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 ) 36 kmh , b ) 40 kmh , c ) 42 kmh , d ) 44 kmh , e ) 34 kmh | e | divide(const_3, add(add(divide(const_1, 80), divide(const_1, 24)), divide(const_1, 30))) | if a car went the first third of the distance at 80 kmh , the second third at 24 kmh , and the last third at 30 kmh , what was the average speed of the car for the entire trip ? | "assume d / 3 = 240 ( this number is convenient because it is divisible by 80 , 24 and 30 ) so : 240 = 80 * t 1 = 3 hrs 240 = 24 * t 2 = 10 hrs 240 = 30 * t 3 = 8 hrs t = t 1 + t 2 + t 3 = 21 hrs d = rt ( 240 * 3 ) = r * 21 r = 34 answer : e" | a = 1 / 80
b = 1 / 24
c = a + b
d = 1 / 30
e = c + d
f = 3 / e
|
a ) 22 sec , b ) 77 sec , c ) 25 sec , d ) 14.3 sec , e ) 17 sec | d | divide(add(150, 100), multiply(63, const_0_2778)) | how many seconds will a train 100 meters long take to cross a bridge 150 meters long if the speed of the train is 63 kmph ? | "d = 100 + 150 = 250 s = 63 * 5 / 18 = 17.5 mps t = 250 / 17.5 = 14.3 sec answer : d" | a = 150 + 100
b = 63 * const_0_2778
c = a / b
|
a ) 2 % , b ) 17 % , c ) 20 % , d ) 65 % , e ) 83 % | c | multiply(divide(subtract(385, 320), 320), const_100) | a store reported total sales of $ 385 million for february of this year . if the total sales for the same month last year was $ 320 million , approximately what was the percent increase i in sales ? | "last year ' s sales = $ 320 million ; this year ' s sales = $ 385 million ; increase i = $ 65 million . now , 20 % of $ 320 million is $ 64 million , which is very close to actual increase of $ 65 million . answer : c ." | a = 385 - 320
b = a / 320
c = b * 100
|
a ) 3600 , b ) 3607 , c ) 3608 , d ) 3602 , e ) 3000 | e | subtract(15000, multiply(divide(4, 5), 15000)) | income and expenditure of a person are in the ratio 5 : 4 . if the income of the person is rs . 15000 , then find his savings ? | "let the income and the expenditure of the person be rs . 5 x and rs . 4 x respectively . income , 5 x = 15000 = > x = 3000 savings = income - expenditure = 5 x - 4 x = x so , savings = rs . 3000 . answer : e" | a = 4 / 5
b = a * 15000
c = 15000 - b
|
a ) $ 5000 , b ) $ 7000 , c ) $ 9000 , d ) $ 11,000 , e ) $ 13,000 | d | divide(add(divide(subtract(280, multiply(divide(6, const_100), 1,000)), subtract(divide(8, const_100), divide(6, const_100))), divide(subtract(280, multiply(divide(6, const_100), 1,000)), subtract(divide(8, const_100), divide(6, const_100)))), 1,000) | salesperson a ' s compensation for any week is $ 280 plus 6 percent of the portion of a ' s total sales above $ 1,000 for that week . salesperson b ' s compensation for any week is 8 percent of b ' s total sales for that week . for what amount of total weekly sales would both salespeople earn the same compensation ? | "280 + 0.06 ( x - 1000 ) = 0.08 x 0.02 x = 220 x = $ 11,000 the answer is d ." | a = 6 / 100
b = a * 1
c = 280 - b
d = 8 / 100
e = 6 / 100
f = d - e
g = c / f
h = 6 / 100
i = h * 1
j = 280 - i
k = 8 / 100
l = 6 / 100
m = k - l
n = j / m
o = g + n
p = o / 1
|
a ) 13 , b ) 12 , c ) 11 , d ) 10 , e ) 9 | d | divide(add(10, 10), multiply(const_60, subtract(divide(const_1, 5), divide(const_1, 6)))) | rashmi walks to her bus stop at 5 kmph and reaches there late by 10 minutes . on the next day , she increases her speed to 6 kmph and reaches the bus stop 10 minutes early . how far is the bus stop ? | let the distance to the bus stop is ` ` d ' ' km time taken by her ( at 5 kmph ) = d / 5 hour , and time taken by her ( at 6 kmph ) = d / 6 hour as per given condition , d / 5 - d / 6 = 20 / 60 d = 10 kms answer : d | a = 10 + 10
b = 1 / 5
c = 1 / 6
d = b - c
e = const_60 * d
f = a / e
|
a ) s . 9621 , b ) s . 6921 , c ) s . 8820 , d ) s . 6261 , e ) s . 6361 | c | multiply(8000, power(add(const_1, divide(5, const_100)), 2)) | the amount of principal rs . 8000 at compound interest at the ratio of 5 % p . a . for 2 years is | "c . i = p ( 1 + r / 100 ) ^ n = 8000 ( 1 + 5 / 100 ) ^ 2 = rs 8820 answer : c" | a = 5 / 100
b = 1 + a
c = b ** 2
d = 8000 * c
|
a ) 49 , b ) 50 , c ) 53.5 , d ) 55 , e ) 57 | a | add(multiply(const_2, add(multiply(add(divide(25, const_100), 1), 4), multiply(add(divide(25, const_100), 1), 6))), multiply(4, 6)) | a cistern 6 m long and 4 m wide contains water up to a depth of 1 m 25 cm . the total area of the wet surface is : | "explanation : area of the wet surface = [ 2 ( lb + bh + lh ) - lb ] = 2 ( bh + lh ) + lb = [ 2 ( 4 x 1.25 + 6 x 1.25 ) + 6 x 4 ] m 2 = 49 m 2 answer : a" | a = 25 / 100
b = a + 1
c = b * 4
d = 25 / 100
e = d + 1
f = e * 6
g = c + f
h = 2 * g
i = 4 * 6
j = h + i
|
a ) 1216 , b ) 1723 , c ) 1129 , d ) 2613 , e ) 1372 | a | divide(152, subtract(inverse(const_2), subtract(const_1, divide(62.5, const_100)))) | all the milk in container a which was filled to its brim was poured into two containers b and c . the quantity of milk in container b was 62.5 % less than the capacity of container a . if 152 liters was now transferred from c to b , then both the containers would have equal quantities of milk . what was the initial quantity of milk in container a ? | a b has 62.5 % or ( 5 / 8 ) of the milk in a . therefore , let the quantity of milk in container a ( initially ) be 8 k . quantity of milk in b = 8 k - 5 k = 3 k . quantity of milk in container c = 8 k - 3 k = 5 k container : a b c quantity of milk : 8 k 3 k 5 k it is given that if 152 liters was transferred from container c to container b , then both the containers would have equal quantities of milk . 5 k - 152 = 3 k + 152 = > 2 k = 304 = > k = 152 the initial quantity of milk in a = 8 k = 8 * 152 = 1216 liters . | a = 1/(2)
b = 62 / 5
c = 1 - b
d = a - c
e = 152 / d
|
a ) 70.11 , b ) 72.02 , c ) 86.5 , d ) 78.2 , e ) none of the above | c | divide(add(add(multiply(85, 7), multiply(60, 4)), 30), add(7, 4)) | the average expenditure of a labourer for 7 months was 85 and he fell into debt . in the next 4 months by reducing his monthly expenses to 60 he not only cleared off his debt but also saved 30 . his monthly income is | "income of 7 months = ( 7 Γ 85 ) β debt = 595 β debt income of the man for next 4 months = 4 Γ 60 + debt + 30 = 270 + debt β΄ income of 10 months = 865 average monthly income = 865 Γ· 10 = 86.5 answer c" | a = 85 * 7
b = 60 * 4
c = a + b
d = c + 30
e = 7 + 4
f = d / e
|
a ) 2 a . m , b ) 6 a . m , c ) 9 a . m , d ) 5 a . m , e ) 1 a . m | c | add(6, subtract(const_4, const_1)) | a man leaves a point p at 6 a . m . and reaches the point q at 10 a . m . another man leaves the point give at 8 a . m . and reaches the point p at 12 noon . at what time do they meet ? | 9 a . m . answer : c | a = 4 - 1
b = 6 + a
|
a ) 25 % , b ) 35 % , c ) 45 % , d ) 55 % , e ) 65 % | b | multiply(35, const_1) | 35 % of major airline companies equip their planes with wireless internet access . 70 % of major airlines offer passengers free on - board snacks . what is the greatest possible percentage of major airline companies that offer both wireless internet and free on - board snacks ? | "to maximize the percentage of companies offering both , let ' s assume that all 35 % of companies which offer wireless internet also offer snacks . the answer is b ." | a = 35 * 1
|
