Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 164 | correct stringclasses 5 values | annotated_formula stringlengths 7 1.65k | linear_formula stringlengths 8 925 | category stringclasses 6 values | answer stringclasses 5 values |
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a is twice as good a work man as b and together they finish the work in 6 days . in how many days a alone can finish the work ? | "wc = 2 : 1 2 x + x = 1 / 6 = > x = 1 / 18 2 x = 1 / 9 a can do the work in 9 days . answer : b" | a ) 3 , b ) 9 , c ) 7 , d ) 2 , e ) 6 | b | inverse(divide(inverse(6), add(const_2, const_1))) | add(const_1,const_2)|inverse(n0)|divide(#1,#0)|inverse(#2)| | physics | B |
if money is invested at r percent interest , compounded annually , the amount of investment will double in approximately 70 / r years . if pat ' s parents invested $ 10000 in a long term bond that pays 8 percent interest , compounded annually , what will be the approximate total amount of investment 18 years later , when pat is ready for college ? | since investment doubles in 70 / r years then for r = 8 it ' ll double in 70 / 8 = ~ 9 years ( we are not asked about the exact amount so such an approximation will do ) . thus in 18 years investment will double twice and become ( $ 10,000 * 2 ) * 2 = $ 40,000 ( after 9 years investment will become $ 10,000 * 2 = $ 20,000 and in another 9 years it ' ll become $ 20,000 * 2 = $ 40,000 ) . answer : a . | a ) $ 40000 , b ) $ 15000 , c ) $ 12000 , d ) $ 10000 , e ) $ 9000 | a | multiply(multiply(10000, const_2), const_2) | multiply(n1,const_2)|multiply(#0,const_2) | general | A |
a sum of money is to be distributed among a , b , c , d in the proportion of 5 : 2 : 4 : 3 . if c gets rs . 500 more than d , what is b ' s share ? | "let the shares of a , b , c and d be 5 x , 2 x , 4 x and 3 x rs . respectively . then , 4 x - 3 x = 500 = > x = 500 . b ' s share = rs . 2 x = 2 * 500 = rs . 1000 . answer : c" | a ) a ) 8239 , b ) b ) 2900 , c ) c ) 1000 , d ) d ) 2393 , e ) e ) 2009 | c | multiply(multiply(subtract(4, 3), 500), 3) | subtract(n2,n3)|multiply(n4,#0)|multiply(n3,#1)| | general | C |
108 . triangle a ’ s base is 20 % greater than the base of triangle b , and a ’ s height is 20 % less than the height of triangle b . the area of triangle a is what percent less or more than the area of triangle b ? | wish the question specified that we are talking about corresponding height . base of a = 21 / 20 * base of b height of a = 19 / 20 * height of b area of a = ( 1 / 2 ) * base of a * height of a = 21 / 20 * 19 / 20 * area of b = 399 / 400 * area of b area of a is 0.25 % less than the area of b . answer ( a ) | ['a ) 0.25 % less', 'b ) 1 % less', 'c ) equal to each other', 'd ) 1 % more', 'e ) 9 % more'] | a | divide(const_100, subtract(multiply(const_100, const_100), multiply(add(const_100, 20), subtract(const_100, 20)))) | add(n1,const_100)|multiply(const_100,const_100)|subtract(const_100,n1)|multiply(#0,#2)|subtract(#1,#3)|divide(const_100,#4) | geometry | A |
at a certain committee meeting only associate professors and assistant professors are present . each associate professor has brought 2 pencils and 1 chart to the meeting , while each assistant professor has brought 1 pencil and 2 charts . if a total of 10 pencils and 14 charts have been brought to the meeting , how many people are present ? | say there are ' a ' associate professors . so we have 2 a pencils and a charts . say there are ' b ' assistant professors . so we have b pencils and 2 b charts . total pencils are 10 so 2 a + b = 10 total charts are 11 so a + 2 b = 14 add both : 3 a + 3 b = 24 so a + b = 8 total number of people = 8 c | a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | c | divide(add(14, 10), add(2, 1)) | add(n4,n5)|add(n0,n1)|divide(#0,#1) | general | C |
12 is 6 % of a , and 6 is 12 % of b . c equals b / a . what is the value of c ? | 6 a / 100 = 12 a = 200 12 b / 100 = 6 b = 50 c = b / a = 50 / 200 = 1 / 4 the answer is c . | a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 6 , e ) 1 / 12 | c | divide(multiply(divide(6, 12), const_100), multiply(divide(12, 6), const_100)) | divide(n1,n0)|divide(n0,n1)|multiply(#0,const_100)|multiply(#1,const_100)|divide(#2,#3) | general | C |
the ratio of type of pen and type of notebook is 5 : 4 . and in 80 ways one can start writing . tell me the no . of pen and no . of notebook . | let the ratio constant be x . no . of pen = 5 x , no . of notebook = 4 x . total no . of ways of start writing = 5 x * 4 x = 20 ( square of x ) = 80 x = â ˆ š 4 = 2 pen = 10 , notebook = 8 answer c | a ) 6,12 , b ) 10,6 , c ) 10,8 , d ) 8,4 , e ) 6,8 | c | divide(divide(multiply(5, 80), 4), const_10) | multiply(n0,n2)|divide(#0,n1)|divide(#1,const_10) | other | C |
a train covers a distance of 9 km in 6 min . if it takes 5 sec to pass a telegraph post , then the length of the train is ? | "speed = ( 9 / 6 * 60 ) km / hr = ( 90 * 5 / 18 ) m / sec = 25 m / sec . length of the train = 25 * 5 = 125 m . answer : option e" | a ) 100 , b ) 105 , c ) 110 , d ) 120 , e ) 125 | e | divide(9, subtract(divide(9, 6), 5)) | divide(n0,n1)|subtract(#0,n2)|divide(n0,#1)| | physics | E |
if x and y are integers such that x ^ 2 = 2 y and xy = 256 , then x – y = ? | "here x and y are integers . x ^ 2 = 2 y , xy = 256 . substitute ( x ^ 2 ) / 2 = y in xy = > x ^ 3 = 256 * 2 = > x ^ 3 = 512 . here x ^ 3 is positive , x is also positive . x = 8 then y = 32 . x - y = - 24 so option b is correct" | a ) - 30 , b ) - 24 , c ) - 5 , d ) 5 , e ) 20 | b | subtract(power(multiply(256, 2), const_0_33), divide(256, power(multiply(256, 2), const_0_33))) | multiply(n1,n2)|power(#0,const_0_33)|divide(n2,#1)|subtract(#1,#2)| | general | B |
because he ’ s taxed by his home planet , mork pays a tax rate of 40 % on his income , while mindy pays a rate of only 30 % on hers . if mindy earned 2 times as much as mork did , what was their combined tax rate ? | "say morks income is - 100 so tax paid will be 40 say mindys income is 2 * 100 = 200 so tax paid is 30 % * 200 = 60 total tax paid = 40 + 60 = 100 . combined tax % will be 100 / 100 + 200 = 33.3 %" | a ) 32.5 % , b ) 34 % , c ) 22.5 % , d ) 33.3 % , e ) 37.5 % | d | multiply(const_100, divide(add(divide(40, const_100), multiply(2, divide(30, const_100))), add(const_1, 2))) | add(n2,const_1)|divide(n0,const_100)|divide(n1,const_100)|multiply(n2,#2)|add(#1,#3)|divide(#4,#0)|multiply(#5,const_100)| | gain | D |
find the product of the localvalue and absolutevalue of 7 in 20768 ? | "local value of 7 = 7 x 100 = 700 place value of 7 = 7 there fore = 7 x 700 = 4900 c" | a ) 9200 , b ) 1100 , c ) 4900 , d ) 3200 , e ) 1600 | c | multiply(multiply(7, const_1000), 7) | multiply(n0,const_1000)|multiply(n0,#0)| | general | C |
company x provides bottled water to its employees in 5 liter bottles , $ 3 each . at least how many 5 l bottles must company x buy monthly , so that new contract with $ 50 fixed monthly rate and $ 1 for 20 l bottle each paid off ? ( assume that no other costs apply ) | let the no . of 5 liter bottles be x , so the no . of 20 l bottles will be x / 4 ( to equate the vol . ) since the total cost will be equal , 3 x = 50 + x / 4 so x = 18.18 or 19 . answer is ( c ) . | a ) 15 , b ) 17 , c ) 19 , d ) 20 , e ) 21 | c | divide(50, subtract(5, divide(const_1, 1))) | divide(const_1,n4)|subtract(n0,#0)|divide(n3,#1)| | general | C |
