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a man can row a distance of 5 km in 60 min with the help of the tide . the direction of the tide reverses with the same speed . now he travels a further 20 km in 20 hours . how much time he would have saved if the direction of tide has not changed ? | "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 ) 49 , b ) 1.745 % , c ) 0 , d ) 1000 , e ) 16 | e | subtract(20, divide(20, 5)) | divide(n2,n0)|subtract(n3,#0)| | physics |
one pipe can fill a tank three times as fast as another pipe . if together the two pipes can fill the tank in 36 minutes , then the slower pipe alone will be able to fill the tank in ? | "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 ) 2 , b ) 150 , c ) 144 min , d ) 9620 , e ) 600 | c | multiply(add(const_1, const_4), 36) | add(const_1,const_4)|multiply(n0,#0)| | physics |
a man can do a job in 15 days . his father takes 20 days and his son finishes it in 15 days . how long will they take to complete the job if they all work together ? | "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 ) 605.03 , b ) 10 , c ) 9 , d ) 5.5 , e ) 8 | d | 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 |
how long does a train 250 meters long running at the rate of 72 km / hr take to cross a bridge 150 meters in length ? | "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 ) ( 40 , b ) 20 sec , c ) 0.4 , d ) 80 % , e ) 44 seconds | b | divide(add(250, 150), multiply(72, const_0_2778)) | add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)| | gain |
the price of commodity x increases by 45 cents every year , while the price of commodity y increases by 20 cents every year . in 2001 , the price of commodity x was $ 5.20 and the price of commodity y was $ 7.30 . in which year will the price of commodity x be 10 cents less than the price of commodity y ? | "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 ) 45 % , b ) 10 kmph , c ) 350.5 , d ) 2009 , e ) 1 / 2 | d | 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 a wholesaler distributes items to several outlets ( a , b , c and d ) in the ratio of 1 / 12 : 1 / 13 : 1 / 15 : 1 / 2 , then find the total number of items the wholesaler distributes ? | 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 ) 1700 , c ) 3.2 , d ) 2 , e ) $ 280 | 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 |
john makes $ 50 a week from his job . he earns a raise and now makes $ 60 a week . what is the % increase ? | "increase = ( 10 / 50 ) * 100 = ( 1 / 5 ) * 100 = 20 % . e" | a ) 8.33 km , b ) 20 % , c ) 110 , d ) 190 , e ) 40 | b | multiply(divide(subtract(60, 50), 50), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
in the fifth grade at parkway elementary school there are 420 students . 312 students are boys and 250 students are playing soccer . 90 % of the students that play soccer are boys . how many girl student are in parkway that is not playing soccer ? | "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 ) 4.17 % , b ) 4 % , c ) 425 , d ) $ 54,000 , e ) 83 . | e | 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 |
john bought a shirt on sale for 25 % off the original price and another 25 % off the discounted price . if the final price was $ 14 , what was the price before the first discount ? | "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 ) 25 % , b ) 1.8 % , c ) $ 24.88 , d ) 50 liters , e ) 52500 | 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 length of a rectangular plot is 10 mtr more than its width . the cost of fencing the plot along its perimeter at the rate of rs . 6.5 mtr is rs . 1690 . the perimeter of the plot is ? | "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 ) 80 , b ) $ 0.40 , c ) 260 , d ) 5 , e ) 1 | c | 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 |
the average weight of 4 person ' s increases by 1.5 kg when a new person comes in place of one of them weighing 95 kg . what might be the weight of the new person ? | "total weight increased = ( 4 x 1.5 ) kg = 6 kg . weight of new person = ( 95 + 6 ) kg = 101 kg . answer : option a" | a ) rs . 8500 , b ) 2.5 sec , c ) 50 , d ) 101 kg , e ) 38 kg | d | add(multiply(4, 1.5), 95) | multiply(n0,n1)|add(n2,#0)| | general |
