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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if ( a + b ) = 11 , ( b + c ) = 9 and ( c + d ) = 3 , what is the value of ( a + d ) ? | "given a + b = 11 = > a = 11 - b - - > eq 1 b + c = 9 c + d = 3 = > d = 3 - c - - > eq 2 then eqs 1 + 2 = > a + d = 11 - b + 3 - c = > 14 - ( b + c ) = > 14 - 9 = 5 . option c . . ." | a ) 16 . , b ) 8 . , c ) 5 . , d ) 2 . , e ) - 2 . | c | subtract(add(11, 3), 9) | add(n0,n2)|subtract(#0,n1)| | general |
from a vessel on the first day , 1 / 3 rd of the liquid evaporates . on the second day 3 / 4 th of the remaining liquid evaporates . what fraction of the volume is present at the end of the 2 day | let x be the volume . . . after 1 st day volume remaining = ( x - x / 3 ) = 2 x / 3 after 2 st day volume remaining = ( 2 x / 3 ) - ( ( 2 x / 3 ) * ( 3 / 4 ) ) = ( 2 x / 3 ) ( 1 - 3 / 4 ) = ( 2 x / 3 ) * ( 1 / 4 ) = x / 6 answer : d | a ) 1 / 3 , b ) 1 / 4 , c ) 1 / 5 , d ) 1 / 6 , e ) 1 / 7 | d | subtract(1, add(divide(1, 3), multiply(divide(3, 4), subtract(const_1, divide(1, 3))))) | divide(n0,n1)|divide(n1,n3)|subtract(const_1,#0)|multiply(#1,#2)|add(#0,#3)|subtract(n0,#4) | physics |
the ratio of the incomes of uma and bala is 8 : 7 and the ratio of their expenditure is 7 : 6 . if at the end of the year , each saves $ 2000 then the income of uma is ? | let the income of uma and bala be $ 8 x and $ 7 x let their expenditures be $ 7 y and $ 6 y 8 x - 7 y = 2000 - - - - - - - 1 ) 7 x - 6 y = 2000 - - - - - - - 2 ) from 1 ) and 2 ) x = 1000 uma ' s income = 8 x = 8 * 2000 = $ 16000 answer is c | a ) $ 16800 , b ) $ 16500 , c ) $ 16000 , d ) $ 16300 , e ) $ 16200 | c | multiply(8, 2000) | multiply(n0,n4) | other |
the average of 6 numbers is 30 . if the average of first 4 is 25 and that of last 3 is 35 , the fourth number is : | let the six numbers be , a , b , c , d , e , f . a + b + c + d + e + f = 30 × 6 = 180 - - - - ( 1 ) a + b + c + d = 25 × 4 = 100 - - - - ( 2 ) d + e + f = 35 × 3 = 105 - - - - ( 3 ) add 2 nd and 3 rd equations and subtract 1 st equation from this . d = 25 answer : a | a ) 25 , b ) 26 , c ) 18 , d ) 19 , e ) 10 | a | subtract(multiply(3, 35), subtract(multiply(6, 30), multiply(4, 25))) | multiply(n4,n5)|multiply(n0,n1)|multiply(n2,n3)|subtract(#1,#2)|subtract(#0,#3) | general |
how many of the positive factors of 15 , 45 and how many common factors are there in numbers ? | factors of 15 - 1 , 3 , 5 , and 15 factors of 45 - 1 , 3 , 9 , 15 and 45 comparing both , we have three common factors of 45,16 - 3 answer c | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | divide(45, 15) | divide(n1,n0) | other |
divide rs . 32000 in the ratio 1 : 9 ? | "1 / 10 * 32000 = 3200 9 / 10 * 32000 = 28800 answer : d" | a ) 12000 , 20000 , b ) 12000 , 200098 , c ) 12000 , 20007 , d ) 3200 , 28800 , e ) 12000 , 20001 | d | multiply(subtract(9, const_2), divide(32000, add(1, subtract(9, const_2)))) | subtract(n2,const_2)|add(n1,#0)|divide(n0,#1)|multiply(#2,#0)| | other |
if a car had traveled 25 kmh faster than it actually did , the trip would have lasted 30 minutes less . if the car went exactly 75 km , at what speed did it travel ? | time = distance / speed difference in time = 1 / 2 hrs 75 / x - 75 / ( x + 25 ) = 1 / 2 substitute the value of x from the options . - - > x = 50 - - > 75 / 50 - 75 / 75 = 3 / 2 - 1 = 1 / 2 answer : d | a ) 30 kmh , b ) 40 kmh , c ) 45 kmh , d ) 50 kmh , e ) 55 kmh | d | divide(subtract(sqrt(add(multiply(multiply(const_2, multiply(75, 25)), const_4), power(25, const_2))), 25), const_2) | multiply(n0,n2)|power(n0,const_2)|multiply(#0,const_2)|multiply(#2,const_4)|add(#3,#1)|sqrt(#4)|subtract(#5,n0)|divide(#6,const_2) | physics |
in the faculty of reverse - engineering , 240 second year students study numeric methods , 423 second year students study automatic control of airborne vehicles and 134 second year students study them both . how many students are there in the faculty if the second year students are approximately 80 % of the total ? | "answer is c : 661 solution : total number of students studying both are 423 + 240 - 134 = 529 ( subtracting the 134 since they were included in the both the other numbers already ) . so 80 % of total is 529 , so 100 % is approx . 661 ." | a ) 515 . , b ) 545 . , c ) 661 . , d ) 644 . , e ) 666 . | c | add(240, 423) | add(n0,n1)| | general |
machine a and machine b are each used to manufacture 770 sprockets . it takes machine a 10 hours longer to produce 770 sprockets than machine b . machine b produces 10 % more sprockets per hour than machine a . how many sprockets per hour does machineaproduce ? | "time taken by b = t time taken by a = t + 10 qty produced by a = q qty produced by b = 1.1 q for b : t ( 1.1 q ) = 770 qt = 700 for a : ( t + 10 ) ( q ) = 770 qt + 10 q = 770 700 + 10 q = 770 q = 7 so a can produce 7 / hour . then b can produce = 7 ( 1.1 ) = 7.7 / hour . e" | a ) 6 , b ) 6.6 , c ) 60 , d ) 100 , e ) 7.7 | e | divide(770, divide(multiply(multiply(10, 770), divide(add(const_100, 10), const_100)), subtract(multiply(770, divide(add(const_100, 10), const_100)), 770))) | add(n1,const_100)|multiply(n0,n1)|divide(#0,const_100)|multiply(#2,#1)|multiply(n0,#2)|subtract(#4,n0)|divide(#3,#5)|divide(n0,#6)| | gain |
for any integer p , * p is equal to the product of all the integers between 1 and p , inclusive . how many prime numbers are there between * 5 + 3 and * 5 + 5 , inclusive ? | generally * p or p ! will be divisible by all numbers from 1 to p . therefore , * 5 would be divisible by all numbers from 1 to 5 . = > * 5 + 3 would give me a number which is a multiple of 3 and therefore divisible ( since * 5 is divisible by 3 ) in fact adding anyprimenumber between 1 to 5 to * 5 will definitely be d... | a ) none , b ) one , c ) two , d ) three , e ) four | a | subtract(subtract(add(multiply(multiply(multiply(5, const_3), const_2), const_4), 5), add(multiply(multiply(multiply(5, const_3), const_2), const_4), 3)), 1) | multiply(n1,const_3)|multiply(#0,const_2)|multiply(#1,const_4)|add(n1,#2)|add(n2,#2)|subtract(#3,#4)|subtract(#5,n0) | general |
the total marks obtained by a student in mathematics and physics is 60 and his score in chemistry is 20 marks more than that in physics . find the average marks scored in mathamatics and chemistry together . | "let the marks obtained by the student in mathematics , physics and chemistry be m , p and c respectively . given , m + c = 60 and c - p = 20 m + c / 2 = [ ( m + p ) + ( c - p ) ] / 2 = ( 60 + 20 ) / 2 = 40 . answer : a" | a ) 40 , b ) 30 , c ) 25 , d ) data inadequate , e ) none of these . | a | divide(add(60, 20), const_2) | add(n0,n1)|divide(#0,const_2)| | general |
two numbers are respectively 40 % and 30 % more than a third number . the second number expressed in terms of percentage of the first is ? | "here , x = 40 and y = 30 ; therefore second number = [ [ ( 100 + y ) / ( 100 + x ) ] x 100 ] % of first number = [ [ ( 100 + 30 ) / ( 100 + 40 ) ] x 100 ] % of first number = 92.8 % of the first answer : b" | a ) 95 % , b ) 93 % , c ) 90 % , d ) 80 % , e ) none of these | b | subtract(const_100, multiply(divide(add(40, const_100), add(30, const_100)), const_100)) | add(n0,const_100)|add(n1,const_100)|divide(#0,#1)|multiply(#2,const_100)|subtract(const_100,#3)| | general |
