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 , b ) 12 , c ) 17 , d ) 18 , e ) 20 | c | add(16, 1) | seller selling an apple for rs . 16 , a seller loses 1 / 6 th of what it costs him . the cp of the apple is ? | "sp = 16 loss = cp 17 loss = cp − sp = cp − 16 ⇒ cp 17 = cp − 16 ⇒ 16 cp 17 = 16 ⇒ cp 17 = 1 ⇒ cp = 17 c" | a = 16 + 1
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a ) 2927 , b ) 3147 , c ) 3387 , d ) 3567 , e ) 3797 | b | add(lcm(lcm(9, 70), lcm(25, 21)), 3) | what is the smallest number which when increased by 3 is divisible by 9 , 70 , 25 and 21 ? | "when increased by 3 , the number must include at least 2 * 3 ^ 2 * 5 ^ 2 * 7 = 3150 the answer is b ." | a = math.lcm(9, 70)
b = math.lcm(25, 21)
c = math.lcm(a, b)
d = c + 3
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a ) 3.1 feet , b ) 3.2 feet , c ) 3.3 feet , d ) 3.4 feet , e ) 3.5 feet | e | divide(add(add(multiply(add(2, 10), 2), 10), 8), add(2, 10)) | carmen made a sculpture from small pieces of wood . the sculpture is 2 feet 10 inches tall . carmen places her sculpture on a base that is 8 inches tall . how tall are the sculpture andbase together ? | "we know 1 feet = 12 inch then 2 feet = 24 inch 24 + 10 = 34 then 34 + 8 = 42 42 / 12 = 3.5 feet answer : e" | a = 2 + 10
b = a * 2
c = b + 10
d = c + 8
e = 2 + 10
f = d / e
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a ) 9 , b ) 13 , c ) 16 , d ) 20 , e ) 25 | d | subtract(const_100, divide(const_100, divide(add(25, const_100), const_100))) | an employee whose hourly wage was increased by 25 percent decided to reduce the number of hours worked per week so that the employee ' s total weekly income would remain unchanged . by what percent should the number of hours worked be reduced ? | let original hourly wage be x and let the no of hours worked be y total wage will be = x * y after the increment the wage will be = 1.25 x now we need to find number of hours worked so that x * y = 1.25 x * z i . e z = 4 / 5 y % decrease = ( y - 4 / 5 y ) / y * 100 = 100 / 5 = 20 % . thus my answer is d | a = 25 + 100
b = a / 100
c = 100 / b
d = 100 - c
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a ) 10 , b ) 20 , c ) 30 , d ) 40 , e ) 50 | c | subtract(divide(multiply(divide(multiply(45, 8), 40), 50), 6), 45) | 45 men working 8 hours per day dig 40 m deep . how many extra men should be put to dig to a depth of 50 m working 6 hours per day ? | "( 45 * 8 ) / 40 = ( x * 6 ) / 50 = > x = 75 75 – 45 = 30 answer : c" | a = 45 * 8
b = a / 40
c = b * 50
d = c / 6
e = d - 45
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a ) 11 , b ) 37 , c ) 36 , d ) 76 , e ) 33 | d | add(divide(150, const_2), const_1) | there are 150 ' zeroes ' and 151 ' ones ' written on a black board . a step involves choosing 3 digits and replacing them with a single digit . if all the 3 digits are identical , they are replaced with the same digit . otherwise , the digit that appears twice replaces the 3 chosen digits . what is the minimum number of steps after which there will be no zero written on the blackboard ? | if we strike out 3 zeros , we are in effect reducing two zeroes . so we need 74 steps to reduce 148 zeroes . there are 2 zeroes and 151 ones left now . the remaining zeroes can be reduced in 2 steps . hence the answer is 76 . answer : d | a = 150 / 2
b = a + 1
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a ) 6 days , b ) 8 days , c ) 5 days , d ) 7 days , e ) 15 days | a | inverse(add(inverse(9), inverse(18))) | a and b complete a work in 9 days . a alone can do it in 18 days . if both together can do the work in how many days ? | "1 / 9 + 1 / 18 = 1 / 6 6 / 1 = 6 days answer : a" | a = 1/(9)
b = 1/(18)
c = a + b
d = 1/(c)
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a ) 140 , b ) 141 , c ) 142 , d ) 143 , e ) 146 | e | power(add(power(multiply(subtract(1, divide(1, 5)), 150), 2), power(subtract(power(add(power(150, 2), power(90, 2)), divide(1, 2)), multiply(subtract(1, divide(1, 5)), 150)), 2)), divide(1, 2)) | in a village there are 150 men and 90 women in present time . if in next year population will be p = ( a ^ 2 + b ^ 2 ) ^ 1 / 2 , and in every year men are reduces 5 % . what is population of after 2 year . | "next year total population = [ 150 ^ 2 + 90 ^ 2 ] ^ . 5 = 174.92 = 175 man decreased by 5 % so total man = 150 * . 95 = 142.5 women will be = 175 - 142.5 = 32.5 so population after two years = [ 135 ^ 2 + 32.5 ^ 2 ] ^ . 5 = 146.15 = 146 so population after two year = 146 answer : e" | a = 1 / 5
b = 1 - a
c = b * 150
d = c ** 2
e = 150 ** 2
f = 90 ** 2
g = e + f
h = 1 / 2
i = g ** h
j = 1 / 5
k = 1 - j
l = k * 150
m = i - l
n = m ** 2
o = d + n
p = 1 / 2
q = o ** p
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a ) 36 , b ) 34 , c ) 356 , d ) 12 , e ) 45 | b | divide(divide(200, 5280), multiply(4, divide(const_1, const_3600))) | if an object travels 200 feet in 4 seconds , what is the object ' s approximate speed in miles per hour ? ( note : 1 mile = 5280 feet ) | 1 mile = 5280 feet = > 1 feet = 1 / 5280 miles if the object travels 200 feet in 4 sec then it travels 200 / 4 * 60 * 60 feet in 1 hour ( 1 hr = 60 min * 60 sec ) = 3600 * 50 feet in 1 hour = 180000 feet in 1 hr = 180000 / 5280 miles in 1 hour = 18000 / 528 miles / hr ~ 34 miles / hr answer - b | a = 200 / 5280
b = 1 / 3600
c = 4 * b
d = a / c
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a ) 8078 , b ) 1040 , c ) 1058 , d ) 1021 , e ) 8000 | e | divide(power(power(3, const_2), const_3), power(3, const_3)) | a cube of side 3 meter length is cut into small cubes of side 15 cm each . how many such small cubes can be obtained ? | "along one edge , the number of small cubes that can be cut = 300 / 15 = 20 along each edge 20 cubes can be cut . ( along length , breadth and height ) . total number of small cubes that can be cut = 20 * 20 * 20 = 8000 answer : e" | a = 3 ** 2
b = a ** 3
c = 3 ** 3
d = b / c
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a ) 3348 , b ) 3898 , c ) 1200 , d ) 2881 , e ) 7881 | c | multiply(divide(multiply(15, 12), add(15, 12)), multiply(15, 12)) | a hall is 15 m long and 12 m broad . if the sum of the areas of the floor and the ceiling is equal to the sum of the areas of 4 walls , the volume of the hall is : | answer : c ) 1200 | a = 15 * 12
b = 15 + 12
c = a / b
d = 15 * 12
e = c * d
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a ) 120 cm 2 , b ) 112 cm 2 , c ) 776 cm 2 , d ) 666 cm 2 , e ) 168 cm 2 | e | divide(multiply(28, 12), const_2) | if the sides of a triangle are 30 cm , 28 cm and 12 cm , what is its area ? | "the triangle with sides 30 cm , 28 cm and 12 cm is right angled , where the hypotenuse is 30 cm . area of the triangle = 1 / 2 * 28 * 12 = 168 cm 2 answer : e" | a = 28 * 12
b = a / 2
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a ) 45 , b ) 65 , c ) 75 , d ) 89 , e ) 90 | c | subtract(multiply(80, const_4), subtract(multiply(79, const_4), add(3, subtract(multiply(80, const_4), multiply(84, 3))))) | the avg weight of a , b & c is 84 kg . if d joins the group , the avg weight of the group becomes 80 kg . if another man e who weights is 3 kg more than d replaces a , then the avgof b , c , d & e becomes 79 kg . what is the weight of a ? | "a + b + c = 3 * 84 = 252 a + b + c + d = 4 * 80 = 320 - - - - ( i ) so , d = 68 & e = 68 + 3 = 71 b + c + d + e = 79 * 4 = 316 - - - ( ii ) from eq . ( i ) & ( ii ) a - e = 320 – 316 = 4 a = e + 4 = 71 + 4 = 75 c" | a = 80 * 4
