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 |
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a ) a . 10 , b ) b . 12 , c ) c . 14 , d ) d . 18 , e ) e . 24 | d | add(add(add(add(add(add(add(add(add(add(add(add(const_1, add(const_3, const_2)), const_1), const_1), const_1), const_1), const_2), const_1), const_1), const_1), const_1), const_1), const_1) | working at constant rate , pump x pumped out half of the water in a flooded basement in 2 hours . the pump y was started and the two pumps , working independently at their respective constant rates , pumped out rest of the water in 3 hours . how many hours would it have taken pump y , operating alone at its own constant rate , to pump out all of the water that was pumped out of the basement ? | rate of x = 1 / 8 rate of x + y = 1 / 6 rate of y = 1 / 6 - 1 / 8 = 1 / 24 18 hours d | a = 3 + 2
b = 1 + a
c = b + 1
d = c + 1
e = d + 1
f = e + 1
g = f + 2
h = g + 1
i = h + 1
j = i + 1
k = j + 1
l = k + 1
m = l + 1
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a ) 100 , b ) 120 , c ) 130 , d ) 140 , e ) 150 | b | divide(6, subtract(1, add(divide(1, 5), divide(3, 4)))) | a cement mixture is composed of 3 elements . by weight , 1 / 5 of the mixture is sand , 3 / 4 of the mixture is water , and the remaining 6 pounds of the mixture is gravel . what is the weight of the entire mixture in pounds ? | "let the total weight be x . sand content = ( 1 / 5 ) x water content = ( 3 / 4 ) x gravel = x - ( 1 / 5 ) x - ( 3 / 4 ) x = ( 1 / 20 ) x = 6 x = 120 then answer will be b = 120" | a = 1 / 5
b = 3 / 4
c = a + b
d = 1 - c
e = 6 / d
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a ) 8 / 5 , b ) 5 / 2 , c ) 3 , d ) 11 / 3 , e ) 22 | e | divide(multiply(subtract(const_12, const_1), subtract(inverse(subtract(const_12, const_3)), inverse(const_10))), subtract(inverse(add(const_1, const_4)), inverse(subtract(const_12, const_3)))) | for each month of a given year except december , a worker earned the same monthly salary and donated one - tenth of that salary to charity . in december , the worker earned n times his usual monthly salary and donated one - fifth of his earnings to charity . if the worker ' s charitable contributions totaled one - nineth of his earnings for the entire year , what is the value of n ? | let monthly salary for each of the 11 months except december was x , then 11 x * 1 / 10 + nx * 1 / 5 = 1 / 9 ( 11 x + nx ) ; 11 / 10 + n / 5 = 1 / 9 ( 11 + n ) = > 11 + 2 n / 10 = 11 + n / 9 = > 66 + 12 n = 110 + 10 n = > 2 n = 44 n = 22 answer : e | a = 12 - 1
b = 12 - 3
c = 1/(b)
d = 1/(10)
e = c - d
f = a * e
g = 1 + 4
h = 1/(g)
i = 12 - 3
j = 1/(i)
k = h - j
l = f / k
|
a ) 5 hrs , b ) 16.6 hrs , c ) 15.6 hrs , d ) 20.4 hrs , e ) 30 hrs | b | divide(50, 3) | ajay can walk 3 km in 1 hour . in how many hours he can walk 50 km ? | "1 hour he walk 3 km he walk 50 km in = 50 / 3 * 1 = 16.6 hours answer is b" | a = 50 / 3
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a ) 8 % , b ) 1 % , c ) 20 % , d ) 80 % , e ) none | c | multiply(subtract(const_1, divide(divide(4, const_100), divide(5, const_100))), const_100) | the income of a broker remains unchanged though the rate of commission is increased from 4 % to 5 % . the percentage of slump in business is ? | answer let the business value change from a to b . then , 4 % of a = 5 % of b β 4 a / 100 = 5 b / 100 β b = 4 a / 5 β΄ change in business = ( a - 4 a / 5 ) = a / 5 percentage slump in business = { ( a / 5 ) / a } x 100 % = 20 % correct option : c | a = 4 / 100
b = 5 / 100
c = a / b
d = 1 - c
e = d * 100
|
a ) 96 , b ) 105.6 , c ) 86 , d ) 74 , e ) 110 | b | divide(44, 240) | find 44 % of 240 | "we know that r % of m is equal to r / 100 Γ m . so , we have 44 % of 240 44 / 100 Γ 240 = 105.6 answer : b" | a = 44 / 240
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a ) 456 , b ) 837 , c ) 912 , d ) 1200 , e ) 1400 | b | divide(multiply(divide(324, divide(subtract(62, subtract(const_100, 62)), const_100)), 62), const_100) | there were two candidates in an election . winner candidate received 62 % of votes and won the election by 324 votes . find the number of votes casted to the winning candidate ? | "w = 62 % l = 38 % 62 % - 38 % = 24 % 24 % - - - - - - - - 324 62 % - - - - - - - - ? = > 837 answer : b" | a = 100 - 62
b = 62 - a
c = b / 100
d = 324 / c
e = d * 62
f = e / 100
|
a ) none , b ) three , c ) four , d ) five , e ) seven | b | subtract(const_4, const_1) | r is the set of positive even integers less than 51 , and s is the set of the squares of the integers in r . how many elements does the intersection of r and s contain ? | "r is the set of positive even integers less than 51 , and s is the set of the squares of the integers in r . how many elements does the intersection of r and s contain ? r = 2,4 , 6,8 , 10,12 . . . s = 4,16 , 36,64 . . . numbers : 4 , 16 , and 36 are even integers ( less than 51 ) that are in both sets . solution : three answer : b" | a = 4 - 1
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a ) 2 / 9 , b ) 5 / 7 , c ) 1 / 7 , d ) 1 / 9 , e ) 3 / 7 | d | divide(const_1, 9) | a die is rolled twice . what is the probability of getting a sum equal to 9 ? | total number of outcomes possible when a die is rolled = 6 ( β΅ any one face out of the 6 faces ) hence , total number of outcomes possible when a die is rolled twice , n ( s ) = 6 Γ 6 = 36 e = getting a sum of 9 when the two dice fall = { ( 3 , 6 ) , { 4 , 5 } , { 5 , 4 } , ( 6 , 3 ) } hence , n ( e ) = 4 p ( e ) = n ( e ) / n ( s ) = 4 / 36 = 1 / 9 answer : option d | a = 1 / 9
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a ) 1 / 16 , b ) 1 / 6 , c ) 1 / 5 , d ) 5 , e ) 6 | a | divide(divide(2, 8), 8) | if xy > 0 , 1 / x + 1 / y = 2 , and 1 / xy = 8 , then ( x + y ) / 4 = ? | "( 1 / x + 1 / y ) = 2 canbe solved as { ( x + y ) / xy } = 8 . substituting for 1 / xy = 8 , we get x + y = 2 / 8 = = > ( x + y ) / 4 = 2 / ( 8 * 4 ) = 1 / 16 . a" | a = 2 / 8
b = a / 8
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a ) 40 / 3 , b ) 40 , c ) 120 , d ) 360 , e ) 420 | d | add(add(40, multiply(3, const_100)), multiply(2, const_10)) | the difference in compound interest earned on a deposit ( compounded annually ) in year 1 and year 2 is $ 40 . had the interest rate been 3 times its present value , the difference a would have been how much ? | case 1 : deposit = $ x ; rate of increase = r . interest yearned in 1 year = xr . deposit in 1 year = x + xr . interest yearned in 2 year = ( x + xr ) r . the difference a = ( x + xr ) r - xr = xr ^ 2 = 40 . case 2 : deposit = $ x ; rate of increase = 3 r . interest yearned in 1 year = x ( 3 r ) . deposit in 1 year = x + 3 xr . interest yearned in 2 year = ( x + 3 xr ) 3 r . the difference = ( x + 3 xr ) 3 r - 3 xr = 9 xr ^ 2 . since from case 1 we know that xr ^ 2 = 40 , then 9 xr ^ 2 = 9 * 40 = 360 . answer : d . | a = 3 * 100
b = 40 + a
c = 2 * 10
d = b + c
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a ) 1320 , b ) 1196 , c ) 1210 , d ) 1440 , e ) 1050 | d | multiply(divide(3120, add(add(2, 3), 4)), 4) | an amount of rs . 3120 was divided among a , b and c , in the ratio 1 / 2 : 1 / 3 : 1 / 4 . find the share of a ? | "let the shares of a , b and c be a , b and c respectively . a : b : c = 1 / 2 : 1 / 3 : 1 / 4 let us express each term with a common denominator which is the last number divisible by the denominators of each term i . e . , 12 . a : b : c = 6 / 12 : 4 / 12 : 3 / 12 = 6 : 4 : 3 . share of a = 6 / 13 * 3120 = rs . 1440 answer : d" | a = 2 + 3
b = a + 4
c = 3120 / b
d = c * 4
|
a ) 74 , b ) k = 75 , c ) k = 175 , d ) k = 680 , e ) 690 | b | add(multiply(34, const_2), divide(subtract(multiply(34, const_2), multiply(8, 5)), subtract(5, const_1))) | a number when divided by 5 gives a number which is 8 more than the remainder obtained on dividing the same number by 34 . such a least possible number k is | "i solved this question by plugging in numbers from the answer choices . a . ) 74 starting with answer choice a , i immediately eliminated it because 74 is not even divisible by 5 . b . ) 75 i divide 75 / 5 and get 15 as an answer . i divide 75 / 34 and get a remainder of 7 . 15 - 7 = 8 so i know the correct answer isb" | a = 34 * 2
b = 34 * 2
c = 8 * 5
d = b - c
e = 5 - 1
