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
values |
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find out the c . i on rs . 6000 at 4 % p . a . compound half - yearly for 1 1 / 2 years | a = 6000 ( 51 / 50 ) 3 = 6367.248 6000 - - - - - - - - - - - 367.25 answer : a | a ) 367.25 , b ) 306.07 , c ) 306.04 , d ) 306.03 , e ) 306.01 | a | subtract(multiply(6000, power(add(const_1, divide(4, const_100)), divide(const_3, const_2))), 6000) | divide(n1,const_100)|divide(const_3,const_2)|add(#0,const_1)|power(#2,#1)|multiply(n0,#3)|subtract(#4,n0) | general |
a card is drawn from a pack of 52 cards . the probability of getting a queen club or a king of heart ? | "n ( s ) = 52 let e = event of getting a queen of club or a king of heart . n ( e ) = 2 p ( e ) = n ( e ) / n ( s ) = 2 / 52 = 1 / 26 option : b" | a ) 3 / 26 , b ) 1 / 26 , c ) 4 / 26 , d ) 6 / 26 , e ) 10 / 26 | b | divide(subtract(52, multiply(const_4, const_4)), 52) | multiply(const_4,const_4)|subtract(n0,#0)|divide(#1,n0)| | probability |
a , b and c run around a circular track starting from the same point simultaneously and in the same direction at speeds of 4 kmph , 6 kmph and 8 kmph respectively . if the length of the track is 400 meters , when will a , b and c meet at the starting point for the first time after they started the race ? | 4 kmph , 6 kmph and 8 kmph is equal to 4000 / 60 mtrs / min , 6000 / 60 mtrs / min and 8000 / 60 mtrs / min or 200 / 3 , 100 , 400 / 3 mtrs / min respectively . seeing this we can infer that the answer should be at least divisible by 3 . ( minutes in options are integers . so if the answer is not divisible by 3 , we wi... | a ) 1 minute , b ) 5 minutes , c ) 12 minutes , d ) 20 minutes , e ) 24 minutes | c | multiply(divide(const_60, divide(6, divide(400, const_1000))), divide(const_60, divide(8, divide(400, const_1000)))) | divide(n3,const_1000)|divide(n1,#0)|divide(n2,#0)|divide(const_60,#1)|divide(const_60,#2)|multiply(#3,#4) | physics |
if 15 workers can build a wall in 48 hours , how many workers will be required to do the same work in 30 hours ? | let the number of workers employed to build the wall in 30 hours be y 48 * 15 = 30 * y ( 48 * 15 ) / 30 = y y = 24 answer : b | a ) 22 , b ) 24 , c ) 26 , d ) 25 , e ) 23 | b | divide(multiply(15, 48), 30) | multiply(n0,n1)|divide(#0,n2) | physics |
the least number of 4 digits which is divisible by 15 , 25 , 40 and 75 is : | greatest number of 4 digits is 9999 . l . c . m of 15 , 25 , 40 and 75 is 600 . on dividing 9999 by 600 , the remainder is 399 . required number = 9999 - 399 = 9600 answer : c | a ) 9000 , b ) 9400 , c ) 9600 , d ) 9800 , e ) 9700 | c | multiply(multiply(multiply(75, multiply(4, const_2)), multiply(4, const_2)), const_2) | multiply(n0,const_2)|multiply(n4,#0)|multiply(#1,#0)|multiply(#2,const_2) | general |
find the probability that a leap year chosen at random will have 53 sundays | "a leap year has 366 day which is 52 full weeks + 2 odd days . now these two odd days may be ( sun + mon ) , ( mon + tue ) , . . . . ( sat + sun ) . now there are total 7 ways . of which sunday appeared two times . so answer 2 / 7 answer : a" | a ) 2 / 7 , b ) 2 / 6 , c ) 2 / 5 , d ) 2 / 3 , e ) 2 / 4 | a | divide(const_2, add(const_3, const_4)) | add(const_3,const_4)|divide(const_2,#0)| | probability |
three pipes a , b and c can fill a tank from empty to full in 20 minutes , 20 minutes and 30 minutes respectively . when the tank is empty , all the three pipes are opened . a , b and c discharge chemical solutions p , q and r respectively . what is the proportion of solution r in the liquid in the tank after 3 minutes... | part filled by ( a + b + c ) in 3 minutes = 3 ( 1 / 20 + 1 / 20 + 1 / 30 ) = 6 / 15 part filled by c in 3 minutes = 3 / 30 required ratio = 3 / 30 * 15 / 6 = 1 / 4 answer : d | a ) 1 / 2 , b ) 1 / 5 , c ) 1 / 3 , d ) 1 / 4 , e ) 1 / 6 | d | multiply(divide(3, 30), divide(const_1, multiply(3, add(divide(const_1, 30), add(divide(const_1, 20), divide(const_1, 20)))))) | divide(n3,n2)|divide(const_1,n0)|divide(const_1,n2)|add(#1,#1)|add(#3,#2)|multiply(n3,#4)|divide(const_1,#5)|multiply(#0,#6) | physics |
a man ' s speed with the current is 25 km / hr and the speed of the current is 2.5 km / hr . the man ' s speed against the current is ? | "man ' s speed with the current = 25 km / hr = > speed of the man + speed of the current = 25 km / hr speed of the current is 2.5 km / hr hence , speed of the man = 25 - 2.5 = 22.5 km / hr man ' s speed against the current = speed of the man - speed of the current = 22.5 - 2.5 = 20 km / hr answer is e ." | a ) 10 , b ) 80 , c ) 50 , d ) 30 , e ) 20 | e | subtract(subtract(25, 2.5), 2.5) | subtract(n0,n1)|subtract(#0,n1)| | gain |
in a fraction , if numerator is added by 2 and denominator by 1 it results 1 , and if numerator added again by 4 and denominator by 2 it results 1 / 2 . find the fraction ? | x + 2 / y + 1 = 1 i . e , x - y + 1 = 0 ; - - - - - - - - ( 1 ) x + 4 / y + 2 = 1 / 2 i . e . , 2 x - y + 6 = 0 ; - - - - - - - - - ( 2 ) solving 1 and 2 . . . we get x = - 5 and y = - 4 answer is 5 / 4 . . . answer : b | a ) 3 / 4 , b ) 5 / 4 , c ) 7 / 4 , d ) 9 / 4 , e ) 11 / 4 | b | divide(add(4, 1), 4) | add(n1,n3)|divide(#0,n3) | general |
a rectangular room has the rectangular shaped rug shown as above figure such that the rug β s area is 180 square feet and its length is 8 feet longer than its width . if the uniform width between the rug and room is 8 feet , what is the area of the region uncovered by the rug ( shaded region ) , in square feet ? | rug ' s area = 180 which is ( x ) x ( 8 + x ) = 180 so x = 10 rug maintains a uniform distance of 8 feet so room has dimension 10 + 16 and 18 + 16 i . e . 26 and 34 area of room 26 x 34 = 884 area covered is 180 so uncovered area is 884 - 180 = 704 ( answer c ) | ['a ) 32', 'b ) 36', 'c ) 704', 'd ) 46', 'e ) 104'] | c | subtract(multiply(add(subtract(divide(const_180, const_10), const_2), const_10), add(divide(const_180, const_10), subtract(divide(const_180, const_10), const_2))), 180) | divide(const_180,const_10)|subtract(#0,const_2)|add(#1,const_10)|add(#0,#1)|multiply(#2,#3)|subtract(#4,n0) | geometry |
7 people average age is 30 . youngest person age is 4 . find average of the people when youngest was born . | "average age of people = 30 so have total age = 210 before 7 years we have to deduct each person age by seven years 210 - 32 = 182 so average age would be 182 / 7 = 26 answer : d" | a ) 23 , b ) 24 , c ) 25 , d ) 26 , e ) 27 | d | divide(subtract(multiply(30, 7), multiply(4, 7)), 7) | multiply(n0,n1)|multiply(n0,n2)|subtract(#0,#1)|divide(#2,n0)| | general |
in a group of cows and chickens , the number of legs was 10 more than twice the number of heads . the number of cows was : | "let the number of cows be x and their legs be 4 x . let the number of chicken be y and their legs be 2 x . total number of legs = 4 x + 2 y . total number of heads = x + y . the number of legs was 10 more than twice the number of heads . therefore , 2 Γ ( x + y ) + 10 = 4 x + 2 y . or , 2 x + 2 y + 10 = 4 x + 2 y . or... | a ) 5 , b ) 7 , c ) 10 , d ) 12 , e ) 14 | a | divide(10, subtract(const_4, const_2)) | subtract(const_4,const_2)|divide(n0,#0)| | general |
