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a train 150 m long passes a km stone in 15 seconds and another train of the same length travelling in opposite direction in 8 seconds . the speed of the second train is | "the 5 inch side should be aligned to the 15 inch side ( 3 layer ) 2 inch side should be aligned to the 20 inch side . ( 10 layer ) 7 inch side should be aligned to the 35 inch side . ( 5 layer ) maximum number of rectangles = 3 * 10 * 5 = 150 answer is d" | a ) 200 , b ) 350 , c ) 100 , d ) 150 , e ) 120 | d | divide(multiply(multiply(15, 20), 35), multiply(multiply(5, 2), 7)) | multiply(n3,n4)|multiply(n0,n1)|multiply(n5,#0)|multiply(n2,#1)|divide(#2,#3)| | geometry |
what is the max number of rectangular boxes , each measuring 5 inches by 2 inches by 7 inches , that can be packed into a rectangular packing box measuring 15 inches by 20 inches by 35 inches , if all boxes are aligned in the same direction ? | "p = 288 ( 100 / 5 ) ^ 2 = > p = 7200 answer : d" | a ) 2277 , b ) 2667 , c ) 3600 , d ) 7200 , e ) 1811 | d | divide(288, subtract(power(add(divide(20, const_100), const_1), 2), add(multiply(divide(20, const_100), 2), const_1))) | divide(n1,const_100)|add(#0,const_1)|multiply(n0,#0)|add(#2,const_1)|power(#1,n0)|subtract(#4,#3)|divide(n2,#5)| | gain |
the difference between the compound interest compounded annually and simple interest for 2 years at 20 % per annum is rs . 288 . find the principal ? | "we assume that 867 is 100 % assume ' x ' is value we looking for here , 867 = 100 % and x % = 63.2 therefore , 100 / x = 867 / 63.2 100 / x = 13.71 x = 7.29 c" | a ) 6.9 , b ) 8.99 , c ) 7.29 , d ) 7.98 , e ) 9.21 | c | multiply(divide(63.2, 867), const_100) | divide(n0,n1)|multiply(#0,const_100)| | gain |
63.2 is what percent of 867 ? | "work done by p and q in 1 day = 1 / 10 work done by r in 1 day = 1 / 15 work done by p , q and r in 1 day = 1 / 10 + 1 / 15 = 1 / 6 but work done by p in 1 day = work done by q and r in 1 day . hence the above equation can be written as work done by p in 1 day Γ£ β 2 = 1 / 6 = > work done by p in 1 day = 1 / 12 = > work done by q and r in 1 day = 1 / 12 hence work done by q in 1 day = 1 / 12 Γ’ β¬ β 1 / 15 = 1 / 60 so q alone can do the work in 60 days answer is e ." | a ) 20 , b ) 22 , c ) 25 , d ) 27 , e ) 60 | e | divide(const_1, subtract(divide(add(divide(const_1, 10), divide(const_1, 15)), const_2), divide(const_1, 15))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(#2,const_2)|subtract(#3,#1)|divide(const_1,#4)| | physics |
p can do a work in the same time in which q and r together can do it . if p and q work together , the work can be completed in 10 days . r alone needs 15 days to complete the same work . then q alone can do it in | "lets work with the data given to us . we know that there ratio of cats to dogs is 6 : 12 or cats 6 dogs 12 we can write number of cats as 6 x and number of dogs as 12 x and we know that 12 x - 6 x = 24 ( therefore 6 x = 24 = > x = 4 ) then # of dogs = 12 x 4 = 48 answer is b" | a ) 50 , b ) 48 , c ) 52 , d ) 54 , e ) 56 | b | multiply(24, 12) | multiply(n1,n2)| | other |
local kennel has cats and dogs in the ratio of 6 : 12 . if there are 24 fewer cats than dogs , how many dogs are in the kennel ? | "1 / 80 + 1 / 40 = 0.0375 days answer : d" | a ) 1.0875 days , b ) 0.1875 days , c ) 0.0675 days , d ) 0.0375 days , e ) 0.0775 days | d | inverse(add(inverse(80), inverse(40))) | inverse(n0)|inverse(n1)|add(#0,#1)|inverse(#2)| | physics |
a and b complete a work in 80 days . a alone can do it in 40 days . if both together can do the work in how many days ? | for simplicity , let ' s assume the customer bought 1 pound of ground beef for $ 1.80 . let x be the price per pound for the steak . then 0.8 x = 180 x = 180 / 0.8 = $ 2.25 the answer is c . | a ) $ 2.05 , b ) $ 2.15 , c ) $ 2.25 , d ) $ 2.35 , e ) $ 2.45 | c | divide(1.8, add(multiply(const_0_25, const_2), multiply(const_0_33, const_1))) | multiply(const_0_25,const_2)|multiply(const_0_33,const_1)|add(#0,#1)|divide(n0,#2) | general |
a customer purchased a package of ground beef at a cost of $ 1.80 per pound . for the same amount of money , the customer could have purchased a piece of steak that weighed 20 percent less than the package of ground beef . what was the cost per pound of the steak ? | "the number of exhaustive outcomes is 36 . let e be the event of getting a prime number on both the dice . p ( e ) = 8 / 36 = 2 / 9 . c )" | a ) 5 / 9 , b ) 1 / 9 , c ) 2 / 9 , d ) 4 / 9 , e ) 7 / 9 | c | divide(const_6, multiply(const_6, const_6)) | multiply(const_6,const_6)|divide(const_6,#0)| | other |
if two dice are thrown together , the probability of getting prime number on both dice is | say x guests ate both types of meat . ( 125 - x ) + ( 75 - x ) = 100 - - > x = 50 . answer : e . | a ) 5 , b ) 100 , c ) 7 , d ) 4 , e ) 50 | e | add(subtract(125, 100), subtract(100, 75)) | subtract(n0,n2)|subtract(n2,n1)|add(#0,#1) | other |
at a wedding reception , 125 guests ate chicken and 75 guests ate beef . if exactly 100 guests ate only one of the two types of meat , how many guests ate both types of meat ? | explanation : mon + tue + wed temperature = 3 x 36.3 = 108.9 tue + wed temperature = 108.9 β 39 = 69.9 tue + wed + thu temperature = 3 x 36.7 = 110.1 so , thursday β s temperature = 110.1 β 69.9 = 40.2 degrees c answer : c | a ) 60.2 degrees c , b ) 50.2 degrees c , c ) 40.2 degrees c , d ) 70.2 degrees c , e ) none of these | c | subtract(multiply(36.7, const_3), subtract(multiply(36.3, const_3), 39)) | multiply(n1,const_3)|multiply(n0,const_3)|subtract(#1,n2)|subtract(#0,#2) | general |
the average temperature for monday , tuesday and wednsday is 36.3 degrees c . the average temperature for tuesday , wednesday and thursday is 36.7 degrees c . if monday β s temperature recorded as 39 degrees c , find the thursday β s temperature ? | "originally ( 3 / 9 ) * 1350 = 450 acres were planted with tobacco . in the new system ( 5 / 9 ) * 1350 = 750 acres were planted with tobacco . thus 750 - 450 = 300 more acres were planted with tobacco . the answer is d ." | a ) 90 , b ) 150 , c ) 270 , d ) 300 , e ) 450 | d | subtract(multiply(add(add(multiply(const_100, const_10), multiply(const_3, const_100)), multiply(4, const_10)), divide(4, add(add(4, 2), 2))), multiply(add(add(multiply(const_100, const_10), multiply(const_3, const_100)), multiply(4, const_10)), divide(2, add(add(4, 2), 2)))) | add(n1,n2)|multiply(const_10,const_100)|multiply(const_100,const_3)|multiply(n1,const_10)|add(#1,#2)|add(n2,#0)|add(#4,#3)|divide(n1,#5)|divide(n2,#5)|multiply(#6,#7)|multiply(#6,#8)|subtract(#9,#10)| | other |
a farmer with 1,350 acres of land had planted his fields with corn , sugar cane , and tobacco in the ratio of 4 : 2 : 3 , respectively , but he wanted to make more money , so he shifted the ratio to 2 : 2 : 5 , respectively . how many more acres of land were planted with tobacco under the new system ? | "first recognize you only need to consider the first three digits ( because the second three are just the first three flipped ) there are 900 possibilities for the first three digits of a 6 digit number , 100 - 999 inclusive . everything starting with a 1 , 3,5 , 7,9 will be odd , which is 5 / 9 ths of the combinations . 5 / 9 * 900 = 500 answer : c" | a ) 400 , b ) 450 , c ) 500 , d ) 900 , e ) 2500 | c | divide(power(const_10, divide(6, const_2)), const_2) | divide(n1,const_2)|power(const_10,#0)|divide(#1,const_2)| | general |
