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a ) 45 , b ) 50 , c ) 55 , d ) 60 , e ) 75
e
multiply(divide(multiply(divide(1, 12), const_3), divide(divide(1, 30), const_2)), 5)
in the standard formulation of a flavored drink the ratio by volume of flavoring to corn syrup to water is 1 : 12 : 30 . in the sport formulation , the ratio of flavoring to corn syrup is three times as great as in the standard formulation , and the ratio of flavoring to water is half that of the standard formulation . if a large bottle of the sport formulation contains 5 ounces of corn syrup , how many ounces of water does it contain ?
"f : c : w 1 : 12 : 30 sport version : f : c 3 : 12 f : w 1 : 60 or 3 : 180 so c : f : w = 12 : 3 : 180 c / w = 12 / 180 = 3 ounces / x ounces x = 5 * 180 / 12 = 75 ounces of water e"
a = 1 / 12 b = a * 3 c = 1 / 30 d = c / 2 e = b / d f = e * 5
a ) 20 , b ) 24 , c ) 30 , d ) 12 , e ) 48
d
multiply(multiply(multiply(negate(2), 1), 1), subtract(negate(2), 1))
what is the value of x ^ 2 yz βˆ’ xyz ^ 2 , if x = - 1 , y = 1 , and z = 3 ?
"1 * 1 * 3 - ( - 1 * 1 * 9 ) = 3 + 9 = 12 ans : d"
a = negate * ( b = a * 1 c = b * 1
a ) 58 , b ) 60 , c ) 62 , d ) 64 , e ) 66
d
add(subtract(100, multiply(2, 19)), 2)
a batsman in his 19 th inning makes a score of 100 and their by increasing his average by 2 . what is his average after the 19 th inning ?
"18 x + 100 = 19 ( x + 2 ) x = 62 + 2 = 64 answer : d"
a = 2 * 19 b = 100 - a c = b + 2
a ) 48 days , b ) 36 days , c ) 28 days , d ) 32 days , e ) 22 days
b
add(divide(multiply(12, const_10), add(const_4, const_1)), 12)
b takes 12 more days than a to finish a task . b and a start this task and a leaves the task 12 days before the task is finished . b completes 60 % of the overall task . how long would b have taken to finish the task if he had worked independently ?
detailed solution let us say a and b split their share of the task and started doing their respective shares simultaneously . let ’ s say a takes a days to finish the task . therefore , b takes a + 12 days to finish the entire task . a has to finish 40 % of the task , since b is doing the rest . so a will only take 2 a / 5 number of days . b only has to finish 60 % of the task , so b will take ( 3 ( a + 12 ) / 5 ) number of days . but as we know , b starts working along with a and finishes 12 days after a stops working . so , ( 3 ( a + 12 ) / 5 ) = ( ( 2 a ) / 5 + 12 ) 3 a + 36 = 2 a + 60 a = 24 ; b = 36 days . answer choice b
a = 12 * 10 b = 4 + 1 c = a / b d = c + 12
a ) 6 sec . , b ) 7 sec . , c ) 9 sec . , d ) 11 sec . , e ) none
a
divide(100, multiply(add(68, 8), const_0_2778))
a train 100 m long is running with a speed of 68 kmph . in what time will it pass a man who is running at 8 kmph in the same direction in which the train is going ?
"solution speed of the train relative to man = ( 68 - 8 ) = 60 kmph = 60 x 5 / 18 = 50 / 3 m / sec . time taken by it to cover 100 m at ( 50 / 3 ) m / sec = ( 100 x 3 / 50 ) sec = 6 sec . answer a"
a = 68 + 8 b = a * const_0_2778 c = 100 / b
a ) 347 , b ) 480 , c ) 288 , d ) 278 , e ) 131
b
divide(1, 2)
with a uniform speed a car covers the distance in 8 hours . had the speed been increased by 4 km / hr , the same distance could have been covered in 7 1 / 2 hours . what is the distance covered ?
"let the distance be x km . then , x / ( 7 1 / 2 ) - x / 8 = 4 2 x / 15 - x / 8 = 4 = > x = 480 km . answer : b"
a = 1 / 2
a ) 7.25 , b ) 8.5 , c ) 8 , d ) 9.15 , e ) 10
c
divide(subtract(324, multiply(20, 4.2)), 30)
in the first 20 overs of a cricket game , the run rate was only 4.2 . what should be the run rate in the remaining 30 overs to reach the target of 324 runs ?
"explanation : required run rate = ( 324 - 4.2 20 ) = 240 / 30 = 8 . answer : c"
a = 20 * 4 b = 324 - a c = b / 30
a ) 1 , b ) 2 , c ) 4 , d ) - 2 , e ) - 4
c
subtract(divide(add(subtract(add(11, 6), const_4), const_1), const_2), const_3)
x and y are integers . a + b < 11 , and a > 6 . what is the smallest possible value of a - b ?
focus on the transition points and plug in the values . a > 6 so a could be 7 , 8 , 9 , 10 , 11 etc look at a = 7 7 + b < 11 b < 4 b could be 3 , 2 , 1 , 0 , - 1 etc when b is 3 , we get a - b = 4 this must be the smallest value because as a increases , b reduces so a - b increases . take another example . a = 11 so b < 0 . a - b will give a value greater than 11 . answer ( c ) .
a = 11 + 6 b = a - 4 c = b + 1 d = c / 2 e = d - 3
a ) 145 , b ) 150 , c ) 195 , d ) 160 , e ) 130
c
multiply(divide(65, 20), const_60)
a mobile battery in 1 hour charges to 20 percent . how much time ( in minute ) will it require more to charge to 65 percent .
1 hr = 20 percent . thus 15 min = 5 percent . now to charge 65 percent 195 min . answer : c
a = 65 / 20 b = a * const_60
a ) 4 , b ) 7 , c ) 8 , d ) 11 , e ) 12
d
add(add(add(7, 2), divide(5, 5)), divide(3, 3))
if x is the product of the positive integers from 1 to 8 , inclusive , and if i , k , m , and p are positive integers such that x = 2 ^ i * 3 ^ k * 5 ^ m * 7 ^ p , then i + k + m + p =
given that x = 8 ! = 2 ^ 7 βˆ— 3 ^ 2 βˆ— 5 βˆ— 7 x . hence , x = 2 ^ 7 βˆ— 3 ^ 2 βˆ— 5 ^ 1 βˆ— 7 ^ 1 = 2 ^ i βˆ— 3 ^ k βˆ— 5 ^ m βˆ— 7 p , since i , k , m , and p are positive integers , then we can equate the exponents , so we have that i = 7 , k = 2 , m = 1 , and p = 1 therefore , i + k + m + p = 7 + 2 + 1 + 1 = 11 answer : d .
a = 7 + 2 b = 5 / 5 c = a + b d = 3 / 3 e = c + d
['a ) 4', 'b ) 5', 'c ) 6', 'd ) 7', 'e ) 8']
d
divide(subtract(sqrt(subtract(const_1, multiply(const_4, divide(divide(subtract(const_4, add(const_4, add(multiply(const_2, 8), multiply(power(8, const_2), divide(const_10, const_2))))), const_3), const_2)))), const_1), const_2)
find the base k of the number system , if ( 524 ) 8 = ( 664 ) k ?
5 * 8 ^ 2 + 2 * 8 ^ 1 + 4 * 8 ^ 0 = 6 * k ^ 2 + 6 * k ^ 1 + 4 * k ^ 0 320 + 16 + 4 = 6 ( k ^ 2 ) + 6 k + 4 340 = 6 ( k ^ 2 ) + 6 k + 4 6 ( k ^ 2 ) + 6 k + 4 - 340 = 0 6 ( k ^ 2 ) + 6 k - 336 = 0 dividing by 6 ; ( k ^ 2 ) + k - 56 = 0 solving this quadratic equation we get , k = 7 and k = - 8 thus k = 7 answer : d
a = 2 * 8 b = 8 ** 2 c = 10 / 2 d = b * c e = a + d f = 4 + e g = 4 - f h = g / 3 i = h / 2 j = 4 * i k = 1 - j l = math.sqrt(k) m = l - 1 n = m / 2
['a ) rs . 4000', 'b ) rs . 4450', 'c ) rs . 4457', 'd ) rs . 4560', 'e ) rs . 5457']
c
multiply(circumface(multiply(sqrt(divide(17, const_pi)), const_100)), 3)
an area of a circularplace is 17 hectares . find the cost of fencing it at the rate of rs . 3 permeter approximately .
area = ( 17.56 x 10000 ) m 2 = 175600 m 2 . Ο€ r 2 = 175600 ⇔ ( r ) 2 = ( 175600 x ( 7 / 22 ) ) ⇔ r = 236.37 m . circumference = 2 Ο€ r = ( 2 x ( 22 / 7 ) x 236.37 ) m = 1485.78 m . cost of fencing = rs . ( 1485.78 x 3 ) = rs . 4457 . c
a = 17 / math.pi b = math.sqrt(a) c = b * 100 d = circumface * (
a ) $ 2.40 , b ) $ 3.00 , c ) $ 3.40 , d ) $ 3.60 , e ) $ 4.80
c
divide(0.90, subtract(const_1, add(divide(3, 5), multiply(divide(const_1, 3), subtract(const_1, divide(3, 5))))))
having received his weekly allowance , john spent 3 / 5 of his allowance at the arcade . the next day he spent one third of his remaining allowance at the toy store , and then spent his last $ 0.90 at the candy store . what is john ’ s weekly allowance ?