b = 79 * 4
c = 80 * 4
d = 84 * 3
e = c - d
f = 3 + e
g = b - f
h = a - g
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a ) 540 , b ) 570 , c ) 619 , d ) 649 , e ) 580 | e | multiply(divide(377, divide(add(const_100, 30), const_100)), 2) | if the price of a certain computer increased 30 percent from d dollars to 377 dollars , then 2 d = | "before price increase price = d after 30 % price increase price = d + ( 30 / 100 ) * d = 1.3 d = 377 ( given ) i . e . d = 377 / 1.3 = $ 290 i . e . 2 d = 2 * 290 = 580 answer : option e" | a = 100 + 30
b = a / 100
c = 377 / b
d = c * 2
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a ) 178 , b ) 180 , c ) 181 , d ) 182 , e ) 183 | e | add(add(add(50, 48), 44), 46) | you have been given a physical balance and 7 weights of 52 , 50 , 48 , 44 , 45 , 46 , and 78 kgs . . keeping weights on one pan and object on the other , what is the maximum you can weigh less than 183 kgs . ? | "balance = 52 + 50 + 48 + 44 + 45 + 46 + 78 - 183 answer : e" | a = 50 + 48
b = a + 44
c = b + 46
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a ) 4 % , b ) 5 % , c ) 6 % , d ) 7 % , e ) 8 % | b | divide(multiply(const_100, subtract(615, 600)), divide(600, 2)) | on a sum of money , the s . i . for 2 years is $ 600 , while the c . i . is $ 615 , the rate of interest being the same in both the cases . the rate of interest is ? | "difference in c . i . and s . i for 2 years = $ 615 - $ 600 = $ 15 s . i for one year = $ 300 s . i . on $ 300 for 1 year = $ 15 rate = ( 100 * 15 ) / ( 300 ) = 5 % the answer is b ." | a = 615 - 600
b = 100 * a
c = 600 / 2
d = b / c
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a ) 1978 , b ) 2707 , c ) 7728 , d ) 4416 , e ) 7291 | d | subtract(9600, multiply(multiply(9600, subtract(const_1, divide(10, const_100))), divide(3000, add(2000, 3000)))) | a is a working partner and b is a sleeping partner in the business . a puts in rs . 2000 and b rs . 3000 , a receives 10 % of the profit for managing the business the rest being divided in proportion of their capitals . out of a total profit of rs . 9600 , money received by a is ? | "2 : 3 = > 2 : 3 9600 * 10 / 100 = 960 9600 - 960 = 8640 8640 * 2 / 5 = 3456 + 960 = 4416 answer : d" | a = 10 / 100
b = 1 - a
c = 9600 * b
d = 2000 + 3000
e = 3000 / d
f = c * e
g = 9600 - f
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a ) 20 , b ) 40 , c ) 60 , d ) 80 , e ) 100 | e | multiply(divide(subtract(divide(12, multiply(subtract(15, 12), divide(40, const_60))), multiply(subtract(15, 12), divide(40, const_60))), subtract(15, 12)), const_60) | annie and sam set out together on bicycles traveling at 15 and 12 km per hour respectively . after 40 minutes , annie stops to fix a flat tire . if it takes annie 35 minutes to fix the flat tire and sam continues to ride during this time , how many minutes will it take annie to catch up with sam assuming that annie resumes riding at 15 km per hour ? | "annie gains 3 km per hour ( or 1 km every 20 minutes ) on sam . after 40 minutes annie is 2 km ahead . sam rides 1 km every 5 minutes . in the next 35 minutes , sam rides 7 km so sam will be 5 km ahead . it will take annie 100 minutes to catch sam . the answer is e ." | a = 15 - 12
b = 40 / const_60
c = a * b
d = 12 / c
e = 15 - 12
f = 40 / const_60
g = e * f
h = d - g
i = 15 - 12
j = h / i
k = j * const_60
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a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 9 | a | subtract(divide(12, 4), const_1) | find the unknown term 8 , 86 , x , - 4 , - 12 | 8 , 86 , x , - 4 , - 12 i guess each differ in d range of 2 8 - 8 = 0 8 - 6 = 2 6 - 2 = 4 2 - 6 = - 4 - 4 - 8 = - 12 answer : a | a = 12 / 4
b = a - 1
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a ) 381 m , b ) 112 m , c ) 500 m , d ) 125 m , e ) 433 m | c | multiply(multiply(360, const_0_2778), 5) | if a train , travelling at a speed of 360 kmph , crosses a pole in 5 sec , then the length of train is ? | "c c = 360 * 5 / 18 * 5 = 500 m" | a = 360 * const_0_2778
b = a * 5
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a ) 25 , b ) 35 , c ) 60 , d ) 75 , e ) 140 | b | multiply(add(const_2, const_3), const_2) | if y is the smallest positive integer such that 1260 multiplied by y is the square of an integer , then y must be | "1260 = 2 ^ 2 * 3 ^ 2 * 5 * 7 to be perfect square , we need to multiply by at least 5 * 7 = 35 . the answer is b ." | a = 2 + 3
b = a * 2
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a ) 2003 , b ) 2004 , c ) 2005 , d ) 2006 , e ) 2007 | e | add(1988, divide(subtract(70, 32), const_2)) | a certain company retirement plan has arule of 70 provision that allows an employee to retire when the employee ' s age plus years of employment with the company total at least 70 . in what year could a female employee hired in 1988 on her 32 nd birthday first be eligible to retire under this provision ? | "she must gain at least 70 points , now she has 32 and every year gives her two more points : one for age and one for additional year of employment , so 32 + 2 * ( # of years ) = 70 - - > ( # of years ) = 19 - - > 1988 + 19 = 2007 . answer : e ." | a = 70 - 32
b = a / 2
c = 1988 + b
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a ) 1 / 2 , b ) 1 / 4 , c ) 3 / 8 , d ) 5 / 8 , e ) 3 / 16 | c | multiply(power(divide(const_1, const_2), 2), multiply(choose(3, 2), divide(const_1, const_2))) | a coin is tossed 3 times . what is the probability of getting exactly 2 heads ? | "the number of possible outcomes is 2 ^ 3 = 8 there are 3 ways to get exactly 2 heads . p ( exactly 2 heads ) = 3 / 8 the answer is c ." | a = 1 / 2
b = a ** 2
c = math.comb(3, 2)
d = 1 / 2
e = c * d
f = b * e
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a ) 60 , b ) 60.75 , c ) 22.78 , d ) 32.52 , e ) 88 | c | subtract(negate(10.13), multiply(subtract(4.5, 6.75), divide(subtract(4.5, 6.75), subtract(3, 4.5)))) | 3 , 4.5 , 6.75 , 10.13 , 15.18 , ( . . . ) | "3 ( 3 ã — 3 ) ã · 2 = 4.5 ( 4.5 ã — 3 ) ã · 2 = 6.75 ( 6.75 ã — 3 ) ã · 2 = 10.13 ( 10.18 ã — 3 ) ã · 2 = 15.18 ( 15.18 ã — 3 ) ã · 2 = 60.75 answer is c" | a = negate - (
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a ) 25 % , b ) 33 1 / 3 % , c ) 50 % , d ) 66 2 / 3 % , e ) 75 % | b | multiply(divide(subtract(60, 45), 45), const_100) | a certain protective additive increases from 45 days to 60 days the time between required maintenance checks on an industrial vehicle . by what percent is the time between maintenance checks increased by using the additive ? | "general formula for percent increase or decrease , ( percent change ) : percent = change / original ∗ 100 so , the time between maintenance checks increased by 60 − 45 / 45 ∗ 100 = 33 1 / 3 answer : b ." | a = 60 - 45
b = a / 45
c = b * 100
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a ) 2250 m , b ) 1000 m , c ) 1250 m , d ) 1800 m , e ) 2000 m | a | multiply(add(multiply(1, const_60), 30), divide(multiply(4.5, const_1000), multiply(3, const_60))) | a can run 4.5 km distance in 1 min 30 seconds , while b can run this distance in 3 min . by how much distance can a beat b ? | a takes time 1.30 minutes = 90 sec b takes time 3 minutes = 180 sec diffrence = 180 - 90 = 90 sec now we are to find distance covered in 90 sec by b 180 sec = 4500 m 1 sec = 25 m 90 sec = 25 x 90 = 2250 m answer : a | a = 1 * const_60
b = a + 30
c = 4 * 5
d = 3 * const_60
e = c / d
f = b * e
