Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
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in an election a candidate who gets 60 % of the votes is elected by a majority of 1040 votes . what is the total number of votes polled ? | "let the total number of votes polled be x then , votes polled by other candidate = ( 100 - 60 ) % of x = 40 % of x 60 % of x - 40 % of x = 1040 20 x / 100 = 1040 x = 1040 * 100 / 20 = 5200 answer is b" | a ) a ) 4500 , b ) b ) 5200 , c ) c ) 6900 , d ) d ) 7520 , e ) e ) 6000 | b | divide(1040, divide(subtract(60, subtract(const_100, 60)), const_100)) | subtract(const_100,n0)|subtract(n0,#0)|divide(#1,const_100)|divide(n1,#2)| | gain |
a person is traveling at 40 km / hr and reached his destiny in 6 hr then find the distance ? | "t = 6 hrs d = t * s = 40 * 6 = 240 km answer is b" | a ) 260 km , b ) 240 km , c ) 280 km , d ) 340 km , e ) 350 km | b | multiply(40, 6) | multiply(n0,n1)| | physics |
a certain library assesses fines for overdue books as follows . on the first day that a book is overdue , the total fine is $ 0.07 . for each additional day that the book is overdue , the total fine is either increased by $ 0.30 or doubled , whichever results in the lesser amount . what is the total for a book on the f... | "1 st day fine - 0.07 2 nd day fine - 0.07 * 2 = 0.14 ( as doubling gives lower value ) 3 rd day fine - 0.14 * 2 = 0.28 ( as doubling gives lower value ) 4 rd day fine - 0.28 * 2 = 0.56 ( as doubling gives lower value ) 5 th day fine - 0.56 + 0.3 = 0.86 ( as doubling gives higher value we add 0.3 this time ) answer : c... | a ) $ 0.60 , b ) $ 0.70 , c ) $ 0.86 , d ) $ 0.90 , e ) $ 1.00 | c | add(multiply(multiply(multiply(0.07, const_2), const_2), const_2), 0.30) | multiply(n0,const_2)|multiply(#0,const_2)|multiply(#1,const_2)|add(n1,#2)| | general |
7 , 26 , 63 , 124 , ( . . . . ) | "explanation : numbers are 2 ^ 3 - 1 = 7 3 ^ 3 - 1 = 26 4 ^ 3 - 1 = 63 5 ^ 3 - 1 = 124 6 ^ 3 - 1 = 342 answer : a" | a ) 342 , b ) 125 , c ) 289 , d ) 564 , e ) 36 | a | subtract(negate(124), multiply(subtract(26, 63), divide(subtract(26, 63), subtract(7, 26)))) | negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)| | general |
by selling 100 pens , a trader gains the cost of 40 pens . find his gain percentage ? | "let the cp of each pen be rs . 1 . cp of 100 pens = rs . 100 profit = cost of 40 pens = rs . 40 profit % = 40 / 100 * 100 = 40 % answer : a" | a ) 40 % , b ) 50 % , c ) 60 % , d ) 70 % , e ) 30 % | a | multiply(divide(40, 100), const_100) | divide(n1,n0)|multiply(#0,const_100)| | gain |
a sum of rs . 1870 has been divided among a , b and c such that a gets of what b gets and b gets of what c gets . b β s share is : | "explanation let c β s share = rs . x then , b β s share = rs . x / 4 , a β s share = rs . ( 2 / 3 x x / 4 ) = rs . x / 6 = x / 6 + x / 4 + x = 1870 = > 17 x / 12 = 1870 = > 1870 x 12 / 17 = rs . 1320 hence , b β s share = rs . ( 1320 / 4 ) = rs . 330 . answer d" | a ) rs . 120 , b ) rs . 160 , c ) rs . 240 , d ) rs . 330 , e ) none | d | subtract(subtract(multiply(divide(1870, const_10), const_2), const_12), const_12) | divide(n0,const_10)|multiply(#0,const_2)|subtract(#1,const_12)|subtract(#2,const_12)| | general |
a circular logo is enlarged to fit the lid of a jar . the new diameter is 30 per cent larger than the original . by what percentage has the area of the logo increased ? | "let old diameter be 4 , so radius is 2 old area = 4 Ο new diameter is 5.2 , so radius is 2.6 new area = 6.76 Ο increase in area is 2.76 Ο % increase in area = 2.76 / 4 * 100 so , % increase is 69 % answer will be ( c )" | a ) 50 , b ) 80 , c ) 69 , d ) 125 , e ) 250 | c | multiply(divide(subtract(power(add(divide(multiply(30, const_2), const_1000), const_3), const_2), const_4), const_4), const_100) | multiply(n0,const_2)|divide(#0,const_1000)|add(#1,const_3)|power(#2,const_2)|subtract(#3,const_4)|divide(#4,const_4)|multiply(#5,const_100)| | geometry |
a is a working partner and b is a sleeping partner in the business . a puts in rs . 15000 and b rs . 25000 , a receives 10 % of the profit for managing the business the rest being divided in proportion of their capitals . out of a total profit of rs . 9600 , money received by a is ? | "15 : 25 = > 3 : 5 9600 * 10 / 100 = 960 9600 - 960 = 8640 8640 * 3 / 8 = 3240 + 960 = 4200 answer : d" | a ) 1978 , b ) 2707 , c ) 7728 , d ) 4200 , e ) 7291 | d | subtract(9600, multiply(multiply(9600, subtract(const_1, divide(10, const_100))), divide(25000, add(15000, 25000)))) | add(n0,n1)|divide(n2,const_100)|divide(n1,#0)|subtract(const_1,#1)|multiply(n3,#3)|multiply(#2,#4)|subtract(n3,#5)| | gain |
how many numbers from 19 to 79 are exactly divisible by 11 ? | "19 / 11 = 1 and 79 / 11 = 7 = = > 7 - 1 = 6 numbers answer : b" | a ) 5 , b ) 6 , c ) 9 , d ) 11 , e ) 12 | b | add(divide(subtract(multiply(floor(divide(79, 11)), 11), multiply(add(floor(divide(19, 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 |
two pipes can fill a tank in 10 minutes and 15 minutes . an outlet pipe can empty the tank in 30 minutes . if all the pipes are opened when the tank is empty , then how many minutes will it take to fill the tank ? | "let v be the volume of the tank . the rate per minute at which the tank is filled is : v / 10 + v / 15 - v / 30 = 2 v / 15 per minute the tank will be filled in 15 / 2 = 7.5 minutes . the answer is b ." | a ) 7.0 , b ) 7.5 , c ) 8.0 , d ) 8.5 , e ) 9.0 | b | subtract(add(divide(const_1, 10), divide(const_1, 15)), divide(const_1, 30)) | divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|subtract(#3,#2)| | physics |
the length of a rectangle is four times its width . if the area is 100 m 2 what is the length of the rectangle ? | "let l be the length and w be the width of the rectangle . hence l = 4 w we now use the area to write 100 = l ? w substitute l by 4 w in the equation above 100 = 4 w ? w = 4 w 2 solve for w and find l 4 w 2 = 100 w 2 = 25 , w = 5 and l = 4 w = 20 m correct answer d" | a ) 50 m , b ) 40 m , c ) 80 m , d ) 20 m , e ) 70 m | d | rectangle_area(2, multiply(const_2.0, 2)) | multiply(const_2.0,n1)|rectangle_area(n1,#0)| | geometry |
exactly 20 % of the reporters for a certain wire service cover local politics in country x . if 20 % of the reporters who cover politics for the wire service do not cover local politics in country x , what percent of the reporters for the wire service do not cover politics ? | "you are correct . people who cover local politics are a subset of people who cover politics . 20 % of reporters who cover politics do not cover local politics so 80 % do cover local politics . reporters covering local politics = 80 % of reporters covering politics = 20 % ofall reporters reporters covering politics / a... | a ) 20 % , b ) 42 % , c ) 44 % , d ) 75 % , e ) 84 % | d | multiply(subtract(const_1, divide(20, subtract(const_100, 20))), const_100) | subtract(const_100,n1)|divide(n0,#0)|subtract(const_1,#1)|multiply(#2,const_100)| | gain |
