Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
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two trains of length 140 m and 280 m are running towards each other on parallel lines at 42 kmph and 30 kmph respectively . in what time will they be clear of each other from the moment they meet ? | "relative speed = ( 42 + 30 ) * 5 / 18 = 4 * 5 = 20 mps . distance covered in passing each other = 140 + 280 = 420 m . the time required = d / s = 420 / 20 = 21 sec . answer : e" | a ) 28 , b ) 266 , c ) 990 , d ) 20 , e ) 21 | e | divide(add(140, 280), multiply(add(42, 30), const_0_2778)) | add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)| | physics |
the tax on a commodity is diminished by 28 % but its consumption is increased by 12 % . find the decrease percent in the revenue derived from it ? | "100 * 100 = 10000 72 * 112 = 8064 10000 - - - - - - - 1936 100 - - - - - - - ? = 19.36 % answer : b" | a ) 18 % , b ) 19 % , c ) 32 % , d ) 12 % , e ) 52 % | b | subtract(const_100, divide(multiply(add(const_100, 12), subtract(const_100, 28)), const_100)) | add(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|subtract(const_100,#3)| | general |
two mechanics were working on your car . one can complete the given job in 5 hours , but the new guy takes eight hours . they worked together for the first two hours , but then the first guy left to help another mechanic on a different job . how long will it take the new guy to finish your car ? | "rate ( 1 ) = 1 / 5 rate ( 2 ) = 1 / 8 combined = 13 / 40 work done in 2 days = 13 / 20 work left = 7 / 20 rate * time = work left 1 / 8 * time = 7 / 20 time = 14 / 5 c" | a ) 7 / 4 , b ) 4 / 3 , c ) 14 / 5 , d ) 10 / 3 , e ) 17 / 5 | c | max(divide(subtract(const_1, multiply(add(divide(const_1, add(const_4, const_2)), divide(const_1, 5)), const_2)), divide(const_1, 5)), const_3) | add(const_2,const_4)|divide(const_1,n0)|divide(const_1,#0)|add(#2,#1)|multiply(#3,const_2)|subtract(const_1,#4)|divide(#5,#1)|max(#6,const_3)| | physics |
how many multiples of 4 are there between 20 and 196 ? | "it should be mentioned whether 20 and 196 are inclusive . if 20 and 196 are inclusive , then the answer is ( 196 - 20 ) / 4 + 1 = 45 . if 20 and 196 are not inclusive , then the answer is ( 192 - 24 ) / 4 + 1 = 43 . since oa is d , then we have not inclusive case ." | a ) 45 , b ) 42 , c ) 52 , d ) 43 , e ) 51 | d | add(divide(subtract(196, 20), 4), const_1) | subtract(n2,n1)|divide(#0,n0)|add(#1,const_1)| | general |
a group of men decided to do a work in 6 days , but 4 of them became absent . if the rest of the group did the work in 12 days , find the original number of men ? | "original number of men = 4 * 12 / ( 12 - 6 ) = 8 answer is e" | a ) 15 , b ) 6 , c ) 40 , d ) 10 , e ) 8 | e | divide(multiply(4, 12), subtract(12, 6)) | multiply(n1,n2)|subtract(n2,n0)|divide(#0,#1)| | physics |
after 4 games , team b had an average of 61.5 points per game . if it got only 47 points in game 5 , how many more points does it need to score to get its total above 500 ? | "( 4 * 61.5 ) + 47 + x > 500 246 + 47 + x > 500 293 + x > 500 = > x > 207 option e" | a ) 85 , b ) 74 , c ) 67 , d ) 53 , e ) 207 | e | subtract(500, add(multiply(4, 61.5), 47)) | multiply(n0,n1)|add(n2,#0)|subtract(n4,#1)| | general |
what is the square root of 151,321 ? | "1 ) checking the last digits in the answer options a , c , e are out since last digit of the square is known to be 1 . 2 ) b = 389 ^ 2 and d = 411 ^ 2 b = ( 400 - 11 ) ^ 2 and d = ( 400 + 11 ) ^ 2 since we need the answer less than 160000 = > d is out . thus we have b as the ans ." | a ) 476 , b ) 389 , c ) 497 , d ) 411 , e ) 524 | b | circle_area(divide(151,321, multiply(const_2, const_pi))) | multiply(const_2,const_pi)|divide(n0,#0)|circle_area(#1)| | other |
the jogging track in a sports complex is 726 m in circumference . deepak and his wife start from the same point and walk in opposite directions at 4.5 km / hr and 3.75 km / hr respectively . they will meet for the first time in ? | "clearly , the two will meet when they are 726 m apart . to be ( 4.5 + 3.75 ) = 8.25 km apart , they take 1 hour . to be 726 m apart , they take ( 100 / 825 * 726 / 1000 ) hrs = ( 242 / 2750 * 60 ) min = 5.28 min . answer : b" | a ) 5.29 min , b ) 5.28 min , c ) 5.08 min , d ) 9.28 min , e ) 5.988 min | b | multiply(divide(divide(726, const_1000), add(4.5, 3.75)), const_60) | add(n1,n2)|divide(n0,const_1000)|divide(#1,#0)|multiply(#2,const_60)| | general |
two trains running in opposite directions cross a man standing on the platform in 35 seconds and 28 seconds respectively and they cross each other in 32 seconds . the ratio of their speeds is : | "let the speeds of the two trains be x m / sec and y m / sec respectively . then , length of the first train = 35 x meters , and length of the second train = 28 y meters . ( 35 x + 28 y ) / ( x + y ) = 32 = = > 35 x + 28 y = 32 x + 32 y = = > 3 x = 4 y = = > x / y = 4 / 3 answer : option b" | a ) 3 : 1 , b ) 4 : 3 , c ) 3 : 8 , d ) 3 : 25 , e ) 3 : 4 | b | divide(subtract(35, 32), subtract(32, 28)) | subtract(n0,n2)|subtract(n2,n1)|divide(#0,#1)| | physics |
40 % of a number is added to 120 , the result is the same number . find the number ? | ": ( 40 / 100 ) * x + 120 = x 2 x = 400 x = 200 answer : d" | a ) 300 , b ) 277 , c ) 266 , d ) 200 , e ) 121 | d | divide(120, divide(120, const_100)) | divide(n1,const_100)|divide(n1,#0)| | gain |
16.02 ã — 0.01 = ? | "16.02 ã — 0.01 = 0.1602 the answer is a ." | a ) 0.1602 , b ) 0.001602 , c ) 1.6021 , d ) 0.01602 , e ) none of these | a | multiply(divide(16.02, 0.01), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
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 . 2415 ? | 2415 * 1 / 3 = 805 option c | a ) s . 800 , b ) s . 810 , c ) s . 805 , d ) s . 900 , e ) s . 920 | c | multiply(2415, divide(1, 3)) | divide(n0,n1)|multiply(n6,#0) | general |
2 students appeared at an examination . 1 of them secured 9 marks more than the other one and hismarks was 56 % of the sum of their marks . find the marks obtained by them are ? | let their marks be ( x + 9 ) and x . then , x + 9 = 56 / 100 ( x + 9 + x ) = > 25 ( x + 9 ) = > 14 ( 2 x + 9 ) = > 3 x = 99 = > x = 33 . so , their marks are 42 and 33 c | a ) 39 and 31 , b ) 40 and 33 , c ) 42 and 33 , d ) 37 and 31 , e ) 35 and 32 | c | multiply(divide(multiply(9, const_100), subtract(56, subtract(const_100, 56))), 56) | multiply(n2,const_100)|subtract(const_100,n3)|subtract(n3,#1)|divide(#0,#2)|multiply(n3,#3) | general |
