Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 164 | correct stringclasses 5 values | annotated_formula stringlengths 7 1.65k | linear_formula stringlengths 8 925 | category stringclasses 6 values | answer stringclasses 5 values |
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if the length of the side of a square is doubled , what is the ratio of the areas of the original square to the area of the new square ? | solution if x be the side of the original square , then its area is equal to x 2 if x is doubled to 2 x , then the new area is equal to ( 2 x ) 2 = 4 x 2 the ratio of the areas of the original square to the area of the new square x 2 / ( 4 x 2 ) = 1 / 4 or 1 : 4 answer is a | ['a ) 1 : 4', 'b ) 2 : 4', 'c ) 3 : 4', 'd ) 4 : 4', 'e ) 5 : 4'] | a | divide(multiply(const_1, const_1), power(multiply(const_2, const_1), const_2)) | multiply(const_1,const_1)|multiply(const_1,const_2)|power(#1,const_2)|divide(#0,#2) | geometry | A |
a jogger running at 9 km / hr along side a railway track is 150 m ahead of the engine of a 100 m long train running at 45 km / hr in the same direction . in how much time will the train pass the jogger ? | "speed of train relative to jogger = 45 - 9 = 36 km / hr . = 36 * 5 / 18 = 10 m / sec . distance to be covered = 150 + 100 = 250 m . time taken = 250 / 10 = 25 sec . answer : e" | a ) 88 , b ) 27 , c ) 36 , d ) 80 , e ) 25 | e | divide(add(150, 100), multiply(subtract(45, 9), divide(divide(const_10, const_2), divide(subtract(45, 9), const_2)))) | add(n1,n2)|divide(const_10,const_2)|subtract(n3,n0)|divide(#2,const_2)|divide(#1,#3)|multiply(#4,#2)|divide(#0,#5)| | general | E |
a started a business with an investment of rs . 70000 and after 3 months b joined him investing rs . 120000 . if the profit at the end of a year is rs . 43000 , then the share of b is ? | ratio of investments of a and b is ( 70000 * 12 ) : ( 120000 * 9 ) = 7 : 36 total profit = rs . 43000 share of b = 36 / 43 ( 43000 ) = rs . 36000 answer : b | a ) 33888 , b ) 36000 , c ) 27778 , d ) 27772 , e ) 81122 | b | subtract(43000, multiply(const_60, const_100)) | multiply(const_100,const_60)|subtract(n3,#0) | gain | B |
in a company of 16 employees , 8 employees earn $ 38,000 , 4 employees earn $ 42,000 , and the 4 highest - paid employees earn the same amount . if the average annual salary for the 16 employees is $ 44,000 , what is the annual salary for each of the highest - paid employees ? | "8 * 38,000 + 4 * 42,000 + 4 x = 16 * 44,000 4 x = 704,000 - 304,000 - 168,000 4 x = 232,000 x = 58,000 the answer is e ." | a ) $ 54,000 , b ) $ 55,000 , c ) $ 56,000 , d ) $ 57,000 , e ) $ 58,000 | e | subtract(divide(multiply(divide(subtract(subtract(multiply(add(add(multiply(multiply(4, 4), add(4, const_1)), const_2), divide(const_1, const_2)), 16), multiply(multiply(16, 4), add(4, const_1))), multiply(multiply(multiply(4, 4), add(4, const_1)), 4)), 4), const_1000), const_1000), 8) | add(const_1,n3)|divide(const_1,const_2)|multiply(n5,n5)|multiply(n0,n5)|multiply(#0,#2)|multiply(#0,#3)|add(#4,const_2)|multiply(n3,#4)|add(#6,#1)|multiply(n0,#8)|subtract(#9,#5)|subtract(#10,#7)|divide(#11,n5)|multiply(#12,const_1000)|divide(#13,const_1000)|subtract(#14,n1)| | general | E |
the salary of a typist was at first raised by 10 % and then the same was reduced by 5 % . if he presently draws rs . 6270 . what was his original salary ? | x * ( 110 / 100 ) * ( 95 / 100 ) = 6270 x * ( 11 / 10 ) * ( 1 / 100 ) = 66 x = 6000 answer : a | a ) 6000 , b ) 2999 , c ) 1000 , d ) 2651 , e ) 1971 | a | divide(6270, multiply(add(const_1, divide(10, const_100)), subtract(const_1, divide(5, const_100)))) | divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|divide(n2,#4) | gain | A |
jack and jill work at a hospital with 3 other workers . for an internal review , 2 of the 5 workers will be randomly chosen to be interviewed . what is the probability that jack and jill will both be chosen ? | 1 / 5 c 2 = 1 / 10 . answer : b . | a ) 1 / 3 , b ) 1 / 10 , c ) 1 / 15 , d ) 3 / 8 , e ) 2 / 3 | b | inverse(divide(factorial(5), multiply(factorial(2), factorial(3)))) | factorial(n2)|factorial(n1)|factorial(n0)|multiply(#1,#2)|divide(#0,#3)|inverse(#4) | physics | B |
the sides of a triangle are in the ratio 5 : 12 : 13 and its perimeter is 240 m , its area is ? | "5 x + 12 x + 13 x = 240 = > x = 8 a = 40 , b = 96 , c = 104 s = ( 40 + 96 + 104 ) / 2 = 120 answer : e" | a ) 150 , b ) 882 , c ) 277 , d ) 261 , e ) 120 | e | multiply(240, divide(240, add(add(5, 12), 13))) | add(n0,n1)|add(n2,#0)|divide(n3,#1)|multiply(n3,#2)| | geometry | E |
in a class of 20 students , 3 students did not borrow any books from the library , 9 students each borrowed 1 book , 4 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 total number of books the students borrowed is 20 * 2 = 40 . the students who borrowed zero , one , or two books borrowed 9 * 1 + 4 * 2 = 17 books . the 4 students who borrowed at least three books borrowed 40 - 17 = 23 books . if 3 of these students borrowed exactly 3 books , then the maximum that one student could have borrowed is 23 - 9 = 14 books . the answer is b ." | a ) 13 , b ) 14 , c ) 15 , d ) 16 , e ) 17 | b | subtract(multiply(20, const_2.0), add(multiply(subtract(subtract(20, add(add(multiply(9, 1), 4), 3)), 1), 3), add(multiply(9, 1), multiply(4, 3)))) | multiply(n0,n5)|multiply(n2,n3)|multiply(n5,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 | B |
a and b together can do a piece of work in 4 days . if a alone can do the same work in 20 days , then b alone can do the same work in ? | "b = 1 / 4 β 1 / 20 = 0.2 days answer : e" | a ) 0.35 days , b ) 0.45 days , c ) 0.55 days , d ) 0.25 days , e ) 0.2 days | e | inverse(subtract(inverse(4), inverse(20))) | inverse(n0)|inverse(n1)|subtract(#0,#1)|inverse(#2)| | physics | E |
a cricket player whose bowling average was 24.85 runs per wicket , takes 5 wicket for 52 runs in a match . due to this his average decreases by 0.85 . what will be the number of wickets taken by him till the last match ? | "average = total runs / total wickets total runs after last match = 24.85 w + 52 total wickets after last match = w + 5 ( 24.85 w + 52 ) / ( w + 5 ) = 24.85 - 0.85 = 24 w = 80 so total wickets aftr last match = w + 5 = 85 answer : c" | a ) 64 , b ) 72 , c ) 85 , d ) 96 , e ) 108 | c | divide(subtract(multiply(subtract(24.85, 0.85), 5), 52), 0.85) | subtract(n0,n3)|multiply(n1,#0)|subtract(#1,n2)|divide(#2,n3)| | general | C |
a certain bacteria colony doubles in size every day for 20 days , at which point it reaches the limit of its habitat and can no longer grow . if two bacteria colonies start growing simultaneously , how many days will it take them to reach the habitat β s limit ? | "simultaneous grow = same amount of time in 20 days we should expect to have p ( 2 ) ^ 20 so if we have two things doing the job of getting us there then ; 2 ^ x + 2 ^ x = 20 2 ( 2 ) ^ x = 2 2 ^ ( x + 1 ) = 2 ^ 20 x + 1 = 20 x = 19 ; answer : e" | a ) 6.33 , b ) 7.5 , c ) 10 , d ) 15 , e ) 19 | e | subtract(20, divide(20, 20)) | divide(n0,n0)|subtract(n0,#0)| | physics | E |
the prices of tea and coffee per kg were the same in june . in july the price of coffee shot up by 20 % and that of tea dropped by 10 % . if in july , a mixture containing equal quantities of tea and coffee costs 50 / kg . how much did a kg of coffee cost in june ? | let the price of tea and coffee be x per kg in june . price of tea in july = 1.2 x price of coffee in july = 0.9 x . in july the price of 1 / 2 kg ( 500 gm ) of tea and 1 / 2 kg ( 500 gm ) of coffee ( equal quantities ) = 50 1.2 x ( 1 / 2 ) + 0.9 x ( 1 / 2 ) = 50 = > x = 47.61 d | a ) 45 , b ) 40 , c ) 35 , d ) 47.61 , e ) 30 | d | divide(50, multiply(subtract(const_1, divide(10, const_100)), add(divide(20, const_100), const_1))) | divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|divide(n2,#4) | general | D |
