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Shang Data Lab @UCSD 7

how many odd factors does 250 have ?

"start with the prime factorization : 250 = 2 * 5 * 7 for odd factors , we put aside the factor of two , and look at the other prime factors . set of exponents = { 1 , 1 } plus 1 to each = { 2 , 2 } product = 2 * 2 = 4 therefore , there are 4 odd factors of 250 . answer : b ."

a ) 113 m 2 , b ) 4 , c ) 450 , d ) 20 , e ) 11 / 30

b

add(add(add(const_4, const_2), const_1), const_1)

add(const_2,const_4)|add(#0,const_1)|add(#1,const_1)|

other

in kaya ' s teacher ' s desk there are 24 pink highlighters , 28 yellow highlighters , and 25 blue highlighters . how many highlighters are there in all ?

"add the numbers of highlighters . 24 + 28 + 25 = 77 . answer is c ."

a ) 77 , b ) 550 sq . units', ' , c ) 791 , d ) 16 , e ) 37

a

add(add(24, 28), 25)

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

general

a train speeds past a pole in 10 seconds and a platform 50 m long in 20 seconds . its length is :

"let the length of the train be x meters and its speed be y m / sec . they , x / y = 10 = > y = x / 10 x + 50 / 20 = x / 10 x = 50 m . answer : option d"

a ) $ 2,350 , b ) 80 % , c ) 50 m . , d ) rs . 126 , e ) $ 0.40

c

multiply(50, subtract(const_2, const_1))

subtract(const_2,const_1)|multiply(n1,#0)|

physics

the sum of two numbers is 528 and their h . c . f is 33 . the number of pairs of numbers satisfying the above condition is

"let the required numbers be 33 a and 33 b . then 33 a + 33 b = 528 \ inline \ fn _ jvn \ rightarrow a + b = 16 . now , co - primes with sum 16 are ( 1,15 ) , ( 3,13 ) , ( 5,11 ) and ( 7,9 ) . \ inline \ fn _ jvn \ therefore required numbers are ( 33 x 1 , 33 x 15 ) , ( 33 x 3 , 33 x 13 ) , ( 33 x 5 , 33 x 11 ) , ( 33 x 7 , 33 x 9 ) the number of such pairs is 4 answer : a"

a ) 37 , b ) 7 , c ) 2 , d ) 4 , e ) 9 minutes

d

multiply(divide(add(528, 33), add(const_1, const_1)), subtract(divide(add(528, 33), add(const_1, const_1)), 33))

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

general

the average age of 50 students in a class is 10 years . if teacher ' s age is also included then average increases 1 year then find the teacher ' s age ?

"total age of 50 students = 50 * 10 = 500 total age of 51 persons = 51 * 11 = 561 age of teacher = 561 - 500 = 61 years answer is c"

a ) 61 , b ) 128 , c ) 4 / 7 , d ) 35 % , e ) 1000

a

subtract(add(add(multiply(50, 10), 1), 50), multiply(50, 10))

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

general

the denominator of a fraction is 6 greater than the numerator . if the numerator and the denominator are increased by 1 , the resulting fraction is equal to 4 ⠁ „ 5 . what is the value of the original fraction ?

"let the numerator be x . then the denominator is x + 6 . x + 1 / x + 7 = 4 / 5 . 5 x + 5 = 4 x + 28 . x = 23 . the original fraction is 23 / 29 . the answer is b ."

a ) 1 / 2 , b ) 23 / 29 , c ) 135 % , d ) 2 , e ) 5184

b

divide(divide(subtract(multiply(4, add(1, 6)), 5), subtract(5, 4)), add(divide(subtract(multiply(4, add(1, 6)), 5), subtract(5, 4)), 6))

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

general

consider the sets tn = { n , n + 1 , n + 2 , n + 3 , n + 4 ) , where n = 1 , 2 , 3 , … , 96 . how many of these sets contain 6 or any integral multiple thereof ( i . e . , any one of the numbers 6 , 12 , 18 , … ) ?

explanation : if n = 1 , then the set t 1 = { 1 , 23 , 45 } , and it does not have 6 or any multiples . n = 2 to n = 6 has 6 in the set . n = 7 , has the set t 7 = { 7 , 89 , 1011 } , and no 6 or multiples . so 1 in every 6 members do not have 6 or multiples of 6 . so , till n = 96 , there are 16 sets of “ 6 members ” ( 16 * 6 = 96 ) and 16 sets do not have 6 or its multiples , while the remaining 80 sets have . answer : a

a ) 9 / 14 , b ) 5 , c ) 600 , d ) 80 , e ) 16

d

multiply(divide(add(const_2, const_3), 6), 96)

add(const_2,const_3)|divide(#0,n8)|multiply(n7,#1)

general

car z travels 55 miles per gallon of gasoline when driven at a constant rate of 45 miles per hour , but travels 20 percent fewer miles per gallon of gasoline when driven at a constant rate of 60 miles per hour . how many miles does car z travel on 10 gallons of gasoline when driven at a constant rate of 60 miles per hour ?

the question stem asks us for the distance possible with 10 gallons of fuel at a constant speed of 60 miles per hour . we therefore first calculate the fuel efficiency at that speed . the stem tells us that at 45 miles / hour , the car will run 55 miles / gallon and at 60 miles / hour , that distance decreases by 20 % . we can therefore conclude that the car will travel 44 miles / gallon at a constant speed of 60 miles / gallon . with 10 gallons of fuel , the car can therefore travel 44 miles / gallon * 10 gallons = 440 miles . answer e .

a ) 1453 , b ) 440 , c ) 1954404 , d ) 1856 , e ) 62

b

multiply(multiply(subtract(const_1, divide(20, const_100)), 55), 10)

divide(n2,const_100)|subtract(const_1,#0)|multiply(n0,#1)|multiply(n4,#2)

gain

a certain farmer pays $ 70 per acre per month to rent farmland . how much does the farmer pay per month to rent a rectangular plot of farmland that is 360 feet by 605 feet ? ( 43,560 square feet = 1 acre )

basically the question an error . 1 acre = 43,560 square feet and if it is then the answer is 1050 ( e )

a ) 35 % , b ) 20 , c ) 320 m , d ) 5 / 26 , e ) $ 1050

e

multiply(70, divide(multiply(360, 605), divide(multiply(360, 605), const_10)))

multiply(n1,n2)|divide(#0,const_10)|divide(#0,#1)|multiply(n0,#2)|

geometry

if a square mirror has a 20 - inch diagonal , what is the approximate perimeter t 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 t 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 ) $ 96 , b ) 6 , c ) c : 37.5 , d ) 60 , e ) 15 sec .

d

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

a clock shows the time as 9 a . m . if the minute hand gains 6 minutes every hour , how many minutes will the clock gain by 6 p . m . ?

"there are 9 hours in between 9 a . m . to 6 p . m . 9 * 6 = 54 minutes . answer : e"

a ) 54 min , b ) 1.2 , c ) 6.24 , d ) 12 cm , e ) 50 %

a

multiply(add(const_3, 6), 6)

add(const_3,n2)|multiply(n1,#0)|

physics

20 people went to a hotel for combine dinner party 12 of them spent rs . 70 each on their dinner and rest spent 4 more than the average expenditure of all the 20 . what was the total money spent by them .

"solution : let average expenditure of 20 people be x . then , 20 x = 12 * 70 + 8 * ( x + 4 ) ; or , 20 x = 12 * 70 + 8 x + 32 ; or , x = 72.667 ; so , total money spent = 72.67 * 20 = rs . 1453.4 . answer : option c"

a ) 1453 , b ) 79860 , c ) 26 ° , d ) 6000 , e ) 200

a

multiply(divide(add(multiply(12, 70), multiply(subtract(20, 12), 4)), subtract(20, subtract(20, 12))), 20)

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

general

how many integers k greater than 100 and less than 600 are there such that if the hundreds and the unit digits of k are reversed , the resulting integer is k + 99 ?

