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 ) 10 hours , b ) 40 hours , c ) 20 hours , d ) 60 hours , e ) none of them | d | divide(multiply(36, 25), 15) | if 36 men can do a piece of work in 25 hours , in how many hours will 15 men do it ? | "let the required number of hours be x . then , less men , more hours ( indirect proportion ) 15 : 36 : : 25 : x = ( 15 x x ) = ( 36 x 25 ) = ( 36 x 25 ) / 15 = 60 hence , 15 men can do it in 60 hours . answer is d ." | a = 36 * 25
b = a / 15
|
a ) 40 sec , b ) 36 sec , c ) 60 sec , d ) 48 sec , e ) 18 sec | b | divide(420, multiply(subtract(48, 60), const_0_2778)) | a train 420 m long is running at a speed of 48 km / hr . in what time will it pass a bridge 60 m long ? | "total length = 420 + 60 = 480 speed = 48 km / h = ( 48 * 5 ) / 18 = 40 / 3 m / s time = ( 480 * 3 ) / 40 = 36 sec . answer : b" | a = 48 - 60
b = a * const_0_2778
c = 420 / b
|
a ) $ 22 , b ) $ 24 , c ) $ 30 , d ) $ 36 , e ) $ 40 | b | multiply(divide(multiply(1, 8), subtract(8, 6)), 6) | nina has exactly enough money to purchase 6 widgets . if the cost of each widget were reduced by $ 1 , then nina would have exactly enough money to purchase 8 widgets . how much money does nina have ? | "b its is . let price = x ( x - 1 ) 8 = 6 x x = 4 hence total money = 6 * 4 = 24" | a = 1 * 8
b = 8 - 6
c = a / b
d = c * 6
|
a ) 1 / 2 , b ) 3 / 4 , c ) 2 / 3 , d ) 1 / 5 , e ) 3 / 7 | c | divide(add(divide(divide(factorial(20), factorial(subtract(20, const_2))), factorial(const_2)), divide(divide(factorial(10), factorial(subtract(10, const_2))), factorial(const_2))), divide(divide(factorial(add(10, 20)), factorial(subtract(add(10, 20), const_2))), factorial(const_2))) | a bag contains 10 black and 20 white balls . one ball is drawn at random . what is the probability that the ball drawn is white ? | "total number of balls = 10 + 20 = 30 number of white balls = 20 probability of drawing a white ball = 20 / 30 = 2 / 3 answer is c" | a = math.factorial(20)
b = 20 - 2
c = math.factorial(b)
d = a / c
e = math.factorial(2)
f = d / e
g = math.factorial(10)
h = 10 - 2
i = math.factorial(h)
j = g / i
k = math.factorial(2)
l = j / k
m = f + l
n = 10 + 20
o = math.factorial(n)
p = 10 + 20
q = p - 2
r = math.factorial(q)
s = o / r
t = math.factorial(2)
u = s / t
v = m / u
|
a ) rs . 1200 , b ) rs . 1300 , c ) rs . 3500 , d ) rs . 2000 , e ) none of these | c | multiply(multiply(8, const_1000), divide(subtract(6, 5), subtract(8, 6))) | a man invests rs . 7,000 at the rate of 5 % per annum . how much more should he invest at the rate of 8 % , so that he can earn a total of 6 % per annum ? | "explanation : interest on rs . 7000 at 5 % per annum = ( 7000 Γ 5 Γ 1 ) / 100 = rs . 350 let his additional investment at 8 % = x interest on rs . x at 8 % per annum = ( x Γ 8 Γ 1 ) / 100 = 2 x / 25 . to earn 6 % per annum for the total , interest = ( 7000 + x ) Γ 6 Γ 1 / 100 . = > 350 + 2 x / 25 = ( 7000 + x ) Γ 6 Γ 1 / 100 . = > 35000 + 8 x = ( 7000 + x ) Γ 6 . = > 35000 + 8 x = 42000 + 6 x . = > 2 x = 7000 . = > x = 3500 . answer : c" | a = 8 * 1000
b = 6 - 5
c = 8 - 6
d = b / c
e = a * d
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a ) 2 , b ) 3 , c ) 6 , d ) 5 , e ) 0 | e | divide(multiply(5, 2), const_10) | if 16 ^ 5 * 15 ^ 4 * 14 ^ 3 * 13 ^ 2 = x , what is the units digit of x ? | 16 ^ 5 * 15 ^ 4 * 14 ^ 3 * 13 ^ 2 = x units digit of x will be 6 ^ 5 * 5 ^ 4 * 4 ^ 3 * 3 ^ 2 = 6 * 5 * 4 * 9 = 1080 therefore , units digit of x = 0 answer e | a = 5 * 2
b = a / 10
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a ) 54 , b ) 45 , c ) 25 , d ) 15 , e ) 55 | a | multiply(210, divide(multiply(15, 3), add(add(multiply(10, 7), multiply(12, 5)), multiply(15, 3)))) | a , b and c rent a pasture . if a puts 10 oxen for 7 months , b puts 12 oxen for 5 months and c puts 15 oxen for 3 months for grazing and the rent of the pasture is rs . 210 , then how much amount should c pay as his share of rent ? | "a : b : c = 10 Γ 7 : 12 Γ 5 : 15 Γ 3 = 2 Γ 7 : 12 Γ 1 : 3 Γ 3 = 14 : 12 : 9 amount that c should pay = 210 Γ 9 / 35 = 6 Γ 9 = 54 answer is a" | a = 15 * 3
b = 10 * 7
c = 12 * 5
d = b + c
e = 15 * 3
f = d + e
g = a / f
h = 210 * g
|
a ) 1 , b ) 3.5 , c ) 20 , d ) 49 , e ) 50 | e | divide(1, divide(1, 50)) | if 50 honey bees make 50 grams of honey in 50 days , then 1 honey bee will make 1 gram of honey in how many days ? | "explanation : let the required number days be x . less honey bees , more days ( indirect proportion ) less honey , less days ( direct proportion ) honey bees 1 : 50 : : 50 : x honey 50 : 1 = > 1 x 50 x x = 50 x 1 x 50 = > x = 50 . answer : e" | a = 1 / 50
b = 1 / a
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a ) 50 % , b ) 60 % , c ) 75 % , d ) 80 % , e ) 100 % | e | multiply(divide(subtract(const_100, multiply(divide(subtract(const_100, 55), subtract(const_100, 10)), const_100)), multiply(divide(subtract(const_100, 55), subtract(const_100, 10)), const_100)), const_100) | the charge for a single room at hotel p is 55 percent less than the charge for a single room at hotel r and 10 percent less than the charge for a single room at hotel g . the charge for a single room at hotel r is what percent greater than the charge for a single room at hotel g ? | "p = 0.45 r = 0.9 g r = 0.9 g / 0.45 = 2 g thus r is 100 % greater than g . the answer is e ." | a = 100 - 55
b = 100 - 10
c = a / b
d = c * 100
e = 100 - d
f = 100 - 55
g = 100 - 10
h = f / g
i = h * 100
j = e / i
k = j * 100
|
a ) 20 , b ) 23 , c ) 24 , d ) 25 , e ) 26 | b | divide(multiply(multiply(69, 12), 5), multiply(30, 6)) | 69 persons can repair a road in 12 days , working 5 hours a day . in how many days will 30 persons , working 6 hours a day , complete the work ? | "according to the chain rule , m 1 x t 1 = m 2 x t 2 therefore , 69 x 12 x 5 = 30 x 6 x x x = 23 hence , the number of days = 23 . answer : b" | a = 69 * 12
b = a * 5
c = 30 * 6
d = b / c
|
a ) 63 % , b ) 24 % , c ) 50 % , d ) 55 % , e ) 61 % | e | subtract(multiply(add(const_100, 40), divide(add(const_100, 15), const_100)), const_100) | a certain college ' s enrollment at the beginning of 1992 was 40 percent greater than it was at the beginning of 1991 , and its enrollment at the beginning of 1993 was 15 percent greater than it was at the beginning of 1992 . the college ' s enrollment at the beginning of 1993 was what percent greater than its enrollment at the beginning of 1991 ? | "suppose enrollment in 1991 was 100 then enrollment in 1992 will be 140 and enrollment in 1993 will be 140 * 1.15 = 161 increase in 1993 from 1991 = 161 - 100 = 61 answer : e" | a = 100 + 40
b = 100 + 15
c = b / 100
d = a * c
e = d - 100
|
a ) 5 % , b ) 11 % , c ) 13 % , d ) 21 % , e ) 33 % | e | multiply(divide(subtract(const_100, 75), 75), const_100) | if the cost price is 75 % of the selling price , then what is the profit percent ? | "let s . p . = $ 100 c . p . = $ 75 profit = $ 25 profit % = 25 / 75 * 100 = 25 / 6 = 33 % approximately answer is e" | a = 100 - 75
b = a / 75
c = b * 100
|
a ) 0.003 , b ) 0.015 , c ) 0.25 , d ) 0.005 , e ) none of these | b | multiply(divide(divide(3, const_100), 3), const_2) | double of quarter of 3 percent written as a decimal is : | "explanation : solution : ( 2 ) * ( 1 / 4 ) * 3 % = 2 * ( 1 / 4 * 3 / 100 ) = 0.015 . answer : b" | a = 3 / 100
b = a / 3
c = b * 2
|
a ) 2 / 35 , b ) 2 / 3 , c ) 7 / 35 , d ) 1 / 7 , e ) 7 / 5 | d | multiply(divide(5, 7), divide(1, 5)) | two brothers ram and ravi appeared for an exam . the probability of selection of ram is 5 / 7 and that of ravi is 1 / 5 . find the probability that both of them are selected . | "let a be the event that ram is selected and b is the event that ravi is selected . p ( a ) = 5 / 7 p ( b ) = 1 / 5 let c be the event that both are selected . p ( c ) = p ( a ) x p ( b ) as a and b are independent events : = 5 / 7 x 1 / 5 = 1 / 7 answer : d" | a = 5 / 7