a ) 156 , b ) 208 , c ) 210 , d ) 223 , e ) 229 | a | divide(add(150, subtract(multiply(59, 46), multiply(58, subtract(46, const_2)))), const_2) | the batting average of a particular batsman is 59 runs in 46 innings . if the difference in his highest and lowest score is 150 runs and his average excluding these two innings is 58 runs , find his highest score . | "explanation : total runs scored by the batsman = 59 * 46 = 2714 runs now excluding the two innings the runs scored = 58 * 44 = 2552 runs hence the runs scored in the two innings = 2714 β 2552 = 162 runs . let the highest score be x , hence the lowest score = x β 150 x + ( x - 150 ) = 162 2 x = 312 x = 156 runs answer a" | a = 59 * 46
b = 46 - 2
c = 58 * b
d = a - c
e = 150 + d
f = e / 2
|
a ) $ 900 , b ) $ 300 , c ) $ 500 , d ) $ 600 , e ) $ 1000 | a | multiply(3000, divide(3, const_100)) | find the simple interest on $ 3000 for 3 years at 10 % per annum ? | "si = ptr / 100 = 3000 * 3 * 10 / 100 = $ 900 answer is a" | a = 3 / 100
b = 3000 * a
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | e | divide(12, const_2) | if 9 ^ y = 3 ^ 12 , what is y ? | "3 ^ 2 y = 3 ^ 12 2 y = 12 , therefore y = 6 e" | a = 12 / 2
|
a ) 280 , b ) 320 , c ) 360 , d ) 400 , e ) 480 | d | divide(150, subtract(1, add(divide(1, 2), multiply(divide(1, 2), divide(1, 4))))) | during a car trip , maria stopped to rest after she traveled 1 / 2 of the total distance to her destination . she stopped again after she traveled 1 / 4 of the distance remaining between her first stop and her destination , and then she drove the remaining 150 miles to her detination . what was the total distance , in miles from maria ' s starting point to her destination ? | "easy 400 is the answer . 3 / 4 ( x / 2 ) = 150 x = 150 * 8 / 3 = 400 . answer d" | a = 1 / 2
b = 1 / 2
c = 1 / 4
d = b * c
e = a + d
f = 1 - e
g = 150 / f
|
a ) 187 km , b ) 480 km , c ) 278 km , d ) 120 km , e ) 671 km | d | divide(1, 2) | with a uniform speed a car covers the distance in 8 hours . had the speed been increased by 1 km / hr , the same distance could have been covered in 7 1 / 2 hours . what is the distance covered ? | "let the distance be x km . then , x / ( 7 1 / 2 ) - x / 8 = 1 2 x / 15 - x / 8 = 1 = > x = 120 km . answer : d" | a = 1 / 2
|
a ) 18 , b ) 24 , c ) 30 , d ) 36 , e ) 42 | d | subtract(divide(multiply(subtract(multiply(5, add(const_10, 6)), multiply(2, add(const_10, 6))), 7), add(2, 2)), 6) | the ratio of the present age of sandy to that of molly is 7 : 2 . eight years from now , the ratio of the ages of sandy to molly will be 5 : 2 . what was sandy ' s age 6 years ago ? | "let the present age of sandy be 7 x years and that of molly be 2 x years . ( 7 x + 8 ) / ( 2 x + 8 ) = 5 / 2 4 x = 24 x = 6 six years ago , sandy ' s age was 7 ( 6 ) - 6 = 36 the answer is d ." | a = 10 + 6
b = 5 * a
c = 10 + 6
d = 2 * c
e = b - d
f = e * 7
g = 2 + 2
h = f / g
i = h - 6
|
a ) 233 , b ) 600 , c ) 287 , d ) 780 , e ) 191 | d | subtract(multiply(420, 9), subtract(multiply(430, 9), 870)) | the average monthly salary of 8 workers and one supervisor in a factory was 430 . @ sswhen @ ssthe @ sssupervisor @ cc @ sswhose @ sssalary @ sswas @ ss 430 . @ sswhen @ ssthe @ sssupervisor @ cc @ sswhose @ sssalary @ sswas @ ss 430 . whenthesupervisor , whosesalarywas 430 . when the supervisor , whose salary was 870 per month , retired , a new person was appointed and then the average salary of 9 people was $ $ 420 per month . the salary of the new supervisor is : | "explanation : total salary of 8 workers and supervisor together = 9 Γ£ β 430 = 3870 now total salary of 8 workers = 3870 Γ’ Λ β 870 = 3000 total salary of 9 workers including the new supervisor = 9 Γ£ β 420 = 3780 salary of the new supervisor = 3780 Γ’ Λ β 3000 = 780 answer : d" | a = 420 * 9
b = 430 * 9
c = b - 870
d = a - c
|
a ) 4.2 , b ) 4.4 , c ) 4.6 , d ) 5.6 , e ) 5.7 | b | divide(subtract(multiply(6, 3.95), add(multiply(2, 3.6), multiply(2, 3.85))), 2) | the average of 6 no . ' s is 3.95 . the average of 2 of them is 3.6 , while the average of the other 2 is 3.85 . what is the average of the remaining 2 no ' s ? | "sum of the remaining two numbers = ( 3.95 * 6 ) - [ ( 3.6 * 2 ) + ( 3.85 * 2 ) ] = 8.8 . required average = ( 8.8 / 2 ) = 4.4 . b" | a = 6 * 3
b = 2 * 3
c = 2 * 3
d = b + c
e = a - d
f = e / 2
|
a ) 7.19 , b ) 7.17 , c ) 7.62 , d ) 7.15 , e ) 7.11 | c | divide(add(121, 165), multiply(add(80, 55), const_0_2778)) | two trains 121 meters and 165 meters in length respectively are running in opposite directions , one at the rate of 80 km and the other at the rate of 55 kmph . in what time will they be completely clear of each other from the moment they meet ? | "t = ( 121 + 165 ) / ( 80 + 55 ) * 18 / 5 t = 7.62 answer : c" | a = 121 + 165
b = 80 + 55
c = b * const_0_2778
d = a / c
|
a ) 5 , b ) 10 , c ) 15 , d ) 20 , e ) 25 | d | multiply(divide(add(24, 25), const_2), divide(4, add(4, 1))) | 24 oz of juice p and 25 oz of juice q are mixed to make smothies x and y . the ratio of p to q in smothie x is 4 is to 1 and that in y is 1 is to 5 . how many ounces of juice p are contained in the smothie x ? | easy way to solve this question is start from the answer and then conform the information provided in the question . we can start from option d i . e 20 . . . as a quantity of juice p in x because it is the only one option that gets divided by 4 is 20 . . . since in the x the juice p to q ratio is 4 : 1 this gives us that quantity of juice p in x = 20 therefore quantity of juice v will be 5 . . . hence ratio = 4 : 1 this will lead to quantity of juice p in x = 4 and quantity of juice v = 20 . . . hence ratio 1 : 5 if we calculate total juice p = 24 and total of juice q = 25 it fits because totals are same as what mentioned in the question . . . thus ans is d | a = 24 + 25
b = a / 2
c = 4 + 1
d = 4 / c
e = b * d
|
a ) 72 % , b ) 81 % , c ) 52 % , d ) 50 % , e ) 28 % | b | subtract(90, multiply(divide(90, const_100), 10)) | a shirt goes on sale for 90 % of its original price . one week later , the sale price is marked down 10 % . the final price is what percent of the original price ? | "just assume original price is 100 . sale price = 90 then it is marked down by 10 % = 90 - 9 = 81 . hence it is 81 % od the original price . hence answer is b ." | a = 90 / 100
b = a * 10
c = 90 - b
|
a ) 4 % , b ) 6 % , c ) 8 % , d ) 10 % , e ) 12 % | d | divide(subtract(const_1, divide(80, const_100)), divide(2, const_100)) | in a certain parking lot , 2 % of the cars are towed for parking illegally . however 80 % of the cars which are parked illegally are not towed . what percentage of cars in the parking lot are parked illegally ? | let x be the number of cars and let y be the number of cars parked illegally . 2 % * x = 20 % * y y / x = 1 / 10 = 10 % the answer is d . | a = 80 / 100
b = 1 - a
c = 2 / 100
d = b / c
|
a ) 27 , b ) 48 , c ) 45 , d ) 72 , e ) 18 | b | subtract(multiply(multiply(5, 5), divide(145, add(add(multiply(3, 3), multiply(3, 5)), multiply(5, 5)))), multiply(multiply(3, 3), divide(145, add(add(multiply(3, 3), multiply(3, 5)), multiply(5, 5))))) | the ages of patrick and michael are in the ratio of 3 : 5 and that of michael and monica are in the ratio of 3 : 5 . if the sum of their ages is 145 , what is the difference between the ages of patrick and monica ? | "ages of p and mi = 3 x + 5 x ages of mi and mo = 3 x : 5 x rationalizing their ages . ratio of their ages will be 9 x : 15 x : 25 x sum = 47 x = 145 x = 3 difference if ages of pa and mo = 25 x - 9 x = 16 x = 16 * 3 = 48 answer b" | a = 5 * 5
