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 average age of a family of 6 members is 28 years . if the age of the youngest member is 7 years , what was the average age of the family at the birth of the youngest member ? | "present age of total members = 6 x 28 = 168 7 yrs back their ages were = 6 x 7 = 42 ages at the birth of youngest member = 168 - 42 = 126 therefore , avg age at the birth of youngest member = 126 / 6 = 21 . answer : c" | a ) 15 , b ) 18 , c ) 21 , d ) 12 , e ) 19 | c | subtract(divide(subtract(multiply(6, 28), 7), const_4), 7) | multiply(n0,n1)|subtract(#0,n2)|divide(#1,const_4)|subtract(#2,n2)| | general |
there are 84 supermarkets in the fgh chain . all of them are either in the us or canada . if there are 14 more fgh supermarkets in the us than in canada , how many fgh supermarkets are there in the us ? | x + ( x - 14 ) = 84 - - > x = 49 . answer : d . | a ) 20 , b ) 31 , c ) 42 , d ) 49 , e ) 64 | d | divide(add(84, 14), const_2) | add(n0,n1)|divide(#0,const_2) | general |
find out the square of a number which when doubled exceeds its one eighth by 15 ? | "let the number be p , then the square will be p ^ 2 according to question : 2 p = ( p / 8 ) + 15 = > 16 p = p + 120 = > p = 8 p ^ 2 = 8 ^ 2 = 64 . answer : d" | a ) 16 , b ) 25 , c ) 19 , d ) 64 , e ) 17 | d | power(divide(15, subtract(const_2, divide(const_1, add(const_4, const_1)))), const_2) | add(const_1,const_4)|divide(const_1,#0)|subtract(const_2,#1)|divide(n0,#2)|power(#3,const_2)| | general |
the h . c . f . of two numbers is 5 and their l . c . m . is 150 . if one of the numbers is 25 , then the other is : | "product of two nos = product of hcf and lcm let one number = x = > 25 * x = 5 * 150 x = 5 * 150 / 25 = 30 answer a" | a ) 30 , b ) 28 , c ) 24 , d ) 20 , e ) 35 | a | multiply(5, 25) | multiply(n0,n2)| | physics |
a and b are two partially filled buckets of water . if 4 liters are transferred from a to b , then a would contain one - third of the amount of water in b . alternatively , if 5 liters are transferred from b to a , b would contain one - half of the amount of water in a . bucket a contains how many liters of water ? | "let a contains a , b contains b liters so , ( a - 4 ) / ( b + 4 ) = 1 / 3 . . . . . . . ( 1 ) again , ( b - 5 ) / ( a + 5 ) = 1 / 2 . . . . . . . . . . ( 2 ) from ( 1 ) ( 2 ) we find a = 47 / 5 ans : ( a )" | a ) 47 / 5 , b ) 13 , c ) 17 , d ) 21 , e ) 23 | a | divide(multiply(add(add(4, const_3), const_2), divide(4, const_2)), add(const_2, divide(const_1, const_2))) | add(n0,const_3)|divide(n0,const_2)|divide(const_1,const_2)|add(#0,const_2)|add(#2,const_2)|multiply(#3,#1)|divide(#5,#4)| | general |
a sporting good store sells one type of baseball bat and one type of baseball . the cost for 2 bats and 4 balls is $ 220 . the cost for 1 bat and 6 balls is $ 190 , as well . if someone were to buy an equal number of bats and balls , at most how many bats can he purchase if he has a budget of $ 270 for the purchase ? | "imo it should be c that is 4 reason : formed an equation . . . bat = b ball = c 2 b + 4 c = 220 1 b + 6 c = 190 solving both we get b that is bat = 70 and c that is ball = 20 new equation 270 to be divided in equal 3 b + 3 c = 270 3 * 70 + 3 * 20 = 270 210 + 60 = 270" | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | divide(270, add(subtract(190, multiply(6, divide(subtract(190, divide(220, 2)), 4))), divide(subtract(190, divide(220, 2)), 4))) | divide(n2,n0)|subtract(n5,#0)|divide(#1,n1)|multiply(n4,#2)|subtract(n5,#3)|add(#2,#4)|divide(n6,#5)| | geometry |
in a coconut grove , ( x + 2 ) trees yield 30 nuts per year , x trees yield 120 nuts per year and ( x – 2 ) trees yield 180 nuts per year . if the average yield per year per tree be 100 , find x . | "( x + 2 ) × 30 + x × 120 + ( x − 2 ) × 180 / ( x + 2 ) + x + ( x − 2 ) = 100 ⇒ 330 x − 300 / 3 x = 100 ⇒ 30 x = 300 ⇒ x = 10 answer e" | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 10 | e | divide(subtract(multiply(180, 2), multiply(30, 2)), subtract(add(add(30, 120), 180), multiply(100, const_3))) | add(n1,n2)|multiply(n0,n4)|multiply(n0,n1)|multiply(n5,const_3)|add(n4,#0)|subtract(#1,#2)|subtract(#4,#3)|divide(#5,#6)| | general |
a football player scores 2 goals in his fifth match thus increasing his average goals score by 0.3 . the total number of goals in his 5 matches would be | "while this question can be solved with a rather straight - forward algebra approach ( as the other posters have noted ) , it can also be solved by testing the answers . one of those numbers must be the total number of goals . . . from a tactical standpoint , it ' s best to test either answer b or answer d , so if the ... | a ) 4 , b ) 6 , c ) 8 , d ) 10 , e ) 12 | a | add(subtract(multiply(const_4, 2), multiply(multiply(const_4, 5), 0.3)), 2) | multiply(n0,const_4)|multiply(n2,const_4)|multiply(n1,#1)|subtract(#0,#2)|add(n0,#3)| | general |
a salt solution contains 15 grams of salt per 1000 cubic centimeters of solution . if 25 cubic centimeters of the solution were poured into an empty container , how many grams of salt would be in the container ? | we are given that a salt solution contains 15 grams of salt per 1000 cubic centimeters of solution . since we are dealing with a solution , we know that the grams of salt is proportional to the number of cubic centimeters of solution . thus , to determine how many grams of salt would be in the container when we have 25... | ['a ) 0.550', 'b ) 0.350', 'c ) 0.450', 'd ) 0.700', 'e ) 0.375'] | e | divide(multiply(15, 25), 1000) | multiply(n0,n2)|divide(#0,n1) | physics |
jennifer had $ 120 to spend on herself . she spent 1 / 5 of the money on a sandwich , 1 / 6 for a ticket to a museum , and 1 / 2 of it on a book . how much money does jennifer have left over ? | "1 / 5 x $ 120 = $ 24 for sandwich 1 / 6 x $ 120 = $ 20 for museum 1 / 2 x $ 120 = $ 60 for book $ 24 + $ 20 + $ 60 = $ 104 spent $ 120 - $ 104 = $ 16 left over correct answer a" | a ) $ 16 , b ) $ 14 , c ) $ 5 , d ) $ 15 , e ) $ 0 | a | subtract(120, add(add(multiply(120, divide(1, 5)), multiply(120, divide(1, 6))), multiply(120, divide(1, 2)))) | divide(n1,n2)|divide(n1,n4)|divide(n1,n6)|multiply(n0,#0)|multiply(n0,#1)|multiply(n0,#2)|add(#3,#4)|add(#6,#5)|subtract(n0,#7)| | general |
if the sum of two numbers is 55 and the h . c . f . and l . c . m of these numbers are 5 and 120 respectively , then the sum of the reciprocals of the numbers is equal to : | "given a + b = 55 , hcm and lcm is 5 and 120 axb = lcmx hcf 600 = 120 x 5 1 / a + 1 / b = a + b / ab = > 55 / 600 = > 11 / 120 answer : c" | a ) 55 / 601 , b ) 601 / 55 , c ) 11 / 120 , d ) 120 / 11 , e ) 120 / 13 | c | divide(divide(55, 5), 120) | divide(n0,n1)|divide(#0,n2)| | general |
the banker ' s gain on a bill due due 1 year hence at 12 % per annum is rs . 7.8 . the true discount is | "solution t . d = [ b . g x 100 / r x t ] = rs . ( 7.8 x 100 / 12 x 1 ) = rs . 65 . answer b" | a ) rs . 72 , b ) rs . 65 , c ) rs . 54 , d ) rs . 50 , e ) none | b | divide(7.8, divide(12, const_100)) | divide(n1,const_100)|divide(n2,#0)| | gain |
a man engaged a servant on the condition that he would pay him rs . 900 and a uniform after 1 year service . he served only for 9 months and receiveduniform and rs . 650 , find the priceof the uniform ? | "9 / 12 = 3 / 4 * 900 = 675 650 - - - - - - - - - - - - - 25 1 / 4 - - - - - - - - 25 1 - - - - - - - - - ? = > rs . 100 b" | a ) rs . 90 , b ) rs . 100 , c ) rs . 120 , d ) rs . 150 , e ) rs . 160 | b | multiply(divide(subtract(multiply(1, 900), multiply(multiply(const_3, const_4), 9)), multiply(multiply(const_3, const_4), const_1)), const_4) | multiply(n0,n1)|multiply(const_3,const_4)|multiply(n2,#1)|multiply(#1,const_1)|subtract(#0,#2)|divide(#4,#3)|multiply(#5,const_4)| | general |
