Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
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two trains 100 m and 60 m long run at the speed of 60 km / hr and 40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? | "relative speed = 60 + 40 = 100 km / hr . = 100 * 5 / 18 = 250 / 9 m / sec . distance covered in crossing each other = 100 + 60 = 160 m . required time = 160 * 9 / 250 = 5.76 = 5.76 sec . answer : b" | a ) 10.9 sec , b ) 5.76 sec , c ) 10.6 sec , d ) 10.8 sec , e ) 20.8 sec | b | divide(add(100, 60), multiply(add(60, 40), const_0_2778)) | add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)| | physics |
the price of lunch for 15 people was $ 207.00 , including a 15 percent gratuity for service . what was the average price per person , excluding the gratuity ? | "if the amount spent excluding the gratuity was x , then x ( 1.15 ) = 207 = > x = 207 / 1.15 = 180 then the average price per person = 180 / 15 = 12 option ( b )" | a ) $ 11.73 , b ) $ 12.00 , c ) $ 13.80 , d ) $ 14.00 , e ) $ 15.87 | b | multiply(multiply(divide(207.00, add(const_100, 15)), const_100), divide(const_1, 15)) | add(n0,const_100)|divide(const_1,n0)|divide(n1,#0)|multiply(#2,const_100)|multiply(#1,#3)| | general |
find the missing figures : 0.50 % of ? = 0.12 | "let 0.50 % of x = 0.12 . then , 0.50 * x / 100 = 0.12 x = [ ( 0.12 * 100 ) / 0.50 ] = 24 . answer is e ." | a ) 16 , b ) 18 , c ) 20 , d ) 22 , e ) 24 | e | divide(0.12, divide(0.50, const_100)) | divide(n0,const_100)|divide(n1,#0)| | gain |
a no . when divided by 220 gives a remainder 43 , what remainder will beobtained by dividingthe same no . 17 ? | "220 + 43 = 263 / 17 = 8 ( remainder ) b" | a ) 2 , b ) 8 , c ) 9 , d ) 11 , e ) 15 | b | divide(add(220, 43), 17) | add(n0,n1)|divide(#0,n2)| | general |
shipment - - - - - - no . of defective chips / shipment - - - total chips in shipment s 2 - - - - - - - - - - - - - - 5 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 12000 s 3 - - - - - - - - - - - - - - 6 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ... | i agree with your solution = 20 . but the question is : there are different combination to get 60000 chips . for example : 1 * s 3 + 2 * s 4 + 2 * s 2 . in this way , we ship 60000 chips with only 6 + 4 * 2 + 2 * 2 = 18 defective chips , better than the average of 20 . the question is to find the expected number q of d... | a ) 14 , b ) 20 , c ) 22 , d ) 24 , e ) 25 | b | divide(subtract(16000, subtract(subtract(subtract(60000, 18000), 16000), 12000)), const_100) | subtract(n13,n5)|subtract(#0,n8)|subtract(#1,n2)|subtract(n8,#2)|divide(#3,const_100) | general |
if 2 x + y = 7 and x + 2 y = 5 , then 8 xy / 3 = ? | "2 * ( x + 2 y = 5 ) equals 2 x + 4 y = 10 2 x + 4 y = 10 - 2 x + y = 7 = 3 y = 3 therefore y = 1 plug and solve . . . 2 x + 1 = 7 2 x = 6 x = 3 ( 8 * 3 * 1 ) / 3 = 24 / 3 = 8 b" | a ) a ) 1 , b ) b ) 8 , c ) c ) 17 / 5 , d ) d ) 18 / 5 , e ) e ) 4 | b | divide(add(divide(subtract(multiply(7, 2), 5), subtract(multiply(2, 2), const_1)), subtract(7, multiply(2, divide(subtract(multiply(7, 2), 5), subtract(multiply(2, 2), const_1))))), 8) | multiply(n0,n1)|multiply(n0,n0)|subtract(#0,n3)|subtract(#1,const_1)|divide(#2,#3)|multiply(n0,#4)|subtract(n1,#5)|add(#4,#6)|divide(#7,n4)| | general |
the maximum number of students among them 640 pens and 520 pencils can be distributed in such a way that each student get the same number of pens and same number of pencils ? | "number of pens = 640 number of pencils = 520 required number of students = h . c . f . of 640 and 520 = 40 answer is a" | a ) 40 , b ) 45 , c ) 50 , d ) 55 , e ) 60 | a | gcd(640, 520) | gcd(n0,n1)| | general |
a certain airline ' s fleet consisted of 80 type a planes at the beginning of 1980 . at the end of each year , starting with 1980 , the airline retired 3 of the type a planes and acquired 4 new type b plans . how many years did it take before the number of type a planes left in the airline ' s fleet was less than 50 pe... | "let x be the number of years . 4 x > 80 - 3 x 7 x > 80 x > 11 + 3 / 7 the answer is d ." | a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 13 | d | divide(80, add(3, 4)) | add(n3,n4)|divide(n0,#0)| | gain |
if the numbers 1 to 96 are written on 96 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 , 70 composite numbers from 1 to 96 . the number which is neither prime nor composite is 1 . therefore , required probability = 1 / 96 . answer : c | a ) 1 / 50 , b ) 1 / 25 , c ) 1 / 96 , d ) 1 , e ) 2 | c | divide(1, 96) | divide(n0,n1) | other |
sandy walked 25 meters towards south . then sandy turned to her left and walked 20 meters . she then turned to her left and walked 25 meters . she then turned to her right and walked 12 meters . what distance is she from the starting point and in which direction ? | "the net distance is 20 + 12 = 32 meters to the east . the answer is a ." | a ) 32 meters east , b ) 32 meters north , c ) 32 meters west , d ) 42 meters east , e ) 52 meters west | a | add(25, 12) | add(n0,n3)| | physics |
the length of the bridge , which a train 120 meters long and travelling at 45 km / hr can cross in 30 seconds , is : | "speed = ( 45 * 5 / 18 ) m / sec = ( 25 / 2 ) m / sec . time = 30 sec . let the length of bridge be x meters . then , ( 120 + x ) / 30 = 25 / 2 = = > 2 ( 120 + x ) = 750 = = > x = 255 m . answer : c" | a ) 239 , b ) 277 , c ) 255 , d ) 88 , e ) 232 | c | subtract(multiply(divide(multiply(45, speed(const_1000, const_1)), speed(const_3600, const_1)), 30), 120) | speed(const_1000,const_1)|speed(const_3600,const_1)|multiply(n1,#0)|divide(#2,#1)|multiply(n2,#3)|subtract(#4,n0)| | physics |
how many liters of water must be evaporated from 50 liters of a 1 - percent sugar solution to get a 10 - percent solution ? | 1 % of a 50 liter solution is 0.5 l which is 10 % of the solution at the end . the solution at the end must be 5 l . we need to evaporate 45 liters . the answer is d . | a ) 35 , b ) 40 , c ) 42 , d ) 45 , e ) 48 | d | subtract(50, multiply(divide(50, const_100), 10)) | divide(n0,const_100)|multiply(n2,#0)|subtract(n0,#1) | gain |
raman mixed 27 kg of butter at rs . 150 per kg with 36 kg butter at the rate of rs . 125 per kg . at what price per kg should he sell the mixture to make a profit of 40 % in the transaction ? | "explanation : cp per kg of mixture = [ 27 ( 150 ) + 36 ( 125 ) ] / ( 27 + 36 ) = rs . 135.71 sp = cp [ ( 100 + profit % ) / 100 ] = 135.71 * [ ( 100 + 40 ) / 100 ] = rs . 190 . answer : c" | a ) 129 , b ) 287 , c ) 190 , d ) 188 , e ) 112 | c | add(divide(add(multiply(27, 150), multiply(36, 125)), add(36, 27)), multiply(divide(add(multiply(27, 150), multiply(36, 125)), add(36, 27)), divide(40, const_100))) | add(n0,n2)|divide(n4,const_100)|multiply(n0,n1)|multiply(n2,n3)|add(#2,#3)|divide(#4,#0)|multiply(#5,#1)|add(#5,#6)| | gain |