the cash difference between the selling prices of an article at a profit of 4 % and 6 % is rs . 3 . the ratio of the two selling prices is : | "let c . p . of the article be rs . x . then , required ratio = 104 % of x / 106 % of x = 104 / 106 = 52 / 53 = 52 : 53 answer : b" | a ) 51 : 52 , b ) 52 : 53 , c ) 51 : 53 , d ) 52 : 55 , e ) 52 : 56 | b | divide(add(const_100, 4), add(const_100, 6)) | add(n0,const_100)|add(n1,const_100)|divide(#0,#1)| | gain |
the area of a circle is increased by 300 % . by what percent has the radius of the circle increased ? | "the area of the circle is increased by 300 % , thus the area is increased 4 times . the area of a circle it proportional to the square of the radius ( area = Ο r ^ 2 ) , therefore the radius must increase 2 times ( diameter increase 2 times = area increase 4 times ) , which is increase by 100 % . answer : c" | a ) 400 % , b ) 200 % , c ) 100 % , d ) 600 % , e ) 800 % | c | multiply(const_100, divide(const_2, const_2)) | divide(const_2,const_2)|multiply(#0,const_100)| | gain |
the total number of digits used in numbering the pages of a book having 356 pages is | "total number of digits = ( no . of digits in 1 - digit page nos . + no . of digits in 2 - digit page nos . + no . of digits in 3 - digit page nos . ) = ( 1 x 9 + 2 x 90 + 3 x 257 ) = ( 9 + 180 + 771 ) = 960 . answer : d" | a ) 732 , b ) 990 , c ) 1098 , d ) 960 , e ) 1405 | d | subtract(subtract(multiply(356, const_3), subtract(const_100, const_1)), subtract(const_10, const_1)) | multiply(n0,const_3)|subtract(const_100,const_1)|subtract(const_10,const_1)|subtract(#0,#1)|subtract(#3,#2)| | general |
working alone at their respective constant rates , a can complete a task in β a β days and b in β b β days . they take turns in doing the task with each working 2 days at a time . if a starts they finish the task in exact 8 days . if b starts , they take a day more . how long does it take to complete the task if they b... | "work done by ab in a day = xy respectively . when a starts : no . of days when a works = 4 no . of days when b works = 4 β 4 x + 4 y = 1 when b starts : no . of days when a works = 5 no . of days when b works = 3 β 5 x + 3 y = 1 solving the above two equations for xy x = 1 / 8 y = 1 / 8 β total work done by ab in a da... | a ) 6 , b ) 7 , c ) 8 , d ) 4 , e ) 2 | d | inverse(add(inverse(subtract(8, const_1)), inverse(add(8, 2)))) | add(n0,n1)|subtract(n1,const_1)|inverse(#1)|inverse(#0)|add(#2,#3)|inverse(#4)| | physics |
a thief goes away with a santro car at a speed of 40 kmph . the theft has been discovered after half an hour and the owner sets off in a bike at 45 kmph when will the owner over take the thief from the start ? | "explanation : | - - - - - - - - - - - 20 - - - - - - - - - - - - - - - - - - - - | 45 40 d = 20 rs = 45 β 40 = 5 t = 20 / 5 = 4 hours answer : option d" | a ) a ) 2 , b ) b ) 5 , c ) c ) 7 , d ) d ) 4 , e ) e ) 8 | d | subtract(divide(multiply(divide(const_1, const_2), 40), subtract(45, 40)), divide(const_1, const_2)) | divide(const_1,const_2)|subtract(n1,n0)|multiply(n0,#0)|divide(#2,#1)|subtract(#3,#0)| | physics |
the area of a right triangle pcm is 18 . length of smallest side cm is 3 units . find the length of the other two sides . | cm = 3 area = pm * cm / 2 , thus pm = 12 by pythagorean theorem , we can now find the value of pc . d is the correct option . | ['a ) 12 , 9', 'b ) 18 , 3', 'c ) 15.16 , 6', 'd ) 12 , 12.37', 'e ) 12 , 15.16'] | d | power(add(power(divide(multiply(18, const_2), 3), const_2), power(3, const_2)), inverse(const_2)) | inverse(const_2)|multiply(n0,const_2)|power(n1,const_2)|divide(#1,n1)|power(#3,const_2)|add(#4,#2)|power(#5,#0) | geometry |
each machine of type a has 2 steel parts and 2 chrome parts . each machine of type b has 6 steel parts and 5 chrome parts . if a certain group of type a and type b machines has a total of 20 steel parts and 22 chrome parts , how many machines are in the group | "look at the below representation of the problem : steel chrome total a 2 2 20 > > no . of type a machines = 20 / 4 = 5 b 6 5 22 > > no . of type b machines = 22 / 11 = 2 so the answer is 7 i . e e . hope its clear ." | a ) 2 , b ) 3 , c ) 4 , d ) 6 , e ) 7 | e | add(divide(22, add(5, 2)), divide(20, add(2, 2))) | add(n1,n3)|add(n0,n1)|divide(n5,#0)|divide(n4,#1)|add(#2,#3)| | general |
there is a square with sides of 9 . what is the area of the biggest circle that can be cut out of this square ? | "area of a circle = area = Γ― β¬ r ^ 2 square is 9 wide , so circle ' s diameter would be 9 , and radius would be 4.5 area = Γ― β¬ 4.5 ^ 2 which is approximately 63.62 answer is b" | a ) 72.83 , b ) 63.62 , c ) 81 , d ) 254.47 , e ) 100 | b | circumface(divide(9, const_2)) | divide(n0,const_2)|circumface(#0)| | geometry |
a boat can travel with aspeed of 30 km / hr in still water . if the speed of the stream is 5 km / hr , find the time taken by the boat to go 70 km downstream . | "speed downstream = ( 30 + 5 ) km / hr = 35 km / hr . time taken to travel 68 km downstream = 70 / 35 hrs = 2 hrs . b" | a ) 1 hr , b ) 2 hrs , c ) 3 hrs , d ) 4 hrs , e ) 5 hrs | b | divide(70, add(30, 5)) | add(n0,n1)|divide(n2,#0)| | physics |
the average age of a group of 10 persons was decreased by 3 years when one person , whose age was 46 years , was replaced by a new person . find the age of the new person ? | "initial average age of the 10 persons be p . age of the new person q . sum of the ages of the initial 10 persons = 10 p new average = ( p - 3 ) 10 ( p - 3 ) = 10 p - 46 + q = > q = 16 answer : c" | a ) 18 , b ) 56 , c ) 16 , d ) 17 , e ) 14 | c | subtract(46, multiply(10, 3)) | multiply(n0,n1)|subtract(n2,#0)| | general |
a number whose fifth part increased by 4 is equal to its fourth part diminished by 10 , is : | let the number be x . then , ( ( 1 / 5 ) x + 4 ) = ( ( 1 / 4 ) x β 10 ) < = > x / 20 < = > x = 14 * 20 = 280 . answer : d | a ) 240 , b ) 260 , c ) 270 , d ) 280 , e ) 290 | d | divide(add(10, 4), subtract(divide(const_1, const_4), divide(const_1, add(const_1, const_4)))) | add(n0,n1)|add(const_1,const_4)|divide(const_1,const_4)|divide(const_1,#1)|subtract(#2,#3)|divide(#0,#4) | general |
if 0.60 : x : : 6 : 2 , then x is equal to | "sol . ( x Γ 6 ) = ( 0.60 Γ 2 ) β x = 0.60 / 6 = 0.2 . answer a" | a ) 0.2 , b ) 0.5 , c ) 2.5 , d ) 1.8 , e ) none | a | divide(multiply(0.60, 2), 6) | multiply(n0,n2)|divide(#0,n1)| | general |
a and b together can do a piece of work in 6 days and a alone can do it in 12 days . in how many days can b alone can do it ? | "explanation : a and b can do work 1 / 6 in 1 day a alone can do 1 / 12 work in 1 day b alone can do ( 1 / 6 - 1 / 12 ) = 1 / 12 work in 1 day = > complete work can be done in 12 days by b answer : option c" | a ) 11 days , b ) 15 days , c ) 12 days , d ) 21 days , e ) 22 days | c | inverse(subtract(inverse(6), inverse(12))) | inverse(n0)|inverse(n1)|subtract(#0,#1)|inverse(#2)| | physics |