"total allowance = x amount spent at the arcade = 3 / 5 x amount remaining = 2 / 5 x amount spent at the toy store = 2 / 5 * 1 / 3 x = 2 / 15 x amount remaining = 2 / 5 x - 2 / 15 x = 4 / 15 x now , 4 / 15 x = $ 0.9 therefore , x = $ 3.40 . answer c"
a = 3 / 5 b = 1 / 3 c = 3 / 5 d = 1 - c e = b * d f = a + e g = 1 - f h = 0 / 90
a ) 7 meters , b ) 9 meters , c ) 4 meters , d ) 5 meters , e ) 3 meters
c
multiply(divide(96, multiply(multiply(multiply(const_2, const_3), const_10), 400)), const_1000)
if a tire rotates at 400 revolutions per minute when the car is traveling 96 km / h , what is the circumference of the tire ?
"400 rev / minute = 400 * 60 rev / 60 minutes = 24,000 rev / hour 24,000 * c = 96,000 m : c is the circumference c = 4 meters correct answer c"
a = 2 * 3 b = a * 10 c = b * 400 d = 96 / c e = d * 1000
a ) 16 % , b ) 32 % , c ) 48 % , d ) 84 % , e ) 92 %
d
subtract(const_100, divide(subtract(const_100, 68), const_2))
a certain characteristic in a large population has a distribution that is symmetric about the mean m . if 68 % of the distribution lies one standard deviation f of the mean , what percent of the distribution is less than m + f ?
"16 % ________________________________________________ m + f 34 % ________________________________________________ m 34 % ________________________________________________ m - f 16 % since 68 % lies one standard deviation from mean m , = > 50 % of 68 % lies on either side as it is symmetric about m . thus 16 % lie below m - f and 16 % lie above m + f now below m + f = 16 + 34 + 34 = 84 % hence d"
a = 100 - 68 b = a / 2 c = 100 - b
a ) 70 , b ) 65 , c ) 80 , d ) 140 , e ) 60
a
add(divide(subtract(100, 40), const_2), 40)
the result when a number subtracted from 100 is the same as the number added to 40 . what is the number ?
"answer let the number be x . then , x - 40 = 100 - x Γ’ ‑ ’ 2 x = 100 + 40 = 140 Γ’ ‑ ’ 2 x = 140 Γ’ Λ† Β΄ x = 70 correct option : a"
a = 100 - 40 b = a / 2 c = b + 40
a ) 4 , b ) 7 , c ) 5 , d ) 3 , e ) 6
c
divide(add(multiply(5, 6), multiply(4, 5)), subtract(multiply(5, 6), multiply(4, 5)))
if a / b = 6 / 5 , then ( 5 a + 4 b ) / ( 5 a - 4 b ) = ?
"answer dividing numerator as well as denominator by b , we get given exp . = ( 5 a + 4 b ) / ( 5 a - 4 b ) = ( 5 a / b + 4 ) / ( 5 a / b - 4 ) since a / b = 6 / 5 this implies that = [ ( 5 * 6 ) / 5 + 4 ] / [ ( 5 * 6 ) / 5 - 4 ) ] = ( 6 + 4 ) / ( 6 - 4 ) = 5 option : c"
a = 5 * 6 b = 4 * 5 c = a + b d = 5 * 6 e = 4 * 5 f = d - e g = c / f
a ) 4 , b ) 6 , c ) 8 , d ) 10 , e ) 12
b
multiply(multiply(1, 2), 3)
running at their respective constant rate , machine x takes 2 days longer to produce w widgets than machines y . at these rates , if the two machines together produce 5 w / 4 widgets in 3 days , how many days would it take machine x alone to produce 1 w widgets .
"i am getting 12 . e . hope havent done any calculation errors . . approach . . let y = no . of days taken by y to do w widgets . then x will take y + 2 days . 1 / ( y + 2 ) + 1 / y = 5 / 12 ( 5 / 12 is because ( 5 / 4 ) w widgets are done in 3 days . so , x widgets will be done in 12 / 5 days or 5 / 12 th of a widget in a day ) solving , we have y = 4 = > x takes 6 days to doing x widgets . answer : b"
a = 1 * 2 b = a * 3
a ) 4 , b ) 5 , c ) 3 , d ) 8 , e ) 9
c
subtract(add(add(add(add(add(add(add(const_4, const_3), add(const_3, const_3)), const_1), add(const_4, const_4)), const_2), 9), const_3), add(add(add(add(add(add(const_4, const_3), add(const_3, const_3)), const_1), add(const_4, const_4)), const_2), 9))
if the number 761 * 829 is completely divisible by 9 , then the smallest whole number in place of * will b
sum of digits = ( 7 + 6 + 1 + x + 8 + 2 + 9 ) = ( 33 + x ) , which must be divisible by 9 . x = 3 . c )
a = 4 + 3 b = 3 + 3 c = a + b d = c + 1 e = 4 + 4 f = d + e g = f + 2 h = g + 9 i = h + 3 j = 4 + 3 k = 3 + 3 l = j + k m = l + 1 n = 4 + 4 o = m + n p = o + 2 q = p + 9 r = i - q
a ) rs . 147.50 , b ) rs . 785.50 , c ) rs . 179.50 , d ) rs . 258.50 , e ) none of these
c
divide(subtract(multiply(157, add(add(1, 1), 2)), add(126, 135)), 2)
teas worth rs . 126 per kg and rs . 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2 . if the mixture is worth rs 157 per kg , the price of the third variety per kg will be ?
explanation : since first and second varieties are mixed in equal proportions . so , their average price = rs . ( 126 + 135 ) / 2 . = > rs . 130.50 . so , the mixture is formed by mixing two varieties , one at rs . 130.50 per kg and the other at say , rs . x per kg in the ratio 2 : 2 , i . e . , 1 : 1 . we have to find x . by the rule of alligation , we have : cost of 1 kg cost of 1 kg of 1 st kind of 2 nd kind ( rs . 130.50 ) ( rs . x ) \ / mean price ( rs . 157 ) / \ x Γ’ Λ† ’ 157 22.50 = > x Γ’ Λ† ’ ( 157 / 22.50 ) = 1 . = > x Γ’ Λ† ’ 157 = 22.50 . = > x = 179.50 rs . answer : c
a = 1 + 1 b = a + 2 c = 157 * b d = 126 + 135 e = c - d f = e / 2
a ) 4 : 3 , b ) 1 : 2 , c ) 3 : 5 , d ) 1 : 4 , e ) 2 : 3
c
divide(sqrt(9), sqrt(25))
the sub - duplicate ratio of 9 : 25 is
root ( 9 ) : root ( 25 ) = 3 : 5 answer : c
a = math.sqrt(9) b = math.sqrt(25) c = a / b
a ) 18 , b ) 68 , c ) 67 , d ) 55 , e ) 56
a
add(add(add(multiply(3, const_3), 3), multiply(multiply(3, const_3), add(multiply(3, const_3), 3))), 3)
sushil got thrice as many marks in english as in science . his total marks in english , science and maths are 162 . if the ratio of his marks in english and maths is 3 : 5 , find his marks in science ?
"s : e = 1 : 3 e : m = 3 : 5 - - - - - - - - - - - - s : e : m = 3 : 9 : 15 3 / 27 * 162 = 18 answer : a"
a = 3 * 3 b = a + 3 c = 3 * 3 d = 3 * 3 e = d + 3 f = c * e g = b + f h = g + 3
a ) 79.2 km / hr , b ) 69 km / hr , c ) 74 km / hr , d ) 61 km / hr , e ) none of these
b
multiply(speed(264, 8), const_2)
a train moves past a post and a platform 264 m long in 8 seconds and 20 seconds respectively . what is the speed of the train ?
let x is the length of the train and v is the speed time taken to move the post = 8 s = > x / v = 8 = > x = 8 v - - - ( 1 ) time taken to cross the platform 264 m long = 20 s ( x + 264 ) / v = 20 = > x + 264 = 20 v - - - ( 2 ) substituting equation 1 in equation 2 , we get 8 v + 264 = 20 v = > v = 264 / 12 = 22 m / s = 22 Γ— 36 / 10 km / hr = 79.2 km / hr answer : b
a = speed * (
a ) 10 , b ) 15 , c ) 22 , d ) 20 , e ) 24
d
add(subtract(add(56, 72), subtract(120, 12)), subtract(72, 56))
of 120 applicants for a job , 56 had at least 4 years ' experience , 72 had degrees , and 12 had less than 4 years ' experience and did not have a degree . how many of the applicants had at least 4 years ' experience and a degree ?
