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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the price of a t . v . set worth rs . 10000 is to be paid in 20 installments of rs . 1000 each . if the rate of interest be 6 % per annum , and the first installment be paid at the time of purchase , then the value of the last installment covering the interest as well will be ? | "money paid in cash = rs . 1000 balance payment = ( 10000 - 1000 ) = rs . 9000 answer : a" | a ) 9000 , b ) 26699 , c ) 96000 , d ) 19000 , e ) 26711 | a | subtract(10000, 1000) | subtract(n0,n2)| | gain |
find the principle on a certain sum of money at 5 % per annum for 3 years if the amount being rs . 1120 ? | "1120 = p [ 1 + ( 5 * 3 ) / 100 ] p = 974 answer : a" | a ) 974 , b ) 1067 , c ) 1977 , d ) 1056 , e ) 1097 | a | divide(1120, add(const_1, divide(multiply(5, 3), const_100))) | multiply(n0,n1)|divide(#0,const_100)|add(#1,const_1)|divide(n2,#2)| | gain |
a water tank is two - fifth full . pipe a can fill a tank in 10 minutes and pipe b can empty it in 6 minutes . if both the pipes are open , how long will it take to empty or fill the tank completely ? | "clearly , pipe b is faster than pipe a and so , the tank will be emptied . part to be emptied = 2 / 5 part emptied by ( a + b ) in 1 minute = ( 1 / 6 - 1 / 10 ) = 1 / 15 1 / 15 : 2 / 5 : : 1 : x x = ( 2 / 5 * 1 * 15 ) = 6 min . so , the tank will be emptied in 6 min . answer : a" | a ) 6 min , b ) 4 min , c ) 2 min , d ) 5 min , e ) 8 min | a | multiply(divide(const_1, add(const_1, const_4)), divide(const_1, subtract(divide(const_1, 6), divide(const_1, 10)))) | add(const_1,const_4)|divide(const_1,n1)|divide(const_1,n0)|divide(const_1,#0)|subtract(#1,#2)|divide(const_1,#4)|multiply(#3,#5)| | physics |
how many two digit prime numbers are there in which both digits are prime numbers ? ( for example , 23 is one of these numbers but 31 is not , since 1 is not a prime number . ) | the second digit can only be 3 or 7 , so the choice quickly narrows down to 23 , 27 , 33 , 37 , 53 , 57 , 73 , and 77 . of these , 27 , 33 , and 57 are divisible by 3 , and 77 by 7 , leaving 23 , 37 , 53 , and 73 . it is easy to see that none of these is divisible by 2 , 3 , 5 , or 7 , and there is no need to look at g... | a ) 3 , b ) 4 , c ) 5 , d ) 8 , e ) 15 | b | divide(reminder(31, 23), const_2) | reminder(n1,n0)|divide(#0,const_2) | other |
a candidate got 35 % of the votes polled and he lost to his rival by 2460 votes . how many votes were cast ? | "35 % - - - - - - - - - - - l 65 % - - - - - - - - - - - w - - - - - - - - - - - - - - - - - - 30 % - - - - - - - - - - 2460 100 % - - - - - - - - - ? = > 8200 answer : d" | a ) 7500 , b ) 3388 , c ) 2665 , d ) 8200 , e ) 2661 | d | divide(2460, subtract(subtract(const_1, divide(35, const_100)), divide(35, const_100))) | divide(n0,const_100)|subtract(const_1,#0)|subtract(#1,#0)|divide(n1,#2)| | gain |
{ 8 , 46 , 53 , 127 } what number should be inserted in the set of 7 values to get a median of 9 ? | median is the middle number of a sorted distribution . in case of even number of items , median will be the average of middle two values . the question asks fora new number that will cause the median of new set at 9 . mentionable that given set is already sorted and number of items of new set will be 7 + 1 = 8 . to get... | a ) 20 , b ) 14 , c ) 16 , d ) 5 , e ) 8 | b | divide(127, 9) | divide(n3,n5) | general |
the ratio 2 : 3 expressed as a percent equals | "solution 2 : 3 = 2 / 3 = ( 2 / 3 x 100 ) % . = 66.6 % . answer c" | a ) 12.5 % , b ) 40 % , c ) 67 % , d ) 125 % , e ) none | c | multiply(divide(2, 3), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
two brother x and y appeared for an exam . the probability of selection of x is 1 / 7 and that of b is 2 / 5 . find the probability that both of them are selected . | "explanation : let a be the event that x is selected and b is the event that y is selected . p ( a ) = 1 / 7 , p ( b ) = 2 / 5 . let c be the event that both are selected . p ( c ) = p ( a ) ã — p ( b ) as a and b are independent events : = ( 1 / 7 ) ã — ( 2 / 5 ) = 2 / 35 answer : d ) 2 / 35" | a ) 2 / 69 , b ) 2 / 60 , c ) 2 / 63 , d ) 2 / 35 , e ) 2 / 38 | d | multiply(divide(1, 7), divide(2, 5)) | divide(n0,n1)|divide(n2,n3)|multiply(#0,#1)| | general |
3 consecutive odd integers are in increasing order such that the sum of the last two integers is 17 more than the first integer . find the 3 integers ? | explanation : let the three consecutive odd integers be x , x + 2 and x + 4 respectively . x + 4 + x + 2 = x + 17 = > x = 11 hence three consecutive odd integers are 11 , 13 and 15 answer : a | a ) 11 , 1315 , b ) 7 , 2 , 10 , c ) 7 , 9 , 10 , d ) 7 , 9 , 11 , e ) 7 , 9 , 29 | a | subtract(multiply(divide(subtract(subtract(17, const_1), add(3, const_1)), const_2), const_2), const_1) | add(n0,const_1)|subtract(n1,const_1)|subtract(#1,#0)|divide(#2,const_2)|multiply(#3,const_2)|subtract(#4,const_1) | general |
p is able to do a piece of work in 5 days and q can do the same work in 10 days . if they can work together for 3 days , what is the fraction of work completed ? | "explanation : amount of work p can do in 1 day = 1 / 5 amount of work q can do in 1 day = 1 / 10 amount of work p and q can do in 1 day = 1 / 5 + 1 / 10 = 3 / 10 amount of work p and q can together do in 3 days = 3 × ( 3 / 10 ) = 9 / 10 answer : option e" | a ) 13 / 10 , b ) 4 / 5 , c ) 11 / 10 , d ) 7 / 10 , e ) 9 / 10 | e | subtract(const_1, multiply(add(divide(const_1, 10), divide(const_1, 5)), 3)) | divide(const_1,n1)|divide(const_1,n0)|add(#0,#1)|multiply(n2,#2)|subtract(const_1,#3)| | physics |
the sum of three consecutive multiples of 3 is 81 . what is the largest number ? | "let the numbers be 3 x , 3 x + 3 and 3 x + 6 . then , 3 x + ( 3 x + 3 ) + ( 3 x + 6 ) = 81 9 x = 72 x = 8 largest number = 3 x + 6 = 30 . answer : c" | a ) 24 , b ) 28 , c ) 30 , d ) 32 , e ) 35 | c | add(add(power(add(add(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(3, const_10), const_2), const_4), const_2), power(add(di... | subtract(n0,const_10)|subtract(#0,const_2)|divide(#1,const_4)|add(#2,const_2)|power(#2,const_2)|add(#3,const_2)|power(#3,const_2)|add(#5,const_2)|add(#4,#6)|power(#5,const_2)|power(#7,const_2)|add(#9,#10)|add(#11,#8)| | general |
what is the area of square field whose side of length 15 m ? | 15 * 15 = 225 sq m answer : a | ['a ) 225 sq m', 'b ) 186 sq m', 'c ) 586 sq m', 'd ) 287 sq m', 'e ) 296 sq m'] | a | square_area(15) | square_area(n0) | geometry |
how many liters of water must be evaporated from 50 liters of a 4 percent sugar solution to get a 10 percent sugar solution ? | "let x be the amount that needs to be evaporated . 0.04 ( 50 ) = 0.1 ( 50 - x ) 0.1 x = 5 - 2 x = 3 / 0.1 = 30 liters the answer is e ." | a ) 15 , b ) 20 , c ) 25 , d ) 28 , e ) 30 | e | subtract(50, multiply(divide(50, const_100), 10)) | divide(n0,const_100)|multiply(n2,#0)|subtract(n0,#1)| | gain |
