Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
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a certain electric - company plan offers customers reduced rates for electricity used between 8 p . m . and 8 a . m . weekdays and 24 hours a day saturdays and sundays . under this plan , the reduced rates w apply to what fraction of a week ? | "number of hours between 8 pm to 8 am = 12 number of hours with reduced rates = ( 12 * 5 ) + ( 24 * 2 ) hours with reduced rates w / total number of hours in a week = ( 12 * 5 ) + ( 24 * 2 ) / ( 24 * 7 ) = 108 / ( 24 * 7 ) = 9 / 14 answer : c" | a ) 1 / 2 , b ) 5 / 8 , c ) 9 / 14 , d ) 16 / 21 , e ) 9 / 10 | c | divide(add(multiply(divide(24, const_2), add(const_2, const_3)), multiply(24, const_2)), multiply(24, add(const_3, const_4))) | add(const_2,const_3)|add(const_3,const_4)|divide(n2,const_2)|multiply(n2,const_2)|multiply(#0,#2)|multiply(n2,#1)|add(#4,#3)|divide(#6,#5)| | physics |
the sum of two consecutive number is 115 . which is the larger number ? | "let consecutive number be x , x + 1 therefore sum of the consecutive number is x + x + 1 = 115 2 x + 1 = 115 2 x = 114 x = 57 therefore larger number is x + 1 = 58 answer : b" | a ) 42 , b ) 58 , c ) 44 , d ) 45 , e ) 46 | b | add(add(power(add(add(divide(subtract(subtract(115, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(115, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(115, const_10), const_2), const_4), const_2), power(... | 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 |
three pipes a , b and c can fill a tank from empty to full in 20 minutes , 20 minutes and 30 minutes respectively . when the tank is empty , all the three pipes are opened . a , b and c discharge chemical solutions p , q and r respectively . what is the proportion of solution p in the liquid in the tank after 3 minutes... | "part filled by ( a + b + c ) in 3 minutes = 3 ( 1 / 20 + 1 / 20 + 1 / 30 ) = 6 / 15 part filled by a in 3 minutes = 3 / 20 required ratio = 3 / 20 * 15 / 6 = 3 / 8 answer : a" | a ) 3 / 8 , b ) 2 / 8 , c ) 6 / 7 , d ) 3 / 5 , e ) 2 / 5 | a | multiply(divide(3, 30), divide(const_1, multiply(3, add(divide(const_1, 30), add(divide(const_1, 20), divide(const_1, 20)))))) | divide(n3,n2)|divide(const_1,n0)|divide(const_1,n2)|add(#1,#1)|add(#3,#2)|multiply(n3,#4)|divide(const_1,#5)|multiply(#0,#6)| | physics |
if difference between compound interest and simple interest on a sum at 10 % p . a . for 2 years is rs . 36 then sum is | "p ( r / 100 ) ^ 2 = c . i - s . i p ( 10 / 100 ) ^ 2 = 36 3600 answer : e" | a ) s . 5000 , b ) s . 5100 , c ) s . 5800 , d ) s . 6000 , e ) s . 3600 | e | divide(36, multiply(divide(10, const_100), divide(10, const_100))) | divide(n0,const_100)|multiply(#0,#0)|divide(n2,#1)| | gain |
the ratio of ducks and frogs in a pond is 31 : 39 respectively . the average number of ducks and frogs in the pond is 152 . what is the number of frogs in the pond ? | "solution : ratio of ducks and frogs in pond , = 31 : 39 . average of ducks and frogs in pond , = 152 . so , total number of ducks and frogs in the pond , = 2 * 152 = 304 . therefore , number of frogs , = ( 304 * 39 ) / 70 = 169 . answer : option d" | a ) 148 , b ) 152 , c ) 156 , d ) 169 , e ) none | d | multiply(39, divide(152, divide(add(add(31, 39), const_2), const_2))) | add(n0,n1)|add(#0,const_2)|divide(#1,const_2)|divide(n2,#2)|multiply(n1,#3)| | general |
x is a positive number which when increased by 17 is equal to 60 times the reciprocal of the number . find value of x ? | let the number be x . then , x + 17 = 60 x x 2 + 17 x - 60 = 0 ( x + 20 ) ( x - 3 ) = 0 x = 3 a | a ) 3 , b ) 4 , c ) 6 , d ) 7 , e ) 8 | a | divide(add(negate(17), sqrt(subtract(power(17, const_2), multiply(const_4, multiply(const_1, negate(60)))))), const_2) | negate(n0)|negate(n1)|power(n0,const_2)|multiply(#1,const_1)|multiply(#3,const_4)|subtract(#2,#4)|sqrt(#5)|add(#0,#6)|divide(#7,const_2) | general |
a man invests rs . 6,000 at the rate of 5 % per annum . how much more should he invest at the rate of 8 % , so that he can earn a total of 6 % per annum ? | "explanation : interest on rs . 6000 at 5 % per annum = ( 6000 Γ 5 Γ 1 ) / 100 = rs . 300 let his additional investment at 8 % = x interest on rs . x at 8 % per annum = ( x Γ 8 Γ 1 ) / 100 = 2 x / 25 . to earn 6 % per annum for the total , interest = ( 6000 + x ) Γ 6 Γ 1 / 100 . = > 300 + 2 x / 25 = ( 6000 + x ) Γ 6 Γ ... | a ) rs . 1200 , b ) rs . 1300 , c ) rs . 1500 , d ) rs . 3000 , e ) none of these | d | multiply(multiply(8, const_1000), divide(subtract(6, 5), subtract(8, 6))) | multiply(n2,const_1000)|subtract(n3,n1)|subtract(n2,n3)|divide(#1,#2)|multiply(#3,#0)| | gain |
the lcm of two numbers is 2310 and hcf is 47 . if one of the numbers is 210 . then what is the other number ? | first number * second number = lcm * hcf other number = 2310 * 47 / 210 = 11 * 47 = 517 answer : e | a ) 715 , b ) 825 , c ) 330 , d ) 582 , e ) 517 | e | divide(multiply(2310, 47), 210) | multiply(n0,n1)|divide(#0,n2) | physics |
a retailer purchases shirts from a wholesaler and then sells the shirts in her store at a retail price that is 70 percent greater than the wholesale price . if the retailer decreases the retail price by 30 percent this will have the same effect as increasing the wholesale price by what percent ? | "answer : d = 19 . assume rs . 100 to be the price at which the retailer buys from wholesaler . 70 % increase makes retail price = 170 . now 30 % decrease - > ( 1 - 30 / 100 ) * 170 = 119 . now compared to the wholesale price of 100 , 19 % increase is what will have the same effect as increasing the wholesale price ." | a ) 26 , b ) 37.5 , c ) 42 , d ) 19 , e ) 50 | d | multiply(subtract(multiply(add(const_1, divide(70, const_100)), subtract(const_1, divide(30, const_100))), const_1), const_100) | divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|subtract(#4,const_1)|multiply(#5,const_100)| | gain |
machine p and machine q are each used to manufacture 220 sprockets . it takes machine p 10 hours longer to produce 220 sprockets than machine q . machine q produces 10 % more sprockets per hour than machine a . how many sprockets per hour does machine a produce ? | "p makes x sprockets per hour . then q makes 1.1 x sprockets per hour . 220 / x = 220 / 1.1 x + 10 1.1 ( 220 ) = 220 + 11 x 11 x = 22 x = 2 the answer is b ." | a ) 5 , b ) 2 , c ) 55 , d ) 95 , e ) 125 | b | divide(subtract(220, divide(220, add(divide(10, const_100), const_1))), 10) | divide(n1,const_100)|add(#0,const_1)|divide(n0,#1)|subtract(n0,#2)|divide(#3,n1)| | gain |
the diagonal of a square is twice the side of equilateral triangle then the ratio of area of the triangle to the area of square is ? | let the side of equilateral triangle = 1 unit . we know that area of an equilateral triangle = 3 β β 4 a 234 a 2 as side = 1 unit area of the equilateral triangle = 3 β β 434 now diagonal of the square = 2 ( side of the equilateral triangle ) = 2 we know that area of the square = 12 d 212 d 2 where d = diagonal so area... | ['a ) 3 β 9', 'b ) 3 β 8', 'c ) 3 β 7', 'd ) 3 β 5', 'e ) 3 β 1'] | b | divide(triangle_area_three_edges(const_1, const_1, const_1), square_area(sqrt(divide(power(const_2, const_2), add(const_1, const_1))))) | add(const_1,const_1)|power(const_2,const_2)|triangle_area_three_edges(const_1,const_1,const_1)|divide(#1,#0)|sqrt(#3)|square_area(#4)|divide(#2,#5) | geometry |
