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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if x dollars is invested at 10 percent for one year and y dollars is invested at 8 percent for one year , the annual income from the 10 percent investment will exceed the annual income from the 8 percent investment by $ 92 . if $ 2000 is the total amount invested , how much is invested at 8 percent ? | 0.1 x = 0.08 ( 2000 - x ) + 92 0.18 x = 252 x = 1400 then the amount invested at 8 % is $ 2000 - $ 1400 = $ 600 the answer is a . | a ) $ 600 , b ) $ 700 , c ) $ 800 , d ) $ 900 , e ) $ 1000 | a | divide(subtract(multiply(divide(10, const_100), 2000), 92), add(divide(8, const_100), divide(10, const_100))) | divide(n0,const_100)|divide(n1,const_100)|add(#1,#0)|multiply(n5,#0)|subtract(#3,n4)|divide(#4,#2) | general |
four 6 faced dice are thrown together . the probability that all the four show the same number on them is ? | "it all 4 numbers have to be same basically we want quadruplets . 1111 , 2222 , 3333 , 4444 , 5555 and 6666 . those are six in number . further the four dice can fall in 6 * 6 * 6 * 6 = 1296 ways . hence the probability is 6 / 1296 = 1 / 216 answer : b" | a ) 1 / 248 , b ) 1 / 216 , c ) 1 / 144 , d ) 1 / 200 , e ) 1 / 242 | b | multiply(multiply(multiply(divide(const_1, 6), divide(const_1, 6)), divide(const_1, 6)), divide(const_1, 6)) | divide(const_1,n0)|multiply(#0,#0)|multiply(#0,#1)|multiply(#0,#2)| | probability |
the tax on a commodity is diminished by 10 % and its consumption increased by 25 % . the effect on revenue is ? | "100 * 100 = 10000 90 * 125 = 11250 - - - - - - - - - - - 10000 - - - - - - - - - - - 1250 100 - - - - - - - - - - - ? = > 12.5 % decrease answer : e" | a ) 12 % decrease , b ) 18 % decrease , c ) 19 % decrease , d ) 13 % decrease , e ) 12.5 % increase | e | subtract(const_100, multiply(multiply(add(const_1, divide(25, const_100)), subtract(const_1, divide(10, const_100))), const_100)) | divide(n1,const_100)|divide(n0,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|multiply(#4,const_100)|subtract(const_100,#5)| | general |
a man ’ s wages were decreased by 50 % . again , the reduced wages were increased by 50 % . he has a loss of ? | here , x = - 50 and y = 50 therefore , the net % change in value = ( x + y + xy / 100 ) % = [ - 50 + 50 + ( - 50 x 50 ) / 100 ] % or - 25 % since the sign is negative , there is loss of 25 % answer : b | a ) 35 % , b ) 25 % , c ) 20 % , d ) 30 % , e ) none of these | b | subtract(const_100, multiply(divide(add(const_100, 50), const_100), 50)) | add(n0,const_100)|divide(#0,const_100)|multiply(n0,#1)|subtract(const_100,#2) | gain |
a cubical block of metal weighs 7 pounds . how much will another cube of the same metal weigh if its sides are twice as long ? | "for example our cube have a side 1 meter , so we have 1 cubical meter in this cube and this cubical meter weigth 7 pounds if we take cube with side 2 meters we will have 8 cubical meters in this cube 8 meters * 7 pounds = 56 pounds so answer is d and similar but more theoretical approach : if we have sides a and b tha... | a ) 48 , b ) 32 , c ) 24 , d ) 56 , e ) 12 | d | multiply(7, multiply(const_2, const_4)) | multiply(const_2,const_4)|multiply(n0,#0)| | geometry |
two cubes of their volumes in the ratio 8 : 125 . the ratio of their surface area is : | the ratio of their surface area is 8 : 125 2 : 5 answer is b . | ['a ) 1.5 : 5', 'b ) 2 : 5', 'c ) 3 : 5', 'd ) 1 : 5', 'e ) 4 : 5'] | b | divide(power(8, const_0_33), power(125, const_0_33)) | power(n0,const_0_33)|power(n1,const_0_33)|divide(#0,#1) | geometry |
find the compound ratio of ( 4 : 3 ) , ( 1 : 3 ) and ( 2 : 3 ) is | required ratio = 4 / 3 * 1 / 3 * 2 / 3 = 16 / 27 = 16 : 27 answer is a | a ) 16 : 27 , b ) 12 : 13 , c ) 13 : 14 , d ) 14 : 15 , e ) 31 : 27 | a | multiply(divide(4, 3), multiply(divide(4, 3), divide(1, 3))) | divide(n0,n1)|divide(n2,n1)|multiply(#0,#1)|multiply(#0,#2) | other |
a train traveling at 72 kmph crosses a platform in 30 seconds and a man standing on the platform in 19 seconds . what is the length of the platform in meters ? | "speed of the train in metres / sec = 72000 / 3600 = 20 distance travelled by train to cross the platform = 30 * 20 = 600 = length of train + length of platform distance travelled by train to cross the man = 19 * 20 = 380 = length of train length of platform = 600 - 380 = 220 answer : a" | a ) 220 meters , b ) 360 meters , c ) 420 meters , d ) 600 meters , e ) can not be determined | a | subtract(multiply(divide(multiply(72, const_1000), const_3600), 30), multiply(divide(multiply(72, const_1000), const_3600), 19)) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)|multiply(n2,#1)|subtract(#2,#3)| | physics |
a batsman scored 140 runs which included 3 boundaries and 8 sixes . what percent of his total score did he make by running between the wickets ? | "number of runs made by running = 140 - ( 3 x 4 + 8 x 6 ) = 140 - ( 60 ) = 80 now , we need to calculate 80 is what percent of 140 . = > 80 / 140 x 100 = 57 % answer : d" | a ) 50 % , b ) 40 % , c ) 60 % , d ) 57 % , e ) 80 % | d | multiply(divide(subtract(140, add(multiply(3, 8), multiply(8, 3))), 140), const_100) | multiply(n1,n2)|multiply(n1,n2)|add(#0,#1)|subtract(n0,#2)|divide(#3,n0)|multiply(#4,const_100)| | general |
an escalator moves towards the top level at the rate of 11 ft . sec and its length is 140 feet . if a person walks on the moving escalator at the rate of 3 feet per second towards the top level , how much time does he take to cover the entire length . | explanation : time taken to cover the entire length = tot . dist / resultant speed = 140 / ( 11 + 3 ) = 10 sec answer : b | a ) 14 sec , b ) 10 sec , c ) 12 sec , d ) 8 sec , e ) 5 sec | b | divide(140, add(11, 3)) | add(n0,n2)|divide(n1,#0) | gain |
for how many unique pairs of nonnegative integers { a , b } is the equation a ^ 2 - b ^ 2 = 25 true ? | answer d ( a + b ) ( a - b ) = 25 5 cases for ( a + b ) , ( a - b ) 25 , 1 5 , 5 answer d | a ) 1 , b ) 3 , c ) 5 , d ) 2 , e ) 9 | d | divide(log(25), log(add(const_4, const_1))) | add(const_1,const_4)|log(n2)|log(#0)|divide(#1,#2) | general |
set s contains exactly 10 numbers and has an average ( arithmetic mean ) of 6.2 . if one of the numbers in set s is increased by 8 , while all other numbers remain the same , what is the new average of set s ? | old set s - total is avg * no of elements = 6.2 * 10 = 62 if one number is increased by 8 then total increased to 62 + 8 = 70 new avg - 70 / 10 = 7.0 . hence answer is d . | a ) 6.6 , b ) 6.7 , c ) 6.8 , d ) 7.0 , e ) 6.9 | d | divide(add(multiply(10, 6.2), 8), 10) | multiply(n0,n1)|add(n2,#0)|divide(#1,n0) | general |
14 men can complete a piece of work in 32 days . in how many days can 16 men complete that piece of work ? | "14 * 32 = 16 * x = > x = 27 1 / 2 days answer : d" | a ) 23 , b ) 27 3 / 4 , c ) 20 1 / 2 , d ) 27 1 / 2 , e ) 11 | d | divide(multiply(32, 14), 16) | multiply(n0,n1)|divide(#0,n2)| | physics |
