Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
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a particular store purchased a stock of turtleneck sweaters and marked up its cost by 20 % . during the new year season , it further marked up its prices by 25 % of the original retail price . in february , the store then offered a discount of 8 % . what was its profit on the items sold in february ? | "assume the total price = 100 x price after 20 % markup = 120 x price after 25 % further markup = 1.25 * 120 x = 150 x price after the discount = 0.92 * 150 x = 138 x hence total profit = 38 % option e" | a ) 27.5 % , b ) 30 % , c ) 35 % , d ) 37.5 % , e ) 38 % | e | subtract(multiply(divide(subtract(const_100, 8), const_100), multiply(add(const_100, 20), divide(add(const_100, 25), const_100))), const_100) | add(n0,const_100)|add(n1,const_100)|subtract(const_100,n2)|divide(#2,const_100)|divide(#1,const_100)|multiply(#0,#4)|multiply(#3,#5)|subtract(#6,const_100)| | gain |
the diagonals of a rhombus are 15 cm and 12 cm . find its area ? | "1 / 2 * 15 * 12 = 90 answer : b" | a ) 70 , b ) 90 , c ) 60 , d ) 80 , e ) 292 | b | rhombus_area(15, 12) | rhombus_area(n0,n1)| | geometry |
( ? ) + 3699 + 1985 - 2047 = 31111 | c 27474 x + 3699 + 1985 - 2047 = 31111 x + 3699 + 1985 = 31111 + 2047 x + 5684 = 33158 x = 33158 - 5684 = 27474 . | a ) 27488 , b ) 27424 , c ) 27474 , d ) 27784 , e ) 27468 | c | subtract(31111, subtract(add(3699, 1985), 2047)) | add(n0,n1)|subtract(#0,n2)|subtract(n3,#1) | general |
pens , pencils and markers in a jenna ' s desk are in the ratio of 2 : 2 : 5 . if there are 10 pens , the number of markers in jenna ' s desk is : | explanation : let pens = 2 x , pencils = 2 x & markers = 5 x . now , 2 x = 10 hence x = 5 . number of markers = 5 x which is 25 . answer : d | a ) 30 , b ) 22 , c ) 20 , d ) 25 , e ) 100 | d | divide(multiply(10, 5), 2) | multiply(n2,n3)|divide(#0,n0) | other |
carl is facing very difficult financial times and can only pay the interest on a $ 20,000 loan he has taken . the bank charges him a quarterly compound rate of 5 % . what is the approximate interest he pays annually ? | "usually , you are given the annual rate of interest and it is mentioned that it is annual rate . the bank charges him a quarterly compounded annual rate of 20 % . here you find per quarter rate as ( 20 / 4 ) % = 5 % i have actually never seen a question with quarter rate given but since this question did not mentionan... | a ) $ 1200 , b ) $ 2000 , c ) $ 2150 , d ) $ 4000 , e ) $ 12000 | d | subtract(multiply(multiply(const_100, const_100), power(add(const_1, divide(5, const_100)), const_4)), multiply(const_100, const_100)) | divide(n1,const_100)|multiply(const_100,const_100)|add(#0,const_1)|power(#2,const_4)|multiply(#1,#3)|subtract(#4,#1)| | gain |
how many digits are in ( 8 Γ 10 ^ 10 ) ( 10 Γ 10 ^ 5 ) ? | the question simplfies to ( 8 Γ 10 ^ 10 ) ( 10 ^ 6 ) = > 8 * 10 ^ 16 = > will contain 16 zeros + 1 digit 8 = > 17 ans a | a ) 17 , b ) 25 , c ) 26 , d ) 27 , e ) 28 | a | add(const_10, add(const_3, const_4)) | add(const_3,const_4)|add(#0,const_10) | general |
a certain car dealership sells economy cars , luxury cars , and sport utility vehicles . the ratio of economy to luxury cars is 3 : 4 . the ratio of economy cars to sport utility vehicles is 4 : 3 . what is the ratio of luxury cars to sport utility vehicles ? | "the ratio of economy to luxury cars is 3 : 4 - - > e : l = 3 : 4 = 12 : 16 . the ratio of economy cars to sport utility vehicles is 4 : 3 - - > e : s = 4 : 3 = 12 : 9 . thus , l : s = 16 : 9 . answer : e ." | a ) 9 : 8 , b ) 8 : 9 , c ) 3 : 2 , d ) 2 : 3 , e ) 16 : 9 | e | divide(divide(multiply(const_4, 3), multiply(3, 3)), divide(multiply(3, const_4), multiply(4, const_4))) | multiply(n3,const_4)|multiply(n3,n3)|multiply(n1,const_4)|divide(#0,#1)|divide(#0,#2)|divide(#3,#4)| | other |
a sum of rs . 1530 has been divided among a , b and c such that a gets of what b gets and b gets of what c gets . b β s share is : | "explanation let c β s share = rs . x then , b β s share = rs . x / 4 , a β s share = rs . ( 2 / 3 x x / 4 ) = rs . x / 6 = x / 6 + x / 4 + x = 1530 = > 17 x / 12 = 1530 = > 1530 x 12 / 17 = rs . 1080 hence , b β s share = rs . ( 1080 / 4 ) = rs . 270 . answer c" | a ) rs . 120 , b ) rs . 160 , c ) rs . 270 , d ) rs . 300 , e ) none | c | subtract(subtract(multiply(divide(1530, const_10), const_2), const_12), const_12) | divide(n0,const_10)|multiply(#0,const_2)|subtract(#1,const_12)|subtract(#2,const_12)| | general |
two pipes can fill the cistern in 10 hr and 12 hr respectively , while the third empty it in 40 hr . if all pipes are opened simultaneously , then the cistern will be filled in | "solution : work done by all the tanks working together in 1 hour . 1 / 10 + 1 / 12 β 1 / 40 = 3 / 19 hence , tank will be filled in 19 / 3 = 6.32 hour option ( a )" | a ) 6.32 hr , b ) 8 hr , c ) 8.5 hr , d ) 10 hr , e ) none of these | a | inverse(subtract(add(divide(const_1, 10), divide(const_1, 12)), divide(const_1, 40))) | divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|subtract(#3,#2)|inverse(#4)| | physics |
if 5 men undertook a piece of construction work and finished half the job in 15 days . if two men drop out , then the job will be completed in | that is , half the work done = 5 * 15 * 1 / 2 then , 5 * 15 * 1 / 2 = 3 * ? * 1 / 2 i . e . 5 * 15 = 3 * ? = 5 * 15 / 3 = 25 day d | a ) 5 days , b ) 15 days , c ) 10 days , d ) 25 days , e ) 20 days | d | divide(multiply(15, 5), subtract(5, const_2)) | multiply(n0,n1)|subtract(n0,const_2)|divide(#0,#1) | physics |
two trains are moving in the same direction at 144 kmph and 72 kmph . the faster train crosses a man in the slower train in 19 seconds . find the length of the faster train ? | "relative speed = ( 144 - 72 ) * 5 / 18 = 4 * 5 = 20 mps . distance covered in 19 sec = 19 * 20 = 380 m . the length of the faster train = 380 m . answer : d" | a ) 320 , b ) 340 , c ) 360 , d ) 380 , e ) 400 | d | multiply(divide(subtract(144, 72), const_3_6), 19) | subtract(n0,n1)|divide(#0,const_3_6)|multiply(n2,#1)| | physics |
a no . when divided by 225 gives a remainder 43 , what remainder will beobtained by dividingthe same no . 17 ? | "225 + 43 = 268 / 17 = 13 ( remainder ) e" | a ) 2 , b ) 7 , c ) 9 , d ) 11 , e ) 13 | e | divide(add(225, 43), 17) | add(n0,n1)|divide(#0,n2)| | general |
the radius of a cone is 14 m , slant height is 20 m . find the curved surface area ? | "cone curved surface area = Ο rl 22 / 7 Γ 14 Γ 120 = 44 Γ 20 = 880 m ( power 2 ) answer is d ." | a ) 770 , b ) 550 , c ) 110 , d ) 880 , e ) 330 | d | volume_cone(14, 20) | volume_cone(n0,n1)| | geometry |
a child want to buy 3 new toys . if there are 10 toys that satisfy the child ' s criteria , in how many different ways can the child select the 3 new toys if the order of selection does n ' t matter . | "the number of ways of choosing 3 toys out of 10 are counted by = 10 c 3 ways = 120 answer : option b" | a ) 20 , b ) 120 , c ) 210 , d ) 15 , e ) 45 | b | choose(10, 3) | choose(n1,n0)| | probability |
