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Shang Data Lab @UCSD 7

noelle walks from point a to point b at an average speed of 3 kilometers per hour . at what speed , in kilometers per hour , must noelle walk from point b to point a so that her average speed for the entire trip is 4 kilometers per hour ?

"let ' s suppose that speed while returning was xkm / h since the distance is same , we can apply the formula of avg speed avg speed = 2 s 1 s 2 / s 1 + s 2 4 = 2 * 3 * x / 3 + x x = 6 b is the answer"

a ) 35 , b ) 9.2 miles , c ) 120 , d ) 6 , e ) $ 480

d

divide(multiply(4, 3), subtract(multiply(const_2, 3), 4))

multiply(n0,n1)|multiply(n0,const_2)|subtract(#1,n1)|divide(#0,#2)|

physics

the product of x and y is a constant . if the value of x is increased by 40 % , by what percentage must the value of y be decreased ?

x * y = constt . let x = y = 100 in beginning i . e . x * y = 100 * 100 = 10000 x ( 100 ) - - - becomes - - - > 1.4 x ( 140 ) i . e . 140 * new ' y ' = 10000 i . e . new ' y ' = 10000 / 140 = 71.43 i . e . y decreases from 100 to 71.43 i . e . decrease of 28.57 % c

a ) 62 , b ) 3 , c ) 6 hours , d ) 28.57 % , e ) 87.3 %

d

multiply(subtract(const_1, divide(const_100, add(const_100, 40))), const_100)

add(n0,const_100)|divide(const_100,#0)|subtract(const_1,#1)|multiply(#2,const_100)

general

if a and b are the roots of the equation x 2 - 6 x + 6 = 0 , then the value of a 2 + b 2 is :

"sol . ( b ) the sum of roots = a + b = 6 product of roots = ab = 6 now , a 2 + b 2 = ( a + b ) 2 - 2 ab = 36 - 12 = 24 answer b"

a ) 2 , b ) 24 , c ) 360 , d ) 89 , e ) 3 , 012,121

b

add(power(divide(subtract(6, sqrt(subtract(power(6, 2), multiply(const_4, 6)))), 2), 2), power(divide(add(6, sqrt(subtract(power(6, 2), multiply(const_4, 6)))), 2), 2))

multiply(n1,const_4)|power(n1,n0)|subtract(#1,#0)|sqrt(#2)|add(n1,#3)|subtract(n1,#3)|divide(#5,n0)|divide(#4,n0)|power(#6,n0)|power(#7,n0)|add(#8,#9)|

general

a honey bee flies for 10 seconds , from a daisy to a rose . it immediately continues to a poppy , flying for 6 additional seconds . the distance the bee passed , flying from the daisy to the rose is 3 meters longer than the distance it passed flying from the rose to the poppy . the bee flies to the poppy at 3 meters per second faster than her speed flying to the rose . the bee flies how many meters per second from the daisy to the rose ?

"let the speed be ' s ' and let the distance between rose and poppy be ' x ' the problem boils down to : rose to poppy : s + 3 = x / 6 - - - - - - - 1 daisy to rose : s = ( x + 3 ) / 10 - - - - - - 2 so from 1 we can re write x as x = 6 s + 18 substitute the value of x in 2 gives us s = 4.2 m / s a"

a ) 49.9 kg , b ) 49 , c ) 310 , d ) 144 min , e ) 4.2

e

divide(add(multiply(6, 3), 3), 10)

multiply(n1,n3)|add(n2,#0)|divide(#1,n0)|

physics

6 persons in an organization including x and y were to be divided in two groups of 3 members each . the total number of groups containing both x and y is what fraction of the total number of groups which can be formed ?

the fraction is nothing but the probability . . number to choose 3 out of 6 = 6 c 3 number to choose x and y and 2 from remaining 4 = 4 c 2 . . prob of a and b choosen = 4 c 2 / 6 c 3 = 3 / 10 answer : d

a ) 481 , b ) it can not be determined from the information given , c ) 225 , d ) 3 / 10 , e ) rs . 1500

d

divide(factorial(3), multiply(subtract(6, const_1), const_4))

factorial(n1)|subtract(n0,const_1)|multiply(#1,const_4)|divide(#0,#2)

general

having scored 94 runs in the 19 th inning , a cricketer increases his average score by 4 . what will be his average score after 19 innings ?

"explanation : let the average score of the first 18 innings be n 18 n + 94 = 19 ( n + 4 ) = > n = 18 so , average score after 19 th innings = x + 4 = 22 . answer : d"

a ) 7 , b ) 240 meters , c ) 40 minutes , d ) 22 , e ) 1070

d

add(subtract(94, multiply(19, 4)), 4)

multiply(n1,n2)|subtract(n0,#0)|add(n2,#1)|

general

ratio between rahul and deepak is 4 : 2 , after 6 years rahul age will be 22 years . what is deepak present age

"explanation : present age is 4 x and 2 x , = > 4 x + 22 = 26 = > x = 1 so deepak age is = 2 ( 1 ) = 2 answer : option d"

a ) 510 , b ) 24 , c ) 11 1 / 9 , d ) 10 , e ) 2

e

divide(multiply(subtract(22, 6), 2), 4)

subtract(n3,n2)|multiply(n1,#0)|divide(#1,n0)|

other

a train 150 m long running at 72 kmph crosses a platform in 25 sec . what is the length of the platform ?

d = 72 * 5 / 18 = 25 = 500 – 150 = 299 . answer : c

a ) 180 m , b ) 5 , c ) 2500 , d ) 9 , e ) 350

e

subtract(multiply(25, multiply(72, const_0_2778)), 150)

multiply(n1,const_0_2778)|multiply(n2,#0)|subtract(#1,n0)|

physics

the present worth of a certain bill due sometime hence is rs . 1296 and the true discount is rs . 72 . what is the banker ' s discount ?

explanation : bg = ( td ) 2 / pw = 722 / 1296 = 72 × 72 / 1296 = 12 × 12 / 36 = 12 / 3 = rs . 4 bg = bd – td = > 4 = bd - 72 = > bd = 72 + 4 = rs . 76 answer : option a

a ) rs . 76 , b ) 40 , c ) 90 , d ) 3200 , e ) 26250

a

add(72, divide(power(72, const_2), 1296))

power(n1,const_2)|divide(#0,n0)|add(n1,#1)

gain

the grade point average of the entire class is 88 . if the average of one third of the class is 94 , what is the average of the rest of the class ?

"let x be the number of students in the class . let p be the average of the rest of the class . 88 x = ( 1 / 3 ) 94 x + ( 2 / 3 ) ( p ) x 264 = 94 + 2 p 2 p = 170 p = 85 . the answer is d ."

a ) 2.14 , b ) 85 , c ) 315 , d ) 75 kg , e ) 96

b

divide(subtract(multiply(88, const_4), 94), subtract(const_4, const_1))

multiply(n0,const_4)|subtract(const_4,const_1)|subtract(#0,n1)|divide(#2,#1)|

general

a truck covers a distance of 376 km at a certain speed in 8 hours . how much time would a car take at an average speed which is 18 kmph more than that of the speed of the truck to cover a distance which is 14 km more than that travelled by the truck ?