a work crew of 7 men takes 10 days to complete one - half of a job . if 8 men are then added to the crew and the men continue to work at the same rate , how many days will it take the enlarged crew to do the rest of the job ? | suppose 1 man can do work in x days . . so 7 men will do in . . 7 / x = 1 / 10 * 1 / 2 as half job is done x = 140 now 8 more are added then 15 / 140 = 1 / 2 * 1 / d for remaining half job d = 4 2 / 3 number of days e | a ) 2 , b ) 3 , c ) 3 1 / 3 , d ) 4 , e ) 4 2 / 3 | e | add(10, divide(multiply(7, 10), add(7, 8))) | add(n0,n2)|multiply(n0,n1)|divide(#1,#0)|add(n1,#2) | physics |
a train running at the speed of 90 km / hr crosses a pole in 6 seconds . what is the length of the train ? | "speed = ( 90 * 5 / 18 ) m / sec = ( 25 ) m / sec length of the train = ( speed x time ) = ( 25 * 9 ) m = 225 m . answer : d" | a ) 255 , b ) 205 , c ) 502 , d ) 225 , e ) 235 | d | multiply(divide(multiply(90, const_1000), const_3600), 6) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics |
the angle of elevation of a ladder leaning against a wall is 60 ° and the foot of the ladder is 4.6 m away from the wall . the length of the ladder is | solution let ab be the wall and bc be the ladder . then , < abc = 60 ° ac = 4.6 m . ; ac / bc = cos 60 ° = 1 / 2 ‹ = › bc = 2 × ac = ( 2 × 4.6 ) m = 9.2 m answer d | a ) 2.3 m , b ) 4.6 m , c ) 7.8 m , d ) 9.2 m , e ) none | d | multiply(4.6, const_2) | multiply(n1,const_2) | physics |
in a certain egg - processing plant , every egg must be inspected , and is either accepted for processing or rejected . for every 96 eggs accepted for processing , 4 eggs are rejected . if , on a particular day , 12 additional eggs were accepted , but the overall number of eggs inspected remained the same , the ratio o... | "straight pluggin in for me . as usual , i started with c and got the answer . lets ' back calculate and see what we get let us consider eggs processed each day to be 400 so initial ratio of eggs processed and rejected is 96 : 4 or 24 : 1 so out of 400 eggs , there will be 384 eggs processed and 16 rejected . now if th... | a ) 100 , b ) 300 , c ) 400 , d ) 3,000 , e ) 4,000 | c | multiply(divide(12, subtract(99, 96)), const_100) | subtract(n3,n0)|divide(n2,#0)|multiply(#1,const_100)| | general |
the sale price sarees listed for rs . 510 after successive discount is 12 % and 15 % is ? | 510 * ( 88 / 100 ) * ( 85 / 100 ) = 381 answer : d | a ) 298 , b ) 237 , c ) 342 , d ) 381 , e ) 291 | d | subtract(subtract(510, divide(multiply(510, 12), const_100)), divide(multiply(subtract(510, divide(multiply(510, 12), const_100)), 15), const_100)) | multiply(n0,n1)|divide(#0,const_100)|subtract(n0,#1)|multiply(n2,#2)|divide(#3,const_100)|subtract(#2,#4) | gain |
the present population of a town is 3456 . population increase rate is 20 % p . a . find the population of town before 2 years ? | "p = 3456 r = 20 % required population of town = p / ( 1 + r / 100 ) ^ t = 3456 / ( 1 + 20 / 100 ) ^ 2 = 3456 / ( 6 / 5 ) ^ 2 = 2400 ( approximately ) answer is a" | a ) 2400 , b ) 2100 , c ) 3500 , d ) 3600 , e ) 2050 | a | add(3456, divide(multiply(3456, 20), const_100)) | multiply(n0,n1)|divide(#0,const_100)|add(n0,#1)| | gain |
the distance from city a to city b is 150 miles . while driving from city a to city b , cara drives at a constant speed of 30 miles per hour . dan leaves city a 90 minutes after cara . what is the minimum constant speed in miles per hour that dan must exceed in order to arrive in city b before cara ? | "the time it takes cara to drive to city b is 150 / 30 = 5 hours . dan needs to take less than 3.5 hours for the trip . dan needs to exceed a constant speed of 150 / 3.5 = 42.85 miles per hour . the answer is a ." | a ) 42.85 , b ) 44 , c ) 46 , d ) 48 , e ) 50 | a | divide(150, subtract(divide(150, 30), divide(90, 90))) | divide(n0,n1)|divide(n2,n2)|subtract(#0,#1)|divide(n0,#2)| | physics |
the interest on a certain deposit at 4.5 % p . a . is rs . 202.50 in one year . how much will the additional interest in one year be on the same deposit at 5 % p . a . ? | "solution s . i . = rs . 202.50 , r = 4.5 % , t = 1 year . principal = rs . ( 100 x 202.50 / 4.5 x 1 ) = rs . 4500 . now , p = rs . 4500 , r = 5 % , t = 1 year . s . i . = rs . ( 4500 x 5 x 1 / 1000 = rs . 225 . ∴ difference in interest = rs . ( 225 - 202.50 ) = rs . 22.50 . answer b" | a ) rs . 20.25 , b ) rs . 22.50 , c ) rs . 25 , d ) rs . 42.75 , e ) none | b | subtract(divide(multiply(divide(202.50, divide(4.5, const_100)), 5), const_100), 202.50) | divide(n0,const_100)|divide(n1,#0)|multiply(n2,#1)|divide(#2,const_100)|subtract(#3,n1)| | gain |
if a ( a - 4 ) = 12 and b ( b - 4 ) = 12 , where a ≠ b , then a + b = | i . e . if a = 6 then b = - 2 or if a = - 2 then b = 6 but in each case a + b = - 2 + 6 = 4 answer : option c | a ) − 48 , b ) − 5 , c ) 4 , d ) 46 , e ) 48 | c | subtract(subtract(subtract(subtract(add(add(4, 12), subtract(4, 12)), const_1), const_1), const_1), const_1) | add(n0,n1)|subtract(n0,n1)|add(#0,#1)|subtract(#2,const_1)|subtract(#3,const_1)|subtract(#4,const_1)|subtract(#5,const_1) | general |
what percent of 60 is 120 ? | "% of 60 is 120 ? = 120 / 60 = 21 = 200 % thus a is the correct answer ." | a ) 200 , b ) 75 , c ) 60 , d ) 33 1 ⁄ 3 , e ) 25 | a | multiply(divide(60, 120), const_100) | divide(n0,n1)|multiply(#0,const_100)| | gain |
the price of commodity x increases by 45 cents every year , while the price of commodity y increases by 20 cents every year . in 2001 , the price of commodity x was $ 4.20 and the price of commodity y was $ 6.30 . in which year will the price of commodity x be 65 cents more than the price of commodity y ? | "the price of commodity x increases 25 cents each year relative to commodity y . the price difference is $ 2.10 and commodity x needs to be 65 cents more than commodity y . $ 2.75 / 25 cents = 11 years the answer is 2001 + 11 years = 2012 . the answer is c ." | a ) 2010 , b ) 2011 , c ) 2012 , d ) 2013 , e ) 2014 | c | add(2001, divide(add(divide(65, const_100), subtract(6.30, 4.20)), subtract(divide(45, const_100), subtract(6.30, 4.20)))) | divide(n5,const_100)|divide(n0,const_100)|subtract(n4,n3)|add(#0,#2)|subtract(#1,#2)|divide(#3,#4)|add(n2,#5)| | general |
the number of timeshare condos available at sunset beach is 3 / 5 the number of timeshare condos available at playa del mar . if the total number of timeshare condos available at the two beaches combined is 400 , what is the difference between the number of condos available at sunset beach and the number of condos avai... | "let x be the number of timeshare condos available at playa del mar . then number of timeshare condos available at sunset beach = 3 / 5 x we know , x + 3 / 5 x = 400 hence , x = 250 . so , number of timeshare condos available at playa del mar = 250 the difference between the number of condos available at sunset beach a... | a ) 60 , b ) 90 , c ) 120 , d ) 100 , e ) 240 | d | add(divide(multiply(400, 3), 5), multiply(3, 5)) | multiply(n0,n2)|multiply(n0,n1)|divide(#0,n1)|add(#2,#1)| | general |