a certain company that sells only cars and trucks reported that revenues from car sales in 1997 were down 11 percent from 1996 and revenues from truck sales were up 7 percent from 1996 . if total revenues from car sales and truck sales in 1997 were up 1 percent from 1996 , what is the ratio of revenue from car sales in... | a . . i have probably solved this question 3 - 4 times by now . . remember the answer . . 1 : 2 | a ) 1 : 2 , b ) 4 : 5 , c ) 1 : 1 , d ) 3 : 2 , e ) 5 : 3 | a | divide(subtract(add(const_100, 7), add(const_100, 1)), subtract(add(const_100, 1), subtract(const_100, 11))) | add(n3,const_100)|add(n6,const_100)|subtract(const_100,n1)|subtract(#0,#1)|subtract(#1,#2)|divide(#3,#4)| | other |
what is the hcf of 2 / 3 , 4 / 9 and 6 / 5 | "explanation : hcf of fractions = hcf of numerators / lcm of denominators = ( hcf of 2 , 4 , 6 ) / ( lcm of 3 , 9 , 5 ) = 2 / 45 answer : option b" | a ) 7 / 45 , b ) 2 / 45 , c ) 4 / 15 , d ) 8 / 45 , e ) 9 / 45 | b | divide(2, 5) | divide(n0,n5)| | general |
3 pumps , working 8 hours a day , can empty a tank in 2 days . how many hours a day must 8 pumps work to empty the tank in 1 day ? | "3 pumps take 16 hrs total ( 8 hrs a day ) if 1 pump will be working then , it will need 16 * 3 = 48 hrs 1 pump need 48 hrs if i contribute 8 pumps then 48 / 8 = 6 hrs . answer : a" | a ) 6 , b ) 10 , c ) 11 , d ) 12 , e ) 13 | a | divide(multiply(multiply(3, 8), 2), 8) | multiply(n0,n1)|multiply(n2,#0)|divide(#1,n3)| | physics |
a , b and c can do a piece of work in 4 days , 6 days and 8 days respectively . how long will they taken , if all the three work together ? | "1 / 4 + 1 / 6 + 1 / 8 = 13 / 24 = > 24 / 13 = > 1 11 / 13 days answer : d" | a ) 2 , b ) 4 , c ) 5 , d ) 1 11 / 13 , e ) 9 | d | inverse(add(inverse(8), add(inverse(4), inverse(6)))) | inverse(n0)|inverse(n1)|inverse(n2)|add(#0,#1)|add(#3,#2)|inverse(#4)| | physics |
a student has to obtain 45 % of the total marks to pass . he got 150 marks and failed by 30 marks . the maximum marks are ? | "let the maximum marks be x then , 45 % of x = 150 + 30 45 x / 100 = 180 45 x = 180 * 100 = 18000 x = 400 answer is e" | a ) 300 , b ) 250 , c ) 280 , d ) 320 , e ) 400 | e | divide(add(150, 30), divide(45, const_100)) | add(n1,n2)|divide(n0,const_100)|divide(#0,#1)| | general |
the units digit of ( 35 ) ^ ( 87 ) + ( 93 ) ^ ( 46 ) is : | "step 1 : ( 35 ) ^ ( 87 ) can be broken up into ( 5 ) ^ 87 x ( 7 ) ^ 87 . each power of 7 ends in a units digit of either a 7 , 9,3 or 1 . each power of 5 ends in a 5 . when you multiply 5 by any odd number you will end up with a units digit of 5 . step 2 : ( 93 ) ^ ( 46 ) can be broken up into ( 3 ) ^ 46 x ( 31 ) ^ 46... | a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 0 | b | add(reminder(multiply(reminder(46, const_4), 93), const_10), reminder(35, const_10)) | reminder(n3,const_4)|reminder(n0,const_10)|multiply(n2,#0)|reminder(#2,const_10)|add(#3,#1)| | general |
a survey of employers found that during 1993 employment costs rose 3.5 percent , where employment costs consist of salary costs and fringe - benefit costs . if salary costs rose 3 percent and fringe - benefit costs rose 5.5 percent during 1993 , then fringe - benefit costs represented what percent of employment costs a... | "in my view easiest method would be , eliminating the percentage sign let 3.5 be 35 5.5 be 55 and 3 be 30 then first eq : s + f = e 130 s + 155 f = 135 e 5 s = 20 f therefore , s = 4 f and then e = 5 f f / e = f / 5 f = 20 % therefore , answer is b" | a ) 16.5 % , b ) 20 % , c ) 35 % , d ) 55 % , e ) 65 % | b | multiply(divide(subtract(3.5, 3), subtract(5.5, 3)), const_100) | subtract(n1,n2)|subtract(n3,n2)|divide(#0,#1)|multiply(#2,const_100)| | gain |
if f ( x ) = 2 x ^ 2 + y , and f ( 2 ) = 30 , what is the value of f ( 10 ) ? | f ( x ) = 2 x ^ 2 + y f ( 2 ) = 30 = > 2 * ( 2 ) ^ 2 + y = 30 = > 8 + y = 30 = > y = 22 f ( 10 ) = 2 * ( 10 ) ^ 2 + 22 = 222 answer e | a ) 104 , b ) 60 , c ) 52 , d ) 50 , e ) 222 | e | add(subtract(30, multiply(power(2, 2), 2)), multiply(2, power(10, 2))) | power(n4,n0)|power(n0,n0)|multiply(n0,#0)|multiply(n0,#1)|subtract(n3,#3)|add(#2,#4) | general |
if a certain coin is flipped , the probability that the coin will land heads is 1 / 2 . if the coin is flipped 5 times , what is the probability that it will land heads up on the first 2 flips but not on the last 3 flips ? | "p ( hhttt ) = 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 * 1 / 2 = 1 / 32 the answer is e ." | a ) 1 / 2 , b ) 1 / 4 , c ) 1 / 8 , d ) 1 / 16 , e ) 1 / 32 | e | power(divide(1, 2), 5) | divide(n0,n1)|power(#0,n2)| | probability |
by weight , liquid x makes up 0.5 percent of solution p and 1.5 percent of solution q . if 200 grams of solution p are mixed with 800 grams of solution q , then liquid x accounts for what percent of the weight of the resulting solution ? | the number of grams of liquid x is 0.5 ( 200 ) / 100 + 1.5 ( 800 ) / 100 = 1 + 12 = 13 grams . 13 / 1000 = 1.3 % the answer is a . | a ) 1.3 % , b ) 1.2 % , c ) 1.1 % , d ) 1.0 % , e ) 0.9 % | a | multiply(divide(add(const_1, divide(multiply(1.5, 800), const_100)), const_1000), const_100) | multiply(n1,n3)|divide(#0,const_100)|add(#1,const_1)|divide(#2,const_1000)|multiply(#3,const_100) | gain |
there 3 kinds of books in the library physics , chemistry and biology . ratio of physics to chemistry is 3 to 2 ; ratio of chemistry to biology is 4 to 3 , and the total of the books is more than 3000 . which one of following can be the total q of the book ? | first , you have to find the common ratio for all 3 books . you have : p : c : b 3 : 2 - - > multiply by 2 ( gives you row 3 ) 4 : 6 6 : 4 : 3 hence : p : c : b : t ( total ) = q 6 : 4 : 3 : 13 - - - - > this means , the total number must be a multiple of 13 . answer a is correct since 299 is divisible by 13 , hence is... | a ) 3003 , b ) 3027 , c ) 3024 , d ) 3021 , e ) 3018 | a | add(3000, 3) | add(n0,n5) | other |
the difference between the place values of 6 and 1 in the number 527631 is | sol . = ( place value of 6 ) – ( place value of 1 ) = ( 600 - 1 ) = 599 answer e | a ) 596 , b ) 587 , c ) 533 , d ) 568 , e ) 599 | e | subtract(multiply(const_10, 6), 6) | multiply(n0,const_10)|subtract(#0,n0)| | general |
when tossed , a certain coin has equal probability of landing on either side . if the coin is tossed 3 times , what is the probability that it will land twice on heads and once on tails ? | "must be once on twice and once on tails 1 / 2 * 1 / 2 * 1 / 2 = 1 / 8 answer : e" | a ) 1 / 16 , b ) 1 / 3 , c ) 1 / 6 , d ) 1 / 2 , e ) 1 / 8 | e | divide(const_1, power(const_2, 3)) | power(const_2,n0)|divide(const_1,#0)| | general |
