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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jerry ’ s average ( arithmetic mean ) score on the first 3 of 4 tests is 85 . if jerry wants to raise his average by 2 points , what score must he earn on the fourth test ? | "total score on 3 tests = 85 * 3 = 255 jerry wants the average to be = 87 hence total score on 4 tests should be = 87 * 4 = 348 score required on the fourth test = 348 - 255 = 93 option d" | a ) 87 , b ) 89 , c ) 90 , d ) 93 , e ) 95 | d | subtract(multiply(4, add(85, 2)), multiply(85, 3)) | add(n2,n3)|multiply(n0,n2)|multiply(n1,#0)|subtract(#2,#1)| | general |
a number exceeds by 20 from its 3 / 8 part . then the number is ? | "x – 3 / 8 x = 20 x = 32 answer : a" | a ) a ) 32 , b ) b ) 35 , c ) c ) 39 , d ) d ) 40 , e ) e ) 45 | a | divide(multiply(20, 8), subtract(8, 3)) | multiply(n0,n2)|subtract(n2,n1)|divide(#0,#1)| | general |
the radius of the two circular fields is in the ratio 3 : 5 the area of the first field is what percent less than the area of the second ? | "r = 3 π r 2 = 9 r = 5 π r 2 = 25 25 π – 16 π 100 - - - - ? = > 64 % answer : d" | a ) 23 % , b ) 62 % , c ) 92 % , d ) 64 % , e ) 22 % | d | subtract(power(5, const_2), power(3, const_2)) | power(n1,const_2)|power(n0,const_2)|subtract(#0,#1)| | geometry |
find the lowest common multiple of 25 , 35 and 50 . | lcm = 2 * 5 * 5 * 7 = 350 . answer is a | a ) 350 , b ) 420 , c ) 510 , d ) 320 , e ) 280 | a | lcm(lcm(25, 35), 50) | lcm(n0,n1)|lcm(n2,#0) | general |
a side of beef lost 35 percent of its weight in processing . if the side of beef weighed 580 pounds after processing , how many pounds did it weigh before processing ? | "let weight of side of beef before processing = x ( 65 / 100 ) * x = 580 = > x = ( 580 * 100 ) / 65 = 892 answer d" | a ) 191 , b ) 355 , c ) 737 , d ) 892 , e ) 1,560 | d | divide(multiply(580, const_100), subtract(const_100, 35)) | multiply(n1,const_100)|subtract(const_100,n0)|divide(#0,#1)| | gain |
country x imported approximately $ 1.20 billion of goods in 1996 . if country x imported $ 288 million of goods in the first two months of 1997 and continued to import goods at the same rate for the rest of the year , by how much would country xs 1997 imports exceed those of 1996 ? | "convert units to millions as answer is in millions 1996 imports = $ 1.20 bill = $ 1200 mill i . e . 1200 / 12 = $ 100 mill / month 1997 imports = $ 288 mill / 2 month i . e . $ 144 mill / month difference / month = 144 - 100 = 44 difference / year = $ 44 mill * 12 = $ 528 mill answer : e" | a ) $ 24 million , b ) $ 120 million , c ) $ 144 million , d ) $ 240 million , e ) $ 528 million | e | subtract(multiply(288, divide(const_12, const_2)), multiply(1.20, const_1000)) | divide(const_12,const_2)|multiply(n0,const_1000)|multiply(n2,#0)|subtract(#2,#1)| | general |
a man goes from a to b at a speed of 21 kmph and comes back to a at a speed of 21 kmph . find his average speed for the entire journey ? | "distance from a and b be ' d ' average speed = total distance / total time average speed = ( 2 d ) / [ ( d / 21 ) + ( d / 24 ] = ( 2 d ) / [ 15 d / 168 ) = > 22.3 kmph . answer : d" | a ) 23.3 kmph , b ) 25.3 kmph , c ) 22.5 kmph , d ) 22.3 kmph , e ) 22.9 kmph | d | divide(add(21, 21), const_2) | add(n0,n1)|divide(#0,const_2)| | physics |
a man rows his boat 72 km downstream and 30 km upstream , taking 3 hours each time . find the speed of the stream ? | "speed downstream = d / t = 72 / ( 3 ) = 24 kmph speed upstream = d / t = 30 / ( 3 ) = 10 kmph the speed of the stream = ( 24 - 10 ) / 2 = 7 kmph answer : e" | a ) 6 kmph , b ) 5 kmph , c ) 2 kmph , d ) 8 kmph , e ) 7 kmph | e | divide(subtract(divide(72, 3), divide(30, 3)), const_2) | divide(n0,n2)|divide(n1,n2)|subtract(#0,#1)|divide(#2,const_2)| | physics |
a certain fruit stand sold apples for $ 0.70 each and bananas for $ 0.60 each . if a customer purchased both apples and bananas from the stand for a total of $ 5.60 , what total number of apples and bananas did the customer purchase ? | "let ' s start with 1 apple for $ 0.70 . let ' s subtract $ 0.70 from $ 5.60 until we get a multiple of $ 0.60 . $ 5.60 , $ 4.90 , $ 4.20 = 7 * $ 0.60 the customer purchased 7 bananas and 2 apples . the answer is b ." | a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | b | add(divide(subtract(5.60, multiply(0.70, const_2)), 0.60), const_2) | multiply(n0,const_2)|subtract(n2,#0)|divide(#1,n1)|add(#2,const_2)| | other |
25 liters of a mixture is created by mixing liquid p and liquid q in the ratio 3 : 2 . how many liters of liquid q must be added to make the ratio 5 : 4 ? | "let x be the amount of liquid q to be added . ( 2 / 5 ) * 25 + x = ( 4 / 9 ) * ( 25 + x ) 450 + 45 x = 500 + 20 x 25 x = 50 x = 2 the answer is b ." | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | b | subtract(divide(multiply(divide(multiply(add(5, 4), divide(multiply(25, 3), add(3, 2))), 5), 4), add(5, 4)), divide(multiply(25, 2), add(3, 2))) | add(n3,n4)|add(n1,n2)|multiply(n0,n1)|multiply(n0,n2)|divide(#2,#1)|divide(#3,#1)|multiply(#0,#4)|divide(#6,n3)|multiply(n4,#7)|divide(#8,#0)|subtract(#9,#5)| | general |
an amount of rs . 1560 was divided among a , b and c in the ratio 1 / 2 : 1 / 3 : 1 / 4 . find the share of c ? | "let the shares of a , b and c be a , b and c respectively . a : b : c = 1 / 2 : 1 / 3 : 1 / 4 a : b : c = 6 / 12 : 4 / 12 : 3 : 12 = 6 : 4 : 3 share of c = 3 / 13 * 1560 = rs . 360 . answer : d" | a ) rs . 278 , b ) rs . 361 , c ) rs . 369 , d ) rs . 360 , e ) rs . 363 | d | multiply(divide(1560, add(add(2, 3), 4)), 4) | add(n2,n4)|add(n6,#0)|divide(n0,#1)|multiply(n6,#2)| | general |
elena purchased brand x pens for $ 4.00 apiece and brand y for $ 2.20 apiece . if elena purchased a total of 12 of these pens for $ 42.00 , how many brand x pens did she purchase ? | 4 x + 2.8 y = 42 - - > multiply by 2.5 ( to get the integers ) - - > 10 x + 7 y = 105 - - > only one positive integers solutions x = 6 and y = 5 ( how to solve : 7 y must have the last digit of 5 in order the last digit of the sum to be 5 ) . answer : c . | a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | c | subtract(multiply(4, 12), 42) | multiply(n0,n2)|subtract(#0,n3) | general |
walking at 5 / 6 th of its usual speed a cab is 6 mnts late . find its usual time to cover the journey ? | "new speed = 5 / 6 th of usual speed new time = 6 / 5 th of usual time 6 / 5 ut - ut = 6 m ut / 5 = 6 m ut = 30 m answer is b" | a ) 25 m , b ) 30 m , c ) 32 m , d ) 50 m , e ) 62 m | b | multiply(5, 6) | multiply(n0,n2)| | physics |
on the number line , if x is halfway between - 6 and 4 , and if y is halfway between - 2 and 6 , what number is halfway between x and y ? | "x = - 1 and y = 2 . the answer is d ." | a ) - 2.5 , b ) - 1.5 , c ) - 0.5 , d ) 0.5 , e ) 1.5 | d | add(negate(6), 4) | negate(n0)|add(n1,#0)| | general |
