options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 727 | problem stringlengths 5 967 | rationale stringlengths 1 2.74k | program stringlengths 10 646 |
|---|---|---|---|---|---|
a ) 5.2 , b ) 1.0 , c ) 0.1 , d ) 2.0 , e ) 5.0 | c | divide(multiply(multiply(sqrt(3), sqrt(3)), 3), const_100) | β 3 percent of 3 β 3 = | "the other solutions are great , but here ' s one more approach . first , we scan the answer choices to see that they are quite spread apart . this means we can also use someapproximation all test - takers must memorize the following approximations : β 2 β 1.4 β 3 β 1.7 so , β 3 percent of 3 β 3 β 1.7 percent of ( 3 ) ( 1.7 ) β ( 1.7 / 100 ) ( 5.1 ) β 8.67 / 100 β 0.08 check the answer choices to see that c is the only answer that ' s close . answer : c" | a = math.sqrt(3)
b = math.sqrt(3)
c = a * b
d = c * 3
e = d / 100
|
a ) 20 , b ) 50 , c ) 60 , d ) 40 , e ) 30 | b | multiply(multiply(divide(100, 40), const_2), 10) | machine a produces 100 parts twice as fast as machine b does . machine b produces 100 parts in 40 minutes . if each machine produces parts at a constant rate , how many parts does machine a produce in 10 minutes ? | "machine b produces 100 part in 40 minutes . machine a produces 100 parts twice as fast as b , so machine a produces 100 parts in 40 / 2 = 20 minutes . now , machine a produces 100 parts in 20 minutes which is 100 / 20 = 5 parts / minute . 5 parts x a total of 10 minutes = 50 b" | a = 100 / 40
b = a * 2
c = b * 10
|
a ) 8 : 6 , b ) 8 : 7 , c ) 8 : 9 , d ) 8 : 4 , e ) 8 : 1 | b | divide(multiply(multiply(3, 2), const_4), multiply(7, 3)) | an order was placed for the supply of a carper whose length and breadth were in the ratio of 3 : 2 . subsequently , the dimensions of the carpet were altered such that its length and breadth were in the ratio 7 : 3 but were was no change in its parameter . find the ratio of the areas of the carpets in both the cases ? | let the length and breadth of the carpet in the first case be 3 x units and 2 x units respectively . let the dimensions of the carpet in the second case be 7 y , 3 y units respectively . from the data , . 2 ( 3 x + 2 x ) = 2 ( 7 y + 3 y ) = > 5 x = 10 y = > x = 2 y required ratio of the areas of the carpet in both the cases = 3 x * 2 x : 7 y : 3 y = 6 x 2 : 21 y 2 = 6 * ( 2 y ) 2 : 21 y 2 = 6 * 4 y 2 : 21 y 2 = 8 : 7 answer : b | a = 3 * 2
b = a * 4
c = 7 * 3
d = b / c
|
a ) 70 , b ) 96 , c ) 108 , d ) 120 , e ) 150 | d | divide(subtract(32, 20), divide(10, const_100)) | of the diplomats who attended a summit conference : 20 spoke french , 32 did not speak english and 20 % of the diplomats spoke neither french nor english . if 10 % of the diplomats spoke both french and english , then how many diplomats attended the conference ? | 2 x 2 matrix will be the easiest way to calculate this . text in black : given statements text in red : calculated values thus d = 120 is the correct answer | a = 32 - 20
b = 10 / 100
c = a / b
|
a ) $ 400 , b ) $ 1400 , c ) $ 3400 , d ) $ 2400 , e ) $ 4400 | d | add(multiply(divide(1400, subtract(1, add(add(divide(2, 5), divide(3, 10)), divide(1, 8)))), divide(3, 10)), multiply(divide(1400, subtract(1, add(add(divide(2, 5), divide(3, 10)), divide(1, 8)))), divide(1, 8))) | a man spends 2 / 5 of his salary on house rent , 3 / 10 of his salary on food and 1 / 8 of his salary on conveyance . if he has $ 1400 left with him , find his expenditure on food . | "part of salary left = 1 - ( 2 / 5 + 3 / 10 + 1 / 8 ) = 7 / 40 let the monthly salary be $ x then ( 7 / 40 ) of x = 1400 x = 8000 expenditure on food = $ 2400 option d" | a = 2 / 5
b = 3 / 10
c = a + b
d = 1 / 8
e = c + d
f = 1 - e
g = 1400 / f
h = 3 / 10
i = g * h
j = 2 / 5
k = 3 / 10
l = j + k
m = 1 / 8
n = l + m
o = 1 - n
p = 1400 / o
q = 1 / 8
r = p * q
s = i + r
|
a ) 63 . , b ) 73 . , c ) 81 , d ) 91 , e ) 108 | a | subtract(subtract(420, 312), subtract(250, divide(multiply(250, 82), const_100))) | in the fifth grade at parkway elementary school there are 420 students . 312 students are boys and 250 students are playing soccer . 82 % of the students that play soccer are boys . how many girl student are in parkway that is not playing soccer ? | "total students = 420 boys = 312 , girls = 108 total playing soccer = 250 82 % of 250 = 205 are boys who play soccer . girls who play soccer = 45 . total girls who do not play soccer = 108 - 45 = 63 . correct option : a" | a = 420 - 312
b = 250 * 82
c = b / 100
d = 250 - c
e = a - d
|
a ) 36 , b ) 33 , c ) 44 , d ) 64 , e ) 56 | b | add(add(subtract(add(multiply(5, 5), multiply(5, 10)), add(multiply(5, 10), 5)), 10), const_3) | x , a , z , and b are single digit positive integers . x = 1 / 5 a . z = 1 / 5 b . ( 10 a + b ) β ( 10 x + z ) could not equal | a = 5 x , b = 5 z therefore ( 5 x * 10 + 5 z ) - ( 10 x + z ) = ( 5 - 1 ) ( 10 x + z ) = . ( 10 x + z ) number should be divisible by 4 b | a = 5 * 5
b = 5 * 10
c = a + b
d = 5 * 10
e = d + 5
f = c - e
g = f + 10
h = g + 3
|
a ) 25 liters , b ) 27 liters , c ) 30 liters , d ) 32 liters , e ) 35 liters | c | subtract(divide(multiply(const_2, const_100), subtract(20, multiply(divide(75, const_100), 20))), 10) | how much water should be added to 10 liters of a 20 % - solution of alcohol to reduce the concentration of alcohol in the solution by 75 % ? | "let x ltr water to be added 2 ltr alcohol to be represented as ( 20 ( 1 - 3 / 4 ( new soln . = 10 + x ) ) ) 2 = 5 % * ( 10 + x ) - - - - - - - - > x = 30 answer : c" | a = 2 * 100
b = 75 / 100
c = b * 20
d = 20 - c
e = a / d
f = e - 10
|
a ) 1 / 9 , b ) 4 / 6 , c ) 3 / 6 , d ) 1 / 2 , e ) 1 / 6 | e | divide(const_1, factorial(3)) | on a game show , a contestant is given 3 keys , each of which opens exactly one of 3 identical boxes . the first box contains $ 5 , the second $ 500 , and the third $ 5000 . the contestant assigns each key to one of the boxes and wins the amount of money contained in any box that is opened by the key assigned to it . what is the probability that a contestant will win more than $ 5000 ? | let ' s call the boxes that contain $ 5 , $ 500 , and $ 5000 , respectively , box a , box b , box c . these are opened , respectively , by key a , key b , and key c . we want to know the probability of winning more than $ 5000 . notice that if the distribution of keys is : box a = key b box b = key a box c = key c then the contestant wins exactly $ 5000 , not more than $ 5000 . the only configuration that leads to winning more than $ 1000 is : box a = key a box b = key b box c = key c i . e . , getting all three keys correct . that ' s the only way to be more than $ 5000 . so , really , the question can be rephrased : what is the probability of guessing the order of keys so that each key matches the correct box ? well , for a set of three items , the number of possible permutations is 3 ! = 3 * 2 * 1 = 6 . of those 6 possible permutations , only one of them leads to all three keys being paired with the right box . so , the answer is probability = 1 / 6 answer : e | a = math.factorial(3)
b = 1 / a
|
a ) 15 , b ) 16 , c ) 18 , d ) 25 , e ) 10 | b | divide(20, add(const_1, divide(25, const_100))) | sakshi can do a piece of work in 20 days . tanya is 25 % more efficient than sakshi . the number of days taken by tanya to do the same piece of work : | "solution ratio of times taken by sakshi and tanya = 125 : 100 = 5 : 4 . suppose tanya taken x days to do the work . 5 : 4 : : 20 : x β x = ( 20 x 4 / 5 ) β x = 16 days . hence , tanya takes 16 days is complete the work . answer b" | a = 25 / 100
b = 1 + a
c = 20 / b
|