b = 1 / 5
c = a * b
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a ) 2 , b ) 3 , c ) 4 , d ) 7 , e ) 8 | a | subtract(11070, multiply(floor(divide(11070, 22)), 22)) | find the least number must be subtracted from 11070 so that remaining number is divisible by 22 . | on dividing 11070 by 22 we get the remainder 4 , so 4 should be subtracted . answer is a . | a = 11070 / 22
b = math.floor(a)
c = b * 22
d = 11070 - c
|
a ) 2333 , b ) 2777 , c ) 2688 , d ) 500 , e ) 2871 | d | divide(subtract(145, divide(multiply(6, 2500), const_100)), subtract(divide(5, const_100), divide(6, const_100))) | rs . 2500 is divided into two parts such that if one part be put out at 5 % simple interest and the other at 6 % , the yearly annual income may be rs . 145 . how much was lent at 5 % ? | "( x * 5 * 1 ) / 100 + [ ( 2500 - x ) * 6 * 1 ] / 100 = 145 x = 500 answer : d" | a = 6 * 2500
b = a / 100
c = 145 - b
d = 5 / 100
e = 6 / 100
f = d - e
g = c / f
|
a ) a ) 59 , b ) b ) 61 , c ) c ) 56 , d ) d ) 65 , e ) e ) 67 | c | divide(add(49, 64), const_2) | find the mean proportional between 49 & 64 ? | "formula = β a Γ b a = 49 and b = 64 β 49 Γ 64 = 7 Γ 8 = 56 c" | a = 49 + 64
b = a / 2
|
a ) 70 min , b ) 94 min , c ) 110 min , d ) 112 min , e ) 120 min | a | add(inverse(add(add(inverse(1), inverse(2)), inverse(2))), inverse(add(inverse(1), inverse(2)))) | one robot builds a robot in 1 hour , another one builds a robot in 2 hours . the 2 robots work together and when a new robot is complete , it joins the others , working at a constant rate of one robot in 2 hours . how much time will it take until there are 4 robots altogether , if the robots build only one robot at a time ? | first the 2 robots work at the rate of 1 + 1 / 2 = 3 / 2 so they complete one robot in 2 / 3 rd of an hour = 40 minutes - ( 1 ) now the 3 robots work together at the rate of 1 + 1 / 2 + 1 / 2 = 4 / 2 = 2 / 1 so they complete one robot in 1 / 2 an hour , i . e 30 minutes - ( 2 ) and now we have 4 robots so total = ( 1 ) + ( 2 ) = 70 minutes answer - a | a = 1/(1)
b = 1/(2)
c = a + b
d = 1/(2)
e = c + d
f = 1/(e)
g = 1/(1)
h = 1/(2)
i = g + h
j = 1/(i)
k = f + j
|
a ) 100 , b ) 50 , c ) 150 , d ) 200 , e ) 75 | a | divide(multiply(multiply(10, 5), divide(const_100, 5)), subtract(multiply(divide(const_100, 5), 3), 50)) | 50 % of a number is 10 less than 3 / 5 th of that number . what is the number ? | "let the number be x then , 3 x / 5 - ( 50 % of x ) = 10 3 x / 5 - 50 x / 100 = 10 x = 100 answer is a" | a = 10 * 5
b = 100 / 5
c = a * b
d = 100 / 5
e = d * 3
f = e - 50
g = c / f
|
a ) 80 , b ) 40 , c ) 50 , d ) 60 , e ) 20 | b | divide(200, 5) | 17 + 18 = 35 . how many such 2 consecutive numbers are there less than 200 when added gives a sum which is divisible by 5 ? | since 2 + 3 = 5 & 7 + 8 = 15 any combination with these no be will give u desirable result . . . so total no in 100 will be 20 & that ' s why in 200 , it will be 40 answer : b | a = 200 / 5
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a ) 960 , b ) 990 , c ) 1,200 , d ) 920 , e ) none of these | b | multiply(divide(add(872, 448), const_2), add(const_1, divide(50, const_100))) | the profit earned by selling an article for 872 is equal to the loss incurred when the same article is sold for 448 . what should be the sale price of the article for making 50 per cent profit ? | "let the profit or loss be x and 872 β x = 448 + x or , x = 424 β 2 = 212 \ cost price of the article = 872 β x = 448 + x = 660 \ sp of the article = 660 Γ 150 β 100 = 990 answer b" | a = 872 + 448
b = a / 2
c = 50 / 100
d = 1 + c
e = b * d
|
a ) 10000 , b ) 2028 , c ) 2775 , d ) 5496 , e ) 6851 | a | divide(2000, subtract(subtract(const_1, divide(40, const_100)), divide(40, const_100))) | a candidate got 40 % of the votes polled and he lost to his rival by 2000 votes . how many votes were cast ? | "40 % - - - - - - - - - - - l 60 % - - - - - - - - - - - w - - - - - - - - - - - - - - - - - - 20 % - - - - - - - - - - 2000 100 % - - - - - - - - - ? = > 10000 answer : a" | a = 40 / 100
b = 1 - a
c = 40 / 100
d = b - c
e = 2000 / d
|
a ) 400 , b ) 420 , c ) 450 , d ) 480 , e ) 500 | e | divide(divide(170, const_2), divide(17, const_100)) | one night 17 percent of the female officers on a police force were on duty . if 170 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 85 . 0.17 x = 85 x = 500 the answer is e . | a = 170 / 2
b = 17 / 100
c = a / b
|
a ) 29960 sq m , b ) 26950 sq m , c ) 43120 sq m , d ) 27680 sq m , e ) 27786 sq m | b | subtract(circle_area(divide(704, multiply(const_2, const_pi))), circle_area(divide(396, multiply(const_2, const_pi)))) | the circumferences of two circles are 396 meters and 704 meters . find the difference between the areas of the larger and the smaller circles ? | "let the radii of the smaller and the larger circles be s m and l m respectively . 2 β s = 396 and 2 β l = 704 s = 396 / 2 β and l = 704 / 2 β difference between the areas = β l ^ 2 - β s ^ 2 = β { 198 ^ 2 / β ^ 2 - 352 ^ 2 / β ^ 2 } = 198 ^ 2 / β - 352 ^ 2 / β = ( 198 - 352 ) ( 198 + 352 ) / β = ( 154 ) ( 550 ) / ( 22 / 7 ) = 26950 sq m answer : b" | a = 2 * math.pi
b = 704 / a
c = circle_area - (
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a ) 39 , b ) 40 , c ) 41 , d ) 42 , e ) 43 | d | divide(495, multiply(subtract(63, 240), const_0_2778)) | a train 495 m long is running at a speed of 63 km / hr . in what time will it pass a bridge 240 m long ? | "speed = 63 * 5 / 18 = 35 / 2 m / sec total distance covered = 495 + 240 = 735 m required time = 700 * 2 / 35 = 42 sec answer : d" | a = 63 - 240
b = a * const_0_2778
c = 495 / b
|
a ) 12 , b ) 15 , c ) 10 , d ) 18 , e ) 9 | e | multiply(multiply(add(3, divide(subtract(sqrt(27), 3), 2)), divide(subtract(sqrt(27), 3), 2)), 2) | if a - b = 3 and a ^ 2 + b ^ 2 = 27 , find the value of ab . | "2 ab = ( a ^ 2 + b ^ 2 ) - ( a - b ) ^ 2 = 27 - 9 = 18 ab = 9 . answer is e ." | a = math.sqrt(27)
b = a - 3
c = b / 2
d = 3 + c
e = math.sqrt(27)
f = e - 3
g = f / 2
h = d * g
i = h * 2
|
a ) 75 , b ) 78 , c ) 80 , d ) 82 , e ) 86 | e | add(multiply(add(multiply(6, const_3), 2), divide(add(multiply(6, const_3), 2), 5)), 6) | in a division sum , the remainder is 6 and the divisor is 5 times the quotient and is obtained by adding 2 to the thrice of the remainder . the dividend is ? | "divisor = ( 6 * 3 ) + 2 = 20 5 * quotient = 20 quotient = 4 . dividend = ( divisor * quotient ) + remainder dividend = ( 20 * 4 ) + 6 = 86 e )" | a = 6 * 3
b = a + 2
c = 6 * 3
d = c + 2
e = d / 5
f = b * e
g = f + 6
|
a ) 3277 , b ) 2977 , c ) 7119 , d ) 6077 , e ) 17112 | c | multiply(multiply(4, multiply(const_100, const_100)), power(divide(3, 4), 6)) | the value of a scooter depreciates in such a way that its value of the end of each year is 3 / 4 of its value of the beginning of the same year . if the initial value of the scooter is rs . 40,000 , what is the value at the end of 6 years ? | "explanation : 40,000 * ( 3 / 4 ) ^ 6 answer : c" | a = 100 * 100
b = 4 * a
c = 3 / 4
d = c ** 6
e = b * d
|
