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
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in the third grade of windblown school there are 108 students , one third of them failed the math test and 1 / 6 failed that literature test . at least how many students failed both tests ? | "one third of them failed the math test = 1 / 3 of 108 = 36 . passed in math test = 82 . and 1 / 6 failed that literature test = 18 . it is a possible scenario that these 18 ( failed in literature ) were the ones who passed the math test . so in this case , no one failed in both . hence a ." | a ) 0 . , b ) 6 . , c ) 8 . , d ) 10 , e ) 12 | a | min(divide(108, const_3), divide(108, 6)) | divide(n0,const_3)|divide(n0,n2)|min(#0,#1)| | other |
if a ( a - 4 ) = 21 and b ( b - 4 ) = 21 , where a ≠ b , then a + b = | "a ( a - 4 ) = 21 and b ( b - 4 ) = 21 = > a , b must be integers and if a is - 3 or 7 , b will be 7 and - 3 respectively = > a + b = 4 ans : c" | a ) − 48 , b ) − 2 , c ) 4 , d ) 46 , e ) 48 | c | subtract(subtract(subtract(subtract(add(add(4, 21), subtract(4, 21)), const_1), const_1), const_1), const_1) | add(n0,n1)|subtract(n0,n1)|add(#0,#1)|subtract(#2,const_1)|subtract(#3,const_1)|subtract(#4,const_1)|subtract(#5,const_1)| | general |
a housing society has been allotted a square piece of land measuring 2550.25 sq . m . what is the side of the plot ? | solution side = √ 2550.25 = √ 255025 / 100 = 505 / 10 = 50.5 m . answer b | ['a ) 50.25 m', 'b ) 50.5 m', 'c ) 50.65 m', 'd ) 50.85', 'e ) none'] | b | sqrt(2550.25) | sqrt(n0) | physics |
in a certificate by mistake a candidate gave his height as 10 % more than actual height . in the interview panel , he clarified that his height was 5 feet 5 nches . find the % correction made by the candidate from his stated height to his actual height ? | "his height was = 5 feet 5 inch = 5 + 60 = 65 inch . required % correction = 65 * ( 1.10 - 1 ) = 6.5 a" | a ) 6.5 , b ) 7 , c ) 8 , d ) 9 , e ) 5 | a | multiply(divide(subtract(add(10, add(const_0_25, const_0_25)), 5), add(const_100, subtract(add(10, add(const_0_25, const_0_25)), 5))), const_100) | add(const_0_25,const_0_25)|add(n0,#0)|subtract(#1,n1)|add(#2,const_100)|divide(#2,#3)|multiply(#4,const_100)| | gain |
rs . 900 becomes rs . 956 in 3 years at a certain rate of simple interest . if the rate of interest is increased by 4 % , what amount will rs . 900 become in 3 years ? | "solution s . i . = rs . ( 956 - 900 ) = rs . 56 rate = ( 100 x 56 / 900 x 3 ) = 56 / 27 % new rate = ( 56 / 27 + 4 ) % = 164 / 27 % new s . i . = rs . ( 900 x 164 / 27 x 3 / 100 ) rs . 164 . ∴ new amount = rs . ( 900 + 164 ) = rs . 1064 . answer c" | a ) rs . 1020.80 , b ) rs . 1025 , c ) rs . 1064 , d ) data inadequate , e ) none of these | c | add(900, divide(multiply(multiply(900, add(divide(multiply(subtract(956, 900), const_100), multiply(900, 3)), 4)), 3), const_100)) | multiply(n0,n2)|subtract(n1,n0)|multiply(#1,const_100)|divide(#2,#0)|add(n3,#3)|multiply(n0,#4)|multiply(n2,#5)|divide(#6,const_100)|add(n0,#7)| | gain |
if the population of a certain country increases at the rate of one person every 20 seconds , by how many persons does the population increase in 30 minutes ? | answer = 3 * 30 = 90 answer = d | a ) 55 , b ) 20 , c ) 96 , d ) 90 , e ) 40 | d | multiply(divide(const_60, 20), 30) | divide(const_60,n0)|multiply(n1,#0) | physics |
jake took a bus from home to market , that travels at 10 kmph . while walking back at 2 kmph , halfway through , he suddenly realized he was getting late and he cycled back the remaining distance in 50 kmph . find the average speed . | let the distance be 2 x ( one way ) time taken by bus = 2 x / 10 , by walking = x / 2 , by cycling = x / 50 hours : . average speed = total distance / total time = 2 x / x / 5 + x / 2 + x / 50 = 2 * 60 / 12 + 30 + 1.2 = 1.67 answer : b | a ) 1.17 kmph , b ) 1.67 kmph , c ) 0.69 kmph , d ) 1.45 kmph , e ) none of these | b | divide(multiply(50, 2), add(add(10, 2), 50)) | add(n0,n1)|multiply(n1,n2)|add(n2,#0)|divide(#1,#2) | physics |
a survey reveals that the average income of a company ’ s customers is $ 45000 per year . if 50 customers respond to the survey and the average income of the wealthiest 10 of those customers is $ 55000 , what is the average income of the other 40 customers ? is there a way to solve this using weighted average concept i... | let x be the average of 40 customers 40 * x + 10 * 55000 = 50 * 45000 solving this we have x = 42500 answer is d . | a ) $ 27,500 , b ) $ 35,000 , c ) $ 37,500 , d ) $ 42,500 , e ) $ 50,000 | d | divide(subtract(multiply(45000, 50), multiply(10, 55000)), 40) | multiply(n0,n1)|multiply(n2,n3)|subtract(#0,#1)|divide(#2,n4) | general |
hcf of two numbers is 11 and the product of these numbers is 363 , what is the the greater number ? | the numbers be 11 a and 11 b 11 a × 11 b = 363 ab = 3 co - primes with product 3 are ( 1 , 3 ) hence the numbers with hcf 11 and product 363 = ( 11 × 1 , 11 × 3 ) = ( 11 , 33 ) numbers are 11 and 33 the greater number = 33 | a ) 28 , b ) 31 , c ) 33 , d ) 35 , e ) 38 | c | divide(363, 11) | divide(n1,n0) | general |
if n divided by 7 has a remainder of 2 , what is the remainder when 4 times n is divided by 7 ? | "as per question = > n = 7 p + 2 for some integer p hence 4 n = > 28 q + 8 = > remainder = > 6 for some integer q alternatively = > n = 2 > 3 n = > 8 = > 8 divided by 7 will leave a remainder 1 hence a" | a ) 1 , b ) 2 , c ) 3 , d ) 5 , e ) 8 | a | multiply(4, 2) | multiply(n1,n2)| | general |
running at the same rate , 8 identical machines can produce 560 paperclips a minute . at this rate , how many paperclips could 12 machines produce in 6 minutes ? | "8 machines produce 560 in 1 min 8 machines produce 560 * 6 in 6 min 12 machine produce 560 * 6 * ( 12 / 8 ) in 6 minutes 560 * 6 * 12 / 8 = 5040 answer is d ." | a ) 1344 , b ) 3360 , c ) 8400 , d ) 5040 , e ) 67200 | d | multiply(multiply(560, divide(12, 8)), 6) | divide(n2,n0)|multiply(n1,#0)|multiply(n3,#1)| | gain |
if 8 workers can build 8 cars in 8 days , then how many days would it take 5 workers to build 5 cars ? | "8 workers can build 1 car per day on average . 1 worker can build 1 / 8 of a car per day . 5 workers can build 5 / 8 car per day . the time required to build 5 cars is 5 / ( 5 / 8 ) = 8 days the answer is c ." | a ) 4 , b ) 5 , c ) 8 , d ) 10 , e ) 12 | c | multiply(divide(multiply(8, 8), 8), divide(5, 5)) | divide(n3,n3)|multiply(n0,n0)|divide(#1,n0)|multiply(#2,#0)| | physics |
square m is inscribed in circle n . if the perimeter of m is 20 , what is the circumference of n ? | square forms two right angled triangles . any time we have a right angle triangle inside a circle , the hypotenuse is the diameter . hypotenuse here = diagonal of the square = 5 sqrt ( 2 ) = diameter = > radius = 2.5 sqrt ( 2 ) circumference of the circle = 2 pi r = 5 pi sqrt ( 2 ) answer is e . | ['a ) 33 √ 2 π', 'b ) 2 √ 2 π', 'c ) 6 √ 2 π', 'd ) 7 √ 2 π', 'e ) 5 √ 2 π'] | e | circumface(divide(sqrt(multiply(power(divide(20, const_4), const_2), const_2)), const_2)) | divide(n0,const_4)|power(#0,const_2)|multiply(#1,const_2)|sqrt(#2)|divide(#3,const_2)|circumface(#4) | geometry |