a ) 987 , b ) 177 , c ) 357 , d ) 267 , e ) 380 | e | multiply(add(18, const_1), add(add(18, const_1), const_1)) | there are 18 stations between hyderabad and bangalore . how many second class tickets have to be printed , so that a passenger can travel from any station to any other station ? | "the total number of stations = 20 from 20 stations we have to choose any two stations and the direction of travel ( i . e . , hyderabad to bangalore is different from bangalore to hyderabad ) in Β² β° p β ways . Β² β° p β = 20 * 19 = 380 . answer : e" | a = 18 + 1
b = 18 + 1
c = b + 1
d = a * c
|
a ) 2.5 , b ) 2.4 , c ) 2.3 , d ) 2.2 , e ) 2.1 | a | divide(90, add(sqrt(divide(subtract(add(multiply(90, 3), multiply(90, 3)), multiply(divide(const_1, const_2), multiply(3, 3))), divide(const_1, const_2))), 3)) | a boat traveled upstream 90 miles at an average speed of ( v - 3 ) miles per hour and then traveled the same distance downstream at an average speed of ( v + 3 ) miles per hour . if the trip upstream took a half hour longer than the trip downstream , then how many hours did it take the boat to travel downstream ? | 90 / ( v - 3 ) = 90 / ( v + 3 ) + 0.5 hours = > 90 ( v + 3 ) = ( v - 3 ) [ 90 + 0.5 v + 1.5 ] 90 v + 270 = ( v - 3 ) ( 183 / 2 + v / 2 ) 180 v + 540 = ( v - 3 ) ( 183 + v ) = 183 v - 549 + v ^ 2 - 3 v v ^ 2 - 1089 = 0 = > v = 33 miles / hr = > time downstream = 90 / 36 = 30 / 12 = 5 / 2 = 2.5 hrs answer a 2.5 | a = 90 * 3
b = 90 * 3
c = a + b
d = 1 / 2
e = 3 * 3
f = d * e
g = c - f
h = 1 / 2
i = g / h
j = math.sqrt(i)
k = j + 3
l = 90 / k
|
a ) 14 , b ) 15 , c ) 20 , d ) 22 , e ) 25 | b | multiply(divide(3, 4), 20) | in a certain country 1 / 3 of 4 = 6 . assuming the same proportion , what would be the value of 1 / 6 of 20 ? | "b 15" | a = 3 / 4
b = a * 20
|
a ) 4 days , b ) 5 days , c ) 8 days , d ) 9 days , e ) 10 days | b | inverse(divide(add(add(inverse(6), inverse(10)), inverse(7.5)), const_2)) | working together , a and b can do a job in 6 days . b and c can do the same job in 10 days , while c and a can do it in 7.5 days . how long will it take if all a , b and c work together to complete the job ? | a and b 1 day work = 1 / 6 b and c 1 day work = 1 / 10 c and a 1 day work = 1 / 7.5 a , b , c 1 day work = ( 1 / 6 + 1 / 10 + 1 / 7.5 ) * 1 / 2 = 1 / 5 so a . b . c completes work in 5 days answer : b | a = 1/(6)
b = 1/(10)
c = a + b
d = 1/(7)
e = c + d
f = e / 2
g = 1/(f)
|
a ) a ) 2 , b ) b ) 3 , c ) c ) 4 , d ) d ) 5 , e ) e ) 7 | e | sqrt(49) | from below option 49 is divisible by which one ? | 49 / 7 = 7 e | a = math.sqrt(49)
|
a ) 17608 , b ) 17920 , c ) 17604 , d ) 17600 , e ) 117601 | b | divide(multiply(add(const_100, 12), add(divide(multiply(12500, const_100), subtract(const_100, 20)), add(125, 250))), const_100) | ramesh purchased a refrigerator for rs . 12500 after getting a discount of 20 % on the labelled price . he spent rs . 125 on transport and rs . 250 on installation . at what price should it be sold so that the profit earned would be 12 % if no discount was offered ? | "price at which the tv set is bought = rs . 12,500 discount offered = 20 % marked price = 12500 * 100 / 80 = rs . 15625 the total amount spent on transport and installation = 125 + 250 = rs . 375 \ total price of tv set = 15625 + 375 = rs . 16000 the price at which the tv should be sold to get a profit of 10 % if no discount was offered = 16000 * 112 / 100 = rs . 17920 . answer : b" | a = 100 + 12
b = 12500 * 100
c = 100 - 20
d = b / c
e = 125 + 250
f = d + e
g = a * f
h = g / 100
|
a ) 22 , b ) 33 , c ) 44 , d ) 48 , e ) 99 | d | multiply(multiply(12, const_2), const_2) | if x / ( 12 p ) is an even prime number , where x is a positive integer and p is a prime number , what is the least value of x ? | x / ( 12 p ) = even prime number x = even prime number * 11 p least value of x = lowest even prime number * 12 * lowest value of p = 2 * 12 * 2 = 48 answer d | a = 12 * 2
b = a * 2
|
a ) 10 , b ) 30 , c ) 28 , d ) 60 , e ) 90 | c | divide(multiply(8, subtract(8, const_1)), const_2) | there are 8 players in a chess group , and each player plays each of the others once . given that each game is played by two players , how many total games will be played ? | "10 players are there . two players play one game with one another . so 8 c 2 = 8 * 7 / 2 = 28 so option c is correct" | a = 8 - 1
b = 8 * a
c = b / 2
|
a ) 1.3 , b ) 1.7 , c ) 2.1 , d ) 2.5 , e ) 2.9 | d | divide(multiply(45, 5), subtract(const_100, 10)) | a driver just filled the car ' s gas tank with 45 liters of gasohol , a mixture consisting of 5 % ethanol and 95 % gasoline . if the car runs best on a mixture consisting of 10 % ethanol and 90 % gasoline , how many liters of ethanol must be added into the gas tank for the car to achieve optimum performance ? | "let x be the number of liters of ethanol added to the gas tank . 0.05 ( 45 ) + x = 0.1 ( 45 + x ) 0.9 x = 4.5 - 2.25 = 2.25 x = 2.5 liters the answer is d ." | a = 45 * 5
b = 100 - 10
c = a / b
|
a ) 24 , b ) 77 , c ) 297 , d ) 267 , e ) 95 | e | multiply(subtract(42, 23), 5) | think of a number , divide it by 5 and add 23 to it . the result is 42 . what is the number thought of ? | "explanation : 42 - 23 = 19 19 x 5 = 95 answer : e" | a = 42 - 23
b = a * 5
|
a ) $ 118.00 , b ) $ 120.00 , c ) $ 121.00 , d ) $ 122.00 , e ) $ 140.00 | c | multiply(100, power(add(const_1, divide(divide(20, const_100), const_2)), const_2)) | ms . lopez deposits $ 100 in an account that pays 20 % interest , compounded semiannually . how much money will there be in the account at the end of one year ? | "et me explain the concept for those who are not familiar with compounded semi annually / quarterly etc . . it basically means , here , that the interest will be paid twice , after 6 months and at the end of year . . so twice . . but interest is 20 % , which will get distributed semiannually that is 10 % . . . so the q literally means . . after 6 months . . 100 * 10 / 100 + 100 = 110 . . at the end of year 110 * 10 / 100 + 110 = 121 . . or we put it in compound interest formula . . where ci = 10 % , t is 2 and p = 100 . . so amount = 100 ( 1 + 10 / 100 ) ^ 2 = 100 * 1.1 * 1.1 = 121 . . answer : c" | a = 20 / 100
b = a / 2
c = 1 + b
d = c ** 2
e = 100 * d
|
a ) a ) 6 , b ) b ) 4 , c ) c ) 8 , d ) d ) 10 , e ) e ) 12 | b | divide(subtract(multiply(86, 12), multiply(84, 12)), subtract(92, 86)) | here ' s an easy question of averages , but let ' s try to see innovative ways of solving this . a class has 12 boys and x girls . average score of boys and girls is 84 and 92 respectively . the average of the whole class is 86 , what is the value of x ? | 12 ( 84 ) + 92 x / 12 + x = 86 1008 + 92 x / 12 + x = 86 1008 + 92 x = 86 ( 12 + x ) 1008 + 92 x = 1032 + 86 x x ' s one side , numbers one side we get , 92 x - 86 x = 1032 - 1008 6 x = 24 hence , x = 4 answer b | a = 86 * 12
b = 84 * 12
c = a - b
d = 92 - 86
e = c / d
|
a ) 1440 , b ) 1220 , c ) 1400 , d ) 1500 , e ) 1620 | a | add(1000, divide(multiply(1000, 20), const_100)) | the present population of a town is 1000 . population increase rate is 20 % p . a . find the population of town after 2 years ? | "p = 1000 r = 20 % required population of town = p * ( 1 + r / 100 ) ^ t = 1000 * ( 1 + 20 / 100 ) ^ 2 = 1000 * ( 6 / 5 ) ^ 2 = 1440 ( approximately ) answer is a" | a = 1000 * 20
b = a / 100
c = 1000 + b
|
a ) 3377 , b ) 2688 , c ) 2688 , d ) 8774 , e ) 1268 | d | multiply(7500, multiply(divide(add(const_100, 4), const_100), divide(add(const_100, 4), const_100))) | if rs . 7500 are borrowed at c . i at the rate of 4 % per annum , then after 4 years the amount to be paid is ? | "a = 7500 ( 26 / 25 ) ^ 4 = 8774 answer : d" | a = 100 + 4
b = a / 100
c = 100 + 4
d = c / 100
e = b * d
f = 7500 * e
|
