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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there are 7 thieves . they stole diamonds from a diamond merchant and ran away . while running , night sets in and they decide to rest in the jungle . when everybody was sleeping , two of them woke up and decided to divide the diamonds equally among themselves . but when they divided the diamonds equally , one diamond ... | we need a number that is a multiple of 7 that will give a remainder of 1 when divided by 2 , 3 , 4 , 5 , and 6 . the least common multiple of these numbers is 60 . so , we need a multiple of 7 that is 1 greater than a multiple of 60 . answer b | a ) 1560 , b ) 1561 , c ) 1559 , d ) 1557 , e ) none of the above | b | multiply(add(add(multiply(multiply(4, 5), add(7, 3)), multiply(4, 5)), 3), 7) | add(n0,n1)|multiply(n2,n3)|multiply(#0,#1)|add(#2,#1)|add(n1,#3)|multiply(n0,#4) | general |
in a race where 15 cars are running , the chance that car x will win is 1 / 4 , that y will win is 1 / 8 and that z will win is 1 / 12 . 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 / 4 + 1 / 8 + 1 / 12 = 11 / 24 answer : c" | a ) 1 / 140 , b ) 1 / 180 , c ) 11 / 24 , d ) 10 / 15 , e ) 57 / 120 | c | add(add(divide(1, 4), divide(1, 8)), divide(1, 12)) | divide(n1,n2)|divide(n1,n4)|divide(n1,n6)|add(#0,#1)|add(#3,#2)| | general |
find the compound interest on $ 26,000 at 20 % per annum for 9 months , compounded quarterly | "principal = $ 26000 ; time = 9 months = 3 quarters ; rate = 20 % per annum = 5 % per quarter . amount = $ [ 26000 x ( 1 + ( 5 / 100 ) ) ^ 3 ] = $ 30098.25 ci . = $ ( 30098.25 - 26000 ) = $ 4098.25 answer d ." | a ) 2532.93 , b ) 2552.26 , c ) 2524.23 , d ) 4098.25 , e ) 2512.23 | d | subtract(multiply(26,000, power(add(const_1, divide(9, const_100)), 20)), 26,000) | divide(n2,const_100)|add(#0,const_1)|power(#1,n1)|multiply(n0,#2)|subtract(#3,n0)| | gain |
some of 50 % - intensity red paint is replaced with 20 % solution of red paint such that the new paint intensity is 40 % . what fraction of the original paint was replaced ? | "40 % is 20 % - points above 20 % and 10 % - points below 50 % . thus the ratio of 20 % - solution to 50 % - solution is 1 : 2 . 1 / 3 of the original paint was replaced . the answer is a ." | a ) 1 / 3 , b ) 1 / 5 , c ) 2 / 3 , d ) 3 / 4 , e ) 4 / 5 | a | divide(subtract(divide(40, const_100), divide(50, const_100)), subtract(divide(20, const_100), divide(50, const_100))) | divide(n2,const_100)|divide(n0,const_100)|divide(n1,const_100)|subtract(#0,#1)|subtract(#2,#1)|divide(#3,#4)| | gain |
shannon and maxine work in the same building and leave work at the same time . shannon lives due north of work and maxine lives due south . the distance between maxine ' s house and shannon ' s house is 50 miles . if they both drive home at the rate 2 r miles per hour , maxine arrives home 40 minutes after shannon . if... | "nice question + 1 we have that x / 24 - ( 60 - x ) / 2 r = 40 also x / r - ( 60 - x ) / 2 r = 120 so we get that 2 x - 60 = 80 r 3 x - 60 = 240 r get rid of r 120 = 3 x x = 46 hence answer is e" | a ) 20 , b ) 34 , c ) 38 , d ) 40 , e ) 46 | e | divide(subtract(multiply(const_3, 50), 50), const_3) | multiply(n0,const_3)|subtract(#0,n0)|divide(#1,const_3)| | physics |
how many three digit numbers r are divisible by 78 or 91 ? | "the answer will be 19 . explanation : 78 = 2 * 3 * 13 now multiples of 78 , 156 . . . . 780 , now 1000 - 780 = 220 only two more muktiples of 78 can exists . so total number of 3 digit multiples of 78 are 9 + 2 = 11 91 = 13 * 7 - - total number of three digit multiples - - 9 no remember we have a common multiples as w... | a ) 17 , b ) 19 , c ) 20 , d ) 21 , e ) 22 | b | add(subtract(subtract(const_1000, const_10), multiply(multiply(const_10, multiply(78, 78)), multiply(const_4, const_2))), const_10) | multiply(n0,n0)|multiply(const_2,const_4)|subtract(const_1000,const_10)|multiply(#0,const_10)|multiply(#3,#1)|subtract(#2,#4)|add(#5,const_10)| | general |
the malibu country club needs to drain its pool for refinishing . the hose they use to drain it can remove 60 cubic feet of water per minute . if the pool is 60 feet wide by 150 feet long by 10 feet deep and is currently at 80 % capacity , how long will it take to drain the pool ? | "volume of pool = 60 * 150 * 10 cu . ft , 80 % full = 60 * 150 * 10 * 0.8 cu . ft water is available to drain . draining capacity = 60 cu . ft / min therefore time taken = 60 * 150 * 10 * 0.8 / 60 min = 1200 min c" | a ) 1800 , b ) 1000 , c ) 1200 , d ) 1400 , e ) 1600 | c | divide(multiply(divide(80, const_100), multiply(multiply(60, 150), 10)), 60) | divide(n4,const_100)|multiply(n1,n2)|multiply(n3,#1)|multiply(#0,#2)|divide(#3,n0)| | gain |
a shopkeeper labeled the price of his articles so as to earn a profit of 50 % on the cost price . he then sold the articles by offering a discount of 10 % on the labeled price . what is the actual percent profit earned in the deal ? | "explanation : let the cp of the article = rs . 100 . then labeled price = rs . 150 . sp = rs . 150 - 10 % of 150 = rs . 150 - 15 = rs . 135 . gain = rs . 135 â € “ rs . 100 = rs . 35 therefore , gain / profit percent = 35 % . answer : option a" | a ) 35 % , b ) 20 % , c ) 17 % , d ) 18 % , e ) none of these | a | subtract(subtract(add(const_100, 50), multiply(add(const_100, 50), divide(10, const_100))), const_100) | add(n0,const_100)|divide(n1,const_100)|multiply(#0,#1)|subtract(#0,#2)|subtract(#3,const_100)| | gain |
to be considered for “ movie of the year , ” a film must appear in at least 1 / 4 of the top - 10 - movies lists submitted by the cinematic academy ’ s 770 members . what is the smallest number of top - 10 lists a film can appear on and still be considered for “ movie of the year ” ? | "total movies submitted are 770 . as per question we need to take 1 / 4 of 770 to be considered for top 10 movies = 192.25 approximate the value we 193 imo option b is the correct answer . . ." | a ) 191 , b ) 193 , c ) 195 , d ) 212 , e ) 213 | b | divide(770, 4) | divide(n3,n1)| | general |
1 / 3 of girls , 1 / 2 of boys go to canteen . what factor and total number of classmates go to canteen . | let total girls be 3 . . . and total boys be 2 . . . so total girls going to canteen = 1 boys going to canteen = 1 total no . of classmates = 5 ( only suppose . . . . . . to find out the factor of students going to canteen . . . ) factor of classmates going to canteen = 2 / 5 . answer : d | a ) 1 / 5 , b ) 1 / 4 , c ) 1 / 3 , d ) 2 / 5 , e ) 2 / 3 | d | divide(add(divide(const_12, 3), divide(const_12, 2)), add(const_12, const_12)) | add(const_12,const_12)|divide(const_12,n1)|divide(const_12,n3)|add(#1,#2)|divide(#3,#0) | general |
suraj has a certain average of runs for 9 innings . in the 10 th innings he scores 200 runs thereby increasing his average by 8 runs . what is his average after the 10 th innings ? | "to improve his average by 8 runs per innings he has to contribute 9 x 8 = 72 runs for the previous 8 innings . thus , the average after the 9 th innings = 200 - 72 = 128 . answer : c" | a ) 149 , b ) 190 , c ) 128 , d ) 178 , e ) 190 | c | divide(subtract(200, multiply(9, 8)), subtract(10, 9)) | multiply(n0,n3)|subtract(n1,n0)|subtract(n2,#0)|divide(#2,#1)| | general |