f = d / e
g = a + f
|
['a ) 5.25 m', 'b ) 7 m', 'c ) 10.5 m', 'd ) 21 m', 'e ) 22 m'] | c | divide(66, multiply(const_2, const_pi)) | a circular road runs round a circular ground . if the difference between the circumferences of the outer circle and inner circle is 66 metres , the width of the road is : | 2 Ο ( r - r ) = 60 = > 2 * 22 / 7 * ( r - r ) = 60 . therefore , ( r - r ) = ( 66 * 7 / 44 ) = 10.5 m answer : c | a = 2 * math.pi
b = 66 / a
|
a ) 238 , b ) 228 , c ) 208 , d ) 277 , e ) 101 | b | divide(12, 19) | find 12 Γ Γ 19 | "mentally imagine this number as ( 10 + 2 ) Γ Γ 19 = 190 + 38 = 228 . answer : b" | a = 12 / 19
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a ) 10,300 , b ) 10,030 , c ) 1,353 , d ) 1,352 , e ) 1,239 | e | subtract(458,600, add(add(multiply(const_2, const_100), multiply(add(const_3, const_4), const_10)), const_2)) | how many integers between 334,700 and 458,600 have tens digit 1 and units digit 3 ? | "there is one number in hundred with 1 in the tens digit and 3 in the units digit : 13 , 113 , 213 , 313 , . . . the difference between 334,700 and 458,600 is 458,600 - 334,700 = 123,900 - one number per each hundred gives 123,900 / 100 = 1,239 numbers . answer : e ." | a = 2 * 100
b = 3 + 4
c = b * 10
d = a + c
e = d + 2
f = 458 - 600
|
a ) 4 , b ) 5 , c ) 8 , d ) 6 , e ) 9 | d | multiply(multiply(3, 2), const_1) | if number divisible by both 3 and 2 , then the same number divisible by which number ? | obviously , the answer is 6 option d | a = 3 * 2
b = a * 1
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['a ) 1 : 4', 'b ) 1 : 3', 'c ) 1 : 2', 'd ) 1 : 5', 'e ) none of these'] | a | divide(1, power(2, 2)) | the radii of two cones are in ratio 2 : 1 , their volumes are equal . find the ratio of their heights . | explanation : let their radii be 2 x , x and their heights be h and h resp . then , volume of cone = 1 / 3 Ο r 2 h 1 / 3 β Ο β ( 2 x ) 2 β h / 1 / 3 β Ο β x 2 β h = > hh = 14 = > h : h = 1 : 4 option a | a = 2 ** 2
b = 1 / a
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a ) 11.5 , b ) 25.5 , c ) 22 , d ) 25 , e ) 27 | b | add(50, const_1) | the average of first 50 non - zero positive integers is | "explanation : sum of first n non - zero positive integers n ( n + 1 ) / 2 so , average of first n non - zero positive integers n ( n + 1 ) / 2 n = ( n + 1 ) / 2 = > ( 50 + 1 ) / 2 = 25.5 answer : b" | a = 50 + 1
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a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 20 | b | subtract(const_60, multiply(const_60, divide(45, 54))) | excluding stoppages , the speed of the bus is 54 kms / hr and with stoppages , it is 45 kmph . for how many minutes does the bus stop per hour ? | by including stoppages , bus will cover 9 km less . . ( 54 km - 45 km = 9 km ) excluding stoppages , x = 54 kmph including stoppages , y = 45 kmph difference , z = 9 km time taken to cover that 9 km ( bus stops per hour ( 60 minutes ) ) = z / x ( 60 minutes ) = 9 / 54 ( 60 ) = 1 / 6 ( 60 ) = 10 minutes . answer : b | a = 45 / 54
b = const_60 * a
c = const_60 - b
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a ) 560 , b ) 561 , c ) 562 , d ) 563 , e ) 673 | e | subtract(multiply(8, 526), multiply(subtract(8, 1), 505)) | sanoop bought 8 t - shirts at an average price ( arithmetic mean ) of rs . 526 . if sanoop returned 1 t - shirts to the retailer , and the average price of the remaining t - shirts was rs . 505 , then what is the average price , of the 3 returned t - shirts ? | total price of 8 t - shirts = 8 * 526 = 4208 total price of 7 t - shirts = 7 * 505 = 3535 total price of 1 t - shirts = 4208 - 3535 = 673 average price of 3 t - shirts = 673 correct option answer : e | a = 8 * 526
b = 8 - 1
c = b * 505
d = a - c
|
['a ) 56', 'b ) 80', 'c ) 100', 'd ) 120', 'e ) 144'] | e | multiply(divide(factorial(6), factorial(5)), factorial(subtract(5, const_1))) | in how many q ways can 5 people from a group of 6 people be seated around a circular table | q = 6 c 5 * ( 5 - 1 ) ! ( select 5 out of 6 and arrange them in circular manner ) = 6 * 4 ! = 6 * 24 = 144 answer - e | a = math.factorial(6)
b = math.factorial(5)
c = a / b
d = 5 - 1
e = math.factorial(d)
f = c * e
|
a ) 8 , b ) 10 , c ) 9 , d ) 14 , e ) 16 | c | divide(subtract(72, multiply(const_3, 6)), multiply(const_3, const_2)) | a number is doubled and 6 is added . if resultant is trebled , it becomes 72 . what is that number | "explanation : = > 3 ( 2 x + 6 ) = 72 = > 2 x + 6 = 24 = > x = 9 option c" | a = 3 * 6
b = 72 - a
c = 3 * 2
d = b / c
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a ) 0.1 % , b ) 1 % , c ) 7 % , d ) 10 % , e ) noneof these | b | divide(multiply(divide(70, const_100), const_100), 70) | i gain 70 paise on rs . 70 . my gain percent is : | solution 11.1 gain % = ( 0.70 / 70 x 100 ) % = 1 % . answer b | a = 70 / 100
b = a * 100
c = b / 70
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a ) 420 , b ) 700 , c ) 220 , d ) 500 , e ) none of these | d | add(400, multiply(400, divide(20, const_100))) | a fruit seller had some oranges . he sells 20 % oranges and still has 400 oranges . how many oranges he had originally ? | "explanation : he sells 20 % of oranges and still there are 400 oranges remaining = > 80 % of oranges = 400 β ( 80 Γ total oranges ) / 100 = 400 β total oranges / 100 = 5 β total oranges = 5 Γ 100 = 500 answer : option d" | a = 20 / 100
b = 400 * a
c = 400 + b
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a ) 2 / 125 , b ) 12 / 25 , c ) c ) 2 / 25 , d ) 3 / 25 , e ) 1 / 5 | b | divide(subtract(divide(multiply(800, 3), 5), 96), 800) | in a group of 800 people , 3 / 5 play at least one instrument , 96 play two or more . what is the probability that one student play exactly one instrument ? | p ( playing 2 or more instruments ) = 96 / 800 = 3 / 25 . then , the probability of playing exactly one instrument is given by : p ( playing 1 or more instruments ) - p ( playing 2 or more instruments ) = 3 / 5 - 3 / 25 = 12 / 25 . answer b . | a = 800 * 3
b = a / 5
c = b - 96
d = c / 800
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a ) 12 : 18 , b ) 144 : 169 , c ) 44 : 22 , d ) 28 : 144 , e ) 96 : 12 | b | power(divide(1728, 2197), divide(const_1, const_3)) | the ratio of the volumes of two cubes is 1728 : 2197 . what is the ratio of their total surface areas ? | "ratio of the sides = Γ’ Β³ Γ’ Λ Ε‘ 1728 : Γ’ Β³ Γ’ Λ Ε‘ 2197 = 12 : 13 ratio of surface areas = 144 : 169 answer : b" | a = 1728 / 2197
b = 1 / 3
c = a ** b
|
a ) 400 , b ) 600 , c ) 800 , d ) 1000 , e ) 1200 | b | divide(multiply(240, const_100), subtract(const_100, 60)) | in an election between two candidates , the first candidate got 60 % of the votes and the second candidate got 240 votes . what was the total number of votes ? | let v be the total number of votes . 0.4 v = 240 v = 600 the answer is b . | a = 240 * 100
b = 100 - 60
c = a / b
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a ) 58 , b ) 62 , c ) 74 , d ) 81 , e ) 65 | b | subtract(multiply(divide(280, 9), const_3_6), 50) | a man sitting in a train which is travelling at 50 kmph observes that a goods train travelling in a opposite direction , takes 9 seconds to pass him . if the goods train is 280 m long , find its speed . | "relative speed = ( 280 / 9 ) m / s = ( 280 / 9 ) * ( 18 / 5 ) = 112 kmph speed of goods train = 112 - 50 = 62 kmph answer is b" | a = 280 / 9
b = a * const_3_6
c = b - 50
|
a ) 100010 , b ) 100012 , c ) 100011 , d ) 100013 , e ) 120012 | c | divide(6, power(const_2, const_3)) | what is the smallest 6 digit number exactly divisible by 111 ? | "smallest 6 digit number = 100000 100000 / 111 = 900 , remainder = 100 . hence 11 more should be added to 100000 to get the smallest 6 digit number exactly divisible by 111 = > smallest 6 digit number exactly divisible by 111 = 100000 + 11 = 100011 answer is c" | a = 2 ** 3
b = 6 / a
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a ) 8 / 15 , b ) 2 / 5 , c ) 3 / 5 , d ) 11 / 15 , e ) 17 / 33 | e | add(multiply(divide(8, add(4, 8)), divide(subtract(8, const_1), subtract(add(4, 8), const_1))), multiply(divide(4, add(4, 8)), divide(subtract(4, const_1), subtract(add(4, 8), const_1)))) | a bag contains 4 green and 8 white balls . if two balls are drawn simultaneously , the probability that both are of the same colour is | "drawing two balls of same color from four green balls can be done in 4 c β ways . similarly from eight white balls two can be drawn in βΈ c β ways . p = 4 c β / ΒΉ β΅ c β + βΈ c β / ΒΉ β΅ c β = 17 / 33 answer : e" | a = 4 + 8
