Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
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in a rectangular coordinate system , what is the area of a rhombus whose vertices have the coordinates ( 0 , 3.5 ) , ( 9 , 0 ) , ( 0 , - 3.5 ) , ( - 9 , 0 ) ? | "ares of rhombus = 1 / 2 * d 1 * d 2 length of 1 st diagonal = 9 + 9 = 18 length of 2 nd diagonal = 3.5 + 3.5 = 7 area = 1 / 2 * 18 * 7 = 63 a is the answer" | a ) 63 , b ) 88 , c ) 112 , d ) 116 , e ) 120 | a | rhombus_area(multiply(9, const_2), multiply(3.5, const_2)) | multiply(n2,const_2)|multiply(n1,const_2)|rhombus_area(#0,#1)| | geometry |
each of the 35 points is placed either inside or on the surface of a perfect sphere . if 88 % or fewer of the points touch the surface , what is the maximum number of segments which , if connected from those points to form chords , could be the diameter of the sphere ? | "maximum number of points on the surface is 88 % * 35 = 30.8 . . . or 30 since it has to be an integer now note that if two points form a diameter , they can not be part of any other diameter . so in the best case we can pair up the points we have 30 points , so at best we can form 15 pairs ( 30 ) . so , answer is ( e ... | a ) 7 , b ) 11 , c ) 13 , d ) 14 , e ) 15 | e | divide(multiply(35, divide(88, const_100)), const_2) | divide(n1,const_100)|multiply(n0,#0)|divide(#1,const_2)| | geometry |
6 years ago , p was half of q in age . if the ratio of their present ages is 3 : 4 , what will be the total of their present ages | explanation : let the present age of p and q be 3 x and 4 x respectively . six years ago , p was half of q in age = > 2 ( 3 x β 6 ) = ( 4 x β 6 ) = > 6 x β 12 = 4 x β 6 = > 2 x = 6 = > x = 3 7 x = 7 * 3 = 21 answer : option a | a ) a ) 21 , b ) b ) 34 , c ) c ) 45 , d ) d ) 25 , e ) e ) 26 | a | add(multiply(const_2, 6), divide(multiply(3, multiply(const_2, 6)), 4)) | multiply(n0,const_2)|multiply(n1,#0)|divide(#1,n2)|add(#2,#0) | general |
the sale price of an article including the sales tax is rs . 616 . the rate of sales tax is 10 % . if the shopkeeper has made a profit of 12 % , then the cost price of the article is : | "110 % of s . p . = 616 s . p . = ( 616 * 100 ) / 110 = rs . 560 c . p = ( 110 * 560 ) / 112 = rs . 500 . answer : a" | a ) 500 , b ) 277 , c ) 266 , d ) 288 , e ) 261 | a | divide(multiply(divide(multiply(616, const_100), add(const_100, 10)), add(const_100, 10)), add(const_100, 12)) | add(n1,const_100)|add(n2,const_100)|multiply(n0,const_100)|divide(#2,#0)|multiply(#0,#3)|divide(#4,#1)| | gain |
if 2 log ( 4 * 5 ^ 2 ) = x , find x | "2 ( log 2 ^ 2 * 5 ^ 2 ) = x 2 log ( 5 * 2 ) ^ 2 = x 2 * 2 log ( 5 * 2 ) = x 4 log 10 = x log 10 base 10 = 1 so 4 * 1 = x x = 4 answer : d" | a ) 5 , b ) 6 , c ) 7 , d ) 4 , e ) 8 | d | power(multiply(4, power(5, 2)), 2) | power(n2,n3)|multiply(n1,#0)|power(#1,n0)| | general |
a certain automobile company β s best - selling model is the speedster . the speedster , like all of their other models , comes in coupe and convertible styles . 3 / 4 of the current inventory is speedsters , of which 3 / 5 are convertibles . if there are 30 vehicles that are not speedsters , how many speedster convert... | "total vehicle = 3 / 4 of speedster + 1 / 4 of others . speedster convertibles = 3 / 4 total vehicle * 3 / 5 given : 1 / 4 constitutes 30 vehicles . hence 3 / 4 constitutes 90 speedster convertibls = 90 * 3 / 5 = 54 b" | a ) 30 , b ) 54 , c ) 60 , d ) 70 , e ) 90 | b | multiply(multiply(divide(30, subtract(const_1, divide(3, 4))), divide(3, 4)), divide(3, 5)) | divide(n2,n3)|divide(n0,n1)|subtract(const_1,#1)|divide(n4,#2)|multiply(#3,#1)|multiply(#0,#4)| | physics |
a rower can row 6 km / h in still water . when the river is running at 1 km / h , it takes the rower 1 hour to row to big rock and back . how many kilometers is it to big rock ? | "let x be the distance to big rock . time = x / 5 + x / 7 = 1 x = 35 / 12 = 2.92 the answer is e ." | a ) 2.66 , b ) 2.78 , c ) 2.8 , d ) 2.87 , e ) 2.92 | e | multiply(divide(subtract(6, 1), add(add(6, 1), subtract(6, 1))), add(6, 1)) | add(n0,n1)|subtract(n0,n1)|add(#0,#1)|divide(#1,#2)|multiply(#0,#3)| | physics |
in a stockpile of products produced by 3 machines m 1 , m 2 and m 3 , 40 % and 30 % were manufactured by m 1 and m 2 respectively . 3 % of the products of m 1 are defective , 1 % of products of m 2 defective , while 93 % of the products of m 3 iii are not defective . what is the percentage of defective in the stockpile... | explanation : let there be 100 products in the stockpile . 40 % and 30 % were manufactured by m 1 and m 2 respectively . so , 30 % are manufactured by m 3 . products from m 1 = 40 , from m 2 = 30 and from m 3 = 30 . number of defective products from m 1 = 0.03 x 40 = 1.2 , from m 2 = 0.01 x 30 = 0.3 and from m 3 = 0.07... | a ) 3 % , b ) 5 % , c ) 2.5 % , d ) 4 % , e ) 3.6 % | e | add(add(multiply(divide(subtract(const_100, 93), const_100), subtract(const_100, add(40, 30))), multiply(divide(1, const_100), 30)), multiply(divide(3, const_100), 40)) | add(n4,n5)|divide(n1,const_100)|divide(n0,const_100)|subtract(const_100,n12)|divide(#3,const_100)|multiply(n5,#1)|multiply(n4,#2)|subtract(const_100,#0)|multiply(#4,#7)|add(#8,#5)|add(#9,#6) | gain |
the parameter of a square is equal to the perimeter of a rectangle of length 18 cm and breadth 10 cm . find the circumference of a semicircle whose diameter is equal to the side of the square . | "let the side of the square be a cm . parameter of the rectangle = 2 ( 18 + 10 ) = 56 cm parameter of the square = 56 cm i . e . 4 a = 56 a = 14 diameter of the semicircle = 14 cm circimference of the semicircle = 1 / 2 ( Γ’ Λ Β ) ( 14 ) = 22 / 7 * 14 = 44 cm answer : a" | a ) 44.0 , b ) 23.59 , c ) 23.55 , d ) 23.53 , e ) 23.57 | a | divide(circumface(divide(square_edge_by_perimeter(rectangle_perimeter(18, 10)), const_2)), const_2) | rectangle_perimeter(n0,n1)|square_edge_by_perimeter(#0)|divide(#1,const_2)|circumface(#2)|divide(#3,const_2)| | geometry |
the average mark obtained by 22 candidates in an examination is 48 . the average of the first 10 is 55 while the last 11 is 40 . the marks obtained by the 11 th candidate is ? | it is clear that 22 x 48 = 10 x 55 + k + 11 x 40 β β k = 66 answer : e | a ) 22 , b ) 0 , c ) 49 , d ) 58 , e ) 66 | e | subtract(multiply(22, 48), add(multiply(10, 55), multiply(11, 40))) | multiply(n0,n1)|multiply(n2,n3)|multiply(n4,n5)|add(#1,#2)|subtract(#0,#3) | general |
' a ' and ' b ' are positive integers such that their lcm is 20 and their hcf is 1 . what is the difference between the minimum and maximum possible values of ' a + b ' ? | "possible values of a and b can be 5,4 ; 4,5 ( which are same for a + b ) and 1,20 ; 20,1 ( same result for a + b ) so 9 - 21 = - 12 . ans e ." | a ) 0 , b ) - 24 , c ) - 32 , d ) - 20 , e ) - 12 | e | subtract(add(divide(20, const_4), const_4), add(divide(20, const_4), const_4)) | divide(n0,const_4)|add(#0,const_4)|subtract(#1,#1)| | general |
