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a ) $ 1010.46 , b ) $ 505.23 , c ) $ 2020.92 , d ) $ 800.46 , e ) $ 1515.69 | a | divide(multiply(150, divide(multiply(add(150, 90), divide(add(150, 30), const_2)), 150)), subtract(divide(add(150, 30), const_2), 30)) | a clock store sold a certain clock to a collector for 30 percent more than the store had originally paid for the clock . when the collector tried to resell the clock to the store , the store bought it back at 60 percent of what the collector had paid . the shop then sold the clock again at a profit of 90 percent on its buy - back price . if the difference between the clock ' s original cost to the shop and the clock ' s buy - back price was $ 150 , for how much did the shop sell the clock the second time ? | "now , in the question above , lets say the original cost of the clock to store was c $ and then it sold the same to the collector at 30 % profit . this means the clocks ' selling price was c ( 1.3 ) and this becomes cost price for the collector . now , when the collector tries to sell the same clock to the store , the store buys it for 60 % the price at which the collector bought it . thus , you get = 1.3 * 0.6 * c = 0.78 c furthermore , the store sells the clock for the second time for 90 % profit and thus the selling price of the clock becomes = cost price of the clock for the store at buy - back * 1.9 = 1.9 * 0.78 c finally given that c - 0.78 c = 150 - - - - > c = 681.82 $ thus , the cost of the clock the second time around = 1.9 * 0.78 c = 1.9 * 0.78 * 681.82 = 1010.46 $ . hence a is the correct answer ." | a = 150 + 90
b = 150 + 30
c = b / 2
d = a * c
e = d / 150
f = 150 * e
g = 150 + 30
h = g / 2
i = h - 30
j = f / i
|
a ) 3 / 40000 , b ) 2 / 30000 , c ) 9 / 2000 , d ) 1 / 60 , e ) 1 / 15 | b | divide(1, const_3) | a certain junior class has 1000 students and a certain senior class has 900 students . among these students , there are 60 siblings pairs each consisting of 1 junior and 1 senior . if 1 student is to be selected at random from each class , what is the probability that the 2 students selected will be a sibling pair ? | "let ' s see pick 60 / 1000 first then we can only pick 1 other pair from the 800 so total will be 60 / 900 * 1000 simplify and you get 2 / 30000 answer is b" | a = 1 / 3
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a ) 3.5 , b ) 3.75 , c ) 4 , d ) 4.25 , e ) 4.86 | e | divide(350, add(divide(multiply(3, 1000), const_100), divide(multiply(1400, 3), const_100))) | a money lender lent rs . 1000 at 3 % per year and rs . 1400 at 3 % per year . the amount should be returned to him when the total interest comes to rs . 350 . find the number of years . | ( 1000 xtx 3 / 100 ) + ( 1400 xtx 3 / 100 ) = 350 Γ’ β β t = 4.86 answer e | a = 3 * 1000
b = a / 100
c = 1400 * 3
d = c / 100
e = b + d
f = 350 / e
|
a ) 1.58 % , b ) 1.98 % , c ) 10 % , d ) 15 % , e ) 19 % | a | divide(add(multiply(400, 0.8), multiply(700, 1.8)), const_1000) | by weight , liquid x makes up 0.8 percent of solution a and 1.8 percent of solution b . if 400 grams of solution a are mixed with 700 grams of solution b , then liquid x accounts for what percent of the weight of the resulting solution ? | "i think there is a typo in question . it should have been ` ` by weight liquid ' x ' makes up . . . . . ` ` weight of liquid x = 0.8 % of weight of a + 1.8 % of weight of b when 400 gms of a and 700 gms of b is mixed : weight of liquid x = ( 0.8 * 400 ) / 100 + ( 1.8 * 700 ) / 100 = 15.8 gms % of liquid x in resultant mixture = ( 15.8 / 1000 ) * 100 = 1.58 % a" | a = 400 * 0
b = 700 * 1
c = a + b
d = c / 1000
|
a ) 5 , b ) 8 , c ) 4 , d ) 6 , e ) 3 | d | divide(subtract(21, power(3, 2)), 2) | if a - b = 3 and a ( power 2 ) + b ( power 2 ) = 21 , find the value of ab . | "2 ab = ( a ( power 2 ) + b ( power 2 ) - ( a - b ) ( power 2 ) = 21 - 9 = 12 ab = 6 . answer is d ." | a = 3 ** 2
b = 21 - a
c = b / 2
|
a ) 120 , b ) 150 , c ) 100 , d ) 180 , e ) 90 | e | subtract(add(175, 325), subtract(460, 50)) | out of 460 students of a school , 325 play football , 175 play cricket and 50 neither play football nor cricket . how many students play both football and cricket ? | "n ( a ) = 325 , n ( b ) = 175 , n ( aub ) = 460 - 50 = 410 . required number = n ( anb ) = n ( a ) + n ( b ) - n ( aub ) = 325 + 175 - 410 = 90 . answer is e" | a = 175 + 325
b = 460 - 50
c = a - b
|
a ) 238 , b ) 204 , c ) 678 , d ) 169 , e ) 161 | b | divide(multiply(multiply(8, add(8, const_1)), add(multiply(8, const_2), const_1)), subtract(8, const_2)) | in a 8 x 8 chess board what is the total number of squares . | ans : the total number of squares in a n x n chess board is equal to ` ` the sum of first n natural number squares ' ' i . e . , n ( n + 1 ) ( 2 n + 1 ) 6 n ( n + 1 ) ( 2 n + 1 ) 6 so substituting 8 in the above formula we get 204 answer : b | a = 8 + 1
b = 8 * a
c = 8 * 2
d = c + 1
e = b * d
f = 8 - 2
g = e / f
|
a ) 1 / 4 , b ) 1 / 10 , c ) 7 / 15 , d ) 8 / 15 , e ) 9 / 15 | d | subtract(const_1, multiply(add(divide(const_1, 15), divide(const_1, 20)), 4)) | a can do a work in 15 days and b in 20 days . if they work on it together for 4 days , then the fraction of the work that is left is : | "explanation : a ' s 1 day ' s work = 1 / 15 b ' s 1 day ' s work = 1 / 20 ( a + b ) ' s 1 day ' s work = ( 1 / 15 + 1 / 20 ) = 7 / 60 ( a + b ) ' s 4 day ' s work = 7 / 60 x 4 = 7 / 15 therefore , remaining work = 1 - 7 / 15 = 8 / 15 answer is d" | a = 1 / 15
b = 1 / 20
c = a + b
d = c * 4
e = 1 - d
|
a ) rs . 3500 , b ) rs . 4000 , c ) rs . 4500 , d ) rs . 5000 , e ) none | b | divide(2380, multiply(divide(subtract(const_100, 15), const_100), divide(subtract(const_100, 30), const_100))) | rakesh credits 15 % of his salary in his fixed deposit account and spends 30 % of the remaining amount on groceries . if the cash in hand is rs . 2380 , what is his salary ? | explanation : let salary be rs . x . then , x - 15 % of x - 30 % of 85 % of x = 2380 or x - 15 x / 100 β 30 Γ 85 Γ x / 100 Γ 100 = 2380 or 200 x - 30 x - 51 x = 2380 Γ 2002380 Γ 200 or 119 x = 2380 Γ 2002380 Γ 200 or x 2380 Γ 200 / 119 = 4000 correct option : b | a = 100 - 15
b = a / 100
c = 100 - 30
d = c / 100
e = b * d
f = 2380 / e
|
a ) 12000 , b ) 14000 , c ) 15000 , d ) 20000 , e ) 60000 | a | subtract(multiply(5, const_4), const_12) | a salesman Γ’ β¬ β’ s terms were changed from a flat commission of 5 % on all his sales to a fixed salary of rs . 1000 plus 2.5 % commission on all sales exceeding rs . 4,000 . if his remuneration as per new scheme was rs . 600 more than that by the previous schema , his sales were worth | "[ 1000 + ( x - 4000 ) * ( 2.5 / 100 ) ] - x * ( 5 / 100 ) = 600 x = 12000 answer a" | a = 5 * 4
b = a - 12
|
a ) 451 times , b ) 638 times , c ) 838 times , d ) 436 times , e ) 435 times | a | divide(multiply(divide(const_3600, const_4), const_3), 6) | if a light flashes every 6 seconds , how many times will it flash in ΒΎ of an hour ? | there are 60 minutes in an hour . in ΒΎ of an hour there are ( 60 * ΒΎ ) minutes = 45 minutes . in ΒΎ of an hour there are ( 60 * 45 ) seconds = 2700 seconds . light flashed for every 6 seconds . in 2700 seconds 2700 / 6 = 450 times . the count start after the first flash , the light will flashes 451 times in ΒΎ of an hour . answer : a | a = 3600 / 4
b = a * 3
c = b / 6
|
a ) 228 , b ) 267 , c ) 287 , d ) 250 , e ) 811 | d | divide(multiply(400, const_100), add(const_100, 60)) | by selling an article at rs . 400 , a profit of 60 % is made . find its cost price ? | "sp = 400 cp = ( sp ) * [ 100 / ( 100 + p ) ] = 400 * [ 100 / ( 100 + 60 ) ] = 400 * [ 100 / 160 ] = rs . 250 answer : d" | a = 400 * 100
b = 100 + 60
c = a / b
|