the average of 50 numbers id 56 . if two numbers , namely 45 and 55 are discarded , the average of the remaining numbers is : | explanation : total of 50 numbers = ( 50 × 56 ) = 2800 total of 48 numbers = ( 2800 - ( 45 + 55 ) ] = 2700 required average = 2700 / 48 = 56.25 answer : e | a ) 22.35 , b ) 33.25 , c ) 22.25 , d ) 51.35 , e ) 56.25 | e | divide(subtract(multiply(50, 56), add(45, 55)), subtract(50, const_2)) | add(n2,n3)|multiply(n0,n1)|subtract(n0,const_2)|subtract(#1,#0)|divide(#3,#2) | general |
a set consists of 12 numbers , all are even or multiple of 5 . if 7 numbers are even and 14 numbers are multiple of 5 , how many numbers is multiple of 10 ? | "{ total } = { even } + { multiple of 5 } - { both } + { nether } . since { neither } = 0 ( allare even or multiple of 5 ) then : 12 = 7 + 14 - { both } + 0 ; { both } = 9 ( so 1 number is both even and multiple of 5 , so it must be a multiple of 10 ) . answer : e ." | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 9 | e | subtract(12, 14) | subtract(n0,n3)| | general |
sum of the squares of 3 no . is 149 and the sum of their products taken two at a time is 70 . find the sum ? | "( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 ( ab + bc + ca ) = 149 + 2 * 70 a + b + c = â ˆ š 289 = 17 answer a" | a ) 17 , b ) 19 , c ) 21 , d ) 13 , e ) 18 | a | sqrt(add(149, multiply(70, const_2))) | multiply(n2,const_2)|add(n1,#0)|sqrt(#1)| | general |
if x and y are positive integers and x = 5 y + 2 , what is the remainder when x is divided by 5 ? | "this question asks what is . . . ( the answer ) , so we know that the answer will be consistent . as such , we can test values to quickly get the solution . we ' re told that x and y are positive integers and x = 5 y + 2 . we ' re asked for the remainder when x is divided by 5 . if . . . . y = 1 x = 7 7 / 5 = 1 remain... | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | c | divide(add(2, const_1), 5) | add(n1,const_1)|divide(#0,n0)| | general |
if , 1 * 3 * 5 = 16 3 * 5 * 7 = 38 5 * 7 * 9 = 68 then find , 7 * 9 * 11 = ? | a 106 ( 11 * 9 ) + 7 = 106 | a ) 106 , b ) 49 , c ) 68 , d ) 38 , e ) 55 | a | add(multiply(9, 11), 7) | multiply(n10,n14)|add(n6,#0) | general |
a is twice as good a workman as b and they took 7 days together to do the work b alone can do it in ? | "wc = 2 : 1 2 x + x = 1 / 7 x = 1 / 21 = > 21 days answer : c" | a ) 25 days , b ) 88 days , c ) 21 days , d ) 11 days , e ) 13 days | c | multiply(divide(multiply(7, add(const_2, const_1)), const_2), const_2) | add(const_1,const_2)|multiply(n0,#0)|divide(#1,const_2)|multiply(#2,const_2)| | physics |
points a , b , c , and d , in that order , lie on a line . if ab = 2 cm , ac = 4 cm , and bd = 6 cm , what is cd , in centimeters ? | "putting a value to each point , lets use the following : a - 0 b - 2 ( ab = 2 ) c - 4 ( ac = 4 ) d - 8 ( bd = 6 ) cd is 8 - 4 = 4 . ans d" | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | d | subtract(4, 2) | subtract(n1,n0)| | physics |
a sum of money at simple interest amounts to rs . 815 in 3 years and to rs . 844 in 4 years . the sum is : | "s . i . for 1 year = rs . ( 844 - 815 ) = rs . 29 . s . i . for 3 years = rs . ( 29 x 3 ) = rs . 87 . principal = rs . ( 815 - 87 ) = rs . 728 . answer : option d" | a ) s . 650 , b ) s . 690 , c ) s . 698 , d ) s . 728 , e ) s . 760 | d | subtract(815, divide(multiply(subtract(844, 815), 3), 4)) | subtract(n2,n0)|multiply(n1,#0)|divide(#1,n3)|subtract(n0,#2)| | gain |
a side of beef lost 40 percent of its weight in processing . if the side of beef weighed 240 pounds after processing , how many pounds did it weigh before processing ? | "let weight of side of beef before processing = x ( 60 / 100 ) * x = 240 = > x = ( 240 * 100 ) / 60 = 400 answer b" | a ) 191 , b ) 400 , c ) 737 , d ) 840 , e ) 1,560 | b | divide(multiply(240, const_100), subtract(const_100, 40)) | multiply(n1,const_100)|subtract(const_100,n0)|divide(#0,#1)| | gain |
36 men can complete a piece of work in 18 days . in how days will 27 men can complete the work ? | "let the required number of days be x less men , more days ( indirect proportion ) 27 : 36 : : 18 : x 27 x = 36 * 18 x = ( 36 * 18 ) / 27 x = 24 days option is d" | a ) 18 , b ) 20 , c ) 17 , d ) 24 , e ) 21 | d | divide(multiply(18, 36), 27) | multiply(n0,n1)|divide(#0,n2)| | physics |
a trader bought a car at 50 % discount on its original price . he sold it at a 80 % increase on the price he bought it . what percent of profit did he make on the original price ? | "original price = 100 cp = 95 s = 95 * ( 180 / 100 ) = 112 100 - 171 = 71 % answer : e" | a ) 17 % , b ) 62 % , c ) 12 % , d ) 19 % , e ) 71 % | e | multiply(subtract(divide(divide(multiply(subtract(const_100, 50), add(const_100, 80)), const_100), const_100), const_1), const_100) | add(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|divide(#3,const_100)|subtract(#4,const_1)|multiply(#5,const_100)| | gain |
renu can do a piece of work in 8 days , but with the help of her friend suma , she can do it in 3 days . in what time suma can do it alone ? | "renu â € ™ s one day â € ™ s work = 1 / 8 suma â € ™ s one day â € ™ s work = 1 / 3 - 1 / 8 = 5 / 24 suma can do it alone in 4.8 days . answer : a" | a ) 4.8 , b ) 1.2 , c ) 4.4 , d ) 4.6 , e ) 4.7 | a | inverse(subtract(divide(const_1, 3), divide(const_1, 8))) | divide(const_1,n1)|divide(const_1,n0)|subtract(#0,#1)|inverse(#2)| | physics |
what is the sum of the odd integers from 35 to 65 , inclusive ? | "the mean is 50 . sum = mean ( # of elements ) there are 16 odd numbers between 35 - 65 inclusive . 16 * 50 = 800 a" | a ) 800 , b ) 550 , c ) 555 , d ) 600 , e ) 605 | a | divide(multiply(35, 65), const_4) | multiply(n0,n1)|divide(#0,const_4)| | general |
excluding stoppages , the speed of a bus is 54 kmph and including stoppages , it is 36 kmph . for how many minutes does the bus stop per hour ? | "due to stoppages , it covers 18 km less . time taken to cover 18 km = ( 18 / 54 ) x 60 = 20 min answer : a" | a ) 20 min , b ) 8 min , c ) 9 min , d ) 10 min , e ) 11 min | a | multiply(const_60, divide(subtract(54, 36), 54)) | subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_60)| | physics |
a worker can load one truck in 6 hours . a second worker can load the same truck in 5 hours . if both workers load one truck simultaneously while maintaining their constant rates , approximately how long , in hours , will it take them to fill one truck ? | "the workers fill the truck at a rate of 1 / 6 + 1 / 5 = 11 / 30 of the truck per hour . then the time to fill one truck is 30 / 11 which is about 2.7 hours . the answer is a ." | a ) 2.7 , b ) 2.9 , c ) 3.1 , d ) 3.3 , e ) 3.5 | a | inverse(add(divide(const_1, 6), divide(const_1, 5))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|inverse(#2)| | physics |
