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a certain debt will be paid in 52 installments from january 1 to december 31 of a certain year . each of the first 8 payments is to be $ 410 ; each of the remaining payments is to be $ 65 more than each of the first 8 payments . what is the average ( arithmetic mean ) payment that will be made on the debt for the year ?
"total number of installments = 52 payment per installment for the first 8 installments = 410 payment per installment for the remaining 32 installments = 410 + 65 = 475 average = ( 8 * 410 + 44 * 475 ) / 52 = 465 answer c"
a ) 443 , b ) 450 , c ) 465 , d ) 468 , e ) 475
c
divide(add(multiply(8, 410), multiply(add(410, 65), subtract(52, 8))), 52)
add(n4,n5)|multiply(n3,n4)|subtract(n0,n3)|multiply(#0,#2)|add(#1,#3)|divide(#4,n0)|
general
the sum of the fourth and twelfth term of an arithmetic progression is 20 . what is the sum of the first 20 terms of the arithmetic progression ?
"n th term of a . p . is given by a + ( n - 1 ) d 4 th term = a + 3 d 12 th term = a + 11 d given a + 3 d + a + 11 d = 20 - - > 2 a + 14 d = 20 - - > a + 7 d = 10 sum of n term of a . p = n / 2 [ 2 a + ( n - 1 ) d ] subsitiuing n = 20 . . . we get 20 / 2 [ 2 a + 14 d ] = 20 [ a + 7 d ] = 20 * 10 = 200 . . . answer is e . . ."
a ) 300 , b ) 120 , c ) 150 , d ) 170 , e ) 200
e
divide(multiply(20, 20), const_2)
multiply(n0,n1)|divide(#0,const_2)|
general
a alone can finish a work in 10 days which b alone can finish in 15 days . if they work together and finish it , then out of a total wages of rs . 3100 , a will get :
explanation : ratio of working days of a : b = 10 : 15 therefore , their wages ratio = reverse ratio = 15 : 10 therefore , a will get 15 units of ratio total ratio = 25 1 unit of ratio = 3100 / 25 = 124 so , a ’ s amount = 124 Γ— 15 = rs . 1860 . answer : option c
a ) rs . 1200 , b ) rs . 1500 , c ) rs . 1860 , d ) rs . 2000 , e ) none of these
c
multiply(divide(divide(multiply(15, const_2), 10), add(divide(multiply(15, const_2), 10), divide(multiply(15, const_2), 15))), 3100)
multiply(n1,const_2)|divide(#0,n0)|divide(#0,n1)|add(#1,#2)|divide(#1,#3)|multiply(n2,#4)
physics
crazy eddie has a key chain factory . eddie managed to decrease the cost of manufacturing his key chains while keeping the same selling price , and thus increased the profit from the sale of each key chain from 40 % of the selling price to 50 % of the selling price . if the manufacturing cost is now $ 50 , what was it before the decrease ?
"deargoodyear 2013 , i ' m happy to help . this is a relatively straightforward problem , not very challenging . btw , crazy eddiewas the actually name of an electronics chain on the east coast of the usa back in the 1970 s . manufacturing now is $ 50 . they now are making a 50 % profit , so the selling price must be $ 100 . they had this same selling price , $ 100 , before they made the change , and had a profit of 40 % , so the manufacturing must have been $ 60 . answer = ( e ) ."
a ) $ 20 , b ) $ 40 , c ) $ 50 , d ) $ 80 , e ) $ 60
e
subtract(divide(50, divide(50, const_100)), multiply(divide(50, divide(50, const_100)), divide(40, const_100)))
divide(n1,const_100)|divide(n0,const_100)|divide(n1,#0)|multiply(#2,#1)|subtract(#2,#3)|
general
while driving from a - ville to b - town , harriet drove at a constant speed of 110 kilometers per hour . upon arriving in b - town , harriet immediately turned and drove back to a - ville at a constant speed of 140 kilometers per hour . if the entire trip took 5 hours , how many minutes did it take harriet to drive from a - ville to b - town ?
"5 hr = 300 min . if harriet spend equal hrs on each leg she will spend 150 min on each . since speed a - b is less than speed b - a and distance on each leg is the same , time spent on a - b is more than 150 min , which mean we can eliminate ans . a , b and c . now let plug in ans . d or e and verify which one give same distance on each leg . e . t = 168 min * leg a - b - - - > d = 110.168 / 60 = 18480 / 60 * leg b - a - - - - > d = 140 * 132 / 60 = 18480 / 60 so the correct ans . ise"
a ) 138 , b ) 148 , c ) 150 , d ) 162 , e ) 168
e
multiply(subtract(5, divide(multiply(110, 5), add(110, 140))), const_60)
add(n0,n1)|multiply(n0,n2)|divide(#1,#0)|subtract(n2,#2)|multiply(#3,const_60)|
physics
if the average of 1 , 2 , 4 , 5 , 6 , 9 , 9 , 10 , 12 and x is 7 , what is the value of x ?
"sum of the deviations of the numbers in the set from the mean is always zero 1 , 2 , 4 , 5 , 6 , 9 , 9 , 10 , 12 mean is 7 so the list is - 6 - 5 - 3 - 2 - 1 + 2 + 2 + 3 + 5 . . . this shud total to zero but this is - 5 , hence we need a number that is 5 more than the mean to get a + 5 and make it zero hence the answer is 7 + 5 = 12 d"
a ) 7 , b ) 9 , c ) 11 , d ) 12 , e ) 13
d
subtract(multiply(7, const_10), add(add(add(add(add(add(add(add(1, 2), 4), 5), 6), 9), 9), 10), 12))
add(n0,n1)|multiply(n9,const_10)|add(n2,#0)|add(n3,#2)|add(n4,#3)|add(n5,#4)|add(n5,#5)|add(n7,#6)|add(n8,#7)|subtract(#1,#8)|
general
a jogger running at 9 km / hr along side a railway track is 200 m ahead of the engine of a 200 m long train running at 45 km / hr in the same direction . in how much time will the train pass the jogger ?
"speed of train relative to jogger = 45 - 9 = 36 km / hr . = 36 * 5 / 18 = 10 m / sec . distance to be covered = 200 + 200 = 360 m . time taken = 400 / 10 = 40 sec . answer : option c"
a ) 89 , b ) 20 , c ) 40 , d ) 88 , e ) 34
c
divide(add(200, 200), multiply(subtract(45, 9), divide(divide(const_10, const_2), divide(subtract(45, 9), const_2))))
add(n1,n2)|divide(const_10,const_2)|subtract(n3,n0)|divide(#2,const_2)|divide(#1,#3)|multiply(#4,#2)|divide(#0,#5)|
general
yesterday ' s closing prices of 2,200 different stocks listed on a certain stock exchange were all different from today ' s closing prices . the number of stocks that closed at a higher price today than yesterday was 20 percent greater than the number that closed at a lower price . how many of the stocks closed at a higher price today than yesterday ?
"lets consider the below - the number of stocks that closed at a higher price = h the number of stocks that closed at a lower price = l we understand from first statement - > h + l = 2200 - - - - ( 1 ) we understand from second statement - > h = ( 120 / 100 ) l = > h = 1.2 l - - - - ( 2 ) solve eq ( 1 ) ( 2 ) to get h = 1200 . c is my answer ."
a ) 484 , b ) 726 , c ) 1,200 , d ) 1,320 , e ) 1,694
c
multiply(divide(subtract(subtract(multiply(20, const_100), const_10), const_10), add(add(const_1, divide(20, const_100)), const_1)), add(const_1, divide(20, const_100)))
divide(n1,const_100)|multiply(n1,const_100)|add(#0,const_1)|subtract(#1,const_10)|add(#2,const_1)|subtract(#3,const_10)|divide(#5,#4)|multiply(#2,#6)|
gain
the dimensions of an open box are 50 cm , 40 cm and 23 cm . its thickness is 2 cm . if 1 cubic cm of metal used in the box weighs 0.5 gms , find the weight of the box .
explanation : volume of the metal used in the box = external volume - internal volume = [ ( 50 x 40 x 23 ) - ( 44 x 34 x 20 ) ] cm 3 = 16080 cm 3 weight of the metal = [ ( 16080 x 0.5 ) / 1000 ] kg = 8.04 kg . answer : a
['a ) 8.04 kg', 'b ) 8.14 kg', 'c ) 8.24 kg', 'd ) 9.04 kg', 'e ) none of these']
a
divide(multiply(subtract(multiply(multiply(50, 40), 23), multiply(multiply(subtract(50, multiply(2, const_3)), subtract(40, multiply(2, const_3))), subtract(23, const_3))), 0.5), const_1000)
multiply(n0,n1)|multiply(n3,const_3)|subtract(n2,const_3)|multiply(n2,#0)|subtract(n0,#1)|subtract(n1,#1)|multiply(#4,#5)|multiply(#6,#2)|subtract(#3,#7)|multiply(n5,#8)|divide(#9,const_1000)
physics
two goods trains each 680 m long are running in opposite directions on parallel tracks . their speeds are 45 km / hr and 30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ?
"relative speed = 45 + 30 = 75 km / hr . 75 * 5 / 18 = 125 / 6 m / sec . distance covered = 680 + 680 = 1360 m . required time = 1360 * 6 / 125 = 65.28 sec . answer : b"
a ) 228 , b ) 65.28 , c ) 48.32 , d ) 27 , e ) 21.1
b
add(45, 30)
add(n1,n2)|
physics
mike needs 30 % to pass . if he scored 212 marks and falls short by 25 marks , what was the maximum marks he could have got ?