a ) 1.6 , b ) 1.8 , c ) 2 , d ) 2.2 , e ) 2.4 | e | divide(240, add(divide(240, 12), divide(240, 3))) | while working alone at their constant rates , computer x can process 240 files in 12 hours , and computer y can process 240 files in 3 hours . if all files processed by these computers are the same size , how many hours would it take the two computers , working at the same time at their respective constant rates , to process a total of 240 files ? | "both computers together process files at a rate of 240 / 12 + 240 / 3 = 20 + 80 = 100 files per hour . the time required to process 240 files is 240 / 100 = 2.4 hours the answer is e ." | a = 240 / 12
b = 240 / 3
c = a + b
d = 240 / c
|
a ) 10 , b ) 12 , c ) 5 , d ) 25 , e ) 85 | c | multiply(subtract(divide(power(39, const_2), 357), floor(divide(power(39, const_2), 357))), 357) | on dividing a number by 357 , we get 39 as remainder . on dividing the same number 17 , what will be the remainder ? | "explanation : let x be the number and y be the quotient . then , x = 357 x y + 39 = ( 17 x 21 x y ) + ( 17 x 2 ) + 5 = 17 x ( 21 y + 2 ) + 5 ) required remainder = 5 . c )" | a = 39 ** 2
b = a / 357
c = 39 ** 2
d = c / 357
e = math.floor(d)
f = b - e
g = f * 357
|
a ) 12 , b ) 15 , c ) 18 , d ) 20 , e ) 25 | d | divide(divide(300, 5), 3) | two dogsled teams raced across a 300 mile course in wyoming . team a finished the course in 3 fewer hours than team r . if team a ' s average speed was 5 mph greater than team r ' s , what was team r ' s average mph ? | "this is a very specific format that has appeared in a handful of real gmat questions , and you may wish to learn to recognize it : here we have a * fixed * distance , and we are given the difference between the times and speeds of two things that have traveled that distance . this is one of the very small number of question formats where backsolving is typically easier than solving directly , since the direct approach normally produces a quadratic equation . say team r ' s speed was s . then team r ' s time is 300 / s . team a ' s speed was then s + 5 , and team a ' s time was then 300 / ( s + 5 ) . we need to find an answer choice for s so that the time of team a is 3 less than the time of team r . that is , we need an answer choice so that 300 / ( s + 5 ) = ( 300 / s ) - 3 . you can now immediately use number properties to zero in on promising answer choices : the times in these questions will always work out to be integers , and we need to divide 300 by s , and by s + 5 . so we want an answer choice s which is a factor of 300 , and for which s + 5 is also a factor of 300 . so you can rule out answers a and c immediately , since s + 5 wo n ' t be a divisor of 300 in those cases ( sometimes using number properties you get to the correct answer without doing any other work , but unfortunately that ' s not the case here ) . testing the other answer choices , if you try answer d , you find the time for team r is 15 hours , and for team a is 12 hours , and since these differ by 3 , as desired , d is correct ." | a = 300 / 5
b = a / 3
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a ) 52.2 , b ) 59.5 , c ) 52.8 , d ) 53.5 , e ) 52.1 | d | divide(add(multiply(24, 40), multiply(50, 60)), add(24, 50)) | the average marks of a class of 24 students is 40 and that of another class of 50 students is 60 . find the average marks of all the students ? | "sum of the marks for the class of 24 students = 24 * 40 = 960 sum of the marks for the class of 50 students = 50 * 60 = 3000 sum of the marks for the class of 74 students = 960 + 3000 = 3960 average marks of all the students = 3960 / 74 = 53.5 answer : d" | a = 24 * 40
b = 50 * 60
c = a + b
d = 24 + 50
e = c / d
|
a ) 191 , b ) 213 , c ) 221 , d ) 223 , e ) 226 | d | divide(multiply(100, 30), subtract(30, divide(multiply(add(divide(multiply(100, 30), 10), 30), 5), 100))) | each month a retailer sells 100 identical items . on each item he makes a profit of $ 30 that constitutes 10 % of the item ' s price to the retailer . if the retailer contemplates giving a 5 % discount on the items he sells , what is the least number of items he will have to sell each month to justify the policy of the discount ? | "for this question , we ' ll need the following formula : sell price = cost + profit we ' re told that the profit on 1 item is $ 20 and that this represents 10 % of the cost : sell price = cost + $ 30 sell price = $ 300 + $ 30 thus , the sell price is $ 330 for each item . selling all 100 items gives the retailer . . . 100 ( $ 30 ) = $ 2,000 of profit if the retailer offers a 5 % discount on the sell price , then the equation changes . . . 5 % ( 330 ) = $ 16.5 discount $ 313.5 = $ 300 + $ 13.5 now , the retailer makes a profit of just $ 13.5 per item sold . to earn $ 2,000 in profit , the retailer must sell . . . . $ 13.5 ( x ) = $ 2,000 x = 2,000 / 13.5 x = 222.222222 items you ' ll notice that this is not among the answer choices . . . . 221 and 223 are . selling 221 items would get us 9 ( 221 ) = $ 1989 which is not enough money . to get back to at least $ 2,000 , we need to sell 223 items . final answer : d" | a = 100 * 30
b = 100 * 30
c = b / 10
d = c + 30
e = d * 5
f = e / 100
g = 30 - f
h = a / g
|
a ) rs . 300 , b ) rs . 200 , c ) rs . 240 , d ) rs . 150 , e ) none of these | d | divide(555, add(add(multiply(add(const_1, divide(25, const_100)), add(const_1, divide(20, const_100))), add(const_1, divide(20, const_100))), const_1)) | if x gets 25 % more than y and y gets 20 % more than z , the share of z out of rs . 555 will be : | "z share = z , y = 1.2 z x = 1.25 Γ£ β 1.2 z , x + y + z = 555 ( 1.25 Γ£ β 1.2 + 1.2 + 1 ) z = 55 3.7 z = 555 , z = 150 answer : . d" | a = 25 / 100
b = 1 + a
c = 20 / 100
d = 1 + c
e = b * d
f = 20 / 100
g = 1 + f
h = e + g
i = h + 1
j = 555 / i
|
a ) undefined , b ) 0 , c ) 3 , d ) 5 , e ) 7 | b | subtract(4, 3) | for integers x , y , and z , if ( 3 ^ x ) ( 4 ^ y ) ( 5 ^ z ) = 13107 , 200000 and x + y + z = 16 , what is the value of xy / z ? | the number 13,107 , 200,000 is not divisible by 3 . ( we can verify this by adding the digits of the number to see that the sum is not a multiple of 3 . ) thus x = 0 and xy / z = 0 . the answer is b . | a = 4 - 3
|
a ) 16 , b ) 17 , c ) 18 , d ) 19 , e ) 14 | c | sqrt(divide(972, const_3)) | the length of a rectangular plot is thrice its breadth . if the area of the rectangular plot is 972 sq m , then what is the breadth of the rectangular plot ? | "let the breadth of the plot be b m . length of the plot = 3 b m ( 3 b ) ( b ) = 972 3 b 2 = 972 b 2 = 324 b = 18 m . answer : option c" | a = 972 / 3
b = math.sqrt(a)
|
a ) 20 , b ) 30 , c ) 15 , d ) 18 , e ) 26 | e | divide(add(multiply(25, 15), subtract(subtract(15, const_2), 25)), 15) | the mean of 15 observations was 25 . it was found later that an observation 40 was wrongly taken as 25 . the corrected new mean is | "explanation : correct sum = ( 25 * 15 + 40 - 25 ) = 390 correct mean = = 390 / 15 = 26 answer : e" | a = 25 * 15
b = 15 - 2
c = b - 25
d = a + c
e = d / 15
|
a ) 240 sq m , b ) 250 sq m , c ) 200 sq m , d ) 400 sq m , e ) 800 sq m | a | divide(square_area(24), const_2) | what is the area of a square field whose diagonal of length 24 m ? | "d 2 / 2 = ( 24 * 24 ) / 2 = 240 answer : a" | a = square_area / (
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a ) 7 : 3 , b ) 3 : 7 , c ) 2 : 5 , d ) 5 : 3 , e ) none of these | d | divide(divide(const_1, const_4), divide(40, const_100)) | if 40 % of a number is equal to two - thirds of another number , what is the ratio of the first number to the second ? | "suppose the first number is x and the second numbery . therefore , 40 % of x = 2 β 3 of y β΄ x β y = 2 β 3 Γ 100 β 40 = 5 β 3 answer d" | a = 1 / 4
b = 40 / 100
c = a / b
