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 |
|---|---|---|---|---|---|---|
if ( 4 - x ) / ( 4 + x ) = x , what is the value of x ^ 2 + 5 x - 4 ? | ( 4 - x ) = x * ( 4 + x ) ( 4 - x ) = 4 x + x ^ 2 0 = x ^ 2 + 5 x - 4 the answer is a . | a ) 0 , b ) 2 , c ) 4 , d ) 6 , e ) 8 | a | subtract(multiply(const_4, const_2), 4) | multiply(const_2,const_4)|subtract(#0,n0) | general |
walking with 4 / 5 of my usual speed , i miss the bus by 5 minutes . what is my usual time ? | "speed ratio = 1 : 4 / 5 = 5 : 4 time ratio = 4 : 51 - - - - - - - - 5 4 - - - - - - - - - ? Γ¨ 20 answer : d" | a ) 145 , b ) 5617 , c ) 15 , d ) 20 , e ) 12 | d | multiply(divide(5, divide(5, 4)), 5) | divide(n1,n0)|divide(n2,#0)|multiply(n1,#1)| | physics |
the average weight of a , b and c is 45 kg . if the average weight of a and b be 40 kg and that of b and c be 45 kg , then the weight of b is : | "explanation : let a , b , c represent their respective weights . then , we have : a + b + c = ( 45 x 3 ) = 135 β¦ . ( i ) a + b = ( 40 x 2 ) = 80 β¦ . ( ii ) b + c = ( 45 x 2 ) = 90 β¦ . ( iii ) adding ( ii ) and ( iii ) , we get : a + 2 b + c = 170 β¦ . ( iv ) subtracting ( i ) from ( iv ) , we get : b = 35 . b β s weigh... | a ) 31 kg , b ) 32 kg , c ) 33 kg , d ) 35 kg , e ) none of these | d | subtract(add(multiply(40, const_2), multiply(45, const_2)), multiply(45, const_3)) | multiply(n1,const_2)|multiply(n2,const_2)|multiply(n0,const_3)|add(#0,#1)|subtract(#3,#2)| | general |
evaluate : 35 - 12 * 3 * 2 = ? | "according to order of operations , 12 ? 3 ? 2 ( division and multiplication ) is done first from left to right 12 * * 2 = 4 * 2 = 8 hence 35 - 12 * 3 * 2 = 35 - 8 = 27 correct answer d" | a ) 62 , b ) 52 , c ) 32 , d ) 27 , e ) 22 | d | subtract(35, multiply(multiply(12, 3), 2)) | multiply(n1,n2)|multiply(n3,#0)|subtract(n0,#1)| | general |
a waiter ' s salary consists of his salary and tips . during one week his tips were 5 / 4 of his salary . what fraction of his income came from tips ? | "income = salary ( s ) + tips = s + s * 5 / 4 = s * 9 / 4 tips = s * 5 / 4 fraction of his income came from tips = ( s * 5 / 4 ) / ( s * 9 / 4 ) = 5 / 9 answer : d" | a ) 4 / 9 , b ) 5 / 4 , c ) 5 / 8 , d ) 5 / 9 , e ) 6 / 9 | d | divide(divide(5, 4), add(divide(5, 4), const_1)) | divide(n0,n1)|add(#0,const_1)|divide(#0,#1)| | general |
in a fort , there are 1200 soldiers . if each soldier consumes 3 kg per day , the provisions available in the fort will last for 30 days . if some more soldiers join , the provisions available will last for 25 days given each soldier consumes 2.5 kg per day . find the number of soldiers joining the fort in that case ? | "assume x soldiers join the fort . 1200 soldiers have provision for 1200 ( days for which provisions last them ) ( rate of consumption of each soldier ) = 1200 ( 30 ) ( 3 ) kg also provisions available for ( 1200 + x ) soldiers is ( 1200 + x ) ( 25 ) ( 2.5 ) kg as the same provisions are available = > 1200 ( 30 ) ( 3 )... | a ) 273 , b ) 528 , c ) 539 , d ) 538 , e ) 723 | b | subtract(divide(divide(multiply(multiply(1200, 3), 30), 25), 2.5), 1200) | multiply(n0,n1)|multiply(n2,#0)|divide(#1,n3)|divide(#2,n4)|subtract(#3,n0)| | general |
19 people went to a hotel for combine dinner party 13 of them spent rs . 79 each on their dinner and rest spent 4 more than the average expenditure of all the 19 . what was the total money spent by them . | solution : let average expenditure of 19 people be x . then , 19 x = 13 * 79 + 6 * ( x + 4 ) ; or , 19 x = 13 * 79 + 6 x + 24 ; or , x = 80.84 ; so , total money spent = 80.84 * 19 = rs . 1536.07 . answer : option e | a ) 1628.4 , b ) 1534 , c ) 1492 , d ) 1496 , e ) none of these | e | multiply(divide(add(multiply(13, 79), multiply(subtract(19, 13), 4)), subtract(19, subtract(19, 13))), 19) | multiply(n1,n2)|subtract(n0,n1)|multiply(n3,#1)|subtract(n0,#1)|add(#0,#2)|divide(#4,#3)|multiply(n0,#5) | general |
in a competitive examination in state a , 6 % candidates got selected from the total appeared candidates . state b had an equal number of candidates appeared and 7 % candidates got selected with 81 more candidates got selected than a . what was the number of candidates appeared from each state ? | state a and state b had an equal number of candidates appeared . in state a , 6 % candidates got selected from the total appeared candidates in state b , 7 % candidates got selected from the total appeared candidates but in state b , 81 more candidates got selected than state a from these , it is clear that 1 % of the ... | a ) 7000 , b ) 8100 , c ) 6000 , d ) 5000 , e ) 4000 | b | divide(81, divide(subtract(7, 6), const_100)) | subtract(n1,n0)|divide(#0,const_100)|divide(n2,#1) | general |
find the compound ratio of ( 3 : 2 ) , ( 4 : 3 ) and ( 1 : 4 ) is | "required ratio = 3 / 2 * 4 / 3 * 1 / 4 = 1 / 2 = 1 : 2 answer is a" | a ) 1 : 2 , b ) 2 : 3 , c ) 3 : 4 , d ) 4 : 5 , e ) 3 : 2 | a | multiply(divide(3, 4), multiply(divide(const_4.0, 2), divide(4, 2))) | divide(n0,n5)|divide(n2,n5)|multiply(#0,#1)|multiply(#0,#2)| | other |
a ( 5 , w ^ 2 ) is the ( x , y ) coordinate of point located on the parabola y = x ^ 2 + 11 . what is the value of w ? | "y = x ^ 2 + 11 w ^ 2 = 5 ^ 2 + 11 w ^ 2 = 36 w = 6 answer d" | a ) 3 . , b ) 4 . , c ) 5 . , d ) 6 . , e ) 9 . | d | power(subtract(power(5, 2), 11), divide(11, 5)) | divide(n3,n0)|power(n0,n2)|subtract(#1,n3)|power(#2,#0)| | general |
how many positive integers less than 5,000 are evenly divisible by neither 16 nor 21 ? | "integers less than 5000 divisible by 16 5000 / 16 = 333 . something , so 333 integers less than 5000 divisible by 21 5000 / 21 = 238 . # # , so 238 we have double counted some , so take lcm of 16 and 21 = 105 and divide by 5000 , we get 47 . so all numbers divisible by 16 and 21 = 333 + 238 - 47 = 524 now subtract tha... | a ) 4,514 , b ) 4,475 , c ) 4,521 , d ) 4,428 , e ) 4,349 | a | divide(factorial(subtract(add(const_4, 16), const_1)), multiply(factorial(16), factorial(subtract(const_4, const_1)))) | add(n1,const_4)|factorial(n1)|subtract(const_4,const_1)|factorial(#2)|subtract(#0,const_1)|factorial(#4)|multiply(#1,#3)|divide(#5,#6)| | general |
a girl walks at a speed of 5 kmph . how much time does he take to walk a distance of 30 km ? | time = distance / speed = 30 / 5 = 6 hours . answer is e | a ) 10 hrs , b ) 2 hrs , c ) 3 hrs , d ) 5 hrs , e ) 6 hrs | e | divide(30, 5) | divide(n1,n0) | physics |
a man Γ’ β¬ β’ s current age is ( 2 / 5 ) of the age of his father . after 10 years , he will be ( 1 / 2 ) of the age of his father . what is the age of father at now ? | "let , father Γ’ β¬ β’ s current age is a years . then , man Γ’ β¬ β’ s current age = [ ( 2 / 5 ) a ] years . therefore , [ ( 2 / 5 ) a + 10 ] = ( 1 / 2 ) ( a + 10 ) 2 ( 2 a + 50 ) = 5 ( a + 10 ) a = 50 d" | a ) 40 , b ) 45 , c ) 38 , d ) 50 , e ) 39 | d | divide(subtract(10, multiply(10, divide(1, 2))), subtract(divide(1, 2), divide(2, 5))) | divide(n3,n0)|divide(n0,n1)|multiply(n2,#0)|subtract(#0,#1)|subtract(n2,#2)|divide(#4,#3)| | general |
