Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
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if the population of a certain country increases at the rate of one person every 15 seconds , by how many persons does the population increase in 1 hour ? | "answer = 4 * 60 = 240 answer is e" | a ) 100 , b ) 120 , c ) 150 , d ) 180 , e ) 240 | e | multiply(divide(const_60, 15), 1) | divide(const_60,n0)|multiply(n1,#0)| | physics |
the difference of two numbers is 8 and one - fourth of their sum is 6 . find the numbers . | "let numbers be x and y . equation ( i ) : x - y = 8 equation ( ii ) : ( x + y ) / 4 = 6 solve system of equations : x = 16 , y = 8 a is the correct answer ." | a ) 16,8 , b ) 10,2 , c ) 17,6 , d ) 14,10 , e ) 12,4 | a | add(subtract(multiply(divide(const_10, const_2), 6), divide(add(8, multiply(divide(const_10, const_2), 6)), const_2)), divide(const_10, const_2)) | divide(const_10,const_2)|multiply(n1,#0)|add(n0,#1)|divide(#2,const_2)|subtract(#1,#3)|add(#0,#4)| | general |
a retailer bought a shirt at wholesale and marked it up 80 % to its initial price of $ 45 . by how many more dollars does he need to increase the price to achieve a 100 % markup ? | "x is the buy price . x + 80 % x = 1,8 x is the sale price = 45 so x , the buy price becomes 45 / 1,8 = 25 % 100 mark up means x + 100 % x = 2 x so 50 . 50 - 45 = 5 answer : e" | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | e | subtract(multiply(divide(45, add(const_1, divide(80, 100))), const_2), 45) | divide(n0,n2)|add(#0,const_1)|divide(n1,#1)|multiply(#2,const_2)|subtract(#3,n1)| | general |
indu gave bindu rs . 1250 on compound interest for 2 years at 8 % per annum . how much loss would indu has suffered had she given it to bindu for 2 years at 4 % per annum simple interest ? | "1250 = d ( 100 / 4 ) 8 d = 8 answer : c" | a ) 1 , b ) 2 , c ) 8 , d ) 4 , e ) 5 | c | subtract(subtract(multiply(1250, power(add(const_1, divide(8, const_100)), 2)), 1250), multiply(multiply(1250, divide(8, const_100)), 2)) | divide(n2,const_100)|add(#0,const_1)|multiply(n0,#0)|multiply(n1,#2)|power(#1,n1)|multiply(n0,#4)|subtract(#5,n0)|subtract(#6,#3)| | gain |
the ratio between the length and the breadth of a rectangular park is 1 : 2 . if a man cycling along the boundary of the park at the speed of 6 km / hr completes one round in 6 min , then the area of the park ( in sq . m ) is ? | "perimeter = distance covered in 6 min . = ( 6000 x 6 ) / 60 m = 600 m . let length = 1 x metres and breadth = 2 x metres . then , 2 ( 1 x + 2 x ) = 600 or x = 100 . length = 100 m and breadth = 200 m . area = ( 100 x 200 ) m 2 = 20000 m c" | a ) 600 m , b ) 6000 m , c ) 20000 m , d ) 100 m , e ) 2000 m | c | rectangle_area(divide(divide(multiply(multiply(divide(6, multiply(const_10, multiply(const_3, const_2))), 6), const_1000), add(1, 2)), const_2), multiply(divide(divide(multiply(multiply(divide(6, multiply(const_10, multiply(const_3, const_2))), 6), const_1000), add(1, 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 |
at a certain company , each employee has a salary grade s that is at least 1 and at most 5 . each employee receives an hourly wage p , in dollars , determined by the formula p = 7.50 + 0.25 ( s – 1 ) . an employee with a salary grade of 5 receives how many more dollars per hour than an employee with a salary grade of 1... | oa is definitely wrong . the answer should be c . | a ) $ 0.50 , b ) $ 1.00 , c ) $ 1.25 , d ) $ 1.50 , e ) $ 1.75 | c | add(multiply(0.25, subtract(5, 1)), 0.25) | subtract(n1,n0)|multiply(n3,#0)|add(n3,#1) | general |
the lateral surface area of cuboid length 12 m , breadth 8 m and height 6 m . | "cuboid lateral surface = 2 h ( l + b ) = 2 × 6 ( 20 ) = 240 m ( power 2 ) answer is b" | a ) 540 , b ) 240 , c ) 440 , d ) 140 , e ) 340 | b | power(sqrt(divide(12, const_4)), const_3) | divide(n0,const_4)|sqrt(#0)|power(#1,const_3)| | geometry |
running at the same constant rate , 6 identical machines can produce a total of 270 bottles per minute . at this rate , how many bottles could 12 such machines produce in 4 minutes ? | "solution let the required number of bottles be x . more machines , more bottles ( direct proportion ) more minutes , more bottles ( direct proportion ) â ˆ ´ 6 ã — 1 ã — x = 12 ã — 4 ã — 270 â ‡ ” x = 12 x 4 x 270 / 6 = 2160 . answer d" | a ) 648 , b ) 1800 , c ) 2700 , d ) 2160 , e ) none of these | d | multiply(multiply(divide(270, 6), 4), 12) | divide(n1,n0)|multiply(n3,#0)|multiply(n2,#1)| | gain |
of the 400 employees in a certain company , 25 percent will be relocated to city x and the remaining 75 percent will be relocated to city y . however , 40 percent of the employees prefer city y and 60 percent prefer city x . what is the highest possible number of employees who will be relocated to the city they prefer ... | "240 prefer x ( group 1 ) ; 160 prefer y ( group 2 ) . city y needs 300 people : letall 160 who prefer y ( entire group 2 ) be relocated there , the rest 140 will be those who prefer x from group 1 ; city x needs 100 people : 240 - 140 = 100 from group 1 will be relocated to x , which they prefer . so , the highest pos... | a ) 65 , b ) 100 , c ) 115 , d ) 130 , e ) 260 | e | add(multiply(40, const_2), 60) | multiply(n3,const_2)|add(n4,#0)| | gain |
two friends plan to walk along a 22 - km trail , starting at opposite ends of the trail at the same time . if friend p ' s rate is 20 % faster than friend q ' s , how many kilometers will friend p have walked when they pass each other ? | "if q complete x kilometers , then p completes 1.2 x kilometers . x + 1.2 x = 22 2.2 x = 22 x = 10 then p will have have walked 1.2 * 10 = 12 km . the answer is c ." | a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | c | multiply(divide(22, add(add(divide(20, const_100), const_1), const_1)), add(divide(20, const_100), const_1)) | divide(n1,const_100)|add(#0,const_1)|add(#1,const_1)|divide(n0,#2)|multiply(#1,#3)| | gain |
a car traveled 40 % of the way from town x to town y at an average speed of 40 mph . the car traveled at an average speed of v mph for the remaining part of the trip . the average speed for the entire trip was 50 mph . what is v in mph ? | "assume total distance = 100 miles time taken for 40 miles = 40 / 40 = 1 hour time taken for the rest of the 60 miles = 60 / v hours . average speed = 50 therefore the total time needed = 2 hours . 2 = 1 + 60 / v hence v = 60 mph answer : d" | a ) 50 , b ) 45 , c ) 55 , d ) 60 , e ) 65 | d | divide(subtract(const_100, 40), divide(40, 40)) | divide(n0,n0)|subtract(const_100,n0)|divide(#1,#0)| | general |
the average weight of a , b and c is 45 kg . if the average weight of a and b be 40 kg and that of b and c be 43 kg , what is the weight of b ? | "let the weight of a , b and c are a , b and c respectively . average weight of a , b and c = 45 a + b + c = 45 × 3 = 135 - - - equation ( 1 ) average weight of a and b = 40 a + b = 40 × 2 = 80 - - - equation ( 2 ) average weight of b and c = 43 b + c = 43 × 2 = 86 - - - equation ( 3 ) equation ( 2 ) + equation ( 3 ) -... | a ) 28 , b ) 30 , c ) 32 , d ) 34 , e ) 31 | e | 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 |