a ) 11 , b ) 5 , c ) 9 , d ) 8 , e ) 10 | a | subtract(50, subtract(subtract(add(add(22, 14), 15), 10), multiply(1, const_2))) | there are 50 employees in the office of abc company . of these , 22 have taken an accounting course , 14 have taken a course in finance and 15 have taken a marketing course . 10 of the employees have taken exactly two of the courses and 1 employee has taken all 3 of the courses . how many of the 50 employees have taken none of the courses ? | 50 employees . counting every different attendand to the courses we have : accounting : 22 finance : 14 marketing : 15 which would add up to 51 different attendands , which is not possible . now 10 have taken exactly 2 courses , which means that there are 10 less different attendands . say that 10 of the finance attentands also attended accounting . 51 - 10 = 41 1 person has taken all three courses . as above , we subtract him from the number of different attendands . since this time the person took all three courses , we have to substract him two times . 41 - 2 = 39 . answer : a | a = 22 + 14
b = a + 15
c = b - 10
d = 1 * 2
e = c - d
f = 50 - e
|
a ) 21.21 , b ) 24.24 , c ) 27.27 , d ) 30.35 , e ) 33.33 | c | divide(divide(80, 5280), multiply(2, divide(1, const_3600))) | if an object travels 80 feet in 2 seconds , what is the object β s approximate speed in miles per hour ? ( note : 1 mile = 5280 feet ) | "80 feet / 2 seconds = 40 feet / second ( 40 feet / second ) * ( 3600 seconds / hour ) * ( 1 mile / 5280 feet ) = 27.27 miles / hour ( approximately ) the answer is c ." | a = 80 / 5280
b = 1 / 3600
c = 2 * b
d = a / c
|
a ) 1210 , b ) 1221 , c ) 1332 , d ) 1443 , e ) 1554 | e | multiply(add(add(multiply(add(add(4, 2), 1), const_100), multiply(add(add(4, 2), 1), const_10)), add(add(4, 2), 1)), const_2) | what is the sum of all possible 3 - digit numbers that can be constructed using the digits 1 , 2 , and 4 if each digit can be used only once in each number ? | there are 6 possible arrangements of the three numbers . then each number will be in the hundreds , tens , and ones place two times each . the sum is 2 ( 111 ) + 2 ( 222 ) + 2 ( 444 ) = 1554 the answer is e . | a = 4 + 2
b = a + 1
c = b * 100
d = 4 + 2
e = d + 1
f = e * 10
g = c + f
h = 4 + 2
i = h + 1
j = g + i
k = j * 2
|
a ) 16.125 , b ) 20.125 , c ) 24.125 , d ) 28.125 , e ) 32.125 | d | subtract(add(multiply(11, divide(multiply(2, 2), 11)), 13), multiply(2, power(divide(multiply(2, 2), 11), 2))) | if x is real , find the maximum value of the expression - 2 x ^ 2 + 11 x + 13 . | "this is an equation of a downward facing parabola . the maximum value is the top point of the parabola . - 2 x ^ 2 + 11 x + 13 = ( - 2 x + 13 ) ( x + 1 ) the roots are 13 / 2 and - 1 . the maximum value must be when x is halfway between these two points . x = 2.75 the maximum value is - 2 ( 2.75 ) ^ 2 + 11 ( 2.75 ) + 13 = 28.125 the answer is d ." | a = 2 * 2
b = a / 11
c = 11 * b
d = c + 13
e = 2 * 2
f = e / 11
g = f ** 2
h = 2 * g
i = d - h
|
a ) 74 , b ) x = 75 , c ) x = 175 , d ) x = 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 x 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 ) 1 , b ) 22 , c ) 23 , d ) 10 , e ) 25 | d | subtract(divide(multiply(multiply(2, 6), 2), const_2), const_2) | if 1 = x . 2 = 2 x 3 = 4 x . . . . what is 2 x * 6 x + 5 x / 4 x - 56 x | let 1 = x 2 = 2 x 3 = 4 x 4 = 8 x 5 = 16 x 6 = 32 x 7 = 64 x now given 2 x * 6 x + 5 x / 4 x - 56 x i can write it as = 2 x * ( 2 x + 4 x ) + ( x + 4 x ) / 4 x - ( 64 x - 8 x ) now by putting the values i get = 2 * ( 2 + 3 ) + ( 8 + 1 ) / 3 - ( 7 - 4 ) = 2 * 5 + 9 / 3 - 3 = 10 + 9 / 3 - 3 = 10 + 3 - 3 = 10 answer : d | a = 2 * 6
b = a * 2
c = b / 2
d = c - 2
|
a ) 0.092 , b ) 0.92 , c ) 9.2 , d ) 92 , e ) 920 | c | multiply(divide(0.004, 0.03), 69.28) | the closest approximation of a ( 69.28 Γ 0.004 ) / 0.03 is | "a ( 69.28 Γ 0.004 ) / 0.03 1 . 0.004 = 4 Γ 10 ^ ( - 3 ) 2 . 0.03 = 3 Γ 10 ^ ( - 2 ) 3 . ( a Γ b ) / c = a Γ ( b / c ) 4 . 0.004 / 0.03 = 4 Γ 10 ^ ( - 3 ) / ( 3 Γ 10 ^ ( - 2 ) ) = 4 Γ 10 ^ ( - 3 - ( - 2 ) ) / 3 = 4 Γ 10 ^ ( - 1 ) / 3 = ( 4 / 3 ) Γ 10 ^ ( - 1 ) = 1.333 Γ 10 ^ ( - 1 ) therefore , ( 69.28 Γ 0.004 ) / 0.03 = 69.28 Γ ( 0.004 / 0.03 ) = 69.28 Γ 1.33 Γ 10 ^ ( - 1 ) = 69.28 Γ 1.33 / 10 = 6.928 * 1.33 now , 7 Γ 2 = 14 7 Γ 1 = 7 or better : 6.9 Γ 1 = 6.9 6.9 Γ 2 = 13.8 hence , 6.9 < 6.928 Γ 1.33 < 13.8 9.2 is the only answer that satisfies this condition . c" | a = 0 / 4
b = a * 69
|
a ) 50 , b ) 60 , c ) 80 , d ) 320 , e ) none | a | divide(40, subtract(const_1, divide(20, const_100))) | a number exceeds 20 % of itself by 40 . the number is ? | answer β΅ a - 20 % of a = 40 β a - a / 5 = 40 β 4 a / 5 = 40 β΄ a = ( 5 x 40 ) / 4 = 50 . correct option : a | a = 20 / 100
b = 1 - a
c = 40 / b
|
a ) 1.08 , b ) 1.4 , c ) 1.8018 , d ) 1.4014 , e ) 1.8 | e | multiply(divide(9.009, 5.005), const_100) | 9.009 / 5.005 = | "9.009 / 5.005 = 9009 / 5005 = 9 ( 1001 ) / 5 ( 1001 ) = 9 / 5 = 1.8 the answer is e ." | a = 9 / 9
b = a * 100
|
a ) 19 , b ) 20 , c ) 21 , d ) 22 , e ) 23 | c | divide(294, const_10) | the ratio between the perimeter and the width of a rectangle is 5 : 1 . if the area of the rectangle is 294 square centimeters , what is the length of the rectangle in centimeters ? | "perimeter = 2 ( w + l ) = 5 w 3 w = 2 l w = 2 l / 3 wl = 294 2 l ^ 2 / 3 = 294 l ^ 2 = 441 l = 21 cm the answer is c ." | a = 294 / 10
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a ) 36,100 , b ) 38,210 , c ) 40,320 , d ) 42,430 , e ) 44,540 | c | factorial(6) | if each digit in the set a = { 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 } is used exactly once , in how many ways can the digits be arranged ? | "8 ! = 40,320 the answer is c ." | a = math.factorial(6)
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a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | e | add(8, const_1) | mark and ann together were allocated n boxes of cookies to sell for a club project . mark sold 8 boxes less than n and ann sold 2 boxes less than n . if mark and ann have each sold at least one box of cookies , but together they have sold less than n boxes , what is the value of n ? | "if n = 9 mark sold 1 box and ann sold 7 boxes total 8 < 9 answer : e" | a = 8 + 1
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a ) 80.0 , b ) 80.9 , c ) 81.0 , d ) 81.1 , e ) 81.9 | e | multiply(multiply(divide(subtract(const_100, 9), const_100), divide(subtract(const_100, 10), const_100)), const_100) | a store reduced the price of all items in the store by 9 % on the first day and by another 10 % on the second day . the price of items on the second day was what percent of the price before the first reduction took place ? | "consider price of the all items as $ 100 after a initial reduction of 9 % price becomes = 0.91 * 100 = $ 91 after the final reduction of 10 % price becomes = 0.9 * 91 = $ 81.9 price of all items on second day is 81.9 % of price on first day correct answer option e" | a = 100 - 9
b = a / 100
c = 100 - 10
d = c / 100
e = b * d
f = e * 100
|
a ) 10 , b ) 910 , c ) 1001 , d ) 1911 , e ) none of these | a | gcd(890, 630) | the maximum number of students among them 890 pens and 630 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 890 and 630 = 10 . answer : a" | a = math.gcd(890, 630)
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a ) 6.33 , b ) 7.5 , c ) 21 , d ) 15 , e ) 19 | c | subtract(22, divide(22, 22)) | a certain bacteria colony doubles in size every day for 22 days , at which point it reaches the limit of its habitat and can no longer grow . if two bacteria colonies start growing simultaneously , how many days will it take them to reach the habitat β s limit ? | "if there is one bacteria colony , then it will reach the limit of its habitat in 20 days . if there are two bacteria colonies , then in order to reach the limit of habitat they would need to double one time less than in case with one colony . thus colonies need to double 21 times . answer : c . similar questions to practice : hope it helps ." | a = 22 / 22
b = 22 - a
