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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a can complete a project in 20 days and b can complete the same project in 30 days . if a and b start working on the project together and a quits 5 days before the project is completed , in how many days will the project be completed ? | "let x = the number of days taken to complete the project . the amount of work done by a is ( x - 10 ) * ( 1 / 20 ) . the amount of work done by b is ( x ) * ( 1 / 30 ) . ( 1 / 20 ) * ( x - 5 ) + ( 1 / 30 ) * ( x ) = 1 ( x / 20 ) + ( x / 30 ) - ( 5 / 20 ) = 1 5 x / 60 = 5 / 4 x = 60 / 4 x = 15 therefore , the answer is... | a ) 18 days , b ) 27 days , c ) 26.67 days , d ) 15 days , e ) 12 days | d | add(divide(subtract(const_1, multiply(divide(const_1, 30), 5)), add(divide(const_1, 20), divide(const_1, 30))), 5) | divide(const_1,n1)|divide(const_1,n0)|add(#1,#0)|multiply(n2,#0)|subtract(const_1,#3)|divide(#4,#2)|add(n2,#5)| | physics |
sides of a rectangular park are in the ratio 3 : 2 and its area is 3750 sq m , the cost of fencing it at 90 ps per meter is ? | "3 x * 2 x = 3750 = > x = 25 2 ( 75 + 50 ) = 250 m 250 * 0.9 = rs . 225 answer : c" | a ) 287 , b ) 369 , c ) 225 , d ) 279 , e ) 361 | c | divide(multiply(90, rectangle_perimeter(sqrt(divide(multiply(3750, 2), 3)), divide(3750, sqrt(divide(multiply(3750, 2), 3))))), const_100) | multiply(n1,n2)|divide(#0,n0)|sqrt(#1)|divide(n2,#2)|rectangle_perimeter(#3,#2)|multiply(n3,#4)|divide(#5,const_100)| | physics |
if 30 % of a number is equal to one - third of another number , what is the ratio of first number to the second number ? | "explanation : let 30 % of a = 1 / 3 b then , 30 a / 100 = 1 b / 3 = > 3 a / 10 = b / 3 = > a : b = 10 : 9 answer : c" | a ) 2 : 5 , b ) 3 : 7 , c ) 10 : 9 , d ) 7 : 3 , e ) none of these | c | divide(divide(const_1, const_4), divide(30, const_100)) | divide(const_1,const_4)|divide(n0,const_100)|divide(#0,#1)| | general |
every year an amount increases by 1 / 8 th of itself . how much will it be after two years if its present value is rs . 32000 ? | "32000 * 9 / 8 * 9 / 8 = 40500 answer : e" | a ) 81000 , b ) 33888 , c ) 77678 , d ) 200988 , e ) 40500 | e | add(add(32000, multiply(divide(1, 8), 32000)), multiply(divide(1, 8), add(32000, multiply(divide(1, 8), 32000)))) | divide(n0,n1)|multiply(n2,#0)|add(n2,#1)|multiply(#2,#0)|add(#2,#3)| | general |
out of 450 students of a school , 325 play football , 175 play cricket and 50 neither play football nor cricket . how many students play both football and cricket ? | "explanation : students who play cricket , n ( a ) = 325 students who play football , n ( b ) = 175 total students who play either or both games , = n ( a βͺ b ) = 450 β 50 = 400 required number , n ( a β© b ) = n ( a ) + n ( b ) β n ( a βͺ b ) = 325 + 175 β 400 = 100 option b" | a ) 75 , b ) 100 , c ) 125 , d ) 150 , e ) none of these | b | subtract(add(175, 325), subtract(450, 50)) | add(n1,n2)|subtract(n0,n3)|subtract(#0,#1)| | other |
a number is doubled and 7 is added . if resultant is trebled , it becomes 99 . what is that number | "explanation : = > 3 ( 2 x + 7 ) = 99 = > 2 x + 7 = 33 = > x = 13 option d" | a ) 8 , b ) 10 , c ) 14 , d ) 13 , e ) 16 | d | divide(subtract(99, multiply(const_3, 7)), multiply(const_3, const_2)) | multiply(n0,const_3)|multiply(const_2,const_3)|subtract(n1,#0)|divide(#2,#1)| | general |
a bag holds 3 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 green ? | "given : 3 r and 3 g marbles required : probability that 2 marbles removed without replacement are both red initially we have to pick one red from a total of 3 red and 3 green marbles after one green has been picked , we need to pick 1 green from a total of 2 greenand 3 red marbles . p ( both green ) = ( 3 / 6 ) * ( 2 ... | a ) 1 / 10 , b ) 1 / 6 , c ) 3 / 10 , d ) 1 / 5 , e ) 1 / 2 | d | divide(choose(3, 3), choose(add(const_2.0, 3), const_2)) | add(n0,n1)|choose(const_2.0,n0)|choose(#0,const_2)|divide(#1,#2)| | other |
in a 200 m race , a covers the distance in 45 seconds and b in 60 second . in this race a beats b by : | "solution distance covered by b in 15 sec . = ( 200 / 60 x 15 ) m = 50 m . β΄ a beats b by 50 metres . answer b" | a ) 20 m , b ) 50 m , c ) 45 m , d ) 60 m , e ) none of these | b | multiply(divide(200, 60), subtract(60, 45)) | divide(n0,n2)|subtract(n2,n1)|multiply(#0,#1)| | physics |
find the value of a / b + b / a , if a and b are the roots of the quadratic equation x 2 + 8 x + 4 = 0 ? | "a / b + b / a = ( a 2 ] / ab a + b = + b 2 ) / ab = ( a 2 + b 2 + a + b ) / ab = [ ( a + b ) 2 - 2 ab - 8 / 1 = - 8 ab = 4 / 1 = 4 hence a / b + b / a = [ ( - 8 ) 2 - 2 ( 4 ) ] / 4 = 56 / 4 = 14 . answer : b" | a ) 15 , b ) 14 , c ) 13 , d ) 12 , e ) 11 | b | subtract(divide(power(negate(8), 2), 4), 2) | negate(n1)|power(#0,n0)|divide(#1,n2)|subtract(#2,n0)| | general |
20 ^ 10 / 120 ^ 5 = ? | "20 ^ 10 / 120 ^ 5 = ? a . 6 ^ 5 b . 5 ^ 6 c . ( 10 / 3 ) ^ 5 d . 6 ^ 3 e . 15 ^ 3 - > 20 ^ 10 / 120 ^ 5 = ( 20 ^ 10 ) / ( 6 ^ 5 ) ( 20 ^ 5 ) = ( 20 ^ 5 ) / ( 6 ^ 5 ) = ( 2 ^ 5 ) ( 10 ^ 5 ) / ( 2 ^ 5 ) ( 3 ^ 5 ) = ( 10 / 3 ) ^ 5 . thus , c is the answer ." | a ) 6 ^ 5 , b ) 5 ^ 6 , c ) ( 10 / 3 ) ^ 5 , d ) 6 ^ 3 , e ) 15 ^ 3 | c | divide(power(20, 10), power(20, 5)) | power(n0,n1)|power(n0,n3)|divide(#0,#1)| | general |
darren bought 5 packs of white t - shirts and 3 packs of blue t - shirts for his basketball team . the white t - shirts come in packs of 6 , and the blue t - shirts come in packs of 9 . how many t - shirts did darren buy in all ? | step 1 : find the number of white t - shirts . 5 Γ 6 = 30 step 2 : find the number of blue t - shirts . 3 Γ 9 = 27 step 3 : find the total number of t - shirts . 30 + 27 = 57 darren bought 57 t - shirts . answer is b . | a ) 77 , b ) 57 , c ) 34 , d ) 65 , e ) 21 | b | add(multiply(5, 6), multiply(3, 9)) | multiply(n0,n2)|multiply(n1,n3)|add(#0,#1) | general |
car a is 10 miles behind car b , which is traveling in the same direction along the same route as car a . car a is traveling at a constant speed of 58 miles per hour and car bis traveling at a constant speed of 50 miles per hour . how many hours will it take for car a to overtake and drive 8 miles ahead of car b ? | "relative speed of car a is 58 - 50 = 8 miles per hour , to catch up 10 miles and drive 8 miles ahead so to drive 18 miles it ' ll need 18 / 8 = 2.25 hours . answer : b" | a ) 1.5 , b ) 2.25 , c ) 2.75 , d ) 2.5 , e ) 3.5 | b | divide(add(10, 8), subtract(58, 50)) | add(n0,n3)|subtract(n1,n2)|divide(#0,#1)| | physics |
in a 400 m race , a covers the distance in 60 seconds and b in 100 second . in this race a beats b by : | "solution distance covered by b in 40 sec . = ( 400 / 100 x 40 ) m = 160 m . β΄ a beats b by 160 metres . answer d" | a ) 20 m , b ) 125 m , c ) 120 m , d ) 160 m , e ) none of these | d | multiply(divide(400, 100), subtract(100, 60)) | divide(n0,n2)|subtract(n2,n1)|multiply(#0,#1)| | physics |
