Datasets:

math_qa

like 68

if a person buy radio worth rs 2468 and pay 7 % sales . how much price of radio should reduce to pay only rs 2468 .

( x + . 07 x ) = 2468 1.07 x = 2468 x = 2306 reduce = 2468 - 2306 = 162 answer : a

a ) 162 , b ) 158 , c ) 159 , d ) 130 , e ) 160

a

subtract(2468, divide(2468, add(const_1, divide(7, const_100))))

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

gain

3 buffaloes eat as much as 4 cows or 2 oxen . at a farm , there are 15 buffaloes , 8 oxen , and 24 cows . the fodder for these cattle is available for 48 days . if 60 more cows and 30 more buffaloes are brought in , how many days will the fodder last ?

2 oxen = 3 buffaloes = 4 cows also : 15 buffaloes + 8 oxen + 24 cows = 10 oxen + 8 oxen + 12 oxen = 30 oxen there is enough fodder for 1 oxen for 30 * 48 days . 60 cows + 30 buffaloes = 30 oxen + 20 oxen = 50 oxen the new total is equal to 80 oxen instead of 30 oxen . 30 * 48 / 80 oxen = 18 days the answer is b .

a ) 15 days , b ) 18 days , c ) 21 days , d ) 24 days , e ) 27 days

b

divide(multiply(48, add(add(15, multiply(divide(24, 4), 3)), multiply(divide(8, 2), 3))), add(add(add(15, 30), multiply(divide(add(24, 60), 4), 3)), multiply(divide(8, 2), 3)))

add(n3,n8)|add(n5,n7)|divide(n5,n1)|divide(n4,n2)|divide(#1,n1)|multiply(n0,#2)|multiply(n0,#3)|add(n3,#5)|multiply(n0,#4)|add(#7,#6)|add(#0,#8)|add(#10,#6)|multiply(n6,#9)|divide(#12,#11)

general

the difference between a number and its three - fifth is 50 . what is the number ?

"let the number be x . then , x - 3 / 5 x = 50 = > 2 / 5 x = 50 x = ( 50 * 5 ) / 2 = 125 . answer : e"

a ) 185 , b ) 190 , c ) 160 , d ) 120 , e ) 125

e

power(add(50, const_4), const_4)

add(n0,const_4)|power(#0,const_4)|

general

two thirds of the roads from a to b are at least 5 miles long , and 3 / 4 of the roads from b to c are at least 5 miles long . if you randomly pick a road from a to b and then randomly pick a road from b to c , what is the probability that at least one of the roads you pick is at least 5 miles long ?

so please : find the probability of the event thatnoneof the roads you pick will be at least 5 miles long and subtract from 1 to get the probability thatat least oneof the roads you pick will be at least 5 miles long : p = 1 - 1 / 3 * 1 / 4 = 11 / 12 . answer : e .

a ) 1 / 6 , b ) 1 / 4 , c ) 2 / 3 , d ) 3 / 4 , e ) 11 / 12

e

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

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

physics

if x = 6 and y = − 2 , what is the value of ( x − 2 y ) ^ y ?

"( x − 2 y ) ^ y = ( 6 − 2 ( − 2 ) ) ^ − 2 = 1 / ( 6 + 4 ) ^ 2 = 1 / 100 = 0.01 answer : b"

a ) − 100 , b ) 0.01 , c ) 0.25 , d ) 4 , e ) 8

b

divide(divide(2, 2), const_1000)

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

general

a rectangular swimming pool is 8 feet by 20 feet . a deck that has uniform width surrounds the pool . the total area of the pool and deck is 540 square feet . what is the width of the deck ?

"let the width = w total area of the pool and deck = ( 2 w + 8 ) ( 2 w + 20 ) we can test the answer choices along with unit digit method a ) 2 feet . . . . . . . . . . . 12 * 24 has unit digit 8 . . . . . . . . . . eliminate b ) 2.5 feet . . . . . . . . . 13 * 25 has unit digit 5 . . . . . . . . . . eliminate c ) 3 feet . . . . . . . . . . . . 14 * 26 has unit digit 4 . . . . . . . . . . . hold d ) 4 feet . . . . . . . . . . . . 16 * 28 has unit digit 8 . . . . . . . . . . . eliminate e ) 5 feet . . . . . . . . . . . . 18 * 30 has unit digit 0 . . . . . . . . . . . eliminate answer : e w = 5"

a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5

e

divide(subtract(sqrt(add(power(subtract(20, const_1), const_2), subtract(540, rectangle_area(8, 20)))), subtract(20, const_1)), const_2)

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

geometry

a 270 m long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds . what is the length of the other train ?

"speed of first train = ( 120 * 5 / 18 ) m / sec = 100 / 3 m / sec speed of other train = ( 80 * 5 / 18 ) m / sec = 200 / 9 m / sec time to cross each other = 9 sec let x be the length of 2 nd train therefore 9 = ( 270 + x ) / ( 100 / 3 + 200 / 9 ) 500 = 270 + x 230 m = x answer : a"

a ) 230 m , b ) 240 m , c ) 260 m , d ) 320 m , e ) 350 m

a

subtract(multiply(divide(add(120, 80), const_3_6), 9), 270)

add(n1,n2)|divide(#0,const_3_6)|multiply(n3,#1)|subtract(#2,n0)|

physics

a sum of money is to be divided among ann , bob and chloe . first , ann receives $ 4 plus one - half of what remains . next , bob receives $ 4 plus one - third of what remains . finally , chloe receives the remaining $ 32 . how much money t did bob receive ?

"notice that we need not consider ann ' s portion in the solution . we can just let k = the money remaining after ann has received her portion and go from there . our equation will use the fact that , once we remove bob ' s portion , we have $ 32 for chloe . so , we getk - bob ' s $ = 32 bob received 4 dollars plus one - third of what remained once bob receives $ 4 , the amount remaining is k - 4 dollars . so , bob gets a 1 / 3 of that as well . 1 / 3 of k - 4 is ( k - 4 ) / 3 so altogether , bob receives 4 + ( k - 4 ) / 3 so , our equation becomes : k - [ 4 + ( k - 4 ) / 3 ] = 32 simplify to get : k - 4 - ( k - 4 ) / 3 = 32 multiply both sides by 3 to get : 3 k - 12 - k + 4 = 96 simplify : 2 k - 8 = 96 solve : k = 52 plug this k - value intok - bob ' s $ = 32 to get : 52 - bob ' s $ = 32 so , bob ' s $ t = 20 answer : b"

a ) 20 , b ) 22 , c ) 24 , d ) 26 , e ) 52

b

divide(multiply(32, const_2), const_3)

multiply(n2,const_2)|divide(#0,const_3)|

general

maxwell leaves his home and walks toward brad ' s house at the same time that brad leaves his home and runs toward maxwell ' s house . if the distance between their homes is 40 kilometers , maxwell ' s walking speed is 3 km / h , and brad ' s running speed is 5 km / h . what is the distance traveled by maxwell when they meet in the middle ?

"consider max starts from point a and brad starts from point b and move towards each other . assume they shall meet at point o after time ' t ' . the question asks us to find oa . from the question stem we can make out : - distance oa = 50 km - distance ob = > 3 xt = 40 - 5 xt ( i . e distance = speed x time ) = > 8 t = 40 hence t = 5 oa = 3 x 5 = 15 km answer : d"

a ) 16 , b ) 17 , c ) 18 , d ) 15 , e ) 14

d

multiply(3, divide(40, add(3, 5)))

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

physics

in a urban village of india named ` ` owlna ' ' , 70 % people have refrigerator , 75 % people have television , 65 % people got computers and 95 % got air - conditionor . how many people ( minimum ) got all these luxury .

a 10 % 100 - [ ( 100 - 65 ) + ( 100 - 95 ) + ( 100 - 75 ) + ( 100 - 70 ) ] = 100 - ( 30 + 25 + 35 + 5 ) = 100 - 95

a ) 5 % , b ) 7 % , c ) 3 % , d ) 9 % , e ) 15 %

a

subtract(const_100, add(add(add(subtract(const_100, 70), subtract(const_100, 75)), subtract(const_100, 65)), subtract(const_100, 95)))

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

general

in a group of 12 engineers , 4 engineers have a phd degree ; the others have only an ms or bs degree . a team of 4 engineers is to be chosen which must have at least 1 engineer with a phd , how many different teams can be chosen ?

