Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 164 | correct stringclasses 5 values | annotated_formula stringlengths 7 1.65k | linear_formula stringlengths 8 925 | category stringclasses 6 values | answer stringclasses 5 values |
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on rainy mornings , mo drinks exactly n cups of hot chocolate ( assume that n is an integer ) . on mornings that are not rainy , mo drinks exactly 2 cups of tea . last week mo drank a total of 16 cups of tea and hot chocolate together . if during that week mo drank 4 more tea cups than hot chocolate cups , then how many rainy days were there last week ? | "t = the number of cups of tea c = the number of cups of hot chocolate t + c = 16 t - c = 4 - > t = 10 . c = 6 . mo drinks 5 cups of tea a day then number of days that are not rainy = 10 / 2 = 5 so number of rainy days = 7 - 5 = 2 e is the answer ." | a ) 6 , b ) 3 , c ) 4 , d ) 5 , e ) 2 | e | subtract(add(const_4, 2), divide(divide(add(16, 4), const_2), 2)) | add(n0,const_4)|add(n1,n2)|divide(#1,const_2)|divide(#2,n0)|subtract(#0,#3)| | general | E |
simplify : ( 7 + 2 ) – ( 5 + 3 + 1 ) - 1 . | "solution : ( 7 + 2 ) – ( 5 + 3 + 1 ) - 1 = 9 - 5 - 3 + 1 - 1 = 9 - 8 + 1 - 1 = 2 - 1 = 1 answer : ( c )" | a ) - 1 , b ) – 2 , c ) 1 , d ) 2 , e ) 0 | c | add(add(7, const_3.0), subtract(subtract(5, 2), 1)) | add(n0,n1)|subtract(n2,n3)|subtract(#1,n4)|add(#0,#2)| | general | C |
the instructions state that cheryl needs 4 / 9 square yards of one type of material and 2 / 3 square yards of another type of material for a project . she buys exactly that amount . after finishing the project , however , she has 6 / 12 square yards left that she did not use . what is the total amount of square yards of material cheryl used ? | "total bought = 4 / 9 + 2 / 3 left part 6 / 12 - - - > 2 / 3 so used part 4 / 9 + 2 / 3 - 2 / 3 = 4 / 9 ans e" | a ) 1 / 12 , b ) 1 / 9 , c ) 2 / 3 , d ) 1 1 / 9 , e ) 4 / 9 | e | subtract(add(divide(4, 9), divide(2, 3)), divide(6, 12)) | divide(n0,n1)|divide(n2,n3)|divide(n4,n5)|add(#0,#1)|subtract(#3,#2)| | general | E |
simple interest on a certain sum of money for 3 years at 10 % per annum is half the compound interest on rs . 6500 for 2 years at 14 % per annum . the sum placed on simple interest is | "solution c . i . = rs [ 6500 x ( 1 + 14 / 100 ) â ² - 6500 ] rs . ( 6500 x 114 / 100 x 114 / 100 - 6500 ) = rs . 1947.4 . sum = rs . [ 973.7 x 100 / 3 x 10 ] = rs . 3245.67 . answer e" | a ) rs . 6500 , b ) rs . 1947.4 , c ) rs . 973.7 , d ) rs . 2000 , e ) none | e | divide(multiply(divide(divide(add(divide(multiply(6500, 14), const_100), divide(multiply(add(6500, divide(multiply(6500, 14), const_100)), 14), const_100)), 2), 3), const_100), 10) | multiply(n2,n4)|divide(#0,const_100)|add(n2,#1)|multiply(n4,#2)|divide(#3,const_100)|add(#1,#4)|divide(#5,n3)|divide(#6,n0)|multiply(#7,const_100)|divide(#8,n1)| | gain | E |
a cistern 7 m long and 5 m wide contains water up to a breadth of 1 m 40 cm . find the total area of the wet surface . | "explanation : area of the wet surface = 2 [ lb + bh + hl ] - lb = 2 [ bh + hl ] + lb = 2 [ ( 5 * 1.40 + 7 * 1.40 ) ] + 7 * 5 = 69 m square option a" | a ) 69 m sqaure , b ) 49 m sqaure , c ) 52 m sqaure , d ) 64 m sqaure , e ) none of these | a | add(multiply(const_2, add(multiply(add(divide(40, const_100), 1), 5), multiply(add(divide(40, const_100), 1), 7))), multiply(5, 7)) | divide(n3,const_100)|multiply(n0,n1)|add(n2,#0)|multiply(n1,#2)|multiply(n0,#2)|add(#3,#4)|multiply(#5,const_2)|add(#6,#1)| | physics | A |
a boy rides his bicycle 10 km at an average sped of 12 km / hr and again travels 12 km at an average speed of 10 km / hr . his average speed for the entire trip is approximately . | sol . total distance travelled = ( 10 + 12 ) km / hr = 22 km / hr . total time taken = [ 10 / 12 + 12 / 10 ] hrs = 61 / 30 hrs ∴ average speed = [ 22 * 30 / 61 ] km / hr = 10.8 km / hr . answer c | a ) 10.2 kmph , b ) 10.4 kmph , c ) 10.8 kmph , d ) 11.8 kmph , e ) none | c | divide(add(10, 12), add(divide(10, 12), divide(12, 10))) | add(n0,n1)|divide(n0,n1)|divide(n1,n0)|add(#1,#2)|divide(#0,#3) | general | C |
how many terminating zeroes t does 200 ! have ? | you have 40 multiples of 5 , 8 of 25 and 1 of 125 . this will give 49 zeros . c | a ) 40 , b ) 48 , c ) 49 , d ) 55 , e ) 64 | c | add(divide(200, add(const_4, const_1)), divide(200, multiply(add(const_4, const_1), add(const_4, const_1)))) | add(const_1,const_4)|divide(n0,#0)|multiply(#0,#0)|divide(n0,#2)|add(#1,#3)| | other | C |
two goods trains each 500 m long are running in opposite directions on parallel tracks . their speeds are 30 km / hr and 30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? | "relative speed = 30 + 30 = 60 km / hr . 60 * 5 / 18 = 50 / 3 m / sec . distance covered = 500 + 500 = 1000 m . required time = 1000 * 3 / 50 = 60 sec . answer : d" | a ) 12 sec , b ) 24 sec , c ) 48 sec , d ) 60 sec , e ) 62 sec | d | add(30, 30) | add(n1,n2)| | physics | D |
x starts a business with rs . 45000 . y joins in the business after 8 months with rs . 30000 . what will be the ratio in which they should share the profit at the end of the year ? | "explanation : ratio in which they should share the profit = ratio of the investments multiplied by the time period = 45000 ã — 12 : 30000 ã — 4 = 45 ã — 12 : 30 ã — 4 = 3 ã — 12 : 2 ã — 4 = 9 : 2 answer : option b" | a ) 1 : 2 , b ) 9 : 2 , c ) 1 : 3 , d ) 3 : 1 , e ) 1 : 1 | b | divide(multiply(45000, const_12), multiply(30000, add(const_4, const_3))) | add(const_3,const_4)|multiply(n0,const_12)|multiply(n2,#0)|divide(#1,#2)| | other | B |
beginning in town a , biker bob rides his bike 20 miles west , 6 miles north , 10 miles east , and then 18 miles north , to town b . how far apart are town a and town b ? ( ignore the curvature of the earth . ) | using pythagoras we have one side i , e total distance traveled in north direction = 18 + 6 = 24 m other being the base ie distance traveled west - distance traveled eat = 20 - 10 = 10 m now this third side or the distance between town a and town b = 24 ^ 2 + 10 ^ 2 = sq root 676 = 26 m answer : c | a ) 28 miles , b ) 30 miles , c ) 26 miles , d ) 18 miles , e ) 22 miles | c | sqrt(add(power(add(6, 18), const_2), power(10, const_2))) | add(n1,n3)|power(n2,const_2)|power(#0,const_2)|add(#2,#1)|sqrt(#3) | physics | C |
a car takes 4 hours to cover a distance of 540 km . how much should the speed in kmph be maintained to cover the same direction in 3 / 2 th of the previous time ? | "time = 4 distence = 540 3 / 2 of 4 hours = 4 * 3 / 2 = 6 hours required speed = 540 / 6 = 90 kmph c )" | a ) 92 kmph , b ) 98 kmph , c ) 90 kmph , d ) 80 kmph , e ) 82 kmph | c | divide(540, divide(multiply(4, 3), 2)) | multiply(n0,n2)|divide(#0,n3)|divide(n1,#1)| | physics | C |
what is the rate percent when the simple interest on rs . 800 amount to rs . 160 in 3 years ? | "160 = ( 800 * 3 * r ) / 100 r = 6.6 % answer : b" | a ) 5 % , b ) 6.6 % , c ) 9 % , d ) 2 % , e ) 4 % | b | divide(multiply(const_100, 160), multiply(800, 3)) | multiply(n1,const_100)|multiply(n0,n2)|divide(#0,#1)| | gain | B |
nicky and cristina are running a 300 meter race . since cristina is faster than nicky , she gives him a 12 second head start . if cristina runs at a pace of 5 meters per second and nicky runs at a pace of only 3 meters per second , how many seconds will nicky have run before cristina catches up to him ? | "the distance traveled by both of them is the same at the time of overtaking . 3 ( t + 12 ) = 5 t t = 18 . cristina will catch up nicky in 18 seconds . so in 18 seconds cristina would cover = 18 * 5 = 90 meter . now time taken my nicky to cover 90 meter = 90 / 3 = 30 seconds . c" | a ) 40 , b ) 50 , c ) 30 , d ) 60 , e ) 45 | c | add(divide(multiply(3, 12), 3), divide(multiply(3, 12), subtract(5, 3))) | multiply(n1,n3)|subtract(n2,n3)|divide(#0,n3)|divide(#0,#1)|add(#2,#3)| | physics | C |
