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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in an office in singapore there are 60 % female employees . 50 % of all the male employees are computer literate . if there are total 62 % employees computer literate out of total 1400 employees , then the no . of female employees who are computer literate ? | "solution : total employees , = 1400 female employees , 60 % of 1400 . = ( 60 * 1400 ) / 100 = 840 . then male employees , = 560 50 % of male are computer literate , = 280 male computer literate . 62 % of total employees are computer literate , = ( 62 * 1400 ) / 100 = 868 computer literate . thus , female computer lite... | a ) 588 , b ) 674 , c ) 672 , d ) 960 , e ) none | a | multiply(subtract(divide(62, const_100), multiply(subtract(const_1, divide(60, const_100)), divide(50, const_100))), 1400) | divide(n2,const_100)|divide(n1,const_100)|divide(n0,const_100)|subtract(const_1,#2)|multiply(#1,#3)|subtract(#0,#4)|multiply(n3,#5)| | gain |
sales price is $ 81 , gross profit is 170 % of cost , what is the value of gross profit ? | "cost + profit = sales cost + ( 170 / 100 ) cost = 81 cost = 30 profit = 81 - 30 = 51 answer ( e )" | a ) 32 , b ) 33 , c ) 39 , d ) 40 , e ) 51 | e | subtract(81, divide(81, add(const_1, divide(170, const_100)))) | divide(n1,const_100)|add(#0,const_1)|divide(n0,#1)|subtract(n0,#2)| | gain |
if the product of two numbers is 17820 and their h . c . f . is 12 , find their l . c . m . | "explanation : hcf * lcm = 17820 , because we know product of two numbers = product of hcf and lcm lcm = 17820 / 12 = 1485 option c" | a ) 2574 , b ) 2500 , c ) 1485 , d ) 1574 , e ) none of these | c | divide(17820, 12) | divide(n0,n1)| | physics |
thabo owns exactly 160 books , and each book is either paperback fiction , paperback nonfiction , or hardcover nonfiction . if he owns 20 more paperback nonfiction books than hardcover nonfiction books , and twice as many paperback fiction books as paperback nonfiction books , how many hardcover books nonfiction books ... | i think we can use double - matrix method and solve using only one variable . our goal is to find the number of hardcover nonfiction books . let that number be x . we are given that all 140 books are either paperback fiction , paperback nonfiction , or hardcover nonfiction . this implies that number of hardcover fictio... | a ) 10 , b ) 25 , c ) 35 , d ) 46 , e ) 50 | b | subtract(divide(add(160, 20), const_4), 20) | add(n0,n1)|divide(#0,const_4)|subtract(#1,n1) | general |
if x + y = 2 x - 2 z , x - 2 y = 4 z and x + y + z = 21 , what is the value of y / z ? | given : x + y = 2 x - 2 z , - - > eq 1 x - 2 y = 4 z - - > eq 2 x + y + z = 21 - - > eq 3 . now we are asked to find the value of y / z . now let choose some option randomly , take c , which is 3 , i randomly chose this value as it is positive : ) y / z = 3 , this is the value we need to get . then y = 3 z - - eq 4 sub... | a ) - 4.5 . , b ) - 2 . , c ) - 1.7 . , d ) 3 . , e ) 2.5 . | c | multiply(21, 2) | multiply(n0,n4) | general |
evaluate 100 / . 25 | explanation : 100 / . 25 = 10000 / 25 = 400 option a | a ) 400 , b ) 510 , c ) 500 , d ) none of these , e ) 506 | a | divide(100, divide(25, const_100)) | divide(n1,const_100)|divide(n0,#0) | general |
the price of rice falls by 20 % . how much rice can be bought now with the money that was sufficient to buy 10 kg of rice previously ? | "solution : let rs . 100 be spend on rice initially for 10 kg . as the price falls by 20 % , new price for 10 kg rice , = ( 100 - 20 % of 100 ) = 80 new price of rice = 80 / 10 = rs . 8 per kg . rice can bought now at = 100 / 8 = 12,5 kg . answer : option b" | a ) 5 kg , b ) 12,5 kg , c ) 25 kg , d ) 30 kg , e ) none | b | divide(const_100, divide(subtract(const_100, 20), 10)) | subtract(const_100,n0)|divide(#0,n1)|divide(const_100,#1)| | gain |
what is the smallest integer that is the sum of 2 distinct prime integers , each greater than 70 ? | a prime number ( or a prime ) is a natural number greater than 1 that has no positive divisors other than 1 and itself . here are the first few prime numbers : 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 , 41 , 43 , 47 , 53 , 59 , 61 , 67 , 71 , 73 , 79 , 83 , 89 , 97 , 101 , 103 , 107 , 109 , 113 , 127 , 131... | a ) 147 , b ) 142 , c ) 173 , d ) 144 , e ) 186 | d | add(add(70, const_1), add(70, const_3)) | add(n1,const_1)|add(n1,const_3)|add(#0,#1) | general |
for any integer k > 1 , the term “ length of an integer ” refers to the number of positive prime factors , not necessarily distinct , whose product is equal to k . for example , if k = 24 , the length of k is equal to 4 , since 24 = 2 × 2 × 2 × 3 . if x and y are positive integers such that x > 1 , y > 1 , and x + 3 y ... | "we know that : x > 1 , y > 1 , and x + 3 y < 940 , and it is given that length means no of factors . for any value of x and y , the max no of factors can be obtained only if factor is smallest noall factors are equal . hence , lets start with smallest no 2 . 2 ^ 1 = 2 2 ^ 2 = 4 2 ^ 3 = 8 2 ^ 4 = 16 2 ^ 5 = 32 2 ^ 6 = ... | a ) 24 , b ) 22 , c ) 16 , d ) 18 , e ) 20 | c | add(add(4, 3), add(add(4, 4), 1)) | add(n2,n7)|add(n2,n2)|add(n0,#1)|add(#0,#2)| | general |
how many 7 in between 1 to 100 ? | "7 , 17,27 , 37,47 , 57,67 , 70,71 , 72,73 , 74,75 , 76,77 ( two 7 ' s ) , 78,79 , 87,97 20 7 ' s between 1 to 100 answer : b" | a ) 18 , b ) 20 , c ) 22 , d ) 23 , e ) 24 | b | divide(100, multiply(7, const_3.0)) | multiply(n0,const_3.0)|divide(n2,#0)| | general |
the perimeter of a semi circle is 162 cm then the radius is ? | "36 / 7 r = 162 = > r = 31.5 answer : a" | a ) 31.5 , b ) 28 , c ) 98 , d ) 37 , e ) 13 | a | divide(162, add(const_2, const_pi)) | add(const_2,const_pi)|divide(n0,#0)| | physics |
together , 15 type a machines and 7 type b machines can complete a certain job in 4 hours . together 8 type b machines and 15 type c machines can complete the same job in 11 hours . how many hours r would it take one type a machine , one type b machine , and one type c machine working together to complete the job ( ass... | "say the rates of machines a , b and c are a , b , and c , respectively . together 15 type a machines and 7 type b machines can complete a certain job in 4 hours - - > 15 a + 7 b = 1 / 4 ; together 8 type b machines and 15 type c machines can complete the same job in 11 hours - - > 8 b + 15 c = 1 / 11 . sum the above :... | a ) 22 hours , b ) 30 hours , c ) 44 hours , d ) 60 hours , e ) it can not be determined from the information above . | c | divide(const_1, divide(add(divide(const_1, 4), divide(const_1, 11)), 15)) | divide(const_1,n2)|divide(const_1,n5)|add(#0,#1)|divide(#2,n0)|divide(const_1,#3)| | physics |