"not sure if this is the shortest . . but this is how i did this there are 8 sets of integers with hundreds and units digits exchanged that satisfies k + 99 . 1 . 102 | 201 ( satisfies k + 99 , where k = 102 ) 2 . 203 | 302 ( satisfies k + 99 , where k = 203 ) 3 . . . . 4 . 405 | 504 each set has 10 such numbers . 1 . 102 | 201 ( still k + 99 holds good ) 2 . 112 | 211 3 . 122 | 221 4 . 132 | 231 5 . . . . 6 . . . . 7 . . . . 8 . . . . 9 . 182 | 281 10 . 192 | 291 therefore , 4 sets with 10 such number in each set will give 4 x 10 = 40 integers . a"

a ) 14.3 % , b ) 40 , c ) 7 , d ) 64.29 % , e ) $ 9640

b

multiply(const_10, subtract(const_10, const_2))

subtract(const_10,const_2)|multiply(#0,const_10)|

general

the ratio of the two natural numbers is 5 : 6 . if a certain number is added to both the numbers , the ratio becomes 7 : 8 . if the larger number exceeds the smaller number by 10 , find the number added ?

"let the two numbers be 5 x and 6 x . let the numbers added to both so that their ratio becomes 7 : 8 be k . ( 5 x + k ) / ( 6 x + k ) = 7 / 8 = > 40 x + 8 k = 42 x + 7 k = > k = 2 x . 6 x - 5 x = 10 = > x = 10 k = 2 x = 20 . answer : c"

a ) rs . 76 , b ) 75 , c ) 174 , d ) 10 , e ) 4

d

subtract(multiply(multiply(6, 6), 6), multiply(add(multiply(7, 6), 6), 8))

multiply(n1,n1)|add(n1,#0)|multiply(n2,#0)|multiply(n3,#1)|subtract(#2,#3)|

other

what annual payment will discharge a debt of rs . 1060 due in 2 years at the rate of 5 % compound interest ?

"explanation : let each installment be rs . x . then , x / ( 1 + 5 / 100 ) + x / ( 1 + 5 / 100 ) 2 = 1060 820 x + 1060 * 441 x = 570.07 so , value of each installment = rs . 570.07 answer : option c"

a ) 12 , b ) 4.37 sec , c ) $ 24.88 , d ) 570.07 , e ) $ 1600

d

divide(multiply(power(add(divide(5, const_100), const_1), 2), 1060), 2)

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

gain

50 % of the population of a village is 23040 . the total population of the village is ?

"answer ∵ 50 % of p = 23040 ∴ p = ( 23040 x 100 ) / 50 = 46080 correct option : d"

a ) 200 sq feet , b ) 1 / 32 , c ) 50 days , d ) 46080 , e ) 5

d

multiply(divide(const_100, 50), 23040)

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

general

if the perimeter of a rectangular house is 1400 m , its length when its breadth is 300 m is ?

2 ( l + 300 ) = 1400 = > l = 400 m answer : b

a ) 400 , b ) 24 , c ) 4 , d ) $ 370,000 , e ) 15 / 11

a

subtract(divide(1400, const_2), 300)

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

physics

the price of an item is discounted 4 percent on day 1 of a sale . on day 2 , the item is discounted another 4 percent , and on day 3 , it is discounted an additional 10 percent . the price of the item on day 3 is what percentage of the sale price on day 1 ?

"let initial price be 100 price in day 1 after 4 % discount = 96 price in day 2 after 4 % discount = 92.16 price in day 3 after 10 % discount = 82.94 so , price in day 3 as percentage of the sale price on day 1 will be = 82.94 / 96 * 100 = > 86.4 % answer will definitely be ( d )"

a ) 65500 $ , b ) 11 years , c ) 555681 , d ) 5 ⁄ 3 , e ) 86.4 %

e

add(multiply(divide(divide(10, const_100), subtract(1, divide(1, 4))), const_100), 2)

divide(n5,const_100)|divide(n1,n0)|subtract(n1,#1)|divide(#0,#2)|multiply(#3,const_100)|add(n2,#4)|

gain

if a # b = ab – b + b ^ 2 , then 3 # 4 =

"solution - simply substitute 3 and 4 in equation in the place of a and b respectively . 3 # 4 = 3 * 4 - 4 + 4 ^ 2 = 12 - 4 + 16 = 24 . ans d"

a ) 19 , 956.732 , b ) 24 , c ) 3200 , d ) 130 cm , e ) 8

b

add(subtract(multiply(3, 4), 4), power(4, 2))

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

general

what number has a 4 : 1 ratio to the number 100 ?

"4 : 1 = x : 100 x = 4 * 100 x = 400 answer : c"

a ) 7 , b ) $ 900 , c ) 400 , d ) 40 , e ) 135 %

c

multiply(100, 4)

multiply(n0,n2)|

other

after 10 % of the inhabitants of a village disappeared , a panic set in during which 25 % of the remaining inhabitants left the village . at that time , the population was reduced to 5535 . what was the number of original inhabitants ?

"let the total number of original inhabitants be x . ( 75 / 100 ) * ( 90 / 100 ) * x = 5535 ( 27 / 40 ) * x = 5535 x = 5535 * 40 / 27 = 8200 the answer is b ."

a ) 15 , b ) 8200 , c ) 50', ' , d ) 94 , e ) $ 6.80

b

divide(5535, subtract(subtract(const_1, divide(10, const_100)), multiply(subtract(const_1, divide(10, const_100)), divide(25, const_100))))

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

gain

is two years older than b who is twice as old as c . if the total of the ages of a , b and c be 42 , then how old is b ?

"let c ' s age be x years . then , b ' s age = 2 x years . a ' s age = ( 2 x + 2 ) years . ( 2 x + 2 ) + 2 x + x = 42 5 x = 40 = > x = 8 hence , b ' s age = 2 x = 16 years . answer : a"

a ) 128 , b ) 4.37 sec , c ) 16 , d ) 50 % , e ) 27 days

c

multiply(divide(subtract(42, const_2), add(const_3, const_2)), const_2)

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

general

today joelle opened an interest - bearing savings account and deposited $ 7,000 . if the annual interest rate is 5 percent compounded interest , and she neither deposits nor withdraws money for exactly 2 years , how much money will she have in the account ?

"interest for 1 st year = 7000 * 5 / 100 = 350 interest for 2 nd year = 7350 * 5 / 100 = 367.50 total = 7000 + 350 + 367.50 = 7717.50 answer : b"

a ) 25 % , b ) 20 , c ) 8 / 3 , d ) $ 7717.50 , e ) 2068

d

add(add(multiply(multiply(multiply(const_3, 2), const_100), const_10), divide(multiply(multiply(multiply(multiply(const_3, 2), const_100), const_10), 5), const_100)), divide(multiply(add(multiply(multiply(multiply(const_3, 2), const_100), const_10), divide(multiply(multiply(multiply(multiply(const_3, 2), const_100), const_10), 5), const_100)), 5), const_100))

multiply(n2,const_3)|multiply(#0,const_100)|multiply(#1,const_10)|multiply(n1,#2)|divide(#3,const_100)|add(#4,#2)|multiply(n1,#5)|divide(#6,const_100)|add(#5,#7)|

gain

last year the range of the annual bonus of the 100 employees at company x was $ 20000 . if the annual bonus of each of the 100 employees this year is 10 percent greater than it was last year , what is the range of the annual bonus of the 100 employees this year ?

let the lowest bonus be x . therefore , highest bonus is x + 20000 . now bonus of each employee is increased by 10 % . therefore the bonus will remain arranged in the same order as before . or lowest bonus = 1.1 x and highest = 1.1 * ( x + 20000 ) or range = highest - lowest = 1.1 * ( x + 20000 ) - 1.1 x = 22000 , hence , b

a ) 75 % , b ) $ 22000 , c ) 1200 , d ) 10 , e ) 28 / 9

b

multiply(20000, add(const_1, divide(10, const_100)))

divide(n3,const_100)|add(#0,const_1)|multiply(n1,#1)

general

the list price of an article is rs . 65 . a customer pays rs . 56.16 for it . he was given two successive discounts , one of them being 10 % . the other discount is ?

