Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 164 | correct stringclasses 5 values | annotated_formula stringlengths 7 1.65k | linear_formula stringlengths 8 925 | category stringclasses 6 values | answer stringclasses 5 values |
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jane makes toy bears . when she works with an assistant , she makes 70 percent more bears per week and works 10 percent fewer hours each week . having an assistant increases jane ’ s output of toy bears per hour by what percent ? | "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 80 % b" | a ) 20 % , b ) 80 % , c ) 100 % , d ) 180 % , e ) 200 % | b | multiply(divide(10, subtract(subtract(const_100, 70), 10)), const_100) | subtract(const_100,n0)|subtract(#0,n1)|divide(n1,#1)|multiply(#2,const_100)| | physics | B |
what could be the range of a set consisting of odd multiples of 3 ? | "range = the difference between the greatest and the smallest numbers in the sequence . our sequence is odd and is a multiple of 7 . every number in that sequence can be represented like this : 3 * ( 2 n + 1 ) where n is any positive integer . range = 3 * ( 2 m + 1 ) - 3 * ( 2 n + 1 ) = 3 * 2 * ( m - n ) = 6 * ( m - n ) . m , n - any positive integers the answer must be divisible by 6 , which is only 24 . the correct answer is b" | a ) 21 , b ) 24 , c ) 35 , d ) 62 , e ) 70 | b | multiply(3, const_4) | multiply(n0,const_4)| | general | B |
a sun is divided among x , y and z in such a way that for each rupee x gets , y gets 50 paisa and z gets 40 paisa . if the share of x is rs . 30 , what is the total amount ? | "x : y : z = 100 : 50 : 40 10 : 5 : 4 10 - - - 30 19 - - - ? = > 57 answer : e" | a ) 67 , b ) 55 , c ) 62 , d ) 47 , e ) 57 | e | add(add(multiply(divide(const_100, 50), 30), multiply(divide(40, 50), 30)), 30) | divide(const_100,n0)|divide(n1,n0)|multiply(n2,#0)|multiply(n2,#1)|add(#2,#3)|add(n2,#4)| | general | E |
a shopkeeper fixes the marked price of an item 30 % above its cost price . the percentage of discount allowed to gain 8 % is | "explanation : let the cost price = rs 100 then , marked price = rs 130 required gain = 8 % , so selling price = rs 108 discount = 130 - 108 = 22 discount % = ( 22 / 130 ) * 100 = 16.92 % option b" | a ) 15.92 % , b ) 16.92 % , c ) 17.92 % , d ) 18.92 % , e ) none of these | b | subtract(const_100, multiply(divide(add(8, const_100), add(30, const_100)), const_100)) | add(n1,const_100)|add(n0,const_100)|divide(#0,#1)|multiply(#2,const_100)|subtract(const_100,#3)| | gain | B |
if the operation € is defined for all x and y by the equation x € y = 3 * x * y , then 3 € ( 4 € 5 ) = | "working inside out , ( 4 € 5 ) = 3 * 4 * 5 = 60 3 € 60 = 3 * 3 * 60 = 540 hence , answer is d" | a ) 80 , b ) 120 , c ) 160 , d ) 540 , e ) 660 | d | multiply(multiply(3, 3), multiply(multiply(3, 4), 5)) | multiply(n0,n1)|multiply(n0,n2)|multiply(n3,#1)|multiply(#0,#2)| | general | D |
the ratio between the length and the breadth of a rectangular park is 3 : 2 . if a man cycling alongthe oundary of the park at the speed of 12 km / hr completes one round in 8 minutes , then the area of the park ( in sq . m ) is | "perimeter = distance covered in 8 min . = 12000 x 8 m = 1600 m . 60 let length = 3 x metres and breadth = 2 x metres . then , 2 ( 3 x + 2 x ) = 1600 or x = 160 . length = 480 m and breadth = 320 m . area = ( 480 x 320 ) m 2 = 153600 m c" | a ) 124345 m , b ) 233234 m , c ) 153600 m , d ) 123567 m , e ) 168999 m | c | rectangle_area(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 8), const_1000), add(3, 2)), const_2), multiply(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 8), 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 | C |
a man goes from a to b at a speed of 35 kmph and comes back to a at a speed of 28 kmph . find his average speed for the entire journey ? | "distance from a and b be ' d ' average speed = total distance / total time average speed = ( 2 d ) / [ ( d / 35 ) + ( d / 28 ] = ( 2 d ) / [ 11 d / 140 ) = > 38.1 kmph . answer : c" | a ) 38.9 kmph , b ) 38.8 kmph , c ) 38.1 kmph , d ) 37.1 kmph , e ) 37.8 kmph | c | divide(add(35, 28), const_2) | add(n0,n1)|divide(#0,const_2)| | physics | C |
what is the smallest number which when increased by 3 is divisible by 18 , 70 , 100 and 21 ? | "when increased by 3 , the number must include at least 2 ^ 2 * 3 ^ 2 * 5 ^ 2 * 7 = 6300 the answer is d ." | a ) 5037 , b ) 5447 , c ) 5867 , d ) 6297 , e ) 6617 | d | add(lcm(lcm(18, 70), lcm(100, 21)), 3) | lcm(n1,n2)|lcm(n3,n4)|lcm(#0,#1)|add(n0,#2)| | general | D |
in a kilometer race , a beats b by 50 meters or 10 seconds . what time does a take to complete the race ? | "explanation : time taken by b run 1000 meters = ( 1000 * 10 ) / 50 = 200 sec . time taken by a = 200 - 10 = 190 sec . answer b" | a ) 50 sec , b ) 190 sec , c ) 287 sec , d ) 74 sec , e ) 697 sec | b | subtract(divide(multiply(const_1, const_1000), divide(50, 10)), 10) | divide(n0,n1)|multiply(const_1,const_1000)|divide(#1,#0)|subtract(#2,n1)| | physics | B |
the ratio of the ages of mini and minakshi is 4 : 3 . the sum of their ages is 28 years . the ratio of their ages after 5 years will be | "let mini ’ s age = 4 x and minakshi ’ s age = 3 x then 4 x + 3 x = 28 x = 4 mini ’ s age = 16 years and minakshi ’ s age = 12 years ratio of their ages after 8 years = ( 16 + 5 ) : ( 12 + 5 ) = 22 : 17 answer : a" | a ) 22 : 17 , b ) 9 : 11 , c ) 5 : 4 , d ) 4 : 5 , e ) 8 : 11 | a | divide(add(multiply(4, divide(28, add(4, 3))), 5), add(multiply(3, divide(28, add(4, 3))), 5)) | add(n0,n1)|divide(n2,#0)|multiply(n0,#1)|multiply(n1,#1)|add(n3,#2)|add(n3,#3)|divide(#4,#5)| | general | A |
a dishonest milkman wants to make a profit on the selling of milk . he would like to mix water ( costing nothing ) with milk costing rs . 42 per litre so as to make a profit of 20 % on cost when he sells the resulting milk and water mixture for rs . 48 in what ratio should he mix the water and milk ? | "water = w ( liter ) milk = m ( liter ) = = > cost = price x quantity = 0.42 m = = > revenue = price x quantity = 0.48 ( m + w ) = = > profit = 0.48 ( m + w ) - 0.42 m = 0.2 * ( 0.42 m ) [ 20 % of cost ] = = > 0.48 m + 0.48 w - 0.42 m = 0.084 m = = > 0.024 m = 0.48 w = = > m / w = 0.48 / 0.024 = 20 - - or - - w / m = 1 / 20 a is correct ." | a ) 1 : 20 , b ) 1 : 10 , c ) 1 : 8 , d ) 1 : 4 , e ) 6 : 11 | a | divide(const_1, divide(20, const_2)) | divide(n1,const_2)|divide(const_1,#0)| | gain | A |
a car gets 40 kilometers per gallon of gasoline . how many gallons of gasoline would the car need to travel 180 kilometers ? | "each 40 kilometers , 1 gallon is needed . we need to know how many 40 kilometers are there in 180 kilometers ? 180 ÷ 40 = 4.5 × 1 gallon = 4.5 gallons correct answer is d ) 4.5 gallons" | a ) 3.5 gallons , b ) 2.7 gallons , c ) 5.7 gallons , d ) 4.5 gallons , e ) 7.5 gallons | d | divide(180, 40) | divide(n1,n0)| | physics | D |
if 2 ^ 98 = 256 l + n , where land n are integers and 0 ≤ n ≤ 4 , what is the value of n ? | if 2 ^ 98 = 256 l + n and 0 < = n < = 4 , what is the value of n ? to solve this , note that 2 ^ 8 = 256 , so 2 ^ 98 = 2 ^ [ ( 8 * 12 ) + 2 ] = 4 * [ 256 * 256 * . . . . . . 12 times ] as this is a multiple of 256 , n can only be 0 . answer : a | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | a | reminder(power(2, 98), 256) | power(n0,n1)|reminder(#0,n2) | general | A |