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a ) 9.5 % , b ) 10.5 % , c ) 11.5 % , d ) 9.1 % , e ) 12.5 % | d | multiply(divide(subtract(add(multiply(10, 12), 10), multiply(10, 12)), 110), const_100) | on a purchase of $ 110 , a store offered a payment plan consisting of a $ 10 down payment and 12 monthly payments of $ 10 each . what percent of the purchase price , to the nearest tenth of a percent , did the customer pay in interest by using this plan ? | "12 * 10 + 10 = 130 ( 10 / 110 ) * 100 = 9.1 answer : d" | a = 10 * 12
b = a + 10
c = 10 * 12
d = b - c
e = d / 110
f = e * 100
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a ) $ 2.10 , b ) $ 37.10 , c ) $ 37.16 , d ) $ 38.10 , e ) $ 38.15 | e | add(multiply(divide(9, const_100), 35), 35) | laura took out a charge account at the general store and agreed to pay 9 % simple annual interest . if she charges $ 35 on her account in january , how much will she owe a year later , assuming she does not make any additional charges or payments ? | principal that is amount taken by laura at year beginning = 35 $ rate of interest = 9 % interest = ( 9 / 100 ) * 35 = 3.15 $ total amount that laura owes a year later = 35 + 3.15 = 38.15 $ answer e | a = 9 / 100
b = a * 35
c = b + 35
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a ) 342 km , b ) 352 km , c ) 642 km , d ) 742 km , e ) 382 km | a | divide(add(add(30, multiply(2, 9)), 30), 2) | a car started running at a speed of 30 km / hr and the speed of the car was increased by 2 km / hr at the end of every hour . find the total distance covered by the car in the first 9 hours of the journey . | "the total distance covered by the car in the first 10 hours = 30 + 32 + 34 + 36 + 38 + 40 + 42 + 44 + 46 = sum of 10 terms in ap whose first term is 30 and last term is 46 = 9 / 2 [ 30 + 46 ] = 342 km . answer : a" | a = 2 * 9
b = 30 + a
c = b + 30
d = c / 2
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a ) 36 - 48 , b ) 50 - 34 , c ) 60 - 24 , d ) 30 - 42 , e ) 21 - 63 | d | divide(subtract(72, 12), const_2) | the sum of two numbers is 72 , and one of them is 12 more than the other . what are the two numbers ? | "in this problem , we are asked to find two numbers . therefore , we must let x be one of them . let x , then , be the first number . we are told that the other number is 12 more , x + 12 . the problem states that their sum is 72 : word problem = 72 the line over x + 12 is a grouping symbol called a vinculum . it saves us writing parentheses . we have : 2 x = 72 Γ’ Λ β 12 = 60 . x = 60 / 2 = 30 . this is the first number . therefore the other number is x + 12 = 30 + 12 = 42 . the sum of 30 + 42 is 72 . d" | a = 72 - 12
b = a / 2
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a ) 3 % , b ) 5 % , c ) 7 % , d ) 10 % , e ) 12 % | c | multiply(multiply(10, 10), subtract(const_1, divide(add(multiply(9, const_60), 36), add(multiply(10, const_60), 20)))) | bob wants to run a mile in the same time as his sister . if bob β s time for a mile is currently 10 minutes 20 seconds and his sister β s time is currently 9 minutes 36 seconds , by what percent does bob need to improve his time in order run a mile in the same time as his sister ? | "bob ' s time = 620 secs . his sis ' time = 576 secs . percent increase needed = ( 620 - 576 / 620 ) * 100 = 44 / 640 * 100 = 7 % . ans ( c ) ." | a = 10 * 10
b = 9 * const_60
c = b + 36
d = 10 * const_60
e = d + 20
f = c / e
g = 1 - f
h = a * g
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a ) 8,5 , b ) 4,0 , c ) 1,4 , d ) 0,8 , e ) 8,0 | b | divide(divide(multiply(lcm(8, 5), const_2), const_10), const_2) | which digits should come in place of * and $ if the number 62684 * $ is divisible by both 8 and 5 ? | "since the given number is divisible by 5 , so 0 or 5 must come in place of $ . but , a number ending with 5 is never divisible by 8 . so , 0 will replace $ . now , the number formed by the last three digits is 4 * 0 , which becomes divisible by 8 , if * is replaced by 4 . hence , digits in place of * and $ are 4 and 0 respectively . answer b 4,0" | a = math.lcm(8, 5)
b = a * 2
c = b / 10
d = c / 2
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a ) 2788 , b ) 4000 , c ) 7282 , d ) 2782 , e ) 2729 | b | multiply(subtract(12, 7), divide(3200, subtract(7, 3))) | an amount of money is to be divided between p , q and r in the ratio of 3 : 7 : 12 . if the difference between the shares of p and q is rs . 3200 , what will be the difference between q and r ' s share ? | "4 - - - 3200 5 - - - ? = > 4000 answer : b" | a = 12 - 7
b = 7 - 3
c = 3200 / b
d = a * c
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a ) 2 , 7 , b ) 7 , 4 , c ) 8 , 7 , d ) 4 , 7 , e ) 17 , 7 | e | divide(divide(add(24, 10), const_2), const_2) | a man can row downstream at 24 kmph and upstream at 10 kmph . find the speed of the man in still water and the speed of stream respectively ? | "let the speed of the man in still water and speed of stream be x kmph and y kmph respectively . given x + y = 24 - - - ( 1 ) and x - y = 10 - - - ( 2 ) from ( 1 ) & ( 2 ) 2 x = 34 = > x = 17 , y = 7 answer : e" | a = 24 + 10
b = a / 2
c = b / 2
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a ) s . 90 , b ) s . 620 , c ) s . 600 , d ) s . 650 , e ) s . 690 | c | subtract(multiply(15, divide(8000, const_100)), multiply(4, divide(15000, const_100))) | john purchased a grinder and a mobile for rs . 15000 & rs . 8000 respectively . he sold the grinder at a loss of 4 % and the mobile phone at a profit of 15 % . overall how much he make a profit . | "let the sp of the refrigerator and the mobile phone be rs . r and rs . m respectively . r = 15000 ( 1 - 4 / 100 ) = 15000 - 600 m = 8000 ( 1 + 15 / 100 ) = 8000 + 120 total sp - total cp = r + m - ( 15000 + 8000 ) = - 600 + 1200 = rs . 600 as this is positive , an overall profit of rs . 600 was made . c" | a = 8000 / 100
b = 15 * a
c = 15000 / 100
d = 4 * c
e = b - d
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a ) 380 sec , b ) 190 sec , c ) 726 sec , d ) 127 sec , e ) 128 sec | a | subtract(divide(multiply(const_1, const_1000), divide(50, 20)), 20) | in a kilometer race , a beats b by 50 meters or 20 seconds . what time does a take to complete the race ? | "time taken by b run 1000 meters = ( 1000 * 20 ) / 50 = 400 sec . time taken by a = 400 - 20 = 380 sec . answer : a" | a = 1 * 1000
b = 50 / 20
c = a / b
d = c - 20
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a ) 2 , b ) 0 , c ) 1 , d ) 3 , e ) 4 | b | divide(divide(divide(lcm(24, 57), 57), const_4), const_4) | what is the least value of x . so that 24 x 57 is divisible by 3 . | "explanation : the sum of the digits of the number is divisible by 3 , then the number is divisible by 3 . 2 + 4 + x + 5 + 7 = 18 + x least value of x may be 0 therefore 18 + 0 = 18 is divisible by 3 . answer : option b" | a = math.lcm(24, 57)
b = a / 57
c = b / 4
d = c / 4
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a ) 11.0 km , b ) 13.8 km , c ) 11.5 km , d ) 12.5 km , e ) 13.5 km | e | multiply(divide(multiply(5, 6), subtract(6, 5)), divide(add(12, 15), const_60)) | if i walk at 5 km / h , i miss the bus by 12 minutes . if i walk at 6 km / h , i reach 15 minutes before the arrival of the bus . how far i walk to reach the bus stand ? | "d = product of speed difference of time / difference of speed d = 5 x 6 / 60 [ 12 Γ’ Λ β ( Γ’ Λ β 15 ) / 6 - 5 ] [ here , Γ’ β¬ β ve sign indicates before the schedule time ] Γ’ β‘ β d = 13.5 km answer e" | a = 5 * 6
b = 6 - 5
c = a / b
d = 12 + 15
e = d / const_60
f = c * e
