options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 727 | problem stringlengths 5 967 | rationale stringlengths 1 2.74k | program stringlengths 10 646 |
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a ) a ) 10 , b ) b ) 12 , c ) c ) 15 , d ) d ) 14 , e ) e ) 9 | b | sqrt(divide(4320, add(power(5, 2), add(power(1, 2), power(2, 2))))) | the ratio of three numbers is 1 : 2 : 5 and the sum of their squares is 4320 . the sum of the numbers is ? | "let the numbers be x , 2 x , 5 x then , x ^ 2 + 4 x ^ 2 + 25 x ^ 2 = 4320 30 x ^ 2 = 4320 x ^ 2 = 144 x = 12 answer is b" | a = 5 ** 2
b = 1 ** 2
c = 2 ** 2
d = b + c
e = a + d
f = 4320 / e
g = math.sqrt(f)
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a ) 1 , b ) 8 , c ) 7 , d ) 6 , e ) 2 | a | divide(90, add(84, add(const_3, const_3))) | the average salary per head of the entire staff of an office including the officers and clerks is rs . 90 . the average salary of officers is rs . 600 and that of the clerks is rs . 84 . if the number of officers is 2 , find the number of officers in the office ? | 6 * 22 = 132 7 * 19 = 133 - - - - - - - - - - - - - - 1 . answer : a | a = 3 + 3
b = 84 + a
c = 90 / b
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a ) 28.28 cm , b ) 25.45 cm , c ) 26.45 cm , d ) 20.18 cm , e ) 32.14 cm | a | divide(circumface(divide(square_edge_by_perimeter(rectangle_perimeter(20, 16)), const_2)), const_2) | the parameter of a square is equal to the perimeter of a rectangle of length 20 cm and breadth 16 cm . find the circumference of a semicircle whose diameter is equal to the side of the square . ( round off your answer to two decimal places ) ? | "let the side of the square be a cm . parameter of the rectangle = 2 ( 20 + 16 ) = 72 cm parameter of the square = 72 cm i . e . 4 a = 72 a = 18 diameter of the semicircle = 18 cm circimference of the semicircle = 1 / 2 ( ∏ ) ( 18 ) = 1 / 2 ( 22 / 7 ) ( 18 ) = 396 / 14 = 28.28 cm to two decimal places answer : a" | a = square_edge_by_perimeter / (
b = circumface / (
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a ) 29 , b ) 780 , c ) 920 , d ) 979 , e ) none of these | c | multiply(100, 10) | 100 × 10 – 100 + 2000 ÷ 100 = ? | "solution given exp . = 100 × 10 - 100 + 20 = 1000 - 100 + 20 = 1020 - 100 = 920 . answer c" | a = 100 * 10
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a ) 0.1 % , b ) 1 % , c ) 1.5 % , d ) 2 % , e ) 3 % | a | multiply(divide(1, 1), const_100) | what percent is 1 gm of 1 kg ? | "1 kg = 1000 gm 1 / 1000 × 100 = 100 / 1000 = 1 / 10 = 0.1 % a )" | a = 1 / 1
b = a * 100
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a ) s . 3700 , b ) s . 4570 , c ) s . 4500 , d ) s . 4550 , e ) s . 2500 | a | subtract(multiply(8000, const_4), subtract(multiply(8700, const_4), 6500)) | the average salary of a person for the months of january , february , march and april is rs . 8000 and that for the months february , march , april and may is rs . 8700 . if his salary for the month of may is rs . 6500 , find his salary for the month of january ? | "sum of the salaries of the person for the months of january , february , march and april = 4 * 8000 = 32000 - - - - ( 1 ) sum of the salaries of the person for the months of february , march , april and may = 4 * 8700 = 34800 - - - - ( 2 ) ( 2 ) - ( 1 ) i . e . may - jan = 2800 salary of may is rs . 6500 salary of january = rs . 3700 answer : a" | a = 8000 * 4
b = 8700 * 4
c = b - 6500
d = a - c
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a ) 1 hour , b ) 1.2 hour , c ) 3 hours , d ) 5 hours , e ) 1.1 hours | e | divide(const_1, subtract(const_1, divide(const_1, multiply(5, const_2)))) | one pump drains one - half of a pond in 5 hours , and then a second pump starts draining the pond . the two pumps working together finish emptying the pond in one - half hour . how long would it take the second pump to drain the pond if it had to do the job alone ? | "the tricky part here , i believed is one half hour = 1 / 2 . then everything would be easy . we have the 1 st pump working rate / hour = 1 / 2 : 5 = 1 / 10 working rate of 2 pumps : 1 / 2 : 1 / 2 = 1 . working rate of 2 nd pump : 1 - 1 / 10 = 9 / 10 - - > time taken for the 2 nd pump to finish : 1 : 9 / 10 = 10 / 9 = 1.1 hours . e" | a = 5 * 2
b = 1 / a
c = 1 - b
d = 1 / c
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a ) 50 km , b ) 40 km , c ) 30 km , d ) 20 km , e ) none of these | c | multiply(12, divide(10, subtract(16, 12))) | if a jogger jogs 16 km / hr instead of 12 km / hr , he would have jogged 10 km more . the actual distance he jogged is | explanation : the actual distance jogged be d . = > d / 12 = ( d + 10 ) / 16 = > 16 d = 12 d + 120 = > 4 d = 120 = > d = 30 answer : c | a = 16 - 12
b = 10 / a
c = 12 * b
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a ) 2 , b ) 8 , c ) 4 , d ) 5 , e ) 10 | e | add(divide(subtract(1000, 100), 100), const_1) | how many multiples of 100 are there between 100 and 1000 ( both are inclusive ) ? | "the answer is ( 1000 - 100 ) / 100 + 1 = 10 answer is e" | a = 1000 - 100
b = a / 100
c = b + 1
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a ) 750 , b ) 882 , c ) 772 , d ) 652 , e ) 271 | a | add(500, multiply(500, divide(50, const_100))) | a person buys an article at rs . 500 . at what price should he sell the article so as to make a profit of 50 % ? | "cost price = rs . 500 profit = 50 % of 500 = rs . 250 selling price = cost price + profit = 500 + 250 = 750 answer : a" | a = 50 / 100
b = 500 * a
c = 500 + b
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a ) 2800 , b ) 2700 , c ) 2100 , d ) 2500 , e ) none of these | b | multiply(multiply(subtract(const_1, divide(20, const_100)), subtract(const_1, divide(55, const_100))), 7500) | in an election between two candidates , one got 55 % of the total valid votes , 20 % of the votes were invalid . if the total number of votes was 7500 , what was the number of valid votes that the other candidate got ? | "explanation : total number of votes = 7500 given that 20 % of percentage votes were invalid = > valid votes = 80 % total valid votes = 7500 × ( 80 / 100 ) 1 st candidate got 55 % of the total valid votes . hence the 2 nd candidate should have got 45 % of the total valid votes = > valid votes that 2 nd candidate got = total valid votes × ( 45 / 100 ) = 7500 × ( 80 / 100 ) × ( 45 / 100 ) = 75 × 45 × 45 = 75 × 4 × 9 = 300 × 9 = 2700 answer : option b" | a = 20 / 100
b = 1 - a
c = 55 / 100
d = 1 - c
e = b * d
f = e * 7500
|
a ) 18 days , b ) 24 days , c ) 30 days , d ) 36 days , e ) 35 days | a | divide(subtract(const_1, multiply(divide(add(add(divide(const_1, 30), divide(const_1, 24)), divide(const_1, 20)), const_2), 10)), subtract(divide(add(add(divide(const_1, 30), divide(const_1, 24)), divide(const_1, 20)), const_2), divide(const_1, 24))) | a and b can do a piece of work in 30 days , while b and c can do the same work in 24 days and c and a in 20 days . they all work together for 10 days when b and c leave . how many days more will a take to finish the work ? | explanation : 2 ( a + b + c ) ' s 1 day ' s work = ( 1 / 30 + 1 / 24 + 1 / 20 ) = 15 / 120 = 1 / 8 work done by a , b , c in 10 days = 10 / 6 = 5 / 8 . a ' s 1 day ' s work = ( 1 / 16 - 1 / 24 ) = 1 / 48 now , 1 / 48 work is done by a in 1 day so , 3 / 8 work will be done by a in ( 48 x 3 / 8 ) = 18 days . answer is a | a = 1 / 30
b = 1 / 24
c = a + b
d = 1 / 20
e = c + d
f = e / 2
g = f * 10
h = 1 - g
i = 1 / 30
j = 1 / 24
k = i + j
l = 1 / 20
m = k + l
n = m / 2
o = 1 / 24
p = n - o
q = h / p
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a ) 32 , b ) 37 , c ) 84 , d ) 29 , e ) 21 | c | multiply(sqrt(divide(multiply(84, 21), 4)), 4) | the h . c . f and l . c . m of two numbers are 84 and 21 respectively . if the ratio of the two numbers is 1 : 4 , then the larger of the two numbers is | explanation : let the numbers be x and 4 x . then , \ inline \ fn _ jvn x \ times 4 x = 84 \ times 21 \ : \ : \ leftrightarrow \ : \ : x ^ { 2 } = \ frac { 84 \ times 21 } { 4 } \ : \ : \ leftrightarrow \ : \ : x = 21 hence larger number = 4 x = 84 answer : c ) 84 | a = 84 * 21