"explanation : speed of the truck = distance / time = 376 / 8 = 47 kmph now , speed of car = ( speed of truck + 18 ) kmph = ( 47 + 18 ) = 65 kmph distance travelled by car = 376 + 14 = 390 km time taken by car = distance / speed = 390 / 65 = 6 hours . answer – a"

a ) 273054 , b ) 3 : 2 , c ) 6 hours , d ) 4264 , e ) 40

c

divide(add(376, 14), add(divide(376, 8), 18))

add(n0,n3)|divide(n0,n1)|add(n2,#1)|divide(#0,#2)|

physics

the ratio of the ages of maala and kala is 3 : 5 . the total of their ages is 3.2 decades . the proportion of their ages after 0.8 decades will be [ 1 decade = 10 years ]

let , maala ’ s age = 3 a and kala ’ s age = 5 a then 3 a + 5 a = 32 a = 4 maala ’ s age = 12 years and kala ’ s age = 20 years proportion of their ages after 8 is = ( 12 + 8 ) : ( 20 + 8 ) = 20 : 28 = 5 : 7 answer : b

a ) 24 , b ) . 2 , c ) 1652 , d ) 5 : 7 , e ) 260 m

d

divide(add(multiply(divide(multiply(3.2, 10), add(3, 5)), 3), multiply(0.8, 10)), add(multiply(5, divide(multiply(3.2, 10), add(3, 5))), multiply(0.8, 10)))

add(n0,n1)|multiply(n2,n5)|multiply(n3,n5)|divide(#1,#0)|multiply(n0,#3)|multiply(n1,#3)|add(#4,#2)|add(#5,#2)|divide(#6,#7)

general

mike took a taxi to the airport and paid $ 2.50 to start plus $ 0.25 per mile . annie took a different route to the airport and paid $ 2.50 plus $ 5.00 in bridge toll fees plus $ 0.25 per mile . if each was charged exactly the same amount , and annie ' s ride was 18 miles , how many miles was mike ' s ride ?

"the cost of annie ' s ride was 2.5 + 5 + ( 0.25 * 18 ) = $ 12 let x be the distance of mike ' s ride . the cost of mike ' s ride is 2.5 + ( 0.25 * x ) = 12 0.25 * x = 9.5 x = 38 miles the answer is c ."

a ) 5 , b ) 38 , c ) 245 , d ) 20 , e ) s . 7000

b

divide(subtract(add(add(2.50, 5.00), multiply(0.25, 18)), 2.50), 0.25)

add(n0,n3)|multiply(n1,n5)|add(#0,#1)|subtract(#2,n0)|divide(#3,n1)|

general

a mobile battery in 1 hour charges to 20 percent . how much time ( in minute ) will it require more to charge to 55 percent .

1 hr = 20 percent . thus 15 min = 5 percent . now to charge 55 percent 165 min . answer : d

a ) 6 / 11 , b ) 2 , c ) 165 , d ) 4 / 12 , e ) 331 times

c

multiply(divide(55, 20), const_60)

divide(n2,n1)|multiply(#0,const_60)|

physics

in a box of 11 pens , a total of 3 are defective . if a customer buys 2 pens selected at random from the box , what is the probability that neither pen will be defective ?

"p ( neither pen is defective ) = 8 / 11 * 7 / 10 = 28 / 55 the answer is a ."

a ) 21 years , b ) 43 , c ) 23 , d ) 10 , e ) 28 / 55

e

multiply(divide(subtract(11, 3), 11), divide(subtract(subtract(11, 3), const_1), subtract(11, const_1)))

subtract(n0,n1)|subtract(n0,const_1)|divide(#0,n0)|subtract(#0,const_1)|divide(#3,#1)|multiply(#2,#4)|

general

an engineer designed a ball so that when it was dropped , it rose with each bounce exactly one - half as high as it had fallen . the engineer dropped the ball from a 32 - meter platform and caught it after it had traveled 46.5 meters . how many times did the ball bounce ?

"going down = 32 m going up = 16 - - > total = 48 going down = 16 - - > total = 64 going up = 8 - - > total = 72 going down = 8 - - > total = 80 going up = 4 - - > total = 84 going down = 4 - - > total = 86 going up = 2 - - > total = 88 ( caught ) no of bounces = 4 answer : a"

a ) 28.57 % , b ) 14 kmph , c ) 4 , d ) 19 , e ) 5 %

c

divide(divide(32, const_2), const_2)

divide(n0,const_2)|divide(#0,const_2)|

general

there are two cars . one is 90 miles north of the other . simultaneously , the car to the north is driven westward at 25 miles per hour and the other car is driven eastward at 35 miles per hour . how many miles apart are the cars after 2 hours ?

"here , drawing a quick sketch of the ' actions ' described will end in a diagonal line that you canbuilda right triangle around : the right triangle will have a base of 90 and a height of 120 . the hidden pattern here is a 3 / 4 / 5 right triangle ( the 90 lines up with the ' 3 ' and the 120 lines up with the ' 4 ' ) . in this way , you can deduce that each side is ' 30 times ' bigger than it ' s corresponding side : 3 / 4 / 5 becomes 90 / 120 / 150 thus the distance between the two cars is the length of the hypotenuse of this larger right triangle . . . final answer : d"

a ) 40 , b ) 265 miles . , c ) 149 , d ) 150 , e ) 1000

d

sqrt(add(power(add(multiply(35, 2), multiply(25, 2)), const_2), power(multiply(35, 2), const_2)))

multiply(n2,n3)|multiply(n1,n3)|add(#0,#1)|power(#0,const_2)|power(#2,const_2)|add(#4,#3)|sqrt(#5)|

physics

a rectangular box measures internally 1.6 m long , 1 m broad and 60 cm deep . the number of cubical box each of edge 20 cm that can be packed inside the box is :

"explanation : number of blocks = ( 160 x 100 x 60 / 20 x 20 x 20 ) = 120 answer : d"

a ) 120 , b ) 1 % , c ) 416 , d ) rs 10123.20 , e ) 2

a

volume_rectangular_prism(divide(1.6, divide(20, const_100)), divide(1, divide(20, const_100)), divide(divide(60, const_100), divide(20, const_100)))

divide(n3,const_100)|divide(n2,const_100)|divide(n0,#0)|divide(n1,#0)|divide(#1,#0)|volume_rectangular_prism(#2,#3,#4)|

physics

a certain meter records voltage between 0 and 10 volts inclusive . if the average value of 3 recordings on the meter was 9 volts , what was the smallest possible recording in volts ?