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a ) $ 1220 , b ) $ 1330 , c ) $ 1550 , d ) $ 1770 , e ) $ 1880 | e | multiply(surface_rectangular_prism(4, 5, 3), 20) | a rectangular tank needs to be coated with insulation . the tank has dimensions of 4 feet , 5 feet , and 3 feet . each square foot of insulation costs $ 20 . how much will it cost to cover the surface of the tank with insulation ? | "the total surface area is 2 ( 3 * 4 + 4 * 5 + 3 * 5 ) = 94 square feet the total cost is 94 * $ 20 = $ 1880 the answer is e ." | a = surface_rectangular_prism * (
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a ) 10 % , b ) 15 % , c ) 20 % , d ) 25 % , e ) 30 % | a | multiply(divide(subtract(4015, 3650), 3650), const_100) | a sum of money deposited at c . i . amounts to rs . 3650 in 2 years and to rs . 4015 in 3 years . find the rate percent ? | "explanation : 3650 - - - - - - - - 365 100 - - - - - - - - ? ( 10 % ) answer : option a" | a = 4015 - 3650
b = a / 3650
c = b * 100
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a ) 1 , b ) 9 , c ) - 1 , d ) 5 / 4 , e ) - 4 / 5 | b | divide(add(4, 5), subtract(5, 4)) | if 5 - 5 / x = 4 + 4 / x , then x = | "we ' re given the equation 5 - 5 / x = 4 + 4 / x . we ' re asked for the value of x . the common - denominator of these 4 numbers is x , so we need to multiply both sides of the equation by x , giving us . . . 5 x - 5 x / x = 4 x + 4 x / x we can then eliminate that denominator , which gives us . . . . 5 x - 5 = 4 x + 4 9 = x b" | a = 4 + 5
b = 5 - 4
c = a / b
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a ) 35 and 46 , b ) 35 and 49 , c ) 35 and 42 , d ) 35 and 41 , e ) 35 and 48 | a | add(multiply(sqrt(divide(subtract(3341, 891), const_2)), const_100), sqrt(subtract(3341, divide(subtract(3341, 891), const_2)))) | the sum of the squares of two numbers is 3341 and the diference of their squares is 891 . the numbers are : | explanation : let the numbers be x and y . then , adding ( i ) and ( ii ) , we get : subtracting ( ii ) from ( i ) , we get : so , the numbers are 35 and 46 answer : a | a = 3341 - 891
b = a / 2
c = math.sqrt(b)
d = c * 100
e = 3341 - 891
f = e / 2
g = 3341 - f
h = math.sqrt(g)
i = d + h
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a ) 36 , b ) 45 , c ) 48 , d ) 49 , e ) 50 | c | multiply(multiply(const_1000, add(1.1, 0.9)), inverse(multiply(add(60, 90), divide(const_1000, const_3600)))) | two trains are moving in opposite direction @ 60 kmph and 90 kmph . their lengths are 1.10 km and 0.9 km respectively . the time taken by the slower train to cross the faster train in seconds is | r . s = ( 60 + 90 ) kmph = > ( 150 * 5 / 18 ) m / sec = > ( 125 / 3 ) m / sec distance covered = ( 1.10 + 0.9 ) km = ( 2 * 1000 ) = 2000 m . required time = ( 2000 * 3 / 125 ) = 48 sec . answer : c | a = 1 + 1
b = 1000 * a
c = 60 + 90
d = 1000 / 3600
e = c * d
f = 1/(e)
g = b * f
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a ) s . 4000 , b ) s . 5000 , c ) s . 4500 , d ) s . 4800 , e ) s . 8000 | e | divide(8820, power(add(divide(5, const_100), const_1), 2)) | a sum amounts to rs . 8820 in 2 years at the rate of 5 % p . a . if interest was compounded yearly then what was the principal ? | "ci = 8820 , r = 5 , n = 2 ci = p [ 1 + r / 100 ] ^ 2 = p [ 1 + 5 / 100 ] ^ 2 8820 = p [ 21 / 20 ] ^ 2 8820 [ 20 / 21 ] ^ 2 8000 answer : e" | a = 5 / 100
b = a + 1
c = b ** 2
d = 8820 / c
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a ) 21 , b ) 28 , c ) 34 , d ) 32 , e ) 76 | b | divide(multiply(24, 42), add(24, 12)) | a gardener wants to plant trees in his garden in rows in such away that the number of trees in each row to be the same . if there are 24 rows the number of trees in each row is 42 if there are 12 more rows find the number of trees in each row ? | required number of trees = 24 / 36 * 42 = 28 . answer : b | a = 24 * 42
b = 24 + 12
c = a / b
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a ) 21.6 , b ) 23.6 , c ) 22.6 , d ) 21.8 , e ) 21.3 | a | divide(multiply(13, 5), subtract(13, 10)) | a contractor undertook to do a piece of work in 10 days . he employed certain number of laboures but 5 of them were absent from the very first day and the rest could finish the work in only 13 days . find the number of men originally employed ? | "let the number of men originally employed be x . 10 x = 13 ( x Γ’ β¬ β 5 ) or x = 21.6 answer a" | a = 13 * 5
b = 13 - 10
c = a / b
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a ) 0.06 , b ) ( 0.06 ) ^ 25 , c ) 1 - ( 0.06 ) ^ 25 , d ) 1 - ( 0.94 ) ^ 25 , e ) ( 0.94 ) ^ 25 | d | subtract(const_1, power(subtract(const_1, 0.06), 25)) | a string of 25 light bulbs is wired in such a way that if any individual light bulb fails , the entire string fails . if for each individual light bulb the probability of failing during time period t is 0.06 , what is the probability that the string of light bulbs will fail during the time period t ? | "the string of light bulbs will fail if at least one light bulb fails . so , let ' s find the probability of the opposite event and subtract that value from 1 . the opposite event is when none of the 25 light bulbs fails , since the probability of each light bulb not to fail is 1 - 0.06 = 0.94 the probability that none of the 25 light bulbs fails is 0.94 ^ 25 . hence , the probability that at least one light bulb fails is 1 - 0.94 ^ 25 . answer : d . now , you should have spotted that your reasoning was not right because of one simple thing , consider the case when we have 100 light bulbs instead of 25 , then according to your logic the probability that the string of light bulbs will fail would be 100 * 0.06 = 6 , which is not possible since the probability of an event can not be more than 1 ( 100 % ) ." | a = 1 - 0
b = a ** 25
c = 1 - b
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a ) 1 / 140 , b ) 10 / 21 , c ) 27 / 410 , d ) 11 / 21 , e ) 57 / 120 | b | add(add(divide(1, 4), divide(1, 12)), divide(1, 7)) | in a race where 16 cars are running , the chance that car x will win is 1 / 4 , that y will win is 1 / 12 and that z will win is 1 / 7 . assuming that a dead heat is impossible , find the chance that one of them will win . | "required probability = p ( x ) + p ( y ) + p ( z ) ( all the events are mutually exclusive ) . = 1 / 4 + 1 / 12 + 1 / 7 = 10 / 21 answer : b" | a = 1 / 4
b = 1 / 12
c = a + b
d = 1 / 7
e = c + d
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a ) 7,000 , b ) 24,000 , c ) 40,000 , d ) 109,200 , e ) 168,000 | d | multiply(const_4, const_10) | a certain machine produces 650 units of product p per hour . working continuously at this constant rate , this machine will produce how many units of product p in 7 days ? | "since 7 days consist of 24 * 7 hours the total hours of a week is 168 hours . since every hour the machine produces 650 units of product p the total product during 168 hours is 650 * 168 = 109,200 correct option : d" | a = 4 * 10
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a ) 83.33 , b ) 110 , c ) 200 , d ) 120 , e ) 160 | c | multiply(divide(const_100, 10), 20) | a 20 % stock yielding 10 % is quoted at : | "solution to earn rs . 10 , money invested = rs . 100 . to earn rs . 20 , money invested = rs . ( 100 / 10 x 20 ) = rs . 200 . Γ’ Λ Β΄ market value of rs . 100 stock = rs . 200 answer c" | a = 100 / 10
b = a * 20
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a ) 64 , b ) 72 , c ) 141 , d ) 96 , e ) 108 | c | divide(subtract(multiply(subtract(24.50, 0.50), 5), 52), 0.50) | a cricket player whose bowling average was 24.50 runs per wicket , takes 5 wicket for 52 runs in a match . due to this his average decreases by 0.50 . what will be the number of wickets taken by him till the last match ? | "average = total runs / total wickets total runs after last match = 24.50 w + 52 total wickets after last match = w + 5 ( 24.50 w + 52 ) / ( w + 5 ) = 24.5 - 0.50 = 24 w = 136 so total wickets aftr last match = w + 5 = 141 answer : c" | a = 24 - 50