at a particular graduation party with 220 guests , 70 % of the guests brought gifts , and 40 % of the female guests brought gifts . if 12 males did not bring gifts to the party , how many females did bring gifts ? | "the correct method total = 220 . . 70 % of 220 = 154 got gifts . . 66 did not get gift , out of which 12 are males , so remaining 60 - 12 = 54 are females . . but 40 % females brought gift , so 60 % did not get it . . so 60 % = 54 , 100 % = 54 * 100 / 60 = 90 . . ans 40 % of 90 = 36 b" | a ) 18 , b ) 36 , c ) 42 , d ) 68 , e ) 70 | b | divide(multiply(divide(multiply(subtract(subtract(220, divide(multiply(70, 220), const_100)), 12), const_100), subtract(const_100, 40)), 40), const_100) | multiply(n0,n1)|subtract(const_100,n2)|divide(#0,const_100)|subtract(n0,#2)|subtract(#3,n3)|multiply(#4,const_100)|divide(#5,#1)|multiply(n2,#6)|divide(#7,const_100)| | gain |
if ( 1 / 5 ) ^ m * ( 1 / 4 ) ^ 3 = 1 / ( 2 * ( 10 ) ^ 3 ) , then m = | "1 / 5 ^ m * 1 / 4 ^ 3 = 2 * 1 / 2 ^ 3 * 5 ^ 3 * 8 2 ^ 3 * 5 ^ 3 * 8 / 4 ^ 3 = 5 ^ m or 5 ^ 3 = 5 ^ m m = 3 b" | a ) 4 , b ) 3 , c ) 1 , d ) 6 , e ) 8 | b | subtract(add(multiply(divide(1, 4), 2), multiply(divide(1, 5), 2)), divide(5, 2)) | divide(n0,n3)|divide(n0,n1)|divide(n1,n6)|multiply(n6,#0)|multiply(n6,#1)|add(#3,#4)|subtract(#5,#2)| | general |
in town x , 64 percent of the population are employed , and 42 percent of the population are employed males . what percent of the employed people in town x are females ? | "we are asked to find the percentage of females in employed people . total employed people 64 % , out of which 42 are employed males , hence 22 % are employed females . ( employed females ) / ( total employed people ) = 22 / 64 = 34 % answer : d ." | a ) 16 % , b ) 25 % , c ) 32 % , d ) 34 % , e ) 52 % | d | multiply(divide(subtract(64, 42), 64), const_100) | subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)| | gain |
if n divided by 7 has a remainder of 1 , what is the remainder when 3 times n is divided by 7 ? | "as per question = > n = 7 p + 1 for some integer p hence 3 n = > 21 q + 3 = > remainder = > 6 for some integer q alternatively = > n = 2 > 3 n = > 3 = > 3 divided by 7 will leave a remainder 3 hence c" | a ) 1 , b ) 2 , c ) 3 , d ) 5 , e ) 6 | c | multiply(3, 1) | multiply(n1,n2)| | general |
how many different pairs of positive integers ( a , b ) satisfy the equation 1 / a + 1 / b = 34 / 53 ? | "there is no certain way to solve 2 unknown with 1 equation . the best way is to look at the question and retrospect the most efficient way . in this question , a and b are only positive integers . so that is a big relief . now , we can start with putting a = 1,2 , . . and so on till the time we are confident about one... | a ) 6 , b ) 3 , c ) 2 , d ) 1 , e ) 0 | c | divide(34, 53) | divide(n2,n3)| | general |
in a house , there are 4 birds , 3 dogs , and 18 cats living with the human family . if the total number of feet in the house are 74 more than the number of heads in the house , how many family members are there ? | explanation : let number of family members be x . then , total number of feet = 2 x 4 + 4 x 3 + 4 x 18 + 2 x = 2 x + 92 . total number of heads = 4 + 3 + 18 + x = 25 + x . therefore ( 2 x + 92 ) = ( 25 + x ) + 74 or x = 7 . answer : d | a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | d | add(divide(18, 3), const_1) | divide(n2,n1)|add(#0,const_1) | general |
what will be the fraction of 33.33 % | "explanation : it will 33.33 * 1 / 100 = 1 / 3 answer : option c" | a ) 1 / 4 , b ) 1 / 5 , c ) 1 / 3 , d ) 1 / 8 , e ) none of above | c | divide(circle_area(divide(33.33, const_2)), const_2) | divide(n0,const_2)|circle_area(#0)|divide(#1,const_2)| | gain |
the ratio of the adjacent angles of a parallelogram is 6 : 9 . also , the ratio of the angles of quadrilateral is 5 : 6 : 7 : 12 . what is the sum of the smaller angle of the parallelogram and the second largest angle of the quadrilateral ? | "the measures of the adjacent angles of a parallelogram add up to be 180 ° given so , 6 x + 9 x = 180 ° or , 15 x = 180 ° or , x = 12 ° hence the angles of the parallelogram are 72 ° and 108 ° further it is given we know sum of all the four angles of a quadrilateral is 360 ° so , 5 y + 6 y + 7 y + 12 y = 360 ° or , 5 y... | a ) 168 ° , b ) 228 ° , c ) 156 ° , d ) 224 ° , e ) none of these | c | multiply(divide(multiply(add(multiply(multiply(6, const_2), const_10), const_100), const_2), add(add(add(5, 6), 7), 12)), 9) | add(n2,n3)|multiply(const_2,n0)|add(n4,#0)|multiply(#1,const_10)|add(#3,const_100)|add(n5,#2)|multiply(#4,const_2)|divide(#6,#5)|multiply(n1,#7)| | other |
the h . c . f . of two numbers is 12 and their difference is 12 . the numbers are | "solution out of the given numbers , the two with h . c . f . 12 and difference 12 are 84 and 96 answer d" | a ) 66 , 78 , b ) 70 , 82 , c ) 94 , 106 , d ) 84 , 96 , e ) none | d | multiply(12, add(const_3, const_4)) | add(const_3,const_4)|multiply(n0,#0)| | general |
if the length and breadth of a room are increased by y feet each , the perimeter increases by 16 feet . find y | 2 ( l + b ) = x ; 2 ( l + y + b + y ) = x + 16 ; 2 ( l + b ) + 4 y = x + 16 ; x + 4 y = x + 16 ; y = 4 answer : d | ['a ) 8', 'b ) 2', 'c ) 6', 'd ) 4', 'e ) 3'] | d | divide(16, const_4) | divide(n0,const_4) | physics |
jack and paul are running 1000 m . jack runs 1.5 km / hr faster than paul but also finishes the race 1.5 sec earlier . what is the speed of jack ? ? | distance ( given ) 1000 m paul speed ( let ) = x m / s time ( paul take ) = 1000 / x sec jack speed = x + 1.5 * ( 5 / 18 ) m / s tym ( jack ) = 1000 / ( x + 7.5 / 18 ) s tym ( paul ) - tym ( jack ) = 1.5 solving for x we get 20.20 speed of jack is 20.20 + 7.5 / 18 = 20.61 answer : b | a ) 19.6 , b ) 20.6 , c ) 21.6 , d ) 22.6 , e ) 23.6 | b | subtract(divide(multiply(multiply(1.5, const_0_2778), const_100), const_2), divide(multiply(1.5, const_0_2778), const_2)) | multiply(n1,const_0_2778)|divide(#0,const_2)|multiply(#0,const_100)|divide(#2,const_2)|subtract(#3,#1) | physics |
the speed of boat in still water is 20 km / hr . if it travels 26 km downstream and 14 km upstream in same time , what is the speed of the stream ? | explanation : let x be speed of stream speed of boat downstream = ( 20 + x ) speed of boat upstream = ( 20 – x ) it is given that boat takes same time to travel 26 km downstream and 14 km upstream [ 26 / ( 20 + x ) ] = [ 14 / ( 20 – x ) ] 520 – 26 x = 280 + 14 x 40 x = 240 x = 6 km / hr answer : a | a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 2 | a | divide(subtract(multiply(20, 26), multiply(20, 14)), add(26, 14)) | add(n1,n2)|multiply(n0,n1)|multiply(n0,n2)|subtract(#1,#2)|divide(#3,#0) | physics |
on my sister ' s birthday , she was 147 cm in height , having grown 5 % since the year before . how tall was she the previous year ? | "let the previous year ' s height be x . 1.05 x = 147 x = 140 the answer is b ." | a ) 145 cm , b ) 140 cm , c ) 142 cm , d ) 138 cm , e ) 135 cm | b | subtract(147, divide(multiply(147, 5), const_100)) | multiply(n0,n1)|divide(#0,const_100)|subtract(n0,#1)| | physics |