each of the three people individually can complete a certain job in 3 , 5 , and 6 hours , respectively . what is the lowest fraction of the job that can be done in 1 hour by 2 of the people working together at their respective rates ? | "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 ) 20 , b ) 9.56 % , c ) 11 / 30 , d ) 68 , e ) 100775 | c | add(divide(1, 5), divide(1, 6)) | divide(n3,n1)|divide(n3,n2)|add(#0,#1)| | physics |
the average of 5 consecutive odd numbers a , b , c , d and e is 33 . what percent of a is d ? | 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 ) 120 % , b ) 5.5 , c ) $ 192 , d ) 88.6 , e ) 210 | d | 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 |
find the area of trapezium whose parallel sides are 20 cm and 18 cm long , and the distance between them is 30 cm ? | "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 ) 40 , b ) 1000 , c ) 475 , d ) 216 , e ) 570 cm 2 | e | quadrilateral_area(30, 18, 20) | quadrilateral_area(n2,n1,n0)| | physics |
daniel went to a shop and bought things worth rs . 25 , out of which 60 paise went on sales tax on taxable purchases . if the tax rate was 6 % , then what was the cost of the tax free items ? | "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 ) 1943236 , b ) 6 , c ) 4.17 % , d ) 8915 , e ) 14.4 | e | subtract(subtract(25, divide(60, const_100)), divide(60, 6)) | divide(n1,const_100)|divide(n1,n2)|subtract(n0,#0)|subtract(#2,#1)| | gain |
the volumes of two cubes are in the ratio 27 : 125 , what shall be the ratio of their surface areas ? | 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 ) 600 , b ) 9 : 25', ' , c ) 760 , d ) 42 , e ) 6336000 | b | 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 |
when positive integer n is divided by positive integer j , the remainder is 15 . if n / j = 134.08 , what is value of j ? | "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 ) 24 , b ) 190 , c ) 187.5 , d ) 12 cm', ' , e ) 40 | c | 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 |
bag contains 7 green and 8 white balls . if two balls are drawn simultaneously , the probability that both are of the same colour is - . | 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 ) 20 % loss , b ) $ 1,354 , c ) 7 / 15 , d ) 25 % , e ) 60 | c | 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 |
two trains of equal length , running with the speeds of 60 and 40 kmph , take 75 seconds to cross each other while they are running in the same direction . what time will they take to cross each other if they are running in opposite directions ? | "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 ) 40 % , b ) 60000 , c ) 130 , d ) 16 hrs , e ) 15 sec | e | 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 |
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 ? | "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 ) 672 , b ) 11 sec , c ) 60 , d ) 38.9 % , e ) 3400 | e | 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 |
how many figures are required to number the pages the pages of a book containing 223 pages ? | "1 to 9 = 9 * 1 = 9 10 to 99 = 90 * 2 = 180 100 to 223 = 124 * 3 = 372 - - - - - - - - - - - 561 answer : c" | a ) 1 / 81 , b ) 210 , c ) $ 120 , d ) 1 / 3 , e ) 561 | e | 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 |
if 40 % of a certain number is 160 , then what is 90 % of that number ? | "explanation : 40 % = 40 * 4 = 160 90 % = 90 * 4 = 360 answer : option d" | a ) 923 , b ) $ 900 , c ) 0.6 , d ) 360 , e ) 252 sec | 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 |
if x / y = 8 / 7 , then ( 7 x + 6 y ) / ( 7 x Γ’ β¬ β 6 y ) = ? | "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 ) 30 % , b ) 7 , c ) 45 km , d ) 80 , e ) 3 / 4 | b | divide(add(8, 7), subtract(8, 7)) | add(n0,n1)|subtract(n0,n1)|divide(#0,#1)| | general |
subtracting 30 from a number , the remainder is one fourth of the number . find the number ? | explanation : 3 / 4 x = 30 = > x = 40 answer : c | a ) 75 % , b ) 2340 , c ) 3.1 sec , d ) 40 , e ) 55 sec | d | divide(30, subtract(const_1, divide(const_1, const_4))) | divide(const_1,const_4)|subtract(const_1,#0)|divide(n0,#1) | general |
the sum of three consecutive multiples of 3 is 108 . what is the largest number ? | "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 ) 9 , b ) 39 , c ) 10,000 , d ) 40000 , e ) 3000000 | 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 |