"the problem asks for a combination , since order does n ' t matter . now , selecting r items from a set of n gives the combination formula : ncr = n ! / r ! ( n - r ) ! n = 12 r = 4 so , total teams is 12 c 4 = 12 ! / ( 4 ! ( 12 - 4 ) ! ) = 495 , and n = 12 - 4 = 8 r = 4 for teams without a phd is 8 c 4 = 8 ! / ( 4 ! ( 8 - 4 ) ! ) = 70 , so , teams with at least 1 phd = 495 - 70 = 425 answer : d"

a ) 495 , b ) 70 , c ) 245 , d ) 425 , e ) 555

d

add(1, multiply(multiply(subtract(12, 4), 4), 4))

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

other

what will be the ci on a sum of rs . 25000 after 3 years at the rate of 12 per year ?

amount = rs . 25000 x 1 + 12 3 100 = rs . 25000 x 28 x 28 x 28 25 25 25 = rs . 35123.20 c . i . = rs . ( 35123.20 - 25000 ) = rs . 10123.20 c

a ) rs . 10100 , b ) rs . 10110 , c ) rs . 10123.20 , d ) rs . 10135.50 , e ) rs . 10167.20

c

subtract(multiply(power(add(divide(12, const_100), const_1), 3), 25000), 25000)

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

gain

if a person walks at 15 km / hr instead of 10 km / hr , he would have walked 10 km more . the actual distance traveled by him is ?

"let the actual distance traveled be x km . then , x / 10 = ( x + 10 ) / 15 5 x - 150 = > x = 30 km . answer : b"

a ) 50 km , b ) 30 km , c ) 18 km , d ) 16 km , e ) 97 km

b

multiply(10, divide(10, subtract(15, 10)))

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

general

a rectangular grass field is 85 m * 55 m , it has a path of 2.5 m wide all round it on the outside . find the area of the path and the cost of constructing it at rs . 2 per sq m ?

"area = ( l + b + 2 d ) 2 d = ( 85 + 55 + 2.5 * 2 ) 2 * 2.5 = > 725 725 * 2 = rs . 1450 answer : a"

a ) 1450 , b ) 1971 , c ) 9676 , d ) 1679 , e ) 2691

a

multiply(subtract(rectangle_area(add(85, multiply(2.5, 2)), add(55, multiply(2.5, 2))), rectangle_area(85, 55)), 2)

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

geometry

the speed at which a man can row a boat in still water is 18 kmph . if he rows downstream , where the speed of current is 3 kmph , what time will he take to cover 60 metres ?

"speed of the boat downstream = 18 + 3 = 21 kmph = 21 * 5 / 18 = 35 / 9 m / s hence time taken to cover 60 m = 60 * 9 / 35 = 15.4 seconds . answer : b"

a ) 16 seconds , b ) 15.4 seconds , c ) 26 seconds , d ) 12 seconds , e ) 18 seconds

b

divide(60, multiply(add(18, 3), const_0_2778))

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

physics

two trains are moving at 50 kmph and 70 kmph in opposite directions . their lengths are 170 m and 100 m respectively . the time they will take to pass each other completely is ?

"70 + 50 = 120 * 5 / 18 = 100 / 3 mps d = 170 + 100 = 270 m t = 270 * 3 / 100 = 8 sec answer : b"

a ) 3 sec , b ) 8 sec , c ) 9 sec , d ) 5 sec , e ) 7 sec

b

divide(add(170, 100), multiply(add(50, 70), const_0_2778))

add(n2,n3)|add(n0,n1)|multiply(#1,const_0_2778)|divide(#0,#2)|

physics

having scored 96 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 + 96 = 19 ( n + 4 ) = > n = 20 so , average score after 19 th innings = x + 4 = 24 . answer : e"

a ) 28 , b ) 27 , c ) 26 , d ) 22 , e ) 24

e

add(subtract(96, multiply(19, 4)), 4)

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

general

what is the greatest possible length which can be used to measure exactly the lengths 8 m , 4 m 20 cm and 12 m 20 cm ?

"required length = hcf of 800 cm , 420 cm , 1220 cm = 20 cm answer : option d"

a ) 10 cm , b ) 30 cm , c ) 25 cm , d ) 20 cm , e ) 35 cm

d

multiply(12, const_4)

multiply(n3,const_4)|

physics

what least number should be added to 1019 , so that the sum is completely divisible by 25 ?

1019 ã · 25 = 40 with remainder = 19 19 + 6 = 25 . hence 6 should be added to 1019 so that the sum will be divisible by 25 answer : option e

a ) 4 , b ) 3 , c ) 2 , d ) 0 , e ) 6

e

subtract(25, reminder(1019, 25))

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

general

in a school having roll strength 286 , the ratio of boys and girls is 8 : 5 . if 22 more girls get admitted into the school , the ratio of boys and girls becomes

solution : boys : girls = 8 : 5 ; ( let the boys = 8 x ; girl = 5 x ) total strength = 286 ; 8 x + 5 x = 286 ; 13 x = 286 ; or , x = 286 / 13 = 22 ; boys = 176 and girls = 110 ; 22 more girls get admitted then number of girls become , ( 5 x + 22 ) = 110 + 22 = 132 ; now , new ratio of boys and girls = 176 : 132 = 4 : 3 . answer : option d

['a ) 12 : 7', 'b ) 10 : 7', 'c ) 8 : 7', 'd ) 4 : 3', 'e ) none']

d

divide(multiply(divide(286, add(8, 5)), 8), add(multiply(divide(286, add(8, 5)), 5), 22))

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

other

if x = 1 + √ 2 , then what is the value of x 4 - 4 x 3 + 4 x 2 + 5 ?

"answer x = 1 + √ 2 ∴ x 4 - 4 x 3 + 4 x 2 + 5 = x 2 ( x 2 - 4 x + 4 ) + 5 = x 2 ( x - 2 ) 2 + 5 = ( 1 + √ 2 ) 2 ( 1 + √ 2 - 2 ) 2 + 5 = ( √ 2 + 1 ) 2 ( √ 2 - 1 ) 2 + 5 = [ ( √ 2 ) 2 - ( 1 ) 2 ] 2 + 5 = ( 2 - 1 ) 2 = 1 + 5 = 6 correct option : c"

a ) - 1 , b ) 0 , c ) 6 , d ) 2 , e ) 3

c

add(multiply(power(add(1, sqrt(2)), 2), power(subtract(add(1, sqrt(2)), 2), 2)), 4)

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

general

the ratio of sum of squares of first n natural numbers to square of sum of first n natural numbers is 17 : 325 . the value of n is

sum of 1 st n natural no . s is n ( n + 1 ) / 2 , sum of sqaures of 1 st n natural no . s is n ( n + 1 ) ( 2 n + 1 ) / 6 , the ratio of sum of squares of first n natural numbers to square of sum of first n natural numbers = n ( n + 1 ) ( 2 n + 1 ) * 2 * 2 / ( n * n * ( n + 1 ) * ( n + 1 ) ) = 2 ( 2 n + 1 ) / ( 3 n ( n + 1 ) ) now by hit and trial for the given options , if we put 25 then this will give 17 / 325 therefore the answer is 25 answer : b

['a ) 15', 'b ) 25', 'c ) 35', 'd ) 30', 'e ) none of these']

b

divide(add(325, const_100), 17)

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

other

a man is walking at a speed of 10 km per hour . after every kilometre , he takes rest for 5 minutes . how much time will be take to cover a distance of 5 kilometres ?

rest time = number of rest × time for each rest = 4 × 5 = 20 minutes total time to cover 5 km = ( 5 ⁄ 10 × 60 ) minutes + 20 minutes = 50 minutes answer b

a ) 48 min . , b ) 50 min . , c ) 45 min . , d ) 55 min . , e ) none of these

b

add(multiply(divide(5, 10), speed(const_60, const_1)), multiply(const_4, 5))

divide(n1,n0)|multiply(n1,const_4)|speed(const_60,const_1)|multiply(#0,#2)|add(#3,#1)

physics

an error 4 % in excess is made while measuring the side of a square . the percentage of error in the calculated area of the square is

"100 cm is read as 102 cm . a 1 = ( 100 x 100 ) cm 2 and a 2 ( 102 x 102 ) cm 2 . ( a 2 - a 1 ) = [ ( 104 ) 2 - ( 100 ) 2 ] = ( 104 + 100 ) x ( 104 - 100 ) = 816 cm 2 . percentage error = 8.16 d"

a ) 8.04 % , b ) 8.14 % , c ) 8.23 % , d ) 8.16 % , e ) 8.5 %

d

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

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

gain

harkamal purchased 8 kg of grapes at the rate of 70 per kg and 9 kg of mangoes at the rate of 55 per kg . how much amount did he pay to the shopkeeper ?