if both 7 ^ 2 and 3 ^ 4 are factors of the number a * 4 ^ 3 * 6 ^ 2 * 13 ^ 11 , then what is the smallest possible value of a ? | "the number a must include at least 3 ^ 2 * 7 ^ 2 = 441 the answer is d ." | a ) 120 , b ) 256 , c ) 343 , d ) 441 , e ) 512 | d | add(add(add(add(add(multiply(multiply(7, 2), 2), multiply(multiply(7, 2), 2)), multiply(multiply(7, 2), 2)), 2), const_4), const_4) | multiply(n0,n7)|multiply(n1,#0)|add(#1,#1)|add(#2,#1)|add(n7,#3)|add(#4,const_4)|add(#5,const_4)| | other |
how many times digit 6 is used while writing numbers from 100 to 1400 ? | "there are 100 numbers which begin with 600 next , in every 10 numbers such as 100 to 110 , 110 to 120 , 120 to 130 6 comes at least once . number of such intervals = end limit - first no . / interval . our range of numbers is 100 - 1000 1000 - 100 = 900 / 10 = 90 number of 10 s interval in this is 90 . so 90 ' 6 s ' s... | a ) 648 , b ) 350 , c ) 252 , d ) 225 , e ) 26 | b | add(add(divide(subtract(1400, 100), const_10), multiply(add(const_10, const_1), add(const_10, const_1))), multiply(6, const_2)) | add(const_1,const_10)|multiply(n0,const_2)|subtract(n2,n1)|divide(#2,const_10)|multiply(#0,#0)|add(#3,#4)|add(#5,#1)| | general |
a confectioner decides to sell all of his pastry due to the coming holiday . his pastry goods are equally divided among a group of 25 regular customers . if only 49 customers come to the bakery , each one will receive 6 less pastry goods . how much pastry does the confectioner needs to sell ? | "pastry is divided in 25 customers equally . so , total number of pastry must be a multiple of 25 only option a satisfies the condition , and hence is the answer" | a ) 300 . , b ) 412 . , c ) 432 . , d ) 502 . , e ) 522 . | a | multiply(divide(multiply(49, 6), subtract(49, 25)), 25) | multiply(n1,n2)|subtract(n1,n0)|divide(#0,#1)|multiply(n0,#2)| | general |
the sum of the non - prime numbers between 40 and 50 , non - inclusive , is | "the sum of the non - prime numbers between 40 and 50 , non - inclusive , are { 42 , 44 , 45 , 46 , 48 , 49 } . 6 times 40 something is approx north of 240 hence the answer is ( c ) 274" | a ) 131 , b ) 176 , c ) 274 , d ) 405 , e ) 495 | c | add(add(add(add(add(add(add(40, const_1), add(add(40, const_1), const_1)), add(add(add(40, const_1), const_1), const_2)), add(add(add(add(40, const_1), const_1), const_2), const_1)), add(add(add(add(add(40, const_1), const_1), const_2), const_1), const_1)), add(add(add(add(add(add(40, const_1), const_1), const_2), cons... | add(n0,const_1)|add(#0,const_1)|add(#0,#1)|add(#1,const_2)|add(#2,#3)|add(#3,const_1)|add(#4,#5)|add(#5,const_1)|add(#6,#7)|add(#7,const_1)|add(#8,#9)|add(#9,const_1)|add(#10,#11)| | general |
fresh grapes contain 90 % by weight while dried grapes contain 20 % water by weight . what is the weight of dry grapes available from 20 kg of fresh grapes ? | "the weight of non - water in 20 kg of fresh grapes ( which is 100 - 90 = 10 % of whole weight ) will be the same as the weight of non - water in x kg of dried grapes ( which is 100 - 20 = 80 % of whole weight ) , so 20 ∗ 0.1 = x ∗ 0.8 - - > x = 2.5 answer : c ." | a ) 2 kg , b ) 2.4 kg , c ) 2.5 kg , d ) 10 kg , e ) none of these | c | multiply(divide(divide(multiply(subtract(const_100, 90), 20), const_100), subtract(const_100, 20)), const_100) | subtract(const_100,n0)|subtract(const_100,n1)|multiply(n2,#0)|divide(#2,const_100)|divide(#3,#1)|multiply(#4,const_100)| | gain |
there are 9 players in a chess group , and each player plays each of the others once . given that each game is played by two players , how many total games will be played ? | "9 players are there . two players play one game with one another . so 9 c 2 = 9 x 8 / 2 = 36 so option c is correct" | a ) 10 , b ) 30 , c ) 36 , d ) 60 , e ) 90 | c | divide(multiply(9, subtract(9, const_1)), const_2) | subtract(n0,const_1)|multiply(n0,#0)|divide(#1,const_2)| | general |
in order to obtain an income of rs . 680 from 10 % stock at rs . 120 , one must make an investment of | "to obtain rs . 10 , investment = rs . 120 . to obtain rs . 680 , investment = = rs . 8160 . answer : b" | a ) 5363 , b ) 8160 , c ) 2368 , d ) 1987 , e ) 2732 | b | multiply(divide(120, 10), 680) | divide(n2,n1)|multiply(n0,#0)| | gain |
in a party every person shakes hands with every other person . if there are 105 hands shakes , find the number of person in the party | "let n be the number of persons in the party . number of hands shake = 105 ; total number of hands shake is given by nc 2 . now , according to the question , nc 2 = 105 ; or , n ! / [ 2 ! * ( n - 2 ) ! ] = 105 ; or , n * ( n - 1 ) / 2 = 105 ; or , n 2 - n = 210 ; or , n 2 - n - 210 = 0 ; or , n = 15 , - 14 ; but , we c... | a ) 14 , b ) 12 , c ) 13 , d ) 15 , e ) 16 | d | divide(add(sqrt(add(multiply(multiply(105, const_2), const_4), const_1)), const_1), const_2) | multiply(n0,const_2)|multiply(#0,const_4)|add(#1,const_1)|sqrt(#2)|add(#3,const_1)|divide(#4,const_2)| | general |
the price of stock decreased by 8 % last year and rose by 10 % this year . what is the net percentage change in the price of the stock ? | ( 100 % - 8 % ) * ( 100 % + 10 % ) = 0.92 * 1.10 = 1.012 = 101.2 % . the net percentage change in the price of the stock is ( + ) 1.2 % the answer is c | a ) 0.2 % , b ) 0.8 % , c ) 1.2 % , d ) 1.6 % , e ) 2 % | c | multiply(subtract(multiply(add(divide(10, const_100), const_1), subtract(const_1, divide(8, const_100))), const_1), const_100) | divide(n1,const_100)|divide(n0,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|subtract(#4,const_1)|multiply(#5,const_100) | gain |
two men and three women working 7 hours a day finish a work in 5 days . four men and four women working 3 hours a day complete the work in 7 days . the number of days in which only 7 men working 4 hours a day will finish the work is ? | answer : option a 2 m + 3 w - - - - - 35 h 4 m + 4 w - - - - - - - 21 h 7 m - - - - - - - ? d 70 m + 105 w = 84 m + 84 m 21 w = 14 m = > 2 m = 3 w 4 * 35 = 7 * x = > x = 20 hours 20 / 4 = 5 days | a ) 5 days , b ) 3 days , c ) 2 days , d ) 9 days , e ) 10 days | a | divide(const_1, multiply(multiply(4, 7), divide(subtract(divide(3, multiply(3, 7)), divide(4, multiply(5, 7))), 4))) | multiply(n0,n2)|multiply(n0,n1)|multiply(n0,n5)|divide(n2,#0)|divide(n5,#1)|subtract(#3,#4)|divide(#5,n5)|multiply(#6,#2)|divide(const_1,#7) | physics |
a train is running at a speed of 40 km / hr and it crosses a post in 19.8 seconds . what is the length of the train ? | "speed of the train , v = 40 km / hr = 40000 / 3600 m / s = 400 / 36 m / s time taken to cross , t = 19.8 s distance covered , d = vt = ( 400 / 36 ) ã — 19.8 = 220 m distance covered is equal to the length of the train = 220 m correct answer is 220 metres d" | a ) 190 metres , b ) 160 metres , c ) 200 metres , d ) 220 metres , e ) 250 metres | d | multiply(19.8, multiply(40, const_0_2778)) | multiply(n0,const_0_2778)|multiply(n1,#0)| | physics |
the average age of 20 persons in a office is 15 years . out of these , the average age of 5 of them is 14 years and that of the other 9 persons is 16 years . the age of the 15 th person is ? | "age of the 15 th student = 20 * 15 - ( 14 * 5 + 16 * 9 ) = 300 - 214 = 86 years answer is b" | a ) 79 , b ) 86 , c ) 95 , d ) 72 , e ) 80 | b | subtract(subtract(multiply(20, 15), multiply(5, 14)), multiply(9, 16)) | multiply(n0,n1)|multiply(n2,n3)|multiply(n4,n5)|subtract(#0,#1)|subtract(#3,#2)| | general |