"explanation : 65 * ( 90 / 100 ) * ( ( 100 - x ) / 100 ) = 56.16 x = 4 % answer : option b"

a ) 1453 , b ) 49 , c ) 4 % , d ) 8 , e ) $ 1260

c

multiply(divide(subtract(subtract(65, multiply(65, divide(10, const_100))), 56.16), subtract(65, multiply(65, divide(10, const_100)))), const_100)

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

gain

a spirit and water solution is sold in a market . the cost per liter of the solution is directly proportional to the part ( fraction ) of spirit ( by volume ) the solution has . a solution of 1 liter of spirit and 1 liter of water costs 50 cents . how many cents does a solution of 1 liter of spirit and 3 liters of water cost ?

c . 50 cents yes , ensure that you understand the relation thoroughly ! cost per liter = k * fraction of spirit 50 cents is the cost of 2 liters of solution ( 1 part water , 1 part spirit ) . so cost per liter is 25 cents . fraction of spirit is 1 / 2 . 25 = k * ( 1 / 2 ) k = 50 cost per liter = 50 * ( 1 / 4 ) ( 1 part spirit , 3 parts water ) cost for 4 liters = 50 * ( 1 / 4 ) * 4 = 50 cents d . 50 cents

a ) 40 , b ) 50', ' , c ) $ 22000 , d ) 5 cm , e ) 274

b

multiply(multiply(50, divide(1, add(1, 3))), add(1, 3))

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

geometry

two trains are running in opposite directions in the same speed . the length of each train is 120 meter . if they cross each other in 12 seconds , the speed of each train ( in km / hr ) is

"explanation : distance covered = 120 + 120 = 240 m time = 12 s let the speed of each train = x . then relative velocity = x + x = 2 x 2 x = distance / time = 240 / 12 = 20 m / s speed of each train = x = 20 / 2 = 10 m / s = 10 * 18 / 5 km / hr = 36 km / hr option b"

a ) 9 : 5 , b ) 21 % , c ) 270 m , d ) 55 , e ) 36 km / hr

e

multiply(const_3_6, divide(divide(add(120, 120), 12), const_2))

add(n0,n0)|divide(#0,n1)|divide(#1,const_2)|multiply(#2,const_3_6)|

physics

what is the measure of the radius of the circle inscribed in a triangle whose sides measure 8 , 15 and 21 units ?

"sides are 8 , 15 and 21 . . . thus it is right angle triangle since 21 ^ 2 = 8 ^ 2 + 15 ^ 2 therefore , area = 1 / 2 * 15 * 8 = 60 we have to find in - radius therefore , area of triangle = s * r . . . . where s = semi - perimeter and r = in - radius now s = semi - perimeter = 21 + 15 + 8 / 2 = 22 thus , 60 = 22 * r and hence r = in - radius = 2.6 option b"

a ) 2.6 units , b ) 180 , c ) 158 , d ) 1040 , e ) 1800

a

divide(triangle_area_three_edges(8, 15, 21), divide(triangle_perimeter(8, 15, 21), const_2))

triangle_area_three_edges(n0,n1,n2)|triangle_perimeter(n0,n1,n2)|divide(#1,const_2)|divide(#0,#2)|

geometry

if ( 4 - x ) / ( 5 + x ) = x , what is the value of x ^ 2 + 6 x - 4 ?

"( 4 - x ) = x * ( 5 + x ) ( 4 - x ) = 5 x + x ^ 2 0 = x ^ 2 + 6 x - 4 the answer is b ."

a ) 9 , b ) 0 , c ) 6 hours , d ) 15 / 11 , e ) 9560

b

subtract(multiply(4, 2), 4)

multiply(n2,n0)|subtract(#0,n0)|

general

a certain bus driver is paid a regular rate of $ 16 per hour for any number of hours that does not exceed 40 hours per week . for any overtime hours worked in excess of 40 hours per week , the bus driver is paid a rate that is 75 % higher than his regular rate . if last week the bus driver earned $ 1004 in total compensation , how many total hours did he work that week ?

"for 40 hrs = 40 * 16 = 640 excess = 1004 - 640 = 364 for extra hours = . 75 ( 16 ) = 12 + 16 = 28 number of extra hrs = 364 / 28 = 13 total hrs = 40 + 13 = 53 answer e 53"

a ) 17 th , b ) 14 , c ) 53 , d ) 120 % , e ) 16

c

add(40, divide(subtract(1004, multiply(16, 40)), divide(multiply(16, add(const_100, 75)), const_100)))

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

general

you have to send 3000 grapes 1000 kilometers from grapecity to appleland . your truck can carry 1000 grapes at a time . every time you travel a kilometer towards appleland you must pay a tax of 1 grape but you pay nothing when going in the other direction ( towards grapecity ) . what is highest number of grapes you can get to appleland ?

step one : first you want to make 3 trips of 1,000 grapes 333 kilometers . you will be left with 2,001 grapes and 667 kilometers to go . step two : next you want to take 2 trips of 1,000 grapes 500 kilometers . you will be left with 1,000 grapes and 167 kilometers to go ( you have to leave a grape behind ) . step three : finally , you travel the last 167 kilometers with one load of 1,000 grapes and are left with 833 grapes in appleland . correct answer is a ) 833

a ) 50 km / hr , b ) 10000 , c ) 833 , d ) 33 , e ) 4

c

subtract(1000, subtract(subtract(1000, floor(divide(1000, const_3))), divide(1000, const_2)))

divide(n1,const_3)|divide(n1,const_2)|floor(#0)|subtract(n1,#2)|subtract(#3,#1)|subtract(n1,#4)

physics

the sum of all consecutive odd integers from − 19 to 29 , inclusive , is

"the sum of the odd numbers from - 19 to + 19 is 0 . let ' s add the remaining numbers . 21 + 23 + 25 + 27 + 29 = 5 ( 25 ) = 125 the answer is a ."

a ) 28.57 % , b ) 125 , c ) 9 , d ) 25 % , e ) 7 days

b

add(add(add(add(19, const_2), add(add(19, const_2), const_2)), add(add(add(19, const_2), const_2), const_2)), 29)

add(n0,const_2)|add(#0,const_2)|add(#0,#1)|add(#1,const_2)|add(#2,#3)|add(n1,#4)|

physics

the contents of a certain box consist of 14 apples and 25 oranges . how many oranges must be removed from the box so that 70 percent of the pieces of fruit in the box will be apples ?

"the objective here is that 70 % of the fruit in the box should be apples . now , there are 14 apples at start and there is no talk of removing any apples , so number of apples should remain 14 and they should constitute 70 % of total fruit , so total fruit = 14 / 0.7 = 20 so we should have 20 - 14 = 6 oranges . right now , there are 25 oranges , so to get to 6 oranges , we should remove 25 - 6 = 19 oranges . answer d"

a ) 800 , b ) 19 , c ) 1.745 % , d ) 52 % , e ) 3400

b

subtract(add(14, 25), divide(14, divide(70, const_100)))

add(n0,n1)|divide(n2,const_100)|divide(n0,#1)|subtract(#0,#2)|

general

jo ' s collection contains us , indian and british stamps . if the ratio of us to indian stamps is 7 to 2 and the ratio of indian to british stamps is 5 to 1 , what is the ratio of us to british stamps ?

u / i = 7 / 2 i / b = 5 / 1 since i is multiple of both 2 ( as per first ratio ) and 5 ( as per second ratio ) so let ' s assume that i = 10 i . e . multiplying teh first ratio by 5 and second ration by 2 in each numerator and denominator then , u : i : b = 35 : 21 : 2 i . e . u : b = 35 : 2 answer : option e

a ) 7200 , b ) 35 : 2 , c ) 6923 , d ) 28 , e ) 36.5

b

divide(multiply(7, 5), multiply(1, 2))

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

other

how many prime numbers are between 13 / 3 and 83 / 6 ?

"13 / 3 = 4 . xxx 83 / 6 = 13 . xxx so we need to find prime numbers between 4 ( exclusive ) - 12 ( inclusive ) there are 2 prime numbers 7 11 hence answer will be ( b ) 2 b"

a ) 1000 , b ) 14 , c ) 30 % , d ) $ 2000 , e ) 2

e

floor(const_2)

floor(const_2)|

general

john makes $ 60 a week from his job . he earns a raise andnow makes $ 80 a week . what is the % increase ?