b = 8 / a
c = 8 - 1
d = 4 + 8
e = d - 1
f = c / e
g = b * f
h = 4 + 8
i = 4 / h
j = 4 - 1
k = 4 + 8
l = k - 1
m = j / l
n = i * m
o = g + n
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a ) 3 / 4 , b ) 5 / 6 , c ) 7 / 8 , d ) 5 / 8 , e ) 7 / 12 | a | add(divide(1, 2), multiply(divide(1, 2), divide(1, 2))) | jar x is 1 / 2 full of water . jar y , which has half the capacity of jar x , is 1 / 2 full of water . if the water in jar y is poured into jar x , then jar x will be filled to what fraction of its capacity ? | let p be the capacity of jar x . the amount of water in jar y is 1 / 2 * p / 2 = p / 4 then the total amount in jar x is p / 2 + p / 4 = 3 p / 4 the answer is a . | a = 1 / 2
b = 1 / 2
c = 1 / 2
d = b * c
e = a + d
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a ) 63 / 64 , b ) 7 / 8 , c ) 1 / 4 , d ) 1 / 8 , e ) 1 / 16 | a | add(add(add(add(divide(1, 2), divide(divide(1, 2), 2)), divide(divide(divide(1, 2), 2), 2)), divide(divide(divide(divide(1, 2), 2), 2), 2)), divide(divide(divide(divide(divide(1, 2), 2), 2), 2), 2)) | if 1 / 2 of the air in a tank is removed with each stroke of a vacuum pump , what fraction of the original amount of air has been removed after 6 strokes ? | "left after 1 st stroke = 1 / 2 left after 2 nd stroke = 1 / 2 * 1 / 2 = 1 / 4 left after 3 rd stroke = 1 / 2 * 1 / 4 = 1 / 8 left after 4 th stroke = 1 / 2 * 1 / 8 = 1 / 16 left after 5 th stroke = 1 / 2 * 1 / 16 = 1 / 32 left after 6 th stroke = 1 / 2 * 1 / 32 = 1 / 64 so removed = 1 - 1 / 64 = 63 / 64" | a = 1 / 2
b = 1 / 2
c = b / 2
d = a + c
e = 1 / 2
f = e / 2
g = f / 2
h = d + g
i = 1 / 2
j = i / 2
k = j / 2
l = k / 2
m = h + l
n = 1 / 2
o = n / 2
p = o / 2
q = p / 2
r = q / 2
s = m + r
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a ) 1 / 5 , b ) 2 / 3 , c ) 5 / 7 , d ) 7 / 5 , e ) 3 / 2 | a | divide(subtract(divide(const_1, const_2), subtract(subtract(const_1, divide(const_1, const_3)), multiply(divide(const_1, const_3), subtract(const_1, divide(const_1, const_3))))), subtract(divide(const_1, const_2), multiply(divide(const_1, const_3), subtract(const_1, divide(const_1, const_3))))) | a certain ball team has an equal number of right - and left - handed players . on a certain day , one - third of the players were absent from practice . of the players at practice that day , one - third were left handed . what is the ratio of the number of right - handed players who were not at practice that day to the number of lefthanded players who were not at practice ? | say the total number of players is 18 , 9 right - handed and 9 left - handed . on a certain day , two - thirds of the players were absent from practice - - > 6 absent and 12 present . of the players at practice that day , one - third were left - handed - - > 12 * 1 / 3 = 4 were left - handed and 8 right - handed . the number of right - handed players who were not at practice that day is 9 - 8 = 1 . the number of left - handed players who were not at practice that days is 9 - 4 = 5 . the ratio = 1 / 5 . answer : a . | a = 1 / 2
b = 1 / 3
c = 1 - b
d = 1 / 3
e = 1 / 3
f = 1 - e
g = d * f
h = c - g
i = a - h
j = 1 / 2
k = 1 / 3
l = 1 / 3
m = 1 - l
n = k * m
o = j - n
p = i / o
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a ) 4 , b ) 6 , c ) 8 , d ) 12 , e ) 16 | d | multiply(multiply(4, const_2), divide(const_3, const_2)) | a group of hikers is planning a trip that will take them up a mountain using one route and back down using another route . they plan to travel down the mountain at a rate of one and a half times the rate they will use on the way up , but the time each route will take is the same . if they will go up the mountain at a rate of 4 miles per day and it will take them two days , how many miles long is the route down the mountain ? | "d = s * t given condition is for up the mountain hikers takes 4 miles per day and total 2 days . . . miles = 8 miles . . . for down speed they take = 3 / 2 ( 4 ) = 6 miles per day and for the same 2 days . . . miles = 12 miles . . answer : option d is correct answer" | a = 4 * 2
b = 3 / 2
c = a * b
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a ) $ 1360 , b ) $ 1620 , c ) $ 1650 , d ) $ 680 , e ) $ 770 | a | add(add(multiply(200, 5), multiply(80, 3)), multiply(multiply(20, 3), const_2)) | rates for having a manuscript typed at a certain typing service are $ 5 per page for the first time a page is typed and $ 3 per page each time a page is revised . if a certain manuscript has 200 pages , of which 80 were revised only once , 20 were revised twice , and the rest required no revisions , what was the total cost of having the manuscript typed ? | "for 200 - 80 - 20 = 100 pages only cost is 5 $ per page for the first time page is typed - 100 * 5 = 500 $ ; for 80 pages the cost is : first time 5 $ + 3 $ of the first revision - 80 * ( 5 + 3 ) = 640 $ ; for 20 pages the cost is : first time 5 $ + 3 $ of the first revision + 3 $ of the second revision - 20 ( 5 + 3 + 3 ) = 220 $ ; total : 500 + 640 + 220 = 1360 $ . answer : a ." | a = 200 * 5
b = 80 * 3
c = a + b
d = 20 * 3
e = d * 2
f = c + e
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a ) 3 , b ) β 5 / 3 , c ) β 7 / 5 , d ) 16 / 3 , e ) β 38 / 3 | e | subtract(negate(1), negate(divide(multiply(5, 7), 3))) | the line y = 5 x / 3 + b goes through the point ( 7 , β 1 ) . what is the value of b ? | β finding the equation β procedure . plug ( x , y ) = ( 7 , β 1 ) into this equation : { - 1 } = { { 5 ( 7 ) } / 3 } + b = { 35 / 3 } + b b = { - 35 / 3 } - 1 = { - 35 / 3 } - { 3 / 3 } = - { 38 / 3 } answer = e | a = negate - (
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a ) rs . 6003 , b ) rs . 3800 , c ) rs . 1288 , d ) rs . 6108 , e ) rs . 6011 | b | subtract(multiply(add(1700, 100), add(20, const_1)), multiply(1700, 20)) | the average monthly salary of 20 employees in an organisation is rs . 1700 . if the manager ' s salary is added , then the average salary increases by rs . 100 . what is the manager ' s monthly salary ? | "explanation : manager ' s monthly salary rs . ( 1800 * 21 - 1700 * 20 ) = rs . 3800 . answer : b" | a = 1700 + 100
b = 20 + 1
c = a * b
d = 1700 * 20
e = c - d
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a ) 199 , b ) 154 , c ) 852 , d ) 741 , e ) 785 | a | divide(subtract(17698, 14), 89) | on dividing 17698 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 ( 17698 - 14 ) / 89 = 199 answer ( a )" | a = 17698 - 14
b = a / 89
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a ) 270 mts , b ) 190 mts , c ) 100 mts , d ) 500 mts , e ) 110 mts | d | sqrt(add(power(multiply(multiply(36, 30), const_0_2778), const_2), power(multiply(multiply(48, 30), const_0_2778), const_2))) | two trains start from same place at same time at right angles to each other . their speeds are 36 km / hr and 48 km / hr respectively . after 30 seconds the distance between them will be ? | explanation : using pythagarous theorem , distance travelled by first train = 36 x 5 / 18 x 30 = 300 m distance travelled by second train = 48 x 5 / 18 x 30 = 400 m so distance between them = Γ’ Λ Ε‘ ( 90000 + 160000 ) = Γ’ Λ Ε‘ 250000 = 500 mts . answer : d | a = 36 * 30
b = a * const_0_2778
c = b ** 2
d = 48 * 30
e = d * const_0_2778
f = e ** 2
g = c + f
h = math.sqrt(g)
|
