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
values |
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ravi can do a piece of work in 24 days while prakash can do it in 40 days . in how many days will they finish it together ? | "1 / 24 + 1 / 40 = 1 / 15 15 days answer : a" | a ) 15 , b ) 16 , c ) 17 , d ) 18 , e ) 10 | a | divide(const_1, add(inverse(24), inverse(40))) | inverse(n0)|inverse(n1)|add(#0,#1)|divide(const_1,#2)| | physics |
the tax on a commodity is diminished by 30 % but its consumption is increased by 20 % . find the decrease percent in the revenue derived from it ? | 100 * 100 = 10000 70 * 120 = 8400 10000 - - - - - - - 1600 100 - - - - - - - ? = 16 % answer : d | a ) 18 % , b ) 72 % , c ) 32 % , d ) 16 % , e ) 52 % | d | subtract(const_100, divide(multiply(add(const_100, 20), subtract(const_100, 30)), const_100)) | add(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|subtract(const_100,#3) | general |
when throwing 2 dices , and looking at the sum of numbers on the dices - what is the probability to have a sum which is smaller than 5 ? | a dice is composed of 6 numbers - 12 , 34 , 56 . when trowing 2 dices - there are 36 options ( 6 x 6 = 36 ) . if we want the sum of two dices to be < 5 there are those options : ( 11 ) , ( 12 ) , ( 21 ) , ( 13 ) , ( 31 ) , ( 22 ) - total of 6 options . therefore - the probability to have a sum which is smaller than 5 i... | a ) 1 / 9 , b ) 1 / 6 , c ) 1 / 2 , d ) 1 / 3 , e ) 32 / 36 | b | divide(divide(2, const_2), add(5, const_1)) | add(n1,const_1)|divide(n0,const_2)|divide(#1,#0) | general |
a man is 24 years older than his son . in two years , his age will be twice the age of his son . the present age of the son is : | "let the son ' s present age be x years . then , man ' s present age = ( x + 24 ) years . ( x + 24 ) + 2 = 2 ( x + 2 ) x + 26 = 2 x + 4 = > x = 22 . answer is a ." | a ) 22 , b ) 28 , c ) 32 , d ) 48 , e ) 92 | a | divide(subtract(24, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | multiply(const_2,const_2)|subtract(const_2,const_1)|subtract(#0,const_2)|subtract(n0,#2)|divide(#3,#1)| | general |
in a mixed college 160 students are there in one class . out of this 160 students 1 / 4 students are girls . how many boys are there ? | total number of students : 160 total girls : 160 * 1 / 4 = 40 total boys : 160 - 40 = 120 answer is d | a ) a ) 40 , b ) b ) 60 , c ) c ) 80 , d ) d ) 120 , e ) e ) 140 | d | multiply(divide(160, 4), const_3) | divide(n0,n3)|multiply(#0,const_3) | general |
a and b put in rs . 300 and rs . 500 respectively into a business . a reinvests into the business his share of the first year ' s profit of rs . 200 where as b does not . in what ratio should they divide the second year ' s profit ? | explanation : 3 : 5 a = 3 / 8 * 200 = 75 375 : 400 39 : 40 answer : c | a ) 39 : 40 , b ) 39 : 49 , c ) 15 : 16 , d ) 14 : 18 , e ) 39 : 41 | c | divide(add(300, multiply(divide(300, add(subtract(500, const_100), 300)), add(200, multiply(const_2, add(const_3, const_2))))), subtract(500, const_100)) | add(const_2,const_3)|subtract(n1,const_100)|add(n0,#1)|multiply(#0,const_2)|add(n2,#3)|divide(n0,#2)|multiply(#4,#5)|add(n0,#6)|divide(#7,#1) | gain |
if 42.18 = k ( 14 + m / 50 ) , where k and m are positive integers and m < 50 , then what is the value of k + m ? | "42.18 = 14 k + km / 50 . . . we can rewrite the number as follows : 42 + 0.18 = 14 k + km / 50 . . . . . . . . since k is integer , then 42 = 14 k . . . . . . . . . . k = 3 0.18 = km / 50 . . . . . . 18 / 100 = 3 m / 50 . . . . . . m = 3 k + m = 3 + 3 = 6 answer : a" | a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | a | add(divide(multiply(subtract(const_0_25, divide(const_1, const_100)), 50), divide(subtract(42.18, subtract(const_0_25, divide(const_1, const_100))), 14)), divide(subtract(42.18, subtract(const_0_25, divide(const_1, const_100))), 14)) | divide(const_1,const_100)|subtract(const_0_25,#0)|multiply(n2,#1)|subtract(n0,#1)|divide(#3,n1)|divide(#2,#4)|add(#5,#4)| | general |
it takes joey the postman 1 hours to run a 5 mile long route every day . he delivers packages and then returns to the post office along the same path . if the average speed of the round trip is 8 mile / hour , what is the speed with which joey returns ? | let his speed for one half of the journey be 5 miles an hour let the other half be x miles an hour now , avg speed = 8 mile an hour 2 * 5 * x / 5 + x = 8 10 x = 8 x + 40 = > 2 x = 40 = > x = 20 e | a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 20 | e | divide(5, subtract(divide(multiply(const_2, 5), 8), 1)) | multiply(n1,const_2)|divide(#0,n2)|subtract(#1,n0)|divide(n1,#2) | physics |
a β s speed is 20 / 16 times that of b . if a and b run a race , what part of the length of the race should a give b as a head start , so that the race ends in a dead heat ? | "we have the ratio of a β s speed and b β s speed . this means , we know how much distance a covers compared with b in the same time . this is what the beginning of the race will look like : ( start ) a _________ b ______________________________ if a covers 20 meters , b covers 16 meters in that time . so if the race i... | a ) 1 / 17 , b ) 3 / 17 , c ) 1 / 10 , d ) 4 / 20 , e ) 3 / 10 | d | divide(subtract(20, 16), 20) | subtract(n0,n1)|divide(#0,n0)| | general |
for any integer k > 1 , the term β length of an integer β refers to the number of positive prime factors , not necessarily distinct , whose product is equal to k . for example , if k = 24 , the length of k is equal to 4 , since 24 = 2 Γ 2 Γ 2 Γ 3 . if x and y are positive integers such that x > 1 , y > 1 , and x + 3 y ... | "we know that : x > 1 , y > 1 , and x + 3 y < 980 , and it is given that length means no of factors . for any value of x and y , the max no of factors can be obtained only if factor is smallest no all factors are equal . hence , lets start with smallest no 2 . 2 ^ 1 = 2 2 ^ 2 = 4 2 ^ 3 = 8 2 ^ 4 = 16 2 ^ 5 = 32 2 ^ 6 =... | a ) 16 , b ) 14 , c ) 18 , d ) 10 , e ) 20 | a | add(add(4, 3), add(add(4, 4), 1)) | add(n2,n7)|add(n2,n2)|add(n0,#1)|add(#0,#2)| | general |
how long will a boy take to run round a square field of side 55 meters , if he runs at the rate of 9 km / hr ? | "speed = 9 km / hr = 9 * 5 / 18 = 5 / 2 m / sec distance = 55 * 4 = 220 m time taken = 220 * 2 / 5 = 88 sec answer is a" | a ) 88 sec , b ) 45 sec , c ) 1 min , d ) 32 sec , e ) 25 sec | a | divide(multiply(55, const_4), multiply(9, divide(const_1000, const_3600))) | divide(const_1000,const_3600)|multiply(n0,const_4)|multiply(n1,#0)|divide(#1,#2)| | gain |
a contractor undertook to do a piece of work in 15 days . he employed certain number of laboures but 5 of them were absent from the very first day and the rest could finish the work in only 20 days . find the number of men originally employed ? | "let the number of men originally employed be x . 15 x = 20 ( x Γ’ β¬ β 5 ) or x = 20 answer b" | a ) 25 , b ) 20 , c ) 19 , d ) 23 , e ) 18 | b | divide(multiply(20, 5), subtract(20, 15)) | multiply(n1,n2)|subtract(n2,n0)|divide(#0,#1)| | physics |
