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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find the compound interest on $ 500 in 5 years at 5 % per annum , the interest being compounded yearly ? | "principle = $ 500 rate = 5 % yearly amount = 500 * ( 1 + 5 / 100 ) ^ 5 = 500 * 21 / 20 * 21 / 20 * 21 / 20 * 21 / 20 * 21 / 20 = $ 638.14 c . i . = 638.14 - 500 = $ 138.14 answer is a" | a ) $ 138.14 , b ) $ 150.12 , c ) $ 142.65 , d ) $ 152 , e ) $ 164.12 | a | subtract(multiply(500, power(add(const_1, divide(5, const_100)), 5)), 500) | divide(n2,const_100)|add(#0,const_1)|power(#1,n1)|multiply(n0,#2)|subtract(#3,n0)| | gain |
during 2003 , a company produced an average of 1,000 products per month . how many products will the company need to produce from 2004 through 2007 in order to increase its monthly average for the period from 2003 through 2007 by 200 % over its 2003 average ? | "company produced 12 * 1000 = 12,000 products in 2003 . if company produces x products from 2004 to 2007 , then total amount of product produced in 4 years ( 2003 through 2007 ) is x + 12,000 . the gives the average of ( x + 12,000 ) / 4 . this average needs to be 200 % higher than that in 2003 . in math terms , 12,000... | a ) 132,000 , b ) 235,000 , c ) 175,000 , d ) 200,000 , e ) 250,000 | a | multiply(200, divide(subtract(2007, 2003), const_2)) | subtract(n3,n0)|divide(#0,const_2)|multiply(n6,#1)| | general |
the sum of two numbers is 45 . their difference is 5 . the bigger number is | x + y = 45 x - y = 5 adding 2 x = 50 = = > x = 25 and y = 45 - 25 = 20 bigger number is x = 25 answer b . | a ) 20 , b ) 25 , c ) 30 , d ) 35 , e ) 45 | b | divide(add(45, 5), const_2) | add(n0,n1)|divide(#0,const_2) | general |
the present worth of rs . 338 due in 2 years at 4 % per annum compound interest is | "solution present worth = rs . [ 338 / ( 1 + 4 / 100 ) ² ] = rs . ( 338 x 25 / 26 x 25 / 26 ) = rs . 312.5 answer a" | a ) rs . 312.5 , b ) rs . 154.75 , c ) rs . 156.25 , d ) rs . 158 , e ) none | a | divide(338, power(add(divide(4, const_100), const_1), 2)) | divide(n2,const_100)|add(#0,const_1)|power(#1,n1)|divide(n0,#2)| | gain |
the average of 6 observations is 11 . a new observation is included and the new average is decreased by 1 . the seventh observation is ? | "let seventh observation = x . then , according to the question we have = > ( 66 + x ) / 7 = 10 = > x = 4 . hence , the seventh observation is 4 . answer : a" | a ) 4 , b ) 3 , c ) 5 , d ) 6 , e ) 7 | a | subtract(multiply(subtract(11, 1), add(6, 1)), multiply(11, 6)) | add(n0,n2)|multiply(n0,n1)|subtract(n1,n2)|multiply(#0,#2)|subtract(#3,#1)| | general |
a company wants to spend equal amounts of money for the purchase of two types of computer printers costing $ 350 and $ 200 per unit , respectively . what is the fewest number of computer printers that the company can purchase ? | "the smallest amount that the company can spend is the lcm of 350 and 200 , which is 1400 for each , which is total 2800 . the number of 1 st type of computers which costing $ 350 = 1400 / 350 = 4 . the number of 2 nd type of computers which costing $ 200 = 1400 / 200 = 7 . total = 4 + 7 = 11 answer is c ." | a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 13 | c | add(divide(lcm(350, 200), 350), divide(lcm(350, 200), 200)) | lcm(n0,n1)|divide(#0,n0)|divide(#0,n1)|add(#1,#2)| | general |
if the height of a cone is increased by 120 % then its volume is increased by ? | "120 % answer : e" | a ) 100 , b ) 777 , c ) 998 , d ) 729 , e ) 120 | e | rectangle_area(120, const_1) | rectangle_area(n0,const_1)| | geometry |
after decreasing 40 % in the price of an article costs rs . 1050 . find the actual cost of an article ? | "cp * ( 60 / 100 ) = 1050 cp = 17.5 * 100 = > cp = 1750 answer : b" | a ) 1200 , b ) 1750 , c ) 1050 , d ) 1570 , e ) 1005 | b | divide(1050, subtract(const_1, divide(40, const_100))) | divide(n0,const_100)|subtract(const_1,#0)|divide(n1,#1)| | gain |
a fair sided dice labeled 1 to 6 is tossed 2 times . what is the probability the sum of the 2 throws is 5 ? | the total possible number of combination if the dice is thrown 2 times is 36 . there are 4 possible outcomes that the sum of the number is 5 which is ( 2 + 3 ) ( 3 + 2 ) ( 1 + 4 ) and ( 4 + 1 ) answer : c | a ) 1 / 6 , b ) 7 / 216 , c ) 4 / 36 , d ) 9 / 216 , e ) 11 / 216 | c | divide(subtract(5, const_1), power(6, 2)) | power(n1,n2)|subtract(n4,const_1)|divide(#1,#0) | general |
a merchant marks his goods up by 50 % and then offers a discount of 10 % on the marked price . what % profit does the merchant make after the discount ? | "let the price be 100 . the price becomes 150 after a 50 % markup . now a discount of 10 % on 150 . profit = 135 - 100 35 % answer d" | a ) 21 % , b ) 25 % , c ) 69 % , d ) 35 % , e ) 19 % | d | subtract(subtract(add(50, const_100), divide(multiply(add(50, const_100), 10), const_100)), const_100) | add(n0,const_100)|multiply(n1,#0)|divide(#1,const_100)|subtract(#0,#2)|subtract(#3,const_100)| | gain |
i bought two books ; for rs . 460 . i sold one at a loss of 15 % and other at a gain of 19 % and then i found each book was sold at the same price . find the cost of the book sold at a loss ? | "x * ( 85 / 100 ) = ( 460 - x ) 119 / 100 x = 280 answer : d" | a ) 337.33 , b ) 280.33 , c ) 299.33 , d ) 268.33 , e ) 289.33 | d | divide(multiply(460, add(const_100, 19)), add(subtract(const_100, 15), add(const_100, 19))) | add(n2,const_100)|subtract(const_100,n1)|add(#0,#1)|multiply(n0,#0)|divide(#3,#2)| | gain |
what is the least number of squares tiles required to pave the floor of a room 6 m 75 cm long and 4 m 5 cm broad ? | "length of largest tile = h . c . f . of 675 cm and 405 cm = 135 cm . area of each tile = ( 135 x 135 ) cm 2 . required number of tiles = 675 x 405 / ( 135 ^ 2 ) = 15 answer : a" | a ) 15 , b ) 20 , c ) 40 , d ) 44 , e ) 54 | a | divide(rectangle_area(multiply(6, const_100), multiply(75, const_100)), square_area(add(multiply(const_4, const_10), const_1))) | multiply(n0,const_100)|multiply(n1,const_100)|multiply(const_10,const_4)|add(#2,const_1)|rectangle_area(#0,#1)|square_area(#3)|divide(#4,#5)| | physics |
in how many years will a sum of money doubles itself at 5 % per annum simple interest | "let the initial sum be x then , amount = 2 x ( because sum will be double ) amount = principal + s . i 2 x = x + s . i s . i = x s . i = ( p * r * t ) / 100 x = ( x * 5 * t ) / 100 therefore , t = 20 years answer : d" | a ) 25 years , b ) 28 years , c ) 10 years , d ) 20 years , e ) 21 years | d | divide(const_100, 5) | divide(const_100,n0)| | gain |
for any positive integer n , the sum of the first n positive integers equals n ( n + 1 ) / 2 . what is the sum of odd all the integers between 1 and 50 ? | "numbers are 1 , 3,5 - - - - - - - , 47,49 average of the set : ( largest + smallest ) / 2 = ( 49 + 1 ) / 2 = 25 ; # of terms : ( largest - smallest ) / 2 + 1 = ( 49 - 1 ) / 2 + 1 = 25 the sum = 25 * 25 = 625 answer : b" | a ) 500 , b ) 625 , c ) 750 , d ) 550 , e ) 600 | b | add(divide(subtract(subtract(50, 1), add(1, 1)), 2), 1) | add(n2,n0)|subtract(n3,n0)|subtract(#1,#0)|divide(#2,n1)|add(n0,#3)| | general |
