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find the value of 72519 x 9999 = y ? | "72519 x 9999 = 72519 x ( 10000 - 1 ) = 72519 x 10000 - 72519 x 1 = 725190000 - 72519 = 725117481 d" | a ) 23,200 , b ) 257 , c ) 58.65 ft , d ) 16 , e ) 725117481 | e | multiply(subtract(9999, const_4), 72519) | subtract(n1,const_4)|multiply(#0,n0)| | general |
a bag contains 12 red marbles . if someone were to remove 2 marbles from the bag , one at a time , and replace the first marble after it was removed , the probability that neither marble would be red is 49 / 64 . how many marbles are in the bag ? | "ok let me see if i can explain what went on in the previous post lets say i have x marbles in the bag in total - - > out of them 12 are red so the probability of pulling a non - red marble is ( x - 12 ) / x now the marble is placed back in the bag and we have x marbles again , of which again 12 are red . so the probability of pulling a non - red marble out is ( x - 12 ) / x probability theorm states that if the probability of event a occuring is m and the probability of event b occuring is n then the probability of both a and b occuring is m * n so therefore the probability of 2 non - red marbles getting pulled out is [ ( x - 12 ) / x ] * [ ( x - 12 ) / x ] this is given as 49 / 64 - - > ( x - 12 ) ^ 2 = 49 / 64 x ^ 2 square rooting u have x - 12 / x = 7 / 8 or x = 96 d" | a ) 12 , b ) 89 % , c ) 96 , d ) 10 , e ) 16 | c | divide(12, subtract(const_1, sqrt(divide(49, 64)))) | divide(n2,n3)|sqrt(#0)|subtract(const_1,#1)|divide(n0,#2)| | other |
average of 15 results is 43 . if the average of first 7 results is 41 and average of last 7 results is 45 then find the eighth result ? | option ' c ' | a ) 60 liters , b ) 2 , 3,4 , c ) 43 , d ) 30 , e ) 625 | c | subtract(multiply(15, 43), add(multiply(7, 41), multiply(7, 45))) | multiply(n0,n1)|multiply(n2,n3)|multiply(n2,n5)|add(#1,#2)|subtract(#0,#3) | general |
the difference between the compound interest compounded annually and simple interest for 2 years at 20 % per annum is rs . 432 . find the principal ? | p = 432 ( 100 / 5 ) ^ 2 = > p = 10800 answer : e | a ) $ 25 , b ) 350 , c ) $ 37.8 , d ) 2500 , e ) 10800 | e | divide(432, subtract(power(add(divide(20, const_100), const_1), 2), add(multiply(divide(20, const_100), 2), const_1))) | divide(n1,const_100)|add(#0,const_1)|multiply(n0,#0)|add(#2,const_1)|power(#1,n0)|subtract(#4,#3)|divide(n2,#5) | gain |
a sprinter starts running on a circular path of radius r metres . her average speed ( in metres / minute ) is Ο r during the first 30 seconds , Ο r / 2 during next one minute , Ο r / 4 during next 2 minutes , Ο r / 8 during next 4 minutes , and so on . what is the ratio of the time taken for the nth round to that for the previous round ? | explanation : there is more than 1 way to approach the solution ; however , i will detail the easiest way to go about it here . we want to find the ratio of time taken for nth round : time taken for ( n - 1 ) th round it will be same as finding the ratio of time taken for 2 nd round : time taken for 1 st round . 1 round = circumference of the circle = 2 Ο r 1 st round : speed = Ο r for 30 seconds . so , total distance travelled = Ο r / 2 . speed = Ο r / 2 for 1 minute . so , total distance travelled = Ο r / 2 . speed = Ο r / 4 for 2 minutes . so , total distance travelled = Ο r / 2 . speed = Ο r / 8 for 4 minutes . so , total distance travelled = Ο r / 2 . so , for a distance of 2 Ο r , time taken is 7.5 minutes . 2 nd round : speed = Ο r / 16 for 8 minutes . so , total distance travelled = Ο r / 2 . speed = Ο r / 32 for 16 minutes . so , total distance travelled = Ο r / 2 . speed = Ο r / 64 for 32 minutes . so , total distance travelled = Ο r / 2 . speed = Ο r / 128 for 64 minutes . so , total distance travelled = Ο r / 2 . so , for a distance of 2 Ο r , time taken is 120 minutes . ratio is 120 : 7.5 = 16 : 1 . answer : c | a ) 18 , b ) 16 , c ) - 4 , d ) 26 Β° , e ) $ 300,000 | b | power(2, 4) | power(n1,n2) | physics |
. a car covers a distance of 1028 km in 4 hours . find its speed ? | "1028 / 4 = 257 kmph answer : d" | a ) 16 hrs , b ) 6 , c ) 150 , d ) 2.5 , e ) 257 | e | divide(1028, 4) | divide(n0,n1)| | physics |
the cost price of a radio is rs . 2550 and it was sold for rs . 2130 , find the loss % ? | "2550 - - - - 400 100 - - - - ? = > 15 % answer : c" | a ) 15 , b ) 10750 , c ) 4 cm', ' , d ) 1 , e ) 9000 | a | multiply(divide(subtract(2550, 2130), 2550), const_100) | subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)| | gain |
if x * y = xy β 2 ( x + y ) for all integers x and y , then 4 * ( β 3 ) = | "4 * ( - 3 ) = 4 * ( - 3 ) - 2 ( 4 + ( - 3 ) ) = - 12 - 2 = - 14 option ( a )" | a ) 8 .', ' , b ) 20.9 , c ) β 14 , d ) 990 , e ) 37 | c | add(negate(multiply(add(negate(3), 4), 2)), multiply(negate(3), 4)) | negate(n2)|add(n1,#0)|multiply(n1,#0)|multiply(#1,n0)|negate(#3)|add(#2,#4)| | general |
the weight of a hollow sphere is directly dependent on its surface area . the surface area of a sphere is 4 Ο Β· r ^ 2 , where r is the radius of the sphere . if a hollow sphere of radius 0.15 cm made of a certain metal weighs 8 grams , a hollow sphere of radius 0.3 cm made of the same metal would weigh how many grams t ? | "weight directly proportional to 4 pi r ^ 2 now , 4 pi is constant , so , weight is directly proportional to r ^ 2 . when radius = 0.15 , weight = 8 , so ( 0.15 ) ^ 2 proportional to 8 ; ( 0.15 ) ^ 2 * 4 proportional to 8 * 4 , solving further ( 0.15 ) ^ 2 * 2 ^ 2 = ( 0.15 * 2 ) ^ 2 = 0.3 ^ 2 ; so answer = 32 ( b )" | a ) 10,800 , b ) 844.03 m , c ) t = 32 , d ) 420 gallons', ' , e ) 23 / 29 | c | multiply(8, 4) | multiply(n0,n3)| | geometry |
an enterprising businessman earns an income of re 5 on the first day of his business . on every subsequent day , he earns an income which is just thrice of that made on the previous day . on the 10 th day of business , he earns an income of : | 2 nd day he earns = 3 ( 2 β 5 ) 3 rd day he earns = 3 ( 3 β 5 ) on 20 th day he earns 3 ( 20 - 5 ) = 45 rupees answer : d | a ) 9 years , b ) 4 % , c ) 500 , d ) 9620 , e ) 45 | e | subtract(multiply(5, 10), 5) | multiply(n0,n1)|subtract(#0,n0) | physics |
a student chose a number , multiplied it by 3 , then subtracted 138 from the result and got 102 . what was the number he chose ? | "solution : let xx be the number he chose , then 3 β
x β 138 = 102 3 x = 240 x = 80 answer a" | a ) 120 , b ) 1173.98 , c ) 475 , d ) 80 , e ) 9000 | d | divide(add(102, 138), 3) | add(n1,n2)|divide(#0,n0)| | general |
two goods train each 500 m long , are running in opposite directions on parallel tracks . their speeds are 45 km / hr and 30 km / hr respectively find the time taken by the slower train to pass the driver of the faster one . | solution relative speed = ( 45 + 30 ) km / hr = ( 75 x 5 / 18 ) m / sec = ( 125 / 6 ) m / sec total distance covered = ( 500 + 500 ) m = 1000 m required time = ( 1000 x 6 / 125 ) sec = 48 sec answer c | a ) 48 sec , b ) 1 / 12 , c ) 50 , d ) 15 , e ) 25 | a | multiply(divide(500, divide(multiply(const_1000, add(45, 30)), const_3600)), const_2) | add(n1,n2)|multiply(#0,const_1000)|divide(#1,const_3600)|divide(n0,#2)|multiply(#3,const_2) | physics |
the population of a village is 14300 . it increases annually at the rate of 15 % p . a . what will be its population after 2 years ? | formula : ( after = 100 denominator ago = 100 numerator ) 14300 Γ 115 / 100 Γ 115 / 100 = 18911 a | a ) 18911 , b ) 257 , c ) $ 300 , d ) 24000 , e ) 22.8 | a | multiply(14300, power(add(const_1, divide(15, const_100)), 2)) | divide(n1,const_100)|add(#0,const_1)|power(#1,n2)|multiply(n0,#2) | gain |
