Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 164 | correct stringclasses 5 values | annotated_formula stringlengths 7 1.65k | linear_formula stringlengths 8 925 | category stringclasses 6 values | answer stringclasses 5 values |
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there are 6 people in the elevator . their average weight is 154 lbs . another person enters the elevator , and increases the average weight to 151 lbs . what is the weight of the 7 th person . | "solution average of 7 people after the last one enters = 151 . â ˆ ´ required weight = ( 7 x 151 ) - ( 6 x 154 ) = 1057 - 924 = 133 . answer a" | a ) 133 , b ) 168 , c ) 189 , d ) 190 , e ) 200 | a | subtract(multiply(151, 7), multiply(6, 154)) | multiply(n2,n3)|multiply(n0,n1)|subtract(#0,#1)| | general | A |
there are 2 available positions and 50 candidates , one half of whom are democrats and another half are republicans . if it was decided that the positions would be filled at random , then what is the probability e that the both positions will be taken by members of just one party ? | e probability of one party having both spots : ( 1 / 2 ) * ( 24 / 49 ) = 12 / 49 ( 1 / 2 ) or ( 25 / 50 ) because it does not matter which party or which person gets the first spot . ( 24 / 49 ) because after one person from a particular party is chosen , there are 24 members of the same party left out of 49 total candidates . since this result can happen for both parties , ( 12 / 49 ) + ( 12 / 49 ) = ( 24 / 49 ) answer : d | a ) 1 / 25 , b ) 12 / 49 , c ) 1 / 4 , d ) 24 / 49 , e ) 1 / 2 | d | multiply(multiply(divide(subtract(divide(50, const_2), const_1), subtract(50, const_1)), divide(divide(50, const_2), 50)), 2) | divide(n1,const_2)|subtract(n1,const_1)|divide(#0,n1)|subtract(#0,const_1)|divide(#3,#1)|multiply(#4,#2)|multiply(n0,#5) | other | D |
two goods trains each 750 m long are running in opposite directions on parallel tracks . their speeds are 45 km / hr and 30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? | "relative speed = 45 + 30 = 75 km / hr . 75 * 5 / 18 = 125 / 6 m / sec . distance covered = 750 + 750 = 1500 m . required time = 1500 * 6 / 125 = 72 sec . answer : option a" | a ) 72 , b ) 45 , c ) 48 , d ) 51 , e ) 44 | a | add(45, 30) | add(n1,n2)| | physics | A |
when working alone , painter w can paint a room in 2 hours , and working alone , painter x can paint the same room in a hours . when the two painters work together and independently , they can paint the room in 3 / 4 of an hour . what is the value of a ? | "rate * time = work let painter w ' s rate be w and painter x ' s rate be x r * t = work w * 2 = 1 ( if the work done is same throughout the question then the work done can be taken as 1 ) = > w = 1 / 2 x * a = 1 = > x = 1 / a when they both work together then their rates get added up combined rate = ( w + x ) r * t = work ( w + x ) * 3 / 4 = 1 = > w + x = 4 / 3 = > 1 / 2 + 1 / a = 4 / 3 = > 1 / a = ( 8 - 3 ) / 6 = 5 / 6 = > a = 6 / 5 = 1 [ 1 / 5 ] answer b" | a ) 3 / 4 , b ) 1 [ 1 / 5 ] , c ) 1 [ 2 / 5 ] , d ) 1 [ 3 / 4 ] , e ) 2 | b | add(subtract(4, 2), divide(const_1, add(2, 3))) | add(n0,n1)|subtract(n2,n0)|divide(const_1,#0)|add(#2,#1)| | physics | B |
the length of a rectangle is reduced by 30 % . by what % would the width have to be increased to maintain the original area ? | "sol . required change = ( 30 * 100 ) / ( 100 - 30 ) = 42.8 % a" | a ) 42.8 % , b ) 20 % , c ) 25 % , d ) 30 % , e ) 35 % | a | multiply(divide(subtract(const_1, divide(subtract(const_100, 30), const_100)), divide(subtract(const_100, 30), const_100)), const_100) | subtract(const_100,n0)|divide(#0,const_100)|subtract(const_1,#1)|divide(#2,#1)|multiply(#3,const_100)| | geometry | A |
annie and sam set out together on bicycles traveling at 15 and 12 km per hour respectively . after 40 minutes , annie stops to fix a flat tire . if it takes annie 35 minutes to fix the flat tire and sam continues to ride during this time , how many minutes will it take annie to catch up with sam assuming that annie resumes riding at 15 km per hour ? | "annie gains 3 km per hour ( or 1 km every 20 minutes ) on sam . after 40 minutes annie is 2 km ahead . sam rides 1 km every 5 minutes . in the next 35 minutes , sam rides 7 km so sam will be 5 km ahead . it will take annie 100 minutes to catch sam . the answer is e ." | a ) 20 , b ) 40 , c ) 60 , d ) 80 , e ) 100 | e | multiply(divide(subtract(divide(12, multiply(subtract(15, 12), divide(40, const_60))), multiply(subtract(15, 12), divide(40, const_60))), subtract(15, 12)), const_60) | divide(n2,const_60)|subtract(n0,n1)|multiply(#0,#1)|divide(n1,#2)|subtract(#3,#2)|divide(#4,#1)|multiply(#5,const_60)| | physics | E |
a courtyard is 25 meter long and 16 meter board is to be paved with bricks of dimensions 20 cm by 10 cm . the total number of bricks required is ? | "number of bricks = courtyard area / 1 brick area = ( 2500 × 1600 / 20 × 10 ) = 20000 answer : c" | a ) 22877 , b ) 27778 , c ) 20000 , d ) 27999 , e ) 17799 | c | divide(multiply(multiply(25, const_100), multiply(16, const_100)), multiply(20, 10)) | multiply(n0,const_100)|multiply(n1,const_100)|multiply(n2,n3)|multiply(#0,#1)|divide(#3,#2)| | physics | C |
calculate the average of first 18 even numbers is ? | explanation : sum of 10 even numbers = 18 * 19 = 342 average = 342 / 18 = 19 answer : option c | a ) 21 , b ) 29 , c ) 19 , d ) 14 , e ) 10 | c | add(18, const_1) | add(n0,const_1) | general | C |
the market value of a certain machine decreased by 20 percent of its purchase price each year . if the machine was purchased in 1982 for its market value of $ 8,000 , what was its market value two years later ? | "e . market value in 1982 = $ 8000 market value in 1983 = $ 8000 - ( $ 8000 x 20 / 100 ) = 8000 - 1600 = $ 6400 market value in 1984 = market value in 1983 - ( 20 % of $ 8000 ) = 6400 - 1600 = $ 4800" | a ) $ 8,000 , b ) $ 5,600 , c ) $ 3,200 , d ) $ 2,400 , e ) $ 4,800 | e | subtract(multiply(multiply(const_100, const_10), multiply(const_2, const_4)), multiply(multiply(multiply(const_100, 20), multiply(const_2, const_4)), multiply(divide(20, const_100), const_2))) | divide(n0,const_100)|multiply(const_10,const_100)|multiply(const_2,const_4)|multiply(#1,#2)|multiply(#0,const_2)|multiply(#3,#4)|subtract(#3,#5)| | gain | E |
if taxi fares were $ 2.00 for the first 1 / 5 mile and $ 0.60 for each 1 / 5 mile there after , then the taxi fare for a 8 - mile ride was | "in 8 miles , initial 1 / 5 mile charge is $ 2 rest of the distance = 8 - ( 1 / 5 ) = 39 / 5 rest of the distance charge = 39 ( 0.6 ) = $ 23.4 ( as the charge is 0.6 for every 1 / 5 mile ) = > total charge for 4 miles = 2 + 23.4 = 25.4 answer is c" | a ) $ 2.40 , b ) $ 25.20 , c ) $ 25.40 , d ) $ 25.60 , e ) $ 25.50 | c | add(2.00, multiply(subtract(divide(2.00, divide(1, 5)), 1), 0.60)) | divide(n1,n2)|divide(n6,#0)|subtract(#1,n1)|multiply(n3,#2)|add(n0,#3)| | general | C |
before leaving home for the town of madison , pete checks a map which shows that madison is 5 inches from his current location , gardensquare . pete arrives in madison 5 hours later and drove at an average speed of 60 miles per hour . at what scale , in inches per mile , is the map drawn ? | "pete covered 5 * 60 = 300 miles which correspond to 5 inches on the map - - > scale in inches per mile is 5 / 300 = 1 / 60 . answer : e ." | a ) 1 / 3 , b ) 1 / 30 , c ) 1 / 10 , d ) 1 / 45 , e ) 1 / 60 | e | divide(const_1, multiply(divide(5, 5), 60)) | divide(n1,n0)|multiply(#0,n2)|divide(const_1,#1)| | physics | E |