b = 3 * 3
c = 3 * 5
d = b + c
e = 5 * 5
f = d + e
g = 145 / f
h = a * g
i = 3 * 3
j = 3 * 3
k = 3 * 5
l = j + k
m = 5 * 5
n = l + m
o = 145 / n
p = i * o
q = h - p
|
a ) 1 / 4 , b ) 1 / 2 , c ) 2 / 5 , d ) 1 , e ) 5 / 4 | c | subtract(divide(subtract(41, const_1), add(39, const_1)), divide(add(29, const_1), subtract(51, const_1))) | a is an integer greater than 29 but less than 41 , b is an integer greater than 39 but less than 51 , what is the range of a / b ? | "min value of a / b will be when b is highest and a is lowest - - - > a = 30 and b = 50 so , a / b = 3 / 5 max value of a / b will be when b is lowest and a is highest - - - > a = 40 and b = 40 so , a / b = 1 range is 1 - ( 3 / 5 ) = 2 / 5 answer should be c ." | a = 41 - 1
b = 39 + 1
c = a / b
d = 29 + 1
e = 51 - 1
f = d / e
g = c - f
|
a ) 7787 , b ) 10000 , c ) 1988 , d ) 1277 , e ) 2081 | b | divide(divide(9975, subtract(const_1, divide(5, const_100))), add(const_1, divide(5, const_100))) | in one year , the population , of a village increased by 5 % and in the next year , it decreased by 5 % . if at the end of 2 nd year , the population was 9975 , what was it in the beginning ? | "x * 105 / 100 * 95 / 100 = 9975 x * 0.9975 = 9975 x = 9975 / 0.9975 = > 10000 answer : b" | a = 5 / 100
b = 1 - a
c = 9975 / b
d = 5 / 100
e = 1 + d
f = c / e
|
a ) 3 , b ) 6 , c ) 9 , d ) 12 , e ) 15 | a | divide(multiply(const_12, log(2)), log(2)) | if 2 ^ ( 2 w ) = 8 ^ ( w β 1 ) , what is the value of w ? | "2 ^ ( 2 w ) = 8 ^ ( w β 1 ) 2 ^ ( 2 w ) = 2 ^ ( 3 * ( w β 1 ) ) 2 ^ ( 2 w ) = 2 ^ ( 3 w - 3 ) let ' s equate the exponents as the bases are equal . 2 w = 3 w - 3 w = 3 the answer is a ." | a = math.log(2)
b = 12 * a
c = math.log(2)
d = b / c
|
a ) 6 km , b ) 3 km , c ) 7 km , d ) 9 km , e ) 2 km | a | multiply(multiply(divide(divide(47, const_60), add(add(divide(const_1, 7), divide(const_1, 8)), divide(const_1, 9))), const_3), const_1000) | a person travels equal distances with speeds of 7 km / hr , 8 km / hr and 9 km / hr and takes a total time of 47 minutes . the total distance is ? | "let the total distance be 3 x km . then , x / 7 + x / 8 + x / 9 = 47 / 60 36 x / 95 = 47 / 60 = > x = 2.07 . total distance = 3 * 2.07 = 6.2 km . answer : a" | a = 47 / const_60
b = 1 / 7
c = 1 / 8
d = b + c
e = 1 / 9
f = d + e
g = a / f
h = g * 3
i = h * 1000
|
a ) 7580 , b ) 12960 , c ) 8290 , d ) 8640 , e ) none | b | divide(multiply(6, multiply(12, const_60)), subtract(divide(multiply(12, const_60), multiply(9, const_60)), const_1)) | a leak in the bottom of a tank can empty the full tank in 9 hours . an inlet pipe fills water at the rate of 6 litres a minute . when the tank is full , the inlet is opened and due to the leak , the tank is empty in 12 hours . how many litres does the cistern hold ? | "solution work done by the inlet in 1 hour = ( 1 / 9 - 1 / 12 ) = 1 / 36 work done by the inlet in 1 min . = ( 1 / 36 Γ 1 / 60 ) = 0.000463 volume of 0.000463 part = 6 litres . therefore , volume of whole = 1 / 0.000463 x 6 βΉ = βΊ 12960 litres . answer b" | a = 12 * const_60
b = 6 * a
c = 12 * const_60
d = 9 * const_60
e = c / d
f = e - 1
g = b / f
|
a ) 60 , b ) 30 , c ) 12 , d ) 90 , e ) 120 | a | divide(divide(240, const_2), const_2) | if k ^ 2 is divisible by 240 what is the least possible value of integer k ? | "240 can be written as ( 2 ^ 4 ) * 3 * 5 . for k ^ 2 to be divisible by 240 it should contain at least 2 ^ 4 and 3 and 5 in its factors . we can leave out option c because 12 doesnt have 5 as one of its factor . now if we check for option b , 30 can be written as 2 * 3 * 5 , hence 30 ^ 2 will have 2 as the maximum power of 2 , so we can leave out this option too . option c is the right answer if we follow the same method as we followed for other two previous options . 60 = ( 2 ^ 2 ) * 3 * 5 ; 60 ^ 2 = ( 2 ^ 4 ) * ( 3 ^ 2 ) * ( 5 ^ 2 ) . so it shows that 60 ^ 2 is divisible by 240 and hence the answer . answer : a" | a = 240 / 2
b = a / 2
|
a ) 13 % , b ) 9.22 % , c ) 9 % , d ) 14 % , e ) 12 % | b | multiply(subtract(const_1, multiply(subtract(const_1, divide(11, const_100)), add(const_1, divide(2, const_100)))), const_100) | you enter a weight loss challenge game and manage to lose 11 % of your body weight . for the final weigh in you are forced to wear clothes that add 2 % to your weight . what percentage of weight loss is measured at the final weigh in ? | "( 100 % - 11 % ) * ( 100 % + 2 % ) = 0.89 * 1.02 = 9.22 % the weigh in records your weight loss at 9.22 % ! the answer is b" | a = 11 / 100
b = 1 - a
c = 2 / 100
d = 1 + c
e = b * d
f = 1 - e
g = f * 100
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | a | subtract(5, const_2) | in a certain game , a large bag is filled with blue , green , purple and red chips worth 1 , 5 , x and 11 points each , respectively . the purple chips are worth more than the green chips , but less than the red chips . a certain number of chips are then selected from the bag . if the product of the point values of the selected chips is 2200 , how many purple chips were selected ? | "2200 = 1 * 5 ^ 2 * 8 * 11 the factor of 8 must come from the purple point value , so there is 1 purple chip . the answer is a ." | a = 5 - 2
|
a ) 6 : 4 , b ) 6 : 14 , c ) 4 : 4 , d ) 4 : 6 , e ) 3 : 1 | e | divide(divide(subtract(multiply(50, divide(80, const_100)), multiply(50, divide(50, const_100))), subtract(divide(80, const_100), divide(40, const_100))), subtract(50, divide(subtract(multiply(50, divide(80, const_100)), multiply(50, divide(50, const_100))), subtract(divide(80, const_100), divide(40, const_100))))) | solution a is 40 % salt and solution b is 80 % salt . if you have 30 ounces of solution a and 60 ounces of solution b , in what ratio could you mix solution a with solution b to produce 50 ounces of a 50 % salt solution ? | forget the volumes for the time being . you have to mix 20 % and 80 % solutions to get 50 % . this is very straight forward since 50 is int he middle of 20 and 80 so we need both solutions in equal quantities . if this does n ' t strike , use w 1 / w 2 = ( a 2 - aavg ) / ( aavg - a 1 ) w 1 / w 2 = ( 80 - 50 ) / ( 50 - 40 ) = 3 / 1 so the volume of the two solutions will be equal . answer has to be 3 : 1 . e | a = 80 / 100
b = 50 * a
c = 50 / 100
d = 50 * c
e = b - d
f = 80 / 100
g = 40 / 100
h = f - g
i = e / h
j = 80 / 100
k = 50 * j
l = 50 / 100
m = 50 * l
n = k - m
o = 80 / 100
p = 40 / 100
q = o - p
r = n / q
s = 50 - r
t = i / s
|
a ) 220044 , b ) 145778 , c ) 220400 , d ) 220014 , e ) 220040 | e | add(multiply(multiply(add(555, 445), 2), subtract(555, 445)), 40) | a no . when divided by the sum of 555 and 445 gives 2 times their difference as quotient & 40 as remainder . find the no . is ? | "( 555 + 445 ) * 2 * 110 + 30 = 220000 + 30 = 220040 e" | a = 555 + 445
b = a * 2
c = 555 - 445
d = b * c
e = d + 40
|