the sum of four consecutive even integers is 1284 . the greatest of them is : | "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 ) 40 , c ) 1 : 4', ' , d ) 334 , e ) 176 | 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 area of a square field is 24200 sq m . how long will a lady take to cross the field diagonally at the rate of 6.6 km / hr ? | "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 ) 23.57 , b ) 39.8 , c ) 24 days , d ) $ 24 , e ) 2 minutes | e | divide(24200, multiply(6.6, const_1000)) | multiply(n1,const_1000)|divide(n0,#0)| | geometry |
anne bought doughnuts for a class breakfast party . she bought 12 chocolate doughnuts , 6 coconut doughnuts , and 8 jam - filled doughnuts . how many doughnuts did anne buy in all ? | "add the numbers of doughnuts . 12 + 6 + 8 = 26 . answer is b ." | a ) 165 Β° , b ) 67 kg , c ) 26 , d ) 3 / 4000 , e ) 8985 | c | add(add(12, 6), 8) | add(n0,n1)|add(n2,#0)| | general |
in what time will a train 100 m long cross an electric pole , it its speed be 90 km / hr ? | "speed = 90 * 5 / 18 = 25 m / sec time taken = 100 / 25 = 4 sec . answer : c" | a ) 314.3 m , b ) 1 / 10 , c ) 868 cm ^ 2', ' , d ) 6 o kmph , e ) 4 sec | e | divide(100, multiply(90, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics |
if the number is divided by 3 , it reduced by 34 . the number is | explanation : let the number be x . then , x - ( x / 3 ) = 34 = > 2 x / 3 = 34 = > x = 51 answer : option a | a ) 5 / 8 , b ) 4 , c ) 51 , d ) 135 % , e ) 2 | c | divide(multiply(34, 3), subtract(3, const_1)) | multiply(n0,n1)|subtract(n0,const_1)|divide(#0,#1) | general |
if a train runs at 40 kmph , it reach its destination late by 11 minutes but if it runs at 50 kmph it is late by 5 minutes only . the correct time for a train to complete its journey is ? 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 | 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 ) 400 , b ) 0 , c ) 20 , d ) 19 min , e ) none | d | 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 |
how many cubes of 8 cm edge can be cut out of a cube of 16 cm edge | "explanation : number of cubes = ( 16 x 16 x 16 ) / ( 8 x 8 x 8 ) = 8 answer : c" | a ) 227.04 mtrs , b ) 49 , c ) 5 % , d ) 8 , e ) 104 meters', ' | d | divide(volume_cube(16), volume_cube(divide(8, const_100))) | divide(n0,const_100)|volume_cube(n1)|volume_cube(#0)|divide(#1,#2)| | probability |
share rs . 5400 among john , jose & binoy in the ration 2 : 4 : 6 . find the amount received by john ? | "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 ) 2 kmph , c ) 11.25 , d ) 30 , e ) 6 km | a | subtract(divide(5400, 2), divide(5400, 6)) | divide(n0,n1)|divide(n0,n3)|subtract(#0,#1)| | other |
a room is a square of side 50 feet . a second room is of area 100 square yards . a third room is of area 200 square feet . which of these can seat maximum people ? ( hint : 1 yard = 3 feet ) | 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 ) 2 / 3 , c ) 28000 , d ) s . 800 , e ) 2972 | a | multiply(100, const_2) | multiply(n1,const_2) | geometry |
alex and brian start a business with rs . 7000 each , and after 8 months , brian withdraws half of his capital . how should they share the profits at the end of the 18 months ? | 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 ) 18 : 13 , b ) 240 , c ) 3 , d ) 1 / 13 , e ) 40.2 degrees c | a | divide(18, add(const_12, const_1)) | add(const_1,const_12)|divide(n2,#0) | gain |
of the people who responded to a market survey , 240 preferred brand x and the rest preferred brand y . if the respondents indicated a preference for brand x over brand y by ratio of 6 to 1 , how many people responded to the survey ? | "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 ) 20 sec , b ) 256 , c ) 280 , d ) 8 , e ) 158 | c | add(divide(240, 6), 240) | divide(n0,n1)|add(n0,#0)| | other |
a block of wood has dimensions 10 cm x 10 cm x 40 cm . the block is painted red and then cut evenly at the 20 cm mark , parallel to the sides , to form two rectangular solids of equal volume . what percentage of the surface area of each of the new solids is not painted red ? | "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 ) 19 , b ) β 3 , c ) 250 , d ) 7.4 , e ) 10 % | e | 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 |
two trains 119 meters and 165 meters in length respectively are running in opposite directions , one at the rate of 80 km and the other at the rate of 65 kmph . in what time will they be completely clear of each other from the moment they meet ? | "t = ( 119 + 165 ) / ( 80 + 65 ) * 18 / 5 t = 7.05 answer : e" | a ) 86 , b ) 12 hours , c ) 7.05 , d ) 1 / 104 , e ) 75 % | c | 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 |
