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a lawn is in the form of a rectangle having its sides in the ratio 2 : 3 . the area of the lawn is ( 1 / 6 ) hectares . find the length and breadth of the lawn . | "area = ( l + b + 2 d ) 2 d = ( 70 + 55 + 2.5 * 2 ) 2 * 2.5 = > 650 650 * 2 = rs . 1300 answer : b" | a ) s . 1350 , b ) s . 1300 , c ) s . 1328 , d ) s . 1397 , e ) s . 1927 | b | multiply(subtract(rectangle_area(add(70, multiply(2.5, 2)), add(55, multiply(2.5, 2))), rectangle_area(70, 55)), 2) | multiply(n2,n3)|rectangle_area(n0,n1)|add(n0,#0)|add(n1,#0)|rectangle_area(#2,#3)|subtract(#4,#1)|multiply(n3,#5)| | geometry |
the population of a town is 8000 . it decreases annually at the rate of 20 % p . a . what will be its population after 3 years ? | "the triangle with sides 39 cm , 36 cm and 15 is right angled , where the hypotenuse is 39 cm . area of the triangle = 1 / 2 * 36 * 15 = 270 cm 2 answer : e" | a ) 570 cm 2 , b ) 370 cm 2 , c ) 170 cm 2 , d ) 271 cm 2 , e ) 270 cm 2 | e | multiply(divide(36, const_2), 15) | divide(n1,const_2)|multiply(n2,#0)| | geometry |
if henry were to add 6 gallons of water to a tank that is already 3 / 4 full of water , the tank would be 7 / 8 full . how many gallons of water would the tank hold if it were full ? | "explanation : he covered 5 km in 1 hour , so he might cover 20 km in 4 hours . but he took 20 hours . he would have saved 20 Γ’ β¬ β 4 = 16 hours . answer : e" | a ) 2 , b ) 8 , c ) 1 , d ) 6 , e ) 16 | e | subtract(20, divide(20, 5)) | divide(n2,n0)|subtract(n3,#0)| | physics |
a honey bee flies for 10 seconds , from a daisy to a rose . it immediately continues to a poppy , flying for 6 additional seconds . the distance the bee passed , flying from the daisy to the rose is 14 meters longer than the distance it passed flying from the rose to the poppy . the bee flies to the poppy at 3 meters per second faster than her speed flying to the rose . the bee flies how many meters per second from the daisy to the rose ? | "let the slower pipe alone fill the tank in x minutes then , faster pipe will fill it in x / 3 minutes 1 / x + 3 / x = 1 / 36 4 / x = 1 / 36 x = 144 min answer is a" | a ) 144 min , b ) 250 min , c ) 196 min , d ) 100 min , e ) 112 min | a | multiply(add(const_1, const_4), 36) | add(const_1,const_4)|multiply(n0,#0)| | physics |
a certain university will select 1 of 8 candidates eligible to fill a position in the mathematics department and 2 of 12 candidates eligible to fill 2 identical positions in the computer science department . if none of the candidates is eligible for a position in both departments , how many different sets of 3 candidates are there to fill the 3 positions ? | "1 day work of the three persons = ( 1 / 15 + 1 / 20 + 1 / 15 ) = 11 / 60 so , all three together will complete the work in 300 / 47 = 5.5 days . answer : c" | a ) 6.3 , b ) 6.9 , c ) 5.5 , d ) 6.1 , e ) 6.2 | c | divide(const_1, add(divide(const_1, 15), add(divide(const_1, 15), divide(const_1, 20)))) | divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|add(#3,#2)|divide(const_1,#4)| | physics |
in the set of positive integers from 1 to 100 , what is the sum of all the odd multiples of 5 ? | "distance = length of train + length of bridge = 250 + 150 = 400 speed = 72 km / hr = 72 * 5 / 18 = 20 m / s required time = 400 / 20 = 20 seconds answer is b" | a ) 10 sec , b ) 20 sec , c ) 25 sec , d ) 30 sec , e ) 35 sec | b | divide(add(250, 150), multiply(72, const_0_2778)) | add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)| | gain |
a man cycles round the boundary of a rectangular park at the rate of 12 kmph and completes one full round in 8 minutes . if the ratio between the length and breadth of the park be 3 : 2 , then its area is : | "the price of commodity x increases 25 cents each year relative to commodity y . the price difference is $ 2.10 and commodity x needs to be 10 cents less than commodity y . $ 2.00 / 25 cents = 8 years the answer is 2001 + 8 years = 2009 . the answer is b ." | a ) 2008 , b ) 2009 , c ) 2010 , d ) 2011 , e ) 2012 | b | add(2001, divide(add(divide(10, const_100), subtract(7.30, 5.20)), subtract(divide(45, const_100), subtract(7.30, 5.20)))) | divide(n5,const_100)|divide(n0,const_100)|subtract(n4,n3)|add(#0,#2)|subtract(#1,#2)|divide(#3,#4)|add(n2,#5)| | general |
if log 8 x + log 8 1 / 6 = 1 / 3 , then the value of x is : | here , a : b : c : d = 1 / 12 : 1 / 13 : 1 / 15 : 1 / 2 1 ) l . c . m of 12 : 13 : 15 : 2 is 780 2 ) find the number of books each friend received - - - - - - - - - ( to find no . of books each friend has , multiply the ratio with the l . c . m . calculated ) a = ( 1 / 12 ) x 780 = 65 b = ( 1 / 13 ) x 780 = 60 c = ( 1 / 15 ) x 780 = 52 d = ( 1 / 2 ) x 780 = 390 3 ) total number of toys = ( 65 x + 60 x + 52 x + 390 x ) = 567 x minimum number of pens ( x ) = 1 therefore , total number of items = 567 items . correct option : a | a ) 567 , b ) 167 , c ) 267 , d ) 467 , e ) 667 | a | add(add(multiply(const_100, const_4), const_100), add(multiply(15, const_4), add(const_4, const_3))) | add(const_3,const_4)|multiply(const_100,const_4)|multiply(n5,const_4)|add(#1,const_100)|add(#0,#2)|add(#3,#4) | general |
if the average marks of 3 batches of 55 , 60 and 45 students respectively is 40 , 62 , 58 , then the average marks of all the students is | "increase = ( 10 / 50 ) * 100 = ( 1 / 5 ) * 100 = 20 % . e" | a ) 16 % , b ) 16.66 % , c ) 17.9 % , d ) 18.12 % , e ) 20 % | e | multiply(divide(subtract(60, 50), 50), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
the average salary of all the workers in a workshop is rs . 8000 . the average salary of 7 technicians is rs . 12000 and the average salary of the rest is rs . 6000 . the total number of workers in the workshop is ? | "total students = 420 boys = 312 , girls = 108 total playing soccer = 250 90 % of 250 = 225 are boys who play soccer . girls who play soccer = 25 . total girls who do not play soccer = 108 - 25 = 83 . correct option : b" | a ) 69 . , b ) 83 . , c ) 81 , d ) 91 , e ) 108 | b | subtract(subtract(420, 312), subtract(250, divide(multiply(250, 90), const_100))) | multiply(n2,n3)|subtract(n0,n1)|divide(#0,const_100)|subtract(n2,#2)|subtract(#1,#3)| | gain |
the radius of the two circular fields is in the ratio 4 : 5 the area of the first field is what percent less than the area of the second ? | "let x be the price before the first discount . the price after the first discount is x - 25 % x ( price after first discount ) a second discount of 25 % of the discounted price after which the final price is 14 ( x - 25 % x ) - 25 % ( x - 25 % x ) = 14 solve for x x = $ 24.88 correct answer c" | a ) $ 45.10 , b ) $ 34.31 , c ) $ 24.88 , d ) $ 67.54 , e ) $ 65.23 | c | divide(multiply(multiply(const_100, const_100), 14), subtract(multiply(subtract(const_100, 25), const_100), multiply(subtract(const_100, 25), 25))) | multiply(const_100,const_100)|subtract(const_100,n0)|multiply(n2,#0)|multiply(#1,const_100)|multiply(n0,#1)|subtract(#3,#4)|divide(#2,#5)| | gain |
the difference of 2 digit number & the number obtained by interchanging the digits is 36 . what is the difference the sum and the number if the ratio between the digits of the number is 1 : 2 ? | "sol . let width = x , length = ( 10 + x ) perimeter = 2 ( x + ( 10 + x ) ) = 2 ( 2 x = 10 ) & 2 ( 2 x + 10 ) * 6.5 = 1650 x = 60 required perimeter = 2 ( 60 + 70 ) = 260 e" | a ) 126 , b ) 156 , c ) 190 , d ) 321 , e ) 260 | e | multiply(add(divide(subtract(divide(divide(1690, 6.5), const_2), 10), const_2), add(divide(subtract(divide(divide(1690, 6.5), const_2), 10), const_2), 10)), const_2) | divide(n2,n1)|divide(#0,const_2)|subtract(#1,n0)|divide(#2,const_2)|add(#3,n0)|add(#4,#3)|multiply(#5,const_2)| | geometry |