a can do a piece of work in 15 days and b alone can do it in 10 days . b works at it for 5 days and then leaves . a alone can finish the remaining work in | "explanation : b ' s 5 days work = 1 / 10 * 5 = 1 / 2 remaining work = 1 â ˆ ’ 1 / 2 = 1 / 2 a can finish work = 15 â ˆ — 1 / 2 = 7.5 days answer is c" | a ) 5 days , b ) 6 days , c ) 7.5 days , d ) 8.5 days , e ) 8 days | c | divide(const_1, add(divide(const_1, 15), divide(const_1, 10))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)| | physics | C |
5 cats can catch 5 mice in 5 minutes . how many cats does it take to catch 100 mice in 100 minutes ? | let x be no . of cat 5 cat catch 5 mice in 5 min 5 cat catch 1 mice in ( 5 / 5 = 1 min ) 1 cat catch 1 mice in 1 * 5 = 5 min 1 cat catch 100 mice in 5 * 100 min x cat catch 100 mice in ( 5 * 100 ) / x min a / q ( 5 * 100 ) / x = 100 = > 5 * 100 = x * 100 = > x = 5 answer : e | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | e | divide(100, multiply(5, const_4)) | multiply(n0,const_4)|divide(n3,#0) | physics | E |
a can complete a work in 18 days and b can do the same work in 10 days . if a after doing 3 days , leaves the work , find in how many days b will do the remaining work ? | "the required answer = ( 18 - 3 ) * 5 / 18 = 150 / 18 = 8 days answer is b" | a ) 2 days , b ) 8 days , c ) 6 days , d ) 7 days , e ) 10 days | b | add(multiply(18, const_2.0), divide(18, 3)) | divide(n0,const_2.0)|multiply(n0,n2)|add(#0,#1)| | physics | B |
a train 1000 m long can cross an electric pole in 200 sec and then find the speed of the train ? | "length = speed * time speed = l / t s = 1000 / 200 s = 5 m / sec speed = 5 * 18 / 5 ( to convert m / sec in to kmph multiply by 18 / 5 ) speed = 18 kmph answer : d" | a ) 12 , b ) 14 , c ) 16 , d ) 18 , e ) 20 | d | divide(divide(1000, const_1000), divide(200, const_3600)) | divide(n0,const_1000)|divide(n1,const_3600)|divide(#0,#1)| | physics | D |
of the 65 house in a development , 50 have a two - car garage , 40 have an in - the - ground swimming pool , and 35 have both a two - car garage and an in - the - ground swimming pool . how many houses in the development have neither a two - car garage nor an in - the - ground swimming pool ? | neither car nor garage = total - garage - ( swim - common ) = 65 - 50 - ( 40 - 35 ) = 65 - 55 = 10 answer a | a ) 10 , b ) 15 , c ) 20 , d ) 25 , e ) 30 | a | subtract(65, add(add(subtract(50, 35), subtract(40, 35)), 35)) | subtract(n1,n3)|subtract(n2,n3)|add(#0,#1)|add(n3,#2)|subtract(n0,#3) | other | A |
a man can do a piece of work in 5 days , but with the help of his son he can do it in 4 days . in what time can the son do it alone ? | "explanation : in this type of question , where we have one person work and together work done . then we can easily get the other person work just by subtracting them . as son ' s one day work = ( 1 / 4 − 1 / 5 ) = ( 5 − 4 ) / 20 = 1 / 20 so son will do whole work in 20 days answer : d" | a ) 2 days , b ) 6 days , c ) 12 days , d ) 20 days , e ) none of these | d | divide(multiply(5, 4), subtract(5, 4)) | multiply(n0,n1)|subtract(n0,n1)|divide(#0,#1)| | physics | D |
the length of a bridge in meters , which a train 100 - meters long and traveling at 45 km / hr can cross in 30 seconds is ? | "45 km / h = 45000 m / 3600 s = 12.5 m / s in 30 seconds , the train can go 30 ( 12.5 ) = 375 meters let x be the length of the bridge . x + 100 = 375 meters x = 275 meters the answer is c ." | a ) 225 , b ) 245 , c ) 275 , d ) 325 , e ) 375 | c | subtract(multiply(divide(multiply(45, const_1000), const_3600), 30), 100) | multiply(n1,const_1000)|divide(#0,const_3600)|multiply(n2,#1)|subtract(#2,n0)| | physics | C |
a tourist does not have to pay tax on the first $ 600 of goods he purchases in country b , but does have to pay a 10 percent tax on the portion of the total value that is in excess of $ 600 . what tax must be paid by a tourist if he buys goods with a total value of $ 1720 ? | "correct answer : a the tourist must pay tax on $ 1720 - $ 600 = $ 1120 . thus , the amount of tax he has to pay is 0.1 ( $ 1120 ) = $ 112 . the correct answer is a ." | a ) $ 112.00 , b ) $ 64.80 , c ) $ 90.00 , d ) $ 100.80 , e ) $ 154.80 | a | divide(multiply(subtract(1720, 600), 10), const_100) | subtract(n3,n0)|multiply(n1,#0)|divide(#1,const_100)| | general | A |
in how many q ways can a 4 - letter password be chosen , using the letters a , b , c , d , e , and / or f , such that at least one letter is repeated within the password ? | total number of four letter passwords = 6 * 6 * 6 * 6 = 1296 - - - - - - ( 1 ) total number of passwords in which no letter repeats = 6 c 4 * 4 ! = 15 * 24 = 360 - - - - - - ( 2 ) therefore required value q = ( 1 ) - ( 2 ) = 1296 - 360 = 936 . d | a ) 720 , b ) 864 , c ) 900 , d ) 936 , e ) 1296 | d | multiply(multiply(divide(divide(factorial(4), factorial(const_2)), factorial(const_2)), divide(divide(factorial(4), factorial(const_2)), factorial(const_2))), multiply(subtract(multiply(const_2, const_3), const_1), subtract(multiply(const_2, const_3), const_1))) | factorial(n0)|factorial(const_2)|multiply(const_2,const_3)|divide(#0,#1)|subtract(#2,const_1)|divide(#3,#1)|multiply(#4,#4)|multiply(#5,#5)|multiply(#7,#6) | general | D |
a certain characteristic in a large population has a distribution that is symmetric about the mean m . if 84 percent of the distribution lies within one standard deviation d of the mean , what percent of the distribution is less than m + d ? | "this is easiest to solve with a bell - curve histogram . m here is equal to µ in the gaussian normal distribution and thus m = 50 % of the total population . so , if 84 % is one st . dev , then on either side of m we have 84 / 2 = 42 % . so , 84 % are to the right and left of m ( = 50 % ) . in other words , our value m + d = 50 + 42 = 92 % goingfrom the mean m , to the right of the distributionin the bell shaped histogram . . this means that 92 % of the values are below m + d . like i said , doing it on a bell - curve histogram is much easier to fullygethow this works , or you could apply gmat percentile jargon / theory to it e" | a ) 16 % , b ) 32 % , c ) 48 % , d ) 84 % , e ) 92 % | e | subtract(const_100, divide(subtract(const_100, 84), const_2)) | subtract(const_100,n0)|divide(#0,const_2)|subtract(const_100,#1)| | general | E |
in a 500 m race , the ratio of the speeds of two contestants a and b is 3 : 4 . a has a start of 140 m . then , a wins by | "solution to reach the winning post a will have to cover a distance of ( 500 - 140 ) m , i . e , 360 m . while a covers 3 m , b covers 4 m . while a covers 360 m , b covers ( 4 / 3 x 360 ) m = 480 m thus , when a reaches the winning post , b covers 480 m and therefore remains 20 m behind . ∴ a wins by 20 m . answer c" | a ) 60 m , b ) 40 m , c ) 20 m , d ) 10 m , e ) 50 m | c | subtract(500, divide(multiply(subtract(500, 140), 4), 3)) | subtract(n0,n3)|multiply(n2,#0)|divide(#1,n1)|subtract(n0,#2)| | physics | C |
a grocer has a sale of rs . 5921 , rs . 5468 , rs . 5568 , rs . 6088 and rs . 6433 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . 5900 ? | "total sale for 5 months = rs . ( 5921 + 5468 + 5568 + 6088 + 6433 ) = rs . 29478 . required sale = rs . [ ( 5900 x 6 ) - 29478 ] = rs . ( 35400 - 29478 ) = rs . 5922 . answer : c" | a ) 4902 , b ) 4922 , c ) 5922 , d ) 5924 , e ) 5928 | c | subtract(multiply(add(5, const_1), 5900), add(add(add(add(5921, 5468), 5568), 6088), 6433)) | add(n5,const_1)|add(n0,n1)|add(n2,#1)|multiply(n6,#0)|add(n3,#2)|add(n4,#4)|subtract(#3,#5)| | general | C |
right triangle pqr is the base of the prism in the figure above . if pq = pr = â ˆ š 14 and the height of the prism is 8 , what is the volume of the prism ? | "volume of prism = area of base * height = 1 / 2 * ( square root of 14 ) * ( square root of 14 ) * 8 = 56 answer : d" | a ) 12 , b ) 45 , c ) 87 , d ) 56 , e ) 23 | d | multiply(triangle_area(sqrt(14), sqrt(14)), 8) | sqrt(n0)|triangle_area(#0,#0)|multiply(n1,#1)| | geometry | D |