b = a / 4
c = math.sqrt(b)
d = c * 4
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a ) 48 seconds , b ) 1 minute , 12 seconds , c ) 1 minute , 50 seconds , d ) 2 minutes , 14 seconds , e ) 4 minutes , 12 seconds | a | add(subtract(const_1, divide(5, 7)), divide(5, 7)) | if it takes a tub 2 minutes to drain 5 / 7 of its content , how much more time will it take for the tub to be empty ? | "if 5 / 7 of tub ' s content is drained 2 / 7 th of tub still needs to be drained . if it takes 2 minutes to drain 5 / 7 th of tub it takes 2 * ( 7 / 5 ) minutes to drain the entire tub and 2 * ( 7 / 5 ) * ( 2 / 7 ) min to drain 2 / 7 th of the tub which is 4 / 5 minutes or 48 minutes so answer is a" | a = 5 / 7
b = 1 - a
c = 5 / 7
d = b + c
|
a ) 1200 , b ) 1960 , c ) 1600 , d ) 1354 , e ) none of these | b | add(divide(210, divide(multiply(divide(9, multiply(const_4, const_3)), 16), const_100)), 210) | the true discount on a bill due 9 months hence at 16 % per annum is rs . 210 . the amount of the bill is | "explanation : let p . w . be rs . x . then , s . i . on rs . x at 16 % for 9 months = rs . 210 . x ã — 16 ã — ( 9 / 12 ) ã — ( 1 / 100 ) = 210 or x = 1750 p . w . = rs . 1750 sum due = p . w . + t . d . = rs . ( 1750 210 ) = rs . 1960 . answer : b" | a = 4 * 3
b = 9 / a
c = b * 16
d = c / 100
e = 210 / d
f = e + 210
|
a ) 112 , b ) 77 , c ) 267 , d ) 13 , e ) 99 | d | subtract(divide(630, multiply(multiply(2, 3), 7)), divide(100, multiply(multiply(2, 3), 7))) | how many numbers between 100 and 630 are divisible by 2 , 3 , and 7 together ? | "explanation : as the division is by 2 , 3 , 7 together , the numbers are to be divisible by : 2 * 3 * 7 = 42 the limits are 100 and 600 the first number divisible is 42 * 3 = 126 to find out the last number divisible by 42 within 630 : 630 / 42 = 13 hence , 42 * 15 = 630 is the last number divisible by 42 within 630 hence , total numbers divisible by 2 , 3 , 7 together are ( 15 â € “ 2 ) = 13 answer : d" | a = 2 * 3
b = a * 7
c = 630 / b
d = 2 * 3
e = d * 7
f = 100 / e
g = c - f
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a ) 10 days , b ) 11 days , c ) 15 days , d ) 20 days , e ) none of these | c | divide(inverse(subtract(inverse(12), inverse(20))), const_2) | a and b together can complete a work in 12 days . a alone can complete it in 20 days . if b does the work only for half a day daily , then in how many days a and b together will complete the work ? | b ' s 1 day ' s work = ( 1 / 12 - 1 / 20 ) = 2 / 60 = 1 / 30 now , ( a + b ) ' s 1 day ' s work = ( 1 / 20 + 1 / 60 ) = 4 / 60 = 1 / 15 so , a and b together will complete the work in 15 days . correct option : c | a = 1/(12)
b = 1/(20)
c = a - b
d = 1/(c)
e = d / 2
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a ) 53 % , b ) 58 % , c ) 62.5 % , d ) 64 % , e ) 75 % | c | multiply(divide(subtract(const_100, 25), add(20, const_100)), const_100) | marketing executives for a certain chewing gum company projected a 20 percent increase in revenue this year over that of last year , but revenue this year actually decreased by 25 % . what percent of the projected revenue was the actual revenue ? | "last year revenue = 100 ( assume ) ; this year revenue = 75 ; projected revenue = 120 . actual / projected * 100 = 75 / 120 * 100 = 62.5 % . answer : c" | a = 100 - 25
b = 20 + 100
c = a / b
d = c * 100
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a ) 0 , b ) 1 , c ) 0 , d ) 3 , e ) 4 | c | floor(divide(reminder(power(7, reminder(1033, add(const_4, const_1))), const_100), const_10)) | what is the tens digit of 7 ^ 1033 ? | "7 ^ 1 = 7 7 ^ 2 = 49 7 ^ 3 = 343 7 ^ 4 = 2401 7 ^ 5 = 16807 7 ^ 6 = 117649 we should see this as pattern recognition . we have a cycle of 4 . ( we can multiply the last 2 digits only as we care about ten ' s digit ) 0 , 4 , 4 , 0 . 1033 = 4 * 258 + 1 the ten ' s digit will be 0 . answer c" | a = 4 + 1
b = 7 ** reminder
c = reminder / (
d = math.floor(c, 100)
|
a ) $ 24 million , b ) $ 120 million , c ) $ 644 million , d ) $ 540 million , e ) $ 720 million | e | subtract(multiply(360, divide(const_12, const_2)), multiply(1.44, const_1000)) | country x imported approximately $ 1.44 billion of goods in 1996 . if country x imported $ 360 million of goods in the first two months of 1997 and continued to import goods at the same rate for the rest of the year , by how much would country xs 1997 imports exceed those of 1996 ? | "convert units to millions as answer is in millions 1996 imports = $ 1.44 bill = $ 1440 mill i . e . 1440 / 12 = $ 120 mill / month 1997 imports = $ 360 mill / 2 month i . e . $ 180 mill / month difference / month = 180 - 120 = 60 difference / year = $ 60 mill * 12 = $ 720 mill answer : e" | a = 12 / 2
b = 360 * a
c = 1 * 44
d = b - c
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a ) 1000 meters , b ) 1050 meters , c ) 1200 meters , d ) 1500 meters , e ) none of these | d | multiply(multiply(6, divide(15, const_60)), const_1000) | a man is walking at the rate of 6 km / hr crosses a bridge in 15 minutes . the length of the bridge is | "explanation : we need to get the answer in meters . so we will first of change distance from km / hour to meter / sec by multiplying it with 5 / 18 and also change 15 minutes to seconds by multiplying it with 60 . speed = 6 â ˆ — 5 / 18 = 5 / 3 m / sec time = 15 â ˆ — 60 seconds = 900 seconds distance = time â ˆ — speed / distance = 5 / 3 â ˆ — 900 = 1500 meter option d" | a = 15 / const_60
b = 6 * a
c = b * 1000
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['a ) 25 are', 'b ) 30 are', 'c ) 50 are', 'd ) 60 are', 'e ) 45 are'] | c | divide(divide(power(multiply(multiply(2, multiply(multiply(2, 3), const_10)), divide(multiply(3, const_1000), multiply(multiply(const_180, const_10), const_2))), const_2), 2), const_100) | 2 minute is taken to travel the diagonal of square field at 3 km / hour . calculate the area of field ? | speed of man = [ 3 * 5 / 18 ] m / sec . = 5 / 6 m / sec . , distance travel in 2 minute = { 5 / 6 * 2 * 60 } m . = 100 m . diagonal of square field = 100 m area of field = 1 / 2 * diagonal ^ 2 = [ 1 / 2 * 100 * 100 ] sq . m = 5000 sq . m = 5000 / 100 = 50 are ( 1 are = 100 sq . m ) answer c | a = 2 * 3
b = a * 10
c = 2 * b
d = 3 * 1000
e = const_180 * 10
f = e * 2
g = d / f
h = c * g
i = h ** 2
j = i / 2
k = j / 100
|
a ) 200 , b ) 400 , c ) 350 , d ) 1800 , e ) 45 | a | subtract(multiply(divide(400, 30), 45), 400) | a 400 meter long train crosses a platform in 45 seconds while it crosses a signal pole in 30 seconds . what is the length of the platform ? | "speed = [ 400 / 30 ] m / sec = 40 / 3 m / sec . let the length of the platform be x meters . then , x + 400 / 45 = 40 / 3 3 ( x + 400 ) = 1800 è x = 200 m . answer : a" | a = 400 / 30
b = a * 45
c = b - 400
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a ) 5 days , b ) 8 days , c ) 6 days , d ) 3.15 days , e ) 2.75 days | a | inverse(add(inverse(8), inverse(14))) | a and b complete a work in 8 days . a alone can do it in 14 days . if both together can do the work in how many days ? | "1 / 8 + 1 / 14 = 11 / 56 56 / 11 = 5 days answer : a" | a = 1/(8)
b = 1/(14)
c = a + b
d = 1/(c)
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a ) 20 , b ) 10 , c ) 34 , d ) 40 , e ) 5 | c | subtract(multiply(27, const_2), multiply(10, const_2)) | if the arithmetic mean of p and q is 10 and the arithmetic mean of q and r is 27 , what is the value of r - p ? | "arithmetic mean expression for p and q : ( p + q ) / 2 = 10 ; p + q = 20 - - - - eq 1 arithmetic mean expression for q and r : ( q + r ) / 2 = 20 ; q + r = 54 - - - - eq 2 subtracting eq 1 from eq 2 we get : r - p = 34 hence , the correct answer is c" | a = 27 * 2
b = 10 * 2
c = a - b
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a ) 3 km , b ) 4 km , c ) 6 km , d ) 9 km , e ) 5 km | a | multiply(const_3, divide(divide(47, const_60), add(add(divide(const_1, 3), divide(const_1, 4)), divide(const_1, 5)))) | a person travels equal distances with speed of 3 km / hr , 4 km / hr and 5 km / hr and takes a total of 47 minutes . find the total distane | explanation : let the distance be 3 x km , then , x / 3 + x / 4 + x / 5 = 47 / 60 47 x / 60 = 47 / 60 x = 1 so total distance = 3 * 1 = 3 km option a | a = 47 / const_60
