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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country x imported approximately $ 1.20 billion of goods in 1996 . if country x imported $ 288 million of goods in the first two months of 1997 and continued to import goods at the same rate for the rest of the year , by how much would country xs 1997 imports exceed those of 1996 ? | "convert units to millions as answer is in millions 1996 imports = $ 1.20 bill = $ 1200 mill i . e . 1200 / 12 = $ 100 mill / month 1997 imports = $ 288 mill / 2 month i . e . $ 144 mill / month difference / month = 144 - 100 = 44 difference / year = $ 44 mill * 12 = $ 528 mill answer : e" | a ) $ 24 million , b ) $ 120 million , c ) $ 144 million , d ) $ 240 million , e ) $ 528 million | e | subtract(multiply(288, divide(const_12, const_2)), multiply(1.20, const_1000)) | divide(const_12,const_2)|multiply(n0,const_1000)|multiply(n2,#0)|subtract(#2,#1)| | general | E |
the average weight of 19 students is 15 kg . by the admission of a new student the average weight is reduced to 14.4 kg . the weight of the new student is ? | "answer weight of new student = total weight of all 20 students - total weight of initial 19 students = ( 20 x 14.4 - 19 x 15 ) kg = 3 kg . correct option : b" | a ) 10.6 kg , b ) 3 kg , c ) 11 kg , d ) 14.9 kg , e ) none | b | subtract(multiply(add(19, const_1), 14.4), multiply(19, 15)) | add(n0,const_1)|multiply(n0,n1)|multiply(n2,#0)|subtract(#2,#1)| | general | B |
if the radius of a circle is decreased 10 % , what happens to the area ? | "area of square = pi * radius ^ 2 new radius = 0.9 * old radius so new area = ( 0.9 ) ^ 2 old area = > 0.81 of old area = > 81 % old area ans : b" | a ) 10 % decrease , b ) 19 % decrease , c ) 36 % decrease , d ) 40 % decrease , e ) 50 % decrease | b | subtract(const_100, multiply(power(divide(10, const_100), const_2), const_100)) | divide(n0,const_100)|power(#0,const_2)|multiply(#1,const_100)|subtract(const_100,#2)| | geometry | B |
if 41 / 88 = 0.46590 , what is the 77 th digit to the right of the decimal point of the fraction ? | "we are not concerned what 41 / 88 means . . we have to look at the decimal . . 0.6590 means 0.465909090 . . . . so leaving girst , second and third digit to the right of decimal , all odd numbered are 0 and all even numbered are 9 . . here 77 is odd , so ans is 0 answer is d" | a ) 6 , b ) 2 , c ) 5 , d ) 0 , e ) 9 | d | add(const_2, const_3) | add(const_2,const_3)| | general | D |
what is the last digit in ( 7 ^ 95 - 3 ^ 58 ) ? | explanation : unit digit in 795 = unit digit in [ ( 74 ) 23 x 73 ] = unit digit in [ ( unit digit in ( 2401 ) ) 23 x ( 343 ) ] = unit digit in ( 123 x 343 ) = unit digit in ( 343 ) = 3 unit digit in 358 = unit digit in [ ( 34 ) 14 x 32 ] = unit digit in [ unit digit in ( 81 ) 14 x 32 ] = unit digit in [ ( 1 ) 14 x 32 ] = unit digit in ( 1 x 9 ) = unit digit in ( 9 ) = 9 unit digit in ( 795 - 358 ) = unit digit in ( 343 - 9 ) = unit digit in ( 334 ) = 4 . so , answer c | a ) 5 , b ) 6 , c ) 4 , d ) 9 , e ) 5.5 | c | divide(divide(multiply(const_4, const_4), const_2), const_2) | multiply(const_4,const_4)|divide(#0,const_2)|divide(#1,const_2) | general | C |
a man is 24 years older than his son . in two years , his age will be twice the age of his son . the present age of this son is | "explanation : let ' s son age is x , then father age is x + 24 . = > 2 ( x + 2 ) = ( x + 24 + 2 ) = > 2 x + 4 = x + 26 = > x = 22 years option b" | a ) 21 years , b ) 22 years , c ) 23 years , d ) 24 years , e ) 26 years | b | divide(subtract(24, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | multiply(const_2,const_2)|subtract(const_2,const_1)|subtract(#0,const_2)|subtract(n0,#2)|divide(#3,#1)| | general | B |
a can finish a work in 24 days , b in 9 days and c in 12 days . b and c start the work but are forced to leave after 3 days . when a done the work ? | "b + c = = > 1 / 9 + 1 / 12 = 7 / 36 b , c = in 3 days = 7 / 36 * 3 = 7 / 12 remaining work = 1 - 7 / 12 = 5 / 12 1 / 24 work is done by a in 1 day 5 / 12 work is done a 24 * 5 / 12 = 10 days answer a" | a ) 10 days , b ) 12 days , c ) 13 days , d ) 9 days , e ) 14 days | a | multiply(divide(const_1, add(divide(const_1, 9), divide(const_1, 12))), 3) | divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|divide(const_1,#2)|multiply(n3,#3)| | physics | A |
two employees m and n are paid a total of $ 594 per week by their employer . if m is paid 120 percent of the salary paid to n , how much is n paid per week ? | 1.2 n + n = 594 2.2 n = 594 n = 270 the answer is d . | a ) $ 240 , b ) $ 250 , c ) $ 260 , d ) $ 270 , e ) $ 280 | d | divide(594, add(divide(120, const_100), const_1)) | divide(n1,const_100)|add(#0,const_1)|divide(n0,#1) | general | D |
if $ 5,000 is invested in an account at a simple annual rate of r percent , the interest is $ 250 . when $ 20,000 is invested at the same interest rate , what is the interest from the investment ? | - > 250 / 5,000 = 5 % and 20,000 * 5 % = 1000 . thus , d is the answer . | a ) $ 700 , b ) $ 750 , c ) $ 800 , d ) $ 1000 , e ) $ 900 | d | divide(multiply(250, multiply(multiply(const_2, const_100), const_100)), divide(multiply(multiply(const_2, const_100), const_100), const_4)) | multiply(const_100,const_2)|multiply(#0,const_100)|divide(#1,const_4)|multiply(n1,#1)|divide(#3,#2) | gain | D |
eddy and freddy start simultaneously from city a and they travel to city b and city c respectively . eddy takes 3 hours and freddy takes 4 hours to complete the journey . if the distance between city a and city b is 570 kms and city a and city c is 300 kms . what is the ratio of their average speed of travel ? ( eddy : freddy ) | "distance traveled by eddy = 570 km time taken by eddy = 3 hours average speed of eddy = 570 / 3 = 190 km / hour distance traveled by freddy = 300 km time taken by freddy = 4 hours average speed of freddy = 300 / 4 = 75 km / hour ratio of average speed of eddy to freddy = 190 / 75 = 38 / 15 answer c" | a ) 8 / 3 , b ) 3 / 8 , c ) 38 / 15 , d ) 5 / 8 , e ) 5 / 3 | c | divide(divide(570, 3), divide(300, 4)) | divide(n2,n0)|divide(n3,n1)|divide(#0,#1)| | physics | C |
a can do a piece of work in 14 days . when he had worked for 2 days b joins him . if the complete work was finished in 8 days . in how many days b alone can finish the work ? | "8 / 14 + 6 / x = 1 x = 14 days answer : e" | a ) 18 , b ) 77 , c ) 66 , d ) 55 , e ) 14 | e | subtract(inverse(subtract(multiply(divide(const_1, 8), subtract(const_1, multiply(2, divide(const_1, 14)))), divide(const_1, 14))), add(14, 8)) | add(n0,n2)|divide(const_1,n2)|divide(const_1,n0)|multiply(n1,#2)|subtract(const_1,#3)|multiply(#1,#4)|subtract(#5,#2)|inverse(#6)|subtract(#7,#0)| | physics | E |
the grade point average of the entire class is 87 . if the average of one third of the class is 95 , what is the average of the rest of the class ? | "let x be the number of students in the class . let p be the average of the rest of the class . 87 x = ( 1 / 3 ) 95 x + ( 2 / 3 ) ( p ) x 261 = 95 + 2 p 2 p = 166 p = 83 . the answer is c ." | a ) 81 , b ) 82 , c ) 83 , d ) 84 , e ) 85 | c | divide(subtract(multiply(87, const_4), 95), subtract(const_4, const_1)) | multiply(n0,const_4)|subtract(const_4,const_1)|subtract(#0,n1)|divide(#2,#1)| | general | C |
