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pipe a can fill the tank in 30 minutes and pipe b can empty the tank in 90 minutes . how long it will take to fill the tank if both pipes are operating together ? | work done by the leak in 1 hour = 1 / 4 - 1 / 8 = 1 / 8 the leak will empty the tank in 8 hours answer is c | a ) 10 hours , b ) 12 hours , c ) 8 hours , d ) 5 hours , e ) 15 hours | c | divide(8, const_1) | divide(n1,const_1) | physics |
an electric pump can fill a tank in 4 hours . because of a leak in the tank , it took 8 hours to fill the tank . if the tank is full , how much time will the leak take to empty it ? | "i think it should be d . i can buy 8 250 - pack for rs 22.95 * 8 = $ 183.60 now , i can buy 6 20 - pack for 3.05 * 5 = $ 18.30 now , i am left with only $ 1.15 . i can not but anything with this . hence total hotdogs = 250 * 8 + 20 * 5 = 2120" | a ) 1,108 , b ) 2,100 , c ) 2,108 , d ) 2,120 , e ) 2,256 | d | multiply(divide(203, 22.95), 250) | divide(n6,n5)|multiply(n4,#0)| | general |
at the wholesale store you can buy an 8 - pack of hot dogs for $ 1.55 , a 20 - pack for $ 3.05 , and a 250 - pack for $ 22.95 . what is the greatest number of hot dogs you can buy at this store with $ 203 ? | "both sari and ken climb in the same direction . speed of sari = 700 / 2 = 350 meters / hr ( since she covers 700 meters in 2 hrs ) speed of ken = 500 meters / hr at 8 : 00 , distance between ken and sari is 700 meters . ken needs to cover this and another 50 meters . time he will take = total distance to be covered / relative speed = ( 700 + 50 ) / ( 500 - 350 ) = 5 hrs starting from 9 : 00 , in 5 hrs , the time will be 14 : 00 answer ( c )" | a ) 13 : 00 , b ) 13 : 30 , c ) 14 : 00 , d ) 15 : 00 , e ) 15 : 30 | c | add(divide(add(700, 50), subtract(500, divide(700, const_2))), 09) | add(n4,n6)|divide(n4,const_2)|subtract(n5,#1)|divide(#0,#2)|add(n2,#3)| | physics |
sari and ken climb up a mountain . at night , they camp together . on the day they are supposed to reach the summit , sari wakes up at 07 : 00 and starts climbing at a constant pace . ken starts climbing only at 09 : 00 , when sari is already 700 meters ahead of him . nevertheless , ken climbs at a constant pace of 500 meters per hour , and reaches the summit before sari . if sari is 50 meters behind ken when he reaches the summit , at what time did ken reach the summit ? | answer : option ' d ' that is , 1 girl can do one time of the work in 3 days . therefore , 7 girls can do 7 times work in the same 3 days itself . | a ) 1 1 / 5 days , b ) 2 days , c ) 2 1 / 5 days , d ) 3 days , e ) 4 days | d | multiply(divide(3, 3), 3) | divide(n0,n0)|multiply(n0,#0) | physics |
if 3 girls can do 3 times of a particular work in 3 days , then , 7 girls can do 7 times of that work in | "principal = ( 100 x 4043.25 ) / ( 9 x 5 ) = 404325 / 45 = 8985 . answer b" | a ) 5768 , b ) 8985 , c ) 2345 , d ) 6474 , e ) 8723 | b | divide(divide(multiply(4043.25, const_100), 9), 5) | multiply(n0,const_100)|divide(#0,n1)|divide(#1,n2)| | gain |
a sum fetched a total simple interest of 4043.25 at the rate of 9 % . p . a . in 5 years . what is the sum ? | "for the 1 st no . there are 2 digits after decimal for the 2 nd no . there are 2 digits after decimal total no . of decimals = 4 req . no . of digits = ( n - 1 ) = ( 4 - 1 ) = 3 answer : d" | a ) 5 , b ) 6 , c ) 4 , d ) 3 , e ) 8 | d | divide(52.416, 0.68) | divide(n3,n4)| | general |
when 52416 is divided by 312 , the quotient is 168 . what will be the quotient when 52.416 is divided by 0.68 ? | let distance be x km travelling at 10 kmph reach at 2 pm travelling at 15 kmph reach at 12 noon = > time taken when travelling at 10 km - time taken when travelling at 15 km = 2 hrs x / 10 - x / 15 = 2 3 x - 2 x * 30 x = 60 time needed if travelled at 10 kmph = 60 / 10 = 6 hrs = > reach at 1 pm = > ( 6 - 1 ) = 5 hrs req speed = 60 / 5 = 12 kmph answer b | a ) 8 kmph , b ) 12 kmph , c ) 10 kmph , d ) 14 kmph , e ) 15 kmph | b | divide(add(15, 10), 2) | add(n1,n3)|divide(#0,n0) | physics |
arun is travelling on his cycle and has calculated to reach point a at 2 pm if he travels at 10 kmph . he will reach there at 12 noon if he travels at 15 kmph . at what speed must he travel to reach a at 1 pm ? | "( 600 * 3 * 4 ) / 100 = 72 600 + 72 = 672 answer : a" | a ) 672 , b ) 246 , c ) 258 , d ) 856 , e ) 653 | a | multiply(power(add(const_1, divide(4, const_100)), 3), 600) | divide(n3,const_100)|add(#0,const_1)|power(#1,n2)|multiply(n0,#2)| | gain |
rs . 600 amounts to rs . 900 in 3 years at simple interest . if the interest is increased by 4 % , it would amount to how much ? | "in one resolution , the distance covered by the wheel is its own circumference . distance covered in 600 resolutions . = 600 * 2 * 22 / 7 * 22.4 = 84403 cm = 844.03 m answer : b" | a ) 843.03 m , b ) 844.03 m , c ) 845.03 m , d ) 846.03 m , e ) 847.03 m | b | divide(multiply(multiply(multiply(divide(add(multiply(add(const_3, const_4), const_3), const_1), add(const_3, const_4)), 22.4), const_2), 600), const_100) | add(const_3,const_4)|multiply(#0,const_3)|add(#1,const_1)|divide(#2,#0)|multiply(n0,#3)|multiply(#4,const_2)|multiply(n1,#5)|divide(#6,const_100)| | physics |
the radius of a wheel is 22.4 cm . what is the distance covered by the wheel in making 600 resolutions . | "angle between hands of a clock when the minute hand is behind the hour hand , the angle between the two hands at m minutes past h ' o clock = 30 ( h − m / 5 ) + m / 2 degree when the minute hand is ahead of the hour hand , the angle between the two hands at m minutes past h ' o clock = 30 ( m / 5 − h ) − m / 2 degree here h = 7 , m = 30 and minute hand is behind the hour hand . hence the angle = 30 ( h − m / 5 ) + m / 2 = 30 ( 7 − 30 / 5 ) + 30 / 2 = 30 ( 7 − 6 ) + 15 = 30 × 1 + 15 = 45 ° answer is d ." | a ) 35 ° , b ) 65 ° , c ) 55 ° , d ) 45 ° , e ) 95 ° | d | divide(multiply(subtract(multiply(divide(multiply(const_3, const_4), subtract(multiply(const_3, const_4), const_1)), multiply(add(const_4, const_1), subtract(multiply(const_3, const_4), const_1))), divide(const_60, const_2)), subtract(multiply(const_3, const_4), const_1)), const_2) | add(const_1,const_4)|divide(const_60,const_2)|multiply(const_3,const_4)|subtract(#2,const_1)|divide(#2,#3)|multiply(#0,#3)|multiply(#4,#5)|subtract(#6,#1)|multiply(#7,#3)|divide(#8,const_2)| | physics |
the angle between the minute hand and the hour hand of a clock when the time is 7.30 , is | "explanation : the difference between any divisor and the corresponding remainder is 14 , l . c . m of 26 , 36,46 - 14 = 10764 - 14 = 10750 answer : option b" | a ) 10570 , b ) 10750 , c ) 17050 , d ) 10075 , e ) 10085 | b | add(46, lcm(26, 36)) | lcm(n0,n1)|add(n2,#0)| | general |
find the least number which when divided by 26 , 36 and 46 leaves the remainders 12 , 22 and 32 respectively . | according to the stem the ball can be white , green or yellow , so the probability is ( white + green + yellow ) / ( total ) = ( 50 + 25 + 10 ) / 100 = 85 / 100 = 0.85 . answer is b | a ) 0.9 , b ) 0.85 , c ) 0.6 , d ) 0.8 , e ) 0.5 | b | divide(subtract(100, add(7, 8)), 100) | add(n4,n5)|subtract(n0,#0)|divide(#1,n0) | other |
a certain bag contains 100 balls â € ” 50 white , 25 green , 10 yellow , 7 red , and 8 purple . if a ball is to be chosen at random , what is the probability that the ball will be neither red nor purple ? | length after removing pole is 12 / 3 = 4 then before removing pole is 2 ( ' coz | 2 | 2 | is | 4 | ) i . e . gap between two poles is 2 m 1 km = 1000 m then split 1000 m by 2 m = > we have 500 sections or gaps then no . of poles is 500 + 1 st pole = 501 poles therefore n = 501 . answer : b | a ) 500 , b ) 501 , c ) 502 , d ) 503 , e ) 504 | b | subtract(add(add(add(add(multiply(multiply(12, 3), const_12), const_10), multiply(const_10, const_4)), const_10), const_10), 1) | multiply(n1,n2)|multiply(const_10,const_4)|multiply(#0,const_12)|add(#2,const_10)|add(#3,#1)|add(#4,const_10)|add(#5,const_10)|subtract(#6,n0) | physics |