the temperature of a certain cup of coffee 2 minutes after it was poured was 120 degrees fahrenheit . if the temperature f of the coffee t minutes after it was poured can be determined by the formula f = 120 * 2 ^ ( - at ) + 60 , where f is in degrees fahrenheit and a is a constant . then the temperature of the coffee ... | "first , we have to find a . we know that after t = 2 minutes the temperature f = 120 degrees . hence : 120 = 120 * ( 2 ^ - 2 a ) + 60 60 = 120 * ( 2 ^ - 2 a ) 60 / 120 = 2 ^ - 2 a 1 / 2 = 2 ^ - 2 a 2 ^ - 1 = 2 ^ - 2 a - 1 = - 2 a 1 / 2 = a now we need to find f after t = 30 minutes : f = 120 * ( 2 ^ - 1 / 2 * 30 ) + 6... | a ) 65 , b ) 60.003662 , c ) 80.2 , d ) 85 , e ) 90 | b | add(multiply(power(2, multiply(divide(60, 2), subtract(const_1, 2))), 120), 60) | divide(n4,n0)|subtract(const_1,n3)|multiply(#0,#1)|power(n3,#2)|multiply(n1,#3)|add(n4,#4)| | general |
10 % of ( 50 % of $ 500 ) is ? | "10 % of ( 50 % of $ 500 ) = 10 / 100 ( 50 / 100 * 500 ) = $ 25 answer is c" | a ) $ 15 , b ) $ 20 , c ) $ 25 , d ) $ 30 , e ) $ 22 | c | divide(multiply(divide(multiply(500, 50), const_100), 10), const_100) | multiply(n1,n2)|divide(#0,const_100)|multiply(n0,#1)|divide(#2,const_100)| | gain |
a chemical supply company has 60 liters of a 20 % hno 3 solution . how many liters of pure undiluted hno 3 must the chemists add so that the resultant solution is a 50 % solution ? | "60 liters of a 20 % hno 3 solution means hno 3 = 12 liters in 60 liters of the solution . now , let x be the pure hno 3 added . as per question , 12 + x = 50 % of ( 60 + x ) or x = 36 . hence , e" | a ) 12 , b ) 15 , c ) 20 , d ) 24 , e ) 36 | e | multiply(subtract(divide(60, const_2), divide(multiply(60, 20), const_100)), const_2) | divide(n0,const_2)|multiply(n0,n1)|divide(#1,const_100)|subtract(#0,#2)|multiply(#3,const_2)| | gain |
a can finish a piece of work in 5 days . b can do it in 25 days . they work together for four days and then a goes away . in how many days will b finish the work ? | "4 / 5 + ( 4 + x ) / 25 = 1 = > x = 1 day answer : b" | a ) 3 days , b ) 1 day , c ) 2 days , d ) 4 days , e ) 5 days | b | divide(subtract(const_1, add(multiply(divide(const_1, 5), const_2), multiply(divide(const_1, 25), const_2))), divide(const_1, 25)) | divide(const_1,n0)|divide(const_1,n1)|multiply(#0,const_2)|multiply(#1,const_2)|add(#2,#3)|subtract(const_1,#4)|divide(#5,#1)| | physics |
4 women can complete a work in 7 days and 10 children take 14 days to complete the work . how many days will 5 women and 10 children take to complete the work ? | "1 women ' s 1 day work = 1 / 28 1 child ' s 1 day work = 1 / 140 ( 5 women + 10 children ) ' s 1 day work = ( 5 / 28 + 10 / 140 ) = 1 / 4 5 women and 10 children will complete the work in 4 days . answer : a" | a ) 4 days , b ) 6 days , c ) 7 days , d ) 9 days , e ) 1 days | a | inverse(add(divide(5, multiply(4, 7)), divide(4, multiply(4, 14)))) | multiply(n0,n1)|multiply(n0,n3)|divide(n4,#0)|divide(n0,#1)|add(#2,#3)|inverse(#4)| | physics |
if p is a prime number greater than 3 , find the remainder when p ^ 2 + 13 is divided by 12 . | "every prime number greater than 3 can be written 6 n + 1 or 6 n - 1 . if p = 6 n + 1 , then p ^ 2 + 13 = 36 n ^ 2 + 12 n + 1 + 13 = 36 n ^ 2 + 12 n + 12 + 2 if p = 6 n - 1 , then p ^ 2 + 13 = 36 n ^ 2 - 12 n + 1 + 13 = 36 n ^ 2 - 12 n + 12 + 2 when divided by 12 , it must leave a remainder of 2 . the answer is d ." | a ) 6 , b ) 1 , c ) 0 , d ) 2 , e ) 7 | d | subtract(add(13, power(add(const_1, const_4), 2)), multiply(12, 3)) | add(const_1,const_4)|multiply(n0,n3)|power(#0,n1)|add(n2,#2)|subtract(#3,#1)| | general |
yesterday ' s closing prices of 2,640 different stocks listed on a certain stock exchange were all different from today ' s closing prices . the number of stocks that closed at a higher price today than yesterday was 20 percent greater than the number that closed at a lower price . how many of the stocks closed at a hi... | "lets consider the below - the number of stocks that closed at a higher price = h the number of stocks that closed at a lower price = l we understand from first statement - > h + l = 2640 - - - - ( 1 ) we understand from second statement - > h = ( 120 / 100 ) l = > h = 1.2 l - - - - ( 2 ) solve eq ( 1 ) ( 2 ) to get h ... | a ) 484 , b ) 726 , c ) 1,100 , d ) 1,320 , e ) 1,440 | e | multiply(divide(subtract(subtract(multiply(20, const_100), const_10), const_10), add(add(const_1, divide(20, const_100)), const_1)), add(const_1, divide(20, const_100))) | divide(n1,const_100)|multiply(n1,const_100)|add(#0,const_1)|subtract(#1,const_10)|add(#2,const_1)|subtract(#3,const_10)|divide(#5,#4)|multiply(#2,#6)| | gain |
a bag consists of 20 marbles , of which 6 are blue , 9 are red , and the remainder are white . if lisa is to select a marble from the bag at random , what is the probability that the marble will be red or white ? | "bag consists of 20 marbles , of which 6 are blue , 9 are red remainder are white . so , white = 20 - 6 - 9 = 5 . probability that the marble will be red or white = probability that the marble will be red + probability that the marble will be white probability that the marble will be red or white = 9 / 20 + 5 / 20 = 14... | a ) 7 / 10 , b ) 2 / 4 , c ) 1 / 4 , d ) 1 / 8 , e ) 1 / 16 | a | divide(add(subtract(20, add(6, 9)), 9), 20) | add(n1,n2)|subtract(n0,#0)|add(n2,#1)|divide(#2,n0)| | probability |
suppose 12 monkeys take 12 minutes to eat 12 bananas . how many monkeys would it take to eat 72 bananas in 72 minutes ? | "one monkey takes 12 min to eat 1 banana , so in 72 mins 1 monkey will eat 6 bananas , so for 72 bananas in 72 min we need 72 / 6 = 12 monkeys answer : d" | a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 13 | d | divide(72, divide(72, 12)) | divide(n3,n0)|divide(n3,#0)| | physics |
a farmer with 1,350 acres of land had planted his fields with corn , sugar cane , and tobacco in the ratio of 5 : 2 : 2 , respectively , but he wanted to make more money , so he shifted the ratio to 2 : 2 : 5 , respectively . how many more acres of land were planted with tobacco under the new system ? | originally ( 2 / 9 ) * 1350 = 300 acres were planted with tobacco . in the new system ( 5 / 9 ) * 1350 = 750 acres were planted with tobacco . thus 750 - 300 = 450 more acres were planted with tobacco . the answer is e . | a ) 90 , b ) 150 , c ) 270 , d ) 300 , e ) 450 | e | subtract(multiply(add(add(multiply(const_100, const_10), multiply(const_3, const_100)), multiply(5, const_10)), divide(5, add(add(5, 2), 2))), multiply(add(add(multiply(const_100, const_10), multiply(const_3, const_100)), multiply(5, const_10)), divide(2, add(add(5, 2), 2)))) | add(n1,n2)|multiply(const_10,const_100)|multiply(const_100,const_3)|multiply(n1,const_10)|add(#1,#2)|add(n2,#0)|add(#4,#3)|divide(n1,#5)|divide(n2,#5)|multiply(#6,#7)|multiply(#6,#8)|subtract(#9,#10) | other |
the ratio of ages of aman , bren , and charlie are in the ratio 5 : 8 : 7 respectively . if 8 years ago , the sum of their ages was 76 , what will be the age of charlie 9 years from now ? | "let the present ages of aman , bren , and charlie be 5 x , 8 x and 7 x respectively . 5 x - 8 + 8 x - 8 + 7 x - 8 = 76 x = 5 present age of charlie = 7 * 5 = 35 charlie ' s age 9 years hence = 35 + 9 = 44 answer = a" | a ) 44 , b ) 52 , c ) 57 , d ) 65 , e ) 80 | a | add(multiply(divide(add(multiply(8, const_3), 76), add(add(5, 8), 7)), 8), 9) | add(n0,n1)|multiply(n1,const_3)|add(n4,#1)|add(n2,#0)|divide(#2,#3)|multiply(n1,#4)|add(n5,#5)| | general |