bert and rebecca were looking at the price of a condominium . the price of the condominium was 80 % more than bert had in savings , and separately , the same price was also 20 % more than rebecca had in savings . what is the ratio of what bert has in savings to what rebecca has in savings . | "suppose bert had 100 so price becomes 180 , this 180 = 1.2 times r ' s saving . . so r ' s saving becomes 150 so required ratio is 100 : 150 = 2 : 3 answer : c" | a ) 1 : 4 , b ) 4 : 1 , c ) 2 : 3 , d ) 3 : 2 , e ) 3 : 4 | c | divide(divide(const_100, add(const_100, 80)), divide(const_100, add(const_100, 20))) | add(n0,const_100)|add(n1,const_100)|divide(const_100,#0)|divide(const_100,#1)|divide(#2,#3)| | gain |
of the 1000 inhabitants of a town , 60 % are males of whom 120 % are literate . if , of all the inhabitants , 25 % are literate , then what percent of the females of the town are literate ? | "explanation : number of males = 60 % of 1000 = 600 . number of females = ( 1000 - 600 ) = 400 . number of literates = 25 % of 1000 = 250 . number of literate males = 20 % of 600 = 120 . number of literate females = ( 250 - 120 ) = 130 . required pecentage = ( 130 / 400 * 100 ) % = 32.5 % . answer : a ) 32.5 %" | a ) 32.5 % , b ) 32.9 % , c ) 92.5 % , d ) 38.5 % , e ) 37.5 % | a | multiply(divide(subtract(multiply(divide(25, const_100), 1000), multiply(divide(120, const_100), multiply(1000, divide(60, const_100)))), subtract(1000, multiply(1000, divide(60, const_100)))), const_100) | divide(n3,const_100)|divide(n2,const_100)|divide(n1,const_100)|multiply(n0,#0)|multiply(n0,#2)|multiply(#1,#4)|subtract(n0,#4)|subtract(#3,#5)|divide(#7,#6)|multiply(#8,const_100)| | gain |
if 4 < x < 6 < y < 10 , then what is the greatest possible positive integer difference of x and y ? | "4 < x < 6 < y < 10 ; 4 < x y < 10 4 + y < x + 10 y - x < 6 . positive integer difference is 5 ( for example y = 9.5 and x = 4.5 ) answer : c ." | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | c | subtract(subtract(10, 4), const_1) | subtract(n2,n0)|subtract(#0,const_1)| | general |
a car traveling at a certain constant speed takes 15 seconds longer to travel 1 kilometer than it would take to travel 1 kilometer at 48 kilometers per hour . at what speed , in kilometers per hour , is the car traveling ? | "48 * t = 1 km = > t = 1 / 48 km / h v * ( t + 15 / 3600 ) = 1 v ( 1 / 48 + 15 / 3600 ) = 1 v ( 90 / 3600 ) = 1 v = 40 km / h the answer is b ." | a ) 39 , b ) 40 , c ) 41 , d ) 42 , e ) 43 | b | divide(1, divide(add(multiply(const_3600, divide(1, 48)), 15), const_3600)) | divide(n1,n3)|multiply(#0,const_3600)|add(n0,#1)|divide(#2,const_3600)|divide(n1,#3)| | physics |
in a fuel station the service costs $ 1.75 per car , every liter of fuel costs 0.75 $ . assuming that a company owns 12 cars and that every fuel tank contains 55 liters and they are all empty , how much money total will it cost to fuel all cars ? | 12 * 1.75 + 0.75 * 12 * 55 = 516 hence - d | a ) 320 $ , b ) 420 $ , c ) 490 $ , d ) 516 $ , e ) 680 $ | d | multiply(multiply(0.75, 55), 12) | multiply(n1,n3)|multiply(n2,#0)| | general |
the sum of the first n positive perfect squares , where n is a positive integer , is given by the formula n 3 / 3 + cn 2 + n / 6 , where c is a constant . what is the sum of the first 15 positive perfect squares ? | "for n = 1 , 1 = 1 / 3 + c + 1 / 6 1 = 1 / 2 + c = > c = 1 / 2 15 * 15 * 15 / 3 + 1 / 2 * 15 * 15 + 15 / 6 = 1240 answer : c" | a ) 1,010 , b ) 1,164 , c ) 1,240 , d ) 1,316 , e ) 1,476 | c | divide(divide(multiply(multiply(15, add(15, const_1)), add(multiply(15, 2), const_1)), 6), divide(multiply(multiply(15, add(15, const_1)), add(multiply(15, 2), const_1)), 6)) | add(n4,const_1)|multiply(n4,n2)|add(#1,const_1)|multiply(n4,#0)|multiply(#2,#3)|divide(#4,n3)|divide(#5,#5)| | general |
a rectangular farm has to be fenced one long side , one short side and the diagonal . if the cost of fencing is rs . 15 per meter . the area of farm is 1200 m 2 and the short side is 30 m long . how much would the job cost ? | explanation : l * 30 = 1200 è l = 40 40 + 30 + 50 = 120 120 * 15 = 1800 answer : option d | ['a ) 1276', 'b ) 1200', 'c ) 2832', 'd ) 1800', 'e ) 1236'] | d | multiply(add(add(30, divide(1200, 30)), sqrt(add(power(30, const_2), power(divide(1200, 30), const_2)))), 15) | divide(n1,n3)|power(n3,const_2)|add(n3,#0)|power(#0,const_2)|add(#1,#3)|sqrt(#4)|add(#2,#5)|multiply(n0,#6) | geometry |
there are 7 players in a chess group , and each player plays each of the others once . given that each game is played by two players , how many total games will be played ? | "7 players are there . two players play one game with one another . so 7 c 2 = 7 * 6 / 2 = 21 so option a is correct" | a ) 21 , b ) 30 , c ) 45 , d ) 60 , e ) 90 | a | divide(multiply(7, subtract(7, const_1)), const_2) | subtract(n0,const_1)|multiply(n0,#0)|divide(#1,const_2)| | general |
in a graduating class of 238 students , 144 took geometry and 119 took biology . what is the difference between the greatest possible number and the smallest possible number of students that could have taken both geometry and biology ? | "greatest possible number taken both should be 144 ( as it is maximum for one ) smallest possible number taken both should be given by total - neither = a + b - both both = a + b + neither - total ( neither must be 0 to minimize the both ) so 144 + 119 + 0 - 238 = 25 greatest - smallest is 144 - 25 = 119 so answer must... | a ) 144 , b ) 119 , c ) 113 , d ) 88 , e ) 31 | b | subtract(119, subtract(add(144, 119), 238)) | add(n1,n2)|subtract(#0,n0)|subtract(n2,#1)| | other |
the diagonals of a rhombus are 12 cm and 15 cm . find its area ? | "area = 0.5 x 12 x 15 = 90 cm â ² answer : c" | a ) 30 , b ) 45 , c ) 90 , d ) 120 , e ) 180 | c | rhombus_area(12, 15) | rhombus_area(n0,n1)| | geometry |
there are 50 students in a class . if 10 % are absent on a particular day , find the number of students present in the class . | "number of students absent on a particular day = 10 % of 50 i . e . , 10 / 100 × 50 = 5 therefore , the number of students present = 50 - 5 = 45 students . answer : d" | a ) 43 , b ) 36 , c ) 28 , d ) 45 , e ) 11 | d | multiply(subtract(const_1, divide(10, const_100.0)), 50) | divide(n1,const_100.0)|subtract(const_1,#0)|multiply(#1,n0)| | gain |
the speed at which a woman can row a boat in still water is 300 kmph . if he rows downstream , where the speed of current is 42 kmph , what time will he take to cover 190 metres ? | "speed of the boat downstream = 300 + 42 = 342 kmph = 342 * 5 / 18 = 95 m / s hence time taken to cover 190 m = 190 / 95 = 2 seconds . answer : b" | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | b | divide(190, multiply(add(300, 42), const_0_2778)) | add(n0,n1)|multiply(#0,const_0_2778)|divide(n2,#1)| | physics |
in a circle with a radius of 6 , what is the area of the biggest rectangle that can be cut out of this circle ? | the biggest rectangle would be a square , whose diagonal is 12 ( 6 + 6 ) if side of square is a , 144 = a ^ 2 + a ^ 2 = 2 a ^ 2 so area of square = a ^ 2 = 144 / 2 = 72 answer : d | ['a ) 12', 'b ) 96', 'c ) 50', 'd ) 72', 'e ) 144'] | d | power(sqrt(divide(power(multiply(6, const_2), const_2), const_2)), const_2) | multiply(n0,const_2)|power(#0,const_2)|divide(#1,const_2)|sqrt(#2)|power(#3,const_2) | geometry |