"120 - 12 = 108 108 - 56 - 72 = - 20 then 20 are in the intersection between 4 years experience and degree . answer : d"
a = 56 + 72 b = 120 - 12 c = a - b d = 72 - 56 e = c + d
a ) a ) 34500 , b ) b ) 24000 , c ) c ) 24098 , d ) d ) 19007 , e ) e ) 18000
e
subtract(39000, multiply(const_60, const_100))
a started a business with an investment of rs . 70000 and after 6 months b joined him investing rs . 120000 . if the profit at the end of a year is rs . 39000 , then the share of b is ?
"ratio of investments of a and b is ( 70000 * 12 ) : ( 120000 * 6 ) = 7 : 6 total profit = rs . 39000 share of b = 6 / 13 ( 39000 ) = rs . 18000 answer : e"
a = const_60 * 100 b = 39000 - a
a ) 12000 , 20000 , b ) 12000 , 200098 , c ) 12000 , 20007 , d ) 12000 , 20006 , e ) 12000 , 20001
a
multiply(subtract(5, const_2), divide(32000, add(3, subtract(5, const_2))))
divide rs . 32000 in the ratio 3 : 5 ?
"3 / 8 * 32000 = 12000 5 / 8 * 32000 = 20000 answer : a"
a = 5 - 2 b = 5 - 2 c = 3 + b d = 32000 / c e = a * d
a ) 7 / 45 , b ) 2 / 45 , c ) 4 / 15 , d ) 8 / 45 , e ) 2 / 15
e
divide(2, 5)
what is the hcf of 2 / 3 , 4 / 15 and 6 / 5
"explanation : hcf of fractions = hcf of numerators / lcm of denominators = ( hcf of 2 , 4 , 6 ) / ( lcm of 3 , 15 , 5 ) = 2 / 15 answer : option e"
a = 2 / 5
a ) 127.6 , b ) 127.5 , c ) 127.0 , d ) 127.2 , e ) 127.3
b
subtract(600, divide(multiply(divide(multiply(subtract(1000, 100), 600), 1000), subtract(800, 100)), 800))
n a race of 1000 m , a can beat by 100 m , in a race of 800 m , b can beat c by 100 m . by how many meters will a beat c in a race of 600 m ?
when a runs 1000 m , b runs 900 m and when b runs 800 m , c runs 700 m . when b runs 900 m , distance that c runs = ( 900 * 700 ) / 800 = 6300 / 8 = 787.5 m . in a race of 1000 m , a beats c by ( 1000 - 787.5 ) = 212.5 m to c . in a race of 600 m , the number of meters by which a beats c = ( 600 * 212.5 ) / 1000 = 127.5 m . answer : ob
a = 1000 - 100 b = a * 600 c = b / 1000 d = 800 - 100 e = c * d f = e / 800 g = 600 - f
a ) 81 , b ) 82 , c ) 83 , d ) 84 , e ) 85
d
multiply(divide(3, 14,39), const_100)
3 , 14,39 , _ ?
"n ^ 3 + n ^ 2 + n soln : 1 ^ 3 + 1 ^ 2 + 1 = 3 2 ^ 3 + 2 ^ 2 + 2 = 8 + 4 + 2 = 14 3 ^ 3 + 3 ^ 2 + 3 = 27 + 9 + 3 = 39 4 ^ 3 + 4 ^ 2 + 4 = 64 + 16 + 4 = 84 answer : d"
a = 3 / 14 b = a * 100
a ) 29 : 7 , b ) 22 : 7 , c ) 26 : 7 , d ) 20 : 7 , e ) 26 : 5
c
divide(add(power(2, 4), 4), add(add(3, 5), 4))
if x : y = 3 : 5 , find the value of ( 2 x + 4 y ) : ( 4 x – y )
"explanation : given : a / b = 3 / 5 ( 2 * 3 + 4 * 5 ) : ( 4 * 3 – 5 ) = 26 : 7 answer : c"
a = 2 ** 4 b = a + 4 c = 3 + 5 d = c + 4 e = b / d
a ) 235 cm 2 , b ) 260 cm 2 , c ) 266 cm 2 , d ) 270 cm 2 , e ) 280 cm 2
c
divide(multiply(14, add(20, 18)), const_2)
nd the area of trapezium whose parallel sides are 20 cm and 18 cm long , and the distance between them is 14 cm ?
"area of a trapezium = 1 / 2 ( sum of parallel sides ) * ( perpendicular distance between them ) = 1 / 2 ( 20 + 18 ) * ( 14 ) = 266 cm 2 answer : c"
a = 20 + 18 b = 14 * a c = b / 2
a ) 150 , b ) 200 , c ) 240 , d ) 300 , e ) 480
b
multiply(multiply(subtract(6, const_4), const_10), const_10)
a β€œ palindromic integer ” is an integer that remains the same when its digits are reversed . so , for example , 43334 and 516615 are both examples of palindromic integers . how many 6 - digit palindromic integers are both even and greater than 600,000 ?
"the first digit and last digit are the same so the 2 possibilities are 6 or 8 . the second and third digits can be any number from 0 to 9 . the total number of palindromic integers is 2 * 10 * 10 = 200 the answer is b ."
a = 6 - 4 b = a * 10 c = b * 10
a ) 6.16 , b ) 6.21 , c ) 6.25 , d ) 2.4 , e ) 2.7
c
multiply(divide(divide(multiply(divide(20, const_100), 100), 4), multiply(divide(20, const_100), 100)), const_100)
a reduction of 20 % in the price of salt enables a lady to obtain 4 kgs more for rs . 100 , find the original price per kg ?
"100 * ( 20 / 100 ) = 20 - - - 4 ? - - - 1 = > rs . 5 100 - - - 80 ? - - - 5 = > rs . 6.25 answer : c"
a = 20 / 100 b = a * 100 c = b / 4 d = 20 / 100 e = d * 100 f = c / e g = f * 100
a ) 3 / 80 , b ) 21 / 80 , c ) 5 / 16 , d ) 25 / 76 , e ) 9 / 80
b
divide(multiply(3, 3), add(multiply(const_60, 8), multiply(8, 8)))
kyle , david , and catherine each try independently to solve a problem . if their individual probabilities for success are 1 / 8 , 1 / 4 and 3 / 5 , respectively , what is the probability no one will solve the problem ?
"p ( kyle will not solve ) = 1 - 1 / 8 = 7 / 8 p ( david will not solve ) = 1 - 1 / 4 = 3 / 4 p ( catherine will not solve ) = 1 - 3 / 5 = 2 / 5 p = ( 7 / 8 ) * ( 3 / 4 ) * ( 2 / 5 ) = 42 / 160 = 21 / 80 answer : b"
a = 3 * 3 b = const_60 * 8 c = 8 * 8 d = b + c e = a / d
a ) 40 , b ) 56 , c ) 60 , d ) 62 , e ) 70
b
multiply(add(6, const_1), add(7, const_1))
if p and q are prime numbers , how many divisors does the product p ^ 6 * q ^ 7 have ?
"when a number n = a ^ x * b ^ y , where a and b are prime numbers , and x , y are positive integers , the number of divisors of n = ( x + 1 ) ( y + 1 ) therefore , the answer is b . 7 * 8 = 56"
a = 6 + 1 b = 7 + 1 c = a * b
a ) 1 / 3 , b ) 2 / 3 , c ) 1 / 4 , d ) 1 / 2 , e ) 3 / 5
d
divide(const_2, 4)
if there is an equal probability of a child being born a boy or a girl , what is the probability that a couple who have 4 children have two children of the same sex and one of the opposite sex ?
no of ways of selecting a gender - 2 c 1 no of ways of selecting any 2 children out of 4 = 4 c 2 total possible outcomes - 2 ^ 4 ( each child can be either a girl or a boy ) probability = 2 c 1 * 4 c 2 / 2 ^ 4 = 2 * 4 / 2 * 2 * 2 * 2 = 8 / 16 = 1 / 2 ans = d
a = 2 / 4
a ) 288 , b ) 48 , c ) 72 , d ) 864 , e ) 964
a
multiply(divide(const_4, 3), power(3, 3))
the measurement of a rectangular box with lid is 25 cmx 12 cmx 18 cm . find the volume of the largest sphere that can be inscribed in the box ( in terms of Ο€ cm 3 ) . ( hint : the lowest measure of rectangular box represents the diameter of the largest sphere )
"d = 12 , r = 6 ; volume of the largest sphere = 4 / 3 Ο€ r 3 = 4 / 3 * Ο€ * 6 * 6 * 6 = 288 Ο€ cm 3 answer : a"
a = 4 / 3 b = 3 ** 3 c = a * b
a ) 600 , b ) 450 , c ) 350 , d ) 550 , e ) 400
e
divide(subtract(multiply(2000, 3), 4800), const_3)
in a tournament prize of 1 st , 2 nd , and 3 rd is given in same interval . if total amount of prize rs 4800 and prize of 1 st is rs 2000 , find the interval ? ?
let the interval is x . prize of 2 nd and 3 rd is 2000 - x and 2000 - 2 x 2000 + 2000 - x + 2000 - 2 x = 4800 x = 400 answer : e
a = 2000 * 3 b = a - 4800 c = b / 3
a ) 3 , b ) 1 , c ) 0 , d ) 6 , e ) 4
a
divide(log(1), log(power(8, 11)))
if n = 8 ^ 11 – 1 , what is the units digit of n ?