if the sum of the 4 th term and the 12 th term of an arithmetic progression is 10 , what is the sum of the first 15 terms of the progression ? | "4 th term + 12 th term = 10 i . e . , ( a + 3 d ) + ( a + 11 d ) = 10 now , sum of first 15 terms = ( 15 / 2 ) * [ 2 a + ( 15 - 1 ) d ] = ( 15 / 2 ) * [ 2 a + 14 d ] = ( 15 / 2 ) * 10 - - - - - - - - - - - - - - - from ( 1 ) = 75 answer : b" | a ) 65 , b ) 75 , c ) 60 , d ) 55 , e ) 50 | b | multiply(divide(15, const_2), 10) | divide(n3,const_2)|multiply(n2,#0)| | general |
during a certain season , a team won 65 percent of its first 100 games and 50 percent of its remaining games . if the team won 70 percent of its games for the entire season , what was the total number of games that the team played ? | "we are first given that a team won 65 percent of its first 100 games . this means the team won 0.65 x 100 = 65 games out of its first 100 games . we are next given that the team won 50 percent of its remaining games . if we use variable t to represent the total number of games in the season , then we can say t – 100 e... | a ) 80 , b ) 75 , c ) 56 , d ) 50 , e ) 105 | b | add(100, divide(subtract(65, 70), subtract(divide(70, 100), divide(50, 100)))) | divide(n3,n1)|divide(n2,n1)|subtract(n0,n3)|subtract(#0,#1)|divide(#2,#3)|add(n1,#4)| | gain |
worker a takes 8 hours to do a job . worker b takes 10 hours to do a job . how long should it take both a and b , working together to do same job . | "explanation : in this type of questions , first we need to calculate 1 hours work , then their collective work as , a ' s 1 hour work is 1 / 8 b ' s 1 hour work is 1 / 10 ( a + b ) ' s 1 hour work = 1 / 8 + 1 / 10 = 9 / 40 so both will finish the work in 40 / 9 hours = 4 4 / 9 option d" | a ) 4 / 9 , b ) 2 4 / 9 , c ) 3 4 / 9 , d ) 4 4 / 9 , e ) none of these | d | divide(const_1, add(divide(const_1, 8), divide(const_1, 10))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)| | physics |
a man can row downstream at 18 kmph and upstream at 12 kmph . find the speed of the man in still water and the speed of stream respectively ? | "let the speed of the man in still water and speed of stream be x kmph and y kmph respectively . given x + y = 18 - - - ( 1 ) and x - y = 12 - - - ( 2 ) from ( 1 ) & ( 2 ) 2 x = 30 = > x = 15 , y = 3 . answer : a" | a ) 3 , b ) 5 , c ) 6 , d ) 4 , e ) 9 | a | divide(divide(add(18, 12), const_2), const_2) | add(n0,n1)|divide(#0,const_2)|divide(#1,const_2)| | physics |
the weight of a glass of jar is 25 % of the weight of the jar filled with coffee beans . after some of the beans have been removed , the weight of the jar and the remaining beans is 60 % of the original total weight . what fraction part of the beans remain in the jar ? | "let weight of jar filled with beans = 100 g weight of jar = 25 g weight of coffee beans = 75 g weight of jar and remaining beans = 60 g weight of remaining beans = 35 g fraction remaining = 35 / 75 = 7 / 15 answer is e ." | a ) 1 / 5 , b ) 1 / 3 , c ) 2 / 5 , d ) 1 / 2 , e ) 7 / 15 | e | divide(subtract(60, 25), subtract(const_100, 25)) | subtract(n1,n0)|subtract(const_100,n0)|divide(#0,#1)| | gain |
one ton has 2500 pounds , and one pound has 16 ounces . how many packets containing wheat weighing 16 pounds and 4 ounces each would totally fill a gunny bag of capacity 13 tons ? | "16 pounds and 4 ounces = 16 * 16 + 4 = 260 ounces . 13 tons = 13 * 2500 pound = 13 * 2500 * 16 ounces . hence the answer is ( 13 * 2500 * 16 ) / 260 = 2000 . answer : a ." | a ) 2000 , b ) 1700 , c ) 2350 , d ) 2500 , e ) 8000 | a | divide(multiply(multiply(13, 2500), 16), add(multiply(16, 16), 4)) | multiply(n0,n4)|multiply(n1,n1)|add(n3,#1)|multiply(n1,#0)|divide(#3,#2)| | general |
in an examination , there were 2,000 candidates , out of which 900 candidates were girls and rest were boys . if 28 % of the boys and 32 % of the girls passed , then the total percentage of failed candidates is ? | girls = 900 , boys = 1100 passed = ( 28 % of 1100 ) + ( 32 % of 900 ) = 308 + 288 = 596 failed = 2000 - 596 = 1404 failed % = [ ( 1404 / 2000 ) x 100 ] % = 70.2 % . answer : d | a ) 35.67 % , b ) 64.75 % , c ) 68.57 % , d ) 70.20 % , e ) none of these | d | multiply(divide(subtract(subtract(multiply(const_2, multiply(const_100, const_10)), multiply(divide(28, const_100), subtract(multiply(const_2, multiply(const_100, const_10)), 900))), multiply(divide(32, const_100), 900)), multiply(const_2, multiply(const_100, const_10))), const_100) | divide(n2,const_100)|divide(n3,const_100)|multiply(const_10,const_100)|multiply(#2,const_2)|multiply(n1,#1)|subtract(#3,n1)|multiply(#0,#5)|subtract(#3,#6)|subtract(#7,#4)|divide(#8,#3)|multiply(#9,const_100) | general |
for what value of x , is | x – 5 | + | x + 4 | + | x | = 14 ? | "for what value of x , is | x – 5 | + | x + 4 | + | x | = 14 ? it ' s easiest just to plug in answer choices : ( e ) : 5 | x – 5 | + | x + 4 | + | x | = 14 ? | 5 - 5 | + | 5 + 4 | + | 5 | = 14 ? | 0 | + | 9 | + | 5 | = 14 ( e )" | a ) 0 , b ) 3 , c ) 4 , d ) - 4 , e ) 5 | e | negate(5) | negate(n0)| | general |
an amount of money is to be distributed among faruk , vasim and ranjith in the ratio 3 : 3 : 7 . if vasims share is rs . 1500 , what is the difference between faruk ' s and ranjith ' s shares ? | "explanation : let p = faruk , q = vasim , r = ranjith let p = 3 x , q = 3 x and r = 7 x . then , 5 x = 1500 ? x = 500 . p = 1500 , q = 1500 and r = 3500 . hence , ( r - p ) = ( 3500 - 1500 ) = 2000 answer : d" | a ) s 1200 , b ) s 1500 , c ) s 1600 , d ) s 2000 , e ) s 1700 | d | multiply(divide(1500, 3), subtract(7, 3)) | divide(n3,n1)|subtract(n2,n0)|multiply(#0,#1)| | other |
a , b and c are employed to do apiece of work for rs . 529 . a and c are supposed to finish 19 / 23 of the work together . how much shall be paid to b ? | c 12 12 men can complete a work in 18 days . six days after they started working , 4 more men joined them . how many days will all of them together complete the remaining work ? | a ) 10 , b ) 11 , c ) 12 , d ) 14 , e ) 16 | c | divide(add(subtract(529, multiply(divide(19, 23), 529)), const_4), add(const_4, const_4)) | add(const_4,const_4)|divide(n1,n2)|multiply(n0,#1)|subtract(n0,#2)|add(#3,const_4)|divide(#4,#0) | general |
a train 990 m long is running at a speed of 78 km / hr . if it crosses a tunnel in 1 min , then the length of the tunnel is ? | "speed = 78 * 5 / 18 = 65 / 3 m / sec . time = 1 min = 60 sec . let the length of the train be x meters . then , ( 990 + x ) / 60 = 65 / 3 x = 310 m . answer : c" | a ) 2898 , b ) 277 , c ) 310 , d ) 297 , e ) 435 | c | divide(990, multiply(subtract(78, 1), const_0_2778)) | subtract(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
calculate how many days it will take for 5 boys to paint a 30 m long fence if 2 boys can paint a 20 m long fence in 3 days , | "the length of wall painted by one boy in one day = 20 / 2 * 1 / 3 = 3.33 m no . of days required to paint 50 m cloth by 8 boys = 30 / 5 * 1 / 3.33 = 1.80 days . e" | a ) 6.80 days , b ) 4.80 days , c ) 3.80 days , d ) 2 days , e ) 1.80 days | e | divide(multiply(multiply(2, 3), 30), multiply(20, 5)) | multiply(n2,n4)|multiply(n0,n3)|multiply(n1,#0)|divide(#2,#1)| | physics |