a 1 k . m . long wire is held by n poles . if one pole is removed , the length of the gap becomes 12 / 3 m . what is the number of poles initially ? | 10 grams of combined mixture and 40 % blue pigment means that the mixtures were mixed 50 % each . thus 5 grams a piece . out of the 5 grams of the dark blue paint , 60 % is red . therefore , 5 * . 55 = 2.75 grams of red pigment | a ) 1.5 , b ) 2.5 , c ) 3.5 , d ) 2.75 , e ) 4.5 | d | multiply(divide(55, multiply(const_100, const_2)), 10) | multiply(const_100,const_2)|divide(n1,#0)|multiply(n5,#1) | gain |
a certain deep blue paint contains 45 percent blue pigment and 55 percent red pigment by weight . a certain green paint contains 35 percent blue pigment and 65 percent yellow pigment . when these paints are mixed to produce a brown paint , the brown paint contains 40 percent blue pigment . if the brown paint weighs 10 grams , then the red pigment contributes how many grams of that weight ? | "solution angle traced by hour hand in 13 / 3 hrs = ( 360 / 12 x 13 / 3 ) ° = 130 ° angle traced by min . hand in 20 min = ( 360 / 60 x 20 ) ° = 120 ° required angle = ( 130 - 120 ) ° = 10 ° . answer c" | a ) 0 ° , b ) 5 ° , c ) 10 ° , d ) 20 ° , e ) none | c | divide(multiply(subtract(multiply(divide(multiply(const_3, const_4), subtract(multiply(const_3, const_4), const_1)), multiply(add(const_4, const_1), subtract(multiply(const_3, const_4), const_1))), divide(const_60, const_2)), subtract(multiply(const_3, const_4), const_1)), const_2) | add(const_1,const_4)|divide(const_60,const_2)|multiply(const_3,const_4)|subtract(#2,const_1)|divide(#2,#3)|multiply(#0,#3)|multiply(#4,#5)|subtract(#6,#1)|multiply(#7,#3)|divide(#8,const_2)| | physics |
the angle between the minute hand and the hour hand of a clock when the time is 4.20 , is | "p : q = 48000 : 24000 = 2 : 1 . answer : c" | a ) 2 : 6 , b ) 2 : 3 , c ) 2 : 1 , d ) 17 : 9 , e ) 17 : 4 | c | divide(add(multiply(add(add(2, const_3), const_3), multiply(add(2, const_3), 2)), add(2, const_3)), add(multiply(const_3, multiply(add(2, const_3), 2)), add(2, const_3))) | add(n2,const_3)|add(#0,const_3)|multiply(n2,#0)|multiply(#1,#2)|multiply(#2,const_3)|add(#0,#3)|add(#0,#4)|divide(#5,#6)| | gain |
p and q started a business investing rs . 48,000 and rs . 24,000 respectively . in what ratio the profit earned after 2 years be divided between p and q respectively ? | "income of 6 months = ( 6 × 95 ) – debt = 570 – debt income of the man for next 4 months = 4 × 60 + debt + 30 = 270 + debt ∴ income of 10 months = 840 average monthly income = 840 ÷ 10 = 84 answer e" | a ) 70 , b ) 72 , c ) 75 , d ) 78 , e ) 84 | e | divide(add(add(multiply(95, 6), multiply(60, 4)), 30), add(6, 4)) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|add(n4,#3)|divide(#4,#0)| | general |
the average expenditure of a labourer for 6 months was 95 and he fell into debt . in the next 4 months by reducing his monthly expenses to 60 he not only cleared off his debt but also saved 30 . his monthly income i | "on dividing 6709 by 9 , we get remainder = 4 therefore , required number to be subtracted = 4 answer : c" | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | c | subtract(6709, multiply(add(multiply(add(const_4, const_1), const_10), add(const_4, const_2)), 9)) | add(const_2,const_4)|add(const_1,const_4)|multiply(#1,const_10)|add(#0,#2)|multiply(n1,#3)|subtract(n0,#4)| | general |
the least number which must be subtracted from 6709 to make it exactly divisible by 9 is : | "first 12 min = 90 * 12 / 60 = 18 km 2 nd 12 min = 110 * 12 / 60 = 22 km 3 rd 12 min = 130 * 12 / 60 = 26 km total time 12.3 = 36 min e" | a ) 52 , b ) 48 , c ) 44 , d ) 40 , e ) 36 | e | add(add(add(12, 12), 12), 12) | add(n1,n1)|add(n1,#0)|add(n1,#1)| | physics |
xavier starts from p towards q at a speed of 90 kmph and after every 12 mins increases his speed by 20 kmph . if the distance between p and q is 61 km , then how much time does he take to cover the distance ? | "let the number be x . then 60 % of 3 / 5 of x = 18 60 / 100 * 3 / 5 * x = 18 x = ( 18 * 25 / 9 ) = 50 required number = 50 . correct option : d" | a ) 80 , b ) 100 , c ) 75 , d ) 50 , e ) none of these | d | divide(18, multiply(divide(60, const_100), divide(3, 5))) | divide(n0,const_100)|divide(n1,n2)|multiply(#0,#1)|divide(n3,#2)| | gain |
if 60 % of 3 / 5 of a number is 18 , then the number is ? | "let x be the number of liters of ethanol added to the gas tank . 0.05 ( 18 ) + x = 0.1 ( 18 + x ) 0.9 x = 1.8 - 0.9 x = 1 liter the answer is a ." | a ) 1 , b ) 1.5 , c ) 1.8 , d ) 2.4 , e ) 3 | a | divide(multiply(18, 5), subtract(const_100, 10)) | multiply(n0,n1)|subtract(const_100,n3)|divide(#0,#1)| | general |
a driver just filled the car ' s gas tank with 18 liters of gasohol , a mixture consisting of 5 % ethanol and 95 % gasoline . if the car runs best on a mixture consisting of 10 % ethanol and 90 % gasoline , how many liters of ethanol must be added into the gas tank for the car to achieve optimum performance ? | explanation : if arun doubles his speed , he needs 3 hour less . double speed means half time . hence , half of the time required by arun to cover 30 km = 3 hour i . e . , time required by arun to cover 30 km = 6 hour arun ' s speed = 30 / 6 = 5 kmph answer is b | a ) 8 kmph , b ) 5 kmph , c ) 4 kmph , d ) 7 kmph , e ) 9 kmph | b | divide(30, multiply(add(1, const_2), const_2)) | add(n2,const_2)|multiply(#0,const_2)|divide(n0,#1) | physics |
in covering a distance of 30 km , arun takes 22 hours more than anil . if arun doubles his speed , then he would take 1 hour less than anil . what is arun ' s speed ? | reduce in consumption = r / ( 100 + r ) * 100 % = 8 / 108 * 100 = 7.41 % answer is b | a ) 7.5 % , b ) 7.41 % , c ) 10.9 % , d ) 12.6 % , e ) 15 % | b | multiply(divide(divide(8, const_100), add(divide(8, const_100), const_1)), const_100) | divide(n0,const_100)|add(#0,const_1)|divide(#0,#1)|multiply(#2,const_100) | general |
in the new budget the price of wheat rose by 8 % . by how much percent must a person reduce his consumption so that his expenditure on it does not increase ? | "a 2 = 104976 = > a = 324 4 a = 1296 1296 * 12 = 15552 answer : e" | a ) 15840 , b ) 3388 , c ) 2667 , d ) 8766 , e ) 15552 | e | multiply(square_perimeter(square_edge_by_area(104976)), 12) | square_edge_by_area(n1)|square_perimeter(#0)|multiply(n0,#1)| | physics |
find the length of the wire required to go 12 times round a square field containing 104976 m 2 . | "explanation : let the numbers be 5 x , 4 x and 3 x , then , ( 5 x + 4 x + 3 x ) / 3 = 360 = > 12 x = 360 * 3 = > x = 90 largest number 5 x = 5 * 90 = 450 answer : b" | a ) 30 , b ) 450 , c ) 27 , d ) 21 , e ) 22 | b | add(multiply(multiply(5, 3), const_100), multiply(4, 3)) | multiply(n0,n2)|multiply(n1,n2)|multiply(#0,const_100)|add(#2,#1)| | general |