a jar contains only red , yellow , and orange marbles . if there are 3 red , 7 yellow , and 4 orange marbles , and 3 marbles are chosen from the jar at random without replacing any of them , what is the probability that 2 yellow , 1 red , and no orange marbles will be chosen ? | "i started by finding the 2 probabilities , without calculation , like this : p ( yyr ) p ( yry ) p ( ryy ) i calculated the first one and ended in 1 / 22 . i looked at the answer choices at this point and saw answer d : 5 / 22 . this helped me realise that for the 3 possible orderings the probabbility is the same . so , it should be ( 1 / 22 ) * ( 5 ) , which indeed is 5 / 22 . e" | a ) 1 / 60 , b ) 1 / 45 , c ) 2 / 45 , d ) 3 / 22 , e ) 5 / 22 | e | divide(multiply(choose(7, 2), choose(3, 1)), choose(add(add(7, 4), 3), 3)) | add(n1,n2)|choose(n1,n4)|choose(n0,n5)|add(n0,#0)|multiply(#1,#2)|choose(#3,n0)|divide(#4,#5)| | probability | E |
father is aged 3 times more than his son ronit . after 8 years , he would be two and half times if ronit ' s age . after further 8 years , how many times would he be of ronit ' s age ? | ronit ' s present age be x years . then , father ' s present age = ( x + 3 x ) years = 4 x years therefore ( 4 x + 8 ) = 5 / 2 ( x + 8 ) 8 x + 16 = 5 x + 40 3 x = 24 , x = 8 , ratio = ( 4 x + 16 ) / ( x + 16 ) = 48 / 24 = 2 , correct answer ( a ) | a ) 2 times , b ) 2.5 times , c ) 3 times , d ) 3.8 times , e ) 4 times | a | divide(add(multiply(const_4, 8), add(8, 8)), add(8, add(8, 8))) | add(n1,n1)|multiply(n1,const_4)|add(#0,#1)|add(n1,#0)|divide(#2,#3) | general | A |
in a rectangular coordinate system , what is the area of a rectangle whose vertices have the coordinates ( - 8 , 1 ) , ( 1 , 1 ) , ( 1 , - 7 and ( - 8 , - 7 ) ? | "length of side 1 = 8 + 1 = 9 length of side 2 = 7 + 1 = 8 area of rectangle = 9 * 8 = 72 c is the answer" | a ) 144 , b ) 36 , c ) 72 , d ) 56 , e ) 112 | c | multiply(add(8, 1), add(1, 7)) | add(n0,n1)|add(n1,n5)|multiply(#0,#1)| | geometry | C |
the average expenditure of a labourer for 6 months was 85 and he fell into debt . in the next 4 months by reducing his monthly expenses to 60 he not only cleared off his debt but also saved 30 . his monthly income i | "income of 6 months = ( 6 × 85 ) – debt = 510 – debt income of the man for next 4 months = 4 × 60 + debt + 30 = 270 + debt ∴ income of 10 months = 780 average monthly income = 780 ÷ 10 = 78 answer d" | a ) 70 , b ) 72 , c ) 75 , d ) 78 , e ) 80 | d | divide(add(add(multiply(85, 6), multiply(60, 4)), 30), add(6, 4)) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|add(n4,#3)|divide(#4,#0)| | general | D |
how many 3 - digit even numbers are possible such that if one of the digits is 5 , the next / succeeding digit to it should be 6 | 560 , 562 , 564 , 566 , and 568 , so total 5 . hence option a . | a ) 5 , b ) 305 , c ) 365 , d ) 405 , e ) 495 | a | add(add(3, 6), 6) | add(n0,n2)|add(n2,#0)| | general | A |
pipe a can fill a tank in 10 hrs and pipe b can fill it in 8 hrs . if both the pipes are opened in the empty tank . there is an outlet pipe in 3 / 4 th of the tank . in how many hours will it be fill 3 / 4 th of that tank ? | part filled a in 1 hr = ( 1 / 10 ) part filled b in 1 hr = ( 1 / 8 ) part filled by ( a + b ) together in 1 hr = ( 1 / 10 ) + ( 1 / 8 ) = 18 / 80 so , the tank will be full in 80 / 18 hrs . time taken to fill exact 3 / 4 th of the tank = ( 80 / 18 ) * ( 3 / 4 ) = 3.20 hrs answer : d | a ) 3 hr , b ) 3.10 hr , c ) 3.15 hr , d ) 3.20 hr , e ) 3.30 hr | d | multiply(divide(3, 4), inverse(add(divide(const_1, 10), divide(const_1, 8)))) | divide(n2,n3)|divide(const_1,n0)|divide(const_1,n1)|add(#1,#2)|inverse(#3)|multiply(#0,#4) | physics | D |
bruce purchased 7 kg of grapes at the rate of 70 per kg and 9 kg of mangoes at the rate of 55 per kg . how much amount did he pay to the shopkeeper ? | cost of 7 kg grapes = 70 × 7 = 490 . cost of 9 kg of mangoes = 55 × 9 = 495 . total cost he has to pay = 490 + 495 = 985 a | a ) a ) 985 , b ) b ) 1050 , c ) c ) 1055 , d ) d ) 1065 , e ) e ) 1075 | a | add(multiply(7, 70), multiply(9, 55)) | multiply(n0,n1)|multiply(n2,n3)|add(#0,#1) | gain | A |
the area of sector of a circle whose radius is 12 metro and whose angle at the center is 38 ° is ? | "38 / 360 * 22 / 7 * 12 * 12 = 47.7 m 2 answer : c" | a ) 30 m 2 , b ) 40 m 2 , c ) 47.7 m 2 , d ) 50 m 2 , e ) 55 m 2 | c | multiply(multiply(power(12, const_2), divide(add(multiply(const_2, const_10), const_2), add(const_4, const_3))), divide(38, divide(const_3600, const_10))) | add(const_3,const_4)|divide(const_3600,const_10)|multiply(const_10,const_2)|power(n0,const_2)|add(#2,const_2)|divide(n1,#1)|divide(#4,#0)|multiply(#6,#3)|multiply(#5,#7)| | geometry | C |
in a group of 68 students , each student is registered for at least one of 3 classes – history , math and english . 20 - one students are registered for history , 20 - 6 students are registered for math , and thirty - 4 students are registered for english . if only 3 students are registered for all 3 classes , how many students are registered for exactly two classes ? | a u b u c = a + b + c - ab - bc - ac + abc 68 = 21 + 26 + 34 - ab - bc - ac + 3 = > ab + bc + ac = 16 exactly two classes = ab + bc + ac - 3 abc = 16 - 3 * 3 = 7 hence e | a ) 13 , b ) 10 , c ) 9 , d ) 8 , e ) 7 | e | subtract(subtract(add(add(add(20, const_1), add(20, 6)), add(add(20, const_10), 4)), 68), multiply(const_2, 3)) | add(n2,const_1)|add(n2,n4)|add(n2,const_10)|multiply(n1,const_2)|add(#0,#1)|add(n5,#2)|add(#4,#5)|subtract(#6,n0)|subtract(#7,#3) | general | E |
following an increase in prices , the price of a candy box was 15 pounds and the price of a can of soda was 6 pounds . if the price of a candy box was raised by 25 % , and the price of a can of soda was raised by 50 % . what was the price of a box of candy plus a can of soda before prices were raised ? | "price of candy before price increase = 15 / 1.25 = 12 price of soda before price increase = 6 / 1.5 = 4 total price = 12 + 4 = 16 e is the answer" | a ) 11 . , b ) 12 . , c ) 13 . , d ) 14 . , e ) 16 | e | add(divide(multiply(15, const_100), add(const_100, 25)), divide(multiply(6, const_100), add(const_100, 50))) | add(n2,const_100)|add(n3,const_100)|multiply(n0,const_100)|multiply(n1,const_100)|divide(#2,#0)|divide(#3,#1)|add(#4,#5)| | general | E |
if a lends rs . 3500 to b at 10 % per annum and b lends the same sum to c at 11.5 % per annum then the gain of b in a period of 3 years is ? | "( 3500 * 1.5 * 3 ) / 100 = > 157.50 answer : c" | a ) 157.20 , b ) 157.29 , c ) 157.50 , d ) 157.30 , e ) 157.23 | c | subtract(divide(multiply(multiply(3500, 11.5), 3), const_100), divide(multiply(multiply(3500, 10), 3), const_100)) | multiply(n0,n2)|multiply(n0,n1)|multiply(#0,n3)|multiply(n3,#1)|divide(#2,const_100)|divide(#3,const_100)|subtract(#4,#5)| | gain | C |
income and expenditure of a person are in the ratio 5 : 4 . if the income of the person is rs . 20000 , then find his savings ? | "let the income and the expenditure of the person be rs . 5 x and rs . 4 x respectively . income , 5 x = 20000 = > x = 4000 savings = income - expenditure = 5 x - 4 x = x so , savings = rs . 4000 answer : b" | a ) rs . 3600 , b ) rs . 4000 , c ) rs . 3639 , d ) rs . 3632 , e ) rs . 3602 | b | subtract(20000, multiply(divide(4, 5), 20000)) | divide(n1,n0)|multiply(n2,#0)|subtract(n2,#1)| | other | B |
