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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the amount of an investment will double in approximately 70 / p years , where p is the percent interest , compounded annually . if thelma invests $ 50,000 in a long - term cd that pays 5 percent interest , compounded annually , what will be the approximate total value of the investment when thelma is ready to retire 42... | "the amount of an investment will double in approximately 70 / p years , where p is the percent interest , compounded annually . if thelma invests $ 50,000 in a long - term cd that pays 5 percent interest , compounded annually , what will be the approximate total value of the investment when thelma is ready to retire 4... | a ) $ 280,000 , b ) $ 320,000 , c ) $ 360,000 , d ) $ 400,000 , e ) $ 540,000 | d | divide(const_3600, const_10) | divide(const_3600,const_10)| | general |
a man buys an article for $ 110 . and sells it for $ 125 . find the gain percent ? | "c . p . = $ 110 s . p . = $ 125 gain = $ 15 gain % = 15 / 110 * 100 = 13.6 % answer is b" | a ) 10 % , b ) 13.6 % , c ) 25 % , d ) 20 % , e ) 30 % | b | subtract(divide(125, divide(110, const_100)), const_100) | divide(n0,const_100)|divide(n1,#0)|subtract(#1,const_100)| | gain |
the slant height of a cone is 10 cm and radius of the base is 5 cm , find the curved surface of the cone ? | "π * 10 * 5 = 157 answer : d" | a ) 150 , b ) 155 , c ) 158 , d ) 157 , e ) 190 | d | multiply(multiply(const_pi, 5), 10) | multiply(n1,const_pi)|multiply(n0,#0)| | geometry |
excluding stoppages , the speed of a train is 54 kmph and including stoppages it is 36 kmph . of how many minutes does the train stop per hour ? | "t = 18 / 54 * 60 = 20 answer : a" | a ) 20 , b ) 17 , c ) 15 , d ) 18 , e ) 12 | a | subtract(const_60, multiply(const_60, divide(36, 54))) | divide(n1,n0)|multiply(#0,const_60)|subtract(const_60,#1)| | physics |
a rectangular region has a fence along three sides and a wall along the fourth side . the fenced side opposite the wall is twice the length of each of the other two fenced sides . if the area of the rectangular region is 200 square feet , what is the total length of the fence , in feet ? | two sides each = x the third = 2 x and the wall length is thus 2 x too x * 2 x = 2 x ^ 2 = 200 ie x ^ 2 = 100 ie x = 10 l = 20 w = 10 total lenght of fence = 2 * 10 + 20 = 40 my answer is d | ['a ) 10', 'b ) 20', 'c ) 30', 'd ) 40', 'e ) 80'] | d | add(add(multiply(const_2, add(multiply(add(const_1, const_1), const_2), const_1)), add(multiply(const_2, add(multiply(add(const_1, const_1), const_2), const_1)), multiply(const_2, add(multiply(add(const_1, const_1), const_2), const_1)))), divide(200, add(add(add(multiply(add(const_1, const_1), const_2), const_1), multi... | add(const_1,const_1)|multiply(#0,const_2)|add(#1,const_1)|multiply(#2,const_2)|add(#3,#3)|add(#2,#3)|add(#4,#3)|add(#5,#2)|divide(n0,#7)|add(#6,#8) | geometry |
if x ¤ y = ( x + y ) ^ 2 - ( x - y ) ^ 2 . then √ 3 ¤ √ 3 = | "x = √ 3 and y also = √ 3 applying the function ( √ 3 + √ 3 ) ^ 2 - ( √ 3 - √ 3 ) ^ 2 = ( 2 √ 3 ) ^ 2 - 0 = 4 x 3 = 12 . note : alternative approach is the entire function is represented as x ^ 2 - y ^ 2 = ( x + y ) ( x - y ) which can be simplified as ( x + y + x - y ) ( x + y - ( x - y ) ) = ( 2 x ) ( 2 y ) = 4 xy . ... | a ) 0 , b ) 5 , c ) 12 , d ) 15 , e ) 20 | c | power(add(sqrt(3), sqrt(3)), 2) | sqrt(n2)|add(#0,#0)|power(#1,n0)| | general |
a palindrome is a number that reads the same forward and backward , such as 124 . how many odd , 4 - digit numbers are palindromes ? | "a palindrome is a number that reads the same forward and backward . examples of four digit palindromes are 1221 , 4334 , 2222 etc you basically get to choose the first two digits and you repeat them in opposite order . say , you choose 45 as your first two digits . the next two digits are 54 and the number is 4554 . a... | a ) 40 , b ) 45 , c ) 50 , d ) 60 , e ) 2500 | d | divide(power(const_10, divide(4, const_2)), const_2) | divide(n1,const_2)|power(const_10,#0)|divide(#1,const_2)| | general |
hillary and eddy are climbing to the summit of mt . everest from a base camp 5,000 ft from the summit . when they depart for the summit at 06 : 00 , hillary climbs at a rate of 800 ft / hr with eddy lagging behind at a slower rate of 500 ft / hr . if hillary stops 1000 ft short of the summit and then descends at a rate... | solution : h stopped 1000 ft before reaching the final point , time taken to reach 4000 ft = 4000 / 800 = 5 hrs . this means she reached there at 11 : 00 . speed difference between them is 800 - 500 = 300 ft / hr so by the time h stops they have 1500 ft of distance so now here we use relative speed formula they both ar... | a ) 11.0 , b ) 12.0 , c ) 13.0 , d ) 14.0 , e ) 15.0 | b | add(6, add(divide(multiply(divide(subtract(divide(500, const_100), const_1), divide(800, const_1000)), subtract(divide(800, const_1000), divide(500, const_1000))), add(divide(500, const_1000), const_1)), divide(subtract(divide(500, const_100), const_1), divide(800, const_1000)))) | divide(n4,const_100)|divide(n3,const_1000)|divide(n4,const_1000)|add(#2,const_1)|subtract(#0,const_1)|subtract(#1,#2)|divide(#4,#1)|multiply(#6,#5)|divide(#7,#3)|add(#8,#6)|add(n1,#9) | general |
a car drives 60 miles on local roads at 20 mph , and 120 miles on the highway at 60 mph , what is the average speed of the entire trip ? | "so the answer is plainly a . . . . we have a general relation for speed , time and distance : v ( velocity ) * t ( time ) = d ( distance ) for first part we have d = 60 miles , and v = 20 mph so we can obtain time : 20 * t = 60 or t = 60 / 20 = 3 hours the needed time to cover 60 miles in the same way we should divide... | a ) 36 mph , b ) 40 mph , c ) 44 mph , d ) 52 mph , e ) 58 mph | a | divide(add(60, 120), add(divide(60, 20), divide(120, 60))) | add(n0,n2)|divide(n0,n1)|divide(n2,n3)|add(#1,#2)|divide(#0,#3)| | physics |
how many kg of pure salt must be added to 100 kg of 10 % solution of salt and water to increase it to a 20 % solution ? | "amount salt in 100 kg solution = 10 * 100 / 100 = 10 kg let x kg of pure salt be added then ( 10 + x ) / ( 100 + x ) = 20 / 100 100 + 10 x = 200 + 2 x 8 x = 100 x = 12.5 answer is d" | a ) 6.7 , b ) 1.3 , c ) 9.6 , d ) 12.5 , e ) 7.9 | d | divide(subtract(20, 10), subtract(const_1, divide(20, 100))) | divide(n2,n0)|subtract(n2,n1)|subtract(const_1,#0)|divide(#1,#2)| | general |
a train 200 m long passes a man , running at 6 km / hr in the same direction in which the train is going , in 10 seconds . the speed of the train is ? | "speed of the train relative to man = ( 200 / 10 ) m / sec = 20 m / sec . [ 20 * ( 18 / 5 ) ] km / hr = 72 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 6 ) km / hr . x - 6 = 72 = = > x = 78 km / hr . answer : c" | a ) 36 , b ) 50 , c ) 78 , d ) 66 , e ) 22 | c | divide(divide(subtract(200, multiply(multiply(6, const_0_2778), 6)), 6), const_0_2778) | multiply(n1,const_0_2778)|multiply(n1,#0)|subtract(n0,#1)|divide(#2,n1)|divide(#3,const_0_2778)| | physics |