a ) 8867 , b ) 299 , c ) 279 , d ) 600 , e ) 2612 | d | multiply(subtract(divide(12000, 10000), divide(8000, 10000)), 1500) | a , b and c started a business with capitals of rs . 8000 , rs . 10000 and rs . 12000 respectively . at the end of the year , the profit share of b is rs . 1500 . the difference between the profit shares of a and c is ? | "ratio of investments of a , b and c is 8000 : 10000 : 12000 = 4 : 5 : 6 and also given that , profit share of b is rs . 1500 = > 5 parts out of 15 parts is rs . 1500 now , required difference is 6 - 4 = 2 parts required difference = 2 / 5 ( 1500 ) = rs . 600 . answer : d" | a = 12000 / 10000
b = 8000 / 10000
c = a - b
d = c * 1500
|
a ) 30 , b ) 36 , c ) 42 , d ) 46 , e ) 48 | e | divide(8, subtract(1, add(divide(1, 3), divide(1, 2)))) | a cement mixture is composed of 3 elements . by weight , 1 / 3 of the mixture is sand , 1 / 2 of the mixture is water , and the remaining 8 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 / 3 ) x water content = ( 1 / 2 ) x gravel = x - ( 1 / 3 ) x - ( 1 / 2 ) x = ( 1 / 6 ) x = 8 x = 48 then answer will be e = 48" | a = 1 / 3
b = 1 / 2
c = a + b
d = 1 - c
e = 8 / d
|
a ) 3.3 , b ) 6.6 , c ) 60 , d ) 100 , e ) 110 | a | divide(330, divide(multiply(multiply(10, 330), divide(add(const_100, 10), const_100)), subtract(multiply(330, divide(add(const_100, 10), const_100)), 330))) | machine a and machine b are each used to manufacture 330 sprockets . it takes machine a 10 hours longer to produce 330 sprockets than machine b . machine b produces 10 % more sprockets per hour than machine a . how many sprockets per hour does machineaproduce ? | "time taken by b = t time taken by a = t + 10 qty produced by a = q qty produced by b = 1.1 q for b : t ( 1.1 q ) = 330 qt = 300 for a : ( t + 10 ) ( q ) = 330 qt + 10 q = 330 300 + 10 q = 330 q = 3 so a can produce 3 / hour . then b can produce = 3 ( 1.1 ) = 3.3 / hour . a" | a = 10 * 330
b = 100 + 10
c = b / 100
d = a * c
e = 100 + 10
f = e / 100
g = 330 * f
h = g - 330
i = d / h
j = 330 / i
|
a ) 5 , b ) 7 , c ) 10 , d ) 15 , e ) 18 | b | divide(multiply(2, 14), 4) | albert is 2 times mary β s age and 4 times as old as betty . mary is 14 years younger than albert . how old is betty ? | "a = 2 m = m + 14 m = 14 a = 28 a = 4 b , and so b = 7 the answer is b ." | a = 2 * 14
b = a / 4
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a ) a ) 6.7 , b ) b ) 38.5 , c ) c ) 9.6 , d ) d ) 12.5 , e ) e ) 7.9 | b | divide(subtract(35, 10), subtract(const_1, divide(35, 100))) | how many kg of pure salt must be added to 100 kg of 10 % solution of salt and water to increase it to a 35 % solution ? | "amount salt in 100 kg solution = 10 * 100 / 100 = 10 kg let x kg of pure salt be added then ( 10 + x ) / ( 100 + x ) = 35 / 100 200 + 20 x = 700 + 7 x 13 x = 500 x = 28.5 answer is b" | a = 35 - 10
b = 35 / 100
c = 1 - b
d = a / c
|
a ) 2 , b ) 3 , c ) 4 , d ) 6 , e ) 8 | a | subtract(200, subtract(398, 200)) | when 200 is divided by positive integer x , the remainder is 2 . what is the remainder when 398 is divided by x ? | "f 200 / x leaves a reminder 2 then ( 200 - 2 ) i . e . 198 is divisible by x so ( 200 + 198 ) / x leaves a reminder rem ( 200 / x ) + rem ( 198 / x ) = > 2 + 0 = 2 answer : a" | a = 398 - 200
b = 200 - a
|
a ) 4 liters , b ) 10 liters , c ) 15 liters , d ) 8 liters , e ) 6 liters | a | subtract(12, divide(multiply(12, 40), 60)) | what quantity of water should taken out to concentrate 12 liters of 40 % acidic liquid to 60 % acidic liquid ? | "required answer is = 12 ( 60 - 40 ) / 60 = 4 liters answer is a" | a = 12 * 40
b = a / 60
c = 12 - b
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a ) 26.5 % , b ) 50 % , c ) 46.5 % , d ) 56.5 % , e ) 60 % | e | multiply(divide(subtract(700, add(add(add(50, 82), add(60, 48)), 40)), 700), const_100) | john had a stock of 700 books in his bookshop . he sold 50 on monday , 82 on tuesday , 60 on wednesday , 48 on thursday and 40 on friday . what percentage of the books were not sold ? | "let n be the total number of books sold . hence n = 50 + 82 + 64 + 78 + 135 = 280 let m be the books not sold m = 700 - n = 700 - 280 = 420 percentage books not sold / total number of books = 420 / 700 = 0.60 = 60 % correct answer e" | a = 50 + 82
b = 60 + 48
c = a + b
d = c + 40
e = 700 - d
f = e / 700
g = f * 100
|
a ) 48 seconds , b ) 1 minute , 12 seconds , c ) 1 minute , 36 seconds , d ) 2 minutes , 14 seconds , e ) 4 minutes , 12 seconds | c | add(subtract(const_1, divide(5, 7)), divide(5, 7)) | if it takes a tub 4 minutes to drain 5 / 7 of its content , how much more time will it take for the tub to be empty ? | "if 5 / 7 of tub ' s content is drained 2 / 7 th of tub still needs to be drained . if it takes 4 minutes to drain 5 / 7 th of tub it takes 4 * ( 7 / 5 ) minutes to drain the entire tub and 4 * ( 7 / 5 ) * ( 2 / 7 ) min to drain 2 / 7 th of the tub which is 8 / 5 minutes or 1 3 / 5 minutes or 1 minute 36 seconds so answer is c" | a = 5 / 7
b = 1 - a
c = 5 / 7
d = b + c
|
a ) 92 , b ) 90 , c ) 82 , d ) 78 , e ) 75 | e | divide(add(add(multiply(power(power(1000, divide(const_1, const_3)), const_2), const_4), power(power(1000, divide(const_1, const_3)), const_2)), power(power(1000, divide(const_1, const_3)), const_2)), multiply(4, 2)) | a welder received an order to make a 1 million liter cube - shaped tank . if he has only 4 x 2 meter sheets of metal that can be cut , how many metal sheets will be required for this ordert ? ( 1 cubic meter = 1000 liters ) | i get 75 . a cube with 1 million liters cube would be a cube with the dimensions of 100 * 100 * 100 . 4 * 2 covers 8 sq liters so 100 / 8 = 12.5 . t = 12.5 * 6 = 75 . e | a = 1 / 3
b = 1000 ** a
c = b ** 2
d = c * 4
e = 1 / 3
f = 1000 ** e
g = f ** 2
h = d + g
i = 1 / 3
j = 1000 ** i
k = j ** 2
l = h + k
m = 4 * 2
n = l / m
|
a ) $ 1,150 , b ) $ 1,350 , c ) $ 1,550 , d ) $ 1,750 , e ) $ 1,950 | e | floor(divide(add(divide(85.50, divide(9, const_100)), 1,000), 1,000)) | when a merchant imported a certain item , he paid a 9 percent import tax on the portion of the total value of the item in excess of $ 1,000 . if the amount of the import tax that the merchant paid was $ 85.50 , what was the total value of the item ? | "let x be the value in excess of $ 1,000 . 0.09 x = 85.5 x = $ 950 the total value was $ 950 + $ 1,000 = $ 1,950 . the answer is e ." | a = 9 / 100
b = 85 / 50
c = b + 1
d = c / 1
e = math.floor(d)
|
a ) 3 , b ) 6 , c ) 10 , d ) 12 , e ) 18 | b | divide(subtract(multiply(7, 4), 4), 4) | how many different 3 - digit multiples of 5 can be composed of digits 2 , 7 , 4 , and 5 if none of the digits is repeated ? | = 3 c 2 x 2 = 6 can you explain your method . i solved it differently xyz , where z must be 5 . therefore 1 variation of digit in z . y can be any of the 3 possible choices . x can be any of the 2 possible choices . 2 + 3 + 1 = 6 b | a = 7 * 4
b = a - 4
c = b / 4
|
a ) 11 , b ) 12 , c ) 14 , d ) 15 , e ) 19 | b | subtract(add(19, 21), subtract(38, 10)) | in a class of 38 children , 19 play tennis , 21 play squash and 10 play neither sport . ( 1 ) how many children play both sports ? | 28 play some sport 19 + 21 - both = 28 both = 12 answer is b | a = 19 + 21
b = 38 - 10
c = a - b
|