jim ’ s taxi service charges an initial fee of $ 2.35 at the beginning of a trip and an additional charge of $ 0.35 for each 2 / 5 of a mile traveled . what is the total charge for a trip of 3.6 miles ? | "let the fixed charge of jim ’ s taxi service = 2.35 $ and charge per 2 / 5 mile ( . 4 mile ) = . 35 $ total charge for a trip of 3.6 miles = 2.35 + ( 3.6 / . 4 ) * . 35 = 2.35 + 9 * . 35 = 5.5 $ answer d" | a ) $ 3.15 , b ) $ 4.45 , c ) $ 4.80 , d ) $ 5.5 , e ) $ 5.40 | d | add(2.35, multiply(0.35, divide(3.6, divide(2, 5)))) | divide(n2,n3)|divide(n4,#0)|multiply(n1,#1)|add(n0,#2)| | general | D |
the average of runs of a cricket player of 10 innings was 35 . how many runs must he make in his next innings so as to increase his average of runs by 4 ? | "explanation : average = total runs / no . of innings = 35 so , total = average x no . of innings = 35 x 10 = 350 . now increase in avg = 4 runs . so , new avg = 35 + 4 = 39 runs total runs = new avg x new no . of innings = 39 x 11 = 429 runs made in the 11 th inning = 429 - 350 = 79 answer : b" | a ) 76 , b ) 79 , c ) 85 , d ) 87 , e ) 89 | b | subtract(multiply(add(10, const_1), add(4, 35)), multiply(10, 35)) | add(n0,const_1)|add(n1,n2)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2)| | general | B |
150 ml of 30 % sulphuric acid was added to approximate 400 ml of 12 % sulphuric acid solution . find the approximate concentration e of the acid in the mixture ? | "do not need any computation 30 % - - - - - - - - - - - 21 % - - - - - - - - - 12 % if volume of both sol . were equal the concentration e would be 21 % = 1 / 5 , but 12 % is more than 3 times only possibility is 1 / 6 d" | a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 6 , e ) 1 / 5 | d | divide(add(divide(multiply(150, 30), const_100), divide(multiply(400, 12), const_100)), add(150, 400)) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|divide(#1,const_100)|divide(#2,const_100)|add(#3,#4)|divide(#5,#0)| | general | D |
what is the area of a square field whose diagonal of length 18 m ? | "d 2 / 2 = ( 18 * 18 ) / 2 = 180 answer : d" | a ) 120 sq m , b ) 250 sq m , c ) 200 sq m , d ) 180 sq m , e ) 100 sq m | d | divide(square_area(18), const_2) | square_area(n0)|divide(#0,const_2)| | geometry | D |
a man invests rs . 8,000 at the rate of 5 % per annum . how much more should he invest at the rate of 8 % , so that he can earn a total of 6 % per annum ? | explanation : interest on rs . 8000 at 5 % per annum = ( 8000 × 5 × 1 ) / 100 = rs . 400 let his additional investment at 8 % = x interest on rs . x at 8 % per annum = ( x × 8 × 1 ) / 100 = 2 x / 25 . to earn 6 % per annum for the total , interest = ( 8000 + x ) × 6 × 1 / 100 . = > 400 + 2 x / 25 = ( 8000 + x ) × 6 × 1 / 100 . = > 40000 + 8 x = ( 8000 + x ) × 6 . = > 40000 + 8 x = 48000 + 6 x . = > 2 x = 8000 . = > x = 4000 . answer : d | a ) rs . 1200 , b ) rs . 1300 , c ) rs . 1500 , d ) rs . 4000 , e ) none of these | d | multiply(multiply(8, const_1000), divide(subtract(6, 5), subtract(8, 6))) | multiply(n2,const_1000)|subtract(n3,n1)|subtract(n2,n3)|divide(#1,#2)|multiply(#3,#0) | gain | D |
vijay sells a cupboard at 12 % below cost price . had he got rs . 1650 more , he would have made a profit of 12 % . what is the cost price of the cupboard ? | "explanation : cost price = 1650 / ( 0.12 + 0.12 ) = 2086 / 0.24 = rs . 6875 answer d" | a ) 7450 , b ) 14900 , c ) 7400 , d ) 6875 , e ) none of these | d | divide(1650, divide(subtract(add(const_100, 12), subtract(const_100, 12)), const_100)) | add(n0,const_100)|subtract(const_100,n0)|subtract(#0,#1)|divide(#2,const_100)|divide(n1,#3)| | gain | D |
money invested at x % , compounded annually , triples in value in approximately every 112 / x years . if $ 1500 is invested at a rate of 8 % , compounded annually , what will be its approximate worth in 28 years ? | "x = 8 % 112 / x years = 112 / 8 = 14 years now , money triples every 14 years therefore , in 14 yrs , if $ 1500 triples to $ 4500 , in 28 years , it will again triple to $ 4500 * 3 = $ 13,500 answer c" | a ) $ 3,750 , b ) $ 5,600 , c ) $ 13,500 , d ) $ 15,000 , e ) $ 22,500 | c | multiply(1500, power(const_3, divide(28, divide(112, 8)))) | divide(n0,n2)|divide(n3,#0)|power(const_3,#1)|multiply(n1,#2)| | gain | C |
milk contains 5 % water . what quantity of pure milk should be added to 10 litres of milk to reduce this 2 % ? | present ration of milk to water is 95 : 5 amount of milk is 10 litres hence , pure milk content is 9.5 litre and water content is 0.5 litres now , we should not disturb the water content so , 0.5 litres of water is equivalent to 2 % so 100 % mixture is : ( 0.5 / 2 ) * 100 = 25 litre thus , milk should be added = 25 - 10 = 15 litres of pure milk answer : a | a ) 15 litres , b ) 16 litres , c ) 17 litres , d ) 25 litres , e ) 35 litres | a | subtract(divide(multiply(divide(5, const_100), 10), divide(2, const_100)), 10) | divide(n0,const_100)|divide(n2,const_100)|multiply(n1,#0)|divide(#2,#1)|subtract(#3,n1) | gain | A |
a train 360 m long is running at a speed of 45 km / hr . in what time will it pass a bridge 160 m long ? | "speed = 45 * 5 / 18 = 25 / 2 m / sec total distance covered = 360 + 160 = 520 m required time = 520 * 2 / 25 = 41.6 sec answer : a" | a ) 41.6 sec , b ) 11 sec , c ) 88 sec , d ) 19 sec , e ) 10 sec | a | divide(360, multiply(subtract(45, 160), const_0_2778)) | subtract(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics | A |
( 500 + 200 + 100 ) ã — 4 ã — ( 3 + 4 ) = ? | "( 500 + 200 + 100 ) ã — 4 ã — ( 3 + 4 ) = ? or , ? = 800 ã — 4 ã — 7 = 22400 answer c" | a ) 52000 , b ) 32400 , c ) 22400 , d ) 22800 , e ) 24200 | c | subtract(divide(multiply(4, add(4, const_1)), const_2), divide(multiply(subtract(500, const_1), 500), const_2)) | add(n3,const_1)|subtract(n0,const_1)|multiply(n3,#0)|multiply(n0,#1)|divide(#2,const_2)|divide(#3,const_2)|subtract(#4,#5)| | general | C |
the speed of a boat in still water in 42 km / hr and the rate of current is 7 km / hr . the distance travelled downstream in 44 minutes is : | "speed downstream = ( 42 + 7 ) = 49 kmph time = 44 minutes = 44 / 60 hour = 11 / 15 hour distance travelled = time × speed = 11 / 15 × 49 = 35.9 km answer : c" | a ) 86.6 km , b ) 46.6 km , c ) 35.9 km , d ) 35.6 km , e ) 26.6 km | c | multiply(add(42, 7), divide(44, const_60)) | add(n0,n1)|divide(n2,const_60)|multiply(#0,#1)| | physics | C |
a cricket bat is sold for $ 900 , making a profit of $ 100 . the profit percentage would be | "100 / ( 900 - 100 ) = 100 / 800 = 0.125 = 12.5 % . answer : b" | a ) 24 % , b ) 12.5 % , c ) 30 % , d ) 36 % , e ) 40 % | b | multiply(divide(100, subtract(900, 100)), const_100) | subtract(n0,n1)|divide(n1,#0)|multiply(#1,const_100)| | gain | B |
the h . c . f of two numbers is 17 and their l . c . m is 5134 . if one of the numbers is 289 , then the other is ? | "other number = ( 17 * 5134 ) / 289 = 302 . answer : a" | a ) 302 , b ) 304 , c ) 306 , d ) 308 , e ) 310 | a | multiply(17, 289) | multiply(n0,n2)| | physics | A |