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a ) 76601 , b ) 76600 , c ) 76655 , d ) 76313 , e ) 76723 | a | multiply(subtract(subtract(add(90189, 6149), 16125), 3612), divide(const_60, const_2)) | ( x ) + 3612 + 16125 - 6149 = 90189 . calculate the value of x | "x + 3612 + 16125 - 6149 = 90189 = x + 3612 + 16125 = 90189 + 6149 = x + 19737 = 96338 = x = 96338 - 19737 = 76601 answer is a" | a = 90189 + 6149
b = a - 16125
c = b - 3612
d = const_60 / 2
e = c * d
|
['a ) decreases by 25.44 %', 'b ) increases by 25.44 %', 'c ) no change in area', 'd ) decreases by 12 %', 'e ) none'] | b | multiply(subtract(power(add(divide(12, const_100), const_1), const_2), const_1), const_100) | if the radius of a circle is increased by 12 % , then the area of the circle | a 1 = pi ( 100 ) ^ 2 a 2 = pi ( 112 ) ^ 2 so ( a 2 - a 1 ) / a 1 * 100 = 25.44 answer : b | a = 12 / 100
b = a + 1
c = b ** 2
d = c - 1
e = d * 100
|
a ) s . 49180 , b ) s . 49910 , c ) s . 11915 , d ) s 6997 , e ) s . 5000 | c | subtract(multiply(add(5, const_1), 7500), add(add(add(add(2500, 6500), 9855), 7230), 7000)) | a grocer has a sale of rs . 2500 , rs . 6500 , rs . 9855 , rs . 7230 and rs . 7000 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . 7500 ? | total sale for 5 months = rs . ( 2500 + 6500 + 9855 + 7230 + 7000 ) = rs . 33085 . required sale = rs . [ ( 7500 x 6 ) - 33085 ] = rs . ( 45000 - 33085 ) = rs . 11915 . c | a = 5 + 1
b = a * 7500
c = 2500 + 6500
d = c + 9855
e = d + 7230
f = e + 7000
g = b - f
|
a ) 17.6 , b ) 17.3 , c ) 8.6 , d ) 17.2 , e ) 9.0 | a | divide(add(20, 16), const_2) | a cyclist bikes x distance at 20 miles per hour and returns over the same path at 16 miles per hour . what is the cyclist ' s average rate for the round trip in miles per hour ? | "distance = d 1 = x miles speed = s 1 = 20 miles per hour time = t 1 = distance / speed = x / 20 2 . going from b to a distance = d 2 = x miles speed = s 2 = 16 miles per hour time = t 2 = distance / speed = x / 16 3 . average speed = total distance / total time total distance = x + x = 2 x total time = x / 20 + x / 16 = 9 x / 80 speed = 2 x / ( 9 x / 80 ) = 160 / 9 = 17.6 answer : a" | a = 20 + 16
b = a / 2
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a ) 15 / 4 , b ) 13 / 2 , c ) 9 / 7 , d ) 1 / 5 , e ) 7 / 6 | a | subtract(divide(subtract(17, const_1), add(3, const_1)), divide(add(6, const_1), subtract(29, const_1))) | if a is an integer greater than 6 but less than 17 and b is an integer greater than 3 but less than 29 , what is the range of a / b ? | "the way to approach this problem is 6 < a < 17 and 3 < b < 29 minimum possible value of a is 7 and maximum is 16 minimum possible value of b is 4 and maximum is 28 range = max a / min b - min a / max b ( highest - lowest ) 16 / 4 - 7 / 28 = 15 / 4 hence a" | a = 17 - 1
b = 3 + 1
c = a / b
d = 6 + 1
e = 29 - 1
f = d / e
g = c - f
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a ) 31.25 , b ) 37.5 , c ) 50.0 , d ) 52.5 , e ) 56.25 | e | multiply(subtract(multiply(add(divide(25, const_100), const_1), add(divide(25, const_100), const_1)), const_1), const_100) | increasing the original price of a certain item by 25 percent and then increasing the new price by 25 percent is equivalent to increasing the original price by what percent ? | "soln : - x * 1.25 * 1.25 = 1.5625 x so there is a net increase of 56.25 % . answer e ." | a = 25 / 100
b = a + 1
c = 25 / 100
d = c + 1
e = b * d
f = e - 1
g = f * 100
|
a ) 22 , b ) 77 , c ) 18 , d ) 99 , e ) 12.5 | e | multiply(divide(subtract(61900, add(42000, 13000)), add(42000, 13000)), const_100) | ramu bought an old car for rs . 42000 . he spent rs . 13000 on repairs and sold it for rs . 61900 . what is his profit percent ? | "total cp = rs . 42000 + rs . 13000 = rs . 55000 and sp = rs . 61900 profit ( % ) = ( 61900 - 55000 ) / 55000 * 100 = 12.5 % answer : e" | a = 42000 + 13000
b = 61900 - a
c = 42000 + 13000
d = b / c
e = d * 100
|
a ) 10.25 % , b ) 12.5 % , c ) 14.75 % , d ) 15.15 % , e ) 16.25 % | a | multiply(const_100, subtract(multiply(add(const_1, divide(5, const_100)), add(const_1, divide(5, const_100))), const_1)) | increasing the original price of an article by 5 percent and then increasing the new price by 5 percent is equivalent to increasing the original price by | "1.05 * 1.05 * x = 1.1025 * x the answer is a ." | a = 5 / 100
b = 1 + a
c = 5 / 100
d = 1 + c
e = b * d
f = e - 1
g = 100 * f
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a ) 650 , b ) 700 , c ) 750 , d ) 800 , e ) 850 | d | subtract(1200, divide(add(1200, 400), const_4)) | there are 1200 jelly beans divided between two jars , jar x and jar y . if there are 400 fewer jelly beans in jar x than 3 times the number of beans in jar y , how many beans are in jar x ? | x + y = 1200 so y = 1200 - x x = 3 y - 400 x = 3 ( 1200 - x ) - 400 4 x = 3200 x = 800 the answer is d . | a = 1200 + 400
b = a / 4
c = 1200 - b
|
a ) 500 % , b ) 250 % , c ) 500 / 3 % , d ) 36 % , e ) 60 % | d | multiply(add(const_1, divide(20, const_100)), subtract(const_100, 70)) | if x is 20 percent more than y and y is 70 percent less than z , then x is what percent of z ? | z = 100 ; y = 30 so x = 36 x as % of z = 36 / 100 * 100 = > 36 % answer will be ( d ) | a = 20 / 100
b = 1 + a
c = 100 - 70
d = b * c
|
a ) 321 , b ) 65 , c ) 120 , d ) 100 , e ) 80 | d | divide(multiply(multiply(1, const_1000), const_1000), multiply(const_10, const_1000)) | how many alphabets need to be there in a language if one were to make 1 million distinct 3 digit initials using the alphabets of the language ? | 1 million distinct 3 digit initials are needed . let the number of required alphabets in the language be β n β . therefore , using β n β alphabets we can form n * n * n = n 3 distinct 3 digit initials . note distinct initials is different from initials where the digits are different . for instance , aaa and bbb are acceptable combinations in the case of distinct initials while they are not permitted when the digits of the initials need to be different . this n 3 different initials = 1 million i . e . n 3 = 106 ( 1 million = 106 ) = > n 3 = ( 102 ) 3 = > n = 102 = 100 hence , the language needs to have a minimum of 100 alphabets to achieve the objective . ans : d | a = 1 * 1000
b = a * 1000
c = 10 * 1000
d = b / c
|
a ) 100 , b ) 110 , c ) 160 , d ) 130 , e ) 140 | c | divide(12, subtract(divide(12, 10), 8)) | a train covers a distance of 12 km in 10 min . if it takes 8 sec to pass a telegraph post , then the length of the train is ? | "speed = ( 12 / 10 * 60 ) km / hr = ( 72 * 5 / 18 ) m / sec = 20 m / sec . length of the train = 20 * 8 = 160 m . answer : option c" | a = 12 / 10
b = a - 8
c = 12 / b
|
a ) 4 , b ) 6 , c ) 9 , d ) 12 , e ) 15 | d | multiply(divide(12, 14), divide(12, 14)) | at a certain restaurant , the ratio of the number of cooks to the number of waiters is 3 to 11 . when 12 more waiters are hired , the ratio of the number of cooks to the number of waiters changes to 3 to 14 . how many cooks does the restaurant have ? | "originally there were 3 k cooks and 11 k waiters . 14 k = 11 k + 12 k = 4 there are 12 cooks . the answer is d ." | a = 12 / 14
b = 12 / 14
c = a * b
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a ) a ) 1040 , b ) b ) 1165 , c ) c ) 1055 , d ) d ) 1065 , e ) e ) 1075 | b | add(multiply(8, 70), multiply(11, 55)) | bruce purchased 8 kg of grapes at the rate of 70 per kg and 11 kg of mangoes at the rate of 55 per kg . how much amount did he pay to the shopkeeper ? | "cost of 8 kg grapes = 70 Γ 8 = 560 . cost of 11 kg of mangoes = 55 Γ 11 = 605 . total cost he has to pay = 560 + 605 = 1165 b" | a = 8 * 70
b = 11 * 55
c = a + b
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a ) 8 , b ) 12 , c ) 13 , d ) 14 , e ) 16 | d | divide(multiply(56, subtract(10, 8)), 8) | a 56 gallon solution of salt and water is 10 % salt . how many gallons of water must be added to the solution in order to decrease the salt to 8 % of the volume ? | "amount of salt = 5.6 assume x gallons of water are added . 5.6 / 56 + x = 8 / 100 560 = 8 x + 448 8 x = 112 x = 14 correct option : d" | a = 10 - 8