if an integer e is to be chosen at random from the integers 1 to 96 , inclusive , what is the probability that e ( e + 1 ) ( e + 2 ) will be divisible by 8 ? | 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 ) 64 , b ) $ 130 , c ) 36 , d ) 5 / 8 , e ) 3584 | 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 |
if 20 men can build a wall 66 metres long in 10 days , what length of a similar can be built by 86 men in 8 days ? | "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 ) 22 , c ) 22 Β½ days , d ) 1575 , e ) 270 cm 2 | a | multiply(66, divide(multiply(86, 8), multiply(20, 10))) | multiply(n3,n4)|multiply(n0,n2)|divide(#0,#1)|multiply(n1,#2)| | physics |
if ( 2 to the x ) - ( 2 to the ( x - 2 ) ) = 3 ( 2 to the 9 ) , what is the value of x ? | "( 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 ) 11 , b ) 50 , c ) 21 % , d ) 33 1 / 3 % , e ) 1 / 6 | a | add(9, 2) | add(n0,n5)| | general |
by travelling at 60 kmph , a person reaches his destination on time . he covered two - third the total distance in one - third of the total time . what speed should he maintain for the remaining distance to reach his destination on time ? | "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 , b ) 64 , c ) $ 900 , d ) 30 kmph , e ) 26 years | d | 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 candidate got 35 % of the votes polled and he lost to his rival by 2430 votes . how many votes were cast ? | "35 % - - - - - - - - - - - l 65 % - - - - - - - - - - - w - - - - - - - - - - - - - - - - - - 30 % - - - - - - - - - - 2430 100 % - - - - - - - - - ? = > 8100 answer : c" | a ) 72 , b ) 5 , c ) 2 , d ) 220010 , e ) 8100 | e | 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 |
if the average of 5 positive integers is 65 and the difference between the largest and the smallest of these 5 numbers is 10 , what is the maximum value possible for the largest of these 5 integers ? | "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 ) 265 , b ) 4.8 , c ) 73 , d ) 50 % , e ) 122 | c | add(divide(subtract(multiply(65, 5), 10), 5), 10) | multiply(n0,n1)|subtract(#0,n3)|divide(#1,n0)|add(n3,#2)| | general |
p can do a work in the same time in which q and r together can do it . if p and q work together , the work can be completed in 10 days . r alone needs 20 days to complete the same work . then q alone can do it in | "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 ) 40 , b ) 70 , c ) 0 , d ) 75 , e ) $ 55.55 | a | 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 |
sale of rs 6835 , rs . 9927 , rs . 6855 , rs . 7230 and rs . 6562 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs , 6500 ? | 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 ) 66 , c ) 400,200 , d ) 4 / 9 , e ) 41.4 | 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 area of a parallelogram with base 15 cm and height 40 cm ? | "area of a parallelogram = base * height = 15 * 40 = 600 cm 2 answer : d" | a ) none of these , b ) 100 , c ) 5 , d ) 600 cm 2 , e ) 50 | d | multiply(15, 40) | multiply(n0,n1)| | geometry |
the number 70 can be written as the sum of the squares of 3 different positive integers . what is the sum of these 3 integers ? | "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 ) 20,160 , b ) 400 % , c ) 1534 , d ) 14 , e ) 16.36 % | 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 5 machines can produce 20 units in 10 hours , how long would it take 25 to produce 100 units ? | "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 ) 130 , b ) 10 , c ) 12 kmph , d ) 3.75 , e ) 4 | b | divide(100, multiply(divide(divide(20, 10), 5), 20)) | divide(n1,n2)|divide(#0,n0)|multiply(n1,#1)|divide(n4,#2)| | physics |
population of a city in 20004 was 1000000 . if in 2005 there isan increment of 15 % , in 2006 there is a decrements of 35 % and in 2007 there is an increment of 45 % , then find the population of city atthe end of the year 2007 | "required population = p ( 1 + r 1 / 100 ) ( 1 - r 2 / 100 ) ( 1 + r 3 / 100 ) = p ( 1 + 15 / 100 ) ( 1 - 35 / 100 ) ( 1 + 45 / 100 ) = 1083875 e" | a ) 1083875 , b ) 63 , c ) 11988 , d ) 40 , e ) 2991 | a | 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 sum of money is distributed among a , b , c , d in the proportion of 1 : 3 : 4 : 2 . if c gets $ 500 more than d , what is the b ' s share ? | 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 ) 36 % , b ) $ 750 , c ) 12 , d ) 16 : 15 , e ) 6.24 | b | 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 length of a rectangular plot is thrice its breadth . if the area of the rectangular plot is 507 sq m , then what is the breadth of the rectangular plot ? | "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 ) 80 / 243 , b ) m , c ) 6 : 00 , d ) 23 , e ) 13 | e | sqrt(divide(507, const_3)) | divide(n0,const_3)|sqrt(#0)| | geometry |
