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 |
|---|---|---|---|---|---|---|---|
the ratio of the volumes of two cubes is 9261 : 12167 . what is the ratio of their total surface areas ? | "ratio of the sides = ³ √ 9261 : ³ √ 12167 = 21 : 23 ratio of surface areas = 212 : 232 = 53 : 58 answer : e" | a ) 1 : 12 , b ) 85 : 31 , c ) 51 : 45 , d ) 58 : 67 , e ) 53 : 58 | e | power(divide(9261, 12167), divide(const_1, const_3)) | divide(n0,n1)|divide(const_1,const_3)|power(#0,#1)| | geometry | E |
two friends decide to get together ; so they start riding bikes towards each other . they plan to meet halfway . each is riding at 6 mph . they live 36 miles apart . one of them has a pet carrier pigeon and it starts flying the instant the friends start traveling . the pigeon flies back and forth at 12 mph between the 2 friends until the friends meet . how many miles does the pigeon travel ? | d 36 it takes 3 hours for the friends to meet ; so the pigeon flies for 3 hours at 12 mph = 36 miles | a ) 54 , b ) 66 , c ) 80 , d ) 36 , e ) 96 | d | multiply(12, divide(divide(36, const_2), 6)) | divide(n1,const_2)|divide(#0,n0)|multiply(n2,#1) | physics | D |
find the perimeter and area of a square of side 15 cm . | "we know that the perimeter of square = 4 ã — side side = 15 cm therefore , perimeter = 4 ã — 15 cm = 60 cm now , area of the square = ( side ã — side ) sq . units = 15 ã — 15 cm â ² = 225 cm â ² answer : e" | a ) 144 , b ) 169 , c ) 196 , d ) 121 , e ) 225 | e | square_area(15) | square_area(n0)| | geometry | E |
40 + 5 * 12 / ( 180 / 3 ) = ? | "explanation : 40 + 5 * 12 / ( 180 / 3 ) = 40 + 5 * 12 / ( 60 ) = 40 + ( 5 * 12 ) / 60 = 40 + 1 = 41 . answer : e" | a ) 23 , b ) 78 , c ) 27 , d ) 61 , e ) 41 | e | add(40, divide(multiply(5, 12), divide(180, 3))) | divide(n3,n4)|multiply(n1,n2)|divide(#1,#0)|add(n0,#2)| | general | E |
ifaequals the sum of the even integers from 2 to 60 , inclusive , andbequals the sum of the odd integers from 1 to 59 , inclusive , what is the value of a - b ? | "this is a solution from beatthegmat : even numbers : ( 60 - 2 ) / 2 + 1 = 30 even integers . ( 60 + 2 ) / 2 = 31 is the average of the even set . sum = avg * ( # of elements ) = 31 * 30 = 930 = a odd numbers : ( 59 - 1 ) / 2 + 1 = 30 odd integers . ( 59 + 1 ) / 2 = 30 is the average of the odd set . sum = avg * ( # of elements ) = 30 * 30 = 900 = b a - b = 930 - 900 = 30 . answer : c" | a ) 1 , b ) 10 , c ) 30 , d ) 20 , e ) 21 | c | subtract(multiply(divide(60, 2), add(divide(60, 2), 1)), multiply(divide(add(59, 1), 2), add(divide(subtract(59, 1), 2), 1))) | add(n2,n3)|divide(n1,n0)|subtract(n3,n2)|add(n2,#1)|divide(#2,n0)|divide(#0,n0)|add(n2,#4)|multiply(#3,#1)|multiply(#6,#5)|subtract(#7,#8)| | general | C |
the number 0.127 is how much greater than 1 / 8 ? | "explanation : 0.127 expressed as a fraction = 127 / 1000 1 / 8 can also be expressed as ( 1 x 125 ) / ( 8 x 125 ) = 125 / 1000 the difference is 2 / 1000 , which is 1 / 500 answer is d" | a ) 1 / 100 , b ) 1 / 1500 , c ) 1 / 2 , d ) 1 / 500 , e ) 1 / 600 | d | subtract(0.127, divide(1, 8)) | divide(n1,n2)|subtract(n0,#0)| | general | D |
calculate the average of all the numbers between 18 and 57 which are divisible by 7 . | "explanation : numbers divisible by 7 are 21,28 , 35,42 , 49,56 , average = ( 21 + 28 + 35 + 42 + 49 + 56 , ) / 6 = 231 / 6 = 38.5 answer : b" | a ) 36.9 , b ) 38.5 , c ) 18.5 , d ) 32.5 , e ) 28.5 | b | multiply(divide(add(add(floor(divide(18, 7)), const_1), floor(divide(57, 7))), const_2), 7) | divide(n0,n2)|divide(n1,n2)|floor(#0)|floor(#1)|add(#2,const_1)|add(#4,#3)|divide(#5,const_2)|multiply(n2,#6)| | general | B |
a fill pipe can fill 1 / 6 of cistern in 90 minutes in how many minutes , it can fill 5 / 6 of the cistern ? | "1 / 6 of the cistern can fill in 90 min 5 / 6 of the cistern can fill in = 90 * 6 * 5 / 6 = 450 min answer is a" | a ) 450 min , b ) 360 min , c ) 240 min , d ) 306 min , e ) 500 min | a | divide(90, 1) | divide(n2,n0)| | physics | A |
12 + 6 = 792 10 + 2 = 110 1 + 9 = 9 2 + 7 = 16 11 + 4 = ? ? solve it ? | x + y = x [ y + ( x - 1 ) ] = x ^ 2 + xy - x 12 + 6 = 12 [ 6 + ( 12 - 1 ) ] = 792 10 + 2 = 10 [ 2 + ( 10 - 1 ) ] = 110 1 + 9 = 1 [ 9 + ( 1 - 1 ) ] = 9 2 + 7 = 2 [ 7 + ( 2 - 1 ) ] = 16 11 + 4 = 11 [ 4 + ( 11 - 1 ) ] = 154 answer : e | a ) 100 , b ) 120 , c ) 190 , d ) 160 , e ) 154 | e | add(multiply(add(11, 4), 10), 4) | add(n12,n13)|multiply(n3,#0)|add(n13,#1) | general | E |
what is the rate percent when the simple interest on rs . 800 amount to rs . 192 in 4 years ? | "192 = ( 800 * 4 * r ) / 100 r = 6 % answer : b" | a ) 5 % , b ) 6 % , c ) 3 % , d ) 9 % , e ) 1 % | b | divide(multiply(const_100, 192), multiply(800, 4)) | multiply(n1,const_100)|multiply(n0,n2)|divide(#0,#1)| | gain | B |
if the mean of numbers 28 , x , 42 , 78 and 104 is 62 , what is the mean of 48 , 62 , 98 , 124 and x ? | "x is common to both the series . so , x is not going to make a difference to the average . only the remaining 4 numbers will contribute to the difference in average between the two series . sum of the 4 numbers , excluding x , of the first series is 28 + 42 + 78 + 104 = 252 sum of the 4 numbers , excluding x , of the second series is 48 + 62 + 98 + 124 = 332 the difference between the sum of the two sets of numbers = 332 - 252 = 80 the sum of the second series is 80 more than the sum of the first series . if the sum of the second series is 80 more , the average of the second series will be 80 / 5 = 16 more than the first series . therefore , the average of the second series = 62 + 16 = 78 . answer a" | a ) 78 , b ) 58 , c ) 390 , d ) 310 , e ) 66 | a | divide(add(add(add(add(subtract(multiply(62, add(const_4, const_1)), add(add(add(28, 42), 78), 104)), 48), 62), 98), 124), add(const_4, const_1)) | add(const_1,const_4)|add(n0,n1)|add(n2,#1)|multiply(n4,#0)|add(n3,#2)|subtract(#3,#4)|add(n5,#5)|add(n6,#6)|add(n7,#7)|add(n8,#8)|divide(#9,#0)| | general | A |
if 7800 / 3.25 = 2400 , then 780.0 / 32.5 is equal to ? | "answer given expression 780.0 / 32.5 = 7800 / 325 = 7800 / ( 3.25 x 100 ) = 2400 / 100 = 24 correct option : e" | a ) 21 , b ) 28 , c ) 26 , d ) 25 , e ) 24 | e | divide(multiply(2400, 3.25), 7800) | multiply(n1,n2)|divide(#0,n0)| | general | E |
find the amount on rs . 5000 in 2 years , the rate of interest being 10 % per first year and 12 % for the second year ? | 5000 * 110 / 100 * 112 / 100 = > 6160 answer : b | a ) 3377 , b ) 6160 , c ) 5460 , d ) 1976 , e ) 1671 | b | divide(multiply(divide(multiply(5000, add(const_100, 10)), const_100), add(const_100, 12)), const_100) | add(n3,const_100)|add(n2,const_100)|multiply(n0,#1)|divide(#2,const_100)|multiply(#0,#3)|divide(#4,const_100) | gain | B |
155 liters of a mixture of milk and water contains in the ratio 3 : 2 . how much water should now be added so that the ratio of milk and water becomes 3 : 4 ? | milk = 3 / 5 * 155 = 93 liters water = 62 liters 93 : ( 62 + p ) = 3 : 4 186 + 3 p = 372 = > p = 62 62 liters of water are to be added for the ratio become 3 : 4 . answer : b | a ) 12 liters , b ) 62 liters , c ) 41 liters , d ) 50 liters , e ) 34 liters | b | multiply(divide(155, add(3, 2)), 2) | add(n1,n2)|divide(n0,#0)|multiply(n2,#1) | general | B |
joe went on a diet 4 months ago when he weighed 212 pounds . if he now weighs 188 pounds and continues to lose at the same average monthly rate , in approximately how many months will he weigh 160 pounds ? | "212 - 188 = 24 pounds lost in 4 months 24 / 4 = 6 , so joe is losing weight at a rate of 6 pounds per month . . . . in approximately how many months will he weigh 160 pounds ? a simple approach is to just list the weights . now : 188 lbs in 1 month : 182 lbs in 2 months : 176 lbs in 3 months : 170 lbs in 4 months : 164 lbs in 5 months : 158 lbs since 160 pounds is halfway between 164 and 158 , the correct answer must be 4.5 months . answer : d" | a ) 3 , b ) 3.5 , c ) 4 , d ) 4.5 , e ) 5 | d | divide(subtract(188, 160), divide(subtract(212, 188), 4)) | subtract(n2,n3)|subtract(n1,n2)|divide(#1,n0)|divide(#0,#2)| | general | D |