what is the sum of all possible solutions to | x - 3 | ^ 2 + | x - 3 | = 20 ? | "first of all | x - 3 | ^ 2 = ( x - 3 ) ^ 2 , so we have : ( x - 3 ) ^ 2 + | x - 3 | = 20 . when x < 3 , x - 3 is negative , thus | x - 3 | = - ( x - 3 ) . in this case we ' ll have ( x - 3 ) ^ 2 - ( x - 3 ) = 20 - - > x = - 1 or x = 8 . discard x = 8 because it ' s not in the range we consider ( < 3 ) . when x > = 3 , x - 3 is non - negative , thus | x - 3 | = x - 3 . in this case we ' ll have ( x - 3 ) ^ 2 + ( x - 3 ) = 20 - - > x = - 2 or x = 7 . discard x = - 2 because it ' s not in the range we consider ( > = 3 ) . thus there are two solutions : x = - 1 and x = 7 - - > the sum = 6 . answer : b ." | a ) 6 % , b ) 11 , c ) 0.01616 , d ) 6 , e ) 5.6 | d | add(add(const_4, 3), subtract(3, const_4)) | add(n0,const_4)|subtract(n0,const_4)|add(#0,#1)| | general |
if x / 5 + 9 / x = 14 / 5 , what are the values of 3 x - 7 ? | i got the same thing b is the answer 8 or 20 | a ) 3 / 4 , b ) 40.46 , c ) 8 and 20 , d ) 81 , e ) 228.623 | c | add(multiply(subtract(add(subtract(9, 5), sqrt(subtract(power(subtract(9, 5), 5), multiply(5, multiply(9, 5))))), 3), const_10), subtract(subtract(subtract(9, 5), sqrt(subtract(power(subtract(9, 5), 5), multiply(5, multiply(9, 5))))), 3)) | multiply(n0,n1)|subtract(n1,n0)|multiply(n0,#0)|power(#1,n3)|subtract(#3,#2)|sqrt(#4)|add(#5,#1)|subtract(#1,#5)|subtract(#6,n4)|subtract(#7,n4)|multiply(#8,const_10)|add(#10,#9)| | general |
two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively . if they cross each other in 23 seconds , what is the ratio of their speeds ? | "speed = x and y resp length of first = 27 x length of second train = 17 y related speed = x + y time taken to cross each other = 23 s 27 x + 17 y / x + y = 23 27 x + 17 y = 23 ( x + y ) 4 x = 6 y x / y = 6 / 4 = 3 / 2 answer b" | a ) 36 , b ) 3 : 2 , c ) 28 m , d ) 51 , e ) 600 | b | divide(subtract(27, 23), subtract(23, 17)) | subtract(n0,n2)|subtract(n2,n1)|divide(#0,#1)| | physics |
a closed cylindrical tank contains 36 pie cubic feet of water and its filled to half its capacity . when the tank is placed upright on its circular base on level ground , the height of water in the tank is 4 feet . when the tank is placed on its side on level ground , what is the height , in feet , of the surface of the water above the ground ? | 36 pie cubic feet of water and its filled to half tank ' s capacity . . . volume of tank = 72 pie cubic feet height of tank = 4 * 2 = 8 feet ( since tank is placed upright on its circular base on level ground , the height of water in the tank is 4 feet . ) 72 pie = pie * r 2 * 8 r 2 = 9 r = 3 feet answer : d | a ) 1804 , b ) 1400 , c ) 3 feet', ' , d ) 9 , e ) x + 2 x + 2 | c | sqrt(divide(divide(multiply(36, const_pi), 4), const_pi)) | multiply(n0,const_pi)|divide(#0,n1)|divide(#1,const_pi)|sqrt(#2) | geometry |
the ratio of two speeds of two trains is 3 to 4 . if each of the trains slows its speed 5 km / hr , what will be the ratio of these two train speeds ? | 3 / 4 = 3 x / 4 x we need to find out ( 3 x + 5 ) / ( 4 x + 5 ) off course we can not solve this to arrive at any rational number hence e . | a ) 2599980 , b ) 11 / 48,000 , c ) it can not be determined from the information given , d ) 500 , e ) 8 | c | divide(3, 4) | divide(n0,n1) | other |
shopkeeper rise price by 33 % and gives successive discount of 10 % and 15 % . what is overall % gain or loss ? | "let d initial price be 100 33 % rise now price = 133 / 100 * 100 = 133 10 % discount then price = 133 * 90 / 100 = 119.7 15 % discount then price = 119.7 * 85 / 100 = 101.745 so gain = 101.745 - 100 = 1.745 gain % = gain * 100 / cp = = > 1.745 * 100 / 100 = 1.745 % answer : a" | a ) $ 8 , b ) 2200 , c ) 24 , d ) 30 , e ) 1.745 % | e | subtract(multiply(multiply(add(const_100, 33), divide(subtract(const_100, 10), const_100)), divide(subtract(const_100, 15), const_100)), const_100) | add(n0,const_100)|subtract(const_100,n2)|subtract(const_100,n1)|divide(#1,const_100)|divide(#2,const_100)|multiply(#0,#4)|multiply(#3,#5)|subtract(#6,const_100)| | gain |
in a certain pet shop , the ratio of dogs to cats to bunnies in stock is 3 : 7 : 13 . if the shop carries 352 dogs and bunnies total in stock , how many dogs are there ? | "let us assume the number of dogs , cats and bunnies to be 3 x , 7 x and 13 x total dogs and bunnies = 16 x . and we are given that 16 x = 352 . hence x = 22 . dogs = 3 x = 3 * 22 = 66 ( option b )" | a ) 1260 , b ) 11 , c ) 4 : 5 , d ) 66 , e ) rs . 3800 | d | multiply(divide(352, add(3, 13)), 3) | add(n0,n2)|divide(n3,#0)|multiply(n0,#1)| | other |
a box contain the number of balls which is as much times greater than 15 as much times lesser than 240 . the no . of ball is ? | "answer let the number be x . x / 15 = 240 / x x ^ 2 = 240 * 15 = 3600 x = β 3600 = 60 correct option : c" | a ) $ 875 , b ) 60 , c ) 1600 , d ) 48 , e ) 2500 | b | divide(add(240, 15), const_2) | add(n0,n1)|divide(#0,const_2)| | general |
when the price of an article was reduced by 25 % its sale increased by 80 % . what was the net effect on the sale ? | "if n items are sold for $ p each , revenue is $ np . if we reduce the price by 25 % , the new price is 0.75 p . if we increase the number sold by 80 % , the new number sold is 1.8 n . so the new revenue is ( 0.75 p ) ( 1.8 n ) = 1.35 np , which is 1.35 times the old revenue , so is 35 % greater . answer : a" | a ) 3 , b ) 16 , c ) 1.33 , d ) 35 : 44 , e ) 35 % increase | e | subtract(divide(multiply(add(80, const_100), subtract(const_100, 25)), const_100), const_100) | add(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|subtract(#3,const_100)| | gain |
5 years ago , the average age of a and b was 15 years . average age of a , b and c today is 20 years . how old will c be after 14 years ? | explanation : ( a + b ) , five years ago = ( 15 * 2 ) = 30 years . ( a + b ) , now = ( 30 + 5 * 2 ) years = 40 years . ( a + b + c ) , now = ( 20 x 3 ) years = 60 years . c , now = ( 60 - 40 ) years = 20 years . c , after 14 years = ( 20 + 14 ) years = 34 years . answer : b | a ) 16.2 % , b ) 158.256 m , c ) 3 : 4 , d ) 34 , e ) 33 % | d | add(subtract(multiply(20, const_3), add(add(multiply(15, const_2), 5), 5)), 14) | multiply(n2,const_3)|multiply(n1,const_2)|add(n0,#1)|add(n0,#2)|subtract(#0,#3)|add(n3,#4) | general |
one fourth of a solution that was 10 % sugar by weight was replaced with by a second solution resulting in a solution that was 16 percent sugar by weight . the second solution was what percent sugar by weight ? | "say the second solution ( which was 1 / 4 th of total ) was x % sugar , then 3 / 4 * 0.1 + 1 / 4 * x = 1 * 0.16 - - > x = 0.34 . alternately you can consider total solution to be 100 liters and in this case you ' ll have : 75 * 0.1 + 25 * x = 100 * 0.16 - - > x = 0.34 . answer : a ." | a ) 3 , b ) 351 , c ) 34 % , d ) 10.5 , e ) 1 / 25 | c | multiply(divide(subtract(multiply(const_100, divide(16, const_100)), multiply(subtract(const_100, multiply(divide(const_1, const_4), const_100)), divide(10, const_100))), multiply(divide(const_1, const_4), const_100)), const_100) | divide(n1,const_100)|divide(n0,const_100)|divide(const_1,const_4)|multiply(#0,const_100)|multiply(#2,const_100)|subtract(const_100,#4)|multiply(#1,#5)|subtract(#3,#6)|divide(#7,#4)|multiply(#8,const_100)| | gain |
from a pack of cards , two cards are drawn one after the other , with replacement . what is the probability that the first card is a black card and the second card is a king or queen ? | "p ( black card ) = 1 / 2 p ( king or queen ) = 2 / 13 p ( black card then a king / queen ) = 1 / 2 * 2 / 13 = 1 / 13 the answer is b ." | a ) 1 / 13 , b ) 227.04 mtrs , c ) 87 % , d ) 7 , e ) 7.2 hr | a | multiply(divide(add(multiply(const_3, const_4), const_1), const_52), divide(const_2, const_52)) | divide(const_2,const_52)|multiply(const_3,const_4)|add(#1,const_1)|divide(#2,const_52)|multiply(#3,#0)| | probability |
