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what is the units digit of 222 ^ ( 333 ) * 333 ^ ( 222 ) ? | "each of the other explanations to this question has properly explained that you need to break down the calculation into pieces and figure out the repeatingpatternof the units digits . here ' s another way to organize the information . we ' re given [ ( 2222 ) ^ 333 ] [ ( 3333 ) ^ 222 ] we can ' combine ' some of the pieces and rewrite this product as . . . . ( [ ( 2222 ) ( 3333 ) ] ^ 222 ) [ ( 2222 ) ^ 111 ] ( 2222 ) ( 3333 ) = a big number that ends in a 6 taking a number that ends in a 6 and raising it to a power creates a nice pattern : 6 ^ 1 = 6 6 ^ 2 = 36 6 ^ 3 = 216 etc . thus , we know that ( [ ( 2222 ) ( 3333 ) ] ^ 222 ) will be a gigantic number that ends in a 6 . 2 ^ 111 requires us to figure out thecycleof the units digit . . . 2 ^ 1 = 2 2 ^ 2 = 4 2 ^ 3 = 8 2 ^ 4 = 16 2 ^ 5 = 32 2 ^ 6 = 64 2 ^ 7 = 128 2 ^ 8 = 256 so , every 4 powers , the pattern of the units digits repeats ( 2 , 4 , 8 , 6 . . . . . 2 , 4 , 8 , 6 . . . . ) . 111 = 27 sets of 4 with a remainder of 3 . . . . this means that 2 ^ 111 = a big number that ends in an 8 so we have to multiply a big number that ends in a 6 and a big number that ends in an 8 . ( 6 ) ( 8 ) = 48 , so the final product will be a gigantic number that ends in an 4 . final answer : c" | a ) 270 , b ) 3024 , c ) 25 days , d ) 4 , e ) 60 | d | add(add(const_4, const_3), const_2) | add(const_3,const_4)|add(#0,const_2)| | general |
the length and breadth of a rectangular courtyard is 75 m and 32 m . find the cost of leveling it at the rate of $ 3 per m 2 . also , find the distance covered by a boy to take 4 rounds of the courtyard . | length of the courtyard = 75 m breadth of the courtyard = 32 m perimeter of the courtyard = 2 ( 75 + 32 ) m = 2 Γ 107 m = 214 m distance covered by the boy in taking 4 rounds = 4 Γ perimeter of courtyard = 4 Γ 214 = 856 m we know that area of the courtyard = length Γ breadth = 75 Γ 32 m 2 = 2400 m 2 for 1 m 2 , the cost of levelling = $ 3 for 2400 m 2 , the cost of levelling = $ 3 Γ 2400 = $ 7200 answer : e | a ) 7200 , b ) 274 , c ) 14 , d ) 62 , e ) 3 , 012,121 | a | multiply(3, multiply(75, 32)) | multiply(n0,n1)|multiply(n2,#0) | physics |
city a and city b are 140 miles apart . train c departs city a , heading towards city b , at 4 : 00 and travels at 40 miles per hour . train d departs city b , heading towards city a , at 4 : 40 and travels at 20 miles per hour . the trains travel on parallel tracks . at what time do the two trains meet ? | "train c has traveled 20 mi in the half hour before train d has started its journey . 140 - 20 = 120 40 + 20 = 60 mph 120 mi / 60 mph = 2 hrs 4 : 40 pm + 2 hrs = 6 : 40 pm answer : c . 6 : 40" | a ) 30 , 10 , b ) 6 : 40 , c ) 30 , d ) 18 , e ) 2011 . | b | divide(add(4, const_2), 40) | add(n4,const_2)|divide(#0,n5)| | physics |
a corporation that had $ 2 billion in profits for the year paid out $ 100 million in employee benefits . approximately what percent of the profits were the employee benefits ? ( note : 1 billion = 10 ^ 9 ) | "required answer = [ employee benefit / profit ] * 100 = [ ( 100 million ) / ( 2 billion ) ] * 100 = [ ( 100 * 10 ^ 6 ) / ( 2 * 10 ^ 9 ) ] * 100 = ( 50 / 1000 ) * 100 = 5 % so answer is ( c )" | a ) 150 , b ) 870 , c ) 5 % , d ) 6970 , e ) - 5 | c | multiply(divide(multiply(100, power(10, add(const_3, const_3))), multiply(2, power(10, 9))), const_100) | add(const_3,const_3)|power(n3,n4)|multiply(n0,#1)|power(n3,#0)|multiply(n1,#3)|divide(#4,#2)|multiply(#5,const_100)| | general |
a man and a boy complete a work together in 24 days . if for the last 6 days man alone does the work then it is completed in 26 days . how long the boy will take to complete the work alone ? | explanation : ( man + boy ) β s 1 day β s work = 1 / 24 their 20 day β s work = 1 / 24 Γ 20 = 5 / 6 the remaining 1 / 6 work is done by the man in 6 days therefore , the man alone will finish the work in 6 Γ 6 days = 36 days man β s 1 day β s work = 1 / 36 therefore , boy β s 1 day β s work = 1 / 24 β 1 / 36 = 3 β 2 / 72 = 1 / 72 therefore , the boy alone will finish the work in 72 days . answer : option a | a ) 600 , b ) 72 days , c ) 62 , d ) 14', ' , e ) 31 | b | add(subtract(26, 6), multiply(26, const_2)) | multiply(n2,const_2)|subtract(n2,n1)|add(#0,#1) | physics |
mathew is planning a vacation trip to london next year from today for 5 days , he has calculated he would need about $ 3000 for expenses , including a round - trip plane ticket from l . a to london . he nets around $ 1500 monthly in gross income , after all bills are paid , he is left with about $ 350 each month free for whatever he desires . how much money would mathew need to evenly save from his $ 350 to have $ 3000 in his bank within 12 months ? | answer is ( d ) . if mathew is left with about $ 350 after all expenses each month , he would need to divide the total expense budget to london ( $ 3000 ) by 12 months to determine how much he would need to put away every single month to hit his target . $ 3000 / 12 = $ 250 . | a ) $ 250 , b ) 1 / 12 , c ) 18 , d ) 3200 , e ) 123 | a | divide(3000, 12) | divide(n1,n6) | general |
three pipes a , b and c can fill a tank from empty to full in 30 minutes , 20 minutes and 10 minutes respectively . when the tank is empty , all the three pipes are opened . a , b and c discharge chemical solutions p , q and r respectively . what is the proportion of solution r in the liquid in the tank after 3 minutes ? | "part filled by ( a + b + c ) in 3 minutes = 3 ( 1 / 30 + 1 / 20 + 1 / 10 ) = 11 / 20 part filled by c in 3 minutes = 3 / 10 required ratio = 3 / 10 * 20 / 11 = 6 / 11 answer : b" | a ) 6 / 11 , b ) 97 , c ) 270 , d ) 12.50 % , e ) 10 kmph | a | multiply(divide(3, 10), divide(const_1, multiply(3, add(divide(const_1, 10), add(divide(const_1, 30), divide(const_1, 30)))))) | divide(n3,n2)|divide(const_1,n0)|divide(const_1,n2)|add(#1,#1)|add(#3,#2)|multiply(n3,#4)|divide(const_1,#5)|multiply(#0,#6)| | physics |
the present worth of rs . 1014 due in 2 years at 4 % per annum compound interest is | "solution present worth = rs . [ 1014 / ( 1 + 4 / 100 ) Β² ] = rs . ( 1014 x 25 / 26 x 25 / 26 ) = rs . 937.5 answer b" | a ) 16 days , b ) 65500 $ , c ) rs . 937.5 , d ) 720 , e ) 78 + ( 4 / | c | divide(1014, power(add(divide(4, const_100), const_1), 2)) | divide(n2,const_100)|add(#0,const_1)|power(#1,n1)|divide(n0,#2)| | gain |
if ( a β b ) is 15 more than ( c + d ) and ( a + b ) is 3 less than ( c β d ) , then ( a β c ) is : | "( a β b ) β ( c + d ) = 15 and ( c β d ) β ( a + b ) = 3 = > ( a β c ) β ( b + d ) = 15 and ( c β a ) β ( b + d ) = 3 = > ( b + d ) = ( a β c ) β 15 and ( b + d ) = ( c β a ) β 3 = > ( a β c ) β 15 = ( c β a ) β 3 = > 2 ( a β c ) = 12 = > ( a β c ) = 6 answer : a" | a ) $ 5.6 , b ) 0.05 , c ) 6 , d ) 7.5 sec , e ) 18 | c | divide(15, 3) | divide(n0,n1)| | general |
if jake loses 8 pounds , he will weigh twice as much as his sister kendra . together they now weigh 281 pounds . what is jake β s present weight , in pounds ? | "j + k = 281 and so k = 281 - j j - 8 = 2 k j - 8 = 2 ( 281 - j ) 3 j = 570 j = 190 the answer is e ." | a ) 190 , b ) 16 , c ) 22.5 , d ) 384 cm 2 , e ) 2 | a | add(multiply(divide(subtract(281, 8), const_3), const_2), 8) | subtract(n1,n0)|divide(#0,const_3)|multiply(#1,const_2)|add(n0,#2)| | general |
the total cost of 100 paper plates and 200 paper cups is $ 8.00 at the same rates what is the total cost of 20 of the plates and 40 of the cups ? | "u dont need to go through all this what u have with u is 100 p + 200 c = $ 8.00 just divide the equation by 5 and you will get what u are looking for 20 p + 40 c = $ 1.60 therefore oa is e" | a ) 12 / 5 , b ) $ 1.60 , c ) 6 , d ) 158 , e ) 29 | b | multiply(divide(20, 100), 8.00) | divide(n3,n0)|multiply(n2,#0)| | gain |