a man can row downstream at 22 kmph and upstream at 14 kmph . find the speed of the man in still water and the speed of stream respectively ? | "explanation : let the speed of the man in still water and speed of stream be x kmph and y kmph respectively . given x + y = 22 - - - ( 1 ) and x - y = 14 - - - ( 2 ) from ( 1 ) & ( 2 ) 2 x = 36 = > x = 18 , y = 4 . answer : option b" | a ) 13 , 3 , b ) 18 , 4 , c ) 15 , 3 , d ) 14 , 4 , e ) none of these | b | divide(divide(add(22, 14), const_2), const_2) | add(n0,n1)|divide(#0,const_2)|divide(#1,const_2)| | physics |
8.5 × 6.4 + 4.5 × 11.6 = ? ÷ 4 | "explanation : ( 54.4 + 52.2 ) × 4 = 426.4 answer : option b" | a ) 426.2 , b ) 426.4 , c ) 106.6 , d ) 422.4 , e ) 522.5 | b | multiply(8.5, 6.4) | multiply(n0,n1)| | general |
a cubical tank is filled with water to a level of 2 feet . if the water in the tank occupies 50 cubic feet , to what fraction of its capacity is the tank filled with water ? | the volume of water in the tank is h * l * b = 50 cubic feet . since h = 2 , then l * b = 25 and l = b = 5 . since the tank is cubical , the capacity of the tank is 5 * 5 * 5 = 125 . the ratio of the water in the tank to the capacity is 50 / 125 = 2 / 5 the answer is d . | ['a ) 1 / 3', 'b ) 2 / 3', 'c ) 3 / 4', 'd ) 2 / 5', 'e ) 5 / 6'] | d | divide(2, divide(50, const_10)) | divide(n1,const_10)|divide(n0,#0) | physics |
a jogger running at 9 km / hr along side a railway track is 240 m ahead of the engine of a 210 m long train running at 45 km / hr in the same direction . in how much time will the train pass the jogger ? | "speed of train relative to jogger = 45 - 9 = 36 km / hr . = 36 * 5 / 18 = 10 m / sec . distance to be covered = 240 + 210 = 450 m . time taken = 450 / 10 = 45 sec . answer : b" | a ) 88 , b ) 45 , c ) 36 , d ) 80 , e ) 12 | b | divide(add(240, 210), multiply(subtract(45, 9), divide(divide(const_10, const_2), divide(subtract(45, 9), const_2)))) | add(n1,n2)|divide(const_10,const_2)|subtract(n3,n0)|divide(#2,const_2)|divide(#1,#3)|multiply(#4,#2)|divide(#0,#5)| | general |
a cistern is normally filled in 12 hrs , but takes 2 hrs longer to fill because of a leak on its bottom , if cistern is full , how much time citern would empty ? | "if leakage / hour = 1 / x , then 1 / 12 - 1 / x = 1 / 14 , solving 1 / x = 1 / 84 so in 84 hours full cistern will be empty . answer : b" | a ) 10 hours , b ) 84 hours , c ) 30 hours , d ) 40 hours , e ) 50 hours | b | inverse(subtract(divide(const_1, 12), divide(const_1, add(12, 2)))) | add(n0,n1)|divide(const_1,n0)|divide(const_1,#0)|subtract(#1,#2)|inverse(#3)| | physics |
what is the units digit of the product of the first 100 odd numbers ? | 1 * 3 * 5 * 7 . . . . . . . . . . . . . . . . . . . will end up in 5 in the units place answer : d | a ) 1 , b ) 3 , c ) 0 , d ) 5 , e ) 7 | d | divide(divide(lcm(100, const_1), const_10), const_2) | lcm(n0,const_1)|divide(#0,const_10)|divide(#1,const_2) | general |
how many positive integers between 20 and 2000 ( both are inclusive ) are there such that they are multiples of 10 ? | "multiples of 10 = 20 , 30,40 - - - - - , 1990,2000 number of multiples of 10 = > 2000 - 20 / 10 + 1 = 199 answer is d" | a ) 201 , b ) 193 , c ) 200 , d ) 199 , e ) 195 | d | divide(subtract(2000, 20), 10) | subtract(n1,n0)|divide(#0,n2)| | other |
what is the cost of leveling the field in the form of parallelogram at the rate of rs . 50 / 10 sq . metre , whose base & perpendicular distance from the other side being 54 m & 24 m respectively ? | area of the parallelogram = length of the base * perpendicular height = 54 * 24 = 1296 m . total cost of levelling = rs . 6480 d | a ) rs . 5280 , b ) rs . 5500 , c ) rs . 6000 , d ) rs . 6480 , e ) rs . 7680 | d | multiply(multiply(54, 24), divide(50, 10)) | divide(n0,n1)|multiply(n2,n3)|multiply(#0,#1) | physics |
a and b together can complete a piece of work in 4 days . if a alone can complete the same work in 12 days , in how many days , in how many days can b alone complete that work ? | "( a + b ) ' s 1 day ' s work = 1 / 4 a ' s 1 day ' s work = 1 / 12 b ' s 1 days ' s work = ( 1 / 4 ) - ( 1 / 12 ) = 1 / 6 hence , b alone can complete the work in 6 days . answer is a" | a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 5 | a | divide(const_1, subtract(divide(const_1, 4), divide(const_1, 12))) | divide(const_1,n0)|divide(const_1,n1)|subtract(#0,#1)|divide(const_1,#2)| | physics |
the area of a square is added to one of its sides , and the perimeter is then subtracted from this total , the result is 4 . what is the length of one side ? | the equation is ; side + area - perimeter = s + a - p = s + s ^ 2 - 4 s = s ( 1 + s - 4 ) . by plugging in the answers we can test the answers quickly ; then , 4 is the only possible answer . answer : b | ['a ) 5', 'b ) 4', 'c ) 3', 'd ) 2', 'e ) 1'] | b | power(divide(4, const_2), const_2) | divide(n0,const_2)|power(#0,const_2) | geometry |
a , b and c rents a pasture for rs . 870 . a put in 12 horses for 8 months , b 16 horses for 9 months and 18 horses for 6 months . how much should a pay ? | "12 * 8 : 16 * 9 = 18 * 6 8 : 12 : 9 8 / 29 * 870 = 240 answer : a" | a ) 240 , b ) 270 , c ) 276 , d ) 271 , e ) 272 | a | multiply(divide(870, add(add(multiply(12, 8), multiply(16, 9)), multiply(18, 6))), multiply(16, 9)) | multiply(n1,n2)|multiply(n3,n4)|multiply(n5,n6)|add(#0,#1)|add(#3,#2)|divide(n0,#4)|multiply(#5,#1)| | general |
caleb spends $ 68.50 on 50 hamburgers for the marching band . if single burgers cost $ 1.00 each and double burgers cost $ 1.50 each , how many double burgers did he buy ? | solution - lets say , single hamburgersxand double hamburgersy given that , x + y = 50 and 1 x + 1.5 y = 68.50 . by solving the equations y = 37 . ans c . | a ) 5 , b ) 10 , c ) 37 , d ) 40 , e ) 45 | c | divide(subtract(68.5, 50), subtract(1.5, 1)) | subtract(n0,n1)|subtract(n3,n2)|divide(#0,#1) | general |
in a graduating class of 234 students , 144 took geometry and 119 took biology . what is the difference between the greatest possible number and the smallest possible number of students that could have taken both geometry and biology ? | "greatest possible number taken both should be 144 ( as it is maximum for one ) smallest possible number taken both should be given by total - neither = a + b - both both = a + b + neither - total ( neither must be 0 to minimize the both ) so 144 + 119 + 0 - 234 = 29 greatest - smallest is 144 - 29 = 115 so answer must... | a ) 144 , b ) 115 , c ) 113 , d ) 88 , e ) 31 | b | subtract(119, subtract(add(144, 119), 234)) | add(n1,n2)|subtract(#0,n0)|subtract(n2,#1)| | other |
two trains are moving at 60 kmph and 70 kmph in opposite directions . their lengths are 150 m and 100 m respectively . the time they will take to pass each other completely is ? | "70 + 60 = 130 * 5 / 18 = 325 / 9 mps d = 150 + 100 = 250 m t = 250 * 9 / 325 = 7 sec answer : c" | a ) 5 sec , b ) 6 sec , c ) 7 sec , d ) 8 sec , e ) 9 sec | c | divide(add(150, 100), multiply(add(60, 70), const_0_2778)) | add(n2,n3)|add(n0,n1)|multiply(#1,const_0_2778)|divide(#0,#2)| | physics |