on dividing a number by 357 , we get 41 as remainder . on dividing the same number by 17 , what will be the remainder ? | "let x be the number and y be the quotient . then , x = 357 * y + 41 = ( 17 * 21 * y ) + ( 17 * 2 ) + 7 = 17 * ( 21 y + 2 ) + 7 . required number = 7 . answer is d" | a ) 4 , b ) 5 , c ) 8 , d ) 7 , e ) 2 | d | multiply(subtract(divide(power(41, const_2), 357), floor(divide(power(41, const_2), 357))), 357) | power(n1,const_2)|divide(#0,n0)|floor(#1)|subtract(#1,#2)|multiply(n0,#3)| | general |
machines x and y produce bottles at their respective constant rates . machine x produces k bottles in 6 hours and machine y produces k bottles in 12 hours . how many hours does it take machines x and y , working simultaneously , to produce 12 k bottles ? | "x rate = k / 6 y rate = k / 12 k / 6 + k / 12 = 12 k / t solving t = 48 answer d" | a ) 8 , b ) 12 , c ) 15 , d ) 48 , e ) 24 | d | divide(multiply(12, divide(add(6, 12), const_2.0)), 12) | add(n0,n1)|divide(#0,const_2.0)|multiply(n2,#1)|divide(#2,n1)| | physics |
a certain music store stocks 800 cellos and 600 violas . of these instruments , there are 80 cello - viola pairs , such that a cello and a viola were both made with wood from the same tree ( each tree can make at most one viola and one cello , so there are no pairs other than these 90 ) . if one viola and one cello are... | "solution provided by stanford 2012 is correct : 80 / 800 choosing one of the cellos which has a pair viola , 1 / 600 choosing the viola which is the pair of chosen cello - - > p = 80 / 800 * 1 / 600 = 1 / 6,000 . answer : c ." | a ) 3 / 16,000 , b ) 1 / 8,100 , c ) 1 / 600 , d ) 1 / 90 , e ) 2 / 45 | c | multiply(divide(80, 800), divide(const_1, 600)) | divide(n2,n0)|divide(const_1,n1)|multiply(#0,#1)| | other |
in a race where 8 cars are running , the chance that car x will win is 1 / 2 , that y will win is 1 / 4 and that z will win is 1 / 3 . assuming that a dead heat is impossible , find the chance that one of them will win . | required probability = p ( x ) + p ( y ) + p ( z ) ( all the events are mutually exclusive ) . = 1 / 2 + 1 / 4 + 1 / 3 = 13 / 12 answer : d | a ) 1 / 140 , b ) 11 / 12 , c ) 14 / 8 , d ) 13 / 12 , e ) 57 / 120 | d | add(add(divide(1, 2), divide(1, 4)), divide(1, 3)) | divide(n1,n2)|divide(n1,n4)|divide(n1,n6)|add(#0,#1)|add(#3,#2) | general |
at a tanning salon , customers are charged $ 10 for their first visit in a calendar month and $ 4 for each visit after that in the same calendar month . in the last calendar month , 100 customers visited the salon , of which 30 made a second visit , and 10 made a third visit . all other customers made only one visit . ... | "i get b . this question seems too straightforward for 600 + . am i missing something ? 100 first - time visits - - > 100 ( 10 ) = $ 1000 30 + 10 = 40 subsequent visits - - > 40 ( 4 ) = $ 160 total revenue : 1000 + 160 = $ 1160 the answer is b ." | a ) $ 1220 , b ) $ 1160 , c ) $ 1300 , d ) $ 1340 , e ) $ 1880 | b | add(multiply(add(10, 4), 30), multiply(subtract(100, 30), 10)) | add(n0,n1)|subtract(n2,n3)|multiply(n3,#0)|multiply(n0,#1)|add(#2,#3)| | physics |
3 pumps , working 8 hours a day , can empty a tank in 2 days . how many hours a day should 4 pumps work in order to empty the tank in 1 day ? | "explanation : let the required hours needed be x more pumps , less hours ( indirect proportion ) more days , less hours ( indirect proportion ) hence we can write as ( pumps ) 3 : 4 ( days ) 2 : 1 } : : x : 8 ⇒ 3 × 2 × 8 = 4 × 1 × x ⇒ 3 × 2 × 2 = x ⇒ x = 12 answer : option b" | a ) 10 , b ) 12 , c ) 8 , d ) 15 , e ) 18 | b | divide(multiply(multiply(3, 8), 2), 4) | multiply(n0,n1)|multiply(n2,#0)|divide(#1,n3)| | physics |
in what ratio must rice of rs . 54 per kg be mixed with rice of rs . 35 per kg so that cost of mixture is rs . 42 per kg ? | ( 42 - 35 ) / ( 54 - 42 ) = 7 / 12 = 7 : 12 answer : b | a ) 9 : 7 , b ) 7 : 12 , c ) 12 : 7 , d ) 5 : 6 , e ) 7 : 9 | b | divide(divide(subtract(42, 35), subtract(54, 35)), subtract(const_1, divide(subtract(42, 35), subtract(54, 35)))) | subtract(n2,n1)|subtract(n0,n1)|divide(#0,#1)|subtract(const_1,#2)|divide(#2,#3) | other |
the sum of the numbers is 264 . if the first number be twice the second and third number be one - third of the first , then the second number is : | let the second number be x . then , first number = 2 x and third number = 2 x / 3 . 2 x + x + 2 x / 3 = 264 11 x / 3 = 264 x = 72 answer : c | a ) 48 , b ) 54 , c ) 72 , d ) 84 , e ) 27 | c | divide(multiply(264, const_3), add(const_10, const_1)) | add(const_1,const_10)|multiply(n0,const_3)|divide(#1,#0) | general |
if rs . 64 amount to rs . 83.20 in 2 years , what will rs . 86 amount to in 4 years at the same rate percent per annum ? | "solution p = rs . 64 , s . i . = rs . ( 83.20 - 64 ) = rs . 19.20 , t = 2 years . so , rate = ( 100 x 19.20 / 64 x 2 ) = 15 % . now , p = rs . 86 , r = 15 % , t = 4 years . ∴ s . i . = rs . ( 86 x 15 x 4 / 100 ) rs . 51.60 . answer d" | a ) rs . 114.80 , b ) rs . 86.70 , c ) rs . 55.40 , d ) rs . 51.60 , e ) none | d | subtract(multiply(subtract(83.20, 64), 86), subtract(83.20, 64)) | subtract(n1,n0)|multiply(n3,#0)|subtract(#1,#0)| | gain |
nd the area of trapezium whose parallel sides are 20 cm and 18 cm long , and the distance between them is 12 cm ? | "area of a trapezium = 1 / 2 ( sum of parallel sides ) * ( perpendicular distance between them ) = 1 / 2 ( 20 + 18 ) * ( 12 ) = 228 cm 2 answer : e" | a ) 243 cm 2 , b ) 239 cm 2 , c ) 267 cm 2 , d ) 230 cm 2 , e ) 228 cm 2 | e | divide(multiply(12, add(20, 18)), const_2) | add(n0,n1)|multiply(n2,#0)|divide(#1,const_2)| | physics |
if x = 6 ^ 36 and x ^ x = 6 ^ k , what is k ? | "solution : we know that x = 6 ^ 36 which implies x ^ x = ( 6 ^ 36 ) ^ ( 6 ^ 36 ) = 6 ^ ( 36 * 6 ^ 36 ) [ because ( x ^ y ) ^ z = x ^ ( y * z ) ) ] so 6 ^ ( 6 ^ 2 * 6 ^ 36 ) = 6 ^ ( 6 ^ ( 2 + 36 ) ) [ because x ^ a * x ^ b = x ^ ( a + b ) ] therefore x ^ x = 6 ^ ( 6 ^ 38 ) given that x ^ x = 6 ^ k so 6 ^ ( 6 ^ 38 ) = 6... | a ) 6 ^ 36 , b ) 6 ^ 38 , c ) 6 ^ 72 , d ) 6 ^ 216 , e ) 6 ^ 432 | b | multiply(power(6, 36), 36) | power(n0,n1)|multiply(n1,#0)| | general |
a cylindrical can has a radius of 6 centimeters and a height of 12 centimeters . what is the area , in square centimeters , of a rectangular label that completely covers the curved surface of the can without over - lapping ? | "they are asking the lateral surface area of the cylinder . the lateral surface area = 2 * pi * r * h = 2 * pi * 6 * 12 = 144 pi answer is d ." | a ) 16 pi , b ) 64 pi , c ) 96 pi , d ) 144 pi , e ) 576 pi | d | multiply(multiply(6, 12), multiply(const_2, const_pi)) | multiply(n0,n1)|multiply(const_2,const_pi)|multiply(#0,#1)| | geometry |
a digital wristwatch was set accurately at 8.30 a . m and then lost 2 seconds every 5 minutes . what time was indicated on the watch at 6.30 p . m of the same day if the watch operated continuously that time ? | for every 5 min watch looses 2 secs , so for 60 ms it looses 24 sec , i . e . , for 1 hr it looses 24 secs 8.30 amto 6.30 pm = 10 hr it looses 240 secs i . e . , 4 mins , hence d watch will show 6.26 answer : e | a ) 5.56 , b ) 5.58 , c ) 6.0 , d ) 6.23 , e ) 6.26 | e | subtract(6.3, divide(divide(multiply(divide(multiply(subtract(add(multiply(const_3, const_4), 6.3), 8.3), const_60), 5), 2), const_60), const_100)) | multiply(const_3,const_4)|add(n3,#0)|subtract(#1,n0)|multiply(#2,const_60)|divide(#3,n2)|multiply(n1,#4)|divide(#5,const_60)|divide(#6,const_100)|subtract(n3,#7) | physics |