the speed of a boat in still water in 42 km / hr and the rate of current is 3 km / hr . the distance travelled downstream in 44 minutes is : | "speed downstream = ( 42 + 3 ) = 45 kmph time = 44 minutes = 44 / 60 hour = 11 / 15 hour distance travelled = time × speed = 11 / 15 × 45 = 33 km answer : c" | a ) 86.6 km , b ) 46.6 km , c ) 33 km , d ) 35.6 km , e ) 26.6 km | c | multiply(add(42, 3), divide(44, const_60)) | add(n0,n1)|divide(n2,const_60)|multiply(#0,#1)| | physics |
find the total average marks of all the students in 2 separate classes , if the average marks of students in the first class of 39 students is 45 and that of another class of 35 students is 70 . | sum of the marks for the class of 39 students = 39 * 45 = 1755 sum of the marks for the class of 35 students = 35 * 70 = 2450 sum of the marks for the class of 74 students = 1755 + 2450 = 4205 average marks of all the students = 4205 / 74 = 56.8 answer : d | a ) 55.8 , b ) 51.8 , c ) 53.8 , d ) 56.8 , e ) 52.8 | d | subtract(add(subtract(subtract(add(add(multiply(45, const_1), const_4), const_12), const_4), const_0_25), const_1), const_1) | multiply(n2,const_1)|add(#0,const_4)|add(#1,const_12)|subtract(#2,const_4)|subtract(#3,const_0_25)|add(#4,const_1)|subtract(#5,const_1) | general |
on an order of 4 dozen boxes of a consumer product , a retailer receives an extra dozen free . this is equivalent to allowing him a discount of : | "clearly , the retailer gets 1 dozen out of 5 dozens free . equivalent discount = 1 / 5 * 100 = 20 % . answer b ) 20 %" | a ) 10 % , b ) 20 % , c ) 30 % , d ) 40 % , e ) 50 % | b | subtract(const_100, multiply(divide(const_3.0, const_4), const_100)) | divide(const_3.0,const_4)|multiply(#0,const_100)|subtract(const_100,#1)| | general |
in an examination , 300 students appeared . out of these students ; 27 % got first division , 54 % got second division and the remaining just passed . assuming that no student failed ; find the number of students who just passed . | "the number of students with first division = 27 % of 300 = 27 / 100 × 300 = 8100 / 100 = 81 and , the number of students with second division = 54 % of 300 = 54 / 100 × 300 = 16200 / 100 = 162 therefore , the number of students who just passed = 300 – ( 81 + 162 ) = 57 answer : b" | a ) 33 , b ) 57 , c ) 54 , d ) 99 , e ) 01 | b | divide(multiply(300, subtract(const_100, add(27, 54))), const_100) | add(n1,n2)|subtract(const_100,#0)|multiply(n0,#1)|divide(#2,const_100)| | general |
x starts a business with rs . 77000 . y joins in the business after 5 months with rs . 42000 . 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 = 77000 * 12 : 42000 * 7 = 77 * 12 : 42 * 7 = 11 * 2 : 7 = 22 : 7 . answer : option c" | a ) 21 : 23 , b ) 23 : 45 , c ) 22 : 7 , d ) 25 : 29 , e ) none of these | c | divide(multiply(77000, const_12), multiply(42000, add(const_4, const_3))) | add(const_3,const_4)|multiply(n0,const_12)|multiply(n2,#0)|divide(#1,#2)| | other |
what is the % change in the area of a rectangle when its length increases by 10 % and its width decreases by 10 % ? | "explanatory answer whenever you encounter problems like this , use a numerical example and then proceed . for ease of computation , it is safe in most cases , to assume the length to be 100 units and the width to be 100 units . ( remember , a square is a rectangle too and the problem works the same way when you assume... | a ) 0 % , b ) 20 % increase , c ) 20 % decrease , d ) 1 % decrease , e ) insufficient data | d | subtract(const_100, divide(multiply(add(const_100, 10), subtract(const_100, 10)), const_100)) | add(n0,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|subtract(const_100,#3)| | geometry |
on a scale of map , 0.6 cm represent 6.6 km . if the distance between the points on the map is 80.5 cm , the actual distance between these points is : | solution let the actual distance be x km . then , more distance on the map , more is the actual distance ( direct proportion ) ∴ 0.6 : 80.5 : : 6.6 : x ⇔ 0.6 x = 80.5 × 6.6 ⇔ x = ⇔ x = 80.5 x 6.6 / 0.6 = x = 885.5 . answer d | a ) 9 km , b ) 72.5 km , c ) 190.75 km , d ) 885.5 km , e ) none of these | d | divide(multiply(80.5, 6.6), 0.6) | multiply(n1,n2)|divide(#0,n0) | physics |
there are 9 cities numbered 1 to 9 . from how many cities the flight can start so as to reach the city 8 either directly or indirectly such that the path formed is divisible by 3 . | start from 1 : 1368 , start from 2 : 2358 , start from 3 : 378 , start from 4 : 48 , start from 5 : 528 , start from 6 : 678 , start from 7 : 78 , why start from 8 ? ? ? , start from 9 : 918 . so flights can start from all 8 other cities . answer : b | a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | b | subtract(9, const_1) | subtract(n0,const_1) | general |
find the sum lend at c . i . at 5 p . c per annum will amount to rs . 740 in 2 years ? | "explanation : 740 = p ( 21 / 20 ) 2 p = 671 answer : e" | a ) 221 , b ) 287 , c ) 400 , d ) 288 , e ) 671 | e | divide(740, power(add(const_1, divide(5, const_100)), 2)) | divide(n0,const_100)|add(#0,const_1)|power(#1,n2)|divide(n1,#2)| | general |
what will be the compound interest on a sum of rs . 25,000 after 3 years at the rate of 12 % p . a . ? | "amount = [ 25000 * ( 1 + 12 / 100 ) 3 ] = 25000 * 28 / 25 * 28 / 25 * 28 / 25 = rs . 35123.20 c . i . = ( 35123.20 - 25000 ) = rs . 10123.20 answer : c" | a ) rs . 10123.77 , b ) rs . 10123.21 , c ) rs . 10123.20 , d ) rs . 10123.28 , e ) rs . 10123.21 | c | subtract(multiply(multiply(multiply(const_4, const_100), const_100), power(add(const_1, divide(12, const_100)), 3)), multiply(multiply(const_4, const_100), const_100)) | divide(n2,const_100)|multiply(const_100,const_4)|add(#0,const_1)|multiply(#1,const_100)|power(#2,n1)|multiply(#3,#4)|subtract(#5,#3)| | gain |
what is the value of 5 x ^ 2 − 1.9 x − 3.7 for x = − 0.3 ? | "5 x ^ 2 - 1.9 x - 3.7 , x = - 0.3 firstly , 5 x ^ 2 can be written as 5 * x * x - substituting one of the x with - 0.3 we have - 1.5 x - 1.9 x - 3.7 - - - > - 3.4 x - 3.7 - - - > - 3.4 ( - 0.3 ) - 3.7 - - - > 1.02 - 3.7 = - 2.68 answer : d" | a ) − 4.72 , b ) − 3.82 , c ) − 3.58 , d ) − 2.68 , e ) 0.57 | d | subtract(multiply(divide(divide(subtract(power(5, 2), power(1.9, 3.7)), const_1000), const_1000), 5), divide(divide(subtract(power(5, 2), power(1.9, 3.7)), const_1000), const_1000)) | power(n0,n1)|power(n2,n3)|subtract(#0,#1)|divide(#2,const_1000)|divide(#3,const_1000)|multiply(n0,#4)|subtract(#5,#4)| | general |
the income of a man increase consistently by 2 % per year . if his present income is rs . 12000 then what will his income be after 1 year ? | explanation : income = 12000 x 1.02 = 12240 answer : option c | a ) 12040 , b ) 12140 , c ) 12240 , d ) 12340 , e ) 12440 | c | add(multiply(divide(2, const_100), 12000), 12000) | divide(n0,const_100)|multiply(n1,#0)|add(n1,#1) | gain |
find the largest 2 digit number which is exactly divisible by 88 ? | "largest 2 digit number is 99 after doing 99 ÷ 88 we get remainder 11 hence largest 2 digit number exactly divisible by 88 = 99 - 11 = 88 a" | a ) 88 , b ) 89 , c ) 99 , d ) 54 , e ) 90 | a | multiply(add(add(add(add(multiply(const_100, const_100), multiply(const_100, const_10)), multiply(const_100, const_3)), multiply(2, const_10)), const_3), 88) | multiply(const_100,const_100)|multiply(const_10,const_100)|multiply(const_100,const_3)|multiply(n0,const_10)|add(#0,#1)|add(#4,#2)|add(#5,#3)|add(#6,const_3)|multiply(n1,#7)| | general |