a ) 72 sec , b ) 24 sec , c ) 44 sec , d ) 62 sec , e ) 21 sec | b | multiply(const_3600, divide(divide(440, const_1000), add(60, 6))) | a train 440 meters long is running with a speed of 60 kmph . in what time will it pass a man who is running at 6 kmph in the direction opposite to that in which the train is going ? | "speed of train relative to man = ( 60 + 6 ) km / hr = 66 km / hr [ 66 * 5 / 18 ] m / sec = [ 55 / 3 ] m / sec . time taken to pass the man = [ 440 * 3 / 55 ] sec = 8 sec answer : b" | a = 440 / 1000
b = 60 + 6
c = a / b
d = 3600 * c
|
a ) 20 % , b ) 30 % , c ) 60 % , d ) 50 % , e ) 33 % | b | divide(multiply(divide(multiply(subtract(13.27, 8.77), 1000), subtract(multiply(subtract(13.27, 8.77), 2000), 1500)), const_100), const_2) | firm a ' s revenues are described by the equation ra = 17.12 x and its costs by ca = 12.87 x + 2400 . firm b ' s revenues are described by the equation rb = 13.27 x and its costs by cb = 8.77 x + 1500 . ( x denotes the quantity of sales ) . if a ' s volume of sales increases from 1000 to 3000 over the same period that b ' s volume increases from 2000 to 3000 , the percent increase in profits for the firm with the biggest dollar increase in profits is | ok , i do not know whether it is exactly what you want to hear but i did it that way : first look at difference between revenue and variable cost = > take the bigger one ( b ) which gives you : 4.5 x - 1500 just plug in 2000 and 3000 . you get 7500 and 12000 ( difference 4500 ) . 4500 > 0.5 * 7500 so 30 % b | a = 13 - 27
b = a * 1000
c = 13 - 27
d = c * 2000
e = d - 1500
f = b / e
g = f * 100
h = g / 2
|
a ) 33 , b ) 87 , c ) 30 , d ) 88 , e ) 77 | c | divide(500, multiply(subtract(63, 3), const_0_2778)) | how many seconds will a 500 meter long train moving with a speed of 63 km / hr take to cross a man walking with a speed of 3 km / hr in the direction of the train ? | "explanation : distance = 500 m speed of the train relative to man = ( 63 - 3 ) km / hr = 60 km / hr = m / s = m / s time taken to pass the man = distance / speed = sec = 30 sec answer : c ) 30" | a = 63 - 3
b = a * const_0_2778
c = 500 / b
|
a ) 12 days , b ) 15 days , c ) 18 days , d ) 21 days , e ) 22 days | c | inverse(subtract(inverse(6), inverse(9))) | a and b together can do a piece of work in 6 days and a alone can do it in 9 days . in how many days can b alone can do it ? | "explanation : a and b can do work 1 / 6 in 1 day a alone can do 1 / 9 work in 1 day b alone can do ( 1 / 6 - 1 / 9 ) = 1 / 18 work in 1 day = > complete work can be done in 18 days by b answer : option c" | a = 1/(6)
b = 1/(9)
c = a - b
d = 1/(c)
|
a ) 0.3 , b ) 0.25 , c ) 0.1 , d ) 0.95 , e ) 0.35 | e | divide(multiply(multiply(sqrt(6), sqrt(6)), 6), const_100) | β 6 percent of 6 β 6 = | "the other solutions are great , but here ' s one more approach . first , we scan the answer choices to see that they are quite spread apart . this means we can also use someapproximation all test - takers must memorize the following approximations : β 2 β 1.4 β 3 β 1.7 β 5 β 2.2 β 6 β 2.4 so , β 6 percent of 6 β 6 β 2.4 percent of ( 6 ) ( 2.4 ) β ( 2.4 / 100 ) ( 14.4 ) β 34.56 / 100 β 0.34 check the answer choices to see that e is the only answer that ' s close . answer : e" | a = math.sqrt(6)
b = math.sqrt(6)
c = a * b
d = c * 6
e = d / 100
|
a ) 24 , b ) 18 , c ) 17 , d ) 16 , e ) 10 | a | divide(divide(1600, 20), divide(1400, multiply(20, 21))) | 20 binders can bind 1400 books in 21 days . how many binders will be required to bind 1600 books in 20 days ? | "binders books days 20 1400 21 x 1600 20 x / 20 = ( 1600 / 1400 ) * ( 21 / 20 ) = > x = 24 answer : a" | a = 1600 / 20
b = 20 * 21
c = 1400 / b
d = a / c
|
a ) 6.3 , b ) 6.9 , c ) 7.1 , d ) 6.1 , e ) 6.2 | c | divide(const_1, add(divide(const_1, 25), add(divide(const_1, 20), divide(const_1, 20)))) | a man can do a job in 20 days . his father takes 20 days and his son finishes it in 25 days . how long will they take to complete the job if they all work together ? | "1 day work of the three persons = ( 1 / 20 + 1 / 20 + 1 / 25 ) = 7 / 50 so , all three together will complete the work in 300 / 47 = 7.1 days . answer : c" | a = 1 / 25
b = 1 / 20
c = 1 / 20
d = b + c
e = a + d
f = 1 / e
|
a ) 3 / 5 , b ) 2 / 5 , c ) 3 / 2 , d ) 3 / 7 , e ) 3 / 8 | b | divide(circle_area(divide(0.4, const_2)), const_2) | what will be the vulgar fraction of 0.4 | "explanation : 0.4 = 40 / 100 = 2 / 5 option b" | a = 0 / 4
b = circle_area / (
|
a ) 5 : 2 , b ) 2 : 3 , c ) 1 : 4 , d ) 3 : 7 , e ) 2 : 5 | a | divide(divide(10, const_100), divide(1, 4)) | in a school 10 % of the boys are same in number as 1 / 4 th of the girls . what is the ratio of boys to the girls in the school ? | "10 % of b = 1 / 4 g 10 b / 100 = g / 4 b = 5 g / 2 b / g = 5 / 2 b : g = 5 : 2 answer is a" | a = 10 / 100
b = 1 / 4
c = a / b
|
['a ) 0.125', 'b ) 0.25', 'c ) 0.5', 'd ) 0.75', 'e ) not enough information to determine the rate'] | e | divide(divide(multiply(840, const_1000), multiply(0.5, const_1000)), const_1000) | the volume of a cistern is 840 cubic meters and water is flowing into the cistern . if the surface level of the water is rising at the rate of 0.5 meters per minute , what is the rate , in cubic meters per minutes , at which the water is flowing into the cistern ? | there are not enough info to answer the question . a 840 cubic meters rectangle is built from : height * length * width . from the question we know the volume of the pool and the filling rate . a pool can have a a height of 10 * width 8.4 * length 10 and have a volume of 840 cubic meters , and it can have a height of 1 meter , width of 100 meters and length of 8.4 . in both cases the pool will fill up in a different rate . the correct answer is e . | a = 840 * 1000
b = 0 * 5
c = a / b
d = c / 1000
|
a ) 31 : 1 , b ) 31 : 4 , c ) 31 : 8 , d ) 31 : 9 , e ) 31 : 2 | d | divide(add(divide(multiply(62.5, 4), const_100), divide(multiply(87.5, 6), const_100)), add(subtract(4, divide(multiply(62.5, 4), const_100)), subtract(6, divide(multiply(87.5, 6), 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 6 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 * 6 = 21 / 4 litres quantity of alcohol in the mixture formed = 5 / 2 + 21 / 4 = 31 / 4 = 7.75 litres as 10 litres of mixture is formed , ratio of alcohol and water in the mixture formed = 7.75 : 2.25 = 31 : 9 . 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 = 6 - j
l = h + k
m = e / l
|
a ) 160 , b ) 80 , c ) 200 , d ) 220 , e ) none | b | divide(add(2, 2), subtract(divide(const_1, 2), divide(const_1, add(const_1, 2)))) | a number whose fifth part increased by 2 is equal to its fourth part diminished by 2 is ? | "answer let the number be n . then , ( n / 5 ) + 2 = ( n / 4 ) - 2 Γ’ β‘ β ( n / 4 ) - ( n / 5 ) = 4 Γ’ β‘ β ( 5 n - 4 n ) / 20 = 4 Γ’ Λ Β΄ n = 80 option : b" | a = 2 + 2
b = 1 / 2
c = 1 + 2
d = 1 / c
e = b - d
f = a / e
|
a ) 30 , b ) 300 , c ) 720 , d ) 14400 , e ) 18000 | d | multiply(multiply(const_3, const_60), const_60) | if an object travels at four feet per second , how many feet does it travel in one hour ? | "speed = 4 feet per second . 1 hour = 60 x 60 seconds = 3600 seconds . total no of feet traveled in 1 hour = 3600 x 4 = 1800 x 10 = 14400 answer d" | a = 3 * const_60