what is the remainder when 14,452 × 15,652 × 16,781 is divided by 5 ? | "only the unit ' s digit of the product will decide the remainder when divided by 5 . hence , 2 * 2 * 1 = will give units digit as 4 so , whatever be the number , if it ends in 4 , the remainder after dividing with 5 will be 4 . optione" | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | e | reminder(multiply(15,652, 14,452), 16,781) | multiply(n0,n1)|reminder(#0,n2)| | general | E |
the megatek corporation is displaying its distribution of employees by department in a circle graph . the size of each sector of the graph representing a department is proportional to the percentage of total employees in that department . if the section of the circle graph representing the manufacturing department takes up 252 ° of the circle , what percentage of megatek employees are in manufacturing ? | "answer : e 252 ° divided by 360 ° equals 0.7 , therefore the sector is equal to 70 % of the total" | a ) 20 % , b ) 25 % , c ) 30 % , d ) 35 % , e ) 70 % | e | multiply(divide(252, divide(const_3600, const_10)), const_100) | divide(const_3600,const_10)|divide(n0,#0)|multiply(#1,const_100)| | physics | E |
the area of a side of a box is 120 sq cm . the area of the other side of the box is 72 sq cm . if the area of the upper surface of the box is 60 sq cm then find the volume of the box . | volume of the box = √ 120 × 72 × 60 = 720 cm 3 answer c | ['a ) 259200 cm 3', 'b ) 86400 cm 3', 'c ) 720 cm 3', 'd ) can not be determined', 'e ) none of these'] | c | volume_rectangular_prism(sqrt(divide(multiply(60, 72), 120)), divide(60, sqrt(divide(multiply(60, 72), 120))), multiply(divide(120, 60), sqrt(divide(multiply(60, 72), 120)))) | divide(n0,n2)|multiply(n1,n2)|divide(#1,n0)|sqrt(#2)|divide(n2,#3)|multiply(#0,#3)|volume_rectangular_prism(#4,#5,#3) | geometry | C |
how many positive integer solutions does the equation 2 x + 5 y = 100 have ? | "formula : ( constant ) / ( lcm of two nos ) = 100 / ( 2 * 5 ) = 10 answer : e" | a ) 50 , b ) 33 , c ) 16 , d ) 35 , e ) 10 | e | divide(5, 2) | divide(n1,n0)| | general | E |
a jar full of whisky contains 40 % alcohol . a part of this whisky is replaced by another containg 19 % alcohol and now the percentage of alcohol was found to be 26 % . what quantity of whisky is replaced ? | let us assume the total original amount of whiskey = 10 ml - - - > 4 ml alcohol and 6 ml non - alcohol . let x ml be the amount removed - - - > total alcohol left = 4 - 0.4 x new quantity of whiskey added = x ml out of which 0.19 is the alcohol . thus , the final quantity of alcohol = 4 - 0.4 x + 0.19 x - - - - > ( 4 - 0.21 x ) / 10 = 0.26 - - - > x = 20 / 3 ml . per the question , you need to find the x ml removed as a ratio of the initial volume - - - > ( 20 / 3 ) / 10 = 2 / 3 . hence , b is the correct answer . | a ) 1 / 3 , b ) 2 / 3 , c ) 2 / 5 , d ) 3 / 5 , e ) 4 / 5 | b | divide(subtract(40, 26), subtract(40, 19)) | subtract(n0,n2)|subtract(n0,n1)|divide(#0,#1) | gain | B |
for the past n days , the average ( arithmetic mean ) daily production at a company was 50 units . if today ' s production of 90 units raises the average to 52 units per day , what is the value of n ? | "( average production for n days ) * n = ( total production for n days ) - - > 50 n = ( total production for n days ) ; ( total production for n days ) + 90 = ( average production for n + 1 days ) * ( n + 1 ) - - > 50 n + 90 = 52 * ( n + 1 ) - - > n = 19 . or as 40 extra units increased the average for n + 1 days by 2 units per day then 40 / ( n + 1 ) = 2 - - > n = 19 . answer : b ." | a ) 30 , b ) 19 , c ) 10 , d ) 9 , e ) 7 | b | subtract(divide(subtract(90, 50), subtract(52, 50)), const_1) | subtract(n1,n0)|subtract(n2,n0)|divide(#0,#1)|subtract(#2,const_1)| | general | B |
a rectangular lawn of dimensions 80 m * 40 m has two roads each 10 m wide running in the middle of the lawn , one parallel to the length and the other parallel to the breadth . what is the cost of traveling the two roads at rs . 3 per sq m ? | "area = ( l + b â € “ d ) d ( 80 + 40 â € “ 10 ) 10 = > 1100 m 2 1100 * 3 = rs . 3300 answer : c" | a ) 2288 , b ) 2779 , c ) 3300 , d ) 3900 , e ) 2781 | c | multiply(multiply(subtract(add(80, 40), 10), 10), 3) | add(n0,n1)|subtract(#0,n2)|multiply(n2,#1)|multiply(n3,#2)| | geometry | C |
find the area of trapezium whose parallel sides are 20 cm and 18 cm long , and the distance between them is 13 cm | "area of a trapezium = 1 / 2 ( sum of parallel sides ) * ( perpendicular distance between them ) = 1 / 2 ( 20 + 18 ) * ( 13 ) = 247 cm 2 answer : d" | a ) 178 cm 2 , b ) 179 cm 2 , c ) 285 cm 2 , d ) 247 cm 2 , e ) 197 cm 2 | d | quadrilateral_area(13, 18, 20) | quadrilateral_area(n2,n1,n0)| | physics | D |
how much space , in cubic units , is left vacant when maximum number of 7 x 7 x 7 cubes are fitted in a rectangular box measuring 14 x 21 x 17 ? | "no of cubes that can be accommodated in box = ( 14 * 21 * 17 ) / ( 7 * 7 * 7 ) 12 * 16 in numerator can be perfectly divided by 7 * 7 in denominator . side with length 17 ca n ' t be perfectly divided by 7 and hence is the limiting factor . closet multiple of 7 less that 17 is 14 . so vacant area in cube = = 14 * 21 * ( 17 - 14 ) = 14 * 21 * 3 = 882 ans - e" | a ) 878 , b ) 879 , c ) 880 , d ) 881 , e ) 882 | e | multiply(subtract(14, multiply(7, const_2)), multiply(21, 17)) | multiply(n4,n5)|multiply(n0,const_2)|subtract(n3,#1)|multiply(#0,#2)| | geometry | E |
two trains of length 210 m and 120 m are 160 m apart . they start moving towards each other on parallel tracks , at speeds 74 kmph and 92 kmph . after how much time will the trains meet ? | they are moving in opposite directions , relative speed is equal to the sum of their speeds . relative speed = ( 74 + 92 ) * 5 / 18 = 46.1 mps . the time required = d / s = 160 / 46.1 = 35 / 10 sec . answer : a | a ) 35 / 10 , b ) 31 / 10 , c ) 31 / 10 , d ) 36 / 10 , e ) 37 / 10 | a | divide(160, multiply(add(74, 92), const_0_2778)) | add(n3,n4)|multiply(#0,const_0_2778)|divide(n2,#1) | physics | A |
the cost price of 12 articles is equal to the selling price of 10 articles . what is the profit percent ? | "10 * sp = 12 * cp sp = 1.2 * cp the profit percent is 20 % . the answer is b ." | a ) 10 % , b ) 20 % , c ) 30 % , d ) 40 % , e ) 50 % | b | divide(multiply(12, const_4), add(const_4, const_1)) | add(const_1,const_4)|multiply(n0,const_4)|divide(#1,#0)| | gain | B |
a sells a cricket bat to b at a profit of 20 % . b sells it to c at a profit of 25 % . if c pays $ 228 for it , the cost price of the cricket bat for a is : | "a 152 125 % of 120 % of a = 228 125 / 100 * 120 / 100 * a = 228 a = 228 * 2 / 3 = 152 ." | a ) 152 , b ) 120 , c ) 130 , d ) 160 , e ) 210 | a | divide(228, multiply(add(const_1, divide(20, const_100)), add(const_1, divide(25, const_100)))) | divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|add(#1,const_1)|multiply(#2,#3)|divide(n2,#4)| | gain | A |