in a rectangular coordinate system , what is the area of a rectangle whose vertices have the coordinates ( - 8 , 1 ) , ( 1 , 1 ) , ( 1 , - 7 and ( - 8 , - 7 ) ? | "length of side 1 = 8 + 1 = 9 length of side 2 = 7 + 1 = 8 area of rectangle = 9 * 8 = 72 c is the answer" | a ) 144 , b ) 36 , c ) 72 , d ) 56 , e ) 112 | c | multiply(add(8, 1), add(1, 7)) | add(n0,n1)|add(n1,n5)|multiply(#0,#1)| | geometry |
on a certain farm the ratio of horses to cows is 6 : 1 . if the farm were to sell 15 horses and buy 15 cows , the ratio of horses to cows would then be 3 : 1 . after the transaction , how many more horses than cows would the farm own ? | "originally , there were 6 k horses and k cows . ( 6 k - 15 ) = 3 ( k + 15 ) 6 k - 3 k = 45 + 15 3 k = 60 k = 20 the difference between horses and cows is ( 6 k - 15 ) - ( k + 15 ) = 5 k - 30 = 70 the answer is d ." | a ) 40 , b ) 50 , c ) 60 , d ) 70 , e ) 80 | d | subtract(subtract(multiply(6, divide(add(multiply(15, 3), multiply(15, 1)), subtract(multiply(6, 1), 3))), 15), add(divide(add(multiply(15, 3), multiply(15, 1)), subtract(multiply(6, 1), 3)), 15)) | multiply(n2,n4)|multiply(n2,n5)|multiply(n0,n5)|add(#0,#1)|subtract(#2,n4)|divide(#3,#4)|add(n2,#5)|multiply(n0,#5)|subtract(#7,n2)|subtract(#8,#6)| | other |
if a * b * c = 195 , b * c * d = 65 , c * d * e = 1000 and d * e * f = 250 the ( a * f ) / ( c * d ) = ? | explanation : a â ˆ — b â ˆ — c / b â ˆ — c â ˆ — d = 195 / 65 = > a / d = 3 d â ˆ — e â ˆ — f / c â ˆ — d â ˆ — e = 250 / 1000 = > f / c = 1 / 4 a / d * f / c = 3 * 1 / 4 = 3 / 4 answer : c | a ) 1 / 2 , b ) 1 / 4 , c ) 3 / 4 , d ) 2 / 3 , e ) none of these | c | divide(multiply(195, 250), multiply(1000, 65)) | multiply(n0,n3)|multiply(n1,n2)|divide(#0,#1) | general |
a circular well with a diameter of 2 metres , is dug to a depth of 4 metres . what is the volume of the earth dug out ? | "sol . volume = ∏ r ² h = [ 22 / 7 * 1 * 1 * 14 ] m ³ = 44 m ³ answer c" | a ) 26 m ³ , b ) 28 m ³ , c ) 44 m ³ , d ) 46 m ³ , e ) none | c | volume_cylinder(divide(2, 2), 4) | divide(n0,n0)|volume_cylinder(#0,n1)| | geometry |
what is the least value of x , so that 2 x 5478 is divisible by 9 | "explanation : the sum of the digits of the number is divisible by 9 . then the number is divisible by 9 . 2 + x + 5 + 4 + 7 + 8 = 26 + x least value of x may be ' 1 ' , so that the total 26 + 1 = 27 is divisible by 9 . answer : option d" | a ) 7 , b ) 8 , c ) 9 , d ) 1 , e ) 2 | d | divide(divide(divide(lcm(2, 5478), 5478), const_4), const_4) | lcm(n0,n1)|divide(#0,n1)|divide(#1,const_4)|divide(#2,const_4)| | general |
the jogging track in a sports complex is 594 m in circumference . deepak and his wife start from the same point and walk in opposite directions at 4.5 km / hr and 3.75 km / hr respectively . they will meet for the first time in ? | "clearly , the two will meet when they are 594 m apart . to be ( 4.5 + 3.75 ) = 8.25 km apart , they take 1 hour . to be 594 m apart , they take ( 100 / 825 * 594 / 1000 ) hrs = ( 594 / 8250 * 60 ) min = 4.32 min . answer : b" | a ) 5.29 min , b ) 4.32 min , c ) 5.08 min , d ) 9.28 min , e ) 5.988 min | b | multiply(divide(divide(594, const_1000), add(4.5, 3.75)), const_60) | add(n1,n2)|divide(n0,const_1000)|divide(#1,#0)|multiply(#2,const_60)| | general |
a can do a job in 45 days and b can do it in 30 days . a and b working together will finish 4 times the amount of work in - - - - - - - days ? | 1 / 45 + 1 / 30 = 5 / 90 = 1 / 18 18 / 1 = 18 * 4 = 72 days answer : e | a ) 30 days , b ) 45 days , c ) 80 days , d ) 72 1 / 2 days , e ) 72 days | e | multiply(inverse(add(divide(const_1, 45), divide(const_1, 30))), 4) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|inverse(#2)|multiply(n2,#3) | physics |
on dividing a number by 357 , we get 37 as remainder . on dividing the same number by 17 , what will be the remainder ? | "let x be the number and y be the quotient . then , x = 357 * y + 37 = ( 17 * 21 * y ) + ( 17 * 2 ) + 3 = 17 * ( 21 y + 2 ) + 3 . required number = 3 . answer is b" | a ) 4 , b ) 3 , c ) 8 , d ) 7 , e ) 2 | b | multiply(subtract(divide(power(37, const_2), 357), floor(divide(power(37, const_2), 357))), 357) | power(n1,const_2)|divide(#0,n0)|floor(#1)|subtract(#1,#2)|multiply(n0,#3)| | general |
a certain social security recipient will receive an annual benefit of $ 12,000 provided he has annual earnings of $ 9,360 or less , but the benefit will be reduced by $ 1 for every $ 3 of annual earnings over $ 9,360 . what amount of total annual earnings would result in a 55 percent reduction in the recipient ' s annu... | "for every $ 3 earn above $ 9360 , the recipient loses $ 1 of benefit . or for every $ 1 loss in the benefit , the recipient earns $ 3 above $ 9360 if earning is ; 9360 + 3 x benefit = 12000 - x or the vice versa if benefit is 12000 - x , the earning becomes 9360 + 3 x he lost 50 % of the benefit ; benefit received = 1... | a ) $ 15,360 , b ) $ 17,360 , c ) $ 25,560 , d ) $ 21,360 , e ) $ 27,360 | c | add(multiply(const_100, 3), const_60) | multiply(const_100,n3)|add(#0,const_60)| | general |
a vessel of capacity 45 litres is fully filled with pure milk . nine litres of milk is removed from the vessel and replaced with water . nine litres of the solution thus formed is removed and replaced with water . find the quantity of pure milk in the final milk solution ? | "explanation : let the initial quantity of milk in vessel be t litres . let us say y litres of the mixture is taken out and replaced by water for n times , alternatively . quantity of milk finally in the vessel is then given by [ ( t - y ) / t ] ^ n * t for the given problem , t = 45 , y = 9 and n = 2 . hence , quantit... | a ) 28.8 , b ) 72.9 , c ) 38.3 , d ) 78.3 , e ) 79.3 | a | subtract(subtract(45, multiply(const_3, const_3)), multiply(divide(subtract(45, multiply(const_3, const_3)), 45), multiply(const_3, const_3))) | multiply(const_3,const_3)|subtract(n0,#0)|divide(#1,n0)|multiply(#2,#0)|subtract(#1,#3)| | physics |
if circles x and y have the same area and circle x has a circumference of 12 π , half of the radius of circle y is : | "x be radius of circle x y be radius of circle y given : pi * x ^ 2 = pi * y ^ 2 also , 2 * pi * x = 12 * pi x = 6 thus y = 6 y / 2 = 3 ans : c" | a ) 16 , b ) 8 , c ) 3 , d ) 2 , e ) 1 | c | divide(divide(12, const_2), const_2) | divide(n0,const_2)|divide(#0,const_2)| | geometry |
convert the 18 / 42 m / s into kilometers per hour ? | "18 / 42 m / s = 18 / 42 * 18 / 5 = 1 ( 27 / 50 ) = 1.54 kmph . answer : d" | a ) 3.5 kmph . , b ) 2.5 kmph . , c ) 1.2 kmph . , d ) 1.54 kmph . , e ) 1.9 kmph . | d | multiply(const_3_6, divide(18, 42)) | divide(n0,n1)|multiply(#0,const_3_6)| | physics |
if a man lost 1 % by selling oranges at the rate of 16 a rupee at how many a rupee must he sell them to gain 44 % ? | "99 % - - - - 16 144 % - - - - ? 99 / 144 * 16 = 11 answer : d" | a ) 12 , b ) 14 , c ) 7 , d ) 11 , e ) 10 | d | divide(multiply(subtract(const_100, 1), 16), add(const_100, 44)) | add(n2,const_100)|subtract(const_100,n0)|multiply(n1,#1)|divide(#2,#0)| | gain |