which is equal to 11 ? | "11 / 1 is equal to 11 . answer : d" | a ) 1.1 , b ) - 11 , c ) 11 / 11 , d ) 11 / 1 , e ) 1 / 11 | d | divide(divide(divide(divide(divide(divide(11, const_4), const_3), const_4), const_3), const_2), const_2) | divide(n0,const_4)|divide(#0,const_3)|divide(#1,const_4)|divide(#2,const_3)|divide(#3,const_2)|divide(#4,const_2)| | general |
the speed of a train is 102 kmph . what is the distance covered by it in 30 minutes ? | "102 * 30 / 60 = 51 kmph answer : b" | a ) 50 kmph , b ) 51 kmph , c ) 41 kmph , d ) 61 kmph , e ) 21 kmph | b | multiply(divide(30, const_60), 102) | divide(n1,const_60)|multiply(n0,#0)| | physics |
a rectangular of certain dimensions is chopped off from one corner of a larger rectangle as shown . ab = 8 cm and bc = 4 cm . the perimeter of the figure abcpqra ( in cm ) is : | solution required perimeter = ( ab + bc + cp + pq + qr + ra ) = ab + bc + ( cp + qr ) + ( pq + ra ) = ab + bc + ab + bc = 2 ( ab + bc ) = [ 2 ( 8 + 4 ) ] cm = 24 cm . answer a | ['a ) 24', 'b ) 28', 'c ) 36', 'd ) 48', 'e ) none of these'] | a | multiply(add(8, 4), const_2) | add(n0,n1)|multiply(#0,const_2) | geometry |
20 machines can do a work in 10 days . how many machines are needed to complete the work in 40 days ? | "required number of machines = 20 * 10 / 40 = 5 answer is e" | a ) 10 , b ) 6 , c ) 4 , d ) 7 , e ) 5 | e | divide(multiply(20, 10), 40) | multiply(n0,n1)|divide(#0,n2)| | physics |
if a 10 percent deposit that has been paid toward the purchase of a certain product is $ 110 , how much more remains to be paid ? | "10 / 100 p = 110 > > p = 110 * 100 / 10 = 1100 1100 - 110 = 990 answer : b" | a ) $ 880 , b ) $ 990 , c ) $ 1,000 , d ) $ 1,100 , e ) $ 1,210 | b | subtract(multiply(110, divide(const_100, 10)), 110) | divide(const_100,n0)|multiply(n1,#0)|subtract(#1,n1)| | general |
let f ( x ) = x ^ 2 + bx + c . if f ( 1 ) = 0 and f ( - 9 ) = 0 , then f ( x ) crosses the y - axis at what y - coordinate ? | "when x = 1 and when x = - 9 , the expression f ( x ) = x ² + bx + c equals 0 . then f ( x ) = ( x - 1 ) ( x + 9 ) f ( 0 ) = - 9 the answer is a ." | a ) - 9 , b ) - 1 , c ) 0 , d ) 1 , e ) 9 | a | negate(divide(subtract(power(9, 2), 1), add(9, 1))) | add(n3,n1)|power(n3,n0)|subtract(#1,n1)|divide(#2,#0)|negate(#3)| | general |
in a trapezium abcd , ab is parallel to dc , ab = 3 * dc , and the diagonals of the trapezium intersect at o . the ratio of the area of ∆ ocd to the area of ∆ oab is : | triangles are similar , so ratio will be 1 : 9 a | ['a ) 1 : 9', 'b ) 1 : 3', 'c ) 3 : 1', 'd ) 1 : 10', 'e ) can not be determined'] | a | multiply(divide(const_1, 3), divide(const_1, 3)) | divide(const_1,n0)|multiply(#0,#0) | geometry |
when 1 / 10 percent of 6,000 is subtracted from 1 / 10 of 6,000 , the difference is | "we can break this problem into two parts : 1 ) what is 1 / 10 percent of 6,000 ? 2 ) what is 1 / 10 of 6,000 ? to calculate 1 / 10 percent of 6,000 we must first remember to divide 1 / 10 by 100 . so we have : ( 1 / 10 ) / ( 100 ) to divide a number by 100 means to multiply it by 1 / 100 , so we have : 1 / 10 x 1 / 10... | a ) 0 , b ) 50 , c ) 450 , d ) 495 , e ) 694 | e | subtract(multiply(multiply(add(const_3, const_4), const_1000), divide(1, 10)), multiply(divide(divide(1, 10), const_100), multiply(add(const_3, const_4), const_1000))) | add(const_3,const_4)|divide(n0,n1)|divide(#1,const_100)|multiply(#0,const_1000)|multiply(#1,#3)|multiply(#2,#3)|subtract(#4,#5)| | general |
if there are 200 questions in a 3 hr examination . among these questions are 10 type a problems , which requires twice as much as time be spent than the rest of the type b problems . how many minutes should be spent on type a problems ? | x = time for type b prolems 2 x = time for type a problem total time = 3 hrs = 180 min 190 x + 10 * 2 x = 180 x = 180 / 210 x = 0.72 time taken for type a problem = 10 * 2 * 0.857143 = 17.14286 min answer : a | a ) 17.14286 min , b ) 62 min , c ) 70.1346 min , d ) 74 min , e ) 76.1346 min | a | multiply(multiply(const_2, divide(multiply(3, const_60), add(subtract(200, 10), multiply(const_2, 10)))), 10) | multiply(n1,const_60)|multiply(n2,const_2)|subtract(n0,n2)|add(#1,#2)|divide(#0,#3)|multiply(#4,const_2)|multiply(n2,#5) | general |
working individually , jane can bake cakes for 4 hours and roy can bake cakes in 5 hours . if jane and roy work together but independently at the task for 2 hours , at which point roy leaves , how many remaining hours will it take jane to complete the task alone ? | in first 2 hrs roy will finish 2 / 5 = 2 / 5 of work and jane will finish 2 / 4 work so total 2 / 5 + 1 / 2 = 9 / 10 work is finished and 1 - 9 / 10 = 1 / 10 work remaining . now jane will take ( 1 / 10 ) * 4 = 2 / 5 hrs to finish it . so answer is b . | a ) 1 / 2 , b ) 2 / 5 , c ) 3 / 4 , d ) 1 / 4 , e ) 1 / 5 | b | divide(4, const_10) | divide(n0,const_10) | physics |
the arithmetic mean of set x is zero . if y = { - 2.22 ; - 1.96 ; - 1.68 ; 1.62 ; 1.94 ; 2.16 } is the subset of x consisting of all those elements in x which are more than two but less than 3 standard deviations away from the arithmetic mean of x , what could be equal to the standard deviation of x ? | this is a fairly straightforward question that can be solved quickly by just applying the options we are provided with mean , m = 0 and y = { - 2.22 ; - 1.96 ; - 1.68 ; 1.62 ; 1.94 ; 2.16 } is the subset of s y consists of all those elements in x that are more than 2 but less than 3 xds away from the arithmetic mean of... | a ) 0.54 , b ) 0.77 , c ) 0.82 , d ) 0.97 , e ) 1.62 | b | divide(2.22, const_3) | divide(n0,const_3) | general |
what is the sum of odd integers from 25 to 85 , inclusive ? | number of odd integers = ( 85 - 25 ) / 2 + 1 = 60 / 2 + 1 = 31 sum of odd integers = ( 25 + 85 ) / 2 * 31 = 1705 answer a | a ) 1,705 , b ) 1,500 , c ) 1,240 , d ) 1,120 , e ) 1,100 | a | divide(85, 85) | divide(n1,n1) | general |
carrie likes to buy t - shirts at the local clothing store . they cost $ 9.95 each . one day , she bought 25 t - shirts . how much money did she spend ? | $ 9.95 * 25 = $ 248.75 . answer is b . | a ) $ 150 , b ) $ 248.75 , c ) $ 200 , d ) $ 171.6 , e ) $ 190 | b | floor(multiply(25, 9.95)) | multiply(n0,n1)|floor(#0)| | general |
evaluate the sum 1 - 2 + 3 - 4 + 5 - 6 + . . . + 997 - 998 + 999 - 1000 | one groups the first two terms and each successive two terms to obtain a sum of 500 expressions each of which is - 1 . the answer is - 500 correct answer a | a ) - 500 , b ) - 1000 , c ) - 999 , d ) - 1001 , e ) 500500 | a | divide(1000, 2) | divide(n9,n1) | general |