how many boxes do we need if we have to carry 250 apples into boxes that each hold 25 apples ? | "total weight increased = ( 4 x 1.5 ) kg = 6 kg . weight of new person = ( 95 + 6 ) kg = 101 kg . answer : option a" | a ) 101 kg , b ) 103.4 kg , c ) 105 kg , d ) data inadequate , e ) none of these | a | add(multiply(4, 1.5), 95) | multiply(n0,n1)|add(n2,#0)| | general |
in a certain quiz that consists of 10 questions , each question after the first is worth 4 points more than the preceding question . if the 10 questions on the quiz are worth a total of 300 points , how many points is the third question worth ? | "the two slowest people work at rates of 1 / 5 and 1 / 6 of the job per hour . the sum of these rates is 1 / 5 + 1 / 6 = 11 / 30 of the job per hour . the answer is c ." | a ) 4 / 15 , b ) 7 / 30 , c ) 11 / 30 , d ) 7 / 18 , e ) 5 / 18 | c | add(divide(1, 5), divide(1, 6)) | divide(n3,n1)|divide(n3,n2)|add(#0,#1)| | physics |
the ratio of spinsters to cats is 2 to 7 . if there are 40 more cats than spinsters , how many spinsters are there ? | explanation : in such a case the middle number ( c ) is the average β΄ c = 33 and a = 31 and d = 35 required percentage = 31 / 35 x 100 = 88.6 answer : option b | a ) 86.8 , b ) 88.6 , c ) 89.2 , d ) 90.1 , e ) 92.2 | b | multiply(const_100, divide(divide(multiply(33, 5), 5), add(add(add(divide(multiply(33, 5), 5), const_2), const_2), const_2))) | multiply(n0,n1)|divide(#0,n0)|add(#1,const_2)|add(#2,const_2)|add(#3,const_2)|divide(#1,#4)|multiply(#5,const_100) | general |
the manufacturer β s suggested retail price ( msrp ) of a certain item is $ 60 . store a sells the item for 20 percent more than the msrp . the regular price of the item at store b is 30 percent more than the msrp , but the item is currently on sale for 10 percent less than the regular price . if sales tax is 5 percent of the purchase price at both stores , what is the result when the total cost of the item at store b is subtracted from the total cost of the item at store a ? | "area of a trapezium = 1 / 2 ( sum of parallel sides ) * ( perpendicular distance between them ) = 1 / 2 ( 20 + 18 ) * ( 30 ) = 570 cm 2 answer : b" | a ) 827 cm 2 , b ) 570 cm 2 , c ) 285 cm 2 , d ) 178 cm 2 , e ) 176 cm 2 | b | quadrilateral_area(30, 18, 20) | quadrilateral_area(n2,n1,n0)| | physics |
if the average ( arithmetic mean ) of a and b is 45 and the average of b and c is 80 , what is the value of c Γ’ Λ β a ? | "total cost of the items he purchased = rs . 25 given that out of this rs . 25 , 30 paise is given as tax = > total tax incurred = 60 paise = rs . 60 / 100 let the cost of the tax free items = x given that tax rate = 6 % β΄ ( 25 β 60 / 100 β x ) 6 / 100 = 60 / 100 β 6 ( 25 β 0.6 β x ) = 60 β ( 25 β 0.6 β x ) = 10 β x = 25 β 0.6 β 10 = 14.4 a" | a ) a ) 14.4 , b ) b ) 20 , c ) c ) 21.3 , d ) d ) 21.5 , e ) e ) 22 | a | subtract(subtract(25, divide(60, const_100)), divide(60, 6)) | divide(n1,const_100)|divide(n1,n2)|subtract(n0,#0)|subtract(#2,#1)| | gain |
what is the largest number of 4 digits which is divisible by 15 , 25 , 40 and 75 ? | a 13 : a 23 = 27 : 125 a 1 : a 2 = 3 : 5 6 a 12 : 6 a 22 a 12 : a 22 = 9 : 25 answer : c | ['a ) 6 : 25', 'b ) 3 : 5', 'c ) 9 : 25', 'd ) 16 : 25', 'e ) 19 : 25'] | c | divide(surface_cube(divide(divide(27, const_3), const_3)), surface_cube(divide(125, divide(125, add(const_4, const_1))))) | add(const_1,const_4)|divide(n0,const_3)|divide(#1,const_3)|divide(n1,#0)|divide(n1,#3)|surface_cube(#2)|surface_cube(#4)|divide(#5,#6) | geometry |
a corporation 5 times its annual bonus to 10 of its employees . what percent of the employees β new bonus is the increase ? | "when a number is divided by another number , we can represent it as : dividend = quotient * divisor + remainder so , dividend / divisor = quotient + remainder / divisor given that n / j = 134.08 here 134 is the quotient . given that remainder = 15 so , 134.08 = 134 + 15 / j so , j = 187.5 ans e" | a ) 22 , b ) 56 , c ) 78 , d ) 112 , e ) 187.5 | e | divide(15, subtract(134.08, add(const_100, add(multiply(const_4, const_10), const_2)))) | multiply(const_10,const_4)|add(#0,const_2)|add(#1,const_100)|subtract(n1,#2)|divide(n0,#3)| | general |
at what rate percent per annum will a sum of money double in 9 years . | explanation : drawing two balls of same color from seven green balls can be done in Γ’ Β Β· c Γ’ β β ways . similarly from eight white balls two can be drawn in Γ’ Β ΒΈ c Γ’ β β ways . p = Γ’ Β Β· c Γ’ β β / Γ’ ΒΉ Γ’ Β Β΅ c Γ’ β β + Γ’ Β ΒΈ c Γ’ β β / Γ’ ΒΉ Γ’ Β Β΅ c Γ’ β β = 7 / 15 a | a ) 7 / 15 , b ) 2 / 8 , c ) 7 / 11 , d ) 13 / 5 , e ) 87 | a | divide(add(divide(factorial(7), multiply(factorial(subtract(7, const_2)), factorial(const_2))), divide(factorial(8), multiply(factorial(subtract(8, const_2)), factorial(const_2)))), divide(factorial(add(7, 8)), multiply(factorial(subtract(add(7, 8), const_2)), factorial(const_2)))) | add(n0,n1)|factorial(n0)|factorial(const_2)|factorial(n1)|subtract(n0,const_2)|subtract(n1,const_2)|factorial(#4)|factorial(#5)|factorial(#0)|subtract(#0,const_2)|factorial(#9)|multiply(#6,#2)|multiply(#7,#2)|divide(#1,#11)|divide(#3,#12)|multiply(#10,#2)|add(#13,#14)|divide(#8,#15)|divide(#16,#17) | other |
what least number should be subtracted from 13601 such that the remainder is divisible by 87 ? | "rs = 60 - 40 = 20 * 5 / 18 = 100 / 18 t = 75 d = 75 * 100 / 18 = 1250 / 3 rs = 60 + 50 = 100 * 5 / 18 t = 1250 / 3 * 18 / 500 = 15 sec answer : a" | a ) 15 sec , b ) 16 sec , c ) 14 sec , d ) 67 sec , e ) 13 sec | a | multiply(multiply(multiply(const_0_2778, subtract(60, 40)), 75), inverse(multiply(const_0_2778, add(60, 40)))) | add(n0,n1)|subtract(n0,n1)|multiply(#0,const_0_2778)|multiply(#1,const_0_2778)|inverse(#2)|multiply(n2,#3)|multiply(#4,#5)| | physics |
in a 400 member association consisting of men and women , exactly 20 % of men and exactly 25 % women are homeowners . what is the least number of members who are homeowners ? | "let the total number of trucks = 3 x total number of sedans = 7 x total number of motorcycles = 2 x total number of sedans = 11900 = > 7 x = 11900 = > x = 1700 total number of motorcycles = 2 x = 2 * 1700 = 3400 answer c" | a ) 1260 , b ) 2100 , c ) 3400 , d ) 4200 , e ) 5200 | c | multiply(divide(add(multiply(multiply(3, 3), const_1000), const_100), 7), 2) | multiply(n0,n0)|multiply(#0,const_1000)|add(#1,const_100)|divide(#2,n1)|multiply(n2,#3)| | other |
given that 100.48 = x , 100.70 = y and xz = y 2 , then the value of z is close to : | "1 to 9 = 9 * 1 = 9 10 to 99 = 90 * 2 = 180 100 to 223 = 124 * 3 = 372 - - - - - - - - - - - 561 answer : c" | a ) 372 , b ) 661 , c ) 561 , d ) 467 , e ) 761 | c | add(add(subtract(divide(divide(223, const_10), const_10), const_1), subtract(subtract(divide(223, const_10), const_1), subtract(divide(divide(223, const_10), const_10), const_1))), multiply(subtract(subtract(223, const_1), subtract(divide(223, const_10), const_1)), const_3)) | divide(n0,const_10)|subtract(n0,const_1)|divide(#0,const_10)|subtract(#0,const_1)|subtract(#2,const_1)|subtract(#1,#3)|multiply(#5,const_3)|subtract(#3,#4)|add(#4,#7)|add(#8,#6)| | general |
3251 + 587 + 369 - ? = 3007 | "explanation : 40 % = 40 * 4 = 160 90 % = 90 * 4 = 360 answer : option d" | a ) 270 , b ) 380 , c ) 260 , d ) 360 , e ) 290 | d | multiply(divide(160, divide(40, const_100)), divide(90, const_100)) | divide(n0,const_100)|divide(n2,const_100)|divide(n1,#0)|multiply(#2,#1)| | gain |