1 , 13 , 35 , 67 , 109 , ____ | "1 , 13 , 35 , 67 , 109 , . . . . . 13 = 1 + 12 35 = 13 + 22 67 = 35 + 32 109 = 67 + 42 so 109 + 52 = 161 answer : a" | a ) 161 , b ) 154 , c ) 216 , d ) 158 , e ) none | a | subtract(negate(67), multiply(subtract(13, 35), divide(subtract(13, 35), subtract(1, 13)))) | negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)| | general | A |
set j consists of 18 consecutive even numbers . if the smallest term in the set is - 22 , what is the range of the positive integers in set j ? | "since there are only 18 integers , another approach is the just list all 18 . we get : - 22 , - 20 , - 18 , - 16 , - 14 , - 12 , - 10 , - 8 , - 6 , - 4 , - 2 , 0 , 2 , 4 , 6 , 8 , 10 , 12 range of positive integers = 12 - 2 = 10 answer : e" | a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | e | subtract(add(negate(22), multiply(subtract(18, const_1), 22)), 22) | negate(n1)|subtract(n0,const_1)|multiply(n1,#1)|add(#2,#0)|subtract(#3,n1)| | general | E |
harkamal purchased 8 kg of grapes at the rate of 70 per kg and 9 kg of mangoes at the rate of 65 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 = 65 × 9 = 585 . total cost he has to pay = 560 + 585 = 1145 . c )" | a ) a ) 1055 , b ) b ) 1075 , c ) c ) 1145 , d ) d ) 1070 , e ) e ) 1080 | c | add(multiply(8, 70), multiply(9, 65)) | multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)| | gain | C |
a person crosses a 1080 m long street in 14 minutes . what is his speed in km per hour ? | "speed = 1080 / ( 12 x 60 ) m / sec = 1.3 m / sec . converting m / sec to km / hr = 1.3 x ( 18 / 5 ) km / hr = 4.6 km / hr . answer : c" | a ) 4.1 , b ) 4.5 , c ) 4.6 , d ) 5.4 , e ) 5.5 | c | divide(divide(1080, const_1000), divide(multiply(14, const_60), const_3600)) | divide(n0,const_1000)|multiply(n1,const_60)|divide(#1,const_3600)|divide(#0,#2)| | physics | C |
two consultants can type up a report in 12.5 hours and edit it in 7.5 hours . if mary needs 30 hours to type the report and jim needs 12 hours to edit it alone , how many e hours will it take if jim types the report and mary edits it immediately after he is done ? | "break down the problem into two pieces : typing and editing . mary needs 30 hours to type the report - - > mary ' s typing rate = 1 / 30 ( rate reciprocal of time ) ( point 1 in theory below ) ; mary and jim can type up a report in 12.5 and - - > 1 / 30 + 1 / x = 1 / 12.5 = 2 / 25 ( where x is the time needed for jim to type the report alone ) ( point 23 in theory below ) - - > x = 150 / 7 ; jim needs 12 hours to edit the report - - > jim ' s editing rate = 1 / 12 ; mary and jim can edit a report in 7.5 and - - > 1 / y + 1 / 12 = 1 / 7.5 = 2 / 15 ( where y is the time needed for mary to edit the report alone ) - - > y = 20 ; how many e hours will it take if jim types the report and mary edits it immediately after he is done - - > x + y = 150 / 7 + 20 = ~ 41.4 answer : a ." | a ) 41.4 , b ) 34.1 , c ) 13.4 , d ) 12.4 , e ) 10.8 | a | add(inverse(subtract(divide(const_1, 12.5), divide(const_1, 30))), inverse(subtract(divide(const_1, 7.5), divide(const_1, 12)))) | divide(const_1,n0)|divide(const_1,n2)|divide(const_1,n1)|divide(const_1,n3)|subtract(#0,#1)|subtract(#2,#3)|inverse(#4)|inverse(#5)|add(#6,#7)| | physics | A |
if m : n is 3 : 9 and n : p is 9 : 12 then m : p is equal to | "the two ratios given are having the same number 9 for n in both the ratios . hence - m : n = 3 : 9 n : p = 9 : 12 = > m : p = 3 : 12 = > 1 : 4 answer b" | a ) 1 : 5 , b ) 1 : 4 , c ) 1 : 6 , d ) 1 : 8 , e ) 1 : 7 | b | divide(3, 12) | divide(n0,n3)| | physics | B |
the sum of the ages of 5 children born at the intervals of 3 year each is 40 year . what is the age of the eldest child ? | "solution let the ages of the children be x , ( x + 3 ) , ( x + 6 ) , ( x + 9 ) and ( x + 12 ) year . then , x + ( x + 3 ) + ( x + 6 ) + ( x + 9 ) + ( x + 12 ) = 40 â ‡ ” 5 x = 10 â ‡ ” x = 2 . â ˆ ´ age of the youngest child = x + 12 = 2 + 12 = 14 years . answer a" | a ) 14 year , b ) 8 year , c ) 10 year , d ) none of these , e ) 9 year | a | divide(subtract(divide(40, divide(5, 3)), multiply(subtract(5, const_1), 3)), 3) | divide(n0,n1)|subtract(n0,const_1)|divide(n2,#0)|multiply(n1,#1)|subtract(#2,#3)|divide(#4,n1)| | general | A |
in the junior basketball league there are 15 teams , 2 / 3 of them are bad and ½ are rich . what ca n ' t be the number of teams that are rich and bad ? | "total teams = 16 bad teams = ( 2 / 3 ) * 15 = 10 rich teams = 8 so maximum value that the both rich and bad can take will be 8 . so e = 9 can not be that value . ans d ." | a ) 4 . , b ) 6 . , c ) 7 . , d ) 9 . , e ) 5 . | d | add(multiply(15, divide(const_1, 2)), const_1) | divide(const_1,n1)|multiply(n0,#0)|add(#1,const_1)| | general | D |
45 workers work 8 hours to dig a hole 30 meters deep . how many extra workers should be hired to dig another hole 30 meters deep by working for 6 hours ? | "45 workers * 8 hours / 30 meters = x * 6 / 30 x = 60 total workers 60 - 45 = 15 new workers the answer is b ." | a ) 10 , b ) 15 , c ) 20 , d ) 25 , e ) 30 | b | subtract(multiply(multiply(45, divide(8, 6)), divide(30, 30)), 45) | divide(n3,n2)|divide(n1,n4)|multiply(n0,#1)|multiply(#0,#2)|subtract(#3,n0)| | physics | B |
there are 70 supermarkets in the fgh chain . all of them are either in the us or canada . if there are 14 more fgh supermarkets in the us than in canada , how many fgh supermarkets are there in the us ? | "x + ( x - 14 ) = 70 - - > x = 42 . answer : c ." | a ) 20 , b ) 31 , c ) 42 , d ) 53 , e ) 64 | c | divide(add(70, 14), const_2) | add(n0,n1)|divide(#0,const_2)| | general | C |
in the first 7 overs of a cricket game , the run rate was only 4.2 . what should be the rate in the remaining 30 overs to reach the target of 282 runs ? | "required run rate = [ 282 - ( 4.2 * 7 ) ] / 30 = 252.60 / 40 = 8.42 answer : b" | a ) 6.25 , b ) 8.42 , c ) 8.3 , d ) 8.1 , e ) 6.21 | b | divide(subtract(282, multiply(7, 4.2)), 30) | multiply(n0,n1)|subtract(n3,#0)|divide(#1,n2)| | gain | B |
find the greatest number which on dividing 1255 and 1490 , leaves a reminder of 8 and 11 respectively | "explanation : in this type of question , its obvious we need to calculate the hcf , trick is hcf of ( 1255 - 8 ) and ( 1490 - 11 ) = hcf ( 1247 , 1479 ) = 29 option b" | a ) 27 , b ) 29 , c ) 33 , d ) 39 , e ) 41 | b | gcd(subtract(1255, 8), subtract(1490, 11)) | subtract(n0,n2)|subtract(n1,n3)|gcd(#0,#1)| | general | B |
two men start from opposite banks of a river . they meet 340 meters away from one of the banks on forward journey . after that they meet at 170 meters from the other bank of the river on their backward journey . what will be the width of the river ( in meters ) ? | let p , q are two persons then their speeds be a , b - > m / hr x - - - - 340 - - - - - - - - - - - - - - - - - - - - - - | - - - ( d - 340 ) - - - - - - - - - y in forward journey time taken by p = 340 / a ; time taken by q = d - 340 / b ; in forward journey , time taken by p = time taken by q so 340 / a = d - 340 / b a / b = 340 / d - 340 - - - - - - - - - - - - ( 1 ) after that they continue their journeyand reach other banks now in backward journey , x - - - - - - - - - - - - - - - ( d - 170 ) - - - - - - - - - - - - - - - - - - - - - - - - - - | - - - - - 170 - - - - - - y distance travelled by p from meet 1 to meet 2 = ( d - 340 ) + 170 distance travelled by p from meet 1 to meet 2 = 340 + ( d - 170 ) now time taken by p = time taken by q ( d - 340 ) + 170 / a = 340 + ( d - 170 ) / b ( d - 340 ) + 170 / 340 + ( d - 170 ) = a / b - - - - - - - - - ( 2 ) from 1 & 2 340 / ( d - 340 ) = ( d - 340 ) + 170 / 340 + ( d - 170 ) 340 / ( d - 340 ) = d - 170 / d + 170 340 d + 57800 = d ^ 2 - 340 d - 170 d + 57800 340 d = d ^ 2 - 510 d d ^ 2 - 850 d = 0 d ( d - 850 ) = 0 d = 0 ord = 850 so width of the river is 850 answer : b | a ) 750 , b ) 850 , c ) 900 , d ) 980 , e ) 950 | b | add(add(add(340, 170), 170), 170) | add(n0,n1)|add(n1,#0)|add(n1,#1) | physics | B |