the average mark of the students of a class in a particular exam is 65 . if 5 students whose average mark in that exam is 20 are excluded , the average mark of the remaining will be 90 . find the number of students who wrote the exam . | "let the number of students who wrote the exam be x . total marks of students = 75 x . total marks of ( x - 5 ) students = 90 ( x - 5 ) 65 x - ( 5 * 20 ) = 90 ( x - 5 ) 350 = 25 x = > x = 14 answer : a" | a ) 14 , b ) 25 , c ) 35 , d ) 45 , e ) 55 | a | divide(subtract(multiply(90, 5), multiply(5, 20)), subtract(90, 65)) | multiply(n1,n3)|multiply(n1,n2)|subtract(n3,n0)|subtract(#0,#1)|divide(#3,#2)| | general |
there are 24 students in a class . in how many different ways can a committee of 3 students be formed ? | "24 c 3 = 24 * 23 * 22 / 6 = 2024 the answer is c ." | a ) 1250 , b ) 1800 , c ) 2024 , d ) 3256 , e ) 2864 | c | multiply(subtract(const_1, divide(3, 24)), 24) | divide(n1,n0)|subtract(const_1,#0)|multiply(#1,n0)| | probability |
fred and sam are standing 35 miles apart and they start walking in a straight line toward each other at the same time . if fred walks at a constant speed of 2 miles per hour and sam walks at a constant speed of 5 miles per hour , how many miles has sam walked when they meet ? | relative distance = 35 miles relative speed = 2 + 5 = 7 miles per hour time taken = 35 / 7 = 5 hours distance travelled by sam = 5 * 5 = 25 miles = c | a ) 5 , b ) 9 , c ) 25 , d ) 30 , e ) 45 | c | multiply(5, divide(35, add(2, 5))) | add(n1,n2)|divide(n0,#0)|multiply(n2,#1) | physics |
a camera lens filter kit containing 5 filters sells for $ 72.50 . if the filters are purchased individually , 2 of them are priced at $ 12.45 each , 2 at $ 14.05 each , 1 at $ 11.50 . the amount saved by purchasing the kit is what percent of the total price of the 5 filters purchased individually ? | cost of kit = $ 72.50 if filters are purchased individually - $ 12.45 * 2 + $ 14.05 * 2 + $ 11.50 = $ 64.50 amount saved = $ 72.50 - $ 64.50 = $ 8 required % age = ( $ 8 / $ 72.50 ) * 100 = 11.03 % so , the correct answer is b . | a ) 10.03 % , b ) 11.03 % , c ) 12.03 % , d ) 13.03 % , e ) 11 % | b | divide(multiply(subtract(72.5, add(11.5, add(multiply(2, 12.45), multiply(2, 14.05)))), const_100), 72.5) | multiply(n2,n3)|multiply(n2,n5)|add(#0,#1)|add(n7,#2)|subtract(n1,#3)|multiply(#4,const_100)|divide(#5,n1) | general |
every student in a room is either a junior or a senior . there is at least one junior and at least one senior in the room . if 1 / 2 of the juniors is equal to 2 / 3 of the seniors , what fraction of the students in the room are juniors ? | let total number of juniors = j total number of seniors = s ( 1 / 2 ) j = ( 2 / 3 ) s = > s = 3 / 4 j total number of students = j + s = ( 7 / 4 ) j fraction of the students in the room are juniors = j / ( j + s ) = j / [ ( 7 / 4 ) j ] = 4 / 7 answer a | a ) 4 / 7 , b ) 1 / 3 , c ) 5 / 3 , d ) 12 / 7 , e ) 17 / 20 | a | divide(multiply(2, 2), add(3, multiply(2, 2))) | multiply(n1,n1)|add(n3,#0)|divide(#0,#1) | general |
in a certain alphabet , 10 letters contain a dot and a straight line . 24 letters contain a straight line but do not contain a dot . if that alphabet has 40 letters , all of which contain either a dot or a straight line or both , how many letters contain a dot but do not contain a straight line ? | we are told that all of the letters contain either a dot or a straight line or both , which implies that there are no letters without a dot and a line ( no line / no dot box = 0 ) . first we find the total # of letters with lines : 10 + 24 = 34 ; next , we find the total # of letters without line : 40 - 34 = 6 ; finall... | a ) 6 , b ) 8 , c ) 14 , d ) 20 , e ) 28 | a | subtract(40, add(10, 24)) | add(n0,n1)|subtract(n2,#0) | other |
a certain manufacturer increased its gross profit on a product from 10 percent of the cost of the product to 15 percent of the cost by changing the selling price . if the new selling price was $ 92.00 and the cost of the product remained the same , what was the old selling price ? | given that { cost of the product } * 1.15 = $ 92 - - > { cost of the product } = $ 80 . the old price was $ 80 * 1.1 = $ 88 . answer : e . | a ) $ 77.40 , b ) $ 80.00 , c ) $ 83.64 , d ) $ 87.40 , e ) $ 88.00 | e | multiply(divide(add(10, const_100), add(15, const_100)), 92) | add(n0,const_100)|add(n1,const_100)|divide(#0,#1)|multiply(n2,#2) | gain |
in a certain city , 60 percent of the registered voters are democrats and the rest are republicans . in a mayoral race , if 65 percent of the registered voters who are democrats and 20 percent of the registered voters who are republicans are expected to vote for candidate a , what percent of the registered voters are e... | "say there are total of 100 registered voters in that city . thus 60 are democrats and 40 are republicans . 60 * 0.65 = 39 democrats are expected to vote for candidate a ; 40 * 0.20 = 8 republicans are expected to vote for candidate a . thus total of 39 + 8 = 47 registered voters are expected to vote for candidate a , ... | a ) 50 % , b ) 53 % , c ) 54 % , d ) 55 % , e ) 47 % | e | add(multiply(60, divide(65, const_100)), multiply(subtract(const_100, 60), divide(20, const_100))) | divide(n1,const_100)|divide(n2,const_100)|subtract(const_100,n0)|multiply(n0,#0)|multiply(#1,#2)|add(#3,#4)| | gain |
the smallest number when increased by ` ` 2 ` ` is exactly divisible by 12 , 30 , 48 , 74 and 100 is : | "lcm = 44400 44400 - 2 = 44398 answer : b" | a ) 14439 , b ) 44398 , c ) 44400 , d ) 44402 , e ) 15005 | b | add(lcm(lcm(lcm(12, 30), lcm(48, 74)), 100), 2) | lcm(n1,n2)|lcm(n3,n4)|lcm(#0,#1)|lcm(n5,#2)|add(n0,#3)| | general |
how many integers between 1 and 1400 are divisible by 25 , and 35 ? | lcm of the given numbers = 175 therefore , number of integers = 1400 / 175 = 8 answer is option e | a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | e | divide(1400, lcm(25, 35)) | lcm(n2,n3)|divide(n1,#0) | general |
a semicircular cubicle has a radius of 14 . what is the approximate perimeter of the cubicle ? | perimeter of a circle = 2 pi * r perimeter of a semicircle = pi * r + 2 r aprox perimiter = 3.14 * 14 + 2 * 14 = 71.96 approximately 72 answer d | ['a ) 55', 'b ) 86', 'c ) 25', 'd ) 72', 'e ) 35'] | d | add(multiply(const_pi, 14), multiply(const_2, 14)) | multiply(n0,const_pi)|multiply(n0,const_2)|add(#0,#1) | geometry |
a works thrice as much as b . if a takes 60 days less than b to do a work then find the number of days it would take to complete the work if both work together ? | if a finishes the job in x days b finishes it in 3 x days 3 x = x + 60 thus x = 30 in one hour 1 / t = 1 / 30 + 1 / 90 = 4 / 90 where t is the number of days they finish the job together . t = 90 / 4 = 22.5 days answer : a | a ) 22.5 days , b ) 21.5 days , c ) 23.5 days , d ) 24.5 days , e ) 25.5 days | a | inverse(add(inverse(multiply(const_3, divide(60, subtract(const_3, const_1)))), inverse(divide(60, subtract(const_3, const_1))))) | subtract(const_3,const_1)|divide(n0,#0)|inverse(#1)|multiply(#1,const_3)|inverse(#3)|add(#4,#2)|inverse(#5) | physics |