two brothers took the gmat exam , the higher score is x and the lower one is y . if the difference between the two scores 1 / 4 , what is the value of y / x ? | "answer is c : 1 / 5 x - y = ( x + y ) / 4 solving for y / x = 1 / 5" | a ) 3 . , b ) 2 . , c ) 1 / 5 , d ) 1 / 3 . , e ) there is n ' t enough data to answer the question . | c | add(4, divide(multiply(4, 1), 4)) | multiply(n0,n1)|divide(#0,n1)|add(n1,#1)| | general | C |
convert the 9 / 36 m / s into kilometers per hour ? | "9 / 36 m / s = 9 / 36 * 18 / 5 = 9 / 10 = 0.9 kmph . answer : a" | a ) 0.9 kmph , b ) 9.9 kmph , c ) 1.3 kmph , d ) 1.2 kmph , e ) 5.7 kmph | a | multiply(const_3_6, divide(9, 36)) | divide(n0,n1)|multiply(#0,const_3_6)| | physics | A |
if the difference between compound interest ( interest compounded yearly ) and simple interest on a sum for 3 years at 20 % p . a . is rs . 132 , then the sum is | "c . i - s . i = 132 c . i = a - p a = p ( 1 + ( r / 100 ) ) ^ n a = x ( 1 + ( 20 / 100 ) ) ^ 3 a = x ( 6 / 5 ) ^ 3 a = 216 / 125 ( x ) c . i = ( 91 / 125 ) x s . i = ( pnr ) / 100 = ( x * 3 * 20 ) / 100 = ( 6 / 10 ) x diff = 132 diff = ( 91 / 25 ) x - ( 6 / 10 ) x = ( 182 x - 150 x ) / 250 = 32 x / 250 32 x / 250 = 132 x = ( 132 * 250 ) / 32 x = 1031.25 answer : e" | a ) rs . 3500 , b ) rs . 1500 , c ) rs . 2500 , d ) rs . 3000 , e ) none | e | multiply(multiply(132, 20), 20) | multiply(n1,n2)|multiply(n1,#0)| | gain | E |
in a 3 digit number , the 100 digit is 2 more than the tens digit and the units digit is 2 less than the tens digit . if the sum of the digits is 27 , find the number ? | e 1197 let the three digit numbers be 100 a + 10 b + c a = b + 2 c = b - 2 a + b + c = 3 b = 27 = > b = 9 so a = 11 and b = 7 hence the three digit number is : 1197 | a ) 867 , b ) 1179 , c ) 1196 , d ) 1098 , e ) 1197 | e | add(add(multiply(add(divide(27, 3), 2), const_100), multiply(divide(27, 3), const_10)), subtract(divide(27, 3), 2)) | divide(n4,n0)|add(n2,#0)|multiply(#0,const_10)|subtract(#0,n2)|multiply(#1,const_100)|add(#4,#2)|add(#5,#3) | general | E |
in a race of 1000 m , a can beat by 100 m , in a race of 800 m , b can beat c by 100 m . by how many meters will a beat c in a race of 1000 m ? | when a runs 1000 m , b runs 900 m and when b runs 800 m , c runs 700 m . when b runs 900 m , distance that c runs = ( 900 * 700 ) / 800 = 6300 / 8 = 787.5 m . in a race of 1000 m , a beats c by ( 1000 - 787.5 ) = 212.5 m to c . in a race of 1000 m , the number of meters by which a beats c = ( 1000 * 212.5 ) / 1000 = 212.5 m . answer : e | a ) 122.9 m , b ) 127.5 m . , c ) 122.2 m , d ) 222.9 m , e ) 212.5 m | e | subtract(1000, divide(multiply(subtract(1000, 100), subtract(800, 100)), 800)) | subtract(n0,n1)|subtract(n2,n1)|multiply(#0,#1)|divide(#2,n2)|subtract(n0,#3) | physics | E |
population of a city in 20004 was 1000000 . if in 2005 there isan increment of 20 % , in 2006 there is a decrements of 35 % and in 2007 there is an increment of 45 % , then find the population of city atthe end of the year 2007 | "required population = p ( 1 + r 1 / 100 ) ( 1 - r 2 / 100 ) ( 1 + r 3 / 100 ) = p ( 1 + 20 / 100 ) ( 1 - 35 / 100 ) ( 1 + 45 / 100 ) = 1131000 a" | a ) 1131000 , b ) 979923 , c ) 980241 , d ) 2356677 , e ) 1083875 | a | multiply(1000000, multiply(multiply(add(const_1, divide(20, const_100)), subtract(const_1, divide(35, const_100))), add(const_1, divide(35, const_100)))) | divide(n5,const_100)|divide(n3,const_100)|add(#0,const_1)|add(#1,const_1)|subtract(const_1,#0)|multiply(#3,#4)|multiply(#2,#5)|multiply(n1,#6)| | gain | A |
the output of a factory is increased by 10 % to keep up with rising demand . to handle the holiday rush , this new output is increased by 60 % . by approximately what percent would the output of the factory now have to be decreased in order to restore the original output ? | "take it as original output = 100 . to meet demand increase by 10 % , then output = 110 . to meet holiday demand , new output increase by 60 % then output equals 176 to restore new holidy demand output to original 100 . final - initial / final * 100 = 76 / 176 * 100 = 8 / 33 * 100 = 43 % approxiamately . option d is correct ." | a ) 20 % , b ) 24 % , c ) 30 % , d ) 43 % , e ) 79 % | d | multiply(divide(subtract(multiply(divide(add(const_100, 10), const_100), divide(add(const_100, 60), const_100)), const_1), multiply(divide(add(const_100, 10), const_100), divide(add(const_100, 60), const_100))), const_100) | add(n0,const_100)|add(n1,const_100)|divide(#0,const_100)|divide(#1,const_100)|multiply(#2,#3)|subtract(#4,const_1)|divide(#5,#4)|multiply(#6,const_100)| | general | D |
? % of 360 = 115.2 | ? % of 360 = 115.2 or , ? = 115.2 × 100 / 360 = 32 answer d | a ) 277 , b ) 36 , c ) 64 , d ) 32 , e ) none of these | d | divide(multiply(115.2, const_100), 360) | multiply(n1,const_100)|divide(#0,n0) | gain | D |
what percentage of numbers from 1 to 95 have squares that end in the digit 0 ? | clearly , the numbers which have 1 or 9 in the unit ' s digit , have squares that end in the digit 1 . such numbers from 1 to 95 are 10 , 20,30 , 40,50 , 60,70 , 80,90 . number of such numbers = 14 . required percentage = ( 9 / 95 * 100 ) = 9.47 % answer : c | a ) 1 , b ) 14 , c ) 9.47 , d ) 21 , e ) 22 | c | multiply(divide(floor(divide(95, const_10)), 95), const_100) | divide(n1,const_10)|floor(#0)|divide(#1,n1)|multiply(#2,const_100) | gain | C |
what is the measure of the radius of the circle that circumscribes a triangle whose sides measure 11 , 60 and 61 ? | "some of pyhtagron triplets we need to keep it in mind . like { ( 2 , 3,5 ) , ( 5 , 12,13 ) , ( 7 , 24,25 ) , ( 11 , 60,61 ) . so now we know the triangle is an right angle triangle . the circle circumscribes the triangle . the circumraduis of the circle that circumscribes the right angle triangle = hypotanse / 2 = 61 / 2 = 30.5 ans . d" | a ) 6.5 , b ) 12.5 , c ) 28 , d ) 30.5 , e ) 40 | d | divide(triangle_area_three_edges(11, 60, 61), divide(triangle_perimeter(11, 60, 61), const_2)) | triangle_area_three_edges(n0,n1,n2)|triangle_perimeter(n0,n1,n2)|divide(#1,const_2)|divide(#0,#2)| | geometry | D |
when a number is divided by 13 , the remainder is 6 . when the same number is divided by 7 , then remainder is 1 . what is the number ? | "explanation : take 243 243 ÷ 7 = 34 , remainder = 5 hence this is not the answer take 312 312 ÷ 7 = 44 , remainder = 4 hence this is not the answer take 253 253 ÷ 7 = 36 , remainder = 1 . 253 ÷ 13 = 19 , remainder = 6 this satisfies both the conditions given in the question . hence it is the answer . answer : b" | a ) 243 , b ) 253 , c ) 312 , d ) 432 , e ) none of these | b | add(multiply(13, add(add(const_10, const_12), const_4)), 6) | add(const_10,const_12)|add(#0,const_4)|multiply(n0,#1)|add(n1,#2)| | general | B |
a is half good a work man as b and together they finish a job in 18 days . in how many days working alone b finish the job ? | "c 27 wc = 1 : 2 2 x + x = 1 / 18 = > x = 1 / 54 2 x = 1 / 27 = > 27 days" | a ) 23 , b ) 22 , c ) 27 , d ) 36 , e ) 48 | c | multiply(18, divide(const_3, const_2)) | divide(const_3,const_2)|multiply(n0,#0)| | physics | C |
12 times a number gives 156 . the number is | "explanation : let the number be ' n ' 12 × n = 156 ⇒ n = 13 correct option : c" | a ) 11 , b ) 12 , c ) 13 , d ) none , e ) can not be determined | c | divide(156, divide(156, 12)) | divide(n1,n0)|divide(n1,#0)| | general | C |
