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 / 10 , b ) 9 / 10 , c ) 7 / 10 , d ) 6 / 10 , e ) 4 / 10 | c | subtract(const_1, multiply(add(divide(const_1, 10), divide(const_1, 20)), 2)) | p is able to do a piece of work in 20 days and q can do the same work in 10 days . if they can work together for 2 days , what is the fraction of work left ? | "explanation : amount of work p can do in 1 day = 1 / 20 amount of work q can do in 1 day = 1 / 10 amount of work p and q can do in 1 day = 1 / 20 + 1 / 10 = 3 / 20 amount of work p and q can together do in 2 days = 2 Γ ( 3 / 20 ) = 3 / 10 fraction of work left = 1 β 3 / 10 = 7 / 10 answer : option c" | a = 1 / 10
b = 1 / 20
c = a + b
d = c * 2
e = 1 - d
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a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | c | divide(subtract(70, multiply(5, 2)), add(10, 5)) | carina has 70 ounces of coffee divided into 5 - and 10 - ounce packages . if she has 2 more 5 - ounce packages than 10 - ounce packages , how many 10 - ounce packages does she have ? | "lets say 5 and 10 ounce packages be x and y respectively . given that , 5 x + 10 y = 70 and x = y + 2 . what is the value of y . substituting the x in first equation , 5 y + 10 + 10 y = 70 - > y = 60 / 15 . = 4 c" | a = 5 * 2
b = 70 - a
c = 10 + 5
d = b / c
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a ) 33 kg , b ) 31 kg , c ) 32 kg , d ) 36 kg , e ) 35 kg | e | subtract(add(multiply(42, const_2), multiply(43, const_2)), multiply(45, const_3)) | the average weight of a , b and c is 45 kg . if the average weight of a and b be 42 kg and that of b and c be 43 kg , then the weight of b is : | "let a , b , c represent their respective weights . then , we have : a + b + c = ( 45 x 3 ) = 135 . . . . ( i ) a + b = ( 42 x 2 ) = 84 . . . . ( ii ) b + c = ( 43 x 2 ) = 86 . . . . ( iii ) adding ( ii ) and ( iii ) , we get : a + 2 b + c = 170 . . . . ( iv ) subtracting ( i ) from ( iv ) , we get : b = 35 b ' s weight = 35 kg . e" | a = 42 * 2
b = 43 * 2
c = a + b
d = 45 * 3
e = c - d
|
a ) 51 , b ) 45 , c ) 65 , d ) 78 , e ) 64 | a | add(40, 11) | john found that the average of 15 numbers is 40 . if 11 is added to each number then the mean of number is ? | "( x + x 1 + . . . x 14 ) / 15 = 40 51 option a" | a = 40 + 11
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a ) 3 , b ) 6.5 , c ) 5.5 , d ) 4 , e ) 3.75 | e | divide(subtract(divide(100, 8), divide(75, 15)), const_2) | a boat goes 100 km downstream in 8 hours , and 75 km upstream in 15 hours . the speed of the stream is ? | "100 - - - 10 ds = 12.5 ? - - - - 1 75 - - - - 15 us = 5 ? - - - - - 1 s = ( 12.5 - 5 ) / 2 = 3.75 kmph . answer : e" | a = 100 / 8
b = 75 / 15
c = a - b
d = c / 2
|
a ) 884 , b ) 890 , c ) 892 , d ) 910 , e ) 945 | a | subtract(1000, subtract(add(divide(1000, 11), divide(1000, 30)), divide(1000, multiply(11, 30)))) | what is the number of integers from 1 to 1000 ( inclusive ) that are divisible by neither 11 nor by 30 ? | "normally , i would use the method used by bunuel . it ' s the most accurate . but if you are looking for a speedy solution , you can use another method which will sometimes give you an estimate . looking at the options ( most of them are spread out ) , i wont mind trying it . ( mind you , the method is accurate here since the numbers start from 1 . ) in 1000 consecutive numbers , number of multiples of 11 = 1000 / 11 = 90 ( ignore decimals ) in 1000 consecutive numbers , number of multiples of 35 = 1000 / 35 = 28 number of multiples of 11 * 35 i . e . 385 = 1000 / 385 = 2 number of integers from 1 to 1000 that are divisible by neither 11 nor by 35 = 1000 - ( 90 + 28 - 2 ) { using the concept of sets here ) = 884 think : why did i say the method is approximate in some cases ? think what happens if the given range is 11 to 1010 both inclusive ( again 1000 numbers ) what is the number of multiples in this case ? a" | a = 1000 / 11
b = 1000 / 30
c = a + b
d = 11 * 30
e = 1000 / d
f = c - e
g = 1000 - f
|
a ) 288 , b ) 564 , c ) 877 , d ) 278 , e ) 178 | b | add(multiply(divide(60, subtract(21, 26)), 26), multiply(divide(60, subtract(21, 26)), 21)) | two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of 26 kmph and 21 kmph respectively . when they meet , it is found that one train has traveled 60 km more than the other one . the distance between the two stations is ? | "1 h - - - - - 5 ? - - - - - - 60 12 h rs = 26 + 21 = 47 t = 12 d = 47 * 12 = 564 answer : b" | a = 21 - 26
b = 60 / a
c = b * 26
d = 21 - 26
e = 60 / d
f = e * 21
g = c + f
|
a ) 5 , b ) 10 , c ) 11 , d ) 20 , e ) 30 | b | divide(multiply(25268, 11), 25268) | what is the smallest no . which must be added to 25268 so as to obtain a sum which is divisible by 11 ? | "for divisibility by 11 , the difference of sums of digits at even and odd places must be either zero or divisible by 11 . for 25268 , difference = ( 2 + 2 + 8 ) - ( 5 + 6 ) = 12 - 11 = 1 . the units digit is at odd place . so we add 10 to the number = > 25268 + 10 = 25278 now , ( 2 + 2 + 8 ) - ( 5 + 7 ) = 12 - 12 = 0 = > 25268 is also divisible by 11 b" | a = 25268 * 11
b = a / 25268
|
a ) 22 sec , b ) 16 sec , c ) 17 sec , d ) 39 sec , e ) 12 sec | d | multiply(divide(455, multiply(63, const_1000)), const_3600) | a train 455 m long , running with a speed of 63 km / hr will pass a tree in ? | "speed = 63 * 5 / 18 = 35 / 2 m / sec time taken = 455 * 2 / 35 = 39 sec answer : d" | a = 63 * 1000
b = 455 / a
c = b * 3600
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a ) 28 % , b ) 40 % , c ) 64.8 % , d ) 76.5 % , e ) 72 % | d | add(multiply(divide(divide(15, const_100), subtract(1, divide(1, 10))), const_100), 2) | the price of an item is discounted 10 percent on day 1 of a sale . on day 2 , the item is discounted another 10 percent , and on day 3 , it is discounted an additional 15 percent . the price of the item on day 3 is what percentage of the sale price on day 1 ? | "original price = 100 day 1 discount = 10 % , price = 100 - 10 = 90 day 2 discount = 10 % , price = 90 - 9 = 81 day 3 discount = 15 % , price = 81 - 12.15 = 68.85 which is 68.85 / 90 * 100 of the sale price on day 1 = ~ 76.5 % answer d" | a = 15 / 100
b = 1 / 10
c = 1 - b
d = a / c
e = d * 100
f = e + 2
|
a ) 520 , b ) 288 , c ) 820 , d ) 740 , e ) 720 | b | divide(divide(multiply(120, 4), add(const_1, divide(const_2, const_3))), const_2) | an aeroplane covers a certain distance at a speed of 120 kmph in 4 hours . to cover the same distance in 1 2 / 3 hours , it must travel at a speed of : | "distance = ( 240 x 5 ) = 480 km . speed = distance / time speed = 480 / ( 5 / 3 ) km / hr . [ we can write 1 2 / 3 hours as 5 / 3 hours ] required speed = ( 480 x 3 / 5 ) km / hr = 288 km / hr answer b ) 288 km / hr" | a = 120 * 4
b = 2 / 3
c = 1 + b
d = a / c
e = d / 2
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a ) 33 , b ) 38 , c ) 27 , d ) 40 , e ) 48 | c | subtract(subtract(add(3, multiply(3, subtract(subtract(add(multiply(const_2, 5), 10), 5), 3))), const_10), const_1) | the present age of a father is 3 years more than 3 times the age of his son . 5 years hence , father ' s age will be 10 years more than twice the age of the son . find the present age of the father . | if the present age be x years . father ' s will be ( 3 x + 3 ) years . . so , ( 3 x + 3 + 5 ) = 2 ( x + 3 ) + 10 or , x = 8 so the fathers present age = ( 3 x + 3 ) = ( 3 * 8 + 3 ) years = 27 years . . answer : option c | a = 2 * 5
b = a + 10
c = b - 5
d = c - 3
e = 3 * d
f = 3 + e
g = f - 10
h = g - 1
|
a ) 100 , b ) 360 , c ) 120 , d ) 200 , e ) 150 | b | multiply(12, 20) | the h . c . f . of two numbers is 12 and their l . c . m . is 600 . if one of the number is 20 , find the other ? | "other number = 12 * 600 / 20 = 360 answer is b" | a = 12 * 20