a certain company reported that the revenue on sales increased 50 % from 2000 to 2003 , and increased 80 % from 2000 to 2005 . what was the approximate percent increase in revenue for this store from 2003 to 2005 ? | "assume the revenue in 2000 to be 100 . then in 2003 it would be 150 and and in 2005 180 , so from 2003 to 2005 it increased by ( 180 - 150 ) / 150 = 30 / 150 = ~ 20 % . answer : a" | a ) 20 % , b ) 40 % , c ) 35 % , d ) 32 % , e ) 29 % | a | multiply(divide(subtract(add(const_1, divide(80, const_100)), add(const_1, divide(50, const_100))), add(const_1, divide(50, const_100))), const_100) | divide(n3,const_100)|divide(n0,const_100)|add(#0,const_1)|add(#1,const_1)|subtract(#2,#3)|divide(#4,#3)|multiply(#5,const_100)| | gain |
if 21 lines are drawn in a plane such that no two of them are parallel and no three are concurrent , then in how many points do they intersect ? | "if two lines are not parallel , then they will intersect in exactly one point . lines can be extended infinitely on both ends so somewhere they will intersect with each other if they are not parallel . we are also given that no three lines are concurrent . this means that no three lines intersect at the same point . s... | a ) 150 , b ) 170 , c ) 190 , d ) 210 , e ) 230 | d | divide(factorial(21), multiply(factorial(subtract(21, const_2)), factorial(const_2))) | factorial(n0)|factorial(const_2)|subtract(n0,const_2)|factorial(#2)|multiply(#3,#1)|divide(#0,#4)| | physics |
three rugs have a combined area of 200 square meters . by overlapping the rugs to cover floor area of 140 square meters , the area that is covered by exactly two layers of rug is 22 square meters . what is the area that is covered with three layers of rug ? | total = rug 1 + rug 2 + rug 3 - { overlap of exactly 2 rugs } - 2 * { overlap of exactly 3 rugs } 140 = 200 - 22 - 2 * { overlap of exactly 2 rugs } - - > { overlap of exactly 3 rugs } = 19 . answer : d . | ['a ) 18 square meters', 'b ) 20 square meters', 'c ) 24 square meters', 'd ) 19 square meters', 'e ) 30 square meters'] | d | divide(subtract(subtract(200, 140), 22), const_2) | subtract(n0,n1)|subtract(#0,n2)|divide(#1,const_2) | geometry |
find the simple interest on $ 3200 for 3 years at 10 % per annum ? | "si = ptr / 100 = 3200 * 3 * 10 / 100 = $ 960 answer is e" | a ) $ 250 , b ) $ 300 , c ) $ 500 , d ) $ 600 , e ) $ 960 | e | multiply(3200, divide(3, const_100)) | divide(n1,const_100)|multiply(n0,#0)| | gain |
if a person walks at 12 km / hr instead of 8 km / hr , he would have walked 20 km more . the actual distance traveled by him is ? | "let the actual distance traveled be x km . then , x / 8 = ( x + 20 ) / 12 3 x - 2 x = 40 = > x = 40 km . answer : d" | a ) 50 km , b ) 58 km , c ) 60 km , d ) 40 km , e ) 65 km | d | multiply(8, divide(20, subtract(12, 8))) | subtract(n0,n1)|divide(n2,#0)|multiply(n1,#1)| | general |
two different primes may be said torhymearound an integer if they are the same distance from the integer on the number line . for instance , 3 and 7 rhyme around 5 . what integer e between 1 and 20 , inclusive , has the greatest number of distinct rhyming primes around it ? | "since we are concerned with integers between 1 and 20 , write down the primes till 40 . 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 ( you should be very comfortable with the first few primes . . . ) 2 , 3 , 5 , 7 , 11,12 , 13 , 17 , 19 , 23 , 29 , 31 , 37 - three pairs ( 11,13 ) , ( 7,17 ) , ( 5 , 19 ) 2 , 3... | a ) 12 , b ) 15 , c ) 17 , d ) e = 18 , e ) 20 | d | multiply(multiply(3, const_2), 3) | multiply(const_2,n0)|multiply(n0,#0)| | physics |
a merchant purchased a jacket for $ 60 and then determined a selling price that equalled the purchase price of the jacket plus a markup that was 25 percent of the selling price . during a sale , the merchant discounted the selling price by 20 percent and sold the jacket . what was the merchant β s gross profit on this ... | "markup = x x = 0.25 ( 60 + x ) = > 3 x / 4 = 15 = > x = 20 so sp = 80 but actual sp = 0.8 * 80 = 64 so profit = $ 4 answer - c" | a ) $ 0 , b ) $ 3 , c ) $ 4 , d ) $ 12 , e ) $ 15 | c | subtract(multiply(divide(60, subtract(const_1, divide(25, const_100))), subtract(const_1, divide(20, const_100))), 60) | divide(n1,const_100)|divide(n2,const_100)|subtract(const_1,#0)|subtract(const_1,#1)|divide(n0,#2)|multiply(#4,#3)|subtract(#5,n0)| | gain |
a car travels at a speed of 65 miles per hour . how far will it travel in 7 hours ? | "during each hour , the car travels 65 miles . for 7 hours it will travel 65 + 65 + 65 + 65 + 65 + 65 + 65 = 7 Γ 65 = 455 miles correct answer is c ) 455 miles" | a ) 125 miles , b ) 225 miles , c ) 455 miles , d ) 425 miles , e ) 525 miles | c | multiply(65, 7) | multiply(n0,n1)| | physics |
the percentage profit earned by selling an article for rs . 1720 is equal to the percentage loss incurred by selling the same article for rs . 1280 . at what price should the article be sold to make 25 % profit ? | "then , ( 1720 - x ) / x * 100 = ( x - 1280 ) / x * 100 1720 - x = x - 1280 2 x = 3000 = > x = 1500 required s . p . = 125 % of rs . 1500 = 125 / 100 * 1500 = rs . 1875 . answer c" | a ) 3000 , b ) 1230 , c ) 1875 , d ) 5600 , e ) 3400 | c | multiply(divide(add(const_100, 25), const_100), divide(add(1720, 1280), const_2)) | add(n2,const_100)|add(n0,n1)|divide(#0,const_100)|divide(#1,const_2)|multiply(#2,#3)| | gain |
find the surface area of a 10 cm x 4 cm x 2 cm brick . | "sol . surface area = [ 2 ( 10 x 4 + 4 x 2 + 10 x 2 ) ] = ( 2 x 68 ) = 136 cu cm answer a" | a ) 136 cu cm , b ) 124 cu cm , c ) 164 cu cm , d ) 180 cu cm , e ) none | a | surface_rectangular_prism(2, 4, 10) | surface_rectangular_prism(n0,n1,n2)| | geometry |
a motorcyclist goes from bombay to pune , a distance of 192 kms at an average of 32 kmph speed . another man starts from bombay by car 2 Γ’ Β½ hours after the first , and reaches pune Γ’ Β½ hour earlier . what is the ratio of the speed of the motorcycle and the car ? | "explanation : t = 192 / 32 = 6 h t = 6 - 3 = 3 time ratio = 6 : 3 = 2 : 1 speed ratio = 1 : 2 answer is a" | a ) 1 : 2 , b ) 1 : 3 , c ) 10 : 27 , d ) 5 : 4 , e ) 10 : 40 | a | divide(divide(192, divide(192, 32)), divide(192, subtract(divide(192, 32), const_3))) | divide(n0,n1)|divide(n0,#0)|subtract(#0,const_3)|divide(n0,#2)|divide(#1,#3)| | physics |
a certain country is divided into 4 provinces . each province consists entirely of progressives and traditionalists . if each province contains the same number of traditionalists and the number of traditionalists in any given province is 1 / 12 the total number of progressives in the entire country , what fraction of t... | "let p be the number of progressives in the country as a whole . in each province , the number of traditionalists is p / 12 the total number of traditionalists is 4 p / 12 = p / 3 . the total population is p + p / 3 = 4 p / 3 p / ( 4 p / 3 ) = 3 / 4 the answer is e ." | a ) 1 / 5 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 3 / 4 | e | subtract(1, divide(divide(4, 12), add(divide(4, 12), 1))) | divide(n0,n2)|add(n1,#0)|divide(#0,#1)|subtract(n1,#2)| | general |
12.5 * 3.2 * 8.8 = ? | "a 352 ? = 12.5 * 3.2 * 8.8 = 352" | a ) 352 , b ) 435 , c ) 224 , d ) 646 , e ) 742 | a | multiply(12.5, 3.2) | multiply(n0,n1)| | general |