"if average of 3 is 9 so sum of 3 should be 27 3 recording can be from 0 - 10 inclusive to find one smallest other two should be highest so , lets assume three var are a , b , c say a is smallest and give b and c greatest readings say 10 and 10 so a has to be 7 e"

a ) 7 , b ) s . 180 , c ) 70 , d ) 1,350 , e ) 250

a

subtract(multiply(3, 9), 10)

multiply(n2,n3)|subtract(#0,n1)|

general

bucket p has thrice the capacity as bucket q . it takes 60 turns for bucket p to fill the empty drum . how many turns it will take for both the buckets p & q , having each turn together to fill the empty drum ?

if caoacity of q is x units , then capacity of p is 3 x and capacity of drum is 60 * 3 x = 180 x . it will take 180 x / 4 x = 45 turns it will take for both the buckets p & q , having each turn together to fill the empty drum . answer : a

a ) 45 , b ) 10800 , c ) 38 , d ) 9', ' , e ) 1 / 12

a

divide(const_1, add(divide(const_1, 60), divide(const_1, multiply(60, const_3))))

divide(const_1,n0)|multiply(n0,const_3)|divide(const_1,#1)|add(#0,#2)|divide(const_1,#3)

other

a father said his son , ` ` i was as old as you are at present at the time of your birth . ` ` if the father age is 40 now , the son age 5 years back was

"let the son ' s present age be x years . then , ( 40 - x ) = x x = 20 . son ' s age 5 years back = ( 20 - 5 ) = 15 years answer : c"

a ) 22 , b ) 15 , c ) 3.2 , d ) 1 / 24 , e ) $ 290

b

subtract(divide(40, const_2), 5)

divide(n0,const_2)|subtract(#0,n1)|

general

the h . c . f . of two numbers is 30 and the other two factors of their l . c . m . are 13 and 14 . the larger of the two numbers is :

"clearly , the numbers are ( 30 x 13 ) and ( 30 x 14 ) . larger number = ( 30 x 14 ) = 420 . answer : option c"

a ) $ 32500 , b ) 95 , c ) 36.92 mph , d ) 420 , e ) 60

d

multiply(30, 14)

multiply(n0,n2)|

other

find compound interest on rs . 7500 at 4 % per year for 2 years , compounded annually .

"amount = rs [ 7500 * ( 1 + ( 4 / 100 ) 2 ] = rs ( 7500 * ( 26 / 25 ) * ( 26 / 25 ) ) = rs . 8112 . therefore , compound interest = rs . ( 8112 - 7500 ) = rs . 612 . answer is e ."

a ) 612 , b ) 100 , c ) 24 days , d ) 20,160 , e ) 1070

a

subtract(add(add(7500, divide(multiply(7500, 4), const_100)), divide(multiply(add(7500, divide(multiply(7500, 4), const_100)), 4), const_100)), 7500)

multiply(n0,n1)|divide(#0,const_100)|add(n0,#1)|multiply(n1,#2)|divide(#3,const_100)|add(#2,#4)|subtract(#5,n0)|

gain

a man whose speed is 5 kmph in still water rows to a certain upstream point and back to the starting point in a river which flows at 1.5 kmph , find his average speed for the total journey ?

"m = 5.0 s = 1.5 ds = 6.5 us = 3.5 as = ( 2 * 6 * 3 ) / 10 = 4.55 answer : e"

a ) 762 , b ) 420 , c ) 4.55 , d ) 432 . , e ) 237

c

divide(add(5, subtract(5, 1.5)), const_2)

subtract(n0,n1)|add(n0,#0)|divide(#1,const_2)|

general

in an election only two candidates contested . a candidate secured 70 % of the valid votes and won by a majority of 178 votes . find the total number of valid votes ?

let the total number of valid votes be x . 70 % of x = 70 / 100 * x = 7 x / 10 number of votes secured by the other candidate = x - 7 x / 100 = 3 x / 10 given , 7 x / 10 - 3 x / 10 = 178 = > 4 x / 10 = 178 = > 4 x = 1780 = > x = 445 . answer : a

a ) 11 : 9 , b ) 6.8 % , c ) 315 , d ) 445 , e ) 9 / 25

d

divide(178, divide(subtract(70, subtract(const_100, 70)), const_100))

subtract(const_100,n0)|subtract(n0,#0)|divide(#1,const_100)|divide(n1,#2)

gain

what is the angle between the hands of a clock when time is 10 : 30 ?

"angle between two hands = 30 h - 11 / 2 m = 30 * 10 - 30 * 11 / 2 = 300 - 165 = 135 deg answer : e"

a ) 4 / 13 , b ) 135 deg , c ) 8 : 5 , d ) 9 / 20 , e ) 25 days

b

subtract(multiply(30, multiply(const_3, const_2)), 10)

multiply(const_2,const_3)|multiply(n1,#0)|subtract(#1,n0)|

geometry

at a certain company , each employee has a salary grade s that is at least 1 and at most 5 . each employee receives an hourly wage p , in dollars , determined by the formula p = 10.50 + 0.25 ( s – 1 ) . an employee with a salary grade of 5 receives how many more dollars per hour than an employee with a salary grade of 1 ?

oa is definitely wrong . the answer should be d .

a ) 4 , b ) $ 1.50 , c ) 64 , d ) 1 , e ) 84

b

add(multiply(0.25, subtract(5, 1)), 0.25)

subtract(n1,n0)|multiply(n3,#0)|add(n3,#1)|

general

if jake loses 5 pounds , he will weigh thrice as much as his sister . together they now weigh 285 pounds . what is jake ' s present weight , in pounds ?

"this problem can be solved as a simple word problem in which we must convert words to math . before we create our equations , we want to define some variables . j = jake ’ s current weight , in pounds s = sister ’ s current weight , in pounds we are told that “ if jake loses 8 pounds , he will weigh twice as much as his sister . we put this into an equation : j – 5 = 3 s j = 3 s + 5 ( equation 1 ) next , we are told that “ together they now weigh 275 pounds . ” we can also put this into an equation . j + s = 285 ( equation 2 ) to solve this equation , we can substitute 2 s + 8 from equation 1 for the variable j in equation 2 : 3 s + 5 = 285 - s 4 s = 280 s = 70 j + 70 = 285 j = 215 answer : e"

a ) 60 , b ) 215 , c ) 35 , d ) 176 , e ) 108 kmph

b

divide(add(multiply(const_2, 285), 5), const_3)

multiply(n1,const_2)|add(n0,#0)|divide(#1,const_3)|

general

6 computers , each working at the same constant rate , together can process a certain amount of data in 9 days . how many additional computers , each working at the same constant rate , will be needed to process the same amount of data in 6 days ?

explanation : if six computers require 9 days to process the data , thats a total of 54 computer - days the product of 6 and 9 . if you change the number of computers or the number of days , 54 will have to remain the product , whether that means 54 days of one computer or one day with 54 computers . in 6 days , the number of computers is : 6 c = 54 c = 9 9 computers is 3 more than the 6 that it took to do the job in 9 days , so the correct choice is ( a ) .

a ) 1920 , b ) 3 , c ) 2 / 3 , d ) 120 , e ) 58.65 ft

b

subtract(divide(multiply(6, divide(const_1, 6)), divide(const_1, 9)), 6)

divide(const_1,n0)|divide(const_1,n1)|multiply(n0,#0)|divide(#2,#1)|subtract(#3,n0)

physics

the radius of a cylindrical vessel is 7 cm and height is 4 cm . find the whole surface of the cylinder ?

"r = 7 h = 4 2 π r ( h + r ) = 2 * 22 / 7 * 7 ( 11 ) = 484 answer : a"

a ) 484 , b ) 4327 , c ) 55 , d ) 16 , e ) 49

a

surface_cylinder(7, 4)

surface_cylinder(n0,n1)|

geometry

mark bought a set of 6 flower pots of different sizes at a total cost of $ 8.00 . each pot cost 0.25 more than the next one below it in size . what was the cost , in dollars , of the largest pot ?