a retailer bought a machine at a wholesale price of $ 90 and later on sold it after a 10 % discount of the retail price . if the retailer made a profit equivalent to 20 % of the whole price , what is the retail price t of the machine ? | "since the wholesale price was $ 90 and the profit was 20 % of the wholesale price ( [ . 2 ] [ 90 ] = $ 18 ) , the retail price would have to be above $ 108 , but not that much greater than that . let ' s start by testing answer c : $ 120 . . . . if . . . . . retail price t = $ 120 10 % discount off = $ 120 - ( . 1 ) (... | a ) 81 , b ) 100 , c ) 120 , d ) 135 , e ) 160 | c | divide(multiply(add(90, divide(multiply(90, 20), const_100)), const_100), multiply(multiply(const_3, const_3), 10)) | multiply(n0,n2)|multiply(const_3,const_3)|divide(#0,const_100)|multiply(n1,#1)|add(n0,#2)|multiply(#4,const_100)|divide(#5,#3)| | gain |
in a certain company , a third of the workers do not have a retirement plan . 20 % of the workers who do not have a retirement plan are women , and 40 % of the workers who do have a retirement plan are men . if 128 of the workers of that company are men , how many of the workers are women ? | "set up equation : x = total number of workers 128 = 0,4 * 2 / 3 * x + 0,8 * 1 / 3 * x 128 = 16 / 30 x x = 240 240 - 128 = 112 answer d" | a ) 80 , b ) 95 , c ) 105 , d ) 112 , e ) 210 | d | multiply(divide(128, add(subtract(divide(const_1, const_3), multiply(divide(const_1, const_3), divide(20, const_100))), multiply(subtract(const_1, divide(const_1, const_3)), divide(40, const_100)))), add(multiply(divide(const_1, const_3), divide(20, const_100)), subtract(subtract(const_1, divide(const_1, const_3)), mul... | divide(const_1,const_3)|divide(n0,const_100)|divide(n1,const_100)|multiply(#0,#1)|subtract(const_1,#0)|multiply(#2,#4)|subtract(#0,#3)|add(#5,#6)|subtract(#4,#5)|add(#3,#8)|divide(n2,#7)|multiply(#9,#10)| | gain |
calculate the number of urns of water needed to fill a tank of dimensions of 10 x 10 x 5 m to 90 % full , if the volume of the urn is 0.8 m 3 and can only be filled to 80 % . | volume of the tank that needed to be filled = 10 x 10 x 5 x 0.9 = 450 volume of the urn that can be filled = 0.8 x 0.8 = 0.64 no . of urns of water needed to fill the tank = 450 / 0.64 = 703.1 answer : d | a ) 520 , b ) 650.2 , c ) 685.5 , d ) 703.1 , e ) 712.2 | d | divide(divide(multiply(multiply(multiply(10, 10), 5), 90), const_100), divide(multiply(0.8, 80), const_100)) | multiply(n0,n0)|multiply(n4,n6)|divide(#1,const_100)|multiply(n2,#0)|multiply(n3,#3)|divide(#4,const_100)|divide(#5,#2) | physics |
a batsman makes a score of 92 runs in the 17 th inning and thus increases his average by 3 . find his average after 17 th inning . | "let the average after 17 th inning = x . then , average after 16 th inning = ( x β 3 ) . β΄ 16 ( x β 3 ) + 92 = 17 x or x = ( 92 β 48 ) = 44 . answer d" | a ) 36 , b ) 39 , c ) 42 , d ) 44 , e ) none of the above | d | add(subtract(92, multiply(17, 3)), 3) | multiply(n1,n2)|subtract(n0,#0)|add(n2,#1)| | general |
a tank can be filled by two pipes a and b in 60 minutes and 40 minutes respectively . how many minutes will it take to fill the tank from empty state if b is used for the first half time and then a and b fill it together for the other half . | explanation : let the total time be x mins . part filled in first half means in x / 2 = 1 / 40 part filled in second half means in x / 2 = 1 / 60 + 1 / 40 = 1 / 24 total = x / 2 x 1 / 40 + x / 2 x 1 / 24 = 1 x = 30 mins answer is d | a ) 15 mins , b ) 20 mins , c ) 25 mins , d ) 30 mins , e ) 35 mins | d | divide(multiply(multiply(40, const_3), const_2), add(const_4, const_4)) | add(const_4,const_4)|multiply(n1,const_3)|multiply(#1,const_2)|divide(#2,#0) | physics |
in a division sum , the quotient is 65 , the divisor 24 and the remainder 5 , find the dividend ? | "explanation : 65 * 24 + 5 = 1565 answer : e" | a ) 1595 , b ) 1569 , c ) 1265 , d ) 1555 , e ) 1565 | e | add(multiply(65, 24), 5) | multiply(n0,n1)|add(n2,#0)| | general |
the principal that amounts to rs . 3913 in 3 years at 6 1 / 4 % per annum c . i . compounded annually , is ? | "principal = [ 4913 / ( 1 + 25 / ( 4 * 100 ) ) 3 ] = 3913 * 16 / 17 * 16 / 17 * 16 / 17 = rs . 3096 . answer : a" | a ) s . 3096 , b ) s . 4076 , c ) s . 4085 , d ) s . 4096 , e ) s . 5096 | a | divide(3913, power(add(1, divide(add(6, divide(1, 4)), const_100)), 3)) | divide(n3,n4)|add(n2,#0)|divide(#1,const_100)|add(#2,n3)|power(#3,n1)|divide(n0,#4)| | gain |
the length of a rectangular field is 7 / 5 its width . if the perimeter of the field is 240 meters , what is the width of the field ? | "let l be the length and w be the width . l = ( 7 / 5 ) w perimeter : 2 l + 2 w = 240 , 2 ( 7 / 5 ) w + 2 w = 240 solve the above equation to find : w = 50 m and l = 70 m . correct answer a ) 50" | a ) 50 , b ) 60 , c ) 70 , d ) 80 , e ) 90 | a | divide(240, add(add(divide(7, 5), divide(7, 5)), const_2)) | divide(n0,n1)|add(#0,#0)|add(#1,const_2)|divide(n2,#2)| | geometry |
the sum of the present ages of two persons a and b is 60 . if the age of a is twice that of b , find the sum of their ages 10 years hence ? | "explanation : a + b = 60 , a = 2 b 2 b + b = 60 = > b = 20 then a = 40 . 10 years , their ages will be 50 and 30 . sum of their ages = 50 + 30 = 80 . e )" | a ) 58 , b ) 59 , c ) 62 , d ) 70 , e ) 80 | e | add(add(multiply(divide(60, 10), const_2), 10), add(divide(60, 10), 10)) | divide(n0,n1)|add(#0,n1)|multiply(#0,const_2)|add(#2,n1)|add(#3,#1)| | general |
at a certain committee meeting only associate professors and assistant professors are present . each associate professor has brought 2 pencils and 1 chart to the meeting , while each assistant professor has brought 1 pencil and 2 charts . if a total of 10 pencils and 8 charts have been brought to the meeting , how many... | "say there are ' a ' associate professors . so we have 2 a pencils and a charts . say there are ' b ' assistant professors . so we have b pencils and 2 b charts . total pencils are 10 so 2 a + b = 10 total charts are 11 so a + 2 b = 8 add both : 3 a + 3 b = 18 so a + b = 6 total number of people = 6 a" | a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | a | divide(add(8, 10), add(2, 1)) | add(n4,n5)|add(n0,n1)|divide(#0,#1)| | general |
it is the new year and mandy has made a resolution to lose weight this year . she plans to exercise and do yoga . for exercise she plans to workout at the gym and ride her bicycle in the ratio of 2 : 3 everyday . she will also do yoga in the ratio , yoga : exercise = 2 : 3 . if she rides her bike for 14 minutes , how m... | "the ratio is 2 : 3 = gym : ride , so ( 14 ) ( 3 / 2 ) = 21 minutes at the gym , and 21 + 14 = 35 minutes exercise , so ( 2 / 3 ) ( 35 ) = 23 minutes yoga . answer : d" | a ) 10 min . , b ) 41 min . , c ) 17 min . , d ) 23 min . , e ) 25 min . | d | divide(multiply(14, divide(3, add(2, 3))), multiply(divide(3, add(2, 3)), divide(3, add(2, 3)))) | add(n0,n1)|divide(n1,#0)|multiply(n4,#1)|multiply(#1,#1)|divide(#2,#3)| | physics |