2 men and 4 boys can complete a work in 4 days . 5 men and 6 boys can complete the same work in 3 days . the work done by 2 boys is equal to the work of how many men ? | exp : ( 2 m + 4 b ) ’ s one day ’ s work = 1 / 4 ( 5 m + 6 b ) ’ s one day ’ s work = 1 / 3 = > ( 8 m + 16 b ) ’ s one day ’ s work = ( 15 m + 18 b ) ’ s one day ’ s work = > 7 men ’ s work = 2 boy ’ s work so , we should be employ 5 more men to be completed the work . answer : b | a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | b | multiply(5, divide(subtract(multiply(5, 4), multiply(3, 4)), subtract(multiply(5, 4), multiply(3, 4)))) | multiply(n1,n3)|multiply(n1,n5)|subtract(#0,#1)|divide(#2,#2)|multiply(n3,#3) | physics |
the sides of a rectangular field are in the ratio 3 : 4 . if the area of the field is 10092 sq . m , the cost of fencing the field @ 25 paise per metre is | "solution let length = ( 3 x ) metres and breadth = ( 4 x ) metres . then , 3 x × 4 x = 10092 ⇔ 12 x 2 = 10092 ⇔ x 2 = 841 ⇔ x = 29 . so , length = 87 m and breadth = 116 m . perimeter = [ 2 ( 87 + 116 ) ] m = 406 m . ∴ cost of fencing = rs . ( 0.25 × 406 ) = rs . 101.50 . answer d" | a ) rs . 55.50 , b ) rs . 67.50 , c ) rs . 86.50 , d ) rs . 101.50 , e ) none of these | d | divide(multiply(rectangle_perimeter(multiply(3, sqrt(divide(10092, multiply(3, 4)))), multiply(4, sqrt(divide(10092, multiply(3, 4))))), 25), const_100) | multiply(n0,n1)|divide(n2,#0)|sqrt(#1)|multiply(n0,#2)|multiply(n1,#2)|rectangle_perimeter(#3,#4)|multiply(n3,#5)|divide(#6,const_100)| | physics |
of 1000 surveyed students , 20 % of those who read book a also read book b and 25 % of those who read book b also read book a . if each student read at least one of the books , what is the difference between the number of students who read only book a and the number of students who read only book b ? | "say the number of students who read book a is a and the number of students who read book b is b . given that 20 % of those who read book a also read book b and 25 % of those who read book b also read book a , so the number of students who read both books is 0.2 a = 0.25 b - - > a = 1.25 b . since each student read at ... | a ) 120 , b ) 125 , c ) 130 , d ) 145 , e ) 155 | b | subtract(multiply(multiply(divide(1000, subtract(add(divide(divide(25, const_100), divide(20, const_100)), const_1), divide(25, const_100))), divide(divide(25, const_100), divide(20, const_100))), subtract(const_1, divide(20, const_100))), multiply(divide(1000, subtract(add(divide(divide(25, const_100), divide(20, cons... | divide(n2,const_100)|divide(n1,const_100)|divide(#0,#1)|subtract(const_1,#1)|subtract(const_1,#0)|add(#2,const_1)|subtract(#5,#0)|divide(n0,#6)|multiply(#7,#2)|multiply(#7,#4)|multiply(#8,#3)|subtract(#10,#9)| | gain |
a and b finish the job in 15 days . while a , b and c can finish it in 3 days . c alone will finish the job in | "explanation : 3 = ( 15 * x ) / ( 15 + x ) 45 + 3 x = 15 x 12 x = 45 x = 3.75 answer : option b" | a ) 4.75 days , b ) 3.75 days , c ) 6.50 days , d ) 7.50 days , e ) 5.50 days | b | divide(multiply(3, 15), subtract(15, 3)) | multiply(n0,n1)|subtract(n0,n1)|divide(#0,#1)| | physics |
the sum of 3 integers a , b and c is 60 . a is one third of the sum of b and c and b is one fifth of the sum of a and c . what is c ? | a is one third of the sum of b and c . . or a = 1 / 3 ( b + c ) . . . but a + b + c = 60 so 1 / 3 ( b + c ) + b + c = 60 4 ( b + c ) / 3 = 60 b + c = 60 * 3 / 4 = 45 a = 60 - 45 = 15 2 ) b is one fifth of the sum of a and c or b = 1 / 5 ( a + c ) . . . but a + b + c = 60 so 1 / 5 ( a + c ) + a + c = 60 a + c = 60 * 5 /... | a ) 40 , b ) 45 , c ) 35 , d ) 50 , e ) 55 | c | add(add(multiply(inverse(const_3), 60), multiply(inverse(add(const_1, const_4)), 60)), 3) | add(const_1,const_4)|inverse(const_3)|inverse(#0)|multiply(n1,#1)|multiply(n1,#2)|add(#3,#4)|add(n0,#5) | general |
the instructions state that cheryl needs 5 / 11 square yards of one type of material and 2 / 3 square yards of another type of material for a project . she buys exactly that amount . after finishing the project , however , she has 25 / 55 square yards left that she did not use . what is the total amount of square yards... | "total bought = 5 / 11 + 2 / 3 left part 25 / 55 - - - > 5 / 11 so used part 5 / 11 + 2 / 3 - 5 / 11 = 2 / 3 ans d" | a ) 1 / 12 , b ) 1 / 9 , c ) 1 2 / 3 , d ) 2 / 3 , e ) 2 1 / 9 | d | subtract(add(divide(5, 11), divide(2, 3)), divide(25, 55)) | divide(n0,n1)|divide(n2,n3)|divide(n4,n5)|add(#0,#1)|subtract(#3,#2)| | general |
in a class of 87 students 41 are taking french , 22 are taking german . of the students taking french or german , 9 are taking both courses . how many students are not enrolled in either course ? | formula for calculating two overlapping sets : a + b - both + not ( a or b ) = total so in our task we have equation : 41 ( french ) + 22 ( german ) - 9 ( both ) + not = 87 54 + not = 87 not = 87 - 54 = 33 so answer is d | a ) 6 , b ) 15 , c ) 24 , d ) 33 , e ) 54 | d | subtract(87, subtract(add(41, 22), 9)) | add(n1,n2)|subtract(#0,n3)|subtract(n0,#1) | other |
what is the least positive integer x for which ( x ^ 2 ) + 164 is divisible by 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , and 11 ? | 2 = 2 3 = 3 4 = 2 * 2 5 = 5 6 = 2 * 3 7 = 7 8 = 2 * 2 * 2 9 = 3 * 3 10 = 2 * 5 11 = 11 so ( x ^ 2 ) + 164 = ( 2 * 2 * 2 * 3 * 3 * 5 * 7 * 11 ) = 27720 . then ( x ^ 2 ) = 27556 . then x = 166 , which is option e . | a ) 27556 , b ) 800 , c ) 796 , d ) 87 , e ) 166 | e | sqrt(subtract(lcm(lcm(lcm(lcm(lcm(lcm(lcm(lcm(2, 3), 4), 5), 6), 7), 8), 9), 11), 164)) | lcm(n0,n3)|lcm(n4,#0)|lcm(n5,#1)|lcm(n6,#2)|lcm(n7,#3)|lcm(n8,#4)|lcm(n9,#5)|lcm(n11,#6)|subtract(#7,n1)|sqrt(#8) | general |
the cross - section of a cannel is a trapezium in shape . if the cannel is 16 m wide at the top and 4 m wide at the bottom and the area of cross - section is 700 sq m , the depth of cannel is ? | "1 / 2 * d ( 16 + 4 ) = 700 d = 70 answer : e" | a ) 76 , b ) 28 , c ) 27 , d ) 80 , e ) 70 | e | divide(divide(divide(700, divide(add(16, 4), const_2)), 4), const_2) | add(n0,n1)|divide(#0,const_2)|divide(n2,#1)|divide(#2,n1)|divide(#3,const_2)| | physics |
two equal triangles constitute rhombus a . if the diagonals of rhombus are 15 cm and 20 cm long . what is the area of each triangle ? | area of rhombus = 0.5 x 15 x 20 = 150 cm â ² area of triangle = 150 / 2 = 75 cm â ² answer : c | ['a ) 300', 'b ) 150', 'c ) 75', 'd ) 40', 'e ) 20'] | c | divide(multiply(15, 20), const_4) | multiply(n0,n1)|divide(#0,const_4) | geometry |
there is a total of 84 marbles in a box , each of which is red , green , blue , or white . if one marble is drawn from the box at random , the probability that it will be white is 1 / 4 and the probability that it will be green is 1 / 7 . what is the probability that the marble will be either red or blue ? | "p ( red or blue ) = 1 - p ( white ) - p ( green ) = 28 / 28 - 7 / 28 - 4 / 28 = 17 / 28 the answer is d ." | a ) 4 / 7 , b ) 9 / 14 , c ) 11 / 14 , d ) 17 / 28 , e ) 23 / 28 | d | subtract(1, add(divide(1, 4), divide(1, 7))) | divide(n1,n2)|divide(n3,n4)|add(#0,#1)|subtract(n1,#2)| | general |