4242 Γ 9999 = ? | "a 42415758 4242 Γ 9999 = 4242 Γ ( 10000 - 1 ) = 4242 Γ 10000 - 4242 Γ 1 = 42420000 - 4242 = 42415758" | a ) 17 th , b ) $ 1600 , c ) 48 , d ) 26 m , e ) 42415758 | e | multiply(divide(4242, 9999), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
2 cow β s and 5 goats are brought for rs . 1050 . if the average price of a goat be rs . 90 . what is the average price of a cow . | "explanation : average price of a goat = rs . 90 total price of 5 goats = 5 * 90 = rs . 450 but total price of 2 cows and 5 goats = rs . 1050 total price of 2 cows is = 1050 - 450 = 600 average price of a cow = 600 / 2 = rs . 300 answer : a" | a ) 7 , b ) 11 , c ) 30 , d ) 300 , e ) 10 % | d | divide(subtract(1050, multiply(5, 90)), 2) | multiply(n1,n3)|subtract(n2,#0)|divide(#1,n0)| | general |
24 oz of juice p and 25 oz of juice v are mixed to make smothies x and y . the ratio of p to v in smothie x is 4 is to 1 and that in y is 1 is to 5 . how many ounces of juice p are contained in the smothie x ? | let us now solve for x : ( 4 / 5 ) x + ( 1 / 6 ) ( 49 - x ) = 24 24 x + 5 ( 49 - x ) = ( 24 ) ( 30 ) 24 x + 245 - 5 x = ( 24 ) ( 30 ) 19 x = 720 - 245 19 x = 475 x = 25 answer : e | a ) 1997 , b ) 106 , c ) 3 , d ) 855 , e ) 25 | e | multiply(25, 1) | multiply(n1,n3) | other |
a certain industrial loom weaves 0.13 meters of cloth every second . approximately how many seconds will it take for the loom to weave 15 meters of cloth ? | "let the required number of seconds be x more cloth , more time , ( direct proportion ) hence we can write as ( cloth ) 0.13 : 15 : : 1 : x = > 0.13 * x = 15 = > x = 15 / 0.13 = > x = 115 answer : b" | a ) 4 , b ) 115 , c ) 6 days , d ) 12 / 7 , e ) 4000 | b | divide(15, 0.13) | divide(n1,n0)| | physics |
in a garden , there are yellow and green flowers which are straight and curved . if the probability of picking a green flower is 1 / 8 and picking a straight flower is 1 / 2 , then what is the probability of picking a flower which is yellow and straight | "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 / 8 of the garden is green , so , since all the flowers must be either green or yellow , we know that 7 / 8 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 ) = 7 / 8 * 1 / 2 = 4 / 9 , or e ." | a ) 25 % , b ) 4 / 9 , c ) 3584 , d ) 45 , e ) 26 | b | multiply(subtract(1, divide(1, 8)), divide(1, 2)) | divide(n2,n3)|divide(n0,n1)|subtract(n2,#1)|multiply(#0,#2)| | probability |
how many 3 digit positive integers with distinct digits are there , which are not multiples of 10 ? | a number not to be a multiple of 10 should not have the units digit of 0 . xxx 9 options for the first digit ( from 1 to 9 inclusive ) . 8 options for the third digit ( from 1 to 9 inclusive minus the one we used for the first digit ) . 8 options for the second digit ( from 0 to 9 inclusive minus 2 digits we used for the first and the third digits ) 9 * 8 * 8 = 576 . answer : a . | a ) 4966 , b ) 10,800 , c ) 481 , d ) 300 , e ) 576 | e | multiply(multiply(multiply(3, 3), subtract(multiply(3, 3), const_1)), subtract(multiply(3, 3), const_1)) | multiply(n0,n0)|subtract(#0,const_1)|multiply(#0,#1)|multiply(#2,#1) | general |
3 pounds of 05 grass seed contain 1 percent herbicide . a different type of grass seed , 20 , which contains 20 percent herbicide , will be mixed with 3 pounds of 05 grass seed . how much grass seed of type 20 should be added to the 3 pounds of 05 grass seed so that the mixture contains 15 percent herbicide ? | 05 grass seed contains 5 % herbicide and its amount is 3 pound 20 grass seed contains 20 % herbicide and its amount is x when these two types of grass seeds are mixed , their average becomes 15 % thus we have 3 ( 1 ) + x ( 20 ) / ( x + 3 ) = 15 3 + 20 x = 15 x + 45 5 x = 42 or x = 8.4 d | a ) 28.57 % , b ) 8.4 , c ) 80 minutes , d ) 1095 , e ) 3.5 | b | divide(subtract(multiply(15, 3), 3), subtract(20, 15)) | multiply(n0,n10)|subtract(n3,n10)|subtract(#0,n0)|divide(#2,#1) | general |
if f ( f ( n ) ) + f ( n ) = 2 n + 3 and f ( 0 ) = 1 , what is the value of f ( 2012 ) ? | "put n = 0 then f ( f ( 0 ) ) + f ( 0 ) = 2 ( 0 ) + 3 β β f ( 1 ) + 1 = 3 β β f ( 1 ) = 2 put n = 1 f ( f ( 1 ) ) + f ( 1 ) = 2 ( 1 ) + 3 β β f ( 2 ) + 2 = 5 β β f ( 2 ) = 3 put n = 2 f ( f ( 2 ) ) + f ( 2 ) = 2 ( 2 ) + 3 β β f ( 3 ) + 3 = 7 β β f ( 3 ) = 4 . . . . . . f ( 2012 ) = 2013 answer : c" | a ) 760 , b ) 96 , c ) 1 / 2 , d ) 2013 , e ) 25 % | d | add(1, 2012) | add(n3,n4)| | general |
a certain barrel , which is a right circular cylinder , is filled to capacity with 60 gallons of oil . the first barrel is poured into a second barrel , also a right circular cylinder , which is empty . the second barrel is twice as tall as the first barrel and has twice the diameter of the first barrel . if all of the oil in the first barrel is poured into the second barrel , how much empty capacity , in gallons , is left in the second barrel ? | radius of first cylinder = r , diameter = 2 r , height = h radius of second cylinder = 2 r , diamter = 2 d and height = 2 h volume of first cylinder = pie ( r ^ 2 ) * h = 60 volume of second cylinder = pie ( 2 r ^ 2 ) 2 h put the value of pie ( r ^ 2 ) * h = 60 in the second cylinder , volume = pie ( r ^ 2 ) * 4 * 2 = 60 * 8 = 480 gallons empty capacity = 420 gallons answer d | a ) 270 , b ) 4.1 hr , c ) 420 gallons', ' , d ) 27 / 128 , e ) 1600 | c | subtract(multiply(60, power(const_2, const_3)), 60) | power(const_2,const_3)|multiply(n0,#0)|subtract(#1,n0) | geometry |
concentrated apples juice comes inside a cylinder tube with a radius of 2.5 inches and a height of 15 inches . the tubes are packed into wooden boxes , each with dimensions of 11 inches by 10 inches by 31 inches . how many tubes of concentrated apples juice , at the most , can fit into 3 wooden boxes ? | concentrated apples juice comes inside a cylinder tube since height of a tube is 15 inches , the tubes can fit only in one way now , diameter of each tube = 5 inches therefore , 4 * 2 can be put in each wooden box in 3 boxes 3 * 4 * 2 can be accommodated = 24 = a | a ) 900 , b ) 41 , c ) 24 . , d ) 77 , e ) 40 m | c | subtract(divide(multiply(multiply(multiply(11, 10), 31), 3), multiply(multiply(divide(multiply(add(const_10, const_1), const_2), add(const_3, const_4)), power(2.5, const_2)), 15)), 10) | add(const_1,const_10)|add(const_3,const_4)|multiply(n2,n3)|power(n0,const_2)|multiply(n4,#2)|multiply(#0,const_2)|divide(#5,#1)|multiply(n5,#4)|multiply(#6,#3)|multiply(n1,#8)|divide(#7,#9)|subtract(#10,n3) | gain |
a group of students decided to collect as many paise from each member of group as is the number of members . if the total collection amounts to rs . 51.84 , the number of the member is the group is : | "explanation : money collected = ( 51.84 x 100 ) paise = 5184 paise . β΄ number of members = β ( 5184 ) = 72 . answer : c" | a ) 425 , b ) 12 / 5 , c ) 5.33 kmph , d ) 3.2 , e ) 72 | e | sqrt(multiply(51.84, const_100)) | multiply(n0,const_100)|sqrt(#0)| | general |
john purchased some shirts and trousers for $ 1550 . he paid $ 250 less for the shirts than he did for the trousers . if he bought 5 shirts and the cost of a shirt is $ 20 less than that of a trouser , how many trousers did he buy ? | "given that the total purchase of two items cost 1550 . so the average purchase of one item will cost 1550 / 2 = 775 . its given as total shirt cost 250 $ less . hence total shirt cost = 775 - 125 and total trouser cost = 775 + 125 5 shirts = 650 $ = = > one shirt = 130 $ one trouser = 130 + 20 = 150 $ total trousers = 900 / 150 = 6 . c" | a ) 45 Β° , b ) 80 , c ) 40 % , d ) $ 250 , e ) 6 | e | divide(subtract(1550, multiply(5, add(20, 20))), add(add(20, 20), 20)) | add(n3,n3)|add(n3,#0)|multiply(n2,#0)|subtract(n0,#2)|divide(#3,#1)| | general |