the distance between two cities a and b is 465 km . a train starts from a at 8 a . m . and travel towards b at 60 km / hr . another train starts from b at 9 a . m and travels towards a at 75 km / hr . at what time do they meet ? | "explanation : suppose they meet x hrs after 8 a . m then , [ distance moved by first in x hrs ] + [ distance moved by second in ( x - 1 ) hrs ] = 465 therefore , 60 x + 75 ( x - 1 ) = 465 = > x = 3 . so , they meet at ( 8 + 4 ) i . e , 12 noon answer : c )" | a ) 09 , b ) 07 , c ) 12 , d ) 05 , e ) 03 | c | add(divide(add(465, 75), add(60, 75)), 8) | add(n0,n4)|add(n2,n4)|divide(#0,#1)|add(n1,#2)| | physics | C |
how many positive integers less than 248 are multiple of 4 but not multiples of 6 ? | "244 / 4 = 61 multiples of 4 which are a multiple of 6 will be of the form 2 * 2 * 3 = 12 n where n > 0 240 / 12 = 20 61 - 20 = 41 answer : b" | a ) 20 , b ) 41 , c ) 42 , d ) 53 , e ) 64 | b | divide(factorial(subtract(add(const_4, 4), const_1)), multiply(factorial(4), 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 | B |
hammers and wrenches are manufactured at a uniform weight per hammer and a uniform weight per wrench . if the total weight of two hammers and two wrenches is one - third that of 8 hammers and 5 wrenches , then the total weight of one wrench is how many times that of one hammer ? | x be the weight of a hammer and y be the weight of a wrench . ( 2 x + 2 y ) = 1 / 3 * ( 8 x + 5 y ) 3 ( 2 x + 2 y ) = ( 8 x + 5 y ) 6 x + 6 y = 8 x + 5 y y = 2 x ans - e | a ) 1 / 2 , b ) 2 / 3 , c ) 1 , d ) 3 / 2 , e ) 2 | e | divide(subtract(8, multiply(const_3, const_2)), subtract(multiply(const_3, const_2), 5)) | multiply(const_2,const_3)|subtract(n0,#0)|subtract(#0,n1)|divide(#1,#2) | general | E |
vlaudia can choose any two of 4 different candles and any 8 of 9 different flowers for a centerpiece arrangement . given these choices , how many candle + flower groupings can she select ? | vlaudia can choose any two of four different candles and any 8 of 9 different flowers for a centerpiece arrangement . 2 candles from 4 candles : 4 c 2 : 6 8 flowers from 9 flowers : 9 c 8 : 9 so total grouping 9 * 6 = 54 . answer is a . | a ) 54 , b ) 72 , c ) 96 , d ) 144 , e ) 432 | a | multiply(divide(factorial(9), factorial(8)), divide(factorial(4), multiply(const_2, const_2))) | factorial(n2)|factorial(n1)|factorial(n0)|multiply(const_2,const_2)|divide(#0,#1)|divide(#2,#3)|multiply(#4,#5) | general | A |
2 / 3 rd of the boys and 3 / 4 th of the girls of a school participate in a function . if the no . of participating students is 550 , out of which 150 are girls , what is the total no . of students in the school ? | let total number of boys be x and total number of girls be y . y = 400 = > x = 600 and ^ = i 50 = > y = 200 now , x + y = 800 c | a ) 500 , b ) 600 , c ) 800 , d ) 900 , e ) 1200 | c | add(divide(multiply(subtract(550, 150), 3), 2), divide(multiply(150, 4), 3)) | multiply(n3,n5)|subtract(n4,n5)|divide(#0,n1)|multiply(n1,#1)|divide(#3,n0)|add(#4,#2) | general | C |
the value of a machine depreciates at 20 % per annum . if its present value is rs . 1 , 50,000 , at what price should it be sold after two years such that a profit of rs . 22,000 is made ? | the value of the machine after two years = 0.8 * 0.8 * 1 , 50,000 = rs . 96,000 sp such that a profit of rs . 22,000 is made = 96,000 + 24,000 = rs . 1 , 18,000 answer : c | a ) 328897 , b ) 120000 , c ) 118000 , d ) 277768 , e ) 188871 | c | add(multiply(multiply(subtract(const_1, divide(20, const_100)), subtract(const_1, divide(20, const_100))), add(multiply(multiply(const_100, const_100), sqrt(const_100)), multiply(multiply(divide(sqrt(const_100), const_2), const_100), const_100))), multiply(multiply(add(20, const_2), const_100), sqrt(const_100))) | add(n0,const_2)|divide(n0,const_100)|multiply(const_100,const_100)|sqrt(const_100)|divide(#3,const_2)|multiply(#2,#3)|multiply(#0,const_100)|subtract(const_1,#1)|multiply(#4,const_100)|multiply(#7,#7)|multiply(#6,#3)|multiply(#8,const_100)|add(#5,#11)|multiply(#12,#9)|add(#13,#10) | gain | C |
the ages of two persons differ by 32 years . if 5 years ago , the elder one be 5 times as old as the younger one , their present ages ( in years ) are respectively | "explanation : let their ages be x and ( x + 32 ) years . 5 ( x - 5 ) = ( x + 32 - 5 ) or 4 x = 52 or x = 14 . their present ages are 46 years and 14 years option b" | a ) 20,20 , b ) 46,14 , c ) 25,15 , d ) 30,10 , e ) none of these | b | subtract(add(divide(multiply(32, 5), subtract(5, const_1)), 5), 32) | multiply(n0,n1)|subtract(n1,const_1)|divide(#0,#1)|add(n1,#2)|subtract(#3,n0)| | general | B |
a person saved $ 10 in buying an item on sale . if he spent $ 600 for the item , approximately how much percent he saved in the transaction ? | "actual price = 600 + 10 = $ 610 saving = 10 / 610 * 100 = 100 / 61 = 1.6 % approximately answer is b" | a ) 8 % , b ) 1.6 % , c ) 2 % , d ) 5 % , e ) 6 % | b | add(floor(multiply(divide(10, add(10, 600)), 600)), const_1) | add(n0,n1)|divide(n0,#0)|multiply(#1,n1)|floor(#2)|add(#3,const_1)| | general | B |
in an election , candidate a got 80 % of the total valid votes . if 15 % of the total votes were declared invalid and the total numbers of votes is 560000 , find the number of valid vote polled in favor of candidate ? | "total number of invalid votes = 15 % of 560000 = 15 / 100 × 560000 = 8400000 / 100 = 84000 total number of valid votes 560000 – 84000 = 476000 percentage of votes polled in favour of candidate a = 80 % therefore , the number of valid votes polled in favour of candidate a = 80 % of 476000 = 80 / 100 × 476000 = 38080000 / 100 = 380800 b )" | a ) 355600 , b ) 380800 , c ) 356500 , d ) 356800 , e ) 357000 | b | multiply(multiply(560000, subtract(const_1, divide(15, const_100))), divide(80, const_100)) | divide(n0,const_100)|divide(n1,const_100)|subtract(const_1,#1)|multiply(n2,#2)|multiply(#0,#3)| | gain | B |
what is the least number of squares tiles required to pave the floor of a room 6 m 75 cm long and 4 m 5 cm broad ? | "length of largest tile = h . c . f . of 675 cm and 405 cm = 135 cm . area of each tile = ( 135 x 135 ) cm 2 . required number of tiles = 675 x 405 / ( 135 ^ 2 ) = 15 answer : a" | a ) 15 , b ) 20 , c ) 40 , d ) 44 , e ) 54 | a | divide(rectangle_area(multiply(6, const_100), multiply(75, const_100)), square_area(add(multiply(const_4, const_10), const_1))) | multiply(n0,const_100)|multiply(n1,const_100)|multiply(const_10,const_4)|add(#2,const_1)|rectangle_area(#0,#1)|square_area(#3)|divide(#4,#5)| | physics | A |
suresh started a business , investing rs . 18000 . after 3 months and 4 months respectively , rohan and sudhir joined him with capitals of 12000 and 9000 . at the end of the year the total profit was rs . 4048 . what is the difference between rohan ’ s and sudhir ’ s share in the profit ? | "suresh : rohan : sudhir ratio of their investments = 18000 × 12 : 12000 × 9 : 9000 × 8 = 6 : 3 : 2 the difference between rohan ’ s and sudhir ’ s share = 1 share : . i . e . = rs . 4048 × 1 / 11 = rs . 368 . e" | a ) s . 345 , b ) s . 350 , c ) s . 352 , d ) s . 362 , e ) s . 368 | e | multiply(subtract(multiply(12000, divide(subtract(multiply(3, 4), 3), multiply(3, 4))), multiply(9000, divide(subtract(multiply(3, 4), 4), multiply(3, 4)))), divide(4048, add(add(18000, multiply(12000, divide(subtract(multiply(3, 4), 3), multiply(3, 4)))), multiply(9000, divide(subtract(multiply(3, 4), 4), multiply(3, 4)))))) | multiply(n1,n2)|subtract(#0,n1)|subtract(#0,n2)|divide(#1,#0)|divide(#2,#0)|multiply(n3,#3)|multiply(n4,#4)|add(n0,#5)|subtract(#5,#6)|add(#7,#6)|divide(n5,#9)|multiply(#10,#8)| | gain | E |
