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 ) 16 pi , b ) 64 pi , c ) 96 pi , d ) 144 pi , e ) 576 pi | d | multiply(multiply(6, 12), multiply(const_2, const_pi)) | a cylindrical can has a radius of 6 centimeters and a height of 12 centimeters . what is the area , in square centimeters , of a rectangular label that completely covers the curved surface of the can without over - lapping ? | "they are asking the lateral surface area of the cylinder . the lateral surface area = 2 * pi * r * h = 2 * pi * 6 * 12 = 144 pi answer is d ." | a = 6 * 12
b = 2 * math.pi
c = a * b
|
a ) 5.56 , b ) 5.58 , c ) 6.0 , d ) 6.23 , e ) 6.26 | e | subtract(6.3, divide(divide(multiply(divide(multiply(subtract(add(multiply(const_3, const_4), 6.3), 8.3), const_60), 5), 2), const_60), const_100)) | a digital wristwatch was set accurately at 8.30 a . m and then lost 2 seconds every 5 minutes . what time was indicated on the watch at 6.30 p . m of the same day if the watch operated continuously that time ? | for every 5 min watch looses 2 secs , so for 60 ms it looses 24 sec , i . e . , for 1 hr it looses 24 secs 8.30 amto 6.30 pm = 10 hr it looses 240 secs i . e . , 4 mins , hence d watch will show 6.26 answer : e | a = 3 * 4
b = a + 6
c = b - 8
d = c * const_60
e = d / 5
f = e * 2
g = f / const_60
h = g / 100
i = 6 - 3
|
['a ) 33.1', 'b ) 33.5', 'c ) 33.7', 'd ) 33.2', 'e ) 33.9'] | a | multiply(const_100, divide(subtract(power(add(const_100, 10), const_3), power(const_100, const_3)), power(const_100, const_3))) | if each edge of cube increased by 10 % , the percentage increase in surface area is : | 100 Γ ( 110 ) / 100 Γ ( 110 ) / 100 Γ ( 110 ) / 100 = > 1331 / 100 = 33.1 % answer is a . | a = 100 + 10
b = a ** 3
c = 100 ** 3
d = b - c
e = 100 ** 3
f = d / e
g = 100 * f
|
a ) 5 , b ) 6.4 , c ) 7 , d ) 8 , e ) 9 | b | divide(110, multiply(add(56, 6), const_0_2778)) | a train 110 m long is running with a speed of 56 km / hr . in what time will it pass a man who is running at 6 km / hr in the direction opposite to that in which the train is going ? | "speed of train relative to man = 56 + 6 = 62 km / hr . = 62 * 5 / 18 = 155 / 9 m / sec . time taken to pass the men = 110 * 9 / 155 = 6.4 sec . answer : option b" | a = 56 + 6
b = a * const_0_2778
c = 110 / b
|
a ) 8 , b ) 16 , c ) 162 , d ) 144 , e ) 100 | c | multiply(divide(48, power(const_2, 5)), const_2) | if a and b are integers and ( a * b ) ^ 5 = 48 y , y could be : | "distribute the exponent . a ^ 5 * b ^ 5 = 48 y find the prime factorization of 48 . this is 2 ^ 4 * 3 ^ 1 . we need 2 ^ 1 * 3 ^ 4 ( or some other power of 3 that will give us a multiple of 3 ^ 5 as our second term ) . 3 ^ 4 * 2 = 81 * 2 = 162 the answer is c ." | a = 2 ** 5
b = 48 / a
c = b * 2
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a ) 6 rs , b ) 7 rs , c ) 8 rs , d ) 4 rs , e ) 1 rs | b | divide(105, multiply(const_3, 5)) | 5 men are equal to as many women as are equal to 8 boys . all of them earn rs . 105 only . men Γ’ β¬ β’ s wages are ? | "5 m = xw = 8 b 5 m + xw + 8 b - - - - - 105 rs . 5 m + 5 m + 5 m - - - - - 105 rs . 15 m - - - - - - 105 rs . = > 1 m = 7 rs . answer : b" | a = 3 * 5
b = 105 / a
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a ) 5 , b ) 4 , c ) 12 , d ) 40 , e ) 160 | a | subtract(divide(520, 40), 8) | the pilot of a small aircraft with a 40 - gallon fuel tank wants to fly to cleveland , which is 520 miles away . the pilot recognizes that the current engine , which can fly only 8 miles per gallon , will not get him there . by how many miles per gallon must the aircraft β s fuel efficiency be improved to make the flight to cleveland possible ? | "actual miles / gallon is = 520 / 40 = 13 miles / gallon . current engine miles / gallon is 8 miles / gallon . additional 5 miles / gallon is required to match the actual mileage . imo option a ." | a = 520 / 40
b = a - 8
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a ) 37 , b ) 42.5 , c ) 44 , d ) 50 , e ) 52.5 | a | divide(add(8, 20), add(add(divide(8, 40), divide(11, 60)), divide(20, 60))) | jerry travels 8 miles at an average speed of 40 miles per hour , stops for 11 minutes , and then travels another 20 miles at an average speed of 60 miles per hour . what is jerry β s average speed , in miles per hour , for this trip ? | "total time taken by jerry = ( 8 / 40 ) * 60 minutes + 11 minutes + ( 20 / 60 ) * 60 minutes = 45 minutes average speed = total distance / total time = ( 8 + 20 ) miles / ( 45 / 60 ) hours = 28 * 60 / 45 = 37 miles per hour answer : option a" | a = 8 + 20
b = 8 / 40
c = 11 / 60
d = b + c
e = 20 / 60
f = d + e
g = a / f
|
a ) 26 : 15 , b ) 12 : 7 , c ) 13 : 7 , d ) 15 : 6 , e ) 13 : 6 | a | divide(multiply(65, 8), multiply(75, 4)) | car a runs at the speed of 65 km / hr & reaches its destination in 8 hr . car b runs at the speed of 75 km / h & reaches its destination in 4 h . what is the respective ratio of distances covered by car a & car b ? | "sol . distance travelled by car a = 65 Γ£ β 8 = 520 km distance travelled by car b = 75 Γ£ β 4 = 300 km ratio = 520 / 300 = 26 : 15 a" | a = 65 * 8
b = 75 * 4
c = a / b
|
a ) 6 ^ 5 , b ) 5 ^ 6 , c ) 3 ^ 6 , d ) 10 ^ 5 , e ) 15 ^ 3 | d | divide(power(30, 10), power(30, 5)) | 30 ^ 10 / 90 ^ 5 = ? | "30 ^ 10 / 90 ^ 5 = ? a . 6 ^ 5 b . 5 ^ 6 c . 3 ^ 6 d . 10 ^ 5 e . 15 ^ 3 - > 30 ^ 10 / 90 ^ 5 = ( 30 ^ 10 ) / ( 3 ^ 5 ) ( 30 ^ 5 ) = ( 30 ^ 5 ) / ( 3 ^ 5 ) = ( 3 ^ 5 ) ( 10 ^ 5 ) / ( 3 ^ 5 ) = 10 ^ 5 . thus , d is the answer ." | a = 30 ** 10
b = 30 ** 5
c = a / b
|
a ) 1 : 2 , b ) 2 : 1 , c ) 1 : 1 , d ) 3 : 1 , e ) 2 : 3 | c | divide(multiply(inverse(const_2), const_4), inverse(const_2)) | the ratio of the areas of a square and rhombus whose base is same is : | "solution : the square and the rhombus are equal in area if they have a common base . answer is c" | a = 1/(2)
b = a * 4
c = 1/(2)
d = b / c
|
a ) 1 hr , b ) 2 hrs , c ) 3 hrs , d ) 4 hrs , e ) 5 hrs | d | divide(120, add(25, 5)) | a boat can travel with a speed of 25 km / hr in still water . if the speed of the stream is 5 km / hr , find the time taken by the boat to go 120 km downstream . | "speed downstream = ( 25 + 5 ) km / hr = 30 km / hr . time taken to travel 120 km downstream = 120 / 30 hrs = 4 hrs . answer : d" | a = 25 + 5
b = 120 / a
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a ) rs . 960 , b ) rs . 840 , c ) rs . 296 , d ) rs . 760 , e ) none | c | add(divide(multiply(100, const_100), multiply(3, 17)), 100) | the banker Γ’ β¬ β’ s gain on a sum due 3 years hence at 17 % per annum is rs . 100 . the banker Γ’ β¬ β’ s discount is | "solution t . d = ( b . g x 100 / r x t ) = rs . ( 100 x 100 / 17 x 3 ) = rs . 196 . b . d = rs ( 196 + 100 ) = rs . 296 . answer c" | a = 100 * 100
b = 3 * 17
c = a / b
d = c + 100
|
a ) 6.5 % , b ) 8.5 % , c ) 10.5 % , d ) 12.5 % , e ) 14.5 % | d | add(8, multiply(divide(5, const_100), 90)) | a 90 - liter solution of cool - drink is made from 5 % jasmine water . if 8 liters of jasmine and 2 liters of water were added to the solution , what percent of the solution is jasmine ? | the percent of jasmine in the resulting solution is : ( amount of jasmine ) / ( total volume ) ( 0.05 ( 90 ) + 8 ) / 100 = 12.5 / 100 = 12.5 % the answer is d . | a = 5 / 100
b = a * 90
c = 8 + b
|
a ) 12 , b ) 20 , c ) 25 , d ) 30 , e ) 35 | a | divide(280, multiply(4, 6)) | if the ratio of two number is 4 : 6 and lcm of the number is 280 then what is the number . | product of two no = lcm * hcf 4 x * 6 x = 280 * x x = 12 answer : a | a = 4 * 6