a cube has a volume of 125 cubic feet . if a similar cube is twice as long , twice as wide , and twice as high , then the volume , in cubic feet of such cube is ? | "volume = 125 = side ^ 3 i . e . side of cube = 5 new cube has dimensions 10 , 10 , and 10 as all sides are twice of teh side of first cube volume = 10 * 10 * 10 = 1000 square feet answer : option e" | a ) 7 / 15 , b ) 1000 , c ) 40 , d ) 24 days , e ) $ 20 | b | volume_cube(multiply(const_2, cube_edge_by_volume(125))) | cube_edge_by_volume(n0)|multiply(#0,const_2)|volume_cube(#1)| | geometry |
a car is running at a speed of 110 kmph . what distance will it cover in 9 sec ? | "speed = 110 kmph = 110 * 5 / 18 = 31 m / s distance covered in 9 sec = 31 * 9 = 279 m answer is b" | a ) four , b ) 184 , c ) 279 m , d ) 2 : 1 , e ) 62 | c | multiply(divide(110, const_3_6), 9) | divide(n0,const_3_6)|multiply(n1,#0)| | physics |
in an election , candidate a got 65 % of the total valid votes . if 15 % of the total votes were declared invalid and the total numbers of votes is 560000 , find the number of valid vote polled in favor of candidate ? | "total number of invalid votes = 15 % of 560000 = 15 / 100 Γ 560000 = 8400000 / 100 = 84000 total number of valid votes 560000 β 84000 = 476000 percentage of votes polled in favour of candidate a = 65 % therefore , the number of valid votes polled in favour of candidate a = 65 % of 476000 = 65 / 100 Γ 476000 = 30940000 / 100 = 309400 d )" | a ) 8 kmph , b ) 5 , c ) 309400 , d ) 65500 $ , e ) 280 | c | multiply(multiply(560000, subtract(const_1, divide(15, const_100))), divide(65, const_100)) | divide(n0,const_100)|divide(n1,const_100)|subtract(const_1,#1)|multiply(n2,#2)|multiply(#0,#3)| | gain |
if a coin has an equal probability of landing heads up or tails up each time it is flipped , what is the probability that the coin will land heads up exactly twice in 2 consecutive flips ? | "total number of ways in which h or t can appear in 3 tosses of coin is = 2 * 2 = 4 ways for 2 h hh , thus probability is = p ( hh ) = 1 / 4 = . 25 answer : c" | a ) 0.25 , b ) 324 , c ) 4966 , d ) 9216 , e ) 375 | a | multiply(power(divide(const_1, const_2), 2), 2) | divide(const_1,const_2)|power(#0,n0)|multiply(n0,#1)| | general |
in a manufacturing plant , it takes 36 machines 4 hours of continuous work to fill 6 standard orders . at this rate , how many hours of continuous work by 72 machines are required to fill 12 standard orders ? | the choices give away the answer . . 36 machines take 4 hours to fill 8 standard orders . . in next eq we aredoubling the machines from 36 to 72 , but thework is not doubling ( only 1 1 / 2 times ) , = 4 * 48 / 72 * 12 / 6 = 4 ans a | a ) 6 , b ) 9 / 25 , c ) $ 12,500 , d ) 14 , e ) 4 | e | divide(divide(multiply(multiply(36, 12), const_4), 72), 6) | multiply(n0,n4)|multiply(#0,const_4)|divide(#1,n3)|divide(#2,n2) | physics |
a car is running at a speed of 90 kmph . what distance will it cover in 10 sec ? | "speed = 90 kmph = 90 * 5 / 18 = 25 m / s distance covered in 10 sec = 25 * 10 = 250 m answer is e" | a ) 250 m , b ) 3 / 10 , c ) 3 : 2 , d ) 1 : 2 , e ) 720 | a | multiply(divide(90, const_3_6), 10) | divide(n0,const_3_6)|multiply(n1,#0)| | physics |
how many 3 digit number formed by using 23 , 45 , 67 once such that number is divisible by 15 . | 4 * 2 * 1 = 8 at one ' s place only 5 will come and at ten ' s place 4 and 7 can be placed , and at 100 th place rest of the 4 digits can come . . . so the answer is 8 answer : a | a ) 81 , b ) 8 , c ) 1 / 13 , d ) 10 , e ) 5390 | b | add(divide(divide(45, 3), 3), const_3) | divide(n2,n0)|divide(#0,n0)|add(#1,const_3) | general |
how many 3 digit positive integers t exist that when divided by 7 leave a remainder of 5 ? | "minimum three digit number is 100 and maximum three digit number is 999 . the first three digit number that leaves remainder 5 when divided by 7 is 103 . 14 * 7 = 98 + 5 = 103 the second three digit number that leaves remainder 5 when divided by 7 is 110 . 15 * 7 = 105 + 5 = 110 the third three digit number that leaves remainder 5 when divided by 7 is 117 and so on the last three digit number that leaves remainder 5 when divided by 7 is 999 142 * 7 = 994 + 5 = 999 therefore , we identify the sequence 103 , 110,117 . . . . . 999 use the formula of last term last term = first term + ( n - 1 ) * common difference you will get the answer 129 that is definitely e ." | a ) 15 kmph , b ) 4.0 , c ) 12 , d ) 384 , e ) 129 | e | divide(subtract(subtract(multiply(const_100, const_10), const_1), add(multiply(add(const_10, const_4), 7), 5)), 7) | add(const_10,const_4)|multiply(const_10,const_100)|multiply(n1,#0)|subtract(#1,const_1)|add(n2,#2)|subtract(#3,#4)|divide(#5,n1)| | general |
in a 100 member association consisting of men and women , exactly 10 % of men and exactly 20 % women are homeowners . what is the maximum number of members who are homeowners ? | "solution simple out of 100 10 % are male i . e 10 and 20 % are female i . e 20 , so total homeowner is 30 . now min number homeowner is 10 and max is 30 so question ask us to find maximum and 29 has maximum value among all option . so ans is 29 . ans : a" | a ) 29 , b ) 27 days , c ) 157 , d ) 96 , e ) 56 | a | add(multiply(multiply(divide(20, const_100), 10), multiply(divide(20, const_100), 10)), divide(subtract(100, 10), 10)) | divide(n2,const_100)|subtract(n0,n1)|divide(#1,n1)|multiply(n1,#0)|multiply(#3,#3)|add(#2,#4)| | gain |
how many single - digit whole numbers yield a remainder of 1 when divided by 10 and also yield a remainder of 1 when divided by 6 ? | the possible number n can be written as follow : n = multiple of lcm ( 610 ) + 1 st such number n = 30 x + 1 possible values = 1 answer : b | a ) 0 % , b ) 280 , c ) 99 kmph , d ) one , e ) 80 | d | multiply(1, 1) | multiply(n0,n0) | general |
the average of 10 numbers is calculated as 16 . it is discovered later on that while calculating the average , one number namely 36 was wrongly read as 26 . the correct average is ? | "explanation : 10 * 16 + 36 β 26 = 170 = > 170 / 10 = 17 a )" | a ) 17 , b ) 10000 , c ) 1453 , d ) 100775 , e ) 7.94 | a | add(16, divide(subtract(36, 26), 10)) | subtract(n2,n3)|divide(#0,n0)|add(n1,#1)| | general |
48 persons can repair a road in 12 days , working 5 hours a day . in how many days will 30 persons , working 6 hours a day , complete the work ? | "let the required number of days be x . less persons , more days ( indirect proportion ) more working hours per day , less days ( indirect proportion ) persons 30 : 48 : : 12 : x working hours / day 6 : 5 30 x 6 x x = 48 x 5 x 12 x = ( 48 x 5 x 12 ) / ( 30 x 6 ) x = 16 answer b" | a ) 12 , b ) 16 , c ) 14 , d ) 3 / 5 , e ) 1.46 % | b | divide(multiply(multiply(48, 12), 5), multiply(30, 6)) | multiply(n0,n1)|multiply(n3,n4)|multiply(n2,#0)|divide(#2,#1)| | physics |
in the coordinate plane , a circle centered on point ( - 3 , 4 ) passes through point ( 1 , 1 ) . what is the area of the circle ? | "r ^ 2 = ( - 3 - 1 ) ^ 2 + ( 4 - 1 ) ^ 2 = 16 + 9 = 25 area of circle = Ο r ^ 2 = 25 Ο answer : c" | a ) 25 Ο , b ) 4.1 hr , c ) 2 , d ) 90 , e ) 5390 | a | circle_area(sqrt(add(power(subtract(3, 1), const_2), power(add(1, 4), const_2)))) | add(n1,n2)|subtract(n0,n2)|power(#1,const_2)|power(#0,const_2)|add(#2,#3)|sqrt(#4)|circle_area(#5)| | geometry |
a cistern has a leak which would empty the cistern in 20 minutes . a tap is turned on which admits 7 liters a minute into the cistern , and it is emptied in 24 minutes . how many liters does the cistern hold ? | "1 / x - 1 / 20 = - 1 / 24 x = 120 120 * 7 = 840 answer : d" | a ) $ 320 , b ) 1 : 729 , c ) 8925 , d ) 840 , e ) 305 | d | multiply(24, 20) | multiply(n0,n2)| | physics |