2 , 6 , 12 , 20 , 30 , 42 , 56 , ( . . . ) | "answer dividing numerator as well as denominator by y , we get given exp . = ( 7 x + 6 y ) / ( 7 x Γ’ β¬ β 6 y ) = ( 7 x / y + 6 ) / ( 7 x / y Γ’ β¬ β 6 ) since x / y = 8 / 7 this implies that = [ ( 7 * 8 ) / 7 + 6 ] / [ ( 7 * 8 ) / 7 - 6 ) ] = ( 8 + 6 ) / ( 8 - 6 ) = 7 option : d" | a ) 11 , b ) 8 , c ) 9 , d ) 7 , e ) 6 | d | divide(add(8, 7), subtract(8, 7)) | add(n0,n1)|subtract(n0,n1)|divide(#0,#1)| | general |
a person buys an article at $ 380 . at what price should he sell the article so as to make a profit of 25 % ? | explanation : 3 / 4 x = 30 = > x = 40 answer : c | a ) 29 , b ) 88 , c ) 40 , d ) 28 , e ) 27 | c | divide(30, subtract(const_1, divide(const_1, const_4))) | divide(const_1,const_4)|subtract(const_1,#0)|divide(n0,#1) | general |
each digit 1 through 5 is used exactly once to create a 5 - digit integer . if the 3 and the 24 can not be adjacent digits in the integer , how many 5 - digit integers are possible ? | "let the numbers be 3 x , 3 x + 3 and 3 x + 6 . then , 3 x + ( 3 x + 3 ) + ( 3 x + 6 ) = 108 9 x = 99 x = 11 largest number = 3 x + 6 = 39 answer : b" | a ) 36 , b ) 39 , c ) 33 , d ) 30 , e ) 42 | b | add(add(power(add(add(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), const_2))) | 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)| | general |
when greenville state university decided to move its fine arts collection to a new library , it had to package the collection in 20 - inch by 20 - inch by 15 - inch boxes . if the university pays $ 0.60 for every box , and if the university needs 3.06 million cubic inches to package the collection , what is the minimum amount the university must spend on boxes ? | "sol . let the four integers be x , x + 2 , x + 4 and x + 6 then , x + ( x + 2 ) + ( x + 4 ) + ( x + 6 ) = 1284 β 4 x = 1272 β x = 318 β΄ greatest integer = x + 6 = 324 . answer a" | a ) 324 , b ) 342 , c ) 364 , d ) 382 , e ) none | a | add(add(power(add(add(divide(subtract(subtract(1284, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(1284, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(1284, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(1284, const_10), const_2), const_4), const_2), const_2))) | 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 |
the greatest number which on dividing 1657 and 2037 leaves remainders 9 and 5 respectively , is : | "explanatory answer step 1 : compute the length of the diagonal of the square let ' a ' meters be the length of a side of the square field . therefore , its area = a 2 square meters . - - - ( 1 ) the length of the diagonal ' d ' of a square whose side is ' a ' meters = β 2 a - - - ( 2 ) from ( 1 ) and ( 2 ) , we can deduce that the square of the diagonal = d 2 = 2 a 2 = 2 ( area of the square ) or d = β 2 * area meters . d = β 2 β 24200 = 48400 = 220 m . step 2 : compute the time taken to cross the field the time taken to cross a distance of 220 meters while traveling at 6.6 kmph = 220 m / 6.6 kmph convert unit of speed from kmph to m / min 1 km = 1000 meters and 1 hour = 60 minutes . so , 6.6 kmph = 6.6 β 1000 / 60 m / min = 110 m / min β΄ time taken = 220 / 110 = 2 minutes choice c" | a ) 3 minutes , b ) 0.04 hours , c ) 2 minutes , d ) 2.4 minutes , e ) 2 minutes 40 seconds | c | divide(24200, multiply(6.6, const_1000)) | multiply(n1,const_1000)|divide(n0,#0)| | geometry |
if x / 5 + 9 / x = 14 / 5 , what are the values of 3 x - 7 ? | "add the numbers of doughnuts . 12 + 6 + 8 = 26 . answer is b ." | a ) 25 , b ) 26 , c ) 39 , d ) 21 , e ) 11 | b | add(add(12, 6), 8) | add(n0,n1)|add(n2,#0)| | general |
prints a page 40 pg per min . if the printed for 2 hours except 20 min . where there was an paper jam , how many page did it print | "speed = 90 * 5 / 18 = 25 m / sec time taken = 100 / 25 = 4 sec . answer : c" | a ) 2.5 , b ) 2.9 , c ) 4 sec , d ) 2.8 , e ) 2.1 | c | divide(100, multiply(90, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics |
a shop owner professes to sell his articles at certain cost price but he uses false weights with which he cheats by 50 % while buying and by 10 % while selling . what is his percentage profit ? | explanation : let the number be x . then , x - ( x / 3 ) = 34 = > 2 x / 3 = 34 = > x = 51 answer : option a | a ) 51 , b ) 50 , c ) 45 , d ) 40 , e ) 36 | a | divide(multiply(34, 3), subtract(3, const_1)) | multiply(n0,n1)|subtract(n0,const_1)|divide(#0,#1) | general |
gold is 19 times as heavy as water and copper is 9 times as heavy as water . in what ratio should these be mixed to get an alloy 13 times as heavy as water ? | let the correct time to complete the journey be x min distance covered in ( x + 11 ) min . at 40 kmph distance covered in ( x + 5 ) min . at 50 kmph ( x + 11 ) / 60 * 40 = ( x + 5 ) / 60 * 50 x = 19 min answer ( a ) | a ) 19 min , b ) 19 hrs , c ) 52 min , d ) 126 min , e ) 52 min | a | divide(subtract(multiply(multiply(60, 40), 11), multiply(multiply(60, 50), 5)), subtract(multiply(60, 50), multiply(60, 40))) | multiply(n0,n9)|multiply(n2,n9)|multiply(n1,#0)|multiply(n3,#1)|subtract(#1,#0)|subtract(#2,#3)|divide(#5,#4) | general |
a marketing survey of anytown found that the ratio of trucks to sedans to motorcycles was 3 : 7 : 2 , respectively . given that there are 11,900 sedans in anytown , how many motorcycles are there ? | "explanation : number of cubes = ( 16 x 16 x 16 ) / ( 8 x 8 x 8 ) = 8 answer : c" | a ) 36 , b ) 2 , c ) 8 , d ) 48 , e ) none of these | c | divide(volume_cube(16), volume_cube(divide(8, const_100))) | divide(n0,const_100)|volume_cube(n1)|volume_cube(#0)|divide(#1,#2)| | probability |
on dividing 21 by a number , the quotient is 10 and the remainder is 1 . find the divisor . | "amount received by sanjay . 4 / 12 x 5400 = 1800 = ( related ratio / sum of ratio ) x total amount so , the amount received by sanjay is 1800 . a" | a ) 1800 , b ) 980 , c ) 1200 , d ) 1240 , e ) 1400 | a | subtract(divide(5400, 2), divide(5400, 6)) | divide(n0,n1)|divide(n0,n3)|subtract(#0,#1)| | other |
in a basketball game , dhoni scored 30 points more than dravid , but only half as many points as shewag . if the 3 players scored a combined total of 150 points , how many points did dhoni score ? | first room because area 50 * 50 = 2500 sq feet second room area 100 sq yard in feet 300 sq feet third room area 200 sq feet answer : a | a ) 200 sq feet , b ) 300 sq feet , c ) 400 sq feet , d ) 500 sq feet , e ) 600 sq feet | a | multiply(100, const_2) | multiply(n1,const_2) | geometry |
( 0.0066 ) ( 3.6 ) / ( 0.04 ) ( 0.1 ) ( 0.006 ) = | alex invests rs . 7000 for 18 months , but brian invests rs . 7000 for the first 8 months and then withdraws rs . 3500 . so , the investment of brian for remaining 10 months is rs . 3500 only . alex : brian 7000 * 18 : ( 7000 * 8 ) + ( 3500 * 10 ) 126000 : 91000 alex : brian = 18 : 13 answer : e | a ) 5 : 4 , b ) 4 : 3 , c ) 18 : 11 , d ) 3 : 2 , e ) 18 : 13 | e | divide(18, add(const_12, const_1)) | add(const_1,const_12)|divide(n2,#0) | gain |
in covering a distance of 30 km , arun takes 22 hours more than anil . if arun doubles his speed , then he would take 1 hour less than anil . what is arun ' s speed ? | "ratio = 6 : 1 = > 6 x respondents preferred brand x and x preferred brand y since , no . of respondents who preferred brand x = 240 = > 6 x = 240 = > x = 40 hence total no . of respondents = 240 + 40 = 280 hence c is the answer ." | a ) 80 , b ) 160 , c ) 280 , d ) 360 , e ) 480 | c | add(divide(240, 6), 240) | divide(n0,n1)|add(n0,#0)| | other |