the length of a rectangular garden is three times its width . if the area of the rectangular garden is 432 square meters , then what is the width of the rectangular garden ? | "let x be the width of the garden . 3 x ^ 2 = 432 x ^ 2 = 144 x = 12 the answer is e ." | a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | e | sqrt(divide(432, const_3)) | divide(n0,const_3)|sqrt(#0)| | geometry | E |
in a group of 25 , 13 can speak latin , 15 can speak french , and 6 do n ' t speak either . how many of these speak both latin and french ? | let no . of persons who speeks booth latin and french = x so , no . of persons who speeks only latin = 13 - x no . of persons who speeks only french = 15 - x no . of persons who speeks dont speek any lag = 6 therefore ( 13 - x ) + ( 15 - x ) + x + 6 = 25 34 - x = 25 x = 9 answer : a | a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 8 | a | subtract(add(add(13, 15), 6), 25) | add(n1,n2)|add(n3,#0)|subtract(#1,n0) | other | A |
two positive integers differ by 4 , and sum of their reciprocals is 1 . then one of the numbers is | algebraic approach : let n be the smaller integer = > 1 / n + 1 / ( n + 4 ) = 1 or ( ( n + 4 ) + n ) / n ( n + 4 ) = 1 or n ^ 2 + 4 n = 2 n + 4 or n = 2 as n can not be - negative solve for n = > n = 2 . hence , a | a ) 2 , b ) b ) 1 , c ) c ) 5 , d ) d ) 21 , e ) e ) 28 | a | add(1, const_1) | add(n1,const_1) | general | A |
the purchase price of an article is $ 48 . in order to include 25 % of cost for overhead and to provide $ 12 of net profit , the markup should be | cost price of article = 48 $ % of overhead cost = 25 net profit = 12 $ we need to calculate % markup net profit as % of cost price = ( 12 / 48 ) * 100 = 25 % total markup should be = 25 + 25 = 50 % answer a | a ) 50 % , b ) 25 % , c ) 35 % , d ) 40 % , e ) 45 % | a | add(multiply(divide(12, 48), const_100), 25) | divide(n2,n0)|multiply(#0,const_100)|add(n1,#1) | gain | A |
if two integers x , y ( x > y ) are selected from - 10 to 11 ( inclusive ) , how many possible cases are there ? | "if two integers x , y ( x > y ) are selected from - 10 to 9 ( inclusive ) , how many possible cases are there ? a . 150 b . 180 c . 190 d . 210 e . 240 - - > 22 c 2 = 22 * 21 / 2 = 231 . therefore , the answer is e ." | a ) 150 , b ) 180 , c ) 190 , d ) 210 , e ) 231 | e | add(add(add(add(add(add(add(11, 10), add(11, const_2)), add(11, const_1)), 11), 10), const_2), const_1) | add(n0,n1)|add(n1,const_2)|add(n1,const_1)|add(#0,#1)|add(#3,#2)|add(#4,n1)|add(#5,n0)|add(#6,const_2)|add(#7,const_1)| | probability | E |
there are 5 guys who have equal number of bullets . all of them shoot 4 bullets . the remaining bullets are equal to the total number of bullets remaining is equal to the bullets each had after division . what was the original number of bullets each one had ? | let initially each was having x bullets they shot 4 bullets so 5 x - 20 = x x = 5 so each was having 5 bullets . total was 25 . answer : e | a ) 18 , b ) 19 , c ) 20 , d ) 21 , e ) 25 | e | multiply(divide(multiply(4, 5), subtract(5, const_1)), 5) | multiply(n0,n1)|subtract(n0,const_1)|divide(#0,#1)|multiply(n0,#2) | general | E |
sam invested rs . 3000 @ 10 % per annum for one year . if the interest is compounded half - yearly , then the amount received by sam at the end of the year will be ? | "p = rs . 3000 ; r = 10 % p . a . = 5 % per half - year ; t = 1 year = 2 half - year amount = [ 3000 * ( 1 + 5 / 100 ) 2 ] = ( 3000 * 21 / 20 * 21 / 20 ) = rs . 3307.50 answer : d" | a ) 2554.0 , b ) 3387.0 , c ) 2503.0 , d ) 3307.5 , e ) 16537.11 | d | multiply(power(add(divide(divide(10, const_2), const_100), const_1), const_2), 3000) | divide(n1,const_2)|divide(#0,const_100)|add(#1,const_1)|power(#2,const_2)|multiply(n0,#3)| | gain | D |
johnny makes $ 7.35 per hour at his work . if he works 6 hours , how much money will he earn ? | 4.75 * 6 = 28.50 . answer is c . | a ) $ 30 , b ) $ 54 , c ) $ 48.32 , d ) $ 44.10 , e ) $ 9.60 | c | multiply(7.35, 6) | multiply(n0,n1)| | physics | C |
the area of a rectangular plot is 460 square metres . if the length is 15 % more than the breadth , what is the breadth of the plot ? | "lb = 460 m 2 let breath = b l = b * ( 100 + 15 ) / 100 = 115 b / 100 from these 115 b / 100 * b = 460 b 2 = 46000 / 115 = 400 b = root of 400 = 20 m answer b" | a ) 14 m , b ) 20 m , c ) 18 m , d ) 12 m , e ) 15 m | b | sqrt(divide(460, subtract(const_1, divide(15, const_100)))) | divide(n1,const_100)|subtract(const_1,#0)|divide(n0,#1)|sqrt(#2)| | geometry | B |
two trains of equal length , running with the speeds of 60 and 40 kmph , take 55 seconds to cross each other while they are running in the same direction . what time will they take to cross each other if they are running in opposite directions ? | "rs = 60 - 40 = 20 * 5 / 18 = 100 / 18 t = 55 d = 55 * 100 / 18 = 2750 / 9 rs = 60 + 40 = 100 * 5 / 18 t = 2750 / 9 * 18 / 500 = 11 sec answer : e" | a ) 10 sec , b ) 16 sec , c ) 13 sec , d ) 67 sec , e ) 11 sec | e | multiply(multiply(multiply(const_0_2778, subtract(60, 40)), 55), inverse(multiply(const_0_2778, add(60, 40)))) | add(n0,n1)|subtract(n0,n1)|multiply(#0,const_0_2778)|multiply(#1,const_0_2778)|inverse(#2)|multiply(n2,#3)|multiply(#4,#5)| | physics | E |
it takes 1.5 hours for tim to mow the lawn . linda can mow the same lawn in 2 hours . how long will it take john and linda , work together , to mow the lawn ? | we first calculate the rate of work of john and linda john : 1 / 1.5 and linda 1 / 2 let t be the time for john and linda to mow the lawn . the work done by john alone is given by t * ( 1 / 1.5 ) the work done by linda alone is given by t * ( 1 / 2 ) when the two work together , their work will be added . hence t * ( 1 / 1.5 ) + t * ( 1 / 2 ) = 1 multiply all terms by 6 6 ( t * ( 1 / 1.5 ) + t * ( 1 / 2 ) ) = 6 and simplify 4 t + 3 t = 6 solve for t t = 6 / 7 hours = 51.5 minutes . answer c | a ) 51.1 minutes , b ) 51.3 minutes , c ) 51.5 minutes , d ) 51.7 minutes , e ) 51.9 minutes | c | inverse(add(divide(const_1, multiply(1.5, const_60)), divide(const_1, multiply(2, const_60)))) | multiply(n0,const_60)|multiply(n1,const_60)|divide(const_1,#0)|divide(const_1,#1)|add(#2,#3)|inverse(#4) | physics | C |
a cistern can be filled by a tap in 7 hours while it can be emptied by another tap in 9 hours . if both the taps are opened simultaneously , then after how much time will the cistern get filled ? | "net part filled in 1 hour = 1 / 7 - 1 / 9 = 2 / 63 therefore the cistern will be filled in 63 / 2 hours or 31.5 hours . answer : e" | a ) 4.5 hrs , b ) 5 hrs , c ) 6.5 hrs , d ) 17.2 hrs , e ) 31.5 hrs | e | divide(const_1, subtract(divide(const_1, 7), divide(const_1, 9))) | divide(const_1,n0)|divide(const_1,n1)|subtract(#0,#1)|divide(const_1,#2)| | physics | E |
how many positive factors do 120 and 80 have in common ? | "the number of common factors will be same as number of factors of the highest common factor ( hcf ) hcf of 120 and 80 is 40 number of factors of 40 = 8 answer : a" | a ) 8 , b ) 12 , c ) 16 , d ) 18 , e ) 24 | a | divide(subtract(80, const_10), const_10) | subtract(n1,const_10)|divide(#0,const_10)| | other | A |
30 men can do a work in 40 days . when should 12 men leave the work so that the entire work is completed in 40 days after they leave the work ? | "total work to be done = 30 * 40 = 1200 let 12 men leave the work after ' p ' days , so that the remaining work is completed in 40 days after they leave the work . 40 p + ( 12 * 40 ) = 1200 40 p = 720 = > p = 18 days answer : a" | a ) 18 days , b ) 10 days , c ) 55 days , d ) 44 days , e ) 22 days | a | divide(subtract(multiply(30, 40), multiply(40, 12)), 40) | multiply(n0,n1)|multiply(n1,n2)|subtract(#0,#1)|divide(#2,n1)| | physics | A |