a man whose bowling average is 22.2 , takes 4 wickets for 36 runs and thereby decreases his average by 1.2 . the number of wickets , taken by him before his last match is : | "explanation : let the number of wickets taken before the last match is x . then , ( 22.2 x + 36 ) / ( x + 4 ) = 21 = > 22.2 x + 36 = 21 x + 84 = > 1.2 x = 48 = > x = 48 / 1.2 = 40 answer : d" | a ) $ 110 , b ) 24 , c ) 150 , d ) 7 , e ) 40 | e | divide(subtract(multiply(floor(22.2), 4), 36), subtract(22.2, floor(22.2))) | floor(n0)|multiply(n1,#0)|subtract(n0,#0)|subtract(#1,n2)|divide(#3,#2)| | general |
a can give b 100 meters start and c 120 meters start in a kilometer race . how much start can b give c in a kilometer race ? | "explanation : a runs 1000 meters while b runs 900 meters and c runs 880 meters . therefore , b runs 900 meters while c runs 880 meters . so , the number of meters that c runs when b runs 1000 meters = ( 1000 x 880 ) / 900 = 977.778 meters thus , b can give c ( 1000 - 977.77 ) = 22.22 meters start answer : c" | a ) 2700 , b ) $ 370,000 , c ) 108 , d ) 22.22 meters , e ) 54 | d | subtract(multiply(const_100, const_10), divide(multiply(multiply(const_100, const_10), subtract(multiply(const_100, const_10), 120)), subtract(multiply(const_100, const_10), 100))) | multiply(const_10,const_100)|subtract(#0,n1)|subtract(#0,n0)|multiply(#0,#1)|divide(#3,#2)|subtract(#0,#4)| | physics |
a train 450 metres long is moving at a speed of 25 kmph . it will cross a man coming from the opposite direction at a speed of 2 km per hour in : | "relative speed = ( 25 + 2 ) km / hr = 27 km / hr = ( 27 Γ 5 / 18 ) m / sec = 15 / 2 m / sec . time taken by the train to pass the man = ( 450 Γ 2 / 15 ) sec = 60 sec answer : e" | a ) 60 sec , b ) 300 , c ) 3.2 . , d ) 3024 , e ) rs . 20985 | a | multiply(const_3600, divide(divide(450, const_1000), add(25, 2))) | add(n1,n2)|divide(n0,const_1000)|divide(#1,#0)|multiply(#2,const_3600)| | physics |
a rectangular with dimensions 35 inches by 45 inches is to be divided into squares of equal size . which of the following could be a length of a side of the squares ? | you need to divide the width and length of the rectangular to equal pieces where l = w you can solve this using gcf 35 = 5 * 7 45 = 3 * 3 * 5 gcf = 5 p . s you can make squares with side of 5 answer : c | a ) 20 , b ) 2 , c ) $ 96 , d ) 5 inches', ' , e ) 62 | d | divide(subtract(45, 35), const_2) | subtract(n1,n0)|divide(#0,const_2) | geometry |
find the greatest number which leaves the same remainder when it divides 25 , 50 and 120 . | "120 - 50 = 70 50 - 25 = 25 120 - 25 = 95 the h . c . f of 25 , 70 and 95 is 5 . answer : b" | a ) 25 , b ) 20 , c ) 21600 , d ) 5 , e ) 180 ares . | d | divide(subtract(50, 25), const_2) | subtract(n1,n0)|divide(#0,const_2)| | general |
o ( x ) represents the least odd integer greater than x , whereas o ( x ) represents the greatest odd integer less than x . likewise , e ( x ) represents the least even integer greater than x , whereas e ( x ) represents the greatest even integer less than x . according to these definitions , the value of o ( 11.6 ) + e ( β 10.4 ) + o ( β 9.2 ) + e ( 9.5 ) is : | "o ( 11.6 ) + e ( β 10.4 ) + o ( β 9.2 ) + e ( 9.5 ) = 13 + ( - 10 ) + ( - 11 ) + 8 = 0 the answer is c ." | a ) 200 % , b ) 220 , c ) 3 , d ) s . 180 , e ) 0 | e | add(add(add(multiply(11.6, const_1), multiply(negate(10.4), const_1)), negate(9.2)), multiply(9.5, const_1)) | multiply(n0,const_1)|multiply(n3,const_1)|negate(n1)|negate(n2)|multiply(#2,const_1)|add(#0,#4)|add(#5,#3)|add(#6,#1)| | general |
what is the smallest positive integer k such that the product of 2205 x k is a perfect square ? | "a perfect square , is just an integer that can be written as the square of some other integer . for example 16 = 4 ^ 2 , is a perfect square . now , 2205 = 3 ^ 2 * 7 ^ 7 * 5 , so if k = 5 then 2205 k = ( 3 * 7 * 5 ) ^ 2 , which is a perfect square ( basically the least positive value of k must complete only the power of 7 to even power as powers of other primes are already even ) . answer : a ." | a ) 9 / 7 , b ) 5 , c ) 11.25 , d ) 81 , e ) 990 | b | add(const_3, const_4) | add(const_3,const_4)| | general |
the function f ( p ) represents the number of ways that prime numbers can be uniquely summed to form a certain number p such that p = a + b + c + d β¦ where those summed variables are each prime and a β€ b β€ c β€ d . . . for instance f ( 8 ) = 3 and the unique ways are 2 + 2 + 2 + 2 and 2 + 3 + 3 and 3 + 5 . what is f ( 12 ) ? | so we can start with 2 and check whether sum of two primes is primes is even . 1 ) 2 ( 6 times ) 2 ) 2 ( 3 times ) + 3 ( 2 times ) 3 ) 2 ( 2 times ) + 3 + 5 4 ) 2 + 3 + 7 5 ) 2 + 5 + 5 6 ) 3 ( 4 times ) 7 ) 5 + 7 answer : d | a ) 40 , b ) 7 , c ) 35 % , d ) 55 % , e ) 200 | b | add(divide(12, const_4), const_4) | divide(n11,const_4)|add(#0,const_4) | general |
in what time will a train 175 m long cross an electric pole , it its speed be 144 km / hr ? | "speed = 144 * 5 / 18 = 40 m / sec time taken = 175 / 40 = 4.37 sec . answer : c" | a ) 4 years , b ) 67 % , c ) 4.37 sec , d ) 5120 , e ) 6 | c | divide(175, multiply(144, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics |
a student chose a number , multiplied it by 5 , then subtracted 275 from the result and got 135 . what was the number he chose ? | "let x be the number he chose , then 5 β
x β 275 = 135 5 x = 410 x = 82 correct answer b" | a ) 110 , b ) 70 , c ) 40 , d ) 2.56 % , e ) 82 | e | divide(add(135, 275), 5) | add(n1,n2)|divide(#0,n0)| | general |
in a can , there is a mixture of milk and water in the ratio 3 : 2 . if the can is filled with an additional 6 liters of milk , the can would be full and the ratio of milk and water would become 2 : 1 . find the capacity of the can ? | "let c be the capacity of the can . ( 3 / 5 ) * ( c - 6 ) + 6 = ( 2 / 3 ) * c 9 c - 54 + 90 = 10 c c = 36 the answer is a ." | a ) 36 , b ) 4 sec , c ) 280 , d ) 36 days , e ) $ 0.40 | a | add(add(multiply(divide(multiply(1, 6), subtract(multiply(2, 2), multiply(1, 3))), 3), 6), multiply(divide(multiply(1, 6), subtract(multiply(2, 2), multiply(1, 3))), 2)) | multiply(n2,n4)|multiply(n1,n3)|multiply(n0,n4)|subtract(#1,#2)|divide(#0,#3)|multiply(n0,#4)|multiply(n1,#4)|add(n2,#5)|add(#7,#6)| | general |
susan drives from city a to city b . after two hours of driving she noticed that she covered 80 km and calculated that , if she continued driving at the same speed , she would end up been 15 minutes late . so she increased her speed by 10 km / hr and she arrived at city b 36 minutes earlier than she planned . find the distance between cities a and b . | let xx be the distance between a and b . since susan covered 80 km in 2 hours , her speed was v = 802 = 40 v = 802 = 40 km / hr . if she continued at the same speed she would be 1515 minutes late , i . e . the planned time on the road is x 40 β 1560 x 40 β 1560 hr . the rest of the distance is ( x β 80 ) ( x β 80 ) km . v = 40 + 10 = 50 v = 40 + 10 = 50 km / hr . so , she covered the distance between a and b in 2 + x β 80502 + x β 8050 hr , and it was 36 min less than planned . therefore , the planned time was 2 + x β 8050 + 36602 + x β 8050 + 3660 . when we equalize the expressions for the scheduled time , we get the equation : x 40 β 1560 = 2 + x β 8050 + 3660 x 40 β 1560 = 2 + x β 8050 + 3660 x β 1040 = 100 + x β 80 + 3050 x β 1040 = 100 + x β 80 + 3050 x β 104 = x + 505 x β 104 = x + 505 5 x β 50 = 4 x + 2005 x β 50 = 4 x + 200 x = 250 x = 250 so , the distance between cities a and b is 250 km . answer : c | a ) 250 , b ) . 2 , c ) 10.28 % , d ) 16 , e ) 16.2 % | a | add(divide(subtract(add(subtract(divide(36, const_60), divide(80, add(divide(80, const_2), 10))), const_2), divide(15, const_60)), subtract(divide(const_1, divide(80, const_2)), divide(const_1, add(divide(80, const_2), 10)))), const_100) | divide(n3,const_60)|divide(n0,const_2)|divide(n1,const_60)|add(n2,#1)|divide(const_1,#1)|divide(n0,#3)|divide(const_1,#3)|subtract(#0,#5)|subtract(#4,#6)|add(#7,const_2)|subtract(#9,#2)|divide(#10,#8)|add(#11,const_100) | physics |