a palindrome is a number that reads the same forward and backward , such as 343 . how many odd , 6 - digit numbers are palindromes ? | "a 2400 let the amount with r be $ r r = 2 / 3 ( total amount with p and q ) r = 2 / 3 ( 6000 - r ) = > 3 r = 12000 - 2 r = > 5 r = 12000 = > r = 2400 ." | a ) 2400 , b ) 2403 , c ) 3998 , d ) 2539 , e ) 1930 | a | divide(multiply(6000, multiply(const_2, const_2)), add(add(multiply(divide(multiply(const_2, const_2), const_3), const_3), multiply(const_1, const_3)), multiply(const_1, const_3))) | multiply(const_2,const_2)|multiply(const_1,const_3)|divide(#0,const_3)|multiply(n0,#0)|multiply(#2,const_3)|add(#4,#1)|add(#5,#1)|divide(#3,#6)| | general |
p , q and r have $ 6000 among themselves . r has two - thirds of the total amount with p and q . find the amount with r ? | "time difference is 1 hr , 22 min , 49 sec = 4969 sec . so , light glows floor ( 4969 / 15 ) = 331 times . answer : e" | a ) 380 times , b ) 381 times , c ) 382 times , d ) 392 times , e ) 331 times | e | divide(add(add(const_2, 47), multiply(add(20, add(const_2, const_60)), const_60)), 15) | add(n6,const_2)|add(const_2,const_60)|add(n5,#1)|multiply(#2,const_60)|add(#0,#3)|divide(#4,n0)| | general |
light glows for every 15 seconds . how many max . times did it glow between 1 : 57 : 58 and 3 : 20 : 47 am . | "sol . average speed = [ 2 * 3 * 2 / 3 + 2 ] km / hr = 12 / 5 km / hr . distance travelled = [ 12 / 5 * 5 ] km = 12 km . β΄ distance between house and school = [ 12 / 2 ] km = 6 km . answer c" | a ) 4.5 km , b ) 5.5 km , c ) 6 km , d ) 7 km , e ) none | c | multiply(divide(5, add(divide(3, 2), const_1)), 3) | divide(n0,n1)|add(#0,const_1)|divide(n2,#1)|multiply(n0,#2)| | physics |
a boy goes to his school from his house at a speed of 3 km / hr and returns at a speed of 2 km / hr . if he takes 5 hours in going and coming . the distance between his house and school is : | "number of passengers on flight = 120 number of female passengers = . 6 * 120 = 72 number of passengers in first class = ( 10 / 100 ) * 120 = 12 number of passengers in coach class = ( 90 / 100 ) * 120 = 108 number of male passengers in first class = 1 / 3 * 12 = 4 number of female passengers in first class = 12 - 4 = 8 number of female passengers in coach class = 72 - 8 = 64 answer d" | a ) 44 , b ) 48 , c ) 50 , d ) 64 , e ) 56 | d | subtract(multiply(120, divide(60, const_100)), subtract(multiply(120, divide(10, const_100)), divide(multiply(120, divide(10, const_100)), 3))) | divide(n2,const_100)|divide(n3,const_100)|multiply(n0,#0)|multiply(n0,#1)|divide(#3,n5)|subtract(#3,#4)|subtract(#2,#5)| | gain |
of the 120 passengers on flight 750 , 60 % are female . 10 % of the passengers sit in first class , and the rest of the passengers sit in coach class . if 1 / 3 of the passengers in first class are male , how many females are there in coach class ? | "radius / side = 2 / 7 Γ’ β‘ β area of circle / area of square = 4 / 49 answer : d" | a ) 2 : 1 , b ) 4 : 7 , c ) 8 : 77 , d ) 4 : 49 , e ) none | d | power(divide(2, 7), 2) | divide(n0,n1)|power(#0,n0)| | geometry |
the ratio of radius of a circle and the side of a square is 2 : 7 . find the ratio of their areas : | by the rule of alligation : c . p . of 1 kg sugar of 1 st kind c . p . of 1 kg sugar of 2 nd kind { \ color { blue } \ therefore } ratio of quantities of 1 st and 2 nd kind = 14 : 6 = 7 : 3 . let x kg of sugar of 1 st kind be mixed with 27 kg of 2 nd kind . then , 7 : 3 = x : 27 or x = ( 7 x 27 / 3 ) = 63 kg . answer : d ) 63 kg | a ) 33 , b ) 39 , c ) 38 , d ) 63 , e ) 01 | d | divide(subtract(multiply(27, divide(9.24, add(divide(10, const_100), const_1))), multiply(27, 7)), subtract(9, divide(9.24, add(divide(10, const_100), const_1)))) | divide(n3,const_100)|multiply(n1,n2)|add(#0,const_1)|divide(n4,#2)|multiply(n1,#3)|subtract(n0,#3)|subtract(#4,#1)|divide(#6,#5) | gain |
how many kilograms of sugar costing rs . 9 per kg must be mixed with 27 kg of sugar costing rs . 7 per kg so that there may be a gain of 10 % by selling the mixture at rs . 9.24 per kg ? | "s . i for 5 years = ( 1020 - 820 ) = rs . 200 . s . i . for 2 years = 200 / 5 * 2 = rs . 80 . principal = ( 820 - 80 ) = rs . 740 . answer : d" | a ) rs . 440 , b ) rs . 500 , c ) rs . 540 , d ) rs . 740 , e ) rs . 840 | d | subtract(820, multiply(divide(subtract(1020, 820), 5), 2)) | subtract(n2,n0)|divide(#0,n3)|multiply(n1,#1)|subtract(n0,#2)| | general |
a sum of money lent out at s . i . amounts to rs . 820 after 2 years and to rs . 1020 after a further period of 5 years . the sum is ? | "let x is the cost of the food 1.07 x is the gross bill after including sales tax 1.15 * 1.07 x = 60 x = 48.7 hence , the correct option is e" | a ) 39.55 $ , b ) 40.63 $ , c ) 41.63 $ , d ) 42.15 $ , e ) 48.7 $ | e | divide(60, add(divide(add(7, 15), const_100), const_1)) | add(n1,n2)|divide(#0,const_100)|add(#1,const_1)|divide(n0,#2)| | general |
a business executive and his client are charging their dinner tab on the executive ' s expense account . the company will only allow them to spend a total of 60 $ for the meal . assuming that they will pay 7 % in sales tax for the meal and leave a 15 % tip , what is the most their food can cost ? | sol . income in first year = * x 42000 = rs . 24000 expenses in second year = \ x 21000 = rs . 35000 total savings = total income - total expenses = ( 42000 + 24000 ) - ( 21000 + 35000 ) = 66000 - s 6000 = rs . 10000 e | a ) rs . 8000 , b ) rs . 9000 , c ) rs . 9800 , d ) rs . 9900 , e ) rs . 10000 | e | add(subtract(42000, divide(multiply(21000, 5), 3)), subtract(divide(multiply(42000, 4), 7), 21000)) | multiply(n4,n7)|multiply(n1,n6)|divide(#0,n3)|divide(#1,n2)|subtract(n6,#2)|subtract(#3,n7)|add(#4,#5) | general |
for 2 consecutive yrs , my incomes are in the ratio of 4 : 7 and expenses in the ratio of 3 : 5 . if my income in the 2 nd yr is rs . 42000 & my expenses in the first yr in rs . 21000 , my total savings for the two - year is | "solution required numbers are 102 , 018,114 , . . . . 996 . this is an a . p with a = 102 , d = 6 . let the number of its terms be n . then a + ( n - 1 ) d βΉ = βΊ 102 + ( n - 1 ) Γ 6 = 996 βΉ = βΊ n = 150 . answer b" | a ) 149 , b ) 150 , c ) 151 , d ) 166 , e ) none | b | add(multiply(const_100, 6), const_100) | multiply(n0,const_100)|add(#0,const_100)| | general |
how many three - digit numbers are divisible by 6 in all ? | "speed of boat in still water = 12 km / hr speed of the stream = 4 km / hr speed downstream = ( 12 + 4 ) = 16 km / hr time taken to travel 68 km downstream = 68 β 16 = 17 β 4 = 4.25 hours answer is a" | a ) 4.25 hr , b ) 5.25 hr , c ) 8.25 hr , d ) 2.25 hr , e ) 2.50 hr | a | divide(68, add(12, 4)) | add(n0,n1)|divide(n2,#0)| | physics |
a boat can travel with a speed of 12 km / hr in still water . if the speed of the stream is 4 km / hr , find the time taken by the boat to go 68 km downstream . | if saran and david submit separate orders , each would be smaller than 100 photocopies , so no discount . each would pay ( 80 ) * ( $ 0.02 ) = $ 1.60 , or together , a cost of $ 3.20 - - - that ' s the combinedno discount cost . if they submit things together as one big order , they get a discount off of that $ 3.20 price - - - - 25 % or 1 / 4 of that is $ 0.80 , the discount on the combined sale . they each effective save half that amount , or $ 0.40 . answer = ( b ) . | a ) $ 0.32 , b ) $ 0.40 , c ) $ 0.45 , d ) $ 0.48 , e ) $ 0.54 | b | divide(subtract(multiply(const_2, multiply(80, 0.02)), multiply(multiply(160, divide(subtract(100, 25), 100)), 0.02)), const_2) | multiply(n0,n3)|subtract(n2,n1)|divide(#1,n2)|multiply(#0,const_2)|multiply(n4,#2)|multiply(n0,#4)|subtract(#3,#5)|divide(#6,const_2) | gain |
