Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
values |
|---|---|---|---|---|---|---|
in a certain pond , 30 fish were caught , tagged , and returned to the pond . a few days later , 30 fish were caught again , of which 2 were found to have been tagged . if the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond , what ` s the approximate number of fish in the ... | "if x is total number of fish in the pond : 4 = 30 / x * 100 = > x = 750 so answer is b" | a ) 400 , b ) 750 , c ) 1250 , d ) 2500 , e ) 10 000 | b | divide(30, divide(2, 30)) | divide(n2,n1)|divide(n0,#0)| | gain |
a 300 m long train crosses a platform in 39 sec while it crosses a signal pole in 9 sec . what is the length of the platform ? | speed = 300 / 9 = 100 / 3 m / sec . let the length of the platform be x meters . then , ( x + 300 ) / 39 = 100 / 3 = > x = 1000 m . answer : e | a ) 389 m , b ) 350 m , c ) 289 m , d ) 799 m , e ) 1000 m | e | subtract(multiply(speed(300, 9), 39), 300) | speed(n0,n2)|multiply(n1,#0)|subtract(#1,n0) | physics |
the largest 4 digit number exactly divisible by 88 is : | "largest 4 - digit number = 9999 88 ) 9999 ( 113 88 - - - - 119 88 - - - - 319 264 - - - 55 - - - required number = ( 9999 - 55 ) = 9944 . answer : a" | a ) 9944 , b ) 9955 , c ) 9966 , d ) 6677 , e ) 9958 | a | square_area(const_pi) | square_area(const_pi)| | general |
a and b β s salaries together amount to rs . 4,000 . a spends 95 % of his salary and b spends 85 % of his . if now their savings are the same , what is a β s salary ? | ( 5 / 100 ) a = ( 15 / 100 ) b a = 3 b a + b = 4000 4 b = 4000 = > b = 1000 a = 3000 answer c | a ) 2777 , b ) 1500 , c ) 3000 , d ) 2776 , e ) 2881 | c | divide(multiply(multiply(multiply(const_2, multiply(const_4, add(const_2, const_3))), const_100), subtract(const_1, divide(85, const_100))), add(subtract(const_1, divide(95, const_100)), subtract(const_1, divide(85, const_100)))) | add(const_2,const_3)|divide(n2,const_100)|divide(n1,const_100)|multiply(#0,const_4)|subtract(const_1,#1)|subtract(const_1,#2)|add(#5,#4)|multiply(#3,const_2)|multiply(#7,const_100)|multiply(#8,#4)|divide(#9,#6) | gain |
the telephone company wants to add an area code composed of 2 letters to every phone number . in order to do so , the company chose a special sign language containing 324 different signs . if the company used 322 of the signs fully and two remained unused , how many additional area codes can be created if the company u... | "# of 2 - letter codes possible from 324 different signs = 324 * 324 . # of 2 - letter codes possible from 322 different signs = 322 * 322 . the difference = 324 ^ 2 - 322 ^ 2 = ( 324 - 322 ) ( 324 + 322 ) = 1292 . answer : c ." | a ) 246 , b ) 248 , c ) 1292 , d ) 15,128 , e ) 30,256 | c | multiply(multiply(322, subtract(324, 322)), subtract(324, 322)) | subtract(n1,n2)|multiply(n2,#0)|multiply(#1,#0)| | general |
the toll t , in dollars , for a truck using a certain bridge is given by the formula t = 3.50 + 0.50 ( x β 2 ) , where x is the number of axles on the truck . what is the toll for an 18 - wheel truck that has 2 wheels on its front axle and 4 wheels on each of its other axles ? | number of wheels in truck = 18 number of wheels on its front axle = 2 number of wheels remaining = 16 number of axles remaining axles = 16 / 4 = 4 total number of axles = 5 t = 3.50 + 0.50 ( x β 2 ) = 3.50 + . 5 * 3 = 3.5 + 1.5 = 5 $ answer e | a ) $ 2.50 , b ) $ 3.00 , c ) $ 3.50 , d ) $ 4.00 , e ) $ 5.00 | e | add(3.5, multiply(0.5, subtract(add(divide(subtract(18, 2), 4), const_1), 2))) | subtract(n3,n2)|divide(#0,n5)|add(#1,const_1)|subtract(#2,n2)|multiply(n1,#3)|add(n0,#4) | general |
a train of length l is traveling at a constant velocity and passes a pole in t seconds . if the same train travelling at the same velocity passes a platform in 4 t seconds , then what is the length of the platform ? | "the train passes a pole in t seconds , so velocity v = l / t ( l + p ) / v = 4 t ( l + p ) / ( l / t ) = 4 t p = 3 l the answer is e ." | a ) 0.5 l , b ) l , c ) 1.5 l , d ) 2 l , e ) 3 l | e | subtract(4, const_1) | subtract(n0,const_1)| | physics |
a 45 Β° - 45 Β° - 90 Β° right triangle has hypotenuse of length h . what is the area of the triangle e in terms of h ? | if . . . each of the two shorter sides = 3 , then the hypotenuse = h = 3 ( root 2 ) . the area e = ( 1 / 2 ) ( base ) ( height ) = ( 1 / 2 ) ( 3 ) ( 3 ) = 9 / 2 . so we ' re looking for an answer that = 9 / 2 when h = 3 ( root 2 ) . there ' s only one answer that matches . . . e | ['a ) h / β 2', 'b ) h / 2', 'c ) h / 4', 'd ) ( h ) ^ 2', 'e ) ( h ) ^ 2 / 4'] | e | triangle_perimeter(45, 45, 90) | triangle_perimeter(n0,n0,n2) | geometry |
the length of a rectangular plot is thrice its width . if the area of the rectangular plot is 675 sq meters , then what is the width ( in meters ) of the rectangular plot ? | "area = l * w = 3 w ^ 2 = 675 w ^ 2 = 225 w = 15 the answer is b ." | a ) 12 , b ) 15 , c ) 18 , d ) 21 , e ) 24 | b | divide(divide(divide(675, const_3), const_3), const_4) | divide(n0,const_3)|divide(#0,const_3)|divide(#1,const_4)| | geometry |
in feb mrs bil ' s earnings were 70 percent of the bil family ' s total income . in june mrs bil earned 10 percent more than in may . if the rest of the family ' s income was the same both months , then , in june , mrs bil ' s earnings were approximately what percent of the bil family ' s total income ? | lets say the family income is 100 in may , bil earned 70 family income is 30 in june , bil earned 10 % more than may , so it is ( 70 + 10 * 70 / 100 = 77 ) family income is same 30 in june bil ' s income percent is 77 * 100 / 107 ~ 72 ans is a | a ) 72 , b ) 34 , c ) 78 , d ) 37 , e ) 27 | a | multiply(divide(add(70, divide(70, 10)), add(const_100, divide(70, 10))), const_100) | divide(n0,n1)|add(n0,#0)|add(#0,const_100)|divide(#1,#2)|multiply(#3,const_100) | general |
find the number of zeroes at the end of 50 ! | "calculate division by 5 and 25 50 / 5 = 10 50 / 25 = 2 total = 12 answer : c" | a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | c | add(const_2, const_2) | add(const_2,const_2)| | other |
a box contains 20 electric bulbs , out of which 4 are defective . two bulbs are chosen at random form this box . the probability that at least one of these is defective is : | "p ( none is defective ) = 16 c 2 / 20 c 2 = 12 / 19 p ( at least one is defective ) = 1 - 12 / 19 = 7 / 19 answer b" | a ) 8 / 19 , b ) 7 / 19 , c ) 10 / 19 , d ) 20 / 19 , e ) 22 / 19 | b | subtract(const_1, divide(choose(subtract(20, 4), const_2), choose(20, const_2))) | choose(n0,const_2)|subtract(n0,n1)|choose(#1,const_2)|divide(#2,#0)|subtract(const_1,#3)| | probability |
john and amanda stand at opposite ends of a straight road and start running towards each other at the same moment . their rates are randomly selected in advance so that john runs at a constant rate of 2 , 3 , 4 , or 5 miles per hour and amanda runs at a constant rate of 3 , 4 , 5 , 6 , or 7 miles per hour . what is the... | john will run farther if he runs at 5 mph and amanda runs at 4 mph or 3 mph . in this case , p ( john runs farther ) = 1 / 4 * 2 / 5 = 2 / 20 john will run farther if he runs at 4 mph and amanda runs at 3 mph . in this case , p ( john runs farther ) = 1 / 4 * 1 / 5 = 1 / 20 p ( john runs farther ) = 2 / 20 + 1 / 20 = 3... | a ) 1 / 5 , b ) 2 / 5 , c ) 3 / 10 , d ) 3 / 20 , e ) 7 / 20 | d | divide(const_3, multiply(4, 5)) | multiply(n2,n3)|divide(const_3,#0) | physics |