in a certain town , the ratio of ny yankees fans to ny mets fans is 2 : 1 , and the ratio of ny mets fans to boston red sox fans is 3 : 7 . if there are 352 baseball fans in the town , each of whom is a fan of exactly one of those three teams , how many ny mets fans are there in this town ? | "the ratio of yankees : mets : red sox = 6 : 3 : 7 the mets fans are 3 / 16 of the population . ( 3 / 16 ) * 352 = 66 the answer is b ." | a ) 58 , b ) 66 , c ) 74 , d ) 82 , e ) 90 | b | multiply(divide(352, add(add(multiply(2, 1), 3), 7)), 3) | multiply(n0,n1)|add(n2,#0)|add(n3,#1)|divide(n4,#2)|multiply(n2,#3)| | other |
in a certain neighborhood , 60 percent of the student are going to school ( a ) , and the rest are going to school ( b ) . a new school ( c ) is opened , if 30 percent of student of school ( a ) and 40 percent of student of school ( b ) are expected to go to the new school ( c ) , what percent of student are expected t... | say there are total of 100 students in the neighborhood , 60 students are going to school ( a ) and 40 student are going to school ( b ) 60 * 0.3 = 18 student of school ( a ) are expected to go to the new school ( c ) 40 * 0.4 = 16 student of school ( b ) are expected to go to the new school ( c ) thus total of 18 + 16... | a ) 30 % , b ) 34 % , c ) 38 % , d ) 42 % , e ) 46 % | b | add(multiply(divide(30, const_100), 60), multiply(divide(40, const_100), subtract(const_100, 60))) | divide(n1,const_100)|divide(n2,const_100)|subtract(const_100,n0)|multiply(n0,#0)|multiply(#1,#2)|add(#3,#4) | gain |
two trains of equal length , running with the speeds of 60 and 100 kmph , take 50 seconds to cross each other while they are running in the same direction . what time will they take to cross each other if they are running in opposite directions ? | "rs = 60 - 40 = 20 * 5 / 18 = 100 / 18 t = 50 d = 50 * 100 / 18 = 2500 / 9 rs = 60 + 100 = 160 * 5 / 18 t = 2500 / 9 * 18 / 800 = 6.25 sec . answer : b" | a ) 10 , b ) 6.25 , c ) 7.35 , d ) 2.62 , e ) 2.28 | b | multiply(multiply(multiply(const_0_2778, subtract(60, 100)), 50), inverse(multiply(const_0_2778, add(60, 100)))) | add(n0,n1)|subtract(n0,n1)|multiply(#0,const_0_2778)|multiply(#1,const_0_2778)|inverse(#2)|multiply(n2,#3)|multiply(#4,#5)| | physics |
find the number of elements in the power set of { 1,2 } | "power set is the set of subsets of { 1,2 } that is { { 1,2 } , { 1 } , { 2 } , f } answer : a" | a ) 4 , b ) 0 , c ) 2 , d ) 3 , e ) 5 | a | add(const_2, const_2) | add(const_2,const_2)| | general |
64 + 5 * 12 / ( 180 / 3 ) = ? | 64 + 5 * 12 / ( 180 / 3 ) = 64 + 5 * 12 / ( 60 ) = 64 + ( 5 * 12 ) / 60 = 64 + 1 = 65 . answer : d | a ) 22 , b ) 77 , c ) 29 , d ) 65 , e ) 21 | d | add(64, divide(multiply(5, 12), divide(180, 3))) | divide(n3,n4)|multiply(n1,n2)|divide(#1,#0)|add(n0,#2) | general |
alok ordered 16 chapatis , 5 plates of rice , 7 plates of mixed vegetable and 6 ice - cream cups . the cost of each chapati is rs . 6 , that of each plate of rice is rs . 45 and that of mixed vegetable is rs . 70 . the amount that alok paid the cashier was rs . 1021 . find the cost of each ice - cream cup ? | "let the cost of each ice - cream cup be rs . x 16 ( 6 ) + 5 ( 45 ) + 7 ( 70 ) + 6 ( x ) = 1021 96 + 225 + 490 + 6 x = 1021 6 x = 210 = > x = 35 . answer : a" | a ) 35 , b ) 66 , c ) 77 , d ) 99 , e ) 91 | a | divide(subtract(subtract(subtract(1021, multiply(16, 6)), multiply(5, 45)), multiply(7, 70)), 6) | multiply(n0,n3)|multiply(n1,n5)|multiply(n2,n6)|subtract(n7,#0)|subtract(#3,#1)|subtract(#4,#2)|divide(#5,n3)| | general |
the length of a train and that of a platform are equal . if with a speed of 54 k / hr , the train crosses the platform in one minute , then the length of the train ( in meters ) is ? | "speed = [ 54 * 5 / 18 ] m / sec = 15 m / sec ; time = 1 min . = 60 sec . let the length of the train and that of the platform be x meters . then , 2 x / 60 = 15 Γ¨ x = 15 * 60 / 2 = 450 answer : c" | a ) 227 , b ) 299 , c ) 450 , d ) 750 , e ) 211 | c | divide(divide(multiply(54, const_1000), divide(const_60, const_1)), const_2) | divide(const_60,const_1)|multiply(n0,const_1000)|divide(#1,#0)|divide(#2,const_2)| | physics |
jack went on a diet 6 months ago when he weighed 222 pounds . if he now weighs 198 pounds and continues to lose at the same average monthly rate , in approximately how many months will he weigh 180 pounds ? | joe lost his weigh 24 pounds during 6 months . that means that he lost 4 pounds per month . he should lose more 18 pounds to be 180 pounds . so it would take 18 / 4 = 4.5 months . correct answer is d | a ) 3 , b ) 35 , c ) 4 , d ) 45 , e ) 5 | d | multiply(multiply(subtract(198, 180), divide(6, subtract(222, 198))), const_10) | subtract(n1,n2)|subtract(n2,n3)|divide(n0,#0)|multiply(#2,#1)|multiply(#3,const_10) | general |
if 2 x = 3 y = 10 , then 3 xy = ? | "2 x = 10 ; x = 5 3 y = 10 ; y = 10 / 3 multiply : 3 xy = 3 * 5 * 10 / 3 = 50 answer : d ." | a ) 100 , b ) 200 , c ) 120 , d ) 50 , e ) 20 | d | divide(subtract(divide(multiply(2, const_100), const_2), const_2), add(divide(multiply(2, const_100), const_2), const_2)) | multiply(n0,const_100)|divide(#0,const_2)|add(#1,const_2)|subtract(#1,const_2)|divide(#3,#2)| | general |
b completes a work in 7 days . a alone can do it in 10 days . if both work together , the work can be completed in how many days ? | "1 / 7 + 1 / 10 = 17 / 70 70 / 17 = 4.11 days answer : b" | a ) 3.75 days , b ) 4.11 days , c ) 5.11 days , d ) 6.75 days , e ) 7.33 days | b | divide(multiply(10, 7), add(7, 10)) | add(n0,n1)|multiply(n0,n1)|divide(#1,#0)| | physics |
excluding the stoppages , the speed of a bus is 48 km / hr and including the stoppages the speed of the bus is 24 km / hr . for how many minutes does the bus stop per hour ? | "speed of the bus without stoppage = 48 km / hr speed of the bus with stoppage = 24 km / hr difference in speed = 24 km / hr so , the time taken in the stoppages = time taken to cover 24 km = ( 24 / 48 ) hr = 1 / 2 hr = 30 min answer : a" | a ) 30 min , b ) 10 min , c ) 12 min , d ) 20 min , e ) 18 min | a | subtract(const_60, multiply(divide(24, 48), const_60)) | divide(n1,n0)|multiply(#0,const_60)|subtract(const_60,#1)| | physics |
what is the % change in the area of a rectangle when its length increases by 20 % and its width decreases by 20 % ? | "( 12 / 10 ) * ( 8 / 10 ) = 96 / 100 of original area 96 / 100 is a 4 % decrease from 100 / 100 - > d" | a ) 0 % , b ) 20 % increase , c ) 20 % decrease , d ) 4 % decrease , e ) insufficient data | d | subtract(const_100, divide(multiply(add(const_100, 20), subtract(const_100, 20)), const_100)) | add(n0,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|subtract(const_100,#3)| | geometry |