b = a * 5
c = b - 52
d = c / 0
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a ) 24 , b ) 30 , c ) 36 , d ) 90 , e ) 96 | a | divide(subtract(300, add(120, 60)), add(4, const_1)) | a polling company found that , of 300 households surveyed , 120 spent at least $ 100 per month on both gasoline and electricity , 60 spent at least $ 100 per month on gasoline but not on electricity , and for every household that did not spend at least $ 100 per month on gasoline or electricity , 4 spent at least $ 100 per month on electricity but not on gasoline . how many of the 300 households did not spend at least $ 100 per month on either gasoline or electricity ? | 120 spend $ 100 on gas & electricity 60 spend $ 100 on gas but not on electricity for every one that spent < $ 100 on both there are 4 that spent $ 100 on electricity not gas let the ones that spent on neither be x the ones that spent on just electricity will be 4 x total households = those who spend > $ 100 on both + those who spend > $ 100 on just one + those who spend < $ 100 on both = 120 + 60 + 4 x + x 300 = 180 + 5 x implies x = 24 answer is ( a ) | a = 120 + 60
b = 300 - a
c = 4 + 1
d = b / c
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a ) 366 , b ) 107 , c ) 102 , d ) 192 , e ) 122 | b | subtract(107.25, divide(1, 4)) | the cash realised on selling a 14 % stock is rs . 107.25 , brokerage being 1 / 4 % is ? | "cash realised = rs . ( 107.25 - 0.25 ) = rs . 107 . answer : b" | a = 1 / 4
b = 107 - 25
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a ) 28.5 , b ) 46 , c ) 30 , d ) data inadequate , e ) none of these | b | subtract(divide(multiply(add(const_100, 38.7), const_100), subtract(const_100, 5)), const_100) | a shopkeeper sold an article offering a discount of 5 % and earned a profit of 38.7 % . what would have been the percentage of profit earned if no discount had been offered ? | "giving no discount to customer implies selling the product on printed price . suppose the cost price of the article is 100 . then printed price = 100 Γ£ β ( 100 + 38.7 ) / ( 100 Γ’ Λ β 5 ) = 146 hence , required % profit = 146 Γ’ β¬ β 100 = 46 % answer b" | a = 100 + 38
b = a * 100
c = 100 - 5
d = b / c
e = d - 100
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a ) 48 kg , b ) 49 kg , c ) 50 kg , d ) 51 kg , e ) none | c | divide(multiply(const_100, 49), subtract(const_100, const_2)) | the price of rice is reduced by 2 % . how many kilograms of rice can now be bought for the money which was sufficient to buy 49 kg of rice earlier ? | sol . let the original price be rs . 100 per kg . money required to buy 49 kg of rice = rs . ( 100 x 49 ) = rs . 4900 . new price = rs . 98 per kg . β΄ quantity of rice bought = [ 4900 / 98 ] kg = 50 kg . answer c | a = 100 * 49
b = 100 - 2
c = a / b
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a ) 2100 , b ) 2550 , c ) 5050 , d ) 6275 , e ) 11325 | a | multiply(subtract(110, 10), add(divide(subtract(50, 10), const_2), const_1)) | set a contains all the even numbers between 10 and 50 inclusive . set b contains all the even numbers between 110 and 150 inclusive . what is the difference between the sum of elements of set b and the sum of the elements of set a ? | "set a contains 10 , 12,14 , 16 . . . 50 set b contains 110 , 112 , 114 . . . 150 number of terms in each set = 21 difference between corresponding terms in set a and b = 100 difference between sum of set b and set a = 100 * 21 = 2100 answer a" | a = 110 - 10
b = 50 - 10
c = b / 2
d = c + 1
e = a * d
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a ) 887 , b ) 678 , c ) 765 , d ) 345 , e ) 256 | a | subtract(1000, subtract(add(divide(1000, 15), divide(1000, 20)), divide(1000, multiply(15, 20)))) | what is the number of integers from 1 to 1000 ( inclusive ) that are divisible by neither 15 nor by 20 ? | "in 1000 consecutive numbers , number of multiples of 15 = 1000 / 15 = 66 ( ignore decimals ) in 1000 consecutive numbers , number of multiples of 20 = 1000 / 20 = 50 number of multiples of 15 * 20 i . e . 300 = 1000 / 300 = 3 number of integers from 1 to 1000 that are divisible by neither 15 nor by 20 = 1000 - ( 66 + 50 - 3 ) { using the concept of sets here ) = 887 answer is a" | a = 1000 / 15
b = 1000 / 20
c = a + b
d = 15 * 20
e = 1000 / d
f = c - e
g = 1000 - f
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a ) 3 , b ) 5 , c ) 7 , d ) 9 , e ) 11 | e | subtract(subtract(13, 1), 1) | if x is an integer such that 5 < x < 21 , 7 < x < 18 , 13 > x > 2 , 12 > x > 9 , and x + 1 < 13 , then x is | 5 < x < 21 , 7 < x < 18 , 2 < x < 13 9 < x < 12 x < 13 from above : 10 < x < 12 - - > x = 11 . answer : e . | a = 13 - 1
b = a - 1
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a ) 700 , b ) 800 , c ) 1,300 , d ) 1,500 , e ) 400 | e | multiply(const_1, const_1) | the cost of registration at a professional association meeting was $ 50 per person ; a lunch for registrants only was available for an additional $ 22 per person . if the number of registrants who paid for lunch was 20 more than the number who did not , and if receipts for registration and lunch totaled $ 50,240 , how many people paid just for registration at the meeting ? | hope this might be useful to you . let the number of people who have opted only to register = x now since the registration cost is 50 $ per person , the total amount sums to = 50 x $ as per the information given in the question , the number of registrants who paid for lunch was 20 more than the number who did not . that means , total number of people who registered and paid for lunch = 20 + x . for the people who registered for lunch the cost is 50 $ ( for the event registration ) + 22 $ ( for lunch ) = 72 $ . total amount in this case sums to = 72 ( 20 + x ) = 1440 + 72 x now , total amount received was 50240 . thus , from the above data , 50 x + 1440 + 72 x = 50240 122 x = 50240 - 1440 122 x = 48800 x = 400 . hence the correct ans is e | a = 1 * 1
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a ) 53 , b ) 47 , c ) 48 , d ) 59 , e ) 45 | a | subtract(negate(37), multiply(subtract(29, 31), divide(subtract(29, 31), subtract(23, 29)))) | 23 , 29 , 31 , 37 , 41 , 43 , 47 , ( . . . ) | "explanation : all are prime numbers in their order , starting from 23 hence , next number is 53 answer : a" | a = negate - (
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a ) 23 / 38 , b ) 29 / 44 , c ) 33 / 48 , d ) 38 / 53 , e ) 42 / 57 | e | divide(divide(subtract(multiply(3, add(3, 15)), 4), subtract(4, 3)), add(divide(subtract(multiply(3, add(3, 15)), 4), subtract(4, 3)), 15)) | the denominator of a fraction is 15 greater than the numerator . if the numerator and the denominator are increased by 3 , the resulting fraction is equal to 3 Γ’ Β β 4 . what is the value of the original fraction ? | "let the numerator be x . then the denominator is x + 15 . x + 3 / x + 18 = 3 / 4 . 4 x + 12 = 3 x + 54 . x = 42 . the original fraction is 42 / 57 . the answer is e ." | a = 3 + 15
b = 3 * a
c = b - 4
d = 4 - 3
e = c / d
f = 3 + 15
g = 3 * f
h = g - 4
i = 4 - 3
j = h / i
k = j + 15
l = e / k
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a ) 5 : 2 , b ) 9 : 2 , c ) 18 : 20 , d ) 1 : 4 , e ) 18 : 4 | a | 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 . 50,000 and rs 20,000 respectively . in what ratio the profit earned after 2 years be divided between a and b respectively ? | "a : b = 50000 : 20000 = 5 : 2 answer : a" | 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 ) rs . 3000 , b ) rs . 4991 , c ) rs . 5991 , d ) rs . 6001 , e ) none | b | subtract(multiply(add(5, const_1), 6500), add(add(add(add(6345, 6927), 6855), 7230), 6562)) | a grocer has a sale of rs . 6345 , rs . 6927 , rs . 6855 , rs . 7230 and rs . 6562 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . 6500 ? | "sol . total sale for 5 months = rs . ( 6435 + 6927 + 6855 + 7230 + 6562 ) = rs . 34009 . β΄ required sale = rs . [ ( 6500 x 6 ) - 34009 ] = rs . ( 39000 - 34009 ) = rs . 4991 . answer b" | a = 5 + 1
