options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 727 | problem stringlengths 5 967 | rationale stringlengths 1 2.74k | program stringlengths 10 646 |
|---|---|---|---|---|---|
a ) 22 % , b ) 25 % , c ) 28 % , d ) 31 % , e ) 34 % | b | multiply(const_100, divide(add(divide(45, const_100), multiply(4, divide(20, const_100))), add(const_1, 4))) | because he ’ s taxed by his home planet , mork pays a tax rate of 45 % on his income , while mindy pays a rate of only 20 % on hers . if mindy earned 4 times as much as mork did , what was their combined tax rate ? | "let x be mork ' s income , then mindy ' s income is 4 x . the total tax paid is 0.45 x + 0.8 x = 1.25 x 1.25 x / 5 x = 0.25 the answer is b ." | a = 45 / 100
b = 20 / 100
c = 4 * b
d = a + c
e = 1 + 4
f = d / e
g = 100 * f
|
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | a | subtract(2, const_1) | let p be a prime number greater than 2 , and let n = 32 p . how many odd divisors does n have ? | the answer is a , there is exactly one . since is the largest ( and only ) even prime number , any p larger than 2 is odd . then since 32 = 2 * 2 * 2 * 2 , all of which are even , this means that p must be the only odd prime factor . | a = 2 - 1
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a ) 39 , b ) 28 , c ) 27 , d ) 40 , e ) 71 | d | divide(divide(divide(344, divide(add(12, 5), const_2)), 5), const_2) | the cross - section of a cannel is a trapezium in shape . if the cannel is 12 m wide at the top and 5 m wide at the bottom and the area of cross - section is 344 sq m , the depth of cannel is ? | "1 / 2 * d ( 12 + 5 ) = 344 d = 40 answer : d" | a = 12 + 5
b = a / 2
c = 344 / b
d = c / 5
e = d / 2
|
a ) 0 , b ) 1 , c ) 2 , d ) 5 , e ) 9 | a | subtract(209, 209) | when the number 209 y 913 is exactly divisible by 11 , then the smallest whole number that can replace y ? | "the given number = 209 y 913 sum of the odd places = 3 + 9 + 9 + 2 = 23 sum of the even places = 1 + y + 0 ( sum of the odd places ) - ( sum of even places ) = number ( exactly divisible by 11 ) 23 - ( 1 + y ) = divisible by 11 22 � y = divisible by 11 . y must be 0 , to make given number divisible by 11 . a" | a = 209 - 209
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a ) 9 / 29 , b ) 8 / 23 , c ) 9 / 25 , d ) 17 / 29 , e ) 3 / 4 | c | divide(multiply(3, 3), add(multiply(multiply(4, 3), const_2), multiply(3, 3))) | pipe p can drain the liquid from a tank in 3 / 4 the time that it takes pipe q to drain it and in 1 / 3 the time that it takes pipe r to do it . if all 3 pipes operating simultaneously but independently are used to drain liquid from the tank , then pipe q drains what portion of the liquid from the tank ? | "suppose q can drain in 1 hr . so , rq = 1 / 1 = 1 so , rp = 1 / [ ( 3 / 4 ) rq ] = 4 / 3 also , rp = rr / ( 1 / 3 ) = > 4 / 3 = rr / ( 1 / 3 ) = > rr = 4 / 9 let h is the time it takes to drain by running all 3 pipes simultaneously so combined rate = rc = 1 / h = 1 + 4 / 3 + 4 / 9 = 25 / 9 = 1 / ( 9 / 25 ) thus runnin... | a = 3 * 3
b = 4 * 3
c = b * 2
d = 3 * 3
e = c + d
f = a / e
|
a ) l , b ) 2 l , c ) 3 l , d ) 4 l , e ) 5 l | d | subtract(5, const_1) | a train of length l is traveling at a constant velocity and passes a pole in t seconds . if the same train travelling at the same velocity passes a platform in 5 t seconds , then what is the length of the platform ? | the train passes a pole in t seconds , so velocity v = l / t ( l + p ) / v = 5 t ( l + p ) / ( l / t ) = 5 t p = 4 l the answer is d . | a = 5 - 1
|
a ) 276 , b ) 299 , c ) 312 , d ) 460 , e ) none | d | multiply(20, 23) | the h . c . f . of two numbers is 20 and the other two factors of their l . c . m . are 21 and 23 . the larger of the two numbers is | "solution clearly , the numbers are ( 20 x 21 ) and ( 20 x 23 ) . larger number = ( 20 x 23 ) = 460 . answer d" | a = 20 * 23
|
a ) 150,170 , b ) 150,270 , c ) 50,270 , d ) 180,270 , e ) 150,290 | b | multiply(5, 25) | two numbers are in the ratio of 5 : 9 . if 25 be subtracted from each , they are in the ratio of 35 : 59 . find the numbers ? | "( 5 x - 25 ) : ( 9 x - 25 ) = 35 : 59 x = 30 = > 150,270 answer : b" | a = 5 * 25
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['a ) 4', 'b ) 6', 'c ) 8', 'd ) 9', 'e ) 10'] | e | add(multiply(power(divide(128, multiply(1, multiply(1, const_2))), inverse(const_3)), multiply(1, const_2)), multiply(1, const_2)) | there is a rectangular prism made of 1 in cubes that has been covered in tin foil . there are exactly 128 cubes that are not touching any tin foil on any of their sides . if the width of the figure created by these 128 cubes is twice the length and twice the height , what is the measure e in inches of the width of the ... | if the width is w , then length and height would be w / 2 . so , w * w / 2 * w / 2 = 128 = > w ^ 3 = ( 2 ^ 3 ) * 64 = ( 2 ^ 3 ) * ( 4 ^ 3 ) = > w = 2 * 4 = 8 in . along the width of the cuboid , 8 cubes do n ' t touch the tin foil . so the actual width will be non - touching cubes + touching cubes = 8 + 2 = e = 10 ans ... | a = 1 * 2
b = 1 * a
c = 128 / b
d = 1/(3)
e = c ** d
f = 1 * 2
g = e * f
h = 1 * 2
i = g + h
|
a ) 7 sec , b ) 6 sec , c ) 8 sec , d ) 14 sec , e ) 13.2 sec | e | divide(110, multiply(add(24, 6), const_0_2778)) | a train 110 m long is running with a speed of 24 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 = 24 + 6 = 30 km / hr . = 30 * 5 / 18 = 8.33 m / sec . time taken to pass the men = 110 / 8.33 = 13.2 sec . answer : e" | a = 24 + 6
b = a * const_0_2778
c = 110 / b
|
a ) 40 days , b ) 40 / 9 days , c ) 60 / 11 days , d ) 30 / 9 days , e ) 60 / 9 days | c | divide(const_1, add(divide(const_1, 10), divide(const_1, 12))) | worker a takes 10 hours to do a job . worker b takes 12 hours to do the same job . how long it take both a & b , working together but independently , to do the same job ? | "a ' s one hour work = 1 / 10 . b ' s one hour work = 1 / 12 . ( a + b ) ' s one hour work = 1 / 10 + 1 / 12 = 11 / 60 . both a & b can finish the work in 60 / 11 days c" | a = 1 / 10
b = 1 / 12
c = a + b
d = 1 / c
|
a ) a ) 400 , b ) b ) 450 , c ) c ) 500 , d ) d ) 550 , e ) e ) 600 | d | subtract(subtract(multiply(add(add(3, 5), const_2), 160), multiply(5, 150)), multiply(3, 100)) | a women purchased 3 towels @ rs . 100 each , 5 towels @ rs . 150 each and two towels at a certain rate which is now slipped off from his memory . but she remembers that the average price of the towels was rs . 160 . find the unknown rate of two towels ? | "10 * 160 = 1600 3 * 100 + 5 * 150 = 1050 1600 – 1050 = 550 d" | a = 3 + 5
b = a + 2
c = b * 160
d = 5 * 150
e = c - d
f = 3 * 100
g = e - f
|
a ) 228 , b ) 299 , c ) 266 , d ) 500 , e ) 1000 | e | multiply(add(5, 6), const_100) | rs . 1500 is divided into two parts such that if one part is invested at 6 % and the other at 5 % the whole annual interest from both the sum is rs . 80 . how much was lent at 5 % ? | "( x * 5 * 1 ) / 100 + [ ( 1500 - x ) * 6 * 1 ] / 100 = 80 5 x / 100 + 90 â € “ 6 x / 100 = 80 x / 100 = 10 = > x = 1000 answer : e" | a = 5 + 6
b = a * 100
|