b = 56 * a
c = b / 8
|
a ) 2348 , b ) 3480 , c ) 2767 , d ) 1998 , e ) 2771 | b | multiply(add(add(multiply(3, 6), multiply(4, 5)), multiply(5, 4)), divide(120, subtract(multiply(4, 5), multiply(3, 6)))) | a , b and c invest in the ratio of 3 : 4 : 5 . the percentage of return on their investments are in the ratio of 6 : 5 : 4 . find the total earnings , if b earns rs . 120 more than a : | "explanation : a b c investment 3 x 4 x 5 x rate of return 6 y % 5 y % 4 y % return \ inline \ frac { 18 xy } { 100 } \ inline \ frac { 20 xy } { 100 } \ inline \ frac { 20 xy } { 100 } total = ( 18 + 20 + 20 ) = \ inline \ frac { 58 xy } { 100 } b ' s earnings - a ' s earnings = \ inline \ frac { 2 xy } { 100 } = 120 total earning = \ inline \ frac { 58 xy } { 100 } = 3480 answer : b ) rs . 3480" | a = 3 * 6
b = 4 * 5
c = a + b
d = 5 * 4
e = c + d
f = 4 * 5
g = 3 * 6
h = f - g
i = 120 / h
j = e * i
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a ) 3.876 , b ) 2.967 , c ) 2.87 , d ) 3.912 , e ) 1.9 | a | multiply(0.3010, divide(0.3010, 0.4771)) | if log 2 = 0.3010 and log 3 = 0.4771 , the value of log 5 ( 512 ) | "log 5 ( 512 ) = log ( 512 ) / log 5 = log 2 ^ 9 / log ( 10 / 2 ) = 9 log 2 / ( log 10 - log 2 ) = ( 9 x 0.3010 ) / ( 1 - 0.3010 ) = 2.709 / 0.699 = 2709 / 699 = 3.876 answer is a ." | a = 0 / 3010
b = 0 * 3010
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a ) 15 , b ) 5 , c ) 50 , d ) 2 , e ) 4 | b | add(const_3, const_4) | what is the smallest positive integer x such that 507 + x is the cube of a positive integer ? | "given 507 + x is a perfect cube so we will take 512 = 8 * 8 * 8 507 + x = 512 x = 512 - 507 = 5 correct option is b" | a = 3 + 4
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a ) 56 sec , b ) 44 sec , c ) 36 sec , d ) 29 sec . , e ) 19 sec . | b | divide(440, subtract(divide(45, const_3_6), divide(divide(9, const_2), const_3_6))) | a train which has 440 m long , is running 45 kmph . in what time will it cross a person moving at 9 kmph in same direction ? | "time taken to cross a moving person = length of train / relative speed time taken = 440 / ( ( 45 - 9 ) ( 5 / 18 ) = 440 / 36 * ( 5 / 18 ) = 440 / 10 = 44 sec answer : b" | a = 45 / const_3_6
b = 9 / 2
c = b / const_3_6
d = a - c
e = 440 / d
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a ) 140 , b ) 150 , c ) 160 , d ) 170 , e ) 180 | c | divide(2560, 16) | two numbers have a h . c . f of 16 and a product of two numbers is 2560 . find the l . c . m of the two numbers ? | "l . c . m of two numbers is given by ( product of the two numbers ) / ( h . c . f of the two numbers ) = 2560 / 16 = 160 . answer : c" | a = 2560 / 16
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a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | c | floor(divide(reminder(power(36, reminder(4, add(const_4, const_1))), const_100), const_10)) | what is the tens digit of 36 ^ 4 ? | "36 ^ 10 = 6 ^ 20 ( 6 ^ 2 ) = 6 * 6 = 36 ( 6 ^ 3 ) = 36 * 6 = . 16 ( 6 ^ 4 ) = . 16 * 6 = . . 96 ( 6 ^ 5 ) = . . 96 * 6 = . . 76 ( 6 ^ 6 ) = . . 76 * 6 = . . . 56 ( 6 ^ 7 ) = . . . . 56 * 6 = . . . . 36 if you see there is a pattern here in tens digits 3 , 1,9 , 7,5 , 3,1 and so on . . . continue the pattern up to 6 ^ 6 ( dont actually calculate full values ) and answer is c : 5" | a = 4 + 1
b = 36 ** reminder
c = reminder / (
d = math.floor(c, 100)
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a ) 35.67 % , b ) 64.75 % , c ) 68.57 % , d ) 69.57 % , e ) none of these | b | multiply(divide(subtract(subtract(multiply(const_2, multiply(const_100, const_10)), multiply(divide(38, const_100), subtract(multiply(const_2, multiply(const_100, const_10)), 900))), multiply(divide(32, const_100), 900)), multiply(const_2, multiply(const_100, const_10))), const_100) | in an examination , there were 2,000 candidates , out of which 900 candidates were girls and rest were boys . if 38 % of the boys and 32 % of the girls passed , then the total percentage of failed candidates is ? | "girls = 900 , boys = 1100 passed = ( 38 % of 1100 ) + ( 32 % of 900 ) = 418 + 288 = 706 failed = 2000 - 706 = 1294 failed % = [ ( 1294 / 2000 ) x 100 ] % = 64.7 % . answer : b" | a = 100 * 10
b = 2 * a
c = 38 / 100
d = 100 * 10
e = 2 * d
f = e - 900
g = c * f
h = b - g
i = 32 / 100
j = i * 900
k = h - j
l = 100 * 10
m = 2 * l
n = k / m
o = n * 100
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a ) 16 % , b ) 18 % , c ) 19 % , d ) 20 % , e ) 25 % | e | subtract(const_100, divide(multiply(450, const_100), 600)) | an article is bought for rs . 600 and sold for rs . 450 , find the loss percent ? | "600 - - - - 150 100 - - - - ? = > 16 2 / 3 % answer : e" | a = 450 * 100
b = a / 600
c = 100 - b
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a ) 0.00015 , b ) 0.0015 , c ) 0.015 , d ) 0.15 , e ) none of these | b | multiply(divide(0.003, 0.5), const_100) | 0.003 * 0.5 = ? | "explanation : 3 * 5 = 15 . sum of decimal places = 4 . 0.003 * 0.5 = 0.0015 = 0.0015 answer b" | a = 0 / 3
b = a * 100
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a ) 5 kmph , b ) 6 kmph , c ) 8 kmph , d ) 1 kmph , e ) 2 kmph | a | divide(subtract(divide(60, 3), divide(30, 3)), const_2) | a man rows his boat 60 km downstream and 30 km upstream taking 3 hrs each time . find the speed of the stream ? | explanation : speed of the boat downstream = speed of the boat upstream \ small \ therefore the speed of the stream = answer : a | a = 60 / 3
b = 30 / 3
c = a - b
d = c / 2
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a ) $ 60 , b ) $ 80 , c ) $ 100 , d ) $ 120 , e ) $ 140 | a | divide(10, subtract(divide(5, 3), divide(3, 2))) | the ratio of the amount of the oil bill for the month of february to the amount of the oil bill for the month of january was 3 : 2 . if the oil bill for february had been $ 10 more , the corresponding ratio would have been 5 : 3 . how much was the oil bill for january ? | "3 : 2 = 9 : 6 and 5 : 3 = 10 : 6 . an increase in $ 10 increases the ratio by 1 : 6 . therefore , january ' s bill was 6 ( $ 10 ) = $ 60 . the answer is a ." | a = 5 / 3
b = 3 / 2
c = a - b
d = 10 / c
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a ) 190 , b ) 195 , c ) 305 , d ) 205 , e ) 210 | c | divide(divide(multiply(add(10, 600), add(divide(subtract(600, 10), 10), const_1)), const_2), add(divide(subtract(600, 10), 10), const_1)) | what is the average ( arithmetic mean ) of all multiples of 10 from 10 to 600 inclusive ? | "this question can be solved with the average formula and ' bunching . ' we ' re asked for the average of all of the multiples of 10 from 10 to 600 , inclusive . to start , we can figure out the total number of terms rather easily : 1 ( 10 ) = 10 2 ( 10 ) = 20 . . . 60 ( 10 ) = 600 so we know that there are 60 total numbers . we can now figure out the sum of those numbers with ' bunching ' : 10 + 600 = 610 20 + 590 = 610 30 + 580 = 610 etc . since there are 60 total terms , this pattern will create 30 ' pairs ' of 610 . thus , since the average = ( sum of terms ) / ( number of terms ) , we have . . . ( 30 ) ( 610 ) / ( 60 ) = 305 final answer : c" | a = 10 + 600
b = 600 - 10
c = b / 10
d = c + 1
e = a * d
f = e / 2
g = 600 - 10
h = g / 10
i = h + 1
j = f / i
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a ) 15 % , b ) 20 % , c ) 25 % , d ) 60 % , e ) 80 % | b | multiply(divide(subtract(74.95, 59.95), 74.95), const_100) | a $ 74.95 lawn chair was sold for $ 59.95 at a special sale . by approximately what percent was the price decreased ? | "listed selling price of chair = 74.95 $ discounted selling price of chair = 59.95 $ discount = 74.95 - 59.95 = 15 $ % decrease in price of chair = ( 15 / 74.95 ) * 100 % = 20 % approx answer b" | a = 74 - 95
b = a / 74
c = b * 100