a certain car uses one gallon of gasoline every 34 miles when it travels on highway , and one gallon of gasoline every 20 miles when it travels in the city . when a car travels 4 miles on highway and 4 additional miles in the city , it uses what percent more gasoline than if it travels 8 miles on the highway ? | 4 miles on the highway = 4 / 34 gallons ; 4 miles in the city = 4 / 20 gallons ; total = 4 / 34 + 4 / 20 = 27 / 85 gallons . 8 miles on the highway = 4 / 17 gallons . the % change = ( 27 / 85 - 4 / 17 ) / ( 4 / 17 ) = 0.35 . answer : d . | a ) 15 % , b ) 20 % , c ) 22.5 % , d ) 35 % , e ) 50 % | d | multiply(divide(subtract(add(multiply(divide(const_1, 20), 4), multiply(4, divide(const_1, 34))), multiply(8, divide(const_1, 34))), multiply(8, divide(const_1, 34))), const_100) | divide(const_1,n1)|divide(const_1,n0)|multiply(n2,#0)|multiply(n2,#1)|multiply(n4,#1)|add(#2,#3)|subtract(#5,#4)|divide(#6,#4)|multiply(#7,const_100) | general |
in measuring the sides of a rectangle , one side is taken 9 % in excess , and the other 8 % in deficit . find the error percent in the area calculated from these measurements . | "let x and y be the sides of the rectangle . then , correct area = xy . calculated area = ( 12 / 11 ) x ( 23 / 25 ) y = ( 358 / 357 ) ( xy ) error in measurement = ( 358 / 357 ) xy - xy = ( 1 / 357 ) xy error percentage = [ ( 1 / 357 ) xy ( 1 / xy ) 100 ] % = ( 7 / 25 ) % = 0.28 % . answer is e ." | a ) 0.11 % , b ) 0.7 % , c ) 0.4 % , d ) 0.6 % , e ) 0.28 % | e | subtract(subtract(9, 8), divide(multiply(9, 8), const_100)) | multiply(n0,n1)|subtract(n0,n1)|divide(#0,const_100)|subtract(#1,#2)| | geometry |
there is 50 % increase in an amount in 6 years at s . i . what will be the c . i . of rs . 12,000 after 3 years at the same rate ? | "let p = rs . 100 . then , s . i . rs . 50 and t = 6 years . r = ( 100 * 50 ) / ( 100 * 6 ) = 10 % p . a . now , p = rs . 12000 , t = 3 years and r = 8.33 % p . a . c . i . = [ 12000 * { ( 1 + 8.33 / 100 ) 3 - 1 } ] = rs . 17988 answer : c" | a ) 2160 , b ) 3120 , c ) 17988 , d ) 6240 , e ) 5240 | c | subtract(multiply(add(multiply(const_100, const_100), multiply(multiply(const_100, divide(50, 6)), 3)), multiply(multiply(add(const_1, divide(divide(50, 6), const_100)), add(const_1, divide(divide(50, 6), const_100))), add(const_1, divide(divide(50, 6), const_100)))), add(multiply(const_100, const_100), multiply(multip... | divide(n0,n1)|multiply(const_100,const_100)|divide(#0,const_100)|multiply(#0,const_100)|add(#2,const_1)|multiply(#3,n3)|add(#1,#5)|multiply(#4,#4)|multiply(#4,#7)|multiply(#6,#8)|subtract(#9,#6)| | gain |
if the cost price is 86 % of the selling price , then what is the profit percent ? | "let s . p . = $ 100 c . p . = $ 86 profit = $ 14 profit % = 14 / 86 * 100 = 25 / 6 = 16 % approximately answer is c" | a ) 5 % , b ) 11 % , c ) 16 % , d ) 21 % , e ) 19 % | c | multiply(divide(subtract(const_100, 86), 86), const_100) | subtract(const_100,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
find the length of the wire required to go 15 times round a square field containing 69696 m ^ 2 . | "explanation : a 2 = 69696 = > a = 264 4 a = 1056 1056 * 15 = 15840 answer : d" | a ) 1250 , b ) 2489 , c ) 6398 , d ) 15840 , e ) 2487 | d | multiply(square_perimeter(square_edge_by_area(69696)), 15) | square_edge_by_area(n1)|square_perimeter(#0)|multiply(n0,#1)| | general |
the average of 5 quantities is 8 . the average of 3 of them is 4 . what is the average of remaining 2 numbers ? | "answer : a ( 5 x 8 - 3 x 4 ) / 2 = 14" | a ) 14 , b ) 10 , c ) 8 , d ) 9.5 , e ) none of these | a | divide(subtract(multiply(5, 8), multiply(3, 4)), 2) | multiply(n0,n1)|multiply(n2,n3)|subtract(#0,#1)|divide(#2,n4)| | general |
the perimeter of one square is 48 cm and that of another is 20 cm . find the perimeter and the diagonal of a square which is equal in area to these two combined ? | explanation : 4 a = 48 4 a = 20 a = 12 a = 5 a ^ 2 = 144 a ^ 2 = 25 combined area = a ^ 2 = 169 = > a = 13 d = 13 √ 2 cm answer : b | ['a ) 5 cm', 'b ) 13 √ 2 cm', 'c ) 12 cm', 'd ) 6 cm', 'e ) 8 cm'] | b | sqrt(multiply(add(power(divide(48, const_4), const_2), power(divide(20, const_4), const_2)), const_2)) | divide(n0,const_4)|divide(n1,const_4)|power(#0,const_2)|power(#1,const_2)|add(#2,#3)|multiply(#4,const_2)|sqrt(#5) | geometry |
in a class of 40 students , 2 students did not borrow any books from the library , 12 students each borrowed 1 book , 14 students each borrowed 2 books , and the rest borrowed at least 3 books . if the average number of books per student was 2 , what is the maximum number of books any single student could have borrowed... | the class borrowed a total of 40 * 2 = 80 books . the 28 students who borrowed 0 , 1 , or 2 books borrowed a total of 12 + 14 * 2 = 40 . to maximize the number of books borrowed by 1 student , let ' s assume that 11 students borrowed 3 books and 1 student borrowed the rest . 80 - 40 - 3 * 11 = 7 the maximum number of b... | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | e | subtract(multiply(40, 2), add(multiply(subtract(subtract(40, add(add(multiply(12, 1), 14), 2)), 1), 3), add(multiply(12, 1), multiply(14, 2)))) | multiply(n0,n1)|multiply(n2,n3)|multiply(n1,n4)|add(#1,#2)|add(n4,#1)|add(n1,#4)|subtract(n0,#5)|subtract(#6,n3)|multiply(n6,#7)|add(#3,#8)|subtract(#0,#9) | general |
a picnic attracts 240 persons . there are 30 more men than women , and 30 more adults than children . how many men are at this picnic ? | "adult + children = 240 let , children = y then , adult = y + 30 i . e . y + ( y + 30 ) = 210 i . e . y = 115 i . e . adult = 115 + 30 = 145 adults include only men and women i . e . men + women = 145 let women , w = x then men , m = x + 30 i . e . x + ( x + 30 ) = 2 x + 30 = 145 i . e . x = 55 i . e . men , m = 55 + 3... | a ) 240 , b ) 75 , c ) 85 , d ) 130 , e ) 200 | c | add(divide(subtract(add(divide(subtract(240, 30), const_2), 30), 30), const_2), 30) | subtract(n0,n1)|divide(#0,const_2)|add(n1,#1)|subtract(#2,n1)|divide(#3,const_2)|add(n1,#4)| | general |
find 80 / ? = ? / 20 | "answer let 80 / y = y / 20 then , y 2 = 80 x 20 = 1600 ∴ y = √ 1600 = 40 . option : a" | a ) 40 , b ) 400 , c ) 800 , d ) 1600 , e ) none | a | divide(80, 20) | divide(n0,n1)| | general |
what is the greatest positive integer r such that 3 ^ r is a factor of 9 ^ 10 ? | "what is the greatest positive integer r such that 3 ^ r is a factor of 9 ^ 10 ? 9 ^ 10 = ( 3 ^ 2 ) ^ 10 = 3 ^ 20 d . 20" | a ) r = 5 , b ) r = 9 , c ) r = 10 , d ) r = 20 , e ) 30 | d | multiply(subtract(9, 10), 10) | subtract(n1,n2)|multiply(n2,#0)| | other |
find the average of all the numbers between 40 and 80 which are divisible by 3 . | "sol . average = ( 42 + 45 + 48 + 51 + 54 + 57 + 60 + 63 + 66 + 69 + 72 + 75 + 78 / 13 ) = 780 / 5 = 60 . answer d" | a ) 15 , b ) 70 , c ) 20 , d ) 60 , e ) none | d | divide(add(add(40, const_4), subtract(80, const_4)), const_2) | add(n0,const_4)|subtract(n1,const_4)|add(#0,#1)|divide(#2,const_2)| | general |
a can do a job in 18 days and b can do it in 30 days . in how many days will they finish it together ? | explanation : 1 / 18 + 1 / 30 = 8 / 90 = 4 / 45 so , both together will finish the work in 4 / 45 days . answer : b | a ) 1 / 9 , b ) 4 / 45 , c ) 2 / 7 , d ) 6 / 3 , e ) 5 / 2 | b | add(divide(const_1, 18), divide(const_1, 30)) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1) | physics |