a train leaves delhi at 9 a . m . at a speed of 30 kmph . another train leaves at 3 p . m . at a speed of 40 kmph on the same day and in the same direction . how far from delhi , will the two trains meet ? | "d = 30 * 6 = 180 rs = 40 – 30 = 10 t = 180 / 10 = 18 d = 40 * 18 = 720 km answer : d" | a ) 229 , b ) 288 , c ) 600 , d ) 720 , e ) 121 | d | multiply(divide(multiply(30, add(const_3, 3)), subtract(40, 30)), 40) | add(n2,const_3)|subtract(n3,n1)|multiply(n1,#0)|divide(#2,#1)|multiply(n3,#3)| | physics |
when a merchant imported a certain item , she paid a 7 percent import tax on the portion of the total value of the item in excess of $ 1,000 . if the amount of the import tax that the merchant paid was $ 112.70 , what was the total value of the item ? | "let x be the value of the item . 0.07 * ( x - 1000 ) = 112.70 x = 2610 the answer is b ." | a ) $ 2420 , b ) $ 2610 , c ) $ 2860 , d ) $ 3050 , e ) $ 3270 | b | add(1,000, divide(112.70, divide(7, const_100))) | divide(n0,const_100)|divide(n2,#0)|add(#1,n1)| | general |
compound x contains elements a and b at an approximate ratio , by weight , of 2 : 10 . approximately how many grams of element b are there in 300 grams of compound x ? | "total number of fractions = 2 + 10 = 12 element b constitutes = 10 out of 12 parts of x so in 300 gms of x have 300 * 10 / 12 = 250 gms of b and 300 - 250 = 50 gms of a . cross check : - a / b = 50 / 250 = 2 / 10 ( as given ) ans c" | a ) 54 , b ) 162 , c ) 250 , d ) 270 , e ) 322 | c | divide(multiply(300, 10), add(2, 10)) | add(n0,n1)|multiply(n1,n2)|divide(#1,#0)| | other |
find the cost of fencing around a circular field of diameter 26 m at the rate of rs . 1.50 a meter ? | "2 * 22 / 7 * 13 = 81.7 81.7 * 1 1 / 2 = rs . 122.55 answer : a" | a ) 122.55 , b ) 132 , c ) 156 , d ) 158 , e ) 267 | a | multiply(circumface(divide(26, const_2)), 1.50) | divide(n0,const_2)|circumface(#0)|multiply(n1,#1)| | physics |
a can give b a start of 50 metres or 10 seconds in a kilometer race . how long does a take to complete the race ? | solution : a can give b a start of 50 metres or 10 seconds in a 1000 m race . that is , b takes 10 seconds to run 50 metres . therefore , b will take ( 10 / 50 ) * 1000 = 200 seconds to run 1000 metres . a who can give b a start of 10 seconds will take 10 seconds lesser to run the 1000 m . hence , the time taken by a =... | a ) 200 seconds , b ) 140 seconds , c ) 220 seconds , d ) 190 seconds , e ) none | d | subtract(multiply(divide(10, 50), const_1000), 10) | divide(n1,n0)|multiply(#0,const_1000)|subtract(#1,n1) | physics |
what is the remainder when 43717 ^ ( 43628233 ) is divided by 5 ? | "we need to find the units digit of the number . the units digit of powers of seven repeats 7 , 9 , 3 , and 1 cyclically . since 43628233 has the form 4 a + 1 , the units digit is 7 . then the remainder when dividing by 5 is 2 . the answer is c ." | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | c | reminder(multiply(43628233, 43717), 5) | multiply(n0,n1)|reminder(#0,n2)| | general |
last year sandy saved 6 % of her annual salary . this year , she made 10 % more money than last year , and she saved 10 % of her salary . the amount saved this year was what percent of the amount she saved last year ? | "let last year ' s salary be x . last year , sandy save 0.06 x this year , sandy saved 0.1 * 1.1 x = 0.11 x 0.11 x / 0.06 x = 11 / 6 = 1.83 = 183 % the answer is e ." | a ) 133 % , b ) 145 % , c ) 158 % , d ) 170 % , e ) 183 % | e | multiply(divide(multiply(divide(10, const_100), add(const_100, 10)), divide(multiply(6, const_100), const_100)), const_100) | add(n1,const_100)|divide(n2,const_100)|multiply(n0,const_100)|divide(#2,const_100)|multiply(#0,#1)|divide(#4,#3)|multiply(#5,const_100)| | gain |
if 10 litres of an oil of rs . 50 per litres be mixed with 5 litres of another oil of rs . 68 per litre then what is the rate of mixed oil per litre ? | "50 * 10 = 500 68 * 5 = 340 840 / 15 = 56 answer : c" | a ) rs . 49.17 , b ) rs . 51.03 , c ) rs . 56 , d ) rs . 55.33 , e ) none of the above | c | divide(add(multiply(10, 50), multiply(5, 68)), add(10, 5)) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0)| | gain |
a broker invested her own money in the stock market . during the first year , she increased her stock market wealth by 90 percent . in the second year , largely as a result of a slump in the stock market , she suffered a 50 percent decrease in the value of her stock investments . what was the net increase or decrease o... | "the actual answer is obtained by multiplying 140 % by 70 % and subtracting 100 % from this total . that is : 190 % × 50 % = 95 % ; 95 % − 100 % = - 5 % . answer : a" | a ) − 5 % , b ) 5 % , c ) 15 % , d ) 20 % , e ) 80 % | a | multiply(subtract(multiply(add(const_1, divide(90, const_100)), subtract(const_1, divide(50, const_100))), const_1), const_100) | divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|subtract(#4,const_1)|multiply(#5,const_100)| | gain |
what least number should be added to 1056 , so that the sum is completely divisible by 23 | "explanation : ( 1056 / 23 ) gives remainder 21 21 + 2 = 23 , so we need to add 2 answer : option c" | a ) 4 , b ) 1 , c ) 2 , d ) 3 , e ) 5 | c | subtract(23, reminder(1056, 23)) | reminder(n0,n1)|subtract(n1,#0)| | general |
the average temperature for tuesday , wednesday and thursday was 52 ° c . the average temperature for wednesday , thursday and friday was 54 ° c . if the temperature on friday be 53 ° c , what was the temperature on tuesday ? | explanation : t + w + t = 52 × 3 = 156 ° c w + t + f = 54 × 3 = 162 ° c also , temperature on friday = 53 ° c temperature on tuesday = 156 + 53 - 162 = 47 ° c answer : option d | a ) 39 ° c , b ) 44 ° c , c ) 37 ° c , d ) 47 ° c , e ) none of these | d | subtract(multiply(52, const_3), subtract(multiply(54, const_3), 53)) | multiply(n0,const_3)|multiply(n1,const_3)|subtract(#1,n2)|subtract(#0,#2) | physics |
the smallest number which when diminished by 6 , is divisible by 12 , 16 , 18 , 21 and 28 is | "required number = ( l . c . m of 12 , 16 , 18 , 21,28 ) + 6 = 1008 + 6 = 1014 answer : b" | a ) 1008 , b ) 1014 , c ) 1022 , d ) 1032 , e ) 1043 | b | add(multiply(multiply(power(const_3, const_2.0), power(const_2.0, const_4)), add(const_3, const_4)), 6) | add(const_3,const_4)|power(const_3,const_2.0)|power(const_2.0,const_4)|multiply(#1,#2)|multiply(#0,#3)|add(#4,n0)| | general |
the true discount on a bill due 9 months hence at 16 % per annum is rs . 240 . the amount of the bill is | "explanation : let p . w . be rs . x . then , s . i . on rs . x at 16 % for 9 months = rs . 240 . x ã — 16 ã — ( 9 / 12 ) ã — ( 1 / 100 ) = 240 or x = 2000 . p . w . = rs . 2000 . sum due = p . w . + t . d . = rs . ( 2000 240 ) = rs . 2240 . answer : b" | a ) 2200 , b ) 2240 , c ) 1600 , d ) 1354 , e ) none of these | b | add(divide(240, divide(multiply(divide(9, multiply(const_4, const_3)), 16), const_100)), 240) | multiply(const_3,const_4)|divide(n0,#0)|multiply(n1,#1)|divide(#2,const_100)|divide(n2,#3)|add(n2,#4)| | gain |