a line has a slope of 3 / 4 and intersects the point t ( - 12 , - 39 ) . at which point does this line intersect the x - axis ? | "assume that the equation of the line is y = mx + c , where m and c are the slope and y - intercept . you are also given that the line crosses the point ( - 12 , - 39 ) , this means that this point will also lie on the line above . thus you get - 39 = m * ( - 12 ) + c , with m = 3 / 4 as the slope is given to be 3 / 4 ... | a ) ( 40,0 ) , b ) ( 30,0 ) , c ) ( 0,40 ) , d ) ( 40,30 ) , e ) ( 0,30 ) | a | multiply(negate(divide(subtract(negate(39), multiply(negate(12), divide(3, 4))), divide(3, 4))), const_10) | divide(n0,n1)|negate(n3)|negate(n2)|multiply(#0,#2)|subtract(#1,#3)|divide(#4,#0)|negate(#5)|multiply(#6,const_10)| | general |
the owner of a furniture shop charges his customer 20 % more than the cost price . if a customer paid rs . 8337 for a computer table , then what was the cost price of the computer table ? | "cp = sp * ( 100 / ( 100 + profit % ) ) = 8337 ( 100 / 120 ) = rs . 6947.5 answer : e" | a ) rs . 6725 , b ) rs . 6727 , c ) rs . 6908 , d ) rs . 6725 , e ) rs . 6947.5 | e | divide(8337, add(const_1, divide(20, const_100))) | divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)| | gain |
two drivers began their journey with the same amount of petrol in their cars at the same time . the only difference is that the first driver � s car could drive 4 hours in that amount of petrol and the second one could drive 5 hours . however , they only drove for some time and found that the amount of petrol that was ... | a 3.75 hours while you can solve it as you like , a simple mathematical equation can be used to find out . let m be the amount of petrol initially . let n be the time for which they drove . according to the question , the amount of petrol used by first car in n hours = mn / 4 the amount of petrol used by second car in ... | a ) 3.75 hours , b ) 5.45 hours , c ) 7.10 hours , d ) 4.75 hours , e ) 3.33 hours | a | divide(subtract(4, const_1), subtract(multiply(divide(const_1, 4), 4), divide(const_1, 5))) | divide(const_1,n0)|divide(const_1,n1)|subtract(n0,const_1)|multiply(n0,#0)|subtract(#3,#1)|divide(#2,#4) | physics |
the ratio of two numbers is 5 : 9 . if each number is decreased by 5 , the ratio becomes 5 : 11 . find the numbers . | "explanation : let the two numbers be 5 x and 9 x . ( 5 x - 5 ) / ( 9 x - 5 ) = 5 : 11 ( 5 x - 5 ) * 11 = ( 9 x - 5 ) * 5 55 x – 55 = 45 x – 25 10 x = 30 x = 3 therefore , the numbers are 15 and 27 . answer : d" | a ) 30 , 19 , b ) 21 , 37 , c ) 15 , 34 , d ) 15 , 27 , e ) none of these | d | sqrt(divide(5, add(power(5, 9), add(power(5, const_2), power(9, 9))))) | power(n0,n1)|power(n1,const_2)|power(n2,n1)|add(#0,#1)|add(#3,#2)|divide(n3,#4)|sqrt(#5)| | other |
tim came second in math . when his mother asked him how much he had scored , he answered that he got the sum of the first 11 even numbers . his mother immediately worked out the answer . how much had he scored in math ? | "c 132 sum = ( n x n ) + n hence , 11 x 11 = 121 + 11 = 132" | a ) 80 , b ) 190 , c ) 132 , d ) 170 , e ) 60 | c | multiply(add(11, const_1), 11) | add(n0,const_1)|multiply(n0,#0)| | physics |
the average height of 12 girls out of a class of 20 is 161 cm . and that of the remaining girls is 158 cm . the average height of the whole class is : | "explanation : average height of the whole class = ( 12 × 161 + 8 × 158 / 20 ) = 159.8 cms answer e" | a ) 142.3 cms , b ) 162.3 cms , c ) 160.5 cms , d ) 142.8 cms , e ) 159.8 cms | e | divide(add(multiply(161, 12), multiply(158, const_10)), 20) | multiply(n0,n2)|multiply(n3,const_10)|add(#0,#1)|divide(#2,n1)| | general |
find out the c . i on rs . 9000 at 4 % p . a . compound half - yearly for 1 1 / 2 years | "a = 9000 ( 51 / 50 ) 3 = 9550.87 9000 - - - - - - - - - - - 550.87 answer : a" | a ) 550.87 , b ) 506.07 , c ) 506.04 , d ) 506.03 , e ) 306.01 | a | subtract(multiply(9000, power(add(1, divide(4, const_100)), divide(const_3, 2))), 9000) | divide(n1,const_100)|divide(const_3,n4)|add(#0,n2)|power(#2,#1)|multiply(n0,#3)|subtract(#4,n0)| | general |
the length of rectangle is thrice its breadth and its perimeter is 56 m , find the area of the rectangle ? | "2 ( 3 x + x ) = 56 l = 21 b = 7 lb = 21 * 7 = 147 answer : e" | a ) 432 , b ) 212 , c ) 252 , d ) 992 , e ) 147 | e | multiply(multiply(divide(56, add(multiply(const_3, const_2), multiply(const_1, const_2))), const_3), divide(56, add(multiply(const_3, const_2), multiply(const_1, const_2)))) | multiply(const_2,const_3)|multiply(const_1,const_2)|add(#0,#1)|divide(n0,#2)|multiply(#3,const_3)|multiply(#3,#4)| | geometry |
in a flight of 600 km , an aircraft was slowed down due to bad weather . its average speed for the trip was reduced by 200 km / hr and the time of flight increased by 30 minutes . the duration of the flight is | "e 1 hr . let the duration of the flight be x hours . then , 600 / x - 600 / 0.5 x = 200 - - > 600 / x - 1200 / ( 2 x + 1 ) = 200 - - > x ( 2 x + 1 ) = 3 - - > 2 x ^ 2 + x - 3 = 0 - - > ( 2 x + 3 ) ( x - 1 ) = 0 - - > x = 1 hr ." | a ) 3 hr . , b ) 5 hr . , c ) 2 hr . , d ) 7 hr . , e ) 1 hr . | e | divide(600, divide(multiply(200, 30), const_10)) | multiply(n1,n2)|divide(#0,const_10)|divide(n0,#1)| | general |
money invested at x % , compounded annually , triples in value in approximately every 112 / x years . if $ 2200 is invested at a rate of 8 % , compounded annually , what will be its approximate worth in 28 years ? | "x = 8 % 112 / x years = 112 / 8 = 14 years now , money triples every 14 years therefore , in 14 yrs , if $ 2200 triples to $ 6600 , in 28 years , it will again triple to $ 6600 * 3 = $ 19,800 answer b" | a ) $ 3,750 , b ) $ 19,800 , c ) $ 8,100 , d ) $ 15,000 , e ) $ 22,500 | b | multiply(2200, power(const_3, divide(28, divide(112, 8)))) | divide(n0,n2)|divide(n3,#0)|power(const_3,#1)|multiply(n1,#2)| | gain |
a rocket soars for 12 seconds at 150 meters per second . it then plummets 600 meters in 3 seconds . what is the average speed of the rocket in meters per second ? | when soaring , the rocket travels a total distance of 1800 m ( 150 m / s for 12 s ) while plummeting the rockets travel distance of 600 m in 3 s the total distance traveled is 2400 m ( 1800 + 600 ) the time taken to travel this distance is 15 s ( 12 + 3 ) average speed = 2400 / 15 = 160 ans : ( option d ) | a ) 140 , b ) 150 , c ) 155 , d ) 160 , e ) 175 | d | divide(add(multiply(12, 150), 600), add(12, 3)) | add(n0,n3)|multiply(n0,n1)|add(n2,#1)|divide(#2,#0) | physics |
for any integer n greater than 1 , # n denotes the product of all the integers from 1 to n , inclusive . how many prime numbers w are there between # 6 + 2 and # 6 + 6 , inclusive ? | none is the answer . a . because for every k 6 ! + k : : k , because 6 ! : : k , since k is between 2 and 6 . a | a ) none , b ) one , c ) two , d ) three , e ) four | a | add(1, 1) | add(n0,n0) | general |
a no . when divided by 224 gives a remainder 43 , what remainder will beobtained by dividingthe same no . 17 ? | "224 + 43 = 267 / 17 = 12 ( remainder ) e" | a ) 2 , b ) 7 , c ) 9 , d ) 11 , e ) 12 | e | divide(add(224, 43), 17) | add(n0,n1)|divide(#0,n2)| | general |
