Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
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66 cubic centimetres of silver is drawn into a wire 1 mm in diameter . the length if the wire in meters will be : | explanation : let the length of the wire be h radius = 1 / 2 mm = 120 cm Ο r 2 h = 66 22 / 7 β 1 / 20 β 1 / 20 β h = 66 = > h = 66 β 20 β 20 β 7 / 22 = 8400 cm = 84 m option c | ['a ) 76 m', 'b ) 80 m', 'c ) 84 m', 'd ) 88 m', 'e ) none of these'] | c | divide(66, multiply(power(divide(1, const_2), const_2), const_pi)) | divide(n1,const_2)|power(#0,const_2)|multiply(#1,const_pi)|divide(n0,#2) | physics |
a certain sum of money is divided among a , b and c so that for each rs . a has , b has 65 paisa and c 40 paisa . if c ' s share is rs . 48 , find the sum of money ? | "a : b : c = 100 : 65 : 40 = 20 : 13 : 8 8 - - - - 48 41 - - - - ? = > rs . 246 answer : c" | a ) 288 , b ) 262 , c ) 246 , d ) 205 , e ) 267 | c | multiply(divide(48, 40), add(add(const_100, 65), 40)) | add(n0,const_100)|divide(n2,n1)|add(n1,#0)|multiply(#2,#1)| | general |
a is four times as fast as b . if b alone can do a piece of work in 60 days , in what time can a and b together complete the work ? | a can do the work in 60 / 4 i . e . , 15 days . a and b ' s one day ' s work = 1 / 15 + 1 / 60 = ( 4 + 1 ) / 60 = 1 / 12 so a and b together can do the work in 12 days . answer : b | a ) 10 , b ) 12 , c ) 22 , d ) 28 , e ) 20 | b | inverse(add(inverse(divide(60, const_4)), inverse(60))) | divide(n0,const_4)|inverse(n0)|inverse(#0)|add(#2,#1)|inverse(#3) | physics |
8 men can do a piece of work in 12 days . 4 women can do it in 48 days and 10 children can do it in 24 days . in how many days can 10 men , 4 women and 10 children together complete the piece of work ? | explanation : 1 man β s 1 day β s work = 1 / 8 Γ 12 = 1 / 96 10 men β s 1 day β s work = 1 Γ 10 / 96 = 5 / 48 1 woman β s 1 day β s work = 1 / 192 4 women β s 1 day β s work = 1 / 192 Γ 4 = 1 / 48 1 child β s 1 day β s work = 1 / 240 10 children β s 1 day β s work = 1 / 24 therefore , ( 10 men + 4 women + 10 children )... | a ) 5 days , b ) 15 days , c ) 28 days , d ) 6 days , e ) 7 days | d | inverse(add(multiply(10, inverse(multiply(24, 10))), add(multiply(inverse(multiply(12, 8)), 10), multiply(inverse(multiply(48, 4)), 4)))) | multiply(n0,n1)|multiply(n2,n3)|multiply(n4,n5)|inverse(#0)|inverse(#1)|inverse(#2)|multiply(n4,#3)|multiply(n2,#4)|multiply(n4,#5)|add(#6,#7)|add(#9,#8)|inverse(#10) | physics |
if a positive integer n has 211 factors , then how many prime factors does n have ? | let the factor be a ^ x * b ^ y . . . . . so ( x + 1 ) ( y + 1 ) . . . . . = 211 . . now 211 is a prime number . . . so only one variable x or y is 210 and the others are 0 . . ( 210 + 1 ) ( 0 + 1 ) . . . so the number becomes a ^ 210 . . thus only one prime factor is there . | a ) 3 , b ) 2 , c ) can not be determined , d ) 4 , e ) 1 | e | divide(211, 211) | divide(n0,n0) | other |
rahim bought 65 books for rs . 1150 from one shop and 50 books for rs . 920 from another . what is the average price he paid per book ? | "average price per book = ( 1150 + 920 ) / ( 65 + 50 ) = 2070 / 115 = rs . 18 answer : b" | a ) rs . 17 , b ) rs . 18 , c ) rs . 12 , d ) rs . 11 , e ) rs . 10 | b | divide(add(1150, 920), add(65, 50)) | add(n1,n3)|add(n0,n2)|divide(#0,#1)| | general |
a cistern 10 m long and 6 m wide contains water up to a breadth of 1 m 35 cm . find the total area of the wet surface . | "explanation : area of the wet surface = 2 [ lb + bh + hl ] - lb = 2 [ bh + hl ] + lb = 2 [ ( 6 * 1.35 + 10 * 1.35 ) ] + 10 * 6 = 103 m square option b" | a ) 100 m sqaure , b ) 103 m sqaure , c ) 152 m sqaure , d ) 164 m sqaure , e ) none of these | b | add(multiply(const_2, add(multiply(add(divide(35, const_100), 1), 6), multiply(add(divide(35, const_100), 1), 10))), multiply(6, 10)) | divide(n3,const_100)|multiply(n0,n1)|add(n2,#0)|multiply(n1,#2)|multiply(n0,#2)|add(#3,#4)|multiply(#5,const_2)|add(#6,#1)| | physics |
a city with a population of 173,460 is to be divided into 9 voting districts , and no district is to have a population that is more than 10 percent greater than the population of any other district . what is the minimum possible population that the least populated district could have ? | "the minimum possible population occurs when all the other districts have a population that is 10 % greater than the least populated district . let p be the population of the least populated district . then 173,460 = p + 8 ( 1.1 ) p 9.8 p = 173,460 p = 17,700 the answer is d ." | a ) 17,400 , b ) 17,500 , c ) 17,600 , d ) 17,700 , e ) 17,800 | d | divide(divide(subtract(multiply(const_1000, const_100), subtract(subtract(const_3600, const_100), const_1000)), const_1000), add(multiply(add(const_1, divide(10, const_100)), subtract(9, const_1)), const_1)) | divide(n2,const_100)|multiply(const_100,const_1000)|subtract(const_3600,const_100)|subtract(n1,const_1)|add(#0,const_1)|subtract(#2,const_1000)|multiply(#4,#3)|subtract(#1,#5)|add(#6,const_1)|divide(#7,const_1000)|divide(#9,#8)| | general |
car q and car y traveled the same 80 - mile route . if car q took 2 hours and car y traveled at an average speed that was 50 percent faster than the average speed of car q , how many hours did it take car y to travel the route ? | the speed of car q is ( distance ) / ( time ) = 80 / 2 = 40 miles per hour . the speed of car y = 3 / 2 * 40 = 60 miles per hour - - > ( time ) = ( distance ) / ( speed ) = 80 / 60 = 4 / 3 hours . answer : c . or : to cover the same distance at 3 / 2 as fast rate 2 / 3 as much time is needed - - > ( time ) * 2 / 3 = 2 ... | a ) 2 / 3 , b ) 1 , c ) 4 / 3 , d ) 8 / 5 , e ) 3 | c | divide(80, add(divide(multiply(divide(80, 2), 50), const_100), divide(80, 2))) | divide(n0,n1)|multiply(n2,#0)|divide(#1,const_100)|add(#2,#0)|divide(n0,#3) | physics |
3 pig β s and 10 hens are brought for rs . 1200 . if the average price of a hen be rs . 30 . what is the average price of a pig . | explanation : average price of a hen = rs . 30 total price of 10 hens = 10 * 30 = rs . 300 but total price of 3 pigs and 10 hens = rs . 1200 total price of 3 pigs is = 1200 - 300 = 900 average price of a pig = 900 / 3 = rs . 300 answer : a | a ) 300 , b ) 350 , c ) 400 , d ) 425 , e ) 275 | a | divide(subtract(1200, multiply(10, 30)), 3) | multiply(n1,n3)|subtract(n2,#0)|divide(#1,n0) | general |
how many factors of 240 are also multiples of 3 ? | "take factors of 240 240 = 2 ^ 4 * 3 ^ 1 * 5 ^ 1 total factors of 240 = ( 4 + 1 ) * ( 1 + 1 ) * ( 1 + 1 ) = 5 * 2 * 2 = 20 out of total factors of 20 , half will have 0 as power of 3 and half will have 1 as power of 3 . it is because we have considered only 0 and 1 as power of 3 to compute total factors of 240 . theref... | a ) 5 , b ) 8 , c ) 9 , d ) 10 , e ) 20 | d | divide(240, multiply(const_10, const_2)) | multiply(const_10,const_2)|divide(n0,#0)| | other |