when positive integer n is divided by 5 , the remainder is 1 . when n is divided by 7 , the remainder is 3 . what is the smallest positive integer k such that k + n is a multiple of 34 ? | "n = 5 p + 1 = 6,11 , 16,21 , 26,31 n = 7 q + 3 = 3 , 10,17 , 24,31 = > n = 34 m + 31 to get this , we need to take lcm of co - efficients of p and q and first common number in series . so we need to add 3 more to make it 34 m + 34 answer - a" | a ) 3 , b ) 4 , c ) 12 , d ) 32 , e ) 35 | a | subtract(34, reminder(3, 7)) | reminder(n3,n2)|subtract(n4,#0)| | general |
a cricketer whose bowling average is 12.4 runs per wicket takes 5 wickets for 26 runs and thereby decreases his average by 0.4 . the number of wickets taken by him till the last match was : | "let the number of wickets taken till the last match be x . then , 12.4 x + 26 / x + 5 = 12 β 12.4 x + 26 = 12 x + 60 β 0.4 x = 34 β x = 34 β 0.4 = 340 β 4 = 85 . answer d" | a ) 64 , b ) 72 , c ) 80 , d ) 85 , e ) none of the above | d | divide(subtract(multiply(5, subtract(12.4, 0.4)), 26), 0.4) | subtract(n0,n3)|multiply(n1,#0)|subtract(#1,n2)|divide(#2,n3)| | general |
working together , tim and tom can type 12 pages in one hour . if they would be able to type 14 pages in one hour if tom increases his typing speed by 25 % , what is the ratio of tom ' s normal typing speed to that of tim ? | "lets say tim types x pages an hour and tom types y pages an hour . we know that x + y = 12 tom increase speed by 25 % means he will type 1.25 y pages an hour . so we get x + 1.25 y = 14 we need to know the ratio of tom ' s speed to tim ' s speed . this is going to be proportional to the number of pages each can type i... | a ) 2 / 1 , b ) 1 / 2 , c ) 1 / 3 , d ) 4 / 1 , e ) 1 / 4 | a | divide(divide(subtract(14, 12), divide(25, multiply(const_100, const_1))), subtract(12, divide(subtract(14, 12), divide(25, multiply(const_100, const_1))))) | multiply(const_1,const_100)|subtract(n1,n0)|divide(n2,#0)|divide(#1,#2)|subtract(n0,#3)|divide(#3,#4)| | physics |
in a bus left side are 15 seats available , 3 few seats in right side because in rear exit door . each seat hold 3 people . in addition , there is a seat back can sit 9 people all together . how many people can sit in a bus ? | "right side = 15 seat left side = 15 - 3 ( 3 few seat in right side ) = 12 seat total = 15 + 12 = 27 people can seat in 27 seat = 27 * 3 = 81 people can seat in last seat = 9 total people can seat = 81 + 9 = 90 answer : d" | a ) 52 , b ) 49 , c ) 95 , d ) 90 , e ) 66 | d | add(multiply(add(15, subtract(15, 3)), 3), 9) | subtract(n0,n1)|add(n0,#0)|multiply(n1,#1)|add(n3,#2)| | general |
the average of temperatures at noontime from monday to friday is 25 ; the lowest one is 15 , what is the possible maximum range of the temperatures ? | average = 25 , sum of temperatures = 25 * 5 = 125 as the min temperature is 15 , max would be 125 - 4 * 15 = 65 - - > the range = 65 ( max ) - 15 ( min ) = 50 answer : c | a ) 20 , b ) 25 , c ) 50 , d ) 45 , e ) 75 | c | subtract(subtract(multiply(25, add(const_2, const_3)), multiply(15, const_4)), 15) | add(const_2,const_3)|multiply(n1,const_4)|multiply(n0,#0)|subtract(#2,#1)|subtract(#3,n1) | general |
what will be the compound interest on a sum of rs . 27,000 after 3 years at the rate of 12 % p . a . ? | "amount = [ 27000 * ( 1 + 12 / 100 ) 3 ] = 27000 * 28 / 25 * 28 / 25 * 28 / 25 = rs . 37933.0 c . i . = ( 37933.056 - 27000 ) = rs . 10933.05 answer : c" | a ) s . 10123.77 , b ) s . 10123.21 , c ) s . 10933.05 , d ) s . 10123.28 , e ) s . 10123.21 | c | subtract(multiply(multiply(multiply(const_4, const_100), const_100), power(add(const_1, divide(12, const_100)), 3)), multiply(multiply(const_4, const_100), const_100)) | divide(n2,const_100)|multiply(const_100,const_4)|add(#0,const_1)|multiply(#1,const_100)|power(#2,n1)|multiply(#3,#4)|subtract(#5,#3)| | gain |
a merchant marks goods up by 75 % and then offers a discount on the marked price . the profit that the merchant makes after offering the discount is 22.5 % . what % discount did the merchant offer ? | let p be the original price of the goods and let x be the rate after the markup . ( 1.75 p ) * x = 1.225 p x = 1.225 / 1.75 = 0.7 which is a discount of 30 % . the answer is c . | a ) 20 % , b ) 25 % , c ) 30 % , d ) 35 % , e ) 40 % | c | multiply(divide(subtract(subtract(add(const_1, divide(75, const_100)), const_1), divide(22.5, const_100)), add(const_1, divide(75, const_100))), const_100) | divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|subtract(#2,const_1)|subtract(#3,#1)|divide(#4,#2)|multiply(#5,const_100) | gain |
a , b and c rents a pasture for rs . 1130 . a put in 12 horses for 8 months , b 16 horses for 9 months and 18 horses for 6 months . how much should c pay ? | "12 * 8 : 16 * 9 = 18 * 6 8 : 12 : 9 9 / 29 * 1130 = 351 answer : c" | a ) 270 , b ) 199 , c ) 351 , d ) 156 , e ) 122 | c | multiply(divide(1130, add(add(multiply(12, 8), multiply(16, 9)), multiply(18, 6))), multiply(16, 9)) | multiply(n1,n2)|multiply(n3,n4)|multiply(n5,n6)|add(#0,#1)|add(#3,#2)|divide(n0,#4)|multiply(#5,#1)| | general |
how many multiples of 2 are there between 1 and 54 , exclusive ? | "26 multiples of 2 between 1 and 54 exclusive . from 2 * 1 upto 2 * 26 , ( 1,2 , 3,4 , . . . , 26 ) . hence , 26 multiples ! correct option is d" | a ) 21 , b ) 22 , c ) 24 , d ) 26 , e ) 28 | d | add(divide(subtract(54, 1), 2), const_1) | subtract(n2,n1)|divide(#0,n0)|add(#1,const_1)| | general |
at what rate percent per annum will a sum of money double in 2 years . | "let principal = p , then , s . i . = p and time = 8 years rate = [ ( 100 x p ) / ( p x 2 ) ] % = 50 % per annum . answer : b" | a ) 34 % , b ) 50 % , c ) 11.5 % , d ) 20 % , e ) 21.5 % | b | divide(const_100, 2) | divide(const_100,n0)| | gain |
the sum of 2 consecutive numbers is definitely : | "if 1 st term is x : x + ( x + 1 ) = 2 x + 1 - - - > positive if 2 nd term is x : ( x - 1 ) + x = 2 x - 1 - - - > positive a" | a ) positive . , b ) divisible by 2 . , c ) divisible by 3 . , d ) divisible by 4 . , e ) divisible by 5 . | a | add(add(power(add(add(divide(subtract(subtract(2, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(2, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(2, const_10), const_2), const_4), const_2), power(add(di... | subtract(n0,const_10)|subtract(#0,const_2)|divide(#1,const_4)|add(#2,const_2)|power(#2,const_2)|add(#3,const_2)|power(#3,const_2)|add(#5,const_2)|add(#4,#6)|power(#5,const_2)|power(#7,const_2)|add(#9,#10)|add(#11,#8)| | physics |
four concentric circles r given . the radius of 1 st circle is x . next is 2 x , then 3 x and 4 x . given that area b / w 2 nd and 1 st is a and 4 th and 3 rd circles being b . find ratio of a to b | area of first circle = pi * ( x ) ^ 2 = pi * x ^ 2 area of second circle = pi * ( 2 x ) ^ 2 = 4 * pi * x ^ 2 area of third circle = pi * ( 3 x ) ^ 2 = 9 * pi * x ^ 2 area of fourth circle = pi * ( 4 x ) ^ 2 = 16 * pi * x ^ 2 area b / w 1 st and 2 nd circles = 3 * pi * x ^ 2 area b / w 3 rd and 4 th circles = 7 * pi * x... | ['a ) 2 : 7', 'b ) 1 : 7', 'c ) 3 : 5', 'd ) 3 : 7', 'e ) 3 : 4'] | d | divide(3, add(3, 4)) | add(n2,n3)|divide(n2,#0) | geometry |