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a ) 270 , b ) 289 , c ) 297 , d ) 255 , e ) 552 | c | multiply(divide(957, add(add(multiply(12, 8), multiply(16, 9)), multiply(18, 6))), multiply(16, 9)) | a , b and c rents a pasture for rs . 957 . a put in 12 horses for 8 months , b 16 horses for 9 months and 18 horses for 6 months . how much should c pay ? | "12 * 8 : 16 * 9 = 18 * 6 8 : 12 : 9 9 / 29 * 957 = 297 answer : c" | a = 12 * 8
b = 16 * 9
c = a + b
d = 18 * 6
e = c + d
f = 957 / e
g = 16 * 9
h = f * g
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a ) 61 % , b ) 63 % , c ) 65 % , d ) 67 % , e ) 69 % | d | subtract(multiply(const_3, 75), add(74, 84)) | a student got 74 % in math and 84 % in history . to get an overall average of 75 % , how much should the student get in the third subject ? | 74 + 84 + x = 3 * 75 x = 67 the answer is d . | a = 3 * 75
b = 74 + 84
c = a - b
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a ) 91.5 cm , b ) 92.2 cm , c ) 28.9 cm , d ) 30.5 cm , e ) 98.2 cm | d | multiply(multiply(const_2, divide(multiply(subtract(7, const_3), const_2), add(const_4, const_3))), 7) | the sector of a circle has radius of 7 cm and central angle 135 o . find its perimeter ? | "perimeter of the sector = length of the arc + 2 ( radius ) = ( 135 / 360 * 2 * 22 / 7 * 7 ) + 2 ( 7 ) = 16.5 + 14 = 30.5 cm answer : d" | a = 7 - 3
b = a * 2
c = 4 + 3
d = b / c
e = 2 * d
f = e * 7
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a ) 8 m , b ) 10 m , c ) 12 m , d ) 15 m , e ) 17 m | a | divide(sqrt(divide(128, divide(const_1, const_2))), const_2) | the width of a rectangular hall is ½ of its length . if the area of the hall is 128 sq . m , what is the difference between its length and breadth ? | "let the length of the hall be x m breadth of the hall = 1 x / 2 m area of the hall = length * breadth 128 = x * 1 x / 2 x ² = 256 x = 16 difference between the length and breadth of the hall = x - 1 x / 2 = x / 2 = 16 / 2 = 8 m answer : a" | a = 1 / 2
b = 128 / a
c = math.sqrt(b)
d = c / 2
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a ) 3 : 1 , b ) 3 : 2 , c ) 2 : 7 , d ) 3 : 25 , e ) 3 : 4 | c | divide(subtract(47, 40), subtract(40, 38)) | two trains running in opposite directions cross a man standing on the platform in 47 seconds and 38 seconds respectively and they cross each other in 40 seconds . the ratio of their speeds is : | "let the speeds of the two trains be x m / sec and y m / sec respectively . then , length of the first train = 47 x meters , and length of the second train = 38 y meters . ( 47 x + 37 y ) / ( x + y ) = 40 = = > 47 x + 38 y = 40 x + 40 y = = > 7 x = 2 y = = > x / y = 2 / 7 answer : option c" | a = 47 - 40
b = 40 - 38
c = a / b
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a ) 338 , b ) 277 , c ) 350 , d ) 288 , e ) 271 | d | subtract(multiply(22, multiply(80, const_0_2778)), 200) | a train 200 m long running at 80 kmph crosses a platform in 22 sec . what is the length of the platform ? | "d = 80 * 5 / 18 = 22 = 488 – 200 = 288 answer : d" | a = 80 * const_0_2778
b = 22 * a
c = b - 200
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a ) rs . 11.81 , b ) rs . 12 , c ) rs . 12.25 , d ) rs . 12.31 , e ) none | a | divide(multiply(9, add(const_100, 5)), subtract(const_100, 20)) | a fruit seller sells mangoes at the rate of rs . 9 per kg and thereby loses 20 % . at what price per kg , he should have sold them to make a profit of 5 % ? | "solution 85 : 9 = 105 : x x = ( 9 × 105 / 80 ) = rs . 11.81 hence , s . p per kg = rs . 11.81 answer a" | a = 100 + 5
b = 9 * a
c = 100 - 20
d = b / c
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a ) 4.1 % , b ) 7 % , c ) 2.5 % , d ) 2.8 % , e ) 3 % | c | divide(divide(const_100, add(const_1, const_4)), 8) | if the simple interest on a certain sum of money for 8 years is one – fifth of the sum , then the rate of interest per annum is | explanation : let the principal ( p ) be x then , simple interest ( si ) = x / 5 time ( t ) = 8 years rate of interest per annum ( r ) = ( 100 × si ) / pt = ( 100 × ( x / 5 ) / ( x × 8 ) = 20 / 8 = 2.5 % answer : option c | a = 1 + 4
b = 100 / a
c = b / 8
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a ) 12 min , b ) 15 min , c ) 25 min , d ) 50 min , e ) 55 min | a | add(multiply(20, const_100), multiply(multiply(subtract(const_1, multiply(add(divide(const_1, 20), divide(const_1, 30)), const_2)), 20), const_60)) | two pipes a and b can fill a tank in 20 and 30 minutes respectively . if both the pipes are used together , then how long will it take to fill the tank ? | "part filled by a in 1 min = 1 / 20 part filled by b in 1 min = 1 / 30 part filled by ( a + b ) in 1 min = ( 1 / 20 + 1 / 30 ) = 1 / 12 so both pipes can fill the tank in 12 minutes . answer : a" | a = 20 * 100
b = 1 / 20
c = 1 / 30
d = b + c
e = d * 2
f = 1 - e
g = f * 20
h = g * const_60
i = a + h
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a ) 20 years , b ) 21 years , c ) 22 years , d ) 23 years , e ) 24 years | b | divide(subtract(add(24, add(24, 3)), multiply(3, 3)), const_2) | the captain of a cricket team of 11 members is 24 years old and the wicket keeper is 3 years older . if the ages of these two are excluded , the average age of the remaining players is one year less than the average age of the whole team . what is the average age of the team ? | "explanation : let the average age of the whole team by x years . 11 x â € “ ( 24 + 27 ) = 9 ( x - 1 ) 11 x â € “ 9 x = 42 2 x = 42 x = 21 . so , average age of the team is 21 years . answer b" | a = 24 + 3
b = 24 + a
c = 3 * 3
d = b - c
e = d / 2
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a ) 148 , b ) 150 , c ) 152 , d ) 154 , e ) 156 | e | add(add(power(add(add(divide(subtract(subtract(111, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(111, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(111, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(111, const_10), const_2), const_4), const_2), const_2))) | the sum of 111 consecutive integers is 11211 . what is the greatest integer in the set ? | "let x be the first integer in the set , then x + 110 is the largest integer . the sum is : x + ( x + 1 ) + ( x + 2 ) + . . . + ( x + 110 ) = 111 x + 110 * 111 / 2 = 111 ( x + 55 ) then x + 55 = 101 x = 46 the largest integer in the set is 46 + 110 = 156 the answer is e ." | a = 111 - 10
b = a - 2
c = b / 4
d = c + 2
e = d + 2
f = e ** 2
g = 111 - 10
h = g - 2
i = h / 4
j = i + 2
k = j + 2
l = k + 2
m = l ** 2
n = f + m
o = 111 - 10
p = o - 2
q = p / 4
r = q ** 2
s = 111 - 10
t = s - 2
u = t / 4
v = u + 2
w = v ** 2
x = r + w
y = n + x
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a ) 9 , b ) 15.55 , c ) 47 , d ) 48 3 / 5 , e ) 59 | b | divide(multiply(28, 5), 9) | according to the formula f = 9 / 5 ( c ) + 32 , if the temperature in degrees farenheit ( f ) increases by 28 , by how much does the temperature in degrees celsius ( c ) increase ? | you can plug in values . c = 5 / 9 * ( f - 32 ) f = 32 - - > c = 0 ; f = 32 + 28 = 60 - - > c = 5 / 9 * 28 = 15.55 . increase = 15.55 degrees . answer : b . | a = 28 * 5
b = a / 9