in the xy - plane the point ( - 2 - 3 ) is the centre of a circle , the point ( - 2 , 1 ) lies inside the circle and the point ( 4 , - 3 ) lies outside the circle . if the radius r of the circle r is an integer then diameter ? | "can be solved without much calculations . you are given that ( - 2 , - 3 ) is the center of the circle . point ( 4 , - 3 ) lies inside the circle - - - > the radius is lesser than distance of ( - 2 , - 3 ) from ( 4 , - 3 ) - - - > lesser than 6 units but the radius will also be greater than the distance of ( - 2 , - 3... | a ) 6 , b ) 10 , c ) 4 , d ) 3 , e ) 2 | b | divide(add(subtract(4, negate(2)), 4), 2) | negate(n0)|subtract(n4,#0)|add(n4,#1)|divide(#2,n0)| | physics |
the mass of 1 cubic meter of a substance is 300 kilograms under certain conditions . what is the volume , in cubic centimeters , of 1 gram of this substance under these conditions ? ( 1 kilogram = 1,000 grams and 1 cubic meter = 1 , 000,000 cubic centimeters ) | density is mass divided by volume . so density of the given substance will be mass / volume = 800 kg / 1 m ^ 3 = 300 kg / m ^ 3 or 1 g / 1.33 cm ^ 3 = 0.3 g / cm ^ 3 . next , ask yourself if 300,000 g is equivalent to 1 , 000,000 cubic centimeters then 1 g is equivalent to how many cubic centimeters ? - - > 1 g - 1 , 0... | ['a ) 1.33', 'b ) 1.25', 'c ) 8.0', 'd ) 12.5', 'e ) 80.0'] | a | subtract(divide(power(const_100, const_3), multiply(300, const_1000)), const_2) | multiply(n1,const_1000)|power(const_100,const_3)|divide(#1,#0)|subtract(#2,const_2) | geometry |
in a 300 member association consisting of men and women , exactly 80 % of men and exactly 15 % women are homeowners . what is the least number of members who are homeowners ? | "solution simple out of 300 80 % are male i . e 240 and 15 % are female i . e 45 , so total homeowner is 295 . now min number homeowner is 45 and max is 240 so question ask us to find least and 46 has least value among all option . so ans is 46 . ans : a" | a ) 46 , b ) 47 , c ) 48 , d ) 49 , e ) 50 | a | add(multiply(multiply(divide(15, const_100), 80), multiply(divide(15, const_100), 80)), divide(subtract(300, 80), 80)) | divide(n2,const_100)|subtract(n0,n1)|divide(#1,n1)|multiply(n1,#0)|multiply(#3,#3)|add(#2,#4)| | general |
mark has a rectangular driveway with a line painted diagonally across . the length of the driveway is 5 feet longer than its width . if the total area of the driveway is 36 square feet , what is the length of the painted line ( in feet ) ? | to solve this , we first need to set up an equation for the area of the room . if x is the width , then we have x ( x + 5 ) = 36 . by putting the equation in standard form , we get x ^ 2 + 5 x - 36 = 0 . by using the quadratic formula , we get roots of 4 and - 9 . we know that x is the width , and x + 5 is the length ,... | ['a ) between 9 and 10', 'b ) between 10 and 11', 'c ) between 11 and 12', 'd ) between 12 and 13', 'e ) between 13 and 14'] | a | sqrt(add(power(divide(add(5, sqrt(add(power(5, const_2), multiply(36, const_4)))), const_2), const_2), power(subtract(divide(add(5, sqrt(add(power(5, const_2), multiply(36, const_4)))), const_2), 5), const_2))) | multiply(n1,const_4)|power(n0,const_2)|add(#0,#1)|sqrt(#2)|add(n0,#3)|divide(#4,const_2)|power(#5,const_2)|subtract(#5,n0)|power(#7,const_2)|add(#6,#8)|sqrt(#9) | geometry |
tom reads at an average rate of 30 pages per hour , while jan reads at an average rate of 41 pages per hour . if tom starts reading a novel at 3 : 20 , and jan begins reading an identical copy of the same book at 4 : 26 , at what time will they be reading the same page ? | since tom reads an average of 1 page every 2 minutes , tom will read 33 pages in the first 66 minutes . jan can catch tom at a rate of 11 pages per hour , so it will take 3 hours to catch tom . the answer is c . | a ) 6 : 26 , b ) 6 : 56 , c ) 7 : 26 , d ) 7 : 56 , e ) 8 : 26 | c | divide(add(4, divide(multiply(add(subtract(4, 3), divide(subtract(26, 20), const_60)), 30), subtract(41, 30))), 26) | subtract(n5,n3)|subtract(n4,n2)|subtract(n1,n0)|divide(#0,const_60)|add(#3,#1)|multiply(n0,#4)|divide(#5,#2)|add(n4,#6)|divide(#7,n5) | general |
a theater charges $ 12 for seats in the orchestra and $ 8 for seats in the balcony . on a certain night , a total of 355 tickets were sold for a total cost of $ 3,320 . how many more tickets were sold that night for seats in the balcony than for seats in the orchestra ? | "orchestra seats - a balcony seats - b a + b = 355 and 12 a + 8 b = 3320 solving equations simultaneously ( multiply equation 1 with 8 and subtract from second equation ) 4 a = 3320 - 8 * 355 = 3320 - 2840 = 480 i . e . a = 120 and b = 355 - 120 = 235 more seats in balcony than orchestra = b - a = 235 - 120 = 115 answe... | a ) 90 , b ) 115 , c ) 120 , d ) 130 , e ) 220 | b | subtract(divide(subtract(multiply(12, 355), add(add(multiply(const_3, const_1000), multiply(const_3, const_100)), multiply(const_2, const_10))), subtract(12, 8)), subtract(355, divide(subtract(multiply(12, 355), add(add(multiply(const_3, const_1000), multiply(const_3, const_100)), multiply(const_2, const_10))), subtrac... | multiply(n0,n2)|multiply(const_1000,const_3)|multiply(const_100,const_3)|multiply(const_10,const_2)|subtract(n0,n1)|add(#1,#2)|add(#5,#3)|subtract(#0,#6)|divide(#7,#4)|subtract(n2,#8)|subtract(#8,#9)| | general |
a certain family has 3 sons : richard is 6 years older than david , and david is 8 years older than scott . if in 8 years , richard will be twice as old as scott , then how old was david 6 years ago ? | "let ' s say age of richard isr age of david isd age of scott iss now richard is 6 years older than david , i . e . r = d + 6 david is 8 years older than scott i . e . d = s + 8 if in 8 years , richard will be twice as old as scott i . e . r + 8 = 2 x ( s + 8 ) i . e . r + 8 = 2 s + 16 i . e . r = 2 s + 8 but r = d + 6... | a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 16 | a | subtract(add(6, 8), 6) | add(n1,n2)|subtract(#0,n4)| | general |
a type t machine can complete a job in 5 hours and a type b machine can complete the job in 7 hours . how many hours will it take 2 type t machines and 3 type b machines working together and independently to complete the job ? | now d should be the answer . t need 5 hours to complete and b needs 7 hours to compete so 2 t + 3 b will complete 2 / 5 + 3 / 7 or 29 / 35 portion of the job in 1 hour so the whole job will take 35 / 29 hours . . . . = d | a ) 1 / 5 , b ) 29 / 35 , c ) 5 / 6 , d ) 35 / 29 , e ) 35 / 12 | d | inverse(add(inverse(divide(5, 2)), inverse(divide(7, 3)))) | divide(n0,n2)|divide(n1,n3)|inverse(#0)|inverse(#1)|add(#2,#3)|inverse(#4) | physics |
on a certain date , pat invested $ 10,000 at x percent annual interest , compounded annually . if the total value of the investment plus interest at the end of 12 years will be $ 40,000 , in how many years , the total value of the investment plus interest will increase to $ 120,000 ? | "if i were to choose during the test , would go for 18 or 20 . probably 18 cuz it wont take too long to get the value doubled . . . . i found a method : rule of 72 . given an x % return , it takes 10,000 to quadralope 12 years . so according to the rule : 72 / x is the no of years 10 , 000.00 took to double 20 , 000.00... | a ) 15 , b ) 21 , c ) 18 , d ) 20 , e ) 24 | b | divide(log(divide(multiply(const_3, const_10), add(const_4, const_1))), log(power(divide(multiply(const_2, const_10), add(const_4, const_1)), divide(const_1, 12)))) | add(const_1,const_4)|divide(const_1,n1)|multiply(const_10,const_3)|multiply(const_10,const_2)|divide(#2,#0)|divide(#3,#0)|log(#4)|power(#5,#1)|log(#7)|divide(#6,#8)| | general |
