Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
values |
|---|---|---|---|---|---|---|
the average of first 6 natural numbers is ? | "sum of 6 natural no . = 42 / 2 = 21 average = 21 / 6 = 3.5 answer : b" | a ) 5.2 , b ) 3.5 , c ) 5.3 , d ) 5.9 , e ) 5.1 | b | add(6, const_1) | add(n0,const_1)| | general |
a part - time employee ’ s hourly wage was increased by 20 % . she decided to decrease the number of hours worked per week so that her total income did not change . by approximately what percent should the number of hours worked be decreased ? | "let ' s plug in somenicenumbers and see what ' s needed . let ' s say the employee used to make $ 1 / hour and worked 100 hours / week so , the total weekly income was $ 100 / week after the 20 % wage increase , the employee makes $ 1.20 / hour we want the employee ' s income to remain at $ 100 / week . so , we want (... | a ) 9 % , b ) 17 % , c ) 25 % , d ) 50 % , e ) 100 % | b | multiply(divide(divide(20, const_100), divide(add(20, const_100), const_100)), const_100) | add(n0,const_100)|divide(n0,const_100)|divide(#0,const_100)|divide(#1,#2)|multiply(#3,const_100)| | general |
the ratio of three numbers is 1 : 2 : 4 and the sum of their squares is 4725 . the sum of the numbers is ? | let the numbers be x , 2 x , 4 x then , x ^ 2 + 4 x ^ 2 + 16 x ^ 2 = 4725 21 x ^ 2 = 4725 x ^ 2 = 225 x = 15 answer is c | ['a ) a ) 10', 'b ) b ) 12', 'c ) c ) 15', 'd ) d ) 14', 'e ) e ) 9'] | c | sqrt(divide(4725, add(power(4, const_2), add(power(1, const_2), power(2, const_2))))) | power(n0,const_2)|power(n1,const_2)|power(n2,const_2)|add(#0,#1)|add(#3,#2)|divide(n3,#4)|sqrt(#5) | other |
madhav ranks seventeenth in a class of thirtyone . what is his rank from the last ? | ( rank from top ) + ( rank from bottom ) - 1 = total students 17 + x - 1 = 31 x = 15 answer : c | a ) 13 , b ) 14 , c ) 15 , d ) 16 , e ) 17 | c | add(subtract(add(add(multiply(const_3, const_4), add(const_1, add(multiply(const_3, const_4), const_4))), const_2), add(const_1, add(multiply(const_3, const_4), const_4))), const_1) | multiply(const_3,const_4)|add(#0,const_4)|add(#1,const_1)|add(#2,#0)|add(#3,const_2)|subtract(#4,#2)|add(#5,const_1) | other |
what positive number , when squared , is equal to the cube of the positive square root of 17 ? | let the positive number be x x ^ 2 = ( ( 17 ) ^ ( 1 / 2 ) ) ^ 3 = > x ^ 2 = 4 ^ 3 = 64 = > x = 8 answer c | ['a ) 64', 'b ) 32', 'c ) 8', 'd ) 4', 'e ) 2'] | c | sqrt(power(power(17, divide(const_1, const_2)), const_3)) | divide(const_1,const_2)|power(n0,#0)|power(#1,const_3)|sqrt(#2) | geometry |
the cost of an article was rs . 75 . the cost was first increased by 20 % and later on it was reduced by 20 % . the present cost of the article is : | solution : initial cost = rs . 75 after 20 % increase in the cost , it becomes , ( 75 + 20 % of 75 ) = rs . 90 now , cost is decreased by 20 % , so cost will become , ( 90 - 20 % of 90 ) = rs . 72 . so , present cost is rs . 72 . mind calculation method : 75 - - - - - 20 % ↑ - - → 90 - - - - - 20 % ↓ - - - - - → 72 . a... | a ) rs . 72 , b ) rs . 60 , c ) rs . 75 , d ) rs . 76 , e ) none | a | multiply(multiply(75, add(const_1, divide(20, const_100))), subtract(const_1, divide(20, const_100))) | divide(n1,const_100)|add(#0,const_1)|subtract(const_1,#0)|multiply(n0,#1)|multiply(#3,#2) | gain |
two trains 111 meters and 165 meters in length respectively are running in opposite directions , one at the rate of 100 km and the other at the rate of 120 kmph . in what time will they be completely clear of each other from the moment they meet ? | "t = ( 111 + 165 ) / ( 100 + 120 ) * 18 / 5 t = 4.51 answer : c" | a ) 4.85 , b ) 7.85 , c ) 4.51 , d ) 5.85 , e ) 6.15 | c | divide(add(111, 165), multiply(add(100, 120), const_0_2778)) | add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)| | physics |
in how many years will a sum of money be doubled given that the annual interest on it is 10 % ? | p = ( p * 10 * r ) / 100 r = 10 % answer : e | a ) 11 , b ) 14 , c ) 13 , d ) 15 , e ) 10 | e | divide(const_100, 10) | divide(const_100,n0) | gain |
rectangle a has sides a and b , and rectangle b has sides c and d . if a / c = b / d = 2 / 5 , what is the ratio of rectangle a ’ s area to rectangle b ’ s area ? | "the area of rectangle a is ab . c = 5 a / 2 and d = 5 b / 2 . the area of rectangle b is cd = 25 ab / 4 . the ratio of rectangle a ' s area to rectangle b ' s area is ab / ( 25 ab / 4 ) = 4 / 25 . the answer is e ." | a ) 2 / 5 , b ) 2 / 25 , c ) 5 / 2 , d ) 5 / 4 , e ) 4 / 25 | e | divide(power(2, const_2), power(5, const_2)) | power(n0,const_2)|power(n1,const_2)|divide(#0,#1)| | geometry |
if a boat is rowed downstream for 600 km in 20 hours and upstream for 48 km in 4 hours , what is the speed of the boat and the river ? | "explanation : if x : speed of boats man in still water y : speed of the river downstream speed ( ds ) = x + y upstream speed ( us ) = x â € “ y x = ( ds + us ) / 2 y = ( ds â € “ us ) / 2 in the above problem ds = 30 ; us = 12 x = ( 30 + 12 ) / 2 = 42 / 2 = 21 km / hr y = ( 30 - 12 ) / 2 = 18 / 2 = 9 km / hr answer : ... | a ) 21 , 21 , b ) 9 , 21 , c ) 21 , 9 , d ) 9 , 9 , e ) 20 , 8 | c | divide(add(divide(48, 4), divide(600, 20)), const_2) | divide(n2,n3)|divide(n0,n1)|add(#0,#1)|divide(#2,const_2)| | physics |
a 1200 m long train crosses a tree in 120 sec , how much time will i take to pass a platform 700 m long ? | "l = s * t s = 1200 / 120 s = 10 m / sec . total length ( d ) = 1900 m t = d / s t = 1900 / 10 t = 190 sec answer : b" | a ) 333 sec , b ) 190 sec , c ) 176 sec , d ) 2687 sec , e ) 1876 sec | b | divide(add(1200, 700), divide(1200, 120)) | add(n0,n2)|divide(n0,n1)|divide(#0,#1)| | physics |
each of the integers from 0 to 9 , inclusive , is written on a separate slip of blank paper and the ten slips are dropped into a hat . if 4 of the slips are the drawn , without replacement , what is the probability that all 4 have a even number written on it ? | key is that there is no replacement , so each successive choice will become more skewed towards picking a neg ( i . e . the pool of positives decreases , while the pool of negatives stay the same ) p ( + on 1 st pick ) = 5 / 10 p ( + on 2 nd pick ) = 4 / 9 p ( + on 3 rd pick ) = 3 / 8 p ( + on 4 rd pick ) = 2 / 7 5 / 1... | a ) 1 / 12 , b ) 1 / 10 , c ) 1 / 8 , d ) 1 / 42 , e ) 5 / 9 | d | multiply(multiply(multiply(divide(add(const_4, const_1), add(9, const_1)), divide(subtract(add(const_4, const_1), const_1), subtract(add(9, const_1), const_1))), divide(subtract(subtract(add(const_4, const_1), const_1), const_1), subtract(subtract(add(9, const_1), const_1), const_1))), divide(subtract(subtract(subtract... | add(const_1,const_4)|add(n1,const_1)|divide(#0,#1)|subtract(#0,const_1)|subtract(#1,const_1)|divide(#3,#4)|subtract(#3,const_1)|subtract(#4,const_1)|divide(#6,#7)|multiply(#2,#5)|subtract(#6,const_1)|subtract(#7,const_1)|divide(#10,#11)|multiply(#8,#9)|multiply(#12,#13) | gain |