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a ) 19 , b ) 25 , c ) 24 , d ) 22 , e ) 20 | a | subtract(subtract(subtract(24, 2), const_1), const_1) | how many positive integers less than 24 are prime numbers , odd multiples of 5 , or the sum of a positive multiple of 2 and a positive multiple of 4 ? | "9 prime numbers less than 28 : { 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 } 2 odd multiples of 5 : { 5 , 15 } 9 numbers which are the sum of a positive multiple of 2 and a positive multiple of 4 : { 6 , 8 , 10 , 12 , 14 , 16 , 18 , 20 , 22 } notice , that 5 is in two sets , thus total # of integers satisfying the given conditions is 9 + 2 + 9 - 1 = 19 . answer : a ." | a = 24 - 2
b = a - 1
c = b - 1
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a ) 25 , b ) 30 , c ) 35 , d ) 40 , e ) 45 | b | subtract(const_60, multiply(divide(125, 250), const_60)) | without stoppages , a train travels certain distance with an average speed of 250 km / h , and with stoppages , it covers the same distance with an average speed of 125 km / h . how many minutes per hour the train stops ? | "due to stoppages , it covers 125 km less . time taken to cover 125 km = 125 â „ 250 h = 1 â „ 2 h = 1 â „ 2 ã — 60 min = 30 min answer b" | a = 125 / 250
b = a * const_60
c = const_60 - b
|
a ) $ 55.70 , b ) $ 56.20 , c ) $ 56.70 , d ) $ 57.20 , e ) $ 57.70 | c | multiply(add(const_1, divide(5, const_100)), 54) | cara took out a charge account at the general store and agreed to pay 5 % simple annual interest . if she charges $ 54 on her account in january , how much will she owe a year later , assuming she does not make any additional charges or payments ? | 1.05 * $ 54 = $ 56.70 the answer is c . | a = 5 / 100
b = 1 + a
c = b * 54
|
a ) 10 , b ) 20 , c ) 21 , d ) 25 , e ) 27 | c | multiply(7, divide(multiply(add(7, 9), subtract(9, multiply(divide(5, add(7, 5)), 9))), subtract(multiply(9, 7), multiply(7, 5)))) | a can contains a mixture of liquids a and b is the ratio 7 : 5 . when 9 litres of mixture are drawn off and the can is filled with b , the ratio of a and b becomes 7 : 9 . how many liter t of liquid a was contained by the can initially ? | "as a : b : : 7 : 5 - - - > only option c is a multiple of 7 and hence it is a good place to start . also a : b : : 7 : 5 means that , a = ( 712 ) * total and b = ( 5 / 12 ) * total if a = 21 , b = 15 - - - > remove 9 litres - - - > you remove ( 7 / 12 ) * 9 of a - - - > a remaining = 21 - ( 7 / 12 ) * 9 = 63 / 4 similarly , for b , you remove ( 5 / 12 ) * 9 - - - > b remaining = 15 - ( 5 / 12 ) * 9 = 45 / 4 and then add 9 more litres of b - - - > 9 + 45 / 4 = 81 / 4 thus a / b ( final ratio ) = ( 45 / 4 ) / ( 81 / 4 ) = 7 : 9 , the same as the final ratio mentioned in the question . hence c is the correct answer . a / b = 7 / 9 = ( 7 x - ( 7 / 12 ) * 9 ) / ( 5 x - ( 5 / 12 ) * 9 + 9 ) , where 7 x and 5 x are initial quantities of a and b respectively . thus , 7 / 9 = ( 7 x - ( 7 / 12 ) * 9 ) / ( 5 x - ( 5 / 12 ) * 9 + 9 ) - - - > giving you x = 3 . thus a ( original ) t = 7 * 3 = 21 . c" | a = 7 + 9
b = 7 + 5
c = 5 / b
d = c * 9
e = 9 - d
f = a * e
g = 9 * 7
h = 7 * 5
i = g - h
j = f / i
k = 7 * j
|
['a ) 20', 'b ) 30', 'c ) 40', 'd ) 56', 'e ) 65'] | c | multiply(const_2, divide(add(multiply(5, 5), 75), 5)) | the length of a rectangle is double its width . if the length is diminished by 5 cm and the width is increased by 5 cm then its area is increased by 75 cm square . what is the length of the rectangle ? | sol . according to question ( l - 5 ) ( b + 5 ) - lb = 75 and l = 2 b so , b = 20 , l = 40 ans . ( c ) | a = 5 * 5
b = a + 75
c = b / 5
d = 2 * c
|
a ) 120 sec , b ) 165 sec , c ) 300 sec , d ) 167 sec , e ) 168 sec | c | divide(1500, subtract(multiply(54, const_0_2778), multiply(36, const_0_2778))) | a and b go around a circular track of length 1500 m on a cycle at speeds of 36 kmph and 54 kmph . after how much time will they meet for the first time at the starting point ? | "time taken to meet for the first time at the starting point = lcm { length of the track / speed of a , length of the track / speed of b } = lcm { 1500 / ( 36 * 5 / 18 ) , 1500 / ( 54 * 5 / 18 ) } = lcm ( 150 , 100 ) = 300 sec . answer : c" | a = 54 * const_0_2778
b = 36 * const_0_2778
c = a - b
d = 1500 / c
|
a ) 100 kmph , b ) 90 kmph , c ) 120 kmph , d ) 117 kmph , e ) 125 kmph | d | subtract(divide(divide(300, 9), const_0_2778), 3) | a train 300 m long takes 9 sec to cross a man walking at 3 kmph in a direction opposite to that of the train . find the speed of the train ? | "let the speed of the train be x kmph speed of the train relative to man = x + 3 = ( x + 3 ) * 5 / 18 m / sec 300 / [ ( x + 3 ) * 5 / 18 ] = 9 9 ( x + 3 ) = 1080 x = 117 kmph answer is d" | a = 300 / 9
b = a / const_0_2778
c = b - 3
|
a ) 94.3 % , b ) 95.3 % , c ) 93.3 % , d ) 92.3 % , e ) 91.3 % | a | multiply(divide(198, 210), const_100) | if there are 210 workers in a factory , and on a certain day , 198 were present . calculate the percentage that showed up for work ? ( round to the nearest tenth ) . | 198 / 210 * 100 = 94.28 94.3 % correct answer a | a = 198 / 210
b = a * 100
|
a ) 8.25 gallons , b ) 7.5 gallons , c ) 6.55 gallons , d ) 5.25 gallons , e ) 4.5 gallons | d | divide(210, 40) | a car gets 40 kilometers per gallon of gasoline . how many gallons of gasoline would the car need to travel 210 kilometers ? | "each 40 kilometers , 1 gallon is needed . we need to know how many 40 kilometers are there in 210 kilometers ? 210 / 40 = 5.25 * 1 gallon = 5.25 gallons correct answer d" | a = 210 / 40
|
a ) 20 , b ) 40 , c ) 50 , d ) 60 , e ) 80 | d | subtract(subtract(multiply(4, divide(add(multiply(15, 7), multiply(15, 3)), subtract(multiply(4, 3), 7))), 15), add(divide(add(multiply(15, 7), multiply(15, 3)), subtract(multiply(4, 3), 7)), 15)) | on a certain farm the ratio of horses to cows is 4 : 1 . if the farm were to sell 15 horses and buy 15 cows , the ratio of horses to cows would then be 7 : 3 . after the transaction , how many more horses than cows would the farm own ? | "originally , there were 4 k horses and k cows . 3 ( 4 k - 15 ) = 7 ( k + 15 ) 12 k - 7 k = 105 + 45 5 k = 150 k = 30 the difference between horses and cows is ( 4 k - 15 ) - ( k + 15 ) = 3 k - 30 = 60 the answer is d ." | a = 15 * 7
b = 15 * 3
c = a + b
d = 4 * 3
e = d - 7
f = c / e
g = 4 * f
h = g - 15
i = 15 * 7
j = 15 * 3
k = i + j
l = 4 * 3
m = l - 7
n = k / m
o = n + 15
p = h - o
|
a ) 1 / 8 , b ) 1 / 4 , c ) 2 / 4 , d ) 10 / 11 , e ) 1 / 4 | d | multiply(subtract(const_1, multiply(add(divide(const_1, 11), divide(const_1, 15)), 3)), 15) | two pipes p and q can fill a cistern in 11 and 15 minutes respectively . both are opened together , but at the end of 3 minutes the first is turned off . how much longer will the cistern take to fill ? | "3 / 11 + x / 15 = 1 x = 10 10 / 11 answer : d" | a = 1 / 11
b = 1 / 15
c = a + b
d = c * 3
e = 1 - d
f = e * 15
|
a ) 16.8 % , b ) 17.4 % , c ) 17.9 % , d ) 18.5 % , e ) 19.1 % | b | multiply(divide(subtract(653, add(multiply(multiply(const_2, 10), add(multiply(const_2, 10), const_1)), multiply(divide(10, add(divide(25, const_100), const_1)), add(15, const_2)))), add(multiply(multiply(const_2, 10), add(multiply(const_2, 10), const_1)), multiply(divide(10, add(divide(25, const_100), const_1)), add(15, const_2)))), const_100) | rani bought more apples than oranges . she sells apples at ₹ 23 apiece and makes 15 % profit . she sells oranges at ₹ 10 apiece and makes 25 % profit . if she gets ₹ 653 after selling all the apples and oranges , find her profit percentage q . | given : selling price of an apple = 23 - - > cost price = 23 / 1.15 = 20 selling price of an orange = 10 - - > cost price = 10 / 1.25 = 8 a > o 23 * ( a ) + 10 * ( o ) = 653 653 - 23 * ( a ) has to be divisible by 10 - - > units digit has to be 0 values of a can be 1 , 11 , 21 , 31 , . . . . - - > 1 can not be the value between 11 and 21 , if a = 11 , o = 30 - - > not possible if a = 21 , o = 17 - - > possible cost price = 20 * 21 + 8 * 17 = 420 + 136 = 556 profit = 653 - 556 = 97 profit % q = ( 97 / 556 ) * 100 = 17.4 % answer : b | a = 2 * 10
