Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 164 | correct stringclasses 5 values | annotated_formula stringlengths 7 1.65k | linear_formula stringlengths 8 925 | category stringclasses 6 values | answer stringclasses 5 values |
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in how many years will a sum of money doubles itself at 18 % per annum on simple interest ? | "p = ( p * 18 * r ) / 100 r = 5.5 % answer : b" | a ) 5.9 % , b ) 5.5 % , c ) 6 % , d ) 6.4 % , e ) 7.5 % | b | divide(const_100, 18) | divide(const_100,n0)| | gain | B |
what is the 8 th digit to the right of the decimal point in the decimal equivalent of 73 / 11 ? | "73 / 11 = 6.6363 . . . . 63 is non - terminating repeating decimal . the 8 th digit to the right of decimal point will be 3 . answer b" | a ) 1 , b ) 3 , c ) 4 , d ) 5 , e ) 2 | b | divide(73, 11) | divide(n1,n2)| | general | B |
if x ^ 2 + ( 1 / x ^ 2 ) = 8 , x ^ 4 + ( 1 / x ^ 4 ) = ? | "- > x ^ 4 + ( 1 / x ^ 4 ) = ( x ^ 2 ) ^ 2 + ( 1 / x ^ 2 ) ^ 2 = ( x ^ 2 + 1 / x ^ 2 ) ^ 2 - 2 x ^ 2 ( 1 / x ^ 2 ) = 8 ^ 2 - 2 = 62 . thus , the answer is c ." | a ) 10 , b ) 11 , c ) 62 , d ) 14 , e ) 15 | c | subtract(power(2, 2), 2) | power(n0,n0)|subtract(#0,n0)| | general | C |
a house wife saved $ 4 in buying an item on sale . if she spent $ 28 for the item , approximately how much percent she saved in the transaction ? | "actual price = 28 + 4 = $ 32 saving = 4 / 32 * 100 = 13 % approximately answer is b" | a ) 8 % , b ) 13 % , c ) 10 % , d ) 11 % , e ) 12 % | b | multiply(divide(4, add(28, 4)), const_100) | add(n0,n1)|divide(n0,#0)|multiply(#1,const_100)| | general | B |
we bought orange juice and apple juice at the store . a bottle of orange juice costs 70 cents and a bottle of apple juice costs 60 cents . we bought a total of 70 bottles for $ 46.20 . how many bottles of orange juice did we buy ? | let oj be the bottles of orange juice and let aj be the bottles of apple juice . ( oj ) + aj = 70 . aj = 70 - ( oj ) . 70 ( oj ) + 60 aj = 4620 . 70 ( oj ) + 60 ( 70 - oj ) = 4620 . 10 ( oj ) + 4200 = 4620 . oj = 42 . the answer is d . | a ) 32 , b ) 36 , c ) 38 , d ) 42 , e ) 48 | d | divide(subtract(multiply(46.2, const_100), multiply(60, 70)), subtract(70, 60)) | multiply(n3,const_100)|multiply(n0,n1)|subtract(n0,n1)|subtract(#0,#1)|divide(#3,#2) | general | D |
line m lies in the xy - plane . the y - intercept of line m is - 2 , and line m passes through the midpoint of the line segment whose endpoints are ( 2 , 4 ) and ( 6 , - 6 ) . what is the slope of line m ? | "ans : b solution : line m goes through midpoint of ( 2 , 4 ) and ( 6 , - 8 ) . midpoint is ( 4 , - 1 ) as we can see that the y axis of intercept point is ( 0 , - 2 ) means line m is parallel to x axis slope m = - 1 ans : b" | a ) - 3 , b ) - 1 , c ) - 1 / 3 , d ) 0 , e ) undefined | b | divide(add(divide(subtract(4, 6), 2), 2), divide(add(2, 6), 2)) | add(n1,n3)|subtract(n2,n4)|divide(#1,n1)|divide(#0,n1)|add(#2,n0)|divide(#4,#3)| | general | B |
two employees x and y are paid a total of rs . 590 per week by their employer . if x is paid 120 percent of the sum paid to y , how much is y paid per week ? | "let the amount paid to x per week = x and the amount paid to y per week = y then x + y = 590 but x = 120 % of y = 120 y / 100 = 12 y / 10 ∴ 12 y / 10 + y = 590 ⇒ y [ 12 / 10 + 1 ] = 590 ⇒ 22 y / 10 = 590 ⇒ 22 y = 5900 ⇒ y = 5900 / 22 = 500 / 2 = rs . 268 a )" | a ) s . 268 , b ) s . 288 , c ) s . 298 , d ) s . 300 , e ) s . 388 | a | divide(multiply(590, multiply(add(const_1, const_4), const_2)), multiply(add(multiply(add(const_1, const_4), const_2), const_1), const_2)) | add(const_1,const_4)|multiply(#0,const_2)|add(#1,const_1)|multiply(n0,#1)|multiply(#2,const_2)|divide(#3,#4)| | general | A |
on thursday mabel handled 90 transactions . anthony handled 10 % more transactions than mabel , cal handled 2 / 3 rds of the transactions that anthony handled , and jade handled 17 more transactions than cal . how much transactions did jade handled ? | "solution : mabel handled 90 transactions anthony handled 10 % more transactions than mabel anthony = 90 + 90 × 10 % = 90 + 90 × 0.10 = 90 + 9 = 99 cal handled 2 / 3 rds of the transactions than anthony handled cal = 2 / 3 × 99 = 66 jade handled 16 more transactions than cal . jade = 66 + 17 = 83 jade handled = 83 transactions . answer : d" | a ) 80 , b ) 81 , c ) 82 , d ) 83 , e ) 84 | d | add(divide(multiply(multiply(divide(90, const_100), add(10, const_100)), 2), 3), 17) | add(n1,const_100)|divide(n0,const_100)|multiply(#0,#1)|multiply(n2,#2)|divide(#3,n3)|add(n4,#4)| | general | D |
the average age of 42 students in a group is 16 years . when teacher â € ™ s age is included to it , the average increases by one . what is the teacher â € ™ s age in years ? | "sol . age of the teacher = ( 43 ã — 17 â € “ 42 ã — 16 ) years = 59 years . answer d" | a ) 31 , b ) 56 , c ) 41 , d ) 59 , e ) none | d | add(42, const_1) | add(n0,const_1)| | general | D |
4,18 , 100,294 , ___ | "2 ^ 3 - 2 ^ 2 = 4 ; 3 ^ 3 - 3 ^ 2 = 18 ; 5 ^ 3 - 5 ^ 2 = 100 ; 7 ^ 3 - 7 ^ 2 = 294 ; so , 11 ^ 3 - 11 ^ 2 = 1210 answer : c" | a ) 1000 , b ) 1100 , c ) 1210 , d ) 1452 , e ) 1552 | c | multiply(divide(4,18, 100,294), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general | C |
the hcf and lcm of two numbers m and n are respectively 6 and 210 . if m + n = 58 , then 1 / m + 1 / n is equal to | "answer we have , m x n = 6 x 210 = 1260 â ˆ ´ 1 / m + 1 / n = ( m + n ) / mn = 58 / 1260 = 17 / 620 correct option : e" | a ) 1 / 35 , b ) 3 / 35 , c ) 5 / 37 , d ) 2 / 35 , e ) none | e | divide(58, multiply(6, 210)) | multiply(n0,n1)|divide(n2,#0)| | general | E |
a man buys a cycle for rs . 1400 and sells it at a loss of 18 % . what is the selling price of the cycle ? | "solution s . p = 82 % of rs . 1400 = rs . ( 82 / 100 × 1400 ) rs . 1148 . answer a" | a ) rs . 1148 , b ) rs . 1160 , c ) rs . 1190 , d ) rs . 1202 , e ) none | a | divide(multiply(subtract(const_100, 18), 1400), const_100) | subtract(const_100,n1)|multiply(n0,#0)|divide(#1,const_100)| | gain | A |
at a summer camp with 1,800 participants , 1 / 2 of the campers are aged 8 to 12 . next year , the number of campers aged 8 to 12 will increase by 1 / 3 . after this change , what percentage of the total 1,800 campers will the 8 - to 12 - year - olds represent ? | total - 1,800 participants campers are aged 8 to 12 = ( 1 / 2 ) * 1800 = 900 next year , campers are aged 8 to 12 = ( 4 / 3 ) * 900 = 1200 percentage = ( 1200 / 1800 ) * 100 = 66 2 / 3 % answer : option c | a ) 68 % , b ) 66 % , c ) 66 2 / 3 % , d ) 60 % , e ) 70 % | c | add(multiply(multiply(divide(multiply(divide(add(const_1000, multiply(8, const_100)), const_2), add(divide(const_1, const_3), const_1)), add(const_1000, multiply(8, const_100))), const_100), const_3), divide(multiply(multiply(divide(multiply(divide(add(const_1000, multiply(8, const_100)), const_2), add(divide(const_1, const_3), const_1)), add(const_1000, multiply(8, const_100))), const_100), const_3), const_10)) | divide(const_1,const_3)|multiply(n3,const_100)|add(#0,const_1)|add(#1,const_1000)|divide(#3,const_2)|multiply(#2,#4)|divide(#5,#3)|multiply(#6,const_100)|multiply(#7,const_3)|divide(#8,const_10)|add(#9,#8) | general | C |