mr yadav spends 60 % of his monthly salary on consumable items and 50 % of the remaining on clothes and transport . he saves the remaining amount . if his savings at the end of the year were 46800 , how much amount per month would he have spent on clothes and transport ? | "β΅ amount , he have spent in 1 month on clothes transport = amount spent on saving per month β΅ amount , spent on clothes and transport = 46800 β 12 = 3900 answer e" | a ) 4038 , b ) 8076 , c ) 9691.2 , d ) 4845.6 , e ) 3900 | e | multiply(divide(divide(46800, divide(divide(multiply(subtract(const_100, 60), 50), const_100), const_100)), multiply(const_3, const_4)), divide(divide(multiply(subtract(const_100, 60), 50), const_100), const_100)) | multiply(const_3,const_4)|subtract(const_100,n0)|multiply(n1,#1)|divide(#2,const_100)|divide(#3,const_100)|divide(n2,#4)|divide(#5,#0)|multiply(#6,#4)| | general |
a train 605 m long is running with a speed of 60 km / hr . in what time will it pass a man who is running at 6 km / hr in the direction opposite to that in which the train is going ? | "speed of train relative to man = 60 + 6 = 66 km / hr . = 66 * 5 / 18 = 55 / 3 m / sec . time taken to pass the men = 605 * 3 / 55 = 33 sec . answer : c" | a ) 81 , b ) 16 , c ) 33 , d ) 54 , e ) 12 | c | divide(605, multiply(add(60, 6), const_0_2778)) | add(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
tom opened a shop investing rs . 3000 . jose joined him 2 months later , investing rs . 4500 . they earned a profit of rs . 6300 after completion of one year . what will be jose ' s share of profit ? | "sol = ~ s - so anju β s share = [ 5 / 9 ] x 6300 = 3500 a" | a ) 3500 , b ) 2800 , c ) 3251 , d ) 4251 , e ) 3508 | a | multiply(6300, subtract(const_1, divide(multiply(3000, multiply(2, multiply(2, const_3))), add(multiply(4500, subtract(multiply(2, multiply(2, const_3)), 2)), multiply(3000, multiply(2, multiply(2, const_3))))))) | multiply(n1,const_3)|multiply(#0,n1)|multiply(n0,#1)|subtract(#1,n1)|multiply(n2,#3)|add(#4,#2)|divide(#2,#5)|subtract(const_1,#6)|multiply(n3,#7)| | gain |
the simple form of the ratio 2 / 3 : 2 / 5 is | "2 / 3 : 2 / 5 = 10 : 6 = 5 : 3 answer : d" | a ) 5 : 7 , b ) 5 : 2 , c ) 5 : 9 , d ) 5 : 3 , e ) 5 : 4 | d | divide(2, 5) | divide(n2,n3)| | other |
rs . 1500 is divided into two parts such that if one part is invested at 6 % and the other at 5 % the whole annual interest from both the sum is rs . 83 . how much was lent at 5 % ? | "( x * 5 * 1 ) / 100 + [ ( 1500 - x ) * 6 * 1 ] / 100 = 83 5 x / 100 + 90 β 6 x / 100 = 83 x / 100 = 7 = > x = 700 answer : d" | a ) 299 , b ) 566 , c ) 678 , d ) 700 , e ) 8277 | d | multiply(add(5, 6), const_100) | add(n1,n2)|multiply(#0,const_100)| | gain |
a towel , when bleached , was found to have lost 20 % of its length and 10 % of its breadth . the percentage of decrease in area is ? | "explanation : let original length = x and original width = y decrease in area will be = xy β ( 80 x / 100 Γ 90 y / 100 ) = ( xy β 18 / 25 xy ) = 7 / 25 xy decrease = ( 7 xy / 25 xy Γ 100 ) % = 28 option d" | a ) 25 % , b ) 26 % , c ) 27 % , d ) 28 % , e ) 29 % | d | multiply(subtract(const_1, multiply(divide(subtract(const_100, 20), const_100), divide(subtract(const_100, 10), const_100))), const_100) | subtract(const_100,n0)|subtract(const_100,n1)|divide(#0,const_100)|divide(#1,const_100)|multiply(#2,#3)|subtract(const_1,#4)|multiply(#5,const_100)| | gain |
gold is 10 times as heavy as water and copper is 5 times as heavy as water . in what ratio should these be mixed to get an alloy 6 times as heavy as water ? | g = 10 w c = 5 w let 1 gm of gold mixed with x gm of copper to get 1 + x gm of the alloy 1 gm gold + x gm copper = x + 1 gm of alloy 10 w + 5 wx = x + 1 * 6 w 10 + 5 x = 6 ( x + 1 ) x = 4 ratio of gold with copper = 1 : 4 = 1 : 4 answer is b | a ) 3 : 2 , b ) 1 : 4 , c ) 3 : 1 , d ) 5 : 2 , e ) 4 : 3 | b | divide(subtract(6, 5), subtract(10, 6)) | subtract(n2,n1)|subtract(n0,n2)|divide(#0,#1) | general |
calculate how many days it will take for 8 boys to paint a 50 m long wall if 6 boys can paint a 60 m long wall in 5 days , | the length of wall painted by one boy in one day = 60 / 6 * 1 / 5 = 2 m no . of days required to paint 50 m cloth by 8 boys = 50 / 8 * 1 / 2 = 3.13 days . b | a ) 4.13 days , b ) 3.13 days , c ) 513 days , d ) 9.13 days , e ) 2 days | b | divide(multiply(multiply(6, 5), 50), multiply(60, 8)) | multiply(n2,n4)|multiply(n0,n3)|multiply(n1,#0)|divide(#2,#1) | physics |
a fair price shopkeeper takes 10 % profit on his goods . he lost 20 % goods during theft . his loss percent is : | "c 12 % suppose he has 100 items . let c . p . of each item be $ 1 . total cost = $ 100 . number of items left after theft = 80 . s . p . of each item = $ 1.10 total sale = 1.10 * 80 = $ 88 hence , loss % = 12 / 100 * 100 = 12 %" | a ) 10 , b ) 13 , c ) 12 , d ) 18 , e ) 19 | c | subtract(const_100, subtract(add(const_100, 10), divide(multiply(add(const_100, 10), 20), const_100))) | add(n0,const_100)|multiply(n1,#0)|divide(#1,const_100)|subtract(#0,#2)|subtract(const_100,#3)| | gain |
a customer went to a shop and paid a total of $ 40 , out of which 90 cents was for sales tax on taxable purchases . if the tax rate was 6 % , then what was the cost of the tax free items ? | "the total cost was $ 40 . the tax was $ 0.90 let the original price of the taxable items = x given that tax rate = 6 % 0.06 x = 0.90 x = $ 15 the cost of the tax free items was $ 40 - $ 15 - $ 0.90 = $ 24.10 the answer is c ." | a ) $ 22.70 , b ) $ 23.20 , c ) $ 24.10 , d ) $ 25.50 , e ) $ 26.90 | c | subtract(subtract(40, 90), multiply(6, const_2)) | multiply(n2,const_2)|subtract(n0,n1)|subtract(#1,#0)| | gain |
albert buys 4 horses and 9 cows for rs . 13,400 . if he sells the horses at 10 % profit and the cows at 20 % profit , then he earns a total profit of rs . 1880 . the cost of a horse is : | "let c . p . of each horse be rs . x and c . p . of each cow be rs . y . then , 4 x + 9 y = 13400 - - ( i ) and , 10 % of 4 x + 20 % of 9 y = 1880 2 / 5 x + 9 / 5 y = 1880 = > 2 x + 9 y = 9400 - - ( ii ) solving ( i ) and ( ii ) , we get : x = 2000 and y = 600 . cost price of each horse = rs . 2000 . answer : b" | a ) rs . 2007 , b ) rs . 2000 , c ) rs . 2089 , d ) rs . 2067 , e ) rs . 2098 | b | multiply(20, const_100) | multiply(n4,const_100)| | gain |
two motor bikes cover the same distance at the speed of 60 and 64 kmps respectively . find the distance traveled by them if the slower bike takes 1 hour more than the faster bike ? | explanation : 60 ( x + 1 ) = 64 x x = 15 60 x 16 = 960 km answer : c | a ) 860 km , b ) 870 km , c ) 960 km , d ) 260 km , e ) 840 km | c | multiply(60, add(divide(60, subtract(64, 60)), 1)) | subtract(n1,n0)|divide(n0,#0)|add(n2,#1)|multiply(n0,#2) | physics |