b = 2 * 10
c = b + 1
d = a * c
e = 25 / 100
f = e + 1
g = 10 / f
h = 15 + 2
i = g * h
j = d + i
k = 653 - j
l = 2 * 10
m = 2 * 10
n = m + 1
o = l * n
p = 25 / 100
q = p + 1
r = 10 / q
s = 15 + 2
t = r * s
u = o + t
v = k / u
w = v * 100
|
a ) 68 % , b ) 66 % , c ) 66 2 / 3 % , d ) 60 % , e ) 70 % | c | add(multiply(multiply(divide(multiply(divide(add(const_1000, multiply(8, const_100)), const_2), add(divide(const_1, const_3), const_1)), add(const_1000, multiply(8, const_100))), const_100), const_3), divide(multiply(multiply(divide(multiply(divide(add(const_1000, multiply(8, const_100)), const_2), add(divide(const_1, const_3), const_1)), add(const_1000, multiply(8, const_100))), const_100), const_3), const_10)) | at a summer camp with 1,800 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 1,800 campers will the 8 - to 12 - year - olds represent ? | total - 1,800 participants campers are aged 8 to 12 = ( 1 / 2 ) * 1800 = 900 next year , campers are aged 8 to 12 = ( 4 / 3 ) * 900 = 1200 percentage = ( 1200 / 1800 ) * 100 = 66 2 / 3 % answer : option c | 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 ) 1100 , b ) 1150 , c ) 1200 , d ) 1250 , e ) 1300 | b | divide(add(800, multiply(500, 3)), const_3) | village p ’ s population is 800 greater than village q ' s population . if village q ’ s population were reduced by 500 people , then village p ’ s population would be 3 times as large as village q ' s population . what is village q ' s current population ? | "p = q + 800 . p = 3 ( q - 500 ) . 3 ( q - 500 ) = q + 800 . 2 q = 2300 . q = 1150 . the answer is b ." | a = 500 * 3
b = 800 + a
c = b / 3
|
a ) rs 625 , b ) rs 300 , c ) rs 350 , d ) rs 375 , e ) none of these | b | divide(subtract(420, multiply(420, divide(12, const_100))), add(divide(25, const_100), const_1)) | the sale price of a trolley bag including the sale tax is rs . 420 . the rate of sale tax is 12 % . if the shopkeeper has made a profit of 25 % , the cost price of the trolley bag is : | explanation : 112 % of s . p . = 420 s . p . = rs . ( 420 x 100 / 112 ) = rs . 375 . c . p . = rs ( 100 / 125 x 375 ) = rs 300 answer : b | a = 12 / 100
b = 420 * a
c = 420 - b
d = 25 / 100
e = d + 1
f = c / e
|
a ) 5600 , b ) 4600 , c ) 4094 , d ) 7200 , e ) none of these | c | divide(multiply(multiply(7, 8), 15.5), divide(divide(multiply(multiply(20, 13.25), 8), const_100), const_100)) | how many bricks each measuring 20 cm x 13.25 cm x 8 cm , will be needed to build a wall 7 m x 8 m x 15.5 m | "explanation : no . of bricks = volume of the wall / volume of 1 brick = ( 700 x 800 x 15.5 ) / ( 20 x 13.25 x 8 ) = 4094 answer : c" | a = 7 * 8
b = a * 15
c = 20 * 13
d = c * 8
e = d / 100
f = e / 100
g = b / f
|
a ) 150 , b ) 300 , c ) 450 , d ) 490 , e ) 620 | b | multiply(subtract(50, 20), const_10) | one - tenth of the students at a nursery school are 4 years old or older . if 20 students have not yet reached their third birthday , and a total of 50 students are not between 3 years old and 4 years old , how many children are in the nursery school ? | x / 10 students are > 4 yrs 20 students are < 3 yrs x / 10 + 20 = 50 x / 10 = 30 x = 300 answer : b | a = 50 - 20
b = a * 10
|
a ) 25 / 4 , b ) 25 / 2 , c ) 20 / 4 , d ) 35 / 4 , e ) 25 / 3 | a | divide(add(multiply(multiply(3, divide(4, 3)), multiply(3, divide(4, 3))), multiply(3, 3)), multiply(3, divide(4, 3))) | if p / q = 3 / 4 then 3 p + 4 q = ? | 3 p + 4 q = ? divided by q , 3 ( p / q ) + 4 = x 3 * ( 3 / 4 ) + 4 = 25 / 4 answer : a | a = 4 / 3
b = 3 * a
c = 4 / 3
d = 3 * c
e = b * d
f = 3 * 3
g = e + f
h = 4 / 3
i = 3 * h
j = g / i
|
a ) 3200 , b ) 4064 , c ) 3250 , d ) 3825 , e ) 3985 | b | multiply(multiply(48000, subtract(multiply(const_3, 4), 4)), divide(13970, add(add(multiply(36000, multiply(const_3, 4)), multiply(42000, multiply(const_3, 4))), multiply(48000, subtract(multiply(const_3, 4), 4))))) | x and y started a business by investing rs . 36000 and rs . 42000 respectively after 4 months z joined in the business with an investment of rs . 48000 , then find share of z in the profit of rs . 13970 ? | "ratio of investment , as investments is for different time . investment x number of units of time . ratio of investments x : y : z = 36000 : 42000 : 48000 = > 6 : 7 : 8 . x = 6 x 12 months = 72 , y = 7 x 12 = 84 , z = 8 x 8 = 64 = > 18 : 21 : 16 . ratio of investments = > x : y : z = 18 : 21 : 16 . investment ratio = profit sharing ratio . z = 13970 ã — 16 / 55 = rs . 4064 . share of z in the profit is rs . 4064 . option b" | a = 3 * 4
b = a - 4
c = 48000 * b
d = 3 * 4
e = 36000 * d
f = 3 * 4
g = 42000 * f
h = e + g
i = 3 * 4
j = i - 4
k = 48000 * j
l = h + k
m = 13970 / l
n = c * m
|
a ) 252 , b ) 280 , c ) 360 , d ) 450 , e ) none | a | divide(multiply(42, 36), subtract(42, 36)) | the banker â € ™ s discount of a certain sum of money is rs . 42 and the true discount on the same sum for the same time is rs . 36 . the sum due is | "sol . sum = b . d . * t . d . / b . d . - t . d . = rs . [ 42 * 36 / 42 - 36 ] = rs . [ 42 * 36 / 6 ] = rs . 252 answer a" | a = 42 * 36
b = 42 - 36
c = a / b
|
a ) rs . 50 , b ) rs . 100 , c ) rs . 150 , d ) rs . 25 , e ) rs . 125 | a | multiply(multiply(const_3, const_4), divide(400, add(20, 4))) | the cost of 20 pens and 12 pencils is rs . 400 and the cost of 6 pens and 4 pencils is rs . 100 . find the cost of each pen ? | "20 p + 12 q = 400 - - - ( 1 ) 6 p + 4 q = 100 8 p + 8 q = 192 - - - ( 2 ) ( 1 ) - ( 2 ) = > 8 p = 160 = > p = 50 answer : a" | a = 3 * 4
b = 20 + 4
c = 400 / b
d = a * c
|
a ) 17 sec , b ) 12 sec , c ) 18 sec , d ) 19 sec , e ) 52 sec | b | divide(add(120, 120), add(divide(120, 10), divide(120, 15))) | two trains of equal lengths take 10 sec and 15 sec , respectively , to cross a telegraph post . if the length of each train is 120 m , in what time will they cross each other , travelling in opposite directions ? | speed of the first train = 120 / 10 = 12 m / sec . speed of the second train = 120 / 5 = 8 m / sec . relative speed = 12 + 8 = 20 m / sec . required time = ( 120 + 120 ) / 20 = 12 sec . answer : b | a = 120 + 120
b = 120 / 10
c = 120 / 15
d = b + c
e = a / d
|
a ) 1 / 3 , b ) 2 / 3 , c ) 1 / 4 , d ) 5 / 16 , e ) 3 / 5 | d | divide(const_2, 5) | if there is an equal probability of a child being born a boy or a girl , what is the probability that a couple who have 5 children have two children of the same sex and one of the opposite sex ? | "no of ways of selecting a gender - 2 c 1 no of ways of selecting any 2 children out of 5 = 5 c 2 total possible outcomes - 2 ^ 5 ( each child can be either a girl or a boy ) probability = 2 c 1 * 5 c 2 / 2 ^ 5 = 2 * 5 / 2 * 2 * 2 * 2 * 2 = 10 / 32 = 5 / 16 ans = d" | a = 2 / 5
|
a ) 12 , b ) 15 , c ) 18 , d ) 21 , e ) 24 | a | multiply(factorial(2), factorial(3)) | 2 men and 3 women are lined up in a row . what is the number of cases where they stand with each other in turn ? ( the number of cases in which men ( or women ) do not stand next to each other ) | "first arrange 3 women . 3 women can be arranged in 3 places in 3 ! ways . now w blank w blankw we have to fill the two blanks with 2 men . we can arrange 2 persons in 2 places in 2 ! ways . 3 ! * 2 ! = 12 ways . a is the answer" | a = math.factorial(2)
