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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john is going with 10 friends on a trip to sidney for spring break . airfare and hotel costs a total of $ 12100.00 for the group of 11 friends . how much does each person have to pay for their hotel and airfare ? | answer = b the total cost of the trip ( $ 12100.00 ) divided by 11 equals $ 1100.00 . | a ) $ 1010 , b ) $ 1100 , c ) $ 1110 , d ) $ 1101 , e ) $ 1200 | b | divide(12100, 11) | divide(n1,n2) | general | B |
by selling an article at rs . 800 , a profit of 25 % is made . find its cost price ? | sp = 800 cp = ( sp ) * [ 100 / ( 100 + p ) ] = 800 * [ 100 / ( 100 + 25 ) ] = 800 * [ 100 / 125 ] = rs . 640 answer : c | a ) s . 486 , b ) s . 455 , c ) s . 640 , d ) s . 480 , e ) s . 489 | c | divide(multiply(800, const_100), add(const_100, 25)) | add(n1,const_100)|multiply(n0,const_100)|divide(#1,#0) | gain | C |
natasha climbs up a hill , and descends along the same way she went up . it takes her 3 hours to reach the top and 2 hours to come back down . if her average speed along the whole journey is 3 kilometers per hour , what was her average speed ( in kilometers per hour ) while climbing to the top ? | "lets assume distance to top as x , so the total distance travelled by natasha = 2 x total time taken = 3 + 2 = 5 hrs avg speed = total dist / total time taken = 2 x / 5 avg speed of complete journey is given as = 3 hrs 2 x / 5 = 3 x = 7.5 miles avg speed while climbing = distance / time = 7.5 / 3 = 2.5 option b" | a ) 1.5 , b ) 2.5 , c ) 3.75 , d ) 5 , e ) 7.5 | b | divide(divide(multiply(add(3, 2), 3), 2), 3) | add(n0,n1)|multiply(n2,#0)|divide(#1,n1)|divide(#2,n0)| | physics | B |
a crow leaves its nest , and flies back and forth from its nest to a nearby ditch to gather worms . the distance between the nest and the ditch is 200 meters . in one and a half hours , the crow manages to bring worms to its nest 15 times . what is the speed of the crow in kilometers per hour ? | "the distance between the nest and the ditch is 200 meters . 15 times mean = a crow leaves its nest , and flies back ( going and coming back ) i . e . 2 times we get total 30 rounds . so the distance is 30 * 200 = 6000 . d = st 6000 / 1.5 = t , i think we can take 6000 meters as 6 km , then only we get t = 4 . ( 1000 meters = 1 km ) d )" | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 6 | d | divide(divide(multiply(200, multiply(15, const_2)), const_1000), divide(15, const_10)) | divide(n1,const_10)|multiply(n1,const_2)|multiply(n0,#1)|divide(#2,const_1000)|divide(#3,#0)| | physics | D |
if tim had lunch at $ 50.50 and he gave 15 % tip , how much did he spend ? | "the tip is 20 % of what he paid for lunch . hence tip = 15 % of 50.50 = ( 15 / 100 ) * 50.50 = $ 7.575 total spent 50.50 + 7.575 = $ 58 correct answer a" | a ) $ 58 , b ) $ 60.60 , c ) $ 70.60 , d ) $ 40.60 , e ) $ 50.60 | a | add(50.50, divide(multiply(50.50, 15), const_100)) | multiply(n0,n1)|divide(#0,const_100)|add(n0,#1)| | gain | A |
a towel , when bleached , lost 10 % of its length and 20 % of its breadth . what is the percentage decrease in area ? | "percentage change in area = ( − 10 − 20 + ( 10 × 20 ) / 100 ) % = − 28 % i . e . , area is decreased by 28 % answer : a" | a ) 28 % , b ) 30 % , c ) 44 % , d ) 54 % , e ) 64 % | a | divide(subtract(multiply(const_100, const_100), multiply(subtract(const_100, 10), subtract(const_100, 20))), const_100) | multiply(const_100,const_100)|subtract(const_100,n0)|subtract(const_100,n1)|multiply(#1,#2)|subtract(#0,#3)|divide(#4,const_100)| | gain | A |
a certain bag contains 60 balls — 22 white , 18 green , 8 yellow , 5 red , and 7 purple . if a ball is to be chosen at random , what is the probability that the ball will be neither red nor purple ? | "according to the stem the ball can be white , green or yellow , so the probability is ( white + green + yellow ) / ( total ) = ( 22 + 18 + 8 ) / 60 = 48 / 60 = 0.8 . answer : d ." | a ) 0.09 , b ) 0.15 , c ) 0.54 , d ) 0.8 , e ) 0.91 | d | divide(add(add(22, 18), 8), 60) | add(n1,n2)|add(n3,#0)|divide(#1,n0)| | other | D |
in a rectangular coordinate system , what is the area of a quadrilateral whose vertices have the coordinates ( 4 , - 3 ) , ( 4 , 7 ) , ( 12 , 2 ) , ( 12 , - 7 ) ? | by graphing the points , we can see that this figure is a trapezoid . a trapezoid is any quadrilateral that has one set of parallel sides , and the formula for the area of a trapezoid is : area = ( 1 / 2 ) × ( base 1 + base 2 ) × ( height ) , where the bases are the parallel sides . we can now determine the area of the quadrilateral : area = 1 / 2 × ( 10 + 9 ) × 8 = 76 . the answer is d . | ['a ) 64', 'b ) 68', 'c ) 72', 'd ) 76', 'e ) 80'] | d | divide(multiply(add(add(3, 7), add(2, 7)), subtract(12, 4)), const_2) | add(n1,n3)|add(n3,n5)|subtract(n4,n0)|add(#0,#1)|multiply(#3,#2)|divide(#4,const_2) | geometry | D |
find 40 % of 320 | "we know that r % of m is equal to r / 100 × m . so , we have 40 % of 320 40 / 100 × 320 = 128 answer : c" | a ) 96 , b ) 94 , c ) 128 , d ) 74 , e ) 110 | c | divide(40, 320) | divide(n0,n1)| | gain | C |
the speed of the boat in still water in 12 kmph . it can travel downstream through 42 kms in 3 hrs . in what time would it cover the same distance upstream ? | "still water = 12 km / hr downstream = 42 / 3 = 14 km / hr upstream = > > still water = ( u + v / 2 ) = > > 12 = u + 14 / 2 = 10 km / hr so time taken in upstream = 42 / 10 = 4.2 hrs answer : c" | a ) 8 hours , b ) 6 hours , c ) 4.2 hours , d ) 5 hours , e ) 6 hours | c | divide(42, subtract(12, subtract(divide(42, 3), 12))) | divide(n1,n2)|subtract(#0,n0)|subtract(n0,#1)|divide(n1,#2)| | physics | C |
sanoop bought 8 t - shirts at an average price ( arithmetic mean ) of rs . 526 . if sanoop returned 1 t - shirts to the retailer , and the average price of the remaining t - shirts was rs . 505 , then what is the average price , of the 3 returned t - shirts ? | total price of 8 t - shirts = 8 * 526 = 4208 total price of 7 t - shirts = 7 * 505 = 3535 total price of 1 t - shirts = 4208 - 3535 = 673 average price of 3 t - shirts = 673 correct option answer : e | a ) 560 , b ) 561 , c ) 562 , d ) 563 , e ) 673 | e | subtract(multiply(8, 526), multiply(subtract(8, 1), 505)) | multiply(n0,n1)|subtract(n0,n2)|multiply(n3,#1)|subtract(#0,#2) | general | E |
a and b finish the job in 15 days . while a , b and c can finish it in 10 days . c alone will finish the job in | "explanation : 10 = ( 15 * x ) / ( 15 + x ) 150 + 10 x = 15 x 5 x = 150 x = 30 answer : option b" | a ) 40 days , b ) 30 days , c ) 60 days , d ) 70 days , e ) 50 days | b | divide(multiply(10, 15), subtract(15, 10)) | multiply(n0,n1)|subtract(n0,n1)|divide(#0,#1)| | physics | B |
the length of the bridge , which a train 160 meters long and travelling at 45 km / hr can cross in 30 seconds , is ? | "speed = ( 45 * 5 / 18 ) m / sec = ( 25 / 2 ) m / sec . time = 30 sec . let the length of bridge be x meters . then , ( 160 + x ) / 30 = 25 / 2 = = > 2 ( 160 + x ) = 750 = = > x = 215 m . answer : c" | a ) 766 m , b ) 156 m , c ) 215 m , d ) 156 m , e ) 156 m | c | subtract(multiply(divide(multiply(45, speed(const_1000, const_1)), speed(const_3600, const_1)), 30), 160) | speed(const_1000,const_1)|speed(const_3600,const_1)|multiply(n1,#0)|divide(#2,#1)|multiply(n2,#3)|subtract(#4,n0)| | physics | C |
a train running at the speed of 30 km / hr crosses a pole in 12 seconds . what is the length of the train ? | "speed = ( 30 x ( 5 / 18 ) m / sec = ( 25 / 3 ) m / sec . length of the train = ( speed x time ) . length of the train = ( ( 25 / 3 ) x 12 ) m = 100 m d" | a ) 70 , b ) 80 , c ) 90 , d ) 100 , e ) 110 | d | multiply(divide(multiply(30, const_1000), const_3600), 12) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics | D |
a , b , c and d enter into partnership . a subscribes 1 / 3 of the capital b 1 / 4 , c 1 / 5 and d the rest . how much share did a get in a profit of rs . 2475 ? | "2475 * 1 / 3 = 825 option c" | a ) s . 800 , b ) s . 810 , c ) s . 825 , d ) s . 900 , e ) s . 920 | c | multiply(2475, divide(1, 3)) | divide(n0,n1)|multiply(n6,#0)| | general | C |