a car covers a distance of 624 km in 2 2 / 5 hours . find its speed ? | "624 / 2 2 / 5 = 260 kmph answer : d" | a ) 104 kmph , b ) 194 kmph , c ) 109 kmph , d ) 260 kmph , e ) 271 kmph | d | divide(624, add(2, divide(2, 5))) | divide(n2,n3)|add(n1,#0)|divide(n0,#1)| | physics | D |
riya and priya set on a journey . riya moves eastward at a speed of 24 kmph and priya moves westward at a speed of 35 kmph . how far will be priya from riya after 45 minutes | "total eastward distance = 24 kmph * 3 / 4 hr = 18 km total westward distance = 35 kmph * 3 / 4 hr = 26 km total distn betn them = 18 + 26 = 44 km ans 44 km answer : e" | a ) 25 kms , b ) 10 kms , c ) 50 kms , d ) 30 kms , e ) 44 kms | e | multiply(speed(add(24, 35), const_60), 45) | add(n0,n1)|speed(#0,const_60)|multiply(n2,#1)| | physics | E |
david and andrew can finish the work 12 days if they work together . they worked together for 8 days and then andrew left . david finished the remaining work in another 8 days . in how many days david alone can finish the work ? | amount of work done by david and andrew in 1 day = 1 / 12 amount of work done by david and andrew in 8 days = 8 ã — ( 1 / 12 ) = 2 / 3 remaining work â € “ 1 â € “ 2 / 3 = 1 / 3 david completes 1 / 3 work in 8 days amount of work david can do in 1 day = ( 1 / 3 ) / 8 = 1 / 24 = > david can complete the work in 24 days answer : e | a ) 30 days , b ) 60 days , c ) 70 days , d ) 80 days , e ) 24 days | e | multiply(divide(8, subtract(12, 8)), 12) | subtract(n0,n1)|divide(n1,#0)|multiply(n0,#1) | physics | E |
a circular swimming pool is surrounded by a concrete wall 4 ft wide . if the area of the concrete wall surrounding the pool is 11 / 25 that of the pool , then the radius of the pool is ? | "let the radius of the pool be rft radius of the pool including the wall = ( r + 4 ) ft area of the concrete wall = { \ color { black } \ pi \ left [ ( r + 4 ) ^ { 2 } - r ^ { 2 } \ right ] } = { \ color { black } \ pi \ left [ ( r + 4 + r ) ( r + 4 - r ) \ right ] = 8 \ pi ( r + 2 ) } sq feet { \ color { black } \ rightarrow 8 \ pi ( r + 2 ) = \ frac { 11 } { 25 } \ pi r ^ { 2 } \ rightarrow 11 r ^ { 2 } = 200 ( r + 2 ) } radius of the pool r = 20 ft answer : b ) 20 ft" | a ) 22 , b ) 20 , c ) 88 , d ) 387 , e ) 19 | b | divide(divide(add(multiply(const_2, 4), sqrt(add(power(multiply(const_2, 4), const_2), multiply(multiply(divide(11, 25), 4), power(4, const_2))))), divide(11, 25)), const_2) | divide(n1,n2)|multiply(n0,const_2)|power(n0,const_2)|multiply(#0,n0)|power(#1,const_2)|multiply(#3,#2)|add(#5,#4)|sqrt(#6)|add(#1,#7)|divide(#8,#0)|divide(#9,const_2)| | geometry | B |
a small table has a length of 12 inches and a breadth of b inches . cubes are placed on the surface of the table so as to cover the entire surface . the maximum side of such cubes is found to be 4 inches . also , a few such tables are arranged to form a square . the minimum length of side possible for such a square is 24 inches . find b . | "from the info that the maximum sides of the cubes is 4 , we know that the gcf of 12 ( = 2 ^ 2 * 3 ) andbis 4 ( = 2 ^ 2 ) , sob = 2 ^ x , where x > = 2 . from the second premise , we know that the lcm of 12 ( 2 ^ 2 * 3 ) andbis 24 ( 2 ^ 3 * 3 ) , sob = 2 ^ 3 or 2 ^ 3 * 3 ( 8 or 24 ) . combining 2 premises shows the answer is a ( 8 ) ." | a ) 8 , b ) 16 , c ) 24 , d ) 32 , e ) 48 | a | sqrt(subtract(power(divide(24, 4), const_2), power(12, const_2))) | divide(n2,n1)|power(n0,const_2)|power(#0,const_2)|subtract(#2,#1)|sqrt(#3)| | geometry | A |
a reduction of 25 % in the price of oil enables a house wife to obtain 5 kgs more for rs . 1200 , what is the reduced price for kg ? | "1200 * ( 25 / 100 ) = 300 - - - - 5 ? - - - - 1 = > rs . 60 answer : e" | a ) s . 40 , b ) s . 46 , c ) s . 49 , d ) s . 41 , e ) s . 60 | e | divide(divide(multiply(1200, 25), const_100), 5) | multiply(n0,n2)|divide(#0,const_100)|divide(#1,n1)| | gain | E |
a student has to obtain 33 % of the total marks to pass . he got 59 marks and failed by 40 marks . the maximum marks are ? | "let the maximum marks be x then , 33 % of x = 59 + 40 33 x / 100 = 99 x = 300 answer is b" | a ) 450 , b ) 300 , c ) 500 , d ) 610 , e ) 175 | b | divide(add(59, 40), divide(33, const_100)) | add(n1,n2)|divide(n0,const_100)|divide(#0,#1)| | general | B |
ravi and sunil are partners in a business . ravi invests rs . 20,000 for 8 months and sunil invested rs . 16000 for 10 months then after one year ratio of their profits will be | "= ( 15000 * 8 ) : ( 8000 * 10 ) = 160000 : 160000 = 1 : 1 answer : a" | a ) 1 : 1 , b ) 2 : 3 , c ) 3 : 2 , d ) 3 : 1 , e ) 3 : 4 | a | divide(multiply(multiply(multiply(multiply(const_4, const_3), multiply(10, const_2)), const_100), 8), multiply(16000, 10)) | multiply(const_3,const_4)|multiply(n3,const_2)|multiply(n2,n3)|multiply(#0,#1)|multiply(#3,const_100)|multiply(n1,#4)|divide(#5,#2)| | gain | A |
a one - foot stick is marked in 1 / 2 and 1 / 4 portion . how many total markings will there be , including the end points ? | "lcm of 8 = 4 1 / 2 marking are ( table of 2 ) 0 . . . . . . 2 . . . . . . . . . . . 4 ( total = 3 ) 1 / 4 marking are ( table of 1 ) 0 . . . . . . . 1 . . . . . . 2 . . . . . . 3 . . . . . . . . 4 ( total = 5 ) overlapping markings are 0 . . . . . . . . 2 . . . . . . . . . 4 ( total = 3 ) total markings = 3 + 5 - 3 = 5 answer = c" | a ) 8 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | c | add(2, 4) | add(n1,n3)| | general | C |
tickets numbered from 1 to 21 are mixed and then a ticket is selected randomly . what is the probability that the selected ticket bears a number which is a multiple of 3 ? | "here , s = [ 1 , 2 , 3 , 4 , … . , 19 , 20 , 21 ] let e = event of getting a multiple of 3 = [ 3 , 6 , 9 , 12 , 15 , 18 , 21 ] p ( e ) = n ( e ) / n ( s ) = 7 / 21 = 1 / 3 the answer is a ." | a ) 1 / 3 , b ) 2 / 5 , c ) 3 / 10 , d ) 3 / 7 , e ) 1 / 7 | a | divide(divide(21, 3), 21) | divide(n1,n2)|divide(#0,n1)| | general | A |
in a bus left side are 15 seats available , 3 few seats in right side because in rear exit door . each seat hold 3 people . in addition , there is a seat back can sit 8 people all together . how many people can sit in a bus ? | "right side = 15 seat left side = 15 - 3 ( 3 few seat in right side ) = 12 seat total = 15 + 12 = 27 people can seat in 27 seat = 27 * 3 = 81 people can seat in last seat = 8 total people can seat = 81 + 8 = 89 answer : e" | a ) 52 , b ) 49 , c ) 95 , d ) 88 , e ) 89 | e | add(multiply(add(15, subtract(15, 3)), 3), 8) | subtract(n0,n1)|add(n0,#0)|multiply(n1,#1)|add(n3,#2)| | general | E |
what is the area of a square field whose diagonal of length 26 m ? | "d 2 / 2 = ( 26 * 26 ) / 2 = 338 answer : c" | a ) 287 , b ) 269 , c ) 338 , d ) 200 , e ) 230 | c | divide(square_area(26), const_2) | square_area(n0)|divide(#0,const_2)| | geometry | C |
the ages of 2 persons differ by 18 years . if 12 years ago the elder one be 6 times as old as the younger one , find the present age of elder person . | "age of the younger person = x age of the elder person = x + 18 6 ( x - 12 ) = x + 18 - 12 x = 15.6 age of elder person = 15.6 + 18 = 33.6 answer is a" | a ) 33.6 , b ) 47 , c ) 50.4 , d ) 52.4 , e ) 47.9 | a | subtract(add(divide(multiply(2, 18), subtract(18, const_1)), 18), 2) | multiply(n0,n1)|subtract(n1,const_1)|divide(#0,#1)|add(n1,#2)|subtract(#3,n0)| | general | A |