b = 280 / a
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a ) 5 / 2 , b ) 10 / 3 , c ) 7 / 2 , d ) 14 / 3 , e ) 11 / 2 | e | subtract(divide(add(5, sqrt(add(power(5, 2), multiply(multiply(2, 12), const_4)))), multiply(2, 2)), divide(subtract(5, sqrt(add(power(5, 2), multiply(multiply(2, 12), const_4)))), multiply(2, 2))) | by how much does the larger root of the equation 2 b ^ 2 + 5 b = 12 exceed the smaller root ? | "for 2 b ^ 2 + 5 b = 12 roots are [ - 5 + sqrt ( 25 + 96 ) ] / 4 or [ - 5 - sqrt ( 25 + 96 ) ] / 4 = 1.5 or - 4 hence larger root 1.5 is 1.5 - ( - 4 ) = 5.5 = 11 / 2 greater than smaller root ( - 4 ) . hence option ( e ) ." | a = 5 ** 2
b = 2 * 12
c = b * 4
d = a + c
e = math.sqrt(d)
f = 5 + e
g = 2 * 2
h = f / g
i = 5 ** 2
j = 2 * 12
k = j * 4
l = i + k
m = math.sqrt(l)
n = 5 - m
o = 2 * 2
p = n / o
q = h - p
|
a ) 77 , b ) 78 , c ) 79 , d ) 80 , e ) 81 | c | subtract(subtract(100, floor(divide(divide(100, const_2), const_3))), const_2) | between 100 to 200 how many no are divisible by both 3 and 2 and 100,200 are inclusive ? | prime numbers are 21 so 100 - 21 = 79 answer : c | a = 100 / 2
b = a / 3
c = math.floor(b)
d = 100 - c
e = d - 2
|
a ) 100 , b ) 110 , c ) 105 , d ) 20 , e ) 215 | c | subtract(multiply(divide(20, 2), add(divide(20, 2), 1)), multiply(divide(add(19, 1), 2), add(divide(subtract(19, 1), 2), 1))) | ifaequals the sum of the even integers from 2 to 20 , inclusive , andbequals the sum of the odd integers from 1 to 19 , inclusive , what is the value of ( a + b ) / 2 ? | "yes ! there is really a faster way to solve it . sum of consecutive odd or even integers = ( no . of odd or even ints ) * ( first int + last int ) / 2 here a = sum of even ints from 2 to 20 , inclusive number of even ints = 10 , first int + last int = 2 + 20 = 22 a = 10 * 22 / 2 = 110 b = sum of odd ints from 1 to 19 , inclusive number of odd ints = 10 , first int + last int = 1 + 19 = 20 a = 10 * 20 / 2 = 100 ( a + b ) / 2 = 110 + 100 = 210 / 2 = 105 ans : c" | a = 20 / 2
b = 20 / 2
c = b + 1
d = a * c
e = 19 + 1
f = e / 2
g = 19 - 1
h = g / 2
i = h + 1
j = f * i
k = d - j
|
a ) 500 , b ) 505 , c ) 510 , d ) 515 , e ) 525 | a | divide(add(200, 200), divide(80, const_100)) | martin has to secure 80 % marks to clear his exam of class 9 th . he got 200 marks and failed by 200 marks . what is the maximum marks ? | a 500 to pass the exam ravish needs 200 + 200 = 400 marks . = > ( 400 / 80 ) * 100 = 500 | a = 200 + 200
b = 80 / 100
c = a / b
|
a ) 75 % , b ) 80 % , c ) 104.8 % , d ) 150.5 % , e ) 180 % | c | multiply(divide(subtract(subtract(square_area(add(add(const_1, divide(100, 100)), divide(multiply(add(const_1, divide(100, 100)), 60), 100))), const_1), const_4), add(const_1, square_area(add(const_1, divide(100, 100))))), 100) | the length of each side of square a is increased by 100 percent to make square b . if the length of the side of square b is increased by 60 percent to make square c , by what percent is the area of square c greater than the sum of the areas of squares a and b ? | "let length of each side of square a be 10 area of a = 10 ^ 2 = 100 since , length of each side of square a is increased by 100 percent to make square b length of each side of square b = 2 * 10 = 20 area of b = 20 ^ 2 = 400 since , length of the side of square b is increased by 60 percent to make square c length of each side of square c = 1.6 * 20 = 32 area of c = 32 ^ 2 = 1024 difference in areas of c and cummulative areas of a and b = 1024 - ( 400 + 100 ) = 524 percent is the area of square c greater than the sum of the areas of squares a and b = ( 524 / 500 ) * 100 % = 104.8 % answer c" | a = 100 / 100
b = 1 + a
c = 100 / 100
d = 1 + c
e = d * 60
f = e / 100
g = b + f
h = square_area - (
i = h - 1
j = i / 4
k = 100 / 100
l = 1 + k
m = 1 + square_area
n = j * m
|
a ) 19 kg , b ) 13 kg , c ) 10 kg , d ) 23 kg , e ) 15 kg | a | multiply(divide(divide(subtract(696.05, 504.35), subtract(887.75, 504.35)), subtract(const_1, divide(subtract(696.05, 504.35), subtract(887.75, 504.35)))), 19) | a pizzeria in castroville creates its own special blend of shredded cheese at a cost of $ 696.05 per kilogram . the blend includes mozzarella , at $ 504.35 per kilogram , and romano cheese , at $ 887.75 per kilogram . if the employees have already shredded 19 kilograms of mozzarella cheese , how many kilograms of romano cheese will they need to shred now ? | 504.35 ( 19 ) + 887.75 x = 696.05 ( 19 + x ) solve the equation . 504.35 ( 19 ) + 887.75 x = 696.05 ( 19 + x ) 9 , 582.65 + 887.75 x = 13 , 224.95 + 696.05 x 9 , 582.65 + 191.7 x = 13 , 224.95 191.7 x = 3 , 642.3 x = 19 . answer is a . | a = 696 - 5
b = 887 - 75
c = a / b
d = 696 - 5
e = 887 - 75
f = d / e
g = 1 - f
h = c / g
i = h * 19
|
a ) 290 / 289 , b ) 1 / 289 , c ) 290 / 90 , d ) 290 / 19 , e ) none of these | a | add(power(divide(const_1, const_1), const_2), power(divide(const_1, 17), const_2)) | product of two natural numbers is 17 . then , the sum of reciprocals of their squares is | "explanation : if the numbers are a , b , then ab = 17 , as 17 is a prime number , so a = 1 , b = 17 . 1 / a 2 + 1 / b 2 = 1 / 1 ( 2 ) + 1 / 17 ( 2 ) = 290 / 289 option a" | a = 1 / 1
b = a ** 2
c = 1 / 17
d = c ** 2
e = b + d
|
a ) 5 , b ) 6 , c ) 7.07 , d ) 8 , e ) 9 | c | divide(multiply(10, sqrt(2)), 2) | the perimeter of an isosceles right triangle is 10 + 10 sq rt 2 . what is the length of the hypotenuse of the triangle ? | "side of triangle is a then perimeter = a + a + a . sqrt 2 ( right angle and pythagorus ) = 2 a + a . sqrt 2 = 10 + 10 sqrt 2 or , a . ( 2 + sqrt 2 ) = 10 ( 1 + sqrt 2 ) , a = 10 * ( 1 + sqrt 2 ) / 2 + sqrt 2 = 10 * 2.414 / 3.414 = 0.707 * 10 then hypotenuse = 7.07 c" | a = math.sqrt(2)
b = 10 * a
c = b / 2
|
a ) 40.2 , b ) 40.4 , c ) 40.5 , d ) 40.8 , e ) none of the above | c | divide(subtract(add(multiply(40.2, 10), add(13, 15)), 31), 10) | the average of 10 numbers is 40.2 . later it is found that two numbers have been wrongly copied . the first is 15 greater than the actual number and the second number added is 13 instead of 31 . find the correct average . | "sum of 10 numbers = 402 corrected sum of 10 numbers = 402 β 13 + 31 β 15 = 405 hence , new average = 405 β 10 = 40.5 answer c" | a = 40 * 2
b = 13 + 15
c = a + b
d = c - 31
e = d / 10
|
a ) 249 , b ) 498 , c ) 676 , d ) 823 , e ) 1,002 | b | divide(subtract(998, 2), 2) | a straight line in the xy - plane has a slope of 2 and a y - intercept of 2 . on this line , what is the x - coordinate of the point whose y - coordinate is 998 ? | "slope of 2 and a y - intercept of 2 y - coordinate is 998 y = 2 x + 2 996 = 2 x x = 498 answer : b . 498" | a = 998 - 2
b = a / 2
|
a ) 3 minutes , b ) 70 seconds , c ) 60 seconds , d ) 100 seconds , e ) 120 seconds | a | multiply(divide(2, const_2), const_3) | if two - third of a bucket is filled in 2 minutes then the time taken to fill the bucket completely will be . | "2 / 3 filled in 2 mint 1 / 3 filled in 1 mint then 2 / 3 + 1 / 3 = 2 + 1 = 3 minutes answer : a" | a = 2 / 2
b = a * 3
|