how long does a train 110 m long running at the speed of 72 km / hr takes to cross a bridge 132 m length ? | "the area of each half is 100 + 4 ( 200 ) + 100 = 1000 the area that is not painted is 100 . the fraction that is not painted is 100 / 1000 = 1 / 10 = 10 % the answer is b ." | a ) 5 % , b ) 10 % , c ) 15 % , d ) 20 % , e ) 25 % | b | multiply(divide(const_100, add(add(multiply(multiply(const_4, const_100), const_4), const_100), const_100)), const_100) | multiply(const_100,const_4)|multiply(#0,const_4)|add(#1,const_100)|add(#2,const_100)|divide(const_100,#3)|multiply(#4,const_100)| | geometry |
the area of a triangle is with base 2 m and height 5 m ? | "t = ( 119 + 165 ) / ( 80 + 65 ) * 18 / 5 t = 7.05 answer : e" | a ) 7.19 , b ) 7.17 , c ) 7.2 , d ) 7.15 , e ) 7.05 | e | divide(add(119, 165), multiply(add(80, 65), const_0_2778)) | add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)| | physics |
there are 3 prizes to be distributed among 10 students . if no students gets more than one prize , then this can be done in ? | for e total numbers 8 * 12 there are 12 numbers divisible by 8 - > 3 * 12 ( if 8 is an example - ( 6 , 78 ) , ( 7 , 89 ) , ( 8 , 910 ) ) and 12 numbers divisible by 4 but not divisible by 8 - > 2 * 12 ( if 4 is an example ( 2 , 34 ) and ( 4 , 56 ) ) the answer 5 / 8 - > d | a ) 1 / 4 , b ) 3 / 8 , c ) 1 / 2 , d ) 5 / 8 , e ) 3 / 4 | d | divide(add(multiply(divide(divide(96, 8), 8), 2), 2), 8) | divide(n1,n4)|divide(#0,n4)|multiply(n3,#1)|add(n3,#2)|divide(#3,n4) | general |
john makes $ 40 a week from his job . he earns a raise andnow makes $ 70 a week . what is the % increase ? | "if 20 men can build a wall 66 metres long in 10 days , length of a similar wall that can be built by 86 men in 8 days = ( 66 * 86 * 8 ) / ( 10 * 20 ) = 227.04 mtrs answer : a" | a ) 227.04 mtrs , b ) 378.4 mtrs , c ) 478.4 mtrs , d ) 488.4 mtrs , e ) 578.4 mtrs | a | multiply(66, divide(multiply(86, 8), multiply(20, 10))) | multiply(n3,n4)|multiply(n0,n2)|divide(#0,#1)|multiply(n1,#2)| | physics |
a store sells 2 kinds of jelly beans mixes ( a and b ) both made up of red and yellow beans . if b contains 20 % more red beans than a but 10 % fewer yellow beans . and jar a contains twice as many red beans as yellow by what percent is the number of beans in jar b larger than the number in jar a | "( 2 to the power x ) - ( 2 to the power ( x - 2 ) ) = 3 ( 2 to the power 9 ) 2 ^ x - 2 ^ ( x - 2 ) = 3 . 2 ^ 9 hence x = 11 . answer is b" | a ) 9 , b ) 11 , c ) 13 , d ) 15 , e ) 17 | b | add(9, 2) | add(n0,n5)| | general |
which greatest possible length can be used to measure exactly 12 meter 65 cm , 15 meter 25 cm and 10 meter 65 cm | "let the time taken to reach the destination be 3 x hours . total distance = 60 * 3 x = 180 x km he covered 2 / 3 * 180 x = 120 x km in 1 / 3 * 3 x = x hours so , the remaining 60 x km , he has to cover in 2 x hours . required speed = 60 x / 2 x = 30 kmph . answer : a" | a ) 30 kmph , b ) 28 kmph , c ) 26 kmph , d ) 24 kmph , e ) 22 kmph | a | divide(subtract(multiply(60, const_3), divide(multiply(multiply(60, const_3), const_2), const_3)), subtract(const_3, const_1)) | multiply(n0,const_3)|subtract(const_3,const_1)|multiply(#0,const_2)|divide(#2,const_3)|subtract(#0,#3)|divide(#4,#1)| | physics |
a boat can travel with a speed of 15 km / hr in still water . if the speed of the stream is 6 km / hr , find the time taken by the boat to go 86 km downstream . | "35 % - - - - - - - - - - - l 65 % - - - - - - - - - - - w - - - - - - - - - - - - - - - - - - 30 % - - - - - - - - - - 2430 100 % - - - - - - - - - ? = > 8100 answer : c" | a ) 7500 , b ) 3388 , c ) 8100 , d ) 2888 , e ) 2661 | c | divide(2430, subtract(subtract(const_1, divide(35, const_100)), divide(35, const_100))) | divide(n0,const_100)|subtract(const_1,#0)|subtract(#1,#0)|divide(n1,#2)| | gain |
what is x if x + 5 y = 24 and y = 2 ? | "sum of 5 integer ( a , b , c , d , e ) = 5 * 65 = 325 e - a = 10 i . e . e = a + 10 for e to be maximum remaining 4 must be as small as possible since smallest of 5 numbers is a so to minimize other numbers we can take them equal to the smallest of 5 numbers i . e . a + a + a + a + ( a + 10 ) = 325 i . e . 5 a = 315 i . e . a = 63 i . e . largest e = 63 + 10 = 73 answer : option e" | a ) 50 , b ) 52 , c ) 59 , d ) 68 , e ) 73 | e | add(divide(subtract(multiply(65, 5), 10), 5), 10) | multiply(n0,n1)|subtract(#0,n3)|divide(#1,n0)|add(n3,#2)| | general |
a producer of tea blends two varieties of tea from two tea gardens one costing rs 18 per kg and another rs 20 per kg in the ratio 5 : 3 . if he sells the blended variety at rs 26 per kg , then his gain percent is | "work done by p and q in 1 day = 1 / 10 work done by r in 1 day = 1 / 20 work done by p , q and r in 1 day = 1 / 10 + 1 / 20 = 3 / 20 but work done by p in 1 day = work done by q and r in 1 day . hence the above equation can be written as work done by p in 1 day Γ£ β 2 = 3 / 20 = > work done by p in 1 day = 3 / 40 = > work done by q and r in 1 day = 3 / 40 hence work done by q in 1 day = 3 / 40 Γ’ β¬ β 1 / 20 = 1 / 40 so q alone can do the work in 40 days answer is e ." | a ) 20 , b ) 22 , c ) 25 , d ) 27 , e ) 40 | e | divide(const_1, subtract(divide(add(divide(const_1, 10), divide(const_1, 20)), const_2), divide(const_1, 20))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(#2,const_2)|subtract(#3,#1)|divide(const_1,#4)| | physics |
find the area of a parallelogram with base 24 cm and height 12 cm ? | total sale for 5 months = rs . ( 6435 + 6927 + 6855 + 7230 + 6562 ) = rs . 34009 . required sale = rs . [ ( 6500 x 6 ) - 34009 ] = rs . ( 39000 - 34009 ) = rs . 4966 answer : a | a ) 4966 , b ) 2477 , c ) 2877 , d ) 2676 , e ) 1881 | a | multiply(subtract(divide(add(add(add(add(6835, 9927), 6855), 7230), 6562), 5), 6500), 5) | add(n0,n1)|add(n2,#0)|add(n3,#1)|add(n4,#2)|divide(#3,n5)|subtract(#4,n6)|multiply(n5,#5) | general |
find the ratio in which rice at rs . 7.20 a kg be mixed with rice at rs . 5.70 a kg to produce a mixture worth rs . 6.30 a kg ? | "area of a parallelogram = base * height = 15 * 40 = 600 cm 2 answer : d" | a ) 200 cm 2 , b ) 100 cm 2 , c ) 42 cm 2 , d ) 600 cm 2 , e ) 230 cm 2 | d | multiply(15, 40) | multiply(n0,n1)| | geometry |
how many 5 digit nos are there if the 2 leftmost digits are odd and the digit 4 ca n ' t appear more than once in the number ? could someone please provide a solution using a approach other than ( 1 - x ( none ) ) approach ? | "i think brute force with some common sense should be used to solve this problem . write down all perfect squares less than 70 : 1 , 4 , 9 , 16 , 25 , 36 , 49 , 64 . now , 70 should be the sum of 3 of those 8 numbers . also to simplify a little bit trial and error , we can notice that as 70 is an odd numbers then either all three numbers must be odd ( odd + odd + odd = odd ) or two must be even and one odd ( even + even + odd = odd ) . we can find that 60 equals to 9 + 25 + 36 = 3 ^ 2 + 5 ^ 2 + 6 ^ 2 = 70 - - > 3 + 5 + 6 = 14 . answer : d ." | a ) 17 , b ) 16 , c ) 15 , d ) 14 , e ) 13 | d | add(add(add(const_4, 3), add(3, const_2)), 3) | add(n1,const_4)|add(const_2,n1)|add(#0,#1)|add(n1,#2)| | geometry |
if x is equal to the sum of the integers from 40 to 50 , inclusive , and y is the number of even integers from 40 to 50 , inclusive , what is the value of x + y ? | "5 machines would produce 100 units in 50 hours . increasing the amount of machines by 5 would mean dividing 50 hours by 5 . 50 / 5 = 10 answer : d" | a ) 5 , b ) 9 , c ) 6 , d ) 10 , e ) 2 | d | divide(100, multiply(divide(divide(20, 10), 5), 20)) | divide(n1,n2)|divide(#0,n0)|multiply(n1,#1)|divide(n4,#2)| | physics |