0.363 * 0.522 + 0.363 * 0.478 = ? | "given expression = 0.363 * ( 0.522 + 0.478 ) = 0.363 * 1 = 0.363 answer : c" | a ) 0.522 , b ) 0.845 , c ) 0.363 , d ) 0.985 , e ) 0.885 | c | multiply(0.363, 0.522) | multiply(n0,n1)| | general | C |
the present ratio of students to teachers at a certain school is 70 to 1 . if the student enrollment were to increase by 50 students and the number of teachers were to increase by 5 , the ratio of students to teachers would then be 25 to 1 . what is the present number of teachers ? | "we are given that the ratio of students to teacher is 70 to 1 . we can rewrite this using variable multipliers . students : teachers = 70 x : x we are next given that student enrollment increases by 50 and the number of teachers increases by 5 . with this change the new ratio becomes 25 to 1 . we can put all this into an equation : students / teachers 25 / 1 = ( 70 x + 50 ) / ( x + 5 ) if we cross multiply we have : 25 ( x + 5 ) = 70 x + 50 25 x + 125 = 70 x + 50 1.667 = x x ~ 2 since x is the present number of teachers , currently there are 2 teachers . answer a ." | a ) 2 , b ) 8 , c ) 10 , d ) 12 , e ) 15 | a | divide(add(70, 25), 25) | add(n0,n4)|divide(#0,n4)| | other | A |
the ratio of money with ram and gopal is 7 : 17 and that with gopal and krishan is 7 : 17 . if ram has rs . 686 , krishan has ? | "ram : gopal = 7 : 17 = 49 : 119 gopal : krishan = 7 : 17 = 119 : 289 ram : gopal : krishan = 49 : 119 : 289 ram : krishan = 49 : 289 thus , 49 : 289 = 686 : n & there n = 289 x 686 / 49 = rs . 4046 answer : b" | a ) s . 2890 , b ) s . 4046 , c ) s . 1190 , d ) s . 1620 , e ) s . 2680 | b | multiply(divide(multiply(divide(686, 7), 17), 7), 17) | divide(n4,n0)|multiply(n1,#0)|divide(#1,n0)|multiply(n1,#2)| | other | B |
how many pounds of salt at 50 cents / lb must be mixed with 40 lbs of salt that costs 38 cents / lb so that a merchant will get 20 % profit by selling the mixture at 48 cents / lb ? | selling price is 48 cents / lb for a 20 % profit , cost price should be 40 cents / lb ( cp * 6 / 5 = 48 ) basically , you need to mix 38 cents / lb ( salt 1 ) with 50 cents / lb ( salt 2 ) to get a mixture costing 40 cents / lb ( salt avg ) weight of salt 1 / weight of salt 2 = ( salt 2 - saltavg ) / ( saltavg - salt 1 ) = ( 50 - 40 ) / ( 40 - 38 ) = 5 / 1 we know that weight of salt 1 is 40 lbs . weight of salt 2 must be 8 lbs . answer ( c ) | a ) 2 , b ) 5 , c ) 8 , d ) 15 , e ) 25 | c | divide(subtract(multiply(48, 40), multiply(divide(add(const_100, 20), const_100), multiply(38, 40))), subtract(multiply(50, divide(add(const_100, 20), const_100)), 48)) | add(n3,const_100)|multiply(n1,n4)|multiply(n1,n2)|divide(#0,const_100)|multiply(#3,#2)|multiply(n0,#3)|subtract(#1,#4)|subtract(#5,n4)|divide(#6,#7) | gain | C |
a sum of money at simple interest amounts to rs . 800 in 3 years and to rs . 850 in 4 years . the sum is : | "explanation : s . i . for 1 year = rs . ( 850 - 800 ) = rs . 50 . s . i . for 3 years = rs . ( 50 x 3 ) = rs . 150 . principal = rs . ( 800 - 150 ) = rs . 600 . answer : option c" | a ) rs . 670 , b ) rs . 690 , c ) rs . 600 , d ) rs . 625 , e ) rs . 654 | c | subtract(800, divide(multiply(subtract(850, 800), 3), 4)) | subtract(n2,n0)|multiply(n1,#0)|divide(#1,n3)|subtract(n0,#2)| | gain | C |
the population of a town is 10000 . it increases annually at the rate of 27 % p . a . what will be its population after 2 years ? | "formula : 10000 × 127 / 100 × 127 / 100 = 16129 answer : c" | a ) 14000 , b ) 14400 , c ) 16129 , d ) 14600 , e ) 14700 | c | add(10000, multiply(divide(multiply(10000, 27), const_100), 2)) | multiply(n0,n1)|divide(#0,const_100)|multiply(#1,n2)|add(n0,#2)| | gain | C |
the price of an article is cut by 10 % . to restore it to the former value , the new price must be increased by | "explanation : let original price = rs . 100 then , new price = rs . 90 increase on rs . 90 = rs . 100 increase % = ( 10 / 90 × 100 ) % = 11 1 / 9 % correct option : c" | a ) 10 % , b ) 9 1 / 11 % , c ) 11 1 / 9 % , d ) 11 % , e ) none | c | add(subtract(const_100, subtract(const_100, 10)), const_2) | subtract(const_100,n0)|subtract(const_100,#0)|add(#1,const_2)| | gain | C |
marketing executives for a certain chewing gum company projected a 30 percent increase in revenue this year over that of last year , but revenue this year actually decreased by 25 % . what percent of the projected revenue was the actual revenue ? | "last year revenue = 100 ( assume ) ; this year revenue = 75 ; projected revenue = 130 . actual / projected * 100 = 75 / 130 * 100 = 57.7 % . answer : b ." | a ) 53 % , b ) 57.7 % , c ) 62.5 % , d ) 64 % , e ) 75 % | b | multiply(divide(subtract(const_100, 25), add(30, const_100)), const_100) | add(n0,const_100)|subtract(const_100,n1)|divide(#1,#0)|multiply(#2,const_100)| | general | B |
the first year , two cows produced 8100 litres of milk . the second year their production increased by 15 % and 10 % respectively , and the total amount of milk increased to 9100 litres a year . how many litres were milked from one cow ? | "let x be the amount of milk the first cow produced during the first year . then the second cow produced ( 8100 − x ) litres of milk that year . the second year , each cow produced the same amount of milk as they did the first year plus the increase of 15 % 15 % or 10 % so 8100 + 15100 ⋅ x + 10100 ⋅ ( 8100 − x ) = 9100 therefore 8100 + 320 x + 110 ( 8100 − x ) = 9100 120 x = 190 x = 3800 therefore , the cows produced 3800 and 4300 litres of milk the first year , and 4370 and 4730 litres of milk the second year , = > one cow produced 4370 litres correct answer is d ) 4370 lt" | a ) 2178 lt , b ) 3697 lt , c ) 6583 lt , d ) 4370 lt , e ) 5548 lt | d | subtract(8100, divide(subtract(9100, multiply(add(const_1, divide(10, const_100)), 8100)), subtract(add(const_1, divide(15, const_100)), add(const_1, divide(10, const_100))))) | divide(n2,const_100)|divide(n1,const_100)|add(#0,const_1)|add(#1,const_1)|multiply(n0,#2)|subtract(#3,#2)|subtract(n3,#4)|divide(#6,#5)|subtract(n0,#7)| | general | D |
the h . c . f . of two numbers is 23 and the other two factors of their l . c . m . are 13 and 14 . the larger of the two numbers is : | "clearly , the numbers are ( 23 x 13 ) and ( 23 x 14 ) . larger number = ( 23 x 14 ) = 322 . answer : option c" | a ) 276 , b ) 299 , c ) 322 , d ) 345 , e ) 355 | c | multiply(23, 14) | multiply(n0,n2)| | other | C |
on increasing the number of lines in a page by 60 , they become 240 . what is the % of increase in the no . of lines in the page ? | "explanation : number of pages increased = 60 now , the number of pages of book = 240 number of pages of the books before increase = 240 – 60 = 180 % increase in the number of pages in the book = 60 / 180 x 100 % = 33.3 % a" | a ) 33.3 % , b ) 305 , c ) 50 % , d ) 55 % , e ) 60 % | a | subtract(multiply(divide(240, subtract(240, 60)), const_100), const_100) | subtract(n1,n0)|divide(n1,#0)|multiply(#1,const_100)|subtract(#2,const_100)| | general | A |
two nascar stock cars take off from the starting line at the exact same time , heading in opposite directions . the budweiser car travels at 145 miles per hour , while the stella artois car travels at 150 miles per hour . at this rate , and ignoring other variable , how long will the cars have to drive in order to be 500 miles , in total , from each other ? | the cars travel ( 145 + 150 = 295 ) miles in one hour . 500 miles / 295 miles / hour = 1.69 hours answer is e | a ) 1.5 hours , b ) 1.27 hours , c ) 1.73 hours , d ) 2 hours , e ) 1.69 hours | e | divide(500, add(145, 150)) | add(n0,n1)|divide(n2,#0) | general | E |