instead of multiplying a number by 3 , the number is divided by 5 . what is the percentage of error obtained ? | "let the number be x the right number is 3 x the wrong number is x / 5 error is ( 3 x - x / 5 ) = 14 x / 5 percentage of error is ( ( 14 x / 5 ) / 3 x ) * 100 = 93.33 % answer : d" | a ) 92.33 % , b ) 91.33 % , c ) 95.33 % , d ) 93.33 % , e ) 94.33 % | d | subtract(const_100, multiply(divide(divide(5, 5), multiply(3, 5)), const_100)) | divide(n1,n1)|multiply(n0,n1)|divide(#0,#1)|multiply(#2,const_100)|subtract(const_100,#3)| | general |
a man sold 18 toys for rs . 23100 , gaining thereby the cost price of 3 toy find the cost price of a toy | "let the cost of one toy = x . then , cost of 18 toys = 18 x . gain = 3 x . sp of 18 toys = rs . 23100 . gain = sp â € “ cp 3 x = 23100 â € “ 18 x 21 x = 23100 x = rs . 1100 . answer : option e" | a ) s . 600 , b ) s . 800 , c ) s . 500 , d ) s . 900 , e ) s . 1100 | e | divide(23100, add(18, 3)) | add(n0,n2)|divide(n1,#0)| | gain |
if ( a – b ) is 17 more than ( c + d ) and ( a + b ) is 3 less than ( c – d ) , then ( a – c ) is : | "( a – b ) – ( c + d ) = 17 and ( c – d ) – ( a + b ) = 3 = > ( a – c ) – ( b + d ) = 17 and ( c – a ) – ( b + d ) = 3 = > ( b + d ) = ( a – c ) – 17 and ( b + d ) = ( c – a ) – 3 = > ( a – c ) – 17 = ( c – a ) – 3 = > 2 ( a – c ) = 14 = > ( a – c ) = 7 answer : d" | a ) 6 , b ) 2 , c ) 3 , d ) 7 , e ) 5 | d | divide(17, 3) | divide(n0,n1)| | general |
in an objective competitive exam , a correct answer score 4 marks and on a wrong answer 2 marks are negatively added . a student scores 420 marks from 150 question . how many answers were correct ? | let x be the correct answer and y be the wrong answer so the total number of questions is ( x + y ) = 150 . . . . . ( 1 ) = > 4 x - 2 y = 420 . . . . . ( 2 ) by solving ( 1 ) & ( 2 ) , we get = > 6 x = 720 hence x = 120 therefore , number of correct answers are 120 . answer d | a ) 150 , b ) 155 , c ) 140 , d ) 120 , e ) 165 | d | divide(add(multiply(150, const_2), 420), add(4, 2)) | add(n0,n1)|multiply(n3,const_2)|add(n2,#1)|divide(#2,#0) | general |
in a division , divident is 690 , divisior is 36 and quotient is 19 . find the remainder . | "explanation : 690 = 36 x 19 + r 690 = 684 + r r = 690 - 684 = 6 answer : option e" | a ) a ) 4 , b ) b ) 3 , c ) c ) 2 , d ) d ) 5 , e ) e ) 6 | e | reminder(690, 36) | reminder(n0,n1)| | general |
running at their respective constant rates , machine x takes 2 days longer to produce w widgets than machine y . at these rates , if the two machines together produce 5 / 4 w widgets in 3 days , how many days would it take machine x alone to produce 4 w widgets ? | "let y produce w widgets in y days hence , in 1 day y will produce w / y widgets . also , x will produce w widgets in y + 2 days ( given , x takes two more days ) hence , in 1 day x will produce w / y + 2 widgets . hence together x and y in 1 day will produce { w / y + w / y + 2 } widgets . together x and y in 3 days w... | a ) 4 , b ) 6 , c ) 8 , d ) 24 , e ) 12 | d | divide(subtract(multiply(multiply(3, 4), 2), multiply(3, 4)), 2) | multiply(n2,n3)|multiply(n0,#0)|subtract(#1,#0)|divide(#2,n0)| | general |
a car traveled 75 % of the way from town a to town b at an average speed of 50 miles per hour . the car travels at an average speed of s miles per hour for the remaining part of the trip . the average speed for the entire trip was 50 miles per hour . what is s ? | "total distance = 100 miles ( easier to work with % ) 75 % of the distance = 75 miles 25 % of the distance = 25 miles 1 st part of the trip → 75 / 50 = 1.5 2 nd part of the trip → 25 / s = t total trip → ( 75 + 25 ) / 50 = 1.5 + t » 100 / 50 = 1.5 + t » 2 = 1.5 + t » t = 0.5 back to 2 nd part of the trip formula : 25 /... | a ) 10 , b ) 20 , c ) 25 , d ) 30 , e ) 50 | e | divide(subtract(add(50, 50), 75), subtract(divide(add(50, 50), 50), divide(75, 50))) | add(n1,n2)|divide(n0,n1)|divide(#0,n2)|subtract(#0,n0)|subtract(#2,#1)|divide(#3,#4)| | physics |
excluding stoppages , the average speed of a bus is 60 km / hr and including stoppages , the average speed of the bus is 30 km / hr . for how many minutes does the bus stop per hour ? | "in 1 hr , the bus covers 60 km without stoppages and 30 km with stoppages . stoppage time = time take to travel ( 60 - 30 ) km i . e 30 km at 60 km / hr . stoppage time = 30 / 60 hrs = 30 min . answer : b" | a ) 22 , b ) 30 , c ) 77 , d ) 20 , e ) 99 | b | subtract(multiply(const_1, const_60), multiply(divide(30, 60), const_60)) | divide(n1,n0)|multiply(const_1,const_60)|multiply(#0,const_60)|subtract(#1,#2)| | general |
if a + b + c = 13 , then find the ab + bc + ca : | answer : c ) 50 | a ) 22 , b ) 99 , c ) 50 , d ) 29 , e ) 11 | c | divide(divide(factorial(add(13, const_2)), factorial(13)), const_2) | add(n0,const_2)|factorial(n0)|factorial(#0)|divide(#2,#1)|divide(#3,const_2)| | general |
indu gave bindu rs . 6250 on compound interest for 2 years at 4 % per annum . how much loss would indu has suffered had she given it to bindu for 2 years at 4 % per annum simple interest ? | 6250 = d ( 100 / 4 ) 2 d = 10 answer : c | a ) 5 , b ) 2 , c ) 10 , d ) 5 , e ) 1 | c | subtract(subtract(multiply(6250, power(add(const_1, divide(4, const_100)), 2)), 6250), multiply(multiply(6250, divide(4, const_100)), 2)) | divide(n2,const_100)|add(#0,const_1)|multiply(n0,#0)|multiply(n1,#2)|power(#1,n1)|multiply(n0,#4)|subtract(#5,n0)|subtract(#6,#3) | gain |
if log 102 = 0.3010 , what is the value of log 101600 ? | explanation : log 101600 = log 10 ( 16 × 100 ) = log 10 ( 16 ) + log 10 ( 100 ) = log 10 ( 24 ) + log 10 ( 102 ) = 4 log 10 ( 2 ) + 2 = ( 4 × 0.3010 ) + 2 = 1.204 + 2 = 3.204 answer : option d | a ) 3.208 , b ) 3.202 , c ) 3.209 , d ) 3.204 , e ) 3.2112 | d | add(multiply(const_4, 0.301), divide(log(const_100), log(const_10))) | log(const_100)|log(const_10)|multiply(n1,const_4)|divide(#0,#1)|add(#3,#2) | other |
a man can row his boat with the stream at 16 km / h and against the stream in 12 km / h . the man ' s rate is ? | "ds = 16 us = 14 s = ? s = ( 16 - 2 ) / 2 = 2 kmph answer : c" | a ) 8 kmph , b ) 9 kmph , c ) 2 kmph , d ) 4 kmph , e ) 1 kmph | c | divide(subtract(16, 12), const_2) | subtract(n0,n1)|divide(#0,const_2)| | gain |
if 125 % of j is equal to 25 % of k , 150 % of k is equal to 50 % of l , and 175 % of l is equal to 75 % of m , then 20 % of m is equal to what percent of 200 % of j ? | first of all , let us write the information in form of equations and numbers 125 j = 5 k or 5 j = k ( i ) 150 k = 50 l or 3 k = l ( ii ) 175 l = 75 m or 7 l = 3 m ( iii ) we need to find a relation between j and m from ( i ) and ( ii ) , 15 j = l multiplying this by 7 , 105 j = 7 l = 3 m hence , 35 j = m now , we are a... | a ) 0.35 , b ) 3.5 , c ) 35 , d ) 350 , e ) 3500 | d | multiply(divide(multiply(divide(multiply(multiply(125, 150), 175), multiply(multiply(25, 50), 75)), 20), 200), const_100) | multiply(n0,n2)|multiply(n1,n3)|multiply(n4,#0)|multiply(n5,#1)|divide(#2,#3)|multiply(n6,#4)|divide(#5,n7)|multiply(#6,const_100) | gain |