"this question can be solved with a handful of different algebra approaches ( as has been shown in the various posts ) . since the question asks for the price of the largest pot , and the answers are prices , we can test the answers . we ' re told that there are 6 pots and that each pot costs 25 cents more than the next . the total price of the pots is $ 8.25 . we ' re asked for the price of the largest ( most expensive ) pot . since the total price is $ 8.00 ( a 25 - cent increment ) and the the difference in sequential prices of the pots is 25 cents , the largest pot probably has a price that is a 25 - cent increment . from the answer choices , i would then test answer c first ( since answers b and d are not in 25 - cent increments ) . if . . . . the largest pot = $ 1.958 0.708 0.958 1.208 1.458 1.708 1.958 total = $ 8.00 so this must be the answer . b"

a ) 8985 , b ) 5 , c ) 1 / 4 , d ) 4 / 7 , e ) $ 1.96

e

add(divide(subtract(8.00, multiply(divide(multiply(subtract(6, const_1), 6), const_2), 0.25)), 6), multiply(subtract(6, const_1), 0.25))

subtract(n0,const_1)|multiply(n0,#0)|multiply(n2,#0)|divide(#1,const_2)|multiply(n2,#3)|subtract(n1,#4)|divide(#5,n0)|add(#6,#2)|

general

determine the value of ( 27 / 31 * 31 / 27 ) * 3

"solution : both fractions should be reduced before performing arithmetic operations . we get ( 27 / 31 * 31 / 27 ) 3 = 1 * 3 = 3 answer d"

a ) 9 , b ) 8 , c ) 20 , d ) 3 , e ) 1 / 2

d

divide(add(subtract(add(27, multiply(31, 27)), subtract(3, 27)), const_1), 31)

multiply(n0,n1)|subtract(n4,n3)|add(n3,#0)|subtract(#2,#1)|add(#3,const_1)|divide(#4,n2)|

general

a rectangular grassy plot 110 m by 65 cm has a gravel path . 5 cm wide all round it on the inside . find the cost of gravelling the path at 80 paise per sq . mt

"explanation : area of theplot = 110 * 65 = 7150 sq m area of the plot excluding the path = ( 110 - 5 ) * ( 65 - 5 ) = 6300 sq m area of the path = 7150 - 6300 = 850 sq m cost of gravelling the path = 850 * 80 / 100 = 680 rs answer : a ) 680 rs"

a ) 25 , b ) 11.25 , c ) 55 , d ) 8 , e ) 680

e

multiply(divide(80, const_100), subtract(multiply(110, 65), multiply(subtract(110, 5), subtract(65, 5))))

divide(n3,const_100)|multiply(n0,n1)|subtract(n0,n2)|subtract(n1,n2)|multiply(#2,#3)|subtract(#1,#4)|multiply(#0,#5)|

physics

in a dairy farm , 16 cows eat 16 bags of husk in 16 days . in how many days one cow will eat one bag of husk ?

explanation : one bag of husk = 16 cows per day ⇒ 16 × 1 × 16 = 1 × 16 × x for one cow = 16 days answer : a

a ) 49 hr , b ) 5 , c ) 3 , d ) 16 , e ) $ 94,000

d

multiply(divide(16, 16), 16)

divide(n0,n0)|multiply(n0,#0)

physics

in 12 pumps can raise 1218 tons of water in 11 days of 9 hrs each , how many pumps will raise 2030 tons of water in 12 days of 11 hrs each ?

explanation : pumps work time 12 1218 99 x 2030 132 = > 1218 / ( 912 * 99 ) = 2020 / ( x × 132 ) = > x = 15 pumps answer : option b

a ) 9,600 , b ) 15 , c ) 4 / 9 , d ) 17 , e ) 19 / 27

b

divide(multiply(multiply(multiply(12, 11), 9), 2030), multiply(multiply(12, 11), 1218))

multiply(n0,n2)|multiply(n3,#0)|multiply(n1,#0)|multiply(n4,#1)|divide(#3,#2)

physics

if the probability of rain on any given day in chicago during the summer is 20 % , independent of what happens on any other day , what is the probability of having exactly 3 rainy days from july 4 through july 7 inclusive ?

"one possible case is : rainy - rainy - rainy - not rainy . the probability of this case is 1 / 5 * 1 / 5 * 1 / 5 * 4 / 5 = 4 / 625 the number of possible cases is 4 c 3 = 4 . p ( exactly 3 rainy days ) = 4 * 4 / 625 = 16 / 625 the answer is e ."

a ) 96 , b ) 15 days , c ) 16 / 625 , d ) 23 % . , e ) 2 / 42

c

multiply(4, power(divide(const_1, const_2), 4))

divide(const_1,const_2)|power(#0,n2)|multiply(n2,#1)|

general

find the product of the local value and absolute value of 2 in 564823

"explanation : place value = local value face value = absolute value the place value of 2 in 564823 is 2 x 10 = 20 the face value of 2 in 564823 is nothing but 2 . = > 20 x 2 = 40 answer : option a"

a ) 9 kmph , b ) 40 , c ) 129.5 , d ) 180 km , e ) 1300

b

multiply(multiply(2, const_1000), 2)

multiply(n0,const_1000)|multiply(n0,#0)|

general

the present population of a town is 60,000 . the population increases annually at 10 % . find the population after 3 years .

"explanation : solution : the required population is 60000 ( 1 + 10 / 100 ) ^ 3 = 60000 * 110 / 100 * 110 / 100 * 110 / 100 = 79,860 answer : b"

a ) 79860 , b ) 1200 , c ) 1600 , d ) s . 200 , e ) 4

a

add(60,000, divide(multiply(60,000, 10), const_100))

multiply(n0,n1)|divide(#0,const_100)|add(n0,#1)|

gain

32.32 / 2000 is equal to :

"25.25 / 2000 = 2525 / 200000 = 0.01616 answer : d"

a ) 24 , b ) 0.01616 , c ) $ 175 , d ) 5 % , e ) 14 years and 30 years

b

divide(32.32, 2000)

divide(n0,n1)|

general

george went to a fruit market with certain amount of money . with this money he can buy either 50 oranges or 40 mangoes . he retains 5 % of the money for taxi fare and buys 25 mangoes . how many oranges can he buy ?

let the amount of money be 200 let cost of 1 orange be 4 let cost of 1 mango be 5 he decides to retain 5 % of 200 = 10 for taxi fare , so he is left with 190 he buys 20 mangoes ( @ 5 ) so he spends 100 money left is 90 ( 190 - 100 ) no of oranges he can buy = 90 / 4 = > 22,5 so , george can buy 20 oranges . d

a ) 180 m , b ) 220 , c ) 350 m , d ) 22.5 , e ) 2599980

d

multiply(subtract(subtract(const_1, divide(5, const_100)), divide(25, 50)), 50)

divide(n2,const_100)|divide(n3,n0)|subtract(const_1,#0)|subtract(#2,#1)|multiply(n0,#3)

general

if 4 spiders make 4 webs in 4 days , then 1 spider will make 1 web in how many days ?