b = 1 / 3
c = 1 / 4
d = b + c
e = 1 / 5
f = d + e
g = a / f
h = 3 * g
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a ) 60 , b ) 50 , c ) 75 , d ) 100 , e ) 120 | b | multiply(divide(multiply(60, const_1000), const_3600), 3) | a train running at the speed of 60 km / hr crosses a pole in 3 seconds . find the length of the train . | ": speed = 60 * ( 5 / 18 ) m / sec = 50 / 3 m / sec length of train ( distance ) = speed * time ( 50 / 3 ) * 3 = 50 meter answer : b" | a = 60 * 1000
b = a / 3600
c = b * 3
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a ) $ 16537.50 , b ) $ 24512.56 , c ) $ 18475.89 , d ) $ 15478.56 , e ) $ 12456.35 | a | multiply(multiply(add(divide(add(const_4, const_1), const_100), const_1), 15000), add(divide(add(const_4, const_1), const_100), const_1)) | sam invested $ 15000 @ 10 % per annum for one year . if the interest is compounded half yearly , then the amount received by sam at the end of the year will be ? | p = $ 15000 r = 10 % p . a . = 5 % t = 2 half years amount = 15000 * ( 1 + 5 / 100 ) ^ 2 = 15000 * 21 / 20 * 21 / 20 = $ 16537.50 answer is a | a = 4 + 1
b = a / 100
c = b + 1
d = c * 15000
e = 4 + 1
f = e / 100
g = f + 1
h = d * g
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a ) 1 , b ) 1.2 , c ) 1.25 , d ) 1.35 , e ) 1.5 | e | divide(45, add(14, 16)) | two cars start at the same time from opposite ends of a highway that is 45 miles long . one car is riding at 14 mph and the second car is riding at 16 mph . how long after they begin will they meet ? | "as cars are moving in opposite directions their speeds will be added . so their relative speeds : 16 + 14 = 30 mph total distance to be covered = 45 miles . time taken would be : 45 miles / 30 mph = 1.5 hours e is the answer ." | a = 14 + 16
b = 45 / a
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a ) $ 230.00 , b ) $ 145.60 , c ) $ 168.80 , d ) $ 182.00 , e ) $ 210.00 | c | multiply(300, divide(add(30, 20), const_100)) | a discount electronics store normally sells all merchandise at a discount of 10 percent to 30 percent off the suggested retail price . if , during a special sale , an additional 20 percent were to be deducted from the discount price , what would be the lowest possible price of an item costing $ 300 before any discount ? | "since the question is essentially just about multiplication , you can do the various mathstepsin a variety of ways ( depending on whichever method you find easiest ) . we ' re told that the first discount is 10 % to 30 % , inclusive . we ' re told that the next discount is 20 % off of the discounted price . . . . we ' re told to maximize the discount ( thus , 30 % off the original price and then 20 % off of the discounted price ) . thatmathcan be written in a number of different ways ( fractions , decimals , etc . ) : 30 % off = ( 1 - . 3 ) = ( 1 - 30 / 100 ) = ( . 7 ) and the same can be done with the 20 % additional discount . . . the final price of an item that originally cost $ 300 would be . . . . . ( $ 300 ) ( . 7 ) ( . 8 ) = ( $ 300 ) ( . 56 ) = 168 final answer : c" | a = 30 + 20
b = a / 100
c = 300 * b
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a ) 130 miles . , b ) 135 miles . , c ) 145 miles . , d ) 120 miles . , e ) 165 miles . | d | multiply(divide(10, add(5, 10)), 180) | tim and é lan are 180 miles away from one another . they are starting to move towards each other simultaneously , tim at a speed of 10 mph and é lan at a speed of 5 mph . if every hour they double their speeds , what is the distance that tim will pass until he meets é lan ? | "tim and elan will meet at the same time while their ratio of speed is 2 : 1 respectively . so their individual distance traveled ratio will be same . plugging in the answer choice only answer choice d meet the 2 : 1 ( tim : elan = 120 : 60 ) ratio of maintaining total distance traveled 180 miles socorrect answer d" | a = 5 + 10
b = 10 / a
c = b * 180
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a ) 10 , b ) 16 , c ) 12 , d ) 14 , e ) 19 | b | add(add(4, 3), add(add(4, 4), 1)) | for any integer k > 1 , the term “ length of an integer ” refers to the number of positive prime factors , not necessarily distinct , whose product is equal to k . for example , if k = 24 , the length of k is equal to 4 , since 24 = 2 × 2 × 2 × 3 . if x and y are positive integers such that x > 1 , y > 1 , and x + 3 y < 960 , what is the maximum possible sum of the length of x and the length of y ? | "we know that : x > 1 , y > 1 , and x + 3 y < 960 , and it is given that length means no of factors . for any value of x and y , the max no of factors can be obtained only if factor is smallest no all factors are equal . hence , lets start with smallest no 2 . 2 ^ 1 = 2 2 ^ 2 = 4 2 ^ 3 = 8 2 ^ 4 = 16 2 ^ 5 = 32 2 ^ 6 = 64 2 ^ 7 = 128 2 ^ 8 = 256 2 ^ 9 = 512 2 ^ 10 = 1024 ( it exceeds 960 , so , x ca n ' t be 2 ^ 10 ) so , max value that x can take is 2 ^ 9 , for which has length of integer is 9 . now , since x = 512 , x + 3 y < 960 so , 3 y < 448 = = > y < 448 / 3 so , y can take any value which is less than 448 / 3 . and to get the maximum no of factors of smallest integer , we can say y = 2 ^ 7 for 2 ^ 7 has length of integer is 7 . so , combined together : 9 + 7 = 16 . b" | a = 4 + 3
b = 4 + 4
c = b + 1
d = a + c
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a ) 9 kmph , b ) 6 kmph , c ) 5 kmph , d ) 8 kmph , e ) 7 kmph | a | divide(subtract(divide(90, 2), divide(45, 2)), const_2) | a man rows his boat 90 km downstream and 45 km upstream , taking 2 1 / 2 hours each time . find the speed of the stream ? | "speed downstream = d / t = 90 / ( 2 1 / 2 ) = 36 kmph speed upstream = d / t = 45 / ( 2 1 / 2 ) = 18 kmph the speed of the stream = ( 36 - 18 ) / 2 = 9 kmph answer : a" | a = 90 / 2
b = 45 / 2
c = a - b
d = c / 2
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a ) 7 , b ) 9 , c ) 15 , d ) 25 , e ) 63 | a | add(const_3, const_4) | what is the smallest positive integer k such that the product of 1575 x k is a perfect square ? | "perfect square , is just an integer that can be written as the square of some other integer . for example 16 = 4 ^ 2 , is a perfect square . now , 1575 = 3 ^ 2 * 5 ^ 2 * 7 , so if k = 7 then 1575 k = ( 3 * 5 * 7 ) ^ 2 , which is a perfect square ( basically the least positive value of k must complete only the power of 7 to even power as powers of other primes are already even ) . answer : a ." | a = 3 + 4
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a ) 100 , b ) 110 , c ) 120 , d ) 140 , e ) 150 | e | divide(multiply(600, subtract(const_100, 75)), const_100) | in a public show 75 % of the seats were filled . if there were 600 seats in the hall , how many seats were vacant ? | "75 % of 600 = 75 / 100 × 600 = 450 therefore , the number of vacant seats = 600 - 450 = 150 . answer : e" | a = 100 - 75
b = 600 * a
c = b / 100
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a ) 1200 , b ) 1764 , c ) 1680 , d ) 1354 , e ) none of these | c | add(divide(180, divide(multiply(divide(9, multiply(const_4, const_3)), 16), const_100)), 180) | the true discount on a bill due 9 months hence at 16 % per annum is rs . 180 . the amount of the bill is | "explanation : let p . w . be rs . x . then , s . i . on rs . x at 16 % for 9 months = rs . 180 . x ã — 16 ã — ( 9 / 12 ) ã — ( 1 / 100 ) = 180 or x = 1500 . p . w . = rs . 1500 . sum due = p . w . + t . d . = rs . ( 1500 180 ) = rs . 1680 . answer : c" | a = 4 * 3
b = 9 / a
c = b * 16
d = c / 100
e = 180 / d
f = e + 180