"cost of 8 kg grapes = 70 × 8 = 560 . cost of 9 kg of mangoes = 55 × 9 = 490 . total cost he has to pay = 560 + 490 = 1055 . b )"

a ) 1000 , b ) 1055 , c ) 1065 , d ) 1075 , e ) 1080

b

add(multiply(8, 70), multiply(9, 55))

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

gain

if x and y are integers such that ( x + 1 ) ^ 2 is less than or equal to 9 and ( y - 1 ) ^ 2 is less than 64 , what is the sum of the maximum possible value of xy and the minimum possible value of xy ?

"( x + 1 ) ^ 2 < = 9 x < = 2 x > = - 4 ( y - 1 ) ^ 2 < 64 y < 9 y > - 7 max possible value of xy is - 4 × - 6 = 24 minimum possible value of xy is - 4 × 8 = - 32 - 32 + 24 = - 8 answer : b"

a ) - 16 , b ) - 8 , c ) 0 , d ) 14 , e ) 16

b

add(sqrt(9), sqrt(64))

sqrt(n2)|sqrt(n5)|add(#0,#1)|

general

the difference between a two - digit number and the number obtained by interchanging the two digits is 63 . which is the smaller of the two numbers ?

"explanation : let the ten ' s digit be x and units digit by y . then , ( 10 x + y ) - ( 10 y + x ) = 63 9 ( x - y ) = 63 x - y = 7 thus , none of the numbers can be determined . answer is d"

a ) 29 , b ) 70 , c ) 92 , d ) can not be determined , e ) none of these

d

divide(63, subtract(const_10, const_1))

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

general

a man started driving at a constant speed , from the site of a blast , the moment he heard the blast . he heard a second blast after a time of 30 mins and 12 seconds . if the second blast occurred exactly 30 mins after the first , how many meters was he from the site when he heard the second blast ? ( speed of sound = 330 m / s )

"the distance the sound traveled to the man is 12 * 330 = 3960 meters the answer is e ."

a ) 3120 , b ) 3330 , c ) 3540 , d ) 3750 , e ) 3960

e

multiply(330, 12)

multiply(n1,n3)|

physics

how many positive integers less than 5,000 are evenly divisible by neither 13 nor 21 ?

"integers less than 5000 divisible by 13 5000 / 13 = 333 . something , so 333 integers less than 5000 divisible by 21 5000 / 21 = 238 . # # , so 238 we have double counted some , so take lcm of 13 and 21 = 105 and divide by 5000 , we get 47 . so all numbers divisible by 13 and 21 = 333 + 238 - 47 = 524 now subtract that from 4999 . 4999 - 524 = 4349 answer e ."

a ) 4,514 , b ) 4,475 , c ) 4,521 , d ) 4,428 , e ) 4,349

e

divide(factorial(subtract(add(const_4, 13), const_1)), multiply(factorial(13), factorial(subtract(const_4, const_1))))

add(n1,const_4)|factorial(n1)|subtract(const_4,const_1)|factorial(#2)|subtract(#0,const_1)|factorial(#4)|multiply(#1,#3)|divide(#5,#6)|

general

a train leaves delhi at 9 a . m . at a speed of 30 kmph . another train leaves at 2 p . m . at a speed of 40 kmph on the same day and in the same direction . how far from delhi , will the two trains meet ?

explanation : d = 30 * 5 = 150 rs = 40 – 30 = 10 t = 150 / 10 = 15 d = 40 * 15 = 600 km answer : option c

a ) 900 km , b ) 800 km , c ) 600 km , d ) 690 km , e ) 609 km

c

multiply(divide(multiply(30, add(const_3, const_2)), subtract(40, 30)), 40)

add(const_2,const_3)|subtract(n3,n1)|multiply(n1,#0)|divide(#2,#1)|multiply(n3,#3)

physics

what is difference between biggest and smallest fraction among 2 / 5 , 3 / 4 , 4 / 5 and 5 / 6

"explanation : 2 / 5 = . 4 , 3 / 4 = . 75 , 4 / 5 = . 8 and 5 / 6 = . 833 so biggest is 5 / 6 and smallest is 2 / 5 their difference is 5 / 6 - 2 / 5 = 13 / 30 option e"

a ) 2 / 5 , b ) 3 / 5 , c ) 1 / 6 , d ) 1 / 7 , e ) none of these

e

subtract(divide(4, 5), divide(2, 5))

divide(n3,n5)|divide(n0,n1)|subtract(#0,#1)|

general

a 5 % stock yields 10 % . the market value of the stock is :

"explanation : for an income of rs . 10 , investment = rs . 100 . for an income of rs 5 , investment = rs . 100 / 10 x 5 = rs 50 market value of rs . 100 stock = rs . 50 answer is c"

a ) rs . 56 , b ) rs . 55 , c ) rs . 50 , d ) rs . 90 , e ) rs . 75

c

multiply(divide(const_100, 10), 5)

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

gain

a train 120 m long passes a man , running at 6 kmph in the direction opposite to that of the train , in 6 seconds . the speed of the train is

"speed of train relative to man : 120 / 6 * 18 / 5 km / hr = 72 km / hr let speed of train = x therefore x + 6 = 72 x = 72 - 6 x = 66 km / hr answer : c"

a ) 54 kmph , b ) 60 kmph , c ) 66 kmph , d ) 72 kmph , e ) 82 kmph

c

divide(divide(subtract(120, multiply(multiply(6, const_0_2778), 6)), 6), const_0_2778)

multiply(n1,const_0_2778)|multiply(n1,#0)|subtract(n0,#1)|divide(#2,n1)|divide(#3,const_0_2778)|

physics

every day daniel drives 64 miles back from work . on sunday , daniel drove all the way back from work at a constant speed of x miles per hour . on monday , daniel drove the first 32 miles back from work at ( 2 x ) miles per hour , and the rest of the way at ( x / 2 ) miles per hour . the time it took daniel to drive back from work on monday is longer than the time it took him to drive back from work on sunday by what percent ?

"let ' s test x = 4 . . . . on sunday , daniel drove 64 miles at 4 miles / hour . d = ( r ) ( t ) 64 = ( 4 ) ( t ) 64 / 4 = 16 = t it takes 16 hours to drive home on monday , daniel drove the first 32 miles at ( 2 ) ( 4 ) = 8 miles / hour and the rest of the way ( 32 miles ) at 4 / 2 = 2 miles / hour d = ( r ) ( t ) 32 = ( 8 ) ( t ) 32 / 8 = 4 = t it takes 4 hours for the first part d = ( r ) ( t ) 32 = ( 2 ) ( t ) 32 / 2 = 16 = t it takes 16 hours for the second part total time to drive home on monday = 4 + 16 = 20 hours we ' re asked by what percent 20 hours is greater than 16 hours . 20 / 16 = 1.25 , so it is 25 % greater . b"

a ) 15 % , b ) 25 % , c ) 35 % , d ) 45 % , e ) 55 %

b

multiply(divide(subtract(add(divide(32, 2), multiply(subtract(64, 32), 2)), 64), 64), const_100)

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

physics

the diagonal of a rhombus are 25 m and 50 m . its area is :

"area of the rhombus = 1 / 2 d 1 d 2 = ( 1 / 2 ã — 25 ã — 50 ) cm ( power ) 2 = 25 ã — 25 = 625 cm ( power ) 2 answer is d ."

a ) 900 , b ) 800 , c ) 700 , d ) 625 , e ) 650

d

rhombus_area(25, 50)

rhombus_area(n0,n1)|

geometry

what is the angle between the hands of a clock when time is 10 : 20 ?