robert is travelling on his cycle and has calculated to reach point a at 2 p . m . if he travels at 10 kmph , he will reach there at 12 noon if he travels at 15 kmph . at what speed must he travel to reach a at 1 p . m . ? | "explanation : let the distance travelled by x km . then , = > ( x / 10 ) - ( x / 15 ) = 2 = > 3 x - 2 x = 60 = > x = 60 = > time taken to travel 60 km at 10 km / hr = 60 / 10 = 6 hours . = > so , robert started 6 hours before 2 p . m . i . e . , at 8 a . m . therefore , speed required = ( 60 / 5 ) km / hr = 12 km / hr answer : b" | a ) 8 , b ) 12 , c ) 11 , d ) 16 , e ) 18 | b | divide(divide(2, subtract(divide(1, 10), divide(1, 15))), add(const_4, 1)) | add(n4,const_4)|divide(n4,n1)|divide(n4,n3)|subtract(#1,#2)|divide(n0,#3)|divide(#4,#0)| | physics | B |
a train 140 m long is running with a speed of 50 km / hr . in what time will it pass a man who is running at 4 km / hr in the direction opposite to that in which the train is going ? | "speed of train relative to man = 50 + 4 = 54 km / hr . = 54 * 5 / 18 = 15 m / sec . time taken to pass the men = 140 * 1 / 15 = 9.33 sec . answer : option c" | a ) 8.1 , b ) 7.0 , c ) 9.33 , d ) 8 , e ) 9 | c | divide(140, multiply(add(50, 4), const_0_2778)) | add(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics | C |
a man is 24 years older than his son . in two years , his age will be twice the age of his son . the present age of his son is | "explanation : let the son ' s present age be x years . then , man ' s present age = ( x + 24 ) years = > ( x + 24 ) + 2 = 2 ( x + 2 ) = > x + 26 = 2 x + 4 so , x = 22 option c" | a ) 20 years , b ) 21 years , c ) 22 years , d ) 24 years , e ) none of these | c | divide(subtract(24, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | multiply(const_2,const_2)|subtract(const_2,const_1)|subtract(#0,const_2)|subtract(n0,#2)|divide(#3,#1)| | general | C |
if a ã — b = 2 a - 3 b + ab , then 4 ã — 5 + 5 ã — 4 is equal to : | "explanation : 4 ã — 5 + 5 ã — 4 = ( 2 ã — 4 - 3 ã — 5 + 4 ã — 5 ) + ( 2 ã — 5 - 3 ã — 4 + 5 ã — 4 ) = ( 8 - 15 + 20 + 10 - 12 + 20 ) = 31 . answer : a" | a ) 31 , b ) 32 , c ) 43 , d ) 45 , e ) 26 | a | add(add(subtract(multiply(5, 4), multiply(3, const_2.0)), multiply(4, const_2.0)), add(subtract(multiply(2, 2), multiply(3, 4)), multiply(2, 4))) | multiply(n0,n2)|multiply(n0,n1)|multiply(n3,n3)|multiply(n1,n2)|subtract(#0,#1)|subtract(#2,#3)|add(#0,#4)|add(#0,#5)|add(#6,#7)| | general | A |
a sum was put at simple interest at certain rate for 3 years . had it been put at 1 % higher rate it would have fetched rs . 78 more . the sum is : a . rs . 2,400 b . rs . 2,100 c . rs . 2,200 d . rs . 2,480 | "1 percent for 3 years = 78 1 percent for 1 year = 26 = > 100 percent = 2600 answer : b" | a ) 2000 , b ) 2600 , c ) 2200 , d ) 2300 , e ) 2400 | b | multiply(divide(78, 3), const_100) | divide(n2,n0)|multiply(#0,const_100)| | gain | B |
a is four times as fast as b . if b alone can do a piece of work in 60 days , in what time can a and b together complete the work ? | a can do the work in 60 / 4 i . e . , 15 days . a and b ' s one day ' s work = 1 / 15 + 1 / 60 = ( 4 + 1 ) / 60 = 1 / 12 so a and b together can do the work in 12 days . answer : b | a ) 10 , b ) 12 , c ) 22 , d ) 28 , e ) 20 | b | inverse(add(inverse(divide(60, const_4)), inverse(60))) | divide(n0,const_4)|inverse(n0)|inverse(#0)|add(#2,#1)|inverse(#3) | physics | B |
a train 120 m in length crosses a telegraph post in 16 seconds . the speed of the train is ? | "s = 120 / 16 * 18 / 5 = 27 kmph answer : e" | a ) 16 kmph , b ) 88 kmph , c ) 54 kmph , d ) 18 kmph , e ) 27 kmph | e | multiply(const_3_6, divide(120, 16)) | divide(n0,n1)|multiply(#0,const_3_6)| | physics | E |
a parking garage rents parking spaces for $ 10 per week or $ 40 per month . how much does a person save in a year by renting by the month rather than by the week ? | "10 $ per week ! an year has 52 weeks . annual charges per year = 52 * 10 = 520 $ 30 $ per month ! an year has 12 months . annual charges per year = 12 * 40 = 480 $ 520 - 480 = 40 ans a" | a ) $ 40 , b ) $ 160 , c ) $ 220 , d ) $ 240 , e ) $ 260 | a | subtract(multiply(add(multiply(10, add(const_3, const_2)), const_2), 10), multiply(40, const_12)) | add(const_2,const_3)|multiply(n1,const_12)|multiply(#0,n0)|add(#2,const_2)|multiply(n0,#3)|subtract(#4,#1)| | general | A |
the probability that a number selected at random from the first 50 natural numbers is a composite number is - . | "the number of exhaustive events = ⁵ ⁰ c ₁ = 50 . we have 15 primes from 1 to 50 . number of favourable cases are 34 . required probability = 34 / 50 = 17 / 25 . answer : a" | a ) 17 / 25 , b ) 17 / 27 , c ) 17 / 18 , d ) 17 / 22 , e ) 17 / 09 | a | divide(multiply(subtract(multiply(const_6, const_3), const_1), const_2), 50) | multiply(const_3,const_6)|subtract(#0,const_1)|multiply(#1,const_2)|divide(#2,n0)| | probability | A |
ajay and vijay have some marbles with them . ajay told vijay ` ` if you give me ' x ' marbles , both of us will have equal number of marbles ' ' . vijay then told ajay ` ` if you give me twice as many marbles , i will have 30 more marbles than you would ' ' . find ' x ' ? | if vijay gives ' x ' marbles to ajay then vijay and ajay would have v - x and a + x marbles . v - x = a + x - - - ( 1 ) if ajay gives 2 x marbles to vijay then ajay and vijay would have a - 2 x and v + 2 x marbles . v + 2 x - ( a - 2 x ) = 30 = > v - a + 4 x = 30 - - - ( 2 ) from ( 1 ) we have v - a = 2 x substituting v - a = 2 x in ( 2 ) 6 x = 30 = > x = 5 . answer : b | a ) 7 , b ) 5 , c ) 6 , d ) 4 , e ) 3 | b | divide(30, multiply(const_2, const_3)) | multiply(const_2,const_3)|divide(n0,#0) | general | B |
find the value of ( √ 1.21 ) / ( √ 0.64 ) + ( √ 1.44 ) / ( √ 0.49 ) is | "( √ 1.21 ) / ( √ 0.64 ) + ( √ 1.44 ) / ( √ 0.49 ) 11 / 8 + 12 / 7 = > 3.089 answer is d" | a ) 195 / 63 , b ) 145 / 63 , c ) 155 / 63 , d ) 3.089 , e ) 185 / 63 | d | add(divide(sqrt(1.21), sqrt(0.64)), divide(sqrt(1.44), sqrt(0.49))) | sqrt(n0)|sqrt(n1)|sqrt(n2)|sqrt(n3)|divide(#0,#1)|divide(#2,#3)|add(#4,#5)| | general | D |
the average salary of a person for the months of january , february , march and april is rs . 8000 and that for the months february , march , april and may is rs . 8300 . if his salary for the month of may is rs . 6500 , find his salary for the month of january ? | sum of the salaries of the person for the months of january , february , march and april = 4 * 8000 = 32000 - - - - ( 1 ) sum of the salaries of the person for the months of february , march , april and may = 4 * 8300 = 33200 - - - - ( 2 ) ( 2 ) - ( 1 ) i . e . may - jan = 1200 salary of may is rs . 6500 salary of january = rs . 5300 . answer : d | a ) 2177 , b ) 2876 , c ) 4500 , d ) 5300 , e ) 6711 | d | subtract(multiply(8000, const_4), subtract(multiply(8300, const_4), 6500)) | multiply(n0,const_4)|multiply(n1,const_4)|subtract(#1,n2)|subtract(#0,#2) | general | D |