"if mike had scored 25 marks more , he could have scored 30 % therefore , mike required 212 + 25 = 237 marks let the maximum marks be m . then 30 % of m = 237 ( 30 / 100 ) Γ— m = 237 m = ( 237 Γ— 100 ) / 30 m = 23700 / 30 m = 790 answer : c"
a ) 343 , b ) 677 , c ) 790 , d ) 867 , e ) 932
c
divide(add(212, 25), divide(30, const_100))
add(n1,n2)|divide(n0,const_100)|divide(#0,#1)|
gain
if r is the least positive integer that is divisible by every integer from 12 to 16 inclusive , then r / 13 is =
the integer should be divisible by : 12 , 13 , 14 , 15 and 16 , that is 3 * 2 ^ 2 , 13 , 2 * 7 , 3 * 5 and 2 ^ 4 . the least common multiple of these integers is the product of 13 , 7 , 5 , 3 and 2 ^ 4 . then , r / 13 is ( 13 * 7 * 5 * 3 * 2 ^ 4 ) / 13 = 7 * 5 * 3 * 16 = 1680 . answer : c .
a ) 40320 , b ) 21840 , c ) 1680 , d ) 840 , e ) 240
c
reminder(13, const_10)
reminder(n2,const_10)
general
a number is selected at random from the first 30 natural numbers . what is the probability that the number is a multiple of either 5 or 12 ?
"number of multiples of 5 from 1 through 30 = 30 / 5 = 6 number of multiples of 12 from 1 through 30 = 30 / 12 = 2 number of multiples of 5 and 12 both from 1 through 30 = number of multiples of 12 * 5 ( = 60 ) = 0 total favourable cases = 6 + 2 - 0 = 8 probability = 8 / 30 = 4 / 15 answer : option d"
a ) 17 / 30 , b ) 2 / 5 , c ) 7 / 15 , d ) 4 / 15 , e ) 11 / 30
d
divide(add(floor(divide(30, 5)), divide(30, 12)), 30)
divide(n0,n2)|divide(n0,n1)|floor(#1)|add(#0,#2)|divide(#3,n0)|
probability
an aeroplane covers a certain distance of 420 kmph in 6 hours . to cover the same distance in 4 2 / 3 hours , it must travel at a speed of ?
"speed of aeroplane = 420 kmph distance travelled in 6 hours = 420 * 6 = 2520 km speed of aeroplane to acver 2520 km in 14 / 3 = 2520 * 3 / 14 = 540 km answer : option ' d '"
a ) 450 km , b ) 480 km , c ) 500 km , d ) 540 km , e ) 590 km
d
divide(multiply(420, 6), divide(add(multiply(3, 4), 2), 3))
multiply(n0,n1)|multiply(n2,n4)|add(n3,#1)|divide(#2,n4)|divide(#0,#3)|
physics
for what value of x between βˆ’ 5 and 5 , inclusive , is the value of x ^ 2 βˆ’ 10 x + 16 the greatest ?
we can see from the statement that two terms containing x , x ^ 2 will always be positive and - 10 x will be positive if x is - ive . . so the equation will have greatest value if x is - ive , and lower the value of x , greater is the equation . so - 5 will give the greatest value . . ans a
a ) βˆ’ 5 , b ) βˆ’ 2 , c ) 0 , d ) 2 , e ) 4
a
subtract(add(power(negate(5), 2), 16), multiply(10, negate(5)))
negate(n0)|multiply(n3,#0)|power(#0,n2)|add(n4,#2)|subtract(#3,#1)
general
what profit percent is made by selling an article at a certain price , if by selling at 2 / 3 rd of that price , there would be a loss of 15 % ?
"sp 2 = 2 / 3 sp 1 cp = 100 sp 2 = 85 2 / 3 sp 1 = 85 sp 1 = 127.5 100 - - - 27.5 = > 27.5 % answer : b"
a ) 20 % , b ) 27.5 % , c ) 10 % , d ) 80 % , e ) 90 %
b
subtract(divide(subtract(const_100, 15), divide(2, 3)), const_100)
divide(n0,n1)|subtract(const_100,n2)|divide(#1,#0)|subtract(#2,const_100)|
gain
a boat running up stram takes 5 hours to cover a certain distance , while it takes 8 hours to cover the same distance running down stream . what is the ratio between the speed of the boat and the speed of water current respectively ?
"explanation : let speed of boat is x km / h and speed stream is y km / hr 5 ( x + y ) = 8 ( x - y ) 5 x + 5 y = 8 x - 8 y 13 y = 3 x x / y = 13 / 3 13 : 3 answer : option e"
a ) 2 : 3 , b ) 5 : 6 , c ) 4 : 5 , d ) 7 : 1 , e ) 13 : 3
e
subtract(8, 5)
subtract(n1,n0)|
physics
machine p and machine q are each used to manufacture 880 sprockets . it takes machine p 10 hours longer to produce 880 sprockets than machine q . machine q produces 10 % more sprockets per hour than machine a . how many sprockets per hour does machine a produce ?
"p makes x sprockets per hour . then q makes 1.1 x sprockets per hour . 880 / x = 880 / 1.1 x + 10 1.1 ( 880 ) = 880 + 11 x 11 x = 88 x = 8 the answer is d ."
a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10
d
divide(subtract(880, divide(880, add(divide(10, const_100), const_1))), 10)
divide(n1,const_100)|add(#0,const_1)|divide(n0,#1)|subtract(n0,#2)|divide(#3,n1)|
gain
in an examination 35 % of the students passed and 351 failed . how many students appeared for the examination ?
let the number of students appeared be x then , 65 % of x = 351 65 x / 100 = 351 x = 351 * 100 / 65 = 540 answer is a
a ) a ) 540 , b ) b ) 400 , c ) c ) 700 , d ) d ) 650 , e ) e ) 840
a
divide(multiply(351, const_100), subtract(const_100, 35))
multiply(n1,const_100)|subtract(const_100,n0)|divide(#0,#1)
gain
what is the average of 1200 , 1300 , 1400 , 1510 , 1520 , 1530 , and 1200 ?
add 1200 , 1300 , 1400 , 1510 , 1520 , 1530 , and 1200 grouping numbers together may quicken the addition sum = 9660 9660 / 7 = 1380 c
a ) 1360.47 , b ) 1355.8 , c ) 1380 , d ) 1350 , e ) 1333
c
divide(add(add(add(add(add(add(1200, 1300), 1400), 1510), 1520), 1530), 1200), add(const_3, const_4))
add(n0,n1)|add(const_3,const_4)|add(n2,#0)|add(n3,#2)|add(n4,#3)|add(n5,#4)|add(n0,#5)|divide(#6,#1)
general
if w is 40 percent less than q , q is 40 percent less than y , and z is 46 percent less than y , then z is greater than w by what percent of w ?
given w = 0.6 q , q = 0.6 y , z = 0.54 y , substituting , w = 2 / 3 z - - - - > z = 1.5 w and thus z is 50 % greater than w . e is the correct answer .
a ) 4 % , b ) 18 % , c ) 36 % , d ) 40 % , e ) 50 %
e
multiply(divide(subtract(subtract(const_100, 46), subtract(subtract(const_100, 40), divide(multiply(40, subtract(const_100, 40)), const_100))), subtract(subtract(const_100, 40), divide(multiply(40, subtract(const_100, 40)), const_100))), const_100)
subtract(const_100,n2)|subtract(const_100,n0)|multiply(n0,#1)|divide(#2,const_100)|subtract(#1,#3)|subtract(#0,#4)|divide(#5,#4)|multiply(#6,const_100)
gain
a man rows his boat 70 km downstream and 45 km upstream , taking 2 1 / 2 hours each time . find the speed of the stream ?
"speed downstream = d / t = 75 / ( 2 1 / 2 ) = 28 kmph speed upstream = d / t = 45 / ( 2 1 / 2 ) = 18 kmph the speed of the stream = ( 28 - 18 ) / 2 = 5 kmph answer : c"
a ) 1 kmph , b ) 6 kmph , c ) 5 kmph , d ) 8 kmph , e ) 7 kmph
c
divide(subtract(divide(const_60.0, 2), divide(45, 2)), const_2)
divide(const_60.0,n2)|divide(n1,n2)|subtract(#0,#1)|divide(#2,const_2)|
physics
a train passes a station platform in 48 sec and a man standing on the platform in 40 sec . if the speed of the train is 36 km / hr . what is the length of the platform ?
"speed = 36 * 5 / 18 = 10 m / sec . length of the train = 10 * 40 = 400 m . let the length of the platform be x m . then , ( x + 400 ) / 48 = 10 = > x = 80 m . answer : a"
a ) 80 , b ) 120 , c ) 60 , d ) 90 , e ) 50
a
multiply(40, multiply(36, const_0_2778))
multiply(n2,const_0_2778)|multiply(n1,#0)|
physics
a particular store purchased a stock of turtleneck sweaters and marked up its cost by 20 % . during the new year season , it further marked up its prices by 25 % of the original retail price . in february , the store then offered a discount of 12 % . what was its profit on the items sold in february ?
