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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a mixture of 40 liters of milk and water contains 10 % water . how much water should be added to this so that water may be 20 % in the new mixture ? | 4 ltr - water 36 ltr - milk let x water be added then the new mixture must have 20 % water then 4 + x = 20 / 100 ( 40 + x ) x = 5 answer : c | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | c | divide(subtract(multiply(40, divide(20, const_100)), multiply(40, divide(10, const_100))), divide(subtract(const_100, 20), const_100)) | divide(n2,const_100)|divide(n1,const_100)|subtract(const_100,n2)|divide(#2,const_100)|multiply(n0,#0)|multiply(n0,#1)|subtract(#4,#5)|divide(#6,#3) | general |
in the number 11,0 ab , a and b represent the tens and units digits , respectively . if 11,0 ab is divisible by 50 , what is the greatest possible value of b × a ? | "you should notice that 50 * 2 = 110 so 11,000 is divisible by 55 : 55 * 200 = 11,000 ( or you can notice that 11,000 is obviously divisible by both 5 and 11 so by 55 ) - - > b * a = 0 * 0 = 0 . next number divisible by 55 is 11,000 + 55 = 11,055 : b * a = 5 * 4 = 20 ( next number wo n ' t have 110 as the first 3 digit... | a ) 0 , b ) 5 , c ) 20 , d ) 15 , e ) 25 | c | subtract(multiply(reminder(reminder(multiply(add(divide(add(power(11,0, const_2), const_1000), 50), const_1), 50), 11,0), const_10), divide(subtract(reminder(multiply(add(divide(add(power(11,0, const_2), const_1000), 50), const_1), 50), 11,0), reminder(reminder(multiply(add(divide(add(power(11,0, const_2), const_1000),... | power(n0,const_2)|add(#0,const_1000)|divide(#1,n2)|add(#2,const_1)|multiply(n2,#3)|reminder(#4,n0)|reminder(#5,const_10)|subtract(#5,#6)|divide(#7,const_10)|multiply(#8,#6)|subtract(#9,const_10)| | general |
a trained covered x km at 40 kmph and another 2 x km at 20 kmph . find the average speed of the train in covering the entire 5 x km . | "total time taken = x / 40 + 2 x / 20 hours = 5 x / 40 = x / 8 hours average speed = 5 x / ( x / 8 ) = 40 kmph answer : d" | a ) 16 , b ) 18 , c ) 24 , d ) 40 , e ) 12 | d | divide(multiply(40, 5), add(divide(40, 40), divide(multiply(2, 40), 20))) | divide(n0,n0)|multiply(n0,n3)|multiply(n0,n1)|divide(#2,n2)|add(#0,#3)|divide(#1,#4)| | general |
riya and priya set on a journey . riya moves eastward at a speed of 21 kmph and priya moves westward at a speed of 22 kmph . how far will be priya from riya after 60 minutes | "total eastward distance = 21 kmph * 1 hr = 21 km total westward distance = 22 kmph * 1 hr = 22 km total distn betn them = 21 + 22 = 43 km ans 43 km answer : a" | a ) 43 kms , b ) 45 kms , c ) 50 kms , d ) 30 kms , e ) 40 kms | a | multiply(speed(add(21, 22), const_60), 60) | add(n0,n1)|speed(#0,const_60)|multiply(n2,#1)| | physics |
a family had dinner in a restaurant and paid $ 30 for food . they also had to pay 9.5 % sale tax and 10 % for the tip . how much did they pay for the dinner ? | solution they paid for food , sales tax and tip , hence total paid = $ 30 + 9.5 % * 30 + 10 % * 30 = $ 35.85 answer c | a ) $ 35.83 , b ) $ 35.84 , c ) $ 35.85 , d ) $ 35.86 , e ) $ 35.87 | c | subtract(subtract(subtract(add(subtract(multiply(subtract(9.5, const_3), 9.5), const_0_25), const_0_25), const_12), const_12), const_2) | subtract(n1,const_3)|multiply(n1,#0)|subtract(#1,const_0_25)|add(#2,const_0_25)|subtract(#3,const_12)|subtract(#4,const_12)|subtract(#5,const_2) | general |
a man is 26 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 | "let ' s son age is x , then father age is x + 26 . = > 2 ( x + 2 ) = ( x + 26 + 2 ) = > 2 x + 4 = x + 28 = > x = 24 years answer : e" | a ) 78 years , b ) 22 years , c ) 88 years , d ) 66 years , e ) 24 years | e | divide(subtract(26, 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 |
the parameter of a square is equal to the perimeter of a rectangle of length 40 cm and breadth 20 cm . find the circumference of a semicircle whose diameter is equal to the side of the square . ( round off your answer to two decimal places ) ? | "let the side of the square be a cm . parameter of the rectangle = 2 ( 40 + 20 ) = 120 cm parameter of the square = 120 cm i . e . 4 a = 120 a = 30 diameter of the semicircle = 30 cm circimference of the semicircle = 1 / 2 ( â ˆ ) ( 30 ) = 1 / 2 ( 22 / 7 ) ( 30 ) = 660 / 14 = 47.14 cm to two decimal places answer : a... | a ) 47.14 , b ) 45.15 , c ) 43.23 , d ) 41.44 , e ) 41.51 | a | divide(circumface(divide(square_edge_by_perimeter(rectangle_perimeter(40, 20)), const_2)), const_2) | rectangle_perimeter(n0,n1)|square_edge_by_perimeter(#0)|divide(#1,const_2)|circumface(#2)|divide(#3,const_2)| | geometry |
two trains 200 m and 300 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 = 200 + 300 = 500 m . required time = 500 * 9 / 250 = 18 secs answer : a" | a ) 18 , b ) 10.9 , c ) 10.4 , d ) 10.8 , e ) 10.1 | a | divide(add(200, 300), multiply(add(60, 40), const_0_2778)) | add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)| | physics |
the manager at a health foods store mixes a unique superfruit juice cocktail that costs $ 1399.45 per litre to make . the cocktail includes mixed fruit juice and a ç ai berry juice , which cost $ 262.85 per litre and $ 3104.35 per litre , respectively . the manager has already opened 35 litres of the mixed fruit juice ... | "262.85 ( 35 ) + 3 , 104.35 x = 1 , 399.45 ( 35 + x ) solve the equation . 262.85 ( 35 ) + 3 , 104.35 x = 1 , 399.45 ( 35 + x ) 9 , 199.75 + 3 , 104.35 x = 48980.75 + 1 , 399.45 x 9 , 199.75 + 1 , 704.9 x = 48980.75 1 , 704.9 x = 39,781 x ≈ 23.3 answer is b ." | a ) 17 litres , b ) 23.3 litres , c ) 11 litres , d ) 07 litres , e ) 38 litres | b | divide(subtract(multiply(35, 1399.45), multiply(35, 262.85)), subtract(3104.35, 1399.45)) | multiply(n0,n3)|multiply(n1,n3)|subtract(n2,n0)|subtract(#0,#1)|divide(#3,#2)| | general |
when n is divided by 15 , the remainder is 8 . what is the remainder when 5 n is divided by 6 ? | "let n = 8 ( leaves a remainder of 8 when divided by 48 ) 5 n = 5 ( 8 ) = 40 , which leaves a remainder of 4 when divided by 6 . answer d" | a ) 2 , b ) 6 , c ) 3 , d ) 4 , e ) 1 | d | subtract(8, reminder(5, 6)) | reminder(n2,n3)|subtract(n1,#0)| | general |
a number x is 18 times another number y . the percentage that y is less than x is | "say y = 1 and x = 18 . then y = 1 is less than x = 18 by ( 18 - 1 ) / 18 * 100 = 17 / 18 * 100 = 94.4 % . answer : d ." | a ) 12.5 % , b ) 87.5 % , c ) 80 % , d ) 94.4 % , e ) 1 % | d | multiply(divide(subtract(18, const_1), 18), const_100) | subtract(n0,const_1)|divide(#0,n0)|multiply(#1,const_100)| | general |
a dozen eggs and 5 pounds of oranges are currently at the same price . if the price of a dozen eggs rises by 3 percent and the price of oranges rises by 6 % . how much more will it cost to buy a dozen eggs and 10 pounds of oranges . | say currently both a dozen eggs and 5 pounds of oranges cost $ 100 ( they are at the same price ) . so , to buy a dozen eggs and 5 pounds of oranges we need $ 100 . after the increase , the price of a dozen eggs will be $ 103 and the price of 5 pounds of oranges will be $ 106 . so after the increase , to buy a dozen eg... | a ) 8 % , b ) 9 % , c ) 10 % , d ) 11 % , e ) 5 % | b | multiply(divide(multiply(subtract(add(add(10, multiply(divide(3, const_100), const_10)), add(10, multiply(divide(6, const_100), 10))), add(10, 10)), const_100), add(10, 10)), const_2) | add(n3,n3)|divide(n1,const_100)|divide(n2,const_100)|multiply(#1,const_10)|multiply(n3,#2)|add(n3,#3)|add(n3,#4)|add(#5,#6)|subtract(#7,#0)|multiply(#8,const_100)|divide(#9,#0)|multiply(#10,const_2) | general |