jim Γ’ β¬ β’ s taxi service charges an initial fee of $ 2.25 at the beginning of a trip and an additional charge of $ 0.4 for each 2 / 5 of a mile traveled . what is the total charge for a trip of 3.6 miles ? | "let the fixed charge of jim Γ’ β¬ β’ s taxi service = 2.25 $ and charge per 2 / 5 mile ( . 4 mile ) = . 4 $ total charge for a trip of 3.6 miles = 2.25 + ( 3.6 / . 4 ) * . 4 = 2.25 + 9 * . 4 = 5.85 $ answer e" | a ) $ 3.15 , b ) $ 4.45 , c ) $ 4.80 , d ) $ 5.05 , e ) $ 5.85 | e | add(multiply(divide(3.6, divide(2, 5)), 0.4), 2.25) | divide(n2,n3)|divide(n4,#0)|multiply(n1,#1)|add(n0,#2)| | general |
in a market , a dozen eggs cost as much as a pound of rice , and a half - liter of kerosene costs as much as 8 eggs . if the cost of each pound of rice is $ 0.33 , then how many w cents does a liter of kerosene cost ? [ one dollar has 100 cents . ] | "main thing to remember is answer is asked in cents , however when we calculate , it comes up as 0.44 $ just multiply by 100 , answer w = 44 . d" | a ) 0.33 , b ) 0.44 , c ) 0.55 , d ) 44 , e ) 55 | d | multiply(divide(divide(8, divide(const_1, const_2)), const_12), multiply(0.33, 100)) | divide(const_1,const_2)|multiply(n1,n2)|divide(n0,#0)|divide(#2,const_12)|multiply(#3,#1)| | general |
a train requires 5 seconds to pass a pole while it requires 25 seconds to cross a stationary train which is 360 mtrs long . find the speed of the train . | "in 5 s the train crosses the pole and in 25 sec the train crosses one more stationary train in 20 sec the train travels a distance of 360 mtrs speed = 360 / 20 = 18 m / s = 18 ( 3600 / 1000 ) = 18 * 18 / 5 = 64.8 kmph answer : d" | a ) 66.8 kmph , b ) 65.8 kmph , c ) 54.8 kmph , d ) 64.8 kmph , e ) 44.8 kmph | d | multiply(divide(360, subtract(25, 5)), const_3_6) | subtract(n1,n0)|divide(n2,#0)|multiply(#1,const_3_6)| | physics |
a two digit number is 18 less than the square of the sum of its digits . how many such numbers are there ? | option 2 take n = 10 a + b . given that , ( 10 a + b ) + 18 = k 2 = ( a + b ) 2 given number = k 2 - 18 = ( 10 a + b ) that means , when we add 18 to the given number it should be a perfect square . so k 2 takes the following values . 1 , 4 , 9 , 16 , 25 , 36 , 49 , 64 , 81 , 100 , 121 , . . . . 1 to 16 are ruled out a... | a ) 2233 , b ) 2222 , c ) 2211 , d ) 6382 , e ) 23,17 | d | add(multiply(divide(const_100, const_2), const_10), power(18, const_3)) | divide(const_100,const_2)|power(n0,const_3)|multiply(#0,const_10)|add(#2,#1) | general |
what is the average ( arithmetic mean ) of 10 , 20,30 , 40,50 , 60,70 , 80,90 ? | "so addition of all term - 10 , 20 , 30 , . . . . . . . 90 so average = ( 10 + 20 + 30 + 40 + 50 + 60 + 70 + 80 + 90 ) / 9 = ( 450 ) / 9 = 50 hence , the correct answer is e ." | a ) 90 , b ) 95 , c ) 70 , d ) 85 , e ) 50 | e | divide(divide(multiply(add(10, 40,50), add(divide(subtract(40,50, 10), 20,30), const_1)), const_2), add(divide(subtract(40,50, 10), 10), const_1)) | add(n0,n2)|subtract(n2,n0)|divide(#1,n1)|add(#2,const_1)|multiply(#0,#3)|divide(#4,const_2)|divide(#5,#3)| | general |
if the average of 6 x and 8 y is greater than 100 , and x is twice y , what is the least integer value of x ? | "substitution can be used in the following way : always start with the equation : x = 2 y . it is more straight forward to manage as compared to the inequality . substitute y = x / 2 , not the other way because you need to find the minimum value of x . so you can get rid of y . now go on to the inequality . so 8 y = 8 ... | a ) 22 , b ) 20 , c ) 21 , d ) 23 , e ) 24 | b | multiply(6, const_2) | multiply(n0,const_2)| | general |
two cubes of their volumes in the ratio 343 : 729 . the ratio of their surface area is : | "the ratio of their surface area is 343 : 729 7 : 9 answer is a ." | a ) 7 : 9 , b ) 2 : 5 , c ) 3 : 5 , d ) 1 : 5 , e ) 4 : 5 | a | divide(power(343, const_0_33), power(729, const_0_33)) | power(n0,const_0_33)|power(n1,const_0_33)|divide(#0,#1)| | geometry |
two pipes p and q can fill a cistern in 12 and 14 minutes respectively . both are opened together , but at the end of 3 minutes the first is turned off . how much longer will the cistern take to fill ? | "3 / 12 + x / 14 = 1 x = 10 1 / 2 answer : e" | a ) 11 1 / 2 min , b ) 11 1 / 4 min , c ) 11 2 / 4 min , d ) 11 1 / 5 min , e ) 10 1 / 2 min | e | multiply(subtract(const_1, multiply(add(divide(const_1, 12), divide(const_1, 14)), 3)), 14) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|multiply(#2,n2)|subtract(const_1,#3)|multiply(n1,#4)| | physics |
a can complete a project in 20 days and b can complete the same project in 20 days . if a and b start working on the project together and a quits 10 days before the project is completed , in how many days will the project be completed ? | "let x = the number of days taken to complete the project . the amount of work done by a is ( x - 10 ) * ( 1 / 20 ) . the amount of work done by b is ( x ) * ( 1 / 30 ) . ( 1 / 20 ) * ( x - 10 ) + ( 1 / 20 ) * ( x ) = 1 ( x / 20 ) + ( x / 20 ) - ( 10 / 20 ) = 1 x / 10 = 3 / 2 x = 15 therefore , the answer is e : 15 ." | a ) 18 days , b ) 27 days , c ) 26.67 days , d ) 16 days , e ) 15 days | e | add(divide(subtract(const_1, multiply(divide(const_1, 20), 10)), add(divide(const_1, 20), divide(const_1, 20))), 10) | divide(const_1,n1)|divide(const_1,n0)|add(#1,#0)|multiply(n2,#0)|subtract(const_1,#3)|divide(#4,#2)|add(n2,#5)| | physics |
a work crew of 4 men takes 5 days to complete one - half of a job . if 11 men are then added to the crew and the men continue to work at the same rate , how many days will it take the enlarged crew to do the rest of the job ? | "suppose 1 man can do work in x days . . so 4 men will do in . . 4 / x = 1 / 5 * 1 / 2 as half job is done x = 40 now 11 more are added then 15 / 40 = 1 / 2 * 1 / d for remaining half job d = 1 1 / 3 number of days c" | a ) 2 , b ) 3 , c ) 1 1 / 3 , d ) 4 , e ) 4 4 / 5 | c | add(5, divide(multiply(4, 5), add(4, 11))) | add(n0,n2)|multiply(n0,n1)|divide(#1,#0)|add(n1,#2)| | physics |
a train 120 meters long takes 6 seconds to cross a man walking at 5 kmph in the direction opposite to that of the train . find the speed of the train . | "explanation : let the speed of the train be x kmph . speed of the train relative to man = ( x + 5 ) kmph = ( x + 5 ) Γ 5 / 18 m / sec . therefore 120 / ( ( x + 5 ) Γ 5 / 18 ) = 6 < = > 30 ( x + 5 ) = 2160 < = > x = 67 speed of the train is 67 kmph . answer : option a" | a ) 67 kmph , b ) 50 kmph , c ) 55 kmph , d ) 60 kmph , e ) 70 kmph | a | subtract(divide(120, multiply(6, const_0_2778)), 5) | multiply(n1,const_0_2778)|divide(n0,#0)|subtract(#1,n2)| | physics |
a small company reduced its faculty by approximately 13 percent to 195 employees . what was the original number of employees ? | "if x is the original number of employees , then after 13 % reduction in employees number is . 87 x but we are given . 87 x = 195 x = 224 so the original number of employees is 224 correct answer - d" | a ) 182 , b ) 208 , c ) 220 , d ) 224 , e ) 302 | d | divide(195, divide(subtract(const_100, 13), const_100)) | subtract(const_100,n0)|divide(#0,const_100)|divide(n1,#1)| | gain |
population is 22000 . population increases by 10 % every year , then the population after 3 years is ? | "population after 1 st year = 22000 * 10 / 100 = 2200 = = = > 22000 + 2200 = 24000 population after 2 nd year = 24000 * 10 / 100 = 2400 = = = > 24000 + 2400 = 26620 population after 3 rd year = 26620 * 10 / 100 = 2662 = = = > 26620 + 2662 = 29282 answer : b" | a ) 26630 , b ) 29282 , c ) 36620 , d ) 26620 , e ) 26820 | b | multiply(22000, add(const_1, divide(multiply(3, 10), const_100))) | multiply(n1,n2)|divide(#0,const_100)|add(#1,const_1)|multiply(n0,#2)| | gain |