what is the remainder of w = 3 ^ 19 when divided by 10 ? | "i looked for patterns : ^ 2 - units digit 9 ^ 3 - units digit 7 ^ 4 - units digit 1 ^ 5 - units digit 3 hence , we can see that when raised to a power which is multiple of 4 , the units digit is 1 , and when to an even power not multiple of 4 , the units digit is 9 and we can then see : ^ 16 - units digit 1 , or ^ 18 ... | a ) 0 , b ) 1 , c ) 5 , d ) 7 , e ) 9 | d | reminder(power(3, reminder(19, const_4)), 10) | reminder(n1,const_4)|power(n0,#0)|reminder(#1,n2)| | general |
if each edge of cube increased by 10 % , the percentage increase in surface area is : | 100 × ( 110 ) / 100 × ( 110 ) / 100 × ( 110 ) / 100 = > 1331 / 100 = 33.1 % answer is a . | ['a ) 33.1', 'b ) 33.5', 'c ) 33.7', 'd ) 33.2', 'e ) 33.9'] | a | multiply(const_100, divide(subtract(power(add(const_100, 10), const_3), power(const_100, const_3)), power(const_100, const_3))) | add(n0,const_100)|power(const_100,const_3)|power(#0,const_3)|subtract(#2,#1)|divide(#3,#1)|multiply(#4,const_100) | geometry |
a train 110 m long is running with a speed of 56 km / hr . in what time will it pass a man who is running at 6 km / hr in the direction opposite to that in which the train is going ? | "speed of train relative to man = 56 + 6 = 62 km / hr . = 62 * 5 / 18 = 155 / 9 m / sec . time taken to pass the men = 110 * 9 / 155 = 6.4 sec . answer : option b" | a ) 5 , b ) 6.4 , c ) 7 , d ) 8 , e ) 9 | b | divide(110, multiply(add(56, 6), const_0_2778)) | add(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
if a and b are integers and ( a * b ) ^ 5 = 48 y , y could be : | "distribute the exponent . a ^ 5 * b ^ 5 = 48 y find the prime factorization of 48 . this is 2 ^ 4 * 3 ^ 1 . we need 2 ^ 1 * 3 ^ 4 ( or some other power of 3 that will give us a multiple of 3 ^ 5 as our second term ) . 3 ^ 4 * 2 = 81 * 2 = 162 the answer is c ." | a ) 8 , b ) 16 , c ) 162 , d ) 144 , e ) 100 | c | multiply(divide(48, power(const_2, 5)), const_2) | power(const_2,n0)|divide(n1,#0)|multiply(#1,const_2)| | general |
5 men are equal to as many women as are equal to 8 boys . all of them earn rs . 105 only . men â € ™ s wages are ? | "5 m = xw = 8 b 5 m + xw + 8 b - - - - - 105 rs . 5 m + 5 m + 5 m - - - - - 105 rs . 15 m - - - - - - 105 rs . = > 1 m = 7 rs . answer : b" | a ) 6 rs , b ) 7 rs , c ) 8 rs , d ) 4 rs , e ) 1 rs | b | divide(105, multiply(const_3, 5)) | multiply(n0,const_3)|divide(n2,#0)| | general |
the pilot of a small aircraft with a 40 - gallon fuel tank wants to fly to cleveland , which is 520 miles away . the pilot recognizes that the current engine , which can fly only 8 miles per gallon , will not get him there . by how many miles per gallon must the aircraft ’ s fuel efficiency be improved to make the flig... | "actual miles / gallon is = 520 / 40 = 13 miles / gallon . current engine miles / gallon is 8 miles / gallon . additional 5 miles / gallon is required to match the actual mileage . imo option a ." | a ) 5 , b ) 4 , c ) 12 , d ) 40 , e ) 160 | a | subtract(divide(520, 40), 8) | divide(n1,n0)|subtract(#0,n2)| | physics |
jerry travels 8 miles at an average speed of 40 miles per hour , stops for 11 minutes , and then travels another 20 miles at an average speed of 60 miles per hour . what is jerry ’ s average speed , in miles per hour , for this trip ? | "total time taken by jerry = ( 8 / 40 ) * 60 minutes + 11 minutes + ( 20 / 60 ) * 60 minutes = 45 minutes average speed = total distance / total time = ( 8 + 20 ) miles / ( 45 / 60 ) hours = 28 * 60 / 45 = 37 miles per hour answer : option a" | a ) 37 , b ) 42.5 , c ) 44 , d ) 50 , e ) 52.5 | a | divide(add(8, 20), add(add(divide(8, 40), divide(11, 60)), divide(20, 60))) | add(n0,n3)|divide(n0,n1)|divide(n2,n4)|divide(n3,n4)|add(#1,#2)|add(#4,#3)|divide(#0,#5)| | physics |
car a runs at the speed of 65 km / hr & reaches its destination in 8 hr . car b runs at the speed of 75 km / h & reaches its destination in 4 h . what is the respective ratio of distances covered by car a & car b ? | "sol . distance travelled by car a = 65 ã — 8 = 520 km distance travelled by car b = 75 ã — 4 = 300 km ratio = 520 / 300 = 26 : 15 a" | a ) 26 : 15 , b ) 12 : 7 , c ) 13 : 7 , d ) 15 : 6 , e ) 13 : 6 | a | divide(multiply(65, 8), multiply(75, 4)) | multiply(n0,n1)|multiply(n2,n3)|divide(#0,#1)| | physics |
30 ^ 10 / 90 ^ 5 = ? | "30 ^ 10 / 90 ^ 5 = ? a . 6 ^ 5 b . 5 ^ 6 c . 3 ^ 6 d . 10 ^ 5 e . 15 ^ 3 - > 30 ^ 10 / 90 ^ 5 = ( 30 ^ 10 ) / ( 3 ^ 5 ) ( 30 ^ 5 ) = ( 30 ^ 5 ) / ( 3 ^ 5 ) = ( 3 ^ 5 ) ( 10 ^ 5 ) / ( 3 ^ 5 ) = 10 ^ 5 . thus , d is the answer ." | a ) 6 ^ 5 , b ) 5 ^ 6 , c ) 3 ^ 6 , d ) 10 ^ 5 , e ) 15 ^ 3 | d | divide(power(30, 10), power(30, 5)) | power(n0,n1)|power(n0,n3)|divide(#0,#1)| | general |
the ratio of the areas of a square and rhombus whose base is same is : | "solution : the square and the rhombus are equal in area if they have a common base . answer is c" | a ) 1 : 2 , b ) 2 : 1 , c ) 1 : 1 , d ) 3 : 1 , e ) 2 : 3 | c | divide(multiply(inverse(const_2), const_4), inverse(const_2)) | inverse(const_2)|multiply(#0,const_4)|divide(#1,#0)| | geometry |
a boat can travel with a speed of 25 km / hr in still water . if the speed of the stream is 5 km / hr , find the time taken by the boat to go 120 km downstream . | "speed downstream = ( 25 + 5 ) km / hr = 30 km / hr . time taken to travel 120 km downstream = 120 / 30 hrs = 4 hrs . answer : d" | a ) 1 hr , b ) 2 hrs , c ) 3 hrs , d ) 4 hrs , e ) 5 hrs | d | divide(120, add(25, 5)) | add(n0,n1)|divide(n2,#0)| | physics |
the banker â € ™ s gain on a sum due 3 years hence at 17 % per annum is rs . 100 . the banker â € ™ s discount is | "solution t . d = ( b . g x 100 / r x t ) = rs . ( 100 x 100 / 17 x 3 ) = rs . 196 . b . d = rs ( 196 + 100 ) = rs . 296 . answer c" | a ) rs . 960 , b ) rs . 840 , c ) rs . 296 , d ) rs . 760 , e ) none | c | add(divide(multiply(100, const_100), multiply(3, 17)), 100) | multiply(n2,const_100)|multiply(n0,n1)|divide(#0,#1)|add(n2,#2)| | gain |
an error 4 % in excess is made while measuring the side of a square . the percentage of error in the calculated area of the square is : | "explanation : 100 cm is read as 104 cm . a 1 = ( 100 × 100 ) cm 2 = 10000 and a 2 = ( 104 × 104 ) cm 2 = 10816 ( a 2 - a 1 ) = 10816 - 10000 = 816 = > 816 / 10000 * 100 = 8.16 answer : d" | a ) 4.16 , b ) 5.16 , c ) 7.16 , d ) 8.16 , e ) 9.16 | d | divide(multiply(subtract(square_area(add(const_100, 4)), square_area(const_100)), const_100), square_area(const_100)) | add(n0,const_100)|square_area(const_100)|square_area(#0)|subtract(#2,#1)|multiply(#3,const_100)|divide(#4,#1)| | gain |
