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 β s speed is 17 / 15 times that of b . if a and b run a race , what part of the length of the race should a give b as a head start , so that the race ends in a dead heat ? | "let x be the fraction of the distance that b runs . let v be the speed at which b runs . the time should be the same for both runners . time = d / ( 17 v / 15 ) = xd / v ( 15 / 17 ) * d / v = x * d / v x = 15 / 17 b should have a head start of 2 / 17 of the full distance . the answer is c ." | a ) 1 / 16 , b ) 2 / 15 , c ) 2 / 17 , d ) 1 / 8 , e ) 1 / 7 | c | divide(subtract(17, 15), 17) | subtract(n0,n1)|divide(#0,n0)| | general | C |
the c . p of 5 books is equal to the s . p of 20 books . find his gain % or loss % ? | "explanation : 5 cp = 20 sp 20 - - - 15 cp loss 100 - - - ? = > 75 % loss answer : e" | a ) 35 % , b ) 20 % , c ) 5 % , d ) 25 % , e ) 75 % | e | multiply(subtract(const_1, divide(5, 20)), const_100) | divide(n0,n1)|subtract(const_1,#0)|multiply(#1,const_100)| | gain | E |
a manufacturer is using glass as the surface for the multi - touch screen of its smartphone . the glass on the manufactured phone has a 4 % probability of not passing quality control tests . the quality control manager bundles the smartphone in groups of 10 . if that bundle has any smartphone that does not pass the quality control test , the entire bundle of 10 is rejected . what is the probability that a smartphone bundle that will be rejected by quality control ? | find the probability of the opposite event and subtract from 1 . the opposite event is that bundle will not be rejected by quality control , which will happen if all 10 phones pass the test , so p ( all 10 phones pass test ) = 0.96 ^ 10 . p ( at least one phone do not pass the test ) = 1 - p ( all 10 phones pass test ) = 1 - 0.96 ^ 10 . answer : e . | a ) 0.25 , b ) . 05 ^ 10 , c ) 1 - 0.95 ^ 10 , d ) 1 - 0.05 ^ 10 , e ) 1 - 0.96 ^ 10 | e | subtract(const_1, power(subtract(const_1, divide(4, const_100)), 10)) | divide(n0,const_100)|subtract(const_1,#0)|power(#1,n1)|subtract(const_1,#2) | general | E |
a man can ride on motorcycle at 50 kmph upward road and 100 kmph downward road . he takes 12 hours to ride motorcycle uphill from lower point a to upper point b and back to a . what is the total distance traveled by him in 12 hours ? he must return back to base point a in remaining time . | upward distance traveled per hour - 50 km , distance traveled after 8 hours = 400 km . he remained with 4 hours travel downward , i . e . distance traveled in remaining 4 hours downward = 400 km so total distance traveled from a to b and back to point a = 400 km upward + 400 km downhill = 800 km ( answer : c ) | a ) 1000 , b ) 900 , c ) 800 , d ) 400 , e ) 1200 | c | multiply(divide(multiply(multiply(50, 100), const_2), add(50, 100)), 12) | add(n0,n1)|multiply(n0,n1)|multiply(#1,const_2)|divide(#2,#0)|multiply(n2,#3) | physics | C |
pavan travelled for 15 hours . he covered the first half of the distance at 30 kmph and remaining half of the distance at 25 kmph . find the distance travelled by pavan . | let the distance travelled be x km . total time = ( x / 2 ) / 30 + ( x / 2 ) / 25 = 15 = > x / 60 + x / 50 = 15 = > ( 5 x + 6 x ) / 300 = 15 = > x = 409 km answer : b | a ) 408 km , b ) 409 km , c ) 410 km , d ) 412 km , e ) 419 km | b | multiply(inverse(add(inverse(multiply(const_2, 30)), inverse(multiply(25, const_2)))), 15) | multiply(n1,const_2)|multiply(n2,const_2)|inverse(#0)|inverse(#1)|add(#2,#3)|inverse(#4)|multiply(n0,#5) | physics | B |
the food in a camp lasts for 20 men for 50 days . if 20 more men join , how many days will the food last ? | one man can consume the same food in 20 * 50 = 1000 days . 20 more men join , the total number of men = 40 the number of days the food will last = 1000 / 40 = 25 days . answer : b | a ) 22 days , b ) 25 days , c ) 28 days , d ) 16 days , e ) 27 days | b | divide(multiply(20, 50), add(20, 20)) | add(n0,n0)|multiply(n0,n1)|divide(#1,#0) | physics | B |
a person x working alone can complete a work in 5 days . a person y completes the same amount of work in 20 days , and a person z when working alone can complete the same amount of work in 30 days . all 3 people work together for 2 days but then x and y leave . how many more days does z need to complete the work after x and y leave ? | the fraction of work completed in two days is 2 / 5 + 2 / 20 + 2 / 30 = 17 / 30 the fraction of work remaining is 1 - 17 / 30 = 13 / 30 the number of days required for z is 13 days . the answer is a . | a ) 13 , b ) 14 , c ) 15 , d ) 16 , e ) 17 | a | divide(subtract(const_1, multiply(add(add(divide(const_1, 5), divide(const_1, 20)), divide(const_1, 30)), 2)), divide(const_1, 30)) | divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|add(#3,#2)|multiply(n4,#4)|subtract(const_1,#5)|divide(#6,#2) | physics | A |
which number need to add to 956734 to get a number exactly divisible by 412 ? | "956734 / 412 = 2322 and reminder = 70 . 412 - 70 = 342 so , the next number divisible by 412 is 342 places in front of 956734 which means 342 + 956734 = 957076 342 should be added to 956734 b" | a ) 345 , b ) 342 , c ) 343 , d ) 342 , e ) 341 | b | multiply(412, subtract(add(floor(divide(956734, 412)), const_1), divide(956734, 412))) | divide(n0,n1)|floor(#0)|add(#1,const_1)|subtract(#2,#0)|multiply(n1,#3)| | general | B |
a person is traveling at 65 km / hr and reached his destiny in 3 hr then find the distance ? | "t = 3 hrs d = t * s = 65 * 3 = 195 km answer is b" | a ) 205 km , b ) 195 km , c ) 295 km , d ) 115 km , e ) 215 km | b | multiply(65, 3) | multiply(n0,n1)| | physics | B |
bhanu spends 30 % of his income on petrol on scooter 20 % of the remaining on house rent and the balance on food . if he spends rs . 300 on petrol then what is the expenditure on house rent ? | "given 30 % ( income ) = 300 β β income = 1000 after having spent rs . 300 on petrol , he left with rs . 700 . his spending on house rent = 20 % ( 700 ) = rs . 140 answer : b" | a ) 2287 , b ) 140 , c ) 128 , d ) 797 , e ) 123 | b | multiply(subtract(divide(300, divide(30, const_100)), 300), divide(20, const_100)) | divide(n1,const_100)|divide(n0,const_100)|divide(n2,#1)|subtract(#2,n2)|multiply(#0,#3)| | gain | B |
63 men working 8 hours per day dig 30 m deep . how many extra men should be put to dig to a depth of 50 m working 6 hours per day ? | "( 63 * 8 ) / 30 = ( x * 6 ) / 50 = > x = 140 140 β 63 = 77 answer : b" | a ) a ) 87 , b ) b ) 77 , c ) c ) 66 , d ) d ) 55 , e ) e ) 97 | b | subtract(divide(multiply(divide(multiply(63, 8), 30), 50), 6), 63) | multiply(n0,n1)|divide(#0,n2)|multiply(n3,#1)|divide(#2,n4)|subtract(#3,n0)| | physics | B |
working individually , emma can wrap presents for 6 hours and troy can wrap presents in 8 hours . if emma and troy work together but independently at the task for 2 hours , at which point troy leaves , how many remaining hours will it take emma to complete the task alone ? | in first 2 hrs troy will finish 2 / 8 = 1 / 4 of work and emma will finish 2 / 6 work so total 1 / 4 + 1 / 3 = 7 / 12 work is finished and 1 - 7 / 12 = 5 / 12 work remaining . now emma will take ( 5 / 12 ) * 6 = 30 / 12 hrs to finish it . so answer is a . | a ) 30 / 12 , b ) 5 / 12 , c ) 3 / 4 , d ) 2 / 12 , e ) 3 / 5 | a | divide(subtract(const_1, add(multiply(2, inverse(6)), multiply(2, inverse(8)))), inverse(6)) | inverse(n0)|inverse(n1)|multiply(n2,#0)|multiply(n2,#1)|add(#2,#3)|subtract(const_1,#4)|divide(#5,#0) | physics | A |