"increase = ( 20 / 60 ) * 100 = ( 1 / 3 ) * 100 = 33.33 % . b"

a ) 4 , b ) 12 hours , c ) 33.33 % , d ) 10 , e ) 5

c

multiply(divide(subtract(80, 60), 60), const_100)

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

gain

three numbers are in the ratio 3 : 4 : 5 and their l . c . m . is 600 . their h . c . f is ?

"let the numbers be 3 x , 4 x and 5 x their l . c . m . = 60 x 60 x = 600 x = 10 the numbers are 3 * 10 , 4 * 10 , 5 * 10 hence required h . c . f . = 10 answer is a"

a ) 8985 , b ) 0 , c ) 996004 , d ) 10 , e ) 57 %

d

add(multiply(multiply(3, 5), const_100), multiply(4, 5))

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

other

a boat takes 19 hours for travelling downstream from point a to point b and coming back to a point c which is at midway between a and b . if the velocity of the stream is 4 kmph and the speed of the boat in still water is 14 kmph , what is the distance between a and b ?

"explanation : speed in downstream = ( 14 + 4 ) km / hr = 18 km / hr ; speed in upstream = ( 14 â € “ 4 ) km / hr = 10 km / hr . let the distance between a and b be x km . then , x / 18 + ( x / 2 ) / 10 = 19 â ‡ ” x / 18 + x / 20 = 19 â ‡ ’ x = 180 km . answer : a"

a ) 54 min , b ) 7.2 hr , c ) 180 km , d ) 1 , e ) 108

c

divide(19, add(divide(const_1, add(14, 4)), divide(const_1, multiply(subtract(14, 4), const_2))))

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

physics

in a division sum , the quotient is 18 , the divisor 43 and the remainder 12 , find the dividend ?

"explanation : 18 * 43 + 12 = 786 answer : d"

a ) 30 , b ) 786 , c ) 1 : 4', ' , d ) 32 square meters , e ) 18

b

add(multiply(18, 43), 12)

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

general

x and y are both integers . if x / y = 50.60 , then what is the sum of all the possible two digit remainders of x / y ?

"remainder = 0.60 - - > 60 / 100 - - > can be written as ( 60 / 4 ) / ( 100 / 4 ) = 15 / 25 so remainders can be 15 , 30 , 45 , 60 , . . . . . 90 . we need the sum of only 2 digit remainders - - > 15 + 30 + 45 + 60 + 75 + 90 = 315 answer : a"

a ) 10.5 , b ) s 255 , c ) 73.43 , d ) 1000 , e ) 315

e

add(multiply(divide(const_3, const_2), const_100), add(multiply(add(const_2, const_3), 50.60), const_3))

add(const_2,const_3)|divide(const_3,const_2)|multiply(n0,#0)|multiply(#1,const_100)|add(#2,const_3)|add(#4,#3)|

general

the average weight of 10 men is increased by 1 ½ kg when one of the men who weighs 60 kg is replaced by a new man . what is the weight of the new man ?

"since the average has increased by 1.5 kg , the weight of the man who stepped in must be equal to 60 + 10 x 1.5 60 + 15 = 75 kg ans : ' d '"

a ) 240 , b ) 1 / 4 , c ) - 3 , d ) 75 kg , e ) 11.5 sec

d

add(60, multiply(10, add(1, divide(const_1, 1))))

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

general

a shopkeeper sold an article at $ 1050 and gained a 20 % profit . what was the cost price ?

"let x be the cost price . 1.2 x = 1050 x = 1050 / 1.2 = 875 the answer is e ."

a ) $ 875 , b ) 19 , c ) 88 % , d ) 1943236 , e ) 375

a

multiply(const_100.0, divide(const_100, add(1050, 20)))

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

gain

niall ' s income is 60 % less than rex ' s income , and sam ' s income is 25 % less than niall ' s income . if rex gave 60 % of his income to sam and 40 % of his income to niall , niall ' s new income would be what fraction of sam ' s new income ?

we can take some easy numbers and make calculations simpler . let r ( rex ' s income ) = 100 q ( niall ' s income ) = 40 % r = 40 s ( sam ' s income ) = 75 % q = ( 3 / 4 ) * 40 = 30 now , if rex gives 40 % to niall - - > q = 40 + 40 = 80 60 % given to sam - - > s = 30 + 60 = 90 the ratio is : q / s = 80 / 90 = 8 / 9 = a

a ) 8 / 9 , b ) 144 , c ) 4 , d ) 13 seconds , e ) 55

a

divide(add(40, 40), add(add(40, 40), const_10))

add(n3,n3)|add(#0,const_10)|divide(#0,#1)

general

if 2 a = 4 b = 10 , then 40 ab =

2 a * 4 b = 10 * 10 = 100 8 ab = 100 i . e . 40 ab = 500 answer : option e

a ) 500 , b ) 52500 , c ) 40 % , d ) 129.5 , e ) 130

a

multiply(40, multiply(divide(10, 2), divide(10, 4)))

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

general

rahim bought 52 books for rs . 1200 from one shop and 32 books for rs . 480 from another . what is the average price he paid per book ?

"average price per book = ( 1200 + 480 ) / ( 52 + 32 ) = 1680 / 84 = rs . 20 answer : d"

a ) 10 , b ) s . 20 , c ) 252 sec , d ) 175.57 , e ) 2 / 7

b

divide(add(1200, 480), add(52, 32))

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

general

a rower whose speed is 4 km / hr in still water rows to a certain point upstream and back to the starting point in a river which flows at 2 km / hr . what is the rower ' s average speed ( in km / hr ) for the total journey ?

"time upstream = d / 2 time downstream = d / 6 total time = d / 2 + d / 6 = 2 d / 3 average speed = 2 d / ( 2 d / 3 ) = 3 km / hr the answer is d ."

a ) 80 / 243 , b ) 3 , c ) 2 : 1 , d ) 120 km , e ) 217.5 cm 2

b

divide(add(subtract(4, const_0.5), add(4, 2)), const_2)

add(n0,const_0.5)|subtract(n0,n1)|add(#0,#1)|divide(#2,const_2)|

general

if 2 ^ 5 , 4 ^ 3 , and 13 ^ 2 are all factors of the product of 936 and w where w is a positive integer , what is the smallest possible value of w ?

"here 156 has three two ' s two three ' s and one 13 rest of them must be in w so w = 13 * 4 * 4 = 208 smash d"

a ) c : 37.5 , b ) 50 liters , c ) 7 , d ) 208 , e ) - 4

d

multiply(multiply(multiply(power(2, 2), 4), divide(13, 2)), 2)

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

general

if 3 < x < 6 < y < 11 , then what is the greatest possible positive integer difference of x and y ?

"3 < x < 6 < y < 11 ; 3 < x y < 11 3 + y < x + 11 y - x < 8 . positive integer difference is 7 ( for example y = 10.5 and x = 3.5 ) answer : e ."

a ) $ 290 , b ) 6 , c ) 7 , d ) 450 sq . m', ' , e ) 2 6 / 7 %

c

subtract(subtract(11, 3), const_1)

subtract(n2,n0)|subtract(#0,const_1)|

general

chris age after 20 years will be 5 times his age 5 years back . what is the present age of chris ?

chris present age = x after 20 years = x + 20 5 years back = x - 5 x + 20 = 5 ( x - 5 ) x = 5 answer is e

a ) 642 , b ) 83 % , c ) 10 , d ) 5 , e ) 17 min .

d

subtract(divide(add(multiply(5, 5), 20), subtract(5, const_1)), subtract(divide(add(multiply(5, 5), 20), subtract(5, const_1)), 5))

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

general

if x is an integer and 2.134 × 10 ^ x is less than 2 , 200,000 , what is the greatest possible value for x ?

"x is an integer and 2.134 × 10 x is less than 2 , 200,000 , what is the greatest possible value for x ? for 2.134 × 10 x is less than 2 , 200,000 to remain true , the greatest number is 2 , 134,000 , which makes x = 6 b . 6"

a ) 22.5 % , b ) 39 , c ) 189 , d ) 6 , e ) s 255

d

floor(divide(log(divide(2, 2.134)), log(10)))

divide(n2,n0)|log(n1)|log(#0)|divide(#2,#1)|floor(#3)|

general

the cross - section of a cannel is a trapezium in shape . if the cannel is 15 m wide at the top and 9 m wide at the bottom and the area of cross - section is 636 sq m , the depth of cannel is ?