a ) 2480 , b ) 3490 , c ) 6785 , d ) 8265 , e ) 9255 | d | subtract(divide(multiply(multiply(30, add(30, const_1)), add(multiply(const_2, 30), const_1)), add(const_3, const_3)), 1290) | the sum of the squares of the first 15 positive integers ( 1 ^ 2 + 2 ^ 2 + 3 ^ 2 + . . . + 15 ^ 2 ) is equal to 1290 . what is the sum of the squares of the second 15 positive integers ( 16 ^ 2 + 17 ^ 2 + 18 ^ 2 + . . . + 30 ^ 2 ) ? | you ' ll never need a formula for the sums of squares on the actual gmat . you do n ' t need to use that formula here , though it ' s not all that straightforward to solve without one . two different approaches : 16 ^ 2 + 17 ^ 2 + 18 ^ 2 + . . . + 30 ^ 2 = ( 15 + 1 ) ^ 2 + ( 15 + 2 ) ^ 2 + ( 15 + 3 ) ^ 2 + . . . + ( 15 + 15 ) ^ 2 now we can expand each square ; they are all in the ( x + y ) ^ 2 = x ^ 2 + 2 xy + y ^ 2 pattern . = ( 15 ^ 2 + 2 * 15 + 1 ^ 2 ) + ( 15 ^ 2 + 4 * 15 + 2 ^ 2 ) + ( 15 ^ 2 + 6 * 15 + 3 ^ 2 ) + . . . + ( 15 ^ 2 + 30 * 15 + 15 ^ 2 ) now we have fifteen 15 ^ 2 terms , so adding these gives 15 * 15 ^ 2 = 15 ^ 3 = 3375 . we also have the entire sum 1 ^ 2 + 2 ^ 2 + . . . + 15 ^ 2 , which we know is equal to 1240 . finally adding the middle terms , we have : 2 * 15 + 4 * 15 + 6 * 15 + . . . + 30 * 15 = 15 ( 2 + 4 + 6 + . . . . + 30 ) = 15 * 2 * ( 1 + 2 + 3 + . . . + 15 ) = 15 * 2 * 8 * 15 = 3600 so the sum must be 3375 + 1240 + 3600 = 8215 alternatively , we can use a different factoring pattern . we want to find the value of 30 ^ 2 + 29 ^ 2 + . . . + 17 ^ 2 + 16 ^ 2 . well if we subtract 15 ^ 2 + 14 ^ 2 + . . . . + 2 ^ 2 + 1 ^ 2 from this , the answer will be 1240 less than what we want to find . so if we can find the value of 30 ^ 2 + 29 ^ 2 + . . . + 17 ^ 2 + 16 ^ 2 - ( 15 ^ 2 + 14 ^ 2 + . . . . + 2 ^ 2 + 1 ^ 2 ) then we can add 1240 to get the answer . now grouping the terms above to get differences of squares , we have = ( 30 ^ 2 - 15 ^ 2 ) + ( 29 ^ 2 - 14 ^ 2 ) + . . . + ( 16 ^ 2 - 1 ^ 2 ) and factoring each of these using x ^ 2 - y ^ 2 = ( x + y ) ( x - y ) , we have = 45 * 15 + 43 * 15 + 41 * 15 + . . . + 17 * 15 = 15 ( 45 + 43 + 41 + . . . + 17 ) in brackets we have an equally spaced sum with fifteen terms , which we can evaluate using the familiar formula . so the above equals 15 * 15 * 62 / 2 = 6975 and adding back the 1290 , we get the answer of 8265 . ( ans d ) | a = 30 + 1
b = 30 * a
c = 2 * 30
d = c + 1
e = b * d
f = 3 + 3
g = e / f
h = g - 1290
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a ) 100 , b ) 120 , c ) 150 , d ) 180 , e ) 160 | b | multiply(divide(const_60, 30), 1) | if the population of a certain country increases at the rate of one person every 30 seconds , by how many persons does the population increase in 1 hour ? | "answer = 2 * 60 = 120 answer is b" | a = const_60 / 30
b = a * 1
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a ) 14 years , b ) 12 years , c ) 56 years , d ) 10 years , e ) 55 years | d | multiply(10, const_1) | the total age of a and b is 10 years more than the total age of b and c . c is how many year younger than | "given that a + b = 10 + b + c = > a Γ’ β¬ β c = 10 + b Γ’ β¬ β b = 10 = > c is younger than a by 10 years answer : d" | a = 10 * 1
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a ) 14.4 , b ) 14.7 , c ) 15.0 , d ) 15.3 , e ) 15.6 | c | multiply(sqrt(divide(22.5, 10.0)), 10.0) | at 1 : 00 pm , there were 10.0 grams of bacteria . the bacteria increased to x grams at 4 : 00 pm , and 22.5 grams at 7 : 00 pm . if the amount of bacteria present increased by the same fraction during each of the 3 - hour periods , how many grams of bacteria were present at 4 : 00 pm ? | "let x be the factor by which the bacteria increases every three hours . at 4 : 00 pm , the amount of bacteria was 10 x and at 7 : 00 pm it was 10 x ^ 2 . 10 x ^ 2 = 22.5 x ^ 2 = 2.25 x = 1.5 at 4 : 00 pm , the amount of bacteria was 10 ( 1.5 ) = 15 grams . the answer is c ." | a = 22 / 5
b = math.sqrt(a)
c = b * 10
|
a ) 10 , b ) 12 , c ) 15 , d ) 18 , e ) 20 | b | add(add(add(divide(6, const_3), divide(12, const_4)), divide(12, const_4)), const_4) | if 6 x ^ 2 + x - 12 = ( ax + w ) ( cx + d ) , then | a | + | w | + | c | + | d | = for a complete solution and more practice problems , see this blog : http : / / magoosh . com / gmat / 2012 / algebra - on . . . to - factor / | 6 x ^ 2 + x - 12 = 6 x ^ 2 + 9 x - 8 x - 12 = > 3 x ( 2 x + 3 ) - 4 ( 2 x + 3 ) = > ( 2 x + 3 ) ( 3 x - 4 ) = ( ax + w ) ( cx + d ) hence a = 2 , w = c = 3 , d = - 4 so , 2 + 3 + 3 + | - 4 | = 2 + 3 + 3 + 4 = 12 answer b . | a = 6 / 3
b = 12 / 4
c = a + b
d = 12 / 4
e = c + d
f = e + 4
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a ) 10 , b ) 20 , c ) 15 , d ) 40 , e ) 50 | c | add(10, const_1) | the average of first four prime numbers greater than 10 is ? | "11 + 13 + 17 + 19 = 60 / 4 = 15 answer : c" | a = 10 + 1
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a ) 3 : 2 , b ) 4 : 3 , c ) 18 : 20 , d ) 1 : 4 , e ) 18 : 4 | b | divide(add(multiply(multiply(const_3, const_3), add(multiply(const_3, const_3), const_1)), 2), multiply(2, add(multiply(const_3, const_3), const_1))) | a and b started a business investing rs . 60,000 and rs 45,000 respectively . in what ratio the profit earned after 2 years be divided between a and b respectively ? | "a : b = 60000 : 45000 = 4 : 3 answer : b" | a = 3 * 3
b = 3 * 3
c = b + 1
d = a * c
e = d + 2
f = 3 * 3
g = f + 1
h = 2 * g
i = e / h
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a ) 561.68 m , b ) 562.68 m , c ) 563.68 m , d ) 564.68 m , e ) 565.68 m | b | divide(multiply(multiply(multiply(divide(add(multiply(add(const_3, const_4), const_3), const_1), add(const_3, const_4)), 22.4), const_2), 400), const_100) | the radius of a wheel is 22.4 cm . what is the distance covered by the wheel in making 400 resolutions . | "in one resolution , the distance covered by the wheel is its own circumference . distance covered in 400 resolutions . = 500 * 2 * 22 / 7 * 22.4 = 56268 cm = 562.68 m answer : b" | a = 3 + 4
b = a * 3
c = b + 1
d = 3 + 4
e = c / d
f = e * 22
g = f * 2
h = g * 400
i = h / 100
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a ) 4400 , b ) 3500 , c ) 8700 , d ) 9000 , e ) 8000 | a | floor(divide(3553, multiply(divide(subtract(const_100, 5), const_100), divide(subtract(const_100, 15), const_100)))) | 5 % people of a village in sri lanka died by bombardment , 15 % of the remainder left the village on account of fear . if now the population is reduced to 3553 , how much was it in the beginning ? | "x * ( 95 / 100 ) * ( 85 / 100 ) = 3553 x = 4400 answer a" | a = 100 - 5
b = a / 100
c = 100 - 15
d = c / 100
e = b * d
f = 3553 / e
g = math.floor(f)
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a ) 60229 , b ) 62887 , c ) 60000 , d ) 61298 , e ) 61291 | c | divide(multiply(multiply(add(const_1, const_4), const_1000), 2), 3) | x and y invested in a business . they earned some profit which they divided in the ratio of 2 : 3 . if x invested rs . 40,000 , the amount invested by y is ? | "suppose y invested rs . y . then 40000 / y = 2 / 3 or y = 60000 . answer : c" | a = 1 + 4
b = a * 1000
c = b * 2
d = c / 3
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a ) 1.5 : 5 , b ) 2 : 5 , c ) 3 : 5 , d ) 1 : 5 , e ) 4 : 5 | c | divide(power(27, const_0_33), power(125, const_0_33)) | two cubes of their volumes in the ratio 27 : 125 . the ratio of their surface area is : | "the ratio of their surface area is 27 : 125 3 : 5 answer is c ." | a = 27 ** const_0_33
b = 125 ** const_0_33
c = a / b
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a ) 500 , b ) 650 , c ) 800 , d ) 950 , e ) 1100 | a | divide(divide(160, const_2), divide(16, const_100)) | one night 16 percent of the female officers on a police force were on duty . if 160 police officers were on duty that night and half of these were female officers , how many female officers were on the police force ? | "let x be the number of female police officers on the police force . the number of female police officers on duty was 80 . 0.16 x = 80 x = 500 the answer is a ." | a = 160 / 2
b = 16 / 100
c = a / b
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a ) 60 mph , b ) 56.67 mph , c ) 53.33 mph , d ) 64 mph , e ) 66.67 mph | a | add(divide(add(multiply(80, 3), multiply(30, 2)), add(3, 2)), subtract(divide(const_100, 3), const_0_33)) | steve traveled the first 2 hours of his journey at 30 mph and the remaining 3 hours of his journey at 80 mph . what is his average speed for the entire journey ? | "distance traveled in 2 hours = 2 * 30 = 60 m distance traveled in 3 hours = 3 * 80 = 240 m total distance covered = 240 + 60 = 300 m total time = 2 + 3 = 5 h hence avg speed = total distance covered / total time taken = 300 / 5 = 60 mph answer : a" | a = 80 * 3
b = 30 * 2
c = a + b
d = 3 + 2
e = c / d
f = 100 / 3
g = f - const_0_33
h = e + g
|