a business executive and his client are charging their dinner tab on the executive ' s expense account . the company will only allow them to spend a total of 50 $ for the meal . assuming that they will pay 7 % in sales tax for the meal and leave a 20 % tip , what is the most their food can cost ? | let x is the cost of the food 1.07 x is the gross bill after including sales tax 1.20 * 1.07 x = 50 x = 37.88 hence , the correct option is c | a ) 39.55 $ , b ) 40.63 $ , c ) 38.63 $ , d ) 37.88 $ , e ) 35.15 $ | c | divide(50, add(divide(add(7, 20), const_100), const_1)) | add(n1,n2)|divide(#0,const_100)|add(#1,const_1)|divide(n0,#2) | general |
the h . c . f . of two numbers is 11 and their l . c . m . is 693 . if one of the number is 77 , find the other ? | "other number = 11 * 693 / 77 = 99 answer is b" | a ) 88 , b ) 99 , c ) 76 , d ) 55 , e ) 66 | b | multiply(11, 77) | multiply(n0,n2)| | physics |
p , q and r can do a work in 6 , 9 and 12 days respectively . they completed the work and got rs . 195 . what is the share of p ? | "the ratio of their working rates = 1 / 6 : 1 / 9 : 1 / 12 = 6 : 4 : 3 . since , they work together , the share of p = 6 / 13 * 195 = rs . 90 answer : d" | a ) 80 , b ) 70 , c ) 85 , d ) 90 , e ) 95 | d | add(multiply(const_100, const_100), divide(multiply(195, 9), add(add(divide(multiply(9, 12), 6), 9), 12))) | multiply(n1,n3)|multiply(n1,n2)|multiply(const_100,const_100)|divide(#1,n0)|add(n1,#3)|add(n2,#4)|divide(#0,#5)|add(#6,#2)| | physics |
machine a and machine b are each used to manufacture 220 sprockets . it takes machine a 10 hours longer to produce 220 sprockets than machine b . machine b produces 10 percent more sprockets per hour than machine a . how many sprockets per hour does machine a produces ? | "machine b : takes x hours to produce 220 sprockets machine a : takes ( x + 10 ) hours to produce 220 sprockets machine b : in 1 hour , b makes 220 / x sprockets machine a : in 1 hour , a makes 220 / ( x + 10 ) sprockets equating : 1.1 ( 220 / ( x + 10 ) ) = 220 / x 242 / ( x + 10 ) = 220 / x 242 x = 220 x + 2200 22 x ... | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | a | divide(220, divide(multiply(multiply(10, 220), divide(add(const_100, 10), const_100)), subtract(multiply(220, divide(add(const_100, 10), const_100)), 220))) | add(n1,const_100)|multiply(n0,n1)|divide(#0,const_100)|multiply(#2,#1)|multiply(n0,#2)|subtract(#4,n0)|divide(#3,#5)|divide(n0,#6)| | gain |
a shop owner professes to sell his articles at certain cost price but he uses false weights with which he cheats by 30 % while buying and by 10 % while selling . what is his percentage profit ? | "the owner buys 100 kg but actually gets 130 kg ; the owner sells 100 kg but actually gives 90 kg ; profit : ( 130 - 90 ) / 90 * 100 = 30 % answer : d ." | a ) 10.22 % , b ) 20.22 % , c ) 21.22 % , d ) 30 % , e ) ca n ' t be calculated | d | divide(multiply(subtract(add(const_100, 30), subtract(const_100, 10)), const_100), subtract(const_100, 10)) | add(n0,const_100)|subtract(const_100,n1)|subtract(#0,#1)|multiply(#2,const_100)|divide(#3,#1)| | gain |
a cricket bat is sold for $ 900 , making a profit of $ 300 . the profit percentage would be | "300 / ( 900 - 300 ) = 50 % . answer : d" | a ) 24 % , b ) 25 % , c ) 30 % , d ) 50 % , e ) 40 % | d | multiply(divide(300, subtract(900, 300)), const_100) | subtract(n0,n1)|divide(n1,#0)|multiply(#1,const_100)| | gain |
each of the integers from 1 to 15 is written on the a seperate index card and placed in a box . if the cards are drawn from the box at random without replecement , how many cards must be drawn to ensure that the product of all the integers drawn is even ? | "out of the 15 integers : 8 are odd and 7 are even . if we need to make sure that the product of all the integers withdrawn is even then we need to make sure that we have at least one even number . in the worst case : 1 . we will end up picking odd numbers one by one , so we will pick all 8 odd numbers first 2 . 9 th n... | a ) 9 , b ) 12 , c ) 11 , d ) 10 , e ) 3 | a | add(divide(15, const_2), 1) | divide(n1,const_2)|add(n0,#0)| | general |
a student chose a number , multiplied it by 7 , then subtracted 150 from the result and got 130 . what was the number he chose ? | "let x be the number he chose , then 7 β
x β 150 = 130 7 x = 280 x = 40 correct answer a" | a ) 40 , b ) 42 , c ) 44 , d ) 46 , e ) 48 | a | divide(add(130, 150), 7) | add(n1,n2)|divide(#0,n0)| | general |
average monthly income of a family of 4 earning members was rs . 735 . one of the earning members died and therefore , the average income came down to rs 550 . the income of the deceased was ? | "answer income of the deceased = total income of 4 members - total income of remaining 3 members . = 735 x 4 - 550 x 3 rs . = 1290 rs . correct option : e" | a ) rs . 692.80 , b ) rs . 820 , c ) rs . 990 , d ) rs . 1385 , e ) none | e | subtract(multiply(735, 4), multiply(550, subtract(4, const_1))) | multiply(n0,n1)|subtract(n0,const_1)|multiply(n2,#1)|subtract(#0,#2)| | general |
evaluate : ( 1 - 1 / 10 ) ( 1 - 1 / 11 ) ( 1 - 1 / 12 ) . . . ( 1 - 1 / 99 ) ( 1 - 1 / 100 ) | ( 1 - 1 / 10 ) ( 1 - 1 / 11 ) ( 1 - 1 / 12 ) . . . ( 1 - 1 / 99 ) ( 1 - 1 / 100 ) = ( 9 / 10 ) ( 10 / 11 ) ( 11 / 12 ) . . . ( 98 / 99 ) ( 99 / 100 ) = 9 / 100 : simplify correct answer e | a ) 5 / 100 , b ) 6 / 100 , c ) 7 / 100 , d ) 8 / 100 , e ) 9 / 100 | e | divide(subtract(10, 1), 100) | subtract(n2,n0)|divide(#0,n14) | general |
when a person aged 39 is added to a group of n people , the average age increases by 2 . when a person aged 15 is added instead , the average age decreases by 1 . what is the value of w ? | "a simple and elegant solution . as addition of 39 , shifts mean by 2 , and addition of 15 , shifts mean by 1 to the other side , we have the mean lying between 3915 , and in a ratio of 2 : 1 39 - 15 = 24 24 divide by 3 is 8 . meaning mean of the n terms is 15 + 8 = 39 - 16 = 23 now , from first statement , when a pers... | a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | a | subtract(divide(subtract(39, 15), add(2, 1)), 1) | add(n1,n3)|subtract(n0,n2)|divide(#1,#0)|subtract(#2,n3)| | general |
210 college students were asked in a survey if they preferred windows or mac brand computers . 90 students claimed that they preferred mac to windows brand computers . one third as many of the students who preferred mac to windows , equally preferred both brands . 60 of the students had no preference . how many of the ... | "210 = 90 ( mac ) + x ( window ) + 60 ( both ) = > x = 60 answer : d" | a ) 25 , b ) 40 , c ) 50 , d ) 60 , e ) 75 | d | subtract(subtract(subtract(210, 60), 90), divide(subtract(subtract(210, 60), 90), const_3)) | subtract(n0,n2)|subtract(#0,n1)|divide(#1,const_3)|subtract(#1,#2)| | other |