if 7 ^ k = 2 , then 7 ^ ( 4 k + 2 ) = | "7 ^ k = 2 7 ^ 4 k = 2 ^ 4 7 ^ 4 k = 16 7 ^ ( 4 k + 2 ) = 7 ^ 4 k * 7 ^ 2 = 16 * 49 = 784 answer : c" | a ) 729 , b ) 754 , c ) 784 , d ) 783 , e ) 108 | c | multiply(power(2, 4), power(7, 2)) | power(n1,n3)|power(n0,n4)|multiply(#0,#1)| | general | C |
the sum of the squares of three numbers is 100 , while the sum of their products taken two at a time is 48 . their sum is : | "x ^ + y ^ 2 + z ^ 2 = 100 xy + yz + zx = 48 as we know . . ( x + y + z ) ^ 2 = x ^ 2 + y ^ 2 + z ^ 2 + 2 ( xy + yz + zx ) so ( x + y + z ) ^ 2 = 100 + ( 2 * 48 ) ( x + y + z ) ^ 2 = 196 so x + y + z = 14 answer : e" | a ) 20 , b ) 30 , c ) 40 , d ) 50 , e ) none of these | e | add(multiply(sqrt(divide(subtract(100, 48), const_2)), const_100), sqrt(subtract(100, divide(subtract(100, 48), const_2)))) | subtract(n0,n1)|divide(#0,const_2)|sqrt(#1)|subtract(n0,#1)|multiply(#2,const_100)|sqrt(#3)|add(#4,#5)| | general | E |
a basket contains 8 apples , of which 1 is spoiled and the rest are good . if we select 2 apples from the basket simultaneously and at random , what is the probability that the 2 apples selected will include the spoiled apple ? | "the total number of ways to choose 2 apples is 8 c 2 = 28 the number of ways that include the spoiled apple is 7 c 1 = 7 p ( the spoiled apple is included ) = 7 / 28 = 1 / 4 the answer is c ." | a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 5 , e ) 1 / 6 | c | divide(choose(subtract(8, 1), 1), choose(8, 2)) | choose(n0,n2)|subtract(n0,n1)|choose(#1,n1)|divide(#2,#0)| | probability | C |
after decreasing 50 % in the price of an article costs rs . 620 . find the actual cost of an article ? | "cp * ( 50 / 100 ) = 620 cp = 12.4 * 100 = > cp = 1240 answer : b" | a ) 1400 , b ) 1240 , c ) 1200 , d ) 1100 , e ) 1500 | b | divide(620, subtract(const_1, divide(50, const_100))) | divide(n0,const_100)|subtract(const_1,#0)|divide(n1,#1)| | gain | B |
if two students starting from same point , walking in the opposite directions with 6 km / hr and 9 km / hr as average speeds respectively . then the distance between them after 4 hours is ? | explanation : total distance = distance traveled by person a + distance traveled by person b = ( 6 × 4 ) + ( 9 × 4 ) = 24 + 36 = 60 km answer : d | a ) 65 km , b ) 55 km , c ) 15 km , d ) 60 km , e ) 75 km | d | add(multiply(6, 4), multiply(9, 4)) | multiply(n0,n2)|multiply(n1,n2)|add(#0,#1) | physics | D |
find the least number must be subtracted from 5474827 so that remaining no . is divisible by 12 ? | "on dividing 5474827 by 12 we get the remainder 7 , so 7 should be subtracted d" | a ) 3 , b ) 5 , c ) 4 , d ) 7 , e ) 2 | d | subtract(5474827, multiply(floor(divide(5474827, 12)), 12)) | divide(n0,n1)|floor(#0)|multiply(n1,#1)|subtract(n0,#2)| | general | D |
a car was driving at 50 km / h for 30 minutes , and then at 90 km / h for another 40 minutes . what was its average speed ? | "driving at 50 km / h for 30 minutes , distance covered = 50 * 1 / 2 = 25 km driving at 90 km / h for 40 minutes , distance covered = 90 * 2 / 3 = 60 km average speed = total distance / total time = 85 / 5 / 6 = 102 km / h answer : a" | a ) 102 , b ) 80 , c ) 75 , d ) 70 , e ) 65 | a | divide(add(multiply(divide(30, add(30, 40)), 50), multiply(divide(40, add(30, 40)), 90)), divide(add(30, 40), const_60)) | add(n1,n3)|divide(n1,#0)|divide(n3,#0)|divide(#0,const_60)|multiply(n0,#1)|multiply(n2,#2)|add(#4,#5)|divide(#6,#3)| | general | A |
a reduction of 12 % in the price of oil enables a house wife to obtain 6 kgs more for rs . 1200 , what is the reduced price for kg ? | "1200 * ( 12 / 100 ) = 144 - - - - 6 ? - - - - 1 = > rs . 24 answer : a" | a ) 24 , b ) 27 , c ) 40 , d ) 28 , e ) 20 | a | divide(divide(multiply(1200, 12), const_100), 6) | multiply(n0,n2)|divide(#0,const_100)|divide(#1,n1)| | gain | A |
if 0.4 of a number is equal to 0.07 of another number , the ratio of the numbers i | "sol . 0.4 a = 0.08 b â ‡ ” a / b = 0.07 / 0.40 = 7 / 40 . â ˆ ´ a : b = 7 : 40 . answer b" | a ) 2 : 3 , b ) 7 : 40 , c ) 3 : 20 , d ) 20 : 3 , e ) none | b | divide(multiply(0.07, const_100), multiply(0.4, const_100)) | multiply(n1,const_100)|multiply(n0,const_100)|divide(#0,#1)| | other | B |
the average of 10 numbers is 40.2 . later it is found that two numbers have been wrongly copied . the first is 18 greater than the actual number and the second number added is 13 instead of 31 . find the correct average . | sum of 10 numbers = 402 corrected sum of 10 numbers = 402 – 13 + 31 – 18 = 402 hence , new average = 402 ⁄ 10 = 40.2 answer a | a ) 40.2 , b ) 40.4 , c ) 40.6 , d ) 40.8 , e ) none of the above | a | divide(subtract(add(multiply(40.2, 10), add(13, 18)), 31), 10) | add(n2,n3)|multiply(n0,n1)|add(#0,#1)|subtract(#2,n4)|divide(#3,n0) | general | A |
4 men and 6 women finish a job in 8 days , while 3 men and 7 women finish it in 10 days . in how many days will 10 women working together finish it ? | explanation : let 1 man ' s 1 day work = x and 1 woman ' s 1 days work = y . then , 4 x + 6 y = 1 / 8 and 3 x + 7 y = 1 / 10 solving , we get y = 1 / 400 [ means work done by a woman in 1 day ] 10 women 1 day work = 10 / 400 = 1 / 40 10 women will finish the work in 40 days option b | a ) 30 days , b ) 40 days , c ) 50 days , d ) 60 days , e ) 70 days | b | divide(add(multiply(divide(subtract(multiply(10, 7), multiply(8, 6)), subtract(multiply(8, 4), multiply(10, 3))), multiply(8, 4)), multiply(8, 6)), 10) | multiply(n4,n5)|multiply(n1,n2)|multiply(n0,n2)|multiply(n3,n5)|subtract(#0,#1)|subtract(#2,#3)|divide(#4,#5)|multiply(#6,#2)|add(#7,#1)|divide(#8,n5) | physics | B |
a room 14 m 21 cm long and 7 m 77 cm broad is to be paved with square tiles . find the least number of square tiles required to cover the floor . | "explanation : area of the room = ( 1421 x 777 ) cm 2 . size of largest square tile = h . c . f . of 1421 cm and 777 cm = 7 cm . area of 1 tile = ( 7 x 7 ) cm 2 . number of tiles required = ( 1147 × 777 ) / ( 7 × 7 ) = 22533 answer : option d" | a ) 22636 , b ) 22640 , c ) 22647 , d ) 22533 , e ) 22675 | d | divide(multiply(add(multiply(14, const_100), 21), add(multiply(7, const_100), 77)), multiply(subtract(21, add(multiply(const_2, const_4), const_2)), subtract(21, add(multiply(const_2, const_4), const_2)))) | multiply(n0,const_100)|multiply(n2,const_100)|multiply(const_2,const_4)|add(n1,#0)|add(n3,#1)|add(#2,const_2)|multiply(#3,#4)|subtract(n1,#5)|multiply(#7,#7)|divide(#6,#8)| | physics | D |
if two of the 4 expressions x + y , x + 5 y , x - y , 5 x + y are chosen at random , what is the probability that their product will be of the form of x ^ 2 - ( by ) ^ 2 , where b is an integer ? | there are total 4 c 2 = 64 c 2 = 6 ways to choose 2 equations out of 4 at random . we want the product of 2 equations in format x 2 − ( by ) 2 x 2 − ( by ) 2 , where b is a constant coefficient of term y 2 y 2 . also note equation does not contain any xyxy term . there are only 2 equations , ( x + y ) & ( x - y ) which gives the product x 2 − y 2 x 2 − y 2 . product of all other equations will contain term xyxy so answer = 1 / 6 ans : e | a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 5 , e ) 1 / 6 | e | divide(subtract(5, 4), add(4, 2)) | add(n0,n3)|subtract(n1,n0)|divide(#1,#0) | general | E |