the radius of a semi circle is 6.4 cm then its perimeter is ? | 36 / 7 r = 6.4 = 32.9 answer : a | ['a ) 32.9', 'b ) 32.4', 'c ) 22.4', 'd ) 32.8', 'e ) 32.1'] | a | add(divide(circumface(6.4), const_2), multiply(6.4, const_2)) | circumface(n0)|multiply(n0,const_2)|divide(#0,const_2)|add(#2,#1) | physics |
the grade point average of the entire class is 87 . if the average of one third of the class is 95 , what is the average of the rest of the class ? | "let x be the number of students in the class . let p be the average of the rest of the class . 87 x = ( 1 / 3 ) 95 x + ( 2 / 3 ) ( p ) x 261 = 95 + 2 p 2 p = 166 p = 83 . the answer is c ." | a ) 81 , b ) 82 , c ) 83 , d ) 84 , e ) 85 | c | divide(subtract(multiply(87, const_4), 95), subtract(const_4, const_1)) | multiply(n0,const_4)|subtract(const_4,const_1)|subtract(#0,n1)|divide(#2,#1)| | general |
an equilateral triangle t 2 is formed by joining the mid points of the sides of another equilateral triangle t 1 . a third equilateral triangle t 3 is formed by joining the mid - points of t 2 and this process is continued indefinitely . if each side of t 1 is 30 cm , find the sum of the perimeters of all the triangles... | "we have 30 for first triangle , when we join mid - points of first triangle we get the second equilateral triangle then the length of second one is 15 and continues . so we have 30,15 , 7.5 , . . . we have ratio = 1 / 2 , and it is gp type . sum of infinite triangle is a / 1 - r = 30 / 1 - ( 1 / 2 ) = 60 equilateral t... | a ) 180 cm , b ) 220 cm , c ) 240 cm , d ) 270 cm , e ) 300 cm | a | add(triangle_perimeter(30, 30, 30), triangle_perimeter(30, 30, 30)) | triangle_perimeter(n5,n5,n5)|add(#0,#0)| | geometry |
john and amanda stand at opposite ends of a straight road and start running towards each other at the same moment . their rates are randomly selected in advance so that john runs at a constant rate of 2 , 3 , 4 , 5 , or 6 miles per hour and amanda runs at a constant rate of 3 , 4 , 5 , 6 , or 7 miles per hour . what is... | "john will run farther if he runs at 6 mph and amanda runs at 5 mph , 4 mph , or 3 mph . in this case , p ( john runs farther ) = 1 / 5 * 3 / 5 = 3 / 25 john will run farther if he runs at 5 mph and amanda runs at 4 mph or 3 mph . in this case , p ( john runs farther ) = 1 / 5 * 2 / 5 = 2 / 25 john will run farther if ... | a ) 1 / 3 , b ) 2 / 5 , c ) 3 / 5 , d ) 4 / 25 , e ) 6 / 25 | e | divide(6, multiply(4, 5)) | multiply(n5,n3)|divide(n1,#0)| | physics |
the sum of four consecutive even numbers is 292 . what would be the largest number ? | "let the four consecutive even numbers be 2 ( x - 2 ) , 2 ( x - 1 ) , 2 x , 2 ( x + 1 ) their sum = 8 x - 4 = 292 = > x = 37 smallest number is : 2 ( x + 1 ) = 76 . answer : c" | a ) 33 , b ) 88 , c ) 76 , d ) 123 , e ) 12 | c | add(add(power(add(add(divide(subtract(subtract(292, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(292, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(292, 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 |
on dividing 127 by a number , the quotient is 5 and the remainder is 2 . find the divisor . | "d = ( d - r ) / q = ( 127 - 2 ) / 5 = 125 / 5 = 25 a" | a ) 25 , b ) 30 , c ) 35 , d ) 40 , e ) 45 | a | floor(divide(127, 5)) | divide(n0,n1)|floor(#0)| | general |
two pipes a and b can separately fill a cistern in 80 minutes and 160 minutes respectively . there is a third pipe in the bottom of the cistern to empty it . if all the three pipes are simultaneously opened , then the cistern is full in 40 minutes . in how much time , the third pipe alone can empty the cistern ? | "1 / 40 - ( 1 / 80 + 1 / 160 ) = - 1 / 160 third pipe can empty in 160 minutes answer : b" | a ) 90 min , b ) 160 min , c ) 110 min , d ) 120 min , e ) 130 min | b | inverse(subtract(add(inverse(80), inverse(160)), inverse(40))) | inverse(n0)|inverse(n1)|inverse(n2)|add(#0,#1)|subtract(#3,#2)|inverse(#4)| | physics |
if 15 students in a class average 70 % on an exam and 10 students average 95 % on the same exam , what is the average in percent for all 25 students ? | "( 15 * 70 + 10 * 95 ) / 25 = 80 % the answer is d ." | a ) 77 % , b ) 78 % , c ) 79 % , d ) 80 % , e ) 81 % | d | divide(add(multiply(15, 70), multiply(10, 95)), 25) | multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)|divide(#2,n4)| | general |
a reduction of 30 % in the price of oil enables a house wife to obtain 9 kgs more for rs . 1800 , what is the reduced price for kg ? | "explanation : 1800 * ( 30 / 100 ) = 540 - - - - 9 ? - - - - 1 = > rs . 60 answer : b" | a ) rs . 55 , b ) rs . 60 , c ) rs . 65 , d ) rs . 70 , e ) rs . 75 | b | divide(divide(multiply(1800, 30), const_100), 9) | multiply(n0,n2)|divide(#0,const_100)|divide(#1,n1)| | gain |
a train passes a man standing on a platform in 8 seconds and also crosses the platform which is 273 metres long in 20 seconds . the length of the train ( in metres ) is : | "explanation : let the length of train be l m . acc . to question ( 273 + l ) / 20 = l / 8 2184 + 8 l = 20 l l = 2184 / 12 = 182 m answer a" | a ) 182 , b ) 176 , c ) 175 , d ) 96 , e ) none of these | a | multiply(divide(273, subtract(20, 8)), 8) | subtract(n2,n0)|divide(n1,#0)|multiply(n0,#1)| | physics |
a certain manufacturer produces items for which the production costs consist of annual fixed costs totaling $ 130000 and variables costs averaging $ 10 per item . if the manufacturer β s selling price per item is $ 15 , how many items the manufacturer produce and sell to earn an annual profit of $ 150000 ? | let the items manufactured or sold bex 130000 + 10 x = 15 x - 150000 5 x = 280000 x = 56000 ans : e | a ) 2,858 , b ) 18,667 , c ) 21,429 , d ) 35,000 , e ) 56,000 | e | add(divide(divide(divide(add(150000, 130000), subtract(15, 10)), const_1000), const_3), const_3) | add(n0,n3)|subtract(n2,n1)|divide(#0,#1)|divide(#2,const_1000)|divide(#3,const_3)|add(#4,const_3) | general |
robert is travelling on his cycle and has calculated to reach point a at 2 p . m . if he travels at 10 km / hr ; he will reach there at 12 noon if he travels at 15 km / hr . at what speed must he travel to reach a at 1 p . m . ? | d 12 kmph let the distance traveled be x km . then , x / 10 - x / 15 = 2 3 x - 2 x = 60 = > x = 60 km . time taken to travel 60 km at 10 km / hr = 60 / 10 = 6 hrs . so , robert started 6 hours before 2 . p . m . i . e . , at 8 a . m . required speed = 60 / 5 = 12 kmph . | a ) 17 kmph , b ) 19 kmph , c ) 15 kmph , d ) 12 kmph , e ) 16 kmph | d | divide(divide(2, subtract(divide(const_1, 10), divide(const_1, 15))), add(const_4, const_1)) | add(const_1,const_4)|divide(const_1,n1)|divide(const_1,n3)|subtract(#1,#2)|divide(n0,#3)|divide(#4,#0) | physics |