if greg buys 3 shirts , 6 trousers and 4 ties , the total cost is $ 80 . if greg buys 7 shirts , 2 trousers and 4 ties , the total cost is $ 80 . how much will it cost him to buy 5 trousers , 4 shirts and 4 ties ? | "solution : 3 x + 6 y + 4 z = 80 7 x + 2 y + 4 z = 80 adding both the equations = 10 x + 8 y + 8 z = 160 5 x + 4 y + 4 z = 80 ans c" | a ) $ 60 , b ) $ 64 , c ) $ 80 , d ) $ 96 , e ) can not be determined | c | divide(add(80, 80), 4) | add(n3,n7)|divide(#0,n2)| | general |
when a number is divided by 6 & then multiply by 12 the answer is 18 what is the no . ? | "if $ x $ is the number , x / 6 * 12 = 18 = > 2 x = 18 = > x = 9 e" | a ) 4.5 , b ) 5 , c ) 5.6 , d ) 5.7 , e ) 9 | e | multiply(divide(18, 12), 6) | divide(n2,n1)|multiply(n0,#0)| | general |
donovan and michael are racing around a circular 400 - meter track . if donovan runs each lap in 48 seconds and michael runs each lap in 40 seconds , how many laps will michael have to complete in order to pass donovan , assuming they start at the same time ? | "one way of approaching this question is by relative speed method 1 . speed / rate of donovan = distance / time = > 400 / 48 = > 50 / 6 2 . speed / rate of michael = distance / time = > 400 / 40 = > 10 relative speed between them = 10 - 50 / 6 = > 10 / 6 ( we subtract the rates if moving in the same direction and add t... | a ) 5 , b ) 7 , c ) 8 , d ) 9 , e ) 6 | e | divide(divide(400, subtract(divide(400, 40), divide(400, 48))), 40) | divide(n0,n2)|divide(n0,n1)|subtract(#0,#1)|divide(n0,#2)|divide(#3,n2)| | physics |
a cistern 6 m long and 4 m wide contains water up to a breadth of 1 m 25 cm . find the total area of the wet surface . | "area of the wet surface = 2 [ lb + bh + hl ] - lb = 2 [ bh + hl ] + lb = 2 [ ( 4 * 1.25 + 6 * 1.25 ) ] + 6 * 4 = 49 m square answer : b" | a ) 66 m square , b ) 49 m square , c ) 77 m square , d ) 44 m square , e ) 33 m square | b | add(multiply(const_2, add(multiply(add(divide(25, const_100), 1), 4), multiply(add(divide(25, const_100), 1), 6))), multiply(4, 6)) | divide(n3,const_100)|multiply(n0,n1)|add(n2,#0)|multiply(n1,#2)|multiply(n0,#2)|add(#3,#4)|multiply(#5,const_2)|add(#6,#1)| | physics |
65 % of the employees of a company are men . 60 % of the men in the company speak french and 40 % of the employees of the company speak french . what is % of the women in the company who do not speak french ? | "no of employees = 100 ( say ) men = 65 women = 35 men speaking french = 0.60 * 65 = 39 employees speaking french = 0.4 * 100 = 40 therefore women speaking french = 40 - 39 = 1 and women not speaking french = 35 - 1 = 34 % of women not speaking french = 34 / 35 * 100 = 97.14 % answer c" | a ) 4 % , b ) 10 % , c ) 97.14 % , d ) 90.14 % , e ) 20 % | c | multiply(divide(subtract(divide(subtract(const_100, 65), const_100), subtract(divide(40, const_100), multiply(divide(65, const_100), divide(60, const_100)))), divide(subtract(const_100, 65), const_100)), const_100) | divide(n2,const_100)|divide(n0,const_100)|divide(n1,const_100)|subtract(const_100,n0)|divide(#3,const_100)|multiply(#1,#2)|subtract(#0,#5)|subtract(#4,#6)|divide(#7,#4)|multiply(#8,const_100)| | gain |
a can do a piece of work in 6 days . b can do it in 6 days . with the assistance of c they completed the work in 2 days . find in how many days can c alone do it ? | "c = 1 / 2 - 1 / 6 - 1 / 6 = 1 / 6 = > 6 days answer : c" | a ) 8 days , b ) 2 days , c ) 6 days , d ) 7 days , e ) 4 days | c | divide(multiply(6, 6), divide(subtract(multiply(6, 6), multiply(add(divide(multiply(6, 6), 6), divide(multiply(6, 6), 6)), 2)), 2)) | multiply(n0,n1)|divide(#0,n0)|divide(#0,n1)|add(#1,#2)|multiply(n2,#3)|subtract(#0,#4)|divide(#5,n2)|divide(#0,#6)| | physics |
what is the median of a set of consecutive integers if the sum of nth number from the beginning and nth number from the end is 200 ? | "surprisingly no one answered this easy one . property of a set of consecutive integerz . mean = median = ( first element + last element ) / 2 = ( second element + last but one element ) / 2 = ( third element + third last element ) / 2 etc . etc . so mean = median = 200 / 2 = 100 answer is e" | a ) 10 , b ) 25 , c ) 50 , d ) 75 , e ) 100 | e | divide(200, const_2) | divide(n0,const_2)| | general |
57 + 58 = 115 . how many such 2 consecutive numbers are there less than 1500 when added gives a sum which is divisible by 5 ? | "since 2 + 3 = 5 & 7 + 8 = 15 any combination with these no be will give u desirable result . . . so total no in 100 will be 20 & that ' s why in 1000 , it will be 20 x 15 = 300 . answer : c" | a ) 100 , b ) 200 , c ) 300 , d ) 400 , e ) 500 | c | divide(1500, 5) | divide(n4,n5)| | general |
a is a working partner and b is a sleeping partner in a business . a puts in 60,000 and b 40,000 . a gets 20 % of the profit for managing the business , and the rest is divided in proportion to their capitals . find the share of a in profit of 10000 . | "the amount a gets for managing = 20 % of rs . 10000 = 2000 remaining profit = 10000 – 2000 = 8000 this is to be divided in the ratio 3 : 2 . share of a = 3 / 5 of 8000 = 4800 ⇒ total share of a = 4800 + 2000 = 6800 . answer a" | a ) 6800 , b ) 5800 , c ) 4800 , d ) 6500 , e ) none of these | a | add(divide(multiply(subtract(10000, divide(multiply(20, 10000), const_100)), add(const_2, const_3)), add(add(const_2, const_3), add(const_2, const_4))), divide(multiply(20, 10000), const_100)) | add(const_2,const_3)|add(const_2,const_4)|multiply(n2,n3)|add(#0,#1)|divide(#2,const_100)|subtract(n3,#4)|multiply(#0,#5)|divide(#6,#3)|add(#7,#4)| | gain |
find the compound interest on $ 15,624 for 9 months at 16 % per annum compounded quarterly . | "p = $ 15625 , n = 9 months = 3 quarters , r = 16 % p . a = 4 % per quarter . amount = $ [ 15625 * ( 1 + ( 4 / 100 ) ^ 3 ) ] = $ ( 15625 * ( 26 / 25 ) * ( 26 / 25 ) * ( 26 / 25 ) ) = $ 17576 . c . i = $ ( 17576 - 15625 ) = $ 1951 . answer ( b )" | a ) $ 4152 , b ) $ 1951 , c ) $ 2258 , d ) $ 8978 , e ) $ 8875 | b | subtract(multiply(15,624, power(add(const_1, divide(16, const_100)), 9)), 15,624) | divide(n2,const_100)|add(#0,const_1)|power(#1,n1)|multiply(n0,#2)|subtract(#3,n0)| | gain |
find the probability that a year chosen at random has 53 mondays . | "explanation : there are 2 kinds of year leap year - probability = 1 / 4 ( out of 4 years one is a leap year ) non leap year - probability = 3 / 4 so it ' s either this or that now in non leap year , there are 365 days meaning 52 weeks with one day extra that one day can be monday out of 7 possible days . hence the pro... | a ) 5 / 28 , b ) 3 / 28 , c ) 1 / 28 , d ) 3 / 28 , e ) 3 / 27 | a | divide(const_2, add(const_3, const_4)) | add(const_3,const_4)|divide(const_2,#0)| | probability |
a pair of articles was bought for $ 360 at a discount of 10 % . what must be the marked price of each of the article ? | "s . p . of each of the article = 360 / 2 = $ 180 let m . p = $ x 90 % of x = 360 x = 180 * 100 / 90 = $ 200 answer is a" | a ) $ 200 , b ) $ 500 , c ) $ 350 , d ) $ 400 , e ) $ 600 | a | divide(multiply(subtract(const_100, 10), divide(360, const_2)), const_100) | divide(n0,const_2)|subtract(const_100,n1)|multiply(#0,#1)|divide(#2,const_100)| | gain |