b = a * const_60
|
a ) 230 m , b ) 240 m , c ) 250 m , d ) 260 m , e ) 270 m | e | subtract(multiply(multiply(add(120, 80), const_0_2778), 9), 230) | a 230 metres long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds . what is the length of the other train ? | "speed = ( 120 + 80 ) km / h ( because direction is opposite hence relative velocity is added ) = 500 / 9 m / s time = 9 sec let the lenght of second train is x total distance covered = 230 + x therefore , d = speed * time thus 230 + x = 500 / 9 * 9 x = 500 - 230 = 270 m answer : e" | a = 120 + 80
b = a * const_0_2778
c = b * 9
d = c - 230
|
a ) 25 , b ) 90 , c ) 100 , d ) 140 , e ) it can not be determined from the information given | b | subtract(multiply(90, const_2), multiply(45, const_2)) | if the average ( arithmetic mean ) of a and b is 45 and the average of b and c is 90 , what is the value of c Γ’ Λ β a ? | "the arithmetic mean of a and b = ( a + b ) / 2 = 45 - - a + b = 90 - - 1 similarly for b + c = 180 - - 2 subtracting 1 from 2 we have c - a = 90 ; answer : b" | a = 90 * 2
b = 45 * 2
c = a - b
|
a ) 173 , b ) 685 , c ) 865 , d ) 495 , e ) 534 | a | divide(multiply(173, 240), 240) | ? x 240 = 173 x 240 | "let y x 240 = 173 x 240 then y = ( 173 x 240 ) / 240 = 173 answer : a" | a = 173 * 240
b = a / 240
|
a ) 64 % , b ) 66 2 / 3 % , c ) 62 % , d ) 60 % , e ) 56 % | b | add(multiply(multiply(divide(multiply(divide(add(const_1000, multiply(8, const_100)), 2), add(divide(1, 3), 1)), add(const_1000, multiply(8, const_100))), const_100), 3), divide(multiply(multiply(divide(multiply(divide(add(const_1000, multiply(8, const_100)), 2), add(divide(1, 3), 1)), add(const_1000, multiply(8, const_100))), const_100), 3), const_10)) | at a summer camp with 900 participants , 1 / 2 of the campers are aged 8 to 12 . next year , the number of campers aged 8 to 12 will increase by 1 / 3 . after this change , what percentage of the total 900 campers will the 8 - to 12 - year - olds represent ? | "total - 900 participants campers are aged 8 to 12 = ( 1 / 2 ) * 900 = 450 next year , campers are aged 8 to 12 = ( 4 / 3 ) * 450 = 600 percentage = ( 600 / 900 ) * 100 = 66 2 / 3 % answer : option b" | a = 8 * 100
b = 1000 + a
c = b / 2
d = 1 / 3
e = d + 1
f = c * e
g = 8 * 100
h = 1000 + g
i = f / h
j = i * 100
k = j * 3
l = 8 * 100
m = 1000 + l
n = m / 2
o = 1 / 3
p = o + 1
q = n * p
r = 8 * 100
s = 1000 + r
t = q / s
u = t * 100
v = u * 3
w = v / 10
x = k + w
|
['a ) 425', 'b ) 430', 'c ) 440', 'd ) 450', 'e ) 460'] | d | divide(multiply(multiply(3.78, const_100), multiply(5.25, const_100)), power(multiply(add(const_3, const_4), const_3), const_2)) | a rectangular courtyard 3.78 m lang and 5.25 m broad is to be paved exactly with square tiles , all of the same size . the minimum number of such tiles is : | l = 378 cm and b = 525 cm maximum length of a square tile = hcf of ( 378,525 ) = 21 cm number of tiles = ( 378 Γ 525 ) / ( 21 Γ 21 ) = ( 18 Γ 25 ) = 450 answer is d . | a = 3 * 78
b = 5 * 25
c = a * b
d = 3 + 4
e = d * 3
f = e ** 2
g = c / f
|
a ) a ) 35 , b ) b ) 34 , c ) c ) 42 , d ) d ) 25 , e ) e ) 26 | c | multiply(add(3, 4), divide(subtract(multiply(12, const_2), subtract(multiply(12, const_2), 12)), subtract(multiply(3, const_2), 4))) | 12 years ago , p was half of q in age . if the ratio of their present ages is 3 : 4 , what will be the total of their present ages | explanation : let the present age of p and q be 3 x and 4 x respectively . twelve years ago , p was half of q in age = > 2 ( 3 x β 12 ) = ( 4 x β 12 ) = > 6 x β 24 = 4 x β 12 = > 2 x = 12 = > x = 6 7 x = 7 * 6 = 42 answer : option c | a = 3 + 4
b = 12 * 2
c = 12 * 2
d = c - 12
e = b - d
f = 3 * 2
g = f - 4
h = e / g
i = a * h
|
a ) 6 / 11 , b ) 1 / 2 , c ) 3 / 4 , d ) 9 / 11 , e ) 10 / 11 | b | divide(const_1, add(divide(const_1, 6), divide(const_1, 6))) | a can do a piece of work in 6 days and b can do it in 6 days how long will they both work together to complete the work ? | "explanation : a β s one day work = 1 / 5 b β s one day work = 1 / 6 ( a + b ) β s one day work = 1 / 6 + 1 / 6 = 2 / 3 = = > time = 3 / 2 = 1 1 / 2 days answer : option b" | a = 1 / 6
b = 1 / 6
c = a + b
d = 1 / c
|
a ) 1 , b ) 3 , c ) 4 , d ) 6 , e ) 8 | c | divide(20, add(multiply(3, 2), 4)) | bag a contains red , white and blue marbles such that the red to white marble ratio is 1 : 3 and the white to blue marble ratio is 2 : 3 . bag b contains red and white marbles in the ratio of 1 : 4 . together , the two bags contain 20 white marbles . how many red marbles could be in bag a ? | "6 is the answer . bag a - r : w : b = 2 : 6 : 9 let w in bag a be 6 k bab b - r : w = 1 : 4 let w in bag b be 4 k w = 20 = 6 k + 4 k k = 2 total red ' s in bag a will be 2 k = 4 c" | a = 3 * 2
b = a + 4
c = 20 / b
|
a ) 74 , b ) n = 75 , c ) n = 175 , d ) n = 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 n 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 ) $ 11 , b ) $ 5 , c ) $ 45 , d ) $ 400 , e ) $ 4.2 | e | divide(subtract(190, multiply(subtract(const_1, divide(10, const_100)), 190)), subtract(5, divide(const_1, const_2))) | a reduction in the price of petrol by 10 % enables a motorist to buy 5 gallons more for $ 190 . find the original price of petrol ? | "price decreased by 10 % , so 9 / 10 times , which means that original gallons bought increased 10 / 9 times . since this increase equals to 5 gallons then 45 gallons were bought originally ( 45 * 10 / 9 = 50 - - > increase 5 gallons ) . hence original price was 190 / 45 = $ 4.2 answer : e ." | a = 10 / 100
b = 1 - a
c = b * 190
d = 190 - c
e = 1 / 2
f = 5 - e
g = d / f
|
a ) 400 m , b ) 500 m , c ) 600 m , d ) 550 m , e ) 700 m | a | subtract(multiply(multiply(45, 48), const_0_2778), divide(multiply(multiply(add(48, 42), const_0_2778), 12), divide(const_3, const_2))) | a train is travelling at 48 kmph . it crosses another train having half of its length , travelling in opposite direction at 42 kmph , in 12 seconds . it also passes a railway platform in 45 seconds . what is the length of the platform ? | speed of i train = 48 kmph length of i train = 2 x m speed of sec train = 42 kmph length of seconf train = x m dist = ( 2 x + x ) = 3 x m time = 12 sec = 90 * 5 / 18 = 25 m / s x = 100 length of first train = 2 x = 200 m time taken = 45 sec speed of first train = 48 kmph = 48 * 5 / 18 = 40 / 3 m / s length of the platform = y metre 200 + y = 45 * 45 / 3 = 600 y = 400 answer a | a = 45 * 48
b = a * const_0_2778
c = 48 + 42
d = c * const_0_2778
e = d * 12
f = 3 / 2
g = e / f
h = b - g
|
a ) 3 / 19 , b ) 3 / 11 , c ) 3 / 17 , d ) 3 / 10 , e ) 3 / 13 | d | subtract(divide(3, 6), divide(3, 15)) | if 3 / p = 6 and 3 / q = 15 then p β q = ? | explanation : 3 p 3 p = 6 , p = 3636 = 1212 and 3 q 3 q = 15 , q = 315315 = 1515 then p β q = 1 / 2 β 1 / 5 = 3 / 10 answer : d | a = 3 / 6
b = 3 / 15
c = a - b
|
a ) 380 times , b ) 381 times , c ) 382 times , d ) 392 times , e ) 482 times | c | divide(add(add(const_2, 47), multiply(add(20, add(const_2, const_60)), const_60)), 13) | light glows for every 13 seconds . how many max . times did it glow between 1 : 57 : 58 and 3 : 20 : 47 am . | "time difference is 1 hr , 22 min , 49 sec = 4969 sec . so , light glows floor ( 4969 / 13 ) = 382 times . answer : c" | a = 2 + 47