if a : b = 4 : 1 , then find ( a - 3 b ) / ( 2 a - b ) ? | answer : option b a / b = 4 / 1 = > a = 4 b ( a - 3 b ) / ( 2 a - b ) = ( 4 b - 3 b ) / ( 8 b - b ) = b / 7 b = > 1 / 7 | a ) 6 / 7 , b ) 1 / 7 , c ) 5 / 7 , d ) 3 / 2 , e ) 4 / 7 | b | divide(subtract(4, 3), subtract(multiply(2, 4), 1)) | multiply(n0,n3)|subtract(n0,n2)|subtract(#0,n1)|divide(#1,#2) | general | B |
a company producing fruit juice changed its packaging from boxes measuring 5 x 10 x 20 centimeters to boxes measuring 6 x 10 x 20 centimeters . if the price of a box did not change and all boxes are full of juice , by approximately what percent did the price of the juice decrease ? | suppose when v = 5 x 10 x 20 = 1000 cm , the price is $ 1200 per cm price = $ 1.2 the price is $ 1200 if v = 6 x 10 x 20 = 1000 cm . per cm price = $ 1.0 so the price is decreased by $ 0.2 so the % price decreased by $ 0.2 / 1.20 = 16.67 % answer : b | a ) 12.00 % , b ) 16.67 % , c ) 18.33 % , d ) 20.00 % , e ) 21.50 % | b | multiply(subtract(const_1, divide(multiply(multiply(5, 10), 20), multiply(multiply(6, 10), 20))), const_100) | multiply(n0,n1)|multiply(n1,n3)|multiply(n2,#0)|multiply(n2,#1)|divide(#2,#3)|subtract(const_1,#4)|multiply(#5,const_100) | general | B |
how many internal diagonals does a heptagon ( seven sided polygon ) have ? | "number of diagonals in any polygon can be found using this formula : n ( n - 3 ) / 2 here n = 7 no . of diagonals = 7 ( 7 - 3 ) / 2 = 14 ans c" | a ) 7 , b ) 9 , c ) 14 , d ) 20 , e ) 35 | c | multiply(subtract(multiply(const_2, const_4), const_3), divide(multiply(const_2, const_4), const_2)) | multiply(const_2,const_4)|divide(#0,const_2)|subtract(#0,const_3)|multiply(#1,#2)| | geometry | C |
a man buys a cycle for rs . 1600 and sells it at a loss of 10 % . what is the selling price of the cycle ? | "s . p . = 90 % of rs . 1600 = 90 / 100 x 1600 = rs . 1440 answer : d" | a ) 1410 , b ) 1420 , c ) 1430 , d ) 1440 , e ) 1540 | d | divide(multiply(subtract(const_100, 10), 1600), const_100) | subtract(const_100,n1)|multiply(n0,#0)|divide(#1,const_100)| | gain | D |
the wages earned by robin is 35 % more than that earned by erica . the wages earned by charles is 60 % more than that earned by erica . how much % is the wages earned by charles more than that earned by robin ? | "explanatory answer let the wages earned by erica be $ 100 then , wages earned by robin and charles will be $ 135 and $ 160 respectively . charles earns $ 35 more than robin who earns $ 135 . therefore , charles ' wage is 35 / 135 * 100 = 25.92 % . the correct choice is ( a )" | a ) 25.92 % , b ) 18.75 % , c ) 30 % , d ) 50 % , e ) 100 % | a | multiply(divide(subtract(add(const_100, 60), add(const_100, 35)), add(const_100, 35)), const_100) | add(n1,const_100)|add(n0,const_100)|subtract(#0,#1)|divide(#2,#1)|multiply(#3,const_100)| | general | A |
the total marks obtained by a student in mathematics and physics is 60 and his score in chemistry is 20 marks more than that in physics . find the average marks scored in mathamatics and chemistry together . | "let the marks obtained by the student in mathematics , physics and chemistry be m , p and c respectively . given , m + c = 60 and c - p = 20 m + c / 2 = [ ( m + p ) + ( c - p ) ] / 2 = ( 60 + 20 ) / 2 = 40 . answer : a" | a ) 40 , b ) 30 , c ) 25 , d ) data inadequate , e ) none of these . | a | divide(add(60, 20), const_2) | add(n0,n1)|divide(#0,const_2)| | general | A |
( ( 1 ^ 5 m ) / ( 5 ^ 5 m ) ) ( ( 1 ^ 18 ) / ( 4 ^ 18 ) ) = 1 / ( 2 ( 10 ) ^ 35 ) what is m ? | ( ( 1 ^ 5 m ) / ( 5 ^ 5 m ) ) ( ( 1 ^ 18 ) / ( 4 ^ 18 ) ) = 1 / ( 2 ( 10 ) ^ 35 ) ( ( 1 / 5 ) ^ 5 m ) * ( ( 1 / 2 ) ^ 36 ) = 1 / ( 2 * ( 2 * 5 ) ^ 35 ) ) 2 ^ 36 will cancel out , since 1 can be written as 1 ^ 35 , so ( 1 / 5 ) ^ 5 m = ( 1 / 5 ) ^ 35 ( ( 1 / 5 ) ^ 5 m ) * ( ( 1 / 2 ) ^ 36 ) = 1 / [ ( 2 ^ 36 ) * ( 5 ^ 35 ) ] so , m = 7 answer a | a ) 7 , b ) 18 , c ) 34 , d ) 35 , e ) 36 | a | divide(log(divide(multiply(power(1, 18), multiply(power(10, 35), 2)), power(4, 18))), log(power(5, 5))) | power(n10,n11)|power(n0,n5)|power(n6,n5)|power(n1,n1)|log(#3)|multiply(n9,#0)|multiply(#5,#1)|divide(#6,#2)|log(#7)|divide(#8,#4) | general | A |
a sum was put at si at certain rate for 2 years . had it been put at 3 % higher rate , it would have fetched rs . 300 more . find the sum | let p be the sum . and x be the interest rate . ( 2 * ( x + 3 ) * p / 100 ) - ( 2 * x * p / 100 ) = 300 therefore 6 p / 100 = 300 p = rs 5000 answer : b | a ) rs 4000 , b ) rs 5000 , c ) rs 6000 , d ) rs 7000 , e ) rs 8000 | b | divide(300, multiply(divide(3, const_100), 2)) | divide(n1,const_100)|multiply(n0,#0)|divide(n2,#1) | gain | B |
harry started a 4 - mile hike with a full 10 - cup canteen of water and finished the hike in 2 hours with 2 cup of water remaining in the canteen . if the canteen leaked at the rate of 1 cup per hour and harry drank 3 cups of water during the last mile , how many cups did he drink per mile during the first 3 miles of the hike ? | "no of cups leaked during the trip = 2 cups . no of cups harry drank = 6 cups . no of cups harry drank during the first 3 miles = 3 . drink / mile = 3 / 3 answer : c" | a ) 1 / 3 , b ) 2 / 3 , c ) 3 / 3 , d ) 3 / 1 , e ) 3 / 2 | c | divide(subtract(subtract(subtract(10, 2), 3), 2), 3) | subtract(n1,n2)|subtract(#0,n5)|subtract(#1,n3)|divide(#2,n6)| | physics | C |
jamshid and irwin decided to make up their own version of basketball . in addition to being able to score 2 - point baskets , they created ways to score 5 - , 11 - , and 13 - point baskets . if at halftime the product of the point values of irwin β s baskets was 2420 , how many 11 - point baskets e did irwin score ? | first thing i noticed was that all these numbers are prime . . . prime factorization gives the following step 1 : 2 * 1210 step 2 : 2 * 11 * 110 step 4 : 2 * 11 * 11 * 10 step 5 : 2 * 11 * 11 * 2 * 5 so he scored e = 2 11 - point baskets in my opinion . answer c | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | c | divide(divide(divide(2420, 11), 11), const_10) | divide(n4,n2)|divide(#0,n2)|divide(#1,const_10) | general | C |
if [ [ x ] ] = x ^ 2 + 2 x + 4 , what is the value of [ [ 5 ] ] ? | "these functions questions might look intimidating , but they just test your knowledge about how well you can substitute values [ [ x ] ] = x ^ 2 + 2 x + 4 [ [ 5 ] ] = 5 ^ 2 + 2 * 5 + 4 = 39 . option a" | a ) 39 , b ) 9 , c ) 15 , d ) 19 , e ) 25 | a | add(add(power(5, const_2.0), multiply(2, 2)), 4) | multiply(n3,n3)|power(n0,n0)|add(#0,#1)|add(n2,#2)| | general | A |
the cost of 2 chairs and 3 tables is rs . 1300 . the cost of 3 chairs and 2 tables is rs . 1200 . the cost of each table is more than that of each chair by ? | "c 100 2 c + 3 t = 1300 - - - ( 1 ) 3 c + 3 t = 1200 - - - ( 2 ) subtracting 2 nd from 1 st , we get - c + t = 100 = > t - c = 100" | a ) 50 , b ) 60 , c ) none of these , d ) 70 , e ) 100 | c | subtract(divide(subtract(multiply(3, 1300), multiply(2, 3)), subtract(multiply(3, 3), multiply(2, 2))), divide(subtract(3, multiply(2, divide(subtract(multiply(3, 1300), multiply(2, 3)), subtract(multiply(3, 3), multiply(2, 2))))), 3)) | multiply(n2,const_3)|multiply(n3,const_2)|multiply(n1,const_3)|multiply(n0,const_2)|subtract(#0,#1)|subtract(#2,#3)|divide(#4,#5)|multiply(n0,#6)|subtract(n3,#7)|divide(#8,n1)|subtract(#6,#9)| | general | C |