"assume the total price = 100 x price after 20 % markup = 120 x price after 25 % further markup = 1.25 * 120 x = 150 x price after the discount = 0.88 * 150 x = 132 x hence total profit = 32 % option b"
a ) 27.5 % , b ) 32 % , c ) 35 % , d ) 37.5 % , e ) 40 %
b
subtract(multiply(divide(subtract(const_100, 12), const_100), multiply(add(const_100, 20), divide(add(const_100, 25), const_100))), const_100)
add(n0,const_100)|add(n1,const_100)|subtract(const_100,n2)|divide(#2,const_100)|divide(#1,const_100)|multiply(#0,#4)|multiply(#3,#5)|subtract(#6,const_100)|
gain
a , b and c enter into partnership . a invests some money at the beginning , b invests double the amount after 6 months , and c invests thrice the amount after 8 months . if the annual gain be rs . 18600 . a ' s share is ?
"x * 12 : 2 x * 6 : 3 x * 4 1 : 1 : 1 1 / 3 * 18600 = 6200 answer : d"
a ) 8876 , b ) 2765 , c ) 6000 , d ) 6200 , e ) 1261
d
multiply(multiply(const_1, const_12), divide(18600, add(add(multiply(const_1, const_12), multiply(subtract(const_12, 6), const_2)), multiply(subtract(const_12, 8), const_3))))
multiply(const_1,const_12)|subtract(const_12,n0)|subtract(const_12,n1)|multiply(#1,const_2)|multiply(#2,const_3)|add(#0,#3)|add(#5,#4)|divide(n2,#6)|multiply(#7,#0)|
gain
a student took 5 papers in an examination , where the full marks were the same for each paper . his marks in these papers were in the proportion of 5 : 6 : 7 : 8 : 9 . in all papers together , the candidate obtained 60 % of the total marks . then the number of papers in which he got more than 50 % marks is
here we can assume that one subject is of 100 marks so total there are 5 subjects = > 100 * 5 = 500 . now according to the question he secured 60 % of those which is 60 % of 500 = 300 marks in total . the ratio between the marks is given as 5 : 6 : 7 : 8 : 9 , so now we can distribute 300 marks according to the ratio . total ratio = 35 for 6 : ( 300 / 35 ) * 6 = 51.4 similarly , we will get for others as 60 , 68.57 , 77.14 hence , there are 5 subject where he secured more that 50 % . the answer is 5 . d
a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 1
d
add(const_1, const_4)
add(const_1,const_4)
general
if x € y = ( x + y ) ^ 2 - ( x - y ) ^ 2 . then √ 6 € √ 6 =
"x = √ 6 and y also = √ 6 applying the function ( √ 6 + √ 6 ) ^ 2 - ( √ 6 - √ 6 ) ^ 2 = ( 2 √ 6 ) ^ 2 - 0 = 4 x 6 = 24 . note : alternative approach is the entire function is represented as x ^ 2 - y ^ 2 = ( x + y ) ( x - y ) which can be simplified as ( x + y + x - y ) ( x + y - ( x - y ) ) = ( 2 x ) ( 2 y ) = 4 xy . substituting x = √ 6 and y = √ 6 you get the answer 24 . answer d"
a ) 0 , b ) 5 , c ) 10 , d ) 24 , e ) 20
d
power(add(sqrt(6), sqrt(6)), 2)
sqrt(n2)|add(#0,#0)|power(#1,n0)|
general
an investment yields an interest payment of $ 240 each month . if the simple annual interest rate is 9 % , what is the amount of the investment ?
"let the principal amount = p simple annual interest = 9 % simple monthly interest = ( 9 / 12 ) = ( 3 / 4 ) % ( 3 / 4 ) * ( p / 100 ) = 240 = > p = ( 240 * 4 * 10 ^ 2 ) / 3 = 80 * 4 * 10 ^ 2 = 320 * 10 ^ 2 = 32000 answer e"
a ) $ 28,300 , b ) $ 30,400 , c ) $ 31,300 , d ) $ 32,500 , e ) $ 32,000
e
multiply(divide(240, divide(9, multiply(const_3, const_4))), const_100)
multiply(const_3,const_4)|divide(n1,#0)|divide(n0,#1)|multiply(#2,const_100)|
gain
the inner circumference of a circular race track , 14 m wide , is 440 m . find radius of the outer circle ?
"let inner radius be r metres . then , 2 r = 440 ; r = = 70 m . radius of outer circle = ( 70 + 14 ) m = 84 m . answer : d"
a ) 22 , b ) 37 , c ) 87 , d ) 84 , e ) 82
d
add(14, divide(divide(440, const_pi), const_2))
divide(n1,const_pi)|divide(#0,const_2)|add(n0,#1)|
physics
the average speed of a car is 1 4 / 5 times the avg speed of a bike . a tractor covers 575 km in 25 hrs . how much distance will the car cover in 4 hrs if the speed of the bike is twice speed of the tractor ?
"sol . average speed of a tractor = 23 km / h the speed of a bus in an hour = 23 Γ— 2 = 46 km the speed of a car in an hour = 9 / 5 * 46 = 82.8 km so , the distance covered by car in 4 h is 82.8 Γ— 4 = 331.2 km ans . ( c )"
a ) 400 km , b ) 500 km , c ) 331.2 km , d ) 550 km , e ) 600 km
c
multiply(multiply(add(1, divide(4, 5)), multiply(const_2, divide(575, 25))), 4)
divide(n1,n2)|divide(n3,n4)|add(n0,#0)|multiply(#1,const_2)|multiply(#2,#3)|multiply(n5,#4)|
general
there are 56 lights which are functional and each is controlled by a separate on / off switch . two children x and y start playing with the switches . x starts by pressing every third switch till he reaches the end . y , thereafter , presses every fifth switch till he too reaches the end . if all switches were in off position at the beggining , how many lights are switched on by the end of this operation ?
editing my solution : number of switches = 56 number of switches turned on by x : 3 , 6 , . . . 54 = 18 number of switches turned on by y : 5 , 10 , . . . . 55 = 11 few switches are turned on by x and later turned off by y : lcm ( 35 ) = 15 x = 15 , 30 , . . . . 90 = 6 . subtract the above 6 switches from both x and y as they are turned off . number of switches that are turned on = ( 18 - 6 ) + ( 11 - 6 ) = 17 answer : c
a ) 18 , b ) 19 , c ) 17 , d ) 15 , e ) 16
c
add(subtract(56, multiply(const_4, const_10)), const_1)
multiply(const_10,const_4)|subtract(n0,#0)|add(#1,const_1)
general
how many multiples of 5 are there between 5 to 95 ?
"as you know , multiples of 5 are integers having 0 or 5 in the digit to the extreme right ( i . e . the unit ’ s place ) . so the numbers are 10 , 15 , 20 , 25 , 30 , 35 , 40 , 45 , 50 , 55 , 60 , 65 , 70 , 75 , 80 , 85 , 90 . answer c"
a ) 9 , b ) 18 , c ) 17 , d ) 16 , e ) 15
c
add(divide(subtract(95, 5), 5), const_1)
subtract(n2,n1)|divide(#0,n0)|add(#1,const_1)|
general
the difference between the local value and face value of 7 in the numeral 657903 is :
( local value ) - ( face value ) = ( 7000 - 7 ) = 6993 . answer : c
a ) 0 , b ) 7896 , c ) 6993 , d ) 903 , e ) 803
c
subtract(multiply(7, const_1000), 7)
multiply(n0,const_1000)|subtract(#0,n0)
general
find the value of b from ( 15 ) ^ 2 x 9 ^ 2 Γ£ Β· 356 = b .
given exp . = ( 15 ) ^ 2 x 9 ^ 2 Γ£ Β· 356 = b = 225 x 81 Γ£ Β· 356 = 51.2 c
a ) 53.2 , b ) 52.2 , c ) 51.2 , d ) 8.2 , e ) 54.2
c
divide(multiply(power(15, 2), power(9, 2)), 356)
power(n0,n1)|power(n2,n1)|multiply(#0,#1)|divide(#2,n4)
general
an order was placed for a carpet whose length and width were in the ratio of 3 : 2 . subsequently , the dimensions of the carpet were altered such that its length and width were in the ratio 1 : 1 but were was no change in its perimeter . what is the ratio of the areas of the carpets ?
"let the length and width of one carpet be 3 x and 2 x . let the length and width of the other carpet be y and y . 2 ( 3 x + 2 x ) = 2 ( y + y ) 5 x = 2 y ( 5 / 2 ) * x = y the ratio of the areas of the carpet in both cases : = 3 x * 2 x : y * y = 6 x ^ 2 : y ^ 2 = 6 x ^ 2 : ( 25 / 4 ) * x ^ 2 = 24 : 25 the answer is e ."
a ) 4 : 5 , b ) 9 : 10 , c ) 14 : 15 , d ) 19 : 20 , e ) 24 : 25
e
divide(rectangle_area(3, 2), rectangle_area(divide(divide(rectangle_perimeter(3, 2), 2), add(1, 1)), multiply(divide(divide(rectangle_perimeter(3, 2), 2), add(1, 1)), 1)))
add(n2,n3)|rectangle_area(n0,n1)|rectangle_perimeter(n0,n1)|divide(#2,n1)|divide(#3,#0)|multiply(n2,#4)|rectangle_area(#4,#5)|divide(#1,#6)|
geometry
how many cuboids of length 5 m , width 2 m and height 3 m can be farmed from a cuboid of 18 m length , 15 m width and 2 m height .