if a triangle in the xy - coordinate system has vertices at ( - 2 , - 3 ) , ( 4 , - 3 ) and ( 28 , 7 ) , what is the area of the triangle ? | ( - 2 , - 3 ) is in 3 rd quadrant , ( 4 , - 3 ) is in 4 th quadrant while ( 28,7 ) is in 1 st quadrant . also , y coordinate of ( - 2 , - 3 ) and ( 4 , - 3 ) is same so they lie on same horizontal line . so base of triangle = sqrt [ ( - 3 + 3 ) ^ 2 + ( 4 + 2 ) ^ 2 ] = 6 and height of triangle = 7 - ( - 3 ) = 10 so area... | ['a ) 30', 'b ) 36', 'c ) 48', 'd ) 60', 'e ) 65'] | a | triangle_area_three_edges(sqrt(add(power(add(28, 2), const_2), power(add(7, 3), const_2))), sqrt(add(power(subtract(28, 4), const_2), power(add(7, 3), const_2))), sqrt(power(add(4, 2), const_2))) | add(n0,n4)|add(n1,n5)|add(n0,n2)|subtract(n4,n2)|power(#0,const_2)|power(#1,const_2)|power(#3,const_2)|power(#2,const_2)|add(#4,#5)|add(#6,#5)|sqrt(#7)|sqrt(#8)|sqrt(#9)|triangle_area_three_edges(#11,#12,#10) | geometry |
a man can row upstream at 20 kmph and downstream at 80 kmph , and then find the speed of the man in still water ? | "us = 20 ds = 80 m = ( 20 + 80 ) / 2 = 50 answer : b" | a ) 32 kmph , b ) 50 kmph , c ) 30 kmph , d ) 45 kmph , e ) 65 kmph | b | divide(add(20, 80), const_2) | add(n0,n1)|divide(#0,const_2)| | physics |
in a tree , 1 / 5 of the birds are robins while the rest are bluejays . if 1 / 2 of the robins are female and 2 / 5 of the bluejays are female , what fraction of the birds in the tree are male ? | "the fraction of birds that are male robins is ( 1 / 2 ) ( 1 / 5 ) = 1 / 10 . the fraction of birds that are male bluejays is ( 3 / 5 ) ( 4 / 5 ) = 12 / 25 . the total fraction of male birds is 1 / 10 + 12 / 25 = 29 / 50 . the answer is d ." | a ) 9 / 25 , b ) 12 / 25 , c ) 17 / 25 , d ) 29 / 50 , e ) 33 / 50 | d | add(multiply(divide(1, 5), divide(const_2.0, 2)), multiply(divide(2, 5), divide(1, 2))) | divide(n4,n1)|divide(n4,n3)|divide(n3,n1)|divide(n2,n3)|multiply(#0,#1)|multiply(#2,#3)|add(#4,#5)| | general |
salesperson a ' s compensation for any week is $ 360 plus 6 percent of the portion of a ' s total sales above $ 1,000 for that week . salesperson b ' s compensation for any week is 8 percent of b ' s total sales for that week . for what amount of total weekly sales would both salespeople earn the same compensation ? | "360 + 0.06 ( x - 1000 ) = 0.08 x or 360 - 60 = 0.08 x or 300 / 0.02 = x or 15000 = x answer : c" | a ) $ 21,000 , b ) $ 18,000 , c ) $ 15,000 , d ) $ 4,500 , e ) $ 4,000 | c | divide(add(divide(subtract(360, multiply(divide(6, const_100), 1,000)), subtract(divide(8, const_100), divide(6, const_100))), divide(subtract(360, multiply(divide(6, const_100), 1,000)), subtract(divide(8, const_100), divide(6, const_100)))), 1,000) | divide(n1,const_100)|divide(n3,const_100)|multiply(#0,n2)|subtract(#1,#0)|subtract(n0,#2)|divide(#4,#3)|add(#5,#5)|divide(#6,n2)| | general |
a crate measures 7 feet by 8 feet by 12 feet on the inside . a stone pillar in the shape of a right circular cylinder must fit into the crate for shipping so that it rests upright when the crate sits on at least one of its six sides . what is the radius , in feet , of the pillar with the largest volume that could still... | we can find the radius of all the three cases of cylinders . the only crux to find the answer faster is that : voulme is pi * r ^ 2 * h . the volume is a function of r ^ 2 . so r has to be the highest to find the largest volume . so r = 7 for the surface 8 * 12 face . volume = 343 pi answer d | ['a ) 2', 'b ) 4', 'c ) 6', 'd ) 7', 'e ) 12'] | d | divide(divide(multiply(multiply(8, 12), 7), 12), 8) | multiply(n1,n2)|multiply(n0,#0)|divide(#1,n2)|divide(#2,n1) | geometry |
in a fuel station the service costs $ 2.30 per vehicle and every liter of fuel costs $ 0.70 . assuming that you fill up 4 mini - vans and 2 trucks , what will be the total cost , if a mini - van ' s tank is 65 liters and a truck ' s tank is 120 % bigger and they are all empty ? | "the service cost of 4 vans and 2 trucks is 6 * 2.30 $ 13.80 the fuel in 4 vans is 4 * 65 = 260 liters the fuel in 2 trucks is 2 * 65 * 2.2 = 286 liters the total fuel ( vans + trucks ) = 546 liters the total fuel cost is 546 * 0.7 = $ 382.20 the total cost is $ 382.20 + $ 13.80 = $ 396 the answer is a ." | a ) $ 396 , b ) $ 400 , c ) $ 404 , d ) $ 408 , e ) $ 412 | a | add(add(multiply(multiply(add(divide(multiply(120, 65), const_100), 65), 2), 0.70), multiply(multiply(4, 65), 0.70)), multiply(2.30, add(4, 2))) | add(n2,n3)|multiply(n4,n5)|multiply(n2,n4)|divide(#1,const_100)|multiply(n1,#2)|multiply(n0,#0)|add(n4,#3)|multiply(n3,#6)|multiply(n1,#7)|add(#8,#4)|add(#9,#5)| | general |
evaluate : 6502.5 + 650.25 + 65.025 + 6.5025 + 0.65025 | "6502.5 650.25 65.025 6.5025 + 0.65025 - - - - - - - - - - - - - - - 7224.92775 answer is c ." | a ) 6819.59775 , b ) 6981.59775 , c ) 7224.92775 , d ) 6198.59775 , e ) 6891.59775 | c | subtract(6502.5, multiply(multiply(650.25, 65.025), 6.5025)) | multiply(n1,n2)|multiply(n3,#0)|subtract(n0,#1)| | general |
what is the smallest positive integer k such that the product of 450 x k is a perfect square ? | "a perfect square , is just an integer that can be written as the square of some other integer . for example 16 = 4 ^ 2 , is a perfect square . now , 450 = 3 ^ 2 * 5 ^ 2 * 3 , so if k = 3 then 450 k = ( 3 * 5 * 3 ) ^ 2 , which is a perfect square ( basically the least positive value of k must complete only the power of... | a ) 3 , b ) 9 , c ) 15 , d ) 25 , e ) 63 | a | add(const_3, const_4) | add(const_3,const_4)| | general |
log 3 n + log 5 n what is 3 digit number n that will be whole number | "no of values n can take is 1 5 ^ 3 = 125 answer : d" | a ) 629 , b ) 729 , c ) 829 , d ) 125 , e ) 727 | d | power(5, 3) | power(n1,n0)| | other |
the perimeter of a rectangular yard is completely surrounded by a fence that measures 14 meters . what is the length of the yard if the area of the yard is 6 meters squared ? | "perimeter of rectangular yard = 2 ( l + b ) = 14 - - > l + b = 7 area = l * b = 6 b = 7 - l l ( 7 - l ) = 6 7 l - l ^ 2 = 6 l ^ 2 - 7 l + 6 = 0 upon simplifying we get l = 1 or 6 . only 6 is there in the answer choice . answer : b" | a ) 8 , b ) 6 , c ) 2 , d ) 4 , e ) 3 | b | subtract(const_4, const_3) | subtract(const_4,const_3)| | geometry |
if a and b are positive integers , and a = 4 b + 16 , the greatest common divisor of a and b can not be | "if b is 2 , 4 , 8 , or 16 , then gcd of a and b is 2 , 4 , 8 , and 16 respectively . so , by poe the answer must be b . still : if b is a multiple of 9 , then a is 16 greater than a multiple of 9 , so not a multiple of 9 , so both of them can not be divisive by 9 . answer : b ." | a ) 2 , b ) 9 , c ) 4 , d ) 8 , e ) 16 | b | add(divide(16, 4), const_2) | divide(n1,n0)|add(#0,const_2)| | general |