a train running at the speed of 40 km / hr crosses a pole in 9 sec . what is the length of the train ? | "speed = 40 * 5 / 18 = 100 / 9 m / sec length of the train = speed * time = 100 / 9 * 9 = 100 m answer : a" | a ) 16 , b ) 66 , c ) 54 , d ) 100 m , e ) 22 | d | multiply(divide(multiply(40, const_1000), const_3600), 9) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics |
what is the remainder if 7 ^ 16 is divided by 100 ? | "7 ^ 16 can be written as ( 7 ^ 4 ) ^ 4 if we divide 7 ^ 4 by 100 the reminder is 1 so , ( 7 ^ 4 ) ^ 4 by 100 , the reminder is 1 ^ 4 = 1 answer : d" | a ) 90 , b ) 5940 , c ) 29 , d ) 2 ^ 11 - 1 , e ) 1 | e | subtract(divide(100, const_2), multiply(7, 7)) | divide(n2,const_2)|multiply(n0,n0)|subtract(#0,#1)| | general |
a large box contains 17 small boxes and each small box contains 25 chocolate bars . how many chocolate bars are in the large box ? | "the number of chocolate bars is equal to 17 * 25 = 425 correct answer c" | a ) 1 , b ) $ 32500 , c ) 18 days , d ) 425 , e ) 400 | d | multiply(17, 25) | multiply(n0,n1)| | general |
what is the product of all the possible values of x if x ^ 2 + 5 x + 6 ? | explanation : = > y = x ^ 2 + 5 x + 6 = > y = ( x + 2 ) ( x + 3 ) = > x = - 2 , x = - 3 product x = ( - 2 ) ( - 3 ) = 6 answer option 6 answer : d | a ) 3.6 km , b ) 11.25 sec , c ) 3000000 , d ) 6 , e ) 21 hours | d | divide(6, const_1) | divide(n2,const_1) | general |
jo ' s collection contains us , indian and british stamps . if the ratio of us to indian stamps is 6 to 2 and the ratio of indian to british stamps is 5 to 1 , what is the ratio of us to british stamps ? | "u / i = 6 / 2 i / b = 5 / 1 since i is multiple of both 2 ( as per first ratio ) and 5 ( as per second ratio ) so let ' s assume that i = 10 i . e . multiplying teh first ratio by 5 and second ration by 2 in each numerator and denominator then , u : i : b = 30 : 18 : 2 i . e . u : b = 30 : 2 answer : option b" | a ) 43983 , b ) 1800 , c ) 70 , d ) 5 inches', ' , e ) 30 : 2 | e | divide(multiply(6, 5), multiply(1, 2)) | multiply(n0,n2)|multiply(n1,n3)|divide(#0,#1)| | other |
a man walking at a constant rate of 9 miles per hour is passed by a woman traveling in the same direction along the same path at a constant rate of 15 miles per hour . the woman stops to wait for the man 3 minutes after passing him , while the man continues to walk at his constant rate . how many minutes must the woman wait until the man catches up ? | when the woman passes the man , they are aligned ( m and w ) . they are moving in the same direction . after 5 minutes , the woman ( w ) will be ahead the man ( m ) : m - - - - - - m - - - - - - - - - - - - - - - w w in the 5 minutes , after passing the man , the woman walks the distance mw = ww , which is 3 * 15 / 60 = 15 / 20 miles and the man walks the distance mm , which is 3 * 9 / 60 = 9 / 20 mile . the difference of 15 / 20 - 9 / 20 = 3 / 10 miles ( mw ) will be covered by the man in ( 3 / 10 ) / 9 = 1 / 30 of an hour , which is 2 minutes . answer b . | a ) 63 , b ) 2 , c ) 3200 , d ) 50', ' , e ) 450 | b | multiply(const_60, divide(multiply(divide(3, const_60), subtract(15, 9)), 9)) | divide(n2,const_60)|subtract(n1,n0)|multiply(#0,#1)|divide(#2,n0)|multiply(#3,const_60) | physics |
in a certain state , the ratio of registered republicans to registered democrats is 3 to 2 , and every registered voter is either a republican or a democrat . if 80 percent of the republicans and 20 percent of the democrats are expected to vote for candidate x , and everyone else is expected to vote for candidate y , by what percent is candidate x expected to win the election ? | "since we were expected to find a percentage figure - it thought that it might be easier to pick a ' smart number ' to represent the total number of voters ( republicans and democrats ) . therefore , i picked 100 ( as the total number of voters ) and thus 30 : 20 represents the number ratio of republicans : democrats . if 80 % of republicans ( which is ( 60 * 0.8 ) = 48 ) and 20 % of democrats ( 40 * 0.2 = 8 ) voted for candidate x , means that out of total of 100 voters ; 56 ( 48 + 8 ) voters voted for candidate x and 44 voted for candidate y . thus we can infer that candidate x is expected to win the election by 12 ( 56 - 44 ) votes . therefore candidate x is expected to win the election by ( 12 / 100 ) votes which is equivalent to 12 % . i think the answer is e ." | a ) 12 % , b ) 113 m 2 , c ) 9 / 25 , d ) 44 . , e ) 420 gallons', ' | a | multiply(divide(subtract(add(multiply(divide(20, const_100), 2), multiply(divide(80, const_100), 3)), add(subtract(3, multiply(divide(80, const_100), 3)), subtract(2, multiply(divide(20, const_100), 2)))), add(3, 2)), const_100) | add(n0,n1)|divide(n3,const_100)|divide(n2,const_100)|multiply(n1,#1)|multiply(n0,#2)|add(#3,#4)|subtract(n0,#4)|subtract(n1,#3)|add(#6,#7)|subtract(#5,#8)|divide(#9,#0)|multiply(#10,const_100)| | other |
a school has received 50 % of the amount it needs for a new building by receiving a donation of $ 400 each from people already solicited . people already solicited represent 40 % of the people from whom the school will solicit donations . how much average contribution is requited from the remaining targeted people to complete the fund raising exercise ? | "let us suppose there are 100 people . 40 % of them donated $ 16000 ( 400 * 40 ) $ 16000 is 50 % of total amount . so total amount = 16000 * 100 / 50 remaining amount is 50 % of total amount . 50 % of total amount = 16000 * ( 100 / 50 ) * ( 50 / 100 ) = 16000 this amount has to be divided by 50 ( remaining people are 50 ) so per head amount is 16000 / 50 = 320 answer : c" | a ) $ 320 , b ) 80 % , c ) 49 , d ) 6 , e ) 70 | a | divide(multiply(divide(multiply(divide(40, const_100), 400), divide(50, const_100)), divide(40, const_100)), divide(50, const_100)) | divide(n2,const_100)|divide(n0,const_100)|multiply(n1,#0)|divide(#2,#1)|multiply(#3,#0)|divide(#4,#1)| | general |
the sides of a square region , measured to the nearest centimeter , are 10 centimeters long . the least possible value of the actual area of the square region is | "though there might be some technicalities concerning the termnearest ( as 9.5 is equidistant from both 9 and 10 ) the answer still should be : 9.5 ^ 2 = 90.25 . answer : e" | a ) 90.25 sq cm , b ) 1000 m , c ) 100 days , d ) 89 % , e ) 16 / 625 | a | power(subtract(subtract(10, const_0_25), const_0_25), const_2) | subtract(n0,const_0_25)|subtract(#0,const_0_25)|power(#1,const_2)| | geometry |
p software has coding line 5 % more than n , n software has coding line 1 / 2 more than m . m software has 100 lines of coding . find p lines . | "m s / w has 100 line of code n s / w has = 100 + 100 * 1 / 2 = 150 line of code p s / w 5 % more n ' code 150 + 7.5 = 157.5 or 158 line of code answer : e" | a ) 158 , b ) 23 , c ) 1.46 % , d ) 44800 , e ) 26250 | a | add(100, divide(100, 2)) | divide(n3,n2)|add(n3,#0)| | general |
in a garden , there are 10 rows and 15 columns of mango trees . the distance between the two trees is 2 metres and a distance of one metre is left from all sides of the boundary of the garden . the length of the garden is | "explanation : each row contains 15 plants . there are 14 gapes between the two corner trees ( 14 x 2 ) metres and 1 metre on each side is left . therefore length = ( 28 + 2 ) m = 30 m . answer : b" | a ) $ 600 , b ) rs . 5.10 , c ) 30 m , d ) 16 , e ) 27 / 20 | c | add(add(multiply(subtract(15, const_1), 2), divide(10, 2)), divide(10, 2)) | divide(n0,n2)|subtract(n1,const_1)|multiply(n2,#1)|add(#0,#2)|add(#3,#0)| | physics |