36 men can complete a piece of work in 18 days . in how many days will 72 men complete the same work ? | "explanation : less men , means more days { indirect proportion } let the number of days be x then , 72 : 36 : : 18 : x x = 9 answer : d ) 9 days" | a ) 24 , b ) 17 , c ) 18 , d ) 9 , e ) 11 | d | divide(multiply(18, 36), 72) | multiply(n0,n1)|divide(#0,n2)| | physics | D |
every letter in the alphabet has a number value that is equal to its place in the alphabet . thus , the letter a has a value of 1 , the letter b has a value of 2 , the letter c has a value of 3 , etc . . . the number value of a word is obtained by adding up the value of the letters in the word and then multiplying that sum by the length of the word . what is the number value of the word ` ` dog ' ' ? | "` ` dog ' ' = ( 4 + 15 + 7 ) * 3 = 78 . the answer is c ." | a ) 72 , b ) 75 , c ) 78 , d ) 81 , e ) 84 | c | multiply(add(add(subtract(multiply(2, const_10), 2), 1), multiply(2, const_10)), 3) | multiply(n1,const_10)|subtract(#0,n1)|add(n0,#1)|add(#2,#0)|multiply(n2,#3)| | general | C |
if the simple interest on a certain sum of money for 7 years is one – fifth of the sum , then the rate of interest per annum is | "explanation : let the principal ( p ) be x then , simple interest ( si ) = x / 5 time ( t ) = 7 years rate of interest per annum ( r ) = ( 100 × si ) / pt = ( 100 × ( x / 5 ) / ( x × 7 ) = 20 / 7 = 2.85 % answer : option e" | a ) 4 % , b ) 2.50 % , c ) 6.20 % , d ) 5 % , e ) 2.85 % | e | divide(divide(const_100, add(const_1, const_4)), 7) | add(const_1,const_4)|divide(const_100,#0)|divide(#1,n0)| | gain | E |
the cross - section of a cannel is a trapezium in shape . if the cannel is 16 m wide at the top and 4 m wide at the bottom and the area of cross - section is 700 sq m , the depth of cannel is ? | "1 / 2 * d ( 16 + 4 ) = 700 d = 70 answer : e" | a ) 76 , b ) 28 , c ) 27 , d ) 80 , e ) 70 | e | divide(divide(divide(700, divide(add(16, 4), const_2)), 4), const_2) | add(n0,n1)|divide(#0,const_2)|divide(n2,#1)|divide(#2,n1)|divide(#3,const_2)| | physics | E |
a man two flats for $ 675958 each . on one he gains 12 % while on the other he loses 12 % . how much does he gain or lose in the whole transaction ? | "in such a case there is always a loss loss % = ( 12 / 10 ) ^ 2 = 36 / 25 = 1.44 % answer is a" | a ) 1.44 % , b ) 2.56 % , c ) 3.12 % , d ) 4.65 % , e ) 5.12 % | a | multiply(divide(subtract(add(multiply(divide(const_100, add(const_100, 12)), 675958), multiply(divide(const_100, subtract(const_100, 12)), 675958)), add(675958, 675958)), add(multiply(divide(const_100, add(const_100, 12)), 675958), multiply(divide(const_100, subtract(const_100, 12)), 675958))), const_100) | add(n1,const_100)|add(n0,n0)|subtract(const_100,n1)|divide(const_100,#0)|divide(const_100,#2)|multiply(n0,#3)|multiply(n0,#4)|add(#5,#6)|subtract(#7,#1)|divide(#8,#7)|multiply(#9,const_100)| | gain | A |
a working mom wakes up every day at 7 am and goes to sleep at 11 pm . she works 8 hours a day . spends 2 hours working out at the gym . she spends 1.5 hours cooking dinner and doing dishes . she spends . 5 of an hour giving her daughter a bath . she spends 1 hour helping with homework and getting her daughter ready for bed . she spends another . 5 of an hour packing lunches for the family for the next day . she spends . 5 of an hour cleaning the house and 2 hours taking a shower and reading or watching t . v . before she goes to sleep . what percent of her day does she spend at work ? | 8 + 2 + 1.5 + . 5 + 1 + . 5 + . 5 + 2 = 16 = total number of hours 8 / 16 = . 05 = 50 percemt the answer is b | a ) 60 , b ) 50 , c ) 70 , d ) 55 , e ) 40 | b | multiply(divide(8, add(subtract(multiply(const_3, const_4), 7), 11)), const_100) | multiply(const_3,const_4)|subtract(#0,n0)|add(n1,#1)|divide(n2,#2)|multiply(#3,const_100) | physics | B |
the sum of the present ages of two persons a and b is 54 . if the age of a is twice that of b , find the sum of their ages 5 years hence ? | "a + b = 54 , a = 2 b 2 b + b = 54 = > b = 18 then a = 36 . 5 years , their ages will be 41 and 23 . sum of their ages = 41 + 23 = 64 . answer : b" | a ) 50 , b ) 64 , c ) 70 , d ) 80 , e ) 90 | b | add(add(multiply(divide(54, 5), const_2), 5), add(divide(54, 5), 5)) | divide(n0,n1)|add(#0,n1)|multiply(#0,const_2)|add(#2,n1)|add(#3,#1)| | general | B |
how many liters of water must be evaporated from 50 liters of a 3 - percent sugar solution to get a 2 - percent solution ? | "how many liters of water must be evaporated from 50 liters of a 3 - percent sugar solution to get a 2 - percent solution ? 3 % of a 50 liter solution is 1.5 l . so you are trying to determine how many liters must a solution be for the 1.5 l to represent 2 % of the solution . set up an inequality and solve for x : 1.5 / x = 1 / 2 x = 3 since you need a 15 l solution , you must evaporate 18 of the original 50 l solution to get a 2 % solution . answer is e ." | a ) 35 , b ) 33 1 / 3 , c ) 27 , d ) 16 2 / 3 , e ) 18 | e | subtract(50, multiply(divide(50, const_100), 2)) | divide(n0,const_100)|multiply(n2,#0)|subtract(n0,#1)| | gain | E |
in a certain business school class , p students are accounting majors , q students are finance majors , r students are marketing majors , and s students are strategy majors . if pqrs = 1365 , and if 1 < p < q < r < s , how many students in the class are accounting majors ? | pqrs = 1365 = 3 * 5 * 7 * 13 since 1 < p < q < r < s , the number of students who are accounting majors is p = 3 . the answer is a . | a ) 3 , b ) 5 , c ) 8 , d ) 11 , e ) 17 | a | divide(1365, multiply(multiply(add(const_2, const_3), add(add(const_2, const_3), const_2)), add(const_10, const_3))) | add(const_10,const_3)|add(const_2,const_3)|add(#1,const_2)|multiply(#1,#2)|multiply(#0,#3)|divide(n0,#4) | gain | A |
rates for having a manuscript typed at a certain typing service are $ 5 per page for the first time a page is typed and $ 3 per page each time a page is revised . if a certain manuscript has 200 pages , of which 80 were revised only once , 20 were revised twice , and the rest required no revisions , what was the total cost of having the manuscript typed ? | "for 200 - 80 - 20 = 100 pages only cost is 5 $ per page for the first time page is typed - 100 * 5 = 500 $ ; for 80 pages the cost is : first time 5 $ + 3 $ of the first revision - 80 * ( 5 + 3 ) = 640 $ ; for 20 pages the cost is : first time 5 $ + 3 $ of the first revision + 3 $ of the second revision - 20 ( 5 + 3 + 3 ) = 220 $ ; total : 500 + 640 + 220 = 1360 $ . answer : a ." | a ) $ 1360 , b ) $ 1620 , c ) $ 1650 , d ) $ 680 , e ) $ 770 | a | add(add(multiply(200, 5), multiply(80, 3)), multiply(multiply(20, 3), const_2)) | multiply(n0,n2)|multiply(n1,n3)|multiply(n1,n4)|add(#0,#1)|multiply(#2,const_2)|add(#3,#4)| | general | A |
a cyclist bikes x distance at 8 miles per hour and returns over the same path at 812 miles per hour . what is the cyclist ' s average rate for the round trip in miles per hour ? | "distance = d 1 = x miles speed = s 1 = 8 miles per hour time = t 1 = distance / speed = x / 8 2 . going from b to a distance = d 2 = x miles speed = s 2 = 12 miles per hour time = t 2 = distance / speed = x / 12 3 . average speed = total distance / total time total distance = x + x = 2 x total time = x / 12 + x / 8 = x ( 1 / 12 + 1 / 8 ) = = 5 x / 24 speed = 2 x / ( 5 x / 24 ) = 48 / 5 = 9.6 answer : d" | a ) 8.1 , b ) 8.3 , c ) 8.6 , d ) 9.6 , e ) 9.0 | d | divide(add(8, 812), const_2) | add(n0,n1)|divide(#0,const_2)| | physics | D |
the compounded ratio of ( 2 : 3 ) , ( 6 : 11 ) and ( 11 : 2 ) is : | answer : option c 2 / 3 : 6 / 11 : 11 / 2 = 2 : 1 | a ) 1 : 2 , b ) 5 : 9 , c ) 2 : 1 , d ) 11 : 24 , e ) none | c | multiply(multiply(divide(2, 3), divide(6, 11)), divide(11, 2)) | divide(n3,n0)|divide(n0,n1)|divide(n2,n3)|multiply(#1,#2)|multiply(#0,#3) | other | C |