one fourth of a solution that was 11 % salt by weight was replaced by a second solution resulting in a solution that was 16 percent sugar by weight . the second solution was what percent salt by weight ? | "consider total solution to be 100 liters and in this case you ' ll have : 75 * 0.11 + 25 * x = 100 * 0.16 - - > x = 0.31 . answer : d ." | a ) 24 % , b ) 34 % , c ) 22 % , d ) 31 % , e ) 8.5 % | d | multiply(subtract(multiply(divide(16, const_100), const_4), subtract(multiply(divide(11, const_100), const_4), divide(11, const_100))), const_100) | divide(n1,const_100)|divide(n0,const_100)|multiply(#0,const_4)|multiply(#1,const_4)|subtract(#3,#1)|subtract(#2,#4)|multiply(#5,const_100)| | gain |
1297 x 1297 = ? | "1297 x 1297 = ( 1297 ) 2 = ( 1300 - 3 ) 2 = ( 1300 ) 2 + ( 3 ) 2 - ( 2 x 1300 x 3 ) = 1690000 + 9 - 7800 = 1690009 - 7800 = 1682209 . answer : a" | a ) a ) 1682209 , b ) b ) 1951601 , c ) c ) 1951602 , d ) d ) 1951603 , e ) e ) 1951604 | a | multiply(divide(1297, 1297), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
80 percent of the members of a study group are women , and 40 percent of those women are lawyers . if one member of the study group is to be selected at random , what is the probability that the member selected is a woman lawyer ? | say there are 100 people in that group , then there would be 0.8 * 0.40 * 100 = 32 women lawyers , which means that the probability that the member selected is a woman lawyer is favorable / total = 32 / 100 . answer : d | a ) 0.45 , b ) 0.55 , c ) 0.65 , d ) 0.32 , e ) 0.35 | d | multiply(divide(80, multiply(multiply(const_5, const_5), const_4)), divide(40, multiply(multiply(const_5, const_5), const_4))) | multiply(const_5,const_5)|multiply(#0,const_4)|divide(n0,#1)|divide(n1,#1)|multiply(#2,#3) | gain |
if a and b are positive integers and ( 2 ^ a ) ^ b = 2 ^ 5 , what is the value of 2 ^ a * 2 ^ b ? | "2 ^ ab = 2 ^ 5 therefore ab = 5 either a = 1 or 5 or b = 5 or 1 therefore 2 ^ a * 2 ^ b = 2 ^ ( a + b ) = 2 ^ 6 = 64 b" | a ) 8 , b ) 64 , c ) 16 , d ) 32 , e ) 4 | b | multiply(power(2, 2), 2) | power(n0,n0)|multiply(n0,#0)| | general |
if money is invested at r percent interest , compounded annually , the amount of investment will double in approximately 70 / r years . if pat ' s parents invested $ 8000 in a long term bond that pays 8 percent interest , compounded annually , what will be the approximate total amount of investment 18 years later , whe... | "since investment doubles in 70 / r years then for r = 8 it ' ll double in 70 / 8 = ~ 9 years ( we are not asked about the exact amount so such an approximation will do ) . thus in 18 years investment will double twice and become ( $ 8,000 * 2 ) * 2 = $ 32,000 ( after 9 years investment will become $ 8,000 * 2 = $ 16,0... | a ) $ 20000 , b ) $ 15000 , c ) $ 32000 , d ) $ 10000 , e ) $ 9000 | c | multiply(multiply(8000, const_2), const_2) | multiply(n1,const_2)|multiply(#0,const_2)| | general |
the average ( arithmetic mean ) of the even integers from 0 to 120 inclusive is how much greater than the average ( arithmetic mean ) of the even integers from 0 to 60 inclusive ? | "the sum of even numbers from 0 to n is 2 + 4 + . . . + n = 2 ( 1 + 2 + . . . + n / 2 ) = 2 ( n / 2 ) ( n / 2 + 1 ) / 2 = ( n / 2 ) ( n / 2 + 1 ) the average is ( n / 2 ) ( n / 2 + 1 ) / ( n / 2 + 1 ) = n / 2 the average of the even numbers from 0 to 120 is 120 / 2 = 60 the average of the even numbers from 0 to 60 is 6... | a ) 10 , b ) 15 , c ) 20 , d ) 30 , e ) 60 | d | subtract(divide(add(60, 0), const_2), divide(add(120, 0), const_2)) | add(n2,n3)|add(n0,n1)|divide(#0,const_2)|divide(#1,const_2)|subtract(#2,#3)| | general |
a batsman scored 120 runs which included 3 boundaries and 8 sixes . what percent of his total score did he make by running between the wickets ? | "explanation : total runs scored = 120 total runs scored from boundaries and sixes = 3 x 4 + 8 x 6 = 60 total runs scored by running between the wickets = 120 - 60 = 60 required % = ( 60 / 120 ) × 100 = 50 % answer : option b" | a ) 45 ( 4 / 11 ) % , b ) 50 % , c ) 45 ( 5 / 11 ) % , d ) 44 ( 5 / 11 ) % , e ) none of these | b | multiply(divide(subtract(120, add(multiply(3, 8), multiply(8, 3))), 120), const_100) | multiply(n1,n2)|multiply(n1,n2)|add(#0,#1)|subtract(n0,#2)|divide(#3,n0)|multiply(#4,const_100)| | general |
the average age of 6 men increases by 3 years when two women are included in place of two men of ages 24 and 26 years . find the average age of the women ? | explanation : 24 + 26 + 6 * 3 = 68 / 2 = 34 answer : c | a ) 37 , b ) 26 , c ) 34 , d ) 18 , e ) 11 | c | divide(add(add(24, 26), multiply(6, 3)), const_2) | add(n2,n3)|multiply(n0,n1)|add(#0,#1)|divide(#2,const_2) | general |
| 9 - 4 | - | 12 - 14 | = ? | "| 9 - 4 | - | 12 - 14 | = | 5 | - | - 2 | = 5 - 2 = 3 correct answer a" | a ) 3 , b ) 2 , c ) 1 , d ) 0 , e ) 4 | a | subtract(subtract(9, 4), subtract(14, 12)) | subtract(n0,n1)|subtract(n3,n2)|subtract(#0,#1)| | general |
a school has received 80 % of the amount it needs for a new building by receiving a donation of $ 600 each from people already solicited . people already solicited represent 40 % of the people from whom the school will solicit donations . how much average contribution is requited from the remaining targeted people to c... | "let us suppose there are 100 people . 40 % of them donated $ 24000 ( 600 * 40 ) $ 24000 is 80 % of total amount . so total amount = 24000 * 100 / 80 remaining amount is 20 % of total amount . 20 % of total amount = 24000 * ( 100 / 80 ) * ( 20 / 100 ) = 6000 this amount has to be divided by 60 ( remaining people are 60... | a ) $ 200 , b ) 100 , c ) $ 105 , d ) $ 250 , e ) $ 300 | b | divide(multiply(divide(multiply(divide(40, const_100), 600), divide(80, const_100)), divide(40, const_100)), divide(80, const_100)) | divide(n2,const_100)|divide(n0,const_100)|multiply(n1,#0)|divide(#2,#1)|multiply(#3,#0)|divide(#4,#1)| | general |
if 4 ^ k = 5 , then 4 ^ ( 2 k + 2 ) = | "4 ^ k = 5 4 ^ 2 k = 5 ^ 2 4 ^ 2 k = 25 4 ^ ( 2 k + 2 ) = 4 ^ 2 k * 4 ^ 2 = 25 * 16 = 400 answer : a" | a ) 400 , b ) 540 , c ) 100 , d ) 830 , e ) 420 | a | multiply(power(5, 2), power(4, 2)) | power(n1,n3)|power(n0,n4)|multiply(#0,#1)| | general |
there are 16 stations between hyderabad and bangalore . how many second class tickets have to be printed , so that a passenger can travel from any station to any other station ? | the total number of stations = 18 from 18 stations we have to choose any two stations and the direction of travel ( i . e . , hyderabad to bangalore is different from bangalore to hyderabad ) in 18 p ₂ ways . 18 p ₂ = 18 * 17 = 306 . answer : e | a ) 288 , b ) 267 , c ) 261 , d ) 211 , e ) 306 | e | multiply(add(16, const_1), add(add(16, const_1), const_1)) | add(n0,const_1)|add(#0,const_1)|multiply(#0,#1) | physics |