find the area of a parallelogram with base 18 cm and height 16 cm ? | "area of a parallelogram = base * height = 18 * 16 = 288 cm 2 answer : e" | a ) 423 cm 2 , b ) 122 cm 2 , c ) 420 cm 2 , d ) 251 cm 2 , e ) 288 cm 2 | e | multiply(18, 16) | multiply(n0,n1)| | geometry | E |
if a : b = 2 : 3 , b : c = 4 : 5 and c : d = 5 : 9 then a : d is equal to : | "solution : a / d = ( a / b ) * ( b * c ) * ( c / d ) = ( 2 / 3 ) * ( 4 / 5 ) * ( 5 / 9 ) = ( 2 * 4 * 5 ) / ( 3 * 5 * 9 ) = 8 / 27 . answer : option b" | a ) 11 : 17 , b ) 8 : 27 , c ) 5 : 9 , d ) 2 : 9 , e ) none | b | divide(multiply(multiply(2, 4), 5), multiply(multiply(3, 5), 9)) | multiply(n0,n2)|multiply(n1,n3)|multiply(n4,#0)|multiply(n5,#1)|divide(#2,#3)| | general | B |
along a yard 273 metres long , 14 trees are palnted at equal distances , one tree being at each end of the yard . what is the distance between two consecutive trees | "explanation : 14 trees have 13 gaps between them , required distance ( 273 / 13 ) = 21 option d" | a ) 18 , b ) 19 , c ) 10 , d ) 21 , e ) 12 | d | divide(273, subtract(14, const_1)) | subtract(n1,const_1)|divide(n0,#0)| | physics | D |
. 3 + 33 + 333 + 3.33 = ? | . 3 33 333 3.33 - - - - - - - - - - 369.63 - - - - - - - - - - answer is a | a ) 369.63 , b ) 372.33 , c ) 702.33 , d ) 702 , e ) none of them | a | add(add(add(33, 333), 3.33), divide(const_3, const_10)) | add(n1,n2)|divide(const_3,const_10)|add(n3,#0)|add(#2,#1) | general | A |
excluding stoppages , the speed of a bus is 75 kmph and including stoppages , it is 45 kmph . for how many minutes does the bus stop per hour ? | "due to stoppages , it covers 30 km less . time taken to cover 30 km = ( ( 30 / 75 ) ã — 60 ) = 24 min . option ( b ) is correct" | a ) 15 , b ) 24 , c ) 17 , d ) 19 , e ) none of these | b | multiply(const_60, divide(subtract(75, 45), 75)) | subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_60)| | physics | B |
how many three digit numbers w are divisible by 78 or 91 ? | "the answer will be 19 . explanation : 78 = 2 * 3 * 13 now multiples of 78 , 156 . . . . 780 , now 1000 - 780 = 220 only two more muktiples of 78 can exists . so total number of 3 digit multiples of 78 are 9 + 2 = 11 91 = 13 * 7 - - total number of three digit multiples - - 9 no remember we have a common multiples as well - - 13 * 7 * 6 = 91 * 6 = 546 so total number of multiples w - - 11 + 9 - 1 = 19 . hence answer is 19 . b" | a ) 17 , b ) 19 , c ) 20 , d ) 21 , e ) 22 | b | add(subtract(subtract(const_1000, const_10), multiply(multiply(const_10, multiply(78, 78)), multiply(const_4, const_2))), const_10) | multiply(n0,n0)|multiply(const_2,const_4)|subtract(const_1000,const_10)|multiply(#0,const_10)|multiply(#3,#1)|subtract(#2,#4)|add(#5,const_10)| | general | B |
a certain car ' s price decreased by 2.5 % ( from the original price ) each year from 1996 to 2002 , during that time the owner of the car invested in a new carburetor and a new audio system for the car , which increased car ' s price by $ 1,000 . if the price of the car in 1996 was $ 22,000 , what is the car ' s price in 2002 ? | "price in 96 = 22000 price decrease each year = 2.5 / 100 * 22000 = 550 price in 97 = 22000 - 550 price in 98 = 22000 - 2 * 550 price in 99 = 22000 - 3 * 550 price in 00 = 22000 - 4 * 550 price in 01 = 22000 - 5 * 550 price in 02 = 22000 - 6 * 550 = 18700 investment in the car = 1000 net price of the car in 02 = 18700 + 1000 = $ 19700 correct option : b" | a ) $ 18,400 , b ) $ 19,700 , c ) $ 20,200 , d ) $ 20,400 , e ) $ 21,100 | b | multiply(const_2, const_10) | multiply(const_10,const_2)| | gain | B |
two tracks are parallel . the first track has 6 checkpoints and the second one has 10 checkpoints . in how many ways can the 6 checkpoints of first track be joined with the 10 checkpoints of the second to form a triangle ? | to make a triangle , you need 2 checkpoints from one track and 1 from the other . you can not have all 3 from the same track since then the points will be in a line ( assuming straight line of track ) you select 2 checkpoints from the first track and one from the second or two from the second track and one from the first . 6 c 2 * 10 c 1 + 10 c 2 * 6 c 1 = 150 + 270 = 420 answer ( e ) | ['a ) 120', 'b ) 150', 'c ) 200', 'd ) 270', 'e ) 420'] | e | add(multiply(divide(divide(factorial(10), factorial(subtract(10, const_2))), factorial(const_2)), divide(factorial(6), factorial(subtract(6, const_1)))), multiply(divide(factorial(10), factorial(subtract(const_10, const_1))), divide(divide(factorial(6), const_2), factorial(subtract(6, const_2))))) | factorial(n1)|factorial(const_2)|factorial(n0)|subtract(n1,const_2)|subtract(n0,const_1)|subtract(const_10,const_1)|subtract(n0,const_2)|divide(#2,const_2)|factorial(#3)|factorial(#4)|factorial(#5)|factorial(#6)|divide(#0,#8)|divide(#2,#9)|divide(#0,#10)|divide(#7,#11)|divide(#12,#1)|multiply(#14,#15)|multiply(#16,#13)|add(#18,#17) | geometry | E |
p and q started a business investing rs 85000 and rs 10000 resp . in what ratio the profit earned after 2 years be divided between p and q respectively . | "explanation : in this type of question as time frame for both investors is equal then just get the ratio of their investments . p : q = 85000 : 10000 = 85 : 10 = 17 : 2 option d" | a ) 17 : 5 , b ) 17 : 3 , c ) 17 : 6 , d ) 17 : 2 , e ) 17 : 8 | d | divide(85000, 10000) | divide(n0,n1)| | gain | D |
what is the greatest prime factor of ( 11 ! × 10 ! + 10 ! × 9 ! ) / 111 ? | ( 11 ! × 10 ! + 10 ! × 9 ! ) / 111 in numerator take out 10 ! * 9 ! as common term 10 ! ∗ 9 ! ( 11 × 10 + 1 ) / 111 10 ! ∗ 9 ! ∗ 111 / 111 10 ! ∗ 9 ! 10 ! has 7 as the greatest prime factor , so ans is 7 answer : d | a ) 2 , b ) 3 , c ) 5 , d ) 7 , e ) 11 | d | add(add(divide(add(multiply(11, 10), const_1), 111), const_3), const_3) | multiply(n0,n1)|add(#0,const_1)|divide(#1,n4)|add(#2,const_3)|add(#3,const_3) | general | D |
a circular mat with radius 10 inches is placed on a square tabletop , each of whose sides is 24 inches long . which of the following is closest to the fraction of the tabletop covered by the mat ? | c . it is a circle inscribed in a square . square side = 24 - - - > square ( table ) area = 24 ^ 2 circle diameter = 20 - - - > circle area = pir ^ 2 = 100 pi ( where pi = ~ 3.14 ) covered fraction = 100 * 3.14 / 24 * 24 = ~ 314 / 24 * 24 = 0.5451 c | ['a ) 5 / 12', 'b ) 2 / 5', 'c ) 0.5451', 'd ) 3 / 4', 'e ) 5 / 6'] | c | divide(circle_area(10), square_area(24)) | circle_area(n0)|square_area(n1)|divide(#0,#1) | geometry | C |
the sum of two numbers is 40 and their product is 375 . what will be the sum of their reciprocals ? | ( 1 / a ) + ( 1 / b ) = ( a + b ) / ab = 40 / 375 = 8 / 75 answer : b | a ) 1 / 40 , b ) 8 / 75 , c ) 75 / 4 , d ) 75 / 8 , e ) 75 / 6 | b | divide(40, 375) | divide(n0,n1) | general | B |
how long will a boy take to run round a square field of side 45 meters , if he runs at the rate of 9 km / hr ? | "speed = 9 km / hr = 9 * 5 / 18 = 5 / 2 m / sec distance = 45 * 4 = 180 m time taken = 180 * 2 / 5 = 72 sec answer is a" | a ) 72 sec , b ) 45 sec , c ) 1 min , d ) 32 sec , e ) 25 sec | a | divide(multiply(45, const_4), multiply(9, divide(const_1000, const_3600))) | divide(const_1000,const_3600)|multiply(n0,const_4)|multiply(n1,#0)|divide(#1,#2)| | gain | A |