"8 ^ 11 - 8 = 8 ( 8 ^ 10 - 1 ) = = > 8 ( 2 ^ 30 - 1 ) last digit of 2 ^ 30 is 4 based on what explanation livestronger is saying . 2 ^ 30 - 1 yields 4 - 1 = 3 as the unit digit . now on multiply this with 1 , we get unit digit as 3 . answer : a"
a = math.log(1) b = 8 ** 11 c = math.log(b) d = a / c
a ) 10 % , b ) 15 % , c ) 5 % , d ) 7 % , e ) 20 %
c
divide(subtract(720, 600), multiply(600, divide(4, const_100)))
$ 600 will become $ 720 in 4 years find the rate of interest ?
"si = simple interest = a - p = 720 - 600 = $ 120 r = 100 si / pt = 100 * 120 / 600 * 4 = 5 % answer is c"
a = 720 - 600 b = 4 / 100 c = 600 * b d = a / c
['a ) 400 %', 'b ) 200 %', 'c ) 300 %', 'd ) 100 %', 'e ) 800 %']
d
multiply(const_100, divide(const_2, const_2))
the area of a circle is increased by 300 % . by what percent has the diameter of the circle increased ?
the area of the circle is increased by 300 % , thus the area is increased 4 times . the area of a circle it proportional to the square of the diameter ( area = Ο€ d ^ 2 / 4 ) , therefore the diameter must increase 2 times ( diameter increase 2 times = area increase 4 times ) , which is increase by 100 % . answer : d .
a = 2 / 2 b = 100 * a
a ) 32 , b ) 87 , c ) 36 , d ) 99 , e ) 77
c
add(floor(divide(multiply(multiply(21, 8), multiply(18, 3)), multiply(multiply(21, 2), 6))), const_1)
18 men take 21 days of 8 hours each to do a piece of work . how many days of 6 hours each would 21 women take to do the same . if 3 women do as much work as 2 men ?
"3 w = 2 m 18 m - - - - - - 21 * 8 hours 21 w - - - - - - x * 6 hours 14 m - - - - - - x * 6 18 * 21 * 8 = 14 * x * 6 x = 36 answer : c"
a = 21 * 8 b = 18 * 3 c = a * b d = 21 * 2 e = d * 6 f = c / e g = math.floor(f) h = g + 1
a ) rs . 1075 , b ) rs . 975 , c ) rs . 875 , d ) rs . 775 , e ) none of these
c
multiply(7000, divide(50, const_100))
find the simple interest on rs 7000 at 50 / 3 % for 9 months
"explanation : s . i . = p Γ— r Γ— t 100 so , by pu Ζ« ing the values in the above formula , our result will be . required result = 7000 Γ— 50 Γ— 93 Γ— 12 Γ— 100 = 875 [ please note that we have divided by 12 as we converted 9 months in a year format ] answer : c"
a = 50 / 100 b = 7000 * a
a ) p ^ 2 , b ) 4 p , c ) p ^ 2 / 4 , d ) p / 16 , e ) p ^ 2 / 16
a
multiply(multiply(multiply(divide(divide(multiply(sqrt(4), const_4), const_pi), const_2), divide(divide(multiply(sqrt(4), const_4), const_pi), const_2)), const_pi), const_2)
what is the area of a square with perimeter 4 p ?
"each side is p a = ( p ) ^ 2 answer a"
a = math.sqrt(4) b = a * 4 c = b / math.pi d = c / 2 e = math.sqrt(4) f = e * 4 g = f / math.pi h = g / 2 i = d * h j = i * math.pi k = j * 2
a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11
e
divide(add(add(9, 7), 6), 2)
a school has 9 english 7 history and 6 geography teachers each teacher can teach 2 subjects max what is he minimum number of teachers required
total subjects = 9 + 7 + 6 = 22 max subjects by 1 teacher = 2 so , min of teachers required = 22 / 2 = 11 answer : e
a = 9 + 7 b = a + 6 c = b / 2
a ) 230 m , b ) 270 m , c ) 643 m , d ) 832 m , e ) 290 m
e
multiply(subtract(26, divide(230, multiply(const_0_2778, 72))), multiply(const_0_2778, 72))
a goods train runs at the speed of 72 kmph and crosses a 230 m long platform in 26 seconds . what is the length of the goods train ?
"speed = ( 72 x 5 / 18 ) m / sec = 20 m / sec . time = 26 sec . let the length of the train be x metres . then , x + 230 / 26 = 20 x + 230 = 520 x = 290 . answer : e"
a = const_0_2778 * 72 b = 230 / a c = 26 - b d = const_0_2778 * 72 e = c * d
a ) 9 , b ) 12 , c ) 13 , d ) 14 , e ) 15
a
divide(2, subtract(divide(multiply(const_2, 2), 6), 1))
it takes joey the postman 1 hours to run a 2 mile long route every day . he delivers packages and then returns to the post office along the same path . if the average speed of the round trip is 6 mile / hour , what is the speed with which joey returns ?
"let his speed for one half of the journey be 3 miles an hour let the other half be x miles an hour now , avg speed = 5 mile an hour 2 * 2 * x / 3 + x = 6 4 x = 6 x + 18 = > 2 x = 18 x = 9 a"
a = 2 * 2 b = a / 6 c = b - 1 d = 2 / c
a ) 18 , b ) 12 , c ) 9 , d ) 4.5 , e ) 3
a
multiply(9, const_2)
the distance from the x - axis to point p is half the distance from the y - axis to point p . if the coordinates of p are ( x , - 9 ) , how many units is p from the y - axis ?
"the x - axis is 9 units from the point p . thus the y - axis is 18 units from the point p . the answer is a ."
a = 9 * 2
a ) 66.66 % , b ) 81.81 % , c ) 66.766 % , d ) 86.66 % , e ) 66.65 %
b
multiply(divide(45, subtract(const_100, 45)), const_100)
if a ' s height is 45 % less than that of b , how much percent b ' s height is more than that of a ?
"excess of b ' s height over a ' s = [ ( 45 / ( 100 - 45 ) ] x 100 % = 81.81 % answer : b )"
a = 100 - 45 b = 45 / a c = b * 100
a ) 50 , b ) 90 , c ) 100 , d ) 105 , e ) 140
b
subtract(subtract(490, divide(subtract(multiply(4, divide(490, 7)), multiply(4, divide(multiply(4, divide(490, 7)), add(4, 1)))), divide(1, 4))), multiply(4, divide(multiply(4, divide(490, 7)), add(4, 1))))
in an apartment building that has 490 units , 4 out of every 7 units are currently rented , including 1 / 4 of the one - bedroom units . if , of the rented apartments , there is a 6 : 1 ratio of two - bedroom units to one - bedroom units , and the building only consists of two - bedroom and one - bedroom units , how many two - bedroom units are not rented ?
"rented = 4 / 7 * 490 = 280 . rented two - bedroom = 6 / 7 * 280 = 240 ; rented one - bedroom = 1 / 7 * 280 = 40 ( or 280 - 240 = 40 ) . rented one - bedroom units are 1 / 4 of all one - bedroom : 1 / 4 * { all one bedroom } = 40 - - > { all one - bedroom } = 160 { all two - bedroom } = 490 - 160 = 330 two - bedroom not rented = 330 - 240 = 90 b"
a = 490 / 7 b = 4 * a c = 490 / 7 d = 4 * c e = 4 + 1 f = d / e g = 4 * f h = b - g i = 1 / 4 j = h / i k = 490 - j l = 490 / 7 m = 4 * l n = 4 + 1 o = m / n p = 4 * o q = k - p
a ) 1063800 , b ) 1083600 , c ) 1036800 , d ) 1053800 , e ) 1073600
c
multiply(factorial(6), factorial(6))
in how many ways 6 boys and 6 girls can be seated in a row so that boys and girls are alternate ?
"number of ways = 2 * 6 ! * 6 ! = 1036800 answer is c"
a = math.factorial(6) b = math.factorial(6) c = a * b
a ) 10 % , b ) 20 % , c ) 30 % , d ) 40 % , e ) 50 %
b
divide(multiply(12, const_4), add(const_4, const_1))
the cost price of 12 articles is equal to the selling price of 10 articles . what is the profit percent ?