the average age of a group of 10 students is 14 years . if 5 more students join the group , the average age rises by 1 year . the average age of the new students is : | explanation : total age of the 10 students = 10 × 14 = 140 total age of 15 students including the newly joined 5 students = 15 × 15 = 225 total age of the new students = 225 − 140 = 85 average age = 85 / 5 = 17 years answer : d | a ) 22 , b ) 38 , c ) 11 , d ) 17 , e ) 91 | d | divide(subtract(multiply(add(14, 1), add(14, 1)), multiply(14, 10)), 5) | add(n1,n3)|multiply(n0,n1)|multiply(#0,#0)|subtract(#2,#1)|divide(#3,n2) | general | D |
in the hillside summer camp there are 50 children . 80 % of the children are boys and the rest are girls . the camp administrator decided to make the number of girls only 10 % of the total number of children in the camp . how many more boys must she bring to make that happen ? | "given there are 50 students , 80 % of 50 = 40 boys and remaining 10 girls . now here 80 % are boys and 20 % are girls . now question is asking about how many boys do we need to add , to make the girls percentage to 20 or 20 % . . if we add 50 to existing 40 then the count will be 90 and the girls number will be 10 as it . now boys are 90 % and girls are 10 % . ( out of 100 students = 90 boys + 10 girls ) . imo option b is correct ." | a ) 55 , b ) 50 , c ) 40 . , d ) 30 . , e ) 25 . | b | add(multiply(divide(subtract(const_100, 10), const_100), 50), multiply(divide(10, const_100), 50)) | divide(n2,const_100)|subtract(const_100,n2)|divide(#1,const_100)|multiply(n0,#0)|multiply(n0,#2)|add(#4,#3)| | general | B |
p can do a work in the same time in which q and r together can do it . if p and q work together , the work can be completed in 10 days . r alone needs 30 days to complete the same work . then q alone can do it in | work done by p and q in 1 day = 1 / 10 work done by r in 1 day = 1 / 30 work done by p , q and r in 1 day = 1 / 10 + 1 / 30 = 4 / 30 but work done by p in 1 day = work done by q and r in 1 day . hence the above equation can be written as work done by p in 1 day ã — 2 = 4 / 30 = > work done by p in 1 day = 4 / 60 = > work done by q and r in 1 day = 4 / 60 hence work done by q in 1 day = 4 / 60 â € “ 1 / 30 = 2 / 60 = 1 / 30 so q alone can do the work in 30 days answer is e . | a ) 10 , b ) 22 , c ) 25 , d ) 27 , e ) 30 | e | divide(const_1, subtract(divide(add(divide(const_1, 10), divide(const_1, 30)), const_2), divide(const_1, 30))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(#2,const_2)|subtract(#3,#1)|divide(const_1,#4) | physics | E |
the monthly rent of a shop of dimension 20 feet × 18 feet is rs . 1440 . what is the annual rent per square foot of the shop ? | sol . monthly rent per square feet = 1440 / ( 20 * 18 ) = 4 & annual rent per square feet = 12 * 4 = 48 answer : a | ['a ) 48', 'b ) 56', 'c ) 68', 'd ) 87', 'e ) 92'] | a | multiply(add(const_10, const_2), divide(1440, rectangle_area(20, 18))) | add(const_10,const_2)|rectangle_area(n0,n1)|divide(n2,#1)|multiply(#0,#2) | geometry | A |
an urn contains 5 red , 6 blue and 8 green balls . 3 balls are randomly selected from the urn , find the probability that the drawn ball are 2 blue and 1 red ? | "sample space = no . of ways 3 balls were drawn from urn = 19 c 3 = 969 no . ways 2 blue balls and 1 red were drawn from bag = 6 c 2 * 5 c 1 = 75 probability = 75 / 969 = 25 / 323 ans - a" | a ) 25 / 323 , b ) 21 / 969 , c ) 28 / 989 , d ) 74 / 879 , e ) 23 / 589 | a | divide(multiply(choose(6, 2), choose(5, 1)), choose(add(add(5, 6), 8), 3)) | add(n0,n1)|choose(n1,n4)|choose(n0,n5)|add(n2,#0)|multiply(#1,#2)|choose(#3,n3)|divide(#4,#5)| | probability | A |
the least number which when increased by 6 each divisible by each one of 24 , 32 , 36 and 54 is : | "solution required number = ( l . c . m . of 24 , 32 , 36 , 54 ) - 5 = 864 - 6 = 858 . answer a" | a ) 858 , b ) 859 , c ) 869 , d ) 4320 , e ) none of these | a | subtract(lcm(lcm(lcm(24, 32), 36), 54), 6) | lcm(n1,n2)|lcm(n3,#0)|lcm(n4,#1)|subtract(#2,n0)| | general | A |
in a kilometer race , a beats b by 48 meters or 6 seconds . what time does a take to complete the race ? | "time taken by b run 1000 meters = ( 1000 * 6 ) / 48 = 125 sec . time taken by a = 125 - 6 = 119 sec . answer : b" | a ) 22 , b ) 119 , c ) 110 , d ) 109 , e ) 12 | b | subtract(divide(multiply(const_1, const_1000), divide(48, 6)), 6) | divide(n0,n1)|multiply(const_1,const_1000)|divide(#1,#0)|subtract(#2,n1)| | physics | B |
if 5 + 3 + 2 = 151022 , 9 + 2 + 4 = 183652 , then 7 + 2 + 5 = ? | if the given number is a + b + c then a . b | a . c | a . b + a . c - b ⇒ ⇒ 5 + 3 + 2 = 5.3 | 5.2 | 5.3 + 5.2 - 3 = 151022 ⇒ ⇒ 9 + 2 + 4 = 9.2 | 9.4 | 9.2 + 9.4 - 2 = 183652 7 + 2 + 5 = 7.2 | 7.5 | 7.2 + 7.5 - 2 = 143547 answer : b | a ) 223888 , b ) 143547 , c ) 2607778 , d ) 126997 , e ) 127811 | b | add(multiply(add(multiply(multiply(7, 2), const_100), multiply(7, 5)), const_100), subtract(add(multiply(7, 2), multiply(7, 5)), 2)) | multiply(n2,n8)|multiply(n0,n8)|add(#0,#1)|multiply(#0,const_100)|add(#3,#1)|subtract(#2,n2)|multiply(#4,const_100)|add(#6,#5) | general | B |
if the price of petrol increases by 30 , by how much must a user cut down his consumption so that his expenditure on petrol remains constant ? | "explanation : let us assume before increase the petrol will be rs . 100 . after increase it will be rs ( 100 + 30 ) i . e 130 . now , his consumption should be reduced to : - = ( 130 − 100 ) / 130 ∗ 100 . hence , the consumption should be reduced to 23 % . answer : b" | a ) 25 % , b ) 23 % , c ) 16.67 % , d ) 33.33 % , e ) none of these | b | multiply(subtract(const_1, divide(const_100, add(const_100, 30))), const_100) | add(n0,const_100)|divide(const_100,#0)|subtract(const_1,#1)|multiply(#2,const_100)| | general | B |
a car averages 30 miles per hour for the first 6 hours of a trip and averages 46 miles per hour for each additional hour of travel time . if the average speed for the entire trip is 34 miles per hour , how many hours long is the trip ? | "let t be the total time of the trip . 30 * 6 + 46 ( t - 6 ) = 34 t 12 t = 276 - 180 t = 8 the answer is a ." | a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 16 | a | add(divide(subtract(multiply(34, 6), multiply(30, 6)), subtract(46, 34)), 6) | multiply(n1,n3)|multiply(n0,n1)|subtract(n2,n3)|subtract(#0,#1)|divide(#3,#2)|add(n1,#4)| | physics | A |
if ( a + b ) = 14 , ( b + c ) = 9 and ( c + d ) = 3 , what is the value of ( a + d ) ? | "given a + b = 14 = > a = 14 - b - - > eq 1 b + c = 9 c + d = 3 = > d = 3 - c - - > eq 2 then eqs 1 + 2 = > a + d = 14 - b + 3 - c = > 17 - ( b + c ) = > 17 - 9 = 8 . option b . . ." | a ) 16 . , b ) 8 . , c ) 7 . , d ) 2 . , e ) - 2 . | b | subtract(add(14, 3), 9) | add(n0,n2)|subtract(#0,n1)| | general | B |
in an electric circuit , two resistors with resistances x and y are connected in parallel . if r is the combined resistance of these two resistors , then the reciprocal of r is equal to the sum of the reciprocals of x and y . what is r if x is 5 ohms and y is 7 ohms ? | "1 / r = 1 / x + 1 / y 1 / r = 1 / 5 + 1 / 7 = 12 / 35 r = 35 / 12 the answer is e ." | a ) 3 / 5 , b ) 1 / 7 , c ) 12 / 35 , d ) 31 / 12 , e ) 35 / 12 | e | divide(multiply(7, 5), add(5, 7)) | add(n0,n1)|multiply(n0,n1)|divide(#1,#0)| | general | E |