b = math.factorial(3)
c = a * b
|
a ) 2 , b ) 8 , c ) 1 , d ) 3 , e ) 4 | d | divide(const_1, multiply(divide(const_1, multiply(4, 3)), 4)) | a constructor estimates that 3 people can paint mr khans house in 4 days . if he uses 4 people instead of 3 , how long will they take to complete the job ? | explanation : use formula for a work members × days = constant 3 × 4 = 4 × a a = 3 so answer is 3 days answer : d | a = 4 * 3
b = 1 / a
c = b * 4
d = 1 / c
|
a ) 0.86 , b ) 0.68 , c ) 0.96 , d ) 0.69 , e ) 0.72 | e | divide(subtract(power(0.82, 3), power(0.1, 3)), add(add(power(0.82, 2), 0.082), power(0.1, 2))) | ( 0.82 ) ( power 3 ) - ( 0.1 ) ( power 3 ) / ( 0.82 ) ( power 2 ) + 0.082 + ( 0.1 ) ( power 2 ) is : | "given expression = ( 0.82 ) ( power 3 ) - ( 0.1 ) ( power 3 ) / ( 0.82 ) ( power 2 ) + ( 0.82 x 0.1 ) + ( 0.1 ) ( power 2 ) = a ( power 3 ) - b ( power 3 ) / a ( power 2 ) + ab + b ( power 2 ) = ( a - b ) = ( 0.82 - 0.1 ) = 0.72 answer is e" | a = 0 ** 82
b = 0 ** 1
c = a - b
d = 0 ** 82
e = d + 0
f = 0 ** 1
g = e + f
h = c / g
|
a ) 12.6 . , b ) 14.4 . , c ) 15.8 . , d ) 16.2 . , e ) 16.4 | c | subtract(add(multiply(2, 7.2), subtract(8.4, divide(const_4, const_10))), 6.6) | for every x , the action [ x ] is defined : [ x ] is the greatest integer less than or equal to x . what is the value of [ 6.5 ] x [ 2 / 3 ] + [ 2 ] x 7.2 + [ 8.4 ] - 6.6 ? | [ 6.5 ] x [ 2 / 3 ] + [ 2 ] x 7.2 + [ 8.4 ] - 6.6 = 6 * 0 + 2 * 7.2 + 8 - 6.6 = 0 + 14.4 + 1.4 15.8 answer c | a = 2 * 7
b = 4 / 10
c = 8 - 4
d = a + c
e = d - 6
|
a ) 240 meters , b ) 360 meters , c ) 480 meters , d ) 600 meters , e ) can not be determined | c | subtract(multiply(divide(multiply(144, const_1000), const_3600), 30), multiply(divide(multiply(144, const_1000), const_3600), 12)) | a train traveling at 144 kmph crosses a platform in 30 seconds and a man standing on the platform in 12 seconds . what is the length of the platform in meters ? | "answer distance covered by the train when crossing a man and when crossing a platform when a train crosses a man standing on a platform , the distance covered by the train is equal to the length of the train . however , when the same train crosses a platform , the distance covered by the train is equal to the length of the train plus the length of the platform . the extra time that the train takes when crossing the platform is on account of the extra distance that it has to cover . i . e . , length of the platform . compute length of platform length of the platform = speed of train * extra time taken to cross the platform . length of platform = 144 kmph * 12 seconds convert 144 kmph into m / sec 1 kmph = 5 / 18 m / s ( this can be easily derived . but if you can remember this conversion , it saves a good 30 seconds ) . ∴ 144 kmph = 5 / 18 ∗ 144 = 40 m / sec therefore , length of the platform = 40 m / s * 12 sec = 480 meters . choice c" | a = 144 * 1000
b = a / 3600
c = b * 30
d = 144 * 1000
e = d / 3600
f = e * 12
g = c - f
|
a ) 700 , b ) 800 , c ) 300 , d ) 1,500 , e ) 1,800 | c | 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 40 more than the number who did not , and if receipts for registration and lunch totaled $ 39,480 , 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 40 more than the number who did not . that means , total number of people who registered and paid for lunch = 40 + 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 ( 40 + x ) = 2880 + 72 x now , total amount received was 39480 thus , from the above data , 50 x + 2880 + 72 x = 39480 122 x = 39480 - 2880 122 x = 36600 x = 300 . hence the correct ans is c" | a = 1 * 1
|
a ) 15.1 , b ) 15.4 , c ) 15.7 , d ) 16.0 , e ) 16.3 | d | multiply(sqrt(divide(25.6, const_10)), const_10) | at 1 : 00 pm , there were 10.0 grams of bacteria . the bacteria increased to x grams at 4 : 00 pm , and 25.6 grams at 7 : 00 pm . if the amount of bacteria present increased by the same fraction during each of the 3 - hour periods , how many grams of bacteria were present at 4 : 00 pm ? | let x be the factor by which the bacteria increases every three hours . at 4 : 00 pm , the amount of bacteria was 10 x and at 7 : 00 pm it was 10 x ^ 2 . 10 x ^ 2 = 25.6 x ^ 2 = 2.56 x = 1.6 at 4 : 00 pm , the amount of bacteria was 10 ( 1.6 ) = 16 grams . the answer is d . | a = 25 / 6
b = math.sqrt(a)
c = b * 10
|
a ) 1 / 13 , b ) 2 / 23 , c ) 5 / 26 , d ) 5 / 22 , e ) 3 / 23 | d | divide(choose(6, 2), choose(add(add(6, 4), 2), 2)) | a bag contains 6 red , 4 blue and 2 green balls . if 2 ballsare picked at random , what is the probability that both are red ? | "p ( both are red ) , = 6 c 212 c 2 = 6 c 212 c 2 = 15 / 66 = 5 / 22 d" | a = math.comb(6, 2)
b = 6 + 4
c = b + 2
d = math.comb(c, 2)
e = a / d
|
a ) rs . 432 , b ) rs . 422 , c ) rs . 419 , d ) rs . 442 , e ) none of these | c | multiply(divide(360, subtract(2560, 360)), 2560) | the true discount on a bill of rs . 2560 is rs . 360 . what is the banker ' s discount ? | "explanation : f = rs . 2560 td = rs . 360 pw = f - td = 2560 - 360 = rs . 2200 true discount is the simple interest on the present value for unexpired time = > simple interest on rs . 2200 for unexpired time = rs . 360 banker ' s discount is the simple interest on the face value of the bill for unexpired time = simple interest on rs . 2160 for unexpired time = 360 / 2200 × 2560 = 0.16 × 2560 = rs . 419 answer : option c" | a = 2560 - 360
b = 360 / a
c = b * 2560
|
a ) 5000 , b ) 4500 , c ) 3200 , d ) 6000 , e ) 1592 | a | divide(divide(multiply(2000, const_100), 10), 4) | a man took loan from a bank at the rate of 4 % p . a . s . i . after 10 years he had to pay rs . 2000 interest only for the period . the principal amount borrowed by him was ? | "principal = ( 100 * 2000 ) / ( 4 * 10 ) = rs . 5000 answer : a" | a = 2000 * 100
b = a / 10
c = b / 4
|
a ) 35.24 , b ) 36.16 , c ) 36.24 , d ) 36.46 , e ) none | d | divide(add(multiply(36, 50), subtract(subtract(50, const_2), 23)), 50) | the mean of 50 observations was 36 . it was found later that an observation 46 was wrongly taken as 23 . the corrected new mean is | "solution correct sum = ( 36 x 50 + 46 - 23 ) = 1823 . â ˆ ´ correct mean = 1823 / 50 = 36.46 . answer d" | a = 36 * 50
b = 50 - 2
c = b - 23
d = a + c
e = d / 50
|
a ) 420 , b ) 700 , c ) 220 , d ) 500 , e ) none of these | d | add(300, multiply(300, divide(40, const_100))) | a fruit seller had some oranges . he sells 40 % oranges and still has 300 oranges . how many oranges he had originally ? | "explanation : he sells 40 % of oranges and still there are 300 oranges remaining = > 60 % of oranges = 300 ⇒ ( 60 × total oranges ) / 100 = 300 ⇒ total oranges / 100 = 5 ⇒ total oranges = 5 × 100 = 500 answer : option d" | a = 40 / 100
b = 300 * a
c = 300 + b
|
a ) 45 , b ) 47.3 , c ) 39.8 , d ) 45.5 , e ) 40.5 | e | divide(subtract(multiply(25, 45), multiply(15, 48)), const_10) | average weight of 15 boys in a class is 48 kgs . the average weight of the class of 25 students is 45 kgs . what is the average weight of the 15 girls in the class ? | total weight of boys in class 15 x 48 = 720 kg total weight of all student in the class room 25 x 45 = 1125 kg total no of girls in the class 1125 - 720 = 405 kg average weight of girls 405 / 10 = 40.5 kg correct ans is e | a = 25 * 45