john began driving from home on a trip averaging 30 miles per hour . how many miles per hour must carla drive on average to catch up to him in exactly 3 hours if she leaves 30 minutes after john ? | carla starts 30 minutes later and it takes 3 hr for carla to meet john so john total time travelled = 3 hr + 30 minutes john distance = 30 * ( 3 1 / 2 ) = 105 so carla need to travle 105 to meet john in 3 hrs speed of carla = 105 / 3 = 35 miles per hour answer is a | a ) 35 , b ) 55 , c ) 39 , d ) 40 , e ) 60 | a | divide(multiply(30, add(3, divide(const_1, const_2))), 3) | divide(const_1,const_2)|add(n1,#0)|multiply(n0,#1)|divide(#2,n1) | physics | A |
a cycle is bought for rs . 900 and sold for rs . 1125 , find the gain percent ? | "900 - - - - 225 100 - - - - ? = > 25 % answer : e" | a ) 27 % , b ) 20 % , c ) 80 % , d ) 30 % , e ) 25 % | e | multiply(divide(subtract(1125, 900), 900), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain | E |
the average runs scored by a batsman in 20 matches is 30 . in the next 10 matches the batsman scored an average of 15 runs . find his average in all the 30 matches ? | "total score of the batsman in 20 matches = 600 . total score of the batsman in the next 10 matches = 150 . total score of the batsman in the 30 matches = 750 . average score of the batsman = 750 / 30 = 25 . answer : a" | a ) 25 , b ) 46 , c ) 88 , d ) 13 , e ) 12 | a | divide(add(multiply(30, 20), multiply(15, 10)), add(20, 10)) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0)| | general | A |
if 20 men take 15 days to to complete a job , in how many days can 20 men finish that work ? | ans . 15 days | a ) 15 , b ) 16 , c ) 17 , d ) 18 , e ) 19 | a | divide(multiply(20, 15), 20) | multiply(n0,n1)|divide(#0,n2)| | physics | A |
a bag contains 6 green and 8 white balls . if two balls are drawn simultaneously , the probability that both are of the same colour is | "drawing two balls of same color from six green balls can be done in 6 c ₂ ways . similarly from eight white balls two can be drawn in ⁸ c ₂ ways . p = 6 c ₂ / ¹ ⁵ c ₂ + ⁸ c ₂ / ¹ ⁵ c ₂ = 43 / 91 answer : e" | a ) 8 / 15 , b ) 2 / 5 , c ) 3 / 5 , d ) 11 / 15 , e ) 43 / 91 | e | add(multiply(divide(8, add(6, 8)), divide(subtract(8, const_1), subtract(add(6, 8), const_1))), multiply(divide(6, add(6, 8)), divide(subtract(6, const_1), subtract(add(6, 8), const_1)))) | add(n0,n1)|subtract(n1,const_1)|subtract(n0,const_1)|divide(n1,#0)|divide(n0,#0)|subtract(#0,const_1)|divide(#1,#5)|divide(#2,#5)|multiply(#3,#6)|multiply(#4,#7)|add(#8,#9)| | other | E |
a family consists of two grandparents , two parents , and 3 grandchildren . the average age of the grandparents is 64 years , the average age of the parents is 39 years , and the average age of the grandchildren is 6 years . what is the average age ( in years ) of the family ? | total age of the grandparents = 64 ã — 2 = 128 total age of the parents = 39 ã — 2 = 78 total age of the grandchildren = 6 ã — 3 = 18 average age of the family = ( 128 + 78 + 18 ) / 7 = 224 / 7 = 32 years the answer is b . | a ) 31 , b ) 32 , c ) 33 , d ) 34 , e ) 35 | b | divide(add(add(multiply(64, const_2), multiply(39, const_2)), multiply(6, 3)), add(3, const_4)) | add(n0,const_4)|multiply(n1,const_2)|multiply(n2,const_2)|multiply(n0,n3)|add(#1,#2)|add(#4,#3)|divide(#5,#0) | general | B |
the percentage increase in the area of rectangle , if each of its side is increased by 10 % is ? | "let original length = x meters original breadth = y meters original area = xy m ^ 2 new length = 110 x / 100 new breadth = 110 y / 100 = 11 y / 10 new area = 11 x / 10 * 11 y / 10 = 121 xy / 100 m ^ 2 increase percent = 21 xy / 100 * 1 / xy * 100 = 21 % answer is a" | a ) 21 % , b ) 30 % , c ) 25 % , d ) 40 % , e ) 50 % | a | multiply(subtract(multiply(divide(add(const_100, 10), const_100), divide(add(const_100, 10), const_100)), const_1), const_100) | add(n0,const_100)|divide(#0,const_100)|multiply(#1,#1)|subtract(#2,const_1)|multiply(#3,const_100)| | geometry | A |
the average age of 19 students in a group is 17 years . when teacher ’ s age is included to it , the average increases by one . what is the teacher ’ s age in years ? | "age of the teacher = ( 20 × 18 – 19 × 17 ) years = 37 years . answer b" | a ) 36 , b ) 37 , c ) 38 , d ) can not be determined , e ) none of these | b | add(19, const_1) | add(n0,const_1)| | general | B |
a , b and c enter into partnership . a invests some money at the beginning , b invests double the amount after 6 months , and c invests thrice the amount after 8 months . if the annual gain be rs . 18000 . a ' s share is ? | "x * 12 : 2 x * 6 : 3 x * 4 1 : 1 : 1 1 / 3 * 18000 = 6000 answer : c" | a ) rs . 7500 , b ) rs . 7200 , c ) rs . 6000 , d ) rs . 5750 , e ) rs . 6750 | c | multiply(multiply(const_1, const_12), divide(18000, add(add(multiply(const_1, const_12), multiply(subtract(const_12, 6), const_2)), multiply(subtract(const_12, 8), const_3)))) | multiply(const_1,const_12)|subtract(const_12,n0)|subtract(const_12,n1)|multiply(#1,const_2)|multiply(#2,const_3)|add(#0,#3)|add(#5,#4)|divide(n2,#6)|multiply(#7,#0)| | gain | C |
john has 10 pairs of matched socks . if he loses 11 individual socks , what is the greatest number of pairs of matched socks he can have left ? | "because we have to maximize the pair of matched socks , we will remove 5 pairs ( 10 socks ) out of 10 pairs 1 sock from the 6 th pair . thus the no of matching socks pair remaining = 10 - 6 = 4 answer d" | a ) 7 , b ) 6 , c ) 5 , d ) 4 , e ) 3 | d | subtract(10, const_3) | subtract(n0,const_3)| | general | D |
if n is an integer , f ( n ) = f ( n - 1 ) - n and f ( 4 ) = 15 . what is the value of f ( 6 ) ? | since f ( n ) = f ( n - 1 ) - n then : f ( 6 ) = f ( 5 ) - 6 and f ( 5 ) = f ( 4 ) - 5 . as given that f ( 4 ) = 15 then f ( 5 ) = 15 - 5 = 10 - - > substitute the value of f ( 5 ) back into the first equation : f ( 6 ) = f ( 5 ) - 6 = 10 - 6 = 4 . answer : a . questions on funtions to practice : | a ) 4 , b ) 0 , c ) 1 , d ) 2 , e ) 24 | a | subtract(subtract(15, add(1, 4)), 6) | add(n0,n1)|subtract(n2,#0)|subtract(#1,n3) | general | A |
a 160 meter long train crosses a man standing on the platform in 16 sec . what is the speed of the train ? | "s = 160 / 16 * 18 / 5 = 36 kmph answer : a" | a ) 36 kmph , b ) 54 kmph , c ) 92 kmph , d ) 86 kmph , e ) 76 kmph | a | multiply(divide(160, 16), const_3_6) | divide(n0,n1)|multiply(#0,const_3_6)| | physics | A |
divide rs . 32000 in the ratio 1 : 7 ? | "1 / 8 * 32000 = 4000 7 / 8 * 32000 = 28000 answer : c" | a ) 12000 , 20000 , b ) 12000 , 200098 , c ) 4000 , 28000 , d ) 12000 , 20006 , e ) 12000 , 20001 | c | multiply(subtract(7, const_2), divide(32000, add(1, subtract(7, const_2)))) | subtract(n2,const_2)|add(n1,#0)|divide(n0,#1)|multiply(#2,#0)| | other | C |
each factor of 210 is inscribed on its own plastic ball , and all of the balls are placed in a jar . if a ball is randomly selected from the jar , what is the probability that the ball is inscribed with a multiple of 42 ? | "210 = 2 * 3 * 5 * 7 , so the # of factors 210 has is ( 1 + 1 ) ( 1 + 1 ) ( 1 + 1 ) ( 1 + 1 ) = 16 ( see below ) ; 42 = 2 * 3 * 7 , so out of 16 factors only two are multiples of 42 : 42 and 210 , itself ; so , the probability is 2 / 16 = 1 / 8 . answer : c ." | a ) 1 / 16 , b ) 5 / 42 , c ) 1 / 8 , d ) 3 / 16 , e ) 1 / 4 | c | divide(divide(choose(42, const_1), 42), power(const_3, const_2)) | choose(n1,const_1)|power(const_3,const_2)|divide(#0,n1)|divide(#2,#1)| | other | C |