a ) 10 , b ) 24 , c ) 30 , d ) 36 , e ) 48 | a | multiply(divide(divide(600, 1000), 216), const_3600) | if a truck is traveling at a constant rate of 216 kilometers per hour , how many seconds will it take the truck to travel a distance of 600 meters ? ( 1 kilometer = 1000 meters ) | "speed = 216 km / hr = > 216,000 m / hr in one minute = > 216000 / 60 = 3600 meters in one sec = > 3600 / 60 = 60 meters time = total distance need to be covered / avg . speed = > 600 / 60 = 10 and hence the answer : a" | a = 600 / 1000
b = a / 216
c = b * 3600
|
a ) 28 , b ) 30 , c ) 40 , d ) 50 , e ) 60 | a | divide(multiply(4, 48), 3) | the l . c . m of two numbers is 48 . the numbers are in the ratio 4 : 3 . the sum of numbers is : | "let the numbers be 4 x and 3 x . then , their l . c . m = 12 x . so , 12 x = 48 or x = 4 . the numbers are 16 and 12 . hence , required sum = ( 16 + 12 ) = 28 . answer : a" | a = 4 * 48
b = a / 3
|
a ) 2 , b ) 3 , c ) 4 , d ) 1 , e ) 5 | d | divide(divide(6, 3), const_2) | a number a is squared and then multiplied by negative 6 . the result of this operation is equal to 3 times the sum of fourth times a and two . what is one possible value of a ? | - 6 * a ^ 2 = 3 ( 4 a + 2 ) a = - 1 or - 1 a = - 1 = b answer : d | a = 6 / 3
b = a / 2
|
a ) 63 , b ) 72 , c ) 144 , d ) 171 , e ) 400 | d | multiply(divide(multiply(38, const_3), 20), 30) | if 20 typists can type 38 letters in 20 minutes , then how many letters will 30 typists working at the same rate complete in 1 hour ? | "20 typists can type 38 letters , so 30 typists can type = 38 * 30 / 20 38 * 30 / 20 letters can be typed in 20 mins . in 60 mins typist can type = 38 * 30 * 60 / 20 * 20 = 171 d is the answer" | a = 38 * 3
b = a / 20
c = b * 30
|
a ) 25 / 32 , b ) 13 / 4 , c ) 17 / 8 , d ) 57 / 64 , e ) 15 / 16 | a | subtract(const_1, add(multiply(inverse(power(2, 6)), 6), add(inverse(power(2, 6)), inverse(power(2, 6))))) | a fair 2 sided coin is flipped 6 times . what is the probability that tails will be the result at least twice , but not more than 4 times ? | "at least twice , but not more than 5 timesmeans exactly 2 times , 3 times , 4 times the probability of getting exactly k results out of n flips is nck / 2 ^ n 6 c 2 / 2 ^ 6 + 6 c 3 / 2 ^ 6 + 6 c 4 / 2 ^ 6 = ( 20 + 15 + 15 ) / 2 ^ 6 = 25 / 32 option : a" | a = 2 ** 6
b = 1/(a)
c = b * 6
d = 2 ** 6
e = 1/(d)
f = 2 ** 6
g = 1/(f)
h = e + g
i = c + h
j = 1 - i
|
a ) 1 hr , b ) 2 hrs , c ) 3 hrs , d ) 5 hrs , e ) 6 hrs | a | divide(50, add(10, 40)) | two cyclist start from the same places in opposite directions . one is going towards north at 10 kmph and the other is going towards south 40 kmph . what time will they take to be 50 km apart ? | to be ( 10 + 40 ) km apart , they take 1 hour to be 50 km apart , they take 1 / 50 * 50 = 1 hrs answer is a | a = 10 + 40
b = 50 / a
|
a ) 1 , b ) 21 , c ) 26 , d ) 52 , e ) 1014 | b | multiply(add(subtract(13, const_10), const_2), add(subtract(13, const_10), const_2)) | if x is a sum of all even integers on the interval 13 . . . 55 and y is their number , what is the gcd ( x , y ) ? | "x = 14 + 16 + . . . + 54 = ( largest + smallest ) / 2 * ( # of terms ) = ( 14 + 54 ) / 2 * 21 = 34 * 21 . gcd of 21 and 34 * 21 is 21 answer : b ." | a = 13 - 10
b = a + 2
c = 13 - 10
d = c + 2
e = b * d
|
a ) 100 / 7 , b ) 5 , c ) 20 / 3 , d ) 8 , e ) 39 / 4 | a | divide(100, add(const_4, const_3)) | how many liters of pure alcohol must be added to a 100 - liter solution that is 20 percent alcohol in order to produce a solution that is 30 percent alcohol ? | 20 % alcohol solution means ; in the 100 liter solution , 20 liters of solution is alcohol and 80 liters other solvents . if we addxliters of alcohol to the solution , the solution becomes 100 + xliters and alcohol , which was 20 liters , becomes 20 + x liters . according to the statement ; 20 + x = 30 % of ( 100 + x )... | a = 4 + 3
b = 100 / a
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a ) rs . 480 , b ) rs . 520 , c ) rs . 600 , d ) rs . 960 , e ) none | c | divide(multiply(const_100, divide(multiply(24, const_100), multiply(2, 10))), multiply(2, 10)) | the banker ' s gain of a certain sum due 2 years hence at 10 % per annum is rs . 24 . the present worth is | solution t . d = ( b . g x 100 / rate x time ) = rs . ( 24 x 100 / 10 x 2 ) = rs . 120 . p . w = ( 100 x t . d / rate x time ) = rs . ( 100 x 120 / 10 x 2 ) = rs . 600 . answer c | a = 24 * 100
b = 2 * 10
c = a / b
d = 100 * c
e = 2 * 10
f = d / e
|
a ) 300 , b ) 150 , c ) 180 , d ) 109 , e ) 200 | e | divide(power(multiply(const_3, power(3000, const_2)), divide(const_1, const_3)), 1.5) | danny is sitting on a rectangular box . the area of the front face of the box is half the area of the top face , and the area of the top face is 1.5 times the area of the side face . if the volume of the box is 3000 , what is the area of the side face of the box ? | "lets suppose length = l , breadth = b , depth = d front face area = l * w = 1 / 2 w * d ( l = 1 / 2 d or d = 2 l ) top face area = w * d side face area = w * d = 1.5 d * l ( w = 1.5 l ) volume = l * w * d = 3000 l * 1.5 l * 2 l = 3000 l = 10 side face area = l * d = l * 2 l = 10 * 2 * 10 = 200 e is the answer" | a = 3000 ** 2
b = 3 * a
c = 1 / 3
d = b ** c
e = d / 1
|
a ) 50 % , b ) 32 % , c ) 25 % , d ) 43 % , e ) 29 % | c | multiply(divide(multiply(const_100, divide(20, const_100)), subtract(const_100, multiply(const_100, divide(20, const_100)))), const_100) | the salary of a person was reduced by 20 % . by what percent should his reduced salary be raised so as to bring it at par with his original salary ? | "let the original salary be $ 100 new salary = $ 80 increase on 80 = 20 increase on 100 = 20 / 80 * 100 = 25 % answer is c" | a = 20 / 100
b = 100 * a
c = 20 / 100
d = 100 * c
e = 100 - d
f = b / e
g = f * 100
|
a ) 4000 , b ) 6000 , c ) 4400 , d ) 4500 , e ) none of these | b | divide(90, multiply(multiply(divide(50, const_100), divide(30, const_100)), divide(10, const_100))) | if 10 % of 30 % of 50 % of a number is 90 , then what is the number ? | "let the number be a given , 10 / 100 * 30 / 100 * 50 / 100 * a = 90 = > 1 / 10 * 3 / 10 * 1 / 2 * a = 90 = > a = 10 * 20 * 10 * 2 = 6000 . answer : b" | a = 50 / 100
b = 30 / 100
c = a * b
d = 10 / 100
e = c * d
f = 90 / e
|
a ) 9 , b ) 8 , c ) 6 , d ) 4 , e ) 3 | a | subtract(multiply(5, const_2), const_1) | the average of non - zero number and its square is 5 times the number . the number is ? | "let the number be x . then , ( x + x 2 ) / 2 = 5 x = > x 2 - 9 x = 0 = > x ( x - 9 ) = 0 = > x = 0 or x = 9 so , the number is 9 . answer : a" | a = 5 * 2
b = a - 1
|
a ) 26 , b ) 36 , c ) 64 , d ) 67 , e ) 63 | e | add(add(add(add(divide(factorial(6), factorial(5)), divide(factorial(6), multiply(const_2, factorial(4)))), divide(factorial(6), multiply(factorial(3), factorial(3)))), divide(factorial(6), multiply(const_2, factorial(4)))), divide(factorial(6), factorial(5))) | preethi has 6 flavors of ice cream in his parlor . how many options are there for dorathy to pick a one - flavor , two - flavor , 3 - flavor , 4 - flavor , 5 flavor or 6 flavor order ? | 6 c 1 + 6 c 2 + 6 c 3 + 6 c 4 + 6 c 5 + 6 c 6 = 63 . answer : e | a = math.factorial(6)
b = math.factorial(5)
c = a / b
d = math.factorial(6)
e = math.factorial(4)
f = 2 * e
g = d / f
h = c + g
i = math.factorial(6)
j = math.factorial(3)
k = math.factorial(3)
l = j * k
m = i / l
n = h + m
o = math.factorial(6)
p = math.factorial(4)
q = 2 * p
r = o / q
s = n + r
t = math.factorial(... |