find the remainder of the division ( 3 ^ 15 ) / 7 . | "find the pattern of the remainders after each power : ( 3 ^ 1 ) / 7 remainder 3 ( 3 ^ 2 ) / 7 remainder 2 ( 3 ^ 3 ) / 7 remainder 6 ( 3 ^ 4 ) / 7 remainder 4 ( 3 ^ 5 ) / 7 remainder 5 ( 3 ^ 6 ) / 7 remainder 1 - - > this is where the cycle ends ( 3 ^ 7 ) / 7 remainder 3 - - > this is where the cycle begins again ( 3 ^... | a ) 7 , b ) 6 , c ) 15 , d ) 8 , e ) 2 | b | reminder(power(3, 15), 7) | power(n0,n1)|reminder(#0,n2)| | general |
in what ratio mental a at rs . 68 per kg be mixed with another metal at rs . 96 per kg so that cost of alloy ( mixture ) is rs . 76 per kg ? | "( 96 - 76 ) / ( 76 - 68 ) = 20 / 8 = 5 / 2 answer : e" | a ) 5 : 8 , b ) 4 : 7 , c ) 3 : 7 , d ) 9 : 5 , e ) 5 : 2 | e | divide(divide(subtract(96, 76), subtract(96, 68)), subtract(const_1, divide(subtract(96, 76), subtract(96, 68)))) | subtract(n1,n2)|subtract(n1,n0)|divide(#0,#1)|subtract(const_1,#2)|divide(#2,#3)| | other |
what is the ratio between perimeters of two squares one having 3 times the diagonal then the other ? | d = 3 d d = d a β 2 = 3 d a β 2 = d a = 3 d / β 2 a = d / β 2 = > 3 : 1 answer : d | ['a ) 3 : 2', 'b ) 3 : 0', 'c ) 4 : 1', 'd ) 3 : 1', 'e ) 2 : 2'] | d | divide(3, divide(3, const_3)) | divide(n0,const_3)|divide(n0,#0) | geometry |
the shopkeeper increased the price of a product by 25 % so that customer finds it difficult to purchase the required amount . but somehow the customer managed to purchase only 72 % of the required amount . what is the net difference in the expenditure on that product ? | "quantity x rate = price 1 x 1 = 1 0.72 x 1.25 = 0.9 decrease in price = ( 0.1 / 1 ) Γ 100 = 10 % b )" | a ) 12.5 % , b ) 10 % , c ) 15 % , d ) 17 % , e ) 19 % | b | divide(multiply(subtract(multiply(const_100, const_100), multiply(add(const_100, 25), 72)), const_100), multiply(const_100, const_100)) | add(n0,const_100)|multiply(const_100,const_100)|multiply(n1,#0)|subtract(#1,#2)|multiply(#3,const_100)|divide(#4,#1)| | general |
two trains 140 m and 150 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 = 140 + 150 = 290 m . required time = 290 * 9 / 250 = 10.44 sec . answer : d :" | a ) 10.9 sec , b ) 13.8 sec , c ) 53.8 sec , d ) 10.44 sec , e ) 10.4 sec | d | divide(add(140, 150), multiply(add(60, 40), const_0_2778)) | add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)| | physics |
a , b and c start a business each investing 20000 . after 5 months a withdrew 10000 , b withdrew 15000 and c invests 5000 more . at the end of the year , a total profit of 58000 was recorded . find the share of b . | ratio of the capitals of a , b and c = 20000 Γ£ β 5 + 10000 Γ£ β 7 : 20000 Γ£ β 5 + 5000 Γ£ β 7 : 20000 Γ£ β 5 + 25000 Γ£ β 7 = 170000 : 135000 : 275000 = 170 : 135 : 275 . b Γ’ β¬ β’ s share = ( 58000 Γ£ β 135 Γ’ Β β 580 ) = 13500 answer d | a ) 20,000 , b ) 21,200 , c ) 28,200 , d ) 13,500 , e ) none of these | d | divide(multiply(58000, divide(add(multiply(20000, 5), multiply(subtract(20000, 15000), subtract(const_12, 5))), add(add(add(multiply(20000, 5), multiply(subtract(20000, 10000), subtract(const_12, 5))), add(multiply(20000, 5), multiply(subtract(20000, 15000), subtract(const_12, 5)))), add(multiply(20000, 5), multiply(ad... | add(n0,n4)|add(n1,const_60)|multiply(n0,n1)|subtract(n0,n3)|subtract(const_12,n1)|subtract(n0,n2)|multiply(#3,#4)|multiply(#5,#4)|multiply(#0,#4)|add(#2,#6)|add(#2,#7)|add(#2,#8)|add(#10,#9)|add(#12,#11)|divide(#9,#13)|multiply(n5,#14)|divide(#15,#1) | gain |
a glass was filled with 10 ounces of water , and 0.06 ounce of the water evaporated each day during a 20 - day period . what percent of the original amount of water evaporated during this period ? | "we are given that 0.06 ounces of water evaporated each day . furthermore , we know that this process happened over a 20 - day period . to calculate the total amount of water that evaporated during this time frame we need to multiply 0.06 by 20 . this gives us : 0.06 x 20 = 1.2 ounces finally , we are asked for β what ... | a ) 0.002 % , b ) 0.02 % , c ) 0.2 % , d ) 12 % , e ) 20 % | d | multiply(divide(multiply(0.06, 20), 10), const_100) | multiply(n1,n2)|divide(#0,n0)|multiply(#1,const_100)| | gain |
in a class of 25 students , 3 students did not borrow any books from the library , 11 students each borrowed 1 book , 6 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 25 * 2 = 50 . the students who borrowed zero , one , or two books borrowed 11 * 1 + 6 * 2 = 23 books . the 5 students who borrowed at least three books borrowed 50 - 23 = 27 books . if 4 of these students borrowed exactly 3 books , then the maximum that one student co... | a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | d | subtract(multiply(25, const_2.0), add(multiply(subtract(subtract(25, add(add(multiply(11, 1), 6), 3)), 1), 3), add(multiply(11, 1), multiply(6, 3)))) | multiply(n0,n5)|multiply(n2,n3)|multiply(n5,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 |
if two painters can complete two rooms in two hours , how many painters would it take to do 18 rooms in 6 hours ? | "explanation : two painters can complete two rooms in two hours . so 18 rooms can be painted in 6 hrs by 6 painters answer : b ) 6 painters" | a ) 8 , b ) 6 , c ) 5 , d ) 4 , e ) 1 | b | divide(multiply(multiply(6, const_3), const_3), multiply(const_3, const_3)) | multiply(n1,const_3)|multiply(const_3,const_3)|multiply(#0,const_3)|divide(#2,#1)| | physics |
how much space , in cubic units , is left vacant when maximum number of 7 x 7 x 7 cubes are fitted in a rectangular box measuring 14 x 21 x 17 ? | "no of cubes that can be accommodated in box = ( 14 * 21 * 17 ) / ( 7 * 7 * 7 ) 12 * 16 in numerator can be perfectly divided by 7 * 7 in denominator . side with length 17 ca n ' t be perfectly divided by 7 and hence is the limiting factor . closet multiple of 7 less that 17 is 14 . so vacant area in cube = = 14 * 21 *... | a ) 878 , b ) 879 , c ) 880 , d ) 881 , e ) 882 | e | multiply(subtract(14, multiply(7, const_2)), multiply(21, 17)) | multiply(n4,n5)|multiply(n0,const_2)|subtract(n3,#1)|multiply(#0,#2)| | geometry |
according to a recent survey report issued by the commerce ministry , government of india , 30 % of the total fdi goes to gujarat and 20 % of this goes to rural areas . if the fdi in gujarat , which goes to urban areas , is $ 72 m , then find the size of fdi in rural andhra pradesh , which attracts 50 % of the fdi that... | 30 % of total fdi given to gujarat is $ 90 m . so , 100 % is $ 300 m . 20 % of 300 m = 60 m . 50 % of 60 m = 30 m . answer : a | a ) 30 m , b ) 40 m , c ) 35 m , d ) 38 m , e ) 50 m | a | multiply(divide(72, multiply(divide(30, const_100), divide(subtract(const_100, 20), const_100))), multiply(divide(50, const_100), divide(20, const_100))) | divide(n0,const_100)|divide(n3,const_100)|divide(n1,const_100)|subtract(const_100,n1)|divide(#3,const_100)|multiply(#1,#2)|multiply(#0,#4)|divide(n2,#6)|multiply(#7,#5) | gain |