at what rate percent per annum will a sum of money double in 16 years ? | "let principle = p s . i . = p t = 16 yrs rate = 100 * p / p * 16 = 25 / 4 % = 6.25 % answer is c" | a ) 10.6 % , b ) 3.65 % , c ) 6.25 % , d ) 8.32 % , e ) 11 % | c | divide(const_100, 16) | divide(const_100,n0)| | gain |
a hat company ships its hats , individually wrapped , in 8 - inch by 10 - inch by 12 - inch boxes . each hat is valued at $ 7.50 . if the company ’ s latest order required a truck with at least 240,000 cubic inches of storage space in which to ship the hats in their boxes , what was the minimum value of the order ? | "number of boxes = total volume / volume of one box = 240,000 / ( 8 * 10 * 12 ) = 250 one box costs 7.50 , so 250 box will cost = 250 * 7.5 = 1875 b is the answer" | a ) $ 960 , b ) $ 1,875 , c ) $ 1,725 , d ) $ 2,050 , e ) $ 2,250 | b | divide(multiply(divide(multiply(add(add(multiply(const_3, const_100), multiply(8, 10)), const_4), const_1000), multiply(multiply(8, 10), 12)), 7.50), const_1000) | multiply(const_100,const_3)|multiply(n0,n1)|multiply(n0,n1)|add(#0,#1)|multiply(n2,#2)|add(#3,const_4)|multiply(#5,const_1000)|divide(#6,#4)|multiply(n3,#7)|divide(#8,const_1000)| | general |
10 business executives and 7 chairmen meet at a conference . if each business executive shakes the hand of every other business executive and every chairman once , and each chairman shakes the hand of each of the business executives but not the other chairmen , how many handshakes would take place ? | "there are 10 business exec and in each handshake 2 business execs are involved . hence 10 c 2 = 45 also , each of 10 exec will shake hand with every 7 other chairmen for total of 70 handshake . total = 45 + 70 = 115 ans : c" | a ) 144 , b ) 131 , c ) 115 , d ) 90 , e ) 45 | c | add(divide(multiply(10, subtract(10, const_1)), const_2), multiply(10, 7)) | multiply(n0,n1)|subtract(n0,const_1)|multiply(n0,#1)|divide(#2,const_2)|add(#3,#0)| | geometry |
in what ratio must water be mixed with milk to gain 16 ( 2 / 3 ) % on selling the mixture at cost price ? | "explanation : let c . p . of 1 litre milk be re . 1 . s . p . of 1 litre of mixture = re . 1 , gain = 50 / 3 % c . p . of 1 litre of mixture = ( 100 x 3 / 350 x 1 ) = 6 / 7 by the rule of alligation , we have : ratio of water and milk = 1 / 7 : 6 / 7 = 1 : 6 . answer is a" | a ) 1 : 6 , b ) 6 : 1 , c ) 2 : 3 , d ) 4 : 3 , e ) 3 : 2 | a | divide(add(16, divide(2, 3)), const_100) | divide(n1,n2)|add(n0,#0)|divide(#1,const_100)| | gain |
two pipes a and b can fill a tank in 2 and 6 minutes respectively . if both the pipes are used together , then how long will it take to fill the tank ? | "part filled by the first pipe in 1 minute = 1 / 2 part filled by the second pipe in 1 minute = 1 / 6 net part filled by pipe a and pipe b in 1 minute = ( 1 / 2 ) + ( 1 / 6 ) = 2 / 3 i . e , pipe a and b together can fill the tank in 3 / 2 minutes = 1.5 minutes answer is c" | a ) 2.5 min , b ) 1.8 min , c ) 1.5 min , d ) 3.5 min , e ) 2 min | c | add(multiply(2, const_100), multiply(multiply(subtract(const_1, multiply(add(divide(const_1, 2), divide(const_1, 6)), const_2)), 2), const_60)) | divide(const_1,n0)|divide(const_1,n1)|multiply(n0,const_100)|add(#0,#1)|multiply(#3,const_2)|subtract(const_1,#4)|multiply(n0,#5)|multiply(#6,const_60)|add(#2,#7)| | physics |
a train requires 8 seconds to pass a pole while it requires 18 seconds to cross a stationary train which is 400 mtrs long . find the speed of the train . | in 8 s the train crosses the pole and in 18 sec the train crosses one more stationary train in 10 sec the train travels a distance of 400 mtrs speed = 400 / 10 = 40 m / s = 40 ( 3600 / 1000 ) = 40 * 18 / 5 = 144 kmph answer : a | a ) 144 kmph , b ) 124 kmph , c ) 134 kmph , d ) 150 kmph , e ) 160 kmph | a | multiply(divide(400, subtract(18, 8)), const_3_6) | subtract(n1,n0)|divide(n2,#0)|multiply(#1,const_3_6) | physics |
there are 6 people in the elevator . their average weight is 154 lbs . another person enters the elevator , and increases the average weight to 151 lbs . what is the weight of the 7 th person . | "solution average of 7 people after the last one enters = 151 . â ˆ ´ required weight = ( 7 x 151 ) - ( 6 x 154 ) = 1057 - 924 = 133 . answer a" | a ) 133 , b ) 168 , c ) 189 , d ) 190 , e ) 200 | a | subtract(multiply(151, 7), multiply(6, 154)) | multiply(n2,n3)|multiply(n0,n1)|subtract(#0,#1)| | general |
for any number z , z * is defined as the greatest positive even integer less than or equal to y . what is the value of 6.25 – 6.25 * ? | "since z * is defined as the greatest positive even integer less than or equal to z , then 6.25 * = 6 ( the greatest positive even integer less than or equal to 6.25 is 6 ) . hence , 6.25 – 6.25 * = 6.25 - 6 = 0.25 answer : d ." | a ) 1.5 , b ) 0.5 , c ) 6.25 , d ) 0.25 , e ) 6.0 | d | subtract(6.25, floor(6.25)) | floor(n0)|subtract(n0,#0)| | general |
a river boat leaves silver town and travels upstream to gold town at an average speed of 3 kilometers per hour . it returns by the same route at an average speed of 7 kilometers per hour . what is the average speed for the round - trip in kilometers per hour ? | pick a number which is lcm of 7 and 3 = 21 . upstream time = 21 / 3 = 7 hrs downstream time = 21 / 7 = 3 hrs total time = 10 hrs total distance = 42 average speed = 42 / 10 = 4.2 km / hr | a ) 4.2 , b ) 7.1 , c ) 7.2 , d ) 7.5 , e ) 8.0 | a | divide(multiply(multiply(3, 7), const_2), add(3, 7)) | add(n0,n1)|multiply(n0,n1)|multiply(#1,const_2)|divide(#2,#0) | physics |
6 friends visited a book stall and brought 6 books , and everyone decided to contribute equally to the total bill of $ 400 . if one of the friends had a coupon for 5 % off the total bill , and if each friend still contributed equally after the coupon was applied to the bill , how much did each friend pay ? | at the non - discounted price , each friend would pay $ 66.66 , as $ 400 divided by 6 friends is $ 66.66 per friend . but if the bill is 5 % off , then each friend would pay 5 % less . 5 % of $ 66.66 is $ 3.33 , so each friend saves $ 3.33 and pays the remaining $ 63.33 correct option : option c | a ) 62.12 , b ) 63 , c ) 63.33 , d ) 64 , e ) 65 | c | subtract(subtract(multiply(divide(400, add(add(6, 6), 5)), const_3), const_4), const_3) | add(n0,n0)|add(n3,#0)|divide(n2,#1)|multiply(#2,const_3)|subtract(#3,const_4)|subtract(#4,const_3) | general |
the ages of two persons differ by 32 years . if 5 years ago , the elder one be 5 times as old as the younger one , their present ages ( in years ) are respectively | "explanation : let their ages be x and ( x + 32 ) years . 5 ( x - 5 ) = ( x + 32 - 5 ) or 4 x = 52 or x = 14 . their present ages are 46 years and 14 years option b" | a ) 20,20 , b ) 46,14 , c ) 25,15 , d ) 30,10 , e ) none of these | b | subtract(add(divide(multiply(32, 5), subtract(5, const_1)), 5), 32) | multiply(n0,n1)|subtract(n1,const_1)|divide(#0,#1)|add(n1,#2)|subtract(#3,n0)| | general |