b = 15 * 48
c = a - b
d = c / 10
|
a ) 48 , 36 , b ) 36 , 48 , c ) 32 , 48 , d ) 48 , 32 , e ) 32 , 42 | b | divide(subtract(84, 12), const_2) | the sum of two numbers is 84 , and one of them is 12 more than the other . what are the two numbers ? | 2 x = 84 − 12 = 72 . x = 72 / 2 = 36 . this is the first number . therefore the other number is x + 12 = 36 + 12 = 48 . the sum of 36 + 48 is 84 . answer is b . | a = 84 - 12
b = a / 2
|
a ) 352 , b ) 272 , c ) 224 , d ) 646 , e ) 742 | b | multiply(12.5, 3.2) | 12.5 * 3.2 * 6.8 = ? | "b 272 ? = 12.5 * 3.2 * 6.8 = 352" | a = 12 * 5
|
a ) 8250 , b ) 9350 , c ) 10,450 , d ) 11,550 , e ) 12,650 | a | multiply(330, 25) | a man started driving at a constant speed , from the site of a blast , the moment he heard the blast . he heard a second blast after a time of 30 mins and 25 seconds . if the second blast occurred exactly 30 mins after the first , how many meters was he from the site when he heard the second blast ? ( speed of sound = 330 m / s ) | "the distance the sound traveled to the man is 25 * 330 = 8250 meters the answer is a ." | a = 330 * 25
|
a ) 81.57 % , b ) 36.5 % , c ) 80.67 % , d ) 56.5 % , e ) 80.57 % | e | multiply(divide(subtract(1400, add(add(add(62, 62), add(60, 48)), 40)), 1400), const_100) | john had a stock of 1400 books in his bookshop . he sold 62 on monday , 62 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 = 62 + 62 + 60 + 48 + 40 = 272 let m be the books not sold m = 1400 - n = 1400 - 272 = 1128 percentage books not sold / total number of books = 1128 / 1200 = 0.81 = 80.57 % correct answer e" | a = 62 + 62
b = 60 + 48
c = a + b
d = c + 40
e = 1400 - d
f = e / 1400
g = f * 100
|
a ) 25869 , b ) 48586 , c ) 40320 , d ) 58964 , e ) 45698 | c | multiply(const_1, subtract(const_1, const_1)) | if you multiply all the numbers on your mobile phone except 0 and 9 , what is the answer ? | "we have to multiply 1 to 8 to find the answer . therefore 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 = 40320 answer is c" | a = 1 - 1
b = 1 * a
|
a ) 15 sec , b ) 9 sec , c ) 12 sec , d ) 15 sec , e ) 18 sec | a | divide(250, multiply(subtract(68, 8), const_0_2778)) | a train 250 m long is running at a speed of 68 kmph . how long does it take to pass a man who is running at 8 kmph in the same direction as the train ? | "speed of the train relative to man = ( 68 - 8 ) kmph = ( 60 * 5 / 18 ) m / sec = ( 50 / 3 ) m / sec time taken by the train to cross the man = time taken by it to cover 250 m at 50 / 3 m / sec = 250 * 3 / 50 sec = 15 sec answer : a ." | a = 68 - 8
b = a * const_0_2778
c = 250 / b
|
a ) 80 % , b ) 105 % , c ) 140 % , d ) 124.2 % , e ) 138 % | c | multiply(divide(multiply(subtract(const_1, divide(30, const_100)), divide(20, const_100)), divide(10, const_100)), const_100) | in 1998 the profits of company n were 10 percent of revenues . in 1999 , the revenues of company n fell by 30 percent , but profits were 20 percent of revenues . the profits in 1999 were what percent of the profits in 1998 ? | "0,14 r = x / 100 * 0.1 r answer c" | a = 30 / 100
b = 1 - a
c = 20 / 100
d = b * c
e = 10 / 100
f = d / e
g = f * 100
|
a ) 36 - 48 , b ) 50 - 34 , c ) 60 - 24 , d ) 42 - 54 , e ) 21 - 63 | d | divide(subtract(96, 12), const_2) | the sum of two numbers is 96 , 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 96 : word problem = 96 the line over x + 12 is a grouping symbol called a vinculum . it saves us writing parentheses . we have : 2 x = 96 â ˆ ’ 12 = 84 . x = 84 / 2 = 42 . this is the first number . therefore the other number is x + 12 = 42 + 12 = 54 . the sum of 42 + 54 is 96 . d" | a = 96 - 12
b = a / 2
|
a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | b | subtract(5, 2) | in a certain game , each player scores either 2 points or 5 points . if n players score 2 points and m players score 5 points , and the total number of points scored is 50 , what is the least possible positive difference q between n and m ? | "we have equation 2 n + 5 m = 50 we have factor 2 in first number and we have factor 5 in second number . lcm ( 2 , 5 ) = 10 so we can try some numbers and we should start from 5 because it will be less list than for 2 2 * 5 = 10 and n should be equal 20 4 * 5 = 20 and n should be equal 15 6 * 5 = 30 and n should be equal 10 8 * 5 = 40 and n should be equal 5 10 * 5 = 50 and n should be equal 0 third variant give us the mininal difference n - m = 10 - 6 = 4 and there is some mistake in my way of thinking because we do n ' t have such answer ) if we change the task and will seek for difference between m and n than minimal result q will be 8 - 5 = 3 and answer b" | a = 5 - 2
|
a ) 40 , b ) 135 , c ) 50 , d ) 55 , e ) 60 | b | multiply(divide(divide(add(divide(120, const_2), 120), 20), const_4), divide(120, const_2)) | a motorcyclist started riding at highway marker a , drove 120 miles to highway marker b , and then , without pausing , continued to highway marker c , where she stopped . the average speed of the motorcyclist , over the course of the entire trip , was 20 miles per hour . if the ride from marker a to marker b lasted 3 times as many hours as the rest of the ride , and the distance from marker b to marker c was half of the distance from marker a to marker b , what was the average speed , in miles per hour , of the motorcyclist while driving from marker b to marker c ? | a - b = 120 miles b - c = 60 miles avg speed = 20 miles time taken for a - b 3 t and b - c be t avg speed = ( 120 + 60 ) / total time 20 = 180 / 4 t t = 135 b - c = 135 mph answer b | a = 120 / 2
b = a + 120
c = b / 20
d = c / 4
e = 120 / 2
f = d * e
|
a ) 8500 , b ) 8900 , c ) 8600 , d ) 6970 , e ) none | c | subtract(multiply(const_10, 9), 9) | the difference between the place values of 9 and 4 in the number 529435 is | sol . = ( place value of 9 ) – ( place value of 4 ) = ( 9000 - 400 ) = 8600 answer c | a = 10 * 9
b = a - 9
|
a ) 1,108 , b ) 3,700 , c ) 2,108 , d ) 2,124 , e ) 2,256 | b | multiply(divide(200, 22.95), 450) | at the wholesale store you can buy an 8 - pack of hot dogs for $ 1.55 , a 20 - pack for $ 3.05 , and a 450 - pack for $ 22.95 . what is the greatest number of hot dogs you can buy at this store with $ 200 ? | "we have $ 200 and we have to maximize the number of hot dogs that we can buy with this amount . let ' s try to find out what is the maximum number of hot dogs that we can buy for a lesser amount of money , which in this case is 450 for $ 22.95 . for the sake of calculation , let ' s take $ 23 . 23 x 8 gives 184 , i . e . a total of 450 x 8 = 3600 hot dogs . we are left with ~ $ 16 . similarly , let ' s use $ 3 for calculation . we can buy 5 20 - pack hot dogs ( 3 x 5 ) , a total of 20 x 5 = 100 hot dogs . so we have 3700 hot dogs . 2108 looks far - fetched ( since we are not likely to be left with > $ 1.55 ) . hence , ( b ) 3700 ( answer b )" | a = 200 / 22
b = a * 450
|
a ) 18 , b ) 24 , c ) 60 , d ) 120 , e ) 240 | a | multiply(multiply(const_3, 3), multiply(3, 2)) | from a total of 3 boys and 4 girls , how many 4 - person committees can be selected if the committee must have exactly 2 boys and 2 girls ? | "answer = c = 18 no of 4 person committees that can be formed = 3 c 2 * 4 c 2 = 18 answer a" | a = 3 * 3
b = 3 * 2
c = a * b
|