210 school students were asked in a survey if they preferred windows or mac brand computers . 60 students claimed that they preferred mac to windows brand computers . one third as many of the students who preferred mac to windows , equally preferred both brands . 90 of the students had no preference . how many of the s... | we are told that 60 students claimed that they preferred mac to windows , which means that 60 preferred mac but not windows , so # of students who preferred mac ( p ( a ) as you wrote ) , does not equal to 60 , it equals to 60 + 20 ( 20 is # of students who equally preferred both brands ) . also we are asked to find # ... | a ) 25 , b ) 40 , c ) 50 , d ) 60 , e ) 75 | b | subtract(210, add(add(add(90, 60), const_10), const_10)) | add(n1,n2)|add(#0,const_10)|add(#1,const_10)|subtract(n0,#2) | other |
speed of a boat in standing water is 11 kmph and speed of the stream is 2.5 kmph . a man can rows to a place at a distance of 105 km and comes back to the starting point . the total time taken by him is ? | "speed upstream = 8.5 kmph speed downstream = 13.5 kmph total time taken = 105 / 8.5 + 105 / 13.5 = 20.12 hours answer is b" | a ) 21.12 hours , b ) 20.12 hours , c ) 19.12 hours , d ) 18.12 hours , e ) 17.12 hours | b | add(multiply(add(add(11, 2.5), subtract(11, 2.5)), 105), multiply(subtract(add(divide(105, add(11, 2.5)), divide(105, subtract(11, 2.5))), add(add(11, 2.5), subtract(11, 2.5))), const_60)) | add(n0,n1)|subtract(n0,n1)|add(#0,#1)|divide(n2,#0)|divide(n2,#1)|add(#3,#4)|multiply(n2,#2)|subtract(#5,#2)|multiply(#7,const_60)|add(#6,#8)| | physics |
a train 650 m long is running at a speed of 81 km / hr . in what time will it pass a bridge 340 m long ? | "speed = 81 * 5 / 18 = 45 / 2 m / sec total distance covered = 650 + 340 = 990 m required time = 990 * 2 / 45 = 44 sec answer : c" | a ) 42 , b ) 43 , c ) 44 , d ) 45 , e ) 46 | c | divide(650, multiply(subtract(81, 340), const_0_2778)) | subtract(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
what is the probability that carol and bernie will get selected for a job they both applied for in a company , given that their chances of getting selected is 4 / 5 and 3 / 5 respectively ? | explanation : p ( carol ) = 4 / 5 p ( bernie ) = 3 / 5 e = { carol and bernie both get selected } p ( e ) = p ( carol ) * p ( bernie ) = 4 / 5 * 3 / 5 = 12 / 25 answer : d | a ) 8 / 25 , b ) 11 / 24 , c ) 12 / 25 , d ) 12 / 25 , e ) 12 / 23 | d | multiply(divide(4, 5), divide(3, 5)) | divide(n0,n1)|divide(n2,n1)|multiply(#0,#1) | probability |
the weight of every type a widget is the same , the weight of every type b widget is the same , and the weight of every type c widget is the same . if the weight of 7 type a widgets is equal to the weight of 2 type b widgets , and the weight of 5 type b widgets is equal to the weight of 7 type c widgets . what is the r... | "5 b = 7 c and so b = 7 c / 5 7 a = 2 b and so a = 2 b / 7 = 2 c / 5 a + b = 2 c / 5 + 7 c / 5 = 9 c / 5 b + c = 7 c / 5 + c = 12 c / 5 the ratio of a + b : b + c = 9 : 12 = 3 : 4 the answer is c ." | a ) 1 : 2 , b ) 2 : 3 , c ) 3 : 4 , d ) 4 : 5 , e ) 5 : 6 | c | divide(add(1, divide(7, 2)), add(divide(7, 2), divide(5, 2))) | divide(n0,n1)|divide(n2,n1)|add(n4,#0)|add(#0,#1)|divide(#2,#3)| | general |
a certain bag contains 100 balls Γ’ β¬ β 10 white , 30 green , 10 yellow , 47 red , and 3 purple . if a ball is to be chosen at random , what is the probability that the ball will be neither red nor purple ? | "according to the stem the ball can be white , green or yellow , so the probability is ( white + green + yellow ) / ( total ) = ( 10 + 30 + 10 ) / 100 = 50 / 100 = 0.5 . answer is e" | a ) 0.9 , b ) 0.75 , c ) 0.6 , d ) 0.8 , e ) 0.5 | e | divide(subtract(100, add(47, 3)), 100) | add(n4,n5)|subtract(n0,#0)|divide(#1,n0)| | other |
how many bricks , each measuring 25 cm x 11 cm x 6 cm , will be needed to build a wall of 8 m x 1 m x 5 cm ? | "number of bricks = volume of the wall / volume of 1 brick = ( 800 x 100 x 5 ) / ( 25 x 11 x 6 ) = 242 . answer : option c" | a ) 5600 , b ) 6000 , c ) 242 , d ) 7200 , e ) 8600 | c | divide(multiply(multiply(multiply(8, const_100), multiply(1, const_100)), 5), multiply(multiply(25, 11), 6)) | multiply(n3,const_100)|multiply(n4,const_100)|multiply(n0,n1)|multiply(#0,#1)|multiply(n2,#2)|multiply(n5,#3)|divide(#5,#4)| | physics |
the breadth of a rectangular field is 60 % of its length . if the perimeter of the field is 800 m , find out the area of the field . | "given that breadth of the rectangular field is 60 % of its length . β b = 60 l 100 = 3 l 5 perimeter of the field = 800 m β 2 ( l + b ) = 800 β 2 ( l + 3 l 5 ) = 800 β l + 3 l 5 = 400 β 8 l 5 = 400 β l 5 = 50 β l = 5 Γ 50 = 250 m b = 3 l 5 = 3 Γ 2505 = 3 Γ 50 = 150 m area = lb = 250 Γ 150 = 37500 m 2 answer is a ." | a ) 37500 , b ) 38000 , c ) 38500 , d ) 39000 , e ) 39500 | a | multiply(divide(divide(800, const_2), add(divide(60, const_100), const_1)), subtract(divide(800, const_2), divide(divide(800, const_2), add(divide(60, const_100), const_1)))) | divide(n1,const_2)|divide(n0,const_100)|add(#1,const_1)|divide(#0,#2)|subtract(#0,#3)|multiply(#3,#4)| | geometry |
if the number is divided by 3 , it reduced by 30 . the number is | "explanation : let the number be x . then , x - ( x / 3 ) = 30 = > 2 x / 3 = 30 = > x = 45 answer : option c" | a ) a ) 51 , b ) b ) 50 , c ) c ) 45 , d ) d ) 40 , e ) e ) 36 | c | divide(multiply(30, 3), subtract(3, const_1)) | multiply(n0,n1)|subtract(n0,const_1)|divide(#0,#1)| | general |
among 400 students , 51 % study sociology , 46 % study mathematics and 40 % study biology . if 34 % of students study both mathematics and sociology , what is the largest possible number of students who study biology but do not study either mathematics or sociology ? | i would just like to add a bit of explanation after the step where you calculate that the number of students studying both m and s = 136 using your analysis : we see that the total number of students who study either maths or sociology = 184 + 204 - 136 = 252 so , in the image we know that the number of students in the... | a ) 30 , b ) 90 , c ) 148 , d ) 172 , e ) 188 | c | subtract(400, subtract(add(multiply(divide(400, const_100), 51), multiply(divide(400, const_100), 46)), multiply(divide(400, const_100), 34))) | divide(n0,const_100)|multiply(n1,#0)|multiply(n2,#0)|multiply(n4,#0)|add(#1,#2)|subtract(#4,#3)|subtract(n0,#5) | other |