what is the sum of the digits of integer k , if k = ( 10 ^ 50 - 50 ) | "there are 51 digits in 10 ^ 50 when we subtract 50 from it , there will be 50 digits left . 10 ^ 50 can be written as 9999999 . . . . ( 50 times ) + 1 so , 10 ^ 50 - 50 = 9999999 . . . . ( 50 times ) + 1 - 50 = 9999999 . . . . ( 50 times ) - 49 consider the last 2 digits , 99 - 49 = 50 the last 2 digits will be 50 . a... | a ) 337 , b ) 437 , c ) 457 , d ) 537 , e ) 412 | b | add(multiply(subtract(10, const_1), subtract(50, const_2)), 10) | subtract(n0,const_1)|subtract(n1,const_2)|multiply(#0,#1)|add(#2,n0)| | general |
what is the least number which when divided by 6 , 9 , 12 and 18 leaves remainder 4 in each care ? | "explanation : lcm of 6 , 9 , 12 and 18 is 36 required number = 36 + 4 = 40 answer : option b" | a ) 30 , b ) 40 , c ) 36 , d ) 56 , e ) 66 | b | add(lcm(lcm(6, 9), lcm(12, 18)), 4) | lcm(n0,n1)|lcm(n2,n3)|lcm(#0,#1)|add(n4,#2)| | general |
the population of a town increased from 1 , 75,000 to 2 , 80,000 in a decade . the average percent increase of population per year is | "solution increase in 10 years = ( 280000 - 175000 ) = 105000 . increase % = ( 105000 / 175000 ã — 100 ) % = 60 % . required average = ( 60 / 10 ) % = 6 % . answer c" | a ) 4.37 % , b ) 5 % , c ) 6 % , d ) 8.75 % , e ) none | c | add(multiply(divide(subtract(divide(subtract(subtract(subtract(multiply(multiply(const_10, const_1000), const_10), const_1000), const_1000), multiply(add(2, const_3), const_100)), multiply(add(multiply(add(const_3, const_4), const_10), add(2, const_3)), const_1000)), 1), const_10), const_100), const_4) | add(n2,const_3)|add(const_3,const_4)|multiply(const_10,const_1000)|multiply(#2,const_10)|multiply(#0,const_100)|multiply(#1,const_10)|add(#0,#5)|subtract(#3,const_1000)|multiply(#6,const_1000)|subtract(#7,const_1000)|subtract(#9,#4)|divide(#10,#8)|subtract(#11,n0)|divide(#12,const_10)|multiply(#13,const_100)|add(#14,co... | general |
a bullet train 200 m long is running with a speed of 69 kmph . in what time will it pass a man who is running at 3 kmph in the direction opposite to that in which the bullet train is going ? | "a 10 sec speed of the bullet train relative to man = ( 69 + 3 ) kmph = 72 * 5 / 18 m / sec = 60 / 3 m / sec . time taken by the bullet train to cross the man = time taken by it to cover 200 m at ( 60 / 3 ) m / sec = ( 200 * 3 / 60 ) sec = 10 sec" | a ) 10 sec , b ) 15 sec , c ) 12 sec , d ) 11 sec , e ) 16 sec | a | divide(200, divide(multiply(add(69, 3), const_1000), const_3600)) | add(n1,n2)|multiply(#0,const_1000)|divide(#1,const_3600)|divide(n0,#2)| | physics |
in what time will a railway train 90 m long moving at the rate of 36 kmph pass a telegraph post on its way ? | "t = 90 / 36 * 18 / 5 = 9 sec answer : c" | a ) 8 sec , b ) 1 sec , c ) 9 sec , d ) 6 sec , e ) 2 sec | c | divide(90, multiply(36, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics |
after successive discounts of 20 % , 10 % and 5 % a certain good is sold for rs . 6800 . find the actual price of the good . | "let actual price was 100 . after three successive discount this will become , 100 = = 20 % discount = > 80 = = 10 % discount = > 72 = = 5 % discount = 68.4 now compare , 68.4 = 6800 1 = 6800 / 68.4 100 = ( 6800 * 100 ) / 68.4 = rs . 9942 . answer : option d" | a ) s . 6000 , b ) s . 9000 , c ) s . 10800 , d ) s . 9942 , e ) s . 9980 | d | divide(multiply(6800, const_100), subtract(subtract(subtract(const_100, 20), divide(multiply(subtract(const_100, 20), 10), const_100)), divide(multiply(subtract(subtract(const_100, 20), divide(multiply(subtract(const_100, 20), 10), const_100)), 5), const_100))) | multiply(n3,const_100)|subtract(const_100,n0)|multiply(n1,#1)|divide(#2,const_100)|subtract(#1,#3)|multiply(n2,#4)|divide(#5,const_100)|subtract(#4,#6)|divide(#0,#7)| | gain |
a can run 160 metre in 28 seconds and b in 32 seconds . by what distance a beat b ? | "clearly , a beats b by 4 seconds now find out how much b will run in these 4 seconds speed of b = distance / time taken by b = 160 / 32 = 5 m / s distance covered by b in 4 seconds = speed ã — time = 5 ã — 4 = 20 metre i . e . , a beat b by 20 metre answer is c" | a ) 38 metre , b ) 28 metre , c ) 20 metre , d ) 15 metre , e ) 28 metre | c | subtract(160, multiply(divide(160, 32), 28)) | divide(n0,n2)|multiply(n1,#0)|subtract(n0,#1)| | physics |
a club wants to mix 20 pounds of candy worth $ 8.00 per pound with candy worth $ 5.00 per pound to reduce the cost of the mixture to $ 6.00 per pound . how many pounds of the $ 5.00 per pound candy should be used ? | let number of pounds of 5 $ candy to be used be w 6 = ( 20 * 8 + 5 * w ) / ( 20 + w ) = > 120 + 6 w = 160 + 5 w = > w = 40 answer c | a ) 20 , b ) 30 , c ) 40 , d ) 50 , e ) 60 | c | subtract(multiply(20, 8), multiply(6, 20)) | multiply(n0,n1)|multiply(n0,n3)|subtract(#0,#1) | general |
by selling an article at rs . 600 , a profit of 30 % is made . find its cost price ? | "sp = 600 cp = ( sp ) * [ 100 / ( 100 + p ) ] = 600 * [ 100 / ( 100 + 50 ) ] = 600 * [ 100 / 150 ] = rs . 400 answer : a" | a ) 400 , b ) 267 , c ) 287 , d ) 480 , e ) 811 | a | divide(multiply(600, const_100), add(const_100, 30)) | add(n1,const_100)|multiply(n0,const_100)|divide(#1,#0)| | gain |
a student gets 60 % in one subject , 70 % in the other . to get an overall of 70 % how much should get in third subject . | "let the 3 rd subject % = x 60 + 70 + x = 3 * 70 130 + x = 210 x = 210 - 130 = 80 answer : b" | a ) 75 % , b ) 80 % , c ) 45 % , d ) 55 % , e ) 65 % | b | subtract(multiply(70, const_3), add(60, 70)) | add(n0,n1)|multiply(n2,const_3)|subtract(#1,#0)| | gain |
if a ' s height is 30 % less than that of b , how much percent b ' s height is more than that of a ? | excess of b ' s height over a ' s = [ ( 30 / ( 100 - 30 ) ] x 100 % = 42.85 % answer : a ) | a ) 42.85 % , b ) 66.68 % , c ) 66.766 % , d ) 86.66 % , e ) 66.65 % | a | multiply(divide(30, subtract(const_100, 30)), const_100) | subtract(const_100,n0)|divide(n0,#0)|multiply(#1,const_100) | general |
if 13 = 13 w / ( 1 - w ) , then ( 2 w ) 2 = | "13 - 13 w = 13 w 26 w = 13 w = 1 / 2 2 w = 1 2 w * 2 = 1 * 2 = 2 answer : d" | a ) 1 / 4 , b ) 1 / 2 , c ) 1 , d ) 2 , e ) 3 | d | multiply(divide(13, add(13, 13)), 2) | add(n0,n0)|divide(n0,#0)|multiply(n3,#1)| | general |
what is the perimeter of a square field whose diagonal is 4 √ 2 ? | "4 a = 16 m answer : a" | a ) 16 m , b ) 32 m , c ) 27 m , d ) 26 m , e ) 76 m | a | divide(add(add(sqrt(subtract(power(4, const_2), power(2, const_2))), 2), add(sqrt(subtract(power(4, const_2), power(2, const_2))), 2)), 2) | power(n0,const_2)|power(n1,const_2)|subtract(#0,#1)|sqrt(#2)|add(n1,#3)|add(#4,#4)|divide(#5,n1)| | geometry |
if the sides of a triangle are 4 cm , 9 cm and 13 cm , what is its area ? | "the triangle with sides 4 cm , 9 cm and 13 cm is right angled , where the hypotenuse is 13 cm . area of the triangle = 1 / 2 * 4 * 9 = 18 cm 2 answer : option a" | a ) 18 , b ) 20 , c ) 22 , d ) 26 , e ) 28 | a | divide(multiply(9, 13), const_2) | multiply(n1,n2)|divide(#0,const_2)| | geometry |