three numbers are in the ratio 5 : 4 : 3 and their average is 360 . the largest number is : | explanation : let c ' s investment = rs . x b ' s investment = rs . ( x - 3000 ) a ' s investment = rs . ( x - 3000 + 6000 ) = rs . ( x + 3000 ) now , ( a + b + c ) ' s investment = rs . 72000 = > x + ( x - 3000 ) + ( x + 3000 ) = 72000 = > 3 x = 72000 = > x = 24000 hence , a ' s investment = rs . 27000 b ' s investment = rs . 21000 c ' s investment = rs . 24000 ratio of the capitals of a , b and c = 27000 : 21000 : 24000 = 9 : 7 : 8 a ' s share = rs . [ ( 9 / 24 ) × 8640 ] = rs . 3240 answer : option a | a ) rs . 3240 , b ) rs . 2520 , c ) rs . 2880 , d ) rs . 3360 , e ) none of these | a | multiply(8640, divide(add(divide(subtract(72000, add(6000, 3000)), const_3), 6000), 72000)) | add(n1,n2)|subtract(n0,#0)|divide(#1,const_3)|add(n1,#2)|divide(#3,n0)|multiply(n3,#4) | general |
a , b and c started a business with a total investment of rs . 72000 . a invests rs . 6000 more than b and b invests rs . 3000 less than c . if the total profit at the end of a year is rs . 8640 , find a ' s share . | "this is how i used to calculate which i think works pretty well : if you let the average of the 20 other numbers equal a , can you write this equation for sum of the list ( s ) n + 20 a = s the question tells us that n = 4 a plug this back into the first equation and you get that the sum is 24 a 4 a + 20 a = 24 a therefore fraction e of n to the total would be 4 a / 24 a or 1 / 6 answer b" | a ) 1 / 20 , b ) 1 / 6 , c ) 1 / 5 , d ) 4 / 21 , e ) 5 / 21 | b | divide(multiply(const_1, const_1), subtract(subtract(multiply(divide(add(divide(20, 4), 21), 4), const_2), 4), const_3)) | divide(n2,n1)|multiply(const_1,const_1)|add(n0,#0)|divide(#2,n1)|multiply(#3,const_2)|subtract(#4,n1)|subtract(#5,const_3)|divide(#1,#6)| | general |
a certain list consists of 21 different numbers . if n is in the list and n is 4 times the average ( arithmetic mean ) of the other 20 numbers in the list , then n is what fraction e of the sum of the 21 numbers in the list ? | "let 4300331 - x = 2535618 then x = 4300331 - 2535618 = 1764713 answer is c" | a ) 1865113 , b ) 1775123 , c ) 1764713 , d ) 1675123 , e ) none of them | c | multiply(4300331, power(add(const_4, const_1), const_4)) | add(const_1,const_4)|power(#0,const_4)|multiply(n0,#1)| | general |
( 4300331 ) - ? = 2535618 | "e euro 2000 let the shares of a , b , c and d be euro 5 x , euro 2 x , euro 4 x and euro 3 x respectively . then , 4 x - 3 x = 1000 x = 1000 . b ' s share = euro 2 x = euro ( 2 x 1000 ) = euro 2000 ." | a ) euro 1000 , b ) euro 3000 , c ) euro 5000 , d ) euro 4000 , e ) euro 2000 | e | multiply(multiply(subtract(4, 3), 1000), 3) | subtract(n2,n3)|multiply(n4,#0)|multiply(n3,#1)| | general |
a sum of money is to be distributed among a , b , c , d in the proportion of 5 : 2 : 4 : 3 . if c gets euro 1000 more than d , what is b ' s share ? | "4 / 3 : 4 / 5 = 20 : 12 = 5 : 3 answer : d" | a ) 10 : 6 , b ) 10 : 3 , c ) 15 : 3 , d ) 5 : 3 , e ) 30 : 3 | d | divide(4, 5) | divide(n2,n3)| | other |
the simple form of the ratio 4 / 3 : 4 / 5 is ? | c . i . = [ 3000 * ( 1 + 15 / 100 ) 2 - 3000 ] = ( 3000 * 23 / 20 * 23 / 20 - 3000 ) = rs . 967.5 . sum = ( 483.75 * 100 ) / ( 5 * 6 ) = rs . 1612.5 answer : c | a ) 1525.2 , b ) 1256.3 , c ) 1612.5 , d ) 1548.5 , e ) 1254.5 | c | divide(divide(subtract(multiply(3000, power(add(const_1, divide(15, const_100)), 2)), 3000), const_2), multiply(5, divide(6, const_100))) | divide(n4,const_100)|divide(n1,const_100)|add(#0,const_1)|multiply(n0,#1)|power(#2,n3)|multiply(n2,#4)|subtract(#5,n2)|divide(#6,const_2)|divide(#7,#3) | gain |
the s . i . on a certain sum of money for 5 years at 6 % per annum is half the c . i . on rs . 3000 for 2 years at 15 % per annum . the sum placed on s . i . is ? | "if b is 1 , 3 , 5 , or 15 , then gcd of a and b is 1 , 3 , 5 , and 15 respectively . so , by poe the answer must be c . still : if b is a multiple of 18 , then a is 15 smaller than a multiple of 18 , so not a multiple of 18 , so both of them can not be divisive by 18 . answer : c ." | a ) 1 , b ) 3 , c ) 18 , d ) 15 , e ) 5 | c | add(divide(15, 20), const_2) | divide(n1,n0)|add(#0,const_2)| | general |
if a and b are positive integers , and a = 20 b - 15 , the greatest common divisor of a and b can not be | "0.1 m = 0.60 e = > e / m = 1 / 6 * 100 = 16 % so answer is e . m - # of motorists e - # of motorists exceeding speed" | a ) 10.5 % , b ) 12.5 % , c ) 15 % , d ) 22 % , e ) 16 % | e | divide(const_100, multiply(multiply(divide(10, const_100), divide(40, const_100)), const_100)) | divide(n0,const_100)|divide(n1,const_100)|multiply(#0,#1)|multiply(#2,const_100)|divide(const_100,#3)| | gain |
on a certain road 10 % of the motorists exceed the posted speed limit and receive speeding tickets , but 40 % of the motorists who exceed the posted speed limit do not receive speeding tickets . what percent of the motorists on the road exceed the posted speed limit ? | "r = 4 ï € r 2 = 16 r = 5 ï € r 2 = 25 25 ï € â € “ 9 ï € 100 - - - - ? = > 36 % . answer : d" | a ) 42 % , b ) 39 % , c ) 38 % , d ) 36 % , e ) 26 % | d | subtract(power(5, const_2), power(4, const_2)) | power(n1,const_2)|power(n0,const_2)|subtract(#0,#1)| | geometry |
the radius of the two circular fields is in the ratio 4 : 5 the area of the first field is what percent less than the area of the second ? | "s . i = ( p * r * t ) / 100 80 = 800 r / 100 r = 80 / 8 = 10 % answer a" | a ) 10 , b ) 12.5 , c ) 25 , d ) 12 , e ) 14.5 | a | multiply(divide(80, multiply(400, 2)), const_100) | multiply(n0,n2)|divide(n1,#0)|multiply(#1,const_100)| | gain |
what is rate of interest if principal . amount be 400 , simple interest 80 and time 2 year . | "sol . ( x × 4 ) = ( 0.20 × 2 ) ⇒ x = 0.4 / 4 = 0.1 . answer c" | a ) 0.2 , b ) 0.3 , c ) 0.1 , d ) 0.5 , e ) none | c | divide(multiply(0.20, 2), 4) | multiply(n0,n2)|divide(#0,n1)| | general |
if 0.20 : x : : 4 : 2 , then x is equal to | "explanation : 26 trees have 25 gaps between them , required distance ( 225 / 25 ) = 10 option b" | a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | b | divide(225, subtract(26, const_1)) | subtract(n1,const_1)|divide(n0,#0)| | physics |
along a yard 225 metres long , 26 trees are palnted at equal distances , one tree being at each end of the yard . what is the distance between two consecutive trees | let cost price = x profit = y selling price = x + y 50 y = 20 ( x + y ) 30 y = 20 x percentage profit = y / x ∗ 100 = 20 / 30 ∗ 100 = 66.667 answer = a | a ) 66.67 % , b ) 33.33 % , c ) 40 % , d ) 25 % , e ) 20 % | a | multiply(subtract(divide(50, subtract(50, 20)), const_1), const_100) | subtract(n0,n1)|divide(n0,#0)|subtract(#1,const_1)|multiply(#2,const_100) | gain |
a retailer marks her goods in such a way that the profit made by selling 50 articles is equal to the selling price of 20 articles . what is the percentage of profit made by the retailer ? | "the total number of ways to choose 2 apples is 10 c 2 = 45 the number of ways that include the spoiled apple is 9 c 1 = 9 p ( the spoiled apple is included ) = 9 / 45 = 1 / 5 the answer is e ." | a ) 2 / 9 , b ) 3 / 8 , c ) 2 / 7 , d ) 1 / 6 , e ) 1 / 5 | e | divide(choose(subtract(10, 1), 1), choose(10, 2)) | choose(n0,n2)|subtract(n0,n1)|choose(#1,n1)|divide(#2,#0)| | probability |
a basket contains 10 apples , of which 1 is spoiled and the rest are good . if we select 2 apples from the basket simultaneously and at random , what is the probability that the 2 apples selected will include the spoiled apple ? | "last term = first term + ( total no . of terms - 1 ) consecutive difference s is a set of 85 consecutive multiples of 5 . if the smallest number in s is 90 , then the greatest number in s is first term = 90 ; total terms = 85 ; difference = 5 90 + ( 84 ) 5 = 510 ans e" | a ) 158 , b ) 597 , c ) 599 , d ) 402 , e ) 510 | e | add(90, multiply(subtract(85, const_1), 5)) | subtract(n0,const_1)|multiply(n1,#0)|add(n2,#1)| | general |