the ratio between x and y is 8 / 3 ; x is multiplied by x and y is multiplied by y , what is the ratio between the new values of x and y ? | "ratio = 7 k / 9 k = 7 / 9 , 14 / 18 , etc . x is multiplied by x and y is multiplied by y - - > ( 7 k * 7 k ) / ( 9 k * 9 k ) = 49 k ^ 2 / 81 k ^ 2 = 49 / 81 = 7 / 9 answer : a" | a ) 6 , b ) 6970 , c ) 8 / 3 , d ) 63 , e ) 33.33 % | c | divide(multiply(8, 3), multiply(3, 8)) | multiply(n0,n1)|divide(#0,#0)| | general |
a , b and c play a cricket match . the ratio of the runs scored by them in the match is a : b = 2 : 3 and b : c = 2 : 5 . if the total runs scored by all of them are 100 , the runs scored by a are ? | "a : b = 2 : 3 b : c = 2 : 5 a : b : c = 4 : 6 : 15 4 / 25 * 100 = 16 answer : d" | a ) $ 320 , b ) 6970 , c ) $ 250 , d ) 300 , e ) 16 | e | multiply(divide(100, add(add(divide(2, 3), divide(5, 2)), 2)), 5) | divide(n0,n1)|divide(n3,n0)|add(#0,#1)|add(#2,n0)|divide(n4,#3)|multiply(n3,#4)| | general |
1 , 3,5 , 7,9 , . . 50 find term of sequnce | "this is an arithmetic progression , and we can write down a = 1 a = 1 , d = 2 d = 2 , n = 50 n = 50 . we now use the formula , so that sn = 12 n ( 2 a + ( n β 1 ) l ) sn = 12 n ( 2 a + ( n β 1 ) l ) s 50 = 12 Γ 50 Γ ( 2 Γ 1 + ( 50 β 1 ) Γ 2 ) s 50 = 12 Γ 50 Γ ( 2 Γ 1 + ( 50 β 1 ) Γ 2 ) = 25 Γ ( 2 + 49 Γ 2 ) = 25 Γ ( 2 + 49 Γ 2 ) = 25 Γ ( 2 + 98 ) = 25 Γ ( 2 + 98 ) = 2500 = 2500 . e" | a ) 6 , b ) 1 , c ) 2500 , d ) 23 / 29 , e ) 60 | c | subtract(negate(50), multiply(subtract(3,5, 7,9), divide(subtract(3,5, 7,9), subtract(1, 3,5)))) | negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)| | general |
a palindrome is a number that reads the same forward and backward , such as 120 . how many odd , 4 - digit numbers are palindromes ? | "a palindrome is a number that reads the same forward and backward . examples of four digit palindromes are 1221 , 4334 , 2222 etc you basically get to choose the first two digits and you repeat them in opposite order . say , you choose 45 as your first two digits . the next two digits are 54 and the number is 4554 . also , you need only odd palindromes . this means that you need an odd digit at the end i . e . 1 / 3 / 5 / 7 / 9 . this means that you need to start the number with an odd digit . only then will it end with an odd digit . in how many ways can you pick two digits such that the first one is an odd digit ? the first digit can be selected in 5 ways . ( 1 / 3 / 5 / 7 / 9 ) the second digit can be selected in 10 ways . ( 0 / 1 / 2 / 3 . . . 8 / 9 ) total = 5 * 11 = 55 ways b" | a ) 55 , b ) 9 , c ) s . 9360 , d ) 14 m , e ) 10780 | a | divide(power(const_10, divide(4, const_2)), const_2) | divide(n1,const_2)|power(const_10,#0)|divide(#1,const_2)| | general |
the operation is defined for all integers a and b by the equation ab = ( a - 1 ) ( b - 1 ) . if x 20 = 190 , what is the value of x ? | "ab = ( a - 1 ) ( b - 1 ) x 20 = ( x - 1 ) ( 20 - 1 ) = 190 - - > x - 1 = 10 - - > x = 11 answer : c" | a ) 81 , b ) 1943236 , c ) 22 , d ) 11 , e ) 3 / 10 | d | add(divide(190, subtract(20, 1)), 1) | subtract(n2,n0)|divide(n3,#0)|add(n0,#1)| | general |
how many 5 digit nos are there if the 2 leftmost digits are odd and the digit 4 ca n ' t appear more than once in the number ? could someone please provide a solution using a approach other than ( 1 - x ( none ) ) approach ? | a . 4 is used once : oo * * 4 - - > ( 5 * 5 * 9 * 9 ) * 3 : 5 choices for the first digit as there are 5 odd numbers , 5 choices for the second digit for the same reason , 9 choices for one of the two * ( not - 4 digit ) , 9 choices for another * ( not - 4 digit ) , multiplied by 3 as 4 can take the place of any of the three last digits ( oo * * 4 , oo * 4 * , oo 4 * * ) ; b . 4 is not used : oo * * * - - > 5 * 5 * 9 * 9 * 9 : the same logic as above . 5 * 5 * 9 * 9 * 3 + 5 * 5 * 9 * 9 * 9 = 24300 . answer : a | a ) 24300 , b ) 3 / 10 , c ) 1 / 3 , d ) 27 days , e ) 50 km / hr | a | multiply(multiply(multiply(multiply(subtract(multiply(4, 5), 1), 5), 5), 5), const_10) | multiply(n0,n2)|subtract(#0,n3)|multiply(n0,#1)|multiply(n0,#2)|multiply(n0,#3)|multiply(#4,const_10) | general |
the population of a bacteria colony doubles every day . if it was started 8 days ago with 3 bacteria and each bacteria lives for 12 days , how large is the colony today ? | "3 ^ 8 * ( 2 ) = 13122 the answer is c" | a ) 3 : 4 , b ) 12.36 % , c ) 1083875 , d ) 18 , e ) 13122 | e | subtract(power(3, add(8, const_1)), const_1) | add(n0,const_1)|power(n1,#0)|subtract(#1,const_1)| | physics |
a cat leaps 6 leaps for every 5 leaps of a dog , but 2 leaps of the dog are equal to 3 leaps of the cat . what is the ratio of the speed of the cat to that of the dog ? | "given ; 2 dog = 3 cat ; or , dog / cat = 3 / 2 ; let cat ' s 1 leap = 2 meter and dogs 1 leap = 3 meter . then , ratio of speed of cat and dog = 2 * 6 / 3 * 5 = 4 : 5 ' ' answer : 4 : 5 ;" | a ) 4 : 5 , b ) 4 , c ) $ 110 , d ) 32.5 % , e ) 6 % | a | divide(multiply(divide(2, 3), 6), 5) | divide(n2,n3)|multiply(n0,#0)|divide(#1,n1)| | other |
a horse is tethered to one corner of a rectangular grassy field 36 m by 20 m with a rope 18 m long . over how much area of the field can it graze ? | "area of the shaded portion = 1 β 4 Γ Ο Γ ( 18 ) 2 = 254 m 2 answer c" | a ) 254 m 2 , b ) 36 , c ) 10 , d ) 3024 , e ) 5 / 6 | a | divide(multiply(power(18, const_2), const_pi), const_4) | power(n2,const_2)|multiply(#0,const_pi)|divide(#1,const_4)| | geometry |
if the arithmetic mean of seventy 5 numbers is calculated , it is 35 . if each number is increased by 5 , then mean of new number is ? | a . m . of 75 numbers = 35 sum of 75 numbers = 75 * 35 = 2625 total increase = 75 * 5 = 375 increased sum = 2625 + 375 = 3000 increased average = 3000 / 75 = 40 . answer : b | a ) 100 % , b ) 9 , c ) 153600 m 2', ' , d ) 120 , e ) 40 | e | add(35, 5) | add(n0,n1) | general |
8873 + x = 13200 , then x is ? | answer x = 13200 - 8873 = 4327 option : b | a ) $ 120 , b ) 4327 , c ) 296 , d ) 60 , e ) 192 | b | subtract(13200, 8873) | subtract(n1,n0) | general |
a box contains 8 pairs of shoes ( 16 shoes in total ) . if two shoes are selected at random , what it is the probability that they are matching shoes ? | "the problem with your solution is that we do n ' t choose 1 shoe from 16 , but rather choose the needed one after we just took one and need the second to be the pair of it . so , the probability would simply be : 1 / 1 * 1 / 15 ( as after taking one at random there are 15 shoes left and only one is the pair of the first one ) = 1 / 15 answer : e ." | a ) 2011 , b ) 60 , c ) 22.22 meters , d ) 1 / 15 , e ) c : 37.5 | d | divide(const_1, subtract(16, const_1)) | subtract(n1,const_1)|divide(const_1,#0)| | general |
sales price is $ 60 , gross profit is 140 % of cost , what is the value of gross profit ? | "cost + profit = sales cost + ( 140 / 100 ) cost = 60 cost = 25 profit = 60 - 25 = 35 answer ( d )" | a ) 5 , b ) 12.0 , c ) 27.2 % . , d ) 35 , e ) 36 | d | subtract(60, divide(60, add(const_1, divide(140, const_100)))) | divide(n1,const_100)|add(#0,const_1)|divide(n0,#1)|subtract(n0,#2)| | gain |