"( 18 Γ£ β€” 15 Γ£ β€” 12 ) / ( 5 Γ£ β€” 3 Γ£ β€” 2 ) = 18 answer is e ."
a ) 10 , b ) 12 , c ) 15 , d ) 16 , e ) 18
e
divide(multiply(multiply(18, 15), 2), multiply(multiply(5, 2), 3))
multiply(n3,n4)|multiply(n0,n1)|multiply(n5,#0)|multiply(n2,#1)|divide(#2,#3)|
physics
an engineer undertakes a project to build a road 10 km long in 60 days and employs 30 men for the purpose . after 20 days , he finds only 2 km of the road has been completed . find the ( approximate ) number of extra men he must employ to finish the work in time .
"30 workers working already let x be the total men required to finish the task in next 40 days 2 km done hence remaining is 8 km also , work has to be completed in next 40 days ( 60 - 20 = 40 ) we know that , proportion of men to distance is direct proportion and , proportion of men to days is inverse proportion hence , x = ( 30 * 8 * 20 ) / ( 2 * 40 ) thus , x = 60 thus , more men needed to finish the task = 60 - 30 = 30 answer : c"
a ) 20 , b ) 15 , c ) 30 , d ) 45 , e ) 60
c
subtract(divide(multiply(multiply(30, subtract(10, 2)), 20), multiply(2, subtract(60, 20))), 30)
subtract(n0,n4)|subtract(n1,n3)|multiply(n2,#0)|multiply(n4,#1)|multiply(n3,#2)|divide(#4,#3)|subtract(#5,n2)|
physics
a retailer bought a machine at a wholesale price of $ 81 and later on sold it after a 10 % discount of the retail price . if the retailer made a profit equivalent to 20 % of the whole price , what is the retail price of the machine ?
"my solution : wholesale price = 81 retail price , be = x he provides 10 % discount on retail price = x - 10 x / 100 this retail price = 20 % profit on wholesale price x - 10 x / 100 = 81 + 1 / 5 ( 81 ) x = 108 ; answer : b"
a ) 81 , b ) 108 , c ) 120 , d ) 135 , e ) 160
b
divide(multiply(add(81, divide(multiply(81, 20), const_100)), const_100), multiply(multiply(const_3, const_3), 10))
multiply(n0,n2)|multiply(const_3,const_3)|divide(#0,const_100)|multiply(n1,#1)|add(n0,#2)|multiply(#4,const_100)|divide(#5,#3)|
gain
1.5 , 2.3 , 3.1 , 3.9 , . . ?
"1.5 + 0.8 = 2.3 2.3 + 0.8 = 3.1 3.1 + 0.8 = 3.9 3.9 + 0.8 = 4.7 answer : b"
a ) 0.6 , b ) 0.8 , c ) 0.4 , d ) 0.24 , e ) 0.15
b
subtract(negate(3.9), multiply(subtract(2.3, 3.1), divide(subtract(2.3, 3.1), subtract(1.5, 2.3))))
negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)|
general
at steel factory , each employee working the second shift produced 2 / 3 as many widgets as each employee working the first shift . if the first shift has 3 / 4 as many employees , what fraction of the total widgets did the second shift produce ?
at steel factory , let the first shift have 3 employee and each produce 3 widgets , so the total number of widgets produced by the first shift is 3 * 3 = 9 ; then the second shift would have 4 employees and each second shift employee would produce 3 * 2 / 3 = 2 widgets , so the total number of widgets produced by the second shift employees would be 4 * 2 = 8 ; the ratio of the second shift production to the total is 8 / ( 9 + 8 ) = 8 / 17 . answer : a .
a ) 8 / 17 , b ) 1 / 2 , c ) 8 / 9 , d ) 9 / 8 , e ) 17 / 8
a
divide(multiply(divide(2, 3), 4), add(3, multiply(divide(2, 3), 4)))
divide(n0,n1)|multiply(n3,#0)|add(n1,#1)|divide(#1,#2)
physics
a circular swimming pool is surrounded by a concrete wall 4 ft wide . if the area of the concrete wall surrounding the pool is 11 / 25 that of the pool , then the radius of the pool is ?
"let the radius of the pool be rft radius of the pool including the wall = ( r + 4 ) ft area of the concrete wall = { \ color { black } \ pi \ left [ ( r + 4 ) ^ { 2 } - r ^ { 2 } \ right ] } = { \ color { black } \ pi \ left [ ( r + 4 + r ) ( r + 4 - r ) \ right ] = 8 \ pi ( r + 2 ) } sq feet { \ color { black } \ rightarrow 8 \ pi ( r + 2 ) = \ frac { 11 } { 25 } \ pi r ^ { 2 } \ rightarrow 11 r ^ { 2 } = 200 ( r + 2 ) } radius of the pool r = 20 ft answer : b ) 20 ft"
a ) 22 , b ) 20 , c ) 88 , d ) 387 , e ) 19
b
divide(divide(add(multiply(const_2, 4), sqrt(add(power(multiply(const_2, 4), const_2), multiply(multiply(divide(11, 25), 4), power(4, const_2))))), divide(11, 25)), const_2)
divide(n1,n2)|multiply(n0,const_2)|power(n0,const_2)|multiply(#0,n0)|power(#1,const_2)|multiply(#3,#2)|add(#5,#4)|sqrt(#6)|add(#1,#7)|divide(#8,#0)|divide(#9,const_2)|
geometry
36 men can complete a piece of work in 18 days . in how many days will 81 men complete the same work ?
"explanation : less men , means more days { indirect proportion } let the number of days be x then , 81 : 36 : : 18 : x x = 8 answer : a ) 8 days"
a ) 8 , b ) 77 , c ) 88 , d ) 29 , e ) 21
a
divide(multiply(18, 36), 81)
multiply(n0,n1)|divide(#0,n2)|
physics
simplify : 0.2 * 0.4 + 0.6 * 0.8
"given exp . = 0.2 * 0.4 + ( 0.6 * 0.8 ) = 0.08 + 0.48 = 0.56 answer is e ."
a ) 0.52 , b ) 0.42 , c ) 0.48 , d ) 0.64 , e ) 0.56
e
add(multiply(0.2, 0.4), multiply(0.6, 0.8))
multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)|
general
what is 0.05 percent of 12,356 ?
since , percent = 1 / 100 , what = something ( s ) , and is : = . we can write the question as s = 0.05 ( 1 / 100 ) 12,356 . the answer is 6.178 . hence , the correct answer is c .
a ) 0.6178 , b ) 61.78 , c ) 6.178 , d ) 0.06178 , e ) 0.006178
c
divide(multiply(0.05, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100)
add(const_2,const_3)|add(const_3,const_4)|add(const_3,const_3)|multiply(const_3,const_4)|multiply(#0,const_2)|multiply(#3,const_100)|multiply(#1,#0)|multiply(#4,#5)|multiply(#6,#4)|add(#7,#8)|add(#9,#2)|multiply(n0,#10)|divide(#11,const_100)
gain
15 lts are taken of from a container full of liquid a and replaced with liquid b . again 15 more lts of the mixture is taken and replaced with liquid b . after this process , if the container contains liquid a and b in the ratio 9 : 16 , what is the capacity of the container s ?
"if you have a 37.5 liter capacity , you start with 37.5 l of a and 0 l of b . 1 st replacement after the first replacement you have 37.5 - 15 = 22.5 l of a and 15 l of b . the key is figuring out how many liters of a and b , respectively , are contained in the next 15 liters of mixture to be removed . the current ratio of a to total mixture is 22.5 / 37.5 ; expressed as a fraction this becomes ( 45 / 2 ) / ( 75 / 2 ) , or 45 / 2 * 2 / 75 . canceling the 2 s and factoring out a 5 leaves the ratio as 9 / 15 . note , no need to reduce further as we ' re trying to figure out the amount of a and b in 15 l of solution . 9 / 15 of a means there must be 6 / 15 of b . multiply each respective ratio by 15 to get 9 l of a and 6 l of b in the next 15 l removal . final replacement the next 15 l removal means 9 liters of a and 6 liters of b are removed and replaced with 15 liters of b . 22.5 - 9 = 13.5 liters of a . 15 liters of b - 6 liters + 15 more liters = 24 liters of b . test to the see if the final ratio = 9 / 16 ; 13.5 / 24 = ( 27 / 2 ) * ( 1 / 24 ) = 9 / 16 . choice c is correct ."
a ) a : 45 , b ) b : 25 , c ) c : 37.5 , d ) d : 36 , e ) e : 42
c
divide(15, subtract(const_1, sqrt(divide(9, add(9, 16)))))
add(n2,n3)|divide(n2,#0)|sqrt(#1)|subtract(const_1,#2)|divide(n0,#3)|
general
a 20 litre mixture of milk and water contains milk and water in the ratio 3 : 2 . 10 litres of the mixture is removed and replaced with pure milk and the operation is repeated once more . at the end of the two removal and replacement , what is the ratio e of milk and water in the resultant mixture ?