of the 65 cars on a car lot , 45 have air - bag , 30 have power windows , and 12 have both air - bag and power windows . how many of the cars on the lot have neither air - bag nor power windows ? | total - neither = all air bag + all power windows - both or 65 - neither = 45 + 30 - 12 = 63 . = > neither = 2 , hence b . answer : b | a ) 8 , b ) 2 , c ) 10 , d ) 15 , e ) 18 | b | subtract(65, subtract(add(45, 30), 12)) | add(n1,n2)|subtract(#0,n3)|subtract(n0,#1) | other |
a large field of 700 hectares is divided into two parts . the difference of the areas of the two parts is one - fifth of the average of the two areas . what is the area of the smaller part in hectares ? | explanation : let the areas of the two parts be x and ( 700 - x ) hectares therefore , so , the two parts are 385 and 315 . hence , area of the smaller = 315 hectares answer : a ) 315 | a ) 315 , b ) 339 , c ) 288 , d ) 299 , e ) 111 | a | divide(subtract(700, multiply(divide(700, const_2), divide(const_1, add(const_1, const_4)))), const_2) | add(const_1,const_4)|divide(n0,const_2)|divide(const_1,#0)|multiply(#1,#2)|subtract(n0,#3)|divide(#4,const_2) | general |
if 2994 ã · 14.5 = 177 , then 29.94 ã · 1.45 = ? | "29.94 / 1.45 = 299.4 / 14.5 = ( 2994 / 14.5 ) x 1 / 10 ) [ here , substitute 177 in the place of 2994 / 14.5 ] = 177 / 10 = 17.7 answer is d ." | a ) 17.1 , b ) 17.3 , c ) 17.5 , d ) 17.7 , e ) 17.2 | d | divide(177, divide(14.5, 1.45)) | divide(n1,n4)|divide(n2,#0)| | general |
a part - time employee whose hourly wage was decreased by 20 percent decided to increase the number of hours worked per week so that the employee ' s total income did not change . by what percent t should the number of hours worked be increased ? | correct answer : c solution : c . we can set up equations for income before and after the wage reduction . initially , the employee earns w wage and works h hours per week . after the reduction , the employee earns . 8 w wage and works x hours . by setting these equations equal to each other , we can determine the incr... | a ) 12.5 % , b ) 20 % , c ) 25 % , d ) 50 % , e ) 100 % | c | multiply(divide(subtract(divide(multiply(const_10, const_4), multiply(divide(subtract(const_100, 20), const_100), const_10)), const_4), const_4), const_100) | multiply(const_10,const_4)|subtract(const_100,n0)|divide(#1,const_100)|multiply(#2,const_10)|divide(#0,#3)|subtract(#4,const_4)|divide(#5,const_4)|multiply(#6,const_100) | general |
tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random . what is the probability that the ticket drawn has a number which is a multiple of 4 or 15 ? | "explanation : here , s = { 1 , 2 , 3 , 4 , . . . . , 19 , 20 } = > n ( s ) = 20 let e = event of getting a multiple of 4 or 15 = multiples od 4 are { 4 , 8 , 12 , 16 , 20 } and multiples of 15 means multiples of 3 and 5 = { 3 , 6 , 9 , 12 , 15 , 18 , 5 , 10 , 15 , 20 } . = the common multiple is only ( 15 ) . = > e = ... | a ) 9 / 10 , b ) 6 / 10 , c ) 2 / 10 , d ) 3 / 10 , e ) 5 / 10 | d | divide(const_3.0, const_10) | divide(const_3.0,const_10)| | general |
cole drove from home to work at an average speed of 75 kmh . he then returned home at an average speed of 105 kmh . if the round trip took a total of 4 hours , how many minutes did it take cole to drive to work ? | first round distance travelled ( say ) = d speed = 75 k / h time taken , t 2 = d / 75 hr second round distance traveled = d ( same distance ) speed = 105 k / h time taken , t 2 = d / 105 hr total time taken = 4 hrs therefore , 4 = d / 75 + d / 105 lcm of 75 and 105 = 525 4 = d / 75 + d / 105 = > 4 = 7 d / 525 + 5 d / 5... | a ) 66 , b ) 70 , c ) 95 , d ) 112 , e ) 140 | e | multiply(divide(multiply(105, 4), add(75, 105)), const_60) | add(n0,n1)|multiply(n1,n2)|divide(#1,#0)|multiply(#2,const_60) | physics |
what could be the range of a set consisting of odd multiples of ( 6 + 1 ) ? | "range = the difference between the greatest and the smallest numbers in the sequence . our sequence is odd and is a multiple of 7 . every number in that sequence can be represented like this : 7 * ( 2 n + 1 ) where n is any positive integer . range = 7 * ( 2 m + 1 ) - 7 * ( 2 n + 1 ) = 7 * 2 * ( m - n ) = 14 * ( m - n... | a ) 70 , b ) 24 , c ) 35 , d ) 62 , e ) 21 | a | multiply(6, const_4) | multiply(n0,const_4)| | general |
the apartment on king - williams street is an asset that its value is tramping about . from the year 1973 to 1983 it ' s value decreased by 16 % and from 1983 to 1993 it ' s value increased by 16 % . what is the value of the asset in 1993 if in 1973 it was worth $ 55000 ? | 55,000 * 1.16 * . 84 = 53,592 answer e | a ) $ 38,796 . , b ) $ 40,000 . , c ) $ 38,976 . , d ) $ 39,679 . , e ) $ 53,592 . | e | subtract(multiply(log(multiply(55000, multiply(divide(subtract(const_100, 16), const_100), divide(add(const_100, 16), const_100)))), const_3), multiply(const_3, const_3)) | add(n2,const_100)|multiply(const_3,const_3)|subtract(const_100,n2)|divide(#2,const_100)|divide(#0,const_100)|multiply(#3,#4)|multiply(n8,#5)|log(#6)|multiply(#7,const_3)|subtract(#8,#1) | general |
in the xy - coordinate system , if ( m , n ) and ( m + 2 , n + k ) are two points on the line with the equation x = 2 y + 5 , then k = | "the equation of the line can be re written as y = ( 1 / 2 ) x - 5 / 2 . slope = 1 / 2 , which means for every 1 unit increase in the y co - ordinate , x will increase by 2 units . hence k = 1 . answer : b" | a ) 1 / 2 , b ) 1 , c ) 2 , d ) 5 / 2 , e ) 4 | b | divide(subtract(add(5, 2), 5), 2) | add(n0,n2)|subtract(#0,n2)|divide(#1,n1)| | general |
the difference between c . i . and s . i . on an amount of $ 15,000 for 2 years is $ 54 . what is the rate of interest per annum ? | "$ 54 is the interest on the first year of interest . let x be the interest rate . the interest after the first year is 15000 * x . the interest on the first year ' s interest is 15000 * x * x 15000 * x ^ 2 = 54 x = 0.06 the answer is b ." | a ) 5 % , b ) 6 % , c ) 7 % , d ) 8 % , e ) 9 % | b | sqrt(54) | sqrt(n2)| | gain |
$ 600 is divided amongst a , b and c so that a may get 2 / 3 as much as b and c together , b may get 6 / 9 as much as a and c together , then the share of a is | "a : ( b + c ) = 2 : 3 a ' s share = 600 * 2 / 5 = $ 240 answer is e" | a ) $ 100 , b ) $ 150 , c ) $ 125 , d ) $ 200 , e ) $ 240 | e | multiply(divide(600, add(divide(2, 3), const_1)), divide(2, 3)) | divide(n1,n2)|add(#0,const_1)|divide(n0,#1)|multiply(#2,#0)| | general |
what is the smallest no . which must be added to 532869 so as to obtain a sum which is divisible by 9 ? | "for 532869 , 5 + 3 + 2 + 8 + 6 + 9 = 33 3 must be added to 532869 to make it divisible by 9 . now , 5 + 3 + 2 + 8 + 7 + 2 = 27 = > 27 is a multiple of 9 and hence 532869 is also divisible by 9 d" | a ) 9 , b ) 2 , c ) 1 , d ) 3 , e ) 6 | d | divide(multiply(532869, 9), 532869) | multiply(n0,n1)|divide(#0,n0)| | general |