find large number from below question the difference of two numbers is 1375 . on dividing the larger number by the smaller , we get 6 as quotient and the 15 as remainder | "let the smaller number be x . then larger number = ( x + 1375 ) . x + 1375 = 6 x + 15 5 x = 1360 x = 272 large number = 272 + 1375 = 1647 e" | a ) 1234 , b ) 1265 , c ) 1350 , d ) 1467 , e ) 1647 | e | multiply(divide(subtract(1375, 15), subtract(6, const_1)), 6) | subtract(n0,n2)|subtract(n1,const_1)|divide(#0,#1)|multiply(n1,#2)| | general |
3 pumps , working 8 hours a day , can empty a tank in 2 days . how many hours a day must 6 pumps work to empty the tank in 1 day ? | "3 pumps take 16 hrs total ( 8 hrs a day ) if 1 pump will be working then , it will need 16 * 3 = 48 hrs 1 pump need 48 hrs if i contribute 6 pumps then 48 / 6 = 8 hrs . answer : a" | a ) 8 , b ) 10 , c ) 11 , d ) 12 , e ) 13 | a | divide(multiply(multiply(3, 8), 2), 6) | multiply(n0,n1)|multiply(n2,#0)|divide(#1,n3)| | physics |
at a garage sale , all of the items were sold at different prices . if the price of a radio sold at the garage sale was both the 18 th highest price and the 25 th lowest price among the prices of the items sold , how many items were sold at the garage sale ? | "there were 17 items sold at a higher price than the radio and 24 items sold at a lower price than the radio . including the radio , there were 17 + 24 + 1 = 42 items sold . the answer is b ." | a ) 41 , b ) 42 , c ) 43 , d ) 44 , e ) 45 | b | subtract(add(25, 18), const_1) | add(n0,n1)|subtract(#0,const_1)| | other |
tom , working alone , can paint a room in 6 hours . peter and john , working independently , can paint the same room in 3 hours and 3 hours , respectively . tom starts painting the room and works on his own for one hour . he is then joined by peter and they work together for an hour . finally , john joins them and the ... | "tom paints 1 / 6 of the room in the first hour . tom and peter paint 1 / 6 + 1 / 3 = 1 / 2 of the room in the next hour for a total of 4 / 6 . the three people then paint the remaining 2 / 6 in a time of ( 2 / 6 ) / ( 5 / 6 ) = 2 / 5 hours peter worked for 7 / 5 hours so he painted 7 / 5 * 1 / 3 = 7 / 15 of the room .... | a ) 1 / 3 , b ) 2 / 5 , c ) 3 / 10 , d ) 7 / 15 , e ) 9 / 25 | d | divide(const_4, add(multiply(const_4, 3), const_1)) | multiply(n2,const_4)|add(#0,const_1)|divide(const_4,#1)| | physics |
if x ^ 2 + 1 / x ^ 2 = 3 , what is the value of x ^ 4 + 1 / x ^ 4 ? | "important : i notice that if we square x ² , we get x ⁴ , and if we square 1 / x ² , we get 1 / x ⁴ , so let ' s see what happens if we take the equation x ² + 1 / x ² = 3 andsquareboth sides : ( x ² + 1 / x ² ) ² = 9 so , ( x ² + 1 / x ² ) ( x ² + 1 / x ² ) = 9 expand to get : x ⁴ + 1 + 1 + 1 / x ⁴ = 9 simplify : x ⁴... | a ) 7 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | a | subtract(power(2, 2), 2) | power(n0,n0)|subtract(#0,n0)| | general |
find the value of 3.5 x [ ( 3.6 x 0.48 x 2.50 ) / ( 0.12 x 0.09 x 0.5 ) ] | answer 3.5 x [ ( 3.6 x 0.48 x 2.50 ) / ( 0.12 x 0.09 x 0.5 ) ] = 3.5 x [ ( 36 x 48 x 250 ) / ( 12 x 9 x 5 ) ] = 3.5 x 4 x 4 x 50 = 2800 correct option : c | a ) 800 , b ) 500 , c ) 2800 , d ) 1600 , e ) none | c | multiply(divide(multiply(multiply(3.6, 0.48), 2.5), multiply(multiply(0.12, 0.09), 0.5)), 3.5) | multiply(n1,n2)|multiply(n4,n5)|multiply(n3,#0)|multiply(n6,#1)|divide(#2,#3)|multiply(n0,#4) | general |
a and b can do a piece of work in 3 days . with the help of c they finish the work in 2 days . c alone can do that piece of work in ? | "c = 1 / 2 – 1 / 3 = 1 / 6 = > 6 days answer : a" | a ) 6 days , b ) 5 days , c ) 4 days , d ) 3 days , e ) 2 days | a | inverse(subtract(2, divide(2, 3))) | divide(n1,n0)|subtract(n1,#0)|inverse(#1)| | physics |
dacid obtained 91 , 65 , 82 , 67 and 85 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology . what are his average marks ? | "average = ( 91 + 65 + 82 + 67 + 85 ) / 5 = 78 answer : b" | a ) 29 , b ) 78 , c ) 39 , d ) 37 , e ) 75 | b | divide(add(add(add(add(91, 65), 82), 67), 85), divide(const_10, const_2)) | add(n0,n1)|divide(const_10,const_2)|add(n2,#0)|add(n3,#2)|add(n4,#3)|divide(#4,#1)| | general |
harkamal purchased 8 kg of grapes at the rate of 80 per kg and 9 kg of mangoes at the rate of 55 per kg . how much amount did he pay to the shopkeeper ? | "cost of 8 kg grapes = 80 × 8 = 640 . cost of 9 kg of mangoes = 55 × 9 = 495 . total cost he has to pay = 640 + 495 = 1135 . d )" | a ) a ) 1000 , b ) b ) 1055 , c ) c ) 1065 , d ) d ) 1135 , e ) e ) 1080 | d | add(multiply(8, 80), multiply(9, 55)) | multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)| | gain |
the average of 9 observations was 4 , that of the 1 st of 5 being 10 and that of the last 5 being 8 . what was the 5 th observation ? | "explanation : 1 to 9 = 9 * 4 = 36 1 to 5 = 5 * 10 = 50 5 to 9 = 5 * 8 = 40 5 th = 50 + 40 = 90 – 36 = 54 option a" | a ) 54 , b ) 12 , c ) 15 , d ) 17 , e ) 18 | a | subtract(add(multiply(10, 5), multiply(4, 5)), multiply(4, 9)) | multiply(n3,n4)|multiply(n1,n3)|multiply(n0,n1)|add(#0,#1)|subtract(#3,#2)| | general |
60 percent of movie theatres in town x have 1 screens or less . 20 % of those theatres sell an average of more than $ 300 worth of popcorn per showing . 56 percent of all the movie theatres in town x sell $ 300 or less of popcorn per showing . what percent of all the stores on the street have 4 or more screens and sell... | "lets take numbers here . assume that the total number of movie theaters in the town = 100 then number of movie theaters with 3 screens or less = 60 = > number of movie theaters with 4 screens or more = 40 movie theaters with 3 screens or less selling popcorn at more than $ 300 = 20 % of 60 = 12 number of movie theater... | a ) 12 , b ) 18 , c ) 32 , d ) 40 , e ) 44 | d | multiply(divide(20, 300), multiply(divide(20, 300), 300)) | divide(n2,n7)|multiply(n3,#0)|multiply(#0,#1)| | general |
a man can row his boat with the stream at 26 km / h and against the stream in 12 km / h . the man ' s rate is ? | "ds = 26 us = 12 s = ? s = ( 26 - 12 ) / 2 = 7 kmph answer : d" | a ) 1 kmph , b ) 4 kmph , c ) 5 kmph , d ) 7 kmph , e ) 8 kmph | d | divide(subtract(26, 12), const_2) | subtract(n0,n1)|divide(#0,const_2)| | gain |
in n is a positive integer less than 200 , and 10 n / 60 is an integer , then n has how many different positive prime factors ? | "( a ) . 10 n / 60 must be an integer . = > 1 n / 6 must be an integer . hence n must be a multiple of 2 * 3 . = > n has 2 different prime integers ." | a ) 2 , b ) 3 , c ) 5 , d ) 6 , e ) 8 | a | add(divide(add(const_1, const_4), divide(divide(divide(60, const_2), const_2), const_3)), const_2) | add(const_1,const_4)|divide(n2,const_2)|divide(#1,const_2)|divide(#2,const_3)|divide(#0,#3)|add(#4,const_2)| | general |