20 beavers , working together in a constant pace , can build a dam in 3 hours . how many hours x will it take 12 beavers that work at the same pace , to build the same dam ? | "c . 5 hrs if there were 10 beavers it qould have taken double x = 6 hrs . . so closest to that option is 5 ." | a ) 2 . , b ) 4 . , c ) x = 5 . , d ) x = 6 . , e ) 8 . | c | divide(multiply(3, 20), 12) | multiply(n0,n1)|divide(#0,n2)| | physics |
the product of two numbers is 168 and the sum of these two numbers is 26 . what is the largest of these two numbers ? | "let the numbers be x and ( 26 – x ) . therefore , x ( 26 – x ) = 168 x 2 – 26 x + 168 = 0 ( x – 14 ) ( x – 12 ) = 0 x = 14 or x = 12 so , the numbers are 14 and 12 . answer : d" | a ) 22 , b ) 77 , c ) 88 , d ) 12 , e ) 62 | d | sqrt(add(power(sqrt(subtract(26, multiply(const_2, 168))), const_2), multiply(const_4, 168))) | multiply(n0,const_4)|multiply(n0,const_2)|subtract(n1,#1)|sqrt(#2)|power(#3,const_2)|add(#0,#4)|sqrt(#5)| | general |
a small company is planning to rent either computer a or computer b to print customer mailing lists . both computer a and computer b must be rented on an hourly basis . the rental fee is based only on the amount of time the computer is turned on . it will cost 40 percent more per hour to rent computer a than to rent co... | "pa = price of a pb = price of b ta = time for a to complete the job tb = time for b to complete the job given pa = 1.4 pb ta + 20 = tb pa * ta = pb * tb = 700 1.4 pb * ( tb - 20 ) = pb * tb 1.4 pb tb - pb tb = 1.4 pb * 20 0.4 pbtb = 28 pb tb = 28 / 0.4 = 70 pb = 700 / 70 = 10 b" | a ) $ 9.40 , b ) $ 10.00 , c ) $ 13.20 , d ) $ 17.80 , e ) $ 22.10 | b | add(divide(40, const_100), add(add(const_4, const_3), add(divide(40, const_100), divide(divide(20, const_4), const_100)))) | add(const_3,const_4)|divide(n0,const_100)|divide(n1,const_4)|divide(#2,const_100)|add(#1,#3)|add(#0,#4)|add(#5,#1)| | general |
in a survey of 330 employees , 104 of them are uninsured , 54 work part time , and 12.5 percent of employees who are uninsured work part time . if a person is to be randomly selected from those surveyed , what is the probability that the person will neither work part time nor be uninsured ? | "- - - - - - - - - ui - - - - - - - - - - - - - - - - nui - - - - - - - total pt - - - - ( 12.5 / 100 ) * 104 = 13 - - - - - - - - - - - - - 54 npt - - - 104 - 13 - - - - - - - - - - - - - - x - - - - - - - - 276 total - - 104 - - - - - - - - - - - - - - - - - - - - - - - - - - - - 330 we have to find not part time and... | a ) 37 / 66 , b ) 8 / 41 , c ) 9 / 348 , d ) 1 / 8 , e ) 41 / 91 | a | divide(subtract(330, add(subtract(54, divide(multiply(12.5, 104), multiply(const_1, const_100))), 104)), 330) | multiply(n1,n3)|multiply(const_1,const_100)|divide(#0,#1)|subtract(n2,#2)|add(n1,#3)|subtract(n0,#4)|divide(#5,n0)| | physics |
a dishonest dealer professes to sell goods at the cost price but uses a weight of 896 grams per kg , what is his percent ? | "896 - - - 104 100 - - - ? = > 11.6 % answer : e" | a ) 22 % , b ) 25 % , c ) 77 % , d ) 99 % , e ) 11.6 % | e | subtract(multiply(divide(const_100, 896), multiply(const_100, multiply(add(const_3, const_2), const_2))), const_100) | add(const_2,const_3)|divide(const_100,n0)|multiply(#0,const_2)|multiply(#2,const_100)|multiply(#1,#3)|subtract(#4,const_100)| | gain |
jack has two dice , one has 6 equally probable sides , labeled 1 , 2 , 3 , 4 , 5 , 6 , and the other has 7 equally probable sides , labeled 1 , 2 , 3 , 4 , 5 , 6 , 7 . if jack rolls both dice what is the probability that both of the numbers will be even ? | probability that the number on first die is even = 3 / 6 [ because 3 out of 6 faces are even ] probability that the number on second die is even = 3 / 7 [ because 3 out of 7 faces are even ] probability that both dice result in odd numbers = ( 3 / 6 ) * ( 3 / 7 ) = 3 / 14 answer : option a | a ) 3 / 14 , b ) 2 / 7 , c ) 1 / 3 , d ) 1 / 2 , e ) 12 / 21 | a | divide(3, multiply(7, 2)) | multiply(n2,n7)|divide(n3,#0) | other |
what will be the compound interest on a sum of rs . 26000 after 3 years at the rate of 12 % p . a . ? | "explanation : amount = [ 26000 * ( 1 + 12 / 100 ) 3 ] = 26000 * 28 / 25 * 28 / 25 * 28 / 25 = rs . 35123.20 c . i . = ( 36528.13 - 26000 ) = rs . 10528.13 answer : e" | a ) s . 10123.22 , b ) s . 10823.20 , c ) s . 10123.20 , d ) s . 10123.29 , e ) s . 10528.13 | e | subtract(multiply(multiply(multiply(const_4, const_100), const_100), power(add(const_1, divide(12, const_100)), 3)), multiply(multiply(const_4, const_100), const_100)) | divide(n2,const_100)|multiply(const_100,const_4)|add(#0,const_1)|multiply(#1,const_100)|power(#2,n1)|multiply(#3,#4)|subtract(#5,#3)| | gain |
which number need to add to 859622 to get a no . exactly divisible by 456 ? | "dividend = quotient * divisor + reminder 859622 / 456 gives quotient = 1885 and reminder = 62 . so , the next number divisible by 456 is 456 places infront of 456 * 1885 which means 456 – 62 = 394 should be added to 859622 . e" | a ) 345466 , b ) 465767 , c ) 565676 , d ) 645469 , e ) 859622 | e | multiply(456, subtract(add(floor(divide(859622, 456)), const_1), divide(859622, 456))) | divide(n0,n1)|floor(#0)|add(#1,const_1)|subtract(#2,#0)|multiply(n1,#3)| | general |
a case of 12 rolls of paper towels sells for $ 9 . the cost of one roll sold individually is $ 1 . what is the percent r of savings per roll for the 12 - roll package over the cost of 12 rolls purchased individually ? | "cost of 12 paper towels individually = 1 * 12 = 12 cost of a set of 12 paper towels = 9 cost of one roll = 9 / 12 = 3 / 4 = 0.75 savings per roll = 1 - . 75 = 0.25 % of savings is r = . 25 / 1 * 100 = 25 % d is the answer ." | a ) 9 % , b ) 11 % , c ) 15 % , d ) 25 % , e ) 90 % | d | subtract(const_100, multiply(divide(9, 12), const_100)) | divide(n1,n0)|multiply(#0,const_100)|subtract(const_100,#1)| | general |
a cistern is filled by pipe a in 10 hours and the full cistern can be leaked out by an exhaust pipe b in 12 hours . if both the pipes are opened , in what time the cistern is full ? | "time taken to full the cistern = ( 1 / 10 - 1 / 12 ) hrs = 1 / 60 = 60 hrs answer : b" | a ) 50 hrs , b ) 60 hrs , c ) 70 hrs , d ) 80 hrs , e ) 90 hrs | b | divide(const_1, subtract(divide(const_1, 10), divide(const_1, 12))) | divide(const_1,n0)|divide(const_1,n1)|subtract(#0,#1)|divide(const_1,#2)| | physics |
here is 15 dots . if you select 3 dots randomly , what is the probability that 3 dots make a triangle ? | total ways of selecting 3 dots out of 15 is 15 c 3 = 455 if 3 dots are collinear then triangle may not be formed . now look at the above diagram . if we select any 3 dots from the red lines they may not form a triangle . they are 3 x 5 c 3 = 30 . if we select the three letters from blue lines , they may not form a tria... | a ) 440 / 455 , b ) 412 / 455 , c ) 434 / 455 , d ) 449 / 455 , e ) 438 / 455 | b | negate_prob(divide(add(multiply(choose(const_5, 3), const_3), multiply(choose(const_3, 3), const_5)), choose(15, 3))) | choose(const_5,n1)|choose(const_3,n1)|choose(n0,n1)|multiply(#0,const_3)|multiply(#1,const_5)|add(#3,#4)|divide(#5,#2)|negate_prob(#6) | geometry |