a cistern can be filled by a tap in 5 hours while it can be emptied by another tap in 6 hours . if both the taps are opened simultaneously , then after how much time will the cistern get filled ? | "net part filled in 1 hour = 1 / 5 - 1 / 6 = 1 / 30 therefore the cistern will be filled in 30 hours . answer : e" | a ) 4.5 hrs , b ) 5 hrs , c ) 6.5 hrs , d ) 7.2 hrs , e ) 30 hrs | e | divide(const_1, subtract(divide(const_1, 5), divide(const_1, 6))) | divide(const_1,n0)|divide(const_1,n1)|subtract(#0,#1)|divide(const_1,#2)| | physics | E |
1397 x 1397 = ? | "1397 x 1397 = ( 1397 ) 2 = ( 1400 - 3 ) 2 = ( 1400 ) 2 + ( 3 ) 2 - ( 2 x 1400 x 3 ) = 1960000 + 9 - 8400 = 1960009 - 8400 = 1951609 . b )" | a ) 1951607 , b ) 1951609 , c ) 1951610 , d ) 1951611 , e ) 1951613 | b | multiply(divide(1397, 1397), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general | B |
if 21 lines are drawn in a plane such that no two of them are parallel and no three are concurrent , then in how many points do they intersect ? | "if two lines are not parallel , then they will intersect in exactly one point . lines can be extended infinitely on both ends so somewhere they will intersect with each other if they are not parallel . we are also given that no three lines are concurrent . this means that no three lines intersect at the same point . so every pair of two lines we select will have a unique point of intersection which they will not share with any third line . the number of ways to select 2 lines from 21 lines is 21 c 2 = 210 the answer is d ." | a ) 150 , b ) 170 , c ) 190 , d ) 210 , e ) 230 | d | divide(factorial(21), multiply(factorial(subtract(21, const_2)), factorial(const_2))) | factorial(n0)|factorial(const_2)|subtract(n0,const_2)|factorial(#2)|multiply(#3,#1)|divide(#0,#4)| | physics | D |
how much water must be added to 60 litres of milk at 1.5 litres for rs . 20 so as to have a mixture worth rs . 32 / 3 a litre ? | c . p . of 1 litre of milk = rs . 20 Γ 2 / 3 = rs . 40 / 3 . c . p . of 1 litre of water = rs 0 . mean price = rs . 32 / 3 . by the rule of alligation , we have : c . p of 1 litre c . p of 1 litre of water of milk ( 0 ) ( rs . 40 / 3 ) \ / mean price ( rs . 32 / 3 ) / \ 40 / 3 β 32 / 3 32 / ( 3 β 0 ) 8 / 3 32 / 3 the ratio of water and milk = 8 / 3 : 32 / 3 . = 8 : 32 = 1 : 4 . thus , quantity of water to be added to 60 litres of milk : = ( 1 / 4 ) Γ 60 litres . = 15 litres . answer : c | a ) 10 litres , b ) 12 litres , c ) 15 litres , d ) 18 litres , e ) 21 litres | c | divide(60, divide(divide(32, 3), subtract(divide(20, 1.5), divide(32, 3)))) | divide(n3,n4)|divide(n2,n1)|subtract(#1,#0)|divide(#0,#2)|divide(n0,#3) | general | C |
the youngest of 4 children has siblings who are 1 , 4 , and 7 years older than she is . if the average ( arithmetic mean ) age of the 4 siblings is 17 , what is the age of the youngest sibling ? | "x + ( x + 1 ) + ( x + 4 ) + ( x + 8 ) = 68 4 x + 12 = 68 4 x = 56 x = 14 the answer is e ." | a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | e | divide(subtract(multiply(4, 17), add(add(4, 4), 7)), 4) | add(n0,n2)|multiply(n0,n5)|add(n3,#0)|subtract(#1,#2)|divide(#3,n0)| | general | E |
the speed at which a girl can row a boat in still water is 90 kmph . if she rows downstream , where the speed of current is 18 kmph , what time will he take to cover 330 metres ? | "speed of the boat downstream = 90 + 18 = 108 kmph = 108 * 5 / 18 = 30 m / s hence time taken to cover 330 m = 330 / 30 = 11 seconds . answer : a" | a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | a | divide(multiply(330, const_3_6), add(90, 18)) | add(n0,n1)|multiply(n2,const_3_6)|divide(#1,#0)| | physics | A |
the average salary of workers in an industry is rs . 300 the average salary of technicians being rs . 400 and that of non - technicians being rs . 125 . what is the total number of workers ? | 400 125 \ / 300 / \ 175 100 7 : 4 7 - > 150 11 - > ? = > 234 answer : b | a ) 230 , b ) 234 , c ) 250 , d ) 547 , e ) 484 | b | floor(add(multiply(subtract(300, 125), divide(400, 300)), const_1)) | divide(n1,n0)|subtract(n0,n2)|multiply(#0,#1)|add(#2,const_1)|floor(#3) | general | B |
a one - foot stick is marked in 1 / 4 and 1 / 6 portion . how many total markings will there be , including the end points ? | "lcm of 24 = 12 1 / 4 marking are ( table of 3 ) 0 . . . . . . 3 . . . . . . . . . . . 6 . . . . . . . . . . 9 . . . . . . . . . . 12 ( total = 5 ) 1 / 6 marking are ( table of 2 ) 0 . . . . . . . 2 . . . . . . 4 . . . . . . 6 . . . . . . . . 8 . . . . . . . . . 12 ( ( total = 6 ) overlapping markings are 0 . . . . . . . . 6 . . . . . . . . . 12 ( total = 3 ) total markings = 5 + 6 - 3 = 8 answer = a" | a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 16 | a | add(4, 6) | add(n1,n3)| | general | A |
set s contains exactly 10 numbers and has an average ( arithmetic mean ) of 6.2 . if one of the numbers in set s is increased by 8 , while all other numbers remain the same , what is the new average of set s ? | old set s - total is avg * no of elements = 6.2 * 10 = 62 if one number is increased by 8 then total increased to 62 + 8 = 70 new avg - 70 / 10 = 7.0 . hence answer is d . | a ) 6.6 , b ) 6.7 , c ) 6.8 , d ) 7.0 , e ) 6.9 | d | divide(add(multiply(10, 6.2), 8), 10) | multiply(n0,n1)|add(n2,#0)|divide(#1,n0) | general | D |
the average of 11 results is 42 , if the average of first 5 results is 49 and that of the last 7 is 52 . find the fifth result ? | 1 to 11 = 11 * 42 = 462 1 to 5 = 5 * 49 = 245 5 to 11 = 7 * 52 = 364 5 th = 245 + 364 β 462 = 147 answer : c | a ) 145 , b ) 146 , c ) 147 , d ) 148 , e ) 149 | c | subtract(add(multiply(5, 49), multiply(7, 52)), multiply(11, 42)) | multiply(n2,n3)|multiply(n4,n5)|multiply(n0,n1)|add(#0,#1)|subtract(#3,#2) | general | C |
the product z of two prime numbers is between 10 and 30 . if one of the prime numbers is greater than 2 but less than 6 and the other prime number is greater than 6 but less than 24 , then what is z ? | "the smallest possible product is 21 which is 3 * 7 . all other products are too big . the answer is a ." | a ) 21 , b ) 15 , c ) 14 , d ) 10 , e ) 6 | a | multiply(subtract(10, const_4), const_3) | subtract(n0,const_4)|multiply(#0,const_3)| | general | A |
a right circular cylinder has a height of 22 and a radius of 5 . a rectangular solid with a height of 15 and a square base , is placed in the cylinder such that each of the corners of the solid is tangent to the cylinder wall . liquid is then poured into the cylinder such that it reaches the rim . what is the volume of the liquid ? | [ quote = bunuel ] a right circular cylinder has a height of 20 and a radius of 5 . a rectangular solid with a height of 15 and a square base , is placed in the cylinder such that each of the corners of the solid is tangent to the cylinder wall . liquid is then poured into the cylinder such that it reaches the rim . what is the volume of the liquid ? the square base has sides of sqrt ( 50 ) due to the 45 - 45 - 90 triangle 22 * 25 * pi - 15 * sqrt ( 50 ) ^ 2 = 550 ( Ο β 1.5 ) d . 500 ( Ο β 1.5 ) | ['a ) 500 ( Ο β 3 )', 'b ) 500 ( Ο β 2.5 )', 'c ) 500 ( Ο β 2 )', 'd ) 550 ( Ο β 1.5 )', 'e ) 500 ( Ο β 1 )'] | d | divide(volume_cylinder(5, 22), const_2) | volume_cylinder(n1,n0)|divide(#0,const_2) | geometry | D |
in the set of positive integers from 1 to 80 , what is the sum of all the odd multiples of 5 ? | "reduce to 1 - 80 5 - 15 - 25 - 35 - 45 - - 55 - - 65 - - 75 are valid multiples . add them - - > 320 c" | a ) 180 , b ) 245 , c ) 320 , d ) 260 , e ) 270 | c | divide(multiply(80, multiply(const_3, 5)), 5) | multiply(n2,const_3)|multiply(n1,#0)|divide(#1,n2)| | general | C |