a retailer bought a machine at a wholesale price of $ 117 and later on sold it after a 10 % discount of the retail price . if the retailer made a profit equivalent to 20 % of the whole price , what is the retail price of the machine ? | "my solution : wholesale price = 117 retail price , be = x he provides 10 % discount on retail price = x - 10 x / 100 this retail price = 20 % profit on wholesale price x - 10 x / 100 = 117 + 1 / 5 ( 117 ) x = 156 ; answer : e" | a ) 81 , b ) 100 , c ) 120 , d ) 135 , e ) 159 | e | divide(multiply(add(117, divide(multiply(117, 20), const_100)), const_100), multiply(multiply(const_3, const_3), 10)) | multiply(n0,n2)|multiply(const_3,const_3)|divide(#0,const_100)|multiply(n1,#1)|add(n0,#2)|multiply(#4,const_100)|divide(#5,#3)| | gain |
what sum of money put at c . i amounts in 2 years to rs . 8000 and in 3 years to rs . 9261 ? | 8000 - - - - 1261 100 - - - - ? = > 15.76 % x * 115.76 / 100 * 115.76 / 100 = 8000 x * 1.34 = 8000 x = 8000 / 1.34 = > 5969.72 answer : a | a ) 5969.72 , b ) 8877.72 , c ) 2877.72 , d ) 2678.72 , e ) 1011.72 | a | divide(8000, power(add(subtract(divide(9261, 8000), const_1), const_1), const_2)) | divide(n3,n1)|subtract(#0,const_1)|add(#1,const_1)|power(#2,const_2)|divide(n1,#3) | general |
carol spends 1 / 4 of her savings on a stereo and 1 / 6 less than she spent on the stereo for a television . what fraction of her savings did she spend on the stereo and television ? | "total savings = s amount spent on stereo = ( 1 / 4 ) s amount spent on television = ( 1 - 1 / 6 ) ( 1 / 4 ) s = ( 5 / 6 ) * ( 1 / 4 ) * s = ( 5 / 24 ) s ( stereo + tv ) / total savings = s ( 1 / 4 + 5 / 24 ) / s = 11 / 24 answer : e" | a ) 1 / 4 , b ) 2 / 7 , c ) 5 / 12 , d ) 1 / 2 , e ) 11 / 24 | e | divide(1, 4) | divide(n0,n1)| | general |
if x and y are both odd prime numbers and x < y , how many distinct positive integer t factors does 2 xy have ? | since 2 xy prime t factors are x ^ 1 * y ^ 1 * 2 ^ 1 , its total number or factors must be ( 1 + 1 ) ( 1 + 1 ) ( 1 + 1 ) = 2 ^ 3 = 8 . thus , i think d would be the correct answer . | a ) 3 , b ) 4 , c ) 6 , d ) 8 , e ) 12 | d | multiply(multiply(2, add(const_1, const_1)), add(const_1, const_1)) | add(const_1,const_1)|multiply(n0,#0)|multiply(#0,#1)| | other |
how many kg of pure salt must be added to 30 kg of 2 % solution of salt and water to increase it to 10 % solution ? | "amount of salt in 30 kg solution = [ ( 20 / 100 ) * 30 ] kg = 0.6 kg let x kg of pure salt be added then , ( 0.6 + x ) / ( 30 + x ) = 10 / 100 60 + 100 x = 300 + 10 x 90 x = 240 x = 8 / 3 . answer is e ." | a ) 3 / 5 , b ) 4 / 3 , c ) 5 / 7 , d ) 2 / 3 , e ) 8 / 3 | e | divide(subtract(10, 2), subtract(const_1, divide(10, 30))) | divide(n2,n0)|subtract(n2,n1)|subtract(const_1,#0)|divide(#1,#2)| | general |
lionel left his house and walked towards walt ' s house , 48 miles away . two hours later , walt left his house and ran towards lionel ' s house . if lionel ' s speed was 3 miles per hour and walt ' s 3 miles per hour , how many miles had lionel walked when he met walt ? | "in the first 2 hours lionel at the rate of 3 miles per hour covered distance = rate * time = 3 * 2 = 6 miles . so , the distance between him and walt was 48 - 6 = 42 miles when walt left his house . now , their combined rate to cover this distance was 3 + 3 = 6 miles per hour , hence they will meet ( they will cover t... | a ) 12 , b ) 16 , c ) 20 , d ) 27 , e ) 28 | d | multiply(3, add(divide(subtract(48, multiply(3, 3)), add(3, const_2.0)), 3)) | add(n1,n2)|multiply(n1,n1)|subtract(n0,#1)|divide(#2,#0)|add(#3,const_2)|multiply(n1,#4)| | physics |
calculate the value of y : 5 y ^ 2 - 8 y + 55 = 59 | using the elimination method substitute options for x and find the correct option . answer : c | a ) 7 , b ) 1 , c ) 2 , d ) 9 , e ) 8 | c | divide(add(sqrt(subtract(power(8, const_2), multiply(multiply(5, subtract(55, 59)), const_4))), 8), multiply(5, 2)) | multiply(n0,n1)|power(n2,const_2)|subtract(n3,n4)|multiply(n0,#2)|multiply(#3,const_4)|subtract(#1,#4)|sqrt(#5)|add(n2,#6)|divide(#7,#0) | general |
the average ( arithmetic mean ) of all scores on a certain algebra test was 90 . if the average of the 8 male students ’ grades was 85 , and the average of the female students ’ grades was 92 , how many female students took the test ? | "total marks of male = m total marks of female = f number of males = 8 number of females = f given : ( m + f ) / ( 8 + f ) = 90 - - - - - - - - - - - - - 1 also given , m / 8 = 85 thus m = 680 - - - - - - - - - 2 also , f / f = 92 thus f = 92 f - - - - - - - - - 3 put 2 and 3 in 1 : we get ( 680 + 92 f ) / ( 8 + f ) = ... | a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 20 | e | divide(subtract(multiply(90, 8), multiply(85, 8)), subtract(92, 90)) | multiply(n0,n1)|multiply(n1,n2)|subtract(n3,n0)|subtract(#0,#1)|divide(#3,#2)| | general |
a pipe can fill a cistern in 55 minutes . find the time in which 1 / 11 part of the cistern will be filled ? | "full cistern filled in = 55 minutes 1 / 11 part filled in = 55 * 1 / 11 = 5 minutes answer is a" | a ) 5 min , b ) 2 min , c ) 3 min , d ) 1 min , e ) 10 min | a | multiply(55, divide(1, 11)) | divide(n1,n2)|multiply(n0,#0)| | physics |
if the perimeter of a rectangular garden is 600 m , its length when its breadth is 95 m is ? | 2 ( l + 95 ) = 600 = > l = 205 m answer : d | ['a ) 286 m', 'b ) 899 m', 'c ) 200 m', 'd ) 205 m', 'e ) 187 m'] | d | subtract(divide(600, const_2), 95) | divide(n0,const_2)|subtract(#0,n1) | physics |
two trains of equal are running on parallel lines in the same direction at 46 km / hr and 36 km / hr . the faster train passes the slower train in 72 sec . the length of each train is ? | "let the length of each train be x m . then , distance covered = 2 x m . relative speed = 46 - 36 = 10 km / hr . = 10 * 5 / 18 = 25 / 9 m / sec . 2 x / 72 = 25 / 9 = > x = 100 . answer : e" | a ) 50 , b ) 26 , c ) 27 , d ) 28 , e ) 100 | e | divide(multiply(72, divide(multiply(subtract(46, 36), const_1000), const_3600)), const_2) | subtract(n0,n1)|multiply(#0,const_1000)|divide(#1,const_3600)|multiply(n2,#2)|divide(#3,const_2)| | general |
6 workers should finish a job in 8 days . after 3 days came 4 workers join them . how many days d do they need to finish the same job ? | "let rate of one worker be r = > ( 6 * r ) * 8 = 1 ( rate * time = work ) = > r = 1 / 48 = > work remaining after 3 days 1 - ( 3 * 6 ) / 48 = 30 / 48 after 4 ppl joined in ( ( 6 + 4 ) * time ) / 48 = 30 / 48 time d = 3 days to finish the task imo a" | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | divide(subtract(multiply(6, 8), multiply(3, 6)), add(6, 4)) | add(n0,n3)|multiply(n0,n1)|multiply(n0,n2)|subtract(#1,#2)|divide(#3,#0)| | physics |
what is the ratio between perimeters of two squares one having 11 times the diagonal then the other ? | "d = 11 d d = d a √ 2 = 11 d a √ 2 = d a = 11 d / √ 2 a = d / √ 2 = > 11 : 1 answer : d" | a ) 3 : 8 , b ) 3 : 6 , c ) 3 : 7 , d ) 11 : 1 , e ) 3 : 3 | d | divide(11, divide(11, 11)) | divide(n0,n0)|divide(n0,#0)| | geometry |