in the third grade of windblown school there are 102 students , one third of them failed the math test and 1 / 6 failed that literature test . at least how many students failed both tests ? | "total = 102 failed in math = 102 / 3 = 34 failed in literature = 108 / 6 = 17 the least failed in both can be 0 while max can be 17 answer a" | a ) 0 . , b ) 6 . , c ) 8 . , d ) 10 . , e ) 12 . | a | min(divide(102, const_3), divide(102, 6)) | divide(n0,const_3)|divide(n0,n2)|min(#0,#1)| | other |
the principal that amounts to rs . 5913 in 3 years at 6 1 / 4 % per annum c . i . compounded annually , is ? | "principal = [ 4913 / ( 1 + 25 / ( 4 * 100 ) ) 3 ] = 5913 * 16 / 17 * 16 / 17 * 16 / 17 = rs . 5096 . answer : e" | a ) s . 3096 , b ) s . 4076 , c ) s . 4085 , d ) s . 4096 , e ) s . 5096 | e | divide(5913, power(add(1, divide(add(6, divide(1, 4)), const_100)), 3)) | divide(n3,n4)|add(n2,#0)|divide(#1,const_100)|add(#2,n3)|power(#3,n1)|divide(n0,#4)| | gain |
if 2805 / 2.55 = 1100 , then 280.5 / 25.5 is equal to ? | "answer given expression 280.5 / 25.5 = 2805 / 255 = 2805 / ( 2.55 x 100 ) = 1100 / 100 = 11 correct option : d" | a ) 1.01 , b ) 1.1 , c ) 0.11 , d ) 11 , e ) none | d | divide(multiply(1100, 2.55), 2805) | multiply(n1,n2)|divide(#0,n0)| | general |
a and b complete a work in 50 days . a alone can do it in 40 days . if both together can do the work in how many days ? | "1 / 50 + 1 / 40 = 0.045 days answer : d" | a ) 1.075 days , b ) 0.185 days , c ) 0.065 days , d ) 0.045 days , e ) 0.0775 days | d | inverse(add(inverse(50), inverse(40))) | inverse(n0)|inverse(n1)|add(#0,#1)|inverse(#2)| | physics |
a sum of rs . 2769 is lent into two parts so that the interest on the first part for 8 years at 3 % per annum may be equal to the interest on the second part for 3 years at 5 % per annum . find the second sum ? | "( x * 8 * 3 ) / 100 = ( ( 2769 - x ) * 3 * 5 ) / 100 24 x / 100 = 41535 / 100 - 15 x / 100 39 x = 41535 = > x = 1065 second sum = 2769 β 1065 = 1704 answer : e" | a ) 1642 , b ) 1640 , c ) 1632 , d ) 2789 , e ) 1704 | e | subtract(2769, divide(multiply(multiply(3, 5), 2769), add(multiply(3, 5), multiply(8, 3)))) | multiply(n2,n4)|multiply(n1,n2)|add(#0,#1)|multiply(n0,#0)|divide(#3,#2)|subtract(n0,#4)| | gain |
( 0.0077 ) ( 3.6 ) / ( 0.04 ) ( 0.1 ) ( 0.007 ) = | "( 0.0077 ) ( 3.6 ) / ( 0.04 ) ( 0.1 ) ( 0.007 ) = 0.0077 * 360 / 4 * ( 0.1 ) ( 0.007 ) = 0.077 * 90 / 1 * 0.007 = 77 * 90 / 7 = 11 * 90 = 990 answer : d" | a ) 0.099 , b ) 0.0099 , c ) 9.9 , d ) 990 , e ) 99 | d | divide(multiply(0.0077, 3.6), multiply(multiply(0.04, 0.1), 0.007)) | multiply(n0,n1)|multiply(n2,n3)|multiply(n4,#1)|divide(#0,#2)| | general |
in a certain candy store , 22 % of the customers are caught sampling the candy and are charged a small fine , but 12 % of the customers who sample the candy are not caught . what is the total percent of all customers who sample candy ? | "since 12 % of the customers who sample the candy are not caught , then 88 % of the customers who sample the candy are caught : { % of customers who sample candy } * 0.88 = 0.22 ; { % of customers who sample candy } = 0.25 . answer : d ." | a ) 22 % , b ) 23 % , c ) 24 % , d ) 25 % , e ) 34 % | d | divide(22, divide(subtract(const_100, 12), const_100)) | subtract(const_100,n1)|divide(#0,const_100)|divide(n0,#1)| | gain |
a lent rs . 5000 to b for 2 years and rs . 3000 to c for 4 years on simple interest at the same rate of interest and received rs . 1760 in all from both of them as interest . the rate of interest per annum is ? | "let the rate be r % p . a . then , ( 5000 * r * 2 ) / 100 + ( 3000 * r * 4 ) / 100 = 1760 100 r + 120 r = 1760 r = 8 % answer : e" | a ) 16 % , b ) 12 % , c ) 74 % , d ) 10 % , e ) 8 % | e | multiply(divide(1760, add(multiply(5000, 2), multiply(3000, 4))), const_100) | multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)|divide(n4,#2)|multiply(#3,const_100)| | gain |
out of 4 numbers , the average of first 3 is 20 and that of the last 3 is 15 . if the last number is 18 , the first number is : | explanation : let the numbers be a , b , c , d given , a + b + c = 60 , b + c + d = 45 now , d = 18 thus , b + c + 18 = 45 β b + c = 27 putting the value of b + c in a + b + c = 60 a + 27 = 60 β a = 33 answer : b | a ) 22 , b ) 33 , c ) 77 , d ) 99 , e ) 27 | b | subtract(multiply(20, 3), subtract(multiply(15, 3), 18)) | multiply(n1,n2)|multiply(n1,n4)|subtract(#1,n5)|subtract(#0,#2) | general |
4 car rental agencies a , b , c and d rented a plot for parking their cars during the night . a parked 15 cars for 12 days , b parked 12 cars for 20 days , c parked 18 cars for 18 days and d parked 16 cars for 15 days . if a paid rs . 1125 as rent for parking his cars , what is the total rent paid by all the 4 agencies... | the ratio in which the four agencies will be paying the rents = 15 * 12 : 12 * 20 : 18 * 18 : 16 * 15 = 180 : 240 : 324 : 240 = 45 : 60 : 81 : 60 let us consider the four amounts to be 45 k , 60 k , 81 k and 60 k respectively . the total rent paid by the four agencies = 45 k + 60 k + 81 k + 60 k = 246 k it is given tha... | a ) rs . 6152 , b ) rs . 6159 , c ) rs . 6156 , d ) rs . 6150 , e ) rs . 6196 | d | multiply(subtract(add(add(add(divide(multiply(divide(16, 15), 1125), divide(15, 12)), divide(multiply(divide(18, 18), 1125), divide(15, 12))), divide(multiply(divide(12, 20), 1125), divide(15, 12))), 1125), multiply(4, const_100)), const_2) | divide(n7,n1)|divide(n1,n2)|divide(n5,n5)|divide(n2,n4)|multiply(n0,const_100)|multiply(n9,#0)|multiply(n9,#2)|multiply(n9,#3)|divide(#5,#1)|divide(#6,#1)|divide(#7,#1)|add(#8,#9)|add(#11,#10)|add(n9,#12)|subtract(#13,#4)|multiply(#14,const_2) | general |
harry started a 7 - mile hike with a full 9 - cup canteen of water and finished the hike in 2 hours with 3 cup of water remaining in the canteen . if the canteen leaked at the rate of 1 cup per hour and harry drank 2 cups of water during the last mile , how many cups did he drink per mile during the first 6 miles of th... | "no of cups leaked during the trip = 2 cups . no of cups harry drank = 6 cups . no of cups harry drank during the first 6 miles = 4 . drink / mile = 4 / 6 answer d" | a ) 1 / 6 , b ) 2 / 6 , c ) 3 / 6 , d ) 4 / 6 , e ) 5 / 6 | d | divide(subtract(subtract(subtract(9, 2), 2), 1), 6) | subtract(n1,n2)|subtract(#0,n5)|subtract(#1,n4)|divide(#2,n6)| | physics |
the banker ' s gain on a sum due 6 years hence at 12 % per annum is rs . 900 . what is the banker ' s discount ? | "explanation : td = ( bg Γ 100 ) / tr = ( 900 Γ 100 ) / ( 6 Γ 12 ) = rs . 1250 bg = bd β td = > 900 = bd - 1250 = > bd = 2150 answer : option d" | a ) 1240 , b ) 1120 , c ) 1190 , d ) 2150 , e ) none of these | d | add(900, divide(multiply(900, const_100), multiply(12, 6))) | multiply(n2,const_100)|multiply(n0,n1)|divide(#0,#1)|add(n2,#2)| | gain |