a , b , k start from the same place and travel in the same direction at speeds of 30 km / hr , 40 km / hr , 60 km / hr respectively . b starts 6 hours after a . if b and k overtake a at the same instant , how many hours after a did k start ? | the table you made does n ' t make sense to me . all three meet at the same point means the distance they cover is the same . we know their rates are 30 , 40 and 60 . say the time taken by b is t hrs . then a takes 6 + t hrs . and we need to find the time taken by k . distance covered by a = distance covered by b 30 * ( 6 + t ) = 40 * t t = 18 hrs distance covered by b = distance covered by k 40 * t = 60 * time taken by k time taken by k = 40 * 18 / 60 = 12 hrs time taken by a = 6 + t = 6 + 18 = 24 hrs time taken by k = 12 hrs so k starts 24 - 12 = 12 hrs after a . ( answer d ) | a ) $ 10,570 , b ) 1 / 12 , c ) 492 , d ) 180 , e ) 12 | e | divide(multiply(30, add(6, divide(multiply(30, 6), subtract(40, 30)))), 60) | multiply(n0,n3)|subtract(n1,n0)|divide(#0,#1)|add(n3,#2)|multiply(n0,#3)|divide(#4,n2) | physics |
in an exam 80 % of the boys and 40 % of the girls passed . the number of girls who passed is 120 , which is 2 / 3 rd of the number of boys who failed . what is the total number of students who appeared for the exam ? | let the number of boys = x , number of girls = y 40 y / 100 = 120 y = 300 120 = 2 / 3 * 20 x / 100 = 2 x / 15 x = 900 total = x + y = 300 + 900 = 1200 answer : a | a ) 1200 , b ) 52500 , c ) 7028 , d ) 450 , e ) 3 : 4 | a | add(divide(120, multiply(divide(subtract(const_100, 80), const_100), divide(2, 3))), divide(120, divide(40, const_100))) | divide(n3,n4)|divide(n1,const_100)|subtract(const_100,n0)|divide(#2,const_100)|divide(n2,#1)|multiply(#3,#0)|divide(n2,#5)|add(#6,#4) | general |
when 4 is added to half of one - third of one - fifth of a number , the result is one - fifteenth of the number . find the number ? | "explanation : let the number be 4 + 1 / 2 [ 1 / 3 ( a / 5 ) ] = a / 15 = > 4 = a / 30 = > a = 120 answer : d" | a ) 820 , b ) 120 , c ) - 1 , d ) 35 : 44 , e ) 90.25 sq cm | b | divide(4, divide(divide(const_1, multiply(4, add(const_2, 4))), const_2)) | add(const_2,n0)|multiply(#0,n0)|divide(const_1,#1)|divide(#2,const_2)|divide(n0,#3)| | general |
a certain lab experiments with white and brown mice only . in one experiment , 2 / 3 of the mice are white . if there are 15 brown mice in the experiment , how many mice in total are in the experiment ? | "let total number of mice = m number of white mice = 2 / 3 m number of brown mice = 1 / 3 m = 15 = > m = 45 answer a" | a ) rs . 189 , b ) 7 / 3 , c ) 129 , d ) 384 , e ) 45 | e | subtract(divide(15, divide(2, 3)), 15) | divide(n0,n1)|divide(n2,#0)|subtract(#1,n2)| | general |
a merchant sells an item at a 20 % discount , but still makes a gross profit of 30 percent of the cost . what percent of the cost would the gross profit on the item have been if it had been sold without the discount ? | "original sp = x cost = c current selling price = . 8 x ( 20 % discount ) . 8 x = 1.3 c ( 30 % profit ) x = 1.3 / . 8 * c x = 13 / 8 c original selling price is 1.625 c which is 62.5 % profit answer d" | a ) 53 , b ) $ 11000 , c ) 62.5 % , d ) 1000 , e ) 3 | c | subtract(const_100, subtract(subtract(const_100, 20), 30)) | subtract(const_100,n0)|subtract(#0,n1)|subtract(const_100,#1)| | gain |
pipe a fills a tank in 42 minutes . pipe b can fill the same tank 6 times as fast as pipe a . if both the pipes are kept open when the tank is empty , how many minutes will it take to fill the tank ? | "a ' s rate is 1 / 42 and b ' s rate is 1 / 7 . the combined rate is 1 / 42 + 1 / 7 = 1 / 6 the pipes will fill the tank in 6 minutes . the answer is d ." | a ) $ 590 , b ) 6 , c ) 1 / 27 , d ) 80 % , e ) 2.25 | b | inverse(add(divide(const_1, 42), divide(6, 42))) | divide(const_1,n0)|divide(n1,n0)|add(#0,#1)|inverse(#2)| | physics |
by weight , liquid x makes up 1.5 percent of solution p and 6.5 percent of solution q . if 200 grams of solution p are mixed with 800 grams of solution q , then liquid x accounts for what percent of the weight of the resulting solution ? | "the number of grams of liquid x is 1.5 ( 200 ) / 100 + 6.5 ( 800 ) / 100 = 3 + 52 = 55 grams . 55 / 1000 = 5.5 % the answer is c ." | a ) 5.5 % , b ) 170 m , c ) 16 , d ) 162 , e ) 855220 | a | multiply(divide(add(const_1, divide(multiply(6.5, 800), const_100)), const_1000), const_100) | multiply(n1,n3)|divide(#0,const_100)|add(#1,const_1)|divide(#2,const_1000)|multiply(#3,const_100)| | gain |
900 men have provisions for 15 days . if 200 more men join them , for how many days will the provisions last now ? | "900 * 15 = 1100 * x x = 12.27 . answer : e" | a ) 18 , b ) 12.27 , c ) 10,000 , d ) rs 66.66 , e ) 21.5 sec | b | divide(multiply(15, 900), add(900, 200)) | add(n0,n2)|multiply(n0,n1)|divide(#1,#0)| | physics |
set a { 3 , 3,3 , 4,5 , 5,5 } has a standard deviation of 1 . what will the standard deviation be if every number in the set is multiplied by 2 ? | points to remember - 1 . if oneadd / subtractthe same amont from every term in a set , sd does n ' t change . 2 . if onemultiply / divideevery term by the same number in a set , sd changes by same number . hence the answer to the above question is b | a ) 2 , b ) 120 % , c ) 200 , d ) 1653 , e ) 54 | a | multiply(2, 1) | multiply(n4,n5) | general |
a snail , climbing a 24 feet high wall , climbs up 4 feet on the first day but slides down 2 feet on the second . it climbs 4 feet on the third day and slides down again 2 feet on the fourth day . if this pattern continues , how many days will it take the snail to reach the top of the wall ? | total transaction in two days = 4 - 2 = 2 feet in 20 days it will climb 20 feet on the 21 st day , the snail will climb 4 feet , thus reaching the top therefore , total no of days required = 21 e | a ) $ 110 , b ) none , c ) 10 , d ) 55 , e ) 21 | e | subtract(24, 4) | subtract(n0,n1) | physics |
a β s speed is 25 / 18 times that of b . if a and b run a race , what part of the length of the race should a give b as a head start , so that the race ends in a dead heat ? | "we have the ratio of a β s speed and b β s speed . this means , we know how much distance a covers compared with b in the same time . this is what the beginning of the race will look like : ( start ) a _________ b ______________________________ if a covers 25 meters , b covers 18 meters in that time . so if the race is 25 meters long , when a reaches the finish line , b would be 7 meters behind him . if we want the race to end in a dead heat , we want b to be at the finish line too at the same time . this means b should get a head start of 7 meters so that he doesn β t need to cover that . in that case , the time required by a ( to cover 25 meters ) would be the same as the time required by b ( to cover 18 meters ) to reach the finish line . so b should get a head start of 7 / 25 th of the race . answer ( c )" | a ) 78 , b ) 30 , c ) 12 , d ) 7 / 25 , e ) 19 | d | divide(subtract(25, 18), 25) | subtract(n0,n1)|divide(#0,n0)| | general |
the marked price of a book is 20 % more than the cost price . after the book is sold , the vendor realizes that he had wrongly raised the cost price by a margin of 25 % . if the marked price of the book is rs . 30 , what is the original cost price of the book ? | let the incorrect cost price be c 1 and let the original cost price be c 2 . marked price of book is rs . 30 . it is 20 % more than c 1 . therefore , ( 120 / 100 ) x c 1 = 30 or c 1 = 25 . c 1 is more than c 2 margin of 25 % . or c 1 = ( 125 / 100 ) c 2 therefore , c 2 = ( 100 / 125 ) x 25 = rs 20 answer : d | a ) 16 , b ) 77.77 meters , c ) s . 9360 , d ) rs . 20 , e ) 1 | d | divide(divide(30, add(const_1, divide(20, const_100))), add(const_1, divide(25, const_100))) | divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|add(#1,const_1)|divide(n2,#2)|divide(#4,#3) | gain |
the h . c . f . of two numbers is 18 and the other two factors of their l . c . m . are 11 and 15 . the larger of the two numbers is : | "the numbers are ( 18 x 11 ) and ( 18 x 15 ) . larger number = ( 18 x 15 ) = 270 . answer : a" | a ) rs . 5845 , b ) 4.8 , c ) 0.05 , d ) 270 , e ) 2 | d | multiply(18, 15) | multiply(n0,n2)| | other |