the cost of a one - family home was $ 120,000 in 1980 . in 1988 , the price had increased to $ 186,000 . what was the percent increase in the cost of the home ? | "increase = 186000 - 120000 = 66000 % increase = 66000 * 100 / 120000 = 55 % answer : option c" | a ) 60 % , b ) 50 % , c ) 55 % , d ) 40 % , e ) 33.3 % | c | multiply(divide(subtract(multiply(multiply(const_12, multiply(const_4, const_4)), const_1000), multiply(multiply(const_12, const_1000), const_10)), multiply(multiply(const_12, const_1000), const_10)), const_100) | multiply(const_4,const_4)|multiply(const_1000,const_12)|multiply(#0,const_12)|multiply(#1,const_10)|multiply(#2,const_1000)|subtract(#4,#3)|divide(#5,#3)|multiply(#6,const_100)| | general |
[ ( 5 / 2 ÷ 1 / 2 × 5 / 2 ) ] / [ ( 5 / 2 × 1 / 2 ÷ 5 / 2 ) ] = ? | explanation : [ ( 5 / 2 ÷ 1 / 2 × 5 / 2 ) ] / [ ( 5 / 2 × 1 / 2 ÷ 5 / 2 ) ] = [ 5 / 2 × 2 / 1 × 5 / 2 ] / [ 5 / 2 × 1 / 2 × 2 / 5 ] = 25 / 2 / 1 / 2 = 25 / 2 x 2 / 1 = 25 answer : option a | a ) 25 , b ) 20 , c ) 18 , d ) 16 , e ) 14 | a | divide(multiply(divide(divide(5, 2), divide(1, 2)), divide(5, 2)), divide(multiply(divide(5, 2), divide(1, 2)), divide(5, 2))) | divide(n0,n1)|divide(n2,n1)|divide(#0,#1)|multiply(#0,#1)|divide(#3,#0)|multiply(#2,#0)|divide(#5,#4) | general |
the length of a room is 5.5 m and width is 3.75 m . what is the cost of paying the floor by slabs at the rate of $ 700 per sq . metre . | "c area = 5.5 × 3.75 sq . metre . cost for 1 sq . metre . = $ 700 hence , total cost = 5.5 × 3.75 × 700 = $ 14437.50" | a ) $ 12537.50 , b ) $ 13537.50 , c ) $ 14437.50 , d ) $ 16437.50 , e ) $ 18537.50 | c | multiply(700, multiply(5.5, 3.75)) | multiply(n0,n1)|multiply(n2,#0)| | physics |
the h . c . f . of two numbers is 12 and their l . c . m . is 600 . if one of the numbers is 480 , then the other is : | "other number = ( 12 x 600 ) / 480 = 15 . answer : d" | a ) 11 , b ) 10 , c ) 8 , d ) 15 , e ) 12 | d | multiply(12, 480) | multiply(n0,n2)| | physics |
find the value of ( 85 + 32 / 113 ) × 113 | "= ( 85 + 32 / 113 ) × 113 = ( 9605 + 32 ) / 113 × 113 = 9637 / 113 × 113 = 9637 answer is e ." | a ) 9263 , b ) 9244 , c ) 9636 , d ) 9623 , e ) 9637 | e | multiply(add(divide(32, 113), 85), 113) | divide(n1,n2)|add(n0,#0)|multiply(#1,n2)| | general |
you collect baseball cards . suppose you start out with 19 . maria takes half of one more than the number of baseball cards you have . since you ' re nice , you give peter 1 baseball card . since his father makes baseball cards , paul decides to triple your baseball cards . how many baseball cards do you have at the en... | "solution start with 19 baseball cards . maria takes half of one more than the number of baseball cards you have . so maria takes half of 19 + 1 which is 10 , so you ' re left with 19 - 10 = 9 . peter takes 1 baseball card from you : 9 - 1 = 8 baseball cards . paul triples the number of baseball cards you have : 8 ã — ... | a ) 17 , b ) 18 , c ) 19 , d ) 20 , e ) 24 | e | multiply(subtract(subtract(subtract(19, const_4), const_4), 1), const_3) | subtract(n0,const_4)|subtract(#0,const_4)|subtract(#1,n1)|multiply(#2,const_3)| | general |
in township k , 1 / 5 of the housing units are equiped with cable tv . if 1 / 12 of the housing units , including 1 / 3 of those that are equiped with cable tv , are equipped with videocassette recorders , what fraction of the housing units have neither cable tv nor videocassette recorders ? | "1 / 5 - - cable tv ( this includes some data from video cassette recorder ) 1 / 10 - - video cassette recorder including 1 / 3 ( equiped with cable tv ) i . e . 1 / 3 ( 1 / 5 ) = 1 / 15 therefore only video cassette recorder = 1 / 12 - 1 / 15 = 1 / 60 total = 1 / 5 + 1 / 60 + neither cable tv nor videocassette recorde... | a ) 23 / 30 , b ) 11 / 15 , c ) 47 / 60 , d ) 1 / 6 , e ) 2 / 15 | c | divide(add(12, 5), multiply(12, const_2)) | add(n1,n3)|multiply(n3,const_2)|divide(#0,#1)| | other |
the average salary of all the workers in a workshop is rs . 8000 . the average salary of 7 technicians is rs . 20000 and the average salary of the rest is rs . 6000 . the total number of workers in the workshop is : | "let the total number of workers be x . then , 8000 x = ( 20000 * 7 ) + 6000 ( x - 7 ) = 2000 x = 98000 = x = 49 . answer : b" | a ) 23 , b ) 49 , c ) 22 , d ) 20 , e ) 21 | b | add(7, divide(multiply(7, subtract(20000, 8000)), subtract(8000, 6000))) | subtract(n2,n0)|subtract(n0,n3)|multiply(n1,#0)|divide(#2,#1)|add(n1,#3)| | general |
the two lines y = x and x = - 4 intersect on the coordinate plane . if z represents the area of the figure formed by the intersecting lines and the x - axis , what is the side length t of a cube whose surface area is equal to 6 z ? | "800 score official solution : the first step to solving this problem is to actually graph the two lines . the lines intersect at the point ( - 4 , - 4 ) and form a right triangle whose base length and height are both equal to 4 . as you know , the area of a triangle is equal to one half the product of its base length ... | a ) t = 16 , b ) t = 8 √ 2 , c ) t = 8 , d ) t = 2 √ 2 , e ) ( √ 2 ) / 3 | d | sqrt(divide(multiply(4, 4), const_2)) | multiply(n0,n0)|divide(#0,const_2)|sqrt(#1)| | general |
find the value of x from the below equation ? : 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 ) - 2 , c ) 1 , d ) 2 , e ) 3 | a | divide(subtract(2, sqrt(subtract(power(2, 3), multiply(5, 5)))), 3) | multiply(n2,n2)|power(n1,n0)|subtract(#1,#0)|sqrt(#2)|subtract(n1,#3)|divide(#4,n0)| | general |
the h . c . f of two numbers is 23 and the other two factors of their l . c . m are 12 and 13 . the larger of the two numbers is : | "clearly , the numbers are ( 23 * 12 ) and ( 23 * 13 ) . larger number = ( 23 * 13 ) = 299 . answer : a" | a ) 299 , b ) 278 , c ) 322 , d ) 231 , e ) 121 | a | multiply(23, 13) | multiply(n0,n2)| | other |
two sides of a triangle have lengths x and t and meet at a right angle . if the perimeter of the triangle is 4 x , what is the ratio of x to t ? | ahhhh . . . . my bad . thank you , you ' re correct . 8 x = 6 t - - > x / t = 3 / 4 . actually b is the right answer . | ['a ) a ) 2 : 3', 'b ) b ) 3 : 4', 'c ) c ) 4 : 3', 'd ) d ) 3 : 2', 'e ) e ) 2 : 1'] | b | divide(const_3, 4) | divide(const_3,n0) | geometry |