let f ( x ) = x ^ 2 + bx + c . if f ( 5 ) = 0 and f ( - 3 ) = 0 , then b + c = | "f ( x ) = x ^ 2 + bx + c . if f ( 5 ) = 0 and f ( - 3 ) = 0 , then b + c = f ( 5 ) = 0 = 25 + 5 b + c - - - taking 25 to the other side - > 5 b + c = - 25 f ( - 3 ) = 0 = 9 - 3 b + c - - - taking - 3 b + c to the other side - > 3 b - c = 9 when we add these 2 equations , we get 8 b = - 16 - - - > b = - 2 and while sub... | a ) 18 , b ) - 17 , c ) - 15 , d ) - 21 , e ) - 24 | b | negate(divide(subtract(power(3, 2), 5), add(3, 5))) | add(n3,n1)|power(n3,n0)|subtract(#1,n1)|divide(#2,#0)|negate(#3)| | general |
the average weight of 8 person ' s increases by 5 kg when a new person comes in place of one of them weighing 35 kg . what might be the weight of the new person ? | "total weight increased = ( 8 x 5 ) kg = 40 kg . weight of new person = ( 35 + 40 ) kg = 75 kg . answer : c" | a ) 80 kg , b ) 85 kg , c ) 75 kg , d ) 100 kg , e ) 110 kg | c | add(multiply(8, 5), 35) | multiply(n0,n1)|add(n2,#0)| | general |
the average marks of a class of 22 students is 40 and that of another class of 28 students is 60 . find the average marks of all the students ? | "sum of the marks for the class of 22 students = 22 * 40 = 880 sum of the marks for the class of 28 students = 28 * 60 = 1680 sum of the marks for the class of 50 students = 880 + 1680 = 2560 average marks of all the students = 2560 / 50 = 51.2 answer : a" | a ) 51.2 , b ) 59.5 , c ) 52.8 , d ) 52.5 , e ) 52.1 | a | divide(add(multiply(22, 40), multiply(28, 60)), add(22, 28)) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0)| | general |
y = x ^ 2 + bx + 512 cuts the x axis at ( h , 0 ) and ( k , 0 ) . if h and k are integers , what is the least value of b ? | as the curve cuts the x - axis at ( h , 0 ) and ( k , 0 ) . therefore h , k are the roots of the quadratic equation . for the quadratic equation is in the form of ax ^ 2 + bx + c = 0 , the product of the roots = c / a = 512 / 1 = 256 and the sum of the roots = - b / a = - b 512 can be expressed as product of two number... | a ) - 512 , b ) - 257 , c ) - 256 , d ) - 513 , e ) 128 | d | subtract(negate(512), const_1) | negate(n1)|subtract(#0,const_1) | general |
the slant height of a cone is 10 cm and radius of the base is 5 cm , find the curved surface of the cone . | "π * 5 * 10 = 157 answer : d" | a ) 250 , b ) 170 , c ) 148 , d ) 157 , e ) 150 | d | multiply(multiply(const_pi, 5), 10) | multiply(n1,const_pi)|multiply(n0,#0)| | geometry |
find the area of a parallelogram with base 30 cm and height 12 cm ? | "area of a parallelogram = base * height = 30 * 12 = 360 cm 2 answer : b" | a ) 290 cm 2 , b ) 360 cm 2 , c ) 270 cm 2 , d ) 280 cm 2 , e ) 260 cm 2 | b | multiply(30, 12) | multiply(n0,n1)| | geometry |
if 45 - [ 28 - { 37 - ( 15 - * ) } ] = 57 , then * is equal to : | "45 - [ 28 - { 37 - ( 15 - * ) } ] = 57 = > 45 - [ 28 - { 37 - 15 + * } ] = 57 45 - [ 28 - 37 + 15 - * ] = 57 = > 45 [ 43 - 37 - * ] = 57 45 - [ 6 - * ] = 57 = > 45 - 6 + * = 57 39 + * = 57 = > * = 57 - 39 = 18 answer : c" | a ) - 29 , b ) - 19 , c ) 18 , d ) 29 , e ) 39 | c | subtract(57, subtract(45, add(subtract(28, 37), 15))) | subtract(n1,n2)|add(n3,#0)|subtract(n0,#1)|subtract(n4,#2)| | general |
in a kilometer race , a beats b by 51 meters or 11 seconds . what time does a take to complete the race ? | "time taken by b run 1000 meters = ( 1000 * 11 ) / 51 = 215 sec . time taken by a = 215 - 11 = 204 sec . answer : d" | a ) 22 , b ) 190 , c ) 277 , d ) 204 , e ) 208 | d | subtract(divide(multiply(const_1, const_1000), divide(51, 11)), 11) | divide(n0,n1)|multiply(const_1,const_1000)|divide(#1,#0)|subtract(#2,n1)| | physics |
how many three - digit numbers that do not contain the digits 2 or 5 are there ? | "we can have 7 digits ( 1 , 3,4 , 6,7 , 8,9 ) for the first place ( hundred ' s place ) . and similarly 8 digits for tenth ' s and unit digit . ( 0,1 , 3,4 , 6,7 , 8,9 ) so in total 7 * 8 * 8 = 488 hence b" | a ) 488 , b ) 996 , c ) 228 , d ) 456 , e ) 1986 | b | multiply(multiply(multiply(const_10, const_10), subtract(const_10, const_1)), 5) | multiply(const_10,const_10)|subtract(const_10,const_1)|multiply(#0,#1)|multiply(n1,#2)| | general |
a person borrows rs . 5000 for 2 years at 4 % p . a . simple interest . he immediately lends it to another person at 6 % p . a for 2 years . find his gain in the transaction per year . | "explanation : the person borrows rs . 5000 for 2 years at 4 % p . a . simple interest simple interest that he needs to pay = prt / 100 = 5000 × 4 × 2 / 100 = 400 he also lends it at 6 % p . a for 2 years simple interest that he gets = prt / 100 = 5000 × 6 × 2 / 100 = 600 his overall gain in 2 years = rs . 600 - rs . 4... | a ) 100 , b ) 150 , c ) 225 , d ) 112.5 , e ) 212.5 | a | divide(subtract(divide(multiply(multiply(5000, 6), 2), const_100), divide(multiply(multiply(5000, 4), 2), const_100)), 2) | multiply(n0,n3)|multiply(n0,n2)|multiply(n1,#0)|multiply(n1,#1)|divide(#2,const_100)|divide(#3,const_100)|subtract(#4,#5)|divide(#6,n1)| | gain |
shop offered 25 % offer for every shirt , smith bought a shirt at rs . 560 . and what was the shop ' s original selling price ? | sp * ( 75 / 100 ) = 560 sp = 7.46 * 100 = > cp = 746 answer : c | a ) 500 , b ) 550 , c ) 746 , d ) 700 , e ) 750 | c | divide(560, subtract(const_1, divide(25, const_100))) | divide(n0,const_100)|subtract(const_1,#0)|divide(n1,#1) | gain |
what is the least number of square tiles required to pave the floor of a room 10 m 17 cm long and 9 m 9 cm broad ? | "solution length of largest tile = h . c . f . of 1017 cm & 909 cm = 9 cm . area of each tile = ( 9 x 9 ) cm 2 ∴ required number of tiles = [ 1017 x 909 / 9 x 9 ] = 11413 . answer c" | a ) 724 , b ) 804 , c ) 11413 , d ) 844 , e ) none | c | divide(multiply(add(multiply(9, const_100), 9), multiply(10, const_100)), power(9, const_2)) | multiply(n2,const_100)|multiply(n0,const_100)|power(n3,const_2)|add(n3,#0)|multiply(#3,#1)|divide(#4,#2)| | physics |
if a person walks at 15 km / hr instead of 9 km / hr , he would have walked 20 km more . the time traveled by him is ? | "let the actual distance traveled be x km . then , x / 9 = ( x + 20 ) / 15 5 x - 3 x = 60 = > x = 30 km . travel time is = 30 / 9 hours = 10 / 3 hours answer : c" | a ) 8 hours , b ) 9 / 5 hours , c ) 10 / 3 hours , d ) 6 hours , e ) 11 / 3 hours | c | multiply(9, divide(20, subtract(15, 9))) | subtract(n0,n1)|divide(n2,#0)|multiply(n1,#1)| | general |
the sum of 3 consecutive odd natural numbers , each divisible by 3 is 72 . what is the largest among them ? | let the 3 consecutive odd numbers each divisible by 3 be x , x + 3 , x + 6 given , x + x + 3 + x + 6 = 72 3 x = 63 x = 21 the numbers are 21 , 24 , 27 therefore the largest number is 27 answer c . | a ) 21 , b ) 24 , c ) 27 , d ) 36 , e ) 45 | c | add(divide(subtract(72, add(add(const_3, const_3), 3)), 3), add(const_3, const_3)) | add(const_3,const_3)|add(n0,#0)|subtract(n2,#1)|divide(#2,n0)|add(#0,#3) | general |
if sum of two no . is 12 and subtraction to that is 4 . so how much min . should added to smaller no . to make it even ? | let smaller no . be x . bigger no . = 12 - x . 12 - x - x = 4 . 12 - 2 x = 4 2 x = 8 , x = 4 . so 4 is itself even no . so , zero should be added . answer e | a ) 1 , b ) 2 , c ) 0.5 , d ) 1.5 , e ) 0 | e | multiply(divide(subtract(12, 4), const_2), const_1) | subtract(n0,n1)|divide(#0,const_2)|multiply(#1,const_1) | general |