b = a * 6500
c = 6345 + 6927
d = c + 6855
e = d + 7230
f = e + 6562
g = b - f
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a ) 9 , b ) 12 , c ) 15 , d ) 16 , e ) 24 | e | subtract(multiply(divide(multiply(24, divide(4, add(4, 2))), divide(3, add(4, 2))), divide(6, add(4, 2))), subtract(24, multiply(24, divide(4, add(4, 2))))) | at deluxe paint store , fuchsia paint is made by mixing 4 parts of red paint with 2 parts of blue paint . mauve paint is made by mixing 3 parts of red paint with 6 parts blue paint . how many liters of blue paint must be added to 24 liters of fuchsia to change it to mauve paint ? | "in 24 liters , red = 4 / 6 * 24 = 16 and blue = 8 so , 16 / ( 8 + x ) = 3 / 6 or , x = 24 ( answer e )" | a = 4 + 2
b = 4 / a
c = 24 * b
d = 4 + 2
e = 3 / d
f = c / e
g = 4 + 2
h = 6 / g
i = f * h
j = 4 + 2
k = 4 / j
l = 24 * k
m = 24 - l
n = i - m
|
a ) 154 , b ) 248 , c ) 360 , d ) 424 , e ) 168 | e | multiply(multiply(divide(multiply(8, multiply(multiply(multiply(power(2, 4), power(3, 2)), 17), 7)), multiply(136, 144)), 8), const_3) | h . c . f of 136 , 144 and a third number is 8 and their l . c . m is 2 ^ 4 * 3 ^ 2 * 17 * 7 . the third number is : | 136 = 2 ^ 3 * 17 ; 144 = 2 ^ 4 * 3 ^ 2 h . c . f = 2 ^ 3 = 8 since h . c . f is the product of lowest powers of common factors , so the third number must have ( 2 ^ 3 ) as its factor . since l . c . m is the product of highest powers of common prime factors , so the third number must have 3 and 7 as its factors . third number = 2 ^ 3 * 3 * 7 = 168 answer : e | a = 2 ** 4
b = 3 ** 2
c = a * b
d = c * 17
e = d * 7
f = 8 * e
g = 136 * 144
h = f / g
i = h * 8
j = i * 3
|
a ) 80 , b ) 90 , c ) 95 , d ) 100 , e ) 112.5 | e | multiply(multiply(multiply(const_100, divide(add(const_100, 20), const_100)), divide(subtract(const_100, 25), const_100)), divide(add(const_100, 25), const_100)) | from the beginning to the end of 2007 , the price of a stock rose 20 percent . in 2008 , it dropped 25 percent . in 2009 , it rose 25 percent . what percent of the stock Γ’ β¬ β’ s 2007 starting price was the price of the stock at the end of 2009 ? | "assume a value at the beginning of 2007 . as this is a % question , assume p = 100 . at the end of 2007 it becmae = 1.2 * 100 = 120 at the end of 2008 it decreased by 25 % = 120 * . 75 = 90 at the end of 2009 it increased by 25 % = 90 * 1.25 = 112.5 thus ratio = 112.5 / 100 = 1.125 ( in % terms = 12.5 % ) . thus e is the correct answer ." | a = 100 + 20
b = a / 100
c = 100 * b
d = 100 - 25
e = d / 100
f = c * e
g = 100 + 25
h = g / 100
i = f * h
|
a ) 10 mps , b ) 19.44 mps , c ) 97.44 mps , d ) 17.44 mps , e ) 18.44 mps | b | multiply(const_0_2778, 70) | express a speed of 70 kmph in meters per second ? | "70 * 5 / 18 = 19.44 mps answer : b" | a = const_0_2778 * 70
|
a ) 0 , b ) 2 , c ) 4 , d ) 6 , e ) 8 | b | subtract(reminder(add(power(reminder(33, const_10), reminder(43, const_4)), power(reminder(43, const_10), const_4)), const_100), multiply(const_2, const_3)) | if n = ( 33 ) ^ 43 + ( 43 ) ^ 32 what is the units digit of n ? | first of all , the units digit of ( 33 ) ^ 43 is the same as that of 3 ^ 43 and the units digit of ( 43 ) ^ 32 is the same as that of 3 ^ 32 . so , we need to find the units digit of 3 ^ 43 + 3 ^ 33 . next , the units digit of 3 in positive integer power repeats in blocks of four { 3 , 9 , 7 , 1 } : 3 ^ 1 = 3 ( the units digit is 3 ) 3 ^ 2 = 9 ( the units digit is 9 ) 3 ^ 3 = 27 ( the units digit is 7 ) 3 ^ 4 = 81 ( the units digit is 1 ) 3 ^ 5 = 243 ( the units digit is 3 again ! ) . . . thus : the units digit of 3 ^ 43 is the same as the units digit of 3 ^ 3 , so 7 ( 43 divided by the cyclicity of 4 gives the remainder of 3 ) . the units digit of 3 ^ 33 is the same as the units digit of 3 ^ 1 , so 3 ( 33 divided by the cyclicity of 4 gives the remainder of 1 ) . therefore the units digit of ( 33 ) ^ 43 + ( 43 ) ^ 32 is 7 + 5 = 2 . answer : b . | a = reminder ** (
b = a + reminder
c = reminder - (
|
a ) 20 % , b ) 25 % , c ) 30 % , d ) 35 % , e ) 40 % | b | divide(20, subtract(const_1, divide(20, const_100))) | on a certain transatlantic crossing , 20 percent of a ship β s passengers held round - trip tickets and also took their cars abroad the ship . if 20 percent of the passengers with round - trip tickets did not take their cars abroad the ship , what percent of the ship β s passengers held round - trip tickets ? | "let t be the total number of passengers . let x be the number of people with round trip tickets . 0.2 t had round trip tickets and took their cars . 0.8 x had round trip tickets and took their cars . 0.8 x = 0.2 t x = 0.25 t the answer is b ." | a = 20 / 100
b = 1 - a
c = 20 / b
|
a ) 150 meter , b ) 286 meter , c ) 186 meter , d ) 315 meter , e ) 265 meter | d | multiply(divide(multiply(126, const_1000), const_3600), 9) | a train running at the speed of 126 km / hr crosses a pole in 9 seconds . find the length of the train . | "speed = 126 * ( 5 / 18 ) m / sec = 35 m / sec length of train ( distance ) = speed * time ( 35 ) * 9 = 315 meter answer : d" | a = 126 * 1000
b = a / 3600
c = b * 9
|
a ) 200 , b ) 220 , c ) 240 , d ) 322 , e ) 342 | a | add(lcm(lcm(15, 30), lcm(45, 60)), 20) | what is the smallest number which when diminished by 20 , is divisible 15 , 30 , 45 and 60 ? | "required number = ( lcm of 15 , 30 , 45 and 60 ) + 20 = 180 + 20 = 200 option a" | a = math.lcm(15, 30)
b = math.lcm(45, 60)
c = math.lcm(a, b)
d = c + 20
|
a ) 30 , b ) 20 , c ) 23 , d ) 17 , e ) 19 | b | subtract(add(add(22, 20), 3), 25) | mrs . evans gave a test to her freshmen economics class , which has 25 students enrolled and 22 of them answered question 1 correctly . if 20 answered question 2 correctly and 3 did not take the test then how many answered both questions correctly ? | "total number of enrolled students = 25 number of students who did not take test = 3 hence , number of students who took test = 25 - 3 = 22 number of students who answered q 2 correctly = 20 , therefore , all students who took test answered q 2 correctly . so , number of students who answered q 1 correctly , also answered q 2 correctly = 20 . number of students who answered both q 1 & q 2 correctly = 20 . answer : b" | a = 22 + 20
b = a + 3
c = b - 25
|
a ) 34 kg , b ) 35 kg , c ) 36 kg , d ) 30 kg , e ) 32 kg | b | add(add(10, 16), subtract(divide(multiply(add(multiply(16, divide(3, 4)), divide(10, add(4, 1))), 3), 2), multiply(16, divide(3, 4)))) | in two alloys , copper and tin are related in the ratios of 4 : 1 and 1 : 3 . 10 kg of ist alloy , 16 kg of 2 nd alloy and some pure copper melted together . an alloy obtained in which the ratio of copper and tin was 3 : 2 . find the weight of the new alloy . | suppose x kg of pure copper is melted with 10 kg of ist alloy and 16 kg of 2 nd alloy , then pure cooper in new alloy = 10 * 4 / 5 + 16 * 1 / 4 + x = 12 + x kg pure tin in new alloy = 10 * 1 / 5 + 16 * 3 / 4 = 14 kg as per given condition ( 12 + x ) / 14 = 3 / 2 24 + 2 x = 42 x = 9 kg total weight of new alloy = 10 + 16 + 9 = 35 kg answer : b | a = 10 + 16
b = 3 / 4
c = 16 * b
d = 4 + 1
e = 10 / d
f = c + e
g = f * 3
h = g / 2
i = 3 / 4
j = 16 * i
k = h - j
l = a + k
|