a grocery shop has a sale of rs . 6335 , rs . 6927 , rs . 6855 , rs . 7230 and rs . 6562 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . 6500 ? | "let the sale in the sixth month = x then ( 6335 + 6927 + 6855 + 7230 + 6562 + x ) / 6 = 6500 = > 6335 + 6927 + 6855 + 7230 + 6562 + x = 6 Γ 6500 = > 33909 + x = 39000 = > x = 39000 β 33909 = 5091 answer : d" | a ) 4857 , b ) 4184 , c ) 4012 , d ) 5091 , e ) 5291 | d | subtract(multiply(6500, add(5, const_1)), add(add(add(6855, add(6335, 6927)), 7230), 6562)) | add(n5,const_1)|add(n0,n1)|add(n2,#1)|multiply(n6,#0)|add(n3,#2)|add(n4,#4)|subtract(#3,#5)| | general |
two numbers have a h . c . f of 9 and a product of two numbers is 1800 . find the l . c . m of the two numbers ? | "l . c . m of two numbers is given by ( product of the two numbers ) / ( h . c . f of the two numbers ) = 1800 / 9 = 200 . answer : a" | a ) 200 , b ) 150 , c ) 160 , d ) 170 , e ) 180 | a | divide(1800, 9) | divide(n1,n0)| | physics |
in a market , a dozen eggs cost as much as a pound of rice , and a half - liter of kerosene costs as much as 4 eggs . if the cost of each pound of rice is $ 0.33 , then how many cents does a liter of kerosene cost ? [ one dollar has 100 cents . ] | a dozen eggs cost as much as a pound of rice - - > 12 eggs = 1 pound of rice = 33 cents ; a half - liter of kerosene costs as much as 8 eggs - - > 4 eggs = 1 / 2 liters of kerosene . how many cents does a liter of kerosene cost - - > 1 liter of kerosene = 8 eggs = 8 / 12 * 33 = 22 cents . answer : d . | a ) 0.22 , b ) 0.44 , c ) 0.55 , d ) 22 , e ) 55 | d | multiply(divide(divide(4, divide(const_1, const_2)), const_12), multiply(0.33, 100)) | divide(const_1,const_2)|multiply(n1,n2)|divide(n0,#0)|divide(#2,const_12)|multiply(#3,#1) | general |
the average salary of all the workers in a workshop is rs . 8000 . the average salary of 7 technicians is rs . 14000 and the average salary of the rest is rs . 6000 . the total number of workers in the workshop is | "solution let the toatl number of workers be x . then 8000 x = ( 14000 x 7 ) + 6000 ( x - 7 ) x = 28 . answer e" | a ) 20 , b ) 21 , c ) 22 , d ) 23 , e ) 28 | e | add(7, divide(multiply(7, subtract(14000, 8000)), subtract(8000, 6000))) | subtract(n2,n0)|subtract(n0,n3)|multiply(n1,#0)|divide(#2,#1)|add(n1,#3)| | general |
a man buys rs . 48 shares paying 9 % dividend . the man wants to have an interest of 12 % on his money . the market value of each share is : | "dividend on rs . 48 = rs . 9 / 100 x 48 = rs . 4.32 . rs . 12 is an income on rs . 100 . rs . 4.32 is an income on rs . 100 / 12 x 4.32 = rs . 36 . answer : option e" | a ) s . 12 , b ) s . 15 , c ) s . 18 , d ) s . 21 , e ) s . 36 | e | multiply(divide(const_100, 12), multiply(divide(9, const_100), 48)) | divide(const_100,n2)|divide(n1,const_100)|multiply(n0,#1)|multiply(#0,#2)| | gain |
a can do a piece of work in 20 days . when he had worked for 2 days b joins him . if the complete work was finished in 8 days . in how many days b alone can finish the work ? | "8 / 20 + 6 / x = 1 x = 10 days answer : e" | a ) 18 , b ) 77 , c ) 66 , d ) 55 , e ) 10 | e | subtract(inverse(subtract(multiply(divide(const_1, 8), subtract(const_1, multiply(2, divide(const_1, 20)))), divide(const_1, 20))), add(20, 8)) | add(n0,n2)|divide(const_1,n2)|divide(const_1,n0)|multiply(n1,#2)|subtract(const_1,#3)|multiply(#1,#4)|subtract(#5,#2)|inverse(#6)|subtract(#7,#0)| | physics |
the ratio of three numbers is 3 : 4 : 5 and the sum of their squares is 1250 . the sum of the numbers is : | "let the numbers be 3 x , 4 x and 5 x then , 9 x 2 + 16 x 2 + 25 x 2 = 1250 50 x 2 = 1250 x = 5 sum of numbers = 3 x + 4 x + 5 x = 12 x = 60 answer c 60" | a ) 30 , b ) 50 , c ) 60 , d ) 70 , e ) 80 | c | sqrt(divide(1250, add(power(5, 4), add(power(3, 4), power(4, 4))))) | power(n0,n1)|power(n1,n1)|power(n2,n1)|add(#0,#1)|add(#3,#2)|divide(n3,#4)|sqrt(#5)| | other |
elena β s bread recipe calls for 8 ounces of butter for each 14 cups of flour used . she needs to make 4 times the original recipe . if 12 ounces of butter is used , then how many cups of flour are needed ? | "solving through algebra route : 8 b + 14 f = x amount if we multiply this equation with 4 we get : 32 b + 56 f = 4 x therefore , we got 32 ounces of butter and 4 x amount of quantity when we use 56 ounces of floor . ans : e" | a ) 1 , b ) 4 , c ) 9 , d ) 13 , e ) 56 | e | multiply(14, 4) | multiply(n1,n2)| | general |
solution a is 20 % salt and solution b is 80 % salt . if you have 30 ounces of solution a and 60 ounces of solution b , in what ratio could you mix solution a with solution b to produce 50 ounces of a 50 % salt solution ? | "let : x = ounces taken from solution a ( 20 % salt ) y = ounces taken from solution b ( 80 % salt ) to prepare 50 ounce 50 % salt . first equation is simple : x + y = 50 to get another equation so as to be able to solve , compute salt contents . 20 % of x + 80 % of y = 50 % of 50 or x / 5 + 4 / 5 * y = 25 or x + 4 y =... | a ) 6 : 4 , b ) 6 : 14 , c ) 4 : 4 , d ) 4 : 6 , e ) 3 : 7 | c | divide(divide(subtract(multiply(50, divide(80, const_100)), multiply(50, divide(50, const_100))), subtract(divide(80, const_100), divide(20, const_100))), subtract(50, divide(subtract(multiply(50, divide(80, const_100)), multiply(50, divide(50, const_100))), subtract(divide(80, const_100), divide(20, const_100))))) | divide(n1,const_100)|divide(n4,const_100)|divide(n0,const_100)|multiply(n4,#0)|multiply(n4,#1)|subtract(#0,#2)|subtract(#3,#4)|divide(#6,#5)|subtract(n4,#7)|divide(#7,#8)| | other |
a man can row at 6 kmph in still water . if the velocity of current is 1 kmph and it takes him 1 hour to row to a place and come back , how far is the place ? | "explanation : speed in still water = 6 kmph speed of the current = 1 kmph speed downstream = ( 6 + 1 ) = 7 kmph speed upstream = ( 6 - 1 ) = 5 kmph let the required distance be x km total time taken = 1 hour Γ’ β‘ β x / 7 + x / 5 = 1 Γ’ β‘ β 5 x + 7 x = 35 Γ’ β‘ β 12 x = 35 Γ’ β‘ β x = 2.9 km answer : option c" | a ) 2.4 km , b ) 2.6 km , c ) 2.9 km , d ) 4.4 km , e ) 5.6 km | c | divide(multiply(subtract(6, 1), const_3), 6) | subtract(n0,n1)|multiply(#0,const_3)|divide(#1,n0)| | physics |
5 kilograms of oranges contained 95 % of water . if the next day the concentration of water decreased by 5 % , what was the new weight of the oranges , in kilograms ? | five kilograms of oranges contained 95 % of water . if the next day the concentration of water decreased by 5 % , what was the new weight of the oranges , in kilograms ? ( a ) 30 ( b ) 25 ( c ) 15 ( d ) 10 ( e ) 5 if b is the answer , then the question means that the concentration of water the newt day became 90 % . ou... | a ) 30 , b ) 25 , c ) 15 , d ) 10 , e ) 5 | b | multiply(5, 5) | multiply(n0,n0) | general |
the mean of 20 values was 150 . it was detected on rechecking that one value 160 was wrongly copied as 135 for the computation of the mean . find the correct mean . | corrected mean = 150 Γ 20 β 135 + 160 / 20 = 3000 β 135 + 160 / 20 = 3025 / 20 = 151.25 answer c | a ) 151 , b ) 149 , c ) 151.25 , d ) 148 , e ) none of the above | c | divide(add(multiply(20, 150), subtract(160, 135)), 20) | multiply(n0,n1)|subtract(n2,n3)|add(#0,#1)|divide(#2,n0) | general |