5 pieces of wood have an average length of 128 cm and a median length of 140 cm . what is the maximum possible length , in cm , of the shortest piece of wood ? | c . 110 sum of all lengths of all 5 pieces of wood = 128 * 5 = 640 3 rd piece ( sorted in increasing length ) length = 140 ( median ) for sum of first 2 wood length to become max , last two should be least . let 4 th , 5 th wood also have length 140 each . total of last 3 = 140 * 3 = 420 sum of first 2 = 640 - 420 = 22... | a ) a ) 90 , b ) b ) 100 , c ) c ) 110 , d ) d ) 130 , e ) e ) 140 | c | divide(subtract(multiply(5, 128), multiply(140, const_3)), const_2) | multiply(n0,n1)|multiply(n2,const_3)|subtract(#0,#1)|divide(#2,const_2) | general |
if a certain toy store ' s revenue in november was 4 / 5 of its revenue in december and its revenue in january was 3 / 4 of its revenue in november , then the store ' s revenue in december was how many times the average ( arithmetic mean ) of its revenues in november and january ? | "let dec rev = 100 then nov rev is 4 / 5 ( 100 ) = > 80 therefore jan rev = 3 / 4 ( nov rev ) = 3 / 4 ( 80 ) = > 60 hence dec rev = x * ( nov rev + jan rev ) / 2 100 = x * ( 80 + 60 ) / 2 x = 100 / 70 = > 1.42 = 2 ans ) d" | a ) 1 / 4 , b ) 1 / 2 , c ) 2 / 3 , d ) 2 , e ) 4 | d | divide(3, divide(add(divide(4, 5), multiply(divide(4, 5), divide(3, 4))), const_2)) | divide(n0,n1)|divide(n2,n3)|multiply(#0,#1)|add(#0,#2)|divide(#3,const_2)|divide(n2,#4)| | general |
a car started running at a speed of 34 km / hr and the speed of the car was increased by 2 km / hr at the end of every hour . find the total distance covered by the car in the first 10 hours of the journey . | "a 43 km the total distance covered by the car in the first 10 hours = 34 + 36 + 38 + 40 + 42 + 44 + 46 + 48 + 50 + 52 = sum of 10 terms in ap whose first term is 34 and last term is 52 = 10 / 2 [ 34 + 52 ] = 430 km ." | a ) 43 km , b ) 76 km , c ) 25 km , d ) 15 km , e ) 30 km | a | divide(add(add(34, multiply(2, 10)), 34), 2) | multiply(n1,n2)|add(n0,#0)|add(n0,#1)|divide(#2,n1)| | physics |
two boys starts from the same place walking at the rate of 4 kmph and 5.5 kmph respectively in the same direction . what time will they take to be 10.5 km apart ? | "explanation : relative speed = 5.5 - 4 = 1.5 kmph ( because they walk in the same direction ) distance = 10.5 km time = distance / speed = 10.5 / 1.5 = 7 hr answer : c" | a ) 8 , b ) 5 , c ) 7 , d ) 9 , e ) 6 | c | divide(10.5, subtract(5.5, 4)) | subtract(n1,n0)|divide(n2,#0)| | gain |
mr . kramer , the losing candidate in a two - candidate election , received 942568 votes , which was exactly 30 percent of all votes cast . approximately what percent of the remaining votes would he need to have received in order to have won at least 50 percent of all the votes cast ? | let me try a simpler one . lets assume that candidate got 30 % votes and total votes is 100 . candidate won = 30 remaining = 70 to get 50 % , candidate requires 20 votes from 100 which is 20 % and 20 votes from 70 . 20 / 70 = 2 / 7 = . 285 = 28.5 % which is approx 29 % . hence the answer is d . | a ) 10 % , b ) 12 % , c ) 15 % , d ) 29 % , e ) 20 % | d | multiply(divide(subtract(multiply(divide(multiply(942568, const_100), 30), divide(50, const_100)), 942568), subtract(divide(multiply(942568, const_100), 30), 942568)), const_100) | divide(n2,const_100)|multiply(n0,const_100)|divide(#1,n1)|multiply(#2,#0)|subtract(#2,n0)|subtract(#3,n0)|divide(#5,#4)|multiply(#6,const_100) | general |
how many pieces of 0.40 meteres can be cut from a rod 47.5 meteres long | "explanation : we need so simple divide 47.5 / 0.40 , = ( 4750 / 40 ) = 118 option b" | a ) 100 , b ) 118 , c ) 120 , d ) 125 , e ) 70 | b | divide(47.5, 0.40) | divide(n1,n0)| | physics |
y = x ^ 2 + bx + 128 cuts the x axis at ( h , 0 ) and ( k , 0 ) . if h and k are integers , what is the least value of b ? | "as the curve cuts the x - axis at ( h , 0 ) and ( k , 0 ) . therefore h , k are the roots of the quadratic equation . for the quadratic equation is in the form of ax ^ 2 + bx + c = 0 , the product of the roots = c / a = 128 / 1 = 64 and the sum of the roots = - b / a = - b / 1 128 can be expressed as product of two nu... | a ) - 129 , b ) - 132 , c ) 32 , d ) 64 , e ) 128 | a | subtract(negate(128), const_1) | negate(n1)|subtract(#0,const_1)| | general |
3 years ago , the average age of a family of 5 members was 17 . a baby having been born , the average age of the family is the same today . what is the age of the child ? | present age of 5 members = 5 x 17 + 3 x 5 = 100 years also , present ages of 5 members + age of the baby = 6 x 17 = 102 years age of the baby = 102 β 100 = 2 years . answer : b | a ) 3 , b ) 2 , c ) 4 , d ) 1 , e ) 5 | b | subtract(17, multiply(3, 5)) | multiply(n0,n1)|subtract(n2,#0) | general |
car a runs at the speed of 33 km / hr & reaches its destination in 7 hr . car b runs at the speed of 44 km / h & reaches its destination in 5 h . what is the respective ratio of distances covered by car a & car b ? | "sol . distance travelled by car a = 33 Γ 7 = 231 km distance travelled by car b = 44 Γ 5 = 220 km ratio = 231 / 220 = 21 : 20 c" | a ) 20 : 23 , b ) 25 : 27 , c ) 21 : 20 , d ) 29 : 23 , e ) 24 : 25 | c | divide(multiply(33, 7), multiply(44, 5)) | multiply(n0,n1)|multiply(n2,n3)|divide(#0,#1)| | physics |
a and b undertake to do a piece of work for $ 120 . a alone can do it in 4 days while b alone can do it in 6 days . with the help of c , they finish it in 2 days . find the share of b ? | "c ' s 1 day work = ( 1 / 2 ) - ( 1 / 4 + 1 / 6 ) = 1 / 12 a : b : c = 1 / 4 : 1 / 6 : 1 / 12 = 3 : 2 : 1 b ' s share = 120 * 2 / 6 = $ 40 answer is e" | a ) $ 150 , b ) $ 100 , c ) $ 200 , d ) $ 50 , e ) $ 40 | e | multiply(divide(multiply(multiply(2, 6), inverse(4)), add(add(multiply(multiply(2, 6), subtract(inverse(2), add(inverse(4), inverse(6)))), multiply(multiply(2, 6), inverse(4))), multiply(multiply(2, 6), inverse(6)))), 120) | inverse(n1)|inverse(n3)|inverse(n2)|multiply(n2,n3)|add(#0,#2)|multiply(#0,#3)|multiply(#2,#3)|subtract(#1,#4)|multiply(#3,#7)|add(#8,#5)|add(#9,#6)|divide(#5,#10)|multiply(n0,#11)| | physics |
a , b , and c were to be paid in proportion to the part of work they did while working on the same piece of work . a and b individually can finish the piece of work in 12 days and 15 days respectively . they worked together for 5 days and then c completed the remaining work all alone . if $ 810 was the net sum to be pa... | the correct answer is d . | a ) $ 144 , b ) $ 90 , c ) $ 60 , d ) $ 54 , e ) $ 48 | d | divide(multiply(multiply(divide(const_1, 15), 5), 810), 5) | divide(const_1,n1)|multiply(n2,#0)|multiply(n3,#1)|divide(#2,n2) | general |
an aeroplane covers a certain distance at a speed of 240 kmph in 4 hours . to cover the same distance in 1 2 / 3 hours , it must travel at a speed of : | "distance = ( 240 x 4 ) = 960 km . speed = distance / time speed = 960 / ( 5 / 3 ) km / hr . [ we can write 1 2 / 3 hours as 5 / 3 hours ] required speed = ( 960 x 3 / 5 ) km / hr = 576 km / hr answer c ) 576 km / hr" | a ) 520 , b ) 620 , c ) 576 , d ) 740 , e ) 720 | c | divide(divide(multiply(240, 4), add(const_1, divide(const_2, const_3))), const_2) | divide(const_2,const_3)|multiply(n0,n1)|add(#0,const_1)|divide(#1,#2)|divide(#3,const_2)| | physics |