in a class of 20 students , 2 students did not borrow any books from the library , 8 students each borrowed 1 book , 3 students each borrowed 2 books , and the rest borrowed at least 3 books . if the average number of books per student was 2 , what is the maximum number of books any single student could have borrowed ? | "the total number of books the students borrowed is 20 * 2 = 40 . the students who borrowed zero , one , or two books borrowed 8 * 1 + 3 * 2 = 14 books . the 7 students who borrowed at least three books borrowed 40 - 14 = 26 books . if 6 of these students borrowed exactly 3 books , then the maximum that one student cou... | a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | a | subtract(multiply(20, 2), add(multiply(subtract(subtract(20, add(add(multiply(8, 1), 3), 2)), 1), 3), add(multiply(8, 1), multiply(3, 2)))) | multiply(n0,n1)|multiply(n2,n3)|multiply(n1,n4)|add(#1,#2)|add(n4,#1)|add(n1,#4)|subtract(n0,#5)|subtract(#6,n3)|multiply(n6,#7)|add(#3,#8)|subtract(#0,#9)| | general |
a manufacturer produces a certain men ' s athletic shoe in integer sizes from 8 to 17 . for this particular shoe , each unit increase in size corresponds to a 1 / 4 - inch increase in the length of the shoe . if the largest size of this shoe is 30 % longer than the smallest size , how long , in inches , is the shoe in ... | "let x be the length of the size 8 shoe . then 0.3 x = 9 / 4 x = 90 / 12 = 15 / 2 inches the size 15 shoe has a length of 15 / 2 + 7 / 4 = 37 / 4 = 9.25 inches the answer is a ." | a ) 9.25 , b ) 11.75 , c ) 13.25 , d ) 15.75 , e ) 17.25 | a | add(divide(multiply(divide(1, 4), subtract(17, 8)), divide(30, const_100)), multiply(subtract(15, 8), divide(1, 4))) | divide(n2,n3)|divide(n4,const_100)|subtract(n1,n0)|subtract(n5,n0)|multiply(#0,#2)|multiply(#0,#3)|divide(#4,#1)|add(#6,#5)| | general |
what is the measure of the radius of the circle that circumscribes a triangle whose sides measure 11 , 60 and 61 ? | "some of pyhtagron triplets we need to keep it in mind . like { ( 2 , 3,5 ) , ( 5 , 12,13 ) , ( 7 , 24,25 ) , ( 11 , 60,61 ) . so now we know the triangle is an right angle triangle . the circle circumscribes the triangle . the circumraduis of the circle that circumscribes the right angle triangle = hypotanse / 2 = 61 ... | a ) 6.5 , b ) 12.5 , c ) 28 , d ) 30.5 , e ) 40 | d | divide(triangle_area_three_edges(11, 60, 61), divide(triangle_perimeter(11, 60, 61), const_2)) | triangle_area_three_edges(n0,n1,n2)|triangle_perimeter(n0,n1,n2)|divide(#1,const_2)|divide(#0,#2)| | geometry |
a man spends 50 % of his income on food , 15 % on children education and 50 % of the remaining on house rent . what percent of his income he is left with ? | "let the total income be x then , income left = ( 100 - 50 ) % of x - [ 100 - ( 50 + 15 ) ] % of x = 50 % of 35 % of x = 17.5 % of x answer is a" | a ) 17.5 % , b ) 16 % , c ) 15.4 % , d ) 12.6 % , e ) 13.5 % | a | divide(multiply(subtract(subtract(const_100, 50), 15), subtract(const_100, 50)), const_100) | subtract(const_100,n0)|subtract(const_100,n2)|subtract(#0,n1)|multiply(#2,#1)|divide(#3,const_100)| | gain |
a car covers a distance of 624 km in 3 hours . find its speed ? | 624 / 3 = 208 kmph answer : e | a ) 104 , b ) 277 , c ) 298 , d ) 269 , e ) 208 | e | divide(624, 3) | divide(n0,n1) | physics |
it was calculated that 50 men could complete a piece of work in 10 days . when work was scheduled to commence , it was found necessary to send 25 men to another project . how much longer will it take to complete the work ? | "one day work = 1 / 10 one man ’ s one day work = 1 / ( 10 * 50 ) now : no . of workers = 25 one day work = 25 * 1 / ( 10 * 50 ) the total no . of days required to complete the work = ( 50 * 10 ) / 25 = 20 answer : a" | a ) 20 days . , b ) 15 days . , c ) 22 days . , d ) 30 days . , e ) 18 days . | a | multiply(10, divide(50, 25)) | divide(n0,n2)|multiply(n1,#0)| | physics |
at a supermarket , the ratio of the number of shampoos to the number of soaps is 8 to 5 , and the ratio of the number of drinks to the number of shampoos is 3 to 4 . if the ratio of the soaps to the number of cereals is 9 to 2 , what is the ratio of the number of drinks to the number of cereals ? | sh : soaps = 8 : 5 drinks : sh = 3 : 4 = > 6 : 8 drinks : sh : soaps = 6 : 8 : 5 soaps : cer = 9 : 2 = > 9 * 5 : 2 * 5 = 45 : 10 now , drinks : sh : soaps = > 6 * 9 : 8 * 9 : 5 * 9 = 54 : 72 : 45 drinks : sh : soaps : cer = 54 : 72 : 45 : 10 drinks : cer = 54 : 10 = > 27 : 5 e | a ) 4 : 3 , b ) 15 : 6 , c ) 45 : 16 , d ) 15 : 4 , e ) 27 : 5 | e | divide(add(8, 3), const_2) | add(n0,n2)|divide(#0,const_2) | other |
xy = 1 then what is ( 5 ^ ( x + y ) ^ 2 ) / ( 5 ^ ( x - y ) ^ 2 ) | "( x + y ) ^ 2 - ( x - y ) ^ 2 ( x + y + x - y ) ( x + y - x + y ) ( 2 x ) ( 2 y ) 4 xy 4 5 ^ 4 = 625 answer b" | a ) 2 , b ) 625 , c ) 8 , d ) 16 , e ) 32 | b | power(5, multiply(const_4, 1)) | multiply(n0,const_4)|power(n1,#0)| | general |
gold is 10 times as heavy as water and copper is 5 times as heavy as water . in what ratio should these be mixed to get an alloy 8 times as heavy as water ? | "g = 10 w c = 5 w let 1 gm of gold mixed with x gm of copper to get 1 + x gm of the alloy 1 gm gold + x gm copper = x + 1 gm of alloy 10 w + 5 wx = x + 1 * 8 w 10 + 5 x = 8 ( x + 1 ) x = 2 / 3 ratio of gold with copper = 1 : 2 / 3 = 3 : 2 answer is a" | a ) 3 : 2 , b ) 1 : 2 , c ) 3 : 1 , d ) 5 : 2 , e ) 4 : 3 | a | divide(subtract(8, 5), subtract(10, 8)) | subtract(n2,n1)|subtract(n0,n2)|divide(#0,#1)| | general |
the length of a rectangular floor is more than its breadth by 200 % . if rs . 400 is required to paint the floor at the rate of rs . 3 / sq m , what would be the length of the floor ? | "let the length and the breadth of the floor be l m and b m respectively . l = b + 200 % of b = l + 2 b = 3 b area of the floor = 400 / 3 = 133.33 sq m l b = 133.33 i . e . , l * l / 3 = 133.33 l ^ 2 = 400 > l = 20 . c" | a ) 12 , b ) 18 , c ) 20 , d ) 22 , e ) 24 | c | multiply(sqrt(divide(divide(400, 3), const_3)), const_3) | divide(n1,n2)|divide(#0,const_3)|sqrt(#1)|multiply(#2,const_3)| | gain |