the present population of a town is 200 . population increase rate is 10 % p . a . find the population of town after 1 years ? | "p = 200 r = 10 % required population of town = p * ( 1 + r / 100 ) ^ t = 200 * ( 1 + 10 / 100 ) = 200 * ( 11 / 10 ) = 220 ( approximately ) answer is d" | a ) 100 , b ) 120 , c ) 200 , d ) 220 , e ) 250 | d | add(200, divide(multiply(200, 10), const_100)) | multiply(n0,n1)|divide(#0,const_100)|add(n0,#1)| | gain |
at a certain resort , each of the 39 food service employees is trained to work in a minimum of 1 restaurant and a maximum of 3 restaurants . the 3 restaurants are the family buffet , the dining room , and the snack bar . exactly 19 employees are trained to work in the family buffet , 18 are trained to work in the dinin... | "39 = 19 + 18 + 12 - 2 - 2 x 2 x = 19 + 18 + 12 - 2 - 39 = 45 - 37 = 8 x = 4 c" | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | c | divide(subtract(subtract(add(add(19, 18), 12), 2), 39), 2) | add(n4,n5)|add(n6,#0)|subtract(#1,n7)|subtract(#2,n0)|divide(#3,n8)| | physics |
if n is a prime number greater than 7 , what is the remainder when n ^ 2 is divided by 12 ? | "there are several algebraic ways to solve this question , but the easiest way is as follows : since we can not have two correct answers just pick a prime greater than 7 , square it and see what would be the remainder upon division of it by 12 . n = 11 - - > n ^ 2 = 121 - - > remainder upon division 121 by 12 is 1 . an... | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 5 | b | subtract(power(add(7, 2), 2), multiply(12, const_4)) | add(n0,n1)|multiply(n2,const_4)|power(#0,n1)|subtract(#2,#1)| | general |
if sushi rows 27 km upstream and 81 km down steam taking 9 hours each , then the speed of the stream | speed upstream = 27 / 9 = 3 kmph speed down stream = 81 / 3 = 27 kmph speed of stream = Β½ ( 27 - 3 ) = 12 kmph answer : a | a ) 12 kmph , b ) 13 kmph , c ) 22 kmph , d ) 20 kmph , e ) 15 kmph | a | add(divide(81, 9), divide(27, 9)) | divide(n1,n2)|divide(n0,n2)|add(#0,#1) | physics |
a rich aristocrat decided that he would give every man rs . 45 and every women rs . 60 . only one nineth of the man and only one twelfth of the women collected their dues . can u tell me how much money thath aristocrat spent if their were 3552 people in total | iman = 45 and 1 women = 60 then 1 / 9 man = 1 / 9 * 45 = 5 and 1 / 12 of women = 1 / 12 * 60 = 5 total people 3552 = 3552 * 5 = 17760 answer : a | a ) 17760 , b ) 18540 , c ) 16320 , d ) 14640 , e ) 15640 | a | multiply(add(multiply(const_2, const_10), add(add(multiply(multiply(const_4, const_10), const_10), add(add(multiply(const_100, const_100), multiply(const_3600, const_2)), multiply(const_4, const_10))), const_100)), const_1) | multiply(const_100,const_100)|multiply(const_2,const_3600)|multiply(const_10,const_4)|multiply(const_10,const_2)|add(#0,#1)|multiply(#2,const_10)|add(#4,#2)|add(#6,#5)|add(#7,const_100)|add(#8,#3)|multiply(#9,const_1) | general |
9 . on level farmland , two runners leave at the same time from the intersection of two country roads . one runner jogs due north at a constant rate of 8 miles per hour while the second runner jogs due east at a constant rate that is 2 miles per hour slower than the first runner ' s rate . how far apart , to the neares... | if runner 1 is going north and runner 2 is going east they are like two sides of a 90 degree triangle . side 1 = 8 m / h - - > 4 m in 1 / 2 hr side 2 = 6 m / h - - > 3 m in 1 / 2 hr to complete this right angle triangle d ^ 2 = 4 ^ 2 + 3 ^ 2 d ^ 2 = 25 = 5 answer option a | a ) 5 , b ) 7 , c ) 8 , d ) 12 , e ) 14 | a | sqrt(add(power(multiply(8, divide(1, 2)), const_2), power(multiply(subtract(8, 2), divide(1, 2)), const_2))) | divide(n3,n2)|subtract(n1,n2)|multiply(n1,#0)|multiply(#0,#1)|power(#2,const_2)|power(#3,const_2)|add(#4,#5)|sqrt(#6) | physics |
a fill pipe can fill 1 / 3 of cistern in 20 minutes in how many minutes , it can fill 3 / 4 of the cistern ? | "1 / 3 of the cistern can fill in 20 min 3 / 4 of the cistern can fill in = 20 * 3 * 3 / 4 = 45 min answer is b" | a ) 15 min , b ) 45 min , c ) 25 min , d ) 40 min , e ) 50 min | b | divide(20, 1) | divide(n2,n0)| | physics |
the length of the bridge , which a train 145 metres long and travelling at 45 km / hr can cross in 30 seconds , is ? | "speed = [ 45 x 5 / 18 ] m / sec = [ 25 / 2 ] m / sec time = 30 sec let the length of bridge be x metres . then , ( 145 + x ) / 30 = 25 / 2 = > 2 ( 145 + x ) = 750 = > x = 230 m . answer : c" | a ) 76 m , b ) 178 m , c ) 230 m , d ) 187 m , e ) 176 m | c | subtract(multiply(divide(multiply(45, speed(const_1000, const_1)), speed(const_3600, const_1)), 30), 145) | speed(const_1000,const_1)|speed(const_3600,const_1)|multiply(n1,#0)|divide(#2,#1)|multiply(n2,#3)|subtract(#4,n0)| | physics |
a cube is divided into 125 identical cubelets . each cut is made parallel to some surface of the cube . but before doing that , the cube is painted with green on one set of opposite faces , red on another set of opposite faces , and blue on the third set of opposite faces . how many cubelets are painted with exactly on... | "each side of the cube has 5 x 5 = 25 cubelets . only the interior cubelets are painted one colour . on each side , 3 x 3 = 9 cubelets are painted one colour . since the cube has six sides , the number of cubes with one colour is 6 * 9 = 54 the answer is d ." | a ) 36 , b ) 42 , c ) 48 , d ) 54 , e ) 60 | d | divide(subtract(125, multiply(multiply(const_4, const_2), const_3)), const_2) | multiply(const_2,const_4)|multiply(#0,const_3)|subtract(n0,#1)|divide(#2,const_2)| | geometry |
indu gave bindu rs . 3750 on compound interest for 2 years at 4 % per annum . how much loss would indu has suffered had she given it to bindu for 2 years at 4 % per annum simple interest ? | "3750 = d ( 100 / 4 ) 2 d = 6 answer : c" | a ) 1 , b ) 2 , c ) 6 , d ) 4 , e ) 9 | c | subtract(subtract(multiply(3750, power(add(const_1, divide(4, const_100)), 2)), 3750), multiply(multiply(3750, divide(4, const_100)), 2)) | divide(n2,const_100)|add(#0,const_1)|multiply(n0,#0)|multiply(n1,#2)|power(#1,n1)|multiply(n0,#4)|subtract(#5,n0)|subtract(#6,#3)| | gain |
for the positive integers x , x + 2 , x + 4 , x + 7 , and x + 37 , the mean is how much greater than the median ? | "mean = ( x + x + 2 + x + 4 + x + 7 + x + 37 ) / 5 = ( 5 x + 50 ) / 5 = x + 10 median = x + 4 thus mean - median = x + 10 - ( x + 4 ) = 6 answer = a" | a ) 6 , b ) 1 , c ) 2 , d ) 4 , e ) 7 | a | subtract(divide(add(add(add(2, 4), 7), 37), add(4, const_1)), 4) | add(n0,n1)|add(const_1,n1)|add(n2,#0)|add(n3,#2)|divide(#3,#1)|subtract(#4,n1)| | general |