"angle between two hands = 20 h - 11 / 2 m = 20 * 10 - 20 * 11 / 2 = 200 - 110 = 90 deg answer : a"

a ) 90 deg , b ) 75 deg , c ) 45 deg , d ) 15 deg , e ) 30 deg

a

subtract(multiply(20, multiply(const_3, const_2)), 10)

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

geometry

the diameter of a cylindrical tin is 14 cm and height is 5 cm . find the volume of the cylinder ?

"r = 7 h = 5 π * 7 * 7 * 5 = 45 π cc answer : d"

a ) 230 , b ) 141 , c ) 66 , d ) 245 , e ) 21

d

divide(volume_cylinder(divide(14, const_2), 5), const_pi)

divide(n0,const_2)|volume_cylinder(#0,n1)|divide(#1,const_pi)|

geometry

if 40 men take 15 days to to complete a job , in how many days can 25 men finish that work ?

ans . 24 days

a ) 24 , b ) 25 , c ) 26 , d ) 27 , e ) 28

a

divide(multiply(40, 15), 25)

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

physics

it takes printer a 4 more minutes more than printer b to print 40 pages . working together , the two printers can print 50 pages in 6 minutes . how long will it take printer a to print 160 pages ?

if it takes 4 more minutes for a to print 40 pages than it takes b , it takes 5 more minutes for a to print 50 pages than it takes b . thus if b is the number of minutes than b takes to print 50 pages , we can write : 1 / b + 1 / ( b + 5 ) = 1 / 6 ( since in 1 minute , they print 1 / 6 th of the 50 page job ) 6 ( 2 b + 5 ) = b ( b + 5 ) b ^ 2 - 7 b - 30 = 0 ( b - 10 ) ( b + 3 ) = 0 b = 10 thus it takes a 15 minutes to print 50 pages and 15 * 160 / 50 = 48 minutes to print 160 pages ( answer c )

a ) 12 , b ) 18 , c ) 48 , d ) 20 , e ) 30

c

divide(multiply(160, add(multiply(4, const_2), 4)), 40)

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

physics

find the cost of fencing around a circular field of diameter 22 m at the rate of rs . 3 a meter ?

"2 * 22 / 7 * 11 = 69.1 69.1 * 3 = rs . 207.3 answer : d"

a ) 288 , b ) 132 , c ) 772 , d ) 207.3 , e ) 261

d

multiply(circumface(divide(22, const_2)), 3)

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

physics

there are 6 people in the elevator . their average weight is 156 lbs . another person enters the elevator , and increases the average weight to 151 lbs . what is the weight of the 7 th person .

solution average of 7 people after the last one enters = 151 . â ˆ ´ required weight = ( 7 x 151 ) - ( 6 x 156 ) = 1057 - 936 = 121 . answer a

a ) 121 , b ) 168 , c ) 189 , d ) 190 , e ) 200

a

subtract(multiply(151, 7), multiply(6, 156))

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

general

each child has 2 pencils and 13 skittles . if there are 15 children , how many pencils are there in total ?

2 * 15 = 30 . answer is a .

a ) 30 , b ) 12 , c ) 18 , d ) 22 , e ) 08

a

multiply(2, 15)

multiply(n0,n2)

general

if a certain number x is divided by 62 , the reminder is 7 . what is the reminder when x + 11 is divided by 31 ?

x can be written as 62 k + 7 or x = 7 , 69,131 , etc . x + 11 = 62 k + 7 + 11 = 62 k + 18 or x + 11 = 18 , 80,142 etc . when divided by 31 , we will get the remainder 18 . c

a ) 3 , b ) 5 , c ) 18 , d ) 6 , e ) 8

c

add(7, 11)

add(n1,n2)

general

if 7 ^ w is a factor of the product of the integers from 1 to 100 , inclusive , what is the largest value of w ?

so the question just means that we have to find all the multiples w of 7 between 1 to 100 so there are 14 multiples of 7 ( 7 - 98 ) but 49 and 98 contain two 7 ' s as factors so we have to add 14 + 2 = 16 e

a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16

e

floor(add(divide(100, 7), divide(100, power(7, const_2))))

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

general

a women purchased 3 towels @ rs . 100 each , 5 towels @ rs . 150 each and two towels at a certain rate which is now slipped off from his memory . but she remembers that the average price of the towels was rs . 145 . find the unknown rate of two towels ?

"10 * 145 = 1450 3 * 100 + 5 * 150 = 1050 1450 – 1050 = 400 a"

a ) a ) 400 , b ) b ) 450 , c ) c ) 500 , d ) d ) 550 , e ) e ) 600

a

subtract(subtract(multiply(add(add(3, 5), const_2), 145), multiply(5, 150)), multiply(3, 100))

add(n0,n2)|multiply(n2,n3)|multiply(n0,n1)|add(#0,const_2)|multiply(n4,#3)|subtract(#4,#1)|subtract(#5,#2)|

general

if 2 pipes take an hour to fill a tank , then how long should 8 pipes take to fill the same tank ?

if 2 pipes take an hour to fill a tank , then 8 pipes will take 2 * 60 / 8 = 15 mins to dig a ditch of the same type . answer : c

a ) 20 min , b ) 10 min , c ) 15 min , d ) 25 min , e ) 30 min

c

divide(multiply(2, const_60), 8)

multiply(n0,const_60)|divide(#0,n1)

physics

there are 6 chess amateurs playing in villa ' s chess club tournament . if each chess amateur plays with exactly 4 other amateurs , what is the total number of chess games possible to be played in the tournament ?

method 2 : each person is one participant of 4 games . so there are in all 4 * 6 = 24 instances of one participant games . but each game has 2 participants so total number of games = 24 / 2 = 12 e

a ) 10 , b ) 20 , c ) 40 , d ) 60 , e ) 12

e

divide(multiply(6, 4), const_2)

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

general

a jar full of whisky contains 30 % alcohol . a part of this whisky is replaced by another containg 19 % alcohol and now the percentage of alcohol was found to be 26 % . what quantity of whisky is replaced ?

"let us assume the total original amount of whiskey = 10 ml - - - > 4 ml alcohol and 6 ml non - alcohol . let x ml be the amount removed - - - > total alcohol left = 4 - 0.4 x new quantity of whiskey added = x ml out of which 0.19 is the alcohol . thus , the final quantity of alcohol = 4 - 0.4 x + 0.19 x - - - - > ( 4 - 0.21 x ) / 10 = 0.26 - - - > x = 20 / 3 ml . per the question , you need to find the x ml removed as a ratio of the initial volume - - - > ( 20 / 3 ) / 10 = 1 / 3 . hence , a is the correct answer ."

a ) 1 / 3 , b ) 2 / 3 , c ) 2 / 5 , d ) 3 / 5 , e ) 4 / 5

a

divide(subtract(30, 26), subtract(30, 19))

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

gain

worker a takes 4 hours to do a job . worker b takes 10 hours to do the same job . how long it take both a & b , working together but independently , to do the same job ?

one day work of a = 1 / 4 one day work of b = 1 / 10 so one day work of a and b together = 1 / 4 + 1 / 10 = 14 / 40 so total days required = 40 / 14 answer : c

a ) 20 / 9 , b ) 40 / 9 , c ) 40 / 14 , d ) 60 / 9 , e ) 80 / 9

c

divide(const_1, add(divide(const_1, 4), divide(const_1, 10)))

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

physics

express a speed of 30 kmph in meters per second ?

"30 * 5 / 18 = 8 mps answer : b"

a ) 10 mps , b ) 8 mps , c ) 9 mps , d ) 7 mps , e ) 12 mps

b

multiply(const_0_2778, 30)

multiply(n0,const_0_2778)|

physics

in a maths test , students were asked to find 5 / 16 of a certain number . one of the students by mistake found 5 / 6 th of that number and his answer was 100 more than the correct answer . find the number .