an auction house charges a commission of 15 % on the first $ 50,000 of the sale price of an item , plus 10 % on the amount of of the sale price in excess of $ 50,000 . what was the price of a painting for which the house charged a total commission of $ 24,000 ? | say the price of the house was $ x , then 0.15 * 50,000 + 0.1 * ( x - 50,000 ) = 24,000 - - > x = $ 215,000 ( 15 % of $ 50,000 plus 10 % of the amount in excess of $ 50,000 , which is x - 50,000 , should equal to total commission of $ 24,000 ) . answer : c | a ) $ 115,000 , b ) $ 160,000 , c ) $ 215,000 , d ) $ 240,000 , e ) $ 365,000 | c | add(multiply(15, 10), 10) | multiply(n0,n2)|add(n2,#0) | general |
a man travelled for 13 hours . he covered the first half of the distance at 20 kmph and remaining half of the distance at 25 kmph . find the distance travelled by the man ? | let the distance travelled be x km . total time = ( x / 2 ) / 20 + ( x / 2 ) / 25 = 13 = > x / 40 + x / 50 = 13 = > ( 5 x + 4 x ) / 200 = 13 = > x = 200 km answer : c | a ) 168 km , b ) 864 km , c ) 200 km , d ) 240 km , e ) 460 km | c | add(multiply(divide(13, const_3), 20), multiply(divide(13, const_3), 25)) | divide(n0,const_3)|multiply(n1,#0)|multiply(n2,#0)|add(#1,#2) | physics |
in the storage room of a certain bakery , the ratio of sugar to flour is 5 to 4 , and the ratio of flour to baking soda is 10 to 1 . if there were 60 more pounds of baking soda in the room , the ratio of flour to baking soda would be 8 to 1 . how many pounds of sugar are stored in the room ? | sugar : flour = 5 : 4 = 25 : 20 ; flour : soda = 10 : 1 = 20 : 2 ; thus we have that sugar : flour : soda = 25 x : 20 x : 2 x . also given that 20 x / ( 2 x + 60 ) = 8 / 1 - - > x = 120 - - > sugar = 25 x = 3,000 answer : e . | a ) 600 , b ) 1200 , c ) 1500 , d ) 1600 , e ) 3000 | e | multiply(divide(5, 4), multiply(10, divide(multiply(8, 60), subtract(10, 8)))) | divide(n0,n1)|multiply(n4,n5)|subtract(n2,n5)|divide(#1,#2)|multiply(n2,#3)|multiply(#0,#4) | general |
a can do a work in 2 days . b can do the same work in 3 days . both a & b together will finish the work and they got $ 500 from that work . find their shares ? | "ratio of their works a : b = 2 : 3 ratio of their wages a : b = 3 : 2 a ' s share = ( 3 / 5 ) 500 = 300 b ' s share = ( 2 / 5 ) 500 = 200 correct option is d" | a ) 100,200 , b ) 400,100 , c ) 300,150 , d ) 300,200 , e ) 250,250 | d | divide(multiply(2, 3), add(2, 3)) | add(n0,n1)|multiply(n0,n1)|divide(#1,#0)| | physics |
a fruit seller had some apples . he sells 40 % apples and still has 420 apples . originally , he had | "solution suppose originally he had x apples . then , ( 100 - 40 ) % of x = 420 . ‹ = › 60 / 100 × x = 420 x ‹ = › ( 420 × 100 / 60 ‹ = › 700 . answer d" | a ) 588 apples , b ) 600 apples , c ) 672 apples , d ) 700 apples , e ) none | d | original_price_before_loss(40, 420) | original_price_before_loss(n0,n1)| | gain |
the cost price of a radio is rs . 1890 and it was sold for rs . 1500 , find the loss % ? | "1890 - - - - 390 100 - - - - ? = > 20 % answer : b" | a ) 18 , b ) 20 , c ) 77 , d ) 66 , e ) 41 | b | multiply(divide(subtract(1890, 1500), 1890), const_100) | subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)| | gain |
if the price of petrol increases by 44 , by how much must a user cut down his consumption so that his expenditure on petrol remains constant ? | "explanation : let us assume before increase the petrol will be rs . 100 . after increase it will be rs ( 100 + 44 ) i . e 144 . now , his consumption should be reduced to : - = ( 144 − 100 ) / 144 ∗ 100 . hence , the consumption should be reduced to 30.6 % . answer : e" | a ) 25 % , b ) 20 % , c ) 16.67 % , d ) 33.33 % , e ) none of these | e | multiply(subtract(const_1, divide(const_100, add(const_100, 44))), const_100) | add(n0,const_100)|divide(const_100,#0)|subtract(const_1,#1)|multiply(#2,const_100)| | general |
a certain sum is invested at simple interest at 15 % p . a . for two years instead of investing at 12 % p . a . for the same time period . therefore the interest received is more by rs . 840 . find the sum ? | let the sum be rs . x . ( x * 15 * 2 ) / 100 - ( x * 12 * 2 ) / 100 = 840 = > 30 x / 100 - 24 x / 100 = 840 = > 6 x / 100 = 840 = > x = 14000 . answer : c | a ) rs . 7000 , b ) rs . 9000 , c ) rs . 14000 , d ) rs . 17000 , e ) rs . 27000 | c | divide(840, divide(multiply(subtract(15, 12), const_2), const_100)) | subtract(n0,n1)|multiply(#0,const_2)|divide(#1,const_100)|divide(n2,#2) | gain |
what is 35 % of 4 / 13 of 715 ? | "this problem can be solved easily if we just use approximation : 35 % is a little over 1 / 3 , while 4 / 13 is a little less than 4 / 12 , which is 1 / 3 . thus , the answer is about 1 / 3 of 1 / 3 of 715 , or 1 / 9 of 715 . since the first 1 / 3 is a slight underestimate and the second 1 / 3 is a slight overestimate ... | a ) 44 , b ) 55 , c ) 66 , d ) 77 , e ) 88 | d | divide(multiply(35, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100) | add(const_2,const_3)|add(const_3,const_4)|add(const_3,const_3)|multiply(const_3,const_4)|multiply(#0,const_2)|multiply(#3,const_100)|multiply(#1,#0)|multiply(#4,#5)|multiply(#6,#4)|add(#7,#8)|add(#9,#2)|multiply(n0,#10)|divide(#11,const_100)| | gain |
16 welders work at a constant rate they complete an order in 8 days . if after the first day , 9 welders start to work on the other project , how many more days the remaining welders will need to complete the rest of the order ? | "1 . we need to find out the time taken by 7 workers after day 1 . 2 . total no . of wokers * total time taken = time taken by 1 worker 3 . time taken by 1 worker = 16 * 8 = 128 days 4 . but on day 1 sixteen workers had already worked finishing 1 / 8 of the job . so 7 workers have to finish only 7 / 8 of the job . 5 . ... | a ) 5 , b ) 2 , c ) 8 , d ) 4 , e ) 16 | e | divide(divide(subtract(const_1, multiply(divide(const_1, multiply(16, 8)), 16)), subtract(16, 9)), divide(const_1, multiply(16, 8))) | multiply(n0,n1)|subtract(n0,n2)|divide(const_1,#0)|multiply(n0,#2)|subtract(const_1,#3)|divide(#4,#1)|divide(#5,#2)| | physics |
if the radius of a circle is decreased by 50 % , find the percentage decrease in its area . | "let original radius = r . new radius = ( 50 / 100 ) r = ( r / 2 ) original area = 22 / 7 ( r ) 2 = and new area = ( ( r / 2 ) ) 2 = ( ( r ) 2 ) / 4 decrease in area = ( ( 3 ( 22 / 7 ) ( r ) 2 ) / 4 x ( 1 / ( 22 / 7 ) ( r ) 2 ) x 100 ) % = 75 % answer is c ." | a ) 25 % , b ) 50 % , c ) 75 % , d ) 95 % , e ) none of them | c | multiply(subtract(divide(const_100, const_100), power(subtract(divide(const_100, const_100), divide(50, const_100)), const_2)), const_100) | divide(const_100,const_100)|divide(n0,const_100)|subtract(#0,#1)|power(#2,const_2)|subtract(#0,#3)|multiply(#4,const_100)| | gain |
what is the least number of squares tiles required to pave the floor of a room 8 m 82 cm long and 2 m 52 cm broad ? | "length of largest tile = h . c . f . of 882 cm and 252 cm = 126 cm . area of each tile = ( 126 x 126 ) cm 2 . required number of tiles = 882 x 252 / ( 126 ^ 2 ) = 14 . answer : a" | a ) 14 , b ) 20 , c ) 40 , d ) 44 , e ) 48 | a | divide(rectangle_area(multiply(8, const_100), multiply(82, const_100)), square_area(add(multiply(const_4, const_10), const_1))) | multiply(n0,const_100)|multiply(n1,const_100)|multiply(const_10,const_4)|add(#2,const_1)|rectangle_area(#0,#1)|square_area(#3)|divide(#4,#5)| | physics |