how many numbers from 29 to 79 are exactly divisible by 11 ? | "option ' a ' 29 / 11 = 2 and 79 / 11 = 7 = = > 7 - 2 = 5 numbers" | a ) 5 , b ) 7 , c ) 9 , d ) 11 , e ) 12 | a | add(divide(subtract(multiply(floor(divide(79, 11)), 11), multiply(add(floor(divide(29, 11)), const_1), 11)), 11), const_1) | divide(n1,n2)|divide(n0,n2)|floor(#0)|floor(#1)|add(#3,const_1)|multiply(n2,#2)|multiply(n2,#4)|subtract(#5,#6)|divide(#7,n2)|add(#8,const_1)| | general |
the racing magic takes 60 seconds to circle the racing track once . the charging bull makes 40 rounds of the track in an hour . if they left the starting point together , how many minutes will it take for them to meet at the starting point for the second time ? | time taken by racing magic to make one circle = 60 seconds time taken bycharging bullto make one circle = 60 mins / 40 = 1.5 mins = 90 seconds lcm of 90 and 60 seconds = 180 seconds time taken for them to meet at the starting point for the second time = 180 * 2 = 360 seconds = 6 mins answer b | a ) 3 , b ) 6 , c ) 9 , d ) 12 , e ) 15 | b | multiply(divide(multiply(multiply(const_60, divide(60, 40)), const_2), const_60), const_2) | divide(n0,n1)|multiply(#0,const_60)|multiply(#1,const_2)|divide(#2,const_60)|multiply(#3,const_2) | physics |
how many liters of water must be evaporated from 50 liters of a 2 - percent sugar solution to get a 10 - percent solution ? | "2 % of a 50 liter solution is 1 l which is 10 % of the solution at the end . the solution at the end must be 1 l . we need to evaporate 40 liters . the answer is d ." | a ) 25 , b ) 30 , c ) 35 , d ) 40 , e ) 45 | d | subtract(50, multiply(divide(50, const_100), 10)) | divide(n0,const_100)|multiply(n2,#0)|subtract(n0,#1)| | gain |
what is the probability of always having x + y > 0 , where y < 0 | "total value of x can be : 1 . x < 0 ; 2 . x = 0 ; 3 . x > 0 but less than y in magnitude 4 . x > 0 as well as | y | so probability of having x + y > 0 would be 1 / 4 answer : d" | a ) 1 / 2 , b ) 1 / 8 , c ) 3 / 4 , d ) 1 / 4 , e ) none | d | divide(const_2, choose(add(const_3, const_3), const_3)) | add(const_3,const_3)|choose(#0,const_3)|divide(const_2,#1)| | general |
the surface area of a sphere is 4 Ο r 2 , where r is the radius of the sphere . if the area of the base of a hemisphere is 3 , what is the surface area w of that hemisphere ? | "given area of the base of a hemisphere is 3 = pi * r ^ 2 thus r = sqrt ( 3 / pi ) . surface area of whole sphere = 4 * pi * r ^ 2 . = 4 * pi * 3 / pi = 12 . since the hemisphere is half of a sphere the surface area of the hemisphere = 12 / 2 = 6 ( curved part , not including the flat rounded base ) . but the total sur... | a ) 6 / Ο , b ) 9 / Ο , c ) 6 , d ) 9 , e ) 12 | d | add(divide(multiply(multiply(4, const_pi), divide(3, const_pi)), 2), multiply(const_pi, divide(3, const_pi))) | divide(n2,const_pi)|multiply(n0,const_pi)|multiply(#0,#1)|multiply(#0,const_pi)|divide(#2,n1)|add(#4,#3)| | geometry |
factory x ' s bulbs work for over 5000 hours in 37 % of cases , whereas factory y ' s bulbs work for over 5000 hours in 75 % of cases . it is known that factory x supplies 40 % of the total bulbs available . what is the chance that a purchased bulb will work for longer than 5000 hours ? | "for x , 40 % of 37 % will work . for y , 60 % of 75 % will work . * 60 % is the rest of the bulb supply in the market . so , the probability that a purchased bulb will work is : 0.40 ( 0.37 ) = . 148 0.60 ( 0.75 ) = 0.45 the combined probability then is 14.8 + 45 = 59.8 % ans e" | a ) 56.4 % , b ) 58 % , c ) 57.4 % , d ) 55 % , e ) 59.8 % | e | divide(add(37, 75), const_2) | add(n1,n3)|divide(#0,const_2)| | physics |
a certain shade of gray paint is obtained by mixing 3 parts of white paint with 5 parts of black paint . if 2 gallons of the mixture is needed and the individual colors can be purchased only in one gallon or half gallon cans , what is the least amount of paint e , in gallons , that must be purchased in order to measure... | "given w : b = 3 : 5 that means say 3 gallons of white paint + 5 gallons of black paint = 8 gallons of paint mixture . but we want least amount of whiteblack paints for minimum of 2 gallons of mixture , so lets reduce keeping same ratio , 1.5 : 2.5 gives 1.5 + 2.5 = 4 gallons of mixture , but we want only 2 gallons , l... | a ) 2 , b ) 2 1 / 2 , c ) 3 , d ) 3 1 / 2 , e ) 4 | b | add(multiply(const_0_25, 2), multiply(2, 5)) | multiply(const_0_25,n2)|multiply(n1,n2)|add(#0,#1)| | general |
a and b complete a work in 4 days . a alone can do it in 8 days . if both together can do the work in how many days ? | "1 / 4 + 1 / 8 = 3 / 8 8 / 3 = 2.67 days answer : c" | a ) 3.75 days , b ) 3.99 days , c ) 2.67 days , d ) 2.98 days , e ) 2.44 days | c | inverse(add(inverse(4), inverse(8))) | inverse(n0)|inverse(n1)|add(#0,#1)|inverse(#2)| | physics |
roja and pooja start moving in the opposite directions from a pole . they are moving at the speeds of 8 km / hr and 3 km / hr respectively . after 4 hours what will be the distance between them ? | "distance = relative speed * time = ( 8 + 3 ) * 4 = 44 km [ they are travelling in the opposite direction , relative speed = sum of the speeds ] . answer : c" | a ) 22 km , b ) 20 km , c ) 44 km , d ) 18 km , e ) 16 km | c | multiply(add(8, 3), 4) | add(n0,n1)|multiply(n2,#0)| | physics |
in a company with 48 employees , some part - time and some full - time , exactly ( 1 / 3 ) of the part - time employees and ( 1 / 4 ) of the full - time employees take the subway to work . what is the greatest possible number q of employees who take the subway to work ? | "p / 3 + f / 4 = p / 3 + ( 48 - p ) / 4 = 12 + p / 2 p / 3 + f / 3 = ( p + f ) / 3 = 48 / 3 = 16 p / 4 + f / 4 = 12 p / 3 + f / 3 > p / 3 + f / 4 > p / 4 + f / 4 - - > 16 > 12 + p / 12 > 12 greatest possible q : 12 + p / 12 = 15 - - > p = 36 ( integer - - > good ) 15 or d is the answer" | a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | d | add(divide(multiply(48, 3), multiply(3, 4)), divide(subtract(48, multiply(3, 4)), divide(multiply(48, 3), multiply(3, 4)))) | multiply(n0,n2)|multiply(n2,n4)|divide(#0,#1)|subtract(n0,#1)|divide(#3,#2)|add(#2,#4)| | general |
how many multiples of 8 are there between 100 and 200 ( both are inclusive ) ? | "the answer is ( 200 - 100 ) / 8 + 1 = 13.5 answer is d" | a ) 27 , b ) 4 , c ) 6 , d ) 13.5 , e ) 8 | d | add(divide(subtract(200, 100), 8), const_1) | subtract(n2,n1)|divide(#0,n0)|add(#1,const_1)| | other |
the salary of a typist was at first raised by 10 % and then the same was reduced by 5 % . if he presently draws rs . 1045 . what was his original salary ? | "x * ( 110 / 100 ) * ( 95 / 100 ) = 1045 x * ( 11 / 10 ) * ( 1 / 100 ) = 11 x = 1000 answer : c" | a ) 1009 , b ) 1018 , c ) 1000 , d ) 1082 , e ) 1029 | c | divide(1045, multiply(add(const_1, divide(10, const_100)), subtract(const_1, divide(5, const_100)))) | divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|divide(n2,#4)| | gain |