the speed of a boat in still water is 20 km / hr and the rate of current is 5 km / hr . the distance travelled downstream in 12 minutes is : | "explanation : speed downstream = ( 20 + 5 ) kmph = 25 kmph distance travelled = ( 25 * ( 12 / 60 ) ) km = 5 km . answer : b" | a ) 9 , b ) 5 , c ) 7 , d ) 6 , e ) 51 | b | multiply(divide(12, const_60), add(20, 5)) | add(n0,n1)|divide(n2,const_60)|multiply(#0,#1)| | physics |
a sum was put at simple interest at a certain rate for 6 years had it been put at 4 % higher rate , it would have fetched 144 more . find the sum . | "difference in s . i . = p × t / 100 ( r 1 − r 2 ) ⇒ 144 = p × 6 x 4 / 100 ( ∵ r 1 - r 2 = 2 ) ⇒ p = 144 × 100 / 6 × 4 = 600 answer b" | a ) 500 , b ) 600 , c ) 700 , d ) 800 , e ) 900 | b | divide(144, multiply(divide(4, const_100), 6)) | divide(n1,const_100)|multiply(n0,#0)|divide(n2,#1)| | gain |
what least no . must be subtracted from 427398 so that remaining no . is divisible by 10 ? | "explanation : on dividing 427398 by 10 we get the remainder 8 , so 8 should be subtracted option a" | a ) 8 , b ) 9 , c ) 7 , d ) 6 , e ) 3 | a | subtract(427398, multiply(floor(divide(427398, 10)), 10)) | divide(n0,n1)|floor(#0)|multiply(n1,#1)|subtract(n0,#2)| | general |
the number of students enrolled at school xx this year is 7 percent more than it was last year . the number of students enrolled at school yy this year is 3 percent more than it was last year . if school xx grew by 40 more students than school yy did , and if there were 4000 total enrolled students last year at both sc... | given info : no . of students enrolled in school x is = 7 % more than previous year ' s strength = 7 x / 100 no . of students enrolled in school y is = 3 % more than previous year ' s strength = 3 y / 100 given that ` ` this year ' ' school x have 40 students more than school y . so the increase in strength can be writ... | a ) 480 , b ) 1600 , c ) 1920 , d ) 2080 , e ) 2400 | e | divide(subtract(multiply(7, 4000), 4000), add(7, 3)) | add(n0,n1)|multiply(n0,n3)|subtract(#1,n3)|divide(#2,#0) | general |
a 20 litre mixture of milk and water contains milk and water in the ratio 3 : 2 . 10 litres of the mixture is removed and replaced with pure milk and the operation is repeated once more . at the end of the two removal and replacement , what is the ratio t of milk and water in the resultant mixture ? | "he 20 litre mixture contains milk and water in the ratio of 3 : 2 . therefore , there will be 12 litres of milk in the mixture and 8 litres of water in the mixture . step 1 . when 10 litres of the mixture is removed , 6 litres of milk is removed and 4 litres of water is removed . therefore , there will be 6 litres of ... | a ) 17 : 3 , b ) 9 : 1 , c ) 3 : 17 , d ) 5 : 3 , e ) 11 : 2 | b | divide(add(multiply(divide(add(multiply(divide(3, add(3, 2)), subtract(20, 10)), 10), 20), subtract(20, 10)), 10), multiply(divide(multiply(divide(2, add(3, 2)), subtract(20, 10)), 20), subtract(20, 10))) | add(n1,n2)|subtract(n0,n3)|divide(n1,#0)|divide(n2,#0)|multiply(#2,#1)|multiply(#3,#1)|add(n3,#4)|divide(#5,n0)|divide(#6,n0)|multiply(#7,#1)|multiply(#8,#1)|add(n3,#10)|divide(#11,#9)| | general |
a collection of books went on sale , and 2 / 3 of them were sold for $ 3.50 each . if none of the 36 remaining books were sold , what was the total amount received for the books that were sold ? | "since 2 / 3 of the books in the collection were sold , 1 / 3 were not sold . the 36 unsold books represent 1 / 3 of the total number of books in the collection , and 2 / 3 of the total number of books equals 2 ( 36 ) or 72 . the total proceeds of the sale was 72 ( $ 3.50 ) or $ 252 . the best answer is therefore a . a... | a ) $ 252 , b ) $ 135 , c ) $ 90 , d ) $ 60 , e ) $ 54 | a | multiply(const_2.0, divide(multiply(36, divide(2, 3)), divide(const_1, 3))) | divide(n2,n1)|divide(const_1,n1)|multiply(n3,#0)|divide(#2,#1)|multiply(n0,#3)| | general |
the average runs scored by a batsman in 35 matches is 36 . in the next 14 matches the batsman scored an average of 15 runs . find his average in all the 30 matches ? | total score of the batsman in 35 matches = 1260 . total score of the batsman in the next 14 matches = 210 . total score of the batsman in the 30 matches = 1470 . average score of the batsman = 1470 / 49 = 30 . answer : e | a ) 31 , b ) 35 , c ) 29 , d ) 25 , e ) 30 | e | divide(add(multiply(36, 35), multiply(15, 14)), add(35, 14)) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0)| | general |
the average ( arithmetic mean ) of the 5 positive integers k , m , r , s , and t is 18 , and k < m < r < s < t . if t is 40 , what is the greatest possible value of the median of the 5 integers ? | "we need to find the median which is the third value when the numbers are in increasing order . since k < m < r < s < t , the median would be r . the average of the positive integers is 18 which means that in effect , all numbers are equal to 18 . if the largest number is 40 , it is 22 more than 18 . we need r to be ma... | a ) 16 , b ) 23 , c ) 19 , d ) 20 , e ) 22 | b | subtract(divide(subtract(multiply(18, 5), 40), const_2), const_2) | multiply(n0,n1)|subtract(#0,n2)|divide(#1,const_2)|subtract(#2,const_2)| | general |
how many positive integers less than 7,000 are there in which the sum of the digits equals 5 ? | "basically , the question asks how many 4 digit numbers ( including those in the form 0 xxx , 00 xx , and 000 x ) have digits which add up to 5 . think about the question this way : we know that there is a total of 5 to be spread among the 4 digits , we just have to determine the number of ways it can be spread . let x... | a ) 52 , b ) 54 , c ) 56 , d ) 58 , e ) 60 | c | divide(factorial(subtract(add(const_4, 5), const_1)), multiply(factorial(5), factorial(subtract(const_4, const_1)))) | add(n1,const_4)|factorial(n1)|subtract(const_4,const_1)|factorial(#2)|subtract(#0,const_1)|factorial(#4)|multiply(#1,#3)|divide(#5,#6)| | general |
4 / [ ( 1 / 0.03 ) + ( 1 / 0.37 ) ] = ? | "approximate . 1 / . 03 = 100 / 3 = 33 1 / . 37 = 100 / 37 = 3 denominator becomes 33 + 3 = 36 4 / 36 = . 1111 answer ( b )" | a ) 0.004 , b ) 0.11111 , c ) 2.775 , d ) 3.6036 , e ) 36.036 | b | inverse(add(divide(4, 0.03), divide(4, 0.37))) | divide(n0,n2)|divide(n0,n4)|add(#0,#1)|inverse(#2)| | general |
by selling 16 pencils for a rupee a man loses 28 % . how many for a rupee should he sell in order to gain 28 % ? | "72 % - - - 16 128 % - - - ? 72 / 128 * 16 = 9 answer : b" | a ) 8 , b ) 9 , c ) 7 , d ) 6 , e ) 4 | b | multiply(divide(const_1, multiply(add(const_100, 28), divide(const_1, subtract(const_100, 28)))), 16) | add(n2,const_100)|subtract(const_100,n1)|divide(const_1,#1)|multiply(#0,#2)|divide(const_1,#3)|multiply(n0,#4)| | gain |