a ) 74 kg , b ) 76.5 kg , c ) 85 kg , d ) 78 , e ) none of these | a | add(multiply(2, 4.5), 65) | the average weight of 2 person ' s increases by 4.5 kg when a new person comes in place of one of them weighing 65 kg . what might be the weight of the new person ? | "explanation : total weight increased = ( 2 x 4.5 ) kg = 9 kg . weight of new person = ( 65 + 9 ) kg = 74 kg . answer : a" | a = 2 * 4
b = a + 65
|
a ) s . 340 , b ) s . 362 , c ) s . 370 , d ) s . 382 , e ) s . 390 | c | subtract(divide(multiply(4070, divide(multiply(12000, subtract(multiply(3, 4), 3)), multiply(const_10, const_100))), add(add(divide(multiply(multiply(3, 4), 18000), multiply(const_10, const_100)), divide(multiply(12000, subtract(multiply(3, 4), 3)), multiply(const_10, const_100))), divide(multiply(9000, subtract(multip... | john started a business , investing rs . 18000 . after 3 months and 4 months respectively , rose and tom joined him with capitals of 12000 and 9000 . at the end of the year the total profit was rs . 4070 . what is the difference between rose ’ s and tom ’ s share in the profit ? | john : rose : tom ratio of their investments = 18000 × 12 : 12000 × 9 : 9000 × 8 = 6 : 3 : 2 the difference between rose ’ s and tom ’ s share = 1 share : . i . e . = rs . 4070 × 1 / 11 = rs . 370 . c ) | a = 3 * 4
b = a - 3
c = 12000 * b
d = 10 * 100
e = c / d
f = 4070 * e
g = 3 * 4
h = g * 18000
i = 10 * 100
j = h / i
k = 3 * 4
l = k - 3
m = 12000 * l
n = 10 * 100
o = m / n
p = j + o
q = 3 * 4
r = q - 4
s = 9000 * r
t = 10 * 100
u = s / t
v = p + u
w = f / v
x = 3 * 4
y = x - 4
z = 9000 * y
A = 10 * 100
B = z / A
C ... |
a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | b | add(const_4, const_4) | if each year the population of the country grows by 20 % , how many years will elapse before the population of the country doubles ? | "till year 2000 , population is 100 . year 2001 : population becomes 120 . . . . . . . . . . . . . 1 year elapsed year 2002 : population becomes 144 . . . . . . . . . . . . . 2 year elapsed year 2003 : population becomes 172 . . . . . . . . . . . . . 3 year elapsed year 2004 : population > 200 . . . . . . . . . . . . .... | a = 4 + 4
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a ) 670 , b ) 664 , c ) 698 , d ) 744 , e ) 700 | c | subtract(815, multiply(divide(subtract(854, 815), subtract(divide(4, const_100), divide(3, const_100))), divide(3, const_100))) | peter invests a sum of money and gets back an amount of $ 815 in 3 years . david invests an equal amount of money and gets an amount of $ 854 in 4 years . if both amounts were invested at the same rate ( simple interest ) what was the sum of money invested ? | "since both peter and david invested the same amount of money at the same rate , they would earn same interest per year . david invested for one year more than peter and hence he got interest amount for one more year . interest earned per year = amount received by david - amount received by peter = 854 - 815 = 39 inter... | a = 854 - 815
b = 4 / 100
c = 3 / 100
d = b - c
e = a / d
f = 3 / 100
g = e * f
h = 815 - g
|
a ) 1000 , b ) 1055 , c ) 1378 , d ) 1075 , e ) 1080 | c | add(multiply(10, 82), multiply(6, 62)) | andrew purchased 10 kg of grapes at the rate of 82 per kg and 6 kg of mangoes at the rate of 62 per kg . how much amount did he pay to the shopkeeper ? | "cost of 10 kg grapes = 82 × 10 = 820 . cost of 6 kg of mangoes = 62 × 6 = 558 . total cost he has to pay = 820 + 558 = 1378 c" | a = 10 * 82
b = 6 * 62
c = a + b
|
a ) 1 / 10 , b ) 3 / 20 , c ) 1 / 5 , d ) 1 / 4 , e ) 9 / 20 | e | subtract(add(divide(const_2, add(add(const_2, add(const_2, 4)), 12)), divide(add(const_2, add(const_2, 4)), add(add(const_2, add(const_2, 4)), 12))), multiply(divide(const_2, add(add(const_2, add(const_2, 4)), 12)), divide(add(const_2, add(const_2, 4)), add(add(const_2, add(const_2, 4)), 12)))) | maths , physics and chemistry books are stored on a library shelf that can accommodate 25 books . currently , 20 % of the shelf spots remain empty . there are twice as many maths books as physics books and the number of physics books is 4 greater than that of chemistry books . among all the books , 12 books are soft co... | "first phase of this problem requires you to determine how many mathematics and chemistry books are even on the shelf . to do so , you have the equations : m + p + c = 20 ( since 4 / 5 of the 25 spots are full of books ) m = 2 p p = 4 + c from that , you can use substitution to get everything down to one variable . c =... | a = 2 + 4
b = 2 + a
c = b + 12
d = 2 / c
e = 2 + 4
f = 2 + e
g = 2 + 4
h = 2 + g
i = h + 12
j = f / i
k = d + j
l = 2 + 4
m = 2 + l
n = m + 12
o = 2 / n
p = 2 + 4
q = 2 + p
r = 2 + 4
s = 2 + r
t = s + 12
u = q / t
v = o * u
w = k - v
|
a ) 60 , b ) 56 , c ) 84 , d ) 90 , e ) 108 | b | multiply(subtract(divide(multiply(const_2, const_2), subtract(7, multiply(const_2, 2))), divide(const_2, subtract(7, 2))), const_60) | tom and linda stand at point a . linda begins to walk in a straight line away from tom at a constant rate of 2 miles per hour . one hour later , tom begins to jog in a straight line in the exact opposite direction at a constant rate of 7 miles per hour . if both tom and linda travel indefinitely , what is the positive ... | "b is the answer . . . . d = ts where d = distance , t = time and s = speed to travel half distance , ( 2 + 2 t ) = 7 t = = > t = 2 / 5 = = > 24 minutes to travel double distance , 2 ( 2 + 2 t ) = 7 t = = > 2 = = > 80 minutes difference , 56 minutes b" | a = 2 * 2
b = 2 * 2
c = 7 - b
d = a / c
e = 7 - 2
f = 2 / e
g = d - f
h = g * const_60
|
a ) 66 , b ) 26 , c ) 42 , d ) 28 , e ) 11 | d | divide(multiply(divide(add(const_4, const_3), add(add(const_4, const_3), const_2)), 64), const_2) | 64 is divided into two parts in such a way that seventh part of first and ninth part of second are equal . find the smallest part ? | "x / 7 = y / 9 = > x : y = 7 : 9 7 / 16 * 64 = 28 answer : d" | a = 4 + 3
b = 4 + 3
c = b + 2
d = a / c
e = d * 64
f = e / 2
|
a ) 1000 m , b ) 1250 m , c ) 1500 m , d ) 1800 m , e ) 2300 m | c | multiply(divide(multiply(6, const_1000), const_60), 15) | a man walking at the rate of 6 km / hr crosses a bridge in 15 minutes . the length of the bridge is ______ . | "hint : to find the answer in meter , we will first convert distance from km / hour to meter / sec by multiplying it with 5 / 18 . also , change 15 minutes to seconds by multiplying it with 60 . distance = speed x time 1 . convert speed into m / sec : 6 x 5 / 18 m / s = 1.66 m / s 2 . convert time from minutes into sec... | a = 6 * 1000
b = a / const_60
c = b * 15
|
a ) 0 and 3 , b ) 3 and 6 , c ) 6 and 9 , d ) 9 and 12 , e ) 12 and 15 | b | add(multiply(floor(power(100, divide(const_1, 4))), const_10), multiply(floor(power(100, divide(const_1, 4))), const_2)) | if w = x ^ 4 + y ^ 4 = 100 , then the greatest possible value of x is between | my attempt : if w = x ^ 4 + y ^ 4 = 100 , then the greatest possible value of x would be when y is minimum . let y ^ 4 be 0 . now x ^ 4 = 100 . x should be definitely greater than 3 but less than 4 . the only option that fits this range is b hence answer is - - b ) 3 and 6 . | a = 1 / 4
b = 100 ** a
c = math.floor(b)
d = c * 10
e = 1 / 4
f = 100 ** e
g = math.floor(f)
h = g * 2
i = d + h
|