two trains start at same time from two stations and proceed towards each other at the rate of 20 km / hr and 25 km / hr respectively . when they meet , it is found that one train has traveled 70 km more than the other . what is the distance between the two stations ? | "explanation : let us assume that trains meet after ' x ' hours distance = speed * time distance traveled by two trains = 20 x km and 25 x km resp . as one train travels 70 km more than the other , 25 x â € “ 20 x = 70 5 x = 70 x = 14 hours as the two trains are moving towards each other , relative speed = 20 + 25 = 45... | a ) 630 km , b ) 540 km , c ) 276 km , d ) 178 km , e ) 176 km | a | multiply(add(20, 25), divide(70, subtract(25, 20))) | add(n0,n1)|subtract(n1,n0)|divide(n2,#1)|multiply(#0,#2)| | general |
the average age of 10 men increases by 2 years when two women are included in place of two men of ages 10 and 12 years . find the average age of the women ? | "10 + 12 + 10 * 2 = 42 / 2 = 21 answer : a" | a ) 21 , b ) 32 , c ) 30 , d ) 41 , e ) 37 | a | divide(add(add(10, 12), multiply(10, 2)), const_2) | add(n2,n3)|multiply(n0,n1)|add(#0,#1)|divide(#2,const_2)| | general |
a , b , c subscribe rs . 50000 for a business , a subscribes rs . 4000 more than b and b rs . 5000 more than c . out of a total profit of rs . 35000 , c receives : | explanation : 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 * 12 / 50 ) = rs . 8,400 . answer : a ) | a ) 8400 , b ) 14703 , c ) 37798 , d ) 13777 , e ) 14778 | a | multiply(35000, divide(divide(subtract(50000, add(add(4000, 5000), 5000)), const_3), 50000)) | add(n1,n2)|add(n2,#0)|subtract(n0,#1)|divide(#2,const_3)|divide(#3,n0)|multiply(n3,#4) | general |
due to construction , the speed limit along an 6 - mile section of highway is reduced from 55 miles per hour to 35 miles per hour . approximately how many minutes more will it take to travel along this section of highway at the new speed limit than it would have taken at the old speed limit ? | 6 / 35 - 6 / 55 = 6 / 5 * ( 11 - 7 ) / 77 = 6 / 5 * 4 / 77 * 60 min = 6 * 12 * 4 / 77 = 288 / 77 ~ 3.7 answer - a | a ) 4 , b ) 8 , c ) 10 , d ) 15 , e ) 24 | a | max(multiply(subtract(add(55, 6), const_1), subtract(divide(6, 35), divide(6, 55))), const_4) | add(n0,n1)|divide(n0,n2)|divide(n0,n1)|subtract(#0,const_1)|subtract(#1,#2)|multiply(#3,#4)|max(#5,const_4) | physics |
a , b and c invested rs . 600 , rs . 800 and rs . 1200 respectively , in a partnership business . find the share of b in profit of rs . 900 after a year ? | "explanation : 600 : 800 : 1200 3 : 4 : 6 4 / 13 * 900 = 276.92 answer : a" | a ) 276.92 , b ) 267.92 , c ) 266.27 , d ) 900 , e ) 237.8 | a | multiply(divide(600, add(add(600, 800), 1200)), 900) | add(n0,n1)|add(n2,#0)|divide(n0,#1)|multiply(n3,#2)| | gain |
a driver would have reduced the time it took to drive from home to the store by 1 / 3 if the average speed had been increased by 21 miles per hour . what was the actual average speed , in miles per hour , when the driver drove from home to the store ? | "let r be the original speed and let t be the original time . since the distance remains the same ( we ' re just changing the rate and time ) , any increase in rate or time is met with a decrease in the other term . decreasing the time by 1 / 3 would give us : d = ( r ) ( t ) = ( 2 t / 3 ) ( x * r ) x = 3 / 2 since ( 2... | a ) 36 , b ) 42 , c ) 45 , d ) 48 , e ) 51 | b | divide(multiply(subtract(1, divide(1, 3)), 21), subtract(1, subtract(1, divide(1, 3)))) | divide(n0,n1)|subtract(n0,#0)|multiply(#1,n2)|subtract(n0,#1)|divide(#2,#3)| | general |
p works 25 % more efficiently than q and q works 50 % more efficiently than r . to complete a certain project , p alone takes 50 days less than q alone . if , in this project p alone works for 60 days and then q alone works for 135 days , in how many days can r alone complete the remaining work ? | "p works 25 % more efficiently than q : something that takes q 5 days , takes p 4 days q works 50 % more efficiently than r : something that takes r 7.5 days , takes q 5 days p alone takes 50 days less than q : for every 4 days p works , q has to work an extra day . hence p alone can do it in 200 days and q alone in 25... | a ) 50 days , b ) 92.4 days , c ) 100 days , d ) 150 days , e ) 80 days | b | subtract(multiply(const_4, 50), multiply(divide(135, const_100), 60)) | divide(n4,const_100)|multiply(n1,const_4)|multiply(n3,#0)|subtract(#1,#2)| | physics |
find a number such that when 15 is subtracted from 7 times the number , the result is 10 more than twice the number . | let the number be x . then , 7 x - 15 = 2 x + 10 = > 5 x = 25 = > x = 5 . hence , the required number is 5 . answer is c . | a ) 1 , b ) 4 , c ) 5 , d ) 0 , e ) 3 | c | divide(add(15, 10), subtract(7, const_2)) | add(n0,n2)|subtract(n1,const_2)|divide(#0,#1) | general |
harkamal purchased 8 kg of grapes at the rate of 70 per kg and 9 kg of mangoes at the rate of 45 per kg . how much amount did he pay to the shopkeeper ? | "cost of 8 kg grapes = 70 × 8 = 560 . cost of 9 kg of mangoes = 45 × 9 = 405 . total cost he has to pay = 560 + 405 = 965 . c )" | a ) a ) 1000 , b ) b ) 1055 , c ) c ) 965 , d ) d ) 1075 , e ) e ) 1080 | c | add(multiply(8, 70), multiply(9, 45)) | multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)| | gain |
the average monthly income of a and b is rs . 4050 . the average monthly income of b and c is rs . 5250 and the average monthly income of a and c is rs . 4200 . what is the monthly income of a ? | let monthly income of a = x monthly income of b = y monthly income of c = z x + y = 2 × 4050 . . . . ( equation 1 ) y + z = 2 × 5250 . . . . ( equation 2 ) z + x = 2 × 4200 . . . . ( equation 3 ) ( equation 1 ) + ( equation 3 ) - ( equation 2 ) = > x + y + x + z - ( y + z ) = ( 2 × 4050 ) + ( 2 × 4200 ) - ( 2 × 5250 ) ... | a ) 2000 , b ) 2500 , c ) 3000 , d ) 3500 , e ) 4000 | c | subtract(multiply(const_2, 4050), divide(add(multiply(const_2, 4050), subtract(multiply(const_2, 5250), multiply(const_2, 4200))), const_2)) | multiply(n0,const_2)|multiply(n1,const_2)|multiply(n2,const_2)|subtract(#1,#2)|add(#0,#3)|divide(#4,const_2)|subtract(#0,#5) | general |
a merchant marks his goods up by 40 % and then offers a discount of 15 % on the marked price . what % profit does the merchant make after the discount ? | let the original price be 100 . after 40 % markup , price = 140 after 15 % discount on this marked - up price , price = 140 - ( 15 % of 140 ) = 140 - 21 = 119 final profit = ( ( 119 - 100 ) / 100 ) * 100 = 19 % option e | a ) 21 % , b ) 25 % , c ) 69 % , d ) 31 % , e ) 19 % | e | subtract(subtract(add(40, const_100), divide(multiply(add(40, const_100), 15), const_100)), const_100) | add(n0,const_100)|multiply(n1,#0)|divide(#1,const_100)|subtract(#0,#2)|subtract(#3,const_100) | gain |
a rectangular grassy plot 100 m . by 70 m has a gravel path 2.5 m wide all round it on the inside . find the cost of gravelling the path at 90 paise per sq . metre . | "explanation : area of the plot = ( 100 x 70 ) m 2 = 7000 m 2 area of the plot excluding the path = [ ( 100 - 5 ) * ( 70 - 5 ) ] m 2 = 6175 m 2 . area of the path = ( 7000 - 6175 ) m 2 = 825 m 2 . cost of gravelling the path = rs . 825 * ( 90 / 100 ) = rs . 742.50 answer : option d" | a ) rs . 700 , b ) rs . 708.50 , c ) rs . 732.50 , d ) rs . 742.50 , e ) none of these | d | divide(multiply(subtract(multiply(100, 70), multiply(subtract(100, multiply(2.5, const_2)), subtract(70, multiply(2.5, const_2)))), 90), const_100) | multiply(n0,n1)|multiply(n2,const_2)|subtract(n0,#1)|subtract(n1,#1)|multiply(#2,#3)|subtract(#0,#4)|multiply(n3,#5)|divide(#6,const_100)| | physics |