find the value of ( 28 + 48 / 69 ) × 69 | "= ( 28 + 48 / 69 ) × 69 = ( 1932 + 48 ) / 69 × 69 = 1980 / 69 × 69 = 1980 answer is a ." | a ) 1980 , b ) 1982 , c ) 1930 , d ) 1988 , e ) 1680 | a | multiply(add(divide(48, 69), 28), 69) | divide(n1,n2)|add(n0,#0)|multiply(#1,n2)| | general |
a soccer store typically sells replica jerseys at a discount of 20 percent to 50 percent off list price . during the annual summer sale , everything in the store is an additional 20 percent off the original list price . if a replica jersey ' s list price is $ 80 , approximately what percent of the list price is the low... | "let the list price be 2 x for min sale price , the first discount given should be 50 % , 2 x becomes x here now , during summer sale additional 20 % off is given ie sale price becomes 0.8 x it is given lise price is $ 80 = > 2 x = 80 = > x = 50 and 0.8 x = 32 so lowest sale price is 32 , which is 40 % of 80 hence , e ... | a ) 20 , b ) 25 , c ) 30 , d ) 40 , e ) 50 | e | divide(80, const_2) | divide(n3,const_2)| | general |
a new tower has just been built at the verbico military hospital ; the number of beds available for patients at the hospital is now 4 times the number available before the new tower was built . currently , 1 / 3 of the hospital ' s original beds , as well as 1 / 5 of the beds in the new tower , are occupied . for the p... | "i think d - 31 / 45 is the correct answer . here goes : lets assume originally the number of beds = x after the new tower , the total combined no of beds = 4 x so old = x , new = 3 x now 1 / 3 of x are occupied and 1 / 5 of 3 x are occupied which simplifies to ( 3 / 5 ) x we are shifting 1 / 3 of x to the new ward so ... | a ) 11 / 30 , b ) 29 / 60 , c ) 17 / 30 , d ) 31 / 45 , e ) 11 / 15 | d | multiply(add(multiply(4, 3), 1), multiply(divide(1, 3), divide(1, 5))) | divide(n1,n2)|divide(n1,n4)|multiply(n0,n2)|add(n1,#2)|multiply(#0,#1)|multiply(#3,#4)| | general |
if jake loses 32 pounds , he will weigh twice as much as his sister . together they now weigh 212 pounds . what is jake ’ s present weight , in pounds ? | lets say j is the weight of jack and s is the wt of his sister . if he loses 32 pounds , he s twice as heavy as his sister . j - 32 = 2 * s also , together they weight 212 pounds j + s = 212 solvong the 2 equation , we get j = 152 pounds ! d | a ) 131 , b ) 135 , c ) 139 , d ) 152 , e ) 188 | d | add(multiply(divide(subtract(212, 32), const_3), const_2), 32) | subtract(n1,n0)|divide(#0,const_3)|multiply(#1,const_2)|add(n0,#2) | general |
a train 125 m long passes a man , running at 5 km / hr in the same direction in which the train is going , in 10 sec . the speed of the train is ? | "speed of the train relative to man = 125 / 10 = 25 / 2 m / sec . = 25 / 2 * 18 / 5 = 45 km / hr let the speed of the train be x km / hr . then , relative speed = ( x - 5 ) km / hr . x - 5 = 45 = > x = 50 km / hr . answer : b" | a ) 225 km / hr , b ) 50 km / hr , c ) 99 km / hr , d ) 88 km / hr , e ) 66 km / hr | b | divide(divide(subtract(125, multiply(multiply(5, const_0_2778), 5)), 5), const_0_2778) | multiply(n1,const_0_2778)|multiply(n1,#0)|subtract(n0,#1)|divide(#2,n1)|divide(#3,const_0_2778)| | physics |
how many cuboids of length 6 m , width 4 m and height 3 m can be farmed from a cuboid of 18 m length , 15 m width and 2 m height . | "( 18 ã — 15 ã — 2 ) / ( 6 ã — 4 ã — 3 ) = 7.5 answer is d ." | a ) 2.6 , b ) 5.8 , c ) 6.2 , d ) 7.5 , e ) 1.2 | d | divide(multiply(multiply(18, 15), 2), multiply(multiply(6, 4), 3)) | multiply(n3,n4)|multiply(n0,n1)|multiply(n5,#0)|multiply(n2,#1)|divide(#2,#3)| | physics |
how many seconds will a train 100 meters long take to cross a bridge 180 meters long if the speed of the train is 36 kmph ? | "explanation : d = 100 + 180 = 280 s = 36 * 5 / 18 = 10 mps t = 280 / 10 = 28 sec answer : option c" | a ) 26 , b ) 72 , c ) 28 , d ) 82 , e ) 27 | c | divide(add(180, 100), multiply(36, const_0_2778)) | add(n0,n1)|multiply(n2,const_0_2778)|divide(#0,#1)| | physics |
there are 36 students in a certain geometry class . if one third of the students are boys and one fourth of the boys are under 6 feet tall , how many boys in the class are under 6 feet tall ? | total students = 36 one third of the students are boys = 1 / 3 * 36 = 12 one fourth of the boys are under six feet tall = 1 / 4 * 12 = 3 . . . therefore number of boys in the class under six feet tall = 3 . . . answer a . . . . alternatively . . . number of boys in the class under six feet tall = 1 / 4 of 1 / 3 of tota... | a ) 3 , b ) 12 , c ) 18 , d ) 24 , e ) 27 | a | divide(divide(36, const_3), const_4) | divide(n0,const_3)|divide(#0,const_4) | general |
find the area of the quadrilateral of one of its diagonals is 30 cm and its off sets 9 cm and 6 cm ? | "1 / 2 * 30 ( 9 + 6 ) = 225 cm 2 answer : a" | a ) 225 cm 2 , b ) 150 cm 2 , c ) 127 cm 2 , d ) 177 cm 2 , e ) 187 cm 2 | a | multiply(multiply(divide(const_1, const_2), add(6, 9)), 30) | add(n1,n2)|divide(const_1,const_2)|multiply(#0,#1)|multiply(n0,#2)| | geometry |
a cricketer scored 138 runs which included 12 boundaries and 2 sixes . what percent of his total score did he make by running between the wickets . | explanation : number of runs made by running = 138 - ( 12 x 4 + 2 x 6 ) = 138 - ( 60 ) = 78 now , we need to calculate 72 is what percent of 138 . = > 78 / 138 * 100 = 56.52 % answer : d | a ) 46.2 % , b ) 54.54 % , c ) 63 % , d ) 56.52 % , e ) none of these | d | multiply(divide(subtract(138, add(multiply(12, const_4), multiply(2, multiply(const_2, const_3)))), 138), const_100) | multiply(n1,const_4)|multiply(const_2,const_3)|multiply(n2,#1)|add(#0,#2)|subtract(n0,#3)|divide(#4,n0)|multiply(#5,const_100) | general |
if the height of a triangle is decreased by 40 % , land its base is increased by 40 % , what will be the effect on its area ? | regd effect = ∣ ∣ 40 − 40 − 40 × 40100 ∣ ∣ % = − 16 % | 40 − 40 − 40 × 40100 | % = − 16 % i . e . , the area will decrease by 16 % answer d | ['a ) no change', 'b ) 16 % increase', 'c ) 8 % decrease', 'd ) 16 % decrease', 'e ) none of thes'] | d | subtract(const_100, divide(multiply(subtract(const_100, 40), add(const_100, 40)), const_100)) | add(n0,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|subtract(const_100,#3) | geometry |
p and q invested in a shop . the profits were divided in the ratio of 2 : 3 respectively . if p invested rs . 40,000 , the amount invested by q is : | "suppose q invested rs . y . then , 40000 / y = 2 / 3 or y = [ 40000 x 3 / 2 ] = 60000 . answer d" | a ) 35000 , b ) 40000 , c ) 50000 , d ) 60000 , e ) none | d | multiply(divide(40,000, 2), 3) | divide(n2,n0)|multiply(n1,#0)| | gain |
if taxi fares were $ 8.00 for the first 1 / 5 mile and $ 0.8 for each 1 / 5 mile there after , then the taxi fare for a 8 - mile ride was | in 8 miles , initial 1 / 5 mile charge is $ 8 rest of the distance = 8 - ( 1 / 5 ) = 39 / 5 rest of the distance charge = 39 ( 0.8 ) = $ 31.2 ( as the charge is 0.8 for every 1 / 5 mile ) = > total charge for 4 miles = 8 + 31.2 = 39.2 answer is e | a ) $ 39.00 , b ) $ 39.10 , c ) $ 39.25 , d ) $ 39.30 , e ) $ 39.20 | e | add(8, multiply(subtract(divide(8, divide(1, 5)), const_1), 0.8)) | divide(n1,n2)|divide(n0,#0)|subtract(#1,const_1)|multiply(n3,#2)|add(n0,#3) | general |