triangle atriangle b are similar triangles with areas 1792 units square and 2527 units square respectively . the ratio of there corresponding height would be | "let x be the height of triangle a and y be the height of triangle of b . since triangles are similar , ratio of area of a and b is in the ratio of x ^ 2 / y ^ 2 therefore , ( x ^ 2 / y ^ 2 ) = 1792 / 2527 ( x ^ 2 / y ^ 2 ) = ( 16 * 16 * 7 ) / ( 19 * 19 * 7 ) ( x ^ 2 / y ^ 2 ) = 16 ^ 2 / 19 ^ 2 x / y = 16 / 19 ans = d" | a ) 9 : 10 , b ) 17 : 19 , c ) 23 : 27 , d ) 16 : 19 , e ) 15 : 23 | d | sqrt(divide(1792, 2527)) | divide(n0,n1)|sqrt(#0)| | geometry |
a and b together have $ 1210 . if 4 / 15 of a ' s amount is equal to 2 / 5 of b ' s amount , how much amount does b have ? | 4 / 15 a = 2 / 5 b a = 2 / 5 * 15 / 4 b a = 3 / 2 b a / b = 3 / 2 a : b = 3 : 2 b ' s share = 1210 * 2 / 5 = $ 484 answer is c | a ) $ 310 , b ) $ 395 , c ) $ 484 , d ) $ 512 , e ) $ 542 | c | divide(1210, add(divide(divide(2, 5), divide(4, 15)), const_1)) | divide(n3,n4)|divide(n1,n2)|divide(#0,#1)|add(#2,const_1)|divide(n0,#3) | general |
there are two circles of different radii . the are of a square is 784 sq cm and its side is twice the radius of the larger circle . the radius of the larger circle is seven - third that of the smaller circle . find the circumference of the smaller circle . ? | "let the radii of the larger and the smaller circles be l cm and s cm respectively . let the side of the square be a cm . a 2 = 784 = ( 4 ) ( 196 ) = ( 22 ) . ( 142 ) a = ( 2 ) ( 14 ) = 28 a = 2 l , l = a / 2 = 14 l = ( 7 / 3 ) s therefore s = ( 3 / 7 ) ( l ) = 6 circumference of the smaller circle = 2 β s = 12 β cm . ... | a ) 87 , b ) 98 , c ) 12 , d ) 27 , e ) 28 | c | add(divide(divide(square_edge_by_area(784), const_2), divide(add(const_3, const_4), const_3)), const_2) | add(const_3,const_4)|square_edge_by_area(n0)|divide(#1,const_2)|divide(#0,const_3)|divide(#2,#3)|add(#4,const_2)| | geometry |
the speed of a boat in still water is 12 km / hr and the rate of current is 4 km / hr . the distance travelled downstream in 18 minutes is | "explanation : speed downstreams = ( 12 + 4 ) kmph = 16 kmph . distance travelled = ( 16 x 18 / 60 ) km = 4.8 km option d" | a ) 1.6 km , b ) 2 km , c ) 3.6 km , d ) 4.8 km , e ) none of these | d | multiply(divide(18, const_60), add(12, 4)) | add(n0,n1)|divide(n2,const_60)|multiply(#0,#1)| | physics |
a gambler bought $ 3000 worth of chips at a casino in denominations of $ 20 and $ 100 . that evening , the gambler lost 16 chips , and then cashed in the remainder . if the number of $ 20 chips lost was 2 more or 2 less than the number of $ 100 chips lost , what is the largest amount of money that the gambler could hav... | in order to maximize the amount of money that the gambler kept , we should maximize # of $ 20 chips lost and minimize # of $ 100 chips lost , which means that # of $ 20 chips lost must be 2 more than # of $ 100 chips lost . so , if # of $ 20 chips lost is x then # of $ 100 chips lost should be x - 2 . now , given that ... | a ) $ 2,040 , b ) $ 2,120 , c ) $ 1,960 , d ) $ 1,920 , e ) $ 1,400 | b | subtract(divide(add(multiply(add(const_4, const_3), 100), multiply(subtract(const_10, const_1), 20)), 20), add(2, 2)) | add(const_3,const_4)|add(n5,n5)|subtract(const_10,const_1)|multiply(n2,#0)|multiply(n1,#2)|add(#3,#4)|divide(#5,n1)|subtract(#6,#1) | general |
a boat covers a certain distance downstream in 1 hour , while it comes back in 1 Β½ hours . if the speed of the stream be 3 kmph , what is the speed of the boat in still water ? | sol . let the speed of the boat in still water be x kmph . then , speed downstream = ( x + 3 ) kmph , speed upstream = ( x - 3 ) kmph . β΄ ( x + 3 ) * 1 = ( x - 3 ) * 3 / 2 β 2 x + 6 = 3 x - 9 β x = 15 kmph . answer d | a ) 12 kmph , b ) 13 kmph , c ) 14 kmph , d ) 15 kmph , e ) none | d | divide(add(multiply(divide(const_3, const_2), 3), 3), subtract(divide(const_3, const_2), 1)) | divide(const_3,const_2)|multiply(n2,#0)|subtract(#0,n0)|add(n2,#1)|divide(#3,#2) | physics |
simplify : 0.3 * 0.8 + 0.1 * 0.5 | "given exp . = 0.3 * 0.8 + ( 0.1 * 0.5 ) = 0.24 + 0.05 = 0.29 answer is b ." | a ) 0.52 , b ) 0.29 , c ) 0.48 , d ) 0.64 , e ) 0.46 | b | add(multiply(0.3, 0.8), multiply(0.1, 0.5)) | multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)| | general |
working alone , pump a can empty a pool in 4 hours . working alone , pump b can empty the same pool in 2 hours . working together , how many minutes will it take pump a and pump b to empty the pool ? | "pump a can empty ( 1 / 4 ) of the pool per hour . pump b can empty ( 1 / 2 ) of the pool per hour . together the pumps can empty 1 / 4 + 1 / 2 = 3 / 4 of the pool per hour . 1 pool / ( 3 / 4 ) pool per hour = 4 / 3 hours = 80 minutes . the answer is d ." | a ) 65 , b ) 70 , c ) 75 , d ) 80 , e ) 85 | d | multiply(inverse(add(divide(const_1, 4), divide(const_1, 2))), const_60) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|inverse(#2)|multiply(#3,const_60)| | physics |
a garrison of 2000 men has provisions for 62 days . at the end of 15 days , a reinforcement arrives , and it is now found that the provisions will last only for 20 days more . what is the reinforcement ? | "2000 - - - - 62 2000 - - - - 47 x - - - - - 20 x * 20 = 2000 * 47 x = 4700 2000 - - - - - - - 2700 answer : c" | a ) 1977 , b ) 1893 , c ) 2700 , d ) 1900 , e ) 1278 | c | subtract(divide(subtract(multiply(2000, 62), multiply(2000, 15)), 20), 2000) | multiply(n0,n1)|multiply(n0,n2)|subtract(#0,#1)|divide(#2,n3)|subtract(#3,n0)| | physics |
the diagonal of a square is the height in an equilateral triangle . if the side of the square is x , what is the side of the equilateral triangle in terms of x ? | the diagonal of the square x * sqrt ( 2 ) if that ' s the altitude of the equilateral triangle , the formula for which is side * sqrt ( 3 ) / 2 side is 2 sqrt ( 2 ) * x / sqrt ( 3 ) answer : ( option d ) | ['a ) β ( 2 / 3 ) Γ x', 'b ) 2 / β 3 Γ x', 'c ) β 3 x', 'd ) β ( 8 / 3 ) Γ x', 'e ) 2 β 3 Γ x'] | d | divide(multiply(power(add(const_1, const_1), inverse(const_2)), const_2), power(const_3, inverse(const_2))) | add(const_1,const_1)|inverse(const_2)|power(#0,#1)|power(const_3,#1)|multiply(#2,const_2)|divide(#4,#3) | geometry |
if p ( a ) = 6 / 17 , p ( b ) = 5 / 17 , and p ( a βͺ b ) = 4 / 17 find p ( b | a ) ? | "p ( b | a ) = p ( a βͺ b ) / p ( a ) p ( b | a ) = ( 4 / 17 ) / ( 6 / 17 ) = 4 / 6 = 2 / 3 . a" | a ) 2 / 3 , b ) 3 / 5 , c ) 2 / 7 , d ) 3 / 7 , e ) 4 / 9 | a | divide(divide(4, 17), divide(6, 17)) | divide(n4,n1)|divide(n0,n1)|divide(#0,#1)| | general |
last year elaine spent 20 % of her annual earnings on rent . this year she earned 15 % more than last year and she spent 25 % of her annual earnings on rent . the amount she spent on rent this year is what percent of the amount spent on rent last year ? | for this it is easiest to use simple numbers . let ' s assume that elaine ' s annual earnings last year were $ 100 . she would ' ve spent $ 20 of this on rent . this year she earned 15 % more , or $ 115 . she would ' ve spent 25 % of this on rent , or $ 28.75 do $ 34.5 / $ 20 this will give you 143.75 % b is the correc... | a ) 152.5 , b ) 143.75 , c ) 155.5 , d ) 165.5 , e ) 117.5 | b | multiply(divide(multiply(add(divide(15, const_100), const_1), 25), 20), const_100) | divide(n1,const_100)|add(#0,const_1)|multiply(n2,#1)|divide(#2,n0)|multiply(#3,const_100) | gain |