a can do a work in 6 days . b can do the same work in 8 days . both a & b together will finish the work and they got $ 1000 from that work . find their shares ? | "ratio of their works a : b = 6 : 8 ratio of their wages a : b = 3 : 4 a ' s share = ( 3 / 5 ) 1000 = 600 b ' s share = ( 4 / 5 ) 1000 = 800 correct option is a" | a ) 600,800 , b ) 500,500 , c ) 300,700 , d ) 800,200 , e ) 550,450 | a | divide(multiply(6, 8), add(6, 8)) | add(n0,n1)|multiply(n0,n1)|divide(#1,#0)| | physics | A |
a man can do a job in 15 days . his father takes 20 days and his son finishes it in 25 days . how long will they take to complete the job if they all work together ? | a 6.4 days 1 day work of the three persons = ( 1 / 15 + 1 / 20 + 1 / 25 ) = 47 / 300 so , all three together will complete the work in 300 / 47 = 6.4 days . - | a ) 6.4 days , b ) 4.4 days , c ) 5.4 days , d ) 8.4 days , e ) 2.4 days | a | divide(const_1, add(divide(const_1, 25), add(divide(const_1, 15), divide(const_1, 20)))) | divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|add(#3,#2)|divide(const_1,#4) | physics | A |
5 , 7 , 11 , 18 , 33 , ( . . . ) | "explanation : 5 5 × 2 - 3 = 7 7 × 2 - 3 = 11 11 × 2 - 3 = 18 18 × 2 - 3 = 33 33 × 2 - 3 = 63 answer : option d" | a ) 22 , b ) 35 , c ) 27 , d ) 63 , e ) 25 | d | subtract(negate(18), multiply(subtract(7, 11), divide(subtract(7, 11), subtract(5, 7)))) | negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)| | general | D |
in a 100 m race , a beats b by 10 m and c by 13 m . in a race of 180 m , b will beat c by : | a : b = 100 : 90 . a : c = 100 : 87 . \ inline \ fn _ jvn { \ color { black } \ therefore \ frac { b } { c } = \ left ( \ frac { b } { a } \ times \ frac { a } { c } \ right ) = \ left ( \ frac { 90 } { 100 } \ times \ frac { 100 } { 87 } \ right ) = \ frac { 30 } { 29 } } when b runs 30 m , c runs 29 m . when b runs 180 m , c runs \ inline \ fn _ jvn { \ color { black } \ left ( \ frac { 29 } { 30 } \ times 180 \ right ) m } = 174 m \ inline \ fn _ jvn { \ color { blue } \ therefore } b beats c by ( 180 - 174 ) m = 6 m . answer : d ) 6 m | a ) 3 , b ) 7 , c ) 8 , d ) 6 , e ) 0 | d | subtract(180, multiply(inverse(multiply(divide(100, subtract(100, 13)), divide(subtract(100, 10), 100))), 180)) | subtract(n0,n2)|subtract(n0,n1)|divide(n0,#0)|divide(#1,n0)|multiply(#2,#3)|inverse(#4)|multiply(n3,#5)|subtract(n3,#6) | physics | D |
the temperature of a certain cup of coffee 30 minutes after it was poured was 120 degrees fahrenheit . if the temperature f of the coffee t minutes after it was poured can be determined by the formula f = 120 * 2 ^ ( - at ) + 60 , where f is in degrees fahrenheit and a is a constant . then the temperature of the coffee 30 minutes after it was poured was how many degrees fahrenheit ? | "first , we have to find a . we know that after t = 30 minutes the temperature f = 120 degrees . hence : 120 = 120 * ( 2 ^ - 30 a ) + 60 60 = 120 * ( 2 ^ - 30 a ) 60 / 120 = 2 ^ - 30 a 1 / 2 = 2 ^ - 30 a 2 ^ - 1 = 2 ^ - 30 a - 1 = - 30 a 1 / 30 = a now we need to find f after t = 30 minutes : f = 120 * ( 2 ^ - 1 / 30 * 30 ) + 60 f = 120 * ( 2 ^ - 1 ) + 60 f = 120 * ( 1 / 2 ^ 1 ) + 60 f = 120 * 1 / 2 + 60 f = 60 + 60 = 120 answer b !" | a ) 65 , b ) 120 , c ) 80 , d ) 85 , e ) 90 | b | add(multiply(power(2, multiply(divide(60, 30), subtract(const_1, 2))), 120), 60) | divide(n4,n0)|subtract(const_1,n3)|multiply(#0,#1)|power(n3,#2)|multiply(n1,#3)|add(n4,#4)| | general | B |
gopi gives rs . 90 plus one turban as salary to his servant for one year . the servant leaves after 9 months and receives rs . 55 and the turban . find the price of the turban . | let the price of turban be x . thus , for one year the salary = ( 90 + x ) for 9 months he should earn 3434 ( 90 + x ) . now he gets one turban and rs . 55 . thus , 3434 ( 90 + x ) = 55 + x or 270 + 3 x = 220 + 4 x or x = 50 answer : d | a ) 27 , b ) 36 , c ) 29 , d ) 50 , e ) 11 | d | subtract(multiply(divide(subtract(90, 55), subtract(const_12, 9)), const_12), 90) | subtract(n0,n2)|subtract(const_12,n1)|divide(#0,#1)|multiply(#2,const_12)|subtract(#3,n0) | general | D |
a miniature roulette wheel is divided into 9 equal sectors , each bearing a distinct integer from 1 to 9 , inclusive . each time the wheel is spun , a ball randomly determines the winning sector by settling in that sector . if the wheel is spun three times , approximately what is the probability that the product of the three winning sectors ’ integers will be even ? | "the only way to have an odd product is if all 3 integers are odd . p ( odd product ) = 5 / 9 * 5 / 9 * 5 / 9 = 125 / 729 p ( even product ) = 1 - 125 / 729 = 604 / 729 which is about 83 % the answer is c ." | a ) 50 % , b ) 67 % , c ) 83 % , d ) 90 % , e ) 96 % | c | multiply(subtract(1, power(divide(divide(9, const_2), 9), const_2)), const_100) | divide(n0,const_2)|divide(#0,n0)|power(#1,const_2)|subtract(n1,#2)|multiply(#3,const_100)| | general | C |
a retailer bought a machine at a wholesale price of $ 80 and later on sold it after a 20 % discount of the retail price . if the retailer made a profit equivalent to 20 % of the wholesale price , what is the retail price of the machine ? | "the price after the discount was 1.2 * 80 = $ 96 let x be the retail price . 0.8 x = $ 96 x = 96 / 0.8 = $ 120 the answer is c ." | a ) $ 80 , b ) $ 100 , c ) $ 120 , d ) 135 , e ) 160 | c | divide(multiply(add(80, divide(multiply(80, 20), const_100)), const_100), multiply(multiply(const_3, const_3), 20)) | multiply(n0,n2)|multiply(const_3,const_3)|divide(#0,const_100)|multiply(n1,#1)|add(n0,#2)|multiply(#4,const_100)|divide(#5,#3)| | gain | C |
| x + 3 | – | 2 - x | = | 8 + x | how many solutions will this equation have ? | "you have | x + 3 | - | 4 - x | = | 8 + x | first , look at the three values independently of their absolute value sign , in other words : | x + 3 | - | 4 - x | = | 8 + x | ( x + 3 ) - ( 4 - x ) = ( 8 + x ) now , you ' re looking at x < - 8 , so x is a number less than - 8 . let ' s pretend x = - 10 here to make things a bit easier to understand . when x = - 10 i . ) ( x + 3 ) ( - 10 + 3 ) ( - 7 ) ii . ) ( 4 - x ) ( 4 - [ - 10 ] ) ( double negative , so it becomes positive ) ( 4 + 10 ) ( 14 ) iii . ) ( 8 + x ) ( 8 + - 10 ) ( - 2 ) in other words , when x < - 8 , ( x + 3 ) and ( 8 + x ) are negative . to solve problems like this , we need to check for the sign change . here is how i do it step by step . i . ) | x + 3 | - | 4 - x | = | 8 + x | ii . ) ignore absolute value signs ( for now ) and find the values of x which make ( x + 3 ) , ( 4 - x ) and ( 8 + x ) = to zero as follows : ( x + 3 ) x = - 3 ( - 3 + 3 ) = 0 ( 4 - x ) x = 4 ( 4 - 4 ) = 0 ( 8 + x ) x = - 8 ( 8 + - 8 ) = 2 c" | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | c | divide(multiply(add(2, const_3.0), const_2), 8) | add(n0,const_4.0)|multiply(#0,const_2)|divide(#1,n2)| | general | C |
what is the next no . 4 12 84 | "3 ^ 0 + 3 = 4 3 ^ 2 + 3 = 12 3 ^ 4 + 3 = 84 3 ^ 6 + 3 = 732 answer : b" | a ) 632 , b ) 732 , c ) 832 , d ) 850 , e ) 902 | b | add(4, reminder(4, 12)) | reminder(n0,n1)|add(n0,#0)| | general | B |
two trucks each 250 m long are running in opposite directions on parallel paths . their speeds are 30 km / hr and 20 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? | relative speed = 20 + 30 = 50 km / hr . 50 * 5 / 18 = 125 / 9 m / sec . distance covered = 250 + 250 = 500 m . required time = 500 * 9 / 125 = 36 sec . answer : e | a ) 77 sec , b ) 66 sec , c ) 48 sec , d ) 55 sec , e ) 36 sec | e | divide(multiply(250, const_2), multiply(add(30, 20), const_0_2778)) | add(n1,n2)|multiply(n0,const_2)|multiply(#0,const_0_2778)|divide(#1,#2) | physics | E |
the captain of a cricket team of 11 members is 25 years old and the wicket keeper is 5 years older . if the ages of these two are excluded , the average age of the remaining players is one year less than the average age of the whole team . what is the average age of the team ? | "let the average age of the whole team be x years . 11 x - ( 25 + 30 ) = 9 ( x - 1 ) 11 x - 9 x = 46 2 x = 46 x = 23 . the average age of the team is 23 years . the answer is c ." | a ) 20 , b ) 21 , c ) 23 , d ) 25 , e ) 26 | c | divide(subtract(add(25, add(25, 5)), multiply(const_3.0, const_3.0)), const_2) | add(n1,n2)|multiply(const_3.0,const_3.0)|add(n1,#0)|subtract(#2,#1)|divide(#3,const_2)| | general | C |