with both valves open , the pool will be filled with water in 48 minutes . the first valve alone would fill the pool in 2 hours . if the second valve emits 50 cubic meters of water more than the first every minute , then what is the capacity q of the pool ? | "d . 12000 cubic meters . if both hte valves fill the pool in 48 minutes and valve 1 only fills in 120 minutes then valve 2 alone will fill the pool in ( 48 * 120 ) / ( 120 - 48 ) = 80 minutes . now , if valve 1 admits x cubic meter of water per minute then the capacity of pool will be 120 x and also 80 ( x + 50 ) . or... | a ) 9000 cubic meters , b ) 10500 cubic meters , c ) 11750 cubic meters , d ) 12000 cubic meters , e ) 12500 cubic meters | d | multiply(50, multiply(const_60, multiply(divide(48, const_60), add(const_4, const_1)))) | add(const_1,const_4)|divide(n0,const_60)|multiply(#0,#1)|multiply(#2,const_60)|multiply(n2,#3)| | physics |
how many diagonals does a 63 - sided convex polygon have ? | "a 63 - sided convex polygon has 63 vertices . if we examine a single vertex , we can see that we can connect it with 60 other vertices to create a diagonal . note : there are 60 options because we ca n ' t connect the vertex to itself , and we ca n ' t connect it to its adjacent vertices , since this would not create ... | a ) 1890 , b ) 1953 , c ) 3780 , d ) 3843 , e ) 3906 | a | divide(factorial(63), multiply(factorial(subtract(63, const_2)), factorial(const_2))) | factorial(n0)|factorial(const_2)|subtract(n0,const_2)|factorial(#2)|multiply(#3,#1)|divide(#0,#4)| | general |
in a graduating class , the difference between the highest and lowest salaries is $ 100000 . the median salary is $ 50000 higher than the lowest salary and the average salary is $ 20000 higher than the median . what is the minimum number of students t in the class ? | the difference between the highest and lowest salaries is $ 100000 . so there are at least 2 people - say one with salary 0 and the other with 100 k . no salary will be outside this range . median = 50 k more than lowest . so median is right in the center of lowest and highest since lowest and highest differ by 100 k .... | a ) a ) 10 , b ) b ) 12 , c ) c ) 15 , d ) d ) 20 , e ) e ) 25 | c | add(multiply(divide(add(50000, 20000), subtract(subtract(100000, add(50000, 20000)), 20000)), const_2), const_1) | add(n1,n2)|subtract(n0,#0)|subtract(#1,n2)|divide(#0,#2)|multiply(#3,const_2)|add(#4,const_1) | general |
uncle bruce is baking chocolate chip cookies . he has 36 ounces of dough ( with no chocolate ) and 20 ounces of chocolate . how much chocolate is left over if he uses all the dough but only wants the cookies to consist of 10 % chocolate ? | "first , you must find the total weight of the mixture given that 80 % of it will be dough . 90 % * total = 36 = > ( 9 / 10 ) total = 36 = > total = 360 / 9 = > total = 40 oz , from there , you must find 10 % of the total 40 oz of the mixture . 10 % * total = > ( 1 / 10 ) ( 40 ) = 4 oz choclate used , not forgetting th... | a ) 18 , b ) 20 , c ) 16 , d ) 22 , e ) 24 | c | multiply(divide(10, const_100), 10) | divide(n2,const_100)|multiply(n2,#0)| | gain |
the area of one square is x ^ 2 + 10 x + 25 and the area of another square is 4 x ^ 2 − 12 x + 9 . if the sum of the perimeters of both squares is 64 , what is the value of x ? | "spotting the pattern of equations both are in form of ( x + c ) ^ 2 so a 1 = ( x + 5 ) ^ 2 a 2 = ( 2 x - 3 ) ^ 2 l 1 = x + 5 l 2 = 2 x - 3 p 1 = 4 ( x + 5 ) p 2 = 4 ( 2 x - 3 ) p 1 + p 2 = 64 4 ( x + 5 ) + 4 ( 2 x - 3 ) = 64 . . . . . . . . . . . . > x = 4.8 answer : c" | a ) 0 , b ) 2 , c ) 4.8 , d ) 4.67 , e ) 10 | c | divide(subtract(64, subtract(multiply(4, divide(10, 2)), 10)), 10) | divide(n1,n0)|multiply(#0,n3)|subtract(#1,n1)|subtract(n7,#2)|divide(#3,n1)| | general |
how long does a train 100 m long traveling at 72 kmph takes to cross a tunnel of 1400 m in length ? | d = 1400 + 100 = 1500 m s = 72 * 5 / 18 = 20 t = 1500 * 1 / 20 = 75 sec answer : d | a ) 70 sec , b ) 60 sec , c ) 82 sec , d ) 75 sec , e ) 62 sec | d | divide(add(100, 1400), multiply(72, const_0_2778)) | add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1) | physics |
a person walks from one end to the other of a 60 - meter long moving walkway at a constant rate in 30 seconds , assisted by the walkway . when this person reaches the end , they reverse direction and continue walking with the same speed , but this time it takes 120 seconds because the person is traveling against the di... | "let v be the speed of the person and let x be the speed of the walkway . 30 ( v + x ) = 60 then 120 ( v + x ) = 240 120 ( v - x ) = 60 when we add the two equations : 240 v = 300 v = 5 / 4 time = 60 / ( 5 / 4 ) = 48 seconds the answer is a ." | a ) 48 , b ) 60 , c ) 72 , d ) 84 , e ) 96 | a | divide(60, divide(add(divide(60, 30), divide(60, 120)), const_2)) | divide(n0,n1)|divide(n0,n2)|add(#0,#1)|divide(#2,const_2)|divide(n0,#3)| | physics |
mike took a taxi to the airport and paid $ 2.50 to start plus $ 0.25 per mile . annie took a different route to the airport and paid $ 2.50 plus $ 5.00 in bridge toll fees plus $ 0.25 per mile . if each was charged exactly the same amount , and annie ' s ride was 22 miles , how many miles was mike ' s ride ? | "the cost of annie ' s ride was 2.5 + 5 + ( 0.25 * 22 ) = $ 13 let x be the distance of mike ' s ride . the cost of mike ' s ride is 2.5 + ( 0.25 * x ) = 13 0.25 * x = 10.5 x = 42 miles the answer is c ." | a ) 30 , b ) 36 , c ) 42 , d ) 48 , e ) 54 | c | divide(subtract(add(add(2.50, 5.00), multiply(0.25, 22)), 2.50), 0.25) | add(n0,n3)|multiply(n1,n5)|add(#0,#1)|subtract(#2,n0)|divide(#3,n1)| | general |
if g and d are both integers , g > d , and - 3 g > 19 , then the largest value of d would be ? | no , your thinking is incorrect . when we know that g > d and g < - 6.33 , the largest value of g can be - 7 while if g = - 7 , then largest value of d < - 7 will be - 8 . for negative numbers , - 7 > - 8 and - 8 > - 10 . you are right in saying that d can take any value less than - 7 - - - > d could be - 8 , - 9 , - 1... | a ) - 5 , b ) - 6 , c ) - 7 , d ) - 8 , e ) - 10 | d | subtract(divide(19, negate(3)), divide(const_2, const_3)) | divide(const_2,const_3)|negate(n0)|divide(n1,#1)|subtract(#2,#0) | other |
what is the dividend from tony ' s stock , if he invested rs . 3200 at 85 to obtain an income of rs . 250 . | by investing rs . 3200 , income = rs . 250 by investing rs . 85 , income = 250 ã — 85 / 3200 = 6.6 ie , dividend = 6.6 % answer is c . | a ) 4 , b ) 7.6 , c ) 6.6 , d ) 7.6 , e ) 9.6 | c | divide(multiply(250, 85), 3200) | multiply(n1,n2)|divide(#0,n0) | gain |
a is twice as good a workman as b and together they finish a piece of work in 14 days . the number of days taken by a alone to finish the work is : | "( a ' s 1 day ' s work ) : ( b ' s 1 day ' s work ) = 2 : 1 ( a + b ) ' s 1 day ' s work = 1 / 14 divide 1 / 14 in the ratio 2 : 1 a ' s 1 day ' s work = ( 1 / 14 x 2 / 3 ) = 1 / 21 hence , a alone can finish the work in 21 days . correct option : b" | a ) 11 , b ) 21 , c ) 28 , d ) 42 , e ) none of these | b | multiply(add(const_1, const_2), 14) | add(const_1,const_2)|multiply(n0,#0)| | physics |