how many pounds of salt at 50 cents / lb must be mixed with 45 lbs of salt that costs 35 cents / lb so that a merchant will get 20 % profit by selling the mixture at 48 cents / lb ? | "selling price is 48 cents / lb for a 20 % profit , cost price should be 40 cents / lb ( cp * 6 / 5 = 48 ) basically , you need to mix 35 cents / lb ( salt 1 ) with 50 cents / lb ( salt 2 ) to get a mixture costing 45 cents / lb ( salt avg ) weight of salt 1 / weight of salt 2 = ( salt 2 - saltavg ) / ( saltavg - salt ... | a ) 20 , b ) 15 , c ) 40 , d ) 50 , e ) 90 | e | divide(subtract(multiply(48, 45), multiply(divide(add(const_100, 20), const_100), multiply(35, 45))), subtract(multiply(50, divide(add(const_100, 20), const_100)), 48)) | add(n3,const_100)|multiply(n1,n4)|multiply(n1,n2)|divide(#0,const_100)|multiply(#3,#2)|multiply(n0,#3)|subtract(#1,#4)|subtract(#5,n4)|divide(#6,#7)| | gain |
if the sum of a number and its square is 132 , then what is the number ? | "explanation : let the number be x . then , x 2 + x = 132 = > x 2 + x - 132 = 0 = > ( x + 12 ) ( x - 11 ) = 0 = > x = 11 answer : option b" | a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | b | floor(sqrt(132)) | sqrt(n0)|floor(#0)| | geometry |
0.363 * 0.522 + 0.363 * 0.478 = ? | "given expression = 0.363 * ( 0.522 + 0.478 ) = 0.363 * 1 = 0.363 answer : c" | a ) 0.522 , b ) 0.845 , c ) 0.363 , d ) 0.985 , e ) 0.885 | c | multiply(0.363, 0.522) | multiply(n0,n1)| | general |
lilly has 10 fish and rosy has 8 fish . in total , how many fish do they have in all ? | "10 + 8 = 18 the answer is d ." | a ) 12 , b ) 14 , c ) 16 , d ) 18 , e ) 20 | d | add(10, 8) | add(n0,n1)| | general |
a certain clock marks every hour by striking a number of times equal to the hour , and the time required for a stroke is exactly equal to the time interval between strokes . at 6 : 00 the time lapse between the beginning of the first stroke and the end of the last stroke is 55 seconds . at 12 : 00 , how many seconds el... | "at 6 : 00 it ' ll chime 6 times . if we assume that the time taken to chime is x , then time between chimes is also x . so you have 6 chimes , which is 6 x and 5 time intervals between chimes . this means that 11 x = 55 seconds . thus x = 5 seconds . by a similar logic , at 12 : 00 , there are 12 chimes and 11 interva... | a ) a . 72 , b ) b . 115 , c ) c . 48 , d ) d . 46 , e ) e . 44 | b | multiply(divide(55, add(6, add(const_1, const_4))), add(12, add(6, add(const_1, const_4)))) | add(const_1,const_4)|add(n0,#0)|add(n3,#1)|divide(n2,#1)|multiply(#2,#3)| | general |
two numbers n and 10 have lcm = 36 and gcf = 5 . find n . | "the product of two integers is equal to the product of their lcm and gcf . hence . 10 * n = 36 * 5 n = 36 * 5 / 10 = 18 correct answer d" | a ) 35 , b ) 56 , c ) 76 , d ) 18 , e ) 24 | d | divide(multiply(36, 5), 10) | multiply(n1,n2)|divide(#0,n0)| | physics |
the difference between c . i . and s . i . on an amount of rs . 15,000 for 2 years is rs . 216 . what is the rate of interest per annum ? | "explanation : [ 15000 * ( 1 + r / 100 ) 2 - 15000 ] - ( 15000 * r * 2 ) / 100 = 216 15000 [ ( 1 + r / 100 ) 2 - 1 - 2 r / 100 ] = 216 15000 [ ( 100 + r ) 2 - 10000 - 200 r ] / 10000 = 216 r 2 = ( 216 * 2 ) / 3 = 144 = > r = 12 rate = 12 % answer : option e" | a ) 8 , b ) 2 , c ) 9 , d ) 4 , e ) 12 | e | sqrt(216) | sqrt(n2)| | gain |
employees of a certain company are each to receive a unique 7 - digit identification code consisting of the digits 0 , 1 , 2 , 3 , 4 , 5 , and 6 such that no digit is used more than once in any given code . in valid codes , the second digit in the code is exactly twice the first digit . how many valid codes are there ? | "one can use slot method to solve this . 7 digit identification code has 7 slots to fill . for the first slot there are three possible options . ( 1,2 and 3 because first digit has to be half of second ) for the second slot there are three possible options . ( 2,4 and 6 because second digit has to be twice the second )... | a ) 42 , b ) 120 , c ) 210 , d ) 360 , e ) 840 | d | multiply(multiply(2, 3), 2) | multiply(n3,n4)|multiply(n3,#0)| | general |
it is the new year and mandy has made a resolution to lose weight this year . she plans to exercise and do yoga . for exercise she plans to workout at the gym and ride her bicycle in the ratio of 2 : 3 everyday . she will also do yoga in the ratio , yoga : exercise = 2 : 3 . if she rides her bike for 20 minutes , how m... | "the ratio is 2 : 3 = gym : ride , so ( 20 ) ( 3 / 2 ) = 30 minutes at the gym , and 30 + 20 = 50 minutes exercise , so ( 2 / 3 ) ( 50 ) = 33 minutes yoga . answer : b" | a ) 10 min . , b ) 33 min . , c ) 17 min . , d ) 23 min . , e ) 25 min . | b | divide(multiply(20, divide(3, add(2, 3))), multiply(divide(3, add(2, 3)), divide(3, add(2, 3)))) | add(n0,n1)|divide(n1,#0)|multiply(n4,#1)|multiply(#1,#1)|divide(#2,#3)| | physics |
two whales are moving in the same direction at 18 mps and 15 mps . the faster whale crosses the slow whale in 15 seconds . what is the length of the slower whale in meters ? | relative speed = ( 18 - 15 ) = 3 mps . distance covered in 15 sec = 15 * 3 = 45 m . the length of the faster train = 45 m . answer : a | a ) 45 m , b ) 56 m , c ) 39 m , d ) 33 m , e ) 62 m | a | multiply(subtract(18, 15), 15) | subtract(n0,n1)|multiply(n1,#0) | physics |
a circular path of 16 m radius has marginal walk 3 m wide all round it . find the cost of leveling the walk at rs . 2 per m 2 ? | explanation : Ο ( 19 ^ 2 - 16 ^ 2 ) = 22 / 7 * ( 361 - 256 ) = 330 330 * 2 = rs . 660 answer : option a | ['a ) rs . 660', 'b ) rs . 40', 'c ) rs . 44', 'd ) rs . 42', 'e ) rs . 43'] | a | multiply(2, subtract(circle_area(add(16, 3)), circle_area(16))) | add(n0,n1)|circle_area(n0)|circle_area(#0)|subtract(#2,#1)|multiply(n2,#3) | physics |
speed of a boat in standing water is 14 kmph and the speed of the stream is 1.2 kmph . a man rows to a place at a distance of 4864 km and comes back to the starting point . the total time taken by him is | explanation : speed downstream = ( 14 + 1.2 ) = 15.2 kmph speed upstream = ( 14 - 1.2 ) = 12.8 kmph total time taken = ( 4864 / 15.2 ) + ( 4864 / 12.8 ) = 320 + 380 = 700 hours . answer : option a | a ) 700 hours , b ) 350 hours , c ) 1400 hours , d ) 1010 hours , e ) none of these | a | add(divide(4864, add(14, 1.2)), divide(4864, subtract(14, 1.2))) | add(n0,n1)|subtract(n0,n1)|divide(n2,#0)|divide(n2,#1)|add(#2,#3) | physics |
the avearge score of a cricketer for 10 matches is 38.9 runs . if the average for the first 6 matches is 41 . then find the average for the last 4 matches ? | solution required average = ( 38.9 x 10 ) - ( 41 x 6 ) / 4 = 143 / 4 . = 35.75 answer b | a ) 33.25 , b ) 35.75 , c ) 34.25 , d ) 35 , e ) none | b | divide(subtract(multiply(10, 38.9), multiply(6, 41)), 4) | multiply(n0,n1)|multiply(n2,n3)|subtract(#0,#1)|divide(#2,n4) | general |