a precious stone weighing 35 grams worth rs . 12,250 is accidentally dropped and gets broken into two pieces having weights in the ratio of 2 : 5 . if the price varies as the square of the weight then find the loss incurred . | the price varies as the square of the weight , p = k x w 2 12250 = k x 352 k = 10 thus , p = 10 w 2 when the two pieces are in the ratio 2 : 5 ( weight wise ) then we know that their weights must be 10 g and 25 g respectively . their values should be : 10 g piece : 10 x 102 = rs . 1000 ; 25 g piece : 10 x 252 = rs . 6250 . total value of stone now , = 1000 + 6250 = rs . 7250 loss incurred = 12250 - 7250 = rs . 5000 answer : d | a ) rs . 5750 , b ) rs . 6000 , c ) rs . 5500 , d ) rs . 5000 , e ) rs . 7000 | d | subtract(subtract(subtract(subtract(add(multiply(const_100, const_100), divide(divide(multiply(const_100, const_100), const_2), 2)), const_100), const_100), divide(const_100, const_2)), add(multiply(divide(subtract(subtract(subtract(add(multiply(const_100, const_100), divide(divide(multiply(const_100, const_100), const_2), 2)), const_100), const_100), divide(const_100, const_2)), power(35, const_2)), power(multiply(5, 2), const_2)), multiply(divide(subtract(subtract(subtract(add(multiply(const_100, const_100), divide(divide(multiply(const_100, const_100), const_2), 2)), const_100), const_100), divide(const_100, const_2)), power(35, const_2)), power(multiply(5, 5), const_2)))) | divide(const_100,const_2)|multiply(const_100,const_100)|multiply(n2,n3)|multiply(n3,n3)|power(n0,const_2)|divide(#1,const_2)|power(#2,const_2)|power(#3,const_2)|divide(#5,n2)|add(#8,#1)|subtract(#9,const_100)|subtract(#10,const_100)|subtract(#11,#0)|divide(#12,#4)|multiply(#13,#6)|multiply(#13,#7)|add(#14,#15)|subtract(#12,#16) | other | D |
what is 0.01 percent of 12,356 ? | "soln : - 0.01 % of 12,356 = 0 . 011000.01100 x 12,356 = 1100 β 1001100 β 100 x 12,356 = 12,356100 β 10012,356100 β 100 = 1.2356 answer : b" | a ) 0.12356 , b ) 1.2356 , c ) 12.356 , d ) 0.012356 , e ) 0.0012356 | b | divide(multiply(0.01, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100) | add(const_2,const_3)|add(const_3,const_4)|add(const_3,const_3)|multiply(const_3,const_4)|multiply(#0,const_2)|multiply(#3,const_100)|multiply(#1,#0)|multiply(#4,#5)|multiply(#6,#4)|add(#7,#8)|add(#9,#2)|multiply(n0,#10)|divide(#11,const_100)| | gain | B |
the average of first 18 natural numbers is ? | "sum of 18 natural no . = 342 / 2 = 171 average = 171 / 18 = 9.5 answer : b" | a ) 5.6 , b ) 9.5 , c ) 9.1 , d ) 9.8 , e ) 5.2 | b | add(18, const_1) | add(n0,const_1)| | general | B |
a can do a work in 6 days , b can do a work in 8 days and c can do it in 12 days . b left work after 4 days . for how many number of days should a and c should work together to complete the remaining work ? | "b work 1 / 8 * 4 = 1 / 2 remaining work = 1 - 1 / 2 = 1 / 2 a and c work together = 1 / 6 + 1 / 12 = 3 / 12 = 1 / 4 take reciprocal 4 * remaining work = 2 * 1 / 4 = 0.5 answer : e" | a ) 0.1 , b ) 0.2 , c ) 0.3 , d ) 0.4 , e ) 0.5 | e | add(divide(4, 6), divide(4, 8)) | divide(n3,n0)|divide(n3,n1)|add(#0,#1)| | physics | E |
for each 6 - month period during a light bulb ' s life span , the odds of it not burning out from over - use are half what they were in the previous 6 - month period . if the odds of a light bulb burning out during the first 6 - month period following its purchase are 2 / 7 , what are the odds of it burning out during the period from 6 months to 1 year following its purchase ? | "p ( of not burning out in a six mnth period ) = 1 / 2 of p ( of not burning out in prev 6 mnth period ) p ( of burning out in 1 st 6 mnth ) = 2 / 7 - - - > p ( of not burning out in 1 st 6 mnth ) = 1 - 2 / 7 = 5 / 7 - - - - > p ( of not burning out in a six mnth period ) = 1 / 2 * 5 / 7 = 1 / 3 - - - > p ( of burning out in a six mnth period ) = 1 - 1 / 3 = 2 / 3 now p ( of burning out in 2 nd six mnth period ) = p ( of not burning out in 1 st six mnth ) * p ( of burning out in a six mnth ) = 5 / 7 * 2 / 3 = 1 / 2 ans c" | a ) 5 / 27 , b ) 2 / 9 , c ) 1 / 2 , d ) 4 / 9 , e ) 2 / 3 | c | multiply(subtract(1, divide(2, 7)), subtract(1, divide(subtract(1, divide(2, 7)), 2))) | divide(n3,n4)|subtract(n6,#0)|divide(#1,n3)|subtract(n6,#2)|multiply(#1,#3)| | general | C |
3 men and 8 women complete a task in same time as 6 men and 2 women do . how much fraction of work will be finished in same time if 5 men and 2 women will do that task . | "3 m + 8 w = 6 m + 2 w 3 m = 6 w 1 m = 2 w therefore 3 m + 8 w = 14 w 5 m + 2 w = 12 w answer is 12 / 14 = 6 / 7 answer : a" | a ) 6 / 7 , b ) 3 / 10 , c ) 3 / 18 , d ) 3 / 16 , e ) 3 / 11 | a | divide(add(multiply(divide(subtract(8, 2), subtract(6, 3)), 5), 2), add(multiply(6, divide(subtract(8, 2), subtract(6, 3))), 2)) | subtract(n1,n3)|subtract(n2,n0)|divide(#0,#1)|multiply(n4,#2)|multiply(n2,#2)|add(n5,#3)|add(n3,#4)|divide(#5,#6)| | physics | A |
the cost of 8 pens and 4 pencils is rs . 200 and the cost of 2 pens and 2 pencils is rs . 48 . find the cost of each pen ? | "let the cost of each pen and pencil be ' p ' and ' q ' respectively . 8 p + 4 q = 200 - - - ( 1 ) 2 p + 2 q = 48 4 p + 4 q = 96 - - - ( 2 ) ( 1 ) - ( 2 ) = > 4 p = 104 = > p = 26 answer : d" | a ) rs . 32 , b ) rs . 36 , c ) rs . 42 , d ) rs . 26 , e ) rs . 104 | d | multiply(multiply(8, const_4), divide(200, add(8, 2))) | add(n0,n4)|multiply(n0,const_4)|divide(n2,#0)|multiply(#2,#1)| | general | D |
a company that ships boxes to a total of 14 distribution centers uses color coding to identify each center . if either a single color or a pair of two different colors is chosen to represent each center and if each center is uniquely represented by that choice of one or two colors , what is the minimum number of colors needed for the coding ? ( assume that the order of the colors in a pair does not matter ) | "back - solving is the best way to solve this problem . you basically need 14 combinations ( including single colors ) if we start from option 1 - > 1 = > 4 c 2 + 4 = 10 ( not enough ) 2 = > 5 c 2 + 5 = 15 ( enough ) since the minimum number is asked . it should be 5 . answer - c" | a ) 7 , b ) 8 , c ) 5 , d ) 9 , e ) 6 | c | subtract(divide(factorial(subtract(divide(14, const_2), const_1)), multiply(factorial(const_3), factorial(const_2))), subtract(divide(14, const_2), const_1)) | divide(n0,const_2)|factorial(const_3)|factorial(const_2)|multiply(#1,#2)|subtract(#0,const_1)|factorial(#4)|divide(#5,#3)|subtract(#6,#4)| | general | C |
what is the smallest integer c for which 27 ^ c > 3 ^ 24 ? | "27 ^ c > 3 ^ 24 converting into the same bases : 27 ^ c > 27 ^ 8 therefore for the equation to hold true , c > 8 or c = 9 option c" | a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 12 | c | add(floor(divide(24, const_3)), const_1) | divide(n2,const_3)|floor(#0)|add(#1,const_1)| | general | C |
the cost price of 7 articles is equal to the selling price of 5 articles . what is the profit percent ? | "5 * sp = 7 * cp sp = 1.4 * cp the profit percent is 40 % . the answer is d ." | a ) 10 % , b ) 20 % , c ) 30 % , d ) 40 % , e ) 50 % | d | divide(multiply(7, const_4), add(const_4, const_1)) | add(const_1,const_4)|multiply(n0,const_4)|divide(#1,#0)| | gain | D |
an assembly line produces 90 cogs per hour until an initial order of 60 cogs is completed . the speed of the assembly line is then immediately decreased so that it only produces 60 cogs per hour until another 60 cogs are produced . what is the overall average output , in cogs per hour , for the assembly line during this whole time ? | "the time to produce the first 60 cogs is 60 / 90 = 2 / 3 hours . the time to produce the next 60 cogs is 60 / 60 = 1 hour . the average output is 120 cogs / ( 5 / 3 ) hours = 72 cogs per hour . the answer is e ." | a ) 68 , b ) 69 , c ) 70 , d ) 71 , e ) 72 | e | divide(multiply(60, const_2), add(divide(60, 90), const_1)) | divide(n1,n0)|multiply(n1,const_2)|add(#0,const_1)|divide(#1,#2)| | general | E |