in 70 liters of a mixture of milk and water , the quantity of water is 10 % . how much water should be added so that new mixture may 25 % water . | there are 10 % of water in 70 l - - > there are 7 l of waters ; after x liters of water is added amount of water becomes ( 7 + x ) liters , which on the other hand is 1 / 4 th ( 25 % ) of new ( 70 + x ) liters of mixture : 7 + x = 1 / 4 * ( 70 + x ) - - > 28 + 4 x = 70 + x - - > x = 14 answer : a | a ) 14 , b ) 15 , c ) 16 , d ) 17 , e ) 18 | a | divide(70, divide(subtract(const_100, 25), subtract(25, 10))) | subtract(const_100,n2)|subtract(n2,n1)|divide(#0,#1)|divide(n0,#2) | general |
an athlete runs 150 metres race in 24 seconds . what is his speed ? | "speed = distance / time = 150 / 20 = 7.5 m / s = 7.5 * 18 / 5 = 27 km / hr answer : a" | a ) 27 , b ) 38 , c ) 37 , d ) 30 , e ) 28 | a | multiply(divide(150, 24), const_3_6) | divide(n0,n1)|multiply(#0,const_3_6)| | physics |
a can contains a mixture of liquids a and b is the ratio 7 : 5 . when 15 litres of mixture are drawn off and the can is filled with b , the ratio of a and b becomes 7 : 9 . how many liter of liquid a was contained by the can initially ? | "ci * vi = cf * vf ( 7 / 12 ) * ( v 1 - 15 ) = ( 7 / 16 ) * v 1 ( v 1 - 15 ) / v 1 = 3 / 4 15 accounts for the difference of 1 on ratio scale so initial volume = v 1 = 4 * 15 = 60 litres . 7 / 12 of the initial mixture was liquid a so liquid a was ( 7 / 12 ) * 60 = 35 litres . answer : b" | a ) 30 , b ) 35 , c ) 40 , d ) 45 , e ) 50 | b | multiply(7, divide(multiply(add(7, 15), subtract(15, multiply(divide(5, add(7, 5)), 15))), subtract(multiply(15, 7), multiply(7, 5)))) | add(n0,n2)|add(n0,n1)|multiply(n0,n2)|multiply(n0,n1)|divide(n1,#1)|subtract(#2,#3)|multiply(n2,#4)|subtract(n2,#6)|multiply(#0,#7)|divide(#8,#5)|multiply(n0,#9)| | other |
find the smallest number that must be added to 1780 to make it a perfect square . | "4 1780 ( 42 16 82 180 164 16 number to be added = ( 43 ) 2 - 1780 = 1849 - 1780 = 69 . answer c" | a ) 17 , b ) 42 , c ) 69 , d ) 47 , e ) 52 | c | multiply(divide(divide(divide(divide(1780, const_1000), const_3), const_3), const_3), divide(divide(divide(divide(1780, const_1000), const_3), const_3), const_3)) | divide(n0,const_1000)|divide(#0,const_3)|divide(#1,const_3)|divide(#2,const_3)|multiply(#3,#3)| | geometry |
two trains , one from howrah to patna and the other from patna to howrah , start simultaneously . after they meet , the trains reach their destinations after 9 hours and 25 hours respectively . the ratio of their speeds is ? | "let us name the trains a and b . then , ( a ' s speed ) : ( b ' s speed ) = √ b : √ a = √ 25 : √ 9 = 5 : 3 answer : b" | a ) 4 : 5 , b ) 5 : 3 , c ) 4 : 4 , d ) 4 : 8 , e ) 4 : 1 | b | divide(sqrt(25), sqrt(9)) | sqrt(n1)|sqrt(n0)|divide(#0,#1)| | physics |
a thief goes away with a santro car at a speed of 40 kmph . the theft has been discovered after half an hour and the owner sets off in a bike at 55 kmph when will the owner over take the thief from the start ? | "explanation : | - - - - - - - - - - - 20 - - - - - - - - - - - - - - - - - - - - | 55 40 d = 20 rs = 55 – 40 = 15 t = 20 / 15 = 1.33 hours answer : option a" | a ) a ) 1.33 , b ) b ) 5 , c ) c ) 7 , d ) d ) 5 , e ) e ) 8 | a | subtract(divide(multiply(divide(const_1, const_2), 40), subtract(55, 40)), divide(const_1, const_2)) | divide(const_1,const_2)|subtract(n1,n0)|multiply(n0,#0)|divide(#2,#1)|subtract(#3,#0)| | physics |
a newspaper manufacturer produces 800 units of a certain model each month at a cost to the manufacturer of £ 50 per unit and all of the produced units are sold each month . what is the minimum selling price per unit that will ensure that the monthly profit ( revenue from sales minus production costs ) on the sales of t... | 800 ( x - 50 ) ≥ 80000 x - 50 ≥ 800 x ≥ 750 answer : option d | a ) 115 , b ) 520 , c ) 550 , d ) 750 , e ) 796 | d | add(50, divide(80000, 800)) | divide(n2,n0)|add(n1,#0) | general |
p and q started a business investing rs . 85,000 and rs . 20,000 respectively . in what ratio the profit earned after 2 years be divided between p and q respectively ? | "p : q = 85000 : 20000 = 17 : 4 . answer : c" | a ) 17 : 6 , b ) 17 : 0 , c ) 17 : 4 , d ) 17 : 2 , e ) 17 : 3 | c | divide(add(multiply(add(add(2, const_3), const_3), multiply(add(2, const_3), 2)), add(2, const_3)), add(multiply(const_3, multiply(add(2, const_3), 2)), add(2, const_3))) | add(n2,const_3)|add(#0,const_3)|multiply(n2,#0)|multiply(#1,#2)|multiply(#2,const_3)|add(#0,#3)|add(#0,#4)|divide(#5,#6)| | gain |
x does a work in 10 days . y does the same work in 15 days . in how many days they together will do the same work ? | "x ' s 1 day ' s work = 1 / 10 y ' s 1 day ' s work = 1 / 15 ( x + y ) ' s 1 day ' s work = ( 1 / 10 + 1 / 15 ) = 1 / 6 both together will finish the work in 6 days . correct option is d" | a ) 10 , b ) 15 , c ) 3 , d ) 6 , e ) 12 | d | inverse(add(divide(const_1, 10), divide(const_1, 15))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|inverse(#2)| | physics |
a rectangular lawn of length 200 m by 120 m has two roads running along its center , one along the length and the other along the width . if the width of the roads is 5 m what is the area y covered by the two roads ? | "area covered by road along the length = 5 * 200 = 1000 square meter area covered by road along the width = 5 * 120 = 600 square meter common area in both roads ( where the roads intersect ) = square with side 5 meter = 5 * 5 = 25 total area of the roads y = 1000 + 600 - 25 = 1575 answer : option c" | a ) 400 , b ) 1550 , c ) 1575 , d ) 1600 , e ) 1625 | c | add(rectangle_area(200, 5), rectangle_area(120, 5)) | rectangle_area(n0,n2)|rectangle_area(n1,n2)|add(#0,#1)| | geometry |
which expression has the greatest value ? | "the difference between each numerator / denominator pair is three . the largest expression ( which in this case will also be the closest to one ) is the expression where ' 3 ' is the smallest fraction of the numerator ( or the denominator ) . the largest numerator is 1982487 , or c , and this will give an expression w... | a ) 1977452 / 1977455 , b ) 1982466 / 1982469 , c ) 1982487 / 1982490 , d ) 1982456 / 1982459 , e ) 1982465 / 1982468 | c | divide(subtract(const_3600, multiply(const_3, const_3)), subtract(const_3600, const_4)) | multiply(const_3,const_3)|subtract(const_3600,const_4)|subtract(const_3600,#0)|divide(#2,#1)| | general |
a train 108 m long moving at a speed of 50 kmph crosses a train 112 m long coming from opp direction in 6 seconds . the speed of the second train is | both trains are moving in opposite direction means we need to add their speeds . i . e : 1 st train speed is = 50 kmph let the 2 nd train speed = x kmph relative speed = ( 50 + x ) kmph = ( 50 + x ) * 5 / 18 = ( 250 + 5 x ) / 18 m / sec distance covered by two trains = 108 + 112 = 220 m therefore , 220 / ( ( 250 + 5 x ... | a ) 48 kmph , b ) 54 kmph , c ) 66 kmph , d ) 82 kmph , e ) 69 kmph | d | subtract(multiply(const_3_6, divide(add(108, 112), 6)), 50) | add(n0,n2)|divide(#0,n3)|multiply(#1,const_3_6)|subtract(#2,n1) | physics |