what is remainder of the division ( 1125 * 1127 * 1129 ) / 12 ? | "remainder will be number / 100 here as the divisor is two digit number = 12 . hence checking for the last two digits = 5 * 7 * 9 = 15 thus remainder = 3 . answer : b" | a ) 0 , b ) 3 , c ) 1 , d ) 2 , e ) 4 | b | subtract(multiply(multiply(1125, 1127), 1129), subtract(multiply(multiply(1125, 1127), 1129), const_3)) | multiply(n0,n1)|multiply(n2,#0)|subtract(#1,const_3)|subtract(#1,#2)| | general | B |
the number of positive integers n in the range 12 ≤ n ≤ 40 such that the product ( n − 1 ) ( n − 2 ) . . . 3 . 2.1 is not divisible by n is | explanation : ( n - 1 ) ! is always divisible by n unless and untill n is a prime number e . g . ( 5 - 1 ) ! = 4 ! which is not divisible by 5 bcuz 5 is a prime number therefore , find all prime numbers in the range of 12 to 40 i . e 13 , 17 , 19 , 23 , 29 , 31 and 37 hence , the number of positive integers n is 7 . answer : b | a ) 5 , b ) 7 , c ) 13 , d ) 14 , e ) 16 | b | add(add(3, 2), 2) | add(n3,n4)|add(n3,#0) | general | B |
p and q started a business investing rs 75000 and rs 15000 resp . in what ratio the profit earned after 2 years be divided between p and q respectively . | "explanation : in this type of question as time frame for both investors is equal then just get the ratio of their investments . p : q = 75000 : 15000 = 75 : 15 = 5 : 1 option a" | a ) 5 : 1 , b ) 17 : 3 , c ) 5 : 6 , d ) 17 : 7 , e ) 3 : 8 | a | divide(75000, 15000) | divide(n0,n1)| | gain | A |
if a student loses 5 kilograms , he will weigh twice as much as his sister . together they now weigh 104 kilograms . what is the student ' s present weight in kilograms ? | "let x be the weight of the sister . then the student ' s weight is 2 x + 5 . x + ( 2 x + 5 ) = 104 3 x = 99 x = 33 kg then the student ' s weight is 71 kg . the answer is b ." | a ) 70 , b ) 71 , c ) 72 , d ) 73 , e ) 74 | b | subtract(104, divide(subtract(104, 5), const_3)) | subtract(n1,n0)|divide(#0,const_3)|subtract(n1,#1)| | other | B |
if 40 % of 3 / 5 of a number is 36 , then the number is ? | "let the number be x . then 40 % of 3 / 5 of x = 36 40 / 100 * 3 / 5 * x = 36 x = ( 36 * 50 / 12 ) = 150 required number = 150 . correct option : b" | a ) 80 , b ) 150 , c ) 75 , d ) 90 , e ) none of these | b | divide(36, multiply(divide(40, const_100), divide(3, 5))) | divide(n0,const_100)|divide(n1,n2)|multiply(#0,#1)|divide(n3,#2)| | gain | B |
positive integer y is 50 percent of 50 percent of positive integer x , and y percent of x equals 50 . what is the value of x ? | "y = 50 % of 50 % 0 f x = x / 4 and y / 100 of x = 50 y / 100 * 4 y = 50 y = 35 and x = 140 answer - d" | a ) 50 , b ) 100 , c ) 200 , d ) 140 , e ) 2,000 | d | multiply(multiply(divide(50, 50), divide(50, 50)), const_1000) | divide(n0,n2)|divide(n1,n2)|multiply(#0,#1)|multiply(#2,const_1000)| | general | D |
two trains are moving in opposite directions at 60 km / hr and 90 km / hr . their lengths are 1.50 km and 1.25 km respectively . the time taken by the slower train to cross the faster train in seconds is ? | "relative speed = 60 + 90 = 150 km / hr . = 150 * 5 / 18 = 125 / 3 m / sec . distance covered = 1.50 + 1.25 = 2.75 km = 2750 m . required time = 2750 * 3 / 125 = 66 sec . answer : c" | a ) 12 , b ) 77 , c ) 66 , d ) 99 , e ) 11 | c | subtract(divide(multiply(1.50, const_1000), divide(multiply(60, const_1000), const_3600)), divide(multiply(1.25, const_1000), divide(multiply(90, const_1000), const_3600))) | multiply(n2,const_1000)|multiply(n0,const_1000)|multiply(n3,const_1000)|multiply(n1,const_1000)|divide(#1,const_3600)|divide(#3,const_3600)|divide(#0,#4)|divide(#2,#5)|subtract(#6,#7)| | physics | C |
two vessels having volumes in the ratio 3 : 5 are filled with water and milk solutions . the ratio of milk and water in the two vessels are 1 : 2 and 3 : 2 respectively . if the contents of both the vessel are empties into a larger vessel , find the ratio of milk and water in the larger vessel . | vessel a = 300 gallons - - > milk = 100 , water = 200 ; vessel b = 500 gallons - - > milk = 300 , water = 200 ; vessel a + b = 800 gallons - - > milk = 400 , water 400 . the ratio = 400 / 400 - - > 1 : 1 answer : b | ['a ) 99 : 61', 'b ) 1 : 1', 'c ) 61 : 160', 'd ) 61 : 99', 'e ) 99 : 160'] | b | power(1, 3) | power(n2,n0) | other | B |
a fill pipe can fill 1 / 2 of cistern in 35 minutes . in how many minutes , it can fill 1 / 2 of the cistern ? | "required time = 35 * 2 * 1 / 2 = 20 minutes answer is d" | a ) 20 min , b ) 25 min , c ) 30 min , d ) 35 min , e ) 40 min | d | divide(35, 1) | divide(n2,n0)| | physics | D |
if x - y - z = 03 x + 4 y + 3 z = 45 x + 2 y + 7 z = 22 , what is the value of z ? | explanation : 4 x + 2 y - 5 z = - 21 â € “ - - - - - - - i 2 x - 2 y + z = 7 â € “ - - - - - - ii 4 x + 3 y - z = - 1 - - - - - - - - iii solve the first equation for x . x - y - z = 0 x = y + z substitute the solution for x into the second and third equations . ii - - - > 3 * ( y + z ) + 4 y + 3 z = 4 3 y + 3 z + 4 y + 3 z = 4 7 y + 6 z = 4 iii - - - - > 5 * ( y + z ) + 2 y + 7 z = 22 5 y + 5 z + 2 y + 7 z = 22 7 y + 12 z = 22 subtract the new second equation from the new third equation and solve for z . 7 y + 12 z = 22 - ( 7 y + 6 z = 4 ) 6 z = 18 z = 3 answer is a | a ) 3 , b ) 4 , c ) 5 , d ) 2 , e ) 1 | a | divide(divide(multiply(divide(add(multiply(add(const_1, const_4), add(22, multiply(const_4, 22))), add(22, multiply(const_2, 22))), add(add(45, 2), multiply(add(const_1, const_4), 7))), 7), 22), add(const_1, const_4)) | add(const_1,const_4)|add(n3,n4)|multiply(n6,const_2)|multiply(n6,const_4)|add(n6,#2)|add(n6,#3)|multiply(n5,#0)|add(#1,#6)|multiply(#0,#5)|add(#4,#8)|divide(#9,#7)|multiply(n5,#10)|divide(#11,n6)|divide(#12,#0) | general | A |
chris age after 15 years will be 5 times his age 5 years back . what is the present age of chris ? | "chris present age = x after 15 years = x + 15 5 years back = x - 5 x + 15 = 5 ( x - 5 ) x = 10 answer is e" | a ) 20 , b ) 25 , c ) 15 , d ) 22 , e ) 10 | e | subtract(divide(add(multiply(5, 5), 15), subtract(5, const_1)), subtract(divide(add(multiply(5, 5), 15), subtract(5, const_1)), 5)) | multiply(n1,n1)|subtract(n1,const_1)|add(n0,#0)|divide(#2,#1)|subtract(#3,n1)|subtract(#3,#4)| | general | E |
the average salary of all the workers in a workshop is rs . 8000 . the average salary of 7 technicians is rs . 18000 and the average salary of the rest is rs . 6000 . the total number of workers in the workshop is | "solution let the toatl number of workers be x . then 8000 x = ( 18000 x 7 ) + 6000 ( x - 7 ) x = 42 answer c" | a ) 20 , b ) 21 , c ) 42 , d ) 23 , e ) 24 | c | add(7, divide(multiply(7, subtract(18000, 8000)), subtract(8000, 6000))) | subtract(n2,n0)|subtract(n0,n3)|multiply(n1,#0)|divide(#2,#1)|add(n1,#3)| | general | C |
the sum of three consecutive multiples of 3 is 90 . what is the largest number ? | "let the numbers be 3 x , 3 x + 3 and 3 x + 6 . then , 3 x + ( 3 x + 3 ) + ( 3 x + 6 ) = 90 9 x = 81 x = 9 largest number = 3 x + 6 = 33 answer : b" | a ) 30 , b ) 33 , c ) 36 , d ) 39 , e ) 42 | b | add(add(power(add(add(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), const_2))) | subtract(n0,const_10)|subtract(#0,const_2)|divide(#1,const_4)|add(#2,const_2)|power(#2,const_2)|add(#3,const_2)|power(#3,const_2)|add(#5,const_2)|add(#4,#6)|power(#5,const_2)|power(#7,const_2)|add(#9,#10)|add(#11,#8)| | general | B |