['a ) between 16 and 17', 'b ) between 17 and 18', 'c ) between 18 and 19', 'd ) between 19 and 20', 'e ) between 20 and 21'] | a | sqrt(add(power(subtract(divide(add(3, sqrt(add(multiply(const_4, 130), power(3, const_2)))), const_2), 3), const_2), power(divide(add(3, sqrt(add(multiply(const_4, 130), power(3, const_2)))), const_2), const_2))) | marla is hanging a banner diagonally across a rectangular room . the length of the room is 3 feet shorter than its width . if the total area of the room is 130 square feet , what is the length of the banner ( in feet ) ? | to solve this , we first need to set up an equation for the area of the room . if x is the width , then we have x ( x - 3 ) = 130 . by putting the equation in standard form , we get x ^ 2 - 3 x - 130 = 0 . by using the quadratic formula , we get roots of - 10 and 13 . we know that x is the width , and x - 3 is the length , so by using the roots , we get 13 as the width ( x ) , and 10 as the length ( x - 3 ) . once we have this , we can use the pythagorean theorem to solve for the diagonal . plugging in the length and width , we will get d ^ 2 = 13 ^ 2 + 10 ^ 2 = 169 + 100 = 269 the square root of 269 is less than 17 , since 17 ^ 2 = 289 . going down a number , we can compute 16 ^ 2 = 256 < 269 . therefore , the length of the diagonal must be between 16 and 17 . the correct answer is a . | a = 4 * 130
b = 3 ** 2
c = a + b
d = math.sqrt(c)
e = 3 + d
f = e / 2
g = f - 3
h = g ** 2
i = 4 * 130
j = 3 ** 2
k = i + j
l = math.sqrt(k)
m = 3 + l
n = m / 2
o = n ** 2
p = h + o
q = math.sqrt(p)
|
a ) 45 , b ) 34 , c ) 20 , d ) 76 , e ) 54 | c | multiply(divide(divide(25, const_100), add(divide(25, const_100), const_1)), const_100) | if p gets 25 % more than q , then by what percent does q get less than p ? | let q get 1 unit . 25 % of 1 unit = 1 / 4 = 0.25 p gets ( 1 + 0.25 ) i . e . , 1.25 units . difference = ( 1.25 - 1 ) = 0.25 required percentage = 0.25 / 1.25 * 100 = 1 / 5 * 100 = 20 answer : c | a = 25 / 100
b = 25 / 100
c = b + 1
d = a / c
e = d * 100
|
a ) 36 , b ) 42 , c ) 58 , d ) 60 , e ) 64 | c | subtract(subtract(multiply(4, add(subtract(multiply(4, 4), 4), subtract(multiply(4, 5), 5))), subtract(multiply(4, 5), 5)), add(subtract(multiply(4, 4), 4), subtract(multiply(4, 5), 5))) | before 5 years , dog a β s age was 4 times of dog b β s age and after 4 years , dog a β s age will be 3 times of dog b β s age . what is the difference of dog a β s age and dog b β s now ? | "a - 5 = 4 ( b - 5 ) - - > a - 4 b = - 15 . . . . . . . . . . . . . 1 a + 4 = 3 ( b + 4 ) - - > a - 3 b = 8 . . . . . . . . . . . . . 2 ( 2 ) - ( 1 ) - - > b = 23 - - > a = 3 ( 27 ) = 81 a - b = 81 - 23 = 58 answer : c" | a = 4 * 4
b = a - 4
c = 4 * 5
d = c - 5
e = b + d
f = 4 * e
g = 4 * 5
h = g - 5
i = f - h
j = 4 * 4
k = j - 4
l = 4 * 5
m = l - 5
n = k + m
o = i - n
|
a ) 250 , b ) 500 , c ) 750 , d ) 1000 , e ) 1500 | d | subtract(multiply(add(500, divide(500, const_2)), const_2), 500) | the number of people who purchased book a is twice the number of people who purchased book b . the number of people who purchased both books a and b is 500 , which is twice the number of people who purchased only book b . what is the number of people b who purchased only book a ? | this is best solved using overlapping sets or a venn diagram . we know that a = 2 b , and that 500 people purchased both a and b . further , those purchasing both was double those purchasing b only . this gives us 250 people purchasing b only . with the 500 that pruchased both , we have a total of 750 that purchased b and this is 1 / 2 of those that purchased a . so , 1500 purchased a . less the 500 that purchased both , b = 1000 purchased a only . ( this is much simpler to solve using the venn diagram ) . correct answer is d . 1000 | a = 500 / 2
b = 500 + a
c = b * 2
d = c - 500
|
a ) 54 , b ) 59 , c ) 63 , d ) 71 , e ) 92 | b | add(multiply(subtract(const_1, divide(const_1, const_3)), subtract(70, 37)), 37) | of 70 players on a football team , 37 are throwers . the rest of the team is divided so one third are left - handed and the rest are right handed . assuming that all throwers are right handed , how many right - handed players are there total ? | "37 are throwers and right handed 1 / 3 of remaining 33 , i . e 11 are left handed and remaining are right handed . basically , out of 70 , only 11 are left handed and remaining 70 - 11 = 59 are right handed . b is the answer" | a = 1 / 3
b = 1 - a
c = 70 - 37
d = b * c
e = d + 37
|
a ) 11 / 12 , b ) 12 / 11 , c ) 13 / 19 , d ) 12 / 19 , e ) 11 / 19 | b | divide(add(multiply(40, const_100), multiply(40, subtract(const_100, 20))), add(multiply(40, const_100), multiply(add(40, 20), 40))) | paul ' s income is 40 % less than rex ' s income , quentin ' s income is 20 % less than paul ' s income , and sam ' s income is 40 % less than paul ' s income . if rex gave 60 % of his income to sam and 40 % of his income to quentin , sam ' s new income would be what fraction of quentin ' s new income ? | "make r = 10 p = 0.6 r = 6 q = 0.8 p = 4.8 s = 0.6 p = 3.6 for that we get s = 9.6 and q 8.8 so 9.6 / 8.8 = 1.2 / 1.1 ans : b" | a = 40 * 100
b = 100 - 20
c = 40 * b
d = a + c
e = 40 * 100
f = 40 + 20
g = f * 40
h = e + g
i = d / h
|
a ) 50 , b ) 60 , c ) 70 , d ) 80 , e ) 90 | c | multiply(const_10, subtract(const_10, const_2)) | how many integers k greater than 100 and less than 900 are there such that if the hundreds and the unit digits of k are reversed , the resulting integer is k + 99 ? | "not sure if this is the shortest . . but this is how i did this there are 7 sets of integers with hundreds and units digits exchanged that satisfies k + 99 . 1 . 102 | 201 ( satisfies k + 99 , where k = 102 ) 2 . 203 | 302 ( satisfies k + 99 , where k = 203 ) 3 . . . . 4 . . . . 5 . . . . 6 . . . . 7 . 708 | 807 each set has 10 such numbers . 1 . 102 | 201 ( still k + 99 holds good ) 2 . 112 | 211 3 . 122 | 221 4 . 132 | 231 5 . . . . 6 . . . . 7 . . . . 8 . . . . 9 . 182 | 281 10 . 192 | 291 therefore , 7 sets with 10 such number in each set will give 7 x 10 = 70 integers . c" | a = 10 - 2
b = 10 * a
|
a ) q = 300 , b ) q = 400 , c ) q = 500 , d ) q = 600 , e ) none of these | c | divide(200, divide(2, 5)) | there are 200 female managers in a certain company . find the total number of female employees q in the company , if 2 / 5 of all the employees are managers and 2 / 5 of all male employees are managers . | { managers } = { female managers } + { male managers } ; we are told that the total number of managers in the company is 2 / 5 of all the employees , thus { managers } = 2 / 5 ( m + f ) , where m and f are number of female and male employees , respectively . also , we know that 2 / 5 of all male employees are managers : { male managers } = 2 / 5 * mas well as there are total of 200 female managers : { female managers } = 200 ; thus : 2 / 5 ( m + f ) = 200 + 2 / 5 * m - - > f = 500 . answer : c . | a = 2 / 5
b = 200 / a
|
a ) 10000 , b ) 12000 , c ) 16160.8 , d ) 14000 , e ) 15000 | c | divide(subtract(multiply(multiply(multiply(4, const_100), const_100), power(add(1, divide(divide(4, const_100), 4)), 4)), multiply(multiply(4, const_100), const_100)), subtract(power(add(1, divide(divide(2, const_100), 2)), 2), 1)) | john invests $ x at the semi - annual constant compounded rate of 2 percent and also does $ 8,000 at the quarterly constant compounded rate of 4 percent . if the interests are the same after 1 year , what is the value of x ? ? | "a = p ( 1 + r / n ) ^ nt a = total amount accrued p = principal deposited r = rate of interest in decimal form n = number of times per year , interest compounded t = time in number of years . . x ( 1 + 0.02 / 2 ) ^ 2 - x = 8,000 ( 1 + 0.04 / 4 ) ^ 4 - 8,000 [ when the principal is subtracted from the total amount accrued , the resulting difference is the interest portion and question states interests are equal ) = > x [ ( 1.01 ) ^ 2 - 1 ] = 8,000 [ ( 1.01 ) ^ 4 - 1 ] = > x [ ( 1.01 ) ^ 2 - 1 ] = 8,000 [ ( 1.01 ) ^ 2 + 1 ] [ ( 1.01 ) ^ 2 - 1 ] - - > using a ^ 2 - b ^ 2 = a + b x a - b formula and cancel common expression on both sides = > x = 8,000 ( 1.0201 + 1 ) = 16160.8 hence answer is c ." | a = 4 * 100