8 ^ 100 is divisible by 17 . find the remainder for this ? | "this is an extremely difficult problem to solve with out fermat ' s little theorem . by applying fermat ' s little theorem , we know that 816 when divided by 17 , the remainder is 1 . so divide 100 by 16 and find the remainder . remainder = 4 therefore , 100 = ( 16 Γ 6 ) + 4 now this problem can be written as 810017 =... | a ) 10 , b ) 12 , c ) 16 , d ) 17 , e ) 19 | c | subtract(17, 8) | subtract(n2,n0)| | general |
a shopkeeper sold an book offering a discount of 5 % and earned a profit of 42.5 % . what would have been the percentage of profit earned if no discount was offered ? | "let c . p . be $ 100 . then , s . p . = $ 142.50 let marked price be $ x . then , 95 / 100 x = 142.5 x = 14250 / 95 = $ 150 now , s . p . = $ 150 , c . p . = $ 100 profit % = 50 % . e" | a ) 140 , b ) 120 , c ) 130 , d ) 110 , e ) 150 | e | multiply(const_100, divide(add(const_100, 42.5), subtract(const_100, 5))) | add(n1,const_100)|subtract(const_100,n0)|divide(#0,#1)|multiply(#2,const_100)| | gain |
a driver would have reduced the time it took to drive from home to the store by 1 / 4 if the average speed had been increased by 10 miles per hour . what was the actual average speed , in miles per hour , when the driver drove from home to the store ? | since the distance remains the same ( we ' re just changing the rate and time ) , any increase in rate or time is met with a decrease in the other term . decreasing the time by 1 / 4 would give us : d = ( r ) ( t ) = ( 3 t / 4 ) ( x * r ) x = 4 / 3 since ( 3 t / 4 ) ( 4 r / 3 ) = ( r ) ( t ) = d 4 r / 3 = r + 10 r / 3 ... | a ) 12 , b ) 15 , c ) 20 , d ) 25 , e ) 30 | e | divide(multiply(subtract(1, divide(1, 4)), const_10), subtract(1, subtract(1, divide(1, 4)))) | divide(n0,n1)|subtract(n0,#0)|multiply(#1,const_10)|subtract(n0,#1)|divide(#2,#3) | general |
a can do a job in 15 days and b can do it in 30 days . a and b working together will finish twice the amount of work in - - - - - - - days ? | "c 1 / 15 + 1 / 30 = 1 / 10 10 * 2 = 20 days" | a ) 14 days , b ) 16 days , c ) 20 days , d ) 11 days , e ) 19 days | c | add(divide(const_1, 15), divide(const_1, 30)) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)| | physics |
at a supermarket , john spent 1 / 2 of his money on fresh fruits and vegetables , 1 / 3 on meat products , and 1 / 10 on bakery products . if he spent the remaining $ 10 on candy , how much did john spend at the supermarket ? | "let ' s let t = total number of dollars spent at the supermarket . with this variable we can set up an equation and determine t . we are given that john spent 1 / 2 of his money on fresh fruits and vegetables , or ( 1 / 2 ) t , 1 / 3 on meat products , or ( 1 / 3 ) t , and 1 / 10 on bakery products , or ( 1 / 10 ) t .... | a ) $ 60 , b ) $ 80 , c ) $ 90 , d ) $ 120 , e ) $ 150 | e | divide(10, subtract(1, add(add(divide(1, 10), divide(1, 3)), divide(1, 2)))) | divide(n0,n5)|divide(n0,n3)|divide(n0,n1)|add(#0,#1)|add(#3,#2)|subtract(n0,#4)|divide(n6,#5)| | general |
the h . c . f . of two numbers is 12 and their l . c . m . is 600 . if one of the number is 120 , find the other ? | "other number = 12 * 600 / 120 = 60 answer is b" | a ) 100 , b ) 60 , c ) 120 , d ) 200 , e ) 150 | b | multiply(12, 120) | multiply(n0,n2)| | physics |
66.2 is what percent of 1000 ? | "we assume that 1000 is 100 % assume ' x ' is value we looking for here , 1000 = 100 % and x % = 66.2 therefore , 100 / x = 1000 / 66.2 100 / x = 15.105 x = 6.62 a" | a ) 6.62 , b ) 66.2 , c ) 662 , d ) 0.662 , e ) 0.0662 | a | multiply(divide(66.2, 1000), const_100) | divide(n0,n1)|multiply(#0,const_100)| | gain |
if the sum and difference of two numbers are 5 and 10 respectively , then the difference of their square is : | "let the numbers be x and y . then , x + y = 5 and x - y = 10 x 2 - y 2 = ( x + y ) ( x - y ) = 5 * 10 = 50 . answer : a" | a ) 50 , b ) 28 , c ) 160 , d ) 180 , e ) 18 | a | subtract(power(divide(add(5, 10), const_2), const_2), power(subtract(5, divide(add(5, 10), const_2)), const_2)) | add(n0,n1)|divide(#0,const_2)|power(#1,const_2)|subtract(n0,#1)|power(#3,const_2)|subtract(#2,#4)| | general |
find the least number which when divided by 56 , 78 leaves a remainder 3 but when divided by 9 leaves no remainder | l . c . m of 5,6 , 7,8 = 840 required number is of the form of 840 k + 3 least value of k for which ( 840 k + 3 ) is divided by 9 is k = 2 required number = ( 840 * 2 + 3 ) = 1683 answer ( d ) | a ) 9632 , b ) 7896 , c ) 8741 , d ) 1683 , e ) 8523 | d | multiply(multiply(multiply(add(const_2, const_3), add(3, 3)), add(3, const_4)), add(const_4, const_4)) | add(const_4,const_4)|add(n2,const_4)|add(const_2,const_3)|add(n2,n2)|multiply(#2,#3)|multiply(#1,#4)|multiply(#0,#5) | general |
what is the average of first 21 multiples of 8 ? | "required average = 8 ( 1 + 2 + . . . . + 21 ) / 21 ( 8 / 21 ) x ( ( 21 x 22 ) / 2 ) ( because sum of first 21 natural numbers ) = 88 e" | a ) a ) 70 , b ) b ) 77 , c ) c ) 79 , d ) d ) 81 , e ) e ) 88 | e | multiply(divide(divide(multiply(21, add(21, const_1)), const_2), 21), 8) | add(n0,const_1)|multiply(n0,#0)|divide(#1,const_2)|divide(#2,n0)|multiply(n1,#3)| | general |
a invested $ 150 in a business after 6 months b invested $ 200 in the business . end of the year if they got $ 100 as profit . find a shares ? | "a : b = 150 * 12 : 200 * 6 a : b = 3 : 2 a ' s share = 100 * 3 / 5 = $ 60 answer is e" | a ) $ 100 , b ) $ 75 , c ) $ 20 , d ) $ 120 , e ) $ 60 | e | multiply(100, subtract(const_1, divide(divide(200, const_2), add(150, divide(200, const_2))))) | divide(n2,const_2)|add(n0,#0)|divide(#0,#1)|subtract(const_1,#2)|multiply(n3,#3)| | gain |
for any number y , y * is defined as the greatest positive even integer less than or equal to y . what is the value of 9.2 β 9.2 * ? | "since y * is defined as the greatest positive even integer less than or equal to y , then 9.2 * = 8 ( the greatest positive even integer less than or equal to 9.2 is 4 ) . hence , 9.2 β 9.2 * = 9.2 - 8 = 1.2 answer : b ." | a ) 0.2 , b ) 1.2 , c ) 1.8 , d ) 2.2 , e ) 4.0 | b | subtract(9.2, subtract(floor(9.2), const_1)) | floor(n0)|subtract(#0,const_1)|subtract(n0,#1)| | general |
the area of a square field 3136 sq m , if the length of cost of drawing barbed wire 3 m around the field at the rate of rs . 1.40 per meter . two gates of 1 m width each are to be left for entrance . what is the total cost ? | "answer : option c explanation : a 2 = 3136 = > a = 56 56 * 4 * 3 = 672 Γ’ β¬ β 6 = 666 * 1.4 = 932.4 answer : c" | a ) 399 , b ) 272 , c ) 932.4 , d ) 277 , e ) 311 | c | multiply(multiply(subtract(multiply(sqrt(3136), const_4), multiply(const_2, 1)), 1.40), 3) | multiply(n3,const_2)|sqrt(n0)|multiply(#1,const_4)|subtract(#2,#0)|multiply(n2,#3)|multiply(#4,n1)| | physics |