a container holding 12 ounces of a solution that is 1 part alcohol to 2 parts water is added to a container holding 9 ounces of a solution that is 1 part alcohol to 2 parts water . what is the ratio of alcohol to water in the resulting solution ? | "container 1 has 12 ounces in the ratio 1 : 2 or , x + 2 x = 12 gives x ( alcohol ) = 4 and remaining water = 8 container 2 has 9 ounces in the ratio 1 : 2 or , x + 2 x = 9 gives x ( alcohol ) = 3 and remaining water = 6 mixing both we have alcohol = 4 + 3 and water = 8 + 6 ratio thus alcohol / water = 7 / 14 = 1 / 2 answer a" | a ) 1 : 2 , b ) 3 : 7 , c ) 3 : 5 , d ) 4 : 7 , e ) 7 : 3 | a | divide(add(multiply(12, divide(1, add(1, 2))), multiply(9, divide(1, add(1, 2)))), subtract(add(12, 9), add(multiply(12, divide(1, add(1, 2))), multiply(9, divide(1, add(1, 2)))))) | add(n1,n2)|add(n1,n5)|add(n0,n3)|divide(n1,#0)|divide(n1,#1)|multiply(n0,#3)|multiply(n3,#4)|add(#5,#6)|subtract(#2,#7)|divide(#7,#8)| | other | A |
the diagonals of a rhombus are 16 cm and 20 cm . find its area ? | "1 / 2 * 16 * 20 = 160 answer : a" | a ) 160 , b ) 288 , c ) 150 , d ) 238 , e ) 31 | a | rhombus_area(16, 20) | rhombus_area(n0,n1)| | geometry | A |
mr . evans will states that each of his children will receive an equal share of his estate and that his grandchildren will split a portion of the estate that is equal to the share received by each of his children . if mr . evans has 5 children and 6 grandchildren , then approximately what percentage of mr . evans estate will each grandchild receive ? | "share of each child ( 5 no ) and ( total share of grand children together ) 1 no = 1 / ( 5 + 1 ) = 1 / 6 as the grand children again equally share the part given to them : share of each grand child = [ ( 1 / 6 ) / 6 ] = 1 / 36 = 2.8 % answer : e" | a ) 20 % , b ) 17 % , c ) 4.0 % , d ) 3.3 % , e ) 2.8 % | e | multiply(divide(divide(divide(const_100, const_3), 6), const_100), const_100) | divide(const_100,const_3)|divide(#0,n1)|divide(#1,const_100)|multiply(#2,const_100)| | general | E |
if x is an integer such that 0 < x < 7 , 0 < x < 15 , 5 > x > – 1 , 3 > x > 0 , and x + 2 < 4 , then x is | "0 < x < 7 , 0 < x < 15 , 5 > x > – 1 3 > x > 0 x < 2 from above : 0 < x < 2 - - > x = 1 . answer : a ." | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | a | subtract(subtract(5, 2), 2) | subtract(n4,n8)|subtract(#0,n8)| | general | A |
pipe a fills a tank in 20 minutes . pipe b can fill the same tank 4 times as fast as pipe a . if both the pipes are kept open when the tank is empty , how many minutes will it take to fill the tank ? | "a ' s rate is 1 / 20 and b ' s rate is 1 / 5 . the combined rate is 1 / 20 + 1 / 5 = 1 / 4 the pipes will fill the tank in 4 minutes . the answer is c ." | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | c | inverse(add(divide(const_1, 20), divide(4, 20))) | divide(const_1,n0)|divide(n1,n0)|add(#0,#1)|inverse(#2)| | physics | C |
in triangle pqr , the angle q = 90 degree , pq = 7 cm , qr = 8 cm . x is a variable point on pq . the line through x parallel to qr , intersects pr at y and the line through y , parallel to pq , intersects qr at z . find the least possible length of xz | look at the diagram below : now , in case when qy is perpendicular to pr , two right triangles pqr and pqy are similar : qy : qp = qr : pr - - > qy : 7 = 8 : 10 - - > qy = 5.6 . answer : a . | ['a ) 5.6 cm', 'b ) 2.4 cm', 'c ) 4.8 cm', 'd ) 2.16 cm', 'e ) 3.2 cm'] | a | divide(multiply(7, 8), const_10) | multiply(n1,n2)|divide(#0,const_10) | geometry | A |
two trains 140 m and 180 m long run at the speed of 60 km / hr and 40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? | "relative speed = 60 + 40 = 100 km / hr . = 100 * 5 / 18 = 250 / 9 m / sec . distance covered in crossing each other = 140 + 180 = 320 m . required time = 320 * 9 / 250 = 11.52 sec . answer : a" | a ) 11.52 sec , b ) 10.1 sec , c ) 10.6 sec , d ) 10.8 sec , e ) 10.2 sec | a | divide(add(140, 180), multiply(add(60, 40), const_0_2778)) | add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)| | physics | A |
tough and tricky questions : number properties . if ( z + 3 ) / 9 is an integer , what is remainder when z is divided by 9 ? | assume the answer choices as the value of z 9 / 9 = 1 ans e | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | e | subtract(9, 3) | subtract(n1,n0) | general | E |
if the cost price of 15 articles is same as the selling price of 25 articles . find the gain or loss percentage ? | "explanation : 15 cp = 25 sp cost price cp = 25 selling price sp = 15 formula = ( sp - cp ) / cp * 100 = ( 15 - 25 ) / 25 * 100 = 40 % loss answer : option d" | a ) 30 % gain , b ) 30 % loss , c ) 40 % gain , d ) 40 % loss , e ) 50 % loss | d | multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 25), 15)), divide(multiply(const_100, 25), 15))) | multiply(n1,const_100)|divide(#0,n0)|subtract(const_100,#1)|divide(#2,#1)|multiply(#3,const_100)| | gain | D |
two ants , arthur and amy , have discovered a picnic and are bringing crumbs back to the anthill . amy makes twice as many trips and carries one and a half times as many crumbs per trip as arthur . if arthur carries a total of z crumbs to the anthill , how many crumbs will amy bring to the anthill , in terms of z ? | "lets do it by picking up numbers . let arthur carry 2 crumbs per trip , this means amy carries 3 crumbs per trip . also let arthur make 2 trips and so amy makes 4 trips . thus total crumbs carried by arthur ( z ) = 2 x 2 = 4 , total crumbs carried by amy = 3 x 4 = 12 . 12 is 3 times 4 , so e" | a ) z / 2 , b ) z , c ) 3 z / 2 , d ) 2 z , e ) 3 z | e | multiply(const_2, add(const_1, divide(const_1, const_2))) | divide(const_1,const_2)|add(#0,const_1)|multiply(#1,const_2)| | general | E |
the shopkeeper increased the price of a product by 25 % so that customer finds it difficult to purchase the required amount . but somehow the customer managed to purchase only 64 % of the required amount . what is the net difference in the expenditure on that product ? | "quantity x rate = price 1 x 1 = 1 0.64 x 1.25 = 0.8 decrease in price = ( 0.2 / 1 ) × 100 = 20 % e )" | a ) 12.5 % , b ) 13 % , c ) 15 % , d ) 17 % , e ) 20 % | e | divide(multiply(subtract(multiply(const_100, const_100), multiply(add(const_100, 25), 64)), const_100), multiply(const_100, const_100)) | add(n0,const_100)|multiply(const_100,const_100)|multiply(n1,#0)|subtract(#1,#2)|multiply(#3,const_100)|divide(#4,#1)| | general | E |
8 men and 2 boys working together can do 4 times as much work as a man and a boy . working capacity of man and boy is in the ratio | explanation : let 1 man 1 day work = x 1 boy 1 day work = y then 8 x + 2 y = 4 ( x + y ) = > 4 x = 2 y = > x / y = 2 / 4 = > x : y = 1 : 2 option a | a ) 1 : 2 , b ) 1 : 3 , c ) 2 : 1 , d ) 2 : 3 , e ) none of these | a | divide(subtract(4, 2), subtract(8, 4)) | subtract(n2,n1)|subtract(n0,n2)|divide(#0,#1) | other | A |
arnold and danny are two twin brothers that are celebrating their birthday . the product of their ages today is smaller by 17 from the product of their ages a year from today . what is their age today ? | "ad = ( a + 1 ) ( d + 1 ) - 17 0 = a + d - 16 a + d = 16 a = d ( as they are twin brothers ) a = d = 8 e is the answer" | a ) 2 . , b ) 4 . , c ) 5 . , d ) 7 . , e ) 8 . | e | divide(subtract(17, const_1), const_2) | subtract(n0,const_1)|divide(#0,const_2)| | general | E |