a student has to obtain 40 % of the total marks to pass . he got 150 marks and failed by 50 marks . the maximum marks are ? | "let the maximum marks be x then , 40 % of x = 150 + 50 40 x / 100 = 200 x = 500 answer is a" | a ) 500 , b ) 650 , c ) 400 , d ) 575 , e ) 610 | a | divide(add(150, 50), divide(40, const_100)) | add(n1,n2)|divide(n0,const_100)|divide(#0,#1)| | general |
david has $ 1,800 at the beginning of his trip , after spending money , he still has exactly $ 800 less than he spent on the trip . how much money does john still have ? | suppose total money spent = x not spend ( money he still has ) = x - 800 x + x - 800 = 1800 x = 1300 money not spend = 1300 - 800 = 500 answer : b | a ) $ 200 , b ) $ 500 , c ) $ 600 , d ) $ 800 , e ) $ 1,200 | b | subtract(divide(add(add(multiply(const_100, const_10), 800), 800), const_2), 800) | multiply(const_10,const_100)|add(n1,#0)|add(n1,#1)|divide(#2,const_2)|subtract(#3,n1) | general |
what number is 75 more than three - fourth of itself ? | "3 / 4 x + 75 = x that means 75 = 1 / 4 x x = 75 * 4 = 300 c is the answer" | a ) 138 , b ) 225 , c ) 300 , d ) 324 , e ) 336 | c | divide(75, subtract(const_1, divide(const_1, const_4))) | divide(const_1,const_4)|subtract(const_1,#0)|divide(n0,#1)| | general |
kamal was 4 times as old as his son 8 years ago . after 8 years , kamal will be twice as old as his son . find out the present age of kamal . | let age of the son before 8 years = x then , age of kamal before 8 years ago = 4 x after 8 years , kamal will be twice as old as his son ⇒ 4 x + 16 = 2 ( x + 16 ) ⇒ x = 8 present age of kamal = 4 x + 8 = 4 × 8 + 8 = 40 answer is c . | a ) 30 , b ) 35 , c ) 40 , d ) 45 , e ) 50 | c | add(multiply(divide(multiply(4, 4), const_2), const_4), 8) | multiply(n0,n0)|divide(#0,const_2)|multiply(#1,const_4)|add(n1,#2) | general |
80 % of a number is added to 120 , the result is the same number . find the number ? | "( 80 / 100 ) * x + 120 = x 2 x = 1200 x = 600 answer : a" | a ) 600 , b ) 288 , c ) 1687 , d ) 997 , e ) 121 | a | divide(120, divide(120, const_100)) | divide(n1,const_100)|divide(n1,#0)| | gain |
find the least number of complete years in which a sum of money put out at 13 % compound interest will be more than double of itself ? | "6 years answer : b" | a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | b | floor(add(divide(log(const_2), log(add(const_1, divide(13, const_100)))), const_1)) | divide(n0,const_100)|log(const_2)|add(#0,const_1)|log(#2)|divide(#1,#3)|add(#4,const_1)|floor(#5)| | general |
kiran has 85 currency notes in all , some of which were of rs . 100 denomination and the remaining of rs . 50 denomination . the total amount of all these currency notes was rs . 5000 . how much amount did she have in the denomination of rs . 50 ? a . b . c . none of these d . answer : option b explanation : | let the number of 50 – rupee notes be x . then , the number of 100 - rupee notes = ( 85 – x ) 50 x + 100 ( 85 – x ) = 5000 x + 2 ( 85 – x ) = 100 = x = 70 so , required amount = rs . ( 50 x 70 ) = rs . 3500 answer : b | a ) 1900 , b ) 3500 , c ) 4000 , d ) 2000 , e ) 2500 | b | multiply(subtract(85, divide(subtract(5000, multiply(50, 85)), 50)), 50) | multiply(n0,n2)|subtract(n3,#0)|divide(#1,n2)|subtract(n0,#2)|multiply(n2,#3) | general |
a man sells a car to his friend at 12 % loss . if the friend sells it for rs . 54000 and gains 20 % , the original c . p . of the car was : | "explanation : s . p = rs . 54,000 . gain earned = 20 % c . p = rs . [ 100 / 120 ã — 54000 ] = rs . 45000 this is the price the first person sold to the second at at loss of 12 % . now s . p = rs . 45000 and loss = 12 % c . p . rs . [ 100 / 88 ã — 45000 ] = rs . 51136.36 correct option : c" | a ) rs . 21136.36 , b ) rs . 31136.36 , c ) rs . 51136.36 , d ) rs . 61136.36 , e ) none of these | c | divide(multiply(divide(multiply(54000, const_100), add(const_100, 20)), const_100), subtract(const_100, 12)) | add(n2,const_100)|multiply(n1,const_100)|subtract(const_100,n0)|divide(#1,#0)|multiply(#3,const_100)|divide(#4,#2)| | gain |
the breath of a rectangle is three - fifths of the radius of a circle . the radius of the circle is equal to the side of the square , whose area is 2025 sq . units . what is the area ( in sq . units ) of the rectangle if the rectangle if the length is 10 units ? | given that the area of the square = 2025 sq . units = > side of square = √ 2025 = 45 units the radius of the circle = side of the square = 45 units breath of the rectangle = 3 / 5 * 45 = 27 units given that length = 10 units area of the rectangle = lb = 27 * 10 = 270 sq . units answer : b | ['a ) 370 sq . units', 'b ) 270 sq . units', 'c ) 170 sq . units', 'd ) 470 sq . units', 'e ) 570 sq . units'] | b | rectangle_area(10, divide(multiply(square_edge_by_area(2025), const_3), add(const_4, const_1))) | add(const_1,const_4)|square_edge_by_area(n0)|multiply(#1,const_3)|divide(#2,#0)|rectangle_area(n1,#3) | geometry |
there are 3 workers in the factory a , b , c . the age of a is 4 more than the age of b and the age of b is 4 more than the age of c . if the sum of the square of the ages is 1202 . what is the sum of their ages ? | let age of b = x ; age of a = x + 4 ; age of c = x - 4 ; sum of squares of ages = > 3 x ^ 2 + 32 = > x = 19.7 = > 20 sum of ages = 60 answer : c | a ) 50 , b ) 55 , c ) 60 , d ) 65 , e ) 70 | c | multiply(sqrt(divide(subtract(1202, add(power(4, const_2), power(4, const_2))), 3)), 3) | power(n1,const_2)|add(#0,#0)|subtract(n3,#1)|divide(#2,n0)|sqrt(#3)|multiply(n0,#4) | general |
if 12 men can reap 120 acres of land in 36 days , how many acres of land can 54 men reap in 54 days ? | "12 men 120 acres 36 days 54 men ? 54 days 120 * 54 / 12 * 54 / 36 10 * 54 * 3 / 2 54 * 15 = 810 answer : c" | a ) 186 , b ) 188 , c ) 810 , d ) 167 , e ) 861 | c | multiply(120, multiply(divide(54, 12), divide(54, 36))) | divide(n3,n0)|divide(n4,n2)|multiply(#0,#1)|multiply(n1,#2)| | physics |
how many cubes of 4 cm edge can be cut out of a cube of 28 cm edge | "explanation : number of cubes = ( 28 x 28 x 28 ) / ( 4 x 4 x 4 ) = 343 answer : a" | a ) 343 , b ) 232 , c ) 216 , d ) 484 , e ) none of these | a | divide(volume_cube(28), volume_cube(divide(4, const_100))) | divide(n0,const_100)|volume_cube(n1)|volume_cube(#0)|divide(#1,#2)| | probability |
the area of a square is equal to four times the area of a rectangle of dimensions 52 cm * 13 cm . what is the perimeter of the square ? | "area of the square = s * s = 4 ( 52 * 13 ) = > s = 52 cm perimeter of the square = 4 * 52 = 208 cm . answer : option b" | a ) 200 , b ) 208 , c ) 210 , d ) 220 , e ) 228 | b | multiply(sqrt(multiply(rectangle_area(52, 13), divide(13, const_2))), const_4) | divide(n1,const_2)|rectangle_area(n0,n1)|multiply(#0,#1)|sqrt(#2)|multiply(#3,const_4)| | geometry |