the end of a blade on an airplane propeller is 20 feet from the center . if the propeller spins at the rate of 1,320 revolutions per second , how many miles will the tip of the blade travel in one minute ? ( 1 mile = 5,280 feet ) | "distance traveled in 1 revolution = 2 π r = 2 π 20 / 5280 revolutions in one second = 1320 revolutions in 60 seconds ( one minute ) = 1320 * 60 total distance traveled = total revolutions * distance traveled in one revolution 1320 * 60 * 2 π 20 / 5280 = 600 π c is the answer" | a ) 200 π , b ) 240 π , c ) 600 π , d ) 480 π , e ) 1,200 π | c | multiply(multiply(multiply(multiply(divide(20, add(multiply(const_2, const_100), multiply(add(const_2, const_3), const_1000))), const_2), divide(add(const_2, multiply(const_2, const_10)), add(const_3, const_4))), 1,320), const_60) | add(const_3,const_4)|add(const_2,const_3)|multiply(const_10,const_2)|multiply(const_100,const_2)|add(#2,const_2)|multiply(#1,const_1000)|add(#3,#5)|divide(#4,#0)|divide(n0,#6)|multiply(#8,const_2)|multiply(#7,#9)|multiply(n1,#10)|multiply(#11,const_60)| | physics | C |
a person buys an article at $ 800 . at what price should he sell the article so as to make a profit of 12 % ? | "a 896 cost price = $ 800 profit = 12 % of 800 = $ 96 selling price = cost price + profit = 800 + 96 = 896" | a ) 896 , b ) 890 , c ) 990 , d ) 789 , e ) 740 | c | add(800, multiply(800, divide(12, const_100))) | divide(n1,const_100)|multiply(n0,#0)|add(n0,#1)| | gain | C |
car a runs at the speed of 50 km / hr & reaches its destination in 8 hr . car b runs at the speed of 25 km / h & reaches its destination in 4 h . what is the respective ratio of distances covered by car a & car b ? | "sol . distance travelled by car a = 50 × 8 = 400 km distance travelled by car b = 25 × 4 = 100 km ratio = 400 / 100 = 4 : 1 answer : a" | a ) 4 : 1 , b ) 11 : 8 , c ) 13 : 7 , d ) 15 : 7 , e ) 16 : 9 | a | divide(multiply(50, 8), multiply(25, 4)) | multiply(n0,n1)|multiply(n2,n3)|divide(#0,#1)| | physics | A |
what is the product of all the possible values of x if x ^ 2 - 2 x - 8 ? | "explanation : = > y = x ^ 2 - 2 x - 8 = > y = ( x + 2 ) ( x - 4 ) = > x = - 2 , y = 4 product x = ( - 2 ) ( 4 ) = - 8 answer option - 8 answer : c" | a ) – 29 , b ) – 12 , c ) - 8 , d ) 29 , e ) 168 | c | divide(8, const_1) | divide(n2,const_1)| | general | C |
a contractor undertakes to complete the construction of a tunnel 720 meters long in 240 days and employs 60 men for the purpose . after 120 days , he finds that only 240 meters of the tunnel is complete . how many more men should be employ in order to complete the work in time ? | in 120 days , only 240 m of the tunnel is constructed by 60 men . the remaining 120 days , 480 m of the tunnel can be constructed by 120 men . additional number of men required = 120 - 60 = 60 men . answer : b | a ) 22 , b ) 27 , c ) 60 , d ) 88 , e ) 12 | b | subtract(120, 60) | subtract(n3,n2) | physics | B |
a metallic sheet is of rectangular shape with dimensions 48 m x 36 m . from each of its corners , a square is cut off so as to make an open box . if the length of the square is 8 m , the volume of the box ( in m 3 ) is : | "explanation clearly , l = ( 48 – 16 ) m = 32 m , b = ( 36 - 16 ) m = 20 m , h = 8 m . volume of the box = ( 32 x 20 x 8 ) m 3 = 5120 m 3 . answer b" | a ) 4830 , b ) 5120 , c ) 6420 , d ) 8960 , e ) none | b | volume_rectangular_prism(subtract(48, multiply(8, const_2)), subtract(36, multiply(8, const_2)), 8) | multiply(n2,const_2)|subtract(n0,#0)|subtract(n1,#0)|volume_rectangular_prism(n2,#1,#2)| | geometry | B |
a fruit seller had some apples . he sells 40 % apples and still has 420 apples . originally , he had | "solution suppose originally he had x apples . then , ( 100 - 40 ) % of x = 420 . ‹ = › 60 / 100 × x = 420 x ‹ = › ( 420 × 100 / 60 ‹ = › 700 . answer d" | a ) 588 apples , b ) 600 apples , c ) 672 apples , d ) 700 apples , e ) none | d | original_price_before_loss(40, 420) | original_price_before_loss(n0,n1)| | gain | D |
john invests $ x at the semi - annual constant compounded rate of 2 percent and also does $ 40,000 at the quarterly constant compounded rate of 4 percent . if the interests are the same after 1 year , what is the value of x ? ? | a = p ( 1 + r / n ) ^ nt a = total amount accrued p = principal deposited r = rate of interest in decimal form n = number of times per year , interest compounded t = time in number of years . . x ( 1 + 0.02 / 2 ) ^ 2 - x = 40,000 ( 1 + 0.04 / 4 ) ^ 4 - 40,000 [ when the principal is subtracted from the total amount accrued , the resulting difference is the interest portion and question states interests are equal ) = > x [ ( 1.01 ) ^ 2 - 1 ] = 40,000 [ ( 1.01 ) ^ 4 - 1 ] = > x [ ( 1.01 ) ^ 2 - 1 ] = 40,000 [ ( 1.01 ) ^ 2 + 1 ] [ ( 1.01 ) ^ 2 - 1 ] - - > using a ^ 2 - b ^ 2 = a + b x a - b formula and cancel common expression on both sides = > x = 40,000 ( 1.0201 + 1 ) = 80804 . hence answer is e . | a ) 50000 , b ) 55000 , c ) 60000 , d ) 70000 , e ) 80804 | e | divide(subtract(multiply(multiply(multiply(const_4, const_100), const_100), power(add(const_1, divide(divide(4, const_100), const_4)), const_4)), multiply(multiply(const_4, const_100), const_100)), subtract(power(add(const_1, divide(divide(2, const_100), const_2)), const_2), const_1)) | divide(n2,const_100)|divide(n0,const_100)|multiply(const_100,const_4)|divide(#0,const_4)|divide(#1,const_2)|multiply(#2,const_100)|add(#3,const_1)|add(#4,const_1)|power(#6,const_4)|power(#7,const_2)|multiply(#5,#8)|subtract(#9,const_1)|subtract(#10,#5)|divide(#12,#11) | gain | E |
in a new housing development , trees are to be planted along the sidewalk of a certain street . each tree takes up one square foot of sidewalk space , and there are to be 10 feet between each tree . how many trees can be planted if the road is 166 feet long ? | "let t be the number of trees . then the length required for trees on the sidewalk will be 1 * t = t to maximize the number of trees , the number of 10 feet spaces between trees should be 1 less than total number of trees . for example , if there are 3 trees , then there should be 2 spaces between them . so the number of 10 feet spaces will be t - 1 . then , the length of sidewalk required for 10 feet spaces will be 10 * ( t - 1 ) it is given that total lenght of sidewalk is 166 feet . or 10 ( t - 1 ) + t = 166 or 10 t - 10 + t = 166 or t = 16 answer : - e" | a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 16 | e | add(divide(subtract(166, const_1), add(10, const_1)), const_1) | add(n0,const_1)|subtract(n1,const_1)|divide(#1,#0)|add(#2,const_1)| | geometry | E |
cubes with each side one inch long are glued together to form a larger cube . the larger cube ' s face is painted with red color and the entire assembly is taken apart . 26 small cubes are found with no paints on them . how many of unit cubes have at least one face that is painted red ? | "use the options . the options which after getting added to 26 shows a cube of a number could be right . here 64 + 26 = 90 72 + 26 = 98 86 + 26 = 112 98 + 26 + 124 99 + 26 = 125 - - - ( 5 * 5 * 5 ) so we have 99 as the answer ! e" | a ) 64 , b ) 72 , c ) 86 , d ) 98 , e ) 99 | e | subtract(power(add(const_1, add(const_1, const_3)), const_3), 26) | add(const_1,const_3)|add(#0,const_1)|power(#1,const_3)|subtract(#2,n0)| | geometry | E |
how many digits will be there to the right of the decimal point in the product of 95 and . 02554 ? | product of 95 and . 02554 is 2.4263 . therefore number of digits to right of decimal point is 4 answer is c . | a ) 6 , b ) 7 , c ) 4 , d ) 5 , e ) 8 | c | subtract(subtract(const_100, 95), const_1) | subtract(const_100,n0)|subtract(#0,const_1) | general | C |