two assembly line inspectors , lauren and steven , inspect widgets as they come off the assembly line . if lauren inspects every fifth widget , starting with the fifth , and steven inspects every third , starting with the third , how many of the 98 widgets produced in the first hour of operation are not inspected by either inspector ? | "widgets inspected by lauren : ( ( 95 - 5 ) / 5 ) + 1 = 18 + 1 = 19 widgets inspected by steven : ( ( 96 - 3 ) / 3 ) + 1 = 31 + 1 = 32 widgets inspected by both : ( ( 90 - 15 ) + 1 = 6 total : 19 + 32 - 6 = 45 hence , widgets not inspected : 98 - 45 = 53 answer : c" | a ) 91 , b ) 59 , c ) 53 , d ) 47 , e ) 45 | c | subtract(98, subtract(add(floor(divide(98, add(const_4, const_1))), floor(divide(98, const_4))), floor(divide(98, add(const_10, add(const_4, const_1)))))) | add(const_1,const_4)|divide(n0,const_4)|add(#0,const_10)|divide(n0,#0)|floor(#1)|divide(n0,#2)|floor(#3)|add(#6,#4)|floor(#5)|subtract(#7,#8)|subtract(n0,#9)| | other | C |
at an international conference , “ red ” world countries and “ blue ” world countries are the only participants . the ratio of “ red ” world participants to “ blue ” world participants is 5 : 5 . if one - third of “ red ” world participants are left - handed and two - thirds of “ blue ” world participants are left - handed , then what is the fraction of the participants who are left - handed ? | "red : blue = 5 : 5 let red = 5 x and blue = 5 x 1 / 3 of red are left handed = > 1 / 3 * 5 x = 5 x / 3 red left handed 2 / 3 of blue are left handed = > 2 / 3 * 5 x = 10 x / 3 blue left handed fraction of participants who are left handed = total left handed / total participants = ( red left handed + blue left handed ) / total participants = ( 5 x / 3 + 10 x / 3 ) / ( 5 x + 5 x ) = ( 15 x / 30 x ) = 1 / 2 answer : d" | a ) 1 / 4 , b ) 1 / 8 , c ) 1 / 6 , d ) 1 / 2 , e ) 1 | d | divide(add(multiply(5, const_0_33), multiply(5, divide(const_2, const_3))), add(5, 5)) | add(n0,n1)|divide(const_2,const_3)|multiply(n0,const_0_33)|multiply(n1,#1)|add(#2,#3)|divide(#4,#0)| | other | D |
96 % of 5 / 8 = | "should be simple . 0.96 * 5 / 8 = 4.8 / 8 = 0.6 correct option : c" | a ) 0.2 , b ) 0.5 , c ) 0.6 , d ) 0.75 , e ) 1.0 | c | divide(multiply(divide(multiply(8, 5), const_100), 96), const_100) | multiply(n1,n2)|divide(#0,const_100)|multiply(n0,#1)|divide(#2,const_100)| | general | C |
a manufacturing company has 15 % cobalt , 25 % led and 60 % of copper . if 5 kg of led is used in a mixture how much copper we need to use | 25 % = 5 kg 100 % = 20 kg 60 % of 20 kg = 12 kg 12 kg answer : a | a ) 12 kg , b ) 13 kg , c ) 22 kg , d ) 14 kg , e ) 15 kg | a | multiply(divide(60, const_100), multiply(5, const_4)) | divide(n2,const_100)|multiply(n3,const_4)|multiply(#0,#1) | general | A |
a bag contains 8 red , 6 blue and 4 green balls . if 2 ballsare picked at random , what is the probability that both are red ? | "p ( both are red ) , = 8 c 216 c 2 = 8 c 216 c 2 = 28 / 120 = 7 / 30 d" | a ) 1 / 13 , b ) 2 / 23 , c ) 5 / 26 , d ) 7 / 30 , e ) 3 / 23 | d | divide(choose(8, 2), choose(add(add(8, 6), 4), 2)) | add(n0,n1)|choose(n0,n3)|add(n2,#0)|choose(#2,n3)|divide(#1,#3)| | other | D |
john invests $ x at the semi - annual constant compounded rate of 2 percent and also does $ 8,000 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 = 8,000 ( 1 + 0.04 / 4 ) ^ 4 - 8,000 [ 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 ] = 8,000 [ ( 1.01 ) ^ 4 - 1 ] = > x [ ( 1.01 ) ^ 2 - 1 ] = 8,000 [ ( 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 = 8,000 ( 1.0201 + 1 ) = 16160.8 hence answer is c ." | a ) 10000 , b ) 12000 , c ) 16160.8 , d ) 14000 , e ) 15000 | c | divide(subtract(multiply(multiply(multiply(4, const_100), const_100), power(add(1, divide(divide(4, const_100), 4)), 4)), multiply(multiply(4, const_100), const_100)), subtract(power(add(1, divide(divide(2, const_100), 2)), 2), 1)) | divide(n2,const_100)|divide(n0,const_100)|multiply(const_100,n2)|divide(#0,n2)|divide(#1,n0)|multiply(#2,const_100)|add(#3,n3)|add(#4,n3)|power(#6,n2)|power(#7,n0)|multiply(#5,#8)|subtract(#9,n3)|subtract(#10,#5)|divide(#12,#11)| | gain | C |
in a certain country 1 / 3 of 4 = 6 . assuming the same proportion , what would be the value of 1 / 6 of 20 ? | "b 15" | a ) 14 , b ) 15 , c ) 20 , d ) 22 , e ) 25 | b | multiply(divide(3, 4), 20) | divide(n1,n2)|multiply(n6,#0)| | general | B |
if sharon ' s weekly salary increased by 16 percent , she would earn $ 348 per week . if instead , her weekly salary were to increase by 25 percent , how much would she earn per week ? | "( 348 / 116 ) 125 = 375 in this case long division does not take much time . ( 348 / 116 ) = 3 3 * 125 = 375 ( 300 + 75 ) answer e" | a ) $ 374 , b ) $ 382 , c ) $ 385 , d ) $ 392 , e ) $ 375 | e | add(divide(348, add(const_1, divide(16, const_100))), multiply(divide(25, const_100), divide(348, add(const_1, divide(16, const_100))))) | divide(n0,const_100)|divide(n2,const_100)|add(#0,const_1)|divide(n1,#2)|multiply(#1,#3)|add(#3,#4)| | general | E |
a bottle contains a certain solution . in the bottled solution , the ratio of water to soap is 3 : 2 , and the ratio of soap to salt is three times this ratio . the solution is poured into an open container , and after some time , the ratio of water to soap in the open container is halved by water evaporation . at that time , what is the ratio of water to salt in the solution ? | "water : soap = 3 : 2 soap : salt = 9 : 2 = > for 9 soap , salt = 2 = > for 2 soap , salt = ( 2 / 9 ) * 2 = 4 / 9 so , water : soap : salt = 3 : 2 : 4 / 9 = 27 : 18 : 4 after open container , water : soap : salt = 13.5 : 18 : 4 so , water : salt = 13.5 : 4 = 27 : 8 answer : e" | a ) 1 : 1 , b ) 2 : 3 , c ) 3 : 2 , d ) 9 : 4 , e ) 27 : 8 | e | divide(multiply(multiply(2, 3), 3), multiply(multiply(2, 2), 2)) | multiply(n0,n1)|multiply(n1,n1)|multiply(n0,#0)|multiply(n1,#1)|divide(#2,#3)| | other | E |
a ferry can transport 85 tons of vehicles . automobiles range in weight from 1,800 to 3,200 pounds . what is the greatest number of automobiles that can be loaded onto the ferry ? | "to get maximum vehicles we must take into consideration the minimum weight i . e 1800 pounds here since , 1 ton = 2000 pounds 85 tons will be 170,000 pounds from the answer choices : let max number of vehicles be 86 total weight will be = 90 * 1800 = 162000 pounds , which is lesser than the maximum weight allowed . ans : d" | a ) 23 , b ) 41 , c ) 48 , d ) 90 , e ) 86 | d | divide(multiply(multiply(85, const_2), const_1000), add(add(add(add(add(add(const_1000, const_100), const_100), const_100), const_100), const_100), const_100)) | add(const_100,const_1000)|multiply(n0,const_2)|add(#0,const_100)|multiply(#1,const_1000)|add(#2,const_100)|add(#4,const_100)|add(#5,const_100)|add(#6,const_100)|divide(#3,#7)| | general | D |
find value of x : 3 x ^ 2 + 5 x + 2 = 0 | "a = 3 , b = 5 , c = 2 x 1,2 = ( - 5 ± √ ( 52 - 4 × 3 × 2 ) ) / ( 2 × 3 ) = ( - 5 ± √ ( 25 - 24 ) ) / 6 = ( - 5 ± 1 ) / 6 x 1 = ( - 5 + 1 ) / 6 = - 4 / 6 = - 2 / 3 x 2 = ( - 5 - 1 ) / 6 = - 6 / 6 = - 1 a" | a ) - 1 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | a | multiply(add(add(3, 3), divide(multiply(3, 3), const_2)), const_10) | add(n0,n0)|multiply(n0,n0)|divide(#1,const_2)|add(#0,#2)|multiply(#3,const_10)| | general | A |
in one hour , a boat goes 11 km along the stream and 5 km against it . find the speed of the boat in still water | "explanation : we know we can calculate it by 1 / 2 ( a + b ) = > 1 / 2 ( 11 + 5 ) = 1 / 2 ( 16 ) = 8 km / hr answer is c" | a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | c | divide(add(11, 5), const_2) | add(n0,n1)|divide(#0,const_2)| | physics | C |
a train overtakes two person who are walking in the same direction in which the train is going , at the rate of 2 kmph and 4 kmph and passes them completely in 9 and 10 seconds respectively . the length of the train is : | let actual speed of train = s m / sec and length of train = l m . then , s - 2 × 5 / 18 = l 9 ⇒ 9 s = l + 5 . . . … ( i ) and s - 4 × 5 ⁄ 18 = l ⁄ 10 ⇒ 90 s = 9 l + 100 . . . . . ( ii ) by ( i ) & ( ii ) , we get l = 50 m . answer b | a ) 45 m , b ) 50 m , c ) 54 m , d ) 72 m , e ) none of these | b | multiply(9, subtract(subtract(multiply(multiply(4, const_0_2778), 10), multiply(9, multiply(2, const_0_2778))), multiply(2, const_0_2778))) | multiply(n1,const_0_2778)|multiply(n0,const_0_2778)|multiply(n3,#0)|multiply(n2,#1)|subtract(#2,#3)|subtract(#4,#1)|multiply(n2,#5) | physics | B |