a coin is tossed 8 times . what is the probability of getting exactly 7 heads ? | "the number of possible outcomes is 2 ^ 8 = 256 there are 8 ways to get exactly 7 heads . p ( exactly 7 heads ) = 8 / 256 = 1 / 32 the answer is c ." | a ) 1 / 8 , b ) 1 / 16 , c ) 1 / 32 , d ) 1 / 64 , e ) 1 / 128 | c | multiply(power(divide(const_1, const_2), 7), multiply(choose(8, 7), divide(const_1, const_2))) | choose(n0,n1)|divide(const_1,const_2)|multiply(#0,#1)|power(#1,n1)|multiply(#2,#3)| | probability |
the price of a jacket is reduced by 25 % . during a special sale the price of the jacket is reduced another 30 % . by approximately what percent must the price of the jacket now be increased in order to restore it to its original amount ? | "1 ) let the price of jacket initially be $ 100 . 2 ) then it is decreased by 25 % , therefore bringing down the price to $ 75 . 3 ) again it is further discounted by 30 % , therefore bringing down the price to $ 52.5 . 4 ) now 52.5 has to be added byx % in order to equal the original price . 52.5 + ( x % ) 52.5 = 100 ... | a ) 105 , b ) 45 , c ) 85 , d ) 95 , e ) 90.5 | e | multiply(const_100, divide(subtract(const_100, subtract(subtract(const_100, 25), multiply(subtract(const_100, 25), divide(30, const_100)))), subtract(subtract(const_100, 25), multiply(subtract(const_100, 25), divide(30, const_100))))) | divide(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|subtract(#1,#2)|subtract(const_100,#3)|divide(#4,#3)|multiply(#5,const_100)| | gain |
tough and tricky questions : word problems . mike , sarah and david decided to club together to buy a present . each of them gave equal amount of money . luckily sarah negotiated a 20 % discount for the present so that each of them paid 4 dollars less . how much did they pay for a present ? | answer c . we know that sarah negotiated a discount of 20 % , so each of them paid $ 4 less . since there are three people , the 20 % of the original price amounts to $ 12 . 5 times 12 $ is 60 $ , so the original price , before sarah negotiated the discount , had been $ 60 . they paid $ 12 less than the base price , so... | a ) 20 , b ) 36 , c ) 48 , d ) 60 , e ) 72 | c | divide(multiply(divide(multiply(multiply(4, const_3), const_100), 20), multiply(20, 4)), const_100) | multiply(n1,const_3)|multiply(n0,n1)|multiply(#0,const_100)|divide(#2,n0)|multiply(#3,#1)|divide(#4,const_100) | general |
if each digit in the set a = { 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 } is used exactly once , in how many ways can the digits be arranged ? | "9 ! = 362,880 the answer is d ." | a ) 356,580 , b ) 358,680 , c ) 360,780 , d ) 362,880 , e ) 364,980 | d | factorial(6) | factorial(n5)| | general |
8 ^ 100 is divisible by 17 . find the remainder ? | "this is an extremely difficult problem to solve with out fermat ' s little theorem . by applying fermat ' s little theorem , we know that 816 when divided by 17 , the remainder is 1 . so divide 100 by 16 and find the remainder . remainder = 4 therefore , 100 = ( 16 × 6 ) + 4 now this problem can be written as 810017 =... | a ) 10 , b ) 12 , c ) 16 , d ) 18 , e ) 20 | c | subtract(17, 8) | subtract(n2,n0)| | general |
in a competitive examination in a state a , 6 % candidates got selected from the total appeared candidates . state b had an equal number of candidates appeared and 7 % candidates got selected with 80 more candidates got selected than a . what was the number of candidates appeared from each state ? | let the number of candidates appeared be x then , 7 % of x - 6 % of x = 80 1 % of x = 80 x = 80 * 100 = 8000 answer is a | a ) 8000 , b ) 7540 , c ) 6500 , d ) 9100 , e ) 6000 | a | multiply(multiply(subtract(7, 6), const_100), 80) | subtract(n1,n0)|multiply(#0,const_100)|multiply(n2,#1) | general |
how many natural numbers are there between 23 and 100 which are exactly divisible by 6 ? | explanation : required numbers are 24 , 30 , 36 , 42 , . . . , 96 this is an a . p . in which a = 24 , d = 6 and l = 96 let the number of terms in it be n . then tn = 96 a + ( n - 1 ) d = 96 24 + ( n - 1 ) x 6 = 96 ( n - 1 ) x 6 = 72 ( n - 1 ) = 12 n = 13 required number of numbers = 13 . a ) | a ) 13 , b ) 25 , c ) 87 , d ) 49 , e ) 63 | a | add(divide(subtract(subtract(100, const_4), add(23, const_1)), 6), const_1) | add(n0,const_1)|subtract(n1,const_4)|subtract(#1,#0)|divide(#2,n2)|add(#3,const_1) | general |
at a loading dock , each worker on the night crew loaded 3 / 4 as many boxes as each worker on the day crew . if the night crew has 4 / 9 as many workers as the day crew , what fraction of all the boxes loaded by the two crews did the day crew load ? | "method : x = no . of boxes loaded by day crew . boxes by night crew = 3 / 4 * 4 / 9 x = 1 / 3 x % loaded by day crew = x / ( x + 1 / 3 x ) = 3 / 4 answer c" | a ) 1 / 2 , b ) 2 / 5 , c ) 3 / 4 , d ) 4 / 5 , e ) 5 / 8 | c | divide(multiply(9, 4), add(multiply(9, 4), multiply(3, 4))) | multiply(n1,n3)|multiply(n0,n2)|add(#0,#1)|divide(#0,#2)| | physics |
a bowl was filled with 10 ounces of water , and 0.0008 ounce of the water evaporated each day during a 50 - day period . what percent of the original amount of water evaporated during this period ? | total amount of water evaporated each day during a 50 - day period = . 0008 * 50 = . 0008 * 100 / 2 = . 08 / 2 = . 04 percent of the original amount of water evaporated during this period = ( . 04 / 10 ) * 100 % = 0.4 % answer c | a ) 0.004 % , b ) 0.04 % , c ) 0.40 % , d ) 4 % , e ) 40 % | c | multiply(divide(multiply(50, 0.0008), 10), const_100) | multiply(n1,n2)|divide(#0,n0)|multiply(#1,const_100) | gain |
if x ¤ y = ( x + y ) ^ 2 - ( x - y ) ^ 2 . then √ 5 ¤ √ 5 = | "x = √ 5 and y also = √ 5 applying the function ( √ 5 + √ 5 ) ^ 2 - ( √ 5 - √ 5 ) ^ 2 = ( 2 √ 5 ) ^ 2 - 0 = 4 x 5 = 20 . note : alternative approach is the entire function is represented as x ^ 2 - y ^ 2 = ( x + y ) ( x - y ) which can be simplified as ( x + y + x - y ) ( x + y - ( x - y ) ) = ( 2 x ) ( 2 y ) = 4 xy . ... | a ) 0 , b ) 5 , c ) 10 , d ) 15 , e ) 20 | e | power(add(sqrt(5), sqrt(5)), 2) | sqrt(n2)|add(#0,#0)|power(#1,n0)| | general |
31 of the scientists that attended a certain workshop were wolf prize laureates , and 16 of these 31 were also nobel prize laureates . of the scientists that attended that workshop and had not received the wolf prize , the number of scientists that had received the nobel prize was 3 greater than the number of scientist... | lets solve by creating equation . . w = 31 . . total = 50 . . not w = 50 - 31 = 19 . . now let people who were neither be x , so out of 19 who won nobel = x + 3 . . so x + x + 3 = 19 or x = 8 . . so who won nobel but not wolf = x + 3 = 11 . . but people who won both w and n = 13 . . so total who won n = 11 + 16 = 27 . ... | a ) a ) 11 , b ) b ) 18 , c ) c ) 24 , d ) d ) 27 , e ) d ) 36 | d | add(add(3, divide(subtract(subtract(50, 31), 3), const_2)), 16) | subtract(n4,n0)|subtract(#0,n3)|divide(#1,const_2)|add(n3,#2)|add(n1,#3) | physics |