"1 / 2 * d ( 15 + 9 ) = 636 d = 53 answer : a"

a ) 21 , b ) 113 m 2 , c ) 10.87 km , d ) 50000 , e ) 53

e

divide(divide(divide(636, divide(add(15, 9), const_2)), 9), const_2)

add(n0,n1)|divide(#0,const_2)|divide(n2,#1)|divide(#2,n1)|divide(#3,const_2)|

physics

a train passes a station platform in 36 sec and a man standing on the platform in 24 sec . if the speed of the train is 54 km / hr . what is the length of the platform ?

"speed = 54 * 5 / 18 = 15 m / sec . length of the train = 15 * 24 = 360 m . let the length of the platform be x m . then , ( x + 360 ) / 36 = 15 = > x = 180 m . answer : c"

a ) 180 m , b ) 22.37 , c ) 1 / 3 , d ) 240 , e ) 34 %

a

multiply(24, multiply(54, const_0_2778))

multiply(n2,const_0_2778)|multiply(n1,#0)|

physics

the weight of a glass of jar is 30 % of the weight of the jar filled with coffee beans . after some of the beans have been removed , the weight of the jar and the remaining beans is 60 % of the original total weight . what fraction part of the beans remain in the jar ?

let weight of jar filled with beans = 100 g weight of jar = 30 g weight of coffee beans = 70 g weight of jar and remaining beans = 60 g weight of remaining beans = 30 g fraction remaining = 30 / 70 = 3 / 7 answer is e .

a ) 9 / 49 , b ) 3 / 7 , c ) 315 , d ) 36.5 , e ) 21 hours

b

divide(subtract(60, 30), subtract(const_100, 30))

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

gain

what will be the fraction of 12.5 %

"explanation : it will 12.5 * 1 / 100 = 1 / 8 answer : option d"

a ) 2000 , b ) 16 days , c ) 22.5 % , d ) 1 / 8 , e ) 2

d

divide(circle_area(divide(12.5, const_2)), const_2)

divide(n0,const_2)|circle_area(#0)|divide(#1,const_2)|

gain

a number increased by 15 % gives 1150 . the number is

"formula = total = 100 % , increse = ` ` + ' ' decrease = ` ` - ' ' a number means = 100 % that same number increased by 15 % = 115 % 115 % - - - - - - - > 1150 ( 115 × 100 = 1150 ) 100 % - - - - - - - > 1000 ( 100 × 100 = 1000 ) b )"

a ) 30 m , b ) 3 : 4 , c ) 1000 , d ) 30 , e ) 25 %

c

divide(1150, add(const_1, divide(15, const_100)))

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

gain

a train running at a speed of 36 kmph crosses an electric pole in 12 seconds . in how much time will it cross a 390 m long platform ?

"let the length of the train be x m . when a train crosses an electric pole , the distance covered is its own length . so , x = 12 * 36 * 5 / 18 m = 120 m . time taken to cross the platform = ( 120 + 390 ) / 36 * 5 / 18 = 51 min . answer : c"

a ) 22 , b ) 51 , c ) 132 , d ) 0.7 , e ) 50

b

divide(add(390, multiply(multiply(const_0_2778, 36), 12)), multiply(const_0_2778, 36))

multiply(n0,const_0_2778)|multiply(n1,#0)|add(n2,#1)|divide(#2,#0)|

physics

the average of 7 numbers is 23 . if each number be multiplied by 5 . find the average of new set of numbers ?

"explanation : average of new numbers = 23 * 5 = 115 answer : option d"

a ) 115 , b ) rs 20 , c ) 3 , d ) 40 , e ) 6

a

multiply(23, 5)

multiply(n1,n2)|

general

if taxi fares were $ 3.00 for the first 1 / 5 mile and $ 0.20 for each 1 / 5 mile there after , then the taxi fare for a 4 - mile ride was

"in 4 miles , initial 1 / 5 mile charge is $ 3 rest of the distance = 4 - ( 1 / 5 ) = 19 / 5 rest of the distance charge = 19 ( 0.2 ) = $ 3.8 ( as the charge is 0.2 for every 1 / 5 mile ) = > total charge for 4 miles = 3 + 3.8 = 6.8 answer is a"

a ) 5 , b ) 16 days , c ) 60 km , d ) $ 6.80 , e ) 20

d

add(3.00, multiply(subtract(divide(3.00, divide(1, 5)), 1), 0.20))

divide(n1,n2)|divide(n6,#0)|subtract(#1,n1)|multiply(n3,#2)|add(n0,#3)|

general

if 8 men or 12 women can do a piece of work in 35 days , in how many days can the same work be done by 6 men and 11 women ?

"8 men = 12 women ( i . e 2 men = 3 women ) 12 women 1 day work = 1 / 35 soln : 6 men ( 9 women ) + 11 women = 20 women = ? 1 women 1 day work = 12 * 35 = 1 / 420 so , 20 women work = 20 / 420 = 1 / 21 ans : 21 days answer : e"

a ) s . 4400 , b ) 21 days , c ) 49 : 64 , d ) 49 , e ) 5 / 26

b

inverse(add(divide(6, multiply(8, 35)), divide(11, multiply(12, 35))))

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

physics

the total age of a and b is 18 years more than the total age of b and c . c is how many year younger than

"given that a + b = 182 + b + c = > a â € “ c = 18 + b â € “ b = 18 = > c is younger than a by 18 years answer : e"

a ) 12 , b ) 12 hours , c ) 870 , d ) 18 years , e ) 40

d

multiply(18, const_1)

multiply(n0,const_1)|

general

one pump drains one - half of a pond in 1 hours , and then a second pump starts draining the pond . the two pumps working together finish emptying the pond in one - half hour . how long would it take the second pump to drain the pond if it had to do the job alone ?

"the tricky part here , i believed is one half hour = 1 / 2 . then everything would be easy . we have the 1 st pump working rate / hour = 1 / 2 : 1 = 1 / 2 working rate of 2 pumps : 1 / 2 : 1 / 2 = 1 . working rate of 2 nd pump : 1 - 1 / 2 = 1 / 2 - - > time taken for the 2 nd pump to finish : 1 : 1 / 2 = 2 / 1 = 2 hours . c"

a ) 259,200 , b ) 3 hours , c ) $ 1.96 , d ) 50 % , e ) 50

b

divide(const_1, subtract(const_1, divide(const_1, multiply(1, const_2))))

multiply(n0,const_2)|divide(const_1,#0)|subtract(const_1,#1)|divide(const_1,#2)|

physics

the greatest number which on dividing 1657 and 2037 leaves remainders 9 and 5 respectively , is :

"explanation : required number = h . c . f . of ( 1657 - 9 ) and ( 2037 - 5 ) = h . c . f . of 1648 and 2032 = 16 . answer : a"

a ) 33 1 / 3 % , b ) $ 17 , c ) 24 , d ) 16 , e ) 14

d

gcd(subtract(2037, 5), subtract(1657, 9))

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

general

a rectangular farm has to be fenced one long side , one short side and the diagonal . if the cost of fencing is rs . 10 per meter . the area of farm is 1200 m 2 and the short side is 30 m long . how much would the job cost ?

"l * 30 = 1200 l = 40 40 + 30 + 50 = 120 120 * 10 = 1200 e"

a ) 30 % , b ) 16 , c ) 275 , d ) 1200 , e ) 162

d

multiply(add(add(30, divide(1200, 30)), sqrt(add(power(30, 2), power(divide(1200, 30), 2)))), 10)

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

geometry

the length of a rectangle is 2 times its width . if the width of the rectangle is 4 inches , what is the rectangle ' s area , in square inches ?

"if the width is 4 in and the length is 2 times the width , then the length is 2 * 4 = 8 in the area is given by 4 * 8 = 32 square inches correct answer e"

a ) 8.12 % , b ) 32 square inches , c ) 43.33 % , d ) 3 , e ) 169

b

rectangle_area(4, multiply(2, 4))

multiply(n0,n1)|rectangle_area(n1,#0)|

geometry

the ratio between the sale price and the cost price of an article is 7 : 4 . what is the ratio between the profit and the cost price of that article ?