consider a lady took a loan from a bank at the rate of 12 % p . a . simple interest . after 3 years she had to pay rs . 5400 interest only for the period . the principal amount borrowed by her was | "explanation : principal = rs . ( 100 × 5400 / 12 × 3 ) = > rs . 15,000 . answer : c" | a ) rs . 2000 , b ) rs . 10000 , c ) rs . 15000 , d ) rs . 20000 , e ) none of these | c | divide(5400, divide(multiply(3, 12), const_100)) | multiply(n0,n1)|divide(#0,const_100)|divide(n2,#1)| | gain |
a store reported total sales of $ 385 million for february of this year . if the total sales for the same month last year was $ 320 million , approximately what was the percent increase q in sales ? | "last year ' s sales = $ 320 million ; this year ' s sales = $ 385 million ; increase q = $ 65 million . now , 20 % of $ 320 million is $ 64 million , which is very close to actual increase of $ 65 million . answer : c ." | a ) 2 % , b ) 17 % , c ) 20 % , d ) 65 % , e ) 83 % | c | multiply(divide(subtract(385, 320), 320), const_100) | subtract(n0,n1)|divide(#0,n1)|multiply(#1,const_100)| | general |
one computer can upload 120 megabytes worth of data in 6 seconds . two computers , including this one , working together , can upload 1300 megabytes worth of data in 42 seconds . how long would it take for the second computer , working on its own , to upload 120 megabytes of data ? | "since the first computer can upload 120 megabytes worth of data in 6 seconds then in 6 * 7 = 42 seconds it can upload 7 * 120 = 840 megabytes worth of data , hence the second compute in 42 seconds uploads 1300 - 840 = 460 megabytes worth of data . the second computer can upload 120 megabytes of data in 9.13 seconds . ... | a ) 6 , b ) 7 , c ) 9.13 , d ) 11 , e ) 13 | c | subtract(divide(1300, 42), divide(120, 6)) | divide(n2,n3)|divide(n0,n1)|subtract(#0,#1)| | physics |
set k contains every multiple of 6 from 18 to 306 inclusive . if w is the median of set k , and x is the average ( arithmetic mean ) of set k , what is the value of w - x ? | set k = { 18 , 24,30 , . . . . 306 } let the no . of terms be n . so we know the ap formulae to calculate the the tn term : tn = a + ( n - 1 ) * d where d = common difference of the terms . 306 = 18 + ( n - 1 ) * 6 n = 49 . so the set k consist of n terms . the median of the set having odd nos of elements is ( n + 1 ) ... | a ) - 6 , b ) - 3 , c ) 0 , d ) 3 , e ) 6 | c | subtract(add(add(multiply(divide(divide(subtract(306, 18), 6), const_2), 6), 18), 6), add(multiply(divide(divide(subtract(306, 18), 6), const_2), 6), 18)) | subtract(n2,n1)|divide(#0,n0)|divide(#1,const_2)|multiply(n0,#2)|add(n1,#3)|add(n0,#4)|subtract(#5,#4) | general |
# p is defined as 2 p + 20 for any number p . what is p , if # ( # ( # p ) ) = 4 ? | # p = 2 p + 20 - - - > # ( # p ) = 2 ( 2 p + 20 ) + 20 = 4 p + 60 and thus # ( 4 p + 60 ) = 2 ( 4 p + 60 ) + 20 = 8 p + 140 = 4 - - - > 8 p = - 136 - - - > p = - 17 , d is the correct answer . | a ) – 108 , b ) – 44 , c ) 10 , d ) - 17 , e ) 18 | d | divide(add(add(add(multiply(multiply(20, 2), 2), multiply(20, 2)), 20), 4), multiply(4, 2)) | multiply(n1,n0)|multiply(n2,n0)|multiply(n0,#0)|add(#2,#0)|add(n1,#3)|add(n2,#4)|divide(#5,#1)| | general |
if n = 3 ^ 0.15 and n ^ b = 9 , b must equal | "15 / 100 = 3 / 20 n = 3 ^ 3 / 20 n ^ b = 3 ^ 2 ( 3 ^ 3 / 20 ) ^ b = 3 ^ 2 b = 40 / 3 answer : e" | a ) 3 / 80 , b ) 3 / 5 , c ) 4 , d ) 5 / 3 , e ) 40 / 3 | e | divide(log(9), log(power(3, 0.15))) | log(n2)|power(n0,n1)|log(#1)|divide(#0,#2)| | general |
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 36 hours and 16 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 = √ 16 : √ 36 = 4 : 6 answer : c" | a ) 4 : 5 , b ) 4 : 3 , c ) 4 : 6 , d ) 4 : 9 , e ) 4 : 2 | c | divide(sqrt(16), sqrt(36)) | sqrt(n1)|sqrt(n0)|divide(#0,#1)| | physics |
john makes $ 55 a week from his job . he earns a raise andnow makes $ 60 a week . what is the % increase ? | "increase = ( 5 / 55 ) * 100 = ( 1 / 11 ) * 100 = 9.09 % . c" | a ) 16 % , b ) 9.66 % , c ) 9.09 % , d ) 9.98 % , e ) 17 % | c | multiply(divide(subtract(60, 55), 55), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
what is the sum of all the prime numbers greater than 10 but less than 40 ? | "required sum = ( 11 + 13 + 17 + 19 + 23 + 29 + 31 + 37 ) = 180 note : 1 is not a prime number answer c" | a ) 190 , b ) 189 , c ) 180 , d ) 199 , e ) 1981 | c | add(add(add(add(add(add(const_12, const_2), const_1), add(add(const_12, const_2), add(add(add(add(add(const_2, const_4), const_4), subtract(const_10, const_1)), add(add(const_2, const_4), const_4)), add(const_10, const_2)))), add(add(add(const_12, const_2), const_1), const_1)), 10), add(const_2, const_4)) | add(const_12,const_2)|add(const_2,const_4)|add(const_10,const_2)|subtract(const_10,const_1)|add(#0,const_1)|add(#1,const_4)|add(#5,#3)|add(#4,const_1)|add(#6,#5)|add(#8,#2)|add(#0,#9)|add(#4,#10)|add(#11,#7)|add(n0,#12)|add(#13,#1)| | general |
the value of a machine depreciates at the rate of 10 % every year . it was purchased 3 years ago . if its present value is rs . 8748 , its purchase price was | explanation : = rs . 12000 answer : b | a ) rs . 12008 , b ) rs . 12000 , c ) rs . 12002 , d ) rs . 1229 , e ) rs . 12021 | b | divide(8748, subtract(const_1, multiply(divide(10, const_100), 3))) | divide(n0,const_100)|multiply(n1,#0)|subtract(const_1,#1)|divide(n2,#2) | gain |
find 95 × × 97 | "here both numbers are less than 100 . so they are deficient of - 5 and - 3 compared with 100 . so answer : e" | a ) 92 / 198 , b ) 92 / 12 , c ) 92 / 13 , d ) 92 / 10 , e ) 92 / 15 | e | divide(95, 97) | divide(n0,n1)| | general |
what is the units digit of 17 ^ 73 × 13 ^ 82 × 11 ^ 87 ? | "to find : the units digit of 17 ^ 83 × 13 ^ 82 × 11 ^ 87 let ' s reduce the clutter and simplify the product ( 7 ^ 83 ) ( 3 ^ 82 ) ( 1 ^ 87 ) 7 has a cyclicity of 4 : the last digit of any positive power of 7 repeats itself after every 4 th power so 7 ^ 5 has the same last digit as 7 ^ 1 , 7 ^ 9 , 7 ^ 13 thus , 7 ^ 83... | a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | e | divide(add(multiply(factorial(17), factorial(const_4.0)), multiply(factorial(17), factorial(82))), 17) | factorial(n0)|factorial(const_4.0)|factorial(n3)|multiply(#0,#1)|multiply(#0,#2)|add(#3,#4)|divide(#5,n0)| | general |
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 70 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 + 70 = 90 cents on commodity y , which will take 9 years . the answer is a ." | a ) 2010 , b ) 2011 , c ) 2012 , d ) 2013 , e ) 2014 | a | add(2001, divide(add(divide(70, const_100), subtract(4.40, 4.20)), subtract(divide(30, const_100), subtract(4.40, 4.20)))) | divide(n5,const_100)|divide(n0,const_100)|subtract(n4,n3)|add(#0,#2)|subtract(#1,#2)|divide(#3,#4)|add(n2,#5)| | general |