reb normally drives to work in 45 minutes at an average speed of 40 miles per hour . this week , however , she plans to bike to work along a route that decreases the total distance she usually travels when driving by 20 % . if reb averages between 12 and 16 miles per hour when biking , how many minutes earlier will she need to leave in the morning in order to ensure she arrives at work at the same time as when she drives ? | reb normally drives to work in 45 minutes at an average speed of 40 miles per hour . use formula d = rt car : t 1 : 45 min r 1 : 40 mph d 1 : [ ( 40 * 45 ) / 60 ] = 30 miles bike : t 1 : ? r 2 : 12 - 16 mph d 2 : 08 * d 1 = 24 miles t 1 : [ ( 24 * 60 ) / 12 ] = 120 min ( only 12 mph speed yields an answer given in the choices ) therefore , deb has to leave 120 min - 45 min = 75 min early answer : d | a ) 135 , b ) 105 , c ) 95 , d ) 75 , e ) 45 | d | subtract(multiply(divide(multiply(multiply(40, divide(45, const_60)), divide(subtract(const_100, 20), const_100)), 12), const_60), 45) | divide(n0,const_60)|subtract(const_100,n2)|divide(#1,const_100)|multiply(n1,#0)|multiply(#2,#3)|divide(#4,n3)|multiply(#5,const_60)|subtract(#6,n0) | physics | D |
a man â € ™ s current age is ( 2 / 5 ) of the age of his father . after 6 years , he will be ( 1 / 2 ) of the age of his father . what is the age of father at now ? | "let , father â € ™ s current age is a years . then , man â € ™ s current age = [ ( 2 / 5 ) a ] years . therefore , [ ( 2 / 5 ) a + 6 ] = ( 1 / 2 ) ( a + 6 ) 2 ( 2 a + 30 ) = 5 ( a + 6 ) a = 30 e" | a ) 40 , b ) 45 , c ) 38 , d ) 50 , e ) 30 | e | divide(subtract(6, multiply(6, divide(1, 2))), subtract(divide(1, 2), divide(2, 5))) | divide(n3,n0)|divide(n0,n1)|multiply(n2,#0)|subtract(#0,#1)|subtract(n2,#2)|divide(#4,#3)| | general | E |
what least number should be added to 1052 , so that the sum is completely divisible by 23 | "explanation : ( 1052 / 23 ) gives remainder 17 17 + 6 = 23 , so we need to add 6 answer : option c" | a ) a ) 4 , b ) b ) 1 , c ) c ) 6 , d ) d ) 3 , e ) e ) 5 | c | subtract(23, reminder(1052, 23)) | reminder(n0,n1)|subtract(n1,#0)| | general | C |
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 | D |
if a , b , c and d are positive integers less than 4 , and 4 ^ a + 3 ^ b + 2 ^ c + 1 ^ d = 78 then what is the value of b / c ? | since a , b and c are positive integers less than 4 we have the following possibilities : 4 ^ a = 416 , or 64 3 ^ b = 39 , or 27 2 ^ c = 24 , or 8 trial and error gives us quite quick the solution of 4 ^ a = 64 3 ^ b = 9 2 ^ c = 4 64 + 9 + 4 = 77 i . e . c = 2 and b = 2 - - - - > b / c = 1 / 1 = 1 the correct answer is c | a ) 3 , b ) 2 , c ) 1 , d ) 1 / 2 , e ) 1 / 3 | c | divide(2, 2) | divide(n3,n3) | general | C |
to produce an annual income of rs . 1200 from a 12 % stock at 90 , the amount of stock needed is : | solution for an income of rs . 12 , stock needed = rs . 100 . for an income of rs . 1200 , stock needed = rs . 100 / 12 x 1200 = 10000 answer a | a ) rs . 10,000 , b ) rs . 10,800 , c ) rs . 14,400 , d ) rs . 16,000 , e ) none of these | a | subtract(const_100, 90) | subtract(const_100,n2) | gain | A |
a man purchased 3 blankets @ rs . 200 each , 5 blankets @ rs . 150 each and two blankets at a certain rate which is now slipped off from his memory . but he remembers that the average price of the blankets was rs . 150 . find the unknown rate of two blankets ? | "10 * 150 = 1500 3 * 200 + 5 * 150 = 1350 1350 – 1050 = 300 answer : c" | a ) 100 , b ) 200 , c ) 300 , d ) 400 , e ) 500 | c | subtract(multiply(const_10, 150), add(multiply(3, const_100.0), multiply(5, 150))) | multiply(n3,const_10)|multiply(n0,const_100.0)|multiply(n2,n3)|add(#1,#2)|subtract(#0,#3)| | general | C |
how many seconds will it take for a car that is traveling at a constant rate of 77 miles per hour to travel a distance of 21 yards ? ( 1 mile = 1,160 yards ) | "speed = 77 miles / hr = 37.64 yard / s distance = 21 yards time = distance / speed = 21 / 37.64 = 0.6 sec ans - c" | a ) . 1 , b ) 0.3 , c ) 0.6 , d ) 2 , e ) 1.6 | c | divide(21, multiply(divide(77, const_3600), multiply(subtract(add(77, 77), const_4), const_10))) | add(n0,n0)|divide(n0,const_3600)|subtract(#0,const_4)|multiply(#2,const_10)|multiply(#1,#3)|divide(n1,#4)| | physics | C |
in a survey of parents , exactly 7 / 8 of the mothers and 3 / 4 of the fathers held full - time jobs . if 60 percent of the parents surveyed were women , what percent of the parents did not hold full - time jobs ? | "fathers without full - time jobs are 1 / 4 * 2 / 5 = 2 / 20 of all the parents surveyed . mothers without full - time jobs are 1 / 8 * 3 / 5 = 3 / 40 of all the parents surveyed . the percent of parents without full - time jobs is 2 / 20 + 3 / 40 = 7 / 40 = 17.5 % the answer is c ." | a ) 25.5 % , b ) 21.5 % , c ) 17.5 % , d ) 13.5 % , e ) 9.5 % | c | add(subtract(subtract(const_100, 60), multiply(divide(3, 4), subtract(const_100, 60))), subtract(60, multiply(divide(7, 8), 60))) | divide(n2,n3)|divide(n0,n1)|subtract(const_100,n4)|multiply(#0,#2)|multiply(n4,#1)|subtract(#2,#3)|subtract(n4,#4)|add(#5,#6)| | general | C |
a rectangular plot measuring 10 meters by 50 meters is to be enclosed by wire fencing . if the poles of the fence are kept 5 meters apart . how many poles will be needed ? | "perimeter of the plot = 2 ( 10 + 50 ) = 120 m no of poles = 120 / 5 = 24 m answer : d" | a ) 46 m , b ) 66 m , c ) 26 m , d ) 24 m , e ) 25 m | d | divide(multiply(add(10, 50), const_2), 5) | add(n0,n1)|multiply(#0,const_2)|divide(#1,n2)| | physics | D |
in the number 11,0 ab , a and b represent the tens and units digits , respectively . if 11,0 ab is divisible by 45 , what is the greatest possible value of b × a ? | "you should notice that 55 * 2 = 110 so 11,000 is divisible by 55 : 55 * 200 = 11,000 ( or you can notice that 11,000 is obviously divisible by both 5 and 11 so by 55 ) - - > b * a = 0 * 0 = 0 . next number divisible by 55 is 11,000 + 55 = 11,055 : b * a = 5 * 3 = 15 ( next number wo n ' t have 110 as the first 3 digits so we have only two options 0 and 25 ) . answer : d . ! please post ps questions in the ps subforum : gmat - problem - solving - ps - 140 / please post ds questions in the ds subforum : gmat - data - sufficiency - ds - 141 / no posting of ps / ds questions is allowed in the mainmath forum . d" | a ) 0 , b ) 5 , c ) 10 , d ) 15 , e ) 25 | d | subtract(multiply(reminder(reminder(multiply(add(divide(add(power(11,0, const_2), const_1000), 45), const_1), 45), 11,0), const_10), divide(subtract(reminder(multiply(add(divide(add(power(11,0, const_2), const_1000), 45), const_1), 45), 11,0), reminder(reminder(multiply(add(divide(add(power(11,0, const_2), const_1000), 45), const_1), 45), 11,0), const_10)), const_10)), const_10) | power(n0,const_2)|add(#0,const_1000)|divide(#1,n2)|add(#2,const_1)|multiply(n2,#3)|reminder(#4,n0)|reminder(#5,const_10)|subtract(#5,#6)|divide(#7,const_10)|multiply(#8,#6)|subtract(#9,const_10)| | general | D |