the captain of a cricket team of 11 members is 28 years old and the wicket keeper is 3 years older . if the ages of these two are excluded , the average age of the remaining players is one year less than the average age of the whole team . what is the average age of the team ? | "explanation : let the average age of the whole team by x years . 11 x â € “ ( 28 + 31 ) = 9 ( x - 1 ) 11 x â € “ 9 x = 50 2 x = 50 x = 25 so , average age of the team is 25 years . answer e" | a ) 20 years , b ) 21 years , c ) 22 years , d ) 23 years , e ) 25 years | e | divide(subtract(add(28, add(28, 3)), multiply(3, 3)), const_2) | add(n1,n2)|multiply(n2,n2)|add(n1,#0)|subtract(#2,#1)|divide(#3,const_2)| | general | E |
the difference between the place values of 9 and 4 in the number 529435 is | sol . = ( place value of 9 ) – ( place value of 4 ) = ( 9000 - 400 ) = 8600 answer c | a ) 8500 , b ) 8900 , c ) 8600 , d ) 6970 , e ) none | c | subtract(multiply(const_10, 9), 9) | multiply(n0,const_10)|subtract(#0,n0)| | general | C |
a box contains 6 pairs of shoes ( 12 shoes in total ) . if two shoes are selected at random , what it is the probability that they are matching shoes ? | the problem with your solution is that we do n ' t choose 1 shoe from 12 , but rather choose the needed one after we just took one and need the second to be the pair of it . so , the probability would simply be : 1 / 1 * 1 / 11 as after taking one at random there are 11 shoes left and only one is the pair of the first one ) = 1 / 11 answer : a . | a ) 1 / 11 , b ) 1 / 20 , c ) 1 / 19 , d ) 1 / 10 , e ) 1 / 9 | a | divide(const_1, subtract(12, const_1)) | subtract(n1,const_1)|divide(const_1,#0) | general | A |
a man buys an article for $ 100 . and sells it for $ 110 . find the gain percent ? | "c . p . = $ 100 s . p . = $ 110 gain = $ 10 gain % = 10 / 100 * 100 = 10 % answer is a" | a ) 10 % , b ) 15 % , c ) 25 % , d ) 20 % , e ) 30 % | a | subtract(divide(110, divide(100, const_100)), const_100) | divide(n0,const_100)|divide(n1,#0)|subtract(#1,const_100)| | gain | A |
what is the smallest positive integer x , such that 5000 x is a perfect cube ? | "take out the factors of 5000 that will come 10 ^ 3 * 5 . for perfect cube you need every no . raise to the power 3 . for 5000 x to be a perfect cube , you need two 5 that means 25 . d is the answer ." | a ) 4 , b ) 6 , c ) 8 , d ) 25 , e ) 18 | d | add(const_3, const_4) | add(const_3,const_4)| | geometry | D |
a box contains 20 electric bulbs , out of which 4 are defective . two bulbs are chosen at random from this box . the probability that at least one of these is defective is | "explanation : please remember that maximum portability is 1 . so we can get total probability of non defective bulbs and subtract it form 1 to get total probability of defective bulbs . so here we go , total cases of non defective bulbs 16 c 2 = 16 ∗ 15 / 2 ∗ 1 = 120 total cases = 20 c 2 = 20 ∗ 19 / 2 ∗ 1 = 190 probability = 120 / 190 = 12 / 19 p of at least one defective = 1 − 12 / 19 = 7 / 19 option a" | a ) 7 / 19 , b ) 6 / 19 , c ) 5 / 19 , d ) 4 / 19 , e ) none of these | a | subtract(const_1, divide(choose(subtract(20, 4), const_2), choose(20, const_2))) | choose(n0,const_2)|subtract(n0,n1)|choose(#1,const_2)|divide(#2,#0)|subtract(const_1,#3)| | probability | A |
if the number is decreased by 5 and divided by 7 the result is 7 . what would be the result if 2 is subtracted and divided by 13 ? | "explanation : let the number be x . then , ( x - 5 ) / 7 = 7 = > x - 5 = 49 x = 54 . : ( x - 2 ) / 13 = ( 54 - 2 ) / 13 = 4 answer : option a" | a ) 4 , b ) 7 , c ) 8 , d ) 5 , e ) 3 | a | divide(subtract(add(multiply(7, 7), 5), 2), 13) | multiply(n1,n1)|add(n0,#0)|subtract(#1,n3)|divide(#2,n4)| | general | A |
bella is taking a trip in her car from new york ( point a ) chicago , illinois ( point b ) . she is traveling at an average speed of 50 miles per hour , if the total distance from point a to point b is 790 miles , in approximately how long will bella reach her destination ? | answer is ( c ) . if she is traveling at a speed of 50 miles per hour and her trip is a total of 790 miles , 790 miles divided by 50 miles per hours would equal 15.80 hours , just shy of 16 hours . | a ) 10 hours , b ) 12.50 hours , c ) 15.80 hours , d ) 25 hours , e ) 1 day | c | divide(790, 50) | divide(n1,n0) | physics | C |
the dimensions of a room are 25 feet * 15 feet * 12 feet . what is the cost of white washing the four walls of the room at rs . 5 per square feet if there is one door of dimensions 6 feet * 3 feet and three windows of dimensions 4 feet * 3 feet each ? | "area of the four walls = 2 h ( l + b ) since there are doors and windows , area of the walls = 2 * 12 ( 15 + 25 ) - ( 6 * 3 ) - 3 ( 4 * 3 ) = 906 sq . ft . total cost = 906 * 5 = rs . 4530 answer : d" | a ) 2277 , b ) 2977 , c ) 2677 , d ) 4530 , e ) 1971 | d | multiply(subtract(subtract(multiply(multiply(const_2, 12), add(15, 25)), multiply(6, 3)), multiply(3, multiply(4, 3))), 5) | add(n0,n1)|multiply(n2,const_2)|multiply(n4,n5)|multiply(n5,n6)|multiply(#0,#1)|multiply(n5,#3)|subtract(#4,#2)|subtract(#6,#5)|multiply(n3,#7)| | geometry | D |
the amount of an investment will double in approximately 70 / p years , where p is the percent interest , compounded annually . if thelma invests $ 40,000 in a long - term cd that pays 5 percent interest , compounded annually , what will be the approximate total value of the investment when thelma is ready to retire 28 years later ? | "the amount of an investment will double in approximately 70 / p years , where p is the percent interest , compounded annually . if thelma invests $ 40,000 in a long - term cd that pays 5 percent interest , compounded annually , what will be the approximate total value of the investment when thelma is ready to retire 42 years later ? the investment gets doubled in 70 / p years . therefore , the investment gets doubled in 70 / 5 = every 14 years . after 28 years , the investment will get doubled 28 / 14 = 2 times . so the amount invested will get doubled thrice . so , 40000 * 2 = 80000 80000 * 2 = 160000 hence , the answer is a ." | a ) $ 160,000 , b ) $ 320,000 , c ) $ 360,000 , d ) $ 450,000 , e ) $ 540,000 | a | divide(const_3600, const_10) | divide(const_3600,const_10)| | general | A |
in a simultaneous throw of a pair of dice , find the probability of getting a total more than 11 | total number of cases = 10 * 10 = 100 favourable cases = [ ( 2,10 ) , ( 3,9 ) , ( 3,10 ) , ( 4,8 ) , ( 4,9 ) , ( 4,10 ) , ( 5,7 ) , ( 5,8 ) , ( 5,9 ) , ( 5,10 ) , ( 6,6 ) , ( 6,7 ) , ( 6,8 ) , ( 6,9 ) , ( 6,10 ) , ( 7,5 ) , ( 7,6 ) , ( 7,7 ) , ( 7,8 ) , ( 7,9 ) , ( 7,10 ) , ( 8,4 ) , ( 8,5 ) , ( 8,6 ) , ( 8,7 ) , ( 8,8 ) , ( 8,9 ) , ( 8,10 ) , ( 9,3 ) , ( 9,4 ) , ( 9,5 ) , ( 9,6 ) , ( 9,7 ) , ( 9,8 ) , ( 9,10 ) , ( 10,2 ) , ( 10,3 ) , ( 10,4 ) , ( 10,5 ) , ( 10,6 ) , ( 10,7 ) , ( 10,8 ) , ( 10,9 ) , ( 10,10 ) ] = 44 so probability = 44 / 100 = 11 / 25 answer is a | a ) 11 / 25 , b ) 7 / 12 , c ) 5 / 13 , d ) 5 / 12 , e ) 6 / 17 | a | divide(subtract(11, multiply(const_2, const_3)), 11) | multiply(const_2,const_3)|subtract(n0,#0)|divide(#1,n0) | general | A |
a , b , c subscribe rs . 50,000 for a business . a subscribes rs . 4000 more than b and b rs . 5000 more than c . out of a total profit of rs . 35,000 , c receives : | "let c = x . then , b = x + 5000 and a = x + 5000 + 4000 = x + 9000 . so , x + x + 5000 + x + 9000 = 50000 3 x = 36000 x = 12000 a : b : c = 21000 : 17000 : 12000 = 21 : 17 : 12 . c ' s share = rs . ( 35000 x 12 / 50 ) = rs . 8,400 . d" | a ) s . 14,000 , b ) s . 14,200 , c ) s . 4,400 , d ) s . 8,400 , e ) s . 4,800 | d | subtract(floor(divide(multiply(divide(add(divide(subtract(subtract(multiply(const_10, 5000), 5000), add(4000, 5000)), const_3), add(4000, 5000)), multiply(const_10, 5000)), multiply(add(const_3, const_4), 5000)), const_1000)), const_1) | add(n1,n2)|add(const_3,const_4)|multiply(n2,const_10)|multiply(n2,#1)|subtract(#2,n2)|subtract(#4,#0)|divide(#5,const_3)|add(#0,#6)|divide(#7,#2)|multiply(#8,#3)|divide(#9,const_1000)|floor(#10)|subtract(#11,const_1)| | general | D |