a baker charges n dollars for a birthday cake . next week , she intends to earn 320 dollars selling cakes . if she were to reduce the price of the cake by 20 percent , she would have to sell two more cakes to earn the same amount of revenue . how many cakes does she intend to sell next week ? | we ' re told that by selling cakes at n dollars / cake , a baker will make $ 320 . reducing the price by 20 % and selling 2 more cakes will also make her $ 320 . we ' re asked for the original number of cakes that the baker intends to sell . $ 320 is a rather interesting number . based on the answer choices , we ' re a... | a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | d | divide(320, divide(subtract(320, multiply(320, divide(subtract(const_100, 20), const_100))), multiply(divide(subtract(const_100, 20), const_100), const_2))) | subtract(const_100,n1)|divide(#0,const_100)|multiply(n0,#1)|multiply(#1,const_2)|subtract(n0,#2)|divide(#4,#3)|divide(n0,#5) | general |
a sum of money deposited at c . i . amounts to rs . 1820 in 4 years and to rs . 2402 in 5 years . find the rate percent ? | "1820 - - - 582 100 - - - ? = > 31 % answer : b" | a ) 30 % , b ) 31 % , c ) 34 % , d ) 19 % , e ) 50 % | b | multiply(divide(subtract(2402, 1820), 1820), const_100) | subtract(n2,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
an assembly line produces 30 cogs per hour until an initial order of 60 cogs is completed . the speed of the assembly line is then immediately increased so that it can produce 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 ... | "the time to produce the first 60 cogs is 60 / 30 = 2 hours . the time to produce the next 60 cogs is 60 / 60 = 1 hour . the average output is 120 cogs / 3 hours = 40 cogs per hour . the answer is a ." | a ) 40 , b ) 44 , c ) 48 , d ) 52 , e ) 56 | a | divide(multiply(60, const_2), add(divide(60, 30), const_1)) | divide(n1,n0)|multiply(n1,const_2)|add(#0,const_1)|divide(#1,#2)| | general |
if the cost price of 44 chocolates is equal to the selling price of 24 chocolates , the gain percent is : | explanation : solution : let c . p . of each chocolate be re . 1 . then , c . p . of 24 chocolates = rs . 24 ; s . p . of 24 chocolates = rs . 44 . . ' . gain % = 20 * 100 / 24 = 83.33 % answer : b | a ) 83.55 % , b ) 83.33 % , c ) 80 % , d ) 83.39 % , e ) 84.33 % | b | divide(const_100, divide(24, subtract(44, 24))) | subtract(n0,n1)|divide(n1,#0)|divide(const_100,#1) | gain |
90 percent of your marbles are a solid color . 5 percent of your marbles are solid yellow . what percentage of your marbles are a solid color other than yellow ? | 90 percent are solid colors including solid yellow . 5 percent are solid yellow . 90 % - 5 % = 85 % , so 85 percent are a solid color other than yellow = > ( c ) | a ) 5 , b ) 15 , c ) 85 , d ) 90 , e ) 95 | c | subtract(90, 5) | subtract(n0,n1) | gain |
when m divided by 288 , the remainder is 47 . find the remainder when the same m is divided by 24 ? | 23 option d | a ) 20 , b ) 21 , c ) 22 , d ) 23 , e ) 26 | d | reminder(47, 24) | reminder(n1,n2) | physics |
the length of rectangle is thrice its breadth and its perimeter is 112 m , find the area of the rectangle ? | "2 ( 3 x + x ) = 112 l = 42 b = 14 lb = 42 * 14 = 588 answer : d" | a ) 432 sq m , b ) 356 sq m , c ) 452 sq m , d ) 588 sq m , e ) 525 sq m | d | multiply(multiply(divide(112, add(multiply(const_3, const_2), multiply(const_1, const_2))), const_3), divide(112, add(multiply(const_3, const_2), multiply(const_1, const_2)))) | multiply(const_2,const_3)|multiply(const_1,const_2)|add(#0,#1)|divide(n0,#2)|multiply(#3,const_3)|multiply(#3,#4)| | geometry |
in measuring the sides of a rectangle , one side is taken 5 % in excess and the other 4 % in deficit . find the error percent in the area , calculate from the those measurements . | "explanation : let x and y be the sides of the rectangle then correct area = = error % = answer : b ) 0.8 %" | a ) 0.9 % , b ) 0.8 % , c ) 0.3 % , d ) 0.2 % , e ) 0.5 % | b | subtract(subtract(5, 4), divide(multiply(5, 4), const_100)) | multiply(n0,n1)|subtract(n0,n1)|divide(#0,const_100)|subtract(#1,#2)| | geometry |
we have a rectangular metallic piece of paper that covers exactly the area of a cube . the length of the piece of paper is 144 inches and the width is 72 inches . what is the volume of the cube in cubic feet is 1 feet is 12 inches ? | "l = 144 / 12 = 12 ft w = 72 / 12 = 6 ft area of paper = 72 area of cube = 12 * side ^ 2 side of cube = 6 v of cube = 216" | a ) a 216 , b ) b 196 , c ) c 170 , d ) d 140 , e ) e 121 | a | add(cube_edge_by_volume(144), 144) | cube_edge_by_volume(n0)|add(n0,#0)| | geometry |
working at constant rate , pump x pumped out half of the water in a flooded basement in 7 hours . the pump y was started and the two pumps , working independently at their respective constant rates , pumped out rest of the water in 3 hours . how many hours would it have taken pump y , operating alone at its own constan... | "rate of x = 1 / 8 rate of x + y = 1 / 6 rate of y = 1 / 6 - 1 / 8 = 1 / 24 34 hours c" | a ) a . 10 , b ) b . 12 , c ) c . 34 , d ) d . 18 , e ) e . 24 | c | add(add(add(add(add(add(add(add(add(add(add(add(const_1, add(3, 7)), const_1), const_1), const_1), const_1), 7), const_1), const_1), const_1), const_1), const_1), const_1) | add(n0,n1)|add(#0,const_1)|add(#1,const_1)|add(#2,const_1)|add(#3,const_1)|add(#4,const_1)|add(#5,n0)|add(#6,const_1)|add(#7,const_1)|add(#8,const_1)|add(#9,const_1)|add(#10,const_1)|add(#11,const_1)| | physics |
if k is an integer and 0.00010101 x 10 ^ k is greater than 1000 , what is the least possible value of k ? | "0.00010101 * 10 ^ k > 1000 we need to move the decimal point to the right 7 places to get 1010.1 this is equivalent to multiplying by 10 ^ 7 . the answer is b ." | a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | b | divide(log(divide(1000, 0.00010101)), log(10)) | divide(n2,n0)|log(n1)|log(#0)|divide(#2,#1)| | general |
the average weight of a , b and c is 45 kg . if the average weight of a and b be 40 kg and that of b and c be 43 kg , then the weight of b is : | "let a , b , c represent their respective weights . then , we have : a + b + c = ( 45 x 3 ) = 135 . . . . ( i ) a + b = ( 40 x 2 ) = 80 . . . . ( ii ) b + c = ( 43 x 2 ) = 86 . . . . ( iii ) adding ( ii ) and ( iii ) , we get : a + 2 b + c = 166 . . . . ( iv ) subtracting ( i ) from ( iv ) , we get : b = 31 . b ' s wei... | a ) 30 kg , b ) 31 kg , c ) 37 kg , d ) 39 kg , e ) 41 kg | b | subtract(add(multiply(40, const_2), multiply(43, const_2)), multiply(45, const_3)) | multiply(n1,const_2)|multiply(n2,const_2)|multiply(n0,const_3)|add(#0,#1)|subtract(#3,#2)| | general |
a car is purchased on hire - purchase . the cash price is $ 22 000 and the terms are a deposit of 10 % of the price , then the balance to be paid off over 60 equal monthly installments . interest is charged at 12 % p . a . what is the monthly installment ? | "explanation : cash price = $ 22 000 deposit = 10 % ã — $ 22 000 = $ 2200 loan amount = $ 22000 â ˆ ’ $ 2200 number of payments = 60 = $ 19800 i = p * r * t / 100 i = 11880 total amount = 19800 + 11880 = $ 31680 regular payment = total amount / number of payments = 528 answer : d" | a ) $ 503 , b ) $ 504 , c ) $ 515 , d ) $ 528 , e ) $ 537 | d | add(divide(multiply(multiply(22, const_1000), subtract(const_1, divide(10, const_100))), 60), multiply(divide(divide(12, const_100), 12), multiply(multiply(22, const_1000), subtract(const_1, divide(10, const_100))))) | divide(n2,const_100)|divide(n4,const_100)|multiply(n0,const_1000)|divide(#1,n4)|subtract(const_1,#0)|multiply(#2,#4)|divide(#5,n3)|multiply(#3,#5)|add(#6,#7)| | gain |