the captain of a cricket team of 11 members is 26 years old and the wicket keeper is 7 years older . if the ages of these two are excluded , the average age of the remaining players is one year less than the average age of the whole team . what is the average age of the team ? | "explanation let the average age of the whole team by x years . 11 x Γ’ β¬ β ( 26 + 33 ) = 9 ( x - 1 ) 11 x Γ’ β¬ β 9 x = 50 2 x = 50 x = 25 . so , average age of the team is 25 years . answer c" | a ) 23 years , b ) 24 years , c ) 25 years , d ) 26 years , e ) 27 years | c | divide(subtract(add(26, add(26, 7)), multiply(7, 7)), const_2) | add(n1,n2)|multiply(n2,n2)|add(n1,#0)|subtract(#2,#1)|divide(#3,const_2)| | general | C |
50 persons like apple . 7 like orange and mango dislike apple . 10 like mango and apple and dislike orange . 4 like all . how many people like apple ? | "orange + mango - apple = 7 mango + apple - orange = 10 apple = 50 orange + mango + apple = 4 50 + 10 + 4 - 7 = 57 like apple answer : d" | a ) 47 , b ) 46 , c ) 54 , d ) 57 , e ) 62 | d | add(add(subtract(50, 4), 7), subtract(10, 7)) | subtract(n0,n3)|subtract(n2,n1)|add(n1,#0)|add(#2,#1)| | general | D |
there were two candidates in an election . winner candidate received 62 % of votes and won the election by 360 votes . find the number of votes casted to the winning candidate ? | "w = 62 % l = 38 % 62 % - 38 % = 24 % 24 % - - - - - - - - 360 62 % - - - - - - - - ? = > 930 answer : e" | a ) 288 , b ) 744 , c ) 788 , d ) 298 , e ) 930 | e | divide(multiply(divide(360, divide(subtract(62, subtract(const_100, 62)), const_100)), 62), const_100) | subtract(const_100,n0)|subtract(n0,#0)|divide(#1,const_100)|divide(n1,#2)|multiply(n0,#3)|divide(#4,const_100)| | gain | E |
find the simple interest on rs . 567 for 7 months at 9 paisa per month ? | "explanation : i = ( 567 * 7 * 9 ) / 100 = 357.21 answer : option a" | a ) s . 357.21 , b ) s . 322.12 , c ) s . 400 , d ) s . 278.9 , e ) s . 300 | a | multiply(567, divide(7, const_100)) | divide(n1,const_100)|multiply(n0,#0)| | gain | A |
robert is travelling on his cycle andhas calculated to reach point a at 2 pm . if he travels at 10 kmph , he will reach there at 12 pm if he travels at 15 kmph . at what speed must he travel to reach a at 1 pm ? | let the distance travelled by x km . then , x - x = 2 10 15 3 x - 2 x = 60 x = 60 km . time taken to travel 60 km at 10 km / hr = 60 hrs = 6 hrs . 10 so , robert started 6 hours before 2 p . m . i . e . , at 8 a . m . required speed = 60 kmph . = 12 kmph . 5 c | a ) 8 kmph , b ) 9 kmph , c ) 12 kmph , d ) 14 kmph , e ) 16 kmph | c | divide(divide(2, subtract(divide(const_1, 10), divide(const_1, 15))), subtract(divide(divide(2, subtract(divide(const_1, 10), divide(const_1, 15))), 10), 1)) | divide(const_1,n1)|divide(const_1,n3)|subtract(#0,#1)|divide(n0,#2)|divide(#3,n1)|subtract(#4,n4)|divide(#3,#5) | physics | C |
3 friends a , b , c went for week end party to mcdonald β s restaurant and there they measure there weights in some order in 7 rounds . a , b , c , ab , bc , ac , abc . final round measure is 140 kg then find the average weight of all the 7 rounds ? | "average weight = [ ( a + b + c + ( a + b ) + ( b + c ) + ( c + a ) + ( a + b + c ) ] / 7 = 4 ( a + b + c ) / 7 = 4 x 140 / 7 = 80 kgs answer : b" | a ) 98.5 kgs , b ) 80 kgs , c ) 76.5 kgs , d ) 67.5 kgs , e ) 58.2 kgs | b | divide(multiply(add(const_1, 3), 140), 7) | add(const_1,n0)|multiply(n2,#0)|divide(#1,n1)| | general | B |
the arithmetic mean and standard deviation of a certain normal distribution are 15.5 and 1.5 , respectively . what value is exactly 2 standard deviations less than the mean ? | "the value which isexactlytwo sd less than the mean is : mean - 2 * sd = 15.5 - 2 * 1.5 = 12.5 . answer : e ." | a ) 10.5 , b ) 11 , c ) 11.5 , d ) 12 , e ) 12.5 | e | subtract(15.5, multiply(2, 1.5)) | multiply(n1,n2)|subtract(n0,#0)| | general | E |
a single reservoir supplies the petrol to the whole city , while the reservoir is fed by a single pipeline filling the reservoir with the stream of uniform volume . when the reservoir is full and if 40,000 liters of petrol is used daily , the suply fails in 90 days . if 32,000 liters of petrol is used daily , it fails in 60 days . how much petrol can be used daily with out the supply ever failing ? | explanation : let x liter be the per day filling and v litr be the capacity of the reservoir , then 90 x + v = 40000 Γ 90 - - - - - ( 1 ) 60 x + v = 32000 Γ 60 - - - - - - ( 2 ) solving eq . ( 1 ) and ( 2 ) , we get x = 56000 hence , 56000 liters per day can be used without the failure of supply . answer : b | ['a ) 64000 liters', 'b ) 56000 liters', 'c ) 78000 liters', 'd ) 60000 liters', 'e ) none of these'] | b | add(add(add(multiply(60, 90), surface_rectangular_prism(const_180, 60, 60)), const_180), multiply(const_2, const_10)) | multiply(n1,n3)|multiply(const_10,const_2)|surface_rectangular_prism(n3,n3,const_180)|add(#0,#2)|add(#3,const_180)|add(#4,#1) | physics | B |
how many points ( x , y ) lie on the line segment between ( 32 , 12 2 / 3 ) and ( 7 , 17 2 / 3 ) such that x and y are both integers ? | "slope = ( 17 2 / 3 - 12 2 / 3 ) / ( 7 - 32 ) = - 1 / 5 y = mx + b = > 12 2 / 3 = - 22 / 3 + b = > b = 19 y = - x / 5 + 19 only integer values work , and the only multiples of 5 between 7 and 32 for x values are 10 , 15 , 20 , 25 and 30 , thus 5 points . c" | a ) one , b ) 4 , c ) 5 , d ) 8 , e ) 9 | c | divide(7, 7) | divide(n4,n4)| | general | C |
a car travels uphill at 30 km / hr and downhill at 90 km / hr . it goes 100 km uphill and 50 km downhill . find the average speed of the car ? | "avg speed = total distance / total time . total distance traveled = 100 + 50 = 150 km ; time taken for uphill journey = 100 / 30 = 10 / 3 ; time taken for down hill journey = 50 / 90 = 5 / 9 ; avg speed = 150 / ( 10 / 3 + 5 / 9 ) = 39 kmph answer : b" | a ) 32 kmph , b ) 39 kmph , c ) 34 kmph , d ) 35 kmph , e ) 36 kmph | b | divide(add(100, 50), add(divide(100, 30), divide(50, 90))) | add(n2,n3)|divide(n2,n0)|divide(n3,n1)|add(#1,#2)|divide(#0,#3)| | general | B |
an item is being sold for $ 10 each . however , if a customer will β buy at least 3 β they have a promo discount of 15 % . also , if a customer will β buy at least 10 β items they will deduct an additional 28 % to their β buy at least 3 β promo price . if sam buys 10 pcs of that item how much should he pay ? | without any discount sam should pay 10 * 10 = $ 100 . now , the overall discount would be slightly less than 43 % , thus he must pay slightly more than $ 57 . only answer choice e fits . answer : e . | a ) $ 92.00 , b ) $ 88.00 , c ) $ 87.04 , d ) $ 80.96 , e ) $ 60.00 | e | multiply(subtract(10, divide(multiply(15, 28), const_100)), const_10) | multiply(n2,n4)|divide(#0,const_100)|subtract(n0,#1)|multiply(#2,const_10) | gain | E |
if 1 tic equals 4 tacs and 5 tacs equal 8 tocs , what is the ratio of one tic to one toc ? | "tic = 4 * tac and 5 * tac = 8 * toc ; 5 * tic = 20 * tac and 20 * tac = 32 * toc - - > 5 * tic = 32 * toc - - > tic / toc = 32 / 5 . answer : d ." | a ) 15 / 2 , b ) 6 / 5 , c ) 5 / 6 , d ) 32 / 5 , e ) 1 / 15 | d | add(divide(8, 1), divide(4, 1)) | divide(n3,n0)|divide(n1,n0)|add(#0,#1)| | other | D |