john invests $ x at the semi - annual constant compounded rate of 2 percent and also does $ 40,000 at the quarterly constant compounded rate of 4 percent . if the interests are the same after 1 year , what is the value of x ? ? | a = p ( 1 + r / n ) ^ nt a = total amount accrued p = principal deposited r = rate of interest in decimal form n = number of times per year , interest compounded t = time in number of years . . x ( 1 + 0.02 / 2 ) ^ 2 - x = 40,000 ( 1 + 0.04 / 4 ) ^ 4 - 40,000 [ when the principal is subtracted from the total amount acc... | a ) 50000 , b ) 55000 , c ) 60000 , d ) 70000 , e ) 80804 | e | divide(subtract(multiply(multiply(multiply(const_4, const_100), const_100), power(add(const_1, divide(divide(4, const_100), const_4)), const_4)), multiply(multiply(const_4, const_100), const_100)), subtract(power(add(const_1, divide(divide(2, const_100), const_2)), const_2), const_1)) | divide(n2,const_100)|divide(n0,const_100)|multiply(const_100,const_4)|divide(#0,const_4)|divide(#1,const_2)|multiply(#2,const_100)|add(#3,const_1)|add(#4,const_1)|power(#6,const_4)|power(#7,const_2)|multiply(#5,#8)|subtract(#9,const_1)|subtract(#10,#5)|divide(#12,#11) | gain |
one side of a rectangle is 3 cm shorter than the other side . if we increase the length of each side by 1 cm , then the area of the rectangle will increase by 14 cm 2 . find the lengths of all sides . | "let x be the length of the longer side x > 3 , then the other side ' s length is x β 3 cm . then the area is s 1 = x ( x - 3 ) cm 2 . after we increase the lengths of the sides they will become ( x + 1 ) and ( x β 3 + 1 ) = ( x β 2 ) cm long . hence the area of the new rectangle will be a 2 = ( x + 1 ) β
( x β 2 ) cm ... | a ) 10 and 3 , b ) 7 and 10 , c ) 10 and 7 , d ) 3 and 10 , e ) 8 and 5 | e | subtract(add(divide(14, 2), 1), 3) | divide(n2,n3)|add(#0,n1)|subtract(#1,n0)| | geometry |
if two positive numbers are in the ratio 1 / 10 : 1 / 7 , then by what percent is the second number more than the first ? | "given ratio = 1 / 10 : 1 / 7 = 7 : 10 let first number be 7 x and the second number be 10 x . the second number is more than first number by 3 x . required percentage = 3 x / 7 x * 100 = 42.8 % . answer : b" | a ) 67 % . , b ) 42.8 % . , c ) 60 % . , d ) 68 % . , e ) 80 % . | b | multiply(divide(1, 7), const_100) | divide(n0,n3)|multiply(#0,const_100)| | general |
if 7125 ΒΈ 1.25 = 5700 < the value of 712.5 Γ· 12.5 is : | given 7125 / 1.25 = 5700 712.5 / 12.5 = 71.25 / 1.25 = 7125 * 1 / 1.25 * 100 = 5700 / 100 = 57 answer : b | a ) 5.7 , b ) 57 , c ) 570 , d ) 0.57 , e ) 0.057 | b | divide(712.5, 12.5) | divide(n3,n4) | general |
how many of the positive factors of 33 are not factors of 45 ? | "factors of 33 - 1 , 3 , 11 , 33 , factors of 45 - 1 , 3 , 5 , 9 , 15 , 45 , comparing both , we have three factors of 33 which are not factors of 45 - 11 , 33 , the answer is c" | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | c | divide(45, 33) | divide(n1,n0)| | other |
find the principle on a certain sum of money at 5 % per annum for 6 years if the amount being rs . 1120 ? | 1120 = p [ 1 + ( 5 * 6 ) / 100 ] p = 862 answer : a | a ) 862 , b ) 1067 , c ) 1977 , d ) 1056 , e ) 1097 | a | divide(1120, add(const_1, divide(multiply(5, 6), const_100))) | multiply(n0,n1)|divide(#0,const_100)|add(#1,const_1)|divide(n2,#2) | gain |
a certain high school has 500 students . of these students , 40 are taking music , 20 are taking art , and 10 are taking both music and art . how many students are taking neither music nor art ? | we ' re given a series of facts to work with : 1 ) a certain high school has 500 students . 2 ) of these students : x are taking music , y are taking art , and z are taking both music and art . we ' re asked how many students are taking neither music nor art ? let ' s test x = 40 y = 20 z = 10 so , we have 40 students ... | a ) 430 , b ) 440 , c ) 450 , d ) 460 , e ) 470 | c | subtract(500, subtract(add(40, 20), 10)) | add(n1,n2)|subtract(#0,n3)|subtract(n0,#1) | other |
when 1 / 10 percent of 7,000 is subtracted from 1 / 10 of 7,000 , the difference is | we can break this problem into two parts : 1 ) what is 1 / 10 percent of 7,000 ? 2 ) what is 1 / 10 of 7,000 ? to calculate 1 / 10 percent of 7,000 we must first remember to divide 1 / 10 by 100 . so we have : ( 1 / 10 ) / ( 100 ) to divide a number by 100 means to multiply it by 1 / 100 , so we have : 1 / 10 x 1 / 100... | a ) 0 , b ) 50 , c ) 450 , d ) 693 , e ) 500 | d | subtract(multiply(multiply(add(const_3, const_4), const_1000), divide(1, 10)), multiply(divide(divide(1, 10), const_100), multiply(add(const_3, const_4), const_1000))) | add(const_3,const_4)|divide(n0,n1)|divide(#1,const_100)|multiply(#0,const_1000)|multiply(#1,#3)|multiply(#2,#3)|subtract(#4,#5) | general |
what sum of money will produce rs . 210 as simple interest in 4 years at 3 1 / 2 percent ? | 210 = ( p * 4 * 7 / 2 ) / 100 p = 1500 answer : b | a ) rs . 2500 , b ) rs . 1500 , c ) rs . 1400 , d ) rs . 1600 , e ) rs . 1300 | b | divide(210, divide(multiply(4, add(3, divide(1, 2))), const_100)) | divide(n3,n4)|add(n2,#0)|multiply(n1,#1)|divide(#2,const_100)|divide(n0,#3) | gain |
while driving from a - ville to b - town , harriet drove at a constant speed of 90 kilometers per hour . upon arriving in b - town , harriet immediately turned and drove back to a - ville at a constant speed of 160 kilometers per hour . if the entire trip took 5 hours , how many minutes did it take harriet to drive fro... | 5 hr = 300 min . if harriet spend equal hrs on each leg she will spend 150 min on each . since speed a - b is less than speed b - a and distance on each leg is the same , time spent on a - b is more than 150 min , which mean we can eliminate ans . a , b and c . now let plug in ans . d or e and verify which one give sam... | a ) 138 , b ) 148 , c ) 150 , d ) 162 , e ) 192 | e | multiply(subtract(5, divide(multiply(90, 5), add(90, 160))), const_60) | add(n0,n1)|multiply(n0,n2)|divide(#1,#0)|subtract(n2,#2)|multiply(#3,const_60) | physics |
the milk level in a rectangular box measuring 64 feet by 25 feet is to be lowered by 6 inches . how many gallons of milk must be removed ? ( 1 cu ft = 7.5 gallons ) | 6 inches = 1 / 2 feet ( there are 12 inches in a foot . ) , so 64 * 25 * 1 / 2 = 800 feet ^ 3 of milk must be removed , which equals to 800 * 7.5 = 6000 gallons . answer : e . | a ) 100 , b ) 250 , c ) 750 , d ) 5625 , e ) 6000 | e | multiply(multiply(multiply(64, 25), divide(1, const_2)), 7.5) | divide(n3,const_2)|multiply(n0,n1)|multiply(#0,#1)|multiply(n4,#2) | general |
a computer store offers employees a 20 % discount off the retail price . if the store purchased a computer from the manufacturer for $ 1000 dollars and marked up the price 20 % to the final retail price , how much would an employee save if he purchased the computer at the employee discount ( 20 % off retail price ) as ... | "cost price = 1000 profit = 20 % = 20 % of 1000 = 200 selling price = cp + profit sp = 1200 a discount of 20 % to employees means 20 % off on 1200 so 20 % of 1200 = 240 so a employee thinks he saved 240 . . answer : c" | a ) $ 1,000 , b ) $ 960 , c ) $ 240 , d ) $ 200 , e ) $ 150 | c | divide(add(divide(multiply(1000, 20), const_100), 1000), multiply(divide(1000, const_100), const_2)) | divide(n1,const_100)|multiply(n0,n1)|divide(#1,const_100)|multiply(#0,const_2)|add(n1,#2)|divide(#4,#3)| | gain |