a , b and c play a cricket match . the ratio of the runs scored by them in the match is a : b = 4 : 3 and b : c = 4 : 5 . if the total runs scored by all of them are 75 , the runs scored by b are ? | "a : b = 4 : 3 b : c = 4 : 5 a : b : c = 16 : 12 : 15 12 / 43 * 75 = 20.93 answer : d" | a ) 20.23 , b ) 20.13 , c ) 30.93 , d ) 20.93 , e ) 10.93 | d | multiply(divide(75, add(add(divide(4, 3), divide(5, 4)), const_1)), 5) | divide(n0,n1)|divide(n3,n0)|add(#0,#1)|add(#2,const_1)|divide(n4,#3)|multiply(n3,#4)| | general |
the length of a room is 5.5 m and width is 3.75 m . find the cost of paying the floor by slabs at the rate of rs . 800 per sq . metre . | "area = 5.5 × 3.75 sq . metre . cost for 1 sq . metre . = rs . 800 hence total cost = 5.5 × 3.75 × 800 = 5.5 × 3000 = rs . 16500 answer : b" | a ) rs . 15500 , b ) rs . 16500 , c ) rs . 17500 , d ) rs . 18500 , e ) rs . 19500 | b | multiply(800, multiply(5.5, 3.75)) | multiply(n0,n1)|multiply(n2,#0)| | physics |
the ages of 2 persons differ by 38 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 + 38 6 ( x - 12 ) = x + 38 - 12 x = 19.6 age of elder person = 19.6 + 38 = 57.6 answer is e" | a ) 30.5 , b ) 48.5 , c ) 50.4 , d ) 62.6 , e ) 57.6 | e | subtract(add(divide(multiply(2, 38), subtract(38, const_1)), 38), 2) | multiply(n0,n1)|subtract(n1,const_1)|divide(#0,#1)|add(n1,#2)|subtract(#3,n0)| | general |
the product x of two prime numbers is between 15 and 70 . if one of the prime numbers is greater than 2 but less than 6 and the other is greater than 13 but less than 25 , then x = | option bc can be ruled out as they themselves are prime numbers 18 = 2 * 9 = 3 * 6 > > ignore 44 = 2 * 22 = 4 * 11 > > ignore 69 = 3 * 23 > > answer answer = e | a ) 18 , b ) 29 , c ) 37 , d ) 44 , e ) 69 | e | multiply(add(2, const_1), subtract(25, 2)) | add(n2,const_1)|subtract(n5,n2)|multiply(#0,#1) | general |
we define that k @ j is the product of j number from k in increasing order for positive integers k , j . for example , 6 @ 4 = 6 * 7 * 8 * 9 . if a = 2020 and b = 2120 , what is the value r of the ratio a / b ? | "r - > a / b = 20 * 21 * … … * 39 / 21 * 22 * … . * 39 * 40 = 20 / 40 = 1 / 2 . therefore , the answer is a ." | a ) 1 / 2 , b ) 1 / 3 , c ) 2 / 3 , d ) 1 / 4 , e ) 1 / 5 | a | divide(divide(2020, 2020), add(divide(2020, 2020), divide(2020, 2020))) | divide(n6,n6)|add(#0,#0)|divide(#0,#1)| | general |
a candidate got 40 % of the votes polled and he lost to his rival by 5000 votes . how many votes were cast ? | "40 % - - - - - - - - - - - l 60 % - - - - - - - - - - - w - - - - - - - - - - - - - - - - - - 20 % - - - - - - - - - - 5000 100 % - - - - - - - - - ? = > 25000 answer : a" | a ) 25000 , b ) 2028 , c ) 2775 , d ) 5496 , e ) 6851 | a | divide(5000, subtract(subtract(const_1, divide(40, const_100)), divide(40, const_100))) | divide(n0,const_100)|subtract(const_1,#0)|subtract(#1,#0)|divide(n1,#2)| | gain |
two employees a and b are paid a total of rs . 560 per week by their employer . if a is paid 150 percent of the sum paid to b , how much is b paid per week ? | "let the amount paid to a per week = x and the amount paid to b per week = y then x + y = 560 but x = 150 % of y = 150 y / 100 = 15 y / 10 ∴ 15 y / 10 + y = 560 ⇒ y [ 15 / 10 + 1 ] = 560 ⇒ 25 y / 10 = 560 ⇒ 25 y = 5600 ⇒ y = 5600 / 25 = rs . 224 d )" | a ) s . 130 , b ) s . 140 , c ) s . 150 , d ) s . 224 , e ) s . 280 | d | divide(560, add(divide(150, const_100), const_1)) | divide(n1,const_100)|add(#0,const_1)|divide(n0,#1)| | general |
what is the units digit of 17 ^ 83 × 13 ^ 82 × 11 ^ 87 ? | "cyclicity of 7 is 4 - - > 7 , 9 , 3 , 1 cyclicity of 3 is 4 - - > 3 , 9 , 7 , 1 11 raised to any positive power has a units digit 1 17 ^ 83 × 13 ^ 82 × 11 ^ 87 = units digit of 3 * 9 * 1 = 7 answer : d" | a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | d | divide(add(multiply(factorial(17), factorial(83)), multiply(factorial(17), factorial(82))), 17) | factorial(n0)|factorial(n1)|factorial(n3)|multiply(#0,#1)|multiply(#0,#2)|add(#3,#4)|divide(#5,n0)| | general |
find the lowest common multiple of 10 , 14 and 20 . | "lcm = 2 * 2 * 5 * 7 = 140 . answer is e" | a ) 360 , b ) 420 , c ) 510 , d ) 320 , e ) 140 | e | lcm(lcm(10, 14), 20) | lcm(n0,n1)|lcm(n2,#0)| | general |
3034 - ( 1002 / 200.4 ) = ? | "3034 - 5 = 3029 answer : b" | a ) 2984 , b ) 3029 , c ) 2982 , d ) 2981 , e ) none of these | b | subtract(3034, divide(1002, 200.4)) | divide(n1,n2)|subtract(n0,#0)| | general |
if 2 x = 3 y = 10 , then 6 xy = ? | "2 x = 10 ; x = 5 3 y = 10 ; y = 10 / 3 multiply : 6 xy = 6 * 5 * 10 / 3 = 100 answer : a ." | a ) 100 , b ) 200 , c ) 120 , d ) 40 , e ) 20 | a | divide(subtract(divide(multiply(2, const_100), const_2), const_2), add(divide(multiply(2, const_100), const_2), const_2)) | multiply(n0,const_100)|divide(#0,const_2)|add(#1,const_2)|subtract(#1,const_2)|divide(#3,#2)| | general |
a cistern of capacity 8000 litres measures externally 3.3 m by 2.6 m by 1.6 m and its walls are 5 cm thick . the thickness of the bottom is : | explanation : let the thickness of the bottom be x cm . then , [ ( 330 - 10 ) × ( 260 - 10 ) × ( 160 - x ) ] = 8000 × 1000 = > 320 × 250 × ( 160 - x ) = 8000 × 1000 = > ( 160 - x ) = 8000 × 1000 / 320 = 100 = > x = 60 cm = 6 dm . answer : b | a ) 90 cm , b ) 6 dm , c ) 1 m , d ) 1.1 cm , e ) none of these | b | subtract(multiply(multiply(3.3, 2.6), 1.6), divide(8000, const_1000)) | divide(n0,const_1000)|multiply(n1,n2)|multiply(n3,#1)|subtract(#2,#0) | physics |
the length of minute hand of a clock is 5.5 cm . what is the area covered by this in 10 minutes | "area of circle is pi * r ^ 2 but in 10 minutes area covered is ( 10 / 60 ) * 360 = 60 degree so formula is pi * r ^ 2 * ( angle / 360 ) = 3.14 * ( 5.5 ^ 2 ) * ( 60 / 360 ) = 15.83 cm ^ 2 answer : e" | a ) 15.27 , b ) 16.27 , c ) 17.27 , d ) 19.27 , e ) 15.83 | e | multiply(divide(add(multiply(const_2, 10), const_2), add(const_3, const_4)), multiply(multiply(5.5, 5.5), divide(multiply(const_1, const_60), multiply(const_100, const_3_6)))) | add(const_3,const_4)|multiply(n1,const_2)|multiply(const_1,const_60)|multiply(const_100,const_3_6)|multiply(n0,n0)|add(#1,const_2)|divide(#2,#3)|divide(#5,#0)|multiply(#6,#4)|multiply(#7,#8)| | physics |
if jake loses 8 pounds , he will weigh twice as much as his sister kendra . together they now weigh 290 pounds . what is jake ’ s present weight , in pounds ? | "j + k = 290 and so k = 290 - j j - 8 = 2 k j - 8 = 2 ( 290 - j ) 3 j = 588 j = 196 the answer is e ." | a ) 180 , b ) 184 , c ) 188 , d ) 192 , e ) 196 | e | add(multiply(divide(subtract(290, 8), const_3), const_2), 8) | subtract(n1,n0)|divide(#0,const_3)|multiply(#1,const_2)|add(n0,#2)| | general |
a , b , k start from the same place and travel in the same direction at speeds of 30 km / hr , 40 km / hr , 80 km / hr respectively . b starts two hours after a . if b and k overtake a at the same instant , how many hours after a did k start ? | "in 2 hours , a travels 60 km . b can catch a at a rate of 10 km / hr , so b catches a 6 hours after b starts . so a and b both travel a distance of 240 km . c needs 3 hours to travel 240 km , so c leaves 5 hours after a . the answer is d ." | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | d | subtract(40, 30) | subtract(n1,n0)| | physics |