s is a set of 85 consecutive multiples of 5 . if the smallest number in s is 90 , then the greatest number in s is | "a square with an area of 4 has a perimeter of 8 . for the area to be > 4 , the longer piece must be > 8 . the wire must be cut within 6 meters from either end . the probability of this is 12 / 14 = 6 / 7 . the answer is d ." | a ) 5 / 14 , b ) 3 / 14 , c ) 8 / 21 , d ) 6 / 7 , e ) 2 / 5 | d | multiply(const_2, divide(const_2, 14)) | divide(const_2,n0)|multiply(#0,const_2)| | geometry |
a 14 meter long wire is cut into two pieces . if the longer piece is then used to form a perimeter of a square , what is the probability that the area of the square will be more than 4 if the original wire was cut at an arbitrary point ? | "let the side for growing cabbages this year be x ft . thus the area is x ^ 2 . let the side for growing cabbages last year be y ft . thus , the area was y ^ 2 . the area would have increased by 191 sq ft as each cabbage takes 1 sq ft space . x ^ 2 - y ^ 2 = 191 ( x + y ) ( x - y ) = 191 191 is a prime number and thus it will be ( 96 + 95 ) * ( 96 - 95 ) . thus x = 96 and y = 95 x ^ 2 = 96 ^ 2 = 9216 the answer is c ." | a ) 7,251 , b ) 8406 , c ) 9216 , d ) 10,348 , e ) can not be determined | c | power(add(divide(191, const_2), add(const_0_25, const_0_25)), const_2) | add(const_0_25,const_0_25)|divide(n1,const_2)|add(#0,#1)|power(#2,const_2)| | geometry |
a gardener grows cabbages in her garden that is in the shape of a square . each cabbage takes 1 square feet of area in her garden . this year , she has increased her output by 191 cabbages as compared to last year . the shape of the area used for growing the cabbages has remained a square in both these years . how many cabbages did she produce this year ? | "( 3 * 8 + 2 * 4 ) : ( 4 * 8 + 5 * 4 ) 8 : 13 8 / 21 * 630 = 240 answer : a" | a ) 240 , b ) 288 , c ) 277 , d ) 877 , e ) 361 | a | multiply(divide(630, add(add(multiply(3000, 8), multiply(subtract(3000, 1000), subtract(const_12, 8))), add(multiply(4000, 8), multiply(add(4000, 1000), subtract(const_12, 8))))), add(multiply(3000, 8), multiply(subtract(3000, 1000), subtract(const_12, 8)))) | add(n1,n3)|multiply(n0,n2)|multiply(n1,n2)|subtract(n0,n3)|subtract(const_12,n2)|multiply(#3,#4)|multiply(#0,#4)|add(#1,#5)|add(#2,#6)|add(#7,#8)|divide(n5,#9)|multiply(#7,#10)| | gain |
a and b began business with rs . 3000 and rs . 4000 after 8 months , a withdraws rs . 1000 and b advances rs . 1000 more . at the end of the year , their profits amounted to rs . 630 find the share of a . | "compounded annually means that the interest is applied once per year . one can have 10 % annual interest compounded monthly - in this case 10 % / 12 would be applied each month , or 10 % annual interest compounded daily etc . with respect to the problem at hand , at the end of two years , tim would have 1,000 ( 1.10 ) ^ 2 = 1,000 ( 1.21 ) = 1,210 and lana would have 2,000 ( 1.05 ) ^ 2 = 2,000 ( 1.1025 ) = 2,205 thus , tim earned 210 dollars , while lana earned 205 dollars the difference is $ 5 and the answer is a ." | a ) $ 5 , b ) $ 15 , c ) $ 50 , d ) $ 100 , e ) $ 105 | a | subtract(subtract(multiply(1,000, power(add(const_1, divide(10, const_100)), 2)), 1,000), subtract(multiply(power(add(const_1, divide(5, const_100)), 2), 2,000), 2,000)) | divide(n1,const_100)|divide(n3,const_100)|add(#0,const_1)|add(#1,const_1)|power(#2,n4)|power(#3,n4)|multiply(n0,#4)|multiply(n2,#5)|subtract(#6,n0)|subtract(#7,n2)|subtract(#8,#9)| | gain |
on a certain day , tim invested $ 1,000 at 10 percent annual interest , compounded annually , and lana invested 2,000 at 5 percent annual interest , compounded annually . the total amount of interest earned by tim ’ s investment in the first 2 years was how much greater than the total amount of interest earned by lana ’ s investment in the first 2 years ? | at 12 : 24 - minute hand will be at 24 * 6 = 144 degrees from position of 12 . - hour hand will move by 2 * 6 = 12 degree during the same time so the difference between the two hands will be 144 - 12 = 132 degrees . answer : e | a ) 115 , b ) 120 , c ) 124 , d ) 130 , e ) 132 | e | subtract(multiply(24, multiply(const_3, const_2)), 12) | multiply(const_2,const_3)|multiply(n1,#0)|subtract(#1,n0) | physics |
what is the angle between the minute and the hour hand of the clock which shows 12 : 24 ? | "3 x ^ 2 - 1.8 x + 0.7 for x = 0.6 = 3 ( 0.6 * 0.6 ) - 3 * 0.6 * ( 0.6 ) + 0.7 = 0 + 0.7 = 0.7 answer : d" | a ) − 0.3 , b ) 0 , c ) 0.3 , d ) 0.7 , e ) 2.46 | d | subtract(multiply(divide(divide(subtract(power(3, 2), power(1.8, 0.7)), const_1000), const_1000), 3), divide(divide(subtract(power(3, 2), power(1.8, 0.7)), const_1000), const_1000)) | power(n0,n1)|power(n2,n3)|subtract(#0,#1)|divide(#2,const_1000)|divide(#3,const_1000)|multiply(n0,#4)|subtract(#5,#4)| | general |
what is the value of 3 x ^ 2 − 1.8 x + 0.7 for x = 0.6 ? | ( 7000 + 28 ) / 100 * 100 = 7028 answer : a | a ) 7028 , b ) 4028 , c ) 3128 , d ) 3256 , e ) 5264 | a | multiply(add(divide(28, 100), 70), const_100) | divide(n1,n2)|add(n0,#0)|multiply(#1,const_100) | general |
find the value of ( 70 + 28 / 100 ) × 100 | let the length of the train be l metres and speeds of the train arun and sriram be r , a and s respectively , then - - - - - - - - - - ( i ) and - - - - - - - - - ( ii ) from eq . ( i ) and ( ii ) 3 ( r - a ) = 2 ( r + k ) r = 3 a + 2 k in 30 minutes ( i . e 1800 seconds ) , the train covers 1800 r ( distance ) but the arun also covers 1800 a ( distance ) in the same time . therefore distance between arun and sriram , when the train has just crossed sriram = 1800 ( r - a ) - 24 ( a + k ) time required = = ( 3600 - 24 ) = 3576 s e | a ) 2534 , b ) 3545 , c ) 3521 , d ) 4564 , e ) 3576 | e | subtract(divide(multiply(subtract(divide(add(36, 24), subtract(36, 24)), const_1), multiply(multiply(const_10, const_3), const_60)), const_2), 24) | add(n1,n2)|multiply(const_10,const_3)|subtract(n1,n2)|divide(#0,#2)|multiply(#1,const_60)|subtract(#3,const_1)|multiply(#4,#5)|divide(#6,const_2)|subtract(#7,n2) | physics |
a train with 120 wagons crosses john who is going in the same direction , in 36 seconds . it travels for half an hour from the time it starts ove ( who is also riding on his horse ) coming from the opposite direction in 24 seconds . in how much time after the train has crossed the mike do the john meets to mike ? rtaking the john ( he is riding on the horse ) before it starts overtaking the mike | "assume : jose does 1 job in x days , so jose does 1 / x job in a day jane does 1 job in y days , so jane does 1 / y job in a day together , they does ( x + y ) / xy job in a day . this is equals to 1 / 20 . so ( x + y ) / xy = 1 / 10 10 ( x + y ) = xy next , we ' re told 1 job takes 25 days to complete if jose and jane each does half the work . so since jose does 1 job in x days , he wil need x / 2 days to do half the job . jane similarly will need y / 2 days to do the other half . x / 2 + y / 2 = 25 x + y = 50 so xy = 500 the answer choices are : 25 days 30 days 60 days 65 days 36 days from the answer choices , so i ' ll go for 25 days for jose and 20 days for jane . answer : a" | a ) 25 days , b ) 30 days , c ) 60 days , d ) 65 days , e ) 36 days | a | multiply(const_3, 10) | multiply(n0,const_3)| | physics |