if the radius of a circle is diminished by 10 % , then the area is diminished by : | "explanation : let the original radius be r cm . new radius = 2 r area = π r 2 new area = π 2 r 2 = 4 π r 2 increase in area = ( 4 π r 2 − π r 2 ) = 3 π r 2 increase percent = 3 π r 2 / π r 2 ∗ 100 = 300 % option c" | a ) 200 % , b ) 210 % , c ) 300 % , d ) 310 % , e ) none of these | c | multiply(subtract(divide(const_100, const_100), power(subtract(divide(const_100, const_100), divide(10, const_100)), const_2)), const_100) | divide(const_100,const_100)|divide(n0,const_100)|subtract(#0,#1)|power(#2,const_2)|subtract(#0,#3)|multiply(#4,const_100)| | geometry |
a parking garage rents parking spaces for $ 10 per week or $ 24 per month . how much does a person save in a year by renting by the month rather than by the week ? | "10 $ per week ! an year has 52 weeks . annual charges per year = 52 * 10 = 520 $ 30 $ per month ! an year has 12 months . annual charges per year = 12 * 24 = 288 $ 520 - 288 = 232 ans d" | a ) $ 140 , b ) $ 160 , c ) $ 220 , d ) $ 232 , e ) $ 260 | d | subtract(multiply(add(multiply(10, add(const_3, const_2)), const_2), 10), multiply(24, const_12)) | add(const_2,const_3)|multiply(n1,const_12)|multiply(#0,n0)|add(#2,const_2)|multiply(n0,#3)|subtract(#4,#1)| | general |
a boat running downstream covers a distance of 12 km in 2 hours while for covering the same distance upstream , it takes 3 hours . what is the speed of the boat in still water ? | "speed downstream = 12 / 2 = 6 kmph speed upstream = 12 / 3 = 4 kmph speed of the boat in still water = 6 + 4 / 2 = 5 kmph answer : b" | a ) 8 kmph , b ) 5 kmph , c ) 4 kmph , d ) 6 kmph , e ) 7 kmph | b | divide(add(divide(12, 2), divide(12, 3)), 2) | divide(n0,n1)|divide(n0,n2)|add(#0,#1)|divide(#2,n1)| | physics |
what is the area of a square field whose diagonal of length 20 m ? | d 2 / 2 = ( 20 * 20 ) / 2 = 200 answer : c | ['a ) 288', 'b ) 289', 'c ) 200', 'd ) 112', 'e ) 178'] | c | divide(square_area(20), const_2) | square_area(n0)|divide(#0,const_2) | geometry |
a sum of rs . 12,500 amounts to rs . 15,500 in 4 years at the rate of simple interest . what is the rate of interest ? | s . i . = rs . ( 15500 - 12500 ) = rs . 3000 . rate = ( 100 x 3000 ) / ( 12500 x 4 ) % = 6 % answer : d | a ) 3 % , b ) 4 % , c ) 5 % , d ) 6 % , e ) 7 % | d | multiply(divide(divide(subtract(add(multiply(multiply(add(const_2, const_3), const_3), multiply(multiply(add(const_2, const_3), const_2), const_100)), multiply(add(const_2, const_3), const_100)), add(multiply(multiply(const_3, const_4), multiply(multiply(add(const_2, const_3), const_2), const_100)), multiply(add(const_... | add(const_2,const_3)|multiply(const_3,const_4)|multiply(#0,const_3)|multiply(#0,const_2)|multiply(#0,const_100)|multiply(#3,const_100)|multiply(#2,#5)|multiply(#1,#5)|add(#6,#4)|add(#7,#4)|subtract(#8,#9)|divide(#10,#9)|divide(#11,n2)|multiply(#12,const_100) | gain |
ravi purchased a refrigerator and a mobile phone for rs . 15000 and rs . 8000 respectively . he sold the refrigerator at a loss of 4 percent and the mobile phone at a profit of 10 percent . overall he make a . | "let the sp of the refrigerator and the mobile phone be rs . r and rs . m respectively . r = 15000 ( 1 - 4 / 100 ) = 15000 - 600 m = 8000 ( 1 + 10 / 100 ) = 8000 + 800 total sp - total cp = r + m - ( 15000 + 8000 ) = - 600 + 800 = rs . 200 as this is positive , an overall profit of rs . 200 was made . answer : d" | a ) loss of rs . 200 , b ) loss of rs . 100 , c ) profit of rs . 100 , d ) profit of rs . 200 , e ) none of these | d | subtract(add(multiply(15000, subtract(const_1, divide(4, const_100))), multiply(8000, add(const_1, divide(10, const_100)))), add(15000, 8000)) | add(n0,n1)|divide(n2,const_100)|divide(n3,const_100)|add(#2,const_1)|subtract(const_1,#1)|multiply(n0,#4)|multiply(n1,#3)|add(#5,#6)|subtract(#7,#0)| | gain |
tim came second in math . when his mother asked him how much he had scored , he answered that he got the sum of the first 8 even numbers . his mother immediately worked out the answer . how much had he scored in math ? | d 72 sum = ( n x n ) + n hence , 8 x 8 = 64 + 8 = 72 | a ) 80 , b ) 90 , c ) 30 , d ) 72 , e ) 60 | d | multiply(add(8, const_1), 8) | add(n0,const_1)|multiply(n0,#0) | physics |
a and b can do a work in 5 days , b and c in 15 days and c and a in 25 days . in how many days will the work be completed , if all three of them work together ? | "one day work of a and b = 1 / 5 one day work of b and c = 1 / 15 one day work of c and a = 1 / 25 2 ( a + b + c ) = 1 / 5 + 1 / 15 + 1 / 25 2 ( a + b + c ) = 23 / 75 ( a + b + c ) = 23 / 150 number of days required = 150 / 23 days . answer : e" | a ) 26 / 18 , b ) 42 / 15 , c ) 48 / 63 , d ) 12 / 36 , e ) 150 / 23 | e | divide(const_1, divide(add(add(inverse(5), inverse(15)), inverse(25)), 5)) | inverse(n0)|inverse(n1)|inverse(n2)|add(#0,#1)|add(#3,#2)|divide(#4,n0)|divide(const_1,#5)| | physics |
22 percent of 300 | 1 % of 300 = 3 22 % of 300 = 22 * 3 = 66 answer : e | a ) 50 , b ) 55 , c ) 62 , d ) 77 , e ) 66 | e | divide(multiply(22, 300), const_100) | multiply(n0,n1)|divide(#0,const_100)| | gain |
working individually , allison can sew dresses for 9 hours and al can sew dresses in 12 hours . if allison and al work together but independently at the task for 3 hours , at which point al leaves , how many remaining hours will it take allison to complete the task alone ? | in first 3 hrs al will finish 3 / 12 = 1 / 4 of work and allison will finish 3 / 9 = 1 / 3 work so total 1 / 4 + 1 / 3 = 7 / 12 work is finished and 1 - 7 / 12 = 5 / 12 work remaining . now allison will take ( 5 / 12 ) * 9 = 45 / 12 hrs to finish it . so answer is a . | a ) 45 / 12 , b ) 23 / 12 , c ) 40 / 33 , d ) 34 / 12 , e ) 43 / 5 | a | divide(subtract(const_1, multiply(add(divide(const_1, 9), divide(const_1, 12)), 3)), divide(const_1, 9)) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|multiply(n2,#2)|subtract(const_1,#3)|divide(#4,#0) | physics |
3 friends are planning to attend a concert . if another friend also goes with them , they will get a group discount on their tickets . if the new total price of the tickets with the group discount is the same as the original total price of the tickets without the group discount , how much is the discount ? | let x be the original price of one ticket . the total original cost is 3 x . the new cost is 4 y , where y is the discounted price of one ticket . 3 x = 4 y y = 3 x / 4 = 0.75 x which is a discount of 25 % . the answer is d . | a ) 5 % , b ) 10 % , c ) 20 % , d ) 25 % , e ) 33 % | d | multiply(subtract(const_1, divide(const_3, add(3, const_1))), const_100) | add(n0,const_1)|divide(const_3,#0)|subtract(const_1,#1)|multiply(#2,const_100) | general |