what is the greatest prime factor of 4 ^ 17 - 2 ^ 29 ? | i ' m happy to help with this . we know 4 = 2 ^ 2 , so 4 ^ 17 = ( 2 ^ 2 ) ^ 17 = 2 ^ ( 2 * 17 ) = 2 ^ 34 that takes advantage of a law of exponents that says ( a ^ n ) ^ m = a ^ ( n * m ) so , 4 ^ 17 - 2 ^ 29 = 2 ^ 34 - 2 ^ 29 = 2 ^ ( 29 + 5 ) - 2 ^ 29 = ( 2 ^ 29 ) * ( 2 * 6 ) - 2 ^ 29 = ( 2 ^ 6 - 1 ) * ( 2 ^ 29 ) = ( ... | a ) 2 , b ) 3 , c ) 5 , d ) 7 , e ) 31 | e | add(2, 29) | add(n2,n3) | general |
jo ' s collection contains us , indian and british stamps . if the ratio of us to indian stamps is 5 to 2 and the ratio of indian to british stamps is 4 to 1 , what is the ratio of us to british stamps ? | "u / i = 5 / 2 i / b = 4 / 1 since i is multiple of both 2 ( as per first ratio ) and 5 ( as per second ratio ) so let ' s assume that i = 10 i . e . multiplying teh first ratio by 5 and second ration by 2 in each numerator and denominator then , u : i : b = 20 : 8 : 2 i . e . u : b = 20 : 2 answer : option d" | a ) 5 : 1 , b ) 10 : 5 , c ) 15 : 2 , d ) 20 : 2 , e ) 25 : 2 | d | divide(multiply(5, 4), multiply(1, 2)) | multiply(n0,n2)|multiply(n1,n3)|divide(#0,#1)| | other |
by investing in 16 % stock at 64 , one earns rs . 1500 . the investment made is ? | "to earn rs . , investment = rs . 64 . to earn rs . 1500 , investment = = rs . 5760 . answer : c" | a ) 2288 , b ) 2877 , c ) 5760 , d ) 7636 , e ) 5728 | c | multiply(divide(multiply(multiply(multiply(const_4, const_2), multiply(const_4, const_2)), const_3), divide(const_100, const_2)), subtract(1500, multiply(const_4, const_100))) | divide(const_100,const_2)|multiply(const_2,const_4)|multiply(const_100,const_4)|multiply(#1,#1)|subtract(n2,#2)|multiply(#3,const_3)|divide(#5,#0)|multiply(#6,#4)| | gain |
a student , after his teacher ' s question , chose a number , multiplied it by 2 , then subtracted 138 from the result and got 102 . what was the number he chose ? | let x be the number he chose , then 2 ⋅ x − 138 = 102 2 x = 240 x = 120 correct answer is e ) 120 | a ) 344 , b ) 251 , c ) 411 , d ) 212 , e ) 120 | e | divide(add(138, 102), 2) | add(n1,n2)|divide(#0,n0) | general |
what is the sum of the greatest common factor and the lowest common multiple of 36 and 16 ? | prime factorization of 36 = 2 x 2 x 3 x 3 prime factorization of 16 = 2 x 2 x 2 x 2 gcf = 4 lcm = 144 sum = 148 . answer a . | a ) 148 , b ) 210 , c ) 314 , d ) 320 , e ) 344 | a | divide(multiply(36, 16), const_4) | multiply(n0,n1)|divide(#0,const_4) | general |
lisa and robert have taken the same number of photos on their school trip . lisa has taken 3 times as many photos as claire and robert has taken 28 more photos than claire . how many photos has claire taken ? | "l = r l = 3 c r = c + 28 3 c = c + 28 c = 14 the answer is e ." | a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 14 | e | divide(28, subtract(3, const_1)) | subtract(n0,const_1)|divide(n1,#0)| | general |
a fruit seller sells mangoes at the rate of rs . 5 per kg and thereby loses 20 % . at what price per kg , he should have sold them to make a profit of 5 % | "explanation : 80 : 5 = 105 : x x = ( 5 × 105 / 80 ) = rs 6.56 option a" | a ) rs 6.56 , b ) rs 9.56 , c ) rs 10.56 , d ) rs 11.56 , e ) none of these | a | divide(multiply(5, add(const_100, 5)), subtract(const_100, 20)) | add(n2,const_100)|subtract(const_100,n1)|multiply(n0,#0)|divide(#2,#1)| | gain |
a train covers a distance of 15 km in 10 min . if it takes 6 sec to pass a telegraph post , then the length of the train is ? | "speed = ( 15 / 10 * 60 ) km / hr = ( 90 * 5 / 18 ) m / sec = 25 m / sec . length of the train = 25 * 6 = 150 m . answer : c" | a ) m , b ) m , c ) m , d ) m , e ) m | c | divide(15, subtract(divide(15, 10), 6)) | divide(n0,n1)|subtract(#0,n2)|divide(n0,#1)| | physics |
when a certain number x is divided by 66 , the remainder is 14 . what is the remainder when x is divided by 11 ? | "let possible value of x is 80 least possible value of x / 11 is 80 / 11 = > 7 quotient with remainder 3 thus answer is ( b ) 3" | a ) 2 , b ) 3 , c ) 7 , d ) 9 , e ) 10 | b | reminder(14, 11) | reminder(n1,n2)| | general |
a special municipal payroll tax charges not tax on a payroll less than $ 200,000 and only 0.1 % on a company ’ s payroll above $ 200,000 . if belfried industries paid $ 400 in this special municipal payroll tax , then they must have had a payroll of ? | "answer : e , ( with different approach ) : the 400 paid is 0.1 % of the additional amount above 200,000 . let it be x now 0.1 % of x = 400 therefore x = 400,000 total = 200,000 + x = 600,000" | a ) $ 180,000 , b ) $ 202,000 , c ) $ 220,000 , d ) $ 400,000 , e ) $ 600,000 | e | multiply(multiply(400, const_100), const_10) | multiply(n3,const_100)|multiply(#0,const_10)| | general |
there are 7 red shoes & 3 green shoes . if two of red shoes are drawn what is the probability of getting red shoes | "taking 2 red shoe the probablity is 7 c 2 from 10 shoes probablity of taking 2 red shoe is 7 c 2 / 10 c 2 = 7 / 15 answer : d" | a ) 1 / 13 , b ) 1 / 14 , c ) 1 / 15 , d ) 7 / 15 , e ) 7 / 16 | d | divide(choose(7, const_2), choose(add(7, 3), const_2)) | add(n0,n1)|choose(n0,const_2)|choose(#0,const_2)|divide(#1,#2)| | probability |
a can do a piece of work in 8 days . b can do it in 16 days . with the assistance of c they completed the work in 4 days . find in how many days can c alone do it ? | "c = 1 / 4 - 1 / 8 - 1 / 16 = 1 / 16 = > 16 days answer : e" | a ) 87 days , b ) 20 days , c ) 26 days , d ) 19 days , e ) 16 days | e | divide(multiply(8, 16), divide(subtract(multiply(8, 16), multiply(add(divide(multiply(8, 16), 8), divide(multiply(8, 16), 16)), 4)), 4)) | multiply(n0,n1)|divide(#0,n0)|divide(#0,n1)|add(#1,#2)|multiply(n2,#3)|subtract(#0,#4)|divide(#5,n2)|divide(#0,#6)| | physics |
the radius of a cylinder is 10 m , height 15 m . the volume of the cylinder is : | "cylinder volume = ï € r ( power 2 ) h = 22 / 7 ã — 10 ã — 10 ã — 15 = 4714.3 m ( power 3 ) answer is e ." | a ) 2200 , b ) 5500 , c ) 3300 , d ) 1100 , e ) 4714.3 | e | multiply(circumface(10), 15) | circumface(n0)|multiply(n1,#0)| | geometry |