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a ) $ 380 , b ) $ 540 , c ) $ 280 , d ) $ 1380 , e ) $ 1430 | c | subtract(6580, multiply(divide(12600, 12), add(const_2, const_4))) | a project scheduled to be carried out over a single fiscal year has a budget of $ 12600 , divided into 12 equal monthly allocations . at the end of the sixth month of that fiscal year , the total amount actually spent on the project was $ 6580 . by how much was the project over its budget ? | difficulty level : 600 each month ' s budget = 12600 / 12 = 1050 budget for 6 months = 6 * 1050 = 6300 actual amount spent = 6580 amount spent over the budget = 6580 - 6300 = 280 answer ( c ) , regards , | a = 12600 / 12
b = 2 + 4
c = a * b
d = 6580 - c
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a ) 6 , b ) 12 , c ) 5 , d ) 8 , e ) 9 | b | divide(subtract(divide(85, 2), divide(25, 2)), const_2) | a man rows his boat 85 km downstream and 25 km upstream , taking 2 1 / 2 hours each time . find the speed of the stream ? | "speed downstream = d / t = 85 / ( 2 1 / 2 ) = 34 kmph speed upstream = d / t = 25 / ( 2 1 / 2 ) = 10 kmph the speed of the stream = ( 34 - 10 ) / 2 = 12 kmph answer : b" | a = 85 / 2
b = 25 / 2
c = a - b
d = c / 2
|
a ) 9 , b ) 14 , c ) 28 , d ) 40 , e ) 60 | d | divide(14, subtract(const_1, add(divide(1, 4), divide(2, 5)))) | a cement mixture is composed of 3 elements . by weight , 1 / 4 of the mixture is sand , 2 / 5 of the mixture is water , and the remaining 14 pounds of the mixture is gravel . what is the weight of the entire mixture in pounds ? | let the total weight be x . sand content = ( 1 / 4 ) x water content = ( 2 / 5 ) x gravel = x - ( 1 / 4 ) x - ( 2 / 5 ) x = ( 7 / 20 ) x = 14 x = 40 then answer will be d = 40 | a = 1 / 4
b = 2 / 5
c = a + b
d = 1 - c
e = 14 / d
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a ) 30 , b ) 15 , c ) 12 , d ) 23 , e ) 17 | a | divide(5880, multiply(power(add(const_3, const_4), const_2), power(const_2, const_2))) | if a is the smallest positive integer such that 5880 multiplied by a is the square of an integer , then ya must be | 5880 = 2 * 2 * 7 * 7 * 5 * 6 , so we need one 5 and one 6 to make it a square of a number . so 5 * 6 = 30 ans : a | a = 3 + 4
b = a ** 2
c = 2 ** 2
d = b * c
e = 5880 / d
|
a ) 50 , b ) 18 , c ) 16 , d ) 12 , e ) 8 | a | multiply(multiply(6, 5), divide(10, 6)) | working simultaneously and independently at an identical constant rate , 6 machines of a certain type can produce a total of x units of product p in 10 days . how many of these machines , working simultaneously and independently at this constant rate , can produce a total of 5 x units of product p in 6 days ? | "6 machines do x units in 10 days we have x / 10 = > rate of the 6 machines we know that we need to have 5 x units in 6 days therefore , we need to get to 5 x / 6 rate of the machines . rate of one machine is x / 10 * 1 / 6 = x / 60 . now , we need to know how many machines need to work simultaneously , to get 5 x done in 6 days . 5 x / 6 work needs to be done by machines that work at x / 60 rate . let ' s assign a constant y for the number of machines : ( x / 60 ) * y = 5 x / 6 y = 5 x / 6 * 60 / x cancel 6 with 60 , and x with x and get - > 50 . answer choice a" | a = 6 * 5
b = 10 / 6
c = a * b
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a ) 200 / 15 , b ) 200 / 17 , c ) 220 / 17 , d ) 240 / 17 , e ) 260 / 17 | b | divide(multiply(72, multiply(const_100, const_100)), power(12, const_2)) | if the difference between the compound interest & simple interest of a certain sum of money is rs . 72 at 12 % p . a . per two year . find the sum ( in rs . ) ? | 200 / 17 c . i = a - p a = p ( 1 + 12 / 100 ) ^ 2 s . i = p * 2 * 12 / 100 c . i - s . i = 72 given substitute and solve p = 200 / 17 answer : b | a = 100 * 100
b = 72 * a
c = 12 ** 2
d = b / c
|
a ) 1287 , b ) 1285 , c ) 1283 , d ) 1280 , e ) 1281 | d | divide(3125, power(add(divide(1, 4), 1), 4)) | each year for 4 years , a farmer increased the number of trees in a certain orchard by 1 / 4 of the number of trees in the orchard of the preceding year . if all of the trees thrived and there were 3125 trees in the orchard at the end of 4 year period , how many trees were in the orchard at the beginning of the 4 year period . | "trees increase by 1 / 4 the number of trees in preceding year . hence , correct answer must be divisible by 4 . based on divisibility rules , if last 2 digits are divisible by 4 then the number is divisible by 4 . thus , we can eliminate a , b , c , e the answer to be d again , trees increase by 1 / 4 the number of trees in preceding year . hence , the number of trees increase by 5 / 4 times the number of trees the preceding year . if x = initial number of trees = 1280 year 1 = 5 / 4 x year 2 = ( 5 / 4 ) ( 5 / 4 ) x year 3 = ( 5 / 4 ) ( 5 / 4 ) ( 5 / 4 ) x year 4 = ( 5 / 4 ) ( 5 / 4 ) ( 5 / 4 ) ( 5 / 4 ) x only for answer d : ( 5 / 4 ) ( 5 / 4 ) ( 5 / 4 ) ( 5 / 4 ) 1280 = 3125 hence , correct answer = d" | a = 1 / 4
b = a + 1
c = b ** 4
d = 3125 / c
|
a ) 6 only , b ) 6 and 12 , c ) 12 only , d ) 18 only , e ) 20 only | b | add(multiply(6, const_100), multiply(2, 6)) | if n is a natural number , then 6 n ^ 2 + 6 n is always divisible by ? | "6 n ^ 2 + 6 n = 6 n ( n + 1 ) , which is always divisible by 6 and 12 both , since n ( n + 1 ) is always even . answer is b" | a = 6 * 100
b = 2 * 6
c = a + b
|
a ) 22 sec , b ) 27 sec , c ) 25 sec , d ) 24 sec , e ) 11 sec | a | divide(add(120, 100), multiply(36, const_0_2778)) | how many seconds will a train 100 meters long take to cross a bridge 120 meters long if the speed of the train is 36 kmph ? | "explanation : d = 100 + 120 = 220 s = 36 * 5 / 18 = 10 mps t = 250 / 10 = 22 sec answer : option a" | a = 120 + 100
b = 36 * const_0_2778
c = a / b
|
a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | c | add(divide(subtract(6.30, multiply(0.70, const_2)), 0.60), const_2) | a certain fruit stand sold apples for $ 0.70 each and bananas for $ 0.60 each . if a customer purchased both apples and bananas from the stand for a total of $ 6.30 , what total number of apples and bananas did the customer purchase ? | "let ' s start with 1 apple for $ 0.70 . let ' s subtract $ 0.70 from $ 6.30 until we get a multiple of $ 0.60 . $ 6.30 , $ 5.60 , $ 4.90 , $ 4.20 = 7 * $ 0.60 the customer purchased 7 bananas and 3 apples . the answer is c ." | a = 0 * 70
b = 6 - 30
c = b / 0
d = c + 2
|
a ) 8.5 km / hr , b ) 10 km / hr , c ) 12.5 km / hr , d ) 9 km / hr , e ) none of these | b | subtract(subtract(15, 2.5), 2.5) | a man ' s speed with the current is 15 km / hr and the speed of the current is 2.5 km / hr . the man ' s speed against the current is : | "explanation : man ' s speed with the current = 15 km / hr = > speed of the man + speed of the current = 15 km / hr speed of the current is 2.5 km / hr hence , speed of the man = 15 - 2.5 = 12.5 km / hr man ' s speed against the current = speed of the man - speed of the current = 12.5 - 2.5 = 10 km / hr answer is b" | a = 15 - 2
b = a - 2
|
a ) $ 1000 , b ) $ 1020 , c ) $ 1040 , d ) $ 1060 , e ) $ 1080 | a | multiply(const_100.0, divide(const_100, add(1200, 20))) | a shopkeeper sold an article at $ 1200 and gained a 20 % profit . what was the cost price ? | "let x be the cost price . 1.2 x = 1200 x = 1200 / 1.2 = 1000 the answer is a ." | a = 1200 + 20
b = 100 / a
c = 100 * 0
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a ) 10 % , b ) 25 % , c ) 33 % , d ) 50 % , e ) 67 % | c | multiply(add(const_1, const_10), subtract(subtract(20, 15), const_1)) | at company x , senior sales representatives visit the home office once every 20 days , and junior sales representatives visit the home office once every 15 days . the number of visits that a junior sales representative makes in a 2 - year period is approximately what percent greater than the number of visits that a senior representative makes in the same period ? | "each 60 - day period , senior representatives visit the home office 3 times while junior representatives visit 4 times , thus 33 % more . the answer is c ." | a = 1 + 10
b = 20 - 15
c = b - 1
d = a * c
|