if $ 5,000 is invested in an account at a simple annual rate of r percent , the interest is $ 250 . when $ 20,000 is invested at the same interest rate , what is the interest from the investment ? | - > 250 / 5,000 = 5 % and 20,000 * 5 % = 1000 . thus , d is the answer . | a ) $ 700 , b ) $ 750 , c ) $ 800 , d ) $ 1000 , e ) $ 900 | d | divide(multiply(250, multiply(multiply(const_2, const_100), const_100)), divide(multiply(multiply(const_2, const_100), const_100), const_4)) | multiply(const_100,const_2)|multiply(#0,const_100)|divide(#1,const_4)|multiply(n1,#1)|divide(#3,#2) | gain |
a train running at the speed of 180 km / hr crosses a pole in 18 seconds . what is the length of the train ? | "speed = ( 180 x ( 5 / 18 ) m / sec = ( 50 ) m / sec . length of the train = ( speed x time ) . length of the train = ( ( 50 ) x 18 ) m = 900 m b" | a ) 800 , b ) 900 , c ) 950 , d ) 1000 , e ) 1050 | b | multiply(divide(multiply(180, const_1000), const_3600), 18) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics |
a 18 % stock yielding 8 % is quoted at ? | "assume that face value = rs . 100 as it is not given to earn rs . 8 , money invested = rs . 100 to earn rs . 18 , money invested = 100 Γ 18 / 8 = rs . 225 ie , market value of the stock = rs . 175 answer is e ." | a ) 165 , b ) 170 , c ) 175 , d ) 180 , e ) 225 | e | multiply(divide(const_100, 8), 18) | divide(const_100,n1)|multiply(n0,#0)| | gain |
p works 25 % more efficiently than q and q works 50 % more efficiently than r . to complete a certain project , p alone takes 50 days less than q alone . if , in this project p alone works for 60 days and then q alone works for 145 days , in how many days can r alone complete the remaining work ? | p works 25 % more efficiently than q : something that takes q 5 days , takes p 4 days q works 50 % more efficiently than r : something that takes r 7.5 days , takes q 5 days p alone takes 50 days less than q : for every 4 days p works , q has to work an extra day . hence p alone can do it in 200 days and q alone in 250... | a ) 50 days , b ) 110.6 days , c ) 100 days , d ) 150 days , e ) 80 days | b | subtract(multiply(const_4, 50), multiply(divide(145, const_100), 60)) | divide(n4,const_100)|multiply(n1,const_4)|multiply(n3,#0)|subtract(#1,#2) | physics |
water consists of hydrogen and oxygen , and the approximate ratio , by mass , of hydrogen to oxygen is 2 : 16 . approximately how many grams of oxygen are there in 180 grams of water ? | "solution : we are given that the ratio of hydrogen to oxygen in water , by mass , is 2 : 16 . using our ratio multiplier we can re - write this as 2 x : 16 x . we can now use these expressions to determine how much oxygen is 180 18 x = 180 x = 10 since x is 10 , we know that there are 16 x 10 = 160 grams of oxygen in ... | a ) 160 , b ) 72 , c ) 112 , d ) 128 , e ) 142 | a | multiply(2, divide(180, add(2, 16))) | add(n0,n1)|divide(n2,#0)|multiply(n0,#1)| | other |
a jar full of whisky contains 40 % alcohol . a part of this whisky is replaced by another containing 19 % alcohol and now the percentage of alcohol was found to be 26 % . the quantity of whisky replaced is : | "explanation : by the rule of alligation , we have : so , ratio of 1 st and 2 nd quantities = 7 : 14 = 1 : 2 required quantity replaced = 2 / 3 . answer is b" | a ) 1 / 3 , b ) 2 / 3 , c ) 2 / 5 , d ) 3 / 5 , e ) 4 / 5 | b | divide(subtract(40, 26), subtract(40, 19)) | subtract(n0,n2)|subtract(n0,n1)|divide(#0,#1)| | gain |
the speed of a train is 60 kmph . what is the distance covered by it in 20 minutes ? | "60 * 20 / 60 = 20 kmph answer : c" | a ) 21 kmph , b ) 22 kmph , c ) 20 kmph , d ) 23 kmph , e ) 24 kmph | c | multiply(divide(20, const_60), 60) | divide(n1,const_60)|multiply(n0,#0)| | physics |
find the average of all the numbers between 9 and 26 which are divisible by 5 . | "sol . average = ( 10 + 15 + 20 + 25 / 4 ) = 70 / 4 = 17.5 answer c" | a ) 15 , b ) 19 , c ) 18 , d ) 22 , e ) none | c | divide(add(add(9, const_4), subtract(26, const_4)), const_2) | add(n0,const_4)|subtract(n1,const_4)|add(#0,#1)|divide(#2,const_2)| | general |
in one hour , a boat goes 15 km / hr along the stream and 9 km / hr against the stream . the speed of the boat in still water ( in km / hr ) is : | "explanation : let the speed downstream be a km / hr and the speed upstream be b km / hr , then speed in still water = 1 / 2 ( a + b ) km / hr rate of stream = 1 / 2 ( a β b ) km / hr speed in still water = 1 / 2 ( 15 + 9 ) kmph = 12 kmph . answer : option a" | a ) 12 kmph , b ) 13 kmph , c ) 14 kmph , d ) 15 kmph , e ) 16 kmph | a | stream_speed(15, 9) | stream_speed(n0,n1)| | physics |
what is the least common multiple of 12,15 , 18,27 | "factors of 12 = 2 * 2 * 3 factors of 27 = 3 * 3 * 3 factors of 15 = 3 * 5 factors of 18 = 2 * 3 * 3 thus the l . c . m is 2 * 3 * 2 * 3 * * 1 * 5 * 3 = 540 the answer is b" | a ) 640 , b ) 540 , c ) 691 , d ) 385 , e ) 789 | b | multiply(add(12,15, const_1), const_2) | add(n0,const_1)|multiply(#0,const_2)| | general |
given a spinner with 4 sections of equal size labeled e , b , c , and d , what is the probability of not getting an e after spinning the spinner two times ? | the probability of not getting an e after spinning the spinner two times is 3 / 4 * 3 / 4 = 9 / 16 ( so getting any of the remaining 3 letters out of 4 ) . answer : b . | a ) 15 / 16 , b ) 9 / 16 , c ) 1 / 2 , d ) 1 / 4 , e ) 1 / 8 | b | power(divide(const_3, 4), const_2) | divide(const_3,n0)|power(#0,const_2) | general |
two carpenters , working in the same pace , can build 2 desks in two hours and a half . how many desks can 3 carpenters build in 4 hours ? | "w = 2 desks t = 2.5 hrs rate of 2 carpenters = 2 Γ r rate = work done / time 2 xr = 2 / 2.5 r = 1 / 2.5 = 2 / 5 ( this is the rate of each carpenter ) work done by 3 carpenters in 4 hrs = 3 Γ rate of each carpenter x time = 3 Γ 2 / 5 Γ 4 = 4.8 desks b is the correct answer ." | a ) 2.4 . , b ) 4.8 . , c ) 4.2 . , d ) 5.5 . , e ) 6.4 | b | multiply(multiply(divide(divide(2, divide(4, 2)), 2), 3), 4) | divide(n2,n0)|divide(n0,#0)|divide(#1,n0)|multiply(n1,#2)|multiply(n2,#3)| | physics |
p and q started a business investing rs . 45,000 and rs . 10,000 respectively . in what ratio the profit earned after 2 years be divided between p and q respectively ? | "p : q = 45000 : 10000 = 9 : 2 answer : a" | a ) 9 : 2 , b ) 4 : 5 , c ) 9 : 3 , d ) 17 : 9 , e ) 17 : 4 | a | divide(add(multiply(add(add(2, const_3), const_3), multiply(add(2, const_3), 2)), add(2, const_3)), add(multiply(const_3, multiply(add(2, const_3), 2)), add(2, const_3))) | add(n2,const_3)|add(#0,const_3)|multiply(n2,#0)|multiply(#1,#2)|multiply(#2,const_3)|add(#0,#3)|add(#0,#4)|divide(#5,#6)| | gain |
