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
|
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
|
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 running simultaneously , pipe q will drain 9 / 25 of the liquid . thus answer = c ." | 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
|
['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 foil covered prism ? | 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 e . | 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 ) or 20 + x = ( 100 + x ) 3 / 10 200 + 10 x = 300 + 3 x 7 x = 100 x = 100 / 7 ans : a | a = 4 + 3
b = 100 / a
|
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(6)
u = math.factorial(5)
v = t / u
w = s + v
|
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(multiply(3, 4), 4)), multiply(const_10, const_100)))), divide(multiply(4070, divide(multiply(9000, subtract(multiply(3, 4), 4)), 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(multiply(3, 4), 4)), multiply(const_10, const_100))))) | 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 = 4070 * B
D = 3 * 4
E = D * 18000
F = 10 * 100
G = E / F
H = 3 * 4
I = H - 3
J = 12000 * I
K = 10 * 100
L = J / K
M = G + L
N = 3 * 4
O = N - 4
P = 9000 * O
Q = 10 * 100
R = P / Q
S = M + R
T = C / S
U = w - T
|
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 . . . . . . . . . . . . . . . . . . 4 year elapsed answer : b" | a = 4 + 4
|
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 interest earned for 3 years = 39 * 3 = 117 amount invested = 815 - 117 = 698 answer : c" | 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 cover and the remaining are hard - cover . if there are a total of 7 hard - cover books among the maths and physics books . what is the probability r , that a book selected at random is either a hard cover book or a chemistry book ? | "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 = p - 4 m = 2 p p = p then ( p - 4 ) + 2 p + p = 20 , so 4 p = 24 and p = 6 . that means that there are 12 math , 6 physics , and 2 chemistry books on the shelf . with those numbers , you also know that there are 8 total hardcovers , 1 of which is chemistry . so if your goal is to get either a hardcover or a chemistry , there are 9 ways towin - either one of the 7 hardcovers that are n ' t chemistry or the two chemistry books . so out of the 20 total , r = 9 provide the desired outcome , making the answer e ." | 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 difference , in minutes , between the amount of time it takes tom to cover half of the distance that linda has covered and the amount of time it takes tom to cover twice the distance that linda has covered ? | "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 seconds = 15 x 60 s = 900 sec 3 . calculate : distance = 1.66 x 900 = 1500 m answer is c" | 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 way . 1 ^ 2 = 1 2 ^ 2 = 4 3 ^ 2 = 9 4 ^ 2 = 16 5 ^ 2 = 25 6 ^ 2 = 36 7 ^ 2 = 49 8 ^ 2 = 64 the only pair that matches is 8 ^ 2 + 5 ^ 2 = 89 . so 8 ^ 3 + 5 ^ 3 = 637 . answer e . )" | 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)
|
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
|
['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
|
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
|
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
|
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
|
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
|
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
|
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
|
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
|
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)), divide(560, multiply(divide(8, const_100), 4))) | 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
|
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 ( 4 / 4 = 1 ) but in this case we must select its matching pair from set b ( the number of the same color ) and since there are only one matching pair of this particular number in b then the probability of this is 1 / 5 . so , overall : q = 1 * 1 / 5 . answer : b . | a = 4 / 5
b = 1 / 4
c = a * b
|
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
|
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 used each of 10 digits equal # of times , thus we used each digit ( including 3 ) 3000 / 10 = 300 times . answer : c ." | 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 / (
|
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 driveway ? | "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 correct answer is a ." | 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
|
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 number of points of 11 players is 7 * 11 = 77 . therefore , the maximum number of points t for 12 th player is 100 - 77 = 23 . answer : e ." | a = 12 * 7
b = 100 - a
c = b + 7
|
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
|
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
|
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
|
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
|
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
|
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
|
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 . solving for only last two digits and multiplying them we get : ( ( 06 * 04 * 46 ) ( 47 * 01 ) ) ^ 2 = ( 44 * 47 ) ^ 2 = 68 ^ 2 = 24 hence answer is e" | 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
|
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 the remaining 10 will be females who are married . total females = 30 married females = 10 then single females = 30 - 10 = 20 we need to find the fraction of female students who are single i . e single female students / total female student = 10 / 30 = 1 / 3 [ b ]" | 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 so ( 150 + l ) / 7.50 = ( l ) / 2.5 i . e . 3 l = l + 150 i . e . 2 l = 150 i . e . l = 75 answer : option b | 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 - 20 ) = 1 / 3 so the volume of the two solutions will be equal . answer has to be 1 : 3 e ." | 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 answer 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 overlap , what is the area of the entire wall ? | "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 ^ 2 ) = 4.5 * 40 = 180 . total area = 40 + 180 = 220 . answer : b ." | 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 answer is ( d )" | 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 ended up paying $ 1440 in taxes , what is the first tier ' s price level ? | "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
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.