a ) a ) 18.7 , b ) b ) 20 , c ) c ) 21.3 , d ) d ) 21.5 , e ) e ) 22 | a | subtract(subtract(25, divide(30, const_100)), divide(30, 5)) | daniel went to a shop and bought things worth rs . 25 , out of which 30 paise went on sales tax on taxable purchases . if the tax rate was 5 % , then what was the cost of the tax free items ? | "total cost of the items he purchased = rs . 25 given that out of this rs . 25 , 30 paise is given as tax = > total tax incurred = 30 paise = rs . 30 / 100 let the cost of the tax free items = x given that tax rate = 5 % ∴ ( 25 − 30 / 100 − x ) 5 / 100 = 30 / 100 ⇒ 5 ( 25 − 0.3 − x ) = 30 ⇒ ( 25 − 0.3 − x ) = 6 ⇒ x = 25 − 0.3 − 6 = 18.7 a" | a = 30 / 100
b = 25 - a
c = 30 / 5
d = b - c
|
a ) 3.5 gallons , b ) 2.7 gallons , c ) 5.7 gallons , d ) 4.75 gallons , e ) 7.5 gallons | d | divide(190, 40) | a car gets 40 kilometers per gallon of gasoline . how many gallons of gasoline would the car need to travel 190 kilometers ? | "each 40 kilometers , 1 gallon is needed . we need to know how many 40 kilometers are there in 190 kilometers ? 190 ã · 40 = 4.75 ã — 1 gallon = 4.75 gallons correct answer is d ) 4.75 gallons" | a = 190 / 40
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a ) 750 , b ) 800 , c ) 900 , d ) 950 , e ) 1000 | e | add(300, 600) | in the faculty of reverse - engineering , 300 second year students study numeric methods , 600 second year students study automatic control of airborne vehicles and 200 second year students study them both . how many students are there in the faculty if the second year students are approximately 70 % of the total ? | "total number of students studying both are 300 + 600 - 200 = 700 ( subtracting the 200 since they were included in the both the other numbers already ) . so 70 % of total is 700 , so 100 % is 1000 answer is e" | a = 300 + 600
|
a ) 4.8 hr , b ) 2.4 hr , c ) 3.6 hr , d ) 1.8 hr , e ) 2.0 hr | a | divide(const_1, subtract(divide(const_1, 3), divide(const_1, 8))) | a cistern can be filled by a tap in 3 hours while it can be emptied by another tap in 8 hours . if both the taps are opened simultaneously , then after how much time will the cistern get filled ? | explanation : part filled by one tap in 1 hour = 1 / 4 part emptied by second tap 1 hour = 1 / 8 net part filled by both these taps in 1 hour = 1 / 3 - 1 / 8 = 5 / 24 i . e , the cistern gets filled in 24 / 5 hours = 4.8 hrs answer : option a | a = 1 / 3
b = 1 / 8
c = a - b
d = 1 / c
|
a ) 114 , b ) 119 , c ) 113 , d ) 88 , e ) 31 | a | subtract(119, subtract(add(144, 119), 233)) | in a graduating class of 233 students , 144 took geometry and 119 took biology . what is the difference between the greatest possible number and the smallest possible number of students that could have taken both geometry and biology ? | "greatest possible number taken both should be 144 ( as it is maximum for one ) smallest possible number taken both should be given by total - neither = a + b - both both = a + b + neither - total ( neither must be 0 to minimize the both ) so 144 + 119 + 0 - 233 = 30 greatest - smallest is 144 - 30 = 114 so answer must be a . 114" | a = 144 + 119
b = a - 233
c = 119 - b
|
a ) 174.2 , b ) 212 , c ) 288.1 , d ) 296 , e ) 302 | e | multiply(divide(48, 2.54), divide(24, 1.5)) | on a map , 1.5 inches represent 24 miles . how many miles approximately is the distance if you measured 48 centimeters assuming that 1 - inch is 2.54 centimeters ? | "1.5 inch = 2.54 * 1.5 cm . so , 2.54 * 1.5 represents 24 miles . so for 48 cm . : 48 / ( 2.54 * 1.5 ) = x / 24 - - - > x = 24 * 47 / ( 3.81 ) = 302 answer will be e ." | a = 48 / 2
b = 24 / 1
c = a * b
|
a ) 2 , b ) - 3 , c ) 4 , d ) - 5 , e ) 6 | b | divide(subtract(42, 6), 4) | if | 4 x + 6 | = 42 , what is the sum of all the possible values of x ? | "there will be two cases 4 x + 6 = 42 or 4 x + 6 = - 42 = > x = 9 or x = - 12 sum of both the values will be - 12 + 9 = - 3 answer is b" | a = 42 - 6
b = a / 4
|
a ) 2 / 3 , b ) 3 / 7 , c ) 4 / 9 , d ) 1 / 6 , e ) 4 / 7 | d | divide(multiply(divide(1, 4), 8), 12) | a pipe can empty 1 / 4 th of a cistern in 12 mins . in 8 mins , what part of the cistern will be empty ? | "1 / 4 - - - - 12 ? - - - - - 8 = = > 1 / 6 d" | a = 1 / 4
b = a * 8
c = b / 12
|
a ) 25 , b ) 19 , c ) 39 , d ) 61 , e ) 64 | e | multiply(multiply(4, divide(16, 4)), divide(16, 4)) | 4 mat - weavers can weave 4 mats in 4 days . at the same rate , how many mats would be woven by 16 mat - weavers in 16 days ? | "let the required number of bottles be x . more weavers , more mats ( direct proportion ) more days , more mats ( direct proportion ) wavers 4 : 16 : : 4 : x days 4 : 16 4 * 4 * x = 16 * 16 * 4 x = ( 16 * 16 * 4 ) / ( 4 x 4 ) x = 64 . answer is e ." | a = 16 / 4
b = 4 * a
c = 16 / 4
d = b * c
|
a ) 5 min , b ) 6 min , c ) 7 min , d ) 8 min , e ) 9 min | d | divide(3, divide(3, 8)) | suppose 8 monkeys take 8 minutes to eat 8 banana ' s . how many minutes would it take 3 monkeys to eat 3 banana ' s ? | there are equal no . of monkeys and equal no . of bananas and they take equal time and the time is 8 mint to eat a banana so each monkey take 8 mints to eat a banana ' s so 3 monkeys will take 8 mints to eat 3 banana ' s . answer : d | a = 3 / 8
b = 3 / a
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a ) 1 , b ) - 1 , c ) 3 , d ) 5 , e ) 13 | c | multiply(3, const_2) | if x + | x | + y = 4 and x + | y | - y = 3 what is x + y = ? | "if x < 0 and y < 0 , then we ' ll have x - x + y = 7 and x - y - y = 6 . from the first equation y = 7 , so we can discard this case since y is not less than 0 . if x > = 0 and y < 0 , then we ' ll have x + x + y = 7 and x - y - y = 6 . solving gives x = 4 > 0 and y = - 1 < 0 - - > x + y = 3 . since in ps questions only one answer choice can be correct , then the answer is c ( so , we can stop here and not even consider other two cases ) . answer : c . adding both eqn we get 2 x + ixi + iyi = 13 now considering x < 0 and y > 0 2 x - x + y = 13 we get x + y = 3 hence answer should be c" | a = 3 * 2
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a ) rs . 480 , b ) rs . 520 , c ) rs . 420 , d ) rs . 460 , e ) none of these | c | multiply(divide(add(42, 42), subtract(42, 35)), 35) | there were 35 students in a hostel . if the number of students increases by 7 , the expenses of the mess increase by rs . 42 per day while the average expenditure per head diminishes by re 1 . find the original expenditure of the mess . | let d be the average daily expenditure original expenditure = 35 × d new expenditure = 35 × d + 42 new average expenditure will be : ( 35 × d + 42 ) / 42 = d - 1 on solving , we get d = 12 therefore original expenditure = 35 × 12 = 420 answer : c | a = 42 + 42
b = 42 - 35
c = a / b
d = c * 35
|
a ) 20 , b ) 24 , c ) 28 , d ) 32 , e ) 36 | d | multiply(1, 8) | two numbers are in the ratio of 1 : 2 . if 8 be added to both , their ratio changes to 3 : 5 . the greater number is | "let the ratio be x : y , given x / y = 1 / 2 , ( x + 8 ) / ( y + 8 ) = 3 / 5 = > x = 16 and y = 32 answer : d" | a = 1 * 8
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a ) $ 125 , b ) $ 106 , c ) $ 135 , d ) $ 90 , e ) $ 100 | b | multiply(subtract(1000, 200), multiply(subtract(const_1, divide(const_1, 10)), divide(const_1, 10))) | if xerox paper costs 5 cents a sheet and a buyer gets 10 % discount on all xerox paper one buys after the first 200 papers and 20 % discount after first 1000 papers , how much will it cost to buy 2500 sheets of xerox paper ? | "30 sec approach - solve it using approximation 2500 sheet at full price , 5 cent = 125 2500 sheet at max discount price , 4 cent = 100 your ans got to be between these two . ans b it is ." | a = 1000 - 200
b = 1 / 10
c = 1 - b
d = 1 / 10
e = c * d
f = a * e
|
a ) 6 days , b ) 2 days , c ) 8 days , d ) 4 1 / 5 days , e ) 9 days | d | subtract(add(inverse(add(inverse(15), inverse(10))), 10), const_3) | a and b can do a work in 10 days and 15 days respectively . a starts the work and b joins him after 3 days . in how many days can they complete the remaining work ? | "work done by a in 3 days = 3 / 10 remaining work = 7 / 10 work done by both a and b in one day = 1 / 10 + 1 / 15 = 5 / 30 = 1 / 6 remaining work = 7 / 10 * 6 / 1 = 21 / 5 = 4 1 / 5 days . answer : d" | a = 1/(15)
b = 1/(10)
c = a + b
d = 1/(c)
e = d + 10
f = e - 3
|