the compound interest earned by sunil on a certain amount at the end of two years at the rate of 8 % p . a . was rs . 2828.80 . find the total amount that sunil got back at the end of two years in the form of principal plus interest earned | "let the sum be rs . p p { [ 1 + 8 / 100 ] 2 - 1 } = 2828.80 p ( 8 / 100 ) ( 2 + 8 / 100 ) = 2828.80 [ a 2 - b 2 = ( a - b ) ( a + b ) ] p = 2828.80 / ( 0.08 ) ( 2.08 ) = 1360 / 0.08 = 17000 principal + interest = rs . 19828.80 answer : b" | a ) 19828.88 , b ) 19828.8 , c ) 19828.87 , d ) 19828.81 , e ) 19828.82 | b | add(divide(2828.80, subtract(power(add(const_1, divide(8, const_100)), const_2), const_1)), 2828.80) | divide(n0,const_100)|add(#0,const_1)|power(#1,const_2)|subtract(#2,const_1)|divide(n1,#3)|add(n1,#4)| | gain |
if x ^ 2 + y ^ 2 = 18 and xy = 3 , then ( x − y ) ^ 2 = | "but you can not take xy + 3 to mean xy = - 3 . . only if xy + 3 = 0 , it will mean xy = - 3 . . rest your solution is perfect and you will get your correct answer as 18 - 2 * 3 = 12 . . answer d" | a ) 8 , b ) 11 , c ) 14 , d ) 12 , e ) 20 | d | power(3, 2) | power(n3,n0)| | general |
the diameter of a garden roller is 1.4 m and it is 2 m long . how much area will it cover in 5 revolutions ? ( use π = 22 ⁄ 7 ) | "required area covered in 5 revolutions = 5 × 2 π rh = 5 × 2 × 22 ⁄ 7 × 0.7 × 2 = 44 m 2 answer b" | a ) 40 m 2 , b ) 44 m 2 , c ) 48 m 2 , d ) 36 m 2 , e ) none of these | b | multiply(5, multiply(multiply(multiply(2, divide(22, 7)), divide(1.4, 2)), 2)) | divide(n0,n1)|divide(n3,n4)|multiply(#1,n1)|multiply(#0,#2)|multiply(n1,#3)|multiply(n2,#4)| | general |
what is the dividend . divisor 16 , the quotient is 8 and the remainder is 4 | "e = d * q + r e = 16 * 8 + 4 e = 128 + 4 e = 132" | a ) 127 , b ) 128 , c ) 130 , d ) 131 , e ) 132 | e | add(multiply(16, 8), 4) | multiply(n0,n1)|add(n2,#0)| | general |
harold and millicent are getting married and need to combine their already - full libraries . if harold , who has 1 / 2 as many books as millicent , brings 1 / 3 of his books to their new home , then millicent will have enough room to bring 1 / 4 of her books to their new home . what fraction of millicent ' s old libra... | "because we see h willbring 1 / 3 of his booksto the new home - - > try to pick a number that isdivisible by 3 . before : assume h = 30 books h = 1 / 2 m - - > m = 60 books after : h ' = 1 / 3 h = 10 books m ' = 1 / 4 m = 15 books total = 25 books m ' = 25 = 5 / 12 * 60 ratio : 5 / 12 ans : e" | a ) 1 / 2 , b ) 2 / 3 , c ) 3 / 4 , d ) 4 / 5 , e ) 5 / 12 | e | subtract(1, multiply(divide(1, 3), divide(1, 2))) | divide(n0,n3)|divide(n0,n1)|multiply(#0,#1)|subtract(n0,#2)| | general |
if a , b and c together can finish a piece of work in 4 days . a alone in 36 days and b in 18 days , then c alone can do it in ? | "c = 1 / 4 - 1 / 36 – 1 / 18 = 1 / 6 = > 6 days ' answer : e" | a ) 1 , b ) 7 , c ) 8 , d ) 9 , e ) 6 | e | divide(const_1, subtract(subtract(const_0_25, divide(const_1, 36)), divide(const_1, 18))) | divide(const_1,n1)|divide(const_1,n2)|subtract(const_0_25,#0)|subtract(#2,#1)|divide(const_1,#3)| | physics |
in a coconut grove , ( x + 1 ) trees yield 60 nuts per year , x trees yield 120 nuts per year and ( x – 1 ) trees yield 180 nuts per year . if the average yield per year per tree be 100 , find x . | "( x + 1 ) × 60 + x × 120 + ( x − 1 ) × 180 / ( x + 1 ) + x + ( x − 1 ) = 100 ⇒ 360 x − 120 / 3 x = 100 ⇒ 60 x = 120 ⇒ x = 2 answer a" | a ) 2 , b ) 4 , c ) 5 , d ) 6 , e ) none of the above | a | divide(subtract(multiply(180, 1), multiply(60, 1)), subtract(add(add(60, 120), 180), multiply(100, const_3))) | add(n1,n2)|multiply(n0,n4)|multiply(n0,n1)|multiply(n5,const_3)|add(n4,#0)|subtract(#1,#2)|subtract(#4,#3)|divide(#5,#6)| | general |
three numbers are in the ratio 3 : 4 : 5 and their l . c . m . is 2400 . their h . c . f is ? | "let the numbers be 3 x , 4 x and 5 x their l . c . m . = 60 x 60 x = 2400 x = 40 the numbers are 3 * 40 , 4 * 40 , 5 * 40 hence required h . c . f . = 40 answer is c" | a ) 20 , b ) 30 , c ) 40 , d ) 50 , e ) 60 | c | add(multiply(multiply(3, 5), const_100), multiply(4, 5)) | multiply(n0,n2)|multiply(n1,n2)|multiply(#0,const_100)|add(#2,#1)| | other |
the area of a circular field is 17.56 hectares . find the cost of fencing it at the rate of rs . 3 per metre approximately | explanation : area = ( 17.56 x 10000 ) m 2 = 175600 m 2 . π r 2 = 175600 ⇔ ( r ) 2 = ( 175600 x ( 7 / 22 ) ) ⇔ r = 236.37 m . circumference = 2 π r = ( 2 x ( 22 / 7 ) x 236.37 ) m = 1485.78 m . cost of fencing = rs . ( 1485.78 x 3 ) = rs . 4457 . answer : option a | ['a ) 4457', 'b ) 4567', 'c ) 4235', 'd ) 4547', 'e ) 4675'] | a | multiply(circumface(multiply(sqrt(divide(17.56, const_pi)), const_100)), const_3) | divide(n0,const_pi)|sqrt(#0)|multiply(#1,const_100)|circumface(#2)|multiply(#3,const_3) | geometry |
a certain no . whendivided by 35 leaves a remainder 25 , what is the remainder if the same no . be dividedby 15 ? | explanation : 35 + 25 = 60 / 15 = 4 ( remainder ) b | a ) 2 , b ) 4 , c ) 7 , d ) 8 , e ) 9 | b | divide(add(35, 25), 15) | add(n0,n1)|divide(#0,n2) | general |
. on dividing a number by 357 , we get 36 as remainder . on dividing the same number 17 , what will be the remainder ? | let x be the number and y be the quotient . then , x = 357 x y + 36 = ( 17 x 21 x y ) + ( 17 x 2 ) + 2 = 17 x ( 21 y + 2 ) + 2 ) required remainder = 2 . answer : option e | a ) 0 , b ) 3 , c ) 5 , d ) 8 , e ) 2 | e | factorial(divide(36, 17)) | divide(n1,n2)|factorial(#0) | general |
there is a point p on the circle . a and b started running in two constant different speeds . a in clockwise and b in anti - clockwise . first time 500 m in clockwise from p then 400 anti - clockwise . if b is yet to complete one round , what is the circumference of the circle ? | let the speeds of a & b be u and v resp . and let circumference be c . let their first meeting be after time t then ut + vt = c and ut = 500 ∴ vt = c − 500 and u / v = 500 / ( c − 500 ) for their second meeting after time t , ut + vt = 2 c and vt = 400 ∴ ut = 2 c − 400 and u / v = ( 2 c − 400 ) / 400 hence 500 / ( c − ... | a ) 700 , b ) 800 , c ) 900 , d ) 1000 , e ) 1100 | a | divide(add(multiply(multiply(const_1, const_2), 500), 400), const_2) | multiply(const_1,const_2)|multiply(n0,#0)|add(n1,#1)|divide(#2,const_2) | physics |
carina has 55 ounces of coffee divided into 5 - and 10 - ounce packages . if she has 2 more 5 - ounce packages than 10 - ounce packages , how many 10 - ounce packages does she have ? | lets say 5 and 10 ounce packages be x and y respectively . given that , 5 x + 10 y = 55 and x = y + 2 . what is the value of y . substituting the x in first equation , 5 y + 10 + 10 y = 55 - > y = 45 / 15 . = 3 e | a ) 7 , b ) 6 , c ) 5 , d ) 4 , e ) 3 | e | divide(subtract(55, multiply(5, 2)), add(10, 5)) | add(n1,n2)|multiply(n1,n3)|subtract(n0,#1)|divide(#2,#0) | general |