the cricket team of 11 members is 28 yrs old & the wicket keeper is 3 yrs older . if the ages ofthese 2 are excluded , the average age of theremaining players is 1 year less than the average age of the whole team . what is the average age of the team ? | "let the average age of the whole team be x years . 11 x - ( 28 + 31 ) = 9 ( x - 1 ) = > 11 x - 9 x = 50 = > 2 x = 50 = > x = 25 . so , average age of the team is 25 years . d" | a ) 21 , b ) 22 , c ) 23 , d ) 25 , e ) 28 | d | divide(subtract(add(add(28, 3), 28), subtract(11, 2)), subtract(11, subtract(11, 2))) | add(n1,n2)|subtract(n0,n3)|add(n1,#0)|subtract(n0,#1)|subtract(#2,#1)|divide(#4,#3)| | general |
a , b and c are partners . a receives 2 / 3 of profits , b and c dividing the remainder equally . a ' s income is increased by rs . 400 when the rate to profit rises from 5 to 7 percent . find the capital of b ? | "a : b : c = 2 / 3 : 1 / 6 : 1 / 6 = 4 : 1 : 1 x * 2 / 100 * 2 / 3 = 400 b capital = 30000 * 1 / 6 = 5000 . answer : d" | a ) 3999 , b ) 7799 , c ) 2500 , d ) 5000 , e ) 2912 | d | divide(multiply(400, const_100), 2) | multiply(n2,const_100)|divide(#0,n0)| | general |
a reduction of 30 % in the price of apples would enable a man to obtain 54 more for rs . 30 , what is reduced price per dozen ? | "b 30 * ( 30 / 100 ) = 9 - - - 54 ? - - - 12 = > rs . 2" | a ) 1 , b ) 2 , c ) 6 , d ) 7 , e ) 10 | b | multiply(divide(divide(multiply(30, 30), const_100), 54), const_12) | multiply(n0,n2)|divide(#0,const_100)|divide(#1,n1)|multiply(#2,const_12)| | gain |
a certain psychologist charges $ 35 more for the first hour of therapy than for each additional hour . if the total charge to a patient who receives 5 hours of therapy is $ 350 , what is the total charge to a patient who receives only 2 hours of therapy ? | "let the charge for first hour = x + 35 then charge for each additional hour = x x + 35 + 4 x = 350 = > 5 x = 315 = > x = 63 total charge for patient for 3 hours of therapy = x + 35 + x = 2 x + 35 = 161 $ answer c" | a ) $ 120 , b ) $ 135 , c ) $ 161 , d ) $ 165 , e ) $ 192 | c | add(multiply(divide(subtract(350, 35), 5), 2), 35) | subtract(n2,n0)|divide(#0,n1)|multiply(n3,#1)|add(n0,#2)| | general |
the average length of 6 ropes is 80 cm . if the average length of one third of the ropes measure to 70 cm , what is the average of the other ropes ? | edit : given ( x 1 + x 2 . . . + x 6 ) / 6 = 80 ( x 1 + x 2 . . . + x 6 ) = 480 - - > eq 1 . now given avg length of one third ropes is 70 . that means out 6 / 3 = 2 ropes . let the avg length of two ropes be ( x 1 + x 2 ) / 2 = 70 . ( x 1 + x 2 ) = 140 . - - > eq 2 . now we are asked to find the average of the remaini... | a ) 75 . , b ) 85 . , c ) 90 . , d ) 94 . , e ) 100 . | b | multiply(subtract(80, divide(multiply(70, const_2), 6)), divide(6, const_4)) | divide(n0,const_4)|multiply(n2,const_2)|divide(#1,n0)|subtract(n1,#2)|multiply(#0,#3) | general |
x and y are two towns . ganesh covers the distance from x to y at an average speed of 44 km / hr . however , he covers the distance from y to x at an average speed of 36 km / hr . his average speed during the whole journey in km / hr . is : | "solution : average speed = 2 xy / x + y = 2 * 44 * 36 / 44 + 36 = 39.6 answer : a" | a ) 39.6 , b ) 43 , c ) 40 , d ) 38 , e ) 29 | a | divide(add(add(subtract(44, const_10), const_100), add(subtract(44, const_10), const_100)), add(divide(add(subtract(44, const_10), const_100), 44), divide(add(subtract(44, const_10), const_100), 36))) | subtract(n0,const_10)|add(#0,const_100)|add(#1,#1)|divide(#1,n0)|divide(#1,n1)|add(#3,#4)|divide(#2,#5)| | general |
if the cost price is 95 % of selling price then what is the profit percentage . | "selling price = rs 100 : then cost price = rs 95 : profit = rs 5 . profit = { ( 5 / 95 ) * 100 } % = 5.26 % answer is a ." | a ) 5.26 , b ) 5.16 , c ) 5.61 , d ) 5.52 , e ) 5 | a | multiply(divide(subtract(const_100, 95), 95), const_100) | subtract(const_100,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
one fourth of a solution that was 10 % sugar by weight was replaced by a second solution resulting in a solution that was 20 percent sugar by weight . the second solution was what percent sugar by weight ? | "say the second solution ( which was 1 / 4 th of total ) was x % sugar , then 3 / 4 * 0.1 + 1 / 4 * x = 1 * 0.20 - - > x = 0.5 . alternately you can consider total solution to be 100 liters and in this case you ' ll have : 75 * 0.1 + 25 * x = 100 * 0.20 - - > x = 0.5 . answer : b ." | a ) 65 % , b ) 50 % , c ) 22 % , d ) 18 % , e ) 8.5 % | b | multiply(divide(subtract(multiply(const_100, divide(20, const_100)), multiply(subtract(const_100, multiply(divide(const_1, const_4), const_100)), divide(10, const_100))), multiply(divide(const_1, const_4), const_100)), const_100) | divide(n1,const_100)|divide(n0,const_100)|divide(const_1,const_4)|multiply(#0,const_100)|multiply(#2,const_100)|subtract(const_100,#4)|multiply(#1,#5)|subtract(#3,#6)|divide(#7,#4)|multiply(#8,const_100)| | gain |
a circular logo is enlarged to fit the lid of a jar . the new diameter is 20 per cent larger than the original . by what percentage has the area of the logo increased ? | "let old diameter be 4 , so radius is 2 old area = 4 π new diameter is 4.8 , so radius is 2.4 new area = 5.76 π increase in area is 1.76 π % increase in area = 1.76 / 4 * 100 so , % increase is 44 % answer will be ( d )" | a ) 50 , b ) 80 , c ) 100 , d ) 44 , e ) 250 | d | multiply(divide(subtract(power(add(divide(multiply(20, const_2), const_1000), const_3), const_2), const_4), const_4), const_100) | multiply(n0,const_2)|divide(#0,const_1000)|add(#1,const_3)|power(#2,const_2)|subtract(#3,const_4)|divide(#4,const_4)|multiply(#5,const_100)| | geometry |
what is the perimeter of a square field whose diagonal is 8 √ 2 ? | "4 a = 32 m answer : b" | a ) 82 m , b ) 32 m , c ) 27 m , d ) 26 m , e ) 76 m | b | divide(add(add(sqrt(subtract(power(8, const_2), power(2, const_2))), 2), add(sqrt(subtract(power(8, const_2), power(2, const_2))), 2)), 2) | power(n0,const_2)|power(n1,const_2)|subtract(#0,#1)|sqrt(#2)|add(n1,#3)|add(#4,#4)|divide(#5,n1)| | geometry |
in a division sum , the quotient is 120 , the divisor 456 and the remainder 333 , find the dividend ? | "explanation : 120 * 456 + 333 = 55053 answer : d" | a ) 55287 , b ) 29887 , c ) 27768 , d ) 55053 , e ) 17191 | d | add(multiply(120, 456), 333) | multiply(n0,n1)|add(n2,#0)| | general |
what is the remainder when ( 71 ) ( 73 ) is divided by 9 ? | "( 71 ) ( 73 ) = ( 72 - 1 ) ( 72 + 1 ) = 72 ^ 2 - 1 which is 1 less than a multiple of 9 . then the remainder will be 8 . the answer is e ." | a ) 1 , b ) 2 , c ) 4 , d ) 6 , e ) 8 | e | reminder(multiply(73, 71), 9) | multiply(n0,n1)|reminder(#0,n2)| | general |