the edge of a cube is 7 a cm . find its surface ? | 6 a 2 = 6 * 7 a * 7 a = 294 a 2 answer : e | ['a ) 24 a 8', 'b ) 24 a 4', 'c ) 24 a 1', 'd ) 24 a 2', 'e ) 294 a 2'] | e | surface_cube(7) | surface_cube(n0) | geometry |
if x < y < z and y - x > 5 , where x is an even integer and y and z are odd integers , what is the least possible value s of z - x ? | "x < y < z to find the least possible value for z - x ; we need to find the values for z and x that can be closest to each other . if x is some even number , then what could be minimum possible odd z . if x is some even number y - x > 5 ; y > x + 5 ; minimum value for y = x + 5 + 2 = x + 7 [ note : x + 5 is as even + o... | a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | d | add(add(5, const_2), const_2) | add(n0,const_2)|add(#0,const_2)| | general |
if renee earns a raise of between 5 % and 10 % , non - inclusive , by what fraction could her salary have increased ? | 5 % is 5 / 100 = 1 / 20 10 % is 10 / 100 = 1 / 10 the increase must be greater than 1 / 20 and less than 1 / 10 . 1 / 4 , 1 / 5 , and 1 / 10 are not less than 1 / 10 . 1 / 20 is not greater than 1 / 20 . the only option left is 2 / 25 , which is in the specified range . answer : d | a ) 1 / 4 , b ) 1 / 5 , c ) 1 / 10 , d ) 2 / 25 , e ) 1 / 20 | d | divide(subtract(5, const_1), multiply(5, const_10)) | multiply(n0,const_10)|subtract(n0,const_1)|divide(#1,#0) | general |
a merchant sells an item at a 20 % discount , but still makes a gross profit of 20 percent of the cost . what percent r of the cost would the gross profit on the item have been if it had been sold without the discount ? | "let the market price of the product is mp . let the original cost price of the product is cp . selling price ( discounted price ) = 100 % of mp - 20 % mp = 80 % of mp . - - - - - - - - - - - - - - - - ( 1 ) profit made by selling at discounted price = 20 % of cp - - - - - - - - - - - - - - ( 2 ) apply the formula : pr... | a ) 20 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 75 % | c | subtract(const_100, subtract(subtract(const_100, 20), 20)) | subtract(const_100,n0)|subtract(#0,n1)|subtract(const_100,#1)| | gain |
in Ξ΄ pqs above , if pq = 5 and ps = 6 , then | "there are two ways to calculate area of pqs . area remains same , so both are equal . 5 * 6 / 2 = pr * 7 / 2 pr = 30 / 7 d" | a ) 9 / 4 , b ) 12 / 5 , c ) 16 / 5 , d ) 30 / 7 , e ) 50 / 3 | d | divide(6, 5) | divide(n1,n0)| | general |
in a graduate physics course , 70 percent of the students are male and 30 percent of the students are married . if two - sevenths of the male students are married , what fraction of the male students is single ? | "let assume there are 100 students of which 70 are male and 30 are females if 30 are married then 70 will be single . now its given that two - sevenths of the male students are married that means 2 / 7 of 70 = 20 males are married if 30 is the total number of students who are married and out of that 20 are males then t... | a ) 2 / 7 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 5 / 7 | e | divide(const_10, 30) | divide(const_10,n1)| | gain |
johnny travels a total of one hour to and from school . on the way there he jogs at 4 miles per hour and on the return trip he gets picked up by the bus and returns home at 20 miles per hour . how far is it to the school ? | "answer : b ) 6.6 miles . average speed for round trip = 2 * a * b / ( a + b ) , where a , b are speeds so , average speed was = 2 * 4 * 20 / ( 4 + 20 ) = 6.6 m / hr the distance between schoolhome should be half of that . ie . 6.6 miles answer b" | a ) 2 miles , b ) 6.6 miles , c ) 4.8 miles , d ) 8 miles , e ) 10 miles | b | multiply(divide(const_1, add(divide(const_1, 4), divide(const_1, 20))), const_1_6) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)|multiply(#3,const_1_6)| | physics |
an engine moves at the speed of 90 kmph without any coaches attached to it . speed of the train reduces at the rate that varies directly as the square root of the number of coaches attached . when 4 coaches are attached speed decreases to 78 kmph . what will be the speed of train when 16 coaches are attached . | "1 . no . of coaches = 4 sqr root = 2 speed decreases by 12 12 = k * 2 k = 6 no . of coaches = 16 swr root = 4 decrease = 4 * 6 = 24 new speed = 90 - 24 = 66 a" | a ) 66 , b ) 72 , c ) 80 , d ) 68 , e ) 73 | a | subtract(90, multiply(sqrt(16), divide(subtract(90, 78), sqrt(4)))) | sqrt(n1)|sqrt(n3)|subtract(n0,n2)|divide(#2,#0)|multiply(#3,#1)|subtract(n0,#4)| | physics |
at what rate percent of simple interest will a sum of money double itself in 77 years ? | let sum = x . then , s . i . = x . rate = ( 100 * s . i . ) / ( p * t ) = ( 100 * x ) / ( x * 77 ) = 100 / 77 = 1.29 % answer : e | a ) 6.54 % , b ) 2.54 % , c ) 8 . 2 % , d ) 4.94 % , e ) 1.29 % | e | divide(const_100, 77) | divide(const_100,n0) | gain |
what is the value of 10 ^ 7 - 6 ^ 5 ? | "as 10 ^ n will always have last digit as 0 and 6 ^ n will always as last digit 6 . . hence difference of such sum should always be ending with 4 and there is only on option . . answer b" | a ) 99 , 91,223 , b ) 99 , 92,224 , c ) 99 , 94,265 , d ) 99 , 95,300 , e ) 99 , 96,307 | b | divide(multiply(add(add(const_100, const_60), const_1), 10), const_100) | add(const_100,const_60)|add(#0,const_1)|multiply(n0,#1)|divide(#2,const_100)| | general |
dacid obtained 73 , 69 , 92 , 64 and 82 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology . what are his average marks ? | "average = ( 73 + 69 + 92 + 64 + 82 ) / 5 = 380 / 5 = 76 answer : e" | a ) 79 , b ) 99 , c ) 88 , d ) 88 , e ) 76 | e | divide(add(add(add(add(73, 69), 92), 64), 82), divide(const_10, const_2)) | add(n0,n1)|divide(const_10,const_2)|add(n2,#0)|add(n3,#2)|add(n4,#3)|divide(#4,#1)| | general |
in may mrs lee ' s earnings were 60 percent of the lee family ' s total income . in june mrs lee earned 30 percent more than in may . if the rest of the family ' s income was the same both months , then , in june , mrs lee ' s earnings were approximately what percent of the lee family ' s total income ? | "let in may lee family ' s total income = 100 in may mrs lee ' s income = 60 in may rest of the family ' s income = 40 in june mrs lees income = 60 * 130 / 100 = 78 in june total income = 78 + 40 = 118 % of mrs lee ' s income = 72 / 112 = 66.10 ( b )" | a ) 64 % , b ) 66 % , c ) 72 % , d ) 76 % , e ) 80 % | b | multiply(divide(add(const_100, 30), add(add(const_100, 30), const_100)), const_100) | add(n1,const_100)|add(#0,const_100)|divide(#0,#1)|multiply(#2,const_100)| | general |