$ 400 is divided amongst a , b and c so that a may get 2 / 3 as much as b and c together , b may get 6 / 9 as much as a and c together , then the share of a is | "a : ( b + c ) = 2 : 3 a ' s share = 400 * 2 / 5 = $ 160 answer is c" | a ) $ 100 , b ) $ 150 , c ) $ 160 , d ) $ 200 , e ) $ 250 | c | multiply(divide(400, add(divide(2, 3), const_1)), divide(2, 3)) | divide(n1,n2)|add(#0,const_1)|divide(n0,#1)|multiply(#2,#0)| | general |
2 . the value of x + x ( xx ) when x = 2 is : | x + x ( xx ) put the value of x = 2 in the above expression we get , 2 + 2 ( 22 ) = 2 + 2 ( 2 Γ 2 ) = 2 + 2 ( 4 ) = 2 + 8 = 10 answer : b | a ) 5 , b ) 10 , c ) 8 , d ) 15 , e ) 20 | b | add(multiply(multiply(2, 2), 2), 2) | multiply(n0,n0)|multiply(n0,#0)|add(n0,#1) | general |
find the value of x from the below equation ? : x ^ 2 + 5 x + 25 = 0 | "a = 1 , b = 5 , c = 25 x 1,2 = ( - 5 Γ’ Β± Γ’ Λ Ε‘ ( 5 ^ 2 - 4 Γ£ β 1 Γ£ β 25 ) ) / ( 2 Γ£ β 1 ) = ( - 5 Γ’ Β± Γ’ Λ Ε‘ ( 25 - 100 ) ) / 2 = ( - 5 Γ’ Β± Γ’ Λ Ε‘ - 75 ) / 2 no real roots a" | a ) no answer , b ) 25 , c ) 1 , d ) 0 , e ) - 2 | a | divide(subtract(5, sqrt(subtract(power(5, 2), multiply(25, 25)))), 2) | multiply(n2,n2)|power(n1,n0)|subtract(#1,#0)|sqrt(#2)|subtract(n1,#3)|divide(#4,n0)| | general |
a doctor prescribed 18 cubic centimeters of a certain drug to a patient whose body weight was 75 pounds . if the typical dosage is 2 cubic centimeters per 15 pounds of the body weight , by what percent was the prescribed dosage greater than the typical dosage ? | "typical dosage is dose : weight : : 2 : 15 . now if weight is 75 : ( 75 / 15 ) ) then typical dosage would be 2 * 5 = 10 cc . dosage = 18 cc . dosage is greater by 2 cc . % dosage is greater : ( 2 / 10 ) * 100 = 20 % e is the answer ." | a ) 8 % , b ) 9 % , c ) 11 % , d ) 12.5 % , e ) 20 % | e | multiply(divide(subtract(multiply(divide(2, 15), 75), 18), multiply(divide(2, 15), 75)), const_100) | divide(n2,n3)|multiply(n1,#0)|subtract(#1,n0)|divide(#2,#1)|multiply(#3,const_100)| | gain |
if two positive numbers are in the ratio 1 / 7 : 1 / 5 , then by what percent is the second number more than the first ? | "given ratio = 1 / 7 : 1 / 5 = 5 : 7 let first number be 5 x and the second number be 7 x . the second number is more than first number by 2 x . required percentage = 2 x / 5 x * 100 = 40 % . answer : d" | a ) 70 % , b ) 90 % , c ) 60 % , d ) 40 % , e ) 65 % | d | multiply(divide(1, 5), const_100) | divide(n0,n3)|multiply(#0,const_100)| | general |
30.5 % of 2000 | 1 % of 2000 = 20 30 % of 2000 = 30 * 20 = 600 0.5 % of 2000 = 0.5 * 2000 = 10 total = 610 answer : d | a ) 556 , b ) 612.5 , c ) 756.8 , d ) 610 , e ) 820 | d | multiply(divide(30.5, 2000), const_100) | divide(n0,n1)|multiply(#0,const_100)| | gain |
if the operation x is defined by a x b = ( b - a ) ^ 2 / a ^ 2 for all numbers a and b , and a β 0 , then β 1 x ( 1 x β 1 ) = | on solving the inner bracket we get 4 . then we get equation as - 1 x 4 where b = 4 and a = - 1 answer d | a ) β 1 , b ) 0 , c ) 1 , d ) 25 , e ) 29 | d | power(subtract(power(subtract(negate(1), 1), 2), negate(1)), 2) | negate(n3)|subtract(#0,n3)|power(#1,n0)|subtract(#2,#0)|power(#3,n0) | general |
the two lines y = x and x = - 3 intersect on the coordinate plane . if z represents the area of the figure formed by the intersecting lines and the x - axis , what is the side length of a cube whose surface area is equal to 6 z ? | 800 score official solution : the first step to solving this problem is to actually graph the two lines . the lines intersect at the point ( - 3 , - 3 ) and form a right triangle whose base length and height are both equal to 4 . as you know , the area of a triangle is equal to one half the product of its base length a... | a ) 16 , b ) 3 / β 2 , c ) 8 , d ) 2 β 2 , e ) ( β 2 ) / 3 | b | sqrt(divide(multiply(3, 3), const_2)) | multiply(n0,n0)|divide(#0,const_2)|sqrt(#1) | general |
subtracting 4 % of a from a is equivalent to multiplying a by how much ? | "answer let a - 4 % of a = ab . β ( 96 x a ) / 100 = ab β΄ b = 0.96 correct option : a" | a ) 0.96 , b ) 9.4 , c ) 0.094 , d ) 94 , e ) none | a | divide(subtract(const_100, 4), const_100) | subtract(const_100,n0)|divide(#0,const_100)| | general |
each of the integers from 0 to 9 , inclusive , is written on a separate slip of blank paper and the ten slips are dropped into a hat . if 5 of the slips are the drawn , without replacement , what is the probability that all 5 have a even number written on it ? | "key is that there is no replacement , so each successive choice will become more skewed towards picking a neg ( i . e . the pool of positives decreases , while the pool of negatives stay the same ) p ( + on 1 st pick ) = 5 / 10 p ( + on 2 nd pick ) = 4 / 9 p ( + on 3 rd pick ) = 3 / 8 p ( + on 4 rd pick ) = 2 / 7 p ( ... | a ) 1 / 252 , b ) 1 / 10 , c ) 1 / 8 , d ) 1 / 2 , e ) 5 / 9 | a | multiply(multiply(multiply(divide(add(const_4, const_1), add(9, const_1)), divide(subtract(add(5, const_1), const_1), subtract(add(9, const_1), const_1))), divide(subtract(subtract(add(5, const_1), const_1), const_1), subtract(subtract(add(9, const_1), const_1), const_1))), divide(subtract(subtract(subtract(add(5, cons... | add(const_1,const_4)|add(n1,const_1)|divide(#0,#1)|subtract(#0,const_1)|subtract(#1,const_1)|divide(#3,#4)|subtract(#3,const_1)|subtract(#4,const_1)|divide(#6,#7)|multiply(#2,#5)|subtract(#6,const_1)|subtract(#7,const_1)|divide(#10,#11)|multiply(#8,#9)|multiply(#12,#13)| | gain |
if a and b are integers and ( a * b ) ^ 5 = 32 y , y could be : | "distribute the exponent . a ^ 5 * b ^ 5 = 32 y find the prime factorization of 32 . this is 2 ^ 5 * 1 . we need 1 the answer is b ." | a ) 16 , b ) 1 , c ) 8 , d ) 12 , e ) 24 | b | multiply(divide(32, power(const_2, 5)), const_2) | power(const_2,n0)|divide(n1,#0)|multiply(#1,const_2)| | general |
it was calculated that 100 men could complete a piece of work in 20 days . when work was scheduled to commence , it was found necessary to send 50 men to another project . how much longer will it take to complete the work ? | one day work = 1 / 20 one man β s one day work = 1 / ( 20 * 100 ) now : no . of workers = 50 one day work = 50 * 1 / ( 20 * 100 ) the total no . of days required to complete the work = ( 100 * 20 ) / 50 = 40 answer : b | a ) 45 days . , b ) 40 days . , c ) 38 days . , d ) 35 days . , e ) 30 days . | b | multiply(20, divide(100, 50)) | divide(n0,n2)|multiply(n1,#0) | physics |
if a is thrice as fast as b and together can do a work in 15 days . in how many days a alone can do the work ? | "a β s one day β s work = 1 / x b β s one day β s work = 1 / 3 x a + b β s one day β s work = 1 / x + 1 / 3 x = 1 / 15 = 3 + 1 / 3 x = 4 / 3 x = 1 / 15 x = 15 * 4 / 3 = 20 answer : b" | a ) 36 , b ) 20 , c ) 28 , d ) 54 , e ) 45 | b | inverse(divide(const_3, multiply(15, const_4))) | multiply(n0,const_4)|divide(const_3,#0)|inverse(#1)| | physics |