a certain sum of money is divided among a , b and c such that a gets one - third of what b and c together get and b gets two - seventh of what a and c together get . if the amount received by a is $ 35 more than that received by b , find the total amount shared by a , b and c . | "a = 1 / 3 ( b + c ) = > c = 3 a - b - - - ( 1 ) b = 2 / 7 ( a + c ) = > c = 3.5 b - a - - ( b ) a - b = $ 35 a = 35 + b ( 1 ) = = = > c = 105 + 3 b - b = 2 b + 105 = = > 2 b - c = - 105 - - - ( 3 ) ( 2 ) = = = > c = 3.5 b - b - 35 = 2.5 b - 35 = = > 2.5 b - c = 35 - - - ( 4 ) from ( 4 ) and ( 3 ) 0.5 b = 140 b = $ 280 a = $ 315 c = 945 - 280 = $ 665 total amount = 315 + 280 + 665 = $ 1260 answer : c" | a ) $ 1360 , b ) $ 1250 , c ) $ 1260 , d ) $ 1270 , e ) $ 1280 | c | add(add(add(multiply(multiply(const_3, 35), const_2), multiply(const_2, 35)), add(add(multiply(multiply(const_3, 35), const_2), multiply(const_2, 35)), 35)), subtract(multiply(add(add(multiply(multiply(const_3, 35), const_2), multiply(const_2, 35)), 35), const_3), add(multiply(multiply(const_3, 35), const_2), multiply(const_2, 35)))) | multiply(n0,const_3)|multiply(n0,const_2)|multiply(#0,const_2)|add(#2,#1)|add(n0,#3)|add(#3,#4)|multiply(#4,const_3)|subtract(#6,#3)|add(#5,#7)| | general | C |
two positive integers differ by 4 , and sum of their reciprocals is 6 . then one of the numbers is | "algebraic approach : let n be the smaller integer = > 1 / n + 1 / ( n + 4 ) = 6 or ( ( n + 4 ) + n ) / n ( n + 4 ) = 6 or ( n ^ 2 + 4 n ) * 6 = 2 n + 4 or n = 2 as n can not be - negative solve for n = > n = 4 . hence , d" | a ) a ) 3 , b ) b ) 1 , c ) c ) 5 , d ) 4 , e ) e ) 28 | d | add(6, 6) | add(n1,n1)| | general | D |
if x = 1 + β 2 , then what is the value of x 4 - 4 x 3 + 4 x 2 + 4 ? | answer x = 1 + β 2 β΄ x 4 - 4 x 3 + 4 x 2 + 5 = x 2 ( x 2 - 4 x + 4 ) + 4 = x 2 ( x - 2 ) 2 + 4 = ( 1 + β 2 ) 2 ( 1 + β 2 - 2 ) 2 + 4 = ( β 2 + 1 ) 2 ( β 2 - 1 ) 2 + 4 = [ ( β 2 ) 2 - ( 1 ) 2 ] 2 + 4 = ( 2 - 1 ) 2 = 1 + 4 = 5 correct option : c | a ) - 1 , b ) 0 , c ) 5 , d ) 2 , e ) 3 | c | add(multiply(power(add(1, sqrt(2)), const_2), power(subtract(add(1, sqrt(2)), 2), const_2)), 4) | sqrt(n1)|add(n0,#0)|power(#1,const_2)|subtract(#1,n1)|power(#3,const_2)|multiply(#2,#4)|add(n2,#5) | general | C |
two friends a , b decided to share a lot of apples . each of them had half of the total plus half an apple in order . after each of them took their share 4 time , no apples were left . how many apples were there ? | whenever the rate of reduction is ' half of the total and half of it ' , the answer is always ( 2 ^ n ) - 1 , where ' n ' is the number of times the process is repeated . here , the process is repeated 8 times . so answer is ( 2 ^ 8 ) - 1 = 255 answer : e | a ) 250 , b ) 255 , c ) 254 , d ) 253 , e ) 251 | e | subtract(power(const_2, multiply(4, const_2)), const_1) | multiply(n0,const_2)|power(const_2,#0)|subtract(#1,const_1) | general | E |
what is the units digit of 31 ! + 50 ! + 2 ! + 3 ! ? | "for all n greater than 4 , the units digit of n ! is 0 . the sum of the four units digits is 0 + 0 + 2 + 6 = 8 the units digit is 8 . the answer is e ." | a ) 0 , b ) 2 , c ) 4 , d ) 6 , e ) 8 | e | add(add(const_4, const_3), const_2) | add(const_3,const_4)|add(#0,const_2)| | other | E |
a train crosses a platform of 200 m in 20 sec , same train crosses another platform of length 350 m in 25 sec . then find the length of the train ? | "length of the train be β x β x + 200 / 20 = x + 350 / 25 5 x + 1000 = 4 x + 1400 x = 400 m answer : c" | a ) 150 m , b ) 180 m , c ) 400 m , d ) 250 m , e ) 155 m | c | subtract(multiply(350, divide(20, divide(20, const_3))), multiply(200, divide(25, divide(20, const_3)))) | divide(n1,const_3)|divide(n1,#0)|divide(n3,#0)|multiply(n2,#1)|multiply(n0,#2)|subtract(#3,#4)| | physics | C |
a and b together can do a work in 3 days . a alone can do it in 12 days . what time b will take to do the work alone ? | "explanation : a and b 1 day ' s work = 1 / 3 a alone can do 1 day ' s work = 1 / 12 what time b will take to do the work alone ? b = ( a + b ) - a = ( 1 / 3 - ( 1 / 12 ) = 4 days answer : option a" | a ) 4 days , b ) 8 days , c ) 12 days , d ) 10 days , e ) 5 days | a | add(inverse(subtract(divide(const_1, 3), divide(const_1, 12))), divide(const_2, add(const_2, const_3))) | add(const_2,const_3)|divide(const_1,n0)|divide(const_1,n1)|divide(const_2,#0)|subtract(#1,#2)|inverse(#4)|add(#3,#5)| | physics | A |
if 63 percent of a class answered the first question on a certain test correctly , 49 percent answered the second question on the test correctly , and 20 percent answered neither of the questions correctly , what percent answered both correctly ? | "63 % answered the first question correctly and 20 % answered neither correctly . then 17 % missed the first question but answered the second question correctly . then the percent who answered both correctly is 49 % - 17 % = 32 % . the answer is e ." | a ) 22 % , b ) 24 % , c ) 26 % , d ) 28 % , e ) 32 % | e | subtract(add(add(63, 49), 20), const_100) | add(n0,n1)|add(n2,#0)|subtract(#1,const_100)| | other | E |
jack and jill work at a hospital with 5 other workers . for an internal review , 2 of the 7 workers will be randomly chosen to be interviewed . what is the probability that jack and jill will both be chosen ? | "1 / 7 c 2 = 1 / 21 . answer : c ." | a ) 1 / 3 , b ) 1 / 4 , c ) 1 / 21 , d ) 3 / 8 , e ) 2 / 3 | c | inverse(divide(factorial(7), multiply(factorial(2), factorial(5)))) | factorial(n2)|factorial(n1)|factorial(n0)|multiply(#1,#2)|divide(#0,#3)|inverse(#4)| | physics | C |
if a book is sold at 6 % profit instead of 6 % loss , it would have brought rs 15 more . find out the cost price of the book | "let c . p . of the book be rs . β x β given , 1.06 x - 0.94 x = 15 = > 0.12 x = 15 = 15 / 0.12 = rs 125 answer : e" | a ) 75 , b ) 72 , c ) 60 , d ) 70 , e ) 125 | e | divide(multiply(const_100, divide(6, const_2)), 6) | divide(n0,const_2)|multiply(#0,const_100)|divide(#1,n0)| | gain | E |
a corporation that had $ 120 billion in profits for the year paid out $ 200 million in employee benefits . approximately what percent of the profits were the employee benefits ? ( note : 1 billion = 10 ^ 9 ) | "required answer = [ employee benefit / profit ] * 100 = [ ( 200 million ) / ( 120 billion ) ] * 100 = [ ( 200 * 10 ^ 6 ) / ( 120 * 10 ^ 9 ) ] * 100 = ( 1.7 / 1000 ) * 100 = 0.17 % so answer is ( e )" | a ) 50 % , b ) 20 % , c ) 5 % , d ) 4 % , e ) 0.17 % | e | multiply(divide(multiply(200, power(10, add(const_3, const_3))), multiply(120, power(10, 9))), const_100) | add(const_3,const_3)|power(n3,n4)|multiply(n0,#1)|power(n3,#0)|multiply(n1,#3)|divide(#4,#2)|multiply(#5,const_100)| | general | E |
two cyclist start on a circular track from a given point but in opposite direction with speeds of 7 m / s and 8 m / s . if the circumference of the circle is 180 meters , after what time will they meet at the starting point ? | "they meet every 180 / 7 + 8 = 12 sec answer is b" | a ) 20 sec , b ) 12 sec , c ) 30 sec , d ) 50 sec , e ) 1 min | b | divide(180, add(8, 7)) | add(n0,n1)|divide(n2,#0)| | physics | B |
if 85 % of 3 / 5 of a number is 36 , then the number is ? | "let the number be x . then 85 % of 3 / 5 of x = 36 85 / 100 * 3 / 5 * x = 36 x = ( 36 * 100 / 51 ) = 70.5 required number = 70.5 . correct option : c" | a ) 80 , b ) 100 , c ) 70.5 , d ) 90 , e ) none of these | c | divide(36, multiply(divide(85, const_100), divide(3, 5))) | divide(n0,const_100)|divide(n1,n2)|multiply(#0,#1)|divide(n3,#2)| | gain | C |