veena ranks 44 rd from the top in a class of 182 . what is her rank from the bottom if 22 students have failed the examination ? | "total student = 182 failed = 22 paasd student = 182 - 22 = 160 from bottom her rank is = 160 - 44 + 1 = 117 answer : e" | a ) 88 , b ) 108 , c ) 110 , d ) 90 , e ) 117 | e | subtract(subtract(subtract(182, 22), subtract(44, const_1)), const_4) | subtract(n1,n2)|subtract(n0,const_1)|subtract(#0,#1)|subtract(#2,const_4)| | other | E |
three 6 faced dice are thrown together . the probability that all the three show the same number on them is - . | "it all 3 numbers have to be same basically we want triplets . 111 , 222 , 333 , 444 , 555 and 666 . those are six in number . further the three dice can fall in 6 * 6 * 6 = 216 ways . hence the probability is 6 / 216 = 1 / 36 answer a" | a ) 1 / 36 , b ) 1 / 216 , c ) 1 / 48 , d ) 1 / 55 , e ) 1 / 90 | a | multiply(multiply(multiply(divide(const_1, 6), divide(const_1, 6)), divide(const_1, 6)), divide(const_1, 6)) | divide(const_1,n0)|multiply(#0,#0)|multiply(#0,#1)|multiply(#0,#2)| | probability | A |
andrew purchased 7 kg of grapes at the rate of 68 per kg and 9 kg of mangoes at the rate of 48 per kg . how much amount did he pay to the shopkeeper ? | "cost of 7 kg grapes = 68 × 7 = 476 . cost of 9 kg of mangoes = 48 × 9 = 432 . total cost he has to pay = 476 + 432 = 908 b" | a ) 1000 , b ) 908 , c ) 930 , d ) 1075 , e ) 1080 | b | add(multiply(7, 68), multiply(9, 48)) | multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)| | gain | B |
a certain company had a total annual expenditure of 2.55 ∗ 10 ^ 7 on employee salaries last year . if the company employed 425 people , what was the average employee salary ? | "given : total annual expenditure of 2.55 ∗ 10 ^ 7 on employee salaries total employees = 425 observe that 425 * 6 = 2550 therefore try to bring the numerator in terms of 2550 average salary = ( 2550 * 10 ^ 4 ) / 425 = 6 * 10 ^ 4 = 60,000 option a" | a ) $ 60,000 , b ) $ 25,000 , c ) $ 35,000 , d ) $ 40,000 , e ) $ 45,000 | a | multiply(divide(multiply(2.55, multiply(10, 7)), 425), const_100) | multiply(n1,n2)|multiply(n0,#0)|divide(#1,n3)|multiply(#2,const_100)| | general | A |
the sum of the squares of three numbers is 125 , while the sum of their products taken two at a time is 50 . their sum is : | "x ^ + y ^ 2 + z ^ 2 = 125 xy + yz + zx = 50 as we know . . ( x + y + z ) ^ 2 = x ^ 2 + y ^ 2 + z ^ 2 + 2 ( xy + yz + zx ) so ( x + y + z ) ^ 2 = 125 + ( 2 * 50 ) ( x + y + z ) ^ 2 = 225 so x + y + z = 15 answer : a" | a ) 15 , b ) 30 , c ) 40 , d ) 50 , e ) none of these | a | add(multiply(sqrt(divide(subtract(125, 50), const_2)), const_100), sqrt(subtract(125, divide(subtract(125, 50), const_2)))) | subtract(n0,n1)|divide(#0,const_2)|sqrt(#1)|subtract(n0,#1)|multiply(#2,const_100)|sqrt(#3)|add(#4,#5)| | general | A |
of the 200 stamps in a collection , 90 are foreign and 50 are more than 10 years old . if 20 stamps are both foreign and more than 10 years old , how many stamps are neither foreign nor more than 10 years old ? | "20 stamps are both foreign and more than 10 years old . 70 stamps are foreign only . 30 stamps are 10 years old only . the number of remaining stamps is 200 - ( 20 + 70 + 30 ) = 80 the answer is b ." | a ) 30 , b ) 80 , c ) 100 , d ) 130 , e ) 150 | b | subtract(200, subtract(add(90, 50), 20)) | add(n1,n2)|subtract(#0,n4)|subtract(n0,#1)| | other | B |
in a recent election , geoff received 0.5 percent of the 5,000 votes cast . to win the election , a candidate needed to receive more than x % of the vote . if geoff needed exactly 2,571 more votes to win the election , what is the value of x ? | "word problems are tricky in somehow more than other problem because you have the additional step to translate . breaking the problem : geoff how many votes he receives ? ? 25 votes he needs 2571 more votes so : 25 + 2571 = 2596 now what ' s the problem wants ? ? a x % . . . . . . . . 2596 is what % of total votes 5000 . . . . . . . . translating : 2596 = x / 100 * 5000 - - - x = 52 % . . . . . . . . . . c" | a ) 50 , b ) 54 , c ) 52 , d ) 60 , e ) 63 | c | add(divide(const_100, const_2), 0.5) | divide(const_100,const_2)|add(n0,#0)| | gain | C |
the ratio of the incomes of uma and bala is 8 : 7 and the ratio of their expenditure is 7 : 6 . if at the end of the year , each saves $ 2000 then the income of uma is ? | let the income of uma and bala be $ 8 x and $ 7 x let their expenditures be $ 7 y and $ 6 y 8 x - 7 y = 2000 - - - - - - - 1 ) 7 x - 6 y = 2000 - - - - - - - 2 ) from 1 ) and 2 ) x = 1000 uma ' s income = 8 x = 8 * 2000 = $ 16000 answer is c | a ) $ 16800 , b ) $ 16500 , c ) $ 16000 , d ) $ 16300 , e ) $ 16200 | c | multiply(8, 2000) | multiply(n0,n4) | other | C |
a boat goes 130 km downstream in 10 hours , and 75 km upstream in 15 hours . the speed of the stream is ? | "125 - - - 10 ds = 130 ? - - - - 1 75 - - - - 15 us = 5 ? - - - - - 1 s = ( 13 - 5 ) / 2 = 4 kmph . answer : d" | a ) 3.5 , b ) 3 , c ) 5 , d ) 4 , e ) 8 | d | divide(subtract(divide(130, 10), divide(75, 15)), const_2) | divide(n0,n1)|divide(n2,n3)|subtract(#0,#1)|divide(#2,const_2)| | physics | D |
on her annual road trip to visit her family in seal beach , california , traci stopped to rest after she traveled 1 ⁄ 3 of the total distance and again after she traveled 1 ⁄ 2 of the distance remaining between her first stop and her destination . she then drove the remaining 100 miles and arrived safely at her destination . what was the total distance , in miles , from traci ’ s starting point to seal beach ? | "let d = total distance traci traveled 1 / 3 = d / 3 i . e . remaining distance = 2 d / 3 she traveled 1 / 2 th of 2 d / 3 = d / 3 thus : d = ( d / 3 ) + ( d / 3 ) + 100 d = 300 answer : b" | a ) 250 , b ) 300 , c ) 350 , d ) 400 , e ) 550 | b | multiply(100, const_2) | multiply(n4,const_2)| | physics | B |
34 . the side surface of a cylinder is rolled with a rectangular plate . if the height of a cylinder tank is 10 feet and the perimeter of the circular base is 4 feet , what is the area of the rectangular plate ? | think of a pringles can . if you took off the bottom and top and cut a slit down the length , it would flatten to a rectangle . the dimensions of the rectangle are the height of the can and the circumference of the circle . since you know both you can multiply them together to find the area , 40 . ( correct answer d ) | a ) 24 , b ) 34 , c ) 28 , d ) 40 , e ) 14 | d | sqrt(subtract(power(4, const_2), power(10, const_2))) | power(n2,const_2)|power(n1,const_2)|subtract(#0,#1)|sqrt(#2)| | geometry | D |
in the standard formulation of a flavored drink the ratio by volume of flavoring to corn syrup to water is 1 : 12 : 30 . in the sport formulation , the ratio of flavoring to corn syrup is three times as great as in the standard formulation , and the ratio of flavoring to water is half that of the standard formulation . if a large bottle of the sport formulation contains 4 ounces of corn syrup , how many ounces of water does it contain ? | "f : c : w 1 : 12 : 30 sport version : f : c 3 : 12 f : w 1 : 60 or 3 : 180 so c : f : w = 12 : 3 : 180 c / w = 12 / 180 = 3 ounces / x ounces x = 4 * 180 / 12 = 60 ounces of water d" | a ) 45 , b ) 50 , c ) 55 , d ) 60 , e ) 63 | d | multiply(divide(multiply(divide(1, 12), const_3), divide(divide(1, 30), const_2)), 4) | divide(n0,n1)|divide(n0,n2)|divide(#1,const_2)|multiply(#0,const_3)|divide(#3,#2)|multiply(n3,#4)| | other | D |