in a recent head - to - head run - off election , 10,000 absentee ballets were cast . 2 / 3 of the absentee ballets were thrown out and 2 / 5 of the remaining absentee ballets were cast for candidate a . how many absentee votes did candidate b receive ? | "1 / 3 * 3 / 5 ( total absentee votes ) = 1 / 5 ( total votes ) = 1 / 5 * 10000 = 2000 answer is a" | a ) 2,000 , b ) 3,000 , c ) 6,000 , d ) 8,000 , e ) 9,000 | a | multiply(divide(divide(subtract(subtract(multiply(multiply(2, const_4), const_1000), multiply(multiply(multiply(2, const_4), const_1000), divide(2, 3))), multiply(subtract(multiply(multiply(2, const_4), const_1000), multiply(multiply(multiply(2, const_4), const_1000), divide(2, 3))), divide(2, 5))), const_1000), const_4), 2) | divide(n1,n2)|divide(n3,n4)|multiply(n3,const_4)|multiply(#2,const_1000)|multiply(#0,#3)|subtract(#3,#4)|multiply(#1,#5)|subtract(#5,#6)|divide(#7,const_1000)|divide(#8,const_4)|multiply(#9,n3)| | general | A |
rs . 500 amounts to rs . 620 in 2 years at simple interest . if the interest is increased by 2 % , it would amount to how much ? | "( 500 * 3 * 2 ) / 100 = 30 500 + 30 = 530 answer : d" | a ) 120 , b ) 25 , c ) 614 , d ) 530 , e ) 210 | d | multiply(power(add(const_1, divide(2, const_100)), 2), 500) | divide(n3,const_100)|add(#0,const_1)|power(#1,n2)|multiply(n0,#2)| | gain | D |
let q represent a set of 7 distinct prime numbers . if the sum of the numbers in q is even and x is a member of q , then what is the least possible value that x can be ? | q = p 1 + p 2 + p 3 + p 4 + p 5 + p 6 + p 7 = even ( and all primes are distinct ) if the least prime is 2 then we have sum of q = even . ans . c . 3 | a ) 1 , b ) 2 , c ) 3 , d ) 5 , e ) 7 | c | add(divide(7, 7), const_2) | divide(n0,n0)|add(#0,const_2) | general | C |
of the 90 house in a development , 50 have a two - car garage , 40 have an in - the - ground swimming pool , and 35 have both a two - car garage and an in - the - ground swimming pool . how many houses in the development have neither a two - car garage nor an in - the - ground swimming pool ? | "neither car nor garage = total - garage - ( swim - common ) = 90 - 50 - ( 40 - 35 ) = 90 - 55 = 35 answer e" | a ) 10 , b ) 15 , c ) 20 , d ) 25 , e ) 35 | e | subtract(90, add(add(subtract(50, 35), subtract(40, 35)), 35)) | subtract(n1,n3)|subtract(n2,n3)|add(#0,#1)|add(n3,#2)|subtract(n0,#3)| | other | E |
a sum of rs . 2691 is lent into two parts so that the interest on the first part for 8 years at 3 % per annum may be equal to the interest on the second part for 3 years at 5 % per annum . find the second sum ? | "( x * 8 * 3 ) / 100 = ( ( 2691 - x ) * 3 * 5 ) / 100 24 x / 100 = 40365 / 100 - 15 x / 100 39 x = 40365 = > x = 1035 second sum = 2691 â € “ 1035 = 1656 answer : b" | a ) 1629 , b ) 1656 , c ) 1277 , d ) 6298 , e ) 1279 | b | subtract(2691, divide(multiply(multiply(3, 5), 2691), add(multiply(3, 5), multiply(8, 3)))) | multiply(n2,n4)|multiply(n1,n2)|add(#0,#1)|multiply(n0,#0)|divide(#3,#2)|subtract(n0,#4)| | gain | B |
a good train 800 metres long is running at a speed of 78 km / hr . if it crosses a tunnel in 1 minute , then the length of the tunnel ( in meters ) is : | speed = 78 * 5 / 18 m / sec = = > 65 / 3 length of the tunnel be x then 800 + x / 60 = 65 / 3 3 ( 800 + x ) = 3900 x = 500 answer d | a ) 600 , b ) 300 , c ) 400 , d ) 500 , e ) 200 | d | subtract(multiply(add(const_60, add(const_4, const_1)), divide(const_60, const_3)), 800) | add(const_1,const_4)|divide(const_60,const_3)|add(#0,const_60)|multiply(#2,#1)|subtract(#3,n0) | physics | D |
a machine , working at a constant rate , manufactures 18 dies in 25 minutes . how many dies does it make in 1 hr 15 min ? | "change 1 hr 15 min to 75 min . for this , we need to set up a simple proportion of dies per time 18 / 25 = s / 75 the absolutely worst thing you could do at this point in the problem is to cross - multiply . that would be a supremely unstrategic move . we can cancel the common factor of 25 in the two denominators . 18 / 1 = s / 3 s = 3 * 18 s = 54 the machine would be 54 dies in 1 hr 15 min . answer : c" | a ) 55 , b ) 53 , c ) 54 , d ) 52 , e ) 50 | c | multiply(divide(add(multiply(1, const_60), 15), 25), 18) | multiply(n2,const_60)|add(n3,#0)|divide(#1,n1)|multiply(n0,#2)| | physics | C |
on dividing 109 by a number , the quotient is 9 and the remainder is 1 . find the divisor . | "d = ( d - r ) / q = ( 109 - 1 ) / 9 = 108 / 9 = 12 c" | a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 15 | c | floor(divide(109, 9)) | divide(n0,n1)|floor(#0)| | general | C |
there are 60 slots around a circle , numbered 1 to 60 . a man starts fromthe first slot and jumps to the 5 th slot . from there he jumps to the 9 th slot andso on . in which slot will he land in his 2200 th jump ? | every 15 th jump , he comes back to 1 st position so , 2200 / 15 = remainder 10 hence 10 th position is 41 st slot answer : c | ['a ) 1', 'b ) 5', 'c ) 41', 'd ) 45', 'e ) 46'] | c | multiply(subtract(divide(2200, 60), floor(divide(2200, 60))), 60) | divide(n5,n0)|floor(#0)|subtract(#0,#1)|multiply(n0,#2) | geometry | C |
two trains 190 m and 160 m long run at the speed of 60 km / hr and 40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? | "relative speed = 60 + 40 = 100 km / hr . = 100 * 5 / 18 = 250 / 9 m / sec . distance covered in crossing each other = 190 + 160 = 350 m . required time = 350 * 9 / 250 = 12.6 sec . answer : d" | a ) 10.7 , b ) 10.9 , c ) 10.6 , d ) 12.6 , e ) 18.8 | d | divide(add(190, 160), multiply(add(60, 40), const_0_2778)) | add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)| | physics | D |
117 x 217 + 83 x 83 = ? | "= ( 117 ) ^ 2 + ( 83 ) ^ 2 = ( 100 + 17 ) ^ 2 + ( 100 - 17 ) ^ 2 = 2 [ ( 100 ) ^ 2 + ( 17 ) ^ 2 ] = 2 [ 10000 + 289 ] = 2 x 10289 = 20578 answer is b" | a ) 79698 , b ) 20578 , c ) 80698 , d ) 81268 , e ) none of them | b | multiply(117, power(217, 83)) | power(n1,n2)|multiply(n0,#0)| | general | B |
three numbers are in the ratio 1 : 3 : 6 and their average is 100 . the largest number is : | "explanation : let the numbers be x , 3 x and 6 x , then , ( x + 3 x + 6 x ) / 3 = 100 = > 10 x = 100 * 3 = > x = 30 largest number 6 x = 6 * 30 = 180 answer : a" | a ) 180 , b ) 98 , c ) 27 , d ) 21 , e ) 22 | a | add(multiply(multiply(1, 6), const_100), multiply(3, 6)) | multiply(n0,n2)|multiply(n1,n2)|multiply(#0,const_100)|add(#2,#1)| | general | A |
65 % of x = 20 % of 422.50 . find the value of x ? | "65 % of x = 20 % of 422.50 then , 65 / 100 * x = 20 / 100 * 4225 / 10 x = 845 / 10 * 100 / 65 = 130 answer is b" | a ) 100 , b ) 130 , c ) 150 , d ) 180 , e ) 199 | b | divide(multiply(multiply(divide(422.50, const_100), 20), const_100), 65) | divide(n2,const_100)|multiply(n1,#0)|multiply(#1,const_100)|divide(#2,n0)| | general | B |
which number need to add to 1782452 to get a number exactly divisible by 92 ? | "1782452 / 92 = 19374 and reminder = 44 . 92 - 44 = 48 so , the next number divisible by 92 is 48 places in front of 1782452 which means 48 + 1782452 = 1782500 48 should be added to 1782452 e" | a ) 44 , b ) 43 , c ) 42 , d ) 41 , e ) 48 | e | multiply(92, subtract(add(floor(divide(1782452, 92)), const_1), divide(1782452, 92))) | divide(n0,n1)|floor(#0)|add(#1,const_1)|subtract(#2,#0)|multiply(n1,#3)| | general | E |
according to the directions on the can of frozen orange juice concentrate , 1 can of concentrate is to be mixed with 3 cans of water to make orange juice . how many 12 ounces cans of the concentrate are required to prepare 360 6 ounces servings of orange juice ? | "its a . total juice rquired = 360 * 6 = 2160 ounce 12 ounce concentate makes = 12 * 4 = 48 ounce juice total cans required = 2160 / 48 = 45 . answer a" | a ) a ) 45 , b ) b ) 34 , c ) c ) 50 , d ) d ) 67 , e ) e ) 100 | a | divide(divide(multiply(360, 6), 12), const_4) | multiply(n3,n4)|divide(#0,n2)|divide(#1,const_4)| | general | A |