we had $ 840 left after spending 30 % of the money that we took for shopping . how much money did we start with ? | let x be the amount of money we started with . 0.7 x = 840 x = 1200 the answer is e . | a ) $ 1000 , b ) $ 1050 , c ) $ 1100 , d ) $ 1150 , e ) $ 1200 | e | divide(840, subtract(const_1, divide(30, const_100))) | divide(n1,const_100)|subtract(const_1,#0)|divide(n0,#1) | gain |
if the sum and difference of two numbers are 10 and 8 respectively , then the difference of their square is : | "let the numbers be x and y . then , x + y = 10 and x - y = 8 x 2 - y 2 = ( x + y ) ( x - y ) = 10 * 8 = 80 . answer : e" | a ) 12 , b ) 28 , c ) 160 , d ) 180 , e ) 80 | e | subtract(power(divide(add(10, 8), const_2), const_2), power(subtract(10, divide(add(10, 8), const_2)), const_2)) | add(n0,n1)|divide(#0,const_2)|power(#1,const_2)|subtract(n0,#1)|power(#3,const_2)|subtract(#2,#4)| | general |
how many times will the digit 5 be written when listing the integers from 1 to 1000 ? | "many approaches are possible . for example : consider numbers from 0 to 999 written as follows : 1 . 000 2 . 001 3 . 002 4 . 003 . . . . . . . . . 1000 . 999 we have 1000 numbers . we used 3 digits per number , hence used total of 3 * 1000 = 3000 digits . now , why should any digit have preferences over another ? we u... | a ) 300 , b ) 380 , c ) 180 , d ) 415 , e ) 264 | a | multiply(multiply(multiply(1, const_10), const_10), const_3) | multiply(n1,const_10)|multiply(#0,const_10)|multiply(#1,const_3)| | general |
the average of first six prime numbers which are between 30 and 70 is | "explanation : first six prime numbers which are between 30 and 70 = 31 , 37 , 41 , 43 , 47 , 53 average = ( 31 + 37 + 41 + 43 + 47 + 53 ) / 6 = 42 answer : b" | a ) 35.4 , b ) 42 , c ) 45.7 , d ) 57 , e ) 67 | b | add(30, const_1) | add(n0,const_1)| | general |
what is the units digit of ( 3 ^ 11 ) ( 4 ^ 13 ) ? | "- > the ones place of ( ~ 3 ) ^ n repeats after 4 times like 3 ο 9 ο 7 ο 1 ο 3 . the ones place of ( ~ 4 ) ^ n repeats after 2 times like 4 ο 6 ο 4 . then , 3 ^ 13 = 3 ^ 4 * 3 + 1 ο 3 ^ 1 . , 4 ^ 13 = 4 ^ 2 * 6 + 1 = 4 ^ 1 = ~ 4 which is ( 3 ^ 13 ) ( 4 ^ 13 ) ο ( 3 ^ 1 ) ( ~ 4 ) = ( ~ 3 ) ( ~ 4 ) = ~ 2 . therefore , t... | a ) 2 , b ) 4 , c ) 6 , d ) 7 , e ) 8 | a | add(add(const_4, const_3), const_2) | add(const_3,const_4)|add(#0,const_2)| | general |
the distance from steve ' s house to work is 30 km . on the way back steve drives twice as fast as he did on the way to work . altogether , steve is spending 6 hours a day on the roads . what is steve ' s speed on the way back from work ? | "time is in the ratio 2 : 1 : : to : fro office therefore , 2 x + 1 x = 6 hrs time take to come back - 2 hrs , distance travelled - 30 km = > speed = 15 kmph answer : d" | a ) 5 . , b ) 10 . , c ) 14 . , d ) 15 , e ) 20 . | d | multiply(divide(add(30, divide(30, const_2)), 6), const_2) | divide(n0,const_2)|add(n0,#0)|divide(#1,n1)|multiply(#2,const_2)| | physics |
the sector of a circle has radius of 21 cm and central angle 270 o . find its perimeter ? | "perimeter of the sector = length of the arc + 2 ( radius ) = ( 270 / 360 * 2 * 22 / 7 * 21 ) + 2 ( 21 ) = 99 + 42 = 141 cm answer : d" | a ) 145 cm , b ) 135 cm , c ) 121 cm , d ) 141 cm , e ) 151 cm | d | multiply(multiply(const_2, divide(multiply(subtract(21, const_3), const_2), add(const_4, const_3))), 21) | add(const_3,const_4)|subtract(n0,const_3)|multiply(#1,const_2)|divide(#2,#0)|multiply(#3,const_2)|multiply(n0,#4)| | physics |
how many times in a day the hands of a clock are straight ? | "in 12 hours , the hands coincide or are in oppesite direction 22 times . in 24 hours , they are at right angles 22 times . answer is d ." | a ) 21 , b ) 27 , c ) 24 , d ) 22 , e ) 25 | d | divide(subtract(multiply(const_2, multiply(const_2, const_12)), const_4), const_2) | multiply(const_12,const_2)|multiply(#0,const_2)|subtract(#1,const_4)|divide(#2,const_2)| | physics |
on a ranch , a rancher can place a loop of rope , called a lasso , once in every 2 throws around a cow β s neck . what is the probability that the rancher will be able to place a lasso around a cow β s neck at least once in 3 attempts ? | "p ( missing all 3 ) = ( 1 / 2 ) ^ 3 = 1 / 8 p ( success on at least one attempt ) = 1 - 1 / 8 = 7 / 8 the answer is c ." | a ) 3 / 4 , b ) 5 / 8 , c ) 7 / 8 , d ) 9 / 16 , e ) 15 / 16 | c | subtract(const_1, power(divide(const_1, 2), 3)) | divide(const_1,n0)|power(#0,n1)|subtract(const_1,#1)| | probability |
twelve machines , each working at the same constant rate , together can complete a certain job in 12 days . how many additional machines , each working at the same constant rate , will be needed to complete the job in 8 days ? | another solution which is faster is since each machine works at a constant rate . the time needs to bought down from 12 to 8 . so the new time is 2 / 3 of the original time . thus to achieve this we need the rate to be 3 / 2 of original . so 3 / 2 * 12 = 18 so we need 18 - 12 = 6 more machines . answer : d | a ) 2 , b ) 3 , c ) 4 , d ) 6 , e ) 8 | d | subtract(multiply(12, inverse(divide(8, 12))), 12) | divide(n1,n0)|inverse(#0)|multiply(n0,#1)|subtract(#2,n0) | physics |
if 49 ( 7 ^ x ) = 1 then x = | "7 ^ x = 1 / 49 7 ^ x = 1 / 7 ^ 2 7 ^ x = 7 ^ - 2 x = - 2 a" | a ) β 2 , b ) β 1 , c ) 0 , d ) 1 , e ) 2 | a | divide(log(divide(1, 49)), log(7)) | divide(n2,n0)|log(n1)|log(#0)|divide(#2,#1)| | general |
a cab driver 5 days income was $ 200 , $ 150 , $ 750 , $ 400 , $ 500 . then his average income is ? | "avg = sum of observations / number of observations avg income = ( 200 + 150 + 750 + 400 + 500 ) / 5 = 400 answer is c" | a ) $ 420 , b ) $ 410 , c ) $ 400 , d ) $ 390 , e ) $ 380 | c | divide(add(add(add(add(200, 150), 750), 400), 500), 5) | add(n1,n2)|add(n3,#0)|add(n4,#1)|add(n5,#2)|divide(#3,n0)| | general |
x is able to do a piece of work in 14 days and y can do the same work in 20 days . if they can work together for 5 days , what is the fraction of work completed ? | explanation : amount of work x can do in 1 day = 1 / 14 amount of work y can do in 1 day = 1 / 20 amount of work x and y can do in 1 day = 1 / 14 + 1 / 20 = 17 / 140 amount of work x and y can together do in 5 days = 5 Γ ( 17 / 140 ) = 17 / 28 answer : option b | a ) 15 / 28 , b ) 17 / 28 , c ) 13 / 28 , d ) 19 / 28 , e ) 11 / 28 | b | divide(5, divide(multiply(14, 20), add(14, 20))) | add(n0,n1)|multiply(n0,n1)|divide(#1,#0)|divide(n2,#2) | physics |
a football field is 9600 square yards . if 1200 pounds of fertilizer are spread evenly across the entire field , how many pounds of fertilizer were spread over an area of the field totaling 5600 square yards ? | answer c ) 9600 yards need 1200 lbs 1 yard will need 1200 / 9600 = 1 / 8 lbs 3600 yards will need 1 / 8 * 5600 yards = 700 lbs | ['a ) 450', 'b ) 600', 'c ) 700', 'd ) 2400', 'e ) 3200'] | c | multiply(5600, divide(1200, 9600)) | divide(n1,n0)|multiply(n2,#0) | geometry |