r and s together can plough a field in 10 hours but by himself r requires 15 hours . how long would s take to plough the same field ? | f r and s together can do a piece of work in x days and r alone can do the same work in y days , then s alone can do the same work in x y / y Γ’ β¬ β x days . the no . of hours required by s = 10 Γ£ β 20 / 20 Γ’ β¬ β 10 = 200 / 10 = 20 hours e | a ) 24 hours , b ) 5 hours , c ) 10 hours , d ) 15 hours , e ) 20 hours | e | add(15, subtract(15, 10)) | subtract(n1,n0)|add(n1,#0) | physics |
the area of a square is equal to five times the area of a rectangle of dimensions 60 cm * 12 cm . what is the perimeter of the square ? | "area of the square = s * s = 5 ( 60 * 12 ) = > s = 60 = 60 cm perimeter of the square = 4 * 60 = 240 cm . answer : e" | a ) 289 cm , b ) 800 cm , c ) 829 cm , d ) 288 cm , e ) 240 cm | e | multiply(sqrt(multiply(rectangle_area(60, 12), divide(12, const_2))), const_4) | divide(n1,const_2)|rectangle_area(n0,n1)|multiply(#0,#1)|sqrt(#2)|multiply(#3,const_4)| | geometry |
36 is divided in 2 parts such that 8 times the first part added to 3 times the second part makes 203 . what is the first part ? | explanation : let the first part be x . 8 x + 3 ( 36 β x ) = 203 8 x + 108 β 3 x = 203 5 x + 108 = 203 5 x = 95 x = 19 answer : a | a ) 19 , b ) 88 , c ) 267 , d ) 26 , e ) 28 | a | divide(subtract(203, multiply(36, 3)), subtract(8, 3)) | multiply(n0,n3)|subtract(n2,n3)|subtract(n4,#0)|divide(#2,#1) | general |
roy is now 8 years older than julia and half of that amount older than kelly . if in 2 years , roy will be twice as old as julia , then in 2 years what would be roy β s age multiplied by kelly β s age ? | "r = j + 8 = k + 4 r + 2 = 2 ( j + 2 ) ( j + 8 ) + 2 = 2 j + 4 j = 6 r = 14 k = 10 in 2 years ( r + 2 ) ( k + 2 ) = 16 * 12 = 192 the answer is e ." | a ) 160 , b ) 168 , c ) 176 , d ) 184 , e ) 192 | e | multiply(add(add(8, subtract(8, 2)), 2), add(subtract(add(8, subtract(8, 2)), divide(8, const_2)), 2)) | divide(n0,const_2)|subtract(n0,n1)|add(n0,#1)|add(n1,#2)|subtract(#2,#0)|add(n1,#4)|multiply(#3,#5)| | general |
the radius of a cylindrical vessel is 5 cm and height is 3 cm . find the whole surface of the cylinder ? | "r = 5 h = 3 2 Ο r ( h + r ) = 2 * 22 / 7 * 5 ( 8 ) = 251.4 answer : a" | a ) 251.4 sq cm , b ) 220 sq cm , c ) 440 sq cm , d ) 132 sq cm , e ) 138 sq cm | a | surface_cylinder(5, 3) | surface_cylinder(n0,n1)| | geometry |
mike works at a science lab that conducts experiments on bacteria . the population of the bacteria multiplies at a constant rate , and his job is to notate the population of a certain group of bacteria each hour . at 1 p . m . on a certain day , he noted that the population was 500 and then he left the lab . he returne... | let the rate be x , then population of the bacteria after each hour can be given as 500,500 x , 500 ( x ^ 2 ) , 500 ( x ^ 3 ) now population at 4 pm = 62,500 thus we have 500 ( x ^ 3 ) = 62,500 = 125 thus x = 5 therefore population at 3 pm = 500 ( 25 ) = 12500 answer : b | a ) 13000 , b ) 12500 , c ) 13500 , d ) 14000 , e ) 15000 | b | multiply(power(add(const_2, const_3), 2), 500) | add(const_2,const_3)|power(#0,n4)|multiply(n1,#1) | physics |
the average weight of a group of boys is 20 kg . after a boy of weight 35 kg joins the group , the average weight of the group goes up by 1 kg . find the number of boys in the group originally ? | "let the number off boys in the group originally be x . total weight of the boys = 20 x after the boy weighing 35 kg joins the group , total weight of boys = 20 x + 35 so 20 x + 35 = 21 ( x + 1 ) = > x = 14 . answer : b" | a ) 12 , b ) 14 , c ) 18 , d ) 24 , e ) 10 | b | add(subtract(35, add(20, 1)), 1) | add(n0,n2)|subtract(n1,#0)|add(#1,n2)| | general |
at a certain bowling alley , it costs $ 0.50 to rent bowling shoes for the day and $ 1.75 to bowl 1 game . if a person has $ 12.80 and must rent shoes , what is the greatest number of complete games that person can bowl in one day ? | "after renting bowling shoes the person is left with $ 12.80 - $ 0.5 = $ 12.30 , which is enough for 12.3 / 1.75 = 7.02 - > ~ 7 . answer : a ." | a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | a | divide(subtract(12.80, 0.50), 1.75) | subtract(n3,n0)|divide(#0,n1)| | physics |
the ratio of number of boys and girls in a school is 3 : 5 . if there are 320 students in the school , find the number of girls in the school ? | "let the number of boys and girls be 3 x and 5 x total students = 320 number of girls in the school = 5 * 320 / 8 = 200 answer is d" | a ) 150 , b ) 250 , c ) 300 , d ) 200 , e ) 280 | d | multiply(divide(320, const_3), 5) | divide(n2,const_3)|multiply(n1,#0)| | other |
the arithmetic mean and standard deviation of a certain normal distribution are 14.5 and 1.5 , respectively . what value is exactly 2 standard deviations less than the mean ? | "the value which isexactlytwo sd less than the mean is : mean - 2 * sd = 14.5 - 2 * 1.5 = 11.5 . answer : c ." | a ) 10.5 , b ) 11 , c ) 11.5 , d ) 12 , e ) 12.5 | c | subtract(14.5, multiply(2, 1.5)) | multiply(n1,n2)|subtract(n0,#0)| | general |
a girl sitting in a train which is travelling at 100 kmph observes that a goods train travelling in a opposite direction , takes 6 seconds to pass him . if the goods train is 560 m long , find its speed . | "relative speed = ( 560 / 6 ) m / s = ( 560 / 6 ) * ( 18 / 5 ) = 336 kmph speed of goods train = 336 - 100 = 236 kmph answer is e" | a ) 230 , b ) 232 , c ) 234 , d ) 236 , e ) 238 | e | subtract(divide(divide(560, 6), const_0_2778), 100) | divide(n2,n1)|divide(#0,const_0_2778)|subtract(#1,n0)| | physics |
the average age of 15 students of a class is 15 years . out of these , the average age of 4 students is 14 years and that of the other 9 students is 16 years . tee age of the 15 th student is ? | "age of the 15 th student = [ 15 * 15 - ( 14 * 4 + 16 * 9 ) ] = ( 225 - 200 ) = 25 years . answer : d" | a ) 20 years , b ) 22 years , c ) 24 years , d ) 25 years , e ) 27 years | d | subtract(multiply(15, 15), add(multiply(4, 14), multiply(9, 16))) | multiply(n0,n0)|multiply(n2,n3)|multiply(n4,n5)|add(#1,#2)|subtract(#0,#3)| | general |
a can do a piece of work in 10 days and b alone can do it in 15 days . how much time will both take to finish the work ? | "c 6 days time taken to finish the job = xy / ( x + y ) = 10 x 15 / ( 10 + 15 ) = 150 / 25 = 6 days" | a ) 2 days , b ) 7 days , c ) 6 days , d ) 9 days , e ) 3 days | c | divide(const_1, add(divide(const_1, 10), divide(const_1, 15))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)| | physics |
in a class of 25 students in an examination in maths 5 students scored 95 marks each , 3 get zero each and the average of the rest was 45 . what is the average of the whole class ? | "explanation : total marks obtained by a class of 25 students = 5 * 95 + 3 * 0 + 17 * 45 = 1240 average marks of whole class = 1240 / 25 = 49.6 answer : option e" | a ) a ) 47 , b ) b ) 45.6 , c ) c ) 44 , d ) d ) 48 , e ) e ) 49.6 | e | divide(add(multiply(45, subtract(25, add(3, 5))), multiply(95, 5)), 25) | add(n1,n3)|multiply(n1,n2)|subtract(n0,#0)|multiply(n4,#2)|add(#3,#1)|divide(#4,n0)| | general |