working together , jose and jane can complete an assigned task in 10 days . however , if jose worked alone and complete half the work and then jane takes over the task and completes the second half of the task , the task will be completed in 25 days . how long will jose take to complete the task if he worked alone ? assume that jane is more efficient than jose | "explanation : required difference = ( 5 ( 1 / 3 ) of rs . 5000 ) - ( 3 ( 1 / 2 ) of rs . 5000 ) = ( 16 / 3 – 7 / 2 ) % of rs . 5000 = ( 11 / 6 ) x ( 1 / 100 ) x 5000 = rs . 91.66 answer d" | a ) rs . 156.66 , b ) rs . 111.23 , c ) rs . 120.66 , d ) rs . 91.66 , e ) none of these | d | subtract(multiply(add(divide(add(5, divide(1, 5)), const_100), 1), divide(5000, add(divide(add(divide(1, 2), 1), const_100), 1))), 5000) | divide(n1,n0)|divide(n1,n5)|add(n0,#0)|add(n1,#1)|divide(#2,const_100)|divide(#3,const_100)|add(n1,#4)|add(n1,#5)|divide(n6,#7)|multiply(#6,#8)|subtract(#9,n6)| | general |
if the sales tax be reduced from 5 ( 1 / 3 ) % to 3 ( 1 / 2 ) % , then what difference does it make to a person who purchases a bag with marked price of rs . 5000 ? | "option b 20 + 20 t = 50 t t = 0.6" | a ) 0.1 , b ) 0.6 , c ) 1 , d ) 1.5 , e ) 2 | b | divide(20, subtract(50, 20)) | subtract(n1,n0)|divide(n2,#0)| | physics |
two cars are traveling in the same direction along the same route . the red car travels at a constant speed of 20 miles per hour , and the black car is traveling at a constant speed of 50 miles per hour . if the red car is 20 miles ahead of the black car , how many hours will it take the black car to overtake the red car ? | "number of diagonals in any polygon can be found using this formula : n ( n - 3 ) / 2 here n = 5 no . of diagonals = 5 ( 5 - 3 ) / 2 = 5 ans a" | a ) 5 , b ) 8 , c ) 9 , d ) 10 , e ) 12 | a | multiply(subtract(multiply(const_2, const_4), const_3), divide(multiply(const_2, const_4), const_2)) | multiply(const_2,const_4)|divide(#0,const_2)|subtract(#0,const_3)|multiply(#1,#2)| | geometry |
how many internal diagonals does a pentagon ( five sided polygon ) have ? | "let the length and the width be 4 x and 3 x respectively . area = ( 4 x ) ( 3 x ) = 5808 12 x ^ 2 = 5808 x ^ 2 = 484 x = 22 the ratio of the width and the area is 3 x : 12 x ^ 2 = 1 : 4 x = 1 : 88 the answer is d ." | a ) 1 : 76 , b ) 1 : 80 , c ) 1 : 84 , d ) 1 : 88 , e ) 1 : 92 | d | divide(divide(sqrt(multiply(3, 5808)), const_2), 5808) | multiply(n2,n1)|sqrt(#0)|divide(#1,const_2)|divide(#2,n2)| | geometry |
the ratio of the length and the width of a rectangle is 4 : 3 and the area of the rectangle is 5808 sq cm . what is the ratio of the width and the area of the rectangle ? | "explanation : ( 100 + g ) / ( 100 + x ) = true measure / faulty measure x = 0 true measure = 1000 faulty measure = 990 100 + g / 100 + 0 = 1000 / 990 100 + g = 100 / 99 * 100 g = 1.01 answer : c" | a ) 1.05 , b ) 1.06 , c ) 1.01 , d ) 1.08 , e ) 1.09 | c | multiply(divide(subtract(multiply(add(add(const_4, const_1), add(const_4, const_1)), const_100), 990), 990), const_100) | add(const_1,const_4)|add(#0,#0)|multiply(#1,const_100)|subtract(#2,n0)|divide(#3,n0)|multiply(#4,const_100)| | gain |
a shopkeeper sells his goods at cost price but uses a faulty meter that weighs 990 grams . find the profit percent . | the total number of ways to choose 2 apples is 9 c 2 = 36 the number of ways that include the spoiled apple is 8 c 1 = 8 p ( the spoiled apple is included ) = 8 / 36 = 2 / 9 the answer is d . | a ) 2 / 3 , b ) 2 / 5 , c ) 2 / 7 , d ) 2 / 9 , e ) 2 / 11 | d | divide(choose(subtract(9, 1), 1), choose(9, 2)) | choose(n0,n2)|subtract(n0,n1)|choose(#1,n1)|divide(#2,#0) | probability |
a basket contains 9 apples , of which 1 is spoiled and the rest are good . if we select 2 apples from the basket simultaneously and at random , what is the probability that the 2 apples selected will include the spoiled apple ? | "think of 100 people total : from the first fact , 26 of these are women with fair hair . from the second fact , these 20 women make up 40 % of the total fair haired population . we can then make a ratio of 60 : 40 fair haired men to fair haired women . this means that ( 60 / 40 ) * 26 equals the number of fair haired men , which is 39 men with fair hair . add this 39 to the 26 women and get 65 fair haired men and women out of 100 total men and women . 65 % e" | a ) 25 , b ) 30 , c ) 50 , d ) 55 , e ) 65 | e | multiply(divide(26, 40), const_100) | divide(n0,n1)|multiply(#0,const_100)| | gain |
26 % of employees are women with fair hair . 40 % of fair - haired employees are women . what percent of employees have fair hair ? | "ratio of rates of working of a and b = 2 : 1 ratio of times taken = 1 : 2 a ' s 1 day work = 1 / 12 b ' s 1 day work = 1 / 24 a + b 1 day work = 1 / 12 + 1 / 24 = 3 / 24 = 1 / 8 a and b can finish the work in 8 days answer is c" | a ) 2 days , b ) 3 days , c ) 8 days , d ) 5 days , e ) 6 days | c | inverse(add(inverse(24), multiply(const_2, inverse(24)))) | inverse(n0)|multiply(#0,const_2)|add(#0,#1)|inverse(#2)| | physics |
a work as fast as b . if b can complete a work in 24 days independently , the number of days in which a and b can together finish the work in ? | "2 ^ 2 n + 2 ^ 2 n + 2 ^ 2 n + 2 ^ 2 n = 4 ^ 20 = > 4 x 2 ^ 2 n = 4 ^ 20 = 2 ^ 40 = > 2 ^ 2 x 2 ^ 2 n = 2 ^ 40 = > 2 ^ ( 2 n + 2 ) = 2 ^ 40 = > 2 n + 2 = 40 = > n = 19 so . answer will be c ." | a ) 3 , b ) 6 , c ) 19 , d ) 23 , e ) 24 | c | divide(subtract(multiply(20, 2), 2), 2) | multiply(n0,n9)|subtract(#0,n0)|divide(#1,n0)| | general |
if 2 ^ 2 n + 2 ^ 2 n + 2 ^ 2 n + 2 ^ 2 n = 4 ^ 20 , then n = | "diver = ( 8 * 3 ) + 3 = 27 6 * quotient = 27 quotient = 4.5 dividend = ( divisor * quotient ) + remainder dividend = ( 27 * 4.5 ) + 8 = 129.5 b" | a ) 110.6 , b ) 129.5 , c ) 130.5 , d ) 86 , e ) 88 | b | add(multiply(add(multiply(8, const_3), 3), divide(add(multiply(8, const_3), 3), 6)), 8) | multiply(n0,const_3)|add(n2,#0)|divide(#1,n1)|multiply(#1,#2)|add(n0,#3)| | general |
in a division sum , the remainder is 8 and the divisor is 6 times the quotient and is obtained by adding 3 to the thrice of the remainder . the dividend is : | "let ' s see pick 6 / 100 first then we can only pick 1 other pair from the 800 so total will be 6 / 80 * 100 simplify and you get 3 / 4000 answer is d" | a ) 3 / 40000 , b ) 1 / 3600 , c ) 9 / 2000 , d ) 3 / 4000 , e ) 1 / 15 | d | divide(1, const_3) | divide(n3,const_3)| | probability |
a certain junior class has 100 students and a certain senior class has 80 students . among these students , there are 6 siblings pairs each consisting of 1 junior and 1 senior . if 1 student is to be selected at random from each class , what is the probability that the 2 students selected will be a sibling pair ? | "let the width = w total area of the pool and deck = ( 2 w + 20 ) ( 2 w + 20 ) we can test the answer choices along with unit digit method a ) 2 feet . . . . . . . . . . . 24 * 24 has unit digit 6 . . . . . . . . . . hold b ) 2.5 feet . . . . . . . . . 25 * 25 has unit digit 5 . . . . . . . . . . eliminate c ) 3 feet . . . . . . . . . . . . 26 * 26 has unit digit 6 . . . . . . . . . . . elimate ( area is more than stipulated ) d ) 4 feet . . . . . . . . . . . . 28 * 28 has unit digit 4 . . . . . . . . . . . eliminate e ) 5 feet . . . . . . . . . . . . 30 * 30 has unit digit 0 . . . . . . . . . . . eliminate answer : a" | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | a | divide(subtract(sqrt(add(power(subtract(20, const_1), const_2), subtract(576, rectangle_area(20, 20)))), subtract(20, const_1)), const_2) | rectangle_area(n0,n1)|subtract(n1,const_1)|power(#1,const_2)|subtract(n2,#0)|add(#2,#3)|sqrt(#4)|subtract(#5,#1)|divide(#6,const_2)| | geometry |