average of 5 numbers is - 10 , and the sum of 3 of the numbers is 16 , wat is the average of the other 2 numbers ? | let the five numbers be a , b , c , d , e . then their average is ( a + b + c + d + e 5 ) = 10 . ( a + b + c + d + e 5 ) = 10 . now three of the numbers have a sum of 16 , say , a + b + c = 16 a + b + c = 16 . so substitute 16 for a + b + ca + b + c in the average above : ( 16 + d + e 5 ) = 10 . ( 16 + d + e 5 ) = 10 .... | a ) - 33 , b ) 33 , c ) 35 , d ) 36 , e ) - 35 | a | divide(subtract(multiply(5, 10), 16), 2) | multiply(n0,n1)|subtract(#0,n3)|divide(#1,n4) | general |
a number when divided by 296 leaves 75 as remainder when the same number is divided by 37 then the remainder will be ? | "let x = 296 q + 75 = ( 37 * 8 q + 37 * 2 ) + 1 37 * ( 8 q + 2 ) + 1 thus when the number is divided by 37 the remainder is 1 answer ( e )" | a ) 2 , b ) 9 , c ) 6 , d ) 5 , e ) 1 | e | reminder(add(296, 75), 37) | add(n0,n1)|reminder(#0,n2)| | general |
if | 4 x + 6 | = 50 , what is the sum of all the possible values of x ? | "there will be two cases 4 x + 6 = 50 or 4 x + 6 = - 50 = > x = 11 or x = - 14 sum of both the values will be - 14 + 11 = - 3 answer is d" | a ) 2 , b ) - 2 , c ) 4 , d ) - 3 , e ) 6 | d | divide(subtract(50, 6), 4) | subtract(n2,n1)|divide(#0,n0)| | general |
a merchant sells an item at a 20 % discount , but still makes a gross profit of 20 percent of the cost . what percent e of the cost would the gross profit on the item have been if it had been sold without the discount ? | "let the market price of the product is mp . let the original cost price of the product is cp . selling price ( discounted price ) = 100 % of mp - 20 % mp = 80 % of mp . - - - - - - - - - - - - - - - - ( 1 ) profit made by selling at discounted price = 20 % of cp - - - - - - - - - - - - - - ( 2 ) apply the formula : pr... | a ) 20 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 75 % | c | subtract(const_100, subtract(subtract(const_100, 20), 20)) | subtract(const_100,n0)|subtract(#0,n1)|subtract(const_100,#1)| | gain |
the width of a rectangle is 10 meter and its area is 150 square meter . if length of the rectangle is increased then its new area is 4 / 3 times of the original area . what is the new perimeter of the rectangle ? | let length of new rectangle be x m then x Γ 10 = 150 Γ 4 / 3 = 200 or , x = 20 m hence new perimeter = 2 ( 10 + 20 ) = 60 m answer : d | ['a ) 48 m', 'b ) 52 m', 'c ) 54 m', 'd ) 60 m', 'e ) none of these'] | d | rectangle_perimeter(divide(divide(multiply(150, 4), 3), 10), 10) | multiply(n1,n2)|divide(#0,n3)|divide(#1,n0)|rectangle_perimeter(n0,#2) | geometry |
the cost price of a radio is rs . 2300 and it was sold for rs . 1800 , find the loss % ? | "2300 - - - - 500 100 - - - - ? = > 21 % answer : e" | a ) 18 , b ) 99 , c ) 27 , d ) 26 , e ) 21 | e | multiply(divide(subtract(2300, 1800), 2300), const_100) | subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)| | gain |
a and b can finish a work 30 days if they work together . they worked together for 20 days and then b left . a finished the remaining work in another 20 days . in how many days a alone can finish the work ? | amount of work done by a and b in 1 day = 1 / 30 amount of work done by a and b in 20 days = 20 Γ ( 1 / 30 ) = 20 / 30 = 2 / 3 remaining work β 1 β 2 / 3 = 1 / 3 a completes 1 / 3 work in 20 days amount of work a can do in 1 day = ( 1 / 3 ) / 20 = 1 / 60 = > a can complete the work in 60 days answer is b . | a ) 30 , b ) 60 , c ) 40 , d ) 20 , e ) 50 | b | multiply(30, divide(20, subtract(30, 20))) | subtract(n0,n1)|divide(n1,#0)|multiply(n0,#1) | physics |
0.0004 ? = 0.01 | "explanation : required answer = 0.0004 / 0.01 = 0.04 / 1 = 0.04 . answer : option c" | a ) 4 , b ) 0.4 , c ) 0.04 , d ) 40 , e ) none of these | c | multiply(divide(0.0004, 0.01), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
860 % of 50 + 50 % of 860 = ? | "solution given expression = [ 860 / 100 x 50 + 50 / 100 x 860 ] = 430 + 430 = 860 . answer d" | a ) 430 , b ) 516 , c ) 660 , d ) 860 , e ) none | d | divide(multiply(860, 50), 50) | multiply(n0,n1)|divide(#0,n2)| | general |
in a lake , there is a patch of lily pads . every day , the patch doubles in size . it takes 39 days for the patch to cover the entire lake , how many days would it take the patch to cover half of the lake ? | "working backward from the day it ' s covered : day 39 : fully covered day 38 : half covered so 38 days answer : d" | a ) 36 , b ) 2 ^ 4 * 3 , c ) 24 , d ) 38 , e ) 47 | d | subtract(39, const_1) | subtract(n0,const_1)| | physics |
find the product of the localvalue and absolutevalue of 4 in 20468 ? | "local value of 4 = 4 x 100 = 400 place value of 4 = 4 there fore = 4 x 400 = 1600 e" | a ) 1000 , b ) 1100 , c ) 1200 , d ) 1300 , e ) 1600 | e | multiply(multiply(4, const_1000), 4) | multiply(n0,const_1000)|multiply(n0,#0)| | general |
marcella has 20 pairs of shoes . if she loses 9 individual shoes , what is the greatest number of matching pairs she could have left ? | "marcella has 20 pairs of shoes and loses 9 shoes . to minimize the loss of identical pairs of shoes we want marcella to lose as many identical pairs as possible . this would yield 4 identical pairs and 1 additional shoe ( destroying 5 pairs of shoes ) . the 20 pairs of shoes minus the 5 ' destroyed ' pairs yields 15 p... | a ) 21 , b ) 20 , c ) 19 , d ) 16 , e ) 15 | e | subtract(20, add(floor(divide(9, const_2)), const_1)) | divide(n1,const_2)|floor(#0)|add(#1,const_1)|subtract(n0,#2)| | general |
a miniature roulette wheel is divided into 10 equal sectors , each bearing a distinct integer from 1 to 10 , inclusive . each time the wheel is spun , a ball randomly determines the winning sector by settling in that sector . if the wheel is spun five times , approximately what is the probability that the product of th... | "the only way to have an odd product is if all 5 integers are odd . p ( odd product ) = 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 = 1 / 32 p ( even product ) = 1 - 1 / 32 = 31 / 32 which is about 97 % the answer is e ." | a ) 50 % , b ) 67 % , c ) 85 % , d ) 91 % , e ) 97 % | e | multiply(subtract(1, power(divide(divide(10, const_2), 10), const_2)), const_100) | divide(n0,const_2)|divide(#0,n0)|power(#1,const_2)|subtract(n1,#2)|multiply(#3,const_100)| | general |
on a map , 7 centimeters represents 35 kilometers . two cities 245 kilometers apart would be separated on the map by how many centimeters ? | 1 centimeter represents 5 kilometers ( 35 / 7 ) x = 245 / 5 = 49 answer : c | a ) 15 , b ) 37 , c ) 49 , d ) 110 , e ) 180 | c | multiply(divide(7, 35), 245) | divide(n0,n1)|multiply(n2,#0) | gain |