how many two - digit numbers yield a remainder of 2 when divided by both 5 and 16 ? | "easier to start with numbers that are of the form 16 p + 2 - - - > 18,34 , 50,66 , 82,96 . out of these , there is only one number ( 82 ) is also of the form 5 q + 2 . thus 1 is the answer . b is the correct answer ." | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | b | subtract(add(2, multiply(16, 5)), multiply(16, 5)) | multiply(n1,n2)|add(n0,#0)|subtract(#1,#0)| | general |
a computer manufacturer produces a certain electronic component at a cost of $ 50 per component . shipping costs for delivering the components are $ 8 per unit . further , the manufacturer has costs of $ 16,000 a month related to the electronic component regardless of how many it produces . if the manufacturer produces... | "$ 16000 is a fixed cost each component is $ 58 ( $ 50 to produce , $ 8 to ship ) manufacturer will be producing and selling 200 components so therefore the equation to find price would be 200 * p = 16000 + ( 200 * 50 ) + ( 200 * 8 ) p = ( 16000 + 10000 + 1600 ) / 200 p = 138 answer : d" | a ) 135 , b ) 129 , c ) 131 , d ) 138 , e ) 141 | d | divide(add(add(multiply(multiply(const_4, const_4), const_1000), multiply(8, const_100)), multiply(add(50, 8), 200)), 200) | add(n0,n1)|multiply(const_4,const_4)|multiply(n1,const_100)|multiply(#1,const_1000)|multiply(n3,#0)|add(#3,#2)|add(#5,#4)|divide(#6,n3)| | general |
the area of an isosceles trapezoid with sides of length 5 and bases of length 8 and 14 is ? | "we can divide the trapezoid into two triangles and a rectangle sides of triangle , t 1 - - > h 1 = 5 , b 1 = 3 , s 1 = root ( 25 - 9 ) = 4 = s 2 of triangle t 2 area of triangle = ( 1 / 2 ) * 4 * 3 = 6 side of the rectangle = 8 * 4 = 32 total area of the trapezoid = 32 + 6 + 6 = 44 answer : c" | a ) 22 , b ) 32 , c ) 44 , d ) 55 , e ) 56 | c | quadrilateral_area(const_4, 14, 8) | quadrilateral_area(const_4,n2,n1)| | geometry |
a rope of which a calf is tied is increased from 15 m to 18 m , how much additional grassy ground shall it graze ? | "π ( 182 – 152 ) = 311.14 answer : d" | a ) 1218 , b ) 1212 , c ) 1210 , d ) 3 . 11.14 , e ) 291.1 | d | multiply(subtract(power(18, const_2), power(15, const_2)), divide(add(multiply(15, const_2), const_2), add(const_4, const_3))) | add(const_3,const_4)|multiply(n0,const_2)|power(n1,const_2)|power(n0,const_2)|add(#1,const_2)|subtract(#2,#3)|divide(#4,#0)|multiply(#6,#5)| | general |
the radius of a circle is increased by 1 % . find how much % does its area increases ? | r = 100 r = 101 r 2 = 10000 r 2 = 10201 10000 - - - - 201 100 - - - - ? = > 2.01 % answer : d | ['a ) 2.21 %', 'b ) 2.07 %', 'c ) 2.08 %', 'd ) 2.01 %', 'e ) 2.11 %'] | d | divide(add(multiply(const_2, const_100), 1), const_100) | multiply(const_100,const_2)|add(n0,#0)|divide(#1,const_100) | geometry |
in gabriel ' s sheep herd , every sheep has either fleas or lice or both . half of the sheep have lice , and 84 sheep have both pests . if 94 sheep have lice , how many sheep have only fleas ? | n ( lice ) = n ( only lice ) + n ( both lice and fleas ) n ( both lice and fleas ) = 84 n ( lice ) = 94 n ( only lice ) = 94 - 84 = 10 since , it is also said that exactly half the sheep have lice , it can be inferred that half the number of sheep is 94 . therefore , the total number of sheep having only fleas is 10 an... | a ) 10 , b ) 20 , c ) 84 , d ) 94 , e ) 178 | a | subtract(94, 84) | subtract(n1,n0) | other |
tanks a and b are each in the shape of a right circular cylinder . the interior of tank a has a height of 7 meters and a circumference of 8 meters , and the interior of tank b has a height of 8 meters and a circumference of 10 meters . the capacity of tank a is what percent of the capacity of tank b ? | "for a , r = 8 / 2 pi . its capacity = ( 4 pi ) ^ 2 * 7 = 112 pi for b , r = 10 / pi . its capacity = ( 5 pi ) ^ 2 * 8 = 200 pi a / b = 112 pi / 200 pi = 0.56 a" | a ) 56 % , b ) 80 % , c ) 100 % , d ) 120 % , e ) 125 % | a | multiply(multiply(power(divide(8, 10), const_2), divide(7, 8)), const_100) | divide(n0,n2)|divide(n1,n3)|power(#1,const_2)|multiply(#0,#2)|multiply(#3,const_100)| | physics |
the speed of a subway train is represented by the equation z = s ^ 2 + 2 s for all situations where 0 ≤ s ≤ 7 , where z is the rate of speed in kilometers per hour and s is the time in seconds from the moment the train starts moving . in kilometers per hour , how much faster is the subway train moving after 6 seconds t... | given : z = s ^ 2 + 2 s for 0 ≤ s ≤ 7 z ( 3 ) = 3 ^ 2 + 2 * 3 = 15 z ( 6 ) = 6 ^ 2 + 2 * 6 = 48 therefore z ( 7 ) - z ( 3 ) = 48 - 15 = 33 km / hr option e | a ) 4 , b ) 9 , c ) 15 , d ) 48 , e ) 33 | e | subtract(add(power(6, 2), multiply(6, 2)), add(power(3, const_2), multiply(3, 2))) | multiply(n0,n4)|multiply(n0,n5)|power(n4,n0)|power(n5,const_2)|add(#0,#2)|add(#1,#3)|subtract(#4,#5) | physics |
notebooks are sold in packages of 4 or 7 only . if wilson bought 69 notebooks exactly , what could be the number of large packs wilson bought ? | let number of packs of four = f let number of packs of seven = s 4 f + 7 s = 69 now , we need to test for values of s . since sum 69 is odd and 4 f will always be even , e ca n ' t be even . now , we can test for values e = 3 , 7 and 9 4 * 5 + 7 * 7 = 20 + 49 = 69 answer c | a ) 3 , b ) 4 , c ) 7 , d ) 8 , e ) 9 | c | divide(subtract(subtract(subtract(subtract(subtract(69, 4), 4), 4), 4), 4), 7) | subtract(n2,n0)|subtract(#0,n0)|subtract(#1,n0)|subtract(#2,n0)|subtract(#3,n0)|divide(#4,n1) | general |
a person lent a certain sum of money at 4 % per annum at simple interest and in 8 years the interest amounted to rs . 340 less than the sum lent . what was the sum lent ? | "p - 340 = ( p * 4 * 8 ) / 100 p = 500 answer : d" | a ) 228 , b ) 278 , c ) 289 , d ) 500 , e ) 821 | d | divide(340, subtract(const_1, divide(multiply(4, 8), const_100))) | multiply(n0,n1)|divide(#0,const_100)|subtract(const_1,#1)|divide(n2,#2)| | gain |
a total of 40 percent of the geese included in a certain migration study were male . if some of the geese migrated during the study and 20 percent of the migrating geese were male , what was the ratio of the migration rate for the male geese to the migration rate for the female geese ? [ migration rate for geese of a c... | "let ' take the number of geese to be 100 . male = 40 . female = 60 . now the second part of the q , let ' s take the number migrated to be 20 . so we have 20 geese that migrated and out of that 20 % are male i . e 20 / 100 * 20 = 4 geese ( males ) and now we know out of the total 20 geese , 4 are male , then 16 have t... | a ) 15 / 40 , b ) 7 / 12 , c ) 2 / 3 , d ) 7 / 8 , e ) 8 / 7 | a | divide(divide(divide(20, const_100), divide(40, const_100)), divide(divide(multiply(multiply(const_2, const_4), const_10), const_100), divide(40, const_100))) | divide(n1,const_100)|divide(n0,const_100)|multiply(const_2,const_4)|divide(#0,#1)|multiply(#2,const_10)|divide(#4,const_100)|divide(#5,#1)|divide(#3,#6)| | general |
a contractor isengaged for 30 days on the condition that he receives rs . 25 for eachday he works & fined rs . 7.50 for each day is absent . he gets rs . 425 in all . for how many days was heabsent ? | 30 * 25 = 750 425 - - - - - - - - - - - 325 25 + 7.50 = 32.5 325 / 32.5 = b | a ) 8 , b ) 10 , c ) 15 , d ) 17 , e ) 21 | b | divide(subtract(multiply(25, 30), 425), add(25, 7.5)) | add(n1,n2)|multiply(n0,n1)|subtract(#1,n3)|divide(#2,#0) | physics |