in a college , 50 % of total 400 arts students are locals . 25 % of students from 100 science students are locals and 85 % of total 120 commerce students are locals . what is the total percentage of locals from arts , science and commerce . | locals from arts = 50 % of 400 = 200 locals from science = 25 % of 100 = 25 locals from commerce = 85 % of 120 = 102 total locals = 200 + 25 + 102 = 327 d | a ) 322 , b ) 340 , c ) 333 , d ) 327 , e ) 500 | d | add(add(multiply(400, divide(50, const_100)), multiply(divide(25, const_100), 100)), multiply(divide(85, const_100), 120)) | divide(n0,const_100)|divide(n2,const_100)|divide(n4,const_100)|multiply(n1,#0)|multiply(n3,#1)|multiply(n5,#2)|add(#3,#4)|add(#6,#5) | gain | D |
9 people decided to split the restaurant bill evenly . if the bill was $ 314.16 dollars , how much money did they 1 cent is the smallest unit ? | this is equivalent to finding the first number that is divisible by 9 that occurs after 31416 . in order to divide the sum in 9 parts , the amount must be divisible by 9 divisibility rule of 9 : the sum of the digits must be divisible by 9 sum of digits of 31416 = 15 and 18 is divisible by 9 . hence , we need to add 3 to this number for it to be divisible by 9 correct option : d | a ) $ 314.16 , b ) $ 314.17 , c ) $ 314.18 , d ) $ 314.19 , e ) $ 314.20 | d | add(314.16, divide(const_3, const_100)) | divide(const_3,const_100)|add(n1,#0) | general | D |
if there are 210 workers in a factory , and on a certain day , 198 were present . calculate the percentage that showed up for work ? ( round to the nearest tenth ) . | 198 / 210 * 100 = 94.28 94.3 % correct answer a | a ) 94.3 % , b ) 95.3 % , c ) 93.3 % , d ) 92.3 % , e ) 91.3 % | a | multiply(divide(198, 210), const_100) | divide(n1,n0)|multiply(#0,const_100) | physics | A |
working alone , printers x , y , and z can do a certain printing job , consisting of a large number of pages , in 12 , 30 , and 40 hours , respectively . what is the ratio of the time it takes printer x to do the job , working alone at its rate , to the time it takes printers y and z to do the job , working together at their individual rates ? | "p 1 takes 12 hrs rate for p 2 p 3 together = 1 / 30 + 1 / 40 = 7 / 120 therefore they take 120 / 7 ratio = 120 / 7 = d" | a ) 4 / 11 , b ) 1 / 2 , c ) 15 / 22 , d ) 120 / 7 , e ) 11 / 4 | d | divide(12, divide(const_1, add(divide(const_1, 30), divide(const_1, 40)))) | divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|divide(const_1,#2)|divide(n0,#3)| | general | D |
in a certain store , the profit is 320 % of the cost . if the cost increases by 40 % but the selling price remains constant , approximately what percentage of the selling price is the profit | explanation : let c . p . = rs . 100 . then , profit = rs . 320 , s . p . = rs . 420 . new c . p . = 140 % of rs . 100 = rs . 140 new s . p . = rs . 420 . profit = rs . ( 420 - 140 ) = rs . 280 required percentage = ( 280 / 420 ) * 100 = 66.67 % ( approx ) = 67 % answer : a | a ) 67 % , b ) 80 % , c ) 90 % , d ) 100 % , e ) none of above | a | multiply(divide(subtract(add(const_100, 320), add(const_100, 40)), add(const_100, 320)), const_100) | add(n0,const_100)|add(n1,const_100)|subtract(#0,#1)|divide(#2,#0)|multiply(#3,const_100) | gain | A |
in what time will a train 400 meters long cross an electric pole , if its speed is 144 km / hr | "explanation : first convert speed into m / sec speed = 144 * ( 5 / 18 ) = 40 m / sec time = distance / speed = 400 / 40 = 10 seconds answer : c" | a ) 5 seconds , b ) 4.5 seconds , c ) 10 seconds , d ) 2.5 seconds , e ) none of these | c | divide(400, multiply(144, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics | C |
the average ( arithmetic mean ) of 22 , 33 , and 53 is 8 less than the average of 30 , 56 , and x . what is x ? | "the average of 22 , 33 , and 53 is 108 / 3 = 36 . the average of 30 , 56 and x is 44 . then 30 + 56 + x = 132 . x = 46 . the answer is c ." | a ) 42 , b ) 44 , c ) 46 , d ) 48 , e ) 50 | c | subtract(add(add(22, 33), 53), add(add(multiply(8, const_3), 30), 56)) | add(n0,n1)|multiply(n3,const_3)|add(n2,#0)|add(n4,#1)|add(n5,#3)|subtract(#2,#4)| | general | C |
find the value of x : x Β² + 6 x + 9 . | that quadratic is factored as follows : x Β² + 6 x + 9 = ( x + 3 ) ( x + 3 ) . lesson 17 . now , it is easy to see that that both factors will be 0 when x = - 3 . the solution is : x = - 3 c | a ) 9 , b ) - 1 , c ) - 3 , d ) 3 , e ) - 9 | c | multiply(6, const_3) | multiply(n0,const_3) | general | C |
if jake loses 8 pounds , he will weigh twice as much as his sister kendra . together they now weigh 293 pounds . what is jake β s present weight , in pounds ? | "j + k = 293 and so k = 293 - j j - 8 = 2 k j - 8 = 2 ( 293 - j ) 3 j = 594 j = 198 the answer is e ." | a ) 182 , b ) 186 , c ) 190 , d ) 194 , e ) 198 | e | add(multiply(divide(subtract(293, 8), const_3), const_2), 8) | subtract(n1,n0)|divide(#0,const_3)|multiply(#1,const_2)|add(n0,#2)| | general | E |
fresh grapes contain 80 % by weight while dried grapes contain 20 % water by weight . what is the weight of dry grapes available from 40 kg of fresh grapes ? | "from the question we know : 40 kg * 80 % = 32 kg of water in the fresh grapes 40 kg - 32 kg of water = 8 kg of non - water mass we are looking for the weight of the dry grapes ( x ) . since the question tells us that 20 % of the weight of the dry graps is water and we know that 8 kg is non - water mass we can set up the following equation : x = 1 / 5 ( x ) + 8 kg 4 / 5 ( x ) = 8 kg x = 10 kg answer - a" | a ) 10 kg , b ) 12 kg , c ) 15 kg , d ) 16 kg , e ) 20 kg | a | multiply(divide(divide(multiply(subtract(const_100, 80), 40), const_100), subtract(const_100, 20)), const_100) | subtract(const_100,n0)|subtract(const_100,n1)|multiply(n2,#0)|divide(#2,const_100)|divide(#3,#1)|multiply(#4,const_100)| | gain | A |
a man buys an article for $ 20 . and sells it for $ 35 . find the gain percent ? | "c . p . = $ 20 s . p . = $ 35 gain = $ 15 gain % = 15 / 20 * 100 = 75 % answer is d" | a ) 25 % , b ) 50 % , c ) 20 % , d ) 75 % , e ) 30 % | d | subtract(divide(35, divide(20, const_100)), const_100) | divide(n0,const_100)|divide(n1,#0)|subtract(#1,const_100)| | gain | D |
the average age of 15 men is increased by years when two of them whose ages are 21 years and 23 years are replaced by two new men . the average age of the two new men is | "total age increased = ( 15 * 2 ) years = 30 years . sum of ages of two new men = ( 21 + 23 + 30 ) years = 74 years average age of two new men = ( 74 / 2 ) years = 37 years . answer : b" | a ) 22 , b ) 37 , c ) 99 , d ) 38 , e ) 27 | b | add(divide(add(21, 23), const_2), multiply(const_1, 15)) | add(n1,n2)|multiply(n0,const_1)|divide(#0,const_2)|add(#2,#1)| | general | B |
a can complete a project in 20 days and b can complete the same project in 30 days . if a and b start working on the project together and b quits 5 days before the project is completed , in how many days total will the project be completed ? | "a ' s rate is 1 / 20 of the project per day . b ' s rate is 1 / 30 of the project per day . the combined rate is 1 / 12 of the project per day . in the last 5 days , a can do 1 / 4 of the project . thus a and b must complete 3 / 4 of the project , which takes 9 days . the total number of days is 9 + 5 = 14 . the answer is c ." | a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | c | add(divide(subtract(const_1, multiply(divide(const_1, 30), 5)), add(divide(const_1, 20), divide(const_1, 30))), 5) | divide(const_1,n1)|divide(const_1,n0)|add(#1,#0)|multiply(n2,#0)|subtract(const_1,#3)|divide(#4,#2)|add(n2,#5)| | physics | C |