if a man buys 1 liter of milk for 12 rs . and mixes it with 20 % water and sells it for 15 rs then what is the % age of gain . . . | quantity after adding 20 % water = 1.2 liter sp of 1.2 lit . @ 15 rs . / liter = 15 * 1.2 = 18 rs cp = 12 rs / lit . % profit = 100 * ( 18 - 12 ) / 12 = 50 answer : c | a ) 30 , b ) 40 , c ) 50 , d ) 60 , e ) 70 | c | multiply(divide(subtract(multiply(add(1, divide(20, const_100)), 15), 12), 12), const_100) | divide(n2,const_100)|add(n0,#0)|multiply(n3,#1)|subtract(#2,n1)|divide(#3,n1)|multiply(#4,const_100) | gain |
a tightrope approximately 320 m long is suspended between two poles . during a performance , a break occurs in the line . assume that the line has an equal chance of breaking anywhere along its length . what is the probability that the break happened in the first 50 meters of the rope ? | simly 50 / 320 = 5 / 32 answer will be ( c ) | a ) 27 / 32 , b ) 1 / 2 , c ) 5 / 32 , d ) 5 / 27 , e ) . 2 / 3 | c | divide(50, 320) | divide(n1,n0) | general |
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 18 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 Γ 10 % = 90 + 90 Γ 0.10 = 90 + 9 = 99 cal handled 2 / 3 rds of the transactions than anthony handled cal = 2 / 3 Γ 99 = 66 jade handled 18 more transactions than cal . jade = 66 + 18 = 84 jade handled = 84 tran... | a ) 80 , b ) 81 , c ) 82 , d ) 83 , e ) 84 | e | add(divide(multiply(multiply(divide(90, const_100), add(10, const_100)), 2), 3), 18) | add(n1,const_100)|divide(n0,const_100)|multiply(#0,#1)|multiply(n2,#2)|divide(#3,n3)|add(n4,#4)| | general |
while working alone at their constant rates , computer x can process 240 files in 4 hours , and computer y can process 240 files in 6 hours . if all files processed by these computers are the same size , how many hours would it take the two computers , working at the same time at their respective constant rates , to pr... | both computers together process files at a rate of 240 / 4 + 240 / 6 = 60 + 40 = 100 files per hour . the time required to process 240 files is 240 / 100 = 2.4 hours the answer is c . | a ) 2 , b ) 2.2 , c ) 2.4 , d ) 2.6 , e ) 2.8 | c | divide(240, add(divide(240, 4), divide(240, 6))) | divide(n0,n1)|divide(n0,n3)|add(#0,#1)|divide(n0,#2) | physics |
$ 2,000 is deposited in a savings account that pays 4 % annual interest compounded semiannually . to the nearest dollar , how much is in the account at the end of the year ? | "this is the case of semi - annual compoundingso , multiply time period by 2 and divide rate by 2 so , new time = 2 periods and new rate = 4 / 2 = 2 % now , ca = 2000 ( 1 + 2 / 100 ) ^ 2 = $ 2080.80 hence answer is e" | a ) $ 2125.15 , b ) $ 2083.12 , c ) $ 2184.22 , d ) $ 2150.84 , e ) $ 2080.80 | e | multiply(const_100, const_100) | multiply(const_100,const_100)| | gain |
a , b and c rent a pasture . if a puts 10 oxen for 7 months , b puts 12 oxen for 5 months and c puts 15 oxen for 3 months for grazing and the rent of the pasture is rs . 140 , then how much amount should c pay as his share of rent ? | "a : b : c = 10 Γ 7 : 12 Γ 5 : 15 Γ 3 = 2 Γ 7 : 12 Γ 1 : 3 Γ 3 = 14 : 12 : 9 amount that c should pay = 140 Γ 9 / 35 = 4 Γ 9 = 36 answer is d" | a ) 35 , b ) 45 , c ) 25 , d ) 36 , e ) 55 | d | multiply(140, divide(multiply(15, 3), add(add(multiply(10, 7), multiply(12, 5)), multiply(15, 3)))) | multiply(n4,n5)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|add(#3,#0)|divide(#0,#4)|multiply(n6,#5)| | general |
what is the rate percent when the simple interest on rs . 1000 amount to rs . 400 in 4 years ? | "interest for 4 yrs = 400 interest for 1 yr = 100 interest rate = 100 / 1000 x 100 = 10 % answer : e" | a ) 5 % , b ) 6 % , c ) 2 % , d ) 95 % , e ) 10 % | e | divide(multiply(const_100, 400), multiply(1000, 4)) | multiply(n1,const_100)|multiply(n0,n2)|divide(#0,#1)| | gain |
a cistern can be filled by a tap in 4 hours while it can be emptied by another tap in 7 hours . if both the taps are opened simultaneously , then after how much time will the cistern get filled ? | "net part filled in 1 hour = ( 1 / 4 - 1 / 7 ) = 3 / 28 the cistern will be filled in 28 / 3 hrs i . e . , 9.3 hrs . answer : c" | a ) 9.5 , b ) 9.4 , c ) 9.3 , d ) 9.2 , e ) 9.1 | c | divide(const_1, subtract(divide(const_1, 4), divide(const_1, 7))) | divide(const_1,n0)|divide(const_1,n1)|subtract(#0,#1)|divide(const_1,#2)| | physics |
by how much is 12 % of 24.2 more than 10 % of 14.2 ? | "answer required difference = ( 12 x 24.2 ) / 100 - ( 10 x 14.2 ) / 100 = 2.904 - 1.42 = 1.484 . correct option : c" | a ) 0.1484 , b ) 14.84 , c ) 1.484 , d ) 2.762 , e ) none | c | subtract(multiply(24.2, divide(12, const_100)), multiply(divide(10, const_100), 14.2)) | divide(n0,const_100)|divide(n2,const_100)|multiply(n1,#0)|multiply(n3,#1)|subtract(#2,#3)| | general |
the jogging track in a sports complex is 627 m in circumference . deepak and his wife start from the same point and walk in opposite directions at 4.5 km / hr and 3.75 km / hr respectively . they will meet for the first time in ? | "clearly , the two will meet when they are 627 m apart . to be ( 4.5 + 3.75 ) = 8.25 km apart , they take 1 hour . to be 627 m apart , they take ( 100 / 825 * 627 / 1000 ) hrs = ( 627 / 8250 * 60 ) min = 4.56 min . answer : a" | a ) 4.56 min , b ) 5.28 min , c ) 5.08 min , d ) 9.28 min , e ) 5.988 min | a | multiply(divide(divide(627, const_1000), add(4.5, 3.75)), const_60) | add(n1,n2)|divide(n0,const_1000)|divide(#1,#0)|multiply(#2,const_60)| | general |
a jogger running at 9 km / hr along side a railway track is 240 m ahead of the engine of a 120 m long train running at 27 km / hr in the same direction . in how much time will the train pass the jogger ? | "speed of train relative to jogger = 27 - 9 = 18 km / hr . = 18 * 5 / 18 = 5 m / sec . distance to be covered = 240 + 120 = 360 m . time taken = 360 / 5 = 72 sec . answer : b" | a ) 76 sec , b ) 72 sec , c ) 98 sec , d ) 36 sec , e ) 23 sec | b | divide(add(240, 120), multiply(subtract(27, 9), divide(divide(const_10, const_2), divide(subtract(27, 9), const_2)))) | add(n1,n2)|divide(const_10,const_2)|subtract(n3,n0)|divide(#2,const_2)|divide(#1,#3)|multiply(#4,#2)|divide(#0,#5)| | general |
a sun is divided among x , y and z in such a way that for each rupee x gets , y gets 45 paisa and z gets 50 paisa . if the share of y is rs . 36 , what is the total amount ? | "x : y : z = 100 : 45 : 50 20 : 9 : 10 9 - - - 36 39 - - - ? = > 156 answer : e" | a ) 115 , b ) 116 , c ) 117 , d ) 118 , e ) 156 | e | add(add(multiply(divide(const_100, 45), 36), multiply(divide(50, 45), 36)), 36) | divide(const_100,n0)|divide(n1,n0)|multiply(n2,#0)|multiply(n2,#1)|add(#2,#3)|add(n2,#4)| | general |