if g and f are both odd prime numbers andg < f , then how many different positive integer factors does 2 gfhave ? | g and f are both odd prime numbers - it means either g or f is not 2 and since prime numbers have only two factors - 1 and the number itself g and f each will have ( 1 + 1 ) = 2 factors hence 2 gf will have ( 1 + 1 ) ( 1 + 1 ) ( 1 + 1 ) = 8 factors d is the answer | a ) 3 , b ) 4 , c ) 6 , d ) 8 , e ) 12 | d | multiply(2, multiply(2, 2)) | multiply(n0,n0)|multiply(n0,#0) | other | D |
the area of a sector of a circle of radius 5 cm formed by an arc of length 5.5 cm is ? | "( 5 * 5.5 ) / 2 = 13.75 answer : c" | a ) 13.78 , b ) 13.67 , c ) 13.75 , d ) 13.98 , e ) 13.28 | c | multiply(divide(const_1, const_2), multiply(5, 5.5)) | divide(const_1,const_2)|multiply(n0,n1)|multiply(#0,#1)| | physics | C |
a certain bus driver is paid a regular rate of $ 14 per hour for any number of hours that does not exceed 40 hours per week . for any overtime hours worked in excess of 40 hours per week , the bus driver is paid a rate that is 75 % higher than his regular rate . if last week the bus driver earned $ 998 in total compensation , how many total hours did he work that week ? | "for 40 hrs = 40 * 14 = 560 excess = 998 - 560 = 438 for extra hours = . 75 ( 14 ) = 10.5 + 14 = 24.5 number of extra hrs = 438 / 24.5 = 17.8 = 18 approx . total hrs = 40 + 18 = 58 answer c" | a ) 54 , b ) 51 , c ) 58 , d ) 55 , e ) 52 | c | add(40, divide(subtract(998, multiply(14, 40)), divide(multiply(14, add(const_100, 75)), const_100))) | add(n3,const_100)|multiply(n0,n1)|multiply(n0,#0)|subtract(n4,#1)|divide(#2,const_100)|divide(#3,#4)|add(n1,#5)| | general | C |
in what time will a train 100 meters long cross an electric pole , if its speed is 360 km / hr | "explanation : first convert speed into m / sec speed = 360 * ( 5 / 18 ) = 100 m / sec time = distance / speed = 100 / 100 = 1 second answer : a" | a ) 1 second , b ) 4.5 seconds , c ) 3 seconds , d ) 2.5 seconds , e ) none of these | a | divide(100, multiply(360, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics | A |
seed mixture x is 40 percent ryegrass and 60 percent bluegrass by weight ; seed mixture y is 25 percent ryegrass and 75 % fescue . if a mixture of x and y contains 30 % ryegrass , what percent of the weight of the mixture is x ? | "assuming the weight of the mixture to be 100 g * * , then the weight of ryegrass in the mixture would be 30 g . also , assume the weight mixture x used in the mixture is xg , then the weight of mixture y used in the mixture would be ( 100 - x ) g . so we can now equate the parts of the ryegrass in the mixture as : 0.4 x + 0.25 ( 100 - x ) = 30 0.4 x + 25 - 0.25 x = 30 0.15 x = 5 x = 5 / 0.15 = 500 / 15 = 100 / 3 so the weight of mixture x as a percentage of the weight of the mixture = ( weight of x / weight of mixture ) * 100 % = ( 100 / 3 ) / 100 * 100 % = 33 % answer : b" | a ) 10 % , b ) 33 1 / 3 % , c ) 40 % , d ) 50 % , e ) 66 2 / 3 % | b | divide(subtract(30, 25), subtract(divide(40, const_100), divide(25, const_100))) | divide(n0,const_100)|divide(n2,const_100)|subtract(n4,n2)|subtract(#0,#1)|divide(#2,#3)| | gain | B |
the average of 9 observations was 7 , that of the 1 st of 5 being 10 and that of the last 5 being 8 . what was the 5 th observation ? | "explanation : 1 to 9 = 9 * 7 = 63 1 to 5 = 5 * 10 = 50 5 to 9 = 5 * 8 = 40 5 th = 50 + 40 = 90 – 63 = 27 option a" | a ) 27 , b ) 12 , c ) 15 , d ) 17 , e ) 18 | a | subtract(add(multiply(10, 5), multiply(7, 5)), multiply(7, 9)) | multiply(n3,n4)|multiply(n1,n3)|multiply(n0,n1)|add(#0,#1)|subtract(#3,#2)| | general | A |
in an office in singapore there are 60 % female employees . 50 % of all the male employees are computer literate . if there are total 62 % employees computer literate out of total 1300 employees , then the no . of female employees who are computer literate ? | "solution : total employees , = 1300 female employees , 60 % of 1300 . = ( 60 * 1300 ) / 100 = 780 . then male employees , = 520 50 % of male are computer literate , = 260 male computer literate . 62 % of total employees are computer literate , = ( 62 * 1300 ) / 100 = 806 computer literate . thus , female computer literate = 806 - 260 = 546 . answer : option a" | a ) 546 , b ) 674 , c ) 672 , d ) 960 , e ) none | a | multiply(subtract(divide(62, const_100), multiply(subtract(const_1, divide(60, const_100)), divide(50, const_100))), 1300) | divide(n2,const_100)|divide(n1,const_100)|divide(n0,const_100)|subtract(const_1,#2)|multiply(#1,#3)|subtract(#0,#4)|multiply(n3,#5)| | gain | A |
the population of a town is 30000 . it decreases annually at the rate of 20 % p . a . what will be its population after 2 years ? | "30000 × 80 / 100 × 80 / 100 = 19200 answer ; b" | a ) 4300 , b ) 19200 , c ) 5120 , d ) 5230 , e ) 5366 | b | subtract(subtract(30000, multiply(30000, divide(20, const_100))), multiply(subtract(30000, multiply(30000, divide(20, const_100))), divide(20, const_100))) | divide(n1,const_100)|multiply(n0,#0)|subtract(n0,#1)|multiply(#0,#2)|subtract(#2,#3)| | gain | B |
evaluate 45 / . 05 | "explanation : 45 / . 05 = 4500 / 5 = 900 option b" | a ) 700 , b ) 900 , c ) 705 , d ) none of these , e ) 506 | b | divide(const_100.0, divide(05, 45)) | divide(n1,const_100)|divide(n0,#0)| | general | B |
let s be the set of all positive integers that , when divided by 8 , have a remainder of 5 . what is the 75 th number in this set ? | "the set s = { 5 , 13 , 21 , 29 , . . . . . . . . . . . . . . . . . . . . . } 1 st number = 8 * 0 + 5 = 5 2 nd number = 8 * 1 + 5 = 13 3 rd number = 8 * 2 + 5 = 21 75 th number = 8 * ( 75 - 1 ) + 5 = 597 answer = a" | a ) 597 , b ) 608 , c ) 613 , d ) 616 , e ) 621 | a | add(multiply(subtract(75, const_1), 8), 5) | subtract(n2,const_1)|multiply(n0,#0)|add(n1,#1)| | general | A |
find the compound ratio of ( 4 : 3 ) , ( 1 : 3 ) and ( 2 : 3 ) is | required ratio = 4 / 3 * 1 / 3 * 2 / 3 = 16 / 27 = 16 : 27 answer is a | a ) 16 : 27 , b ) 12 : 13 , c ) 13 : 14 , d ) 14 : 15 , e ) 31 : 27 | a | multiply(divide(4, 3), multiply(divide(4, 3), divide(1, 3))) | divide(n0,n1)|divide(n2,n1)|multiply(#0,#1)|multiply(#0,#2) | other | A |
a clothing store purchased a pair of pants for $ 81 and was selling it at a price that equaled the purchase price of the pants plus a markup that was 25 percent of the selling price . after some time a clothing store owner decided to decrease the selling price by 20 percent . what was the clothing store ' s gross profit on this sale ? | "sale price ( sp ) = 81 + markup ( mp ) - - > mp = sp - 81 and given mp = sp / 4 ( 25 % is 1 / 4 th ) so sp / 4 = sp - 81 3 sp / 4 = 81 sp = 108 now a discount of 20 % is given so new sp is . 8 * 108 = 86.4 profit = 86.4 - 81 = 5.4 $ answer is b" | a ) $ 14 , b ) $ 5.4 , c ) $ 4.4 , d ) $ 1.2 , e ) $ 5.0 | b | subtract(divide(multiply(subtract(const_100, 20), add(divide(81, const_3), 81)), const_100), 81) | divide(n0,const_3)|subtract(const_100,n2)|add(n0,#0)|multiply(#2,#1)|divide(#3,const_100)|subtract(#4,n0)| | general | B |