the number 341 is equal to the sum of the cubes of two integers . what is the product of those integers ? | 5 ^ 3 + 6 ^ 3 = 341 number is 5 * 6 = 30 d | a ) 7 / 66 , b ) 2 , c ) 80 / 243 , d ) 30 , e ) 0 | 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 |
what is the sum of the multiples of 7 from 77 to 91 , inclusive ? | "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 ) 2 , b ) 36 kmph , c ) 252 , d ) 3 , e ) 2250 | c | 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 |
45 x ? = 25 % of 900 | "answer let 45 x a = ( 25 x 900 ) / 100 β΄ a = ( 25 x 9 ) / 45 = 5 correct option : c" | a ) 650 , b ) 4991 , c ) 5.5 % , d ) 5 , e ) 49995 | d | divide(multiply(divide(25, const_100), 900), 45) | divide(n1,const_100)|multiply(n2,#0)|divide(#1,n0)| | general |
if there are thrice as many women as men in a group and an equal number of men and women do not own cars - a group that is 30 % of the total . what fraction of the total is men who own cars ? | 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 ) 1 β 10 , b ) 30 % , c ) 11 , d ) 30.68 , e ) 22 | a | divide(const_1, divide(30, const_3)) | divide(n0,const_3)|divide(const_1,#0) | general |
| x + 3 | β | 4 - x | = | 7 + x | how many solutions will this equation have ? | 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 ) 39.8 , b ) 2 / 9 , c ) 2 , d ) 309400 , e ) 15552 | c | divide(multiply(add(4, 3), const_2), 7) | add(n0,n1)|multiply(#0,const_2)|divide(#1,n2) | general |
a man is 24 years older than his son . in three years , his age will be twice the age of his son . the present age of the son is | "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 ) 21 years , b ) 19 , c ) 2700 , d ) 1 , e ) 30 | a | 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 |
an auction house charges a commission of 17 % on the first $ 50,000 of the sale price of an item , plus 10 % on the amount of of the sale price in excess of $ 50,000 . what was the price of a painting for which the house charged a total commission of $ 24,000 ? | "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 ) $ 205,000 , b ) 1 / 16 , c ) 53 , d ) 10 , e ) 1717.85 | a | add(multiply(17, 10), 10) | multiply(n0,n2)|add(n2,#0)| | general |
a man covers a certain distance q in a train . if the train moved 4 km / hr faster , it would take 30 min less . if it moved 2 km / hr slower , it would take 20 mins more . find the distance ? | 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 ) rs . 600 , b ) 1.1 inches', ' , c ) 60 km , d ) 144 , e ) 1.52 % | c | multiply(divide(subtract(multiply(4, 2), 4), const_2), 30) | multiply(n0,n2)|subtract(#0,n0)|divide(#1,const_2)|multiply(n1,#2) | general |
if the a radio is sold for rs 490 and sold for rs 465.50 . find loss % . | "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 ) $ 306 , b ) 11 and 9 , c ) 10 , d ) 5 % , e ) 9 | d | multiply(divide(subtract(490, 465.50), 490), const_100) | subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)| | gain |
eight identical machines can produce 360 aluminum cans per hour . if all of the machines work at the same constant rate , how many cans could 5 such machines produce in 6 hours ? | 8 machines / 360 cans = 5 machines / x cans 8 x = 1800 x = 225 ( 225 ) ( 6 hours ) = 1350 cans . the answer is d . | a ) 135 Β° , b ) 1,350 , c ) 6 : 49 , d ) 30 , e ) 16 | b | 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 |
a cycle is bought for rs . 900 and sold for rs . 1160 , find the gain percent ? | "900 - - - - 260 100 - - - - ? = > 29 % answer : b" | a ) 2 , b ) 29 , c ) 510 , d ) 9 , e ) 23 years | b | multiply(divide(subtract(1160, 900), 900), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
the perimeter of a rectangular yard is completely surrounded by a fence that measures 12 meters . what is the length of the yard if the area of the yard is 9 meters squared ? | 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 / 3 , b ) 6 , c ) 1', ' , d ) 14.49 % , e ) 50 litres | c | subtract(const_4, const_3) | subtract(const_4,const_3) | geometry |