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a ) s . 420 , b ) s . 360 , c ) s . 389 , d ) s . 368 , e ) s . 323 | a | divide(595, add(add(multiply(divide(2, 3), divide(1, 4)), divide(1, 4)), 1)) | if rs . 595 be divided among a , b , c in such a way that a gets 2 / 3 of what b gets and b gets 1 / 4 of what c gets , then their shares are respectively ? | "( a = 2 / 3 b and b = 1 / 4 c ) = a / b = 2 / 3 and b / c = 1 / 4 a : b = 2 : 3 and b : c = 1 : 4 = 3 : 12 a : b : c = 2 : 3 : 12 a ; s share = 595 * 2 / 17 = rs . 70 b ' s share = 595 * 3 / 17 = rs . 105 c ' s share = 595 * 12 / 17 = rs . 420 . answer : a" | a = 2 / 3
b = 1 / 4
c = a * b
d = 1 / 4
e = c + d
f = e + 1
g = 595 / f
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a ) 9 % , b ) 15 % , c ) 25 % , d ) 33 % , e ) 50 % | d | multiply(divide(divide(50, const_100), divide(add(50, const_100), const_100)), const_100) | a part - time employee β s hourly wage was increased by 50 % . she decided to decrease the number of hours worked per week so that her total income did not change . by approximately what percent should the number of hours worked be decreased ? | let ' s plug in somenicenumbers and see what ' s needed . let ' s say the employee used to make $ 1 / hour and worked 100 hours / week so , the total weekly income was $ 100 / week after the 50 % wage increase , the employee makes $ 1.50 / hour we want the employee ' s income to remain at $ 100 / week . so , we want ( $ 1.50 / hour ) ( new # of hours ) = $ 100 divide both sides by 1.50 to get : new # of hours = 100 / 1.50 β 67 hours so , the number of hours decreases from 100 hours to ( approximately ) 67 hours . this represents a 33 % decrease ( approximately ) . answer : d | a = 50 / 100
b = 50 + 100
c = b / 100
d = a / c
e = d * 100
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a ) 2 : 5 , b ) 2 : 7 , c ) 1 : 6 , d ) 1 : 4 , e ) 1 : 3 | a | divide(subtract(5, 3), subtract(10, 5)) | cereal a is 10 % sugar by weight , whereas healthier but less delicious cereal b is 3 % sugar by weight . to make a delicious and healthy mixture that is 5 % sugar , what should be the ratio of cereal a to cereal b , by weight ? | "ratio of a / ratio of b = ( average wt of mixture - wt of b ) / ( wt of a - average wt of mixture ) = > ratio of a / ratio of b = ( 5 - 3 ) / ( 10 - 5 ) = 2 / 5 so they should be mixed in the ratio 2 : 5 answer - a" | a = 5 - 3
b = 10 - 5
c = a / b
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a ) 1.8 m , b ) 6 m , c ) 7 m , d ) 9 m , e ) 12 m | b | subtract(divide(const_2, subtract(const_1, divide(5, 7))), const_1) | the q train leaves station a moving at a constant speed , and passes by stations b and c , in this order . it takes the q train 7 hours to reach station b , and 5 additional hours to reach station c . the distance between stations a and b is m kilometers longer than the distance between stations b and c . what is the distance between stations a and c in terms of m ? | the reason it is failing for you is that you chose incorrect numbers . if the question says it q took 7 hrs to reach from a to b and 5 hrs to reach from b to c at a constant speed . it shows that distance ab and bc should be in ratio of 7 / 5 . if you take such numbers you can solve problem . ab = 7 , bc = 5 therefore ab - bc = 2 but from question , ab - bc = m = > m = 2 now total distance = ab + bc = 12 substitute 12 to get answer in terms of m total distance = 12 = 6 m ans b | a = 5 / 7
b = 1 - a
c = 2 / b
d = c - 1
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a ) 225 sq m , b ) 126 sq m , c ) 144 sq m , d ) 267 sq m , e ) 231 sq m | c | divide(square_area(12), const_2) | what is the area of a square field whose sides have a length of 12 meters ? | "12 * 12 = 144 sq m the answer is c ." | a = square_area / (
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a ) 3 , b ) 9 , c ) 10 , d ) 12 , e ) 14 | d | subtract(subtract(add(subtract(subtract(subtract(subtract(subtract(subtract(subtract(32, 4), 4), 2), 2), const_1), 4), const_1), 2), 4), 4) | if 4 a β 2 b β 2 c = 32 and β 3 a - β ( 2 b + 2 c ) = 4 , what is the value of a + b + c ? | "when we look at the two equations , we can relize some similarity , so lets work on it . . 3 a β 2 b β 2 c = 32 can be written as β 3 a ^ 2 - β ( 2 b + 2 c ) ^ 2 = 32 { β 3 a - β ( 2 b + 2 c ) } { β 3 a + β ( 2 b + 2 c ) } = 32 . . or 4 * β 3 a + β ( 2 b + 2 c ) = 32 . . or β 3 a + β ( 2 b + 2 c ) = 8 . . now lets work on these two equations 1 ) β 3 a - β ( 2 b + 2 c ) = 4 . . 2 ) β 3 a + β ( 2 b + 2 c ) = 8 . . a ) add the two eq . . β 3 a + β ( 2 b + 2 c ) + β 3 a - β ( 2 b + 2 c ) = 12 . . 2 β 3 a = 12 . . or β 3 a = 6 . . 3 a = 36 . . a = 12 . b ) subtract 1 from 2 . . β 3 a + β ( 2 b + 2 c ) - β 3 a + β ( 2 b + 2 c ) = 4 . . 2 β ( 2 b + 2 c ) = 4 . . β ( 2 b + 2 c ) = 2 . . 2 b + 2 c = 4 . . or b + c = 2 . . from a and b a + b + c = 12 + 2 = 14 . . d" | a = 32 - 4
b = a - 4
c = b - 2
d = c - 2
e = d - 1
f = e - 4
g = f - 1
h = g + 2
i = h - 4
j = i - 4
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a ) 107 , b ) 117 , c ) 127 , d ) 137 , e ) 147 | b | divide(multiply(add(multiply(6, const_100), 24), add(multiply(4, const_100), 32)), multiply(gcd(add(multiply(6, const_100), 24), add(multiply(4, const_100), 32)), gcd(add(multiply(6, const_100), 24), add(multiply(4, const_100), 32)))) | a room is 6 meters 24 centimeters in length and 4 meters 32 centimeters in width . find the least number of square tiles of equal size required to cover the entire floor of the room . | "let us calculate both the length and width of the room in centimeters . length = 6 meters and 24 centimeters = 624 cm width = 4 meters and 32 centimeters = 432 cm as we want the least number of square tiles required , it means the length of each square tile should be as large as possible . further , the length of each square tile should be a factor of both the length and width of the room . hence , the length of each square tile will be equal to the hcf of the length and width of the room = hcf of 624 and 432 = 48 thus , the number of square tiles required = ( 624 x 432 ) / ( 48 x 48 ) = 13 x 9 = 117 answer : b" | a = 6 * 100
b = a + 24
c = 4 * 100
d = c + 32
e = b * d
f = 6 * 100
g = f + 24
h = 4 * 100
i = h + 32
j = math.gcd(g, i)
k = 6 * 100
l = k + 24
m = 4 * 100
n = m + 32
o = math.gcd(l, n)
p = j * o
q = e / p
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a ) 8 , b ) 10 , c ) 12 , d ) 15 , e ) 16 | b | add(subtract(divide(122, 7), 8), const_1) | how many integers are between 8 and 122 / 7 , inclusive ? | 122 / 7 = 17 . xx we are not concerned about the exact value of 122 / 7 as we just need the integers . the different integers between 8 and 122 / 7 would be 8 , 9 , 10 , 11 , 12,13 , 14,15 , 16,17 total number of integers = 10 option b | a = 122 / 7
b = a - 8
c = b + 1
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a ) 20 , b ) 37 , c ) 42 , d ) 53 , e ) 64 | b | divide(add(60, 14), const_2) | there are 60 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 ) = 60 - - > x = 37 . answer : b ." | a = 60 + 14
b = a / 2
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a ) 50 kmph , b ) 58 kmph , c ) 62 kmph , d ) 65 kmph , e ) 75 kmph | c | subtract(multiply(divide(280, 9), const_3_6), 50) | a man sitting in a train which is traveling at 50 kmph observes that a goods train , traveling in opposite direction , takes 9 seconds to pass him . if the goods train is 280 m long , find its speed . ? | relative speed = 280 / 9 m / sec = ( ( 280 / 9 ) * ( 18 / 5 ) ) kmph = 112 kmph . speed of goods train = ( 112 - 50 ) kmph = 62 kmph . answer : c . | a = 280 / 9
b = a * const_3_6
c = b - 50