a 180 meter long train crosses a man standing on the platform in 12 sec . what is the speed of the train ? | "s = 180 / 12 * 18 / 5 = 54 kmph answer : a" | a ) 54 kmph , b ) 108 kmph , c ) 176 kmph , d ) 134 kmph , e ) 161 kmph | a | multiply(divide(180, 12), const_3_6) | divide(n0,n1)|multiply(#0,const_3_6)| | physics |
( 26.3 * 12 * 20 ) / 3 + 125 = ? | b 2229 ? = [ ( 26.3 * 12 * 20 ) / 3 ] + 125 ? = 2104 + 125 = 2229 | a ) 2339 , b ) 2229 , c ) 4429 , d ) 5529 , e ) 2669 | b | add(divide(multiply(multiply(26.3, 12), 20), const_3), 125) | multiply(n0,n1)|multiply(n2,#0)|divide(#1,const_3)|add(n4,#2) | general |
rectangular tile each of size 45 cm by 50 cm must be laid horizontally on a rectangular floor of size 250 cm by 180 cm , such that the tiles do not overlap and they are placed with edges jutting against each other on all edges . a tile can be placed in any orientation so long as its edges are parallel to the edges of f... | "area of tile = 45 * 50 = 2250 area of floor = 250 * 180 = 45000 no of tiles = 45000 / 2250 = 20 so , the no of tile = 20 answer : e" | a ) 35 , b ) 40 , c ) 25 , d ) 18 , e ) 20 | e | divide(multiply(250, 180), multiply(45, 50)) | multiply(n2,n3)|multiply(n0,n1)|divide(#0,#1)| | geometry |
if 15 people contributed a total of $ 30.00 toward a gift and each of them contributed at least $ 1.00 , then the maximum possible amount any one person could have contributed is | "b for me 14 people with 1 $ each - > maximum = 16" | a ) $ 1.00 , b ) $ 16 , c ) $ 5.00 , d ) $ 6.00 , e ) $ 20.00 | b | subtract(30.00, multiply(subtract(15, 1.00), 1.00)) | subtract(n0,n2)|multiply(n2,#0)|subtract(n1,#1)| | general |
a water tank , having the shape of a rectangular prism of base 100 square centimeters , is being filled at the rate of 1 liter per minute . find the rate at which the height of the water in the water tank increases . express your answer in centimeters per minute . | the volume of a rectangular prism is given by area of base ? height the rate at which the tank is being filled is 1 liter per minute which we need to convert 1 liter = 1 dm 3 but 1 dm = 10 cm . hence 1 liter = ( 10 cm ) 3 = 1000 cm 3 the height h of the water is related to the volume by volume = area of base * h in one... | ['a ) 30 cm', 'b ) 10 cm', 'c ) 50 cm', 'd ) 90 cm', 'e ) 70 cm'] | b | divide(const_1000, 100) | divide(const_1000,n0) | geometry |
the difference between a two - digit number and the number obtained by interchanging the positions of its digits is 72 . what is the difference between the two digits of that number ? | "sol . let the ten β s digit be x and unit β s digit be y , then , ( 10 x + y ) - ( 10 y + x ) = 72 β 9 ( x - y ) = 72 β x - y = 8 answer d" | a ) 4 , b ) 5 , c ) 6 , d ) 8 , e ) 9 | d | divide(72, subtract(const_10, const_1)) | subtract(const_10,const_1)|divide(n0,#0)| | general |
machine a can put caps on a set of bottles in 8 hours at a constant rate . if machine a does the job for 6 hours and machine b does the rest of the job , which works at 1 / 4 constant rate of machine a . how long will it take for machine b alone to do the rest of the job ? | machine a will do 6 / 8 in 6 hrs , so a does 3 / 4 of the work . . therefore , b will do the remaining 1 / 4 th work alone . . as the speed of b is 1 / 4 rate of a , b will do the 1 / 4 th work in same time that a takes to complete full job . . . ans 8 c | a ) 4 hrs , b ) 6 hrs , c ) 8 hrs , d ) 10 hrs , e ) 16 hrs | c | multiply(subtract(const_1, divide(6, 8)), inverse(multiply(inverse(8), inverse(4)))) | divide(n1,n0)|inverse(n0)|inverse(n3)|multiply(#1,#2)|subtract(const_1,#0)|inverse(#3)|multiply(#5,#4) | physics |
a β s speed is 16 / 15 times that of b . if a and b run a race , what part of the length of the race should a give b as a head start , so that the race ends in a dead heat ? | "let x be the fraction of the distance that b runs . let v be the speed at which b runs . the time should be the same for both runners . time = d / ( 16 v / 15 ) = xd / v ( 15 / 16 ) * d / v = x * d / v x = 15 / 16 b should have a head start of 1 / 16 of the full distance . the answer is b ." | a ) 1 / 8 , b ) 1 / 16 , c ) 1 / 15 , d ) 1 / 32 , e ) 1 / 31 | b | divide(subtract(16, 15), 16) | subtract(n0,n1)|divide(#0,n0)| | general |
6 women can do 75 unit of work in 8 days by working 5 hr / day in how many days 4 women do 30 units of work by working 8 hr / day | ( 6 x 8 x 5 ) / 75 = ( ax 4 x 8 ) / 30 a = 3 answer : a | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | divide(30, multiply(multiply(divide(divide(divide(75, 6), 8), 5), 8), 4)) | divide(n1,n0)|divide(#0,n2)|divide(#1,n3)|multiply(n2,#2)|multiply(n4,#3)|divide(n5,#4) | physics |
50 % of a number is more than 20 % of 650 by 190 . find the number ? | "( 50 / 100 ) * x Γ’ β¬ β ( 20 / 100 ) * 650 = 190 1 / 2 x = 320 x = 640 answer : a" | a ) 640 , b ) 278 , c ) 800 , d ) 267 , e ) 121 | a | divide(multiply(add(divide(multiply(650, 20), const_100), 190), const_100), 50) | multiply(n1,n2)|divide(#0,const_100)|add(n3,#1)|multiply(#2,const_100)|divide(#3,n0)| | gain |
what is the units digit of the expression 14 ^ 7 β 16 ^ 4 ? | i think answer on this one should be c too . since we know that 14 ^ 7 > 16 ^ 4 , as will said one should always check if the number is positive . | a ) 0 , b ) 3 , c ) 8 , d ) 6 , e ) 4 | c | add(add(const_4, const_3), const_2) | add(const_3,const_4)|add(#0,const_2)| | general |
when xyz co discontinued an agency arrangement with john , management held his commission of rs . 25000 / - for one month . earlier john had availed an advance agency fees of 8280 / - from company . but robert forgot that . after one month john asked his fees and accountant gives rs . 18500 / - to him . what is the inc... | total fees = rs . 25000 / - advance fees = 8280 / - balance fees = 25000 - 8280 = 16720 paid amount = 18500 / - incentive amount = 18500 - 16720 = 1780 / - answer is a | a ) a ) 1780 , b ) b ) 1250 , c ) c ) 10780 , d ) d ) 10500 , e ) e ) 8600 | a | subtract(add(8280, 18500), 25000) | add(n1,n2)|subtract(#0,n0) | general |
if | 5 x - 5 | = 150 , what is the sum of all the possible values of x ? | "there will be two cases 5 x - 5 = 150 or 5 x - 5 = - 150 = > x = 31 or x = - 29 sum of both the values will be - 29 + 31 = 2 answer is a" | a ) 2 , b ) - 2 , c ) 6 , d ) - 3 , e ) 5 | a | subtract(subtract(subtract(150, 5), add(150, 5)), 5) | add(n1,n2)|subtract(n2,n1)|subtract(#1,#0)|subtract(#2,n1)| | general |
x is the product of each integer from 1 to 100 , inclusive and y = 100 ^ k , where k is an integer . what is the greatest value of k for which y is a factor of x ? | "the number of trailing zeros in the decimal representation of n ! , the factorial of a non - negative integer n , can be determined with this formula : n 5 + n 52 + n 53 + . . . + n 5 k , where k must be chosen such that 5 k β€ n x = 1 * 2 * 3 . . . . * 100 = 100 ! no . of trailing zeros in 100 ! = 100 / 5 + 100 / 5 ^ ... | a ) 5 , b ) 6 , c ) 8 , d ) 10 , e ) 12 | e | add(divide(100, const_10), 1) | divide(n1,const_10)|add(#0,n0)| | general |