if tim had lunch at $ 50.50 and he gave 10 % tip , how much did he spend ? | "the tip is 20 % of what he paid for lunch . hence tip = 20 % of 50.50 = ( 10 / 100 ) * 50.50 = $ 5.05 total spent 50.50 + 5.05 = $ 55.55 correct answer d" | a ) 38 , b ) 55 % , c ) 25 % , d ) $ 55.55 , e ) $ 500 | d | add(50.50, divide(multiply(50.50, 10), const_100)) | multiply(n0,n1)|divide(#0,const_100)|add(n0,#1)| | gain |
a 240 meter long train running at the speed of 120 kmph crosses another train running in the opposite direction at the speed of 80 kmph in 9 seconds . what is the lenght of other train . | "relative speeds = ( 120 + 80 ) km / hr = 200 km / hr = ( 200 * 5 / 18 ) m / s = ( 500 / 9 ) m / s let length of train be xm x + 240 / 9 = 500 / 9 x = 260 ans is 260 m answer : a" | a ) 5 / 1 , b ) 32.8 % , c ) 120 , d ) 260 m , e ) 70 | d | subtract(multiply(9, multiply(add(120, 80), const_0_2778)), 240) | add(n1,n2)|multiply(#0,const_0_2778)|multiply(n3,#1)|subtract(#2,n0)| | physics |
company a imported 12,000 widgets made of either brass or aluminum . the widgets are painted blue , red or green . if 10 percent of the widgets are made of brass and of those 20 percent are painted green and 40 percent are painted red how many brass widgets painted blue were imported ? | answer a . we are told that 10 % of all imported widgets are made of brass and of those , 20 % are green and 40 % are red . since we know that there are only three colors , the remaining 40 % must be blue . 40 % blue of 10 % brass widgets leads to 4 % blue brass widgets out of the total 10,550 widgets . 12,000 / 100 * 4 = 480 . answer b . | a ) 8 : 5 , b ) 480 , c ) 80 % , d ) 55 % , e ) 6 | b | multiply(multiply(multiply(multiply(divide(10, const_100), divide(40, const_100)), divide(add(10, const_2), 10)), const_100), const_100) | add(n1,const_2)|divide(n1,const_100)|divide(n3,const_100)|divide(#0,n1)|multiply(#1,#2)|multiply(#3,#4)|multiply(#5,const_100)|multiply(#6,const_100) | gain |
mr . das decided to walk down the escalator of a mall . he found that if he walks down 26 steps , he requires 30 seconds to reach the bottom . however , if he steps down 34 stair she would only require 18 seconds to get to the bottom . if the time is measured from the moment the top step begins to descend to the time he steps off the last step at the bottom , find out the height of the stair way insteps ? | here when he step down 26 steps he has 30 seconds for remaining steps . if he step down 34 stairs he has only 18 sec . 30 - 18 = 12 12 secs for 8 steps . . 18 secs for 12 steps . 12 + 34 = 46 so ans is 46 . . answer : b | a ) 6 , b ) 32.8 % , c ) 3.6 , d ) 5 / 4 , e ) 44800 | a | subtract(add(multiply(divide(subtract(34, 26), subtract(30, 18)), 30), 26), multiply(const_4, const_10)) | multiply(const_10,const_4)|subtract(n2,n0)|subtract(n1,n3)|divide(#1,#2)|multiply(n1,#3)|add(n0,#4)|subtract(#5,#0) | physics |
solution x is 30 % chemical a and 70 % chemical b by volume . solution y is 40 % chemical a and 60 % chemical b by volume . if a mixture of x and y is 36 % chemical a , what percent of the mixture is solution x ? | "the volume of the mixture be x + y . 0.3 x + 0.4 y = 0.36 ( x + y ) x = 2 y / 3 x / ( x + y ) = ( 2 y / 3 ) / ( 5 y / 3 ) = 2 / 5 = 40 % . the answer is c ." | a ) 1.6 . , b ) 40 % , c ) 240 meters , d ) 15 , e ) 600 | b | multiply(divide(divide(subtract(40, 36), subtract(36, 30)), add(divide(subtract(40, 36), subtract(36, 30)), const_1)), const_100) | subtract(n2,n4)|subtract(n4,n0)|divide(#0,#1)|add(#2,const_1)|divide(#2,#3)|multiply(#4,const_100)| | gain |
jim β s taxi service charges an initial fee of $ 2.45 at the beginning of a trip and an additional charge of $ 0.35 for each 2 / 5 of a mile traveled . what is the total charge for a trip of 3.6 miles ? | "let the fixed charge of jim β s taxi service = 2.45 $ and charge per 2 / 5 mile ( . 4 mile ) = . 35 $ total charge for a trip of 3.6 miles = 2.45 + ( 3.6 / . 4 ) * . 35 = 2.45 + 9 * . 35 = 5.6 $ answer e" | a ) $ 5.6 , b ) 25 , c ) 80 , d ) 1 : 1 , e ) 6 | a | add(2.45, multiply(0.35, divide(3.6, divide(2, 5)))) | divide(n2,n3)|divide(n4,#0)|multiply(n1,#1)|add(n0,#2)| | general |
a bowl contains equal numbers of red , orange , green , blue , and yellow candies . kaz eats all of the green candies and half of the orange ones . next , he eats half of the remaining pieces of each color . finally , he eats red and yellow candies in equal proportions until the total number of remaining candies of all colors equals 24 % of the original number . what percent of the red candies remain ? | let x be the original number of each color so there are a total of 5 x candies . kaz eats all of the green candies and half of the orange ones . there are 0 green candies and 0.5 x orange candies remaining . he eats half of the remaining pieces of each color . there are 0.25 x orange candies , and 0.5 x each of red , yellow , and blue candies . he eats red and yellow candies in equal proportions . orange + blue + red + yellow = 0.75 x + red + yellow = 1.2 x red + yellow = 0.45 x red = 0.225 x , since red = yellow . the answer is c . | a ) 43 square meters , b ) 100775 , c ) 3 % , d ) 22.5 % , e ) 308 | d | multiply(divide(divide(subtract(24, add(divide(divide(const_100, add(const_2, const_3)), const_2), divide(divide(divide(const_100, add(const_2, const_3)), const_2), const_2))), const_2), divide(const_100, add(const_2, const_3))), const_100) | add(const_2,const_3)|divide(const_100,#0)|divide(#1,const_2)|divide(#2,const_2)|add(#2,#3)|subtract(n0,#4)|divide(#5,const_2)|divide(#6,#1)|multiply(#7,const_100) | 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 : | "( c . p . of 9 balls ) - ( s . p . of 9 balls ) = ( c . p . of 5 balls ) c . p . of 4 balls = s . p . of 9 balls = rs . 720 . c . p . of 1 ball = rs . 720 / 4 = rs . 180 . answer : option e" | a ) 4 : 5 , b ) 2500 , c ) 151 / 31 , d ) 6 , e ) s . 180 | e | divide(720, subtract(9, 5)) | subtract(n0,n2)|divide(n1,#0)| | gain |
boy sells a book for rs . 630 he gets a loss of 10 % , to gain 10 % , what should be the sp ? | "cost price = 630 / 90 x 100 = 700 to gain 10 % = 700 x 10 / 100 = 70 sp = cp + gain = 700 + 70 = 770 answer : d" | a ) 10000 , b ) 2 . , c ) 770 , d ) 32 , e ) 33.3 % | c | add(divide(630, subtract(const_1, divide(10, const_100))), multiply(divide(630, subtract(const_1, divide(10, const_100))), divide(10, const_100))) | divide(n1,const_100)|divide(n2,const_100)|subtract(const_1,#0)|divide(n0,#2)|multiply(#3,#1)|add(#3,#4)| | gain |
a and b can do a piece of work in 11 days . with the help of c they finish the work in 5 days . c alone can do that piece of work in ? | "c = 1 / 5 β 1 / 11 = 6 / 55 = > 9.2 days answer : d" | a ) 10 / 3 , b ) 31 , c ) 103.4 % , d ) 9.2 days , e ) 40 % | d | inverse(subtract(5, divide(5, 11))) | divide(n1,n0)|subtract(n1,#0)|inverse(#1)| | physics |
a train 310 meters long is running with a speed of 60 kmph . in what time will it pass a man who is running at 6 kmph in the direction opposite to that in which the train is going ? | "speed of train relative to man = ( 60 + 6 ) km / hr = 66 km / hr [ 66 * 5 / 18 ] m / sec = [ 55 / 3 ] m / sec . time taken to pass the man = [ 310 * 3 / 55 ] sec = 17 sec answer : d" | a ) 3 , b ) 17 , c ) 21 , d ) 1 , e ) 4 : 5 | b | multiply(const_3600, divide(divide(310, const_1000), add(60, 6))) | add(n1,n2)|divide(n0,const_1000)|divide(#1,#0)|multiply(#2,const_3600)| | physics |
how many odd factors does 210 have ? | start with the prime factorization : 210 = 2 * 3 * 5 * 7 for odd factors , we put aside the factor of two , and look at the other prime factors . set of exponents = { 1 , 1 , 1 } plus 1 to each = { 2 , 2 , 2 } product = 2 * 2 * 2 = 8 therefore , there are 8 odd factors of 210 . in case you are curious , they are { 1 , 3 , 5 , 7 , 15 , 21 , 35 , and 105 } answer : e . | a ) 32.8 % , b ) 40 , c ) 62.5 % , d ) 8 , e ) 28.57 % | d | add(add(add(const_4, const_2), const_1), const_1) | add(const_2,const_4)|add(#0,const_1)|add(#1,const_1) | other |