x varies inversely as square of y . given that y = 3 for x = 1 . the value of x for y = 7 will be equal to : | 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 ) 23 , b ) 2 / 42 , c ) 18 years , d ) 7.9 % , 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 |
each machine of type a has 3 steel parts and 3 chrome parts . each machine of type b has 6 steel parts and 5 chrome parts . if a certain group of type a and type b machines has a total of 60 steel parts and 44 chrome parts , how many machines are in the group | "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 ) 36 min , b ) 14 , c ) 12.5', ' , d ) $ 864 , e ) 5 pm . | b | 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 rectangular wall is covered entirely with two kinds of decorative tiles : regular and jumbo . 1 / 3 of the tiles are jumbo tiles , which have a length three times that of regular tiles and have the same ratio of length to width as the regular tiles . if regular tiles cover 50 square feet of the wall , and no tiles overlap , what is the area of the entire wall ? | "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 ) 275 , b ) 16 , c ) 19 , d ) rs . 76 , e ) 730 | a | add(50, multiply(divide(multiply(50, 3), const_2), 3)) | multiply(n2,n1)|divide(#0,const_2)|multiply(n1,#1)|add(n2,#2)| | geometry |
a clothing store purchased a pair of pants for $ 90 and was selling it at a price that equaled the purchase price of the pants plus a markup that was 25 percent of the selling price . after some time a clothing store owner decided to decrease the selling price by 20 percent . what was the clothing store ' s gross profit on this sale ? | 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 ) 90 o , b ) $ 6 , c ) 195 m , d ) 40 , e ) 859622 | b | 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 |
a certain social security recipient will receive an annual benefit of $ 12,000 provided he has annual earnings of $ 9,360 or less , but the benefit will be reduced by $ 1 for every $ 3 of annual earnings over $ 9,360 . what amount of total annual earnings would result in a 60 percent reduction in the recipient ' s annual social security benefit ? ( assume social security benefits are not counted as part of annual earnings . ) | "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 ) 0.4 , b ) 26 , c ) $ 23,760 , d ) 5.2 , e ) 4.8 cm', ' | c | add(multiply(const_100, 3), const_60) | multiply(const_100,n3)|add(#0,const_60)| | general |
the area of a circular field is 17.56 hectares . find the cost of fencing it at the rate of rs . 6 per metre approximately | "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 ) 5 % , b ) $ 23,760 , c ) 300 , d ) 8915 , e ) 240 meters | d | 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 |
the annual birth and death rate in a country per 1000 are 39.4 and 19.4 respectively . the number of years q in which the population would be doubled assuming there is no emigration or immigration is | 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 ) 3 : 2 , b ) 3 , c ) q = 35 , d ) 55 , e ) 820 | c | 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 |
a is twice as good a workman as b and they took 8 days together to do the work b alone can do it in . | "wc = 2 : 1 2 x + x = 1 / 8 x = 1 / 24 = > 24 days answer : a" | a ) 24 days , b ) 120 , c ) rs . 15,000 , d ) 150 , e ) 26 | 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 |
5 n + 2 > 12 and 7 n - 5 < 23 ; n must be between which numbers ? | "5 n > 10 - - > n > 2 7 n < 28 - - > n < 4 2 < n < 4 answer : b" | a ) 2 and 4 , b ) s . 20 , c ) 4 / 1 , d ) $ 12.50 , e ) rs . 3800 | a | add(multiply(2, const_10), divide(add(23, 5), 7)) | add(n4,n5)|multiply(const_10,n1)|divide(#0,n3)|add(#2,#1)| | general |
how many unique positive odd integers less than 70 are equal to the product of a positive multiple of 5 and an odd number ? | "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 ) 11 / 3 , b ) rs 66.66 , c ) 31.67 % , d ) 40 , e ) 7 | e | divide(divide(70, 5), const_2) | divide(n0,n1)|divide(#0,const_2)| | general |
find the value of 201834 x 99999 = m ? | "201834 x 99999 = 201834 x ( 100000 - 1 ) = 201834 x 100000 - 201834 x 1 = 20183400000 - 201834 = 20183198166 a" | a ) 252 sec , b ) 900 , c ) 4.8 cm', ' , d ) 29 , e ) 20183198166 | e | multiply(subtract(99999, const_4), 201834) | subtract(n1,const_4)|multiply(#0,n0)| | general |