"he 20 litre mixture contains milk and water in the ratio of 3 : 2 . therefore , there will be 12 litres of milk in the mixture and 8 litres of water in the mixture . step 1 . when 10 litres of the mixture is removed , 6 litres of milk is removed and 4 litres of water is removed . therefore , there will be 6 litres of milk and 4 litres of water left in the container . it is then replaced with pure milk of 10 litres . now the container will have 16 litres of milk and 4 litres of water . step 2 . when 10 litres of the new mixture is removed , 8 litres of milk and 2 litres of water is removed . the container will have 8 litres of milk and 2 litres of water in it . now 10 litres of pure milk is added . therefore , the container will have 18 litres of milk and 2 litres of water in it at the end of the second step . therefore , the ratio of milk and water is 18 : 2 or 9 : 1 . shortcut . we are essentially replacing water in the mixture with pure milk . let w _ o be the amount of water in the mixture originally = 8 litres . let w _ r be the amount of water in the mixture after the replacements have taken place . then , { w _ r } / { w _ o } = ( 1 - r / m ) ^ n where r is the amount of the mixture replaced by milk in each of the steps , m is the total volume of the mixture and n is the number of times the cycle is repeated . hence , { w _ r } / { w _ o } Β  = ( 1 / 2 ) ^ 2 Β  = 1 / 4 therefore e , w _ r Β  = { w _ o } / 4 = 8 / 4 Β  = 2 litres . b"
a ) 17 : 3 , b ) 9 : 1 , c ) 3 : 17 , d ) 5 : 3 , e ) 11 : 2
b
divide(add(multiply(divide(add(multiply(divide(3, add(3, 2)), subtract(20, 10)), 10), 20), subtract(20, 10)), 10), multiply(divide(multiply(divide(2, add(3, 2)), subtract(20, 10)), 20), subtract(20, 10)))
add(n1,n2)|subtract(n0,n3)|divide(n1,#0)|divide(n2,#0)|multiply(#2,#1)|multiply(#3,#1)|add(n3,#4)|divide(#5,n0)|divide(#6,n0)|multiply(#7,#1)|multiply(#8,#1)|add(n3,#10)|divide(#11,#9)|
general
an article is bought for rs . 800 and sold for rs . 1200 , find the loss percent ?
"800 - - - - 400 400 - - - - ? = > 50 % answer : e"
a ) 55 % , b ) 25.25 % , c ) 33.33 % , d ) 45 % , e ) 50 %
e
subtract(const_100, divide(multiply(1200, const_100), 800))
multiply(n1,const_100)|divide(#0,n0)|subtract(const_100,#1)|
gain
when greenville state university decided to move its fine arts collection to a new library , it had to package the collection in 20 - inch by 20 - inch by 15 - inch boxes . if the university pays $ 0.50 for every box , and if the university needs 3.06 million cubic inches to package the collection , what is the minimum amount the university must spend on boxes ?
"total no . of boxes = 3060000 / ( 20 Γ— 20 Γ— 15 ) = 510 total cost = 510 Γ— $ 0.5 = $ 255 answer a"
a ) $ 255 , b ) $ 275 , c ) $ 510 , d ) $ 1,250 , e ) $ 2,550
a
multiply(divide(multiply(3.06, multiply(const_1000, const_1000)), multiply(multiply(20, 20), 15)), 0.50)
multiply(const_1000,const_1000)|multiply(n0,n0)|multiply(n4,#0)|multiply(n2,#1)|divide(#2,#3)|multiply(n3,#4)|
general
diana works 10 hours per day on monday , wednesday and friday , and 15 hours per day on tuesday and thursday . she does not work on saturday and sunday . she earns $ 1800 per week . how much does she earn in dollars per hour ?
so , she works 30 hours in 3 days so , she works 30 hours in 2 days so in a week she works 60 hours ( 30 + 30 ) and earns $ 1800 so , hourly wage is 1800 / 60 = > 30 hence answer will be ( c ) 30
a ) 20 , b ) 22 , c ) 30 , d ) 24 , e ) 25
c
divide(1800, add(multiply(const_2, 15), multiply(10, const_3)))
multiply(n1,const_2)|multiply(n0,const_3)|add(#0,#1)|divide(n2,#2)
physics
a car travels at a speed of 65 miles per hour . how far will it travel in 8 hours ?
"during each hour , the car travels 65 miles . for 8 hours it will travel 65 + 65 + 65 + 65 + 65 + 65 + 65 + 65 = 8 Γ— 65 = 520 miles correct answer is c ) 520 miles"
a ) 125 miles , b ) 225 miles , c ) 520 miles , d ) 425 miles , e ) 525 miles
c
multiply(65, 8)
multiply(n0,n1)|
physics
the mean of 50 observations was 36 . it was found later that an observation 30 was wrongly taken as 23 . the corrected new mean is :
explanation : correct sum = ( 36 * 50 + 30 - 23 ) = 1825 . correct mean = = 1807 / 50 = 36.14 answer : b ) 36.14
a ) 36.02 , b ) 36.14 , c ) 36.12 , d ) 36.11 , e ) 36.8
b
divide(add(multiply(36, 50), subtract(subtract(50, const_2), 23)), 50)
multiply(n0,n1)|subtract(n0,const_2)|subtract(#1,n3)|add(#0,#2)|divide(#3,n0)
general
a , b and c invest in the ratio of 3 : 4 : 5 . the percentage of return on their investments are in the ratio of 6 : 5 : 4 . find the total earnings , if b earns rs . 200 more than a :
explanation : a b c investment 3 x 4 x 5 x rate of return 6 y % 5 y % 4 y % return \ inline \ frac { 18 xy } { 100 } \ inline \ frac { 20 xy } { 100 } \ inline \ frac { 20 xy } { 100 } total = ( 18 + 20 + 20 ) = \ inline \ frac { 58 xy } { 100 } b ' s earnings - a ' s earnings = \ inline \ frac { 2 xy } { 100 } = 200 total earning = \ inline \ frac { 58 xy } { 100 } = 5800 answer : e ) rs . 5800
a ) 2348 , b ) 7250 , c ) 2767 , d ) 1998 , e ) 5800
e
multiply(add(add(multiply(3, 6), multiply(4, 5)), multiply(5, 4)), divide(200, subtract(multiply(4, 5), multiply(3, 6))))
multiply(n0,n3)|multiply(n1,n2)|add(#0,#1)|subtract(#1,#0)|add(#2,#1)|divide(n6,#3)|multiply(#4,#5)
general
the average of first 14 even numbers is ?
"sum of 14 even numbers = 14 * 15 = 210 average = 210 / 14 = 15 answer : e"
a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 15
e
add(14, const_1)
add(n0,const_1)|
general
a certain rectangular crate measures 12 feet by 16 feet by 18 feet . a cylindrical gas tank is to be made for shipment in the crate and will stand upright when the crate is placed on one of its six faces . what should the radius of the tank be if it is to be of the largest possible volume ?
for max volume of cylinder ( pi * r ^ 2 * h ) we need to max out r ^ 2 * h . we do n ' t know what the dimensions of the crate refer to . . therefore for max vol base should be 18 x 16 i . e . of radius 16 / 2 = 8 d
['a ) 4', 'b ) 5', 'c ) 6', 'd ) 8', 'e ) 10']
d
sqrt(divide(volume_cylinder(divide(16, const_2), 12), multiply(const_pi, 12)))
divide(n1,const_2)|multiply(n0,const_pi)|volume_cylinder(#0,n0)|divide(#2,#1)|sqrt(#3)
geometry
for what values of k will the pair of equations 4 ( 3 x + 4 y ) = 48 and kx + 12 y = 30 does not have a unique solution ?
we have 2 equations 1 . 4 ( 3 x + 4 y ) = 48 - - > 3 x + 4 y = 12 - - > 9 x + 12 y = 36 2 . kx + 12 y = 30 substract 1 - 2 , we get ( 9 - k ) x = 6 i . e . x = 6 / ( 9 - k ) then , by looking at options , we get some value of x except for b . when we put k = 9 , x becomes 6 / 0 and hence answer is b
a ) 12 , b ) 9 , c ) 3 , d ) 7.5 , e ) 2.5
b
divide(multiply(12, 3), 4)
multiply(n1,n4)|divide(#0,n0)
general
the sector of a circle has radius of 21 cm and central angle 120 o . find its perimeter ?
"perimeter of the sector = length of the arc + 2 ( radius ) = ( 120 / 360 * 2 * 22 / 7 * 21 ) + 2 ( 21 ) = 44 + 42 = 86 cm answer : d"
a ) 91.5 cm , b ) 11.5 cm , c ) 91.8 cm , d ) 86 cm , e ) 99.5 cm
d
multiply(multiply(const_2, divide(multiply(subtract(21, const_3), const_2), add(const_4, const_3))), 21)
add(const_3,const_4)|subtract(n0,const_3)|multiply(#1,const_2)|divide(#2,#0)|multiply(#3,const_2)|multiply(n0,#4)|
physics
how many different positive integers are factors of 40 ?
2 * 20 4 * 10 8 * 5 answer : d
a ) 3 , b ) 2 , c ) 1 , d ) 6 , e ) 4
d
add(power(const_2, const_2), const_2)
power(const_2,const_2)|add(#0,const_2)
other
a couple married in 1980 had two children , one in 1982 and the other in 1984 . their combined ages will equal the years of the marriage in ?
sum of year = 1 + 9 + 8 + 0 = ( 18 ) 1 + 8 = 9 1 + 9 + 8 + 2 = ( 20 ) 2 + 0 = 2 1 + 9 + 8 + 4 = ( 13 ) 1 + 3 = 4 now 9 + 2 + 4 = ( 15 ) 1 + 5 = 6 nd option a 1 + 9 + 8 + 6 = 24 ( 2 + 4 ) = 6 answer : a
a ) 1986 , b ) 1985 , c ) 1987 , d ) 1988 , e ) 1989
a
add(1980, add(subtract(1982, 1980), subtract(1984, 1980)))
subtract(n1,n0)|subtract(n2,n0)|add(#0,#1)|add(n0,#2)
general
find the expenditure on digging a well 14 m deep and of 3 m diameter at rs . 19 per cubic meter ?