find the area of the quadrilateral of one of its diagonals is 30 cm and its off sets 10 cm and 6 cm ? | "1 / 2 * 30 ( 10 + 6 ) = 240 cm 2 answer : c" | a ) 189 cm 2 , b ) 150 cm 2 , c ) 240 cm 2 , d ) 177 cm 2 , e ) 187 cm 2 | c | multiply(multiply(divide(const_1, const_2), add(6, 10)), 30) | add(n1,n2)|divide(const_1,const_2)|multiply(#0,#1)|multiply(n0,#2)| | geometry |
a certain junior class has 1,200 students and a certain senior class has 800 students . among these students , there are 60 siblings pairs , each consisting of 1 junior and 1 senior . if 1 student is to be selected at random from each class , what is the probability that the 2 students selected at will be a sibling pai... | "total number of ways of choosing one student from each group is = 800 * 1200 number of cases in which a sibling pair will be got is = 60 thus the probability that the 2 students selected will be a sibling pair is = 60 / ( 800 * 1200 ) = 1 / 16,000 e" | a ) 3 / 40,000 , b ) 1 / 3,600 , c ) 9 / 2,000 , d ) 1 / 60 , e ) 1 / 16,000 | e | divide(1, const_3) | divide(n3,const_3)| | probability |
two trains are moving in opposite directions at 60 km / hr and 90 km / hr . their lengths are 1.10 km and 0.65 km respectively . the time taken by the slower train to cross the faster train in seconds is ? | relative speed = 60 + 90 = 150 km / hr . = 150 * 5 / 18 = 125 / 3 m / sec . distance covered = 1.10 + 0.65 = 1.75 km = 1750 m . required time = 1750 * 3 / 125 = 42 sec . answer : a | a ) 42 , b ) 77 , c ) 48 , d ) 99 , e ) 11 | a | subtract(divide(multiply(1.1, const_1000), divide(multiply(60, const_1000), const_3600)), divide(multiply(0.65, const_1000), divide(multiply(90, const_1000), const_3600))) | multiply(n2,const_1000)|multiply(n0,const_1000)|multiply(n3,const_1000)|multiply(n1,const_1000)|divide(#1,const_3600)|divide(#3,const_3600)|divide(#0,#4)|divide(#2,#5)|subtract(#6,#7) | physics |
the sum of the fourth and twelfth term of an arithmetic progression is 20 . what is the sum of the first 10 terms of the arithmetic progression ? | n th term of a . p . is given by a + ( n - 1 ) d 4 th term = a + 3 d 12 th term = a + 11 d given a + 3 d + a + 11 d = 20 - - > 2 a + 14 d = 20 - - > a + 7 d = 10 sum of n term of a . p = n / 2 [ 2 a + ( n - 1 ) d ] subsitiuing n = 10 . . . we get 10 / 2 [ 2 a + 14 d ] = 10 [ a + 7 d ] = 10 * 10 = 100 . . . answer is b ... | a ) 300 , b ) 100 , c ) 150 , d ) 170 , e ) 270 | b | divide(multiply(20, 10), const_2) | multiply(n0,n1)|divide(#0,const_2) | general |
a salt manufacturing company produced a total of 3000 tonnes of salt in january of a particular year . starting from february its production increased by 100 tonnes every month over the previous months until the end of the year . find its ave 66 rage daily production for that year ? | "total production of salt by the company in that year = 3000 + 3100 + 3200 + . . . . + 4100 = 42600 . average monthly production of salt for that year = 42600 / 365 Γ’ β° Λ 116 answer : d" | a ) 119 , b ) 185 , c ) 190 , d ) 116 , e ) 200 | d | divide(divide(multiply(add(multiply(3000, const_2), multiply(subtract(const_12, const_1), 100)), const_12), const_2), add(add(multiply(const_3, 100), multiply(multiply(const_2, const_3), const_10)), add(const_2, const_3))) | add(const_2,const_3)|multiply(n0,const_2)|multiply(n1,const_3)|multiply(const_2,const_3)|subtract(const_12,const_1)|multiply(n1,#4)|multiply(#3,const_10)|add(#1,#5)|add(#2,#6)|add(#8,#0)|multiply(#7,const_12)|divide(#10,const_2)|divide(#11,#9)| | general |
set x consists of the integers from 10 to 15 , inclusive , while set y consists of the integers from 5 to 10 , inclusive . how many distinct integers do belong to the both sets at the same time ? | "x = { 10 , 11 , 12,13 , 14,15 } y = { 5,6 , 7 , 8 , 9 , 10 } common elements = { 10 } = 1 element answer : option d ." | a ) 5 , b ) 8 , c ) 4 , d ) 1 , e ) 9 | d | add(subtract(15, 10), 10) | subtract(n1,n0)|add(#0,n0)| | other |
in a class of boys and girls vikas ' s rank is 9 th and tanvi ' s rank is 17 th . vikas ' s rank among the boys in that class is 4 th from the top and 18 th from the bottom and tanvi ' s rank among the girls is 8 th from top and 21 st from bottom . in the order of rank , how many girls are there between tanvi and vikas... | explanation : vikas ' s rank in the class is 9 . so there are 8 people before him . his rank among boys is 4 . so 3 boys are before him . so there are 8 β 3 = 5 girls before him . tanvi ' s rank among the girls is 8 . so there are 7 girls before her . so number of girls between vikas and tanvi is 7 β 5 = 2 answer : c | a ) 29 , b ) 8 , c ) 2 , d ) 4 , e ) 6 | c | subtract(subtract(8, const_1), subtract(subtract(9, const_1), subtract(4, const_1))) | subtract(n4,const_1)|subtract(n0,const_1)|subtract(n2,const_1)|subtract(#1,#2)|subtract(#0,#3) | other |
arjun started a business with rs . 5,000 and is joined afterwards by anoop with rs . 10 , 000 . after how many months did anoop join if the profits at the end of the year are divided equally ? | "suppose anoop joined after 3 months . then , 5000 * 12 = 10000 * ( 12 β x ) = > x = 6 . answer : d" | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | d | divide(subtract(multiply(multiply(multiply(const_4, const_100), const_100), const_12), multiply(multiply(multiply(const_2, const_100), const_100), const_12)), multiply(multiply(10, const_100), const_10)) | multiply(const_100,const_4)|multiply(const_100,const_2)|multiply(n1,const_100)|multiply(#0,const_100)|multiply(#1,const_100)|multiply(#2,const_10)|multiply(#3,const_12)|multiply(#4,const_12)|subtract(#6,#7)|divide(#8,#5)| | general |
tom purchased 8 kg of apples at the rate of 70 per kg and 910 kg of mangoes at the rate of 55 per kg . how much amount did he pay to the shopkeeper ? | "cost of 8 kg apples = 70 Γ 8 = 560 . cost of 10 kg of mangoes = 55 Γ 10 = 550 . total cost he has to pay = 560 + 550 = 1110 . a )" | a ) a ) 1110 , b ) b ) 1045 , c ) c ) 1055 , d ) d ) 1060 , e ) e ) 1075 | a | add(multiply(8, 70), multiply(910, 55)) | multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)| | gain |
what is the value of β 16 % ? | "explanation : br > β 16 % = β 14 / β 100 = 4 / 10 = 40 / 100 = 40 % correct answer is c ) 40 %" | a ) 20 % , b ) 30 % , c ) 40 % , d ) 50 % , e ) 60 % | c | circle_area(divide(16, multiply(const_2, const_pi))) | multiply(const_2,const_pi)|divide(n0,#0)|circle_area(#1)| | gain |
a certain company that sells only cars and trucks reported that revenues from car sales in 1997 were down 11 percent from 1996 and revenues from truck sales were up 7 percent from 1996 . if total revenues from car sales and truck sales in 1997 were up 1 percent from 1996 , what is the ratio q of revenue from car sales ... | a . . i have probably solved this question 3 - 4 times by now . . remember the answer . . 1 : 2 | a ) 1 : 2 , b ) 4 : 5 , c ) 1 : 1 , d ) 3 : 2 , e ) 5 : 3 | a | divide(subtract(add(const_100, 7), add(const_100, 1)), subtract(add(const_100, 1), subtract(const_100, 11))) | add(n3,const_100)|add(n6,const_100)|subtract(const_100,n1)|subtract(#0,#1)|subtract(#1,#2)|divide(#3,#4) | other |