sarah operated her lemonade stand monday through friday over a two week period and made a total profit of 350 dollars . on hot days she sold cups of lemonade for a price that was 25 percent higher than the regular days . each cup she sold had a total cost of 75 cents and sarah did not incur any other costs . if every day she sold exactly 32 cups and 3 of the days were hot , then what was the price of 1 cup on a hot day ? | "7 regular days - - > sales = 7 * 32 * x = 224 x ; 3 hot days - - > sales = 3 * 32 * ( 1.25 x ) = 120 x ; total sales = 224 x + 120 x = 344 x . total cost = 10 * 32 * 0.75 = 240 . profit = 344 x - 240 = 350 - - > x = 1.715 . 1.25 x = ~ 2.14 . answer : c ." | a ) 24 , b ) $ 2.14 , c ) 300 , d ) 32 kmph , e ) 240000 | b | multiply(divide(add(multiply(multiply(32, divide(75, const_100)), multiply(add(const_4, 1), const_2)), 350), add(multiply(subtract(multiply(add(const_4, 1), const_2), 3), 32), multiply(multiply(divide(add(const_100, 25), const_100), 3), 32))), divide(add(const_100, 25), const_100)) | add(n5,const_4)|add(n1,const_100)|divide(n2,const_100)|divide(#1,const_100)|multiply(n3,#2)|multiply(#0,const_2)|multiply(#4,#5)|multiply(n4,#3)|subtract(#5,n4)|add(n0,#6)|multiply(n3,#8)|multiply(n3,#7)|add(#10,#11)|divide(#9,#12)|multiply(#13,#3)| | gain |
what is the greatest prime factor of 2 ^ 8 - 1 ? | "2 ^ 8 - 1 = ( 2 ^ 4 - 1 ) ( 2 ^ 4 + 1 ) = 15 * 17 the answer is c ." | a ) 16 , b ) 40 % , c ) 12 kmph , d ) 2 , e ) 17 | e | floor(divide(2, divide(8, const_2))) | divide(n1,const_2)|divide(n0,#0)|floor(#1)| | general |
if x , y , and z are positive integers and 2 x = 3 y = 4 z , then the least possible value of x + y + z is | "given 2 x = 3 y = 5 z x + y + z in terms of x = x + ( 2 x / 3 ) + ( 2 x / 5 ) = 31 x / 15 now checking with each of the answers and see which value gives a minimum integer value . a x = 15 / 31 * 40 , not an integer b , c , d can be ruled out similarly . e is minimum value as x = 15 * 31 / 31 = 15 answer is e" | a ) 40 , b ) 48 , c ) 25 , d ) 7 , e ) 31 | e | add(subtract(divide(multiply(multiply(2, 3), 4), 2), divide(multiply(multiply(2, 3), 4), 3)), divide(multiply(multiply(2, 3), 4), 4)) | multiply(n0,n1)|multiply(n2,#0)|divide(#1,n2)|divide(#1,n0)|divide(#1,n1)|subtract(#3,#4)|add(#2,#5)| | general |
right triangle abc is to be drawn in the xy - plane so that the right angle is at a and ab is parallel to the y - axis . if the x - and y - coordinates of a , b , and c are to be integers that are consistent with the inequalities - 7 β€ x β€ 1 and 4 β€ y β€ 9 , then how many different triangles can be drawn that will meet these conditions ? | "we have the rectangle with dimensions 9 * 7 ( 9 horizontal dots and 7 vertical ) . ab is parallel to y - axis and ac is parallel to x - axis . choose the ( x , y ) coordinates for vertex a : 9 c 1 * 7 c 1 ; choose the x coordinate for vertex c ( as y coordinate is fixed by a ) : 8 c 1 , ( 9 - 1 = 8 as 1 horizontal dot is already occupied by a ) ; choose the y coordinate for vertex b ( as x coordinate is fixed by a ) : 6 c 1 , ( 7 - 1 = 6 as 1 vertical dot is already occupied by a ) . 9 c 1 * 7 c 1 * 8 c 1 * 6 c 1 = 3024 answer : d ." | a ) 174 cm , b ) 3 / 5 , c ) 3024 , d ) 50', ' , e ) 150 | c | multiply(multiply(7, subtract(7, const_1)), multiply(9, 7)) | multiply(n0,n3)|subtract(n0,const_1)|multiply(n0,#1)|multiply(#2,#0)| | geometry |
a man has $ 480 in the denominations of one - dollar , 5 - dollar notes and 10 - dollar . the number of dollars of each denomination is equal . what is the total number of dollar that he has ? | c $ 90 let number of notes of each denomination be x . then x + 5 x + 10 x = 480 16 x = 480 x = 30 . hence , total number of notes = 3 x = 90 . | a ) 40 % , b ) 7 , c ) 13 , d ) 2 , e ) 90 | e | add(divide(multiply(480, 10), const_60), 10) | multiply(n0,n2)|divide(#0,const_60)|add(n2,#1) | general |
when n is divided by 48 , the remainder is 6 . what is the remainder when 4 n is divided by 8 ? | "let n = 6 ( leaves a remainder of 6 when divided by 48 ) 4 n = 4 ( 6 ) = 24 , which leaves a remainder of 0 when divided by 8 . answer a" | a ) 100 % , b ) rs . 20985 , c ) 78 , d ) 0 , e ) 0.45 % | d | subtract(6, reminder(4, 8)) | reminder(n2,n3)|subtract(n1,#0)| | general |
find how many positive integers less than 10000 are there such thatthe sum of the digits of the no . is divisible by 3 ? | if sum of the digits is divisible by 3 , the number is divisible by 3 . therefore , required number of non - negative integers is equal to count of numbers less than 10000 which are divisible by 3 . such numbers are ( 3 , 6 , 9 , . . . , 9999 ) ( arithmetic progression with first term = 3 , last term = 9999 , common difference = 3 ) . count of such numbers = 9999 3 = 3333 99993 = 3333 but zero is also divisible by 3 . this makes our total count 3334 d | a ) 1 , b ) 40 , c ) $ 214.16 , d ) 3334 , e ) 50 % | d | add(floor(divide(10000, 3)), const_1) | divide(n0,n1)|floor(#0)|add(#1,const_1) | general |
a , b and c can do a work in 7 , 14 and 21 days respectively . they completed the work and got rs . 242 . what is the share of c ? | "the ratio of their working rates = 1 / 7 : 1 / 14 : 1 / 21 = 6 : 3 : 2 . since , they work together , the share of c = 2 / 11 * 242 = rs . 44 \ answer : b" | a ) 49 hr , b ) 127 , c ) 44 , d ) 4.2 , e ) 855 | c | multiply(242, divide(inverse(14), add(inverse(21), add(inverse(7), inverse(14))))) | inverse(n1)|inverse(n0)|inverse(n2)|add(#1,#0)|add(#3,#2)|divide(#0,#4)|multiply(n3,#5)| | physics |
the present population of a town is 4320 . population increase rate is 20 % p . a . find the population of town before 2 years ? | "p = 4320 r = 20 % required population of town = p / ( 1 + r / 100 ) ^ t = 4320 / ( 1 + 20 / 100 ) ^ 2 = 4320 / ( 6 / 5 ) ^ 2 = 3000 ( approximately ) answer is d" | a ) 6 : 15 , b ) 1 / 6 , c ) 36', ' , d ) 36.3 , e ) 3000 | e | add(4320, divide(multiply(4320, 20), const_100)) | multiply(n0,n1)|divide(#0,const_100)|add(n0,#1)| | gain |
if 12 men and 16 boys can do a piece of work in 7 days and 13 men together will 24 boys can do it in 4 days . compare the daily work done by a man with that of a boy . | "12 m + 16 b - - - - - 7 days 13 m + 24 b - - - - - - - 4 days 84 m + 112 b = 52 m + 96 b 32 m = 16 b = > 2 m = b m : b = 1 : 2 answer : b" | a ) 1 : 2 , b ) 9 , c ) $ 154.1 , d ) 7500 , e ) 32 square inches | a | divide(subtract(multiply(4, 24), multiply(7, 16)), subtract(multiply(7, 12), multiply(4, 13))) | multiply(n4,n5)|multiply(n1,n2)|multiply(n0,n2)|multiply(n3,n5)|subtract(#0,#1)|subtract(#2,#3)|divide(#4,#5)| | physics |
two persons a and b can complete a piece of work in 30 days and 45 days respectively . if they work together , what part of the work will be completed in 6 days ? | "a ' s one day ' s work = 1 / 30 b ' s one day ' s work = 1 / 45 ( a + b ) ' s one day ' s work = 1 / 30 + 1 / 45 = 1 / 18 the part of the work completed in 6 days = 6 ( 1 / 18 ) = 1 / 3 . answer c" | a ) 1 / 12 , b ) 54 min , c ) 50 % , d ) 1405.25 , e ) 1 / 3 | e | multiply(6, add(divide(const_1, 30), divide(const_1, 45))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|multiply(n2,#2)| | physics |
a hat company ships its hats , individually wrapped , in 8 - inch by 10 - inch by 12 - inch boxes . each hat is valued at $ 7.50 . if the company β s latest order required a truck with at least 432,000 cubic inches of storage space in which to ship the hats in their boxes , what was the minimum value of the order ? | "number of boxes = total volume / volume of one box = 432,000 / ( 8 * 10 * 12 ) = 450 one box costs 7.50 , so 450 box will cost = 450 * 7.5 = 3375 e is the answer" | a ) 200 , b ) 100 , c ) 46 , d ) $ 3,375 , e ) 3.5 | d | divide(multiply(divide(multiply(add(add(multiply(const_3, const_100), multiply(8, 10)), const_4), const_1000), multiply(multiply(8, 10), 12)), 7.50), const_1000) | multiply(const_100,const_3)|multiply(n0,n1)|multiply(n0,n1)|add(#0,#1)|multiply(n2,#2)|add(#3,const_4)|multiply(#5,const_1000)|divide(#6,#4)|multiply(n3,#7)|divide(#8,const_1000)| | general |