what is the probability that a number selected from number 1 to 30 is prime number , when each of the given numbers is equally likely to be selected ? | x = { 2,3 , 5,7 , 11,13 , 17,19 , 23,29 } n ( x ) = 10 n ( s ) = 30 hence required probability = n ( x ) / n ( s ) = 10 / 30 option a is answer | a ) 10 / 30 , b ) 20 / 30 , c ) 1 , d ) 5 / 30 , e ) 15 / 30 | a | divide(multiply(const_5, const_2), 30) | multiply(const_2,const_5)|divide(#0,n1) | probability |
a grocer purchased a quantity of bananas at 3 pounds for $ 0.50 and sold the entire quantity at 4 pounds for $ 1.00 . how many pounds did the grocer purchase if the profit from selling the bananas was $ 8.00 ? | "cost price of 1 pound of bananas = 0.5 / 3 = 1 / 6 selling price of 1 pound of bananas = 1 / 4 profit per pound = ( 1 / 4 - 1 / 6 ) = ( 1 / 12 ) total profit is given as 8 ( 1 / 12 ) * x = 8 x = 96 answer : c" | a ) 40 , b ) 60 , c ) 96 , d ) 120 , e ) 240 | c | divide(8.00, subtract(divide(1.00, 4), divide(0.50, 3))) | divide(n3,n2)|divide(n1,n0)|subtract(#0,#1)|divide(n4,#2)| | gain |
the volume of a certain substance is always directly proportional to its weight . if 48 cubic inches of the substance weigh 112 ounces , what is the volume , in cubic inches , of 53 ounces of this substance ? | "112 ounces of a substance has a volume of 48 cubic inches 53 ounces of a substance has a volume of ( 48 / 112 ) * 53 = 22 cubic inches answer d" | a ) 27 , b ) 36 , c ) 42 , d ) 22 , e ) 147 | d | multiply(divide(48, 112), 53) | divide(n0,n1)|multiply(n2,#0)| | geometry |
two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively . if they cross each other in 22 seconds , what is the ratio of their speeds ? | let the speed of the trains be x and y respectively length of train 1 = 27 x length of train 2 = 17 y relative speed = x + y time taken to cross each other = 22 s = ( 27 x + 17 y ) / ( x + y ) = 22 = ( 27 x + 17 y ) / = 22 ( x + y ) = 5 x = 5 y = x / y = 5 / 5 = 1 / 1 i . e 1 : 1 answer : a | a ) 1 : 1 , b ) 3 : 1 , c ) 2 : 3 , d ) 3 : 2 , e ) 3 : 4 | a | divide(subtract(27, 22), subtract(22, 17)) | subtract(n0,n2)|subtract(n2,n1)|divide(#0,#1) | physics |
roses can be purchased individually for $ 5.30 , one dozen for $ 36 , or two dozen for $ 50 . what is the greatest number of roses that can be purchased for $ 680 ? | "buy as many $ 50 deals as possible . we can by 650 / 50 = 13 two dozen roses , thus total of 13 * 24 = 312 roses . we are left with 680 - 650 = $ 30 . we can buy 30 / 5.3 = ~ 5 roses for that amount . total = 312 + 5 = 317 . answer : c ." | a ) 156 , b ) 162 , c ) 317 , d ) 324 , e ) 325 | c | add(multiply(floor(divide(680, 50)), multiply(const_12, const_2)), floor(divide(subtract(680, multiply(floor(divide(680, 50)), 50)), 5.30))) | divide(n3,n2)|multiply(const_12,const_2)|floor(#0)|multiply(n2,#2)|multiply(#2,#1)|subtract(n3,#3)|divide(#5,n0)|floor(#6)|add(#7,#4)| | general |
the salaries of a and b together is rs . 14000 . a spend 80 % of his salary and b spends 85 % of his salary . what is the salary of b if their savings are equal ? | let the salaries of a and b are x and y respectively x + y = 14000 savings of a = 20 x / 100 = savings of b = 15 y / 100 x = ¾ y 3 / 4 y + y = 14000 , 7 y / 4 = 14000 , y = 8000 answer : b | a ) rs . 6000 , b ) rs . 8000 , c ) rs . 7500 , d ) rs . 6500 , e ) rs . 8500 | b | divide(subtract(14000, multiply(14000, divide(80, const_100))), subtract(subtract(add(const_1, const_1), divide(80, const_100)), divide(85, const_100))) | add(const_1,const_1)|divide(n1,const_100)|divide(n2,const_100)|multiply(n0,#1)|subtract(#0,#1)|subtract(n0,#3)|subtract(#4,#2)|divide(#5,#6) | gain |
how many of the positive divisors q of 120 are also multiples of 4 not including 120 ? | "4 , 8,12 , 20,24 , 40,60 . ( 7 ) is the answer other way : factors of 120 = 2 ^ 3 * 3 * 5 separate 2 ^ 2 ( which means 4 ) now , calculate the number of other factors . q = 2 * 3 * 5 = total positive factors are 2 * 2 * 2 = 8 this 8 factors include 120 so subtract 1 from 8 ans is 7 = d" | a ) 3 . , b ) 4 . , c ) 5 . , d ) 7 . , e ) 8 . | d | divide(divide(divide(120, 4), const_2), const_3) | divide(n0,n1)|divide(#0,const_2)|divide(#1,const_3)| | general |
how many integers between 323,700 and 458,600 have tens digit 1 and units digit 3 ? | "there is one number in hundred with 1 in the tens digit and 3 in the units digit : 13 , 113 , 213 , 313 , . . . the difference between 323,700 and 458,600 is 458,600 - 323,700 = 134,900 - one number per each hundred gives 134,900 / 100 = 1,349 numbers . answer : e ." | a ) 10,300 , b ) 10,030 , c ) 1,353 , d ) 1,352 , e ) 1,349 | e | subtract(458,600, add(add(multiply(const_2, const_100), multiply(add(const_3, const_4), const_10)), const_2)) | add(const_3,const_4)|multiply(const_100,const_2)|multiply(#0,const_10)|add(#1,#2)|add(#3,const_2)|subtract(n1,#4)| | general |
two equally sized jugs full of water are each emptied into two separate unequally sized empty jugs , x and y . now , jug x is 1 / 6 full , while jug y is 2 / 3 full . if water is poured from jug x into jug y until jug y is filled , what fraction of jug x then contains water ? | "suppose the water in each jug is l liters cx x ( 1 / 6 ) = l cx = 6 l liters cx is capacity of x cy x ( 2 / 3 ) = l cy = 3 l / 2 liters cy is capacity of y now , y is 3 l / 2 - l empty = l / 2 empty so , we can put only l / 2 water in jug y from jug x jug x ' s remaining water = l - l / 2 = l / 2 fraction of x which c... | a ) 0 , b ) 1 / 12 , c ) 2 / 15 , d ) 1 / 10 , e ) 2 / 10 | b | multiply(divide(1, 6), divide(subtract(1, divide(2, 3)), divide(2, 3))) | divide(n0,n1)|divide(n2,n3)|subtract(n0,#1)|divide(#2,#1)|multiply(#0,#3)| | general |
11 + 12 + 13 + . . . 51 + 52 + 53 = ? | "sum = 11 + 12 + 13 + . . . 51 + 52 + 53 sum of n consecutive positive integers starting from 1 is given as n ( n + 1 ) / 2 sum of first 53 positive integers = 53 * 54 / 2 sum of first 10 positive integers = 11 * 10 / 2 sum = 11 + 12 + 13 + . . . 51 + 52 + 53 = 53 * 54 / 2 - 11 * 10 / 2 = 1376 answer : b" | a ) 1361 , b ) 1376 , c ) 1363 , d ) 1364 , e ) 1365 | b | divide(multiply(51, add(51, 11)), 12) | add(n0,n3)|multiply(n3,#0)|divide(#1,n1)| | general |
a man buys 52 pens at marked price of 46 pens from a whole seller . if he sells these pens giving a discount of 1 % , what is the profit percent ? | "explanation : let marked price be re . 1 each c . p . of 52 pens = rs . 46 s . p . of 52 pens = 99 % of rs . 52 = rs . 51.48 profit % = ( profit / c . p . ) x 100 profit % = ( 5.48 / 46 ) x 100 = 11.91 % answer : d" | a ) 7.6 % , b ) 7.7 % , c ) 11.32 % , d ) 11.91 % , e ) 7.8 % | d | multiply(const_100, divide(subtract(multiply(52, subtract(const_100, 1)), multiply(46, const_100)), multiply(46, const_100))) | multiply(n1,const_100)|subtract(const_100,n2)|multiply(n0,#1)|subtract(#2,#0)|divide(#3,#0)|multiply(#4,const_100)| | gain |
express 50 mps in kmph ? | "25 * 18 / 5 = 180 kmph answer : d" | a ) 172 , b ) 160 , c ) 150 , d ) 180 , e ) 120 | d | multiply(divide(50, const_1000), const_3600) | divide(n0,const_1000)|multiply(#0,const_3600)| | physics |