when greenville state university decided to move its fine arts collection to a new library , it had to package the collection in 20 - inch by 20 - inch by 12 - inch boxes . if the university pays $ 0.40 for every box , and if the university needs 2.16 million cubic inches to package the collection , what is the minimum amount the university must spend on boxes ? | "the volume of each box is 20 * 20 * 12 = 4800 cubic inches . number of boxes = 2 , 160,000 / 4800 = 450 boxes total cost = 450 × $ 0.4 = $ 180 the answer is b ." | a ) $ 120 , b ) $ 180 , c ) $ 310 , d ) $ 450 , e ) $ 640 | b | multiply(divide(multiply(2.16, multiply(const_1000, const_1000)), multiply(multiply(20, 20), 12)), 0.40) | multiply(const_1000,const_1000)|multiply(n0,n0)|multiply(n4,#0)|multiply(n2,#1)|divide(#2,#3)|multiply(n3,#4)| | general | B |
each of the 59 members in lourdes school class is required to sign up for a minimum of one and a maximum of 3 academic clubs . the 3 clubs to choose from are the poetry club , the history club , and the writing club . a total of 22 students sign up for the poetry club , 27 students for the history club , and 28 students for the writing club . if 6 students sign up for exactly two clubs , how many students sign up for all 3 clubs ? | each of the 59 members in lourdes school class is required to sign up for a minimum of one and a maximum of three academic clubs . total = g 1 + g 2 + g 3 - ( # in exactly 2 ) - 2 * ( # in 3 sets ) 59 = 22 + 27 + 28 - ( 6 ) - 2 x so , # in 3 sets = 6 = c | a ) 2 , b ) 5 , c ) 6 , d ) 8 , e ) 9 | c | divide(subtract(subtract(add(add(22, 27), 28), 59), 6), const_2) | add(n3,n4)|add(n5,#0)|subtract(#1,n0)|subtract(#2,n6)|divide(#3,const_2) | general | C |
what is the sum between the place values of three 7 ' s in the numeral 87953.0727 | "required sum = 7000 + 0.0700 + 0.0007 = 7000.0707 answer is b" | a ) 21 , b ) 7000.0707 , c ) 7777 , d ) 7014 , e ) 7000.0077 | b | add(multiply(7, const_10), multiply(7, const_1000)) | multiply(n0,const_10)|multiply(n0,const_1000)|add(#0,#1)| | general | B |
the sale price of an article including the sales tax is rs . 616 . the rate of sales tax is 10 % . if the shopkeeper has made a profit of 16 % , then the cost price of the article is : | "110 % of s . p . = 616 s . p . = ( 616 * 100 ) / 110 = rs . 560 c . p = ( 110 * 560 ) / 116 = rs . 531 answer : option c" | a ) 500 , b ) 334 , c ) 531 , d ) 664 , e ) 5598 | c | divide(multiply(divide(multiply(616, const_100), add(const_100, 10)), add(const_100, 10)), add(const_100, 16)) | add(n1,const_100)|add(n2,const_100)|multiply(n0,const_100)|divide(#2,#0)|multiply(#0,#3)|divide(#4,#1)| | gain | C |
if 10 men do a work in 80 days , in how many days will 20 men do it ? | "10 * 80 = 20 * x x = 40 days answer : c" | a ) 18 days , b ) 38 days , c ) 40 days , d ) 48 days , e ) 50 days | c | divide(multiply(10, 80), 20) | multiply(n0,n1)|divide(#0,n2)| | physics | C |
roses can be purchased individually for $ 2.30 , one dozen for $ 36 , or two dozen for $ 50 . what is the greatest number of roses that can be purchased for $ 680 ? | "buy as many $ 50 deals as possible . we can by 650 / 50 = 13 two dozen roses , thus total of 13 * 24 = 312 roses . we are left with 680 - 650 = $ 30 . we can buy 30 / 2.3 = ~ 13 roses for that amount . total = 312 + 13 = 325 . answer : c ." | a ) 156 , b ) 162 , c ) 325 , d ) 324 , e ) 326 | c | add(multiply(floor(divide(680, 50)), multiply(const_12, const_2)), floor(divide(subtract(680, multiply(floor(divide(680, 50)), 50)), 2.30))) | divide(n3,n2)|multiply(const_12,const_2)|floor(#0)|multiply(n2,#2)|multiply(#2,#1)|subtract(n3,#3)|divide(#5,n0)|floor(#6)|add(#7,#4)| | general | C |
in a certain school , 20 % of students are below 8 years of age . the number of students above 8 years of age is 2 / 3 of the number of students of 8 years of age which is 60 . what is the total number of students in the school ? | explanation : let the number of students be x . then , number of students above 8 years of age = ( 100 - 20 ) % of x = 80 % of x . 80 % of x = 60 + 2 / 3 of 60 80 / 100 x = 100 x = 125 . answer : option c | a ) 72 , b ) 80 , c ) 125 , d ) 150 , e ) 100 | c | divide(add(60, multiply(60, divide(2, 3))), subtract(const_1, divide(20, const_100))) | divide(n3,n4)|divide(n0,const_100)|multiply(n6,#0)|subtract(const_1,#1)|add(n6,#2)|divide(#4,#3) | general | C |
if x = 1 + √ 2 , then what is the value of x 4 - 4 x 3 + 4 x 2 + 5 ? | "answer x = 1 + √ 2 ∴ x 4 - 4 x 3 + 4 x 2 + 5 = x 2 ( x 2 - 4 x + 4 ) + 5 = x 2 ( x - 2 ) 2 + 5 = ( 1 + √ 2 ) 2 ( 1 + √ 2 - 2 ) 2 + 5 = ( √ 2 + 1 ) 2 ( √ 2 - 1 ) 2 + 5 = [ ( √ 2 ) 2 - ( 1 ) 2 ] 2 + 5 = ( 2 - 1 ) 2 = 1 + 5 = 6 correct option : c" | a ) - 1 , b ) 0 , c ) 6 , d ) 2 , e ) 3 | c | add(multiply(power(add(1, sqrt(2)), 2), power(subtract(add(1, sqrt(2)), 2), 2)), 4) | sqrt(n1)|add(n0,#0)|power(#1,n1)|subtract(#1,n1)|power(#3,n1)|multiply(#2,#4)|add(n2,#5)| | general | C |
a bus starts from city x . the number of women in the bus is half of the number of men . in city y , 16 men leave the bus and 8 women enter . now , number of men and women is equal . in the beginning , how many passengers entered the bus ? | explanation : originally , let number of women = x . then , number of men = 2 x . so , in city y , we have : ( 2 x - 16 ) = ( x + 8 ) or x = 24 . therefore total number of passengers in the beginning = ( x + 2 x ) = 3 x = 72 . answer : d | a ) 15 , b ) 30 , c ) 36 , d ) 72 , e ) 46 | d | multiply(const_3, add(16, 8)) | add(n0,n1)|multiply(#0,const_3) | general | D |
a person spent rs . 8,550 from his salary on food and 8,640 on house rent . after that he was left with 40 % of his monthly salary . what is his monthly salary ? | "total money spent on food and house rent = 8,550 + 8,640 = 17,190 which is 100 - 40 = 60 % of his monthly salary ∴ his salary = 17190 x 100 / 60 = 28650 answer : b" | a ) 38,650 , b ) 28,650 , c ) 48,650 , d ) 68,650 , e ) 18,650 | b | divide(add(multiply(add(const_4, const_1), const_100), 40), sqrt(const_100)) | add(const_1,const_4)|sqrt(const_100)|multiply(#0,const_100)|add(n2,#2)|divide(#3,#1)| | gain | B |
a bag has 1 , 3 , 5 , 7 , 9,11 and 13 cm sticks . find the probability that they they will form a triangle if 3 sticks are drawn ? | explanation : total cases : = 7 c 3 = 35 favourable cases : ( 3 , 5,7 ) , ( 3 , 7,9 ) , ( 3 , 9,11 ) , ( 3 , 11,13 ) , ( 5 , 7,9 ) , ( 5 , 7,11 ) , ( 5 , 9,11 ) , ( 5 , 9,13 ) , ( 5 , 11,13 ) , ( 7 , 9,11 ) , ( 7 , 9,13 ) , ( 7 , 11,13 ) , ( 9 , 11,13 ) = total 13 . [ others are rejected because to form a triangle it is necessary that the sum of 2 small sides must be greater than the third greatest side . ] hence ( a ) is correct answer . answer : a | a ) 13 / 35 , b ) 14 / 35 , c ) 1 / 2 , d ) 5 / 6 , e ) 5 / 5 | a | divide(add(7, permutation(3, 7)), choose(7, 3)) | choose(n3,n1)|permutation(n1,n3)|add(n3,#1)|divide(#2,#0) | geometry | A |
a worker ' s daily wage is increased by 40 % and the new wage is $ 28 per day . what was the worker ' s daily wage before the increase ? | "let x be the daily wage before the increase . 1.4 x = $ 28 x = $ 20 the answer is b ." | a ) $ 18 , b ) $ 20 , c ) $ 22 , d ) $ 24 , e ) $ 25 | b | divide(28, add(const_1, divide(40, const_100))) | divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)| | physics | B |