the sum of two numbers is 184 . if one - third of the one exceeds one - seventh of the other by 8 , find the smaller number . | "required population = p ( 1 + r 1 / 100 ) ( 1 - r 2 / 100 ) ( 1 + r 3 / 100 ) = p ( 1 + 15 / 100 ) ( 1 - 35 / 100 ) ( 1 + 45 / 100 ) = 1083875 e" | a ) 1034355 , b ) 1035677 , c ) 1045778 , d ) 1067675 , e ) 1083875 | e | multiply(1000000, multiply(multiply(add(const_1, divide(15, 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 basket contains 10 apples , of which 1 is spoiled and the rest are good . if we select 2 apples from the basket simultaneously and at random , what is the probability that the 2 apples selected will include the spoiled apple ? | let the shares of a , b , c , d are x , 3 x , 4 x , 2 x 4 x - 2 x = 500 x = 250 b ' s share = 3 x = $ 750 answer is c | a ) $ 450 , b ) $ 500 , c ) $ 750 , d ) $ 800 , e ) $ 840 | c | divide(multiply(divide(multiply(add(500, 500), 2), 4), 3), 2) | add(n4,n4)|multiply(n3,#0)|divide(#1,n2)|multiply(n1,#2)|divide(#3,n3) | general |
the average weight of 18 boys in a class is 50.25 kg and that of the remaining 8 boys is 45.15 kg . find the average weights of all the boys in the class . | "let the breadth of the plot be b m . length of the plot = 3 b m ( 3 b ) ( b ) = 507 3 b 2 = 507 b 2 = 169 b = 13 m . answer : option c" | a ) 16 , b ) 17 , c ) 13 , d ) 19 , e ) 14 | c | sqrt(divide(507, const_3)) | divide(n0,const_3)|sqrt(#0)| | geometry |
sum of two numbers prime to each other is 20 and their l . c . m . is 99 . what are the numbers ? | 5 ^ 3 + 6 ^ 3 = 341 number is 5 * 6 = 30 d | a ) 8 , b ) 15 , c ) 21 , d ) 30 , e ) 39 | d | multiply(floor(power(divide(341, const_2), divide(const_1, const_3))), power(subtract(341, power(floor(power(divide(341, const_2), divide(const_1, const_3))), const_3)), divide(const_1, const_3))) | divide(n0,const_2)|divide(const_1,const_3)|power(#0,#1)|floor(#2)|power(#3,const_3)|subtract(n0,#4)|power(#5,#1)|multiply(#3,#6) | general |
on selling 9 balls at rs . 720 , there is a loss equal to the cost price of 5 balls . the cost price of a ball is : | "the formula we want to use in this type of problem is this : average * total numbers = sum first , find the average by taking the sum of the f + l number and divide it by 2 : a = ( f + l ) / 2 second , find the total numbers in our range by dividing our f and l numbers by 7 and add 1 . ( 91 / 7 ) - ( 77 / 7 ) + 1 = 3 multiply these together so what we show average * total numbers = sum ( 91 + 77 ) / 2 * 3 = sum 84 * 3 = 252 e" | a ) 260 , b ) 452 , c ) 352 , d ) 260 , e ) 252 | e | multiply(divide(add(subtract(91, const_3), add(77, const_2)), const_2), add(divide(subtract(subtract(91, const_3), add(77, const_2)), 7), const_1)) | add(n1,const_2)|subtract(n2,const_3)|add(#0,#1)|subtract(#1,#0)|divide(#3,n0)|divide(#2,const_2)|add(#4,const_1)|multiply(#6,#5)| | general |
solution p is 20 percent lemonade and 80 percent carbonated water by volume ; solution q is 45 percent lemonade and 55 percent carbonated water by volume . if a mixture of pq contains 75 percent carbonated water , what percent of the volume of the mixture is p ? | "answer let 45 x a = ( 25 x 900 ) / 100 β΄ a = ( 25 x 9 ) / 45 = 5 correct option : c" | a ) 16.2 , b ) 4 , c ) 5 , d ) 500 , e ) none | c | divide(multiply(divide(25, const_100), 900), 45) | divide(n1,const_100)|multiply(n2,#0)|divide(#1,n0)| | general |
elena purchased brand x pens for $ 4.00 apiece and brand y for $ 2.40 apiece . if elena purchased a total of 12 of these pens for $ 42.00 , how many brand x pens did she purchase ? | consider a group of 100 men and 300 women , a total of 400 people . 30 % of them , which is 120 , form a group of people who do n ' t own a car . half of them are men , and the other half are women , more precisely 60 . it means that there are 100 - 60 = 40 men who own a car , and this represents 40 / 400 = 1 / 10 of the total . answer d | a ) 3 β 20 , b ) 11 β 60 , c ) 9 β 40 , d ) 1 β 10 , e ) 11 β 20 | d | divide(const_1, divide(30, const_3)) | divide(n0,const_3)|divide(const_1,#0) | general |
in an examination , the percentage of students qualified to the students appeared from school ' p ' is 70 % . in school ' q ' , the number of students appeared is 30 % more than the students appeared from school ' p ' and the number of students qualified from school ' q ' is 50 % more than the students qualified from school ' p ' . what is the % of students qualified to the number of students appeared from school ' q ' ? | you have | x + 3 | - | 4 - x | = | 8 + x | first , look at the three values independently of their absolute value sign , in other words : | x + 3 | - | 4 - x | = | 8 + x | ( x + 3 ) - ( 4 - x ) = ( 8 + x ) now , you ' re looking at x < - 8 , s α» x is a number less than - 8 . let ' s pretend x = - 10 here to make things a bit easier to understand . when x = - 10 i . ) ( x + 3 ) ( - 10 + 3 ) ( - 7 ) ii . ) ( 4 - x ) ( 4 - [ - 10 ] ) ( double negative , s α» Γ t becomes positive ) ( 4 + 10 ) ( 14 ) iii . ) ( 8 + x ) ( 8 + - 10 ) ( - 2 ) in other words , when x < - 8 , ( x + 3 ) and ( 8 + x ) are negative . to solve problems like this , we need to check for the sign change . here is how i do it step by step . i . ) | x + 3 | - | 4 - x | = | 8 + x | ii . ) ignore absolute value signs ( for now ) and find the values of x which make ( x + 3 ) , ( 4 - x ) and ( 8 + x ) = to zero as follows : ( x + 3 ) x = - 3 ( - 3 + 3 ) = 0 ( 4 - x ) x = 4 ( 4 - 4 ) = 0 ( 8 + x ) x = - 8 ( 8 + - 8 ) = 1 c | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | c | divide(multiply(add(4, 3), const_2), 7) | add(n0,n1)|multiply(#0,const_2)|divide(#1,n2) | general |
alice and bob drive at constant speeds toward each other on a highway . alice drives at a constant speed of 30 km per hour . at a certain time they pass by each other , and then keep driving away from each other , maintaining their constant speeds . if alice is 100 km away from bob at 7 am , and also 100 km away from bob at 11 am , then how fast is bob driving ( in kilometers per hour ) ? | "solution let the son ' s present age be x years . then , man ' s present age = ( x + 24 ) years . then Γ’ β¬ ΒΉ = Γ’ β¬ ΒΊ ( x + 24 ) + 3 = 2 ( x + 3 ) Γ’ β¬ ΒΉ = Γ’ β¬ ΒΊ x + 27 = 2 x + 6 x = 21 . answer d" | a ) 14 years , b ) 18 years , c ) 20 years , d ) 21 years , e ) none | d | divide(subtract(24, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | multiply(const_2,const_2)|subtract(const_2,const_1)|subtract(#0,const_2)|subtract(n0,#2)|divide(#3,#1)| | general |
the sum of ages of 5 children born at the intervals of 3 years each is 80 years . what is the age of the youngest child ? | "say the price of the house was $ x , then 0.17 * 50,000 + 0.1 * ( x - 50,000 ) = 24,000 - - > x = $ 205,000 ( 17 % of $ 50,000 plus 10 % of the amount in excess of $ 50,000 , which is x - 50,000 , should equal to total commission of $ 24,000 ) . answer : c ." | a ) $ 115,000 , b ) $ 160,000 , c ) $ 205,000 , d ) $ 240,000 , e ) $ 365,000 | c | add(multiply(17, 10), 10) | multiply(n0,n2)|add(n2,#0)| | general |
