that it will be divisible by 16 and , thus , a probability of 1 / 16 that it will be divisible by 32 1 * 1 / 16 = 1 / 16 answer : c" | a ) 1 / 16 , b ) 10 % , c ) 340 , d ) rs . 315 , e ) 1 / 3 | a | divide(const_2, 32) | divide(const_2,n3)| | general |
today jim is twice as old as fred , and sam is 4 years younger than fred . 4 years ago jim was 8 times as old as sam . how old is jim now ? | we ' re asked how old jim is now . we ' re given three facts to work with : 1 ) today , jim is twice as old as fred 2 ) today , sam is 4 years younger than fred 3 ) four years ago , jim was 8 times as old as sam . let ' s test answer d : 20 if . . . . jim is currently 20 years old . . . . fred is 10 years old sam is 6 years old 4 years ago , jim was 16 and sam was 2 , so jim was 8 times sam ' s age . this is an exact match for what we were told , so this must be the answer . d | a ) 95 kg , b ) - 39 , c ) 3 / 4 , d ) 20 , e ) 2 / 9 | d | multiply(divide(subtract(multiply(8, 8), 4), subtract(8, const_2)), const_2) | multiply(n2,n2)|subtract(n2,const_2)|subtract(#0,n0)|divide(#2,#1)|multiply(#3,const_2) | general |
a bullet train 150 m long is running with a speed of 30 kmph . in what time will it pass a man who is running at 6 kmph in the direction opposite to that in which the bullet train is going ? | e 15 sec speed of the bullet train relative to man = ( 30 + 6 ) kmph = 36 * 5 / 18 m / sec = 30 / 3 m / sec . time taken by the bullet train to cross the man = time taken by it to cover 150 m at ( 30 / 3 ) m / sec = ( 150 * 3 / 30 ) sec = 15 sec | a ) 75 % , b ) $ 306 , c ) 9 , d ) 15 sec , e ) 50 % | d | divide(150, divide(multiply(add(30, 6), const_1000), const_3600)) | add(n1,n2)|multiply(#0,const_1000)|divide(#1,const_3600)|divide(n0,#2) | physics |
the sum of all the integers k such that β 21 < k < 24 is | "- 20 - - - - - - - - - - - - - - - - - - 0 - - - - - - - - - - - - - - - - - 23 values upto + 23 cancels outwe are left with only - 20 - 19 sum of which is - 39 . hence option d . d" | a ) - 39 , b ) none of above , c ) 112 , d ) 80 % , e ) 12 | a | add(add(negate(21), const_1), add(add(negate(21), const_1), const_1)) | negate(n0)|add(#0,const_1)|add(#1,const_1)|add(#1,#2)| | general |
a dealer purchases 15 articles for rs . 25 and sells 12 articles for rs . 30 . find the profit percentage ? | "l . c . m of 15 and 12 = 60 cp of 60 articles = rs . 100 ( 25 * 4 ) sp of 60 articles = rs . 150 ( 30 * 5 ) profit percentage = ( 150 - 100 ) / 100 * 100 = 50 % answer : b" | a ) 6 , b ) 4 / 1 , c ) 50 % , d ) 1856 , e ) 13.75 | c | subtract(multiply(30, add(const_4, const_1)), multiply(25, const_4)) | add(const_1,const_4)|multiply(n1,const_4)|multiply(n3,#0)|subtract(#2,#1)| | gain |
the digital sum of a number is the sum of its digits . for how many of the positive integers 24 - 130 inclusive is the digital sum a multiple of 7 ? | "is there other way than just listing ? 25 34 43 52 59 61 68 70 77 86 95 106 115 124 14 ways . . c" | a ) 223 , b ) 13 % , c ) 3 / 4000 , d ) 12,526 , e ) 14 | e | subtract(subtract(24, 7), const_2) | subtract(n0,n2)|subtract(#0,const_2)| | general |
a worker ' s daily wage is increased by 50 % and the new wage is $ 30 per day . what was the worker ' s daily wage before the increase ? | let x be the daily wage before the increase . 1.5 x = $ 30 x = $ 20 the answer is c . | a ) $ 290 , b ) 16 , c ) 77.77 meters , d ) $ 20 , e ) 625 | d | divide(30, add(const_1, divide(50, const_100))) | divide(n0,const_100)|add(#0,const_1)|divide(n1,#1) | physics |
what is 120 % of 13 / 24 of 600 ? | "120 % * 13 / 24 * 360 = 1.2 * 13 * 25 = 390 the answer is b ." | a ) 900 , b ) 15 / 8 ohms', ' , c ) 1,200 , d ) 390 , e ) 15 sec | d | divide(multiply(120, 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 |
if the weight of 12 meters long rod is 13.4 kg . what is the weight of 6 meters long rod ? | "answer β΅ weight of 12 m long rod = 13.4 kg β΄ weight of 1 m long rod = 13.4 / 12 kg β΄ weight of 6 m long rod = 13.4 x 6 / 12 = 6.7 kg option : a" | a ) 4 , b ) 5184 , c ) 6.7 kg . , d ) 5 hours , e ) 3576 | c | divide(multiply(6, 13.4), 12) | multiply(n1,n2)|divide(#0,n0)| | physics |