sushil got thrice as many marks in english as in science . his total marks in english , science and maths are 115 . if the ratio of his marks in english and maths is 5 : 1 , find his marks in science ? | "s : e = 1 : 3 e : m = 5 : 1 - - - - - - - - - - - - s : e : m = 5 : 15 : 3 5 / 23 * 115 = 25 answer : c" | a ) 18 , b ) 77 , c ) 25 , d ) 55 , e ) 31 | c | add(add(add(multiply(5, const_3), 5), multiply(multiply(5, const_3), add(multiply(5, const_3), 5))), 5) | multiply(n1,const_3)|add(n1,#0)|multiply(#1,#0)|add(#1,#2)|add(n1,#3)| | general | C |
the average age of students of a class is 15.8 years . the average age of boys in the class is 16.4 years and that of the girls is 15.4 years . the ration of the number of boys to the number of girls in the class is ? | "let the ratio be k : 1 . then , k * 16.4 + 1 * 15.4 = ( k + 1 ) * 15.8 = ( 16.4 - 15.8 ) k = ( 15.8 - 15.4 ) = k = 0.4 / 0.6 = 2 / 3 required ratio = 2 / 3 : 1 = 2 : 3 . answer : option b explanation : let the ratio be k : 1 . then , k * 16.4 + 1 * 15.4 = ( k + 1 ) * 15.8 = ( 16.4 - 15.8 ) k = ( 15.8 - 15.4 ) = k = 0.4 / 0.6 = 2 / 3 required ratio = 2 / 3 : 1 = 2 : 3 . answer : b" | a ) 2 : 7 , b ) 2 : 3 , c ) 2 : 1 , d ) 2 : 9 , e ) 2 : 2 | b | divide(subtract(15.8, 15.4), subtract(16.4, 15.8)) | subtract(n0,n2)|subtract(n1,n0)|divide(#0,#1)| | general | B |
total 15 cows 5 cow gives each 2 liter milk 5 cow gives each 3 / 4 liter milk 5 cow gives each 1 / 4 liter milk this is split into 3 son per each 5 cows & 5 liter milk how ? | "5 cow 2 liter each = 10 liter 5 cow 3 / 4 liter each = 3 / 4 = 0.75 * 5 = 3.75 5 cow 1 / 4 liter each = 1 / 4 = 0.25 * 5 = 1.25 add 10 + 3.75 + 1.25 = 15 milk split into 3 son each 5 liter then 15 / 3 = 5 answer : b" | a ) 10 , b ) 5 , c ) 15 , d ) 7.5 , e ) 12.5 | b | add(add(divide(3, 4), multiply(divide(3, 4), 5)), multiply(const_0_25, 5)) | divide(n4,n5)|multiply(n1,const_0_25)|multiply(n1,#0)|add(#0,#2)|add(#3,#1)| | general | B |
what will be the ratio between the area of a rectangle and the area of a triangle with one of the sides of rectangle as base and a vertex on the opposite side of rectangle . | "area of δ ebc = 1 ⁄ 2 × bc × ef = 1 ⁄ 2 × bc × ab [ since , ef = ab ] area of δ ebc = 1 ⁄ 2 × area of δ abcd \ required ratio = 2 : 1 . answer b" | a ) 1 : 2 , b ) 2 : 1 , c ) 3 : 1 , d ) data inadequate , e ) none of these | b | divide(rectangle_area(const_1, const_1), triangle_area(rectangle_area(const_1, const_1), rectangle_area(const_1, const_1))) | rectangle_area(const_1,const_1)|triangle_area(#0,#0)|divide(#0,#1)| | geometry | B |
a student chose a number , multiplied it by 2 , then subtracted 140 from the result and got 102 . what was the number he chose ? | "solution : let x be the number he chose , then 2 * x * 140 = 102 2 x = 242 x = 121 correct answer e" | a ) 90 , b ) 100 , c ) 120 , d ) 160 , e ) 121 | e | divide(add(102, 140), 2) | add(n1,n2)|divide(#0,n0)| | general | E |
a garden center sells a certain grass seed in 5 - pound bags at $ 13.85 per bag , 10 - pound bags at $ 20.42 per bag , and 25 - pound bags $ 32.25 per bag . if a customer is to buy at least 65 pounds of the grass seed , but no more than 80 pounds , what is the least possible cost of the grass seed that the customer will buy ? | "there can be 2 cases 1 ) 25 + 25 + 10 + 5 = $ 98.77 or 2 ) 25 + 25 + 25 = $ 96.75 d" | a ) $ 94.03 , b ) $ 96.75 , c ) $ 98.78 , d ) $ 98.77 , e ) $ 105.3 | d | add(add(multiply(const_2, 32.25), multiply(const_1, 20.42)), multiply(const_1, 13.85)) | multiply(n5,const_2)|multiply(n3,const_1)|multiply(n1,const_1)|add(#0,#1)|add(#3,#2)| | general | D |
the vertex of a rectangle are ( 1 , 0 ) , ( 9 , 0 ) , ( 1 , 2 ) and ( 9 , 2 ) respectively . if line l passes through the origin and divided the rectangle into two identical quadrilaterals , what is the slope of line l ? | if line l divides the rectangle into two identical quadrilaterals , then it must pass through the center ( 5 , 1 ) . the slope of a line passing through ( 0,0 ) and ( 5 , 1 ) is 1 / 5 . the answer is d . | a ) 5 , b ) 4 , c ) 1 / 2 , d ) 1 / 5 , e ) 1 / 8 | d | divide(const_1, divide(add(subtract(9, 1), const_2), const_2)) | subtract(n2,n0)|add(#0,const_2)|divide(#1,const_2)|divide(const_1,#2) | general | D |
of all the students in a certain dormitory , 1 / 2 are first - year students and the rest are second - year students . if 4 / 5 of the first - year students have not declared a major and if the fraction of second - year students who have declared a major is 4 times the fraction of first - year students who have declared a major , what fraction of all the students in the dormitory are second - year students who have not declared a major ? | tot students = x 1 st year student = x / 2 - - - - > non majaor = 4 / 5 ( x / 2 ) - - - - - > maj = 1 / 5 ( x / 2 ) 2 nd year student = x / 2 - - - - > maj = 4 ( 1 / 5 ( x / 2 ) ) = 2 / 5 ( x ) - - - > non major = x / 2 - 2 / 5 ( x ) = 1 / 5 ( x ) hence 1 / 5 d | a ) 1 / 15 , b ) 4 / 5 , c ) 4 / 15 , d ) 1 / 5 , e ) 2 / 5 | d | subtract(const_1, divide(4, 5)) | divide(n2,n3)|subtract(const_1,#0) | general | D |
alok ordered 16 chapatis , 5 plates of rice , 7 plates of mixed vegetable and 6 ice - cream cups . the cost of each chapati is rs . 6 , that of each plate of rice is rs . 45 and that of mixed vegetable is rs . 70 . the amount that alok paid the cashier was rs . 1051 . find the cost of each ice - cream cup ? | "let the cost of each ice - cream cup be rs . x 16 ( 6 ) + 5 ( 45 ) + 7 ( 70 ) + 6 ( x ) = 1051 96 + 225 + 490 + 6 x = 1051 6 x = 240 = > x = 40 . answer : b" | a ) 25 , b ) 40 , c ) 77 , d ) 99 , e ) 91 | b | divide(subtract(subtract(subtract(1051, multiply(16, 6)), multiply(5, 45)), multiply(7, 70)), 6) | multiply(n0,n3)|multiply(n1,n5)|multiply(n2,n6)|subtract(n7,#0)|subtract(#3,#1)|subtract(#4,#2)|divide(#5,n3)| | general | B |
in an examination , 20 % of total students failed in hindi , 70 % failed in english and 10 % in both . the percentage of these who passed in both the subjects is : | "pass percentage = 100 - ( 20 + 70 - 10 ) = 100 - 80 = 20 answer : b" | a ) 10 % , b ) 20 % , c ) 30 % , d ) 40 % , e ) 50 % | b | subtract(const_100, subtract(add(20, 70), 10)) | add(n0,n1)|subtract(#0,n2)|subtract(const_100,#1)| | general | B |
raja and ram can together complete a piece of work in 4 days . if raja alone can complete the same work in 12 days , in how many days can ram alone complete that work ? | ( raja + ram ) ' s 1 days work = 1 / 4 raja 1 day work = 1 / 2 ram 1 day work = ( 1 / 4 - 1 / 12 ) = 1 / 6 = = > 6 days answer a | a ) 6 days , b ) 5 days , c ) 7 days , d ) 8 days , e ) 10 days | 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 | A |
an order was placed for a carpet whose length and width were in the ratio of 3 : 2 . subsequently , the dimensions of the carpet were altered such that its length and width were in the ratio 2 : 1 but were was no change in its perimeter . what is the ratio of the areas of the carpets ? | "let the length and width of one carpet be 3 x and 2 x . let the length and width of the other carpet be 2 y and y . 2 ( 3 x + 2 x ) = 2 ( 2 y + y ) 5 x = 3 y ( 5 / 3 ) * x = y the ratio of the areas of the carpet in both cases : = 3 x * 2 x : 2 y * y = 6 x ^ 2 : 2 y ^ 2 = 6 x ^ 2 : 2 * ( 25 / 9 ) * x ^ 2 = 54 : 50 = 27 : 25 the answer is e ." | a ) 7 : 5 , b ) 12 : 11 , c ) 17 : 15 , d ) 22 : 17 , e ) 27 : 25 | e | divide(rectangle_area(3, 2), rectangle_area(divide(divide(rectangle_perimeter(3, 2), 2), add(2, 1)), multiply(divide(divide(rectangle_perimeter(3, 2), 2), add(2, 1)), 2))) | add(n2,n3)|rectangle_area(n0,n1)|rectangle_perimeter(n0,n1)|divide(#2,n1)|divide(#3,#0)|multiply(n2,#4)|rectangle_area(#4,#5)|divide(#1,#6)| | geometry | E |