a sum of money is to be distributed among a , b , c , d in the proportion of 5 : 2 : 4 : 3 . if c gets rs . 1100 more than d , what is b ' s share ? | "let the shares of a , b , c and d be rs . 5 x , rs . 2 x , rs . 4 x and rs . 3 x respectively . then , 4 x - 3 x = 1100 x = 1100 . b ' s share = rs . 2 x = rs . ( 2 x 1100 ) = rs . 2200 . answer = c" | a ) rs . 500 , b ) rs . 1500 , c ) rs . 2200 , d ) rs . 2500 , e ) none of the above | c | multiply(multiply(subtract(4, 3), 1100), 3) | subtract(n2,n3)|multiply(n4,#0)|multiply(n3,#1)| | general |
find the sum lend at c . i . at 5 p . c per annum will amount to rs . 500 in 2 years ? | explanation : 500 = p ( 21 / 20 ) 2 p = 453.51 answer : c | a ) 221 , b ) 287 , c ) 453 , d ) 288 , e ) 171 | c | divide(500, power(add(const_1, divide(5, const_100)), const_2)) | divide(n0,const_100)|add(#0,const_1)|power(#1,const_2)|divide(n1,#2) | general |
two trains are moving in the same direction at 72 kmph and 36 kmph . the faster train crosses a man in the slower train in 20 seconds . find the length of the faster train ? | "relative speed = ( 72 - 36 ) * 5 / 18 = 2 * 5 = 10 mps . distance covered in 27 sec = 20 * 10 = 200 m . the length of the faster train = 200 m . answer : c" | a ) 270 , b ) 277 , c ) 200 , d ) 257 , e ) 271 | c | multiply(divide(subtract(72, 36), const_3_6), 20) | subtract(n0,n1)|divide(#0,const_3_6)|multiply(n2,#1)| | physics |
x varies inversely as square of y . given that y = 3 for x = 1 . the value of x for y = 4 will be equal to : | "explanation : solution : given x = k / y ^ 2 , where k is constant . now , y = 3 and x = 1 gives k = 9 . . ' . x = 9 / y ^ 2 = > x = 9 / 4 ^ 2 = 9 / 16 answer : e" | a ) 3 , b ) 6 , c ) 1 / 9 , d ) 1 / 3 , e ) 9 / 16 | e | divide(multiply(1, power(3, const_2)), power(4, const_2)) | power(n0,const_2)|power(n2,const_2)|multiply(n1,#0)|divide(#2,#1)| | general |
the sum of all solutions for x in the equation x ^ 2 – 4 x + 14 = | x – 4 | + 8 is equal to : | "x ^ 2 - 4 x + 6 = | x - 4 | rhs can be - ve or + ve x ^ 2 - 5 x + 10 = 0 x ^ 2 - 3 x + 2 = 0 x = 5 , 2,1 we test all 3 values in original equation , all ok . thus , sum = 5 + 2 + 1 = 8 ans ( b )" | a ) – 7 , b ) 8 , c ) 10 , d ) 12 , e ) 14 | b | multiply(8, 4) | multiply(n3,n4)| | general |
r is the set of positive odd integers less than 50 , and s is the set of the squares of the integers in r . how many elements does the intersection of r and s contain ? | "squares < 50 { 1 , 4,9 , 16,25 , 36,49 } s = { 1,9 , 25,49 } r = { 1,3 , . . . . . 49 } hence c ." | a ) none , b ) two , c ) four , d ) five , e ) seven | c | subtract(subtract(50, const_4), const_4) | subtract(n0,const_4)|subtract(#0,const_4)| | physics |
in a sports club with 27 members , 17 play badminton and 19 play tennis and 2 do not play either . how many members play both badminton and tennis ? | "let x play both badminton and tennis so 17 - x play only badminton and 19 - x play only tennis . 2 play none and there are total 30 students . hence , ( 17 - x ) + ( 19 - x ) + x + 2 = 27 38 - 2 x + x = 27 38 - x = 27 x = 11 so 8 members play both badminton and tennis . e" | a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | e | subtract(add(add(17, 19), 2), 27) | add(n1,n2)|add(n3,#0)|subtract(#1,n0)| | other |
a & b can separately do a piece of work in 20 & 15 days . they worked together for 6 days , after which b was replaced by c . the work was finished in next 5 days . the no . of days in which c alone could do the work is | b worked for 6 days = > in 6 days b completed = 6 x 1 15 = 2 5 th work a worked for first 6 days and later 4 days = > in 10 days a completed = 10 x 1 20 = 1 2 of the work total work done by a and b = 2 5 + 1 2 = 4 + 5 10 = 9 10 balance work = 1 - 9 10 = 1 10 th of total work balance work was completed by c in 5 days c ... | a ) 10 days , b ) 20 days , c ) 30 days , d ) 40 days , e ) 50 days | e | divide(inverse(subtract(divide(subtract(const_1, multiply(add(divide(const_1, 20), divide(const_1, 15)), 6)), 5), divide(const_1, 20))), const_2) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|multiply(n2,#2)|subtract(const_1,#3)|divide(#4,n3)|subtract(#5,#0)|inverse(#6)|divide(#7,const_2) | physics |
a string of 13 light bulbs is wired in such a way that if any individual light bulb fails , the entire string fails . if for each individual light bulb the probability of failing during time period t is 0.02 , what is the probability that the string of light bulbs will fail during the time period t ? | "the string of light bulbs will fail if at least one light bulb fails . so , let ' s find the probability of the opposite event and subtract that value from 1 . the opposite event is when none of the 13 light bulbs fails , since the probability of each light bulb not to fail is 1 - 0.02 = 0.98 the the probability that ... | a ) 0.02 , b ) ( 0.02 ) ^ 13 , c ) 1 - ( 0.02 ) ^ 13 , d ) ( 0.98 ) ^ 13 , e ) 1 - ( 0.98 ) ^ 13 | e | subtract(const_1, power(subtract(const_1, 0.02), 13)) | subtract(const_1,n1)|power(#0,n0)|subtract(const_1,#1)| | probability |
145 liters of a mixture of milk and water contains in the ratio 3 : 2 . how much water should now be added so that the ratio of milk and water becomes 3 : 4 ? | "milk = 3 / 5 * 145 = 87 liters water = 58 liters 87 : ( 58 + p ) = 3 : 4 174 + 3 p = 348 = > p = 58 58 liters of water are to be added for the ratio become 3 : 4 . answer : e" | a ) 12 liters , b ) 32 liters , c ) 41 liters , d ) 50 liters , e ) 58 liters | e | multiply(divide(145, add(3, 2)), 2) | add(n1,n2)|divide(n0,#0)|multiply(n2,#1)| | general |
edward ' s home have 3 taps , one tap , working alone at its maximum constant rate , can fill a certain sink in 270 seconds . a second tap working alone at its maximum constant rate , can fill a certain sink in 274 seconds . a third tap , working alone at its maximum constant rate , can fill the same sink in 276 second... | tap 1 : 270 secs tap 2 : 274 secs tap 3 : 276 secs considering the average of these 3 taps : 273.33 secs . so 1 tap can fill the tank in 273.33 secs , so 3 taps can fill the tank in 273.33 / 3 = 91.11 secs . closest answer is a . | a ) 91 seconds , b ) 102 seconds , c ) 177 seconds , d ) 200 seconds , e ) 270 seconds | a | divide(add(add(divide(270, 3), divide(274, 3)), divide(276, 3)), 3) | divide(n1,n0)|divide(n2,n0)|divide(n3,n0)|add(#0,#1)|add(#3,#2)|divide(#4,n0) | physics |
the overall age of x and y is 12 year greater than the overall age of y and z . z is how many decades younger that x ? | "c 12 ( x + y ) – ( y + z ) = 12 x – z = 12" | a ) 11 , b ) 15 , c ) 12 , d ) 17 , e ) 19 | c | divide(12, const_1) | divide(n0,const_1)| | general |
the compound ratio of 1 : 3 , 1 : 2 and 1 : 5 ? | 1 / 3 * 1 / 2 * 1 / 5 = 1 / 30 1 : 30 answer : a | a ) 1 : 30 , b ) 1 : 5 , c ) 1 : 7 , d ) 1 : 1 , e ) 1 : 6 | a | divide(divide(multiply(1, 1), multiply(3, 2)), divide(multiply(1, 1), multiply(2, 5))) | multiply(n0,n2)|multiply(n1,n3)|multiply(n2,n4)|multiply(n3,n5)|divide(#0,#1)|divide(#2,#3)|divide(#4,#5)| | other |