evaluate : 60 - 12 * 3 * 2 = ? | "according to order of operations , 12 ? 3 ? 2 ( division and multiplication ) is done first from left to right 12 * * 2 = 4 * 2 = 8 hence 60 - 12 * 3 * 2 = 60 - 8 = 52 correct answer b" | a ) 52 , b ) 1700 , c ) 3 , d ) 105 , e ) 4 | a | subtract(60, multiply(multiply(12, 3), 2)) | multiply(n1,n2)|multiply(n3,#0)|subtract(n0,#1)| | general |
a rectangular photograph is surrounded by a border that is 1 inch wide on each side . the total area of the photograph and the border is m square inches . if the border had been 4 inches wide on each side , the total area would have been ( m + 108 ) square inches . what is the perimeter of the photograph , in inches ? | "let length and breadth of photograph be l and b respectively . perimeter is given by 2 * ( l + b ) - - - - - ( 1 ) according to the question : ( l + 2 ) ( b + 2 ) = m - - - - ( 2 ) and ( l + 8 ) ( b + 8 ) = m + 108 - - - - - - - - - > ( l + 8 ) ( b + 8 ) - 108 = m - - - - - - ( 3 ) equating ( 2 ) and ( 3 ) ( l + 2 ) ( b + 2 ) = ( l + 8 ) ( b + 8 ) - 108 lb + 2 l + 2 b + 4 = lb + 8 l + 8 b + 64 - 108 simplify 6 l + 6 b = 48 - - - - - - > 2 ( l + b ) = 16 ( check eq ( 1 ) ) answer is c" | a ) 44 , b ) 90 , c ) 14 , d ) 30 , e ) 16 | e | divide(subtract(108, subtract(power(multiply(4, const_2), const_2), power(multiply(1, const_2), const_2))), const_2) | multiply(n1,const_2)|multiply(n0,const_2)|power(#0,const_2)|power(#1,const_2)|subtract(#2,#3)|subtract(n2,#4)|divide(#5,const_2)| | geometry |
a store sells 2 kinds of jelly beans mixes ( a and b ) both made up of red and yellow beans . if b contains 20 % more red beans than a but 10 % fewer yellow beans . and jar a contains twice as many red beans as yellow by what percent is the number of beans in jar b larger than the number in jar a | a has 10 yellows 20 reds total = 30 so b has 1.2 x 20 = 24 reds 0.9 x 10 = 9 yellows total = 33 difference = 3 / 30 = 10 % answer : e | a ) 16 , b ) 10 , c ) 24 . , d ) rs . 6720 , e ) 44 hours | b | multiply(subtract(divide(add(add(const_100, 20), subtract(const_100, 10)), const_100), const_2), const_100) | add(n1,const_100)|subtract(const_100,n2)|add(#0,#1)|divide(#2,const_100)|subtract(#3,const_2)|multiply(#4,const_100) | general |
find the greatest common factor ( gfc ) of 24 , 40 and 60 . | we first write the prime factorization of each given number 24 = 2 Γ 2 Γ 2 Γ 3 = 23 * cubic * Γ 3 40 = 2 Γ 2 Γ 2 Γ 5 = 23 * cubic * Γ 5 60 = 2 Γ 2 Γ 3 Γ 5 = 22 * square * Γ 3 Γ 5 gfc = 22 * square * = 4 corect answer is d ) 4 | a ) 40.46 , b ) 35 km , c ) 11 , d ) 4 , e ) 24 % | d | gcd(gcd(24, 40), 60) | gcd(n0,n1)|gcd(n2,#0) | other |
when 242 is divided by a certain divisor the remainder obtained is 12 . when 698 is divided by the same divisor the remainder obtained is 16 . however , when the sum of the two numbers 242 and 698 is divided by the divisor , the remainder obtained is 10 . what is the value of the divisor ? | "let that divisor be x since remainder is 12 or 16 it means divisor is greater than 16 . now 242 - 12 = 230 = kx ( k is an integer and 234 is divisble by x ) similarly 698 - 16 = 682 = lx ( l is an integer and 689 is divisible by x ) adding both 698 and 242 = ( 230 + 682 ) + 12 + 16 = x ( k + l ) + 28 when we divide this number by x then remainder will be equal to remainder of ( 28 divided by x ) = 10 hence x = 28 - 10 = 18 hence b" | a ) $ 41 , b ) 70400 yards , c ) 1954404 , d ) 18 , e ) 32 | d | subtract(add(12, 16), 10) | add(n1,n3)|subtract(#0,n6)| | general |
the cost price of an article is 64 % of the marked price . calculate the gain percent after allowing a discount of 20 % ? | "explanation : let marked price = rs . 100 . then , c . p . = rs . 64 , s . p . = rs . 80 gain % = 16 / 64 * 100 = 25 % . answer : option e" | a ) 78 + ( 4 / , b ) 7.7 % . , c ) 25 % , d ) 800 , e ) 94 | c | multiply(subtract(divide(subtract(const_100, 20), 64), const_1), const_100) | subtract(const_100,n1)|divide(#0,n0)|subtract(#1,const_1)|multiply(#2,const_100)| | gain |
by selling 22 pens for a rupee a woman loses 10 % . how many for a rupee should he sell in order to gain 50 % ? | "d 90 % - - - 22 150 % - - - ? 90 / 150 * 22 = 13" | a ) 18 % , b ) 65500 $ , c ) s . 240 , d ) 50 , e ) 13 | e | inverse(add(divide(divide(const_100, subtract(const_100, 10)), 22), divide(multiply(divide(divide(const_100, subtract(const_100, 10)), 22), 50), const_100))) | subtract(const_100,n1)|divide(const_100,#0)|divide(#1,n0)|multiply(n2,#2)|divide(#3,const_100)|add(#2,#4)|inverse(#5)| | gain |