i sold a book at a profit of 10 % . had i sold it for $ 90 more , 15 % would have been gained . find the cost price ? | "115 % of cost - 110 % of cost = $ 90 5 % of cost = $ 90 cost = 90 * 100 / 5 = $ 1800 answer is a" | a ) $ 1800 , b ) $ 2500 , c ) $ 3000 , d ) $ 3120 , e ) $ 1540 | a | divide(multiply(divide(multiply(90, 10), subtract(15, 10)), const_100), 10) | multiply(n0,n1)|subtract(n2,n0)|divide(#0,#1)|multiply(#2,const_100)|divide(#3,n0)| | gain |
reena took a loan of $ . 1200 with simple interest for as many years as the rate of interest . if she paid $ 972 as interest at the end of the loan period , what was the rate of interest ? | "let rate = r % and time = r years . then , 1200 x r x r / 100 = 972 12 r 2 = 972 r 2 = 81 r = 9 . answer : e" | a ) 3.6 , b ) 6 , c ) 18 , d ) can not be determined , e ) none of these | e | sqrt(divide(multiply(972, const_100), 1200)) | multiply(n1,const_100)|divide(#0,n0)|sqrt(#1)| | gain |
7 , 10 , 8 , 11 , 9 , 12 , ___ | "10 = 7 + 3 8 = 10 - 2 11 = 8 + 3 9 = 11 - 2 12 = 9 + 3 similarly 12 - 2 = 10 answer : a" | a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | a | subtract(negate(11), multiply(subtract(10, 8), divide(subtract(10, 8), subtract(7, 10)))) | negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)| | general |
how much more would rs . 20000 fetch , after two years , if it is put at 20 % p . a . compound interest payable half yearly than if is put at 20 % p . a . compound interest payable yearly ? | "20000 ( 11 / 10 ) 4 - 20000 ( 6 / 5 ) 2 = 6 , 602.50 answer : a" | a ) a ) rs . 6 , 602.50 , b ) b ) rs . 6 , 601.50 , c ) c ) rs . 6 , 603.50 , d ) d ) rs . 6 , 604.50 , e ) e ) rs . 6 , 605.50 | a | subtract(multiply(power(add(const_1, divide(divide(20, const_100), const_2)), const_4), 20000), multiply(power(add(divide(20, const_100), const_1), const_2), 20000)) | divide(n1,const_100)|add(#0,const_1)|divide(#0,const_2)|add(#2,const_1)|power(#1,const_2)|multiply(n0,#4)|power(#3,const_4)|multiply(n0,#6)|subtract(#7,#5)| | gain |
how much interest will $ 10,000 earn in 9 months at an annual rate of 10 % ? | "soln : - 9 months = 3 / 4 of year ; 10 % = 10 / 100 = 1 / 10 ; $ 10,000 ( principal ) * 3 / 50 ( interest rate ) * 1 / 10 ( time ) = $ 60 . answer : d" | a ) $ 250 , b ) $ 350 , c ) $ 450 , d ) $ 60 , e ) $ 650 | d | multiply(multiply(power(const_100, const_2), divide(10, const_100)), divide(const_3, 10)) | divide(const_3,n2)|divide(n2,const_100)|power(const_100,const_2)|multiply(#1,#2)|multiply(#0,#3)| | gain |
a new home buyer pays 4 % annual interest on her first mortgage and 9 % annual interest on her second mortgage . if she borrowed a total of $ 320,000 , 80 % of which was in the first mortgage , what is her approximate monthly interest payment ? | "making the numbers a little easy looking : monthly payment = 320,000 * ( 80 / 100 * 4 / 100 + 20 / 100 * 9 / 100 ) / 12 = 3200 / 12 * ( 3.2 + 1.8 ) = 16000 / 12 = 1333 answer : a" | a ) $ 1,333 , b ) $ 1,733 , c ) $ 3,466 , d ) $ 13,333 , e ) $ 20,796 | a | subtract(9, multiply(4, const_2)) | multiply(n0,const_2)|subtract(n1,#0)| | general |
a dog takes 5 leaps for every 8 leaps of a hare . if one leap of the dog is equal to 4 leaps of the hare , the ratio of the speed of the dog to that of the hare is : | "explanation : dog : hare = ( 5 * 4 ) leaps of hare : 8 leaps of hare = 20 : 8 - - > 10 : 4 - - > 5 : 2 answer : c" | a ) 5 : 8 , b ) 5 : 3 , c ) 5 : 2 , d ) 5 : 9 , e ) 2 : 5 | c | divide(multiply(5, 4), 8) | multiply(n0,n2)|divide(#0,n1)| | other |
by selling 200 apples , a fruit - seller gains the selling price of 50 apples . find the gain percent ? | sp = cp + g 200 sp = 200 cp + 50 sp 150 sp = 200 cp 150 - - - 50 cp 100 - - - ? = > 33.33 % answer : a | a ) 33.33 % , b ) 35 % , c ) 27.75 % , d ) 35.75 % , e ) 32.25 % | a | multiply(divide(50, subtract(200, 50)), const_100) | subtract(n0,n1)|divide(n1,#0)|multiply(#1,const_100) | gain |
a , b and c can do a piece of work in 24 days , 30 days and 40 days respectively . they began the work together but c left 8 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 / 40 = 1 / 10 work done by a and b together in the last 8 days = 8 * ( 1 / 24 + 1 / 30 ) = 3 / 5 remaining work = 2 / 5 the number of days required for this initial work = 4 days . the total number of days required = 8 + 4 = 12 days . answer : a" | a ) 12 days , b ) 16 days , c ) 18 days , d ) 11 days , e ) 38 days | a | add(divide(subtract(const_1, multiply(add(divide(const_1, 24), divide(const_1, 30)), 8)), add(divide(const_1, multiply(add(const_2, const_3), multiply(const_2, 8))), add(divide(const_1, 24), divide(const_1, 30)))), 8) | add(const_2,const_3)|divide(const_1,n0)|divide(const_1,n1)|multiply(const_2,n3)|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 |
a man whose bowling average is 12.4 , takes 4 wickets for 26 runs and there by decreases his average by 0.4 . the number of wickets taken by him before his last match is ? | "12.4 * x + 26 = ( 4 + x ) 12 solve equation x = 55 answer : c" | a ) 53 , b ) 54 , c ) 55 , d ) 56 , e ) 57 | c | divide(subtract(multiply(floor(12.4), 4), 26), subtract(12.4, floor(12.4))) | floor(n0)|multiply(n1,#0)|subtract(n0,#0)|subtract(#1,n2)|divide(#3,#2)| | general |
an engine moves at the speed of 60 kmph without any coaches attached to it . speed of the train reduces at the rate that varies directly as the square root of the number of coaches attached . when 9 coaches are attached speed decreases to 48 kmph . what will be the speed of train when 25 coaches are attached . | 1 . no . of coaches = 9 sqr root = 3 speed decreases by 12 12 = k * 3 k = 4 no . of coaches = 25 swr root = 5 decrease = 5 * 4 = 20 new speed = 60 - 20 = 40 c | a ) 35 , b ) 28 , c ) 40 , d ) 44 , e ) 42 | c | subtract(60, multiply(sqrt(25), divide(subtract(60, 48), sqrt(9)))) | sqrt(n1)|sqrt(n3)|subtract(n0,n2)|divide(#2,#0)|multiply(#3,#1)|subtract(n0,#4) | physics |
a survey was sent to 80 customers , 12 of whom responded . then the survey was redesigned and sent to another 63 customers , 10 of whom responded . by approximately what percent did the response rate increase from the original survey to the redesigned survey ? | "rate of first survey = 12 / 80 rate of second survey = 10 / 63 % response rate increase ( 10 / 63 - 12 / 80 ) / ( 12 / 80 ) = 5 % answer is b = 5 %" | a ) 2 % , b ) 5 % , c ) 14 % , d ) 28 % , e ) 63 % | b | floor(multiply(subtract(divide(10, 63), divide(12, 80)), const_100)) | divide(n3,n2)|divide(n1,n0)|subtract(#0,#1)|multiply(#2,const_100)|floor(#3)| | gain |
a train covers a distance of 12 km in 10 min . if it takes 9 sec to pass a telegraph post , then the length of the train is ? | "speed = ( 12 / 10 * 60 ) km / hr = ( 72 * 5 / 18 ) m / sec = 20 m / sec . length of the train = 20 * 9 = 180 m . answer : e" | a ) 298 , b ) 288 , c ) 120 , d ) 776 , e ) 180 | e | divide(const_100.0, subtract(divide(const_100.0, 10), 9)) | divide(const_100.0,n1)|subtract(#0,n2)|divide(const_100.0,#1)| | physics |