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 2000 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 2000 , loss of pyarelal = 2000 * 9 x / 10 x = 1800 answer : c" | a ) 1600 , b ) 1700 , c ) 1800 , d ) 1900 , e ) 1000 | c | divide(multiply(9, 2000), add(1, 9)) | add(n0,n1)|multiply(n1,n2)|divide(#1,#0)| | gain | C |
you collect pens . suppose you start out with 20 . mike gives you another 22 pens . since her father makes pens , cindy decides to double your pens . since you ' re nice , you give sharon 19 pens . how many pens do you have at the end ? | solution start with 20 pens . mike gives you 22 pens : 20 + 22 = 42 pens . cindy doubles the number of pens you have : 42 ã — 2 = 84 pens . sharon takes 19 pens from you : 84 - 19 = 65 pens . so you have 65 at the end . correct answer : c | a ) 39 , b ) 40 , c ) 65 , d ) 42 , e ) 43 | c | subtract(multiply(add(22, 20), const_2), 19) | add(n0,n1)|multiply(#0,const_2)|subtract(#1,n2) | general | C |
the average of 5 numbers id 27 . if one number is excluded , the average becomes 25 . the excluded number is | solution excluded number = ( 27 x 5 ) - ( 25 x 4 ) = 135 - 100 = 35 . answer d | a ) 25 , b ) 27 , c ) 30 , d ) 35 , e ) none | d | subtract(multiply(27, 5), multiply(25, const_4)) | multiply(n0,n1)|multiply(n2,const_4)|subtract(#0,#1) | general | D |
the sale price sarees listed for rs . 480 after successive discount is 15 % and 25 % is ? | "480 * ( 85 / 100 ) * ( 75 / 100 ) = 306 answer : c" | a ) 298 , b ) 237 , c ) 306 , d ) 876 , e ) 291 | c | subtract(subtract(480, divide(multiply(480, 15), const_100)), divide(multiply(subtract(480, divide(multiply(480, 15), const_100)), 25), const_100)) | multiply(n0,n1)|divide(#0,const_100)|subtract(n0,#1)|multiply(n2,#2)|divide(#3,const_100)|subtract(#2,#4)| | gain | C |
a owes b rs . 50 . he agrees to pay b over a number of consecutive days on a monday , paying single note or rs . 10 or rs . 20 on each day . in how many different ways can a repay b . | he can pay by all 10 rupee notes = 1 way 3 ten rupee + 1 twenty rupee = 4 ! 3 ! × 1 ! 4 ! 3 ! × 1 ! = 4 ways 1 ten rupee + 2 twenty rupee notes = 3 ! 2 ! × 1 ! 3 ! 2 ! × 1 ! = 3 ways total ways = 1 + 4 + 3 = 8 answer : b | a ) 2 , b ) 8 , c ) 9 , d ) 6 , e ) 5 | b | add(add(divide(factorial(const_3), factorial(const_2)), divide(factorial(const_4), factorial(const_3))), divide(factorial(divide(50, 10)), factorial(divide(50, 10)))) | divide(n0,n1)|factorial(const_3)|factorial(const_2)|factorial(const_4)|divide(#1,#2)|divide(#3,#1)|factorial(#0)|add(#4,#5)|divide(#6,#6)|add(#7,#8) | physics | B |
rectangular tile each of size 35 cm by 30 cm must be laid horizontally on a rectangular floor of size 1000 cm by 210 cm , such that the tiles do not overlap and they are placed with edges jutting against each other on all edges . a tile can be placed in any orientation so long as its edges are parallel to the edges of floor . no tile should overshoot any edge of the floor . the maximum number of tiles that can be accommodated on the floor is : | area of tile = 35 * 30 = 1050 area of floor = 1000 * 210 = 210000 no of tiles = 210000 / 1050 = 200 so , the no of tile = 200 answer : c | ['a ) 600', 'b ) 400', 'c ) 200', 'd ) 500', 'e ) 100'] | c | divide(multiply(1000, 210), multiply(35, 30)) | multiply(n2,n3)|multiply(n0,n1)|divide(#0,#1) | geometry | C |
there 3 kinds of books in the library physics , chemistry and biology . ratio of physics to chemistry is 3 to 2 ; ratio of chemistry to biology is 4 to 3 , and the total of the books is more than 3000 . which one of following can be the total t of the book ? | "first , you have to find the common ratio for all 3 books . you have : p : c : b 3 : 2 - - > multiply by 2 ( gives you row 3 ) 4 : 6 6 : 4 : 3 hence : p : c : b : t ( total ) t 6 : 4 : 3 : 13 - - - - > this means , the total number must be a multiple of 13 . answer a is correct since 299 is divisible by 13 , hence is 2990 and so is 3003 ( 2990 + 13 ) ." | a ) 3003 , b ) 3027 , c ) 3024 , d ) 3021 , e ) 3018 | a | add(3000, 3) | add(n0,n5)| | other | A |
obra drove 180 π meters along a circular track . if the area enclosed by the circular track on which she drove is 57,600 π square meters , what percentage of the circular track did obra drive ? | "area enclosed by the circular track on which she drove is 57,600 π square meters so , π ( r ^ 2 ) = 57,600 π - - - > ( r ^ 2 ) = 57,600 - - - > r = 240 circumference of the circular track = 2 π r = 480 π therefore , part of circumference covered = 180 π / 480 π = 37.5 % hence , answer is a ." | a ) 37.5 % , b ) 12.5 % , c ) 18.75 % , d ) 25 % , e ) 33.3 % | a | subtract(subtract(multiply(multiply(divide(circumface(180), 180), const_3), const_2), const_4), const_2) | circumface(n0)|divide(#0,n0)|multiply(#1,const_3)|multiply(#2,const_2)|subtract(#3,const_4)|subtract(#4,const_2)| | geometry | A |
what is the least number of square tiles required to pave the floor of a room 50 m 00 cm long and 11 m 25 cm broad ? | solution length of largest tile = h . c . f . of 5000 cm & 1125 cm = 25 cm . area of each tile = ( 25 x 25 ) cm 2 ∴ required number of tiles = [ 5000 x 1125 / 25 x 25 ] = 9000 . answer d | a ) 724 , b ) 804 , c ) 814 , d ) 9000 , e ) none | d | divide(multiply(add(multiply(11, const_100), 25), multiply(50, const_100)), power(25, const_2)) | multiply(n2,const_100)|multiply(n0,const_100)|power(n3,const_2)|add(n3,#0)|multiply(#3,#1)|divide(#4,#2) | physics | D |
two trains of equal are running on parallel lines in the same direction at 46 km / hr and 36 km / hr . the faster train passes the slower train in 36.00001 sec . the length of each train is ? | "let the length of each train be x m . then , distance covered = 2 x m . relative speed = 46 - 36 = 10 km / hr . = 10 * 5 / 18 = 25 / 9 m / sec . 2 x / 36 = 25 / 9 = > x = 50 . answer : a" | a ) 50 , b ) 99 , c ) 77 , d ) 26 , e ) 23 | a | divide(multiply(36.00001, divide(multiply(subtract(46, 36), const_1000), const_3600)), const_2) | subtract(n0,n1)|multiply(#0,const_1000)|divide(#1,const_3600)|multiply(n2,#2)|divide(#3,const_2)| | general | A |
the area of sector of a circle whose radius is 18 metro and whose angle at the center is 42 â ° is ? | "42 / 360 * 22 / 7 * 18 * 18 = 118.8 m 2 answer : d" | a ) 52.6 , b ) 52.9 , c ) 52.8 , d ) 118.8 , e ) 52.2 | d | multiply(multiply(power(18, const_2), divide(add(multiply(const_2, const_10), const_2), add(const_4, const_3))), divide(42, divide(const_3600, const_10))) | add(const_3,const_4)|divide(const_3600,const_10)|multiply(const_10,const_2)|power(n0,const_2)|add(#2,const_2)|divide(n1,#1)|divide(#4,#0)|multiply(#6,#3)|multiply(#5,#7)| | geometry | D |
the total circumference of two circles is 25 . if the first circle has a circumference that is exactly twice the circumference of the second circle , then what is the approximate sum of their two radii ? | "let r = radius of smaller circle . let r = radius of larger circle therefore : 2 π r + 2 π r = 25 where 2 r = r thus : 2 π r + 4 π r = 25 6 π r = 25 r = approx 1.33 π r + 2 r π = 25 3 π r = 25 r = approx 2.65 r + r = approx 3.98 = 4.0 answer : c" | a ) 5.7 , b ) 6.0 , c ) 4.0 , d ) 9.7 , e ) 18.0 | c | divide(add(divide(divide(25, const_3), const_3), divide(multiply(divide(25, const_3), const_2), const_3)), const_2) | divide(n0,const_3)|divide(#0,const_3)|multiply(#0,const_2)|divide(#2,const_3)|add(#1,#3)|divide(#4,const_2)| | general | C |