a ) 2337 , b ) 2667 , c ) 1440 , d ) 2887 , e ) 2661 | c | add(multiply(120, const_2), multiply(add(120, divide(multiply(120, 25), const_100)), multiply(const_4, const_2))) | a farmer planned to plough a field by doing 120 hectares a day . after two days of work he increased his daily productivity by 25 % and he finished the job two days ahead of schedule . a ) what is the area of the field ? b ) in how many days did the farmer get the job done ? c ) in how many days did the farmer plan to get the job done ? | first of all we will find the new daily productivity of the farmer in hectares per day : 25 % of 120 hectares is 25100 β
120 = 3025100 β
120 = 30 hectares , therefore 120 + 30 = 150120 + 30 = 150 hectares is the new daily productivity . lets x be the planned number of days allotted for the job . then the farm is 120 β
x 120 β
x hectares . on the other hand , we get the same area if we add 120 β
2120 β
2 hectares to 150 ( x β 4 ) 150 ( x β 4 ) hectares . then we get the equation 120 x = 120 β
2 + 150 ( x β 4 ) 120 x = 120 β
2 + 150 ( x β 4 ) x = 12 x = 12 so , the job was initially supposed to take 12 days , but actually the field was ploughed in 12 - 2 = 10 days . the field ' s area is 120 β
12 = 1440120 β
12 = 1440 hectares . answer : c | a = 120 * 2
b = 120 * 25
c = b / 100
d = 120 + c
e = 4 * 2
f = d * e
g = a + f
|
a ) 2 days , b ) 7 days , c ) 8 days , d ) 9 days , e ) 45 days | a | divide(subtract(multiply(12, 9), multiply(12, 6)), add(12, 6)) | 12 men complete a work in 9 days . after they have worked for 6 days , 6 more men join them . how many days will they take to complete the remaining work ? | 1 man ' s 1 day work = 1 / 108 12 men ' s 6 day ' s work = 1 / 9 * 6 = 2 / 3 remaining work = 1 - 2 / 3 = 1 / 3 18 men ' s 1 day work = 1 / 108 * 18 = 1 / 6 1 / 6 work is done by them in 1 day . 1 / 3 work is done by them in 6 * 1 / 3 = 2 days . answer : a | a = 12 * 9
b = 12 * 6
c = a - b
d = 12 + 6
e = c / d
|
a ) $ 120 , b ) $ 130 , c ) $ 140 , d ) $ 150 , e ) $ 160 | e | divide(add(add(200, 20), 20), add(subtract(const_1, divide(const_1, const_4)), subtract(const_1, divide(const_1, const_4)))) | joe has a total of $ 200 in his two pockets . he takes one fourth of the money in his left pocket and puts it in his right pocket . he then takes $ 20 from his left pocket and puts it in his right pocket . if he now has an equal amount of money in each pocket , then how much money did he originally have in his left pocket ? | joe had $ 100 in his left and $ 100 in his right pocket at the end . so , he had $ 120 in his left pocket ( and $ 80 in his right pocket ) before moving $ 20 from his left to his right pocket . thus , he had $ 160 in his left pocket originally ( solve x β x / 4 = 120 ) . correct answer e | a = 200 + 20
b = a + 20
c = 1 / 4
d = 1 - c
e = 1 / 4
f = 1 - e
g = d + f
h = b / g
|
a ) 50 hours , b ) 40 hours , c ) 25 hours , d ) 12 hours , e ) 8 hours | c | divide(200, multiply(divide(divide(20, 10), 5), 20)) | if 5 machines can produce 20 units in 10 hours , how long would it take 20 machines to produce 200 units ? | here , we ' re told that 5 machines can produce 20 units in 10 hours . . . . that means that each machine works for 10 hours apiece . since there are 5 machines ( and we ' re meant to assume that each machine does the same amount of work ) , then the 5 machines equally created the 20 units . 20 units / 5 machines = 4 units are made by each machine every 10 hours now that we know how long it takes each machine to make 4 units , we can break this down further if we choose to . . . 10 hours / 4 units = 2.5 hours per unit when 1 machine is working . the prompt asks us how long would it take 20 machines to produce 200 units . if 20 machines each work for 2.5 hours , then we ' ll have 20 units . since 200 units is ' 10 times ' 20 , we need ' 10 times ' more time . ( 2.5 hours ) ( 10 times ) = 25 hours final answer : [ reveal ] spoiler : c | a = 20 / 10
b = a / 5
c = b * 20
d = 200 / c
|
a ) 45 % , b ) 70 % , c ) 35 % , d ) 32.5 % , e ) 30 % | b | subtract(const_100, multiply(divide(subtract(const_100, 60), const_100), subtract(const_100, 25))) | the price of a cycle is reduced by 25 per cent . the new price is reduced by a further 60 per cent . the two reductions together are equal to a single reduction of | "price = p initially price reduced by 25 % which means new price is 3 / 4 p now on this new price further 60 percent is reduced which means the new price is merely 40 percent of 3 / 4 p = = > ( 3 / 4 ) x ( 2 / 5 ) p = 6 / 20 p is the new price after both deduction which is 30 percent of the original value p . this implies this entire series of deduction is worth having discounted 70 % of p . so answer is b = 70 %" | a = 100 - 60
b = a / 100
c = 100 - 25
d = b * c
e = 100 - d
|
a ) 1767 , b ) 1800 , c ) 1812 , d ) 1890 , e ) none | a | multiply(add(add(const_1, add(const_3, const_4)), divide(divide(1225, add(const_2, const_3)), add(const_2, const_3))), add(add(const_1, add(const_2, const_3)), power(add(const_2, const_3), const_2))) | given a number , there is an algorithm described here to find it ' s sum and number of factors . for example , let us take the number 1225 ? | solving for 1225 = 52 Γ 72 sum of divisors = ( ap + 1 β 1 ) / ( a β 1 ) Γ ( bq + 1 β 1 ) / ( b β 1 ) here a = 5 , b = 7 prime factors p = 2 and q = 2 sum of divisors = ( 53 β 1 ) / ( 5 β 1 ) Γ ( 73 β 1 ) / ( 7 β 1 ) = ( 124 / 4 ) Γ ( 342 / 6 ) = ( 31 Γ 57 ) = 1767 option a | a = 3 + 4
b = 1 + a
c = 2 + 3
d = 1225 / c
e = 2 + 3
f = d / e
g = b + f
h = 2 + 3
i = 1 + h
j = 2 + 3
k = j ** 2
l = i + k
m = g * l
|
a ) 20 , b ) 10 , c ) 15 , d ) 16 , e ) 18 | a | add(add(divide(3, 4), multiply(divide(3, 4), 20)), multiply(const_0_25, 20)) | total 60 cows 20 cow gives each 2 liter milk 20 cow gives each 3 / 4 liter milk 20 cow gives each 1 / 4 liter milk this is split into 3 son per each 20 cows & 20 liter milk how ? | 20 cow 2 liter each = 40 liter 20 cow 3 / 4 liter each = 3 / 4 = 0.75 * 20 = 15 20 cow 1 / 4 liter each = 1 / 4 = 0.25 * 20 = 5 add 40 + 15 + 5 = 60 milk split into 3 son each 20 liter then 60 / 3 = 20 answer : a | a = 3 / 4
b = 3 / 4
c = b * 20
d = a + c
e = const_0_25 * 20
f = d + e
|
a ) 1 : 2 , b ) 2 : 1 , c ) 1 : 3 , d ) 3 : 1 , e ) 18 : 7 | e | divide(multiply(45000, const_12), multiply(30000, add(const_4, const_3))) | x starts a business with rs . 45000 . y joins in the business after 5 months with rs . 30000 . what will be the ratio in which they should share the profit at the end of the year ? | explanation : ratio in which they should share the profit = ratio of the investments multiplied by the time period = 45000 Γ£ β 12 : 30000 Γ£ β 7 = 45 Γ£ β 12 : 30 Γ£ β 7 = 3 Γ£ β 12 : 2 Γ£ β 7 = 18 : 7 answer : option e | a = 45000 * 12
b = 4 + 3
c = 30000 * b
d = a / c
|
a ) 1600 , b ) 1700 , c ) 1800 , d ) 1900 , e ) 1000 | c | divide(multiply(9, 2000), add(1, 9)) | ashok and pyarelal invested money together in a business and share a capital of ashok is 1 / 9 of that of pyarelal . if the incur a loss of rs 2000 then loss of pyarelal ? | "let the capital of pyarelal be x , then capital of ashok = x / 9 so ratio of investment of pyarelal and ashok = x : x / 9 = 9 x : x hence out of the total loss of 2000 , loss of pyarelal = 2000 * 9 x / 10 x = 1800 answer : c" | a = 9 * 2000
b = 1 + 9
c = a / b
|
a ) rs . 500 , b ) rs . 1500 , c ) rs . 2000 , d ) rs . 2400 , e ) none of the above | d | multiply(multiply(subtract(4, 3), 1200), 3) | a sum of money is to be distributed among a , b , c , d in the proportion of 5 : 2 : 4 : 3 . if c gets rs . 1200 more than d , what is b ' s share ? | "let the shares of a , b , c and d be rs . 5 x , rs . 2 x , rs . 4 x and rs . 3 x respectively . then , 4 x - 3 x = 1200 x = 1200 . b ' s share = rs . 2 x = rs . ( 2 x 1200 ) = rs . 2400 . answer = d" | a = 4 - 3
b = a * 1200
c = b * 3
|
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