"explanation : let the required number days be x . less spiders , more days ( indirect proportion ) less webs , less days ( direct proportion ) spiders 1 : 4 | | : : 4 : 1 webs 4 : 1 | = > 1 * 4 * x = 4 * 1 * 4 = > x = 4 answer : d"

a ) five , b ) 4 , c ) 12 , d ) c : 37.5 , e ) - 49

b

multiply(1, 4)

multiply(n0,n3)|

physics

in an examination , there were 2,000 candidates , out of which 900 candidates were girls and rest were boys . if 36 % of the boys and 32 % of the girls passed , then the total percentage of failed candidates is ?

"girls = 900 , boys = 1100 passed = ( 36 % of 1100 ) + ( 32 % of 900 ) = 396 + 288 = 684 failed = 2000 - 684 = 1316 failed % = [ ( 1316 / 2000 ) x 100 ] % = 65.8 % . answer : b"

a ) 44 , b ) 104 miles , c ) 65.80 % , d ) 0.55 , e ) 1320

c

multiply(divide(subtract(subtract(multiply(const_2, multiply(const_100, const_10)), multiply(divide(36, const_100), subtract(multiply(const_2, multiply(const_100, const_10)), 900))), multiply(divide(32, const_100), 900)), multiply(const_2, multiply(const_100, const_10))), const_100)

divide(n2,const_100)|divide(n3,const_100)|multiply(const_10,const_100)|multiply(#2,const_2)|multiply(n1,#1)|subtract(#3,n1)|multiply(#0,#5)|subtract(#3,#6)|subtract(#7,#4)|divide(#8,#3)|multiply(#9,const_100)|

general

john left home and drove at the rate of 50 mph for 2 hours . he stopped for lunch then drove for another 3 hours at the rate of 55 mph to reach his destination . how many miles did john drive ?

"the total distance d traveled by john is given by d = 50 * 2 + 3 * 55 = 265 miles . answer d"

a ) 90 sq . cm' , b ) 265 miles . , c ) 4 , d ) 187.5 , e ) 3

b

add(multiply(50, 2), multiply(3, 55))

multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)|

physics

a boat having a length 3 m and breadth 2 m is floating on a lake . the boat sinks by 1 cm when a man gets on it . the mass of the man is :

"volume of water displaced = ( 3 x 2 x 0.01 ) m 3 = 0.06 m 3 . mass of man = volume of water displaced x density of water = ( 0.06 x 1000 ) kg = 60 kg . answer : option b"

a ) 3 , b ) 378 , c ) 6923 , d ) 60 kg , e ) $ 44.33

d

multiply(multiply(multiply(3, 2), divide(1, const_100)), const_1000)

divide(n2,const_100)|multiply(n0,n1)|multiply(#0,#1)|multiply(#2,const_1000)|

physics

anne earned $ 3 an hour baby - sitting , and $ 4 an hour working in the garden . last week she did baby - sitting for 5 hours and garden work for 3 hours . how much more money does she need to buy a game that costs $ 35 ?

5 x $ 3 = $ 15 for baby - sitting 3 x $ 4 = $ 12 for garden work $ 15 + $ 12 = $ 27 she has $ 35 - $ 27 = $ 8 more needed to buy the game correct answer a

a ) 1954404 , b ) 24 , c ) $ 8 , d ) 1 / 8 , e ) 64

c

subtract(35, add(multiply(5, 3), multiply(3, 4)))

multiply(n0,n2)|multiply(n0,n1)|add(#0,#1)|subtract(n4,#2)

general

if the average ( arithmetic mean ) of a and b is 45 and the average of b and c is 80 , what is the value of c â ˆ ’ a ?

"the arithmetic mean of a and b = ( a + b ) / 2 = 45 - - a + b = 90 - - 1 similarly for b + c = 160 - - 2 subtracting 1 from 2 we have c - a = 70 ; answer : b"

a ) s . 4400 , b ) 49 , c ) 70 , d ) $ 615 , e ) 510

c

subtract(multiply(80, const_2), multiply(45, const_2))

multiply(n1,const_2)|multiply(n0,const_2)|subtract(#0,#1)|

general

a envelop weight 8.2 gm , if 800 of these envelop are sent with an advertisement mail . how much wieght ?

"800 * 8.2 6560.0 gm 6.56 kg answer : a"

a ) 17 th , b ) 7 : 3 , c ) 223 , d ) 6.56 kg , e ) 3

d

divide(multiply(8.2, 800), const_1000)

multiply(n0,n1)|divide(#0,const_1000)|

general

if a coin is flipped , the probability that the coin will land heads is 1 / 2 . if the coin is flipped 5 times , what is the probability that it will land tails up on the first 4 flips and not on the last flip ?

( 1 / 2 ) * ( 1 / 2 ) * ( 1 / 2 ) * ( 1 / 2 ) * ( 1 / 2 ) = 1 / 32 answer : b

a ) 4.37 sec , b ) 8 kmph , c ) 40 days , d ) 1 / 32 , e ) 6000

d

divide(const_1, power(2, 5))

power(n1,n2)|divide(const_1,#0)

probability

the cross - section of a water channel is a trapezium in shape . if the channel is 14 meters wide at the top and 8 meters wide at the bottom and the area of cross - section is 990 square meters , what is the depth of the channel ( in meters ) ?

"1 / 2 * d * ( 14 + 8 ) = 990 d = 90 the answer is e ."

a ) rs . 315 , b ) 90 , c ) 360 , d ) 21.48 hours , e ) 11

b

divide(990, divide(add(14, 8), const_2))

add(n0,n1)|divide(#0,const_2)|divide(n2,#1)|

physics

income and expenditure of a person are in the ratio 5 : 4 . if the income of the person is rs . 14000 , then find his savings ?

"let the income and the expenditure of the person be rs . 5 x and rs . 4 x respectively . income , 5 x = 14000 = > x = 2800 savings = income - expenditure = 5 x - 4 x = x so , savings = rs . 2800 . answer : b"

a ) 0.4 , b ) 2800 , c ) 0.15 , d ) 49995 , e ) 1000

b

subtract(14000, multiply(divide(4, 5), 14000))

divide(n1,n0)|multiply(n2,#0)|subtract(n2,#1)|

other

two employees x and y are paid a total of rs . 750 per week by their employer . if x is paid 120 percent of the sum paid to y , how much is y paid per week ?

"let the amount paid to x per week = x and the amount paid to y per week = y then x + y = 750 but x = 120 % of y = 120 y / 100 = 12 y / 10 ∴ 12 y / 10 + y = 750 ⇒ y [ 12 / 10 + 1 ] = 750 ⇒ 22 y / 10 = 750 ⇒ 22 y = 7500 ⇒ y = 7500 / 22 = rs . 340.90 e )"

a ) 6 , b ) 4 : 5 , c ) 21600 , d ) s . 340.90 , e ) 60

d

divide(multiply(750, multiply(add(const_1, const_4), const_2)), multiply(add(multiply(add(const_1, const_4), const_2), const_1), const_2))

add(const_1,const_4)|multiply(#0,const_2)|add(#1,const_1)|multiply(n0,#1)|multiply(#2,const_2)|divide(#3,#4)|

general

the difference between compound interest and simple interest on a certain amount of money at 5 % per annum for 2 years is 19 . find the sum :

"sol . ( d ) let the sum be 100 . therefore , si = 100 × 5 × 2100 = 10100 × 5 × 2100 = 10 and ci = 100 ( 1 + 5100 ) 2 − 100100 ( 1 + 5100 ) 2 − 100 ∴ = 100 × 21 × 2120 × 20 − 100 = 414 = 100 × 21 × 2120 × 20 − 100 = 414 difference of ci and si = 41 ⁄ 4 - 10 = 1 ⁄ 4 if the difference is 1 ⁄ 4 , the sum = 100 = > if the difference is 19 , the sum = 400 × 19 = 7600 answer b"

a ) 9 , b ) 7600 , c ) 14', ' , d ) 80 % , e ) 350

b

divide(19, subtract(power(add(const_1, divide(5, const_100)), 2), add(const_1, multiply(2, divide(5, const_100)))))

divide(n0,const_100)|add(#0,const_1)|multiply(n1,#0)|add(#2,const_1)|power(#1,n1)|subtract(#4,#3)|divide(n2,#5)|

gain

1398 x 1398 = ?