a boat takes 19 hours for travelling downstream from point a to point b and coming back to a point c midway between a and b . if the velocity of the stream is 4 kmph and the speed of the boat in still water is 14 kmph , what is the distance between a and b ? | down stream speed : ( 14 + 4 ) km / hr = 18 km / hr up stream speed : ( 14 - 4 ) km / hr = 10 km / hr ( x / 18 ) + ( x / 2 / 20 ) = 19 x = 180 km answer : b | a ) 160 km , b ) 180 km , c ) 200 km , d ) 220 km , e ) 240 km | b | divide(19, add(divide(const_1, add(14, 4)), divide(const_1, multiply(subtract(14, 4), const_2)))) | add(n1,n2)|subtract(n2,n1)|divide(const_1,#0)|multiply(#1,const_2)|divide(const_1,#3)|add(#2,#4)|divide(n0,#5) | physics | B |
a bus started its journey from chennai and reached mumbai in 50 min with its average speed of 60 km / hr . if the average speed of the bus is increased by 6 km / hr , how much time will it take to cover the same distance ? | sol . distance between mumbai and pune = ( 60 * 50 ) / 60 = 50 average speed of the bus is increased by 6 km / hr then the speed of the bus = 66 km / hr required time = 50 * 66 / 60 = 55 min . b | a ) 45 min . , b ) 55 min . , c ) 65 min . , d ) 60 min . , e ) 50 min . | b | multiply(divide(multiply(divide(50, 60), 60), subtract(60, 6)), 60) | divide(n0,n1)|subtract(n1,n2)|multiply(n1,#0)|divide(#2,#1)|multiply(n1,#3) | general | B |
a and b can do a piece of work in 30 days , while b and c can do the same work in 24 days and c and a in 20 days . they all work together for 10 days when b and c leave . how many days more will a take to finish the work ? | explanation : 2 ( a + b + c ) ' s 1 day ' s work = ( 1 / 30 + 1 / 24 + 1 / 20 ) = 15 / 120 = 1 / 8 work done by a , b , c in 10 days = 10 / 6 = 5 / 8 . a ' s 1 day ' s work = ( 1 / 16 - 1 / 24 ) = 1 / 48 now , 1 / 48 work is done by a in 1 day so , 3 / 8 work will be done by a in ( 48 x 3 / 8 ) = 18 days . answer is a | a ) 18 days , b ) 24 days , c ) 30 days , d ) 36 days , e ) 35 days | a | divide(subtract(const_1, multiply(divide(add(add(divide(const_1, 30), divide(const_1, 24)), divide(const_1, 20)), const_2), 10)), subtract(divide(add(add(divide(const_1, 30), divide(const_1, 24)), divide(const_1, 20)), const_2), divide(const_1, 24))) | divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|add(#3,#2)|divide(#4,const_2)|multiply(n3,#5)|subtract(#5,#1)|subtract(const_1,#6)|divide(#8,#7) | physics | A |
a train running at the speed of 60 km / hr crosses a pole in 12 seconds . find the length of the train ? | "speed = 60 * ( 5 / 18 ) m / sec = 50 / 3 m / sec length of train ( distance ) = speed * time ( 50 / 3 ) * 12 = 200 meter answer : b" | a ) 150 meter , b ) 200 meter , c ) 156 meter , d ) 168 meter , e ) 154 meter | b | multiply(divide(multiply(60, const_1000), const_3600), 12) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics | B |
one fourth of one third of two fifth of a number is 20 . what will be 40 % of that number | explanation : ( 1 / 4 ) * ( 1 / 3 ) * ( 2 / 5 ) * x = 20 then x = 20 * 30 = 600 40 % of 600 = 240 answer : option d | a ) a ) 140 , b ) b ) 150 , c ) c ) 180 , d ) d ) 240 , e ) e ) 220 | d | divide(multiply(divide(20, multiply(multiply(divide(const_1, const_4), divide(const_1, const_3)), divide(const_2, add(const_2, const_3)))), 40), const_100) | add(const_2,const_3)|divide(const_1,const_4)|divide(const_1,const_3)|divide(const_2,#0)|multiply(#1,#2)|multiply(#3,#4)|divide(n0,#5)|multiply(n1,#6)|divide(#7,const_100) | gain | D |
in a games hour 4 different types of players came to the ground ? cricket 22 , hokey 15 , football 21 , softball 19 . how many players are present in the ground ? | "total number of players = 22 + 15 + 21 + 19 = 77 answer is d" | a ) 70 , b ) 52 , c ) 62 , d ) 77 , e ) 79 | d | add(add(22, 15), add(21, 19)) | add(n1,n2)|add(n3,n4)|add(#0,#1)| | physics | D |
the average age of an adult class is 50 years . 12 new students with an avg age of 32 years join the class . therefore decreasing the average by 4 years . find what was the original average age of the class ? | "let original strength = y then , 50 y + 12 x 32 = ( y + 12 ) x 46 â ‡ ’ 50 y + 384 = 46 y + 552 â ‡ ’ 4 y = 168 â ˆ ´ y = 42 c" | a ) 10 , b ) 12 , c ) 42 , d ) 20 , e ) 22 | c | divide(subtract(multiply(12, subtract(50, 4)), multiply(12, 32)), 4) | multiply(n1,n2)|subtract(n0,n3)|multiply(n1,#1)|subtract(#2,#0)|divide(#3,n3)| | general | C |
a car traveled 75 % of the way from town a to town b at an average speed of 80 miles per hour . the car travels at an average speed of s miles per hour for the remaining part of the trip . the average speed for the entire trip was 40 miles per hour . what is s ? | "total distance = 100 miles ( easier to work with % ) 75 % of the distance = 75 miles 25 % of the distance = 25 miles 1 st part of the trip → 75 / 80 = 0.9375 2 nd part of the trip → 25 / s = t total trip → ( 75 + 25 ) / 40 = 0.9375 + t » 100 / 40 = 0.9375 + t » 2.5 = 0.9375 + t » t = 1.5625 back to 2 nd part of the trip formula : 25 / s = 1.5625 » s = 16 ans a" | a ) 16 , b ) 20 , c ) 25 , d ) 30 , e ) 37.5 | a | divide(subtract(add(40, 80), 75), subtract(divide(add(40, 80), 40), divide(75, 80))) | add(n1,n2)|divide(n0,n1)|divide(#0,n2)|subtract(#0,n0)|subtract(#2,#1)|divide(#3,#4)| | physics | A |
a car traveled from san diego to san francisco at an average speed of 63 miles per hour . if the journey back took twice as long , what was the average speed of the trip ? | "let the time taken be = x one way distance = 63 x total distance traveled = 2 * 63 x = 126 x total time taken = x + 2 x = 3 x average speed = 126 x / 3 x = 42 answer : d" | a ) 24 . , b ) 32 . , c ) 36 . , d ) 42 . , e ) 44 . | d | inverse(add(inverse(63), divide(inverse(63), const_2))) | inverse(n0)|divide(#0,const_2)|add(#1,#0)|inverse(#2)| | physics | D |
how much water must be added to 56 litres of milk at 1 1 ⁄ 2 litres for 20 so as to have a mixture worth 10 2 ⁄ 3 a litre ? | "c . p . of 1 litre of milk = ( 20 × 2 ⁄ 3 ) = 40 ⁄ 3 ∴ ratio of water and milk = 8 ⁄ 3 : 32 ⁄ 3 = 8 : 32 = 1 : 4 ∴ quantity of water to be added to 56 litres of milk = ( 1 ⁄ 4 × 56 ) litres = 14 litres . answer c" | a ) 10 litres , b ) 12 litres , c ) 14 litres , d ) 18 litres , e ) none of these | c | divide(20, add(1, divide(1, 2))) | divide(n1,n3)|add(n1,#0)|divide(n4,#1)| | general | C |
each light bulb at a hotel is either incandescent or fluorescent . at a certain moment , forty percent of the incandescent bulbs are switched on , and eighty percent of the fluorescent bulbs are switched on . if 72 percent of all the bulbs are switched on at this moment , what percent of the bulbs that are switched on are incandescent ? | let i be the number of incandescent bulbs . let f be the number of fluorescent bulbs . 0.4 i + 0.8 f = 0.72 ( i + f ) 0.08 f = 0.32 i f = 4 i this means that for every 1 incandescent bulb , there are 4 fluorescent bulbs . the percent of bulbs that are switched on which are incandescent is : 0.4 i / ( 0.4 i + 0.8 f ) = 0.4 i / ( 0.4 i + 0.8 * 4 i ) = 0.4 i / 3.6 i = 1 / 9 which is about 11 % . the answer is c . | a ) 24 % , b ) 16 % , c ) 11 % , d ) 7 % , e ) 5 % | c | divide(multiply(subtract(const_100, 72), multiply(add(divide(const_2, const_0_25), const_2), const_4)), const_100) | divide(const_2,const_0_25)|subtract(const_100,n0)|add(#0,const_2)|multiply(#2,const_4)|multiply(#3,#1)|divide(#4,const_100) | gain | C |