one week , a certain truck rental lot had a total of 24 trucks , all of which were on the lot monday morning . if 50 % of the trucks that were rented out during the week were returned to the lot on or before saturday morning of that week , and if there were at least 12 trucks on the lot that saturday morning , what is the greatest number of different trucks that could have been rented out during the week ? | n - not rented trucks ; r - rented trucks n + r = 24 n + r / 2 = 12 r = 24 a | a ) 24 , b ) 16 , c ) 12 , d ) 8 , e ) 4 | a | add(12, multiply(24, divide(50, const_100))) | divide(n1,const_100)|multiply(n0,#0)|add(n2,#1) | general | A |
every disk in a bag is either blue , yellow or green . the ratio of blue disks to yellow disks to green disks in this bag is 3 : 7 : 8 . if the total number of disks in the bag is 54 , how many more green disks than blue disks are in the bag ? | "let b : y : g = 3 x : 7 x : 8 x . 3 x + 7 x + 8 x = 18 x = 54 - - > x = 3 . g - b = 8 x - 3 x = 5 x = 15 . the answer is a ." | a ) 15 , b ) 25 , c ) 30 , d ) 35 , e ) 40 | a | multiply(divide(54, add(add(3, 7), 8)), subtract(8, 3)) | add(n0,n1)|subtract(n2,n0)|add(n2,#0)|divide(n3,#2)|multiply(#3,#1)| | general | A |
on a map , 1.5 inches represent 24 miles . how many miles approximately is the distance if you measured 46 centimeters assuming that 1 - inch is 2.54 centimeters ? | "1.5 inch = 2.54 * 1.5 cm . so , 2.54 * 1.5 represents 24 miles . so for 46 cm . : 46 / ( 2.54 * 1.5 ) = x / 24 - - - > x = 24 * 46 / ( 3.81 ) = 290 answer will be d ." | a ) 174.2 , b ) 212 , c ) 288.1 , d ) 290 , e ) 282.4 | d | multiply(divide(46, 2.54), divide(24, 1.5)) | divide(n2,n4)|divide(n1,n0)|multiply(#0,#1)| | physics | D |
when a certain tree was first planted , it was 4 feet tall , and the height of the tree increased by a constant amount each year for the next 6 years . at the end of the 6 th year , the tree was 1 / 3 taller than it was at the end of the 4 th year . by how many feet did the height of the tree increase each year ? | "say , the tree grows by x feet every year . then , 4 + 6 x = ( 1 + 1 / 3 ) ( 4 + 4 x ) or , x = 2 answer b" | a ) 3 / 10 , b ) 2 , c ) 1 / 2 , d ) 2 / 3 , e ) 6 / 5 | b | divide(4, subtract(multiply(subtract(6, 4), 3), 4)) | subtract(n1,n0)|multiply(n4,#0)|subtract(#1,n0)|divide(n0,#2)| | general | B |
on multiplying a number f by 153 , the result obtained was 102325 . however , it is found that both the 2 ' s are wrong . find the correct result . | the only thing you actually know about the correct number f is that it is divisible by 153 and has 5 as a factor . you should immediately try to find the factors of 153 and look for them in the options . 153 = 9 * 17 divisibility by 9 is easy to check . only ( d ) satisfies . | a ) 104345 , b ) 107375 , c ) 108385 , d ) 109395 , e ) 105355 | d | add(add(multiply(add(const_3, const_4), const_10), multiply(multiply(add(const_3, const_4), const_100), const_10)), 102325) | add(const_3,const_4)|multiply(#0,const_10)|multiply(#0,const_100)|multiply(#2,const_10)|add(#1,#3)|add(n1,#4) | other | D |
of the families in city x in 1994 , 40 percent owned a personal computer . the number of families in city x owning a computer in 1998 was 30 percent greater than it was in 1994 , and the total number of families in city x was 4 percent greater in 1998 than it was in 1994 . what percent of the families in city x owned a personal computer in 1998 ? | "say a 100 families existed in 1994 then the number of families owning a computer in 1994 - 40 number of families owning computer in 1998 = 40 * 130 / 100 = 52 number of families in 1998 = 104 the percentage = 52 / 104 * 100 = 50 % . answer : a" | a ) 50 % , b ) 52 % , c ) 56 % , d ) 70 % , e ) 74 % | a | multiply(const_100, divide(divide(multiply(add(30, const_100), 40), const_100), add(const_100, 4))) | add(n3,const_100)|add(n5,const_100)|multiply(n1,#0)|divide(#2,const_100)|divide(#3,#1)|multiply(#4,const_100)| | general | A |
the ages of two person differ by 20 years . if 4 years ago , the elder one be 5 times as old as the younger one , their present ages ( in years ) are respectively | "let their ages be x and ( x + 20 ) years . then , 5 ( x - 4 ) = ( x + 20 - 4 ) = > 4 x = 36 = > x = 9 their present ages are 29 years and 9 year . answer : c" | a ) 30 , 10 , b ) 25 , 5 , c ) 29 , 9 , d ) 50 , 30 , e ) 20,10 | c | add(divide(add(multiply(5, 4), subtract(20, 4)), subtract(5, const_1)), 20) | multiply(n1,n2)|subtract(n0,n1)|subtract(n2,const_1)|add(#0,#1)|divide(#3,#2)|add(n0,#4)| | general | C |
a worker ' s take - home pay last year was the same each month , and she saved the same fraction of her take - home pay each month . the total amount of money that she had saved at the end of the year was 5 times the amount of that portion of her monthly take - home pay that she did not save . if all the money that she saved last year was from her take - home pay , what fraction of her take - home pay did she save each month ? | "let x be the fraction of her take - home pay that the worker saved . let p be the monthly pay . 12 xp = 5 ( 1 - x ) p 12 xp = 5 p - 5 xp 17 xp = 5 p x = 5 / 17 the answer is c ." | a ) 2 / 11 , b ) 3 / 14 , c ) 5 / 17 , d ) 7 / 20 , e ) 11 / 30 | c | inverse(add(divide(const_12, 5), const_1)) | divide(const_12,n0)|add(#0,const_1)|inverse(#1)| | general | C |
$ 406 is divided among a , b , and c so that a receives half as much as b , and b receives half as much as c . how much money is c ' s share ? | let the shares for a , b , and c be x , 2 x , and 4 x respectively . 7 x = 406 x = 58 4 x = 232 the answer is d . | a ) $ 208 , b ) $ 216 , c ) $ 224 , d ) $ 232 , e ) $ 240 | d | multiply(divide(406, add(add(divide(const_1, const_2), const_1), const_2)), const_2) | divide(const_1,const_2)|add(#0,const_1)|add(#1,const_2)|divide(n0,#2)|multiply(#3,const_2) | general | D |
the average weight of 5 students decreases by 12 kg when one of them weighing 72 kg is replaced by a new student . the weight of the student is | "explanation : let the weight of student be x kg . given , difference in average weight = 12 kg = > ( 72 - x ) / 5 = 12 = > x = 12 answer : a" | a ) 12 kg , b ) 10 kg , c ) 20 kg , d ) 72 kg , e ) none of these | a | subtract(72, multiply(5, 12)) | multiply(n0,n1)|subtract(n2,#0)| | general | A |
a cube of sides 9 is first painted red and then cut into smaller cubes of side 3 . how many of the smaller cube have painted on exactly 2 sides ? | n = side of big cube / side of small cube and no . of smaller cubes with two surfaces painted is ( n - 2 ) * 12 ( 3 - 2 ) * 12 = 12 ans answer : c | ['a ) 30', 'b ) 24', 'c ) 12', 'd ) 8', 'e ) 9'] | c | subtract(subtract(multiply(9, 3), add(3, const_4)), multiply(const_4, 2)) | add(n1,const_4)|multiply(n0,n1)|multiply(n2,const_4)|subtract(#1,#0)|subtract(#3,#2) | geometry | C |
mr . karan borrowed a certain amount at 6 % per annum simple interest for 9 years . after 9 years , he returned rs . 8110 / - . find out the amount that he borrowed . | "explanation : let us assume mr . karan borrowed amount is rs . a . ( the principal ) by formula of simple interest , s . i . = prt / 100 where p = the principal , r = rate of interest as a % , t = time in years s . i . = ( p * 6 * 9 ) / 100 = 54 p / 100 amount = principal + s . i . 8110 = p + ( 54 p / 100 ) 8110 = ( 100 p + 54 p ) / 100 8110 = 154 p / 100 p = ( 8110 * 100 ) / 154 = rs . 5266.233 answer d" | a ) rs . 4,900 , b ) rs . 5,000 , c ) rs . 5,100 , d ) rs . 5266 , e ) none of these | d | divide(8110, add(const_1, divide(multiply(6, 9), const_100))) | multiply(n0,n1)|divide(#0,const_100)|add(#1,const_1)|divide(n3,#2)| | gain | D |