what is the difference between the place values of two 1 ' s in the numeral 135.21 | required difference = 100 - 0.01 = 99.99 answer is d | a ) 99.999 , b ) 100.2 , c ) 134 , d ) 99.99 , e ) 99.9 | d | subtract(multiply(1, const_100), divide(1, const_100)) | divide(n0,const_100)|multiply(n0,const_100)|subtract(#1,#0) | general | D |
there are 3 vessels of equal capacity . vessel a contains milk and water in the ratio 7 : 3 ; vessel b contains milk and water in the ratio 2 : 1 and vessel c contains milk and water in the ratio 3 : 2 . if the mixture in all the 3 vessels is mixed up . what will be the ratio of milk and water ? | 7 : 3 = > 7 x + 3 x = 10 x 2 : 1 = > 2 y + 1 y = 3 y 3 : 2 = > 3 z + 2 z = 5 z 10 x = 3 y = 5 z take lcm of 10 , 3,5 = 30 or simply ; x = 3 y = 10 z = 6 so , ratio of milk : water = ( 7 x + 2 y + 3 z ) / ( 3 x + y + 2 z ) = 59 / 31 ans : a | a ) 59 : 31 , b ) 3 : 2 , c ) 118 : 126 , d ) 193 : 122 , e ) 201 : 132 | a | divide(add(add(multiply(7, multiply(add(2, 1), add(3, 2))), multiply(2, multiply(add(7, 3), add(3, 2)))), multiply(3, multiply(add(7, 3), add(2, 1)))), add(add(multiply(3, multiply(add(2, 1), add(3, 2))), multiply(1, multiply(add(7, 3), add(3, 2)))), multiply(2, multiply(add(7, 3), add(2, 1))))) | add(n3,n4)|add(n0,n3)|add(n0,n1)|multiply(#0,#1)|multiply(#2,#1)|multiply(#2,#0)|multiply(n1,#3)|multiply(n3,#4)|multiply(n0,#5)|multiply(n0,#3)|multiply(n4,#4)|multiply(n3,#5)|add(#6,#7)|add(#9,#10)|add(#12,#8)|add(#13,#11)|divide(#14,#15) | general | A |
in a renowned city , the average birth rate is 7 people every two seconds and the death rate is 2 people every two seconds . estimate the size of the population net increase that occurs in one day . | this question can be modified so that the birth rate is given every m seconds and the death rate is given every n seconds . for this particular question : increase in the population every 2 seconds = 7 - 2 = 5 people . total 2 second interval in a day = 24 * 60 * 60 / 2 = 43,200 population increase = 43,200 * 5 = 216,000 . hence b . | a ) 215,000 , b ) 216,000 , c ) 217,000 , d ) 218,000 , e ) 219,000 | b | multiply(multiply(subtract(7, const_2), const_3600), const_12) | subtract(n0,const_2)|multiply(#0,const_3600)|multiply(#1,const_12) | general | B |
a train 225 m in length crosses a telegraph post in 15 seconds . the speed of the train is ? | "s = 225 / 25 * 18 / 5 = 32 kmph answer : c" | a ) 37 kmph , b ) 34 kmph , c ) 32 kmph , d ) 38 kmph , e ) 76 kmph | c | multiply(const_3_6, divide(225, 15)) | divide(n0,n1)|multiply(#0,const_3_6)| | physics | C |
how many 3 - digit even numbers are possible such that if one of the digits is 4 , the next / succeeding digit to it should be 7 ? | 470 , 472 , 474 , 476 , and 478 , so total 5 . hence option b . | a ) 305 , b ) 5 , c ) 365 , d ) 405 , e ) 495 | b | add(add(3, 7), 7) | add(n0,n2)|add(n2,#0)| | general | B |
in a certain animal shelter , the ratio of the number of dogs to the number of cats is 15 to 7 . if 8 additional cats were to be taken in by the shelter , the ratio of the number of dogs to the number of cats would be 15 to 11 . how many dogs are in the shelter ? | "this ratio question can be solved in a couple of different ways . here ' s an algebraic approach . . . we ' re told that the ratio of the number of dogs to the number of cats is 15 : 7 . we ' re then told that 8 more cats are added to this group and the ratio becomes 15 : 11 . we ' re asked for the number of dogs . algebraically , since the number of dogs is a multiple of 15 and the number of cats is a multiple of 7 , we can write this initial relationship as . . . 15 x / 7 x when we add the 18 cats and factor in the ' ending ratio ' , we have an equation . . . . 15 x / ( 7 x + 8 ) = 15 / 11 here we have 1 variable and 1 equation , so we can solve for x . . . . ( 15 x ) ( 11 ) = ( 7 x + 8 ) ( 15 ) ( x ) ( 11 ) = ( 7 x + 8 ) ( 1 ) 11 x = 7 x + 8 4 x = 8 x = 2 with this x , we can figure out the initial number of dogs and cats . . . initial dogs = 15 x = 15 ( 2 ) = 30 final answer : c" | a ) 15 , b ) 25 , c ) 30 , d ) 45 , e ) 60 | c | multiply(divide(divide(multiply(7, 8), subtract(11, 7)), 7), 15) | multiply(n1,n2)|subtract(n4,n1)|divide(#0,#1)|divide(#2,n1)|multiply(n0,#3)| | other | C |
for any integer k > 1 , the term β length of an integer β refers to the number of positive prime factors , not necessarily distinct , whose product is equal to k . for example , if k = 24 , the length of k is equal to 4 , since 24 = 2 Γ 2 Γ 2 Γ 3 . if x and y are positive integers such that x > 1 , y > 1 , and x + 3 y < 940 , what is the maximum possible sum of the length of x and the length of y ? | "we know that : x > 1 , y > 1 , and x + 3 y < 940 , and it is given that length means no of factors . for any value of x and y , the max no of factors can be obtained only if factor is smallest noall factors are equal . hence , lets start with smallest no 2 . 2 ^ 1 = 2 2 ^ 2 = 4 2 ^ 3 = 8 2 ^ 4 = 16 2 ^ 5 = 32 2 ^ 6 = 64 2 ^ 7 = 128 2 ^ 8 = 256 2 ^ 9 = 512 2 ^ 10 = 1024 ( opps / / it exceeds 1000 , so , x ca n ' t be 2 ^ 10 ) so , max value that x can take is 2 ^ 9 , for which haslength of integeris 9 . now , since x = 512 , x + 3 y < 940 so , 3 y < 428 = = > y < 428 / 3 so , y can take any value which is less than 428 / 3 . and to get the maximum no of factors of smallest integer , we can say y = 2 ^ 7 for 2 ^ 7 has length of integer is 7 . so , combined together : 9 + 7 = 16 . c" | a ) 24 , b ) 22 , c ) 16 , d ) 18 , e ) 20 | c | add(add(4, 3), add(add(4, 4), 1)) | add(n2,n7)|add(n2,n2)|add(n0,#1)|add(#0,#2)| | general | C |
the speed of a subway train is represented by the equation z = s ^ 2 + 2 s for all situations where 0 β€ s β€ 7 , where z is the rate of speed in kilometers per hour and s is the time in seconds from the moment the train starts moving . in kilometers per hour , how much faster is the subway train moving after 6 seconds than it was moving after 3 seconds ? | given : z = s ^ 2 + 2 s for 0 β€ s β€ 7 z ( 3 ) = 3 ^ 2 + 2 * 3 = 15 z ( 6 ) = 6 ^ 2 + 2 * 6 = 48 therefore z ( 7 ) - z ( 3 ) = 48 - 15 = 33 km / hr option e | a ) 4 , b ) 9 , c ) 15 , d ) 48 , e ) 33 | e | subtract(add(power(6, 2), multiply(6, 2)), add(power(3, const_2), multiply(3, 2))) | multiply(n0,n4)|multiply(n0,n5)|power(n4,n0)|power(n5,const_2)|add(#0,#2)|add(#1,#3)|subtract(#4,#5) | physics | E |
what is the area of square field whose side of length 17 m ? | "17 * 17 = 289 sq m answer : d" | a ) 225 , b ) 777 , c ) 266 , d ) 289 , e ) 261 | d | square_area(17) | square_area(n0)| | geometry | D |
the ratio between the sale price and the cost price of an article is 10 : 7 . what is the ratio between the profit and the cost price of that article ? | "let c . p . = rs . 7 x and s . p . = rs . 10 x . then , gain = rs . 3 x required ratio = 3 x : 7 x = 3 : 7 answer : c" | a ) 2 : 9 , b ) 2 : 5 , c ) 3 : 7 , d ) 2 : 0 , e ) 2 : 1 | c | divide(subtract(10, 7), 7) | subtract(n0,n1)|divide(#0,n1)| | other | C |
find the smallest number in a gp whose sum is 38 and product 1728 | "in gp : - a / r + a + ar . . . . = 38 a ( 1 + r + r ^ 2 ) = 38 r . . . . . . eq 1 product a ^ 3 = 1728 a = 12 now put this value in eq 1 12 ( 1 + r + r ^ 2 ) = 38 r r = 2 / 3 and 3 / 2 . . so sallest number is 8 answer : c" | a ) 12 , b ) 20 , c ) 8 , d ) all of these , e ) none of these | c | multiply(divide(divide(divide(divide(38, const_1000), const_3), const_3), const_3), divide(divide(divide(divide(38, const_1000), const_3), const_3), const_3)) | divide(n0,const_1000)|divide(#0,const_3)|divide(#1,const_3)|divide(#2,const_3)|multiply(#3,#3)| | general | C |