how many bricks , each measuring 100 cm x 11.25 cm x 6 cm , will be needed to build a wall of 8 m x 6 m x 22.5 cm ? | "number of bricks = volume of the wall / volume of 1 brick = ( 800 x 600 x 22.5 ) / ( 100 x 11.25 x 6 ) = 1600 answer : a" | a ) 1600 , b ) 2400 , c ) 5500 , d ) 7400 , e ) 3400 | a | divide(multiply(multiply(multiply(8, const_100), multiply(6, const_100)), 22.5), multiply(multiply(100, 11.25), 6)) | multiply(n3,const_100)|multiply(n4,const_100)|multiply(n0,n1)|multiply(#0,#1)|multiply(n2,#2)|multiply(n5,#3)|divide(#5,#4)| | physics |
real - estate salesman z is selling a house at a 20 percent discount from its retail price . real - estate salesman x vows to match this price , and then offers an additional 10 percent discount . real - estate salesman y decides to average the prices of salesmen z and x , then offer an additional 25 percent discount .... | "let retail price = $ 100 . z sells with 20 % discount . so z sells at $ 80 . x matches the price and gives additional 10 % . so , x sells at 80 - ( 10 % of 80 ) = 80 - 8 = $ 72 y avgs x and z [ ( 72 + 80 ) / 2 = 76 ] . y gives additional 25 % discount . so , y sells at 76 - ( 25 % of 76 ) = 76 - 19 = $ 57 now , ratio ... | a ) 72 / 80 , b ) 72 / 100 , c ) 57 / 80 , d ) 19 / 24 , e ) 76 / 100 | d | multiply(divide(divide(multiply(divide(add(subtract(const_100, 20), multiply(subtract(const_100, 20), divide(subtract(const_100, 10), const_100))), const_2), subtract(const_100, 25)), const_100), multiply(subtract(const_100, 20), divide(subtract(const_100, 10), const_100))), const_10) | subtract(const_100,n1)|subtract(const_100,n0)|subtract(const_100,n2)|divide(#0,const_100)|multiply(#3,#1)|add(#4,#1)|divide(#5,const_2)|multiply(#6,#2)|divide(#7,const_100)|divide(#8,#4)|multiply(#9,const_10)| | general |
if a - b = 3 and a ^ 2 + b ^ 2 = 28 , find the value of ab . | "2 ab = ( a ^ 2 + b ^ 2 ) - ( a - b ) ^ 2 = 25 - 9 = 16 ab = 8 . answer is d ." | a ) 19 , b ) 15 , c ) 10 , d ) 8 , e ) 13 | d | multiply(multiply(add(3, divide(subtract(sqrt(28), 3), 2)), divide(subtract(sqrt(28), 3), 2)), 2) | sqrt(n3)|subtract(#0,n0)|divide(#1,n1)|add(n0,#2)|multiply(#3,#2)|multiply(n1,#4)| | general |
a group of men decided to do a work in 20 days , but 10 of them became absent . if the rest of the group did the work in 40 days , find the original number of men ? | "original number of men = 10 * 40 / ( 40 - 20 ) = 20 answer is a" | a ) 20 , b ) 50 , c ) 40 , d ) 100 , e ) 25 | a | divide(multiply(10, 40), subtract(40, 20)) | multiply(n1,n2)|subtract(n2,n0)|divide(#0,#1)| | physics |
mr . karan borrowed a certain amount at 6 % per annum simple interest for 9 years . after 9 years , he returned rs . 8310 / - . find out the amount that he borrowed . | "explanation : let us assume mr . karan borrowed amount is rs . a . ( the principal ) by formula of simple interest , s . i . = prt / 100 where p = the principal , r = rate of interest as a % , t = time in years s . i . = ( p * 6 * 9 ) / 100 = 54 p / 100 amount = principal + s . i . 8310 = p + ( 54 p / 100 ) 8310 = ( 1... | a ) s . 5266 , b ) s . 5396 , c ) s . 5228 , d ) s . 5218 , e ) s . 52192 | b | divide(8310, add(const_1, divide(multiply(6, 9), const_100))) | multiply(n0,n1)|divide(#0,const_100)|add(#1,const_1)|divide(n3,#2)| | gain |
an article is bought for rs . 980 and sold for rs . 788 , find the loss percent ? | "980 - - - - 192 100 - - - - ? = > 19 % answer : b" | a ) 16 % , b ) 19 % , c ) 17 % , d ) 13 % , e ) 12 % | b | subtract(const_100, divide(multiply(788, const_100), 980)) | multiply(n1,const_100)|divide(#0,n0)|subtract(const_100,#1)| | gain |
out of 15 students in a class , 7 are from maharashtra , 5 are from karnataka , and 3 are from goa . four students are to be selected at random . what are the chances that at least one is from karnataka ? | total possible ways of selecting 4 students out of 15 = 15 c 4 = ( 15 * 14 * 13 * 12 ) / ( 4 * 3 * 2 * 1 ) = 1365 the number of ways of selecting a 4 students in which no student belongs to karnataka = 10 c 4 number of ways of selecting atleast one student from karnataka = 15 c 4 - 10 c 4 = 1155 probability = 1155 / 13... | a ) 12 / 13 , b ) 11 / 13 , c ) 10 / 15 , d ) 1 / 15 , e ) 4 / 14 | b | divide(subtract(divide(divide(factorial(15), factorial(subtract(15, const_4))), factorial(const_4)), divide(divide(factorial(subtract(15, const_5)), factorial(subtract(subtract(15, const_5), const_4))), factorial(const_4))), divide(divide(factorial(15), factorial(subtract(15, const_4))), factorial(const_4))) | factorial(n0)|factorial(const_4)|subtract(n0,const_4)|subtract(n0,const_5)|factorial(#2)|factorial(#3)|subtract(#3,const_4)|divide(#0,#4)|factorial(#6)|divide(#7,#1)|divide(#5,#8)|divide(#10,#1)|subtract(#9,#11)|divide(#12,#9) | probability |
suppose that f ( x ) is a function such that for every real number x , i ) f ( x ) + f ( 1 - x ) = 10 and ( ii ) f ( 1 + x ) = 3 + f ( x ) . then f ( x ) + f ( - x ) must equal | since ( ii ) holds for every real number x , it will remain valid if we replace x with - x . therefore , f ( 1 - x ) = 3 + f ( - x ) . now , from ( i ) , 10 = f ( x ) + f ( 1 - x ) = f ( x ) + 3 + f ( - x ) so that f ( x ) + f ( - x ) = 10 - 3 = 7 . ( observe that f ( x ) = 3 x + 4 satisfies the conditions in the probl... | a ) 7 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | a | subtract(const_10, 3) | subtract(const_10,n3) | general |
each child has 12 crayons and 24 apples . if there are 18 children , how many crayons are there in total ? | 12 * 18 = 216 . answer is c . | a ) 220 , b ) 65 , c ) 216 , d ) 219 , e ) 230 | c | multiply(18, 12) | multiply(n0,n2)| | general |
in a certain warehouse , 90 percent of the packages weigh less than 75 pounds , and a total of 48 packages weigh less than 25 pounds . if 80 percent of the packages weigh at least 25 pounds , how many of the packages weigh at least 25 pounds but less than 75 pounds ? | "if 80 % of the packages weigh at least 25 pounds this means that 20 % of the packages weigh less than 25 pounds let t = total number of packages so , 20 % of t = # of packages that weigh less than 25 pounds 48 packages weigh less than 25 pounds great . so , 20 % of t = 48 rewrite to get : 0.2 t = 48 solve : t = 240 90... | a ) 8 , b ) 64 , c ) 168 , d ) 102 , e ) 144 | c | subtract(divide(multiply(multiply(divide(48, subtract(const_100, 80)), const_100), 90), const_100), 48) | subtract(const_100,n4)|divide(n2,#0)|multiply(#1,const_100)|multiply(n0,#2)|divide(#3,const_100)|subtract(#4,n2)| | general |
what is the units digit of the expression 14 ^ 7 − 19 ^ 4 ? | i think answer on this one should be e too . since we know that 14 ^ 7 > 19 ^ 4 , as will said one should always check if the number is positive . | a ) 0 , b ) 3 , c ) 4 , d ) 6 , e ) 8 | e | add(add(const_4, const_3), const_2) | add(const_3,const_4)|add(#0,const_2)| | general |