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a ) 632 , b ) 732 , c ) 832 , d ) 850 , e ) 902 | b | add(4, reminder(4, 12)) | what is the next no . 4 12 84 | "3 ^ 0 + 3 = 4 3 ^ 2 + 3 = 12 3 ^ 4 + 3 = 84 3 ^ 6 + 3 = 732 answer : b" | a = 4 + reminder
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a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | c | subtract(add(multiply(11, const_2), 3), add(2, 9)) | x and y are positive integers . when x is divided by 9 , the remainder is 2 , and when x is divided by 7 , the remainder is 4 . when y is divided by 11 , the remainder is 3 , and when y is divided by 13 , the remainder is 12 . what is the least possible value of y - x ? | "when x is divided by 9 , the remainder is 2 : so , the possible values of x are : 2 , 11 , 21 , 29 , etc . when x is divided by 7 , the remainder is 4 : so , the possible values of x are : 4 , 11,18 , . . . stop . since both lists include 11 , the smallest possible value of x is 11 . when y is divided by 11 , the remainder is 3 : so , the possible values of y are : 3 , 14 , 25 , etc . when y is divided by 13 , the remainder is 12 : so , the possible values of y are : 12 , 25 , . . . stop . since both lists include 25 , the smallest possible value of y is 25 since the smallest possible values of x and y are 11 and 25 respectively , the smallest possible value of y - x is 14 . so , c is the correct answer to the original question ." | a = 11 * 2
b = a + 3
c = 2 + 9
d = b - c
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a ) $ 1.40 , b ) $ 2.40 , c ) $ 3.315 , d ) $ 4.40 , e ) $ 5.40 | c | multiply(subtract(const_1, multiply(divide(add(const_100, 22), const_100), divide(const_1, const_2))), 8.50) | at joes steakhouse the hourly wage for a chef is 22 % greater than that of a dishwasher , and the hourly wage of a dishwasher is half as much as the hourly wage of a manager . if a managers wage is $ 8.50 per hour , how much less than a manager does a chef earn each hour ? | "manager wages per hour = $ 8.50 dishwasher wages per hour = half of manager ' s wages . = 1 / 2 ( $ 8.50 ) = = > $ 4.25 chef wages per hour = 22 % greater than dishwasher wages - - > 22 % of $ 4.25 = ( 22 * ( $ 4.25 ) ) / 100 - - > ( $ 93.5 ) / 100 - - > $ 0.935 therefore , chef wages per hour = $ 4.25 + $ 0.935 = = > $ 5.185 difference of wages between manager and chef = $ 8.50 - $ 5.185 = = > $ 3.315 answer : c" | a = 100 + 22
b = a / 100
c = 1 / 2
d = b * c
e = 1 - d
f = e * 8
|
a ) $ 1250 , b ) $ 700 , c ) $ 1350 , d ) $ 900 , e ) $ 1000 | b | multiply(subtract(10000, 2000), multiply(subtract(const_1, divide(const_1, 10)), divide(const_1, 10))) | if xerox paper costs 5 cents a sheet and a buyer gets 10 % discount on all xerox paper one buys after the first 2000 papers and 20 % discount after first 10000 papers , how much will it cost to buy 15000 sheets of xerox paper ? | 30 sec approach - solve it using approximation 15000 sheet at full price , 5 cent = 750 15000 sheet at max discount price , 4 cent = 600 your ans got to be between these two . ans b it is . | a = 10000 - 2000
b = 1 / 10
c = 1 - b
d = 1 / 10
e = c * d
f = a * e
|
a ) 108 , b ) 80 , c ) 128 , d ) 135 , e ) 143 | b | multiply(divide(subtract(18, divide(36, 18)), divide(36, 18)), add(divide(subtract(18, divide(36, 18)), divide(36, 18)), divide(36, 18))) | if the sum of two positive integers is 18 and the difference of their squares is 36 , what is the product of the two integers ? | "let the 2 positive numbers x and y x + y = 18 - - 1 x ^ 2 - y ^ 2 = 36 = > ( x + y ) ( x - y ) = 36 - - 2 using equation 1 in 2 , we get = > x - y = 2 - - 3 solving equation 1 and 3 , we get x = 10 y = 8 product = 10 * 8 = 80 answer b" | a = 36 / 18
b = 18 - a
c = 36 / 18
d = b / c
e = 36 / 18
f = 18 - e
g = 36 / 18
h = f / g
i = 36 / 18
j = h + i
k = d * j
|
a ) 110 , b ) 120 , c ) 186 , d ) 140 , e ) 150 | c | divide(add(42, 23), divide(35, const_100)) | a candidate appearing for an examination has to secure 35 % marks to pass paper i . but he secured only 42 marks and failed by 23 marks . what is the maximum mark for paper i ? | "he secured 42 marks nd fail by 23 marks so total marks for pass the examinatn = 65 let toal marks x x * 35 / 100 = 65 x = 186 answer : c" | a = 42 + 23
b = 35 / 100
c = a / b
|
a ) 13 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | b | subtract(subtract(divide(divide(50, const_2), const_2), const_10), const_10) | how many odd integers from 1 to 50 ( both inclusive ) have odd number of factors ? | "integers having odd number of factors will be perfect squares . odd numbers will have odd perfect squares . thus , the possible values for the perfect squares are : 1,9 , 25,49 and the corresponding integers are 1,3 , 5,7 ( more than 3 ) . thus b is the correct answer ." | a = 50 / 2
b = a / 2
c = b - 10
d = c - 10
|
a ) 55 , b ) 56 , c ) 57 , d ) 52 , e ) none of these | d | divide(rectangle_perimeter(90, 40), 5) | a rectangular plot measuring 90 metres by 40 metres is to be enclosed by wire fencing . if the poles of the fence are kept 5 metres apart , how many poles will be needed ? | "solution perimeter of the plot = 2 ( 90 + 40 ) = 260 m . β΄ number of poles = [ 260 / 5 ] = 52 m answer d" | a = rectangle_perimeter / (
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a ) - 6 , b ) 1 , c ) - 2 , d ) - 3 , e ) 4 | a | multiply(subtract(2, const_4), const_3) | find the value for x from below equation : x / 3 = - 2 ? | 1 . multiply both sides by 3 : x * 3 / 3 = - 2 / 3 2 . simplify both sides : x = - 6 a | a = 2 - 4
b = a * 3
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a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | c | multiply(subtract(const_1, divide(4, 12)), 9) | dan can do a job alone in 12 hours . annie , working alone , can do the same job in just 9 hours . if dan works alone for 4 hours and then stops , how many hours will it take annie , working alone , to complete the job ? | "dan can complete 1 / 12 of the job per hour . in 4 hours , dan completes 4 ( 1 / 12 ) = 1 / 3 of the job . annie can complete 1 / 9 of the job per hour . to complete the job , annie will take 2 / 3 / 1 / 9 = 6 hours . the answer is c ." | a = 4 / 12
b = 1 - a
c = b * 9
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a ) 288 , b ) 48 , c ) 72 , d ) 972 , e ) 964 | d | multiply(divide(const_4, 3), power(3, 3)) | the measurement of a rectangular box with lid is 25 cmx 18 cmx 18 cm . find the volume of the largest sphere that can be inscribed in the box ( in terms of Ο cm 3 ) . ( hint : the lowest measure of rectangular box represents the diameter of the largest sphere ) | "d = 18 , r = 9 ; volume of the largest sphere = 4 / 3 Ο r 3 = 4 / 3 * Ο * 9 * 9 * 9 = 972 Ο cm 3 answer : d" | a = 4 / 3
b = 3 ** 3
c = a * b
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a ) 11 / 30 , b ) 31 / 60 , c ) 41 / 80 , d ) 51 / 120 , e ) 71 / 180 | e | subtract(add(divide(5, 4), subtract(const_1, divide(const_3.0, 4))), divide(9, 5)) | the probability that a computer company will get a computer hardware contract is 3 / 4 and the probability that it will not get a software contract is 5 / 9 . if the probability of getting at least one contract is 4 / 5 , what is the probability that it will get both the contracts ? | "let , a β‘ event of getting hardware contract b β‘ event of getting software contract ab β‘ event of getting both hardware and software contract . p ( a ) = 3 / 4 , p ( ~ b ) = 5 / 9 = > p ( b ) = 1 - ( 5 / 9 ) = 4 / 9 . a and b are not mutually exclusive events but independent events . so , p ( at least one of a and b ) = p ( a ) + p ( b ) - p ( ab ) . = > 4 / 5 = ( 3 / 4 ) + ( 4 / 9 ) - p ( ab ) . = > p ( ab ) = 71 / 180 . hence , the required probability is 71 / 180 . the answer is e ." | a = 5 / 4
b = 3 / 0
c = 1 - b
d = a + c
e = 9 / 5
f = d - e