what is rate of interest if principal . amount be 400 , simple interest 140 and time 2 year . | "s . i = ( p * r * t ) / 100 140 = 800 r / 100 r = 140 / 8 = 17.5 % answer b" | a ) 10 , b ) 17.5 , c ) 25 , d ) 12 , e ) 14.5 | b | multiply(divide(140, multiply(400, 2)), const_100) | multiply(n0,n2)|divide(n1,#0)|multiply(#1,const_100)| | gain |
the population of a village is 12000 . it increases annually at the rate of 24 % p . a . what will be its population after 2 years ? | "formula : ( after = 100 denominator ago = 100 numerator ) 12000 × 124 / 100 × 124 / 100 = 18451 b" | a ) 14567 , b ) 18451 , c ) 18789 , d ) 18786 , e ) 14400 | b | multiply(12000, power(add(const_1, divide(24, const_100)), 2)) | divide(n1,const_100)|add(#0,const_1)|power(#1,n2)|multiply(n0,#2)| | gain |
a school has received 40 % of the amount it needs for a new building by receiving a donation of $ 400 each from people already solicited . people already solicited represent 40 % of the people from whom the school will solicit donations . how much average contribution is requited from the remaining targeted people to c... | "let the amount school needs = x let total people school plans to solicit = t school has received 60 % of x = > ( 3 / 5 ) x people already solicited = 40 % of t = > ( 2 / 5 ) t now , as per the information given in the question : ( 3 / 5 ) x = $ 400 . ( 2 / 5 ) . t - - - - - - - - - - - - - - - - - - - - - - - - - - - ... | a ) $ 200 , b ) $ 177.78 , c ) $ 100 , d ) $ 277.78 , e ) $ 77.78 | e | divide(multiply(divide(multiply(divide(40, const_100), 400), divide(40, const_100)), divide(40, const_100)), divide(40, const_100)) | divide(n2,const_100)|divide(n0,const_100)|multiply(n1,#0)|divide(#2,#1)|multiply(#3,#0)|divide(#4,#1)| | general |
an engineer designed a ball so that when it was dropped , it rose with each bounce exactly one - half as high as it had fallen . the engineer dropped the ball from a 16 - meter platform and caught it after it had traveled 40 meters . how many times did the ball bounce ? | "division of total distance travelled will be 16 + 16 + 8 ans : 2" | a ) 2 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | a | divide(divide(16, const_2), const_2) | divide(n0,const_2)|divide(#0,const_2)| | general |
a trader sold an item at a loss of 35 % . if he had increased the price by $ 80 he would have made a gain of 5 % . what is the cost price of the item ? | "let c . p . be $ x then 105 % of x - 65 % of x = 100 40 % of x = 100 40 / 100 * x = 100 40 x = 10 000 x = 250 answer is e" | a ) $ 245 , b ) $ 150 , c ) $ 260 , d ) $ 220 , e ) $ 250 | e | divide(multiply(80, const_100), multiply(subtract(add(const_1, divide(5, const_100)), subtract(const_1, divide(35, const_100))), const_100)) | divide(n2,const_100)|divide(n0,const_100)|multiply(n1,const_100)|add(#0,const_1)|subtract(const_1,#1)|subtract(#3,#4)|multiply(#5,const_100)|divide(#2,#6)| | gain |
a and b can do a piece of work in 10 days , b and c in 15 days , c and a in 12 days . how long will c take to do it ? | "2 c = 1 / 15 + 1 / 12 – 1 / 10 = 1 / 20 c = 1 / 40 = > 40 days answer : d" | a ) 70 days , b ) 55 days , c ) 24 days , d ) 40 days , e ) 11 days | d | divide(multiply(10, const_3), subtract(divide(add(divide(multiply(10, const_3), 12), add(divide(multiply(10, const_3), 10), divide(multiply(10, const_3), 15))), const_2), divide(multiply(10, const_3), 10))) | multiply(n0,const_3)|divide(#0,n0)|divide(#0,n1)|divide(#0,n2)|add(#1,#2)|add(#4,#3)|divide(#5,const_2)|subtract(#6,#1)|divide(#0,#7)| | physics |
a certain telephone company company charges 3.10 cents for the first 1 / 5 of a minute plus 0.40 cents for each additional 1 / 5 of a minute . what would this company charge for a talk time of 8 minutes long ? | first 1 / 5 minute charge = 3.10 cents rest of the minute = 8 - ( 1 / 5 ) = 39 / 5 charge for the rest of the minute = 39 * 0.4 = 15.6 total charge = 3.10 + 15.6 = 18.7 cents answer is d . | a ) 15.60 , b ) 16.0 , c ) 17.8 , d ) 18.7 , e ) 19.1 | d | add(add(multiply(0.4, subtract(5, 1)), multiply(multiply(0.4, subtract(8, 1)), 5)), 3.1) | subtract(n2,n1)|subtract(n6,n1)|multiply(n3,#0)|multiply(n3,#1)|multiply(n2,#3)|add(#2,#4)|add(n0,#5) | general |
a certain number of men complete a piece of work in 60 days . if there were 8 men more , the work could be finished in 10 days less . how many men were originally there ? | originally 1 et there be x men . more men , less days ( x + 8 ) : x ∷ 60 : 50 so , x + 8 / x = 60 / 50 or x = 40 . answer : d | a ) 30 , b ) 32 , c ) 36 , d ) 40 , e ) 45 | d | divide(multiply(8, subtract(60, 10)), 10) | subtract(n0,n2)|multiply(n1,#0)|divide(#1,n2) | physics |
how many bricks , each measuring 25 cm x 11.25 cm x 6 cm , will be needed to build a wall of 7.5 m x 6 m x 22.5 cm ? | "number of bricks = volume of wall / volume of bricks = 750 x 600 x 22.5 / 25 x 11.25 x 6 = = 6000 answer : a" | a ) 6000 , b ) 6410 , c ) 6440 , d ) 6500 , e ) 6800 | a | divide(multiply(multiply(multiply(7.5, const_100), multiply(6, const_100)), 22.5), multiply(multiply(25, 11.25), 6)) | multiply(n3,const_100)|multiply(n4,const_100)|multiply(n0,n1)|multiply(#0,#1)|multiply(n2,#2)|multiply(n5,#3)|divide(#5,#4)| | physics |
if the function q is defined by the formula q = 5 w / ( 4 vf ( z ^ 2 ) ) , by what factor will q be multiplied if w is quadrupled , f is doubled , and z is tripled ? | "we just need to find the factor thats all , w - > quadrupled - > 4 w f - > doubled - > 2 f z - > tripled - > 3 z hence , z ^ 2 = 9 z ^ 2 w is in numerator , and f * z in denominator . hence , additional factor being introduced = 4 / 2 * 9 = 4 / 18 = 2 / 9 = b" | a ) 1 / 9 , b ) 2 / 9 , c ) 4 / 9 , d ) 3 / 9 , e ) 2 / 27 | b | divide(4, multiply(2, power(const_3, 2))) | power(const_3,n2)|multiply(n2,#0)|divide(n1,#1)| | general |
what is the remainder when 43 ^ 91 is divided by 5 ? | "the units digit of the exponents of 3 cycle in a group of 4 : { 3 , 9 , 7 , 1 } 91 has the form 4 k + 3 so the units digit of 43 ^ 91 is 7 . the remainder when dividing by 5 is 2 . the answer is c ." | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | c | subtract(divide(5, const_2), multiply(43, 43)) | divide(n2,const_2)|multiply(n0,n0)|subtract(#0,#1)| | general |
if a and b are integers and ( a * b ) ^ 5 = 64 y , y could be : | "distribute the exponent . a ^ 5 * b ^ 5 = 64 y find the prime factorization of 64 . this is 2 ^ 5 * 2 ^ 1 we need 2 ^ 4 ( or some other power of 2 that will give us a multiple of 2 ^ 5 as our second term ) . 2 ^ 4 = 16 the answer is a ." | a ) 16 , b ) 32 , c ) 64 , d ) 128 , e ) 256 | a | multiply(divide(64, power(const_2, 5)), const_2) | power(const_2,n0)|divide(n1,#0)|multiply(#1,const_2)| | general |
5 men are equal to as many women as are equal to 8 boys . all of them earn rs . 90 only . men ’ s wages are ? | "5 m = xw = 8 b 5 m + xw + 8 b - - - - - 90 rs . 5 m + 5 m + 5 m - - - - - 90 rs . 15 m - - - - - - 90 rs . = > 1 m = 6 rs . answer : a" | a ) 6 rs , b ) 2 rs , c ) 4 rs , d ) 9 rs , e ) 3 rs | a | divide(90, multiply(const_3, 5)) | multiply(n0,const_3)|divide(n2,#0)| | general |