count the numbers between 10 - 99 which yield a remainder of 3 when divided by 9 and also yield a remainder of 2 when divided by 5 ? | answer = a ) two numbers between 10 - 99 giving remainder 3 when divided by 9 = 12 , 21 , 30 , 39 , 48 , 57 , 66 , 75 , 84 , 93 the numbers giving remainder 2 when divided by 5 = 12 , 57 = 2 | a ) two , b ) five , c ) six , d ) four , e ) one | a | multiply(2, const_1) | multiply(n4,const_1) | general | A |
if x is real , find the maximum value of the expression - 2 x ^ 2 + 9 x + 11 . | "this is an equation of a downward facing parabola . the maximum value is the top point of the parabola . - 2 x ^ 2 + 9 x + 11 = ( - 2 x + 11 ) ( x + 1 ) the roots are 11 / 2 and - 1 . the maximum value must be when x is halfway between these two points . x = 2.25 the maximum value is - 2 ( 2.25 ) ^ 2 + 9 ( 2.25 ) + 11 = 21.125 the answer is d ." | a ) 12.125 , b ) 15.125 , c ) 18.125 , d ) 21.125 , e ) 24.125 | d | subtract(add(multiply(9, divide(multiply(2, 2), 9)), 11), multiply(2, power(divide(multiply(2, 2), 9), 2))) | multiply(n0,n0)|divide(#0,n2)|multiply(n2,#1)|power(#1,n0)|add(n3,#2)|multiply(n0,#3)|subtract(#4,#5)| | general | D |
on a game show , a contestant is given 3 keys , each of which opens exactly one of 3 identical boxes . the first box contains $ 5 , the second $ 500 , and the third $ 5000 . the contestant assigns each key to one of the boxes and wins the amount of money contained in any box that is opened by the key assigned to it . what is the probability that a contestant will win more than $ 5000 ? | let ' s call the boxes that contain $ 5 , $ 500 , and $ 5000 , respectively , box a , box b , box c . these are opened , respectively , by key a , key b , and key c . we want to know the probability of winning more than $ 5000 . notice that if the distribution of keys is : box a = key b box b = key a box c = key c then the contestant wins exactly $ 5000 , not more than $ 5000 . the only configuration that leads to winning more than $ 1000 is : box a = key a box b = key b box c = key c i . e . , getting all three keys correct . that ' s the only way to be more than $ 5000 . so , really , the question can be rephrased : what is the probability of guessing the order of keys so that each key matches the correct box ? well , for a set of three items , the number of possible permutations is 3 ! = 3 * 2 * 1 = 6 . of those 6 possible permutations , only one of them leads to all three keys being paired with the right box . so , the answer is probability = 1 / 6 answer : e | a ) 1 / 9 , b ) 4 / 6 , c ) 3 / 6 , d ) 1 / 2 , e ) 1 / 6 | e | divide(const_1, factorial(3)) | factorial(n0)|divide(const_1,#0) | general | E |
what is x if x + 2 y = 20 and y = 5 ? | "x = 20 - 2 y x = 20 - 10 . x = 10 answer : a" | a ) a ) 10 , b ) b ) 8 , c ) c ) 6 , d ) d ) 4 , e ) e ) 2 | a | subtract(20, multiply(2, 5)) | multiply(n0,n2)|subtract(n1,#0)| | general | A |
jane and thomas are among the 7 people from which a committee of 4 people is to be selected . how many different possible committees of 4 people can be selected from these 7 people if at least one of either jane or thomas is to be selected ? | "the total number of ways to choose 4 people from 7 is 7 c 4 = 35 . the number of committees without jane or thomas is 5 c 4 = 5 . there are 35 - 5 = 30 possible committees which include jane and / or thomas . the answer is a ." | a ) 30 , b ) 45 , c ) 55 , d ) 65 , e ) 70 | a | add(add(choose(7, const_1), choose(7, const_1)), choose(const_4, const_1)) | choose(n0,const_1)|choose(const_4,const_1)|add(#0,#0)|add(#2,#1)| | probability | A |
a can do a work in 15 days and b in 20 days . if they work on it together for 3 days , then the fraction of the work that is left is | "person ( a ) ( b ) ( a + b ) time - ( 15 ) ( 20 ) ( - ) rate - ( 20 ) ( 15 ) ( 35 ) work - ( 300 ) ( 300 ) ( 300 ) therefore a + b requires ( 300 / 35 ) days to complete entire work for 1 st 4 days they work 35 * 3 = 105 remaining work is 300 - 105 = 195 remaining fraction of work is = 195 / 300 = 13 / 20 answer e" | a ) 8 / 17 , b ) 7 / 15 , c ) 3 / 15 , d ) 8 / 15 , e ) 13 / 20 | e | subtract(const_1, multiply(add(divide(const_1, 15), divide(const_1, 20)), 3)) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|multiply(n2,#2)|subtract(const_1,#3)| | physics | E |
walking with 4 / 5 of my usual speed , i miss the bus by 4 minutes . what is my usual time ? | "speed ratio = 1 : 4 / 5 = 5 : 4 time ratio = 4 : 5 1 - - - - - - - - 4 4 - - - - - - - - - ? è 16 answer : a" | a ) 16 min , b ) 26 min , c ) 34 min , d ) 20 min , e ) 12 min | a | multiply(divide(4, divide(5, 4)), 5) | divide(n1,n0)|divide(n2,#0)|multiply(n1,#1)| | physics | A |
the number of diagonals of a polygon of n sides is given by the formula z = n ( n - 3 ) / 2 . if a polygon has twice as many diagonals as sides , how many sides does it have ? | "z = n ( n - 3 ) z = 2 * n 2 n = n ( n - 3 ) = > 2 = n - 3 = > n = 5 answer b" | a ) 3 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | b | add(3, 2) | add(n0,n1)| | general | B |
at an election 2 candidates are participated and a candidate got 30 % of votes and defeated by 1000 . and 100 votes are invalid . find the total polled votes ? | "winner votes = 100 - 30 = 70 polled votes = [ ( 100 * 1000 ) / 2 * 70 - 100 ] + 100 = 2600 answer is b" | a ) 2150 , b ) 2600 , c ) 3120 , d ) 1500 , e ) 1895 | b | divide(add(1000, 100), subtract(const_1, divide(multiply(30, 2), const_100))) | add(n2,n3)|multiply(n0,n1)|divide(#1,const_100)|subtract(const_1,#2)|divide(#0,#3)| | gain | B |
the average age of 35 students in a class is 16 years . the average age of 21 students is 14 . what is the average age of remaining 38 students ? | "solution sum of the ages of 14 students = ( 16 x 35 ) - ( 14 x 21 ) = 560 - 294 . = 266 . ∴ required average = 266 / 38 = 7 years . answer b" | a ) 14 years , b ) 7 years , c ) 19 years , d ) 21 years , e ) none | b | subtract(add(add(multiply(35, 16), 21), 35), multiply(35, 16)) | multiply(n0,n1)|add(n2,#0)|add(n0,#1)|subtract(#2,#0)| | general | B |
a train 130 meters long takes 6 seconds to cross a man walking at 5 kmph in the direction opposite to that of the train . find the speed of the train . | "explanation : let the speed of the train be x kmph . speed of the train relative to man = ( x + 5 ) kmph = ( x + 5 ) × 5 / 18 m / sec . therefore 130 / ( ( x + 5 ) × 5 / 18 ) = 6 < = > 30 ( x + 5 ) = 2340 < = > x = 73 speed of the train is 73 kmph . answer : option e" | a ) 45 kmph , b ) 50 kmph , c ) 55 kmph , d ) 60 kmph , e ) 73 kmph | e | subtract(divide(130, multiply(6, const_0_2778)), 5) | multiply(n1,const_0_2778)|divide(n0,#0)|subtract(#1,n2)| | physics | E |
a towel , when bleached , lost 20 % of its length and 10 % of its breadth . what is the percentage decrease in area ? | "formula for percentage change in area is : = ( − x − y + ( xy ) / 100 ) % = ( − 20 − 10 + ( 20 × 10 ) / 100 ) % = − 28 % i . e . , area is decreased by 28 % answer is b ." | a ) 26 , b ) 28 , c ) 24 , d ) 20 , e ) 22 | b | divide(subtract(multiply(const_100, const_100), multiply(subtract(const_100, 20), subtract(const_100, 10))), 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 | B |