the positive numbers w , x , y , and z are such that x is 30 percent greater than y , y is 20 percent greater than z , and w is 20 percent less than x . what percent greater than z is w ? | "my strategy is same as thedobermanbut instead take z = 100 , which makes life a bit easy . as : z = 100 y = 120 ( 20 % greater than z ) z = 144 ( 20 % greater than y ) now calculate w 20 % less than z = 144 * 80 / 100 = 115.2 now by just looking , relation between w and z : w - z / z * 100 = 20 - answer c" | a ) 15.2 % , b ) 16.0 % , c ) 20.0 % , d ) 23.2 % , e ) 24.8 % | c | multiply(const_100, subtract(multiply(multiply(divide(add(30, const_100), const_100), divide(add(30, const_100), const_100)), divide(subtract(const_100, 30), const_100)), const_1)) | add(n0,const_100)|subtract(const_100,n0)|divide(#1,const_100)|divide(#0,const_100)|multiply(#3,#3)|multiply(#2,#4)|subtract(#5,const_1)|multiply(#6,const_100)| | general | C |
the least number which must be subtracted from 509 to make it exactly divisible by 9 is : | on dividing 509 by 9 , we get remainder = 5 therefore , required number to be subtracted = 5 answer : c | a ) a ) 2 , b ) b ) 3 , c ) c ) 5 , d ) d ) 5 , e ) e ) 6 | c | subtract(509, 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 | C |
the edges of three metal cubes are 1 cm , 3 cm , and 4 cm respectively . a new cube is made by melting these three cubes together . what is the edge of the new cube ( in centimeters ) ? | the total volume is 1 ^ 3 + 3 ^ 3 + 4 ^ 3 = 92 the edge of the new cube is the cube root of 92 which is about 4.5 cm . the answer is c . | a ) 3.9 , b ) 4.2 , c ) 4.5 , d ) 4.8 , e ) 5.1 | c | power(add(power(4, const_3), add(1, power(3, const_3))), const_0_33) | power(n1,const_3)|power(n2,const_3)|add(n0,#0)|add(#2,#1)|power(#3,const_0_33) | physics | C |
pipe a can fill a tank in 4 hours . due to a leak at the bottom , it takes 6 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 / 4 - 1 / x = 1 / 6 = > 1 / x = 1 / 4 - 1 / 6 = ( 3 - 2 ) / 12 = 1 / 12 = > x = 12 . answer : a" | a ) 12 , b ) 67 , c ) 95 , d ) 36 , e ) 66 | a | divide(multiply(6, 4), subtract(6, 4)) | multiply(n0,n1)|subtract(n1,n0)|divide(#0,#1)| | physics | A |
by selling a book for 260 , 20 % profit was earned . what is the cost price of the book ? | "sp = 120 % of cp ; : . cp = 260 × 100 / 120 = 216 option ' b '" | a ) a ) 215 , b ) b ) 216 , c ) c ) 230 , d ) d ) 235 , e ) e ) 240 | b | original_price_before_gain(20, 260) | original_price_before_gain(n1,n0)| | gain | B |
the wages earned by robin is 49 % more than that earned by erica . the wages earned by charles is 60 % more than that earned by erica . how much % is the wages earned by charles more than that earned by robin ? | "explanatory answer let the wages earned by erica be $ 100 then , wages earned by robin and charles will be $ 149 and $ 160 respectively . charles earns $ 49 more than robin who earns $ 149 . therefore , charles ' wage is 49 / 149 * 100 = 32.88 % . the correct choice is ( e )" | a ) 23 % , b ) 18.75 % , c ) 30 % , d ) 50 % , e ) 32.88 % | e | multiply(divide(subtract(add(const_100, 60), add(const_100, 49)), add(const_100, 49)), const_100) | add(n1,const_100)|add(n0,const_100)|subtract(#0,#1)|divide(#2,#1)|multiply(#3,const_100)| | general | E |
if the wheel is 28 cm then the number of revolutions to cover a distance of 1056 cm is ? | "2 * 22 / 7 * 28 * x = 1056 = > x = 6 answer : e" | a ) 3 , b ) 5 , c ) 26 , d ) 12 , e ) 6 | e | divide(1056, multiply(multiply(const_2, divide(add(add(multiply(const_3, const_100), multiply(const_1, const_10)), const_4), const_100)), 28)) | multiply(const_100,const_3)|multiply(const_1,const_10)|add(#0,#1)|add(#2,const_4)|divide(#3,const_100)|multiply(#4,const_2)|multiply(n0,#5)|divide(n1,#6)| | physics | E |
sum of two numbers is 40 . two times of the first exceeds by 5 from the three times of the other . then the numbers will be ? | "explanation : x + y = 15 3 y ã ¢ â ‚ ¬ â € œ 2 x = 5 x = 23 y = 17 answer : b" | a ) 26 , 14 , b ) 23 , 17 , c ) 17 , 23 , d ) 14 , 26 , e ) 15 , 25 | b | subtract(40, divide(subtract(40, divide(5, const_2)), const_2)) | divide(n1,const_2)|subtract(n0,#0)|divide(#1,const_2)|subtract(n0,#2)| | general | B |
what is the remainder when 1201 × 1202 × 1205 × 1210 is divided by 6 ? | "the remainders when dividing each number by six are : 1 , 2 , 5 , and 4 . the product is 1 * 2 * 5 * 4 = 40 the remainder when dividing 40 by 6 is 4 . the answer is d ." | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | d | reminder(multiply(1202, 1201), 1205) | multiply(n0,n1)|reminder(#0,n2)| | general | D |
when positive integer n is divided by 4 , the remainder is 2 . when n is divided by 7 , the remainder is 5 . how many values less than 100 can n take ? | "a quick approac to this q is . . the equation we can form is . . 3 x + 2 = 7 y + 5 . . 3 x - 3 = 7 y . . . 3 ( x - 1 ) = 7 y . . . so ( x - 1 ) has to be a multiple of 7 as y then will take values of multiple of 3 . . here we can see x can be 1 , 8,15 , 22,29 so 5 values till 100 is reached as ( 29 - 1 ) * 3 = 84 and next multiple of 7 will be 84 + 21 > 100 . . ans 3 . . c" | a ) 0 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | subtract(100, reminder(5, 7)) | reminder(n3,n2)|subtract(n4,#0)| | general | C |
java house charges $ 4.25 for a cup of coffee that costs a total of $ 3.85 to make . cup ' o ' joe charges $ 4.80 for a cup that costs $ 4.65 to make . if java house sells 55000 cups of coffee , how many must cup ' o ' joe sell to make at least as much in total gross profit as its competitor does ? | java : profit / cup = 4.25 - 3.85 = 0.4 : no of cups = 55,000 : gross profit = 55,000 * 0.4 = 22,000 joe : profit / cup = 0.15 : gross profit = 22,000 : no of cups = 22,000 / 0.15 = 220,000 / 1.5 ( only closes is 146,667 ) answer e | a ) 7,858 , b ) 8,301 , c ) 14,667 , d ) 63,840 , e ) 146,667 | e | divide(divide(multiply(55000, subtract(4.25, 3.85)), subtract(4.8, 4.65)), multiply(const_10, const_1000)) | multiply(const_10,const_1000)|subtract(n0,n1)|subtract(n2,n3)|multiply(n4,#1)|divide(#3,#2)|divide(#4,#0) | general | E |
10 , 15 , 22.5 , 33.75 , 50.62 , ( . . . ) | "10 ( 10 ã — 3 ) ã · 2 = 15 ( 15 ã — 3 ) ã · 2 = 22.5 ( 22.5 ã — 3 ) ã · 2 = 33.75 ( 33.75 ã — 3 ) ã · 2 = 50.62 ( 50.62 ã — 3 ) ã · 2 = 75.93 answer is c" | a ) 60 , b ) 60.75 , c ) 75.93 , d ) 76.33 , e ) 70.1 | c | subtract(negate(33.75), multiply(subtract(15, 22.5), divide(subtract(15, 22.5), subtract(10, 15)))) | negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)| | general | C |
if the length of the sides of two cubes are in the ratio 4 : 1 , what is the ratio of their total surface area ? | let x be the length of the small cube ' s side . the total surface area of the small cube is 6 x ^ 2 . the total surface area of the large cube is 6 ( 4 x ) ^ 2 = 96 x ^ 2 . the ratio of surface areas is 16 : 1 . the answer is d . | ['a ) 4 : 1', 'b ) 6 : 1', 'c ) 8 : 1', 'd ) 16 : 1', 'e ) 64 : 1'] | d | multiply(4, 4) | multiply(n0,n0) | geometry | D |
a school has received 60 % of the amount it needs for a new building by receiving a donation of $ 300 each from people already solicited . people already solicited represent 40 % of the people from whom the school will solicit donations . how much average contribution is requited from the remaining targeted people to complete the fund raising exercise ? | "let the amount school needs = x let total people school plans to solicit = t school has received 60 % of x = > ( 3 / 5 ) x people already solicited = 40 % of t = > ( 2 / 5 ) t now , as per the information given in the question : ( 3 / 5 ) x = $ 400 . ( 2 / 5 ) . t - - - - - - - - - - - - - - - - - - - - - - - - - - - 1 remaning amount is 40 % i . e ( 2 / 5 ) x - - - - - - because school has already received 60 % and the remaining people are 60 % i . e ( 3 / 5 ) . t - - - - - because 40 % of the people are already solicited so , average contribution required from the remaining targeted people is ( 2 / 5 ) x = ( amount required ) . ( 3 / 5 ) . t - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 2 divide eqn 1 by eqn 2 amount required = $ 177.78 b" | a ) $ 200 , b ) $ 177.78 , c ) $ 100 , d ) $ 277.78 , e ) $ 377.78 | b | divide(multiply(divide(multiply(divide(40, const_100), 300), divide(60, const_100)), divide(40, const_100)), divide(60, const_100)) | divide(n2,const_100)|divide(n0,const_100)|multiply(n1,#0)|divide(#2,#1)|multiply(#3,#0)|divide(#4,#1)| | general | B |