ratio and proportion 215 : 474 : : 537 : ? | 2 + 1 + 5 + 4 + 7 + 4 = 23 5 + 3 + 7 + x = 23 = > x = 8 combination that match 8 is 26 since 2 + 6 = 8 answer : a | a ) 26 , b ) 27 , c ) 25 , d ) 22 , e ) 23 | a | add(add(const_4, const_2), add(const_10, const_10)) | add(const_2,const_4)|add(const_10,const_10)|add(#0,#1) | other | A |
50 square stone slabs of equal size were needed to cover a floor area of 72 sq . m . find the length of each stone slab ? | "area of each slab = 72 / 50 m 2 = 1.44 m 2 length of each slab √ 1.44 = 1.2 m = 120 cm" | a ) 120 cm , b ) 767 cm , c ) 88 cm , d ) 666 cm , e ) 776 cm | a | multiply(sqrt(divide(72, 50)), const_100) | divide(n1,n0)|sqrt(#0)|multiply(#1,const_100)| | geometry | A |
in a fuel station the service costs $ 1.75 per car , every liter of fuel costs 0.75 $ . assuming that a company owns 12 cars and that every fuel tank contains 55 liters and they are all empty , how much money total will it cost to fuel all cars ? | 12 * 1.75 + 0.75 * 12 * 55 = 516 hence - d | a ) 320 $ , b ) 420 $ , c ) 490 $ , d ) 516 $ , e ) 680 $ | d | multiply(multiply(0.75, 55), 12) | multiply(n1,n3)|multiply(n2,#0)| | general | D |
there are 172 lights which are functional and each is controlled by a separate on / off switch . two children a and b start playing with the switches . a starts by pressing every third switch till he reaches the end . b , thereafter , presses every fifth switch till he too reaches the end . if all switches were in off position at the beggining , how many lights are switched on by the end of this operation ? | "editing my solution : number of switches = 172 number of switches turned on by a : 3 , 6 , . . . 171 = 57 number of switches turned on by b : 5 , 10 , . . . . 170 = 34 few switches are turned on by a and later turned off by b : lcm ( 3,5 ) = 15 x = 15 , 30 , . . . . 90 = 6 . subtract the above 6 switches from both a and b as they are turned off . number of switches that are turned on = ( 57 - 6 ) + ( 34 - 6 ) = 79 answer : d" | a ) 83 , b ) 85 , c ) 87 , d ) 79 , e ) 89 | d | subtract(add(floor(divide(172, const_3)), floor(divide(172, add(const_1, const_4)))), multiply(floor(divide(172, multiply(const_3, add(const_1, const_4)))), const_2)) | add(const_1,const_4)|divide(n0,const_3)|divide(n0,#0)|floor(#1)|multiply(#0,const_3)|divide(n0,#4)|floor(#2)|add(#3,#6)|floor(#5)|multiply(#8,const_2)|subtract(#7,#9)| | other | D |
a boy is traveling from his house to school at 10 km / hr and reached school 2 hours late . next day he traveled 20 km / hr and reached 1 hour early . then find the distance between house and school ? | let distance be x s 1 = 10 km / hr s 2 = 20 km / hr t 1 = x / 10 hr t 2 = x / 20 hr difference in time = 2 + 1 = 3 hr ( x / 10 ) - ( x / 20 ) = 3 x = 60 km answer is d | a ) 50 km , b ) 45 km , c ) 33 km , d ) 60 km , e ) 54 km | d | multiply(add(divide(add(multiply(2, 10), 20), subtract(20, 10)), 2), 10) | multiply(n0,n1)|subtract(n2,n0)|add(n2,#0)|divide(#2,#1)|add(n1,#3)|multiply(n0,#4) | physics | D |
on the independence day , bananas were be equally distributed among the children in a school so that each child would get two bananas . on the particular day 305 children were absent and as a result each child got two extra bananas . find the actual number of children in the school ? | "let the number of children in the school be x . since each child gets 2 bananas , total number of bananas = 2 x . 2 x / ( x - 305 ) = 2 + 2 ( extra ) = > 2 x - 610 = x = > x = 610 . answer : b" | a ) 600 , b ) 610 , c ) 500 , d ) 520 , e ) 720 | b | multiply(305, const_2) | multiply(n0,const_2)| | general | B |
by selling an article at rs . 400 , a profit of 25 % is made . find its cost price ? | "sp = 400 cp = ( sp ) * [ 100 / ( 100 + p ) ] = 400 * [ 100 / ( 100 + 25 ) ] = 400 * [ 100 / 125 ] = rs . 320 answer : b" | a ) 228 , b ) 320 , c ) 287 , d ) 480 , e ) 811 | b | divide(multiply(400, const_100), add(const_100, 25)) | add(n1,const_100)|multiply(n0,const_100)|divide(#1,#0)| | gain | B |
a sucrose solution contains 10 grams of sucrose per 100 cubic centimeters of solution . if 60 cubic centimeters of the solution were poured into an empty container , how many grams of sucrose would be in the container ? | we are given that a sucrose solution contains 10 grams of sucrose per 100 cubic centimeters of solution . since we are dealing with a solution , we know that the grams of sucrose is proportional to the number of cubic centimeters of solution . thus , to determine how many grams of sucrose would be in the container when we have 60 cubic centimeters of solution , we can set up a proportion . we can say : “ 10 grams of sucrose is to 100 cubic centimeters of solution as x grams of sucrose is to 60 cubic centimeters of solution . ” let ’ s now set up the proportion and solve for x . 10 / 100 = x / 60 when we cross multiply we obtain : ( 10 ) ( 60 ) = 100 x 600 = 100 x 6.00 = x there are 6.00 grams of sucrose in the solution in the container . the answer is b . | ['a ) 4.00', 'b ) 6.00', 'c ) 5.50', 'd ) 6.50', 'e ) 6.75'] | b | divide(multiply(60, 10), 100) | multiply(n0,n2)|divide(#0,n1) | physics | B |
if the cost price of 55 articles is equal to the selling price of 50 articles , then what is the percent profit ? | "let x be the cost price of one article . let y be the selling price of one article . 50 y = 55 x y = 1.1 x the answer is b ." | a ) 5 % , b ) 10 % , c ) 15 % , d ) 20 % , e ) 25 % | b | multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 50), 55)), divide(multiply(const_100, 50), 55))) | multiply(n1,const_100)|divide(#0,n0)|subtract(const_100,#1)|divide(#2,#1)|multiply(#3,const_100)| | gain | B |
a person walks from one end to the other of a 80 - meter long moving walkway at a constant rate in 40 seconds , assisted by the walkway . when this person reaches the end , they reverse direction and continue walking with the same speed , but this time it takes 120 seconds because the person is traveling against the direction of the moving walkway . if the walkway were to stop moving , how many seconds would it take this person to walk from one end of the walkway to the other ? | "let v be the speed of the person and let x be the speed of the walkway . 40 ( v + x ) = 80 then 120 ( v + x ) = 240 120 ( v - x ) = 80 when we add the two equations : 240 v = 320 v = 4 / 3 time = 80 / ( 4 / 3 ) = 60 seconds the answer is b ." | a ) 56 , b ) 60 , c ) 64 , d ) 68 , e ) 72 | b | divide(80, divide(add(divide(80, 40), divide(80, 120)), const_2)) | divide(n0,n1)|divide(n0,n2)|add(#0,#1)|divide(#2,const_2)|divide(n0,#3)| | physics | B |
if m is a positive integer and m ^ 2 is divisible by 39 , then the largest positive integer that must divide m is ? | "m ^ 2 is divisible by 48 so m ^ 2 must be multiple of 48 . if the value of m is multiples of 13 then it will satisfy the condition . if we if m is 12 or 24 or 36 then it ans is e but if m = 39 then answer should be 16 . is the question right ? or am i missing some thing ? e" | a ) 3 , b ) 6 , c ) 8 , d ) 12 , e ) 13 | e | multiply(const_3, divide(divide(39, const_3), const_3)) | divide(n1,const_3)|divide(#0,const_3)|multiply(#1,const_3)| | general | E |
find the remainder q when 12 ^ 190 is divided by 1729 ? | 12 ^ ( 190 ) can be written as . ( ( 12 ^ 3 ) ^ 63 ) * 12 . 12 ^ 3 when divided by 1729 gives a remainder q - 1 . so in the numerator we have - 12 . now acccording to remainder theorm the answer will be 1729 - 12 = 1717 . d | a ) 12 , b ) 1 , c ) 1728 , d ) 1717 , e ) 4 | d | subtract(1729, 12) | subtract(n2,n0) | general | D |