a company charges a fixed rental of rs . 350 per month . it allows 200 calls free per month . each call is charge at rs . 1.4 when the number of calls exceed 200 per month and it charges rs . 1.6 when the number of calls exceeds 400 per month and so on . a customer made 150 calls in february and 250 calls in march . by how much percent each call is cheaper in march than each call in february . | solution : charge per call in february = 350 / 150 = 7 / 3 = 2.33 charge per call in march = [ 350 + ( 50 * 1.4 ) ] / 250 = 420 / 250 = 42 / 25 = 1.68 % cheaper call rate in march = [ ( 2.33 - 1.68 ) / 2.33 ] * 100 = 28 % . answer : option a | a ) 28 % , b ) 25 % , c ) 18.5 % , d ) 16 % , e ) none of these | a | multiply(divide(subtract(divide(350, 150), divide(add(multiply(subtract(250, 200), 1.4), 350), 250)), divide(350, 150)), const_100) | divide(n0,n6)|subtract(n7,n1)|multiply(n2,#1)|add(n0,#2)|divide(#3,n7)|subtract(#0,#4)|divide(#5,#0)|multiply(#6,const_100) | general | A |
john makes $ 50 a week from his job . he earns a raise and now makes $ 70 a week . what is the % increase ? | "increase = ( 20 / 50 ) * 100 = 40 % . e" | a ) 16 % , b ) 16.66 % , c ) 18 % , d ) 21 % , e ) 40 % | e | multiply(divide(subtract(70, 50), 50), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain | E |
the weight of 4 dogs is determined to be 25 pounds , 31 pounds , 35 pounds and 33 pounds respectively . the weight of a fifth dog is determined to be y pounds . if the average ( arithmetic mean ) weight of the first 4 dogs is the same as that of all 5 dogs what is the value of y ? | total weight of the 4 dogs = ( 25 + 31 + 35 + 33 ) = 124 avg = 124 / 4 = 31 total weight of 5 dogs = 124 + y or 4 ( 31 ) + y average of 5 dogs as per question = 31 equation : 4 ( 31 ) + y = 5 ( 31 ) , or y = 31 . choose a | a ) 31 , b ) 33 , c ) 35 , d ) 37 , e ) 39 | a | divide(add(add(25, 31), add(35, 33)), 4) | add(n1,n2)|add(n3,n4)|add(#0,#1)|divide(#2,n0) | general | A |
what is 2 / 5 of 5 / 9 of 1 / 2 ? | "2 / 5 * 5 / 9 * 1 / 2 = 1 / 9 answer : b" | a ) 1 / 4 , b ) 1 / 9 , c ) 9 / 16 , d ) 5 / 8 , e ) 16 / 9 | b | multiply(divide(2, 5), multiply(divide(5, 9), divide(1, 2))) | divide(n0,n1)|divide(n1,n3)|divide(n4,n0)|multiply(#1,#2)|multiply(#0,#3)| | general | B |
if 18 men take 15 days to to complete a job , in how many days can 24 men finish that work ? | ans . 10 days | a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | a | divide(multiply(18, 15), 24) | multiply(n0,n1)|divide(#0,n2)| | physics | A |
thomas and matt has some money with them in the ratio of 3 : 4 . if their father gives rs . 30 to each of them then the ratio becomes 4 : 5 . what would be the ratio if thomas spend rs . 40 and matt spend rs . 20 from what they have now ? | let money of thomas and matt be x and y respectively . x / y = 3 / 4 = > x = 3 / 4 y ( x + 30 ) / ( y + 30 ) = 4 / 5 = > 5 x + 150 = 4 y + 120 = > 5 x = 4 y - 30 but x = 3 / 4 y 5 * 3 / 4 y = 4 y - 30 15 y = 16 y - 120 = > y = 120 = > x = 3 * 120 / 4 = > x = 90 ratio of their money , if thomas spend rs . 40 and matt spend rs . 20 = > ( 90 - 40 ) / ( 120 - 20 ) = 50 / 100 . = > 1 : 2 answer : a | a ) 1 : 2 , b ) 2 : 3 , c ) 3 : 4 , d ) 4 : 5 , e ) 5 : 6 | a | divide(subtract(multiply(multiply(30, 4), divide(3, 4)), 40), subtract(multiply(30, 4), 20)) | divide(n0,n1)|multiply(n1,n2)|multiply(#0,#1)|subtract(#1,n6)|subtract(#2,n5)|divide(#4,#3) | other | A |
a vendor sells 20 percent of the pears he had and throws away 50 percent of the remainder . the next day , the vendor sells 20 percent of the remaining pears and throws away the rest . in total , what percent of his pears does the vendor throw away ? | let x be the original number of pears . on day one , the vendor throws away ( 0.5 ) ( 0.8 ) x = 0.4 x . the remaining pears are ( 0.5 ) ( 0.8 ) x = 0.4 x . on day two , the vendor throws away ( 0.8 ) ( 0.4 ) x = 0.32 x . the vendor throws away a total of 0.4 x + 0.32 x = 0.72 x . the vendor throws away 72 percent of the pears . the answer is e . | a ) 56 , b ) 60 , c ) 64 , d ) 68 , e ) 72 | e | multiply(const_100, add(subtract(subtract(subtract(const_1, divide(20, const_100)), multiply(divide(50, const_100), subtract(const_1, divide(20, const_100)))), multiply(subtract(subtract(const_1, divide(20, const_100)), multiply(divide(50, const_100), subtract(const_1, divide(20, const_100)))), divide(20, const_100))), multiply(divide(50, const_100), subtract(const_1, divide(20, const_100))))) | divide(n1,const_100)|divide(n0,const_100)|subtract(const_1,#1)|multiply(#0,#2)|subtract(#2,#3)|multiply(#1,#4)|subtract(#4,#5)|add(#3,#6)|multiply(#7,const_100) | general | E |
a money lender finds that due to a fall in the annual rate of interest from 8 % to 7 3 / 4 % his yearly income diminishes by rs . 63.50 , his capital is ? | "let the capital be rs . x . then , ( x * 8 * 1 ) / 100 - ( x * 31 / 4 * 1 / 100 ) = 63.50 32 x - 31 x = 6350 * 4 x = 25,400 . answer : e" | a ) 24,602 , b ) 28,888 , c ) 24,600 , d ) 24,628 , e ) 24,400 | e | divide(63.50, divide(const_4, 4)) | divide(const_4,n3)|divide(n4,#0)| | gain | E |
a sum of money at simple interest amounts to $ 815 in 3 years and to $ 854 in 4 years . the sum is : | "b $ 698 s . i . for 1 year = $ ( 854 - 815 ) = $ 39 . s . i . for 3 years = $ ( 39 x 3 ) = $ 117 . principal = $ ( 815 - 117 ) = $ 698 ." | a ) $ 153 , b ) $ 698 , c ) $ 398 , d ) $ 549 , e ) $ 675 | b | subtract(815, divide(multiply(subtract(854, 815), 3), 4)) | subtract(n2,n0)|multiply(n1,#0)|divide(#1,n3)|subtract(n0,#2)| | gain | B |
a waitress ' s income consists of her salary and tips . during one week , her tips were 2 / 4 of her salary . what fraction of her income for the week came from tips ? | "her tips were 2 / 4 of her salary . let ' s say her salary = $ 4 this mean her tips = ( 2 / 4 ) ( $ 4 ) = $ 2 so , her total income = $ 4 + $ 2 = $ 6 what fraction of her income for the week came from tips $ 2 / $ 6 = 1 / 3 = c" | a ) 1 / 9 , b ) 1 / 6 , c ) 1 / 3 , d ) 4 / 9 , e ) 5 / 9 | c | divide(2, add(2, 4)) | add(n0,n1)|divide(n0,#0)| | general | C |
at what price must an book costing $ 47.50 be marked in order that after deducting 5 % from the list price . it may be sold at a profit of 25 % on the cost price ? | "c $ 62.50 cp = 47.50 sp = 47.50 * ( 125 / 100 ) = 59.375 mp * ( 95 / 100 ) = 59.375 mp = 62.5" | a ) 72.5 , b ) 55.5 , c ) 62.5 , d ) 82.5 , e ) 60.5 | c | multiply(divide(divide(multiply(47.50, add(const_100, 25)), const_100), subtract(const_100, 5)), const_100) | add(n2,const_100)|subtract(const_100,n1)|multiply(n0,#0)|divide(#2,const_100)|divide(#3,#1)|multiply(#4,const_100)| | gain | C |
what is the area of a square with perimeter 12 p ? | each side is 3 p a = ( 3 p ) ^ 2 = 9 p ^ 2 answer c | ['a ) 16 p ^ 2', 'b ) 4 p', 'c ) 9 p ^ 2', 'd ) p / 16', 'e ) p ^ 2 / 16'] | c | square_area(square_edge_by_perimeter(12)) | square_edge_by_perimeter(n0)|square_area(#0) | geometry | C |
working together , tim and tom can type 12 pages in one hour . if they would be able to type 15 pages in one hour if tom increases his typing speed by 30 % , 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 30 % means he will type 1.3 y pages an hour . so we get x + 1.3 y = 15 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 in an hour , hence ( y / x ) . subtracting both : 0.3 y = 3 so y = 10 . . . so x = 2 ( y / x ) = 10 / 2 = 5 / 1 answer is ( b ) | a ) 1 / 5 , b ) 5 / 1 , c ) 4 / 1 , d ) 1 / 4 , e ) 1 / 3 | b | divide(divide(subtract(15, 12), divide(30, multiply(const_100, const_1))), subtract(12, divide(subtract(15, 12), divide(30, 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 | B |