mr . kramer , the losing candidate in a two - candidate election , received 942,568 votes , which was exactly 47 percent of all votes cast . approximately what percent of the remaining votes would he need to have received in order to have won at least 50 percent of all the votes cast ? | "let me try a simpler one . lets assume that candidate got 47 % votes and total votes is 100 . candidate won = 47 remaining = 53 to get 50 % , candidate requires 3 votes from 100 which is 3 % and 3 votes from 53 . 3 / 53 = . 056 = 5.6 % which is approx 6 % . hence the answer is b ." | a ) 10 % , b ) 6 % , c ) 15 % , d ) 17 % , e ) 20 % | b | multiply(divide(subtract(divide(50, const_100), divide(47, const_100)), subtract(const_1, divide(47, const_100))), const_100) | divide(n2,const_100)|divide(n1,const_100)|subtract(#0,#1)|subtract(const_1,#1)|divide(#2,#3)|multiply(#4,const_100)| | general |
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 |
what is the cost of leveling the field in the form of parallelogram at the rate of rs . 30 / 10 sq . metre , whose base & perpendicular distance from the other side being 64 m & 74 m respectively ? | "area of the parallelogram = length of the base * perpendicular height = 64 * 74 = 4736 m . total cost of levelling = rs . 4736 c" | a ) s . 6000 , b ) s . 6300 , c ) s . 4736 , d ) s . 6480 , e ) s . 7000 | c | multiply(multiply(64, 74), divide(30, 10)) | divide(n0,n1)|multiply(n2,n3)|multiply(#0,#1)| | physics |
sides of a rectangular park are in the ratio 3 : 2 and its area is 3750 sq m , the cost of fencing it at 70 ps per meter is ? | "3 x * 2 x = 3750 = > x = 25 2 ( 75 + 50 ) = 250 m 250 * 0.70 = rs . 175 answer : c" | a ) 287 , b ) 369 , c ) 175 , d ) 279 , e ) 361 | c | divide(multiply(70, rectangle_perimeter(sqrt(divide(multiply(3750, 2), 3)), divide(3750, sqrt(divide(multiply(3750, 2), 3))))), const_100) | multiply(n1,n2)|divide(#0,n0)|sqrt(#1)|divide(n2,#2)|rectangle_perimeter(#3,#2)|multiply(n3,#4)|divide(#5,const_100)| | physics |
when 1 / 20 % of 4,000 is subtracted from 1 / 20 of 4,000 , the difference is | "1 / 20 % of 4000 = 2 1 / 20 of 4000 = 200 200 - 2 = 198 ans : b" | a ) 50 , b ) 198 , c ) 380 , d ) 398 , e ) 400 | b | multiply(inverse(20), multiply(multiply(const_100, 20), add(const_4, const_4))) | add(const_4,const_4)|inverse(n4)|multiply(n4,const_100)|multiply(#0,#2)|multiply(#1,#3)| | general |
what percent is 2 minutes 24 seconds of an hour ? | "2 mt 24 s = 144 s ( 144 * 100 ) / ( 60 * 60 ) = 4 % answer : c" | a ) 6 % , b ) 8 % , c ) 4 % , d ) 2 % , e ) 9 % | c | multiply(divide(2, 24), const_100) | divide(n0,n1)|multiply(#0,const_100)| | physics |
what is the remainder when 43 * 49 is divided by 8 ? | "we can make use of the rule : remainder of { ( a * b ) / n } } = remainder of ( a / n ) * remainder of ( b / n ) here remainder of { 43 * 49 ) / 8 } } = remainder of ( 43 / 8 ) * remainder of ( 49 / 8 ) = 3 * 1 = 3 answer : b" | a ) 1 , b ) 3 , c ) 4 , d ) 5 , e ) 7 | b | subtract(divide(8, const_2), multiply(43, 43)) | divide(n2,const_2)|multiply(n0,n0)|subtract(#0,#1)| | general |
the difference between a number and its two - fifth is 510 . what is 60 % of that number ? | "let the number be x . then , x - 2 / 5 x = 510 x = ( 510 * 5 ) / 3 = 850 60 % of 850 = 510 . answer : b" | a ) 512 , b ) 510 , c ) 514 , d ) 516 , e ) 518 | b | divide(multiply(510, add(const_4, const_1)), add(const_1, const_2)) | add(const_1,const_4)|add(const_1,const_2)|multiply(n0,#0)|divide(#2,#1)| | general |
a 40 - liter solution of alcohol and water is 5 percent alcohol . if 5.5 liters of alcohol and 4.5 liters of water are added to this solution , what percent of the solution produced is alcohol ? | "the percent of alcohol in the solution is ( 0.05 ( 40 ) + 5.5 ) / 50 = 7.5 / 50 = 15 % the answer is d ." | a ) 9 % , b ) 11 % , c ) 13 % , d ) 15 % , e ) 17 % | d | multiply(divide(add(multiply(40, divide(5, const_100)), 5.5), add(40, add(5.5, 4.5))), const_100) | add(n2,n3)|divide(n1,const_100)|add(n0,#0)|multiply(n0,#1)|add(n2,#3)|divide(#4,#2)|multiply(#5,const_100)| | general |
in the first round of the elections , the only two candidates got exactly the same number of votes . during the second round , 16000 votes switched from the first candidate to the second one . the total number of votes remained the same in both rounds , and no other votes switched sides . if , in the second round , the... | let a be the first candidate and b be the second one . given x + 16000 = 5 ( x - 16000 ) = > x = 24000 2 x = 48000 imo option d . | a ) 15,000 , b ) 25,000 , c ) 40,000 , d ) 48,000 , e ) 60,000 | d | multiply(divide(add(multiply(5, 16000), 16000), const_4), const_2) | multiply(n0,n1)|add(n0,#0)|divide(#1,const_4)|multiply(#2,const_2) | general |
in an examination , a student scores 4 marks for every correct answer and loses 1 mark for every wrong answer . if he attempts all 80 questions and secures 120 marks , how many questions does he answer correctly ? | "let the number of correct answers be x then , number of wrong answers = ( 80 - x ) 4 x – ( 80 - x ) = 120 = > 4 x – 80 + x = 120 = > 5 x = 200 = > x = 200 / 5 = 40 i . e . , he does 40 questions correctly answer is e ." | a ) 20 , b ) 10 , c ) 30 , d ) 50 , e ) 40 | e | divide(add(120, 80), add(4, 1)) | add(n2,n3)|add(n0,n1)|divide(#0,#1)| | physics |
after striking the floor , a rubber ball rebounds to 4 / 5 th of the height from which it has fallen . find the total distance that it travels before coming to rest if it has been gently dropped from a height of 120 metres . | solution : the first drop is 120 metres . after this the ball will rise by 96 metres and fall by 96 metres . this process will continue in the form of infinite gp with common ratio 0.8 and first term 96 . the required answer is given by the formula : a / ( 1 - r ) now , [ { 120 / ( 1 / 5 ) } + { 96 / ( 1 / 5 ) } ] = 10... | a ) 540 m , b ) 960 m , c ) 1080 m , d ) 1020 m , e ) 1120 m | c | multiply(120, add(4, 5)) | add(n0,n1)|multiply(n2,#0) | physics |
a sum was put a simple interest at a certain rate for 5 years . had it been put at 2 % higher rate , it would have fetched rs . 250 more . the sum is : | "explanation : let the sub be rs . x and the initial rate be r % . then x ã — ( r + 2 ) ã — 5 / 100 â ˆ ’ x ã — r ã — 5 / 100 = 250 â ‡ ’ x ã — 2 ã — 5 / 100 = 250 â ‡ ’ x ã — 5 / 100 = 125 â ‡ ’ 5 x = 12500 â ‡ ’ x = 2500 answer : option b" | a ) s . 2000 , b ) s . 2500 , c ) s . 3000 , d ) s . 3500 , e ) s . 4000 | b | multiply(divide(divide(250, 5), 2), const_100) | divide(n2,n0)|divide(#0,n1)|multiply(#1,const_100)| | gain |