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a ) 18 , b ) 21 , c ) 23 , d ) 25 , e ) 27 | c | divide(subtract(add(add(26, 3), 26), subtract(11, 2)), subtract(11, subtract(11, 2))) | the cricket team of 11 members is 26 yrs old & the wicket keeper is 3 yrs older . if the ages ofthese 2 are excluded , the average age of the remainingplayers is 1 year less than the average age of the whole team . what is the average age of the team ? | "let the average age of the whole team be x years . 11 x - ( 26 + 29 ) = 9 ( x - 1 ) = > 11 x - 9 x = 46 = > 2 x = 46 = > x = 23 . so , average age of the team is 23 years . c" | a = 26 + 3
b = a + 26
c = 11 - 2
d = b - c
e = 11 - 2
f = 11 - e
g = d / f
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a ) rs . 432 , b ) rs . 422 , c ) rs . 412 , d ) rs . 442 , e ) none of these | a | multiply(divide(360, subtract(2160, 360)), 2160) | the true discount on a bill of rs . 2160 is rs . 360 . what is the banker ' s discount ? | "explanation : f = rs . 2160 td = rs . 360 pw = f - td = 2160 - 360 = rs . 1800 true discount is the simple interest on the present value for unexpired time = > simple interest on rs . 1800 for unexpired time = rs . 360 banker ' s discount is the simple interest on the face value of the bill for unexpired time = simple interest on rs . 2160 for unexpired time = 360 / 1800 Γ 2160 = 15 Γ 2160 = rs . 432 answer : option a" | a = 2160 - 360
b = 360 / a
c = b * 2160
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a ) 5 % , b ) 22 % , c ) 13 % , d ) 21 % , e ) 19 % | b | multiply(divide(subtract(const_100, 82), 82), const_100) | if the cost price is 82 % of the selling price , then what is the profit percent ? | "let s . p . = $ 100 c . p . = $ 82 profit = $ 18 profit % = 18 / 82 * 100 = 25 / 6 = 22 % approximately answer is b" | a = 100 - 82
b = a / 82
c = b * 100
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a ) 35.4 , b ) 42 , c ) 45.7 , d ) 64 , e ) 67 | d | add(50, const_1) | the average of first six prime numbers which are between 50 and 90 is | "explanation : first six prime numbers which are between 50 and 90 = 53 , 59 , 61 , 67 , 71 , 73 average = ( 53 + 59 + 61 + 67 + 71 + 73 ) / 6 = 64 answer : d" | a = 50 + 1
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a ) rs . 170 , b ) rs . 17 , c ) rs . 1.70 , d ) rs . 4.25 , e ) none | a | divide(85, 0.5) | if 0.5 % of a = 85 paise , then the value of a is ? | "answer β΅ 0.5 / 100 of a = 85 / 100 β΄ a = rs . ( 85 / 0.5 ) = rs . 170 correct option : a" | a = 85 / 0
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a ) rs . 9471 , b ) rs . 12,628 , c ) rs . 18,040 , d ) rs . 7015.56 , e ) none | d | multiply(multiply(const_0_25, const_100), 10) | suganya and suriya are partners in a business . suganya invests rs . 15,000 for 8 months and suriya invests rs . 42,000 for 10 months . out of a profit of rs . 31,570 . suganya ' s share is | "solution ratio of their shares = ( 15000 Γ£ β 8 ) : ( 42000 Γ£ β 10 ) = 2 : 7 . suganya ' s share = rs . ( 31570 Γ£ β 2 / 9 ) = rs . 7015.56 answer d" | a = const_0_25 * 100
b = a * 10
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a ) $ 120 , b ) $ 150 , c ) $ 1786 , d ) $ 250 , e ) $ 300 | c | subtract(multiply(1200, power(add(const_1, divide(20, const_100)), 5)), 1200) | find the compound interest on $ 1200 for 5 years at 20 % p . a . if ci is component yearly ? | "a = p ( 1 + r / 100 ) ^ t = 1200 ( 1 + 20 / 100 ) ^ 5 = $ 2986 ci = $ 1786 answer is c" | a = 20 / 100
b = 1 + a
c = b ** 5
d = 1200 * c
e = d - 1200
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a ) $ 4000 , b ) $ 2890 , c ) $ 3005 , d ) $ 6952 , e ) $ 6003 | d | multiply(divide(const_3.0, add(add(5, const_3.0), 6)), 15800) | a person want to give his money of $ 15800 to his 4 children a , b , c , d in the ratio 5 : 9 : 6 : 5 . what is the a + c share ? | "a ' s share = 15800 * 5 / 25 = $ 3160 c ' s share = 15800 * 6 / 25 = $ 3792 a + c = $ 6952 answer is d" | a = 5 + 3
b = a + 6
c = 3 / 0
d = c * 15800
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a ) 24 , b ) 36 , c ) 48 , d ) 60 , e ) 72 | a | lcm(12, 24) | a certain class of students is being divided into teams . the class can either be divided into 12 teams with an equal number of players on each team or 24 teams with an equal number of players on each team . what is the lowest possible number of students in the class ? | "let total no of students in the class be n so , we are told that n is divisible by both 12 24 so , lets find the least common multiple of 12 24 , ie 24 so our answer is ( a ) 24" | a = math.lcm(12, 24)
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a ) $ 220 , b ) $ 340 , c ) $ 150 , d ) $ 100 , e ) $ 120 | a | add(multiply(6, 10), multiply(10, 16)) | a group of 6 children and 10 adults are going to the zoo . child tickets cost $ 10 , and adult tickets cost $ 16 . how much will the zoo tickets cost in all ? | step 1 : find the cost of the children ' s tickets . 6 Γ $ 10 = $ 60 step 2 : find the cost of the adults ' tickets . 10 Γ $ 16 = $ 160 step 3 : find the total cost . $ 60 + $ 160 = $ 220 the zoo tickets will cost $ 220 . answer is a . | a = 6 * 10
b = 10 * 16
c = a + b
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a ) 5 kg , b ) 15 kg , c ) 25 kg , d ) 30 kg , e ) none | e | divide(const_100, divide(subtract(const_100, 40), 20)) | the price of rice falls by 40 % . how much rice can be bought now with the money that was sufficient to buy 20 kg of rice previously ? | "solution : let rs . 100 be spend on rice initially for 20 kg . as the price falls by 40 % , new price for 20 kg rice , = ( 100 - 40 % of 100 ) = 60 new price of rice = 60 / 20 = rs . 3 per kg . rice can bought now at = 100 / 3 = 33.33 kg . answer : option e" | a = 100 - 40
b = a / 20
c = 100 / b
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a ) 30 , b ) 15 , c ) 40 , d ) 45 , e ) 20 | e | add(multiply(sqrt(divide(subtract(138, 131), const_2)), const_100), sqrt(subtract(138, divide(subtract(138, 131), const_2)))) | the sum of the squares of three numbers is 138 , while the sum of their products taken two at a time is 131 . find the sum ? | "a 2 + b 2 + c 2 = 138 and ( ab + bc + ca ) = 131 a + b + c ^ 2 = 400 root ( 400 ) = 20 answer e" | a = 138 - 131
b = a / 2
c = math.sqrt(b)
d = c * 100
e = 138 - 131
f = e / 2
g = 138 - f
h = math.sqrt(g)
i = d + h
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a ) s . 800 , b ) s . 400 , c ) s . 600 , d ) s . 500 , e ) s . 300 | e | divide(1200, const_3) | divide rs . 1200 among a , b and c so that a receives 1 / 3 as much as b and c together and b receives 2 / 3 as a and c together . a ' s share is ? | "a + b + c = 1200 a = 1 / 3 ( b + c ) ; b = 2 / 3 ( a + c ) a / ( b + c ) = 1 / 3 a = 1 / 4 * 1200 = > 300 answer : e" | a = 1200 / 3
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a ) 9 % less , b ) 1 % less , c ) equal to each other , d ) 1.3 % more , e ) 9 % more | d | divide(const_100, subtract(multiply(const_100, const_100), multiply(add(const_100, 9), subtract(const_100, 9)))) | 108 . triangle a β s base is 9 % greater than the base of triangle b , and a β s height is 9 % less than the height of triangle b . the area of triangle a is what percent less or more than the area of triangle b ? | "wish the question specified that we are talking about corresponding height . base of a = 10 / 9 * base of b height of a = 8 / 9 * height of b area of a = ( 1 / 2 ) * base of a * height of a = 10 / 9 * 8 / 9 * area of b = 80 / 81 * area of b area of a is 1.3 % more than the area of b . answer ( d )" | a = 100 * 100
b = 100 + 9
c = 100 - 9
d = b * c
e = a - d
f = 100 / e
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a ) 6 , b ) 12 , c ) 15 , d ) 20 , e ) 24 | a | divide(subtract(multiply(62.5, 30), 1800), subtract(62.5, 50)) | tourist purchased a total of 30 travelers checks in $ 50 and $ 100 denominations . the total worth of the travelers checks is $ 1800 . how many checks of $ 50 denominations can he spend so that average amount ( arithmetic mean ) of the remaining travelers checks is $ 62.5 ? | "you could set - up a quick table and brute force the answer . a 6 * 50 300 1800 - 300 1500 24 62.50 b 16 * 50 600 1800 - 600 1200 18 66.67 c 15 * 50 750 1800 - 750 1050 15 70.00 d 20 * 50 1000 1800 - 1000 800 10 80.00 e 24 * 50 1200 1800 - 1200 600 6 100.00 answer is a" | a = 62 * 5