a ) 72 , b ) 224 , c ) 320 , d ) 512 , e ) 637 | e | gcd(89, const_4) | if m and n are positive integers and m ^ 2 + n ^ 2 = 89 , what is the value of m ^ 3 + n ^ 3 ? | "you need to integers which squared are equal 89 . which could it be ? let ' s start with the first integer : 1 ^ 2 = 1 2 ^ 2 = 4 3 ^ 2 = 9 4 ^ 2 = 16 5 ^ 2 = 25 6 ^ 2 = 36 7 ^ 2 = 49 8 ^ 2 = 64 stop . the integers ca n ' t be greater than 8 or we will score above 89 . the second integer need to be picked up the same w... | a = math.gcd(89, 4)
|
a ) 11 , b ) 26 , c ) 24 , d ) 27 , e ) 32 | b | floor(multiply(divide(3, 8), 70)) | the ratio of the number of red cars in a certain parking lot to the number of black cars is 3 to 8 . if there are 70 black cars in the lot , how many red cars are there in the lot ? | "b is correct r / b = 3 / 8 and b = 70 r = 70 * 3 / 8 = 26" | a = 3 / 8
b = a * 70
c = math.floor(b)
|
a ) 14 , b ) 16 , c ) 19 , d ) 21 , e ) none of these | b | floor(sqrt(272)) | if the sum of a number and its square is 272 , what is the number ? | "explanation : let the integer be x . then , x + x 2 = 272 x 2 + x - 272 = 0 ( x + 17 ) ( x – 16 ) = 0 x = 16 answer : b" | a = math.sqrt(272)
b = math.floor(a)
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a ) 28888 , b ) 27789 , c ) 2777 , d ) 14000 , e ) 2881 | d | multiply(2460, divide(1, 3)) | a , b , c and d enter into partnership . a subscribes 1 / 3 of the capital b 1 / 4 , c 1 / 5 and d the rest . how much share did a get in a profit of rs . 2460 ? | "25 * 12 : 30 * 12 : 35 * 8 15 : 18 : 14 14 / 47 * 47000 = 14000 answer : d" | a = 1 / 3
b = 2460 * a
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['a ) 19 cm', 'b ) 20 cm', 'c ) 21 cm', 'd ) 30 cm', 'e ) none'] | c | power(divide(19404, divide(multiply(const_2, const_pi), const_3)), divide(const_1, const_3)) | volume of a hemisphere is 19404 cu . cm . its radius is : | sol . let the radius be r cm . then , 2 / 3 * 22 / 7 * r ³ = 19404 ⇔ r ³ = [ 19404 * 21 / 44 ] = ( 21 ) ³ ⇔ r = 21 cm . answer c | a = 2 * math.pi
b = a / 3
c = 19404 / b
d = 1 / 3
e = c ** d
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a ) 24 , b ) 44 , c ) 58 , d ) 60 , e ) 62 | c | add(add(18, multiply(4, const_4)), multiply(4, 5)) | a hiker walked for 3 days . she walked 18 miles on the first day , walking 3 miles per hour . on the second day she walked for one less hour but she walked one mile per hour , faster than on the first day . on the third day she walked at 4 miles per hour for 5 hours . how many miles in total did she walk ? | "first day - 18 miles with 3 miles per hours then total - 6 hours for that day second day - 4 miles per hour and 5 hours - 20 miles third day - 4 miles per hour and 5 hours - 20 miles total 18 + 20 + 20 = 58 answer : option c" | a = 4 * 4
b = 18 + a
c = 4 * 5
d = b + c
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a ) 27 , b ) 28 , c ) 29 , d ) 30 , e ) 31 | d | subtract(multiply(multiply(multiply(2, 2), multiply(2, 2)), 2), const_2) | 2 corner most boxes of a chess board ( diagonally opposite ) haven been cut out there ' s a rectangular block = 2 sqaures of chess board , how many such blocks can be placed on the chess board ? ” | chess board has 64 squares . so we can place 32 rectangular blocks , but 2 are cut off from the corner . hence 32 - 2 = 30 blocks answer : d | a = 2 * 2
b = 2 * 2
c = a * b
d = c * 2
e = d - 2
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a ) 13 √ 4 , b ) 12.1 √ 2 , c ) 23 √ 2 , d ) 12 √ 4 , e ) 13 √ 9 | b | sqrt(multiply(add(power(divide(44, const_4), const_2), power(divide(20, const_4), const_2)), const_2)) | the perimeter of one square is 44 cm and that of another is 20 cm . find the perimeter and the diagonal of a square which is equal in area to these two combined ? | "4 a = 44 4 a = 20 a = 11 a = 5 a 2 = 121 a 2 = 25 combined area = a 2 = 146 = > a = 12.1 d = 12.1 √ 2 answer : b" | a = 44 / 4
b = a ** 2
c = 20 / 4
d = c ** 2
e = b + d
f = e * 2
g = math.sqrt(f)
|
a ) 45 % , b ) 125 % , c ) 145 % , d ) 150 % , e ) 225 % | b | divide(subtract(19,947, 8,902), 8,902) | in 1970 there were 8,902 women stockbrokers in the united states . by 1978 the number had increased to 19,947 . approximately what was the percent increase ? | "we can use approximation to get the answer quickly . like , 19,947 is approx 20000 8902 is approx 8900 so , total increase = 20000 - 8900 = 11100 hence , % increase = 11100 / 8900 = approx . 11 / 9 = 1.22 = approx . 1.25 so , choice is b ." | a = 19 - 947
b = a / 8
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a ) 18 % , b ) 20 % , c ) 21 % , d ) 23 % , e ) can not be determined | d | add(divide(multiply(30, 4), 2), add(30, 5)) | two kinds of vodka are mixed in the ratio 1 : 2 and 2 : 1 and they are sold fetching the profit 30 % and 20 % respectively . if the vodkas are mixed in equal ratio and the individual profit percent on them are increased by 4 / 3 and 5 / 3 times respectively , then the mixture will fetch the profit of | answer : d . | a = 30 * 4
b = a / 2
c = 30 + 5
d = b + c
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a ) 32 , b ) 1360 , c ) 1000 , d ) 46 , e ) 104 | c | subtract(multiply(add(subtract(divide(200, const_10), const_2), const_10), add(divide(200, const_10), subtract(divide(200, const_10), const_2))), 200) | a rectangular room has the rectangular shaped rug shown as above figure such that the rug ’ s area is 200 square feet and its length is 10 feet longer than its width . if the uniform width between the rug and room is 10 feet , what is the area of the region uncovered by the rug ( shaded region ) , in square feet ? | "rug ' s area = 200 which is ( x ) x ( 10 + x ) = 200 so x = 10 rug maintains a uniform distance of 10 feet so room has dimension 10 + 20 and 20 + 20 i . e . 30 and 40 area of room 30 x 40 = 1200 area covered is 200 so uncovered area is 1200 - 200 = 1000 ( answer c )" | a = 200 / 10
b = a - 2
c = b + 10
d = 200 / 10
e = 200 / 10
f = e - 2
g = d + f
h = c * g
i = h - 200
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a ) 10 % , b ) 5 % , c ) 15 % , d ) 20 % , e ) 25 % | b | multiply(divide(20, const_100), const_100) | the sum of money will be double itself in 20 years and simple interest find rate of interest ? | "t = 20 years p = principle amount = x a = total amount = 2 x si = simple interest = a - p = 2 x - x = x r = 100 si / pt = 100 x / 20 x = 5 % answer is b" | a = 20 / 100
b = a * 100
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a ) 16 , b ) 17 , c ) 18 , d ) 10 , e ) 20 | d | multiply(3, const_2) | how many positive even integers less than 100 contain digits 3 or 7 ? | "two digit numbers : 3 at tens place : 30 , 32,34 , 36,38 7 at tens place : 70 , 72,74 , 76,78 if 3 and 7 is at units place , the number cant be even total : 5 + 5 = 10 answer d" | a = 3 * 2
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a ) rs . 640.86 , b ) rs . 430.86 , c ) rs . 330.86 , d ) rs . 630.86 , e ) rs . 130.86 | d | subtract(add(add(divide(multiply(divide(560, multiply(divide(8, const_100), const_2.0)), 8), const_100), divide(560, multiply(divide(8, const_100), 4))), divide(multiply(add(divide(multiply(divide(560, multiply(divide(8, const_100), 4)), 8), const_100), divide(560, multiply(divide(8, const_100), 4))), 8), const_100)), ... | if the simple interest on a sum of money for 4 years at 8 % per annum is rs . 560 , what is the compound interest on the same sum at the rate and for the same time ? | "sum = ( 50 * 100 ) / ( 2 * 5 ) = rs . 1 , 750.00 c . i . on rs . 1 , 750.00 for 4 years at 8 % = rs . 2 , 380.86 . = rs . 2 , 380.86 - 1 , 750.00 = rs . 630.86 answer : d" | a = 8 / 100