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a ) 3 minutes , b ) 4 minutes , c ) 4.2 minutes , d ) 6 minutes , e ) 12 minutes | c | divide(multiply(3, 10), add(speed(10, 10), speed(multiply(3, 10), 5))) | working alone at its constant rate , machine a produces x boxes in 10 minutes and working alone at its constant rate , machine b produces 3 x boxes in 5 minutes . how many minutes does it take machines a and b , working simultaneously at their respective constant rates , to produce 3 x boxes ? | "rate = work / time given rate of machine a = x / 10 min machine b produces 3 x boxes in 5 min hence , machine b produces 4 x boxes in 10 min . rate of machine b = 6 x / 10 we need tofind the combined time that machines a and b , working simultaneouslytakeat their respective constant rates let ' s first find the combined rate of machine a and b rate of machine a = x / 10 min + rate of machine b = 6 x / 10 = 7 x / 10 now combine time = combine work needs to be done / combine rate = 3 x / 7 x * 10 = 4.2 min ans : c" | a = 3 * 10
b = speed + (
c = 3 * 10
d = a / b
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a ) 60 , b ) 120 , c ) 130 , d ) 140 , e ) 150 | a | divide(add(102, 138), 4) | a student chose a number , multiplied it by 4 , then subtracted 138 from the result and got 102 . what was the number he chose ? | solution : let xx be the number he chose , then 4 β
x β 138 = 102 4 x = 240 x = 60 answer a | a = 102 + 138
b = a / 4
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a ) 1 kmph , b ) 4 kmph , c ) 3 kmph , d ) 2 kmph , e ) 1.9 kmph | b | divide(subtract(12, 4), const_2) | what is the speed of the stream if a canoe rows upstream at 4 km / hr and downstream at 12 km / hr | "sol . speed of stream = 1 / 2 ( 12 - 4 ) kmph = 4 kmph . answer b" | a = 12 - 4
b = a / 2
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a ) 2 pm , b ) 9 pm , c ) 3 pm , d ) 8 pm , e ) 6 pm | c | subtract(multiply(3, const_12), divide(multiply(3, const_12), add(divide(3, 5), const_1))) | when asked what the time is , a person answered that the amount of time left is 3 / 5 of the time already completed . what is the time . | "a day has 24 hrs . assume x hours have passed . remaining time is ( 24 - x ) 24 β x = 3 / 5 x β x = 15 time is 3 pm answer : c" | a = 3 * 12
b = 3 * 12
c = 3 / 5
d = c + 1
e = b / d
f = a - e
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a ) 6 min , b ) 8 min , c ) 7 min , d ) 9 min , e ) 1 min | d | divide(divide(const_4, add(const_2, const_3)), subtract(inverse(6), inverse(10))) | a water tank is three - fifths full . pipe a can fill a tank in 10 minutes and pipe b can empty it in 6 minutes . if both the pipes are open , how long will it take to empty or fill the tank completely ? | "the combined rate of filling / emptying the tank = 1 / 10 - 1 / 6 = - 1 / 15 since the rate is negative , the tank will be emptied . a full tank would take 15 minutes to empty . since the tank is only three - fifths full , the time is ( 3 / 5 ) * 15 = 9 minutes the answer is d ." | a = 2 + 3
b = 4 / a
c = 1/(6)
d = 1/(10)
e = c - d
f = b / e
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a ) 19 : 1 , b ) 19 : 4 , c ) 19 : 8 , d ) 38 : 10 , e ) 38 : 2 | d | divide(add(divide(multiply(62.5, 4), const_100), divide(multiply(87.5, 8), const_100)), add(subtract(4, divide(multiply(62.5, 4), const_100)), subtract(8, divide(multiply(87.5, 8), const_100)))) | two vessels p and q contain 62.5 % and 87.5 % of alcohol respectively . if 4 litres from vessel p is mixed with 8 litres from vessel q , the ratio of alcohol and water in the resulting mixture is ? | "quantity of alcohol in vessel p = 62.5 / 100 * 4 = 5 / 2 litres quantity of alcohol in vessel q = 87.5 / 100 * 8 = 7 / 1 litres quantity of alcohol in the mixture formed = 5 / 2 + 7 / 1 = 19 / 2 = 9.50 litres as 12 litres of mixture is formed , ratio of alcohol and water in the mixture formed = 9.50 : 2.50 = 38 : 10 . answer : d" | a = 62 * 5
b = a / 100
c = 87 * 5
d = c / 100
e = b + d
f = 62 * 5
g = f / 100
h = 4 - g
i = 87 * 5
j = i / 100
k = 8 - j
l = h + k
m = e / l
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a ) a ) 125 , b ) b ) 175 , c ) c ) 481 , d ) d ) 375 , e ) e ) 524 | c | divide(multiply(divide(228, const_100), 1265), 6) | ( 228 % of 1265 ) Γ· 6 = ? | "explanation : ? = ( 228 x 1265 / 100 ) Γ· 6 = 288420 / 600 = 481 answer : option c" | a = 228 / 100
b = a * 1265
c = b / 6
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a ) 2178 lt , b ) 3697 lt , c ) 6583 lt , d ) 4370 lt , e ) 5548 lt | d | subtract(8100, divide(subtract(9100, multiply(add(const_1, divide(10, const_100)), 8100)), subtract(add(const_1, divide(15, const_100)), add(const_1, divide(10, const_100))))) | the first year , two cows produced 8100 litres of milk . the second year their production increased by 15 % and 10 % respectively , and the total amount of milk increased to 9100 litres a year . how many litres were milked from one cow ? | "let x be the amount of milk the first cow produced during the first year . then the second cow produced ( 8100 β x ) litres of milk that year . the second year , each cow produced the same amount of milk as they did the first year plus the increase of 15 % 15 % or 10 % so 8100 + 15100 β
x + 10100 β
( 8100 β x ) = 9100 therefore 8100 + 320 x + 110 ( 8100 β x ) = 9100 120 x = 190 x = 3800 therefore , the cows produced 3800 and 4300 litres of milk the first year , and 4370 and 4730 litres of milk the second year , = > one cow produced 4370 litres correct answer is d ) 4370 lt" | a = 10 / 100
b = 1 + a
c = b * 8100
d = 9100 - c
e = 15 / 100
f = 1 + e
g = 10 / 100
h = 1 + g
i = f - h
j = d / i
k = 8100 - j
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a ) 138.9 cms , b ) 149.2 cms , c ) 168.6 cms , d ) 159.2 cms , e ) 142.5 cms | c | divide(add(multiply(169, 40), multiply(167, const_10)), 50) | the average height of 40 girls out of a class of 50 is 169 cm . and that of the remaining girls is 167 cm . the average height of the whole class is : | "explanation : average height of the whole class = ( 40 Γ 169 + 10 Γ 167 / 50 ) = 168.6 cms answer c" | a = 169 * 40
b = 167 * 10
c = a + b
d = c / 50
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a ) 1 / 7 , b ) 7 / 12 , c ) 1 / 2 , d ) 2 / 7 , e ) 2 / 3 | d | divide(const_2, 7) | equal amount of water were poured into two empty jars of different capacities , which made one jar 1 / 7 full and other jar 1 / 6 full . if the water in the jar with lesser capacity is then poured into the jar with greater capacity , what fraction of the larger jar will be filled with water ? | "same amount of water made bigger jar 1 / 7 full , then the same amount of water ( stored for a while in smaller jar ) were added to bigger jar , so bigger jar is 1 / 7 + 1 / 7 = 2 / 7 full . answer : d ." | a = 2 / 7
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a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | divide(subtract(add(multiply(2, 5), 10), add(multiply(3, 2), 5)), subtract(multiply(2, 3), multiply(3, const_1))) | given f ( x ) = 3 x β 5 , for what value of x does 2 * [ f ( x ) ] β 10 = f ( x β 2 ) ? | "2 ( 3 x - 5 ) - 10 = 3 ( x - 2 ) - 5 3 x = 9 x = 3 the answer is c ." | a = 2 * 5
b = a + 10
c = 3 * 2
d = c + 5
e = b - d
f = 2 * 3
g = 3 * 1
h = f - g
i = e / h
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a ) 14000 , b ) 14400 , c ) 19600 , d ) 14600 , e ) 14700 | c | add(10000, multiply(divide(multiply(10000, 40), const_100), 2)) | the population of a town is 10000 . it increases annually at the rate of 40 % p . a . what will be its population after 2 years ? | "formula : 10000 Γ 140 / 100 Γ 140 / 100 = 19600 answer : c" | a = 10000 * 40
b = a / 100
c = b * 2
d = 10000 + c