"10 * sp = 12 * cp sp = 1.2 * cp the profit percent is 20 % . the answer is b ."
a = 12 * 4 b = 4 + 1 c = a / b
a ) 5 kmph , b ) 4 kmph , c ) 9 kmph , d ) 8 kmph , e ) 1 kmph
b
divide(subtract(divide(const_60.0, 2), divide(45, 2)), const_2)
a man rows his boat 65 km downstream and 45 km upstream , taking 2 1 / 2 hours each time . find the speed of the stream ?
"speed downstream = d / t = 65 / ( 2 1 / 2 ) = 26 kmph speed upstream = d / t = 45 / ( 2 1 / 2 ) = 18 kmph the speed of the stream = ( 26 - 18 ) / 2 = 4 kmph answer : b"
a = const_60 / 0 b = 45 / 2 c = a - b d = c / 2
a ) 2 . , b ) 4 . , c ) 5 . , d ) 7 , e ) 9 .
b
divide(subtract(9, const_1), const_2)
arnold and danny are two twin brothers that are celebrating their birthday . the product of their ages today is smaller by 9 from the product of their ages a year from today . what is their age today ?
"ad = ( a + 1 ) ( d + 1 ) - 9 0 = a + d - 8 a + d = 8 a = d ( as they are twin brothers ) a = d = 4 b is the answer"
a = 9 - 1 b = a / 2
a ) 7 , b ) 12 , c ) 5 , d ) 6 , e ) 8
b
multiply(27, 27)
27 * 27 * 27 * 27 = 3 ^ ?
"3 ^ 3 * 3 ^ 3 * 3 ^ 3 * 3 ^ 3 = 3 ^ ( 3 + 3 + 3 + 3 ) = 3 ^ 12 answer : 12 option : b"
a = 27 * 27
a ) 228 , b ) 720 , c ) 255 , d ) 199 , e ) 231
b
divide(multiply(divide(240, divide(subtract(60, subtract(const_100, 60)), const_100)), 60), const_100)
there were two candidates in an election . winner candidate received 60 % of votes and won the election by 240 votes . find the number of votes casted to the winning candidate ?
"w = 60 % l = 40 % 60 % - 40 % = 20 % 20 % - - - - - - - - 240 60 % - - - - - - - - ? = > 720 answer : b"
a = 100 - 60 b = 60 - a c = b / 100 d = 240 / c e = d * 60 f = e / 100
a ) 2 , b ) 1 , c ) 3 , d ) 5 , e ) 4
c
subtract(25, 20)
a factory producing tennis balls stores them in either big boxes , 25 balls per box , or small boxes , 20 balls per box . if 153 freshly manufactured balls are to be stored , what is the least number of balls that can be left unboxed ?
"we have to work with multiples of 20 and 25 . first , we must know the limits of this multiples , so : 152 / 25 = 6 . . . . so the max is 6 152 / 20 = 7 . . . so the max is 7 153 - 150 = 3 ( 6 big box ) answer : c"
a = 25 - 20
a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5
d
subtract(997, add(multiply(multiply(multiply(5, 9), 11), const_2), 3))
what least number should be subtracted from 997 so that the remainder when divided by 5 , 9 , and 11 will leave in each case the same remainder 3 ?
the lcm of 5,9 and 11 is 495 . the next multiple is 2 * 495 = 990 . 990 + { remainder } = 990 + 3 = 993 , which is 4 less than 997 . answer : d .
a = 5 * 9 b = a * 11 c = b * 2 d = c + 3 e = 997 - d
a ) $ 94.03 , b ) $ 96.75 , c ) $ 98.75 , d ) $ 102.07 , e ) $ 105.3
c
add(add(multiply(const_2, 32.25), multiply(const_1, 20.43)), multiply(const_1, 13.82))
a garden center sells a certain grass seed in 5 - pound bags at $ 13.82 per bag , 10 - pound bags at $ 20.43 per bag , and 25 - pound bags $ 32.25 per bag . if a customer is to buy at least 65 pounds of the grass seed , but no more than 80 pounds , what is the least possible cost of the grass seed that the customer will buy ?
"there can be 2 cases 1 ) 25 + 25 + 10 + 5 = $ 98.75 or 2 ) 25 + 25 + 25 = $ 96.75 c"
a = 2 * 32 b = 1 * 20 c = a + b d = 1 * 13 e = c + d
a ) 3 , b ) 5 , c ) 9 , d ) 7 , e ) 8
d
divide(14, subtract(const_4, const_2))
in a group of cows and chickens , the number of legs was 14 more than twice the number of heads . the number of cows was :
d 7 let the number of cows be x and their legs be 4 x . let the number of chicken be y and their legs be 2 x . total number of legs = 4 x + 2 y . total number of heads = x + y . the number of legs was 14 more than twice the number of heads . therefore , 2 Γ— ( x + y ) + 14 = 4 x + 2 y . or , 2 x + 2 y + 14 = 4 x + 2 y . or , 2 x + 14 = 4 x [ subtracting 2 y from both sides ] . or , 14 = 4 x – 2 x [ subtracting 2 x from both sides ] . or , 14 = 2 x . or , x = 7 [ dividing by 2 on both sides ] . therefore , the number of cows = 7 .
a = 4 - 2 b = 14 / a
a ) 16 , b ) 72 , c ) 112 , d ) 136 , e ) 142
d
multiply(2, divide(153, add(2, 16)))
water consists of hydrogen and oxygen , and the approximate ratio , by mass , of hydrogen to oxygen is 2 : 16 . approximately how many grams of oxygen are there in 153 grams of water ?
"since the ratio , by mass , of hydrogen to oxygen is 2 : 16 , then oxygen is 16 / ( 2 + 16 ) = 8 / 9 of mass of water . therefore there are 153 * 8 / 9 = 136 grams of oxygen in 153 grams of water . answer : d ."
a = 2 + 16 b = 153 / a c = 2 * b
a ) 0 , b ) 2 , c ) 5 , d ) 13 , e ) 17
e
add(multiply(add(const_2, const_3), const_2), divide(135, divide(135, const_2)))
how many zeros does 135 ^ 19 x 42 ^ 17 end with ?
"to know the zeros we need to know how many 10 ' s are there for this need to find number of 2 s and 5 s ( 2 * 5 = 10 ) 135 = ( 5 x 3 ^ 3 ) ^ 19 so number of 5 s are = 19 42 = ( 2 x 7 x 3 ) ^ 17 so number of 2 s = 17 so tot number of 10 s in this number is ( 17 - > 2 s x 17 - > 5 s ) so answer is 17 answer : e"
a = 2 + 3 b = a * 2 c = 135 / 2 d = 135 / c e = b + d
a ) a ) 3.12 , b ) b ) 8 , c ) c ) 10 , d ) d ) 15 , e ) e ) 6.13
e
max(multiply(subtract(add(40, 5), const_1), subtract(divide(5, 22), divide(5, 40))), const_4)
due to construction , the speed limit along an 5 - mile section of highway is reduced from 40 miles per hour to 22 miles per hour . approximately how many minutes more will it take to travel along this section of highway at the new speed limit than it would have taken at the old speed limit ?
"old time in minutes to cross 5 miles stretch = 5 * 60 / 40 = 5 * 3 / 2 = 7.5 new time in minutes to cross 5 miles stretch = 5 * 60 / 22 = 5 * 30 / 11 = 13.63 time difference = 6.13 ans : e"
a = 40 + 5 b = a - 1 c = 5 / 22 d = 5 / 40 e = c - d f = b * e g = max(f)
a ) 10 / 3 , b ) 2 , c ) 3 , d ) 4 , e ) 5
a
divide(divide(4, 3), divide(2, 5))
2 : 5 = 4 / 3 : x . the value of x is ?
x * 2 = 5 * 4 / 3 x * 2 = 20 / 3 x = 10 / 3 answer : a
a = 4 / 3 b = 2 / 5 c = a / b
a ) 26 , b ) 30 , c ) 34 , d ) 38 , e ) 42
b
add(divide(subtract(subtract(200, const_2), add(20, const_2)), add(const_2, 4)), const_1)
how many even integers n , such that 20 < = n < = 200 are of the form 3 k + 4 , where k is any natural number ?
"the first number is 22 = 16 + 6 ( 1 ) . we can continue adding 6 to make a list : 22 , 28 , 34 , . . . the last number is 196 = 16 + 6 ( 30 ) there are 30 numbers in the list . the answer is b ."
a = 200 - 2 b = 20 + 2 c = a - b d = 2 + 4 e = c / d f = e + 1
a ) 25 , b ) 17 , c ) 30 , d ) 34 , e ) 36
b
add(add(add(add(add(add(add(add(const_2, const_3), add(const_2, const_3)), add(add(const_2, const_3), const_2)), add(4, const_2)), add(add(4, const_2), const_2)), add(add(add(4, const_2), const_2), const_4)), add(add(add(add(4, const_2), const_2), const_4), const_2)), add(add(add(add(add(4, const_2), const_2), const_4), const_2), const_4))
find a sum for 1 st 4 prime number ' s ?