a number is said to be prime saturated if the product of all the different positive prime factors of a is less than the square root of a . what is the greatest two digit prime saturated integer ? | "clearly d a number is said to be prime saturated if the product of all the different positive prime factors of a is less than the square root of a . 96 has more number of smaller prime factor thats the clue ! ! = d" | a ) 99 , b ) 98 , c ) 97 , d ) 96 , e ) 95 | d | add(multiply(const_2, const_3), subtract(const_100, const_10)) | multiply(const_2,const_3)|subtract(const_100,const_10)|add(#0,#1)| | other |
the area of a circular field is 13.86 hectares . find the cost of fencing it at the rate of rs . 4.60 per metre . | "explanation : area = ( 13.86 x 10000 ) sq . m = 138600 sq . m circumference = cost of fencing = rs . ( 1320 x 4.60 ) = rs . 6072 . answer : e ) 6072" | a ) 2399 , b ) 3888 , c ) 2999 , d ) 5808 , e ) 6072 | e | multiply(circumface(multiply(sqrt(divide(13.86, const_pi)), const_100)), 4.60) | divide(n0,const_pi)|sqrt(#0)|multiply(#1,const_100)|circumface(#2)|multiply(#3,n1)| | geometry |
3 years ago the average age of a class of 6 members was 19 years . a boy have been added , the average age of the class is the same today . what is the age of the boy ? | 6 * 22 = 132 7 * 19 = 133 ` 1 a | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | a | subtract(multiply(add(6, const_1), 19), add(multiply(6, 19), multiply(6, 3))) | add(n1,const_1)|multiply(n1,n2)|multiply(n0,n1)|add(#1,#2)|multiply(n2,#0)|subtract(#4,#3) | general |
find the number of shares that can be bought for rs . 8200 if the market value is rs . 28 each with brokerage being 2.5 % . | "explanation : cost of each share = ( 28 + 2.5 % of 28 ) = rs . 28.7 therefore , number of shares = 8200 / 28.7 = 285.7 answer : d" | a ) 237 , b ) 270 , c ) 177 , d ) 285.7 , e ) 111 | d | floor(divide(8200, add(28, divide(2.5, const_100)))) | divide(n2,const_100)|add(n1,#0)|divide(n0,#1)|floor(#2)| | gain |
if ( x + 3 ) ^ 2 / ( 3 x + 29 ) = 2 , then the difference between the two possible values of x is : | "( x + 3 ) ^ 2 / ( 3 x + 29 ) = 2 ( x + 3 ) ^ 2 = 2 ( 3 x + 29 ) x ^ 2 + 6 x + 9 = 6 x + 58 x ^ 2 - 49 = 0 ( x - 7 ) ( x + 7 ) = 0 x = 7 or x = - 7 the answer is d ." | a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 16 | d | multiply(sqrt(subtract(multiply(29, const_3.0), multiply(3, 3))), 2) | multiply(n4,n3)|multiply(n0,n0)|subtract(#0,#1)|sqrt(#2)|multiply(#3,n1)| | general |
a train 710 m long is running at a speed of 78 km / hr . if it crosses a tunnel in 1 min , then the length of the tunnel is ? | "speed = 78 * 5 / 18 = 65 / 3 m / sec . time = 1 min = 60 sec . let the length of the train be x meters . then , ( 710 + x ) / 60 = 65 / 3 x = 590 m . answer : e" | a ) 2898 , b ) 277 , c ) 500 , d ) 297 , e ) 590 | e | divide(710, multiply(subtract(78, 1), const_0_2778)) | subtract(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
in a certain corporation , there are 300 male employees and 100 female employees . it is known that 20 % of the male employees have advanced degrees and 40 % of the females have advanced degrees . if one of the 400 employees is chosen at random , what is the probability this employee has an advanced degree or is female... | "in this corporation , there are 400 total employees . there are 100 women . of the 300 men , 20 % have advanced degrees — - 10 % of 300 must be 30 , so 20 % of 300 must be 60 . add the women and the men with advanced degrees : 100 + 60 = 160 . this is the or region , full set of individuals that satisfy the condition ... | a ) 1 / 20 , b ) 1 / 10 , c ) 1 / 5 , d ) 2 / 5 , e ) 3 / 4 | d | add(divide(multiply(subtract(const_1, divide(40, multiply(20, 20))), 100), 400), divide(add(multiply(divide(20, multiply(20, 20)), 300), multiply(divide(40, multiply(20, 20)), 100)), 400)) | multiply(n2,n2)|divide(n3,#0)|divide(n2,#0)|multiply(n0,#2)|multiply(n1,#1)|subtract(const_1,#1)|add(#3,#4)|multiply(n1,#5)|divide(#7,n4)|divide(#6,n4)|add(#8,#9)| | other |
the price of 24 apples is equal to that of 28 oranges . the price of 45 apples and 60 oranges together is rs . 1350 . the total price of 30 apples and 40 oranges is | let the price of one apple = a and price of one orange = b price of 24 apples is equal to that of 28 oranges 24 a = 28 b = > 6 a = 7 b ⇒ b = 6 a / 7 - - - - - ( equation 1 ) price of 45 apples and 60 oranges together is rs . 1350 = > 45 a + 60 b = 1350 = > 3 a + 4 b = 90 ⇒ 3 a + 4 ( 6 a ) / 7 = 90 ( ∵ substituted the v... | a ) 850 , b ) 900 , c ) 750 , d ) 950 , e ) 800 | b | add(multiply(divide(1350, add(45, multiply(60, divide(24, 28)))), 30), multiply(multiply(divide(1350, add(45, multiply(60, divide(24, 28)))), divide(24, 28)), 40)) | divide(n0,n1)|multiply(n3,#0)|add(n2,#1)|divide(n4,#2)|multiply(n5,#3)|multiply(#3,#0)|multiply(n6,#5)|add(#4,#6) | general |
rhoda tosses a coin 5 times . find the probability of getting exactly 4 heads . | exactly 4 heads out of 5 would be : hhhht , hhthh , or many other combinations . therefore , no . of possible combinations are 5 ! / 4 ! * 1 ! = 5 probability = 5 / 2 ^ 5 = 5 / 32 answer : a | a ) 5 / 32 , b ) 6 / 32 , c ) 5 / 16 , d ) 4 / 32 , e ) 8 / 32 | a | divide(choose(5, 4), power(const_2, 5)) | choose(n0,n1)|power(const_2,n0)|divide(#0,#1) | probability |
if two of the 4 expressions x + y , x + 5 y , x - y , and 5 x - y are chosen at random , what is the probability r that their product will be of the form of x ^ 2 - ( by ) ^ 2 , where b is an integer ? | only ( x + y ) ( x - y ) pair will give the form x ^ 2 - ( by ) ^ 2 the probability of selecting these two pairs are 1 / 4 * 1 / 3 = 1 / 12 , assuming x + y is picked first , then x - y , but x - y can be picked first followed by x + y . so the probability r = 1 / 12 * 2 = 1 / 6 ans e | a ) 1 / 2 , b ) 1 / 3 , c ) 1 / 4 , d ) 1 / 5 , e ) r = 1 / 6 | e | multiply(divide(const_1, 2), divide(const_1, const_3)) | divide(const_1,n3)|divide(const_1,const_3)|multiply(#0,#1) | general |
25 binders can bind 1400 books in 21 days . how many binders will be required to bind 1600 books in 20 days ? | "binders books days 25 1400 21 x 1600 20 x / 25 = ( 1600 / 1400 ) * ( 21 / 20 ) = > x = 30 answer : e" | a ) 87 , b ) 18 , c ) 17 , d ) 16 , e ) 30 | e | divide(divide(1600, 20), divide(1400, multiply(25, 21))) | divide(n3,n4)|multiply(n0,n2)|divide(n1,#1)|divide(#0,#2)| | physics |
the amount of principal rs . 8000 at compound interest at the ratio of 5 % p . a . for 3 years is | "c . i = p ( 1 + r / 100 ) ^ n = 8000 ( 1 + 5 / 100 ) ^ 3 = 8000 ( 21 / 20 * 21 / 20 * 21 / 20 ) = rs 9261 answer : c" | a ) rs . 9621 , b ) rs . 6921 , c ) rs . 9261 , d ) rs . 6261 , e ) rs . 6361 | c | multiply(8000, power(add(const_1, divide(5, const_100)), 3)) | divide(n1,const_100)|add(#0,const_1)|power(#1,n2)|multiply(n0,#2)| | gain |