if the side length of square b is four times that of square a , the area of square b is how many times the area of square a ? | let x be the side length of square a . then the area of square a is x ^ 2 . the area of square b is ( 4 x ) ^ 2 = 16 x ^ 2 . the answer is b . | ['a ) 20', 'b ) 16', 'c ) 12', 'd ) 8', 'e ) 4'] | b | multiply(const_4, const_4) | multiply(const_4,const_4) | geometry |
two trains of equal are running on parallel lines in the same direction at 50 km / hr and 36 km / hr . the faster train passes the slower train in 36 sec . the length of each train is ? | "let the length of each train be x m . then , distance covered = 2 x m . relative speed = 50 - 36 = 14 km / hr . = 14 * 5 / 18 = 35 / 9 m / sec . 2 x / 36 = 35 / 9 = > x = 70 . answer : a" | a ) 70 m , b ) 72 m , c ) 80 m , d ) 82 m , e ) 84 m | a | divide(multiply(36, divide(multiply(subtract(50, 36), const_1000), const_3600)), const_2) | subtract(n0,n1)|multiply(#0,const_1000)|divide(#1,const_3600)|multiply(n2,#2)|divide(#3,const_2)| | general |
a dog takes 3 leaps for every 7 leaps of a hare . if one leap of the dog is equal to 3 leaps of the hare , the ratio of the speed of the dog to that of the hare is : | "explanation : dog : hare = ( 3 * 3 ) leaps of hare : 7 leaps of hare = 9 : 7 . answer : a ) 9 : 7" | a ) 9 : 7 , b ) 9 : 9 , c ) 9 : 6 , d ) 9 : 1 , e ) 9 : 2 | a | divide(multiply(3, 3), 7) | multiply(n0,n2)|divide(#0,n1)| | other |
what is the dividend . divisor 17 , the quotient is 9 and the remainder is 5 ? | "d = d * q + r d = 17 * 9 + 5 d = 153 + 5 d = 158" | a ) 150 , b ) 152 , c ) 154 , d ) 158 , e ) 144 | d | add(multiply(17, 9), 5) | multiply(n0,n1)|add(n2,#0)| | general |
circular gears p and q start rotating at the same time at constant speeds . gear p makes 10 revolutions per minute and gear q makes 40 revolutions per minute . how many seconds after the gears start rotating will gear q have made exactly 2 more revolutions than gear p ? | "gear q makes 30 more revolutions each 60 seconds . the time to make 2 = 30 / 15 more revolutions is 60 / 15 = 4 seconds . the answer is a ." | a ) 4 , b ) 8 , c ) 10 , d ) 12 , e ) 15 | a | divide(2, subtract(divide(40, const_60), divide(10, const_60))) | divide(n1,const_60)|divide(n0,const_60)|subtract(#0,#1)|divide(n2,#2)| | physics |
a and b together can do a piece of work in 3 days . if a alone can do the same work in 20 days , then b alone can do the same work in ? | "b = 1 / 2 – 1 / 20 = 0.28 days answer : a" | a ) 0.28 days , b ) 0.45 days , c ) 0.55 days , d ) 0.25 days , e ) 0.15 days | a | inverse(subtract(inverse(3), inverse(20))) | inverse(n0)|inverse(n1)|subtract(#0,#1)|inverse(#2)| | physics |
the hcf of two numbers is 62 and the other two factors of their lcm are 11 and 12 . what is the largest number . | "explanation : hcf of the two numbers = 62 hcf will be always a factor of lcm 62 is factor of lcm other two factors are 11 & 12 then the numbers are ( 62 * 11 ) and ( 62 x 12 ) = 682 and 744 answer : option b" | a ) 462 , b ) 744 , c ) 480 , d ) 504 , e ) 555 | b | multiply(62, 12) | multiply(n0,n2)| | other |
a and b can do a piece of work in 45 days and 40 days respectively . they began to do the work together but a leaves after some days and then b completed the remaining work in 23 days . the number of days after which a left the work was ? | let the total units of work to be done be 360 . the units of work done by a in a single day = 8 similarly , the units of work done by b in a single day = 9 . a and b ’ s one day work = 17 units a and b worked together for some days = 17 x ( assume ) b ’ s work alone for 23 days = 23 x 9 = 207 so , the work done by a an... | a ) 12 , b ) 11 , c ) 10 , d ) 9 , e ) 8 | d | divide(subtract(const_1, divide(23, 40)), add(divide(const_1, 45), divide(const_1, 40))) | divide(n2,n1)|divide(const_1,n0)|divide(const_1,n1)|add(#1,#2)|subtract(const_1,#0)|divide(#4,#3) | physics |
teas worth rs . 126 per kg and rs . 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2 . if the mixture is worth rs 152 per kg , the price of the third variety per kg will be ? | "explanation : since first and second varieties are mixed in equal proportions . so , their average price = rs . ( 126 + 135 ) / 2 . = > rs . 130.50 . so , the mixture is formed by mixing two varieties , one at rs . 130.50 per kg and the other at say , rs . x per kg in the ratio 2 : 2 , i . e . , 1 : 1 . we have to fin... | a ) rs . 147.50 , b ) rs . 785.50 , c ) rs . 174.50 , d ) rs . 258.50 , e ) none of these | c | divide(subtract(multiply(152, add(add(1, 1), 2)), add(126, 135)), 2) | add(n2,n2)|add(n0,n1)|add(n4,#0)|multiply(n5,#2)|subtract(#3,#1)|divide(#4,n4)| | other |
when positive integer n is divided by 3 , the remainder is 1 . when n is divided by 8 , the remainder is 5 . what is the smallest positive integer p , such that ( n + p ) is a multiple of 15 ? | "when positive integer n is divided by 3 , the remainder is 1 i . e . , n = 3 x + 1 values of n can be one of { 1 , 4 , 7 , 10 , 13 , 16 , 19 , 22 . . . . . . . . . . . . . . 49 , 52 , 59 . . . . . . . . . . . . . . . . . . } similarly , when n is divided by 8 , the remainder is 5 . . i . e . , n = 7 y + 5 values of n ... | a ) 1 , b ) 2 , c ) 5 , d ) 19 , e ) 20 | b | subtract(15, reminder(5, 8)) | reminder(n3,n2)|subtract(n4,#0)| | general |
what are the last two digits of ( 301 * 402 * 502 * 604 * 646 * 547 * 448 * 349 ) ^ 2 | "( ( 301 * 402 * 502 * 604 * 646 ) * ( 547 * 448 * 349 ) ) ^ 2 if you observe above digits , last digit are : 1,2 , 2,4 , 6,7 , 8,9 ; 3 & 5 are missing ; so i have rearranged them so that multiplication will be easy for me as initial 4 digits have last two digits as 01 , 02,02 , 04,46 and final three as 47 * 48 * 49 . ... | a ) 96 , b ) 76 , c ) 56 , d ) 36 , e ) 16 | c | multiply(add(divide(const_100, const_4), multiply(multiply(multiply(multiply(2, const_3), subtract(const_1, 2)), subtract(const_1, 2)), subtract(const_1, 2))), const_4) | divide(const_100,const_4)|multiply(n8,const_3)|subtract(const_1,n8)|multiply(#1,#2)|multiply(#3,#2)|multiply(#4,#2)|add(#0,#5)|multiply(#6,const_4)| | general |
p runs 25 % faster than q so p gives q a 60 meter head start . if the race ends in a tie , how far ( in meters ) did p run in the race ? | "let d be the race distance that p ran . let t be the time it took to complete the race . let v be q ' s speed . t = d / 1.25 v = ( d - 60 ) / v d = 1.25 d - 75 0.25 d = 75 d = 300 meters . the answer is c ." | a ) 100 , b ) 200 , c ) 300 , d ) 400 , e ) 500 | c | add(divide(multiply(60, const_100), 25), 60) | multiply(n1,const_100)|divide(#0,n0)|add(n1,#1)| | gain |
joe invested a certain sum of money in a simple interest bond whose value grew to $ 560 at the end of 3 years and to $ 660 at the end of another 5 years . what was the rate of interest in which he invested his sum ? | "in 5 years , the value grew $ 100 , so the simple interest was $ 20 per year . in 3 years , the total interest was 3 * $ 20 = $ 60 the principal is $ 560 - $ 60 = 500 . the interest rate is $ 20 / $ 500 = 4 % the answer is b ." | a ) 3 % , b ) 4 % , c ) 5 % , d ) 6 % , e ) 7 % | b | multiply(divide(divide(subtract(660, 560), 5), subtract(560, multiply(divide(subtract(660, 560), 5), 3))), const_100) | subtract(n2,n0)|divide(#0,n3)|multiply(n1,#1)|subtract(n0,#2)|divide(#1,#3)|multiply(#4,const_100)| | gain |