a ) 77 , b ) 60 , c ) 99 , d ) 26 , e ) 21 | b | multiply(150, divide(const_2, add(2, 3))) | kim can do a work in 3 days while david can do the same work in 2 days . both of them finish the work together and get rs . 150 . what is the share of kim ? | kim ' s wages : david ' s wages = kim ' s 1 day work : david ' s 1 day work = 1 / 3 : 1 / 2 = 2 : 3 kim ' s share = 2 / 5 * 150 = rs . 60 answer : b | a = 2 + 3
b = 2 / a
c = 150 * b
|
a ) 60.5 , b ) 69.5 , c ) 67.5 , d ) 67.7 , e ) 22.5 | d | divide(multiply(add(51.50, divide(multiply(51.50, 25), const_100)), const_100), subtract(const_100, 5)) | at what price must an article costing rs . 51.50 be marked in order that after deducting 5 % from the list price . it may be sold at a profit of 25 % on the cost price ? | "cp = 51.50 sp = 51.50 * ( 125 / 100 ) = 64.375 mp * ( 95 / 100 ) = 64.375 mp = 67.7 answer : d" | a = 51 * 50
b = a / 100
c = 51 + 50
d = c * 100
e = 100 - 5
f = d / e
|
a ) 84 , b ) 102 , c ) 76 , d ) 70 , e ) none of the above | c | subtract(negate(123,199), multiply(subtract(11,18, 29,47), divide(subtract(11,18, 29,47), subtract(4,7, 11,18)))) | 4,7 , 11,18 , 29,47 , . . . ? . . , 123,199 | "the addition of difference between each no . is equal to latter no . so , difference between 47 and 76 is 29 which is previous no . of 47 answer c" | a = negate - (
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a ) 60 , b ) 80 , c ) 100 , d ) 120 , e ) 140 | c | sqrt(add(power(add(multiply(25, 2), multiply(15, 2)), const_2), power(multiply(25, 2), const_2))) | there are two cars . one is 60 miles north of the other . simultaneously , the car to the north is driven westward at 15 miles per hour and the other car is driven eastward at 25 miles per hour . how many miles apart are the cars after 2 hours ? | "here , drawing a quick sketch of the ' actions ' described will end in a diagonal line that you canbuilda right triangle around : the right triangle will have a base of 60 and a height of 80 . the hidden pattern here is a 3 / 4 / 5 right triangle ( the 60 lines up with the ' 3 ' and the 80 lines up with the ' 4 ' ) . in this way , you can deduce that each side is ' 20 times ' bigger than it ' s corresponding side : 3 / 4 / 5 becomes 60 / 80 / 100 thus the distance between the two cars is the length of the hypotenuse of this larger right triangle . . . final answer : c" | a = 25 * 2
b = 15 * 2
c = a + b
d = c ** 2
e = 25 * 2
f = e ** 2
g = d + f
h = math.sqrt(g)
|
a ) 2 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | b | multiply(add(8, 32), subtract(const_1, divide(add(const_3, const_4), 8))) | the diluted wine contains only 8 liters of wine and the rest is water . a new mixture whose concentration is 30 % , is to be formed by replacing wine . how many liters of mixture shall be replaced with pure wine if there was initially 32 liters of water in the mixture ? | explanation : wine water 8 l 32 l 1 : 4 20 % 80 % ( original ratio ) 30 % 70 % ( required ratio ) in ths case , the percentage of water being reduced when the mixture is being replaced with wine . so the ratio of left quantity to the initial quantity is 7 : 8 therefore , \ frac { 7 } { 8 } = \ left [ 1 - \ frac { k } { 40 } \ right ] \ rightarrow \ frac { 7 } { 8 } = \ left [ \ frac { 40 - k } { 40 } \ right ] \ rightarrow k = 5 l answer : b ) 5 | a = 8 + 32
b = 3 + 4
c = b / 8
d = 1 - c
e = a * d
|
a ) 5 , b ) 7 , c ) 4 , d ) 8 , e ) 6 | c | divide(multiply(2, const_4), subtract(multiply(2, const_3), const_4)) | the length of the side of a square is represented by x + 2 . the length of the side of an equilateral triangle is 2 x . if the square and the equilateral triangle have equal perimeter , then the value of x is ? | since the side of the square is x + 2 , its perimeter = 4 ( x + 2 ) = 4 x + 8 since the side of the equilateral triangle is 2 x , its perimeter = 3 * 2 x = 6 x also , the perimeters of both are equal . ( i . e . ) 4 x + 8 = 6 x ( i . e . ) 2 x = 8 Γ¨ x = 4 . answer : c | a = 2 * 4
b = 2 * 3
c = b - 4
d = a / c
|
a ) 30 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 70 % | d | divide(multiply(8, const_4), add(const_4, const_1)) | the cost price of 8 articles is equal to the selling price of 5 articles . what is the profit percent ? | "5 * sp = 8 * cp sp = 1.6 * cp the profit percent is 60 % . the answer is d ." | a = 8 * 4
b = 4 + 1
c = a / b
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a ) 16 , b ) 19 , c ) 15 , d ) 20 , e ) 28 | e | divide(subtract(30, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | a man is 30 years older than his son . in two years , his age will be twice the age of his son . the present age of the son is ? | "let the son ' s present age be x years . then , man ' s present age = ( x + 30 ) years . ( x + 30 ) + 2 = 2 ( x + 2 ) x + 32 = 2 x + 4 = > x = 28 . answer : e" | a = 2 * 2
b = a - 2
c = 30 - b
d = 2 - 1
e = c / d
|
a ) 5 / 6 , b ) 5 / 2 , c ) 5 / 1 , d ) 5 / 3 , e ) 7 / 6 | b | divide(divide(subtract(8.75, 7.50), subtract(8.75, 7)), subtract(const_1, divide(subtract(8.75, 7.50), subtract(8.75, 7)))) | in what ratio should a variety of rice costing rs . 7 per kg be mixed with another variety of rice costing rs . 8.75 per kg to obtain a mixture costing rs . 7.50 per kg ? | "let us say the ratio of the quantities of cheaper and dearer varieties = x : y by the rule of allegation , x / y = ( 8.75 - 7.50 ) / ( 7.50 - 7 ) = 5 / 2 answer : b" | a = 8 - 75
b = 8 - 75
c = a / b
d = 8 - 75
e = 8 - 75
f = d / e
g = 1 - f
h = c / g
|
a ) 15 , b ) 16 , c ) 17 , d ) 18 , e ) 19 | d | multiply(reminder(20, subtract(const_10, const_1)), subtract(const_10, const_1)) | q and r are two - digit positive integers that have the same digits but in reverse order . if the positive difference between q and r is less than 20 , what is the greatest possible value of q minus r ? | "a two - digit integer ` ` ab ' ' can be expressed algebraically as 10 a + b . q - r = ( 10 a + b ) - ( 10 b + a ) = 9 ( a - b ) < 20 . the greatest multiple of 9 which is less than 20 is 18 . the answer is d ." | a = 10 - 1
b = reminder * (
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a ) 24 hours , b ) 20 hours , c ) 30 hours , d ) 40 hours , e ) 50 hours | a | inverse(subtract(divide(const_1, 6), divide(const_1, add(6, 2)))) | a cistern is normally filled in 6 hrs , but takes 2 hrs longer to fill because of a leak on its bottom , if cistern is full , how much time citern would empty ? | "if leakage / hour = 1 / x , then 1 / 6 - 1 / x = 1 / 8 , solving 1 / x = 1 / 24 so in 24 hours full cistern will be empty . answer : a" | a = 1 / 6
b = 6 + 2
c = 1 / b
d = a - c
e = 1/(d)
|
a ) - 4 , b ) - 2 , c ) 2 , d ) 4 , e ) can not be determined . | e | divide(power(8, const_2), 4) | a number x is multiplied with itself and then added to the product of 4 and x . if the result of these two operations is - 8 , what is the value of x ? | "a number x is multiplied with itself - - > x ^ 2 added to the product of 4 and x - - > x ^ 2 + 4 x if the result of these two operations is - 4 - - > x ^ 2 + 4 x = - 8 i . e x ^ 2 + 4 x + 8 = 0 is the quadratic equation which needs to be solved . no solution imo e" | a = 8 ** 2
b = a / 4
|
a ) 108 , b ) 99 , c ) 128 , d ) 135 , e ) 143 | b | multiply(divide(subtract(20, divide(40, 20)), divide(40, 20)), add(divide(subtract(20, divide(40, 20)), divide(40, 20)), divide(40, 20))) | if the sum of two positive integers is 20 and the difference of their squares is 40 , what is the product of the two integers ? | "let the 2 positive numbers x and y x + y = 20 - - 1 x ^ 2 - y ^ 2 = 40 = > ( x + y ) ( x - y ) = 40 - - 2 using equation 1 in 2 , we get = > x - y = 2 - - 3 solving equation 1 and 3 , we get x = 11 y = 9 product = 11 * 9 = 99 answer b" | a = 40 / 20