in a mixture of 60 litres the ratio of milk to water is 5 : 1 . additional 20 litres of water is added to the mixture . find the ratio of milk to water in the resulting mixture . | "given that milk / water = 5 x / x and 5 x + x = 60 - - > x = 10 . thus milk = 5 x = 50 liters and water = x = 10 liters . new ratio = 50 / ( 20 + 10 ) = 50 / 30 = 5 / 3 . answer is b" | a ) 2 : 3 , b ) 5 : 3 , c ) 1 : 4 , d ) 3 : 2 , e ) 3 : 1 | b | divide(subtract(60, divide(60, add(5, 1))), add(divide(60, add(5, 1)), 20)) | add(n1,n2)|divide(n0,#0)|add(n3,#1)|subtract(n0,#1)|divide(#3,#2)| | general |
a train traveling at 72 kmph crosses a platform in 34 seconds and a man standing on the platform in 18 seconds . what is the length of the platform in meters ? | "speed of train = 72 * ( 5 / 18 ) = 20 m / s lets consider the man as a stationery point on the platform . crossing the point gives us the length of the train . lt = 20 * 18 = 360 m . crossing the platform gives us the length of trainlength of platform . l ( t + p ) = 20 * 34 = 680 m . so , length of platform = 680 - 3... | a ) 240 meters , b ) 320 meters , c ) 420 meters , d ) 600 meters , e ) can not be determined | b | subtract(multiply(divide(multiply(72, const_1000), const_3600), 34), multiply(divide(multiply(72, const_1000), const_3600), 18)) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)|multiply(n2,#1)|subtract(#2,#3)| | physics |
square abcd is the base of the cube while square efgh is the cube ' s top face such that point e is above point a , point f is above point b etc . what is the distance between the midpoint of edge ab and the midpoint of edge eh if the area of square abcd is 2 ? | distance from mid point of ab to ad = sqrt [ ( 1 / sqrt 2 ) ^ 2 + ( 1 / sqrt 2 ) ^ 2 ] = 1 the distance between the midpoint of edge ab and the midpoint of edge eh = sqrt [ 1 ^ 2 + ( sqrt 2 ) ^ 2 ] = sqrt 3 . answer : d | ['a ) 1 / sqrt 2', 'b ) 1', 'c ) sqrt 2', 'd ) sqrt 3', 'e ) 2 sqrt 3'] | d | sqrt(add(const_1, 2)) | add(n0,const_1)|sqrt(#0) | geometry |
in an ngo the daily average wages of 20 illiterate employees is decreased from rs . 25 to rs . 10 , thus the average salary of all the literate and illiterate employees is decreased rs . 10 per day . the no . of educated employees working in the ngo is : | explanation : total employees = \ inline \ frac { ( 25 - 10 ) \ times 20 } { 10 } = 30 hence number of educated employees = 30 - 20 = 10 answer : c ) 10 | a ) 12 , b ) 11 , c ) 10 , d ) 27 , e ) 22 | c | subtract(20, 10) | subtract(n0,n2) | general |
you have been given a physical balance and 7 weights of 52 , 50 , 48 , 44 , 45 , 46 and 78 kgs . keeping weights on one pan and object on the other , what is the maximum you can weigh less than 183 kgs . | "add the max weights 52 + 50 + 78 = 180 answer : a" | a ) 180 , b ) 181 , c ) 182 , d ) 178 , e ) 168 | a | add(add(add(50, 48), 44), 46) | add(n2,n3)|add(n4,#0)|add(n6,#1)| | general |
excluding stoppages , the speed of a train is 45 kmph and including stoppages it is 34 kmph . of how many minutes does the train stop per hour ? | "explanation : t = 11 / 45 * 60 = 14.6 answer : option d" | a ) e 982 , b ) 27 , c ) 12 , d ) 14.6 , e ) 28 | d | subtract(const_60, multiply(const_60, divide(34, 45))) | divide(n1,n0)|multiply(#0,const_60)|subtract(const_60,#1)| | physics |
a , b and c start a business each investing 20,000 . after 5 months a withdrew 5000 , b withdrew 4000 and c invests 6000 more . at the end of the year , a total profit of 69,900 was recorded . find the share of b | "ratio of the capitals of a , b and c = 20000 Γ 5 + 15000 Γ 7 : 20000 Γ 5 + 16000 Γ 7 : 20000 Γ 5 + 26000 Γ 7 = 205000 : 212000 : 282000 = 205 : 212 : 282 . b β s share = ( 69900 Γ 212 β 699 ) = 21200 ; answer c" | a ) 20,000 , b ) 21,000 , c ) 28,200 , d ) 20,500 , e ) none of these | c | divide(add(multiply(subtract(const_12, 5), 4000), multiply(multiply(const_2, multiply(const_100, const_100)), 5)), multiply(const_100, const_10)) | multiply(const_100,const_100)|multiply(const_10,const_100)|subtract(const_12,n1)|multiply(n3,#2)|multiply(#0,const_2)|multiply(n1,#4)|add(#3,#5)|divide(#6,#1)| | gain |
5 men or 8 women do equal amount of work in a day . a job requires 3 men and 5 women to finish the job in 10 days how many woman are required to finish the job in 14 days . | 1 man can do the work of 8 / 5 = 1.6 women equivalent of 3 * 1.6 + 5 = 9.8 women take 10 days so women needed to do the work in 14 days = [ 10 / 14 ] * 9.8 = 7 answer : b | a ) 10 , b ) 7 , c ) 6 , d ) 12 , e ) 13 | b | divide(multiply(add(multiply(divide(8, 5), 3), 5), 10), 14) | divide(n1,n0)|multiply(n2,#0)|add(n0,#1)|multiply(n4,#2)|divide(#3,n5) | physics |
ajay can walk 10 km in 1 hour . in how many hours he can walk 50 km ? | 1 hour he walk 10 km he walk 50 km in = 50 / 10 * 1 = 5 hours answer is a | a ) 5 hrs , b ) 10 hrs , c ) 15 hrs , d ) 20 hrs , e ) 30 hrs | a | divide(50, 10) | divide(n2,n0) | physics |
factory x ' s bulbs work for over 5000 hours in 75 % of cases , whereas factory y ' s bulbs work for over 5000 hours in 65 % of cases . it is known that factory x supplies 70 % of the total bulbs available . what is the chance that a purchased bulb will work for longer than 5000 hours ? | "for x , 70 % of 75 % will work . for y , 30 % of 65 % will work . * 30 % is the rest of the bulb supply in the market . so , the probability that a purchased bulb will work is : 0.70 ( 0.75 ) = . 525 0.30 ( 0.65 ) = 0.195 the combined probability then is 52.5 + 19.5 = 72 % ans d" | a ) 76.4 % , b ) 78 % , c ) 77.4 % , d ) 72 % , e ) 73.9 % | d | divide(add(75, 65), const_2) | add(n1,n3)|divide(#0,const_2)| | physics |
the radius of a wheel is 22.4 cm . what is the distance covered by the wheel in making 1500 resolutions ? | "in one resolution , the distance covered by the wheel is its own circumference . distance covered in 1500 resolutions . = 1500 * 2 * 22 / 7 * 22.4 = 211200 cm = 2112 m answer : a" | a ) 2112 m , b ) 2704 m , c ) 2278 m , d ) 2288 m , e ) 2378 m | a | divide(multiply(multiply(multiply(divide(add(multiply(add(const_3, const_4), const_3), const_1), add(const_3, const_4)), 22.4), const_2), 1500), const_100) | add(const_3,const_4)|multiply(#0,const_3)|add(#1,const_1)|divide(#2,#0)|multiply(n0,#3)|multiply(#4,const_2)|multiply(n1,#5)|divide(#6,const_100)| | physics |
a 300 meter long train running at the speed of 120 kmph crosses another train running in the opposite direction at the speed of 80 kmph in 9 seconds . what is the lenght of other train . | relative speeds = ( 120 + 80 ) km / hr = 200 km / hr = ( 200 * 5 / 18 ) m / s = ( 500 / 9 ) m / s let length of train be xm x + 300 / 9 = 500 / 9 x = 200 ans is 200 m answer : e | a ) 210 m , b ) 220 m , c ) 230 m , d ) 240 m , e ) 200 m | e | subtract(multiply(9, multiply(add(120, 80), const_0_2778)), 300) | add(n1,n2)|multiply(#0,const_0_2778)|multiply(n3,#1)|subtract(#2,n0) | physics |
by selling 12 pens for a rupee a woman loses 20 % . how many for a rupee should he sell in order to gain 20 % ? | "d 80 % - - - 12 120 % - - - ? 80 / 120 * 12 = 8" | a ) 12 , b ) 14 , c ) 45 , d ) 8 , e ) 65 | d | inverse(add(divide(divide(const_100, subtract(const_100, 20)), 12), divide(multiply(divide(divide(const_100, subtract(const_100, 20)), 12), 20), const_100))) | subtract(const_100,n1)|divide(const_100,#0)|divide(#1,n0)|multiply(n2,#2)|divide(#3,const_100)|add(#2,#4)|inverse(#5)| | gain |