two trains of equal length are running on parallel lines in the same direction at 46 km / hr and 36 km / hr . the faster train catches and completely passes the slower train in 54 seconds . what is the length of each train ( in meters ) ? | "the relative speed = 46 - 36 = 10 km / hr = 10 * 5 / 18 = 25 / 9 m / s in 54 seconds , the relative difference in distance traveled is 54 * 25 / 9 = 150 meters this distance is twice the length of each train . the length of each train is 150 / 2 = 75 meters the answer is d ." | a ) 45 , b ) 55 , c ) 65 , d ) 75 , e ) 85 | d | divide(multiply(divide(multiply(subtract(46, 36), const_1000), const_3600), 54), const_2) | subtract(n0,n1)|multiply(#0,const_1000)|divide(#1,const_3600)|multiply(n2,#2)|divide(#3,const_2)| | physics | D |
what is the compound interest on rs : 30,000 for 4 months at the rate of 5 % per annum | it is monthly compound rate = 5 / 12 % per month 30000 * ( 1 + 5 / 1200 ) ^ 4 - 30000 = 503.13 answer : c | a ) 501.13 , b ) 502.13 , c ) 503.13 , d ) 504.13 , e ) 505.13 | c | divide(multiply(multiply(multiply(const_3, const_100), const_100), multiply(5, divide(4, multiply(const_4, const_3)))), const_100) | multiply(const_100,const_3)|multiply(const_3,const_4)|divide(n1,#1)|multiply(#0,const_100)|multiply(n2,#2)|multiply(#3,#4)|divide(#5,const_100) | gain | C |
a certain car dealership sells economy cars , luxury cars , and sport utility vehicles . the ratio of economy to luxury cars is 3 : 5 . the ratio of economy cars to sport utility vehicles is 4 : 3 . what is the ratio of luxury cars to sport utility vehicles ? | "the ratio of economy to luxury cars is 3 : 5 - - > e : l = 3 : 5 = 12 : 20 . the ratio of economy cars to sport utility vehicles is 4 : 3 - - > e : s = 4 : 3 = 12 : 9 . thus , l : s = 20 : 9 . answer : d ." | a ) 9 : 8 , b ) 8 : 9 , c ) 3 : 2 , d ) 20 : 9 , e ) 1 : 2 | d | divide(divide(multiply(const_4, 3), multiply(3, 3)), divide(multiply(3, const_4), multiply(5, const_4))) | multiply(n3,const_4)|multiply(n3,n3)|multiply(n1,const_4)|divide(#0,#1)|divide(#0,#2)|divide(#3,#4)| | other | D |
john distributes his pencil among his 4 friends rose , mary , ranjan , and rohit in the ratio 1 / 5 : 1 / 3 : 1 / 4 : 1 / 5 . what is the minimum no . of pencils that the person should have ? | "rakesh : rahul : ranjan : rohit = 1 / 5 : 1 / 3 : 1 / 4 : 1 / 5 step 1 : at first we need to do is lcm of 2 , 3,4 and 5 is 60 . step 2 : then pencil are distributed in ratio among friends , rakesh = ( 1 / 5 x 60 ) = 12 rahul = ( 1 / 3 x 60 ) = 20 . ranjan = ( 1 / 4 x 60 ) = 15 . rohit = ( 1 / 5 x 60 ) = 12 . step 3 : total number of pencils are ( 12 x + 20 x + 15 x + 12 x ) = 59 x . for minimum number of pencils x = 1 . the person should have at least 59 pencils . a )" | a ) 59 , b ) 65 , c ) 70 , d ) 77 , e ) 98 | a | add(add(add(divide(lcm(lcm(lcm(5, 1), 1), 3), 5), divide(lcm(lcm(lcm(5, 1), 1), 3), 1)), divide(lcm(lcm(lcm(5, 1), 1), 3), 1)), divide(lcm(lcm(lcm(5, 1), 1), 3), 3)) | lcm(n2,n3)|lcm(n3,#0)|lcm(n4,#1)|divide(#2,n2)|divide(#2,n3)|divide(#2,n4)|add(#3,#4)|add(#6,#4)|add(#7,#5)| | other | A |
the ratio of the first and second class fares between two stations is 3 : 1 and that of the number of passengers travelling between the two stations by the first and the second class is 1 : 50 . if in a day , rs . 1 , 325 are collected from the passengers travelling between the two stations , then the amount collected from the second class from the second class passengers is | explanation : let the fares be 3 x and x respectively for the first and second class . suppose in a day passengers travelling between the two stations be 1 and 50 . then by given condition , 3 x × 1 + x × 50 = 1,325 = > x = 25 the amount collected from second class = 25 × 50 = rs . 1250 answer : a | a ) 1,250 , b ) 1,000 , c ) 850 , d ) 750 , e ) 350 | a | multiply(1, 1) | multiply(n1,n1) | physics | A |
40 men took a dip in a water tank 40 m long and 20 m broad on a religious day . if the average displacement of water by a man is 4 m 3 , then the rise in the water level in the tank will be : | "explanation : total volume of water displaced = ( 4 x 40 ) m 3 = 160 m 3 rise in water level = 160 / 40 ã — 20 = 0.2 m = 20 cm answer : b" | a ) 25 cm , b ) 20 cm , c ) 35 cm , d ) 50 cm , e ) none of these | b | divide(multiply(4, 40), multiply(40, 20)) | multiply(n0,n3)|multiply(n1,n2)|divide(#0,#1)| | physics | B |
a pair of articles was bought for $ 50 at a discount of 30 % . what must be the marked price of each of the article ? | "s . p . of each of the article = 50 / 2 = $ 25 let m . p = $ x 70 % of x = 25 x = 25 * . 7 = $ 17.50 answer is b" | a ) $ 25 , b ) $ 17.50 , c ) $ 29.65 , d ) $ 35.95 , e ) $ 45.62 | b | divide(multiply(subtract(const_100, 30), divide(50, const_2)), const_100) | divide(n0,const_2)|subtract(const_100,n1)|multiply(#0,#1)|divide(#2,const_100)| | gain | B |
a candidate got 33 % of the votes polled and he lost to his rival by 833 votes . how many votes were cast ? | "let x be the total number of votes . 0.33 x + 833 = 0.67 x 0.34 x = 833 x = 833 / 0.34 = 2450 the answer is b ." | a ) 2250 , b ) 2450 , c ) 2650 , d ) 2850 , e ) 3050 | b | divide(833, subtract(subtract(const_1, divide(33, const_100)), divide(33, const_100))) | divide(n0,const_100)|subtract(const_1,#0)|subtract(#1,#0)|divide(n1,#2)| | gain | B |
working simultaneously and independently at an identical constant rate , 4 machines of a certain type can produce a total of x units of product p in 6 days . how many of these machines , working simultaneously and independently at this constant rate , can produce a total of 2 x units of product p in 6 days ? | "4 machines do x units in 6 days we have x / 6 = > rate of the 4 machines we know that we need to have 2 x units in 4 days therefore , we need to get to 2 x / 4 rate of the machines . rate of one machine is x / 6 * 1 / 4 = x / 24 . now , we need to know how many machines need to work simultaneously , to get 2 x done in 6 days . 2 x / 6 work needs to be done by machines that work at x / 24 rate . let ' s assign a constant y for the number of machines : ( x / 24 ) * y = 2 x / 6 y = 2 x / 6 * 24 / x cancel 6 with 24 , and x with x and get - > 8 . answer choice e" | a ) 24 , b ) 18 , c ) 16 , d ) 12 , e ) 8 | e | multiply(multiply(4, 2), divide(6, 6)) | divide(n1,n3)|multiply(n0,n2)|multiply(#0,#1)| | general | E |
how many even multiples of 25 are there between 249 and 501 ? | "250 = 10 * 25 500 = 20 * 25 the even multiples are 25 multiplied by 10 , 12 , 14 , 16 , 18 , and 20 for a total of 6 . the answer is b ." | a ) 5 , b ) 6 , c ) 9 , d ) 10 , e ) 11 | b | add(divide(subtract(501, 249), multiply(25, const_2)), const_1) | multiply(n0,const_2)|subtract(n2,n1)|divide(#1,#0)|add(#2,const_1)| | general | B |
a company that ships boxes to a total of 12 distribution centers uses color coding to identify each center . if either a single color or a pair of two different colors is chosen to represent each center and if each center is uniquely represented by that choice of one or two colors , what is the minimum number of colors needed for the coding ? ( assume that the order of colors in a pair does not matter ) | let # of colors needed be nn , then it must be true that n + c 2 n ≥ 12 n + cn 2 - # of ways to choose the pair of different colors from nn colors when order does n ' t matter ) - - > n + n ( n − 1 ) / 2 ≥ 12 - - > 2 n + n ( n − 1 ) ≥ 24 - - > n is an integer ( it represents # of colors ) n ≥ 5 - - > nmin = 5 answer : b . | a ) 4 , b ) 5 , c ) 6 , d ) 12 , e ) 24 | b | subtract(divide(factorial(subtract(divide(12, const_2), const_1)), multiply(factorial(const_3), factorial(const_2))), subtract(divide(12, const_2), const_1)) | divide(n0,const_2)|factorial(const_3)|factorial(const_2)|multiply(#1,#2)|subtract(#0,const_1)|factorial(#4)|divide(#5,#3)|subtract(#6,#4) | general | B |