which number should replace both the asterisks in ( * / 20 ) x ( * / 180 ) = 1 ? | "let ( y / 20 ) x ( y / 180 ) = 1 y ^ 2 = 20 x 180 = 20 x 20 x 9 y = ( 20 x 3 ) = 60 the answer is c ." | a ) 20 , b ) 40 , c ) 60 , d ) 90 , e ) 120 | c | sqrt(multiply(180, 20)) | multiply(n0,n1)|sqrt(#0)| | general |
how many seconds will a train 100 meters long take to cross a bridge 150 meters long if the speed of the train is 18 kmph ? | "d = 100 + 150 = 250 s = 18 * 5 / 18 = 5 mps t = 250 / 5 = 50 sec d ) 50 sec" | a ) 80 sec , b ) 20 sec , c ) 40 sec , d ) 50 sec , e ) 60 sec | d | divide(add(150, 100), multiply(18, const_0_2778)) | add(n0,n1)|multiply(n2,const_0_2778)|divide(#0,#1)| | physics |
what is the sum of all even numbers from 1 to 701 ? | "explanation : 700 / 2 = 350 350 * 351 = 122850 answer : d" | a ) 122821 , b ) 281228 , c ) 281199 , d ) 122850 , e ) 128111 | d | divide(multiply(1, 701), const_4) | multiply(n0,n1)|divide(#0,const_4)| | general |
two numbers have a h . c . f of 11 and a product of two numbers is 1991 . find the l . c . m of the two numbers ? | "l . c . m of two numbers is given by ( product of the two numbers ) / ( h . c . f of the two numbers ) = 1991 / 11 = 181 . answer : b" | a ) 140 , b ) 181 , c ) 160 , d ) 170 , e ) 180 | b | divide(1991, 11) | divide(n1,n0)| | physics |
at what price must an book costing $ 47.50 be marked in order that after deducting 15 % from the list price . it may be sold at a profit of 25 % on the cost price ? | "c $ 62.50 cp = 47.50 sp = 47.50 * ( 125 / 100 ) = 59.375 mp * ( 85 / 100 ) = 59.375 mp = 69.85 b" | a ) 72.85 , b ) 69.85 , c ) 62.85 , d ) 82.85 , e ) 60.85 | b | multiply(divide(divide(multiply(47.50, add(const_100, 25)), const_100), subtract(const_100, 15)), const_100) | add(n2,const_100)|subtract(const_100,n1)|multiply(n0,#0)|divide(#2,const_100)|divide(#3,#1)|multiply(#4,const_100)| | gain |
16.02 × 0.001 = ? | "16.02 × 0.001 = ? or , ? = 0.01602 answer d" | a ) 0.1602 , b ) 0.001602 , c ) 1.6021 , d ) 0.01602 , e ) none of these | d | multiply(divide(16.02, 0.001), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
find the numbers which are in the ratio 3 : 2 : 3 such that the sum of the first and the second added to the difference of the third and the second is 24 ? | "let the numbers be a , b and c . a : b : c = 3 : 2 : 3 given , ( a + b ) + ( c - b ) = 24 = > a + c = 24 = > 3 x + 3 x = 24 = > x = 6 a , b , c are 3 x , 2 x , 3 x a , b , c are 18 , 12 , 18 . answer : e" | a ) 4 , 3,22 , b ) 4 , 4,22 , c ) 9 , 3,32 , d ) 9 , 6,12 , e ) 18 , 12,18 | e | divide(multiply(3, 3), 3) | multiply(n2,n0)|divide(#0,n0)| | general |
in one hour , a boat goes 13 km along the stream and 5 km against the stream . the speed of the boat in still water ( in km / hr ) is : | "sol . speed in still water = 1 / 2 ( 13 + 5 ) kmph = 9 kmph . answer c" | a ) 2 , b ) 4 , c ) 9 , d ) 12 , e ) 15 | c | divide(add(13, 5), const_2) | add(n0,n1)|divide(#0,const_2)| | physics |
a library has an average of 510 visitors on sundays and 240 on other days . what is the average number of visitors per day in a month of 30 days beginning with a sunday ? | "in a month of 30 days beginning with a sunday , there will be 4 complete weeks and another two days which will be sunday and monday . hence there will be 5 sundays and 25 other days in a month of 30 days beginning with a sunday average visitors on sundays = 510 total visitors of 5 sundays = 510 × 5 average visitors on... | a ) 290 , b ) 304 , c ) 285 , d ) 270 , e ) 275 | c | divide(add(multiply(add(floor(divide(30, add(const_3, const_4))), const_1), 510), multiply(subtract(30, add(floor(divide(30, add(const_3, const_4))), const_1)), 240)), 30) | add(const_3,const_4)|divide(n2,#0)|floor(#1)|add(#2,const_1)|multiply(n0,#3)|subtract(n2,#3)|multiply(n1,#5)|add(#4,#6)|divide(#7,n2)| | general |
if each participant of a chess tournament plays exactly one game with each of the remaining participants , then 231 games will be played during the tournament . what is the number of participants ? | "let n be the number of participants . the number of games is nc 2 = n * ( n - 1 ) / 2 = 231 n * ( n - 1 ) = 462 = 22 * 21 ( trial and error ) the answer is d ." | a ) 16 , b ) 18 , c ) 20 , d ) 22 , e ) 24 | d | divide(add(sqrt(add(multiply(multiply(231, const_2), const_4), const_1)), const_1), const_2) | multiply(n0,const_2)|multiply(#0,const_4)|add(#1,const_1)|sqrt(#2)|add(#3,const_1)|divide(#4,const_2)| | general |
in a certain archery competition , points were awarded as follows : the first place winner receives 11 points , the second place winner receives 7 points , the third place winner receives 5 points and the fourth place winner receives 2 points . no other points are awarded . john participated several times in the compet... | "15400 = 2 * 2 * 2 * 5 * 5 * 7 * 11 john participated 7 times . the answer is c ." | a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | c | floor(sqrt(divide(15400, multiply(multiply(multiply(11, 7), 5), 2)))) | multiply(n0,n1)|multiply(n2,#0)|multiply(n3,#1)|divide(n4,#2)|sqrt(#3)|floor(#4)| | general |
the owner of a furniture shop charges his customer 30 % more than the cost price . if a customer paid rs . 8450 for a computer table , then what was the cost price of the computer table ? | "cp = sp * ( 100 / ( 100 + profit % ) ) = 8450 ( 100 / 130 ) = rs . 6500 . answer : d" | a ) rs . 5725 , b ) rs . 5275 , c ) rs . 6275 , d ) rs . 6500 , e ) none of these | d | divide(8450, add(const_1, divide(30, const_100))) | divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)| | gain |
on rainy mornings , mo drinks exactly n cups of hot chocolate ( assume that n is an integer ) . on mornings that are not rainy , mo drinks exactly 6 cups of tea . last week mo drank a total of 22 cups of tea and hot chocolate together . if during that week mo drank 14 more tea cups than hot chocolate cups , then how ma... | "t = the number of cups of tea c = the number of cups of hot chocolate t + c = 22 t - c = 14 - > t = 18 . c = 2 . mo drinks 6 cups of tea a day then number of days that are not rainy = 18 / 6 = 3 so number of rainy days = 7 - 3 = 4 c is the answer ." | a ) 0 , b ) 1 , c ) 4 , d ) 5 , e ) 6 | c | subtract(add(const_4, 6), divide(divide(add(22, 14), const_2), 6)) | add(n0,const_4)|add(n1,n2)|divide(#1,const_2)|divide(#2,n0)|subtract(#0,#3)| | general |
find the value of 85 p 3 . | "85 p 3 = 85 ! / ( 85 - 3 ) ! = 85 ! / 82 ! = 85 * 84 * 83 * 82 ! / 82 ! = 85 * 84 * 83 = 595650 answer : b" | a ) 565350 , b ) 595650 , c ) 535950 , d ) 565350 , e ) 575350 | b | multiply(subtract(3, const_4), 85) | subtract(n1,const_4)|multiply(#0,n0)| | general |
a certain city with a population of 144,000 is to be divided into 11 voting districts , and no district is to have a population that is more than 10 percent greater than the population of any other district what is the minimum possible population that the least populated district could have ? | let x = number of people in smallest district x * 1.1 = number of people in largest district x will be minimised when the number of people in largest district is maximised 10 * x * 1.1 = 11 x = total number of people in other districts so we have 11 x + x = 142 k x = 12,000 answer : d | a ) a ) 10,700 , b ) b ) 10,800 , c ) c ) 10,900 , d ) d ) 12,000 , e ) e ) 11,100 | d | multiply(multiply(const_4, const_2), const_100) | multiply(const_2,const_4)|multiply(#0,const_100) | general |