a test has 160 questions . each question has 5 options , but only 1 option is correct . if test - takers mark the correct option , they are awarded 1 point . however , if an answer is incorrectly marked , the test - taker loses 0.25 points . no points are awarded or deducted if a question is not attempted . a certain g... | a correct answers get you 1 point , an incorrect answer gets you minus 1 / 4 point and a skipped question gets you 0 points . since there are 200 total questions , there are a variety of ways to get a total of 40 points . let c be the number of correct answers and let i be the number of incorrect answers . to get 40 po... | a ) 23 , b ) 25 , c ) 27 , d ) 29 , e ) 31 | b | add(add(divide(subtract(160, 40), 5), const_1), const_1) | subtract(n0,n5)|divide(#0,n1)|add(#1,const_1)|add(#2,const_1) | general |
a glucose solution contains 15 grams of glucose per 100 cubic centimeters of solution . if 65 cubic centimeters of the solution were poured into an empty container , how many grams of glucose would be in the container ? | we are given that a glucose solution contains 15 grams of glucose per 100 cubic centimeters of solution . since we are dealing with a solution , we know that the grams of glucose is proportional to the number of cubic centimeters of solution . thus , to determine how many grams of glucose would be in the container when... | ['a ) 7.50', 'b ) 8.75', 'c ) 7.75', 'd ) 9.50', 'e ) 9.75'] | e | multiply(65, divide(15, 100)) | divide(n0,n1)|multiply(n2,#0) | physics |
the price of a coat in a certain store is $ 500 . if the price of the coat is to be reduced by $ 150 , by what percent is the price to be reduced ? | "price of a coat in a certain store = $ 500 the price of the coat is to be reduced by $ 150 % change = ( final value - initial value ) * 100 / initial value % reduction = ( reduction in price ) * 100 / initial value i . e . % reduction = ( 150 ) * 100 / 500 = 30 % answer : option e" | a ) 10 % , b ) 15 % , c ) 20 % , d ) 25 % , e ) 30 % | e | multiply(divide(150, 500), const_100) | divide(n1,n0)|multiply(#0,const_100)| | gain |
if a man lost 4 % by selling oranges at the rate of 60 a rupee at how many a rupee must he sell them to gain 44 % ? | "96 % - - - - 60 144 % - - - - ? 96 / 144 * 60 = 40 answer : d" | a ) 11 , b ) 28 , c ) 39 , d ) 40 , e ) 43 | d | divide(multiply(subtract(const_100, 4), 60), add(const_100, 44)) | add(n2,const_100)|subtract(const_100,n0)|multiply(n1,#1)|divide(#2,#0)| | gain |
a man can row 9 kmph in still water . when the river is running at 1.2 kmph , it takes him 1 hour to row to a place and back . what is the total distance traveled by the man ? | "m = 9 s = 1.2 ds = 10.2 us = 7.8 x / 10.2 + x / 7.8 = 1 x = 4.42 d = 4.42 * 2 = 8.84 answer : c" | a ) 6.24 km , b ) 6 km , c ) 8.84 km , d ) 5.66 km , e ) 10 km | c | multiply(divide(multiply(add(9, 1.2), subtract(9, 1.2)), add(add(9, 1.2), subtract(9, 1.2))), const_2) | add(n0,n1)|subtract(n0,n1)|add(#0,#1)|multiply(#0,#1)|divide(#3,#2)|multiply(#4,const_2)| | physics |
a train speeds past a pole in 15 sec and a platform 100 m long in 25 sec , its length is ? | "let the length of the train be x m and its speed be y m / sec . then , x / y = 15 = > y = x / 15 ( x + 100 ) / 25 = x / 15 = > x = 150 m . answer : b" | a ) 158 m , b ) 150 m , c ) 110 m , d ) 130 m , e ) 157 m | b | multiply(100, subtract(const_2, const_1)) | subtract(const_2,const_1)|multiply(n1,#0)| | physics |
two trains 300 m and 450 m long run at the speed of 60 km / hr and 40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? | "relative speed = 60 + 40 = 100 km / hr . = 100 * 5 / 18 = 250 / 9 m / sec . distance covered in crossing each other = 300 + 450 = 750 m . required time = 700 * 9 / 250 = 27 sec . answer : b" | a ) 10.6 , b ) 27 , c ) 10.4 , d ) 10.8 , e ) 10.1 | b | divide(add(300, 450), multiply(add(60, 40), const_0_2778)) | add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)| | physics |
a shopkeeper buys mangoes at the rate of 6 a rupee and sells them at 3 a rupee . find his net profit or loss percent ? | the total number of mangoes bought by the shopkeeper be 18 . if he buys 6 a rupee , his cp = 3 he selling at 3 a rupee , his sp = 6 profit = sp - cp = 6 - 3 = 3 profit percent = 3 / 6 * 100 = 50 % answer : c | a ) 33 4 / 3 % , b ) 33 1 / 7 % , c ) 50 % , d ) 32 1 / 3 % , e ) 60 % | c | divide(multiply(3, const_100), 6) | multiply(n1,const_100)|divide(#0,n0) | gain |
what percent of 180 is 45 ? | "here ' s an algebraic approach . let p = the required percent . this means p / 100 = the percent as a fraction . so , ` ` what percent of 180 is 45 ? ' ' becomes ` ` ( p / 100 ) of 180 = 45 or : ( p / 100 ) ( 180 ) = 45 divide both sides by 180 to get : p / 100 = 45 / 180 simplify right side : p / 100 = 1 / 4 multiply... | a ) 0.25 % , b ) 4 % , c ) 25 % , d ) 40 % , e ) 250 % | c | multiply(divide(180, 45), const_100) | divide(n0,n1)|multiply(#0,const_100)| | gain |
if 20 men can build a wall 112 metres long in 6 days , what length of a similar wall can be built by 15 men in 3 days ? | "20 men is 6 days can build 112 metres 15 men in 3 days can build = 112 * ( 15 / 20 ) x ( 3 / 6 ) = 42 meters answer : e ." | a ) 65 mtr . , b ) 52 mtr , c ) 70 mtr . , d ) 78 mtr . , e ) 42 mtr . | e | multiply(112, divide(multiply(15, 3), multiply(20, 6))) | multiply(n3,n4)|multiply(n0,n2)|divide(#0,#1)|multiply(n1,#2)| | physics |
b completes a work in 3 days . a alone can do it in 10 days . if both work together , the work can be completed in how many days ? | 1 / 3 + 1 / 10 = 13 / 30 30 / 13 = 2.31 days answer : a | a ) 2.31 days , b ) 4.31 days , c ) 5.31 days , d ) 6.31 days , e ) 7.31 days | a | divide(multiply(10, 3), add(3, 10)) | add(n0,n1)|multiply(n0,n1)|divide(#1,#0) | physics |
a and b started a business in partnership investing rs . 20,000 and rs . 15,000 respectively . after 6 months , c joined them with rs . 20,000 . whatwill be b ' s share in total profit of rs . 20,000 earned at the end of 2 years from the startingof the business ? | a : b : c = ( 20,000 x 24 ) : ( 15,000 x 24 ) : ( 20,000 x 18 ) = 4 : 3 : 3 . b ' s share = rs . 20000 x 3 / 10 = rs . 6,000 . d | a ) s . 5,000 , b ) s . 5,500 , c ) s . 5,700 , d ) s . 6,000 , e ) s . 7,500 | d | multiply(multiply(add(const_4, const_1), const_4), multiply(2, multiply(const_3, const_4))) | add(const_1,const_4)|multiply(const_3,const_4)|multiply(#0,const_4)|multiply(n5,#1)|multiply(#2,#3) | gain |
nick took a bus from home to market , that travels at 80 kmph . while walking back at 8 kmph , halfway through , he suddenly realized he was getting late and he cycled back the remaining distance in 60 kmph . find the average speed . | let the distance be 2 x ( one way ) time taken by bus = 2 x / 80 , by walking = x / 8 , by cycling = x / 60 hours : . average speed = total distance / total time = 8 x / x / 40 + x / 8 + x / 60 = 8 * 60 / 1.5 + 7.5 + 1 = 12.0 answer : d | a ) 49.5 kmph , b ) 49.0 kmph , c ) 48.5 kmph , d ) 48.0 kmph , e ) none of these | d | inverse(add(divide(inverse(60), const_2), inverse(80))) | inverse(n2)|inverse(n0)|divide(#0,const_2)|add(#2,#1)|inverse(#3) | physics |