b = a * 100
c = 4 / 100
d = c / 4
e = 1 + d
f = e ** 4
g = b * f
h = 4 * 100
i = h * 100
j = g - i
k = 2 / 100
l = k / 2
m = 1 + l
n = m ** 2
o = n - 1
p = j / o
|
a ) 40 , b ) 44 , c ) 80 , d ) 86 , e ) 90 | d | divide(subtract(power(24, const_2), 404), const_2) | if the sum of two numbers is 24 and the sum of their squares is 404 , then the product of the numbers is | "sol . let the numbers be x and y . then , ( x + y ) = 24 and x 2 + y 2 = 404 . now , 2 xy = ( x + y ) 2 - ( x 2 + y 2 ) = ( 24 ) 2 - 404 = 576 - 404 = 172 xy = 86 . answer d" | a = 24 ** 2
b = a - 404
c = b / 2
|
a ) 10 / 9 , b ) 1 / 289 , c ) 290 / 90 , d ) 290 / 19 , e ) none of these | a | add(power(divide(const_1, const_1), const_2), power(divide(const_1, 3), const_2)) | product of two natural numbers is 3 . then , the sum of reciprocals of their squares is | "explanation : if the numbers are a , b , then ab = 3 , as 3 is a prime number , so a = 1 , b = 3 . 1 / a 2 + 1 / b 2 = 1 / 1 ( 2 ) + 1 / 3 ( 2 ) = 10 / 9 option a" | a = 1 / 1
b = a ** 2
c = 1 / 3
d = c ** 2
e = b + d
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a ) 7 , b ) 9 , c ) 11 , d ) 12 , e ) 21 | d | add(multiply(divide(10, const_2), 2), const_2) | if the least common multiple of two prime numbers x and y is 10 , where x > y , then the value of 2 x + y is | lcm of two prime numbers will basically be multiple of those numbers . lcm = 10 = 2 * 5 , both of which are prime . now we know x > y , so x = 5 and y = 2 . 2 x + y = 2 * 5 + 2 = 10 + 2 = 12 . answer : d | a = 10 / 2
b = a * 2
c = b + 2
|
a ) 7.14 , b ) 7.41 , c ) 4.71 , d ) 4.17 , e ) none of them | d | multiply(divide(subtract(const_100, 96), 96), const_100) | if the cost price is 96 % of selling price then what is the profit percentage . | "selling price = rs 100 : then cost price = rs 96 : profit = rs 4 . profit = { ( 4 / 96 ) * 100 } % = 4.17 % answer is d ." | a = 100 - 96
b = a / 96
c = b * 100
|
a ) 14 , b ) 15 , c ) 33 , d ) 22 , e ) 23 | c | divide(multiply(subtract(50, 6), 3), 4) | ratio between rahul and deepak is 4 : 3 , after 6 years rahul age will be 50 years . what is deepak present age | "explanation : present age is 4 x and 3 x , = > 4 x + 6 = 50 = > x = 11 so deepak age is = 3 ( 11 ) = 33 answer : option c" | a = 50 - 6
b = a * 3
c = b / 4
|
a ) 12 , b ) 30 , c ) 60 , d ) 90 , e ) 120 | c | divide(divide(240, 3), 3) | if x ^ 3 is divisible by 240 what is the least possible value of integer x ? | "240 can be written as ( 2 ^ 4 ) * 3 * 5 . for k ^ 3 to be divisible by 240 it should contain atleast 2 ^ 4 and 3 and 5 in its factors . we can leave out option a because 12 doesnt have 5 as one of its factor . now if we check for option b , 30 can be written as 2 * 3 * 5 , hence 30 ^ 3 will have 3 as the maximum power of 2 , so we can leave out this option too . option c is the right answer if we follow the same method as we followed for other two previous options . 60 = ( 2 ^ 2 ) * 3 * 5 ; 60 ^ 3 = ( 2 ^ 6 ) * ( 3 ^ 3 ) * ( 5 ^ 3 ) . so it shows that 60 ^ 3 is divisible by 240 and hence the answer . answer : c" | a = 240 / 3
b = a / 3
|
a ) s . 1090 , b ) s . 1610 , c ) s . 1190 , d ) s . 1202 , e ) s . 1092 | b | divide(multiply(subtract(const_100, 8), 1750), const_100) | a man buys a cycle for rs . 1750 and sells it at a loss of 8 % . what is the selling price of the cycle ? | "s . p . = 92 % of rs . 1750 = rs . 92 x 1750 / 100 = rs . 1610 answer : option b" | a = 100 - 8
b = a * 1750
c = b / 100
|
a ) 40 , b ) 44 , c ) 66 , d ) 52 , e ) none of these | c | add(add(multiply(5, divide(12, subtract(multiply(divide(6, 5), 5), 4))), divide(12, subtract(multiply(divide(6, 5), 5), 4))), 12) | in a can , there is a mixture of milk and water in the ratio 4 : 5 . if it is filled with an additional 12 litres of milk the can would be full and ratio of milk and water would become 6 : 5 . find the capacity of the can ? | "let the capacity of the can be t litres . quantity of milk in the mixture before adding milk = 4 / 9 ( t - 12 ) after adding milk , quantity of milk in the mixture = 6 / 11 t . 6 t / 11 - 12 = 4 / 9 ( t - 12 ) 10 t = 1188 - 528 = > t = 66 . answer : c" | a = 6 / 5
b = a * 5
c = b - 4
d = 12 / c
e = 5 * d
f = 6 / 5
g = f * 5
h = g - 4
i = 12 / h
j = e + i
k = j + 12
|
a ) 1218 , b ) 180 , c ) 176 , d ) 200 , e ) 291 | c | multiply(subtract(power(9, const_2), power(5, const_2)), divide(add(multiply(5, const_2), const_2), add(const_4, const_3))) | a rope of which a calf is tied is increased from 5 m to 9 m , how much additional grassy ground shall it graze ? | "Ο ( 92 β 52 ) = 176 answer : c" | a = 9 ** 2
b = 5 ** 2
c = a - b
d = 5 * 2
e = d + 2
f = 4 + 3
g = e / f
h = c * g
|
a ) a - 56 , b ) b - 57 , c ) c - 58 , d ) d - 59 , e ) e - 60 | a | add(multiply(6, 8), 8) | if the number of white balls in a box are 6 times the black balls , which of the below is the total number of balls , if there are 8 black balls in the box ? | let the number of black balls = x and the number of white balls = 6 x then , total number of balls in the box = x + 6 x = 7 x i . e . , the total number of balls must be a multiple of 7 from the given choices , only 56 is the multiple of 7 . hence , 56 are the number of balls in the box . answer is a . | a = 6 * 8
b = a + 8
|
a ) 7.3 % , b ) 10.5 % , c ) 11 % , d ) 12.5 % , e ) none | a | multiply(const_100, divide(add(multiply(add(const_2, const_3), const_2), 1), add(const_100, 50))) | a trader marked the selling price of an article at 50 % above the cost price . at the time of selling , he allows certain discount and suffers a loss of 1 % . he allowed a discount of : | "sol . let c . p . = rs . 100 . then , marked price = rs . 110 , s . p . = rs . 99 . β΄ discount % = [ 11 / 150 * 100 ] % = 7.3 % answer a" | a = 2 + 3
b = a * 2
c = b + 1
d = 100 + 50
e = c / d
f = 100 * e
|
a ) 216 , b ) 217.5 , c ) 236.5 , d ) 245 , e ) 250 | b | divide(multiply(45, 750), 25) | 45 % of 750 - 25 % of 480 = ? | "explanation : given expression = ( 45 / 100 * 750 ) - ( 25 / 100 * 480 ) = ( 337.50 - 120 ) = 217.50 answer is b" | a = 45 * 750
b = a / 25
|
a ) 46 , b ) 42 , c ) 56 , d ) 32 , e ) 40 | e | divide(subtract(divide(30, divide(15, const_100)), 90), 2) | among all sales staff at listco corporation , college graduates and those without college degrees are equally represented . each sales staff member is either a level - 1 or level - 2 employee . level - 1 college graduates account for 15 % of listco ' s sales staff . listco employs 90 level - 1 employees , 30 of whom are college graduates . how many sales staff members without college degrees are level - 2 employees ? | "i ' m going in on this one . so let ' s say that we have the following so we know that l 1 = 90 and that c and l 1 = 0.15 x , we should set up a double set matrix btw but anyways , i ' m just explaining the point with this problem . now we are told that 0.15 x = 30 , therefore the grand total is 200 . now we know that l 2 is 200 - 90 = 110 . we also learn that c and no c are equally represented thus 100 each . therefore no c and no l 2 will be 100 - 60 = 40 . thus e is the correct answer choice" | a = 15 / 100
b = 30 / a
c = b - 90
d = c / 2
|