the cost of one photocopy is $ 0.02 . however , a 25 % discount is offered on orders of more than 100 photocopies . if saran and david have to make 80 copies each , how much will each of them save if they submit a single order of 160 copies ? | "we can use fractional equivalents here to solve the problem 80 % = 4 / 5 ; this means that in 1 st case if she prepares 5 bears , in 2 nd case she prepares 9 bears 10 % = 1 / 10 ; this means that in 1 st case if she needs 10 hours , in 2 nd case she needs 9 hours now we come to productivity based on above fractional values the productivity in 1 st case is 0.5 bears / hour and in the 2 nd case it is 1 bear / hour hence the productivity is double with the assistant i . e . the increase in productivity is 200 % e" | a ) 20 % , b ) 80 % , c ) 100 % , d ) 180 % , e ) 200 % | e | multiply(divide(10, subtract(subtract(const_100, 100), 10)), const_100) | subtract(const_100,n0)|subtract(#0,n1)|divide(n1,#1)|multiply(#2,const_100)| | physics |
jane makes toy bears . when she works with an assistant , she makes 100 percent more bears per week and works 10 percent fewer hours each week . having an assistant increases jane β s output of toy bears per hour by what percent ? | "explanation : amount of work p can do in 1 day = 1 / 10 amount of work q can do in 1 day = 1 / 12 amount of work p and q can do in 1 day = 1 / 10 + 1 / 12 = 11 / 60 amount of work p and q can together do in 5 days = 5 Γ ( 11 / 60 ) = 11 / 12 fraction of work left = 1 β 11 / 12 = 1 / 12 answer : option c" | a ) 7 / 12 , b ) 5 / 12 , c ) 1 / 12 , d ) 3 / 12 , e ) 1 / 2 | c | subtract(const_1, multiply(add(divide(const_1, 12), divide(const_1, 10)), 5)) | divide(const_1,n1)|divide(const_1,n0)|add(#0,#1)|multiply(n2,#2)|subtract(const_1,#3)| | physics |
p is able to do a piece of work in 10 days and q can do the same work in 12 days . if they can work together for 5 days , what is the fraction of work left ? | "perimeter = distance covered in 6 min . = ( 12000 / 60 ) x 6 m = 1200 m . let length = 3 x metres and breadth = 2 x metres . then , 2 ( 3 x + 2 x ) = 1200 or x = 120 . length = 360 m and breadth = 240 m . area = ( 360 x 240 ) m 2 = 86400 m 2 . answer : c" | a ) 153601 , b ) 153600 , c ) 86400 , d ) 153603 , e ) 153604 | c | rectangle_area(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 6), const_1000), add(3, 2)), const_2), multiply(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 6), const_1000), add(3, 2)), const_2), 2)) | add(n0,n1)|multiply(const_2,const_3)|multiply(#1,const_10)|divide(n2,#2)|multiply(n3,#3)|multiply(#4,const_1000)|divide(#5,#0)|divide(#6,const_2)|multiply(n1,#7)|rectangle_area(#7,#8)| | physics |
the ratio between the length and the breadth of a rectangular park is 3 : 2 . if a man cycling along the boundary of the park at the speed of 12 km / hr completes one round in 6 minutes , then the area of the park ( in sq . m ) is : | "area of a parallelogram = base * height = 24 * 12 = 288 cm 2 answer : e" | a ) 297 cm 2 , b ) 384 cm 2 , c ) 672 cm 2 , d ) 267 cm 2 , e ) 288 cm 2 | e | multiply(24, 12) | multiply(n0,n1)| | geometry |
find the area of a parallelogram with base 24 cm and height 12 cm ? | p ( of not burning out in a six mnth period ) = 1 / 2 of p ( of not burning out in prev 6 mnth period ) p ( of burning out in 1 st 6 mnth ) = 2 / 3 - - - > p ( of not burning out in 1 st 6 mnth ) = 1 - 2 / 3 = 1 / 3 - - - - > p ( of not burning out in a six mnth period ) = 1 / 2 * 1 / 3 = 1 / 6 - - - > p ( of burning out in a six mnth period ) = 1 - 1 / 3 = 2 / 3 now p ( of burning out in 2 nd six mnth period ) = p ( of not burning out in 1 st six mnth ) * p ( of burning out in a six mnth ) = 2 / 3 * 1 / 6 = 2 / 7 ans e | a ) 5 / 27 , b ) 2 / 9 , c ) 1 / 3 , d ) 4 / 9 , e ) 2 / 7 | e | multiply(subtract(1, divide(2, 3)), subtract(1, divide(subtract(1, divide(2, 3)), 2))) | divide(n3,n4)|subtract(n6,#0)|divide(#1,n3)|subtract(n6,#2)|multiply(#1,#3) | general |
for each 6 - month period during a light bulb ' s life span , the odds of it not burning out from over - use are half what they were in the previous 6 - month period . if the odds of a light bulb burning out during the first 6 - month period following its purchase are 2 / 3 , what are the odds of it burning out during the period from 6 months to 1 year following its purchase ? | "we are given that 0.05 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 0.05 by 20 . this gives us : 0.05 x 20 = 1 ounces finally , we are asked for β what percent β of the original amount of water evaporated during this period . to determine this percentage , we have to make sure we translate the expression correctly . we can translate it to : ( amount evaporated / original amount ) x 100 % ( 1 / 10 ) x 100 % ( 10 / 100 ) x 100 % = 10 % answer e" | a ) 0.002 % , b ) 0.02 % , c ) 0.2 % , d ) 2 % , e ) 10 % | e | multiply(divide(multiply(0.05, 20), 10), const_100) | multiply(n1,n2)|divide(#0,n0)|multiply(#1,const_100)| | gain |
a glass was filled with 10 ounces of water , and 0.05 ounce of the water evaporated each day during a 20 - day period . what percent of the original amount of water evaporated during this period ? | "explanation : let b contribution is x . 3500 * 12 / 7 x = 2 / 3 = > 14 x = 126000 = > x = rs 9000 option a" | a ) rs 9000 , b ) rs 7000 , c ) rs 5000 , d ) rs 4000 , e ) none of these | a | divide(multiply(multiply(3500, const_12), 3), multiply(subtract(const_12, 5), 2)) | multiply(n0,const_12)|subtract(const_12,n1)|multiply(n3,#0)|multiply(n2,#1)|divide(#2,#3)| | other |
a starts business with rs . 3500 and after 5 months , b joins with a as his partner . after a year , the profit is divided in the ratio 2 : 3 . what is b ' s contribution in the capital | "explanation : required percentage = ( 16 / 5200 * 100 ) % = 3 / 10 % = 0.30 % answer : a ) . 30 %" | a ) 30 , b ) 66 , c ) 58 , d ) 29 , e ) 17 | a | multiply(divide(5.2, 16), const_100) | divide(n0,n1)|multiply(#0,const_100)| | gain |
what percent of 5.2 kg is 16 gms ? | "solution 32.5 let p . w . be rs . x . then , s . i . on rs . x at 16 % for 9 months = rs . 162 . β΄ x 16 x 9 / 12 x 1 / 100 } = 162 or x = 1350 . β΄ p . w . = rs . 1350 . answer b" | a ) rs . 1386 , b ) rs . 1350 , c ) rs . 1575 , d ) rs . 2268 , e ) none of these | b | add(divide(162, divide(multiply(divide(9, multiply(const_4, const_3)), 16), const_100)), 162) | multiply(const_3,const_4)|divide(n0,#0)|multiply(n1,#1)|divide(#2,const_100)|divide(n2,#3)|add(n2,#4)| | gain |
the true discount on a bill due 9 months hence at 16 % per annum is rs . 162 . the amount of the bill is | "net part filled in 1 hour 1 / 5 - 1 / 10 = 1 / 10 the cistern will be filled in 10 hr answer is b" | a ) 20 hr , b ) 10 hr , c ) 5 hr , d ) 4 hr , e ) 15 hr | b | divide(const_1, subtract(divide(const_1, 5), divide(const_1, 10))) | divide(const_1,n0)|divide(const_1,n1)|subtract(#0,#1)|divide(const_1,#2)| | physics |
a cistern can be filled by a tap in 5 hours while it can be emptied by another tap in 10 hours . if both the taps are opened simultaneously then after how much time will the cistern get filled ? | charges for 7 hours = ( first hour @ $ 0.50 ) + ( 3 hours @ $ 3 ) + ( 3.5 hours @ $ 1 ) charges for 7 hours = ( 1 @ $ 0.50 ) + ( 3 hours @ $ 3 ) + ( 3.5 hours @ $ 1 ) charges for 7 hours = ( $ 0.50 ) + ( $ 9 ) + ( $ 3.5 ) charges for 7 hours = ( $ 0.50 ) + ( $ 9 ) + ( $ 3.50 ) charges for 7 hours = $ 13 hence correct answer must be ( c ) | a ) $ 11.5 . , b ) $ 12 . , c ) $ 13 . , d ) $ 14.5 , e ) $ 15 . | c | add(add(multiply(3, 3), multiply(add(subtract(7, 4), divide(50, const_100)), 1)), divide(50, const_100)) | divide(n0,const_100)|multiply(n1,n1)|subtract(n4,n2)|add(#0,#2)|multiply(n3,#3)|add(#1,#4)|add(#5,#0) | physics |