the flow of water through a drainage pipe was monitored for a 3 - hour period . in the second hour , the rate of flow was 36 gallons per hour , which was 50 percent faster than the rate of flow for the first hour . if 25 percent more water flowed through the pipe in the third hour than it did in the second , how many g... | rate of flow of water in second hour = 36 gallons per hour rate of flow of water in first hour = 36 / ( 3 / 2 ) = 24 gallons per hour rate of flow of water in third hour = ( 125 / 100 ) * 36 = ( 5 / 4 ) * 36 = 45 gallons per hour number of gallons of water that flowed through the pipe during the entire 3 hours = 24 + 3... | a ) 106.25 , b ) 105.5 , c ) 105.75 , d ) 106 , e ) 105.0 | e | add(add(divide(36, divide(3, const_2)), add(36, multiply(const_0_25, 36))), 36) | divide(n0,const_2)|multiply(n1,const_0_25)|add(n1,#1)|divide(n1,#0)|add(#2,#3)|add(n1,#4) | general |
running at the same constant rate , 6 identical machines can produce a total of 330 bottles per minute . at this rate , how many bottles could 10 such machines produce in 4 minutes ? | "let the required number of bottles be x . more machines , more bottles ( direct proportion ) more minutes , more bottles ( direct proportion ) machines 6 : 10 : : 330 : x time ( in minutes ) 1 : 4 6 x 1 x x = 10 x 4 x 330 x = ( 10 x 4 x 330 ) / ( 6 ) x = 2200 . answer : c" | a ) 648 , b ) 1800 , c ) 2200 , d ) 10800 , e ) 10900 | c | multiply(multiply(divide(330, 6), 4), 10) | divide(n1,n0)|multiply(n3,#0)|multiply(n2,#1)| | gain |
if a light flashes every 6 seconds , how many times will it flash in 1 / 4 of an hour ? | "in 1 / 4 of an hour there are 15 * 60 = 900 seconds the number of 6 - second intervals = 900 / 6 = 150 after the first flash , there will be 150 more flashes for a total of 151 . the answer is b ." | a ) 51 , b ) 151 , c ) 251 , d ) 351 , e ) 451 | b | divide(multiply(divide(const_3600, const_4), const_3), 6) | divide(const_3600,const_4)|multiply(#0,const_3)|divide(#1,n0)| | physics |
a sum of rs . 1360 has been divided among a , b and c such that a gets of what b gets and b gets of what c gets . b β s share is : | "explanation let c β s share = rs . x then , b β s share = rs . x / 4 , a β s share = rs . ( 2 / 3 x x / 4 ) = rs . x / 6 = x / 6 + x / 4 + x = 13660 = > 17 x / 12 = 13660 = > 13660 x 12 / 17 = rs . 960 hence , b β s share = rs . ( 960 / 4 ) = rs . 240 . answer c" | a ) rs . 120 , b ) rs . 160 , c ) rs . 240 , d ) rs . 300 , e ) none | c | subtract(subtract(multiply(divide(1360, const_10), const_2), const_12), const_12) | divide(n0,const_10)|multiply(#0,const_2)|subtract(#1,const_12)|subtract(#2,const_12)| | general |
find the compound interest on $ 20000 in 2 years at 4 % per annum , the interest being compounded half - yearly ? | "principle = $ 20000 rate = 2 % half yearly = 4 half years amount = 20000 * ( 1 + 2 / 100 ) ^ 4 = 20000 * 51 / 50 * 51 / 50 * 51 / 50 * 51 / 50 = $ 21648.64 c . i . = 21648.64 - 20000 = $ 1648.64 answer is b" | a ) $ 645.56 , b ) $ 1648.64 , c ) $ 954.26 , d ) $ 745.69 , e ) $ 1020.45 | b | subtract(multiply(power(add(divide(divide(4, const_100), 2), const_1), 4), 20000), 20000) | divide(n2,const_100)|divide(#0,n1)|add(#1,const_1)|power(#2,n2)|multiply(n0,#3)|subtract(#4,n0)| | gain |
78 persons can repair a road in 12 days , working 5 hours a day . in how many days will 30 persons , working 6 hours a day , complete the work ? | "let the required number of days be x . less persons , more days ( indirect proportion ) more working hours per day , less days ( indirect proportion ) persons 30 : 78 : : 12 : x working hours / day 6 : 5 30 x 6 x x = 78 x 5 x 12 x = ( 78 x 5 x 12 ) / ( 30 x 6 ) x = 26 . answer : c" | a ) 10 , b ) 13 , c ) 26 , d ) 15 , e ) 16 | c | divide(multiply(multiply(78, 12), 5), multiply(30, 6)) | multiply(n0,n1)|multiply(n3,n4)|multiply(n2,#0)|divide(#2,#1)| | physics |
can you find a 6 digit number under 5 lakh such that the sum of it ' s digits is 43 ? | b the required number is 499449 . the maximum sum of a number under 5 lakh will be of the number 499999 i . e . 49 . thus the required number must be close to 499999 . the root of 499999 ~ 707.106 by calculating the square of 707 , we get 499849 which is the answer . | a ) 588338 , b ) 499449 , c ) 399826 , d ) 788392 , e ) 622975 | b | power(add(multiply(add(6, const_1), const_100), add(6, const_1)), const_2) | add(n0,const_1)|multiply(#0,const_100)|add(#0,#1)|power(#2,const_2) | general |
a real estate agent received a 6 % commission on the selling price of a house . if his commission was $ 8880 , what was the selling price of the house ? | 6 % x = 8,880 : x = selling price of house . x = $ 148,000 : solve for x correct answer c . | a ) $ 234000 , b ) $ 227000 , c ) $ 148000 , d ) $ 479000 , e ) $ 337,000 | c | divide(8880, divide(6, const_100)) | divide(n0,const_100)|divide(n1,#0) | gain |
there is a total of 84 marbles in a box , each of which is red , green , blue , or white . if one marble is drawn from the box at random , the probability that it will be white is 1 / 4 and the probability that it will be green is 2 / 7 . what is the probability that the marble will be either red or blue ? | p ( red or blue ) = 1 - p ( white ) - p ( green ) = 28 / 28 - 7 / 28 - 8 / 28 = 13 / 28 the answer is c . | a ) 3 / 7 , b ) 11 / 14 , c ) 13 / 28 , d ) 19 / 28 , e ) 29 / 42 | c | subtract(1, add(divide(1, 4), divide(2, 7))) | divide(n1,n2)|divide(n3,n4)|add(#0,#1)|subtract(n1,#2) | general |
( x + 7 ) is a factor in x ^ 2 - mx - 35 . what is the value of m ? | "i solved the second degree equation and found it like this : x ^ 2 - mx - 35 ( x - 5 ) ( x + 7 ) x = 5 or x = - 7 substituting both values for x in the equation we find : x ^ 2 - mx - 35 = > ( - 7 ) ^ 2 - m ( - 7 ) = 35 = > 49 + 7 m = 35 = > 7 m = 35 - 49 = - 14 = > m = - 2 and with 5 , using a similar process we end ... | a ) - 3 , b ) - 2 , c ) - 4 , d ) - 5 , e ) - 6 | b | subtract(divide(35, 7), 7) | divide(n2,n0)|subtract(#0,n0)| | general |
the product of two positive integers is 575 and their difference is 2 . what is the bigger number ? | "let ' s use trial and error to find the two numbers . 27 * 25 = 675 ( too high ) 25 * 23 = 575 the answer is c ." | a ) 21 , b ) 23 , c ) 25 , d ) 27 , e ) 29 | c | subtract(add(multiply(divide(575, divide(575, 2)), const_4), const_10), const_1) | divide(n0,n1)|divide(n0,#0)|multiply(#1,const_4)|add(#2,const_10)|subtract(#3,const_1)| | general |
at a certain pizza parlor , the diameter of a large pizza is 40 % larger than the diameter of a small pizza . what is the percent increase in total amount of pizza , from a small to a large ? | a real world problem β - yum ! β - but again , the specifics of the shape don β t matter . diameter increases 40 % , so k = 1.4 in this case . β amount of pizza β just means the area of the circle , so k ^ 2 = ( 1.4 ) ^ 2 = 1.96 , which is the multiplier for a 96 % increase . that β s the percent increase in area . ans... | a ) 20 % , b ) 40 % , c ) 64 % , d ) 80 % , e ) 96 % | e | divide(subtract(power(add(const_100, 40), const_2), power(const_100, const_2)), const_100) | add(n0,const_100)|power(const_100,const_2)|power(#0,const_2)|subtract(#2,#1)|divide(#3,const_100) | general |
on a map , 2.5 inches represent 40 miles . how many miles approximately is the distance if you measured 155 centimeters assuming that 1 - inch is 2.54 centimeters ? | "1 inch = 2.54 cm 2.5 inch = 2.54 * 2.5 cm 6.35 cm = 40 miles 155 cms = 40 / 6.35 * 155 = 976.4 miles answer : d" | a ) 990.4 , b ) 970 , c ) 972 , d ) 976.4 , e ) 975 | d | multiply(divide(155, 2.54), divide(40, 2.5)) | divide(n2,n4)|divide(n1,n0)|multiply(#0,#1)| | physics |