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 | C |
two trains , one from howrah to patna and the other from patna to howrah , start simultaneously . after they meet , the trains reach their destinations after 16 hours and 9 hours respectively . the ratio of their speeds is ? | "let us name the trains a and b . then , ( a ' s speed ) : ( b ' s speed ) = β b : β a = β 9 : β 16 = 3 : 4 answer : e" | a ) 4 : 5 , b ) 4 : 3 , c ) 4 : 4 , d ) 4 : 9 , e ) 3 : 4 | e | divide(sqrt(9), sqrt(16)) | sqrt(n1)|sqrt(n0)|divide(#0,#1)| | physics | E |
a walks at 10 kmph and 6 hours after his start , b cycles after him at 20 kmph . how far from the start does b catch up with a ? | suppose after x km from the start b catches up with a . then , the difference in the time taken by a to cover x km and that taken by b to cover x km is 6 hours . x / 10 - x / 20 = 6 x = 120 km answer is d | a ) 100 km , b ) 150 km , c ) 50 km , d ) 120 km , e ) 200 km | d | multiply(6, 20) | multiply(n1,n2) | physics | D |
a certain rectangular window is two times as long as it is wide . if its perimeter is 30 feet , what are its dimensions in terms of length by width ? | "2 x + 2 y = 30 x + y = 15 x + 2 x = 15 3 x = 15 x = 5 answer d" | a ) 12 by 2 , b ) 11 by 3 , c ) 10.5 by 3.5 , d ) 10 by 5 , e ) 9 by 3 | d | rectangle_area(30, const_4) | rectangle_area(n0,const_4)| | geometry | D |
from january 1 , 2015 , to january 1 , 2017 , the number of people enrolled in health maintenance organizations increased by 12 percent . the enrollment on january 1 , 2017 , was 45 million . how many million people , to the nearest million , were enrolled in health maintenance organizations on january 1 , 2015 ? | "soln : - 12 x = 45 - - > 28 / 25 * x = 45 - - > x = 45 * 25 / 28 = 1125 / 28 = ~ 40 answer : c ." | a ) 38 , b ) 39 , c ) 40 , d ) 41 , e ) 42 | c | multiply(divide(const_100, add(const_100, 12)), 45) | add(n4,const_100)|divide(const_100,#0)|multiply(n7,#1)| | gain | C |
the average marks of 10 students in a class is 100 . but a student mark is wrongly noted as 50 instead of 10 then find the correct average marks ? | "correct avg marks = 100 + ( 10 - 50 ) / 10 avg = 100 - 4 = 96 answer is c" | a ) a ) 78 , b ) b ) 82 , c ) c ) 96 , d ) d ) 91 , e ) e ) 85 | c | divide(add(subtract(multiply(100, 10), 50), 10), 10) | multiply(n0,n1)|subtract(#0,n2)|add(n3,#1)|divide(#2,n0)| | general | C |
if the area of a square with sides of length 5 centimeters is equal to the area of a rectangle with a width of 4 centimeters , what is the length of the rectangle , in centimeters ? | "let length of rectangle = l 5 ^ 2 = l * 4 = > l = 25 / 4 = 7 answer d" | a ) 4 , b ) 8 , c ) 12 , d ) 7 , e ) 18 | d | divide(power(5, const_2), 4) | power(n0,const_2)|divide(#0,n1)| | geometry | D |
the diameter of a garden roller is 1.4 m and it is 2 m long . how much area will it cover in 6 revolutions ? ( use Γ― β¬ = 22 Γ’ Β β 7 ) | "required area covered in 5 revolutions = 6 Γ£ β 2 Γ― β¬ rh = 6 Γ£ β 2 Γ£ β 22 Γ’ Β β 7 Γ£ β 0.7 Γ£ β 2 = 52.8 m 2 answer c" | a ) 40 m 2 , b ) 44 m 2 , c ) 52.8 m 2 , d ) 36 m 2 , e ) none of these | c | multiply(6, multiply(multiply(multiply(2, divide(22, 7)), divide(1.4, 2)), 2)) | divide(n0,n1)|divide(n3,n4)|multiply(#1,n1)|multiply(#0,#2)|multiply(n1,#3)|multiply(n2,#4)| | physics | C |
a certain characteristic in a large population has a distribution that is symmetric about the mean m . if 68 % of the distribution lies one standard deviation g of the mean , what percent of the distribution is less than m + g ? | "16 % ________________________________________________ m + g 34 % ________________________________________________ m 34 % ________________________________________________ m - g 16 % since 68 % lies one standard deviation from mean m , = > 50 % of 68 % lies on either side as it is symmetric about m . thus 16 % lie below m - g and 16 % lie above m + g now below m + g = 16 + 34 + 34 = 84 % hence d" | a ) 16 % , b ) 32 % , c ) 48 % , d ) 84 % , e ) 92 % | d | subtract(const_100, divide(subtract(const_100, 68), const_2)) | subtract(const_100,n0)|divide(#0,const_2)|subtract(const_100,#1)| | general | D |
ratio of ages of 3 persons is 4 : 7 : 9 , 8 years ago , the sum of their ages was 56 . find their present ages | explanation : let the present ages are 4 x , 7 x , 9 x . = > ( 4 x - 8 ) + ( 7 x - 8 ) + ( 9 x - 8 ) = 56 = > 20 x = 80 = > x = 4 so their present ages are : 16 , 28,36 answer : option d | a ) 16 , 35,36 , b ) 12 , 28,36 , c ) 16 , 28,27 , d ) 16 , 28,36 , e ) none of these | d | multiply(divide(add(add(add(56, 8), 8), 8), add(add(4, 7), 9)), 3) | add(n4,n5)|add(n1,n2)|add(n4,#0)|add(n3,#1)|add(n4,#2)|divide(#4,#3)|multiply(n0,#5) | general | D |
when the price of an article was reduced by 35 % its sale increased by 80 % . what was the net effect on the sale ? | "if n items are sold for $ p each , revenue is $ np . if we reduce the price by 35 % , the new price is 0.65 p . if we increase the number sold by 80 % , the new number sold is 1.8 n . so the new revenue is ( 0.65 p ) ( 1.8 n ) = 1.17 np , which is 1.17 times the old revenue , so is 17 % greater . answer : a" | a ) 17 % increase , b ) 44 % decrease , c ) 60 % increase , d ) 66 % increase , e ) 66 % decrease | a | subtract(divide(multiply(add(80, const_100), subtract(const_100, 35)), const_100), const_100) | add(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|subtract(#3,const_100)| | gain | A |
if * stands for / , / stands for - , + stands for * and - stands for + , then 9 / 8 * 7 + 5 - 10 = ? | 9 / 8 * 7 + 5 - 10 this expression becomes 9 - 8 / 7 * 5 + 10 after exchanging the symbols then the values for it would be 9 - 1.142 * 5 + 10 = 13.290 ( app ) = 13.3 answer : a | a ) 13.3 , b ) 10.8 , c ) 10.7 , d ) 11.4 , e ) 12.5 | a | add(10, subtract(9, multiply(divide(8, 7), 5))) | divide(n1,n2)|multiply(n3,#0)|subtract(n0,#1)|add(n4,#2) | general | A |
john has exactly 31 pens , and each pen is either bule , black , or red . if he has 5 more black pens than red pens , and twice as many blue pens as black pens , how many blue pens does john has ? | x = the number of blue pens y = the number of black pens z = the number of red pens from the first sentence we have equation # 1 : x + y + z = 31 . . . he has 5 more black pens than red pens . . . equation # 2 : y = 5 + z . . . twice as many blue pens as black pens . . . equation # 3 : x = 2 y solve equation # 1 for z : z = y - 5 now , we can replace both x and z with y in equation # 1 2 y + y + ( y - 5 ) = 31 4 y - 5 = 31 4 y = 36 y = 9 there are 9 black pens . this is 5 more than the number of red pens , so z = 4 . hence the answer is . x = 18 , and just for check 18 + 9 + 4 = 31 . answer = 18 , ( b ) | a ) 15 , b ) 18 , c ) 20 , d ) 22 , e ) 17 | b | multiply(divide(add(31, 5), const_4), const_2) | add(n0,n1)|divide(#0,const_4)|multiply(#1,const_2) | general | B |
the perimeter of a semi circle is 162 cm then the radius is ? | "36 / 7 r = 162 = > r = 31.5 answer : a" | a ) 31.5 , b ) 28 , c ) 98 , d ) 37 , e ) 13 | a | divide(162, add(const_2, const_pi)) | add(const_2,const_pi)|divide(n0,#0)| | physics | A |