the true discount on a bill of rs . 540 is rs . 90 . the banker ' s discount is : | "p . w . = rs . ( 540 - 90 ) = rs . 450 . s . i . on rs . 450 = rs . 90 . s . i . on rs . 540 = rs . ( 90 / 450 ) x 540 = rs . 108 . b . d . = rs . 108 . answer : b" | a ) rs . 60 , b ) rs . 108 , c ) rs . 110 , d ) rs . 112 , e ) rs . 122 | b | multiply(divide(90, subtract(540, 90)), 540) | subtract(n0,n1)|divide(n1,#0)|multiply(n0,#1)| | gain |
when positive integer x is divided by positive integer y , the remainder is 3 . if x / y = 96.15 , what is the value of y ? | by the definition of a remainder , the remainder here is equal to 3 / y . the remainder in decimal form is given as . 15 therefore , 3 / y = . 15 solve for y and get 20 . e | a ) 96 , b ) 75 , c ) 48 , d ) 25 , e ) 20 | e | divide(3, subtract(96.15, floor(96.15))) | floor(n1)|subtract(n1,#0)|divide(n0,#1) | general |
fetched a total simple interest of rs . 4216.25 at the rate of 9 p . c . p . a . in 5 years . what is the sum ? | explanation : principal = rs . = rs . = rs . 8925 answer : d | a ) rs . 8928 , b ) rs . 89228 , c ) rs . 8911 , d ) rs . 8925 , e ) rs . 8922 | d | divide(multiply(4216.25, const_100), multiply(9, 5)) | multiply(n0,const_100)|multiply(n1,n2)|divide(#0,#1) | gain |
the average of 10 numbers is calculated as 15 . it is discovered later on that while calculating the average , one number namely 36 was wrongly read as 26 . the correct average is ? | "explanation : 10 * 15 + 36 – 26 = 160 / 10 = 16 a )" | a ) 16 , b ) 18 , c ) 19 , d ) 22 , e ) 24 | a | add(15, divide(subtract(36, 26), 10)) | subtract(n2,n3)|divide(#0,n0)|add(n1,#1)| | general |
if x is 20 percent greater than 98 , then x = | "x = 98 * 1.2 = 117.6 so the answer is e ." | a ) 68 , b ) 70.4 , c ) 86 , d ) 105.6 , e ) 117.6 | e | add(98, multiply(divide(20, const_100), 98)) | divide(n0,const_100)|multiply(n1,#0)|add(n1,#1)| | general |
30 pens and 75 pencils were purchased for 450 . if the average price of a pencil was 2.00 , find the average price of a pen . | since average price of a pencil = 2 ∴ price of 75 pencils = 150 ∴ price of 30 pens = ( 450 – 150 ) = 300 ∴ average price of a pen = 360 ⁄ 60 = 10 answer a | a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | a | divide(subtract(450, multiply(75, const_2)), 30) | multiply(n1,const_2)|subtract(n2,#0)|divide(#1,n0) | general |
a person is traveling at 100 km / hr and reached his destiny in 5 then find the distance ? | "t = 5 hr s = 100 km / hr d = t * s = 100 * 5 = 500 km answer is c" | a ) 250 km , b ) 300 km , c ) 500 km , d ) 450 km , e ) 350 km | c | multiply(100, 5) | multiply(n0,n1)| | physics |
if jake loses 29 pounds , he will weigh twice as much as his sister . together they now weigh 99 pounds . what is jake ' s present weight , in pounds ? | "j = jake ’ s current weight , in pounds s = sister ’ s current weight , in pounds we are told that “ if jake loses 8 pounds , he will weigh twice as much as his sister . we put this into an equation : j – 29 = 2 s j = 2 s + 29 ( equation 1 ) next , we are told that “ together they now weigh 89 pounds . ” we can also p... | a ) 69 , b ) 135 , c ) 139 , d ) 147 , e ) 188 | a | add(multiply(divide(subtract(99, 29), const_3), const_2), 29) | subtract(n1,n0)|divide(#0,const_3)|multiply(#1,const_2)|add(n0,#2)| | general |
a man can row downstream at the rate of 18 kmph and upstream at 6 kmph . find the man ’ s rate in still water and rate of current ? | "rate of still water = 1 / 2 ( down stream + upstream ) = 1 / 2 ( 18 + 6 ) = 12 kmph rate of current = 1 / 2 ( down stream - upstream ) = 1 / 2 ( 18 - 6 ) = 1 / 2 ( 12 ) = 6 kmph answer is b ." | a ) 8.5 , b ) 6.0 , c ) 9.5 , d ) 9.0 , e ) 8.25 | b | divide(subtract(18, 6), const_2) | subtract(n0,n1)|divide(#0,const_2)| | gain |
the average ( arithmetic mean ) of the even integers from 0 to 40 inclusive is how much greater than the average ( arithmetic mean ) of the even integers from 0 to 20 inclusive ? | "the sum of even numbers from 0 to n is 2 + 4 + . . . + n = 2 ( 1 + 2 + . . . + n / 2 ) = 2 ( n / 2 ) ( n / 2 + 1 ) / 2 = ( n / 2 ) ( n / 2 + 1 ) the average is ( n / 2 ) ( n / 2 + 1 ) / ( n / 2 + 1 ) = n / 2 the average of the even numbers from 0 to 40 is 40 / 2 = 20 the average of the even numbers from 0 to 20 is 20 ... | a ) 5 , b ) 8 , c ) 10 , d ) 15 , e ) 20 | c | subtract(divide(add(20, 0), const_2), divide(add(40, 0), const_2)) | add(n2,n3)|add(n0,n1)|divide(#0,const_2)|divide(#1,const_2)|subtract(#2,#3)| | general |
what is the smallest number h which when increased by 5 is completely divisible by 8 , 11 and 24 ? | this question includes a number of great number property shortcuts that you can take advantage of : 1 ) the question asks for the smallest number that , when increased by 5 is divisible by 8 , 11 and 24 . since the answers are numbers , we can test the answers . 2 ) any number that is divisible by 24 is also divisible ... | a ) 264 , b ) 259 , c ) 269 , d ) 270 , e ) 282 | b | subtract(lcm(lcm(8, 11), 24), 5) | lcm(n1,n2)|lcm(n3,#0)|subtract(#1,n0) | general |
the cost price of a radio is rs . 1500 and it was sold for rs . 1290 , find the loss % ? | "1500 - - - - 210 100 - - - - ? = > 14 % answer : a" | a ) 14 % , b ) 16 % , c ) 17 % , d ) 78 % , e ) 28 % | a | multiply(divide(subtract(1500, 1290), 1500), const_100) | subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)| | gain |
in an election a candidate who gets 60 % of the votes is elected by a majority of 900 votes . what is the total number of votes polled ? | "let the total number of votes polled be x then , votes polled by other candidate = ( 100 - 60 ) % of x = 40 % of x 60 % of x - 40 % of x = 900 20 x / 100 = 900 x = 900 * 100 / 20 = 4500 answer is a" | a ) a ) 4500 , b ) b ) 5200 , c ) c ) 6900 , d ) d ) 7520 , e ) e ) 6000 | a | divide(900, divide(subtract(60, subtract(const_100, 60)), const_100)) | subtract(const_100,n0)|subtract(n0,#0)|divide(#1,const_100)|divide(n1,#2)| | gain |
( 17 ) 3.75 x ( 17 ) ? = 178 | "solution let ( 17 ) 3.75 * ( 17 ) x = 178 . then , ( 17 ) 3.5 + x = ( 17 ) 8 . ∴ 3.75 + x = 8 ⇔ x = ( 8 - 3.75 ) ⇔ x = 4.25 answer c" | a ) 2.29 , b ) 2.75 , c ) 4.25 , d ) 4.5 , e ) none of these | c | multiply(17, 17) | multiply(n0,n2)| | general |
the speed of a boat in still water in 15 km / hr and the rate of current is 3 km / hr . the distance travelled downstream in 12 minutes is : | "speed downstream = ( 15 + 3 ) kmph = 18 kmph . distance travelled = 18 x 12 / 60 km = 3.6 km . answer : d" | a ) 1.2 km , b ) 2.4 km , c ) 3.2 km , d ) 3.6 km , e ) 4.6 km | d | multiply(add(15, 3), divide(12, const_60)) | add(n0,n1)|divide(n2,const_60)|multiply(#0,#1)| | physics |
if 1000 microns = 1 decimeter , and 1 , 000,000 angstroms = 1 decimeter , how many angstroms equal 1 micron ? | "1000 microns = 1 decimeter , and 1 , 000,000 angstroms = 1 decimeter 1000 microns = 1 , 000,000 angstroms 1 micron = 1 , 000,000 / 1,000 = 1,000 answer : b" | a ) 10 , b ) 1000 , c ) 100 , d ) 0.01 , e ) 0.0001 | b | multiply(divide(1, multiply(const_100, const_100)), multiply(const_100, const_100)) | multiply(const_100,const_100)|divide(n1,#0)|multiply(#1,#0)| | general |