"explanation : 1398 x 1398 = ( 1398 ) 2 = ( 1400 - 2 ) 2 = 14002 - ( 2 × 1400 × 2 ) + 22 = 1960000 - 5600 + 4 = 1954404 . answer : option d"

a ) $ 300 , b ) 16 , c ) 1954404 , d ) 24', ' , e ) 30

c

multiply(divide(1398, 1398), const_100)

divide(n0,n1)|multiply(#0,const_100)|

general

machine a produces 100 parts twice as fast as machine b does . machine b produces 100 parts in 60 minutes . if each machine produces parts at a constant rate , how many parts does machine a produce in 6 minutes ?

"machine b produces 100 part in 60 minutes . machine a produces 100 parts twice as fast as b , so machine a produces 100 parts in 60 / 2 = 30 minutes . now , machine a produces 100 parts in 30 minutes which is 100 / 30 = 10 / 3 parts / minute . 10 / 3 parts x a total of 6 minutes = 20 a"

a ) 14 , b ) 20 , c ) 3000000 , d ) 9560 , e ) 7 / 12

b

multiply(multiply(divide(100, 60), const_2), 6)

divide(n0,n2)|multiply(#0,const_2)|multiply(n3,#1)|

gain

if the area of a square with sides of length 3 centimeters is equal to the area of a rectangle with a width of 4 centimeters , what is the length of the rectangle , in centimeters ?

"let length of rectangle = l 3 ^ 2 = l * 4 = > l = 9 / 4 = 3 answer d"

a ) 24 . , b ) 1717.85 , c ) 3 , d ) 7 / 12 , e ) 879

c

divide(power(3, const_2), 4)

power(n0,const_2)|divide(#0,n1)|

geometry

if n is a prime number greater than 5 , what is the remainder when n ^ 2 is divided by 12 ?

there are several algebraic ways to solve this question , but the easiest way is as follows : since we can not have two correct answers just pick a prime greater than 5 , square it and see what would be the remainder upon division of it by 12 . n = 7 - - > n ^ 2 = 49 - - > remainder upon division 49 by 12 is 1 . answer : b .

a ) 6 , b ) 16.2 % , c ) 1 , d ) 11 , e ) 20

c

subtract(power(add(5, const_2), 2), multiply(12, const_4))

add(n0,const_2)|multiply(n2,const_4)|power(#0,n1)|subtract(#2,#1)

general

set j consists of 5 consecutive even numbers . if the smallest term in the set is - 2 , what is the range of the positive integers in set j ?

since there are only 5 integers , another approach is the just list all 5 . we get : - 2 , 0,2 , 4 , 6 range of positive integers = 6 - 2 = 4 answer : c

a ) 3 , b ) 144 , c ) 5 ⁄ 3 , d ) 4 years , e ) 4

e

subtract(add(negate(2), multiply(subtract(5, const_1), 2)), 2)

negate(n1)|subtract(n0,const_1)|multiply(n1,#1)|add(#2,#0)|subtract(#3,n1)

general

what is the greatest positive integer n such that 3 ^ n is a factor of 36 ^ 450 ?

"36 = 3 ^ 2 * 2 ^ 2 . 36 ^ 450 = 3 ^ 900 * 2 ^ 900 the answer is e ."

a ) 95 , b ) 51 days , c ) 10 , d ) 900 , e ) 4176

d

multiply(subtract(36, 450), 450)

subtract(n1,n2)|multiply(n2,#0)|

other

the sum of all the integers g such that - 26 < g < 24 is

"easy one - - 25 , - 24 , - 23 , - 22 , . . . . . . - 1,0 , 1 , 2 . . . . , 22 , 23 cancel everyhitng and we ' re left with - - 25 and - 24 g = - 49 . d is the answer ."

a ) - 49 , b ) 3 , c ) 6 , d ) $ 0.40 , e ) 5000

a

add(add(negate(26), const_1), add(add(negate(26), const_1), const_1))

negate(n0)|add(#0,const_1)|add(#1,const_1)|add(#1,#2)|

general

a sum of money at simple interest amounts to $ 680 in 3 years and to $ 710 in 4 years . the sum is :

"c $ 590 s . i . for 1 year = $ ( 710 - 680 ) = $ 30 . s . i . for 3 years = $ ( 30 x 3 ) = $ 90 . principal = $ ( 680 - 90 ) = $ 590 ."

a ) 20,160 , b ) 2220 , c ) 70 , d ) 778 , e ) $ 590

e

subtract(680, divide(multiply(subtract(710, 680), 3), 4))

subtract(n2,n0)|multiply(n1,#0)|divide(#1,n3)|subtract(n0,#2)|

gain

a student chose a number , multiplied it by 2 , then subtracted 180 from the result and got 104 . what was the number he chose ?

"solution : let x be the number he chose , then 2 * x * 180 = 104 2 x = 284 x = 142 correct answer d"

a ) 308 , b ) 142 , c ) 174 , d ) 14 , e ) 672

b

divide(add(104, 180), 2)

add(n1,n2)|divide(#0,n0)|

general

two brothers take the same route to school on their bicycles , one gets to school in 25 minutes and the second one gets to school in 36 minutes . the ratio of their speeds is

solution let us name the brothers as a and b . = ( a ' s speed ) : ( b ' s speed ) = â ˆ š b : â ˆ š a = â ˆ š 25 : â ˆ š 36 = 5 : 6 answer d

a ) 0.7 % , b ) 74 , c ) 5 : 6 , d ) 90 , e ) $ 1260

c

divide(sqrt(25), sqrt(36))

sqrt(n0)|sqrt(n1)|divide(#0,#1)

physics

the pinedale bus line travels at an average speed of 60 km / h , and has stops every 5 minutes along its route . yahya wants to go from his house to the pinedale mall , which is 9 stops away . how far away , in kilometers , is pinedale mall away from yahya ' s house ?

"number of stops in an hour : 60 / 5 = 12 distance between stops : 60 / 12 = 5 km distance between yahya ' s house and pinedale mall : 5 x 9 = 45 km imo , correct answer is ` ` c . ' '"

a ) 45 km , b ) 62 , c ) 2 , d ) rs . 27500 , e ) 9 : 25', '

a

multiply(60, divide(multiply(5, 9), 60))

multiply(n1,n2)|divide(#0,n0)|multiply(n0,#1)|

physics

in a certain warehouse , 50 percent of the packages weigh less than 75 pounds , and a total of 48 packages weigh less than 25 pounds . if 80 percent of the packages weigh at least 25 pounds , how many of the packages weigh at least 25 pounds but less than 75 pounds ?