what is the largest 4 digit number exactly divisible by 35 ? | "largest 4 digit number = 9999 9999 ÷ 35 = 285 , remainder = 24 hence largest 4 digit number exactly divisible by 88 = 9999 - 24 = 9975 answer : c" | a ) 4676 , b ) 4678 , c ) 9975 , d ) 9504 , e ) 9936 | c | multiply(floor(divide(power(const_10, 4), 35)), 35) | power(const_10,n0)|divide(#0,n1)|floor(#1)|multiply(n1,#2)| | general |
a boat can travel with a speed of 16 km / hr in still water . if the rate of stream is 5 km / hr , then find the time taken by the boat to cover distance of 105 km downstream . | "explanation : it is very important to check , if the boat speed given is in still water or with water or against water . because if we neglect it we will not reach on right answer . i just mentioned here because mostly mistakes in this chapter are of this kind only . lets see the question now . speed downstream = ( 16... | a ) 4 hours , b ) 5 hours , c ) 6 hours , d ) 7 hours , e ) 8 hours | b | divide(105, add(16, 5)) | add(n0,n1)|divide(n2,#0)| | physics |
the sum of the average ( arithmetic mean ) of the even integers from 0 to 100 inclusive and the average ( arithmetic mean ) of the even integers from 0 to 50 inclusive ? | so , according to a mean of a set of even numbers from 0 to 100 = ( 0 + 100 ) / 2 = 50 and mean of a set of even numbers from 0 to 50 = ( 0 + 50 ) / 2 = 25 sum = 50 + 25 = 75 answer : d . | a ) 15 , b ) 125 , c ) 25 , d ) 75 , e ) 50 | d | add(divide(100, const_2), divide(50, const_2)) | divide(n1,const_2)|divide(n3,const_2)|add(#0,#1) | general |
mr . loyd wants to fence his square shaped land of 150 sqft each side . if a pole is laid every 30 ft how many poles do he need ? | if each side is 120 feet . . then total perimeter is 150 * 4 = 600 poles every 20 feet hence no of poles = 600 / 20 = 30 answer : b | ['a ) 20', 'b ) 30', 'c ) 40', 'd ) 50', 'e ) 60'] | b | divide(divide(multiply(150, const_4), const_10), const_2) | multiply(n0,const_4)|divide(#0,const_10)|divide(#1,const_2) | geometry |
when greenville state university decided to move its fine arts collection to a new library , it had to package the collection in 20 - inch by 20 - inch by 12 - inch boxes . if the university pays $ 0.50 for every box , and if the university needs 1.92 million cubic inches to package the collection , what is the minimum... | "the volume of each box is 20 * 20 * 12 = 4800 cubic inches . number of boxes = 1 , 920,000 / 4800 = 400 boxes total cost = 400 × $ 0.5 = $ 200 the answer is b ." | a ) $ 150 , b ) $ 200 , c ) $ 320 , d ) $ 650 , e ) $ 810 | b | multiply(divide(multiply(1.92, multiply(const_1000, const_1000)), multiply(multiply(20, 20), 12)), 0.50) | multiply(const_1000,const_1000)|multiply(n0,n0)|multiply(n4,#0)|multiply(n2,#1)|divide(#2,#3)|multiply(n3,#4)| | general |
find the simple interest on rs . 68,800 at 16 2 / 3 % per annum for 9 months . | "p = rs . 68800 , r = 50 / 3 % p . a and t = 9 / 12 years = 3 / 4 years . s . i . = ( p * r * t ) / 100 = rs . ( 68,800 * ( 50 / 3 ) * ( 3 / 4 ) * ( 1 / 100 ) ) = rs . 8600 answer is a ." | a ) s . 8600 , b ) s . 8000 , c ) s . 7500 , d ) s . 7000 , e ) s . 6500 | a | multiply(multiply(multiply(add(multiply(multiply(multiply(2, 3), const_100), const_100), multiply(multiply(multiply(3, 3), const_100), multiply(add(3, 2), 2))), divide(add(multiply(16, 3), 2), 3)), divide(multiply(3, 3), multiply(2, multiply(2, 3)))), divide(const_1, const_100)) | add(n2,n3)|divide(const_1,const_100)|multiply(n3,n3)|multiply(n2,n3)|multiply(n1,n3)|add(n2,#4)|multiply(n2,#3)|multiply(#3,const_100)|multiply(#2,const_100)|multiply(#0,n2)|divide(#2,#6)|divide(#5,n3)|multiply(#7,const_100)|multiply(#8,#9)|add(#12,#13)|multiply(#14,#11)|multiply(#10,#15)|multiply(#1,#16)| | gain |
the area of square abcd above is 22 . the shaded region is the intersection of the square and a circular region centered at c . what is the area of the shaded region ? | = square - quarter circle = 22 - pi * 22 / 4 = 22 - pi * 11 / 2 answer : b | ['a ) 18 − 9 / 4 ∗ π', 'b ) 22 − 11 / 2 ∗ π', 'c ) 9 + 9 / 4 ∗ π', 'd ) 9 + 9 / 2 ∗ π', 'e ) 9 / 2 − 9 / 4 ∗ π'] | b | subtract(22, multiply(divide(22, const_4), const_pi)) | divide(n0,const_4)|multiply(#0,const_pi)|subtract(n0,#1) | geometry |
in the list 1 , 2 , 3 , 4 , 5 , 5 , 5 , 5 , 7 , 11 , 21 , what fraction of the data is less than the mode ? | "mode : the mode of any set is the term which has the highest frequency ( occurrence ) highest frequent term in the set is 5 ( with frequency 4 ) hence mode = 5 four terms ( 1 , 2 , 3 , 4 ) out of a total of 11 terms are less than mode of the set . fraction of set that are less than mode of set = 4 / 11 answer : option... | a ) 2 / 9 , b ) 1 / 3 , c ) 2 / 5 , d ) 4 / 11 , e ) 7 / 9 | d | divide(const_4.0, 11) | divide(n3,n9)| | general |
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 |
express a speed of 72 kmph in meters per second ? | "72 * 5 / 18 = 20 mps answer : c" | a ) 10 mps , b ) 76 mps , c ) 20 mps , d ) 97 mps , e ) 16 mps | c | multiply(const_0_2778, 72) | multiply(n0,const_0_2778)| | physics |
there are 10 teams in a soccer league and each team plays each of the other teams once . how many games will be played in total ? | "10 c 2 = 45 the answer is e ." | a ) 33 , b ) 36 , c ) 39 , d ) 42 , e ) 45 | e | divide(factorial(10), multiply(factorial(subtract(10, const_2)), factorial(const_2))) | factorial(n0)|factorial(const_2)|subtract(n0,const_2)|factorial(#2)|multiply(#3,#1)|divide(#0,#4)| | general |
if two of the 4 expressions x + y , x + 5 y , x - y , 5 x + y are chosen at random , what is the probability that their product will be of the form of x ^ 2 - ( by ) ^ 2 , where b is an integer ? | there are total 4 c 2 = 64 c 2 = 6 ways to choose 2 equations out of 4 at random . we want the product of 2 equations in format x 2 − ( by ) 2 x 2 − ( by ) 2 , where b is a constant coefficient of term y 2 y 2 . also note equation does not contain any xyxy term . there are only 2 equations , ( x + y ) & ( x - y ) which... | a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 5 , e ) 1 / 6 | e | divide(subtract(5, 4), add(4, 2)) | add(n0,n3)|subtract(n1,n0)|divide(#1,#0) | general |
bert left the house with n dollars . he spent 1 / 4 of this at the hardware store , then $ 7 at the dry cleaners , and then half of what was left at the grocery store . when he got home , he had $ 19 left in his pocket . what was the value of n ? | "started to test answer d if he had 60 , then he spent 15 at hardware store now he was left with 45 $ he spent 7 dollars on cleaning , thus he remained with 38 $ he then spent 1 / 2 of 38 , or 19 , and was left with 19 . hence , the only option that can be right is a ." | a ) $ 36 , b ) $ 44 , c ) $ 52 , d ) $ 60 , e ) $ 68 | a | multiply(multiply(inverse(subtract(1, divide(1, 4))), add(divide(7, const_2), 19)), const_2) | divide(n2,const_2)|divide(n0,n1)|add(n3,#0)|subtract(n0,#1)|inverse(#3)|multiply(#2,#4)|multiply(#5,const_2)| | general |