a ) 1 : 3 , b ) 1 : 2 , c ) 1 : 1 , d ) 2 : 1 , e ) 3 : 1 | d | divide(0.835, add(0.8, 0.9)) | in what proportion must flour at $ 0.8 per pound be mixed with flour at $ 0.9 per pound so that the mixture costs $ 0.835 per pound ? | "using weighted average method : let x be the proportion in which the $ 0.8 per pound flour is mixed with $ 0.9 per pound flour . thus 0.8 * x + 0.9 * ( 1 - x ) = 0.835 0.9 β 0.1 x = 0.835 x = 0.65 thus ratio of both flours is 2 : 1 d" | a = 0 + 8
b = 0 / 835
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a ) 25 , b ) 27 , c ) 30 , d ) 35 , e ) none | d | subtract(multiply(27, 5), multiply(25, const_4)) | the average of 5 numbers id 27 . if one number is excluded , the average becomes 25 . the excluded number is | solution excluded number = ( 27 x 5 ) - ( 25 x 4 ) = 135 - 100 = 35 . answer d | a = 27 * 5
b = 25 * 4
c = a - b
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a ) 19 , b ) 29 , c ) 38 , d ) 49 , e ) 59 | c | add(subtract(86, multiply(17, 3)), 3) | a batsman makes a score of 86 runs in the 17 th inning and thus increases his averages by 3 . find his average after 17 th inning ? | "let the average after 17 th inning = x then average after 16 th inning = ( x - 3 ) therefore 16 ( x - 3 ) + 86 = 17 x therefore x = 38 answer : c" | a = 17 * 3
b = 86 - a
c = b + 3
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a ) 10 days , b ) 20 days , c ) 25 days , d ) 15 days , e ) 45 days | b | inverse(subtract(4, divide(4, 6))) | a and b can do a piece of work in 6 days . with the help of c they finish the work in 4 days . c alone can do that piece of work in ? | "c = 1 / 5 β 1 / 6 = 1 / 30 = > 30 days answer : b" | a = 4 / 6
b = 4 - a
c = 1/(b)
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a ) 150 , b ) 200 , c ) 250 , d ) 300 , e ) 350 | b | divide(50, subtract(subtract(subtract(1, divide(1, 16)), divide(subtract(1, divide(1, 16)), 15)), multiply(subtract(subtract(1, divide(1, 16)), divide(subtract(1, divide(1, 16)), 15)), divide(5, 7)))) | a person deposits 1 / 16 of his income as provident fund and 1 / 15 of the remaining as insurance premium . if he spends 5 / 7 of the balance on domestic needs and deposits an amount of rs . 50 in the bank , his total income would be | ( 1 - ( 1 / 16 ) ) ( 1 - ( 1 / 15 ) ) ( 1 - ( 5 / 7 ) ) of income = 50 hence income = 200 answer : b | a = 1 / 16
b = 1 - a
c = 1 / 16
d = 1 - c
e = d / 15
f = b - e
g = 1 / 16
h = 1 - g
i = 1 / 16
j = 1 - i
k = j / 15
l = h - k
m = 5 / 7
n = l * m
o = f - n
p = 50 / o
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a ) 2 / 7 , b ) 3 / 8 , c ) 9 / 16 , d ) 5 / 8 , e ) 16 / 9 | a | multiply(divide(3, 5), multiply(divide(5, 7), divide(2, 3))) | what is 3 / 5 of 5 / 7 of 2 / 3 ? | "3 / 5 * 5 / 7 * 2 / 3 = 2 / 7 answer : a" | a = 3 / 5
b = 5 / 7
c = 2 / 3
d = b * c
e = a * d
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a ) 7 / 12 , b ) 8 / 41 , c ) 9 / 348 , d ) 1 / 8 , e ) 41 / 70 | e | divide(subtract(350, add(subtract(54, divide(multiply(12.5, 104), multiply(const_1, const_100))), 104)), 350) | in a survey of 350 employees , 104 of them are uninsured , 54 work part time , and 12.5 percent of employees who are uninsured work part time . if a person is to be randomly selected from those surveyed , what is the probability that the person will neither work part time nor be uninsured ? | "- - - - - - - - - ui - - - - - - - - - - - - - - - - nui - - - - - - - total pt - - - - ( 12.5 / 100 ) * 104 = 13 - - - - - - - - - - - - - 54 npt - - - 104 - 13 - - - - - - - - - - - - - - x - - - - - - - - 296 total - - 104 - - - - - - - - - - - - - - - - - - - - - - - - - - - - 350 we have to find not part time and not uninsured . in other words not part time and insured = x / 350 = ( 296 - 104 + 13 ) / 350 = 41 / 70 answer is e ." | a = 12 * 5
b = 1 * 100
c = a / b
d = 54 - c
e = d + 104
f = 350 - e
g = f / 350
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a ) 1 : 20 , b ) 1 : 10 , c ) 1 : 8 , d ) 1 : 4 , e ) 1 : 2 | b | divide(subtract(33, divide(36, divide(add(const_100, 20), const_100))), divide(36, divide(add(const_100, 20), const_100))) | a dishonest person wants to make a profit on the selling of milk . he would like to mix water ( costing nothing ) with milk costing 33 $ per litre so as to make a profit of 20 % on cost when he sells the resulting milk and water mixture for 36 $ . in what ratio should he mix the water and milk ? | "he needs to make a profit of 20 % , which means he needs to sell the milk + water combination for 39.6 ( 20 % of 33 + 33 ) now , he plans to sell 1 liter of this solution of milk and water for 36 . which means he needs to sell 1.1 liters of this solution to earn 39.6 . ( 36 for 1 liter and so 39.6 for 1.1 liters ) hence the water added would be . 1 liter . therefore the ratio of water to milk is . 1 / 1 = 1 / 10 answer : b" | a = 100 + 20
b = a / 100
c = 36 / b
d = 33 - c
e = 100 + 20
f = e / 100
g = 36 / f
h = d / g
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a ) 200 , b ) 288 , c ) 120 , d ) 776 , e ) 991 | a | divide(12, subtract(divide(12, 10), 10)) | a train covers a distance of 12 km in 10 min . if it takes 10 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 * 10 = 200 m . answer : a" | a = 12 / 10
b = a - 10
c = 12 / b
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a ) 130 , b ) 140 , c ) 150 , d ) 220 , e ) 280 | d | divide(550, add(divide(150, const_100), const_1)) | two employees a and b are paid a total of rs . 550 per week by their employer . if a is paid 150 percent of the sum paid to b , how much is b paid per week ? | let the amount paid to a per week = x and the amount paid to b per week = y then x + y = 550 but x = 150 % of y = 150 y / 100 = 15 y / 10 β΄ 15 y / 10 + y = 550 β y [ 15 / 10 + 1 ] = 550 β 25 y / 10 = 550 β 25 y = 5500 β y = 5500 / 25 = rs . 220 d ) | a = 150 / 100
b = a + 1
c = 550 / b
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a ) 59 % , b ) 55 % , c ) 63 % , d ) 66 % , e ) 68 % | d | multiply(const_100, divide(divide(multiply(add(20, const_100), 60), const_100), add(const_100, 9))) | of the families in city x in 1991 , 60 percent owned a personal computer . the number of families in city x owning a computer in 1993 was 20 percent greater than it was in 1991 , and the total number of families in city x was 9 percent greater in 1993 than it was in 1991 . what percent of the families in city x owned a personal computer in 1993 ? | "say a 100 families existed in 1991 then the number of families owning a computer in 1991 - 60 number of families owning computer in 1993 = 60 * 120 / 100 = 72 number of families in 1993 = 109 the percentage = 72 / 109 * 100 = 66 % . answer : d" | a = 20 + 100
b = a * 60
c = b / 100
d = 100 + 9
e = c / d
f = 100 * e
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a ) 83 % , b ) 80 % , c ) 20 % , d ) 17 % , e ) 9 % | e | subtract(10, const_1) | at a local appliance manufacturing facility , the workers received a 10 % hourly pay raise due to extraordinary performance . if one worker decided to reduce the number of hours that he worked so that his overall pay would remain unchanged , by approximately what percent would he reduce the number of hours that he worked ? | let ' s say he works usually 10 hours and earns 100 per hour . 10 * 100 = 1000 10 * 110 = 1100 ( this are the new earnings after the raise ) to figure out how much he needs to work with the new salary in order to earn the original 1000 : 1000 / 110 = 9.09 so he can reduce his work by 0.91 hours . which is > 9 % . answer e | a = 10 - 1
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a ) 25 / 8 , b ) 25 / 4 , c ) 25 / 6 , d ) 3 , e ) none | a | divide(add(subtract(add(64, multiply(32, 25)), subtract(2, 64)), const_1), 50) | determine the value of ( 25 / 32 + 50 / 64 ) * 2 | "solution : both fractions should be reduced before performing arithmetic operations . we get ( 25 / 32 + 2.25 / 2.32 ) * 2 = ( 25 / 32 + 25 / 32 ) * 2 = 2 * ( 25 / 32 ) * 2 = 4 ( 25 / 32 ) = 25 / 8 answer a" | a = 32 * 25
b = 64 + a
c = 2 - 64
d = b - c
e = d + 1
f = e / 50
|
a ) 4800 , b ) 1200 , c ) 1680 , d ) 1100 , e ) 1300 | b | multiply(24, 20) | a cistern has a leak which would empty the cistern in 20 minutes . a tap is turned on which admits 10 liters a minute into the cistern , and it is emptied in 24 minutes . how many liters does the cistern hold ? | "1 / x - 1 / 20 = - 1 / 24 x = 120 120 * 10 = 1200 answer : b" | a = 24 * 20