"c . p . = rs . 4 x and s . p . = rs . 7 x . then , gain = rs . 3 x required ratio = 3 x : 4 x = 3 : 4 a"

a ) 2 / 3 , b ) 25 , c ) 5 kmph , d ) 975 , e ) 3 : 4

e

divide(subtract(7, 4), 4)

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

other

barbata invests $ 2400 in the national bank at 5 % . how much additional money must she invest at 8 % so that the total annual income will be equal to 6 % of her entire investment ?

"let the additional invested amount for 8 % interest be x ; equation will be ; 2400 + 0.05 * 2400 + x + 0.08 x = 2400 + x + 0.06 ( 2400 + x ) 0.05 * 2400 + 0.08 x = 0.06 x + 0.06 * 2400 0.02 x = 2400 ( 0.06 - 0.05 ) x = 2400 * 0.01 / 0.02 = 1200 ans : ` ` a ' '"

a ) 14 kmph , b ) 6 , c ) 28 , d ) 0.05 , e ) 1200

e

divide(subtract(multiply(divide(6, const_100), 2400), multiply(2400, divide(5, const_100))), subtract(divide(8, const_100), divide(6, const_100)))

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

general

if a speaks the truth 30 % of the times , b speaks the truth 40 % of the times . what is the probability that at least one will tell the truth

"probability of a speaks truth p ( a ) = 3 / 10 ; false = 7 / 10 probability of b speaks truth p ( b ) = 4 / 10 ; false = 6 / 10 . for given qtn ans = 1 - ( neither of them tell truth ) . because a & b are independent events = 1 - [ ( 7 / 10 ) * ( 6 / 10 ) ] = 1 - 42 / 100 = 1 - 0.42 = 0.58 answer : a"

a ) 0.58 , b ) 10 ° , c ) 14 , d ) 9 , e ) . 2

a

multiply(divide(30, multiply(multiply(const_4, const_5), const_5)), divide(40, multiply(multiply(const_4, const_5), const_5)))

multiply(const_4,const_5)|multiply(#0,const_5)|divide(n0,#1)|divide(n1,#1)|multiply(#2,#3)|

gain

a certain farmer pays $ 30 per acre per month to rent farmland . how much does the farmer pay per month to rent a rectangular plot of farmland that is 370 feet by 605 feet ? ( 43,560 square feet = 1 acre )

basically the question an error . 1 acre = 43,560 square feet and if it is then the answer is 154.1 ( e )

a ) 14 : 00 , b ) 200 , c ) $ 154.1 , d ) $ 17 , e ) 7028

c

multiply(30, divide(multiply(370, 605), divide(multiply(370, 605), const_10)))

multiply(n1,n2)|divide(#0,const_10)|divide(#0,#1)|multiply(n0,#2)|

geometry

difference between a two - digit number and the number obtained by interchanging the two digits is 36 , what is the difference between two numbers

explanation : let the ten digit be x , unit digit is y . then ( 10 x + y ) - ( 10 y + x ) = 36 = > 9 x - 9 y = 36 = > x - y = 4 . option b

a ) 996004 , b ) 4 , c ) 3 , d ) 420 gallons', ' , e ) rs . 2700

b

divide(36, subtract(const_10, const_1))

subtract(const_10,const_1)|divide(n0,#0)

general

the average score of a cricketer in 6 matches is 27 and in other 4 matches is 32 . then find the average score in all the 10 matches ?

"average in 10 matches = ( 6 * 27 + 4 * 32 ) / 6 + 4 = 162 + 128 / 10 = 290 / 10 = 29 answer is d"

a ) 29 , b ) 0.3 % , c ) 175.5 cm , d ) 6 liters , e ) 60

a

divide(add(multiply(6, 27), multiply(4, 32)), 10)

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

general

rs . 900 amounts to rs . 920 in 3 years at simple interest . if the interest is increased by 3 % , it would amount to how much ?

"( 900 * 3 * 3 ) / 100 = 81 920 + 81 = 1001 answer : b"

a ) 5 5 / 8 km , b ) 60 , c ) 113 m 2 , d ) 1070 , e ) rs . 1001

e

multiply(power(add(const_1, divide(3, const_100)), 3), 900)

divide(n3,const_100)|add(#0,const_1)|power(#1,n2)|multiply(n0,#2)|

gain

the average expenditure of a labourer for 10 months was 85 and he fell into debt . in the next 4 months by reducing his monthly expenses to 60 he not only cleared off his debt but also saved 30 . his monthly income is

"income of 10 months = ( 10 × 85 ) – debt = 850 – debt income of the man for next 4 months = 4 × 60 + debt + 30 = 270 + debt ∴ income of 10 months = 1120 average monthly income = 1120 ÷ 10 = 112 answer c"

a ) 40.5 , b ) 0 , c ) 100 , d ) five , e ) 112

e

divide(add(add(multiply(85, 10), multiply(60, 4)), 30), add(10, 4))

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

general

a store has 10 bottles of juice , including 5 bottles of apple juice . in the evening , 6 bottles of juice are sold one by one . what is the probability of selling 3 bottles of apple juice among the 6 bottles ? assume that every bottle has an equal chance of being bought .

"the total number of ways to sell 6 bottles from 10 is 10 c 6 = 210 . the number of ways to sell 3 bottles of apple juice is 5 c 3 * 5 c 3 = 10 * 10 = 100 p ( selling 3 bottles of apple juice ) = 100 / 210 = 10 / 21 the answer is d ."

a ) 2 , b ) $ 2,350 , c ) 10 / 21 , d ) 129.5 , e ) 7

c

divide(choose(5, 3), choose(10, 5))

choose(n1,n3)|choose(n0,n1)|divide(#0,#1)|

probability

a farmer has an apple orchard consisting of fuji and gala apple trees . due to high winds this year 10 % of his trees cross pollinated . the number of his trees that are pure fuji plus the cross - pollinated ones totals 136 , while 3 / 4 of all his trees are pure fuji . how many of his trees are pure gala ?

let f = pure fuji , g = pure gala and c - cross pollinated . c = 10 % of x where x is total trees . c = . 1 x also 3 x / 4 = f and c + f = 136 = > . 1 x + 3 / 4 x = 136 = > x = 160 160 - 136 = pure gala = 24 . a

a ) 3400 , b ) 51 days , c ) 60 , d ) 24 , e ) 20

d

subtract(divide(136, add(divide(10, const_100), divide(3, 4))), 136)

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

general

at a monthly meeting , 1 / 3 of the attendees were males and 4 / 5 of the male attendees arrived on time . if 5 / 6 of the female attendees arrived on time , what fraction of the attendees at the monthly meeting did not arrive on time ?

"males who did not arrive on time are 1 / 5 * 1 / 3 = 1 / 15 of the attendees . females who did not arrive on time are 1 / 6 * 2 / 3 = 1 / 9 of the attendees . the fraction of all attendees who did not arrive on time is 1 / 15 + 1 / 9 = 8 / 45 the answer is e ."

a ) 9 , b ) 8 / 45 , c ) 378 , d ) 28 , e ) 10 years

b

add(multiply(subtract(const_1, divide(5, 6)), subtract(const_1, divide(1, 3))), multiply(subtract(const_1, divide(4, 5)), divide(1, 3)))

divide(n4,n5)|divide(n0,n1)|divide(n2,n3)|subtract(const_1,#0)|subtract(const_1,#1)|subtract(const_1,#2)|multiply(#3,#4)|multiply(#1,#5)|add(#6,#7)|

general

a shopkeeper sold an book offering a discount of 5 % and earned a profit of 30 % . what would have been the percentage of profit earned if no discount was offered ?

"let c . p . be $ 100 . then , s . p . = $ 130 let marked price be $ x . then , 95 / 100 x = 130 x = 13000 / 95 = $ 136.8 now , s . p . = $ 136.8 , c . p . = $ 100 profit % = 136.8 % . d"

a ) 180 , b ) 136.8 , c ) 275 , d ) $ 8 , e ) 0

b

multiply(const_100, divide(add(const_100, 30), subtract(const_100, 5)))

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

gain

a company wants to spend equal amounts of money for the purchase of two types of computer printers costing $ 300 and $ 200 per unit , respectively . what is the fewest number of computer printers that the company can purchase ?