the ratio of the capacity to do work of a and b is 3 : 2 . if they together can complete a work in 18 days , then how long does a take to complete the work alone ? | llet a and b take 3 x and 2 x days to complete the work 1 / 3 x + 1 / 2 x = 1 / 18 ⇒ x = 15 so a will take 45 days . answer : a | a ) 45 , b ) 30 , c ) 24 , d ) 40 , e ) none of these | a | divide(multiply(divide(add(multiply(18, 3), multiply(18, 2)), 3), 3), 2) | multiply(n0,n2)|multiply(n1,n2)|add(#0,#1)|divide(#2,n0)|multiply(n0,#3)|divide(#4,n1) | physics |
in a can , there is a mixture of milk and water in the ratio 4 : 5 . if it is filled with an additional 14 litres of milk the can would be full and ratio of milk and water would become 6 : 5 . find the capacity of the can ? | "let the capacity of the can be t litres . quantity of milk in the mixture before adding milk = 4 / 9 ( t - 14 ) after adding milk , quantity of milk in the mixture = 6 / 11 t . 6 t / 11 - 14 = 4 / 9 ( t - 14 ) 10 t = 1386 - 616 = > t = 77 . answer : b" | a ) 40 , b ) 77 , c ) 48 , d ) 52 , e ) none of these | b | add(add(multiply(5, divide(14, subtract(multiply(divide(6, 5), 5), 4))), divide(14, subtract(multiply(divide(6, 5), 5), 4))), 14) | divide(n3,n4)|multiply(n1,#0)|subtract(#1,n0)|divide(n2,#2)|multiply(n1,#3)|add(#3,#4)|add(n2,#5)| | general |
the population of a town increased from 1 , 75000 to 2 , 97500 in a decade . the average percent increase of population per year is | solution increase in 10 years = ( 297500 - 175000 ) = 122500 increase % = ( 122500 / 175000 ã — 100 ) % = 70 % . required average = ( 70 / 10 ) % = 7 % . answer e | a ) 4.37 % , b ) 5 % , c ) 6 % , d ) 8.75 % , e ) 7 % | e | multiply(divide(divide(subtract(add(multiply(multiply(const_1000, const_100), 2), 97500), add(multiply(const_1000, const_100), 75000)), const_10), add(multiply(const_1000, const_100), 75000)), const_100) | multiply(const_100,const_1000)|add(n1,#0)|multiply(n2,#0)|add(n3,#2)|subtract(#3,#1)|divide(#4,const_10)|divide(#5,#1)|multiply(#6,const_100) | general |
the workforce of company x is 40 % female . the company hired 20 additional male workers , and as a result , the percent of female workers dropped to 50 % . how many employees did the company have after hiring the additional male workers ? | "let x be the total worker then 0.4 x = female worker and 0.6 x is male worker then 20 male worker added 04 x / ( 0.6 x + 20 ) = 50 / 100 or 40 x = 50 * ( 0.6 x + 100 ) = 30 x + 5000 or 10 x = 5000 , x = 500 total worker = 500 + 20 = 520 c" | a ) 480 , b ) 500 , c ) 520 , d ) 460 , e ) 440 | c | add(divide(multiply(divide(50, const_100), 20), subtract(divide(40, const_100), divide(50, const_100))), 20) | divide(n2,const_100)|divide(n0,const_100)|multiply(n1,#0)|subtract(#1,#0)|divide(#2,#3)|add(n1,#4)| | gain |
the ratio of kamla and nisha is 6 : 5 and the sum of their ages is 44 . what will be the ratio of their ages after 8 years . | answer : let age of kamla is 6 x and age of nisha is 5 x . then , 6 x + 5 x = 4411 x = 44 , x = 4 after 8 year the ratio of their ages = 6 x + 8 / 5 x + 8 = 8 : 7 answer c | a ) 5 : 6 , b ) 7 : 8 , c ) 8 : 7 , d ) 12 : 11 , e ) 14 : 13 | c | divide(8, add(6, const_1)) | add(n0,const_1)|divide(n3,#0) | general |
two identical cubes if one of them is painted pink on its 4 side and blue on the remaining two side then how many faces painted pink to other cube so that probability of getting cube is 1 / 3 when we roll both the cube . | 0 faces should be painted pink i . e all faces blue answer : a | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | a | multiply(divide(divide(1, 3), divide(4, const_6)), const_6) | divide(n1,n2)|divide(n0,const_6)|divide(#0,#1)|multiply(#2,const_6) | other |
due to construction , the speed limit along an 9 - mile section of highway is reduced from 55 miles per hour to 25 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 9 miles stretch = 9 * 60 / 55 = 9 * 12 / 11 = 9.81 new time in minutes to cross 9 miles stretch = 9 * 60 / 25 = 9 * 12 / 5 = 21.6 time difference = 11.79 ans : a" | a ) a ) 11.79 , b ) b ) 8 , c ) c ) 10 , d ) d ) 15 , e ) e ) 24 | a | max(multiply(subtract(add(55, 9), const_1), subtract(divide(9, 25), divide(9, 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 |
the cost price of a radio is rs . 2400 and it was sold for rs . 1980 , find the loss % ? | "2400 - - - - 420 100 - - - - ? = > 17 % answer : a" | a ) 17 % , b ) 12 % , c ) 22 % , d ) 24 % , e ) 21 | a | multiply(divide(subtract(2400, 1980), 2400), const_100) | subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)| | gain |
a couple spent $ 132 in total while dining out and paid this amount using a credit card . the $ 132 figure included a 20 percent tip which was paid on top of the price which already included a sales tax of 10 percent on top of the price of the food . what was the actual price of the food before tax and tip ? | "let the price of the meal be x . after a 10 % sales tax addition , the price is 1.1 * x after a 20 % tip on this amount , the total is 1.2 * 1.1 * x = 1.32 x 1.32 x = 132 x = 100 the correct answer is a ." | a ) $ 100 , b ) $ 104 , c ) $ 108 , d ) $ 112 , e ) $ 116 | a | divide(multiply(divide(multiply(132, const_100), add(const_100, 20)), const_100), add(const_100, 10)) | add(n2,const_100)|add(n3,const_100)|multiply(n0,const_100)|divide(#2,#0)|multiply(#3,const_100)|divide(#4,#1)| | general |
the average salary of all the workers in a workshop is rs . 8000 . the average salary of 7 technicians is rs . 14000 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 = ( 14000 * 7 ) + 6000 ( x - 7 ) = 2000 x = 56000 = x = 28 . answer : a" | a ) 28 , b ) 19 , c ) 22 , d ) 20 , e ) 21 | a | add(7, divide(multiply(7, subtract(14000, 8000)), subtract(8000, 6000))) | subtract(n2,n0)|subtract(n0,n3)|multiply(n1,#0)|divide(#2,#1)|add(n1,#3)| | general |
a rectangular photograph is surrounded by a border that is 1 inch wide on each side . the total area of the photograph and the border is m square inches . if the border had been 3 inches wide on each side , the total area would have been ( m + 52 ) square inches . what is the perimeter of the photograph , in inches ? | "let length and breadth of photograph be l and b respectively . perimeter is given by 2 * ( l + b ) - - - - - ( 1 ) according to the question : ( l + 2 ) ( b + 2 ) = m - - - - ( 2 ) and ( l + 6 ) ( b + 6 ) = m + 52 - - - - - - - - - > ( l + 6 ) ( b + 6 ) - 52 = m - - - - - - ( 3 ) equating ( 2 ) and ( 3 ) ( l + 2 ) ( b... | a ) 10 , b ) 36 , c ) 38 , d ) 40 , e ) 42 | a | divide(subtract(52, subtract(power(multiply(3, const_2), const_2), power(multiply(1, const_2), const_2))), const_2) | multiply(n1,const_2)|multiply(n0,const_2)|power(#0,const_2)|power(#1,const_2)|subtract(#2,#3)|subtract(n2,#4)|divide(#5,const_2)| | geometry |
uncle bruce is baking chocolate chip cookies . he has 50 ounces of dough ( with no chocolate ) and 40 ounces of chocolate . how much chocolate is left over if he uses all the dough but only wants the cookies to consist of 30 % chocolate ? | "first , you must find the total weight of the mixture given that 80 % of it will be dough . 70 % * total = 40 = > ( 7 / 10 ) total = 40 = > total = 400 / 7 = > total = 57.14 oz , from there , you must find 30 % of the total 57.14 oz of the mixture . 30 % * total = > ( 3 / 10 ) ( 57.14 ) = 17.14 oz choclate used , not ... | a ) 22.86 , b ) 23.86 , c ) 25 , d ) 26 , e ) 27 | a | multiply(divide(30, const_100), 30) | divide(n2,const_100)|multiply(n2,#0)| | gain |