a 90 - liter solution of cool - drink is made from 5 % jasmine water . if 8 liters of jasmine and 2 liters of water were added to the solution , what percent of the solution is jasmine ? | the percent of jasmine in the resulting solution is : ( amount of jasmine ) / ( total volume ) ( 0.05 ( 90 ) + 8 ) / 100 = 12.5 / 100 = 12.5 % the answer is d . | a ) 6.5 % , b ) 8.5 % , c ) 10.5 % , d ) 12.5 % , e ) 14.5 % | d | add(8, multiply(divide(5, const_100), 90)) | divide(n1,const_100)|multiply(n0,#0)|add(n2,#1) | gain |
if the ratio of two number is 4 : 6 and lcm of the number is 280 then what is the number . | product of two no = lcm * hcf 4 x * 6 x = 280 * x x = 12 answer : a | a ) 12 , b ) 20 , c ) 25 , d ) 30 , e ) 35 | a | divide(280, multiply(4, 6)) | multiply(n0,n1)|divide(n2,#0) | other |
by how much does the larger root of the equation 2 b ^ 2 + 5 b = 12 exceed the smaller root ? | "for 2 b ^ 2 + 5 b = 12 roots are [ - 5 + sqrt ( 25 + 96 ) ] / 4 or [ - 5 - sqrt ( 25 + 96 ) ] / 4 = 1.5 or - 4 hence larger root 1.5 is 1.5 - ( - 4 ) = 5.5 = 11 / 2 greater than smaller root ( - 4 ) . hence option ( e ) ." | a ) 5 / 2 , b ) 10 / 3 , c ) 7 / 2 , d ) 14 / 3 , e ) 11 / 2 | e | subtract(divide(add(5, sqrt(add(power(5, 2), multiply(multiply(2, 12), const_4)))), multiply(2, 2)), divide(subtract(5, sqrt(add(power(5, 2), multiply(multiply(2, 12), const_4)))), multiply(2, 2))) | multiply(n0,n3)|multiply(n0,n0)|power(n2,n0)|multiply(#0,const_4)|add(#3,#2)|sqrt(#4)|add(n2,#5)|subtract(n2,#5)|divide(#6,#1)|divide(#7,#1)|subtract(#8,#9)| | general |
between 100 to 200 how many no are divisible by both 3 and 2 and 100,200 are inclusive ? | prime numbers are 21 so 100 - 21 = 79 answer : c | a ) 77 , b ) 78 , c ) 79 , d ) 80 , e ) 81 | c | subtract(subtract(100, floor(divide(divide(100, const_2), const_3))), const_2) | divide(n0,const_2)|divide(#0,const_3)|floor(#1)|subtract(n0,#2)|subtract(#3,const_2) | other |
ifaequals the sum of the even integers from 2 to 20 , inclusive , andbequals the sum of the odd integers from 1 to 19 , inclusive , what is the value of ( a + b ) / 2 ? | "yes ! there is really a faster way to solve it . sum of consecutive odd or even integers = ( no . of odd or even ints ) * ( first int + last int ) / 2 here a = sum of even ints from 2 to 20 , inclusive number of even ints = 10 , first int + last int = 2 + 20 = 22 a = 10 * 22 / 2 = 110 b = sum of odd ints from 1 to 19 ... | a ) 100 , b ) 110 , c ) 105 , d ) 20 , e ) 215 | c | subtract(multiply(divide(20, 2), add(divide(20, 2), 1)), multiply(divide(add(19, 1), 2), add(divide(subtract(19, 1), 2), 1))) | add(n2,n3)|divide(n1,n0)|subtract(n3,n2)|add(n2,#1)|divide(#2,n0)|divide(#0,n0)|add(n2,#4)|multiply(#3,#1)|multiply(#6,#5)|subtract(#7,#8)| | general |
martin has to secure 80 % marks to clear his exam of class 9 th . he got 200 marks and failed by 200 marks . what is the maximum marks ? | a 500 to pass the exam ravish needs 200 + 200 = 400 marks . = > ( 400 / 80 ) * 100 = 500 | a ) 500 , b ) 505 , c ) 510 , d ) 515 , e ) 525 | a | divide(add(200, 200), divide(80, const_100)) | add(n2,n2)|divide(n0,const_100)|divide(#0,#1) | gain |
the length of each side of square a is increased by 100 percent to make square b . if the length of the side of square b is increased by 60 percent to make square c , by what percent is the area of square c greater than the sum of the areas of squares a and b ? | "let length of each side of square a be 10 area of a = 10 ^ 2 = 100 since , length of each side of square a is increased by 100 percent to make square b length of each side of square b = 2 * 10 = 20 area of b = 20 ^ 2 = 400 since , length of the side of square b is increased by 60 percent to make square c length of eac... | a ) 75 % , b ) 80 % , c ) 104.8 % , d ) 150.5 % , e ) 180 % | c | multiply(divide(subtract(subtract(square_area(add(add(const_1, divide(100, 100)), divide(multiply(add(const_1, divide(100, 100)), 60), 100))), const_1), const_4), add(const_1, square_area(add(const_1, divide(100, 100))))), 100) | divide(n0,n0)|add(#0,const_1)|multiply(n1,#1)|square_area(#1)|add(#3,const_1)|divide(#2,n0)|add(#1,#5)|square_area(#6)|subtract(#7,const_1)|subtract(#8,const_4)|divide(#9,#4)|multiply(n0,#10)| | geometry |
find the perimeter and area of a square of side 15 cm . | "we know that the perimeter of square = 4 ã — side side = 15 cm therefore , perimeter = 4 ã — 15 cm = 60 cm now , area of the square = ( side ã — side ) sq . units = 15 ã — 15 cm â ² = 225 cm â ² answer : e" | a ) 144 , b ) 169 , c ) 196 , d ) 121 , e ) 225 | e | square_area(15) | square_area(n0)| | geometry |
what is the remainder when 980 * 1050 * 880 * 1010 is divided by 22 ? | "oa is ' c ' . oe : take the remainder from each of 980 / 22 , 1050 / 22 and so on . . 980 / 22 gives remainder = 12 1050 / 22 gives remainder = 16 880 / 22 gives remainder = 0 1010 / 22 gives remainder = 20 the net remainder is the product of above individual remainders . i . e = 12 * 16 * 0 * 20 break them into pairs... | a ) 15 , b ) 10 , c ) 12 , d ) 8 , e ) 3 | c | reminder(multiply(1050, 980), 880) | multiply(n0,n1)|reminder(#0,n2)| | general |
a pizzeria in castroville creates its own special blend of shredded cheese at a cost of $ 696.05 per kilogram . the blend includes mozzarella , at $ 504.35 per kilogram , and romano cheese , at $ 887.75 per kilogram . if the employees have already shredded 19 kilograms of mozzarella cheese , how many kilograms of roman... | 504.35 ( 19 ) + 887.75 x = 696.05 ( 19 + x ) solve the equation . 504.35 ( 19 ) + 887.75 x = 696.05 ( 19 + x ) 9 , 582.65 + 887.75 x = 13 , 224.95 + 696.05 x 9 , 582.65 + 191.7 x = 13 , 224.95 191.7 x = 3 , 642.3 x = 19 . answer is a . | a ) 19 kg , b ) 13 kg , c ) 10 kg , d ) 23 kg , e ) 15 kg | a | multiply(divide(divide(subtract(696.05, 504.35), subtract(887.75, 504.35)), subtract(const_1, divide(subtract(696.05, 504.35), subtract(887.75, 504.35)))), 19) | subtract(n0,n1)|subtract(n2,n1)|divide(#0,#1)|subtract(const_1,#2)|divide(#2,#3)|multiply(n3,#4) | other |
product of two natural numbers is 17 . then , the sum of reciprocals of their squares is | "explanation : if the numbers are a , b , then ab = 17 , as 17 is a prime number , so a = 1 , b = 17 . 1 / a 2 + 1 / b 2 = 1 / 1 ( 2 ) + 1 / 17 ( 2 ) = 290 / 289 option a" | a ) 290 / 289 , b ) 1 / 289 , c ) 290 / 90 , d ) 290 / 19 , e ) none of these | a | add(power(divide(const_1, const_1), const_2), power(divide(const_1, 17), const_2)) | divide(const_1,const_1)|divide(const_1,n0)|power(#0,const_2)|power(#1,const_2)|add(#2,#3)| | general |
the perimeter of an isosceles right triangle is 10 + 10 sq rt 2 . what is the length of the hypotenuse of the triangle ? | "side of triangle is a then perimeter = a + a + a . sqrt 2 ( right angle and pythagorus ) = 2 a + a . sqrt 2 = 10 + 10 sqrt 2 or , a . ( 2 + sqrt 2 ) = 10 ( 1 + sqrt 2 ) , a = 10 * ( 1 + sqrt 2 ) / 2 + sqrt 2 = 10 * 2.414 / 3.414 = 0.707 * 10 then hypotenuse = 7.07 c" | a ) 5 , b ) 6 , c ) 7.07 , d ) 8 , e ) 9 | c | divide(multiply(10, sqrt(2)), 2) | sqrt(n2)|multiply(n0,#0)|divide(#1,n2)| | geometry |