b = a * 2
c = 560 / b
d = c * 8
e = d / 100
f = 8 / 100
g = f * 4
h = 560 / g
i = e + h
j = 8 / 100
k = j * 4
l = 560 / k
m = l * 8
n = m / 100
o = 8 / 100
p = o * 4
q = 560 / p
r = n + q
s = r * 8
t = s / 100
u = i + t
v = 8 / 100
w = v * 4
x = 560 / w
y = u - x
|
a ) $ 42 , b ) $ 44 , c ) $ 52 , d ) $ 60 , e ) $ 68 | a | multiply(multiply(inverse(subtract(1, divide(1, 3))), add(divide(7, const_2), 10.50)), const_2) | bert left the house with n dollars . he spent 1 / 3 of this at the hardware store , then $ 7 at the dry cleaners , and then half of what was left at the grocery store . when he got home , he had $ 10.50 left in his pocket . what was the value of n ? | "started to test answer a if he had 42 , then he spent 13 2 / 3 at hardware store now he was left with 27 1 / 3 $ he spent 7 dollars on cleaning , thus he remained with 20 1 / 3 $ he then spent 1 / 2 of 20 1 / 3 or 10.5 , hence , right option is a ." | a = 1 / 3
b = 1 - a
c = 1/(b)
d = 7 / 2
e = d + 10
f = c * e
g = f * 2
|
a ) 18 , b ) 56 , c ) 12 , d ) 17 , e ) 14 | e | subtract(44, multiply(10, 3)) | the average age of a group of 10 persons was decreased by 3 years when one person , whose age was 44 years , was replaced by a new person . find the age of the new person ? | "initial average age of the 10 persons be p . age of the new person q . sum of the ages of the initial 10 persons = 10 p new average = ( p - 3 ) 10 ( p - 3 ) = 10 p - 44 + q = > q = 14 answer : e" | a = 10 * 3
b = 44 - a
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a ) 40 sec , b ) 29 sec , c ) 26 sec , d ) 27 sec , e ) 43 sec | e | divide(360, multiply(subtract(42, 140), const_0_2778)) | a train 360 m long is running at a speed of 42 km / hr . in what time will it pass a bridge 140 m long ? | "speed = 42 * 5 / 18 = 35 / 3 m / sec total distance covered = 360 + 140 = 500 m required time = 500 * 3 / 35 = 43 sec answer : e" | a = 42 - 140
b = a * const_0_2778
c = 360 / b
|
a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 16 | b | divide(add(divide(96, 16), sqrt(add(multiply(multiply(divide(divide(96, 16), add(const_1, divide(50, const_100))), 4), 4), power(divide(96, 16), const_2)))), const_2) | pascal has 96 miles remaining to complete his cycling trip . if he reduced his current speed by 4 miles per hour , the remainder of the trip would take him 16 hours longer than it would if he increased his speed by 50 % . what is his current speed t ? | "let the current speed be x miles per hour . time taken if speed is 50 % faster ( i . e . 3 x / 2 = 1.5 x ) = 96 / 1.5 x time taken if speed is reduced by 4 miles / hr ( i . e . ( x - 4 ) ) = 96 / ( x - 4 ) as per question , 96 / ( x - 4 ) - 96 / 1.5 x = 16 solving this t we get x = 8 . b ." | a = 96 / 16
b = 96 / 16
c = 50 / 100
d = 1 + c
e = b / d
f = e * 4
g = f * 4
h = 96 / 16
i = h ** 2
j = g + i
k = math.sqrt(j)
l = a + k
m = l / 2
|
a ) 0.15 , b ) 0.20 , c ) 0.25 , d ) 0.30 , e ) 0.33 | b | multiply(divide(4, 5), divide(const_1, 4)) | a = { 2 , 3 , 4 , 5 } b = { 4 , 5 , 6 , 7 , 8 } two integers will be randomly selected from the sets above , one integer from set a and one integer from set b . what is the probability q that the sum of the two integers will equal 9 ? | rearrange the first set : a = { 5,4 , 3,2 } b = { 4,5 , 6,7 , 8 } as you can see numbers in each column ( the numbers of the same color ) give the sum of 9 . so there are 4 such pares possible , total # of pairs is 4 * 5 = 20 . q = favorable / total = 4 / 20 = 0.2 . answer : b . or : we can select any number from set a... | a = 4 / 5
b = 1 / 4
c = a * b
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a ) 40 , b ) 99 , c ) 77 , d ) 66 , e ) 32 | a | divide(360, multiply(subtract(45, 140), const_0_2778)) | a train 360 m long is running at a speed of 45 km / hr . in what time will it pass a bridge 140 m long ? | "speed = 45 * 5 / 18 = 25 / 2 m / sec total distance covered = 360 + 140 = 500 m required time = 500 * 2 / 25 = 40 sec answer : a" | a = 45 - 140
b = a * const_0_2778
c = 360 / b
|
a ) 30 , b ) 54 , c ) 40 , d ) 36 , e ) 31 | c | divide(600, multiply(subtract(56, 2), const_0_2778)) | how many seconds will a 600 m long train take to cross a man walking with a speed of 2 km / hr in the direction of the moving train if the speed of the train is 56 km / hr ? | "speed of train relative to man = 56 - 2 = 54 km / hr . = 54 * 5 / 18 = 15 m / sec . time taken to pass the man = 600 * 1 / 15 = 40 sec . answer : c" | a = 56 - 2
b = a * const_0_2778
c = 600 / b
|
a ) 4500 m 3 , b ) 4580 m 3 , c ) 9000 m 3 , d ) 4900 m 3 , e ) 4700 m 3 | c | divide(multiply(multiply(2, 45), multiply(6, const_1000)), multiply(const_1, const_60)) | a river 2 m deep and 45 m wide is flowing at the rate of 6 kmph the amount of water that runs into the sea per minute is ? | "explanation : ( 6000 * 2 * 45 ) / 60 = 9000 m 3 answer : option c" | a = 2 * 45
b = 6 * 1000
c = a * b
d = 1 * const_60
e = c / d
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a ) 100 , b ) 200 , c ) 300 , d ) 400 , e ) 500 | c | multiply(multiply(multiply(1, const_10), const_10), const_3) | how many times will the digit 3 be written when listing the integers from 1 to 1000 ? | "many approaches are possible . for example : consider numbers from 0 to 999 written as follows : 1 . 000 2 . 001 3 . 002 4 . 003 . . . . . . . . . 1000 . 999 we have 1000 numbers . we used 3 digits per number , hence used total of 3 * 1000 = 3000 digits . now , why should any digit have preferences over another ? we u... | a = 1 * 10
b = a * 10
c = b * 3
|
a ) 11.72 , b ) 11.52 , c ) 11.97 , d ) 10.91 , e ) 11.79 | e | divide(circumface(divide(square_edge_by_perimeter(rectangle_perimeter(9, 6)), const_2)), const_2) | the parameter of a square is equal to the perimeter of a rectangle of length 9 cm and breadth 6 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 ( 9 + 6 ) = 30 cm parameter of the square = 30 cm i . e . 4 a = 30 a = 7.5 diameter of the semicircle = 7.5 cm circimference of the semicircle = 1 / 2 ( â ˆ ) ( 7.5 ) = 1 / 2 ( 22 / 7 ) ( 7.5 ) = 165 / 14 = 11.79 cm to two decimal places answer : e" | a = square_edge_by_perimeter / (
b = circumface / (
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a ) 2 hours , b ) 4 hours , c ) 5 hours , d ) 6 hours , e ) 7 hours | a | inverse(add(multiply(divide(const_1, 3), add(divide(10, const_100), const_1)), multiply(divide(const_1, 6), divide(20, const_100)))) | working alone , mary can pave a driveway in 3 hours and hillary can pave the same driveway in 6 hours . when they work together , mary thrives on teamwork so her rate increases by 10 % , but hillary becomes distracted and her rate decreases by 20 % . if they both work together , how many hours will it take to pave the ... | "initial working rates : mary = 1 / 3 per hour hillary = 1 / 6 per hour rate when working together : mary = 1 / 3 + ( 1 / 10 * 1 / 3 ) = 3 / 8 per hour hillary = 1 / 6 - ( 1 / 5 * 1 / 6 ) = 2 / 15 per hour together they work 3 / 8 + 2 / 15 = 1 / 2 per hour so they will need 2 hours to complete the driveway . the correc... | a = 1 / 3
b = 10 / 100
c = b + 1