b = 2 + const_60
c = 20 + b
d = c * const_60
e = a + d
f = e / 13
|
a ) 355600 , b ) 380800 , c ) 356500 , d ) 356800 , e ) 357000 | b | multiply(multiply(560000, subtract(const_1, divide(15, const_100))), divide(80, const_100)) | in an election , candidate a got 80 % 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 favor 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 = 80 % therefore , the number of valid votes polled in favour of candidate a = 80 % of 476000 = 80 / 100 Γ 476000 = 38080000 / 100 = 380800 b )" | a = 15 / 100
b = 1 - a
c = 560000 * b
d = 80 / 100
e = c * d
|
a ) 3 / 80 , b ) 3 / 5 , c ) 40 , d ) 5 / 3 , e ) 80 / 3 | c | divide(log(16), log(power(2, 0.1))) | if n = 2 ^ 0.1 and n ^ b = 16 , b must equal | 10 / 100 = 1 / 10 n = 2 ^ 1 / 10 n ^ b = 2 ^ 4 ( 2 ^ 1 / 10 ) ^ b = 2 ^ 4 b = 40 answer : c | a = math.log(16)
b = 2 ** 0
c = math.log(b)
d = a / c
|
a ) 91 , b ) 37 , c ) 99 , d ) 277 , e ) 19 | a | gcd(1001, 910) | the maximum number of students among them 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is : | explanation : required number of students = h . c . f of 1001 and 910 = 9 answer : a ) 91 | a = math.gcd(1001, 910)
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a ) $ 4096 , b ) $ 14043 , c ) $ 7892 , d ) $ 17736 , e ) $ 15336 | b | multiply(multiply(30000, divide(12, const_100)), const_4) | a will states that the estate would be divided among 6 sons with the remainder donated to charity . one stipulation in the will is that no one among the sons can receive an amount within 12 % of another ' s amount . if one of the beneficiaries received $ 30000 , what is the smallest possible range between the highest and lowest amounts ( rounded to the nearest dollar ) among the 6 sons ? | 1 st - 30000 . 2 nd - 0.88 * 30000 = 26400 3 rd - 0.88 * 26400 = 23232 4 th - 0.88 * 23232 = 20 , 444.16 5 th - 0.88 * 20 , 444.16 = ~ 17 , 990.86 6 th - 0.88 * 17 , 990.86 = ~ 15 , 831.96 range = 30000 - 15831 = 14168 answer : b . | a = 12 / 100
b = 30000 * a
c = b * 4
|
a ) 4586970843 , b ) 4686970743 , c ) 4691100843 , d ) 4586870843 , e ) 4690640889 | e | multiply(469111, 9999) | calculate 469111 x 9999 = ? | answer 469111 x 9999 = 469111 x ( 10000 - 1 ) = 4691110000 - 469111 = 4690640889 . option : e | a = 469111 * 9999
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a ) 5 % , b ) 2 % , c ) 6 % , d ) 8 % , e ) 3 % | b | divide(add(multiply(floor(sqrt(multiply(divide(1, 25), const_100))), const_10), 1), const_3) | simple interest on a certain sum at a certain annual rate of interest is 1 / 25 of the sum . if the numbers representing rate percent and time in years be equal , then the rate of interest is : | "explanation : let sum = x . then , s . i . = x / 25 let rate = r % and time = r years . [ x * r * r / 100 ] = x / 25 ? r ^ 2 = 100 / 25 = 4 r = 2 hence , rate of interest = 2 % . answer : b" | a = 1 / 25
b = a * 100
c = math.sqrt(b)
d = math.floor(c)
e = d * 10
f = e + 1
g = f / 3
|
a ) 20 , b ) 24 , c ) 28 , d ) 32 , e ) 36 | e | multiply(1, 9) | two numbers are in the ratio of 1 : 2 . if 9 be added to both , their ratio changes to 3 : 5 . the greater number is | "let the ratio be x : y , given x / y = 1 / 2 , ( x + 9 ) / ( y + 9 ) = 3 / 5 = > x = 18 and y = 36 answer : e" | a = 1 * 9
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a ) 22 , b ) 24 , c ) 26 , d ) 28 , e ) 30 | b | divide(subtract(subtract(add(multiply(subtract(60, 10), 10), multiply(10, 20)), 660), multiply(gcd(60, 10), 20)), 10) | bookman purchased 60 copies of a new book released recently , 10 of which are hardback and sold for $ 20 each , and rest are paperback and sold for $ 10 each . if 14 copies were sold and the total value of the remaining books was 660 , how many paperback copies were sold ? | "the bookman had 10 hardback ad 60 - 10 = 50 paperback copies ; 14 copies were sold , hence 60 - 14 = 46 copies were left . let # of paperback copies left be p then 10 p + 20 ( 46 - p ) = 660 - - > 10 p = 260 - - > p = 26 # of paperback copies sold is 50 - 26 = 24 answer : b" | a = 60 - 10
b = a * 10
c = 10 * 20
d = b + c
e = d - 660
f = math.gcd(60, 10)
g = f * 20
h = e - g
i = h / 10
|
a ) $ 752 , b ) $ 755 , c ) $ 765 , d ) $ 783 , e ) $ 775 | d | divide(add(multiply(add(760, 30), 30), multiply(6, 760)), 26) | last year manfred received 26 paychecks . each of his first 6 paychecks was $ 760 ; each of his remaining paychecks was $ 30 more than each of his first 6 paychecks . to the nearest dollar , what was the average ( arithmetic mean ) amount of his pay checks for the year ? | "= ( 760 * 6 + 790 * 20 ) / 26 = 783 answer is d . posted from my mobile device" | a = 760 + 30
b = a * 30
c = 6 * 760
d = b + c
e = d / 26
|
a ) 15 hours , b ) 6 hours , c ) 8 hours , d ) 10 hours , e ) 12 hours | a | divide(subtract(12, 10), 30) | two men started from the same place walk at the rate of 10 kmph and 12 kmph respectively . what time will they take to be 30 km apart , if they walk in the same direction ? | "to be 2 km apart they take 1 hour to be 10 km apart they take = 1 / 2 * 30 = 15 hours answer is a" | a = 12 - 10
b = a / 30
|
a ) 25 % , b ) 15 % , c ) 30 % , d ) 32 % , e ) 40 % | a | divide(subtract(const_100, subtract(const_100, 20)), divide(subtract(const_100, 20), const_100)) | if a ' s salary is 20 % less than b ' s salary , by how much percentage is b ' s salary more than a ' s ? | "required percentage = [ 100 * 20 / ( 100 - 20 ) ] % = 25 % answer is a" | a = 100 - 20
b = 100 - a
c = 100 - 20
d = c / 100
e = b / d
|
a ) 6 , b ) 1 , c ) 5 , d ) 4 , e ) 7 | c | multiply(divide(divide(multiply(divide(20, const_100), 100), 5), multiply(divide(20, const_100), 100)), const_100) | a reduction of 20 % in the price of salt enables a lady to obtain 5 kgs more for rs . 100 , find the original price per kg ? | "100 * ( 20 / 100 ) = 20 - - - 5 ? - - - 1 = > rs . 4 100 - - - 80 ? - - - 4 = > rs . 5 answer : c" | a = 20 / 100
b = a * 100
c = b / 5
d = 20 / 100
e = d * 100
f = c / e
g = f * 100
|
a ) 106 / 19 , b ) 108 / 12 , c ) 108 / 19 , d ) 108 / 15 , e ) 108 / 11 | d | divide(103, 10) | find 103 Γ Γ 10 | "by taking 100 as a base we can easily get the answer for the above question within seconds . 103 is 3 more than 100 and 105 is 5 more than 100 so we will write add the excess of the second number to first number or excess of first number to second number as show below to get left most digits and multiply the excesses to get the right most two digits . note : always remember if you get a single digit number put zero in the left two make it a two digit number to get the right most two digits . answer : d" | a = 103 / 10
|
a ) 180 , b ) 1180 , c ) 280 , d ) 580 , e ) 2080 | b | add(multiply(10, 100), multiply(60, multiply(10, divide(30, 100)))) | a worker is paid a regular rate of rs . 10 for completing a survey . the worker completes 100 surveys per week . for any survey involving the use of her cellphone , she is paid at a rate of that is 30 % higher than her regular rate . if she completed 60 surveys involving the use of her cellphone , how much did she get that week ? | "amount earned using her cell phone = 60 * 13 = 780 earned for remaining surveys = 40 * 10 = 400 total earning = 1180 answer : b" | a = 10 * 100