there are two concentric circles such that the area of the outer circle is four times the area of the inner circle . let a , b and c be three distinct points on the perimeter of the outer circle such that ab and ac are tangents to the inner circle . if the area of the outer circle is 12 square centimetres then the area... | explanation : outer circle : since π r ² = 12 , r = √ ( 12 / π ) = 2 √ ( 3 / π ) . inner circle : since the area is 1 / 4 the area of the outer circle , π r ² = 3 and r = √ ( 3 / π ) ao is the radius of the outer circle and od is the radius of the inner circle . a line tangent to a circle and a radius drawn to the poin... | ['a ) π √ \u200b 12', 'b ) 9 / π', 'c ) 9 √ 3 / π', 'd ) 6 √ 3 / π', 'e ) none of these'] | c | multiply(const_3, multiply(sqrt(divide(const_3, const_pi)), sqrt(subtract(divide(12, const_pi), divide(const_3, const_pi))))) | divide(const_3,const_pi)|divide(n0,const_pi)|sqrt(#0)|subtract(#1,#0)|sqrt(#3)|multiply(#2,#4)|multiply(#5,const_3) | geometry |
if 425 / 999 = 0.125 , what is the 81 st digit to the right of the decimal point in the decimal equivalent of the fraction ? | 0 . [ u ] 125 [ / u = 0.425425425 . . . . . every 3 rd digit is 5 and every multiple of 3 will be the digit 5 . since 81 is multiple of 3 , the 81 st digit is 5 . answer e | a ) 0 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | e | add(const_2, const_3) | add(const_2,const_3) | general |
a and b start a business , with a investing the total capital of rs . 500000 , on the condition that b pays a interest @ 10 % per annum on his half of the capital . a is a working partner and receives rs . 15000 per month from the total profit and any profit remaining is equally shared by both of them . at the end of t... | "interest received by a from b = 10 % of half of rs . 500000 = 10 % * 250000 = 25000 . amount received by a per annum for being a working partner = 15000 * 12 = rs . 180000 . let ' p ' be the part of the remaining profit that a receives as his share . total income of a = ( 25000 + 180000 + p ) total income of b = only ... | a ) a ) 590280 , b ) b ) 590100 , c ) c ) 591900 , d ) d ) 590000 , e ) e ) 927000 | d | add(multiply(multiply(divide(divide(500000, const_2), 10), const_3), const_4), 15000) | divide(n0,const_2)|divide(#0,n1)|multiply(#1,const_3)|multiply(#2,const_4)|add(n2,#3)| | gain |
if p and q are prime numbers , how many divisors does the product p ^ 2 * q ^ 5 have ? | "when a number n = a ^ x * b ^ y , where a and b are prime numbers , and x , y are positive integers , the number of divisors of n = ( x + 1 ) ( y + 1 ) therefore , the answer is e . 3 * 6 = 18" | a ) 8 , b ) 10 , c ) 12 , d ) 15 , e ) 18 | e | multiply(add(2, const_1), add(5, const_1)) | add(n0,const_1)|add(n1,const_1)|multiply(#0,#1)| | general |
the figure above shows the dimensions of a semicircular cross section of a one - way tunnel . the single traffic lane is 12 feet wide and is equidistant from the sides of the tunnel . if vehicles must clear the top of the tunnel by at least ½ foot when they are inside the traffic lane , what should be the limit l on th... | "let ' s label the midpoint of the circle o . since the base of the semi - circle is 20 , we know that the diameter is 20 and , accordingly , the radius is 10 . we also know that the traffic lane is 12 feet long and there ' s an equal amount of space on either side , so the traffic lane extends 6 feet on either side of... | a ) 5 ½ ft , b ) 7 ½ ft , c ) 8 ½ ft , d ) 9 ½ ft , e ) 10 ft | b | subtract(divide(12, const_2), divide(const_1, const_2)) | divide(n0,const_2)|divide(const_1,const_2)|subtract(#0,#1)| | physics |
a factory has 15 machines of equal efficiency in its factory . the annual manufacturing costs are rs . 42000 & establishment charges are rs . 12000 . the annual output of the company is rs . 70000 . the annual output and manufacturing costs are directly proportional to the number of machines . the shareholders get 12 .... | option ( b ) is correct original profit = 70,000 − 42,000 − 12,000 = 16 , 00070,000 − 42,000 − 12,000 = 16,000 if 7.14 % 7.14 % of 14 i . e . one of the machines closed throughout the year , then change in profit will be : = 1314 × [ 70,000 − 42,000 ] = 1314 × [ 70,000 − 42,000 ] = 14,000 = 14,000 thus , decrease in th... | a ) 12.5 % , b ) 13.5 % , c ) 14 % , d ) 15 % , e ) 16 % | a | divide(divide(subtract(subtract(subtract(70000, 42000), 12000), multiply(divide(const_1, const_2), subtract(70000, 42000))), add(15, const_1)), const_10) | add(n0,const_1)|divide(const_1,const_2)|subtract(n3,n1)|multiply(#1,#2)|subtract(#2,n2)|subtract(#4,#3)|divide(#5,#0)|divide(#6,const_10) | gain |
the h . c . f of two numbers is 23 and the other two factors of their l . c . m are 11 and 12 . the larger of the two numbers is : | "clearly , the numbers are ( 23 * 11 ) and ( 23 * 12 ) . larger number = ( 23 * 12 ) = 276 . answer : b" | a ) 338 , b ) 276 , c ) 322 , d ) 231 , e ) 121 | b | multiply(23, 12) | multiply(n0,n2)| | other |
the present population of a town is 3600 . population increase rate is 20 % p . a . find the population of town before 2 years ? | "p = 3600 r = 20 % required population of town = p / ( 1 + r / 100 ) ^ t = 3600 / ( 1 + 20 / 100 ) ^ 2 = 3600 / ( 6 / 5 ) ^ 2 = 2500 ( approximately ) answer is a" | a ) 2500 , b ) 2100 , c ) 3500 , d ) 3600 , e ) 2050 | a | add(3600, divide(multiply(3600, 20), const_100)) | multiply(n0,n1)|divide(#0,const_100)|add(n0,#1)| | gain |
evaluate : 20 - 12 ÷ 4 × 2 = | "according to order of operations , 12 ÷ 4 × 2 ( division and multiplication ) is done first from left to right 12 ÷ 4 × 2 = 3 × 2 = 6 hence 20 - 12 ÷ 4 × 2 = 20 - 6 = 14 correct answer is b ) 22" | a ) a ) 12 , b ) b ) 24 , c ) c ) 36 , d ) d ) 48 , e ) e ) 60 | b | subtract(20, multiply(multiply(12, const_2.0), 2)) | multiply(n1,const_2.0)|multiply(n3,#0)|subtract(n0,#1)| | general |
the average of a batsman for 40 innings is 50 runs . his highest score exceeds his lowest score by 172 runs . if these two innings are excluded , his average drops by 2 runs . find his highest score . | total runs = 40 × 50 = 2000 let his highest score be = x then his lowest score = x – 172 now 200 − x − ( x − 172 ) / 38 = 48 ⇒ 2 x = 2172 – 1824 ⇒ x = 174 answer c | a ) 172 , b ) 173 , c ) 174 , d ) 175 , e ) none of the above | c | add(172, divide(subtract(subtract(multiply(40, 50), multiply(subtract(40, 2), subtract(50, 2))), 172), 2)) | multiply(n0,n1)|subtract(n0,n3)|subtract(n1,n3)|multiply(#1,#2)|subtract(#0,#3)|subtract(#4,n2)|divide(#5,n3)|add(n2,#6) | general |
a corporation paid $ 9 million in federal taxes on its first $ 50 million of gross profits and then $ 30 million in federal taxes on the next $ 150 million in gross profits . by approximately what percent did the ratio of federal taxes to gross profits increase from the first $ 50 million in profits to the next $ 150 m... | "difference in ratios = ( 30 / 150 ) - ( 9 / 50 ) = ( 1 / 50 ) % change = ( change ( 1 / 50 ) / original ratio ( 7 / 50 ) ) * 100 = 14 % answer - e" | a ) 6 % , b ) 14 % , c ) 20 % , d ) 23 % , e ) 14 % | e | multiply(divide(subtract(divide(add(30, 9), 150), divide(9, 50)), divide(9, 50)), const_100) | add(n0,n2)|divide(n0,n1)|divide(#0,n3)|subtract(#2,#1)|divide(#3,#1)|multiply(#4,const_100)| | gain |