d = a * c
e = 1 / 6
f = 20 / 100
g = e * f
h = d + g
i = 1/(h)
|
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | c | subtract(multiply(11, 2), 20) | if the remainder is 11 when the integer n is divided by 20 , what is the remainder when 2 n is divided by 10 ? | "n = 20 k + 11 2 n = 2 ( 20 k + 11 ) = 4 k * 10 + 22 = 4 k * 10 + 2 * 10 + 2 = 10 j + 2 the answer is c ." | a = 11 * 2
b = a - 20
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a ) 7 , b ) 13 , c ) 16 , d ) 21 , e ) 23 | e | add(subtract(100, multiply(12, 7)), 7) | a basketball team composed of 12 players scored 100 points in a particular contest . if none of the individual players scored fewer than 7 points , what is the greatest number of points t that an individual player might have scored ? | "general rule for such kind of problems : to maximize one quantity , minimize the others ; to minimize one quantity , maximize the others . thus to maximize the number of points of one particular player minimize the number of points of all other 11 players . minimum number of points for a player is 7 , so the minimum n... | a = 12 * 7
b = 100 - a
c = b + 7
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a ) $ 100 , b ) $ 75 , c ) $ 20 , d ) $ 120 , e ) $ 50 | c | multiply(100, subtract(const_1, divide(divide(200, const_2), add(400, divide(200, const_2))))) | a invested $ 400 in a business after 6 months b invested $ 200 in the business . end of the year if they got $ 100 as profit . find b ' s shares ? | "a : b = 400 * 12 : 200 * 6 a : b = 4 : 1 b ' s share = 100 * 1 / 5 = $ 20 answer is c" | a = 200 / 2
b = 200 / 2
c = 400 + b
d = a / c
e = 1 - d
f = 100 * e
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a ) 7207 , b ) 7206 , c ) 7203 , d ) 7200 , e ) 7201 | d | add(add(triangle_area(40, 30), triangle_area(50, 120)), multiply(40, subtract(120, 30))) | plot abcd is as shown in figure , where af = 30 m , ce = 40 m , ed = 50 m , ae = 120 m . find the area of the plot abcd ? | "area of plot abcd = area of ade + area of afb + area of bcef = 1 / 2 * 50 * 120 + 1 / 2 * 40 * 30 + 40 * 90 = 3000 + 600 + 3600 = 7200 sq . m answer : d" | a = triangle_area + (
b = a + triangle_area
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a ) 40 days , b ) 16 days , c ) 6 days , d ) 5 days , e ) 40 days | c | inverse(subtract(3, divide(3, 6))) | a and b can do a piece of work in 6 days . with the help of c they finish the work in 3 days . c alone can do that piece of work in ? | "c 6 days c = 1 / 3 – 1 / 6 = 1 / 6 = > 6 days" | a = 3 / 6
b = 3 - a
c = 1/(b)
|
a ) 2.4 . , b ) 3.6 . , c ) 4.2 , d ) 5.5 , e ) 6.4 | e | multiply(multiply(divide(divide(2, divide(4, 2)), 2), 4), 4) | two carpenters , working in the same pace , can build 2 desks in two hours and a half . how many desks can 4 carpenters build in 4 hours ? | "w = 2 desks t = 2.5 hrs rate of 2 carpenters = 2 × r rate = work done / time 2 xr = 2 / 2.5 r = 1 / 2.5 = 2 / 5 ( this is the rate of each carpenter ) work done by 4 carpenters in 4 hrs = 4 × rate of each carpenter x time = 4 × 2 / 5 × 4 = 6.4 desks e is the correct answer ." | a = 4 / 2
b = 2 / a
c = b / 2
d = c * 4
e = d * 4
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a ) 20 % , b ) 24 % , c ) 30 % , d ) 32 % , e ) 39 % | e | multiply(divide(subtract(multiply(divide(add(const_100, 10), const_100), divide(add(const_100, 50), const_100)), const_1), multiply(divide(add(const_100, 10), const_100), divide(add(const_100, 50), const_100))), const_100) | the output of a factory is increased by 10 % to keep up with rising demand . to handle the holiday rush , this new output is increased by 50 % . by approximately what percent would the output of the factory now have to be decreased in order to restore the original output ? | "take it as original output = 100 . to meet demand increase by 10 % , then output = 110 . to meet holiday demand , new output increase by 50 % then output equals 165 to restore new holidy demand output to original 100 . final - initial / final * 100 = 65 / 165 * 100 = 39 % approxiamately . option e is correct ." | a = 100 + 10
b = a / 100
c = 100 + 50
d = c / 100
e = b * d
f = e - 1
g = 100 + 10
h = g / 100
i = 100 + 50
j = i / 100
k = h * j
l = f / k
m = l * 100
|
a ) 19 , b ) 18 , c ) 16 , d ) 11 , e ) 17 | d | add(10, const_1) | the average of first 10 even numbers is ? | "sum of 10 even numbers = 10 * 11 = 110 average = 110 / 10 = 11 answer : d" | a = 10 + 1
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a ) 20 % , b ) 25 % , c ) 30 % , d ) 35 % , e ) 40 % | b | multiply(divide(subtract(60, 45), 60), const_100) | in town p , 60 percent of the population are employed , and 45 percent of the population are employed males . what percent of the employed people in town p are females ? | "the percent of the population who are employed females is 60 - 45 = 15 % the percent of employed people who are female is 15 % / 60 % = 25 % . the answer is b ." | a = 60 - 45
b = a / 60
c = b * 100
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a ) 35.29 kg , b ) 37.25 kg , c ) 42.45 kg , d ) 38.66 kg , e ) 29.78 kg | d | divide(add(multiply(26, 50), multiply(34, 30)), add(26, 34)) | there are 2 sections a and b in a class , consisting of 26 and 34 students respectively . if the average weight of section a is 50 kg and that of section b is 30 kg , find the average of the whole class ? | "total weight of 26 + 34 students = 26 * 50 + 34 * 30 = 2320 average weight of the class is = 2320 / 60 = 38.66 kg answer is d" | a = 26 * 50
b = 34 * 30
c = a + b
d = 26 + 34
e = c / d
|
a ) 4368 , b ) 2678 , c ) 5460 , d ) 1976 , e ) 1671 | a | divide(multiply(divide(multiply(4000, add(const_100, 4)), const_100), add(const_100, 5)), const_100) | find the amount on rs . 4000 in 2 years , the rate of interest being 4 % per first year and 5 % for the second year ? | "4000 * 104 / 100 * 105 / 100 = > 4368 answer : a" | a = 100 + 4
b = 4000 * a
c = b / 100
d = 100 + 5
e = c * d
f = e / 100
|
a ) 10 % , b ) 9 % , c ) 11.11 % , d ) 12 % , e ) none of these | c | multiply(divide(add(multiply(const_2, const_100), divide(const_100, const_2)), 900), const_100) | a shopkeeper forced to sell at cost price , uses a 900 grams weight for a kilogram . what is his gain percent ? | "shopkeeper sells 900 g instead of 1000 g . so , his gain = 1000 - 900 = 100 g . thus , % gain = ( 100 * 100 ) / 900 = 11.11 % . answer : option c" | a = 2 * 100
b = 100 / 2
c = a + b
d = c / 900
e = d * 100
|
a ) 20 % , b ) 30 % , c ) 40 % , d ) 50 % , e ) 60 % | d | subtract(100, 50) | john want to buy a $ 100 trouser at the store , but he think it â € ™ s too expensive . finally , it goes on sale for $ 50 . what is the percent decrease ? | "the is always the difference between our starting and ending points . in this case , it â € ™ s 100 â € “ 50 = 40 . the â € œ original â € is our starting point ; in this case , it â € ™ s 100 . ( 50 / 100 ) * 100 = ( 0.5 ) * 100 = 50 % . d" | a = 100 - 50
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a ) 96 , b ) 76 , c ) 56 , d ) 36 , e ) 24 | e | multiply(add(divide(const_100, const_4), multiply(multiply(multiply(multiply(2, const_3), subtract(const_1, 2)), subtract(const_1, 2)), subtract(const_1, 2))), const_4) | what are the last two digits of ( 301 * 402 * 503 * 604 * 646 * 547 * 449 * 349 ) ^ 2 | "( ( 301 * 402 * 503 * 604 * 646 ) * ( 547 * 449 * 349 ) ) ^ 2 if you observe above digits , last digit are : 1,2 , 3,4 , 6,7 , 9,9 ; 5 & 9 are missing ; so i have rearranged them so that multiplication will be easy for me as initial 4 digits have last two digits as 01 , 02,03 , 04,46 and final three as 47 * 49 * 49 . ... | a = 100 / 4