in how many ways 4 boys and 4 girls can be seated in a row so that they are alternate . | "solution : let the arrangement be , b g b g b g b g 4 boys can be seated in 4 ! ways . girl can be seated in 4 ! ways . required number of ways , = 4 ! * 4 ! = 576 . answer : option d" | a ) 576 , b ) 1 / 104 , c ) 1 / 216 , d ) 3 , e ) 6 | a | multiply(factorial(4), factorial(4)) | factorial(n0)|factorial(n1)|multiply(#0,#1)| | probability |
two men a and b start from place x walking at 4 Β½ kmph and 5 ΒΎ kmph respectively . how many km apart they are at the end of 4 Β½ hours if they are walking in the same direction ? | "rs = 5 ΒΎ - 4 Β½ = 1 ΒΌ t = 4 Β½ h . d = 5 / 4 * 9 / 2 = 45 / 8 = 5 5 / 8 km answer : c" | a ) 1.8 , b ) 5 5 / 8 km , c ) 130 cm , d ) 40 % , e ) 391 | b | add(multiply(add(4, divide(const_1, const_2)), subtract(add(5, divide(const_3, 4)), add(4, divide(const_1, const_2)))), const_2) | divide(const_1,const_2)|divide(const_3,n0)|add(n2,#0)|add(n1,#1)|add(n0,#0)|subtract(#3,#4)|multiply(#2,#5)|add(#6,const_2)| | physics |
kavi spends 50 % of his monthly salary on food and saves 80 % of the remaining amount . if his monthly salary is rs . 19000 , how much money does he save every month ? | explanation : kavi ' s monthly income = rs . 19,000 he spends 50 % on food . the total money spent on food = 50 / 100 * 19000 = rs . 9500 now , his monthly remaining income = rs . 19000 β rs . 9500 = rs . 9500 out of rs . 9500 , he saves 40 % . amount saved = 40 / 100 * 9500 = rs . 3800 answer : d | a ) $ 0.50 , b ) rs . 3800 , c ) 9 / 20 , d ) 4 , e ) 175.5 cm | b | divide(divide(multiply(divide(multiply(19000, 50), const_100), 80), const_100), const_2) | multiply(n0,n2)|divide(#0,const_100)|multiply(n1,#1)|divide(#2,const_100)|divide(#3,const_2) | gain |
a man can row upstream at 40 kmph and downstream at 52 kmph , and then find the speed of the man in still water ? | "us = 40 ds = 52 m = ( 40 + 52 ) / 2 = 46 answer : d" | a ) 80 % , b ) 44.9091 , c ) 8200 , d ) 46 , e ) 125 | d | divide(add(40, 52), const_2) | add(n0,n1)|divide(#0,const_2)| | physics |
a can do a piece of work in 10 days and b alone can do it in 20 days . how much time will both take to finish the work ? | "this question can be solved by different methods . we need to conserve time in exams so solving this problem using equations is the good idea . time taken to finish the job = xy / ( x + y ) = 10 x 20 / ( 10 + 20 ) = 200 / 30 = 6.666 days answer : c" | a ) 46 seconds , b ) 4 : 5 , c ) 6.666 , d ) 67 % , e ) 125 % | c | divide(const_1, add(divide(const_1, 10), divide(const_1, 20))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)| | physics |
a ferry can transport 100 tons of vehicles . automobiles range in weight from 1,600 to 3,000 pounds . what is the greatest number of automobiles that can be loaded onto the ferry ? | "to get maximum vehicles we must take into consideration the minimum weight i . e 1600 pounds here since , 1 ton = 2000 pounds 100 tons will be 200,000 pounds from the answer choices : let max number of vehicles be 120 total weight will be = 120 * 1600 = 192000 pounds , which is lesser than the maximum weight allowed . ans : e" | a ) 11190 , b ) 1070 , c ) 120 , d ) 227.04 mtrs , e ) $ 120 | c | divide(multiply(multiply(100, const_2), const_1000), add(add(add(add(add(add(const_1000, const_100), const_100), const_100), const_100), const_100), const_100)) | add(const_100,const_1000)|multiply(n0,const_2)|add(#0,const_100)|multiply(#1,const_1000)|add(#2,const_100)|add(#4,const_100)|add(#5,const_100)|add(#6,const_100)|divide(#3,#7)| | general |
the simple interest on rs . 10 for 4 months at the rate of 3 paise per rupeeper month is | "sol . s . i . = rs . [ 10 * 3 / 100 * 4 ] = rs . 1.20 answer a" | a ) 1.2 , b ) 33.33 % , c ) 1528 , d ) 8.12 % , e ) 17 | a | divide(multiply(multiply(10, 4), 3), const_100) | multiply(n0,n1)|multiply(n2,#0)|divide(#1,const_100)| | gain |
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 48 seconds . find the length of the faster train ? | "relative speed = ( 72 - 36 ) * 5 / 18 = 2 * 5 = 10 mps . distance covered in 27 sec = 48 * 10 = 480 m . the length of the faster train = 480 m . answer : e" | a ) 15 % , b ) 480 , c ) 0.85 , d ) 49 , e ) $ 35 | b | multiply(divide(subtract(72, 36), const_3_6), 48) | subtract(n0,n1)|divide(#0,const_3_6)|multiply(n2,#1)| | physics |
a bowl contains equal numbers of red , orange , green , blue , and yellow candies . kaz eats all of the green candies and half of the orange ones . next , he eats half of the remaining pieces of each color . finally , he eats red and yellow candies in equal proportions until the total number of remaining candies of all colors equals 25 % of the original number . what percent of the red candies remain ? | "let x be the original number of each color . kaz eats all of the green candies and half of the orange ones . there are 0 green candies and 0.5 x orange candies remaining . he eats half of the remaining pieces of each color . there are 0.25 x orange candies , and 0.5 x each of red , yellow , and blue candies . he eats red and yellow candies in equal proportions . orange + blue + red + yellow = 0.75 x + red + yellow = 1.25 x red + yellow = 0.5 x red = 0.25 x , since red = yellow . the answer is c ." | a ) 1 / 3 , b ) 14', ' , c ) 25 % , d ) 2000 , e ) 16.5 m 2 | c | multiply(divide(divide(subtract(25, add(divide(divide(const_100, add(const_2, const_3)), const_2), divide(divide(divide(const_100, add(const_2, const_3)), const_2), const_2))), const_2), divide(const_100, add(const_2, const_3))), const_100) | add(const_2,const_3)|divide(const_100,#0)|divide(#1,const_2)|divide(#2,const_2)|add(#2,#3)|subtract(n0,#4)|divide(#5,const_2)|divide(#6,#1)|multiply(#7,const_100)| | general |
what is the units digit of ( 63 ^ 4 ) ( 41 ^ 7 ) ( 99 ^ 9 ) ? | "the units digit of 63 ^ 4 is the units digit of 3 ^ 4 which is 1 . the units digit of 41 ^ 7 is the units digit of 1 ^ 7 which is 1 . the units digit of 99 ^ 9 is the units digit of 9 ^ 9 which is 9 . note the pattern : 9 ^ 1 = 9 , 9 ^ 2 = 81 , 9 ^ 3 = 729 , . . . the units digit alternates between 9 and 1 . the units digit of 1 * 1 * 9 is 9 . the answer is e ." | a ) 144 min , b ) 10780 , c ) 762 , d ) 9 , e ) 4.5 sec | d | divide(add(multiply(factorial(63), factorial(4)), multiply(factorial(63), factorial(7))), 63) | factorial(n0)|factorial(n1)|factorial(n3)|multiply(#0,#1)|multiply(#0,#2)|add(#3,#4)|divide(#5,n0)| | general |
if - 4 and - 8 are negative integers , then - 4 * - 8 + 2 is | answer : c | a ) 40 , b ) 8 , c ) 34 , d ) 10 years , e ) 8 / 25 | c | add(multiply(negate(4), negate(8)), 2) | negate(n0)|negate(n1)|multiply(#0,#1)|add(n4,#2) | general |
the maximum number of students among them 1234 pens and 874 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is : | "explanation : required number of students = h . c . f of 1234 and 874 = 2 . answer : b" | a ) 180 m , b ) 82 , c ) 2 , d ) rs . 440 , e ) 612 | c | gcd(1234, 874) | gcd(n0,n1)| | general |
a boatman selling a boat along river flow . if he sell boat in steal water at 3 m / sec and flow of river is 2 m / sec . how much time he will take to sell 100 m . | net speed = 3 + 2 = 5 m / sec distance = 100 m time = 100 / 5 = 20 sec answer d | a ) 0.225 , b ) 20 , c ) 106 , d ) 30 , e ) 39 | b | divide(100, add(3, 2)) | add(n0,n1)|divide(n2,#0) | physics |
a man can row his boat with the stream at 12 km / h and against the stream in 8 km / h . the man ' s rate is ? | explanation : ds = 12 us = 8 s = ? s = ( 12 - 8 ) / 2 = 2 kmph answer : a | a ) 600 , b ) 45 , c ) - 14 , d ) 2 kmph , e ) 60000 | d | divide(subtract(12, 8), const_2) | subtract(n0,n1)|divide(#0,const_2) | gain |