the total price of a kilogram each of shimla apples and red delicious apples is 250 rupees more than the total price of a kilogram each of red delicious apples and fuji apples . fuji apples is how much cheaper than shimla apples ? | ( shimla + red delicious ) - ( red delicious + fuji ) = 250 shimla - fuji = 250 answer : a | a ) 250 , b ) 125 , c ) 500 , d ) 100 , e ) 80 | a | multiply(250, const_1) | multiply(n0,const_1) | general |
how long does a train 110 m long running at the speed of 72 km / hr takes to cross a bridge 138 m length ? | speed = 72 * 5 / 18 = 20 m / sec total distance covered = 110 + 138 = 248 m . required time = 248 / 20 = 12.4 sec . answer : b | a ) 82.1 sec . , b ) 12.4 sec , c ) 19.1 sec . , d ) 17.1 sec . , e ) 42.1 sec . | b | divide(add(110, 138), multiply(72, const_0_2778)) | add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1) | physics |
what is the 25 th digit to the right of the decimal point in the decimal form of 6 / 11 ? | "explanation : to determine the greatest possible number of members , first recognize that each member had to contribute the lowest amount given . build an inequality for the individual contributions and the total amount collected , with n = the number of members in the club , and solve for n . 12 n ≤ 599 n ≤ 49 11 / 1... | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | c | divide(6, 11) | divide(n1,n2)| | general |
a sum of money lent out at s . i . amounts to a total of $ 590 after 2 years and to $ 815 after a further period of 5 years . what was the initial sum of money that was invested ? | "s . i for 5 years = $ 815 - $ 590 = $ 225 the s . i . is $ 45 / year s . i . for 2 years = $ 90 principal = $ 590 - $ 90 = $ 500 the answer is b ." | a ) $ 490 , b ) $ 500 , c ) $ 510 , d ) $ 520 , e ) $ 530 | b | subtract(590, multiply(subtract(815, 590), divide(2, 5))) | divide(n1,n3)|subtract(n2,n0)|multiply(#0,#1)|subtract(n0,#2)| | general |
if 5 < x < 9 and y = x + 7 , what is the greatest possible integer value of x + y ? | "x + y = x + x + 7 = 2 x + 7 we need to maximize this value and it needs to be an integer . 2 x is an integer when the decimal of x is . 0 or . 5 the largest such value is 8.5 then x + y = 8.5 + 15.5 = 24 . the answer is d ." | a ) 21 , b ) 22 , c ) 23 , d ) 24 , e ) 25 | d | add(add(7, const_10), const_10) | add(n2,const_10)|add(#0,const_10)| | general |
there are 8 students . 4 of them are men and 4 of them are women . if 4 students are selected from the 8 students . what is the probability that the number of men is equal to that of women ? | men = women means 2 men 2 women total such combinations = 4 c 2 * 4 c 2 = 4 ! / 2 ! . 2 ! * 4 ! / 2 ! . 2 ! = 6 * 6 total combinations = 8 c 4 = 8 * 7 * 6 * 5 / 4 * 3 * 2 * 1 = 70 so probability = 36 / 70 = 18 / 35 hence a | a ) 18 / 35 , b ) 16 / 35 , c ) 14 / 35 , d ) 13 / 35 , e ) 12 / 35 | a | divide(multiply(choose(4, divide(4, const_2)), choose(4, divide(4, const_2))), choose(8, 4)) | choose(n0,n1)|divide(n1,const_2)|choose(n1,#1)|multiply(#2,#2)|divide(#3,#0) | probability |
how many seconds will a train 200 meters long running at the rate of 36 kmph take to pass a certain telegraph post ? | distance = length of train = 200 meters speed = 36 kmph = 36 * 5 / 18 = 10 m / s required time = 200 / 10 = 20 sec answer is b | a ) 10 sec , b ) 20 sec , c ) 25 sec , d ) 30 sec , e ) 35 sec | b | divide(200, multiply(const_0_2778, 36)) | multiply(n1,const_0_2778)|divide(n0,#0) | physics |
a circular logo is enlarged to fit the lid of a jar . the new diameter is 40 per cent larger than the original . by what percentage has the area of the logo increased ? | "let old diameter be 4 , so radius is 2 old area = 4 π new diameter is 5.6 , so radius is 2.8 new area = 7.84 π increase in area is 3.84 π % increase in area = 3.84 / 4 * 100 so , % increase is 96 % answer will be ( b )" | a ) 50 , b ) 96 , c ) 100 , d ) 125 , e ) 250 | b | multiply(divide(subtract(power(add(divide(multiply(40, const_2), const_1000), const_3), const_2), const_4), const_4), const_100) | multiply(n0,const_2)|divide(#0,const_1000)|add(#1,const_3)|power(#2,const_2)|subtract(#3,const_4)|divide(#4,const_4)|multiply(#5,const_100)| | geometry |
a , b and c enter into a partnership by investing $ 11000 , $ 15000 and $ 23000 respectively . at the end of 8 months , b receives $ 3315 as his share . find the share of a . | the ratio of capital of a , b and c = 11000 : 15000 : 23000 = 11 : 15 : 23 a ' s share = ( 11 / 15 ) * 3315 = $ 2431 the answer is d . | a ) $ 1865 , b ) $ 2087 , c ) $ 2217 , d ) $ 2431 , e ) $ 2662 | d | divide(multiply(multiply(11000, divide(3315, divide(multiply(15000, 8), const_100))), 8), const_100) | multiply(n1,n3)|divide(#0,const_100)|divide(n4,#1)|multiply(n0,#2)|multiply(n3,#3)|divide(#4,const_100) | gain |
a reduction of 35 % in the price of oil enables a house wife to obtain 5 kgs more for rs . 800 , what is the reduced price for kg ? | 800 * ( 35 / 100 ) = 280 - - - - 5 ? - - - - 1 = > rs . 56 answer : c | a ) 42 , b ) 49 , c ) 56 , d ) 63 , e ) 70 | c | divide(divide(multiply(800, 35), const_100), 5) | multiply(n0,n2)|divide(#0,const_100)|divide(#1,n1) | gain |
water consists of hydrogen and oxygen , and the approximate ratio , by mass , of hydrogen to oxygen is 2 : 16 . approximately how many grams of oxygen are there in 90 grams of water ? | "solution : we are given that the ratio of hydrogen to oxygen in water , by mass , is 2 : 16 . using our ratio multiplier we can re - write this as 2 x : 16 x . we can now use these expressions to determine how much oxygen is in 90 grams of water . 2 x + 16 x = 90 18 x = 90 x = 5 since x is 5 , we know that there are 1... | a ) 16 , b ) 90 , c ) 112 , d ) 128 , e ) 142 | b | multiply(2, divide(90, add(2, 16))) | add(n0,n1)|divide(n2,#0)|multiply(n0,#1)| | other |
suppose 3 monkeys take 3 minutes to eat 3 bananas . how many monkeys would it take to eat 6 bananas in 6 minutes ? | "one monkey takes 3 min to eat 1 banana , so in 6 mins 1 monkey will eat 2 bananas , so for 6 bananas in 6 min we need 6 / 2 = 3 monkeys answer : d" | a ) 9 , b ) 10 , c ) 11 , d ) 3 , e ) 13 | d | divide(6, divide(6, 3)) | divide(n3,n0)|divide(n3,#0)| | physics |
a dessert recipe calls for 50 % melted chocolate and 50 % raspberry puree to make a particular sauce . a chef accidentally makes 15 cups of the sauce with 40 % melted chocolate and 60 % raspberry puree instead . how many cups of the sauce does he need to remove and replace with pure melted chocolate to make the sauce t... | we have 15 cups of sauce with 4040 choc and 6060 rasb cups of choc = 0.4 ∗ 15 = 6 cups of rasb = 0.6 ∗ 15 = 9 now let say we removed x cup of original mix and replaced with x cups of choc . therefore final number of cups of choc = 6 − 0.4 x + x now this number of cup should be 50 % of total = 15 / 2 = 7.5 therefore 6 −... | a ) 1.5 , b ) 2.5 , c ) 3 , d ) 4.5 , e ) 5 | b | divide(divide(subtract(add(50, 50), 50), subtract(50, 40)), const_2) | add(n0,n0)|subtract(n0,n3)|subtract(#0,n0)|divide(#2,#1)|divide(#3,const_2) | gain |