the average weight of 20 persons sitting in a boat had some value . a new person added to them whose weight was 48 kg only . due to his arrival , the average weight of all the persons decreased by 5 kg . find the average weight of first 20 persons ? | 20 x + 48 = 21 ( x β 5 ) x = 57 answer : c | a ) 55 , b ) 56 , c ) 57 , d ) 58 , e ) 59 | c | subtract(multiply(add(20, const_1), 5), 48) | add(n0,const_1)|multiply(n2,#0)|subtract(#1,n1) | general |
if two integers x , y ( x > y ) are selected from - 4 to 4 ( inclusive ) , how many cases are there ? | "there are 9 integers from - 4 to 4 inclusive . 9 c 2 = 36 . the answer is d ." | a ) 27 , b ) 30 , c ) 33 , d ) 36 , e ) 39 | d | add(add(add(add(add(add(add(4, 4), add(4, const_2)), add(4, const_1)), 4), 4), const_2), const_1) | add(n0,n1)|add(n1,const_2)|add(n1,const_1)|add(#0,#1)|add(#3,#2)|add(#4,n1)|add(#5,n0)|add(#6,const_2)|add(#7,const_1)| | general |
a man is 24 years older than his son . in four years , his age will be twice the age of his son . the present age of the son is | "solution let the son ' s present age be x years . then , man ' s present age = ( x + 24 ) years . then Γ’ β¬ ΒΉ = Γ’ β¬ ΒΊ ( x + 24 ) + 4 = 2 ( x + 4 ) Γ’ β¬ ΒΉ = Γ’ β¬ ΒΊ x + 28 = 2 x + 8 x = 20 . answer c" | a ) 14 years , b ) 18 years , c ) 20 years , d ) 22 years , e ) none | c | divide(subtract(24, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | multiply(const_2,const_2)|subtract(const_2,const_1)|subtract(#0,const_2)|subtract(n0,#2)|divide(#3,#1)| | general |
a car is purchased on hire - purchase . the cash price is $ 23 000 and the terms are a deposit of 10 % of the price , then the balance to be paid off over 60 equal monthly instalments . interest is charged at 12 % p . a . what is the monthly instalment ? | "explanation : cash price = $ 23 000 deposit = 10 % Γ£ β $ 23 000 = $ 2300 loan amount = $ 23000 Γ’ Λ β $ 2300 number of payments = 60 = $ 20700 i = p * r * t / 100 i = 12420 total amount = 20700 + 12420 = $ 33120 regular payment = total amount / number of payments = 552 answer : e" | a ) $ 503 , b ) $ 504 , c ) $ 515 , d ) $ 543 , e ) $ 552 | e | add(divide(multiply(multiply(23, const_1000), subtract(const_1, divide(10, const_100))), 60), multiply(divide(divide(12, const_100), 12), multiply(multiply(23, const_1000), subtract(const_1, divide(10, const_100))))) | divide(n2,const_100)|divide(n4,const_100)|multiply(n0,const_1000)|divide(#1,n4)|subtract(const_1,#0)|multiply(#2,#4)|divide(#5,n3)|multiply(#3,#5)|add(#6,#7)| | gain |
on the number line , if x is halfway between - 8 and 4 , and if y is halfway between - 2 and 6 , what number is halfway between x and y ? | x = - 2 and y = 2 . the answer is c . | a ) - 2 , b ) - 1 , c ) 0 , d ) 1 , e ) 2 | c | add(negate(8), 4) | negate(n0)|add(n1,#0) | general |
a survey was sent to 90 customers , 15 of whom responded . then the survey was redesigned and sent to another 50 customers , 9 of whom responded . by approximately what percent did the response rate increase from the original survey to the redesigned survey ? | "rate of first survey = 15 / 90 rate of second survey = 9 / 50 % response rate increase ( 9 / 50 - 15 / 90 ) / ( 15 / 90 ) = 8 % answer is c = 8 %" | a ) 2 % , b ) 5 % , c ) 8 % , d ) 28 % , e ) 63 % | c | floor(multiply(subtract(divide(9, 50), divide(15, 90)), const_100)) | divide(n3,n2)|divide(n1,n0)|subtract(#0,#1)|multiply(#2,const_100)|floor(#3)| | gain |
company kw is being sold , and both company a and company b were considering the purchase . the price of company kw is 50 % more than company a has in assets , and this same price is also 100 % more than company b has in assets . if companies a and b were to merge and combine their assets , the price of company kw woul... | "let the price of company a ' s assets be 100 price of assets of kw is 50 % more than company a ' s assets which is 150 price of assets of kw is 100 % more than company b ' s assets which means price of company b ' s assets is half the price of kw = 75 a + b = 175 kw = 150 kw / ( a + b ) * 100 = 150 / 175 * 100 = 85.7 ... | a ) 66 % , b ) 75 % , c ) 86 % , d ) 116 % , e ) 150 % | c | multiply(divide(add(100, 50), add(100, divide(add(100, 50), const_2))), 100) | add(n0,n1)|divide(#0,const_2)|add(n1,#1)|divide(#0,#2)|multiply(#3,n1)| | gain |
a rectangular field is to be fenced on three sides leaving a side of 80 feet uncovered . if the area of the field is 680 sq . ft , how many feet of fencing will be required ? | "given that length and area , so we can find the breadth . length x breadth = area 80 x breadth = 680 breadth = 8.5 feet area to be fenced = 2 b + l = 2 ( 8.5 ) + 80 = 97 feet answer : b ) 97 ft" | a ) 44 ft , b ) 97 ft , c ) 22 ft , d ) 11 ft , e ) 66 ft | b | add(multiply(divide(680, 80), const_2), 80) | divide(n1,n0)|multiply(#0,const_2)|add(n0,#1)| | geometry |
rs . 120 are divided among a , b , c such that a ' s share is rs . 20 more than b ' s and rs . 20 less than c ' s . what is b ' s share | explanation : let c = x . then a = ( x οΏ½ 20 ) and b = ( x οΏ½ 40 ) . x + x - 20 + x - 40 = 120 or x = 60 . a : b : c = 40 : 20 : 60 = 2 : 1 : 3 . b ' s share = rs . 120 * ( 1 / 6 ) = rs . 20 answer : b | a ) rs 10 , b ) rs 20 , c ) rs 24 , d ) rs 28 , e ) none of these | b | divide(subtract(subtract(120, 20), add(20, 20)), const_3) | add(n1,n1)|subtract(n0,n1)|subtract(#1,#0)|divide(#2,const_3) | general |
a certain meter records voltage between 0 and 10 volts inclusive . if the average value of 3 recordings on the meter was 5 volts , what was the smallest possible recording in volts ? | "if average of 3 is 5 so sum of 3 should be 15 3 recording can be from 0 - 10 inclusive to find one smallest other two should be highest so , lets assume three var are a , b , c say a is smallest and give b and c greatest readings say 6 and 6 so a has to be 3 b" | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | b | subtract(multiply(3, 5), 10) | multiply(n2,n3)|subtract(#0,n1)| | general |
the ratio of investments of two partners p and q is 7 : 5 and the ratio of their profits is 7 : 10 . if p invested the money for 10 months , find for how much time did q invest the money ? | "7 * 5 : 10 * x = 7 : 10 x = 20 answer : c" | a ) 11 , b ) 10 , c ) 20 , d ) 16 , e ) 12 | c | multiply(multiply(divide(7, 5), divide(10, 7)), 10) | divide(n0,n1)|divide(n3,n2)|multiply(#0,#1)|multiply(n4,#2)| | gain |
a jar full of whisky contains 40 % alcohol . a part of this whisky is replaced by another containg 19 % alcohol and now the percentage of alcohol was found to be 25 % . what quantity of whisky is replaced ? | "let us assume the total original amount of whiskey = 10 ml - - - > 4 ml alcohol and 6 ml non - alcohol . let x ml be the amount removed - - - > total alcohol left = 4 - 0.4 x new quantity of whiskey added = x ml out of which 0.19 is the alcohol . thus , the final quantity of alcohol = 4 - 0.4 x + 0.19 x - - - - > ( 4 ... | a ) 1 / 3 , b ) 2 / 3 , c ) 2 / 5 , d ) 3 / 5 , e ) 4 / 5 | c | divide(subtract(40, 25), subtract(40, 19)) | subtract(n0,n2)|subtract(n0,n1)|divide(#0,#1)| | gain |