a sum of rs . 1870 has been divided among a , b and c such that a gets of what b gets and b gets of what c gets . b ’ s share is : | "explanation let c ’ s share = rs . x then , b ’ s share = rs . x / 4 , a ’ s share = rs . ( 2 / 3 x x / 4 ) = rs . x / 6 = x / 6 + x / 4 + x = 1870 = > 17 x / 12 = 1870 = > 1870 x 12 / 17 = rs . 1320 hence , b ’ s share = rs . ( 1320 / 4 ) = rs . 330 . answer d" | a ) rs . 120 , b ) rs . 160 , c ) rs . 240 , d ) rs . 330 , e ) none | d | subtract(subtract(multiply(divide(1870, const_10), const_2), const_12), const_12) | divide(n0,const_10)|multiply(#0,const_2)|subtract(#1,const_12)|subtract(#2,const_12)| | general | D |
pipe a can fill a tank in 8 minutes and pipe b cam empty it in 24 minutes . if both the pipes are opened together after how many minutes should pipe b be closed , so that the tank is filled in 30 minutes ? | let the pipe b be closed after x minutes . 30 / 8 - x / 24 = 1 = > x / 24 = 30 / 8 - 1 = 11 / 4 = > x = 11 / 4 * 24 = 66 . answer : e | a ) 18 , b ) 27 , c ) 98 , d ) 27 , e ) 66 | e | multiply(subtract(divide(30, 8), const_1), 24) | divide(n2,n0)|subtract(#0,const_1)|multiply(n1,#1) | physics | E |
if there is one larger cube with surface area 600 and no . of smaller cubes with surface area 24 . then how many small cubes are required to make a larger cube with surface area of 600 ? | the ratio between larger and smaller is 6 a 12 : 6 a 22 = 600 : 24 a 12 : a 22 = 100 : 4 a 1 : a 2 = 10 : 2 then length of cube is 10 and 2 so make a cube with length 10 , we have to 5 small cubes each in length , breadth and height . so total 125 cubes are required . then the answer is b | ['a ) 5', 'b ) 125', 'c ) 4', 'd ) 100', 'e ) 24'] | b | volume_cube(divide(sqrt(divide(600, divide(600, const_100))), sqrt(divide(24, divide(600, const_100))))) | divide(n0,const_100)|divide(n0,#0)|divide(n1,#0)|sqrt(#1)|sqrt(#2)|divide(#3,#4)|volume_cube(#5) | geometry | B |
a man saves 10 % of his monthly salary . if an account of dearness of things he is to increase his monthly expenses by 10 % , he is only able to save rs . 200 per month . what is his monthly salary ? | "income = rs . 100 expenditure = rs . 90 savings = rs . 10 present expenditure 90 + 90 * ( 10 / 100 ) = rs . 99 present savings = 100 – 99 = rs . 1 if savings is rs . 1 , salary = rs . 100 if savings is rs . 200 , salary = 100 / 1 * 200 = 20000 answer : c" | a ) rs . 22,000 , b ) rs . 21,000 , c ) rs . 20,000 , d ) rs . 23,000 , e ) rs . 24,000 | c | divide(multiply(200, const_100), subtract(const_100, add(subtract(const_100, 10), multiply(subtract(const_100, 10), divide(10, const_100))))) | divide(n1,const_100)|multiply(n2,const_100)|subtract(const_100,n0)|multiply(#0,#2)|add(#3,#2)|subtract(const_100,#4)|divide(#1,#5)| | general | C |
3 / 4 of 1 / 2 of 2 / 5 of 5000 = ? | "d 750 ? = 5000 * ( 2 / 5 ) * ( 1 / 2 ) * ( 3 / 4 ) = 750" | a ) 392 , b ) 229 , c ) 753 , d ) 750 , e ) 540 | d | multiply(multiply(multiply(divide(3, 4), divide(1, 2)), divide(2, 5)), 5000) | divide(n3,n5)|divide(n0,n1)|divide(n2,n3)|multiply(#1,#2)|multiply(#0,#3)|multiply(n6,#4)| | general | D |
if 20 liters of chemical x are added to 80 liters of a mixture that is 30 % chemical x and 70 % chemical y , then what percentage of the resulting mixture is chemical x ? | "the amount of chemical x in the solution is 20 + 0.3 ( 80 ) = 44 liters . 44 liters / 100 liters = 44 % the answer is e ." | a ) 32 % , b ) 35 % , c ) 38 % , d ) 41 % , e ) 44 % | e | add(20, multiply(divide(30, const_100), 80)) | divide(n2,const_100)|multiply(n1,#0)|add(n0,#1)| | general | E |
the speed of a boat in still water in 65 km / hr and the rate of current is 15 km / hr . the distance travelled downstream in 25 minutes is : | "explanation : speed downstream = ( 65 + 15 ) = 80 kmph time = 25 minutes = 25 / 60 hour = 5 / 12 hour distance travelled = time × speed = ( 2 / 5 ) × 80 = 33.33 km answer : option c" | a ) 55.55 km , b ) 44.44 km , c ) 33.33 km , d ) 22.22 km , e ) 11.11 km | c | multiply(add(65, 15), divide(25, const_60)) | add(n0,n1)|divide(n2,const_60)|multiply(#0,#1)| | physics | C |
a committee is reviewing a total of 20 x black - and - white films and 8 y color films for a festival . if the committee selects y / x % of the black - and - white films and all of the color films , what fraction of the selected films are in color ? | "say x = y = 10 . in this case we would have : 20 x = 200 black - and - white films ; 8 y = 80 color films . y / x % = 10 / 10 % = 1 % of the black - and - white films , so 2 black - and - white films and all 80 color films , thus total of 82 films were selected . color films thus compose 80 / 82 = 40 / 41 of the selected films . answer : a" | a ) 40 / 41 , b ) 20 / 41 , c ) 30 / 41 , d ) 60 / 41 , e ) 80 / 41 | a | divide(8, add(divide(20, const_100), 8)) | divide(n0,const_100)|add(n1,#0)|divide(n1,#1)| | general | A |
evaluate : 20 - 12 ÷ 4 × 2 = | "according to order of operations , 12 ÷ 4 × 2 ( division and multiplication ) is done first from left to right 12 ÷ 4 × 2 = 3 × 2 = 6 hence 20 - 12 ÷ 4 × 2 = 20 - 6 = 14 correct answer is b ) 22" | a ) a ) 12 , b ) b ) 24 , c ) c ) 36 , d ) d ) 48 , e ) e ) 60 | b | subtract(20, multiply(multiply(12, const_2.0), 2)) | multiply(n1,const_2.0)|multiply(n3,#0)|subtract(n0,#1)| | general | B |
what is the smallest number which , when increased by 5 , is divisible by 7 , 8 , and 24 ? | "lcm ( 7 , 8,24 ) = 24 x 7 = 168 so the least divisible number is 168 , and the number we are looking for is 168 - 5 = 163 . the answer is d ." | a ) 148 , b ) 153 , c ) 158 , d ) 163 , e ) 168 | d | subtract(lcm(24, 8), 5) | lcm(n2,n3)|subtract(#0,n0)| | general | D |
the marks obtained by polly and sandy are in the ratio 5 : 6 and those obtained by sandy and willy are in the ratio of 3 : 2 . the marks obtained by polly and willy are in the ratio of . . . ? | "polly : sandy = 5 : 6 sandy : willy = 3 : 2 = 6 : 4 polly : sandy : willy = 5 : 6 : 4 polly : willy = 5 : 4 the answer is b ." | a ) 3 : 2 , b ) 5 : 4 , c ) 7 : 6 , d ) 9 : 8 , e ) 11 : 10 | b | divide(multiply(5, 3), multiply(6, 2)) | multiply(n0,n2)|multiply(n1,n3)|divide(#0,#1)| | other | B |
a farm is made up of brown and white goats . the number of male brown goats is 35 % less than the total number of brown goats . the total number of female goats is 20 times more than the number of female brown goats . if the total population of male goats is half the population of the female goats , what percentage of male goats are brown ? | since we are dealing in percentage let us pick 100 as the number of brown male goats . that means that the total number of brown goats is = 135 ( number of male brown goats is 35 % less than the total number of brown goats ) , therefore the number of brown female goats is 35 . the total number of female goats is 15 x the number of brown female goats = 35 * 20 = 700 female goats . male goats total population is half the population of the female goats = 700 / 2 = 350 therefore percentage of male horses goats that are brown = 100 / 350 * 100 = 28.6 correct option is e | a ) 30.1 , b ) 20.5 , c ) 15.2 , d ) 10.2 , e ) 28.6 | e | multiply(multiply(divide(35, const_100), 20), const_4) | divide(n0,const_100)|multiply(n1,#0)|multiply(#1,const_4) | general | E |