vishal invested 10 % more than trishul . trishul invested 10 % less than raghu . if the total sum of their investments is rs . 6069 , how much amount did raghu invest ? | "let money invested by raghu = rs . x money invested by trishul = 9 / 10 x = 0.9 x money invested by vishal = 9 / 10 x * 110 / 100 = 0.99 x also , x + 0.9 x + 0.99 x = 6069 = x = 6069 / 2.89 = 2100 therefore , amount invested by raghu is rs . 2100 . answer : c" | a ) 1287 , b ) 2887 , c ) 2100 , d ) 1129 , e ) 1192 | c | divide(6069, add(add(multiply(subtract(const_1, divide(10, const_100)), add(const_1, divide(10, const_100))), subtract(const_1, divide(10, const_100))), const_1)) | divide(n0,const_100)|add(#0,const_1)|subtract(const_1,#0)|multiply(#1,#2)|add(#3,#2)|add(#4,const_1)|divide(n2,#5)| | general |
a train 125 m long passes a man , running at 7 km / hr in the same direction in which the train is going , in 10 seconds . the speed of the train is ? | "speed of the train relative to man = ( 125 / 10 ) m / sec = ( 25 / 2 ) m / sec . [ ( 25 / 2 ) * ( 18 / 5 ) ] km / hr = 45 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 7 ) km / hr . x - 7 = 45 = = > x = 52 km / hr . answer : c" | a ) 11 , b ) 50 , c ) 52 , d ) 65 , e ) 22 | c | divide(divide(subtract(125, multiply(multiply(7, const_0_2778), 7)), 7), const_0_2778) | multiply(n1,const_0_2778)|multiply(n1,#0)|subtract(n0,#1)|divide(#2,n1)|divide(#3,const_0_2778)| | physics |
one - third of rahul ' s savings in national savings certificate is equal to one - half of his savings in public provident fund . if he has rs . 1 , 25,000 as total savings , how much has he saved in public provident fund ? | "let savings in n . s . c and p . p . f . be rs . x and rs . ( 125000 - x ) respectively . then , = 1 / 3 x = 1 / 2 ( 125000 - x ) = x / 3 + x / 2 = 62500 = 5 x / 6 = 62500 = x = 62500 x 6 / 5 = 75000 savings in public provident fund = rs . ( 125000 - 75000 ) = rs . 50000 answer is c ." | a ) 80000 , b ) 60000 , c ) 50000 , d ) 75000 , e ) 90000 | c | multiply(add(multiply(multiply(const_100, const_10), const_100), subtract(multiply(multiply(const_100, const_10), const_100), multiply(multiply(const_2, const_100), const_100))), divide(1, add(divide(const_3, const_2), 1))) | divide(const_3,const_2)|multiply(const_10,const_100)|multiply(const_100,const_2)|add(n0,#0)|multiply(#1,const_100)|multiply(#2,const_100)|divide(n0,#3)|subtract(#4,#5)|add(#4,#7)|multiply(#8,#6)| | general |
an industrial loom weaves 0.128 metres of cloth every second . approximately , how many seconds will it take for the loom to weave 25 metres of cloth ? | let the required time be x seconds . more metres , more time ( direct proportion ) 0.128 : 25 : : 1 : x 0.128 x = 25 x 1 x = 25 / 0.128 = ( 25 x 1000 ) / 128 x = 195.31 . required time = 195 sec ( approximately ) . answer is c . | a ) 488 , b ) 204 , c ) 195 , d ) 178 , e ) 133 | c | divide(25, 0.128) | divide(n1,n0) | physics |
the sum of two numbers is 45 . the sum of their quotient and is reciprocal is 2.05 , the product of the numbers is ? | a + b = 45 a / b + b / a = 2.05 = > ( a ^ 2 + b ^ 2 ) / ab = 2.05 = > ( ( a + b ) ^ 2 β 2 ab ) / ab = 2.05 = > ( a + b ) ^ 2 = 2.05 ab + 2 ab = 4.05 ab = > ab = 45 ^ 2 / 4.05 = 500 answer : c | a ) 300 , b ) 400 , c ) 500 , d ) 600 , e ) 700 | c | add(add(const_2, const_3), add(divide(multiply(45, const_100), const_10), 45)) | add(const_2,const_3)|multiply(n0,const_100)|divide(#1,const_10)|add(n0,#2)|add(#0,#3) | general |
george went to a fruit market with certain amount of money . with this money he can buy either 50 oranges or 40 mangoes . he retains 15 % of the money for taxi fare and buys 15 mangoes . how many oranges can he buy ? | "let the amount of money be 200 let cost of 1 orange be 4 let cost of 1 mango be 5 he decides to retain 15 % of 200 = 30 for taxi fare , so he is left with 170 he buys 20 mangoes ( @ 5 ) so he spends 100 money left is 70 ( 170 - 100 ) no of oranges he can buy = 70 / 4 = > 17,5 so , george can buy 15 oranges . b" | a ) 25 , b ) 17.5 , c ) 20 , d ) 16 , e ) 12 | b | multiply(subtract(subtract(const_1, divide(15, const_100)), divide(15, 50)), 50) | divide(n2,const_100)|divide(n3,n0)|subtract(const_1,#0)|subtract(#2,#1)|multiply(n0,#3)| | general |
daal is now being sold at rate rs . 20 a kg . during last month its rate was rs 16 per kg . by how reduce percent should a family its consumption so as to keep the expenditure fixed ? | last month rate = 16 per kg this month rate = 20 per kg if 20 = 100 % then 16 = ? therefore % = ( 16 * 100 ) / 20 = 80 % to reduce consumption 100 - 80 = 20 % answer : b | a ) 10 % , b ) 20 % , c ) 30 % , d ) 40 % , e ) 50 % | b | subtract(const_100, multiply(inverse(divide(20, 16)), const_100)) | divide(n0,n1)|inverse(#0)|multiply(#1,const_100)|subtract(const_100,#2) | gain |
a certain maroon paint contains 70 percent blue pigment and 30 percent red pigment by weight . a certain green paint contains 10 percent blue pigment and 90 percent yellow pigment . when these paints are mixed to produce a brown paint , the brown paint contains 40 percent blue pigment . if the brown paint weighs 10 gra... | 10 grams of combined mixture and 40 % blue pigment means that the mixtures were mixed 50 % each . thus 5 grams a piece . out of the 5 grams of the dark blue paint , 60 % is red . therefore , 5 * . 3 = 1.5 grams of red pigment | a ) 1.5 , b ) 2 , c ) 2.5 , d ) 3 , e ) 3.5 | a | multiply(divide(70, add(const_100, const_100)), 10) | add(const_100,const_100)|divide(n0,#0)|multiply(#1,n5)| | gain |
before 4 years , dog a β s age was 4 times of dog b β s age and after 4 years , dog a β s age will be 3 times of dog b β s age . what is the difference of dog a β s age and dog b β s now ? | "a - 4 = 4 ( b - 4 ) - - > a - 4 b = - 12 . . . . . . . . . . . . . 1 a + 4 = 3 ( b + 4 ) - - > a - 3 b = 8 . . . . . . . . . . . . . 2 ( 2 ) - ( 1 ) - - > b = 20 - - > a = 3 ( 24 ) = 72 a - b = 72 - 20 = 52 answer : c" | a ) 36 , b ) 42 , c ) 52 , d ) 60 , e ) 64 | c | subtract(subtract(multiply(4, add(subtract(multiply(4, 4), 4), subtract(multiply(4, 4), 4))), subtract(multiply(4, 4), 4)), add(subtract(multiply(4, 4), 4), subtract(multiply(4, 4), 4))) | multiply(n1,n2)|multiply(n0,n1)|subtract(#0,n1)|subtract(#1,n0)|add(#2,#3)|multiply(n1,#4)|subtract(#5,#3)|subtract(#6,#4)| | general |
a cricket bat is sold for $ 850 , making a profit of $ 225 . the profit percentage would be | "225 / ( 850 - 225 ) = 225 / 625 = 9 / 25 bit more than 33 % . answer : d" | a ) 24 % , b ) 25 % , c ) 30 % , d ) 36 % , e ) 40 % | d | multiply(divide(225, subtract(850, 225)), const_100) | subtract(n0,n1)|divide(n1,#0)|multiply(#1,const_100)| | gain |
what is the least number to be added to 661 to make it a perfect square ? | "the numbers greater than 661 and are square of number is 676 . the least number that should be added 661 to make it perfect square = 676 - 661 = 15 . answer : b" | a ) 13 , b ) 15 , c ) 20 , d ) 16 , e ) 45 | b | subtract(multiply(divide(add(add(const_12, const_4), const_2), const_100), divide(add(add(const_12, const_4), const_2), const_100)), 661) | add(const_12,const_4)|add(#0,const_2)|divide(#1,const_100)|multiply(#2,#2)|subtract(#3,n0)| | general |