b = 30 / 100
c = 10 * b
d = 60 * c
e = a + d
|
a ) 15 % , b ) 20 % , c ) 25 % , d ) 30 % , e ) 60 % | e | multiply(subtract(const_1, divide(multiply(const_1, 2), 5)), const_100) | if the price of sugar rises from rs . 2 per kg to rs . 5 per kg , a person , to have no increase in the expenditure on sugar , will have to reduce his consumption of sugar by | "sol . let the original consumption = 100 kg and new consumption = x kg . so , 100 x 2 = x Γ 5 = x = 40 kg . β΄ reduction in consumption = 60 % . answer e" | a = 1 * 2
b = a / 5
c = 1 - b
d = c * 100
|
a ) 8 , b ) 9 , c ) 6 , d ) 4 , e ) 2 | d | divide(24, 44) | what is the 124 th digit to the right of the decimal point in the decimal form of 24 / 44 ? | "to solve this question , we first have to use some long division . this long division allows us to get 24 / 44 in decimal form , which is 0.545454 β¦ where β 54 β is repeating . we can see that the 1 st , 3 rd , 5 th digit to the right of the decimal point is a 5 and that the 2 nd , 4 th , 6 th digit to the right of the decimal point is a 4 . in other words , each odd - positioned digit is a 5 , and each even - positioned digit is a 4 . then the 124 digit to the right of the decimal point and we see that 124 is even , we know that the 124 th digit is a 4 . answer d ." | a = 24 / 44
|
a ) 40 , b ) 45 , c ) 80 , d ) 90 , e ) 2500 | c | divide(power(const_10, divide(4, const_2)), const_2) | a palindrome is a number that reads the same forward and backward , such as 145 . how many odd , 4 - digit numbers are palindromes ? | "a palindrome is a number that reads the same forward and backward . examples of four digit palindromes are 1221 , 4334 , 2222 etc you basically get to choose the first two digits and you repeat them in opposite order . say , you choose 45 as your first two digits . the next two digits are 54 and the number is 4554 . also , you need only odd palindromes . this means that you need an odd digit at the end i . e . 1 / 3 / 5 / 7 / 9 . this means that you need to start the number with an odd digit . only then will it end with an odd digit . in how many ways can you pick two digits such that the first one is an odd digit ? the first digit can be selected in 5 ways . ( 1 / 3 / 5 / 7 / 9 ) the second digit can be selected in 10 ways . ( 0 / 1 / 2 / 3 . . . 8 / 9 ) total = 5 * 16 = 80 ways c" | a = 4 / 2
b = 10 ** a
c = b / 2
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | c | subtract(4, reminder(15, 5)) | when n is divided by 25 , the remainder is 4 . what is the remainder when n + 15 is divided by 5 ? | "assume n = 29 remainder ( n / 24 ) = 4 n + 15 = 44 remainder ( 44 / 5 ) = 4 option c" | a = 4 - reminder
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a ) s . 50 , b ) s . 70 , c ) s . 100 , d ) s . 125 , e ) s . 60 | d | multiply(500, divide(25, const_100)) | find the 25 % of rs . 500 . | "explanation : 25 % of 500 = > 25 / 100 * 500 = rs . 125 answer : d" | a = 25 / 100
b = 500 * a
|
a ) 190 , b ) 105 , c ) 200 , d ) 205 , e ) 210 | b | divide(divide(multiply(add(10, 200), add(divide(subtract(200, 10), 10), const_1)), const_2), add(divide(subtract(200, 10), 10), const_1)) | what is the average ( arithmetic mean ) of all multiples of 10 from 10 to 200 inclusive ? | "this question can be solved with the average formula and ' bunching . ' we ' re asked for the average of all of the multiples of 10 from 10 to 200 , inclusive . to start , we can figure out the total number of terms rather easily : 1 ( 10 ) = 10 2 ( 10 ) = 20 . . . 20 ( 10 ) = 200 so we know that there are 40 total numbers . we can now figure out the sum of those numbers with ' bunching ' : 10 + 200 = 210 20 + 190 = 210 30 + 180 = 210 etc . since there are 20 total terms , this pattern will create 10 ' pairs ' of 210 . thus , since the average = ( sum of terms ) / ( number of terms ) , we have . . . ( 10 ) ( 210 ) / ( 20 ) = 105 answer : b" | a = 10 + 200
b = 200 - 10
c = b / 10
d = c + 1
e = a * d
f = e / 2
g = 200 - 10
h = g / 10
i = h + 1
j = f / i
|
a ) 20,20 , b ) 20,8 , c ) 25,15 , d ) 30,10 , e ) none of these | b | subtract(add(divide(multiply(12, 5), subtract(5, const_1)), 5), 12) | the ages of two persons differ by 12 years . if 5 years ago , the elder one be 5 times as old as the younger one , their present ages ( in years ) are respectively | "explanation : let their ages be x and ( x + 12 ) years . 5 ( x - 5 ) = ( x + 12 - 5 ) or 4 x = 32 or x = 8 . their present ages are 20 years and 8 years option b" | a = 12 * 5
b = 5 - 1
c = a / b
d = c + 5
e = d - 12
|
a ) 2 , b ) 1 , c ) 6 , d ) 12 , e ) 24 | b | divide(divide(divide(divide(divide(divide(54, const_2), const_2), const_2), const_2), const_3), const_3) | if n is the smallest integer such that 54 times n is the square of an integer , what is the value of n ? | "54 can written as = 2 * 3 * 3 * 3 - - > 2 ^ 1 * 3 ^ 3 - - - ( 1 ) so for 54 * n to be a square of an integer , the integer should have even powers to the prime numbers it composed of . here 2 already has even power - > so n has to be 2 to make the power of 1 in ( 1 ) even option b is correct" | a = 54 / 2
b = a / 2
c = b / 2
d = c / 2
e = d / 3
f = e / 3
|
a ) 10 s , b ) 6 s , c ) 4 s , d ) 8 s , e ) 12 s | e | divide(add(150, 150), add(divide(multiply(54, const_1000), const_3600), divide(multiply(36, const_1000), const_3600))) | two trains a and b are 150 m and 150 m long and are moving at one another at 54 km / hr and 36 km / hr respectively . arun is sitting on coach b 1 of train a . calculate the time taken by arun to completely cross train b . | "detailed solution speed of a = 54 β 1000 / 60 β 60 = 15 m / s speed of b = 36 β 1000 / 60 β 60 = 10 m / s relative speed = s 1 + s 2 = 15 + 10 m / s = 25 m / s the length that needs to be crossed = length of train b = 150 m . therefore time taken = 150 / 25 = 6 s . what is the time taken for trains to completely cross each other ? the length that needs to be crossed = 150 + 150 = 300 m . time taken = 300 / 25 = 12 s . correct answer e ." | a = 150 + 150
b = 54 * 1000
c = b / 3600
d = 36 * 1000
e = d / 3600
f = c + e
g = a / f
|
a ) 900 , b ) 1800 , c ) 2250 , d ) 1650 , e ) 1250 | b | multiply(40, multiply(divide(30, 4), divide(30, 5))) | if 4 a = 5 b = 30 , then 40 ab = | "4 a * 5 b = 30 * 30 = 900 20 ab = 900 i . e . 30 ab = 1800 answer : option b" | a = 30 / 4
b = 30 / 5
c = a * b
d = 40 * c
|
a ) 281 m , b ) 25 m , c ) 50 m , d ) 125 m , e ) 75 m | b | multiply(multiply(18, const_0_2778), 5) | if a train , travelling at a speed of 18 kmph , crosses a pole in 5 sec , then the length of train is ? | "b b = 18 * 5 / 18 * 5 = 25 m" | a = 18 * const_0_2778
b = a * 5
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a ) 100 m , b ) 125 m , c ) 130 m , d ) 150 m , e ) 165 m | e | multiply(110, subtract(const_2, const_1)) | a train speeds past a pole in 15 seconds and a platform 110 meters long in 25 seconds . what is the length of the train ? | "let the length of the train be x meters . the speed of the train is x / 15 . then , x + 110 = 25 * ( x / 15 ) 10 x = 1650 x = 165 meters the answer is e ." | a = 2 - 1
b = 110 * a
|
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