b = 2 * 3
c = 1 - 2
d = b * c
e = 1 - 2
f = d * e
g = 1 - 2
h = f * g
i = a + h
j = i * 4
|
a ) 400 , b ) 500 , c ) 375 , d ) 450 , e ) 600 | e | divide(1800, const_3) | divide rs . 1800 among a , b and c so that a receives 2 / 5 as much as b and c together and b receives 1 / 5 as a and c together . a ' s share is ? | a + b + c = 1800 a = 2 / 5 ( b + c ) ; b = 1 / 5 ( a + c ) a / ( b + c ) = 2 / 5 a = 1 / 6 * 3600 = > 600 answer : e | a = 1800 / 3
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a ) 18 , b ) 28 , c ) 32 , d ) 56 , e ) 58 | c | multiply(divide(divide(12880, 230), 7), 4) | the ratio between the number of sheep and the number of horses at the stewar farm is 4 to 7 . if each of horse is fed 230 ounces of horse food per day and the farm needs a total 12880 ounces of horse food per day . what is number sheep in the form ? ? | "et no of sheep and horses are 4 k and 7 k no of horses = 12880 / 230 = 56 now 7 k = 56 and k = 8 no of sheep = ( 4 * 8 ) = 32 answer : c" | a = 12880 / 230
b = a / 7
c = b * 4
|
a ) 420 , b ) 550 , c ) 490 , d ) 450 , e ) 457 | c | subtract(multiply(const_10, 150), add(multiply(3, 100), multiply(5, 150))) | a man purchased 3 blankets @ rs . 100 each , 5 blankets @ rs . 150 each and two blankets at a certain rate which is now slipped off from his memory . but he remembers that the average price of the blankets was rs . 154 . find the unknown rate of two blankets ? | "10 * 154 = 1540 3 * 100 + 5 * 150 = 1050 1540 – 1050 = 490 answer : c" | a = 10 * 150
b = 3 * 100
c = 5 * 150
d = b + c
e = a - d
|
a ) $ 40,000 , b ) $ 41,667 , c ) $ 42,000 , d ) $ 43,750 , e ) $ 60,000 | d | add(divide(divide(divide(multiply(add(divide(15, const_100), 1), multiply(subtract(add(const_1000, const_60), const_10), const_1000)), add(multiply(subtract(21, 1), add(divide(15, const_100), 1)), 1)), add(divide(15, const_100), 1)), const_100), add(multiply(const_100, const_2), const_3)) | last year , company x paid out a total of $ 1 , 050,000 in salaries to its 21 employees . if no employee earned a salary that is more than 15 % greater than any other employee , what is the lowest possible salary that any one employee earned ? | "employee 1 earned $ x ( say ) employee 2 will not earn more than $ 1.15 x therfore , to minimize the salary of any one employee , we need to maximize the salaries of the other 20 employees ( 1.15 x * 20 ) + x = 1 , 050,000 solving for x = $ 43,750 answer d" | a = 15 / 100
b = a + 1
c = 1000 + const_60
d = c - 10
e = d * 1000
f = b * e
g = 21 - 1
h = 15 / 100
i = h + 1
j = g * i
k = j + 1
l = f / k
m = 15 / 100
n = m + 1
o = l / n
p = o / 100
q = 100 * 2
r = q + 3
s = p + r
|
a ) 2 / 7 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 5 / 7 | b | divide(const_10, 30) | in a graduate physics course , 70 percent of the students are male and 30 percent of the students are married . if two - sevenths of the male students are married , what fraction of the female students is married ? | "let assume there are 100 students of which 70 are male and 30 are females if 30 are married then 70 will be single . now its given that two - sevenths of the male students are married that means 2 / 7 of 70 = 20 males are married if 30 is the total number of students who are married and out of that 20 are males then t... | a = 10 / 30
|
['a ) 9', 'b ) 7', 'c ) - 9', 'd ) 11', 'e ) 6'] | b | divide(subtract(sqrt(add(multiply(63, const_4), power(const_2, const_2))), const_2), const_2) | the area of a rectangle is 63 sq m . the width is two meters shorter than the length . what is the width ? | a = l x w w = l - 2 l = w + 2 a = ( w + 2 ) x w a = w ^ 2 + 2 x w 63 = w ^ 2 + 2 w 0 = w ^ 2 + 2 w - 63 0 = ( w + 9 ) ( w - 7 ) w = - 9 and w = 7 , width can not be negative so w = 7 answer is b | a = 63 * 4
b = 2 ** 2
c = a + b
d = math.sqrt(c)
e = d - 2
f = e / 2
|
a ) 45 , b ) 86 , c ) 30 , d ) 78 , e ) 38 | a | divide(add(25, 65), const_2) | a man can row upstream at 25 kmph and downstream at 65 kmph , and then find the speed of the man in still water ? | "us = 25 ds = 65 m = ( 65 + 25 ) / 2 = 45 answer : a" | a = 25 + 65
b = a / 2
|
a ) 37.5 m , b ) 75 m , c ) 25 m , d ) 80 m , e ) 30 m | b | multiply(divide(150, subtract(7.5, 2.5)), 2.5) | a train crosses a bridge of length 150 m in 7.5 seconds and a lamp post on the bridge in 2.5 seconds . what is the length of the train in metres ? | let length of train = l case - 1 : distance = 150 + l ( while crossing the bridge ) time = 7.5 seconds i . e . speed = distance / time = ( 150 + l ) / 7.5 case - 2 : distance = l ( while passing the lamp post ) time = 2.5 seconds i . e . speed = distance / time = ( l ) / 2.5 but since speed has to be same in both cases... | a = 7 - 5
b = 150 / a
c = b * 2
|
a ) 20 % , b ) 33.3 % , c ) 40 % , d ) 50 % , e ) 75 % | e | divide(multiply(subtract(multiply(12, 12), multiply(3, 12)), const_100), multiply(12, 12)) | at a special sale , 12 tickets can be purchased for the price of 3 tickets . if 12 tickets are purchased at the sale , the amount saved will be what percent of the original price of the 12 tickets ? | "let the price of a ticket be rs . 100 , so 3 tickets cost 300 & 12 tickets cost 1200 12 tickets purchased at price of 3 tickets ie . , for 300 , so amount saved s rs . 900 , % of 5 tickets = ( 900 / 1200 ) * 100 = 75 % answer : e" | a = 12 * 12
b = 3 * 12
c = a - b
d = c * 100
e = 12 * 12
f = d / e
|
a ) 6 % , b ) 25 % , c ) 37 1 / 2 % , d ) 60 % , e ) 75 % | e | multiply(add(divide(const_1, 40), divide(const_1, 20)), const_100) | if x > 0 , x / 40 + x / 20 is what percent of x ? | "just plug and chug . since the question asks for percents , pick 100 . ( but any number will do . ) 100 / 40 + 100 / 20 = 2.5 + 5 = 7.5 7.5 is 75 % of 100 = e" | a = 1 / 40
b = 1 / 20
c = a + b
d = c * 100
|
a ) 1 : 2 , b ) 2 : 3 , c ) 2 : 4 , d ) 2 : 1 , e ) 2 : 9 | a | divide(subtract(15.8, 15.4), subtract(16.6, 15.8)) | the average age of students of a class is 15.8 years . the average age of boys in the class is 16.6 years and that of the girls is 15.4 years . the ration of the number of boys to the number of girls in the class is : | "let the ratio be k : 1 . then , k * 16.6 + 1 * 15.4 = ( k + 1 ) * 15.8 = ( 16.6 - 15.8 ) k = ( 15.8 - 15.4 ) = k = 0.4 / 0.6 = 1 / 2 required ratio = 1 / 1 : 1 = 1 : 2 . answer : a" | a = 15 - 8
b = 16 - 6
c = a / b
|
a ) 6 : 4 , b ) 6 : 14 , c ) 4 : 4 , d ) 4 : 6 , e ) 1 : 3 | e | divide(divide(subtract(multiply(50, divide(60, const_100)), multiply(50, divide(50, const_100))), subtract(divide(60, const_100), divide(20, const_100))), subtract(50, divide(subtract(multiply(50, divide(60, const_100)), multiply(50, divide(50, const_100))), subtract(divide(60, const_100), divide(20, const_100))))) | solution a is 20 % salt and solution b is 60 % salt . if you have 30 ounces of solution a and 60 ounces of solution b , in what ratio could you mix solution a with solution b to produce 50 ounces of a 50 % salt solution ? | "forget the volumes for the time being . you have to mix 20 % and 80 % solutions to get 50 % . this is very straight forward since 50 is int he middle of 20 and 80 so we need both solutions in equal quantities . if this does n ' t strike , use w 1 / w 2 = ( a 2 - aavg ) / ( aavg - a 1 ) w 1 / w 2 = ( 60 - 50 ) / ( 50 -... | a = 60 / 100