what is 120 % of 13 / 24 of 960 ? | "120 % * 13 / 24 * 360 = 1.2 * 13 * 40 = 624 the answer is e ." | a ) 420 , b ) 484 , c ) 526 , d ) 578 , e ) 624 | e | divide(multiply(120, 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 | E |
a man sold 20 articles for $ 60 and gained 20 % . how many articles should he sell for $ 80 to incur a loss 20 % ? | "production cost per article : $ 60 * ( 100 % - 20 % ) / 20 = $ 2.40 required production costs for a loss of 20 % : $ 80 * ( 100 % + 20 % ) = $ 96 number of articles to be sold for $ 96 to incur a 20 % loss : $ 96 / $ 2.40 = 40 thus , solution c is correct ." | a ) 45 , b ) 36 , c ) 40 , d ) 50 , e ) 48 | c | divide(original_price_before_loss(20, 80), divide(original_price_before_gain(20, 60), 20)) | original_price_before_gain(n0,n1)|original_price_before_loss(n0,n3)|divide(#0,n0)|divide(#1,#2)| | gain | C |
a can do a piece of work in 21 days and b in 28 days . together they started the work and b left after 4 days . in how many days can a alone do the remaining work ? | "let a worked for x days . x / 21 + 4 / 28 = 1 = > x / 21 = 6 / 7 = > x = 18 a worked for 18 days . so , a can complete the remaining work in 18 - 4 = 14 days . answer : d" | a ) 22 , b ) 88 , c ) 27 , d ) 14 , e ) 99 | d | add(divide(subtract(const_1, add(multiply(4, divide(const_1, 21)), multiply(4, divide(const_1, 28)))), divide(const_1, 28)), 4) | divide(const_1,n0)|divide(const_1,n1)|multiply(n2,#0)|multiply(n2,#1)|add(#2,#3)|subtract(const_1,#4)|divide(#5,#1)|add(n2,#6)| | physics | D |
find the simple interest on rs . 750 for 9 months at 6 paisa per month ? | "explanation : i = ( 750 * 9 * 6 ) / 100 = 405 answer : option b" | a ) rs . 415 , b ) rs . 405 , c ) rs . 450 , d ) rs . 345 , e ) rs . 564 | b | multiply(750, divide(9, const_100)) | divide(n1,const_100)|multiply(n0,#0)| | gain | B |
if 15 students in a class average 73 % on an exam and 10 students average 88 % on the same exam , what is the average in percent for all 25 students ? | "( 15 * 73 + 10 * 88 ) / 25 = 79 % the answer is c ." | a ) 77 % , b ) 78 % , c ) 79 % , d ) 80 % , e ) 81 % | c | divide(add(multiply(15, 73), multiply(10, 88)), 25) | multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)|divide(#2,n4)| | general | C |
a squirrel runs up a cylindrical post , in a perfect spiral path making one circuit for each rise of 4 feet . how many feet does the squirrel travels if the post is 12 feet tall and 3 feet in circumference ? | "total circuit = 12 / 4 = 3 total feet squirrel travels = 3 * 3 = 9 feet answer : a" | a ) 9 feet , b ) 12 feet , c ) 13 feet , d ) 15 feet , e ) 18 feet | a | multiply(divide(12, 4), 3) | divide(n1,n0)|multiply(n2,#0)| | geometry | A |
if x # y is defined to equal x ^ 2 / y for all x and y , then ( - 1 # 2 ) # - 3 = | ( - 1 ) ^ 2 / 2 = 1 / 2 ( - 1 / 2 ) ^ 2 / - 3 = - 1 / 12 so d is my answer | a ) 1 / 12 , b ) 1 / 3 , c ) 1 / 2 , d ) - 1 / 12 , e ) - 4 / 3 | d | divide(power(divide(power(negate(const_1), 2), 2), 2), negate(3)) | negate(const_1)|negate(n3)|power(#0,n0)|divide(#2,n0)|power(#3,n0)|divide(#4,#1) | general | D |
find the least number must be subtracted from 427398 so that remaining no . is divisible by 15 ? | "on dividing 427398 by 15 we get the remainder 3 , so 3 should be subtracted b" | a ) 2 , b ) 3 , c ) 5 , d ) 7 , e ) 8 | b | subtract(427398, multiply(floor(divide(427398, 15)), 15)) | divide(n0,n1)|floor(#0)|multiply(n1,#1)|subtract(n0,#2)| | general | B |
if a truck is traveling at a constant rate of 36 kilometers per hour , how many seconds will it take the truck to travel a distance of 600 meters ? ( 1 kilometer = 1000 meters ) | "speed = 36 km / hr = > 36,000 m / hr in one minute = > 36000 / 60 = 600 meters in one sec = > 600 / 60 = 10 meters time = total distance need to be covered / avg . speed = > 600 / 10 = 60 and hence the answer : c" | a ) 18 , b ) 24 , c ) 60 , d ) 36 , e ) 48 | c | multiply(divide(divide(600, 1000), 36), const_3600) | divide(n1,n3)|divide(#0,n0)|multiply(#1,const_3600)| | physics | C |
if 11.25 m of a uniform steel rod weighs 42.75 kg . what will be the weight of 9 m of the same rod ? | "explanation : let the required weight be x kg . then , less length , less weight ( direct proportion ) = > 11.25 : 9 : : 42.75 : x = > 11.25 x x = 9 x 42.75 = > x = ( 9 x 42.75 ) / 11.25 = > x = 34.2 answer : b" | a ) 22.8 kg , b ) 34.2 kg , c ) 28 kg , d ) 26.5 kg , e ) none of these | b | divide(multiply(9, 42.75), 11.25) | multiply(n1,n2)|divide(#0,n0)| | physics | B |
mary works in a restaurant a maximum of 50 hours . for the first 20 hours , she is paid $ 8 per hour . for each overtime hour , she is paid at a rate which is 25 % higher than her regular rate . how much mary can earn in a week ? | "mary receives $ 8 ( 20 ) = $ 160 for the first 20 hours . for the 30 overtime hours , she receives $ 8 ( 0.25 ) + $ 8 = $ 10 per hour , that is $ 10 ( 30 ) = $ 300 . the total amount is $ 160 + $ 300 = $ 460 answer c 460 ." | a ) 300 , b ) 420 , c ) 460 , d ) 320 , e ) 400 | c | add(multiply(50, 8), multiply(subtract(50, 20), multiply(8, divide(25, const_100)))) | divide(n3,const_100)|multiply(n0,n2)|subtract(n0,n1)|multiply(n2,#0)|multiply(#3,#2)|add(#1,#4)| | gain | C |
a can run 224 metre in 28 seconds and b in 32 seconds . by what distance a beat b ? | "clearly , a beats b by 4 seconds now find out how much b will run in these 4 seconds speed of b = distance / time taken by b = 224 / 32 = 28 / 4 = 7 m / sdistance covered by b in 4 seconds = speed × time = 7 × 4 = 28 metre i . e . , a beat b by 28 metre answer is b" | a ) 38 metre , b ) 28 metre , c ) 23 metre , d ) 15 metre , e ) 28 metre | b | subtract(224, multiply(divide(224, 32), 28)) | divide(n0,n2)|multiply(n1,#0)|subtract(n0,#1)| | physics | B |
the difference between simple and compound interest on rs . 1500 for one year at 10 % per annum reckoned half - yearly is ? | "s . i . = ( 1500 * 10 * 1 ) / 100 = rs . 150 c . i . = [ 1500 * ( 1 + 5 / 100 ) 2 - 1500 ] = rs . 153.75 difference = ( 153.75 - 150 ) = rs . 3.75 answer : d" | a ) 8.0 , b ) 3.0 , c ) 9.5 , d ) 3.75 , e ) 2.15 | d | multiply(subtract(power(add(divide(divide(10, const_2), const_100), const_1), const_2), add(divide(10, const_100), const_1)), 1500) | divide(n1,const_2)|divide(n1,const_100)|add(#1,const_1)|divide(#0,const_100)|add(#3,const_1)|power(#4,const_2)|subtract(#5,#2)|multiply(n0,#6)| | gain | D |
a part - time employee ’ s hourly wage was increased by 15 % . she decided to decrease the number of hours worked per week so that her total income did not change . by approximately what percent should the number of hours worked be decreased ? | "let ' s plug in somenicenumbers and see what ' s needed . let ' s say the employee used to make $ 1 / hour and worked 100 hours / week so , the total weekly income was $ 100 / week after the 15 % wage increase , the employee makes $ 1.15 / hour we want the employee ' s income to remain at $ 100 / week . so , we want ( $ 1.15 / hour ) ( new # of hours ) = $ 100 divide both sides by 1.15 to get : new # of hours = 100 / 1.15 ≈ 87 hours so , the number of hours decreases from 100 hours to ( approximately ) 87 hours . this represents a 13 % decrease ( approximately ) . answer : a" | a ) 13 % , b ) 15 % , c ) 25 % , d ) 50 % , e ) 100 % | a | multiply(divide(divide(15, const_100), divide(add(15, const_100), const_100)), const_100) | add(n0,const_100)|divide(n0,const_100)|divide(#0,const_100)|divide(#1,#2)|multiply(#3,const_100)| | general | A |