a certain bag contains 60 balls β 22 white , 10 green , 7 yellow , 15 red , and 6 purple . if a ball is to be chosen at random , what is the probability that the ball will be neither red nor purple ? | "according to the stem the ball can be white , green or yellow , so the probability is ( white + green + yellow ) / ( total ) = ( 22 + 10 + 17 ) / 60 = 39 / 60 = 0.65 . answer : b ." | a ) 0.09 , b ) 0.65 , c ) 0.54 , d ) 0.85 , e ) 0.91 | b | divide(add(add(22, 10), 7), 60) | add(n1,n2)|add(n3,#0)|divide(#1,n0)| | other |
points a , b , c , and d , in that order , lie on a line . if ab = 2 cm , ac = 5 cm , and bd = 9 cm , what is cd , in centimeters ? | "putting a value to each point , lets use the following : a - 0 b - 2 ( ab = 2 ) c - 5 ( ac = 5 ) d - 11 ( bd = 9 ) cd is 11 - 5 = 6 . ans d" | a ) 1 , b ) 2 , c ) 3 , d ) 6 , e ) 5 | d | subtract(5, 2) | subtract(n1,n0)| | physics |
if the area of a square with sides of length 4 centimeters is equal to the area of a rectangle with a width of 8 centimeters , what is the length of the rectangle , in centimeters ? | "let length of rectangle = l 4 ^ 2 = l * 8 = > l = 16 / 8 = 2 answer c" | a ) 4 , b ) 8 , c ) 2 , d ) 16 , e ) 18 | c | divide(power(4, const_2), 8) | power(n0,const_2)|divide(#0,n1)| | geometry |
the average of 6 numbers is 6.8 . if one of the numbers is multiplied by a factor of 3 , the average of the numbers increases to 9.2 . what number is multiplied by 3 ? | "the average of 6 numbers is 6.8 the sum of 6 numbers will be 6.8 x 6 = 40.8 the average of 6 number after one of the number is multiplied by 3 is 9.2 the sum of the numbers will now be 9.2 x 6 = 55.2 so the sum has increased by 55.2 - 40.8 = 14.4 let the number multiplied by 3 be n then , 3 n = n + 14.4 or 2 n = 14.4 ... | a ) 9.5 , b ) 8.0 , c ) 7.2 , d ) 5.0 , e ) 4.0 | c | subtract(multiply(9.2, 6), multiply(6.8, 6)) | multiply(n0,n3)|multiply(n0,n1)|subtract(#0,#1)| | general |
two trains 200 m and 800 m long run at the speed of 60 km / hr and 40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? | "relative speed = 60 + 40 = 100 km / hr . = 100 * 5 / 18 = 250 / 9 m / sec . distance covered in crossing each other = 200 + 800 = 1000 m . required time = 1000 * 9 / 250 = 36 sec . answer : c" | a ) 10.6 , b ) 10.9 , c ) 36 , d ) 10.8 , e ) 10.1 | c | divide(add(200, 800), multiply(add(60, 40), const_0_2778)) | add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)| | physics |
if 3 < x < 6 < y < 7 , then what is the greatest possible positive integer difference of x and y ? | "3 < x < 6 < y < 7 ; 3 < x y < 7 3 + y < x + 7 y - x < 4 . positive integer difference is 3 ( for example y = 6.5 and x = 3.5 ) answer : a ." | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | subtract(subtract(7, 3), const_1) | subtract(n2,n0)|subtract(#0,const_1)| | general |
the ratio between the perimeter and the width of a rectangle is 5 : 1 . if the area of the rectangle is 96 square centimeters , what is the length of the rectangle in centimeters ? | "perimeter = 2 ( w + l ) = 5 w 3 w = 2 l w = 2 l / 3 wl = 96 2 l ^ 2 / 3 = 96 l ^ 2 = 144 l = 12 cm the answer is b ." | a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | b | divide(96, const_10) | divide(n2,const_10)| | geometry |
in a group of 90 people , 55 have visited iceland and 33 have visited norway . if 51 people have visited both iceland and norway , how many people have visited neither country ? | "this is an example of a standard overlapping sets question . it has no ' twists ' to it , so you ' ll likely find using the overlapping sets formula to be a fairly easy approach . if you ' re not familiar with it , then here is the formula : 90 = 55 + 33 - 51 + ( # in neither group ) = 53 the prompt gives you all of t... | a ) 50 , b ) 51 , c ) 52 , d ) 53 , e ) 54 | d | subtract(90, subtract(add(55, 33), 51)) | add(n1,n2)|subtract(#0,n3)|subtract(n0,#1)| | other |
a small company reduced its faculty by approximately 14 percent to 195 employees . what was the original number of employees ? | "if x is the original number of employees , then after 14 % reduction in employees number is . 86 x but we are given . 86 x = 195 x = 227 so the original number of employees is 227 correct answer - b" | a ) a ) 182 , b ) b ) 227 , c ) c ) 220 , d ) d ) 224 , e ) e ) 302 | b | divide(195, divide(subtract(const_100, 14), const_100)) | subtract(const_100,n0)|divide(#0,const_100)|divide(n1,#1)| | gain |
in a dairy farm , 40 cows eat 40 bags of husk in 40 days . in how many days one cow will eat one bag of husk ? | "assume that in x days , one cow will eat one bag of husk . more cows , less days ( indirect proportion ) more bags , more days ( direct proportion ) hence we can write as cowsbags 40 : 11 : 40 β« β β¬ : : x : 40 β 40 Γ 1 Γ 40 = 1 Γ 40 Γ x β x = 40 answer : c" | a ) 1 , b ) 1 / 40 , c ) 40 , d ) 80 , e ) 1 / 80 | c | multiply(divide(40, 40), 40) | divide(n0,n0)|multiply(n0,#0)| | physics |
the average age of 4 men is increased by years when two of them whose ages are 21 years and 23 years are replaced by two new men . the average age of the two new men is | total age increased = ( 8 * 2 ) years = 8 years . sum of ages of two new men = ( 21 + 23 + 8 ) years = 52 years average age of two new men = ( 52 / 2 ) years = 26 years . answer : d | a ) 22 , b ) 30 , c ) 99 , d ) 26 , e ) 27 | d | add(divide(add(21, 23), const_2), multiply(const_1, 4)) | add(n1,n2)|multiply(n0,const_1)|divide(#0,const_2)|add(#2,#1) | general |
one type of liquid contains 25 % of kerosene , the other contains 30 % of kerosene . p can is filled with 6 parts of the first liquid and 4 parts of the second liquid . find the percentage of the kerosene in the new mixture . | let p be filled by 60 lts of 1 st liquid and 40 lts . of 2 nd liquid . amount of kerosene = ( 25 * 60 / 100 ) + ( 30 * 40 / 100 ) = 27 lts . % of kerosene = 27 % answer : d | a ) 28 % , b ) 25 % , c ) 30 % , d ) 27 % , e ) 35 % | d | multiply(add(multiply(6, divide(25, const_100)), multiply(4, divide(30, const_100))), add(6, 4)) | add(n2,n3)|divide(n0,const_100)|divide(n1,const_100)|multiply(n2,#1)|multiply(n3,#2)|add(#3,#4)|multiply(#5,#0) | gain |
the ratio of income of anand to that of balu is 5 : 4 and the expenditure of anand to that of balu is 3 : 2 . if at the end of the year , each saves rs , 800 , the income of anand is : | explanation : let the income of anand and balu be 5 x and 4 x and . the expenditures of anand and balu be 3 y and 2 y . then , 5 x 3 y = 800 and 4 x 2 y = 800 . on solving we get : x = 400 . as income = 5 x = rs . 2000 . answer : c | a ) rs 1600 , b ) rs 1800 , c ) rs 2000 , d ) rs 2200 , e ) none of these | c | add(800, multiply(divide(subtract(multiply(800, 5), multiply(800, 4)), subtract(multiply(3, 4), multiply(2, 5))), 3)) | multiply(n0,n4)|multiply(n1,n4)|multiply(n1,n2)|multiply(n0,n3)|subtract(#0,#1)|subtract(#2,#3)|divide(#4,#5)|multiply(n2,#6)|add(n4,#7) | other |
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