in a corporation , 50 percent of the male employees and 40 percent of the female employees are at least 35 years old . if 42 percent of all the employees are at least 35 years old , what fraction of the employees in the corporation are females ? | you can use the weighted averages formula for a 10 sec solution . no of females / no of males = ( 50 - 42 ) / ( 42 - 40 ) = 4 / 1 no of females as a fraction of total employees = 4 / ( 4 + 1 ) = 4 / 5 ; answer : d | a ) 3 / 5 , b ) 2 / 3 , c ) 3 / 4 , d ) 4 / 5 , e ) 5 / 6 | d | divide(subtract(50, 42), subtract(50, 40)) | subtract(n0,n3)|subtract(n0,n1)|divide(#0,#1) | other | D |
find the sum lend at c . i . at 5 p . c per annum will amount to rs . 650 in 2 years ? | "explanation : 650 = p ( 21 / 20 ) 2 p = 589.56 answer : d" | a ) 221 , b ) 287 , c ) 400 , d ) 589 , e ) 171 | d | divide(650, power(add(const_1, divide(5, const_100)), 2)) | divide(n0,const_100)|add(#0,const_1)|power(#1,n2)|divide(n1,#2)| | general | D |
the mall charges 60 cents for the first hour of parking and $ 4 for each additional hour until the customer reaches 4 hours , after that the parking fee is $ 2 per hour . if edward parked his car in the mall for 7 hours and 30 minutes , how much is he going to pay ? | "0.60 + 4 * 4 + 2 * 4 + 0.60 = $ 25.2 answer : $ 25.2" | a ) $ 25.2 , b ) $ 22 . , c ) $ 13 . , d ) $ 14.5 . , e ) $ 15.5 | a | add(add(multiply(4, 4), multiply(add(subtract(7, 4), divide(60, const_100)), 2)), divide(60, const_100)) | divide(n0,const_100)|multiply(n1,n1)|subtract(n4,n2)|add(#0,#2)|multiply(n3,#3)|add(#1,#4)|add(#5,#0)| | physics | A |
what percent of 13 is 13 percent of 1 ? | "13 % of 1 = ( 13 / 100 ) * 1 = 13 / 100 to determine what percentage of 13 this is : [ 13 ] [ / 100 * 13 ] * 100 = 1 % ans : c" | a ) 0.001 , b ) 0.01 , c ) 1 , d ) 10 , e ) 101 . | c | p_after_gain(13, 13) | p_after_gain(n1,n0)| | gain | C |
the malibu country club needs to drain its pool for refinishing . the hose they use to drain it can remove 60 cubic feet of water per minute . if the pool is 50 feet wide by 150 feet long by 10 feet deep and is currently at 80 % capacity , how long will it take to drain the pool ? | volume of pool = 50 * 150 * 10 cu . ft , 80 % full = 50 * 150 * 10 * 0.8 cu . ft water is available to drain . draining capacity = 60 cu . ft / min therefore time taken = 50 * 150 * 10 * 0.8 / 60 min = 1000 min a | a ) 1000 min , b ) 1200 min , c ) 1300 min , d ) 1400 min , e ) 1600 min | a | divide(multiply(divide(80, const_100), multiply(multiply(50, 150), 10)), 60) | divide(n4,const_100)|multiply(n1,n2)|multiply(n3,#1)|multiply(#0,#2)|divide(#3,n0) | gain | A |
a car traveled 462 miles per tankful of gasoline on the highway and 336 miles per tankful of gasoline in the city . if the car traveled 12 fewer miles per gallon in the city than on the highway , how many miles per gallon did the car travel in the city ? | "i treat such problems as work ones . work = rate * time mileage ( m ) = rate ( mpg ) * gallons ( g ) x gallons is a full tank { 462 = rx { 336 = ( r - 12 ) x solve for r , r = 44 44 - 12 = 32 mpg e" | a ) 14 , b ) 16 , c ) 21 , d ) 22 , e ) 32 | e | divide(336, divide(subtract(462, 336), 12)) | subtract(n0,n1)|divide(#0,n2)|divide(n1,#1)| | physics | E |
a train speeds past a pole in 15 sec and a platform 100 m long in 25 sec , its length is ? | "let the length of the train be x m and its speed be y m / sec . then , x / y = 15 = > y = x / 15 ( x + 100 ) / 25 = x / 15 = > x = 150 m . answer : b" | a ) 158 m , b ) 150 m , c ) 110 m , d ) 130 m , e ) 157 m | b | multiply(100, subtract(const_2, const_1)) | subtract(const_2,const_1)|multiply(n1,#0)| | physics | B |
a salesman β s terms were changed from a flat commission of 5 % on all his sales to a fixed salary of rs . 1300 plus 2.5 % commission on all sales exceeding rs . 4,000 . if his remuneration as per new scheme was rs . 600 more than that by the previous schema , his sales were worth ? | [ 1300 + ( x - 4000 ) * ( 2.5 / 100 ) ] - x * ( 5 / 100 ) = 600 x = 18000 answer : c | a ) 12028 , b ) 12000 , c ) 12019 , d ) 12197 , e ) 18000 | c | divide(600, divide(5, const_100)) | divide(n0,const_100)|divide(n4,#0) | general | C |
there is 60 % increase in an amount in 6 years at si . what will be the ci of rs . 11,000 after 3 years at the same rate ? | "let p = rs . 100 . then , s . i . rs . 60 and t = 6 years . r = 100 x 60 = 10 % p . a . 100 x 6 now , p = rs . 11000 . t = 3 years and r = 10 % p . a . c . i . = rs . 11000 x 1 + 10 3 - 1 100 = rs . 11000 x 331 1000 = 3641 . a" | a ) 3641 , b ) 2572 , c ) 4542 , d ) 2343 , e ) 3972 | a | subtract(multiply(add(multiply(const_100, const_100), multiply(multiply(const_100, divide(60, 6)), 3)), multiply(multiply(add(const_1, divide(divide(60, 6), const_100)), add(const_1, divide(divide(60, 6), const_100))), add(const_1, divide(divide(60, 6), const_100)))), add(multiply(const_100, const_100), multiply(multiply(const_100, divide(60, 6)), 3))) | divide(n0,n1)|multiply(const_100,const_100)|divide(#0,const_100)|multiply(#0,const_100)|add(#2,const_1)|multiply(#3,n3)|add(#1,#5)|multiply(#4,#4)|multiply(#4,#7)|multiply(#6,#8)|subtract(#9,#6)| | gain | A |
( 0.0088 ) ( 4.5 ) / ( 0.05 ) ( 0.1 ) ( 0.008 ) = | "( 0.0088 ) ( 4.5 ) / ( 0.05 ) ( 0.1 ) ( 0.008 ) = 0.0088 * 450 / 5 * ( 0.1 ) ( 0.008 ) = 0.088 * 90 / 1 * 0.008 = 88 * 90 / 8 = 11 * 90 = 990 answer : e" | a ) 950 , b ) 940 , c ) 980 , d ) 960 , e ) 990 | e | divide(multiply(0.0088, 4.5), multiply(multiply(0.05, 0.1), 0.008)) | multiply(n0,n1)|multiply(n2,n3)|multiply(n4,#1)|divide(#0,#2)| | general | E |
a train passes a station platform in 50 sec and a man standing on the platform in 15 sec . if the speed of the train is 54 km / hr . what is the length of the platform ? | "speed = 54 * 5 / 18 = 15 m / sec . length of the train = 15 * 15 = 225 m . let the length of the platform be x m . then , ( x + 225 ) / 50 = 15 = > x = 525 m . answer : d" | a ) 227 , b ) 500 , c ) 550 , d ) 525 , e ) 171 | d | multiply(15, multiply(54, const_0_2778)) | multiply(n2,const_0_2778)|multiply(n1,#0)| | physics | D |
in a recent election , james received 1.5 percent of the 2,000 votes cast . to win the election , a candidate needed to receive more than 50 percent of the vote . how many additional votes would james have needed to win the election ? | "james = ( 1.5 / 100 ) * 2000 = 30 votes to win = ( 50 / 100 ) * total votes + 1 = ( 50 / 100 ) * 2000 + 1 = 1001 remaining voted needed to win election = 1001 - 30 = 971 answer : option d" | a ) 901 , b ) 989 , c ) 990 , d ) 971 , e ) 1,001 | d | subtract(add(const_1000, const_1000), multiply(add(const_1000, const_1000), const_0.5)) | add(const_1000,const_1000)|multiply(const_0.5,#0)|subtract(#0,#1)| | general | D |
a man can row downstream at 22 kmph and upstream at 10 kmph . find the speed of the man in still water and the speed of stream respectively ? | "let the speed of the man in still water and speed of stream be x kmph and y kmph respectively . given x + y = 22 - - - ( 1 ) and x - y = 10 - - - ( 2 ) from ( 1 ) & ( 2 ) 2 x = 32 = > x = 16 , y = 6 . answer : d" | a ) 16 , 2 , b ) 16 , 4 , c ) 16 , 8 , d ) 16 , 6 , e ) 16 , 7 | d | divide(divide(add(22, 10), const_2), const_2) | add(n0,n1)|divide(#0,const_2)|divide(#1,const_2)| | physics | D |
( 35423 + 7164 + 41720 ) - ( 317 x 89 ) = ? | "= ( 84307 ) - 317 x 89 = 84307 - ( 317 x ( 90 - 1 ) ) = 84307 - ( 317 x 90 - 317 ) = 84307 - 28213 = 54094 answer is d" | a ) 28213 , b ) 84307 , c ) 50694 , d ) 54094 , e ) none of them | d | divide(add(35423, 7164), 41720) | add(n0,n1)|divide(#0,n2)| | general | D |