if the radius of a circle decreased by 50 % its area is decreased by : | original area = π r ( power ) 2 new area = π ( r / 2 ) ( power ) 2 = ( π r ( power ) 2 ) / 4 reduction in area = [ ( π r ( power ) 2 - ( π r ( power ) 2 ) / 4 ] 2 = ( 3 π r ( power ) 2 ) / 4 reduction percent = ( ( 3 π r ( power ) 2 ) / 4 × 1 / ( π r ( power ) 2 × 100 ) % = 75 % answer is a . | ['a ) 75', 'b ) 79', 'c ) 72', 'd ) 70', 'e ) 80'] | a | subtract(const_100, multiply(power(divide(50, const_100), const_2), const_100)) | divide(n0,const_100)|power(#0,const_2)|multiply(#1,const_100)|subtract(const_100,#2) | geometry | A |
dacid obtained 91 , 65 , 82 , 67 and 85 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology . what are his average marks ? | "average = ( 91 + 65 + 82 + 67 + 85 ) / 5 = 78 answer : b" | a ) 29 , b ) 78 , c ) 39 , d ) 37 , e ) 75 | b | divide(add(add(add(add(91, 65), 82), 67), 85), divide(const_10, const_2)) | add(n0,n1)|divide(const_10,const_2)|add(n2,#0)|add(n3,#2)|add(n4,#3)|divide(#4,#1)| | general | B |
a company sells pens and pencils . the revenue from pens in 2007 was up 5 % from 2006 . the revenue from pencils declined 13 % over the same period . overall revenue was down 1 % from 06 to 07 . what was the ratio of pencil revenue to pen revenue in 2006 | let the revenue of pens be ' x ' let the revenue of pencils be ' y ' the revenue of pens up from 2006 by 5 % = > 105 % the revenue of pencils down from 2006 by 13 % = > 87 % overall revenue down by 1 % = > 99 % hence the ratio of pencils to pens = 6 / 12 = 1 / 2 answer : a | a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 5 , e ) 1 / 6 | a | divide(subtract(add(5, const_100), subtract(const_100, 1)), subtract(subtract(const_100, 1), subtract(const_100, 13))) | add(n1,const_100)|subtract(const_100,n4)|subtract(const_100,n3)|subtract(#0,#1)|subtract(#1,#2)|divide(#3,#4) | other | A |
what is the probability of getting a sum of 12 when rolling 3 dice simultaneously ? | "sum of 12 can be achieved in following ways 6 , 5,1 - - - total cases = 3 ! = 6 6 , 4,2 - - - total cases = 3 ! = 6 6 , 3,3 - - - total cases = 3 ! / 2 ! = 3 5 , 5,2 - - - total cases = 3 ! / 2 ! = 3 5 , 4,3 - - - total cases = 3 ! = 6 4 , 4,4 - - - total cases = 3 ! / 3 ! = 1 total cases = 25 probability = 25 * ( 1 / 6 * 1 / 6 * 1 / 6 ) = 25 / 216 answer : e" | a ) 10 / 216 , b ) 12 / 216 , c ) 21 / 216 , d ) 23 / 216 , e ) 25 / 216 | e | divide(const_2, choose(add(const_3, const_3), const_3)) | add(const_3,const_3)|choose(#0,const_3)|divide(const_2,#1)| | probability | E |
a library has an average of 500 visitors on sundays and 140 on other days . the average number of visitors per day in a month of 30 days beginning with a sunday is : | "explanation : since the month begins with a sunday , so there will be five sundays in the month , required average = ( 500 * 5 + 140 * 25 ) / 30 = 6000 / 30 = 200 answer : b ) 200" | a ) 100 , b ) 200 , c ) 300 , d ) 400 , e ) 550 | b | divide(add(multiply(add(floor(divide(30, add(const_3, const_4))), const_1), 500), multiply(subtract(30, add(floor(divide(30, add(const_3, const_4))), const_1)), 140)), 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 | B |
at company x , senior sales representatives visit the home office once every 18 days , and junior sales representatives visit the home office once every 12 days . the number of visits that a junior sales representative makes in a 2 - year period is approximately what percent greater than the number of visits that a senior representative makes in the same period ? | "each 36 - day period , senior representatives visit the home office 2 times while junior representatives visit 3 times , thus 50 % more . the answer is d ." | a ) 10 % , b ) 25 % , c ) 33 % , d ) 50 % , e ) 67 % | d | multiply(add(const_1, const_10), subtract(subtract(18, 12), const_1)) | add(const_1,const_10)|subtract(n0,n1)|subtract(#1,const_1)|multiply(#0,#2)| | general | D |
the average of 7 consecutive numbers is 33 . the highest of these numbers is : | "the average of 7 consecutive numbers will be the 4 th number which is given to be = 33 5 th number = 34 , 6 th number = 35 , 7 th number = 36 answer : d ." | a ) 28 , b ) 30 , c ) 33 , d ) 36 , e ) 28 | d | add(divide(subtract(multiply(33, 7), add(add(add(add(const_1, add(add(add(add(add(const_1, const_2), const_1), const_1), const_1), const_1)), const_1), const_1), add(add(add(add(const_1, add(add(add(add(add(const_1, const_2), const_1), const_1), const_1), const_1)), const_1), const_1), const_1))), 7), add(add(add(add(const_1, const_2), const_1), const_1), const_1)) | add(const_1,const_2)|multiply(n0,n1)|add(#0,const_1)|add(#2,const_1)|add(#3,const_1)|add(#4,const_1)|add(#5,const_1)|add(#6,const_1)|add(#7,const_1)|add(#8,const_1)|add(#8,#9)|subtract(#1,#10)|divide(#11,n0)|add(#4,#12)| | general | D |
a dishonest dealer claims to sell a product at its cost price . he uses a counterfeit weight which is 20 % less than the real weight . further greed overtook him and he added 30 % impurities to the product . find the net profit percentage of the dealer ? | "the dealer uses weight which is 20 % less than the real weight . or ( 1 - 1 / 5 ) or 4 / 5 of real weight . it means that he is selling $ 4 worth of product for $ 5 . the dealer then further added 30 % impurities to the product . it means that he is selling $ 5 worth of product for $ 5.5 . so his profit is $ 5.5 - $ 4 = $ 1.5 and his profit percent is ( 1.5 / 4 ) * 100 = 37.5 % answer : - e" | a ) 44 % , b ) 40 % , c ) 50 % , d ) 56.25 % , e ) 37.5 % | e | multiply(subtract(divide(add(const_100, 30), subtract(const_100, 20)), const_1), const_100) | add(n1,const_100)|subtract(const_100,n0)|divide(#0,#1)|subtract(#2,const_1)|multiply(#3,const_100)| | general | E |
the average salary of 15 people in the shipping department at a certain firm is $ 21,000 . the salary of 5 of the employees is $ 25,000 each and the salary of 4 of the employees is $ 16,000 each . what is the average salary of the remaining employees ? | "total salary . . . 15 * 21 k = 315 k 5 emp @ 25 k = 125 k 4 emp @ 16 k = 64 k remaing 6 emp sal = 315 k - 125 k - 64 k = 126 k average = 126 k / 6 = 21000 ans : c" | a ) $ 19,250 , b ) $ 18,500 , c ) $ 21,000 , d ) $ 15,850 , e ) $ 12,300 | c | divide(subtract(subtract(multiply(multiply(5, 4), multiply(4, 4)), multiply(multiply(5, 5), 5)), multiply(4, 15)), add(const_2, 5)) | add(n2,const_2)|multiply(n2,n4)|multiply(n4,n4)|multiply(n2,n2)|multiply(n0,n4)|multiply(#1,#2)|multiply(n2,#3)|subtract(#5,#6)|subtract(#7,#4)|divide(#8,#0)| | general | C |
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 × 8 = 560 . cost of 10 kg of mangoes = 55 × 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 | A |
the annual interest rate earned by an investment increased by 11 percent from last year to this year . if the annual interest rate earned by the investment this year was 22 percent , what was the annual interest rate last year ? | "let i = interest rate i ( this year ) = i ( last year ) + 0.11 i ( last year ) = 1.11 i ( last year ) 22 = 1.11 x i ( last year ) i ( last year ) = 22 / 1.1 = 220 / 11 = 20 % answer : d" | a ) 17 % , b ) 19 % , c ) 18 % , d ) 20 % , e ) 22 % | d | divide(multiply(22, const_100), add(22, const_100)) | add(n1,const_100)|multiply(n1,const_100)|divide(#1,#0)| | gain | D |