if x and y are both odd prime numbers andx < y , then how many different positive integer factors does 2 xyhave ? | "x and y are both odd prime numbers - it means either x or y is not 2 and since prime numbers have only two factors - 1 and the number itself x and y each will have ( 1 + 1 ) = 2 factors hence 2 xy will have ( 1 + 1 ) ( 1 + 1 ) ( 1 + 1 ) = 8 factors e is the answer" | a ) 3 , b ) 4 , c ) 6 , d ) 12 , e ) 8 | e | multiply(multiply(2, add(const_1, const_1)), add(const_1, const_1)) | add(const_1,const_1)|multiply(n0,#0)|multiply(#0,#1)| | other |
1 , 6 , 24 , 60 , 120 , 210 , ___ ? | "it follow the pattern like this . . 2 ^ 3 - 2 = 6 3 ^ 3 - 3 = 24 4 ^ 3 - 4 = 60 5 ^ 3 - 5 = 120 6 ^ 3 - 6 = 210 so . . . 7 ^ 3 - 7 = 336 answer : a" | a ) 336 , b ) 366 , c ) 330 , d ) 660 , e ) 770 | a | subtract(negate(60), multiply(subtract(6, 24), divide(subtract(6, 24), subtract(1, 6)))) | negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)| | general |
there are 30 students in a classroom . the ratio of the # of girls to boys was 1 : 2 , how many boys are there ? | # of girls = x # of boys = 2 x x + 2 x = 30 3 x = 30 x = 30 / 3 = 10 boys = 2 x = 2 * 10 = 20 answer : c | a ) 10 , b ) 15 , c ) 20 , d ) 25 , e ) 30 | c | multiply(divide(2, add(1, 2)), 30) | add(n1,n2)|divide(n2,#0)|multiply(n0,#1) | other |
99 ! how many zero ' s ? | 99 ! = 99 / 5 = 19 = 19 / 5 = 3 so 99 ! has ( 19 + 3 = 22 zeroes ) answer : c | a ) 20 , b ) 21 , c ) 22 , d ) 23 , e ) 24 | c | add(floor(divide(99, add(const_1, const_4))), floor(divide(99, power(add(const_1, const_4), const_2)))) | add(const_1,const_4)|divide(n0,#0)|power(#0,const_2)|divide(n0,#2)|floor(#1)|floor(#3)|add(#4,#5) | other |
in a kilometer race , a beats b by 40 meters or 8 seconds . what time does a take to complete the race ? | "time taken by b run 1000 meters = ( 1000 * 8 ) / 40 = 200 sec . time taken by a = 200 - 8 = 192 sec . answer : c" | a ) 221 , b ) 190 , c ) 192 , d ) 191 , e ) 121 | c | subtract(divide(multiply(const_1, const_1000), divide(40, 8)), 8) | divide(n0,n1)|multiply(const_1,const_1000)|divide(#1,#0)|subtract(#2,n1)| | physics |
the volume of a sphere with radius r is ( 4 / 3 ) * pi * r ^ 3 and the surface area is 4 * pi * r ^ 3 . if a sperical balloon has a volume of 4500 pi cubic centimeters , what is hte surface area of the balloon in square centimeters ? | "the surface area is 4 . pi . r ^ 2 ( its area remember not volume ) as 4 / 3 . pi . r ^ 3 = 4500 pi r = 15 so area = 4 . pi . r ^ 2 = 900 . pi = 900 x 3.14 = 2000 ( approx ) e" | a ) a . 40 , b ) b . 100 , c ) c . 400 , d ) d . 1,000 , e ) e . 2,000 | e | divide(const_180, const_1000) | divide(const_180,const_1000)| | geometry |
in what time will a train 300 m long cross an electric pole , it its speed be 144 km / hr ? | "speed = 144 * 5 / 18 = 40 m / sec time taken = 300 / 40 = 7.5 sec . answer : e" | a ) 2.5 sec , b ) 9.7 sec , c ) 3.5 sec , d ) 2.9 sec , e ) 7.5 sec | e | divide(300, multiply(144, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics |
anand and deepak started a business investing rs . 2250 and rs . 3200 respectively . out of a total profit of rs . 1380 , deepak ' s share is : | ratio of their shares = 2250 : 3200 = 45 : 64 deepak ' s share = 13800 * 64 / 109 = rs . 810.27 . answer : b | a ) 3776 , b ) 810.27 , c ) 815.21 , d ) 688.12 , e ) 267.1 | b | multiply(1380, divide(3200, add(2250, 3200))) | add(n0,n1)|divide(n1,#0)|multiply(n2,#1) | gain |
the parameter of a square is equal to the perimeter of a rectangle of length 18 cm and breadth 14 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 ( 18 + 14 ) = 64 cm parameter of the square = 64 cm i . e . 4 a = 64 a = 16 diameter of the semicircle = 16 cm circimference of the semicircle = 1 / 2 ( ∏ ) ( 16 ) = 1 / 2 ( 22 / 7 ) ( 16 ) = 352 / 14 = 25.14 cm to two decimal places answer : b" | a ) 77.14 cm , b ) 25.14 cm , c ) 84.92 cm , d ) 94.94 cm , e ) 23.57 cm | b | divide(circumface(divide(square_edge_by_perimeter(rectangle_perimeter(18, 14)), const_2)), const_2) | rectangle_perimeter(n0,n1)|square_edge_by_perimeter(#0)|divide(#1,const_2)|circumface(#2)|divide(#3,const_2)| | geometry |
find the probability that a leap year selected at random will not have 53 sundays | "there are 366 days in a leap year : 52 weeks and 2 more days . so , 52 sundays and 2 days . these 2 days can be : { mon , tue } , { tue , wed } , { wed , thu } , { thu , fri } , { fri , sat } , { sat , sun } and { sun , mon } ( 7 cases ) . in order to have 53 sundays we should have either { sat , sun } or { sun , mon ... | a ) 6 / 7 , b ) 5 / 7 , c ) 4 / 7 , d ) 3 / 7 , e ) 2 / 7 | b | divide(const_2, add(const_3, const_4)) | add(const_3,const_4)|divide(const_2,#0)| | probability |
a can do a piece of work in 3 hours ; b and c together can do it in 3 hours , while a and c together can do it in 2 hours . how long will b alone take to do it ? | "a ' s 1 hour ' s work = 1 / 3 ; ( b + c ) ' s 1 hour ' s work = 1 / 3 ; ( a + c ) ' s 1 hour ' s work = 1 / 2 . ( a + b + c ) ' s 1 hour ' s work = ( 1 / 3 + 1 / 3 ) = 2 / 3 . b ' s 1 hour ' s work = ( 2 / 3 - 1 / 2 ) = 1 / 6 . therefore a alone will take 6 hours to do the work . d" | a ) 12 , b ) 9 , c ) 8 , d ) 6 , e ) 5 | d | divide(const_1, subtract(divide(const_1, 3), subtract(divide(const_1, 2), divide(const_1, 3)))) | divide(const_1,n1)|divide(const_1,n2)|divide(const_1,n0)|subtract(#1,#2)|subtract(#0,#3)|divide(const_1,#4)| | physics |
the area of the floor of a room is 20 m 2 that of a longer wall 15 m 2 and of the shorter wall 12 m 2 , find the edge of the new cube ? | lb = 20 ; lh = 15 ; fh = 12 ( lbh ) 2 = 20 * 15 * 12 = > lbh = 60 m 3 answer : c | ['a ) 450 m 3', 'b ) 100 m 2', 'c ) 60 m 3', 'd ) 400 m 3', 'e ) 500 m 3'] | c | volume_rectangular_prism(sqrt(15), sqrt(12), sqrt(20)) | sqrt(n2)|sqrt(n4)|sqrt(n0)|volume_rectangular_prism(#0,#1,#2) | geometry |
in a 200 m race , if a gives b a start of 25 metres , then a wins the race by 10 seconds . alternatively , if a gives b a start of 45 metres the race ends in a dead heat . how long does a take to run 200 m ? | explanatory answer a gives b a start of 25 metres and still wins the race by 10 seconds . alternatively , if a gives b a start of 45 metres , then the race ends in a dead heat . therefore , the additional 20 metres start given to b compensates for the 10 seconds . i . e . , b runs 20 metres in 10 seconds . hence , b wi... | a ) 100 seconds , b ) 112.5 seconds , c ) 77.5 seconds , d ) 87.5 seconds , e ) none | c | divide(subtract(200, 45), const_2) | subtract(n0,n3)|divide(#0,const_2) | physics |