the height of the wall is 6 times its width and length of the wall is 7 times its height . if the volume of the wall be 129024 cu . m . its width is | "explanation : let width = x then , height = 6 x and length = 42 x 42 x ã — 6 x ã — x = 129024 x = 8 answer : e" | a ) 4 m , b ) 5 m , c ) 6 m , d ) 7 m , e ) 8 m | e | power(divide(129024, multiply(multiply(6, 7), 6)), divide(const_1, const_3)) | divide(const_1,const_3)|multiply(n0,n1)|multiply(n0,#1)|divide(n2,#2)|power(#3,#0)| | physics |
evaluate : 136.09 + 43.9 | "25 / 0.0005 = ( 25 * 10000 ) / ( 0.0005 * 10000 ) = 25000 / 5 = 5000 answer is d" | a ) 1000 , b ) 3000 , c ) 7000 , d ) 5000 , e ) 2000 | d | divide(136.09, divide(43.9, 136.09)) | divide(n1,n0)|divide(n0,#0)| | general |
16 balls are numbered 1 to 16 . a ball is drawn and then another ball is drawn without replacement . what is the probability that both balls have even numbers ? | "p ( 1 st ball is even ) = 8 / 16 p ( 2 nd ball is also even ) = 7 / 15 p ( both balls are even ) = 8 / 16 * 7 / 15 = 7 / 30 the answer is d ." | a ) 4 / 11 , b ) 5 / 17 , c ) 6 / 25 , d ) 7 / 30 , e ) 8 / 37 | d | multiply(divide(add(const_4, const_4), 16), divide(subtract(add(const_4, const_4), 1), subtract(16, 1))) | add(const_4,const_4)|subtract(n0,n1)|divide(#0,n0)|subtract(#0,n1)|divide(#3,#1)|multiply(#2,#4)| | other |
the number of years of service of the 8 employees in a production department are 15 , 10 , 9 , 17 , 6 , 3 , 14 and 16 . what is the range in the number of years of service of the 8 employees ? | = 17 - 3 = 14 answer e | a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | e | subtract(17, 3) | subtract(n4,n6) | general |
how many factors of 440 are odd numbers greater than 1 ? | when factorized , 440 has 4 prime factors . of these prime factors 2 are odd and 2 are even . hence total number of odd factors is 2 * 2 ( 4 ) , which includes 2 . the total number of odd factors greater than 1 are 2 . ( option a ) | a ) 2 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | divide(const_60.0, multiply(const_10, const_2)) | multiply(const_10,const_2)|divide(const_60.0,#0)| | other |
annika hikes at a constant rate of 10 minutes per kilometer . she has hiked 2.5 kilometers east from the start of a hiking trail when she realizes that she has to be back at the start of the trail in 35 minutes . if annika continues east , then turns around and retraces her path to reach the start of the trail in exact... | set up two r x t = d cases . 1 . 1 / 10 km / min x t = 2.5 from which t = 25 mins . we know total journey time now is 35 + 25 = 60 the rate is the same ie 1 / 10 km / min . set up second r x t = d case . 1 / 10 km / min x 60 = 6 km now the total journey would be halved as distance would be same in each direction . 6 / ... | a ) 3.625 , b ) 3.5 , c ) 3 , d ) 4 , e ) 4.5 | c | add(divide(subtract(divide(35, const_10), 2.5), const_2), 2.5) | divide(n2,const_10)|subtract(#0,n1)|divide(#1,const_2)|add(n1,#2) | physics |
the average temperature of the town in the first 4 days of a month was 58 degrees . the average for the second , third , fourth and fifth days was 59 degrees . if the temperatures of the first and fifth days were in the ratio 7 : 8 , then what is the temperature on the fifth day ? | explanation : sum of temperatures on 1 st , 2 nd , 3 rd and 4 th days = ( 58 * 4 ) = 232 degrees . . . ( 1 ) sum of temperatures on 2 nd , 3 rd , 4 th and 5 th days - ( 59 * 4 ) = 236 degrees . . . . ( 2 ) subtracting ( 1 ) from ( 2 ) , we get : temp , on 5 th day - temp on 1 st day = 4 degrees . let the temperatures o... | a ) 62 degrees , b ) 32 degrees , c ) 65 degrees , d ) 66 degrees , e ) 67 degrees | b | multiply(8, subtract(multiply(4, 59), multiply(4, 58))) | multiply(n0,n2)|multiply(n0,n1)|subtract(#0,#1)|multiply(n4,#2) | physics |
49 ã — 49 ã — 49 ã — 49 x 49 = 7 ? | "49 ã — 49 ã — 49 ã — 49 x 49 = 7 ? or , 7 ( 2 ) ã — 7 ( 2 ) ã — 7 ( 2 ) ã — 7 ( 2 ) x 7 ( 2 ) = 7 ? or 7 ( 10 ) = 7 ? or , ? = 10 answer d" | a ) 4 , b ) 7 , c ) 8 , d ) 10 , e ) none of these | d | subtract(divide(49, 49), const_1) | divide(n0,n3)|subtract(#0,const_1)| | general |
reena took a loan of $ . 1200 with simple interest for as many years as the rate of interest . if she paid $ 108 as interest at the end of the loan period , what was the rate of interest ? | let rate = r % and time = r years . then , 1200 x r x r / 100 = 108 12 r 2 = 108 r 2 = 9 r = 3 . answer : d | a ) 3.6 , b ) 6 , c ) 18 , d ) 3 , e ) none of these | d | sqrt(divide(multiply(108, const_100), 1200)) | multiply(n1,const_100)|divide(#0,n0)|sqrt(#1) | gain |
today joelle opened an interest - bearing savings account and deposited $ 5,000 . if the annual interest rate is 4 percent compounded interest , and she neither deposits nor withdraws money for exactly 2 years , how much money will she have in the account ? | "interest for 1 st year = 5000 * 4 / 100 = 200 interest for 2 nd year = 5200 * 4 / 100 = 208 total = 5000 + 200 + 208 = 5408 answer : b" | a ) $ 5200 , b ) $ 5408 , c ) $ 5208 , d ) $ 5608 , e ) $ 5808 | b | add(add(multiply(multiply(multiply(const_3, 2), const_100), const_10), divide(multiply(multiply(multiply(multiply(const_3, 2), const_100), const_10), 4), const_100)), divide(multiply(add(multiply(multiply(multiply(const_3, 2), const_100), const_10), divide(multiply(multiply(multiply(multiply(const_3, 2), const_100), co... | multiply(n2,const_3)|multiply(#0,const_100)|multiply(#1,const_10)|multiply(n1,#2)|divide(#3,const_100)|add(#4,#2)|multiply(n1,#5)|divide(#6,const_100)|add(#5,#7)| | gain |
ana climbs up stairs 20 floors building and back . she takes the exact same route both ways . on the trip up she climbs an average speed of 2 steps per second . on the trip back she climbs down at an average speed of 4 steps per second . what is her approximate average speed for the round trip in miles per hour ? | average speed = total distance / total time here number of steps to 20 th floor and back is same as she takes the same route . d = 2 t 1 and d = 4 t 2 therefore , t 1 = d / 2 and t 2 = d / 4 t 1 + t 2 = 3 d / 4 therefore , average speed = 2 d / 3 d / 4 = 2 d * 4 / 3 d answer is 1.3 since , the options are far away we n... | a ) 1 , b ) 1.25 , c ) 1.33 , d ) 1.5 , e ) 1.75 | c | divide(20, add(divide(20, 2), divide(20, 4))) | divide(n0,n1)|divide(n0,n2)|add(#0,#1)|divide(n0,#2) | physics |
a cube of edge 6 cm is cut into cubes each of edge 2 cm . the ratio of the total surface area of one of the small cubes to that of the large cube is equal to : | "sol . required ratio = 6 * 2 * 2 / 6 * 6 * 6 = 1 / 9 = 1 : 9 . answer a" | a ) 1 : 9 , b ) 1 : 225 , c ) 1 : 52 , d ) 1 : 522 , e ) none | a | divide(const_4, const_100) | divide(const_4,const_100)| | geometry |
two positive integers differ by 4 , and sum of their reciprocals is 6 . then one of the numbers is | "algebraic approach : let n be the smaller integer = > 1 / n + 1 / ( n + 4 ) = 6 or ( ( n + 4 ) + n ) / n ( n + 4 ) = 6 or ( n ^ 2 + 4 n ) * 6 = 2 n + 4 or n = 2 as n can not be - negative solve for n = > n = 4 . hence , d" | a ) a ) 3 , b ) b ) 1 , c ) c ) 5 , d ) 4 , e ) e ) 28 | d | add(6, 6) | add(n1,n1)| | general |