at a certain company , each employee has a salary grade s that is at least 1 and at most 7 . each employee receives an hourly wage p , in dollars , determined by the formula p = 9.50 + 0.25 ( s – 1 ) . an employee with a salary grade of 7 receives how many more dollars per hour than an employee with a salary grade of 1 ? | "salary grade of 7 is p ( 7 ) = 9.50 + 0.25 ( 7 – 1 ) = 9.50 + 0.25 * 6 ; salary grade of 1 is p ( 1 ) = 9.50 + 0.25 ( 1 – 1 ) = 9.50 ; p ( 7 ) - p ( 1 ) = 9.50 + 0.25 * 6 - 9.50 = 1.5 . answer : d ." | a ) $ 0.50 , b ) $ 1.00 , c ) $ 1.25 , d ) $ 1.50 , e ) $ 1.75 | d | add(multiply(0.25, subtract(7, 1)), 0.25) | subtract(n1,n0)|multiply(n3,#0)|add(n3,#1)| | general | D |
a and b have monthly incomes in the ratio 5 : 6 and monthly expenditures in the ratio 3 : 4 . if they save rs . 1800 and rs . 1600 respectively , find the monthly income of b | explanation : incomes of a and b = 5 x and 6 x and expenses of a and b = 3 y and 4 y then , savings of a = 5 x - 3 y = 1800 — ? ( 1 ) savings of b = 6 x - 4 y = 1600 — ? ( 2 ) by solving equations ( 1 ) and ( 2 ) y = 1400 monthly income of b = expenses of b + savings of b = 4 y + 1600 = 4 ( 1400 ) + 1600 = rs . 7200 answer : d | a ) rs . 3400 , b ) rs . 2700 , c ) rs . 1720 , d ) rs . 7200 , e ) rs . 8200 | d | multiply(divide(subtract(divide(multiply(4, 1800), 3), 1600), subtract(divide(multiply(4, 5), 3), 6)), 6) | multiply(n3,n4)|multiply(n0,n3)|divide(#0,n2)|divide(#1,n2)|subtract(#2,n5)|subtract(#3,n1)|divide(#4,#5)|multiply(n1,#6) | other | D |
the hcf and lcm of two numbers m and n are respectively 5 and 210 . if m + n = 75 , then 1 / m + 1 / n is equal to | "answer we have , m x n = 5 x 210 = 1050 â ˆ ´ 1 / m + 1 / n = ( m + n ) / mn = 75 / 1050 = 1 / 14 correct option : a" | a ) 1 / 14 , b ) 3 / 35 , c ) 5 / 37 , d ) 2 / 35 , e ) none | a | divide(75, multiply(5, 210)) | multiply(n0,n1)|divide(n2,#0)| | general | A |
a card is drawn from a pack of 52 cards . the probability of not getting a face card ? | clearly in the 52 cards other than face cards = 52 - 16 = 36 probability of not getting a face card = 36 / 52 = 9 / 13 correct option is a | a ) 9 / 13 , b ) 5 / 12 , c ) 3 / 13 , d ) 6 / 17 , e ) 1 / 13 | a | divide(subtract(52, multiply(const_4, const_4)), 52) | multiply(const_4,const_4)|subtract(n0,#0)|divide(#1,n0) | probability | A |
a farmer has an apple orchard consisting of fuji and gala apple trees . due to high winds this year 10 % of his trees cross pollinated . the number of his trees that are pure fuji plus the cross - pollinated ones totals 170 , while 3 / 4 of all his trees are pure fuji . how many of his trees are pure gala ? | "let f = pure fuji , g = pure gala and c - cross pollinated . c = 10 % of x where x is total trees . c = . 1 x also 3 x / 4 = f and c + f = 170 = > . 1 x + 3 / 4 x = 170 = > x = 200 200 - 170 = pure gala = 30 . answer a" | a ) 30 , b ) 33 , c ) 55 , d ) 77 , e ) 88 | a | subtract(divide(170, add(divide(10, const_100), divide(3, 4))), 170) | divide(n0,const_100)|divide(n2,n3)|add(#0,#1)|divide(n1,#2)|subtract(#3,n1)| | general | A |
the size of a flat - screen tablet is given as the length of the screen ’ s diagonal . how many square inches greater is the screen of a square 6 - inch flat - screen tablet than a square 5 - inch flat - screen tablet ? | if we take a square with side length x and draw a diagonal , we get two isosceles right triangles . if we focus on one such right triangle , we see that the legs have length x . square 6 - inch flat - screen television the diagonal ( hypotenuse ) = 6 so , we can apply the pythagorean theorem to get x ² + x ² = 6 ² simplify : 2 x ² = 6 ² divide both sides by 2 to get : x ² = 6 ² / 2 since the area of the square = x ² , we can see that the area of this square is 6 ² / 2 square 5 - inch flat - screen television the diagonal ( hypotenuse ) = 5 so , we can apply the pythagorean theorem to get x ² + x ² = 5 ² simplify : 2 x ² = 5 ² divide both sides by 2 to get : x ² = 5 ² / 2 since the area of the square = x ² , we can see that the area of this square is 5 ² / 2 difference in areas = 6 ² / 2 - 5 ² / 2 = ( 6 ² - 5 ² ) / 2 = ( 36 - 25 ) / 2 = 11 / 2 = 5.5 e | ['a ) 5.0', 'b ) 6.0', 'c ) 6.8', 'd ) 5.8', 'e ) 5.5'] | e | subtract(divide(power(6, const_2), const_2), divide(power(5, const_2), const_2)) | power(n0,const_2)|power(n1,const_2)|divide(#0,const_2)|divide(#1,const_2)|subtract(#2,#3) | geometry | E |
how many litres of pure acid are there in 4 litres of a 35 % solution | explanation : question of this type looks a bit typical , but it is too simple , as below . . . it will be 8 * 20 / 100 = 1.4 answer : option e | a ) 1.5 , b ) 1.6 , c ) 1.7 , d ) 1.8 , e ) 1.4 | e | multiply(divide(35, const_100), 4) | divide(n1,const_100)|multiply(n0,#0)| | gain | E |
jerry travels 8 miles at an average speed of 40 miles per hour , stops for 12 minutes , and then travels another 20 miles at an average speed of 60 miles per hour . what is jerry ’ s average speed , in miles per hour , for this trip ? | "total time taken by jerry = ( 8 / 40 ) * 60 minutes + 12 minutes + ( 20 / 60 ) * 60 minutes = 44 minutes average speed = total distance / total time = ( 8 + 20 ) miles / ( 44 / 60 ) hours = 28 * 60 / 44 = 42 miles per hour answer : option a" | a ) 42 , b ) 42.5 , c ) 44 , d ) 50 , e ) 52.5 | a | divide(add(8, 20), add(add(divide(8, 40), divide(12, 60)), divide(20, 60))) | add(n0,n3)|divide(n0,n1)|divide(n2,n4)|divide(n3,n4)|add(#1,#2)|add(#4,#3)|divide(#0,#5)| | physics | A |
a car started running at a speed of 30 km / hr and the speed of the car was increased by 2 km / hr at the end of every hour . find the total distance covered by the car in the first 7 hours of the journey . | "the total distance covered by the car in the first 7 hours = 30 + 32 + 34 + 36 + 38 + 40 + 42 = sum of 7 terms in ap whose first term is 30 and last term is 42 = 7 / 2 [ 30 + 42 ] = 252 km . answer : c" | a ) 342 km , b ) 352 km , c ) 252 km , d ) 742 km , e ) 382 km | c | divide(add(add(30, multiply(2, 7)), 30), 2) | multiply(n1,n2)|add(n0,#0)|add(n0,#1)|divide(#2,n1)| | physics | C |
a tank contains 10,000 gallons of a solution that is 5 percent sodium chloride by volume . if 2,000 gallons of water evaporate from the tank , the remaining solution will be approximately what percent sodium chloride ? | "the amount of sodium chloride is 0.05 * 10,000 = 500 gallons 500 / 8000 = 1 / 16 = 6.25 % the answer is c ." | a ) 1.25 % , b ) 3.75 % , c ) 6.25 % , d ) 6.67 % , e ) 11.7 % | c | multiply(divide(multiply(multiply(const_100, const_100), divide(5, const_100)), subtract(multiply(const_100, const_100), add(multiply(add(const_2, const_3), multiply(multiply(add(const_2, const_3), const_2), const_100)), multiply(add(const_2, const_3), const_100)))), const_100) | add(const_2,const_3)|divide(n1,const_100)|multiply(const_100,const_100)|multiply(#1,#2)|multiply(#0,const_2)|multiply(#0,const_100)|multiply(#4,const_100)|multiply(#0,#6)|add(#7,#5)|subtract(#2,#8)|divide(#3,#9)|multiply(#10,const_100)| | gain | C |