in february wilson ’ s earnings were 40 percent of his family ’ s total income . in march wilson earned 25 percent less than in february . if the rest of his family ’ s income was the same in both months , then , in march , wilson ’ s earnings were approximately what percent x of his family ’ s total income ? | "lets suppose the total family income in feb = 100 x wilson ' s earning in feb = 40 % of 100 x = 40 x earnings of remaining family in feb = 100 x - 40 x = 60 x wilson ' s earning in march = 75 % of wilson ' s feb earnings = 75 % of 40 x = 30 x earnings of remaining family in march = earnings of remaining family in feb ... | a ) 15 % , b ) 17 % , c ) 24 % , d ) 30 % , e ) 33 % | e | multiply(divide(subtract(divide(40, const_100), multiply(divide(25, const_100), divide(40, const_100))), subtract(const_1, multiply(divide(25, const_100), divide(40, const_100)))), const_100) | divide(n0,const_100)|divide(n1,const_100)|multiply(#1,#0)|subtract(#0,#2)|subtract(const_1,#2)|divide(#3,#4)|multiply(#5,const_100)| | general |
at a garage sale , all of the items were sold at different prices . if the price of a radio sold at the garage sale was both the 16 th highest price and the 23 rd lowest price among the prices of the items sold , how many items were sold at the garage sale ? | "there were 15 items sold at a higher price than the radio and 22 items sold at a lower price than the radio . including the radio , there were 15 + 22 + 1 = 38 items sold . the answer is d ." | a ) 35 , b ) 36 , c ) 37 , d ) 38 , e ) 39 | d | subtract(add(23, 16), const_1) | add(n0,n1)|subtract(#0,const_1)| | other |
one hour after yolanda started walking from x to y , a distance of 52 miles , bob started walking along the same road from y to x . if yolanda ' s walking rate was 3 miles per hour and bob т ' s was 4 miles per hour , how many miles had bob walked when they met ? | "when b started walking y already has covered 3 miles out of 52 , hence the distance at that time between them was 52 - 3 = 49 miles . combined rate of b and y was 3 + 4 = 7 miles per hour , hence they would meet each other in 49 / 7 = 7 hours . in 6 hours b walked 7 * 4 = 28 miles . answer : a ." | a ) 28 , b ) 23 , c ) 22 , d ) 21 , e ) 19.5 | a | multiply(divide(subtract(52, 3), add(3, 4)), 4) | add(n1,n2)|subtract(n0,n1)|divide(#1,#0)|multiply(n2,#2)| | physics |
find the sum the difference between the compound and s . i . on a certain sum of money for 2 years at 10 % per annum is rs . 13 of money ? | "p = 13 ( 100 / 10 ) 2 = > p = 1300 answer : a" | a ) 1300 , b ) 1992 , c ) 9921 , d ) 2798 , e ) 2789 | a | multiply(multiply(divide(13, multiply(10, 2)), const_100), multiply(10, 2)) | multiply(n0,n1)|divide(n2,#0)|multiply(#1,const_100)|multiply(#2,#0)| | general |
30 men can do a work in 40 days . when should 10 men leave the work so that the entire work is completed in 40 days after they leave the work ? | "total work to be done = 30 * 40 = 1200 let 10 men leave the work after ' p ' days , so that the remaining work is completed in 40 days after they leave the work . 40 p + ( 10 * 40 ) = 1200 40 p = 800 = > p = 20 days answer : e" | a ) 87 days , b ) 10 days , c ) 55 days , d ) 44 days , e ) 20 days | e | divide(subtract(multiply(30, 40), multiply(40, 10)), 40) | multiply(n0,n1)|multiply(n1,n2)|subtract(#0,#1)|divide(#2,n1)| | physics |
two trains of equal lengths take 10 sec and 15 sec respectively to cross a telegraph post . if the length of each train be 120 m , in what time will they cross other travelling in opposite direction ? | "speed of the first train = 120 / 10 = 12 m / sec . speed of the second train = 120 / 5 = 8 m / sec . relative speed = 12 + 8 = 20 m / sec . required time = ( 120 + 120 ) / 20 = 12 sec . answer : b" | a ) 16 sec , b ) 12 sec , c ) 17 sec , d ) 21 sec , e ) 23 sec | b | divide(multiply(120, const_2), add(speed(120, 15), speed(120, 10))) | multiply(n2,const_2)|speed(n2,n1)|speed(n2,n0)|add(#1,#2)|divide(#0,#3)| | physics |
what is 990 * 990 ? | "if you take a base of 1000 then 990 is 10 less than 1000 to get the product of 990 x 990 write like this 990 - 10 ( as 10 less than base 1000 ) 990 - 10 now 10 x 10 = 100 and 990 - 10 = 980 so 990 x 990 = 980100 . . . ( bingo the answer is d . you can even have a shortcut . . . . . . 10 x 10 = 100 . . . only answer ch... | a ) 974,169 , b ) 974,219 , c ) 974,549 , d ) 980,100 , e ) 985,369 | d | circle_area(divide(990, multiply(const_2, const_pi))) | multiply(const_2,const_pi)|divide(n0,#0)|circle_area(#1)| | general |
jill ' s favorite number is even and has some repeating prime factors . john is guessing jills favorite number , and the only hint she gives is that 7 is a prime factor . what is johns best guess ? | a 84 = 2 * 2 * 2 * 7 b 105 is odd c 54 = 2 * 3 * 3 d 42 = 2 * 3 * 7 e 70 = 2 * 5 * 7 the answer must meet three criteria : odd , have repeating prime factors , and 7 is a prime factor . a is the only answer that meets all criteria | a ) 84 , b ) 105 , c ) 54 , d ) 42 , e ) 70 | a | multiply(power(const_2, const_2), multiply(7, const_3)) | multiply(n0,const_3)|power(const_2,const_2)|multiply(#0,#1) | other |
kamal obtained 76 , 65 , 82 , 67 and 85 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology . what are his average marks ? | "sol . average = 76 + 65 + 82 + 67 + 85 / 5 ) = ( 375 / 5 ) = 75 . answer d" | a ) 65 , b ) 69 , c ) 72 , d ) 75 , e ) none | d | divide(add(add(add(add(76, 65), 82), 67), 85), add(const_1, const_4)) | add(n0,n1)|add(const_1,const_4)|add(n2,#0)|add(n3,#2)|add(n4,#3)|divide(#4,#1)| | general |
find the number of factors of 180 that are in the form ( 4 * k + 2 ) , where k is a non - negative integer ? | make sure to follow posting guidelines ( link in my signatures ) . title of the thread must mention the first few words of the question itself . as for the question , the most straightforward way is to list out the factors ( this is fine for this question as 180 is a relatively small number ) . number of factors of 180... | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 6 | e | divide(divide(divide(180, const_3), 2), add(const_3, const_2)) | add(const_2,const_3)|divide(n0,const_3)|divide(#1,n2)|divide(#2,#0) | general |
every student of a certain school must play an instrument . in last year , 1 / 2 of the students picked a woodwind instrument , 2 / 5 of the students picked a brass instrument , and all of the other students took percussion . in this year , 1 / 2 of the students who play woodwind and 1 / 4 of the students who play bras... | lets pick smart numbers . total number of students : 20 woodwind ( 1 / 2 ) : 10 brass ( 2 / 5 ) : 8 percussion ( 1 / 10 ) : 2 after leaving school woodwind : 5 brass : 6 percussion : 2 new total number of students : 13 woodwind and brass : 11 answer 11 / 13 or d | a ) 7 / 13 , b ) 4 / 5 , c ) 9 / 13 , d ) 11 / 13 , e ) 7 / 10 | d | divide(add(subtract(subtract(1, divide(const_1, const_2)), divide(divide(const_1, const_2), 2)), subtract(divide(const_2, 5), divide(divide(const_2, 5), 4))), subtract(subtract(1, subtract(subtract(1, divide(const_1, const_2)), divide(divide(const_1, const_2), 2))), subtract(subtract(1, divide(const_1, const_2)), divid... | divide(const_1,const_2)|divide(const_2,n3)|divide(#0,n1)|divide(#1,n7)|subtract(n0,#0)|subtract(#4,#2)|subtract(#1,#3)|subtract(#4,#1)|add(#5,#6)|subtract(n0,#5)|subtract(#9,#7)|divide(#8,#10) | general |