"required sum = ( 2 + 3 + 5 + 7 ) = 17 note : 1 is not a prime number option b"
a = 2 + 3 b = 2 + 3 c = a + b d = 2 + 3 e = d + 2 f = c + e g = 4 + 2 h = f + g i = 4 + 2 j = i + 2 k = h + j l = 4 + 2 m = l + 2 n = m + 4 o = k + n p = 4 + 2 q = p + 2 r = q + 4 s = r + 2 t = o + s u = 4 + 2 v = u + 2 w = v + 4 x = w + 2 y = x + 4 z = t + y
a ) 12 : 5 , b ) 11 : 5 , c ) 13 : 5 , d ) 14 : 5 , e ) 9 : 5
b
power(5, 3)
two vessels having volumes in the ratio 3 : 5 are filled with water and milk solutions . the ratio of milk and water in the two vessels are 5 : 1 and 6 : 4 respectively . if the contents of both the vessel are empties into a larger vessel , find the ratio of milk and water in the larger vessel .
"vessel a = 300 gallons - - > milk = 250 , water = 50 ; vessel b = 500 gallons - - > milk = 300 , water = 200 ; vessel a + b = 800 gallons - - > milk = 550 , water 250 the ratio = 550 / 250 - - > 11 : 5 answer : b"
a = 5 ** 3
a ) 48 , b ) 36 , c ) 26 , d ) 11 , e ) 45
e
divide(750, multiply(subtract(63, 3), const_0_2778))
how many seconds will a 750 meter long train moving with a speed of 63 km / hr take to cross a man walking with a speed of 3 km / hr in the direction of the train ?
"explanation : here distance d = 750 mts speed s = 63 - 3 = 60 kmph = 60 x 5 / 18 m / s time t = = 45 sec . answer : e"
a = 63 - 3 b = a * const_0_2778 c = 750 / b
a ) $ 500 , b ) $ 1,000 , c ) $ 2,000 , d ) $ 3,000 , e ) $ 5,000
d
divide(multiply(7.2, power(10, 11)), multiply(240, power(10, add(const_4, const_2))))
a certain country had a total annual expenditure of $ 7.2 x 10 ^ 11 last year . if the population of the country was 240 million last year , what was the per capita expenditure ?
"total expenditure / population = per capita expenditure hence , ( 7,2 x 10 ^ 11 ) / 240 000 000 = ( 7,2 x 10 ^ 11 ) / ( 2,4 x 10 ^ 8 ) = 3 x 10 ^ ( 11 - 8 ) = 3 x 10 ^ 3 = 3000 . answer is d ."
a = 10 ** 11 b = 7 * 2 c = 4 + 2 d = 10 ** c e = 240 * d f = b / e
a ) 2,858 , b ) 18,667 , c ) 28,000 , d ) 35,000 , e ) 40,000
c
divide(multiply(divide(divide(multiply(add(add(const_3, const_10), 15), power(const_100, const_2)), subtract(15, 5)), multiply(const_4, const_2)), const_3), const_1000)
a certain manufacturer produces items for which the production costs consist of annual fixed costs totaling $ 130,000 and variables costs averaging $ 5 per item . if the manufacturer ’ s selling price per item is $ 15 , how many items the manufacturer produce and sell to earn an annual profit of $ 150,000 ?
"let the items manufactured or sold bex 130000 + 5 x = 15 x - 150000 10 x = 280000 x = 28000 ans : c"
a = 3 + 10 b = a + 15 c = 100 ** 2 d = b * c e = 15 - 5 f = d / e g = 4 * 2 h = f / g i = h * 3 j = i / 1000
a ) 94 , b ) 95 , c ) 96 , d ) 97 , e ) 98
b
subtract(multiply(add(add(add(add(const_10, const_10), add(const_10, const_10)), add(const_10, const_10)), add(const_4, const_3)), add(add(add(add(const_10, const_10), add(const_10, const_10)), add(const_10, const_10)), add(const_4, const_3))), 4529)
what should be added to 4529 so that it may become a perfect square ?
"68 x 68 = 4624 4624 - 4529 = 95 if added to 95 get perfect square answer = b"
a = 10 + 10 b = 10 + 10 c = a + b d = 10 + 10 e = c + d f = 4 + 3 g = e + f h = 10 + 10 i = 10 + 10 j = h + i k = 10 + 10 l = j + k m = 4 + 3 n = l + m o = g * n p = o - 4529
a ) 1200 , b ) 3000 , c ) 1000 , d ) 3600 , e ) 2400
d
divide(subtract(multiply(divide(8, const_100), 2400), multiply(2400, divide(5, const_100))), subtract(divide(8, const_100), divide(8, const_100)))
barbata invests $ 2400 in the national bank at 5 % . how much additional money must she invest at 8 % so that the total annual income will be equal to 8 % of her entire investment ?
"let the additional invested amount for 8 % interest be x ; equation will be ; 2400 + 0.05 * 2400 + x + 0.08 x = 2400 + x + 0.08 ( 2400 + x ) 0.05 * 2400 + 0.08 x = 0.06 x + 0.08 * 2400 0.02 x = 2400 ( 0.08 - 0.05 ) x = 2400 * 0.03 / 0.02 = 3600 ans : d"
a = 8 / 100 b = a * 2400 c = 5 / 100 d = 2400 * c e = b - d f = 8 / 100 g = 8 / 100 h = f - g i = e / h
a ) 32 / 7 , b ) 13 / 7 , c ) 2 / 7 , d ) 32 / 17 , e ) 2 / 17
a
divide(add(power(4, 5), 5), add(add(3, 4), 5))
if x : y = 3 : 4 , find ( 4 x + 5 y ) : ( 5 x - 2 y ) .
". x / y = 3 / 4  ( 4 x + 5 y ) / ( 5 x + 2 y ) = ( 4 ( x / y ) + 5 ) / ( 5 ( x / y ) - 2 ) = ( 4 ( 3 / 4 ) + 5 ) / ( 5 ( 3 / 4 ) - 2 ) = ( 3 + 5 ) / ( 7 / 4 ) = 32 / 7 answer a 32 / 7"
a = 4 ** 5 b = a + 5 c = 3 + 4 d = c + 5 e = b / d
a ) 64 and 15,625 , b ) 60 and 15,625 , c ) 64 and 15,620 , d ) 64 and 15,635 , e ) 64 and 16,625
a
divide(multiply(multiply(const_100, const_100), const_100), divide(multiply(multiply(const_100, const_100), const_100), subtract(multiply(const_3, const_12), const_4)))
find two integers , neither of which ends in a zero , and whose product is exactly 1 , 000,000
"1 , 000,000 = 10 ^ 6 = 10 x 10 x 10 x 10 x 10 x 10 = ( 2 x 5 ) x ( 2 x 5 ) x ( 2 x 5 ) x ( 2 x 5 ) x ( 2 x 5 ) x ( 2 x 5 ) = ( 2 ^ 6 ) x ( 5 ^ 6 ) = 64 x 15625 so the numbers are 64 and 15,625 answer : a"
a = 100 * 100 b = a * 100 c = 100 * 100 d = c * 100 e = 3 * 12 f = e - 4 g = d / f h = b / g
a ) 64 , b ) 68 , c ) 65 , d ) 66 , e ) 67
b
subtract(multiply(9, 63), subtract(multiply(17, 59), multiply(56, 9)))
the average of 17 numbers is 59 . average of the first 9 of them is 56 and that of the last 9 is 63 . find the 9 th number ?
sum of all the 17 numbers = 17 * 59 = 1,003 sum of the first 9 of them = 9 * 56 = 504 sum of the last 9 of them = 9 * 63 = 567 so , the 8 th number = 504 + 567 - 1,003 = 68 . answer : b
a = 9 * 63 b = 17 * 59 c = 56 * 9 d = b - c e = a - d
a ) 2 hours , b ) 4 hours , c ) 5 hours , d ) 6 hours , e ) 7 hours
a
divide(56, add(24, 4))
a boat can travel with a speed of 24 km / hr in still water . if the speed of the stream is 4 km / hr , find the time taken by the boat to go 56 km downstream
"explanation : speed of the boat in still water = 24 km / hr speed of the stream = 4 km / hr speed downstream = ( 22 + 5 ) = 28 km / hr distance travelled downstream = 56 km time taken = distance / speed = 56 / 28 = 2 hours answer : option a"
a = 24 + 4 b = 56 / a
a ) 1000 , b ) 1500 , c ) 3000 , d ) 2500 , e ) 2400
c
subtract(multiply(multiply(const_100, const_100), power(add(const_1, divide(7, const_100)), const_4)), multiply(const_100, const_100))
carl is facing very difficult financial times and can only pay the interest on a $ 10,000 loan he has taken . the bank charges him a quarterly compound rate of 7 % . what is the approximate interest he pays annually ?