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a ) none , b ) one , c ) two , d ) three , e ) four | a | add(1, 1) | for any integer n greater than 1 , # n denotes the product of all the integers from 1 to n , inclusive . how many prime numbers e are there between # 6 + 2 and # 6 + 6 , inclusive ? | "none is the answer . a . because for every k 6 ! + k : : k , because 6 ! : : k , since k is between 2 and 6 . a" | a = 1 + 1
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a ) 5 , b ) 9 , c ) 16 , d ) 2 , e ) 42 | d | subtract(multiply(multiply(multiply(784, 618), 917), 463), subtract(multiply(multiply(multiply(784, 618), 917), 463), add(const_4, const_4))) | the unit digit in the product ( 784 x 618 x 917 x 463 ) is : | "explanation : unit digit in the given product = unit digit in ( 4 x 8 x 7 x 3 ) = ( 672 ) = 2 d" | a = 784 * 618
b = a * 917
c = b * 463
d = 784 * 618
e = d * 917
f = e * 463
g = 4 + 4
h = f - g
i = c - h
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a ) 33 , b ) 45 , c ) 66 , d ) 80 , e ) 21 | d | divide(volume_cylinder(divide(8, const_2), 5), const_pi) | the diameter of a cylindrical tin is 8 cm and height is 5 cm . find the volume of the cylinder ? | "r = 4 h = 5 Ο * 4 * 4 * 5 = 80 Ο cc answer : d" | a = 8 / 2
b = volume_cylinder / (
|
a ) 1 , b ) 7 . , c ) 2 . , d ) 3 , e ) 9 | a | add(multiply(const_4, const_2), reminder(add(add(multiply(subtract(const_10, const_1), const_100), multiply(multiply(add(const_3, const_2), const_100), const_10)), multiply(add(const_12, add(const_3, const_2)), add(const_3, const_2))), 2)) | whats the reminder when 54,879 , 856,985 , 421,547 , 895,689 , 874,525 , 826,547 is divided by 2 | "a number ending in a 0 is divisible by 2 . given the obscene number , you should immediately be convinced that you will need to focus on a very small part of it . 54,879 , 856,985 , 421,547 , 895,689 , 874,525 , 826,547 = 54,879 , 856,985 , 421,547 , 895,689 , 874,525 , 826,540 + 7 the first number is divisible by 16 . you just have to find the remainder when you divide 5287 by 16 . that will be the remainder when you divide the original number by 2 . 7 / 16 gives remainder 1 . answer ( a )" | a = 4 * 2
b = 10 - 1
c = b * 100
d = 3 + 2
e = d * 100
f = e * 10
g = c + f
h = 3 + 2
i = 12 + h
j = 3 + 2
k = i * j
l = g + k
m = a + reminder
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a ) 26 , b ) 19 , c ) 11 , d ) 8 , e ) 6 | b | subtract(add(26, 30), subtract(45, 8)) | each of the dogs in a certain kennel is a single color . each of the dogs in the kennel either has long fur or does not . of the 45 dogs in the kennel , 26 have long fur , 30 are brown , and 8 are neither long - furred nor brown . how many long - furred dogs are brown ? | "no of dogs = 45 long fur = 26 brown = 30 neither long fur nor brown = 8 therefore , either long fur or brown = 45 - 8 = 37 37 = 26 + 30 - both both = 19 answer b" | a = 26 + 30
b = 45 - 8
c = a - b
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a ) 40 , b ) 3000 / 11 , c ) 5900 / 247 , d ) 2790 / 11 , e ) 2709 / 8 | c | multiply(divide(subtract(multiply(108, 17), multiply(114, 13)), multiply(114, 13)), const_100) | a person bought 114 glass bowls at a rate of rs . 13 per bowl . he sold 108 of them at rs . 17 and the remaining broke . what is the percentage gain for a ? | "cp = 114 * 13 = 1482 and sp = 108 * 17 = 1836 gain % = 100 * ( 1836 - 1482 ) / 1482 = 5900 / 247 answer : c" | a = 108 * 17
b = 114 * 13
c = a - b
d = 114 * 13
e = c / d
f = e * 100
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a ) 4 hours , b ) 5 hours , c ) 6 hours , d ) 7 hours , e ) 8 hours | d | divide(147, add(16, 5)) | a boat can travel with a speed of 16 km / hr in still water . if the rate of stream is 5 km / hr , then find the time taken by the boat to cover distance of 147 km downstream . | "explanation : it is very important to check , if the boat speed given is in still water or with water or against water . because if we neglect it we will not reach on right answer . i just mentioned here because mostly mistakes in this chapter are of this kind only . lets see the question now . speed downstream = ( 16 + 5 ) = 21 kmph time = distance / speed = 147 / 21 = 7 hours option d" | a = 16 + 5
b = 147 / a
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a ) 31 , b ) 36 , c ) 53 , d ) 58 , e ) none | c | add(38, const_1) | the average age of 38 students in a group is 14 years . when teacher Γ’ β¬ β’ s age is included to it , the average increases by one . what is the teacher Γ’ β¬ β’ s age in years ? | "sol . age of the teacher = ( 39 Γ£ β 15 Γ’ β¬ β 38 Γ£ β 14 ) years = 53 years . answer c" | a = 38 + 1
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a ) 5 : 28 , b ) 5 : 19 , c ) 15 : 12 , d ) 5 : 13 , e ) 15 : 46 | e | divide(subtract(sqrt(4761), 24), multiply(sqrt(4761), const_2)) | the area of a square is 4761 sq cm . find the ratio of the breadth and the length of a rectangle whose length is twice the side of the square and breadth is 24 cm less than the side of the square . | "let the length and the breadth of the rectangle be l cm and b cm respectively . let the side of the square be a cm . a 2 = 4761 a = 69 l = 2 a and b = a - 24 b : l = a - 24 : 2 a = 45 : 138 = 15 : 46 answer : e" | a = math.sqrt(4761)
b = a - 24
c = math.sqrt(4761)
d = c * 2
e = b / d
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a ) 1 / 4 , b ) 56 / 27 , c ) 2 , d ) 3 , e ) 4 | c | multiply(multiply(const_2, divide(const_1, 28)), 28) | a positive integer n is a perfect number provided that the sum of all the positive factors of n , including 1 and n , is equal to 2 n . what is the sum of the reciprocals of all the positive factors of the perfect number 28 ? | soln : 28 = 1 * 28 2 * 14 4 * 7 sum of reciprocals = 1 + 1 / 28 + 1 / 2 + 1 / 14 + 1 / 4 + 1 / 7 = 56 / 28 = 2 answer : c | a = 1 / 28
b = 2 * a
c = b * 28
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a ) 72 sec , b ) 210 sec , c ) 192 sec , d ) 252 sec , e ) none | b | multiply(const_3600, divide(divide(add(200, 150), const_1000), subtract(46, 40))) | two trains 200 m and 150 m long are running on parallel rails at the rate of 40 kmph and 46 kmph respectively . in how much time will they cross each other , if they are running in the same direction ? | "solution relative speed = ( 46 - 40 ) kmph = 6 kmph = ( 6 x 5 / 18 ) m / sec = ( 30 / 18 ) m / sec time taken = ( 350 x 18 / 30 ) sec = 210 sec . answer b" | a = 200 + 150
b = a / 1000
c = 46 - 40
d = b / c
e = 3600 * d
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a ) $ 128,000 , b ) $ 120,000 , c ) $ 110,000 , d ) $ 139,000 , e ) $ 125,000 | b | multiply(subtract(divide(multiply(subtract(const_100, const_10), const_1000), subtract(multiply(subtract(const_100, const_10), const_1000), multiply(multiply(const_0_25, const_100), const_1000))), const_1), const_100) | an employee β s annual salary was increased 50 % . if her old annual salary equals $ 80,000 , what was the new salary ? | "old annual salary = $ 80,000 salary increase = 50 % . original salary = $ 80,000 * 50 / 100 = $ 40,000 new salary = $ 80,000 + $ 40,000 = $ 120,000 hence b ." | a = 100 - 10
b = a * 1000
c = 100 - 10
d = c * 1000
e = const_0_25 * 100
f = e * 1000
g = d - f
h = b / g
i = h - 1
j = i * 100
|
a ) 2 / 15 , b ) 8 / 15 , c ) 3 / 11 , d ) 1 / 12 , e ) 1 / 15 | e | subtract(const_1, multiply(8, add(divide(const_1, 15), divide(const_1, 20)))) | a can do a job in 15 days and b in 20 days . if they work on it together for 8 days , then the fraction of the work that is left is ? | "a ' s 1 day work = 1 / 15 b ' s 1 day work = 1 / 20 a + b 1 day work = 1 / 15 + 1 / 20 = 7 / 60 a + b 8 days work = 7 / 60 * 8 = 14 / 15 remaining work = 1 - 14 / 15 = 1 / 15 answer is e" | a = 1 / 15
b = 1 / 20
c = a + b
d = 8 * c
e = 1 - d