a certain company retirement plan has arule of 70 provision that allows an employee to retire when the employee ' s age plus years of employment with the company total at least 70 . in what year could a female employee hired in 1989 on her 32 nd birthday first be eligible to retire under this provision ? | "she must gain at least 70 points , now she has 32 and every year gives her two more points : one for age and one for additional year of employment , so 32 + 2 * ( # of years ) = 70 - - > ( # of years ) = 19 - - > 1989 + 19 = 2008 . answer : a ." | a ) 2008 , b ) 2004 , c ) 2005 , d ) 2006 , e ) 2007 | a | add(1989, divide(subtract(70, 32), const_2)) | subtract(n0,n3)|divide(#0,const_2)|add(n2,#1)| | general |
carina has 115 ounces of coffee divided into 5 - and 10 - ounce packages . if she has 2 more 5 - ounce packages than 10 - ounce packages , how many 10 - ounce packages does she have ? | "lets say 5 and 10 ounce packages be x and y respectively . given that , 5 x + 10 y = 115 and x = y + 2 . what is the value of y . substituting the x in first equation , 5 y + 10 + 10 y = 85 - > y = 105 / 15 . = 7 b" | a ) 5 , b ) 7 , c ) 6 , d ) 4 , e ) 3 | b | divide(subtract(115, multiply(5, 2)), add(10, 5)) | add(n1,n2)|multiply(n1,n3)|subtract(n0,#1)|divide(#2,#0)| | general |
by weight , liquid x makes up 1.5 percent of solution p and 6.5 percent of solution q . if 400 grams of solution p are mixed with 600 grams of solution q , then liquid x accounts for what percent of the weight of the resulting solution ? | "the number of grams of liquid x is 1.5 ( 400 ) / 100 + 6.5 ( 600 ) / 100 = 6 + 39 = 45 grams . 45 / 1000 = 4.5 % the answer is d ." | a ) 3.6 % , b ) 3.9 % , c ) 4.2 % , d ) 4.5 % , e ) 4.8 % | d | multiply(divide(add(const_1, divide(multiply(6.5, 600), const_100)), const_1000), const_100) | multiply(n1,n3)|divide(#0,const_100)|add(#1,const_1)|divide(#2,const_1000)|multiply(#3,const_100)| | gain |
a trader bought a car at 40 % discount on its original price . he sold it at a 80 % increase on the price he bought it . what percent of profit did he make on the original price ? | "original price = 100 cp = 60 s = 60 * ( 180 / 100 ) = 112 100 - 108 = 8 % answer : a" | a ) 8 % , b ) 62 % , c ) 2 % , d ) 9 % , e ) 22 % | a | multiply(subtract(divide(divide(multiply(subtract(const_100, 40), add(const_100, 80)), const_100), const_100), const_1), const_100) | add(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|divide(#3,const_100)|subtract(#4,const_1)|multiply(#5,const_100)| | gain |
what is the sum of 100 consecutive integers from - 49 inclusive , in a increasing order ? | "from - 49 to - 1 - - > 49 nos . zero - - > 1 number from + 1 to + 49 - - > 49 nos . when we add up nos . from - 49 to + 49 sum will be zero . total 99 nos will be added . 100 th number will be 50 . sum of these 100 nos . = 50 . b is the answer ." | a ) - 29 , b ) 50 , c ) - 30 , d ) 30 , e ) 60 | b | add(49, const_1) | add(n1,const_1)| | general |
a reduction of 40 % in the price of bananas would enable a man to obtain 64 more for rs . 40.00001 , what is reduced price per dozen ? | 40 * ( 40 / 100 ) = 16 - - - 64 ? - - - 12 = > rs . 3 answer : d | a ) 6 , b ) 5 , c ) 4 , d ) 3 , e ) 2 | d | multiply(const_12, divide(multiply(40, divide(40, const_100)), 64)) | divide(n0,const_100)|multiply(n0,#0)|divide(#1,n1)|multiply(#2,const_12) | gain |
an industrial loom weaves 0.128 metres of cloth every second . approximately , how many seconds will it take for the loom to weave 24 metre of cloth ? | "explanation : let the time required by x seconds . then , more cloth means more time ( direct proportion ) so , 0.128 : 1 : : 24 : x = > x = { \ color { blue } \ frac { 24 \ times 1 } { 0.128 } } = > x = 187.5 so time will be approx 188 seconds answer : d" | a ) 175 seconds , b ) 195 seconds , c ) 155 seconds , d ) 188 seconds , e ) 115 seconds | d | divide(24, 0.128) | divide(n1,n0)| | physics |
the perimeter of one face of a cube is 20 cm . its volume will be : | "explanation : edge of cude = 20 / 4 = 5 cm volume = a * a * a = 5 * 5 * 5 = 125 cm cube option a" | a ) 125 cm 3 , b ) 400 cm 3 , c ) 250 cm 3 , d ) 625 cm 3 , e ) none of these | a | volume_cube(square_edge_by_perimeter(20)) | square_edge_by_perimeter(n0)|volume_cube(#0)| | geometry |
if the average ( arithmetic mean ) of ( 2 a + 16 ) and ( 3 a - 8 ) is 84 , what is the value of a ? | "( ( 2 a + 16 ) + ( 3 a - 8 ) ) / 2 = ( 5 a + 8 ) / 2 = 84 a = 32 the answer is d ." | a ) 25 , b ) 30 , c ) 28 , d ) 32 , e ) 42 | d | divide(subtract(multiply(84, 2), subtract(16, 8)), add(2, 3)) | add(n0,n2)|multiply(n0,n4)|subtract(n1,n3)|subtract(#1,#2)|divide(#3,#0)| | general |
a square mirror has exactly half the area of the rectangular wall on which it is hung . if each side of the mirror is 18 inches and the width of the wall is 32 inches , what is the length of the wall , in inches ? | "since the mirror is 42 inches in all sides , it must be a square . area of a square is a = a ^ 2 ; 18 ^ 2 = 324 . area of rectangle is double of that 2 * 324 = 648 . now a = lw and we need find w so a / l = w ; 648 / 32 = 20.25 answer ! answer is d" | a ) 15.25 , b ) 18.25 , c ) 19.25 , d ) 20.25 , e ) 21.25 | d | divide(multiply(const_2, square_area(18)), 32) | square_area(n0)|multiply(#0,const_2)|divide(#1,n1)| | geometry |
a sun is divided among x , y and z in such a way that for each rupee x gets , y gets 50 paisa and z gets 40 paisa . if the share of x is rs . 30 , what is the total amount ? | "x : y : z = 100 : 50 : 40 10 : 5 : 4 10 - - - 30 19 - - - ? = > 57 answer : e" | a ) 67 , b ) 55 , c ) 62 , d ) 47 , e ) 57 | e | add(add(multiply(divide(const_100, 50), 30), multiply(divide(40, 50), 30)), 30) | divide(const_100,n0)|divide(n1,n0)|multiply(n2,#0)|multiply(n2,#1)|add(#2,#3)|add(n2,#4)| | general |
the product of two numbers is 276 and the sum of their squares is 289 . the sum of the number is ? | "let the numbers be x and y . then , xy = 276 and x 2 + y 2 = 289 . ( x + y ) 2 = x 2 + y 2 + 2 xy = 289 + ( 2 x 276 ) = 841 x + y = 29 . option a" | a ) a ) 29 , b ) b ) 25 , c ) c ) 27 , d ) d ) 31 , e ) e ) 35 | a | sqrt(add(power(sqrt(subtract(289, multiply(const_2, 276))), const_2), multiply(const_4, 276))) | multiply(n0,const_4)|multiply(n0,const_2)|subtract(n1,#1)|sqrt(#2)|power(#3,const_2)|add(#0,#4)|sqrt(#5)| | general |
how many kilograms of sugar costing rs . 9 per kg must be mixed with 27 kg of sugar costing rs . 7 per kg so that there may be gain of 10 % by selling the mixture at rs . 9.24 per kg ? | "explanation : let the rate of second quality be rs x per kg . step 1 : s . p of 1 kg of mixture = rs . 9.24 gain = 10 % c . p of 1 kg of mixture = [ 100 / ( 100 + 10 ) Γ 9.24 ] . = > rs . 8.40 . thus , the mean price = rs . 8.40 . step 2 : c . p of 1 kg of sugar of 1 st kind = 900 p c . p of 1 kg of sugar of 2 nd kind... | a ) 60 kg , b ) 63 kg , c ) 50 kg , d ) 77 kg , e ) none of these | b | divide(subtract(multiply(27, divide(9.24, add(divide(10, const_100), const_1))), multiply(27, 7)), subtract(9, divide(9.24, add(divide(10, const_100), const_1)))) | divide(n3,const_100)|multiply(n1,n2)|add(#0,const_1)|divide(n4,#2)|multiply(n1,#3)|subtract(n0,#3)|subtract(#4,#1)|divide(#6,#5)| | gain |