if x is an integer such that 0 < x < 12 , 1 < x < 15 , 10 > x > 5 , 11 > x > 8 , and x + 2 < 12 , then x is | "0 < x < 12 , 1 < x < 15 , - 5 < x < 10 8 < x < 11 x < 10 from above : 8 < x < 10 - - > x = 9 . answer : d ." | a ) 3 , b ) 5 , c ) 7 , d ) 9 , e ) 11 | d | subtract(subtract(12, 2), 2) | subtract(n9,n8)|subtract(#0,n8)| | general |
a sum of money lent out at s . i . amounts to rs . 660 after 2 years and to rs . 1020 after a further period of 5 years . the sum is ? | "s . i for 5 years = ( 1020 - 660 ) = rs . 360 . s . i . for 2 years = 360 / 5 * 2 = rs . 144 . principal = ( 660 - 144 ) = rs . 516 . answer : e" | a ) s . 440 , b ) s . 500 , c ) s . 540 , d ) s . 740 , e ) s . 516 | e | subtract(660, multiply(divide(subtract(1020, 660), 5), 2)) | subtract(n2,n0)|divide(#0,n3)|multiply(n1,#1)|subtract(n0,#2)| | general |
what number has a 5 : 1 ratio to the number 8 ? | 5 : 1 = x : 8 x = 40 answer : b | a ) 22 , b ) 40 , c ) 88 , d ) 52 , e ) 12 | b | multiply(8, 5) | multiply(n0,n2) | other |
for an agricultural experiment , 500 seeds were planted in one plot and 200 were planted in a second plot . if exactly 30 percent of the seeds in the first plot germinated and exactly 50 percent of the seeds in the second plot germinated , what percent of the total number of seeds germinated ? | "from the first plot 0.30 * 500 = 150 ; from the second plot 0.50 * 200 = 100 ; total percent = ( 150 + 100 ) / ( 500 + 200 ) = 0.357 . answer : a ." | a ) 35.7 % , b ) 25 % , c ) 32 % , d ) 30.8 % , e ) 60 % | a | multiply(divide(add(multiply(divide(30, const_100), 500), multiply(divide(50, const_100), 200)), add(500, 200)), const_100) | add(n0,n1)|divide(n2,const_100)|divide(n3,const_100)|multiply(n0,#1)|multiply(n1,#2)|add(#3,#4)|divide(#5,#0)|multiply(#6,const_100)| | general |
what is the total number of prime numbers less than 70 ? | solution prime numbers less than 70 are = 12 . 2,3 , 5,7 , 11,13 , 17,19 , 23,29 , 31,37 , 41,43 , 47,53 , 59,61 , and 67 their number is 19 . answer c | a ) 17 , b ) 18 , c ) 19 , d ) 20 , e ) 21 | c | max(add(add(const_10, const_3), const_4), add(add(add(const_10, const_3), const_4), const_2)) | add(const_10,const_3)|add(#0,const_4)|add(#1,const_2)|max(#1,#2) | general |
3 / 5 of all married couples have more than one child . 1 / 2 of all married couples have more than 3 children . what fraction of all married couples have 2 or 3 children ? | "plug in simple numbers . take 100 couples for example . 3 / 5 of 100 couples have more than one child = 60 couples . 1 / 2 of 100 couples have more than 3 kids = 50 couples . this implies that 50 couples are a subset of 60 couples and the complement of 60 couples within those 100 couples , which equals 40 couples have... | a ) 1 / 5 , b ) 1 / 10 , c ) 7 / 20 , d ) 3 / 5 , e ) it can not be determined from the given information . | b | subtract(divide(3, 5), divide(2, 5)) | divide(n0,n1)|divide(n2,n3)|subtract(#0,#1)| | general |
at a special sale , 6 tickets can be purchased for the price of 3 tickets . if 6 tickets are purchased at the sale , the amount saved will be what percent of the original price of the 6 tickets ? | let the price of a ticket be rs . 100 , so 3 tickets cost 300 & 6 tickets cost 600 6 tickets purchased at price of 3 tickets ie . , for 300 , so amount saved s rs . 300 , % of 6 tickets = ( 300 / 600 ) * 100 = 50 % answer : d | a ) 20 % , b ) 33.3 % , c ) 40 % , d ) 50 % , e ) 66.6 % | d | divide(multiply(subtract(multiply(6, 6), multiply(3, 6)), const_100), multiply(6, 6)) | multiply(n0,n0)|multiply(n0,n1)|subtract(#0,#1)|multiply(#2,const_100)|divide(#3,#0) | gain |
a 4 - digit code can consist of the digits 0 ~ 9 except that it does not contain the digits 1 , 2 , 4 , and 6 at all . if repeated digits are allowed , what is the probability that it has at least one even digit ? | p ( no even digits ) = 2 / 3 * 2 / 3 * 2 / 3 * 2 / 3 = 16 / 81 p ( at least one even digit ) = 1 - 16 / 81 = 65 / 81 the answer is d . | a ) 23 / 24 , b ) 31 / 36 , c ) 55 / 64 , d ) 65 / 81 , e ) 75 / 121 | d | divide(subtract(power(subtract(const_10, const_4), 4), power(const_4, 4)), power(subtract(const_10, const_4), 4)) | power(const_4,n0)|subtract(const_10,const_4)|power(#1,n0)|subtract(#2,#0)|divide(#3,#2) | general |
by selling an article at rs . 800 , a shopkeeper makes a profit of 25 % . at what price should he sell the article so as to make a loss of 15 % ? | "sp = 800 profit = 25 % cp = ( sp ) * [ 100 / ( 100 + p ) ] = 800 * [ 100 / 125 ] = 640 loss = 25 % = 25 % of 640 = rs . 96 sp = cp - loss = 640 - 96 = rs . 544 answer : e" | a ) 228 , b ) 480 , c ) 267 , d ) 288 , e ) 544 | e | subtract(divide(multiply(800, const_100), add(25, const_100)), divide(multiply(divide(multiply(800, const_100), add(25, const_100)), 15), const_100)) | add(n1,const_100)|multiply(n0,const_100)|divide(#1,#0)|multiply(n2,#2)|divide(#3,const_100)|subtract(#2,#4)| | gain |
the speed of a train is 120 kmph . what is the distance covered by it in 10 minutes ? | "120 * 10 / 60 = 20 kmph answer : b" | a ) 26 kmph , b ) 20 kmph , c ) 28 kmph , d ) 30 kmph , e ) 40 kmph | b | multiply(divide(10, const_60), 120) | divide(n1,const_60)|multiply(n0,#0)| | physics |
from a total of 4 boys and 4 girls , how many 4 - person committees can be selected if the committee must have exactly 2 boys and 2 girls ? | "answer = b = 36 no of 4 person committees that can be formed = 4 c 2 * 4 c 2 = 36 answer b" | a ) 16 , b ) 36 , c ) 60 , d ) 120 , e ) 240 | b | multiply(multiply(const_3, 4), multiply(4, 2)) | multiply(n0,const_3)|multiply(n0,n3)|multiply(#0,#1)| | general |
if an article is sold at 20 % profit instead of 14 % profit , then the profit would be rs . 180 more . what is the cost price ? | "explanation : let the cost price of an article be rs . x . ( 20 % of x ) - ( 14 % of x ) = 180 20 x / 100 - 14 x / 100 = 105 = > 6 x = 180 * 100 = > x = 3000 cost price = rs . 3000 answer : c" | a ) rs . 1000 , b ) rs . 2000 , c ) rs . 3000 , d ) rs . 4000 , e ) rs . 5200 | c | multiply(divide(180, 14), const_100) | divide(n2,n1)|multiply(#0,const_100)| | gain |
find the greatest number that exactly divides 35 , 91 and 840 ? | "the greatest number that exactly divides 35 , 91 and 840 is the hcf of the three numbers . so , calculating hcf we get the answer 7 . answer : c" | a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | c | gcd(gcd(subtract(subtract(840, 91), 35), subtract(91, 35)), subtract(840, 91)) | subtract(n2,n1)|subtract(n1,n0)|subtract(#0,n0)|gcd(#2,#1)|gcd(#3,#0)| | general |
joe needs to paint all the airplane hangars at the airport , so he buys 360 gallons of paint to do the job . during the first week , he uses 1 / 3 of all the paint . during the second week , he uses 1 / 5 of the remaining paint . how many gallons of paint has joe used ? | "total paint initially = 360 gallons paint used in the first week = ( 1 / 3 ) * 360 = 120 gallons . remaning paint = 240 gallons paint used in the second week = ( 1 / 5 ) * 240 = 48 gallons total paint used = 168 gallons . option a" | a ) 168 , b ) 144 , c ) 175 , d ) 216 , e ) 250 | a | add(multiply(divide(360, 3), 1), divide(subtract(360, multiply(divide(360, 3), 1)), 5)) | divide(n0,n2)|multiply(n1,#0)|subtract(n0,#1)|divide(#2,n4)|add(#3,#1)| | physics |