a cylinder is being filled with sand weighing 200 kg per cubic foot . the cylinder has a diameter of 1 foot and is 5 feet tall . how much sand is being used ? ? | answer is a . volume of cylinder = pi * r ^ 2 * h - - - - - - - - > pi * ( 1 / 2 ) ^ 2 * 5 - - - - - - - - > pi * ( 1 / 4 ) * 5 - - - - - - - - > ( 5 / 4 ) pi cubic feet . 1 cubic feet = 200 kg . so ( 5 / 4 ) pi cubic feet = ( ( 200 * 5 ) / 4 ) pi - - - - - - - - - > 250 pi regards , abhijit answer is a | ['a ) 250 pi kg', 'b ) 500 pi kg', 'c ) 500 pi ^ 2 kg', 'd ) 1000 pi kg', 'e ) 1000 pi ^ 2 kg'] | a | multiply(200, volume_cylinder(divide(1, const_2), 5)) | divide(n1,const_2)|volume_cylinder(#0,n2)|multiply(n0,#1) | geometry |
a can complete the job in 4 hours and b can complete the same job in 3 hours . a works for 1 hour and then b joins and both complete the job . what fraction of the job did b complete | a = 1 / 7 | a ) 1 / 7 , b ) 3 / 10 , c ) 1 / 2 , d ) 5 / 6 , e ) 8 / 9 | a | divide(subtract(subtract(4, 1), 3), multiply(4, 3)) | multiply(n0,n1)|subtract(n0,n2)|subtract(#1,n1)|divide(#2,#0)| | physics |
there is food for 760 men for 22 days . how many more men should join after two days so that the same food may last for 16 days more ? | "760 - - - - 22 760 - - - - 20 x - - - - - 16 x * 16 = 760 * 20 x = 950 760 - - - - - - - 190 answer : e" | a ) 122 , b ) 140 , c ) 199 , d ) 188 , e ) 190 | e | subtract(divide(multiply(760, subtract(22, const_2)), 16), 760) | subtract(n1,const_2)|multiply(n0,#0)|divide(#1,n2)|subtract(#2,n0)| | physics |
if n = 8 ^ 8 – 8 , what is the units digit of n ? | "8 ^ 8 - 8 = 8 ( 8 ^ 7 - 1 ) = = > 8 ( 2 ^ 21 - 1 ) last digit of 2 ^ 21 is 2 based on what explanation livestronger is saying . 2 ^ 24 - 1 yields 2 - 1 = 1 as the unit digit . now on multiply this with 8 , we get unit digit as 8 answer : c" | a ) 0 , b ) 1 , c ) 8 , d ) 2 , e ) 4 | c | divide(log(8), log(power(8, 8))) | log(n2)|power(n0,n1)|log(#1)|divide(#0,#2)| | general |
if w is the set of all the integers between 19 and 99 , inclusive , that are either multiples of 3 or multiples of 2 or multiples of both , then w contains how many numbers ? | "multiples of 2 from 19 to 99 = multiples of 2 from 1 to 99 - multiples of 2 from 1 to 18 = [ 99 / 2 ] - [ 18 / 2 ] = 49 - 9 = 40 multiples of 3 from 19 to 99 = multiples of 3 from 1 to 99 - multiples of 3 from 1 to 18 = [ 99 / 3 ] - [ 18 / 3 ] = 33 - 6 = 27 multiples of 2 and 3 bothi . e . 6 from 19 to 99 = multiples ... | a ) 26 , b ) 32 , c ) 33 , d ) 34 , e ) 54 | e | subtract(add(floor(divide(subtract(99, 19), 3)), divide(subtract(99, 19), 2)), floor(divide(subtract(99, 19), multiply(2, 3)))) | multiply(n3,n2)|subtract(n1,n0)|divide(#1,n3)|divide(#1,n2)|divide(#1,#0)|floor(#3)|floor(#4)|add(#2,#5)|subtract(#7,#6)| | other |
find the curved surface area , if the radius of a cone is 28 m and slant height is 30 m ? | "cone curved surface area = ï € rl 22 / 7 ã — 28 ã — 30 = 88 ã — 30 = 2640 m ( power 2 ) answer is b ." | a ) 2550 , b ) 2640 , c ) 3960 , d ) 4280 , e ) 5490 | b | multiply(const_pi, multiply(28, 30)) | multiply(n0,n1)|multiply(#0,const_pi)| | geometry |
| 16 - 5 | - | 5 - 12 | = ? | "| 16 - 5 | - | 5 - 12 | = | 11 | - | - 7 | = 11 - 7 = 4 correct answer e" | a ) 3 , b ) 2 , c ) 1 , d ) 0 , e ) 4 | e | subtract(subtract(16, 5), subtract(12, 5)) | subtract(n0,n1)|subtract(n3,n2)|subtract(#0,#1)| | general |
if two - third of a bucket is filled in 90 seconds then the time taken to fill the bucket completely will be . | "2 / 3 filled in 90 seconds 1 / 3 filled in 45 seconds then 2 / 3 + 1 / 3 = 90 + 45 seconds = 135 seconds answer : c" | a ) 90 seconds , b ) 70 seconds , c ) 135 seconds , d ) 100 seconds , e ) 120 seconds | c | multiply(divide(90, const_2), const_3) | divide(n0,const_2)|multiply(#0,const_3)| | physics |
if a tire rotates at 400 revolutions per minute when the car is traveling 120 km / h , what is the circumference of the tire ? | 400 rev / minute = 400 * 60 rev / 60 minutes = 24,000 rev / hour 24,000 * c = 12,0000 m : c is the circumference c = 5 meters correct answer d | ['a ) 7 meters', 'b ) 9 meters', 'c ) 8 meters', 'd ) 5 meters', 'e ) 3 meters'] | d | multiply(divide(120, multiply(multiply(multiply(const_2, const_3), const_10), 400)), const_1000) | multiply(const_2,const_3)|multiply(#0,const_10)|multiply(n0,#1)|divide(n1,#2)|multiply(#3,const_1000) | physics |
the population of a town increased from 1 , 75000 to 2 , 62500 in a decade . what is the average percent increase of population per year ? | explanation : increase in the population in 10 years = 2 , 62500 - 1 , 75000 = 87500 % ncrease in the population in 10 years = ( 87500 / 175000 ) × 100 = 8750 / 175 = 50 % average % increase of population per year = 50 % / 10 = 5 % answer : option c | a ) 4 % , b ) 6 % , c ) 5 % , d ) 50 % , e ) none of these | c | multiply(divide(divide(subtract(add(2, divide(62500, multiply(const_1000, const_100))), add(1, divide(75000, multiply(const_1000, const_100)))), const_10), add(1, divide(75000, multiply(const_1000, const_100)))), const_100) | multiply(const_100,const_1000)|divide(n3,#0)|divide(n1,#0)|add(n2,#1)|add(n0,#2)|subtract(#3,#4)|divide(#5,const_10)|divide(#6,#4)|multiply(#7,const_100) | general |
10 men do a work in 10 days . how many men are needed to finish the work in 2 days ? | "men required to finish the work in 2 days = 10 * 10 / 2 = 50 answer is a" | a ) 50 , b ) 20 , c ) 30 , d ) 10 , e ) 15 | a | divide(multiply(10, 10), 2) | multiply(n0,n1)|divide(#0,n2)| | physics |
the average age of 35 students in a class is 16 years . the average age of 21 students is 14 . what is the average age of remaining 38 students ? | "solution sum of the ages of 14 students = ( 16 x 35 ) - ( 14 x 21 ) = 560 - 294 . = 266 . ∴ required average = 266 / 38 = 7 years . answer b" | a ) 14 years , b ) 7 years , c ) 19 years , d ) 21 years , e ) none | b | subtract(add(add(multiply(35, 16), 21), 35), multiply(35, 16)) | multiply(n0,n1)|add(n2,#0)|add(n0,#1)|subtract(#2,#0)| | general |
a man can row 8 kmph in still water . if the velocity of the current is 2 kmph and it takes him 2 hours to row to a place and come back . how far is the place ? | man ' s rate down stream = 8 + 2 = 10 kmph man ' s rate upstream = 8 - 2 = 6 kmph let the required distance be x km then x / 10 + x / 6 = 2 3 x + 5 x = 60 8 x = 60 x = 7.5 km answer is e . | a ) 5.5 , b ) 8.5 , c ) 6.5 , d ) 9.5 , e ) 7.5 | e | divide(multiply(multiply(subtract(8, 2), add(8, 2)), 2), add(add(8, 2), subtract(8, 2))) | add(n0,n1)|subtract(n0,n1)|add(#0,#1)|multiply(#0,#1)|multiply(n1,#3)|divide(#4,#2) | physics |
if x is equal to the sum of the integers from 20 to 40 , inclusive , and y is the number of even integers from 20 to 40 , inclusive , what is the value of x + y ? | "x = 20 + 21 + . . . + 40 = 21 ( 30 ) = 630 y = 11 x + y = 641 the answer is e ." | a ) 601 , b ) 611 , c ) 621 , d ) 631 , e ) 641 | e | add(multiply(divide(add(20, 40), const_2), add(subtract(40, 20), const_1)), add(divide(subtract(40, 20), const_2), const_1)) | add(n0,n1)|subtract(n1,n0)|add(#1,const_1)|divide(#1,const_2)|divide(#0,const_2)|add(#3,const_1)|multiply(#2,#4)|add(#5,#6)| | general |