the mall charges 50 cents for the first hour of parking and $ 3 for each additional hour until the customer reaches 4 hours , after that the parking fee is $ 1 per hour . if a certain customer parked his in the mall for 7 hours and 30 minutes , how much is he going to pay ? | explanation : suppose y invested rs . y . then 40000 / y = 2 / 3 or y = 60000 . answer : c ) 60000 | a ) 33488 , b ) 63809 , c ) 60000 , d ) 37887 , e ) 77824 | c | multiply(divide(multiply(40000, add(2, 3)), 2), divide(3, add(2, 3))) | add(n0,n1)|divide(n1,#0)|multiply(n2,#0)|divide(#2,n0)|multiply(#3,#1) | gain |
x and y invested in a business . they earned some profit which they divided in the ratio of 2 : 3 . if x invested rs . 40000 , the amount invested by y is | "a square with an area of 4 has a perimeter of 8 . for the area to be > 4 , the longer piece must be > 8 . the wire must be cut within 4 meters from either end . the probability of this is 8 / 12 = 2 / 3 . the answer is b ." | a ) 1 / 6 , b ) 2 / 3 , c ) 3 / 10 , d ) 1 / 3 , e ) 2 / 5 | b | multiply(const_2, divide(const_2, 12)) | divide(const_2,n0)|multiply(#0,const_2)| | geometry |
a 12 meter long wire is cut into two pieces . if the longer piece is then used to form a perimeter of a square , what is the probability that the area of the square will be more than 4 if the original wire was cut at an arbitrary point ? | sp = 1.34 * 245 = 328 answer : a | a ) 328 , b ) 320 , c ) 300 , d ) 207 , e ) 310 | a | add(245, multiply(245, divide(34, const_100))) | divide(n1,const_100)|multiply(n0,#0)|add(n0,#1) | gain |
an article with cost price of 245 is sold at 34 % profit . what is the selling price ? | "assume both trains meet after x hours after 7 am distance covered by train starting from p in x hours = 20 x km distance covered by train starting from q in ( x - 1 ) hours = 25 ( x - 1 ) total distance = 155 = > 20 x + 25 ( x - 1 ) = 155 = > 45 x = 180 = > x = 4 means , they meet after 3 hours after 7 am , ie , they meet at 11 am answer is e ." | a ) 10 am , b ) 12 am , c ) 10.30 am , d ) 12.30 am , e ) 11 am | e | add(divide(add(155, 25), add(20, 25)), 7) | add(n0,n4)|add(n2,n4)|divide(#0,#1)|add(n1,#2)| | physics |
two stations p and q are 155 km apart on a straight track . one train starts from p at 7 a . m . and travels towards q at 20 kmph . another train starts from q at 8 a . m . and travels towards p at a speed of 25 kmph . at what time will they meet ? | "a 23 years let the average age of the whole team by x years . 11 x - ( 26 + 29 ) = 9 ( x - 1 ) 11 x - 9 x = 46 2 x = 46 x = 23 . so , average age of the team is 23 years" | a ) 23 years , b ) 20 years , c ) 25 years , d ) 22 years , e ) 28 years | a | divide(subtract(add(26, add(26, 3)), multiply(3, 3)), const_2) | add(n1,n2)|multiply(n2,n2)|add(n1,#0)|subtract(#2,#1)|divide(#3,const_2)| | general |
the captain of a cricket team of 11 members is 26 years old and the wicket keeper is 3 years older . if the ages of these two are excluded , the average age of the remaining players is one year less than the average age of the whole team . what is the average age of the team ? | original number of sections = 16 - 3 = 13 original number of students = 24 x 13 = 312 present number of students = 21 x 16 = 336 number of new students admitted = 336 - 312 = 24 so the answer is option c ) 24 . | a ) 12 , b ) 42 , c ) 24 , d ) 28 , e ) 26 | c | subtract(multiply(21, 16), multiply(24, subtract(16, 3))) | multiply(n2,n3)|subtract(n2,n1)|multiply(n0,#1)|subtract(#0,#2) | physics |
the number of students in each section of a school is 24 . after admitting new students , 3 new sections were started . now , the total number of sections is 16 and there are 21 students in each section . the number of new students admitted is : | = ( 81 ) ^ 2 + ( 68 ) ^ 2 β 2 x 81 x 68 = a ^ 2 + b ^ 2 β 2 ab , where a = 81 , b = 68 = ( a - b ) ^ 2 = ( 81 β 68 ) ^ 2 = ( 13 ) ^ 2 = 169 . answer is a . | a ) 169 , b ) 159 , c ) 189 , d ) 179 , e ) 219 | a | add(81, 81) | add(n0,n0) | general |
simplify : 81 x 81 + 68 x 68 - 2 x 81 x 68 . | "= > - 24 * ( 30 - 1 ) + 1240 ; = > - ( 24 * 30 ) + 24 + 1240 ; = > - 720 + 1264 = 544 . correct option : c" | a ) - 544 , b ) 584 , c ) 544 , d ) 345 , e ) none of these | c | add(multiply(negate(24), 29), 1240) | negate(n0)|multiply(n1,#0)|add(n2,#1)| | general |
- 24 * 29 + 1240 = ? | "the total amount of alcohol is 0.25 ( 4 ) + 0.11 ( 3 ) = 1.33 liters . the percentage is 1.33 / 7 = 133 / 700 = 19 / 100 which is 19 % the answer is c ." | a ) 18.2 % , b ) 18.6 % , c ) 19.0 % , d ) 19.4 % , e ) 19.8 % | c | multiply(divide(add(multiply(4, divide(25, const_100)), multiply(divide(11, const_100), multiply(4, divide(25, const_100)))), add(4, 3)), const_100) | add(n0,n2)|divide(n1,const_100)|divide(n3,const_100)|multiply(n0,#1)|multiply(#2,#3)|add(#3,#4)|divide(#5,#0)|multiply(#6,const_100)| | general |
4 liters of a 25 percent solution of alcohol in water are mixed with 3 liters of a 11 percent alcohol in water solution . what is the percentage of alcohol in the new solution ? | ratio of investment , as investments is for different time . investment x number of units of time . ratio of investments x : y : z = 36000 : 42000 : 48000 = > 6 : 7 : 8 . x = 6 x 12 months = 72 , y = 7 x 12 = 84 , z = 8 x 8 = 64 = > 18 : 21 : 16 . ratio of investments = > x : y : z = 18 : 21 : 16 . investment ratio = profit sharing ratio . z = 13750 Γ 16 / 55 = rs . 4000 . share of z in the profit is rs . 4000 . option b | a ) 3200 , b ) 4000 , c ) 3250 , d ) 3825 , e ) 3985 | b | multiply(multiply(48000, subtract(multiply(const_3, const_4), const_4)), divide(13750, add(add(multiply(36000, multiply(const_3, const_4)), multiply(42000, multiply(const_3, const_4))), multiply(48000, subtract(multiply(const_3, const_4), const_4))))) | multiply(const_3,const_4)|multiply(n0,#0)|multiply(n1,#0)|subtract(#0,const_4)|add(#1,#2)|multiply(n3,#3)|add(#4,#5)|divide(n4,#6)|multiply(#7,#5) | gain |
x and y started a business by investing rs . 36000 and rs . 42000 respectively after 4 months z joined in the business with an investment of rs . 48000 , then find share of z in the profit of rs . 13750 ? | "machine b : takes x hours to produce 550 sprockets machine a : takes ( x + 10 ) hours to produce 550 sprockets machine b : in 1 hour , b makes 550 / x sprockets machine a : in 1 hour , a makes 550 / ( x + 10 ) sprockets equating : 1.1 ( 550 / ( x + 10 ) ) = 550 / x 605 / ( x + 10 ) = 550 / x 605 x = 550 x + 5500 55 x = 5500 x = 100 a makes 550 / ( 110 ) = 5 sprockets per hour answer : b" | a ) 6 , b ) 5 , c ) 7 , d ) 8 , e ) 9 | b | divide(550, divide(multiply(multiply(10, 550), divide(add(const_100, 10), const_100)), subtract(multiply(550, divide(add(const_100, 10), const_100)), 550))) | add(n1,const_100)|multiply(n0,n1)|divide(#0,const_100)|multiply(#2,#1)|multiply(n0,#2)|subtract(#4,n0)|divide(#3,#5)|divide(n0,#6)| | gain |
machine a and machine b are each used to manufacture 550 sprockets . it takes machine a 10 hours longer to produce 550 sprockets than machine b . machine b produces 10 percent more sprockets per hour than machine a . how many sprockets per hour does machine a produces ? | "b 13.2 sec d = 100 + 120 = 220 m s = 60 * 5 / 18 = 50 / 3 t = 220 * 3 / 50 = 13.2 sec answer is b" | a ) 15.8 sec , b ) 13.2 sec , c ) 12.4 sec , d ) 16.8 sec , e ) 11.8 sec | b | divide(add(100, 120), multiply(60, const_0_2778)) | add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)| | physics |