noelle walks from point a to point b at an average speed of 6 kilometers per hour . at what speed , in kilometers per hour , must noelle walk from point b to point a so that her average speed for the entire trip is 7 kilometers per hour ? | "let ' s suppose that speed while returning was xkm / h since the distance is same , we can apply the formula of avg speed avg speed = 2 s 1 s 2 / s 1 + s 2 7 = 2 * 6 * x / 6 + x x = 8.4 e is the answer" | a ) 6.75 , b ) 7 , c ) 7.25 , d ) 7.5 , e ) 8.4 | e | divide(multiply(7, 6), subtract(multiply(const_2, 6), 7)) | multiply(n0,n1)|multiply(n0,const_2)|subtract(#1,n1)|divide(#0,#2)| | physics |
6 x 0.6 x 0.06 x 0.006 x 60 = ? | "explanation : ? = 6 x 6 / 10 x 6 / 100 x 6 / 1000 x 60 = 77760 / 1000000 = 7776 / 100000 = 0.07776 answer : option d" | a ) 77.76 , b ) 7.776 , c ) 0.7776 , d ) 0.07776 , e ) none otf these | d | divide(subtract(multiply(6, 0.6), multiply(0.06, 0.006)), 60) | multiply(n0,n1)|multiply(n2,n3)|subtract(#0,#1)|divide(#2,n4)| | general |
0.009 / x = 0.02 . find the value of x | "x = 0.009 / 0.045 = 0.9 answer : a" | a ) 0.045 , b ) 0.09 , c ) 0.9 , d ) 9 , e ) 90 | a | divide(0.009, 0.02) | divide(n0,n1)| | general |
if two projectiles are launched at the same moment from 2520 km apart and travel directly towards each other at 432 km per hour and 576 km per hour respectively , how many minutes will it take for them to meet ? | "the projectiles travel a total of 1008 km per hour . the time to meet is 2520 / 1008 = 2.5 hours = 150 minutes the answer is e ." | a ) 130 , b ) 135 , c ) 140 , d ) 145 , e ) 150 | e | multiply(divide(2520, add(432, 576)), const_60) | add(n1,n2)|divide(n0,#0)|multiply(#1,const_60)| | physics |
what is the smallest number by which 3600 be divided to make it a perfect cube ? | solution 3600 = 23 * 52 * 32 * 2 . to make it a perfect cube , it must be divided by 52 * 32 * 2 i . e . , 450 . answer d | ['a ) 9', 'b ) 50', 'c ) 300', 'd ) 450', 'e ) 550'] | d | divide(3600, power(const_2, const_3)) | power(const_2,const_3)|divide(n0,#0) | geometry |
radius of a circle is 5 cm , if we draw a rectangle of maximum size , what is the area of rectangle ? | area of rectangle of max size inside circle will be a square with diagonal equal to 10 cms . then side of square = root ( 100 / 2 ) = root 50 area of square = root 50 * root 50 = 50 sq cms . answer : c | ['a ) 30', 'b ) 40', 'c ) 50', 'd ) 60', 'e ) 70'] | c | divide(multiply(multiply(5, const_2), multiply(5, const_2)), const_2) | multiply(n0,const_2)|multiply(#0,#0)|divide(#1,const_2) | geometry |
each week , harry is paid x dollars per hour for the first 30 hours and 1.5 x dollars for each additional hour worked that week . each week , james is paid x dollars per hour for the first 40 hours and 2 x dollars for each additional hour worked that week . last week james worked a total of 47 hours . if harry and jame... | "amount earned by james = 40 * x + 7 * 2 x = 54 x therefore , amount earned by harry = 54 x but we know the amount harry earned assuming working y hours ( y > 30 ) is 30 * x + ( y - 30 ) * 1.5 x [ [ we know y > 30 because in 30 h the most harry could earn is 30 x , but he has earned 54 x ] ] so x * ( 1.5 y - 45 + 30 ) ... | a ) 35 , b ) 36 , c ) 37 , d ) 38 , e ) 46 | e | add(divide(subtract(add(40, 2), 30), 1.5), 30) | add(n2,n3)|subtract(#0,n0)|divide(#1,n1)|add(n0,#2)| | general |
the greatest number that divides 60 and 190 leaving remainder 6 and 10 respectively is : | explanation : 60 - 6 = 54 , 190 - 10 = 180 highest number that divides 54 and 180 is hcf of numbers hcf of 54 nad 180 = 18 answer : b | a ) 15 , b ) 18 , c ) 19 , d ) 22 , e ) 23 | b | divide(subtract(190, 10), gcd(60, 190)) | gcd(n0,n1)|subtract(n1,n3)|divide(#1,#0) | general |
5 bells begin to toll together at 12 : 00 noon and toll respectively at intervals of 3 , 4 , 5 , 6 , and 7 seconds . not including the toll at 12 : 00 noon , how many more times will all 5 bells toll together before 1 : 00 pm ( i . e . one hour later ) ? | the least common multiple is 2 * 2 * 3 * 5 * 7 = 420 . 3600 seconds / 420 = 8 + remainder . the answer is b . | a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 14 | b | floor(divide(const_3600, multiply(multiply(multiply(3, 4), 5), 7))) | multiply(n3,n4)|multiply(n0,#0)|multiply(n7,#1)|divide(const_3600,#2)|floor(#3) | physics |
9 students bought burgers for $ 143 dollars . if the bill has to be divided among these students , how much amount should be added to the amound to make it divided amoung them equally in dollars ( in whole number of dollars ) | in order to divide the sum in 9 parts , the amount must be divisible by 9 divisibility rule of 9 : the sum of the digits must be divisible by 9 sum of digits of 143 = 8 and 9 is divisible by 9 . hence , we need to add 1 to this number for it to be divisible by 9 correct option : a | a ) $ 1 , b ) $ 2 , c ) $ 3 , d ) $ 4 , e ) $ 5 | a | subtract(9, reminder(143, 9)) | reminder(n1,n0)|subtract(n0,#0) | general |
amit and ananthu can do a work in 15 days and 25 days respectively . amit started the work and left after 3 days . ananthu took over and completed the work . in how many days was the total work completed ? | "amit β s one day β s work = 1 / 15 amit β s 3 day β s work = 1 / 15 * 3 = 1 / 5 work left = 1 - 1 / 5 = 4 / 5 ananthu β s one day β s work = 1 / 25 ananthu can do work in = 4 / 5 * 25 = 20 / 1 days so total days = 25 + 3 = 28 days answer : a" | a ) 28 days , b ) 20 days , c ) 23 days , d ) 25 days , e ) 27 days | a | add(divide(subtract(const_1, multiply(inverse(15), 3)), inverse(25)), 3) | inverse(n0)|inverse(n1)|multiply(#0,n2)|subtract(const_1,#2)|divide(#3,#1)|add(#4,n2)| | physics |
b alone can do piece of work in 10 days . a alone can do it in 15 days . if the total wages for the work is rs 5000 , how much should b be paid if they work together for the entire duration of the work ? | answer : a | a ) 267 , b ) 7 , c ) 76 , d ) 8 , e ) 82 | a | add(divide(5000, add(15, 10)), add(add(add(add(multiply(const_10, const_4), const_10), 10), const_4), const_3)) | add(n0,n1)|multiply(const_10,const_4)|add(#1,const_10)|divide(n2,#0)|add(n0,#2)|add(#4,const_4)|add(#5,const_3)|add(#6,#3) | general |
a , b and c , each working alone can complete a job in 6 , 8 and 12 days respectively . if all three of them work together to complete a job and earn $ 1170 , what will be c ' s share of the earnings ? | "explanatory answer a , b and c will share the amount of $ 1170 in the ratio of the amounts of work done by them . as a takes 6 days to complete the job , if a works alone , a will be able to complete 1 / 6 th of the work in a day . similarly , b will complete 1 / 8 th and c will complete 1 / 12 th of the work . so , t... | a ) $ 1100 , b ) $ 260 , c ) $ 1080 , d ) $ 1170 , e ) $ 630 | b | multiply(1170, divide(inverse(8), add(inverse(12), add(inverse(6), inverse(8))))) | inverse(n1)|inverse(n0)|inverse(n2)|add(#1,#0)|add(#3,#2)|divide(#0,#4)|multiply(n3,#5)| | physics |
express a speed of 180 kmph in meters per second ? | "c 50 mps 180 * 5 / 18 = 50 mps" | a ) 10 mps , b ) 05 mps , c ) 50 mps , d ) 12 mps , e ) 11 mps | c | multiply(const_0_2778, 180) | multiply(n0,const_0_2778)| | physics |