two bullet train s 140 m and 160 m long run at the speed of 60 km / hr and 40 km / hr respectively in opposite directions on parallel tracks . the time ( in seconds ) which they take to cross each other , is : | "d 10.8 sec . relative speed = ( 60 + 40 ) km / hr = 100 x 5 / 18 = 250 / 9 m / sec . distance covered in crossing each other = ( 140 + 160 ) m = 300 m . required time = 300 x 9 / 250 = 54 / 5 = 10.8 sec ." | a ) 15.8 sec . , b ) 12.8 sec . , c ) 11.8 sec . , d ) 10.8 sec . , e ) 08.8 sec . | d | divide(add(140, 160), multiply(add(60, 40), const_0_2778)) | add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)| | physics | D |
a can complete a project in 30 days and b can complete the same project in 30 days . if a and b start working on the project together and a quits 10 days before the project is completed , in how many days will the project be completed ? | "let x = the number of days taken to complete the project . the amount of work done by a is ( x - 10 ) * ( 1 / 20 ) . the amount of work done by b is ( x ) * ( 1 / 30 ) . ( 1 / 30 ) * ( x - 10 ) + ( 1 / 30 ) * ( x ) = 1 ( x / 30 ) + ( x / 30 ) - ( 10 / 30 ) = 1 2 x / 30 = 1 / 3 x = 5 therefore , the answer is d : 5 ." | a ) 18 days , b ) 27 days , c ) 26.67 days , d ) 5 days , e ) 12 days | d | add(divide(subtract(const_1, multiply(divide(const_1, 30), 10)), add(divide(const_1, 30), divide(const_1, 30))), 10) | divide(const_1,n1)|divide(const_1,n0)|add(#1,#0)|multiply(n2,#0)|subtract(const_1,#3)|divide(#4,#2)|add(n2,#5)| | physics | D |
what is the dividend . divisor 18 , the quotient is 9 and the remainder is 4 | "d = d * q + r d = 18 * 9 + 4 d = 166 answer : a" | a ) a ) 166 , b ) b ) 148 , c ) c ) 150 , d ) d ) 153 , e ) e ) 158 | a | add(multiply(18, 9), 4) | multiply(n0,n1)|add(n2,#0)| | general | A |
the manager at a health foods store mixes a unique superfruit juice cocktail that costs $ 1399.45 per litre to make . the cocktail includes mixed fruit juice and a ç ai berry juice , which cost $ 262.85 per litre and $ 3104.35 per litre , respectively . the manager has already opened 36 litres of the mixed fruit juice . how many litres of the a ç ai berry juice does he need to add ? | "262.85 ( 36 ) + 3 , 104.35 x = 1 , 399.45 ( 36 + x ) solve the equation . 262.85 ( 36 ) + 3 , 104.35 x = 1 , 399.45 ( 36 + x ) 9 , 462.6 + 3 , 104.35 x = 50 , 380.2 + 1 , 399.45 x 9 , 462.6 + 1 , 704.9 x = 50 , 380.2 1 , 704.9 x = 40 , 917.6 x = 24 answer is b ." | a ) 17 litres , b ) 24 litres , c ) 11 litres , d ) 07 litres , e ) 38 litres | b | divide(subtract(multiply(36, 1399.45), multiply(36, 262.85)), subtract(3104.35, 1399.45)) | multiply(n0,n3)|multiply(n1,n3)|subtract(n2,n0)|subtract(#0,#1)|divide(#3,#2)| | general | B |
the avg weight of a , b & c is 80 kg . if d joins the group , the avg weight of the group becomes 82 kg . if another man e who weights is 3 kg more than d replaces a , then the avgof b , c , d & e becomes 81 kg . what is the weight of a ? | a + b + c = 3 * 80 = 240 a + b + c + d = 4 * 82 = 328 - - - - ( i ) so , d = 88 & e = 88 + 3 = 91 b + c + d + e = 81 * 4 = 324 - - - ( ii ) from eq . ( i ) & ( ii ) a - e = 328 – 324 = 4 a = e + 4 = 91 + 4 = 95 answer : e | a ) 56 , b ) 65 , c ) 75 , d ) 89 , e ) 95 | e | subtract(multiply(82, const_4), subtract(multiply(81, const_4), add(3, subtract(multiply(82, const_4), multiply(80, const_3))))) | multiply(n1,const_4)|multiply(n3,const_4)|multiply(n0,const_3)|subtract(#0,#2)|add(n2,#3)|subtract(#1,#4)|subtract(#0,#5) | general | E |
a train traveling at 216 kmph crosses a platform in 30 seconds and a man standing on the platform in 12 seconds . what is the length of the platform in meters ? | "answer distance covered by the train when crossing a man and when crossing a platform when a train crosses a man standing on a platform , the distance covered by the train is equal to the length of the train . however , when the same train crosses a platform , the distance covered by the train is equal to the length of the train plus the length of the platform . the extra time that the train takes when crossing the platform is on account of the extra distance that it has to cover . i . e . , length of the platform . compute length of platform length of the platform = speed of train * extra time taken to cross the platform . length of platform = 216 kmph * 12 seconds convert 216 kmph into m / sec 1 kmph = 5 / 18 m / s ( this can be easily derived . but if you can remember this conversion , it saves a good 30 seconds ) . ∴ 216 kmph = 5 / 18 ∗ 216 = 60 m / sec therefore , length of the platform = 60 m / s * 12 sec = 720 meters . choice d" | a ) 240 meters , b ) 360 meters , c ) 420 meters , d ) 720 meters , e ) can not be determined | d | subtract(multiply(divide(multiply(216, const_1000), const_3600), 30), multiply(divide(multiply(216, const_1000), const_3600), 12)) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)|multiply(n2,#1)|subtract(#2,#3)| | physics | D |
nina has exactly enough money to purchase 6 widgets . if the cost of each widget were reduced by $ 2 , then nina would have exactly enough money to purchase 8 widgets . how much money does nina have ? | e its is . let price = x ( x - 2 ) 8 = 6 x x = 8 hence total money = 6 * 8 = 48 | a ) $ 22 , b ) $ 24 , c ) $ 30 , d ) $ 36 , e ) $ 48 | e | multiply(divide(multiply(2, 8), subtract(8, 6)), 6) | multiply(n1,n2)|subtract(n2,n0)|divide(#0,#1)|multiply(n0,#2) | general | E |
a train 150 m long takes 6 sec to cross a man walking at 5 kmph in a direction opposite to that of the train . find the speed of the train ? | let the speed of the train be x kmph speed of the train relative to man = x + 5 = ( x + 5 ) * 5 / 18 m / sec 150 / [ ( x + 5 ) * 5 / 18 ] = 6 30 ( x + 5 ) = 2700 x = 85 kmph answer is b | a ) 75 kmph , b ) 85 kmph , c ) 95 kmph , d ) 105 kmph , e ) 115 kmph | b | subtract(divide(divide(150, 6), const_0_2778), 5) | divide(n0,n1)|divide(#0,const_0_2778)|subtract(#1,n2) | physics | B |
if 25 % of a number is subtracted from a second number , the second number reduces to its 4 - sixth . what is the ratio of the first number to the second number ? | explanation : let the numbers be p & q q - p / 4 = 4 q / 6 = > q - 4 q / 6 = p / 4 = > 2 q / 6 = p / 4 = > p / q = 4 / 3 answer b | a ) 5 : 3 , b ) 4 : 3 , c ) 6 : 3 , d ) 3 : 4 , e ) 5 : 4 | b | divide(subtract(const_1, divide(4, multiply(const_3, const_2))), divide(25, const_100)) | divide(n0,const_100)|multiply(const_2,const_3)|divide(n1,#1)|subtract(const_1,#2)|divide(#3,#0) | general | B |
a cylindrical can has a radius of 6 centimeters and a height of 12 centimeters . what is the area , in square centimeters , of a rectangular label that completely covers the curved surface of the can without over - lapping ? | "they are asking the lateral surface area of the cylinder . the lateral surface area = 2 * pi * r * h = 2 * pi * 6 * 12 = 144 pi answer is d ." | a ) 16 pi , b ) 64 pi , c ) 96 pi , d ) 144 pi , e ) 576 pi | d | multiply(multiply(6, 12), multiply(const_2, const_pi)) | multiply(n0,n1)|multiply(const_2,const_pi)|multiply(#0,#1)| | geometry | D |
one side of a rectangle is 3 cm shorter than the other side . if we increase the length of each side by 1 cm , then the area of the rectangle will increase by 18 cm 2 . find the lengths of all sides . | "let x be the length of the longer side x > 3 , then the other side ' s length is x − 3 cm . then the area is s 1 = x ( x - 3 ) cm 2 . after we increase the lengths of the sides they will become ( x + 1 ) and ( x − 3 + 1 ) = ( x − 2 ) cm long . hence the area of the new rectangle will be a 2 = ( x + 1 ) ⋅ ( x − 2 ) cm 2 , which is 18 cm 2 more than the first area . therefore a 1 + 18 = a 2 x ( x − 3 ) + 18 = ( x + 1 ) ( x − 2 ) x 2 − 3 x + 18 = x 2 + x − 2 x − 2 2 x = 20 x = 10 . so , the sides of the rectangle are 10 cm and ( 10 − 3 ) = 7 cm long . so answer is c ." | a ) 10 and 3 , b ) 7 and 10 , c ) 10 and 7 , d ) 3 and 10 , e ) 10 and 10 | c | subtract(add(divide(18, 2), 1), 3) | divide(n2,n3)|add(#0,n1)|subtract(#1,n0)| | geometry | C |