the average marks of a class of 25 students is 40 and that of another class of 30 students is 60 . find the average marks of all the students ? | "sum of the marks for the class of 25 students = 25 * 40 = 1000 sum of the marks for the class of 30 students = 30 * 60 = 1800 sum of the marks for the class of 55 students = 1000 + 1800 = 2800 average marks of all the students = 2800 / 55 = 50.9 . answer : d" | a ) 52.6 , b ) 52.4 , c ) 52.1 , d ) 50.9 , e ) 52.9 | d | divide(add(multiply(25, 40), multiply(30, 60)), add(25, 30)) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0)| | general |
a lady grows cabbage in her garden that is in the shape of a square . each cabbage takes 1 square foot of area in her garden . this year , she has increased her output by 211 cabbages when compared to last year . the shape of the area used for growing the cabbage has remained a square in both these years . how many cab... | explanatory answer the shape of the area used for growing cabbages has remained a square in both the years . let the side of the square area used for growing cabbages this year be x ft . therefore , the area of the ground used for cultivation this year = x 2 sq . ft . let the side of the square area used for growing ca... | ['a ) 11236', 'b ) 11025', 'c ) 14400', 'd ) 12696', 'e ) can not be determined'] | a | power(add(divide(subtract(211, 1), const_2), 1), const_2) | subtract(n1,n0)|divide(#0,const_2)|add(n0,#1)|power(#2,const_2) | geometry |
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 / 3 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 / 3 ) = l cx = 3 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 / 15 , c ) 2 / 15 , d ) 1 / 6 , e ) 2 / 10 | d | multiply(divide(1, 3), 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 |
find the number of factors of 9321 ? | "9321 = 3 x 13 x 239 = 31 x 13 1 x 239 1 ; add one to all the powers i . e . , ( 1 + 1 ) x ( 1 + 1 ) x ( 1 + 1 ) = 8 ; we get eight factors 1 , 3 , 13 , 39 , 239 , 717 , 3107 , 9321 answer : c" | a ) 3 , b ) 6 , c ) 8 , d ) 16 , e ) 32 | c | add(const_2, const_2) | add(const_2,const_2)| | other |
the average ( arithmetic mean ) of 20 , 40 , and 60 is 5 more than the average of 10 , 60 , and what number ? | "a 1 = 120 / 3 = 40 a 2 = a 1 - 5 = 35 sum of second list = 35 * 3 = 105 therefore the number = 105 - 70 = 35 b" | a ) 30 , b ) 35 , c ) 40 , d ) 45 , e ) 55 | b | subtract(add(add(20, 40), 60), add(add(multiply(5, const_3), 10), 60)) | add(n0,n1)|multiply(n3,const_3)|add(n2,#0)|add(n4,#1)|add(n5,#3)|subtract(#2,#4)| | general |
on dividing a number by 357 , we get 39 as remainder . on dividing the same number by 17 , what will be the remainder ? | "let x be the number and y be the quotient . then , x = 357 * y + 39 = ( 17 * 21 * y ) + ( 17 * 2 ) + 5 = 17 * ( 21 y + 2 ) + 5 . required number = 5 . answer is b" | a ) 4 , b ) 5 , c ) 8 , d ) 7 , e ) 2 | b | multiply(subtract(divide(power(39, const_2), 357), floor(divide(power(39, const_2), 357))), 357) | power(n1,const_2)|divide(#0,n0)|floor(#1)|subtract(#1,#2)|multiply(n0,#3)| | general |
a man â € ™ s current age is ( 2 / 5 ) of the age of his father . after 12 years , he will be ( 1 / 2 ) of the age of his father . what is the age of father at now ? | "let , father â € ™ s current age is a years . then , man â € ™ s current age = [ ( 2 / 5 ) a ] years . therefore , [ ( 2 / 5 ) a + 12 ] = ( 1 / 2 ) ( a + 12 ) 2 ( 2 a + 60 ) = 5 ( a + 12 ) a = 60 a" | a ) 60 , b ) 45 , c ) 38 , d ) 50 , e ) 39 | a | divide(subtract(12, multiply(12, divide(1, 2))), subtract(divide(1, 2), divide(2, 5))) | divide(n3,n0)|divide(n0,n1)|multiply(n2,#0)|subtract(#0,#1)|subtract(n2,#2)|divide(#4,#3)| | general |
potato chips are on sale , buy 2 get 1 free . there are only 9 bags of chips left ( 4 plain and 5 bbq ) . a shopper randomly grabs 3 bags . what is the probability the shopper will grab exactly 3 bags of bbq chips ? | combination probability formula : ncr = n ! / [ r ! ( n - r ) ! ] total possible , select 3 bags from 9 bags = 9 c 3 = 9 ! / [ 3 ! ( 9 - 3 ) ! ] = 84 . to have 3 bbq there must be 0 plain , select 0 plain from 4 = 4 c 0 = 1 . and , select 3 bbq from 5 = 5 c 3 = 10 . 3 bbq and 0 plain = ( 1 ) ( 10 ) = 10 probability = (... | a ) 2 / 3 , b ) 1 / 3 , c ) 1 / 5 , d ) 5 / 42 , e ) 7 / 42 | d | multiply(divide(subtract(subtract(5, const_1), const_1), subtract(subtract(9, const_1), const_1)), multiply(divide(5, 9), divide(subtract(5, const_1), subtract(9, const_1)))) | divide(n4,n2)|subtract(n4,const_1)|subtract(n2,const_1)|divide(#1,#2)|subtract(#1,const_1)|subtract(#2,const_1)|divide(#4,#5)|multiply(#0,#3)|multiply(#6,#7) | probability |
two spherical balls lie on the ground touching . if one of the balls has a radius of 10 cm , and the point of contact is 12 cm above the ground , what is the radius of the other ball ( in centimeters ) ? | "a straight line will join the two centers and the point of contact , thus making similar triangles . 2 / 10 = ( r - 12 ) / r 2 r = 10 r - 120 r = 15 the answer is b ." | a ) 14 , b ) 15 , c ) 16 , d ) 18 , e ) 20 | b | add(add(10, 12), 10) | add(n0,n1)|add(#0,n0)| | physics |
there are two bags a and b . a contains 6 red flowers and 3 pink flowers . where as bag b contains 2 red flowers and 7 pink flowers . one flower is chosen from a bag randomly . what is the probability that the flower chosen is pink ? | bag a 6 red + 3 pink bag b 2 red + 7 pink probability of picking 1 pink from bag a is 3 / 9 probability of picking 1 pink from bag b is 7 / 9 total = 3 / 9 + 7 / 9 = 10 / 9 and probability of selecting a bag is 1 / 2 so finally probability of choosing a pink flower = 1 / 2 ( 10 / 9 ) = 5 / 9 answer : d | a ) 5 / 8 , b ) 5 / 6 , c ) 5 / 1 , d ) 5 / 9 , e ) 5 / 2 | d | multiply(add(divide(const_3, add(6, 3)), divide(7, add(7, 2))), divide(const_1, const_2)) | add(n0,n1)|add(n2,n3)|divide(const_1,const_2)|divide(const_3,#0)|divide(n3,#1)|add(#3,#4)|multiply(#5,#2) | probability |
the length of the bridge , which a train 180 meters long and travelling at 45 km / hr can cross in 30 seconds , is ? | "speed = ( 45 * 5 / 18 ) m / sec = ( 25 / 2 ) m / sec . time = 30 sec . let the length of bridge be x meters . then , ( 180 + x ) / 30 = 25 / 2 = = > 2 ( 180 + x ) = 750 = = > x = 195 m . answer : a" | a ) 195 m , b ) 156 m , c ) 245 m , d ) 156 m , e ) 156 m | a | subtract(multiply(divide(multiply(45, speed(const_1000, const_1)), speed(const_3600, const_1)), 30), 180) | speed(const_1000,const_1)|speed(const_3600,const_1)|multiply(n1,#0)|divide(#2,#1)|multiply(n2,#3)|subtract(#4,n0)| | physics |
a mixture of 50 litres of milk and water contains 10 % water . how many litres of water should be added to the mixture so that the mixture contains 15 % water ? | quantity of milk in the mixture = 90 / 100 ( 50 ) = 45 litres . after adding water , milk would form 60 % of the mixture . hence , if quantity of mixture after adding x liters of water , ( 60 ) / 100 x = 45 = > x = 75 hence 75 - 50 = 25 litres of water must be added . answer : c | a ) 15 , b ) 20 , c ) 25 , d ) 30 , e ) 10 | c | subtract(divide(multiply(50, 15), 10), 50) | multiply(n0,n2)|divide(#0,n1)|subtract(#1,n0) | general |
a train 100 m long crosses a platform 125 m long in 30 sec ; find the speed of the train ? | "d = 100 + 125 = 225 t = 30 s = 225 / 30 * 18 / 5 = 27 kmph answer : c" | a ) 36 , b ) 77 , c ) 27 , d ) 33 , e ) 99 | c | subtract(multiply(30, multiply(const_60.0, const_0_2778)), 100) | multiply(const_60.0,const_0_2778)|multiply(n2,#0)|subtract(#1,n0)| | physics |