the average of 10 numbers is 40.2 . later it is found that two numbers have been wrongly copied . the first is 15 greater than the actual number and the second number added is 13 instead of 31 . find the correct average . | "sum of 10 numbers = 402 corrected sum of 10 numbers = 402 – 13 + 31 – 15 = 405 hence , new average = 405 ⁄ 10 = 40.5 answer c" | a ) 40.2 , b ) 40.4 , c ) 40.5 , d ) 40.8 , e ) none of the above | c | divide(subtract(add(multiply(40.2, 10), add(13, 15)), 31), 10) | add(n2,n3)|multiply(n0,n1)|add(#0,#1)|subtract(#2,n4)|divide(#3,n0)| | general |
a straight line in the xy - plane has a slope of 2 and a y - intercept of 2 . on this line , what is the x - coordinate of the point whose y - coordinate is 998 ? | "slope of 2 and a y - intercept of 2 y - coordinate is 998 y = 2 x + 2 996 = 2 x x = 498 answer : b . 498" | a ) 249 , b ) 498 , c ) 676 , d ) 823 , e ) 1,002 | b | divide(subtract(998, 2), 2) | subtract(n2,n0)|divide(#0,n0)| | general |
if two - third of a bucket is filled in 2 minutes then the time taken to fill the bucket completely will be . | "2 / 3 filled in 2 mint 1 / 3 filled in 1 mint then 2 / 3 + 1 / 3 = 2 + 1 = 3 minutes answer : a" | a ) 3 minutes , b ) 70 seconds , c ) 60 seconds , d ) 100 seconds , e ) 120 seconds | a | multiply(divide(2, const_2), const_3) | divide(n0,const_2)|multiply(#0,const_3)| | physics |
marla is hanging a banner diagonally across a rectangular room . the length of the room is 3 feet shorter than its width . if the total area of the room is 130 square feet , what is the length of the banner ( in feet ) ? | to solve this , we first need to set up an equation for the area of the room . if x is the width , then we have x ( x - 3 ) = 130 . by putting the equation in standard form , we get x ^ 2 - 3 x - 130 = 0 . by using the quadratic formula , we get roots of - 10 and 13 . we know that x is the width , and x - 3 is the leng... | ['a ) between 16 and 17', 'b ) between 17 and 18', 'c ) between 18 and 19', 'd ) between 19 and 20', 'e ) between 20 and 21'] | a | sqrt(add(power(subtract(divide(add(3, sqrt(add(multiply(const_4, 130), power(3, const_2)))), const_2), 3), const_2), power(divide(add(3, sqrt(add(multiply(const_4, 130), power(3, const_2)))), const_2), const_2))) | multiply(n1,const_4)|power(n0,const_2)|add(#0,#1)|sqrt(#2)|add(n0,#3)|divide(#4,const_2)|power(#5,const_2)|subtract(#5,n0)|power(#7,const_2)|add(#8,#6)|sqrt(#9) | geometry |
if p gets 25 % more than q , then by what percent does q get less than p ? | let q get 1 unit . 25 % of 1 unit = 1 / 4 = 0.25 p gets ( 1 + 0.25 ) i . e . , 1.25 units . difference = ( 1.25 - 1 ) = 0.25 required percentage = 0.25 / 1.25 * 100 = 1 / 5 * 100 = 20 answer : c | a ) 45 , b ) 34 , c ) 20 , d ) 76 , e ) 54 | c | multiply(divide(divide(25, const_100), add(divide(25, const_100), const_1)), const_100) | divide(n0,const_100)|add(#0,const_1)|divide(#0,#1)|multiply(#2,const_100) | gain |
before 5 years , dog a ’ s age was 4 times of dog b ’ s age and after 4 years , dog a ’ s age will be 3 times of dog b ’ s age . what is the difference of dog a ’ s age and dog b ’ s now ? | "a - 5 = 4 ( b - 5 ) - - > a - 4 b = - 15 . . . . . . . . . . . . . 1 a + 4 = 3 ( b + 4 ) - - > a - 3 b = 8 . . . . . . . . . . . . . 2 ( 2 ) - ( 1 ) - - > b = 23 - - > a = 3 ( 27 ) = 81 a - b = 81 - 23 = 58 answer : c" | a ) 36 , b ) 42 , c ) 58 , d ) 60 , e ) 64 | c | subtract(subtract(multiply(4, add(subtract(multiply(4, 4), 4), subtract(multiply(4, 5), 5))), subtract(multiply(4, 5), 5)), add(subtract(multiply(4, 4), 4), subtract(multiply(4, 5), 5))) | 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 |
the number of people who purchased book a is twice the number of people who purchased book b . the number of people who purchased both books a and b is 500 , which is twice the number of people who purchased only book b . what is the number of people b who purchased only book a ? | this is best solved using overlapping sets or a venn diagram . we know that a = 2 b , and that 500 people purchased both a and b . further , those purchasing both was double those purchasing b only . this gives us 250 people purchasing b only . with the 500 that pruchased both , we have a total of 750 that purchased b ... | a ) 250 , b ) 500 , c ) 750 , d ) 1000 , e ) 1500 | d | subtract(multiply(add(500, divide(500, const_2)), const_2), 500) | divide(n0,const_2)|add(n0,#0)|multiply(#1,const_2)|subtract(#2,n0) | other |
of 70 players on a football team , 37 are throwers . the rest of the team is divided so one third are left - handed and the rest are right handed . assuming that all throwers are right handed , how many right - handed players are there total ? | "37 are throwers and right handed 1 / 3 of remaining 33 , i . e 11 are left handed and remaining are right handed . basically , out of 70 , only 11 are left handed and remaining 70 - 11 = 59 are right handed . b is the answer" | a ) 54 , b ) 59 , c ) 63 , d ) 71 , e ) 92 | b | add(multiply(subtract(const_1, divide(const_1, const_3)), subtract(70, 37)), 37) | divide(const_1,const_3)|subtract(n0,n1)|subtract(const_1,#0)|multiply(#2,#1)|add(n1,#3)| | general |
the perimeter of a rhombus is 120 feet and one of its diagonal has a length of 40 feet . find the area of the rhombus . | a perimeter of 120 when divided by 4 gives the side of the rhombus 30 feet . the length of the side oc of the right triangle is equal to half the diagonal : 20 feet . let us now consider the right triangle boc and apply pythagora ' s theorem to find the length of side bo . 30 ^ 2 = bo ^ 2 + 20 ^ 2 bo = 10 sqrt ( 5 ) fe... | ['a ) 100 sqrt ( 5 ) ft . sq', 'b ) 400 sqrt ( 5 ) ft . sq', 'c ) 500 sqrt ( 5 ) ft . sq', 'd ) 200 sqrt ( 5 ) ft . sq', 'e ) 150 sqrt ( 5 ) ft . sq'] | b | rhombus_area(multiply(const_2, sqrt(subtract(power(divide(120, const_4), const_2), power(divide(40, const_2), const_2)))), 40) | divide(n0,const_4)|divide(n1,const_2)|power(#0,const_2)|power(#1,const_2)|subtract(#2,#3)|sqrt(#4)|multiply(#5,const_2)|rhombus_area(n1,#6) | geometry |
paul ' s income is 40 % less than rex ' s income , quentin ' s income is 20 % less than paul ' s income , and sam ' s income is 40 % less than paul ' s income . if rex gave 60 % of his income to sam and 40 % of his income to quentin , sam ' s new income would be what fraction of quentin ' s new income ? | "make r = 10 p = 0.6 r = 6 q = 0.8 p = 4.8 s = 0.6 p = 3.6 for that we get s = 9.6 and q 8.8 so 9.6 / 8.8 = 1.2 / 1.1 ans : b" | a ) 11 / 12 , b ) 12 / 11 , c ) 13 / 19 , d ) 12 / 19 , e ) 11 / 19 | b | divide(add(multiply(40, const_100), multiply(40, subtract(const_100, 20))), add(multiply(40, const_100), multiply(add(40, 20), 40))) | add(n0,n1)|multiply(n4,const_100)|multiply(n0,const_100)|subtract(const_100,n1)|multiply(n4,#3)|multiply(n4,#0)|add(#1,#4)|add(#2,#5)|divide(#6,#7)| | general |
how many integers k greater than 100 and less than 900 are there such that if the hundreds and the unit digits of k are reversed , the resulting integer is k + 99 ? | "not sure if this is the shortest . . but this is how i did this there are 7 sets of integers with hundreds and units digits exchanged that satisfies k + 99 . 1 . 102 | 201 ( satisfies k + 99 , where k = 102 ) 2 . 203 | 302 ( satisfies k + 99 , where k = 203 ) 3 . . . . 4 . . . . 5 . . . . 6 . . . . 7 . 708 | 807 each ... | a ) 50 , b ) 60 , c ) 70 , d ) 80 , e ) 90 | c | multiply(const_10, subtract(const_10, const_2)) | subtract(const_10,const_2)|multiply(#0,const_10)| | general |