in arun ' s opinion , his weight is greater than 61 kg but leas than 72 kg . his brother does not agree with arun and he thinks that arun ' s weight is greater than 60 kg but less than 70 kg . his mother ' s view is that his weight can not be greater than 64 kg . if all of them are correct in their estimation , what is... | let arun ' s weight be x kg . according to arun , 61 < x < 72 . according to arun ' s brother , 60 < x < 70 . according to arun ' s mother , x < 64 . the values satisfying all the above conditions are 62 and 63 . required average = ( 62 + 63 ) / 2 = 62.5 kg answer : b | a ) 86.5 kg , b ) 62.5 kg , c ) 46.5 kg , d ) 26.5 kg , e ) 16.5 kg | b | divide(add(64, add(61, const_1)), const_2) | add(n0,const_1)|add(n4,#0)|divide(#1,const_2) | general |
a and b together have $ 1210 . if of a ' s amount is equal to of b ' s amount , how much amount does b have ? | ( 4 / 15 ) a = ( 2 / 5 ) b = > a = ( ( 2 / 5 ) * ( 15 / 4 ) ) b = > a = ( 3 / 2 ) b = > a / b = 3 / 2 a : b = 3 : 2 b ' s share = rs . ( 1210 * ( 2 / 5 ) ) = $ 484 option c | a ) $ 400 , b ) $ 489 , c ) $ 484 , d ) $ 365 , e ) % 342 | c | divide(multiply(multiply(1210, const_2), const_2), const_10) | multiply(n0,const_2)|multiply(#0,const_2)|divide(#1,const_10) | general |
what profit percent is made by selling an article at a certain price , if by selling at 2 / 3 rd of that price , there would be a loss of 20 % ? | "sp 2 = 2 / 3 sp 1 cp = 100 sp 2 = 80 2 / 3 sp 1 = 80 sp 1 = 120 100 - - - 20 = > 20 % answer : a" | a ) 20 % , b ) 28 % , c ) 60 % , d ) 26 % , e ) 30 % | a | subtract(divide(subtract(const_100, 20), divide(2, 3)), const_100) | divide(n0,n1)|subtract(const_100,n2)|divide(#1,#0)|subtract(#2,const_100)| | gain |
what is the minimum value of | x - 4 | + | x + 6 | + | x - 5 | ? | a can not be the answer as all the three terms are in modulus and hence the answer will be non negative . | x - 4 | > = 0 - - > minimum occurs at x = 4 | x + 6 | > = 0 - - > minimum occurs at x = - 6 | x - 5 | > = 0 - - > minimum occurs at x = 5 x = - 6 - - > result = 10 + 0 + 11 = 21 . also any negative value will pus... | a ) - 3 , b ) 3 , c ) 5 , d ) 7 , e ) 11 | e | add(6, 5) | add(n1,n2) | general |
how many five - digit numbers that do not contain the digits 4 or 7 are there ? | "we can have 7 digits ( 1 , 2,3 , 5,6 , 8,9 ) for the first place ( ten thousand ' s place ) . and similarly 8 digits for thousand ' s , hundred ' s , tenth ' s and unit digit . ( 0,1 , 2,3 , 5,6 , 8,9 ) so in total 7 * 8 * 8 * 8 * 8 = 28672 hence c" | a ) 44648 , b ) 27844 , c ) 28642 , d ) 16864 , e ) 32458 | c | multiply(multiply(multiply(const_10, const_10), subtract(const_10, const_1)), 7) | multiply(const_10,const_10)|subtract(const_10,const_1)|multiply(#0,#1)|multiply(n1,#2)| | general |
the larger interior angle of a parallelogram is 90 ° more than its smaller interior angle . what is the measure of the smaller interior angle of the parallelogram ? | let smaller be x , larger becomes x + 90 . . . total of all angles = 360 so 4 x + 180 = 360 and x = 45 answer : d | a ) 90 ° , b ) 75 ° , c ) 60 ° , d ) 45 ° , e ) 30 ° | d | divide(90, const_2) | divide(n0,const_2) | geometry |
a train is 400 meter long is running at a speed of 60 km / hour . in what time will it pass a bridge of 800 meter length ? | "speed = 60 km / hr = 60 * ( 5 / 18 ) m / sec = 150 / 9 m / sec total distance = 400 + 800 = 1200 meter time = distance / speed = 1200 * ( 9 / 150 ) = 72 seconds answer : a" | a ) 72 seconds , b ) 27 seconds , c ) 40 seconds , d ) 128 seconds , e ) 18 seconds | a | divide(add(400, 800), divide(multiply(60, const_1000), const_3600)) | add(n0,n2)|multiply(n1,const_1000)|divide(#1,const_3600)|divide(#0,#2)| | physics |
a man saves 20 % of his monthly salary . if an account of dearness of things he is to increase his monthly expenses by 20 % , he is only able to save rs . 200 per month . what is his monthly salary ? | "income = rs . 100 expenditure = rs . 80 savings = rs . 20 present expenditure 80 * ( 20 / 100 ) = rs . 96 present savings = 100 – 96 = rs . 4 100 - - - - - - 4 ? - - - - - - - - - 200 = > 5000 answer : a" | a ) rs . 5000 , b ) rs . 25000 , c ) rs . 7500 , d ) rs . 8500 , e ) rs . 9500 | a | divide(multiply(200, const_100), subtract(const_100, add(subtract(const_100, 20), multiply(subtract(const_100, 20), divide(20, 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 |
the average height of 20 students is 20 cm and the average height of 11 students is 20 cm . what is the average height of the 31 students ? | the total height of the 31 students = ( 20 * 20 ) + ( 11 * 20 ) = 400 + 220 = 620 cm the required average height = 620 / 31 = 20 cm answer : b | a ) 40 cm , b ) 20 cm , c ) 30 cm , d ) 25 cm , e ) 35 cm | b | divide(add(multiply(20, 20), multiply(11, 20)), 31) | multiply(n0,n0)|multiply(n0,n2)|add(#0,#1)|divide(#2,n4) | general |
a certain ball team has an equal number of right - and left - handed players . on a certain day , two - thirds of the players were absent from practice . of the players at practice that day , two - third were left handed . what is the ratio of the number of right - handed players who were not at practice that day to th... | say the total number of players is 18 , 9 right - handed and 9 left - handed . on a certain day , two - thirds of the players were absent from practice - - > 12 absent and 6 present . of the players at practice that day , one - third were left - handed - - > 6 * 2 / 3 = 4 were left - handed and 2 right - handed . the n... | a ) 1 / 3 , b ) 2 / 3 , c ) 5 / 7 , d ) 7 / 5 , e ) 3 / 2 | d | divide(subtract(divide(const_1, const_2), subtract(subtract(const_1, divide(const_2, const_3)), multiply(subtract(const_1, divide(const_2, const_3)), divide(const_2, const_3)))), subtract(divide(const_1, const_2), multiply(subtract(const_1, divide(const_2, const_3)), divide(const_2, const_3)))) | divide(const_1,const_2)|divide(const_2,const_3)|subtract(const_1,#1)|multiply(#1,#2)|subtract(#2,#3)|subtract(#0,#3)|subtract(#0,#4)|divide(#6,#5) | general |
excluding the stoppages , the speed of a bus is 64 km / hr and including the stoppages the speed o the bus is 48 km / hr . for how many minutes does the bus stop per hour ? | answer : b ) 15 minutes | a ) 22 , b ) 15 , c ) 77 , d ) 28 , e ) 91 | b | subtract(const_60, multiply(divide(48, 64), const_60)) | divide(n1,n0)|multiply(#0,const_60)|subtract(const_60,#1) | physics |
what amount does an investor receive if the investor invests $ 3000 at 10 % p . a . compound interest for two years , compounding done annually ? | "a = ( 1 + r / 100 ) ^ n * p ( 1.1 ) ^ 2 * 5000 = 1.21 * 5000 = 3630 the answer is b ." | a ) $ 3420 , b ) $ 3630 , c ) $ 3870 , d ) $ 4040 , e ) $ 4220 | b | multiply(3000, power(add(const_1, divide(10, const_100)), const_2)) | divide(n1,const_100)|add(#0,const_1)|power(#1,const_2)|multiply(n0,#2)| | gain |