each factor of 200 is inscribed on its own plastic ball , and all of the balls are placed in a jar . if a ball is randomly selected from the jar , what is the probability that the ball is inscribed with a multiple of 40 ? | "200 = the # of factors 200 has is = 12 ( see below ) ; so out of 12 factors only two are multiples of 40 : 40 and 200 , itself ; so , the probability is 2 / 12 = 1 / 6 . answer : a ." | a ) 1 / 6 , b ) 5 / 42 , c ) 1 / 8 , d ) 3 / 16 , e ) 1 / 4 | a | divide(divide(choose(40, const_1), 40), power(const_3, const_2)) | choose(n1,const_1)|power(const_3,const_2)|divide(#0,n1)|divide(#2,#1)| | other |
a certain bag contains 100 balls — 50 white , 30 green , 10 yellow , 7 red , and 3 purple . if a ball is to be chosen at random , what is the probability that the ball will be neither red nor purple ? | "according to the stem the ball can be white , green or yellow , so the probability is ( white + green + yellow ) / ( total ) = ( 50 + 30 + 10 ) / 100 = 90 / 100 = 0.9 . answer is a" | a ) 0.9 , b ) 0.75 , c ) 0.6 , d ) 0.8 , e ) 0.5 | a | divide(add(add(50, 30), 10), 100) | add(n1,n2)|add(n3,#0)|divide(#1,n0)| | other |
if ( n + 2 ) ! = 2550 ( n ! ) ; find ’ n ’ | "( n + 2 ) ! = 2550 ( n ! ) ( n + 2 ) ( n + 1 ) ( n ! ) = 2550 ( n ! ) ( n + 2 ) ( n + 1 ) = 2550 2 n + 3 n + 2 - 2550 = 0 2 n + 3 n - 2548 = 0 2 n + 52 n - 49 n - 2548 = 0 n ( n - 49 ) ( n + 52 ) = 0 n = 49 or n = - 52 answer : c" | a ) 38 , b ) 35 , c ) 49 , d ) 43 , e ) 34 | c | subtract(add(const_4, const_4), const_1) | add(const_4,const_4)|subtract(#0,const_1)| | general |
find the remainder when 32 ^ 33 ^ 34 is divided by 11 | "so 32103210 gives remainder 1 . now we have to write 32 ^ 33 ^ 34 in this format . so we have to find the remainder 33 ^ 34 when divided by 10 . the remainder is nothig but unit digit of the number . click here to learn this concept 33 ^ 34 gives unit digit of 9 . so 33 ^ 34 = 10 k + 9 323334 = 32 ( 10 k + 9 ) = ( 321... | a ) 22 , b ) 10 , c ) 99 , d ) 76 , e ) 28 | b | reminder(power(32, 33), 34) | power(n0,n1)|reminder(#0,n2)| | general |
a heap of coconuts is divided into groups of 2 , 3 and 11 and each time one coconut is left over . the least number of coconuts in the heap is ? a . 31 b . 41 c . 51 d . 61 | lcm = 66 = > 66 + 1 = 67 answer : d | a ) a ) 31 , b ) b ) 41 , c ) c ) 51 , d ) d ) 67 , e ) e ) 71 | d | subtract(lcm(lcm(2, 3), 11), const_1) | lcm(n0,n1)|lcm(n2,#0)|subtract(#1,const_1) | general |
along a yard 441 metres long , 22 trees are palnted at equal distances , one tree being at each end of the yard . what is the distance between two consecutive trees | "explanation : 22 trees have 21 gaps between them , required distance ( 441 / 21 ) = 21 option d" | a ) 18 , b ) 19 , c ) 20 , d ) 21 , e ) 12 | d | divide(441, subtract(22, const_1)) | subtract(n1,const_1)|divide(n0,#0)| | physics |
calculate the area of a triangle , if the sides are 30 cm , 21 cm and 10 cm , what is its area ? | "the triangle with sides 30 cm , 21 cm and 10 cm is right angled , where the hypotenuse is 30 cm . area of the triangle = 1 / 2 * 21 * 10 = 105 cm 2 answer : b" | a ) 145 cm 2 , b ) 105 cm 2 , c ) 125 cm 2 , d ) 115 cm 2 , e ) 135 cm 2 | b | multiply(divide(21, const_2), 10) | divide(n1,const_2)|multiply(n2,#0)| | geometry |
the circumference of the front wheel of a cart is 30 ft long and that of the back wheel is 32 ft long . what is the distance traveled by the cart , when the front wheel has done 5 more revolutions than the rear wheel ? | point to note : both the wheels would have traveled the same distance . now consider , no . of revolutions made by back wheel as x , which implies that the number of revolutions made by the front wheel is ( x + 5 ) . equating the distance traveled by front wheel to back wheel : ( x + 5 ) * 30 = x * 32 . ( formula for c... | a ) 20 ft , b ) 25 ft , c ) 750 ft , d ) 900 ft , e ) 2400 ft | e | multiply(30, add(divide(multiply(30, 5), subtract(32, 30)), 5)) | multiply(n0,n2)|subtract(n1,n0)|divide(#0,#1)|add(n2,#2)|multiply(n0,#3) | physics |
two pipes a and b can fill a cistern in 12 and 15 minutes respectively . both are opened together but after 4 minutes a is turned off . after how much more time will the cistern be filled ? | "4 / 12 + ( 4 + x ) / 15 = 1 x = 6 answer : a" | a ) 6 , b ) 8 1 / 1 , c ) 8 1 / 4 , d ) 8 1 / 2 , e ) 8 3 / 7 | a | add(multiply(12, subtract(const_1, multiply(add(inverse(12), inverse(15)), 4))), 4) | inverse(n0)|inverse(n1)|add(#0,#1)|multiply(#2,n2)|subtract(const_1,#3)|multiply(n0,#4)|add(n2,#5)| | physics |
find the annual income derived by investing $ 6800 in 50 % stock at 136 . | "by investing $ 136 , income obtained = $ 50 . by investing $ 6800 , income obtained = $ [ ( 50 / 136 ) * 6800 ] = $ 2500 . answer b ." | a ) 550 , b ) 2500 , c ) 250 , d ) 3000 , e ) 400 | b | divide(multiply(6800, 50), 136) | multiply(n0,n1)|divide(#0,n2)| | gain |
a boat moves upstream at the rate of 1 km in 20 minutes and down stream 1 km in 15 minutes . then the speed of the current is : | "rate upstream = ( 1 / 20 * 60 ) = 3 kmph rate down stream = 1 / 15 * 60 = 4 kmph rate of the current = ½ ( 4 - 3 ) = 0.5 kmph answer : b" | a ) 1 kmph , b ) 0.5 kmph , c ) 3 kmph , d ) 2.5 kmph , e ) 3.5 kmph | b | divide(subtract(multiply(divide(1, 15), const_60), multiply(divide(1, 20), const_60)), const_2) | divide(n0,n3)|divide(n0,n1)|multiply(#0,const_60)|multiply(#1,const_60)|subtract(#2,#3)|divide(#4,const_2)| | physics |
a motorcyclist x drives along a circular fence at a rate of 2 rounds per hour and another motor cyclist y at a rate of 4 rounds per hour . after how long they will cross each other for the first time ? | explanation : since x and y move in the same direction they cross each other when there is a difference of two round between the two . relative speed of x and y = 4 - 2 = 2 . time taken to complete one round at this speed = 1 / 2 hour = 30 minutes answer : c | a ) 10 min , b ) 20 min , c ) 30 min , d ) 40 min , e ) none of these | c | divide(multiply(const_2, const_60), 4) | multiply(const_2,const_60)|divide(#0,n1) | gain |
a solution contains 8 parts of water for every 7 parts of lemonade syrup . how many parts of the solution should be removed and replaced with water so that the solution will now contain 30 % lemonade syrup ? | "let the total solution is 150 l with 80 l water 70 l syrup . to make 30 % syrup solution , the result solution must have 105 l syrup and 45 l syrup . therefore we are taking 25 l of syrup from initial solution and replacing with water . using urinary method : 70 l syrup in 150 l solution 25 l syrup in 53.6 l solution ... | a ) 1.5 , b ) 1.75 , c ) 2.14 , d ) 2.34 , e ) 2.64 | c | multiply(divide(subtract(divide(7, add(8, 7)), divide(const_2, add(const_2, const_3))), divide(7, add(8, 7))), add(8, 7)) | add(n0,n1)|add(const_2,const_3)|divide(n1,#0)|divide(const_2,#1)|subtract(#2,#3)|divide(#4,#2)|multiply(#0,#5)| | gain |
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