there are 8 stores in town that had a total of 23 visitors on a particular day . however , only 12 people went shopping that day ; some people visited more than one store . if 8 people visited exactly two stores each , and everyone visited at least one store , what is the largest number of stores anyone could have visited ? | "8 people visited 2 stores each for 16 visits . to maximize the number of stores that one person visited , let ' s assume that 3 people visited 1 store each . the number of remaining visits is 23 - 16 - 3 = 4 , which is the maximum that one person could have visited . the answer is c ." | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | c | subtract(subtract(23, multiply(8, const_2)), subtract(subtract(12, 8), const_1)) | multiply(n0,const_2)|subtract(n2,n0)|subtract(n1,#0)|subtract(#1,const_1)|subtract(#2,#3)| | general | C |
at an election 2 candidates are participated a candidate got 30 % of votes and defeated by 5000 votes . 100 votes are invalid . find the total polled votes ? | winners votes = 100 - 30 = 70 % invalid votes = 100 polled votes = [ ( 100 * 5000 ) / ( 2 * 70 - 100 ) ] + 100 = 12600 answer is b | a ) 11520 , b ) 12600 , c ) 15000 , d ) 13560 , e ) 12560 | b | add(const_100, multiply(divide(5000, subtract(subtract(100, 30), 30)), const_100)) | subtract(n3,n1)|subtract(#0,n1)|divide(n2,#1)|multiply(#2,const_100)|add(#3,const_100) | gain | B |
a car going at 40 miles per hour set out on an 90 - mile trip at 9 : 00 a . m . exactly 10 minutes later , a second car left from the same place and followed the same route . how fast , in miles per hour , was the second car going if it caught up with the first car at 10 : 30 a . m . ? | "let car a = car that starts at 9 am car b = car that starts at 9 : 10 am time for which car a travels at speed of 40 m per hour = 1.5 hours distance travelled by car a = 40 * 1.5 = 60 miles since car b catches up car a at 10 : 30 , time = 90 mins = 3 / 2 hour speed of car b = 60 / ( 3 / 2 ) = 90 miles per hour answer e" | a ) 45 , b ) 50 , c ) 53 , d ) 55 , e ) 90 | e | divide(90, divide(add(multiply(subtract(10, 9), const_60), subtract(40, 10)), const_60)) | subtract(n4,n2)|subtract(n0,n4)|multiply(#0,const_60)|add(#2,#1)|divide(#3,const_60)|divide(n1,#4)| | physics | E |
a certain music store stocks 800 cellos and 600 violas . of these instruments , there are 120 cello - viola pairs , such that a cello and a viola were both made with wood from the same tree ( each tree can make at most one viola and one cello , so there are no pairs other than these 90 ) . if one viola and one cello are chosen at random , what is the probability that the two instruments are made with wood from the same tree ? | "solution provided by stanford 2012 is correct : 120 / 800 choosing one of the cellos which has a pair viola , 1 / 600 choosing the viola which is the pair of chosen cello - - > p = 120 / 800 * 1 / 600 = 1 / 8,000 . answer : b ." | a ) 3 / 16,000 , b ) 1 / 8,000 , c ) 3 / 1,600 , d ) 1 / 90 , e ) 2 / 45 | b | multiply(divide(120, 800), divide(const_1, 600)) | divide(n2,n0)|divide(const_1,n1)|multiply(#0,#1)| | other | B |
if 20 typists can type 48 letters in 20 minutes , then how many letters will 30 typists working at the same rate complete in 1 hour ? | "20 typists can type 48 letters , so 30 typists can type = 48 * 30 / 20 48 * 30 / 20 letters can be typed in 20 mins . in 60 mins typist can type = 48 * 30 * 60 / 20 * 20 = 216 d is the answer" | a ) 63 , b ) 72 , c ) 144 , d ) 216 , e ) 400 | d | multiply(divide(multiply(48, const_3), 20), 30) | multiply(n1,const_3)|divide(#0,n0)|multiply(n3,#1)| | physics | D |
a sum of money is sufficient to pay a ' s wages for 21 days and b ' s wages for 28 days . the same money is sufficient to pay the wages of both for ? | let total money be rs . x a ' s 1 day ' s wages = rs . x / 21 , b ' s 1 day ' s wages = rs . x / 28 ( a + b ) ' s 1 day ' s wages = rs . ( x / 21 + x / 28 ) = rs . x / 12 ; money is sufficient to pay the wages of both for 12 days . correct option : a | a ) 12 days , b ) 12 1 / 4 days , c ) 14 days , d ) 24 1 / 2 days , e ) none of these | a | divide(const_1, add(divide(const_1, 21), divide(const_1, 28))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2) | general | A |
ifaequals the sum of the even integers from 2 to 20 , inclusive , andbequals the sum of the odd integers from 1 to 19 , inclusive , what is the value of ( a + b ) / 2 ? | "yes ! there is really a faster way to solve it . sum of consecutive odd or even integers = ( no . of odd or even ints ) * ( first int + last int ) / 2 here a = sum of even ints from 2 to 20 , inclusive number of even ints = 10 , first int + last int = 2 + 20 = 22 a = 10 * 22 / 2 = 110 b = sum of odd ints from 1 to 19 , inclusive number of odd ints = 10 , first int + last int = 1 + 19 = 20 a = 10 * 20 / 2 = 100 ( a + b ) / 2 = 110 + 100 = 210 / 2 = 105 ans : c" | a ) 100 , b ) 110 , c ) 105 , d ) 20 , e ) 215 | c | subtract(multiply(divide(20, 2), add(divide(20, 2), 1)), multiply(divide(add(19, 1), 2), add(divide(subtract(19, 1), 2), 1))) | add(n2,n3)|divide(n1,n0)|subtract(n3,n2)|add(n2,#1)|divide(#2,n0)|divide(#0,n0)|add(n2,#4)|multiply(#3,#1)|multiply(#6,#5)|subtract(#7,#8)| | general | C |
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 $ 8000 in a long term bond that pays 4 percent interest , compounded annually , what will be the approximate total amount of investment 36 years later , when pat is ready for college ? | "since investment doubles in 70 / r years then for r = 4 it ' ll double in 70 / 4 = ~ 18 years ( we are not asked about the exact amount so such an approximation will do ) . thus in 36 years investment will double twice and become ( $ 8,000 * 2 ) * 2 = $ 32,000 ( after 18 years investment will become $ 8,000 * 2 = $ 16,000 and in another 18 years it ' ll become $ 16,000 * 2 = $ 32,000 ) . answer : c ." | a ) $ 20000 , b ) $ 15000 , c ) $ 32000 , d ) $ 10000 , e ) $ 9000 | c | multiply(multiply(8000, const_2), const_2) | multiply(n1,const_2)|multiply(#0,const_2)| | general | C |
wink , inc . follows a certain procedure that requires two tasks to be finished independently in order for a job to be done . on any given day , there is a 2 / 3 probability that task 1 will be completed on time , and a 3 / 5 probability that task 2 will be completed on time . on a certain day , what is the probability that task 1 will be completed on time , but task 2 will not ? | "p ( 1 and not 2 ) = 2 / 3 * ( 1 - 3 / 5 ) = 4 / 15 . answer : a ." | a ) 4 / 15 , b ) 3 / 40 , c ) 13 / 40 , d ) 7 / 20 , e ) 13 / 22 | a | multiply(divide(2, 3), subtract(const_1, divide(3, 2))) | divide(n0,n1)|divide(n3,n4)|subtract(n2,#1)|multiply(#0,#2)| | probability | A |
if the numbers 1 to 99 are written on 99 pieces of paper , ( one on each ) and one piece is picked at random , then what is the probability that the number drawn is neither prime nor composite ? | "there are 25 primes , 73 composite numbers from 1 to 100 . the number which is neither prime nor composite is 1 . therefore , required probability = 1 / 99 . answer : c" | a ) 1 / 50 , b ) 1 / 25 , c ) 1 / 99 , d ) 1 , e ) 2 | c | divide(1, 99) | divide(n0,n1)| | other | C |
one side of a rectangular field is 15 m and one of its diagonal is 17 m . find the area of field ? | "other side = [ ( 17 x 17 ) - ( 15 x 15 ) ] = ( 289 - 225 ) = 8 m area = 15 x 8 = 120 sq . m answer : b" | a ) 110 , b ) 120 , c ) 130 , d ) 140 , e ) 150 | b | rectangle_area(15, sqrt(subtract(power(17, const_2), power(15, const_2)))) | power(n1,const_2)|power(n0,const_2)|subtract(#0,#1)|sqrt(#2)|rectangle_area(n0,#3)| | geometry | B |
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