"an easy way too attempt this is by approximation : for the first quarter , he pays 7 % of 10,000 which is $ 700 . so for the four quarters in the year , he will approximately play 700 * 4 = 2800 over the entire year . because of compound interest it will be more than 2800 . approx . option c is correct . answer : c"
a = 100 * 100 b = 7 / 100 c = 1 + b d = c ** 4 e = a * d f = 100 * 100 g = e - f
a ) rs . 450 , b ) rs . 600 , c ) rs . 800 , d ) rs . 375 , e ) rs . 400
e
multiply(multiply(subtract(inverse(3), add(inverse(8), inverse(6))), 3200), 3)
a alone can do a piece of work in 6 days and b alone in 8 days . a and b undertook to do it for rs . 3200 . with the help of c , they completed the work in 3 days . how much is to be paid to c ?
"c ' s 1 day ' s work = 1 / 3 - ( 1 / 6 + 1 / 8 ) = 1 / 24 . a ' s wages : b ' s wages : c ' s wages = 1 / 6 : 1 / 8 : 1 / 24 = 4 : 3 : 1 . c ' s share ( for 3 days ) = rs . ( 3 x 1 / 24 x 3200 ) = rs . 400 . correct option ` ` e ' ' ."
a = 1/(3) b = 1/(8) c = 1/(6) d = b + c e = a - d f = e * 3200 g = f * 3
a ) 29 / 16 , b ) 19 / 16 , c ) 15 / 16 , d ) 9 / 13 , e ) 0
e
divide(9, 16)
1 / 2 - [ ( 2 / 5 * 5 / 32 ) + 1 ] + 9 / 16 =
"1 / 2 - [ ( 2 / 5 * 5 / 32 ) + 1 ] + 9 / 16 = 1 / 2 - [ ( 1 / 16 ) + 1 ] + 9 / 16 = 1 / 2 - [ 17 / 16 ] + 9 / 16 = 8 / 16 - 17 / 16 + 9 / 16 = 0 the answer is e ."
a = 9 / 16
a ) 520 , b ) 120 , c ) 300 , d ) 30 , e ) 333
a
subtract(subtract(subtract(divide(divide(divide(factorial(20), factorial(subtract(20, 3))), factorial(3)), const_2), 20), 20), const_10)
mariah has decided to hire three workers . to determine whom she will hire , she has selected a group of 20 candidates . she plans to have one working interview with 3 of the 20 candidates every day to see how well they work together . how many days will it take her to have working interviews with all the different combinations of job candidates ?
520 . answer a
a = math.factorial(20) b = 20 - 3 c = math.factorial(b) d = a / c e = math.factorial(3) f = d / e g = f / 2 h = g - 20 i = h - 20 j = i - 10
a ) 2 , b ) - 2 , c ) 3 , d ) - 3 , e ) 4
a
subtract(multiply(11, divide(subtract(multiply(2, 3), 1), subtract(multiply(7, 3), multiply(2, 10)))), add(multiply(10, divide(subtract(multiply(2, 3), 1), subtract(multiply(7, 3), multiply(2, 10)))), 3))
when positive integer x is divided by 10 , the quotient is y and the remainder is 3 . when 2 x is divided by 7 , the quotient is 3 y and the remainder is 1 . what is the value of 11 y – x ?
"( 1 ) x = 10 y + 3 ( 2 ) 2 x = 21 y + 1 ( 2 ) - ( 1 ) : x = 11 y - 2 11 y - x = 2 the answer is a ."
a = 2 * 3 b = a - 1 c = 7 * 3 d = 2 * 10 e = c - d f = b / e g = 11 * f h = 2 * 3 i = h - 1 j = 7 * 3 k = 2 * 10 l = j - k m = i / l n = 10 * m o = n + 3 p = g - o
a ) $ 21 , b ) $ 27 , c ) $ 31 , d ) $ 39 , e ) it can not be determined
d
subtract(add(120, subtract(160.50, 120)), multiply(subtract(160.50, 120), 3))
at an upscale fast - food restaurant , shin can buy 3 burgers , 7 shakes , and one cola for $ 120 . at the same place it would cost $ 160.50 for 4 burgers , 10 shakes , and one cola . how much would it cost for a meal of one burger , one shake , and one cola ?
"let ' s suppose that the price of a burger is bb , of a shake - ss and that of a cola is cc . we can then construct these equations : 3 b + 7 s + c = 120 4 b + 10 s + c = 160.5 subtracting the first equation from the second gives us b + 3 s = 40.5 now if we subtract the new equation two times from first or 3 times from second we will get b + s + c = 39 . in any case , there is no necessity to know each item ' s price , just the sum . answer : d"
a = 160 - 50 b = 120 + a c = 160 - 50 d = c * 3 e = b - d
a ) 75 , b ) 74 , c ) 79 , d ) 77 , e ) 72
d
add(divide(divide(300, add(const_4, const_1)), add(const_4, const_1)), divide(300, add(const_4, const_1)))
if 300 ! / 10 ^ n is an integer , what is the largest possible value of n ?
"the question actually asks the highest power of 10 which divides 300 ! ( for a number to be an integer - without any remainder all the trailing zeroe ' s must be divided by the denominator ) 10 = 2 x 5 300 factorial will have 74 as - 300 / 5 = 60 60 / 5 = 12 12 / 5 = 2 so answer will be ( d ) 74"
a = 4 + 1 b = 300 / a c = 4 + 1 d = b / c e = 4 + 1 f = 300 / e g = d + f
a ) 15060000 , b ) 0.001506 , c ) 0.01506 , d ) 0.1506 , e ) none of these
b
multiply(divide(15.06, 0.0001), const_100)
15.06 * 0.0001 = ?
"explanation : clearly after decimal 6 digits should be there . option b"
a = 15 / 6 b = a * 100
a ) 33.33 , b ) 60 , c ) 80 , d ) 66.67 , e ) 64
d
multiply(divide(add(multiply(divide(60, const_100), 200), multiply(subtract(const_1, divide(75, const_100)), 160)), add(200, 160)), const_100)
addison high school ’ s senior class has 200 boys and 160 girls . if 60 % of the boys and 75 % of the girls attended college , what percentage of the total class that did not attend college ?
"number of boys attending the college : 0.60 * 200 = 120 number of girls attending the college : 0.75 * 160 = 120 total attending = 120 + 120 = 240 total people = 200 + 160 = 360 % of people attending = 240 / 360 = 66.67 % . d is the correct answer"
a = 60 / 100 b = a * 200 c = 75 / 100 d = 1 - c e = d * 160 f = b + e g = 200 + 160 h = f / g i = h * 100
a ) 110 / 1547 , b ) 120 / 1547 , c ) 140 / 1547 , d ) 160 / 1547 , e ) 220 / 1547
b
multiply(divide(divide(5, 34), divide(7, 48)), divide(1, 13))
find the fraction which has the same ratio to 1 / 13 that 5 / 34 has to 7 / 48 .
p : 1 / 13 = 5 / 34 : 7 / 48 as the product of the means is equal to the product of the extremes . p * 7 / 48 = 1 / 13 * 5 / 34 p * 7 / 48 = 5 / 442 p = 240 / 3094 = > p = 120 / 1547 answer : b
a = 5 / 34 b = 7 / 48 c = a / b d = 1 / 13 e = c * d
a ) 6 , 8 , b ) 9 , 12 , c ) 12 , 16 , d ) 16 , 20 , e ) none
a
sqrt(divide(multiply(power(3, const_2), 224), gcd(multiply(power(3, const_2), 224), subtract(multiply(8, power(3, const_2)), power(4, const_2)))))
two numbers are such that the square of one is 224 less than 8 times the square of the other . if the numbers be in the ratio of 3 : 4 , the numbers are
solution let the numbers be 3 x and 4 x . then , ( 4 x ) 2 = 8 x ( 3 x ) 2 - 224 β€Ή = β€Ί 16 x 2 = 72 x 2 - 224 β€Ή = β€Ί 56 x 2 = 224 . β€Ή = β€Ί x 2 = 4 . β€Ή = β€Ί x = 2 . so , the numbers are 6 and 8 . answer a
a = 3 ** 2 b = a * 224 c = 3 ** 2 d = c * 224 e = 3 ** 2 f = 8 * e g = 4 ** 2 h = f - g i = math.gcd(d, h) j = b / i k = math.sqrt(j)
a ) 21 : 5 , b ) 21 : 16 , c ) 21 : 2 , d ) 9 : 9 , e ) 21 : 11
b
divide(add(multiply(3500, 6), multiply(multiply(3500, const_2), 6)), multiply(4000, add(6, 6)))
a and b invests rs . 3500 and rs . 4000 respectively in a business . if a doubles his capital after 6 months . in what ratio should a and b divide that year ' s profit ?
"( 3.5 * 6 + 7 * 6 ) : ( 4 * 12 ) 63 : 48 = > 21 : 16 answer : b"
a = 3500 * 6 b = 3500 * 2 c = b * 6 d = a + c e = 6 + 6 f = 4000 * e g = d / f
a ) 5 / 6 , b ) 3 / 4 , c ) 2 / 3 , d ) 1 / 2 , e ) 3 / 5
c
subtract(const_1, divide(10, multiply(10, const_3)))
a searchlight on top of the watchtower makes 2 revolutions per minute . what is the probability that a man appearing near the tower will stay in the dark for at least 10 seconds ?