"the smallest amount that the company can spend is the lcm of 300 and 200 , which is 600 for each , which is total 1200 . the number of 1 st type of computers which costing $ 300 = 600 / 300 = 2 . the number of 2 nd type of computers which costing $ 200 = 600 / 200 = 3 . total = 2 + 3 = 5 answer is b ."

a ) 5 , b ) 7 , c ) 88.6 , d ) 9 , e ) 2

a

add(divide(lcm(300, 200), 300), divide(lcm(300, 200), 200))

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

general

if p is a prime number greater than 3 , find the remainder when p ^ 2 + 14 is divided by 12 .

"every prime number greater than 3 can be written 6 n + 1 or 6 n - 1 . if p = 6 n + 1 , then p ^ 2 + 14 = 36 n ^ 2 + 12 n + 1 + 14 = 36 n ^ 2 + 12 n + 12 + 3 if p = 6 n - 1 , then p ^ 2 + 14 = 36 n ^ 2 - 12 n + 1 + 14 = 36 n ^ 2 - 12 n + 12 + 3 when divided by 12 , it must leave a remainder of 3 . the answer is c ."

a ) 416 , b ) 3 , c ) 16 , d ) 20 , e ) s . 5100

b

subtract(add(14, power(add(const_1, const_4), 2)), multiply(12, 3))

add(const_1,const_4)|multiply(n0,n3)|power(#0,n1)|add(n2,#2)|subtract(#3,#1)|

general

three unbiased coins are tossed . what is the probability of getting 3 heads and 1 tail ?

"let , h - - > head , t - - > tail here s = { ttt , tth , tht , htt , thh , hth , hht , hhh } let e = event of getting 3 heads then e = { hhh , hth , thh , hht } p ( e ) = n ( e ) / n ( s ) = 4 / 8 = 1 / 2 answer is d"

a ) 8 , b ) 4 , c ) five , d ) 77 sq metres', ' , e ) 1 / 2

e

negate_prob(divide(const_1, power(const_2, const_3)))

power(const_2,const_3)|divide(const_1,#0)|negate_prob(#1)|

probability

find 95 × × 98

"here both numbers are less than 100 . so they are deficient of - 5 and - 2 compared with 100 . so answer : d"

a ) 4 years , b ) 93 / 10 , c ) 9,600 , d ) 16 , e ) 10.8 km / hr

b

divide(95, 98)

divide(n0,n1)|

general

5 + 5

d

a ) 20 % , b ) 4608 , c ) 10 , d ) 600 , e ) 20000

c

multiply(divide(5, 5), const_100)

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

general

a certain ball team has an equal number of right - and left - handed players . on a certain day , one - third of the players were absent from practice . of the players at practice that day , one - third were right handed . what is the ratio of the number of right - handed players who were not at practice that day to the number of left handed players who were not at practice ?

"say the total number of players is 18 , 9 right - handed and 9 left - handed . on a certain day , two - thirds of the players were absent from practice - - > 6 absent and 12 present . of the players at practice that day , one - third were right - handed - - > 12 * 1 / 3 = 4 were right - handed and 8 left - handed . the number of right - handed players who were not at practice that day is 9 - 4 = 5 . the number of left - handed players who were not at practice that days is 9 - 8 = 1 . the ratio = 5 / 1 . answer : b"

a ) 12003 , b ) 5 / 1 , c ) 2 , d ) $ 1.80 , e ) 334

b

divide(subtract(divide(const_1, const_2), subtract(subtract(const_1, divide(const_1, const_3)), multiply(divide(const_1, const_3), subtract(const_1, divide(const_1, const_3))))), subtract(divide(const_1, const_2), multiply(divide(const_1, const_3), subtract(const_1, divide(const_1, const_3)))))

divide(const_1,const_2)|divide(const_1,const_3)|subtract(const_1,#1)|multiply(#1,#2)|subtract(#2,#3)|subtract(#0,#3)|subtract(#0,#4)|divide(#6,#5)|

general

bag a contains red , white and blue marbles such that the red to white marble ratio is 1 : 3 and the white to blue marble ratio is 2 : 3 . bag b contains red and white marbles in the ratio of 1 : 4 . together , the two bags contain 42 white marbles . how many red marbles could be in bag a ?

"6 is the answer . bag a - r : w : b = 2 : 6 : 9 let w in bag a be 6 k bab b - r : w = 1 : 4 let w in bag b be 4 p w = 42 = 6 k + 4 p = > k = 5 , p = 3 total red ' s in bag a will be 2 k = 10 d"

a ) 3720 , b ) 30 , c ) 10 , d ) 44.9091 , e ) 2

c

divide(42, add(multiply(3, 2), 4))

multiply(n1,n2)|add(n5,#0)|divide(n6,#1)|

other

a man has some hens and cows . if the number of heads be 42 and the number of feet equals 124 , then the number of hens will be

"explanation : let number of hens = h and number of cows = c number of heads = 42 = > h + c = 42 - - - ( equation 1 ) number of feet = 124 = > 2 h + 4 c = 124 = > h + 2 c = 62 - - - ( equation 2 ) ( equation 2 ) - ( equation 1 ) gives 2 c - c = 62 - 42 = > c = 20 substituting the value of c in equation 1 , we get h + 22 = 42 = > h = 42 - 20 = 22 i . e . , number of hens = 22 answer : a"

a ) 47 , b ) 45 , c ) 300 , d ) 22 , e ) 120

d

divide(subtract(multiply(42, const_4), 124), const_2)

multiply(n0,const_4)|subtract(#0,n1)|divide(#1,const_2)|

general

find large number from below question the difference of two numbers is 1385 . on dividing the larger number by the smaller , we get 6 as quotient and the 15 as remainder

"let the smaller number be x . then larger number = ( x + 1385 ) . x + 1385 = 6 x + 15 5 x = 1370 x = 274 large number = 274 + 1385 = 1659 e"

a ) 18 , b ) $ 8.45 , c ) 90 kg , d ) 1659 , e ) 13.4

d

multiply(divide(subtract(1385, 15), subtract(6, const_1)), 6)

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

general

the average age of 36 students in a group is 13 years . when teacher ' s age is included to it , the average increases by one . what is the teacher ' s age in years ?

"explanation : age of the teacher = ( 37 * 14 - 36 * 13 ) years = 50 years . answer : e"

a ) 50 years , b ) 272 , c ) 25 , d ) rs 9000 , e ) 1 / 2

a

add(36, const_1)

add(n0,const_1)|

general

convert 0.30 in to a vulgar fraction ?

answer 0.30 = 30 / 100 = 3 / 10 correct option : c

a ) 60 , b ) 200 , c ) 130 , d ) 3 / 10 , e ) 61 cm

d

divide(multiply(multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)), 0.3), multiply(multiply(add(const_3, const_2), const_2), multiply(add(const_3, const_2), const_2)))

add(const_2,const_3)|multiply(#0,const_2)|multiply(#1,#1)|multiply(n0,#2)|divide(#3,#2)

physics

a certain galaxy is known to comprise approximately 5 x 10 ^ 11 stars . of every 50 million of these stars , one is larger in mass than our sun . approximately how many stars in this galaxy are larger than the sun ?