a , b , k start from the same place and travel in the same direction at speeds of 30 km / hr , 40 km / hr , 50 km / hr respectively . b starts two hours after a . if b and k overtake a at the same instant , how many hours after a did k start ? | "in 2 hours , a travels 60 km . b can catch a at a rate of 10 km / hr , so b catches a 6 hours after b starts . so a and b both travel a distance of 240 km . c needs 4.8 hours to travel 240 km , so c leaves 3.2 hours after a . the answer is b ." | a ) 2.4 , b ) 3.2 , c ) 4.0 , d ) 4.8 , e ) 5.6 | b | subtract(40, 30) | subtract(n1,n0)| | physics |
a certain shade of gray paint is obtained by mixing 3 parts of white paint with 5 parts of black paint . if 2 gallons of the mixture is needed and the individual colors can be purchased only in one gallon or half gallon cans , what is the least amount of paint r , in gallons , that must be purchased in order to measure... | given w : b = 3 : 5 that means say 3 gallons of white paint + 5 gallons of black paint = 8 gallons of paint mixture . but we want least amount of whiteblack paints for minimum of 2 gallons of mixture , so lets reduce keeping same ratio , 1.5 : 2.5 gives 1.5 + 2.5 = 4 gallons of mixture , but we want only 2 gallons , le... | a ) 2 , b ) 2 1 / 2 , c ) 3 , d ) 3 1 / 2 , e ) 4 | b | add(multiply(const_0_25, const_2), multiply(2, 5)) | multiply(const_0_25,const_2)|multiply(n1,n2)|add(#0,#1) | general |
what least number must be subtracted from 652543 so that remaining no . is divisible by 12 | "explanation : on dividing 652543 by 12 we get the remainder 7 , so 7 should be subtracted option c" | a ) 3 , b ) 5 , c ) 7 , d ) 8 , e ) 9 | c | subtract(652543, multiply(floor(divide(652543, 12)), 12)) | divide(n0,n1)|floor(#0)|multiply(n1,#1)|subtract(n0,#2)| | general |
if 20 men can build a wall 112 metres long in 6 days , what length of a similar wall can be built by 25 men in 3 days ? | "20 men is 6 days can build 112 metres 25 men in 3 days can build = 112 * ( 25 / 20 ) x ( 3 / 6 ) = 70 meters answer : c ." | a ) 65 mtr . , b ) 52 mtr , c ) 70 mtr . , d ) 78 mtr . , e ) 17 mtr . | c | multiply(112, divide(multiply(25, 3), multiply(20, 6))) | multiply(n3,n4)|multiply(n0,n2)|divide(#0,#1)|multiply(n1,#2)| | physics |
if the sides of a triangle are 26 cm , 24 cm and 12 cm , what is its area ? | "the triangle with sides 26 cm , 24 cm and 12 cm is right angled , where the hypotenuse is 26 cm . area of the triangle = 1 / 2 * 24 * 12 = 144 cm 2 answer : a" | a ) 144 , b ) 772 , c ) 288 , d ) 266 , e ) 2848 | a | divide(multiply(24, 12), const_2) | multiply(n1,n2)|divide(#0,const_2)| | geometry |
in a certain candy store , 22 % of the customers are caught sampling the candy and are charged a small fine , but 10 % of the customers who sample the candy are not caught . what is the total percent of all customers who sample candy ? | "since 10 % of the customers who sample the candyare notcaught , then 90 % of the customers who sample the candyarecaught : { % of customers who sample candy } * 0.90 = 0.22 ; { % of customers who sample candy } = 0.25 . answer : c ." | a ) 22 % , b ) 23 % , c ) 24.4 % , d ) 25 % , e ) 34 % | c | divide(22, divide(subtract(const_100, 10), const_100)) | subtract(const_100,n1)|divide(#0,const_100)|divide(n0,#1)| | gain |
carl can wash all the windows of his house in 7 hours . his wife maggie can wash all the windows in 4 hours . how many hours will it take for both of them working together to wash all the windows ? | work hrs = ab / ( a + b ) = 28 / 11 = 11 6 / 2 answer is c | a ) 2 , b ) 2 1 / 4 , c ) 11 6 / 2 , d ) 4 1 / 2 , e ) 5 | c | multiply(divide(const_60, add(divide(const_1, 7), divide(const_1, 4))), add(divide(const_1, 7), divide(const_1, 4))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_60,#2)|multiply(#2,#3) | physics |
each month , after jill pays for rent , utilities , food , and other necessary expenses , she has one fifth of her net monthly salary left as discretionary income . of this discretionary income , she puts 30 % into a vacation fund , 20 % into savings , and spends 35 % on eating out and socializing . this leaves her wit... | "let x be the monthly salary 15 % of 1 / 5 * x = 108 x = 3600 answer d" | a ) $ 2400 , b ) $ 3200 , c ) $ 6000 , d ) $ 3600 , e ) $ 9600 | d | multiply(multiply(const_100, add(const_3, const_2)), divide(108, subtract(const_100, add(add(30, 20), 35)))) | add(n0,n1)|add(const_2,const_3)|add(n2,#0)|multiply(#1,const_100)|subtract(const_100,#2)|divide(n3,#4)|multiply(#5,#3)| | gain |
if a certain toy store ' s revenue in november was 2 / 3 of its revenue in december and its revenue in january was 1 / 3 of its revenue in november , then the store ' s revenue in december was how many times the average ( arithmetic mean ) of its revenues in november and january ? | "let dec rev = 100 then nov rev is 2 / 3 ( 100 ) = > 67 therefore jan rev = 1 / 3 ( nov rev ) = 1 / 3 ( 67 ) = > 23 hence dec rev = x * ( nov rev + jan rev ) / 2 100 = x * ( 67 + 23 ) / 2 x = 100 / 45 = > 2.22 = 3 ans ) e" | a ) 1 / 4 , b ) 1 / 2 , c ) 2 / 3 , d ) 2 , e ) 3 | e | divide(1, divide(add(divide(2, 3), multiply(divide(2, 3), divide(1, 3))), const_2)) | divide(n0,n1)|divide(n2,n3)|multiply(#0,#1)|add(#0,#2)|divide(#3,const_2)|divide(n2,#4)| | general |
the largest two digit number which is a perfect cube , is : | "explanation : 4 * 4 * 4 = 64 answer : d" | a ) 25 , b ) 36 , c ) 49 , d ) 64 , e ) 81 | d | square_area(const_pi) | square_area(const_pi)| | geometry |
a , b , c and d enter into partnership . a subscribes 1 / 3 of the capital b 1 / 4 , c 1 / 5 and d the rest . how much share did a get in a profit of rs . 2475 ? | "2475 * 1 / 3 = 825 option c" | a ) s . 800 , b ) s . 810 , c ) s . 825 , d ) s . 900 , e ) s . 920 | c | multiply(2475, divide(1, 3)) | divide(n0,n1)|multiply(n6,#0)| | general |
a rectangular tank measuring 5 m Γ 4.5 m Γ 2.1 m is dug in the centre of the field measuring 13.5 m Γ 2.5 . the earth dug out is spread evenly over the remaining portion of a field . how much is the level of the field raised ? | area of the field = 13.5 Γ 2.5 = 33.75 m 2 area covered by the rectangular tank = 5 Γ 4.5 = 22.50 m 2 area of the field on which the earth dug out is to be spread = 33.75 β 22.50 = 11.25 m 2 let the required height be h . then , 11.25 Γ h = 5 Γ 4.5 Γ 2.1 or h = 4.2 m answer c | a ) 4.0 m , b ) 4.1 m , c ) 4.2 m , d ) 4.3 m , e ) none of these | c | divide(multiply(multiply(5, 4.5), 2.1), subtract(multiply(13.5, 2.5), multiply(5, 4.5))) | multiply(n0,n1)|multiply(n3,n4)|multiply(n2,#0)|subtract(#1,#0)|divide(#2,#3) | physics |
a batsman makes a score of 80 runs in the 11 th inning and thus increases his average by 5 . find his average after 11 th inning . | "let the average after 11 th inning = x then , average after 10 th inning = x - 5 10 ( x - 5 ) + 80 = 11 x x = 80 - 50 = 30 answer is d" | a ) 40 , b ) 50 , c ) 62 , d ) 30 , e ) 56 | d | add(subtract(80, multiply(11, 5)), 5) | multiply(n1,n2)|subtract(n0,#0)|add(n2,#1)| | general |