"22 / 7 * 14 * 3 / 2 * 3 / 2 = 99 m 2 99 * 19 = 1881 answer : b"
a ) 2998 , b ) 1881 , c ) 2890 , d ) 1485 , e ) 2780
b
multiply(volume_cylinder(divide(3, const_2), 14), 19)
divide(n1,const_2)|volume_cylinder(#0,n0)|multiply(n2,#1)|
physics
at an election 2 candidates are participated and a candidate got 45 % of votes and defeated by 9000 . and 83 votes are invalid . find the total polled votes ?
winner votes = 100 - 45 = 55 polled votes = [ ( 100 * 9000 ) / 2 * 55 - 100 ] + 83 = 90083 answer is d
a ) 84512 , b ) 78452 , c ) 65893 , d ) 90083 , e ) 74502
d
divide(add(9000, 83), subtract(const_1, divide(multiply(45, 2), const_100)))
add(n2,n3)|multiply(n0,n1)|divide(#1,const_100)|subtract(const_1,#2)|divide(#0,#3)
gain
after working for 6 days , david was joined by moore . together they completed the remaining job in 3 days . how many days will it take both of them to complete the entire job , given that it would have taken david 12 days to complete the job alone ?
explanation : david and moore complete half work in 3 days = > they can complete whole work in 6 days answer : option c
a ) 7 , b ) 4 , c ) 6 , d ) 2 , e ) 7
c
multiply(divide(12, 6), 3)
divide(n2,n0)|multiply(n1,#0)
physics
how long does a train 110 m long running at the speed of 54 km / hr takes to cross a bridge 132 m length ?
"speed = 54 * 5 / 18 = 15 m / sec total distance covered = 110 + 132 = 242 m . required time = 242 / 15 = 16.13 sec . answer : d"
a ) 12.9 sec , b ) 12.1 sec , c ) 17.9 sec , d ) 16.13 sec , e ) 14.9 sec
d
divide(add(110, 132), multiply(54, const_0_2778))
add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)|
physics
by selling an article at rs . 800 , a shopkeeper makes a profit of 25 % . at what price should he sell the article so as to make a loss of 35 % ?
"sp = 800 profit = 25 % cp = ( sp ) * [ 100 / ( 100 + p ) ] = 800 * [ 100 / 125 ] = 640 loss = 25 % = 25 % of 640 = rs . 224 sp = cp - loss = 640 - 224 = rs . 416 answer : d"
a ) 228 , b ) 480 , c ) 267 , d ) 416 , e ) 276
d
subtract(divide(multiply(800, const_100), add(25, const_100)), divide(multiply(divide(multiply(800, const_100), add(25, const_100)), 35), const_100))
add(n1,const_100)|multiply(n0,const_100)|divide(#1,#0)|multiply(n2,#2)|divide(#3,const_100)|subtract(#2,#4)|
gain
the average weight of 8 people increases by 2.5 kg when a new person comes in place of one of them weighing 55 kg . what is the weight of the new person ?
"the total weight increase = ( 8 x 2.5 ) kg = 20 kg weight of new person = ( 55 + 20 ) kg = 75 kg the answer is a ."
a ) 75 kg , b ) 85 kg , c ) 95 kg , d ) 65 kg , e ) 55 kg
a
add(multiply(2.5, 8), 55)
multiply(n0,n1)|add(n2,#0)|
general
the length of a rectangular plot is 10 mtr more than its width . the cost of fencing the plot along its perimeter at the rate of rs . 6.5 mtr is rs . 910 . the perimeter of the plot is ?
"sol . let width = x , length = ( 10 + x ) perimeter = 2 ( x + ( 10 + x ) ) = 2 ( 2 x = 10 ) & 2 ( 2 x + 10 ) * 6.5 = 910 x = 30 required perimeter = 2 ( 30 + 40 ) = 140 d"
a ) 126 , b ) 156 , c ) 190 , d ) 140 , e ) 260
d
multiply(add(divide(subtract(divide(divide(910, 6.5), const_2), 10), const_2), add(divide(subtract(divide(divide(910, 6.5), const_2), 10), const_2), 10)), const_2)
divide(n2,n1)|divide(#0,const_2)|subtract(#1,n0)|divide(#2,const_2)|add(#3,n0)|add(#4,#3)|multiply(#5,const_2)|
geometry
what is the least number which should be added to 5432 so that the sum is exactly divisible by 5 , 6 , 4 , and 3 ?
l . c . m . of 5 , 6 , 4 and 3 = 60 . when dividing 5432 by 60 , the remainder is 32 . the number to be added = 60 - 32 = 28 . the answer is c .
a ) 16 , b ) 22 , c ) 28 , d ) 34 , e ) 40
c
add(3, add(4, add(6, add(5, multiply(5, const_2)))))
multiply(n1,const_2)|add(n1,#0)|add(n2,#1)|add(n3,#2)|add(n4,#3)
general
two airplanes take off from one airfield at noon . one flies due east at 203 miles per hour while the other flies directly northeast at 283 miles per hour . approximately how many miles apart are the airplanes at 2 p . m . ?
"c in two hours : the plane flying east will be 406 miles away from airport . the other plane will be 566 miles away from airport . 566 / 406 = ~ 1.4 = ~ sqrt ( 2 ) this means that planes formed a right isocheles triangle = > sides of such triangles relate as 1 : 1 : sqrt ( 2 ) = > the planes are 406 miles apart . c"
a ) 166 , b ) 332 , c ) 406 , d ) 483 , e ) 566
c
sqrt(subtract(power(multiply(283, 2), 2), power(multiply(203, 2), 2)))
multiply(n1,n2)|multiply(n0,n2)|power(#0,n2)|power(#1,n2)|subtract(#2,#3)|sqrt(#4)|
physics
a man can swim in still water at 6 km / h , but takes twice as long to swim upstream than downstream . the speed of the stream is ?
"m = 6 s = x ds = 6 + x us = 6 - x 6 + x = ( 6 - x ) 2 6 + x = 12 - 2 x 3 x = 6 x = 2 answer : d"
a ) 3 , b ) 7.5 , c ) 2.25 , d ) 2 , e ) 4
d
divide(6, const_3)
divide(n0,const_3)|
general
the population of a town is 10000 . it increases annually at the rate of 20 % p . a . what will be its population after 4 years ?
"formula : ( after = 100 denominator ago = 100 numerator ) 10000 Γ£ β€” 120 / 100 Γ£ β€” 120 / 100 Γ£ β€” 120 / 100 Γ£ β€” 120 / 100 = 20730 c"
a ) 20300 , b ) 20400 , c ) 20730 , d ) 20500 , e ) 20600
c
add(10000, multiply(divide(multiply(10000, 20), const_100), 4))
multiply(n0,n1)|divide(#0,const_100)|multiply(#1,n2)|add(n0,#2)|
gain
a certain roller coaster has 4 cars , and a passenger is equally likely to ride in any 1 of the 4 cars each time that passenger rides the roller coaster . if a certain passenger is to ride the roller coaster 4 times , what is the probability that the passenger will ride in each of the 4 cars ?
"if he is to ride 4 times and since he can choose any of the 4 cars each time , total number of ways is = 4 * 4 * 4 * 4 = 256 now the number of ways if he is to choose a different car each time is = 4 * 3 * 2 * 1 = 24 so the probability is = 24 / 256 = 6 / 64 = 3 / 32 = 3 / 32 answer : d"
a ) 0 , b ) 1 / 9 , c ) 2 / 9 , d ) 3 / 32 , e ) 1
d
multiply(factorial(4), power(divide(1, 4), 4))
divide(n1,n0)|factorial(n0)|power(#0,n0)|multiply(#1,#2)|
general
what will be the ratio between the area of a rectangle and the area of a triangle with one of the sides of rectangle as base and a vertex on the opposite side of rectangle .
"area of Ξ΄ ebc = 1 ⁄ 2 Γ— bc Γ— ef = 1 ⁄ 2 Γ— bc Γ— ab [ since , ef = ab ] area of Ξ΄ ebc = 1 ⁄ 2 Γ— area of Ξ΄ abcd \ required ratio = 2 : 1 . answer b"
a ) 1 : 2 , b ) 2 : 1 , c ) 3 : 1 , d ) data inadequate , e ) none of these
b
divide(rectangle_area(const_1, const_1), triangle_area(rectangle_area(const_1, const_1), rectangle_area(const_1, const_1)))
rectangle_area(const_1,const_1)|triangle_area(#0,#0)|divide(#0,#1)|
geometry
the average age of a , b and c is 25 years . if the average age of a and c is 29 years , what is the age of b in years ?