10 men , working 7 hours a day can complete a work in 18 days . how many hours a day must 15 men work to complete the same work in 12 days ? | the number of hours required to complete the work is 10 * 7 * 18 = 1260 15 × 12 × ( x ) = 1260 x = 7 the answer is b . | a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | b | divide(multiply(multiply(10, 7), 18), multiply(15, 12)) | multiply(n0,n1)|multiply(n3,n4)|multiply(n2,#0)|divide(#2,#1) | physics |
calculate the value of n from the below equation : y ^ 4 â ˆ ’ 20 y + 1 = 22 | use elimination method to find the correct option . you find that of all the options 3 is the correct value for y answer : b | a ) 5 , b ) 3 , c ) 6 , d ) 2 , e ) 1 | b | subtract(4, divide(subtract(22, 1), add(20, 1))) | add(n1,n2)|subtract(n3,n2)|divide(#1,#0)|subtract(n0,#2) | general |
the average of first five prime numbers greater than 37 is ? | "41 + 43 + 47 + 53 + 59 = 243 / 5 = 48.6 answer : d" | a ) 32.2 , b ) 32.98 , c ) 42.3 , d ) 48.6 , e ) 32.4 | d | add(37, const_1) | add(n0,const_1)| | general |
a bag contains 6 black and 6 white balls . one ball is drawn at random . what is the probability that the ball drawn is white ? | "let number of balls = ( 6 + 6 ) = 12 . number of white balls = 6 . p ( drawing a white ball ) = 6 / 12 = 1 / 2 . option b ." | a ) 3 / 4 , b ) 1 / 2 , c ) 1 / 7 , d ) 1 / 8 , e ) 4 / 3 | b | divide(add(divide(divide(factorial(6), factorial(subtract(6, const_2))), factorial(const_2)), divide(divide(factorial(6), factorial(subtract(6, const_2))), factorial(const_2))), divide(divide(factorial(add(6, 6)), factorial(subtract(add(6, 6), const_2))), factorial(const_2))) | add(n0,n1)|factorial(n1)|factorial(const_2)|factorial(n0)|subtract(n1,const_2)|subtract(n0,const_2)|factorial(#4)|factorial(#5)|factorial(#0)|subtract(#0,const_2)|divide(#1,#6)|divide(#3,#7)|factorial(#9)|divide(#10,#2)|divide(#11,#2)|divide(#8,#12)|add(#13,#14)|divide(#15,#2)|divide(#16,#17)| | probability |
the list price of an article is rs . 67 . a customer pays rs . 56.16 for it . he was given two successive discounts , one of them being 10 % . the other discount is ? | 67 * ( 90 / 100 ) * ( ( 100 - x ) / 100 ) = 56.16 x = 6.86 % answer : d | a ) 3.86 % , b ) 4.86 % , c ) 5.86 % , d ) 6.86 % , e ) 7.86 % | d | multiply(divide(subtract(subtract(67, multiply(67, divide(10, const_100))), 56.16), subtract(67, multiply(67, divide(10, const_100)))), const_100) | divide(n2,const_100)|multiply(n0,#0)|subtract(n0,#1)|subtract(#2,n1)|divide(#3,#2)|multiply(#4,const_100) | gain |
karen places a bet with tom that she will beat tom in a car race by 4 miles even if karen starts 4 minutes late . assuming that karen drives at an average speed of 60 mph and tom drives at an average speed of 45 mph , how many w miles will tom drive before karen wins the bet ? | "let k and t be the speeds of karen and tom respectively . t be the time that karen will travel - - - - > t + 4 / 60 will be the total time tom will travel by the time the distance between karen and tom is 4 miles . thus , per the question , k ( t ) - t ( t + 4 / 60 ) = 4 - - - > t = 7 / 15 hours thus the distance trav... | a ) 15 , b ) 18 , c ) 21 , d ) 24 , e ) 27 | d | subtract(divide(add(multiply(divide(45, 60), 4), 4), subtract(divide(60, 60), divide(45, 60))), 4) | divide(n3,n2)|divide(n2,n2)|multiply(n0,#0)|subtract(#1,#0)|add(n0,#2)|divide(#4,#3)|subtract(#5,n0)| | physics |
rectangle a has sides a and b , and rectangle b has sides c and d . if a / c = b / d = 2 / 3 , what is the ratio of rectangle a ’ s area to rectangle b ’ s area ? | "the area of rectangle a is ab . c = 3 a / 2 and d = 3 b / 2 . the area of rectangle b is cd = 9 ab / 4 . the ratio of rectangle a ' s area to rectangle b ' s area is ab / ( 9 ab / 4 ) = 4 / 9 . the answer is a ." | a ) 4 / 9 , b ) 2 / 3 , c ) 3 / 2 , d ) 9 / 2 , e ) 9 / 4 | a | divide(power(2, const_2), power(3, const_2)) | power(n0,const_2)|power(n1,const_2)|divide(#0,#1)| | geometry |
the probability that event a occurs is 0.4 , and the probability that events a and b both occur is 0.25 . if the probability that either event a or event b occurs is 0.8 , what is the probability that event b will occur ? | "p ( a or b ) = p ( a ) + p ( b ) - p ( a n b ) 0.8 = 0.4 + p ( b ) - 0.25 p ( b ) = 0.65 ans : e" | a ) 0.05 , b ) 0.15 , c ) 0.45 , d ) 0.5 , e ) 0.65 | e | subtract(add(0.8, 0.25), 0.4) | add(n1,n2)|subtract(#0,n0)| | other |
a lemonade stand sold only small and large cups of lemonade on tuesday . 1 / 5 of the cups sold were small and the rest were large . if the large cups were sold for 7 / 6 as much as the small cups , what fraction of tuesday ' s total revenue was from the sale of large cups ? | let the total # of cups be 10 . # of small cups sold 1 / 5 * 10 = 2 ; # of large cups sold 10 - 2 = 8 ; let the price of small cup be $ 6 , then the price of larges cup would be 7 / 6 * 6 = $ 7 ; revenue from small cups : 2 * $ 6 = $ 12 ; revenue from large cups cups : 8 * $ 7 = $ 56 ; fraction of total revenue from la... | a ) 7 / 16 , b ) 7 / 15 , c ) 10 / 21 , d ) 14 / 17 , e ) 1 / 2 | d | divide(multiply(subtract(const_10, multiply(divide(1, 5), const_10)), multiply(divide(7, 6), multiply(divide(1, 5), const_10))), add(multiply(multiply(divide(1, 5), const_10), multiply(divide(1, 5), const_10)), multiply(subtract(const_10, multiply(divide(1, 5), const_10)), multiply(divide(7, 6), multiply(divide(1, 5), ... | divide(n2,n3)|divide(n0,n1)|multiply(#1,const_10)|multiply(#0,#2)|multiply(#2,#2)|subtract(const_10,#2)|multiply(#3,#5)|add(#4,#6)|divide(#6,#7) | general |
if money is invested at r percent interest , compounded annually , the amount of the investment will double in approximately 50 / r years . if luke ' s parents invested $ 14,500 in a long term bond that pays 12 percent interest compounded annually , what will be the approximate total amount of the investment 12 years l... | answer equals d in 48 years . i thought by 50 th year it would reach 120,500 . options should have been separated more widely for clarity . | a ) 62,000 , b ) 85,500 , c ) 95,500 , d ) 120,500 , e ) 100,000 | d | divide(multiply(multiply(add(const_2, const_3), const_1000), 12), const_2) | add(const_2,const_3)|multiply(#0,const_1000)|multiply(n2,#1)|divide(#2,const_2)| | general |
12 is 4 % of a , and 4 is 12 % of b . c equals b / a . what is the value of c ? | "4 a / 100 = 12 a = 300 12 b / 100 = 4 b = 100 / 3 c = b / a = 100 / ( 3 * 300 ) = 1 / 9 the answer is d ." | a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 6 , d ) 1 / 9 , e ) 1 / 12 | d | divide(multiply(divide(4, 12), const_100), multiply(divide(12, 4), const_100)) | divide(n1,n0)|divide(n0,n1)|multiply(#0,const_100)|multiply(#1,const_100)|divide(#2,#3)| | general |