calculate 12351 Γ· ? = 69 | "answer let 12351 Γ· ? = 69 then x = 12351 / 69 = 179 . option : a" | a ) 179 , b ) 119 , c ) 129 , d ) 173 , e ) 156 | a | multiply(12351, 69) | multiply(n0,n1)| | general | A |
$ 500 will become $ 1000 in 5 years find the rate of interest ? | si = simple interest = a - p = 1000 - 500 = $ 500 r = 100 si / pt = 100 * 500 / 500 * 5 = 20 % answer is b | a ) 10 % , b ) 20 % , c ) 25 % , d ) 30 % , e ) 50 % | b | divide(subtract(1000, 500), multiply(500, divide(5, const_100))) | divide(n2,const_100)|subtract(n1,n0)|multiply(n0,#0)|divide(#1,#2) | gain | B |
excluding stoppages , the speed of a bus is 50 km / hr and including stoppages , it is 36 km / hr . for how many minutes does the bus stop per hour ? | "due to stoppages , it covers 14 km less . time taken to cover 14 km = 14 / 50 * 60 = 16 min . answer : d" | a ) 118 min , b ) 10 min , c ) 18 min , d ) 16 min , e ) 15 min | d | multiply(const_60, divide(subtract(50, 36), 50)) | subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_60)| | physics | D |
a cylinder is 6 cms in diameter and 6 cms in height . if spheres of the same size are made from the material obtained , what is the diameter of each sphere ? | since the sphere is made out of the cylinder material their volume will be same volume of cylinder = pi * 3 ^ 2 * 6 volume of one sphere = 4 / 3 * pi * r ^ 3 ( where r is radius of sphere ) so if there are n such sphere ' s equating the volumes , pi * r ^ 2 * h = n * ( 4 / 3 ) * pi * r ^ 3 n is the number of spheres , substituting the , r and h 9 * 6 = n * ( 4 / 3 ) * ( d / 2 ) ^ 3 d is the diameter of a sphere , n = ( 9 * 6 * 3 * 8 ) / ( 4 * d ^ 3 ) n = 324 / d ^ 3 now substitute the given answers for d and find the values for n , since a n has to be an integer only value fit for d is 3 , answer : c | ['a ) 5 cms', 'b ) 2 cms', 'c ) 3 cms', 'd ) 4 cms', 'e ) 6 cms'] | c | floor(divide(multiply(multiply(const_3, const_3), 6), multiply(divide(const_4, const_3), multiply(6, const_2)))) | divide(const_4,const_3)|multiply(const_3,const_3)|multiply(n0,const_2)|multiply(n0,#1)|multiply(#0,#2)|divide(#3,#4)|floor(#5) | physics | C |
the value of a machine depreciates at 24 % per annum . if its present value is $ 1 , 50,000 , at what price should it be sold after two years such that a profit of $ 24,000 is made ? | "the value of the machine after two years = 0.76 * 0.76 * 1 , 50,000 = $ 86,640 sp such that a profit of $ 24,000 is made = 86,640 + 24,000 = $ 1 , 10,640 d" | a ) $ 250640 , b ) $ 430640 , c ) $ 120640 , d ) $ 110640 , e ) $ 150640 | d | add(multiply(multiply(subtract(1, divide(24, const_100)), subtract(1, divide(24, const_100))), add(multiply(multiply(const_100, const_100), sqrt(const_100)), multiply(multiply(divide(sqrt(const_100), const_2), const_100), const_100))), multiply(multiply(add(24, const_2), const_100), sqrt(const_100))) | add(n0,const_2)|divide(n0,const_100)|multiply(const_100,const_100)|sqrt(const_100)|divide(#3,const_2)|multiply(#2,#3)|multiply(#0,const_100)|subtract(n1,#1)|multiply(#4,const_100)|multiply(#7,#7)|multiply(#6,#3)|multiply(#8,const_100)|add(#5,#11)|multiply(#12,#9)|add(#13,#10)| | gain | D |
a chemist mixes one liter of pure water with x liters of a 75 % salt solution , and the resulting mixture is a 15 % salt solution . what is the value of x ? | "concentration of salt in pure solution = 0 concentration of salt in salt solution = 75 % concentration of salt in the mixed solution = 15 % the pure solution and the salt solution is mixed in the ratio of - - > ( 75 - 15 ) / ( 15 - 0 ) = 4 / 1 1 / x = 4 / 1 x = 1 / 4 answer : a" | a ) 1 / 4 , b ) 1 / 3 , c ) 1 / 2 , d ) 1 , e ) 3 | a | divide(15, subtract(75, 15)) | subtract(n0,n1)|divide(n1,#0)| | gain | A |
a circular metal plate of even thickness has 12 holes of radius 1 cm drilled into it . as a result the plate lost 1 / 6 th its original weight . the radius of the circular plate is | area of 12 holes = 12 * pi * 1 ^ 2 = 12 * pi as 12 * pi = ( 1 / 6 ) th weight , so total area of plate = 12 * pi / ( 1 / 6 ) = 72 * pi if r = radius of the plate , then its area = pi * r ^ 2 = 72 * pi , so r = sqrt 72 answer : d | ['a ) 16 sqrt 2', 'b ) 8 sqrt 2', 'c ) 32 sqrt 2', 'd ) sqrt 72', 'e ) sqrt 82'] | d | sqrt(divide(multiply(multiply(circle_area(1), 12), 6), const_pi)) | circle_area(n1)|multiply(n0,#0)|multiply(n3,#1)|divide(#2,const_pi)|sqrt(#3) | physics | D |
in a group of 100 people , 60 like volleyball , 50 like hockey . how many like both volleyball and hockey ? | "make a venn diagram , and enter your data . let the number of people who like both volleyball and hockey be x 60 - x + x + 50 - x = 100 x = 10 so number who like both volleyball and hockey = 10 answer c" | a ) 20 , b ) 15 , c ) 10 , d ) 5 , e ) 25 | c | add(add(50, subtract(60, 50)), 50) | subtract(n1,n2)|add(n2,#0)|add(n2,#1)| | other | C |
in a public show 60 % of the seats were filled . if there were 600 seats in the hall , how many seats were vacant ? | "75 % of 600 = 60 / 100 Γ 600 = 360 therefore , the number of vacant seats = 600 - 360 = 240 . answer : e" | a ) 100 , b ) 110 , c ) 120 , d ) 140 , e ) 240 | e | divide(multiply(600, subtract(const_100, 60)), const_100) | subtract(const_100,n0)|multiply(n1,#0)|divide(#1,const_100)| | gain | E |
two trains start from same place at same time at right angles to each other . their speeds are 36 km / hr and 48 km / hr respectively . after 30 seconds the distance between them will be ? | explanation : using pythagarous theorem , distance travelled by first train = 36 x 5 / 18 x 30 = 300 m distance travelled by second train = 48 x 5 / 18 x 30 = 400 m so distance between them = Γ’ Λ Ε‘ ( 90000 + 160000 ) = Γ’ Λ Ε‘ 250000 = 500 mts . answer : d | a ) 270 mts , b ) 190 mts , c ) 100 mts , d ) 500 mts , e ) 110 mts | d | sqrt(add(power(multiply(multiply(36, 30), const_0_2778), const_2), power(multiply(multiply(48, 30), const_0_2778), const_2))) | multiply(n0,n2)|multiply(n1,n2)|multiply(#0,const_0_2778)|multiply(#1,const_0_2778)|power(#2,const_2)|power(#3,const_2)|add(#4,#5)|sqrt(#6) | physics | D |
the average of 10 numbers is calculated as 16 . it is discovered later on that while calculating the average , the number 55 was incorrectly read as 25 , and this incorrect number was used in the calculation . what is the correct average ? | "the total sum of the numbers should be increased by 30 . then the average will increase by 30 / 10 = 3 . the correct average is 19 . the answer is b ." | a ) 17 , b ) 19 , c ) 25 , d ) 31 , e ) 46 | b | divide(add(subtract(multiply(16, 10), 25), 55), 10) | multiply(n0,n1)|subtract(#0,n3)|add(n2,#1)|divide(#2,n0)| | general | B |
the h . c . f . of two numbers is 15 and their l . c . m . is 620 . if one of the numbers is 375 , then the other is : | "other number = ( 15 x 620 ) / 375 = 24.8 . answer : e" | a ) 36.6 , b ) 42.8 , c ) 28.4 , d ) 11.6 , e ) 24.8 | e | multiply(15, 375) | multiply(n0,n2)| | physics | E |
a , b and c have rs . 500 between them , a and c together have rs . 200 and b and c rs . 340 . how much does c have ? | "a + b + c = 500 a + c = 200 b + c = 340 - - - - - - - - - - - - - - a + b + 2 c = 540 a + b + c = 500 - - - - - - - - - - - - - - - - c = 40 answer : b" | a ) 50 , b ) 40 , c ) 267 , d ) 29 , e ) 27 | b | subtract(add(200, 340), 500) | add(n1,n2)|subtract(#0,n0)| | general | B |