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a ) $ 200 , b ) $ 300 , c ) $ 100 , d ) $ 400 , e ) $ 250 | d | divide(multiply(divide(multiply(divide(40, const_100), 600), divide(const_60.0, const_100)), divide(40, const_100)), divide(40, const_100)) | a school has received 40 % of the amount it needs for a new building by receiving a donation of $ 600 each from people already solicited . people already solicited represent 40 % of the people from whom the school will solicit donations . how much average contribution is requited from the remaining targeted people to complete the fund raising exercise ? | "let us suppose there are 100 people . 40 % of them donated $ 16000 ( 400 * 40 ) $ 16000 is 40 % of total amount . so total amount = 16000 * 100 / 40 remaining amount is 60 % of total amount . 60 % of total amount = 16000 * ( 100 / 40 ) * ( 60 / 100 ) = 24000 this amount has to be divided by 60 ( remaining people are 60 ) so per head amount is 24000 / 60 = $ 400 answer : d" | a = 40 / 100
b = a * 600
c = const_60 / 0
d = b / c
e = 40 / 100
f = d * e
g = 40 / 100
h = f / g
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a ) $ 800 , b ) $ 1000 , c ) $ 600 , d ) $ 400 , e ) $ 500 | a | divide(multiply(subtract(multiply(multiply(add(const_3, const_4), const_4), const_1000), multiply(multiply(add(const_4, const_1), const_4), const_1000)), 10), const_100) | in township k each property is taxed at 10 percent of its assessed value . if the assessed value of a property in township k is increased from $ 20,000 to $ 28,000 , by how much will the property tax increase ? | increase in house value = $ 28,000 - $ 20,000 = $ 8000 so , tax increase = 10 % of $ 8000 = $ 800 answer : a | a = 3 + 4
b = a * 4
c = b * 1000
d = 4 + 1
e = d * 4
f = e * 1000
g = c - f
h = g * 10
i = h / 100
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a ) 25 , b ) 26 , c ) 27 , d ) 28 , e ) 29 | b | divide(subtract(60, subtract(8, 2)), const_3) | george is 8 years more than christopher and ford is 2 years younger than christopher . the sum of their ages is 60 . find the ages of george . | "christopher age = x george age , y = x + 8 - - - - - - - - - - > ( 1 ) ford age , z = x - 2 - - - - - - - - - - - - > ( 2 ) sum of their ages , x + y + z = 60 - - - - > ( 3 ) substitute z and y values in equation ( 3 ) therefore , x + ( x + 8 ) + ( x - 2 ) = 60 = > 3 x + 8 - 2 = 60 = > 3 x = 60 - 6 = > 3 x = 54 = > x = 54 / 3 x = 18 ( christopher ' s age ) substitute x value in equation 1 & 2 y = x + 8 y = 18 + 8 y = 26 answer : b" | a = 8 - 2
b = 60 - a
c = b / 3
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a ) 0.65 , b ) 0.25 , c ) 0.75 , d ) 0.85 , e ) none | c | divide(multiply(0.25, 6), 2) | if 0.25 : x : : 2 : 6 , then x is equal to | "sol . ( x Γ 2 ) = ( 0.25 Γ 6 ) β x = 1.5 / 2 = 0.75 . answer c" | a = 0 * 25
b = a / 2
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a ) 1.5 , b ) 1.75 , c ) 2.14 , d ) 2.34 , e ) 2.64 | c | multiply(divide(subtract(divide(7, add(8, 7)), divide(const_2, add(const_2, const_3))), divide(7, add(8, 7))), add(8, 7)) | a solution contains 8 parts of water for every 7 parts of lemonade syrup . how many parts of the solution should be removed and replaced with water so that the solution will now contain 10 % lemonade syrup ? | "let the total solution is 150 l with 80 l water 70 l syrup . to make 10 % syrup solution , the result solution must have 135 l syrup and 15 l syrup . therefore we are taking 55 l of syrup from initial solution and replacing with water . using urinary method : 70 l syrup in 150 l solution 55 l syrup in 117.9 l solution we started by multiplying 10 now to get to the result we need to divide by 55 = > amount of solution to be replaced with water = ( 117.9 / 55 ) = 2.14 . correct option : c" | a = 8 + 7
b = 7 / a
c = 2 + 3
d = 2 / c
e = b - d
f = 8 + 7
g = 7 / f
h = e / g
i = 8 + 7
j = h * i
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a ) 534 , b ) 446 , c ) 354 , d ) 324 , e ) none of them | d | subtract(add(multiply(287, 287), multiply(269, 269)), multiply(multiply(2, 287), 269)) | 287 x 287 + 269 x 269 - 2 x 287 x 269 = ? | = a ^ 2 + b ^ 2 - 2 ab where a = 287 and b = 269 = ( a - b ) ^ 2 = ( 287 - 269 ) ^ 2 = ( 18 ) ^ 2 = 324 answer is d | a = 287 * 287
b = 269 * 269
c = a + b
d = 2 * 287
e = d * 269
f = c - e
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a ) 40 days , b ) 40 / 9 days , c ) 39 / 9 days , d ) 30 / 9 days , e ) 20 / 3 days | e | divide(const_1, add(divide(const_1, 12), divide(const_1, 15))) | worker a takes 12 hours to do a job . worker b takes 15 hours to do the same job . how long it take both a & b , working together but independently , to do the same job ? | "a ' s one hour work = 1 / 12 . b ' s one hour work = 1 / 15 . ( a + b ) ' s one hour work = 1 / 12 + 1 / 15 = 9 / 60 = 3 / 20 . both a & b can finish the work in 20 / 3 days e" | a = 1 / 12
b = 1 / 15
c = a + b
d = 1 / c
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a ) 10 / 28 , b ) 1 / 28 , c ) 12 / 28 , d ) 28 / 1 , e ) 582 / 695 | b | divide(1, add(add(12, 10), 5)) | jamal had a bag of marbles . he had 12 yellow , 10 blue , 5 green and 1 black marbles . he decided to randomly select a ball from the bag . what is the proabilitly that he will draw a black marble ? | b | a = 12 + 10
b = a + 5
c = 1 / b
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a ) 90 , b ) 120 , c ) 160 , d ) 360 , e ) 400 | e | divide(100, subtract(const_1, divide(3, 4))) | the visitors of a modern art museum who watched a certain picasso painting were asked to fill in a short questionnaire indicating whether they had enjoyed looking at the picture and whether they felt they had understood it . according to the results of the survey , all 100 visitors who did not enjoy the painting also did not feel they had understood the painting , and the number of visitors who enjoyed the painting was equal to the number of visitors who felt they had understood the painting . if 3 / 4 of the visitors who answered the questionnaire both enjoyed the painting and felt they had understood the painting , then how many visitors answered the questionnaire ? | "if we exclude those cases and take the question at face value , then it seems straightforward . group # 1 = ( did n ' t like , did n ' t understand ) = 100 group # 2 = ( likeunderstood ) = 3 / 4 ( 1 / 4 ) n = 100 n = 400 answer = ( e )" | a = 3 / 4
b = 1 - a
c = 100 / b
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a ) rs 824.32 , b ) rs 822.32 , c ) rs 700 , d ) rs 900 , e ) rs 624.00 | a | subtract(multiply(power(add(const_1, divide(divide(2, const_4), const_100)), const_3), multiply(multiply(multiply(const_4, const_4), const_100), sqrt(const_100))), multiply(multiply(multiply(const_4, const_4), const_100), sqrt(const_100))) | find the compound interest on rs . 10,000 in 2 years at 4 % per annum , the interest being compounded half - yearly . | "principal = rs . 10000 ; rate = 2 % per half - year ; time = 2 years = 4 half - years . amount = rs [ 10000 * ( 1 + ( 2 / 100 ) ) 4 ] = rs ( 10000 * ( 51 / 50 ) * ( 51 / 50 ) * ( 51 / 50 ) * ( 51 / 50 ) ) = rs . 10824.32 . : . c . i . = rs . ( 10824.32 - 10000 ) = rs . 824.32 . answer a rs 824.32" | a = 2 / 4
b = a / 100
c = 1 + b
d = c ** 3
e = 4 * 4
f = e * 100
g = math.sqrt(100)
h = f * g
i = d * h
j = 4 * 4
k = j * 100
l = math.sqrt(100)
m = k * l
n = i - m
|
a ) 5 % , b ) 10 % , c ) 16 % , d ) 22 % , e ) none of these | d | add(add(multiply(divide(divide(120, const_4), add(const_1, const_4)), const_2), divide(divide(100, add(const_1, const_4)), const_4)), add(const_1, const_4)) | in an acoustics class , 120 students are male and 100 students are female . 25 % of the male students and 20 % of the female students are engineering students . 20 % of the male engineering students and 25 % of the female engineering students passed the final exam . what percentage of engineering students passed the exam ? | explanation : the number of female engineering students in the class is ( 20 / 100 ) Γ 100 = 20 . now , 25 % of the female engineering students passed the final exam : 25 . hence , the number of female engineering students who passed is 5 . there are 120 male students in the class . and 25 % of them are engineering students . hence , the number of male engineering students is ( 1 / 4 ) Γ 120 = 30 . now , 20 % of the male engineering students passed the final exam : - 20 . hence , the number of male engineering students who passed is 6 . hence , the total number of engineering students who passed is : ( female engineering students who passed ) + ( male engineering students who passed ) = > 5 + 6 = 11 . the total number of engineering students in the class is : ( number of female engineering students ) + ( number of male engineering students ) = = > 30 + 20 = 50 hence , the percentage of engineering students who passed is : - = > ( total number of engineering students passed / total number of engineering students ) Γ 100 . = > ( 11 / 50 ) Γ 100 . = > 22 % . answer : d | a = 120 / 4