if the sum of two numbers is 16 and the sum of their squares is 200 , then the product of the numbers is | "according to the given conditions x + y = 16 and x ^ 2 + y ^ 2 = 200 now ( x + y ) ^ 2 = x ^ 2 + y ^ 2 + 2 xy so 16 ^ 2 = 200 + 2 xy so xy = 56 / 2 = 28 answer : e" | a ) 40 , b ) 44 , c ) 80 , d ) 88 , e ) 28 | e | divide(subtract(power(16, const_2), 200), const_2) | power(n0,const_2)|subtract(#0,n1)|divide(#1,const_2)| | general |
a and b can do a piece of work in 12 days . with the help of c they finish the work in 6 days . c alone can do that piece of work in ? | "c = 1 / 6 â € “ 1 / 12 = 1 / 12 = > 12 days answer : e" | a ) 33 , b ) 878 , c ) 30 , d ) 88 , e ) 12 | e | inverse(subtract(6, divide(6, 12))) | divide(n1,n0)|subtract(n1,#0)|inverse(#1)| | physics |
an amount of money is to be divided between p , q and r in the ratio of 3 : 7 : 12 . if the difference between the shares of p and q is rs . 4000 , what will be the difference between q and r ' s share ? | "4 - - - 4000 5 - - - ? = > 5000 answer : b" | a ) 2788 , b ) 5000 , c ) 7282 , d ) 2782 , e ) 2729 | b | multiply(subtract(12, 7), divide(4000, subtract(7, 3))) | subtract(n1,n0)|subtract(n2,n1)|divide(n3,#0)|multiply(#2,#1)| | general |
running at the same constant rate , 6 identical machines can produce a total of 390 bottles per minute . at this rate , how many bottles could 10 such machines produce in 4 minutes ? | "let the required number of bottles be x . more machines , more bottles ( direct proportion ) more minutes , more bottles ( direct proportion ) machines 6 : 10 : : 390 : x time ( in minutes ) 1 : 4 6 x 1 x x = 10 x 4 x 390 x = ( 10 x 4 x 390 ) / ( 6 ) x = 2600 . answer : c" | a ) 648 , b ) 1800 , c ) 2600 , d ) 10800 , e ) 10900 | c | multiply(multiply(divide(390, 6), 4), 10) | divide(n1,n0)|multiply(n3,#0)|multiply(n2,#1)| | gain |
a certain pair of used shoes can be repaired for $ 14.50 and will last for 1 year . a pair of the same kind of shoes can be purchased new for $ 32.00 and will last for 2 years . the average cost per year of the new shoes is what percent greater than the cost of repairing the used shoes ? | 1 ) cost of repairing = 14.5 ( for one year ) , therefore for 2 years it would be $ 29 . 2 ) cost of new pair which will last for 2 years is $ 32 . percentage change formula = ( final value - initial value ) / ( initial value ) * 100 . in this case the final value would be the price of new shoeinitial value would be th... | a ) 16 , b ) 10.34 , c ) 12.46 , d ) 11.2 , e ) 14 | b | multiply(const_100, divide(subtract(32, multiply(14.5, 2)), multiply(14.5, 2))) | multiply(n0,n3)|subtract(n2,#0)|divide(#1,#0)|multiply(#2,const_100) | general |
if length of rectangle is increased by 10 % , and breadth is decreased by 10 % . what will it effect on area of rectangle ? | let initial length = l , and breadth = b area = l * b new length = l * 110 % = [ l * 110 / 100 ] = 11 l / 10 , new breadth = 9 b / 10 new area = 11 l / 10 * 9 b / 10 = 99 / 100 lb decrease in area = [ lb - 99 / 100 lb ] = lb / 100 = % decrease = [ lb / 100 * 1 / lb * 100 ] = 1 % answer c | ['a ) it will have no effect', 'b ) 1 % increase', 'c ) 1 % decrease', 'd ) 2 % increase', 'e ) 2 % decrease'] | c | subtract(add(divide(multiply(10, 10), const_100), 10), 10) | multiply(n0,n0)|divide(#0,const_100)|add(n0,#1)|subtract(#2,n0) | geometry |
what is the ratio of 12 minute to 1 hour ? | c 1 / 5 | a ) 1 / 4 , b ) 3 / 4 , c ) 1 / 5 , d ) 2 / 2 , e ) 3 / 1 | c | divide(12, const_60) | divide(n0,const_60) | physics |
8 * 8 * 8 * 8 = 2 ^ ? | "solution : 2 ^ 3 * 2 ^ 3 * 2 ^ 3 * 2 ^ 3 = 2 ^ ( 3 + 3 + 3 + 3 ) = 2 ^ 12 answer : 12 option : e" | a ) 4 , b ) 6 , c ) 8 , d ) 10 , e ) 12 | e | multiply(8, 8) | multiply(n0,n1)| | general |
what is the smallest no . which must be added to 53403977 so as to obtain a sum which is divisible by 11 ? | "for divisibility by 11 , the difference of sums of digits at even and odd places must be either zero or divisible by 11 . for 53403977 , difference = ( 5 + 4 + 3 + 7 ) - ( 3 + 0 + 9 + 7 ) = 19 - 19 = 0 . = > 53403977 is also divisible by 11 c" | a ) 5 , b ) 4 , c ) 0 , d ) 1 , e ) 2 | c | divide(multiply(53403977, 11), 53403977) | multiply(n0,n1)|divide(#0,n0)| | general |
each child has 8 crayons and 15 apples . if there are 7 children , how many crayons are there in total ? | 8 * 7 = 56 . answer is b . | a ) 22 , b ) 56 , c ) 12 , d ) 36 , e ) 10 | b | multiply(7, 8) | multiply(n0,n2)| | general |
a student took 6 courses last year and received an average ( arithmetic mean ) grade of 100 points . the year before , the student took 5 courses and received an average grade of 60 points . to the nearest tenth of a point , what was the student ’ s average grade for the entire two - year period ? | "let the 6 courses that were taken last year be a 1 , a 2 , a 3 , a 4 , a 5 , a 6 a 1 + a 2 + a 3 + a 4 + a 5 + a 6 = 100 * 6 = 600 the year before , the 5 courses be b 1 , b 2 , b 3 , b 4 , b 5 b 1 + b 2 + b 3 + b 4 + b 5 = 60 * 5 = 300 student ' s average = ( 600 + 300 ) / 11 = 81.81 answer d" | a ) 79 , b ) 89 , c ) 95 , d ) 81.81 , e ) 97.2 | d | floor(divide(add(multiply(6, 100), multiply(5, 60)), add(6, 5))) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0)|floor(#4)| | general |
a person is traveling at 20 km / hr and reached his destiny in 2.5 hr then find the distance ? | "t = 2.5 hrs = 5 / 2 hrs d = t * s = 20 * 5 / 2 = 50 km answer is e" | a ) 53 km , b ) 55 km , c ) 52 km , d ) 60 km , e ) 50 km | e | multiply(20, 2.5) | multiply(n0,n1)| | physics |
the parameter of a square is equal to the perimeter of a rectangle of length 16 cm and breadth 12 cm . find the circumference of a semicircle whose diameter is equal to the side of the square . ( round off your answer to two decimal places ) | "let the side of the square be a cm . parameter of the rectangle = 2 ( 16 + 12 ) = 56 cm parameter of the square = 56 cm i . e . 4 a = 56 a = 14 diameter of the semicircle = 14 cm circimference of the semicircle = 1 / 2 ( ∏ ) ( 14 ) = 1 / 2 ( 22 / 7 ) ( 14 ) = 308 / 14 = 22 cm answer : a" | a ) 22 cm , b ) 47.14 cm , c ) 84.92 cm , d ) 94.94 cm , e ) 23.57 cm | a | divide(circumface(divide(square_edge_by_perimeter(rectangle_perimeter(16, 12)), const_2)), const_2) | rectangle_perimeter(n0,n1)|square_edge_by_perimeter(#0)|divide(#1,const_2)|circumface(#2)|divide(#3,const_2)| | geometry |
a motorcyclist goes from bombay to pune , a distance of 192 kms at an average of 32 kmph speed . another man starts from bombay by car 2 ½ hours after the first , and reaches pune ½ hour earlier . what is the ratio of the speed of the motorcycle and the car ? | b 1 : 2 t = 192 / 32 = 6 h t = 6 - 3 = 3 time ratio = 6 : 3 = 2 : 1 speed ratio = 1 : 2 | a ) 2 : 8 , b ) 1 : 2 , c ) 2 : 5 , d ) 6 : 9 , e ) 6 : 2 | b | divide(divide(192, divide(192, 32)), divide(192, subtract(divide(192, 32), const_3))) | divide(n0,n1)|divide(n0,#0)|subtract(#0,const_3)|divide(n0,#2)|divide(#1,#3) | physics |