if a 5 percent deposit that has been paid toward the purchase of a certain product is $ 70 , how much more remains to be paid ? | "95 % remains to be paid so the remaining amount is 19 * 70 = $ 1330 . the answer is d ." | a ) $ 1330 , b ) 400 m , c ) 34 , d ) 80 minutes , e ) 2.25 | a | subtract(multiply(70, divide(const_100, 5)), 70) | divide(const_100,n0)|multiply(n1,#0)|subtract(#1,n1)| | general |
for all even integers n , h ( n ) is defined to be the sum of the even integers between 4 and n , inclusive . what is the value of h ( 18 ) / h ( 10 ) ? | 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 ) 2 kmph , c ) 50 , d ) 36.5 , e ) $ 920.24 | 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 |
due to construction , the speed limit along an 5 - mile section of highway is reduced from 60 miles per hour to 40 miles per hour . approximately how many minutes more will it take to travel along this section of highway at the new speed limit than it would have taken at the old speed limit ? | "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 ) 40 , b ) 30 hr , c ) 2.5 , d ) 84.6 % , e ) 60 | c | 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 |
a pyramid has a square base of 6 cm , and the four lateral faces are four congruent equilateral triangles . what is the total surface area of the pyramid in square cm ? | 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 + 36 * sqrt ( , 3 ) 180 km , b ) 20 , c ) 1200 , d ) 192 , e ) 1 / 3 | a | 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 |
sonika deposited rs . 8000 which amounted to rs . 9200 after 3 years at simple interest . had the interest been 1.5 % more . she would get how much ? | ( 8000 * 3 * 1.5 ) / 100 = 360 9200 - - - - - - - - 9560 answer : a | a ) 24 seconds , b ) 2 , c ) rs . 30,000 , d ) 9560 , e ) 28 | d | 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 horse is tethered to one corner of a rectangular grassy field 36 m by 20 m with a rope 12 m long . over how much area of the field can it graze ? | "area of the shaded portion = 1 β 4 Γ Ο Γ ( 12 ) 2 = 113 m 2 answer b" | a ) 4 , b ) 0.2 , c ) 113 m 2 , d ) 20,160 , e ) 74 | c | divide(multiply(power(12, const_2), const_pi), const_4) | power(n2,const_2)|multiply(#0,const_pi)|divide(#1,const_4)| | geometry |
there are 6 people in the elevator . their average weight is 170 lbs . another person enters the elevator , and increases the average weight to 151 lbs . what is the weight of the 7 th person . | "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 ) 52 , b ) 43983 , c ) 37 , d ) 7 pm , e ) 90 cc', ' | c | subtract(multiply(151, 7), multiply(6, 170)) | multiply(n2,n3)|multiply(n0,n1)|subtract(#0,#1)| | general |
two trains are moving in the same direction at 72 kmph and 36 kmph . the faster train crosses a man in the slower train in 27 seconds . find the length of the faster 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 ) 108 Β° , b ) 270 m , c ) 160 , d ) 10.5 , e ) 14 | b | multiply(divide(subtract(72, 36), const_3_6), 27) | subtract(n0,n1)|divide(#0,const_3_6)|multiply(n2,#1)| | physics |
find the smallest number in gp whose sum is 38 and product is 1728 | "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 ) 10 , b ) 416 , c ) 8 , d ) 6.6 kmph , e ) 2068 | c | 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 fair coin is tossed 4 times . what is the probability of getting at least 2 tails ? | "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 ) 4 , b ) 93 / 10 , c ) 64.3 , d ) 11 / 16 , e ) s . 352 | 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 |
find the area of trapezium whose parallel sides are 30 cm and 18 cm long , and the distance between them is 15 cm . | "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 ) 360 , b ) 378 , c ) 25 % , d ) $ 22000 , e ) 28 | a | quadrilateral_area(15, 18, 30) | quadrilateral_area(n2,n1,n0)| | physics |
the mean of 50 observations is 100 . but later he found that there is decrements of 13 from each observations . what is the the updated mean is ? | "87 answer is a" | a ) 8 , b ) 14 kmph , c ) 4096 , d ) 87 , e ) 1000 | d | subtract(100, 13) | subtract(n1,n2)| | general |