seed mixture x is 40 percent ryegrass and 60 percent bluegrass by weight ; seed mixture y is 25 percent ryegrass and 75 percent fescue . if a mixture of x and y contains 32 percent ryegrass , what percent of the weight of this mixture is x ? | "- - - - - - - - - - - - - - - - > ryegrass x - - - - - - - - - - - - - - > 40 % y - - - - - - - - - - - - - - > 25 % m ( mixture ) - - - - > 32 % 0.4 x + ( m - x ) 0.25 = 0.32 m 0.15 x = 0.07 m x = 0.4666 m x = 46.66 % of m c" | a ) 10 % , b ) 33.33 % , c ) 46.66 % , d ) 50 % , e ) 66.66 % | c | divide(subtract(32, 25), subtract(divide(40, const_100), divide(25, const_100))) | divide(n0,const_100)|divide(n2,const_100)|subtract(n4,n2)|subtract(#0,#1)|divide(#2,#3)| | gain |
during 2003 , a company produced an average of 2000 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 150 % over its 2003 average ? | company produced 12 * 2000 = 24,000 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 + 24,000 . the gives the average of ( x + 24,000 ) / 4 . this average needs to be 300 % higher than that in 2003 . in math terms , 36,000 ... | a ) 287,000 , b ) 290,000 , c ) 284,000 , d ) 285,000 , e ) 286,000 | c | divide(subtract(multiply(multiply(multiply(add(divide(150, const_100), const_1), 2000), add(subtract(2007, 2003), const_1)), const_12), multiply(2000, const_12)), const_1000) | divide(n6,const_100)|multiply(n1,const_12)|subtract(n3,n0)|add(#2,const_1)|add(#0,const_1)|multiply(n1,#4)|multiply(#3,#5)|multiply(#6,const_12)|subtract(#7,#1)|divide(#8,const_1000) | general |
in how many ways can a 5 - letter password be chosen , using the letters a , b , c , d , e , f , g , and / or h such that at least one letter is repeated within the password ? | total number of four letter passwords = 8 * 8 * 8 * 8 * 8 = 32768 - - - - - - ( 1 ) total number of passwords in which no letter repeats = 8 c 5 * 5 ! = 56 * 120 = 6720 - - - - - - ( 2 ) therefore required value = ( 1 ) - ( 2 ) = 32768 - 6720 = 26048 e | a ) 720 , b ) 864 , c ) 900 , d ) 936 , e ) 26048 | e | subtract(power(add(5, const_3), 5), divide(factorial(add(5, const_3)), factorial(const_3))) | add(n0,const_3)|factorial(const_3)|factorial(#0)|power(#0,n0)|divide(#2,#1)|subtract(#3,#4) | general |
how many seconds will a 700 metre long train take to cross a man walking with a speed of 3 km / hr in the direction of the moving train if the speed of the train is 63 km / hr ? | "relative speed of the train = 63 - 3 = 60 kmph = 60 * 5 / 18 = 50 / 3 m / sec t = 700 * 3 / 50 = 42 sec answer : c" | a ) 25 , b ) 30 , c ) 42 , d ) 45 , e ) 50 | c | divide(700, multiply(const_0_2778, subtract(63, 3))) | subtract(n2,n1)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
a man walking at a constant rate of 5 miles per hour is passed by a woman traveling in the same direction along the same path at a constant rate of 15 miles per hour . the woman stops to wait for the man 2 minutes after passing him , while the man continues to walk at his constant rate . how many minutes must the woman... | "when the woman passes the man , they are aligned ( m and w ) . they are moving in the same direction . after 5 minutes , the woman ( w ) will be ahead the man ( m ) : m - - - - - - m - - - - - - - - - - - - - - - w w in the 5 minutes , after passing the man , the woman walks the distance mw = ww , which is 2 * 15 / 60... | a ) 1 , b ) 2 , c ) 3 , d ) 5 , e ) 4 | e | multiply(const_60, divide(multiply(divide(2, const_60), subtract(15, 5)), 5)) | divide(n2,const_60)|subtract(n1,n0)|multiply(#0,#1)|divide(#2,n0)|multiply(#3,const_60)| | physics |
a certain galaxy is known to comprise approximately 2 x 10 ^ 11 stars . of every 50 million of these stars , one is larger in mass than our sun . approximately how many stars in this galaxy are larger than the sun ? | total no . of stars on galaxy = 2 * 10 ^ 11 of every 50 million stars , 1 is larger than sun . 1 million = 10 ^ 6 therofore , 50 million = 50 * 10 ^ 6 total no . of stars larger than sun = 2 * 10 ^ 11 / 50 * 10 ^ 6 = 20 * 10 ^ 3 / 5 = 4000 therefore answer is d | a ) 800 , b ) 1,250 , c ) 8,000 , d ) 4,000 , e ) 80,000 | d | multiply(divide(multiply(divide(multiply(2, 10), 50), power(10, const_4)), const_1000), 2) | multiply(n0,n1)|power(n1,const_4)|divide(#0,n3)|multiply(#2,#1)|divide(#3,const_1000)|multiply(n0,#4) | general |
a side of beef lost 35 percent of its weight in processing . if the side of beef weighed 550 pounds after processing , how many pounds did it weigh before processing ? | "let weight of side of beef before processing = x ( 65 / 100 ) * x = 550 = > x = ( 550 * 100 ) / 65 = 846 answer d" | a ) 191 , b ) 355 , c ) 737 , d ) 846 , e ) 1,560 | d | divide(multiply(550, const_100), subtract(const_100, 35)) | multiply(n1,const_100)|subtract(const_100,n0)|divide(#0,#1)| | gain |
right triangle abc is to be drawn in the xy - plane so that the right angle is at a and ab is parallel to the y - axis . if the x - and y - coordinates of a , b , and c are to be integers that are consistent with the inequalities - 5 β€ x β€ 2 and 4 β€ y β€ 9 , then how many different triangles can be drawn that will meet ... | "we have the rectangle with dimensions 9 * 5 ( 9 horizontal dots and 5 vertical ) . ab is parallel to y - axis and ac is parallel to x - axis . choose the ( x , y ) coordinates for vertex a : 9 c 1 * 5 c 1 ; choose the x coordinate for vertex c ( as y coordinate is fixed by a ) : 8 c 1 , ( 9 - 1 = 8 as 1 horizontal dot... | a ) 54 , b ) 1440 , c ) 2,160 , d ) 2,916 , e ) 148,824 | b | multiply(multiply(5, subtract(5, const_1)), multiply(9, 5)) | multiply(n0,n3)|subtract(n0,const_1)|multiply(n0,#1)|multiply(#2,#0)| | geometry |
one half of a two digit number exceeds its one third by 5 . what is the sum of the digits of the number ? | "x / 2 β x / 3 = 5 = > x = 30 3 + 0 = 3 answer : a" | a ) a ) 3 , b ) b ) 5 , c ) c ) 7 , d ) d ) 9 , e ) e ) 11 | a | add(reminder(multiply(5, const_4), const_10), const_1) | multiply(n0,const_4)|reminder(#0,const_10)|add(#1,const_1)| | general |
the triplicate ratio of 1 : 8 is ? | "13 : 83 = 1 : 512 answer : d" | a ) 1 : 0 , b ) 1 : 8 , c ) 1 : 7 , d ) 1 : 512 , e ) 1 : 1 | d | divide(power(const_2.0, 8), power(const_3.0, 8)) | power(const_2.0,n1)|power(const_3.0,n1)|divide(#0,#1)| | other |
a white paper 2 inches wide is placed around a rectangular black paper with dimensions 8 inches by 12 inches . what is the area of the white paper , in square inches ? | this question is an example of a ' punch out ' question - we have to find the area of everything , then ' punch out ' the part that we do n ' t want . we ' re told that a white paper 2 inches wide is placed around a rectangular black paper with dimensions 8 inches by 12 inches . we ' re asked for the area of the frame ... | ['a ) 96', 'b ) 196', 'c ) 128', 'd ) 144', 'e ) 168'] | b | multiply(12, multiply(8, 2)) | multiply(n0,n1)|multiply(n2,#0) | geometry |
find the greatest number which on dividing 4351 and 5161 , leaves a reminder of 8 and 10 respectively | "explanation : in this type of question , its obvious we need to calculate the hcf , trick is hcf of ( 4351 - 8 ) and ( 5161 - 10 ) = hcf ( 4343 , 5151 ) = 101 option d" | a ) 125 , b ) 111 , c ) 129 , d ) 101 , e ) 141 | d | gcd(subtract(4351, 8), subtract(5161, 10)) | subtract(n0,n2)|subtract(n1,n3)|gcd(#0,#1)| | general |