the ratio of buses to cars on river road is 1 to 3 . if there are 20 fewer buses than cars on river road , how many cars are on river road ? | "b / c = 1 / 3 c - b = 20 . . . . . . . . . > b = c - 20 ( c - 20 ) / c = 1 / 3 testing answers . clearly eliminate abce put c = 30 . . . . . . . . . > ( 30 - 20 ) / 30 = 10 / 30 = 1 / 3 answer : d" | a ) 30 , b ) 216 , c ) 1,350 , d ) 190 , e ) 2 | a | multiply(divide(20, subtract(3, 1)), 3) | subtract(n1,n0)|divide(n2,#0)|multiply(n1,#1)| | other |
speed of a boat in standing water is 10 kmph and speed of the stream is 2.5 kmph . a man can rows to a place at a distance of 105 km and comes back to the starting point . the total time taken by him is ? | "speed upstream = 7.5 kmph speed downstream = 12.5 kmph total time taken = 105 / 7.5 + 105 / 12.5 = 22.4 hours answer is b" | a ) 1.9 , b ) 1 / 4 , c ) 1 , d ) 22.4 hours , e ) 4 | d | add(multiply(add(add(10, 2.5), subtract(10, 2.5)), 105), multiply(subtract(add(divide(105, add(10, 2.5)), divide(105, subtract(10, 2.5))), add(add(10, 2.5), subtract(10, 2.5))), const_60)) | add(n0,n1)|subtract(n0,n1)|add(#0,#1)|divide(n2,#0)|divide(n2,#1)|add(#3,#4)|multiply(n2,#2)|subtract(#5,#2)|multiply(#7,const_60)|add(#6,#8)| | physics |
the total of 324 of 20 paise and 25 paise make a sum of rs . 71 . the no of 20 paise coins is | "explanation : let the number of 20 paise coins be x . then the no of 25 paise coins = ( 324 - x ) . 0.20 * ( x ) + 0.25 ( 324 - x ) = 71 = > x = 200 . . answer : d ) 200" | a ) 54.4 cm , b ) 1 / 25 , c ) 200 , d ) 576 , e ) 60 kg | c | divide(subtract(multiply(324, 25), multiply(71, const_100)), subtract(25, 20)) | multiply(n0,n2)|multiply(n3,const_100)|subtract(n2,n1)|subtract(#0,#1)|divide(#3,#2)| | general |
if annual decrease in the population of a town is 5 % and the present number of people is 40000 what will the population be in 2 years ? | population in 2 years = 40000 ( 1 - 5 / 100 ) ^ 2 = 40000 * 19 * 19 / 20 * 20 = 36100 answer is c | a ) 45 , b ) 107 kg , c ) 16 , d ) 2 minutes , 00 seconds , e ) 36100 | e | multiply(power(divide(subtract(const_100, 5), const_100), 2), 40000) | subtract(const_100,n0)|divide(#0,const_100)|power(#1,n2)|multiply(n1,#2) | gain |
the price of lunch for 15 people was $ 206.00 , including a 15 percent gratuity for service . what was the average price per person , excluding the gratuity ? | take the initial price before the gratuity is 100 the gratuity is calculated on the final price , so as we assumed the final bill before adding gratuity is 100 so gratuity is 15 % of 100 is 15 so the total price of meals is 115 so the given amount i . e 206 is for 115 then we have to calculate for 100 for 115 206 for 100 x so by cross multiplication we get 115 x = 100 * 206 = > x = 100 * 206 / 110 by simplifying we get x as 187.27 which is the price of lunch before gratuity so the gratuity is 18.73 so as the question ask the average price person excluding gratuity is 187.27 / 15 = 12.48 so our answer is b ) | a ) 14 years and 30 years , b ) $ 12.48 , c ) 45 , d ) $ 130 , e ) 17 | b | multiply(multiply(divide(206, add(const_100, 15)), const_100), divide(const_1, 15)) | add(n0,const_100)|divide(const_1,n0)|divide(n1,#0)|multiply(#2,const_100)|multiply(#1,#3) | general |
a man can do a piece of work in 5 days , but with the help of his son , he can finish it in 3 days . in what time can the son do it alone ? | "son ' s 1 day work = 1 / 3 - 1 / 5 = 2 / 15 son alone can do the work in 15 / 2 days = 7 1 / 2 days answer is c" | a ) 12 , b ) 10 ^ 5 , c ) 7 1 / 2 , d ) 64 , e ) 20 sec | c | divide(multiply(5, 3), subtract(5, 3)) | multiply(n0,n1)|subtract(n0,n1)|divide(#0,#1)| | physics |
joe drives 600 miles at 60 miles per hour , and then he drives the next 120 miles at 40 miles per hour . what is his average speed for the entire trip in miles per hour ? | "t 1 = 600 / 60 = 10 hours t 2 = 120 / 40 = 3 hours t = t 1 + t 2 = 13 hours avg speed = total distance / t = 720 / 13 = 55 mph = b" | a ) 140 , b ) 131.6 , c ) 55 , d ) 0 , e ) 64 | c | divide(add(600, 120), add(divide(600, 60), divide(120, 40))) | add(n0,n2)|divide(n0,n1)|divide(n2,n3)|add(#1,#2)|divide(#0,#3)| | physics |
a pet store regularly sells pet food at a discount of 10 percent to 30 percent from the manufacturer β s suggested retail price . if during a sale , the store discounts an additional 20 percent from the discount price , what would be the lowest possible price of a container of pet food that had a manufacturer β s suggested retail price o f $ 20.00 ? | "for retail price = $ 20 first maximum discounted price = 20 - 30 % of 20 = 20 - 6 = 14 price after additional discount of 20 % = 14 - 20 % of 14 = 14 - 2.8 = 11.2 answer : option b" | a ) $ 11.20 , b ) 10 , c ) 165 , d ) 21 , e ) 36.48 | a | multiply(divide(subtract(const_100, 20), const_100), multiply(divide(subtract(const_100, 30), const_100), 20.00)) | subtract(const_100,n2)|subtract(const_100,n1)|divide(#0,const_100)|divide(#1,const_100)|multiply(n3,#3)|multiply(#2,#4)| | gain |
if 20 liters of chemical x are added to 80 liters of a mixture that is 15 % chemical x and 85 % chemical y , then what percentage of the resulting mixture is chemical x ? | "the amount of chemical x in the solution is 20 + 0.15 ( 80 ) = 32 liters . 32 liters / 100 liters = 32 % the answer is b ." | a ) 2520 , b ) 16.5 m 2 , c ) 5 : 6 , d ) 32 % , e ) $ 920.24 | d | add(20, multiply(divide(15, const_100), 80)) | divide(n2,const_100)|multiply(n1,#0)|add(n0,#1)| | general |
a set consists of 12 numbers , all are even or multiple of 5 . if 4 numbers are even and 10 numbers are multiple of 5 , how many numbers is multiple of 10 ? | "{ total } = { even } + { multiple of 5 } - { both } + { nether } . since { neither } = 0 ( allare even or multiple of 5 ) then : 12 = 4 + 10 - { both } + 0 ; { both } = 2 ( so 1 number is both even and multiple of 5 , so it must be a multiple of 10 ) . answer : c ." | a ) 20 , b ) 2 , c ) 49 , d ) 129 , e ) 2009 | b | subtract(12, 10) | subtract(n0,n3)| | general |
two employees x and y are paid a total of rs . 506 per week by their employer . if x is paid 120 percent of the sum paid to y , how much is y paid per week ? | "let the amount paid to x per week = x and the amount paid to y per week = y then x + y = 506 but x = 120 % of y = 120 y / 100 = 12 y / 10 β΄ 12 y / 10 + y = 506 β y [ 12 / 10 + 1 ] = 506 β 22 y / 10 = 506 β 22 y = 5060 β y = 5060 / 22 = 460 / 2 = rs . 230 e" | a ) s . 230 , b ) 410 m , c ) 40 , d ) 79 kmph , e ) 54.4 cm | a | divide(multiply(506, multiply(add(const_1, const_4), const_2)), multiply(add(multiply(add(const_1, const_4), const_2), const_1), const_2)) | add(const_1,const_4)|multiply(#0,const_2)|add(#1,const_1)|multiply(n0,#1)|multiply(#2,const_2)|divide(#3,#4)| | general |
for any integer k greater than 1 , the symbol k * denotes the product of all the fractions of the form 1 / t , where t is an integer between 1 and k , inclusive . what is the value of 3 * / 4 * ? | "when dealing with ' symbolism ' questions , it often helps to ' play with ' the symbol for a few moments before you attempt to answer the question that ' s asked . by understanding how the symbol ' works ' , you should be able to do the latter calculations faster . here , we ' re told that k * is the product of all the fractions of the form 1 / t , where t is an integer between 1 and k , inclusive . based on this definition . . . . if . . . . k = 2 k * = ( 1 / 1 ) ( 1 / 2 ) = 1 / 2 if . . . . k = 3 k * = ( 1 / 1 ) ( 1 / 2 ) ( 1 / 3 ) = 1 / 6 we ' re asked to find the value of 5 * / 4 * now that we know how the symbol ' works ' , solving this problem should n ' t be too difficult . you can actually choose to do the math in a couple of different ways . . . . 5 * = ( 1 / 1 ) ( 1 / 2 ) ( 1 / 3 ) ( 1 / 4 ) ( 1 / 5 ) do n ' t calculate this just yet though . . . . since we ' re dividing by 4 * , many of those fractions will ' cancel out . ' 4 * = ( 1 / 1 ) ( 1 / 2 ) ( 1 / 3 ) ( 1 / 4 ) we ' re looking for the value of : ( 1 / 1 ) ( 1 / 2 ) ( 1 / 3 ) ( 1 / 4 ) ( 1 / 5 ) / ( 1 / 1 ) ( 1 / 2 ) ( 1 / 3 ) ( 1 / 4 ) since the first four fraction in the numerator and denominator cancel out , we ' re left with just one fraction : 5 / 4 b" | a ) 5 / 4 , b ) 6 , c ) 65500 $ , d ) 2 : 3 , e ) 300 | a | divide(divide(divide(1, const_3), const_3), add(1, const_4)) | add(n0,const_4)|divide(n2,const_3)|divide(#1,const_3)|divide(#2,#0)| | general |