joe β s average ( arithmetic mean ) test score across 4 equally weighted tests was 40 . he was allowed to drop his lowest score . after doing so , his average test score improved to 45 . what is the lowest test score that was dropped ? | "the arithmetic mean of 4 equally weighted tests was 40 . so what we can assume is that we have 4 test scores , each 40 . he dropped his lowest score and the avg went to 45 . this means that the lowest score was not 40 and other three scores had given the lowest score 5 each to make it up to 40 too . when the lowest sc... | a ) 20 , b ) 25 , c ) 55 , d ) 65 , e ) 80 | b | subtract(multiply(40, 4), multiply(45, const_3)) | multiply(n0,n1)|multiply(n2,const_3)|subtract(#0,#1)| | general |
the ages of patrick and michael are in the ratio of 3 : 5 and that of michael and monica are in the ratio of 3 : 5 . if the sum of their ages is 147 , what is the difference between the ages of patrick and monica ? | "let their ages are 9 x , 15 x and 25 x respectively . difference between the ages of patrick and monica = 25 x - 9 x = 16 x where 49 x = 147 or x = 3 so diff in ages = 16 x = 16 * 3 = 48 yrs answer : a" | a ) 48 , b ) 47 , c ) 46 , d ) 49 , e ) 52 | a | subtract(multiply(multiply(5, 5), divide(147, add(add(multiply(3, 3), multiply(3, 5)), multiply(5, 5)))), multiply(multiply(3, 3), divide(147, add(add(multiply(3, 3), multiply(3, 5)), multiply(5, 5))))) | multiply(n0,n0)|multiply(n1,n0)|multiply(n1,n3)|add(#0,#1)|add(#3,#2)|divide(n4,#4)|multiply(#5,#2)|multiply(#5,#0)|subtract(#6,#7)| | general |
if the price of a certain bond on may 1 st was 2 / 3 the price of the bond on june 1 st and the price of the bond on july 1 st was 50 % greater than the price of the bond on may 1 st . then the price of the bond on june 1 st st was what percent of the average ( arithmetic mean ) price of the bond on may 1 st and july 1... | "the price on june 1 st = 12 ( assume ) ; the price on may 1 st = 2 / 3 * 12 = 8 ; the price on july 1 st = 8 * 1.50 = 12 . the average price of the bond on may 1 st and july 1 st = ( 8 + 12 ) / 2 = 10 . the price of the bond on june 1 st ( 12 ) is 6 / 5 times ( 120 % ) the average price of the bond on may 1 st and jul... | a ) 50 % , b ) 75 % , c ) 120 % , d ) 133 1 / 3 % , e ) 150 % | c | add(multiply(50, const_0_33), add(const_100, const_0_33)) | add(const_0_33,const_100)|multiply(n5,const_0_33)|add(#0,#1)| | general |
cricket match is conducted in us . the run rate of a cricket game was only 3.2 in first 10 over . what should be the run rate in the remaining 40 overs to reach the target of 252 runs ? | "required run rate = 252 - ( 3.2 x 10 ) = 220 = 5.5 40 40 d" | a ) 6 , b ) 6.25 , c ) 7.25 , d ) 5.5 , e ) 8 | d | divide(subtract(252, multiply(3.2, 10)), 40) | multiply(n0,n1)|subtract(n3,#0)|divide(#1,n2)| | gain |
the area of a square is equal to nine times the area of a rectangle of dimensions 121 cm * 144 cm . what is the perimeter of the square ? | "area of the square = s * s = 9 ( 121 * 144 ) = > s = 3 * 11 * 12 = 396 cm perimeter of the square = 4 * 396 = 1584 cm . answer : c" | a ) 1500 cm , b ) 1800 cm , c ) 1584 cm , d ) 1718 cm , e ) 1086 cm | c | multiply(sqrt(multiply(rectangle_area(121, 144), divide(144, const_2))), const_4) | divide(n1,const_2)|rectangle_area(n0,n1)|multiply(#0,#1)|sqrt(#2)|multiply(#3,const_4)| | geometry |
the credit card and a global payment processing companies have been suffering losses for some time now . a well known company recently announced its quarterly results . according to the results , the revenue fell to $ 48.0 billion from $ 69.0 billion , a year ago . by what percent did the revenue fall ? | "$ 69 - $ 48 = 21 $ ( 21 / 69 ) * 100 = 30.434 % answer : b" | a ) 20.8 , b ) 30.4 , c ) 31.8 , d ) 32.5 , e ) 33.8 | b | multiply(divide(subtract(69.0, 48.0), 69.0), const_100) | subtract(n1,n0)|divide(#0,n1)|multiply(#1,const_100)| | gain |
a and b started a business investing rs . 90,000 and rs 30,000 respectively . in what ratio the profit earned after 2 years be divided between a and b respectively ? | "a : b = 90000 : 30000 = 9 : 3 = 3 : 1 answer : b" | a ) 9 : 2 , b ) 3 : 1 , c ) 18 : 20 , d ) 18 : 4 , e ) 17 : 4 | b | divide(add(multiply(multiply(const_3, const_3), add(multiply(const_3, const_3), const_1)), 2), multiply(2, add(multiply(const_3, const_3), const_1))) | multiply(const_3,const_3)|add(#0,const_1)|multiply(#1,#0)|multiply(n2,#1)|add(n2,#2)|divide(#4,#3)| | gain |
what is the time required for a bike to cover a distance of 450 km with a speed of 90 kmh ? | time = distance / speed time = 450 / 90 = 5 answer : a | a ) 5 hours , b ) 3 hours , c ) 8 hours , d ) 6 hours , e ) 4 hours | a | divide(450, 90) | divide(n0,n1) | physics |
how many even multiples of 25 are there between 249 and 501 ? | "250 = 10 * 25 500 = 20 * 25 the even multiples are 25 multiplied by 10 , 12 , 14 , 16 , 18 , and 20 for a total of 6 . the answer is b ." | a ) 5 , b ) 6 , c ) 9 , d ) 10 , e ) 11 | b | add(divide(subtract(501, 249), multiply(25, const_2)), const_1) | multiply(n0,const_2)|subtract(n2,n1)|divide(#1,#0)|add(#2,const_1)| | general |
tanks a and b are each in the shape of a right circular cylinder . the interior of tank a has a height of 8 meters and a circumference of 8 meters , and the interior of tank b has a height of 8 meters and a circumference of 10 meters . the capacity of tank a is what percent of the capacity of tank b ? | "for a , r = 8 / 2 pi . its capacity = ( 4 pi ) ^ 2 * 8 = 128 pi for b , r = 10 / pi . its capacity = ( 5 pi ) ^ 2 * 8 = 200 pi a / b = 128 pi / 200 pi = 0.64 a" | a ) 64 % , b ) 80 % , c ) 100 % , d ) 120 % , e ) 125 % | a | multiply(multiply(power(divide(8, 10), const_2), divide(8, 8)), const_100) | divide(n0,n2)|divide(n1,n3)|power(#1,const_2)|multiply(#0,#2)|multiply(#3,const_100)| | physics |
in n is a positive integer less than 200 , and 14 n / 60 is an integer , then n has how many different positive prime factors c ? | "i like to put the numbers in prime factors so it is easier and faster to visualize . 14 * n / 60 if we write the factors of 14 - - > 2 , 7 , and the factors of 60 - - > 2 , 2 , 3 , 5 , we have ( 2 * 7 * n ) / ( 2 ^ 2 * 3 * 5 ) simplifying 7 * n / ( 2 * 3 * 5 ) the only way the equation above has an integer value is if... | a ) 2 , b ) 3 , c ) 5 , d ) 6 , e ) 8 | b | add(divide(add(const_1, const_4), divide(divide(divide(60, const_2), const_2), const_3)), const_2) | add(const_1,const_4)|divide(n2,const_2)|divide(#1,const_2)|divide(#2,const_3)|divide(#0,#3)|add(#4,const_2)| | general |
if the average ( arithmetic mean ) of a and b is 110 , and the average of b and c is 150 , what is the value of a β c ? | "question : a - c = ? ( a + b ) / 2 = 110 = = = > a + b = 220 ( b + c ) / 2 = 150 = = = > b + c = 300 ( a + b ) - ( b + c ) = 220 - 300 = = = > a + b - b - c = - 80 = = = > a - c = - 80 answer : d" | a ) β 220 , b ) β 100 , c ) 100 , d ) β 80 , e ) it can not be determined from the information given | d | subtract(multiply(150, const_2), multiply(110, const_2)) | multiply(n1,const_2)|multiply(n0,const_2)|subtract(#0,#1)| | general |