the difference of two numbers is 1340 . on dividing the larger number by the smaller , we get 6 as quotient and the 15 as remainder . what is the smaller number ? | "let the smaller number be x . then larger number = ( x + 1340 ) . x + 1340 = 6 x + 15 5 x = 1325 x = 265 smaller number = 265 . a )" | a ) a ) 265 , b ) b ) 280 , c ) c ) 290 , d ) d ) 300 , e ) e ) 310 | a | divide(add(1340, 15), subtract(6, const_1)) | add(n0,n2)|subtract(n1,const_1)|divide(#0,#1)| | general |
a jar contains only red , yellow , and orange marbles . if there are 1 red , 5 yellow , and 4 orange marbles , and 3 marbles are chosen from the jar at random without replacing any of them , what is the probability that 2 yellow , 1 red , and no orange marbles will be chosen ? | "i started by finding the 2 probabilities , without calculation , like this : p ( yyr ) p ( yry ) p ( ryy ) i calculated the first one and ended in 1 / 22 . i looked at the answer choices at this point and saw answer d : 1 / 45 . this helped me realise that for the 3 possible orderings the probabbility is the same . so... | a ) 1 / 60 , b ) 1 / 45 , c ) 2 / 45 , d ) 3 / 22 , e ) 5 / 22 | b | divide(multiply(choose(5, 2), choose(1, 1)), choose(add(add(5, 4), 1), 1)) | add(n1,n2)|choose(n1,n4)|choose(n0,n5)|add(n0,#0)|multiply(#1,#2)|choose(#3,n0)|divide(#4,#5)| | probability |
a scale 6 ft . 8 inches long is divided into 4 equal parts . find the length of each part . | "explanation : total length of scale in inches = ( 6 * 12 ) + 8 = 80 inches length of each of the 4 parts = 80 / 4 = 20 inches answer : a" | a ) 20 inches , b ) 77 inches , c ) 66 inches , d ) 97 inches , e ) 66 inches | a | divide(add(multiply(6, const_12), 8), 4) | multiply(n0,const_12)|add(n1,#0)|divide(#1,n2)| | general |
if w is the set of all the integers between 59 and 99 , inclusive , that are either multiples of 3 or multiples of 2 or multiples of both , then w contains how many numbers ? | multiples of 2 from 59 to 99 = multiples of 2 from 1 to 99 - multiples of 2 from 1 to 58 = [ 99 / 2 ] - [ 58 / 2 ] = 49 - 29 = 20 multiples of 3 from 59 to 99 = multiples of 3 from 1 to 99 - multiples of 3 from 1 to 58 = [ 99 / 3 ] - [ 58 / 3 ] = 33 - 169 = 14 multiples of 2 and 3 bothi . e . 6 from 59 to 99 = multiple... | a ) 27 , b ) 32 , c ) 33 , d ) 34 , e ) 35 | a | subtract(add(floor(divide(subtract(99, 59), 3)), divide(subtract(99, 59), 2)), floor(divide(subtract(99, 59), multiply(const_2, const_3)))) | multiply(const_2,const_3)|subtract(n1,n0)|divide(#1,n3)|divide(#1,n2)|divide(#1,#0)|floor(#3)|floor(#4)|add(#2,#5)|subtract(#7,#6) | other |
find the value of a / b + b / a , if a and b are the roots of the quadratic equation x 2 + 4 x + 2 = 0 ? | "a / b + b / a = ( a 2 + b 2 ) / ab = ( a 2 + b 2 + a + b ) / ab = [ ( a + b ) 2 - 2 ab ] / ab a + b = - 4 / 1 = - 4 ab = 2 / 1 = 2 hence a / b + b / a = [ ( - 4 ) 2 - 2 ( 2 ) ] / 2 = 4 / 2 = 2 . c )" | a ) 8 , b ) 10 , c ) 2 , d ) 16 , e ) 24 | c | subtract(divide(power(negate(4), 2), 2), 2) | negate(n1)|power(#0,n0)|divide(#1,n2)|subtract(#2,n0)| | general |
what least no . must be subtracted from 50248 so that remaining no . is divisible by 20 ? | "explanation : on dividing 50248 by 20 we get the remainder 8 , so 8 should be subtracted option d" | a ) 3 , b ) 2 , c ) 4 , d ) 8 , e ) 6 | d | subtract(50248, multiply(floor(divide(50248, 20)), 20)) | divide(n0,n1)|floor(#0)|multiply(n1,#1)|subtract(n0,#2)| | general |
find the area of a parallelogram with base 32 cm and height 15 cm . | "area of a parallelogram = base * height = 32 * 15 = 480 cm 2 answer : option d" | a ) 200 , b ) 384 , c ) 345 , d ) 480 , e ) 242 | d | multiply(32, 15) | multiply(n0,n1)| | geometry |
a clock shows the time as 10 a . m . if the minute hand gains 5 minutes every hour , how many minutes will the clock gain by 4 p . m . ? | "there are 6 hours in between 10 a . m . to 4 p . m . 6 * 5 = 30 minutes . answer : a" | a ) 30 min , b ) 35 min , c ) 45 min , d ) 50 min , e ) 55 min | a | multiply(add(const_3, 4), 5) | add(const_3,n2)|multiply(n1,#0)| | physics |
a space shuttle orbits the earth at about 6 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 6 * 3600 = 21,600 answer : c | a ) 480 , b ) 2,880 , c ) 21,600 , d ) 28,800 , e ) 48,000 | c | multiply(6, const_3600) | multiply(n0,const_3600) | physics |
a father purchased dress for his 3 daughters . the dresses are of same color but diff size and they are kept in dark room . what is probability that all the 3 will not choose there own dress ? | explanation : let 1 st girl come and she choose wrong dress so probability of that girl to choose wrong dress out of 3 is = 2 / 3 . now 2 nd girl come nd she choose wrong dress so probability of that girl to choose wrong dress out of 2 is 1 / 2 . now for 3 rd girl probability is 1 to choose wrong dress . so probability... | a ) 1 / 3 , b ) 2 / 3 , c ) 1 / 2 , d ) 3 / 4 , e ) 3 / 7 | a | divide(choose(const_2, const_1), factorial(const_3)) | choose(const_2,const_1)|factorial(const_3)|divide(#0,#1) | probability |
a man has some hens and cows . if the number of heads be 48 and the number of feet equals 144 , then the number of hens will be : | "let hens be x and cows be y now , feet : x * 2 + y * 4 = 144 heads : x * 1 + y * 1 = 48 implies , 2 x + 4 y = 144 and x + y = 48 solving these two equations , we get x = 24 and y = 24 therefore , hens are 26 . answer : c" | a ) 22 , b ) 23 , c ) 24 , d ) 26 , e ) 28 | c | divide(subtract(multiply(48, const_4), 144), const_2) | multiply(n0,const_4)|subtract(#0,n1)|divide(#1,const_2)| | general |
a batsman in his 10 th innings makes a score of 60 , and thereby increases his average by 3 . what is his average after the 10 th innings ? he had never been β not out β . | "average score before 10 th innings = 60 - 3 Γ 10 = 30 average score after 10 th innings = > 30 + 3 = 33 answer : e" | a ) 47 , b ) 37 , c ) 39 , d ) 43 , e ) 33 | e | add(subtract(60, multiply(3, 10)), 3) | multiply(n0,n2)|subtract(n1,#0)|add(n2,#1)| | general |
the radius of a cylinder is 10 m , height 14 m . the volume of the cylinder is : | "cylinder volume = Ο r ( power 2 ) h = 22 / 7 Γ 10 Γ 10 Γ 14 = 4400 m ( power 3 ) answer is e ." | a ) 2200 , b ) 5500 , c ) 3300 , d ) 1100 , e ) 4400 | e | multiply(circumface(10), 14) | circumface(n0)|multiply(n1,#0)| | geometry |
at what price must an article costing rs . 47.50 be marked in order that after deducting 5 % from the list price . it may be sold at a profit of 30 % on the cost price ? | "cp = 47.50 sp = 47.50 * ( 130 / 100 ) = 61.75 mp * ( 95 / 100 ) = 61.75 mp = 65 answer : c" | a ) 62.5 , b ) 62.3 , c ) 65 , d ) 62.2 , e ) 62.9 | c | divide(multiply(add(47.50, divide(multiply(47.50, 30), const_100)), const_100), subtract(const_100, 5)) | multiply(n0,n2)|subtract(const_100,n1)|divide(#0,const_100)|add(n0,#2)|multiply(#3,const_100)|divide(#4,#1)| | gain |
income and expenditure of a person are in the ratio 5 : 3 . if the income of the person is rs . 10000 , then find his savings ? | "let the income and the expenditure of the person be rs . 5 x and rs . 3 x respectively . income , 5 x = 10000 = > x = 2000 savings = income - expenditure = 5 x - 3 x = 2 x so , savings = rs . 2 * 2000 = rs . 4000 answer : a" | a ) rs . 4000 , b ) rs . 3603 , c ) rs . 3639 , d ) rs . 3632 , e ) rs . 3602 | a | subtract(10000, multiply(divide(3, 5), 10000)) | divide(n1,n0)|multiply(n2,#0)|subtract(n2,#1)| | other |