if a mixture is 3 ⁄ 7 alcohol by volume and 4 ⁄ 7 water by volume , what is the ratio of the volume of alcohol to the volume of water in this mixture ? | "should be a sub - 600 level q . . volume = { 3 / 7 } / { 4 / 7 } = 3 / 4 answer : c" | a ) 3 ⁄ 7 , b ) 4 ⁄ 7 , c ) 3 ⁄ 4 , d ) 4 ⁄ 3 , e ) 7 ⁄ 4 | c | divide(divide(3, 7), divide(4, 7)) | divide(n0,n1)|divide(n2,n1)|divide(#0,#1)| | general |
the ratio of two quantities is 4 : 3 . if each of the quantities is decreased by 8 , their ratio changes to 3 : 1 then the smallest number is ? | "let the numbers be 4 x and 3 x then 4 x - 8 / 3 x - 8 = 3 / 1 4 x - 8 = 9 x - 24 5 x = 16 x = 3.2 smallest number = 3 * 3.2 = 9.6 answer is c" | a ) 5.2 , b ) 6.9 , c ) 9.6 , d ) 7.8 , e ) 4.7 | c | divide(add(4, 8), add(3, 8)) | add(n0,n2)|add(n1,n2)|divide(#0,#1)| | other |
the instructions state that cheryl needs 4 / 19 square yards of one type of material and 2 / 13 square yards of another type of material for a project . she buys exactly that amount . after finishing the project , however , she has 4 / 26 square yards left that she did not use . what is the total amount of square yards... | "total bought = 4 / 19 + 2 / 13 left part 4 / 26 - - - > 2 / 13 so used part 4 / 19 + 2 / 13 - 2 / 13 = 4 / 19 answer : a" | a ) 4 / 19 , b ) 1 / 9 , c ) 2 / 3 , d ) 1 1 / 9 , e ) 2 1 / 9 | a | subtract(add(divide(4, 19), divide(2, 13)), divide(4, 26)) | divide(n0,n1)|divide(n2,n3)|divide(n4,n5)|add(#0,#1)|subtract(#3,#2)| | general |
running at the same constant rate , 6 identical machines can produce a total of 270 bottles per minute . at this rate , how many bottles could 16 such machines produce in 4 minutes ? | "solution let the required number of bottles be x . more machines , more bottles ( direct proportion ) more minutes , more bottles ( direct proportion ) â ˆ ´ 6 ã — 1 ã — x = 16 ã — 4 ã — 270 â ‡ ” x = 16 x 4 x 270 / 6 = 2880 . answer a" | a ) 2880 , b ) 1800 , c ) 2700 , d ) 10800 , e ) none of these | a | multiply(multiply(divide(270, 6), 4), 16) | divide(n1,n0)|multiply(n3,#0)|multiply(n2,#1)| | gain |
all the milk in container a which was filled to its brim was poured into two containers b and c . the quantity of milk in container b was 62.5 % less than the capacity of container a . if 154 liters was now transferred from c to b , then both the containers would have equal quantities of milk . what was the initial qua... | "a b has 62.5 % or ( 5 / 8 ) of the milk in a . therefore , let the quantity of milk in container a ( initially ) be 8 k . quantity of milk in b = 8 k - 5 k = 3 k . quantity of milk in container c = 8 k - 3 k = 5 k container : a b c quantity of milk : 8 k 3 k 5 k it is given that if 154 liters was transferred from cont... | a ) 1232 , b ) 1723 , c ) 1129 , d ) 2613 , e ) 1372 | a | divide(154, subtract(inverse(const_2), subtract(const_1, divide(62.5, const_100)))) | divide(n0,const_100)|inverse(const_2)|subtract(const_1,#0)|subtract(#1,#2)|divide(n1,#3)| | general |
of the 170 people at a party , 70 were women , and 30 women tried the appetizer . if 60 people did not try the appetizer , what is the total number of men who tried the appetizer ? | total people at party = 170 women = 70 so men 170 - 70 = 100 no . of pple who tried appetizer = 170 - 60 ( given info ) = 110 no of women who tried appetizer = 30 so remaining ppl ( men ) who tried the appetizer = 110 - 30 = 80 correct option e | a ) 40 , b ) 50 , c ) 60 , d ) 70 , e ) 80 | e | subtract(subtract(170, 60), 30) | subtract(n0,n3)|subtract(#0,n2) | general |
on dividing a number by 7 , we get 5 as quotient and 0 as remainder . on dividing the same number by 11 , what will be the remainder ? | number = 7 * 5 + 0 = 35 11 ) 35 ( 3 33 - - - - - - - - 2 required number = 2 . answer : a | a ) 2 , b ) 5 , c ) 8 , d ) 1 , e ) 3 | a | reminder(multiply(7, 5), 11) | multiply(n0,n1)|reminder(#0,n3) | general |
if a 10 percent deposit that has been paid toward the purchase of a certain product is $ 130 , how much more remains to be paid ? | 10 / 100 p = 130 > > p = 130 * 100 / 10 = 1300 1300 - 130 = 1170 answer : d | a ) $ 880 , b ) $ 990 , c ) $ 1,000 , d ) $ 1,170 , e ) $ 1,210 | d | subtract(multiply(130, divide(const_100, 10)), 130) | divide(const_100,n0)|multiply(n1,#0)|subtract(#1,n1) | general |
a man swims downstream 36 km and upstream 18 km taking 6 hours each time , what is the speed of the man in still water ? | "36 - - - 6 ds = 6 ? - - - - 1 18 - - - - 6 us = 3 ? - - - - 1 m = ? m = ( 6 + 3 ) / 2 = 4.5 answer : a" | a ) 4.5 , b ) 5.5 , c ) 6.5 , d ) 4.8 , e ) 2 | a | divide(add(divide(18, 6), divide(36, 6)), const_2) | divide(n1,n2)|divide(n0,n2)|add(#0,#1)|divide(#2,const_2)| | physics |
a rectangular block 12 cm by 36 cm by 60 cm is cut into an exact number of equal cubes . find the least possible number of cubes ? | "volume of the block = 12 * 36 * 60 = 25920 cm ^ 3 side of the largest cube = h . c . f of 12 , 36,60 = 12 cm volume of the cube = 12 * 12 * 12 = 1728 cm ^ 3 number of cubes = 25920 / 1728 = 15 answer is c" | a ) 6 , b ) 10 , c ) 15 , d ) 40 , e ) 22 | c | divide(add(subtract(divide(rectangle_area(const_360, const_1000), const_10), multiply(const_1000, multiply(const_3, const_2))), add(multiply(const_3, const_1000), multiply(60, const_10))), divide(add(subtract(divide(rectangle_area(const_360, const_1000), const_10), multiply(const_1000, multiply(const_3, const_2))), add... | multiply(const_1000,const_3)|multiply(n2,const_10)|multiply(const_2,const_3)|rectangle_area(const_1000,const_360)|add(#0,#1)|divide(#3,const_10)|multiply(#2,const_1000)|subtract(#5,#6)|add(#4,#7)|divide(#8,const_10)|divide(#8,#9)| | geometry |
a man swims downstream 65 km and upstream 35 km taking 5 hours each time ; what is the speed of the current ? | "65 - - - 5 ds = 13 ? - - - - 1 35 - - - - 5 us = 7 ? - - - - 1 s = ? s = ( 13 - 7 ) / 2 = 3 answer : b" | a ) 2 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | b | divide(add(divide(35, 5), divide(65, 5)), const_2) | divide(n1,n2)|divide(n0,n2)|add(#0,#1)|divide(#2,const_2)| | physics |
a , b and c started a business with capitals of rs . 8000 , rs . 10000 and rs . 12000 respectively . at the end of the year , the profit share of b is rs . 3500 . the difference between the profit shares of a and c is ? | "explanation : ratio of investments of a , b and c is 8000 : 10000 : 12000 = 4 : 5 : 6 and also given that , profit share of b is rs . 3500 = > 5 parts out of 15 parts is rs . 3500 now , required difference is 6 - 4 = 2 parts required difference = 2 / 5 ( 3500 ) = rs . 1400 answer : b" | a ) 621 , b ) 1400 , c ) 236 , d ) 600 , e ) 211 | b | multiply(subtract(divide(12000, 10000), divide(8000, 10000)), 3500) | divide(n2,n1)|divide(n0,n1)|subtract(#0,#1)|multiply(n3,#2)| | gain |
how many pieces of 75 cm can be cut from a rope 60 meters long ? | "explanation : total pieces of 75 cm that can be cut from a rope of 60 meters long is = ( 60 meters ) / ( 75 cm ) = ( 60 meters ) / ( 0.75 meters ) = 80 answer : a" | a ) 80 , b ) 40 , c ) 60 , d ) none , e ) can not be determined | a | divide(60, 75) | divide(n1,n0)| | physics |
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