"if 80 % of the packages weigh at least 25 pounds this means that 20 % of the packages weigh less than 25 pounds let t = total number of packages so , 20 % of t = # of packages that weigh less than 25 pounds 48 packages weigh less than 25 pounds great . so , 20 % of t = 48 rewrite to get : 0.2 t = 48 solve : t = 240 50 % of the packages weigh less than 75 pounds so , 50 % oft = number of packages that weigh less than 75 pounds 50 % of 240 = 120 , so 120 packages weigh less than 75 pounds of those 120 packages that weigh less than 75 pounds , 48 packages weigh less than 25 pounds . so , the number of packages that weight between 25 and 75 pounds = 120 - 48 = 72 = c"

a ) 72 , b ) 1000 , c ) rs . 10000 , d ) 75 % , e ) 40.5

a

subtract(divide(multiply(multiply(divide(48, subtract(const_100, 80)), const_100), 50), const_100), 48)

subtract(const_100,n4)|divide(n2,#0)|multiply(#1,const_100)|multiply(n0,#2)|divide(#3,const_100)|subtract(#4,n2)|

general

in one hour , a boat goes 11 km along the stream and 5 km against the stream . the speed of the boat in still water ( in km / hr ) is :

"solution speed in still water = 1 / 2 ( 11 + 5 ) kmph . = 8 kmph . answer c"

a ) 25 , b ) 8 , c ) 1173.98 , d ) 68 kmph , e ) 1

b

divide(add(11, 5), const_2)

add(n0,n1)|divide(#0,const_2)|

physics

the ratio of the cost price and the selling price is 4 : 5 . the profit percent is ?

"let c . p . = rs . 4 x . then , s . p . = rs . 5 x gain = ( 5 x - 4 x ) = rs . x gain % = ( x * 100 ) / 4 x = 25 % . answer : c"

a ) 75.5 kg , b ) 21 , c ) 3 , d ) 25 , e ) 30

d

multiply(subtract(divide(5, 4), const_1), const_100)

divide(n1,n0)|subtract(#0,const_1)|multiply(#1,const_100)|

gain

if 45 % of a class averages 100 % on a test , 50 % of the class averages 78 % on the test , and the remainder of the class averages 65 % on the test , what is the overall class average ? ( round final answer to the nearest percent ) .

"this question is a weighted average question with a series of dependent variables . the remaining portion of the class represents 100 % - 45 % - 50 % = 5 % of the class converting the portions of the class population to decimal weights , we find : class average = 0.45 x 100 + 0.50 x 78 + 0.05 x 63 = 87.15 the class average ( rounded ) is 87 % final answer b ) 87 %"

a ) 48 , b ) 14 % , c ) 4000 , d ) 87 % , e ) 5 / 26

d

divide(add(add(multiply(45, 100), multiply(50, 78)), multiply(subtract(const_100, add(45, 50)), 65)), const_100)

add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|subtract(const_100,#0)|multiply(n4,#4)|add(#3,#5)|divide(#6,const_100)|

gain

of the votes cast on a certain proposal , 62 more were in favor of the proposal than were against it . if the number of votes against the proposal was 40 percent of the total vote , what was the total number of votes cast ? ( each vote cast was either in favor of the proposal or against it . )

let x be the total number of votes cast . 0.6 x = 0.4 x + 62 0.2 x = 62 x = 310 the answer is d .

a ) 3 , b ) 3 : 2 , c ) 1764 , d ) 310 , e ) 4.5 sec

d

divide(62, subtract(subtract(const_1, divide(40, const_100)), divide(40, const_100)))

divide(n1,const_100)|subtract(const_1,#0)|subtract(#1,#0)|divide(n0,#2)

gain

how many of the integers between 20 and 80 are even ?

"number start between 20 to 80 is 60 numbers half of them is even . . which is 30 answer : b"

a ) 1200 , b ) rs . 3240 , c ) 30 , d ) 3 , e ) 2

c

divide(subtract(80, 20), const_2)

subtract(n1,n0)|divide(#0,const_2)|

general

if a and b are positive integers , and a = 5 b + 20 , the greatest common divisor of a and b can not be

if b is 2 , 4 , 5 , or 10 , then gcd of a and b is 2 , 4 , 5 , and 10 respectively . so , by poe the answer must be d . still : if b is a multiple of 6 , then a is 20 greater than a multiple of 6 , so not a multiple of 6 , so both of them can not be divisive by 6 . answer : d .

a ) $ 2.25 , b ) 6 , c ) 378 , d ) 1 , e ) 5.5

b

add(divide(20, 5), const_2)

divide(n1,n0)|add(#0,const_2)

general

what is the value of ( 44444445 * 88888885 * 44444442 + 44444438 ) / 44444444 ^ 2

"ans : a let x = 44444444 ( x + 1 ) × ( 2 x − 3 ) × ( x − 2 ) + ( x − 6 ) x 2 ( x + 1 ) × ( 2 x − 3 ) × ( x − 2 ) + ( x − 6 ) x 2 ( x 2 − x − 2 ) × ( 2 x − 3 ) + ( x − 6 ) x 2 ( x 2 − x − 2 ) × ( 2 x − 3 ) + ( x − 6 ) x 2 2 x 3 − 2 x 2 − 4 x − 3 x 2 + 3 x + 6 + x − 6 x 22 x 3 − 2 x 2 − 4 x − 3 x 2 + 3 x + 6 + x − 6 x 2 2 x 3 − 5 x 2 x 2 = 2 x − 52 x 3 − 5 x 2 x 2 = 2 x − 5 substituting the value of x in 2 x - 5 , we get 88888883 answer : a"

a ) 8 kmph , b ) 1 / 25 , c ) 90 cc', ' , d ) 88888883 , e ) rs . 2500

d

power(44444445, negate(88888885))

negate(n1)|power(n0,#0)|

general

find the greatest number which leaves the same remainder when it divides 21 , 57 and 105 .

"105 - 57 = 48 57 - 21 = 36 105 - 21 = 84 the h . c . f of 36 , 48 and 84 is 12 . answer : c"

a ) 12 , b ) 12.0 , c ) 480 , d ) 22 , e ) 7 / 15

a

divide(subtract(57, 21), const_2)

subtract(n1,n0)|divide(#0,const_2)|

general

at a summer camp with 2100 participants , 1 / 2 of the campers are aged 8 to 12 . next year , the number of campers aged 8 to 12 will increase by 1 / 3 . after this change , what percentage of the total 2100 campers will the 8 - to 12 - year - olds represent ?