b = 20 - a
c = 40 / 20
d = b / c
e = 40 / 20
f = 20 - e
g = 40 / 20
h = f / g
i = 40 / 20
j = h + i
k = d * j
|
a ) 5 days , b ) 15 days , c ) 28 days , d ) 4 days , e ) 7 days | d | inverse(add(multiply(10, inverse(multiply(24, 10))), add(multiply(inverse(multiply(12, 8)), 10), multiply(inverse(multiply(48, 4)), 4)))) | 8 men can do a piece of work in 12 days . 4 women can do it in 48 days and 10 children can do it in 24 days . in how many days can 10 men , 20 women and 10 children together complete the piece of work ? | "explanation : 1 man β s 1 day β s work = 1 / 8 Γ 12 = 1 / 96 10 men β s 1 day β s work = 1 Γ 10 / 96 = 5 / 48 1 woman β s 1 day β s work = 1 / 192 20 women β s 1 day β s work = 1 / 192 Γ 20 = 5 / 48 1 child β s 1 day β s work = 1 / 240 10 children β s 1 day β s work = 1 / 24 therefore , ( 10 men + 20 women + 10 children ) β s 1 day β s work = 5 / 48 + 5 / 48 + 1 / 24 = 1 / 4 the required no . of days = 4 days answer : option d" | a = 24 * 10
b = 1/(a)
c = 10 * b
d = 12 * 8
e = 1/(d)
f = e * 10
g = 48 * 4
h = 1/(g)
i = h * 4
j = f + i
k = c + j
l = 1/(k)
|
a ) 1 , b ) 2 , c ) 4 , d ) 0 , e ) 3 | c | subtract(divide(5, const_2), multiply(3, 3)) | what is the remainder when 3 ^ 86 is divided by 5 ? | "3 ^ 0 = 1 3 ^ 1 = 3 3 ^ 2 = 9 3 ^ 3 = 27 3 ^ 4 = . . . 1 3 ^ 5 = . . . . 3 3 ^ 6 = . . . . . . 9 and so on . 86 / 4 = 21 ( with a remainder 2 ) so the last number is a 9 . so when 3 ^ 86 is divided by 5 the remainder is 4 , ans = c" | a = 5 / 2
b = 3 * 3
c = a - b
|
a ) 15 , b ) 20 , c ) 16 , d ) 30 , e ) 35 | c | divide(160, multiply(2, 5)) | if the ratio of two number is 2 : 5 and lcm of the number is 160 then what is the number . | "product of two no = lcm * hcf 2 x * 5 x = 160 * x x = 16 answer : c" | a = 2 * 5
b = 160 / a
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a ) 60 , b ) 61.2 , c ) 62.5 , d ) 3,708 , e ) 3,672 | d | multiply(1.03, const_3600) | the moon revolves around the earth at a speed of approximately 1.03 kilometers per second . this approximate speed is how many kilometers per hour ? | "moon revolves around the earth at a speed of 1.02 kilometers per second . one hour equal to 60 minutes . one minute equals to 60 seconds . so one hour equals to 3600 seconds . so one hour , speed = 1.03 * 3600 = 3708 kilometers per hour . option d is correct" | a = 1 * 3
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a ) 33 , b ) 77 , c ) 60 , d ) 29 , e ) 11 | c | divide(subtract(multiply(100, 50), multiply(20, 100)), subtract(70, 20)) | the average marks obtained by 100 candidates in an examination is 50 . find how many candidates have passed in the examination , if the average marks of candidates who passed is 70 and that of failed candidates is 20 . | explanation : let the number of students who passed the examination be x number of students failed = ( 100 - x ) total marks of students who have passed = 70 x total marks of 100 students = 100 * 50 = 5000 total marks of students who have failed = 20 ( 100 β x ) 20 ( 100 β x ) + 70 x = 5000 2000 β 20 x + 70 x = 5000 50 x = 3000 x = 60 answer : c | a = 100 * 50
b = 20 * 100
c = a - b
d = 70 - 20
e = c / d
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a ) - 2 , b ) - 1 , c ) 0 , d ) 1 , e ) 2 | c | add(subtract(multiply(divide(subtract(19, 3), multiply(divide(divide(26, const_2), divide(26, const_2)), 16)), const_2), 3), subtract(multiply(divide(subtract(19, 3), multiply(divide(divide(26, const_2), divide(26, const_2)), 16)), const_2), 3)) | set v consists of integers { 3 , - 8 , y , 19 , - 6 } and set b consists of integers { k , - 3 , 0 , 16 , - 5 , 9 } . number l represents the median of set v , number m represents mode of set b and number z = l ^ m . if y is an integer greater than 21 , for what value of k will z be a divisor of 26 ? | v = { 3 , - 8 , y , 19 , - 6 } b = { k , - 3 , 016 , - 59 } y > 21 so l = median of v = 3 m = mode of set b z = ( 3 ) ^ m if z is a divisor of 26 , ( 3 ) ^ m = 1 because 26 does not have 3 as a factor = > m = 0 hence k = 0 , as m is mode and 0 will be the most frequently occuring number in set b . answer - c | a = 19 - 3
b = 26 / 2
c = 26 / 2
d = b / c
e = d * 16
f = a / e
g = f * 2
h = g - 3
i = 19 - 3
j = 26 / 2
k = 26 / 2
l = j / k
m = l * 16
n = i / m
o = n * 2
p = o - 3
q = h + p
|
a ) 3 % , b ) 16 2 / 3 % , c ) 25 % , d ) 4 1 / 3 % , e ) 60 % | d | subtract(const_100, divide(multiply(subtract(14.0, 11.0), const_100), subtract(18.8, 15.6))) | in 1982 and 1983 , company b β s operating expenses were $ 11.0 million and $ 14.0 million , respectively , and its revenues were $ 15.6 million and $ 18.8 million , respectively . what was the percent increase in company b β s profit ( revenues minus operating expenses ) from 1982 to 1983 ? | "profit in 1982 = 15.6 - 11 = 4.6 million $ profit in 1983 = 18.8 - 14 = 4.8 million $ percentage increase in profit = ( 4.8 - 4.6 ) / 4.6 * 100 % = 4 1 / 3 % answer d" | a = 14 - 0
b = a * 100
c = 18 - 8
d = b / c
e = 100 - d
|
a ) 14 , b ) 15 , c ) 16 , d ) 17 , e ) 18 | b | subtract(1,800, add(add(multiply(const_2, const_100), multiply(add(const_3, const_4), const_10)), const_2)) | how many integers between 1 and 1,800 are divisible by 20 , 30,40 ? | "lcm of the given numbers = 120 therefore , number of integers = 1800 / 120 = 15 option b is answer" | a = 2 * 100
b = 3 + 4
c = b * 10
d = a + c
e = d + 2
f = 1 - 800
|
a ) 1672 , b ) 1664 , c ) 1677 , d ) 1698 , e ) 1679 | b | subtract(2704, divide(multiply(multiply(3, 5), 2704), add(multiply(3, 5), multiply(8, 3)))) | a sum of rs . 2704 is lent into two parts so that the interest on the first part for 8 years at 3 % per annum may be equal to the interest on the second part for 3 years at 5 % per annum . find the second sum ? | "( x * 8 * 3 ) / 100 = ( ( 2704 - x ) * 3 * 5 ) / 100 24 x / 100 = 40560 / 100 - 15 x / 100 39 x = 40560 = > x = 1040 second sum = 2704 Γ’ β¬ β 1040 = 1664 answer : b" | a = 3 * 5
b = a * 2704
c = 3 * 5
d = 8 * 3
e = c + d
f = b / e
g = 2704 - f
|
a ) 200 , b ) 210 , c ) 180 , d ) 190 , e ) 220 | b | add(150, multiply(150, divide(40, const_100))) | 150 is increased by 40 % . find the final number . | "final number = initial number + 40 % ( original number ) = 150 + 40 % ( 150 ) = 150 + 60 = 210 . answer b" | a = 40 / 100
b = 150 * a
c = 150 + b
|
a ) 25 % , b ) 50 % , c ) 75 % , d ) 85 % , e ) 95 % | a | multiply(subtract(divide(const_1, subtract(const_1, divide(20, const_100))), const_1), const_100) | a person ' s salary is getting reduced by 20 % . what percentage should be added to get back his original salary ? | salary he get after reduced 20 % = 100 - 20 % = 80 % so , percentage should be added to get back his salary = ( 20 / 80 ) * 100 = 25 % answer : a | a = 20 / 100
b = 1 - a
c = 1 / b
d = c - 1
e = d * 100
|
a ) 2 , b ) 3 , c ) 4 , d ) 6 , e ) 8 | a | divide(18, add(multiply(3, 2), 4)) | bag a contains red , white and blue marbles such that the red to white marble ratio is 1 : 3 and the white to blue marble ratio is 2 : 3 . bag b contains red and white marbles in the ratio of 1 : 4 . together , the two bags contain 18 white marbles . how many red marbles could be in bag a ? | "6 is the answer . bag a - r : w : b = 2 : 6 : 9 let w in bag a be 6 k bab b - r : w = 1 : 4 let w in bag b be 4 p w = 18 = 6 k + 4 p = > k = 1 , p = 3 total red ' s in bag a will be 2 k = 2 a" | a = 3 * 2
b = a + 4
c = 18 / b
|
a ) 11.73 , b ) 12 , c ) 13.8 , d ) 14 , e ) 18 | e | divide(divide(multiply(207, subtract(const_100, 15)), const_100), const_10) | the price of lunch for 10 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 ) 10 * 18 = 180 now 180 ( 115 / 100 ) = 207 = > possible answer imo e | a = 100 - 15
b = 207 * a
c = b / 100
d = c / 10