paul has to secure 50 % marks to clear his exam of class 7 th . he got 50 marks and failed by 10 marks . what is the maximum marks ? | c 120 to pass the exam ravish needs 50 + 10 = 60 marks . = > ( 60 / 50 ) * 100 = 120 | a ) 100 , b ) 110 , c ) 120 , d ) 210 , e ) 200 | c | divide(multiply(add(50, 10), const_100), 50) | add(n0,n3)|multiply(#0,const_100)|divide(#1,n0) | gain |
on a two - dimensional coordinate plane , the line s = x ^ 2 - x ^ 3 touches the x - axis in how many places ? | "apparently it ' s s = x ^ 2 - x ^ 3 instead of s = x ^ 2 - s ^ 3 . in this case : the x - intercept is the value ( s ) of x for s = 0 . 0 = x ^ 2 - x ^ 3 ; 0 = x ^ 2 ( 1 - x ) ; x = 0 or x = 1 . answer : c ." | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | c | subtract(max(3, 2), const_1) | max(n0,n1)|subtract(#0,const_1)| | general |
two vertical ladders length of 6 m and 11 m are kept vertically at a distance of 12 m . find the top distance of both ladders ? | so distance between the top points = ad = 122 + 52 β β β β β β β β = 13122 + 52 = 13 so in paper b he did well . answer : c | a ) 22 , b ) 27 , c ) 13 , d ) 19 , e ) 17 | c | sqrt(add(power(subtract(11, 6), const_2), power(12, const_2))) | power(n2,const_2)|subtract(n1,n0)|power(#1,const_2)|add(#2,#0)|sqrt(#3) | physics |
the volume of a certain substance is always directly proportional to its weight . if 48 cubic inches of the substance weigh 112 ounces , what is the volume , in cubic inches , of 84 ounces of this substance ? | 112 ounces of a substance has a volume of 48 cubic inches 84 ounces of a substance has a volume of ( 48 / 112 ) * 84 = 36 cubic inches answer b | ['a ) 27', 'b ) 36', 'c ) 42', 'd ) 64', 'e ) 147'] | b | multiply(divide(48, 112), 84) | divide(n0,n1)|multiply(n2,#0) | geometry |
a batsman makes a score of 95 runs in the 11 th inning and thus increases his average by 5 . find his average after 11 th inning . | "let the average after 11 th inning = x then , average after 10 th inning = x - 5 10 ( x - 5 ) + 95 = 11 x x = 95 - 50 = 45 answer is d" | a ) 40 , b ) 50 , c ) 62 , d ) 45 , e ) 56 | d | add(subtract(95, multiply(11, 5)), 5) | multiply(n1,n2)|subtract(n0,#0)|add(n2,#1)| | general |
the largest six digit number which is a perfect cube , is : | "explanation : 99 * 99 * 99 = 970299 answer : a" | a ) 970299 , b ) 907289 , c ) 992289 , d ) 900279 , e ) 999999 | a | square_area(const_pi) | square_area(const_pi)| | geometry |
of the three - digit positive integers that have no digits equal to zero , how many have two digits that are equal to each other and the remaining digit different from the other two ? | "assume : a , b , c is the digit and a , b , c not = 0 thus a , b , c could be 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , and 9 count the possible way to get three digit number abc . a = b , and c must be different from a , b thus , there are 3 possible ways of digit arrangement : aac , aca , caa case i : aac = > ( digit 1 st ) x... | a ) 24 , b ) 36 , c ) 72 , d ) 1 , e ) 216 | e | multiply(multiply(subtract(const_10, const_1), subtract(const_10, const_2)), const_1) | subtract(const_10,const_1)|subtract(const_10,const_2)|multiply(#0,#1)|multiply(#2,const_1)| | general |
a cistern can be filled by a tap in 7 hours while it can be emptied by another tap in 9 hours . if both the taps are opened simultaneously , then after how much time will the cistern get filled ? | "net part filled in 1 hour = 1 / 7 - 1 / 9 = 2 / 63 therefore the cistern will be filled in 63 / 2 hours or 31.5 hours . answer : e" | a ) 4.5 hrs , b ) 5 hrs , c ) 6.5 hrs , d ) 17.2 hrs , e ) 31.5 hrs | e | divide(const_1, subtract(divide(const_1, 7), divide(const_1, 9))) | divide(const_1,n0)|divide(const_1,n1)|subtract(#0,#1)|divide(const_1,#2)| | physics |
16.036 divided by 0.04 gives : | = 16.036 / 0.04 = 1603.6 / 4 = 400.9 answer is d . | a ) 40.09 , b ) 4.06 , c ) 40.06 , d ) 400.9 , e ) 400.6 | d | divide(16.036, 0.04) | divide(n0,n1) | general |
a canoe has two oars , left and right . each oar either works or breaks . the failure or non - failure of each oar is independent of the failure or non - failure of the other . you can still row the canoe with one oar . the probability that the left oar works is 3 / 5 . the probability that the right oar works is also ... | "simply look at the question from the other side . what is the probability that you can β t row the canoe ? this would be 2 / 5 x 2 / 5 = 4 / 25 . using the idea that the probability of something happening is 1 β the probability that it doesn β t happen , you can use the following equation to reach the right answer r :... | a ) 9 / 25 , b ) 10 / 25 , c ) 6 / 10 , d ) 2 / 3 , e ) 21 / 25 | e | add(multiply(divide(3, 5), divide(3, 5)), add(multiply(divide(3, 5), divide(const_2, 5)), multiply(divide(3, 5), divide(const_2, 5)))) | divide(n0,n1)|divide(const_2,n1)|multiply(#0,#1)|multiply(#0,#0)|add(#2,#2)|add(#4,#3)| | general |
a is the average ( arithmetic mean ) of the first 7 positive multiples of 8 and b is the median of the first 3 positive multiples of positive integer n . if the value of a ^ 2 β b ^ 2 is zero , what is the value of n ? | "if a ^ 2 - b ^ 2 = 0 , then let ' s assume that a = b . a must equal the 4 th positive multiple of 4 , thus a = 32 , which also equals b . b is the second positive multiple of n , thus n = 32 / 2 = 16 . the answer is c ." | a ) 10 , b ) 12 , c ) 16 , d ) 24 , e ) 32 | c | divide(multiply(8, const_4), 2) | multiply(n1,const_4)|divide(#0,n3)| | general |
the population of a bacteria culture doubles every 3 minutes . approximately how many minutes will it take for the population to grow from 1,000 to 500,000 bacteria | this one ' s easy . 1000 * 2 ^ t = 500,000 2 ^ t = 500 now gauging , since 2 ^ 8 = 256 , then 2 ^ 9 = 512 so t = 9 but be careful , ' t ' is in time intervals of 3 minutes so answer is 9 * 3 = 27 minutes answer ( a ) | a ) 27 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | a | multiply(log(divide(multiply(multiply(add(const_4, const_1), const_1000), const_100), const_1000)), 3) | add(const_1,const_4)|multiply(#0,const_1000)|multiply(#1,const_100)|divide(#2,const_1000)|log(#3)|multiply(n0,#4) | general |
shopkeeper rise price by 36 % and gives successive discount of 10 % and 15 % . what is overall % gain or loss ? | "let d initial price be 100 36 % rise now price = 136 / 100 * 100 = 136 10 % discount then price = 136 * 90 / 100 = 122.4 15 % discount then price = 122.4 * 85 / 100 = 104.04 so gain = 104.04 - 100 = 4.04 gain % = gain * 100 / cp = = > 4.04 * 100 / 100 = 4.04 % answer : b" | a ) 5.04 % , b ) 4.04 % , c ) 3.04 % , d ) 6.04 % , e ) 7.04 % | b | subtract(multiply(multiply(add(const_100, 36), divide(subtract(const_100, 10), const_100)), divide(subtract(const_100, 15), const_100)), const_100) | add(n0,const_100)|subtract(const_100,n2)|subtract(const_100,n1)|divide(#1,const_100)|divide(#2,const_100)|multiply(#0,#4)|multiply(#3,#5)|subtract(#6,const_100)| | gain |