how long does a train 100 m long travelling at 60 kmph takes to cross a bridge of 120 m in length ? | after the reservoir is filled to 7.5 gallons the amount of water is at 55 % - which means that 45 % of the reservoir is empty . to figure out what that 45 % is approximate : 7.5 gallons / 55 percent = x gallons / 45 percent , therefore , x = 6.136 gallons , answer choices e , b , c , d are below 6.136 . we know that the reservoir must be short more than 6.136 gallons , therefore , the only possible choice is a . | a ) 6.9 , b ) 1.4 , c ) 2.5 , d ) 3.0 , e ) 4.4 | a | divide(divide(multiply(7.5, const_100), 55), const_2) | multiply(n1,const_100)|divide(#0,n2)|divide(#1,const_2) | general |
a rainstorm increased the amount of water stored in state j reservoirs from 5 billion gallons to 7.5 billion gallons . if the storm increased the amount of water in the reservoirs to 55 percent of total capacity , approximately how many billion gallons of water were the reservoirs short of total capacity prior to the storm ? | "u dont need to go through all this what u have with u is 100 p + 200 c = $ 4.50 just divide the equation by 5 and you will get what u are looking for 20 p + 40 c = $ 0.90 therefore oa is a" | a ) $ . 90 , b ) $ 1.00 , c ) $ 1.20 , d ) $ 1.50 , e ) $ 1.60 | a | multiply(divide(20, 100), 4.50) | divide(n3,n0)|multiply(n2,#0)| | gain |
the total cost of 100 paper plates and 200 paper cups is $ 4.50 at the same rates what is the total cost of 20 of the plates and 40 of the cups ? | "( 1 / 2 + 1 / 3 ) t = 2 t = 12 / 5 answer : c" | a ) 8 / 15 , b ) 4 / 3 , c ) 12 / 5 , d ) 9 / 4 , e ) 15 / 4 | c | multiply(divide(const_1, add(divide(const_1, 2), divide(const_1, 3))), 2) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)|multiply(#3,n0)| | physics |
kathleen can paint a room in 2 hours , and anthony can paint an identical room in 3 hours . how many hours would it take kathleen and anthony to paint both rooms if they work together at their respective rates ? | "a takes time 1.20 minutes = 80 sec b takes time 1.30 minutes = 90 sec diffrence = 90 - 80 = 10 sec now we are to find distance covered in 10 sec by b 90 sec = 30 m 1 sec = 30 m 10 sec = 10 x 30 = 300 m answer : d" | a ) 90 m , b ) 30 m , c ) 120 m , d ) 300 m , e ) 190 m | d | subtract(multiply(const_2.0, const_1000), multiply(divide(multiply(2.7, const_1000), add(multiply(1, const_60), 30)), add(multiply(2.7, const_60), 20))) | multiply(n0,const_1000)|multiply(n1,const_60)|multiply(n3,const_60)|add(n2,#1)|add(n4,#2)|divide(#0,#4)|multiply(#3,#5)|subtract(#0,#6)| | physics |
a can run 2.7 km distance in 1 min 20 seconds , while b can run this distance in 1 min 30 sec . by how much distance can a beat b ? | "5 * 5 answer : a" | a ) 1 , b ) 5 , c ) 4 , d ) 6 , e ) 8 | a | add(power(const_2, const_2), const_2) | power(const_2,const_2)|add(#0,const_2)| | other |
how many different positive integers are factors of 25 ? | 4 x * x = 2304 = > x = 24 answer : c | ['a ) 19', 'b ) 23', 'c ) 24', 'd ) 16', 'e ) 17'] | c | sqrt(divide(add(add(multiply(const_1000, const_2), multiply(const_100, const_3)), const_4), const_4)) | multiply(const_1000,const_2)|multiply(const_100,const_3)|add(#0,#1)|add(#2,const_4)|divide(#3,const_4)|sqrt(#4) | geometry |
a parallelogram has a base that is four time the size of it ' s height . the total area of this parallelogram is 2,304 sq ft . what is the height of the parallelogram ? | "d = 150 + 170 = 320 m s = 60 * 5 / 18 = 50 / 3 t = 320 * 3 / 50 = 19.2 sec answer : b" | a ) 16.5 , b ) 19.2 , c ) 16.4 , d ) 16.8 , e ) 16.1 | b | divide(add(150, 170), multiply(60, const_0_2778)) | add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)| | physics |
how long does a train 150 m long traveling at 60 kmph takes to cross a bridge of 170 m in length ? | "we are selecting from non - negative single digit integers , so from { 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 } . these 10 digits represent the total number of outcomes . hence , the total number of outcomes is 10 . we need to find the probability that the median of the set will increase but the range still remains the same . the median of the set is ( 3 + 4 ) / 2 = 3.5 , thus the number selected must be 4 or greater . for the range to remain the same , the number must be between 2 and 7 inclusive . to satisfy both conditions , the number selected must be 4 , 5 , 6 , or 7 . the probability is 4 / 10 = 0.4 the answer is c ." | a ) 0.2 , b ) 0.3 , c ) 0.4 , d ) 0.5 , e ) 0.6 | c | divide(const_4, const_10) | divide(const_4,const_10)| | general |
if a randomly selected non - negative single digit integer is added to { 2 , 3 , 4 , 7 } . what is the probability that the median of the set will increase but the range still remains the same ? | "14 ^ 2 has units digit 6 15 ^ 8 has units digit 5 thus w = 14 ^ 2 * 15 ^ 8 has units digit 0 and will be divisible by 5 . the remainder will be zero answer : ( a )" | a ) 0 , b ) 1 , c ) 2 , d ) 4 , e ) 5 | a | divide(5, 5) | divide(n4,n4)| | general |
what is the remainder when the number w = 14 ^ 2 * 15 ^ 8 is divided by 5 ? | "let the numbers be 13 a and 13 b . then , 13 a * 13 b = 2028 = > ab = 12 . now , co - primes with product 12 are ( 1 , 12 ) and ( 3 , 4 ) . so , the required numbers are ( 13 * 1 , 13 * 12 ) and ( 13 * 3 , 13 * 4 ) . clearly , there are 2 such pairs . answer : b" | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | b | sqrt(add(power(sqrt(subtract(13, multiply(const_2, 2028))), const_2), multiply(const_4, 2028))) | multiply(n0,const_4)|multiply(n0,const_2)|subtract(n1,#1)|sqrt(#2)|power(#3,const_2)|add(#0,#4)|sqrt(#5)| | general |
the product of two numbers is 2028 and their h . c . f is 13 . the number of such pairs is : | "let x be the number he chose , then 6 β
x β 250 = 122 6 x = 372 x = 62 correct answer d" | a ) 59 , b ) 60 , c ) 61 , d ) 62 , e ) 63 | d | divide(add(122, 250), 6) | add(n1,n2)|divide(#0,n0)| | general |
a student chose a number , multiplied it by 6 , then subtracted 250 from the result and got 122 . what was the number he chose ? | "prime numbers between 30 and 50 are 37 , 41 , 43 , 47 required average = ( 37 + 41 + 43 + 47 ) / 4 = 168 / 4 = 42 answer is b" | a ) 15 , b ) 42 , c ) 45 , d ) 34 , e ) 26 | b | divide(add(add(add(30, const_1), add(add(const_4.0, const_1), const_2)), add(subtract(50, const_4.0), subtract(50, const_2))), 30) | add(n0,const_1)|subtract(n1,const_4.0)|subtract(n1,const_2)|add(#0,const_2)|add(#1,#2)|add(#0,#3)|add(#5,#4)|divide(#6,const_4)| | general |
find the average of all prime numbers between 30 and 50 | explanation : ( 55 Γ£ β 40 ) + ( 60 Γ£ β 62 ) + ( 45 Γ£ β 58 ) / 55 + 60 + 45 8530 / 160 = 53.3 option b | a ) 54.48 , b ) 53.31 , c ) 54.6 , d ) 54.58 , e ) none of these | b | divide(add(add(multiply(55, 40), multiply(60, 62)), multiply(40, 58)), add(add(55, 60), 45)) | add(n1,n2)|multiply(n1,n4)|multiply(n2,n5)|multiply(n4,n6)|add(#1,#2)|add(n3,#0)|add(#4,#3)|divide(#6,#5) | general |
if the average marks of 3 batches of 55 , 60 and 45 students respectively is 40 , 62 , 58 , then the average marks of all the students is | "in an arithmetic progression , the nth term is given by tn = a + ( n - 1 ) d here tn = 65 , a = - 60 , d = 1 hence , 65 = - 60 + ( n - 1 ) or n = 126 sum of n terms can be calculated by sn = n / 2 ( a + l ) a = first term , l = last term , n = no . of terms sn = 126 * ( - 60 + 65 ) / 2 sn = 126 * 5 / 2 = 315 answer : a" | a ) 315 , b ) 215 , c ) 115 , d ) 165 , e ) 765 | a | divide(multiply(60, 65), const_4) | multiply(n0,n1)|divide(#0,const_4)| | general |