jean drew a gumball at random from a jar of pink and blue gumballs . since the gumball she selected was blue and she wanted a pink one , she replaced it and drew another . the second gumball also happened to be blue and she replaced it as well . if the probability of her drawing the two blue gumballs was 9 / 49 , what ... | the probability of drawing a pink gumball both times is the same . the probability that she drew two blue gumballs = 9 / 49 = ( 3 / 7 ) * ( 3 / 7 ) therefore probability that the next one she draws is pink = 4 / 7 option ( b ) | a ) 1 / 49 , b ) 4 / 7 , c ) 3 / 7 , d ) 16 / 49 , e ) 40 / 49 | b | subtract(const_1, sqrt(divide(9, 49))) | divide(n0,n1)|sqrt(#0)|subtract(const_1,#1) | general |
a sum was put a simple interest at a certain rate for 2 years . had it been put at 4 % higher rate , it would have fetched rs . 60 more . the sum is : | "explanation : let the sub be rs . x and the initial rate be r % . then x Γ ( r + 4 ) Γ 2 / 100 β x Γ r Γ 2 / 100 = 60 β x Γ 4 Γ 2 / 100 = 60 β x Γ 2 / 100 = 15 β 2 x = 1500 β x = 750 answer : option a" | a ) rs . 750 , b ) rs . 700 , c ) rs . 820 , d ) rs . 940 , e ) rs . 900 | a | multiply(divide(divide(60, 2), 4), const_100) | divide(n2,n0)|divide(#0,n1)|multiply(#1,const_100)| | gain |
three coins are tossed . find the probability of at least 2 tails ? | "n ( s ) = 2 ^ 3 = 8 let e is the event of getting at least 2 tails n ( e ) = ttt , tth , htt , tht = 4 p ( e ) = n ( e ) / n ( s ) = 4 / 8 = 1 / 2 ans - a" | a ) 1 / 2 , b ) 2 / 3 , c ) 1 / 4 , d ) 1 / 8 , e ) 3 / 8 | a | divide(2, power(const_2, 2)) | power(const_2,n0)|divide(n0,#0)| | probability |
a train can travel 50 % faster than a car . both start from point a at the same time and reach point b at 75 kms away from a at the same time . on the way , however , the train lost about 12.5 minutes while stopping at the stations . the speed of the car is ? | "let speed of the car be x km / hr . then , speed of the train = 150 / 100 x = 3 / 2 x km / hr . 75 / x - 75 / ( 3 / 2 x ) = 125 / ( 10 * 60 ) 75 / x - 50 / x = 5 / 24 = 120 km / hr . answer : c" | a ) 2387 , b ) 267 , c ) 120 , d ) 277 , e ) 271 | c | divide(multiply(subtract(multiply(12.5, const_2), const_1), multiply(12.5, const_2)), add(const_4, const_1)) | add(const_1,const_4)|multiply(n2,const_2)|subtract(#1,const_1)|multiply(#1,#2)|divide(#3,#0)| | physics |
tom traveled the entire 60 miles trip . if he did the first 12 miles of at a constant rate 24 miles per hour and the remaining trip of at a constant rate 48 miles per hour , what is the his average speed , in miles per hour ? | "avg speed = total distance / total time = ( d 1 + d 2 ) / ( t 1 + t 2 ) = ( 12 + 48 ) / ( ( 12 / 24 ) + ( 48 / 48 ) ) = 60 * 2 / 3 = 40 mph d" | a ) 20 mph , b ) 24 mph , c ) 30 mph , d ) 40 mph , e ) 42 mph | d | divide(60, add(divide(48, subtract(60, 12)), divide(12, 24))) | divide(n1,n2)|subtract(n0,n1)|divide(n3,#1)|add(#2,#0)|divide(n0,#3)| | physics |
a grocer has a sale of rs . 2435 , rs . 2920 , rs . 2855 , rs . 3230 and rs . 2560 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . 2500 ? | "total sale for 5 months = rs . ( 2435 + 2920 + 2855 + 3230 + 2560 ) = rs . 14000 . required sale = rs . [ ( 2500 x 6 ) - 14000 ] = rs . ( 15000 - 14000 ) = rs . 1000 answer : option c" | a ) 5000 , b ) 4000 , c ) 1000 , d ) 6000 , e ) 8000 | c | subtract(multiply(add(5, const_1), 2500), add(add(add(add(2435, 2920), 2855), 3230), 2560)) | add(n5,const_1)|add(n0,n1)|add(n2,#1)|multiply(n6,#0)|add(n3,#2)|add(n4,#4)|subtract(#3,#5)| | general |
a mixture contains alcohol and water in the ratio 4 : 3 . if 5 litres of water is added to the mixture , the ratio becomes 4 : 5 . find the quantity of alcohol in the given mixture | "let the quantity of alcohol and water be 4 x litres and 3 x litres respectively 4 x / ( 3 x + 5 ) = 4 / 5 ο³ 20 x = 4 ( 3 x + 5 ) ο³ 8 x = 20 ο³ x = 2.5 quantity of alcohol = ( 4 x 2.5 ) litres = 10 litres . answer a 10 litres" | a ) 10 litres , b ) 20 litres , c ) 30 litres , d ) 5 litres , e ) 14 litres | a | multiply(5, const_1) | multiply(n2,const_1)| | general |
evaluate 28 % of 550 + 45 % of 280 | "explanation : = ( 28 / 100 ) * 550 + ( 45 / 100 ) * 280 = 154 + 126 = 280 answer : option e" | a ) 232 , b ) 242 , c ) 252 , d ) 262 , e ) 280 | e | divide(28, divide(550, 28)) | divide(n1,n0)|divide(n0,#0)| | gain |
what sum of money put at c . i amounts in 2 years to rs . 8840 and in 3 years to rs . 9261 ? | "8840 - - - - 421 100 - - - - ? = > 4.76 % x * 104.76 / 100 * 104.76 / 100 = 8840 x * 1.0975 = 8840 x = 8840 / 1.0975 = > 8054.54 answer : a" | a ) 8054 , b ) 8877 , c ) 2877 , d ) 2678 , e ) 1011 | a | divide(8840, power(add(subtract(divide(9261, 8840), const_1), const_1), 2)) | divide(n3,n1)|subtract(#0,const_1)|add(#1,const_1)|power(#2,n0)|divide(n1,#3)| | general |
if square - root of 15 is 3.87 . then find the value of square - root of ( 5 / 3 ) | = > root 5 / root 3 = ( root 5 / root 3 ) * ( root 3 / root 3 ) = root 15 / root 9 = 3.87 / 3 = 1.29 answer : a | a ) 1.29 , b ) 1.39 , c ) 1.49 , d ) 1.59 , e ) 1.69 | a | sqrt(divide(5, 3)) | divide(n2,n3)|sqrt(#0) | general |
a salesman sold twice as much pears in the afternoon than in the morning . if he sold $ 480 kilograms of pears that day , how many kilograms did he sell in the afternoon ? | "3 x = 480 x = 160 therefore , the salesman sold 160 kg in the morning and 2 β
160 = 320 kg in the afternoon . so answer is e ." | a ) 120 , b ) 180 , c ) 240 , d ) 280 , e ) 320 | e | multiply(divide(480, const_3), const_2) | divide(n0,const_3)|multiply(#0,const_2)| | other |
0.004 Γ 0.5 = ? | "explanation : 0.004 Γ 0.5 = 0.002 . answer : option c" | a ) none of these , b ) 0.02 , c ) 0.002 , d ) 0.0002 , e ) 2 | c | multiply(divide(0.004, 0.5), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
on rainy mornings , mo drinks exactly n cups of hot chocolate ( assume that n is an integer ) . on mornings that are not rainy , mo drinks exactly 3 cups of tea . last week mo drank a total of 20 cups of tea and hot chocolate together . if during that week mo drank 10 more tea cups than hot chocolate cups , then how ma... | t = the number of cups of tea c = the number of cups of hot chocolate t + c = 20 t - c = 10 - > t = 15 . c = 5 . mo drinks 3 cups of tea a day then number of days that are not rainy = 15 / 3 = 5 so number of rainy days = 7 - 5 = 2 a is the answer . | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | a | subtract(add(const_4, const_3), divide(divide(add(20, 10), const_2), 3)) | add(const_3,const_4)|add(n1,n2)|divide(#1,const_2)|divide(#2,n0)|subtract(#0,#3) | general |
how long will a boy take to run round a square field of side 50 meters , if he runs at the rate of 10 km / hr ? | "speed = 10 km / hr = 10 * 5 / 18 = 25 / 9 m / sec distance = 50 * 4 = 200 m time taken = 200 * 9 / 25 = 72 sec answer is c" | a ) 52 sec , b ) 45 sec , c ) 72 sec , d ) 25 sec , e ) 39 sec | c | divide(multiply(50, const_4), multiply(10, divide(const_1000, const_3600))) | divide(const_1000,const_3600)|multiply(n0,const_4)|multiply(n1,#0)|divide(#1,#2)| | gain |