"total - 2100 participants campers are aged 8 to 12 = ( 1 / 2 ) * 2100 = 1050 next year , campers are aged 8 to 12 = ( 4 / 3 ) * 1050 = 1400 percentage = ( 1400 / 2100 ) * 100 = 66 2 / 3 % answer : option e"

a ) 855 , b ) 5 / 12 , c ) 64 , d ) 66 2 / 3 , e ) 2

d

add(multiply(multiply(divide(multiply(divide(add(const_1000, multiply(8, const_100)), 2), add(divide(1, 3), 1)), add(const_1000, multiply(8, const_100))), const_100), 3), divide(multiply(multiply(divide(multiply(divide(add(const_1000, multiply(8, const_100)), 2), add(divide(1, 3), 1)), add(const_1000, multiply(8, const_100))), const_100), 3), const_10))

divide(n1,n8)|multiply(n3,const_100)|add(#0,n1)|add(#1,const_1000)|divide(#3,n2)|multiply(#2,#4)|divide(#5,#3)|multiply(#6,const_100)|multiply(#7,n8)|divide(#8,const_10)|add(#9,#8)|

general

express 35 mps in kmph ?

"35 * 18 / 5 = 126 kmph answer : c"

a ) 12 hours , b ) 360 , c ) 900 , d ) 10 , e ) 126

e

multiply(divide(35, const_1000), const_3600)

divide(n0,const_1000)|multiply(#0,const_3600)|

physics

the area of a square field 3136 sq m , if the length of cost of drawing barbed wire 3 m around the field at the rate of rs . 2.50 per meter . two gates of 1 m width each are to be left for entrance . what is the total cost ?

"a 2 = 3136 = > a = 56 56 * 4 * 3 = 672 – 6 = 666 * 2.5 = 1665 answer : d"

a ) 8 and 20 , b ) 257 , c ) 80.76 % , d ) s . 1665 , e ) 5625

d

multiply(multiply(subtract(multiply(sqrt(3136), const_4), multiply(const_2, 1)), 2.50), 3)

multiply(n3,const_2)|sqrt(n0)|multiply(#1,const_4)|subtract(#2,#0)|multiply(n2,#3)|multiply(#4,n1)|

physics

what is the area inscribed by the lines y = 4 , x = 2 , y = 6 - 2 x on an xy - coordinate plane ?

"first , let ' s graph the lines y = 4 and x = 2 at this point , we need to find the points where the line y = 6 - 2 x intersects the other two lines . for the vertical line , we know that x = 2 , so we ' ll plug x = 2 into the equation y = 6 - 2 x to get y = 6 - 2 ( 2 ) = 2 perfect , when x = 2 , y = 2 , so one point of intersection is ( 2,2 ) for the horizontal line , we know that y = 4 , so we ' ll plug y = 4 into the equation y = 6 - 2 x to get 4 = 6 - 2 x . solve to get : x = 1 so , when y = 4 , x = 1 , so one point of intersection is ( 1,4 ) now add these points to our graph and sketch the line y = 6 - 2 x at this point , we can see that we have the following triangle . the base has length 1 and the height is 2 area = ( 1 / 2 ) ( base ) ( height ) = ( 1 / 2 ) ( 1 ) ( 2 ) = 1 answer : a"

a ) 50 , b ) 1 , c ) 50000 , d ) 8 , e ) rs . 315

b

multiply(subtract(subtract(6, 4), 4), multiply(multiply(4, const_0_25), subtract(subtract(6, 4), 4)))

multiply(const_0_25,n0)|subtract(n2,n0)|subtract(#1,n0)|multiply(#0,#2)|multiply(#3,#2)|

general

there are 690 male and female participants in a meeting . half the female participants and one - quarter of the male participants are democrats . one - third of all the participants are democrats . how many of the democrats are female ?

female = x male = 690 - x x / 2 + 690 - x / 4 = 1 / 3 * ( 690 ) = 230 x = 230 x / 2 = 115 is supposed to be the answer m is missing something correct option c

a ) 14 , b ) 75 , c ) 350 , d ) 52 % , e ) 115

e

divide(subtract(multiply(divide(690, const_3), const_4), 690), const_2)

divide(n0,const_3)|multiply(#0,const_4)|subtract(#1,n0)|divide(#2,const_2)

general

find the area of a parallelogram with base 20 cm and height 40 cm ?

"area of a parallelogram = base * height = 20 * 40 = 800 cm 2 answer : c"

a ) 72.33 , b ) 800 cm 2 , c ) 250 m , d ) 119 , e ) 28

b

multiply(20, 40)

multiply(n0,n1)|

geometry

in a garden , there are 10 rows and 14 columns of mango trees . the distance between the two trees is 2 metres and a distance of one metre is left from all sides of the boundary of the garden . the length of the garden is

"explanation : each row contains 14 plants . there are 13 gapes between the two corner trees ( 13 x 2 ) metres and 1 metre on each side is left . therefore length = ( 26 + 2 ) m = 28 m . answer : e"

a ) 125 , b ) 30 , c ) 200 % , d ) 28 m , e ) 1000 m

d

add(add(multiply(subtract(14, const_1), 2), divide(10, 2)), divide(10, 2))

divide(n0,n2)|subtract(n1,const_1)|multiply(n2,#1)|add(#0,#2)|add(#3,#0)|

physics

a can do a piece of work in 6 hours ; b and c together can do it in 4 hours , which a and c together can do it in 3 hours . how long will b alone take to do it ?

"a ' s 1 hour work = 1 / 6 ; ( b + c ) ' s 1 hour work = 1 / 4 ; ( a + c ) ' s 1 hour work = 1 / 3 ( a + b + c ) ' s 1 hour work = ( 1 / 4 + 1 / 6 ) = 5 / 12 b ' s 1 hour work = ( 5 / 12 - 1 / 3 ) = 1 / 12 b alone will take 12 hours to do the work . answer : d"

a ) 12 hours , b ) 2522 , c ) 40 , d ) 1176000 , e ) 19

a

divide(const_1, subtract(divide(const_1, 4), subtract(divide(const_1, 3), divide(const_1, 6))))

divide(const_1,n1)|divide(const_1,n2)|divide(const_1,n0)|subtract(#1,#2)|subtract(#0,#3)|divide(const_1,#4)|

physics

a train 250 m long running at 72 kmph crosses a platform in 30 sec . what is the length of the platform ?

"d = 72 * 5 / 18 = 30 = 600 â € “ 250 = 350 m answer : a"

a ) 384 , b ) 350 m , c ) 7 , d ) 40.6 , e ) 9.87 %

b

subtract(multiply(30, multiply(72, const_0_2778)), 250)

multiply(n1,const_0_2778)|multiply(n2,#0)|subtract(#1,n0)|

physics

4 shepherds were watching over the flocks and they were commenting on how many sheep they each had . if ram had 3 more sheep than he would have one less than rahul . wheras akar has the same number as the other 3 shepherds put togeher . if john had 3 less sheep he would have exactly trile the number of ram . if they were evenly distributed if they would each have 11 seep how many sheep did ram have ?

akar has = ram + rahul + john after evenly distribution each has 11 . so , total no . is 44 so , akar has = 22 & ram + rahul + john = 22 also ram = rahul - 4 & john - 3 = 3 * ram solving these we get the sol . answer : b

a ) 60 kg , b ) 3 , c ) 12 , d ) 62 , e ) 1000

b

divide(subtract(multiply(11, const_2), add(4, 3)), add(4, const_1))

add(n0,n1)|add(n0,const_1)|multiply(n4,const_2)|subtract(#2,#0)|divide(#3,#1)

general