mr . kramer , the losing candidate in a two - candidate election , received 942,568 votes , which was exactly 30 percent of all votes cast . approximately what percent of the remaining votes would he need to have received in order to have won at least 50 percent of all the votes cast ? | let me try a simpler one . lets assume that candidate got 30 % votes and total votes is 100 . candidate won = 30 remaining = 70 to get 50 % , candidate requires 20 votes from 100 which is 20 % and 20 votes from 70 . 20 / 70 = 2 / 7 = . 285 = 28.5 % which is approx 29 % . hence the answer is d . | a ) 10 % , b ) 12 % , c ) 15 % , d ) 29 % , e ) 20 % | d | multiply(divide(subtract(divide(50, const_100), divide(30, const_100)), subtract(const_1, divide(30, const_100))), const_100) | divide(n2,const_100)|divide(n1,const_100)|subtract(#0,#1)|subtract(const_1,#1)|divide(#2,#3)|multiply(#4,const_100) | general | D |
the true discount on a bill due 9 months hence at 16 % per annum is rs . 153 . the amount of the bill is | "solution 32.5 let p . w . be rs . x . then , s . i . on rs . x at 16 % for 9 months = rs . 153 . ∴ x 16 x 9 / 12 x 1 / 100 } = 153 or x = 1275 . ∴ p . w . = rs . 1275 . answer c" | a ) rs . 1386 , b ) rs . 1764 , c ) rs . 1275 , d ) rs . 2268 , e ) none of these | c | add(divide(153, divide(multiply(divide(9, multiply(const_4, const_3)), 16), const_100)), 153) | multiply(const_3,const_4)|divide(n0,#0)|multiply(n1,#1)|divide(#2,const_100)|divide(n2,#3)|add(n2,#4)| | gain | C |
a boat can travel with a speed of 16 km / hr in still water . if the rate of stream is 5 km / hr , then find the time taken by the boat to cover distance of 147 km downstream . | "explanation : it is very important to check , if the boat speed given is in still water or with water or against water . because if we neglect it we will not reach on right answer . i just mentioned here because mostly mistakes in this chapter are of this kind only . lets see the question now . speed downstream = ( 16 + 5 ) = 21 kmph time = distance / speed = 147 / 21 = 7 hours option d" | a ) 4 hours , b ) 5 hours , c ) 6 hours , d ) 7 hours , e ) 8 hours | d | divide(147, add(16, 5)) | add(n0,n1)|divide(n2,#0)| | physics | D |
at the end of the month , a certain ocean desalination plant ’ s reservoir contained 40 million gallons of water . this amount is one fifth of the normal level . if this amount represents 80 % of the reservoir ’ s total capacity , how many million gallons short of total capacity is the normal level ? | "the q talks of total capacity , normal level , present level , shortage etc . . so it is all about not going wrong in these terms 40 mg = 80 % of total . . total = 40 / . 8 = 50 mg . . normal level = 1 / 5 of 50 = 10 mg . . shortage of normal level = 50 - 10 = 40 mg . . c" | a ) 20 , b ) 30 , c ) 40 , d ) 50 , e ) 60 | c | divide(divide(40, const_2), divide(80, const_100)) | divide(n0,const_2)|divide(n1,const_100)|divide(#0,#1)| | gain | C |
the average weight of a group of boys is 20 kg . after a boy of weight 36 kg joins the group , the average weight of the group goes up by 1 kg . find the number of boys in the group originally ? | "let the number off boys in the group originally be x . total weight of the boys = 20 x after the boy weighing 36 kg joins the group , total weight of boys = 20 x + 36 so 20 x + 36 = 21 ( x + 1 ) = > x = 15 answer : b" | a ) 12 , b ) 15 , c ) 18 , d ) 24 , e ) 10 | b | add(subtract(36, add(20, 1)), 1) | add(n0,n2)|subtract(n1,#0)|add(#1,n2)| | general | B |
a train 200 m long is running with a speed of 60 km / hr . in what time will it pass a man who is running at 10 km / hr in the direction opposite to that in which the train is going ? | "speed of train relative to man = 60 + 10 = 70 km / hr . = 70 * 5 / 18 = 175 / 9 m / sec . time taken to pass the men = 200 * 9 / 175 = 10 sec . answer : c" | a ) 5 , b ) 6 , c ) 10 , d ) 9 , e ) 5 | c | divide(200, multiply(add(60, 10), const_0_2778)) | add(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics | C |
a tourist does not have to pay tax on the first $ 600 of goods he purchases in country b , but does have to pay a 12 percent tax on the portion of the total value that is in excess of $ 600 . what tax must be paid by a tourist if he buys goods with a total value of $ 1720 ? | "correct answer : e the tourist must pay tax on $ 1720 - $ 600 = $ 1120 . thus , the amount of tax he has to pay is 0.12 ( $ 1120 ) = $ 134.40 . the correct answer is e ." | a ) $ 54.00 , b ) $ 64.80 , c ) $ 90.00 , d ) $ 100.80 , e ) $ 134.40 | e | divide(multiply(subtract(1720, 600), 12), const_100) | subtract(n3,n0)|multiply(n1,#0)|divide(#1,const_100)| | general | E |
harish can paint a wall in just 3 hours . however it takes 6 hours for ganpat to complete the same job of painting the wall . how long would it take for both of them to paint the wall , if both work together ? | d two hours . harish who could paint the wall in 3 hours could paint 2 / 3 of the wall in just two hours . ganpat who paint the wall in 6 hours could paint 1 / 3 of the wall in two hours . 2 / 3 + 1 / 3 = 1 . | a ) 3 hours , b ) 1 hours , c ) 3.20 hours , d ) 2 hours , e ) 1.35 hours | d | inverse(add(inverse(3), inverse(6))) | inverse(n0)|inverse(n1)|add(#0,#1)|inverse(#2) | physics | D |
a certain car dealership sells economy cars , luxury cars , and sport utility vehicles . the ratio of economy to luxury cars is 3 : 2 . the ratio of economy cars to sport utility vehicles is 5 : 3 . what is the ratio of luxury cars to sport utility vehicles ? | "the ratio of economy to luxury cars is 3 : 2 - - > e : l = 3 : 2 = 15 : 10 . the ratio of economy cars to sport utility vehicles is 5 : 3 - - > e : s = 5 : 3 = 15 : 9 . thus , l : s = 10 : 9 . answer : e ." | a ) 9 : 8 , b ) 8 : 9 , c ) 3 : 2 , d ) 2 : 3 , e ) 10 : 9 | e | divide(divide(multiply(const_4, 3), multiply(3, 3)), divide(multiply(3, const_4), multiply(2, const_4))) | multiply(n3,const_4)|multiply(n3,n3)|multiply(n1,const_4)|divide(#0,#1)|divide(#0,#2)|divide(#3,#4)| | other | E |
the slant height of a cone is 22 cm and radius of the base is 7 cm , find the curved surface of the cone ? | "π * 22 * 7 = 483 answer : c" | a ) 426 , b ) 284 , c ) 483 , d ) 256 , e ) 428 | c | multiply(multiply(const_pi, 7), 22) | multiply(n1,const_pi)|multiply(n0,#0)| | geometry | C |
if both 5 ^ 2 and 3 ^ 3 are factors of the number a * 4 ^ 3 * 6 ^ 2 * 13 ^ 11 , then what is the smallest possible value of a ? | "the number a must include at least 3 * 5 ^ 2 = 75 the answer is b ." | a ) 49 , b ) 75 , c ) 150 , d ) 225 , e ) 343 | b | add(add(add(add(add(multiply(multiply(5, 2), 2), multiply(multiply(5, 2), 2)), multiply(multiply(5, 2), 2)), 2), const_4), const_4) | multiply(n0,n7)|multiply(n1,#0)|add(#1,#1)|add(#2,#1)|add(n7,#3)|add(#4,const_4)|add(#5,const_4)| | other | B |