water consists of hydrogen and oxygen , and the approximate ratio , by mass , of hydrogen to oxygen is 2 : 16 . approximately how many grams of hydrogen are there in 171 grams of water ? | "( 2 / 18 ) * 171 = 19 grams the answer is e ." | a ) 11 , b ) 13 , c ) 15 , d ) 17 , e ) 19 | e | multiply(2, divide(171, add(2, 16))) | add(n0,n1)|divide(n2,#0)|multiply(n0,#1)| | other |
two whole numbers whose sum is 42 can not be in the ratio | d ) 3 : 8 | a ) 2 : 5 , b ) 1 : 6 , c ) 2 : 4 , d ) 3 : 8 , e ) 2 : 40 | d | divide(divide(subtract(divide(const_100.0, const_2), const_10), const_2), add(divide(42, const_2), const_10)) | divide(const_100.0,const_2)|add(#0,const_10)|subtract(#0,const_10)|divide(#2,const_2)|divide(#3,#1)| | other |
running at their respective constant rate , machine x takes 2 days longer to produce w widgets than machines y . at these rates , if the two machines together produce 5 w / 4 widgets in 3 days , how many days would it take machine x alone to produce 4 w widgets . | "i am getting 12 . e . hope havent done any calculation errors . . approach . . let y = no . of days taken by y to do w widgets . then x will take y + 2 days . 1 / ( y + 2 ) + 1 / y = 5 / 12 ( 5 / 12 is because ( 5 / 4 ) w widgets are done in 3 days . so , x widgets will be done in 12 / 5 days or 5 / 12 th of a widget ... | a ) 4 , b ) 6 , c ) 8 , d ) 10 , e ) 24 | e | multiply(multiply(4, 2), 3) | multiply(n0,n4)|multiply(n3,#0)| | general |
the average of first 32 natural numbers is ? | "sum of 32 natural no . = 1056 / 2 = 55 average = 528 / 32 = 16.5 answer : a" | a ) 16.5 , b ) 17.5 , c ) 18.4 , d ) 15.4 , e ) 15.1 | a | add(32, const_1) | add(n0,const_1)| | general |
a , b and c , each working alone can complete a job in 6 , 8 and 12 days respectively . if all three of them work together to complete a job and earn $ 4680 , what will be a ' s share of the earnings ? | "explanatory answer a , b and c will share the amount of $ 4680 in the ratio of the amounts of work done by them . as a takes 6 days to complete the job , if a works alone , a will be able to complete 1 / 6 th of the work in a day . similarly , b will complete 1 / 8 th and c will complete 1 / 12 th of the work . so , t... | a ) $ 1100 , b ) $ 520 , c ) $ 2080 , d ) $ 1170 , e ) $ 630 | c | multiply(4680, divide(inverse(8), add(inverse(12), add(inverse(6), inverse(8))))) | inverse(n1)|inverse(n0)|inverse(n2)|add(#1,#0)|add(#3,#2)|divide(#0,#4)|multiply(n3,#5)| | physics |
by selling an article at rs . 600 , a profit of 25 % is made . find its cost price ? | "explanation : sp = 600 cp = ( sp ) * [ 100 / ( 100 + p ) ] = 600 * [ 100 / ( 100 + 25 ) ] = 600 * [ 100 / 125 ] = rs . 480 answer : b" | a ) 322 , b ) 480 , c ) 287 , d ) 192 , e ) 107 | b | divide(multiply(600, const_100), add(const_100, 25)) | add(n1,const_100)|multiply(n0,const_100)|divide(#1,#0)| | gain |
ashok and pyarelal invested money together in a business and share a capital of ashok is 1 / 9 of that of pyarelal . if the incur a loss of rs 1200 then loss of pyarelal ? | let the capital of pyarelal be x , then capital of ashok = x / 9 so ratio of investment of pyarelal and ashok = x : x / 9 = 9 x : x hence out of the total loss of 1200 , loss of pyarelal = 1200 * 9 x / 10 x = 1080 answer : c | a ) 600 , b ) 700 , c ) 1080 , d ) 900 , e ) 1000 | c | divide(multiply(9, 1200), add(1, 9)) | add(n0,n1)|multiply(n1,n2)|divide(#1,#0) | gain |
the cross - section of a cannel is a trapezium in shape . if the cannel is 20 m wide at the top and 12 m wide at the bottom and the area of cross - section is 800 sq m , the depth of cannel is ? | "1 / 2 * d ( 20 + 12 ) = 800 d = 50 answer : b" | a ) 76 , b ) 50 , c ) 27 , d ) 80 , e ) 25 | b | divide(divide(divide(800, divide(add(20, 12), const_2)), 12), const_2) | add(n0,n1)|divide(#0,const_2)|divide(n2,#1)|divide(#2,n1)|divide(#3,const_2)| | physics |
the average wages of a worker during a fortnight comprising 15 consecutive working days was $ 90 per day . during the first 7 days , his average wages was $ 87 per day and the average wages during the last 7 days was $ 95 per day . what was his wage on the 8 th day ? | average daily wage of a worker for 15 consecutive working days = 90 $ during the first 7 days , the daily average daily wage = 87 $ during the last 7 days , the daily average daily wage = 95 $ wage on 8 th day = 90 * 15 - ( 87 * 7 + 95 * 7 ) = 1350 - ( 609 + 665 ) = 1350 - 1274 = 76 answer a | a ) $ 76 , b ) $ 90 , c ) $ 92 , d ) $ 97 , e ) $ 104 | a | subtract(multiply(90, 15), add(multiply(87, 7), multiply(95, 7))) | multiply(n0,n1)|multiply(n2,n3)|multiply(n2,n5)|add(#1,#2)|subtract(#0,#3) | physics |
what is the rate percent when the simple interest on rs . 1500 amount to rs . 250 in 5 years ? | "interest for 5 yrs = 250 interest for 1 yr = 50 interest rate = 50 / 1500 x 100 = 3.33 % answer : a" | a ) 3.33 % , b ) 6 % , c ) 2 % , d ) 95 % , e ) 1 % | a | divide(multiply(const_100, 250), multiply(1500, 5)) | multiply(n1,const_100)|multiply(n0,n2)|divide(#0,#1)| | gain |
nd the area of trapezium whose parallel sides are 20 cm and 15 cm long , and the distance between them is 14 cm ? | "area of a trapezium = 1 / 2 ( sum of parallel sides ) * ( perpendicular distance between them ) = 1 / 2 ( 20 + 15 ) * ( 14 ) = 245 cm 2 answer : b" | a ) 230 cm 2 , b ) 245 cm 2 , c ) 255 cm 2 , d ) 260 cm 2 , e ) 280 cm 2 | b | divide(multiply(14, add(20, 15)), const_2) | add(n0,n1)|multiply(n2,#0)|divide(#1,const_2)| | physics |
if 50 apprentices can finish a job in 4 hours , and 30 journeymen can finish the same job in 7 hours , how much of the job should be completed by 10 apprentices and 15 journeymen in one hour ? | "50 apprentices can finish the job in 4 hours , thus : 10 apprentices can finish the job in 4 * 5 = 20 hours ; in 1 hour 10 apprentices can finish 1 / 20 of the job . 30 journeymen can finish the same job in 4,5 hours , thus : 15 journeymen can finish the job in 7 * 2 = 14 hours ; in 1 hour 15 journeymen can finish 1 /... | a ) 17 / 140 , b ) 29 / 180 , c ) 26 / 143 , d ) 1 / 5 , e ) 39 / 121 | a | add(divide(const_1, divide(multiply(50, 4), 10)), divide(const_1, divide(multiply(30, 7), 15))) | multiply(n0,n1)|multiply(n2,n3)|divide(#0,n4)|divide(#1,n5)|divide(const_1,#2)|divide(const_1,#3)|add(#4,#5)| | physics |
a sum of money lent out at s . i . amounts to a total of $ 420 after 2 years and to $ 595 after a further period of 5 years . what was the initial sum of money that was invested ? | "s . i for 5 years = $ 595 - $ 420 = $ 175 the s . i . is $ 35 / year s . i . for 2 years = $ 70 principal = $ 420 - $ 70 = $ 350 the answer is e ." | a ) $ 310 , b ) $ 320 , c ) $ 330 , d ) $ 340 , e ) $ 350 | e | subtract(420, multiply(subtract(595, 420), divide(2, 5))) | divide(n1,n3)|subtract(n2,n0)|multiply(#0,#1)|subtract(n0,#2)| | general |