what is the sum of the integers from - 60 to 65 , inclusive ? | "a = first number l = last number sn = n / 2 [ a + l ] between 60 and 100 numbers = 41 = > 100 - 60 = 40 + 1 = 41 sn = 41 / 2 Γ£ β 160 = 41 Γ£ β 80 = 3280 answer : e" | a ) 4800 , b ) 4860 , c ) 5000 , d ) 5500 , e ) 3280 | e | add(100, const_1) | add(n1,const_1)| | general |
what is the sum of natural numbers between 60 and 100 | explanatory answer let the percentage of the total votes secured by party d be x % then the percentage of total votes secured by party r = ( x - 12 ) % as there are only two parties contesting in the election , the sum total of the votes secured by the two parties should total up to 100 % i . e . , x + x - 12 = 100 2 x - 12 = 100 or 2 x = 112 or x = 56 % . if party d got 56 % of the votes , then party got ( 56 - 12 ) = 44 % of the total votes . 44 % of the total votes = 132,000 i . e . , 44 / 100 * t = 132,000 = > t = 132000 * 100 / 44 = 300,000 votes . the margin by which party r lost the election = 12 % of the total votes = 12 % of 300,000 = 36,000 . the correct choice is ( d ) | a ) 240000 , b ) 300000 , c ) 168000 , d ) 36000 , e ) 24,000 | d | multiply(divide(132000, divide(subtract(const_100, 12), const_2)), 12) | subtract(const_100,n0)|divide(#0,const_2)|divide(n1,#1)|multiply(n0,#2) | general |
in an election contested by two parties , party d secured 12 % of the total votes more than party r . if party r got 132000 votes , by how many votes did it lose the election ? | "let x and y be the sides of the rectangle . then , correct area = xy . calculated area = ( 26 / 25 ) x ( 32 / 33 ) y = ( 344 / 341 ) ( xy ) error in measurement = ( 344 / 341 ) xy - xy = ( 3 / 341 ) xy error percentage = [ ( 3 / 341 ) xy ( 1 / xy ) 100 ] % = ( 22 / 25 ) % = 0.88 % . answer is e ." | a ) 0.11 % , b ) 0.7 % , c ) 0.4 % , d ) 0.6 % , e ) 0.88 % | e | subtract(subtract(4, 3), divide(multiply(4, 3), const_100)) | multiply(n0,n1)|subtract(n0,n1)|divide(#0,const_100)|subtract(#1,#2)| | geometry |
in measuring the sides of a rectangle , one side is taken 4 % in excess , and the other 3 % in deficit . find the error percent in the area calculated from these measurements . | "5 choices for each of the 4 questions , thus total e of 5 * 5 * 5 * 5 = 5 ^ 4 = 625 ways to answer all of them . answer : c ." | a ) 24 , b ) 120 , c ) 625 , d ) 720 , e ) 1024 | c | power(5, 4) | power(n1,n0)| | general |
a multiple choice test consists of 4 questions , and each question has 5 answer choices . in how many e ways can the test be completed if every question is unanswered ? | c ' s 1 day work = ( 1 / 3 ) - ( 1 / 6 + 1 / 8 ) = 1 / 24 a : b : c = 1 / 6 : 1 / 8 : 1 / 24 = 4 : 3 : 1 a ' s share = 600 * 4 / 8 = $ 300 answer is c | a ) $ 100 , b ) $ 150 , c ) $ 300 , d ) $ 250 , e ) $ 350 | c | multiply(divide(multiply(multiply(3, 8), inverse(6)), add(add(multiply(multiply(3, 8), subtract(inverse(3), add(inverse(6), inverse(8)))), multiply(multiply(3, 8), inverse(6))), multiply(multiply(3, 8), inverse(8)))), 600) | inverse(n1)|inverse(n3)|inverse(n2)|multiply(n2,n3)|add(#0,#2)|multiply(#0,#3)|multiply(#2,#3)|subtract(#1,#4)|multiply(#3,#7)|add(#8,#5)|add(#9,#6)|divide(#5,#10)|multiply(n0,#11) | physics |
a and b undertake to do a piece of work for $ 600 . a alone can do it in 6 days while b alone can do it in 8 days . with the help of c , they finish it in 3 days . find the share of a ? | "let the amount paid to x per week = x and the amount paid to y per week = y then x + y = 660 but x = 120 % of y = 120 y / 100 = 12 y / 10 Γ’ Λ Β΄ 12 y / 10 + y = 660 Γ’ β‘ β y [ 12 / 10 + 1 ] = 660 Γ’ β‘ β 22 y / 10 = 660 Γ’ β‘ β 22 y = 6600 Γ’ β‘ β y = 6600 / 22 = 600 / 2 = rs . 300 e" | a ) s . 150 , b ) s . 200 , c ) s . 250 , d ) s . 350 , e ) s . 300 | e | divide(multiply(660, multiply(add(const_1, const_4), const_2)), multiply(add(multiply(add(const_1, const_4), const_2), const_1), const_2)) | add(const_1,const_4)|multiply(#0,const_2)|add(#1,const_1)|multiply(n0,#1)|multiply(#2,const_2)|divide(#3,#4)| | general |
two employees x and y are paid a total of rs . 660 per week by their employer . if x is paid 120 percent of the sum paid to y , how much is y paid per week ? | "c if we calculate we will get 10749" | a ) 2449 , b ) 10449 , c ) 10749 , d ) 10449 , e ) 6468 | c | subtract(multiply(divide(54671, const_100), 10456), multiply(divide(const_1, const_3), multiply(divide(54671, const_100), 10456))) | divide(n0,const_100)|divide(const_1,const_3)|multiply(n1,#0)|multiply(#1,#2)|subtract(#2,#3)| | general |
54671 - 10456 - 33466 = ? | "speed = ( 120 + 80 ) km / h ( because direction is opposite hence relative velocity is added ) = 500 / 9 m / s time = 9 sec let the lenght of second train is x total distance covered = 250 + x therefore , d = speed * time thus 250 + x = 500 / 9 * 9 x = 500 - 250 = 250 m answer : c" | a ) 230 m , b ) 240 m , c ) 250 m , d ) 260 m , e ) 270 m | c | subtract(multiply(multiply(add(120, 80), const_0_2778), 9), 250) | add(n1,n2)|multiply(#0,const_0_2778)|multiply(n3,#1)|subtract(#2,n0)| | physics |
a 250 metres long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds . what is the length of the other train ? | x + x ( xx ) put the value of x = 7 in the above expression we get , 7 + 7 ( 77 ) = 7 + 7 ( 7 Γ£ β 7 ) = 7 + 7 ( 49 ) = 7 + 343 = 350 the answer is ( a ) | a ) a ) 350 , b ) b ) 346 , c ) c ) 358 , d ) d ) 336 , e ) e ) 364 | a | add(multiply(7, multiply(7, 7)), 7) | multiply(n0,n0)|multiply(n0,#0)|add(n0,#1) | general |
the value of x + x ( xx ) when x = 7 | "the class borrowed a total of 40 * 2 = 80 books . the 25 students who borrowed 0 , 1 , or 2 books borrowed a total of 12 + 11 * 2 = 34 . to maximize the number of books borrowed by 1 student , let ' s assume that 14 students borrowed 3 books and 1 student borrowed the rest . 80 - 34 - 3 * 14 = 4 the maximum number of books borrowed by any student is 4 . the answer is b ." | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | b | subtract(multiply(40, 2), add(multiply(subtract(subtract(40, add(add(multiply(12, 1), 11), 2)), 1), 3), add(multiply(12, 1), multiply(11, 2)))) | multiply(n0,n1)|multiply(n2,n3)|multiply(n1,n4)|add(#1,#2)|add(n4,#1)|add(n1,#4)|subtract(n0,#5)|subtract(#6,n3)|multiply(n6,#7)|add(#3,#8)|subtract(#0,#9)| | general |
in a class of 40 students , 2 students did not borrow any books from the library , 12 students each borrowed 1 book , 11 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 ? | t ^ 2 - kt - 48 = ( t - 8 ) ( t + m ) where m is any positive integer . if 48 / 8 = 6 , then we know as a matter of fact that : m = + 6 and thus k = 8 - 6 = 12 t ^ 2 - kt - m = ( t - a ) ( t + m ) where a > m t ^ 2 + kt - m = ( t - a ) ( t + m ) where a < m t ^ 2 - kt + m = ( t - a ) ( t - m ) t ^ 2 + kt + m = ( t + a ) ( t + m ) b | a ) 16 , b ) 12 , c ) 2 , d ) 6 , e ) 14 | b | add(const_10, 2) | add(n1,const_10) | general |
if ( t - 8 ) is a factor of t ^ 2 - kt - 45 , then k = | set a contains 12,14 , 16 . . . 50 set b contains 112 , 114 , 116 . . . 150 number of terms in each set = 20 difference between corresponding terms in set a and b = 100 difference between sum of set b and set a = 100 * 20 = 2000 answer a | a ) 2000 , b ) 2550 , c ) 5050 , d ) 6275 , e ) 11325 | a | multiply(subtract(112, 12), add(divide(subtract(50, 12), const_2), const_1)) | subtract(n1,n0)|subtract(n2,n0)|divide(#0,const_2)|add(#2,const_1)|multiply(#3,#1) | general |