if a farmer wants to plough a farm field on time , he must plough 340 hectares a day . for technical reasons he ploughed only 85 hectares a day , hence he had to plough 2 more days than he planned and he still has 40 hectares left . what is the area of the farm field and how many days the farmer planned to work initial... | "let x be the number of days in the initial plan . therefore , the whole field is 340 â ‹ … x hectares . the farmer had to work for x + 2 days , and he ploughed 85 ( x + 2 ) hectares , leaving 40 hectares unploughed . then we have the equation : 340 x = 85 ( x + 2 ) + 40 255 x = 210 x = 0.823 so the farmer planned to h... | a ) 280 , b ) 490 , c ) 720 , d ) 435 , e ) 679 | a | multiply(divide(add(40, multiply(2, 85)), subtract(340, 85)), 340) | multiply(n1,n2)|subtract(n0,n1)|add(n3,#0)|divide(#2,#1)|multiply(n0,#3)| | physics |
p is 60 % more efficient than q . p can complete a work in 26 days . if p and q work together , how many days will it take to complete the same work ? | the work done by p in 1 day = 1 / 26 let work done by q in 1 day = q q × ( 160 / 100 ) = 1 / 26 q = 100 / ( 26 × 160 ) = 10 / ( 26 × 16 ) the work done by p and q in 1 day = 1 / 26 + 10 / ( 26 × 16 ) = 26 / ( 26 × 16 ) = 1 / 16 p and q together can do the work in 16 days . the answer is c . | a ) 12 , b ) 14 , c ) 16 , d ) 18 , e ) 20 | c | divide(multiply(add(divide(60, const_100), const_1), 26), add(add(divide(60, const_100), const_1), const_1)) | divide(n0,const_100)|add(#0,const_1)|add(#1,const_1)|multiply(n1,#1)|divide(#3,#2) | physics |
if the cost price is 92 % of sp then what is the profit % | "sol . sp = rs 100 : then cp = rs 92 : profit = rs 8 . profit = { ( 8 / 92 ) * 100 } % = 8.70 % answer is a ." | a ) 8.70 % , b ) 8.60 % , c ) 8.10 % , d ) 8.30 % , e ) 8.27 % | a | multiply(divide(subtract(const_100, 92), 92), const_100) | subtract(const_100,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
a sum was put a simple interest at a certain rate for 7 years . had it been put at 2 % higher rate , it would have fetched rs . 350 more . the sum is : | "explanation : let the sub be rs . x and the initial rate be r % . then x ã — ( r + 2 ) ã — 7 / 100 â ˆ ’ x ã — r ã — 7 / 100 = 350 â ‡ ’ x ã — 2 ã — 7 / 100 = 350 â ‡ ’ x ã — 7 / 100 = 175 â ‡ ’ 7 x = 17500 â ‡ ’ x = 2500 answer : option a" | a ) s . 500 , b ) s . 1000 , c ) s . 1500 , d ) s . 2000 , e ) s . 2500 | a | multiply(divide(divide(350, 7), 2), const_100) | divide(n2,n0)|divide(#0,n1)|multiply(#1,const_100)| | gain |
the size of a flat - screen television is given as the length of the screen ’ s diagonal . how many square inches greater is the screen of a square 34 - inch flat - screen television than a square 27 - inch flat - screen television ? | "diagonal of a square = √ 2 * side of square = > side = diagonal / ( √ 2 ) area of square = s * s = ( diagonal ^ 2 ) / 2 difference of both tv ' s = ( 34 ^ 2 - 27 ^ 2 ) / 2 = 427 answer : c" | a ) 106.75 , b ) 213.5 , c ) 427 , d ) 729 , e ) 1,156 | c | subtract(divide(power(34, const_2), const_2), divide(power(27, const_2), const_2)) | power(n0,const_2)|power(n1,const_2)|divide(#0,const_2)|divide(#1,const_2)|subtract(#2,#3)| | geometry |
â ˆ š 1.5625 = ? | explanation : 1 | 1.5625 ( 1.25 | 1 | - - - - - - - 22 | 56 | 44 | - - - - - - - 245 | 1225 | 1225 | - - - - - - - | x | - - - - - - - 1.5625 = 1.25 . answer is b | a ) 1.05 , b ) 1.25 , c ) 1.45 , d ) 1.55 , e ) 1.65 | b | add(divide(1.5625, 1.5625), const_0_25) | divide(n0,n0)|add(#0,const_0_25) | general |
the tax on a commodity is diminished by 20 % but its consumption is increased by 20 % . find the decrease percent in the revenue derived from it ? | "100 * 100 = 10000 80 * 120 = 9600 10000 - - - - - - - 400 100 - - - - - - - ? = 4 % answer : c" | a ) 20 % , b ) 18 % , c ) 4 % , d ) 12 % , e ) 14 % | c | subtract(const_100, divide(multiply(add(const_100, 20), subtract(const_100, 20)), const_100)) | add(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|subtract(const_100,#3)| | general |
three cubes of metal whose edges are 9 , 12 and 18 cm respectively , are melted and one new cube is made . find the edge of the new cube ? | "93 + 123 + 183 = a 3 = > a = 20.2 answer : a" | a ) 20.2 , b ) 77 , c ) 66 , d ) 18 , e ) 99 | a | power(add(add(volume_cube(9), volume_cube(12)), volume_cube(18)), const_0_33) | volume_cube(n0)|volume_cube(n1)|volume_cube(n2)|add(#0,#1)|add(#3,#2)|power(#4,const_0_33)| | physics |
find ( 7 x + 4 y ) / ( x - 2 y ) if x / 2 y = 3 / 2 ? | "x / 2 y = 3 / 2 = > x = 6 y / 2 = 3 y = > ( 7 x + 4 y ) / ( x - 2 y ) = ( ( 7 * ( 3 y ) ) + 4 y ) / ( 3 y - 2 y ) = > 25 y / y = 25 answer : b" | a ) 24 , b ) 25 , c ) 26 , d ) 27 , e ) 29 | b | divide(add(multiply(multiply(divide(3, 2), 2), 7), 4), subtract(multiply(divide(3, 2), 2), 2)) | divide(n4,n2)|multiply(n2,#0)|multiply(n0,#1)|subtract(#1,n2)|add(n1,#2)|divide(#4,#3)| | general |
the price of lunch for 12 people was $ 207 including a 15 % gratuity for service . what was the average price per person , excluding the gratuity ? | "clearly e is the answer i used poe here lets consider option ( e ) 12 * 15 = 180 now 180 ( 115 / 100 ) = 207 = > possible answer imo e" | a ) 11.73 , b ) 12 , c ) 13.8 , d ) 14 , e ) 15 | e | divide(divide(multiply(207, subtract(const_100, 15)), const_100), 12) | subtract(const_100,n2)|multiply(n1,#0)|divide(#1,const_100)|divide(#2,n0)| | general |
the banker ' s gain on a sum due 3 years hence at 12 % per annum is rs . 270 . the banker ' s discount is : | t . d = ( b . gx 100 ) / ( r / t ) = ( 270 x 100 ) / ( 12 x 3 ) = rs . 750 b . d . = rs . ( 750 + 270 ) = rs . 1020 . answer : d | a ) rs . 960 , b ) rs . 760 , c ) rs . 840 , d ) rs . 1020 , e ) rs . 1615 | d | add(270, divide(multiply(270, const_100), multiply(12, 3))) | multiply(n2,const_100)|multiply(n0,n1)|divide(#0,#1)|add(n2,#2) | gain |
a started a business with an investment of rs . 70000 and after 6 months b joined him investing rs . 120000 . if the profit at the end of a year is rs . 26000 , then the share of a is ? | "ratio of investments of a and b is ( 70000 * 12 ) : ( 120000 * 6 ) = 7 : 6 total profit = rs . 26000 share of b = 7 / 13 ( 26000 ) = rs . 14000 answer : b" | a ) s . 14028 , b ) s . 14000 , c ) s . 14003 , d ) s . 14029 , e ) s . 24029 | b | subtract(26000, multiply(const_60, const_100)) | multiply(const_100,const_60)|subtract(n3,#0)| | gain |
by selling 150 apples , a fruit - seller gains the selling price of 30 apples . find the gain percent ? | "option c explanation : sp = cp + g 150 sp = 150 cp + 30 sp 120 sp = 150 cp 120 - - - 30 cp 100 - - - ? = > 25 %" | a ) 30 % , b ) 58 % , c ) 25 % , d ) 60 % , e ) 12 % | c | multiply(divide(30, subtract(150, 30)), const_100) | subtract(n0,n1)|divide(n1,#0)|multiply(#1,const_100)| | gain |