donovan and michael are racing around a circular 300 - meter track . if donovan runs each lap in 45 seconds and michael runs each lap in 40 seconds , how many laps will michael have to complete in order to pass donovan , assuming they start at the same time ? | "one way of approaching this question is by relative speed method 1 . speed / rate of donovan = distance / time = > 300 / 45 = > 60 / 9 2 . speed / rate of michael = distance / time = > 300 / 40 = > 30 / 4 relative speed between them = 30 / 4 - 60 / 9 = > 270 - 240 / 36 = 30 / 36 = 5 / 6 ( we subtract the rates if movi... | a ) 6 , b ) 9 , c ) 8 , d ) 7 , e ) 10 | b | divide(divide(300, subtract(divide(300, 40), divide(300, 45))), 40) | divide(n0,n2)|divide(n0,n1)|subtract(#0,#1)|divide(n0,#2)|divide(#3,n2)| | physics |
the average age of a group of persons going for picnic is years . fifteen new persons with an average age of 15 years join the group on the spot due to which their average age becomes 15.5 years . the number of persons initially going for picnic is | solution let the initial number of persons be x . then 16 x + 15 x 15 = 15.5 ( x + 20 ) = 0.5 x = 7.5 x = 15 . answer c | a ) 5 , b ) 10 , c ) 15 , d ) 40 , e ) 50 | c | divide(multiply(15, 15.5), 15.5) | multiply(n0,n1)|divide(#0,n1) | general |
the average of 10 numbers was calculated as 19 . it is discovered later on that while calculating the average , one number , namely 76 , was incorrectly read as 26 . what is the correct average ? | "10 * 19 - 26 + 76 = 240 240 / 10 = 24 the answer is d ." | a ) 21 , b ) 22 , c ) 23 , d ) 24 , e ) 25 | d | divide(add(multiply(10, 19), subtract(76, 26)), 10) | multiply(n0,n1)|subtract(n2,n3)|add(#0,#1)|divide(#2,n0)| | general |
anand and deepak started a business investing rs . 22,500 and rs . 35,000 respectively . out of a total profit of rs . 13,800 , deepak ’ s share is _____ | explanation : ratio of their investments = 22500 : 35000 = 9 : 14 so deepak ' s share = 923923 × 13800 = rs . 5,400 answer : a | a ) 5400 , b ) 3797 , c ) 27877 , d ) 2772 , e ) 9911 | a | subtract(multiply(add(add(multiply(multiply(add(const_2, const_3), const_2), multiply(const_100, multiply(add(const_2, const_3), const_2))), multiply(const_3, multiply(const_100, multiply(add(const_2, const_3), const_2)))), multiply(multiply(const_2, const_4), const_100)), divide(add(multiply(multiply(multiply(add(cons... | add(const_2,const_3)|multiply(const_2,const_4)|multiply(const_100,const_2)|multiply(#0,const_2)|multiply(#1,const_100)|multiply(#0,const_100)|multiply(#2,const_100)|multiply(#3,const_100)|multiply(#5,#3)|multiply(#3,#7)|multiply(#7,const_3)|multiply(#7,const_2)|add(#9,#10)|add(#6,#11)|multiply(#9,const_3)|add(#12,#4)|a... | gain |
of the 3,600 employees of company x , 3 / 4 are clerical . if the clerical staff were to be reduced by 4 / 5 , what percent of the total number of the remaining employees would then be clerical ? | "let ' s see , the way i did it was 3 / 4 are clerical out of 3600 so 2700 are clerical 2700 reduced by 4 / 5 is 2700 * 4 / 5 so it reduced 2160 people , so there is 540 clerical people left but since 2160 people left , it also reduced from the total of 3600 so there are 1440 people total since 540 clerical left / 1440... | a ) 37 % , b ) 22.2 % , c ) 20 % , d ) 12.5 % , e ) 11.1 % | a | multiply(divide(multiply(divide(3, 4), subtract(4, divide(4, 5))), add(multiply(divide(3, 4), subtract(4, divide(4, 5))), subtract(4, divide(3, 4)))), const_100) | divide(n1,n2)|divide(n3,n4)|subtract(n3,#1)|subtract(n3,#0)|multiply(#0,#2)|add(#4,#3)|divide(#4,#5)|multiply(#6,const_100)| | general |
the sum of the squares of three numbers is 41 , while the sum of their products taken two at a time is 20 . their sum is : | "x ^ + y ^ 2 + z ^ 2 = 138 xy + yz + zx = 131 as we know . . ( x + y + z ) ^ 2 = x ^ 2 + y ^ 2 + z ^ 2 + 2 ( xy + yz + zx ) so ( x + y + z ) ^ 2 = 41 + ( 2 * 20 ) ( x + y + z ) ^ 2 = 81 so x + y + z = 9 answer : c" | a ) 10 , b ) 5 , c ) 9 , d ) 15 , e ) none of these | c | add(multiply(sqrt(divide(subtract(41, 20), const_2)), const_100), sqrt(subtract(41, divide(subtract(41, 20), const_2)))) | subtract(n0,n1)|divide(#0,const_2)|sqrt(#1)|subtract(n0,#1)|multiply(#2,const_100)|sqrt(#3)|add(#4,#5)| | general |
when a merchant imported a certain item , she paid a 7 percent import tax on the portion of the total value of the item in excess of $ 1,000 . if the amount of the import tax that the merchant paid was $ 109.90 , what was the total value of the item ? | let x be the value of the item . 0.07 * ( x - 1000 ) = 109.90 x = 2570 the answer is d . | a ) $ 1940 , b ) $ 2150 , c ) $ 2360 , d ) $ 2570 , e ) $ 2780 | d | add(const_1000, divide(109.9, divide(7, const_100))) | divide(n0,const_100)|divide(n2,#0)|add(#1,const_1000) | general |
a cyclist rides a bicycle 9 km at an average speed of 12 km / hr and again travels 12 km at an average speed of 9 km / hr . what is the average speed for the entire trip ? | "distance = 21 km time = 9 / 12 + 12 / 9 = ( 81 + 144 ) / 108 = 225 / 108 = 25 / 12 hours average speed = ( 21 * 12 ) / 25 = 10.1 km / h the answer is c ." | a ) 9.5 , b ) 9.8 , c ) 10.1 , d ) 10.6 , e ) 11.2 | c | divide(add(9, 12), add(divide(9, 12), divide(12, 9))) | add(n0,n1)|divide(n0,n1)|divide(n1,n0)|add(#1,#2)|divide(#0,#3)| | general |
tea worth rs . 126 per kg are mixed with a third variety in the ratio 1 : 1 : 2 . if the mixture is worth rs . 173 per kg , the price of the third variety per kg will be | "solution since first second varieties are mixed in equal proportions , so their average price = rs . ( 126 + 135 / 2 ) = rs . 130.50 so , the mixture is formed by mixing two varieties , one at rs . 130.50 per kg and the other at say , rs . x per kg in the ratio 2 : 2 , i . e . , 1 : 1 . we have to find x . x - 173 / 2... | a ) rs . 169.50 , b ) rs . 1700 , c ) rs . 195.50 , d ) rs . 180 , e ) none | c | divide(subtract(multiply(173, add(add(1, 1), 2)), add(126, 126)), 2) | add(n1,n1)|add(n0,n0)|add(n3,#0)|multiply(n4,#2)|subtract(#3,#1)|divide(#4,n3)| | other |
110 is increased by 50 % . find the final number . | "final number = initial number + 50 % ( original number ) = 110 + 50 % ( 110 ) = 110 + 55 = 165 . answer e" | a ) 100 , b ) 110 , c ) 150 , d ) 155 , e ) 165 | e | add(110, multiply(110, divide(50, const_100))) | divide(n1,const_100)|multiply(n0,#0)|add(n0,#1)| | gain |
in a manufacturing plant , it takes 36 machines 4 hours of continuous work to fill 8 standard orders . at this rate , how many hours of continuous work by 72 machines are required to fill 24 standard orders ? | "the choices give away the answer . . 36 machines take 4 hours to fill 8 standard orders . . in next eq we aredoubling the machines from 36 to 72 , but thework is not doubling ( only 1 1 / 2 times ) , = 4 * 48 / 72 * 24 / 8 = 6 ans b" | a ) 3 , b ) 6 , c ) 8 , d ) 9 , e ) 12 | b | divide(divide(multiply(multiply(36, 24), 4), 72), 8) | multiply(n0,n4)|multiply(#0,n1)|divide(#1,n3)|divide(#2,n2)| | physics |
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