how many of the positive factors of 26 , 16 and how many common factors are there in numbers ? | "factors of 26 - 1 , 2 , 13 and 26 factors of 16 - 1 , 2 , 4 , 8 and 16 comparing both , we have three common factors of 45,16 - 2 answer : b" | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | b | divide(16, 26) | divide(n1,n0)| | other |
find unit digit of n when n = 63 ^ 1 ! + 2 ! + . . . + 63 ! + 18 ^ 1 ! + 2 ! + . . . + 18 ! + 37 ^ 1 ! + 2 ! + . . . 37 ! | 3,8 and 7 have a power cycle of 4 , i . e . the units digit in each case will repeat after every 4 th power . for eg : 3 ^ 1 = 3 , 3 ^ 2 = 9 , 3 ^ 3 = 27 , 3 ^ 4 = 81 , 3 ^ 5 = 243 all the powers given ( 1 ! + 2 ! + . . . . ) , are multiples of 4 . it is so because the last 2 digits of the total sum will be 00 , for ea... | a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 0 | d | subtract(subtract(add(add(63, 18), 37), const_100), const_10) | add(n0,n4)|add(n8,#0)|subtract(#1,const_100)|subtract(#2,const_10) | general |
6 students are equally divided into 3 groups , then , the 3 groups were assigned to 3 different topics . how many different arrangements w are possible ? | 90 is the number of ways you can assign 3 teams formed out of 12 people to 3 different tasks . but now you can order the 3 tasks in 3 ! ways . t 1 t 2 t 3 or t 2 t 1 t 3 . . . . etc etc . i was confused between 90 and 540 but since question used the wordarrangementsdecided to go with complete arrangements w including t... | a ) 30 , b ) 60 , c ) 90 , d ) 180 , e ) 540 | c | multiply(divide(divide(factorial(6), factorial(subtract(6, const_2))), const_2), factorial(3)) | factorial(n0)|factorial(n1)|subtract(n0,const_2)|factorial(#2)|divide(#0,#3)|divide(#4,const_2)|multiply(#5,#1) | probability |
find the value of ( β 1.5 ) / ( β 0.81 ) + ( β 1.44 ) / ( β 0.49 ) is | "( β 1.5 ) / ( β 0.81 ) + ( β 1.44 ) / ( β 0.49 ) = > 3.075 answer is e ." | a ) 195 / 63 , b ) 145 / 63 , c ) 155 / 63 , d ) 125 / 63 , e ) 3.075 | e | add(divide(sqrt(1.5), sqrt(0.81)), divide(sqrt(1.44), sqrt(0.49))) | sqrt(n0)|sqrt(n1)|sqrt(n2)|sqrt(n3)|divide(#0,#1)|divide(#2,#3)|add(#4,#5)| | general |
a completes a piece of work in 3 / 4 of the time in b does , b takes 4 / 5 of the time in c does . they got a profit of rs . 40000 how much b gets ? | assume c takes 20 days . now b takes 4 / 5 ( 20 ) = 16 days . a takes 3 / 4 ( 16 ) = 12 now their efficiencies ratio = 1 / 20 : 1 / 16 : 1 / 12 = 12 : 15 : 20 b ' s share in the profit of rs . 40000 = 15 / 47 ( 40000 ) = rs . 12765 answer : a | a ) 12765 , b ) 121227 , c ) 26887 , d ) 19977 , e ) 26992 | a | add(add(add(const_1000, const_1000), divide(40000, 4)), add(add(multiply(const_60, const_10), const_100), add(const_60, 5))) | add(const_1000,const_1000)|add(n3,const_60)|divide(n4,n1)|multiply(const_10,const_60)|add(#0,#2)|add(#3,const_100)|add(#5,#1)|add(#4,#6) | general |
the sum of two consecutive number is 87 . which is the larger number ? | "let consecutive number be x , x + 1 therefore sum of the consecutive number is x + x + 1 = 87 2 x + 1 = 87 2 x = 86 x = 43 therefore larger number is x + 1 = 44 answer : c" | a ) 42 , b ) 43 , c ) 44 , d ) 45 , e ) 46 | c | add(add(power(add(add(divide(subtract(subtract(87, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(87, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(87, const_10), const_2), const_4), const_2), power(add... | subtract(n0,const_10)|subtract(#0,const_2)|divide(#1,const_4)|add(#2,const_2)|power(#2,const_2)|add(#3,const_2)|power(#3,const_2)|add(#5,const_2)|add(#4,#6)|power(#5,const_2)|power(#7,const_2)|add(#9,#10)|add(#11,#8)| | physics |
right triangle pqr is to be constructed in the xy - plane so that the right angle is at p and pr is parallel to the x - axis . the x and y coordinates of p , q and r are to be integers that satisfy the inequalitites - 4 β€ x β€ 5 and 6 β€ y β€ 16 . how many different triangles with these properties could be constructed ? | total values for x = 10 ; y = 11 x 1 , y 1 = 10 * 11 . . . . . . . . . . . . . . . . . . . . . . . coordinates of 1 st pnt x 2 , y 2 = 9 * 1 ( y 2 = y 1 ) . . . . . . . . . . . . . . . coordinates of 2 nd pnt y coordinates will be same as that of 1 st pnt bcoz it is parallel to x axis x 3 , y 3 = 1 * 10 ( x 2 = x 3 ) .... | a ) 110 , b ) 1100 , c ) 9900 , d ) 10000 , e ) 12100 | c | multiply(multiply(subtract(const_10, const_1), multiply(add(6, 4), subtract(16, 5))), const_10) | add(n0,n2)|subtract(n3,n1)|subtract(const_10,const_1)|multiply(#0,#1)|multiply(#3,#2)|multiply(#4,const_10) | geometry |
when a student joe , weighing 40 kg , joins a group of students whose average weight is 30 kg , the average weight goes up by 1 kg . subsequently , if two students , excluding joe , leave the group the average weight comes back to 30 kg . what is the difference between the average weight of the two students who left an... | "after two persons leave the group the average remains the same . that means the weight of the two persons = 40 + 30 = 70 so , the average the two persons = 35 that gives the answer 40 - 35 = 5 answer a" | a ) 5 kg , b ) 11 kg , c ) 30 kg , d ) 36.5 kg , e ) 71 kg | a | subtract(40, divide(subtract(add(multiply(30, subtract(40, add(30, 1))), 40), multiply(subtract(subtract(40, add(30, 1)), 1), 30)), const_2)) | add(n1,n2)|subtract(n0,#0)|multiply(n1,#1)|subtract(#1,n2)|add(n0,#2)|multiply(n1,#3)|subtract(#4,#5)|divide(#6,const_2)|subtract(n0,#7)| | general |
a certain clock marks every hour by striking a number of times equal to the hour , and the time required for a stroke is exactly equal to the time interval between strokes . at 6 : 00 the time lapse between the beginning of the first stroke and the end of the last stroke is 33 seconds . at 12 : 00 , how many seconds el... | "at 6 : 00 it ' ll chime 6 times . if we assume that the time taken to chime is x , then time between chimes is also x . so you have 6 chimes , which is 6 x and 5 time intervals between chimes . this means that 11 x = 33 seconds . thus x = 3 seconds . by a similar logic , at 12 : 00 , there are 12 chimes and 11 interva... | a ) a . 72 , b ) b . 50 , c ) c . 48 , d ) d . 69 , e ) e . 44 | d | multiply(divide(33, add(6, add(const_1, const_4))), add(12, add(6, add(const_1, const_4)))) | add(const_1,const_4)|add(n0,#0)|add(n3,#1)|divide(n2,#1)|multiply(#2,#3)| | general |
john bought a total of 20 mangoes and oranges . each mango costs 80 cents and each orange costs 60 cents . if the average price of the 20 mangoes and oranges that john originally purchased was 69 cents , then how many oranges needs to return to raise the average price of his purchase to 72 cents ? | "let number of mangoes be x , number of oranges be 12 - x 0.80 x + ( 20 - x ) 0.60 / 20 = 0.69 solving for x , we get x = 9 - - > mangoes 9 , oranges 11 now , number of oranges to be returned be y 0.80 * 9 + ( 11 - y ) * 0.60 / 20 - y = 0.72 solving for y , y = 5 ans : b" | a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | b | multiply(divide(69, const_100), 20) | divide(n4,const_100)|multiply(n0,#0)| | general |
a goods train runs at a speed of 72 kmph and crosses a 240 m long platform in 26 seconds . what is the length of the goods train ? | "s = 240 + x / t 72 * 5 / 18 = 240 + x / 26 x = 280 answer : e" | a ) 230 m , b ) 240 m , c ) 260 m , d ) 270 m , e ) 280 m | e | subtract(multiply(multiply(72, const_0_2778), 26), 240) | multiply(n0,const_0_2778)|multiply(n2,#0)|subtract(#1,n1)| | physics |