a library has an average of 570 visitors on sundays and 240 on other days . the average number of visitors per day in a month of 30 days beginning with a sunday is : | "since the month begins with sunday , to there will be five sundays in the month average required = ( 570 x 5 + 240 x 25 ) / 30 ) = 295 answer : option d" | a ) 250 , b ) 276 , c ) 280 , d ) 295 , e ) none | d | divide(add(multiply(add(floor(divide(30, add(const_3, const_4))), const_1), 570), multiply(subtract(30, add(floor(divide(30, add(const_3, const_4))), const_1)), 240)), 30) | add(const_3,const_4)|divide(n2,#0)|floor(#1)|add(#2,const_1)|multiply(n0,#3)|subtract(n2,#3)|multiply(n1,#5)|add(#4,#6)|divide(#7,n2)| | general |
in a room filled with 7 people , 1 people have exactly 1 sibling in the room and 6 people have exactly 2 siblings in the room . if two individuals are selected from the room at random , what is the probability that those two individuals are not siblings ? | "there are suppose a b c d e f g members in the room 4 people who have exactly one sibling . . . . a b c d . . . . ( a is bs ∘ sssibl ∈ g ∘ ssand ∘ ssviceversa ) ∘ ss ( c ∘ ssis ∘ ssds ∘ sssibl ∈ g ∘ ssand ∘ ssviceversa ) ∘ ss ( c ∘ ssis ∘ ssdssibl ∈ gandviceversa ) ( cisds sibling and viceversa ) ( c is ds sibling and... | a ) 5 / 21 , b ) 3 / 7 , c ) 4 / 21 , d ) 5 / 7 , e ) 16 / 21 | c | add(multiply(divide(1, 7), divide(subtract(7, 2), subtract(7, 1))), multiply(divide(1, 7), divide(subtract(7, 1), subtract(7, 1)))) | divide(n1,n0)|subtract(n0,n4)|subtract(n0,n2)|subtract(n0,n1)|divide(#1,#2)|divide(#3,#2)|multiply(#0,#4)|multiply(#0,#5)|add(#6,#7)| | general |
the roof of an apartment building is rectangular and its length is 4 times longer than its width . if the area of the roof is 676 feet squared , what is the difference between the length and the width of the roof ? | "let the width = x x * 4 x = 676 x ^ 2 = 169 x = 13 length = 4 * 13 = 52 difference = 52 - 13 = 39 a is the answer" | a ) 39 . , b ) 40 . , c ) 42 . , d ) 44 . , e ) 46 . | a | subtract(multiply(sqrt(divide(676, 4)), 4), sqrt(divide(676, 4))) | divide(n1,n0)|sqrt(#0)|multiply(#1,n0)|subtract(#2,#1)| | geometry |
the area of a rectangular plot is 360 square metres . if the length is 25 % less than the breadth , what is the breadth of the plot ? | length = 75 % of breadth . length × breadth = 360 m 2 ⇒ 75 % of breadth × breadth = 360 m 2 ⇒ 75 / 100 × breadth × breadth = 360 m 2 ⇒ breadth × breadth = 480 m 2 ⇒ breadth = 22 m answer : d | ['a ) 32 m', 'b ) 12 m', 'c ) 20 m', 'd ) 22 m', 'e ) 25 m'] | d | sqrt(divide(360, subtract(const_1, divide(25, const_100)))) | divide(n1,const_100)|subtract(const_1,#0)|divide(n0,#1)|sqrt(#2) | geometry |
a 60 cm long wire is to be cut into two pieces so that one piece will be 2 / 5 th of the other , how many centimeters will the shorter piece be ? | "explanation : 1 : 2 / 5 = 5 : 2 2 / 7 * 60 = 20 answer : option d" | a ) a ) 73 , b ) b ) 20 , c ) c ) 83 , d ) d ) 17.1 , e ) e ) 52 | d | subtract(60, divide(60, add(divide(2, 5), const_1))) | divide(n1,n2)|add(#0,const_1)|divide(n0,#1)|subtract(n0,#2)| | physics |
one week , a certain truck rental lot had a total of 24 trucks , all of which were on the lot monday morning . if 50 % of the trucks that were rented out during the week were returned to the lot on or before saturday morning of that week , and if there were at least 12 trucks on the lot that saturday morning , what is ... | n - not rented trucks ; r - rented trucks n + r = 24 n + r / 2 = 12 r = 24 a | a ) 24 , b ) 16 , c ) 12 , d ) 8 , e ) 4 | a | add(12, multiply(24, divide(50, const_100))) | divide(n1,const_100)|multiply(n0,#0)|add(n2,#1) | general |
if two projectiles are launched at the same moment from 1455 km apart and travel directly towards each other at 470 km per hour and 500 km per hour respectively , how many minutes will it take for them to meet ? | "the projectiles travel a total of 970 km per hour . the time to meet is 1455 / 970 = 1.5 hours = 90 minutes the answer is a ." | a ) 90 , b ) 85 , c ) 80 , d ) 75 , e ) 70 | a | multiply(divide(1455, add(470, 500)), const_60) | add(n1,n2)|divide(n0,#0)|multiply(#1,const_60)| | physics |
if the least common addition of two prime numbers x and y is 36 , where x < y , then the value of 4 x + y is | ( x + y ) = 36 and both x an y are prime . the only values of x and y can be 17 and 19 ( x = 17 and y = 19 ) 4 x + y = 4 * 17 + 19 = 87 correct option : e | a ) 32 , b ) 45 , c ) 59 , d ) 73 , e ) 87 | e | add(add(multiply(36, const_2), divide(36, 4)), add(const_3, const_3)) | add(const_3,const_3)|divide(n0,n1)|multiply(n0,const_2)|add(#1,#2)|add(#3,#0) | general |
a man can do a job in 10 days . his father takes 20 days and his son finishes it in 25 days . how long will they take to complete the job if they all work together ? | "1 day work of the three persons = ( 1 / 10 + 1 / 20 + 1 / 25 ) = 19 / 100 so , all three together will complete the work in 100 / 19 = 5.3 days . answer : c" | a ) 6.3 , b ) 6.9 , c ) 5.3 , d ) 6.1 , e ) 6.2 | c | divide(const_1, add(divide(const_1, 25), add(divide(const_1, 10), divide(const_1, 20)))) | divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|add(#3,#2)|divide(const_1,#4)| | physics |
a train 2500 m long can cross an electric pole in 50 sec and then find the speed of the train ? | "length = speed * time speed = l / t s = 2500 / 50 s = 50 m / sec speed = 50 * 18 / 5 ( to convert m / sec in to kmph multiply by 18 / 5 ) speed = 180 kmph answer : b" | a ) 178 , b ) 180 , c ) 182 , d ) 184 , e ) 186 | b | divide(divide(2500, const_1000), divide(50, const_3600)) | divide(n0,const_1000)|divide(n1,const_3600)|divide(#0,#1)| | physics |
a semicircle has a radius of 14 . what is the approximate perimeter of the semicircle ? | "the perimeter of a circle is 2 * pi * r . the perimeter of a semicircle is 2 * pi * r / 2 + 2 r = pi * r + 2 r the perimeter is pi * 14 + 2 * 14 which is about 72 . the answer is a ." | a ) 72 , b ) 76 , c ) 80 , d ) 84 , e ) 88 | a | add(divide(circumface(14), const_2), multiply(const_2, 14)) | circumface(n0)|multiply(n0,const_2)|divide(#0,const_2)|add(#2,#1)| | geometry |
the average of 25 results is 24 . the average of first 12 of those is 14 and the average of last 12 is 17 . what is the 13 th result ? | "solution : sum of 1 st 12 results = 12 * 14 sum of last 12 results = 12 * 17 13 th result = x ( let ) now , 12 * 14 + 12 * 17 + x = 25 * 24 or , x = 228 . answer : option b" | a ) 740 , b ) 228 , c ) 690 , d ) 780 , e ) 458 | b | subtract(subtract(multiply(25, 24), multiply(12, 17)), multiply(12, 14)) | multiply(n0,n1)|multiply(n2,n5)|multiply(n2,n3)|subtract(#0,#1)|subtract(#3,#2)| | general |