a tailor trims 7 feet from opposite edges of a square piece of cloth , and 5 feet from the other two edges . if 45 square feet of cloth remain , what was the length of a side of the original piece of cloth ? | "let the original side of the square be x . ( x - 14 ) * ( x - 10 ) = 45 = 5 * 9 x = 19 the answer is d ." | a ) 13 , b ) 15 , c ) 17 , d ) 19 , e ) 21 | d | divide(add(add(multiply(const_2, 7), multiply(const_2, const_3.0)), sqrt(add(multiply(7, subtract(45, multiply(multiply(const_2, 5), multiply(const_2, 7)))), power(add(multiply(const_2, 7), multiply(const_2, 5)), const_2)))), const_2) | multiply(n0,const_2)|multiply(const_3.0,const_2)|add(#0,#1)|multiply(#1,#0)|power(#2,const_2)|subtract(n2,#3)|multiply(#5,n0)|add(#6,#4)|sqrt(#7)|add(#2,#8)|divide(#9,const_2)| | geometry | D |
the maximum number of students among them 1340 pens and 1280 pencils can be distributed in such a way that each student get the same number of pens and same number of pencils ? | "number of pens = 1340 number of pencils = 1280 required number of students = h . c . f . of 1340 and 1280 = 20 answer is c" | a ) 25 , b ) 30 , c ) 20 , d ) 45 , e ) 90 | c | gcd(1340, 1280) | gcd(n0,n1)| | general | C |
a train passes a man standing on a platform in 8 seconds and also crosses the platform which is 270 metres long in 20 seconds . the length of the train ( in metres ) is : | "explanation : let the length of train be l m . acc . to question ( 270 + l ) / 20 = l / 8 2160 + 8 l = 20 l l = 2160 / 12 = 180 m answer a" | a ) 180 , b ) 176 , c ) 175 , d ) 96 , e ) none of these | a | multiply(divide(270, subtract(20, 8)), 8) | subtract(n2,n0)|divide(n1,#0)|multiply(n0,#1)| | physics | A |
how many 5 - digit numbers that do not contain the digits 4 or 7 are there ? | we can have 7 digits ( 1 , 2,3 , 5,6 , 8,9 ) for the first place ( ten thousand ' s place ) . and similarly 8 digits for thousand ' s , hundred ' s , tenth ' s and unit digit . ( 0,1 , 2,3 , 5,6 , 8,9 ) so in total 7 * 8 * 8 * 8 * 8 = 28672 hence c | a ) 44648 , b ) 27844 , c ) 28642 , d ) 16864 , e ) 32458 | c | multiply(power(multiply(const_4, const_2), subtract(5, const_1)), 7) | multiply(const_2,const_4)|subtract(n0,const_1)|power(#0,#1)|multiply(n2,#2) | general | C |
the closest approximation of a ( 69.28 × 0.004 ) / 0.03 is | "a ( 69.28 × 0.004 ) / 0.03 1 . 0.004 = 4 × 10 ^ ( - 3 ) 2 . 0.03 = 3 × 10 ^ ( - 2 ) 3 . ( a × b ) / c = a × ( b / c ) 4 . 0.004 / 0.03 = 4 × 10 ^ ( - 3 ) / ( 3 × 10 ^ ( - 2 ) ) = 4 × 10 ^ ( - 3 - ( - 2 ) ) / 3 = 4 × 10 ^ ( - 1 ) / 3 = ( 4 / 3 ) × 10 ^ ( - 1 ) = 1.333 × 10 ^ ( - 1 ) therefore , ( 69.28 × 0.004 ) / 0.03 = 69.28 × ( 0.004 / 0.03 ) = 69.28 × 1.33 × 10 ^ ( - 1 ) = 69.28 × 1.33 / 10 = 6.928 * 1.33 now , 7 × 2 = 14 7 × 1 = 7 or better : 6.9 × 1 = 6.9 6.9 × 2 = 13.8 hence , 6.9 < 6.928 × 1.33 < 13.8 9.2 is the only answer that satisfies this condition . c" | a ) 0.092 , b ) 0.92 , c ) 9.2 , d ) 92 , e ) 920 | c | multiply(divide(0.004, 0.03), 69.28) | divide(n1,n2)|multiply(n0,#0)| | general | C |
sophia finished 2 / 3 of a book . she calculated that she finished 90 more pages than she has yet to read . how long is her book ? | let xx be the total number of pages in the book , then she finished 23 ⋅ x 23 ⋅ x pages . then she has x − 23 ⋅ x = 13 ⋅ xx − 23 ⋅ x = 13 ⋅ x pages left . 23 ⋅ x − 13 ⋅ x = 9023 ⋅ x − 13 ⋅ x = 90 13 ⋅ x = 9013 ⋅ x = 90 x = 270 x = 270 so the book is 270 pages long . answer : b | a ) 229 , b ) 270 , c ) 877 , d ) 266 , e ) 281 | b | divide(90, subtract(const_1, divide(2, 3))) | divide(n0,n1)|subtract(const_1,#0)|divide(n2,#1) | general | B |
a certain bacteria colony doubles in size every day for 22 days , at which point it reaches the limit of its habitat and can no longer grow . if two bacteria colonies start growing simultaneously , how many days will it take them to reach the habitat ’ s limit ? | "if there is one bacteria colony , then it will reach the limit of its habitat in 20 days . if there are two bacteria colonies , then in order to reach the limit of habitat they would need to double one time less than in case with one colony . thus colonies need to double 21 times . answer : c . similar questions to practice : hope it helps ." | a ) 6.33 , b ) 7.5 , c ) 21 , d ) 15 , e ) 19 | c | subtract(22, divide(22, 22)) | divide(n0,n0)|subtract(n0,#0)| | physics | C |
the speed of a boat in still water is 21 km / hr and the rate of current is 8 km / hr . the distance travelled downstream in 13 minutes is | "explanation : speed downstreams = ( 21 + 8 ) kmph = 18 kmph . distance travelled = ( 29 x 13 / 60 ) km = 6.3 km option c" | a ) 1.6 km , b ) 2 km , c ) 6.3 km , d ) 4 km , e ) none of these | c | multiply(divide(13, const_60), add(21, 8)) | add(n0,n1)|divide(n2,const_60)|multiply(#0,#1)| | physics | C |
a certain airline ' s fleet consisted of 100 type a planes at the beginning of 1980 . at the end of each year , starting with 1980 , the airline retired 3 of the type a planes and acquired 4 new type b plans . how many years did it take before the number of type a planes left in the airline ' s fleet was less than 50 percent of the fleet ? | "let x be the number of years . 4 x > 100 - 3 x 7 x > 100 x > 14 + 2 / 7 the answer is b ." | a ) 14 , b ) 15 , c ) 16 , d ) 17 , e ) 18 | b | divide(100, add(3, 4)) | add(n3,n4)|divide(n0,#0)| | gain | B |
( 0.8 ) ( power 3 ) - ( 0.5 ) ( power 3 ) / ( 0.8 ) ( power 2 ) + 0.40 + ( 0.5 ) ( power 2 ) is : | "given expression = ( 0.8 ) ( power 3 ) - ( 0.5 ) ( power 3 ) / ( 0.8 ) ( power 2 ) + ( 0.8 x 0.5 ) + ( 0.5 ) ( power 2 ) = a ( power 3 ) - b ( power 3 ) / a ( power 2 ) + ab + b ( power 2 ) = ( a - b ) = ( 0.8 - 0.5 ) = 0.30 answer is d ." | a ) 0.6 , b ) 0.5 , c ) 0.35 , d ) 0.3 , e ) none of them | d | divide(subtract(power(0.8, 3), power(0.5, 3)), add(add(power(0.8, 2), 0.40), power(0.5, 2))) | power(n0,n1)|power(n2,n1)|power(n0,n5)|power(n2,n5)|add(n6,#2)|subtract(#0,#1)|add(#4,#3)|divide(#5,#6)| | general | D |
working together , printer a and printer b would finish the task in 24 minutes . printer a alone would finish the task in 60 minutes . how many pages does the task contain if printer b prints 7 pages a minute more than printer a ? | answer : b . | a ) 600 , b ) 840 , c ) 1000 , d ) 1200 , e ) 1500 | b | multiply(divide(7, subtract(divide(60, 24), const_1)), 60) | divide(n1,n0)|subtract(#0,const_1)|divide(n2,#1)|multiply(#2,n1)| | physics | B |
the length of a room is 5.5 m and width is 3.75 m . what is the cost of paying the floor by slabs at the rate of $ 300 per sq . metre . | "area = 5.5 × 3.75 sq . metre . cost for 1 sq . metre . = $ 300 hence , total cost = 5.5 × 3.75 × 300 = $ 6187.50 d" | a ) $ 2587.50 , b ) $ 3587.50 , c ) $ 4187.50 , d ) $ 6187.50 , e ) $ 8587.50 | d | multiply(300, multiply(5.5, 3.75)) | multiply(n0,n1)|multiply(n2,#0)| | physics | D |
an exam consists of 2 true / false questions . brian forgets to study , so he must guess blindly on each question . if any score above 70 % is a passing grade , what is the probability that brian passes ? | "if you have 8 t or f and brian is going to guess then each question he has a 50 % chance of getting correct . if a passing score is 70 % it means brian needs to get 6 / 8 = 75 % , 7 / 8 = 87.5 % , or 8 / 8 = 100 % to pass . each is a possibility . if brian gets a 5 / 8 ( = 62.5 % ) or below he fails . so first figure out the number of ways that brian can get 6 out of 8 , 7 out of 8 , and 8 out of 8 questions correct . which is 8 choose 6 , equals is 28 , 8 choose 7 , equals 8 , and 8 choose 8 , equals 1 . this sums to 37 . the number of possible questions outcomes - the sum of 8 choose 8 , 7 choose 8 , 6 choose 8 … . 2 choose 8 , 1 choose 8 , and 0 choose 8 is 256 , so the chance of him passing is 1 / 16 . a" | a ) 1 / 16 , b ) 37 / 256 , c ) 1 / 2 , d ) 219 / 256 , e ) 15 / 16 | a | add(divide(subtract(const_1, add(add(power(divide(const_1, const_2), 2), multiply(2, power(divide(const_1, const_2), 2))), multiply(multiply(2, const_3), power(divide(const_1, const_2), 2)))), const_10), subtract(const_1, add(add(power(divide(const_1, const_2), 2), multiply(2, power(divide(const_1, const_2), 2))), multiply(multiply(2, const_3), power(divide(const_1, const_2), 2))))) | divide(const_1,const_2)|multiply(n0,const_3)|power(#0,n0)|multiply(n0,#2)|multiply(#1,#2)|add(#3,#2)|add(#5,#4)|subtract(const_1,#6)|divide(#7,const_10)|add(#8,#7)| | general | A |