if | 20 x - 10 | = 190 , then find the product of the values of x ? | "| 20 x - 10 | = 190 20 x - 10 = 190 or 20 x - 10 = - 190 20 x = 200 or 20 x = - 180 x = 10 or x = - 9 product = - 9 * 10 = - 90 answer is e" | a ) - 45 , b ) 50 , c ) - 62 , d ) 35 , e ) - 90 | e | subtract(subtract(subtract(190, 10), add(190, 10)), 10) | add(n1,n2)|subtract(n2,n1)|subtract(#1,#0)|subtract(#2,n1)| | general |
how many positive integers between 1 and 400 are there such that they are multiples of 25 ? | "multiples of 25 = 25 , 50,75 , - - - - - 400 number of multiples of 25 = > 25 * 16 = 400 answer is c" | a ) 12 , b ) 13 , c ) 16 , d ) 15 , e ) 11 | c | divide(subtract(400, 1), 25) | subtract(n1,n0)|divide(#0,n2)| | general |
if the range w of the 6 numbers 4 , 314 , 710 and x is 12 , what is the difference between the greatest possible value of x and least possible value of x ? | the range w of a set is the difference between the largest and smallest elements of a set . without x , the difference between the largest and smallest elements of a set is 14 - 3 = 11 < 12 , which means that in order 12 to be the range of the set x must be either the smallest element so that 14 - x = 12 - - - > x = 2 ... | a ) 0 , b ) 2 , c ) 12 , d ) 13 , e ) 15 | d | subtract(add(12, const_3), subtract(add(const_10, 4), 12)) | add(n4,const_3)|add(n1,const_10)|subtract(#1,n4)|subtract(#0,#2) | general |
how many 7 digit number contain number 3 ? | "total 7 digit no . = 9 * 10 * 10 * 10 * 10 * 10 * 10 = 9000000 not containing 3 = 8 * 9 * 9 * 9 * 9 * 9 * 9 = 4251528 total 7 digit number contain 3 = 9000 - 4251528 = 4748472 answer : c" | a ) 2748472 , b ) 3748472 , c ) 4748472 , d ) 5748472 , e ) 6748472 | c | add(subtract(subtract(const_1000, const_10), multiply(multiply(const_10, multiply(7, 7)), multiply(const_4, const_2))), const_10) | multiply(n0,n0)|multiply(const_2,const_4)|subtract(const_1000,const_10)|multiply(#0,const_10)|multiply(#3,#1)|subtract(#2,#4)|add(#5,const_10)| | general |
the 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 18 % , then the cost price of the article is : | "explanation : 110 % of s . p . = 616 s . p . = ( 616 * 100 ) / 110 = rs . 560 c . p = ( 100 * 560 ) / 118 = rs . 474.6 answer : b" | a ) 500 , b ) 474.6 , c ) 222 , d ) 297 , e ) 111 | b | divide(multiply(divide(multiply(616, const_100), add(const_100, 10)), add(const_100, 10)), add(const_100, 18)) | add(n1,const_100)|add(n2,const_100)|multiply(n0,const_100)|divide(#2,#0)|multiply(#0,#3)|divide(#4,#1)| | gain |
how many integers from 10 to 180 , inclusive , are divisible by 3 but not divisible by 7 ? | "we should find # of integers divisible by 3 but not by 3 * 7 = 21 . # of multiples of 21 in the range from 10 to 180 , inclusive is ( 168 - 21 ) / 21 + 1 = 8 ; 57 - 8 = 49 . answer : b ." | a ) 45 , b ) 49 , c ) 50 , d ) 52 , e ) 56 | b | divide(180, const_10) | divide(n1,const_10)| | general |
12 men work 8 hours per day to complete the work in 10 days . to complete the same work in 8 days , working 15 hours a day , the number of men required | "that is , 1 work done = 12 Γ 8 Γ 10 then , 12 8 Γ 10 = ? Γ 15 Γ 8 ? ( i . e . no . of men required ) = 12 Γ 8 Γ 10 / 15 Γ 10 = 8 days . a" | a ) 8 days , b ) 3 days , c ) 7 days , d ) 5 days , e ) 6 days | a | divide(multiply(multiply(12, 10), 8), multiply(8, 15)) | multiply(n0,n2)|multiply(n3,n4)|multiply(n1,#0)|divide(#2,#1)| | physics |
if x is an integer and 2.134 Γ 10 ^ x is less than 220,000 , what is the greatest possible value for x ? | "x is an integer and 2.134 Γ 10 x is less than 220,000 , what is the greatest possible value for x ? for 2.134 Γ 10 x is less than 220,000 to remain true , the greatest number is 213,400 , which makes x = 5 c . 5" | a ) 7 , b ) 6 , c ) 5 , d ) 4 , e ) 3 | c | floor(divide(log(divide(220,000, 2.134)), log(10))) | divide(n2,n0)|log(n1)|log(#0)|divide(#2,#1)|floor(#3)| | general |
calculate 469157 x 9999 = ? | "answer 469157 x 9999 = 469157 x ( 10000 - 1 ) = 4691570000 - 469157 = 4691100843 . option : c" | a ) 4586970843 , b ) 4686970743 , c ) 4691100843 , d ) 4586870843 , e ) none | c | multiply(469157, 9999) | multiply(n0,n1)| | general |
the area of a square field is 400 km 2 . how long will it take for a horse to run around at the speed of 20 km / h ? | explanation area of field = 400 km 2 . then , each side of field = β 400 = 20 km distance covered by the horse = perimeter of square field = 20 Γ 4 = 80 km β΄ time taken by horse = distances / peed = 80 / 20 = 4 h answer c | a ) 12 h , b ) 10 h , c ) 4 h , d ) 6 h , e ) none of these | c | divide(multiply(const_4, sqrt(400)), 20) | sqrt(n0)|multiply(#0,const_4)|divide(#1,n2)| | geometry |
there are 10 dozen mangoes in a box . if there are 36 such boxes , how many mangoes are there in all the boxes together ? | number of mangoes = 12 dozens = 12 Γ 10 = 120 β΄ number of mangoes in 36 boxes = 36 Γ 120 = 4320 answer e | a ) 516 , b ) 3096 , c ) 6192 , d ) 628 , e ) 4320 | e | multiply(multiply(36, 10), add(const_10, const_2)) | add(const_10,const_2)|multiply(n0,n1)|multiply(#0,#1) | general |
if a and b are in the ratio 3 : 4 , and b and c in the ratio 12 : 13 , then a and c will be in the ratio | solution : ( a / b ) * ( b / c ) = ( 3 / 4 ) * ( 12 / 13 ) ; or , a / b = 36 / 39 = 9 : 13 . answer : option b | a ) 3 : 13 , b ) 9 : 13 , c ) 36 : 13 , d ) 13 : 9 , e ) none | b | divide(multiply(3, const_3), 13) | multiply(n0,const_3)|divide(#0,n3) | other |
the sum of a number and the number preceding it is 33 . by how much is two less than 6 times the number ? | two numbers must be 16 and 17 16 + 17 = 33 required number is 17 six times of this number = 6 * 17 = 102 two less than 102 = 102 - 2 = 100 answer : d | a ) 196 , b ) 94 , c ) 90 , d ) 100 , e ) 120 | d | subtract(multiply(subtract(33, multiply(const_4, const_4)), 6), const_2) | multiply(const_4,const_4)|subtract(n0,#0)|multiply(n1,#1)|subtract(#2,const_2) | general |
4,18 , 100,294 , ___ | "2 ^ 3 - 2 ^ 2 = 4 ; 3 ^ 3 - 3 ^ 2 = 18 ; 5 ^ 3 - 5 ^ 2 = 100 ; 7 ^ 3 - 7 ^ 2 = 294 ; so , 11 ^ 3 - 11 ^ 2 = 1210 answer : c" | a ) 1000 , b ) 1100 , c ) 1210 , d ) 1452 , e ) 1552 | c | multiply(divide(4,18, 100,294), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
a bag holds 5 red marbles and 5 green marbles . if you removed two randomly selected marbles from the bag , without replacement , what is the probability that both would be red ? | "probability of selecting first red marble = 5 / 10 probability of selecting second red marble without replacement = 4 / 9 final probability = 5 / 10 * 4 / 9 = 2 / 9 the correct answer is d ." | a ) 1 / 6 , b ) 2 / 5 , c ) 3 / 10 , d ) 2 / 9 , e ) 1 / 2 | d | divide(choose(5, 5), choose(add(const_2.0, 5), const_2)) | add(n0,n1)|choose(const_2.0,n0)|choose(#0,const_2)|divide(#1,#2)| | other |