a ) a ) 77 , b ) b ) 99 , c ) c ) 66 , d ) d ) 55 , e ) e ) 88 | e | subtract(divide(multiply(divide(multiply(72, 8), 30), 50), 6), 72) | 72 men working 8 hours per day dig 30 m deep . how many extra men should be put to dig to a depth of 50 m working 6 hours per day ? | "( 72 * 8 ) / 30 = ( x * 6 ) / 50 = > x = 160 160 β 72 = 88 answer : e" | a = 72 * 8
b = a / 30
c = b * 50
d = c / 6
e = d - 72
|
a ) 3 . , b ) 4 . , c ) 5 . , d ) 7 . , e ) 8 . | c | divide(divide(divide(120, 4), const_2), const_3) | how many of the positive divisors of 120 are also multiples of 4 not including 120 ? | given to find the positive divisors of 120 and condition is that divisors have to be multiple of 4 but not including 120 . consider factors of 120 4 * 30 6 * 20 15 * 8 24 * 5 60 * 2 total we get 5 different ways and all are part of 4 . answer : c is correct option . | a = 120 / 4
b = a / 2
c = b / 3
|
a ) 297 , b ) 879 , c ) 323 , d ) 762 , e ) 269 | c | subtract(subtract(400, divide(multiply(400, 15), const_100)), divide(multiply(subtract(400, divide(multiply(400, 15), const_100)), 5), const_100)) | the sale price sarees listed for rs . 400 after successive discount is 15 % and 5 % is ? | "400 * ( 85 / 100 ) * ( 95 / 100 ) = 323 answer : c" | a = 400 * 15
b = a / 100
c = 400 - b
d = 400 * 15
e = d / 100
f = 400 - e
g = f * 5
h = g / 100
i = c - h
|
a ) 6 , b ) 7 , c ) 8 , d ) 10 , e ) 13 | e | add(sqrt(divide(subtract(power(15, const_2), power(const_2, const_2)), const_2)), const_2) | efrida and frazer who live 15 miles apart , meet at a restaurant that is directly north of efrida ' s home and directly east of frazer ' s home . if the restaurant is two miles closer to efrida ' s home , than to frazer ' s home , how many miles is the restaurant from frazer ' s home ? | it ' s a simple geometry problem . look at the diagram below : answer : e . | a = 15 ** 2
b = 2 ** 2
c = a - b
d = c / 2
e = math.sqrt(d)
f = e + 2
|
a ) 17 , 18 , b ) 7 , 8 , c ) 5 , 6 , d ) 1 , 2 , e ) 8 , 9 | b | add(add(power(add(add(divide(subtract(subtract(15, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(15, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(15, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(15, const_10), const_2), const_4), const_2), const_2))) | the sum of two consecutive integers is 15 . find the numbers . | "n + ( n + 1 ) = 15 2 n + 1 = 15 2 n = 14 n = 7 answer : b" | a = 15 - 10
b = a - 2
c = b / 4
d = c + 2
e = d + 2
f = e ** 2
g = 15 - 10
h = g - 2
i = h / 4
j = i + 2
k = j + 2
l = k + 2
m = l ** 2
n = f + m
o = 15 - 10
p = o - 2
q = p / 4
r = q ** 2
s = 15 - 10
t = s - 2
u = t / 4
v = u + 2
w = v ** 2
x = r + w
y = n + x
|
a ) 11 , b ) 13 , c ) 39 / 5 , d ) 21 , e ) 23 | c | divide(multiply(add(add(3, const_3), const_2), divide(3, const_2)), add(const_2, divide(const_1, const_2))) | a and b are two partially filled buckets of water . if 3 liters are transferred from a to b , then a would contain one - third of the amount of water in b . alternatively , if 5 liters are transferred from b to a , b would contain one - half of the amount of water in a . bucket a contains how many liters of water ? | "let a contains a , b contains b liters so , ( a - 3 ) / ( b + 3 ) = 1 / 3 . . . . . . . ( 1 ) again , ( b - 5 ) / ( a + 5 ) = 1 / 2 . . . . . . . . . . ( 2 ) from ( 1 ) ( 2 ) we find a = 39 / 5 ans : ( c )" | a = 3 + 3
b = a + 2
c = 3 / 2
d = b * c
e = 1 / 2
f = 2 + e
g = d / f
|
a ) 7,200 , b ) 8,880 , c ) 10,800 , d ) 28,800 , e ) 48,000 | a | multiply(2, const_3600) | a space shuttle orbits the earth at about 2 kilometers per second . this speed is equal to how many kilometers per hour ? | "seconds in 1 hours : 60 s in 1 min 60 min in 1 hr 60 * 60 = 3600 sec in 1 hr 2 * 3600 = 28,800 answer : a" | a = 2 * 3600
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a ) 8 sec , b ) 10 sec , c ) 12 sec , d ) 14 sec , e ) 15 sec | c | multiply(divide(divide(220, const_1000), subtract(59, 7)), const_3600) | a train 220 m long is running with a speed of 59 kmph . . in what will it pass a man who is running at 7 kmph in the direction opposite to that in which the train is going | "explanation : speed of the train relative to man = ( 59 + 7 ) kmph = 66 Γ 5 / 18 m / sec = 55 / 3 m / sec . time taken by the train to cross the man = time taken by it to cover 220 m at ( 55 / 3 ) m / sec = ( 220 Γ 3 / 55 ) sec = 12 sec answer : option c" | a = 220 / 1000
b = 59 - 7
c = a / b
d = c * 3600
|
a ) 35 , b ) 36 , c ) 34 , d ) 38 , e ) 39 | c | add(divide(subtract(add(40, 2), 35), 1.5), 35) | each week , harry is paid x dollars per hour for the first 35 hours and 1.5 x dollars for each additional hour worked that week . each week , james is paid x dollars per per hour for the first 40 hours and 2 x dollars for each additional hour worked that week . last week james worked a total of 41 hours if harry and james were paid the same amount last week , how many hours did harry work last week ? | "james worked for 41 hours hence he earned 40 * x + 1 * 2 x = 42 x dollars ; we know that harry also earned the same 42 x dollars , out of which he earned 35 x dollars for thefirst 35 hoursplus 7 x additional dollars . since for each additional hour he gets 1.5 x dollars then he worked for 7 x / 1.5 x = 4 additional hours , so harry worked for total of 30 + 4 = 34 hours . answer : c ." | a = 40 + 2
b = a - 35
c = b / 1
d = c + 35
|
a ) 4 sec , b ) 7 sec , c ) 2 sec , d ) 6 sec , e ) 8 sec | e | divide(80, multiply(36, const_0_2778)) | in what time will a railway train 80 m long moving at the rate of 36 kmph pass a telegraph post on its way ? | "t = 80 / 36 * 18 / 5 = 8 sec answer : e" | a = 36 * const_0_2778
b = 80 / a
|
['a ) 68', 'b ) 83', 'c ) 57', 'd ) 61', 'e ) 62'] | a | subtract(circle_area(divide(multiply(multiply(divide(sqrt(const_3), const_2), 12), const_2), const_3)), multiply(divide(multiply(power(divide(12, const_3), const_2), sqrt(const_3)), const_4), 12)) | two equilateral triangles of side 12 cm are placed one on top of another , such that a 6 pointed star is formed . if the six verticals lie on a circle what is the area of the circle not enclosed by the star ? | area of star = β 3 / 4 * ( 12 ) ^ 2 + 3 * β 3 / 4 * ( 12 / 3 ) ^ 2 = 48 β 3 if r be radius of circle then r = 6 / cos 30 = > r = 12 / β 3 = > r ^ 2 = 48 area of circle = pi * 48 area of the circle not enclosed by the star = 48 * pi - 48 β 3 = 48 ( 3.14 - 1.732 ) = 68 ( approax ) answer : a | a = math.sqrt(3)
b = a / 2
c = b * 12
d = c * 2
e = d / 3
f = circle_area - (
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a ) 712.5 , b ) 1234 , c ) 1289 , d ) 1345 , e ) none of these | a | divide(divide(multiply(multiply(const_100, const_100), 9.5), const_100), const_4) | jaclyn buys $ 30 000 worth of debentures in a company . she earns 9.5 % p . a . simple interest , paid to her quarterly ( that is , every 3 months ) . if the agreed period of the debenture was 18 months : calculate the amount of interest jaclyn will earn for each quarter | "explanation : i = ( p x r x t ) / 100 = 30000 * 9.5 / 100 * ( 18 / 12 ) ^ 1 / 6 = 712.50 answer : a" | a = 100 * 100
b = a * 9
c = b / 100
d = c / 4
|
a ) 1 , b ) 3 , c ) 5 , d ) 6 , e ) 8 | c | subtract(multiply(subtract(12, 1), add(6, 1)), multiply(12, 6)) | the average of 6 observations is 12 . a new seventh observation is included and the new average is decreased by 1 . the seventh observation is : | "explanation : seventh observation = ( 7 Γ Γ 11 - 6 Γ Γ 12 ) = 5 correct option : c" | a = 12 - 1
b = 6 + 1
c = a * b
d = 12 * 6
e = c - d
|