how many numbers are divisible by 4 between 1 and 100 . | "numbers divisible by 4 till 100 = ( 100 / 4 ) = 25 but we should not consider 100 as we are asked to find the number between 1 to 100 which are divisible by 4 . so answer is 24 numbers . option : a" | a ) 24 , b ) 23 , c ) 22 , d ) 21 , e ) 25 | a | divide(100, multiply(4, 1)) | multiply(n0,n1)|divide(n2,#0)| | general | A |
working together , printer a and printer b would finish the task in 30 minutes . printer a alone would finish the task in 50 minutes . how many pages does the task contain if printer b prints 3 pages a minute more than printer a ? | "30 * a + 30 * b = x pages in 30 mins printer a will print = 30 / 50 * x pages = 3 / 5 * x pages thus in 30 mins printer printer b will print x - 3 / 5 * x = 2 / 5 * x pages also it is given that printer b prints 3 more pages per min that printer a . in 30 mins printer b will print 90 more pages than printer a thus 3 / 5 * x - 2 / 5 * x = 90 = > x = 400 pages answer : b" | a ) 200 , b ) 400 , c ) 600 , d ) 800 , e ) 900 | b | multiply(divide(3, subtract(divide(const_60.0, 30), const_1)), const_60) | divide(n1,n0)|subtract(#0,const_1)|divide(const_4.0,#1)|multiply(#2,const_60)| | physics | B |
at what rate percent per annum will a sum of money double in 2 years . | "let principal = p , then , s . i . = p and time = 8 years rate = [ ( 100 x p ) / ( p x 2 ) ] % = 50 % per annum . answer : b" | a ) 34 % , b ) 50 % , c ) 11.5 % , d ) 20 % , e ) 21.5 % | b | divide(const_100, 2) | divide(const_100,n0)| | gain | B |
what is the sum of the greatest common factor and the lowest common multiple of 16 and 36 ? | "prime factorization of the given numbers 16 = 2 ^ 4 36 = 2 ^ 2 * 3 * 2 greatest common factor = 2 ^ 2 = 4 lowest common multiple = 2 ^ 4 * 3 ^ 2 = 144 sum = 4 + 144 = 148 answer a" | a ) 148 , b ) 222 , c ) 144 , d ) 157 , e ) 164 | a | divide(multiply(16, 36), const_4) | multiply(n0,n1)|divide(#0,const_4)| | general | A |
a scale 6 ft . 8 inches long is divided into 4 equal parts . find the length of each part . | "explanation : total length of scale in inches = ( 6 * 12 ) + 8 = 80 inches length of each of the 4 parts = 80 / 4 = 20 inches answer : a" | a ) 20 inches , b ) 77 inches , c ) 66 inches , d ) 97 inches , e ) 66 inches | a | divide(add(multiply(6, const_12), 8), 4) | multiply(n0,const_12)|add(n1,#0)|divide(#1,n2)| | general | A |
how many five - 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(multiply(multiply(const_10, const_10), subtract(const_10, const_1)), 7) | multiply(const_10,const_10)|subtract(const_10,const_1)|multiply(#0,#1)|multiply(n1,#2)| | general | C |
tom and jerry enter into a partnership by investing $ 700 and $ 300 respectively . at the end of one year , they divided their profits such that a third of the profit is divided equally for the efforts they have put into the business and the remaining amount of profit is divided in the ratio of the investments they made in the business . if tom received $ 800 more than jerry did , what was the profit made by their business in that year ? | say the profit was $ x . tom share = x / 6 ( half of the third ) + ( x - x / 3 ) * 0.7 jerry share = x / 6 ( half of the third ) + ( x - x / 3 ) * 0.3 thus ( x - x / 3 ) * 0.7 - ( x - x / 3 ) * 0.3 = 800 - - > x = 3000 . answer is a | a ) 3000 , b ) 4000 , c ) 5000 , d ) 6900 , e ) 6677 | a | divide(800, subtract(add(multiply(700, divide(subtract(const_1, divide(const_1, const_3)), add(700, 300))), divide(divide(const_1, const_3), const_2)), add(multiply(300, divide(subtract(const_1, divide(const_1, const_3)), add(700, 300))), divide(divide(const_1, const_3), const_2)))) | add(n0,n1)|divide(const_1,const_3)|divide(#1,const_2)|subtract(const_1,#1)|divide(#3,#0)|multiply(n0,#4)|multiply(n1,#4)|add(#2,#5)|add(#2,#6)|subtract(#7,#8)|divide(n2,#9) | general | A |
how many trailing zeroes does 49 ! + 50 ! have ? | "49 ! + 50 ! = 51 * 49 ! no of trailing 0 ' s = no of 5 * 2 no of 2 ' s = 49 / 2 + 49 / 4 + 49 / 8 + 49 / 16 + 49 / 32 = d no of 5 ' s = 49 / 5 + 49 / 25 = 10 = p now since p < d no of trailing 0 ' s = 10 answer : b" | a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 22 | b | add(divide(add(50, const_1), add(const_1, const_4)), const_2) | add(n1,const_1)|add(const_1,const_4)|divide(#0,#1)|add(#2,const_2)| | other | B |
23 , 29 , 31 , 37 , 41 , 43 , ( . . . ) | "all are prime numbers in their order , starting from 23 hence , next number is 47 answer is d" | a ) 53 , b ) 48 , c ) 59 , d ) 47 , e ) 50 | d | subtract(negate(37), multiply(subtract(29, 31), divide(subtract(29, 31), subtract(23, 29)))) | negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)| | general | D |
in an election , candidate douglas won 64 percent of the total vote in counties x and y . he won 76 percent of the vote in county x . if the ratio of people who voted in county x to county y is 2 : 1 , what percent of the vote did candidate douglas win in county y ? | "given voters in ratio 2 : 1 let x has 200 votersy has 100 voters for x 76 % voted means 76 * 200 = 152 votes combined for xy has 300 voters and voted 64 % so total votes = 192 balance votes = 192 - 152 = 40 as y has 100 voters so 40 votes means 40 % of votes required ans e" | a ) 10 % , b ) 25 % , c ) 30 % , d ) 35 % , e ) 40 % | e | multiply(divide(subtract(divide(64, const_100), multiply(divide(76, const_100), divide(2, const_3))), divide(1, const_3)), const_100) | divide(n0,const_100)|divide(n1,const_100)|divide(n2,const_3)|divide(n3,const_3)|multiply(#1,#2)|subtract(#0,#4)|divide(#5,#3)|multiply(#6,const_100)| | general | E |
a soft drink company had 6000 small and 14000 big bottles in storage . if 20 % of small 23 % of big bottles have been sold , then the total bottles remaining in storage is | "6000 + 14000 - ( 0.2 * 6000 + 0.22 * 14000 ) = 15720 . answer : b ." | a ) 15360 , b ) 15720 , c ) 15060 , d ) 14930 , e ) 16075 | b | subtract(add(6000, 14000), add(multiply(6000, divide(20, const_100)), multiply(divide(23, const_100), 14000))) | add(n0,n1)|divide(n2,const_100)|divide(n3,const_100)|multiply(n0,#1)|multiply(n1,#2)|add(#3,#4)|subtract(#0,#5)| | general | B |
the average ( arithmetic mean ) of all scores on a certain algebra test was 90 . if the average of the 8 male students ’ grades was 87 , and the average of the female students ’ grades was 92 , how many female students took the test ? | "total marks of male = m total marks of female = f number of males = 8 number of females = f given : ( m + f ) / ( 8 + f ) = 90 - - - - - - - - - - - - - 1 also given , m / 8 = 87 thus m = 696 - - - - - - - - - 2 also , f / f = 92 thus f = 92 f - - - - - - - - - 3 put 2 and 3 in 1 : we get ( 696 + 92 f ) / ( 8 + f ) = 90 solving this we get f = 12 answer : e" | a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | e | divide(subtract(multiply(90, 8), multiply(87, 8)), subtract(92, 90)) | multiply(n0,n1)|multiply(n1,n2)|subtract(n3,n0)|subtract(#0,#1)|divide(#3,#2)| | general | E |
the output of a factory was increased by 10 % to keep up with rising demand . to handle the holiday rush , this new output was increased by 25 % . by approximately what percent would the output now have to be decreased in order to restore the original output ? | "the original output increases by 10 % and then 25 % . total % change = a + b + ab / 100 total % change = 10 + 25 + 10 * 25 / 100 = 37.5 % now , you want to change it to 0 , so , 0 = 37.5 + x + 37.5 x / 100 x = - 37.5 ( 100 ) / 137.5 = 27 % approximately answer is d" | a ) 20 % , b ) 24 % , c ) 30 % , d ) 27 % , e ) 79 % | d | divide(multiply(subtract(add(add(const_100, 10), multiply(add(const_100, 10), divide(25, const_100))), const_100), const_100), add(add(const_100, 10), multiply(add(const_100, 10), divide(25, const_100)))) | add(n0,const_100)|divide(n1,const_100)|multiply(#0,#1)|add(#0,#2)|subtract(#3,const_100)|multiply(#4,const_100)|divide(#5,#3)| | general | D |