jayant opened a shop investing rs . 30,000 . madhu joined him 2 months later , investing rs . 45,000 . they earned a profit of rs . 60,000 after completion of one year . what will be madhu ' s share of profit ? | "30,000 * 12 = 45,000 * 8 1 : 1 madhu ' s share = 1 / 2 * 60,000 i . e . rs . 30,000 answer : c" | a ) rs . 30,000 , b ) 0 , c ) rs . 25,000 , d ) 1 / 12 , e ) 1750 m | a | multiply(add(multiply(multiply(multiply(const_4, 2), multiply(add(2, const_3), 2)), const_100), multiply(multiply(add(2, const_3), const_100), const_100)), divide(divide(multiply(add(2, const_3), 2), 2), multiply(const_4, const_3))) | add(n1,const_3)|multiply(n1,const_4)|multiply(const_3,const_4)|multiply(#0,n1)|multiply(#0,const_100)|divide(#3,n1)|multiply(#1,#3)|multiply(#4,const_100)|divide(#5,#2)|multiply(#6,const_100)|add(#9,#7)|multiply(#10,#8)| | gain |
if the average ( arithmetic mean ) of a and b is 120 , and the average of b and c is 150 , what is the value of a β c ? | "a + b = 240 b + c = 300 a - c = - 60 . imo option a ." | a ) β 60 , b ) rs 66.66 , c ) 25 , d ) 30 , e ) 786858 | a | subtract(multiply(150, const_2), multiply(120, const_2)) | multiply(n1,const_2)|multiply(n0,const_2)|subtract(#0,#1)| | general |
a vessel of capacity 2 litre has 25 % of alcohol and another vessel of capacity 6 litre had 30 % alcohol . the total liquid of 8 litre was poured out in a vessel of capacity 10 litre and thus the rest part of the vessel was filled with the water . what is the new concentration of mixture ? | "25 % of 2 litres = 0.5 litres 30 % of 6 litres = 1.8 litres therefore , total quantity of alcohol is 2.3 litres . this mixture is in a 10 litre vessel . hence , the concentration of alcohol in this 10 litre vessel is 23 % answer : a" | a ) 150 m , b ) 500 , c ) 23 % . , d ) 629 , e ) 0 | c | multiply(divide(add(multiply(divide(25, const_100), 2), multiply(divide(30, const_100), 6)), 10), const_100) | divide(n1,const_100)|divide(n3,const_100)|multiply(n0,#0)|multiply(n2,#1)|add(#2,#3)|divide(#4,n5)|multiply(#5,const_100)| | general |
if x / 4 years ago roger was 14 years old and x / 4 years from now he will be 4 x years old , how old will he be 3 x years from now ? | "assume the current age = a a - x / 4 = 14 ( i ) a + x / 4 = 4 x or a = 15 x / 4 ( ii ) putting the value of a from ( ii ) in ( i ) 15 x / 4 - x / 4 = 14 or 14 x / 4 = 14 therefore x = 4 and a = 15 3 x years from now , age will be 15 + 3 * 4 = 27 option e" | a ) 14 : 00 , b ) 1 / 12 , c ) 27 , d ) 129 , e ) 5625 | c | subtract(multiply(multiply(4, 3), 4), divide(subtract(14, const_1), const_2)) | multiply(n0,n4)|subtract(n1,const_1)|divide(#1,const_2)|multiply(n0,#0)|subtract(#3,#2)| | general |
jack and jill work at a hospital with 4 other workers . for an internal review , 2 of the 6 workers will be randomly chosen to be interviewed . what is the probability that jack and jill will both be chosen ? | "total number of ways to choose 2 out of 6 workers = 6 ! / 2 ! 4 ! = 15 number of ways to choose both jack and jill = 1 probability = 1 / 15 c should be the answer" | a ) 5 : 6 , b ) 16 days , c ) 4991 , d ) 110 , e ) 1 / 15 | e | inverse(divide(factorial(6), multiply(factorial(2), factorial(4)))) | factorial(n2)|factorial(n1)|factorial(n0)|multiply(#1,#2)|divide(#0,#3)|inverse(#4)| | physics |
a fashion designer sold a pair of jeans to a retail store for 40 percent more than it cost to manufacture the pair of jeans . a customer bought the pair of jeans for 35 percent more than the retailer paid for them . the price the customer paid was what percent greater than the cost of manufacturing the jeans ? | "find the product of the two increases : ( 1.4 ) * ( 1.35 ) which is 1.89 and a 89 % increase . d" | a ) 900 , b ) 210 , c ) 8 : 1 , d ) 89 % , e ) 3 : 2 | d | multiply(subtract(divide(multiply(multiply(const_100, add(const_1, divide(40, const_100))), add(const_1, divide(35, const_100))), const_100), const_1), const_100) | divide(n1,const_100)|divide(n0,const_100)|add(#0,const_1)|add(#1,const_1)|multiply(#3,const_100)|multiply(#2,#4)|divide(#5,const_100)|subtract(#6,const_1)|multiply(#7,const_100)| | gain |
what least number must be added to 9879 , so that the sum is completely divisible by 10 ? | "if we divide 9879 by 10 remainder is 9 10 - 9 = 1 answer : c" | a ) 89 % , b ) 1 , c ) $ 3.85 , d ) $ 24.88 , e ) 3 | b | subtract(multiply(add(multiply(const_4, const_10), const_2), 10), 9879) | multiply(const_10,const_4)|add(#0,const_2)|multiply(n1,#1)|subtract(#2,n0)| | general |
solution x is 10 percent alcohol by volume , and solution y is 30 percent alcohol by volume . how many milliliters of solution y must be added to 150 milliliters of solution x to create a solution that is 25 percent alcohol by volume ? | "we know that x is 10 % , y is 30 % and w . avg = 25 % . what does this mean with respect to w . avg technique ? w . avg is 1 portion away from y and 3 portion away from x so for every 1 portion of x we will have to add 3 portions of y . if x = 150 then y = 450 answer : d" | a ) 30 kmph , b ) 450 , c ) 104 meters', ' , d ) 2 : 1 , e ) 28 / 55 | b | multiply(divide(subtract(25, 10), subtract(30, 25)), 150) | subtract(n3,n0)|subtract(n1,n3)|divide(#0,#1)|multiply(n2,#2)| | general |
a cycle is bought for rs . 750 and sold for rs . 1080 , find the gain percent ? | "750 - - - - 180 100 - - - - ? = > 44 % answer : b" | a ) 28 , b ) 44 , c ) 150 , d ) 5.75 , e ) 1076 | b | multiply(divide(subtract(1080, 750), 750), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
a , b and c invested rs . 6500 , rs . 1300 and rs . 7800 respectively , in a partnership business . find the share of b in profit of rs . 11700 after a year ? | "explanation : 6500 : 1300 : 7800 5 : 1 : 6 1 / 12 * 11700 = 975 answer : d" | a ) 975 , b ) 174 , c ) 6 , d ) 36 , e ) $ 480 | a | multiply(divide(6500, add(add(6500, 1300), 7800)), 11700) | add(n0,n1)|add(n2,#0)|divide(n0,#1)|multiply(n3,#2)| | gain |
the average ( arithmetic mean ) of 16 students first quiz scores in a difficult english class is 62.5 . when one student dropped the class , the average of the remaining scores increased to 65.0 . what is the quiz score of the student who dropped the class ? | "total score of 16 students is 16 * 62.50 = 1000 total score of 15 students is 15 * 65 = 975 so , the score of the person who left is 25 ( 1000 - 975 ) answer will be ( b )" | a ) 16 , b ) 25 , c ) 50 days , d ) 6 , e ) 93 / 10 | b | subtract(multiply(16, 62.5), multiply(subtract(16, const_1), 65.0)) | multiply(n0,n1)|subtract(n0,const_1)|multiply(n2,#1)|subtract(#0,#2)| | general |
evaluate 35 % of 450 + 45 % of 350 | "explanation : = ( 35 / 100 ) * 450 + ( 45 / 100 ) * 350 = 315 option c" | a ) 7 / 20 , b ) 315 , c ) 21 , d ) 10 , e ) 31.67 % | b | divide(35, divide(450, 35)) | divide(n1,n0)|divide(n0,#0)| | gain |
when positive integer x is divided by positive integer y , the remainder is 8 . if x / y = 96.12 , what is the value of y ? | "when positive integer x is divided by positive integer y , the remainder is 8 - - > x = qy + 8 ; x / y = 96.12 - - > x = 96 y + 0.12 y ( so q above equals to 96 ) ; 0.12 y = 8 - - > y = 66.7 . answer : c ." | a ) 104 meters', ' , b ) 8.4 , c ) 0.036 , d ) 88 , e ) 66.7 | e | divide(8, subtract(96.12, floor(96.12))) | floor(n1)|subtract(n1,#0)|divide(n0,#1)| | general |
a bullock cart has to cover a distance of 80 km in 10 hrs . if it covers half of the journey in 3 / 5 th time . what should be its speed to cover the remaining distance in the time left . | a 10 kmph time left = 10 - 3 / 5 * 10 = 4 hr 10 km / h speed = 40 km / 4 hr = 10 kmph | a ) 10 kmph , b ) 100 % , c ) 2013 , d ) $ 65.00 , e ) 18 | a | divide(divide(80, const_2), subtract(10, multiply(divide(10, 5), 3))) | divide(n0,const_2)|divide(n1,n3)|multiply(n2,#1)|subtract(n1,#2)|divide(#0,#3) | physics |