the average weight of 8 person ' s increases by 2.5 kg when a new person comes in place of one of them weighing 50 kg . what is the weight of the new person ? | "explanation : total increase in weight = 8 ã — 2.5 = 20 if x is the weight of the new person , total increase in weight = x â ˆ ’ 50 = > 20 = x - 50 = > x = 20 + 50 = 70 answer : option c" | a ) 75 kg , b ) 50 kg , c ) 70 kg , d ) 80 kg , e ) 60 kg | c | add(multiply(8, 2.5), 50) | multiply(n0,n1)|add(n2,#0)| | general | C |
the simple form of the ratio 7 / 6 : 3 / 2 is ? | "7 / 6 : 3 / 2 = 7 : 9 answer : b" | a ) 5 : 8 , b ) 7 : 9 , c ) 5 : 9 , d ) 5 : 3 , e ) 5 : 1 | b | divide(3, 2) | divide(n2,n3)| | other | B |
if | 20 x - 10 | = 90 , then find the product of the values of x ? | "| 20 x - 10 | = 90 20 x - 10 = 90 or 20 x - 10 = - 90 20 x = 100 or 20 x = - 80 x = 5 or x = - 4 product = - 4 * 5 = - 20 answer is e" | a ) - 45 , b ) 50 , c ) - 62 , d ) 35 , e ) - 20 | e | subtract(subtract(subtract(90, 10), add(90, 10)), 10) | add(n1,n2)|subtract(n2,n1)|subtract(#1,#0)|subtract(#2,n1)| | general | E |
a man engaged a servant on the condition that he would pay him rs . 900 and auniform after 1 year service . he served only for 9 months and received uniform and rs . 650 , find the price of the uniform ? | "9 / 12 = 3 / 4 * 900 = 675 650 - - - - - - - - - - - - - 25 1 / 4 - - - - - - - - 25 1 - - - - - - - - - ? = > rs . 100 b" | a ) rs . 80 , b ) rs . 100 , c ) rs . 120 , d ) rs . 145 , e ) rs . 156 | b | multiply(divide(subtract(multiply(1, 900), multiply(multiply(const_3, const_4), 9)), multiply(multiply(const_3, const_4), const_1)), const_4) | multiply(n0,n1)|multiply(const_3,const_4)|multiply(n2,#1)|multiply(#1,const_1)|subtract(#0,#2)|divide(#4,#3)|multiply(#5,const_4)| | general | B |
a cab driver 5 days income was $ 600 , $ 250 , $ 450 , $ 400 , $ 800 . then his average income is ? | "avg = sum of observations / number of observations avg income = ( 600 + 250 + 450 + 400 + 800 ) / 5 = 500 answer is c" | a ) $ 460 , b ) $ 480 , c ) $ 500 , d ) $ 520 , e ) $ 540 | c | divide(add(add(add(add(600, 250), 450), 400), 800), 5) | add(n1,n2)|add(n3,#0)|add(n4,#1)|add(n5,#2)|divide(#3,n0)| | general | C |
in the xy - coordinate system , what is the slope of the line that goes through point ( 2 , 4 ) and is equidistant from the two points p = ( 9 , 3 ) and q = ( 5 , 9 ) ? | "first , get the middle coordinate between ( 9,3 ) and ( 5,9 ) . x = 9 + ( 5 - 9 ) / 2 = 7 y = 3 + ( 9 - 3 ) / 2 = 6 second , get the slope of ( 7,6 ) and ( 2,4 ) . m = 6 - 4 / 7 - 2 = 2 / 5 = 0.4 answer : d" | a ) 0.1 , b ) 0.2 , c ) 0.3 , d ) 0.4 , e ) 0.5 | d | divide(divide(add(4, 3), 3), divide(add(const_4.0, 4), const_2)) | add(n1,n3)|add(n0,n2)|divide(#0,n3)|divide(#1,const_2)|divide(#2,#3)| | general | D |
laxmi and prasanna set on a journey . laxmi moves northwards at a speed of 40 kmph and prasanna moves southward at a speed of 38 kmph . how far will be prasanna from laxmi after 60 minutes ? | "explanation : we know 60 min = 1 hr total northward laxmi ' s distance = 40 kmph x 1 hr = 40 km total southward prasanna ' s distance = 38 kmph x 1 hr = 38 km total distance between prasanna and laxmi is = 40 + 38 = 78 km . answer : c" | a ) 11 , b ) 50 , c ) 78 , d ) 27 , e ) 18 | c | add(40, 38) | add(n0,n1)| | physics | C |
on dividing a number by 357 , we get 39 as remainder . on dividing the same number by 17 , what will be the remainder ? | "let x be the number and y be the quotient . then , x = 357 * y + 39 = ( 17 * 21 * y ) + ( 17 * 2 ) + 5 = 17 * ( 21 y + 2 ) + 5 . required number = 5 . answer is b" | a ) 4 , b ) 5 , c ) 8 , d ) 7 , e ) 2 | b | multiply(subtract(divide(power(39, const_2), 357), floor(divide(power(39, const_2), 357))), 357) | power(n1,const_2)|divide(#0,n0)|floor(#1)|subtract(#1,#2)|multiply(n0,#3)| | general | B |
three numbers are in the ratio of 2 : 3 : 4 . if the sum of the squares of the extremes is 180 , then the middle number is : | "number be = 2 x , 3 x , 4 x , ( 2 x ) 2 + ( 4 x ) 2 = 180 20 x = 180 ⇒ x = 9 ⇒ x = 3 , midline number = 3 × 3 = 9 answer : d" | a ) 6 , b ) 12 , c ) 15 , d ) 9 , e ) 7 | d | multiply(2, 4) | multiply(n0,n2)| | general | D |
the average weight of 24 boys in a class is 50.25 kg and that of the remaining 8 boys is 45.15 kg . find the average weights of all the boys in the class . | "explanation : average weight of 24 boys = 50.25 total weight of 24 boys = 50.25 × 24 average weight of remaining 8 boys = 45.15 total weight of remaining 8 boys = 45.15 × 8 total weight of all boys in the class = ( 50.25 × 24 ) + ( 45.15 × 8 ) total boys = 24 + 8 = 32 average weight of all the boys = ( 50.25 × 24 ) + ( 45.15 × 8 ) / 32 = 48.975 answer : option a" | a ) 48.975 , b ) 42.255 , c ) 50 , d ) 51.255 , e ) 52.253 | a | divide(add(multiply(24, 50.25), multiply(8, 45.15)), add(24, 8)) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0)| | general | A |
a & b can separately do a piece of work in 20 & 15 days . they worked together for 6 days , after which b was replaced by c . the work was finished in next 5 days . the no . of days in which c alone could do the work is | b worked for 6 days = > in 6 days b completed = 6 x 1 15 = 2 5 th work a worked for first 6 days and later 4 days = > in 10 days a completed = 10 x 1 20 = 1 2 of the work total work done by a and b = 2 5 + 1 2 = 4 + 5 10 = 9 10 balance work = 1 - 9 10 = 1 10 th of total work balance work was completed by c in 5 days c can do 1 / 10 of the work in 5 days . therefore c alone can do the work in 50 days . e | a ) 10 days , b ) 20 days , c ) 30 days , d ) 40 days , e ) 50 days | e | divide(inverse(subtract(divide(subtract(const_1, multiply(add(divide(const_1, 20), divide(const_1, 15)), 6)), 5), divide(const_1, 20))), const_2) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|multiply(n2,#2)|subtract(const_1,#3)|divide(#4,n3)|subtract(#5,#0)|inverse(#6)|divide(#7,const_2) | physics | E |
a car going at 20 miles per hour set out on an 80 - mile trip at 9 : 00 a . m . exactly 10 minutes later , a second car left from the same place and followed the same route . how fast , in miles per hour , was the second car going if it caught up with the first car at 10 : 30 a . m . ? | "let car a = car that starts at 9 am car b = car that starts at 9 : 10 am time for which car a travels at speed of 20 m per hour = 1.5 hours distance travelled by car a = 20 * 1.5 = 30 miles since car b catches up car a at 10 : 30 , time = 80 mins = 4 / 3 hour speed of car b = 30 / ( 4 / 3 ) = 40 miles per hour answer a" | a ) 40 , b ) 50 , c ) 53 , d ) 55 , e ) 60 | a | divide(80, divide(add(multiply(subtract(10, 9), const_60), subtract(20, 10)), const_60)) | subtract(n4,n2)|subtract(n0,n4)|multiply(#0,const_60)|add(#2,#1)|divide(#3,const_60)|divide(n1,#4)| | physics | A |
the dimensions of a rectangular solid are 4 inches , 5 inches , and 9 inches . if a cube , a side of which is equal to one of the dimensions of the rectangular solid , is placed entirely within thespherejust large enough to hold the cube , what the ratio of the volume of the cube to the volume within thespherethat is not occupied by the cube ? | answer : b . | a ) 2 : 5 , b ) 5 : 9 , c ) 5 : 16 , d ) 25 : 7 , e ) 32 : 25 | b | divide(const_10, add(add(multiply(9, const_2), const_2), const_1)) | multiply(n2,const_2)|add(#0,const_2)|add(#1,const_1)|divide(const_10,#2)| | geometry | B |