a certain manufacturer of cake , muffin , and bread mixes has 100 buyers , of whom 50 purchases cake mix , 40 purchase muffin mix , and 15 purchase both cake mix and muffin mix . if a buyer is to be selected at random from the 100 buyers , what is the probability that the buyer selected will be one who purchases neithe... | "c + m + b - cm - mb - cb - 2 cmb = 100 c - cake buyers , m - muffin and b - bread buyers . cm , mb , cb and cmb are intersecting regions . the question asks for people who have bought only bread mixes = b - cb - mb - 2 cmb has to be found out . 50 + 40 + b - cb - mb - 15 - 2 cmb = 100 b - cb - mb - 2 cmb = 25 hence th... | a ) 1 / 10 , b ) 3 / 10 , c ) 1 / 4 , d ) 7 / 10 , e ) 9 / 10 | c | divide(subtract(100, subtract(add(50, 40), 15)), 100) | add(n1,n2)|subtract(#0,n3)|subtract(n0,#1)|divide(#2,n0)| | other |
when the price of an article was reduced by 40 % its sale increased by 80 % . what was the net effect on the sale ? | "if n items are sold for $ p each , revenue is $ np . if we reduce the price by 40 % , the new price is 0.6 p . if we increase the number sold by 80 % , the new number sold is 1.8 n . so the new revenue is ( 0.6 p ) ( 1.8 n ) = 1.08 np , which is 1.08 times the old revenue , so is 8 % greater . answer : a" | a ) 8 % increase , b ) 4 % decrease , c ) 6 % increase , d ) 66 % increase , e ) 66 % decrease | a | subtract(divide(multiply(add(80, const_100), subtract(const_100, 40)), const_100), const_100) | add(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|subtract(#3,const_100)| | gain |
how many integers from 101 to 900 , inclusive , remains the value unchanged when the digits were reversed ? | "question is asking for palindrome first digit possibilities - 1 through 8 = 8 9 is not possible here because it would result in a number greater than 9 ( i . e 909 , 919 . . ) second digit possibilities - 0 though 9 = 10 third digit is same as first digit = > total possible number meeting the given conditions = 8 * 10... | a ) 50 , b ) 60 , c ) 70 , d ) 80 , e ) 90 | d | divide(900, const_10) | divide(n1,const_10)| | general |
what is the units digit of 2222 ^ ( 333 ) * 4444 ^ ( 222 ) ? | "each of the other explanations to this question has properly explained that you need to break down the calculation into pieces and figure out the repeatingpatternof the units digits . here ' s another way to organize the information . we ' re given [ ( 2222 ) ^ 333 ] [ ( 4444 ) ^ 222 ] we can ' combine ' some of the p... | a ) 0 , b ) 2 , c ) 4 , d ) 6 , e ) 10 | e | add(add(const_4, const_3), const_2) | add(const_3,const_4)|add(#0,const_2)| | general |
the ratio of petrol and kerosene in the container is 3 : 2 when 10 liters of the mixture is taken out and is replaced by the kerosene , the ratio become 2 : 3 . then total quantity of the mixture in the container is : | explanation : pool : kerosene 3 : 2 ( initially ) 2 : 3 ( after replacement ) \ frac { remaining \ : ( or \ : left ) \ : quantity } { initial \ : quantity } = \ left ( 1 - \ frac { replaced \ : quantity } { total \ : quantity } \ right ) ( for petrol ) \ frac { 2 } { 3 } = \ left ( 1 - \ frac { 10 } { k } \ right ) \ r... | a ) 33 , b ) 30 , c ) 38 , d ) 37 , e ) 27 | b | divide(10, subtract(const_1, divide(2, 3))) | divide(n1,n0)|subtract(const_1,#0)|divide(n2,#1) | general |
abcd is a square . e , f , g , h is the midpoint of the sides of square . j and k is the midpoint of the sides hg and fg . l is a point on sides ef such that lf = 1 / 3 ef . find the ratio of area of triangle jkl to the area of square abcd . | ratio = 5 : 48 answer : d | ['a ) 5 : 46', 'b ) 5 : 44', 'c ) 4 : 48', 'd ) 5 : 48', 'e ) 5 : 4'] | d | triangle_area(divide(const_1, const_2), add(multiply(divide(const_1, const_2), divide(1, 3)), multiply(divide(const_1, const_2), divide(const_1, const_2)))) | divide(const_1,const_2)|divide(n0,n1)|multiply(#0,#1)|multiply(#0,#0)|add(#2,#3)|triangle_area(#4,#0) | geometry |
what is the greatest 6 - digit number when divided by 6 , 7 , 8 , 9 , and 10 leaves a remainder of 3 , 4 , 5 , 6 and 7 respectively ? | when you divide a positive integer by 10 , the remainder will just be the units digit . we know the remainder is 7 when we divide by 10 , so d is the only possible answer . | a ) 456780 , b ) 678910 , c ) 997479 , d ) 997917 , e ) 997920 | d | divide(multiply(const_1000, const_1000), 10) | multiply(const_1000,const_1000)|divide(#0,n5)| | general |
a part - time employee whose hourly wage was decreased by 20 percent decided to increase the number of hours worked per week so that the employee ' s total income did not change . by what percent q should the number of hours worked be increased ? | "correct answer : c solution : c . we can set up equations for income before and after the wage reduction . initially , the employee earns w wage and works h hours per week . after the reduction , the employee earns . 8 w wage and works x hours . by setting these equations equal to each other , we can determine the inc... | a ) 12.5 % , b ) 20 % , c ) 25 % , d ) 50 % , e ) 100 % | c | multiply(divide(subtract(divide(multiply(const_10, const_4), multiply(divide(subtract(const_100, 20), const_100), const_10)), const_4), const_4), const_100) | multiply(const_10,const_4)|subtract(const_100,n0)|divide(#1,const_100)|multiply(#2,const_10)|divide(#0,#3)|subtract(#4,const_4)|divide(#5,const_4)|multiply(#6,const_100)| | general |
if each digit in the set a = { 1 , 2 , 3 , 4 , 5 , 6 , 7 } is used exactly once , in how many ways can the digits be arranged ? | "7 ! = 5040 the answer is b ." | a ) 4830 , b ) 5040 , c ) 5250 , d ) 5460 , e ) 5680 | b | factorial(6) | factorial(n5)| | general |
if the sum of a number and its square is 20 , what is the number ? | "let the number be x . then , x + x 2 = 20 ( x + 5 ) ( x - 4 ) = 0 x = 4 answer : d" | a ) 15 , b ) 26 , c ) 28 , d ) 4 , e ) none of these | d | floor(sqrt(20)) | sqrt(n0)|floor(#0)| | geometry |
roy is now 8 years older than julia and half of that amount older than kelly . if in 4 years , roy will be twice as old as julia , then in 4 years what would be roy β s age multiplied by kelly β s age ? | "r = j + 8 = k + 4 r + 4 = 2 ( j + 4 ) ( j + 8 ) + 4 = 2 j + 8 j = 4 r = 12 k = 8 in 4 years ( r + 4 ) ( k + 4 ) = 16 * 12 = 192 the answer is c ." | a ) 172 , b ) 180 , c ) 192 , d ) 200 , e ) 216 | c | multiply(add(add(8, subtract(8, 4)), 4), add(subtract(add(8, subtract(8, 4)), divide(8, const_2)), 4)) | divide(n0,const_2)|subtract(n0,n1)|add(n0,#1)|add(n1,#2)|subtract(#2,#0)|add(n1,#4)|multiply(#3,#5)| | general |
an investor deposited $ 5,000 to open a new savings account that earned 4 percent annual interest , compounded quarterly . if there were no other transactions in the account , what was the amount of money in the account 6 months after the account was opened ? | the amount in the account after 6 months is 1.01 * 1.01 ( $ 5,000 ) = $ 5100.50 the answer is b . | a ) $ 5050.50 , b ) $ 5100.50 , c ) $ 5200.50 , d ) $ 5300.50 , e ) $ 5400.50 | b | multiply(multiply(multiply(add(const_2, const_3), const_100), multiply(add(const_2, const_3), const_2)), power(add(divide(divide(4, const_4), const_100), const_1), const_2)) | add(const_2,const_3)|divide(n1,const_4)|divide(#1,const_100)|multiply(#0,const_100)|multiply(#0,const_2)|add(#2,const_1)|multiply(#3,#4)|power(#5,const_2)|multiply(#6,#7) | gain |