the hiker walking at a constant rate of 5 miles per hour is passed by a cyclist traveling in the same direction along the same path at 20 miles per hour . the cyclist stops to wait for the hiker 5 minutes after passing her , while the hiker continues to walk at her constant rate , how many minutes must the cyclist wait... | "after passing the hiker the cyclist travels for 5 minutes at a rate of 20 miles / hour . in those 5 mins the cyclist travels a distance of 5 / 3 miles . in those 5 mins the hiker travels a distance of 5 / 12 miles . so the hiker still has to cover 15 / 12 miles to meet the waiting cyclist . the hiker will need 1 / 4 h... | a ) 15 , b ) 10 , c ) 20 , d ) 30 , e ) 35 | a | multiply(divide(multiply(subtract(20, 5), divide(5, const_60)), 5), const_60) | divide(n2,const_60)|subtract(n1,n0)|multiply(#0,#1)|divide(#2,n0)|multiply(#3,const_60)| | physics |
a tank with a volume of 30 cubic feet has one inlet pipe and 2 outlet pipes . the inlet pipe fills water into the tank at the rate of 5 cubic inches / min and the 2 outlet pipes empty it out at the rates of 9 cubic inches / min and 8 cubic inches / min respectively . if all 3 pipes are opened when the tank is full , ho... | "the tank is emptied at this rate : 9 + 8 - 5 = 12 cubic inches / min the tank has a volume of 30 * 12 * 12 * 12 = 51840 cubic inches . the time it takes to empty the tank is 51840 / 12 = 4320 minutes . the answer is d ." | a ) 1440 , b ) 2340 , c ) 3240 , d ) 4320 , e ) 5420 | d | divide(multiply(30, power(9, 5)), subtract(add(12, 8), 5)) | add(n4,n5)|power(n8,n6)|multiply(n0,#1)|subtract(#0,n2)|divide(#2,#3)| | physics |
a train consists of 12 boggies , each boggy 15 metres long . the train crosses a telegraph post in 9 seconds . due to some problem , three boggies were detached . the train now crosses a telegraph post in | "length of train = 12 ã — 15 = 180 m . then , speed of train = 180 â „ 9 = 20 m / s now , length of train = 9 ã — 15 = 135 m â ˆ ´ required time = 135 â „ 20 = 6.75 sec . answer a" | a ) 6.75 sec , b ) 12 sec , c ) 15 sec , d ) 20 sec , e ) none of these | a | divide(subtract(multiply(12, 15), 15), divide(multiply(12, 15), 9)) | multiply(n0,n1)|divide(#0,n2)|subtract(#0,n1)|divide(#2,#1)| | physics |
a 90 metres long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds . what is the length of the other train ? | "explanation : as trains are running in opposite directions so their relative speed will get added so , relative speed = 120 + 80 = 200 kmph = 200 * ( 5 / 18 ) = 500 / 9 m / sec let the length of other train is x meter then x + 90 / 9 = 500 / 9 = > x + 90 = 500 = > x = 410 so the length of the train is 410 meters optio... | a ) 220 meter , b ) 225 meter , c ) 230 meter , d ) 410 meter , e ) none of these | d | subtract(multiply(multiply(add(120, 80), const_0_2778), 9), 90) | add(n1,n2)|multiply(#0,const_0_2778)|multiply(n3,#1)|subtract(#2,n0)| | physics |
with a uniform speed a car covers the distance in 8 hours . had the speed been increased by 2 km / hr , the same distance could have been covered in 7 1 / 2 hours . what is the distance covered ? | "let the distance be x km . then , x / ( 7 1 / 2 ) - x / 8 = 2 2 x / 15 - x / 8 = 2 = > x = 240 km . answer : d" | a ) 187 km , b ) 480 km , c ) 278 km , d ) 240 km , e ) 671 km | d | divide(1, 2) | divide(n3,n4)| | physics |
a store ’ s selling price of $ 2240 for a certain computer would yield a profit of 30 percent of the store ’ s cost for the computer . what selling price would yield a profit of 40 percent of the computer ’ s cost ? | "1.3 x = 2240 x = 2240 / 1.3 so , 1.4 x = 2240 * 1.4 / 1.3 = 2412 answer : - b" | a ) $ 2400 , b ) $ 2412 , c ) $ 2650 , d ) $ 2732 , e ) $ 2800 | b | multiply(2240, divide(add(const_100, 40), add(const_100, 30))) | add(n2,const_100)|add(n1,const_100)|divide(#0,#1)|multiply(n0,#2)| | gain |
three table runners have a combined area of 220 square inches . by overlapping the runners to cover 80 % of a table of area 175 square inches , the area that is covered by exactly two layers of runner is 24 square inches . what is the area of the table that is covered with three layers of runner ? | "total = a + b + c - ( sum of exactly 2 - group overlaps ) - 2 * ( all three ) + neither 80 % * 175 = 220 - 24 - 2 * ( all three ) + 0 2 * ( all three ) = 220 - 24 - 140 all three = 28 answer : d" | a ) 18 square inches , b ) 20 square inches , c ) 24 square inches , d ) 28 square inches , e ) 30 square inches | d | divide(subtract(subtract(220, 24), multiply(175, divide(80, const_100))), const_2) | divide(n1,const_100)|subtract(n0,n3)|multiply(n2,#0)|subtract(#1,#2)|divide(#3,const_2)| | geometry |
the youngest of 4 children has siblings who are 2 , 7 , and 11 years older than she is . if the average ( arithmetic mean ) age of the 4 siblings is 25 , what is the age of the youngest sibling ? | "x + ( x + 2 ) + ( x + 7 ) + ( x + 11 ) = 100 4 x + 20 = 100 4 x = 80 x = 20 the answer is d ." | a ) 17 , b ) 18 , c ) 19 , d ) 20 , e ) 21 | d | divide(subtract(multiply(const_4.0, 25), add(add(4, 7), 11)), 4) | add(n1,n2)|multiply(n0,n5)|add(n3,#0)|subtract(#1,#2)|divide(#3,n0)| | general |
addison high school ’ s senior class has 300 boys and 240 girls . if 60 % of the boys attended the college then 60 % of the total class attended college , what percentage of the girls class attended college ? | "number of boys attending the party : 0.60 * 300 = 180 total students = 300 + 240 = 540 number of students attending the party : 0.60 * 540 = 324 number of girls attending = 324 - 180 = 144 % of girls attending = 144 / 240 = 60 % . a is the correct answer" | a ) 60 , b ) 50 , c ) 46 , d ) 55.56 , e ) 65 | a | multiply(divide(add(multiply(divide(60, const_100), 300), multiply(subtract(const_1, divide(60, const_100)), 240)), add(300, 240)), const_100) | add(n0,n1)|divide(n2,const_100)|divide(n3,const_100)|multiply(n0,#1)|subtract(const_1,#2)|multiply(n1,#4)|add(#3,#5)|divide(#6,#0)|multiply(#7,const_100)| | general |
if 6 ! / 3 ^ x is an integer , what is the greatest possible value of x ? | "6 - 2 * 3 3 - 1 * 3 hence max of 3 ^ 2 is allowed . imo a ." | a ) 2 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | add(divide(6, 3), divide(3, divide(6, 3))) | divide(n0,n1)|divide(n1,#0)|add(#0,#1)| | general |
a person walking at 10 kmph reaches his office 10 minutes late . if he walks at 15 kmph , he reaches there 10 minutes earlier . how far is the office from his house ? | formula = s 1 * s 2 / s 2 - s 1 * t 1 + t 2 / 60 = 10 * 15 / 5 * 20 / 60 = 30 * 20 / 60 = 10 km answer is d | a ) 5 km , b ) 15 km , c ) 8 km , d ) 10 km , e ) 20 km | d | multiply(add(divide(add(multiply(10, divide(10, const_60)), multiply(15, divide(10, const_60))), subtract(15, 10)), divide(10, const_60)), 10) | divide(n0,const_60)|subtract(n2,n0)|multiply(n0,#0)|multiply(n2,#0)|add(#2,#3)|divide(#4,#1)|add(#5,#0)|multiply(n0,#6) | physics |
running at the same rate , 8 identical machines can produce 560 paperclips a minute . at this rate , how many paperclips could 18 machines produce in 6 minutes ? | "8 machines produce 560 in 1 min 8 machines produce 560 * 6 in 6 min 18 machine produce 560 * 6 * ( 18 / 8 ) in 6 minutes 560 * 6 * 18 / 8 = 7560 answer is c ." | a ) 1344 , b ) 3360 , c ) 7560 , d ) 50400 , e ) 67200 | c | multiply(multiply(560, divide(18, 8)), 6) | divide(n2,n0)|multiply(n1,#0)|multiply(n3,#1)| | gain |