a corporation paid $ 8 million in federal taxes on its first $ 50 million of gross profits and then $ 30 million in federal taxes on the next $ 150 million in gross profits . by approximately what percent did the ratio of federal taxes to gross profits increase from the first $ 50 million in profits to the next $ 150 m... | "difference in ratios = ( 30 / 150 ) - ( 8 / 50 ) = ( 6 / 50 ) % change = ( change ( 6 / 50 ) / original ratio ( 7 / 50 ) ) * 100 = 86 % answer - e" | a ) 6 % , b ) 14 % , c ) 20 % , d ) 23 % , e ) 86 % | e | multiply(divide(subtract(divide(add(30, 8), 150), divide(8, 50)), divide(8, 50)), const_100) | add(n0,n2)|divide(n0,n1)|divide(#0,n3)|subtract(#2,#1)|divide(#3,#1)|multiply(#4,const_100)| | gain |
what is the measure of the radius of the circle that circumscribes a triangle whose sides measure 9 , 40 and 41 ? | "explanatory answer 9 , 40 and 41 is pythagorean triplet . so , the triangle is right angled . key property about right triangles in a right angled triangle , the radius of the circle that circumscribes the triangle is half the hypotenuse . in the given triangle , the hypotenuse = 41 . therefore , the radius of the cir... | a ) 6 , b ) 4 , c ) 24.5 , d ) 20.5 , e ) 12.5 | d | divide(triangle_area_three_edges(9, 40, 41), divide(triangle_perimeter(9, 40, 41), const_2)) | triangle_area_three_edges(n0,n1,n2)|triangle_perimeter(n0,n1,n2)|divide(#1,const_2)|divide(#0,#2)| | geometry |
70 , 75 , 80 , 85 , 90 , 105 , 105 , 130 , 130 , 130 the list shown consists of the times , in seconds , that it took each of 10 schoolchildren to run a distance of 400 on of meters . if the standard devastation of the 10 running times is 25.4 seconds , rounded to the nearest tenth of a second , how many of the 10 runn... | the most time consuming part in this question is to define the mean . under exam pressure and time pressure it is very easy to make mistake . it is easier to group numbers : 130 * 3 = 390 ; 105 * 2 = 210 ; 75 + 85 = 160 ; 70 + 80 = 150 ; 90 ; next stage combine results , again using more convenient ways to calculate : ... | a ) one , b ) two , c ) three , d ) four , e ) five | b | add(subtract(75, floor(subtract(const_100, 25.4))), const_1) | subtract(const_100,n13)|floor(#0)|subtract(n1,#1)|add(#2,const_1) | general |
if x is invested in a bank at a rate of simple interest of y % p . a . for two years , then the interest earned is 400 . if x is invested at y % p . a . , for two years when the interest is compounded annually , the interest is 410 . what is the value of x ? | "simple way to solve this question is to use options . from si , we know that x * y = 20,000 . now , put the value of x = 4000 , we will have y = 5 % to calculate ci , now , we know 1 st year amount = 4000 + 5 % of 4000 = 4200 . 2 nd year , amount = 4200 + 5 % of 4200 = 4410 . we can see after 2 years interest = 4410 -... | a ) 8000 , b ) 6000 , c ) 5000 , d ) 4000 , e ) 3000 | d | divide(power(divide(400, const_2), const_2), subtract(410, 400)) | divide(n0,const_2)|subtract(n1,n0)|power(#0,const_2)|divide(#2,#1)| | gain |
a cricket bat is sold for $ 900 , making a profit of $ 225 . the profit percentage would be | "225 / ( 900 - 225 ) = 33.33 % . answer : c" | a ) 24 % , b ) 25.5 % , c ) 33.33 % , d ) 36.3 % , e ) 40 % | c | multiply(divide(225, subtract(900, 225)), const_100) | subtract(n0,n1)|divide(n1,#0)|multiply(#1,const_100)| | gain |
a part of certain sum of money is invested at 16 % per annum and the rest at 12 % per annum , if the interest earned in each case for the same period is equal , then ratio of the sums invested is ? | "12 : 16 = 3 : 4 answer : e" | a ) 5 : 4 , b ) 1 : 4 , c ) 3 : 1 , d ) 3 : 5 , e ) 3 : 4 | e | multiply(divide(12, const_100), 16) | divide(n1,const_100)|multiply(n0,#0)| | gain |
reduce 4128 / 4386 to its lowest terms | explanation : hcf of 4128 and 4386 4128 ) 4386 ( 1 4128 - - - - - 258 ) 4128 ( 16 4128 - - - - 0 hence hcf of 4128 and 4386 is 258 . 4128 / 258 = 16 4386 / 258 = 17 4128 / 4386 = 16 / 17 answer : option c | a ) 14 / 13 , b ) 13 / 17 , c ) 16 / 17 , d ) 13 / 18 , e ) none of these | c | divide(divide(divide(4128, const_2), const_3), divide(divide(4386, const_2), const_3)) | divide(n0,const_2)|divide(n1,const_2)|divide(#0,const_3)|divide(#1,const_3)|divide(#2,#3) | general |
a bag holds 2 red marbles and 3 green marbles . if you removed two randomly selected marbles from the bag , without replacement , what is the probability that both would be red ? | probability of selecting first red marble = 2 / 5 probability of selecting second red marble without replacement = 1 / 4 final probability = 2 / 5 * 1 / 4 = 1 / 10 correct answer - a | a ) 1 / 10 , b ) 1 / 5 , c ) 3 / 10 , d ) 2 / 5 , e ) 1 / 2 | a | divide(choose(2, 2), choose(add(2, 3), const_2)) | add(n0,n1)|choose(n0,n0)|choose(#0,const_2)|divide(#1,#2) | other |
for how many integer values of n will the value of the expression 4 n + 7 be an integer greater than 1 and less than 40 ? | "4 n + 7 > 1 4 n > - 6 n > - ( 3 / 2 ) n > - 1.5 ( n = - 1 , 0 , 1 , 2 3 . . . . . . . . upto infinity ) from second constraint 4 n + 7 < 40 4 n < 33 n < 8 . 25 n = ( - infinity , . . . . . . . - 3 , - 2 , - 1 , 0 , 1 , 2 , . . . . . . . . . upto 8 ) combining the two - 1.5 < n < 8.25 n = 1 to 8 ( 48 integers ) and n =... | a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | a | subtract(floor(divide(subtract(40, 7), 4)), floor(divide(subtract(1, 7), 4))) | subtract(n3,n1)|subtract(n2,n1)|divide(#0,n0)|divide(#1,n0)|floor(#2)|floor(#3)|subtract(#4,#5)| | general |
a contractor undertook to do a piece of work in 9 days . he employed certain number of laboures but 6 of them were absent from the very first day and the rest could finish the work in only 15 days . find the number of men originally employed . | let the number of men originally employed be x . 9 x = 15 ( x – 6 ) or x = 15 answer a | a ) 15 , b ) 6 , c ) 13 , d ) 9 , e ) none of these | a | divide(multiply(15, 6), subtract(15, 9)) | multiply(n1,n2)|subtract(n2,n0)|divide(#0,#1) | physics |
the average of 11 results is 42 , if the average of first 5 results is 49 and that of the last 7 is 52 . find the fifth result ? | 1 to 11 = 11 * 42 = 462 1 to 5 = 5 * 49 = 245 5 to 11 = 7 * 52 = 364 5 th = 245 + 364 – 462 = 147 answer : c | a ) 145 , b ) 146 , c ) 147 , d ) 148 , e ) 149 | c | subtract(add(multiply(5, 49), multiply(7, 52)), multiply(11, 42)) | multiply(n2,n3)|multiply(n4,n5)|multiply(n0,n1)|add(#0,#1)|subtract(#3,#2) | general |
a train travelling at 48 kmph completely crosses another train having half its length and travelling in opposite direction at 42 kmph , in 12 seconds . it also passes a railway platform in 45 seconds . the length of the platform is | length of the ist train = 2 x m and the length of the other train be x m ( 1 / 2 ) the relative speed of the train ( 48 + 42 ) = 90 kmph or 25 m / s time taken to cross one another = 12 s hence , 12 = 2 x + x / 25 solving , x = 100 m and 200 m now to determine the length of the platform 45 = 100 + length of the platfor... | a ) 400 m , b ) 450 m , c ) 560 m , d ) 600 m , e ) 680 m | d | multiply(45, multiply(48, const_0_2778)) | multiply(n0,const_0_2778)|multiply(n3,#0) | physics |
if x is to be chosen at random from the set { 1 , 2 , 3 } and y is to be chosen at random from the set { 5 , 6 , 7 } , what is the probability that xy will be even ? | "probably the best way to solve would be to use 1 - p ( opposite event ) = 1 - p ( odd ) = 1 - p ( odd ) * p ( odd ) = 1 - 2 / 4 * 2 / 3 = 8 / 12 = 5 / 6 . answer : e ." | a ) 1 / 6 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 5 / 6 | e | divide(power(2, 2), multiply(choose(5, 2), choose(2, 1))) | choose(n3,n1)|choose(n1,n0)|power(n1,n1)|multiply(#0,#1)|divide(#2,#3)| | probability |