b = 1 + 4
c = a / b
d = c * 2
e = 1 + 4
f = 100 / e
g = f / 4
h = d + g
i = 1 + 4
j = h + i
|
a ) 14 / 45 , b ) 4 / 9 , c ) 4 / 5 , d ) 4 / 6 , e ) none of these | a | subtract(add(divide(2, 3), subtract(const_1, divide(2, 9))), divide(9, 5)) | the probability that a computer company will get a computer hardware contract is 2 / 3 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 ? | "explanation : 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 ) = 2 / 3 , 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 = ( 2 / 3 ) + ( 4 / 9 ) - p ( ab ) . = > p ( ab ) = 14 / 45 . hence , the required probability is 14 / 45 . answer : a" | a = 2 / 3
b = 2 / 9
c = 1 - b
d = a + c
e = 9 / 5
f = d - e
|
a ) 3 , b ) 2 , c ) 1 , d ) 0 , e ) 4 | a | subtract(subtract(8, 3), subtract(6, 4)) | | 8 - 3 | - | 4 - 6 | = ? | "| 8 - 3 | - | 4 - 6 | = | 5 | - | - 2 | = 5 - 2 = 3 correct answer a" | a = 8 - 3
b = 6 - 4
c = a - b
|
a ) 7 , b ) 6 , c ) 5 , d ) 4 , e ) 3 | c | floor(divide(log(divide(240,000, 2.134)), log(10))) | if x is an integer and 2.134 Γ 10 ^ x is less than 240,000 , what is the greatest possible value for x ? | "x is an integer and 2.134 Γ 10 x is less than 240,000 , what is the greatest possible value for x ? for 2.134 Γ 10 x is less than 240,000 to remain true , the greatest number is 213,400 , which makes x = 5 c . 5" | a = 240 / 0
b = math.log(a)
c = math.log(10)
d = b / c
e = math.floor(d)
|
a ) 47 / 10 , b ) 46 / 10 , c ) 45 / 10 , d ) 42 / 10 , e ) 43 / 10 | c | divide(200, multiply(add(71, 89), const_0_2778)) | two trains of length 90 m and 100 m are 200 m apart . they start moving towards each other on parallel tracks , at speeds 71 kmph and 89 kmph . after how much time will the trains meet ? | "they are moving in opposite directions , relative speed is equal to the sum of their speeds . relative speed = ( 71 + 89 ) * 5 / 18 = 44.4 mps . the time required = d / s = 200 / 44.4 = 45 / 10 sec . answer : c" | a = 71 + 89
b = a * const_0_2778
c = 200 / b
|
a ) 3 , b ) 5 , c ) 8 , d ) 13 , e ) 15 | a | subtract(multiply(40, 2), add(multiply(subtract(subtract(40, add(add(multiply(12, 1), 10), 2)), 1), 3), add(multiply(12, 1), multiply(10, 2)))) | in a class of 40 students , 2 students did not borrow any books from the library , 12 students each borrowed 1 book , 10 students each borrowed 2 books , and the rest borrowed at least 3 books . if the average number of books per student was 2 , what is the maximum number of books any single student could have borrowed ? | "total # of students = 40 avg # of books per student = 2 total # of books = 40 * 2 = 80 # of student borrowed at least 3 books = 40 - 2 - 12 - 10 = 16 # of books borrowed by above 16 students = 80 - ( 12 * 1 ) - ( 10 * 2 ) = 48 considering that 15 out of above 16 students borrowed only 3 books , # of books borrowed = 15 * 3 = 45 so maximum # of books borrowed by any single student = 48 - 45 = 3 option a" | a = 40 * 2
b = 12 * 1
c = b + 10
d = c + 2
e = 40 - d
f = e - 1
g = f * 3
h = 12 * 1
i = 10 * 2
j = h + i
k = g + j
l = a - k
|
a ) 14.3 % , b ) 16.67 % , c ) 33 % , d ) 28.6 % , e ) 49.67 % | e | multiply(multiply(divide(divide(divide(2, 7), 2), add(divide(divide(2, 7), 2), subtract(const_1, divide(2, 7)))), const_100), const_3) | of the 20210 employees of the anvil factory , 2 / 7 are journeymen . if half of the journeymen were laid off , what percentage of the total remaining employees would be journeymen ? | the exam gives us a number that is easily divisible by 7 to pique our curiosity and tempt us into calculating actual numbers ( also because otherwise the ratio would be incorrect ) . since the question is about percentages , the actual numbers will be meaningless , as only the ratio of that number versus others will be meaningful . nonetheless , for those who are curious , each 1 / 7 portion represents ( 14210 / 7 ) 2,030 employees . this in turn means that 4,060 employees are journeymen and the remaining 10,150 are full time workers . if half the journeymen were laid off , that would mean 1 / 7 of the total current workforce would be removed . this statistic is what leads many students to think that since half the journeymen are left , the remaining journeymen would represent half of what they used to be , which means 1 / 7 of the total workforce . if 1 / 7 of the workforce is journeymen , and 1 / 7 is roughly 14.3 % , then answer choice a should be the right answer . in this case , though , it is merely the tempting trap answer choice . what changed between the initial statement and the final tally ? well , you let go of 1 / 7 of the workforce , so the total number of workers went down . the remaining workers are still 1 / 7 of the initial workers , but the group has changed . the new workforce is smaller than the original group , specifically 6 / 7 of it because 1 / 7 was eliminated . the remaining workers now account for 1 / 7 out of 6 / 7 of the force , which if we multiply by 7 gives us 1 out of 6 . this number as a percentage is answer choice b , 49.67 % . using the absolute numbers we calculated before , there were 4,060 journeymen employees out of 14,210 total . if 2,030 of them are laid off , then there are 2,030 journeyman employees left , but now out of a total of ( 14,210 - 2,030 ) 12,180 employees . 2,030 / 12,180 is exactly 1 / 6 , or 16.67 % . the answer will work with either percentages or absolute numbers , but the percentage calculation will be significantly faster and applicable to any similar situation . the underlying principle of percentages ( and , on a related note , ratios ) can be summed up in the brainteaser i like to ask my students : if you β re running a race and you overtake the 2 nd place runner just before the end , what position do you end up in ? the correct answer is 2 nd place . percentages , like ratios and other concepts of relative math , depend entirely on the context . whether 100 % more of something is better than 50 % more of something else depends on the context much more than the percentages quoted . when it comes to percentages on the gmat , the goal is to understand them enough to instinctively not fall into the traps laid out for you . e | a = 2 / 7
b = a / 2
c = 2 / 7
d = c / 2
e = 2 / 7
f = 1 - e
g = d + f
h = b / g
i = h * 100
j = i * 3
|
a ) 26 , b ) 22 , c ) 25 , d ) 27 , e ) 33 | e | add(add(multiply(2, 13), 5), 2) | find the total number of prime factors in the expression ( 4 ) ^ 13 x ( 7 ) ^ 5 x ( 11 ) ^ 2 | ( 4 ) ^ 13 x ( 7 ) ^ 5 x ( 11 ) ^ 2 = ( 2 x 2 ) ^ 13 x ( 7 ) ^ 5 x ( 11 ) ^ 2 = 2 ^ 13 x 2 ^ 13 x 7 ^ 5 x 11 ^ 2 = 2 ^ 26 x 7 ^ 5 x 11 ^ 2 total number of prime factors = ( 26 + 5 + 2 ) = 33 . answer is e . | a = 2 * 13
b = a + 5
c = b + 2
|
a ) 724 m , b ) 704 m , c ) 915.2 m , d ) 278 m , e ) 927 m | c | divide(multiply(multiply(multiply(divide(add(multiply(add(const_3, const_4), const_3), const_1), add(const_3, const_4)), 22.4), const_2), 650), const_100) | the radius of a wheel is 22.4 cm . what is the distance covered by the wheel in making 650 resolutions ? | "in one resolution , the distance covered by the wheel is its own circumference . distance covered in 650 resolutions . = 650 * 2 * 22 / 7 * 22.4 = 91520 cm = 915.2 m answer : c" | a = 3 + 4
b = a * 3
c = b + 1
d = 3 + 4
e = c / d
f = e * 22
g = f * 2
h = g * 650
i = h / 100
|
a ) 20 , b ) 25 , c ) 27 , d ) 22 , e ) 91 | a | divide(subtract(22, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | a man is 22 years older than his son . in two years , his age will be twice the age of his son . the present age of the son is ? | "let the son ' s present age be x years . then , man ' s present age = ( x + 22 ) years . ( x + 22 ) + 2 = 2 ( x + 2 ) x + 24 = 2 x + 4 = > x = 20 . answer : a" | a = 2 * 2