a boat having a length 6 m and breadth 3 m is floating on a lake . the boat sinks by 1 cm when a man gets on it . the mass of the man is : | "volume of water displaced = ( 6 x 3 x 0.01 ) m 3 = 0.18 m 3 . mass of man = volume of water displaced x density of water = ( 0.18 x 1000 ) kg = 180 kg . answer : d" | a ) 100 kg , b ) 120 kg , c ) 89 kg , d ) 180 kg , e ) 110 kg | d | multiply(multiply(multiply(6, 3), divide(1, const_100)), const_1000) | divide(n2,const_100)|multiply(n0,n1)|multiply(#0,#1)|multiply(#2,const_1000)| | physics |
a computer is programmed to multiply consecutive even integers 2 * 4 * 6 * 8 * … * n until the product is divisible by 1551 , what is the value of n ? | "factorise 1551 . . 3 * 11 * 47 . . so n has to be a multiple of largest prime number , 61 . . so n = 2 * 47 = 94 . . ans : d" | a ) 22 , b ) 38 , c ) 62 , d ) 94 , e ) 672 | d | multiply(2, divide(divide(1551, add(const_10, const_1)), const_3)) | add(const_1,const_10)|divide(n4,#0)|divide(#1,const_3)|multiply(n0,#2)| | general |
0.999999 + 0.111111 = ? | "0.999999 + 0.111111 = 0.999999 + 0.11111 + 0.000001 = ( 0.999999 + 0.000001 ) + 0.11111 = 1 + 0.11111 = 1.11111 e" | a ) 1 , b ) 1.0001 , c ) 1.0021 , d ) 1.111 , e ) 1.11111 | e | multiply(divide(0.999999, 0.111111), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
if the cost price of 10 articles is same as the selling price of 25 articles . find the gain or loss percentage ? | "explanation : 10 cp = 25 sp cost price cp = 25 selling price sp = 10 formula = ( sp - cp ) / cp * 100 = ( 10 - 25 ) / 25 * 100 = 60 % loss answer : option d" | a ) 30 % gain , b ) 30 % loss , c ) 40 % gain , d ) 60 % loss , e ) 50 % loss | d | multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 25), 10)), divide(multiply(const_100, 25), 10))) | multiply(n1,const_100)|divide(#0,n0)|subtract(const_100,#1)|divide(#2,#1)|multiply(#3,const_100)| | gain |
what is the unit digit in 7 ^ 105 ? | "unit digit in 7 ^ 105 = unit digit in [ ( 7 ^ 4 ) ^ 26 x 7 ] but , unit digit in ( 7 ^ 4 ) ^ 26 = 1 unit digit in 7 ^ 105 = ( 1 x 7 ) = 7 . answer b ) 7 ." | a ) 9 , b ) 7 , c ) 3 , d ) 8 , e ) 5 | b | reminder(power(const_4.0, const_2), const_10) | power(const_4.0,const_2)|reminder(#0,const_10)| | general |
julie put half of her savings in a savings account that pays an annual simple interest and half in a savings account that pays an annual compound interest . after two years she earned $ 112 and $ 120 from the simple interest account and the compound interest account respectively . if the interest rates for both account... | $ 112 for 2 years = $ 56 per year . extra $ 8 yearned with the compound interest is the percent yearned on percent . so , $ 8 is yearned on $ 56 , which means that the interest = 7 % . this on the other hand means that half of the savings = 56 * 7 = $ 392 . twice of that = $ 784 . answer : b . | a ) 600 , b ) 784 , c ) 1080 , d ) 1200 , e ) 1440 | b | divide(112, divide(multiply(const_2, subtract(120, 112)), 112)) | subtract(n1,n0)|multiply(#0,const_2)|divide(#1,n0)|divide(n0,#2) | gain |
if 70 % of ( x - y ) = 30 % of ( x + y ) then what percent of x is y ? | "70 % of ( x - y ) = 30 % of ( x + y ) ( 70 / 100 ) ( x - y ) = ( 30 / 100 ) ( x + y ) 7 ( x - y ) = 3 ( x + y ) 4 x = 10 y x = 5 / 2 y therefore required percentage = ( ( y / x ) x 100 ) % = ( ( y / ( 5 / 2 ) y ) x 100 ) = 40 % answer is b ." | a ) 2.5 % , b ) 40 % , c ) 5 % , d ) 15 % , e ) 25 % | b | multiply(divide(subtract(70, 30), add(70, 30)), const_100) | add(n0,n1)|subtract(n0,n1)|divide(#1,#0)|multiply(#2,const_100)| | general |
a man spends 1 / 3 rd of his salary on food . he spends 1 / 4 th on rent and 1 / 5 th on cloths . if he is left with 1760 , then who much salary does he earn . | "let salary be x total spent = ( x / 3 + x / 4 + x / 5 ) = ( 47 / 60 ) * x left = ( x - 47 x / 60 ) = ( 13 / 60 ) * x given , ( 13 / 60 ) * x = 1760 x = ( 1760 * 60 ) / 13 = 8123.077 = 8123 ( approax ) answer : b" | a ) 8120 , b ) 8123 , c ) 8222 , d ) 8322 , e ) 8422 | b | divide(1760, subtract(1, add(add(divide(1, 3), divide(1, 4)), divide(1, 5)))) | divide(n0,n1)|divide(n0,n3)|divide(n0,n5)|add(#0,#1)|add(#3,#2)|subtract(n0,#4)|divide(n6,#5)| | general |
what is the least value of x , so that 2 x 5476 is divisible by 9 | "explanation : the sum of the digits of the number is divisible by 9 . then the number is divisible by 9 . 2 + x + 5 + 4 + 7 + 6 = 24 + x least value of x may be ' 3 ' , so that the total 24 + 3 = 27 is divisible by 9 . answer : option d" | a ) 7 , b ) 8 , c ) 9 , d ) 3 , e ) 2 | d | divide(divide(divide(lcm(2, 5476), 5476), const_4), const_4) | lcm(n0,n1)|divide(#0,n1)|divide(#1,const_4)|divide(#2,const_4)| | general |
the length of the ribbon was originally 55 cm . it was reduced in the ratio 11 : 7 . what is its length now ? | explanation : length of ribbon originally = 55 cm let the original length be 11 x and reduced length be 7 x . but 11 x = 55 cm x = 55 / 11 cm = 5 cm therefore , reduced length = 7 x cm = 7 5 cm = 35 cm answer : c | a ) 6 cm , b ) 12 cm , c ) 35 cm , d ) 44 cm , e ) none of these | c | multiply(55, divide(7, 11)) | divide(n2,n1)|multiply(n0,#0) | other |
at 12.00 hours , j starts to walk from his house at 6 kmph . at 13.30 , p follows him from j ' s house on his bicycle at 8 kmph . when will j be 3 km behind p ? | by the time p starts j is 1.5 hr x 6 = 9 km away from his house . j is 3 km behind when p is 3 km ahead of him . ie . , p has to cover 12 km . so he takes 12 / ( 8 - 6 ) = 6 hrs after 13.30 . so the required time is 19.30 hrs answer : b | a ) 19.39 , b ) 19.3 , c ) 19.32 , d ) 19.33 , e ) 19.36 | b | add(13.3, divide(add(multiply(6, divide(const_3, const_2)), 3), subtract(8, 6))) | divide(const_3,const_2)|subtract(n3,n1)|multiply(n1,#0)|add(n4,#2)|divide(#3,#1)|add(n2,#4) | physics |
how many seconds will a 800 m long train take to cross a man walking with a speed of 5 km / hr in the direction of the moving train if the speed of the train is 65 km / hr ? | "speed of train relative to man = 65 - 5 = 60 km / hr . = 60 * 5 / 18 = 50 / 3 m / sec . time taken to pass the man = 800 * 3 / 50 = 48 sec . answer : e" | a ) 30 , b ) 50 , c ) 66 , d ) 44 , e ) 48 | e | divide(800, multiply(subtract(65, 5), const_0_2778)) | subtract(n2,n1)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
the sum of five consecutive even numbers is 500 . find the middle number of the five ? | "middle number = 500 / 5 = 100 ans c" | a ) 140 , b ) 66 , c ) 100 , d ) 99 , e ) 120 | c | add(add(power(add(add(divide(subtract(subtract(500, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(500, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(500, const_10), const_2), const_4), const_2), power(... | 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 together can do a piece of work in 6 days and a alone can do it in 14 days . in how many days can b alone can do it ? | "explanation : a and b can do work 1 / 6 in 1 day a alone can do 1 / 14 work in 1 day b alone can do ( 1 / 6 - 1 / 14 ) = 2 / 21 work in 1 day = > complete work can be done in 21 / 2 days by b answer : option c" | a ) 12 days , b ) 15 days , c ) 21 / 2 days , d ) 21 days , e ) 22 days | c | inverse(subtract(inverse(6), inverse(14))) | inverse(n0)|inverse(n1)|subtract(#0,#1)|inverse(#2)| | physics |