what is the smallest integer x for which 27 ^ x > 3 ^ 24 ? | "27 ^ x > 3 ^ 24 converting into the same bases : 27 ^ x > 27 ^ 8 therefore for the equation to hold true , x > 8 or x = 9 option c" | a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 12 | c | add(floor(divide(24, const_3)), const_1) | divide(n2,const_3)|floor(#0)|add(#1,const_1)| | general |
a man can row downstream at 16 kmph and upstream at 12 kmph . find the speed of the man in still water . | "let the speed of the man in still water and speed of stream be x kmph and y kmph respectively . given x + y = 16 - - - ( 1 ) and x - y = 12 - - - ( 2 ) from ( 1 ) & ( 2 ) 2 x = 28 = > x = 14 , y = 2 . answer : a" | a ) 14 , b ) 13 , c ) 12 , d ) 16 , e ) 15 | a | divide(divide(add(16, 12), const_2), const_2) | add(n0,n1)|divide(#0,const_2)|divide(#1,const_2)| | physics |
a shopkeeper sold an article offering a discount of 4 % and earned a profit of 38 % . what would have been the percentage of profit earned if no discount was offered ? | "let c . p . be rs . 100 . then , s . p . = rs . 138 let marked price be rs . x . then , 96 / 100 x = 138 x = 13800 / 96 = rs . 143.75 now , s . p . = rs . 143.75 , c . p . = rs . 100 profit % = 44 % . answer : e" | a ) 24 , b ) 28 , c ) 30 , d ) 32 , e ) 44 | e | subtract(divide(multiply(add(const_100, 38), const_100), subtract(const_100, 4)), const_100) | add(n1,const_100)|subtract(const_100,n0)|multiply(#0,const_100)|divide(#2,#1)|subtract(#3,const_100)| | gain |
two trains of length 160 m and 320 m are running towards each other on parallel lines at 42 kmph and 30 kmph respectively . in what time will they be clear of each other from the moment they meet ? | "relative speed = ( 42 + 30 ) * 5 / 18 = 4 * 5 = 20 mps . distance covered in passing each other = 160 + 320 = 480 m . the time required = d / s = 480 / 20 = 24 sec . answer : d" | a ) 10 sec , b ) 32 sec , c ) 82 sec , d ) 24 sec , e ) 89 sec | d | divide(add(160, 320), multiply(add(42, 30), const_0_2778)) | add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)| | physics |
solve ( 0.76 Γ 0.76 Γ 0.76 β 0.008 ) / ( 0.76 Γ 0.76 + 0.76 Γ 0.2 + 0.04 ) | 0.56 option ' a ' | a ) 0.56 , b ) 0.62 , c ) 0.5 , d ) 0.48 , e ) 0.52 | a | divide(subtract(multiply(multiply(0.76, 0.76), 0.76), 0.008), add(add(multiply(0.76, 0.76), multiply(0.76, 0.2)), 0.04)) | multiply(n0,n0)|multiply(n0,n7)|add(#0,#1)|multiply(n0,#0)|add(n8,#2)|subtract(#3,n3)|divide(#5,#4) | general |
what is the maximum area that can be enclosed by a wire 44 cm long ? | perimeter of circle = 2 * 22 / 7 * r = 44 cm from this r = 7 then area of circle = 22 / 7 * r ^ 2 = 154 answer : b | ['a ) 77', 'b ) 154', 'c ) 308', 'd ) 318', 'e ) none of these'] | b | circle_area(divide(divide(44, const_2), const_pi)) | divide(n0,const_2)|divide(#0,const_pi)|circle_area(#1) | physics |
if the average ( arithmetic mean ) of 2 x , 4 x , and 8 x is 140 , what is the value of x ? | "am of 2 x , 4 x and 8 x = 2 x + 4 x + 8 x / 3 = 14 x / 3 given that 14 x / 3 = 140 x = 30 answer is b" | a ) 20 , b ) 30 , c ) 15 , d ) 33 , e ) 31 | b | subtract(multiply(4, const_2), multiply(2, const_2)) | multiply(n1,const_2)|multiply(n0,const_2)|subtract(#0,#1)| | general |
if the probability of rain on any given day in chicago during the summer is 50 % , independent of what happens on any other day , what is the probability of having exactly 3 rainy days from july 4 through july 10 inclusive ? | "one possible case is : rainy - rainy - rainy - not rainy - not rainy - not rainy - not rainy . the probability of this case is ( 1 / 2 ) ^ 7 = 1 / 128 the number of possible cases is 7 c 3 = 35 . p ( exactly 3 rainy days ) = 35 * 1 / 128 = 35 / 128 the answer is c ." | a ) 7 / 32 , b ) 11 / 64 , c ) 35 / 128 , d ) 49 / 128 , e ) 65 / 128 | c | multiply(4, power(divide(const_1, const_2), 4)) | divide(const_1,const_2)|power(#0,n2)|multiply(n2,#1)| | general |
what is the positive difference between the sum of the squares of the first 5 positive integers and the sum of the prime numbers between the first square and fourth square ? | "forget conventional ways of solving math questions . in ps , ivy approach is the easiest and quickest way to find the answer . the sum of the squares of the first 4 positive integers = 1 ^ 2 + 2 ^ 2 + 3 ^ 2 + 4 ^ 2 + 5 ^ 2 = 55 the sum of the prime numbers between the first square ( = 1 ) and fourth square ( = 16 ) = ... | a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | d | subtract(add(add(add(add(add(const_1, power(const_2, const_2)), power(const_3, const_2)), power(const_4, const_2)), power(add(const_4, const_1), const_2)), power(5, const_2)), add(add(add(const_4, const_3), 5), add(add(add(add(const_2, const_3), add(const_4, const_1)), add(const_4, const_3)), add(add(const_2, const_3),... | add(const_1,const_4)|add(const_3,const_4)|add(const_2,const_3)|power(const_2,const_2)|power(const_3,const_2)|power(const_4,const_2)|power(n0,const_2)|add(#3,const_1)|add(n0,#1)|add(#2,#0)|add(n0,#2)|power(#0,const_2)|add(#7,#4)|add(#9,#1)|add(#12,#5)|add(#13,#10)|add(#14,#11)|add(#8,#15)|add(#16,#6)|subtract(#18,#17)| | general |
find the value of 80641 x 9999 = m ? | "80641 x 9999 = 80641 x ( 10000 - 1 ) = 80641 x 10000 - 80641 x 1 = 806410000 - 80641 = 806329359 d" | a ) 807518799 , b ) 806436469 , c ) 807538799 , d ) 806329359 , e ) 817431046 | d | multiply(subtract(9999, const_4), 80641) | subtract(n1,const_4)|multiply(#0,n0)| | general |
3889 + 12.952 β ? = 3854.002 | explanation let 3889 + 12.952 β x = 3854.002 . then x = ( 3889 + 12.952 ) β 3854.002 = 3901.952 β 3854.002 = 47.95 . answer d | a ) 47.095 , b ) 47.752 , c ) 47.932 , d ) 47.95 , e ) none | d | subtract(add(3889, 12.952), 3854.002) | add(n0,n1)|subtract(#0,n2) | general |
huey ' s hip pizza sells two sizes of square pizzas : a small pizza that measures 9 inches on a side and costs $ 10 , and a large pizza that measures 14 inches on a side and costs $ 20 . if two friends go to huey ' s with $ 30 apiece , how many more square inches of pizza can they buy if they pool their money than if t... | in the first case each can buy one pizza of $ 10 and one pizza of $ 20 . in square inches that would be ( 9 * 9 = 81 ) for the small pizza and ( 14 * 14 = 196 ) for the large pizza . in total sq inches that would be ( 81 + 196 ) * 2 = 554 sq inches . in the second case if they pool their money together they can buy 3 l... | ['a ) 5 square inches', 'b ) 10 square inches', 'c ) 15 square inches', 'd ) 25 square inches', 'e ) 34 square inches'] | e | subtract(multiply(power(14, const_2), const_3), add(add(power(9, const_2), power(14, const_2)), add(power(9, const_2), power(14, const_2)))) | power(n2,const_2)|power(n0,const_2)|add(#1,#0)|multiply(#0,const_3)|add(#2,#2)|subtract(#3,#4) | geometry |