|
a ) 756 , b ) 750 , c ) 600 , d ) 760 , e ) none of these | a | multiply(divide(multiply(1800, 60), const_100), divide(subtract(const_100, 30), const_100)) | in an office , totally there are 1800 employees and 60 % of the total employees are males . 30 % of the males in the office are at - least 50 years old . find the number of males aged below 50 years ? | "number of male employees = 1800 * 60 / 100 = 1080 required number of male employees who are less than 50 years old = 1080 * ( 100 - 30 ) % = 1080 * 70 / 100 = 756 . answer : a" | a = 1800 * 60
b = a / 100
c = 100 - 30
d = c / 100
e = b * d
|
a ) 2300 , b ) 600 , c ) 250 , d ) 300 , e ) none of these | d | divide(factorial(25), multiply(factorial(subtract(25, const_2)), factorial(const_2))) | if 25 lines are drawn in a plane such that no two of them are parallel and no three are concurrent , then in how many points do they intersect ? | "1 st line - 0 points 2 nd line - new 1 point 3 td line - new 2 points + old 1 point 4 th line - new 3 points + old 2 + 1 points 5 th line - new 4 points + old 3 + 2 + 1 points n - th line - ( n - 1 ) points + ( n - 2 ) . . . . 3 + 2 + 1 therefore , s = 1 + 2 + 3 . . . ( n - 1 ) s = ( n - 1 ) n / 2 = 24 * 25 / 2 = 300 ans ; d" | a = math.factorial(25)
b = 25 - 2
c = math.factorial(b)
d = math.factorial(2)
e = c * d
f = a / e
|
a ) 85,400 , b ) 86,400 , c ) 87,400 , d ) 88,400 , e ) 89,400 | b | multiply(multiply(subtract(5, 3), const_3600), const_12) | in a renowned city , the average birth rate is 5 people every two seconds and the death rate is 3 people every two seconds . estimate the size of the population net increase that occurs in one day . | "this question can be modified so that the birth rate is given every m seconds and the death rate is given every n seconds . for this particular question : increase in the population every 2 seconds = 5 - 3 = 2 people . total 2 second interval in a day = 24 * 60 * 60 / 2 = 43,200 population increase = 43,200 * 2 = 86,400 . hence b ." | a = 5 - 3
b = a * 3600
c = b * 12
|
a ) 26630 , b ) 27951 , c ) 36621 , d ) 26621 , e ) 26821 | b | multiply(21000, add(const_1, divide(multiply(3, 10), const_100))) | population is 21000 . population increases by 10 % every year , then the population after 3 years is ? | "population after 1 st year = 21000 * 10 / 100 = 2100 = = = > 21000 + 2100 = 23100 population after 2 nd year = 23100 * 10 / 100 = 2310 = = = > 23100 + 2310 = 25410 population after 3 rd year = 254100 * 10 / 100 = 2541 = = = > 25410 + 2541 = 27951 answer : b" | a = 3 * 10
b = a / 100
c = 1 + b
d = 21000 * c
|
a ) rs 220 , b ) rs . 225 , c ) rs . 230 , d ) rs . 235 , e ) rs . 255 | b | multiply(720, divide(10, add(add(9, 10), 13))) | a , b , c hired a car for rs . 720 and used it for 9 , 10 and 13 hours respectively . hire charges paid by b are : | "a : b : c = 9 : 10 : 13 . hire charges paid by b = rs . 720 * 10 / 32 = rs . 225 . answer : b" | a = 9 + 10
b = a + 13
c = 10 / b
d = 720 * c
|
a ) 72.5 , b ) 55.5 , c ) 67.5 , d ) 82.5 , e ) 60.5 | c | multiply(divide(divide(multiply(47.50, add(const_100, 25)), const_100), subtract(const_100, 12)), const_100) | at what price must an book costing $ 47.50 be marked in order that after deducting 12 % from the list price . it may be sold at a profit of 25 % on the cost price ? | "c $ 62.50 cp = 47.50 sp = 47.50 * ( 125 / 100 ) = 59.375 mp * ( 88 / 100 ) = 59.375 mp = 67.5 c" | a = 100 + 25
b = 47 * 50
c = b / 100
d = 100 - 12
e = c / d
f = e * 100
|
a ) 600 , b ) 480 , c ) 750 , d ) 650 , e ) 695 | e | subtract(subtract(multiply(45, 13), multiply(22, 20)), multiply(22, 15)) | the average of 45 results is 13 . the average of first 22 of them is 15 and that of last 22 is 20 . find the 23 result ? | "23 th result = sum of 45 results - sum of 44 results 13 * 45 - 15 * 22 + 20 * 22 = 585 - 330 + 440 = 695 answer is e" | a = 45 * 13
b = 22 * 20
c = a - b
d = 22 * 15
e = c - d
|
a ) 30 , b ) 70 , c ) 27 , d ) 26 , e ) 25 | b | add(multiply(const_4, 5.9421), divide(log(const_100), log(const_10))) | if log 1087.5 = 5.9421 , then the number of digits in ( 875 ) 10 is ? | "x = ( 875 ) 10 = ( 87.5 x 10 ) 10 therefore , log 10 x = 10 ( log 1087.5 + 1 ) = 10 ( 5.9421 + 1 ) = 10 ( 6.9421 ) = 69.421 x = antilog ( 69.421 ) therefore , number of digits in x = 70 . answer : b" | a = 4 * 5
b = math.log(100)
c = math.log(10)
d = b / c
e = a + d
|
a ) 24 % , b ) 27 % , c ) 30 % , d ) 33 % , e ) 36 % | d | multiply(subtract(const_1, divide(divide(5, 30), divide(3, 12))), const_100) | a japanese rock band made Β₯ 3 million on their first album with sales of Β₯ 12 million , and made Β₯ 5 million on their second album with sales of Β₯ 30 million . by approximately what percent did the ratio of profit to sales decrease from the first album to second album ? | solution : this is a percent decrease problem . we will use the formula : percent change = ( new β old ) / old x 100 to calculate the final answer . we first set up the ratios of profits to sales . the first ratio will be for the first album , and the second ratio will be for the next album . because all of the sales are in millions , we do not have to express all the trailing zeros in our ratios . first album profit / sales = 3 / 12 = 1 / 4 second album profit / sales = 5 / 30 = 1 / 6 we can simplify each ratio by multiplying each by the lcm of the two denominators , which is 12 . keep in mind that we are able to do this only because our answer choices are expressed in percents . first album profit / sales = ( 1 / 4 ) x 12 = 3 next 2 million profit / sales = ( 1 / 6 ) x 12 = 2 we can plug 3 and 2 into our percent change formula : ( 2 β 3 ) / 3 x 100 = - 1 / 3 x 100 so approximately a 33 % decrease . answer d . | a = 5 / 30
b = 3 / 12
c = a / b
d = 1 - c
e = d * 100
|
a ) a ) 1500 , b ) b ) 2000 , c ) c ) 2500 , d ) d ) 3000 , e ) e ) 4000 | e | divide(multiply(2, const_100), 0.05) | an inspector rejects 0.05 % of the meters as defective . how many will he examine to reject 2 ? | "let the number of meters to be examined be x then , 0.05 % of x = 2 ( 5 / 100 ) * ( ( 1 / 100 ) * x = 2 x = 4000 answer is e" | a = 2 * 100
b = a / 0
|
a ) 7 , b ) 10 , c ) 12 , d ) 14 , e ) 15 | a | divide(subtract(69, multiply(const_3, 9)), multiply(const_3, const_2)) | a number is doubled and 9 is added . if resultant is trebled , it becomes 69 . what is that number | "explanation : = > 3 ( 2 x + 9 ) = 69 = > 2 x + 9 = 23 = > x = 7 answer : option a" | a = 3 * 9
b = 69 - a
c = 3 * 2
d = b / c
|
a ) 26 h , b ) 16 h , c ) 86 h , d ) 96 h , e ) 66 h | b | divide(multiply(const_2, multiply(12, 24)), add(12, 24)) | a motor boat takes 12 hours to go downstream and it takes 24 hours to return the same distance . what is the time taken by boat in still water ? | explanation : if t 1 and t 2 are the upstream and down stream times . then time taken in still water is given by \ inline \ frac { 2 \ times t 1 \ times t 2 } { t 1 + t 2 } = \ frac { 2 \ times 12 \ times 24 } { 36 } = 16 h answer : b | a = 12 * 24
b = 2 * a
c = 12 + 24
d = b / c
|
a ) 5 , b ) 9 , c ) 12 , d ) 20 , e ) 30 | d | divide(100000, multiply(add(const_1, const_4), const_1000)) | if x is a positive integer , then the least value of x for which x ! is divisible by 100000 is ? can someone please explain intuitively what the question is asking ? | in order x ! to be divisible by 1,000 , it should have at least 3 trailing zeros . a trailing 0 in factorial of a number is produced by 2 and 5 in it : 2 * 5 = 10 . so , we need 10 to be in x ! at least in power of 3 . 5 ! = 120 has 1 trailing zeros . 10 ! will have 2 trailing zeros . 15 ! will have 3 trailing zeros . 20 ! will have 5 trailing zeros . answer : d . | a = 1 + 4