5 is added to a certain number , the sum is multiplied by 2 , the product is divided by 5 and 5 is subtracted from the quotient . the remainder left is half of 44 . what is the number ? | "let number is x . when 5 added to it , = ( x + 5 ) 2 multiplied to sum , = 2 * ( x + 5 ) now , = [ { 2 * ( x + 5 ) } / 5 ] and , = [ { 2 * ( x + 5 ) } / 5 ] - 5 according to question , [ { 2 * ( x + 5 ) } / 5 ] - 5 = half of 44 [ ( 2 x + 10 ) / 5 ) = 22 + 5 2 x + 10 = 27 * 5 2 x - 135 - 10 x = 125 / 2 = 62.5 = 63 so ,... | a ) 21 , b ) 20 , c ) 63 , d ) 30 , e ) 45 | c | subtract(divide(multiply(add(divide(44, const_2), 5), 5), 2), 5) | divide(n4,const_2)|add(n0,#0)|multiply(n0,#1)|divide(#2,n1)|subtract(#3,n0)| | general |
the ratio between the sale price and the cost price of an article is 9 : 7 . what is the ratio between the profit and the cost price of that article ? | "let c . p . = rs . 7 x and s . p . = rs . 9 x . then , gain = rs . 2 x required ratio = 2 x : 7 x = 2 : 7 answer : b" | a ) 2 : 9 , b ) 2 : 7 , c ) 3 : 6 , d ) 2 : 0 , e ) 2 : 1 | b | divide(subtract(9, 7), 7) | subtract(n0,n1)|divide(#0,n1)| | other |
find the number which when multiplied by 15 is increased by 196 . | let the number be x . then , 15 x - x = 196 < = > 14 x = 196 < = > x = 14 . answer : a | a ) 14 , b ) 20 , c ) 26 , d ) 28 , e ) 30 | a | divide(196, subtract(15, const_1)) | subtract(n0,const_1)|divide(n1,#0) | general |
one night a certain hotel rented 3 / 4 of its rooms . including 2 / 3 of their air conditioned rooms . if 3 / 5 of its rooms were air conditioned , what percent of the rooms that were not rented were air conditioned ? | consider total # of rooms to be 100 ; as 3 / 5 of the rooms are air conditioned then # of rooms that are air conditioned is 3 / 5 * 100 = 60 ; 3 / 4 rooms were rented - - > 1 / 4 * 100 = 25 were not rented ; 2 / 3 of air conditioned rooms were rented - - > 1 / 3 * 60 = 20 air conditioned room were not rented ; 20 / 25 ... | a ) 20 , b ) 33 1 / 3 , c ) 35 , d ) 40 , e ) 80 | e | multiply(divide(divide(divide(multiply(const_100, 3), 5), 3), subtract(const_100, divide(multiply(const_100, 3), 4))), const_100) | multiply(n0,const_100)|divide(#0,n5)|divide(#0,n1)|divide(#1,n0)|subtract(const_100,#2)|divide(#3,#4)|multiply(#5,const_100) | gain |
two trains of equal length are running on parallel lines in the same direction at 46 km / hr and 36 km / hr . the faster train catches and completely passes the slower train in 72 seconds . what is the length of each train ( in meters ) ? | "the relative speed = 46 - 36 = 10 km / hr = 10 * 5 / 18 = 25 / 9 m / s in 72 seconds , the relative difference in distance traveled is 72 * 25 / 9 = 200 meters this distance is twice the length of each train . the length of each train is 200 / 2 = 100 meters the answer is d ." | a ) 70 , b ) 80 , c ) 90 , d ) 100 , e ) 110 | d | divide(multiply(divide(multiply(subtract(46, 36), const_1000), const_3600), 72), const_2) | subtract(n0,n1)|multiply(#0,const_1000)|divide(#1,const_3600)|multiply(n2,#2)|divide(#3,const_2)| | physics |
80 + 5 * 12 / ( 180 / 3 ) = ? | "explanation : 80 + 5 * 12 / ( 180 / 3 ) = 80 + 5 * 12 / ( 60 ) = 80 + ( 5 * 12 ) / 60 = 80 + 1 = 81 . answer : e" | a ) 23 , b ) 78 , c ) 27 , d ) 61 , e ) 81 | e | add(80, divide(multiply(5, 12), divide(180, 3))) | divide(n3,n4)|multiply(n1,n2)|divide(#1,#0)|add(n0,#2)| | general |
ann and bob drive separately to a meeting . ann ' s average driving speed is greater than bob ' s avergae driving speed by one - third of bob ' s average driving speed , and ann drives twice as many miles as bob . what is the ratio t of the number of hours ann spends driving to the meeting to the number of hours bob sp... | say the rate of bob is 3 mph and he covers 6 miles then he needs 6 / 3 = 2 hours to do that . now , in this case the rate of ann would be 3 + 3 * 1 / 3 = 4 mph and the distance she covers would be 6 * 2 = 12 miles , so she needs 12 / 4 = 3 hours for that . the ratio t of ann ' s time to bob ' s time is 3 : 2 . answer :... | a ) 8 : 3 , b ) 3 : 2 , c ) 4 : 3 , d ) 2 : 3 , e ) 3 : 8 | b | divide(const_2, add(const_1, divide(const_1, const_3))) | divide(const_1,const_3)|add(#0,const_1)|divide(const_2,#1) | general |
if 6 x = 10 y = 14 z , then what is a possible sum of positive integers x , y , and z ? | answer = d = 122 6 x = 10 y = 14 z 3 x = 5 y = 7 z 3 ( 5 * 7 ) = 5 ( 3 * 7 ) = 7 ( 3 * 5 ) addition = 35 + 21 + 15 = 71 answer would be multiple of 71 which is 213 answer : a | a ) 213 , b ) 58 , c ) 84 , d ) 122 , e ) 168 | a | divide(multiply(multiply(10, 14), 6), const_4) | multiply(n1,n2)|multiply(n0,#0)|divide(#1,const_4) | general |
tough and tricky questions : percents . over the course of a year , a certain microbrewery increased its beer output by 20 percent . at the same time , it decreased its total working hours by 30 percent . by what percent did this factory increase its output per hour ? | lets assume the initial production was 100 litres of beer for 100 hr . with the 20 % increase the total amount of beer production will be 120 litres and with 30 % decrease in total hours will be reduced to 70 hr . 100 hr - - - - > 100 lts 1 hr - - - - - > 1 lts 70 hr - - - - - > 120 lts 1 hr - - - - - > 1.714 lts total... | a ) 171.4 , b ) 171.5 , c ) 171.6 , d ) 171.7 , e ) none of the above | a | multiply(divide(add(const_100, 20), subtract(const_100, 30)), const_100) | add(n0,const_100)|subtract(const_100,n1)|divide(#0,#1)|multiply(#2,const_100) | general |
walking with 4 / 5 of my usual speed , i arrive at the bus stop 8 minutes later than normal . how many minutes does it take to walk to the bus stop at my usual speed ? | "let t = usual time = distance / usual speed t + 8 = distance / ( 4 * usual speed / 5 ) = ( 5 * distance ) / ( 4 * usual speed ) = 5 t / 4 t = 32 the answer is d ." | a ) 20 , b ) 24 , c ) 28 , d ) 32 , e ) 36 | d | divide(8, subtract(divide(5, 4), const_1)) | divide(n1,n0)|subtract(#0,const_1)|divide(n2,#1)| | physics |
an inspector rejects 0.08 % of the meters as defective . how many will be examine to project ? | let the number of meters to be examined be x . then , 0.08 % of x = 2 [ ( 8 / 100 ) * ( 1 / 100 ) * x ] = 2 x = [ ( 2 * 100 * 100 ) / 8 ] = 2500 answer is d . | a ) 2000 , b ) 2300 , c ) 2700 , d ) 2500 , e ) 250 | d | divide(multiply(multiply(const_2, const_100), const_100), multiply(0.08, const_100)) | multiply(const_100,const_2)|multiply(n0,const_100)|multiply(#0,const_100)|divide(#2,#1) | gain |
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