there are 680 students in a school . the ratio of boys and girls in this school is 3 : 5 . find the total of girls & boys are there in this school ? | "in order to obtain a ratio of boys to girls equal to 3 : 5 , the number of boys has to be written as 3 x and the number of girls as 5 x where x is a common factor to the number of girls and the number of boys . the total number of boys and girls is 680 . hence 3 x + 5 x = 680 solve for x 8 x = 680 x = 85 number of boys 3 x = 3 × 85 = 255 number of girls 5 x = 5 × 85 = 425 a" | a ) 425 , b ) 345 , c ) 375 , d ) 380 , e ) 400 | a | multiply(divide(680, 5), 3) | divide(n0,n2)|multiply(n1,#0)| | other | A |
a and b can do a work in 15 days , b and c in 30 days , c and a in 45 days . if a , b and c work together , they will complete the work in ? | "a + b 1 day work = 1 / 15 b + c 1 day work = 1 / 30 c + a 1 day work = 1 / 45 adding we get 2 ( a + b + c ) = 1 / 15 + 1 / 30 + 1 / 45 = 11 / 90 a + b + c 1 day work = 11 / 180 a , b , c can finish the work in 180 / 11 days = 16 days approximately answer is b" | a ) 10 days , b ) 16 days , c ) 20 days , d ) 25 days , e ) 22 days | b | divide(const_1, divide(add(add(inverse(15), inverse(30)), inverse(45)), const_2)) | inverse(n0)|inverse(n1)|inverse(n2)|add(#0,#1)|add(#3,#2)|divide(#4,const_2)|divide(const_1,#5)| | physics | B |
a man buys a cycle for rs . 800 and sells it at a loss of 15 % . what is the selling price of the cycle ? | "s . p . = 85 % of rs . 800 = rs . 85 / 100 x 800 = rs . 680 answer : c" | a ) s . 1090 , b ) s . 1160 , c ) s . 680 , d ) s . 520 , e ) s . 700 | c | divide(multiply(subtract(const_100, 15), 800), const_100) | subtract(const_100,n1)|multiply(n0,#0)|divide(#1,const_100)| | gain | C |
after giving a discount of rs . 80 the shopkeeper still gets a profit of 10 % , if the cost price is rs . 200 . find the markup % ? | "cost price = 180 s . p = 200 * 110 / 100 = 220 disc = 80 so . . . mark price = 216 + 80 = 296 . . . . . . mark up % = 261 - 200 / 200 = 61 / 200 = 0.305 or 31 % answer : a" | a ) 31 % , b ) 33 % , c ) 28 % , d ) 30 % , e ) 32 % | a | divide(subtract(add(add(200, divide(200, divide(const_100, 10))), 80), 200), 200) | divide(const_100,n1)|divide(n2,#0)|add(n2,#1)|add(n0,#2)|subtract(#3,n2)|divide(#4,n2)| | gain | A |
the side of a square is increased by 10 % then how much % does its area increases ? | "a = 100 a 2 = 10000 a = 110 a 2 = 12100 - - - - - - - - - - - - - - - - 10000 - - - - - - - - - 2100 100 - - - - - - - ? = > 21 % answer : a" | a ) 21 , b ) 56.25 , c ) 50.75 , d ) 42.75 , e ) 52.75 | a | divide(multiply(subtract(square_area(add(const_100, 10)), square_area(const_100)), const_100), square_area(const_100)) | add(n0,const_100)|square_area(const_100)|square_area(#0)|subtract(#2,#1)|multiply(#3,const_100)|divide(#4,#1)| | geometry | A |
sandy had $ 224 left after spending 30 % of the money she took for shopping . how much money did sandy take along with her ? | "let the money sandy took for shopping be x . 0.7 x = 224 x = 320 the answer is c ." | a ) $ 260 , b ) $ 290 , c ) $ 320 , d ) $ 350 , e ) $ 380 | c | divide(224, divide(subtract(const_100, 30), const_100)) | subtract(const_100,n1)|divide(#0,const_100)|divide(n0,#1)| | gain | C |
the difference between the ages of two persons is 30 years . fifteen years ago , the elder one was twice as old as the younger one . the present age of the younger person is ? | "let their ages be x years and ( x + 30 ) years then , ( x + 30 ) - 15 = 2 ( x - 15 ) x + 15 = 2 x - 30 x = 45 answer is c" | a ) 30 yr , b ) 25 yr , c ) 45 yr , d ) 40 yr , e ) 50 yr | c | add(subtract(30, subtract(30, add(const_3, const_2))), multiply(subtract(30, add(const_3, const_2)), const_2)) | add(const_2,const_3)|subtract(n0,#0)|multiply(#1,const_2)|subtract(n0,#1)|add(#2,#3)| | general | C |
what is the sum of all digits for the number 10 ^ 27 - 48 ? | "10 ^ 27 is a 28 - digit number : 1 followed by 27 zeros . 10 ^ 27 - 48 is a 27 - digit number : 25 9 ' s and 52 at the end . the sum of the digits is 25 * 9 + 5 + 2 = 232 . the answer is c ." | a ) 212 , b ) 222 , c ) 232 , d ) 242 , e ) 252 | c | multiply(add(divide(subtract(subtract(27, 10), const_2), const_2), 10), divide(add(subtract(27, 10), const_2), const_2)) | subtract(n1,n0)|add(#0,const_2)|subtract(#0,const_2)|divide(#2,const_2)|divide(#1,const_2)|add(n0,#3)|multiply(#5,#4)| | general | C |
a train running at 1 / 5 of its own speed reached a place in 20 hours . how much time could be saved if the train would have run at its own speed ? | "time taken if run its own speed = 1 / 5 * 20 = 4 hrs time saved = 20 - 4 = 16 hrs answer : d" | a ) 8 hrs , b ) 10 hrs , c ) 12 hrs , d ) 16 hrs , e ) 6 hrs | d | multiply(divide(1, 5), 20) | divide(n0,n1)|multiply(n2,#0)| | physics | D |
if the average ( arithmetic mean ) of ( 2 a + 16 ) and ( 3 a - 8 ) is 69 , what is the value of a ? | ( ( 2 a + 16 ) + ( 3 a - 8 ) ) / 2 = ( 5 a + 8 ) / 2 = 69 a = 26 the answer is a . | a ) 26 , b ) 30 , c ) 28 , d ) 36 , e ) 42 | a | divide(subtract(multiply(69, 2), subtract(16, 8)), add(2, 3)) | add(n0,n2)|multiply(n0,n4)|subtract(n1,n3)|subtract(#1,#2)|divide(#3,#0) | general | A |
the h . c . f and l . c . m of two numbers are 84 and 21 respectively . if the ratio of the two numbers is 1 : 4 , then the larger of the two numbers is | explanation : let the numbers be x and 4 x . then , \ inline \ fn _ jvn x \ times 4 x = 84 \ times 21 \ : \ : \ leftrightarrow \ : \ : x ^ { 2 } = \ frac { 84 \ times 21 } { 4 } \ : \ : \ leftrightarrow \ : \ : x = 21 hence larger number = 4 x = 84 answer : c ) 84 | a ) 32 , b ) 37 , c ) 84 , d ) 29 , e ) 21 | c | multiply(sqrt(divide(multiply(84, 21), 4)), 4) | multiply(n0,n1)|divide(#0,n3)|sqrt(#1)|multiply(n3,#2) | other | C |
two numbers are in the ratio 3 : 5 . if 9 be subtracted from each , they are in the ratio of 9 : 17 . the first number ? | "( 3 x - 9 ) : ( 5 x - 9 ) = 9 : 17 x = 12 = > 3 x = 36 answer : a" | a ) 36 , b ) 86 , c ) 27 , d ) 86 , e ) 27 | a | add(multiply(3, divide(9, multiply(3, 5))), multiply(5, divide(9, multiply(3, 5)))) | multiply(n0,n1)|divide(n2,#0)|multiply(n0,#1)|multiply(n1,#1)|add(#2,#3)| | other | A |
at the wholesale store you can buy an 8 - pack of hot dogs for $ 1.55 , a 20 - pack for $ 3.05 , and a 300 - pack for $ 22.95 . what is the greatest number of hot dogs you can buy at this store with $ 200 ? | "we have $ 200 and we have to maximize the number of hot dogs that we can buy with this amount . let ' s try to find out what is the maximum number of hot dogs that we can buy for a lesser amount of money , which in this case is 300 for $ 22.95 . for the sake of calculation , let ' s take $ 23 . 23 x 8 gives 184 , i . e . a total of 300 x 8 = 2400 hot dogs . we are left with ~ $ 16 . similarly , let ' s use $ 3 for calculation . we can buy 5 20 - pack hot dogs ( 3 x 5 ) , a total of 20 x 5 = 100 hot dogs . so we have 2500 hot dogs . 2108 looks far - fetched ( since we are not likely to be left with > $ 1.55 ) . hence , ( b ) 2500 ( answer b )" | a ) 1,108 , b ) 2,500 , c ) 2,108 , d ) 2,124 , e ) 2,256 | b | multiply(divide(200, 22.95), 300) | divide(n6,n5)|multiply(n4,#0)| | general | B |