at a loading dock , each worker on the night crew loaded 1 / 4 as many boxes as each worker on the day crew . if the night crew has 4 / 5 as many workers as the day crew , what fraction of all the boxes loaded by the two crews did the day crew load ? | "method : x = no . of boxes loaded by day crew . boxes by night crew = 1 / 4 * 4 / 5 x = 1 / 5 x % loaded by day crew = x / ( x + 1 / 5 x ) = 5 / 6 answer e" | a ) 1 / 2 , b ) 2 / 5 , c ) 3 / 5 , d ) 4 / 5 , e ) 5 / 6 | e | divide(multiply(5, 4), add(multiply(5, 4), multiply(1, 4))) | multiply(n1,n3)|multiply(n0,n2)|add(#0,#1)|divide(#0,#2)| | physics |
a train speeds past a pole in 15 seconds and a platform 150 meters long in 25 seconds . what is the length of the train ( in meters ) ? | "let the length of the train be x meters . the speed of the train is x / 15 . then , x + 150 = 25 * ( x / 15 ) 10 x = 2250 x = 225 meters the answer is c ." | a ) 205 , b ) 215 , c ) 225 , d ) 235 , e ) 245 | c | multiply(150, subtract(const_2, const_1)) | subtract(const_2,const_1)|multiply(n1,#0)| | physics |
a jeep takes 7 hours to cover a distance of 420 km . how much should the speed in kmph be maintained to cover the same direction in 3 / 2 th of the previous time ? | "time = 7 distance = 420 3 / 2 of 7 hours = 7 * 3 / 2 = 10.5 hours required speed = 420 / 10.5 = 40 kmph c )" | a ) 48 kmph , b ) 52 kmph , c ) 40 kmph , d ) 63 kmph , e ) 65 kmph | c | divide(420, multiply(divide(3, 2), 7)) | divide(n2,n3)|multiply(n0,#0)|divide(n1,#1)| | physics |
the end of a blade on an airplane propeller is 5 feet from the center . if the propeller spins at the rate of 1,320 revolutions per second , how many miles will the tip of the blade travel in one minute ? ( 1 mile = 5,280 feet ) | "distance traveled in 1 revolution = 2 Ο r = 2 Ο 5 / 5280 revolutions in one second = 1320 revolutions in 60 seconds ( one minute ) = 1320 * 60 total distance traveled = total revolutions * distance traveled in one revolution 1320 * 60 * 2 Ο 5 / 5280 = 150 Ο e is the answer" | a ) 200 Ο , b ) 240 Ο , c ) 300 Ο , d ) 480 Ο , e ) 150 Ο | e | multiply(multiply(multiply(multiply(divide(5, add(multiply(const_2, const_100), multiply(add(const_2, const_3), const_1000))), const_2), divide(add(const_2, multiply(const_2, const_10)), add(const_3, const_4))), 1,320), const_60) | add(const_3,const_4)|add(const_2,const_3)|multiply(const_10,const_2)|multiply(const_100,const_2)|add(#2,const_2)|multiply(#1,const_1000)|add(#3,#5)|divide(#4,#0)|divide(n0,#6)|multiply(#8,const_2)|multiply(#7,#9)|multiply(n1,#10)|multiply(#11,const_60)| | physics |
what will be the cost of building a fence around a square plot with area equal to 144 sq ft , if the price per foot of building the fence is rs . 58 ? | "let the side of the square plot be a ft . a 2 = 144 = > a = 12 length of the fence = perimeter of the plot = 4 a = 48 ft . cost of building the fence = 48 * 58 = rs . 2784 . answer : a" | a ) 2784 , b ) 2287 , c ) 2977 , d ) 2668 , e ) 1298 | a | multiply(square_perimeter(sqrt(144)), 58) | sqrt(n0)|square_perimeter(#0)|multiply(n1,#1)| | geometry |
what is the 36 th digit to the right of the decimal point in the decimal expansion of 1 / 37 ? | "1 / 37 = 0.027027 . . . so , we have a repeating cycle of 027 . every third digit ( 3 rd , 6 th , 9 th , . . . ) to the right of the decimal point is 7 , thus 36 th digit to the right of the decimal point is also 7 . answer : d ." | a ) 0 , b ) 2 , c ) 4 , d ) 7 , e ) 9 | d | divide(1, 37) | divide(n1,n2)| | general |
if a ( a - 2 ) = 2 and b ( b - 2 ) = 2 , where a β b , then a + b = | "i . e . if a = - 1 then b = 2 or if a = 2 then b = - 1 but in each case a + b = - 1 + 2 = - 1 answer : option a" | a ) β 1 , b ) β 2 , c ) 4 , d ) 46 , e ) 48 | a | subtract(subtract(subtract(subtract(add(add(2, 2), subtract(2, 2)), const_1), const_1), const_1), const_1) | add(n0,n1)|subtract(n0,n1)|add(#0,#1)|subtract(#2,const_1)|subtract(#3,const_1)|subtract(#4,const_1)|subtract(#5,const_1)| | general |
if an object travels 400 feet in 4 seconds , what is the object ' s approximate speed in miles per hour ? ( note : 1 mile = 5280 feet ) | "1 mile = 5280 feet = > 1 feet = 1 / 5280 miles if the object travels 400 feet in 4 sec then it travels 400 / 4 * 60 * 60 feet in 1 hour ( 1 hr = 60 min * 60 sec ) = 3600 * 100 feet in 1 hour = 360000 feet in 1 hr = 360000 / 5280 miles in 1 hour = 36000 / 528 miles / hr ~ 68 miles / hr answer - a" | a ) 68 , b ) 54 , c ) 87 , d ) 96 , e ) 15 | a | divide(divide(400, 5280), multiply(4, divide(1, const_3600))) | divide(n0,n3)|divide(n2,const_3600)|multiply(n1,#1)|divide(#0,#2)| | physics |
if the sum and difference of two numbers are 10 and 19 respectively , then the difference of their square is : | let the numbers be x and y . then , x + y = 10 and x - y = 19 x 2 - y 2 = ( x + y ) ( x - y ) = 10 * 19 = 190 . answer : e | a ) 12 , b ) 28 , c ) 160 , d ) 180 , e ) 190 | e | subtract(power(divide(add(10, 19), const_2), const_2), power(subtract(10, divide(add(10, 19), const_2)), const_2)) | add(n0,n1)|divide(#0,const_2)|power(#1,const_2)|subtract(n0,#1)|power(#3,const_2)|subtract(#2,#4) | general |
if 14 percent of the students at a certain school went to a camping trip and took more than $ 100 , and 75 percent of the students who went to the camping trip did not take more than $ 100 , what percentage of the students at the school went to the camping trip ? | let x be the number of students in the school . 0.14 x students went to the trip and took more than 100 $ . they compose ( 100 - 75 ) = 25 % of all students who went to the trip . therefore the toal of 0.14 x / 0.25 = 0.56 x students went to the camping which is 56 % . the answer is e | a ) 95 , b ) 90 , c ) 85 , d ) 80 , e ) 56 | e | divide(14, divide(subtract(const_100, 75), const_100)) | subtract(const_100,n2)|divide(#0,const_100)|divide(n0,#1) | general |
cole drove from home to work at an average speed of 80 kmh . he then returned home at an average speed of 120 kmh . if the round trip took a total of 3 hours , how many minutes did it take cole to drive to work ? | "first round distance travelled ( say ) = d speed = 80 k / h time taken , t 2 = d / 80 hr second round distance traveled = d ( same distance ) speed = 120 k / h time taken , t 2 = d / 120 hr total time taken = 3 hrs therefore , 3 = d / 80 + d / 120 lcm of 80 and 120 = 240 3 = d / 80 + d / 120 = > 3 = 3 d / 240 + 2 d / ... | a ) 66 , b ) 70 , c ) 72 , d ) 85 , e ) 108 | e | multiply(divide(multiply(120, 3), add(80, 120)), const_60) | add(n0,n1)|multiply(n1,n2)|divide(#1,#0)|multiply(#2,const_60)| | physics |
the reciprocal of the hcf and lcm of two are 1 / 16 and 1 / 312 . if one of the number is 24 then other no . is | "reciprocal of the hcf and lcm of two are 1 / 16 and 1 / 312 so , hcf = 16 , lcm = 312 lcm * hcf = product of two numbers = a * b = > b = lcm * hcf / a so , other = 16 * 312 / 24 = 208 answer : d" | a ) 126 , b ) 136 , c ) 146 , d ) 208 , e ) 266 | d | divide(multiply(16, 312), 24) | multiply(n1,n3)|divide(#0,n4)| | physics |