in a group of people , if 30 people were made to stand in each column , 16 columns could be formed . if 48 people were made to stand in a column , how many columns could be formed ? | 16 * 30 = 48 * n n = 10 the answer is a . | a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | a | divide(30, divide(48, 16)) | divide(n2,n1)|divide(n0,#0) | other |
18 buckets of water fill a tank when the capacity of each bucket is 60 litres . how many buckets will be needed to fill the same tank , if the capacity of each bucket is 5 litres ? | "capacity of the tank = ( 18 Γ£ β 60 ) litre number of buckets required of capacity of each bucket is 17 litre = 18 Γ£ β 60 / 5 = 18 Γ£ β 12 = 216 answer is a" | a ) 216 , b ) 200 , c ) 212 , d ) 220 , e ) 210 | a | divide(multiply(60, 18), 5) | multiply(n0,n1)|divide(#0,n2)| | physics |
a car travels from point a to point b . the average speed of the car is 60 km / hr and it travels the first half of the trip at a speed of 50 km / hr . what is the speed of the car in the second half of the trip ? | "let d be the distance and let v be the speed in the second half . the total time = t 1 + t 2 d / 60 = d / 100 + ( d / 2 ) / v d / 150 = d / 2 v and so v = 75 km / hr the answer is c ." | a ) 55 , b ) 60 , c ) 75 , d ) 80 , e ) 90 | c | divide(multiply(60, divide(multiply(50, 60), multiply(subtract(50, 60), 60))), divide(60, const_10)) | divide(n0,const_10)|multiply(n0,n1)|subtract(n1,n0)|multiply(n0,#2)|divide(#1,#3)|multiply(n0,#4)|divide(#5,#0)| | general |
james took a 3 - hour bike ride . in the second hour he traveled 36 miles , which was 20 percent farther than he traveled the first hour . if he traveled 25 percent farther in the third hour than he did in the second hour , how many miles did jose travel during the entire ride ? | "let the distance travelled in the first hour be x . thus , 1.2 x = 36 , x = 30 . now , the distance travelled in the 3 rd hour = 36 + 1 / 4 Γ’ Λ β 36 = 45 . 36 + 30 + 45 = 111 answer : c" | a ) 54.0 , b ) 54.9 , c ) 111.0 , d ) 157.0 , e ) 163.0 | c | add(add(multiply(divide(subtract(const_100, 20), const_100), 36), 36), multiply(divide(add(const_100, 25), const_100), 36)) | add(n3,const_100)|subtract(const_100,n2)|divide(#1,const_100)|divide(#0,const_100)|multiply(n1,#2)|multiply(n1,#3)|add(n1,#4)|add(#6,#5)| | physics |
a large tanker can be filled by two pipes a and b in 30 and 15 hours respectively . how many hours will it take to fill the tanker from empty state if a and b fill it together ? | part filled by a in 1 hr = 1 / 30 part filled by b in 1 hr = 1 / 15 part filled by a + b = 1 / 30 + 1 / 15 = 1 / 10 both the pipes can fill the tank in = 10 hours answer is b | a ) 8 hr , b ) 10 hr , c ) 12 hr , d ) 9 hr , e ) 5 hr | b | inverse(add(inverse(30), inverse(15))) | inverse(n0)|inverse(n1)|add(#0,#1)|inverse(#2) | physics |
a rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered . if the area of the field is 440 sq . feet , how many feet of fencing will be required ? | "given that length and area , so we can find the breadth . length x breadth = area 20 x breadth = 440 breadth = 22 feet area to be fenced = 2 b + l = 2 ( 22 ) + 20 = 64 feet answer : c" | a ) 34 , b ) 40 , c ) 64 , d ) 88 , e ) 92 | c | add(multiply(divide(440, 20), const_2), 20) | divide(n1,n0)|multiply(#0,const_2)|add(n0,#1)| | geometry |
the length of a rectangle is two - fifths of the radius of a circle . the radius of the circle is equal to the side of the square , whose area is 3600 sq . units . what is the area ( in sq . units ) of the rectangle if the rectangle if the breadth is 10 units ? | "given that the area of the square = 3600 sq . units = > side of square = β 3600 = 60 units the radius of the circle = side of the square = 60 units length of the rectangle = 2 / 5 * 60 = 24 units given that breadth = 10 units area of the rectangle = lb = 24 * 10 = 240 sq . units answer : option a" | a ) 240 , b ) 250 , c ) 260 , d ) 270 , e ) 280 | a | multiply(10, multiply(const_2, divide(sqrt(3600), divide(10, const_2)))) | divide(n1,const_2)|sqrt(n0)|divide(#1,#0)|multiply(#2,const_2)|multiply(n1,#3)| | geometry |
a certain rectangular window is four times as long as it is wide . if its perimeter is 50 feet , what are its dimensions in terms of length by width ? | "2 x + 2 y = 50 x + y = 25 x + 4 x = 25 5 x = 25 x = 5 answer c" | a ) 12 by 2 , b ) 11 by 3 , c ) 20 by 5 , d ) 10 by 4 , e ) 9 by 3 | c | rectangle_area(50, const_4) | rectangle_area(n0,const_4)| | geometry |
if l is the least positive integer that is divisible by every integer from 1 to 9 inclusive , then l / 2 is = | the integer should be divisible by : 2 , 3 , 4 , 5 , 6 , 7 , 8 and 9 , that is 2 , 3 , 2 ^ 2 , 5 , 2 * 3 , 7 , 2 ^ 3 and 3 ^ 2 . the least common multiple of these integers is the product of 7 , 5 , 3 ^ 2 and 2 ^ 3 . then , l / 2 is ( 7 * 5 * 3 ^ 2 * 2 ^ 3 ) / 2 = 7 * 5 * 9 * 4 = 1260 . answer : d . | a ) 5040 , b ) 2520 , c ) 1890 , d ) 1260 , e ) 630 | d | divide(multiply(multiply(multiply(power(2, add(2, 1)), power(add(2, 1), 2)), add(add(add(2, 1), 2), 2)), add(add(2, 1), 2)), 2) | add(n0,n2)|add(n2,#0)|power(n2,#0)|power(#0,n2)|add(n2,#1)|multiply(#2,#3)|multiply(#4,#5)|multiply(#1,#6)|divide(#7,n2) | general |
rs . 1500 is divided into two parts such that if one part is invested at 6 % and the other at 5 % the whole annual interest from both the sum is rs . 82 . how much was lent at 5 % ? | "( x * 5 * 1 ) / 100 + [ ( 1500 - x ) * 6 * 1 ] / 100 = 82 5 x / 100 + 90 β 6 x / 100 = 82 x / 100 = 8 = > x = 800 answer : d" | a ) 899 , b ) 866 , c ) 878 , d ) 800 , e ) 877 | d | multiply(add(5, 6), const_100) | add(n1,n2)|multiply(#0,const_100)| | gain |
an athlete runs 200 metres race in 24 seconds . what is his speed ? | speed = dist / time = 200 / 24 200 / 24 * 18 / 5 = 40 * 3 / 4 = 30 km / hr answer b | a ) 20 km / hr , b ) 30 km / hr , c ) 25 km / hr , d ) 35 km / hr , e ) 40 km / hr | b | multiply(divide(200, 24), const_3_6) | divide(n0,n1)|multiply(#0,const_3_6) | physics |
if the average ( arithmetic mean ) of a and b is 115 , and the average of b and c is 160 , what is the value of a β c ? | "question : a - c = ? ( a + b ) / 2 = 115 = = = > a + b = 230 ( b + c ) / 2 = 160 = = = > b + c = 320 ( a + b ) - ( b + c ) = 230 - 320 = = = > a + b - b - c = - 90 = = = > a - c = - 90 answer : c" | a ) β 220 , b ) β 100 , c ) β 90 , d ) 135 , e ) it can not be determined from the information given | c | subtract(multiply(160, const_2), multiply(115, const_2)) | multiply(n1,const_2)|multiply(n0,const_2)|subtract(#0,#1)| | general |
in a certain parallelogram the degree measure of one angle exceeds that of the other by 50 what is the degree measure of the smaller angle ? | "in a parallelogram opposite angles are equal and the angles at each side are supplementary to each other ( supplementary angles are two angles that add up to 180 Β° ) . given : x + ( x + 50 ) = 180 - - > x = 65 . answer : a ." | a ) 65 , b ) 80 , c ) 85 , d ) 90 , e ) 95 | a | divide(subtract(divide(const_3600, const_10), multiply(50, const_2)), const_4) | divide(const_3600,const_10)|multiply(n0,const_2)|subtract(#0,#1)|divide(#2,const_4)| | geometry |
jim is able to sell a hand - carved statue for $ 550 which was a 35 % profit over his cost . how much did the statue originally cost him ? | "550 = 1.35 * x x = 550 / 1.35 = 407.407407 . . . which rounds to $ 407.41 , which is ( a ) ." | a ) $ 407.41 , b ) $ 412.40 , c ) $ 455.40 , d ) $ 474.90 , e ) $ 488.20 | a | divide(550, add(divide(35, const_100), const_1)) | divide(n1,const_100)|add(#0,const_1)|divide(n0,#1)| | gain |