a certain number of horses and an equal number of men are going somewhere . half of the owners are on their horses ' back while the remaining ones are walking along leading their horses . if the number of legs walking on the ground is 70 , how many horses are there ? | "explanation : let number of horses = number of men = x . then , number of legs = 4 x + 2 x ( x / 2 ) = 5 x . so , 5 x = 70 or x = 14 . answer : c" | a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | c | divide(70, add(const_4, divide(const_2, const_2))) | divide(const_2,const_2)|add(#0,const_4)|divide(n0,#1)| | general |
617 x 617 + 583 x 583 = ? | "= ( 617 ) ^ 2 + ( 583 ) ^ 2 = ( 600 + 17 ) ^ 2 + ( 600 - 17 ) ^ 2 = 2 [ ( 600 ) ^ 2 + ( 17 ) ^ 2 ] = 2 [ 360000 + 289 ] = 2 x 360289 = 720578 answer is a" | a ) 720578 , b ) 80578 , c ) 80698 , d ) 81268 , e ) none of them | a | multiply(617, power(617, 583)) | power(n1,n2)|multiply(n0,#0)| | general |
what positive number , when squared , is equal to the cube of the positive square root of 15 ? | "let the positive number be x x ^ 2 = ( ( 15 ) ^ ( 1 / 2 ) ) ^ 3 = > x ^ 2 = 4 ^ 3 = 36 = > x = 6 answer d" | a ) 64 , b ) 32 , c ) 8 , d ) 6 , e ) 2 | d | sqrt(power(power(15, divide(const_1, const_2)), const_3)) | divide(const_1,const_2)|power(n0,#0)|power(#1,const_3)|sqrt(#2)| | geometry |
a cube with its sides numbered 1 through 6 is rolled twice , first landing on a and then landing on b . if any roll of the cube yields an equal chance of landing on any of the numbers 1 through 6 , what is the probability t that a + b is prime ? | "total # of outcomes is 6 * 6 = 36 ; favorable outcomes : a - b - - > prime 1 - 1 - - > 2 ; 1 - 2 - - > 3 ; 2 - 1 - - > 3 ; 1 - 4 - - > 5 ; 4 - 1 - - > 5 ; 2 - 3 - - > 5 ; 3 - 2 - - > 5 ; 1 - 6 - - > 7 ; 6 - 1 - - > 7 ; 2 - 5 - - > 7 ; 5 - 2 - - > 7 ; 3 - 4 - - > 7 ; 4 - 3 - - > 7 ; 6 - 5 - - > 11 ; 5 - 6 - - > 11 . to... | a ) 0 , b ) 1 / 12 , c ) 5 / 12 , d ) 7 / 18 , e ) 4 / 9 | c | multiply(add(const_12, const_3), power(divide(1, 6), const_2)) | add(const_12,const_3)|divide(n0,n1)|power(#1,const_2)|multiply(#0,#2)| | general |
a man invested rs . 14400 in rs . 100 shares of a company at 20 % premium . if the company declares 7 % dividend at the end of the year , then how much does he get ? | solution number of shares = ( 14400 / 120 ) = 120 . face value = rs . ( 100 x 120 ) = rs . 12000 . annual income = rs . ( 7 / 100 x 12000 ) = rs . 840 . answer b | a ) rs . 500 , b ) rs . 840 , c ) rs . 650 , d ) rs . 720 , e ) none | b | divide(multiply(multiply(divide(14400, add(100, 20)), 100), 7), 100) | add(n1,n2)|divide(n0,#0)|multiply(n1,#1)|multiply(n3,#2)|divide(#3,n1) | gain |
a boy standing idle sounds a whistle to his friend at a distance of 1200 m moving away from him in a speeding car at 108 kms / hr . find the duration after which his friend is going to hear him . ( speed of sound = 330 m / sec ) . | if friend hears sound after t secs , then 330 * t = 1200 + ( 108 * 5 / 18 ) * t 300 t = 1200 t = 4 secs answer : b | a ) 3.6 secs , b ) 4.00 secs , c ) 40 secs , d ) 45 secs , e ) none of these | b | add(divide(1200, 330), divide(multiply(multiply(const_3, const_10), divide(1200, 330)), 330)) | divide(n0,n2)|multiply(const_10,const_3)|multiply(#0,#1)|divide(#2,n2)|add(#0,#3) | physics |
if ' a ' and ' b ' are non - zero numbers such that their sum is 6 times the product , what is the value of 1 / a + 1 / b ? | a + b = 6 ab 1 / a + 1 / b = ( b + a ) / ( ab ) = 6 . the answer is b . | a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 12 | b | divide(6, 1) | divide(n0,n1) | general |
two integers are in the ratio of 1 to 4 . if 12 is added to the smaller number , the ratio becomes 1 to 2 . find the larger integer . | one option is to set up the equations and solve : if the ratio of two integers x and y is 1 to 4 , then 4 x = y , where x is the smaller integer . if adding 12 to the smaller integer makes the ratio 1 to 2 , then 2 ( x + 12 ) = y . substituting y = 4 x into the second equation yields 2 x + 24 = 4 x . so , x = 12 ( smal... | a ) 8 , b ) 16 , c ) 32 , d ) 48 , e ) 54 | d | multiply(divide(multiply(2, 12), subtract(4, 2)), 4) | multiply(n2,n4)|subtract(n1,n4)|divide(#0,#1)|multiply(n1,#2) | other |
if the annual increase in the population of a town is 10 % and the present number of people is 11000 , what will the population be in 2 years ? | "the required population is = 11000 ( 1 + 10 / 100 ) ^ 2 = 11000 * 11 / 10 * 11 / 10 = 13310 answer is e" | a ) 12100 , b ) 15240 , c ) 12456 , d ) 11452 , e ) 13310 | e | multiply(multiply(divide(add(10, const_100), const_100), 11000), divide(add(10, const_100), const_100)) | add(n0,const_100)|divide(#0,const_100)|multiply(n1,#1)|multiply(#1,#2)| | gain |
shannon and maxine work in the same building and leave work at the same time . shannon lives due north of work and maxine lives due south . the distance between maxine ' s house and shannon ' s house is 60 miles . if they both drive home at the rate 2 r miles per hour , maxine arrives home 30 minutes after shannon . if... | "nice question + 1 we have that x / 24 - ( 60 - x ) / 2 r = 30 also x / r - ( 60 - x ) / 2 r = 120 so we get that 2 x - 60 = 80 r 3 x - 60 = 240 r get rid of r 120 = 3 x x = 30 hence answer is c" | a ) 20 , b ) 34 , c ) 30 , d ) 40 , e ) 46 | c | divide(subtract(multiply(const_3, 60), 60), const_3) | multiply(n0,const_3)|subtract(#0,n0)|divide(#1,const_3)| | physics |
the owner of a furniture shop charges his customer 15 % more than the cost price . if a customer paid rs . 6400 for a computer table , then what was the cost price of the computer table ? | "cp = sp * ( 100 / ( 100 + profit % ) ) = 6400 ( 100 / 115 ) = rs . 5565 . answer : a" | a ) rs . 5565 , b ) rs . 6298 , c ) rs . 6290 , d ) rs . 6725 , e ) rs . 6708 | a | divide(6400, add(const_1, divide(15, const_100))) | divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)| | gain |
a train speeds past a pole in 15 seconds and a platform 140 meters long in 25 seconds . what is the length of the train ( in meters ) ? | "let the length of the train be x meters . the speed of the train is x / 15 . then , x + 140 = 25 * ( x / 15 ) 10 x = 2100 x = 210 meters the answer is c ." | a ) 190 , b ) 200 , c ) 210 , d ) 220 , e ) 230 | c | multiply(140, subtract(const_2, const_1)) | subtract(const_2,const_1)|multiply(n1,#0)| | physics |
it takes 3 workers a total of 15 hours to build a giant machine with each worker working at the same rate . if 5 workers start to build the machine at 11 : 00 am , and one extra worker per hour is added beginning at 1 : 00 pm , at what time will the machine be complete ? | "3 workers build 1 / 15 of a machine in one hour . 1 worker builds 1 / 45 of a machine in one hour . in the first 2 hours , 5 workers build 5 * ( 1 / 45 ) * 2 = 10 / 45 of a machine . from 1 : 00 to 2 : 00 , 6 workers build another 6 / 45 . the total is 16 / 45 . from 2 : 00 to 3 : 00 , 7 workers build another 7 / 45 .... | a ) 4 : 30 , b ) 4 : 45 , c ) 5 : 00 , d ) 5 : 15 , e ) 5 : 30 | e | subtract(divide(multiply(3, 15), 15), const_0_33) | multiply(n0,n1)|divide(#0,n1)|subtract(#1,const_0_33)| | general |