there are 200 female managers in a certain company . find the total number of female employees q in the company , if 2 / 5 of all the employees are managers and 2 / 5 of all male employees are managers . | { managers } = { female managers } + { male managers } ; we are told that the total number of managers in the company is 2 / 5 of all the employees , thus { managers } = 2 / 5 ( m + f ) , where m and f are number of female and male employees , respectively . also , we know that 2 / 5 of all male employees are managers ... | a ) q = 300 , b ) q = 400 , c ) q = 500 , d ) q = 600 , e ) none of these | c | divide(200, divide(2, 5)) | divide(n1,n2)|divide(n0,#0) | general |
john invests $ x at the semi - annual constant compounded rate of 2 percent and also does $ 8,000 at the quarterly constant compounded rate of 4 percent . if the interests are the same after 1 year , what is the value of x ? ? | "a = p ( 1 + r / n ) ^ nt a = total amount accrued p = principal deposited r = rate of interest in decimal form n = number of times per year , interest compounded t = time in number of years . . x ( 1 + 0.02 / 2 ) ^ 2 - x = 8,000 ( 1 + 0.04 / 4 ) ^ 4 - 8,000 [ when the principal is subtracted from the total amount accr... | a ) 10000 , b ) 12000 , c ) 16160.8 , d ) 14000 , e ) 15000 | c | divide(subtract(multiply(multiply(multiply(4, const_100), const_100), power(add(1, divide(divide(4, const_100), 4)), 4)), multiply(multiply(4, const_100), const_100)), subtract(power(add(1, divide(divide(2, const_100), 2)), 2), 1)) | divide(n2,const_100)|divide(n0,const_100)|multiply(const_100,n2)|divide(#0,n2)|divide(#1,n0)|multiply(#2,const_100)|add(#3,n3)|add(#4,n3)|power(#6,n2)|power(#7,n0)|multiply(#5,#8)|subtract(#9,n3)|subtract(#10,#5)|divide(#12,#11)| | gain |
if the sum of two numbers is 24 and the sum of their squares is 404 , then the product of the numbers is | "sol . let the numbers be x and y . then , ( x + y ) = 24 and x 2 + y 2 = 404 . now , 2 xy = ( x + y ) 2 - ( x 2 + y 2 ) = ( 24 ) 2 - 404 = 576 - 404 = 172 xy = 86 . answer d" | a ) 40 , b ) 44 , c ) 80 , d ) 86 , e ) 90 | d | divide(subtract(power(24, const_2), 404), const_2) | power(n0,const_2)|subtract(#0,n1)|divide(#1,const_2)| | general |
product of two natural numbers is 3 . then , the sum of reciprocals of their squares is | "explanation : if the numbers are a , b , then ab = 3 , as 3 is a prime number , so a = 1 , b = 3 . 1 / a 2 + 1 / b 2 = 1 / 1 ( 2 ) + 1 / 3 ( 2 ) = 10 / 9 option a" | a ) 10 / 9 , b ) 1 / 289 , c ) 290 / 90 , d ) 290 / 19 , e ) none of these | a | add(power(divide(const_1, const_1), const_2), power(divide(const_1, 3), const_2)) | divide(const_1,const_1)|divide(const_1,n0)|power(#0,const_2)|power(#1,const_2)|add(#2,#3)| | general |
if the least common multiple of two prime numbers x and y is 10 , where x > y , then the value of 2 x + y is | lcm of two prime numbers will basically be multiple of those numbers . lcm = 10 = 2 * 5 , both of which are prime . now we know x > y , so x = 5 and y = 2 . 2 x + y = 2 * 5 + 2 = 10 + 2 = 12 . answer : d | a ) 7 , b ) 9 , c ) 11 , d ) 12 , e ) 21 | d | add(multiply(divide(10, const_2), 2), const_2) | divide(n0,const_2)|multiply(n1,#0)|add(#1,const_2) | general |
if the cost price is 96 % of selling price then what is the profit percentage . | "selling price = rs 100 : then cost price = rs 96 : profit = rs 4 . profit = { ( 4 / 96 ) * 100 } % = 4.17 % answer is d ." | a ) 7.14 , b ) 7.41 , c ) 4.71 , d ) 4.17 , e ) none of them | d | multiply(divide(subtract(const_100, 96), 96), const_100) | subtract(const_100,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
ratio between rahul and deepak is 4 : 3 , after 6 years rahul age will be 50 years . what is deepak present age | "explanation : present age is 4 x and 3 x , = > 4 x + 6 = 50 = > x = 11 so deepak age is = 3 ( 11 ) = 33 answer : option c" | a ) 14 , b ) 15 , c ) 33 , d ) 22 , e ) 23 | c | divide(multiply(subtract(50, 6), 3), 4) | subtract(n3,n2)|multiply(n1,#0)|divide(#1,n0)| | other |
if x ^ 3 is divisible by 240 what is the least possible value of integer x ? | "240 can be written as ( 2 ^ 4 ) * 3 * 5 . for k ^ 3 to be divisible by 240 it should contain atleast 2 ^ 4 and 3 and 5 in its factors . we can leave out option a because 12 doesnt have 5 as one of its factor . now if we check for option b , 30 can be written as 2 * 3 * 5 , hence 30 ^ 3 will have 3 as the maximum power... | a ) 12 , b ) 30 , c ) 60 , d ) 90 , e ) 120 | c | divide(divide(240, 3), 3) | divide(n1,n0)|divide(#0,n0)| | general |
a man buys a cycle for rs . 1750 and sells it at a loss of 8 % . what is the selling price of the cycle ? | "s . p . = 92 % of rs . 1750 = rs . 92 x 1750 / 100 = rs . 1610 answer : option b" | a ) s . 1090 , b ) s . 1610 , c ) s . 1190 , d ) s . 1202 , e ) s . 1092 | b | divide(multiply(subtract(const_100, 8), 1750), const_100) | subtract(const_100,n1)|multiply(n0,#0)|divide(#1,const_100)| | gain |
in a can , there is a mixture of milk and water in the ratio 4 : 5 . if it is filled with an additional 12 litres of milk the can would be full and ratio of milk and water would become 6 : 5 . find the capacity of the can ? | "let the capacity of the can be t litres . quantity of milk in the mixture before adding milk = 4 / 9 ( t - 12 ) after adding milk , quantity of milk in the mixture = 6 / 11 t . 6 t / 11 - 12 = 4 / 9 ( t - 12 ) 10 t = 1188 - 528 = > t = 66 . answer : c" | a ) 40 , b ) 44 , c ) 66 , d ) 52 , e ) none of these | c | add(add(multiply(5, divide(12, subtract(multiply(divide(6, 5), 5), 4))), divide(12, subtract(multiply(divide(6, 5), 5), 4))), 12) | divide(n3,n4)|multiply(n1,#0)|subtract(#1,n0)|divide(n2,#2)|multiply(n1,#3)|add(#3,#4)|add(n2,#5)| | general |
a metallic sheet is of rectangular shape with dimensions 100 m x 50 m . from each of its corners , a square is cut off so as to make an open box . if the length of the square is 10 m , the volume of the box ( in m cube ) is : | "explanation : l = ( 100 - 20 ) m = 80 m , [ because 10 + 10 = 20 ] b = ( 50 - 20 ) m = 30 m , h = 10 m . volume of the box = ( 80 x 30 x 8 ) m cube = 24000 m cube . option d" | a ) 42500 m cube , b ) 20000 m cube , c ) 44140 m cube , d ) 24000 m cube , e ) none of these | d | volume_rectangular_prism(subtract(100, multiply(10, const_2)), subtract(50, multiply(10, const_2)), 10) | multiply(n2,const_2)|subtract(n0,#0)|subtract(n1,#0)|volume_rectangular_prism(n2,#1,#2)| | geometry |
a rope of which a calf is tied is increased from 5 m to 9 m , how much additional grassy ground shall it graze ? | "π ( 92 – 52 ) = 176 answer : c" | a ) 1218 , b ) 180 , c ) 176 , d ) 200 , e ) 291 | c | multiply(subtract(power(9, const_2), power(5, const_2)), divide(add(multiply(5, const_2), const_2), add(const_4, const_3))) | add(const_3,const_4)|multiply(n0,const_2)|power(n1,const_2)|power(n0,const_2)|add(#1,const_2)|subtract(#2,#3)|divide(#4,#0)|multiply(#6,#5)| | general |