pipe b that can fill a tank in an hour and pipe a that can fill the tank in half an hour are opened simultaneously when the tank is empty . pipe a is shut 15 minutes before the tank overflows . when will the tank overflow ? | the last 15 minutes only pipe b was open . since it needs 1 hour to fill the tank , then in 15 minutes it fills 1 / 4 th of the tank , thus 3 / 4 of the tank is filled with both pipes open . the combined rate of two pipes is 1 + 2 = 3 tanks / hour , therefore to fill 3 / 4 th of the tank they need ( time ) = ( work ) /... | a ) 32 mins , b ) 35 mins , c ) 40 mins , d ) 30 mins , e ) 36 mins | d | add(divide(subtract(const_1, multiply(15, divide(const_1, multiply(const_1, const_60)))), add(divide(const_1, multiply(divide(const_1, const_2), const_60)), divide(const_1, multiply(const_1, const_60)))), 15) | divide(const_1,const_2)|multiply(const_1,const_60)|divide(const_1,#1)|multiply(#0,const_60)|divide(const_1,#3)|multiply(n0,#2)|add(#4,#2)|subtract(const_1,#5)|divide(#7,#6)|add(n0,#8) | physics |
working together , tim and tom can type 12 pages in one hour . if they would be able to type 15 pages in one hour if tom increases his typing speed by 30 % , what is the ratio of tom ' s normal typing speed to that of tim ? | lets say tim types x pages an hour and tom types y pages an hour . we know that x + y = 12 tom increase speed by 30 % means he will type 1.3 y pages an hour . so we get x + 1.3 y = 15 we need to know the ratio of tom ' s speed to tim ' s speed . this is going to be proportional to the number of pages each can type in a... | a ) 1 / 5 , b ) 5 / 1 , c ) 4 / 1 , d ) 1 / 4 , e ) 1 / 3 | b | divide(divide(subtract(15, 12), divide(30, multiply(const_100, const_1))), subtract(12, divide(subtract(15, 12), divide(30, multiply(const_100, const_1))))) | multiply(const_1,const_100)|subtract(n1,n0)|divide(n2,#0)|divide(#1,#2)|subtract(n0,#3)|divide(#3,#4) | physics |
the interest on a certain deposit at 5 % per annum is rs . 101.20 in one year . how much will the additional interest in one year be on the same deposit at 6 % per annum ? | explanation : principal , p = 100 Γ si / rt = 100 Γ 101.20 / 5 Γ 1 = 20 Γ 101.20 = rs . 2024 simple interest for rs . 2024 at 6 % per annum for 1 year , si = 2024 Γ 6 Γ 1100 = 121.44 additional interest = rs . 121.44 - rs . 101.20 = rs . 20.24 answer : option c | a ) rs . 20.8 , b ) rs . 19.74 , c ) rs . 20.24 , d ) rs . 19.5 , e ) rs . 19.00 | c | subtract(divide(multiply(divide(multiply(const_100, 101.2), 5), 6), const_100), 101.2) | multiply(n1,const_100)|divide(#0,n0)|multiply(n2,#1)|divide(#2,const_100)|subtract(#3,n1) | gain |
in an election , candidate a got 65 % of the total valid votes . if 15 % of the total votes were declared invalid and the total numbers of votes is 560000 , find the number of valid vote polled in favor of candidate . | "total number of invalid votes = 15 % of 560000 = 15 / 100 Γ 560000 = 8400000 / 100 = 84000 total number of valid votes 560000 β 84000 = 476000 percentage of votes polled in favour of candidate a = 65 % therefore , the number of valid votes polled in favour of candidate a = 65 % of 476000 = 65 / 100 Γ 476000 = 30940000... | a ) 330000 , b ) 309400 , c ) 347000 , d ) 356000 , e ) 357000 | b | multiply(multiply(560000, subtract(const_1, divide(15, const_100))), divide(65, const_100)) | divide(n0,const_100)|divide(n1,const_100)|subtract(const_1,#1)|multiply(n2,#2)|multiply(#0,#3)| | gain |
the profit obtained by selling an article for rs . 86 is the same as the loss obtained by selling it for rs . 42 . what is the cost price of the article ? | "s . p 1 - c . p = c . p β s . p 2 86 - c . p = c . p - 42 2 c . p = 86 + 42 ; c . p = 128 / 2 = 64 answer : e" | a ) rs . 40 , b ) rs . 50 , c ) rs . 49 , d ) rs . 59 , e ) rs . 64 | e | divide(add(86, 42), const_2) | add(n0,n1)|divide(#0,const_2)| | gain |
a squirrel runs up a cylindrical post , in a perfect spiral path making one circuit for each rise of 4 feet . how many feet does the squirrel travels if the post is 16 feet tall and 3 feet in circumference ? | "total circuit = 16 / 4 = 4 total feet squirrel travels = 4 * 3 = 12 feet answer : a" | a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | a | multiply(divide(16, 4), 3) | divide(n1,n0)|multiply(n2,#0)| | geometry |
a batsman in his 12 th innings makes a score of 70 and thereby increases his average by 3 runs . what is his average after the 12 th innings if he had never been β not out β ? | "let β x β be the average score after 12 th innings β 12 x = 11 Γ ( x β 3 ) + 70 β΄ x = 37 answer b" | a ) 42 , b ) 37 , c ) 44 , d ) 45 , e ) 46 | b | add(subtract(70, multiply(12, 3)), 3) | multiply(n0,n2)|subtract(n1,#0)|add(n2,#1)| | general |
a train 125 m long passes a man , running at 2 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 - 2 ) km / hr . x - 2 = 45 = = > x = 47 km / hr answer : d" | a ) 36 , b ) 50 , c ) 28 , d ) 47 , e ) 29 | d | divide(divide(subtract(125, multiply(multiply(2, const_0_2778), 2)), 10), const_0_2778) | multiply(n1,const_0_2778)|multiply(n1,#0)|subtract(n0,#1)|divide(#2,n2)|divide(#3,const_0_2778)| | physics |
50 persons like apple . 7 like orange and mango dislike apple . 10 like mango and apple and dislike orange . 4 like all . how many people like apple ? | "orange + mango - apple = 7 mango + apple - orange = 10 apple = 50 orange + mango + apple = 4 50 + 10 + 4 - 7 = 57 like apple answer : d" | a ) 47 , b ) 46 , c ) 54 , d ) 57 , e ) 62 | d | add(add(subtract(50, 4), 7), subtract(10, 7)) | subtract(n0,n3)|subtract(n2,n1)|add(n1,#0)|add(#2,#1)| | general |
5 ^ 100 is divided by 18 . then what is the remainder ? | "here n = 18 = 2 Γ 32 Ο ( 18 ) = 18 ( 1 β 12 ) ( 1 β 13 ) = 6 so 56 when divided by 18 , remainder is 1 . so we can write the given expression 5100 = ( 56 ) 16 Γ 54 = ( 1 ) 16 Γ 54 = 52 Γ 52 = 7 Γ 7 = 49 now 49 when divided by 18 , remainder is 13 . c" | a ) 10 , b ) 11 , c ) 13 , d ) 15 , e ) 16 | c | subtract(18, 5) | subtract(n2,n0)| | general |
a corporation that had $ 120 billion in profits for the year paid out $ 200 million in employee benefits . approximately what percent of the profits were the employee benefits ? ( note : 1 billion = 10 ^ 9 ) | "required answer = [ employee benefit / profit ] * 100 = [ ( 200 million ) / ( 120 billion ) ] * 100 = [ ( 200 * 10 ^ 6 ) / ( 120 * 10 ^ 9 ) ] * 100 = ( 1.7 / 1000 ) * 100 = 0.17 % so answer is ( e )" | a ) 50 % , b ) 20 % , c ) 5 % , d ) 4 % , e ) 0.17 % | e | multiply(divide(multiply(200, power(10, add(const_3, const_3))), multiply(120, power(10, 9))), const_100) | add(const_3,const_3)|power(n3,n4)|multiply(n0,#1)|power(n3,#0)|multiply(n1,#3)|divide(#4,#2)|multiply(#5,const_100)| | general |
a , b and c have rs . 900 between them , a and c together have rs . 400 and b and c rs . 750 . how much does c have ? | "a + b + c = 900 a + c = 400 b + c = 750 - - - - - - - - - - - - - - a + b + 2 c = 1150 a + b + c = 900 - - - - - - - - - - - - - - - - c = 250 answer : d" | a ) 100 , b ) 150 , c ) 200 , d ) 250 , e ) 350 | d | subtract(add(400, 750), 900) | add(n1,n2)|subtract(#0,n0)| | general |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.