a rectangular swimming pool is 20 feet by 20 feet . a deck that has uniform width surrounds the pool . the total area of the pool and deck is 576 square feet . what is the width of the deck ? | the percent of the budget for transportation is 100 - ( 61 + 10 + 6 + 5 + 3 ) = 15 % 100 % of the circle is 360 degrees . then ( 15 % / 100 % ) * 360 = 54 degrees the answer is c . | ['a ) 18 °', 'b ) 36 °', 'c ) 54 °', 'd ) 72 °', 'e ) 90 °'] | c | divide(multiply(const_360, subtract(const_100, add(add(add(add(61, 10), 6), 5), 3))), const_100) | add(n0,n1)|add(n2,#0)|add(n3,#1)|add(n4,#2)|subtract(const_100,#3)|multiply(#4,const_360)|divide(#5,const_100) | geometry |
a circle graph shows how the budget of a certain company was spent : 61 percent for salaries , 10 percent for research and development , 6 percent for utilities , 5 percent for equipment , 3 percent for supplies , and the remainder for transportation . if the area of each sector of the graph is proportional to the percent of the budget it represents , how many degrees of the circle are used to represent transportation ? | let a be the event ‘ the number on the card drawn is even ’ and b be the event ‘ the number on the card drawn is greater than 3 ’ . we have to find p ( a | b ) . now , the sample space of the experiment is s = { 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 } then a = { 2 , 4 , 6 , 8 , 10 } , b = { 4 , 5 , 6 , 7 , 8 , 9 , 10 } and a n b = { 4 , 6 , 8 , 10 } also p ( a ) = 5 / 2 , p ( b ) = 7 / 10 and p ( a n b ) = 4 / 10 then p ( a | b ) = p ( a n b ) / p ( b ) = ( 4 / 10 ) / ( 7 / 10 ) = 4 / 7 b ) | a ) 3 / 7 , b ) 4 / 7 , c ) 5 / 7 , d ) 7 / 11 , e ) 9 / 11 | b | multiply(divide(const_4, 10), divide(10, subtract(10, 3))) | divide(const_4,n0)|subtract(n0,n3)|divide(n0,#1)|multiply(#0,#2) | general |
10 stickers numbered 1 to 10 are placed in a bowl , mixed up thoroughly and then one sticker is drawn randomly . if it is known that the number on the drawn sticker is more than 3 , what is the probability that it is an even number ? | "a score of 58 was 2 standard deviations below the mean - - > 58 = mean - 2 d a score of 98 was 3 standard deviations above the mean - - > 98 = mean + 3 d solving above for mean w = 74 . answer : a ." | a ) 74 , b ) 76 , c ) 78 , d ) 80 , e ) 82 | a | divide(add(multiply(58, 3), multiply(98, 2)), add(2, 3)) | add(n1,n3)|multiply(n0,n3)|multiply(n1,n2)|add(#1,#2)|divide(#3,#0)| | general |
for a certain exam , a score of 58 was 2 standard deviations below mean and a score of 98 was 3 standard deviations above mean . what was the mean score w for the exam ? | "let the length and the breadth of the floor be l m and b m respectively . l = b + 200 % of b = l + 2 b = 3 b area of the floor = 450 / 3 = 150 sq m l b = 150 i . e . , l * l / 3 = 150 l 2 = 450 = > l = 21.21 answer : c" | a ) 65 , b ) 44 , c ) 21.21 , d ) 16 , e ) 14 | c | multiply(sqrt(divide(divide(450, 5), const_3)), const_3) | divide(n1,n2)|divide(#0,const_3)|sqrt(#1)|multiply(#2,const_3)| | gain |
the length of a rectangular floor is more than its breadth by 200 % . if rs . 450 is required to paint the floor at the rate of rs . 5 per sq m , then what would be the length of the floor ? | "for a , r = 8 / 2 pi . its capacity = ( 4 pi ) ^ 2 * 9 = 144 pi for b , r = 10 / pi . its capacity = ( 5 pi ) ^ 2 * 8 = 200 pi a / b = 144 pi / 200 pi = 0.72 a" | a ) 72 % , b ) 80 % , c ) 100 % , d ) 120 % , e ) 125 % | a | multiply(multiply(power(divide(8, 10), const_2), divide(9, 8)), const_100) | divide(n0,n2)|divide(n1,n3)|power(#1,const_2)|multiply(#0,#2)|multiply(#3,const_100)| | physics |
tanks a and b are each in the shape of a right circular cylinder . the interior of tank a has a height of 9 meters and a circumference of 8 meters , and the interior of tank b has a height of 8 meters and a circumference of 10 meters . the capacity of tank a is what percent of the capacity of tank b ? | "w = 2 desks t = 2.5 hrs rate of 2 carpenters = 2 × r rate = work done / time 2 xr = 2 / 2.5 r = 1 / 2.5 = 2 / 5 ( this is the rate of each carpenter ) work done by 4 carpenters in 1 hrs = 4 × rate of each carpenter x time = 4 × 2 / 5 × 1 = 1.6 desks a is the correct answer ." | a ) 1.6 . , b ) 3.6 . , c ) 4.2 . , d ) 5.5 . , e ) 6.4 | a | multiply(multiply(divide(divide(2, divide(1, 2)), 2), 4), 1) | divide(n2,n0)|divide(n0,#0)|divide(#1,n0)|multiply(n1,#2)|multiply(n2,#3)| | physics |
two carpenters , working in the same pace , can build 2 desks in two hours and a half . how many desks can 4 carpenters build in 1 hours ? | "21 trees have 20 gaps between them , required distance ( 220 / 20 ) = 11 d" | a ) 10 , b ) 12 , c ) 14 , d ) 11 , e ) 17 | d | divide(220, add(subtract(21, 2), const_1)) | subtract(n1,n2)|add(#0,const_1)|divide(n0,#1)| | physics |
220 metres long yard , 21 trees are palnted at equal distances , one tree being at each end of the yard . what is the distance between 2 consecutive trees | "number of stops in an hour : 60 / 5 = 12 distance between stops : 60 / 12 = 5 km distance between yahya ' s house and pinedale mall : 5 x 10 = 50 km imo , correct answer is ` ` d . ' '" | a ) 20 km , b ) 30 km , c ) 40 km , d ) 50 km , e ) 60 km | d | multiply(60, divide(multiply(5, 10), 60)) | multiply(n1,n2)|divide(#0,n0)|multiply(n0,#1)| | physics |
the pinedale bus line travels at an average speed of 60 km / h , and has stops every 5 minutes along its route . yahya wants to go from his house to the pinedale mall , which is 10 stops away . how far away , in kilometers , is pinedale mall away from yahya ' s house ? | let x be the distance from his house to the school . x / 10 = x / 12 + 2 6 x = 5 x + 120 x = 120 km the answer is e . | a ) 100 , b ) 105 , c ) 110 , d ) 115 , e ) 120 | e | multiply(multiply(10, 12), divide(subtract(12, 10), add(1, 1))) | add(n1,n1)|multiply(n0,n2)|subtract(n2,n0)|divide(#2,#0)|multiply(#3,#1) | physics |
a student travels from his house to school at 10 km / hr and reaches school 1 hour late . the next day he travels 12 km / hr and reaches school 1 hour early . what is the distance between his house and the school ? | "ans 25 reverse of 52 answer : b" | a ) 49 , b ) 25 , c ) 36 , d ) 64 , e ) 56 | b | multiply(52, divide(43, 34)) | divide(n0,n1)|multiply(n2,#0)| | general |
43 : 34 : : 52 : ? | "speed of the first bullet train = 120 / 10 m / sec = 12 m / sec . speed of the second bullet train = 120 / 30 m / sec = 4 m / sec . relative speed = ( 12 + 4 ) = 16 m / sec . required time = ( 120 + 120 ) / 16 sec = 15 sec . d" | a ) 13 sec . , b ) 14 sec . , c ) 12 sec . , d ) 15 sec . , e ) 19 sec . | d | divide(add(120, 120), add(speed(120, 10), speed(120, 30))) | add(n2,n2)|speed(n2,n0)|speed(n2,n1)|add(#1,#2)|divide(#0,#3)| | physics |
two bullet trains of equal lengths take 10 seconds and 30 seconds respectively to cross a telegraph post . if the length of each bullet train be 120 metres , in what time ( in seconds ) will they cross each other travelling in opposite direction ? | ( 1 / 2 + 1 / 7 ) t = 2 t = 28 / 9 answer : a | a ) 28 / 9 , b ) 4 / 3 , c ) 15 / 8 , d ) 9 / 4 , e ) 15 / 4 | a | multiply(divide(const_1, add(divide(const_1, 2), divide(const_1, 7))), const_2) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)|multiply(#3,const_2) | physics |
kathleen can paint a room in 2 hours , and anthony can paint an identical room in 7 hours . how many hours would it take kathleen and anthony to paint both rooms if they work together at their respective rates ? | "16 students are interested , 8 are not interested prob = 1 - 8 c 2 / 24 c 2 = 1 - ( 8 * 7 / ( 24 * 23 ) ) = 1 - 7 / 69 = 62 / 69 answer : d" | a ) 1 / 14 , b ) 4 / 49 , c ) 2 / 7 , d ) 62 / 69 , e ) 13 / 14 | d | divide(subtract(choose(24, const_2), choose(subtract(24, multiply(24, divide(4, 6))), const_2)), choose(24, const_2)) | choose(n2,const_2)|divide(n0,n1)|multiply(n2,#1)|subtract(n2,#2)|choose(#3,const_2)|subtract(#0,#4)|divide(#5,#0)| | gain |