the average of marks obtained by 120 boys was 40 . if the average of marks of passed boys was 39 and that of failed boys was 15 , the number of boys who passed the examination is ? | "let the number of boys who passed = x . then , 39 x x + 15 x ( 120 - x ) = 120 x 40 24 x = 4800 - 1800 = > x = 3000 / 24 x = 125 hence , the number of boys passed = 125 . answer : c" | a ) 100 , b ) 110 , c ) 125 , d ) 130 , e ) 145 | c | divide(subtract(multiply(40, 120), multiply(120, 15)), subtract(39, 15)) | multiply(n0,n1)|multiply(n0,n3)|subtract(n2,n3)|subtract(#0,#1)|divide(#3,#2)| | general |
in a group of 27 junior high school students , 7 take french , 10 take spanish , and 4 take both languages . the students taking both french and spanish are not counted with the 7 taking french or the 10 taking spanish . how many students are not taking either french or spanish ? | "a 7 add 7 + 10 + 4 to get 21 . then subtract 21 from the total students β 27 β 21 = 6 . answer is e" | a ) 7 , b ) 6 , c ) 9 , d ) 4 , e ) 6 | e | subtract(27, add(7, 10)) | add(n1,n2)|subtract(n0,#0)| | other |
angelo and isabella are both salespersons . in any given week , angelo makes $ 550 in base salary plus 8 percent of the portion of his sales above $ 4,000 for that week . isabella makes 10 percent of her total sales for any given week . for what amount of weekly sales would angelo and isabella earn the same amount of m... | "official solution : the problem asks for the amount of weekly sales it takes for angelo and isabella to earn the same amount of money . you can write an equation that sets angelo β s and isabella β s weekly earnings equal to each other , with x representing weekly sales . weekly earnings for each salesperson equal bas... | a ) 11,500 , b ) 24,500 , c ) 25,500 , d ) 26,500 , e ) 27,500 | a | floor(divide(divide(subtract(550, multiply(4,000, divide(8, const_100))), subtract(divide(10, const_100), divide(8, const_100))), 4,000)) | divide(n1,const_100)|divide(n3,const_100)|multiply(#0,n2)|subtract(#1,#0)|subtract(n0,#2)|divide(#4,#3)|divide(#5,n2)|floor(#6)| | general |
the points a ( 0 , 0 ) , b ( 0 , 4 a - 4 ) and c ( 2 a + 1 , 2 a + 6 ) form a triangle . if angle abc = 90 , what is the area of triangle abc ? | "1 / 2 bh = 1 / 2 ( 2 a + 1 ) ( 2 a + 6 ) now 4 a - 4 = 2 a + 6 2 a = 10 , a = 5 therefore , a ( 0,0 ) ; b ( 0,16 ) ; c ( 11,16 ) 1 / 2 * 11 * 16 = 88 answer : a" | a ) 88 , b ) 98 , c ) 108 , d ) 78 , e ) 90 | a | divide(multiply(subtract(multiply(divide(add(6, 4), 4), 4), 4), add(multiply(divide(add(6, 4), 4), subtract(4, 4)), 1)), 4) | add(n4,n8)|subtract(n3,n4)|divide(#0,n4)|multiply(#2,#1)|multiply(n3,#2)|add(#3,n6)|subtract(#4,n4)|multiply(#5,#6)|divide(#7,n4)| | geometry |
machine a takes 100 hours to complete a certain job and starts that job at 7 am . after two hour of working alone , machine a is joined by machine b and together they complete the job at 6 pm . how long would it have taken machine b to complete the jobif it had worked alone for the entire job ? | "let us assume total job = 100 units a finishes 100 units in 100 hrs ( given ) hence a ( working rate ) = 1 units / hr now given that a works for 2 hr ( so 2 units done ) then a and b finish total work in 15 hours . hence a and b finish 98 units in 9 hours . of these 1 x 9 = 9 units were done by a . hence b did 89 unit... | a ) 10.11 , b ) 9.99 , c ) 11.1 , d ) 44.11 , e ) 1.01 | a | inverse(subtract(divide(subtract(const_1, multiply(inverse(100), const_2)), subtract(subtract(multiply(6, const_2), 7), const_2)), inverse(100))) | inverse(n0)|multiply(n2,const_2)|multiply(#0,const_2)|subtract(#1,n1)|subtract(const_1,#2)|subtract(#3,const_2)|divide(#4,#5)|subtract(#6,#0)|inverse(#7)| | physics |
average age of students of an adult school is 46 years . 120 new students whose average age is 32 years joined the school . as a result the average age is decreased by 4 years . find the number of students of the school after joining of the new students . | "explanation : let the original no . of students be x . according to situation , 46 x + 120 * 32 = ( x + 120 ) 36 β x = 48 so , required no . of students after joining the new students = x + 120 = 168 . answer : d" | a ) 1200 , b ) 120 , c ) 360 , d ) 168 , e ) none of these | d | multiply(divide(subtract(multiply(add(32, 4), 120), multiply(120, 32)), subtract(46, add(32, 4))), 4) | add(n2,n3)|multiply(n1,n2)|multiply(n1,#0)|subtract(n0,#0)|subtract(#2,#1)|divide(#4,#3)|multiply(n3,#5)| | general |
on a map , 3.5 inches represent 35 miles . how many miles approximately is the distance if you measured 127 centimeters assuming that 1 - inch is 2.54 centimeters ? | "1 inch = 2.54 cm 3.5 inch = 2.54 * 3.5 cm 8.89 cm = 35 miles 127 cms = 35 / 8.89 * 127 = 500 miles answer : b" | a ) 515 , b ) 500 , c ) 525 , d ) 530 , e ) 535 | b | multiply(divide(127, 2.54), divide(35, 3.5)) | divide(n2,n4)|divide(n1,n0)|multiply(#0,#1)| | physics |
how long does a train 110 m long traveling at 60 kmph takes to cross a bridge of 140 m in length ? | "d = 110 + 140 = 250 m s = 60 * 5 / 18 = 50 / 3 t = 250 * 3 / 50 = 15 sec answer : c" | a ) 16.5 , b ) 16.0 , c ) 15 , d ) 16.8 , e ) 16.1 | c | divide(add(110, 140), multiply(60, const_0_2778)) | add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)| | physics |
if the product of two numbers is 51216 and their h . c . f . is 33 , find their l . c . m . | "explanation : hcf * lcm = 51216 , because we know product of two numbers = product of hcf and lcm lcm = 51216 / 88 = 582 option c" | a ) 600 , b ) 680 , c ) 582 , d ) 560 , e ) none of these | c | divide(51216, 33) | divide(n0,n1)| | physics |
at an elementary school , 70 % of the faculty members are women and 40 % of the faculty members are married . if 1 β 3 of the men are single , what fraction of the women are married ? | "- - - - - - - - - - - - - - - - - - - - m - - - - - - w - - - - - - - - total marrried - - - - - - - - - - 20 - - - - - 20 - - - - - - - - - 40 not married - - - - - 10 - - - - - 50 - - - - - - - - - 60 total - - - - - - - - - - - - - 30 - - - - - 70 - - - - - - - - 100 need married woman / total woman , so 20 / 70 = ... | a ) 5 β 7 , b ) 7 β 10 , c ) 2 / 7 , d ) 7 β 30 , e ) 5 β 70 | c | subtract(const_1, divide(1, 3)) | divide(n2,n3)|subtract(const_1,#0)| | gain |
a train which has 300 m long , is running 80 kmph . in what time will it cross a person moving at 16 kmph in same direction ? | time taken to cross a moving person = length of train / relative speed time taken = 300 / ( ( 80 - 16 ) ( 5 / 18 ) = 300 / 64 * ( 5 / 18 ) = 300 / 20 = 15 sec answer : c | a ) 25 sec , b ) 35 sec , c ) 15 sec , d ) 45 sec , e ) 20 sec | c | divide(300, subtract(divide(80, const_3_6), divide(divide(16, const_2), const_3_6))) | divide(n1,const_3_6)|divide(n2,const_2)|divide(#1,const_3_6)|subtract(#0,#2)|divide(n0,#3) | physics |
what is the number of integers from 1 to 500 ( inclusive ) that are divisible by neither 10 nor by 15 ? | in 500 consecutive numbers , number of multiples of 10 = 500 / 10 = 50 ( ignore decimals ) in 500 consecutive numbers , number of multiples of 15 = 500 / 15 = 33 number of multiples of 10 * 15 i . e . 150 = 500 / 150 = 3 number of integers from 1 to 500 that are divisible by neither 10 nor by 15 = 500 - ( 50 + 33 - 3 )... | a ) 170 , b ) 250 , c ) 300 , d ) 560 , e ) 420 | e | subtract(500, subtract(add(divide(500, 10), divide(500, 15)), divide(500, multiply(10, 15)))) | divide(n1,n2)|divide(n1,n3)|multiply(n2,n3)|add(#0,#1)|divide(n1,#2)|subtract(#3,#4)|subtract(n1,#5) | other |