a starts business with rs . 4500 and after 7 months , b joins with a as his partner . after a year , the profit is divided in the ratio 2 : 3 . what is b â € ™ s contribution in the capital ? | "explanation : a invested rs . 3500 for 12 months . let b joined with investment x . and he invested for 12 - 7 = 5 months . so there profit ratio = ( 4500 ã — 12 ) : ( 5 x ) = 2 : 3 â ‡ ’ x = 16200 answer : c" | a ) 24887 , b ) 20778 , c ) 16200 , d ) 9000 , e ) 2811 | c | divide(multiply(multiply(4500, const_12), 3), multiply(subtract(const_12, 7), 2)) | multiply(n0,const_12)|subtract(const_12,n1)|multiply(n3,#0)|multiply(n2,#1)|divide(#2,#3)| | other |
the average of 6 no . ' s is 3.95 . the average of 2 of them is 3.4 , while the average of theother 2 is 3.85 . what is the average of the remaining 2 no ' s ? | "sum of the remaining two numbers = ( 3.95 * 6 ) - [ ( 3.4 * 2 ) + ( 3.85 * 2 ) ] = 23.70 - ( 6.8 + 7.7 ) = 23.70 - 14.5 = 9.20 . required average = ( 9.2 / 2 ) = 4.6 . e" | a ) 2.3 , b ) 2.5 , c ) 3.5 , d ) 3.6 , e ) 4.6 | e | divide(subtract(multiply(6, 3.95), add(multiply(2, 3.4), multiply(2, 3.85))), 2) | multiply(n0,n1)|multiply(n2,n3)|multiply(n2,n5)|add(#1,#2)|subtract(#0,#3)|divide(#4,n2)| | general |
some persons can do a piece of work in 12 days . two times the number of such persons will do half of that work in | "explanation : let x men can do the in 12 days and the required number of days be z more men , less days [ indirect proportion ] less work , less days [ direct proportion ] answer : d ) 3 days" | a ) 9 , b ) 8 , c ) 4 , d ) 3 , e ) 1 | d | multiply(multiply(12, divide(const_1, const_2)), divide(const_1, const_2)) | divide(const_1,const_2)|multiply(n0,#0)|multiply(#0,#1)| | physics |
find the number of zero ’ s in 173 ! ( 173 factorial ) ? | no of zeroes in 173 ! is 173 / 5 = 34 ( quotient ) 34 / 5 = 6 ( quotient ) 6 / 5 = 1 ( quotient ) 34 + 6 + 1 = 41 ans 41 zeroes answer : d | a ) 38 , b ) 39 , c ) 40 , d ) 41 , e ) 42 | d | add(add(floor(divide(173, add(const_1, const_4))), floor(divide(173, power(add(const_1, const_4), const_2)))), floor(divide(173, power(add(const_1, const_4), const_3)))) | add(const_1,const_4)|divide(n0,#0)|power(#0,const_2)|power(#0,const_3)|divide(n0,#2)|divide(n0,#3)|floor(#1)|floor(#4)|floor(#5)|add(#6,#7)|add(#9,#8) | other |
270 ã · ? ã — 15 + 270 = 405 | "explanation : 270 ã · ? ã — 15 = 405 - 270 = 135 ( 270 ã — 15 ) / ? = 135 ? = ( 270 ã — 15 ) / 135 = 30 answer : option c" | a ) 24 , b ) 25 , c ) 30 , d ) 39 , e ) 43 | c | add(add(divide(multiply(add(subtract(405, 270), 15), const_2), const_10), divide(divide(multiply(add(subtract(405, 270), 15), const_2), const_10), const_10)), const_1) | subtract(n3,n2)|add(n1,#0)|multiply(#1,const_2)|divide(#2,const_10)|divide(#3,const_10)|add(#3,#4)|add(#5,const_1)| | general |
what is the sum of the odd integers from 55 to 65 , inclusive ? | "the mean is 60 . sum = mean ( # of elements ) there are 6 odd numbers between 55 - 65 inclusive . 6 * 60 = 360 b" | a ) 495 , b ) 360 , c ) 555 , d ) 600 , e ) 605 | b | divide(multiply(55, 65), const_4) | multiply(n0,n1)|divide(#0,const_4)| | general |
the ratio between the length and the breadth of a rectangular park is 3 : 2 . if a man cycling along the boundary of the park at the speed of 12 km / hr completes one round in 6 minutes , then the area of the park ( in sq . m ) is | "perimeter = distance covered in 6 min . = 12000 x 6 m = 1000 m . 60 let length = 3 x metres and breadth = 2 x metres . then , 2 ( 3 x + 2 x ) = 1000 or x = 100 . length = 300 m and breadth = 200 m . area = ( 300 x 200 ) m 2 = 60000 m 2 . e" | a ) 153200 , b ) 153400 , c ) 153600 , d ) 154000 , e ) 60000 | e | rectangle_area(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 6), const_1000), add(3, 2)), const_2), multiply(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 6), const_1000), add(3, 2)), const_2), 2)) | add(n0,n1)|multiply(const_2,const_3)|multiply(#1,const_10)|divide(n2,#2)|multiply(n3,#3)|multiply(#4,const_1000)|divide(#5,#0)|divide(#6,const_2)|multiply(n1,#7)|rectangle_area(#7,#8)| | physics |
when a person aged 39 is added to a group of n people , the average age increases by 2 . when a person aged 15 is added instead , the average age decreases by 1 . what is the value of e ? | a simple and elegant solution . as addition of 39 , shifts mean by 2 , and addition of 15 , shifts mean by 1 to the other side , we have the mean lying between 3915 , and in a ratio of 2 : 1 39 - 15 = 24 24 divide by 3 is 8 . meaning mean of the n terms is 15 + 8 = 39 - 16 = 23 now , from first statement , when a perso... | a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | a | subtract(divide(subtract(39, 15), add(2, 1)), const_1) | add(n1,n3)|subtract(n0,n2)|divide(#1,#0)|subtract(#2,const_1) | general |
the present age of a father is 3 years more than 3 times the age of his son . 3 years hence , father as age will be 8 years more than twice the age of the son . find the present age of the father . | explanation : let the present age be ' x ' years . then father ' s present age is 3 x + 3 years . three years hence ( 3 x + 3 ) + 3 = 2 ( x + 3 ) + 8 x = 8 hence father ' s present age = 3 x + 3 = [ ( 3 x 8 ) + 3 ] = 27 years . answer : e | a ) 88 , b ) 77 , c ) 33 , d ) 44 , e ) 27 | e | add(multiply(3, 8), 3) | multiply(n0,n3)|add(n0,#0) | general |
the “ length of integer x ” refers to the number of prime factors , not necessarily distinct , that x has . ( if x = 60 , the length of x would be 4 because 60 = 2 × 2 × 3 × 5 . ) what is the greatest possible length of integer z if z < 2500 ? | "to maximize the length of z , we should minimize its prime base . the smallest prime is 2 and since 2 ^ 11 = 2048 < 2500 , then the greatest possible length of integer z is 11 . the answer is c ." | a ) 7 , b ) 9 , c ) 11 , d ) 13 , e ) 15 | c | log(power(2, const_10)) | power(n3,const_10)|log(#0)| | general |
in measuring the sides of a rectangle , one side is taken 6 % in excess , and the other 5 % in deficit . find the error percent in the area calculated from these measurements . | "let x and y be the sides of the rectangle . then , correct area = xy . calculated area = ( 53 / 50 ) x ( 19 / 20 ) y = ( 144 / 143 ) ( xy ) error in measurement = ( 144 / 143 ) xy - xy = ( 1 / 143 ) xy error percentage = [ ( 1 / 143 ) xy ( 1 / xy ) 100 ] % = ( 7 / 10 ) % = 0.7 % . answer is b ." | a ) 0.11 % , b ) 0.7 % , c ) 0.4 % , d ) 0.6 % , e ) 0.8 % | b | subtract(subtract(6, 5), divide(multiply(6, 5), const_100)) | multiply(n0,n1)|subtract(n0,n1)|divide(#0,const_100)|subtract(#1,#2)| | geometry |
a courier charges for packages to a certain destination are 65 cents for the first 250 grams and 10 cents for each additional 100 grams or part thereof . what could be the weight in grams of a package for which the charge is $ 1.95 ? | "the charge is 65 cents for the first 250 grams . this leaves a charge of $ 1.95 - $ 0.65 = $ 1.30 the charge for the next 1200 grams is $ 1.20 which leaves a charge of $ 0.10 the weight is somewhere between 1450 and 1550 . the answer is c ." | a ) 980 , b ) 1170 , c ) 1530 , d ) 1720 , e ) 1960 | c | divide(add(subtract(1.95, divide(65, 100)), multiply(divide(divide(10, 100), 100), 250)), divide(divide(10, 100), 100)) | divide(n2,n3)|divide(n0,n3)|divide(#0,n3)|subtract(n4,#1)|multiply(n1,#2)|add(#4,#3)|divide(#5,#2)| | general |