the roof of an apartment building is rectangular and its length is 3 times longer than its width . if the area of the roof is 675 feet squared , what is the difference between the length and the width of the roof ? | "let the width = x x * 3 x = 675 x ^ 2 = 225 x = 15 length = 3 * 15 = 45 difference = 45 - 15 = 30 b is the answer" | a ) 38 . , b ) 30 . , c ) 42 . , d ) 44 . , e ) 46 . | b | subtract(multiply(sqrt(divide(675, 3)), 3), sqrt(divide(675, 3))) | divide(n1,n0)|sqrt(#0)|multiply(#1,n0)|subtract(#2,#1)| | geometry |
a grocery shop has a sale of rs . 6235 , rs . 6927 , rs . 6855 , rs . 7230 and rs . 6562 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . 6500 ? | "let the sale in the sixth month = x then ( 6235 + 6927 + 6855 + 7230 + 6562 + x ) / 6 = 6500 = > 6235 + 6927 + 6855 + 7230 + 6562 + x = 6 Γ 6500 = > 33809 + x = 39000 = > x = 39000 β 33809 = 5191 answer : d" | a ) 4857 , b ) 4184 , c ) 4012 , d ) 5191 , e ) 5291 | d | subtract(multiply(6500, add(5, const_1)), add(add(add(6855, add(6235, 6927)), 7230), 6562)) | add(n5,const_1)|add(n0,n1)|add(n2,#1)|multiply(n6,#0)|add(n3,#2)|add(n4,#4)|subtract(#3,#5)| | general |
a train 225 m long passes a man , running at 5 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 = ( 225 / 10 ) m / sec = ( 45 / 2 ) m / sec . [ ( 45 / 2 ) * ( 18 / 5 ) ] km / hr = 81 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 5 ) km / hr . x - 5 = 81 = = > x = 86 km / hr . answer : option a" | a ) 86 , b ) 50 , c ) 12 , d ) 13 , e ) 67 | a | divide(divide(subtract(225, multiply(multiply(5, const_0_2778), 5)), 5), const_0_2778) | multiply(n1,const_0_2778)|multiply(n1,#0)|subtract(n0,#1)|divide(#2,n1)|divide(#3,const_0_2778)| | physics |
0.0006688 / 0.0000120 x 17.05 = ? | "explanation : ? = 0.0006688 / 0.0000120 x 17.05 = 950.1 answer : option e" | a ) 450.1 , b ) 910.1 , c ) 930.1 , d ) 920.1 , e ) 950.1 | e | divide(0.0006688, 0.0000120) | divide(n0,n1)| | general |
the average age of a group of 10 persons was decreased by 3 years when one person , whose age was 45 years , was replaced by a new person . find the age of the new person ? | initial average age of the 10 persons be p . age of the new person q . sum of the ages of the initial 10 persons = 10 p new average = ( p - 3 ) 10 ( p - 3 ) = 10 p - 45 + q = > q = 15 answer : d | a ) 18 , b ) 56 , c ) 12 , d ) 15 , e ) 14 | d | subtract(45, multiply(10, 3)) | multiply(n0,n1)|subtract(n2,#0) | general |
a car traveled 75 % of the way from town a to town b at an average speed of 75 mph . the car traveled at an average speed of v mph for the remaining part of the trip . the average speed for the entire trip was 50 mph . what is v in mph ? | "assume total distance = 100 miles time taken for 75 miles = 75 / 75 = 1 hour time taken for the rest of the 25 miles = 25 / v hours . average speed = 50 therefore the total time needed = 2 hours . 2 = 1 + 25 / v hence v = 25 mph answer : c" | a ) 30 , b ) 35 , c ) 25 , d ) 40 , e ) 45 | c | divide(subtract(add(50, 75), 75), subtract(divide(add(50, 75), 50), divide(75, 75))) | add(n1,n2)|divide(n0,n1)|divide(#0,n2)|subtract(#0,n0)|subtract(#2,#1)|divide(#3,#4)| | general |
if 20 men can build a wall 66 metres long in 15 days , what length of a similar can be built by 86 men in 8 days ? | "if 20 men can build a wall 66 metres long in 15 days , length of a similar wall that can be built by 86 men in 8 days = ( 66 * 86 * 8 ) / ( 15 * 20 ) = 151.36 mtrs answer : a" | a ) 151.36 mtrs , b ) 378.4 mtrs , c ) 478.4 mtrs , d ) 488.4 mtrs , e ) 578.4 mtrs | a | multiply(66, divide(multiply(86, 8), multiply(20, 15))) | multiply(n3,n4)|multiply(n0,n2)|divide(#0,#1)|multiply(n1,#2)| | physics |
on a certain day , orangeade was made by mixing a certain amount of orange juice with an equal amount of water . on the next day , orangeade was made by mixing the same amount of orange juice with twice the amount of water . on both days , all the orangeade that was made was sold . if the revenue from selling the orang... | "we are given that orangeade is made on day 1 with an equal amount of water and orange juice . we can set this information up into a ratio using a variable multiplier : w : oj = x : x thus , orangeade quantity = amount of water + amount of oj = x + x = 2 x we are next given that orangeade on day 2 was made by mixing th... | a ) $ 0.15 , b ) $ 0.20 , c ) $ 0.30 , d ) $ 0.40 , e ) $ 0.45 | d | divide(multiply(add(const_1, const_1), 0.60), add(const_1, const_2)) | add(const_1,const_1)|add(const_1,const_2)|multiply(n0,#0)|divide(#2,#1)| | general |
two consultants can type up a report in 12.5 hours and edit it in 7.5 hours . if mary needs 30 hours to type the report and jim needs 12 hours to edit it alone , how many r hours will it take if jim types the report and mary edits it immediately after he is done ? | "break down the problem into two pieces : typing and editing . mary needs 30 hours to type the report - - > mary ' s typing rate = 1 / 30 ( rate reciprocal of time ) ( point 1 in theory below ) ; mary and jim can type up a report in 12.5 and - - > 1 / 30 + 1 / x = 1 / 12.5 = 2 / 25 ( where x is the time needed for jim ... | a ) 41.4 , b ) 34.1 , c ) 13.4 , d ) 12.4 , e ) 10.8 | a | add(inverse(subtract(divide(const_1, 12.5), divide(const_1, 30))), inverse(subtract(divide(const_1, 7.5), divide(const_1, 12)))) | divide(const_1,n0)|divide(const_1,n2)|divide(const_1,n1)|divide(const_1,n3)|subtract(#0,#1)|subtract(#2,#3)|inverse(#4)|inverse(#5)|add(#6,#7)| | physics |
the number 1254 is divisible by which number ? | last 3 digit can not divisible by 8 so , reject option a 1254 is not divisible by 7 last number is not 0 or 5 , so neglect 4 is not divide 1254 divisibility rule of 6 is a number divisible both by 3 and 2 . 1254 attains both divisible . so the number is divisible by 6 option e | a ) 8 , b ) 7 , c ) 4 , d ) 5 , e ) 6 | e | divide(1254, add(add(const_100, const_100), subtract(const_10, const_1))) | add(const_100,const_100)|subtract(const_10,const_1)|add(#0,#1)|divide(n0,#2) | general |
what is the cost of leveling the field in the form of parallelogram at the rate of rs . 40 / 10 sq . metre , whose base & perpendicular distance from the other side being 24 m & 34 m respectively ? | "area of the parallelogram = length of the base * perpendicular height = 24 * 34 = 816 m . total cost of levelling = rs . 816 e" | a ) s . 600 , b ) s . 630 , c ) s . 640 , d ) s . 648 , e ) s . 816 | e | multiply(multiply(24, 34), divide(40, 10)) | divide(n0,n1)|multiply(n2,n3)|multiply(#0,#1)| | physics |
a certain no . when divided by 35 leaves aremainder 25 , what is the remainder if the same no . be divided by 15 ? | explanation : 35 + 25 = 60 / 15 = 4 ( remainder ) c | a ) 1 , b ) 3 , c ) 4 , d ) 6 , e ) 8 | c | subtract(subtract(subtract(35, 25), const_4), const_2) | subtract(n0,n1)|subtract(#0,const_4)|subtract(#1,const_2) | general |
on a certain road , 40 % of the motorists exceed the posted speed limit and receive speeding tickets , but 20 % of the motorists who exceed the posted speed limit do not receive speeding tickets . what percent of the motorists on that road exceed the posted speed limit ? | "suppose there are x motorists . 40 % of them exceeded the speed limit and received the ticket , i . e . 2 x / 5 . again , suppose total no . of motorists who exceeded the speed limit are y . 20 % of y exceeded the speed limit but did n ' t received the ticket , i . e . y / 5 . it means 4 y / 5 received the ticket . he... | a ) 10.5 % , b ) 12.5 % , c ) 15 % , d ) 22 % , e ) 50 % | e | multiply(divide(40, subtract(const_100, 20)), const_100) | subtract(const_100,n1)|divide(n0,#0)|multiply(#1,const_100)| | gain |