b = a * 1000
c = 100000 / b
|
a ) $ 1080 , b ) $ 1180 , c ) $ 1280 , d ) $ 1380 , e ) $ 1480 | a | add(add(add(multiply(multiply(const_3, 30), const_2), multiply(const_2, 30)), add(add(multiply(multiply(const_3, 30), const_2), multiply(const_2, 30)), 30)), subtract(multiply(add(add(multiply(multiply(const_3, 30), const_2), multiply(const_2, 30)), 30), const_3), add(multiply(multiply(const_3, 30), const_2), multiply(const_2, 30)))) | a certain sum of money is divided among a , b and c such that a gets one - third of what b and c together get and b gets two - seventh of what a and c together get . if the amount received by a is $ 30 more than that received by b , find the total amount shared by a , b and c . | a = 1 / 3 ( b + c ) = > c = 3 a - b - - - ( 1 ) b = 2 / 7 ( a + c ) = > c = 3.5 b - a - - ( b ) a - b = $ 30 a = 30 + b ( 1 ) = = = > c = 90 + 3 b - b = 2 b + 90 = = > 2 b - c = - 90 - - - ( 3 ) ( 2 ) = = = > c = 3.5 b - b - 30 = 2.5 b - 30 = = > 2.5 b - c = 30 - - - ( 4 ) from ( 4 ) and ( 3 ) 0.5 b = 120 b = $ 240 a = $ 270 c = 810 - 240 = $ 570 total amount = 270 + 240 + 570 = $ 1080 answer : a | a = 3 * 30
b = a * 2
c = 2 * 30
d = b + c
e = 3 * 30
f = e * 2
g = 2 * 30
h = f + g
i = h + 30
j = d + i
k = 3 * 30
l = k * 2
m = 2 * 30
n = l + m
o = n + 30
p = o * 3
q = 3 * 30
r = q * 2
s = 2 * 30
t = r + s
u = p - t
v = j + u
|
a ) 2 : 5 , b ) 3 : 4 , c ) 1 : 5 , d ) 20 : 5 , e ) 30 : 7 | c | divide(multiply(0.05, const_100), multiply(0.25, const_100)) | if 0.25 of a number is equal to 0.05 of another number , the ratio of the numbers is : | "0.25 a = 0.05 b - > a / b = 0.05 / 0.25 = 5 / 25 = 1 / 5 : . a : b = 1 : 5 answer : c" | a = 0 * 5
b = 0 * 25
c = a / b
|
a ) 1058.24 , b ) 2006.24 , c ) 838.55 , d ) 848.55 , e ) 858.55 | c | divide(multiply(multiply(multiply(const_3, const_100), const_100), multiply(5, divide(4, multiply(4, const_3)))), const_100) | what is the compound interest on rs : 50,000 for 4 months at the rate of 5 % per annum | "it is monthly compound rate = 5 / 12 % per month 50000 * ( 1 + 5 / 1200 ) ^ 4 - 50000 = 838.55 answer : c" | a = 3 * 100
b = a * 100
c = 4 * 3
d = 4 / c
e = 5 * d
f = b * e
g = f / 100
|
['a ) two', 'b ) three', 'c ) four', 'd ) five', 'e ) six'] | b | multiply(const_3, const_1) | if k and w are the dimensions of a rectangle that has area 20 and if k and w are integers such that k > w , what is the total number of possible values of k ? | kw = 20 = 20 * 1 = 10 * 2 = 5 * 4 - - > k can take 3 values , answer : b | a = 3 * 1
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['a ) 32', 'b ) 78', 'c ) 84', 'd ) 90', 'e ) 72'] | e | divide(power(multiply(const_3, power(648, const_2)), divide(const_1, const_3)), 1.5) | danny is sitting on a rectangular box . the area of the front face of the box is half the area of the top face , and the area of the top face is 1.5 times the area of the side face . if the volume of the box is 648 , what is the area of the side face of the box ? | lets suppose length = l , breadth = b , depth = d front face area = l * w = 1 / 2 w * d ( l = 1 / 2 d or d = 2 l ) top face area = w * d side face area = w * d = 1.5 d * l ( w = 1.5 l ) volume = l * w * d = 648 l * 1.5 l * 2 l = 648 l = 6 side face area = l * d = l * 2 l = 6 * 2 * 6 = 72 e is the answer | a = 648 ** 2
b = 3 * a
c = 1 / 3
d = b ** c
e = d / 1
|
a ) 252 kms , b ) 152 kms , c ) 492 kms , d ) 752 kms , e ) 152 kms | c | multiply(add(multiply(2, 30), multiply(subtract(12, const_1), 2)), divide(12, 2)) | the speed of a car increases by 2 kms after every one hour . if the distance travelling in the first one hour was 30 kms . what was the total distance traveled in 12 hours ? | "explanation : total distance travelled in 12 hours = ( 30 + 32 + 34 + . . . . . upto 12 terms ) this is an a . p with first term , a = 30 , number of terms , n = 12 , d = 2 . required distance = 12 / 2 [ 2 x 30 + { 12 - 1 ) x 2 ] = 6 ( 82 ) = 492 kms . answer : c" | a = 2 * 30
b = 12 - 1
c = b * 2
d = a + c
e = 12 / 2
f = d * e
|
a ) 22 sec , b ) 77 sec , c ) 16.7 sec , d ) 18 sec , e ) 17 sec | c | divide(add(150, 100), multiply(54, const_0_2778)) | how many seconds will a train 100 meters long take to cross a bridge 150 meters long if the speed of the train is 54 kmph ? | "d = 100 + 150 = 250 s = 54 * 5 / 18 = 15 mps t = 250 / 15 = 16.7 sec answer : c" | a = 150 + 100
b = 54 * const_0_2778
c = a / b
|
a ) β 5 , b ) 1 , c ) 13 , d ) 17 , e ) 551 | d | subtract(multiply(add(4, 2), const_3), const_1) | if x and g are integers and 2 x β g = 11 , then 4 x + g can not be | 2 x - g = 11 . . . . g = 2 x - 11 4 x + g = 4 x + 2 x - 11 = 6 x - 11 6 x - 11 = - 5 . . . x = 1 6 x - 11 = 1 . . . x = 2 6 x - 11 = 13 . . . x = 4 6 x - 11 = 17 . . x is not integer 6 x - 11 = 551 . . x is not integer i think the choice e is 55 not 551 . otherwise both de can not be solution = d | a = 4 + 2
b = a * 3
c = b - 1
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a ) 57 / 512 , b ) 7 / 512 , c ) 55 / 512 , d ) 9 / 512 , e ) 11 / 512 | a | divide(subtract(subtract(subtract(subtract(multiply(multiply(8, 3), 3), 8), 3), 3), 1), multiply(8, multiply(8, 8))) | a fair 8 - sided dice labeled 1 to 8 is tossed 3 times . what is the probability the value of the dice shown is 3 at least once ? | this is a fair dice with 8 fair sides , so the probability of the dice landing on any sides is equal . the question ask if the value of the dice is shown 3 at least once so the calculation of the first possible outcomes is 1 / 8 * 1 / 8 * 1 / 8 ( all of the dice thrown shown the value of 3 ) this is follow by the second possible outcomes if the dice thrown show the value of 3 2 times ( 1 / 8 * 1 / 8 * 7 / 8 ) * 7 here referring to any number besides 3 the third possible outcomes is that the dice show the value 3 one time ( 1 / 8 * 7 / 8 * 7 / 8 ) so the answer is 1 / 512 + 7 / 512 + 49 / 512 = 57 / 512 answer : a | a = 8 * 3
b = a * 3
c = b - 8
d = c - 3
e = d - 3
f = e - 1
g = 8 * 8
h = 8 * g
i = f / h
|
a ) 2.5 sec , b ) 2.9 sec , c ) 1.7 sec , d ) 7.5 sec , e ) 2.4 sec | c | divide(100, multiply(216, const_0_2778)) | in what time will a train 100 m long cross an electric pole , it its speed be 216 km / hr ? | "speed = 215 * 5 / 18 = 60 m / sec time taken = 100 / 60 = 1.7 sec . answer : c" | a = 216 * const_0_2778
b = 100 / a
|
a ) 12.0 , b ) 12.1 , c ) 12.2 , d ) 12.3 , e ) 11.0 | e | divide(add(10, 12.1), const_2) | consider below info to be a table : amount of bacteria time - amount 1 pm - 10 gms 4 pm - x gms 7 pm - 12.1 gms data for a certain biology experiment are given in the table above . if the amount of bacteria present increased by the same factor during each of the two 3 - hour periods shown , how many grams of bacteria were present at 4 pm ? | let say the factor is x , the at 7 pm quantity = 10 ( i . e . quantity @ 1 pm ) * x ^ 2 = 12.1 = > x ^ 2 = 1.21 or x = 1.1 = > @ 4 pm quantity = 10 x = 10 * 1.1 = 11 . answer e | a = 10 + 12
b = a / 2
|
a ) 10 % , b ) 20 % , c ) 30 % , d ) 40 % , e ) 60 % | e | multiply(divide(subtract(add(divide(32, 2), multiply(subtract(120, 32), const_2)), 120), 120), const_100) | every day daniel drives 120 miles back from work . on sunday , daniel drove all the way back from work at a constant speed of x miles per hour . on monday , daniel drove the first 32 miles back from work at ( 2 x ) miles per hour , and the rest of the way at ( x / 2 ) miles per hour . the time it took daniel to drive back from work on monday is longer than the time it took him to drive back from work on sunday by what percent ? | let ' s test x = 4 . . . . on sunday , daniel drove 120 miles at 4 miles / hour . d = ( r ) ( t ) 120 = ( 4 ) ( t ) 120 / 4 = 30 = t it takes 30 hours to drive home on monday , daniel drove the first 32 miles at ( 2 ) ( 4 ) = 8 miles / hour and the rest of the way ( 88 miles ) at 4 / 2 = 2 miles / hour d = ( r ) ( t ) 32 = ( 8 ) ( t ) 32 / 8 = 4 = t it takes 4 hours for the first part d = ( r ) ( t ) 88 = ( 2 ) ( t ) 88 / 2 = 44 = t it takes 44 hours for the second part total time to drive home on monday = 4 + 44 = 48 hours we ' re asked by what percent 48 hours is greater than 30 hours . 48 / 30 = 1.6 , so it is 60 % greater . answer : e | a = 32 / 2