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a ) 1 : 20 , b ) 1 : 10 , c ) 1 : 8 , d ) 1 : 4 , e ) 6 : 11 | b | divide(const_1, divide(20, const_2)) | a dishonest milkman wants to make a profit on the selling of milk . he would like to mix water ( costing nothing ) with milk costing rs . 33 per litre so as to make a profit of 20 % on cost when he sells the resulting milk and water mixture for rs . 36 in what ratio should he mix the water and milk ? | cost needed to net a 20 % profit : ( 36 - x ) / x = . 2 x = 30 actual cost : 33 solution ( x = liters of water needed to be added to the 1 liter of milk ) : 33 / ( 1 + x ) = 30 x = 1 / 10 so to get the cost down to 30 milk : water 1 : 1 / 10 or ( 10 / 10 ) : ( 1 / 10 ) answer : b | a = 20 / 2
b = 1 / a
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a ) 2.04 % , b ) 5.36 % , c ) 4.26 % , d ) 6.26 % , e ) 7.26 % | a | multiply(subtract(inverse(divide(980, multiply(multiply(add(const_4, const_1), const_2), const_100))), const_1), const_100) | a dishonest shopkeeper professes to sell pulses at the cost price , but he uses a false weight of 980 gm . for a kg . his gain is β¦ % . | "his percentage gain is 100 * 20 / 980 as he is gaining 20 units for his purchase of 980 units . so 2.04 % . . answer : a" | a = 4 + 1
b = a * 2
c = b * 100
d = 980 / c
e = 1/(d)
f = e - 1
g = f * 100
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a ) 20 , b ) 120 , c ) 360 , d ) 2400 , e ) 2820 | d | divide(multiply(120, 400), 20) | if 20 % of a number = 400 , then 120 % of that number will be ? | "let the number x . then , 20 % of x = 400 x = ( 400 * 100 ) / 20 = 2000 120 % of x = ( 120 / 100 * 2000 ) = 2400 . answer : d" | a = 120 * 400
b = a / 20
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a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | a | divide(subtract(subtract(add(add(17, 18), 12), 4), 39), 2) | at a certain resort , each of the 39 food service employees is trained to work in a minimum of 1 restaurant and a maximum of 3 restaurants . the 3 restaurants are the family buffet , the dining room , and the snack bar . exactly 17 employees are trained to work in the family buffet , 18 are trained to work in the dining room , and 12 are trained to work in the snack bar . if 4 employees are trained to work in exactly 2 restaurants , how many employees are trained to work in all 3 restaurants ? | "39 = 17 + 18 + 12 - 4 - 2 x 2 x = 17 + 18 + 12 - 4 - 39 = 43 - 39 = 4 x = 2 a" | a = 17 + 18
b = a + 12
c = b - 4
d = c - 39
e = d / 2
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a ) 50 hrs , b ) 60 hrs , c ) 70 hrs , d ) 80 hrs , e ) 100 hrs | e | divide(const_1, subtract(divide(const_1, 20), divide(const_1, 25))) | a cistern is filled by pipe a in 20 hours and the full cistern can be leaked out by an exhaust pipe b in 25 hours . if both the pipes are opened , in what time the cistern is full ? | "time taken to full the cistern = ( 1 / 20 - 1 / 25 ) hrs = 1 / 100 = 100 hrs answer : e" | a = 1 / 20
b = 1 / 25
c = a - b
d = 1 / c
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a ) 4.37 % , b ) 5 % , c ) 3 % , d ) 8.75 % , e ) none | c | add(multiply(divide(subtract(divide(subtract(subtract(subtract(multiply(multiply(const_10, const_1000), const_10), const_1000), const_1000), multiply(add(2, const_3), const_100)), multiply(add(multiply(add(const_3, const_4), const_10), add(2, const_3)), const_1000)), 1), const_10), const_100), const_4) | the population of a town increased from 1 , 75,000 to 2 , 27,500 in a decade . the average percent increase of population per year is | "solution increase in 10 years = ( 227500 - 175000 ) = 52500 . increase % = ( 52500 / 175000 Γ£ β 100 ) % = 30 % . required average = ( 30 / 10 ) % = 3 % . answer c" | a = 10 * 1000
b = a * 10
c = b - 1000
d = c - 1000
e = 2 + 3
f = e * 100
g = d - f
h = 3 + 4
i = h * 10
j = 2 + 3
k = i + j
l = k * 1000
m = g / l
n = m - 1
o = n / 10
p = o * 100
q = p + 4
|
a ) - 1466 , b ) 2801 , c ) - 2801 , d ) - 2071 , e ) none of them | a | multiply(subtract(const_1, const_2), subtract(multiply(54, 29), 100)) | - 54 x 29 + 100 = ? | given exp . = - 54 x ( 30 - 1 ) + 100 = - ( 54 x 30 ) + 54 + 100 = - 1620 + 154 = - 1466 answer is a | a = 1 - 2
b = 54 * 29
c = b - 100
d = a * c
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a ) 60 , b ) 30 , c ) 15 , d ) 75 , e ) 100 | c | divide(subtract(multiply(divide(50, const_100), 100), 47), divide(20, const_100)) | if 50 % of 100 is greater than 20 % of a number by 47 , what is the number ? | "explanation : 50 / 100 * 100 - 20 / 100 * x = 47 50 - 20 / 100 * x = 47 3 = 20 / 100 * x 3 * 100 / 20 = x 15 = x answer : option c" | a = 50 / 100
b = a * 100
c = b - 47
d = 20 / 100
e = c / d
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a ) 2520 , b ) 1800 , c ) 2700 , d ) 10800 , e ) none of these | a | multiply(multiply(divide(270, 6), 4), 14) | running at the same constant rate , 6 identical machines can produce a total of 270 bottles per minute . at this rate , how many bottles could 14 such machines produce in 4 minutes ? | "solution let the required number of bottles be x . more machines , more bottles ( direct proportion ) more minutes , more bottles ( direct proportion ) Γ’ Λ Β΄ 6 Γ£ β 1 Γ£ β x = 14 Γ£ β 4 Γ£ β 270 Γ’ β‘ β x = 14 x 4 x 270 / 6 = 2520 . answer a" | a = 270 / 6
b = a * 4
c = b * 14
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a ) 37.5 , b ) 75 , c ) 100 , d ) 150 , e ) 90 | e | divide(subtract(multiply(12, const_2), 6), subtract(subtract(subtract(const_1, divide(20, const_100)), multiply(subtract(const_1, divide(20, const_100)), divide(const_1, const_4))), multiply(const_2, divide(20, const_100)))) | a tank holds x gallons of a saltwater solution that is 20 % salt by volume . one fourth of the water is evaporated , leaving all of the salt . when 6 gallons of water and 12 gallons of salt are added , the resulting mixture is 33 1 / 3 % salt by volume . what is the value of x ? | nope , 150 . i can only get it by following pr ' s backsolving explanation . i hate that . original mixture has 20 % salt and 80 % water . total = x out of which salt = 0.2 x and water = 0.8 x now , 1 / 4 water evaporates and all salt remains . so what remains is 0.2 x salt and 0.6 x water . now 12 gallons salt is added and 6 gallons of water is added . so salt now becomes - > ( 0.2 x + 12 ) and water - - > ( 0.6 x + 6 ) amount of salt is 33.33 % of total . so amount of water is 66.66 % . so salt is half of the volume of water . so ( 0.2 x + 12 ) = ( 0.6 x + 6 ) / 2 = > 0.4 x + 24 = 0.6 x + 6 - > 0.2 x = 18 solving , x = 90 answer : e | a = 12 * 2
b = a - 6
c = 20 / 100
d = 1 - c
e = 20 / 100
f = 1 - e
g = 1 / 4
h = f * g
i = d - h
j = 20 / 100
k = 2 * j
l = i - k
m = b / l
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a ) 325 , b ) 425 , c ) 625 , d ) 700 , e ) none of these | c | divide(250, divide(2, 5)) | there are 250 female managers in a certain company . find the total number of female employees in the company , if 2 / 5 of all the employees are managers and 2 / 5 of all male employees are managers . | as per question stem 2 / 5 m ( portion of men employees who are managers ) + 250 ( portion of female employees who are managers ) = 2 / 5 t ( portion of total number of employees who are managers ) , thus we get that 2 / 5 m + 250 = 2 / 5 t , or 2 / 5 ( t - m ) = 250 , from here we get that t - m = 625 , that would be total number of female employees and the answer ( c ) | a = 2 / 5
b = 250 / a
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a ) 6999 , b ) 7000 , c ) 7001 , d ) 7002 , e ) 7003 | b | subtract(15000, multiply(divide(8, 15), 15000)) | income and expenditure of a person are in the ratio 15 : 8 . if the income of the person is rs . 15000 , then find his savings ? | "let the income and the expenditure of the person be rs . 15 x and rs . 8 x respectively . income , 15 x = 15000 = > x = 1000 savings = income - expenditure = 15 x - 8 x = 7 x = 7 ( 1000 ) so , savings = rs . 7000 . answer : b" | a = 8 / 15