in a triangle , one side is 6 cm and another side is 8 cm . which of the following can be the perimeter of the triangle ? | given : one side is 6 cm and another side is 8 cm . so the 3 rd side will be > 3 and < 15 . thus the perimeter will be : 18 < perimeter < 30 . only option satisfying this condition is 22 . hence a . | ['a ) 22 .', 'b ) 25 .', 'c ) 30 .', 'd ) 32 .', 'e ) 34 .'] | a | subtract(triangle_perimeter(6, 8, sqrt(add(power(6, const_2), power(8, const_2)))), const_2) | power(n0,const_2)|power(n1,const_2)|add(#0,#1)|sqrt(#2)|triangle_perimeter(n0,n1,#3)|subtract(#4,const_2) | geometry |
in 50 l can , milk and water in the ratio of 3 : 1 . if the ratio has made 1 : 3 then the quantity of water that has to be added further is ? | 100 litre water is to be added . let ' x ' liter water is to be added more . then 1 / 3 = 37.5 / ( 12.5 + x ) x = 100 answer : a | a ) 100 , b ) 120 , c ) 90 , d ) 80 , e ) 70 | a | subtract(multiply(multiply(divide(50, add(3, 1)), 3), 3), multiply(divide(50, add(3, 1)), 1)) | add(n1,n2)|divide(n0,#0)|multiply(n1,#1)|multiply(n2,#1)|multiply(n1,#2)|subtract(#4,#3) | other |
find the sum of all 4 digit numbers formed using digits 1,2 , 5,6 . | "( n - 1 ) ! * ( 111 . . . n ) * ( sum of the digits ) = ( 4 - 1 ) ! * 1111 * ( 1 + 2 + 5 + 6 ) = 93324 answer : a" | a ) 93324 , b ) 92324 , c ) 93424 , d ) 93424 , e ) 93824 | a | multiply(add(add(const_100, const_4), subtract(multiply(const_100, const_10), 1,2)), divide(add(divide(subtract(subtract(multiply(const_100, const_10), 1,2), add(const_100, const_4)), 1,2), const_1), const_2)) | add(const_100,const_4)|multiply(const_10,const_100)|subtract(#1,n1)|add(#0,#2)|subtract(#2,#0)|divide(#4,n1)|add(#5,const_1)|divide(#6,const_2)|multiply(#3,#7)| | general |
8.036 divided by 0.04 gives : | "= 8.036 / 0.04 = 803.6 / 4 = 200.9 answer is a ." | a ) 200.9 , b ) 2.06 , c ) 20.06 , d ) 100.9 , e ) 200.6 | a | divide(8.036, 0.04) | divide(n0,n1)| | general |
the simple interest on rs . 20 for 6 months at the rate of 5 paise per rupeeper month is | "sol . s . i . = rs . [ 20 * 5 / 100 * 6 ] = rs . 6 answer c" | a ) 1.2 , b ) 1.4 , c ) 6 , d ) 7 , e ) none | c | divide(multiply(multiply(20, 6), 5), const_100) | multiply(n0,n1)|multiply(n2,#0)|divide(#1,const_100)| | gain |
tom has travelling to 7 cities . gasoline prices varied from city to city . what is the median gasoline price ? | ordering the data from least to greatest , we get : $ 1.61 , $ 1.75 , $ 1.79 , $ 1.84 , $ 1.96 , $ 2.09 , $ 2.11 the median gasoline price is $ 1.84 . ( there were 3 states with higher gasoline prices and 3 with lower prices . ) b | a ) $ 1 , b ) $ 1.84 , c ) $ 1.98 , d ) $ 2.34 , e ) $ 2.56 | b | divide(add(add(multiply(add(7, const_1), const_10), const_4), const_100), const_100) | add(n0,const_1)|multiply(#0,const_10)|add(#1,const_4)|add(#2,const_100)|divide(#3,const_100) | general |
if n is an integer and 102 n ^ 2 is less than or equal to 8100 , what is the greatest possible value of n ? | "102 * n ^ 2 < = 8100 n ^ 2 < = 8100 / 102 which will be less than 81 since 8100 / 100 = 81 which is the square of 9 next closest value of n where n ^ 2 < = 81 is 8 ans b" | a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | b | floor(sqrt(divide(8100, 102))) | divide(n2,n0)|sqrt(#0)|floor(#1)| | general |
4 mat - weavers can weave 4 mats in 4 days . at the same rate , how many mats would be woven by 12 mat - weavers in 12 days ? | "let the required number of bottles be x . more weavers , more mats ( direct proportion ) more days , more mats ( direct proportion ) wavers 4 : 12 : : 4 : x days 4 : 12 4 * 4 * x = 12 * 12 * 4 x = ( 12 * 12 * 4 ) / ( 4 x 4 ) x = 36 . answer is e ." | a ) 25 , b ) 19 , c ) 39 , d ) 61 , e ) 36 | e | multiply(multiply(4, divide(12, 4)), divide(12, 4)) | divide(n3,n0)|multiply(n0,#0)|multiply(#0,#1)| | gain |
if 3 people can do 3 times of a particular work in 3 days , then how many days would it take 6 people to do 6 times of that particular work ? | "3 people can do the work one time in one day . 1 person can do 1 / 3 of the work in one day . 6 people can do 6 / 3 of the work in one day . 6 people can do 6 times the work in 3 days . the answer is c ." | a ) 1 , b ) 2 , c ) 3 , d ) 6 , e ) 9 | c | multiply(3, divide(6, 6)) | divide(n3,n3)|multiply(n0,#0)| | physics |
if 4 xz + yw = 7 and xw + yz = 14 , what is the value of the expression ( 2 x + y ) ( 2 z + w ) ? | "( 2 x + y ) * ( 2 z + w ) = 7 + 2 ( 14 ) = 35 answer : c" | a ) 9 . , b ) 12 . , c ) 35 . , d ) 16 . , e ) 18 . | c | add(4, multiply(14, 2)) | multiply(n2,n3)|add(n0,#0)| | general |
a , b , k start from the same place and travel in the same direction at speeds of 30 km / hr , 40 km / hr , 120 km / hr respectively . b starts two hours after a . if b and k overtake a at the same instant , how many hours after a did k start ? | "in 2 hours , a travels 60 km . b can catch a at a rate of 10 km / hr , so b catches a 6 hours after b starts . so a and b both travel a distance of 240 km . c needs 2 hours to travel 240 km , so c leaves 6 hours after a . the answer is c ." | a ) 3 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | c | subtract(40, 30) | subtract(n1,n0)| | physics |
a river 3 m deep and 55 m wide is flowing at the rate of 1 kmph the amount of water that runs into the sea per minute is ? | "rate of water flow - 1 kmph - - 1000 / 60 - - 16.66 m / min depth of river - - 3 m width of river - - 55 m vol of water per min - - 16.66 * 3 * 55 - - - 2748.9 answer b" | a ) 2338.9 , b ) 2748.9 , c ) 2148.9 , d ) 2745.9 , e ) 2718.9 | b | divide(multiply(multiply(3, 55), multiply(1, const_1000)), multiply(const_1, const_60)) | multiply(n0,n1)|multiply(n2,const_1000)|multiply(const_1,const_60)|multiply(#0,#1)|divide(#3,#2)| | physics |
in x - y plane , there is a right triangle abc ( β b = 90 o ) . if the length of ac is 100 and the slope of line segment ac is 4 / 3 , what is the length of ab ? | slope = change in vertical direction / change in horizontal direction = 4 / 3 change in vertical direction = 4 x = ab change in horizontal direction = 3 x = bc ab ^ 2 + bc ^ 2 = 100 ^ 2 16 x ^ 2 + 9 x ^ 2 = 10000 25 x ^ 2 = 400 x ^ 2 = 400 x = 20 therefore ab = 20 * 4 = 80 answer : e | a ) 12 , b ) 18 , c ) 24 , d ) 28 , e ) 80 | e | multiply(4, divide(100, sqrt(add(power(4, const_2), power(3, const_2))))) | power(n2,const_2)|power(n3,const_2)|add(#0,#1)|sqrt(#2)|divide(n1,#3)|multiply(n2,#4) | general |
what is the remainder when 14,451 Γ 15,651 Γ 16,788 is divided by 5 ? | "only the unit ' s digit of the product will decide the remainder when divided by 5 . hence , 1 * 1 * 8 = will give units digit as 8 so , whatever be the number , if it ends in 8 , the remainder after dividing with 5 will be 3 . optiond" | a ) 1 , b ) 2 , c ) 4 , d ) 3 , e ) 5 | d | reminder(multiply(15,651, 14,451), 16,788) | multiply(n0,n1)|reminder(#0,n2)| | general |
a dishonest dealer professes to sell goods at the cost price but uses a weight of 850 grams per kg , what is his percent ? | "850 - - - 150 100 - - - ? = > 17.64 % answer : b" | a ) 25 % , b ) 17 % , c ) 29 % , d ) 55 % , e ) 45 % | b | subtract(multiply(divide(const_100, 850), multiply(const_100, multiply(add(const_3, const_2), const_2))), const_100) | add(const_2,const_3)|divide(const_100,n0)|multiply(#0,const_2)|multiply(#2,const_100)|multiply(#1,#3)|subtract(#4,const_100)| | gain |