495 x 495 = ? | "495 x 495 = ( 495 ) 2 = ( 500 - 5 ) 2 = ( 500 ) 2 + ( 5 ) 2 - ( 2 x 500 x 5 ) = 250000 + 25 - 5000 = 250025 - 5000 = 245025 . d )" | a ) 250000 , b ) 250025 , c ) 5000 , d ) 245025 , e ) 240000 | d | multiply(divide(495, 495), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
the average weight of a group of boys is 30 kg . after a boy of weight 33 kg joins the group , the average weight of the group goes up by 1 kg . find the number of boys in the group originally ? | "let the number off boys in the group originally be x . total weight of the boys = 30 x after the boy weighing 33 kg joins the group , total weight of boys = 30 x + 33 so 30 x + 33 = 31 ( x + 1 ) = > x = 2 . answer : c" | a ) 4 , b ) 5 , c ) 2 , d ) 7 , e ) 8 | c | add(subtract(33, add(30, 1)), 1) | add(n0,n2)|subtract(n1,#0)|add(#1,n2)| | general |
what is the number of integers from 11 to 1010 ( inclusive ) that are divisible by neither 12 nor by 15 ? | "in 1000 consecutive numbers , number of multiples of 12 = 1000 / 12 = 83 ( ignore decimals ) in 1000 consecutive numbers , number of multiples of 15 = 1000 / 15 = 66 number of multiples of 12 * 15 i . e . 180 = 1000 / 180 = 5 number of integers from 11 to 1010 that are divisible by neither 12 nor by 15 = 1000 - ( 83 +... | a ) 565 , b ) 856 , c ) 783 , d ) 590 , e ) 300 | b | subtract(1010, subtract(add(divide(1010, 12), divide(1010, 15)), divide(1010, multiply(12, 15)))) | divide(n1,n2)|divide(n1,n3)|multiply(n2,n3)|add(#0,#1)|divide(n1,#2)|subtract(#3,#4)|subtract(n1,#5)| | other |
the ages of ashley and mary are in the ratio 4 : 7 . the sum of their ages is 22 . find the age of mary . | "a : m = 4 : 7 let the ages of a & m be x & 22 - x x / 22 - x = 4 / 7 x = 8 a = 8 yrs m = 14 yrs answer : d" | a ) 11 yrs , b ) 12 yrs , c ) 13 yrs , d ) 14 yrs , e ) 15 yrs | d | multiply(divide(22, add(4, 7)), 4) | add(n0,n1)|divide(n2,#0)|multiply(n0,#1)| | general |
a rectangular field has a length 10 meters more than it is width . if the area of the field is 171 , what is the length ( in meters ) of the rectangular field ? | area = l * w = ( l ) * ( l - 10 ) = 171 trial and error : 20 * 10 = 200 ( too high ) 19 * 9 = 171 the length is 19 meters . the answer is b . | ['a ) 17', 'b ) 19', 'c ) 21', 'd ) 23', 'e ) 25'] | b | add(10, add(const_0_25, add(const_0_33, divide(divide(171, 10), const_2)))) | divide(n1,n0)|divide(#0,const_2)|add(#1,const_0_33)|add(#2,const_0_25)|add(n0,#3) | geometry |
a man is 26 years older than his son . in two years , his age will be twice the age of his son . the present age of the son is ? | "let the son ' s present age be x years . then , man ' s present age = ( x + 26 ) years . ( x + 26 ) + 2 = 2 ( x + 2 ) x + 28 = 2 x + 4 = > x = 24 . answer : d" | a ) 16 , b ) 19 , c ) 15 , d ) 24 , e ) 10 | d | divide(subtract(26, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | multiply(const_2,const_2)|subtract(const_2,const_1)|subtract(#0,const_2)|subtract(n0,#2)|divide(#3,#1)| | general |
if a 2 + b 2 + c 2 = 250 and ab + bc + ca = 3 , then a + b + c is | "by formula , ( a + b + c ) ^ 2 = a ^ 2 + b ^ 2 + c ^ 2 + 2 ( ab + bc + ca ) , since , a ^ 2 + b ^ 2 + c ^ 2 = 250 and ab + bc + ca = 3 , ( a + b + c ) ^ 2 = 250 + 2 ( 3 ) = 256 = 16 ^ 2 therefore : a + b + c = 16 answer : c" | a ) 4 , b ) 8 , c ) 16 , d ) 32 , e ) 64 | c | sqrt(add(250, multiply(3, 2))) | multiply(n4,n0)|add(n3,#0)|sqrt(#1)| | general |
if x is a positive number and 5 the square root of x is equal to x / 25 , then x = | "5 of sqrt ( x ) = x / 25 , which means that sqrt ( x ) = x / 5 or x = x ^ 2 / 25 - > divide by x 1 = x / 25 x = 25 b ." | a ) 10 , b ) 25 , c ) 5 , d ) 2 , e ) 1 | b | power(25, multiply(25, const_3)) | multiply(n1,const_3)|power(n1,#0)| | general |
if sales tax is reduced from 5 % to 4 % , then what difference it will make if you purchase an item of rs . 1000 | explanation : clue : answer will be 5 % of 1000 - 4 % of 1000 answer : option a | a ) 10 , b ) 20 , c ) 30 , d ) 40 , e ) 50 | a | subtract(divide(multiply(1000, 5), const_100), divide(multiply(1000, 4), const_100)) | multiply(n0,n2)|multiply(n1,n2)|divide(#0,const_100)|divide(#1,const_100)|subtract(#2,#3) | general |
of all the homes on gotham street , 1 / 3 are termite - ridden , and 1 / 4 of these are collapsing . what fraction of the homes are termite - ridden , but not collapsing ? | "the fraction of homes which are termite - infested but not collapsing is 3 / 4 * 1 / 3 = 1 / 4 the answer is c ." | a ) 1 / 3 , b ) 2 / 5 , c ) 1 / 4 , d ) 1 / 6 , e ) 3 / 4 | c | divide(add(1, const_2.0), multiply(3, 4)) | add(n0,const_2.0)|multiply(n1,n3)|divide(#0,#1)| | general |
convert 100 miles into yards ? | "1 mile = 1760 yards 100 miles = 100 * 1760 = 176000 yards answer is e" | a ) 25630 yards , b ) 35200 yards , c ) 39520 yards , d ) 42560 yards , e ) 176000 yards | e | divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 100), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2))) | add(const_2,const_3)|multiply(#0,const_2)|multiply(#1,#1)|multiply(n0,#2)|divide(#3,#2)| | physics |
take number 1000 and then add 20 to it . now add 1000 one more time . now add 30 . now add 1000 one more time . now add 40 . now add 1000 one more time . now add 10 . what is the total ? | a 4100 | a ) 4100 , b ) 3590 , c ) 2800 , d ) 3700 , e ) 3600 | a | add(add(add(add(1000, 1000), 1000), 1000), add(add(add(20, 30), 40), 10)) | add(n0,n0)|add(n1,n3)|add(n0,#0)|add(n5,#1)|add(n0,#2)|add(n7,#3)|add(#4,#5) | general |
elena β s bread recipe calls for 2 ounces of butter for each 5 cups of flour used . she needs to make 4 times the original recipe . if 12 ounces of butter is used , then how many cups of flour are needed ? | "solving through algebra route : 2 b + 5 f = x amount if we multiply this equation with 4 we get : 8 b + 20 f = 4 x therefore , we got 8 ounces of butter and 4 x amount of quantity when we use 20 ounces of floor . ans : e" | a ) 1 , b ) 4 , c ) 9 , d ) 13 , e ) 20 | e | multiply(5, 4) | multiply(n1,n2)| | general |
three walls have wallpaper covering a combined area of 300 square meters . by overlapping the wallpaper to cover a wall with an area of 180 square meters , the area that is covered by exactly two layers of wallpaper is 38 square meters . what is the area that is covered with three layers of wallpaper ? | 300 - 180 = 120 sq m of the wallpaper overlaps ( in either two layers or three layers ) if 36 sq m has two layers , 120 - 38 = 82 sq m of the wallpaper overlaps in three layers . 82 sq m makes two extra layers hence the area over which it makes two extra layers is 41 sq m . answer ( e ) . | ['a ) 5 square meters', 'b ) 36 square meters', 'c ) 42 square meters', 'd ) 83.3 square meters', 'e ) 41 square meters'] | e | divide(subtract(subtract(300, 180), 38), const_2) | subtract(n0,n1)|subtract(#0,n2)|divide(#1,const_2) | geometry |