a garrison of 2000 men has provisions for 54 days . at the end of 15 days , a reinforcement arrives , and it is now found that the provisions will last only for 30 days more . what is the reinforcement ? | "2000 - - - - 54 2000 - - - - 39 x - - - - - 20 x * 30 = 2000 * 39 x = 2600 2000 - - - - - - - 600 answer : c" | a ) 1800 men , b ) 3900 men , c ) 600 men , d ) 3800 men , e ) 900 men | c | subtract(divide(subtract(multiply(2000, 54), multiply(2000, 15)), 30), 2000) | multiply(n0,n1)|multiply(n0,n2)|subtract(#0,#1)|divide(#2,n3)|subtract(#3,n0)| | physics |
the ages of two person differ by 20 years . if 4 years ago , the elder one be 5 times as old as the younger one , their present ages ( in years ) are respectively | "let their ages be x and ( x + 20 ) years . then , 5 ( x - 4 ) = ( x + 20 - 4 ) = > 4 x = 36 = > x = 9 their present ages are 29 years and 9 year . answer : c" | a ) 30 , 10 , b ) 25 , 5 , c ) 29 , 9 , d ) 50 , 30 , e ) 20,10 | c | add(divide(add(multiply(5, 4), subtract(20, 4)), subtract(5, const_1)), 20) | multiply(n1,n2)|subtract(n0,n1)|subtract(n2,const_1)|add(#0,#1)|divide(#3,#2)|add(n0,#4)| | general |
28 machines can do a work in 10 days . how many machines are needed to complete the work in 40 days ? | "required number of machines = 28 * 10 / 40 = 7 answer is d" | a ) 10 , b ) 6 , c ) 4 , d ) 7 , e ) 5 | d | divide(multiply(28, 10), 40) | multiply(n0,n1)|divide(#0,n2)| | physics |
sandy buys an old scooter for $ 800 and spends $ 200 on its repairs . if sandy sells the scooter for $ 1400 , what is the gain percent ? | "selling price / total cost = 1400 / 1000 = 1.4 the gain percent is 40 % . the answer is c ." | a ) 20 % , b ) 30 % , c ) 40 % , d ) 50 % , e ) 60 % | c | multiply(divide(subtract(1400, add(200, 800)), add(200, 800)), const_100) | add(n0,n1)|subtract(n2,#0)|divide(#1,#0)|multiply(#2,const_100)| | gain |
a jogger running at 9 kmph alongside a railway track in 240 meters ahead of the engine of a 120 metres long train running at 45 kmph in the same direction . in how much time will the train pass the jogger ? | total distance ( d ) = 240 + 120 = 360 meter relative speed ( s ) = 45 - 9 = 36 kmph = 36 * 5 / 18 = 10 mps d = s * t 360 = 10 * t t = 36 seconds answer : c | a ) 18 seconds , b ) 26 seconds , c ) 36 seconds , d ) 16 seconds , e ) 46 seconds | c | divide(add(240, 120), multiply(subtract(45, 9), const_0_2778)) | add(n1,n2)|subtract(n3,n0)|multiply(#1,const_0_2778)|divide(#0,#2) | physics |
a woman complete a journey in 15 hours . she travels first half of the journey at the rate of 21 km / hr and second half at the rate of 24 km / hr . find the total journey in km . | "0.5 x / 21 + 0.5 x / 24 = 15 - - > x / 21 + x / 24 = 30 - - > x = 336 km . a" | a ) 336 km . , b ) 216 km . , c ) 314 km . , d ) 224 km . , e ) 544 km . | a | multiply(multiply(divide(multiply(15, 24), add(24, 21)), 21), const_2) | add(n1,n2)|multiply(n0,n2)|divide(#1,#0)|multiply(n1,#2)|multiply(#3,const_2)| | physics |
how many 3 - digit integers exist such that all their digits are odd ? | so last digit can be filled in 5 ways - 1 , 3,5 , 7,9 there are 5 possibilities . the answer is 5 ∗ 5 ∗ 5 = 125 the correct answer is d | a ) 80 , b ) 100 , c ) 120 , d ) 125 , e ) 135 | d | power(add(3, const_2), 3) | add(n0,const_2)|power(#0,n0) | general |
matt gets a $ 1,300 commission on a big sale . this commission alone raises his average commission by $ 150 . if matt ' s new average commission is $ 400 , how many sales has matt made ? | "let , average commission = x no . of items sold = y total commission = xy new commission = xy + 1300 new average = ( xy + 1300 ) / ( y + 1 ) = 150 + x i . e . ( xy + 1300 ) = ( y + 1 ) * ( 150 + x ) i . e . ( xy + 1300 ) = ( xy + x + 150 y + 150 ) i . e . ( 1150 ) = ( x + 150 y ) new commission = 400 = 150 + x i . e .... | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | d | divide(subtract(const_1000, const_100), subtract(400, subtract(400, 150))) | subtract(const_1000,const_100)|subtract(n2,n1)|subtract(n2,#1)|divide(#0,#2)| | general |
a train speeds past a pole in 15 seconds and a platform 100 meter long in 25 seconds . what is length of the train ? | "explanation : let the length of the train is x meter and speed of the train is y meter / second then x / y = 15 [ because distance / speed = time ] = > y = 15 / x x + 100 / 25 = x / 15 x = 150 meters so length of the train is 150 meters answer is c" | a ) 140 meter , b ) 145 meter , c ) 150 meter , d ) 155 meter , e ) 160 meter | c | multiply(100, subtract(const_2, const_1)) | subtract(const_2,const_1)|multiply(n1,#0)| | physics |
25 % of 30 % of 45 % is equal to ? | "( 25 / 100 ) * ( 30 / 100 ) * ( 45 / 100 ) 1 / 4 * 3 / 10 * 9 / 20 27 / 800 = 0.03375 answer : a" | a ) 0.03375 , b ) 0.03372 , c ) 0.0337 , d ) 0.03377 , e ) 0.03376 | a | divide(multiply(25, 30), 45) | multiply(n0,n1)|divide(#0,n2)| | gain |
a student reached his school late by 20 mins by travelling at a speed of 9 kmph . if he had travelled at a speed of 12 kmph , he would have reached his school 20 mins early . what is the distance between house and school ? | distance = distance either student reaches school before time or after time 9 ( t + 20 / 60 ) = 12 ( t - 20 / 60 ) i . e 9 ( t + 1 / 3 ) = 12 ( t - 1 / 3 ) 9 ( 3 t + 1 ) = 12 ( 3 t - 1 ) t = 7 / 3 put t value in any of distance equation put t = 7 / 3 in 9 ( t + 1 / 3 ) we get 24 km answer : a | a ) 24 , b ) 25 , c ) 26 , d ) 27 , e ) 28 | a | multiply(add(divide(20, const_60), divide(20, const_60)), divide(multiply(12, 9), subtract(12, 9))) | divide(n0,const_60)|multiply(n1,n2)|subtract(n2,n1)|add(#0,#0)|divide(#1,#2)|multiply(#3,#4) | physics |
the volume of a rectangular swimming pool is 840 cubic meters and water is flowing into the swimming pool . if the surface level of the water is rising at the rate of 0.5 meters per minute , what is the rate w , in cubic meters per minutes , at which the water is flowing into the swimming pool ? | the correct answer is e . there are not enough info to answer the question . a 840 cubic meters rectangle is built from : height * length * width . from the question we know the volume of the pool and the filling rate . a pool can have a height of 10 * width 8.4 * length 10 and have a volume of 840 cubic meters , and i... | ['a ) 0.125', 'b ) 0.25', 'c ) 0.5', 'd ) 0.75', 'e ) not enough information to determine the rate'] | e | divide(840, 0.5) | divide(n0,n1) | geometry |
the sum of four consecutive even numbers is 292 . what would be the smallest number ? | "e 70 let the four consecutive even numbers be 2 ( x - 2 ) , 2 ( x - 1 ) , 2 x , 2 ( x + 1 ) their sum = 8 x - 4 = 292 = > x = 37 smallest number is : 2 ( x - 2 ) = 70 ." | a ) 49 , b ) 68 , c ) 60 , d ) 57 , e ) 70 | e | add(add(power(add(add(divide(subtract(subtract(292, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(292, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(292, const_10), const_2), const_4), const_2), power(... | subtract(n0,const_10)|subtract(#0,const_2)|divide(#1,const_4)|add(#2,const_2)|power(#2,const_2)|add(#3,const_2)|power(#3,const_2)|add(#5,const_2)|add(#4,#6)|power(#5,const_2)|power(#7,const_2)|add(#9,#10)|add(#11,#8)| | physics |