b = a * 15000
c = 15000 - b
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a ) 66 , b ) 74 , c ) 78 , d ) 98 , e ) none of these | d | subtract(multiply(78, 6), multiply(74, 5)) | ashok secured average of 78 marks in 6 subjects . if the average of marks in 5 subjects is 74 , how many marks did he secure in the 6 th subject ? | "explanation : number of subjects = 6 average of marks in 6 subjects = 78 therefore total marks in 6 subjects = 78 * 6 = 468 now , no . of subjects = 5 total marks in 5 subjects = 74 * 5 = 370 therefore marks in 6 th subject = 468 β 370 = 98 answer d" | a = 78 * 6
b = 74 * 5
c = a - b
|
a ) 16 km , b ) 10 km , c ) 12 km , d ) 24 km , e ) 25 km | d | multiply(3, 4) | a man performs 1 / 2 of the total journey by rail , 1 / 3 by bus and the remaining 4 km on foot . his total journey is | "explanation : let the journey be x km then , 1 x / 2 + 1 x / 3 + 4 = x 5 x + 24 = 6 x x = 24 km answer : option d" | a = 3 * 4
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a ) $ 3,000 , b ) $ 1,350 , c ) $ 1,725 , d ) $ 2,050 , e ) $ 2,250 | a | divide(multiply(divide(multiply(add(add(multiply(const_3, const_100), multiply(8, const_10)), const_4), const_1000), multiply(multiply(8, 10), 12)), 7.5), const_1000) | a hat company ships its hats , individually wrapped , in 8 - inch by 10 - inch by 12 - inch boxes . each hat is valued at $ 7.50 . if the company β s latest order required a truck with at least 384,000 cubic inches of storage space in which to ship the hats in their boxes , what was the minimum value of the order ? | number of boxes = total volume / volume of one box = 384,000 / ( 8 * 10 * 12 ) = 400 one box costs 7.50 , so 400 box will cost = 400 * 7.5 = 3000 a is the answer | a = 3 * 100
b = 8 * 10
c = a + b
d = c + 4
e = d * 1000
f = 8 * 10
g = f * 12
h = e / g
i = h * 7
j = i / 1000
|
a ) 12 , b ) 11 , c ) 13 , d ) 15 , e ) 16 | b | add(multiply(3, 2), 5) | if p / q = 3 / 5 , then 2 p + q = ? | "let p = 3 , q = 5 then 2 * 3 + 5 = 11 so 2 p + q = 11 . answer : b" | a = 3 * 2
b = a + 5
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a ) 5 , b ) 9 , c ) 25 , d ) 30 , e ) 75 | e | multiply(5, divide(100, add(5, 5))) | fred and sam are standing 100 miles apart and they start walking in a straight line toward each other at the same time . if fred walks at a constant speed of 5 miles per hour and sam walks at a constant speed of 5 miles per hour , how many miles has sam walked when they meet ? | "relative distance = 100 miles relative speed = 5 + 5 = 10 miles per hour time taken = 100 / 10 = 15 hours distance travelled by sam = 15 * 5 = 75 miles = e" | a = 5 + 5
b = 100 / a
c = 5 * b
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a ) 4 , b ) 8 , c ) 12 , d ) 16 , e ) 18 | b | multiply(subtract(const_3, 1), divide(36, subtract(add(multiply(const_10, 2), 1), add(const_10, 2)))) | the difference between a two digit number and the number obtained by interchanging the digits is 36 . what is the difference between the sum and the difference of the digits of the number if the ratio between the digits of the number is 1 : 2 ? | sol . since the number is greater than the number obtained on reversing the digits , so the ten ' s is greater than the unit ' s digit . let the ten ' s and units digit be 2 x and x respectively . then , ( 10 Γ 2 x + x ) - ( 10 x + 2 x ) = 36 β 9 x = 36 β x = 4 . β΄ required difference = ( 2 x + x ) - ( 2 x - x ) = 2 x = 8 . answer b | a = 3 - 1
b = 10 * 2
c = b + 1
d = 10 + 2
e = c - d
f = 36 / e
g = a * f
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a ) 23 , b ) 78 , c ) 27 , d ) 61 , e ) 41 | e | add(40, divide(multiply(5, 12), divide(180, 3))) | 40 + 5 * 12 / ( 180 / 3 ) = ? | "explanation : 40 + 5 * 12 / ( 180 / 3 ) = 40 + 5 * 12 / ( 60 ) = 40 + ( 5 * 12 ) / 60 = 40 + 1 = 41 . answer : e" | a = 5 * 12
b = 180 / 3
c = a / b
d = 40 + c
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a ) 743 , b ) 155 , c ) 852 , d ) 741 , e ) 785 | b | divide(subtract(13787, 14), 89) | on dividing 13787 by a certain number , we get 89 as quotient and 14 as remainder . what is the divisor ? | "divisor * quotient + remainder = dividend divisor = ( dividend ) - ( remainder ) / quotient ( 13787 - 14 ) / 89 = 155 answer ( b )" | a = 13787 - 14
b = a / 89
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a ) 20 liters , b ) 30 liters , c ) 50 liters , d ) 60 liters , e ) none of these | d | subtract(multiply(divide(multiply(60, 2), add(2, 1)), 2), subtract(60, divide(multiply(60, 2), add(2, 1)))) | in a mixture 60 litres , the ra Ι΅ o of milk and water 2 : 1 . if the this ra Ι΅ o is to be 1 : 2 , then the quanity of water to be further added is | explanation : quantity of milk = 60 * ( 2 / 3 ) = 40 liters quantity of water = 60 - 40 = 20 liters answer : d | a = 60 * 2
b = 2 + 1
c = a / b
d = c * 2
e = 60 * 2
f = 2 + 1
g = e / f
h = 60 - g
i = d - h
|
a ) 17 inches , b ) 20 inches , c ) 15 inches , d ) 18 inches , e ) 19 inches | b | divide(add(multiply(6, const_12), 8), 4) | a scale 6 ft . 8 inches long is divided into 4 equal parts . find the length of each part | "explanation : total length of scale in inches = ( 6 * 12 ) + 8 = 80 inches length of each of the 4 parts = 80 / 4 = 20 inches answer : b" | a = 6 * 12
b = a + 8
c = b / 4
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a ) 12 % , b ) 16 % , c ) 20.18 % , d ) 82 % , e ) 23 % | c | multiply(divide(subtract(64900, add(42000, 12000)), add(42000, 12000)), const_100) | ramu bought an old car for rs . 42000 . he spent rs . 12000 on repairs and sold it for rs . 64900 . what is his profit percent ? | "total cp = rs . 42000 + rs . 12000 = rs . 54000 and sp = rs . 64900 profit ( % ) = ( 64900 - 54000 ) / 54000 * 100 = 20.18 % answer : c" | a = 42000 + 12000
b = 64900 - a
c = 42000 + 12000
d = b / c
e = d * 100
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a ) 2 , b ) 4 , c ) 8 , d ) 49 , e ) 32 | d | add(subtract(79, add(add(add(15, 10), 5), 3)), 3) | there are 79 people that own pets . 15 people own only dogs , 10 people own only cats , 5 people own only cats and dogs , 3 people own cats , dogs and snakes . how many total snakes are there ? | lets assign variables to all the areas in venn diagram of three . three different units are dog , cat , snake = total = 79 only dog = d = 15 only cat = c = 10 only snake = s exactly dog and cat = 5 exactly dog and snake = x exactly cat and snake = y all three = 3 so 79 = 15 + 10 + 5 + 3 + x + y + s we need to know total snakes = x + y + s + 3 = 49 answer : d | a = 15 + 10
b = a + 5
c = b + 3
d = 79 - c
e = d + 3
|
a ) 240 , b ) 280 , c ) 320 , d ) 360 , e ) 400 | d | divide(multiply(divide(multiply(900, const_2), const_3), const_3), add(const_2, const_3)) | a man traveled a total distance of 900 km . he traveled one - third of the whole trip by plane and the distance traveled by train is two - thirds of the distance traveled by bus . if he traveled by train , plane and bus , how many kilometers did he travel by bus ? | "total distance traveled = 900 km . distance traveled by plane = 300 km . distance traveled by bus = x distance traveled by train = 2 x / 3 x + 2 x / 3 + 300 = 900 5 x / 3 = 600 x = 360 km the answer is d ." | a = 900 * 2
b = a / 3
c = b * 3
d = 2 + 3
e = c / d
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a ) 52 : 56 , b ) 52 : 53 , c ) 52 : 50 , d ) 22 : 56 , e ) 51 : 53 | e | divide(add(const_100, 2), add(const_100, 6)) | the cash difference between the selling prices of an article at a profit of 2 % and 6 % is rs . 3 . the ratio of the two selling prices is ? | "let c . p . of the article be rs . x . then , required ratio = 102 % of x / 106 % of x = 102 / 106 = 51 / 53 = 51 : 53 answer : e" | a = 100 + 2
b = 100 + 6
c = a / b