a shopkeeper sold an article offering a discount of 5 % and earned a profit of 19.7 % . what would have been the percentage of profit earned if no discount was offered ? | "let c . p . be rs . 100 . then , s . p . = rs . 19.70 let marked price be rs . x . then , 95 / 100 x = 119.70 x = 11970 / 95 = rs . 126 now , s . p . = rs . 126 , c . p . = rs . 100 profit % = 26 % . answer : b" | a ) 60 % , b ) 26 % , c ) 30 % , d ) 56 % , e ) 73 % | b | subtract(divide(multiply(add(const_100, 19.7), const_100), subtract(const_100, 5)), const_100) | add(n1,const_100)|subtract(const_100,n0)|multiply(#0,const_100)|divide(#2,#1)|subtract(#3,const_100)| | gain | B |
at a meeting , 25 attendants used a pencil and 15 attendants used a pen . if exactly 20 attendants used only one of the two types of writing tools , how many attendants wrote with both types of writing tools ? | say x attendants wrote with both writing tools . ( 25 - x ) + ( 15 - x ) = 20 - - > x = 10 . answer : a . | a ) 10 , b ) 6 , c ) 7 , d ) 4 , e ) 14 | a | divide(subtract(add(25, 15), 20), const_2) | add(n0,n1)|subtract(#0,n2)|divide(#1,const_2) | other | A |
10 % of a number is more than 20 % of 650 by 190 . find the number ? | "( 10 / 100 ) * x â € “ ( 20 / 100 ) * 650 = 190 1 / 10 x = 320 x = 3200 answer : b" | a ) 288 , b ) 3200 , c ) 800 , d ) 267 , e ) 121 | b | divide(multiply(add(divide(multiply(650, 20), const_100), 190), const_100), 10) | multiply(n1,n2)|divide(#0,const_100)|add(n3,#1)|multiply(#2,const_100)|divide(#3,n0)| | gain | B |
a courtyard is 28 meter long and 13 meter board is to be paved with bricks of dimensions 22 cm by 12 cm . the total number of bricks required is : | explanation : number of bricks = courtyard area / 1 brick area = ( 2800 ã — 1300 / 22 ã — 12 ) = 13787 option c | a ) 16000 , b ) 14567 , c ) 13787 , d ) 13456 , e ) none of these | c | divide(multiply(multiply(28, const_100), multiply(13, const_100)), multiply(22, 12)) | multiply(n0,const_100)|multiply(n1,const_100)|multiply(n2,n3)|multiply(#0,#1)|divide(#3,#2) | physics | C |
a train 125 m long passes a man , running at 4 km / hr in the same direction in which the train is going , in 10 seconds . the speed of the train is ? | "speed of the train relative to man = ( 125 / 10 ) m / sec = ( 25 / 2 ) m / sec . [ ( 25 / 2 ) * ( 18 / 5 ) ] km / hr = 45 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 4 ) km / hr . x - 4 = 45 = = > x = 49 km / hr . answer : b" | a ) 11 , b ) 49 , c ) 88 , d ) 65 , e ) 22 | b | divide(divide(subtract(125, multiply(multiply(4, const_0_2778), 4)), 10), const_0_2778) | multiply(n1,const_0_2778)|multiply(n1,#0)|subtract(n0,#1)|divide(#2,n2)|divide(#3,const_0_2778)| | physics | B |
a sum amounts to rs . 5292 in 2 years at the rate of 5 % p . a . if interest was compounded yearly then what was the principal ? | "ci = 5292 , r = 5 , n = 2 ci = p [ 1 + r / 100 ] ^ 2 = p [ 1 + 5 / 100 ] ^ 2 5292 = p [ 21 / 20 ] ^ 2 5292 [ 20 / 21 ] ^ 2 4800 answer : d" | a ) s . 4000 , b ) s . 5000 , c ) s . 4500 , d ) s . 4800 , e ) s . 5800 | d | divide(5292, power(add(divide(5, const_100), const_1), 2)) | divide(n2,const_100)|add(#0,const_1)|power(#1,n1)|divide(n0,#2)| | gain | D |
34 men can complete a piece of work in 18 days . in how many days will 17 men complete the same work ? | "explanation : let the required number of days be a . then , less men , more days ( indirect proportion ) 17 : 34 : : 18 : a 17 x a = 34 x 18 a = ( 34 x 18 ) / 17 a = 36 answer : a" | a ) 36 , b ) 26 , c ) 97 , d ) 26 , e ) 19 | a | divide(multiply(18, 34), 17) | multiply(n0,n1)|divide(#0,n2)| | physics | A |
if a certain coin is flipped , the probability that the coin will land heads is 1 / 2 . if the coin is flipped 6 times , what is the probability that it will land heads up on the first 3 flips but not on the last 3 flips ? | "p ( hhhttt ) = 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 = 1 / 64 the answer is d ." | a ) 1 / 8 , b ) 1 / 16 , c ) 1 / 32 , d ) 1 / 64 , e ) 1 / 128 | d | power(divide(1, 2), 6) | divide(n0,n1)|power(#0,n2)| | probability | D |
a barrel full of beer has 2 taps one midway , , which draw a litre in 6 minutes and the other at the bottom , which draws a litre in 4 minutes . the lower tap is lower normally used after the level of beer in the barrel is lower than midway . the capacity of the barrel is 36 litres . a new assistant opens the lower tap when the barrel is full and draws out some beer . as a result the lower tap has been used 24 minutes before the usual time . for how long was the beer drawn out by the new assistant ? | sol . the top tab is operational till 18 litres is drawn out . ∴ time after which the lower tap is usually open = 18 × 6 = 108 minutes ∴ time after which it is open now = 108 – 24 = 84 minutes ∴ litres drawn = 84 / 6 = 14 litres ∴ 18 – 14 = 4 litres were drawn by the new assistant . ∴ time = 4 × 4 = 16 minutes answer b | a ) 15 minutes , b ) 16 minutes , c ) 17 minutes , d ) 18 minutes , e ) none of these | b | multiply(4, subtract(divide(36, const_2), divide(subtract(multiply(divide(36, const_2), 6), 24), 6))) | divide(n3,const_2)|multiply(n1,#0)|subtract(#1,n4)|divide(#2,n1)|subtract(#0,#3)|multiply(n2,#4) | physics | B |
a , b and c can do a piece of work in 24 days , 30 days and 60 days respectively . they began the work together but c left 4 days before the completion of the work . in how many days was the work completed ? | one day work of a , b and c = 1 / 24 + 1 / 30 + 1 / 60 = 11 / 120 work done by a and b together in the last 4 days = 4 * ( 1 / 24 + 1 / 30 ) = 3 / 10 remaining work = 7 / 10 the number of days required for this initial work = 7 days . the total number of days required = 4 + 7 = 11 days . answer : b | a ) 10 days , b ) 11 days , c ) 13 days , d ) 21 days , e ) 31 days | b | add(divide(subtract(const_1, multiply(add(divide(const_1, 24), divide(const_1, 30)), 4)), add(divide(const_1, multiply(add(const_2, const_3), multiply(const_2, const_4))), add(divide(const_1, 24), divide(const_1, 30)))), 4) | add(const_2,const_3)|divide(const_1,n0)|divide(const_1,n1)|multiply(const_2,const_4)|add(#1,#2)|multiply(#0,#3)|divide(const_1,#5)|multiply(n3,#4)|add(#4,#6)|subtract(const_1,#7)|divide(#9,#8)|add(n3,#10) | physics | B |
rs . 1200 is divided so that 5 times the first share , 10 times the 2 nd share and fifteen times third share amount to the same . what is the value of the second share ? | "a + b + c = 1200 5 a = 10 b = 15 c = x a : b : c = 1 / 5 : 1 / 10 : 1 / 15 = 3 : 2 : 1 2 / 6 * 1200 = rs 400 answer : d" | a ) s 525 , b ) s 527 , c ) s 598 , d ) s 400 , e ) s 500 | d | multiply(5, divide(1200, add(add(5, 10), const_3))) | add(n1,n2)|add(#0,const_3)|divide(n0,#1)|multiply(n1,#2)| | general | D |