a small company is planning to rent either computer a or computer b to print customer mailing lists . both computer a and computer b must be rented on an hourly basis . the rental fee is based only on the amount of time the computer is turned on . it will cost 40 percent more per hour to rent computer a than to rent co... | "pa = price of a pb = price of b ta = time for a to complete the job tb = time for b to complete the job given pa = 1.4 pb ta + 20 = tb pa * ta = pb * tb = 650 1.4 pb * ( tb - 20 ) = pb * tb 1.4 pb tb - pb tb = 1.4 pb * 20 0.4 pbtb = 28 pb tb = 28 / 0.4 = 70 pb = 650 / 70 ~ 9.28 a" | a ) $ 9.28 , b ) $ 11.30 , c ) $ 13.20 , d ) $ 17.80 , e ) $ 22.10 | a | add(divide(40, const_100), add(add(const_4, const_3), add(divide(40, const_100), divide(divide(20, const_4), const_100)))) | add(const_3,const_4)|divide(n0,const_100)|divide(n1,const_4)|divide(#2,const_100)|add(#1,#3)|add(#0,#4)|add(#5,#1)| | general |
each of the 20 boxes in a certain shipment weighs either 10 pounds or 20 pounds , and the average ( arithmetic mean ) weight of the boxes in the shipment is 18 pounds . if the average weight of the boxes in the shipment is to be reduced to 16 pounds by removing some of the 20 - pound boxes , how many 20 - pound boxes m... | "if the average of 10 - pound and 20 - pound boxes is 18 , the ratio of 10 - pound boxes : 20 - pound boxes is 1 : 4 . so out of 20 boxes , 4 are 10 - pound boxes and 16 are 20 - pound boxes . if the average of 10 and 20 - pound boxes is to be 16 , the ratio of 10 - pound boxes : 20 - pound boxes should be 2 : 3 . the ... | a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 15 | b | divide(subtract(add(multiply(10, divide(subtract(multiply(20, 20), multiply(20, 18)), subtract(20, 10))), multiply(20, subtract(20, divide(subtract(multiply(20, 20), multiply(20, 18)), subtract(20, 10))))), multiply(20, 16)), subtract(20, 16)) | multiply(n0,n2)|multiply(n0,n3)|multiply(n0,n4)|subtract(n2,n1)|subtract(n2,n4)|subtract(#0,#1)|divide(#5,#3)|multiply(n1,#6)|subtract(n0,#6)|multiply(n2,#8)|add(#7,#9)|subtract(#10,#2)|divide(#11,#4)| | general |
a television have side 6 * 5 . its screen size is 5 * 4 . calculate the percentage of unscreened part t . v . | unscreened part = 30 - 20 = 10 . % unscreened part = 10 / 30 * 100 = 33.33 % answer c | a ) 30 , b ) 25 , c ) 33.33 , d ) 44.44 , e ) 28 | c | multiply(divide(subtract(multiply(6, 5), multiply(5, 4)), multiply(6, 5)), const_100) | multiply(n0,n1)|multiply(n1,n3)|subtract(#0,#1)|divide(#2,#0)|multiply(#3,const_100) | general |
how many 5 digit number contain number 3 ? | "total 5 digit no . = 9 * 10 * 10 * 10 * 10 = 90000 not containing 3 = 8 * 9 * 9 * 9 * 9 = 52488 total 5 digit number contain 3 = 90000 - 52488 = 37512 answer : e" | a ) 31512 , b ) 32512 , c ) 33512 , d ) 34512 , e ) 37512 | e | add(subtract(subtract(const_1000, const_10), multiply(multiply(const_10, multiply(5, 5)), multiply(const_4, const_2))), const_10) | multiply(n0,n0)|multiply(const_2,const_4)|subtract(const_1000,const_10)|multiply(#0,const_10)|multiply(#3,#1)|subtract(#2,#4)|add(#5,const_10)| | general |
a train crosses a post in 15 seconds and a platform 100 m long in 25 seconds . its length is | explanation : assume x is the length of the train and v is the speed x / v = 15 = > v = x / 15 ( x + 100 ) / v = 25 = > v = ( x + 100 ) / 25 ie , x / 15 = ( x + 100 ) / 25 = > 5 x = 3 x + 300 = > x = 300 / 2 = 150 answer : option a | a ) 150 m , b ) 300 m , c ) 400 m , d ) 180 m , e ) 160 m | a | divide(multiply(15, 100), subtract(25, 15)) | multiply(n0,n1)|subtract(n2,n0)|divide(#0,#1) | physics |
how many factors does 49 ^ 2 have ? | "36 ^ 2 = 6 * 6 * 6 * 6 = 2 ^ 4 * 3 ^ 4 total factors = ( 4 + 1 ) * ( 4 + 1 ) = 6 * 6 = 36 answer e ." | a ) 2 , b ) 8 , c ) 24 , d ) 25 , e ) 36 | e | divide(49, multiply(const_10, const_2)) | multiply(const_10,const_2)|divide(n0,#0)| | other |
a cube of side one meter length is cut into small cubes of side 25 cm each . how many such small cubes can be obtained ? | "along one edge , the number of small cubes that can be cut = 100 / 25 = 4 along each edge 25 cubes can be cut . ( along length , breadth and height ) . total number of small cubes that can be cut = 4 * 4 * 4 = 64 answer : d" | a ) 56 , b ) 60 , c ) 62 , d ) 64 , e ) 65 | d | divide(power(power(25, const_2), const_3), power(25, const_3)) | power(n0,const_2)|power(n0,const_3)|power(#0,const_3)|divide(#2,#1)| | physics |
last week john spent 10 percent of his wages on recreation . this week , his wages are 10 percent less than last week ʼ s wages and he spent 40 percent of his wages on recreation . the amount he spends on recreation this week is what percent of the amount he spent on recreation last week | "say john ' s wages last week were $ 100 , so he spent 0.10 * 100 = $ 10 on recreation ; this week ' s wages is 0.90 * 100 = $ 90 , so he spends 0.4 * 90 = $ 36 on recreation ; 36 / 10 = 3.60 , hence the amount he spends on recreation this week is 360 % of the amount he spent on recreation last week : 10 * 3.6 = 36 ans... | a ) 180 % , b ) 360 % , c ) 200 % , d ) 220 % , e ) 250 % | b | multiply(divide(multiply(divide(40, const_100), multiply(divide(subtract(const_100, 10), const_100), const_100)), multiply(divide(10, const_100), const_100)), const_100) | divide(n2,const_100)|divide(n0,const_100)|subtract(const_100,n1)|divide(#2,const_100)|multiply(#1,const_100)|multiply(#3,const_100)|multiply(#0,#5)|divide(#6,#4)|multiply(#7,const_100)| | general |
company s produces two kinds of stereos : basic and deluxe . of the stereos produced by company s last month , 2 / 3 were basic and the rest were deluxe . if it takes 8 / 5 as many hours to produce a deluxe stereo as it does to produce a basic stereo , then the number of hours it took to produce the deluxe stereos last... | "# of basic stereos was 2 / 3 of total and # of deluxe stereos was 1 / 3 of total , let ' s assume total = 15 , then basic = 10 and deluxe = 5 . now , if time needed to produce one deluxe stereo is 1 unit than time needed to produce one basic stereo would be 8 / 5 units . total time for basic would be 10 * 1 = 10 and t... | a ) 7 / 17 , b ) 14 / 31 , c ) 4 / 9 , d ) 17 / 35 , e ) 1 / 2 | c | divide(add(3, 2), const_10) | add(n0,n1)|divide(#0,const_10)| | general |
the age of mother one decade age was 3 times the age of her daughter . 10 years hence , mother ’ s age will be two times that of her daughter . the proposition of their current ages is : | a 7 : 3 let , ages of mother and daughter 1 decade ago was 3 a and a years correspondingly then , ( 3 a + 10 ) + 10 = 2 [ ( a + 10 ) + 10 ] 3 a + 20 = 2 a + 40 x = 20 needed ratio = ( 3 a + 10 ) : ( a + 10 ) = 70 : 30 = 7 : 3 | a ) 7 : 3 , b ) 6 : 3 , c ) 8 : 3 , d ) 2 : 3 , e ) 6 : 3 | a | divide(add(multiply(3, divide(subtract(multiply(const_2, add(10, 10)), add(10, 10)), subtract(3, const_2))), 10), add(divide(subtract(multiply(const_2, add(10, 10)), add(10, 10)), subtract(3, const_2)), 10)) | add(n1,n1)|subtract(n0,const_2)|multiply(#0,const_2)|subtract(#2,#0)|divide(#3,#1)|add(n1,#4)|multiply(n0,#4)|add(n1,#6)|divide(#7,#5) | general |
a line with slope 2 intersects a line with slope 6 at the point ( 40 , 30 ) . what is the distance between the x - intercepts of these lines ? | the line with slope 2 passing through ( 40 , 30 ) has x - intercept 40 − ( 30 / 2 ) . the line with slope 6 passing through ( 40 , 30 ) has x - intercept 40 − ( 30 / 6 ) . so the difference in the x - intercepts is ( 30 / 2 ) − ( 30 / 6 ) = 10 . correct answer d | a ) 4 , b ) 6 , c ) 8 , d ) 10 , e ) 12 | d | divide(divide(multiply(30, 40), multiply(6, 2)), const_10) | multiply(n2,n3)|multiply(n0,n1)|divide(#0,#1)|divide(#2,const_10) | general |