triathlete dan runs along a 2 - mile stretch of river and then swims back along the same route . if dan runs at a rate of 8 miles per hour and swims at a rate of 6 miles per hour , what is his average rate for the entire trip in miles per minute ? | "dan travels 4 miles round trip . running part : ( 2 / 8 = 1 / 4 * 60 = 15 minutes ) swimming part : ( 2 / 6 = 1 / 3 * 60 = 20 minutes ) 4 miles in ( 15 + 20 ) minutes 4 / 35 = 1 / 9 mile per minute answer : 1 / 9 mile per minute" | a ) 1 / 9 , b ) 2 / 15 , c ) 3 / 15 , d ) 1 / 4 , e ) 3 / 8 | a | divide(add(2, 2), add(multiply(divide(2, 6), const_60), multiply(divide(2, 8), const_60))) | add(n0,n0)|divide(n0,n2)|divide(n0,n1)|multiply(#1,const_60)|multiply(#2,const_60)|add(#3,#4)|divide(#0,#5)| | physics |
a man engaged a servant on the condition that he would pay him rs . 600 and a uniform after one year service . he served only for 9 months and received uniform and rs . 400 , find the price of the uniform ? | 9 / 12 = 3 / 4 * 600 = 450 400 - - - - - - - - - - - - - 50 1 / 4 - - - - - - - - 50 1 - - - - - - - - - ? = > rs . 200 answer : e | a ) s . 80 , b ) s . 85 , c ) s . 90 , d ) s . 100 , e ) s . 200 | e | multiply(divide(subtract(multiply(9, 600), multiply(multiply(const_3, const_4), 400)), multiply(multiply(const_3, const_4), const_1)), const_4) | multiply(n0,n1)|multiply(const_3,const_4)|multiply(n2,#1)|multiply(#1,const_1)|subtract(#0,#2)|divide(#4,#3)|multiply(#5,const_4) | general |
find the value of y from ( 12 ) ^ 1 x 6 ^ 4 ÷ 432 = y ? | "36 e" | a ) 2134 , b ) 2234 , c ) 2540 , d ) 2560 , e ) 36 | e | divide(multiply(power(12, 1), power(6, 4)), 432) | power(n0,n1)|power(n2,n3)|multiply(#0,#1)|divide(#2,n4)| | general |
find the least number which when divided by 20 , 25,35 and 40 leaves remainders 14 , 19,29 and 34 respectively . | "here , ( 20 - 14 ) = 6 , ( 25 – 19 ) = 6 , ( 35 - 29 ) = 6 and ( 40 - 34 ) = 6 . therefore , required number = ( l . c . m . of 20,25 , 35,40 ) – 6 = 1394 . answer is a ." | a ) 1394 , b ) 1294 , c ) 1194 , d ) 1094 , e ) none of them | a | add(40, lcm(20, 25,35)) | lcm(n0,n1)|add(n2,#0)| | general |
if 1 / 8 of a pencil is black , 1 / 2 of the remaining is white and the remaining 3 1 / 2 cm is blue , find the total length of the pencil . | let the total length of the pencil be x cm . black part is 1 / 8 of the length so x / 8 . so the remaining part is x - x / 8 = = > 8 x / 8 - x / 8 = = > 7 x / 8 cm . white part = 1 / 2 * 7 x / 8 = 7 x / 16 cm . so the remaining part is 7 x / 8 - 7 x / 16 = = > 14 x / 16 - 7 x / 16 = 7 x / 16 which is equal to 3 1 / 2 .... | a ) 12 , b ) 9 , c ) 8 , d ) 6 , e ) 5 | c | multiply(8, 1) | multiply(n0,n1) | general |
marcella has 25 pairs of shoes . if she loses 9 individual shoes , what is the greatest number of matching pairs she could have left ? | "greatest number of paired shoes means greatest number of lost pairs 9 shoes have 4 pairs and 1 extra shoes of a pair . so a otal of 5 pairs she loses leftover pair = 20 answer : b" | a ) 21 , b ) 20 , c ) 19 , d ) 16 , e ) 15 | b | subtract(25, add(floor(divide(9, const_2)), const_1)) | divide(n1,const_2)|floor(#0)|add(#1,const_1)|subtract(n0,#2)| | general |
the radius of a wheel is 22.4 cm . what is the distance covered by the wheel in making 180 resolutions ? | "in one resolution , the distance covered by the wheel is its own circumference . distance covered in 180 resolutions . = 180 * 2 * 22 / 7 * 22.4 = 25344 cm = 253.4 m answer : e" | a ) 724 m , b ) 704 m , c ) 287 m , d ) 278 m , e ) 253.4 m | e | divide(multiply(multiply(multiply(divide(add(multiply(add(const_3, const_4), const_3), const_1), add(const_3, const_4)), 22.4), const_2), 180), const_100) | add(const_3,const_4)|multiply(#0,const_3)|add(#1,const_1)|divide(#2,#0)|multiply(n0,#3)|multiply(#4,const_2)|multiply(n1,#5)|divide(#6,const_100)| | physics |
if a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster , he would have covered 80 more miles than he actually did . how many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that... | "case 1 : let rate = r , time = t , and distance = d so d = rt case 2 : ( d + 80 ) = ( r + 5 ) ( t + 1 ) case 3 : ( d + x ) = ( r + 10 ) ( t + 2 ) x = 160 ans e" | a ) 100 , b ) 120 , c ) 140 , d ) 150 , e ) 160 | e | add(multiply(subtract(80, 5), 2), multiply(2, 10)) | multiply(n3,n4)|subtract(n2,n1)|multiply(n3,#1)|add(#2,#0)| | physics |
a reduction of 20 % in the price of oil enables a house wife to obtain 10 kgs more for rs . 1500 , what is the reduced price for kg ? | "explanation : 1500 * ( 20 / 100 ) = 300 - - - - 10 ? - - - - 1 = > rs . 30 answer : c" | a ) rs . 20 , b ) rs . 25 , c ) rs . 30 , d ) rs . 35 , e ) rs . 40 | c | divide(divide(multiply(1500, 20), const_100), 10) | multiply(n0,n2)|divide(#0,const_100)|divide(#1,n1)| | gain |
a train passes a station platform in 36 sec and a man standing on the platform in 20 sec . if the speed of the train is 99 km / hr . what is the length of the platform ? | "speed = 99 * 5 / 18 = 27.5 m / sec . length of the train = 27.5 * 20 = 550 m . let the length of the platform be x m . then , ( x + 550 ) / 36 = 27.5 = > x = 440 m . answer : a" | a ) 440 , b ) 488 , c ) 477 , d ) 426 , e ) 422 | a | multiply(20, multiply(99, const_0_2778)) | multiply(n2,const_0_2778)|multiply(n1,#0)| | physics |
100 white and black tiles will be used to form a 20 x 20 square pattern . if there must be at least one black tile in every row and at least one white tile in every column , what is the maximum difference between the number of black and white tiles that can be used ? | "answer = b please refer diagram below 90 - 20 = 70" | a ) 75 , b ) 70 , c ) 85 , d ) 90 , e ) 95 | b | subtract(100, 20) | subtract(n0,n1)| | general |
two numbers a and b are such that the sum of 6 % of a and 4 % of b is two - third of the sum of 6 % of a and 8 % of b . find the ratio of a : b . | "explanation : 6 % of a + 4 % of b = 2 / 3 ( 6 % of a + 8 % of b ) 6 a / 100 + 4 b / 100 = 2 / 3 ( 6 a / 100 + 8 b / 100 ) ⇒ 6 a + 4 b = 2 / 3 ( 6 a + 8 b ) ⇒ 18 a + 12 b = 12 a + 16 b ⇒ 6 a = 4 b ⇒ ab = 4 / 6 = 2 / 3 ⇒ a : b = 2 : 3 answer : option a" | a ) 2 : 3 , b ) 1 : 2 , c ) 4 : 3 , d ) 1 : 1 , e ) 3 : 2 | a | divide(subtract(divide(multiply(divide(8, const_100), 4), const_3), divide(const_2.0, const_100)), subtract(divide(6, const_100), divide(multiply(divide(6, const_100), const_2), const_3))) | divide(n3,const_100)|divide(n1,const_100)|divide(n0,const_100)|divide(n2,const_100)|multiply(#0,const_2)|multiply(#3,const_2)|divide(#4,const_3)|divide(#5,const_3)|subtract(#6,#1)|subtract(#2,#7)|divide(#8,#9)| | general |
if the sum of the first n positive integers is s , what is the sum of the first n negative even integers , in terms of s ? | answer is c given sum of - 1 - 2 - 3 . . . . . - n = s sum of - 2 , - 4 , - 6 . . . . - 2 n = 2 [ sum ( - 1 - 2 - 3 . . . . n ) ] = 2 s | a ) s / 2 , b ) s , c ) 2 s , d ) 2 s + 2 , e ) 4 s | c | negate(const_2) | negate(const_2) | general |
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