b = 50 * a
c = 50 / 100
d = 50 * c
e = b - d
f = 60 / 100
g = 20 / 100
h = f - g
i = e / h
j = 60 / 100
k = 50 * j
l = 50 / 100
m = 50 * l
n = k - m
o = 60 / 100
p = 20 / 100
q = o - p
r = n / q
s = 50 - r
t = i / s
|
a ) 22.8 kg , b ) 25.6 kg , c ) 28 kg , d ) 30.4 kg , e ) none of these | d | divide(multiply(8, 42.75), 11.25) | if 11.25 m of a uniform steel rod weighs 42.75 kg . what will be the weight of 8 m of the same rod ? | explanation : let the required weight be x kg . then , less length , less weight ( direct proportion ) = > 11.25 : 8 : : 42.75 : x = > 11.25 x x = 8 x 42.75 = > x = ( 8 x 42.75 ) / 11.25 = > x = 30.4 answer : d | a = 8 * 42
b = a / 11
|
a ) 70 , b ) 125 , c ) 300 , d ) 500 , e ) none of these | a | divide(divide(multiply(const_100, multiply(const_1000, 66)), multiply(const_60, const_1)), multiply(multiply(const_2, 250), add(const_3, divide(add(const_2, multiply(const_3, const_4)), power(add(const_2, multiply(const_4, const_2)), const_2))))) | the radius of the wheel of a bus is 250 cms and the speed of the bus is 66 km / h , then the r . p . m . ( revolutions per minutes ) of the wheel is | "radius of the wheel of bus = 250 cm . then , circumference of wheel = 2 ï € r = 500 ï € = 1571.43440 cm distance covered by bus in 1 minute = 66 â „ 60 ã — 1000 ã — 100 cms distance covered by one revolution of wheel = circumference of wheel = 1571.45 cm â ˆ ´ revolutions per minute = 6600000 / 60 ã — 1571.43 = 70 a... | a = 1000 * 66
b = 100 * a
c = const_60 * 1
d = b / c
e = 2 * 250
f = 3 * 4
g = 2 + f
h = 4 * 2
i = 2 + h
j = i ** 2
k = g / j
l = 3 + k
m = e * l
n = d / m
|
a ) 12.9 , b ) 12.5 , c ) 12.6 , d ) 11.6 , e ) 12.1 | d | divide(multiply(21, 1000), add(1000, 800)) | 1000 men have provisions for 21 days . if 800 more men join them , for how many days will the provisions last now ? | "1000 * 21 = 1800 * x x = 11.6 answer : d" | a = 21 * 1000
b = 1000 + 800
c = a / b
|
a ) 160 , b ) 220 , c ) 360 , d ) 440 , e ) 560 | b | add(40, multiply(divide(multiply(40, 3), const_2), 3)) | a rectangular wall is covered entirely with two kinds of decorative tiles : regular and jumbo . 1 / 3 of the tiles are jumbo tiles , which have a length three times that of regular tiles and have the same ratio of length to width as the regular tiles . if regular tiles cover 40 square feet of the wall , and no tiles ov... | "the number of jumbo tiles = x . the number of regular tiles = 2 x . assume the ratio of the dimensions of a regular tile is a : a - - > area = a ^ 2 . the dimensions of a jumbo tile is 3 a : 3 a - - > area = 9 a ^ 2 . the area of regular tiles = 2 x * a ^ 2 = 40 . the area of jumbo tiles = x * 9 a ^ 2 = 4.5 ( 2 x * a ... | a = 40 * 3
b = a / 2
c = b * 3
d = 40 + c
|
a ) 24 , b ) 30 , c ) 48 , d ) 54 , e ) 72 | d | divide(factorial(subtract(add(const_4, 210), const_1)), multiply(factorial(210), factorial(subtract(const_4, const_1)))) | how many positive integers less than 10,000 are such that the product of their digits is 210 ? | "210 = 2 x 5 x 3 x 7 = 5 x 6 x 7 x 1 = 5 x 6 x 7 those are the only sets of digits we can use to for the numbers ( any other combination of factors will have two digit factors ) . numbers using 2,5 , 3,7 = 4 ! numbers using 5,6 , 7,1 = 4 ! numbers using 5 , 6,7 ( 3 - digit numbers ) = 3 ! answer = 24 + 24 + 6 = 54 answ... | a = 4 + 210
b = a - 1
c = math.factorial(b)
d = math.factorial(210)
e = 4 - 1
f = math.factorial(e)
g = d * f
h = c / g
|
a ) 1 / 3 , b ) 2 / 5 , c ) 1 / 4 , d ) 2 / 3 , e ) 4 / 5 | a | divide(1, 3) | find the last term of a g . p whose first term is 9 and common ratio is ( 1 / 3 ) if the sum of the terms of the g . p is ( 40 / 3 ) | sum of the g . p . = ( first term - r * last term ) / 1 – r 40 / 3 = 9 – 1 / 3 ( last term ) / 2 / 3 last term = ( - 40 / 3 * 2 / 3 + 9 ) * 3 = - 80 / 3 + 27 = 1 / 3 answer : a | a = 1 / 3
|
a ) 105 , b ) 110 , c ) 120 , d ) 130 , e ) 140 | a | divide(subtract(multiply(36, 120), multiply(120, 15)), subtract(39, 15)) | the average of marks obtained by 120 boys was 36 . if the average of marks of passed boys was 39 and that of failed boys was 15 , the number of boys who passed the examination is ? | "let the number of boys who passed = x . then , 39 x x + 15 x ( 120 - x ) = 120 x 36 24 x = 4320 - 1800 = > x = 2520 / 24 x = 105 . hence , the number of boys passed = 105 . answer : a" | a = 36 * 120
b = 120 * 15
c = a - b
d = 39 - 15
e = c / d
|
a ) 74 , b ) 69 , c ) 75 , d ) 85 , e ) 90 | a | divide(add(add(add(add(76, 65), 82), 62), 85), add(const_1, const_4)) | shekar scored 76 , 65 , 82 , 62 and 85 marks in mathematics , science , social studies , english and biology respectively . what are his average marks ? | "explanation : average = ( 76 + 65 + 82 + 62 + 85 ) / 5 = 370 / 5 = 74 hence average = 74 answer : a" | a = 76 + 65
b = a + 82
c = b + 62
d = c + 85
e = 1 + 4
f = d / e
|
a ) 46 m , b ) 66 m , c ) 26 m , d ) 24 m , e ) 25 m | d | divide(multiply(add(10, 50), const_2), 5) | a rectangular plot measuring 10 meters by 50 meters is to be enclosed by wire fencing . if the poles of the fence are kept 5 meters apart . how many poles will be needed ? | "perimeter of the plot = 2 ( 10 + 50 ) = 120 m no of poles = 120 / 5 = 24 m answer : d" | a = 10 + 50
b = a * 2
c = b / 5
|
a ) 12 , b ) 13 , c ) 16 , d ) 14 , e ) 18 | d | divide(subtract(const_1, multiply(5, divide(const_1, 15))), divide(const_1, 21)) | x can finish a work in 21 days . y can finish the same work in 15 days . y worked for 5 days and left the job . how many days does x alone need to finish the remaining work ? | "work done by x in 1 day = 1 / 21 work done by y in 1 day = 1 / 15 work done by y in 5 days = 5 / 15 = 1 / 3 remaining work = 1 – 1 / 3 = 2 / 3 number of days in which x can finish the remaining work = ( 2 / 3 ) / ( 1 / 21 ) = 14 d" | a = 1 / 15
b = 5 * a
c = 1 - b
d = 1 / 21
e = c / d
|
a ) $ 1600 , b ) $ 6000 , c ) $ 6050 , d ) $ 7050 , e ) $ 4000 | e | divide(subtract(1440, multiply(multiply(multiply(const_3, multiply(const_2, const_3)), const_1000), divide(8, const_100))), subtract(divide(16, const_100), divide(8, const_100))) | country c imposes a two - tiered tax on imported cars : the first tier imposes a tax of 16 % of the car ' s price up to a certain price level . if the car ' s price is higher than the first tier ' s level , the tax on the portion of the price that exceeds this value is 8 % . if ron imported a $ 14,000 imported car and ... | "let t be the tier price , p be total price = 14000 per the given conditions : 0.16 t + 0.08 ( p - t ) = 1440 - - - - > t = 8000 . e is the correct answer ." | a = 2 * 3
b = 3 * a
c = b * 1000
d = 8 / 100
e = c * d
f = 1440 - e
g = 16 / 100
h = 8 / 100
i = g - h
j = f / i
|
a ) 0 , b ) 1 , c ) 2 , d ) 4 , e ) 5 | a | divide(5, 5) | what is the remainder when the number r = 14 ^ 2 * 15 ^ 8 is divided by 5 ? | "14 ^ 2 has units digit 6 15 ^ 8 has units digit 5 thus r = 14 ^ 2 * 15 ^ 8 has units digit 0 and will be divisible by 5 . the remainder will be zero answer : ( a )" | a = 5 / 5
|
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