if x dollars is invested at 10 percent for one year and y dollars is invested at 8 percent for one year , the annual income from the 10 percent investment will exceed the annual income from the 8 percent investment by $ 56 . if $ 2,000 is the total amount invested , how much is invested at 8 percent ? | "( a ) $ 280 ( b ) $ 800 ( c ) $ 892 ( d ) $ 1,108 ( e ) $ 1,200 y = 2000 - x . 1 x - . 08 ( 2000 - x ) = 56 . - 160 + . 18 x = 56 x = 1200 and y = 800 answer : b" | a ) $ 280 , b ) $ 800 , c ) $ 892 , d ) $ 1,108 , e ) $ 1,200 | b | subtract(multiply(multiply(const_100, 10), const_2), divide(add(multiply(multiply(10, 8), const_2), 56), add(divide(10, const_100), divide(8, const_100)))) | divide(n0,const_100)|divide(n1,const_100)|multiply(n0,const_100)|multiply(n0,n1)|add(#0,#1)|multiply(#2,const_2)|multiply(#3,const_2)|add(n4,#6)|divide(#7,#4)|subtract(#5,#8)| | general |
if a mixture is 1 β 2 alcohol by volume and 3 β 2 water by volume , what is the ratio of the volume of alcohol to the volume of water in this mixture ? | "should be a sub - 600 level q . . volume = { 1 / 2 } / { 3 / 2 } = 1 / 3 c" | a ) 3 β 7 , b ) 4 β 7 , c ) 3 β 4 , d ) 4 β 3 , e ) 7 β 4 | c | divide(divide(const_2.0, 2), divide(3, 2)) | divide(const_2.0,n1)|divide(n2,n1)|divide(#0,#1)| | general |
a number when divided by 342 gives a remainder 38 , what remainder will be obtained by dividing the same number 21 ? | "explanation : 342 + 38 = 380 / 21 = 2 ( remainder ) answer : d" | a ) 7 , b ) 6 , c ) 9 , d ) 2 , e ) 3 | d | subtract(38, multiply(21, const_2)) | multiply(n2,const_2)|subtract(n1,#0)| | general |
at the beginning of 1992 , maria ' s stock portfolio was worth $ 600000 . during 1992 , the portfolio ' s value increased by 95 % . during the next year , the portfolio increased its worth by only 5 % . what was maria ' s portfolio worth , in dollars , by the end of 1993 ? | by the end of 1992 , the port folio worth is 1.95 * ( $ 600 k ) by 1993 , the portfolio is 1.05 * ( 1.95 * $ 600 k ) = $ 1 , 228,500 answer : d | a ) 1 , 180000 , b ) 1 , 200000 , c ) 1 , 200300 , d ) 1 , 228500 , e ) 1 , 330,000 | d | divide(600000, const_3) | divide(n1,const_3) | general |
amit and ananthu can do a work in 15 days and 45 days respectively . amit started the work and left after 3 days . ananthu took over and completed the work . in how many days was the total work completed ? | "amit β s one day β s work = 1 / 15 amit β s 3 day β s work = 1 / 15 * 3 = 1 / 5 work left = 1 - 1 / 5 = 4 / 5 ananthu β s one day β s work = 1 / 45 ananthu can do work in = 4 / 5 * 45 = 36 days so total days = 36 + 3 = 39 days answer : e" | a ) 28 days , b ) 30 days , c ) 33 days , d ) 35 days , e ) 39 days | e | add(divide(subtract(const_1, multiply(inverse(15), 3)), inverse(45)), 3) | inverse(n0)|inverse(n1)|multiply(#0,n2)|subtract(const_1,#2)|divide(#3,#1)|add(#4,n2)| | physics |
a pet store holds cats and dogs . if the difference between the number of cats and the number of dogs is 5 . what could be the ratio of cats to dogs in the pet store ? | "say theratioof cats to dogs is a / b . then thenumberof cats would be ax and thenumberof dogs bx , for some positive integer x . we are told that ax - bx = 5 - - > x ( a - b ) = 5 . since 5 is a prime number it could be broken into the product of two positive multiples only in one way : x ( a - b ) = 1 * 5 . the above... | a ) 1 : 3 , b ) 20 : 25 , c ) 1 : 5 , d ) 2 : 5 , e ) 4 : 6 | b | divide(multiply(5, const_4), add(multiply(5, const_4), 5)) | multiply(n0,const_4)|add(n0,#0)|divide(#0,#1)| | other |
a number is said to be prime saturated if the product of all the different positive prime factors of w is less than the square root of w . what is the greatest two digit prime saturated integer ? | "w 96 = 3 * 32 = 3 * 2 ^ 5 answer is d ." | a ) 99 , b ) 98 , c ) 97 , d ) 96 , e ) 95 | d | add(multiply(const_2, const_3), subtract(const_100, const_10)) | multiply(const_2,const_3)|subtract(const_100,const_10)|add(#0,#1)| | other |
54 is to be divided into two parts such that the sum of 10 times the first and 22 times the second is 780 . the bigger part is | "solution let the two parts be ( 54 - x ) and x . then , 10 ( 54 - x ) + 22 x = 780 = 12 x = 240 βΉ = βΊ x = 20 . therefore bigger part = ( 54 - x ) = 34 . answer d" | a ) 24 , b ) 30 , c ) 32 , d ) 34 , e ) 36 | d | subtract(54, divide(subtract(780, multiply(54, 10)), subtract(22, 10))) | multiply(n0,n1)|subtract(n2,n1)|subtract(n3,#0)|divide(#2,#1)|subtract(n0,#3)| | general |
a work which could be finished in 9 days was finished 3 days earlier after 10 more men joined . the number of men employed was ? | "explanation : x - - - - - - - 9 ( x + 10 ) - - - - 6 x * 9 = ( x + 10 ) 6 x = 20 answer : b" | a ) 18 , b ) 20 , c ) 22 , d ) 24 , e ) 26 | b | divide(multiply(multiply(3, const_2), 10), subtract(9, multiply(3, const_2))) | multiply(n1,const_2)|multiply(n2,#0)|subtract(n0,#0)|divide(#1,#2)| | physics |
if the compound interest on a certain sum of money for 6 years at 10 % per annum be rs . 993 , what would be the simple interest ? | "let p = principal a - amount we have a = p ( 1 + r / 100 ) 3 and ci = a - p atq 993 = p ( 1 + r / 100 ) 3 - p ? p = 3000 / - now si @ 10 % on 3000 / - for 6 yrs = ( 3000 x 10 x 6 ) / 100 = 1800 / - answer : c ." | a ) rs . 880 , b ) rs . 890 , c ) rs . 1800 , d ) rs . 900 , e ) none | c | divide(multiply(multiply(multiply(multiply(const_3.0, const_100), 10), 10), 6), const_100) | multiply(const_3.0,const_100)|multiply(n1,#0)|multiply(n1,#1)|multiply(n0,#2)|divide(#3,const_100)| | gain |
a bicycle wheel has a diameter of 0.66 m . how many complete revolutions does it make in 1 km ? | "1 revolution = 3.14 * diameter . number of revolutions in 1 km = 1000 m / ( 3.14 * 0.66 m ) = 482.5 . hence 482 complete revolutions . answer d" | a ) 246 , b ) 448 , c ) 1408 , d ) 482 , e ) 223 | d | divide(multiply(const_1000, 1), multiply(add(const_3, divide(add(multiply(const_3, const_4), const_2), power(add(multiply(const_4, const_2), const_2), const_2))), 0.66)) | multiply(n1,const_1000)|multiply(const_3,const_4)|multiply(const_2,const_4)|add(#1,const_2)|add(#2,const_2)|power(#4,const_2)|divide(#3,#5)|add(#6,const_3)|multiply(n0,#7)|divide(#0,#8)| | physics |
a sum of money amounts to rs . 9800 after 5 years and rs . 12015 after 8 years at the same rate of simple interest . the rate of interest per annum is ? | s . i . for 3 years = ( 12015 - 9800 ) = rs . 2215 s . i . for 5 years = rs . 2215 / 3 * 5 = rs . 3691 . principal = ( 9800 - 3675 ) = rs . 6108 hence , rate = ( 100 * 3691 ) / ( 6125 * 5 ) = 12.05 % answer : a | a ) 12.05 % , b ) 5 % , c ) 13.05 % , d ) 22.05 % , e ) 18.05 % | a | multiply(divide(subtract(12015, 9800), subtract(multiply(9800, 8), multiply(5, 12015))), const_100) | multiply(n0,n3)|multiply(n1,n2)|subtract(n2,n0)|subtract(#0,#1)|divide(#2,#3)|multiply(#4,const_100) | gain |
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