set a contains all the even numbers between 12 and 50 inclusive . set b contains all the even numbers between 112 and 150 inclusive . what is the difference between the sum of elements of set b and the sum of the elements of set a ? | "let ' s just look at the dimensions ( no calculation needed ) . with dimension 11 the same , the other dimension 15 is twice 7.5 then the area will be double which means 100 % greater . the answer is c ." | a ) 50 % , b ) 87 % , c ) 100 % , d ) 187 % , e ) 200 % | c | multiply(divide(subtract(multiply(rectangle_area(11, 15), const_2), multiply(rectangle_area(7.5, 11), const_2)), rectangle_area(11, 15)), const_100) | rectangle_area(n0,n1)|rectangle_area(n0,n2)|multiply(#0,const_2)|multiply(#1,const_2)|subtract(#2,#3)|divide(#4,#0)|multiply(#5,const_100)| | geometry |
how much greater is the combined area in square inches of the front and back of a rectangular sheet of paper measuring 11 inches by 15 inches than that of a rectangular sheet of paper measuring 7.5 inches by 11 inches ? | "10 ^ 2 + 7 ^ 2 + 1 ^ 2 = 150 = = = > sum of these 3 integers = 10 + 7 + 1 = 18 b" | a ) 17 , b ) 18 , c ) 15 , d ) 14 , e ) 13 | b | add(add(add(const_4, 3), add(3, const_2)), 3) | add(n1,const_4)|add(const_2,n1)|add(#0,#1)|add(n1,#2)| | geometry |
the number 150 can be written as the sum of the squares of 3 different positive integers . what is the sum of these 3 integers ? | "let other diagonal = 2 x cm . since diagonals of a rhombus bisect each other at right angles , we have : ( 20 ) 2 = ( 12 ) 2 + ( x ) 2 = > x = β ( 20 ) 2 β ( 12 ) 2 = β 256 = 16 cm . so , other diagonal = 32 cm . area of rhombus = ( 1 / 2 ) x ( product of diagonals ) = ( 1 / 2 Γ 24 x 32 ) cm 2 = 384 cm 2 hence c" | a ) 320 cm 2 , b ) 280 cm 2 , c ) 384 cm 2 , d ) 290 cm 2 , e ) 350 cm 2 | c | add(multiply(multiply(divide(const_1, const_2), 24), sqrt(subtract(multiply(multiply(20, 20), const_4), multiply(24, 24)))), 24) | divide(const_1,const_2)|multiply(n0,n0)|multiply(n1,n1)|multiply(n1,#0)|multiply(#1,const_4)|subtract(#4,#2)|sqrt(#5)|multiply(#3,#6)|add(n1,#7)| | geometry |
find the area of a rhombus one side of which measures 20 cm and one diagonal is 24 cm ? | "explanation : the sum of the digits of the number is divisible by 3 , then the number is divisible by 3 . 2 + 2 + x + 5 + 7 = 16 + x least value of x may be 2 therefore 16 + 2 = 18 is divisible by 3 . answer : option a" | a ) 2 , b ) 0 , c ) 1 , d ) 3 , e ) 4 | a | divide(divide(divide(lcm(22, 57), 57), const_4), const_4) | lcm(n0,n1)|divide(#0,n1)|divide(#1,const_4)|divide(#2,const_4)| | general |
what is the least value of x . so that 22 x 57 is divisible by 3 . | "the percentage of tagged deer in the second sample = 5 / 50 * 100 = 10 % . so , 120 tagged deers comprise 10 % of total # of deers - - > total # of deers = 120 * 10 = 1,200 answer : a" | a ) 1,200 , b ) 750 , c ) 1,250 , d ) 1,500 , e ) 2,500 | a | multiply(50, 5) | multiply(n1,n3)| | general |
in a forest 120 deer were caught , tagged with electronic markers , then released . a week later , 50 deer were captured in the same forest . of these 50 deer , it was found that 5 had been tagged with the electronic markers . if the percentage of tagged deer in the second sample approximates the percentage of tagged deer in the forest , and if no deer had either left or entered the forest over the preceding week , what is the approximate number of deer in the forest ? | "let the c . p of each article be re 1 . then , s . p of 7 articles = c . p of 10 articles = rs . 10 now , c . p of 7 articles = rs . 7 , s . p of 7 articles = rs 10 gain = rs ( 10 - 7 ) = rs 3 . gain % = ( 3 / 7 Γ 100 ) % = 43 % answer : d" | a ) 93 % , b ) 23 % , c ) 33 % , d ) 43 % , e ) 53 % | d | subtract(7, 10) | subtract(n0,n1)| | gain |
if the selling price of 7 articles is same as the cost price of 10 articles . find the gain or loss percentage ? | "let the number of boys and girls be 2 x and 7 x total students = 360 number of girls in the school = 7 * 360 / 9 = 280 answer is e" | a ) 150 , b ) 250 , c ) 300 , d ) 370 , e ) 280 | e | multiply(divide(360, const_3), const_2.0) | divide(n2,const_3)|multiply(const_2.0,#0)| | other |
the ratio of number of boys and girls in a school is 2 : 7 . if there are 360 students in the school , find the number of girls in the school ? | "yes , ensure that you understand the relation thoroughly ! cost per liter = k * fraction of spirit 30 cents is the cost of 2 liters of solution ( 1 part water , 1 part spirit ) . so cost per liter is 15 cents . fraction of spirit is 1 / 2 . 15 = k * ( 1 / 2 ) k = 30 cost per liter = 30 * ( 1 / 3 ) ( 1 part spirit , 2 parts water ) cost for 3 liters = 30 * ( 1 / 3 ) * 3 = 50 cents b . 30 cents" | a ) 13 , b ) 30 , c ) 50 , d ) 51 , e ) 52 | b | multiply(multiply(30, divide(1, add(1, 2))), add(1, 2)) | add(n0,n4)|divide(n0,#0)|multiply(n2,#1)|multiply(#0,#2)| | geometry |
a spirit and water solution is sold in a market . the cost per liter of the solution is directly proportional to the part ( fraction ) of spirit ( by volume ) the solution has . a solution of 1 liter of spirit and 1 liter of water costs 30 cents . how many cents does a solution of 1 liter of spirit and 2 liters of water cost ? | let us consider it is a rectangle . so area = 24 * 9 = 216 sq . cm now ( 24 - 3 * 2 ) * ( 9 - 2 * 2 ) = 18 * 5 = 90 sq . cm answer : e | ['a ) 60 sq . cm', 'b ) 70 sq . cm', 'c ) 95 sq . cm', 'd ) 80 sq . cm', 'e ) 90 sq . cm'] | e | multiply(subtract(divide(216, power(const_3, const_2)), multiply(3, const_2)), subtract(power(const_3, const_2), multiply(2, const_2))) | multiply(n1,const_2)|multiply(n2,const_2)|power(const_3,const_2)|divide(n0,#2)|subtract(#2,#1)|subtract(#3,#0)|multiply(#5,#4) | other |
you need to print a document of the area 216 sq cm . condition is 3 cm margin is to be left at both top & bottom and 2 cm at the sides . what is the optimized size of your paper ? | "explanation : average = ( 2 + 4 + 6 + 8 + 10 + 12 ) / 2 = 42 / 2 = 21 option b" | a ) 15 , b ) 21 , c ) 25 , d ) 30 , e ) 35 | b | divide(add(add(add(multiply(2, const_3), add(2, multiply(2, const_2))), multiply(2, const_4)), multiply(add(const_4, const_1), 2)), 2) | add(const_1,const_4)|multiply(n2,const_2)|multiply(n2,const_3)|multiply(n2,const_4)|add(n2,#1)|multiply(n2,#0)|add(#4,#2)|add(#6,#3)|add(#7,#5)|divide(#8,n2)| | general |
find the average of all numbers between 1 and 13 which are divisible by 2 | "any x and y satisfying x / y = 5 / 3 should give the same value for ( x + y ) / ( x - y ) . say x = 5 and y = 3 , then ( x + y ) / ( x - y ) = ( 5 + 3 ) / ( 5 - 3 ) = 4 . answer : a ." | a ) 4 , b ) 1 / 5 , c ) - 1 / 6 , d ) - 1 / 5 , e ) - 5 | a | divide(add(5, 3), subtract(5, 3)) | add(n0,n1)|subtract(n0,n1)|divide(#0,#1)| | general |
if x / y = 5 / 3 , then ( x + y ) / ( x - y ) = ? | explanation : x - 35 = 80 x / 100 = > x = 175 = > 4 x / 5 = 4 x 175 / 5 = 140 . answer d | a ) 130 , b ) 155 , c ) 490 , d ) 140 , e ) 160 | d | multiply(divide(4, add(const_4, const_1)), multiply(35, add(const_4, const_1))) | add(const_1,const_4)|divide(n2,#0)|multiply(n0,#0)|multiply(#1,#2) | general |
a number , when 35 is subtracted from it , reduces to its 80 percent . what is 4 - fifth of that number ? | "the actual answer is obtained by multiplying 140 % by 70 % and subtracting 100 % from this total . that is : 140 % Γ 70 % = 98 % ; 98 % β 100 % = - 2 % . answer : b" | a ) β 5 % , b ) β 2 % , c ) 15 % , d ) 20 % , e ) 80 % | b | multiply(subtract(multiply(add(const_1, divide(40, const_100)), subtract(const_1, divide(30, const_100))), const_1), const_100) | divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|subtract(#4,const_1)|multiply(#5,const_100)| | gain |