if the number of white balls in a box are 6 times the black balls , which of the below is the total number of balls , if there are 8 black balls in the box ? | let the number of black balls = x and the number of white balls = 6 x then , total number of balls in the box = x + 6 x = 7 x i . e . , the total number of balls must be a multiple of 7 from the given choices , only 56 is the multiple of 7 . hence , 56 are the number of balls in the box . answer is a . | a ) a - 56 , b ) b - 57 , c ) c - 58 , d ) d - 59 , e ) e - 60 | a | add(multiply(6, 8), 8) | multiply(n0,n1)|add(n1,#0) | general |
a trader marked the selling price of an article at 50 % above the cost price . at the time of selling , he allows certain discount and suffers a loss of 1 % . he allowed a discount of : | "sol . let c . p . = rs . 100 . then , marked price = rs . 110 , s . p . = rs . 99 . ∴ discount % = [ 11 / 150 * 100 ] % = 7.3 % answer a" | a ) 7.3 % , b ) 10.5 % , c ) 11 % , d ) 12.5 % , e ) none | a | multiply(const_100, divide(add(multiply(add(const_2, const_3), const_2), 1), add(const_100, 50))) | add(const_2,const_3)|add(n0,const_100)|multiply(#0,const_2)|add(#2,n1)|divide(#3,#1)|multiply(#4,const_100)| | gain |
45 % of 750 - 25 % of 480 = ? | "explanation : given expression = ( 45 / 100 * 750 ) - ( 25 / 100 * 480 ) = ( 337.50 - 120 ) = 217.50 answer is b" | a ) 216 , b ) 217.5 , c ) 236.5 , d ) 245 , e ) 250 | b | divide(multiply(45, 750), 25) | multiply(n0,n1)|divide(#0,n2)| | general |
among all sales staff at listco corporation , college graduates and those without college degrees are equally represented . each sales staff member is either a level - 1 or level - 2 employee . level - 1 college graduates account for 15 % of listco ' s sales staff . listco employs 90 level - 1 employees , 30 of whom ar... | "i ' m going in on this one . so let ' s say that we have the following so we know that l 1 = 90 and that c and l 1 = 0.15 x , we should set up a double set matrix btw but anyways , i ' m just explaining the point with this problem . now we are told that 0.15 x = 30 , therefore the grand total is 200 . now we know that... | a ) 46 , b ) 42 , c ) 56 , d ) 32 , e ) 40 | e | divide(subtract(divide(30, divide(15, const_100)), 90), 2) | divide(n3,const_100)|divide(n6,#0)|subtract(#1,n4)|divide(#2,n1)| | general |
72 men working 8 hours per day dig 30 m deep . how many extra men should be put to dig to a depth of 50 m working 6 hours per day ? | "( 72 * 8 ) / 30 = ( x * 6 ) / 50 = > x = 160 160 – 72 = 88 answer : e" | a ) a ) 77 , b ) b ) 99 , c ) c ) 66 , d ) d ) 55 , e ) e ) 88 | e | subtract(divide(multiply(divide(multiply(72, 8), 30), 50), 6), 72) | multiply(n0,n1)|divide(#0,n2)|multiply(n3,#1)|divide(#2,n4)|subtract(#3,n0)| | physics |
how many of the positive divisors of 120 are also multiples of 4 not including 120 ? | given to find the positive divisors of 120 and condition is that divisors have to be multiple of 4 but not including 120 . consider factors of 120 4 * 30 6 * 20 15 * 8 24 * 5 60 * 2 total we get 5 different ways and all are part of 4 . answer : c is correct option . | a ) 3 . , b ) 4 . , c ) 5 . , d ) 7 . , e ) 8 . | c | divide(divide(divide(120, 4), const_2), const_3) | divide(n0,n1)|divide(#0,const_2)|divide(#1,const_3) | general |
the sale price sarees listed for rs . 400 after successive discount is 15 % and 5 % is ? | "400 * ( 85 / 100 ) * ( 95 / 100 ) = 323 answer : c" | a ) 297 , b ) 879 , c ) 323 , d ) 762 , e ) 269 | c | subtract(subtract(400, divide(multiply(400, 15), const_100)), divide(multiply(subtract(400, divide(multiply(400, 15), const_100)), 5), const_100)) | multiply(n0,n1)|divide(#0,const_100)|subtract(n0,#1)|multiply(n2,#2)|divide(#3,const_100)|subtract(#2,#4)| | gain |
efrida and frazer who live 15 miles apart , meet at a restaurant that is directly north of efrida ' s home and directly east of frazer ' s home . if the restaurant is two miles closer to efrida ' s home , than to frazer ' s home , how many miles is the restaurant from frazer ' s home ? | it ' s a simple geometry problem . look at the diagram below : answer : e . | a ) 6 , b ) 7 , c ) 8 , d ) 10 , e ) 13 | e | add(sqrt(divide(subtract(power(15, const_2), power(const_2, const_2)), const_2)), const_2) | power(n0,const_2)|power(const_2,const_2)|subtract(#0,#1)|divide(#2,const_2)|sqrt(#3)|add(#4,const_2) | physics |
the sum of two consecutive integers is 15 . find the numbers . | "n + ( n + 1 ) = 15 2 n + 1 = 15 2 n = 14 n = 7 answer : b" | a ) 17 , 18 , b ) 7 , 8 , c ) 5 , 6 , d ) 1 , 2 , e ) 8 , 9 | b | add(add(power(add(add(divide(subtract(subtract(15, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(15, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(15, const_10), const_2), const_4), const_2), power(add... | subtract(n0,const_10)|subtract(#0,const_2)|divide(#1,const_4)|add(#2,const_2)|power(#2,const_2)|add(#3,const_2)|power(#3,const_2)|add(#5,const_2)|add(#4,#6)|power(#5,const_2)|power(#7,const_2)|add(#9,#10)|add(#11,#8)| | physics |
a and b are two partially filled buckets of water . if 3 liters are transferred from a to b , then a would contain one - third of the amount of water in b . alternatively , if 5 liters are transferred from b to a , b would contain one - half of the amount of water in a . bucket a contains how many liters of water ? | "let a contains a , b contains b liters so , ( a - 3 ) / ( b + 3 ) = 1 / 3 . . . . . . . ( 1 ) again , ( b - 5 ) / ( a + 5 ) = 1 / 2 . . . . . . . . . . ( 2 ) from ( 1 ) ( 2 ) we find a = 39 / 5 ans : ( c )" | a ) 11 , b ) 13 , c ) 39 / 5 , d ) 21 , e ) 23 | c | divide(multiply(add(add(3, const_3), const_2), divide(3, const_2)), add(const_2, divide(const_1, const_2))) | add(n0,const_3)|divide(n0,const_2)|divide(const_1,const_2)|add(#0,const_2)|add(#2,const_2)|multiply(#3,#1)|divide(#5,#4)| | general |
a space shuttle orbits the earth at about 2 kilometers per second . this speed is equal to how many kilometers per hour ? | "seconds in 1 hours : 60 s in 1 min 60 min in 1 hr 60 * 60 = 3600 sec in 1 hr 2 * 3600 = 28,800 answer : a" | a ) 7,200 , b ) 8,880 , c ) 10,800 , d ) 28,800 , e ) 48,000 | a | multiply(2, const_3600) | multiply(n0,const_3600)| | physics |
a train 220 m long is running with a speed of 59 kmph . . in what will it pass a man who is running at 7 kmph in the direction opposite to that in which the train is going | "explanation : speed of the train relative to man = ( 59 + 7 ) kmph = 66 × 5 / 18 m / sec = 55 / 3 m / sec . time taken by the train to cross the man = time taken by it to cover 220 m at ( 55 / 3 ) m / sec = ( 220 × 3 / 55 ) sec = 12 sec answer : option c" | a ) 8 sec , b ) 10 sec , c ) 12 sec , d ) 14 sec , e ) 15 sec | c | multiply(divide(divide(220, const_1000), subtract(59, 7)), const_3600) | divide(n0,const_1000)|subtract(n1,n2)|divide(#0,#1)|multiply(#2,const_3600)| | physics |
each week , harry is paid x dollars per hour for the first 35 hours and 1.5 x dollars for each additional hour worked that week . each week , james is paid x dollars per per hour for the first 40 hours and 2 x dollars for each additional hour worked that week . last week james worked a total of 41 hours if harry and ja... | "james worked for 41 hours hence he earned 40 * x + 1 * 2 x = 42 x dollars ; we know that harry also earned the same 42 x dollars , out of which he earned 35 x dollars for thefirst 35 hoursplus 7 x additional dollars . since for each additional hour he gets 1.5 x dollars then he worked for 7 x / 1.5 x = 4 additional ho... | a ) 35 , b ) 36 , c ) 34 , d ) 38 , e ) 39 | c | add(divide(subtract(add(40, 2), 35), 1.5), 35) | add(n2,n3)|subtract(#0,n0)|divide(#1,n1)|add(n0,#2)| | general |