"the searchlight completes one revolution every 30 seconds . the probability that the man ' s area will be lit up is 10 / 30 = 1 / 3 . the probability that he will stay in the dark is 1 - 1 / 3 = 2 / 3 the answer is c ."
a = 10 * 3 b = 10 / a c = 1 - b
a ) 7 / 15 , b ) 9 / 20 , c ) 11 / 25 , d ) 13 / 27 , e ) 15 / 31
d
divide(const_4, add(multiply(const_4, 4), const_1))
tom , working alone , can paint a room in 6 hours . peter and john , working independently , can paint the same room in 3 hours and 4 hours , respectively . tom starts painting the room and works on his own for one hour . he is then joined by peter and they work together for an hour . finally , john joins them and the three of them work together to finish the room , each one working at his respective rate . what fraction of the whole job was done by peter ?
"tom paints 1 / 6 of the room in the first hour . tom and peter paint 1 / 6 + 1 / 3 = 1 / 2 of the room in the next hour for a total of 4 / 6 . the three people then paint the remaining 2 / 6 in a time of ( 2 / 6 ) / ( 3 / 4 ) = 4 / 9 hours peter worked for 13 / 9 hours so he painted 13 / 9 * 1 / 3 = 13 / 27 of the room . the answer is d ."
a = 4 * 4 b = a + 1 c = 4 / b
a ) 45 , b ) 88 , c ) 120 , d ) 152 , e ) 154
c
divide(30, subtract(divide(30, 40), divide(30, const_60)))
the distance from city a to city b is 30 miles . while driving from city a to city b , bob drives at a constant speed of 40 miles per hour . alice leaves city a 30 minutes after bob . what is the minimum constant speed in miles per hour that alice must exceed in order to arrive in city b before bob ?
"the time it takes bob to drive to city b is 30 / 40 = 0.75 hours . alice needs to take less than 0.25 hours for the trip . alice needs to exceed a constant speed of 30 / 0.25 = 120 miles per hour . the answer is c ."
a = 30 / 40 b = 30 / const_60 c = a - b d = 30 / c
a ) $ 550 , b ) $ 650 , c ) $ 750 , d ) $ 850 , e ) $ 950
b
subtract(multiply(multiply(const_100, 10), const_2), divide(add(multiply(multiply(10, 8), const_2), 83), add(divide(10, const_100), divide(8, const_100))))
if x dollars is invested at 10 percent for one year and y dollars is invested at 8 percent for one year , the annual income from the 10 percent investment will exceed the annual income from the 8 percent investment by $ 83 . if $ 2,000 is the total amount invested , how much is invested at 8 percent ?
"0.1 x = 0.08 ( 2000 - x ) + 83 0.18 x = 243 x = 1350 then the amount invested at 8 % is $ 2000 - $ 1350 = $ 650 the answer is b ."
a = 100 * 10 b = a * 2 c = 10 * 8 d = c * 2 e = d + 83 f = 10 / 100 g = 8 / 100 h = f + g i = e / h j = b - i
a ) 58 , b ) 61 , c ) 64 , d ) 67 , e ) 70
d
add(subtract(divide(90, 5), const_2), add(divide(90, 3), divide(90, 5)))
line q has the equation 5 y – 3 x = 90 . if line s is perpendicular to q , has an integer for its y - intercept , and intersects q in the second quadrant , then how many possible line s ’ s exist ? ( note : intersections on one of the axes do not count . )
5 y - 3 x = 90 and so y = 3 x / 5 + 18 when x = 0 , then y = 18 . when y = 0 , then x = - 30 the slope is 3 / 5 , so the slope of line s is - 5 / 3 . through the point ( - 30 , 0 ) , 0 = - 5 ( - 30 ) / 3 + c the y - intercept is c = - 50 . thus the perpendicular line s can have y - intercepts from - 49 up to 17 . the number of possible lines is 49 + 17 + 1 = 67 the answer is d .
a = 90 / 5 b = a - 2 c = 90 / 3 d = 90 / 5 e = c + d f = b + e
a ) 22 , b ) 21 , c ) 88 , d ) 76 , e ) 29
b
divide(add(multiply(subtract(15, 6), 2), multiply(15, 3)), const_3)
the ratio of expenditure and savings is 3 : 2 . if the income increases by 15 % and the savings increases by 6 % , then by how much percent should his expenditure increases ?
answer : b ) 21
a = 15 - 6 b = a * 2 c = 15 * 3 d = b + c e = d / 3
a ) 750 , b ) 400 , c ) 800 , d ) 850 , e ) none of these
b
divide(multiply(const_100, 160), subtract(70, subtract(const_100, 70)))
in an election between the two candidates , the candidates who gets 70 % of votes polled is winned by 160 vote ’ s majority . what is the total number of votes polled ?
"explanation : note : majority ( 40 % ) = difference in votes polled to win ( 70 % ) & defeated candidates ( 30 % ) 40 % = 70 % - 30 % 40 % - - - - - > 160 ( 40 * 4 = 160 ) 100 % - - - - - > 400 ( 100 * 4 = 400 ) answer : option b"
a = 100 * 160 b = 100 - 70 c = 70 - b d = a / c
a ) 14 , b ) 16 , c ) 18 , d ) 24 , e ) 64
e
add(add(power(add(add(divide(subtract(subtract(48, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(48, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(48, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(48, const_10), const_2), const_4), const_2), const_2)))
the sum of three consecutive even numbers is 48 . find the middle number of the three ?
"three consecutive even numbers ( 2 p - 2 ) , 2 p , ( 2 p + 2 ) . ( 2 p - 2 ) + 2 p + ( 2 p + 2 ) = 48 6 p = 48 = > p = 8 . the middle number is : 8 p = 64 . answer : e"
a = 48 - 10 b = a - 2 c = b / 4 d = c + 2 e = d + 2 f = e ** 2 g = 48 - 10 h = g - 2 i = h / 4 j = i + 2 k = j + 2 l = k + 2 m = l ** 2 n = f + m o = 48 - 10 p = o - 2 q = p / 4 r = q ** 2 s = 48 - 10 t = s - 2 u = t / 4 v = u + 2 w = v ** 2 x = r + w y = n + x
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6
c
subtract(subtract(24, multiply(8, const_2)), subtract(subtract(12, 8), const_1))
there are 8 stores in town that had a total of 24 visitors on a particular day . however , only 12 people went shopping that day ; some people visited more than one store . if 9 people visited exactly two stores each , and everyone visited at least one store , what is the largest number of stores anyone could have visited ?
"9 people visited 2 stores each for 18 visits . to maximize the number of stores that one person visited , let ' s assume that 2 people visited 1 store each . the number of remaining visits is 24 - 18 - 2 = 4 , which is the maximum that one person could have visited . the answer is c ."
a = 8 * 2 b = 24 - a c = 12 - 8 d = c - 1 e = b - d
a ) 110.111 m , b ) 121.111 m , c ) 111.11 m , d ) 141.111 m , e ) 181.111 m
c
multiply(multiply(80, const_0_2778), 5)
if a train , travelling at a speed of 80 kmph , crosses a pole in 5 sec , then the length of train is ?
"d = 80 * 5 / 18 * 5 = 111.11 m answer : c"
a = 80 * const_0_2778 b = a * 5
a ) 10 , b ) 8 , c ) 5 , d ) 4 , e ) 2
c
subtract(divide(divide(120, const_1), const_3), const_3)
in how many ways can the integer 120 be expressed as a product of two different positive integers ?
"120 = ( 2 ^ 3 ) * 3 * 5 since 120 is not a perfect square , no of ways = 5 answer c"
a = 120 / 1 b = a / 3 c = b - 3
a ) 15 , b ) 12 , c ) 8 , d ) 6 , e ) 4
c
subtract(multiply(4, const_2), multiply(2, const_2))
a fruit - salad mixture consists of apples , peaches , and grapes in the ratio 4 : 5 : 2 , respectively , by weight . if 44 pounds of the mixture is prepared , the mixture includes how many more pounds of apples than grapes ?
"we can first set up our ratio using variable multipliers . we are given that a fruit - salad mixture consists of apples , peaches , and grapes , in the ratio of 4 : 5 : 2 , respectively , by weight . thus , we can say : apples : peaches : grapes = 4 x : 5 x : 2 x we are given that 39 pounds of the mixture is prepared so we can set up the following question and determine a value for x : 4 x + 5 x + 2 x = 44 11 x = 44 x = 4 now we can determine the number of pounds of apples and of grapes . pounds of grapes = ( 2 ) ( 4 ) = 8 pounds of apples = ( 4 ) ( 4 ) = 16 thus we know that there are 16 – 8 = 8 more pounds of apples than grapes . answer is c ."
a = 4 * 2 b = 2 * 2 c = a - b