b = a - 1800
c = 62 - 5
d = b / c
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a ) rs 16123.20 , b ) rs 16123.30 , c ) rs 16197.12 , d ) rs 16123.50 , e ) none of these | c | subtract(multiply(multiply(multiply(const_4, const_100), const_100), power(add(const_1, divide(12, const_100)), 3)), multiply(multiply(const_4, const_100), const_100)) | what will be the compound interest on rs . 40000 after 3 years at the rate of 12 % per annum | "explanation : ( 40000 Γ ( 1 + 12 / 100 ) 3 ) = > 40000 Γ 28 / 25 Γ 28 / 25 Γ 28 / 25 = > 56197.12 so compound interest will be 56197.12 - 40000 = rs 16197.12 option c" | a = 4 * 100
b = a * 100
c = 12 / 100
d = 1 + c
e = d ** 3
f = b * e
g = 4 * 100
h = g * 100
i = f - h
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a ) 30 , b ) 40 , c ) 60 , d ) 72 , e ) can not be determined | d | divide(54, 75) | how many pieces of 75 cm can be cut from a rope 54 meters long ? | "explanation : total pieces of 75 cm that can be cut from a rope of 54 meters long is = ( 54 meters ) / ( 75 cm ) = ( 54 meters ) / ( 0.75 meters ) = 72 answer : d" | a = 54 / 75
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a ) 7 / 4 , b ) 10 / 3 , c ) 12 / 5 , d ) 15 / 7 , e ) 18 / 11 | c | divide(16, divide(const_1, add(divide(const_1, 10), divide(const_1, 20)))) | working alone , printers x , y , and z can do a certain printing job , consisting of a large number of pages , in 16 , 10 , and 20 hours , respectively . what is the ratio of the time it takes printer x to do the job , working alone at its rate , to the time it takes printers y and z to do the job , working together at their individual rates ? | "the time it takes printer x is 16 hours . the combined rate of y and z is 1 / 10 + 1 / 20 = 3 / 20 the time it takes y and z is 20 / 3 the ratio of times is 16 / ( 20 / 3 ) = 3 * 16 / 20 = 12 / 5 the answer is c ." | a = 1 / 10
b = 1 / 20
c = a + b
d = 1 / c
e = 16 / d
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a ) 59 kg , b ) 62 kg , c ) 74 kg , d ) 85 kg , e ) 71 kg | e | add(multiply(1.8, 10), 53) | the average weight of 10 oarsmen in a boat is increased by 1.8 kg when one of the crew , who weight 53 kg is replaced by a new man . find the weight of the new man . | "total weight increased = 1.8 * 10 = 18 kg weight of the new man = 53 + 18 = 71 kg answer is e" | a = 1 * 8
b = a + 53
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a ) 30 , b ) 29 , c ) 34 , d ) 32 , e ) 33 | b | add(divide(subtract(16, 6000), 13), const_1) | how many multiples of 13 are less than 6000 , and also multiples of 16 ? | "lcm of 13 & 16 = 208 tried dividing 6000 by 208 got quotient 28.84 ' so b is answer" | a = 16 - 6000
b = a / 13
c = b + 1
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a ) x / 2 , b ) x , c ) 3 m / 2 , d ) mx , e ) mx | e | multiply(const_2, add(const_1, divide(const_1, const_2))) | two ants , arthur and amy , have discovered a picnic and are bringing crumbs back to the anthill . amy makes twice as many trips and carries one and a half times as many crumbs per trip as arthur . if arthur carries a total of m crumbs to the anthill , how many crumbs will amy bring to the anthill , in terms of m ? | "lets do it by picking up numbers . let arthur carry 2 crumbs per trip , this means amy carries 3 crumbs per trip . also let arthur make 2 trips and so amy makes 4 trips . thus total crumbs carried by arthur ( m ) = 2 x 2 = 4 , total crumbs carried by amy = 3 x 4 = 12 . 12 is 3 times 4 , so e" | a = 1 / 2
b = 1 + a
c = 2 * b
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a ) 1000 meters , b ) 1050 meters , c ) 1200 meters , d ) 1750 meters , e ) none of these | d | multiply(multiply(7, divide(15, const_60)), const_1000) | a man is walking at the rate of 7 km / hr crosses a bridge in 15 minutes . the length of the bridge is | "explanation : we need to get the answer in meters . so we will first of change distance from km / hour to meter / sec by multiplying it with 5 / 18 and also change 15 minutes to seconds by multiplying it with 60 . speed = 7 Γ’ Λ β 5 / 18 = 35 / 18 m / sec time = 15 Γ’ Λ β 60 seconds = 900 seconds distance = time Γ’ Λ β speed / distance = 35 / 18 Γ’ Λ β 900 = 1750 meter option d" | a = 15 / const_60
b = 7 * a
c = b * 1000
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['a ) 40', 'b ) 6', 'c ) 17', 'd ) 29', 'e ) 38'] | a | divide(multiply(6, 13), const_2) | if two sides of a triangle are 6 and 13 , respectively , which of the following could not be the area of this triangle ? | for this question it would be helpful to know the largest area that this triangle could be , given the two sides of 6 and 12 . we know that the area of a triangle will be maximized when two sides are perpendicular to each other ( consult bunuel ' s drawing above ) . thus we have a max area being equal to one half the base times the height , either ( . 5 ) 13 * 6 - or - ( . 5 ) 6 * 13 will result in a maximum area of 39 for the triangle ; therefore , the triangle could never have an area of 40 . a | a = 6 * 13
b = a / 2
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a ) 350 , b ) 370 , c ) 390 , d ) 430 , e ) none | c | add(350, divide(multiply(4, subtract(420, 350)), add(3, 4))) | average expenditure of a person for the first 3 days of a week is rs . 350 and for the next 4 days is rs . 420 . average expenditure of the man for the whole week is : | "explanation : assumed mean = rs . 350 total excess than assumed mean = 4 Γ ( rs . 420 - rs . 350 ) = rs . 280 therefore , increase in average expenditure = rs . 280 / 7 = rs . 40 therefore , average expenditure for 7 days = rs . 350 + rs . 40 = rs . 390 correct option : c" | a = 420 - 350
b = 4 * a
c = 3 + 4
d = b / c
e = 350 + d
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a ) rs . 473 , b ) rs . 374 , c ) rs . 495 , d ) rs . 157 , e ) none of the above | d | subtract(multiply(1000, power(add(1, divide(divide(10, 2), const_100)), multiply(add(1, divide(1, 2)), 2))), 1000) | compound interest of rs . 1000 at 10 % per annum for 1 1 / 2 years will be ( interest compounded half yearly ) . | "10 % interest per annum will be 5 % interest half yearly for 3 terms ( 1 1 / 2 years ) so compound interest = 3000 [ 1 + ( 5 / 100 ) ] ^ 3 - 1000 = 1000 [ ( 21 / 20 ) ^ 3 - 1 ] = 1000 ( 9261 - 8000 ) / 8000 = 1 * 1261 / 8 = 157 answer : d" | a = 10 / 2
b = a / 100
c = 1 + b
d = 1 / 2
e = 1 + d
f = e * 2
g = c ** f
h = 1000 * g
i = h - 1000
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a ) 149 , b ) 169 , c ) 841 , d ) 920 , e ) 219 | c | subtract(negate(361,529), multiply(subtract(49,121, 169,289), divide(subtract(49,121, 169,289), subtract(4,25, 49,121)))) | 4,25 , 49,121 , 169,289 , 361,529 , | "29 ^ 2 = 841 because follow sequence of square of the prime numbers answer : c" | a = negate - (
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a ) 3 / 5 , b ) 7 / 10 , c ) 3 / 2 , d ) 3 / 7 , e ) 3 / 8 | b | divide(circle_area(divide(0.70, const_2)), const_2) | what will be the vulgar fraction of 0.70 | "explanation : 0.70 = 70 / 100 = 7 / 10 option b" | a = 0 / 70
b = circle_area / (
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a ) 35 , b ) 20 , c ) 45 , d ) 50 , e ) 55 | b | divide(add(40, 40), add(divide(40, 30), divide(40, 15))) | susan drove an average speed of 30 miles per hour for the first 40 miles of a trip then at a average speed of 15 miles / hr for the remaining 40 miles of the trip if she made no stops during the trip what was susan ' s avg speed in miles / hr for the entire trip | avg . speed = total distance / total time total distance = 80 miles total time = 40 / 30 + 40 / 15 = 4 avg . speed = 20 . answer - b | a = 40 + 40