maxwell leaves his home and walks toward brad ' s house at the same time that brad leaves his home and runs toward maxwell ' s house . if the distance between their homes is 80 kilometers , maxwell ' s walking speed is 4 km / h , and brad ' s running speed is 6 km / h , what is the distance traveled by brad ? | "time taken = total distance / relative speed total distance = 80 kms relative speed ( opposite side ) ( as they are moving towards each other speed would be added ) = 6 + 4 = 10 kms / hr time taken = 80 / 10 = 8 hrs distance traveled by brad = brad ' s speed * time taken = 6 * 8 = 40 kms . . . answer - c" | a ) 16 , b ) 18 , c ) 40 , d ) 24 , e ) 30 | c | multiply(const_3.0, divide(80, add(4, 6))) | add(const_3.0,n2)|divide(n0,#0)|multiply(n1,#1)| | physics | C |
the number of students in each section of a school is 24 . after admitting new students , three new sections were started . now , the total number of sections is 16 and there are 21 students in each section . the number of new students admitted is : | original number of sections = 16 - 3 = 13 original number of students = 24 x 13 = 312 present number of students = 21 x 16 = 336 number of new students admitted = 336 - 312 = 24 so the answer is option c ) 24 . | ['a ) 12', 'b ) 42', 'c ) 24', 'd ) 28', 'e ) 26'] | c | subtract(multiply(16, 21), multiply(24, subtract(16, const_3))) | multiply(n1,n2)|subtract(n1,const_3)|multiply(n0,#1)|subtract(#0,#2) | physics | C |
a , b , c subscribe rs . 50000 for a business . a subscribes rs . 4000 more than b and b rs . 5000 more than c . out of a total profit of rs . 30000 , b receives : | let c = x . then , b = x + 5000 and a = x + 5000 + 4000 = x + 9000 . so , x + x + 5000 + x + 9000 = 50000 3 x = 36000 x = 12000 a : b : c = 21000 : 17000 : 12000 = 21 : 17 : 12 . b ' s share = rs . ( 30000 x 17 / 50 ) = rs . 10,200 . b | a ) s . 10,000 , b ) s . 10,200 , c ) s . 10,400 , d ) s . 10,700 , e ) s . 10,800 | b | multiply(30000, divide(divide(add(50000, subtract(5000, 4000)), const_3), 50000)) | subtract(n2,n1)|add(n0,#0)|divide(#1,const_3)|divide(#2,n0)|multiply(n3,#3) | general | B |
if k ^ 3 is divisible by 84 , what is the least possible value of integer k ? | "k ^ 3 = 84 * x = 2 ^ 2 * 3 * 7 * x the factors of k must at minimum include 2 * 3 * 7 = 42 . the answer is b ." | a ) 36 , b ) 42 , c ) 48 , d ) 56 , e ) 60 | b | divide(divide(84, const_2), const_2) | divide(n1,const_2)|divide(#0,const_2)| | general | B |
jack and jill are marathon runners . jack can finish a marathon ( 41 km ) in 4.5 hours and jill can run a marathon in 4.1 hours . what is the ratio of their average running speed ? ( jack : jill ) | "average speed of jack = distance / time = 41 / ( 9 / 2 ) = 82 / 9 average speed of jill = 41 / ( 4.1 ) = 10 ratio of average speed of jack to jill = ( 82 / 9 ) / 10 = 82 / 90 = 41 / 45 answer a" | a ) 41 / 45 , b ) 15 / 14 , c ) 4 / 5 , d ) 5 / 4 , e ) can not be determined | a | divide(divide(41, 4.5), divide(41, 4.1)) | divide(n0,n1)|divide(n0,n2)|divide(#0,#1)| | physics | A |
having received his weekly allowance , a student spent 3 / 5 of his allowance at the arcade . the next day he spent one third of his remaining allowance at the toy store , and then spent his last $ 1.00 at the candy store . what is this student β s weekly allowance ? | "let x be the value of the weekly allowance . ( 2 / 3 ) ( 2 / 5 ) x = 100 cents ( 4 / 15 ) x = 100 x = $ 3.75 the answer is c ." | a ) $ 2.75 , b ) $ 3.25 , c ) $ 3.75 , d ) $ 4.25 , e ) $ 4.75 | c | divide(multiply(multiply(3, 5), 1.00), const_4) | multiply(n0,n1)|multiply(n2,#0)|divide(#1,const_4)| | general | C |
if a is an integer greater than 2 but less than 11 and b is an integer greater than 11 but less than 21 , what is the range of a / b ? | "the way to approach this problem is 2 < a < 11 and 11 < b < 21 minimum possible value of a is 3 and maximum is 10 minimum possible value of b is 12 and maximum is 20 range = max a / min b - min a / max b ( highest - lowest ) 10 / 12 - 3 / 20 = 41 / 60 hence e" | a ) 52 / 45 , b ) 23 / 47 , c ) 60 / 41 , d ) 51 / 42 , e ) 41 / 60 | e | subtract(divide(subtract(11, const_1), add(11, const_1)), divide(add(2, const_1), subtract(21, const_1))) | add(n2,const_1)|add(n0,const_1)|subtract(n1,const_1)|subtract(n3,const_1)|divide(#2,#0)|divide(#1,#3)|subtract(#4,#5)| | general | E |
what is the smallest positive integer x such that ( x + 1 ) ^ 2 is divisible by 28 , 98 , 242 , and 308 ? | "28 = 2 * 2 * 7 98 = 2 * 7 * 7 242 = 2 * 11 * 11 308 = 2 * 2 * 7 * 11 so ( x + 1 ) ^ 2 = 2 * 2 * 7 * 7 * 11 * 11 , which means ( x + 1 ) = 2 * 7 * 11 = 154 , which means x = 153 , which is option a" | a ) 153 , b ) 308 , c ) 121 , d ) 96 , e ) 511 | a | divide(multiply(1, const_2), 1) | multiply(n0,const_2)|divide(#0,n0)| | general | A |
if a sum of money doubles itself in 5 years at simple interest , the ratepercent per annum is | "explanation : let sum = x then simple interest = x rate = ( 100 * x ) / ( x * 5 ) = 20 option c" | a ) 12 , b ) 12.5 , c ) 20 , d ) 13.5 , e ) 14 | c | divide(divide(const_2, divide(5, const_100)), const_2) | divide(n0,const_100)|divide(const_2,#0)|divide(#1,const_2)| | gain | C |
one - seventh of the light switches produced by a certain factory are defective . 4 - fifths of the defective switches are rejected and 1 / 15 of the non defective switches are rejected by mistake . if all the switches not rejected are sold , what percent of the switches sold by the factory are defective ? | 1 / 7 of the switches are defective . the defective switches that are not rejected are 1 / 5 * 1 / 7 = 1 / 35 = 3 / 105 of all switches . the non defective switches that are sold are 6 / 7 * 14 / 15 = 84 / 105 of all switches . the percent of switches sold that are defective is 3 / 87 which is about 3.4 % . the answer is b . | a ) 2.3 % , b ) 3.4 % , c ) 4.5 % , d ) 5.6 % , e ) 6.7 % | b | multiply(divide(multiply(divide(1, add(const_3, const_4)), inverse(add(const_2, const_3))), add(multiply(divide(subtract(15, 1), 15), divide(add(const_3, const_3), add(const_3, const_4))), multiply(divide(1, add(const_3, const_4)), inverse(add(const_2, const_3))))), const_100) | add(const_3,const_4)|add(const_2,const_3)|add(const_3,const_3)|subtract(n2,n1)|divide(n1,#0)|divide(#3,n2)|divide(#2,#0)|inverse(#1)|multiply(#4,#7)|multiply(#5,#6)|add(#9,#8)|divide(#8,#10)|multiply(#11,const_100) | general | B |
on dividinng 109 by a number , the quotient is 9 and the remainder is 1 . find the divisor . | d = ( d - r ) / q = ( 109 - 1 ) / 9 = 108 / 9 = 18 a | a ) 12 , b ) 14 , c ) 15 , d ) 16 , e ) 17 | a | floor(divide(109, 9)) | divide(n0,n1)|floor(#0) | general | A |
a , b , c hired a car for rs . 520 and used it for 7,8 and 11 hours respectively . hire charges paid by b were : | sol . a : b : c = 7 : 8 : 11 . hire charges paid by b = rs . [ 520 * 8 / 26 ] = rs . 160 answer a | a ) 160 , b ) 180 , c ) 220 , d ) 225 , e ) none | a | multiply(520, divide(multiply(const_4, const_2), add(add(add(const_4, const_3), multiply(const_4, const_2)), 11))) | add(const_3,const_4)|multiply(const_2,const_4)|add(#0,#1)|add(n2,#2)|divide(#1,#3)|multiply(n0,#4) | physics | A |
p , q and r have rs . 4000 among themselves . r has two - thirds of the total amount with p and q . find the amount with r ? | "let the amount with r be rs . r r = 2 / 3 ( total amount with p and q ) r = 2 / 3 ( 4000 - r ) = > 3 r = 8000 - 2 r = > 5 r = 8000 = > r = 1600 . answer : b" | a ) rs . 3000 , b ) rs . 1600 , c ) rs . 2400 , d ) rs . 4000 , e ) none of these | b | multiply(divide(4000, add(const_1, divide(const_2, const_3))), divide(const_2, const_3)) | divide(const_2,const_3)|add(#0,const_1)|divide(n0,#1)|multiply(#2,#0)| | general | B |