find the missing figures : ? % of 25 = 20125 | "let x % of 25 = 2.125 . then , ( x / 100 ) * 25 = 2.125 x = ( 2.125 * 4 ) = 8.5 . answer is e ." | a ) 440 , b ) 3 , c ) 770 , d ) 8.5 , e ) 200 | d | divide(20125, divide(25, const_100)) | divide(n0,const_100)|divide(n1,#0)| | gain |
the average age of 15 students of a class is 15 years . out of these , the average age of 4 students is 14 years and that of the other 10 students is 16 years . the age of the 15 th student is | "solution age of the 15 th student = [ 15 x 15 - ( 14 x 4 + 16 x 10 ) ] = ( 225 - 216 ) = 9 years . answer e" | a ) 450 sq . m', ' , b ) 75 , c ) 150 , d ) 9 years , e ) 6 | d | subtract(multiply(15, 15), add(multiply(4, 14), multiply(10, 16))) | multiply(n0,n0)|multiply(n2,n3)|multiply(n4,n5)|add(#1,#2)|subtract(#0,#3)| | general |
a certain sum of money is divided among a , b and c so that for each rs . a has , b has 65 paisa and c 40 paisa . if c ' s share is rs . 40 , find the sum of money ? | "a : b : c = 100 : 65 : 40 = 20 : 13 : 8 8 - - - - 40 41 - - - - ? = > rs . 205 answer : d" | a ) 12 , b ) 1 , c ) $ 8 , d ) 205 , e ) 57.5 | d | multiply(divide(40, 40), add(add(const_100, 65), 40)) | add(n0,const_100)|divide(n2,n1)|add(n1,#0)|multiply(#2,#1)| | general |
two trains leave the train station at the same time . one train , on the blue line , heads east - while the other , on the red line , heads west . if the train on the blue line averages 40 km / hr and the other train averages 40 km / hr - how long will it take for the trains to be 100 km apart ? | each train is averaging 40 km / hour in an opposite direction . after 1 hour , they will be 80 km apart , and after 1.25 hours , they will be 100 km apart . ( 80 * 1.25 = 100 ) answer is d | a ) 1.25 hours , b ) 28 , c ) 34 % , d ) 300 , e ) 88.2 % | a | divide(divide(100, const_2), 40) | divide(n2,const_2)|divide(#0,n0) | general |
a certain college ' s enrollment at the beginning of 1992 was 30 percent greater than it was at the beginning of 1991 , and its enrollment at the beginning of 1993 was 10 percent greater than it was at the beginning of 1992 . the college ' s enrollment at the beginning of 1993 was what percent greater than its enrollment at the beginning of 1991 ? | "suppose enrollment in 1991 was 100 then enrollment in 1992 will be 130 and enrollment in 1993 will be 130 * 1.1 = 143 increase in 1993 from 1991 = 143 - 100 = 43 answer : a" | a ) 100 % , b ) 43 % , c ) 10 , d ) 77 , e ) 1800 | b | subtract(multiply(add(const_100, 30), divide(add(const_100, 10), const_100)), const_100) | add(n1,const_100)|add(n4,const_100)|divide(#1,const_100)|multiply(#0,#2)|subtract(#3,const_100)| | gain |
what is the greatest possible length which can be used to measure exactly the lengths 10 m 50 cm , 14 m 55 cm and 50 cm ? | "required length = hcf of 1050 cm , 1455 cm , 50 cm = 5 cm answer is e" | a ) 24 days , b ) 9 , c ) 68 kmph , d ) 5 cm , e ) 30 % | d | multiply(55, const_4) | multiply(n3,const_4)| | physics |
for any integer p , * p is equal to the product of all the integers between 1 and p , inclusive . how many prime numbers are there between * 9 + 3 and * 9 + 9 , inclusive ? | "generally * p or p ! will be divisible by all numbers from 1 to p . therefore , * 9 would be divisible by all numbers from 1 to 9 . = > * 9 + 3 would give me a number which is a multiple of 3 and therefore divisible ( since * 9 is divisible by 3 ) in fact adding anyprimenumber between 1 to 9 to * 9 will definitely be divisible . so the answer is none ( a ) ! supposing if the question had asked for prime numbers between * 9 + 3 and * 9 + 11 then the answer would be 1 . for * 9 + 3 and * 9 + 13 , it is 2 and so on . . . a" | a ) 94 , b ) 12 hours , c ) none , d ) 0.125 days , e ) 1 / 3 | c | subtract(subtract(add(multiply(multiply(multiply(9, 3), const_2), const_4), 9), add(multiply(multiply(multiply(9, 3), const_2), const_4), 3)), 1) | multiply(n1,n2)|multiply(#0,const_2)|multiply(#1,const_4)|add(n1,#2)|add(n2,#2)|subtract(#3,#4)|subtract(#5,n0)| | general |
a can give b 120 meters start and c 200 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 880 meters and c runs 800 meters . therefore , b runs 880 meters while c runs 800 meters . so , the number of meters that c runs when b runs 1000 meters = ( 1000 x 800 ) / 880 = 909.09 meters thus , b can give c ( 1000 - 909.09 ) = 90.09 meters start answer : a" | a ) 612 , b ) 3 : 2 , c ) 90.09 meters , d ) 31 % , e ) 21 % | c | subtract(multiply(const_100, const_10), divide(multiply(multiply(const_100, const_10), subtract(multiply(const_100, const_10), 200)), subtract(multiply(const_100, const_10), 120))) | multiply(const_10,const_100)|subtract(#0,n1)|subtract(#0,n0)|multiply(#0,#1)|divide(#3,#2)|subtract(#0,#4)| | physics |
convert 40 miles into yards ? | "1 mile = 1760 yards 40 miles = 40 * 1760 = 70400 yards answer is e" | a ) s . 120 , b ) 8 minutes , c ) 7 , d ) 70400 yards , e ) $ 78.80 | d | divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 40), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2))) | add(const_2,const_3)|multiply(#0,const_2)|multiply(#1,#1)|multiply(n0,#2)|divide(#3,#2)| | physics |
a merchant marks his goods up by 30 % and then offers a discount of 20 % on the marked price . what % profit does the merchant make after the discount ? | "let the price be 100 . the price becomes 130 after a 30 % markup . now a discount of 20 % on 130 . profit = 104 - 100 4 % answer e" | a ) 4 / 5 , b ) s . 200 , c ) 49995 , d ) 9 , e ) 4 % | e | subtract(subtract(add(30, const_100), divide(multiply(add(30, const_100), 20), const_100)), const_100) | add(n0,const_100)|multiply(n1,#0)|divide(#1,const_100)|subtract(#0,#2)|subtract(#3,const_100)| | gain |
the owner of a furniture shop charges his customer 42 % more than the cost price . if a customer paid rs . 8300 for a computer table , then what was the cost price of the computer table ? | cp = sp * ( 100 / ( 100 + profit % ) ) = 8300 ( 100 / 142 ) = rs . 5845 . answer : b | a ) 7 , b ) 6 , c ) rs . 5845 , d ) 5 % , e ) 35 | c | divide(8300, add(const_1, divide(42, const_100))) | divide(n0,const_100)|add(#0,const_1)|divide(n1,#1) | gain |
an article with cost price of 320 is sold at 18 % profit . what is the selling price ? | "sp = 1.18 * 320 = 378 answer : d" | a ) 40 , b ) 3 , c ) 144 , d ) 378 , e ) 36', ' | d | add(320, multiply(320, divide(18, const_100))) | divide(n1,const_100)|multiply(n0,#0)|add(n0,#1)| | gain |
if it takes a tub 5 minutes to drain 5 / 7 of its content , how much more time will it take for the tub to be empty ? | "if 5 / 7 of tub ' s content is drained 2 / 7 th of tub still needs to be drained . if it takes 5 minutes to drain 5 / 7 th of tub it takes 5 * ( 7 / 5 ) minutes to drain the entire tub and 5 * ( 7 / 5 ) * ( 2 / 7 ) min to drain 2 / 7 th of the tub which is 2 minutes so answer is d" | a ) 10 , b ) 519 , c ) 2 / 3 , d ) 432 . , e ) 2 minutes , 00 seconds | e | add(subtract(const_1, divide(5, 7)), divide(5, 7)) | divide(n1,n2)|subtract(const_1,#0)|add(#0,#1)| | general |