a dog is tied to a tree by a long nylon cord . if the dog runs from the due north side of the tree to the due south side of the tree with the cord extended to its full length at all items , and the dog ran approximately 30 feet , what was the approximate length of the nylon cord l in feet ? | "because the cord was extended to its full length at all items , the dog ran along a semi - circular path , from north to south . the circumference of a full circle is 2 * pi * r , but since we only care about the length of half the circle , the semi - circle path is pi * r . l = pi * r = 30 . round pi = 3 , then r = 1... | a ) 30 , b ) 25 , c ) 15 , d ) 10 , e ) 5 | d | divide(30, const_3) | divide(n0,const_3)| | general |
all factors of a positive integer x are multiplied and the product obtained is x ^ 3 . if x is greater than 1 , how many factors does x have ? | the question can be done in seconds if you understand the properties of factors of a number . note that every distinct factor of a number which is not a perfect square has a complementary factor . for example , factors of 6 : 1 , 2 , 3 , 6 1 has a complementary factor 6 such that 1 * 6 = 6 ( the original number ) 2 has... | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | e | add(3, const_3) | add(n0,const_3) | general |
mike needs 30 % to pass . if he scored 212 marks and falls short by 19 marks , what was the maximum marks he could have got ? | "if mike had scored 19 marks more , he could have scored 30 % therefore , mike required 212 + 19 = 231 marks let the maximum marks be m . then 30 % of m = 231 ( 30 / 100 ) Γ m = 231 m = ( 231 Γ 100 ) / 30 m = 23100 / 30 m = 770 answer : c" | a ) 543 , b ) 677 , c ) 770 , d ) 867 , e ) 832 | c | divide(add(212, 19), divide(30, const_100)) | add(n1,n2)|divide(n0,const_100)|divide(#0,#1)| | gain |
mary works in a restaurant a maximum of 40 hours . for the first 20 hours , she is paid $ 8 per hour . for each overtime hour , she is paid at a rate which is 25 % higher than her regular rate . how much mary can earn in a week ? | "mary receives $ 8 ( 20 ) = $ 160 for the first 20 hours . for the 20 overtime hours , she receives $ 8 ( 0.25 ) + $ 8 = $ 10 per hour , that is $ 10 ( 20 ) = $ 200 . the total amount is $ 160 + $ 200 = $ 360 answer c 360 ." | a ) 300 , b ) 420 , c ) 360 , d ) 320 , e ) 400 | c | add(multiply(40, 8), multiply(subtract(40, 20), multiply(8, divide(25, const_100)))) | divide(n3,const_100)|multiply(n0,n2)|subtract(n0,n1)|multiply(n2,#0)|multiply(#3,#2)|add(#1,#4)| | gain |
sonika deposited rs . 13000 which amounted to rs . 10200 after 4 years at simple interest . had the interest been 5 % more . she would get how much ? | "( 13000 * 4 * 5 ) / 100 = 2600 10200 - - - - - - - - 12800 answer : c" | a ) 18990 , b ) 13002 , c ) 12800 , d ) 12300 , e ) 14500 | c | add(multiply(multiply(add(divide(5, const_100), divide(divide(subtract(10200, 13000), 4), 13000)), 13000), 4), 13000) | divide(n3,const_100)|subtract(n1,n0)|divide(#1,n2)|divide(#2,n0)|add(#0,#3)|multiply(n0,#4)|multiply(n2,#5)|add(n0,#6)| | gain |
a is two years older than b who is twice as old as c . if the total of the ages of a , b and c be 47 , the how old is b ? | "let c ' s age be x years . then , b ' s age = 2 x years . a ' s age = ( 2 x + 2 ) years . ( 2 x + 2 ) + 2 x + x = 47 5 x = 45 x = 9 . hence , b ' s age = 2 x = 19 years . d )" | a ) 10 years , b ) 12 years , c ) 14 years , d ) 18 years , e ) 16 years | d | divide(multiply(subtract(47, const_2), const_2), add(const_4, const_1)) | add(const_1,const_4)|subtract(n0,const_2)|multiply(#1,const_2)|divide(#2,#0)| | general |
a shopkeeper sells 400 metres of cloth for rs . 18000 at a loss of rs . 5 per metre . find his cost price for one metre of cloth ? | "sp per metre = 18000 / 400 = rs . 45 loss per metre = rs . 5 cp per metre = 45 + 5 = rs . 50 answer : d" | a ) rs . 59 , b ) rs . 58 , c ) rs . 09 , d ) rs . 50 , e ) rs . 13 | d | add(divide(18000, 400), 5) | divide(n1,n0)|add(n2,#0)| | gain |
the positive integers m and n leave remainders of 2 and 3 , respectively , when divided by 6 . m > n . what is the remainder when m β n is divided by 6 ? | m is 2 more than a multiple of 6 ; n is 3 more than a multiple of 6 . m - n will be 1 less than a multiple of 6 ( 5 , 11 , 17 , . . . ) , therefore m - n will yield a remainder of 5 when divided by 6 . answer : e . | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | e | add(6, subtract(2, 3)) | subtract(n0,n1)|add(n2,#0) | physics |
the mean of 50 observations was 36 . it was found later that an observation 44 was wrongly taken as 23 . the corrected new mean is | "solution correct sum = ( 36 x 50 + 44 - 23 ) = 1821 . Γ’ Λ Β΄ correct mean = 1821 / 50 = 36.42 . answer d" | a ) 35.22 , b ) 36.12 , c ) 36.22 , d ) 36.42 , e ) none | d | divide(add(multiply(36, 50), subtract(subtract(50, const_2), 23)), 50) | multiply(n0,n1)|subtract(n0,const_2)|subtract(#1,n3)|add(#0,#2)|divide(#3,n0)| | general |
the ratio of the length and the width of a rectangle is 4 : 3 and the area of the rectangle is 5292 sq cm . what is the ratio of the width and the area of the rectangle ? | "let the length and the width be 4 x and 3 x respectively . area = ( 4 x ) ( 3 x ) = 5292 12 x ^ 2 = 5292 x ^ 2 = 441 x = 21 the ratio of the width and the area is 3 x : 12 x ^ 2 = 1 : 4 x = 1 : 84 the answer is d ." | a ) 1 : 72 , b ) 1 : 76 , c ) 1 : 80 , d ) 1 : 84 , e ) 1 : 88 | d | divide(divide(sqrt(multiply(3, 5292)), const_2), 5292) | multiply(n2,n1)|sqrt(#0)|divide(#1,const_2)|divide(#2,n2)| | geometry |
is two years older than b who is twice as old as c . if the total of the ages of a , b and c be 27 , then how old is b ? | "let c ' s age be x years . then , b ' s age = 2 x years . a ' s age = ( 2 x + 2 ) years . ( 2 x + 2 ) + 2 x + x = 27 5 x = 25 = > x = 5 hence , b ' s age = 2 x = 10 years . answer : d" | a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | d | multiply(divide(subtract(27, const_2), add(const_3, const_2)), const_2) | add(const_2,const_3)|subtract(n0,const_2)|divide(#1,#0)|multiply(#2,const_2)| | general |
if a * b = 2 a β 3 b + ab , then 5 * 7 + 7 * 5 is equal to : | 5 * 7 + 7 * 5 = ( 2 * 5 β 5 * 7 + 5 * 7 ) + ( 2 * 7 β 5 * 5 + 7 * 5 ) = ( 10 + 14 β 25 + 35 ) = 34 . answer : a | a ) 34 , b ) 18 , c ) 72 , d ) 78 , e ) 82 | a | multiply(5, 7) | multiply(n2,n3) | general |
two family reunions are happening at the park avenue hotel , the oates reunion and the hall reunion . all 100 guests at the hotel attend at least one of the reunions . if 40 people attend the oates reunion and 70 people attend the hall reunion , how many people attend both reunions | no of people in oates reunion = 40 no of people in hall reunion = 70 attending both = x all guests attend at least one . therefore , 100 = 40 + 70 - ( both ) both = 10 answer c | a ) 2 , b ) 5 , c ) 10 , d ) 16 , e ) 22 | c | subtract(add(40, 70), 100) | add(n1,n2)|subtract(#0,n0) | other |
a whale goes on a feeding frenzy that lasts for 9 hours . for the first hour he catches and eats x kilos of plankton . in every hour after the first , it consumes 4 kilos of plankton more than it consumed in the previous hour . if by the end of the frenzy the whale will have consumed a whopping accumulated total 450 ki... | if you list the amount eaten each hour , you ' ll get an equally spaced list , increasing by 4 each hour . in any equally spaced list , the median equals the mean . here , the mean is 450 / 9 = 50 , so the median is also 50 , and that is the amount eaten in the 5 th hour . we need to add 4 to find the total eaten in th... | a ) 38 , b ) 47 , c ) 50 , d ) 53 , e ) 54 | e | add(divide(450, 9), 4) | divide(n2,n0)|add(n1,#0) | general |