360 metres long yard , 31 trees are palnted at equal distances , one tree being at each end of the yard . what is the distance between 2 consecutive trees | "31 trees have 30 gaps between them , required distance ( 360 / 30 ) = 12 b" | a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 17 | b | divide(360, add(subtract(31, 2), const_1)) | subtract(n1,n2)|add(#0,const_1)|divide(n0,#1)| | physics |
a test has 120 questions . each question has 5 options , but only 1 option is correct . if test - takers mark the correct option , they are awarded 1 point . however , if an answer is incorrectly marked , the test - taker loses 0.25 points . no points are awarded or deducted if a question is not attempted . a certain g... | "a correct answers get you 1 point , an incorrect answer gets you minus 1 / 4 point and a skipped question gets you 0 points . since there are 200 total questions , there are a variety of ways to get a total of 40 points . let c be the number of correct answers and let i be the number of incorrect answers . to get 40 p... | a ) 11 , b ) 13 , c ) 15 , d ) 17 , e ) 19 | d | add(add(divide(subtract(120, 40), 5), 1), 1) | subtract(n0,n5)|divide(#0,n1)|add(#1,n2)|add(#2,n2)| | general |
a = 2 ^ 15 - 625 ^ 3 and a / x is an integer , where x is a positive integer greater than 1 , such that it does not have a factor p such that 1 < p < x , then how many different values for x are possible ? | "this is a tricky worded question and i think the answer is should be d not c . . . here is my reason : the stem says that x is a positive integer such that has no factor grater than 2 and less than x itself . the stem wants to say that x is a prime number . because any prime number has no factor grater than 1 and itse... | a ) none , b ) one , c ) two , d ) three , e ) four | b | subtract(15, multiply(3, const_4)) | multiply(n3,const_4)|subtract(n1,#0)| | general |
a train running at the speed of 80 km / hr crosses a pole in 9 sec . what is the length of the train ? | "speed = 80 * 5 / 18 = 200 / 9 m / sec length of the train = speed * time = 200 / 9 * 9 = 200 m answer : e" | a ) 227 m , b ) 150 m , c ) 187 m , d ) 167 m , e ) 200 m | e | multiply(divide(multiply(80, const_1000), const_3600), 9) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics |
a train 180 m long is running with a speed of 60 km / hr . in what time will it pass a man who is running at 6 km / hr in the direction opposite to that in which the train is going ? | "speed of train relative to man = 60 + 6 = 66 km / hr . = 66 * 5 / 18 = 55 / 3 m / sec . time taken to pass the men = 180 * 3 / 55 = 10 sec . answer b" | a ) 7 , b ) 10 , c ) 8 , d ) 2 , e ) 4 | b | divide(180, multiply(add(60, 6), const_0_2778)) | add(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
what is the difference between 38 % of 80 nd 12 % of 160 . | "( 38 / 100 ) * 80 Γ’ β¬ β ( 12 / 100 ) * 160 30.4 - 19.2 = 11.2 answer : d" | a ) 11.3 , b ) 12.4 , c ) 12.7 , d ) 11.2 , e ) 10.3 | d | subtract(divide(multiply(38, 80), const_100), divide(multiply(12, 160), const_100)) | multiply(n0,n1)|multiply(n2,n3)|divide(#0,const_100)|divide(#1,const_100)|subtract(#2,#3)| | gain |
paul completes a piece of work in 80 days , rose completes the same work in 120 days . if both of them work together , then the number of days required to complete the work is ? | if a can complete a work in x days and b can complete the same work in y days , then , both of them together can complete the work in x y / x + y days . that is , the required no . of days = 80 Γ 120 / 200 = 48 days answer is d | a ) 42 , b ) 44 , c ) 46 , d ) 48 , e ) 50 | d | inverse(add(divide(const_1, 80), divide(const_1, 120))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|inverse(#2) | physics |
in a certain pond , 100 fish were caught , tagged , and returned to the pond . a few days later , 100 fish were caught again , of which 2 were found to have been tagged . if the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond , what is the approximate number of fish in the... | "total fish = x percentage of second catch = ( 2 / 100 ) * 100 = 2 % so , x * 2 % = 50 x = 2500 ans . d" | a ) 400 , b ) 625 , c ) 1,250 , d ) 2,500 , e ) 10,000 | d | divide(100, divide(2, 100)) | divide(n2,n1)|divide(n0,#0)| | gain |
the manufacturing cost of a shoe is rs . 210 and the transportation lost is rs . 500 for 100 shoes . what will be the selling price if it is sold at 20 % gains | "explanation : total cost of a watch = 210 + ( 500 / 100 ) = 215 . gain = 20 % = > sp = 1.2 cp = 1.2 x 215 = 258 answer : c" | a ) s 222 , b ) s 216 , c ) s 258 , d ) s 210 , e ) s 217 | c | add(add(210, divide(500, 100)), multiply(divide(20, 100), add(210, divide(500, 100)))) | divide(n1,n2)|divide(n3,n2)|add(n0,#0)|multiply(#2,#1)|add(#2,#3)| | gain |
a vessel of capacity 2 litre has 40 % of alcohol and another vessel of capacity 6 litre had 60 % alcohol . the total liquid of 8 litre was poured out in a vessel of capacity 10 litre and thus the rest part of the vessel was filled with the water . what is the new concentration of mixture ? | "40 % of 2 litres = 0.8 litres 60 % of 6 litres = 3.6 litres therefore , total quantity of alcohol is 4.4 litres . this mixture is in a 10 litre vessel . hence , the concentration of alcohol in this 10 litre vessel is 44 % b" | a ) 31 % . , b ) 44 % . , c ) 49 % . , d ) 29 % . , e ) 51 % . | b | multiply(divide(add(multiply(divide(40, const_100), 2), multiply(divide(60, const_100), 6)), 10), const_100) | divide(n1,const_100)|divide(n3,const_100)|multiply(n0,#0)|multiply(n2,#1)|add(#2,#3)|divide(#4,n5)|multiply(#5,const_100)| | general |
65 % of x = 20 % of 422.50 . find the value of x ? | "65 % of x = 20 % of 422.50 then , 65 / 100 * x = 20 / 100 * 4225 / 10 x = 845 / 10 * 100 / 65 = 130 answer is b" | a ) 100 , b ) 130 , c ) 150 , d ) 180 , e ) 199 | b | divide(multiply(multiply(divide(422.50, const_100), 20), const_100), 65) | divide(n2,const_100)|multiply(n1,#0)|multiply(#1,const_100)|divide(#2,n0)| | general |
on increasing the number of lines in a page by 80 , they become 240 . what is the % of increase in the no . of lines in the page ? | explanation : number of pages increased = 80 now , the number of pages of book = 240 number of pages of the books before increase = 240 β 80 = 160 % increase in the number of pages in the book = 80 160 x 100 % = 50 % c | a ) 20 % , b ) 305 , c ) 50 % , d ) 55 % , e ) 60 % | c | subtract(multiply(divide(240, subtract(240, 80)), const_100), const_100) | subtract(n1,n0)|divide(n1,#0)|multiply(#1,const_100)|subtract(#2,const_100) | general |