find the probability that a non leap year selected at random will have 53 sundays | "there are 365 days in a leap year : 52 weeks and 1 more days . so , 52 sundays and 1 days . this 1 days can be { mon , tue , wed , thu , fri , sat , sun } ( 7 cases ) . in order to have 53 sundays we should have sun case . the probability of that is 1 / 7 . answer : e ." | a ) 6 / 7 , b ) 5 / 7 , c ) 4 / 7 , d ) 3 / 7 , e ) 1 / 7 | e | divide(const_2, add(const_3, const_4)) | add(const_3,const_4)|divide(const_2,#0)| | probability |
when a number is divided by 6 & then multiply by 12 the answer is 11 what is the no . ? | "if $ x $ is the number , x / 6 * 12 = 11 = > 2 x = 11 = > x = 5.5 c" | a ) 4.5 , b ) 5 , c ) 5.5 , d ) 5.8 , e ) 6 | c | multiply(divide(11, 12), 6) | divide(n2,n1)|multiply(n0,#0)| | general |
sides of a rectangular park are in the ratio 3 : 2 and its area is 2400 sq m , the cost of fencing it at 50 ps per meter is ? | "3 x * 2 x = 2400 = > x = 20 2 ( 60 + 40 ) = 200 m 200 * 1 / 2 = rs . 100 answer : b" | a ) 287 , b ) 100 , c ) 125 , d ) 988 , e ) 271 | b | divide(multiply(50, rectangle_perimeter(sqrt(divide(multiply(2400, 2), 3)), divide(2400, sqrt(divide(multiply(2400, 2), 3))))), const_100) | multiply(n1,n2)|divide(#0,n0)|sqrt(#1)|divide(n2,#2)|rectangle_perimeter(#3,#2)|multiply(n3,#4)|divide(#5,const_100)| | physics |
if 3 x + 4 y = 60 and 6 x - 4 y = 12 , what is the product of x and y ? | given 3 x + 4 y = 60 - - - eq 1 6 x - 4 y = 12 - - eq 2 sum both eqns we get 9 x = 72 = > x = 8 sum 6 x in eq 2 = > 48 - 4 y = 12 . = > y = 9 now xy = 8 * 9 = 72 option d is correct answer . | a ) 80 , b ) 54 , c ) 54 , d ) 72 , e ) 12 | d | multiply(divide(add(60, 12), add(6, 3)), divide(subtract(multiply(6, divide(add(60, 12), add(6, 3))), 12), 4)) | add(n2,n5)|add(n0,n3)|divide(#0,#1)|multiply(n3,#2)|subtract(#3,n5)|divide(#4,n1)|multiply(#2,#5) | general |
a rope can make 70 rounds of the circumference of a cylinder whose radius of the base is 14 cm . how many times can it go round a cylinder having radius 20 cm ? | explanation : let the required number of rounds be x more radius , less rounds ( indirect proportion ) hence we can write as radius 14 : 20 } : : x : 70 β 14 Γ 70 = 20 x β 14 Γ 7 = 2 x β x = 7 Γ 7 = 49 . answer : option a | ['a ) 49 rounds', 'b ) 42 rounds', 'c ) 54 rounds', 'd ) 52 rounds', 'e ) none of these'] | a | divide(multiply(circumface(14), 70), circumface(20)) | circumface(n1)|circumface(n2)|multiply(n0,#0)|divide(#2,#1) | geometry |
a seller of used cars has 18 cars to sell and each of his clients selected 3 cars that he liked most . if each car was selected exactly thrice , how many clients visited the garage ? | "ifno caris selected more than once then the number of clients = 18 / 3 = 6 but since every car is being selected three times so no . of clients must be thrice as well = 6 * 3 = 18 answer : option d" | a ) 8 , b ) 10 , c ) 12 , d ) 18 , e ) 16 | d | multiply(divide(18, 3), const_3) | divide(n0,n1)|multiply(#0,const_3)| | general |
what percent of 120 are 90 ? | "explanation : ( ? % / 100 ) * 120 = 90 ? = 75 % answer is c" | a ) 25 % , b ) 50 % , c ) 75 % , d ) 33 % , e ) 30 % | c | multiply(divide(120, 90), const_100) | divide(n0,n1)|multiply(#0,const_100)| | gain |
vijay sells a cupboard at 12 % below cost price . had he got rs . 1650 more , he would have made a profit of 12 % . what is the cost price of the cupboard ? | "explanation : cost price = 1650 / ( 0.12 + 0.12 ) = 2086 / 0.24 = rs . 6875 answer d" | a ) 7450 , b ) 14900 , c ) 7400 , d ) 6875 , e ) none of these | d | divide(1650, divide(subtract(add(const_100, 12), subtract(const_100, 12)), const_100)) | add(n0,const_100)|subtract(const_100,n0)|subtract(#0,#1)|divide(#2,const_100)|divide(n1,#3)| | gain |
a hostel had provisions for 250 men for 48 days . if 50 men left the hostel , how long will the food last at the same rate ? | "a hostel had provisions for 250 men for 48 days if 50 men leaves the hostel , remaining men = 250 - 50 = 200 we need to find out how long the food will last for these 200 men . let the required number of days = x days more men , less days ( indirect proportion ) ( men ) 250 : 200 : : x : 48 250 Γ 48 = 200 x 5 Γ 48 = 4... | a ) 30 , b ) 40 , c ) 50 , d ) 60 , e ) 65 | d | divide(multiply(250, 48), subtract(250, 50)) | multiply(n0,n1)|subtract(n0,n2)|divide(#0,#1)| | gain |
8 identical machines , working alone and at their constant rates , take 6 hours to complete a job lot . how long would it take for 12 such machines to perform the same job ? | "let each machine do 1 unit of work for 1 hour 8 machines - - > 8 units of work in 1 hour for 6 hours = 8 * 6 = 48 units of total work is done . now this 48 units of total work must be done by 3 machines 12 units of work ( 12 machines ) - - - > 1 hour for 48 units of work 12 * 4 - - - > 1 * 4 hours d 4 hours" | a ) 6 hours , b ) 8.75 hours , c ) 12 hours , d ) 4 hours , e ) 16 hours | d | divide(multiply(6, 8), 12) | multiply(n0,n1)|divide(#0,n2)| | physics |
if a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster , he would have covered 70 more miles than he actually did . how many more miles would he have covered than he actually did if he had driven 1 hours longer and at an average rate of 10 miles per hour faster on that... | "case 1 : let rate = r , time = t , and distance = d so d = rt case 2 : ( d + 70 ) = ( r + 5 ) ( t + 1 ) case 3 : ( d + x ) = ( r + 10 ) ( t + 1 ) x = 130 ans c" | a ) 100 , b ) 120 , c ) 130 , d ) 150 , e ) 160 | c | add(multiply(subtract(70, 5), 1), multiply(1, 10)) | multiply(n3,n4)|subtract(n2,n1)|multiply(n3,#1)|add(#2,#0)| | physics |
if a 5 percent deposit that has been paid toward the purchase of a certain product is $ 50 , how much more remains to be paid ? | "95 % remains to be paid so the remaining amount is 19 * 50 = $ 950 . the answer is e ." | a ) $ 550 , b ) $ 650 , c ) $ 750 , d ) $ 850 , e ) $ 950 | e | subtract(multiply(50, divide(const_100, 5)), 50) | divide(const_100,n0)|multiply(n1,#0)|subtract(#1,n1)| | general |
john buys good worth rs . 6650 . he gets a rebate of 6 % on it . after getting the rebate , he pays sales tax @ 10 % . find the amount he will have to pay for the goods . | rebate = 6 % of rs . 6650 = rs . 6 x 6650 = rs . 399 . 100 sales tax = 10 % of rs . ( 6650 - 399 ) = rs . 10 x 6251 = rs . 625.10 100 final amount = rs . ( 6251 + 625.10 ) = rs . 6876.10 correct answer - a | a ) rs . 6876.10 , b ) rs . 4500 , c ) rs . 4600 , d ) rs . 3400 , e ) rs . 7500 | a | add(multiply(divide(10, const_100), subtract(6650, multiply(divide(6, const_100), 6650))), subtract(6650, multiply(divide(6, const_100), 6650))) | divide(n2,const_100)|divide(n1,const_100)|multiply(n0,#1)|subtract(n0,#2)|multiply(#0,#3)|add(#4,#3) | gain |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.