a fruit seller sells mangoes at the rate of rs . 12 per kg and thereby loses 15 % . at what price per kg , he should have sold them to make a profit of 5 % | "explanation : 85 : 12 = 105 : x x = ( 12 × 105 / 85 ) = rs 14.82 option d" | a ) rs 8.82 , b ) rs 9.82 , c ) rs 10.82 , d ) rs 14.82 , e ) none of these | d | divide(multiply(12, add(const_100, 5)), subtract(const_100, 15)) | add(n2,const_100)|subtract(const_100,n1)|multiply(n0,#0)|divide(#2,#1)| | gain |
a sells a bicycle to b and makes a profit of 35 % . b sells the same bicycle to c at a profit of 45 % . if the final s . p . of the bicycle was rs . 225 , find out the cost price of the bicycle for a . | "explanation : let cp be 100 a sells at 35 % profit so sp = 135 b sells at 45 % profit = 135 x ( 1 + 45 / 100 ) = 195.75 cp sp 100 - - - 195.75 x - - - 225 cp = 225 x 100 / 195.75 = 114.94 answer : e" | a ) 237 , b ) 126 , c ) 971 , d ) 611 , e ) 115 | e | divide(divide(225, divide(add(const_100, 45), const_100)), divide(add(35, const_100), const_100)) | add(n1,const_100)|add(n0,const_100)|divide(#0,const_100)|divide(#1,const_100)|divide(n2,#2)|divide(#4,#3)| | gain |
how many zeros does 10 ! end with ? | "according to above 10 ! has 10 / 5 = 2 = 2 trailing zeros . answer : e" | a ) 10 , b ) 4 , c ) 4 , d ) 3 , e ) 2 | e | add(add(divide(10, add(const_4, const_1)), divide(subtract(10, add(const_4, const_1)), power(add(const_4, const_1), const_2))), divide(subtract(10, add(const_4, const_1)), power(add(const_4, const_1), const_3))) | add(const_1,const_4)|divide(n0,#0)|power(#0,const_2)|power(#0,const_3)|subtract(n0,#0)|divide(#4,#2)|divide(#4,#3)|add(#1,#5)|add(#7,#6)| | other |
find the value of y from ( 12 ) ^ 2 x 6 ^ 4 ÷ 432 = y ? | 432 e | a ) 2134 , b ) 2234 , c ) 2540 , d ) 2560 , e ) 432 | e | divide(multiply(power(12, 2), power(6, 4)), 432) | power(n0,n1)|power(n2,n3)|multiply(#0,#1)|divide(#2,n4) | general |
find the simple interest on rs . 5000 at 6 % per annum for the period from 5 th feb to 19 th april , 2015 . | "explanation : given : 1 ) principal = rs . 5000 2 ) rate of interest = 6 % 3 ) time = 5 th feb to 19 th april , 2015 first find the time period 5 th feb to 19 th april , 2015 feb = 28 – 5 = 23 days march = 31 days april = 19 days total days = 23 + 31 + 19 = 73 days convert days into years , by dividing it by 365 time ... | a ) rs . 40 , b ) rs . 50 , c ) rs . 60 , d ) rs . 70 , e ) none of these | c | multiply(multiply(multiply(add(multiply(multiply(multiply(5, 19), const_100), const_100), multiply(multiply(multiply(19, 19), const_100), multiply(add(19, 5), 5))), divide(add(multiply(6, 19), 5), 19)), divide(multiply(19, 19), multiply(5, multiply(5, 19)))), 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 |
a train passes a station platform in 36 sec and a man standing on the platform in 20 sec . if the speed of the train is 36 km / hr . what is the length of the platform ? | "speed = 36 * 5 / 18 = 10 m / sec . length of the train = 10 * 20 = 200 m . let the length of the platform be x m . then , ( x + 200 ) / 36 = 10 = > x = 160 m . answer : a" | a ) 160 , b ) 240 , c ) 288 , d ) 277 , e ) 221 | a | multiply(20, multiply(36, const_0_2778)) | multiply(n2,const_0_2778)|multiply(n1,#0)| | physics |
car a runs at the speed of 42 km / hr & reaches its destination in 6 hr . car b runs at the speed of 52 km / h & reaches its destination in 9 h . what is the respective ratio of distances covered by car a & car b ? | "sol . distance travelled by car a = 42 × 6 = 252 km distance travelled by car b = 52 × 9 = 468 km ratio = 252 / 468 = 7 : 13 b" | a ) 11 : 6 , b ) 7 : 13 , c ) 13 : 7 , d ) 15 : 6 , e ) 13 : 6 | b | divide(multiply(42, 6), multiply(52, 9)) | multiply(n0,n1)|multiply(n2,n3)|divide(#0,#1)| | physics |
the speed of a car is 90 km in the first hour and 55 km in the second hour . what is the average speed of the car ? | "s = ( 90 + 55 ) / 2 = 72.5 kmph answer : a" | a ) 72.5 kmph , b ) 75 kmph , c ) 87 kmph , d ) 56 kmph , e ) 86 kmph | a | divide(add(90, 55), const_2) | add(n0,n1)|divide(#0,const_2)| | physics |
by selling 15 pencils for a rupee a man loses 25 % . how many for a rupee should he sell in order to gain 25 % ? | "75 % - - - 15 125 % - - - ? 75 / 125 * 15 = 9 answer : b" | a ) 8 , b ) 9 , c ) 7 , d ) 6 , e ) 5 | b | multiply(divide(const_1, multiply(add(const_100, 25), divide(const_1, subtract(const_100, 25)))), 15) | add(n2,const_100)|subtract(const_100,n1)|divide(const_1,#1)|multiply(#0,#2)|divide(const_1,#3)|multiply(n0,#4)| | gain |
find the remainder of the division ( 3 ^ 14 ) / 4 . | "find the pattern of the remainders after each power : ( 3 ^ 1 ) / 4 remainder 3 ( 3 ^ 2 ) / 4 remainder 1 - - > this is where the cycle ends ( 3 ^ 3 ) / 4 remainder 3 - - > this is where the cycle begins again ( 3 ^ 4 ) / 4 remainder 1 continuing the pattern to ( 3 ^ 14 ) / 4 gives us a remainder of 1 final answer : a... | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | a | reminder(power(3, 14), 4) | power(n0,n1)|reminder(#0,n2)| | general |
a certain machine produces 850 units of product p per hour . working continuously at this constant rate , this machine will produce how many units of product p in 7 days ? | "since 7 days consist of 24 * 7 hours the total hours of a week is 168 hours . since every hour the machine produces 850 units of product p the total product during 168 hours is 850 * 168 = 142,800 . correct option : d" | a ) 7,000 , b ) 24,000 , c ) 40,000 , d ) 142,800 , e ) 168,000 | d | multiply(const_4, const_10) | multiply(const_10,const_4)| | physics |
the area of a circular field is 17.56 hectares . find the cost of fencing it at the rate of rs . 5 per metre approximately | "explanation : area = ( 17.56 x 10000 ) m 2 = 175600 m 2 . π r 2 = 175600 ⇔ ( r ) 2 = ( 175600 x ( 7 / 22 ) ) ⇔ r = 236.37 m . circumference = 2 π r = ( 2 x ( 22 / 7 ) x 236.37 ) m = 1485.78 m . cost of fencing = rs . ( 1485.78 x 5 ) = rs . 7429 . answer : option e" | a ) 4457 , b ) 4567 , c ) 4235 , d ) 4547 , e ) 7429 | e | multiply(circumface(multiply(sqrt(divide(17.56, const_pi)), const_100)), 5) | divide(n0,const_pi)|sqrt(#0)|multiply(#1,const_100)|circumface(#2)|multiply(#3,n1)| | geometry |
a contractor undertakes to built a walls in 50 days . he employs 70 peoples for the same . however after 25 days he finds that only 40 % of the work is complete . how many more man need to be employed to complete the work in time ? | "70 men complete 0.4 work in 25 days . applying the work rule , m 1 × d 1 × w 2 = m 2 × d 2 × w 1 we have , 70 × 25 × 0.6 = m 2 × 25 × 0.4 or m 2 = 70 × 25 × 0.6 / 25 × 0.4 = 105 men answera" | a ) 105 , b ) 30 , c ) 35 , d ) 20 , e ) none of these | a | divide(multiply(70, divide(subtract(const_100, 40), const_100)), divide(const_4, const_10)) | divide(const_4,const_10)|subtract(const_100,n3)|divide(#1,const_100)|multiply(n1,#2)|divide(#3,#0)| | physics |
a number is said to be prime saturated if the product of all the different positive prime factors of n is less than the square root of n . what is the greatest two digit prime saturated integer ? | "the square roots of all numbers from answer choices are between 9 and 10 , so the product of primes of the number we are looking should be less then 9 , so this number should have only 2 - s , only 3 - s , only 5 - s , or only 2 - s and 3 - s as its primes . only 96 satisfies this . answer : d ." | a ) 99 , b ) 98 , c ) 97 , d ) 96 answer , e ) 95 | d | add(multiply(const_2, const_3), subtract(const_100, const_10)) | multiply(const_2,const_3)|subtract(const_100,const_10)|add(#0,#1)| | other |