b = 120 - 32
c = b * 2
d = a + c
e = d - 120
f = e / 120
g = f * 100
|
a ) 25 , b ) 22 , c ) 23 , d ) 30 , e ) 36 | a | add(divide(subtract(100, 1), 4), const_1) | how many multiples of 4 are there between 1 and 100 , inclusive ? | "25 multiples of 4 between 1 and 100 inclusive . from 4 * 1 upto 4 * 25 , ( 1,2 , 3,4 , . . . , 25 ) . hence , 25 multiples ! correct option is a" | a = 100 - 1
b = a / 4
c = b + 1
|
a ) 5 , b ) 33 , c ) 20 , d ) 40 , e ) 45 | b | divide(subtract(66.50, 50), subtract(1.50, 1.00)) | caleb spends $ 66.50 on 50 hamburgers for the marching band . if single burgers cost $ 1.00 each and double burgers cost $ 1.50 each , how many double burgers did he buy ? | "solution - lets say , single hamburgersxand double hamburgersy given that , x + y = 50 and 1 x + 1.5 y = 66.50 . by solving the equations y = 33 . ans b ." | a = 66 - 50
b = 1 - 50
c = a / b
|
a ) 33 , b ) 37 , c ) 76 , d ) 14 , e ) 29 | d | inverse(subtract(divide(const_1, const_2), inverse(add(2, divide(const_1, const_3))))) | a pump can fill a tank with water in 2 hours . because of a leak , it took hours to fill the tank . the leak can drain all the water of the tank in : | explanation : work done by the leak in 1 hour = { \ color { black } \ left ( \ frac { 1 } { 2 } - \ frac { 3 } { 7 } \ right ) = \ frac { 1 } { 14 } } { \ color { black } \ therefore } leak will empty the tank in 14 hrs answer : d ) 14 hours | a = 1 / 2
b = 1 / 3
c = 2 + b
d = 1/(c)
e = a - d
f = 1/(e)
|
a ) 50 , b ) 70 , c ) 80 , d ) 90 , e ) 100 | e | divide(multiply(900, 600), multiply(70, 30)) | rectangular tile each of size 70 cm by 30 cm must be laid horizontally on a rectangular floor of size 900 cm by 600 cm , such that the tiles do not overlap and they are placed with edges jutting against each other on all edges . a tile can be placed in any orientation so long as its edges are parallel to the edges of floor . no tile should overshoot any edge of the floor . the maximum number of tiles that can be accommodated on the floor is : | "area of tile = 90 * 30 = 2700 area of floor = 900 * 300 = 270000 no of tiles = 270000 / 2700 = 100 so , the no of tile = 100 answer : e" | a = 900 * 600
b = 70 * 30
c = a / b
|
a ) 100 % , b ) 90 % , c ) 80 % , d ) 70 % , e ) 60 % | c | multiply(divide(divide(45, const_100), subtract(const_1, divide(45, const_100))), const_100) | a merchant has selected two items to be placed on sale , one of which currently sells for 45 percent less than the other . if he wishes to raise the price of the cheaper item so that the two items are equally priced , by what percentage must he raise the price of the less expensive item ? | "expensive item = $ 100 ; cheap item = $ 55 ; we must increase $ 55 to $ 100 , so by $ 45 , which is approximately 80 % increase : ( 100 - 55 ) / 55 = 3 / 7 = ~ 0.81 . answer : c ." | a = 45 / 100
b = 45 / 100
c = 1 - b
d = a / c
e = d * 100
|
a ) 1344 , b ) 6300 , c ) 8400 , d ) 50400 , e ) 67200 | b | multiply(multiply(560, divide(15, 8)), 6) | running at the same rate , 8 identical machines can produce 560 paperclips a minute . at this rate , how many paperclips could 15 machines produce in 6 minutes ? | "8 machines produce 560 in 1 min 8 machines produce 560 * 6 in 6 min 15 machine produce 560 * 6 * ( 15 / 8 ) in 6 minutes 560 * 6 * 15 / 8 = 6300 answer is b ." | a = 15 / 8
b = 560 * a
c = b * 6
|
a ) 69 , b ) 70 , c ) 71 , d ) 72 , e ) 73 | d | multiply(factorial(3), factorial(3)) | in how many ways 3 boys and 3 girls can be seated in a row so that boys and girls are alternate ? | "number of ways = 2 * 3 ! * 3 ! = 72 answer is d" | a = math.factorial(3)
b = math.factorial(3)
c = a * b
|
a ) 20 , b ) 24 , c ) 36 , d ) 42 , e ) 60 | d | add(multiply(divide(7, subtract(divide(9, 10), divide(3, 4))), divide(3, 4)), 7) | if 7 gallons of gasoline are added to a tank that is already filled to 3 / 4 of its capacity , the tank is then filled to 9 / 10 of its capacity . how many gallons does the tank hold ? | "let the capacity of the tank = c ( 3 / 4 ) c + 7 = ( 9 / 10 ) c = > ( 9 / 10 ) c - ( 3 / 4 ) c = 7 = > ( 3 / 20 ) c = 7 = > c = ( 7 * 20 ) / 3 = 46.66 number of gallons of gasoline that the tank currently holds = 3 / 4 * c + 7 = 34.99 + 7 = 41.99 = 42 answer d" | a = 9 / 10
b = 3 / 4
c = a - b
d = 7 / c
e = 3 / 4
f = d * e
g = f + 7
|
a ) a . 10 , b ) b . 12 , c ) c . 14 , d ) d . 28 , e ) e . 24 | d | add(add(add(add(add(add(add(add(add(add(add(add(const_1, add(3, 6)), const_1), const_1), const_1), const_1), 6), const_1), const_1), const_1), const_1), const_1), const_1) | working at constant rate , pump x pumped out half of the water in a flooded basement in 6 hours . the pump y was started and the two pumps , working independently at their respective constant rates , pumped out rest of the water in 3 hours . how many hours would it have taken pump y , operating alone at its own constant rate , to pump out all of the water that was pumped out of the basement ? | "rate of x = 1 / 8 rate of x + y = 1 / 6 rate of y = 1 / 6 - 1 / 8 = 1 / 24 28 hours d" | a = 3 + 6
b = 1 + a
c = b + 1
d = c + 1
e = d + 1
f = e + 1
g = f + 6
h = g + 1
i = h + 1
j = i + 1
k = j + 1
l = k + 1
m = l + 1
|
a ) 12 hours , b ) 68 hours , c ) 36 hours , d ) 58 hours , e ) none | b | add(divide(170, add(9, 6)), divide(170, subtract(9, 6))) | speed of a boat in standing water is 9 kmph and the speed of the stream is 6 kmph . a man rows to place at a distance of 170 km and comes back to the starting point . the total time taken by him is : | "sol . speed upstream = 3 kmph ; speed downstream = 15 kmph . β΄ total time taken = [ 105 / 7.5 + 105 / 10.5 ] hours = 68 hours . answer b" | a = 9 + 6
b = 170 / a
c = 9 - 6
d = 170 / c
e = b + d
|
a ) 20 % loss , b ) 6.25 % profit , c ) 33.33 % loss , d ) 30.33 % loss , e ) none of these | b | multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 16), 17)), divide(multiply(const_100, 16), 17))) | if the cost price of 17 articles is equal to the selling price of 16 articles , what is the percentage of profit or loss that the merchant makes ? | "explanation : let cost price of 1 article be re . 1 . therefore , cost price of 17 articles = rs . 17 . selling price of 16 articles = rs . 17 therefore , selling price of 17 articles is : - = > 17 / 16 Γ£ β 17 = > 18.06 therefore , profit = selling price - cost price . = > 18.06 Γ’ Λ β 17 = 1.06 hence , the percentage of profit = profit x 100 / c . p . = > 1.06 / 17 Γ£ β 100 . = > 6.25 % profit . answer : b" | a = 100 * 16
b = a / 17
c = 100 - b
d = 100 * 16
e = d / 17
f = c / e
g = 100 * f
|
a ) 62 / 95 , b ) 52 / 95 , c ) 92 / 95 , d ) 42 / 95 , e ) 32 / 95 | c | divide(subtract(choose(20, const_2), choose(subtract(20, multiply(20, divide(4, 5))), const_2)), choose(20, const_2)) | according to a recent student poll , 4 / 5 out of 20 members of the finance club are interested in a career in investment banking . if two students are chosen at random , what is the probability that at least one of them is interested in investment banking ? | "16 students are interested , 4 are not interested prob = 1 - 4 c 2 / 20 c 2 = 1 - ( 4 * 3 / ( 20 * 19 ) ) = 1 - 3 / 95 = 92 / 95 answer : c" | a = math.comb(20, 2)
b = 4 / 5