if seven machines working at the same rate can do 3 / 4 of a job in 30 minutes , how many minutes would it take two machines working at the same rate to do 3 / 5 of the job ? | "using the std formula m 1 d 1 h 1 / w 1 = m 2 d 2 h 2 / w 2 substituting the values we have 7 * 1 / 2 * 4 / 3 = 2 * 5 / 3 * x ( converted 30 min into hours = 1 / 2 ) 14 / 3 = 10 / 3 * x x = 7 / 5 hour so 84 minutes answer : d" | a ) 45 , b ) 60 , c ) 75 , d ) 84 , e ) 100 | d | divide(divide(3, 5), multiply(divide(divide(divide(3, 4), 5), 30), const_2)) | divide(n0,n4)|divide(n0,n1)|divide(#1,n4)|divide(#2,n2)|multiply(#3,const_2)|divide(#0,#4)| | physics |
what is the % change in the area of a rectangle when its length increases by 60 % and its width decreases by 60 % ? | "( 16 / 10 ) * ( 4 / 10 ) = 64 / 100 of original area 64 / 100 is a 36 % decrease from 100 / 100 - > c" | a ) 0 % , b ) 20 % increase , c ) 36 % decrease , d ) 1 % decrease , e ) insufficient data | c | subtract(const_100, divide(multiply(add(const_100, 60), subtract(const_100, 60)), const_100)) | add(n0,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|subtract(const_100,#3)| | geometry |
initially two cups of same volume are present with milk filled upto 3 / 5 th and 4 / 5 th of their volumes . water is then filled . then two mixtures are mixed . find the ratio of water to milk in the mixture | water content can b in 1 st and second is 2 / 5 and 1 / 5 total : 3 / 5 . the total milk content is 7 / 5 . the ratio is ( ( 3 / 5 ) / ( 7 / 5 ) ) = 3 / 7 . answer : c | a ) 1 / 7 , b ) 2 / 7 , c ) 3 / 7 , d ) 4 / 7 , e ) 5 / 7 | c | divide(add(subtract(const_1, divide(4, 5)), subtract(const_1, divide(3, 5))), add(divide(3, 5), divide(4, 5))) | divide(n2,n1)|divide(n0,n1)|add(#1,#0)|subtract(const_1,#0)|subtract(const_1,#1)|add(#3,#4)|divide(#5,#2) | general |
convert 600 miles into meters ? | "1 mile = 1609.34 meters 600 mile = 600 * 1609.34 = 965604 meters answer is c" | a ) 784596 , b ) 845796 , c ) 965604 , d ) 784596 , e ) 864520 | c | divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 600), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2))) | add(const_2,const_3)|multiply(#0,const_2)|multiply(#1,#1)|multiply(n0,#2)|divide(#3,#2)| | physics |
heinz produces tomato puree by boiling tomato juice . the tomato puree has only 20 % water while the tomato juice has 90 % water . how many liters of tomato puree will be obtained from 30 litres of tomato juice ? | "answer : explanation : in each of the solutions , there is a pure tomato component and some water . so while boiling , water evaporates but tomato not . so we equate tomato part in the both equations . â ‡ ’ â ‡ ’ 10 % ( 30 ) = 80 % ( x ) â ‡ ’ â ‡ ’ x = 3.75 liters . answer : a" | a ) 3.75 liters . , b ) 2.5 liters . , c ) 8.5 liters . , d ) 2.6 liters . , e ) 2.1 liters . | a | divide(multiply(divide(subtract(const_100, 90), const_100), 30), divide(subtract(const_100, 20), const_100)) | subtract(const_100,n1)|subtract(const_100,n0)|divide(#0,const_100)|divide(#1,const_100)|multiply(n2,#2)|divide(#4,#3)| | gain |
if x < y < z and y - x > 3 , where x is an even integer and y and z are odd integers , what is the least possible value of z - x ? | "we have : 1 ) x < y < z 2 ) y - x > 3 3 ) x = 2 k ( x is an even number ) 4 ) y = 2 n + 1 ( y is an odd number ) 5 ) z = 2 p + 1 ( z is an odd number ) 6 ) z - x = ? least value z - x = 2 p + 1 - 2 k = 2 p - 2 k + 1 = 2 ( p - k ) + 1 - that means that z - x must be an odd number . we are asked to find the least value ... | a ) 7 , b ) 6 , c ) 8 , d ) 9 , e ) 10 | a | add(add(3, const_2), const_2) | add(n0,const_2)|add(#0,const_2)| | general |
if an amount of rs 42900 is distributed equally amongst 22 persons , how much amount would each person get ? | required amount = 42900 / 22 = rs 1950 answer a | a ) rs 1950 , b ) rs 2000 , c ) rs 745 , d ) rs 765 , e ) none | a | divide(42900, 22) | divide(n0,n1) | general |
six bells commence tolling together and toll at intervals of 2 , 4 , 6 , 8 10 and 12 seconds respectively . in 50 minutes , how many times do they toll together ? | "l . c . m of 2,4 , 6,8 , 10,12 is 120 . i . e after each 2 min they will toll together . so in 50 min they will toll 25 times . as they have initially tolled once , the answer will be 25 + 1 = 26 . answer : e" | a ) 4 , b ) 10 , c ) 15 , d ) 16 , e ) 26 | e | divide(10, divide(multiply(multiply(2, multiply(multiply(add(2, const_3), 2), const_3)), 2), 10)) | add(n0,const_3)|multiply(n0,#0)|multiply(#1,const_3)|multiply(n0,#2)|multiply(n0,#3)|divide(#4,n4)|divide(n4,#5)| | physics |
what least number must be subtracted from 9671 so that the remaining number is divisible by 5 ? | "on dividing 9671 by 5 , we get remainder = 1 . required number be subtracted = 1 answer : e" | a ) 5 , b ) 4 , c ) 3 , d ) 2 , e ) 1 | e | subtract(9671, multiply(floor(divide(9671, 5)), 5)) | divide(n0,n1)|floor(#0)|multiply(n1,#1)|subtract(n0,#2)| | general |
find the average of all prime numbers between 50 and 70 | "prime numbers between 50 and 70 are 53 , 59 , 61 , 67 required average = ( 53 + 59 + 61 + 67 ) / 4 = 240 / 4 = 60 answer is c" | a ) 90 , b ) 45 , c ) 60 , d ) 90 , e ) 26 | c | divide(add(add(add(50, const_1), add(add(50, const_1), const_2)), add(subtract(70, 50), subtract(70, const_2))), 50) | add(n0,const_1)|subtract(n1,n0)|subtract(n1,const_2)|add(#0,const_2)|add(#1,#2)|add(#0,#3)|add(#5,#4)|divide(#6,n0)| | general |
the sum of ages of 5 children born at the intervals of 3 years each is 70 years . what is the age of the youngest child ? | "let the ages of children be x , ( x + 3 ) , ( x + 6 ) , ( x + 9 ) and ( x + 12 ) years . then , x + ( x + 3 ) + ( x + 6 ) + ( x + 9 ) + ( x + 12 ) = 70 5 x = 40 x = 8 . age of the youngest child = x = 8 years . e )" | a ) 3 years , b ) 4 years , c ) 6 years , d ) 7 years , e ) 8 years | e | subtract(subtract(divide(70, 5), 3), 3) | divide(n2,n0)|subtract(#0,n1)|subtract(#1,n1)| | general |
for how many integer values of n will the value of the expression 4 n + 7 be an integer greater than 1 and less than 120 ? | "4 n + 7 > 1 4 n > - 6 n > - ( 3 / 2 ) n > - 1.5 ( n = - 1 , 0 , 1 , 2 3 . . . . . . . . upto infinity ) from second constraint 4 n + 7 < 120 4 n < 113 n < 28 . 25 n = ( - infinity , . . . . . . . - 3 , - 2 , - 1 , 0 , 1 , 2 , . . . . . . . . . upto 48 ) combining the two - 1.5 < n < 48.25 n = 1 to 28 ( 48 integers ) a... | a ) 22 , b ) 24 , c ) 26 , d ) 28 , e ) 30 | e | subtract(floor(divide(subtract(120, 7), 4)), floor(divide(subtract(1, 7), 4))) | subtract(n3,n1)|subtract(n2,n1)|divide(#0,n0)|divide(#1,n0)|floor(#2)|floor(#3)|subtract(#4,#5)| | general |
the average of thirteen numbers is 9 . the average of first 6 numbers is 5 and the average of last 6 numbers is 7 . what is the middle number ? | the total of thirteen numbers = 13 x 9 = 117 the total of first 6 and last 6 numbers is = 6 x 5 + 6 x 7 = 72 so , the middle number is ( 117 - 72 ) = 45 b | a ) 47 , b ) 45 , c ) 25 , d ) 37 , e ) 44 | b | subtract(multiply(add(add(6, 6), const_1), 9), add(multiply(6, 5), multiply(6, 7))) | add(n1,n1)|multiply(n1,n2)|multiply(n1,n4)|add(#0,const_1)|add(#1,#2)|multiply(n0,#3)|subtract(#5,#4) | general |
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