a contractor undertook to make 15 km of roadway in 40 weeks . in 10 weeks , 3 km was complete by 180 men working 8 hours a day . the men then agreed to work 1 hour a day overtime , and some boys were engaged to assist them , the work was finished in the stipulated time ( 40 weeks ) . how many boys were employed , if th... | explanation : let the capacity of man = 3 units , and boy = 2 units per hour . now total work = 3 × 180 × 8 × 7 × 10 = 3 km . - - - - - - - ( 1 ) let k boys were recruited . now total work = ( 3 × 180 + 2 × k ) × 9 × 7 × 30 = 12 km . - - - - - - ( 2 ) by dividing 2 nd equation by 1 st , ⇒ ( 540 + 2 k ) × 9 × 7 × 303 × ... | a ) 48 , b ) 50 , c ) 22 , d ) 27 , e ) 221 | b | divide(subtract(15, add(3, multiply(180, multiply(multiply(divide(divide(divide(3, 10), 180), 8), add(8, 1)), subtract(40, 10))))), multiply(subtract(40, 10), multiply(divide(2, 3), multiply(divide(divide(divide(3, 10), 180), 8), add(8, 1))))) | add(n5,n6)|divide(n3,n2)|divide(n9,n3)|subtract(n1,n2)|divide(#1,n4)|divide(#4,n5)|multiply(#0,#5)|multiply(#6,#3)|multiply(#2,#6)|multiply(n4,#7)|multiply(#8,#3)|add(n3,#9)|subtract(n0,#11)|divide(#12,#10) | physics |
a man can row 7 kmph in still water . when the river is running at 1.2 kmph , it takes him 1 hour to row to a place and black . how far is the place ? | "m = 7 s = 1.2 ds = 7 + 1.2 = 8.2 us = 7 - 1.2 = 5.8 x / 8.2 + x / 5.8 = 1 x = 3.40 . answer : b" | a ) 3.3 , b ) 3.4 , c ) 3.5 , d ) 3.6 , e ) 3.7 | b | multiply(divide(multiply(add(7, 1.2), subtract(7, 1.2)), add(add(7, 1.2), subtract(7, 1.2))), const_2) | add(n0,n1)|subtract(n0,n1)|add(#0,#1)|multiply(#0,#1)|divide(#3,#2)|multiply(#4,const_2)| | physics |
find the least number of complete years in which a sum of money put out at 33.3 % compound interest will be more than double of itself ? | "3 years answer : a" | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | floor(add(divide(log(const_2), log(add(const_1, divide(33.3, const_100)))), const_1)) | divide(n0,const_100)|log(const_2)|add(#0,const_1)|log(#2)|divide(#1,#3)|add(#4,const_1)|floor(#5)| | general |
the sum of the squares of three numbers is 138 , while the sum of their products taken two at a time is 131 . their sum is | "explanation : let the numbers be a , b and c . then , a 2 + b 2 + c 2 = 138 and ( ab + bc + ca ) = 131 ( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 ( ab + bc + ca ) = 138 + 2 x 131 = 400 = > ( a + b + c ) = 400 − − − √ = 20 . option b" | a ) 15 , b ) 20 , c ) 25 , d ) 35 , e ) 45 | b | add(multiply(sqrt(divide(subtract(138, 131), const_2)), const_100), sqrt(subtract(138, divide(subtract(138, 131), const_2)))) | subtract(n0,n1)|divide(#0,const_2)|sqrt(#1)|subtract(n0,#1)|multiply(#2,const_100)|sqrt(#3)|add(#4,#5)| | general |
by selling a book for 290 , 20 % profit was earned . what is the cost price of the book ? | "sp = 120 % of cp ; : . cp = 290 × 100 / 120 = 241 option ' b '" | a ) a ) 215 , b ) b ) 241 , c ) c ) 230 , d ) d ) 235 , e ) e ) 240 | b | original_price_before_gain(20, 290) | original_price_before_gain(n1,n0)| | gain |
if a number n is chosen at random from the set of two - digit integers whose digits are both prime numbers , what is the probability w that n is divisible by 3 ? | prime digits are : 2 , 3 , 5 , 7 total number of 2 digit # s with both digits prime are : 4 * 4 = 16 out of these numbers divisible by 3 = 33 , 27 , 57 , 72 and 75 . i had to find the numbers manually using the 4 numbers above . = > prob = 5 / 16 . ans d . took me 3 : 20 mins . | a ) 1 / 3 , b ) ¼ , c ) 9 / 25 , d ) 5 / 16 , e ) 0 | d | divide(add(3, const_2), multiply(const_4, const_4)) | add(n0,const_2)|multiply(const_4,const_4)|divide(#0,#1) | general |
a women purchased 3 towels @ rs . 100 each , 5 towels @ rs . 150 each and two towels at a certain rate which is now slipped off from his memory . but she remembers that the average price of the towels was rs . 170 . find the unknown rate of two towels ? | "10 * 170 = 1700 3 * 100 + 5 * 150 = 1050 1700 – 1050 = 650 b" | a ) a ) 400 , b ) b ) 650 , c ) c ) 500 , d ) d ) 550 , e ) e ) 600 | b | subtract(subtract(multiply(add(add(3, 5), const_2), 170), multiply(5, 150)), multiply(3, 100)) | add(n0,n2)|multiply(n2,n3)|multiply(n0,n1)|add(#0,const_2)|multiply(n4,#3)|subtract(#4,#1)|subtract(#5,#2)| | general |
a grocer has a sale of rs . 5266 , rs . 5768 , rs . 5922 , rs . 5678 and rs . 6029 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . 5600 ? | "total sale for 5 months = rs . ( 5266 + 5768 + 5922 + 5678 + 6029 ) = rs . 28663 . required sale = rs . [ ( 5600 x 6 ) - 28663 ] = rs . ( 33600 - 28663 ) = rs . 4937 . answer : c" | a ) 4637 , b ) 4737 , c ) 4937 , d ) 5937 , e ) 5978 | c | subtract(multiply(add(5, const_1), 5600), add(add(add(add(5266, 5768), 5922), 5678), 6029)) | add(n5,const_1)|add(n0,n1)|add(n2,#1)|multiply(n6,#0)|add(n3,#2)|add(n4,#4)|subtract(#3,#5)| | general |
in a certain pond , 40 fish were caught , tagged , and returned to the pond . a few days later , 40 fish were caught again , of which 2 were found to have been tagged . if the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond , what ` s the approximate number of fish in the ... | "the percent of tagged fish in the second catch is 2 / 40 * 100 = 5 % . we are told that 5 % approximates the percent of tagged fish in the pond . since there are 40 tagged fish , then we have 0.05 x = 40 - - > x = 800 . answer : b ." | a ) 400 , b ) 800 , c ) 1250 , d ) 2500 , e ) 10 000 | b | divide(40, divide(2, 40)) | divide(n2,n1)|divide(n0,#0)| | gain |
the guests at a football banquet consumed a total of 327 pounds of food . if no individual guest consumed more than 2 pounds of food , what is the minimum number of guests that could have attended the banquet ? | to minimize one quantity maximize other . 163 * 2 ( max possible amount of food a guest could consume ) = 326 pounds , so there must be more than 163 guests , next integer is 164 . answer : e . | a ) 160 , b ) 161 , c ) 162 , d ) 163 , e ) 164 | e | add(floor(divide(327, 2)), const_1) | divide(n0,n1)|floor(#0)|add(#1,const_1) | general |
a report consists of 20 sheets each of 55 lines and each such line consist of 65 characters . this report is retyped into sheets each of 65 lines such that each line consists of 70 characters . the percentage reduction in number of sheets is closest to | explanation : total no of characters = 20 * 55 * 65 . from the problem it is clear that 20 * 55 * 65 = x * 65 * 70 . we get x = 15.6 ∼ 16 because these are sheets . therefore 4 less sheets . hencde , there is 20 % decrease in number of sheets . answer : a | a ) 20 , b ) 5 , c ) 30 , d ) 35 , e ) 67 | a | divide(multiply(floor(subtract(20, divide(divide(multiply(multiply(20, 55), 65), 70), 65))), const_100), 20) | multiply(n0,n1)|multiply(n2,#0)|divide(#1,n4)|divide(#2,n2)|subtract(n0,#3)|floor(#4)|multiply(#5,const_100)|divide(#6,n0) | gain |
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