b = a - 2
c = 22 - b
d = 2 - 1
e = c / d
|
a ) 150 m , b ) 100 m , c ) 500 m , d ) 180 m , e ) 190 m | c | multiply(subtract(divide(multiply(2, const_1000), multiply(2, const_60)), divide(multiply(2, const_1000), add(multiply(2, const_60), 40))), multiply(2, const_60)) | a can run 2 km distance in 2 min . , while b can run this distance in 2 min 40 sec . by how much distance can a beat b ? | a takes time 2 minutes = 120 sec b takes time 2.40 minutes = 160 sec diffrence = 160 - 120 = 40 sec now we are to find distance covered in 40 sec by b 160 sec = 2000 m 1 sec = 25 / 2 m 40 sec = 40 x 25 / 2 = 500 m answer : c | a = 2 * 1000
b = 2 * const_60
c = a / b
d = 2 * 1000
e = 2 * const_60
f = e + 40
g = d / f
h = c - g
i = 2 * const_60
j = h * i
|
a ) 5 miles , b ) 7 miles , c ) 9 miles , d ) 10 miles , e ) 12 miles | b | multiply(divide(subtract(20, multiply(24, divide(add(2, 5), const_60))), add(5, add(2, 5))), 5) | stacy and heather are 20 miles apart and walk towards each other along the same route . stacy walks at constant rate that is 2 mile per hour faster than heather ' s constant rate of 5 miles / hour . if heather starts her journey 24 minutes after stacy , how far from the original destination has heather walked when the two meet ? | "original distance between s and h = 20 miles . speed of s = 5 + 2 = 7 mph , speed of h = 5 mph . time traveled by h = t hours - - - > time traveled by s = t + 24 / 60 = t + 2 / 5 hours . now , the total distances traveled by s and h = 20 miles - - - > 7 * ( t + 2 / 5 ) + 5 * t = 20 - - - > t = 86 / 60 hours . thus h has traveled for 86 / 60 hours giving you a total distance for h = 5 * 86 / 60 = 7 miles . b is thus the correct answer . p . s . : based on the wording of the question , you should calculatehow far from theoriginal destination has heather walkedwhen the two meet . ' original destination ' for h does not make any sense . original destination for h was situated at a distance of 20 miles ." | a = 2 + 5
b = a / const_60
c = 24 * b
d = 20 - c
e = 2 + 5
f = 5 + e
g = d / f
h = g * 5
|
a ) 82 m , b ) 50 m , c ) 72 m , d ) 80 m , e ) none of these | b | divide(multiply(divide(multiply(subtract(46, 36), const_1000), const_3600), 36), const_2) | two trains of equal length are running on parallel lines in the same directions at 46 km / hr . and 36 km / hr . the faster trains pass the slower train in 36 seconds . the length of each train is : | "explanation : the relative speed of train is 46 - 36 = 10 km / hr = ( 10 x 5 ) / 18 = 25 / 9 m / s 10 Γ 518 = 259 m / s in 36 secs the total distance travelled is 36 x 25 / 9 = 100 m . therefore the length of each train is = 100 / 2 = 50 m . answer b" | a = 46 - 36
b = a * 1000
c = b / 3600
d = c * 36
e = d / 2
|
a ) 346 , b ) 368 , c ) 350 , d ) 337 , e ) 639 | c | divide(1400, const_3) | divide rs . 1400 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 = 1400 a = 1 / 3 ( b + c ) ; b = 2 / 3 ( a + c ) a / ( b + c ) = 1 / 3 a = 1 / 4 * 1400 = > 350 answer : c" | a = 1400 / 3
|
a ) 15 , b ) 26 , c ) 8 , d ) 91 , e ) none of these | c | floor(sqrt(72)) | if the sum of a number and its square is 72 , what is the number ? | "let the number be x . then , x + x 2 = 72 ( x + 9 ) ( x - 8 ) = 0 x = 8 answer : c" | a = math.sqrt(72)
b = math.floor(a)
|
a ) 140 , b ) 175 , c ) 210 , d ) 70 , e ) 280 | d | subtract(30, subtract(20, 30)) | sarah is driving to the airport . after driving at 20 miles per hour for one hour , she realizes that if she continues at that same average rate she will be an hour late for her flight . she then travels 50 miles per hour for the rest of the trip , and arrives 30 minutes before her flight departs . how many miles did she drive in all ? | "after driving at 20 miles per hourfor one hour , this distance left to cover is d - 20 . say this distance is x miles . now , we know that the difference in time between covering this distance at 20 miles per hour and 50 miles per hour is 1 + 1 / 2 = 3 / 2 hours . so , we have that x / 20 - x / 50 = 3 / 2 - - > 5 x / 100 - 2 x / 100 = 3 / 2 - - > 3 x / 100 = 3 / 2 - - > x = 50 . total distance = x + 20 = 70 miles . answer : d" | a = 20 - 30
b = 30 - a
|
a ) 288 , b ) 267 , c ) 261 , d ) 211 , e ) 210 | e | multiply(add(13, const_1), add(add(13, const_1), const_1)) | there are 13 stations between hyderabad and bangalore . how many second class tickets have to be printed , so that a passenger can travel from any station to any other station ? | "the total number of stations = 15 from 15 stations we have to choose any two stations and the direction of travel ( i . e . , hyderabad to bangalore is different from bangalore to hyderabad ) in 15 p β ways . 15 p β = 15 * 14 = 210 . answer : e" | a = 13 + 1
b = 13 + 1
c = b + 1
d = a * c
|
a ) 36 , b ) 63 , c ) 34 , d ) 30 , e ) 33 | a | subtract(multiply(4, const_1000), 4) | the difference between the local value and the face value of 4 in the numeral 98625147 is | "explanation : ( local value of 4 ) - ( face value of 4 ) = ( 40 - 4 ) = 36 a )" | a = 4 * 1000
b = a - 4
|
a ) 1.5 kmph , b ) 2 kmph , c ) 3 kmph , d ) 2.5 kmph , e ) 3.5 kmph | a | divide(subtract(multiply(divide(1, 12), const_60), multiply(divide(1, 30), const_60)), const_2) | a boat moves upstream at the rate of 1 km in 30 minutes and down stream 1 km in 12 minutes . then the speed of the current is : | "rate upstream = ( 1 / 30 * 60 ) = 2 kmph rate down stream = 1 / 12 * 60 = 5 kmph rate of the current = Β½ ( 5 - 2 ) = 1.5 kmph answer : a" | a = 1 / 12
b = a * const_60
c = 1 / 30
d = c * const_60
e = b - d
f = e / 2
|
a ) 45 , b ) 60 , c ) 75 , d ) 80 , e ) 100 | b | divide(divide(3, 5), multiply(divide(divide(divide(3, 4), 5), 30), const_2)) | if five machines working at the same rate can do 3 / 4 of a job in 30 minutes , how many minutes would it take two machines working at the same rate to do 3 / 5 of the job ? | using the std formula m 1 d 1 h 1 / w 1 = m 2 d 2 h 2 / w 2 substituting the values we have 5 * 1 / 2 * 4 / 3 = 2 * 5 / 3 * x ( converted 30 min into hours = 1 / 2 ) 10 / 3 = 10 / 3 * x x = 1 hour so 60 minutes answer : b | a = 3 / 5
b = 3 / 4
c = b / 5
d = c / 30
e = d * 2
f = a / e
|
a ) 1 / 6 , b ) 3 / 10 , c ) 1 / 3 , d ) 1 / 2 , e ) 2 / 3 | c | divide(10, multiply(3, const_10)) | a circular racetrack is 3 miles in length and has signs posted to indicate each 1 / 10 mile increment . if a race car starts at a random location on the track and travels exactly one half mile , what is the probability that the car ends within a half mile of the sign indicating 2 1 / 2 miles ? | the car ends within a half mile of the sign indicating 2 1 / 2 miles means that the car will end in one mile interval , between the signs indicating 2 and 3 miles . now , it does n ' t matter where the car starts or what distance it travels , the probability will be p = ( favorable outcome ) / ( total # of outcomes ) = 1 / 3 ( as the car starts at random point end travels some distance afterwards we can consider its end point as the point where he randomly appeared , so the probability that the car appeared within 1 mile interval out of total 3 miles will be 1 / 3 ) . answer : c . | a = 3 * 10
b = 10 / a
|
a ) 12.5 % , b ) 11 % , c ) 13 % , d ) 15 % , e ) 12.8 % | a | multiply(divide(subtract(2400, 2100), 2400), const_100) | the cost price of a radio is rs . 2400 and it was sold for rs . 2100 , find the loss % ? | "2400 - - - - 300 100 - - - - ? = > 12.5 % answer : a" | a = 2400 - 2100
b = a / 2400
c = b * 100
|
a ) 25 , 5 , b ) 30 , 10 , c ) 35 , 15 , d ) 50 , 30 , e ) none of these | b | subtract(add(divide(multiply(20, 5), subtract(5, const_1)), 5), 20) | the ages of two persons differ by 20 years . if 5 years ago , the older one be 5 times as old as the younger one , then their present ages , in years , are | "let the ages be β x β and β y β years now . then x - y = 20 . . . ( i ) ( x - 5 ) = 5 ( y - 5 ) . . . ( ii ) from ( i ) and ( ii ) , 20 + y - 5 = 5 y - 25 y = 10 and x = 30 . answer : b" | a = 20 * 5
b = 5 - 1
c = a / b
d = c + 5
e = d - 20
|
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