a batsman scored 120 runs whichincluded 3 boundaries and 8 sixes . what % of his total score did he make by running between the wickets | number of runs made by running = 110 - ( 3 x 4 + 8 x 6 ) = 120 - ( 60 ) = 60 now , we need to calculate 60 is what percent of 120 . = > 60 / 120 * 100 = 50 % b | a ) 40 % , b ) 50 % , c ) 55 % , d ) 65 % , e ) 70 % | b | multiply(divide(add(multiply(3, const_4), multiply(add(3, const_3), 8)), 120), const_100) | add(n1,const_3)|multiply(n1,const_4)|multiply(n2,#0)|add(#1,#2)|divide(#3,n0)|multiply(#4,const_100) | general |
in a group of 86 students , each student is registered for at least one of 3 classes β history , math and english . 20 - 8 students are registered for history , 20 - 3 students are registered for math , and forty - 4 students are registered for english . if only 3 students are registered for all 3 classes , how many st... | a u b u c = a + b + c - ab - bc - ac + abc 86 = 28 + 23 + 44 - ab - bc - ac + 3 = > ab + bc + ac = 12 exactly two classes = ab + bc + ac - 3 abc = 12 - 3 * 3 = 3 hence a | a ) 3 , b ) 10 , c ) 9 , d ) 8 , e ) 7 | a | subtract(subtract(add(add(add(20, 3), add(20, 8)), add(multiply(4, const_10), 4)), multiply(3, 3)), subtract(86, 3)) | add(n1,n2)|add(n2,n3)|multiply(n6,const_10)|multiply(n1,n1)|subtract(n0,n1)|add(#0,#1)|add(n6,#2)|add(#5,#6)|subtract(#7,#3)|subtract(#8,#4) | general |
a man can row his boat with the stream at 24 km / h and against the stream in 10 km / h . the man ' s rate is ? | "explanation : ds = 24 us = 10 s = ? s = ( 24 - 10 ) / 2 = 7 kmph answer : c" | a ) 1 kmph , b ) 6 kmph , c ) 7 kmph , d ) 4 kmph , e ) 9 kmph | c | divide(subtract(24, 10), const_2) | subtract(n0,n1)|divide(#0,const_2)| | gain |
there were totally 100 men . 82 are married . 75 have t . v , 85 have radio , 70 have a . c . how many men have t . v , radio , a . c and also married ? | "100 - ( 100 - 82 ) - ( 100 - 75 ) - ( 100 - 85 ) - ( 100 - 70 ) = 100 - 18 - 25 - 15 - 30 = 100 - 88 = 12 answer : b" | a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | b | subtract(100, add(add(add(subtract(100, 82), subtract(100, 75)), subtract(100, 85)), subtract(100, 70))) | subtract(n0,n1)|subtract(n0,n2)|subtract(n0,n3)|subtract(n0,n4)|add(#0,#1)|add(#4,#2)|add(#5,#3)|subtract(n0,#6)| | general |
the roof of an apartment building is rectangular and its length is 4 times longer than its width . if the area of the roof is 900 feet squared , what is the difference between the length and the width of the roof ? | let the width = x x * 4 x = 900 x ^ 2 = 225 x = 15 length = 4 * 15 = 60 difference = 60 - 15 = 45 e is the answer | ['a ) 38 .', 'b ) 40 .', 'c ) 42 .', 'd ) 44 .', 'e ) 45 .'] | e | subtract(multiply(sqrt(divide(900, 4)), const_4), sqrt(divide(900, 4))) | divide(n1,n0)|sqrt(#0)|multiply(#1,const_4)|subtract(#2,#1) | geometry |
sachin is younger than rahul by 18 years . if the ratio of their ages is 7 : 9 , find the age of sachin | "explanation : if rahul age is x , then sachin age is x - 18 , so , 9 x - 162 = 7 x 2 x = 162 x = 81 so sachin age is 81 - 18 = 63 answer : b ) 63" | a ) 24.5 , b ) 63 , c ) 65 , d ) 36 , e ) 24.19 | b | multiply(divide(18, subtract(9, 7)), 7) | subtract(n2,n1)|divide(n0,#0)|multiply(n1,#1)| | other |
average monthly income of a family of 4 earning members was rs . 735 . one of the earning members died and therefore , the average income came down to rs 590 . the income of the deceased was ? | "answer income of the deceased = total income of 4 members - total income of remaining 3 members . = 735 x 4 - 590 x 3 rs . = 1170 rs . correct option : c" | a ) rs . 692.80 , b ) rs . 820 , c ) rs . 1170 , d ) rs . 1385 , e ) none | c | subtract(multiply(735, 4), multiply(590, subtract(4, const_1))) | multiply(n0,n1)|subtract(n0,const_1)|multiply(n2,#1)|subtract(#0,#2)| | general |
two numbers are less than third number by 32 % and 37 % respectively . how much percent is the second number less than by the first | "let the third number is x . then first number = ( 100 - 32 ) % of x = 68 % of x second number is ( 63 x / 100 ) difference = 68 x / 100 - 63 x / 100 = x / 20 so required percentage is , difference is what percent of first number ( x / 20 * 100 / 68 x * 100 ) % = 14 % answer : e" | a ) 8 % , b ) 10 % , c ) 9 % , d ) 11 % , e ) 14 % | e | subtract(multiply(divide(subtract(37, 32), subtract(const_100, 32)), const_100), const_10) | subtract(n1,n0)|subtract(const_100,n0)|divide(#0,#1)|multiply(#2,const_100)|subtract(#3,const_10)| | gain |
peter invested a certain sum of money in a simple interest bond whose value grew to $ 100 at the end of 2 years and further to $ 200 at the end of another 4 years . what was the rate of interest in which he invested his sum ? | "explanatory answer initial amount invested = $ x amount at the end of year 2 = $ 100 amount at the end of year 6 ( another 4 years ) = $ 200 therefore , the interest earned for the 4 year period between the 2 rd year and 6 th year = $ 200 - $ 100 = $ 100 as the simple interest earned for a period of 4 years is $ 100 ,... | a ) 50 % , b ) 25 % , c ) 10 % , d ) 30 % , e ) 18 % | a | multiply(divide(divide(subtract(200, 100), 4), subtract(100, multiply(divide(subtract(200, 100), 4), 2))), const_100) | subtract(n2,n0)|divide(#0,n3)|multiply(n1,#1)|subtract(n0,#2)|divide(#1,#3)|multiply(#4,const_100)| | gain |
two nascar stock cars take off from the starting line at the exact same time , heading in opposite directions . the budweiser car travels at 145 miles per hour , while the stella artois car travels at 150 miles per hour . at this rate , and ignoring other variable , how long will the cars have to drive in order to be 5... | the cars travel ( 145 + 150 = 295 ) miles in one hour . 500 miles / 295 miles / hour = 1.69 hours answer is e | a ) 1.5 hours , b ) 1.27 hours , c ) 1.73 hours , d ) 2 hours , e ) 1.69 hours | e | divide(500, add(145, 150)) | add(n0,n1)|divide(n2,#0) | general |
at a loading dock , each worker on the night crew loaded 3 / 4 as many boxes as each worker on the day crew . if the night crew has 2 / 3 as many workers as the day crew , what fraction of all the boxes loaded by the two crews did the day crew load ? | "let x be the number of workers on the day crew . let y be the number of boxes loaded by each member of the day crew . then the number of boxes loaded by the day crew is xy . the number of boxes loaded by the night crew is ( 2 x / 3 ) ( 3 y / 4 ) = xy / 2 the total number of boxes is xy + xy / 2 = 3 xy / 2 the fraction... | a ) 1 / 2 , b ) 2 / 3 , c ) 3 / 5 , d ) 4 / 5 , e ) 5 / 8 | b | divide(multiply(3, 4), add(multiply(3, 4), multiply(3, 2))) | multiply(n1,n3)|multiply(n0,n2)|add(#0,#1)|divide(#0,#2)| | physics |
convert 400 miles into meters ? | "1 mile = 1609.34 meters 400 mile = 400 * 1609.34 = 643736 meters answer is e" | a ) 784596 , b ) 845796 , c ) 804670 , d ) 784596 , e ) 643736 | e | divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 400), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2))) | add(const_2,const_3)|multiply(#0,const_2)|multiply(#1,#1)|multiply(n0,#2)|divide(#3,#2)| | physics |