a man sells an article at 10 % gain . had be sold at for rs . 60 / - more he could have gained 20 % what is cost price of article | not really . when you solve the 2 equation above , you get , 6 t - 4 / 3 = 5 r / 6 from simplifying equation 1 4 t - 2 = r / 2 from simplifying equation 2 you can now multiply equation 2 by 5 to get 5 ( 4 t - 2 = r / 2 ) = 20 t - 10 = 5 r / 2 and then subtract this new equation from equation 1 to get t = 3 , followed by r = 20 to give you distance q = r * t = 20 * 3 = 60 km . d | a ) 200 km , b ) 50 km , c ) 20 km , d ) 60 km , e ) 80 km | d | multiply(divide(subtract(multiply(4, 2), 4), const_2), 30) | multiply(n0,n2)|subtract(#0,n0)|divide(#1,const_2)|multiply(n1,#2) | general |
the average weight of 20 persons sitting in a boat had some value . a new person added to them whose weight was 46 kg only . due to his arrival , the average weight of all the persons decreased by 5 kg . find the average weight of first 20 persons ? | "sol . cp = rs 490 , sp = 465.50 . loss = rs ( 490 - 465.50 ) = rs 24.50 . loss % = [ ( 24.50 / 490 ) * 100 ] % = 5 % answer is b ." | a ) 4 % , b ) 5 % , c ) 6 % , d ) 3 % , e ) 5.5 % | b | multiply(divide(subtract(490, 465.50), 490), const_100) | subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)| | gain |
a total of 520 players participated in a single tennis knock out tournament . what is the total number of matches played in the tournament ? ( knockout means if a player loses , he is out of the tournament ) . no match ends in a tie . | 8 machines / 360 cans = 5 machines / x cans 8 x = 1800 x = 225 ( 225 ) ( 6 hours ) = 1350 cans . the answer is d . | a ) 675 , b ) 750 , c ) 1,800 , d ) 1,350 , e ) 7,500 | d | subtract(multiply(6, 360), multiply(6, divide(multiply(5, 360), add(const_4, const_4)))) | add(const_4,const_4)|multiply(n0,n2)|multiply(n0,n1)|divide(#2,#0)|multiply(n2,#3)|subtract(#1,#4) | physics |
how many positive integer solutions does the equation 4 x + 5 y = 100 have ? | "900 - - - - 260 100 - - - - ? = > 29 % answer : b" | a ) 11 , b ) 29 , c ) 99 , d ) 77 , e ) 18 | b | multiply(divide(subtract(1160, 900), 900), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
what is the sum of natural numbers between 60 and 100 | perimeter of rectangular yard = 2 ( l + b ) = 12 - - > l + b = 6 area = l * b = 9 b = 6 - l l ( 6 - l ) = 9 6 l - l ^ 2 = 9 l ^ 2 - 6 l + 9 = 0 upon simplifying we get l = 3 . answer : b | ['a ) 8', 'b ) 1', 'c ) 3', 'd ) 4', 'e ) 6'] | b | subtract(const_4, const_3) | subtract(const_4,const_3) | geometry |
a began business with rs . 27000 and was joined afterwards by b with rs . 54000 . when did b join if the profits at the end of the year were divided in the ratio of 2 : 1 ? | explanation : solution : given x = k / y ^ 2 , where k is constant . now , y = 3 and x = 1 gives k = 9 . . ' . x = 9 / y ^ 2 = > x = 9 / 7 ^ 2 = 9 / 49 answer : e | a ) 3 , b ) 6 , c ) 1 / 9 , d ) 1 / 3 , e ) 9 / 49 | e | divide(multiply(1, power(3, const_2)), power(7, const_2)) | power(n0,const_2)|power(n2,const_2)|multiply(n1,#0)|divide(#2,#1) | general |
the difference between compound interest and simple interest on a certain amount of money at 5 % per annum for 2 years is 19 . find the sum : | "look at the below representation of the problem : steel chrome total a 3 3 60 > > no . of type a machines = 60 / 6 = 10 b 6 5 44 > > no . of type b machines = 44 / 11 = 4 so the answer is 14 i . e c . hope its clear ." | a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | c | add(divide(44, add(5, const_3.0)), divide(60, add(3, 3))) | add(n2,n3)|add(n0,n1)|divide(n5,#0)|divide(n4,#1)|add(#2,#3)| | general |
a sum of rs . 100 is lent at simple interest of 3 % p . a . for the first month , 9 % p . a . for the second month , 27 % p . a . for the third month and so on . what is the total amount of interest earned at the end of the year approximately | "the number of jumbo tiles = x . the number of regular tiles = 2 x . assume the ratio of the dimensions of a regular tile is a : a - - > area = a ^ 2 . the dimensions of a jumbo tile is 3 a : 3 a - - > area = 9 a ^ 2 . the area of regular tiles = 2 x * a ^ 2 = 50 . the area of jumbo tiles = x * 9 a ^ 2 = 4.5 ( 2 x * a ^ 2 ) = 4.5 * 50 = 225 . total area = 50 + 225 = 275 . answer : b ." | a ) 160 , b ) 275 , c ) 360 , d ) 440 , e ) 560 | b | add(50, multiply(divide(multiply(50, 3), const_2), 3)) | multiply(n2,n1)|divide(#0,const_2)|multiply(n1,#1)|add(n2,#2)| | geometry |
what quantity of water should be added to reduce 9 liters of 50 % acidic liquid to 30 % acidic liquid ? | sale price ( sp ) = 90 + markup ( mp ) - - > mp = sp - 90 and given mp = sp / 4 ( 25 % is 1 / 4 th ) so sp / 4 = sp - 90 3 sp / 4 = 90 sp = 120 now a discount of 20 % is given so new sp is . 8 * 120 = 96 profit = 96 - 90 = 6.0 $ answer is d | a ) $ 14 , b ) $ 5 , c ) $ 10 , d ) $ 6 , e ) $ 8 | d | subtract(divide(multiply(subtract(const_100, 20), add(divide(90, const_3), 90)), const_100), 90) | divide(n0,const_3)|subtract(const_100,n2)|add(n0,#0)|multiply(#2,#1)|divide(#3,const_100)|subtract(#4,n0) | general |
compute all real solutions to 16 x + 4 x + 1 - 96 = 0 | "for every $ 3 earn above $ 9360 , the recipient loses $ 1 of benefit . or for every $ 1 loss in the benefit , the recipient earns $ 3 above $ 9360 if earning is ; 9360 + 3 x benefit = 12000 - x or the vice versa if benefit is 12000 - x , the earning becomes 9360 + 3 x he lost 50 % of the benefit ; benefit received = 12000 - 0.6 * 12000 = 12000 - 7200 x = 4800 earning becomes 9360 + 3 x = 9360 + 3 * 4800 = 23760 ans : d" | a ) $ 15,360 , b ) $ 17,360 , c ) $ 18,000 , d ) $ 23,760 , e ) $ 27,360 | d | add(multiply(const_100, 3), const_60) | multiply(const_100,n3)|add(#0,const_60)| | general |
find the area of a parallelogram with base 26 cm and height 12 cm ? | "explanation : area = ( 17.56 x 10000 ) m 2 = 175600 m 2 . Ο r 2 = 175600 β ( r ) 2 = ( 175600 x ( 7 / 22 ) ) β r = 236.37 m . circumference = 2 Ο r = ( 2 x ( 22 / 7 ) x 236.37 ) m = 1485.78 m . cost of fencing = rs . ( 1485.78 x 6 ) = rs . 8915 . answer : option e" | a ) 4457 , b ) 4567 , c ) 4235 , d ) 4547 , e ) 8915 | e | multiply(circumface(multiply(sqrt(divide(17.56, const_pi)), const_100)), 6) | divide(n0,const_pi)|sqrt(#0)|multiply(#1,const_100)|circumface(#2)|multiply(#3,n1)| | geometry |
what is the length of a bridge ( in meters ) , which a train 166 meters long and travelling at 45 km / h can cross in 40 seconds ? | suppose the population of the country in current year is 1000 . so annual increase is 1000 + 39.4 - 19.4 = 1020 hence every year there is an increase of 2 % . 2000 = 1000 ( 1 + ( 2 / 100 ) ) ^ n n = 35 answer is d | a ) q = 20 , b ) q = 25 , c ) q = 30 , d ) q = 35 , e ) 40 | d | divide(subtract(const_100, multiply(const_10, const_3)), multiply(divide(subtract(39.4, 19.4), 1000), const_100)) | multiply(const_10,const_3)|subtract(n1,n2)|divide(#1,n0)|subtract(const_100,#0)|multiply(#2,const_100)|divide(#3,#4) | general |
if log 8 x + log 8 1 / 6 = 1 / 3 , then the value of x is : | "wc = 2 : 1 2 x + x = 1 / 8 x = 1 / 24 = > 24 days answer : a" | a ) 24 days , b ) 12 days , c ) 29 days , d ) 25 days , e ) 27 days | a | multiply(divide(multiply(8, add(const_2, const_1)), const_2), const_2) | add(const_1,const_2)|multiply(n0,#0)|divide(#1,const_2)|multiply(#2,const_2)| | physics |
find the l . c . m of 15 , 18 , 28 and 30 . | "5 n > 10 - - > n > 2 7 n < 28 - - > n < 4 2 < n < 4 answer : b" | a ) 1 and 8 , b ) 2 and 4 , c ) 0 and 9 , d ) 2 and 7 , e ) 2 and 9 | b | add(multiply(2, const_10), divide(add(23, 5), 7)) | add(n4,n5)|multiply(const_10,n1)|divide(#0,n3)|add(#2,#1)| | general |
a boatman goes 2 km against the current of the stream in 1 hour and goes 1 km along the current in 15 minutes . how long will it take to go 5 km in stationary water ? | "the question basically asks how many positive odd integers less than 70 are odd multiples of 5 so we have 5,15 , 25,35 , 45,55 and 65 = 7 ans b" | a ) 4 , b ) 7 , c ) 11 , d ) 12 , e ) 15 | b | divide(divide(70, 5), const_2) | divide(n0,n1)|divide(#0,const_2)| | general |