what is the remainder when 1044 * 1047 * 1050 * 1053 is divided by 33 ? | "explanatory answer useful result pertaining to remainders you can solve this problem if you know this rule about remainders . let a number x divide the product of a and b . the remainder will be the product of the remainders when x divides a and when x divides b . using this rule , the remainder when 33 divides 1044 is 21 . the remainder when 33 divides 1047 is 24 . the remainder when 33 divides 1050 is 27 . the remainder when 33 divides 1053 is 30 . ∴ the remainder when 33 divides 1044 * 1047 * 1050 * 1053 is 21 * 24 * 27 * 30 . note : the remainder when a number is divided by a divisor ' d ' will take values from 0 to ( d - 1 ) . it can not be equal to or more than ' d ' . the value of 21 * 24 * 27 * 30 is more than 33 . when the value of the remainder is more than the divisor , final remainder will be the remainder of dividing the product by the divisor . i . e . , the final remainder is the remainder when 33 divides 21 * 24 * 27 * 30 . when 33 divides 21 * 24 * 27 * 30 , the remainder is 30 . choice c" | a ) 3 , b ) 27 , c ) 30 , d ) 21 , e ) 18 | c | reminder(multiply(1047, 1044), 1050) | multiply(n0,n1)|reminder(#0,n2)| | general | C |
a factory has three types of machines , each of which works at its own constant rate . if 7 machine as and 11 machine bs can produce 360 widgets per hour , and if 8 machine as and 22 machine cs can produce 600 widgets per hour , how many widgets could one machine a , one machine b , and one machine c produce in one 8 - hour day ? | "let machine a produce a widgets per hour . b produce b widgets per hour and c produce c widgets per hour . 7 a + 11 b = 360 - - - ( 1 ) 8 a + 22 c = 600 - - - ( 2 ) dividing ( 2 ) by 2 4 a + 11 c = 300 . . . . . ( 3 ) adding ( 1 ) ( 3 ) 11 a + 11 b + 11 c = 660 a + b + c = 60 per hour so for eight hrs = 60 * 8 = 480 = answer = c" | a ) 400 , b ) 475 , c ) 480 , d ) 625 , e ) 700 | c | multiply(divide(600, 11), 8) | divide(n5,n1)|multiply(n3,#0)| | physics | C |
how many times will the digit 8 be written when listing the integers from 1 to 1000 ? | "many approaches are possible . for example : consider numbers from 0 to 999 written as follows : 1 . 000 2 . 001 3 . 002 4 . 003 . . . . . . . . . 1000 . 999 we have 1000 numbers . we used 3 digits per number , hence used total of 3 * 1000 = 3000 digits . now , why should any digit have preferences over another ? we used each of 10 digits equal # of times , thus we used each digit ( including 8 ) 3000 / 10 = 300 times . answer : c ." | a ) 164 , b ) 297 , c ) 300 , d ) 345 , e ) 482 | c | multiply(multiply(multiply(1, const_10), const_10), const_3) | multiply(n1,const_10)|multiply(#0,const_10)|multiply(#1,const_3)| | general | C |
if 7 boys meet at a reunion and each boy shakes hands exactly once with each of the others , then what is the total number of handshakes | "n ( n - 1 ) / 2 = 7 * 6 / 2 = 21 answer : a" | a ) 21 , b ) 42 , c ) 43 , d ) 44 , e ) 45 | a | divide(factorial(7), multiply(factorial(subtract(7, const_2)), factorial(const_2))) | factorial(n0)|factorial(const_2)|subtract(n0,const_2)|factorial(#2)|multiply(#3,#1)|divide(#0,#4)| | general | A |
find a positive number , which when increased by 8 is equal to 128 times reciprocal of the number . | explanation : let the number be x . then , x + 8 = 128 x ( 1 / x ) = > x < sup > 2 < / sup > + 8 x - 128 = 0 = > ( x + 16 ) ( x - 8 ) = 0 = > x = 8 answer : option c | a ) 5 , b ) 6 , c ) 8 , d ) 7 , e ) 9 | c | divide(subtract(sqrt(add(multiply(128, const_4), power(8, const_2))), 8), const_2) | multiply(n1,const_4)|power(n0,const_2)|add(#0,#1)|sqrt(#2)|subtract(#3,n0)|divide(#4,const_2) | general | C |
a and b can do a piece of work in 18 days ; band c can do it in 24 days a and c can do it in 36 days . in how many days will a , band c finish it together ? | sol . ( a + b ) ' s 1 day ' s work = ( 1 / 18 ) ( b + c ) ' s 1 day ' s work = ( 1 / 24 ) and ( a + c ) ' s 1 day ' s work = ( 1 / 36 ) adding , we get : 2 ( a + b + c ) ' s 1 day ' s work = ¬ ( 1 / 18 + 1 / 24 + 1 / 36 ) = 9 / 72 = 1 / 8 ( a + b + c ) ' s 1 day ' s work = 1 / 16 thus , a , band c together can finish the work in 16 days . ans : d | a ) 15 days , b ) 12 days , c ) 8 days , d ) 16 days , e ) 9 days | d | divide(const_1, divide(add(add(inverse(18), inverse(24)), inverse(36)), const_2)) | inverse(n0)|inverse(n1)|inverse(n2)|add(#0,#1)|add(#3,#2)|divide(#4,const_2)|divide(const_1,#5) | physics | D |
a light has a rating of 110 watts , it is replaced with a new light that has 30 % higher wattage . how many watts does the new light have ? | "final number = initial number + 30 % ( original number ) = 110 + 30 % ( 110 ) = 110 + 33 = 143 . answer a" | a ) 143 , b ) 145 , c ) 156 , d ) 134 , e ) 100 | a | multiply(110, add(const_1, divide(30, const_100))) | divide(n1,const_100)|add(#0,const_1)|multiply(n0,#1)| | gain | A |
if the charge of staying in a student youth hostel $ 18.00 / day for the first week , and $ 12.00 / day for each additional week , how much does it cost to stay for 23 days ? | total number of days of stay = 23 charge of staying in first week = 18 * 7 = 126 $ charge of staying for additional days = ( 23 - 7 ) * 12 = 16 * 12 = 192 $ total charge = 126 + 192 = 318 $ answer a | a ) $ 318 , b ) $ 289 , c ) $ 282 , d ) $ 274 , e ) $ 286 | a | add(multiply(18, add(const_3, const_4)), multiply(12, subtract(23, add(const_3, const_4)))) | add(const_3,const_4)|multiply(n0,#0)|subtract(n2,#0)|multiply(n1,#2)|add(#1,#3) | general | A |
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 competition and finished first , second , third , or fourth each time . the product of all the points he received was 107800 . how many times did he participate in the competition ? | "107800 = 2 * 2 * 2 * 5 * 5 * 7 * 7 * 11 john participated 8 times . the answer is e ." | a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | e | floor(sqrt(divide(107800, multiply(multiply(multiply(11, 7), 5), 2)))) | multiply(n0,n1)|multiply(n2,#0)|multiply(n3,#1)|divide(n4,#2)|sqrt(#3)|floor(#4)| | general | E |
if grapes are 93 % water and raisins are 16 % water , then how many kilograms did a quantity of raisins , which currently weighs 8 kilograms , weigh when all the raisins were grapes ? ( assume that the only difference between their raisin - weight and their grape - weight is water that evaporated during their transformation . ) | let x be the original weight . the weight of the grape pulp was 0.07 x . since the grape pulp is 84 % of the raisins , 0.07 x = 0.84 ( 8 ) . then x = 12 * 8 = 96 kg . the answer is b . | a ) 92 , b ) 96 , c ) 100 , d ) 104 , e ) 108 | b | divide(multiply(divide(subtract(const_100, 16), const_100), 8), divide(subtract(const_100, 93), const_100)) | subtract(const_100,n1)|subtract(const_100,n0)|divide(#0,const_100)|divide(#1,const_100)|multiply(n2,#2)|divide(#4,#3) | general | B |
the operation is defined for all integers a and b by the equation ab = ( a - 1 ) ( b - 1 ) . if x 9 = 160 , what is the value of x ? | "ab = ( a - 1 ) ( b - 1 ) x 9 = ( x - 1 ) ( 9 - 1 ) = 160 - - > x - 1 = 20 - - > x = 21 answer : e" | a ) 18 , b ) 15 , c ) 17 , d ) 19 , e ) 21 | e | add(divide(160, subtract(9, 1)), 1) | subtract(n2,n0)|divide(n3,#0)|add(n0,#1)| | general | E |
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