on the number line , if x is halfway between - 11 and 5 , and if y is halfway between - 3 and 7 , what number is halfway between x and y ? | "x = - 3 and y = 2 . the answer is b ." | a ) - 1.5 , b ) - 0.5 , c ) 0.5 , d ) 1.5 , e ) 2.5 | b | add(negate(11), 5) | negate(n0)|add(n1,#0)| | general |
john and david work at a hospital with 4 other workers . for an internal review , 2 of the 4 workers will be randomly chosen to be interviewed . what is the probability that john and david will both be chosen ? | "probability that john and david will both be chosen out of 8 workers = ( 2 / 4 ) * ( 1 / 3 ) = 1 / 6 answer e" | a ) 1 / 14 , b ) 1 / 20 , c ) 1 / 13 , d ) 1 / 7 , e ) 1 / 6 | e | multiply(2, multiply(divide(const_1, subtract(4, const_1)), divide(const_1, 4))) | divide(const_1,n0)|subtract(n0,const_1)|divide(const_1,#1)|multiply(#2,#0)|multiply(n1,#3)| | physics |
if w / x = 5 / 7 and w / y = 4 / 7 , then ( x + y ) / y = | "ratio 1 : 7 w = 5 x ratio 2 : 7 w = 4 y 5 x = 4 y x = 4 y / 3 ( x + y ) / y = ( ( 4 y / 3 ) + y ) / y = y ( 4 / 3 + 1 ) / y = 7 / 3 answer is b" | a ) 1 / 2 , b ) 7 / 3 , c ) 6 / 7 , d ) 1 / 5 , e ) 11 / 13 | b | add(divide(divide(4, 5), divide(7, 7)), const_1) | divide(n2,n0)|divide(n3,n1)|divide(#0,#1)|add(#2,const_1)| | general |
a worker ' s take - home pay last year was the same each month , and she saved the same fraction of her take - home pay each month . the total amount of money that she had saved at the end of the year was 6 times the amount of that portion of her monthly take - home pay that she did not save . if all the money that she... | "let x be the fraction of her take - home pay that the worker saved . let p be the monthly pay . 12 xp = 6 ( 1 - x ) p 12 xp = 6 p - 6 xp 18 xp = 6 p x = 1 / 3 the answer is d ." | a ) 1 / 6 , b ) 1 / 5 , c ) 1 / 4 , d ) 1 / 3 , e ) 1 / 2 | d | inverse(add(divide(const_12, 6), const_1)) | divide(const_12,n0)|add(#0,const_1)|inverse(#1)| | general |
the product of two numbers is 4107 . if the h . c . f of these numbers is 37 , then the greater number is : | "let the numbers be 37 a and 37 b . then , 37 a * 37 b = 4107 = > ab = 3 now , co - primes with product 3 are ( 1 , 3 ) . so , the required numbers are ( 37 * 1 , 37 * 3 ) i . e . , ( 1 , 111 ) . greater number = 111 . answer : c" | a ) 101 , b ) 107 , c ) 111 , d ) 112 , e ) 113 | c | sqrt(add(power(sqrt(subtract(37, multiply(const_2, 4107))), const_2), multiply(const_4, 4107))) | multiply(n0,const_4)|multiply(n0,const_2)|subtract(n1,#1)|sqrt(#2)|power(#3,const_2)|add(#0,#4)|sqrt(#5)| | general |
marts income is 60 percent more than tims income and tims income is 40 percent less than juans income . what percentage of juans income is marts income | "tim = 100 mary ' s income is 60 percent more than tims income ; mary ' s income = t + 0.6 t = 1.6 t = 1.6 * 100 = 160 tims income is 40 percent less than juans income . 100 is 40 % less than juan ' s income 100 = j - 0.4 j 100 = 0.6 j j = 100 / 0.6 = 1000 / 6 what percentage of juans income is marts income ( 160 / ( 1... | a ) 124 % , b ) 120 % , c ) 96 % , d ) 80 % , e ) 64 % | c | multiply(divide(add(const_100, 60), multiply(divide(const_100, subtract(const_100, 40)), const_100)), const_100) | add(n0,const_100)|subtract(const_100,n1)|divide(const_100,#1)|multiply(#2,const_100)|divide(#0,#3)|multiply(#4,const_100)| | general |
on a trip , a cyclist averaged 8 miles per hour for the first 12 miles and 12 miles per hour for the remaining 24 miles . if the cyclist returned immediately via the same route and took a total of 7.5 hours for the round trip , what was the average speed ( in miles per hour ) for the return trip ? | "the time to go 36 miles was 12 / 8 + 24 / 12 = 1.5 + 2 = 3.5 hours . the average speed for the return trip was 36 miles / 4 hours = 9 mph . the answer is a ." | a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 13 | a | divide(add(12, 24), subtract(7.5, add(divide(12, 8), divide(24, 12)))) | add(n1,n3)|divide(n1,n0)|divide(n3,n2)|add(#1,#2)|subtract(n4,#3)|divide(#0,#4)| | physics |
the price of a t . v . set worth rs . 20000 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 = ( 20000 - 1000 ) = rs . 19000 answer : d" | a ) 19002 , b ) 19028 , c ) 27788 , d ) 19000 , e ) 27711 | d | subtract(20000, 1000) | subtract(n0,n2)| | gain |
working alone , pump a can empty a pool in 4 hours . working alone , pump b can empty the same pool in 6 hours . working together , how many minutes will it take pump a and pump b to empty the pool ? | pump a can empty ( 1 / 4 ) of the pool per hour . pump b can empty ( 1 / 6 ) of the pool per hour . together the pumps can empty 1 / 4 + 1 / 6 = 5 / 12 of the pool per hour . 1 pool / ( 5 / 12 ) pool per hour = 12 / 5 hours = 144 minutes . the answer is c . | a ) 132 , b ) 138 , c ) 144 , d ) 150 , e ) 156 | c | multiply(inverse(add(divide(const_1, 4), divide(const_1, 6))), const_60) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|inverse(#2)|multiply(#3,const_60) | physics |
1704 x 1704 = ? | "1704 x 1704 = ( 1704 ) 2 = ( 1700 + 4 ) 2 = ( 1700 ) 2 + ( 4 ) 2 + ( 2 x 1700 x 4 ) = 2890000 + 16 + 13600 . = 2903616 . b )" | a ) a ) 3623216 , b ) b ) 2903616 , c ) c ) 3624316 , d ) d ) 3625116 , e ) e ) 3625216 | b | multiply(divide(1704, 1704), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
the radius of a semi circle is 6.83 cm then its perimeter is ? | "36 / 7 r = 6.3 = 34.97 answer : c" | a ) 32.52 , b ) 32.47 , c ) 34.97 , d ) 32.92 , e ) 32.33 | c | add(divide(circumface(6.83), const_2), multiply(6.83, const_2)) | circumface(n0)|multiply(n0,const_2)|divide(#0,const_2)|add(#2,#1)| | physics |
joan took out a mortgage from hel local bank . each monthly mortgage payment she makes must be triple the amount of the previous month ' s payment . if her first payment is $ 100 , and the total amount she must pay back is $ 109300 , how many months will it take joan to pay back her mortgage ? | "joan starts off with 100 $ . . which is to be tripled every month her monthly payments look like this : 100 , 300 , 900 , 2700 . . . . . . . . . upto 109300 this can be re written as : 100 x 1 , 100 x 3 , 100 x 9 , 100 x 27 . . . . . . 100 x 1093 so we have 1 , 3 , 9 , 27 . . . . . 109300 in gp we know that a = 1 , an... | a ) 7 , b ) 8 , c ) 10 , d ) 11 , e ) 13 | a | divide(log(add(divide(multiply(109300, const_2), 100), const_1)), log(const_3)) | log(const_3)|multiply(n1,const_2)|divide(#1,n0)|add(#2,const_1)|log(#3)|divide(#4,#0)| | general |
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