in how many no . between 10 and 40 exactly two of the digits is 4 ? | "it ' s simple can be solved by elimination of answer choices . option b and c are too large , not possible . even ae are large to have correct choice . ans : d" | a ) 10 , b ) 15 , c ) 35 , d ) 1 , e ) 45 | d | divide(divide(40, 10), 4) | divide(n1,n0)|divide(#0,n2)| | general |
10 people went to a hotel for combine dinner party 8 of them spent rs . 10 each on their dinner and rest spent 4 more than the average expenditure of all the 10 . what was the total money spent by them . | "solution : let average expenditure of 10 people be x . then , 10 x = 8 * 10 + 2 * ( x + 4 ) ; or , 10 x = 10 * 10 + 2 x + 8 ; or , x = 11 ; so , total money spent = 11 * 10 = rs . 110 . answer : option b" | a ) 1628.4 , b ) 110 , c ) 1492 , d ) 1496 , e ) none of these | b | multiply(divide(add(multiply(8, 10), multiply(subtract(10, 8), 4)), subtract(10, subtract(10, 8))), 10) | multiply(n1,n2)|subtract(n0,n1)|multiply(n3,#1)|subtract(n0,#1)|add(#0,#2)|divide(#4,#3)|multiply(n0,#5)| | general |
a carpenter worked alone for 1 day on a job that would take him 6 more days to finish . he and another carpenter completed the job in 5 more days . how many days would it have taken the second carpenter to do the complete job working alone ? | a carpenter worked only 1 day on something that takes him 6 more days . means ; carpenter finishes his work in 7 days . let his buddy finish the same task in x days . respective rates per day : 1 / 7 and 1 / x to complete 1 work : first guy worked for 5 days @ rate = 1 / 7 per day . second one worked for 5 days @ rate ... | a ) 4.5 , b ) 7.5 , c ) 9.5 , d ) 17.5 , e ) 24.5 | d | divide(multiply(5, add(1, 6)), const_2) | add(n0,n1)|multiply(n2,#0)|divide(#1,const_2) | physics |
find the least number of five digits which is exactly divisible by 12 and 18 ? | "the smallest five digit numbers are 10025 , 10080,11080 10025 is not divisible by 12 10080 is divisible by both 12 and 18 answer : b" | a ) a ) 1080 , b ) b ) 10080 , c ) c ) 10025 , d ) d ) 11080 , e ) e ) 12080 | b | add(divide(divide(multiply(floor(add(divide(subtract(multiply(const_10, const_1000), const_1), 18), const_1)), 18), const_4), const_100), const_2) | multiply(const_10,const_1000)|subtract(#0,const_1)|divide(#1,n1)|add(#2,const_1)|floor(#3)|multiply(n1,#4)|divide(#5,const_4)|divide(#6,const_100)|add(#7,const_2)| | general |
how many odd prime numbers are there less than 75 ? | "odd prime number less than 75 : 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 , 41 , 43 , 47 , 53 , 59 , 61 , 67 , 71 , 73 there is 20 the odd prime number answer is a" | a ) 20 , b ) 1 , c ) 2 , d ) 16 , e ) 17 | a | add(subtract(75, const_100), const_4) | subtract(n0,const_100)|add(#0,const_4)| | general |
a is twice as good a workman as b and they took 7 days together to do the work b alone can do it in . | "c 21 days wc = 2 : 1 2 x + x = 1 / 7 x = 1 / 21 = > 21 days" | a ) 23 days , b ) 25 days , c ) 21 days , d ) 26 days , e ) 29 days | c | multiply(divide(multiply(7, add(const_2, const_1)), const_2), const_2) | add(const_1,const_2)|multiply(n0,#0)|divide(#1,const_2)|multiply(#2,const_2)| | physics |
if x is a positive integer with fewer than 3 digits , what is the probability that c x * ( x + 1 ) is a multiple of either 4 or 5 ? | "interesting question ! also one that we should be able to answer very quickly be keeping an eye on our best friends , the answer choices . we know that x belongs to the set { 1 , 2 , 3 , . . . , 99 } . we want to know the probability c that x ( x + 1 ) is a multiple of either 4 or 5 . when will this happen ? if either... | a ) 4 / 99 , b ) 2 / 25 , c ) 8 / 99 , d ) 49 / 100 , e ) 86 / 99 | e | divide(subtract(add(multiply(divide(const_100, 4), const_2), multiply(divide(const_100, 5), const_2)), 4), subtract(const_100, 1)) | divide(const_100,n2)|divide(const_100,n3)|subtract(const_100,n1)|multiply(#0,const_2)|multiply(#1,const_2)|add(#3,#4)|subtract(#5,n2)|divide(#6,#2)| | general |
if the cost price of 65 chocolates is equal to the selling price of 50 chocolates , the gain percent is : | "explanation : solution : let c . p . of each chocolate be re . 1 . then , c . p . of 50 chocolates = rs . 50 ; s . p . of 50 chocolates = rs . 65 . . ' . gain % = 15 * 100 / 50 = 30 % answer : d" | a ) 35 % , b ) 10 % , c ) 20 % , d ) 30 % , e ) 33 % | d | divide(const_100, divide(50, subtract(65, 50))) | subtract(n0,n1)|divide(n1,#0)|divide(const_100,#1)| | gain |
a train , 155 meters long travels at a speed of 45 km / hr crosses a bridge in 30 seconds . the length of the bridge is | "explanation : assume the length of the bridge = x meter total distance covered = 155 + x meter total time taken = 30 s speed = total distance covered / total time taken = ( 155 + x ) / 30 m / s = > 45 ã — ( 10 / 36 ) = ( 155 + x ) / 30 = > 45 ã — 10 ã — 30 / 36 = 155 + x = > 45 ã — 10 ã — 10 / 12 = 155 + x = > 15 ã — ... | a ) 270 m , b ) 245 m , c ) 235 m , d ) 220 m , e ) 240 m | d | subtract(multiply(multiply(45, const_0_2778), 30), 155) | multiply(n1,const_0_2778)|multiply(n2,#0)|subtract(#1,n0)| | physics |
sachin is younger than rahul by 7 years . if the ratio of their ages is 7 : 9 , find the age of sachin | "if rahul age is x , then sachin age is x - 7 , so ( x - 7 ) / x = 7 / 9 = > 9 x - 63 = 7 x = > 2 x = 63 = > x = 31.5 so sachin age is 31.5 - 7 = 24.5 answer : b" | a ) 24.58 , b ) 24.5 , c ) 24.3 , d ) 24.9 , e ) 24.1 | b | multiply(divide(7, subtract(9, 7)), 7) | subtract(n2,n1)|divide(n0,#0)|multiply(n1,#1)| | other |
the present population of a town is 500 . population increase rate is 10 % p . a . find the population of town before 2 years ? | "p = 500 r = 10 % required population of town = p / ( 1 + r / 100 ) ^ t = 500 / ( 1 + 10 / 100 ) ^ 2 = 500 / ( 11 / 10 ) ^ 2 = 413 ( approximately ) answer is d" | a ) 512 , b ) 615 , c ) 810 , d ) 413 , e ) 123 | d | add(500, divide(multiply(500, 10), const_100)) | multiply(n0,n1)|divide(#0,const_100)|add(n0,#1)| | gain |
a perfect square is defined as the square of an integer and a perfect cube is defined as the cube of an integer . how many positive integers n are there such that n is less than 5,000 and at the same time n is a perfect square and a perfect cube ? | "if n is a perfect square and a perfect cube , then n = a ^ 6 for some integer a . the numbers are 1 ^ 6 = 1 , 2 ^ 6 = 64 , 3 ^ 6 = 729 , 4 ^ 6 = 4096 . the answer is c ." | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | c | divide(add(const_4, const_2), const_1) | add(const_2,const_4)|divide(#0,const_1)| | geometry |