"age of b = age of ( a + b + c ) Γ’ € β€œ age of ( a + c ) = 25 Γ£ β€” 3 Γ’ € β€œ 29 Γ£ β€” 2 = 75 Γ’ € β€œ 58 = 17 years a"
a ) 17 , b ) 35 , c ) 20 , d ) 32 , e ) 21
a
subtract(multiply(25, const_3), multiply(29, const_2))
multiply(n0,const_3)|multiply(n1,const_2)|subtract(#0,#1)|
general
a man walking at the rate of 6 km per hour crosses a square field diagonally in 9 seconds the area of the field is
explanation : distance covered in 9 seconds = ( 6 Γ— 1000 / 3600 ) Γ— 9 = 15 m diagonal of square field = 15 m side of square = a then diagonal of that square = √ 2 a hence area of the square = a 2 = ( 152 ) / 2 = 112.5 sq . m answer : option c
['a ) 110 sq . m', 'b ) 111.5 aq . m', 'c ) 112.5 sq . m', 'd ) 114 sq . m', 'e ) none of these']
c
divide(power(multiply(divide(multiply(6, const_1000), multiply(const_360, const_10)), 9), const_2), const_2)
multiply(n0,const_1000)|multiply(const_10,const_360)|divide(#0,#1)|multiply(n1,#2)|power(#3,const_2)|divide(#4,const_2)
geometry
in a college the ratio of the numbers of boys to the girls is 8 : 5 . if there are 175 girls , the total number of students in the college is ?
"let the number of boys and girls be 8 x and 5 x then , 5 x = 175 x = 35 total number of students = 13 x = 13 * 35 = 455 answer is c"
a ) 562 , b ) 356 , c ) 455 , d ) 416 , e ) 512
c
add(multiply(divide(8, 5), 175), 175)
divide(n0,n1)|multiply(n2,#0)|add(n2,#1)|
other
i spend 40 hours a week ( 5 days ) at work and like to organize my time so that i spend an equal number of hours on the two tasks i currently have . currently , i am spending 5 hours a day on task 1 and 3 on task 2 . how many hours a week less do i need to spend on task 1 in order to evenly distribute my time ?
5 x 5 = 25 25 - 5 = 20 the answer is a .
a ) 5 , b ) 3 , c ) 7 , d ) 1 , e ) 6
a
add(subtract(subtract(divide(40, const_2), 3), subtract(divide(40, const_2), 5)), 3)
divide(n0,const_2)|subtract(#0,n4)|subtract(#0,n1)|subtract(#1,#2)|add(n4,#3)
physics
what is the greatest of 3 consecutive integers whose sum is 27 ?
"the sum of three consecutive integers can be written as n + ( n + 1 ) + ( n + 2 ) = 3 n + 3 if the sum is 24 , we need to solve the equation 3 n + 3 = 27 ; = > 3 n = 27 ; = > n = 9 the greatest of the three numbers is therefore 9 + 2 = 11 answer : e"
a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 11
e
add(divide(subtract(27, 3), 3), const_2)
subtract(n1,n0)|divide(#0,n0)|add(#1,const_2)|
physics
the area of a triangle is with base 4 m and height 5 m ?
1 / 2 * 4 * 5 = 10 m 2 answer : b
['a ) 11', 'b ) 10', 'c ) 99', 'd ) 17', 'e ) 12']
b
triangle_area(4, 5)
triangle_area(n0,n1)
geometry
a pharmaceutical company received $ 5 million in royalties on the first $ 20 million in sales of and then $ 10 million in royalties on the next $ 108 million in sales . by approximately what percentage did the ratio of royalties to sales decrease from the first $ 20 million in sales to the next $ 108 million in sales ?
( 10 / 108 ) / ( 5 / 20 ) = 20 / 27 = 74 % it means that 10 / 108 represents only 74 % . therefore a decrease of 26 % . answer b
a ) 8 % , b ) 26 % , c ) 45 % , d ) 52 % , e ) 56 %
b
multiply(divide(5, 20), const_100)
divide(n0,n1)|multiply(#0,const_100)
general
peter brought a scooter for a certain sum of money . he spent 10 % of the cost on repairs and sold the scooter for a profit of $ 1100 . how much did he spend on repairs if he made a profit of 20 % ?
e 500 let the c . p . be $ x . then , 20 % of x = 1100 20 / 100 * x = 1100 = > x = 5500 c . p . = $ 5500 , expenditure on repairs = 10 % actual price = $ ( 100 * 5500 ) / 110 = $ 5000 expenditures on repairs = ( 5500 - 5000 ) = $ 500 .
a ) 100 , b ) 200 , c ) 300 , d ) 400 , e ) 500
e
subtract(divide(multiply(const_100, 1100), 20), multiply(divide(divide(multiply(const_100, 1100), 20), add(const_100, const_10)), const_100))
add(const_10,const_100)|multiply(n1,const_100)|divide(#1,n2)|divide(#2,#0)|multiply(#3,const_100)|subtract(#2,#4)
gain
10 people can write 55 book in 12 days working 8 hour a day . then in how many day 198 can be written by 48 people ?
"work per day epr hour per person = 55 / ( 12 * 8 * 10 ) / / eq - 1 people = 48 ; let suppose day = p ; per day work for 8 hours acc . to condition work per day epr hour per person = 198 / ( p * 8 * 48 ) / / eq - 2 eq - 1 = = eq - 2 ; p = 9 answer : a"
a ) 9 , b ) 13 , c ) 21 , d ) 24 , e ) 23
a
divide(multiply(55, 12), divide(48, const_2))
divide(n5,const_2)|multiply(n1,n2)|divide(#1,#0)|
physics
the current of a stream runs at the rate of 4 kmph . a boat goes 12 km and back to the starting point in 2 hours , then find the speed of the boat in still water ?
"s = 4 m = x ds = x + 4 us = x - 4 12 / ( x + 4 ) + 12 / ( x - 4 ) = 2 x = 13.21 answer : e"
a ) a ) 17.33 , b ) b ) 2 , c ) c ) 18 , d ) d ) 16 , e ) e ) 13.21
e
divide(power(4, 2), 2)
power(n0,n2)|divide(#0,n2)|
physics
a train 200 m long passes a man , running at 5 km / hr in the same direction in which the train is going , in 10 seconds . the speed of the train is :
"speed of the train relative to man = ( 200 / 10 ) m / sec = 20 m / sec . [ 20 * ( 18 / 5 ) ] km / hr = 72 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 5 ) km / hr . x - 5 = 72 = = > x = 77 km / hr . answer : c"
a ) 45 km / hr , b ) 50 km / hr , c ) 77 km / hr , d ) 55 km / hr , e ) 56 km / hr
c
divide(divide(subtract(200, multiply(multiply(5, const_0_2778), 5)), 5), const_0_2778)
multiply(n1,const_0_2778)|multiply(n1,#0)|subtract(n0,#1)|divide(#2,n1)|divide(#3,const_0_2778)|
physics
to fill a tank , 200 buckets of water is required . how many buckets of water will be required to fill the same tank if the capacity of the bucket is reduced to four - fifths of its present ?
"let the capacity of 1 bucket = x . then , the capacity of tank = 200 x . new capacity of bucket = 4 / 5 x therefore , required number of buckets = ( 200 x ) / ( 4 x / 5 ) = ( 200 x ) x 5 / 4 x = 1000 / 4 = 250 answer is e ."
a ) 50 , b ) 100 , c ) 150 , d ) 200 , e ) 250
e
divide(multiply(200, add(const_4, const_1)), const_2)
add(const_1,const_4)|multiply(n0,#0)|divide(#1,const_2)|
physics
if the cost price of 19 articles is equal to the selling price of 16 articles , what is the percentage of profit or loss that the merchant makes ?
"explanation : let cost price of 1 article be re . 1 . therefore , cost price of 19 articles = rs . 19 . selling price of 16 articles = rs . 19 therefore , selling price of 19 articles is : - = > 19 / 16 Γ£ β€” 19 = > 22.56 . therefore , profit = selling price - cost price . = > 22.56 Γ’ Λ† ’ 19 = 3.56 . hence , the percentage of profit = profit x 100 / c . p . = > 3.56 / 19 Γ£ β€” 100 . = > 18.75 % profit . answer : b"
a ) 18.75 % loss , b ) 18.75 % profit , c ) 33.33 % loss , d ) 30.33 % loss , e ) none of these
b
multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 16), 19)), divide(multiply(const_100, 16), 19)))
multiply(n1,const_100)|divide(#0,n0)|subtract(const_100,#1)|divide(#2,#1)|multiply(#3,const_100)|
gain
the mean of 40 observations was 100 . it was found later that an observation 50 was wrongly taken as 75 the corrected new mean is
"explanation : correct sum = ( 100 * 40 + 50 - 75 ) = 3975 . correct mean = = 3975 / 40 = 99.375 answer : c"
a ) 125.8 , b ) 102.8 , c ) 99.375 , d ) 100.52 , e ) 122
c
divide(add(multiply(100, 40), subtract(subtract(40, const_2), 75)), 40)
multiply(n0,n1)|subtract(n0,const_2)|subtract(#1,n3)|add(#0,#2)|divide(#3,n0)|
general
the entrance fee for a fair is $ 5 for persons under the age of 18 , and 20 % more for persons older . each ride at the fair costs $ 0.50 . if joe goes with her 6 years old twin brothers , and they each took 3 rides in total . how much money does joe end up spending at the fair ?
total entrance fee is ( 2 * $ 5 ) + ( 1.20 * 5 ) = $ 16 total rides fee is ( 0.50 * 3 ) * 3 = $ 4.50 total money spent is $ 20.50 answer is b
a ) 16 , b ) 20.5 , c ) 17.5 , d ) 20 , e ) 4.5
b
add(add(multiply(5, const_2), multiply(add(divide(20, const_100), const_1), 5)), multiply(multiply(3, const_3), 0.5))
divide(n2,const_100)|multiply(n0,const_2)|multiply(n5,const_3)|add(#0,const_1)|multiply(n3,#2)|multiply(n0,#3)|add(#1,#5)|add(#6,#4)
general
in the manufacture of a certain product , 7 percent of the units produced are defective and 5 percent of the defective units are shipped for sale . what percent of the units produced are defective units that are shipped for sale ?