a broker invested her own money in the stock market . during the first year , she increased her stock market wealth by 40 percent . in the second year , largely as a result of a slump in the stock market , she suffered a 30 percent decrease in the value of her stock investments . what was the net increase or decrease on her overall stock investment wealth by the end of the second year ? | explanation : since the month begins with a sunday , so there will be five sundays in the month , required average = ( 415 * 5 + 325 * 25 ) / 30 = 10200 / 30 = 340 answer : e ) 340 | a ) 140 , b ) 240 , c ) 260 , d ) 280 , e ) 340 | e | divide(add(multiply(add(floor(divide(30, add(const_3, const_4))), const_1), 425), multiply(subtract(30, add(floor(divide(30, add(const_3, const_4))), const_1)), 325)), 30) | add(const_3,const_4)|divide(n2,#0)|floor(#1)|add(#2,const_1)|multiply(n0,#3)|subtract(n2,#3)|multiply(n1,#5)|add(#4,#6)|divide(#7,n2) | general |
a library has an average of 425 visitors on sundays and 325 on other days . the average number of visitors per day in a month of 30 days beginning with a sunday is : | "solution let s be the sample space . then , n ( s ) = 52 c 3 = 22100 let e = event of getting 1 face card . n ( e ) = number of ways of choosing 1 face card out of 26 = 13 c 1 * 13 c 2 = 13 * 72 = 936 p ( e ) = n ( e ) / n ( s ) = 936 / 22100 = 234 / 5525 = 234 / 5525 . answer d" | a ) 238 / 5525 , b ) 176 / 5534 , c ) 253 / 5523 , d ) 234 / 5525 , e ) 1 / 5525 | d | divide(multiply(divide(52, const_4), divide(52, const_4)), choose(52, const_2)) | choose(n0,const_2)|divide(n0,const_4)|multiply(#1,#1)|divide(#2,#0)| | probability |
from a pack of 52 cards , two cards are drawn together at random . what is the probability that the one is heart and other two is diamond ? | lte central angle = x perimeter of the sector = length of the arc + 2 ( radius ) 50 = ( x / 360 * 2 * 22 / 7 * 14 ) + 2 ( 14 ) 50 = 88 x / 360 + 28 88 x / 360 = 22 88 x = 7920 x = 90 answer : e | a ) 180 o , b ) 225 o , c ) 270 o , d ) 150 o , e ) 90 o | e | multiply(multiply(const_2, divide(multiply(subtract(14, const_3), const_2), add(const_4, const_3))), 14) | add(const_3,const_4)|subtract(n0,const_3)|multiply(#1,const_2)|divide(#2,#0)|multiply(#3,const_2)|multiply(n0,#4) | physics |
the sector of a circle has radius of 14 cm and its perimeter 50 cm . find its central angel ? | this is how i used to calculate which i think works pretty well : if you let the average of the 20 other numbers equal a , can you write this equation for sum of the list ( s ) n + 20 a = s the question tells us that n = 4 a plug this back into the first equation and you get that the sum is 24 a 4 a + 20 a = 24 a therefore fraction q of n to the total would be 4 a / 24 a or 1 / 6 answer b | a ) 1 / 20 , b ) 1 / 6 , c ) 1 / 5 , d ) 4 / 21 , e ) 5 / 21 | b | divide(multiply(const_1, const_1), subtract(subtract(multiply(divide(add(divide(20, 4), 21), 4), const_2), 4), const_3)) | divide(n2,n1)|multiply(const_1,const_1)|add(n0,#0)|divide(#2,n1)|multiply(#3,const_2)|subtract(#4,n1)|subtract(#5,const_3)|divide(#1,#6) | general |
a certain list consists of 21 different numbers . if n is in the list and n is 4 times the average ( arithmetic mean ) of the other 20 numbers in the list , then n is what fraction q of the sum of the 21 numbers in the list ? | 12 men 120 acres 16 days 36 men ? 32 days 120 * 36 / 12 * 32 / 16 120 * 3 * 2 120 * 6 = 720 answer : d | a ) 269 , b ) 512 , c ) 369 , d ) 720 , e ) 450 | d | multiply(120, multiply(divide(36, 12), divide(32, 16))) | divide(n3,n0)|divide(n4,n2)|multiply(#0,#1)|multiply(n1,#2) | physics |
if 12 men can reap 120 acres of land in 16 days , how many acres of land can 36 men reap in 32 days ? | "total fish = x percentage of second catch = ( 4 / 50 ) * 100 = 8 % so , x * 8 % = 50 x = 625 ans d ." | a ) 200 , b ) 325 , c ) 565 , d ) 625 , e ) 700 | d | divide(50, divide(4, 50)) | divide(n2,n1)|divide(n0,#0)| | gain |
in a certain pond , 50 fish were caught , tagged , and returned to the pond . a few days later , 50 fish were caught again , of which 4 were found to have been tagged . if the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond , what is the approximate number of fish in the pond ? | "a : b = 2 : 3 b : c = 2 : 5 a : b : c = 4 : 6 : 15 6 / 25 * 75 = 18 answer : b" | a ) 11 , b ) 18 , c ) 99 , d ) 77 , e ) 51 | b | multiply(divide(75, add(add(divide(2, 3), divide(5, 2)), 2)), 5) | divide(n0,n1)|divide(n3,n0)|add(#0,#1)|add(#2,n0)|divide(n4,#3)|multiply(n3,#4)| | general |
a , b and c play a cricket match . the ratio of the runs scored by them in the match is a : b = 2 : 3 and b : c = 2 : 5 . if the total runs scored by all of them are 75 , the runs scored by b are ? | "answer we have , m x n = 8 x 200 = 1600 Γ’ Λ Β΄ 1 / m + 1 / n = ( m + n ) / mn = 84 / 1600 = 3 / 50 correct option : b" | a ) 1 / 35 , b ) 3 / 50 , c ) 5 / 37 , d ) 2 / 35 , e ) none | b | divide(84, multiply(8, 200)) | multiply(n0,n1)|divide(n2,#0)| | general |
the hcf and lcm of two numbers m and n are respectively 8 and 200 . if m + n = 84 , then 1 / m + 1 / n is equal to | "let the number of meters to be examined be x then , 0.04 % of x = 2 ( 4 / 100 ) * ( ( 1 / 100 ) * x = 2 x = 5000 answer is d" | a ) a ) 1500 , b ) b ) 2000 , c ) c ) 2500 , d ) d ) 5000 , e ) e ) 3100 | d | divide(multiply(2, const_100), 0.04) | multiply(n1,const_100)|divide(#0,n0)| | gain |
an inspector rejects 0.04 % of the meters as defective . how many will he examine to reject 2 ? | "increase = ( 40 / 50 ) * 100 = ( 4 / 5 ) * 100 = 80 % . d" | a ) 15 % , b ) 16.66 % , c ) 17.8 % , d ) 80 % , e ) 21 % | d | multiply(divide(subtract(90, 50), 50), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
john makes $ 50 a week from his job . he earns a raise andnow makes $ 90 a week . what is the % increase ? | "let the son ' s present age be x years . then , ( 48 - x ) = x x = 24 . son ' s age 5 years back = ( 24 - 5 ) = 19 years answer : d" | a ) 14 , b ) 17 , c ) 11 , d ) 19 , e ) 99 | d | subtract(divide(48, const_2), 5) | divide(n0,const_2)|subtract(#0,n1)| | general |
a father said his son , ` ` i was as old as you are at present at the time of your birth . ` ` if the father age is 48 now , the son age 5 years back was | "us = 25 ds = 43 m = ( 43 + 25 ) / 2 = 34 answer : b" | a ) 86 , b ) 34 , c ) 30 , d ) 15 , e ) 17 | b | divide(add(25, 43), const_2) | add(n0,n1)|divide(#0,const_2)| | physics |
a man can row upstream at 25 kmph and downstream at 43 kmph , and then find the speed of the man in still water ? | "there will be two cases x - 20 = 40 and x - 20 = - 40 solve for x = > x = 40 + 20 = > x = 60 or x = - 40 + 20 = > x = - 20 the sum of both values will be 60 + - 20 = 40 answer is d" | a ) 0 , b ) 60 , c ) - 80 , d ) 40 , e ) 80 | d | subtract(add(40, 20), subtract(40, 20)) | add(n0,n1)|subtract(n1,n0)|subtract(#0,#1)| | general |
if | x - 20 | = 40 what is the sum of all the values of x . | "ans is e given d _ ab = 2 * d _ bc let d _ ab = d and d _ bc = x so d = 2 x for average miles per gallon = ( d + x ) / ( ( d / 12 ) + ( x / 14 ) ) = 15 ( formula avg speed = total distance / total time )" | a ) 13 , b ) 13.5 , c ) 14 , d ) 14.5 , e ) 15 | e | divide(add(multiply(14, const_10), divide(multiply(14, const_10), const_2)), add(divide(multiply(14, const_10), 12), divide(divide(multiply(14, const_10), const_2), 14))) | multiply(n1,const_10)|divide(#0,const_2)|divide(#0,n0)|add(#1,#0)|divide(#1,n1)|add(#2,#4)|divide(#3,#5)| | general |