a ) $ 24960 , b ) $ 25920 , c ) $ 28080 , d ) $ 28350 , e ) $ 29160 | d | add(multiply(multiply(divide(multiply(multiply(add(subtract(add(const_12, 5), 9), const_1), const_12), 30), add(3, const_1)), 3), 10), multiply(divide(multiply(multiply(add(subtract(add(const_12, 5), 9), const_1), const_12), 30), add(3, const_1)), 5)) | for a certain art exhibit , a museum sold admission tickets to a group of 30 people every 5 minutes from 9 : 00 in the morning to 5 : 55 in the afternoon , inclusive . the price of a regular admission ticket was $ 10 and the price of a student ticket was $ 5 . if on one day 3 times as many regular admission tickets were sold as student tickets , what was the total revenue from ticket sales that day ? | "from 9 : 00 in the morning to 5 : 55 in the afternoon , inclusive there are 9 * 12 = 108 five - minute intervals , thus total of 108 * 30 tickets were sold . say x student and 3 x regular tickets were sold , then x + 3 x = 108 * 30 - - > x = 27 * 30 and 3 x = 3 * ( 27 * 30 ) = 27 * 90 . therefore , the total revenue from ticket sales that day was 27 * 30 * 5 + 27 * 90 * 10 = $ 28,350 . answer : d ." | a = 12 + 5
b = a - 9
c = b + 1
d = c * 12
e = d * 30
f = 3 + 1
g = e / f
h = g * 3
i = h * 10
j = 12 + 5
k = j - 9
l = k + 1
m = l * 12
n = m * 30
o = 3 + 1
p = n / o
q = p * 5
r = i + q
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | a | divide(const_60.0, multiply(const_10, const_2)) | how many factors of 880 are odd numbers greater than 1 ? | when factorized , 880 has 6 prime factors . of these prime factors 2 are odd and 4 are even . hence total number of odd factors is 2 * 2 ( 4 ) , which includes 4 . the total number of odd factors greater than 1 are 3 . ( option a ) | a = 10 * 2
b = const_60 / 0
|
a ) 12 , b ) 27 , c ) 42 , d ) 61 , e ) 60 | b | add(divide(subtract(270, 18), 9), const_1) | how many multiples of 9 are there between 18 and 270 ? | "it should be mentioned whether 18 and 270 are inclusive . if 18 and 270 are inclusive , then the answer is ( 270 - 18 ) / 9 + 1 = 29 . if 18 and 270 are not inclusive , then the answer is ( 261 - 27 ) / 9 + 1 = 27 . since oa is b , then we have not inclusive case ." | a = 270 - 18
b = a / 9
c = b + 1
|
a ) 52 days , b ) 47 days , c ) 42 days , d ) 37 days , e ) 32 days | c | divide(multiply(70, 30), 50) | if it takes 30 construction workers 70 days to build a skyscraper , how long would it take 50 workers ? | 30 * 70 = 50 * x x = 42 answer : c | a = 70 * 30
b = a / 50
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a ) 50 , b ) 74 , c ) 37 , d ) 77 , e ) 68 | e | add(add(multiply(4, const_10), multiply(const_2, const_10)), multiply(4, const_2)) | there are thirty - 4 lines that are tangent to a circle , and these lines create regions in the plane . can you tell how many of these regions are not enclosed ? | e 68 68 regions . each new tangent increases the non - enclosed areas by two | a = 4 * 10
b = 2 * 10
c = a + b
d = 4 * 2
e = c + d
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a ) rs . 480 , b ) rs . 960 , c ) rs . 960 , d ) rs . 460 , e ) rs . 560 | d | divide(subtract(1400, multiply(8, 60)), 2) | 2 cow β s and 8 goats are brought for rs . 1400 . if the average price of a goat be rs . 60 . what is the average price of a cow . | explanation : average price of a goat = rs . 60 total price of 8 goats = 8 * 60 = rs . 480 but total price of 2 cows and 8 goats = rs . 1400 total price of 2 cows is = 1400 - 480 = 980 average price of a cow = 980 / 2 = rs . 460 answer : option d | a = 8 * 60
b = 1400 - a
c = b / 2
|
a ) 1 , b ) 2 , c ) 345 , d ) 445 , e ) 235 | b | divide(subtract(subtract(365, multiply(multiply(add(const_4, const_2), add(const_4, const_2)), const_10)), 1), const_2) | 1 = 6,2 = 36,3 = 363,4 = 364,5 = 365 , then 36 = ? | 1 = 6,2 = 36,3 = 363,4 = 364,5 = 365 , then 36 = ? 36 = 2 check the second eqn . answer : b | a = 4 + 2
b = 4 + 2
c = a * b
d = c * 10
e = 365 - d
f = e - 1
g = f / 2
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | e | divide(const_1, add(divide(const_1, 6), divide(divide(const_1, 6), const_3))) | p alone can complete a job in 6 days . the work done by q alone in one day is equal to one - fifth of the work done by p alone in one day . in how many days can the work be completed if p and q work together ? | "p ' s rate is 1 / 6 q ' s rate is 1 / 30 the combined rate is 1 / 6 + 1 / 30 = 1 / 5 if they work together , the job will take 5 days . the answer is e ." | a = 1 / 6
b = 1 / 6
c = b / 3
d = a + c
e = 1 / d
|
a ) 375 , b ) 277 , c ) 266 , d ) 99 , e ) 121 | a | divide(150, divide(150, const_100)) | 60 % of a number is added to 150 , the result is the same number . find the number ? | ": ( 60 / 100 ) * x + 150 = x 2 x = 750 x = 375 answer : a" | a = 150 / 100
b = 150 / a
|
a ) 24 , b ) 36 , c ) 48 , d ) 60 , e ) 72 | b | subtract(power(3, const_4), const_4) | x , y , and z are different prime numbers . the product x ^ 3 * y ^ 2 * z ^ 2 is divisible by how many different positive numbers ? | "the exponents of x ^ 3 * y ^ 2 * z ^ 2 are 3 , 2 , and 2 . the number of factors is ( 3 + 1 ) ( 2 + 1 ) ( 2 + 1 ) = 36 the answer is b ." | a = 3 ** 4
b = a - 4
|
a ) 141 , b ) 180 , c ) 130 , d ) 122 , e ) 280 | e | multiply(divide(lcm(87, 20), const_12), const_2) | what is the smallest integer that is multiple of 87 and 20 | correct answer : e it is the lcm of 8,7 and 20 which is 280 | a = math.lcm(87, 20)
b = a / 12
c = b * 2
|
a ) s . 200.72 , b ) s . 250.72 , c ) s . 272.72 , d ) s . 300.72 , e ) s . 310.72 | c | divide(multiply(600, multiply(add(const_1, const_4), const_2)), multiply(add(multiply(add(const_1, const_4), const_2), const_1), const_2)) | two employees x and y are paid a total of rs . 600 per week by their employer . if x is paid 120 percent of the sum paid to y , how much is y paid per week ? | "let the amount paid to x per week = x and the amount paid to y per week = y then x + y = 600 but x = 120 % of y = 120 y / 100 = 12 y / 10 β΄ 12 y / 10 + y = 600 β y [ 12 / 10 + 1 ] = 600 β 22 y / 10 = 600 β 22 y = 6000 β y = 6000 / 22 = rs . 272.72 c" | a = 1 + 4
b = a * 2
c = 600 * b
d = 1 + 4
e = d * 2
f = e + 1
g = f * 2
h = c / g
|
a ) s . 42 , b ) s . 46 , c ) s . 40 , d ) s . 41 , e ) s . 32 | e | divide(divide(multiply(800, 20), const_100), 5) | a reduction of 20 % in the price of oil enables a house wife to obtain 5 kgs more for rs . 800 , what is the reduced price for kg ? | "800 * ( 20 / 100 ) = 160 - - - - 5 ? - - - - 1 = > rs . 32 answer : e" | a = 800 * 20
b = a / 100
c = b / 5
|
a ) 124 % , b ) 120 % , c ) 96 % , d ) 80 % , e ) 84 % | e | multiply(multiply(subtract(const_1, divide(40, const_100)), add(const_1, divide(40, const_100))), const_100) | mary ' s income is 40 percent more than tim ' s income , and tim ' s income is 40 percent less than juan ' s income . what percent of juan ' s income is mary ' s income ? | "juan ' s income = 100 ( assume ) ; tim ' s income = 60 ( 40 percent less than juan ' s income ) ; mary ' s income = 84 ( 40 percent more than tim ' s income ) . thus , mary ' s income ( 84 ) is 84 % of juan ' s income ( 100 ) . answer : e ." | a = 40 / 100
b = 1 - a
c = 40 / 100
d = 1 + c
e = b * d
f = e * 100
|
a ) $ 1.00 , b ) $ 1.12 , c ) $ 1.20 , d ) $ 1.32 , e ) $ 1.40 | b | multiply(divide(1.8, add(const_100, subtract(const_100, 20))), divide(multiply(subtract(const_100, 20), add(const_100, 40)), const_100)) | kim has 40 percent more money than sal and sal has 20 percent less money than phil . if sal and phil have a combined total of $ 1.80 , how much money does kim have ? | phil = x sal = 0.8 x kim = ( 1.4 ) * 0.8 x = 1.12 x x + 0.8 x = 1.8 - - > x = 1 kim = 1.12 answer : b | a = 100 - 20