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a ) 15 , b ) 50 , c ) 55 , d ) 35 , e ) 65 | a | add(subtract(80, 60), subtract(50, 35)) | in an it company , there are a total of 60 employees including 50 programmers . the number of male employees is 80 , including 35 male programmers . how many employees must be selected to guaranty that we have 3 programmers of the same sex ? | "you could pick 10 non - programmers , 2 male programmers , and 2 female programmers , and still not have 3 programmers of the same sex . but if you pick one more person , you must either pick a male or a female programmer , so the answer is 15 . a" | a = 80 - 60
b = 50 - 35
c = a + b
|
a ) 9.99 , b ) 3,615 , c ) 3.62 , d ) 4.62 , e ) 6.14 | d | inverse(subtract(divide(subtract(const_1, multiply(inverse(20), const_2)), subtract(subtract(multiply(1, const_2), 6), const_2)), inverse(20))) | machine a takes 20 hours to complete a certain job and starts that job at 6 am . after four hour of working alone , machine a is joined by machine b and together they complete the job at 1 pm . how long would it have taken machine b to complete the job if it had worked alone for the entire job ? | "let us assume total job = 100 units a finishes 100 units in 20 hrs ( given ) hence a ( working rate ) = 5 units / hr now given that a works for 4 hr ( so 20 units done ) then a and b finish total work in 7 hours . hence a and b finish 80 units in 3 hours . of these 3 x 5 = 15 units were done by a . hence b did 65 units in 3 hours . hence b ( working rate ) = 65 / 3 units / hr hence b takes 100 x 3 / 65 = 4.62 hours to complete the job . answer d ." | a = 1/(20)
b = a * 2
c = 1 - b
d = 1 * 2
e = d - 6
f = e - 2
g = c / f
h = 1/(20)
i = g - h
j = 1/(i)
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a ) 7.5 hr , b ) 8 hr , c ) 8.57 hr , d ) 10 hr , e ) none of these | c | inverse(subtract(add(divide(const_1, 10), divide(const_1, 12)), divide(const_1, 15))) | two pipes can fill the cistern in 10 hr and 12 hr respectively , while the third empty it in 15 hr . if all pipes are opened simultaneously , then the cistern will be filled in | "solution : work done by all the tanks working together in 1 hour . 1 / 10 + 1 / 12 β 1 / 15 = 7 / 60 hence , tank will be filled in 60 / 7 = 8.57 hour option ( c )" | a = 1 / 10
b = 1 / 12
c = a + b
d = 1 / 15
e = c - d
f = 1/(e)
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a ) 100 , b ) 125 , c ) 150 , d ) 175 , e ) 200 | b | divide(multiply(100, add(const_4, const_1)), const_2) | to fill a tank , 100 buckets of water is required . how many buckets of water will be required to fill the same tank if the capacity of the bucket is reduced to four - fifth of its present ? | "let the capacity of 1 bucket = x . then , the capacity of tank = 100 x . new capacity of bucket = 4 / 5 x therefore , required number of buckets = ( 100 x ) / ( 4 x / 5 ) = ( 100 x ) x 5 / 4 x = 500 / 4 = 125 answer is b ." | a = 4 + 1
b = 100 * a
c = b / 2
|
a ) 3 sec , b ) 4 sec , c ) 5 sec , d ) 6 sec , e ) 7 sec | e | divide(65, multiply(32, const_0_2778)) | in what time will a railway train 65 m long moving at the rate of 32 kmph pass a telegraph post on its way ? | "t = 65 / 32 * 18 / 5 = 7 sec answer : e" | a = 32 * const_0_2778
b = 65 / a
|
a ) s . 1518.8 , b ) s . 1327 , c ) s . 1328 , d ) s . 1397 , e ) s . 1927 | a | multiply(subtract(rectangle_area(add(75, multiply(2.8, 2)), add(55, multiply(2.8, 2))), rectangle_area(75, 55)), 2) | a rectangular grass field is 75 m * 55 m , it has a path of 2.8 m wide all round it on the outside . find the area of the path and the cost of constructing it at rs . 2 per sq m ? | "area = ( l + b + 2 d ) 2 d = ( 75 + 55 + 2.8 * 2 ) 2 * 2.8 = > 759.4 759.4 * 2 = rs . 1518.8 answer : a" | a = 2 * 8
b = 75 + a
c = 2 * 8
d = 55 + c
e = rectangle_area - (
f = e * rectangle_area
|
a ) 125 / 990 , b ) 125 / 999 , c ) 125 / 900 , d ) 12 / 999 , e ) 12 / 990 | b | divide(floor(multiply(0.125125, const_1000)), multiply(multiply(add(multiply(add(const_1, const_10), const_10), const_1), const_3), const_3)) | the rational number for recurring decimal 0.125125 . . . . is : | 0.125125 . . . = 0.125 = 125 / 999 answer is b . | a = 0 * 125125
b = math.floor(a)
c = 1 + 10
d = c * 10
e = d + 1
f = e * 3
g = f * 3
h = b / g
|
a ) 3 and 7 , b ) 14 only , c ) 7 only , d ) 7 and 14 , e ) none | d | add(multiply(7, const_100), multiply(2, 7)) | if n is a natural number , then ( 7 ( n 2 ) + 7 n ) is always divisible by : | "solution ( 7 n 2 + 7 n ) = 7 n ( n + 1 ) , which is always divisible by 7 and 14 both , since n ( n + 1 ) is always even . answer d" | a = 7 * 100
b = 2 * 7
c = a + b
|
a ) 8 , b ) 11 , c ) 12 , d ) 15 , e ) 16 | b | divide(divide(divide(360, 4), const_2), const_3) | how many of the positive divisors of 360 are also multiples of 4 not including 360 ? | "360 = 2 ^ 5 * 3 * 5 = ( 4 ) * 2 * 3 ^ 2 * 5 besides ( 4 ) , the exponents of 2 , 3 , and 5 are 1 , 2 , and 1 . there are ( 1 + 1 ) ( 2 + 1 ) ( 1 + 1 ) = 12 ways to make multiples of 4 . we must subtract 1 because one of these multiples is 360 . the answer is b ." | a = 360 / 4
b = a / 2
c = b / 3
|
a ) 40 % , b ) 55.4 % , c ) 57 % , d ) 60 % , e ) 62 % | b | multiply(divide(11628, add(add(1136, 8236), 11628)), const_100) | 3 candidates in an election and received 1136 , 8236 and 11628 votes respectively . what % of the total votes did the winning candidate gotin that election ? | "total number of votes polled = ( 1136 + 8236 + 11628 ) = 21000 so , required percentage = 11628 / 21000 * 100 = 55.4 % b" | a = 1136 + 8236
b = a + 11628
c = 11628 / b
d = c * 100
|
a ) 10 days , b ) 30 days , c ) 20 days , d ) 60 days , e ) 40 days | d | inverse(add(divide(9, multiply(12, 81)), divide(12, multiply(20, 81)))) | if 12 men or 20 women can do a piece of work in 81 days , then in how many days can 9 men and 12 women together do the work ? | "d 60 days given that 12 m = 20 w = > 3 m = 5 w 9 men + 12 women = 15 women + 12 women = 27 women 20 women can do the work in 81 days . so , 27 women can do it in ( 20 * 81 ) / 27 = 60 days ." | a = 12 * 81
b = 9 / a
c = 20 * 81
d = 12 / c
e = b + d
f = 1/(e)
|
a ) 32.25 % , b ) 31.00 % , c ) 30.25 % , d ) 30.00 % , e ) 22.50 % | a | multiply(const_100, subtract(multiply(add(const_1, divide(15, const_100)), add(const_1, divide(15, const_100))), const_1)) | increasing the original price of an article by 15 percent and then increasing the new price by 15 percent is equivalent to increasing the original price by | "you can do this by approximation , or with straight math by calculating 100 * 1.15 ^ 2 . or step by step : if you increase 100 by 15 % you ' ll get 115 , then if you increase 115 = 100 + 15 by 15 % again , 100 will gain 15 again and 15 will gain its 15 % which is 2.25 , so total gain is 15 + 2.25 = 17.25 - - > 115 + 17.25 = 132.25 . answer : a ." | a = 15 / 100
b = 1 + a
c = 15 / 100
d = 1 + c
e = b * d
f = e - 1
g = 100 * f
|
a ) $ 0 , b ) $ 3 , c ) $ 4 , d ) $ 12 , e ) $ 16 | e | subtract(multiply(divide(48, subtract(const_1, divide(40, const_100))), subtract(const_1, divide(20, const_100))), 48) | a merchant purchased a jacket for $ 48 and then determined a selling price that equalled the purchase price of the jacket plus a markup that was 40 percent of the selling price . during a sale , the merchant discounted the selling price by 20 percent and sold the jacket . what was the merchant β s gross profit on this sale ? | "actual cost = $ 48 sp = actual cost + mark up = actual cost + 40 % sp = 48 * 100 / 60 on sale sp = 80 / 100 ( 48 * 100 / 60 ) = 64 gross profit = $ 16 answer is e" | a = 40 / 100
b = 1 - a
c = 48 / b
d = 20 / 100
e = 1 - d
f = c * e
g = f - 48
|