"total no . of stars on galaxy = 5 * 10 ^ 11 of every 50 million stars , 1 is larger than sun . 1 million = 10 ^ 6 therofore , 50 million = 50 * 10 ^ 6 total no . of stars larger than sun = 5 * 10 ^ 11 / 50 * 10 ^ 6 = 50 * 10 ^ 3 / 5 = 10000 therefore answer is e"

a ) 10,000 , b ) 9 , c ) 4 , d ) 5 , e ) 463

a

multiply(divide(multiply(divide(multiply(5, 10), 50), power(10, const_4)), const_1000), 5)

multiply(n0,n1)|power(n1,const_4)|divide(#0,n3)|multiply(#2,#1)|divide(#3,const_1000)|multiply(n0,#4)|

general

a room is 30 m long and 24 m broad . if the sum of the areas of the floor and the ceiling is equal to the sum of the areas of four walls , the volume of the hall is :

let the height be h 2 ( 30 + 24 ) x h – 2 ( 30 - 24 ) h = ( 2 ( 30 x 24 ) ) / ( 2 ( 30 + 24 ) ) = ( 30 x 24 ) / 54 = 40 / 3 m volume = 30 x 24 x 40 / 3 = 9600 m 3 answer : d

a ) 9600 m 3', ' , b ) 30 minutes , c ) 14 , d ) $ 488.9 , e ) 3 / 5

a

volume_rectangular_prism(30, 24, divide(multiply(rectangle_area(30, 24), const_2), rectangle_perimeter(30, 24)))

rectangle_area(n0,n1)|rectangle_perimeter(n0,n1)|multiply(#0,const_2)|divide(#2,#1)|volume_rectangular_prism(n0,n1,#3)

geometry

if $ 120 invested at a certain rate of simple interest amounts to $ 180 at the end of 3 years , how much will $ 150 amount to at the same rate of interest in 6 years ?

"120 amounts to 180 in 3 years . i . e ( principal + interest ) on 120 in 3 years = 180 120 + 120 * ( r / 100 ) * ( 3 ) = 140 = > r = 50 / 3 150 in 6 years = principal + interest = 300 answer is e ."

a ) 5526 , b ) $ 300 , c ) 1,400 , d ) 3 , e ) 1 / 16

b

add(150, divide(multiply(multiply(150, 6), divide(divide(multiply(subtract(180, 120), 120), 120), 3)), 120))

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

gain

find the smallest number which should be multiplied with 520 to make it a perfect square

"explanation : 520 = 26 * 20 = 2 * 13 * 22 * 5 = 23 * 13 * 5 required smallest number = 2 * 13 * 5 = 130 130 is the smallest number which should be multiplied with 520 to make it a perfect square . answer : e"

a ) 25 % , b ) rs . 15,000 , c ) 130 , d ) 131.95 , e ) 100

c

divide(divide(divide(divide(divide(520, const_3), const_3), const_4), const_4), const_4)

divide(n0,const_3)|divide(#0,const_3)|divide(#1,const_4)|divide(#2,const_4)|divide(#3,const_4)|

geometry

the price of a certain product increased by the same percent from 1960 to 1970 as from 1970 to 1980 . if its price of $ 1.20 in 1970 was 150 percent of its price in 1960 , what was its price in 1980 ?

the price in 1970 was 150 percent of its price in 1960 , means that the percent increase was 50 % from 1960 to 1970 ( and from 1970 to 1980 ) . therefore the price in 1980 = $ 1.2 * 1.5 = $ 1.8 . answer : a .

a ) 3 , b ) 544 , c ) 96 , d ) 2 , e ) $ 1.80

e

multiply(divide(150, const_100), 1.2)

divide(n6,const_100)|multiply(n4,#0)

general

a is two years older than b who is twice as old as c . if the total of the ages of a , b and c be 37 , the how old is b ?

explanation : let c ' s age be x years . then , b ' s age = 2 x years . a ' s age = ( 2 x + 2 ) years . ( 2 x + 2 ) + 2 x + x = 37 ⇒ 5 x = 35 ⇒ x = 7 . hence , b ' s age = 2 x = 14 years . answer : d

a ) 14 , b ) $ 32500 , c ) 1320 , d ) $ 23400 , e ) 2 / 42

a

divide(multiply(subtract(37, const_2), const_2), add(const_4, const_1))

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

general

if c and t are positive integers , ct + c + t can not be

let ct + t + c = x add 1 on both sides : ct + t + c + 1 = x + 1 t ( c + 1 ) + c + 1 = x + 1 ( c + 1 ) ( t + 1 ) = x + 1 minimum value of ( c + 1 ) = 2 minimum value of ( t + 1 ) = 2 hence x + 1 can not be prime substitute x from the given options : 6 + 1 = 7 - - > prime - - > ct + t + s can not be 6 answer : b

a ) 720 , b ) 24', ' , c ) 27 / 128 , d ) 7 , e ) 6

e

multiply(const_2, const_3)

multiply(const_2,const_3)

general

the area of a circular field is 17.56 hectares . find the cost of fencing it at the rate of rs . 2 per metre approximately

"explanation : area = ( 17.56 x 10000 ) m 2 = 175600 m 2 . π r 2 = 175600 ⇔ ( r ) 2 = ( 175600 x ( 7 / 22 ) ) ⇔ r = 236.37 m . circumference = 2 π r = ( 2 x ( 22 / 7 ) x 236.37 ) m = 1485.78 m . cost of fencing = rs . ( 1485.78 x 2 ) = rs . 2972 . answer : option a"

a ) 12.0 , b ) 4 , c ) 16', ' , d ) 2972 , e ) $ 5

d

multiply(circumface(multiply(sqrt(divide(17.56, const_pi)), const_100)), 2)

divide(n0,const_pi)|sqrt(#0)|multiply(#1,const_100)|circumface(#2)|multiply(#3,n1)|

geometry

a man swims downstream 30 km and upstream 12 km taking 3 hours each time , what is the speed of the man in still water ?

"30 - - - 3 ds = 10 ? - - - - 1 12 - - - - 3 us = 4 ? - - - - 1 m = ? m = ( 10 + 4 ) / 2 = 7 answer : b"

a ) 5 5 / 8 km , b ) 54 , c ) 2140 , d ) 6000 , e ) 7

e

divide(add(divide(12, 3), divide(30, 3)), const_2)

divide(n1,n2)|divide(n0,n2)|add(#0,#1)|divide(#2,const_2)|

physics

a trained covered x km at 40 kmph and another 2 x km at 20 kmph . find the average speed of the train in covering the entire 7 x km .

"total time taken = x / 40 + 2 x / 20 hours = 5 x / 40 = x / 8 hours average speed = 7 x / ( x / 8 ) = 56 kmph answer : a"

a ) 26 , b ) 27.38 % , c ) 56 , d ) 45 , e ) 8

c

divide(multiply(40, 7), add(divide(40, 40), divide(multiply(2, 40), 20)))

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

general

if the average of 6 digits is 16 and the average of 4 of them is 10 , calculate the average of the remaining 2 numbers ?

"explanation : total of the 6 digits - 6 * 16 = 96 total of the 4 digits - 4 * 10 = 40 total of the remaining 2 digits - 96 - 40 = 56 average of the remaining 2 numbers = 56 / 2 = 28 answer : c"

a ) 125 % , b ) 82 , c ) 85 , d ) 28 , e ) 77

d

divide(subtract(multiply(6, 16), multiply(4, 10)), 2)

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

general

a company pays 20.5 % dividend to its investors . if an investor buys rs . 50 shares and gets 25 % on investment , at what price did the investor buy the shares ?

"explanation : dividend on 1 share = ( 20.5 * 50 ) / 100 = rs . 10.25 rs . 25 is income on an investment of rs . 100 rs . 10.25 is income on an investment of rs . ( 10.25 * 100 ) / 25 = rs . 41 answer : e"

a ) s . 1665 , b ) 40 , c ) 1000 , d ) 41 , e ) 12 days .

d

divide(multiply(divide(multiply(20.5, 50), const_100), const_100), 25)

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

gain

how many diagonals does a 59 - sided convex polygon have ?

a 59 - sided convex polygon has 59 vertices . if we examine a single vertex , we can see that we can connect it with 56 other vertices to create a diagonal . note that we ca n ' t connect the vertex to itself and we ca n ' t connect it to its adjacent vertices , since this would not create a diagonal . if each of the 59 vertices can be connected with 56 vertices to create a diagonal then the total number of diagonals would be ( 59 ) ( 56 ) = 3304 however , we must recognize that we have counted every diagonal twice . to account for counting each diagonal twice , we must divide 3304 by 2 to get 1652 . the answer is b .

a ) 60 kg , b ) 47.4 % , c ) 1652 , d ) 5 , e ) 36 %

c

divide(factorial(59), multiply(factorial(subtract(59, const_2)), factorial(const_2)))

factorial(n0)|factorial(const_2)|subtract(n0,const_2)|factorial(#2)|multiply(#3,#1)|divide(#0,#4)

general