"explanation : let the number be x . 5 * x / 6 = 5 * x / 16 + 100 25 * x / 48 = 100 x = 192 answer a"

a ) 192 , b ) 280 , c ) 384 , d ) 400 , e ) 500

a

divide(multiply(multiply(100, 16), 6), subtract(multiply(5, 16), multiply(5, 6)))

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

general

a boat running up stram takes 6 hours to cover a certain distance , while it takes 9 hours to cover the same distance running down stream . what is the ratio between the speed of the boat and the speed of water current respectively ?

"explanation : let speed of boat is x km / h and speed stream is y km / hr 6 ( x + y ) = 9 ( x - y ) 6 x + 6 y = 9 x - 9 y 15 y = 3 x 5 y = x x / y = 5 / 1 5 : 1 answer : option b"

a ) 2 : 3 , b ) 5 : 1 , c ) 4 : 5 , d ) 7 : 1 , e ) 8 : 1

b

subtract(9, 6)

subtract(n1,n0)|

physics

the cost price of a radio is rs . 1900 and it was sold for rs . 1330 , find the loss % ?

"1900 - - - - 570 100 - - - - ? = > 30 % answer : c"

a ) 18 % , b ) 31 % , c ) 30 % , d ) 45 % , e ) 12 %

c

multiply(divide(subtract(1900, 1330), 1900), const_100)

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

gain

someone on a skateboard is traveling 13 miles per hour . how many feet does she travel in 25 seconds ? ( 1 mile = 5280 feet )

"per second = > 13 * 5280 ft / 60 * 60 = 19.07 ft 25 seconds = > 19.07 * 25 = 476.75 ft answer : a"

a ) 476.75 ft , b ) 450 ft , c ) 480 ft , d ) 490 ft , e ) 500 ft

a

multiply(25, divide(multiply(13, 5280), const_3600))

multiply(n0,n3)|divide(#0,const_3600)|multiply(n1,#1)|

physics

there are 4 more women than there are men on a local co - ed softball team . if there are a total of 14 players on the team , what is the ratio of men to women ?

"w = m + 4 w + m = 14 m + 4 + m = 14 2 m = 10 m = 5 w = 9 ratio : 5 : 9 ans : e"

a ) 10 / 16 , b ) 6 / 16 , c ) 4 / 16 , d ) 6 / 10 , e ) 5 / 9

e

divide(divide(subtract(14, 4), add(const_1, const_1)), add(divide(subtract(14, 4), add(const_1, const_1)), 4))

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

general

8900 ÷ 6 ÷ 4 = ?

explanation : given exp . 8900 * 1 / 6 * 1 / 4 = 370.833 answer : d

a ) 349 , b ) 541.75 , c ) 224.37 , d ) 370.833 , e ) none of these

d

divide(divide(8900, 6), 4)

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

general

the markup on a box of apples is 10 percent of the cost . the markup is what percent of the selling price ? ( markup = selling price - cost )

mp = 0.1 cp sp = cp + 0.1 cp = 1.1 cp hence mp = 0.1 / 1.1 sp = 1 / 1 sp . hence mp is 9.09 % of sp answer a

a ) 9.09 % , b ) 10 % , c ) 12 1 / 2 % , d ) 15 % , e ) 16 2 / 3 %

a

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

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

general

a train speeds past a pole in 15 seconds and a platform 100 m long in 40 seconds . its length is ?

"let the length of the train be x meters and its speed be y m / sec . they , x / y = 15 = > y = x / 15 x + 100 / 40 = x / 15 x = 100 m . answer : d"

a ) 188 m , b ) 876 m , c ) 251 m , d ) 100 m , e ) 145 m

d

multiply(100, subtract(const_2, const_1))

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

physics

a train running at the speed of 30 km / hr crosses a pole in 12 seconds . what is the length of the train ?

"speed = ( 30 x ( 5 / 18 ) m / sec = ( 25 / 3 ) m / sec . length of the train = ( speed x time ) . length of the train = ( ( 25 / 3 ) x 12 ) m = 100 m d"

a ) 70 , b ) 80 , c ) 90 , d ) 100 , e ) 110

d

multiply(divide(multiply(30, const_1000), const_3600), 12)

multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)|

physics

three numbers are in the ratio 3 : 4 : 5 and their l . c . m is 2400 . their h . c . f is :

"let the numbers be 3 x , 4 x and 5 x . then , their l . c . m = 60 x . so , 60 x = 2400 or x = 40 . the numbers are ( 3 * 40 ) , ( 4 * 40 ) and ( 5 * 40 ) . hence , required h . c . f = 40 . answer : a"

a ) 40 , b ) 80 , c ) 120 , d ) 200 , e ) 220

a

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

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

other

how many positive integers less than 9999 are such that the product of their digits is 210 .

"the prime factorization of 210 is 2 * 3 * 5 * 7 . so one way to make the right kind of number is to use those four digits , in any of the 4 ! = 24 orders you can put them in . notice though that we can also get 210 as a product by multiplying 5 , 6 and 7 . so we can make some 3 - digit numbers with the right product : 3 ! = 6 of them to be exact . but we can also get the right product using the digit 1 along with the digits 5 , 6 , and 7 . again we can arrange those digits in 4 ! = 24 orders . a"

a ) 24 , b ) 58 , c ) 26 , d ) 34 , e ) 25

a

divide(factorial(subtract(add(const_4, 210), const_1)), multiply(factorial(210), factorial(subtract(const_4, const_1))))

add(n1,const_4)|factorial(n1)|subtract(const_4,const_1)|factorial(#2)|subtract(#0,const_1)|factorial(#4)|multiply(#1,#3)|divide(#5,#6)|

general

two stations p and q are 65 km apart on a straight track . one train starts from p at 7 a . m . and travels towards q at 20 kmph . another train starts from q at 8 a . m . and travels towards p at a speed of 25 kmph . at what time will they meet ?

assume both trains meet after x hours after 7 am distance covered by train starting from p in x hours = 20 x km distance covered by train starting from q in ( x - 1 ) hours = 25 ( x - 1 ) total distance = 65 = > 20 x + 25 ( x - 1 ) = 65 = > 45 x = 90 = > x = 2 means , they meet after 2 hours after 7 am , ie , they meet at 9 am answer is a .

a ) 9 am , b ) 12 am , c ) 10.30 am , d ) 12.30 am , e ) 11 am

a

add(divide(add(65, 25), add(20, 25)), 7)

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

physics

a store reported total sales of $ 385 million for february of this year . if the total sales for the same month last year was $ 320 million , approximately what was the percent increase w in sales ?

"last year ' s sales = $ 320 million ; this year ' s sales = $ 385 million ; increase w = $ 65 million . now , 20 % of $ 320 million is $ 64 million , which is very close to actual increase of $ 65 million . answer : c ."

a ) 2 % , b ) 17 % , c ) 20 % , d ) 65 % , e ) 83 %

c

multiply(divide(subtract(385, 320), 320), const_100)

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

general

a man saves 10 % of his monthly salary . if an account of dearness of things he is to increase his monthly expenses by 5 % , he is only able to save rs . 400 per month . what is his monthly salary ?

"income = rs . 100 expenditure = rs . 90 savings = rs . 10 present expenditure 90 + 90 * ( 5 / 100 ) = rs . 94.5 present savings = 100 – 94.5 = rs . 5.50 if savings is rs . 5.50 , salary = rs . 100 if savings is rs . 400 , salary = 100 / 5.5 * 400 = 7273 answer : d"

a ) rs . 6500 , b ) rs . 7500 , c ) rs . 7200 , d ) rs . 7273 , e ) rs . 6300

d

divide(multiply(400, const_100), subtract(const_100, add(subtract(const_100, 10), multiply(subtract(const_100, 10), divide(5, const_100)))))

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

general

working individually , emma can wrap presents for 6 hours and troy can wrap presents in 8 hours . if emma and troy work together but independently at the task for 2 hours , at which point troy leaves , how many remaining hours will it take emma to complete the task alone ?

in first 2 hrs troy will finish 2 / 8 = 1 / 4 of work and emma will finish 2 / 6 work so total 1 / 4 + 1 / 3 = 7 / 12 work is finished and 1 - 7 / 12 = 5 / 12 work remaining . now emma will take ( 5 / 12 ) * 6 = 30 / 12 hrs to finish it . so answer is a .

a ) 30 / 12 , b ) 5 / 12 , c ) 3 / 4 , d ) 2 / 12 , e ) 3 / 5

a

divide(subtract(const_1, add(multiply(2, inverse(6)), multiply(2, inverse(8)))), inverse(6))

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

physics

the length of a rectangular plot is 20 metres more than its breadth . if the cost of fencing the plot @ rs . 26.50 per metre is rs . 8480 , what is the length of the plot in metres ?