it takes 40 identical printing presses 9 hours to print 500,000 papers . how many hours would it take 30 of these printing presses to print 500,000 papers ? | "40 printing presses can do 1 / 9 of the job each hour . 30 printing presses can do 3 / 4 * 1 / 9 = 1 / 12 of the job each hour . the answer is d ." | a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 13 | d | divide(multiply(divide(const_1000, const_2), const_1000), multiply(divide(divide(multiply(divide(const_1000, const_2), const_1000), 40), 9), 30)) | divide(const_1000,const_2)|multiply(#0,const_1000)|divide(#1,n0)|divide(#2,n1)|multiply(n3,#3)|divide(#1,#4)| | physics | D |
the cost per pound of green tea and coffee were the same in june . in july , the price of coffee shot up by 100 % and that of green tea dropped by 70 % . if in july , a mixture containing equal quantities of green tea and coffee costs $ 3.45 for 3 lbs , how much did a pound of green tea cost in july ? | lets assume price of coffee in june = 100 x price of green tea in june = 100 x price of coffee in july = 200 x ( because of 100 % increase in price ) price of green tea in july = 30 x ( because of 70 % decrease in price ) price of 1.5 pound of coffee 1.5 pound of green tea in july will be = 300 x + 45 x = 345 x as per question 345 x = 3.45 $ x = 0.01 s so the price of tea in july = 30 x = 30 x 0.01 = 0.3 $ / pound answer b | a ) $ 4 , b ) $ 0.3 , c ) $ 1 , d ) $ 3 , e ) $ 1.65 | b | multiply(divide(add(multiply(divide(add(100, 100), 100), divide(3, const_2)), multiply(divide(3, const_2), divide(subtract(100, 70), 100))), 3.45), divide(subtract(100, 70), 100)) | add(n0,n0)|divide(n3,const_2)|subtract(n0,n1)|divide(#0,n0)|divide(#2,n0)|multiply(#3,#1)|multiply(#1,#4)|add(#5,#6)|divide(#7,n2)|multiply(#8,#4) | general | B |
if g is to be chosen at random from the set { 5,8 , 7,1 } and k is to be chosen at random from the set { 14 , 8,3 } , what is the probability that gk will be even ? | method - 1 gk will be even when 1 ) g is even and k is odd , probability of g even is ( 1 / 4 ) and probability of k odd is ( 1 / 3 ) , so probability of case ( 1 ) = ( 1 / 4 ) * ( 1 / 3 ) = ( 1 / 12 ) 2 ) g is odd and k is even , probability of g odd is ( 3 / 4 ) and probability of k even is ( 2 / 3 ) , so probability of case ( 2 ) = ( 3 / 4 ) * ( 2 / 3 ) = ( 6 / 12 ) 3 ) g is even and k is even , probability of g even is ( 1 / 4 ) and probability of k even is ( 2 / 3 ) , so probability of case ( 1 ) = ( 1 / 4 ) * ( 2 / 3 ) = ( 2 / 12 ) total favorable probability = ( 1 / 12 ) + ( 6 / 12 ) + ( 2 / 12 ) = ( 12 / 12 ) = 1 answer : option a | a ) 1 , b ) 5 , c ) 8 , d ) 4 , e ) 3 | a | divide(multiply(const_4, const_3), multiply(const_4, const_3)) | multiply(const_3,const_4)|divide(#0,#0) | probability | A |
the monthly incomes of a and b are in the ratio 5 : 2 . b ' s monthly income is 12 % more than c ' s monthly income . if c ' s monthly income is rs . 14000 , then find the annual income of a ? | "b ' s monthly income = 14000 * 112 / 100 = rs . 15680 b ' s monthly income = 2 parts - - - - > rs . 15680 a ' s monthly income = 5 parts = 5 / 2 * 15680 = rs . 39200 a ' s annual income = rs . 39200 * 12 = rs . 470400 answer : d" | a ) rs . 420000 , b ) rs . 180000 , c ) rs . 201600 , d ) rs . 470400 , e ) none of these | d | multiply(multiply(multiply(14000, add(const_1, divide(12, const_100))), divide(5, 2)), 12) | divide(n0,n1)|divide(n2,const_100)|add(#1,const_1)|multiply(n3,#2)|multiply(#0,#3)|multiply(n2,#4)| | other | D |
a worker makes a toy in every 2 h . if he works for 100 h , then how many toys will he make ? | "no . of toys = 100 / 2 = 50 answer : c" | a ) 40 , b ) 54 , c ) 50 , d ) 39 , e ) none | c | divide(100, 2) | divide(n1,n0)| | physics | C |
if 4 men can paint 60 m long wall in 3 days , then 5 men can paint 50 m long wall in | the length of wall painted by one man in one day = 60 / 4 ã — 3 = 5 m no . of days required to paint 50 m wall by 5 men = 50 / 5 ã — 5 = 2 day . b | a ) 3 days , b ) 2 days , c ) 5 days , d ) 6 days , e ) 1 day | b | divide(50, multiply(divide(divide(60, 3), 4), 5)) | divide(n1,n2)|divide(#0,n0)|multiply(n3,#1)|divide(n4,#2) | physics | B |
a car is purchased on hire - purchase . the cash price is $ 23 000 and the terms are a deposit of 10 % of the price , then the balance to be paid off over 60 equal monthly instalments . interest is charged at 12 % p . a . what is the monthly instalment ? | "explanation : cash price = $ 23 000 deposit = 10 % ã — $ 23 000 = $ 2300 loan amount = $ 23000 â ˆ ’ $ 2300 number of payments = 60 = $ 20700 i = p * r * t / 100 i = 12420 total amount = 20700 + 12420 = $ 33120 regular payment = total amount / number of payments = 552 answer : e" | a ) $ 503 , b ) $ 504 , c ) $ 515 , d ) $ 543 , e ) $ 552 | e | add(divide(multiply(multiply(23, const_1000), subtract(const_1, divide(10, const_100))), 60), multiply(divide(divide(12, const_100), 12), multiply(multiply(23, const_1000), subtract(const_1, divide(10, const_100))))) | divide(n2,const_100)|divide(n4,const_100)|multiply(n0,const_1000)|divide(#1,n4)|subtract(const_1,#0)|multiply(#2,#4)|divide(#5,n3)|multiply(#3,#5)|add(#6,#7)| | gain | E |
the ratio of two no . addition and subtraction be 4 : 3 . the what is the ratio of numbers ? | ( x + y ) / ( x - y ) = 4 / 3 dividing numerator and denominator by y ( x / y ) + 1 / ( x / y ) - 1 = 4 / 3 let x / y be z z + 1 / z - 1 = 4 / 3 3 z + 3 = 4 z - 4 z = 7 x / y = 7 answer e | a ) 3 : 2 , b ) 4 : 3 , c ) 5 : 1 , d ) 6 : 5 , e ) 7 : 1 | e | add(4, 3) | add(n0,n1) | general | E |
a man is 36 years older than his son . in three years , his age will be thrice the age of his son . the present age of this son is | "let ' s son age is x , then father age is x + 36 . = > 3 ( x + 3 ) = ( x + 36 + 3 ) = > 3 x + 9 = x + 39 = > 2 x = 30 years = 15 years answer : e" | a ) 28 years , b ) 22 years , c ) 18 years , d ) 16 years , e ) 15 years | e | divide(subtract(36, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | multiply(const_2,const_2)|subtract(const_2,const_1)|subtract(#0,const_2)|subtract(n0,#2)|divide(#3,#1)| | general | E |
a , b and c enter into partnership . a invests some money at the beginning , b invests double the amount after 6 months , and c invests thrice the amount after 8 months . if the annual gain be rs . 18900 . a ' s share is ? | "x * 12 : 2 x * 6 : 3 x * 4 1 : 1 : 1 1 / 3 * 18900 = 6300 answer : a" | a ) 6300 , b ) 2765 , c ) 6000 , d ) 1298 , e ) 1261 | a | multiply(multiply(const_1, const_12), divide(18900, add(add(multiply(const_1, const_12), multiply(subtract(const_12, 6), const_2)), multiply(subtract(const_12, 8), const_3)))) | multiply(const_1,const_12)|subtract(const_12,n0)|subtract(const_12,n1)|multiply(#1,const_2)|multiply(#2,const_3)|add(#0,#3)|add(#5,#4)|divide(n2,#6)|multiply(#7,#0)| | gain | A |
a , b , c can do a piece of work in 20 days , 30 days and 40 days respectively , working alone . how soon can the work be done if a is assisted by b and c on alternate days ? | "a + b 1 day work = 1 / 20 + 1 / 30 = 1 / 12 a + c 1 day work = 1 / 20 + 1 / 40 = 3 / 40 work done in 2 days = 1 / 12 + 3 / 40 = 19 / 120 19 / 120 work is done by a in 2 days whole work will be done in 2 * 19 / 120 = 3 days approximately answer is a" | a ) 3 days , b ) 1 day , c ) 5 days , d ) 10 days , e ) 7 days | a | divide(const_2, add(add(divide(const_1, 20), divide(const_1, 30)), add(divide(const_1, 20), divide(const_1, 40)))) | divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|add(#0,#2)|add(#3,#4)|divide(const_2,#5)| | physics | A |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.