in what time will a railway train 80 m long moving at the rate of 36 kmph pass a telegraph post on its way ? | "t = 80 / 36 * 18 / 5 = 8 sec answer : e" | a ) 4 sec , b ) 7 sec , c ) 2 sec , d ) 6 sec , e ) 8 sec | e | divide(80, multiply(36, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics |
two equilateral triangles of side 12 cm are placed one on top of another , such that a 6 pointed star is formed . if the six verticals lie on a circle what is the area of the circle not enclosed by the star ? | area of star = √ 3 / 4 * ( 12 ) ^ 2 + 3 * √ 3 / 4 * ( 12 / 3 ) ^ 2 = 48 √ 3 if r be radius of circle then r = 6 / cos 30 = > r = 12 / √ 3 = > r ^ 2 = 48 area of circle = pi * 48 area of the circle not enclosed by the star = 48 * pi - 48 √ 3 = 48 ( 3.14 - 1.732 ) = 68 ( approax ) answer : a | ['a ) 68', 'b ) 83', 'c ) 57', 'd ) 61', 'e ) 62'] | a | subtract(circle_area(divide(multiply(multiply(divide(sqrt(const_3), const_2), 12), const_2), const_3)), multiply(divide(multiply(power(divide(12, const_3), const_2), sqrt(const_3)), const_4), 12)) | divide(n0,const_3)|sqrt(const_3)|divide(#1,const_2)|power(#0,const_2)|multiply(n0,#2)|multiply(#3,#1)|divide(#5,const_4)|multiply(#4,const_2)|divide(#7,const_3)|multiply(n0,#6)|circle_area(#8)|subtract(#10,#9) | geometry |
jaclyn buys $ 30 000 worth of debentures in a company . she earns 9.5 % p . a . simple interest , paid to her quarterly ( that is , every 3 months ) . if the agreed period of the debenture was 18 months : calculate the amount of interest jaclyn will earn for each quarter | "explanation : i = ( p x r x t ) / 100 = 30000 * 9.5 / 100 * ( 18 / 12 ) ^ 1 / 6 = 712.50 answer : a" | a ) 712.5 , b ) 1234 , c ) 1289 , d ) 1345 , e ) none of these | a | divide(divide(multiply(multiply(const_100, const_100), 9.5), const_100), const_4) | multiply(const_100,const_100)|multiply(n2,#0)|divide(#1,const_100)|divide(#2,const_4)| | gain |
the average of 6 observations is 12 . a new seventh observation is included and the new average is decreased by 1 . the seventh observation is : | "explanation : seventh observation = ( 7 × × 11 - 6 × × 12 ) = 5 correct option : c" | a ) 1 , b ) 3 , c ) 5 , d ) 6 , e ) 8 | c | subtract(multiply(subtract(12, 1), add(6, 1)), multiply(12, 6)) | add(n0,n2)|multiply(n0,n1)|subtract(n1,n2)|multiply(#0,#2)|subtract(#3,#1)| | general |
for a certain art exhibit , a museum sold admission tickets to a group of 30 people every 5 minutes from 9 : 00 in the morning to 5 : 55 in the afternoon , inclusive . the price of a regular admission ticket was $ 10 and the price of a student ticket was $ 5 . if on one day 3 times as many regular admission tickets wer... | "from 9 : 00 in the morning to 5 : 55 in the afternoon , inclusive there are 9 * 12 = 108 five - minute intervals , thus total of 108 * 30 tickets were sold . say x student and 3 x regular tickets were sold , then x + 3 x = 108 * 30 - - > x = 27 * 30 and 3 x = 3 * ( 27 * 30 ) = 27 * 90 . therefore , the total revenue f... | a ) $ 24960 , b ) $ 25920 , c ) $ 28080 , d ) $ 28350 , e ) $ 29160 | d | add(multiply(multiply(divide(multiply(multiply(add(subtract(add(const_12, 5), 9), const_1), const_12), 30), add(3, const_1)), 3), 10), multiply(divide(multiply(multiply(add(subtract(add(const_12, 5), 9), const_1), const_12), 30), add(3, const_1)), 5)) | add(const_12,n4)|add(n8,const_1)|subtract(#0,n2)|add(#2,const_1)|multiply(#3,const_12)|multiply(n0,#4)|divide(#5,#1)|multiply(n8,#6)|multiply(n7,#6)|multiply(n6,#7)|add(#9,#8)| | general |
how many factors of 880 are odd numbers greater than 1 ? | when factorized , 880 has 6 prime factors . of these prime factors 2 are odd and 4 are even . hence total number of odd factors is 2 * 2 ( 4 ) , which includes 4 . the total number of odd factors greater than 1 are 3 . ( option a ) | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | divide(const_60.0, multiply(const_10, const_2)) | multiply(const_10,const_2)|divide(const_60.0,#0)| | other |
how many multiples of 9 are there between 18 and 270 ? | "it should be mentioned whether 18 and 270 are inclusive . if 18 and 270 are inclusive , then the answer is ( 270 - 18 ) / 9 + 1 = 29 . if 18 and 270 are not inclusive , then the answer is ( 261 - 27 ) / 9 + 1 = 27 . since oa is b , then we have not inclusive case ." | a ) 12 , b ) 27 , c ) 42 , d ) 61 , e ) 60 | b | add(divide(subtract(270, 18), 9), const_1) | subtract(n2,n1)|divide(#0,n0)|add(#1,const_1)| | general |
if it takes 30 construction workers 70 days to build a skyscraper , how long would it take 50 workers ? | 30 * 70 = 50 * x x = 42 answer : c | a ) 52 days , b ) 47 days , c ) 42 days , d ) 37 days , e ) 32 days | c | divide(multiply(70, 30), 50) | multiply(n0,n1)|divide(#0,n2) | physics |
there are thirty - 4 lines that are tangent to a circle , and these lines create regions in the plane . can you tell how many of these regions are not enclosed ? | e 68 68 regions . each new tangent increases the non - enclosed areas by two | a ) 50 , b ) 74 , c ) 37 , d ) 77 , e ) 68 | e | add(add(multiply(4, const_10), multiply(const_2, const_10)), multiply(4, const_2)) | multiply(n0,const_10)|multiply(const_10,const_2)|multiply(n0,const_2)|add(#0,#1)|add(#3,#2) | geometry |
2 cow ’ s and 8 goats are brought for rs . 1400 . if the average price of a goat be rs . 60 . what is the average price of a cow . | explanation : average price of a goat = rs . 60 total price of 8 goats = 8 * 60 = rs . 480 but total price of 2 cows and 8 goats = rs . 1400 total price of 2 cows is = 1400 - 480 = 980 average price of a cow = 980 / 2 = rs . 460 answer : option d | a ) rs . 480 , b ) rs . 960 , c ) rs . 960 , d ) rs . 460 , e ) rs . 560 | d | divide(subtract(1400, multiply(8, 60)), 2) | multiply(n1,n3)|subtract(n2,#0)|divide(#1,n0) | general |
1 = 6,2 = 36,3 = 363,4 = 364,5 = 365 , then 36 = ? | 1 = 6,2 = 36,3 = 363,4 = 364,5 = 365 , then 36 = ? 36 = 2 check the second eqn . answer : b | a ) 1 , b ) 2 , c ) 345 , d ) 445 , e ) 235 | b | divide(subtract(subtract(365, multiply(multiply(add(const_4, const_2), add(const_4, const_2)), const_10)), 1), const_2) | add(const_2,const_4)|multiply(#0,#0)|multiply(#1,const_10)|subtract(n5,#2)|subtract(#3,n0)|divide(#4,const_2) | general |
p alone can complete a job in 6 days . the work done by q alone in one day is equal to one - fifth of the work done by p alone in one day . in how many days can the work be completed if p and q work together ? | "p ' s rate is 1 / 6 q ' s rate is 1 / 30 the combined rate is 1 / 6 + 1 / 30 = 1 / 5 if they work together , the job will take 5 days . the answer is e ." | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | e | divide(const_1, add(divide(const_1, 6), divide(divide(const_1, 6), const_3))) | divide(const_1,n0)|divide(#0,const_3)|add(#0,#1)|divide(const_1,#2)| | physics |
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