shekar scored 76 , 65 , 82 , 67 and 95 marks in mathematics , science , social studies , english and biology respectively . what are his average marks ? | explanation : average = ( 76 + 65 + 82 + 67 + 95 ) / 5 = 385 / 5 = 77 hence average = 77 answer : b | a ) 65 , b ) 77 , c ) 75 , d ) 85 , e ) 90 | b | divide(add(add(add(add(76, 65), 82), 67), 95), add(const_1, const_4)) | add(n0,n1)|add(const_1,const_4)|add(n2,#0)|add(n3,#2)|add(n4,#3)|divide(#4,#1) | general | B |
what sum of money will produce rs . 75 as simple interest in 4 years at 3 1 / 2 percent ? | "75 = ( p * 4 * 7 / 2 ) / 100 p = 536 answer : e" | a ) 337 , b ) 500 , c ) 266 , d ) 288 , e ) 536 | e | divide(75, divide(multiply(4, add(3, divide(1, 2))), const_100)) | divide(n3,n4)|add(n2,#0)|multiply(n1,#1)|divide(#2,const_100)|divide(n0,#3)| | gain | E |
100 Γ 10 β 100 + 2000 Γ· 100 = ? | "solution given exp . = 100 Γ 10 - 100 + 20 = 1000 - 100 + 20 = 1020 - 100 = 920 . answer c" | a ) 29 , b ) 780 , c ) 920 , d ) 979 , e ) none of these | c | multiply(100, 10) | multiply(n0,n1)| | general | C |
x , y , and z are different prime numbers . the product x ^ 2 * y * z is divisible by how many different positive numbers ? | the exponents of x ^ 2 * y * z are 2 , 1 , and 1 . the number of factors is ( 2 + 1 ) ( 1 + 1 ) ( 1 + 1 ) = 12 the answer is e . | a ) 4 , b ) 6 , c ) 8 , d ) 9 , e ) 12 | e | subtract(power(2, const_4), const_4) | power(n0,const_4)|subtract(#0,const_4) | general | E |
36 welders work at a constant rate they complete an order in 5 days . if after the first day , 12 welders start to work on the other project , how many more days the remaining welders will need to complete the rest of the order ? | "1 . we need to find out the time taken by 24 workers after day 1 . 2 . total no . of wokers * total time taken = time taken by 1 worker 3 . time taken by 1 worker = 36 * 5 = 180 days 4 . but on day 1 thirty - six workers had already worked finishing 1 / 5 of the job . so 24 workers have to finish only 4 / 5 of the job . 5 . total time taken by 24 workers can be got from formula used at ( 2 ) . i . e . , 24 * total time taken = 180 . total time taken by 6 workers to finish the complete job is 180 / 24 = 7.5 days . 6 . time taken by 24 workers to finish 4 / 5 of the job is 4 / 5 * 7.5 = 6 days . the answer is choice e" | a ) 5 , b ) 2 , c ) 8 , d ) 4 , e ) 6 | e | divide(divide(subtract(const_1, multiply(divide(const_1, multiply(36, 5)), 36)), subtract(36, 12)), divide(const_1, multiply(36, 5))) | multiply(n0,n1)|subtract(n0,n2)|divide(const_1,#0)|multiply(n0,#2)|subtract(const_1,#3)|divide(#4,#1)|divide(#5,#2)| | physics | E |
what least value should be replaced by * in 223 * 431 so the number become divisible by 9 | "explanation : trick : number is divisible by 9 , if sum of all digits is divisible by 9 , so ( 2 + 2 + 3 + * + 4 + 3 + 1 ) = 15 + * should be divisible by 9 , 15 + 3 will be divisible by 9 , so that least number is 3 . answer : option a" | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | subtract(431, subtract(431, 431)) | subtract(n1,n1)|subtract(n1,#0)| | general | A |
if a sum of money doubles itself in 8 years at simple interest , the ratepercent per annum is | explanation : let sum = x then simple interest = x rate = ( 100 * x ) / ( x * 8 ) = 12.5 option b | a ) 12 , b ) 12.5 , c ) 13 , d ) 13.5 , e ) 14 | b | divide(divide(const_2, divide(8, const_100)), const_2) | divide(n0,const_100)|divide(const_2,#0)|divide(#1,const_2) | gain | B |
an owner of a pizza stand sold small slices of pizza for rs . 150 each and large slices for rs . 250 each . one night he sold 5000 slices , for a total of rs . 10.50 lakh . how many small slices were sold ? | explanatory answer this question is a word problem in linear equations . the problem is solved by framing two linear equations from the information given in the question and solving them . let ' s ' be the number of small slices and ' b ' the number of large slices sold on that night . therefore , s + b = 5000 . . . eqn ( 1 ) each small slice was sold for rs . 150 . therefore , ' s ' small slices would have fetched rs . 150 s . each large slice was sold for rs . 250 . therefore , ' b ' large slices would have fetched rs . 250 b . total value of sale = 150 s + 250 b = 10 , 50000 ( note the left hand side is in rupees and hence we have to convert right hand also to rupees from lakhs of rupees ) or 150 s + 250 b = 10 , 50000 . . . eqn ( 2 ) multiplying equation ( 1 ) by 150 , we get 150 s + 150 b = 7 , 50000 . . . eqn ( 3 ) subtracting eqn ( 3 ) from eqn ( 2 ) , we get 100 b = 300000 or b = 3000 we know that s + b = 5000 so , s = 5000 - b = 5000 - 3000 = 2000 . 2000 small slices were sold . answer b | a ) 3000 , b ) 2000 , c ) 4000 , d ) 2500 , e ) 3500 | b | subtract(5000, multiply(const_3, const_1000)) | multiply(const_1000,const_3)|subtract(n2,#0) | general | B |
ramu bought an old car for rs . 42000 . he spent rs . 12000 on repairs and sold it for rs . 64900 . what is his profit percent ? | "total cp = rs . 42000 + rs . 12000 = rs . 54000 and sp = rs . 64900 profit ( % ) = ( 64900 - 54000 ) / 54000 * 100 = 20.18 % answer : c" | a ) 12 % , b ) 16 % , c ) 20.18 % , d ) 82 % , e ) 23 % | c | multiply(divide(subtract(64900, add(42000, 12000)), add(42000, 12000)), const_100) | add(n0,n1)|subtract(n2,#0)|divide(#1,#0)|multiply(#2,const_100)| | gain | C |
of the 13 employees in a certain department , 1 has an annual salary of 38000 , 2 have an annual salary of 45700 each , 2 have an annual salary of 42500 each , 3 have an annual salary of 40000 each and 5 have an annual salary of 48500 each . what is the median annual salary for the 13 employees ? | median is just the value in the middle when you arrange all values in the ascending order in this question , the 7 th value would be the median ( since there are 13 employees ) 38 , 40 , 40 , 40 , 42.5 , 42.5 , 45.7 so , answer is d . | a ) 38,000 , b ) 40,000 , c ) 42,500 , d ) 45,700 , e ) 48,500 | d | multiply(42500, const_1) | multiply(n6,const_1) | general | D |
in assembling a bluetooth device , a factory uses one of two kinds of modules . one module costs $ 10 and the other one , that is cheaper , costs $ 6 . the factory holds a $ 80 worth stock of 12 modules . how many of the modules in the stock are of the cheaper kind ? | "so the number of $ 6 modules must be 10 so that the leftover 2 modules are of $ 10 which will give a total value $ 80 . 10 * 6 + 2 * 10 = 60 + 20 = 80 answer : c" | a ) 20 , b ) 25 , c ) 10 , d ) 30 , e ) 35 | c | divide(subtract(80, 10), 6) | subtract(n2,n0)|divide(#0,n1)| | other | C |
if 15 % of 40 is greater than 25 % of a number by 2 , the number is | "explanation : 15 / 100 * 40 - 25 / 100 * x = 2 or x / 4 = 4 so x = 16 option b" | a ) 14 , b ) 16 , c ) 18 , d ) 20 , e ) 22 | b | divide(subtract(multiply(divide(15, const_100), 40), 2), divide(25, const_100)) | divide(n0,const_100)|divide(n2,const_100)|multiply(n1,#0)|subtract(#2,n3)|divide(#3,#1)| | gain | B |
sakshi can do a piece of work in 10 days . tanya is 25 % more efficient than sakshi . the number of days taken by tanya to do the same piece of work : | "solution ratio of times taken by sakshi and tanya = 125 : 100 = 5 : 4 . suppose tanya taken x days to do the work . 5 : 4 : : 10 : x β x = ( 10 x 4 / 5 ) β x = 8 days . hence , tanya takes 8 days is complete the work . answer a" | a ) 8 , b ) 16 , c ) 18 , d ) 25 , e ) 10 | a | divide(10, add(const_1, divide(25, const_100))) | divide(n1,const_100)|add(#0,const_1)|divide(n0,#1)| | physics | A |