two pipes a and b can fill a tank in 25 hours and 35 hours respectively . if both the pipes are opened simultaneously , how much time will be taken to fill the tank ? | "part filled by a in 1 hour = 1 / 25 part filled by b in 1 hour = 1 / 30 part filled by ( a + b ) in 1 hour = 1 / 25 + 1 / 30 = 11 / 150 both the pipes together fill the tank in 150 / 11 = 14 6 / 11 hours answer is b" | a ) 20 hours , b ) 14 6 / 11 hours , c ) 10 hours , d ) 12 hours , e ) 8 hours | b | divide(const_1, add(divide(const_1, 25), divide(const_1, 35))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)| | physics |
tammy climbed a mountain in two days . she spent a total of 14 hours climbing the mountain . on the second day , she walked at an average speed that was half a kilometer per hour faster , but 2 hours less than what she walked on the first day . if the total distance she climbed during the two days is 52 kilometers , ho... | "ans : c total time = 14 hrs let time traveled during 1 st day = x let time traveled during 2 nd day = x - 2 total time = 14 x + x - 2 = 14 x = 8 speed * time = distance s * 8 + ( s + 0.5 ) ( 8 - 2 ) = 52 solving s = 4.5 now speed for 2 nd day is 0.5 less than the 1 st day which is 4.5 thus speed for 2 nd day = 4 its s... | a ) 3 , b ) 3.5 , c ) 4 , d ) 4.5 , e ) 6 | c | add(divide(subtract(52, divide(subtract(divide(add(14, 2), 2), 2), 2)), add(subtract(divide(add(14, 2), 2), 2), divide(add(14, 2), 2))), divide(const_1, 2)) | add(n0,n1)|divide(const_1,n1)|divide(#0,n1)|subtract(#2,n1)|add(#2,#3)|divide(#3,n1)|subtract(n2,#5)|divide(#6,#4)|add(#7,#1)| | physics |
jolene entered an 14 - month investment contract that guarantees to pay 2 percent interest at the end of 6 months , another 3 percent interest at the end of 12 months , and 4 percent interest at the end of the 18 month contract . if each interest payment is reinvested in the contract , and jolene invested $ 10,000 init... | "if interest were not compounded in every six months ( so if interest were not earned on interest ) then we would have ( 2 + 3 + 4 ) = 9 % simple interest earned on $ 10,000 , which is $ 900 . so , you can rule out a , b and c right away . interest earned after the first time interval : $ 10,000 * 2 % = $ 200 ; interes... | a ) $ 506.00 , b ) $ 726.24 , c ) $ 900.00 , d ) $ 920.24 , e ) $ 926.24 | a | add(multiply(add(multiply(divide(3, const_100), add(multiply(divide(2, const_100), power(const_100, 2)), power(const_100, 2))), add(multiply(divide(2, const_100), power(const_100, 2)), power(const_100, 2))), divide(4, const_100)), multiply(divide(3, const_100), add(multiply(divide(2, const_100), power(const_100, 2)), p... | divide(n1,const_100)|divide(n3,const_100)|divide(n5,const_100)|power(const_100,n1)|multiply(#0,#3)|add(#4,#3)|multiply(#5,#1)|add(#5,#6)|multiply(#7,#2)|add(#8,#6)| | gain |
an isosceles triangle with sides 13 13 10 and there is a circle inscribed it . find the radius of circle ? | height of triangle = √ ( 13 ^ 2 - ( 10 / 2 ) ^ 2 ) = 12 area of triangle = r * s where , r = radius of inscribed circle & s = ( a + b + c ) / 2 area of triangle = 1 / 2 * b * h = 1 / 2 * 10 * 12 = 60 s = ( 13 + 13 + 10 ) / 2 = 18 r = area of triangle / s = 60 / 18 = 10 / 3 = 3.33 answer : b | ['a ) 1.32', 'b ) 3.33', 'c ) 3.58', 'd ) 4.34', 'e ) 5.18'] | b | divide(multiply(triangle_area_three_edges(13, 13, 10), const_2), triangle_perimeter(13, 13, 10)) | triangle_area_three_edges(n0,n0,n2)|triangle_perimeter(n0,n0,n2)|multiply(#0,const_2)|divide(#2,#1) | geometry |
the diagonals of a rhombus are 10 cm and 15 cm . find its area ? | "1 / 2 * 10 * 15 = 75 answer : a" | a ) 75 , b ) 129 , c ) 150 , d ) 123 , e ) 117 | a | rhombus_area(10, 15) | rhombus_area(n0,n1)| | geometry |
in a can , there is a mixture of milk and water in the ratio 1 : 5 . if it is filled with an additional 2 litres of milk the can would be full and ratio of milk and water would become 3 : 5 . find the capacity of the can ? | "let the capacity of the can be t litres . quantity of milk in the mixture before adding milk = 1 / 6 ( t - 2 ) after adding milk , quantity of milk in the mixture = 3 / 8 t . 3 t / 8 - 2 = 1 / 6 = ( t - 2 ) 10 t = 96 - 16 = > t = 8 answer : d" | a ) 40 , b ) 44 , c ) 48 , d ) 8 , e ) 56 | d | add(add(multiply(5, divide(2, subtract(multiply(divide(3, 5), 5), 1))), divide(2, subtract(multiply(divide(3, 5), 5), 1))), 2) | divide(n3,n4)|multiply(n1,#0)|subtract(#1,n0)|divide(n2,#2)|multiply(n1,#3)|add(#3,#4)|add(n2,#5)| | general |
each child has 6 crayons and 12 apples . if there are 12 children , how many crayons are there in total ? | 6 * 12 = 72 . answer is c | a ) 22 , b ) 65 , c ) 72 , d ) 78 , e ) 90 | c | multiply(12, 6) | multiply(n0,n2)| | general |
in a certain school , the ratio of boys to girls is 5 to 17 . if there are 72 more girls than boys , how many boys are there ? | "the ratio of b to g is 5 : 13 and the other data point is g are more than boys by 72 . . . looking at the ratio we can say that the 12 ( 17 - 5 ) extra parts caused this diff of 72 . so 1 part corresponds to 72 / 8 = 9 and so 5 parts correspond to 12 * 9 = 108 . e" | a ) 27 , b ) 36 , c ) 45 , d ) 72 , e ) 108 | e | subtract(divide(72, subtract(const_1, divide(5, 17))), 72) | divide(n0,n1)|subtract(const_1,#0)|divide(n2,#1)|subtract(#2,n2)| | other |
a car takes 8 hours to cover a distance of 540 km . how much should the speed in kmph be maintained to cover the same direction in 3 / 2 th of the previous time ? | "time = 8 distence = 540 3 / 2 of 8 hours = 8 * 3 / 2 = 12 hours required speed = 540 / 12 = 45 kmph b )" | a ) 10 kmph , b ) 45 kmph , c ) 60 kmph , d ) 55 kmph , e ) 30 kmph | b | divide(540, divide(multiply(8, 3), 2)) | multiply(n0,n2)|divide(#0,n3)|divide(n1,#1)| | physics |
in a room with 7 people , 4 people have exactly 1 friend in the room and 3 people have exactly 2 friends in the room ( assuming that friendship is a mutual relationship , i . e . if jane is paul ' s friend , paul is jane ' s friend ) . if two individuals are selected from the room at random , what is the probability th... | "( 4 / 7 ) ( 5 / 6 ) + ( 3 / 7 ) ( 4 / 6 ) if you choose one of the 4 with one other friend , then you have a 5 / 6 chance of not picking their friend 2 nd . if you choose one of the 3 with 2 friends , you have a 4 / 6 chance of not picking one of their friends second . add them up . 20 / 42 + 12 / 42 32 / 42 = 16 / 21... | a ) 5 / 21 , b ) 3 / 7 , c ) 4 / 7 , d ) 5 / 7 , e ) 16 / 21 | e | divide(subtract(subtract(subtract(7, 1), 1), 1), subtract(7, 1)) | subtract(n0,n2)|subtract(#0,n2)|subtract(#1,n2)|divide(#2,#0)| | general |
assisants are needed to prepare for preparation . each helper can make either 2 large cakes or 35 small cakes / hour . the kitchen is available for 3 hours and 20 large cakes and 700 small cakes are needed . how many helpers are required ? | 20 large cakes will require the equivalent of 10 helpers working for one hour . 700 small cakes will require the equivalent of 20 helpers working for one hour . this means if only one hour were available we would need 30 helpers . but since three hours are available we can use 10 helpers . b | a ) 9 , b ) 10 , c ) 13 , d ) 16 , e ) 18 | b | divide(add(divide(20, 2), divide(700, 35)), 3) | divide(n3,n0)|divide(n4,n1)|add(#0,#1)|divide(#2,n2) | other |
if the perimeter of a rectangular park is 1000 m , its length when its breadth is 200 m is ? | 2 ( l + 200 ) = 1000 = > l = 300 m answer : d | a ) 50 , b ) 100 , c ) 200 , d ) 300 , e ) 400 | d | subtract(divide(1000, const_2), 200) | divide(n0,const_2)|subtract(#0,n1)| | physics |