b = 40 / 30
c = 40 / 15
d = b + c
e = a / d
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a ) 720 , b ) 750 , c ) 820 , d ) 840 , e ) 860 | d | divide(subtract(multiply(1500, divide(15, const_100)), 15), divide(25, const_100)) | if 25 % of x is 15 less than 15 % of 1500 , then x is ? | "25 % of x = x / 4 ; 15 % of 1500 = 15 / 100 * 1500 = 225 given that , x / 4 = 225 - 15 = > x / 4 = 210 = > x = 840 . answer : d" | a = 15 / 100
b = 1500 * a
c = b - 15
d = 25 / 100
e = c / d
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a ) 12 sec , b ) 24 sec , c ) 48 sec , d ) 60 sec , e ) none | b | multiply(multiply(250, inverse(multiply(add(45, 30), const_0_2778))), const_2) | two good train each 250 m long , are running in opposite directions on parallel tracks . their speeds are 45 km / hr and 30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one . | "sol . relative speed = ( 45 + 30 ) km / hr = ( 75 x 5 / 18 ) m / sec = ( 125 / 6 ) m / sec . distance covered = ( 250 + 250 ) m = 1000 m . required time = ( 500 x 6 / 125 ) sec = 24 sec . answer b" | a = 45 + 30
b = a * const_0_2778
c = 1/(b)
d = 250 * c
e = d * 2
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a ) 1978 , b ) 2707 , c ) 7728 , d ) 7008 , e ) 7291 | d | subtract(9600, multiply(multiply(9600, subtract(const_1, divide(10, const_100))), divide(1500, add(3500, 1500)))) | a is a working partner and b is a sleeping partner in the business . a puts in rs . 3500 and b rs . 1500 , a receives 10 % of the profit for managing the business the rest being divided in proportion of their capitals . out of a total profit of rs . 9600 , money received by a is ? | "35 : 15 = > 7 : 3 9600 * 10 / 100 = 960 9600 - 960 = 8640 8640 * 7 / 10 = 6048 + 960 = 7008 answer : d" | a = 10 / 100
b = 1 - a
c = 9600 * b
d = 3500 + 1500
e = 1500 / d
f = c * e
g = 9600 - f
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a ) 25 % , b ) 35 % , c ) 45 % , d ) 55 % , e ) 65 % | c | multiply(45, const_1) | 45 % of major airline companies equip their planes with wireless internet access . 70 % of major airlines offer passengers free on - board snacks . what is the greatest possible percentage of major airline companies that offer both wireless internet and free on - board snacks ? | "to maximize the percentage of companies offering both , let ' s assume that all 45 % of companies which offer wireless internet also offer snacks . the answer is c ." | a = 45 * 1
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a ) 5000 , b ) 7000 , c ) 4000 , d ) 8000 , e ) 9000 | b | divide(4725, subtract(subtract(const_1, divide(10, const_100)), multiply(subtract(const_1, divide(10, const_100)), divide(25, const_100)))) | after 10 % of the inhabitants of a village disappeared , a panic set in during which 25 % of the remaining inhabitants left the village . at that time , the population was reduced to 4725 . what was the number of original inhabitants ? | "let the total number of original inhabitants be x . ( 75 / 100 ) * ( 90 / 100 ) * x = 4725 ( 27 / 40 ) * x = 4725 x = 4725 * 40 / 27 = 7000 the answer is b ." | a = 10 / 100
b = 1 - a
c = 10 / 100
d = 1 - c
e = 25 / 100
f = d * e
g = b - f
h = 4725 / g
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a ) 8 : 8 , b ) 9 : 8 , c ) 9 : 6 , d ) 9 : 5 , e ) 9 : 13 | e | divide(add(multiply(3000, 6), multiply(multiply(3000, const_2), 6)), multiply(6500, add(6, 6))) | a and b invests rs . 3000 and rs . 6500 respectively in a business . if a doubles his capital after 6 months . in what ratio should a and b divide that year ' s profit ? | "( 3 * 6 + 6 * 6 ) : ( 6.5 * 12 ) 54 : 78 = > 9 : 13 answer : e" | a = 3000 * 6
b = 3000 * 2
c = b * 6
d = a + c
e = 6 + 6
f = 6500 * e
g = d / f
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a ) 36 kmh , b ) 19 kmh , c ) 42 kmh , d ) 44 kmh , e ) 50 kmh | b | divide(const_3, add(add(divide(const_1, 10), divide(const_1, 24)), divide(const_1, 48))) | if a car went the first third of the distance at 10 kmh , the second third at 24 kmh , and the last third at 48 kmh , what was the average speed of the car for the entire trip ? | "assume d / 3 = 240 ( this number is convenient because it is divisible by 10 , 24 and 48 ) so : 240 = 10 * t 1 = 24 hrs 240 = 24 * t 2 = 10 hrs 240 = 48 * t 3 = 5 hrs t = t 1 + t 2 + t 3 = 39 hrs d = rt ( 240 * 3 ) = r * 39 r = 19 answer : b" | a = 1 / 10
b = 1 / 24
c = a + b
d = 1 / 48
e = c + d
f = 3 / e
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a ) 4 : 7 , b ) 4 : 9 , c ) 4 : 2 , d ) 4 : 1 , e ) 4 : 5 | a | divide(40, 70) | if 70 % of an equal to 40 % of b , then ratio between a and b is ? | explanation : 70 a = 40 b a : b = 4 : 7 answer : a | a = 40 / 70
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a ) 38 min , b ) 41 min , c ) 40 min , d ) 42 min , e ) 52 min | c | divide(subtract(add(40, multiply(const_3, const_60)), multiply(5, const_4)), 5) | a college has classes from 10 : 00 am until 1 : 40 pm . in this duration , there are 5 periods . if 5 minutes are provided between each period to leave one class and enter the next class , how many minutes long is each period ? | the total time is 220 minutes . there are four breaks of five minutes between the classes for a total of 20 minutes . the total class time is 200 minutes . 200 / 5 classes = 40 minutes per class the answer is c . | a = 3 * const_60
b = 40 + a
c = 5 * 4
d = b - c
e = d / 5
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a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | d | divide(multiply(multiply(multiply(const_4, 3), 2), 1), multiply(multiply(3, 2), 1)) | a card game called β high - low β divides a deck of 52 playing cards into 2 types , β high β cards and β low β cards . there are an equal number of β high β cards and β low β cards in the deck and β high β cards are worth 2 points , while β low β cards are worth 1 point . if you draw cards one at a time , how many ways can you draw β high β and β low β cards to earn 5 points if you must draw exactly 3 β low β cards ? | to get a 5 , you need one high and three lows ( you could have had 2 highs and one low , but the constraint is that you must have three low cards ) hlll = 4 ! 3 ! = 4 4 ! is the number of ways you can arrange these four spaces . divide by 3 ! because you you repeat three low cards ans : d | a = 4 * 3
b = a * 2
c = b * 1
d = 3 * 2
e = d * 1
f = c / e
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a ) 41 , b ) 65 , c ) 70 , d ) 77 , e ) 98 | a | add(add(add(divide(lcm(lcm(lcm(2, 3), 3), 5), 2), divide(lcm(lcm(lcm(2, 3), 3), 5), 3)), divide(lcm(lcm(lcm(2, 3), 3), 5), 3)), divide(lcm(lcm(lcm(2, 3), 3), 5), 5)) | john distributes his pencil among his 4 friends rose , mary , ranjan , and rohit in the ratio 1 / 2 : 1 / 3 : 1 / 3 : 1 / 5 . what is the minimum no . of pencils that the person should have ? | rakesh : rahul : ranjan : rohit = 1 / 2 : 1 / 3 : 1 / 3 : 1 / 5 step 1 : at first we need to do is lcm of 2 , 3,3 and 5 is 30 . step 2 : then pencil are distributed in ratio among friends , rakesh = ( 1 / 2 x 30 ) = 15 . rahul = ( 1 / 3 x 30 ) = 10 . ranjan = ( 1 / 3 x 30 ) = 10 . rohit = ( 1 / 5 x 30 ) = 6 . step 3 : total number of pencils are ( 15 x + 10 x + 10 x + 6 x ) = 41 x . for minimum number of pencils x = 1 . the person should have at least 41 pencils . a ) | a = math.lcm(2, 3)
b = math.lcm(a, 3)
c = math.lcm(b, 5)
d = c / 2
e = math.lcm(2, 3)
f = math.lcm(e, 3)
g = math.lcm(f, 5)
h = g / 3
i = d + h
j = math.lcm(2, 3)
k = math.lcm(j, 3)
l = math.lcm(k, 5)
m = l / 3
n = i + m
o = math.lcm(2, 3)
p = math.lcm(o, 3)
q = math.lcm(p, 5)
r = q / 5
s = n + r
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