is two years older than b who is twice as old as c . if the total of the ages of a , b and c be 47 , then how old is b ? | let c ' s age be x years . then , b ' s age = 2 x years . a ' s age = ( 2 x + 2 ) years . ( 2 x + 2 ) + 2 x + x = 47 5 x = 45 = > x = 9 hence , b ' s age = 2 x = 18 years . answer : d | a ) 7 , b ) 8 , c ) 9 , d ) 18 , e ) 11 | d | multiply(divide(subtract(47, const_2), add(const_3, const_2)), const_2) | add(const_2,const_3)|subtract(n0,const_2)|divide(#1,#0)|multiply(#2,const_2) | general |
right now , al and eliot have bank accounts , and al has more money than eliot . the difference between their two accounts is 1 / 12 of the sum of their two accounts . if al ’ s account were to increase by 10 % and eliot ’ s account were to increase by 20 % , then al would have exactly $ 20 more than eliot in his accou... | "lets assume al have amount a in his bank account and eliot ' s bank account got e amount . we can form an equation from the first condition . a - e = 1 / 12 * ( a + e ) = = > 11 a = 13 e - - - - - - - - - - - - ( 1 ) second condition gives two different amounts , al ' s amount = 1.1 a and eliot ' s amount = 1.2 e 1.1 ... | a ) $ 110 , b ) $ 120 , c ) $ 180 , d ) $ 200 , e ) $ 260 | d | divide(multiply(20, const_100), subtract(multiply(add(12, 1), 10), add(const_100, 20))) | add(n0,n1)|add(n3,const_100)|multiply(n4,const_100)|multiply(#0,n2)|subtract(#3,#1)|divide(#2,#4)| | general |
a bus covered a distance of 250 km , partly at an average speed of 40 kmph and partly at 60 kmph . if the total time taken is 6.5 hours , then the distance covered at 40 kmph is | "let the partial distance covered at 40 kmph be x let the another partial distance covered at 60 kmph be ( 250 - x ) thus , x / 40 - ( 250 - x ) / 60 = 6.5 or , x / 40 + ( 250 - x ) / 60 = 6.5 or , ( 3 x + 500 - 2 x ) / / 120 = 6.5 or 500 + x = 780 x = 280 answer : d" | a ) 130 km , b ) 120 km , c ) 100 km , d ) 280 km , e ) none of these | d | multiply(divide(subtract(multiply(60, 6.5), 250), negate(subtract(40, 60))), 40) | multiply(n2,n3)|subtract(n1,n2)|negate(#1)|subtract(#0,n0)|divide(#3,#2)|multiply(n1,#4)| | physics |
at an election 2 candidates are participated a candidate got 45 % of votes and defeated by 9000 votes . 83 votes are invalid . find the total polled votes ? | "winners votes = 100 - 45 = 55 % invalid votes = 83 polled votes = [ ( 100 * 9000 ) / ( 2 * 55 - 100 ) ] + 83 = 90083 answer is d" | a ) 100000 , b ) 85236 , c ) 89562 , d ) 90083 , e ) 95842 | d | add(const_100, multiply(divide(9000, subtract(subtract(const_100.0, 45), 45)), 83)) | subtract(const_100.0,n1)|subtract(#0,n1)|divide(n2,#1)|multiply(#2,const_100)|add(#3,n3)| | gain |
a searchlight on top of the watchtower makes 2 revolutions per minute . what is the probability that a man appearing near the tower will stay in the dark for at least 6 seconds ? | "the searchlight completes one revolution every 30 seconds . the probability that the man ' s area will be lit up is 6 / 30 = 1 / 5 . the probability that he will stay in the dark is 1 - 1 / 5 = 4 / 5 the answer is b ." | a ) 5 / 6 , b ) 4 / 5 , c ) 3 / 4 , d ) 2 / 3 , e ) 1 / 2 | b | subtract(const_1, divide(6, multiply(6, const_3))) | multiply(n1,const_3)|divide(n1,#0)|subtract(const_1,#1)| | physics |
a shop owner professes to sell his articles at certain cost price but he uses false weights with which he cheats by 12 % while buying and by 20 % while selling . what is his percentage profit ? | "the owner buys 100 kg but actually gets 112 kg ; the owner sells 100 kg but actually gives 80 kg ; profit : ( 112 - 80 ) / 80 * 100 = ~ 40 % answer : d ." | a ) 10.22 % , b ) 20.22 % , c ) 21.22 % , d ) 40 % , e ) ca n ' t be calculated | d | divide(multiply(subtract(add(const_100, 12), subtract(const_100, 20)), const_100), subtract(const_100, 20)) | add(n0,const_100)|subtract(const_100,n1)|subtract(#0,#1)|multiply(#2,const_100)|divide(#3,#1)| | gain |
tough and tricky questions : number properties . if ( z + 3 ) / 9 is an integer , what is remainder when z is divided by 9 ? | assume the answer choices as the value of z 9 / 9 = 1 ans e | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | e | subtract(9, 3) | subtract(n1,n0) | general |
if x dollars is invested at 9 percent for one year and y dollars is invested at 8 percent for one year , the annual income from the 9 percent investment will exceed the annual income from the 8 percent investment by $ 48 . if $ 2,000 is the total amount invested , how much is invested at 9 percent ? | "2 equations with 2 unknowns 9 x / 100 - 8 y / 100 = 48 and x + y = 2000 solving these 2 equations , x = 1223.53 and y = 776.47 answer - c" | a ) 300 , b ) 250 , c ) 1223.53 , d ) 1133.24 , e ) 776.47 | c | subtract(multiply(multiply(const_100, 9), const_2), divide(add(multiply(multiply(9, 8), const_2), 48), add(divide(9, const_100), divide(8, const_100)))) | divide(n0,const_100)|divide(n1,const_100)|multiply(n0,const_100)|multiply(n0,n1)|add(#0,#1)|multiply(#2,const_2)|multiply(#3,const_2)|add(n4,#6)|divide(#7,#4)|subtract(#5,#8)| | general |
if the wheel is 8 cm then the number of revolutions to cover a distance of 948 cm is ? | "2 * 22 / 7 * 8 * x = 948 = > x = 18.9 answer : e" | a ) 11 , b ) 14 , c ) 14 , d ) 12 , e ) 19 | e | divide(948, multiply(multiply(const_2, divide(add(add(multiply(const_3, const_100), multiply(const_1, const_10)), const_4), const_100)), 8)) | multiply(const_100,const_3)|multiply(const_1,const_10)|add(#0,#1)|add(#2,const_4)|divide(#3,const_100)|multiply(#4,const_2)|multiply(n0,#5)|divide(n1,#6)| | physics |
the cost price of 11 articles is equal to the selling price of 10 articles . what is the profit percent ? | "10 * sp = 11 * cp sp = 1.1 * cp the profit percent is 10 % . the answer is b ." | a ) 5 % , b ) 10 % , c ) 15 % , d ) 20 % , e ) 25 % | b | divide(multiply(11, const_4), add(const_4, const_1)) | add(const_1,const_4)|multiply(n0,const_4)|divide(#1,#0)| | gain |
if x < y < z and y - x > 5 , where x is an even integer and y and z are odd integers , what is the least possible value d of z - x ? | "x < y < z to find the least possible value for z - x ; we need to find the values for z and x that can be closest to each other . if x is some even number , then what could be minimum possible odd z . if x is some even number y - x > 5 ; y > x + 5 ; minimum value for y = x + 5 + 2 = x + 7 [ note : x + 5 is as even + o... | a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | d | add(add(5, const_2), const_2) | add(n0,const_2)|add(#0,const_2)| | general |
if money is invested at r percent interest , compounded annually , the amount of the investment will double in approximately 70 / r years . if pat ' s parents invested $ 5,000 in a long - term bond that pays 8 percent interest , compounded annually , what will be the approximate total amount of the investment 18 years ... | since investment doubles in 70 / r years , then for r = 8 it ' ll double in 70 / 8 = ~ 9 years ( we are not asked about the exact amount so such an approximation will do ) . thus in 18 years investment will double twice and become ( $ 5,000 * 2 ) * 2 = $ 20,000 ( after 9 years investment will become $ 5,000 * 2 = $ 10,... | a ) $ 20000 , b ) $ 15000 , c ) $ 12000 , d ) $ 10000 , e ) $ 9000 | a | divide(multiply(multiply(add(const_2, const_3), const_1000), 8), const_2) | add(const_2,const_3)|multiply(#0,const_1000)|multiply(n2,#1)|divide(#2,const_2) | general |