the price of commodity x increases by 30 cents every year , while the price of commodity y increases by 20 cents every year . if in 2001 , the price of commodity x was $ 4.20 and the price of commodity y was $ 4.40 , in which year will commodity x cost 90 cents more than the commodity y ? | the cost of commodity x increases by 10 cents per year relative to commodity y . the price of x must gain 20 + 90 = $ 1.10 cents on commodity y , which will take 11 years . the answer is c . | a ) 2010 , b ) 2011 , c ) 2012 , d ) 2013 , e ) 2014 | c | add(2001, divide(add(divide(90, const_100), subtract(4.4, 4.2)), subtract(divide(30, const_100), subtract(4.4, 4.2)))) | divide(n5,const_100)|divide(n0,const_100)|subtract(n4,n3)|add(#0,#2)|subtract(#1,#2)|divide(#3,#4)|add(n2,#5) | general |
shawn invested one half of his savings in a bond that paid simple interest for 2 years and received $ 650 as interest . he invested the remaining in a bond that paid compound interest , interest being compounded annually , for the same 2 years at the same rate of interest and received $ 725 as interest . what was the v... | so , we know that shawn received 20 % of the amount he invested in a year . we also know that in one year shawn received $ 325 , thus 0.2 x = $ 325 - - > x = $ 1625 since , he invested equal sums in his 2 bonds , then his total savings before investing was 2 * $ 1,625 = $ 3,250 answer d | a ) 1000 , b ) 1,250 , c ) 2,250 , d ) 3,250 , e ) 4,250 | d | multiply(divide(multiply(divide(650, 2), divide(650, 2)), subtract(725, 650)), 2) | divide(n1,n0)|subtract(n3,n1)|multiply(#0,#0)|divide(#2,#1)|multiply(n0,#3) | gain |
at joel ’ s bookstore , the current inventory is 40 % historical fiction . of the historical fiction books , 40 % are new releases , while 70 % of the other books are new releases . what fraction of all new releases are the historical fiction new releases ? | "let there be 100 books in all historic fiction books = 40 % of total = 40 other books = 60 new historic fiction = 40 % of 40 = 16 other new books = 70 % of 60 = 42 total new books = 58 fraction = 16 / 58 = 8 / 29 ans : b" | a ) 4 / 25 , b ) 8 / 29 , c ) 2 / 5 , d ) 8 / 15 , e ) 2 / 3 | b | divide(divide(multiply(40, 40), const_100), add(divide(multiply(40, 40), const_100), divide(multiply(70, subtract(const_100, 40)), const_100))) | multiply(n0,n1)|subtract(const_100,n0)|divide(#0,const_100)|multiply(n2,#1)|divide(#3,const_100)|add(#2,#4)|divide(#2,#5)| | gain |
10 years ago , the age of anand was one - third the age of bala at that time . the present age of bala is 10 years more than the present age of anand . find the present age of anand ? | let the present ages of anand and bala be ' a ' and ' b ' respectively . a - 10 = 1 / 3 ( b - 10 ) - - - ( 1 ) b = a + 10 substituting b = a + 12 in first equation , a - 10 = 1 / 3 ( a + 0 ) = > 3 a - 30 = a = > 2 a = 30 = > a = 15 . answer : d | a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | d | divide(multiply(10, const_3), subtract(const_3, const_1)) | multiply(n0,const_3)|subtract(const_3,const_1)|divide(#0,#1) | general |
the average of 10 numbers is 40.2 . later it is found that two numbers have been wrongly copied . the first is 17 greater than the actual number and the second number added is 13 instead of 31 . find the correct average . | "sum of 10 numbers = 402 corrected sum of 10 numbers = 402 – 13 + 31 – 17 = 403 hence , new average = 403 ⁄ 10 = 40.3 answer a" | a ) 40.3 , b ) 40.4 , c ) 40.6 , d ) 40.8 , e ) none of the above | a | divide(subtract(add(multiply(40.2, 10), add(13, 17)), 31), 10) | add(n2,n3)|multiply(n0,n1)|add(#0,#1)|subtract(#2,n4)|divide(#3,n0)| | general |
nails and screws are manufactured at a uniform weight per nail and a uniform weight per screw . if the total weight of one screw and one nail is half that of 6 screws and one nail , then the total weight of 3 screws , and 3 nails is how many times that of 4 screws and 2 nails ? | "let the weight of nail be n and that of screw be s . . so s + w = 1 / 2 * ( 6 s + 1 n ) . . . or 1 n = 4 s . . lets see the weight of 3 s and 3 n = 3 s + 3 * 4 s = 15 s . . and weight of 4 s and 2 n = 4 s + 2 * 4 s = 12 s . . ratio = 15 s / 12 s = 15 / 12 = 5 / 4 answer : e" | a ) 4 , b ) 5 / 2 , c ) 2 , d ) 3 / 2 , e ) 5 / 4 | e | divide(add(multiply(3, const_4), 3), const_10) | multiply(n1,const_4)|add(n1,#0)|divide(#1,const_10)| | general |
the length of a room is 5.5 m and width is 3.75 m . find the cost of paving the floor by slabs at the rate of rs . 1200 per sq . metre . | "solution area of the floor = ( 5.5 × 3.75 ) m 2 = 20.625 m 2 ∴ cost of paving = rs . ( 1200 × 20.625 ) = 24750 . answer c" | a ) rs . 15000 , b ) rs . 15550 , c ) rs . 24750 , d ) rs . 16500 , e ) none of these | c | multiply(1200, multiply(5.5, 3.75)) | multiply(n0,n1)|multiply(n2,#0)| | physics |
15 men take 21 days of 8 hrs . each to do a piece of work . how many days of 6 hrs . each would it take for 21 women if 3 women do as much work as 2 men ? | let 1 man does 1 unit / hr of work 15 m in 21 days of 8 hrs will do ( 15 * 21 * 8 ) units 3 w = 2 m 1 w = ( 2 / 3 ) units / hr 21 w with 6 hrs a day will take ( 15 * 21 * 8 ) / ( 21 * 6 * ( 2 / 3 ) ) days = > 30 days answer : a | a ) 30 , b ) 20 , c ) 19 , d ) 29 , e ) 39 | a | divide(multiply(multiply(15, 21), 8), multiply(multiply(21, 6), divide(2, 3))) | divide(n6,n5)|multiply(n0,n1)|multiply(n1,n3)|multiply(n2,#1)|multiply(#0,#2)|divide(#3,#4) | physics |
4 brothers adam , bill , charles and david together contributed certain amount of money and purchased a car . the contribution made by adam is half that of the total contribution made by the other 3 brothers , contribution made by bill is one - fourth of the total contribution made by the other 3 brothers and the contr... | given that a : ( b + c + d ) = 1 : 2 . hence a contributed 1 / 3 rd of the total price . given b : ( a + c + d ) = 1 : 4 hence b contributed 1 / 5 th of the total price . given c : ( a + b + d ) = 2 : 3 hence c contributed 2 / 5 th of the total price . thus the contribution made by a , b and c = 1 / 3 + 1 / 5 + 2 / 5 =... | a ) $ 540 , b ) $ 580 , c ) $ 600 , d ) $ 650 , e ) $ 680 | d | subtract(9750, add(add(divide(9750, add(const_1, const_2)), divide(9750, add(4, const_1))), divide(9750, add(divide(3, const_2), const_1)))) | add(const_1,const_2)|add(n0,const_1)|divide(n1,const_2)|add(#2,const_1)|divide(n4,#0)|divide(n4,#1)|add(#4,#5)|divide(n4,#3)|add(#6,#7)|subtract(n4,#8) | general |
share rs . 6000 among john , jose & binoy in the ration 2 : 4 : 6 . find the amount received by john ? | "amount received by sanjay . 4 / 12 x 6000 = 2000 = ( related ratio / sum of ratio ) x total amount so , the amount received by sanjay is 2000 . e" | a ) 900 , b ) 980 , c ) 1200 , d ) 1240 , e ) 2000 | e | subtract(divide(6000, 2), divide(6000, 6)) | divide(n0,n1)|divide(n0,n3)|subtract(#0,#1)| | other |
a worker makes a toy in every 2 h . if he works for 120 h , then how many toys will he make ? | "no . of toys = 120 / 2 = 60 answer : b" | a ) 40 , b ) 60 , c ) 45 , d ) 39 , e ) none | b | divide(120, 2) | divide(n1,n0)| | physics |
find the sum of all odd numbers up to 100 . | "the given numbers are 1 , 3 , 5 , 7 , . . . , 99 . this is an a . p . with a = 1 and d = 2 . let it contain n terms . then , 1 + ( n - 1 ) x 2 = 99 or n = 50 . required sum = n / 2 ( first term + last term ) = 50 / 2 ( 1 + 99 ) = 2500 . answer is a ." | a ) 2500 , b ) 2400 , c ) 2300 , d ) 2200 , e ) 2100 | a | add(divide(divide(100, divide(divide(divide(divide(divide(100, const_2), const_2), const_2), const_2), const_2)), const_2), add(const_1, sqrt(divide(divide(100, divide(divide(divide(divide(divide(100, const_2), const_2), const_2), const_2), const_2)), const_2)))) | divide(n0,const_2)|divide(#0,const_2)|divide(#1,const_2)|divide(#2,const_2)|divide(#3,const_2)|divide(n0,#4)|divide(#5,const_2)|sqrt(#6)|add(#7,const_1)|add(#8,#6)| | general |