a bag contains 7 red , 9 blue and 5 green balls . if 3 balls are picked at random , what is the probability that both are red ? | "p ( both are red ) , = 7 c 3 / 21 c 3 = ( 7 * 6 * 5 ) / 21 * 20 * 19 = 1 / 38 d" | a ) 2 / 21 , b ) 3 / 41 , c ) 5 / 26 , d ) 1 / 38 , e ) 5 / 32 | d | divide(choose(7, 3), choose(add(add(7, 9), 5), 3)) | add(n0,n1)|choose(n0,n3)|add(n2,#0)|choose(#2,n3)|divide(#1,#3)| | other |
40 is divided into two parts in such a way that seventh part of first and ninth part of second are equal . find the smallest part ? | "x / 7 = y / 9 = > x : y = 7 : 9 7 / 16 * 40 = 17.5 answer : b" | a ) 13.2 , b ) 17.5 , c ) 12.8 , d ) 34.25 , e ) 31.25 | b | divide(multiply(divide(add(const_4, const_3), add(add(const_4, const_3), const_2)), 40), const_2) | add(const_3,const_4)|add(#0,const_2)|divide(#0,#1)|multiply(n0,#2)|divide(#3,const_2)| | general |
hammers and wrenches are manufactured at a uniform weight per hammer and a uniform weight per wrench . if the total weight of two hammers and three wrenches is one - third that of 8 hammers and 4 wrenches , then the total weight of one wrench is how many times that of one hammer ? | "x b Γ© the weight of a hammer and y be the weight of a wrench . ( 2 x + 3 y ) = 1 / 3 * ( 8 x + 4 y ) 3 ( 2 x + 3 y ) = ( 8 x + 4 y ) 6 x + 9 y = 8 x + 4 y 5 y = 2 x y = 2 x / 5 ans - b" | a ) 1 / 2 , b ) 2 / 5 , c ) 1 , d ) 3 / 2 , e ) 2 | b | divide(subtract(8, multiply(const_3, const_2)), subtract(multiply(const_3, const_2), 4)) | multiply(const_2,const_3)|subtract(n0,#0)|subtract(#0,n1)|divide(#1,#2)| | general |
a man can row at 5 kmph in still water . if the velocity of the current is 1 kmph and it takes him 1 hour to row to a place and come back . how far is that place . | "explanation : let the distance is x km rate downstream = 5 + 1 = 6 kmph rate upstream = 5 - 1 = 4 kmph then x / 6 + x / 4 = 1 [ because distance / speed = time ] = > 2 x + 3 x = 12 = > x = 12 / 5 = 2.4 km option c" | a ) . 4 km , b ) 1.4 km , c ) 2.4 km , d ) 3.4 km , e ) none of these | c | divide(multiply(subtract(5, 1), const_3), 5) | subtract(n0,n1)|multiply(#0,const_3)|divide(#1,n0)| | physics |
in a certain school , the ratio of boys to girls is 5 to 13 . if there are 80 more girls than boys , how many boys are there ? | "the ratio of b to g is 5 : 13 and the other data point is g are more than boys by 80 . . . looking at the ratio we can say that the 8 ( 13 - 5 ) extra parts caused this diff of 80 . so 1 part corresponds to 80 / 8 = 10 and so 5 parts correspond to 5 * 10 = 50 . d" | a ) 27 , b ) 36 , c ) 45 , d ) 50 , e ) 117 | d | subtract(divide(80, subtract(const_1, divide(5, 13))), 80) | divide(n0,n1)|subtract(const_1,#0)|divide(n2,#1)|subtract(#2,n2)| | other |
a reduction in the price of petrol by 10 % enables a motorist to buy 5 gallons more for $ 300 . find the original price of petrol ? | "price decreased by 10 % , so 9 / 10 times , which means that original gallons bought increased 10 / 9 times . since this increase equals to 5 gallons then 45 gallons were bought originally ( 45 * 10 / 9 = 50 - - > increase 5 gallons ) . hence original price was 300 / 45 = $ 6.6 answer : e ." | a ) $ 11 , b ) $ 5 , c ) $ 45 , d ) $ 400 , e ) $ 6.6 | e | divide(subtract(300, multiply(subtract(const_1, divide(10, const_100)), 300)), subtract(5, divide(const_1, const_2))) | divide(n0,const_100)|divide(const_1,const_2)|subtract(const_1,#0)|subtract(n1,#1)|multiply(n2,#2)|subtract(n2,#4)|divide(#5,#3)| | gain |
it costs a publishing company 50000 dollars to make books . the 50000 is a fixed cost or a cost that can not change . to help the publishing company sell the books , a marketing company charges 4 dollars for each book sold . if the company charges 9 dollars per book , how many books should they sell to break even ? | let c be the cost of producing and selling x books let r be the revenue made for selling x books r = selling price of 1 book Γ number of books sold r = 9 x c = fixed cost + variable cost variable cost = fee charged for 1 book Γ number of books sold variable cost = 4 x c = 50000 + 4 x r = c 9 x = 50000 + 4 x 9 x - 4 x =... | a ) 100 , b ) 1000 , c ) 10000 , d ) 100000 , e ) none | c | divide(50000, subtract(9, 4)) | subtract(n3,n2)|divide(n0,#0) | general |
a certain characteristic in a large population has a distribution that is symmetric about the mean m . if 68 % of the distribution lies one standard deviation s of the mean , what percent of the distribution is less than m + s ? | "16 % ________________________________________________ m + s 34 % ________________________________________________ m 34 % ________________________________________________ m - s 16 % since 68 % lies one standard deviation from mean m , = > 50 % of 68 % lies on either side as it is symmetric about m . thus 16 % lie below... | a ) 16 % , b ) 32 % , c ) 48 % , d ) 84 % , e ) 92 % | d | subtract(const_100, divide(subtract(const_100, 68), const_2)) | subtract(const_100,n0)|divide(#0,const_2)|subtract(const_100,#1)| | general |
sam invested rs . 6000 @ 10 % per annum for one year . if the interest is compounded half - yearly , then the amount received by sam at the end of the year will be ? | p = rs . 6000 ; r = 10 % p . a . = 5 % per half - year ; t = 1 year = 2 half - year amount = [ 6000 * ( 1 + 5 / 100 ) 2 ] = ( 6000 * 21 / 20 * 21 / 20 ) = rs . 6615.00 answer : e | a ) 5300.0 , b ) 24580.0 , c ) 16537.5 , d ) 7120.0 , e ) 6615.0 | e | multiply(power(add(divide(divide(10, const_2), const_100), const_1), const_2), 6000) | divide(n1,const_2)|divide(#0,const_100)|add(#1,const_1)|power(#2,const_2)|multiply(n0,#3) | gain |
30 men can do a work in 40 days . when should 20 men leave the work so that the entire work is completed in 40 days after they leave the work ? | "total work to be done = 30 * 40 = 1200 let 20 men leave the work after ' p ' days , so that the remaining work is completed in 40 days after they leave the work . 40 p + ( 20 * 40 ) = 1200 40 p = 400 = > p = 10 days answer : b" | a ) 22 days , b ) 10 days , c ) 77 days , d ) 88 days , e ) 55 days | b | divide(subtract(multiply(30, 40), multiply(40, 20)), 40) | multiply(n0,n1)|multiply(n1,n2)|subtract(#0,#1)|divide(#2,n1)| | physics |
find the area of the quadrilateral of one of its diagonals is 50 cm and its off sets 10 cm and 8 cm ? | "1 / 2 * 50 ( 10 + 8 ) = 450 cm 2 answer : d" | a ) 189 cm 2 , b ) 150 cm 2 , c ) 127 cm 2 , d ) 450 cm 2 , e ) 187 cm 2 | d | multiply(multiply(divide(const_1, const_2), add(8, 10)), 50) | add(n1,n2)|divide(const_1,const_2)|multiply(#0,#1)|multiply(n0,#2)| | geometry |
for any real number x , the operatoris defined as : ( x ) = x ( 2 β x ) if p + 1 = ( p + 1 ) , then p = | ( x ) = x ( 2 β x ) ( p + 1 ) = ( p + 1 ) ( 2 - p - 1 ) = ( 1 - p ) ( p + 1 ) we are given that p + 1 = ( p + 1 ) therefore ( 1 - p ) ( p + 1 ) = ( p + 1 ) or ( p + 1 ) + ( p - 1 ) ( p + 1 ) = 0 ( p + 1 ) ( p - 1 ) = 0 p = - 1 , p = 1 option c | a ) - 2 , b ) 0 , c ) 1 , d ) 2 , e ) 3 | c | negate(divide(negate(2), multiply(const_2, 1))) | multiply(n1,const_2)|negate(n0)|divide(#1,#0)|negate(#2) | general |