a boat running upstream takes 640 min to cover a certain distance , while it takes 280 min to cover the same distance running down stream . what is the ratio between the speed of the boat and speed of water current respectively ? | let the speed of the boat = x and speed of the current = y relative speed in upstream = x - y and in downstream = x + y let the distance ' d ' is to be covered in downstream and upstream , then time taken in upstream = d / ( x - y ) = 640 or d = 640 ( x - y ) - - - - ( i ) and time taken in downstream = d / ( x + y ) =... | a ) 23 : 7 , b ) 22 : 9 , c ) 23 : 8 , d ) 23 : 9 , e ) 23 : 11 | d | divide(divide(add(640, 280), const_2), subtract(640, divide(add(640, 280), const_2))) | add(n0,n1)|divide(#0,const_2)|subtract(n0,#1)|divide(#1,#2) | physics |
f ( 1 ) = 1 , where f ( x + y ) = f ( x ) + f ( y ) + 8 xy - 2 . then f ( 7 ) = ? | f ( 1 ) = 1 f ( 2 ) = f ( 1 + 1 ) = 1 + 1 + 8 - 2 = 8 f ( 3 ) = f ( 2 + 1 ) = 8 + 1 + ( 8 * 2 * 1 ) - 2 = 23 f ( 6 ) = f ( 3 + 3 ) = 23 + 23 + ( 8 * 3 * 3 ) - 2 = 116 f ( 7 ) = f ( 6 + 1 ) = 116 + 1 + ( 8 * 6 * 1 ) - 2 = 163 thus ans is 163 answer : b | a ) 162 , b ) 163 , c ) 164 , d ) 165 , e ) 166 | b | subtract(add(add(subtract(add(add(subtract(add(add(subtract(add(add(subtract(add(add(subtract(add(add(1, 1), 8), 2), 1), multiply(8, const_2)), 2), 1), multiply(8, const_3)), 2), 1), multiply(8, const_4)), 2), 1), multiply(8, add(const_3, const_2))), 2), 1), multiply(8, add(const_3, const_3))), 2) | add(n0,n0)|add(const_2,const_3)|add(const_3,const_3)|multiply(n2,const_2)|multiply(n2,const_3)|multiply(n2,const_4)|add(n2,#0)|multiply(n2,#1)|multiply(n2,#2)|subtract(#6,n3)|add(n0,#9)|add(#10,#3)|subtract(#11,n3)|add(n0,#12)|add(#13,#4)|subtract(#14,n3)|add(n0,#15)|add(#16,#5)|subtract(#17,n3)|add(n0,#18)|add(#19,#7)... | general |
what is the largest possible value of cc if 5 c + ( d − 12 ) ^ 2 = 235 ? | to maximize c , we should minimize ( d - 12 ) ^ 2 . ( d - 12 ) ^ 2 is a square of a number , thus its smallest possible value is 0 ( for d = 12 ) . in this case we ' d have 5 c + 0 = 235 - - > c = 47 . answer : e . | a ) 17 , b ) 25 , c ) 35 , d ) 42 , e ) 47 | e | divide(235, 5) | divide(n3,n0) | general |
find the cost of fencing around a circular field of diameter 35 m at the rate of rs . 1.50 a meter ? | "2 * 22 / 7 * 17.5 = 110 110 * 1 1 / 2 = rs . 115.5 answer : d" | a ) 287 , b ) 132 , c ) 156 , d ) 115.5 , e ) 267 | d | multiply(circumface(divide(35, const_2)), 1.50) | divide(n0,const_2)|circumface(#0)|multiply(n1,#1)| | physics |
the sum of all two digit numbers divisible by 5 is | "required numbers are 10,15 , 20,25 , . . . , 95 this is an a . p . in which a = 10 , d = 5 and l = 95 . let the number of terms in it be n . then t = 95 so a + ( n - 1 ) d = 95 . 10 + ( n - 1 ) * 5 = 95 , then n = 18 . required sum = n / 2 ( a + l ) = 18 / 2 ( 10 + 95 ) = 945 . answer : a ) 945" | a ) 945 , b ) 3487 , c ) 28 , d ) 279 , e ) 1287 | a | multiply(divide(divide(subtract(multiply(5, add(const_10, const_1)), multiply(const_1, 5)), subtract(multiply(5, const_2), multiply(const_1, 5))), const_2), add(multiply(5, const_2), multiply(5, add(const_10, const_1)))) | add(const_1,const_10)|multiply(n0,const_2)|multiply(n0,const_1)|multiply(n0,#0)|subtract(#1,#2)|add(#1,#3)|subtract(#3,#2)|divide(#6,#4)|divide(#7,const_2)|multiply(#5,#8)| | general |
at the faculty of aerospace engineering , 312 students study random - processing methods , 222 students study scramjet rocket engines and 112 students study them both . if every student in the faculty has to study one of the two subjects , how many students are there in the faculty of aerospace engineering ? | "312 + 222 - 112 ( since 112 is counted twice ) = 422 e is the answer" | a ) 404 , b ) 452 , c ) 444 , d ) 468 , e ) 422 | e | add(subtract(312, divide(112, const_2)), subtract(222, divide(112, const_2))) | divide(n2,const_2)|subtract(n0,#0)|subtract(n1,#0)|add(#1,#2)| | other |
a , b and c started a business with capitals of rs . 8000 , rs . 10000 and rs . 12000 respectively . at the end of the year , the profit share of b is rs . 1700 . the difference between the profit shares of a and c is ? | "ratio of investments of a , b and c is 8000 : 10000 : 12000 = 4 : 5 : 6 and also given that , profit share of b is rs . 1700 = > 5 parts out of 15 parts is rs . 1700 now , required difference is 6 - 4 = 2 parts required difference = 2 / 5 ( 1700 ) = rs . 680 answer : d" | a ) 288 , b ) 266 , c ) 155 , d ) 680 , e ) 441 | d | multiply(subtract(divide(12000, 10000), divide(8000, 10000)), 1700) | divide(n2,n1)|divide(n0,n1)|subtract(#0,#1)|multiply(n3,#2)| | gain |
if the operation ø is defined for all positive integers x and w by x ø w = ( 2 ^ x ) / ( 2 ^ w ) then ( 4 ø 2 ) ø 2 = ? | "4 ø 2 = 2 ^ 4 / 2 ^ 2 = 4 4 ø 2 = 2 ^ 4 / 2 ^ 2 = 4 the answer is b ." | a ) 2 , b ) 4 , c ) 8 , d ) 16 , e ) 32 | b | divide(power(2, divide(power(2, 4), power(2, 2))), power(2, 2)) | power(n0,n2)|power(n0,n0)|power(n0,n4)|divide(#0,#1)|power(n0,#3)|divide(#4,#2)| | general |
a rectangular floor is covered by a rug except for a strip 3 meters wide along each of the four edge . if the floor is 12 meters by 10 meters , what is the area of the rug in square meters ? | "a strip of 3 meters is covering the inner rectangular rug for all 4 sides . length of inner rug = 12 - ( 2 * 3 ) breadth of inner rug = 10 - ( 2 * 3 ) area of rug = 6 * 4 = 24 sq . mt â nswer : e" | a ) 30 , b ) 28 , c ) 36 , d ) 42 , e ) 24 | e | rectangle_area(subtract(12, multiply(3, const_2)), subtract(10, multiply(3, const_2))) | multiply(n0,const_2)|subtract(n1,#0)|subtract(n2,#0)|rectangle_area(#1,#2)| | geometry |
if x , y , and z are positive integers , and 2 x = 5 y = 7 z , then the least possible value of x + y + z is | "take lcm of 2,5 and 7 = 70 now 2 x = 70 = > x = 35 5 y = 70 = > y = 14 7 z = 70 = > z = 10 35 + 14 + 10 = 59 . option e ." | a ) 15 , b ) 28 , c ) 37 , d ) 42 , e ) 59 | e | add(add(divide(divide(multiply(multiply(2, 5), 7), const_2), 2), divide(divide(multiply(multiply(2, 5), 7), const_2), 5)), divide(divide(multiply(multiply(2, 5), 7), const_2), 7)) | multiply(n0,n1)|multiply(n2,#0)|divide(#1,const_2)|divide(#2,n0)|divide(#2,n1)|divide(#2,n2)|add(#3,#4)|add(#6,#5)| | general |
on a trip , a cyclist averaged 8 miles per hour for the first 16 miles and 10 miles per hour for the remaining 16 miles . if the cyclist returned immediately via the same route and took a total of 6.8 hours for the round trip , what was the average speed ( in miles per hour ) for the return trip ? | "the time to go 32 miles was 16 / 8 + 16 / 10 = 2 + 1.6 = 3.6 hours . the average speed for the return trip was 32 miles / 3.2 hours = 10 mph . the answer is b ." | a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 13 | b | divide(add(16, 16), subtract(6.8, add(divide(16, 8), divide(16, 10)))) | add(n1,n3)|divide(n1,n0)|divide(n3,n2)|add(#1,#2)|subtract(n4,#3)|divide(#0,#4)| | physics |
for how many unique pairs of nonnegative integers { a , b } is the equation a ^ 2 - b ^ 2 = 235 true ? | "answer b ( a + b ) ( a - b ) = 235 2 cases for ( a + b ) , ( a - b ) 235 , 1 47 , 5 answer b" | a ) 1 , b ) 2 , c ) 5 , d ) 7 , e ) 9 | b | divide(log(235), log(add(const_4, const_1))) | add(const_1,const_4)|log(n2)|log(#0)|divide(#1,#2)| | general |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.