how many multiples of 5 are there between 5 to 90 ? | "explanation : as you know , multiples of 5 are integers having 0 or 5 in the digit to the extreme right ( i . e . the units place ) . so the numbers are 10 , 15 , 20 , 25 , 30 , 35 , 40 , 45 , 50 , 55 , 60 , 65 , 70 , 75 , 80 , 85 . answer : b" | a ) 9 , b ) 16 , c ) 17 , d ) none of these , e ) can not be determined | b | add(divide(subtract(90, 5), 5), const_1) | subtract(n2,n1)|divide(#0,n0)|add(#1,const_1)| | general |
instead of multiplying a number by 9 , the number is divided by 9 . what is the percentage of error obtained ? | "let the number be x the right number is 9 x the wrong number is x / 9 error is ( 9 x - x / 9 ) = 80 x / 9 percentage of error is ( ( 80 x / 9 ) / 9 x ) * 100 = 98.765 % answer : e" | a ) 97.765 % , b ) 99.765 % , c ) 96.765 % , d ) 95.765 % , e ) 98.765 % | e | subtract(const_100, multiply(divide(divide(9, 9), multiply(9, 9)), const_100)) | divide(n1,n1)|multiply(n0,n1)|divide(#0,#1)|multiply(#2,const_100)|subtract(const_100,#3)| | general |
1500 men have provisions for 17 days . if 320 more men join them , for how many days will the provisions last now ? | "1500 * 17 = 1820 * x x = 14 answer : c" | a ) 12.9 , b ) 12.0 , c ) 14 , d ) 15 , e ) 16 | c | divide(multiply(17, 1500), add(1500, 320)) | add(n0,n2)|multiply(n0,n1)|divide(#1,#0)| | physics |
the cost of carpeting a room 15 meters long with a carpet 75 cm wide at 30 paisa per meter is rs . 36 . the breadth of the room is ? | length of carpet = total cost / rate = 3600 / 30 = 120 m area of carpet = ( 120 x 75 ) / 100 m 2 = 90 m 2 ∴ area of the room = 90 m 2 breadth of the room = area / length = 90 / 15 m = 6 m answer : a | a ) 6 meters , b ) 8 meters , c ) 9 meters , d ) 10 meters , e ) 12 meters | a | divide(multiply(divide(multiply(36, const_100), 30), divide(75, const_100)), 15) | divide(n1,const_100)|multiply(n3,const_100)|divide(#1,n2)|multiply(#2,#0)|divide(#3,n0) | physics |
the ratio between the length and the breadth of a rectangular park is 3 : 2 . if a man cycling along the boundary of the park at the speed of 12 km / hr completes one round in 7 minutes , then the area of the park ( in sq . m ) is : | "perimeter = distance covered in 7 min . = ( 12000 / 60 ) x 7 m = 1400 m . let length = 3 x metres and breadth = 2 x metres . then , 2 ( 3 x + 2 x ) = 1400 or x = 140 . length = 420 m and breadth = 280 m . area = ( 420 x 280 ) m 2 = 117600 m 2 . answer : d" | a ) 153601 , b ) 153600 , c ) 153602 , d ) 117600 , e ) 153604 | d | rectangle_area(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 7), const_1000), add(3, 2)), const_2), multiply(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 7), const_1000), add(3, 2)), const_2), 2)) | add(n0,n1)|multiply(const_2,const_3)|multiply(#1,const_10)|divide(n2,#2)|multiply(n3,#3)|multiply(#4,const_1000)|divide(#5,#0)|divide(#6,const_2)|multiply(n1,#7)|rectangle_area(#7,#8)| | physics |
james took a 3 - hour bike ride . in the second hour he traveled 6 miles , which was 20 percent farther than he traveled the first hour . if he traveled 25 percent farther in the third hour than he did in the second hour , how many miles did jose travel during the entire ride ? | "let the distance travelled in the first hour be x . thus , 1.2 x = 6 , x = 5 . now , the distance travelled in the 3 rd hour = 6 + 1 / 4 â ˆ — 6 = 7.5 the only option ending with a 0.5 in the decimal place is c . answer : c" | a ) 54.0 , b ) 54.9 , c ) 18.5 , d ) 57.0 , e ) 63.0 | c | add(add(multiply(divide(subtract(const_100, 20), const_100), 6), 6), multiply(divide(add(const_100, 25), const_100), 6)) | add(n3,const_100)|subtract(const_100,n2)|divide(#1,const_100)|divide(#0,const_100)|multiply(n1,#2)|multiply(n1,#3)|add(n1,#4)|add(#6,#5)| | physics |
what least number must be added to 9599 , so that the sum is completely divisible by 100 ? | "if we divide 9599 by 100 remainder is 99 100 - 99 = 1 answer : a" | a ) 1 , b ) 2 , c ) 4 , d ) 6 , e ) 8 | a | subtract(multiply(add(multiply(const_4, const_10), const_2), 100), 9599) | multiply(const_10,const_4)|add(#0,const_2)|multiply(n1,#1)|subtract(#2,n0)| | general |
in august , a cricket team that played 120 matches won 28 % of the games it played . after a continuous winning streak , this team raised its average to 52 % . how many matches did the team win to attain this average ? | "let the no of matches played more = x so , ( 120 + x ) * 52 / 100 = 33.6 + x by solving we get x = 60 answer : c" | a ) 40 , b ) 52 , c ) 60 , d ) 80 , e ) 98 | c | divide(subtract(multiply(divide(52, const_100), 120), multiply(divide(28, const_100), 120)), subtract(const_1, divide(52, const_100))) | divide(n2,const_100)|divide(n1,const_100)|multiply(n0,#0)|multiply(n0,#1)|subtract(const_1,#0)|subtract(#2,#3)|divide(#5,#4)| | general |
a fish can go upstream at 35 kmph and downstream at 55 kmph , and then find the speed of the fish in still water ? | us = 35 ds = 55 m = ( 35 + 55 ) / 2 = 45 answer : a | a ) 45 , b ) 55 , c ) 35 , d ) 65 , e ) 70 | a | divide(add(35, 55), const_2) | add(n0,n1)|divide(#0,const_2) | physics |
two numbers are respectively 20 % and 25 % more than a third number . so find the percentage that is first of the second is ? | "first - 120 second - 125 third - 100 125 - - - - 120 100 - - - - - 96 answer a" | a ) 96 , b ) 20 , c ) 35 , d ) 78 , e ) 89 | a | subtract(const_100, multiply(divide(add(20, const_100), add(25, const_100)), const_100)) | add(n0,const_100)|add(n1,const_100)|divide(#0,#1)|multiply(#2,const_100)|subtract(const_100,#3)| | gain |
a shopkeeper sells 20 % of his stock at 10 % profit ans sells the remaining at a loss of 5 % . he incurred an overall loss of rs . 450 . find the total worth of the stock ? | "let the total worth of the stock be rs . x . the sp of 20 % of the stock = 1 / 5 * x * 1.1 = 11 x / 50 the sp of 80 % of the stock = 4 / 5 * x * 0.95 = 19 x / 25 = 38 x / 50 total sp = 11 x / 50 + 38 x / 50 = 49 x / 50 overall loss = x - 49 x / 50 = x / 50 x / 50 = 450 = > x = 22500 answer : b" | a ) 20029 , b ) 22500 , c ) 20289 , d ) 20027 , e ) 20026 | b | divide(450, subtract(multiply(divide(5, const_100), divide(subtract(const_100, 20), const_100)), multiply(divide(10, const_100), divide(20, const_100)))) | divide(n2,const_100)|divide(n1,const_100)|divide(n0,const_100)|subtract(const_100,n0)|divide(#3,const_100)|multiply(#1,#2)|multiply(#0,#4)|subtract(#6,#5)|divide(n3,#7)| | gain |
the number which exceeds 16 % of it by 126 is : | "solution solution let the number be x . x - 16 % of x = 126 x - 16 / 100 x = 126 x - 4 / 25 x = 126 21 / 25 x = 126 x = ( 126 x 25 / 21 ) = 150 answer a" | a ) 150 , b ) 52 , c ) 58 , d ) 60 , e ) 62 | a | divide(multiply(126, const_100), subtract(const_100, 16)) | multiply(n1,const_100)|subtract(const_100,n0)|divide(#0,#1)| | gain |
a person ' s present age is two - fifth of the age of his mother . after 8 years , he will be one - half of the age of his mother . how old id the mother at present ? | "let the mother ' s present age be x years . then , the person ' s present age = 2 / 5 x years . ( 2 / 5 x + 8 ) = 1 / 2 ( x + 8 ) 2 ( 2 x + 40 ) = 5 ( x + 8 ) = > x = 40 answer : c" | a ) 46 , b ) 49 , c ) 40 , d ) 21 , e ) 19 | c | divide(subtract(8, add(const_2, const_3)), subtract(divide(const_1, const_2), divide(const_2, add(const_2, const_3)))) | add(const_2,const_3)|divide(const_1,const_2)|divide(const_2,#0)|subtract(n0,#0)|subtract(#1,#2)|divide(#3,#4)| | general |