"let length of plot = l meters , then breadth = l - 20 meters and perimeter = 2 [ l + l - 20 ] = [ 4 l - 40 ] meters [ 4 l - 40 ] * 26.50 = 8480 [ 4 l - 40 ] = 8480 / 26.50 = 320 4 l = 360 l = 360 / 4 = 90 meters . answer : d"

a ) 20 , b ) 200 , c ) 300 , d ) 90 , e ) 140

d

subtract(divide(divide(8480, 26.50), const_2), multiply(const_2, 20))

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

physics

find the last digit of ( 1021 ^ 3921 ) + ( 3081 ^ 3921 )

"last digit of 1 st expression is 1 and second expression is also 1 as , 1 raised to any power will be 1 itself so , 1 + 1 = 2 so the last digit will be 2 answer : c"

a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4

c

reminder(power(reminder(1021, const_10), reminder(3921, const_10)), const_10)

reminder(n0,const_10)|reminder(n1,const_10)|power(#0,#1)|reminder(#2,const_10)|

general

a certain electric - company plan offers customers reduced rates for electricity used between 8 p . m . and 8 a . m . weekdays and 24 hours a day saturdays and sundays . under this plan , the reduced rates z apply to what fraction of a week ?

number of hours between 8 pm to 8 am = 12 number of hours with reduced rates = ( 12 * 5 ) + ( 24 * 2 ) hours with reduced rates z / total number of hours in a week = ( 12 * 5 ) + ( 24 * 2 ) / ( 24 * 7 ) = 108 / ( 24 * 7 ) = 9 / 14 answer : c

a ) 1 / 2 , b ) 5 / 8 , c ) 9 / 14 , d ) 16 / 21 , e ) 9 / 10

c

divide(add(multiply(divide(24, const_2), add(const_2, const_3)), multiply(24, const_2)), multiply(24, add(const_3, const_4)))

add(const_2,const_3)|add(const_3,const_4)|divide(n2,const_2)|multiply(n2,const_2)|multiply(#0,#2)|multiply(n2,#1)|add(#4,#3)|divide(#6,#5)

physics

the ratio of the volumes of two cubes is 125 : 216 . what is the ratio of their total surface areas ?

"ratio of the sides = ³ √ 125 : ³ √ 216 = 5 : 6 ratio of surface areas = 52 : 62 = 13 : 18 answer : a"

a ) 13 : 18 , b ) 8 : 12 , c ) 1 : 12 , d ) 8 : 17 , e ) 1 : 11

a

power(divide(125, 216), divide(const_1, const_3))

divide(n0,n1)|divide(const_1,const_3)|power(#0,#1)|

geometry

the average weight of d , e and f is 42 kg . if the average weight of d and e be 35 kg and that of e and f be 41 kg , what is the weight of e ?

let the weight of d , e and f are a , b and c respectively . average weight of d , e and f = 42 a + b + c = 42 × 3 = 126 - - - equation ( 1 ) average weight of d and e = 35 a + b = 35 × 2 = 70 - - - equation ( 2 ) average weight of e and f = 41 b + c = 41 × 2 = 82 - - - equation ( 3 ) equation ( 2 ) + equation ( 3 ) - equation ( 1 ) = > a + b + b + c - ( a + b + c ) = 70 + 82 - 126 = > b = 152 - 126 = 26 weight of b = 26 kg answer : d

a ) 23 kg , b ) 24 kg , c ) 25 kg , d ) 26 kg , e ) 27 kg

d

subtract(add(multiply(35, const_2), multiply(41, const_2)), multiply(42, const_3))

multiply(n1,const_2)|multiply(n2,const_2)|multiply(n0,const_3)|add(#0,#1)|subtract(#3,#2)

general

two trains 121 meters and 165 meters in length respectively are running in opposite directions , one at the rate of 75 km and the other at the rate of 65 kmph . in what time will they be completely clear of each other from the moment they meet ?

"t = ( 121 + 165 ) / ( 75 + 65 ) * 18 / 5 t = 7.35 answer : d"

a ) 7.19 , b ) 7.17 , c ) 7.2 , d ) 7.35 , e ) 7.11

d

divide(add(121, 165), multiply(add(75, 65), const_0_2778))

add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)|

physics

find the smallest number which when divided by 12 and 15 leaves respective remainders of 8 and 11 .

"let ' n ' is the smallest number which divided by 12 and 15 leaves respective remainders of 8 and 11 . required number = ( lcm of 12 and 15 ) - ( common difference of divisors and remainders ) = ( 60 ) - ( 4 ) = 56 . answer : e"

a ) 87 , b ) 97 , c ) 27 , d ) 21 , e ) 56

e

subtract(multiply(12, 15), add(const_10, const_1))

add(const_1,const_10)|multiply(n0,n1)|subtract(#1,#0)|

general

a merchant marks his goods up by 20 % and then offers a discount of 10 % on the marked price . what % profit does the merchant make after the discount ?

"let the price be 100 . the price becomes 120 after a 20 % markup . now a discount of 10 % on 120 . profit = 108 - 100 8 % answer a"

a ) 8 % , b ) 10 % , c ) 21 % , d ) 15 % , e ) 17 %

a

subtract(subtract(add(20, const_100), divide(multiply(add(20, const_100), 10), const_100)), const_100)

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

gain

a batsman in his 20 th innings makes a score of 90 , and thereby increases his average by 2 . what is his average after the 20 th innings ? he had never been ’ not out ’ .

"average score before 20 th innings = 90 - 2 × 20 = 50 average score after 20 th innings = > 50 + 2 = 52 answer : a"

a ) 52 , b ) 37 , c ) 39 , d ) 43 , e ) 42

a

add(subtract(90, multiply(2, 20)), 2)

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

general

s ( n ) is a n - digit number formed by attaching the first n perfect squares , in order , into one integer . for example , s ( 1 ) = 1 , s ( 2 ) = 14 , s ( 3 ) = 149 , s ( 4 ) = 14916 , s ( 5 ) = 1491625 , etc . how many digits are in s ( 99 ) ?

"focus on the points where the number of digits in squares change : 1 , 2 , 3 - single digit squares . first 2 digit number is 10 . 4 , 5 , . . . 9 - two digit squares . to get 9 , the last number with two digit square , think that first 3 digit number is 100 which is 10 ^ 2 . so 9 ^ 2 must be the last 2 digit square . 10 , 11 , 12 , . . . 31 - three digit squares . to get 31 , think of 1000 - the first 4 digit number . it is not a perfect square but 900 is 30 ^ 2 . 32 ^ 2 = 2 ^ 10 = 1024 , the first 4 digit square . 32 - 99 - four digit squares . to get 99 , think of 10,000 - the first 5 digit number which is 100 ^ 2 . so number of digits in s ( 99 ) = 3 * 1 + 6 * 2 + 22 * 3 + 68 * 4 = 3 + 12 + 66 + 272 = 353 ; answer : b"

a ) 350 , b ) 353 , c ) 354 , d ) 356 , e ) 357

b

add(add(add(add(const_60, 4), 4), multiply(add(add(const_60, 4), 4), 4)), add(multiply(3, 1), multiply(multiply(3, 2), 2)))

add(n6,const_60)|multiply(n0,n4)|multiply(n2,n4)|add(n6,#0)|multiply(n2,#2)|add(#1,#4)|multiply(n6,#3)|add(#3,#6)|add(#7,#5)|

general

the h . c . f . of two numbers is 23 and the other two factors of their l . c . m . are 10 and 12 . the larger of the two numbers is :

"clearly , the numbers are ( 23 x 10 ) and ( 23 x 12 ) . larger number = ( 23 x 12 ) = 276 . answer : option a"

a ) 276 , b ) 295 , c ) 322 , d ) 345 , e ) 354

a

multiply(23, 12)

multiply(n0,n2)|

other

how many figures are required to number the pages the pages of a book containing 1210 pages ?