a man swims downstream 48 km and upstream 32 km taking 8 hours each time ; what is the speed of the current ? | "48 - - - 8 ds = 6 ? - - - - 1 32 - - - - 8 us = 4 ? - - - - 1 s = ? s = ( 6 - 4 ) / 2 = 1 answer : a" | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | a | divide(add(divide(32, 8), divide(48, 8)), const_2) | divide(n1,n2)|divide(n0,n2)|add(#0,#1)|divide(#2,const_2)| | physics |
how many seconds will a 100 m long train take to cross a man walking with a speed of 3 km / hr in the direction of the moving train if the speed of the train is 63 km / hr ? | "speed of train relative to man = 63 - 3 = 60 km / hr . = 60 * 5 / 18 = 50 / 3 m / sec . time taken to pass the man = 100 * 3 / 50 = 6 sec . answer : b" | a ) 11 sec , b ) 6 sec , c ) 77 sec , d ) 14 sec , e ) 12 | b | divide(100, multiply(subtract(63, 3), const_0_2778)) | subtract(n2,n1)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
in a colony of 70 residents , the ratio of the number of men and women is 4 : 3 . among the women , the ratio of the educated to the uneducated is 1 : 4 . if the ratio of the number of educated to uneducated persons is 8 : 27 , then find the ratio of the number of educated to uneducated men in the colony ? | number of men in the colony = 4 / 7 * 70 = 40 . number of women in the colony = 3 / 7 * 70 = 40 . number educated women in the colony = 1 / 5 * 30 = 6 . number of uneducated women in the colony = 4 / 5 * 50 = 24 . number of educated persons in the colony = 8 / 35 * 70 = 16 . as 6 females are educated , remaining 10 edu... | a ) 1 : 7 , b ) 2 : 3 , c ) 1 : 8 , d ) 2 : 3 , e ) 1 : 3 | e | divide(subtract(multiply(divide(70, add(8, 27)), 8), divide(multiply(divide(70, add(4, 3)), 3), add(1, 4))), subtract(multiply(divide(70, add(8, 27)), 27), multiply(divide(multiply(divide(70, add(4, 3)), 3), add(1, 4)), 4))) | add(n5,n6)|add(n1,n2)|add(n1,n3)|divide(n0,#0)|divide(n0,#1)|multiply(n5,#3)|multiply(n2,#4)|multiply(n6,#3)|divide(#6,#2)|multiply(n1,#8)|subtract(#5,#8)|subtract(#7,#9)|divide(#10,#11) | other |
a cube of edge 5 cm is cut into cubes of each edge 1 cm . the ratio of the total surface area of one of the small cubes to that of the large cube is equal to | solution required ratio = ( 6 x 1 x 1 / 6 x 5 x 5 ) ‹ = › 1 / 25 ‹ = › 1 : 125 . answer b | ['a ) 1 : 5', 'b ) 1 : 25', 'c ) 1 : 125', 'd ) 1 : 625', 'e ) none'] | b | divide(surface_cube(1), surface_cube(5)) | surface_cube(n1)|surface_cube(n0)|divide(#0,#1) | geometry |
find large number from below question the difference of two numbers is 1365 . on dividing the larger number by the smaller , we get 7 as quotient and the 15 as remainder | "let the smaller number be x . then larger number = ( x + 1365 ) . x + 1365 = 7 x + 15 6 x = 1350 x = 225 large number = 225 + 1365 = 1590 c" | a ) 1235 , b ) 1346 , c ) 1590 , d ) 1635 , e ) 1489 | c | multiply(divide(subtract(1365, 15), subtract(7, const_1)), 7) | subtract(n0,n2)|subtract(n1,const_1)|divide(#0,#1)|multiply(n1,#2)| | general |
alex age after 13 years will be 10 times his age 10 years back . what is the present age of alex ? | let alex ' s present age be x alex ' s age after 17 years = x + 17 alex ' s age 10 years back = x - 10 x + 17 = 10 ( x - 10 ) x = 13 rajeev ' s present age = 13 years answer is b | a ) 15 , b ) 13 , c ) 20 , d ) 21 , e ) 26 | b | divide(add(multiply(const_10, const_10), 13), subtract(const_12, const_3)) | multiply(const_10,const_10)|subtract(const_12,const_3)|add(n0,#0)|divide(#2,#1) | general |
when a person aged 39 is added to a group of n people , the average age increases by 2 . when a person aged 15 is added instead , the average age decreases by 1 . what is the value of q ? | "a simple and elegant solution . as addition of 39 , shifts mean by 2 , and addition of 15 , shifts mean by 1 to the other side , we have the mean lying between 3915 , and in a ratio of 2 : 1 39 - 15 = 24 24 divide by 3 is 8 . meaning mean of the n terms is 15 + 8 = 39 - 16 = 23 now , from first statement , when a pers... | a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | a | subtract(divide(subtract(39, 15), add(2, 1)), 1) | add(n1,n3)|subtract(n0,n2)|divide(#1,#0)|subtract(#2,n3)| | general |
how many paying stones , each measuring 2 1 / 2 m * 2 m are required to pave a rectangular court yard 60 m long and 16 1 / 2 m board ? | "60 * 33 / 2 = 5 / 2 * 2 * x = > x = 198 answer : d" | a ) 99 , b ) 18 , c ) 16 , d ) 198 , e ) 15 | d | divide(multiply(60, add(16, divide(1, 2))), multiply(add(2, divide(1, 2)), 2)) | divide(n1,n0)|add(n5,#0)|add(n0,#0)|multiply(n4,#1)|multiply(n0,#2)|divide(#3,#4)| | general |
the captain of a cricket team of 11 members is 26 years old and the wicket keeper is 7 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 ? | "explanation let the average age of the whole team by x years . 11 x â € “ ( 26 + 33 ) = 9 ( x - 1 ) 11 x â € “ 9 x = 50 2 x = 50 x = 25 . so , average age of the team is 25 years . answer c" | a ) 23 years , b ) 24 years , c ) 25 years , d ) 26 years , e ) 27 years | c | divide(subtract(add(26, add(26, 7)), multiply(7, 7)), const_2) | add(n1,n2)|multiply(n2,n2)|add(n1,#0)|subtract(#2,#1)|divide(#3,const_2)| | general |
if 6 ^ k + 6 ^ k = ( 6 ^ 9 ) ^ ( 6 ^ 9 ) - 6 ^ k , then k = ? | "6 ^ k + 6 ^ k = ( 6 ^ 9 ) ^ 6 ^ 9 - 6 ^ k 6 * ( 6 ^ k ) = 6 ^ ( 36 * 6 ^ 9 ) = 6 ^ ( 6 ^ 2 * 6 ^ 9 ) = 6 ^ ( 6 ^ 11 ) 6 ^ k + 1 = 6 ^ ( 6 ^ 11 ) so k + 1 = 6 ^ 11 so k = 6 ^ 11 - 1 answer is d" | a ) 11 / 3 , b ) 11 / 2 , c ) 242 , d ) 6 ^ 11 - 1 , e ) 3 ^ 11 - 1 | d | subtract(power(6, add(9, const_2)), const_1) | add(n3,const_2)|power(n0,#0)|subtract(#1,const_1)| | general |
there are 12 stations between ernakulam and chennai . how many second class tickets have to be printed , so that a passenger can travel from one station to any other station ? | "the total number of stations = 12 from 12 stations we have to choose any two stations and the direction of travel ( ernakulam to chennai is different from chennai to ernakulam ) in 12 p 2 ways . 12 p 2 = 12 * 11 = 870 answer : a" | a ) 132 , b ) 145 , c ) 165 ` , d ) 400 , e ) 300 | a | multiply(add(12, const_2), subtract(add(12, const_2), const_1)) | add(n0,const_2)|subtract(#0,const_1)|multiply(#0,#1)| | physics |
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