find the greatest number which leaves the same remainder when it divides 25 , 62 and 105 . | "105 - 62 = 43 62 - 25 = 37 105 - 25 = 80 the h . c . f of 37 , 43 and 80 is 1 . answer : a" | a ) 1 , b ) 8 , c ) 12 , d ) 16 , e ) none of these | a | divide(subtract(62, 25), const_2) | subtract(n1,n0)|divide(#0,const_2)| | general | A |
running at their respective constant rates , machine x takes 2 days longer to produce w widgets than machine y . at these rates , if the two machines together produce 5 / 4 w widgets in 3 days , how many days would it take machine x alone to produce 5 w widgets ? | "let y produce w widgets in y days hence , in 1 day y will produce w / y widgets . also , x will produce w widgets in y + 2 days ( given , x takes two more days ) hence , in 1 day x will produce w / y + 2 widgets . hence together x and y in 1 day will produce { w / y + w / y + 2 } widgets . together x and y in 3 days will produce = 3 * [ { w / y + w / y + 2 } ] widgets . it is given that in 3 days together they produce ( 5 / 4 ) w widgets . equating , 3 * [ { w / y + w / y + 2 } ] = ( 5 / 4 ) w take out w common and move 3 to denominator of rhs w { 1 / y + 1 / ( y + 2 ) } = ( 5 / 12 ) w canceling w from both sides { 1 / y + 1 / ( y + 2 ) } = 5 / 12 2 y + 2 / y ( y + 2 ) = 5 / 12 24 y + 24 = 5 y ^ 2 + 10 y 5 y ^ 2 - 14 y - 24 = 0 5 y ^ 2 - 20 y + 6 y - 24 = 0 5 y ( y - 4 ) + 6 ( y - 4 ) = 0 ( 5 y + 6 ) + ( y - 4 ) = 0 y = - 6 / 5 or y = 4 discarding y = - 6 / 5 as no of days can not be negative y = 4 hence it takes y , 4 days to produce w widgets . therefore , it will take x ( 4 + 2 ) = 6 days to produce w widgets . hence it will take x 5 * 6 = 30 days to produce 2 w widgets . answer : b" | a ) 4 , b ) 30 , c ) 8 , d ) 10 , e ) 12 | b | divide(subtract(multiply(multiply(3, 4), 2), multiply(3, 4)), 2) | multiply(n2,n3)|multiply(n0,#0)|subtract(#1,#0)|divide(#2,n0)| | general | B |
a bag contains 6 white marbles and 6 black marbles . if each of 6 girls and 6 boys randomly selects and keeps a marble , what is the probability that all of the girls select the same colored marble ? | "first , total ways to select for all boys and girls , i . e 12 ! / ( 6 ! * 6 ! ) = 12 * 11 * 10 * 9 * 8 * 7 / 6 * 5 * 4 * 3 * 2 = 924 then there are one two way girls can have all same colors , either white or black . the number of ways in which 6 girls can select 6 white balls = 6 c 6 = 1 the number of ways in which 6 girls can select 6 black balls = 6 c 6 = 1 therefore , total favorable outcomes / total outcomes = 2 / 924 = 1 / 462 d" | a ) 1 / 35 , b ) 1 / 10 , c ) 1 / 3 , d ) 1 / 462 , e ) 1 / 252 | d | divide(const_2, choose(add(6, 6), 6)) | add(n0,n0)|choose(#0,n0)|divide(const_2,#1)| | probability | D |
local kennel has cats and dogs in the ratio of 3 : 4 . if there are 8 fewer cats than dogs , how many dogs are in the kennel ? | "lets work with the data given to us . we know that there ratio of cats to dogs is 3 : 4 or cats 3 dogs 4 we can write number of cats as 3 x and number of dogs as 4 x and we know that 4 x - 3 x = 8 ( therefore x = 8 ) then # of dogs = 4 x 8 = 32 answer is d" | a ) 28 , b ) 26 , c ) 24 , d ) 32 , e ) 30 | d | multiply(8, 4) | multiply(n1,n2)| | other | D |
of the total amount that jill spent on a shopping trip , excluding taxes , she spent 45 percent on clothing , 45 percent on food , and 10 percent on other items . if jill paid a 5 percent tax on the clothing , no tax on the food , and an 10 percent tax on all other items , then the total tax that she paid was what percent of the total amount that she spent , excluding taxes ? | "assume she has $ 200 to spend . tax clothing = 45 % = $ 90 = $ 4.5 food = 45 % = $ 90 = $ 0.00 items = 10 % = $ 20 = $ 2.00 total tax = $ 6.50 % of total amount = 6.5 / 200 * 100 = 3.25 % answer d" | a ) 3.5 , b ) 3.35 , c ) 3.45 , d ) 3.25 , e ) 3.55 | d | multiply(divide(add(multiply(45, divide(5, const_100)), multiply(10, divide(10, const_100))), const_100), const_100) | divide(n3,const_100)|divide(n4,const_100)|multiply(n0,#0)|multiply(n2,#1)|add(#2,#3)|divide(#4,const_100)|multiply(#5,const_100)| | general | D |
if the sides of a square are multiplied by sqrt ( 20 ) , the area of the original square is how many times as large as the area of the resultant square ? | "let x be the original length of one side . then the original area is x ^ 2 . the new square has sides of length sqrt ( 20 ) * x , so the area is 20 x ^ 2 . the area of the original square is 1 / 20 = 5 % times the area of the new square . the answer is b ." | a ) 2 % , b ) 5 % , c ) 20 % , d ) 50 % , e ) 150 % | b | square_perimeter(20) | square_perimeter(n0)| | geometry | B |
on a map , 2.5 inches represent 25 miles . how many miles approximately is the distance if you measured 55 centimeters assuming that 1 - inch is 2.54 centimeters ? | "1 inch = 2.54 cm 2.5 inch = 2.54 * 2.5 cm 6.35 cm = 25 miles 55 cms = 25 / 6.35 * 55 = 216.5 miles answer : e" | a ) 212.5 , b ) 213.5 , c ) 214.5 , d ) 215.5 , e ) 216.5 | e | multiply(divide(55, 2.54), divide(25, 2.5)) | divide(n2,n4)|divide(n1,n0)|multiply(#0,#1)| | physics | E |
a doctor prescribed 24 cubic centimeters of a certain drug to a patient whose body weight was 120 pounds . if the typical dosage is 2 cubic centimeters per 15 pounds of the body weight , by what percent was the prescribed dosage greater than the typical dosage ? | "typical dosage per 15 pound of the body weight = 2 c . c typical dosage per 120 pound of the body weight = 2 * ( 120 / 15 ) = 2 * 8 = 16 c . c dosage prescribed by doctor for 120 pound patient = 24 c . c % prescribed dosage greater than the typical dosage = ( 24 - 16 / 16 ) * 100 % = ( 8 / 16 ) * 100 % = 50 % answer d" | a ) 8 % , b ) 9 % , c ) 50 % , d ) 12.5 % , e ) 14.8 % | d | multiply(divide(subtract(multiply(divide(2, 15), 120), 24), multiply(divide(2, 15), 120)), const_100) | divide(n2,n3)|multiply(n1,#0)|subtract(#1,n0)|divide(#2,#1)|multiply(#3,const_100)| | gain | D |
a can finish a work in 12 days and b can do the same work in 15 days . b worked for 10 days and left the job . in how many days , a alone can finish the remaining work ? | "b ' s 10 day ' s work = ( 1 / 15 * 10 ) = 2 / 3 remaining work = ( 1 - 2 / 3 ) = 1 / 3 now , 1 / 18 work is done by a in 1 day 1 / 3 work is done by a in ( 12 * 1 / 3 ) = 4 days . correct option : a" | a ) 4 , b ) 5 1 / 2 , c ) 6 , d ) 8 , e ) none of these | a | divide(multiply(multiply(divide(const_1, 15), 10), 12), const_2) | divide(const_1,n1)|multiply(n2,#0)|multiply(n0,#1)|divide(#2,const_2)| | physics | A |
the sum of four consecutive even numbers is 292 . what would be the smallest number ? | "e 70 let the four consecutive even numbers be 2 ( x - 2 ) , 2 ( x - 1 ) , 2 x , 2 ( x + 1 ) their sum = 8 x - 4 = 292 = > x = 37 smallest number is : 2 ( x - 2 ) = 70 ." | a ) 49 , b ) 68 , c ) 60 , d ) 57 , e ) 70 | e | add(add(power(add(add(divide(subtract(subtract(292, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(292, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(292, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(292, const_10), const_2), const_4), const_2), const_2))) | subtract(n0,const_10)|subtract(#0,const_2)|divide(#1,const_4)|add(#2,const_2)|power(#2,const_2)|add(#3,const_2)|power(#3,const_2)|add(#5,const_2)|add(#4,#6)|power(#5,const_2)|power(#7,const_2)|add(#9,#10)|add(#11,#8)| | physics | E |
a theater box office sold an average ( arithmetic mean ) of 64 tickets per staff member to a particular movie . among the daytime staff , the average number sold per member was 80 , and among the evening staff , the average number sold was 60 . if there are no other employees , what was the ratio of the number of daytime staff members to the number of evening staff members ? | "deviation from the mean for the daytime staff = 80 - 64 = 16 . deviation from the mean for the evening staff = 64 - 60 = 4 . thus , the ratio of the number of daytime staff members to the number of evening staff members is 4 : 16 = 1 : 4 . the answer is b ." | a ) 1 : 2 , b ) 1 : 4 , c ) 3 : 7 , d ) 7 : 12 , e ) 4 : 5 | b | divide(subtract(64, 60), subtract(80, 64)) | subtract(n0,n2)|subtract(n1,n0)|divide(#0,#1)| | general | B |
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