if the diameter of circle r is 50 % of the diameter of circle s , the area of circle r is what percent of the area of circle s ? | "let diameter of circle r , dr = 50 and diameter of circle s , ds = 100 radius of circle r , rr = 25 radius of circle s , rs = 50 area of circle r / area of circle s = ( pi * rr ^ 2 ) / ( pi * rs ^ 2 ) = ( 25 / 50 ) ^ 2 = ( 5 / 10 ) ^ 2 = 25 % answer : b" | a ) 35 % , b ) 25 % , c ) 20 % , d ) 15 % , e ) 10 % | b | divide(multiply(circle_area(50), const_100), circle_area(const_100)) | circle_area(n0)|circle_area(const_100)|multiply(#0,const_100)|divide(#2,#1)| | geometry |
a can do a piece of work in 9 days . when he had worked for 2 days b joins him . if the complete work was finished in 8 days . in how many days b alone can finish the work ? | "8 / 9 + 6 / x = 1 x = 54 days answer : d" | a ) 18 , b ) 77 , c ) 66 , d ) 54 , e ) 12 | d | subtract(inverse(subtract(multiply(divide(const_1, 8), subtract(const_1, multiply(2, divide(const_1, 9)))), divide(const_1, 9))), add(9, 8)) | add(n0,n2)|divide(const_1,n2)|divide(const_1,n0)|multiply(n1,#2)|subtract(const_1,#3)|multiply(#1,#4)|subtract(#5,#2)|inverse(#6)|subtract(#7,#0)| | physics |
john ' s bank ' s saving amount is decreased 50 % due to loan payment and current balance is rs . 50000 . find the actual balance before deduction ? | "50 % decreased 50 % balance = 50000 100 % = 50000 / 50 * 100 = 100000 answer : a" | a ) 100000 , b ) 8500 , c ) 9000 , d ) 9500 , e ) 10000 | a | add(divide(multiply(50000, 50), const_100), 50000) | multiply(n0,n1)|divide(#0,const_100)|add(n1,#1)| | gain |
a man ' s regular pay is $ 3 per hour up to 40 hours . overtime is twice the payment for regular time . if he was paid $ 192 , how many hours overtime did he work ? | "at $ 3 per hour up to 40 hours , regular pay = $ 3 x 40 = $ 120 if total pay = $ 168 , overtime pay = $ 192 - $ 120 = $ 72 overtime rate ( twice regular ) = 2 x $ 3 = $ 6 per hour = > number of overtime hours = $ 72 / $ 6 = 12 ans is e" | a ) 8 , b ) 5 , c ) 9 , d ) 6 , e ) 12 | e | divide(subtract(192, multiply(3, 40)), multiply(3, const_2)) | multiply(n0,n1)|multiply(n0,const_2)|subtract(n2,#0)|divide(#2,#1)| | physics |
if n is a positive integer and the product of all integers from 1 to n , inclusive , is a multiple of 294 , what is the least possible value of n ? | "294 = 2 * 3 * 7 * 7 , so n must be at least 14 . the answer is d ." | a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 16 | d | divide(divide(divide(divide(294, const_2), const_3), const_4), divide(const_10, const_2)) | divide(n1,const_2)|divide(const_10,const_2)|divide(#0,const_3)|divide(#2,const_4)|divide(#3,#1)| | general |
excluding stoppages , the speed of a train is 55 kmph and including stoppages it is 40 kmph . of how many minutes does the train stop per hour ? | "explanation : t = 15 / 55 * 60 = 16 answer : option b" | a ) 82 , b ) 16 , c ) 12 , d ) 82 , e ) 18 | b | subtract(const_60, multiply(const_60, divide(40, 55))) | divide(n1,n0)|multiply(#0,const_60)|subtract(const_60,#1)| | physics |
a train travels 210 km in 3 hours and 270 km in 4 hours . find the average speed of train . | "as we know that speed = distance / time for average speed = total distance / total time taken thus , total distance = 210 + 270 = 480 km thus , total speed = 7 hrs or , average speed = 480 / 7 or , 69 kmph . answer : b" | a ) 80 kmph , b ) 69 kmph , c ) 70 kmph , d ) 90 kmph , e ) none of these | b | divide(add(210, 270), add(3, 4)) | add(n0,n2)|add(n1,n3)|divide(#0,#1)| | physics |
15 - 2 + 4 ÷ 1 / 2 × 8 = ? | explanation : 15 - 2 + 4 ÷ 1 / 2 × 8 = 15 - 2 + 4 × 2 × 8 = 15 - 2 + 64 = 77 answer : option b | a ) 272 , b ) 77 , c ) 168 , d ) 56 , e ) 59 | b | add(subtract(15, 2), multiply(divide(4, divide(const_1, const_2)), 8)) | divide(const_1,const_2)|subtract(n0,n1)|divide(n2,#0)|multiply(n5,#2)|add(#3,#1) | general |
the difference between the compound interest and simple interest on a certain sum at 20 % per annum for 2 years is $ 500 . find the sum ? | "let the sum be $ x c . i . = x ( 1 + 20 / 100 ) ^ 2 - x = 35 x / 100 s . i . = ( x * 20 * 2 ) / 100 = 2 x / 5 c . i . - s . i . = ( 35 x / 100 ) - ( 2 x / 5 ) = 5 x / 100 5 x / 100 = 500 x = 10000 answer is a" | a ) $ 10000 , b ) $ 50000 , c ) $ 15200 , d ) $ 12500 , e ) $ 15000 | a | divide(multiply(multiply(const_100, const_100), 500), const_100) | multiply(const_100,const_100)|multiply(n2,#0)|divide(#1,const_100)| | gain |
the length of a rectangular garden is three times its width . if the area of the rectangular garden is 432 square meters , then what is the width of the rectangular garden ? | "let x be the width of the garden . 3 x ^ 2 = 432 x ^ 2 = 144 x = 12 the answer is e ." | a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | e | sqrt(divide(432, const_3)) | divide(n0,const_3)|sqrt(#0)| | geometry |
the mean of 50 observations was 39 . it was found later that an observation 48 was wrongly taken as 23 . the corrected new mean is | "sol . therefore correct sum = ( 39 × 50 + 48 – 23 ) = 1975 . therefore correct mean = 1975 / 50 = 39.5 . answer e" | a ) 35.2 , b ) 36.1 , c ) 36.5 , d ) 39.1 , e ) none | e | divide(add(multiply(39, 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 |
when positive integer n is divided by 2 , the remainder is 1 . when n is divided by 7 , the remainder is 5 . what is the smallest positive integer p , such that ( n + p ) is a multiple of 10 ? | "when positive integer n is divided by 2 , the remainder is 1 i . e . , n = 2 x + 1 values of n can be one of { 1 , 3 , 5 , 7 , 9 , 11 , 13 , 15 , 17,19 , . . . . . . . . . . . . . . 31 , 33,35 . . . . . . . . . . . . . . . . . . } similarly , when n is divided by 7 , the remainder is 5 . . i . e . , n = 7 y + 5 values... | a ) 1 , b ) 2 , c ) 5 , d ) 19 , e ) 20 | c | subtract(10, reminder(5, 7)) | reminder(n3,n2)|subtract(n4,#0)| | general |
how many odd factors does 160 have ? | "start with the prime factorization : 160 = 2 * 5 for odd factors , we put aside the factor of two , and look at the other prime factors . set of exponents = { 1 } plus 1 to each = { 2 } product = 2 therefore , there are 2 odd factors of 160 . answer : a ." | a ) 2 , b ) 4 , c ) 5 , d ) 6 , e ) 8 | a | add(add(add(const_4, const_2), const_1), const_1) | add(const_2,const_4)|add(#0,const_1)|add(#1,const_1)| | other |
the length of a rectangle is doubled while its width is tripled . what is the % change in area ? | the original area is l * w the new area is 2 l * 3 w = 6 * l * w = l * w + 5 * l * w the area increased by 500 % . the answer is c . | ['a ) 250 %', 'b ) 300 %', 'c ) 500 %', 'd ) 650 %', 'e ) 700 %'] | c | multiply(subtract(multiply(const_2, const_3), const_1), const_10) | multiply(const_2,const_3)|subtract(#0,const_1)|multiply(#1,const_10) | geometry |