"0.07 * 0.05 = 0.0035 = 0.35 % the answer is b ."
a ) 0.125 % , b ) 0.35 % , c ) 0.8 % , d ) 1.25 % , e ) 2.0 %
b
multiply(7, divide(5, const_100))
divide(n1,const_100)|multiply(n0,#0)|
gain
a bag contains 6 red , 5 blue and 2 green balls . if 2 ballsare picked at random , what is the probability that both are red ?
"p ( both are red ) , = 6 c 213 c 2 = 6 c 213 c 2 = 5 / 26 c"
a ) 2 / 13 , b ) 3 / 17 , c ) 5 / 26 , d ) 5 / 21 , e ) 7 / 15
c
divide(choose(6, 2), choose(add(add(6, 5), 2), 2))
add(n0,n1)|choose(n0,n3)|add(n2,#0)|choose(#2,n3)|divide(#1,#3)|
other
if 1488 / 1.24 = 1200 , then 148.8 / 12.4 is equal to ?
"answer given expression 148.8 / 12.4 = 1488 / 124 = 1488 / ( 1.24 x 100 ) = 1200 / 100 = 12 correct option : d"
a ) 17 , b ) 10 , c ) 14 , d ) 12 , e ) 11
d
divide(multiply(1200, 1.24), 1488)
multiply(n1,n2)|divide(#0,n0)|
general
the cross - section of a cannel is a trapezium in shape . if the cannel is 6 m wide at the top and 4 m wide at the bottom and the area of cross - section is 10290 sq m , the depth of cannel is ?
1 / 2 * d ( 6 + 4 ) = 10290 d = 258 answer : a
a ) 258 , b ) 260 , c ) 262 , d ) 264 , e ) 266
a
divide(divide(divide(10290, divide(add(6, 4), const_2)), const_4), const_2)
add(n0,n1)|divide(#0,const_2)|divide(n2,#1)|divide(#2,const_4)|divide(#3,const_2)
physics
a girl scout was selling boxes of cookies . in a month , she sold both boxes of chocolate chip cookies ( $ 1.25 each ) and boxes of plain cookies ( $ 0.75 each ) . altogether , she sold 1,585 boxes for a combined value of $ 1 , 586.25 . how many boxes of plain cookies did she sell ?
let # plain cookies sold be x then # chocolate cookies = ( total cookies - x ) equating for x ( 0.75 ) * x + ( 1.25 ) * ( 1585 - x ) = 1586.25 = > x = 790 e
a ) 0 , b ) 233 , c ) 500 , d ) 695 , e ) 790
e
divide(add(const_1000, 586.25), const_2)
add(n4,const_1000)|divide(#0,const_2)
other
in what ratio p : q should the mixture p of milk and water in the ratio of 5 : 4 be mixed with another mixture q of milk and water in the ratio 2 : 7 so that the resultant mixture contains equal quantities of milk and water ?
( 5 / 9 ) * p + ( 2 / 9 ) * q = ( 4 / 9 ) * p + ( 7 / 9 ) * q p = 5 q p / q = 5 / 1 the answer is e .
a ) 3 : 2 , b ) 2 : 1 , c ) 4 : 3 , d ) 4 : 1 , e ) 5 : 1
e
add(subtract(add(4, 7), add(5, 2)), const_1)
add(n1,n3)|add(n0,n2)|subtract(#0,#1)|add(#2,const_1)
general
a certain company expects quarterly earnings of $ 0.80 per share of stock , half of which will be distributed as dividends to shareholders while the rest will be used for research and development . if earnings are greater than expected , shareholders will receive an additional $ 0.04 per share for each additional $ 0.10 of per share earnings . if quarterly earnings are $ 1.10 per share , what will be the dividend paid to a person who owns 500 shares of the company ' s stock ?
"eps actual > eps expected . each gets and additional . 12 per share . thus . 52 * 500 - - > $ 260 answer is a"
a ) $ 260 , b ) $ 96 , c ) $ 26 , d ) $ 120 , e ) $ 240
a
multiply(add(divide(0.80, const_2), multiply(multiply(subtract(1.10, 0.80), const_10), 0.04)), 500)
divide(n0,const_2)|subtract(n3,n0)|multiply(#1,const_10)|multiply(n1,#2)|add(#0,#3)|multiply(n4,#4)|
general
the positive integers s and t leave remainders of 2 and 3 , respectively , when divided by 6 . s > t . what is the remainder when s – t is divided by 6 ?
let ' s test out some values of p and q that satisfy the given information . s leaves are remainder of 2 when divided by 6 so , s could equal 8 t leaves are remainder of 3 when divided by 6 so , t could equal 3 what is the remainder when s – t is divided by 6 ? so , s - t = 8 - 3 = 5 , and when we divide 5 by 6 , we get 0 with remainder 5 answer : e
a ) 3 , b ) 1 , c ) 4 , d ) 2 , e ) 5
e
subtract(add(2, 6), 3)
add(n0,n2)|subtract(#0,n1)
general
what is the remainder when 9 ^ 381 is divided by 5 ?
i also agree that the remainder is ' 4 ' ( using the last digit of the powers of 7 ) . could we have the official answer please ? e
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4
e
subtract(divide(5, const_2), multiply(9, 9))
divide(n2,const_2)|multiply(n0,n0)|subtract(#0,#1)|
general
the sum of ages of 5 children born 2 years different each is 40 years . what is the age of the elder child ?
"let the ages of children be x , ( x + 2 ) , ( x + 4 ) , ( x + 6 ) and ( x + 8 ) years . then , x + ( x + 2 ) + ( x + 4 ) + ( x + 6 ) + ( x + 8 ) = 40 5 x = 20 x = 4 . x + 8 = 4 + 8 = 12 answer : a"
a ) 12 , b ) 9 , c ) 10 , d ) 16 , e ) 17
a
divide(add(add(add(add(2, const_4), add(2, const_4)), add(const_4, const_4)), 40), 5)
add(n1,const_4)|add(const_4,const_4)|add(#0,#0)|add(#2,#1)|add(n2,#3)|divide(#4,n0)|
general
find large number from below question the difference of two no . is 1365 . on dividing the larger no . by the smaller , we get 6 as quotient and the 15 as remainder ?
"let the smaller number be x . then larger number = ( x + 1365 ) . x + 1365 = 6 x + 15 5 x = 1350 x = 270 large number = 270 + 1365 = 1635 e"
a ) 1234 , b ) 1456 , c ) 1256 , d ) 1456 , e ) 1635
e
multiply(divide(subtract(1365, 15), subtract(6, const_1)), 6)
subtract(n0,n2)|subtract(n1,const_1)|divide(#0,#1)|multiply(n1,#2)|
general
the compound interest earned by sunil on a certain amount at the end of two years at the rate of 4 % p . a . was rs . 326.40 . find the total amount that sunil got back at the end of two years in the form of principal plus interest earned .
"let the sum be rs . p p { [ 1 + 4 / 100 ] 2 - 1 } = 326.40 p ( 4 / 100 ) ( 2 + 4 / 100 ) = 326.40 [ a 2 - b 2 = ( a - b ) ( a + b ) ] p = 326.40 / ( 0.04 ) ( 2.04 ) = 4000 amount = 4000 + 326.40 = 4326.40 answer : a"
a ) rs . 4326.40 , b ) rs . 4236.40 , c ) rs . 4136.40 , d ) rs . 4316.40 , e ) rs . 4136.40
a
add(divide(326.40, subtract(power(add(const_1, divide(4, const_100)), const_2), const_1)), 326.40)
divide(n0,const_100)|add(#0,const_1)|power(#1,const_2)|subtract(#2,const_1)|divide(n1,#3)|add(n1,#4)|
gain
if a square mirror has a 20 - inch diagonal , what is the approximate perimeter e of the mirror , in inches ?
if you draw the square and diagonal inside the square . u can see square becomes part of two triangles opposite to each other . and we know the property of the triangle , addition of two sides of triangle must be greater than its diagonal in order to complete the triangle . and each side must be less than 20 and perimeter e must be less than 80 , so we can eliminate answer choice c , d and e . so side 1 + side 2 > 20 , that means side 1 or side 2 must be > 10 . so we can eliminate the answer choice a . now we are left with is b
['a ) 40', 'b ) 60', 'c ) 80', 'd ) 100', 'e ) 120']
b
square_perimeter(divide(20, power(add(const_1, const_1), inverse(const_2))))
add(const_1,const_1)|inverse(const_2)|power(#0,#1)|divide(n0,#2)|square_perimeter(#3)
geometry
james and david work at a grocery shop with 6 other workers . for an internal review , 2 of the 6 workers will be randomly chosen to be interviewed . what is the probability that james and david will both be chosen ?
"probability that james and david will both be chosen out of 6 workers = ( 2 / 6 ) * ( 1 / 5 ) = 1 / 15 answer e"
a ) 1 / 20 , b ) 1 / 22 , c ) 1 / 23 , d ) 1 / 25 , e ) 1 / 15
e
multiply(2, multiply(divide(const_1, 6), divide(const_1, subtract(6, const_1))))
divide(const_1,n0)|subtract(n0,const_1)|divide(const_1,#1)|multiply(#0,#2)|multiply(#3,n1)|
physics