sum of two numbers is 25 . two times of the first exceeds by 5 from the three times of the other . then the numbers will be ? | "explanation : x + y = 25 2 x β 3 y = 5 x = 16 y = 9 b )" | a ) a ) 5 , b ) b ) 9 , c ) c ) 11 , d ) d ) 13 , e ) e ) 15 | b | subtract(25, divide(subtract(25, divide(5, const_2)), const_2)) | divide(n1,const_2)|subtract(n0,#0)|divide(#1,const_2)|subtract(n0,#2)| | general |
rs . 1300 is divided so that 4 times the first share , thrice the 2 nd share and twice the third share amount to the same . what is the value of the third share ? | "a + b + c = 1300 4 a = 3 b = 2 c = x a : b : c = 1 / 4 : 1 / 3 : 1 / 2 = 3 : 4 : 6 6 / 13 * 1300 = rs . 600 answer : b" | a ) s . 528 , b ) s . 600 , c ) s . 528 , d ) s . 540 , e ) s . 549 | b | multiply(4, divide(1300, add(add(4, 2), const_3))) | add(n1,n2)|add(#0,const_3)|divide(n0,#1)|multiply(n1,#2)| | general |
joe went on a diet 4 months ago when he weighed 212 pounds . if he now weighs 188 pounds and continues to lose at the same average monthly rate , in approximately how many months will he weigh 160 pounds ? | "212 - 188 = 24 pounds lost in 4 months 24 / 4 = 6 , so joe is losing weight at a rate of 6 pounds per month . . . . in approximately how many months will he weigh 160 pounds ? a simple approach is to just list the weights . now : 188 lbs in 1 month : 182 lbs in 2 months : 176 lbs in 3 months : 170 lbs in 4 months : 16... | a ) 3 , b ) 3.5 , c ) 4 , d ) 4.5 , e ) 5 | d | divide(subtract(188, 160), divide(subtract(212, 188), 4)) | subtract(n2,n3)|subtract(n1,n2)|divide(#1,n0)|divide(#0,#2)| | general |
each day a man meets his wife at the train station after work , and then she drives him home . she always arrives exactly on time to pick him up . one day he catches an earlier train and arrives at the station an hour early . he immediately begins walking home along the same route the wife drives . eventually his wife ... | "as they arrived 12 minutes earlier than usual , they saved 12 minutes on round trip from home to station ( home - station - home ) - - > 6 minutes in each direction ( home - station ) - - > wife meets husband 6 minutes earlier the usual meeting time - - > husband arrived an hour earlier the usual meeting time , so he ... | a ) 45 minutes , b ) 50 minutes , c ) 40 minutes , d ) 54 minutes , e ) 35 minutes | d | subtract(const_60, divide(12, const_2)) | divide(n0,const_2)|subtract(const_60,#0)| | physics |
how many numbers from 29 to 79 are exactly divisible by 11 | "29 / 11 = 2 and 79 / 11 = 7 = = > 7 - 2 = 5 numbers option ' a '" | a ) 5 , b ) 7 , c ) 8 , d ) 9 , e ) 11 | a | add(divide(subtract(multiply(floor(divide(79, 11)), 11), multiply(add(floor(divide(29, 11)), const_1), 11)), 11), const_1) | divide(n1,n2)|divide(n0,n2)|floor(#0)|floor(#1)|add(#3,const_1)|multiply(n2,#2)|multiply(n2,#4)|subtract(#5,#6)|divide(#7,n2)|add(#8,const_1)| | general |
calculate the time it will take for a train that is 200 meter long to pass a bridge of 180 meter length , if the speed of the train is 65 km / hour ? | "speed = 65 km / hr = 65 * ( 5 / 18 ) m / sec = 18.06 m / sec total distance = 200 + 180 = 380 meter time = distance / speed = 380 * ( 100 / 1806 ) = 21.04 seconds answer : e" | a ) 20 seconds , b ) 4 seconds , c ) 31.04 seconds , d ) 11.04 seconds , e ) 21.04 seconds | e | divide(add(200, 180), divide(65, const_3_6)) | add(n0,n1)|divide(n2,const_3_6)|divide(#0,#1)| | physics |
if z is a multiple of 8902 , what is the remainder when z ^ 2 is divided by 4 ? | "2 is a factor of 8902 , so 2 is a factor of z . then 2 ^ 2 = 4 is a factor of z ^ 2 . then the remainder when z ^ 2 is divided by 4 is 0 . the answer is a ." | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) it depends on z . | a | multiply(subtract(divide(power(divide(8902, const_3), 2), 4), add(add(add(multiply(multiply(const_1000, subtract(4, 2)), const_100), multiply(multiply(const_1000, subtract(4, 2)), const_10)), multiply(2, const_100)), subtract(const_100, 2))), 4) | divide(n0,const_3)|multiply(n1,const_100)|subtract(n2,n1)|subtract(const_100,n1)|multiply(#2,const_1000)|power(#0,n1)|divide(#5,n2)|multiply(#4,const_100)|multiply(#4,const_10)|add(#7,#8)|add(#9,#1)|add(#10,#3)|subtract(#6,#11)|multiply(n2,#12)| | general |
a person travels from p to q a speed of 50 km / hr and returns by increasing his speed by 50 % . what is his average speed for both the trips ? | "speed on return trip = 150 % of 50 = 75 km / hr . average speed of trip = 50 + 75 / 2 = 125 / 2 = 62.5 km / hr answer : a" | a ) 62.5 , b ) 77 , c ) 48 , d ) 99 , e ) 21 | a | divide(add(multiply(50, add(const_1, divide(50, const_100))), 50), const_2) | divide(n1,const_100)|add(#0,const_1)|multiply(n0,#1)|add(n0,#2)|divide(#3,const_2)| | general |
the ratio of incomes of two person p 1 and p 2 is 5 : 4 and the ratio of their expenditures is 3 : 2 . if at the end of the year , each saves rs . 1600 , then what is the income of p 1 ? | let the income of p 1 and p 2 be rs . 5 x and rs . 4 x respectively and let their expenditures be rs . 3 y and 2 y respectively . then , 5 x β 3 y = 1600 β¦ ( i ) and 4 x β 2 y = 1600 β¦ β¦ . . ( ii ) on multiplying ( i ) by 2 , ( ii ) by 3 and subtracting , we get : 2 x = 1600 - > x = 800 p 1 β s income = rs 5 * 800 = rs... | a ) rs . 800 , b ) rs . 2400 , c ) rs . 4000 , d ) rs . 3200 , e ) rs . 4200 | c | multiply(subtract(multiply(divide(1600, 2), 3), 1600), 5) | divide(n6,n1)|multiply(n4,#0)|subtract(#1,n6)|multiply(n2,#2) | other |
what is the area of a square field whose diagonal of length 40 m ? | "d 2 / 2 = ( 40 * 40 ) / 2 = 800 answer : e" | a ) 287 , b ) 269 , c ) 750 , d ) 200 , e ) 800 | e | divide(square_area(40), const_2) | square_area(n0)|divide(#0,const_2)| | geometry |
find the area of a rhombus one side of which measures 20 cm and one diagonal is 28 cm . | explanation : let other diagonal = 2 x cm . since diagonals of a rhombus bisect each other at right angles , we have : ( 20 ) 2 = ( 12 ) 2 + ( x ) 2 = > x = β ( 20 ) 2 β ( 12 ) 2 = β 256 = 16 cm . _ i so , other diagonal = 32 cm . area of rhombus = ( 1 / 2 ) x ( product of diagonals ) = ( 1 / 2 Γ 28 x 32 ) cm 2 = 448 c... | ['a ) 200 cm 2', 'b ) 300 cm 2', 'c ) 400 cm 2', 'd ) 448 cm 2', 'e ) 100 cm 2'] | d | add(multiply(multiply(divide(const_1, const_2), 28), sqrt(subtract(multiply(multiply(20, 20), const_4), multiply(28, 28)))), 28) | divide(const_1,const_2)|multiply(n0,n0)|multiply(n1,n1)|multiply(n1,#0)|multiply(#1,const_4)|subtract(#4,#2)|sqrt(#5)|multiply(#3,#6)|add(n1,#7) | geometry |
find the greatest number which is such that when 794 , 858 and 1351 are divided by it , the remainders are all same . | given , the remainders are same i . e . differences of that numbers are exactly divisible . hence , you have to find hcf ( x β y , y β z , z β x ) . 858 β 794 = 64 ; 1351 β 794 = 557 ; 1351 β 858 = 493 . hcf of ( 64 , 557 , 493 ) = 1 . answer : a | a ) 1 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | a | gcd(subtract(1351, 794), subtract(858, 794)) | subtract(n2,n0)|subtract(n1,n0)|gcd(#0,#1) | general |
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