find the average of all numbers between 6 and 36 which are divisible by 7 | explanation : average = ( 7 + 14 + 21 + 28 + 35 ) / 7 = 105 / 7 = 15 option b | a ) 20 , b ) 15 , c ) 25 , d ) 30 , e ) 35 | b | divide(add(add(add(multiply(7, const_3), add(7, multiply(7, const_2))), multiply(7, const_4)), multiply(add(const_4, const_1), 7)), 7) | add(const_1,const_4)|multiply(n2,const_2)|multiply(n2,const_3)|multiply(n2,const_4)|add(n2,#1)|multiply(n2,#0)|add(#4,#2)|add(#6,#3)|add(#7,#5)|divide(#8,n2) | general |
in how many ways 3 boys and 3 girls can be seated in a row so that they are alternate . | "solution : let the arrangement be , b g b g b g b 34 boys can be seated in 3 ! ways . girl can be seated in 3 ! ways . required number of ways , = 3 ! * 3 ! = 36 . answer : option c" | a ) 144 , b ) 288 , c ) 36 , d ) 256 , e ) none | c | multiply(factorial(3), factorial(3)) | factorial(n0)|factorial(n1)|multiply(#0,#1)| | probability |
if you multiply all the numbers on your mobile phone except 0 , what is the answer ? | "we have to multiply 1 to 9 to find the answer . therefore 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 = 362880 answer is b" | a ) 256485 , b ) 362880 , c ) 125425 , d ) 0 , e ) 125826 | b | multiply(const_1, subtract(const_1, const_1)) | subtract(const_1,const_1)|multiply(#0,const_1)| | general |
a straight line in the xy - plane has slope 4 . on this line the x - coordinate of the point is 50 and y - coordinate is 300 then what is the y intercept of the plane ? | eq of line = y = mx + c m = 4 x = 50 y = 50 * 4 + c , substitute y by 300 as given in question . 300 = 200 + c , c = 100 correct option is c | a ) 150 , b ) 110 , c ) 100 , d ) 200 , e ) 50 | c | subtract(300, multiply(50, 4)) | multiply(n0,n1)|subtract(n2,#0) | general |
what two - digit number is less than the sum of the square of its digits by 13 and exceeds their doubled product by 5 ? | let the digits be x and y . the number would be 10 x + y . we are given that 2 xy + 5 = 10 x + y = x ^ 2 y ^ 2 - 13 thus 2 xy + 5 = x ^ 2 + y ^ 2 - 13 x ^ 2 + y ^ 2 - 2 xy = 16 ( x - y ) ^ 2 = 16 ( x - y ) = 4 or - 4 substituting the values of ( x - y ) in the equation 2 xy + 5 = 10 x + y x comes out to be 1 or 9 . . .... | a ) 95 , b ) 99 , c ) 26 , d ) 73 , e ) none of the above | b | add(multiply(const_10, add(subtract(13, 5), const_1)), 5) | subtract(n0,n1)|add(#0,const_1)|multiply(#1,const_10)|add(n1,#2) | 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 q 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 |
a and b invests rs . 10000 each , a investing for 8 months and b investing for all the 12 months in the year . if the total profit at the end of the year is rs . 25000 , find their shares ? | "the ratio of their profits a : b = 8 : 12 = 2 : 3 share of a in the total profit = 2 / 5 * 25000 = rs . 10000 share of a in the total profit = 3 / 5 * 25000 = rs . 15000 answer : d" | a ) 22277 , b ) 26782 , c ) 22882 , d ) 15000 , e ) 28761 | d | divide(multiply(10000, const_1), const_3) | multiply(n0,const_1)|divide(#0,const_3)| | gain |
a river 3 m deep and 36 m wide is flowing at the rate of 2 kmph the amount of water that runs into the sea per minute is ? | "( 2000 * 3 * 36 ) / 60 = 3600 m 3 answer : c" | a ) 4500 , b ) 2678 , c ) 3600 , d ) 3400 , e ) 2500 | c | divide(multiply(multiply(3, 36), multiply(2, const_1000)), multiply(const_1, const_60)) | multiply(n0,n1)|multiply(n2,const_1000)|multiply(const_1,const_60)|multiply(#0,#1)|divide(#3,#2)| | physics |
a train traveling at 100 kmph overtakes a motorbike traveling at 64 kmph in 20 seconds . what is the length of the train in meters ? | "train overtakes a bike means that we are talking about total length of the train . ( train ' s head is close to bike when it started and its tail crosses the bike when it overtakes the bike ) relative speed = 100 - 64 = 36 km / h = 36000 m / h time = 20 seconds distance = speed * time 36000 * 20 / 3600 = 200 meters . ... | a ) 400 meters , b ) 1111 meters , c ) 1777 meters , d ) 200 meters , e ) none of these | d | multiply(multiply(subtract(100, 64), const_0_2778), 20) | subtract(n0,n1)|multiply(#0,const_0_2778)|multiply(n2,#1)| | physics |
how many seconds will a 900 m 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 ? | "speed of train relative to man = 63 - 3 = 60 km / hr . = 60 * 5 / 18 = 50 / 3 m / sec . time taken to pass the man = 900 * 3 / 50 = 54 sec . answer : d" | a ) 23 sec , b ) 30 sec , c ) 49 sec , d ) 54 sec , e ) 59 sec | d | divide(900, multiply(subtract(63, 3), const_0_2778)) | subtract(n2,n1)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
two airplanes take off from one airfield at noon . one flies due east at 201 miles per hour while the other flies directly northeast at 283 miles per hour . approximately how many miles apart are the airplanes at 2 p . m . ? | "d in two hours : the plane flying east will be 402 miles away from airport . the other plane will be 566 miles away from airport . 566 / 402 = ~ 1.4 = ~ sqrt ( 2 ) this means that planes formed a right isocheles triangle = > sides of such triangles relate as 1 : 1 : sqrt ( 2 ) = > the planes are 402 miles apart . d" | a ) 166 , b ) 332 , c ) 400 , d ) 402 , e ) 566 | d | sqrt(subtract(power(multiply(283, 2), 2), power(multiply(201, 2), 2))) | multiply(n1,n2)|multiply(n0,n2)|power(#0,n2)|power(#1,n2)|subtract(#2,#3)|sqrt(#4)| | physics |
what is the sum of all the multiples of 10 between 0 and 120 ? | "the multiples of 10 between 0 and 120 are 10 , 20 , 30 , 40 , 50 , 60 , 70 , 80 , 90 , 100 , 110 and 120 . if these are all added together , the result is 780 . final answer : e" | a ) 500 , b ) 620 , c ) 450 , d ) 340 , e ) 780 | e | add(add(add(add(add(add(const_12, const_2), const_1), add(add(const_12, const_2), add(add(add(add(add(const_2, const_4), const_4), subtract(const_10, const_1)), add(add(const_2, const_4), const_4)), add(const_10, const_2)))), add(add(add(const_12, const_2), const_1), const_1)), 10), add(const_2, const_4)) | add(const_12,const_2)|add(const_2,const_4)|add(const_10,const_2)|subtract(const_10,const_1)|add(#0,const_1)|add(#1,const_4)|add(#5,#3)|add(#4,const_1)|add(#6,#5)|add(#8,#2)|add(#0,#9)|add(#4,#10)|add(#11,#7)|add(n0,#12)|add(#13,#1)| | general |
if money is invested at r percent interest , compounded annually , the amount of the investment will double in approximately 50 / r years . if luke ' s parents invested $ 11,500 in a long term bond that pays 12 percent interest compounded annually , what will be the approximate total amount of the investment 12 years l... | answer equals c in 48 years . i thought by 50 th year it would reach 95,500 . options should have been separated more widely for clarity . | a ) 62,000 , b ) 85,500 , c ) 95,500 , d ) 100,500 , e ) 100,000 | c | divide(multiply(multiply(add(const_2, const_3), const_1000), 12), const_2) | add(const_2,const_3)|multiply(#0,const_1000)|multiply(n2,#1)|divide(#2,const_2)| | general |
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