4.509 / 10.02 = | "4.509 / 10.02 4509 / 1002 = 4.5 move the comma two places to the left as we have 2 decimal places from the divisor : 0.45 . answer : a" | a ) 0.45 , b ) 0.3505 , c ) 0.3509 , d ) 0.351 , e ) 0.3527 | a | multiply(divide(4.509, 10.02), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
the captain of a cricket team of 11 members is 26 years old and the wicket keeper is 5 years older . if the ages of these two are excluded , the average age of the remaining players is one year less than the average age of the whole team . what is the average age of the team ? | "let the average age of the whole team be x years . 11 x - ( 26 + 31 ) = 9 ( x - 1 ) 11 x - 9 x = 48 2 x = 48 x = 24 . the average age of the team is 24 years . the answer is d ." | a ) 21 , b ) 22 , c ) 23 , d ) 24 , e ) 25 | d | divide(subtract(add(26, add(26, 5)), multiply(const_3.0, const_3.0)), const_2) | add(n1,n2)|multiply(const_3.0,const_3.0)|add(n1,#0)|subtract(#2,#1)|divide(#3,const_2)| | general |
the ratio of type of pen and type of notebook is 5 : 4 . and in 80 ways one can start writing . tell me the no . of pen and no . of notebook . | let the ratio constant be x . no . of pen = 5 x , no . of notebook = 4 x . total no . of ways of start writing = 5 x * 4 x = 20 ( square of x ) = 80 x = â ˆ š 4 = 2 pen = 10 , notebook = 8 answer c | a ) 6,12 , b ) 10,6 , c ) 10,8 , d ) 8,4 , e ) 6,8 | c | divide(divide(multiply(5, 80), 4), const_10) | multiply(n0,n2)|divide(#0,n1)|divide(#1,const_10) | other |
joe ’ s average ( arithmetic mean ) test score across 4 equally weighted tests was 50 . he was allowed to drop his lowest score . after doing so , his average test score improved to 55 . what is the lowest test score that was dropped ? | "the arithmetic mean of 4 equally weighted tests was 50 . so what we can assume is that we have 4 test scores , each 50 . he dropped his lowest score and the avg went to 55 . this means that the lowest score was not 50 and other three scores had given the lowest score 5 each to make it up to 50 too . when the lowest sc... | a ) 20 , b ) 35 , c ) 55 , d ) 65 , e ) 80 | b | subtract(multiply(50, 4), multiply(55, const_3)) | multiply(n0,n1)|multiply(n2,const_3)|subtract(#0,#1)| | general |
jamshid and irwin decided to make up their own version of basketball . in addition to being able to score 2 - point baskets , they created ways to score 5 - , 11 - , and 13 - point baskets . if at halftime the product of the point values of irwin ’ s baskets was 2420 , how many 11 - point baskets e did irwin score ? | first thing i noticed was that all these numbers are prime . . . prime factorization gives the following step 1 : 2 * 1210 step 2 : 2 * 11 * 110 step 4 : 2 * 11 * 11 * 10 step 5 : 2 * 11 * 11 * 2 * 5 so he scored e = 2 11 - point baskets in my opinion . answer c | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | c | divide(divide(divide(2420, 11), 11), const_10) | divide(n4,n2)|divide(#0,n2)|divide(#1,const_10) | general |
a train is running at a speed of 40 km / hr and it crosses a post in 18 seconds . what is the length of the train ? | speed of the train , v = 40 km / hr = 40000 / 3600 m / s = 400 / 36 m / s time taken to cross , t = 18 s distance covered , d = vt = ( 400 / 36 ) × 18 = 200 m distance covered is equal to the length of the train = 200 m correct answer is 200 metres | a ) 190 metres , b ) 160 metres , c ) 200 metres , d ) 120 metres , e ) 250 metres | c | multiply(18, multiply(40, const_0_2778)) | multiply(n0,const_0_2778)|multiply(n1,#0) | physics |
70 % of 300 + 25 % of 400 - ? = 256 | "explanation : solution : let 70 % of 300 + 25 % of 400 - x = 256 . then , x = ( 70 / 100 * 300 ) + ( 25 / 100 * 400 ) - 256 = 210 + 100 - 256 = 54 . answer : c" | a ) 48 , b ) 68 , c ) 54 , d ) 60 , e ) none of these | c | divide(multiply(70, 300), 25) | multiply(n0,n1)|divide(#0,n2)| | general |
riya and priya set on a journey . riya moves eastward at a speed of 30 kmph and priya moves westward at a speed of 50 kmph . how far will be priya from riya after 45 minutes | "total eastward distance = 30 kmph * 3 / 4 hr = 22.50 km total westward distance = 50 kmph * 3 / 4 hr = 37.5 km total distn betn them = 22.5 + 37.5 = 60 km ans 60 km answer : b" | a ) 25 kms , b ) 60 kms , c ) 50 kms , d ) 30 kms , e ) 40 kms | b | multiply(speed(add(30, 50), const_60), 45) | add(n0,n1)|speed(#0,const_60)|multiply(n2,#1)| | physics |
let c be defined as the sum of all prime numbers between 0 and 23 . what is c / 3 | "prime numbers between 0 and 30 - 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 sum , c = 77 c / 3 = 25.3 answer d" | a ) 155 , b ) 129 , c ) 61 , d ) 25.3 , e ) 43 | d | add(divide(3, const_10), power(const_2, add(const_2, const_4))) | add(const_2,const_4)|divide(n2,const_10)|power(const_2,#0)|add(#1,#2)| | general |
a cistern which could be filled in 10 hours takes 2 hour more to be filled owing to a leak in its bottom . if the cistern is full , in what time will the leak empty it ? | "let the leak empty the full cistern in x hours 10 x / x - 10 = 10 + 1 x = 110 hours answer is e" | a ) 50 hours , b ) 80 hours , c ) 100 hours , d ) 150 hours , e ) 110 hours | e | inverse(subtract(divide(const_1, 10), divide(const_1, const_10))) | divide(const_1,n0)|divide(const_1,const_10)|subtract(#0,#1)|inverse(#2)| | physics |
when the positive integer x is divided by 9 , the remainder is 5 . what is the remainder when 5 x is divided by 9 ? | "i tried plugging in numbers x = 9 q + 5 x = 14 5 x = 70 5 x / 9 = 63 + 7 remainder is 7 . answer is e ." | a ) 0 , b ) 1 , c ) 3 , d ) 4 , e ) 7 | e | reminder(multiply(5, 5), 9) | multiply(n1,n2)|reminder(#0,n0)| | general |
when positive integer x is divided by positive integer y , the remainder is 7 . if x / y = 86.1 , what is the value of y ? | "when positive integer x is divided by positive integer y , the remainder is 7 - - > x = qy + 7 ; x / y = 86.1 - - > x = 86 y + 0.1 y ( so q above equals to 86 ) ; 0.1 y = 7 - - > y = 70 . answer : c ." | a ) 72 , b ) 71 , c ) 70 , d ) 30 , e ) 45 | c | divide(7, subtract(86.1, floor(86.1))) | floor(n1)|subtract(n1,#0)|divide(n0,#1)| | general |
find the simple interest on $ 10000 at 8 % per annum for 12 months ? | "p = $ 10000 r = 8 % t = 12 / 12 years = 1 year s . i . = p * r * t / 100 = 10000 * 8 * 1 / 100 = $ 800 answer is d" | a ) $ 410 , b ) $ 500 , c ) $ 650 , d ) $ 800 , e ) $ 1000 | d | multiply(10000, divide(8, const_100)) | divide(n1,const_100)|multiply(n0,#0)| | gain |
a fruit - salad mixture consists of apples , peaches , and grapes in the ratio 6 : 5 : 2 , respectively , by weight . if 78 pounds of the mixture is prepared , the mixture includes how many more pounds of apples than grapes ? | "we can first set up our ratio using variable multipliers . we are given that a fruit - salad mixture consists of apples , peaches , and grapes , in the ratio of 6 : 5 : 2 , respectively , by weight . thus , we can say : apples : peaches : grapes = 6 x : 5 x : 2 x we are given that 39 pounds of the mixture is prepared ... | a ) 24 , b ) 12 , c ) 9 , d ) 6 , e ) 4 | a | subtract(multiply(6, const_2), multiply(2, const_2)) | multiply(n0,const_2)|multiply(n2,const_2)|subtract(#0,#1)| | general |
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