"1 to 9 = 9 * 1 = 9 10 to 99 = 90 * 2 = 180 100 to 999 = 900 * 3 = 2700 1000 to 1210 = 211 * 4 = 844 - - - - - - - - - - - 3733 answer : e"

a ) 3533 , b ) 3833 , c ) 3333 , d ) 3633 , e ) 3733

e

add(add(subtract(divide(divide(1210, const_10), const_10), const_1), subtract(subtract(divide(1210, const_10), const_1), subtract(divide(divide(1210, const_10), const_10), const_1))), multiply(subtract(subtract(1210, const_1), subtract(divide(1210, const_10), const_1)), const_3))

divide(n0,const_10)|subtract(n0,const_1)|divide(#0,const_10)|subtract(#0,const_1)|subtract(#2,const_1)|subtract(#1,#3)|multiply(#5,const_3)|subtract(#3,#4)|add(#4,#7)|add(#8,#6)|

general

the average weight of a , b and c is 75 kg . if the average weight of a and b be 88 kg and that of b and c be 72 kg , then the weight of b is :

"explanation let a , b , c represent their respective weights . then , we have : a + b + c = ( 75 x 3 ) = 225 â € ¦ . ( i ) a + b = ( 88 x 2 ) = 176 â € ¦ . ( ii ) b + c = ( 72 x 2 ) = 144 â € ¦ . ( iii ) adding ( ii ) and ( iii ) , we get : a + 2 b + c = 288 â € ¦ . ( iv ) subtracting ( i ) from ( iv ) , we get : b = 63 . b â € ™ s weight = 63 kg . answer d"

a ) 60 kg , b ) 59 kg , c ) 62 kg , d ) 63 kg , e ) 61 kg

d

subtract(add(multiply(88, const_2), multiply(72, const_2)), multiply(75, const_3))

multiply(n1,const_2)|multiply(n2,const_2)|multiply(n0,const_3)|add(#0,#1)|subtract(#3,#2)|

general

a and b are mixed together in the ratio 9 : 11 . what is the weight of mixture , if 26.1 kg of a has been consumed in it ?

explanation : for 9 kg a , mixture = ( 9 + 11 ) kg . for 26.1 kg a , mixture = ( ( 20 / 9 ) x 26.1 ) kg = 58 kg . answer : d

a ) 56 , b ) 45 , c ) 58.2 , d ) 58 , e ) 46

d

add(multiply(11, divide(26.1, 9)), 26.1)

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

general

find the constant k so that : - x 2 - ( k + 12 ) x - 8 = - ( x - 2 ) ( x - 4 )

"- x 2 - ( k + 12 ) x - 8 = - ( x - 2 ) ( x - 4 ) : given - x 2 - ( k + 12 ) x - 8 = - x 2 + 6 x - 8 - ( k + 12 ) = 6 : two polynomials are equal if their corresponding coefficients are equal . k = - 18 : solve the above for k correct answer c"

a ) 11 , b ) 12 , c ) 18 , d ) 14 , e ) 15

c

add(12, add(4, 2))

add(n0,n4)|add(n1,#0)|

general

last year , company x paid out a total of $ 1 , 050,000 in salaries to its 31 employees . if no employee earned a salary that is more than 20 % greater than any other employee , what is the lowest possible salary that any one employee earned ?

"employee 1 earned $ x ( say ) employee 2 will not earn more than $ 1.2 x therfore , to minimize the salary of any one employee , we need to maximize the salaries of the other 30 employees ( 1.2 x * 30 ) + x = 1 , 050,000 solving for x = $ 28 , 378.37 answer b"

a ) $ 10,000 , b ) $ $ 28 , 378.37 , c ) $ 42,000 , d ) $ 50,000 , e ) $ 60,000

b

add(divide(divide(divide(multiply(add(divide(20, const_100), 1), multiply(subtract(add(const_1000, const_60), const_10), const_1000)), add(multiply(subtract(31, 1), add(divide(20, const_100), 1)), 1)), add(divide(20, const_100), 1)), const_100), add(multiply(const_100, const_2), const_3))

add(const_1000,const_60)|divide(n3,const_100)|multiply(const_100,const_2)|subtract(n2,n0)|add(#2,const_3)|add(#1,n0)|subtract(#0,const_10)|multiply(#6,const_1000)|multiply(#5,#3)|add(n0,#8)|multiply(#5,#7)|divide(#10,#9)|divide(#11,#5)|divide(#12,const_100)|add(#4,#13)|

general

if n is a positive integer and the product of all integers from 1 to n , inclusive , is a multiple of 350 , what is the least possible value of n ?

"350 = 2 * 5 * 5 * 7 so the least value forncan be 7 . b"

a ) 5 , b ) 7 , c ) 12 , d ) 13 , e ) 14

b

divide(divide(divide(divide(350, const_2), const_3), const_4), divide(const_10, const_2))

divide(n1,const_2)|divide(const_10,const_2)|divide(#0,const_3)|divide(#2,const_4)|divide(#3,#1)|

general

the duplicate ratio of 2 : 3 is ?

"2 ^ 2 : 3 ^ 2 = 4 : 9 answer : a"

a ) 4 : 9 , b ) 1 : 4 , c ) 1 : 8 , d ) 1 : 18 , e ) 1 : 13

a

divide(power(2, const_2), power(3, const_2))

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

other

the water level in a rectangular swimming pool measuring 40 feet by 25 feet is to be lowered by 6 inches . how many gallons of water must be removed ? ( 1 cu ft = 7.5 gallons )

"6 inches = 1 / 2 feet ( there are 12 inches in a foot . ) , so 40 * 25 * 1 / 2 = 500 feet ^ 3 of water must be removed , which equals to 500 * 7.5 = 3750 gallons . answer : d ."

a ) 100 , b ) 250 , c ) 750 , d ) 3750 , e ) 5625

d

multiply(volume_rectangular_prism(40, 25, divide(6, add(const_10, const_2))), 7.5)

add(const_10,const_2)|divide(n2,#0)|volume_rectangular_prism(n0,n1,#1)|multiply(n4,#2)|

geometry

3 maths classes : a , b and c take an algebra test . the average score of class a is 83 . the average score of class b is 76 . the average score of class c is 85 . the average score of class a and b is 79 and average score of class b and c is 81 . what is the average score of classes a , b , c ?

let the number of students in classes a , b and c be p , q and r respectively . then , total score of a = 83 p , total score of b = 76 q , total score of c = 85 r . also given that , ( 83 p + 76 q ) / ( p + q ) = 79 = > 4 p = 3 q . ( 76 q + 85 r ) / ( q + r ) = 81 = > 4 r = 5 q , = > q = 4 p / 3 and r = 5 p / 3 therefore , average score of a , b , c = ( 83 p + 76 q + 85 r ) / ( p + q + r ) = 978 / 12 = 81.5 answer : d

a ) 81.8 , b ) 81.1 , c ) 81.2 , d ) 81.5 , e ) 81.9

d

add(81, divide(const_1, const_2))

divide(const_1,const_2)|add(n5,#0)

general

at a certain resort , each of the 39 food service employees is trained to work in a minimum of 1 restaurant and a maximum of 3 restaurants . the 3 restaurants are the family buffet , the dining room , and the snack bar . exactly 19 employees are trained to work in the family buffet , 18 are trained to work in the dining room , and 12 are trained to work in the snack bar . if 4 employees are trained to work in exactly 2 restaurants , how many employees are trained to work in all 3 restaurants ?

"39 = 19 + 18 + 12 - 4 - 2 x 2 x = 19 + 18 + 12 - 4 - 39 = 45 - 39 = 6 x = 3 answer : b"

a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6

b

divide(subtract(subtract(add(add(19, 18), 12), 4), 39), 2)

add(n4,n5)|add(n6,#0)|subtract(#1,n7)|subtract(#2,n0)|divide(#3,n8)|

physics

the volume of a cube is 1728 cc . find its surface .