sheela deposits rs . 3800 in bank savings account . if this is 32 % of her monthly income . what is her monthly income in ? | "explanation : 32 % of income = rs . 3800 100 % of income = 3800 x 100 / 32 = rs . 11875 answer : c" | a ) 22000 , b ) 20000 , c ) 11875 , d ) 12340 , e ) none of these | c | divide(multiply(3800, const_100), 32) | multiply(n0,const_100)|divide(#0,n1)| | gain | C |
there is a rectangular prism made of 1 in cubes that has been covered in tin foil . there are exactly 128 cubes that are not touching any tin foil on any of their sides . if the width of the figure created by these 128 cubes is twice the length and twice the height , what is the measure e in inches of the width of the foil covered prism ? | if the width is w , then length and height would be w / 2 . so , w * w / 2 * w / 2 = 128 = > w ^ 3 = ( 2 ^ 3 ) * 64 = ( 2 ^ 3 ) * ( 4 ^ 3 ) = > w = 2 * 4 = 8 in . along the width of the cuboid , 8 cubes do n ' t touch the tin foil . so the actual width will be non - touching cubes + touching cubes = 8 + 2 = e = 10 ans e . | ['a ) 4', 'b ) 6', 'c ) 8', 'd ) 9', 'e ) 10'] | e | add(multiply(power(divide(128, multiply(1, multiply(1, const_2))), inverse(const_3)), multiply(1, const_2)), multiply(1, const_2)) | inverse(const_3)|multiply(n0,const_2)|multiply(n0,#1)|divide(n1,#2)|power(#3,#0)|multiply(#1,#4)|add(#5,#1) | geometry | E |
if log 0.318 = 0.3364 and log 0.317 = 0.33320 then log 0.319 = ? | log 0.318 / log 0.317 = log ( 0.318 - 0.317 ) = log 0.001 = 0.3364 / 0.33320 = 1.0096 now , log 0.318 * log 0.001 = log ( 0.318 + 0.001 ) = log 0.319 = 0.3364 * 1.0096 = 0.3396 so option c answer : c | a ) 0.33365 , b ) 0.3368 , c ) 0.3396 , d ) 0.3369 , e ) 0.3469 | c | multiply(0.3364, divide(0.3364, 0.3332)) | divide(n1,n3)|multiply(n1,#0) | other | C |
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 | C |
the figure above shows the dimensions of a semicircular cross section of a one - way tunnel . the single traffic lane is 12 feet wide and is equidistant from the sides of the tunnel . if vehicles must clear the top of the tunnel by at least ½ foot when they are inside the traffic lane , what should be the limit h on the height of vehicles that are allowed to use the tunnel ? | "let ' s label the midpoint of the circle o . since the base of the semi - circle is 20 , we know that the diameter is 20 and , accordingly , the radius is 10 . we also know that the traffic lane is 12 feet long and there ' s an equal amount of space on either side , so the traffic lane extends 6 feet on either side of o . let ' s call the leftmost point on the base of the traffic lane a . so , the distance oa is 6 . now draw a line straight up from a to the top of the tunnel . let ' s label the point at which the line intersects the circle b . the answer to the question will , therefore , be the height ab - . 5 feet ( we need to leave . 5 feet of clearance ) . here ' s the key to solving the question : if we draw a line from o to b , that line is a radius of the circle and , therefore , has length 10 . we now have right triangle oab ( the right angle is at point a ) , with leg oa = 6 and hypotenuse ob = 10 . we can now solve for leg ab = 8 ( either by applying the pythagorean theorum or by applying the 3 / 4 / 5 special right triangle ratio ) . finally : ab = 8 , so the correct answer h is 8 - . 5 = 7.5 . . . choose ( b ) ! from a strategic guessing point of view , as soon as we realize that the height of the tunnel is 10 in the middle , we should quickly eliminate ( d ) and ( e ) as too big ; worse case you have a 1 / 3 shot at picking up the points . b" | a ) 5 ½ ft , b ) 7 ½ ft , c ) 8 ½ ft , d ) 9 ½ ft , e ) 10 ft | b | subtract(divide(12, const_2), divide(const_1, const_2)) | divide(n0,const_2)|divide(const_1,const_2)|subtract(#0,#1)| | physics | B |
there are two positive numbers in the ratio 5 : 8 . if the larger number exceeds the smaller by 36 , then find the smaller number ? | "let the two positive numbers be 5 x and 8 x respectively . 8 x - 5 x = 36 3 x = 36 = > x = 12 = > smaller number = 5 x = 60 . answer : d" | a ) 25 , b ) 66 , c ) 77 , d ) 60 , e ) 44 | d | divide(multiply(36, 5), const_4) | multiply(n0,n2)|divide(#0,const_4)| | other | D |
on a certain transatlantic crossing , 20 percent of a ship ’ s passengers held round - trip tickets and also took their cars abroad the ship . if 80 percent of the passengers with round - trip tickets did not take their cars abroad the ship , what percent of the ship ’ s passengers held round - trip tickets ? | "let t be the total number of passengers . let x be the number of people with round trip tickets . 0.2 t had round trip tickets and took their cars . 0.2 x had round trip tickets and took their cars . 0.2 x = 0.2 t x = t the answer is e ." | a ) 20 % , b ) 40 % , c ) 60 % , d ) 80 % , e ) 100 % | e | divide(20, subtract(const_1, divide(80, const_100))) | divide(n1,const_100)|subtract(const_1,#0)|divide(n0,#1)| | gain | E |
each of the 9 squares shown is to contain one number chosen from 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , and 9 . no number is to be repeated . suppose the sum of the 5 squares aligned vertically on the right is 32 and that the sum of the 5 squares aligned horizontally on the bottom is 20 . what number goes in the shared corner square ? | if x is the number in the corner square , then the sum of all the numbers in the squares is equal to the sum of the numbers in the five squares aligned vertically plus the sum of the numbers in the five squares aligned horizontally minus x . hence , 1 + 2 + · · · + 9 = 32 + 20 − x . the sum on the left is 45 , so x = 52 − 45 = 7 . correct answer e | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | e | add(4, add(1, 2)) | add(n1,n2)|add(n4,#0) | geometry | E |
the perimeter of an equilateral triangle is 45 . if one of the side is the side of an isoceles triangle of perimeter 40 then what is the lenght of base of isoceles triangle . | "base of isoceles triangle is 40 - 15 - 15 = 10 units . answer : a" | a ) 10 units , b ) 20 units , c ) 30 units , d ) 40 units , e ) 15 units | a | subtract(subtract(40, divide(45, const_3)), divide(45, const_3)) | divide(n0,const_3)|subtract(n1,#0)|subtract(#1,#0)| | geometry | A |
find the simple interest on rs . 235 for 3 months at 8 paisa per month ? | "explanation : i = ( 235 * 3 * 8 ) / 100 = 56.4 answer : option b" | a ) s . 57.8 , b ) s . 56.4 , c ) s . 29.3 , d ) s . 33 , e ) s . 74 | b | multiply(235, divide(3, const_100)) | divide(n1,const_100)|multiply(n0,#0)| | gain | B |
the radius of the two circular fields is in the ratio 5 : 4 the area of the first field is what percent greater than the area of the second ? | "r = 5 ï € r 2 = 25 r = 4 ï € r 2 = 16 25 ï € â € “ 9 ï € 100 - - - - ? = > 36 % . answer : d" | a ) 22 % , b ) 32 % , c ) 34 % , d ) 64 % , e ) 36 % | d | subtract(power(4, const_2), power(5, const_2)) | power(n1,const_2)|power(n0,const_2)|subtract(#0,#1)| | geometry | D |
the h . c . f . of two numbers is 20 and the other two factors of their l . c . m . are 21 and 23 . the larger of the two numbers is | "solution clearly , the numbers are ( 20 x 21 ) and ( 20 x 23 ) . larger number = ( 20 x 23 ) = 460 . answer d" | a ) 276 , b ) 299 , c ) 312 , d ) 460 , e ) none | d | multiply(20, 23) | multiply(n0,n2)| | other | D |
what is the cp of rs 100 stock at 6 discount , with 1 / 5 % brokerage ? | "explanation : use the formula , cp = 100 â € “ discount + brokerage % cp = 100 - 6 + 1 / 5 94.2 thus the cp is rs 94.2 . answer : c" | a ) 96.9 , b ) 96.3 , c ) 94.2 , d ) 96.7 , e ) 96.21 | c | add(subtract(100, 6), divide(1, 5)) | divide(n2,n3)|subtract(n0,n1)|add(#0,#1)| | gain | C |
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