a student is ranked 13 th from right and 8 th from left . how many are there ? | "13 th from right means = x + 12 8 th from left means = 7 + x total = 12 + 7 + 1 = 20 answer : c" | a ) 18 , b ) 19 , c ) 20 , d ) 21 , e ) 22 | c | subtract(add(13, 8), const_1) | add(n0,n1)|subtract(#0,const_1)| | general |
how many 4 - digit odd numbers do not use any digit more than once ? | let a 4 - digit number be represented by abcd here a can have any value between 1 to 9 - so total 9 b can have any value between 0 to 9 , but not a - so total 9 c can have any value between 0 to 9 , but not a or b - so total 8 d can have any value between 0 to 9 , but not a , b or c - so total 7 no . of all possible 4 ... | a ) 1728 , b ) 2160 , c ) 2240 , d ) 2268 , e ) 2520 | d | multiply(multiply(multiply(multiply(const_2, const_4), multiply(const_2, const_4)), add(const_4, const_3)), add(const_1, const_4)) | add(const_1,const_4)|add(const_3,const_4)|multiply(const_2,const_4)|multiply(#2,#2)|multiply(#1,#3)|multiply(#0,#4) | general |
revenues were recorded for store a and store b over a period of 3 months . in the first month , store a ' s revenues were $ 12000 higher than store b ' s revenues . in the second month , store a ' s revenues were $ 8000 higher than store b ' s revenues . if store a ' s average ( arithmetic mean ) monthly revenue for th... | answer : cit may be tempting to come up with a lot of variables , one each for each month ' s revenue for each company . however , focus on the differences . in the first month , the difference was + 12 in favor of a . ( note that we can drop the thousands since every number in the question is in terms of thousands . )... | a ) $ 14,000 , b ) $ 15,000 , c ) $ 42,000 , d ) $ 46,000 , e ) $ 50,000 | a | divide(subtract(add(12000, 8000), multiply(2000, 3)), const_1000) | add(n1,n2)|multiply(n0,n4)|subtract(#0,#1)|divide(#2,const_1000) | general |
a is older than b by 5 years . 7 years hence , thrice a ' s age shall be equal to 4 times that of b . find their present ages . | explanation : let the present ages of a and b be ' a ' and ' b ' respectively . 3 ( a + 7 ) = 4 ( ( b + 7 ) - - - ( 1 ) = > 3 a - 4 b = 7 - - - - - - - ( 1 ) a - b = 5 - - - - - - - ( 2 ) solving ( 1 ) and ( 2 ) we get b = 8 substituting b = 8 in second equation , we get a = 13 . answer : c = > 2 a = 32 = > a = 16 . an... | a ) 14 , 7 , b ) 10 , 5 , c ) 13 , 8 , d ) 12 , 8 , e ) 8 , 12 | c | add(subtract(add(multiply(const_3, 5), multiply(const_3, 7)), multiply(4, 7)), 5) | multiply(n0,const_3)|multiply(n1,const_3)|multiply(n1,n2)|add(#0,#1)|subtract(#3,#2)|add(n0,#4) | general |
three numbers are in the ratio 3 : 4 : 5 and their l . c . m is 2400 . their h . c . f is ? | "let the numbers be 3 x , 4 x and 5 x . then , their l . c . m = 60 x . so , 60 x = 2400 or x = 40 . the numbers are ( 3 * 40 ) , ( 4 * 40 ) and ( 5 * 40 ) . hence , required h . c . f = 40 . answer : a" | a ) 40 , b ) 99 , c ) 88 , d ) 44 , e ) 11 | a | add(multiply(multiply(3, 5), const_100), multiply(4, 5)) | multiply(n0,n2)|multiply(n1,n2)|multiply(#0,const_100)|add(#2,#1)| | other |
what is x if x + 2 y = 20 and y = 5 ? | "x = 20 - 2 y x = 20 - 10 . x = 10 answer : a" | a ) a ) 10 , b ) b ) 8 , c ) c ) 6 , d ) d ) 4 , e ) e ) 2 | a | subtract(20, multiply(2, 5)) | multiply(n0,n2)|subtract(n1,#0)| | general |
a certain experimental mathematics program was tried out in 2 classes in each of 34 elementary schools and involved 40 teachers . each of the classes had 1 teacher and each of the teachers taught at least 1 , but not more than 3 , of the classes . if the number of teachers who taught 3 classes is n , then the least and... | one may notice that greatest possible values differ in each answer choice in contrast to the least values , which repeat . to find out the greatest value you should count the total classes ( 34 * 2 = 68 ) , then subtract the total # of teachers since we know from the question that each teacher taught at least one class... | a ) 0 and 13 , b ) 0 and 14 , c ) 1 and 10 , d ) 1 and 9 , e ) 2 and 8 | b | divide(subtract(multiply(34, 2), 40), 2) | multiply(n0,n1)|subtract(#0,n2)|divide(#1,n0) | general |
nicky and cristina are running a race . since cristina is faster than nicky , she gives him a 48 meter head start . if cristina runs at a pace of 5 meters per second and nicky runs at a pace of only 3 meters per second , how many seconds will nicky have run before cristina catches up to him ? | "used pluging in method say t is the time for cristina to catch up with nicky , the equation will be as under : for nicky = n = 3 * t + 48 for cristina = c = 5 * t @ t = 24 , n = 120 c = 120 right answer ans : c" | a ) 15 seconds , b ) 18 seconds , c ) 24 seconds , d ) 30 seconds , e ) 45 seconds | c | divide(48, subtract(5, 3)) | subtract(n1,n2)|divide(n0,#0)| | physics |
a walks at 10 kmph and 6 hours after his start , b cycles after him at 20 kmph . how far from the start does b catch up with a ? | suppose after x km from the start b catches up with a . then , the difference in the time taken by a to cover x km and that taken by b to cover x km is 6 hours . x / 10 - x / 20 = 6 x = 120 km answer is d | a ) 100 km , b ) 150 km , c ) 50 km , d ) 120 km , e ) 200 km | d | multiply(6, 20) | multiply(n1,n2) | physics |
how many words , with or without meaning , can be formed using all letters of the word sharp using each letter exactly once ? | "the word sharp has exactly 5 letters which are all different . therefore the number of words that can be formed = number of permutations of 5 letters taken all at a time . = p ( 5 , 5 ) = 5 ! = 5 x 4 x 3 x 2 × 1 = 120 answer : d" | a ) 90 , b ) 100 , c ) 110 , d ) 120 , e ) 130 | d | factorial(const_3) | factorial(const_3)| | general |
mary works in a restaurant a maximum of 80 hours . for the first 20 hours , she is paid $ 8 per hour . for each overtime hour , she is paid at a rate which is 25 % higher than her regular rate . how much mary can earn in a week ? | "mary receives $ 8 ( 20 ) = $ 160 for the first 20 hours . for the 60 overtime hours , she receives $ 8 ( 0.25 ) + $ 8 = $ 10 per hour , that is $ 10 ( 60 ) = $ 600 . the total amount is $ 160 + $ 600 = $ 760 answer c 360 ." | a ) 600 , b ) 720 , c ) 760 , d ) 620 , e ) 700 | c | add(multiply(80, 8), multiply(subtract(80, 20), multiply(8, divide(25, const_100)))) | divide(n3,const_100)|multiply(n0,n2)|subtract(n0,n1)|multiply(n2,#0)|multiply(#3,#2)|add(#1,#4)| | gain |
if 2 x + y = 4 and x + 2 y = 5 , then ( x + y ) / 3 = | "we have two equations : 2 x + y = 4 x + 2 y = 5 notice that something nice happens when we add them . we get : 3 x + 3 y = 9 divide both sides by 3 to get : x + y = 3 so , ( x + y ) / 3 = 3 / 3 = 1 answer : a" | a ) 1 , b ) 4 / 3 , c ) 17 / 5 , d ) 18 / 5 , e ) 4 | a | divide(add(divide(subtract(multiply(4, 2), 5), subtract(multiply(2, 2), const_1)), subtract(4, multiply(2, divide(subtract(multiply(4, 2), 5), subtract(multiply(2, 2), const_1))))), 3) | multiply(n0,n1)|multiply(n0,n0)|subtract(#0,n3)|subtract(#1,const_1)|divide(#2,#3)|multiply(n0,#4)|subtract(n1,#5)|add(#4,#6)|divide(#7,n4)| | general |
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