according to a recent student poll , 4 / 6 out of 24 members of the finance club are interested in a career in investment banking . if two students are chosen at random , what is the probability that at least one of them is interested in investment banking ? | "let the two positive numbers be 5 x and 8 x respectively . 8 x - 5 x = 15 3 x = 15 = > x = 5 = > smaller number = 5 x = 25 . answer : a" | a ) 25 , b ) 66 , c ) 88 , d ) 89 , e ) 53 | a | divide(multiply(15, 5), const_4) | multiply(n0,n2)|divide(#0,const_4)| | other |
there are two positive numbers in the ratio 5 : 8 . if the larger number exceeds the smaller by 15 , then find the smaller number ? | "800 score official solution : the first step to solving this problem is to actually graph the two lines . the lines intersect at the point ( - 5 , - 5 ) and form a right triangle whose base length and height are both equal to 4 . as you know , the area of a triangle is equal to one half the product of its base length and height : a = ( 1 / 2 ) bh = ( 1 / 2 ) ( 4 × 4 ) = 25 / 2 ; so z = 25 / 2 . the next step requires us to find the length of a side of a cube that has a face area equal to 25 / 2 . as you know the 6 faces of a cube are squares . so , we can reduce the problem to finding the length of the side of a square that has an area of 25 / 2 . since the area of a square is equal to s ² , where s is the length of one of its side , we can write and solve the equation s ² = 25 / 2 . clearly s = √ 25 / 2 = 5 / √ 2 , oranswer choice ( b ) ." | a ) 16 , b ) 5 / √ 2 , c ) 8 , d ) 2 √ 2 , e ) ( √ 2 ) / 3 | b | sqrt(divide(multiply(5, 5), const_2)) | multiply(n0,n0)|divide(#0,const_2)|sqrt(#1)| | general |
the two lines y = x and x = - 5 intersect on the coordinate plane . if z represents the area of the figure formed by the intersecting lines and the x - axis , what is the side length of a cube whose surface area is equal to 6 z ? | let no . of bicycles be x & no . of wagons be y so , 2 x + 4 y = 190 by solving , we get no . of bicycles = 39 ( wheels = > 2 * 39 = 78 ) no . of wagons = 28 ( wheels = > 4 * 28 = 112 ) answer : e | a ) 35 , b ) 36 , c ) 37 , d ) 38 , e ) 39 | e | multiply(divide(190, add(multiply(4, const_2), const_2)), const_2) | multiply(n0,const_2)|add(#0,const_2)|divide(n1,#1)|multiply(#2,const_2) | physics |
in school there are some bicycles and 4 wheeler wagons . one tuesday there are 190 wheels in the campus . how many bicycles are there ? | "total expenditure = 40 + 25 + 15 + 5 = 85 % saving = ( 100 - 85 ) = 15 % 15 / 100 × salary = 1500 , salary = 10000 rs . answer : d" | a ) 4000 , b ) 6000 , c ) 8000 , d ) 10000 , e ) none of these | d | divide(multiply(1500, const_100), 5) | multiply(n4,const_100)|divide(#0,n3)| | gain |
a person spends 40 % of his salary on food , 25 % on house rent , 15 % on entertainment and 5 % on conveyance . if his savings at the end of the month is rs . 1500 , then his salary per month in rupees is : | "the percent of tagged fish in the second catch is 2 / 90 * 100 = 2.22 % . we are told that 2.22 % approximates the percent of tagged fish in the pond . since there are 90 tagged fish , then we have 0.022 x = 90 - - > x = 4,091 . answer : e ." | a ) 400 , b ) 625 , c ) 1250 , d ) 2500 , e ) 4091 | e | divide(90.00001, divide(2, 90)) | divide(n2,n1)|divide(n0,#0)| | gain |
in a certain pond , 90.00001 fish were caught , tagged , and returned to the pond . a few days later , 90 fish were caught again , of which 2 were found to have been tagged . if the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond , what ` s the approximate number of fish in the pond ? | "good question . so we have a garden where all the flowers have two properties : color ( green or yellow ) and shape ( straight or curved ) . we ' re told that 1 / 5 of the garden is green , so , since all the flowers must be either green or yellow , we know that 4 / 5 are yellow . we ' re also told there is an equal probability of straight or curved , 1 / 2 . we want to find out the probability of something being yellow and straight , pr ( yellow and straight ) . so if we recall , the probability of two unique events occurring simultaneously is the product of the two probabilities , pr ( a and b ) = p ( a ) * p ( b ) . so we multiply the two probabilities , pr ( yellow ) * pr ( straight ) = 4 / 5 * 1 / 2 = 2 / 5 , or c ." | a ) 1 / 7 , b ) 1 / 8 , c ) 2 / 5 , d ) 3 / 4 , e ) 7 / 8 | c | multiply(subtract(1, divide(1, 5)), divide(1, 2)) | divide(n2,n3)|divide(n0,n1)|subtract(n2,#1)|multiply(#0,#2)| | probability |
in a garden , there are yellow and green flowers which are straight and curved . if the probability of picking a green flower is 1 / 5 and picking a straight flower is 1 / 2 , then what is the probability of picking a flower which is yellow and straight | "100 * ( 20 / 100 ) = 20 - - - 2 ? - - - 1 = > rs . 10 100 - - - 80 ? - - - 10 = > rs . 12.5 answer : c" | a ) 12.6 , b ) 12.1 , c ) 12.5 , d ) 12.4 , e ) 12.7 | c | multiply(divide(divide(multiply(divide(20, const_100), 100), 2), multiply(divide(20, const_100), 100)), const_100) | divide(n0,const_100)|multiply(n2,#0)|divide(#1,n1)|divide(#2,#1)|multiply(#3,const_100)| | gain |
a reduction of 20 % in the price of salt enables a lady to obtain 2 kgs more for rs . 100 , find the original price per kg ? | "t = 110 / 36 * 18 / 5 = 11 sec answer : d" | a ) 6 sec , b ) 7 sec , c ) 8 sec , d ) 11 sec , e ) 2 sec | d | divide(110, multiply(36, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics |
in what time will a railway train 110 m long moving at the rate of 36 kmph pass a telegraph post on its way ? | "a = 7 x , b = 7 z therefore ( 7 x . 10 + 7 z ) - ( 10 x + z ) = ( 7 - 1 ) ( 10 x + z ) = 6 . ( 10 x + z ) number should be divisible by 6 d" | a ) 30 , b ) 36 , c ) 42 , d ) 63 , e ) 66 | d | add(add(subtract(add(multiply(7, 7), multiply(7, 10)), add(multiply(7, 10), 7)), 10), const_3) | multiply(n1,n1)|multiply(n1,n4)|add(#0,#1)|add(n1,#1)|subtract(#2,#3)|add(n4,#4)|add(#5,const_3)| | general |
x , a , z , and b are single digit positive integers . x = 1 / 7 a . z = 1 / 7 b . ( 10 a + b ) – ( 10 x + z ) could not equal | "number of students in group a = 45 students who forget homework in group a = 20 % of 45 = 9 students number of students in group b = 55 students who forget homework in group b = 40 % of 55 = 22 students total number of students = 45 + 55 = 100 students who forgot homework = 9 + 22 = 31 students percentage of students who forget homework = 31 / 100 * 100 = 31 % answer : d" | a ) 23 % , b ) 25 % , c ) 29 % , d ) 31 % , e ) 36 % | d | multiply(divide(add(divide(multiply(45, 20), const_100), divide(multiply(55, 40), const_100)), add(45, 55)), const_100) | add(n0,n1)|multiply(n0,n2)|multiply(n1,n3)|divide(#1,const_100)|divide(#2,const_100)|add(#3,#4)|divide(#5,#0)|multiply(#6,const_100)| | gain |
there are two groups of students in the sixth grade . there are 45 students in group a , and 55 students in group b . if , on a particular day , 20 % of the students in group a forget their homework , and 40 % of the students in group b forget their homework , then what percentage of the sixth graders forgot their homework ? | "production during these 5 days = total production in a month - production in first 25 days . = 30 x 43 - 25 x 50 = 40 ∴ average for last 5 days = 40 / 5 = 8 c" | a ) 20 , b ) 36 , c ) 8 , d ) 50 , e ) 59 | c | divide(subtract(multiply(add(25, 5), 43), multiply(25, 50)), 5) | add(n1,n2)|multiply(n0,n1)|multiply(n3,#0)|subtract(#2,#1)|divide(#3,n2)| | general |