a ) 4,629 , b ) 4,639 , c ) 3,946 , d ) 4,926 , e ) 4,936 | e | subtract(758,805, add(add(multiply(const_2, const_100), multiply(add(const_3, const_4), const_10)), const_2)) | how many integers between 265,205 and 758,805 have tens digit 1 and units digit 3 ? | "there is one number in hundred with 1 in the tens digit and 3 in the units digit : 13 , 113 , 213 , 313 , . . . the difference between 265,205 and 758,805 is 758,805 - 265,205 = 493,600 - one number per each hundred gives 133,900 / 100 = 4,936 numbers . answer : e ." | a = 2 * 100
b = 3 + 4
c = b * 10
d = a + c
e = d + 2
f = 758 - 805
|
a ) a ) 12.5 % , b ) b ) 13 % , c ) c ) 13.15 % , d ) d ) 14 % , e ) e ) 15 % | a | divide(multiply(subtract(multiply(const_100, const_100), multiply(add(const_100, 25), 60)), const_100), multiply(const_100, const_100)) | the shopkeeper increased the price of a product by 25 % so that customer finds it difficult to purchase the required amount . but somehow the customer managed to purchase only 60 % of the required amount . what is the net difference in the expenditure on that product ? | "quantity x rate = price 1 x 1 = 1 0.6 x 1.25 = 0.755 decrease in price = ( 0.125 / 1 ) × 100 = 12.5 % a )" | a = 100 * 100
b = 100 + 25
c = b * 60
d = a - c
e = d * 100
f = 100 * 100
g = e / f
|
a ) 64 % , b ) 66 % , c ) 60 % , d ) 68 % , e ) 70 % | d | multiply(divide(subtract(160, add(multiply(3, 8), multiply(8, 3))), 160), const_100) | a batsman scored 160 runs which included 3 boundaries and 8 sixes . what percent of his total score did he make by running between the wickets . | "explanation : number of runs made by running = 160 - ( 3 x 4 + 8 x 6 ) = 160 - ( 60 ) = 100 now , we need to calculate 100 is what percent of 160 . = > 100 / 160 * 100 = 68 option d" | a = 3 * 8
b = 8 * 3
c = a + b
d = 160 - c
e = d / 160
f = e * 100
|
a ) 120 m , b ) 180 m , c ) 240 m , d ) 220 m , e ) 280 m | a | divide(6, subtract(divide(6, 10), 12)) | a train covers a distance of 6 km in 10 min . if it takes 12 sec to pass a telegraph post , then the length of the train is ? | "speed = ( 6 / 10 * 60 ) km / hr = ( 36 * 5 / 18 ) m / sec = 10 m / sec . length of the train = 10 * 12 = 120 m . answer : a" | a = 6 / 10
b = a - 12
c = 6 / b
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a ) 4 % , b ) 10 % , c ) 96 % , d ) 90 % , e ) 65 % | e | multiply(divide(subtract(divide(subtract(const_100, 60), const_100), subtract(divide(50, const_100), multiply(divide(60, const_100), divide(60, const_100)))), divide(subtract(const_100, 60), const_100)), const_100) | 60 % of the employees of a company are men . 60 % of the men in the company speak french and 50 % of the employees of the company speak french . what is % of the women in the company who do not speak french ? | "no of employees = 100 ( say ) men = 60 women = 40 men speaking french = 0.6 * 60 = 36 employees speaking french = 0.5 * 100 = 50 therefore women speaking french = 50 - 36 = 14 and women not speaking french = 40 - 14 = 26 % of women not speaking french = 26 / 40 * 100 = 65 % answer e" | a = 100 - 60
b = a / 100
c = 50 / 100
d = 60 / 100
e = 60 / 100
f = d * e
g = c - f
h = b - g
i = 100 - 60
j = i / 100
k = h / j
l = k * 100
|
a ) 6 , b ) 8 , c ) 11 , d ) 12 , e ) 13 | b | multiply(divide(16, add(const_2, const_2)), const_2) | in a group of donkeys and pigs , the numbers of legs are 16 more than twice the number of heads . the number of donkeys is | let the number of donkeys be x and the number of pigs be y . then , 4 x + 2 y = 2 ( x + y ) = 16 or 2 x + ( 2 x + 2 y ) = ( 2 x + 2 y ) + 16 or 2 x = 16 or x = 8 . answer : b | a = 2 + 2
b = 16 / a
c = b * 2
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a ) $ 40 , b ) $ 60 , c ) $ 160 , d ) $ 90 , e ) $ 100 | c | divide(multiply(subtract(divide(240, subtract(const_1, divide(40, const_100))), 240), const_100), 240) | a furniture dealer purchased a desk for $ 240 and then set the selling price equal to the purchase price plus a markup that was 40 % of the selling price . if the dealer sold the desk at the selling price , what was the amount of the dealer ' s gross profit from the purchase and the sale of the desk ? | "purchase price = 240 selling price = x 240 + 0.4 * x = x 0.6 * x = 240 x = 400 profit = 400 - 240 = 160 answer : c" | a = 40 / 100
b = 1 - a
c = 240 / b
d = c - 240
e = d * 100
f = e / 240
|
a ) rs . 2250 , b ) rs . 3375 , c ) rs . 6750 , d ) rs . 5625 , e ) none of these | b | multiply(12375, inverse(add(add(divide(2, 3), multiply(divide(2, 3), 3)), const_1))) | a , b and c enter into a partnership . a invests 3 times as much as b invests and 2 / 3 of what c invests . at the end of the year , the profit earned is rs . 12375 . what is the share of c ? | "explanation : let the investment of c be rs . x . the inverstment of b = rs . ( 2 x / 3 ) the inverstment of a = rs . ( 3 × ( 2 / 3 ) x ) = rs . ( 2 x ) ratio of capitals of a , b and c = 2 x : 2 x / 3 : x = 6 : 2 : 3 c ' s share = rs . [ ( 3 / 11 ) × 12375 ] = rs . 3375 answer : option b" | a = 2 / 3
b = 2 / 3
c = b * 3
d = a + c
e = d + 1
f = 1/(e)
g = 12375 * f
|
a ) 24 % , b ) 20 % , c ) 26 % , d ) 28 % , e ) 35 % | b | divide(multiply(subtract(add(multiply(divide(multiply(280, 40), const_100), divide(add(const_100, 20), const_100)), multiply(divide(multiply(280, 60), const_100), divide(add(const_100, 20), const_100))), 280), const_100), 280) | a shopkeeper has 280 kg of apples . he sells 40 % of these at 20 % profit and remaining 60 % at 20 % profit . find his % profit on total . | "if the total quantity was 100 then 40 x 20 % + 60 x 20 % = 20 this profit will remain same for any total quantity unless the % of products remains the same . hence ' b ' is the answer" | a = 280 * 40
b = a / 100
c = 100 + 20
d = c / 100
e = b * d
f = 280 * 60
g = f / 100
h = 100 + 20
i = h / 100
j = g * i
k = e + j
l = k - 280
m = l * 100
n = m / 280
|
a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | d | divide(factorial(subtract(add(const_4, 30), const_1)), multiply(factorial(30), factorial(subtract(const_4, const_1)))) | how many positive integers less than 30 have no common prime factor with 30 ? | "30 = 2 * 3 * 5 . so , the number must be less than 30 and not have primes 2 , 3 , or 5 . this means that the number could be : 1 , 7 , 11 , 13 , 17 , 19 , 23 , or 29 . total of 8 numbers . answer : d ." | a = 4 + 30
b = a - 1
c = math.factorial(b)
d = math.factorial(30)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
|
a ) 288 , b ) 350 , c ) 889 , d ) 276 , e ) 333 | e | subtract(multiply(speed(300, 18), 38), 300) | a 300 m long train crosses a platform in 38 sec while it crosses a signal pole in 18 sec . what is the length of the platform ? | "speed = 300 / 18 = 50 / 3 m / sec . let the length of the platform be x meters . then , ( x + 300 ) / 38 = 50 / 3 3 x + 900 = 1900 = > x = 333 m . answer : e" | a = speed * (
b = a - 38
|
a ) 18.4 sec , b ) 77 sec , c ) 76 sec , d ) 20 sec , e ) 66 sec | a | divide(add(120, 280), multiply(add(42, 36), const_0_2778)) | two trains of length 120 m and 280 m are running towards each other on parallel lines at 42 kmph and 36 kmph respectively . in what time will they be clear of each other from the moment they meet ? | "relative speed = ( 42 + 36 ) * 5 / 18 = 21.7 mps . distance covered in passing each other = 120 + 280 = 400 m . the time required = d / s = 400 / 21.7 = 18.4 sec . answer : a" | a = 120 + 280
b = 42 + 36
c = b * const_0_2778
d = a / c
|
a ) 2 : 5 , b ) 5 : 15 , c ) 3 : 5 , d ) 4 : 7 , e ) 14 : 5 | b | divide(add(divide(6, const_3), divide(35, multiply(2, divide(6, const_3)))), multiply(add(divide(6, const_3), divide(35, multiply(2, divide(6, const_3)))), const_3)) | tough and tricky questions : mixture problems . a drink holding 6 ounces of an alcoholic drink that is 1 part rum to 2 parts coke is added to a jug holding 35 ounces of an alcoholic drink that is 1 part rum to 3 parts coke . what is the ratio of rum to coke in the resulting mixture ? | answer = b ) 5 : 15 . . . . . . . . . . . . . . . . . . . . . . . . rum . . . . . . . . . . . . . . . . . . . . . coke . . . . . . . . . . . . . . . . . total first drink . . . . . . . . . . . 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 . . . . . . . . . . . . . . . . . . . . . 6 ( ratio 1 : 2 ) second drink . . . . . . . 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 . . . . . . . . . . . . . . . . . . . 32 ( ratio 1 : 3 ) total . . . . . . . . . . . . . . . . . . 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 . . . . . . . . . . . . . . . . . . . . 38 ratio = 10 : 30 = 5 : 15 b | a = 6 / 3