5359 x 51 = ? | "201834 x 99999 = 201834 x ( 100000 - 1 ) = 201834 x 100000 - 201834 x 1 = 20183400000 - 201834 = 20183198166 a" | a ) 20183198166 , b ) 20194181766 , c ) 20175292556 , d ) 20132191166 , e ) 20153198166 | a | multiply(subtract(99999, const_4), 201834) | subtract(n1,const_4)|multiply(#0,n0)| | general |
how many numbers from 2 to 13 are exactly divisible by 2 ? | "95 % remains to be paid so the remaining amount is 19 * 70 = $ 1330 . the answer is d ." | a ) $ 1120 , b ) $ 1190 , c ) $ 1260 , d ) $ 1330 , e ) $ 1400 | d | subtract(multiply(70, divide(const_100, 5)), 70) | divide(const_100,n0)|multiply(n1,#0)|subtract(#1,n1)| | general |
a train 1000 m long is running at a speed of 78 km / hr . if it crosses a tunnel in 1 min , then the length of the tunnel is ? | concept : when terms are in arithmetic progression ( a . p . ) i . e . terms are equally spaced then mean = median = ( first + last ) / 2 and sum = mean * number of terms h ( 18 ) = [ ( 4 + 18 ) / 2 ] * 8 = 88 h ( 10 ) = ( 4 + 10 ) / 2 ] * 4 = 28 h ( 18 ) / h ( 10 ) = ( 88 ) / ( 28 ) ~ 3 answer : a | a ) 3 , b ) 1.8 , c ) 6 , d ) 18 , e ) 60 | a | divide(divide(multiply(add(18, 4), add(divide(subtract(18, 4), const_2), const_1)), const_2), divide(multiply(add(divide(subtract(10, 4), const_2), const_1), add(4, 10)), const_2)) | add(n0,n1)|add(n0,n2)|subtract(n1,n0)|subtract(n2,n0)|divide(#2,const_2)|divide(#3,const_2)|add(#4,const_1)|add(#5,const_1)|multiply(#0,#6)|multiply(#7,#1)|divide(#8,const_2)|divide(#9,const_2)|divide(#10,#11) | general |
what annual payment will discharge a debt of rs . 1125 due in 2 years at the rate of 5 % compound interest ? | "old time in minutes to cross 5 miles stretch = 5 * 60 / 60 = 5 * 1 / 1 = 5 new time in minutes to cross 5 miles stretch = 5 * 60 / 40 = 5 * 3 / 2 = 7.5 time difference = 2.5 ans : b" | a ) a ) 3.12 , b ) b ) 2.5 , c ) c ) 10 , d ) d ) 15 , e ) e ) 24 | b | max(multiply(subtract(add(60, 5), const_1), subtract(divide(5, 40), divide(5, 60))), const_4) | add(n0,n1)|divide(n0,n2)|divide(n0,n1)|subtract(#0,const_1)|subtract(#1,#2)|multiply(#3,#4)|max(#5,const_4)| | physics |
raman mixed 24 kg of butter at rs . 150 per kg with 36 kg butter at the rate of rs . 125 per kg . at what price per kg should he sell the mixture to make a profit of 40 % in the transaction ? | first of all , of course , the base has an area of 36 . for the lateral surfaces , it would be helpful to remember the formula for the area of an equilateral triangle . the area of one equilateral triangle is a = ( s ^ 2 * sqrt { 3 } ) / 4 . we know the side of the equilateral triangle must be the same as the square : s = 6 . thus , one of these equilateral triangles has an area of a = ( 6 ^ 2 * sqrt { 3 } ) / 4 = 9 * sqrt { 3 } . there are four identical triangles , so their combined area is a = 36 * sqrt { 3 } . now , add the square base , for a total surface area of a = 36 + 36 * sqrt { 3 } . answer = b | ['a ) 36 + 18 * sqrt ( 3 )', 'b ) 36 + 36 * sqrt ( 3 )', 'c ) 72', 'd ) 72 + 36 * sqrt ( 3 )', 'e ) 72 + 72 * sqrt ( 3 )'] | b | add(multiply(divide(multiply(6, sqrt(subtract(square_area(6), power(const_3, const_2)))), const_2), const_4), square_area(6)) | power(const_3,const_2)|square_area(n0)|subtract(#1,#0)|sqrt(#2)|multiply(n0,#3)|divide(#4,const_2)|multiply(#5,const_4)|add(#6,#1) | geometry |
find large number from below question the difference of two numbers is 1380 . on dividing the larger number by the smaller , we get 6 as quotient and the 15 as remainder | ( 8000 * 3 * 1.5 ) / 100 = 360 9200 - - - - - - - - 9560 answer : a | a ) 9560 , b ) 96288 , c ) 26667 , d ) 1662 , e ) 2882 | a | add(multiply(multiply(add(divide(1.5, const_100), divide(divide(subtract(9200, 8000), 3), 8000)), 8000), 3), 8000) | divide(n3,const_100)|subtract(n1,n0)|divide(#1,n2)|divide(#2,n0)|add(#0,#3)|multiply(n0,#4)|multiply(n2,#5)|add(n0,#6) | gain |
a man can row upstream at 25 kmph and downstream at 43 kmph , and then find the speed of the man in still water ? | "area of the shaded portion = 1 β 4 Γ Ο Γ ( 12 ) 2 = 113 m 2 answer b" | a ) 154 cm 2 , b ) 113 m 2 , c ) 123 m 2 , d ) 115 m 2 , e ) none of these | b | divide(multiply(power(12, const_2), const_pi), const_4) | power(n2,const_2)|multiply(#0,const_pi)|divide(#1,const_4)| | geometry |
a boat takes 19 hours for travelling downstream from point a to point b and coming back to a point c which is at 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 ? | "solution average of 7 people after the last one enters = 151 . Γ’ Λ Β΄ required weight = ( 7 x 151 ) - ( 6 x 170 ) = 1057 - 1020 = 37 . answer a" | a ) 37 , b ) 168 , c ) 189 , d ) 190 , e ) 200 | a | subtract(multiply(151, 7), multiply(6, 170)) | multiply(n2,n3)|multiply(n0,n1)|subtract(#0,#1)| | general |
a goods bullet train runs at the speed of 72 km / hr and crosses a 250 m long platform in 26 seconds . what is the length of the goods bullet train ? | "relative speed = ( 72 - 36 ) * 5 / 18 = 2 * 5 = 10 mps . distance covered in 27 sec = 27 * 10 = 270 m . the length of the faster train = 270 m . answer : a" | a ) 270 m , b ) 189 m , c ) 278 m , d ) 279 m , e ) 917 m | a | multiply(divide(subtract(72, 36), const_3_6), 27) | subtract(n0,n1)|divide(#0,const_3_6)|multiply(n2,#1)| | physics |
the ratio of numbers is 5 : 6 and their h . c . f is 4 . their l . c . m is : | "let x , y , z be the numbers in geometric progression ? y ^ 2 = xz x + y + z = 38 xyz = 1728 xyz = xzy = y ^ 2 y = y ^ 3 = 1728 y = 12 y ^ 2 = xz = 144 z = 144 / x x + y + z = x + 12 + 144 / x = 38 x ^ 2 + 12 x + 144 = 38 x x ^ 2 - 26 x + 144 = 0 ( x - 18 ) ( x - 8 ) = 0 x = 8,18 if x = 8 , z = 38 - 8 - 12 = 18 the numbers are 8,12 , 18 their sum is 38 their product is 1,728 the smallest number is 8 answer : d" | a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | d | multiply(divide(divide(divide(divide(38, const_1000), const_3), const_3), const_3), divide(divide(divide(divide(38, const_1000), const_3), const_3), const_3)) | divide(n0,const_1000)|divide(#0,const_3)|divide(#1,const_3)|divide(#2,const_3)|multiply(#3,#3)| | general |
a certain list consists of 21 different numbers . if n is in the list and n is 4 times the average ( arithmetic mean ) of the other 20 numbers in the list , then n is what fraction t of the sum of the 21 numbers in the list ? | "let ' s find the probability of the opposite event and subtract this value from 1 . the opposite event would be getting zero tails ( so all heads ) or 1 tail . p ( hhhh ) = ( 12 ) 4 = 116 p ( hhhh ) = ( 12 ) 4 = 116 . p ( thhh ) = 4 ! 3 ! β ( 12 ) 4 = 416 p ( thhh ) = 4 ! 3 ! β ( 12 ) 4 = 416 , we are multiplying by 4 ! 3 ! 4 ! 3 ! since thhh scenario can occur in number of ways : thhh , hthh , hhth , or hhht ( notice that 4 ! 3 ! 4 ! 3 ! basically gives number of arrangements of 4 letters thhh out of which 3 h ' s are identcal ) . p ( t β₯ 2 ) = 1 β ( 116 + 416 ) = 1116 p ( t β₯ 2 ) = 1 β ( 116 + 416 ) = 1116 . answer : d ." | a ) 1 / 16 , b ) 1 / 2 , c ) 3 / 16 , d ) 11 / 16 , e ) 3 / 8 | d | divide(add(add(add(choose(4, const_2), choose(4, const_3)), choose(4, const_4)), choose(4, 4)), power(const_2, 4)) | choose(n0,const_2)|choose(n0,const_3)|choose(n0,const_4)|choose(n0,n0)|power(const_2,n0)|add(#0,#1)|add(#5,#2)|add(#6,#3)|divide(#7,#4)| | probability |