b = 100 + a
c = 1 / 8
d = 100 - 20
e = 100 + 40
f = d * e
g = f / 100
h = c * g
|
a ) $ 2,240 , b ) $ 2,120 , c ) $ 1,960 , d ) $ 1,920 , e ) $ 1,400 | a | subtract(multiply(const_3, const_1000), add(multiply(divide(add(14, 2), 2), 20), multiply(subtract(divide(add(14, 2), 2), 2), 100))) | a gambler bought $ 3,000 worth of chips at a casino in denominations of $ 20 and $ 100 . that evening , the gambler lost 14 chips , and then cashed in the remainder . if the number of $ 20 chips lost was 2 more or 2 less than the number of $ 100 chips lost , what is the largest amount of money that the gambler could have received back ? | "in order to maximize the amount of money that the gambler kept , we should maximize # of $ 20 chips lost and minimize # of $ 100 chips lost , which means that # of $ 20 chips lost must be 2 more than # of $ 100 chips lost . so , if # of $ 20 chips lost is x then # of $ 100 chips lost should be x - 2 . now , given that total # of chips lost is 14 : x + x - 2 = 14 - - > x = 8 : 8 $ 20 chips were lost and 8 - 2 = 6 $ 100 chips were lost . total worth of chips lost is 8 * 20 + 6 * 100 = $ 760 , so the gambler kept $ 3,000 - $ 760 = $ 2,240 . answer : a ." | a = 3 * 1000
b = 14 + 2
c = b / 2
d = c * 20
e = 14 + 2
f = e / 2
g = f - 2
h = g * 100
i = d + h
j = a - i
|
a ) 0.555 d , b ) 0.765 d , c ) 0.455 d , d ) 0.785 d , e ) 0.805 d | c | subtract(divide(subtract(const_100, 35), const_100), multiply(divide(subtract(const_100, 35), const_100), divide(30, const_100))) | a dress on sale in a shop is marked at $ d . during the discount sale its price is reduced by 35 % . staff are allowed a further 30 % reduction on the discounted price . if a staff member buys the dress what will she have to pay in terms of d ? | "effective discount = a + b + ab / 100 = - 35 - 30 + ( - 35 ) ( - 30 ) / 100 = - 54.5 sale price = d * ( 1 - 54.5 / 100 ) sale price = . 455 * d answer ( c )" | a = 100 - 35
b = a / 100
c = 100 - 35
d = c / 100
e = 30 / 100
f = d * e
g = b - f
|
a ) 27 , b ) 33 , c ) 84 , d ) 81 , e ) 162 | c | multiply(multiply(multiply(multiply(const_2, const_2), const_2), const_3), 6) | the number of boxes in a warehouse can be divided evenly into 6 equal shipments by boat or 28 equal shipments by truck . what is the smallest number of boxes that could be in the warehouse ? | "answer is the lcm of 6 and 28 = 84 answer c" | a = 2 * 2
b = a * 2
c = b * 3
d = c * 6
|
a ) 56 , b ) 70 , c ) 84 , d ) 96 , e ) 120 | c | divide(divide(336, const_2), const_2) | if k ^ 3 is divisible by 336 , what is the least possible value of integer k ? | "k ^ 3 = 336 * x = 2 ^ 4 * 3 * 7 * x the factors of k must at minimum include 2 ^ 2 * 3 * 7 = 84 . the answer is c ." | a = 336 / 2
b = a / 2
|
a ) 5 , b ) 21 , c ) 33 , d ) 60 , e ) 40 | b | add(add(add(add(6, subtract(6, 1)), subtract(subtract(6, 1), 1)), subtract(subtract(subtract(6, 1), 1), 1)), 2) | there are 6 boxes numbered 1 , 2 , . . . 6 . each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered . the total number of ways in which this can be done is | given the constraints of this question , we have to consider 6 possible outcomes . . . 1 green ball : 6 options grrrrr rgrrrr rrgrrr rrrgrr rrrrgr rrrrrg 2 green balls : 5 options ggrrrr rggrrr rrggrr rrrggr rrrrgg 3 green balls : 4 options gggrrr rgggrr rrgggr rrrggg 4 green balls : 3 options ggggrr rggggr rrgggg 5 green balls : 2 options gggggr rggggg 6 green balls : 1 option gggggg 6 + 5 + 4 + 3 + 2 + 1 = 21 options . b | a = 6 - 1
b = 6 + a
c = 6 - 1
d = c - 1
e = b + d
f = 6 - 1
g = f - 1
h = g - 1
i = e + h
j = i + 2
|
['a ) 20 %', 'b ) 30 %', 'c ) 40 %', 'd ) 50 %', 'e ) 55 %'] | b | divide(multiply(subtract(add(power(const_1, const_2), power(const_1, const_2)), sqrt(add(power(const_1, const_2), power(const_1, const_2)))), const_100), add(power(const_1, const_2), power(const_1, const_2))) | a man walked diagonally across a square lot . approximately , what was the percent saved by not walking along the edges ? | let the side of the square ( abcd ) be x metres . then , ab + bc = 2 x metres . ac = 2 x = ( 1.41 x ) m . saving on 2 x metres = ( 0.59 x ) m . saving % = ( 0.59 x ) / 2 x x 100 % = 30 % ( approx . ) b ) | a = 1 ** 2
b = 1 ** 2
c = a + b
d = 1 ** 2
e = 1 ** 2
f = d + e
g = math.sqrt(f)
h = c - g
i = h * 100
j = 1 ** 2
k = 1 ** 2
l = j + k
m = i / l
|
a ) 150 , b ) 550 , c ) 760 , d ) 450 , e ) 260 | b | multiply(power(add(const_1, divide(2, const_100)), 5), 500) | rs . 500 amounts to rs . 670 in 5 years at simple interest . if the interest is increased by 2 % , it would amount to how much ? | ( 500 * 5 * 2 ) / 100 = 50 500 + 50 = 550 answer : b | a = 2 / 100
b = 1 + a
c = b ** 5
d = c * 500
|
a ) 10000 , b ) 562 , c ) 352 , d ) 576 , e ) 594 | a | divide(multiply(10000, 1000), 11000) | the lcm and hcf of two numbers are 10000 and 1000 respectively . find the larger of the two numbers if their sum is 11000 . | "there are 2 approaches in solving this . methode 1 . hcf * lcm = the actual number . 10000 * 1000 = 10000000 so the answer which we are looking for has to be a factor of 10000000 . so among the options shortlist the answers by eliminating those numbers which is not divisible by 10000000 . and then take the highest number as the answer as the question asks abt the highest number . answer is a" | a = 10000 * 1000
b = a / 11000
|
a ) 4.3 hrs , b ) 8.9 hrs , c ) 2.9 hrs , d ) 7.2 hrs , e ) 8.6 hrs | a | divide(const_1, subtract(divide(const_1, 3), divide(const_1, 10))) | a cistern can be filled by a tap in 3 hours while it can be emptied by another tap in 10 hours . if both the taps are opened simultaneously , then after how much time will the cistern get filled ? | "net part filled in 1 hour = ( 1 / 3 - 1 / 10 ) = 7 / 30 the cistern will be filled in 30 / 7 hrs i . e . , 4.3 hrs . answer : a" | a = 1 / 3
b = 1 / 10
c = a - b
d = 1 / c
|
a ) 12 km , b ) 15 km , c ) 18 km , d ) 24 km , e ) data inadequate | a | divide(multiply(subtract(9, 3), subtract(divide(add(add(6, 4), add(4, 3)), const_4), const_0_25)), const_2) | a software engineer starts from home at 3 pm for every walk on a route which has level road for some distance and a then a hillock . he walks speed of 4 kmph on level ground and then at a speed of 3 kmph on the uphill and then down the hill at a speed of 6 kmph to the level ground and then at a speed of 4 kmph to the home if he reaches home at 9 pm . what is the distance one way of his route ? | time = distance / speed so , let the distance to the level road from home be x km and the distance to the hillock from the level road be y km so , ( x / 4 + y / 3 ) + ( y / 6 + x / 4 ) = 6 x + y = 12 km . answer : a | a = 9 - 3
b = 6 + 4
c = 4 + 3
d = b + c
e = d / 4
f = e - const_0_25
g = a * f
h = g / 2
|
a ) 5 % , b ) 10 % , c ) 15 % , d ) 20 % , e ) it can not be determined | d | subtract(add(60, 60), const_100) | a box contains either blue or red flags . the total number of flags in the box is an even number . a group of children are asked to pick up two flags each . if all the flags are used up in the process such that 60 % of the children have blue flags , and 60 % have red flags , what percentage of children have flags of both the colors ? | "solution : let the total number of flags be 100 ( even number ) let the total number of ' blue ' flags alone be ' a ' let the total number of ' red ' flags alone be ' b ' let the total number of ' both ' flags be ' c ' we have given , total number of blue flags = 60 % = 60 = a + c total number of red flags = 55 % = 55 = b + c total number of flags = a + b + c = 100 ( since all the flags have been utilized ) so , substituting for c in the third equation , we have , 60 - c + c + 60 - c = 100 c = 20 option d ." | a = 60 + 60