"a 3 = 1728 = > a = 12 6 a 2 = 6 * 12 * 12 = 864 answer : a"

a ) 864 , b ) 556 , c ) 255 , d ) 287 , e ) 267

a

surface_cube(cube_edge_by_volume(1728))

cube_edge_by_volume(n0)|surface_cube(#0)|

geometry

the length of rectangle is thrice its breadth and its perimeter is 40 m , find the area of the rectangle ?

"2 ( 3 x + x ) = 40 l = 15 b = 5 lb = 15 * 5 = 75 answer : c"

a ) 432 , b ) 212 , c ) 75 , d ) 992 , e ) 212

c

multiply(multiply(divide(40, add(multiply(const_3, const_2), multiply(const_1, const_2))), const_3), divide(40, add(multiply(const_3, const_2), multiply(const_1, const_2))))

multiply(const_2,const_3)|multiply(const_1,const_2)|add(#0,#1)|divide(n0,#2)|multiply(#3,const_3)|multiply(#3,#4)|

geometry

a 25 cm wide path is to be made around a circular garden having a diameter of 4 meters . approximate area of the path is square meters is ?

"area of the path = area of the outer circle - area of the inner circle = ∏ { 4 / 2 + 25 / 100 } 2 - ∏ [ 4 / 2 ] 2 = ∏ [ 2.252 - 22 ] = ∏ ( 0.25 ) ( 4.25 ) { ( a 2 - b 2 = ( a - b ) ( a + b ) } = ( 3.14 ) ( 1 / 4 ) ( 17 / 4 ) = 53.38 / 16 = 3.34 sq m answer : a"

a ) 3.34 sq m , b ) 98.8 sq m , c ) 67.8 sq m , d ) 27.9 sq m , e ) 19.9 sq m

a

subtract(circle_area(add(divide(4, const_2), divide(25, const_100))), circle_area(divide(4, const_2)))

divide(n1,const_2)|divide(n0,const_100)|add(#0,#1)|circle_area(#0)|circle_area(#2)|subtract(#4,#3)|

geometry

a train is 700 meter long is running at a speed of 21 km / hour . in what time will it pass a bridge of 130 meter length ?

"speed = 21 km / hr = 21 * ( 5 / 18 ) m / sec = 35 / 6 m / sec total distance = 700 + 130 = 830 meter time = distance / speed = 830 * ( 6 / 35 ) = 142.286 seconds . answer : a"

a ) 142.286 , b ) 251.228 , c ) 34 , d ) 200 , e ) 150.627

a

divide(add(700, 130), divide(multiply(21, const_1000), const_3600))

add(n0,n2)|multiply(n1,const_1000)|divide(#1,const_3600)|divide(#0,#2)|

physics

a coin is tossed 4 times . what is the probability of getting exactly 3 heads ?

the number of possible outcomes is 2 ^ 4 = 16 there are 4 ways to get exactly 3 heads . p ( exactly 3 heads ) = 4 / 16 = 1 / 4 the answer is b .

a ) 1 / 2 , b ) 1 / 4 , c ) 1 / 8 , d ) 1 / 16 , e ) 3 / 16

b

multiply(power(divide(const_1, const_2), 3), multiply(choose(const_4, const_3), divide(const_1, const_2)))

choose(const_4,const_3)|divide(const_1,const_2)|multiply(#0,#1)|power(#1,n1)|multiply(#2,#3)

probability

the ratio of pens to pencils is 5 to 6 . there are 4 more pencils than pens . how many pencils are there ?

"let the number of pens be 5 x and the number of pencils be 6 x . 6 x - 5 x = 4 x = 4 the number of pencils is 24 . the answer is d ."

a ) 12 , b ) 16 , c ) 20 , d ) 24 , e ) 28

d

add(multiply(4, 5), 4)

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

other

a train passes a station platform in 42 sec and a man standing on the platform in 12 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 * 12 = 180 m . let the length of the platform be x m . then , ( x + 180 ) / 42 = 15 = > x = 450 m . answer : b"

a ) 227 , b ) 450 , c ) 460 , d ) 480 , e ) 171

b

multiply(12, multiply(54, const_0_2778))

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

physics

how many pieces of 75 cm can be cut from a rope 45 meters long ?

"explanation : total pieces of 75 cm that can be cut from a rope of 45 meters long is = ( 45 meters ) / ( 75 cm ) = ( 45 meters ) / ( 0.75 meters ) = 60 answer : c"

a ) 30 , b ) 40 , c ) 60 , d ) none , e ) can not be determined

c

divide(45, 75)

divide(n1,n0)|

physics

in what ratio mental a at rs . 68 per kg be mixed with another metal at rs . 96 per kg so that cost of alloy ( mixture ) is rs . 75 per kg ?

"( 96 - 75 ) / ( 75 - 68 ) = 21 / 7 = 3 / 1 answer : b"

a ) 5 : 8 , b ) 3 : 1 , c ) 3 : 7 , d ) 9 : 5 , e ) 9 : 8

b

divide(divide(subtract(96, 75), subtract(96, 68)), subtract(const_1, divide(subtract(96, 75), subtract(96, 68))))

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

other

a shopkeeper fixes the marked price of an item 30 % above its cost price . the percentage of discount allowed to gain 6 % is

"explanation : let the cost price = rs 100 then , marked price = rs 130 required gain = 6 % , so selling price = rs 106 discount = 130 - 106 = 24 discount % = ( 24 / 130 ) * 100 = 18.46 % option b"

a ) 18.45 % , b ) 18.46 % , c ) 18.47 % , d ) 18.48 % , e ) none of these

b

subtract(const_100, multiply(divide(add(6, const_100), add(30, const_100)), const_100))

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

gain

the product of two numbers is 120 and the sum of their squares is 289 . find the sum ?

"nos a and b ab = 120 and a 2 + b 2 = 289 ( a + b ) 2 = 529 a + b = root ( 529 ) = 23 answer a"

a ) 23 , b ) 20 , c ) 15 , d ) 26 , e ) 28

a

sqrt(add(power(sqrt(subtract(289, multiply(const_2, 120))), const_2), multiply(const_4, 120)))

multiply(n0,const_4)|multiply(n0,const_2)|subtract(n1,#1)|sqrt(#2)|power(#3,const_2)|add(#0,#4)|sqrt(#5)|

general

find the sum of first 26 natural numbers

"explanation : sum of n natural numbers = n ( n + 1 ) / 2 = 26 ( 26 + 1 ) / 2 = 26 ( 27 ) / 2 = 351 answer : option d"

a ) 470 , b ) 468 , c ) 465 , d ) 351 , e ) 485

d

add(divide(divide(26, divide(divide(divide(divide(divide(26, const_2), const_2), const_2), const_2), const_2)), const_2), add(const_1, sqrt(divide(divide(26, divide(divide(divide(divide(divide(26, const_2), const_2), const_2), const_2), const_2)), const_2))))

divide(n0,const_2)|divide(#0,const_2)|divide(#1,const_2)|divide(#2,const_2)|divide(#3,const_2)|divide(n0,#4)|divide(#5,const_2)|sqrt(#6)|add(#7,const_1)|add(#8,#6)|

general

in an electric circuit , two resistors with resistances m and n are connected in parallel . in this case , if p is the combined resistance of these two resistors , then the reciprocal of p is equal to the sum of the reciprocals of m and n . what is p in terms of m and n ?

the wording is a bit confusing , though basically we are told that 1 / p = 1 / m + 1 / n , from which it follows that p = mn / ( m + n ) . answer : b

['a ) ( n - m )', 'b ) mn / ( m + n )', 'c ) ( nm )', 'd ) ( n - m ) / ( m + n )', 'e ) none of these']

b

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

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

geometry

x + ( 1 / x ) = 3 find x ^ 2 + ( 1 / x ^ 2 )

"squaring on both sides ( x + 1 / x ) ^ 2 = 3 ^ 2 x ^ 2 + 1 / x ^ 2 = 9 - 2 x ^ 2 + 1 / x ^ 2 = 7 answer : b"

a ) 2.25 , b ) 7 , c ) 4.25 , d ) 5.25 , e ) 6.25

b

subtract(power(3, 2), 2)

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

general