how many plants will be there in a circular bed whose outer edge measure 31 cms , allowing 4 cm 2 for each plant ? | circumference of circular bed = 31 cm area of circular bed = ( 31 ) 2 â „ 4 ï € space for each plant = 4 cm 2 â ˆ ´ required number of plants = ( 31 ) 2 â „ 4 ï € ã · 4 = 19.11 = 19 ( approx ) answer a | ['a ) 19', 'b ) 750', 'c ) 24', 'd ) 120', 'e ) none of these'] | a | divide(circle_area(divide(31, multiply(2, const_pi))), const_4) | multiply(n2,const_pi)|divide(n0,#0)|circle_area(#1)|divide(#2,const_4) | physics | A |
the malibu country club needs to drain its pool for refinishing . the hose they use to drain it can remove 60 cubic feet of water per minute . if the pool is 60 feet wide by 150 feet long by 10 feet deep and is currently at 80 % capacity , how long will it take to drain the pool ? | "volume of pool = 60 * 150 * 10 cu . ft , 80 % full = 60 * 150 * 10 * 0.8 cu . ft water is available to drain . draining capacity = 60 cu . ft / min therefore time taken = 60 * 150 * 10 * 0.8 / 60 min = 1200 min c" | a ) 1800 , b ) 1000 , c ) 1200 , d ) 1400 , e ) 1600 | c | divide(multiply(divide(80, const_100), multiply(multiply(60, 150), 10)), 60) | divide(n4,const_100)|multiply(n1,n2)|multiply(n3,#1)|multiply(#0,#2)|divide(#3,n0)| | gain | C |
a bowl of fruit contains 12 apples and 23 oranges . how many oranges must be removed so that 60 % of the pieces of fruit in the bowl will be apples ? | "number of apples = 14 number of oranges = 23 let number of oranges that must be removed so that 60 % of pieces of fruit in bowl will be apples = x total number of fruits after x oranges are removed = 12 + ( 23 - x ) = 35 - x 12 / ( 35 - x ) = 6 / 10 = > 20 = 35 - x = > x = 15 answer c" | a ) 3 , b ) 6 , c ) 15 , d ) 17 , e ) 20 | c | divide(subtract(multiply(add(12, 23), divide(60, const_100)), 12), divide(60, const_100)) | add(n0,n1)|divide(n2,const_100)|multiply(#0,#1)|subtract(#2,n0)|divide(#3,#1)| | gain | C |
a 12 by 16 rectangle is inscribed in circle . what is the circumference of the circle ? | "the diagonal of the rectangle will be the diameter of the circle . and perimeter = 2 * pi * r ans : d" | a ) 5 π , b ) 10 π , c ) 15 π , d ) 20 π , e ) 30 π | d | circumface(divide(sqrt(add(power(12, const_2), power(16, 16))), 16)) | power(n0,const_2)|power(n1,n1)|add(#0,#1)|sqrt(#2)|divide(#3,n1)|circumface(#4)| | geometry | D |
bob finishes the first half of an exam in two - sixth the time it takes him to finish the second half . if the whole exam takes him an hour , how many minutes does he spend on the first half of the exam ? | many times , it is easiest to think of problems like this conceptually ( as opposed to formulaically ) . conceptually , if the first half of the exam takes 2 / 6 ' s the time of the second half ( 6 / 6 ' s ) , we can see that the hour the entire exam took to finish can be broken down into eight ' s . ( another way to look at this problem is via ratio ' s - first half is 2 : 8 and the second half is 6 : 8 ) . with each sixth of an hour being 15 / 2 minutes ( 60 / 8 ) , the first half of the exam would have taken 15 minutes . correct answer is a . | a ) 15 , b ) 24 , c ) 27 , d ) 36 , e ) 40 | a | subtract(const_60, multiply(divide(const_2, add(const_2, const_4)), const_60)) | add(const_2,const_4)|divide(const_2,#0)|multiply(#1,const_60)|subtract(const_60,#2) | physics | A |
what profit percent is made by selling an article at a certain price , if by selling at 2 / 3 rd of that price , there would be a loss of 20 % ? | "sp 2 = 2 / 3 sp 1 cp = 100 sp 2 = 80 2 / 3 sp 1 = 80 sp 1 = 120 100 - - - 20 = > 20 % answer : a" | a ) 20 % , b ) 28 % , c ) 60 % , d ) 26 % , e ) 30 % | a | subtract(divide(subtract(const_100, 20), divide(2, 3)), const_100) | divide(n0,n1)|subtract(const_100,n2)|divide(#1,#0)|subtract(#2,const_100)| | gain | A |
x and y invested in a business . they earned some profit which they divided in the ratio of 2 : 3 . if x invested rs . 60,000 . the amount invested by y is | "solution suppose y invested rs . y then , 60000 / y = 2 / 3 â € ¹ = â € º y = ( 60000 ã — 3 / 2 ) . â € ¹ = â € º y = 90000 . answer d" | a ) rs . 45,000 , b ) rs . 50,000 , c ) rs . 60,000 , d ) rs . 90,000 , e ) none | d | divide(multiply(multiply(add(const_1, const_4), const_1000), 2), 3) | add(const_1,const_4)|multiply(#0,const_1000)|multiply(n0,#1)|divide(#2,n1)| | gain | D |
what is the measure of the radius of the circle inscribed in a triangle whose sides measure 8 , 15 and 19 units ? | "sides are 8 , 15 and 19 . . . thus it is right angle triangle since 19 ^ 2 = 8 ^ 2 + 15 ^ 2 therefore , area = 1 / 2 * 15 * 8 = 60 we have to find in - radius therefore , area of triangle = s * r . . . . where s = semi - perimeter and r = in - radius now s = semi - perimeter = 19 + 15 + 8 / 2 = 21 thus , 60 = 21 * r and hence r = in - radius = 2.8 option a" | a ) 2.8 units , b ) 6 units , c ) 3 units , d ) 5 units , e ) 12 units | a | divide(triangle_area_three_edges(8, 15, 19), divide(triangle_perimeter(8, 15, 19), const_2)) | triangle_area_three_edges(n0,n1,n2)|triangle_perimeter(n0,n1,n2)|divide(#1,const_2)|divide(#0,#2)| | geometry | A |
in a group of buffaloes and ducks , the number of legs are 24 more than twice the number of heads . what is the number of buffaloes in the group ? | let no . of buffaloes be x & no . of ducks be y buffaloes have 4 legs while ducks have 2 4 x + 2 y = 2 * ( x + y ) + 24 = > 4 x + 2 y - 2 x - 2 y = 24 = > x = 12 number of buffaloes in the group = 12 answer : a | a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | a | divide(24, const_2) | divide(n0,const_2) | general | A |
if difference between compound interest and simple interest on a sum at 10 % p . a . for 2 years is rs . 41 then sum is | "p ( r / 100 ) ^ 2 = c . i - s . i p ( 10 / 100 ) ^ 2 = 41 4100 answer : e" | a ) s . 5000 , b ) s . 5100 , c ) s . 5800 , d ) s . 6000 , e ) s . 4100 | e | divide(41, multiply(divide(10, const_100), divide(10, const_100))) | divide(n0,const_100)|multiply(#0,#0)|divide(n2,#1)| | gain | E |
36 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 : 36 : : 12 : x working hours / day 6 : 5 30 x 6 x x = 36 x 5 x 12 x = ( 36 x 5 x 12 ) / ( 30 x 6 ) x = 12 answer a" | a ) 12 , b ) 16 , c ) 13 , d ) 18 , e ) 19 | a | divide(multiply(multiply(36, 12), 5), multiply(30, 6)) | multiply(n0,n1)|multiply(n3,n4)|multiply(n2,#0)|divide(#2,#1)| | physics | A |
the center of a circle lies on the origin of the coordinate plane . if a point ( x , y ) is randomly selected inside of the circle , what is the probability that y > 0 or y > x ? | "the line y = x divides the circle into two equal areas . all the points above the line y = x satisfy the condition that y > x . all the points above the x - axis satisfy the condition that y > 0 . the union of these two areas is 5 / 8 of the circle . the answer is a ." | a ) 5 / 8 , b ) 1 / 6 , c ) 3 / 8 , d ) 1 / 2 , e ) 3 / 4 | a | divide(const_3, multiply(const_2, const_4)) | multiply(const_2,const_4)|divide(const_3,#0)| | physics | A |
the area of a square is equal to five times the area of a rectangle of dimensions 25 cm * 5 cm . what is the perimeter of the square ? | "area of the square = s * s = 5 ( 25 * 5 ) = > s = 25 cm perimeter of the square = 4 * 25 = 100 cm . answer : option a" | a ) 100 , b ) 800 , c ) 500 , d ) 600 , e ) 700 | a | multiply(sqrt(multiply(rectangle_area(25, 5), divide(5, const_2))), const_4) | divide(n1,const_2)|rectangle_area(n0,n1)|multiply(#0,#1)|sqrt(#2)|multiply(#3,const_4)| | geometry | A |
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