b = 6 / 3
c = 2 * b
d = 35 / c
e = a + d
f = 6 / 3
g = 6 / 3
h = 2 * g
i = 35 / h
j = f + i
k = j * 3
l = e / k
|
a ) 118 min , b ) 10 min , c ) 18 min , d ) 16 min , e ) 25 min | e | multiply(const_60, divide(subtract(70, 40), 70)) | excluding stoppages , the speed of a bus is 70 km / hr and including stoppages , it is 40 km / hr . for how many minutes does the bus stop per hour ? | "due to stoppages , it covers 30 km less . time taken to cover 30 km = 30 / 70 * 60 = 25 min . answer : e" | a = 70 - 40
b = a / 70
c = const_60 * b
|
a ) 4 , b ) 7 , c ) 8 , d ) 16 , e ) 20 | c | divide(subtract(multiply(49, 49), multiply(49, 49)), 7) | 49 x 49 x 49 x 49 = 7 ? | "solution 49 * 49 * 49 * 49 = ( 72 * 72 * 72 * 72 ) = 7 ( 2 + 2 + 2 + 2 ) = 78 so , the correct answer is 8 answer c" | a = 49 * 49
b = 49 * 49
c = a - b
d = c / 7
|
a ) 179 , b ) 119 , c ) 129 , d ) 173 , e ) 156 | a | multiply(12351, 69) | calculate 12351 ÷ ? = 69 | "answer let 12351 ÷ ? = 69 then x = 12351 / 69 = 179 . option : a" | a = 12351 * 69
|
a ) 23 , b ) 25 , c ) 26 , d ) 28 , e ) 30 | c | multiply(divide(multiply(const_0_25, 24), divide(9, 3)), 13) | if 9 engines consume 24 metric tonnes of coal , when each is working 8 hoursa day , how much coal will be required for 8 engines , each running 13 hours a day , it is being given that 3 engines of former type consume as much as 4 engines of latter type ? | explanation : let 3 engines of former type consume 1 unit in 1 hour . then , 4 engines of latter type consume 1 unit in 1 hour . therefore 1 engine of former type consumes ( 1 / 3 ) unit in 1 hour . 1 engine of latter type consumes ( 1 / 4 ) unit in 1 hour . let the required consumption of coal be x units . less engines , less coal consumed ( direct proportion ) more working hours , more coal consumed ( direct proportion ) less rate of consumption , less coal consumed ( direct proportion ) number of engines 9 : 8 working hours 8 : 13 } : : 24 : x rate of consumption ( 1 / 3 ) : ( 1 / 4 ) ( 9 * 8 * ( 1 / 3 ) * x ) = ( 8 * 13 * ( 1 / 4 ) * 24 ) < = > 24 x = 624 < = > x = 26 . answer : c | a = const_0_25 * 24
b = 9 / 3
c = a / b
d = c * 13
|
a ) 16 , b ) 18 , c ) 20 , d ) 21 , e ) 24 | c | divide(multiply(divide(const_1000, const_2), const_1000), multiply(divide(divide(multiply(divide(const_1000, const_2), const_1000), 40), 15), 30)) | it takes 40 identical printing presses 15 hours to print 500,000 papers . how many hours would it take 30 of these printing presses to print 500,000 papers ? | 40 printing presses can do 1 / 15 of the job each hour . 30 printing presses can do 3 / 4 * 1 / 15 = 1 / 20 of the job each hour . the answer is c . | a = 1000 / 2
b = a * 1000
c = 1000 / 2
d = c * 1000
e = d / 40
f = e / 15
g = f * 30
h = b / g
|
a ) 60 feet , b ) 50 feet , c ) 40 feet , d ) 30 feet , e ) 5 feet | a | add(multiply(divide(400, 20), const_2), 20) | a rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered . if the area of the field is 400 sq . ft , how many feet of fencing will be required ? | "explanation : we are given with length and area , so we can find the breadth . as length * breadth = area = > 20 * breadth = 400 = > breadth = 20 feet area to be fenced = 2 b + l = 2 * 20 + 20 = 60 feet answer : option a" | a = 400 / 20
b = a * 2
c = b + 20
|
a ) 90 , b ) 180 , c ) 250 , d ) 500 , e ) 1,000 | c | divide(divide(100, const_2), divide(20, const_100)) | one night 20 percent of the female officers on a police force were on duty . if 100 officers were on duty that night and half of these were female officers , how many female officers were on the police force ? | "let total number of female officers in the police force = f total number of officers on duty on that night = 100 number of female officers on duty on that night = 100 / 2 = 50 ( 20 / 100 ) * f = 50 = > f = 250 answer c" | a = 100 / 2
b = 20 / 100
c = a / b
|
a ) 12 , b ) 18 , c ) 6 , d ) 1 , e ) 17 | a | divide(multiply(multiply(multiply(100, 25), 30), 12), multiply(multiply(75, 20), 50)) | if a certain number of people can dig earth 100 m deep , 25 m long , and 30 m broad in 12 days , then find the number of days the same number of people will require to dig earth 75 m deep , 20 m long and 50 m broad . | explanation : more number of days means – more length , more depth and more width . hence , it ’ s a direct proportion . ( 100 * 25 * 30 ) : ( 75 * 20 * 50 ) : : 12 : x 75000 : 75000 : : 12 : x x = 12 answer a | a = 100 * 25
b = a * 30
c = b * 12
d = 75 * 20
e = d * 50
f = c / e
|
a ) 10 glas , b ) 8.5 gals , c ) 8 gals , d ) 6.66 gals , e ) 8.33 gals | e | divide(subtract(multiply(divide(multiply(divide(20, const_100), 125), const_100), 125), 25), subtract(const_1, divide(multiply(divide(20, const_100), 125), const_100))) | a mixture of 125 gallons of wine and water contains 20 % water . how much water must be added to the mixture in order to increase the percentage of water to 25 % of the new mixture ? | "in 125 gallons of the solution there are 0.2 ∗ 125 = 250 gallons of water . we want to add ww gallons of water to 125 gallons of solution so that 25 + w gallons of water to be 25 % of new solution : 25 + w = 0.25 ( 125 + w ) - - > w = 25 / 3 ≈ 8.33 answer : e ." | a = 20 / 100
b = a * 125
c = b / 100
d = c * 125
e = d - 25
f = 20 / 100
g = f * 125
h = g / 100
i = 1 - h
j = e / i
|
a ) a ) 10 , b ) b ) 24 , c ) c ) 40 , d ) d ) 55 , e ) e ) 70 | b | subtract(multiply(16, 61.5), multiply(subtract(16, const_1), 64.0)) | the average ( arithmetic mean ) of 16 students first quiz scores in a difficult english class is 61.5 . when one student dropped the class , the average of the remaining scores increased to 64.0 . what is the quiz score of the student who dropped the class ? | "total score of 16 students is 16 * 61.50 = 984 total score of 15 students is 15 * 64 = 960 so , the score of the person who left is 24 ( 984 - 960 ) answer will be ( b )" | a = 16 * 61
b = 16 - 1
c = b * 64
d = a - c
|
['a ) 1300', 'b ) 1500', 'c ) 1600', 'd ) 1700', 'e ) 1800'] | e | divide(divide(divide(divide(divide(volume_rectangular_prism(200, 90, 40), const_3), const_2), const_4), 4), const_4) | the ratio , by volume , of soap to alcohol to water in a certain solution is 4 : 40 : 90 . the solution will be altered so that the ratio of soap to alcohol is doubled while the ratio of soap to water is halved . if the altered solution will contain 200 cubic centimeters of alcohol , how many cubic centimeters of water will it contain ? | soap : alcohol initial ratio soap : alcohol : water - - > 4 : 40 : 90 initial soap : alcohol = 4 / 40 = 4 : 40 after doubled soap : alcohol = 2 * 4 / 40 = 8 : 40 initial soap : water = 4 / 90 = 4 : 90 after halved soap : water : 1 / 2 * 4 / 90 = 2 / 90 = 2 : 90 after soap : alcohol : water - - > 8 : 40 : 360 - - > 1 : 5 : 45 given alcohol 200 cubic centimeter . ratio is 40 : 200 : 1800 ( 1 : 5 : 45 ) for 200 cubic centimeter of alcohol - - - 1800 cubic cm water is required . answer - e | a = volume_rectangular_prism / (
b = a / 3
c = b / 2
d = c / 4
e = d / 4
|
a ) 170 , b ) 180 , c ) 190 , d ) 200 , e ) 210 | d | divide(multiply(2400, 100), 1200) | there is enough provisions for 2400 women in an army camp for 100 days . if there were 1200 men less , how long will the provision last ? | "we have , m 1 d 1 = m 2 d 2 2400 * 100 = 1200 * d 2 d 2 = 2400 * 100 / 1200 = 200 days . answer : d" | a = 2400 * 100
b = a / 1200
|