a waiter ' s salary consists of his salary and tips . during one week his tips were 9 / 4 of his salary . what fraction of his income came from tips ? | "income = salary ( s ) + tips = s + s * 9 / 4 = s * 13 / 4 tips = s * 9 / 4 fraction of his income came from tips = ( s * 9 / 4 ) / ( s * 13 / 4 ) = 9 / 13 answer : c" | a ) 4 / 9 , b ) 5 / 4 , c ) 9 / 13 , d ) 5 / 9 , e ) 6 / 9 | c | divide(divide(9, 4), add(divide(9, 4), const_1)) | divide(n0,n1)|add(#0,const_1)|divide(#0,#1)| | general |
the largest 5 digit number exactly divisible by 91 is : | "explanation : largest 5 - digit number = 99999 91 ) 99999 ( 1098 91 - - - 899 819 - - - - 809 728 - - - 81 - - - required number = ( 99999 - 81 ) = 99918 . answer : d" | a ) 25698 , b ) 581396 , c ) 369741 , d ) 99918 , e ) 1258 | d | square_area(const_pi) | square_area(const_pi)| | general |
the average age of a husband and his wife was 23 years at the time of their marriage . after 5 years they have a one - year old child . the average age of the family now is : | explanation : sum of the present ages of husband , wife and child = ( 23 * 2 + 5 * 2 ) + 1 = 57 years . required average = ( 57 / 3 ) = 19 years . answer : c | a ) 25 , b ) 23 , c ) 19 , d ) 18 , e ) 17 | c | divide(add(add(add(multiply(23, const_2), 5), 5), const_1), const_3) | multiply(n0,const_2)|add(n1,#0)|add(n1,#1)|add(#2,const_1)|divide(#3,const_3) | general |
what sum of money put at c . i amounts in 2 years to rs . 8820 and in 3 years to rs . 9261 ? | "explanation : 8820 - - - - 441 100 - - - - ? = > 5 % x * 105 / 100 * 105 / 100 = 8820 x * 1.1025 = 8820 x = 8820 / 1.1025 = > 8000 answer is a" | a ) rs . 8000 , b ) rs . 8400 , c ) rs . 7500 , d ) rs . 7800 , e ) none of these | a | divide(8820, power(add(subtract(divide(9261, 8820), const_1), const_1), 2)) | divide(n3,n1)|subtract(#0,const_1)|add(#1,const_1)|power(#2,n0)|divide(n1,#3)| | general |
if the length of a rectangular field is 30 metres more than its breadth and the perimeter of the field is 540 metres , what is the area of the field in square metres ? | let breadth is x then length is 30 + x now the perimeter = 2 ( x + 30 + x ) = 540 solving this we get x = 120 now the area is 120 * 150 = 18000 answer : a | ['a ) 18000', 'b ) 40000', 'c ) 31000', 'd ) 19000', 'e ) 16000'] | a | rectangle_area(square_edge_by_perimeter(subtract(540, multiply(30, const_2))), add(square_edge_by_perimeter(subtract(540, multiply(30, const_2))), 30)) | multiply(n0,const_2)|subtract(n1,#0)|square_edge_by_perimeter(#1)|add(n0,#2)|rectangle_area(#3,#2) | geometry |
what is the greater of the two numbers whose product is 2560 , given that the sum of the two numbers exceeds their difference by 32 ? | "let the greater and the smaller number be g and s respectively . gs = 2560 g + s exceeds g - s by 32 i . e . , g + s - ( g - s ) = 32 i . e . , 2 s = 32 = > s = 16 g = 2560 / s = 160 . answer : c" | a ) a ) 90 , b ) b ) 100 , c ) c ) 160 , d ) d ) 70 , e ) of these | c | divide(2560, multiply(power(const_2, const_4), const_2)) | power(const_2,const_4)|multiply(#0,const_2)|divide(n0,#1)| | general |
the profits of qrs company rose 10 % from march to april , then dropped 20 % from april to may , then rose 50 % from may to june . what was the percent increase for the whole quarter , from march to june ? | "we need multipliers . multiplier for a 10 % increases = 1 + 0.10 = 1.1 multiplier for a 20 % decreases = 1 β 0.20 = 0.8 multiplier for a 50 % increases = 1 + 0.50 = 1.5 now , multiply these . first , multiply ( 0.8 ) and ( 1.5 ) , using the doubling & halving trick . half of 0.80 is 0.40 , and twice 1.5 is 3 ( 0.8 ) *... | a ) 15 % , b ) 32 % , c ) 40 % , d ) 62 % , e ) 80 % | b | multiply(const_100, subtract(multiply(add(const_1, divide(50, const_100)), multiply(add(const_1, divide(10, const_100)), subtract(const_1, divide(20, const_100)))), const_1)) | divide(n2,const_100)|divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|add(#1,const_1)|subtract(const_1,#2)|multiply(#4,#5)|multiply(#3,#6)|subtract(#7,const_1)|multiply(#8,const_100)| | gain |
what will be the cost of building a fence around a circular plot with area equal to 289 sq ft , if the price per foot of building the fence is rs . 58 ? | let the side of the circular plot be a ft . pi * ( r ) ^ 2 = 289 = > r = 9.59 length of the fence = perimeter of the plot = 2 * pi * r = 60.26 ft . cost of building the fence = 60.26 * 58 = rs . 3495.28 answer : b | ['a ) a ) rs . 3944', 'b ) b ) rs . 3495.28', 'c ) c ) rs . 3988', 'd ) d ) rs . 3949', 'e ) e ) rs . 3923'] | b | multiply(circumface(sqrt(divide(289, const_pi))), 58) | divide(n0,const_pi)|sqrt(#0)|circumface(#1)|multiply(n1,#2) | geometry |
the amount of principal rs . 4000 at compound interest at the ratio of 10 % p . a . for 3 years is | 5324 answer : c | a ) rs . 5342 , b ) rs . 4325 , c ) rs . 5324 , d ) rs . 3425 , e ) rs . 3525 | c | multiply(4000, power(add(const_1, divide(10, const_100)), const_3)) | divide(n1,const_100)|add(#0,const_1)|power(#1,const_3)|multiply(n0,#2) | gain |
twenty eight meters of wire is available to fence off a flower bed in the form of a circular sector . what must the radius of the circle in meters be , if we wish to have a flower bed with the greatest possible surface area ? | area of sector , a = x / 360 * pi * r ^ 2 circumference of the sector = 28 = > x / 360 * 2 * pi * r + 2 r = 28 = > 2 a / r + 2 r = 28 = > a = r 14 - r ^ 2 = r 14 - r ^ 2 we will now max using derivations max value of a will found at a = 0 i . e 14 - 2 r = 0 r = 7 d | ['a ) 2 β 2', 'b ) 2 β 5', 'c ) 5', 'd ) 7', 'e ) none of these'] | d | divide(divide(multiply(add(const_10, const_4), const_2), const_2), const_2) | add(const_10,const_4)|multiply(#0,const_2)|divide(#1,const_2)|divide(#2,const_2) | geometry |
if 2 x = 3 y = 20 , then 12 xy = ? | "2 x = 20 ; x = 10 3 y = 20 ; y = 20 / 3 multiply : 12 xy = 12 * 10 * 20 / 3 = 800 . answer : b ." | a ) 1,200 , b ) 800 , c ) 120 , d ) 40 , e ) 20 | b | divide(subtract(divide(multiply(2, const_100), const_2), const_2), add(divide(multiply(2, const_100), const_2), const_2)) | multiply(n0,const_100)|divide(#0,const_2)|add(#1,const_2)|subtract(#1,const_2)|divide(#3,#2)| | general |
when a number is divided by 6 & then multiply by 12 the answer is 12 what is the no . ? | "if $ x $ is the number , x / 6 * 12 = 12 = > 2 x = 12 = > x = 6 e" | a ) 4.5 , b ) 5 , c ) 5.5 , d ) 5.8 , e ) 6 | e | multiply(divide(12, 12), 6) | divide(n2,n1)|multiply(n0,#0)| | general |
divide $ 1162 among a , b , c in the ratio 35 : 28 : 20 . | "sum of ratio terms = ( 35 + 28 + 20 ) = 83 . a ' s share = $ ( 1162 x ( 35 / 83 ) ) = $ 490 ; b ' s share = $ ( 1162 x ( 28 / 83 ) ) = $ . 392 ; c ' s share = $ ( 1162 x ( 20 / 83 ) ) = $ 280 . answer c 490 , 392,280" | a ) 490 , 490,490 , b ) 360 , 392,460 , c ) 490 , 392,280 , d ) 260 , 160,60 , e ) 492 , 390,280 | c | divide(1162, 35) | divide(n0,n1)| | other |
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