a collection of books went on sale , and 2 / 3 of them were sold for $ 3.25 each . if none of the 20 remaining books were sold , what was the total amount received for the books that were sold ? | "if 20 books constitute 1 / 3 rd of the total , then 2 / 3 rd of the total = 40 books amount received for sold books = 40 * 3.25 = $ 130 answer : a" | a ) $ 130 , b ) $ 37.8 , c ) 22 , d ) 200 , e ) 30 | a | multiply(2, divide(multiply(20, divide(2, 3)), divide(const_1, 3))) | divide(n0,n1)|divide(const_1,n1)|multiply(n3,#0)|divide(#2,#1)|multiply(n0,#3)| | general |
there are two cars . one is 160 miles north of the other . simultaneously , the car to the north is driven westward at 10 miles per hour and the other car is driven eastward at 30 miles per hour . how many miles apart are the cars after 3 hours ? | "here , drawing a quick sketch of the ' actions ' described will end in a diagonal line that you canbuilda right triangle around : the right triangle will have a base of 120 and a height of 160 . the hidden pattern here is a 3 / 4 / 5 right triangle ( the 120 lines up with the ' 3 ' and the 160 lines up with the ' 4 ' ) . in this way , you can deduce that each side is ' 40 times ' bigger than it ' s corresponding side : 3 / 4 / 5 becomes 120 / 160 / 200 thus the distance between the two cars is the length of the hypotenuse of this larger right triangle . . . final answer : b" | a ) 16 , b ) 54.2 % , c ) 32 , d ) 1800 , e ) 200 | e | sqrt(add(power(add(multiply(30, 3), multiply(10, 3)), const_2), power(multiply(30, 3), const_2))) | multiply(n2,n3)|multiply(n1,n3)|add(#0,#1)|power(#0,const_2)|power(#2,const_2)|add(#4,#3)|sqrt(#5)| | physics |
a batsman scored 120 runs which included 5 boundaries and 8 sixes . what % of his total score did he make by running between the wickets | "number of runs made by running = 120 - ( 5 x 4 + 8 x 6 ) = 120 - ( 68 ) = 52 now , we need to calculate 60 is what percent of 120 . = > 52 / 120 * 100 = 43.33 % a" | a ) 0 , b ) 30 , c ) 12800 , d ) - 2407 , e ) 43.33 % | e | multiply(divide(subtract(120, add(multiply(5, 8), multiply(8, 5))), 120), const_100) | multiply(n1,n2)|multiply(n1,n2)|add(#0,#1)|subtract(n0,#2)|divide(#3,n0)|multiply(#4,const_100)| | general |
given that 100.48 = x , 100.70 = y and xz = y 2 , then the value of z is close to : | "xz = y 2 10 ( 0.48 z ) = 10 ( 2 x 0.70 ) = 101.40 0.48 z = 1.40 z = 140 = 35 = 2.9 ( approx . ) 48 12 answer : c" | a ) 2.9 , b ) 12 / 5 , c ) 189 , d ) 1000', ' , e ) $ 1.83 | a | divide(multiply(subtract(100.70, const_100), const_2), subtract(100.48, const_100)) | subtract(n1,const_100)|subtract(n0,const_100)|multiply(#0,const_2)|divide(#2,#1)| | general |
how many digits are required to number a book containing 250 pages ? | "9 pages from 1 to 9 will require 9 digits . 90 pages from 10 to 99 will require 90 * 2 = 180 digits . 250 - ( 90 + 9 ) = 151 pages will require 151 * 3 = 453 digits . the total number of digits is 9 + 180 + 453 = 642 . the answer is b ." | a ) 642 , b ) 24 % , c ) 44 , d ) 24 metre , e ) 50000 | a | add(add(subtract(const_10, const_1), multiply(multiply(subtract(const_10, const_1), const_10), const_2)), multiply(add(subtract(250, const_100), const_1), const_3)) | subtract(const_10,const_1)|subtract(n0,const_100)|add(#1,const_1)|multiply(#0,const_10)|multiply(#3,const_2)|multiply(#2,const_3)|add(#4,#0)|add(#6,#5)| | general |
at scratch and dents rent - a - car , it costs $ 34.95 a day plus $ 0.23 per mile to rent a car . at rent - a - lemon , the charge is $ 25.00 a day plus $ 1.31 per mile . if you need to rent a car for 3 days , how many miles ( to nearest tenth ) must you drive for a car from both agencies to cost the same amount ? | for sad : saddaily = $ 34.95 / day sadmile = $ 0.23 / mile for ral : raldaily = $ 25.00 / day ralmile = $ 1.31 / mile we want the raltotal = sadtotal , so we get ( raldaily * days ) + ( ralmile * miles ) = ( saddaily * days ) + ( sadmile * miles ) = > miles = ( ( saddaily * days ) - ( raldaily * days ) ) / ( ralmiles - sadmiles ) = ( ( saddaily - raldaily ) * days ) / ( ralmiles - sadmiles ) miles = ( ( $ 34.95 * 3 ) - ( $ 25.00 * 3 ) ) / ( $ 1.31 - $ 0.23 ) = 27.6 miles c . 27.6 miles | a ) 16 , b ) 240 , c ) 8 and 20 , d ) 52 % , e ) 27.6 miles | e | divide(subtract(multiply(34.95, 3), multiply(25, 3)), subtract(1.31, 0.23)) | multiply(n0,n4)|multiply(n2,n4)|subtract(n3,n1)|subtract(#0,#1)|divide(#3,#2) | general |
pipe a can fill a tank in 9 hours . due to a leak at the bottom , it takes 12 hours for the pipe a to fill the tank . in what time can the leak alone empty the full tank ? | "let the leak can empty the full tank in x hours 1 / 9 - 1 / x = 1 / 12 = > 1 / x = 1 / 9 - 1 / 12 = 1 / 12 = > x = 36 . answer : a" | a ) 23 , b ) 36100 , c ) 8985 , d ) 2 / 9 , e ) 36 | e | divide(multiply(12, 9), subtract(12, 9)) | multiply(n0,n1)|subtract(n1,n0)|divide(#0,#1)| | physics |
if p is a prime number greater than 3 , find the remainder when p ^ 2 + 16 is divided by 12 . | every prime number greater than 3 can be written 6 n + 1 or 6 n - 1 . if p = 6 n + 1 , then p ^ 2 + 16 = 36 n ^ 2 + 12 n + 1 + 16 = 36 n ^ 2 + 12 n + 12 + 5 if p = 6 n - 1 , then p ^ 2 + 16 = 36 n ^ 2 - 12 n + 1 + 16 = 36 n ^ 2 - 12 n + 12 + 5 when divided by 12 , it must leave a remainder of 5 . the answer is a . | a ) 5 , b ) 410 m , c ) 60 , d ) 46080 , e ) 8 | a | subtract(add(16, power(add(const_1, const_4), const_2)), multiply(12, 3)) | add(const_1,const_4)|multiply(n0,n3)|power(#0,const_2)|add(n2,#2)|subtract(#3,#1) | general |
a hiker walked for 3 days . she walked 18 miles on the first day , walking 3 miles per hour . on the second day she walked for one less hour but she walked one mile per hour , faster than on the first day . on the third day she walked at 7 miles per hour for 2 hours . how many miles in total did she walk ? | "first day - 18 miles with 3 miles per hours then total - 6 hours for that day second day - 4 miles per hour and 5 hours - 20 miles third day - 7 miles per hour and 2 hours - 14 miles total 18 + 20 + 14 = 52 answer : option b ." | a ) 5 % , b ) 100 kg , c ) 35 , d ) 52 , e ) 30 | d | add(add(18, multiply(7, const_4)), multiply(7, 2)) | multiply(n3,const_4)|multiply(n3,n4)|add(n1,#0)|add(#2,#1)| | physics |
if f ( x ) = 3 x ^ 4 - 4 x ^ 3 - 2 x ^ 2 + 5 x , then f ( - 1 ) = | "f ( - 1 ) = 3 ( - 1 ) ^ 4 - 4 ( - 1 ) ^ 3 - 2 ( - 1 ) ^ 2 + 5 ( - 1 ) = 3 + 4 - 2 - 5 = 0 the answer is c ." | a ) 492 , b ) rs . 937.5 , c ) 7 , d ) 450 , e ) 0 | e | add(subtract(subtract(multiply(3, power(negate(1), 4)), multiply(4, power(negate(1), 3))), multiply(3, power(negate(1), 2))), multiply(5, negate(1))) | negate(n7)|multiply(n6,#0)|power(#0,n1)|power(#0,n0)|power(#0,n5)|multiply(n0,#2)|multiply(n1,#3)|multiply(n0,#4)|subtract(#5,#6)|subtract(#8,#7)|add(#1,#9)| | general |
carol spends 1 / 4 of her savings on a stereo and 1 / 5 less than she spent on the stereo for a television . what fraction of her savings did she spend on the stereo and television ? | "total savings = s amount spent on stereo = ( 1 / 4 ) s amount spent on television = ( 1 - 1 / 5 ) ( 1 / 4 ) s = ( 4 / 5 ) * ( 1 / 4 ) * s = ( 1 / 5 ) s ( stereo + tv ) / total savings = s ( 1 / 4 + 1 / 5 ) / s = 9 / 20 answer : d" | a ) 7 , b ) 9 / 20 , c ) 0.5 , d ) 5 , e ) 13,400 | b | divide(1, 4) | divide(n0,n1)| | general |
if x is 20 percent greater than 55 , then x = | x is 20 % greater than 55 means x is 1.2 times 55 ( in other words 55 + 20 / 100 * 55 = 1.2 * 55 ) therefore , x = 1.2 * 55 = 66 answer : d | a ) 5 / 12 , b ) 21 , c ) 12', ' , d ) 240 , e ) 66 | e | add(55, multiply(divide(20, const_100), 55)) | divide(n0,const_100)|multiply(n1,#0)|add(n1,#1) | general |