a ) 3.3 pm , b ) 5.5 pm , c ) 4.9 pm , d ) 5 pm , e ) 5.2 pm | d | add(3, divide(divide(60, const_2), subtract(75, 60))) | a thief steals a car at 2.30 pm and drives it at 60 kmph . the theft is discovered at 3 pm and the owner sets off in another car at 75 kmph when will he overtake the thief | let the thief is overtaken x hrs after 2.30 pm distance covered by the thief in x hrs = distance covered by the owner in x - 1 / 2 hr 60 x = 75 ( x - ½ ) x = 5 / 2 hr thief is overtaken at 2.30 pm + 2 ½ hr = 5 pm answer is d . | a = 60 / 2
b = 75 - 60
c = a / b
d = 3 + c
|
a ) 1 hour , b ) 1 hour 10 minutes , c ) 2 hours 30 minutes , d ) 1 hour 40 minutes , e ) 2 hours 40 minutes | e | add(multiply(sqrt(divide(100, 90)), 90), 100) | two cars a and b start from boston and new york respectively simultaneously and travel towards each other at constant speeds along the same route . after meeting at a point between boston and new york the two cars a and b proceed to their respective destinations of new york and boston . car a reaches new york 100 minutes after the two cars have met and car b reaches boston 90 minutes after they have met . how long did car a take to cover the distance between boston and new york ? | "both cars leave at the same time both cars travel at constant speed stealing a useful piece of information from paragkan : if two objects a and b start from opposite points and , after having met en route , reach their respective destinations in a and b mins ( or any other measure of time ) respectively , then the ratio of their speeds ratio of speed : ( a / b ) = sq . rt ( b / a ) sq . rt ( b / a ) sq . rt ( 90 / 100 ) sq . rt ( 3 / 2 ) so , for every three units of distance a travels , b travels two . because we know the ratio of speed and the time it took b to travel the distance a has n ' t yet covered , we can find the time it took a to cover the distance b did in 90 minutes . 90 * ( 2 / 3 ) where 2 / 3 represents the lesser amount of time it took a to travel the distance b did in 90 minutes . = 60 minutes . therefore , a took 100 minutes to travel the first portion then 60 minutes to travel the distance b did in 90 minutes . a spent ( 100 + 60 ) = 160 minutes on the road . e . 2 hours 40 minutes" | a = 100 / 90
b = math.sqrt(a)
c = b * 90
d = c + 100
|
a ) 50 km , b ) 56 km , c ) 60 km , d ) 40 km , e ) 80 km | d | multiply(10, divide(20, subtract(15, 10))) | if a person walks at 15 km / hr instead of 10 km / hr , he would have walked 20 km more . the actual distance traveled by him is : | "let the actual distance travelled be x km . x / 10 = ( x + 20 ) / 15 15 x = 10 x + 200 5 x = 200 x = 40 km . answer : d" | a = 15 - 10
b = 20 / a
c = 10 * b
|
a ) $ 22 , b ) $ 24 , c ) $ 30 , d ) $ 36 , e ) $ 40 | d | multiply(divide(multiply(1.5, 8), subtract(8, 6)), 6) | nina has exactly enough money to purchase 6 widgets . if the cost of each widget were reduced by $ 1.5 , then nina would have exactly enough money to purchase 8 widgets . how much money does nina have ? | "d its is . let price = x ( x - 1.5 ) 8 = 6 x x = 6 hence total money = 6 * 6 = 36" | a = 1 * 5
b = 8 - 6
c = a / b
d = c * 6
|
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | d | subtract(10, const_4) | there are 40 doors marked with numbers 1 to 40 . there are 40 individuals marked 1 to 40 . an operation on a door is defined as changing the status of the door from open to closed or vice versa . all the doors are closed to start with . one at a time , one randomly chosen individual goes and operates the doors . the individual however operates only those doors which are a multiple of the number he / she is carrying . for example , the individual marked with number 5 operates the doors marked with 5 , 10 , 15 , 20 , 25 , 30 , 35 , and 40 . if every individual in the group gets one turn , then how many doors are open at the end ? | if a door is closed at the start , it requires an odd number of people to operate to be open at the end . only the perfect squares have an odd number of factors . the doors which are open at the end are : 1 , 4 , 9 , 16 , 25 , 36 for a total of 6 doors . the answer is d . | a = 10 - 4
|
a ) 1.2 , b ) 0.2 , c ) 1.8 , d ) 2.2 , e ) 4.0 | a | subtract(7.2, subtract(floor(7.2), const_1)) | for any number y , y * is defined as the greatest positive even integer less than or equal to y . what is the value of 7.2 – 7.2 * ? | since y * is defined as the greatest positive even integer less than or equal to y , then 7.2 * = 6 ( the greatest positive even integer less than or equal to 7.2 is 6 ) . hence , 7.2 – 7.2 * = 7.2 - 6 = 1.2 answer : a . | a = math.floor(7, 2)
b = a - 1
c = 7 - 2
|
a ) 17 : 5 , b ) 17 : 3 , c ) 17 : 6 , d ) 17 : 2 , e ) 17 : 8 | d | divide(85000, 10000) | p and q started a business investing rs 85000 and rs 10000 resp . in what ratio the profit earned after 2 years be divided between p and q respectively . | "explanation : in this type of question as time frame for both investors is equal then just get the ratio of their investments . p : q = 85000 : 10000 = 85 : 10 = 17 : 2 option d" | a = 85000 / 10000
|
a ) 1 / 126 , b ) 1 / 120 , c ) 1 / 24 , d ) 4 / 25 , e ) 1 / 2 | a | divide(const_2, choose(add(5, 5), 5)) | a bag contains 5 white marbles and 5 black marbles . if each of 5 girls and 5 boys randomly selects and keeps a marble , what is the probability that all of the girls select the same colored marble ? | "consider a case when all girls select black marbles . . . so girl 1 will chose 1 black marble out of 10 . . and black marble will be chosen from 5 black marbles girl 1 will have a probability of picking black marble 5 / 10 girl 2 will have to pick a black marble out of 4 and total remaining no . of marbles 9 = 4 / 9 so we have probability of all girls selecting black marbles as 5 / 10 * 4 / 9 * 3 / 8 * 2 / 7 * 1 / 6 = 1 / 252 since girls can also select white marbles so we will have 2 cases so probability of girls selecting all same colour marbles is 2 * 1 / 252 or 1 / 126 ans is a" | a = 5 + 5
b = math.comb(a, 5)
c = 2 / b
|
a ) 4 , b ) 8 , c ) 12 , d ) 15 , e ) 20 | c | subtract(subtract(multiply(5, 5), add(5, const_4)), power(const_2, const_2)) | if the radius of a circle that centers at the origin is 5 , how many m points on the circle have integer coordinates ? | "i understand this might not be required but i used the equation of a circle . since the origin is at 0 , x ^ 2 + y ^ 2 = 5 ^ 2 . x , y could be + / - ( 0,5 or 5,0 ) - 4 possibilities . x , y could be + / - ( 3,4 or 4,3 ) - 8 possibilities . ans : m = c" | a = 5 * 5
b = 5 + 4
c = a - b
d = 2 ** 2
e = c - d
|
a ) 35 ° , b ) 65 ° , c ) 45 ° , d ) 75 ° , e ) 95 ° | d | divide(multiply(subtract(multiply(divide(multiply(const_3, const_4), subtract(multiply(const_3, const_4), const_1)), multiply(add(const_4, const_1), subtract(multiply(const_3, const_4), const_1))), divide(const_60, const_2)), subtract(multiply(const_3, const_4), const_1)), const_2) | the angle between the minute hand and the hour hand of a clock when the time is 8.30 , is | "angle between hands of a clock when the minute hand is behind the hour hand , the angle between the two hands at m minutes past h ' o clock = 30 ( h − m / 5 ) + m / 2 degree when the minute hand is ahead of the hour hand , the angle between the two hands at m minutes past h ' o clock = 30 ( m / 5 − h ) − m / 2 degree here h = 8 , m = 30 and minute hand is behind the hour hand . hence the angle = 30 ( h − m / 5 ) + m / 2 = 30 ( 8 − 30 / 5 ) + 30 / 2 = 30 ( 8 − 6 ) + 15 = 30 × 2 + 15 = 75 ° answer is d ." | a = 3 * 4
b = 3 * 4
c = b - 1
d = a / c
e = 4 + 1
f = 3 * 4
g = f - 1
h = e * g
i = d * h
j = const_60 / 2
k = i - j
l = 3 * 4
m = l - 1
n = k * m
o = n / 2
|
a ) 1 : 3 , b ) 2 : 3 , c ) 3 : 2 , d ) 3 : 1 , e ) 3 : 4 | b | divide(multiply(multiply(multiply(multiply(8, const_3), multiply(10, const_2)), const_100), 8), multiply(12000, 10)) | ravi and sunil are partners in a business . ravi invests rs . 10,000 for 8 months and sunil invested rs . 12000 for 10 months then after one year ratio of their profits will be | "= ( 15000 * 8 ) : ( 8000 * 10 ) = 80000 : 120000 = 2 : 3 answer : b" | a = 8 * 3
b = 10 * 2
c = a * b
d = c * 100
e = d * 8
f = 12000 * 10
g = e / f
|
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