how many of the positive divisors of 360 are also multiples of 4 not including 360 ? | "360 = 2 ^ 5 * 3 * 5 = ( 4 ) * 2 * 3 ^ 2 * 5 besides ( 4 ) , the exponents of 2 , 3 , and 5 are 1 , 2 , and 1 . there are ( 1 + 1 ) ( 2 + 1 ) ( 1 + 1 ) = 12 ways to make multiples of 4 . we must subtract 1 because one of these multiples is 360 . the answer is b ." | a ) 8 , b ) 11 , c ) 12 , d ) 15 , e ) 16 | b | divide(divide(divide(360, 4), const_2), const_3) | divide(n0,n1)|divide(#0,const_2)|divide(#1,const_3)| | general | B |
a number when divided by 5 gives a number which is 8 more than the remainder obtained on dividing the same number by 34 . such a least possible number n is | "i solved this question by plugging in numbers from the answer choices . a . ) 74 starting with answer choice a , i immediately eliminated it because 74 is not even divisible by 5 . b . ) 75 i divide 75 / 5 and get 15 as an answer . i divide 75 / 34 and get a remainder of 7 . 15 - 7 = 8 so i know the correct answer isb" | a ) 74 , b ) n = 75 , c ) n = 175 , d ) n = 680 , e ) 690 | b | add(multiply(34, const_2), divide(subtract(multiply(34, const_2), multiply(8, 5)), subtract(5, const_1))) | multiply(n2,const_2)|multiply(n0,n1)|subtract(n0,const_1)|subtract(#0,#1)|divide(#3,#2)|add(#4,#0)| | general | B |
a bag contains 5 red , 4 blue and 3 green balls . if 2 ballsare picked at random , what is the probability that both are red ? | "p ( both are red ) , = 5 c 212 c 2 = 5 c 212 c 2 = 10 / 66 = 5 / 33 c" | a ) 1 / 13 , b ) 2 / 23 , c ) 5 / 33 , d ) 4 / 27 , e ) 3 / 23 | c | divide(choose(5, 2), choose(add(add(5, 4), 3), 2)) | add(n0,n1)|choose(n0,n3)|add(n2,#0)|choose(#2,n3)|divide(#1,#3)| | other | C |
the simple interest on rs . 8 for 12 months at the rate of 5 paise per rupeeper month is | "sol . s . i . = rs . [ 8 * 5 / 100 * 12 ] = rs . 4.8 answer b" | a ) 1.2 , b ) 4.8 , c ) 4.4 , d ) 3.21 , e ) none | b | divide(multiply(multiply(8, 12), 5), const_100) | multiply(n0,n1)|multiply(n2,#0)|divide(#1,const_100)| | gain | B |
a person travels equal distances with speeds of 8 km / hr , 9 km / hr and 10 km / hr and takes a total time of 47 minutes . the total distance is ? | "let the total distance be 3 x km . then , x / 8 + x / 9 + x / 10 = 47 / 60 x / 3 = 47 / 60 = > x = 2.33 . total distance = 3 * 2.33 = 6.99 km . answer : c" | a ) 6 km , b ) 3 km , c ) 7 km , d ) 5 km , e ) 2 km | c | multiply(multiply(divide(divide(47, const_60), add(add(divide(const_1, 8), divide(const_1, 9)), divide(const_1, 10))), const_3), const_1000) | divide(n3,const_60)|divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#1,#2)|add(#4,#3)|divide(#0,#5)|multiply(#6,const_3)|multiply(#7,const_1000)| | physics | C |
a car traveling at a certain constant speed takes 2 seconds longer to travel 1 kilometer than it would take to travel 1 kilometer at 450 kilometers per hour . at what speed , in kilometers per hour , is the car traveling ? | "c 450 * t = 1 km = > t = 1 / 450 km / h v * ( t + 2 / 3600 ) = 1 v ( 1 / 450 + 2 / 3600 ) = 1 = > v = 360 km / h" | a ) 370 , b ) 365 , c ) 360 , d ) 380 , e ) 350 | c | divide(1, divide(add(multiply(const_3600, divide(1, 450)), 2), const_3600)) | divide(n1,n3)|multiply(#0,const_3600)|add(n0,#1)|divide(#2,const_3600)|divide(n1,#3)| | physics | C |
amanda sees a sale for 30 % off all items , she sees a dress on sale that originally cost $ 50 . how much will it cost amanda to buy the dress after the sale amount of 30 % has been take off ? | final number = original number - 30 % ( original number ) = 50 - 30 % ( 50 ) = 50 - 15 = $ 35 . answer b | a ) $ 40 , b ) $ 35 , c ) $ 50 , d ) $ 65 , e ) $ 15 | b | subtract(50, multiply(divide(30, const_100), 50)) | divide(n0,const_100)|multiply(n1,#0)|subtract(n1,#1) | gain | B |
300 + 5 Γ 8 = ? | 300 + 5 Γ 8 = ? or , ? = 300 + 40 = 340 answer b | a ) 820 , b ) 340 , c ) 420 , d ) 209 , e ) none of these | b | add(multiply(5, 8), 300) | multiply(n1,n2)|add(n0,#0) | general | B |
the salary of a , b , c , d , e is rs . 8000 , rs . 5000 , rs . 16000 , rs . 7000 , rs . 9000 per month respectively , then the average salary of a , b , c , d , and e per month is | "answer average salary = 8000 + 5000 + 16000 + 7000 + 9000 / 5 = rs . 9000 correct option : d" | a ) rs . 7000 , b ) rs . 8000 , c ) rs . 8500 , d ) rs . 9000 , e ) none | d | divide(add(add(add(add(8000, 5000), 16000), 7000), 9000), add(const_4, const_1)) | add(n0,n1)|add(const_1,const_4)|add(n2,#0)|add(n3,#2)|add(n4,#3)|divide(#4,#1)| | general | D |
a and b go around a circular track of length 1800 m on a cycle at speeds of 36 kmph and 54 kmph . after how much time will they meet for the first time at the starting point ? | time taken to meet for the first time at the starting point = lcm { length of the track / speed of a , length of the track / speed of b } = lcm { 1800 / ( 36 * 5 / 18 ) , 1800 / ( 54 * 5 / 18 ) } = lcm ( 180 , 120 ) = 360 sec . answer : e | a ) 120 sec , b ) 165 sec , c ) 186 sec , d ) 167 sec , e ) 360 sec | e | divide(1800, subtract(multiply(54, const_0_2778), multiply(36, const_0_2778))) | multiply(n2,const_0_2778)|multiply(n1,const_0_2778)|subtract(#0,#1)|divide(n0,#2) | physics | E |
a marching band of 240 musicians are to march in a rectangular formation with s rows of exactly t musicians each . there can be no less than 8 musicians per row and no more than 30 musicians per row . how many different rectangular formations f are possible ? | "the combinations could be f { ( 1,240 ) , ( 2,120 ) , ( 3,80 ) , ( 4,60 ) , ( 5,48 ) , ( 6,40 ) , ( 8,30 ) , ( 10,24 ) , ( 12,20 ) , ) 15,16 ) , ( 16,15 ) , ( 20,12 ) , ( 24,10 ) , ( 30,8 ) , ( 40,6 ) , ( 48,5 ) , ( 60,4 ) , ( 80,3 ) , ( 120,2 ) , ( 240,1 ) } of these we are told 8 < = t < = 30 so we can remove these pairs , and we are left only with . { ( 8,30 , ( 10,24 ) , ( 12,20 ) , ( 15,16 ) , ( 16,15 ) , ( 20,12 ) , ( 24,10 ) , ( 30,8 ) } hence 8 . . e" | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) f = 8 | e | divide(240, 30) | divide(n0,n2)| | general | E |
the guests at a football banquet consumed a total of 406 pounds of food . if no individual guest consumed more than 2.5 pounds of food , what is the minimum number of guests that could have attended the banquet ? | "to minimize one quantity maximize other . 162 * 2.5 ( max possible amount of food a guest could consume ) = 405 pounds , so there must be more than 162 guests , next integer is 163 . answer : d ." | a ) 160 , b ) 161 , c ) 162 , d ) 163 , e ) 164 | d | add(floor(divide(406, 2.5)), const_1) | divide(n0,n1)|floor(#0)|add(#1,const_1)| | general | D |
for any integer p , * p is equal to the product of all the integers between 1 and p , inclusive . how many prime numbers are there between * 5 + 3 and * 5 + 5 , inclusive ? | generally * p or p ! will be divisible by all numbers from 1 to p . therefore , * 5 would be divisible by all numbers from 1 to 5 . = > * 5 + 3 would give me a number which is a multiple of 3 and therefore divisible ( since * 5 is divisible by 3 ) in fact adding anyprimenumber between 1 to 5 to * 5 will definitely be divisible . so the answer is none ( a ) ! supposing if the question had asked for prime numbers between * 5 + 3 and * 5 + 11 then the answer would be 1 . for * 5 + 3 and * 5 + 13 , it is 2 and so on . . . a | a ) none , b ) one , c ) two , d ) three , e ) four | a | subtract(subtract(add(multiply(multiply(multiply(5, const_3), const_2), const_4), 5), add(multiply(multiply(multiply(5, const_3), const_2), const_4), 3)), 1) | multiply(n1,const_3)|multiply(#0,const_2)|multiply(#1,const_4)|add(n1,#2)|add(n2,#2)|subtract(#3,#4)|subtract(#5,n0) | general | A |
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