set j consists of 18 consecutive even numbers . if the smallest term in the set is - 22 , what is the range of the positive integers in set j ? | "since there are only 18 integers , another approach is the just list all 18 . we get : - 22 , - 20 , - 18 , - 16 , - 14 , - 12 , - 10 , - 8 , - 6 , - 4 , - 2 , 0 , 2 , 4 , 6 , 8 , 10 , 12 range of positive integers = 12 - 2 = 10 answer : e" | a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | e | subtract(add(negate(22), multiply(subtract(18, const_1), 22)), 22) | negate(n1)|subtract(n0,const_1)|multiply(n1,#1)|add(#2,#0)|subtract(#3,n1)| | general |
suresh and ramesh started a business investing rs . 24,000 and rs . 40,000 respectively . out of total profit of rs . 18,000 , what is suresh ' s share ? | explanation : ratio of suresh and ramesh ' s share = 24,000 : 40,000 = 3 : 5 suresh ' s share = rs . ( 18000 x 3 / 8 ) = rs . 6750 answer : a | a ) s . 6750 , b ) s . 7750 , c ) s . 8750 , d ) s . 9372 , e ) s . 9750 | a | multiply(multiply(multiply(multiply(const_2, const_3), const_4), multiply(multiply(const_2, add(const_2, const_3)), const_100)), divide(multiply(multiply(const_3, multiply(const_2, const_3)), multiply(multiply(const_2, add(const_2, const_3)), const_100)), add(multiply(multiply(multiply(const_2, const_3), const_4), mult... | add(const_2,const_3)|multiply(const_2,const_3)|multiply(#1,const_3)|multiply(#0,const_2)|multiply(#1,const_4)|multiply(#3,const_100)|multiply(#3,const_4)|add(#6,const_2)|multiply(#2,#5)|multiply(#4,#5)|multiply(#7,#5)|add(#9,#10)|divide(#8,#11)|multiply(#12,#9) | gain |
what is the square root of 36 , divided by 6 ? | "square root is a number times itself square root of 36 = 6 , 6 / 6 = 1 ( e ) 1" | a ) 9 , b ) 36 , c ) 122 , d ) 6 , e ) 1 | e | divide(sqrt(36), 6) | sqrt(n0)|divide(#0,n1)| | other |
a man swims downstream 32 km and upstream 24 km taking 4 hours each time , what is the speed of the man in still water ? | "32 - - - 4 ds = 8 ? - - - - 1 24 - - - - 4 us = 6 ? - - - - 1 m = ? m = ( 8 + 6 ) / 2 = 7 answer : d" | a ) 9 , b ) 8 , c ) 6 , d ) 7 , e ) 2 | d | divide(add(divide(24, 4), divide(32, 4)), const_2) | divide(n1,n2)|divide(n0,n2)|add(#0,#1)|divide(#2,const_2)| | physics |
during a sale , the price of a pair of shoes is marked down 15 % from the regular price . after the sale ends , the price goes back to the original price . what is the percent of increase to the nearest percent from the sale price back to the regular price for the shoes ? | assume the price = 100 price during sale = 85 price after sale = 100 percent increase = 15 / 85 * 100 = 18 % approx . correct option : d | a ) 9 % , b ) 10 % , c ) 11 % , d ) 18 % , e ) 90 % | d | divide(multiply(15, const_100), subtract(const_100, 15)) | multiply(n0,const_100)|subtract(const_100,n0)|divide(#0,#1) | gain |
a and b can do a piece of work in 18 and 9 days respectively . a does the work for a few days and left , then b continued for 6 days . how many days did a and b work together ? | a and b work together for 2 days . a 1 day work 1 / 18 b 1 day work 1 / 9 let a work x days . then equation form is ( x / 18 ) + ( ( x + 6 ) / 9 ) = 1 from this equation x value is 2 . answer : b | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | b | divide(multiply(6, const_3), 9) | multiply(n2,const_3)|divide(#0,n1) | physics |
a and b can finish a work in 10 days while a alone can do the same work in 20 days . in how many days b alone will complete the work ? | "b = 1 / 10 β 1 / 20 = 1 / 20 = > 20 days answer : a" | a ) 20 days , b ) 48 days , c ) 98 days , d ) 31 days , e ) 22 days | a | inverse(subtract(inverse(10), inverse(20))) | inverse(n0)|inverse(n1)|subtract(#0,#1)|inverse(#2)| | physics |
a combustion reaction forms carbon dioxide . a carbon dioxide molecule contains one carbon and two oxygen atoms . if , over a period of 12 minutes , a combustion reaction creates 12,000 molecules of carbon dioxide then approximately how many more atoms of oxygen than carbon are created on average per minute ? | "solution : 12,000 carbon dioxide molecules are created over a period of 12 minutes . therefore 12,000 / 12 = 1,000 carbon dioxide molecules are created on average per minute each carbon dioxide molecule contains one carbon atom and two oxygen atoms . so 1,000 carbon dioxide molecules contain 1 Γ 1,000 = 1,000 carbon a... | a ) 20 , b ) 60 , c ) 600 , d ) 1,000 , e ) 1,200 | d | divide(divide(multiply(subtract(multiply(12, const_2), 12), const_1000), 12), const_1000) | multiply(n0,const_2)|subtract(#0,n0)|multiply(#1,const_1000)|divide(#2,n0)|divide(#3,const_1000)| | general |
8 ! / ( 8 - 2 ) ! = ? | 8 ! / ( 8 - 2 ) ! = 8 ! / 6 ! = 8 * 7 = 56 . hence , the correct answer is c . | a ) 36 , b ) 46 , c ) 56 , d ) 66 , e ) 96 | c | divide(factorial(8), factorial(subtract(8, 2))) | factorial(n0)|subtract(n0,n2)|factorial(#1)|divide(#0,#2)| | general |
sally , mark , sarah , steve and jim are inheriting the money from their grandmother . she left them $ 527500 to split equally between all of them . if the money is split equally , how much money will each person receive ? | there are 5 people inheriting the money . we have to take the total of the inheritance and divide it by 5 for each person receiving money . $ 527500 / 5 = $ 105500 . therefore the correct answer is d . | a ) $ 135875 , b ) $ 52700 , c ) $ 127900 , d ) $ 105500 , e ) $ 105000 | d | divide(527500, add(const_4, const_1)) | add(const_1,const_4)|divide(n0,#0) | general |
if the cost price of 165 chocolates is equal to the selling price of 150 chocolates , the gain percent is | "let c . p . of each chocolate be re . 1 . then , c . p . of 150 chocolates = rs . 150 ; s . p . of 150 chocolates = rs . 165 . gain % = 15 / 150 * 100 = 10 % answer : c" | a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | c | divide(const_100, divide(150, subtract(165, 150))) | subtract(n0,n1)|divide(n1,#0)|divide(const_100,#1)| | gain |
on a trip covering 450 km , a bicyclist travels the first 300 km at 20 km per hour and the remainder of the distance at 15 km per hour . what is the average speed , in km per hour , for the entire trip ? | "time 1 = 300 / 20 = 15 hours time 2 = 150 / 15 = 10 hours total time = 25 hours average speed = 450 / 25 = 18 km / hr the answer is d ." | a ) 16.5 , b ) 17.0 , c ) 17.5 , d ) 18.0 , e ) 18.5 | d | speed(450, add(divide(300, 20), divide(subtract(450, 300), 15))) | divide(n1,n2)|subtract(n0,n1)|divide(#1,n3)|add(#0,#2)|speed(n0,#3)| | physics |
jack and jill work at a hospital with 2 other workers . for an internal review , 2 of the 4 workers will be randomly chosen to be interviewed . what is the probability that jack and jill will both be chosen ? | "1 / 4 c 2 = 1 / 6 . answer : b ." | a ) 1 / 3 , b ) 1 / 6 , c ) 1 / 15 , d ) 3 / 8 , e ) 2 / 3 | b | inverse(divide(factorial(4), multiply(factorial(2), factorial(2)))) | factorial(n2)|factorial(n1)|factorial(n0)|multiply(#1,#2)|divide(#0,#3)|inverse(#4)| | physics |
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