there are , in a certain league , 20 teams , and each team face another team for a total of 4 times . how many games are played in the season ? | "by using the formula , t [ n ( n - 1 ) / 2 ] , where t = no . of games between two teams and n = total no . of teams , we get : 760 option b ." | a ) 450 , b ) 760 , c ) 110 , d ) 180 , e ) 400 | b | multiply(multiply(subtract(20, const_1), 4), divide(20, const_2)) | divide(n0,const_2)|subtract(n0,const_1)|multiply(#1,n1)|multiply(#0,#2)| | general |
if p ( a ) = 2 / 15 , p ( b ) = 4 / 15 , and p ( a â ˆ ª b ) = 12 / 15 find p ( b | a ) | "p ( b | a ) = p ( a â ˆ ª b ) / p ( a ) p ( b | a ) = ( 12 / 15 ) / ( 2 / 15 ) = 6 . e" | a ) 1 / 2 , b ) 2 / 3 , c ) 3 / 2 , d ) 4 / 5 , e ) 6 | e | divide(divide(12, 15), divide(2, 15)) | divide(n4,n1)|divide(n0,n1)|divide(#0,#1)| | general |
the water level in a rectangular swimming pool measuring 20 feet by 25 feet is to be lowered by 6 inches . how many gallons of water must be removed ? ( 1 cu ft = 7.5 gallons ) | 6 inches = 1 / 2 feet ( there are 12 inches in a foot . ) , so 20 * 25 * 1 / 2 = 250 feet ^ 3 of water must be removed , which equals to 250 * 7.5 = 1875 gallons . answer : c . | ['a ) 100', 'b ) 250', 'c ) 1875', 'd ) 1200', 'e ) 5625'] | c | multiply(volume_rectangular_prism(20, 25, divide(6, add(const_10, const_2))), 7.5) | add(const_10,const_2)|divide(n2,#0)|volume_rectangular_prism(n0,n1,#1)|multiply(n4,#2) | geometry |
a man buys rs . 20 shares paying 9 % dividend . the man wants to have an interest of 12 % on his money . the market value of each share is : | dividend on rs 20 = 9 / 100 * 20 = 9 / 5 is 12 income 100 9 / 5 is an income on rs ( 100 / 12 * 9 / 5 ) = rs 15 answer d | a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 18 | d | multiply(divide(const_100, 12), multiply(divide(9, const_100), 20)) | divide(const_100,n2)|divide(n1,const_100)|multiply(n0,#1)|multiply(#0,#2) | gain |
a car travels first 160 km at 75 km / hr and the next 160 km at 90 km / hr . what is the average speed for the first 320 km of the tour ? | "car travels first 160 km at 75 km / hr time taken to travel first 160 km = distancespeed = 160 / 75 car travels next 160 km at 90 km / hr time taken to travel next 160 km = distancespeed = 160 / 90 total distance traveled = 160 + 160 = 2 × 160 total time taken = 160 / 75 + 160 / 90 average speed = total distance trave... | a ) 71.11 , b ) 71.12 , c ) 71.1 , d ) 71.17 , e ) 81.81 | e | multiply(320, divide(const_1, add(divide(160, 75), divide(160, 90)))) | divide(n0,n1)|divide(n0,n3)|add(#0,#1)|divide(const_1,#2)|multiply(n4,#3)| | general |
a , b and c have rs . 500 between them , a and c together have rs . 200 and b and c rs . 360 . how much does c have ? | "a + b + c = 500 a + c = 200 b + c = 360 - - - - - - - - - - - - - - a + b + 2 c = 560 a + b + c = 500 - - - - - - - - - - - - - - - - c = 60 answer : a" | a ) 60 , b ) 78 , c ) 267 , d ) 29 , e ) 27 | a | subtract(add(200, 360), 500) | add(n1,n2)|subtract(#0,n0)| | general |
if each participant of a chess tournament plays exactly one game with each of the remaining participants , then 190 games will be played during the tournament . what is the number of participants ? | "let n be the number of participants . the number of games is nc 2 = n * ( n - 1 ) / 2 = 190 n * ( n - 1 ) = 380 = 20 * 19 ( trial and error ) the answer is d ." | a ) 14 , b ) 16 , c ) 18 , d ) 20 , e ) 22 | d | divide(add(sqrt(add(multiply(multiply(190, const_2), const_4), const_1)), const_1), const_2) | multiply(n0,const_2)|multiply(#0,const_4)|add(#1,const_1)|sqrt(#2)|add(#3,const_1)|divide(#4,const_2)| | general |
a certain number of men can finish a piece of work in 10 days . however , if there were 10 less men it will take 10 days more for the work to be finished . how many men were there originally ? | "let x be the number of men originally thus , 10 x = ( 10 + 10 ) ( x - 10 ) 10 x = 20 ( x - 10 ) x = 2 ( x - 10 ) x = 20 answer : d" | a ) 10 men , b ) 30 men , c ) 40 men , d ) 20 men , e ) none of these | d | divide(10, subtract(const_1, divide(10, add(10, 10)))) | add(n0,n1)|divide(n0,#0)|subtract(const_1,#1)|divide(n1,#2)| | physics |
find the average of all numbers between 4 and 34 which are divisible by 5 | "explanation : average = ( 5 + 10 + 15 + 20 + 25 + 30 ) / 5 = 105 / 5 = 21 option b" | a ) 20 , b ) 21 , c ) 25 , d ) 30 , e ) 55 | b | divide(add(add(add(multiply(5, const_3), add(5, multiply(5, const_2))), multiply(5, const_4)), multiply(add(const_4, const_1), 5)), 5) | add(const_1,const_4)|multiply(n2,const_2)|multiply(n2,const_3)|multiply(n2,const_4)|add(n2,#1)|multiply(n2,#0)|add(#4,#2)|add(#6,#3)|add(#7,#5)|divide(#8,n2)| | general |
how can a cake ( circular ) be cut into 8 pieces by making just 3 cuts ? | explanation : cut the cake using three mutually perpendicular planes . it leaves 8 pieces , one each in each octant . answera | ['a ) 21', 'b ) 787', 'c ) 37', 'd ) 26', 'e ) 01'] | a | multiply(divide(8, 3), 8) | divide(n0,n1)|multiply(n0,#0) | geometry |
the bankers discount of a certain sum of money is rs . 72 and the true discount on the same sum for the same time is rs . 60 . the sum due is : | "sum = ( b . d * t . d ) / ( b . d - t . d ) ( 72 * 60 ) / 72 - 60 ; 360 answer : a" | a ) rs . 360 , b ) rs . 432 , c ) rs . 540 , d ) rs . 1080 , e ) rs . 1280 | a | divide(multiply(60, 72), subtract(72, 60)) | multiply(n0,n1)|subtract(n0,n1)|divide(#0,#1)| | gain |
the length of a room is 5.5 m and width is 3.75 m . find the cost of paying the floor by slabs at the rate of rs . 1000 per sq . metre . | "solution area of the floor = ( 5.5 x 3.75 ) m ² = 20.635 m ² cost of paying = rs . ( 1000 x 20.625 ) = rs . 20625 . answer d" | a ) rs . 15,000 , b ) rs . 15,500 , c ) rs . 15,600 , d ) rs . 20,625 , e ) none | d | multiply(1000, multiply(5.5, 3.75)) | multiply(n0,n1)|multiply(n2,#0)| | physics |
find the value of x in [ ( 17.28 / x ) / ( 3.6 * 0.2 ) ] = 2 | ( 17.28 / x ) = 2 * 3.6 * 0.2 x = ( 17.28 / 1.44 ) = ( 1728 / 14 ) = 12 . option b | a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | b | divide(17.28, multiply(multiply(3.6, 0.2), 2)) | multiply(n1,n2)|multiply(n3,#0)|divide(n0,#1) | general |
set a contains all the even numbers between 22 and 70 inclusive . set b contains all the even numbers between 62 and 110 inclusive . what is the difference between the sum of elements of set b and the sum of the elements of set a ? | "each term in set b is 40 more than the corresponding term in set a . the difference of the sums = 25 * 40 = 1000 . the answer is c ." | a ) 600 , b ) 800 , c ) 1000 , d ) 1200 , e ) 1400 | c | multiply(subtract(62, 22), add(divide(subtract(70, 22), const_2), const_1)) | subtract(n1,n0)|subtract(n2,n0)|divide(#0,const_2)|add(#2,const_1)|multiply(#3,#1)| | general |
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