on a sum of money , the simple interest for 2 years is rs . 660 , while the compound interest is rs . 696.30 , the rate of interest being the same in both the cases . the rate of interest is : | "solution difference in c . i . and s . i . for 2 years = rs . ( 696.30 - 660 ) = rs . 36.30 . s . i . for one year = rs . 330 . ∴ s . i . on rs . 330 for 1 year = rs . 36.30 . ∴ rate ( 100 x 36.30 / 330 x 1 ) % = 11 % . answer d" | a ) 10 % , b ) 10.5 % , c ) 12 % , d ) none of these , e ) can not be determined | d | divide(multiply(const_100, subtract(subtract(696.30, divide(660, 2)), divide(660, 2))), divide(660, 2)) | divide(n1,n0)|subtract(n2,#0)|subtract(#1,#0)|multiply(#2,const_100)|divide(#3,#0)| | gain | D |
a machine , working at a constant rate , manufactures 18 pens in 30 minutes . how many pens does it make in 1 hr 45 min ? | change 1 hr 45 min to 105 min . for this , we need to set up a simple proportion of pens per time 18 / 30 = s / 105 the absolutely worst thing you could do at this point in the problem is to cross - multiply . that would be a supremely unstrategic move . instead , cancel before you multiply . for what we can see this post . we can cancel 18 / 30 as 6 / 10 6 / 10 = s / 105 we can cross - multiply . 10 * s = 6 * 105 = 630 if we divide both sides by 10 , s = 63 the machine would be 63 pens in 1 hr 45 min . answer : a | a ) 63 , b ) 65 , c ) 62 , d ) 60 , e ) 45 | a | multiply(divide(add(multiply(1, const_60), 45), 30), 18) | multiply(n2,const_60)|add(n3,#0)|divide(#1,n1)|multiply(n0,#2) | physics | A |
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 18 % . 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 = 0.82 * 150 x = 123 x hence total profit = 23 % option d" | a ) 27.5 % , b ) 30 % , c ) 35 % , d ) 23 % , e ) 40 % | d | subtract(multiply(divide(subtract(const_100, 18), 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 | D |
a sells a bicycle to b at a profit of 20 % . b sells it to c at a profit of 25 % . if c pays rs . 225 for it , the cost price of the bicycle for a is | explanation : 125 % of 120 % of a = 225 125 / 100 * 120 / 100 * a = 225 a = 225 * 2 / 3 = 150 . answer : d | a ) 627 , b ) 297 , c ) 127 , d ) 150 , e ) 181 | d | divide(multiply(const_100, divide(multiply(const_100, 225), add(25, const_100))), add(20, const_100)) | add(n1,const_100)|add(n0,const_100)|multiply(n2,const_100)|divide(#2,#0)|multiply(#3,const_100)|divide(#4,#1) | gain | D |
the sum of two prime numbers is 85 . what is the product of these two prime numbers ? | since there is only one even prime number ( 2 ) , and the sum of two odd numbers is even , the product of these two numbers must be even . correct answer e | a ) 85 , b ) 91 , c ) 115 , d ) 133 , e ) 166 | e | multiply(subtract(85, const_2), const_2) | subtract(n0,const_2)|multiply(#0,const_2) | general | E |
the average weight of a , b and c is 45 kg . if the average weight of a and b be 40 kg and that of b and c be 43 kg , then the weight of b is : | "let a , b , c represent their respective weights . then , we have : a + b + c = ( 45 x 3 ) = 135 . . . . ( i ) a + b = ( 40 x 2 ) = 80 . . . . ( ii ) b + c = ( 43 x 2 ) = 86 . . . . ( iii ) adding ( ii ) and ( iii ) , we get : a + 2 b + c = 166 . . . . ( iv ) subtracting ( i ) from ( iv ) , we get : b = 31 . b ' s weight = 31 kg b )" | a ) 30 kg , b ) 31 kg , c ) 37 kg , d ) 39 kg , e ) 41 kg | b | subtract(add(multiply(40, const_2), multiply(43, const_2)), multiply(45, const_3)) | multiply(n1,const_2)|multiply(n2,const_2)|multiply(n0,const_3)|add(#0,#1)|subtract(#3,#2)| | general | B |
the speed of a car is 90 km in the first hour and 60 km in the second hour . what is the average speed of the car ? | "s = ( 90 + 60 ) / 2 = 75 kmph answer : b" | a ) 12 , b ) 75 , c ) 88 , d ) 54 , e ) 15 | b | divide(add(90, 60), const_2) | add(n0,n1)|divide(#0,const_2)| | physics | B |
x starts a business with rs . 52000 . y joins in the business after 6 months with rs . 26000 . what will be the ratio in which they should share the profit at the end of the year ? | "explanation : ratio in which they should share the profit = ratio of the investments multiplied by the time period = 52000 * 12 : 26000 * 6 = 52 * 12 : 26 * 6 = 24 : 6 = 4 : 1 . answer : option d" | a ) 1 : 2 , b ) 2 : 3 , c ) 1 : 3 , d ) 4 : 1 , e ) none of these | d | divide(multiply(52000, const_12), multiply(26000, add(const_4, const_3))) | add(const_3,const_4)|multiply(n0,const_12)|multiply(n2,#0)|divide(#1,#2)| | other | D |
the average weight of a class of 20 boys was calculated to be 58.4 kgs and it was later found that one weight was misread as 56 kg instead of 61 kg . what is the correct weight ? | "actual total weight is ( 20 x 58.4 - 56 + 61 ) = 1173 kgs actual average weight is 1177 / 20 = 58.65 kgs b" | a ) 58 kgs , b ) 58.65 kgs , c ) 58.95 kgs , d ) 59 kgs , e ) 59.85 kgs | b | divide(add(multiply(58.4, 20), subtract(61, 56)), 20) | multiply(n0,n1)|subtract(n3,n2)|add(#0,#1)|divide(#2,n0)| | general | B |
a , b and c can do a piece of work in 9 days , 14 days and 28 days respectively . how long will they taken , if all the three work together ? | 1 / 9 + 1 / 14 + 1 / 28 = 55 / 252 = > 252 / 55 days = > 4.52 days answer : b | a ) 4.12 days , b ) 4.52 days , c ) 3.52 days , d ) 2.52 days , e ) 4.92 days | b | inverse(add(inverse(28), add(inverse(9), inverse(14)))) | inverse(n0)|inverse(n1)|inverse(n2)|add(#0,#1)|add(#3,#2)|inverse(#4) | physics | B |
the contents of a certain box consist of 40 apples and 30 oranges . how many oranges must be added to the box so that exactly 40 % of the pieces of fruit in the box will be apples ? | "apple = ( apple + orange + x ) * 0.4 40 = ( 30 + 40 + x ) * 0.4 x = 30 answer : e" | a ) 40 , b ) 50 , c ) 60 , d ) 70 , e ) 30 | e | subtract(add(40, 30), divide(40, divide(40, const_100))) | add(n0,n1)|divide(n2,const_100)|divide(n0,#1)|subtract(#0,#2)| | general | E |
in a college the ratio of the numbers of boys to the girls is 8 : 5 . if there are 175 girls , the total number of students in the college is ? | "let the number of boys and girls be 8 x and 5 x then , 5 x = 175 x = 35 total number of students = 13 x = 13 * 35 = 455 answer is c" | a ) 562 , b ) 356 , c ) 455 , d ) 416 , e ) 512 | c | add(multiply(divide(8, 5), 175), 175) | divide(n0,n1)|multiply(n2,#0)|add(n2,#1)| | other | C |
when the price of an article is reduced by 30 % , the sales increases by 50 % . the percentage change in the total amount of receipts is ? | we have , receipts = price x sales therefore , net % change in receipts = ( x + y + xy / 100 ) % = [ - 30 + 50 + ( - 30 x 50 ) / 100 ] % = 5 % increase . answer : b | a ) 5 % decrease , b ) 5 % increase , c ) 10 % decrease , d ) 15 % decrease , e ) none of these | b | subtract(divide(multiply(subtract(const_100, 30), add(const_100, 50)), const_100), const_100) | add(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|subtract(#3,const_100) | general | B |
a property decreases in value every year at the rate of 6 1 / 4 % of its value at the beginning of the year its value at the end of 3 years was rs . 21093 . find its value at the beginning of the first year ? | 6 1 / 4 % = 1 / 16 x * 15 / 16 * 15 / 16 * 15 / 16 = 21093 x = 25600.24 answer : a | a ) 25600.24 , b ) 25600.27 , c ) 25600.28 , d ) 25600.21 , e ) 25600.29 | a | divide(21093, power(subtract(1, divide(add(divide(1, 4), 6), const_100)), const_3)) | divide(n1,n2)|add(n0,#0)|divide(#1,const_100)|subtract(n1,#2)|power(#3,const_3)|divide(n4,#4) | gain | A |
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