in a factory , an average of 50 tv ' s are produced per day for the fist 25 days of the months . a few workers fell ill for the next 5 days reducing the daily avg for the month to 43 sets / day . the average production per day for day last 5 days is ? | solution : 10 v 2 area of square = 1 / 2 * ( length of diagonal ) 2 area of square 1 = * ( 5 v 2 ) 2 = 25 area of square 2 = 4 * 25 = 100 length of diagonal of square 2 = v 2 * area = v 2 * 100 = v 200 = 10 v 2 cm answer is c | ['a ) 20 v 2', 'b ) 10', 'c ) 10 v 2', 'd ) 20', 'e ) 25'] | c | multiply(const_10, sqrt(2)) | sqrt(n2)|multiply(#0,const_10) | geometry |
what is the length of the diagonal of a square whose area is 4 times of another square with diagonal as 5 v 2 cm ? | "let t be the tier price , p be total price = 12000 per the given conditions : 0.12 t + 0.08 ( p - t ) = 1440 - - - - > t = 12000 . d is the correct answer ." | a ) $ 1600 , b ) $ 6000 , c ) $ 6050 , d ) $ 1200 , e ) $ 8000 | d | divide(subtract(1440, multiply(multiply(multiply(const_3, multiply(const_2, const_3)), const_1000), divide(8, const_100))), subtract(divide(12, const_100), divide(8, const_100))) | divide(n1,const_100)|divide(n0,const_100)|multiply(const_2,const_3)|multiply(#2,const_3)|subtract(#1,#0)|multiply(#3,const_1000)|multiply(#0,#5)|subtract(n3,#6)|divide(#7,#4)| | general |
country c imposes a two - tiered tax on imported cars : the first tier imposes a tax of 12 % of the car ' s price up to a certain price level . if the car ' s price is higher than the first tier ' s level , the tax on the portion of the price that exceeds this value is 8 % . if ron imported a $ 12,000 imported car and ended up paying $ 1440 in taxes , what is the first tier ' s price level ? | "am of x , x + 2 , and x + 4 = x + ( x + 2 ) + ( x + 4 ) / 3 = 3 x + 6 / 3 = x + 2 given that x + 2 = 73 x = 71 answer : a" | a ) 71 , b ) 75 , c ) 85 , d ) 83 , e ) 82 | a | subtract(multiply(4, const_2), multiply(2, const_2)) | multiply(n1,const_2)|multiply(n0,const_2)|subtract(#0,#1)| | general |
if the average ( arithmetic mean ) of x , x + 2 , and x + 4 is 73 , what is the value of x ? | "1904 = p [ 1 + ( 5 * 12 / 5 ) / 100 ] p = 1700 . answer : a" | a ) 1700 , b ) 2777 , c ) 2889 , d ) 27670 , e ) 2771 | a | divide(1904, add(divide(multiply(divide(add(multiply(2, 5), 2), 5), 5), const_100), const_1)) | multiply(n1,n3)|add(n1,#0)|divide(#1,n3)|multiply(n0,#2)|divide(#3,const_100)|add(#4,const_1)|divide(n4,#5)| | general |
find the principle on a certain sum of money at 5 % per annum for 2 2 / 5 years if the amount being rs . 1904 ? | "a 11 ( x + y ) â € “ ( y + z ) = 11 x â € “ z = 11" | a ) 11 , b ) 15 , c ) 12 , d ) 17 , e ) 19 | a | divide(11, const_1) | divide(n0,const_1)| | general |
the overall age of x and y is 11 year greater than the overall age of y and z . z is how many decades younger that x ? | "( place value of 5 ) - ( face value of 5 ) = ( 50000 - 5 ) = 49995 answer : option d" | a ) 973 , b ) 6973 , c ) 5994 , d ) 49995 , e ) none of these | d | subtract(multiply(const_10, 5), 5) | multiply(n0,const_10)|subtract(#0,n0)| | general |
the difference between the place value and the face value of 5 in the numeral 856973 is | "xyt / ( x + y ) x = 5 , y = 10 , t = 3 5 * 10 * 3 / 5 + 10 = 150 / 50 = 10 answer : c" | a ) 15 , b ) 18 , c ) 10 , d ) 12 , e ) 14 | c | divide(multiply(multiply(5, 10), 3), multiply(5, 10)) | multiply(n0,n1)|multiply(n2,#0)|divide(#1,#0)| | physics |
aaron will jog from home at 5 miles per hour and then walk back home by the same route at 10 miles per hour . how many miles from home can aaron jog so that he spends a total of 3 hours jogging and walking ? | "solution correct sum = ( 36 x 50 + 47 - 23 ) = 1824 . â ˆ ´ correct mean = 1824 / 50 = 36.48 . answer d" | a ) 35.24 , b ) 36.16 , c ) 36.22 , d ) 36.48 , e ) none | d | divide(add(multiply(36, 50), subtract(subtract(50, const_2), 23)), 50) | multiply(n0,n1)|subtract(n0,const_2)|subtract(#1,n3)|add(#0,#2)|divide(#3,n0)| | general |
the mean of 50 observations was 36 . it was found later that an observation 47 was wrongly taken as 23 . the corrected new mean is | "let the length and the breadth of the floor be l m and b m respectively . l = b + 200 % of b = l + 2 b = 3 b area of the floor = 324 / 3 = 108 sq m l b = 108 i . e . , l * l / 3 = 108 l 2 = 324 = > l = 18 . answer : c" | a ) 27 m , b ) 24 m , c ) 18 m , d ) 21 m , e ) none of these | c | multiply(sqrt(divide(divide(324, 3), const_3)), const_3) | divide(n1,n2)|divide(#0,const_3)|sqrt(#1)|multiply(#2,const_3)| | gain |
the length of a rectangular floor is more than its breadth by 200 % . if rs . 324 is required to paint the floor at the rate of rs . 3 per sq m , then what would be the length of the floor ? | "90 % 104 % - - - - - - - - 14 % - - - - 182 100 % - - - - ? = > rs : 1300 answer : d" | a ) s : 1000 , b ) s : 1067 , c ) s : 1278 , d ) s : 1300 , e ) s : 1027 | d | divide(multiply(182, const_100), subtract(add(const_100, 4), subtract(const_100, 10))) | add(const_100,n2)|multiply(n1,const_100)|subtract(const_100,n0)|subtract(#0,#2)|divide(#1,#3)| | gain |
a watch was sold at a loss of 10 % . if it was sold for rs . 182 more , there would have been a gain of 4 % . what is the cost price ? | "the incomes of a and b be 3 p and 4 p . expenditures = income - savings ( 3 p - 100 ) and ( 4 p - 100 ) the ratio of their expenditure = 1 : 4 ( 3 p - 100 ) : ( 4 p - 100 ) = 1 : 4 8 p = 300 = > p = 37.5 their incomes = 112.5 , 150 answer : d" | a ) 112.5 , 158.5 , b ) 180.5 , 150 , c ) 100 , 200 , d ) 112.5 , 150 , e ) 122.5 , 150 | d | multiply(3, divide(100, 4)) | divide(n2,n4)|multiply(n0,#0)| | other |
the incomes of two persons a and b are in the ratio 3 : 4 . if each saves rs . 100 per month , the ratio of their expenditures is 1 : 4 . find their incomes ? | "there ' s a direct formula for this . number of diagonals in a regular polygon = [ n * ( n - 3 ) ] / 2 , n = number of sides of the regular polygon . here , n = 10 . plugging it in , we get 35 diagonals ! answer ( b ) ." | a ) 875 , b ) 35 , c ) 1425 , d ) 2025 , e ) 2500 | b | divide(multiply(subtract(10, const_3), 10), const_2) | subtract(n0,const_3)|multiply(n0,#0)|divide(#1,const_2)| | geometry |
a diagonal of a polygon is an segment between two non - adjacent vertices of the polygon . how many diagonals does a regular 10 - sided polygon have ? | "2 3 / 4 - 1 2 / 5 = 11 / 4 - 7 / 5 = ( 55 - 28 ) / 20 = 27 / 20 2 / 3 - 1 / 3 = ( 6 - 3 ) / 3 = 3 / 3 = 1 so 27 / 20 / 1 = 27 - 20 answer - d" | a ) 17 / 36 , b ) 36 / 17 , c ) 7 / 6 , d ) 27 / 20 , e ) 51 / 44 | d | subtract(divide(add(multiply(const_10, 2), 2), 4), divide(add(const_10, 1), 5)) | add(n3,const_10)|multiply(const_10,n0)|add(n0,#1)|divide(#0,n5)|divide(#2,n2)|subtract(#4,#3)| | general |
what is 2 3 / 4 - 1 2 / 5 divided by 2 / 3 - 1 / 3 ? | time of meeting = distance / relative speed = 36 / 8 + 4 = 36 / 12 = 3 hrs after 5 am = 8 am answer is a | a ) 8 am , b ) 6 am , c ) 7 am , d ) 10 am , e ) 8 pm | a | add(5, divide(36, add(4, 8))) | add(n1,n2)|divide(n3,#0)|add(n0,#1) | physics |
a and b start walking towards each other at 5 am at speed of 4 kmph and 8 kmph . they were initially 36 km apart . at what time do they meet ? | n = 4 k + 1 = 7 j + 3 let ' s start at 1 = 4 ( 0 ) + 1 and keep adding 4 until we find a number in the form 7 j + 3 . 1 , 5 , 9 , 13 , 17 = 7 ( 2 ) + 3 the next such number is 17 + 4 * 7 = 45 . 17 + 45 = 62 the answer is c . | a ) 54 , b ) 58 , c ) 62 , d ) 66 , e ) 70 | c | add(add(multiply(7, const_2), 3), add(multiply(7, multiply(const_2, const_3)), 3)) | multiply(n2,const_2)|multiply(const_2,const_3)|add(n3,#0)|multiply(n2,#1)|add(n3,#3)|add(#2,#4) | general |