sachin borrows rs . 5000 for 2 years at 4 % p . a . simple interest . he immediately lends money to rahul at 25 / 4 % p . a . for 2 years . find the gain of one year by sachin . | explanation : two things need to give attention in this question , first we need to calculate gain for 1 year only . gain in 2 year = [ ( 5000 Γ 254 Γ 2100 ) β ( 5000 Γ 4 Γ 2100 ) ] = ( 625 β 400 ) = 225 so gain for 1 year = 2252 = 112.50 answer : c | a ) 110.5 , b ) 111.5 , c ) 112.5 , d ) 113.5 , e ) none of these | c | subtract(p_after_gain(divide(25, 4), 5000), p_after_gain(4, 5000)) | divide(n3,n2)|p_after_gain(n2,n0)|p_after_gain(#0,n0)|subtract(#2,#1) | gain |
this year , mbb consulting fired 8 % of its employees and left remaining employee salaries unchanged . sally , a first - year post - mba consultant , noticed that that the average ( arithmetic mean ) of employee salaries at mbb was 10 % more after the employee headcount reduction than before . the total salary pool all... | "100 employees getting 1000 $ avg , so total salary for 100 ppl = 100000 8 % reduction in employees lead to 92 employees and a salary increase of 10 % of previous avg salary thus the new avg salary is = 10 % ( 1000 ) + 1000 = 1100 so total salary of 92 employees is 92 * 1100 = 101200 now the new salary is more than pre... | a ) 98.5 % , b ) 101.2 % , c ) 102.8 % , d ) 104.5 % , e ) 105.0 % | b | divide(multiply(add(const_100, multiply(const_100, 10)), add(subtract(const_100, 8), const_4)), multiply(const_100, 10)) | multiply(n1,const_100)|subtract(const_100,n0)|add(#0,const_100)|add(#1,const_4)|multiply(#2,#3)|divide(#4,#0)| | general |
on average , the bottle - nosed dolphin comes up for air once every 3 minutes ; the beluga whale , a close relative , comes up for air on average once every 6 minutes . the number of times a bottle - nosed dolphin would come up for air in a 24 hour period is approximately what percent greater than the number of times a... | dolphin once in 3 min ; beluga once in 6 min ; so , dolphin comes up 2 times frequently than beluga , which is 150 % ( 6 - 3 ) / 2 * 100 . answer : e . | a ) 25 % , b ) 50 % , c ) 75 % , d ) 100 % , e ) 150 % | e | multiply(divide(subtract(6, 3), const_2), const_100) | subtract(n1,n0)|divide(#0,const_2)|multiply(#1,const_100) | general |
the circumference of a circle is equal to 72 pi . find the radius of this circle . | the circumference of a circle is given by c = 2 pi r , where r is the radius of the circle . substitute c by 72 pi to obtain the equation 72 pi = 2 pi r simplify and solve for r to obtain r = 36 answer b | ['a ) 35', 'b ) 36', 'c ) 37', 'd ) 38', 'e ) 39'] | b | divide(72, const_2) | divide(n0,const_2) | geometry |
the average of 25 results is 50 . the average of first 12 of those is 14 and the average of last 12 is 17 . what is the 13 th result ? | "solution : sum of 1 st 12 results = 12 * 14 sum of last 12 results = 12 * 17 13 th result = x ( let ) now , 12 * 14 + 12 * 17 + x = 25 * 50 or , x = 878 . answer : option c" | a ) 741 , b ) 752 , c ) 878 , d ) 785 , e ) 458 | c | subtract(subtract(multiply(25, 50), multiply(12, 17)), multiply(12, 14)) | multiply(n0,n1)|multiply(n2,n5)|multiply(n2,n3)|subtract(#0,#1)|subtract(#3,#2)| | general |
two trains each 475 m long are running in opposite directions on parallel tracks . their speeds are 55 km / hr and 40 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? | relative speed = 55 + 40 = 95 km / hr . 95 * 5 / 18 = 475 / 18 m / sec . distance covered = 475 + 475 = 950 m . required time = 950 * 18 / 475 = 36 sec . answer : d | a ) 77 sec , b ) 66 sec , c ) 48 sec , d ) 36 sec , e ) 45 sec | d | multiply(divide(multiply(divide(475, const_1000), const_2), add(55, 40)), const_3600) | add(n1,n2)|divide(n0,const_1000)|multiply(#1,const_2)|divide(#2,#0)|multiply(#3,const_3600) | physics |
mr . loyd wants to fence his square shaped land of 150 sqft each side . if a pole is laid every 10 ft how many poles do he need ? | "if each side is 120 feet . . then total perimeter is 150 * 4 = 600 poles every 10 feet hence no of poles = 600 / 10 = 60 answer : d" | a ) 20 , b ) 30 , c ) 40 , d ) 60 , e ) 65 | d | divide(divide(multiply(150, const_4), const_10), const_2) | multiply(n0,const_4)|divide(#0,const_10)|divide(#1,const_2)| | geometry |
in a city where all streets run east - to - west , all avenues run north - to - south , and all intersections are right angles as shown below , jenn needs to walk from the corner of 1 st street and 1 st avenue to the corner of 6 th street and 3 rd avenue . if her friend amanda is sitting on a bench on 4 th street halfw... | all routes between ( 1,1 ) to ( 3,6 ) = 7 ! / ( 2 ! * 5 ! ) = 21 routes which pass 4 th st . between 1 st and 2 nd avenue = routs between ( 1,1 ) to ( 1,4 ) * routs between ( 2,4 ) to ( 3,6 ) = 1 * ( 3 ! / ( 2 ! * 1 ! ) ) = 3 the probability that jenn will walk down 4 th st . past amanda = 3 / 21 = 1 / 7 c is correct . | a ) 1 / 42 , b ) 1 / 21 , c ) 1 / 7 , d ) 1 / 3 , e ) 1 / 2 | c | divide(3, multiply(7, 3)) | multiply(n3,n7)|divide(n3,#0) | physics |
length of a rectangular plot is 50 mtr more than its breadth . if the cost of fencin g the plot at 26.50 per meter is rs . 5300 , what is the length of the plot in mtr ? | "let breadth = x metres . then , length = ( x + 50 ) metres . perimeter = 5300 / 26.5 m = 200 m . 2 [ ( x + 50 ) + x ] = 200 2 x + 50 = 100 2 x = 50 x = 25 . hence , length = x + 50 = 75 m d" | a ) 46 m , b ) 60 m , c ) 58 m , d ) 75 m , e ) 80 m | d | divide(add(divide(5300, 26.50), multiply(const_2, 50)), const_4) | divide(n2,n1)|multiply(n0,const_2)|add(#0,#1)|divide(#2,const_4)| | physics |
10 camels cost as much as 24 horses , 16 horses cost as much as 4 oxen and 6 oxen as much as 4 elephants . if the cost of 10 elephants is rs . 110000 , find the cost of a camel ? | "cost of the camel = p 10 camels = 24 horses 16 horses = 4 oxen 6 oxen = 4 elephants 10 elephants = rs . 110000 p = rs . [ ( 24 * 4 * 4 * 110000 ) / ( 10 * 16 * 6 * 10 ) ] p = rs . ( 42240000 / 9600 ) = > p = rs . 4400 answer : d" | a ) s . 9800 , b ) s . 3800 , c ) s . 9800 , d ) s . 4400 , e ) s . 6880 | d | divide(multiply(multiply(multiply(24, 4), 4), 110000), multiply(multiply(multiply(10, 16), 6), 10)) | multiply(n1,n3)|multiply(n0,n2)|multiply(n3,#0)|multiply(n4,#1)|multiply(n7,#2)|multiply(n0,#3)|divide(#4,#5)| | general |
kyle , david , and catherine each try independently to solve a problem . if their individual probabilities for success are 1 / 3 , 2 / 7 and 5 / 9 , respectively , what is the probability that kyle and catherine , but not david will solve the problem ? | p ( kyle will solve ) = 1 / 3 p ( david will not solve ) = 1 - 2 / 7 = 5 / 7 p ( catherine will solve ) = 5 / 9 p = ( 1 / 3 ) * ( 5 / 7 ) * ( 5 / 9 ) = 25 / 189 answer : c | a ) 35 / 189 , b ) 33 / 129 , c ) 25 / 189 , d ) 24 / 113 , e ) 20 / 189 | c | divide(multiply(5, 5), add(multiply(const_60, const_3), multiply(const_3, const_3))) | multiply(n4,n4)|multiply(const_3,const_60)|multiply(const_3,const_3)|add(#1,#2)|divide(#0,#3) | general |
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