a ) 13 , b ) 15 , c ) 12 , d ) 10 , e ) 11 | e | subtract(400, add(subtract(subtract(subtract(subtract(divide(divide(divide(divide(819200, const_100), const_2), const_2), const_4), const_10), const_100), const_10), const_4), const_1)) | mary is expected to pay back a total sum of $ 819200 to her bank as mortgage . if she is expected to make a payment which is double of the previous months amount , how long will it take her to fully repay the loan if her initial payment was $ 400 . | first payment - - $ 400 total amount to be repaid - - $ 819200 first month payment - - $ 400 second month payment - - $ 400 * 2 = $ 800 third month payment - - $ 800 * 2 = $ 1600 fourth month payment - - $ 1600 * 2 = $ 3200 fifth month payment - - $ 3200 * 2 = $ 6400 sixth month payment - - $ 6400 * 2 = $ 12800 seventh month payment - - $ 25600 * 2 = $ 51200 eight month payment - - $ 51200 * 2 = $ 102400 ninth month payment - - $ 102400 * 2 = $ 204800 tenth month payment - - $ 204800 * 2 = $ 409600 eleventh month payment - - $ 409600 * 2 = $ 819200 answer is e | a = 819200 / 100
b = a / 2
c = b / 2
d = c / 4
e = d - 10
f = e - 100
g = f - 10
h = g - 4
i = h + 1
j = 400 - i
|
a ) 0.125 % , b ) 1.25 % , c ) 12.5 % , d ) 125 % , e ) 0.152 % | c | multiply(divide(divide(multiply(148, 5.17), add(add(9.18, 5.17), 2.05)), 148), const_100) | irin , ingrid and nell bake chocolate chip cookies in the ratio of 9.18 : 5.17 : 2.05 . if altogether they baked a batch of 148 cookies , what percent of the cookies did nell bake ? | "9.18 x + 5.17 x + 2.05 x = 16.4 x = 148 cookies x = 148 / 16.4 = 9 ( approx ) so , tell baked 9 * 2.05 cookies or 19 cookies ( approx ) % share = 19 / 148 = 12.5 approx hence , answer is c ." | a = 148 * 5
b = 9 + 18
c = b + 2
d = a / c
e = d / 148
f = e * 100
|
a ) 12 , b ) 48 , c ) 16 , d ) 20 , e ) 25 | b | multiply(divide(multiply(add(multiply(3, 2), multiply(2, 2)), divide(30, subtract(multiply(3, 2), 1))), add(4, 1)), 4) | a jar contains a mixture of ab in the ratio 4 : 1 . when 30 l of mixture is replaced with liquid b , ratio becomes 2 : 3 . how many liters of liquid a was present in mixture initially . | "30 litres of mixture that is replaced will contain 24 litres of a and 6 litres of b ( as a : b = 4 : 1 ) let the initial volume of the mixture be 4 k + 1 k = 5 k so by condition , [ 4 k - 24 ] / [ k - 6 + 30 ] = 2 / 3 = > 12 k - 72 = 2 k - 12 + 60 = > 10 k = 120 solve for k which is k = 12 so initial volume of liquid a = 4 k = 48 litres answer : b" | a = 3 * 2
b = 2 * 2
c = a + b
d = 3 * 2
e = d - 1
f = 30 / e
g = c * f
h = 4 + 1
i = g / h
j = i * 4
|
a ) 7 1 / 2 sec , b ) 7 1 / 7 sec , c ) 7 8 / 2 sec , d ) 7 1 / 9 sec , e ) 7 2 / 2 sec | d | divide(add(150, 100), multiply(add(50, 70), const_0_2778)) | two trains are moving at 50 kmph and 70 kmph in opposite directions . their lengths are 150 m and 100 m respectively . the time they will take to pass each other completely is ? | "70 + 50 = 120 * 5 / 18 = 100 / 3 mps d = 150 + 100 = 250 m t = 250 * 3 / 100 = 15 / 2 = 7 1 / 2 sec answer : d" | a = 150 + 100
b = 50 + 70
c = b * const_0_2778
d = a / c
|
a ) 53.6 % , b ) 52.6 % , c ) 54.6 % , d ) 55.6 % , e ) 59.6 % | a | divide(multiply(30, const_100), 56) | if there are 56 laborers in a crew , and on a certain day , 30 were present . calculate the percentage that showed up for work ? ( round to the nearest tenth ) . | 30 / 56 * 100 = 53.57 53.6 % correct answer a | a = 30 * 100
b = a / 56
|
a ) 2 , b ) 3 , c ) 4 , d ) 1 , e ) 5 | c | divide(30, 24) | how many of the positive factors of 24 are not factors of 30 | "factors of 24 - 1 , 2 , 3 , 4 , 6 , 8 , 12,24 factors of 30 - 1 , 2 , 3 , 5 , 6 , 10 , 15 and 30 . comparing both , we have four factors of 24 which are not factors of 30 - 4,8 , 12,24 answer ( c )" | a = 30 / 24
|
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 8 | d | add(divide(power(2, 2), 2), const_1) | if f ( x ) = 12 - x ^ 2 / 2 and f ( 2 k ) = 5 k , what is one possible value for k ? | "first of all , see thisgmat blog postand check the related lesson linked below for some background on function notation . we can plug anything in for x and get a result . you can find f ( 1 ) , for example , by plugging in 1 where x is , and you would get 12 - 1 / 2 = 11.5 . or we could find f ( 2 ) , which would be 12 - 4 / 2 = 10 . so the notation f ( 2 k ) means that we are going to plug a 2 k in for x everywhere in the formula for f ( x ) . that would be : f ( 2 k ) = 12 - ( 2 k ) ^ 2 / 2 = 12 - 2 k ^ 2 . remember that we have to square both the 2 and the k , to get 4 k 2 . now , this expression , the output , we will set equal to 2 k . 12 - 2 k ^ 2 = 2 k - - > k = - 3 or k = 5 . all the answers are positive , so we choose k = 5 . answer = d" | a = 2 ** 2
b = a / 2
c = b + 1
|
a ) 31 , b ) 67 , c ) 88 , d ) 69 , e ) 12 | a | divide(add(multiply(40, 20), multiply(13, 10)), add(20, 10)) | the average runs scored by a batsman in 20 matches is 40 . in the next 10 matches the batsman scored an average of 13 runs . find his average in all the 30 matches ? | total score of the batsman in 20 matches = 800 . total score of the batsman in the next 10 matches = 130 . total score of the batsman in the 30 matches = 930 . average score of the batsman = 930 / 30 = 31 . answer : a | a = 40 * 20
b = 13 * 10
c = a + b
d = 20 + 10
e = c / d
|
a ) rs . 3000 , b ) rs . 3600 , c ) rs . 2400 , d ) rs . 4000 , e ) none of these | b | multiply(divide(9000, add(const_1, divide(const_2, const_3))), divide(const_2, const_3)) | p , q and r have rs . 9000 among themselves . r has two - thirds of the total amount with p and q . find the amount with r ? | "let the amount with r be rs . r r = 2 / 3 ( total amount with p and q ) r = 2 / 3 ( 9000 - r ) = > 3 r = 18000 - 2 r = > 5 r = 18000 = > r = 3600 . answer : b" | a = 2 / 3
b = 1 + a
c = 9000 / b
d = 2 / 3
e = c * d
|
a ) 3.5 gallons , b ) 2.37 gallons , c ) 5.7 gallons , d ) 4.25 gallons , e ) 2.83 gallons | e | divide(170, 60) | a car gets 60 kilometers per gallon of gasoline . how many gallons of gasoline would the car need to travel 170 kilometers ? | "each 60 kilometers , 1 gallon is needed . we need to know how many 60 kilometers are there in 170 kilometers ? 60 ã · 170 = 2.83 ã — 1 gallon = 2.83 gallons correct answer is e ) 2.83 gallons" | a = 170 / 60
|
a ) 182 , b ) 186 , c ) 190 , d ) 194 , e ) 198 | e | add(multiply(divide(subtract(293, 8), const_3), const_2), 8) | if jake loses 8 pounds , he will weigh twice as much as his sister kendra . together they now weigh 293 pounds . what is jake ’ s present weight , in pounds ? | "j + k = 293 and so k = 293 - j j - 8 = 2 k j - 8 = 2 ( 293 - j ) 3 j = 594 j = 198 the answer is e ." | a = 293 - 8
b = a / 3
c = b * 2
d = c + 8
|
['a ) 10', 'b ) 9', 'c ) 8', 'd ) 12', 'e ) 15'] | c | multiply(const_2, multiply(const_pi, divide(4, const_pi))) | the radius of a circle is 4 / π cm then its perimeter is ? | perimeter = 2 π r 2 π * 4 / π = 8 cm answer : c | a = 4 / math.pi
b = math.pi * a
c = 2 * b
|
a ) 6.99 % , b ) 6.89 % , c ) 6.08 % , d ) 6.09 % , e ) 6.19 % | d | add(add(divide(6, const_2), divide(6, const_2)), divide(multiply(divide(6, const_2), divide(6, const_2)), const_100)) | the effective annual rate of interest corresponding to a nominal rate of 6 % per annum payable half - yearly is ? | "amount of rs . 100 for 1 year when compounded half - yearly = [ 100 * ( 1 + 3 / 100 ) 2 ] = rs . 106.09 effective rate = ( 106.09 - 100 ) = 6.09 % answer : d" | a = 6 / 2
b = 6 / 2
c = a + b
d = 6 / 2
e = 6 / 2
f = d * e
g = f / 100
h = c + g
|
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