if 80 % of 90 is greater than 70 % of a number by 30 , what is the number ? | "explanation : 80 / 100 * 90 - 70 / 100 * x = 30 72 - 70 / 100 * x = 30 72 - 30 = 70 / 100 * x 42 = 70 / 100 * x 42 * 100 / 70 = x 60 = x answer : option b" | a ) 70 , b ) 60 , c ) 10 , d ) 75 , e ) 85 | b | divide(subtract(multiply(divide(80, const_100), 90), 30), divide(70, const_100)) | divide(n0,const_100)|divide(n2,const_100)|multiply(n1,#0)|subtract(#2,n3)|divide(#3,#1)| | gain | B |
. 009 / ? = . 01 | "let . 009 / x = . 01 ; then x = . 009 / . 01 = . 9 / 1 = . 9 answer is a" | a ) . 9 , b ) . 09 , c ) . 009 , d ) . 0009 , e ) none of them | a | divide(divide(009, const_1000), divide(01, const_100)) | divide(n0,const_1000)|divide(n1,const_100)|divide(#0,#1)| | general | A |
kanul spent $ 500 in buying raw materials , $ 400 in buying machinery and 10 % of the total amount he had as cash with him . what was the total amount ? | "let the total amount be x then , ( 100 - 10 ) % of x = 500 + 400 90 % of x = 900 90 x / 100 = 9000 / 9 x = $ 1000 answer is e" | a ) $ 1010 , b ) $ 1025 , c ) $ 1125 , d ) $ 1100 , e ) $ 1000 | e | divide(add(500, 400), subtract(const_1, divide(10, const_100))) | add(n0,n1)|divide(n2,const_100)|subtract(const_1,#1)|divide(#0,#2)| | gain | E |
what is the decimal equivalent of ( 1 / 4 ) ^ 6 ? | "( 1 / 4 ) ^ 6 = 1 / 4096 = 0.0002 answer : d" | a ) 0.0016 , b ) 0.0625 , c ) 0.16 , d ) 0.0002 , e ) 0.5 | d | power(divide(1, 4), 6) | divide(n0,n1)|power(#0,n2)| | general | D |
the average ( arithmetic mean ) of 20 , 40 , and 60 is 8 more than the average of 10 , 70 , and what number ? | "a 1 = 120 / 3 = 40 a 2 = a 1 - 8 = 32 sum of second list = 32 * 3 = 96 therefore the number = 96 - 80 = 16 answer : a" | a ) 16 , b ) 25 , c ) 35 , d ) 45 , e ) 55 | a | subtract(add(add(20, 40), 60), add(add(multiply(8, const_3), 10), 70)) | add(n0,n1)|multiply(n3,const_3)|add(n2,#0)|add(n4,#1)|add(n5,#3)|subtract(#2,#4)| | general | A |
what is the dividend . divisor 18 , the quotient is 9 and the remainder is 3 | "d = d * q + r d = 18 * 9 + 3 d = 162 + 3 d = 165 answer : c" | a ) a ) 145 , b ) b ) 148 , c ) c ) 165 , d ) d ) 153 , e ) e ) 158 | c | add(multiply(18, 9), 3) | multiply(n0,n1)|add(n2,#0)| | general | C |
a and b began business with rs . 2000 and rs . 4000 after 8 months , a withdraws rs . 1000 and b advances rs . 1000 more . at the end of the year , their profits amounted to rs . 630 find the share of a . | "( 2 * 8 + 1 * 4 ) : ( 4 * 8 + 5 * 4 ) 5 : 13 5 / 18 * 630 = 175 answer : a" | a ) 175 , b ) 288 , c ) 277 , d ) 877 , e ) 361 | a | multiply(divide(630, add(add(multiply(2000, 8), multiply(subtract(2000, 1000), subtract(const_12, 8))), add(multiply(4000, 8), multiply(add(4000, 1000), subtract(const_12, 8))))), add(multiply(2000, 8), multiply(subtract(2000, 1000), subtract(const_12, 8)))) | add(n1,n3)|multiply(n0,n2)|multiply(n1,n2)|subtract(n0,n3)|subtract(const_12,n2)|multiply(#3,#4)|multiply(#0,#4)|add(#1,#5)|add(#2,#6)|add(#7,#8)|divide(n5,#9)|multiply(#7,#10)| | gain | A |
how many multiples of 9 are there between 1 and 50 , exclusive ? | "5 multiples of 9 between 1 and 50 exclusive . from 9 * 1 upto 9 * 5 , ( 1 , 2,3 , 4,5 ) . hence , 5 multiples ! correct option is a" | a ) 5 , b ) 4 , c ) 6 , d ) 7 , e ) 3 | a | add(divide(subtract(50, 1), 9), const_1) | subtract(n2,n1)|divide(#0,n0)|add(#1,const_1)| | general | A |
a hat company ships its hats , individually wrapped , in 8 - inch by 10 - inch by 12 - inch boxes . each hat is valued at $ 6.0 . if the company ’ s latest order required a truck with at least 288,000 cubic inches of storage space in which to ship the hats in their boxes , what was the minimum value of the order ? | "total volume is 288000 given lbh = 8 * 10 * 12 . the number of hats inside it = 288000 / 10 * 8 * 12 = 300 . price of each hat is 6 $ then total value is 300 * 6.0 = 1800 . imo option e is correct answer . ." | a ) $ 960 , b ) $ 1,350 , c ) $ 1,725 , d ) $ 2,050 , e ) $ 1,800 | e | divide(multiply(divide(multiply(add(add(multiply(const_3, const_100), multiply(8, 10)), const_4), const_1000), multiply(multiply(8, 10), 12)), 6.0), const_1000) | multiply(const_100,const_3)|multiply(n0,n1)|multiply(n0,n1)|add(#0,#1)|multiply(n2,#2)|add(#3,const_4)|multiply(#5,const_1000)|divide(#6,#4)|multiply(n3,#7)|divide(#8,const_1000)| | general | E |
what will be the reminder when ( 43 ^ 43 + 43 ) is divided by 44 ? | "( x ^ n + 1 ) will be divisible by ( x + 1 ) only when n is odd ; ( 43 ^ 43 + 1 ) will be divisible by ( 43 + 1 ) ; ( 43 ^ 43 + 1 ) + 42 when divided by 44 will give 42 as remainder . correct option : b" | a ) 32 , b ) 42 , c ) 83 , d ) 71 , e ) 92 | b | subtract(add(power(43, 43), 43), multiply(44, floor(divide(add(power(43, 43), 43), 44)))) | power(n0,n0)|add(n0,#0)|divide(#1,n3)|floor(#2)|multiply(n3,#3)|subtract(#1,#4)| | general | B |
the marks obtained by vijay and amith are in the ratio 2 : 5 and those obtained by amith and abhishek in the ratio of 3 : 2 . the marks obtained by vijay and abhishek are in the ratio of ? | "2 : 5 3 : 2 - - - - - - - 6 : 15 : 10 6 : 10 3 : 5 answer : e" | a ) 6 : 8 , b ) 3 : 1 , c ) 6 : 5 , d ) 3 : 2 , e ) 3 : 5 | e | divide(multiply(2, 3), multiply(5, 2)) | multiply(n0,n2)|multiply(n1,n3)|divide(#0,#1)| | other | E |
due to construction , the speed limit along an 15 - mile section of highway is reduced from 55 miles per hour to 35 miles per hour . approximately how many minutes more will it take to travel along this section of highway at the new speed limit than it would have taken at the old speed limit ? | "old time in minutes to cross 15 miles stretch = 15 * 60 / 55 = 15 * 12 / 11 = 16.36 new time in minutes to cross 15 miles stretch = 15 * 60 / 35 = 15 * 12 / 7 = 25.71 time difference = 9.36 answer c )" | a ) 5 , b ) 8 , c ) 9.36 , d ) 15 , e ) 24 | c | max(multiply(subtract(add(55, 15), const_1), subtract(divide(15, 35), divide(15, 55))), const_4) | add(n0,n1)|divide(n0,n2)|divide(n0,n1)|subtract(#0,const_1)|subtract(#1,#2)|multiply(#3,#4)|max(#5,const_4)| | physics | C |
60 % of a number is added to 120 , the result is the same number . find the number ? | "( 60 / 100 ) * x + 120 = x 2 x = 600 x = 300 answer : a" | a ) 300 , b ) 288 , c ) 270 , d ) 129 , e ) 281 | a | divide(120, divide(120, const_100)) | divide(n1,const_100)|divide(n1,#0)| | gain | A |
the difference between the ages of two persons is 20 years . fifteen years ago , the elder one was twice as old as the younger one . the present age of the younger person is ? | let their ages be x years and ( x + 20 ) years then , ( x + 20 ) - 15 = 2 ( x - 15 ) x + 5 = 2 x - 30 x = 35 answer is b | a ) 30 , b ) 35 , c ) 25 , d ) 28 , e ) 32 | b | add(subtract(20, subtract(20, add(const_3, const_2))), multiply(subtract(20, add(const_3, const_2)), const_2)) | add(const_2,const_3)|subtract(n0,#0)|multiply(#1,const_2)|subtract(n0,#1)|add(#2,#3) | general | B |
on dividing number by 357 , we get 39 as remainder . on dividing the same number by 17 , what will be the remainder ? | "let the number be x and on dividing x by 5 , we get k as quotient and 3 as remainder . x = 5 k + 3 x 2 = ( 5 k + 3 ) 2 = ( 25 k 2 + 30 k + 9 ) = 5 ( 5 k 2 + 6 k + 1 ) + 4 on dividing x 2 by 5 , we get 4 as remainder . answer : c" | a ) 0 , b ) 3 , c ) 5 , d ) 11 , e ) 13 | c | reminder(39, 17) | reminder(n1,n2)| | general | C |
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