which number is a factor of 20 ? | a whole number that divides exactly into another whole number is called a factor 20 / 5 = 4 answer b | a ) 3 , b ) 4 , c ) 7 , d ) 18 , e ) 9 | b | power(const_2, const_2) | power(const_2,const_2) | other |
there are 2 sections a and b in a class , consisting of 24 and 16 students respectively . if the average weight of section a is 40 kg and that of section b is 35 kg , find the average of the whole class ? | "total weight of 36 + 44 students = 24 * 40 + 16 * 35 = 1520 average weight of the class is = 1520 / 40 = 38 kg answer is d" | a ) 35.29 kg , b ) 37.25 kg , c ) 42 kg , d ) 38 kg , e ) 29.78 kg | d | divide(add(multiply(24, 40), multiply(16, 35)), add(24, 16)) | add(n1,n2)|multiply(n1,n3)|multiply(n2,n4)|add(#1,#2)|divide(#3,#0)| | general |
if n divided by 5 has a remainder of 4 , what is the remainder when 3 times n is divided by 5 ? | "as per question = > n = 5 p + 4 for some integer p hence 3 n = > 15 q + 12 but again , 12 can be divided by 5 to get remainder 2 for some integer q hence d" | a ) 1 , b ) 4 , c ) 3 , d ) 2 , e ) 0 | d | multiply(3, 4) | multiply(n1,n2)| | general |
a man can row downstream at 18 kmph and upstream at 10 kmph . find the speed of the man in still water and the speed of stream respectively ? | "let the speed of the man in still water and speed of stream be x kmph and y kmph respectively . given x + y = 18 - - - ( 1 ) and x - y = 10 - - - ( 2 ) from ( 1 ) & ( 2 ) 2 x = 28 = > x = 14 , y = 4 . answer : d" | a ) 7 , b ) 6 , c ) 9 , d ) 4 , e ) 2 | d | divide(divide(add(18, 10), const_2), const_2) | add(n0,n1)|divide(#0,const_2)|divide(#1,const_2)| | physics |
if four machines working at the same rate can do 3 / 4 of a job in 30 minutes , how many minutes would it take two machines working at the same rate to do 3 / 5 of the job ? | "using the std formula m 1 d 1 h 1 / w 1 = m 2 d 2 h 2 / w 2 substituting the values we have 4 * 1 / 2 * 4 / 3 = 2 * 5 / 3 * x ( converted 30 min into hours = 1 / 2 ) 8 / 3 = 10 / 3 * x x = 4 / 5 hour so 48 minutes answer : a" | a ) 48 , b ) 60 , c ) 75 , d ) 80 , e ) 100 | a | divide(divide(3, 5), multiply(divide(divide(divide(3, 4), 5), 30), const_2)) | divide(n0,n4)|divide(n0,n1)|divide(#1,n4)|divide(#2,n2)|multiply(#3,const_2)|divide(#0,#4)| | physics |
the ratio of number of boys and girls in a school is 2 : 5 . if there are 210 students in the school , find the number of girls in the school ? | "let the number of boys and girls be 2 x and 5 x total students = 210 number of girls in the school = 5 * 210 / 7 = 150 answer is a" | a ) 150 , b ) 250 , c ) 300 , d ) 370 , e ) 280 | a | multiply(divide(210, const_3), 5) | divide(n2,const_3)|multiply(n1,#0)| | other |
what is the lowest positive integer that is divisible by each of the integers 1 through 10 inclusive ? | "we have to find the lcm of 1 , 2 , 3 , 2 ^ 2 , 5 , 2 * 3 , 7 , 2 ^ 3 , 3 ^ 2 , and 2 * 5 . the lcm is 1 * 2 ^ 3 * 3 ^ 2 * 5 * 7 = 2520 the answer is d ." | a ) 420 , b ) 840 , c ) 1260 , d ) 2520 , e ) 5020 | d | lcm(1, 10) | lcm(n0,n1)| | general |
there are two positive integers a and b . what is the probability that a + b is odd / | s = adding two numbers is ( even + even ) , ( even + odd ) , ( odd + odd ) , ( odd + even ) n ( s ) = 4 e = ( even + odd ) , ( odd + even ) are the points in the event . n ( e ) = 2 p ( e ) = n ( e ) / n ( s ) = 2 / 4 = 1 / 2 answer is option c | a ) 3 / 2 , b ) 6 / 7 , c ) 1 / 2 , d ) 7 / 2 , e ) 4 / 5 | c | multiply(divide(const_2, const_4), const_100) | divide(const_2,const_4)|multiply(#0,const_100) | general |
of the 150 employees at company x , 60 are full - time , and 100 have worked at company x for at least a year . there are 20 employees at company x who aren β t full - time and haven β t worked at company x for at least a year . how many full - time employees of company x have worked at the company for at least a year ... | "full time employee who have not worked for at least one year = a full time employee who have worked for at least one year = b non full time employee who have worked for at least one year = c non full time employee who have not worked for at least one year = d a + b + c + d = 150 a + b = 80 i . e . c + d = 70 b + c = 1... | a ) 20 , b ) 30 , c ) 50 , d ) 80 , e ) 100 | d | subtract(subtract(150, 60), 20) | subtract(n0,n1)|subtract(#0,n3)| | general |
if f is the smallest positive integer such that 3,150 multiplied by f is the square of an integer , then f must be | solution : this problem is testing us on the rule that when we express a perfect square by its unique prime factors , every prime factor ' s exponent is an even number . let β s start by prime factorizing 3,150 . 3,150 = 315 x 10 = 5 x 63 x 10 = 5 x 7 x 3 x 3 x 5 x 2 3,150 = 2 ^ 1 x 3 ^ 2 x 5 ^ 2 x 7 ^ 1 ( notice that ... | ['a ) 2', 'b ) 5', 'c ) 6', 'd ) 7', 'e ) 14'] | e | add(const_10, const_4) | add(const_10,const_4) | geometry |
what is the least number . which should be added to 0.0355 to make it a perfect square ? | "0.0355 + 0.0006 = 0.0361 ( 0.19 ) ^ 2 answer : e" | a ) 0.0005 , b ) 0.0016 , c ) 0.0056 , d ) 0.0066 , e ) 0.0006 | e | subtract(multiply(divide(add(add(const_12, const_4), const_2), const_100), divide(add(add(const_12, const_4), const_2), const_100)), 0.0355) | add(const_12,const_4)|add(#0,const_2)|divide(#1,const_100)|multiply(#2,#2)|subtract(#3,n0)| | general |
if x + 4 y = 5 and 5 x + 6 y = 7 , then 3 x + 5 y = ? | add the left terms and right terms of the two equation to obtain a new equation x + 5 x + 4 y + 6 y = 5 + 7 group and simplify 6 x + 10 y = 12 divide all terms of the above equation by 2 to obtain a new equation 3 x + 5 y = 6 correct answer a | a ) 6 , b ) 5 , c ) 4 , d ) 3 , e ) 2 | a | add(multiply(5, divide(subtract(multiply(5, 5), 7), subtract(multiply(5, 4), 6))), multiply(subtract(5, multiply(divide(subtract(multiply(5, 5), 7), subtract(multiply(5, 4), 6)), 4)), 3)) | multiply(n1,n1)|multiply(n0,n1)|subtract(#0,n4)|subtract(#1,n3)|divide(#2,#3)|multiply(n1,#4)|multiply(n0,#4)|subtract(n1,#6)|multiply(n5,#7)|add(#5,#8) | general |
the distance from steve ' s house to work is 10 km . on the way back steve drives twice as fast as he did on the way to work . altogether , steve is spending 6 hours a day on the roads . what is steve ' s speed on the way back from work ? | "time is in the ratio 2 : 1 : : to : fro office therefore , 2 x + 1 x = 6 hrs time take to come back - 2 hrs , distance travelled - 10 km = > speed = 5 kmph a" | a ) 5 . , b ) 10 . , c ) 14 . , d ) 15 . , e ) 20 . | a | multiply(divide(add(10, divide(10, const_2)), 6), const_2) | divide(n0,const_2)|add(n0,#0)|divide(#1,n1)|multiply(#2,const_2)| | physics |