the sum of the fourth and twelfth term of an arithmetic progression is 20 . what is the sum of the first 16 terms of the arithmetic progression ? | "area of a trapezium = 1 / 2 ( sum of parallel sides ) * ( perpendicular distance between them ) = 1 / 2 ( 30 + 18 ) * ( 15 ) = 360 cm 2 answer : c" | a ) 227 , b ) 299 , c ) 360 , d ) 161 , e ) 212 | c | quadrilateral_area(15, 18, 30) | quadrilateral_area(n2,n1,n0)| | physics |
the cross - section of a cannel is a trapezium in shape . if the cannel is 14 m wide at the top and 4 m wide at the bottom and the area of cross - section is 380 sq m , the depth of cannel is ? | "87 answer is a" | a ) 87 , b ) 97 , c ) 67 , d ) 57 , e ) 46 | a | subtract(100, 13) | subtract(n1,n2)| | general |
in a rectangular coordinate system , what is the area of a quadrilateral whose vertices have the coordinates ( 2 , - 2 ) , ( 2 , 6 ) , ( 14 , 2 ) , ( 14 , - 5 ) ? | here x and y are integers . x ^ 2 = y xy = 27 . substitute x ^ 2 = y in xy = > x ^ 3 = 27 . here x 3 is positive , x is also positive . x = 3 then y = 9 . x - y = - 6 so option c is correct | a ) - 30 , b ) - 20 , c ) - 6 , d ) 5 , e ) 20 | c | subtract(power(power(27, divide(const_1, const_3)), const_2), power(27, divide(const_1, const_3))) | divide(const_1,const_3)|power(n1,#0)|power(#1,const_2)|subtract(#2,#1) | general |
a batsman in his 9 th inning makes a score of 75 and their by increasing his average by 7 . what is his average after the 9 th inning ? | "let there be x pupils in the class . total increase in marks = ( x * 1 / 2 ) = x / 2 x / 2 = ( 83 - 63 ) = > x / 2 = 20 = > x = 40 answer : c" | a ) 13 , b ) 18 , c ) 40 , d ) 82 , e ) 43 | c | multiply(subtract(83, 63), const_2) | subtract(n0,n1)|multiply(#0,const_2)| | general |
rs . 6000 is lent out in two parts . one part is lent at 7 % p . a simple interest and the other is lent at 10 % p . a simple interest . the total interest at the end of one year was rs . 450 . find the ratio of the amounts lent at the lower rate and higher rate of interest ? | "sp of 1 m of cloth = 529 / 23 = rs . 23 cp of 1 m of cloth = sp of 1 m of cloth - profit on 1 m of cloth = rs . 23 - rs . 5 = rs . 18 answer : e" | a ) 26 , b ) 88 , c ) 90 , d ) 42 , e ) 18 | e | subtract(divide(529, 23), 5) | divide(n1,n0)|subtract(#0,n2)| | physics |
there are 3000 students in a school and among them 20 % of them attends chess class . 40 % of the students who are in the chess class are also enrolled for swimming . no other students are interested in swimming so how many will attend the swimming class if all enrolled attends ? | 9 % * cost price = $ 54 1 % * cost price = $ 54 / 9 = $ 6 the cost price is $ 600 . the answer is b . | a ) $ 500 , b ) $ 600 , c ) $ 700 , d ) $ 800 , e ) $ 900 | b | multiply(divide(54, 9), const_100) | divide(n2,n1)|multiply(#0,const_100) | gain |
rohan spends 40 % of his salary on food , 20 % on house rent , 10 % on entertainment and 10 % on conveyance . if his savings at the end of a month are rs . 3000 . then his monthly salary is | it will take 8 minutes for 100 cats to kill 100 rats . 1 cat can kill 1 rat in 8 minutes , so 100 cats can kill 100 rats in 8 minutes answer c | a ) 6 minutes , b ) 7 minutes , c ) 8 minutes , d ) 9 minutes , e ) 10 minutes | c | multiply(8, const_1) | multiply(n0,const_1) | physics |
darcy lives 1.5 miles from work . she can walk to work at a constant rate of 3 miles per hour , or she can ride the train to work at a constant rate of 20 miles per hour . if she rides the train , there is an additional x minutes spent walking to the nearest train station , waiting for the train , and walking from the final train station to her work . if it takes darcy a total of 20 more minutes to commute to work by walking than it takes her to commute to work by riding the train , what is the value of x ? | "lets assume the principal amount ( initial amount invested ) to be p rate of interest to berand time as t . we need to find r now after a time of 3 years the principal p amounts to $ 400 and after a time of 5 years ( question says after another 5 years so 3 + 2 ) p becomes $ 600 . formulating the above data amount ( a 1 ) at end of 3 years a 1 = p ( 1 + 3 r / 100 ) = 400 amount ( a 2 ) at end of 8 years a 2 = p ( 1 + 5 r / 100 ) = 600 dividing a 2 by a 1 we get ( 1 + 5 r / 100 ) / ( 1 + 3 r / 100 ) = 6 / 8 after cross multiplication we are left with r = 100 option : a" | a ) 100 % , b ) 12.5 % , c ) 67 % , d ) 25 % , e ) 33 % | a | multiply(divide(divide(subtract(600, 400), 2), subtract(400, multiply(divide(subtract(600, 400), 2), 3))), const_100) | subtract(n2,n0)|divide(#0,n3)|multiply(n1,#1)|subtract(n0,#2)|divide(#1,#3)|multiply(#4,const_100)| | gain |
john had a stock of 1000 books in his bookshop . he sold 75 on monday , 50 on tuesday , 64 on wednesday , 78 on thursday and 135 on friday . what percentage of the books were not sold ? | "e 762 ? = 5080 * ( 2 / 5 ) * ( 1 / 2 ) * ( 3 / 4 ) = 762" | a ) 392 , b ) 229 , c ) 753 , d ) 493 , e ) 762 | e | multiply(multiply(multiply(divide(3, 4), divide(1, 2)), divide(2, 5)), 5080) | divide(n3,n5)|divide(n0,n1)|divide(n2,n3)|multiply(#1,#2)|multiply(#0,#3)|multiply(n6,#4)| | general |
the least number which when increased by 5 each divisible by each one of 24 , 32 , 36 and 54 is : | "not sure if this is the shortest . . but this is how i did this there are 6 sets of integers with hundreds and units digits exchanged that satisfies k + 99 . 1 . 102 | 201 ( satisfies k + 99 , where k = 102 ) 2 . 203 | 302 ( satisfies k + 99 , where k = 203 ) 3 . . . . 4 . . . . 5 . . . . 6 . 607 | 708 each set has 10 such numbers . 1 . 102 | 201 ( still k + 99 holds good ) 2 . 112 | 211 3 . 122 | 221 4 . 132 | 231 5 . . . . 6 . . . . 7 . . . . 8 . . . . 9 . 182 | 281 10 . 192 | 291 therefore , 6 sets with 10 such number in each set will give 6 x 10 = 60 integers . b" | a ) 50 , b ) 60 , c ) 70 , d ) 80 , e ) 90 | b | multiply(const_10, subtract(const_10, const_2)) | subtract(const_10,const_2)|multiply(#0,const_10)| | general |
the mass of 1 cubic meter of a substance is 200 kilograms under certain conditions . what is the volume , in cubic centimeters , of 1 gram of this substance under these conditions ? ( 1 kilogram = 1,000 grams and 1 cubic meter = 1 , 000,000 cubic centimeters ) | β 4 = 2 so , β 4 percent of 4 β 4 = 2 percent of ( 4 ) ( 2 ) = ( 2 / 100 ) ( 8 ) = 16 / 100 = 0.16 answer : a | a ) 0.16 , b ) 0.17 , c ) 0.18 , d ) 0.2 , e ) 0.19 | a | divide(multiply(multiply(sqrt(4), sqrt(4)), 4), const_100) | sqrt(n0)|multiply(#0,#0)|multiply(n0,#1)|divide(#2,const_100) | gain |
find large number from below question the difference of two numbers is 1365 . on dividing the larger number by the smaller , we get 6 as quotient and the 10 as remainder | "first recognize you only need to consider the first two digits ( because the second two are just the first two flipped ) there are 90 possibilities for the first two digits of a 4 digit number , 10 - 99 inclusive . everything starting with a 2,4 , 6,8 will be odd , which is 4 / 9 ths of the combinations . 4 / 9 * 90 = 40 answer : a" | a ) 40 , b ) 45 , c ) 50 , d ) 90 , e ) 2500 | a | divide(power(const_10, divide(4, const_2)), const_2) | divide(n1,const_2)|power(const_10,#0)|divide(#1,const_2)| | general |
23 , 27 , 36 , 52 , . . . | "that is , 1 work done = 12 Γ 8 Γ 10 then , 12 8 Γ 10 = ? Γ 12 Γ 8 ? ( i . e . no . of men required ) = 12 Γ 8 Γ 10 / 12 Γ 8 = 8 days e )" | a ) 11 days , b ) 12 days , c ) 9 days , d ) 8 days , e ) 10 days | e | divide(multiply(multiply(12, 10), 8), multiply(8, 12)) | multiply(n0,n2)|multiply(n3,n4)|multiply(n1,#0)|divide(#2,#1)| | physics |