the age of somu is one - third his father ' s . 5 years back he was one - fifth of his father ' s age . what is his persent age ? | "explanation : let somu ' s age be x and that of his father be 3 x . so , x - 5 = 3 x - 5 / 5 = x = 10 answer : option e" | a ) 11 , b ) 13 , c ) 14 , d ) 12 , e ) 10 | e | divide(subtract(multiply(add(const_4, const_1), 5), 5), subtract(add(const_4, const_1), const_3)) | add(const_1,const_4)|multiply(n0,#0)|subtract(#0,const_3)|subtract(#1,n0)|divide(#3,#2)| | general | E |
a and b can together finish a work in 40 days . they worked together for 10 days and then b left . after another 6 days , a finished the remaining work . in how many days a alone can finish the job ? | "a + b 10 days work = 10 * 1 / 40 = 1 / 4 remaining work = 1 - 1 / 4 = 3 / 4 3 / 4 work is done by a in 6 days whole work will be done by a in 6 * 4 / 3 = 8 days answer is d" | a ) 10 , b ) 25 , c ) 60 , d ) 8 , e ) 20 | d | divide(multiply(6, 40), subtract(40, 10)) | multiply(n0,n2)|subtract(n0,n1)|divide(#0,#1)| | physics | D |
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 | A |
the heights of a wall is 6 times its width and the length of the wall is 7 times its height . if volume of the wall be 16128 cu . m , its width is | solution let the width of the wall be x metres . then , height = ( 6 x ) m and length = ( 42 x ) m . 42 x × x × 6 x = 16128 x 3 ‹ = › ( 16128 / 42 × 6 ) = 64 ‹ = › x = 4 . answer a | a ) 4 m , b ) 4.5 m , c ) 5 m , d ) 6 m , e ) none | a | divide(sqrt(divide(16128, multiply(multiply(6, 7), 6))), const_2) | multiply(n0,n1)|multiply(n0,#0)|divide(n2,#1)|sqrt(#2)|divide(#3,const_2) | physics | A |
a dishonest dealer professes to sell goods at the cost price but uses a weight of 880 grams per kg , what is his percent ? | "explanation : 880 - - - 120 100 - - - ? = > 13.6 % answer : b" | a ) 15 % , b ) 13.6 % , c ) 65 % , d ) 45 % , e ) 35 % | b | subtract(multiply(divide(const_100, 880), multiply(const_100, multiply(add(const_3, const_2), const_2))), const_100) | add(const_2,const_3)|divide(const_100,n0)|multiply(#0,const_2)|multiply(#2,const_100)|multiply(#1,#3)|subtract(#4,const_100)| | gain | B |
how many paying stones , each measuring 2 1 / 2 m * 2 m are required to pave a rectangular court yard 50 m long and 16 1 / 2 m board ? | 50 * 33 / 2 = 5 / 2 * 2 * x = > x = 165 answer : c | a ) 99 , b ) 18 , c ) 165 , d ) 10 , e ) 15 | c | divide(multiply(50, add(16, divide(1, 2))), multiply(add(2, divide(1, 2)), 2)) | divide(n1,n0)|add(n5,#0)|add(n0,#0)|multiply(n4,#1)|multiply(n0,#2)|divide(#3,#4) | general | C |
what is the remainder when 3 ^ 29 is divided by 5 ? | "the units digit of powers of 3 follow a repeating cycle of four : { 3 , 9 , 7 , 1 } 29 has the form 4 k + 1 , so the units digit of 3 ^ 29 is 3 . the remainder when dividing by 5 is 3 . the answer is d ." | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | d | subtract(divide(5, const_2), multiply(3, 3)) | divide(n2,const_2)|multiply(n0,n0)|subtract(#0,#1)| | general | D |
what profit percent is made by selling an article at a certain price , if by selling at 1 / 3 rd of that price , there would be a loss of 20 % ? | "sp 2 = 1 / 3 sp 1 cp = 100 sp 2 = 80 1 / 3 sp 1 = 80 sp 1 = 240 100 - - - 240 = > 140 % answer : a" | a ) 140 % , b ) 29 % , c ) 70 % , d ) 27 % , e ) 28 % | a | subtract(divide(subtract(const_100, 20), divide(1, 3)), const_100) | divide(n0,n1)|subtract(const_100,n2)|divide(#1,#0)|subtract(#2,const_100)| | gain | A |
the area of a triangle is with base 4.5 m and height 6 m ? | "1 / 2 * 4.5 * 6 = 13.5 m 2 answer : c" | a ) 11 m 2 , b ) 10 m 2 , c ) 13.5 m 2 , d ) 19 m 2 , e ) 12.5 m 2 | c | triangle_area(4.5, 6) | triangle_area(n0,n1)| | geometry | C |
the average of first six prime numbers greater than 3 is ? | "5 + 7 + 11 + 13 + 17 + 19 = 72 / 6 = 12 answer : c" | a ) 16 , b ) 10 , c ) 12 , d ) 19 , e ) 17.8 | c | add(3, const_1) | add(n0,const_1)| | general | C |
two trains of equal lengths take 10 sec and 15 sec respectively to cross a telegraph post . if the length of each train be 160 m , in what time will they cross other travelling in opposite direction ? | "speed of the first train = 120 / 10 = 12 m / sec . speed of the second train = 120 / 5 = 8 m / sec . relative speed = 12 + 8 = 20 m / sec . required time = ( 160 + 160 ) / 20 = 16 sec . answer : c" | a ) 17 sec , b ) 12 sec , c ) 16 sec , d ) 15 sec , e ) 18 sec | c | divide(multiply(160, const_2), add(speed(160, 15), speed(160, 10))) | multiply(n2,const_2)|speed(n2,n1)|speed(n2,n0)|add(#1,#2)|divide(#0,#3)| | physics | C |
the edge of a cube is 3 a cm . find its surface ? | "6 a 2 = 6 * 3 a * 3 a = 54 a 2 answer : d" | a ) 24 a 8 , b ) 24 a 4 , c ) 24 a 1 , d ) 54 a 2 , e ) 24 a 7 | d | surface_cube(3) | surface_cube(n0)| | geometry | D |
6 workers should finish a job in 8 days . after 3 days came 4 workers join them . how many days d do they need to finish the same job ? | "let rate of one worker be r = > ( 6 * r ) * 8 = 1 ( rate * time = work ) = > r = 1 / 48 = > work remaining after 3 days 1 - ( 3 * 6 ) / 48 = 30 / 48 after 4 ppl joined in ( ( 6 + 4 ) * time ) / 48 = 30 / 48 time d = 3 days to finish the task imo a" | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | divide(subtract(multiply(6, 8), multiply(3, 6)), add(6, 4)) | add(n0,n3)|multiply(n0,n1)|multiply(n0,n2)|subtract(#1,#2)|divide(#3,#0)| | physics | A |
maxwell leaves his home and walks toward brad ' s house at the same time that brad leaves his home and runs toward maxwell ' s house . if the distance between their homes is 40 kilometers , maxwell ' s walking speed is 3 km / h , and brad ' s running speed is 5 km / h . what is the distance traveled by maxwell when they meet in the middle ? | "consider max starts from point a and brad starts from point b and move towards each other . assume they shall meet at point o after time ' t ' . the question asks us to find oa . from the question stem we can make out : - distance oa = 50 km - distance ob = > 3 xt = 40 - 5 xt ( i . e distance = speed x time ) = > 8 t = 40 hence t = 5 oa = 3 x 5 = 15 km answer : d" | a ) 16 , b ) 17 , c ) 18 , d ) 15 , e ) 14 | d | multiply(3, divide(40, add(3, 5))) | add(n1,n2)|divide(n0,#0)|multiply(n1,#1)| | physics | D |
machine p and machine q are each used to manufacture 110 sprockets . it takes machine p 10 hours longer to produce 110 sprockets than machine q . machine q produces 10 % more sprockets per hour than machine a . how many sprockets per hour does machine a produce ? | p makes x sprockets per hour . then q makes 1.1 x sprockets per hour . 10 / x = 110 / 1.1 x + 10 1.1 ( 110 ) = 110 + 11 x 11 x = 11 x = 1 the answer is e | a ) 5 , b ) 15 , c ) 55 , d ) 95 , e ) 1 | e | divide(subtract(110, divide(110, add(divide(10, const_100), const_1))), 10) | divide(n1,const_100)|add(#0,const_1)|divide(n0,#1)|subtract(n0,#2)|divide(#3,n1) | gain | E |
if x + y = - 4 , and x = 6 / y , what is the value of x ^ 2 + y ^ 2 ? | x ^ 2 + y ^ 2 should make you think of these formulas : ( x + y ) ( x + y ) = x ^ 2 + y ^ 2 + 2 xy we already know ( x + y ) = - 4 and x * y = 6 ( x + y ) ( x + y ) = ( - 4 ) ( - 4 ) = x ^ 2 + y ^ 2 + 2 * ( 6 ) x ^ 2 + y ^ 2 = 16 - 12 = 4 answer : d | a ) 10 , b ) 15 , c ) 6 , d ) 4 , e ) 5 | d | subtract(power(4, const_2), multiply(2, 6)) | multiply(n1,n2)|power(n0,const_2)|subtract(#1,#0) | general | D |