there were totally 100 men . 81 are married . 75 have t . v , 85 have radio , 70 have a . c . how many men have t . v , radio , a . c and also married ? | "100 - ( 100 - 81 ) - ( 100 - 75 ) - ( 100 - 85 ) - ( 100 - 70 ) = 100 - 19 - 25 - 15 - 30 = 100 - 89 = 11 answer : a" | a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | a | subtract(100, add(add(add(subtract(100, 81), subtract(100, 75)), subtract(100, 85)), subtract(100, 70))) | subtract(n0,n1)|subtract(n0,n2)|subtract(n0,n3)|subtract(n0,n4)|add(#0,#1)|add(#4,#2)|add(#5,#3)|subtract(n0,#6)| | general |
nicky and cristina are running a 400 meter race . since cristina is faster than nicky , she gives him a 12 second head start . if cristina runs at a pace of 5 meters per second and nicky runs at a pace of only 3 meters per second , how many seconds will nicky have run before cristina catches up to him ? | "the distance traveled by both of them is the same at the time of overtaking . 3 ( t + 12 ) = 5 t t = 18 . cristina will catch up nicky in 18 seconds . so in 18 seconds cristina would cover = 18 * 5 = 90 meter . now time taken my nicky to cover 90 meter = 90 / 3 = 30 seconds . d" | a ) 40 , b ) 25 , c ) 35 , d ) 30 , e ) 50 | d | add(divide(multiply(3, 12), 3), divide(multiply(3, 12), subtract(5, 3))) | multiply(n1,n3)|subtract(n2,n3)|divide(#0,n3)|divide(#0,#1)|add(#2,#3)| | physics |
a mixture of 70 liters of wine and water contains 10 % water . how much water must be added to make water 12 ½ % of the total mixture ? | "70 * ( 10 / 100 ) = 7 wine water 87 1 / 2 % 12 1 / 2 % 87 1 / 2 % - - - - - - - 63 12 1 / 2 % - - - - - - - ? = > 9 - 7 = 2 answer : d" | a ) 12 liters , b ) 10 liters , c ) 4 liters , d ) 2 liters , e ) 3 liters | d | divide(subtract(multiply(divide(12, const_100), 70), multiply(divide(10, const_100), 70)), subtract(const_1, divide(12, const_100))) | divide(n2,const_100)|divide(n1,const_100)|multiply(n0,#0)|multiply(n0,#1)|subtract(const_1,#0)|subtract(#2,#3)|divide(#5,#4)| | general |
danny and steve are running towards each other , each one from his own house . danny can reach steve ' s house in 35 minutes of running , which is half the time it takes steve to reach danny ' s house . if the two started to run at the same time , how much time longer will it take steve to reach the halfway point betwe... | "danny ' s time is 35 minutes . the time to reach halfway is 17.5 minutes . steve ' s time is 70 minutes . the time to reach halfway is 35 minutes . the time difference is 35 - 17.5 = 17.5 minutes the answer is a ." | a ) 17.5 minutes . , b ) 35 minutes . , c ) 52.5 minutes . , d ) 70 minutes . , e ) 105 minutes . | a | divide(35, const_2) | divide(n0,const_2)| | physics |
a vessel of capacity 2 litre has 20 % of alcohol and another vessel of capacity 6 litre had 40 % alcohol . the total liquid of 8 litre was poured out in a vessel of capacity 10 litre and thus the rest part of the vessel was filled with the water . what is the new concentration of mixture ? | "20 % of 2 litres = 0.4 litres 40 % of 6 litres = 2.4 litres therefore , total quantity of alcohol is 2.8 litres . this mixture is in a 10 litre vessel . hence , the concentration of alcohol in this 10 litre vessel is 28 % answer : e" | a ) 31 % . , b ) 71 % . , c ) 49 % . , d ) 29 % . , e ) 28 % . | e | multiply(divide(add(multiply(divide(20, const_100), 2), multiply(divide(40, const_100), 6)), 10), const_100) | divide(n1,const_100)|divide(n3,const_100)|multiply(n0,#0)|multiply(n2,#1)|add(#2,#3)|divide(#4,n5)|multiply(#5,const_100)| | general |
in a kilometer race , a beats b by 60 meters or 10 seconds . what time does a take to complete the race ? | "time taken by b run 1000 meters = ( 1000 * 10 ) / 60 = 166.7 sec . time taken by a = 166.7 - 10 = 156.7 sec . answer : b" | a ) 199 sec , b ) 156.7 sec , c ) 726 sec , d ) 127 sec , e ) 128 sec | b | subtract(divide(multiply(const_1, const_1000), divide(60, 10)), 10) | divide(n0,n1)|multiply(const_1,const_1000)|divide(#1,#0)|subtract(#2,n1)| | physics |
the speed at which a man can row a boat in still water is 15 kmph . if he rows downstream , where the speed of current is 3 kmph , what time will he take to cover 120 metres ? | "speed of the boat downstream = 15 + 3 = 18 kmph = 18 * 5 / 18 = 5 m / s hence time taken to cover 120 m = 120 / 5 = 24 seconds . answer : b" | a ) 18 , b ) 24 , c ) 17 , d ) 12 , e ) 13 | b | divide(120, multiply(add(15, 3), const_0_2778)) | add(n0,n1)|multiply(#0,const_0_2778)|divide(n2,#1)| | physics |
the perimeter of one face of a cube is 24 cm . its volume will be : | "explanation : edge of cude = 24 / 4 = 6 cm volume = a * a * a = 6 * 6 * 6 = 216 cm cube option a" | a ) 216 cm 3 , b ) 400 cm 3 , c ) 250 cm 3 , d ) 625 cm 3 , e ) none of these | a | volume_cube(square_edge_by_perimeter(24)) | square_edge_by_perimeter(n0)|volume_cube(#0)| | geometry |
when a train moves at an average speed of 100 kmph it reaches its destination on time . when its average speed becomes 80 kmph , then it reaches its destination 20 minutes late . find the length of journey rounded to two decimal places . | sol . difference between timings = 20 min = 1 / 3 hr . let the length of journey be x km . then , x / 80 - x / 100 = 1 / 3 â ‡ ” 5 x - 4 x = 133.33 â ‡ ” x = 133.33 km . answer b | a ) 99.33 km , b ) 133.33 km , c ) 125.33 km , d ) 156.66 km , e ) none | b | divide(divide(20, const_60), divide(subtract(100, 80), multiply(100, 80))) | divide(n2,const_60)|multiply(n0,n1)|subtract(n0,n1)|divide(#2,#1)|divide(#0,#3) | physics |
a certain bacteria colony doubles in size every day for 20 days , at which point it reaches the limit of its habitat and can no longer grow . if two bacteria colonies start growing simultaneously , how many days will it take them to reach the habitat ’ s limit ? | "simultaneous grow = same amount of time in 20 days we should expect to have p ( 2 ) ^ 20 so if we have two things doing the job of getting us there then ; 2 ^ x + 2 ^ x = 20 2 ( 2 ) ^ x = 2 2 ^ ( x + 1 ) = 2 ^ 20 x + 1 = 20 x = 19 ; answer : e" | a ) 6.33 , b ) 7.5 , c ) 10 , d ) 15 , e ) 19 | e | subtract(20, divide(20, 20)) | divide(n0,n0)|subtract(n0,#0)| | physics |
julie put half of her savings in a savings account that pays an annual simple interest and half in a savings account that pays an annual compound interest . after two years she earned $ 120 and $ 126 from the simple interest account and the compound interest account respectively . if the interest rates for both account... | "$ 120 for 2 years = $ 60 per year . extra $ 6 yearned with the compound interest is the percent yearned on percent . so , $ 6 is yearned on $ 60 , which means that the interest = 10 % . this on the other hand means that half of the savings = 60 * 10 = $ 600 . twice of that = $ 1,200 . answer : d ." | a ) 600 , b ) 720 , c ) 1080 , d ) 1200 , e ) 1440 | d | divide(120, divide(multiply(const_2, subtract(126, 120)), 120)) | subtract(n1,n0)|multiply(#0,const_2)|divide(#1,n0)|divide(n0,#2)| | gain |
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