300 first - time customers of a fashion store were surveyed for their shopping experience right after leaving the store . 60 % of the customers in the survey had purchased clothes for less than $ 100 . 40 % of the customers in the survey reported they were overall satisfied with their purchase . 35 % of the customers t... | out of 300 - 180 purchased for less than 100 $ 120 for more out of 300 - 120 responded as satisfied and 180 responded disatisfied out of 180 ( purchased less than 100 $ ) - 35 % = 63 responded as satisfied , so remaining satisfied are 120 - 63 = 57 so 57 is what percentage of 300 - 19 % so the answer should be a | a ) 19 , b ) 25 , c ) 35 , d ) 45 , e ) 75 | a | multiply(divide(subtract(multiply(divide(40, 100), 300), multiply(divide(35, 100), multiply(300, divide(60, 100)))), 300), const_100) | divide(n3,n2)|divide(n4,n2)|divide(n1,n2)|multiply(n0,#0)|multiply(n0,#2)|multiply(#1,#4)|subtract(#3,#5)|divide(#6,n0)|multiply(#7,const_100) | general |
a work can be finished in 30 days by 30 women . the same work can be finished in fifteen days by 15 men . the ratio between the capacity of a man and a woman is | "work done by 30 women in 1 day = 1 / 30 work done by 1 woman in 1 day = 1 / ( 30 Γ 30 ) work done by 15 men in 1 day = 1 / 15 work done by 1 man in 1 day = 1 / ( 15 Γ 15 ) ratio of the capacity of a man and woman = 1 / ( 15 Γ 15 ) : 1 / ( 30 Γ 30 ) ) = 1 / 225 : 1 / 900 = 1 / 1 : 1 / 4 = 4 : 1 option d" | a ) 1 : 3 , b ) 4 : 3 , c ) 2 : 3 , d ) 4 : 1 , e ) 2 : 4 | d | divide(divide(const_1, multiply(add(30, const_2), const_10)), divide(const_1, multiply(30, const_10))) | add(n0,const_2)|multiply(n0,const_10)|divide(const_1,#1)|multiply(#0,const_10)|divide(const_1,#3)|divide(#4,#2)| | physics |
if there is an equal probability of a child being born a boy or a girl , what is the probability that a couple who have 8 children have two children of the same sex and one of the opposite sex ? | "no of ways of selecting a gender - 2 c 1 no of ways of selecting any 2 children out of 8 = 8 c 2 total possible outcomes - 2 ^ 8 ( each child can be either a girl or a boy ) probability = 2 c 1 * 8 c 2 / 2 ^ 8 = 2 * 8 / 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 = 16 / 256 = 1 / 16 ans = d" | a ) 1 / 3 , b ) 2 / 3 , c ) 1 / 4 , d ) 1 / 16 , e ) 3 / 5 | d | divide(const_2, 8) | divide(const_2,n0)| | general |
? % of 360 = 126 | "? % of 360 = 126 or , ? = 126 Γ 100 / 360 = 35 answer b" | a ) 277 , b ) 35 , c ) 64 , d ) 72 , e ) none of these | b | divide(multiply(126, const_100), 360) | multiply(n1,const_100)|divide(#0,n0)| | gain |
the area of a circle of radius 5 is numerically what percent of its circumference ? | solution required % = [ Ο x ( 5 ) 2 / 2 Ο x 5 x 100 ] % βΉ = βΊ 250 % . answer d | ['a ) 200', 'b ) 225', 'c ) 240', 'd ) 250', 'e ) none'] | d | multiply(const_100, divide(circle_area(5), circumface(5))) | circle_area(n0)|circumface(n0)|divide(#0,#1)|multiply(#2,const_100) | geometry |
jony walks along sunrise boulevard daily . he starts walking at 07 : 00 from block 10 and walks to block 90 where he turns around and walks back to block 70 , where he stops at 07 : 40 . the blocks along the boulevard are numbered sequentially ( 1 , 2,3 ) , and each block measures 40 meters . what is jony ' s speed in ... | total distance from 10 to 90 = 80 + from 90 to 70 = 20 so the dist is 100 Γ 40 ( per block dist ) speed = 4000 mts / 40 min = 100 m / min b is the answer | a ) 88 , b ) 100 , c ) 198 , d ) 216 , e ) 252 | b | divide(add(multiply(subtract(90, 10), 40), multiply(subtract(90, 70), 40)), 40) | subtract(n3,n2)|subtract(n3,n4)|multiply(n6,#0)|multiply(n6,#1)|add(#2,#3)|divide(#4,n6) | physics |
car a is 40 miles behind car b , which is traveling in the same direction along the same route as car a . car a is traveling at a constant speed of 58 miles per hour and car bis traveling at a constant speed of 50 miles per hour . how many hours will it take for car a to overtake and drive 8 miles ahead of car b ? | "relative speed of car a is 58 - 50 = 8 miles per hour , to catch up 40 miles and drive 8 miles ahead so to drive 48 miles it ' ll need 48 / 8 = 6 hours . answer : b" | a ) 5 , b ) 6 , c ) 7 , d ) 5.5 , e ) 6.5 | b | divide(add(40, 8), subtract(58, 50)) | add(n0,n3)|subtract(n1,n2)|divide(#0,#1)| | physics |
in the xy - coordinate plane , the graph of y = - x ^ 2 + 9 intersects line l at ( p , 5 ) and ( t , - 8 ) . what is the least possible value of the slope of line l ? | "we need to find out the value of p and l to get to the slope . line l and graph y intersect at point ( p , 5 ) . hence , x = p and y = 5 should sactisfy the graph . soliving 5 = - p 2 + 9 p 2 = 4 p = + or - 2 simillarly point ( t , - 8 ) should satisfy the equation . hence x = t and y = - 8 . - 7 = - t 2 + 9 t = + or ... | a ) - 6.5 , b ) 2 , c ) - 2 , d ) - 6 , e ) - 10 | a | divide(subtract(8, 5), subtract(sqrt(add(9, 8)), sqrt(add(5, 9)))) | add(n1,n3)|add(n1,n2)|subtract(n3,n2)|sqrt(#0)|sqrt(#1)|subtract(#3,#4)|divide(#2,#5)| | general |
two fill taps a and b can separately fill a cistern in 45 and 40 minutes respectively . they started to fill a cistern together but tap a is turned off after few minutes and tap b fills the rest part of cistern in 23 minutes . after how many minutes , was tap a turned - off ? | let a was turned off after x min . then , cistern filled by a in x min + cistern filled by b in ( x + 23 ) min = 1 β x / 45 + x + 23 / 40 = 1 β 17 x + 207 = 360 β x = 9 min . answer a | a ) 9 min , b ) 10 min , c ) 12 min , d ) 7 min , e ) none of these | a | divide(subtract(const_1, multiply(speed(const_1, 40), 23)), add(speed(const_1, 45), speed(const_1, 40))) | speed(const_1,n1)|speed(const_1,n0)|add(#1,#0)|multiply(n2,#0)|subtract(const_1,#3)|divide(#4,#2) | physics |
find the compound interest on $ 1200 for 4 years at 20 % p . a . if ci is component yearly ? | "a = p ( 1 + r / 100 ) ^ t = 1200 ( 1 + 20 / 100 ) ^ 4 = $ 2488 ci = $ 1288 answer is c" | a ) $ 120 , b ) $ 150 , c ) $ 1288 , d ) $ 250 , e ) $ 300 | c | subtract(multiply(1200, power(add(const_1, divide(20, const_100)), 4)), 1200) | divide(n2,const_100)|add(#0,const_1)|power(#1,n1)|multiply(n0,#2)|subtract(#3,n0)| | gain |
in a company 10 % of male staff are same in number as 1 / 4 th of the female staff . what is the ratio of male staff to female staff | 10 % of ms = 1 / 4 th of fs - > 10 ms / 100 = 1 / 4 fs - > ms = 5 / 2 fs : . ms / fs = 5 / 2 = ms : fs = 5 : 2 answer : b | a ) 3 : 2 , b ) 5 : 2 , c ) 2 : 1 , d ) 4 : 3 , e ) 1 : 2 | b | divide(divide(1, 4), divide(10, const_100)) | divide(n1,n2)|divide(n0,const_100)|divide(#0,#1) | other |
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