b = a - 100
|
a ) 14 , b ) 13 , c ) 16 , d ) 17 , e ) 8 | b | subtract(11, const_1) | there are 30 balls which are red , blue or green . if 11 balls are green and the sum of red balls and green balls is less than 25 , at most how many red balls are there ? | "r + g + b = 30 g = 11 r + g < 25 = > r + 11 < 25 = > r < 14 = > at most 13 red balls answer : b" | a = 11 - 1
|
a ) 2187 m , b ) 3704 m , c ) 4179 m , d ) 4224 m , e ) 4297 m | d | divide(multiply(multiply(multiply(divide(add(multiply(add(const_3, const_4), const_3), const_1), add(const_3, const_4)), 22.4), const_2), 3000), const_100) | the radius of a wheel is 22.4 cm . what is the distance covered by the wheel in making 3000 resolutions ? | "in one resolution , the distance covered by the wheel is its own circumference . distance covered in 3000 resolutions . = 3000 * 2 * 22 / 7 * 22.4 = 422400 cm = 4224 m answer : d" | a = 3 + 4
b = a * 3
c = b + 1
d = 3 + 4
e = c / d
f = e * 22
g = f * 2
h = g * 3000
i = h / 100
|
a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | c | multiply(subtract(11, 6), const_2) | half a number plus 6 is 11 . what is the number ? | let x be the number . always replace ` ` is ' ' with an equal sign ( 1 / 2 ) x + 6 = 11 ( 1 / 2 ) x = 11 - 6 ( 1 / 2 ) x = 5 x = 10 correct answer is c | a = 11 - 6
b = a * 2
|
a ) 8 , b ) 12 , c ) 10 , d ) 16 , e ) 18 | c | divide(power(10, 2), 10) | n ^ ( n / 2 ) = 6 is true when n = 6 in the same way what is the value of n if n ^ ( n / 2 ) = 10 ? | n ^ ( n / 2 ) = 10 apply log n / 2 logn = log 10 nlogn = 2 log 10 = log 10 ^ 2 = log 100 logn = log 100 now apply antilog n = 100 / n now n = 10 . answer : c | a = 10 ** 2
b = a / 10
|
a ) 48 , b ) 50 , c ) 56 , d ) 62 , e ) 64 | c | subtract(multiply(6, 52), subtract(multiply(11, 50), multiply(6, 49))) | the average age of 11 boys is 50 , if the average age of first 6 boys is 49 and that of the last 6 is 52 . find the boy age ? | 1 to 11 = 11 * 50 = 550 1 to 6 = 6 * 49 = 294 6 to 11 = 6 * 52 = 312 6 th = 294 + 312 β 550 = 56 c | a = 6 * 52
b = 11 * 50
c = 6 * 49
d = b - c
e = a - d
|
a ) rs . 6000 , b ) rs . 9000 , c ) rs . 10800 , d ) rs . 10000 , e ) rs . 9980 | d | divide(multiply(6840, const_100), subtract(subtract(subtract(const_100, 20), divide(multiply(subtract(const_100, 20), 10), const_100)), divide(multiply(subtract(subtract(const_100, 20), divide(multiply(subtract(const_100, 20), 10), const_100)), 5), const_100))) | after successive discounts of 20 % , 10 % and 5 % a certain good is sold for rs . 6840 . find the actual price of the good . | "let actual price was 100 . after three successive discount this will become , 100 = = 20 % discount = > 80 = = 10 % discount = > 72 = = 5 % discount = 68.4 now compare , 68.4 = 6840 1 = 6840 / 68.4 100 = ( 6840 * 100 ) / 68.4 = rs . 10,000 . answer : option d" | a = 6840 * 100
b = 100 - 20
c = 100 - 20
d = c * 10
e = d / 100
f = b - e
g = 100 - 20
h = 100 - 20
i = h * 10
j = i / 100
k = g - j
l = k * 5
m = l / 100
n = f - m
o = a / n
|
a ) 1218 , b ) 1212 , c ) 1408 , d ) 2707 , e ) 2591 | c | multiply(subtract(power(23, const_2), power(const_10, const_2)), divide(add(multiply(9, const_2), const_2), add(const_4, const_3))) | a rope of which a calf is tied is increased from 9 m to 23 m , how much additional grassy ground shall it graze ? | "Ο ( 232 β 92 ) = 1408 answer : c" | a = 23 ** 2
b = 10 ** 2
c = a - b
d = 9 * 2
e = d + 2
f = 4 + 3
g = e / f
h = c * g
|
a ) β 0.3 , b ) 0 , c ) 0.3 , d ) 0.94 , e ) 2.46 | d | subtract(multiply(divide(divide(subtract(power(3, 2), power(1.6, 0.3)), const_1000), const_1000), 3), divide(divide(subtract(power(3, 2), power(1.6, 0.3)), const_1000), const_1000)) | what is the value of 3 x ^ 2 β 1.6 x + 0.3 for x = 0.8 ? | "3 x ^ 2 - 1.6 x + 0.3 for x = 0.6 = 3 ( 0.8 * 0.8 ) - 2 * 0.8 * ( 0.8 ) + 0.3 = 0.8 * 0.8 + 0.3 = 0.94 correct option : d" | a = 3 ** 2
b = 1 ** 6
c = a - b
d = c / 1000
e = d / 1000
f = e * 3
g = 3 ** 2
h = 1 ** 6
i = g - h
j = i / 1000
k = j / 1000
l = f - k
|
a ) 144 , b ) 153 , c ) 162 , d ) 171 , e ) 180 | c | subtract(900, add(add(multiply(divide(44, const_100), 900), multiply(divide(28, const_100), 900)), multiply(divide(10, const_100), 900))) | in a school of 900 students , 44 % wear blue shirts , 28 % wear red shirts , 10 % wear green shirts , and the remaining students wear other colors . how many students wear other colors ( not blue , not red , not green ) ? | "44 + 28 + 10 = 82 % 100 β 82 = 18 % 900 * 18 / 100 = 162 the answer is c ." | a = 44 / 100
b = a * 900
c = 28 / 100
d = c * 900
e = b + d
f = 10 / 100
g = f * 900
h = e + g
i = 900 - h
|
a ) 2.5 % , b ) 15 % , c ) 25 % , d ) 35 % , e ) 250 % | d | multiply(divide(divide(14, const_100), divide(40, const_100)), const_100) | if c is 14 % of a and 40 % of b , what percent of a is b ? | "answer = d 14 a / 100 = 40 b / 100 b = 14 a / 40 = 35 a / 100 = 35 %" | a = 14 / 100
b = 40 / 100
c = a / b
d = c * 100
|
a ) 11 / 30 , b ) 29 / 60 , c ) 17 / 30 , d ) 19 / 30 , e ) 7 / 15 | e | multiply(add(multiply(2, 3), 1), multiply(divide(1, 3), divide(1, 5))) | a new tower has just been built at the verbico military hospital ; the number of beds available for patients at the hospital is now 2 times the number available before the new tower was built . currently , 1 / 3 of the hospital ' s original beds , as well as 1 / 5 of the beds in the new tower , are occupied . for the purposes of renovating the hospital ' s original wing , all of the patients in the hospital ' s original beds must be transferred to beds in the new tower . if patients are neither admitted nor discharged during the transfer , what fraction of the beds in the new tower will be unoccupied once the transfer is complete ? | i think e - 7 / 15 is the correct answer . here goes : lets assume originally the number of beds = x after the new tower , the total combined no of beds = 2 x so old = x , new = x now 1 / 3 of x are occupied and 1 / 5 of x are occupied which simplifies to ( 1 / 5 ) x we are shifting 1 / 3 of x to the new ward so there will now be : 1 / 3 of x plus 1 / 5 of x occupied in the new ward . add them up to get 8 / 15 of x there are x beds in new tower so ratio is : ( 8 / 15 ) x / x = 8 / 15 of x subtract that from 15 / 15 of x and you get the number of un - occupied beds to total capacity of new towerb = 7 / 15 . e | a = 2 * 3
b = a + 1
c = 1 / 3
d = 1 / 5
e = c * d
f = b * e
|
a ) 6 , b ) 6.25 , c ) 7.25 , d ) 7.5 , e ) 4.75 | e | divide(subtract(222, multiply(3.2, 10)), 40) | cricket match is conducted in us . the run rate of a cricket game was only 3.2 in first 10 over . what should be the run rate in the remaining 40 overs to reach the target of 222 runs ? | "required run rate = 222 - ( 3.2 x 10 ) = 190 = 4.75 40 40 e" | a = 3 * 2
b = 222 - a
c = b / 40
|
a ) 16 , b ) 77 , c ) 25 , d ) 87 , e ) 18 | e | floor(sqrt(342)) | if the sum of a number and its square is 342 , what is the number ? | "explanation : let the integer be x . then , x + x 2 = 342 x 2 + x - 342 = 0 ( x + 19 ) ( x β 18 ) = 0 x = 18 answer : e" | a = math.sqrt(342)
b = math.floor(a)
|
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