the price of food in the area where the sims family lives is scheduled to increase by 50 % next year . since the sims family can not afford an increase in their food bill , how much will they have to reduce consumption to keep their cost the same ? | solution : let the current food expense be represented by rs . 100 . the cost of food rises 50 % . so , to buy same amount of food , they need to increase their expense , = ( 100 + 50 % of 100 ) = rs . 150 . but , they want to keep food expense the same , so they have to cut rs . by 50 to keep it to rs . = 100 . the % decrease in consumption is , ( 50 / 150 ) * 100 = 33.3 % . mental calculation method ; 100 - - - - - 50 % ↑ - - - → 150 - - - - - - x % ↓ - - - → 100 . here , x = ( 50 / 125 ) * 100 = 33.3 % . answer : option d | a ) 17 % , b ) 25.6 % , c ) none % , d ) 33.3 % , e ) 50 % | d | multiply(subtract(const_1, divide(const_1, divide(add(const_100, 50), const_100))), const_100) | add(n0,const_100)|divide(#0,const_100)|divide(const_1,#1)|subtract(const_1,#2)|multiply(#3,const_100) | general | D |
andy solves problems 78 to 125 inclusive in a math exercise . how many problems does he solve ? | "125 - 78 + 1 = 48 ' e ' is the answer" | a ) 53 , b ) 52 , c ) 51 , d ) 50 , e ) 48 | e | add(subtract(125, 78), const_1) | subtract(n1,n0)|add(#0,const_1)| | general | E |
find the missing value : 8597 - ? = 7429 - 4358 | "let 8597 - x = 7429 - 4358 . then , x = ( 8597 + 4358 ) - 7429 = 12955 - 7429 = 5526 . answer is a ." | a ) 5526 , b ) 5426 , c ) 5326 , d ) 5226 , e ) none of them | a | divide(7429, divide(8597, const_100)) | divide(n0,const_100)|divide(n1,#0)| | general | A |
how many seconds will a 500 meter long train take to cross a man walking with a speed of 3 km / hr in the direction of the moving train if the speed of the train is 63 km / hr ? | "let length of tunnel is x meter distance = 800 + x meter time = 1 minute = 60 seconds speed = 78 km / hr = 78 * 5 / 18 m / s = 65 / 3 m / s distance = speed * time 800 + x = ( 65 / 3 ) * 60 800 + x = 20 * 65 = 1300 x = 1300 - 800 = 500 meters answer : option c" | a ) 65 , b ) 69 , c ) 30 , d ) 31 , e ) 32 | c | multiply(multiply(subtract(divide(500, multiply(subtract(63, 3), const_0_2778)), const_1), const_10), const_2) | subtract(n2,n1)|multiply(#0,const_0_2778)|divide(n0,#1)|subtract(#2,const_1)|multiply(#3,const_10)|multiply(#4,const_2)| | physics | C |
a man swims downstream 48 km and upstream 18 km taking 3 hours each time , what is the speed of the man in still water ? | "48 - - - 3 ds = 16 ? - - - - 1 18 - - - - 3 us = 6 ? - - - - 1 m = ? m = ( 16 + 6 ) / 2 = 11 answer : d" | a ) 2 , b ) 8 , c ) 9 , d ) 11 , e ) 14 | d | divide(add(divide(18, 3), divide(48, 3)), const_2) | divide(n1,n2)|divide(n0,n2)|add(#0,#1)|divide(#2,const_2)| | physics | D |
what is the units digit of ( 147 ^ 25 ) ^ 49 ? | the units digit of the exponents of 7 repeat in a cycle of four , which is { 7,9 , 3,1 } . the number 25 has the form 4 n + 1 so the units digit is 7 inside the bracket . the exponent 49 has the form 4 n + 1 , so the units digit is 7 . the answer is d . | a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | d | add(add(const_4, const_3), const_2) | add(const_3,const_4)|add(#0,const_2)| | general | D |
how long does a train 110 m long traveling at 60 kmph takes to cross a bridge of 150 m in length ? | "d = 110 + 150 = 260 m s = 60 * 5 / 18 = 50 / 3 t = 260 * 3 / 50 = 15.6 sec answer : d" | a ) sec , b ) sec , c ) sec , d ) sec , e ) sec | d | divide(add(110, 150), multiply(60, const_0_2778)) | add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)| | physics | D |
65 % of the employees of a company are men . 60 % of the men in the company speak french and 40 % of the employees of the company speak french . what is % of the women in the company who do not speak french ? | "no of employees = 100 ( say ) men = 65 women = 35 men speaking french = 0.60 * 65 = 39 employees speaking french = 0.4 * 100 = 40 therefore women speaking french = 40 - 39 = 1 and women not speaking french = 35 - 1 = 34 % of women not speaking french = 34 / 35 * 100 = 97.14 % answer c" | a ) 4 % , b ) 10 % , c ) 97.14 % , d ) 90.14 % , e ) 20 % | c | multiply(divide(subtract(divide(subtract(const_100, 65), const_100), subtract(divide(40, const_100), multiply(divide(65, const_100), divide(60, const_100)))), divide(subtract(const_100, 65), const_100)), const_100) | divide(n2,const_100)|divide(n0,const_100)|divide(n1,const_100)|subtract(const_100,n0)|divide(#3,const_100)|multiply(#1,#2)|subtract(#0,#5)|subtract(#4,#6)|divide(#7,#4)|multiply(#8,const_100)| | gain | C |
sam invested rs . 8000 @ 10 % per annum for one year . if the interest is compounded half - yearly , then the amount received by sam at the end of the year will be ? | "p = rs . 8000 ; r = 10 % p . a . = 5 % per half - year ; t = 1 year = 2 half - year amount = [ 8000 * ( 1 + 5 / 100 ) 2 ] = ( 8000 * 21 / 20 * 21 / 20 ) = rs . 8820.00 answer : c" | a ) 3200.0 , b ) 11520.0 , c ) 8820.0 , d ) 7354.0 , e ) 16537.11 | c | multiply(power(add(divide(divide(10, const_2), const_100), const_1), const_2), 8000) | divide(n1,const_2)|divide(#0,const_100)|add(#1,const_1)|power(#2,const_2)|multiply(n0,#3)| | gain | C |
the grade point average of one third of the classroom is 60 ; the grade point average of the rest is 66 . what is the grade point average of the whole class ? | "let n = total students in class total points for 1 / 3 class = 60 n / 3 = 20 n total points for 2 / 3 class = 66 * 2 n / 3 = 44 n total points for whole class = 20 n + 44 n = 64 n 64 n total class points / n total students = 64 grade point average for total class answer : c" | a ) 55 , b ) 56 , c ) 64 , d ) 65 , e ) 66 | c | add(multiply(divide(66, const_3), const_2), divide(60, const_3)) | divide(n0,const_3)|divide(n1,const_3)|multiply(#1,const_2)|add(#0,#2)| | general | C |
two unbiased coins are tossed . what is probability of getting at most one tail ? | "explanation : total 4 cases = [ hh , tt , th , ht ] favourable cases = [ hh , th , ht ] please note we need atmost one tail , not atleast one tail . so probability = 3 / 4 answer : d" | a ) 12 , b ) 13 , c ) 32 , d ) 34 , e ) none of these | d | negate_prob(divide(const_1, power(const_2, const_3))) | power(const_2,const_3)|divide(const_1,#0)|negate_prob(#1)| | probability | D |
a number exceeds by 35 from its 3 / 8 part . then the number is ? | "x – 3 / 8 x = 35 x = 56 answer : a" | a ) a ) 56 , b ) b ) 35 , c ) c ) 39 , d ) d ) 40 , e ) e ) 45 | a | divide(multiply(35, 8), subtract(8, 3)) | multiply(n0,n2)|subtract(n2,n1)|divide(#0,#1)| | general | A |
the area of a rectangular plot is 24 times its breadth . if the difference between the length and the breadth is 10 metres , what is its breadth ? | l × b = 24 × b ∴ l = 24 m and l – b = 10 ∴ b = 24 – 10 = 14 m answer a | ['a ) 14 metres', 'b ) 5 metres', 'c ) 7.5 metres', 'd ) data inadequate', 'e ) none of these'] | a | subtract(24, 10) | subtract(n0,n1) | geometry | A |
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