c = 20 * b
d = 20 - c
e = math.comb(d, 2)
f = a - e
g = math.comb(20, 2)
h = f / g
|
a ) 1 , b ) 4 , c ) 5 , d ) 2 , e ) 3 | d | add(divide(const_1, const_2), subtract(divide(8, const_2), const_2)) | let p be a prime number . if n = 8 p , how many distinct even divisors does n have ? | "the correct answer is d . there are two : 2 and 2 p . 8 p = 2 * 2 * 2 p = ( 2 * 2 ) * 2 p . since we care only about distinct divisors , we ignore the fact that 2 appears three times ." | a = 1 / 2
b = 8 / 2
c = b - 2
d = a + c
|
a ) 100 marks , b ) 200 marks , c ) 280 marks , d ) 371 marks , e ) 120 marks | e | add(multiply(divide(add(30, 15), subtract(divide(45, const_100), divide(30, const_100))), divide(30, const_100)), 30) | a candidate who gets 30 % of the marks fails by 30 marks . but another candidate who gets 45 % marks gets 15 marks more than necessary for passing . find the number of marks for passing ? | "30 % - - - - - - - - - - - - 30 45 % - - - - - - - - - - - - 15 - - - - - - - - - - - - - - - - - - - - - - 15 % - - - - - - - - - - - - - 45 30 % - - - - - - - - - - - - - - ? 90 + 30 = 120 marks answer : e" | a = 30 + 15
b = 45 / 100
c = 30 / 100
d = b - c
e = a / d
f = 30 / 100
g = e * f
h = g + 30
|
a ) 150 , b ) 750 , c ) 1,250 , d ) 1,900 , e ) 2,500 | d | multiply(50, 5) | in a forest 190 deer were caught , tagged with electronic markers , then released . a week later , 50 deer were captured in the same forest . of these 50 deer , it was found that 5 had been tagged with the electronic markers . if the percentage of tagged deer in the second sample approximates the percentage of tagged deer in the forest , and if no deer had either left or entered the forest over the preceding week , what is the approximate number of deer in the forest ? | "the percentage of tagged deer in the second sample = 5 / 50 * 100 = 10 % . so , 190 tagged deers comprise 10 % of total # of deers - - > total # of deers = 190 * 10 = 1,900 . answer : d ." | a = 50 * 5
|
a ) a ) 25 , b ) b ) 40 , c ) c ) 32 , d ) d ) 45 , e ) e ) 30 | e | divide(subtract(6, add(const_2, const_3)), subtract(divide(const_1, const_2), divide(const_2, add(const_2, const_3)))) | a person ' s present age is two - fifth of the age of his mother . after 6 years , he will be one - half of the age of his mother . how old is the mother at present ? | "let the mother ' s present age be x years then the person ' s present age = 2 x / 5 ( 2 x / 5 ) + 6 = 1 / 2 ( x + 6 ) 2 ( 2 x + 30 ) = 5 ( x + 6 ) x = 30 answer is e" | a = 2 + 3
b = 6 - a
c = 1 / 2
d = 2 + 3
e = 2 / d
f = c - e
g = b / f
|
a ) 22 , b ) 88 , c ) 48 , d ) 99 , e ) 72 | e | add(20, 30) | two goods trains each 500 m long are running in opposite directions on parallel tracks . their speeds are 20 km / hr and 30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? | "relative speed = 20 + 30 = 50 km / hr . 50 * 5 / 18 = 125 / 9 m / sec . distance covered = 500 + 500 = 1000 m . required time = 1000 * 9 / 125 = 72 sec answer : e" | a = 20 + 30
|
a ) 100 , b ) 110 , c ) 120 , d ) 130 , e ) 140 | b | divide(subtract(multiply(37, 120), multiply(120, 15)), subtract(39, 15)) | the average of marks obtained by 120 boys was 37 . if the average of marks of passed boys was 39 and that of failed boys was 15 , the number of boys who passed the examination is ? | let the number of boys who passed = x . then , 39 x x + 15 x ( 120 - x ) = 120 x 37 24 x = 4440 - 1800 = > x = 2640 / 24 x = 110 . hence , the number of boys passed = 110 . answer : b | a = 37 * 120
b = 120 * 15
c = a - b
d = 39 - 15
e = c / d
|
a ) 220 , b ) 250 , c ) 280 , d ) 300 , e ) 340 | d | add(add(105, divide(multiply(105, 25), subtract(const_100, add(40, 25)))), divide(multiply(105, 40), subtract(const_100, add(40, 25)))) | a fruit drink is made of orange , watermelon , and grape juice , where 25 percent of the drink is orange juice and 40 percent is watermelon juice . if the drink is made with 105 ounces of grape juice , how many ounces is the drink total ? | "let the total number of ounces in the drink be x . % of orange = 25 % % of watermelon = 40 % % of grape = 100 % - 65 % = 35 % 0.35 x = 105 x = 300 therefore there are a total of 300 ounces in the drink . the answer is d ." | a = 105 * 25
b = 40 + 25
c = 100 - b
d = a / c
e = 105 + d
f = 105 * 40
g = 40 + 25
h = 100 - g
i = f / h
j = e + i
|
a ) 1683 , b ) 1684 , c ) 1685 , d ) 1582 , e ) 1584 | a | add(8, lcm(5, 6,7)) | find the least number which when divided by 5 , 6,7 , and 8 leaves a remainder 3 , but when divided by 9 leaves no remainder | "explanation : l . c . m . of 5,6 , 7,8 = 840 . required number is of the form 840 k + 3 least value of k for which ( 840 k + 3 ) is divisible by 9 is k = 2 . required number = ( 840 x 2 + 3 ) = 1683 answer is a" | a = math.lcm(5, 6)
b = 8 + a
|
a ) 35 min , b ) 5 min , c ) 15 min , d ) 25 min , e ) 20 min | d | divide(multiply(25, divide(3, add(2, 3))), multiply(divide(3, add(2, 3)), divide(3, add(2, 3)))) | it is the new year and mandy has made a resolution to lose weight this year . she plans to exercise and do yoga . for exercise she plans to workout at the gym and ride her bicycle in the ratio of 2 : 3 everyday . she will also do yoga in the ratio , yoga : exercise = 2 : 3 . if she does yoga for 25 minutes , how much time will she spend in the gym ? | "the ratio is 2 : 3 = yoga : exer , so ( 25 ) ( 3 / 2 ) = 37.5 minutes for exercise , and 2 : 3 = gym : ride , and ( 2 / 3 ) ( 37.5 ) = 25 minutes in the gym . answer : d" | a = 2 + 3
b = 3 / a
c = 25 * b
d = 2 + 3
e = 3 / d
f = 2 + 3
g = 3 / f
h = e * g
i = c / h
|
a ) 2828 , b ) 2338 , c ) 377 , d ) 4266 , e ) 288 | d | multiply(add(multiply(multiply(add(const_3, const_4), const_2), multiply(const_100, multiply(add(const_2, const_3), const_2))), multiply(subtract(multiply(add(const_2, const_3), const_2), const_2), const_100)), subtract(power(add(const_1, divide(13.5, const_100)), const_2), const_1)) | find the compound interest accrued on an amount of rs . 14,800 at 13.5 % p . a at the end of two years . ( round off your answer to nearest integer ? | "ci = 14800 { [ 1 + 13.5 / 100 ] 2 - 1 } = 14800 { [ 1 + 27 / 200 ] 2 - 1 = 14800 { 2 + 27 / 200 } { 27 / 200 } = ( 74 ) [ 2 + 27 / 200 ] ( 27 ) = 1998 [ 2 + 27 / 200 ] = 3996 + 269.73 = rs . 4266 . answer : d" | a = 3 + 4
b = a * 2
c = 2 + 3
d = c * 2
e = 100 * d
f = b * e
g = 2 + 3
h = g * 2
i = h - 2
j = i * 100
k = f + j
l = 13 / 5
m = 1 + l
n = m ** 2
o = n - 1
p = k * o
|
a ) 15 , b ) 16 , c ) 17 , d ) 18 , e ) 19 | a | divide(150, const_10) | the ratio between the perimeter and the width of a rectangle is 5 : 1 . if the area of the rectangle is 150 square centimeters , what is the length of the rectangle in centimeters ? | "perimeter = 2 ( w + l ) = 5 w 3 w = 2 l w = 2 l / 3 wl = 150 2 l ^ 2 / 3 = 150 l ^ 2 = 225 l = 15 cm the answer is a ." | a = 150 / 10
|
a ) 11 , b ) 14 , c ) 17 , d ) 20 , e ) 23 | c | add(10, const_4) | how many odd numbers between 10 and 1500 are the squares of integers ? | "the numbers are the squares of 5 , 7 , 9 , . . . , 37 which includes 17 numbers . the answer is c ." | a = 10 + 4
|
a ) 3 : 2 , b ) 1 : 1 , c ) 3 : 1 , d ) 5 : 2 , e ) 4 : 3 | b | divide(subtract(8, 5), subtract(11, 8)) | gold is 11 times as heavy as water and copper is 5 times as heavy as water . in what ratio should these be mixed to get an alloy 8 times as heavy as water ? | "g = 11 w c = 5 w let 1 gm of gold mixed with x gm of copper to get 1 + x gm of the alloy 1 gm gold + x gm copper = x + 1 gm of alloy 11 w + 5 wx = x + 1 * 8 w 11 + 5 x = 8 ( x + 1 ) x = 1 / 1 ratio of gold with copper = 1 : 1 = 1 : 1 answer is b" | a = 8 - 5
b = 11 - 8
c = a / b
|
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