a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 16 | b | divide(add(divide(96, 16), sqrt(add(multiply(multiply(divide(divide(96, 16), add(const_1, divide(50, const_100))), 4), const_4), power(divide(96, 16), const_2)))), const_2) | pascal has 96 miles remaining to complete his cycling trip . if he reduced his current speed by 4 miles per hour , the remainder of the trip would take him 16 hours longer than it would if he increased his speed by 50 % . what is his current speed o ? | let the current speed be x miles per hour . time taken if speed is 50 % faster ( i . e . 3 x / 2 = 1.5 x ) = 96 / 1.5 x time taken if speed is reduced by 4 miles / hr ( i . e . ( x - 4 ) ) = 96 / ( x - 4 ) as per question , 96 / ( x - 4 ) - 96 / 1.5 x = 16 solving this o we get x = 8 . b . | a = 96 / 16
b = 96 / 16
c = 50 / 100
d = 1 + c
e = b / d
f = e * 4
g = f * 4
h = 96 / 16
i = h ** 2
j = g + i
k = math.sqrt(j)
l = a + k
m = l / 2
|
a ) 45 , b ) 50 , c ) 55 , d ) 60 , e ) 90 | e | multiply(divide(multiply(divide(1, 12), const_3), divide(divide(1, 30), const_2)), 6) | in the standard formulation of a flavored drink the ratio by volume of flavoring to corn syrup to water is 1 : 12 : 30 . in the sport formulation , the ratio of flavoring to corn syrup is three times as great as in the standard formulation , and the ratio of flavoring to water is half that of the standard formulation . if a large bottle of the sport formulation contains 6 ounces of corn syrup , how many ounces of water does it contain ? | "f : c : w 1 : 12 : 30 sport version : f : c 3 : 12 f : w 1 : 60 or 3 : 180 so c : f : w = 12 : 3 : 180 c / w = 12 / 180 = 3 ounces / x ounces x = 6 * 180 / 12 = 90 ounces of water e" | a = 1 / 12
b = a * 3
c = 1 / 30
d = c / 2
e = b / d
f = e * 6
|
a ) 2 / 3 , b ) 3 / 5 , c ) 3 / 4 , d ) 3 / 2 , e ) 5 / 2 | d | divide(multiply(subtract(multiply(const_6, const_3), const_1), const_2), 100) | the probability that a number selected at random from the first 100 natural numbers is a composite number is ? | "explanation : the number of exhaustive events = 100 c β = 100 . we have 25 primes from 1 to 100 . number of favourable cases are 75 . required probability = 75 / 50 = 3 / 2 . answer is d" | a = 6 * 3
b = a - 1
c = b * 2
d = c / 100
|
['a ) 50', 'b ) 80', 'c ) 100', 'd ) 143.36', 'e ) 250'] | d | multiply(divide(subtract(power(add(divide(multiply(60, const_2), const_1000), const_3), const_2), const_4), const_4), const_100) | a circular logo is enlarged to fit the lid of a jar . the new diameter is 60 per cent larger than the original . by what percentage has the area of the logo increased ? | let old diameter be 4 , so radius is 2 old area = 4 Ο new diameter is 6.24 , so radius is 3.12 new area = 9.7344 Ο increase in area is 5.7344 Ο % increase in area = 5.7344 / 4 * 100 so , % increase is 143.36 % answer will be ( d ) | a = 60 * 2
b = a / 1000
c = b + 3
d = c ** 2
e = d - 4
f = e / 4
g = f * 100
|
a ) rs 840 , b ) rs 1350 , c ) rs 1620 , d ) rs 1680 , e ) none of these | b | multiply(divide(multiply(9, 6), multiply(4, 10)), 1000) | if 4 men working 10 hours a day earn rs . 1000 per week , then 9 men working 6 hours a day will earn how much per week ? | explanation : ( men 4 : 9 ) : ( hrs / day 10 : 6 ) : : 1000 : x hence 4 * 10 * x = 9 * 6 * 1000 or x = 9 * 6 * 1000 / 4 * 10 = 1350 answer : b | a = 9 * 6
b = 4 * 10
c = a / b
d = c * 1000
|
a ) 30 , 10 , b ) 25 , 5 , c ) 29 , 9 , d ) 35 , 15 , e ) 20,10 | d | add(divide(add(multiply(5, 10), subtract(20, 10)), subtract(5, const_1)), 20) | the ages of two person differ by 20 years . if 10 years ago , the elder one be 5 times as old as the younger one , their present ages ( in years ) are respectively | "let their ages be x and ( x + 20 ) years . then , 5 ( x - 10 ) = ( x + 20 - 10 ) = > 4 x = 60 = > x = 15 their present ages are 35 years and 15 year . answer : d" | a = 5 * 10
b = 20 - 10
c = a + b
d = 5 - 1
e = c / d
f = e + 20
|
a ) $ 35,000 , b ) $ 40,000 , c ) $ 55,000 , d ) $ 65,000 , e ) $ 75,000 | b | divide(add(multiply(multiply(multiply(const_3, const_3), const_10), multiply(const_100, const_10)), multiply(multiply(const_3, const_10), multiply(const_100, const_10))), add(75, 15)) | a company has 15 managers and 75 associates . the 15 managers have an average salary of $ 90,000 . the 75 associates have an average salary of $ 30,000 . what is the average salary for the company ? | another method is to get ratios say 30000 = a and we know the # of people are in 1 : 5 ratio average = ( 3 a * 1 + a * 5 ) / 6 = 8 a / 6 = 40000 answer is b . $ 40,000 | a = 3 * 3
b = a * 10
c = 100 * 10
d = b * c
e = 3 * 10
f = 100 * 10
g = e * f
h = d + g
i = 75 + 15
j = h / i
|
a ) 358000 , b ) 368000 , c ) 357000 , d ) 358000 , e ) 367000 | c | multiply(multiply(560000, subtract(const_1, divide(15, const_100))), divide(75, const_100)) | in an election , candidate a got 75 % of the total valid votes . if 15 % of the total votes were declared invalid and the total numbers of votes is 560000 , find the number of valid vote polled in favour of candidate ? | "total number of invalid votes = 15 % of 560000 = 15 / 100 Γ 560000 = 8400000 / 100 = 84000 total number of valid votes 560000 β 84000 = 476000 percentage of votes polled in favour of candidate a = 75 % therefore , the number of valid votes polled in favour of candidate a = 75 % of 476000 = 75 / 100 Γ 476000 = 35700000 / 100 = 357000 answer : c" | a = 15 / 100
b = 1 - a
c = 560000 * b
d = 75 / 100
e = c * d
|
a ) - 7 , b ) - 4 , c ) - 3 , d ) 1 , e ) - 10 | e | divide(subtract(10, add(power(7, 2), multiply(3, 7))), 7) | if 7 is one solution of the equation x ^ 2 + 3 x + k = 10 , where k is a constant , what is the other solution ? | "the phrase β 7 is one solution of the equation β means that one value of x is 7 . thus , we first must plug 7 for x into the given equation to determine the value of k . so we have 7 ^ 2 + ( 3 ) ( 7 ) + k = 10 49 + 21 + k = 10 70 + k = 10 k = - 60 next we plug - 60 into the given equation for k and then solve for x . x ^ 2 + 3 x β 60 = 10 x ^ 2 + 3 x β 70 = 0 ( x + 10 ) ( x - 7 ) = 0 x = - 10 or x = 7 thus , - 10 is the other solution . answer e ." | a = 7 ** 2
b = 3 * 7
c = a + b
d = 10 - c
e = d / 7
|
a ) 121 , b ) 3267 , c ) 363 , d ) 33 , e ) none of the above | c | lcm(power(add(const_10, const_1), const_2), divide(subtract(33, const_3), const_10)) | if both 112 and 33 are factors of the number a * 43 * 62 * 1311 , then what is the smallest possible value of a ? | explanatory answer step 1 : prime factorize the given expression a * 43 * 62 * 1311 can be expressed in terms of its prime factors as a * 28 * 32 * 1311 step 2 : find factors missing after excluding ' a ' to make the number divisible by both 112 and 33 112 is a factor of the given number . if we do not include ' a ' , 11 is not a prime factor of the given number . if 112 is a factor of the number , 112 should have been in ' a ' 33 is a factor of the given number . if we do not include ' a ' , the number has only 32 in it . therefore , if 33 has to be a factor of the given number ' a ' has to contain 31 in it . therefore , ' a ' should be at least 112 * 3 = 363 if the given number has 112 and 33 as its factors . the smallest value that ` ` a ' ' can take is 363 . choice c is the correct answer . | a = 10 + 1
b = a ** 2
c = 33 - 3
d = c / 10
e = math.lcm(b, d)
|
a ) 4 / 5 , b ) 5 / 4 , c ) 3 / 2 , d ) 5 / 7 , e ) 3 / 5 | e | divide(sqrt(9), sqrt(25)) | two isosceles triangles have equal vertical angles and their areas are in the ratio 9 : 25 . find the ratio of their corresponding heights . | "we are basically given that the triangles are similar . in two similar triangles , the ratio of their areas is the square of the ratio of their sides and also , the square of the ratio of their corresponding heights . therefore , area / area = height ^ 2 / height ^ 2 = 9 / 25 - - > height / height = 3 / 5 . answer : e ." | a = math.sqrt(9)
b = math.sqrt(25)
c = a / b
|
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