if x and y are integers such that x ^ 2 = y and xy = 27 , then x β y = ? | 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 ) 21 , b ) $ 28.44 , c ) - 6 , d ) 6 / 5 , e ) 2 | 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 pupil ' s marks were wrongly entered as 83 instead of 63 . due to the average marks for the class got increased by half . the number of pupils in the class is ? | "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 ) 43983 , b ) 1 : 64 .', ' , c ) 5390 , d ) 137 / 216', ' , e ) 40 | e | multiply(subtract(83, 63), const_2) | subtract(n0,n1)|multiply(#0,const_2)| | general |
a trader sells 23 meters of cloth for rs . 529 at the profit of rs . 5 per metre of cloth . what is the cost price of one metre of cloth ? | "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 ) 756 , b ) 18 , c ) 56 , d ) 11.25 Β° , e ) 2 | b | subtract(divide(529, 23), 5) | divide(n1,n0)|subtract(#0,n2)| | physics |
if an article is sold at 18 % profit instead of 9 % profit , then the profit would be $ 54 more . what is the cost price ? | 9 % * cost price = $ 54 1 % * cost price = $ 54 / 9 = $ 6 the cost price is $ 600 . the answer is b . | a ) 10 , b ) $ 600 , c ) 12 , d ) 28 years , e ) 58 and 28 | b | multiply(divide(54, 9), const_100) | divide(n2,n1)|multiply(#0,const_100) | gain |
if 8 cats can kill 8 rats in 8 minutes , how long will it take 100 cats to kill 100 rats ? | 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 ) 2 , b ) 8 minutes , c ) 11 , d ) 36 metre , e ) 24 | b | multiply(8, const_1) | multiply(n0,const_1) | physics |
peter invested a certain sum of money in a simple interest bond whose value grew to $ 400 at the end of 3 years and to $ 600 at the end of another 2 years . what was the rate of interest in which he invested his sum ? | "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 ) 20 , b ) 3584 , c ) 18911 , d ) 3 , e ) 100 % | e | 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 |
3 / 4 of 1 / 2 of 2 / 5 of 5080 = ? | "e 762 ? = 5080 * ( 2 / 5 ) * ( 1 / 2 ) * ( 3 / 4 ) = 762" | a ) 11 , b ) 20 hrs , c ) 226.623 , d ) 2 / 3 , 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 |
how many integers k greater than 100 and less than 800 are there such that if the hundreds and the unit digits of k are reversed , the resulting integer is k + 99 ? | "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 ) 3 : 5 , b ) $ 900 , c ) $ 11.20 , d ) 60 , e ) 12 | d | multiply(const_10, subtract(const_10, const_2)) | subtract(const_10,const_2)|multiply(#0,const_10)| | general |
β 4 percent of 4 β 4 = | 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 ) 840 , c ) w . 100 , d ) 7 1 / 2 , e ) 11 sec | a | divide(multiply(multiply(sqrt(4), sqrt(4)), 4), const_100) | sqrt(n0)|multiply(#0,#0)|multiply(n0,#1)|divide(#2,const_100) | gain |
a palindrome is a number that reads the same forward and backward , such as 616 . how many even , 4 - digit numbers are palindromes ? | "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 ) 54 Β°', ' , b ) 126 , c ) 40 , d ) 3.696 kg , e ) 2500 | c | divide(power(const_10, divide(4, const_2)), const_2) | divide(n1,const_2)|power(const_10,#0)|divide(#1,const_2)| | general |
12 men work 8 hours per day to complete the work in 10 days . to complete the same work in 8 days , working 12 hours a day , the number of men required ? | "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 ) 63 , b ) 36', ' , c ) 642 , d ) 10 days , e ) $ 488.9 | d | divide(multiply(multiply(12, 10), 8), multiply(8, 12)) | multiply(n0,n2)|multiply(n3,n4)|multiply(n1,#0)|divide(#2,#1)| | physics |
8 people decided to split the restaurant bill evenly . if the bill was $ 214.15 dollars , how much money did they 1 cent is the smallest unit ? | "if the last three digits of a whole number are divisible by 8 , then the entire number is divisible by 8 the last 3 digit 415 not divisible by a hence , we need to add 1 to this number for it to be divisible by 8 correct option : a" | a ) $ 9.60 , b ) 5 , c ) $ 54,000 , d ) $ 214.16 , e ) 1.11 % | d | add(214.15, divide(const_3, const_100)) | divide(const_3,const_100)|add(n1,#0)| | general |
in a fuel station the service costs $ 1.50 per car , every liter of fuel costs 0.35 $ . assuming that you own 3 limos and 2 fleet vans and all fuel tanks are empty . how much will it cost to fuel all cars together if a limo tank is 32 liters and an fleet van tank is 75 % bigger ? | "lots of calculations . 1.50 * 4 + 3 * . 35 * 32 + 2 * ( 7 / 4 ) * 32 * . 35 answer = $ 78.80 the correct option is a" | a ) 7 , b ) 32 square inches , c ) $ 78.80 , d ) 49 hr , e ) 54 | c | multiply(multiply(0.35, 2), 3) | multiply(n1,n3)|multiply(n2,#0)| | general |