Datasets:

math_qa

like 68

how long does a train 110 m long traveling at 60 kmph takes to cross a bridge of 390 m in length ?

"d = 110 + 390 = 500 m s = 60 * 5 / 18 = 50 / 3 t = 500 * 3 / 50 = 30 sec answer : d"

a ) 18.9 sec , b ) 88.9 sec , c ) 22.9 sec , d ) 30.00 sec , e ) 72.0 sec

d

divide(add(110, 390), multiply(60, const_0_2778))

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

physics

a can finish a work in 24 days , b in 9 days and c in 12 days . b and c start the work but are forced to leave after 3 days . when a done the work ?

"b + c = = > 1 / 9 + 1 / 12 = 7 / 36 b , c = in 3 days = 7 / 36 * 3 = 7 / 12 remaining work = 1 - 7 / 12 = 5 / 12 1 / 24 work is done by a in 1 day 5 / 12 work is done a 24 * 5 / 12 = 10 days answer a"

a ) 10 days , b ) 12 days , c ) 13 days , d ) 9 days , e ) 14 days

a

multiply(divide(const_1, add(divide(const_1, 9), divide(const_1, 12))), 3)

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

physics

in a recent election , james received 1.5 percent of the 2,000 votes cast . to win the election , a candidate needed to receive more than 50 percent of the vote . how many additional votes would james have needed to win the election ?

"james = ( 1.5 / 100 ) * 2000 = 30 votes to win = ( 50 / 100 ) * total votes + 1 = ( 50 / 100 ) * 2000 + 1 = 1001 remaining voted needed to win election = 1001 - 30 = 971 answer : option d"

a ) 901 , b ) 989 , c ) 990 , d ) 971 , e ) 1,001

d

subtract(add(const_1000, const_1000), multiply(add(const_1000, const_1000), const_0.5))

add(const_1000,const_1000)|multiply(const_0.5,#0)|subtract(#0,#1)|

general

jolene entered an 18 - month investment contract that guarantees to pay 2 percent interest at the end of 6 months , another 3 percent interest at the end of 10 months , and 4 percent interest at the end of the 18 month contract . if each interest payment is reinvested in the contract , and jolene invested $ 10,000 initially , what will be the total amount of interest paid during the 18 - month contract ?

if interest were not compounded in every six months ( so if interest were not earned on interest ) then we would have ( 2 + 3 + 4 ) = 9 % simple interest earned on $ 10,000 , which is $ 900 . so , you can rule out a , b and c right away . interest earned after the first time interval : $ 10,000 * 2 % = $ 200 ; interest earned after the second time interval : ( $ 10,000 + $ 200 ) * 3 % = $ 300 + $ 6 = $ 306 ; interest earned after the third time interval : ( $ 10,000 + $ 200 + $ 306 ) * 4 % = $ 400 + $ 8 + ( ~ $ 12 ) = ~ $ 420 ; total : 200 + 306 + ( ~ 420 ) = ~ $ 726.24 . answer : b .

a ) $ 506.00 , b ) $ 726.24 , c ) $ 900.00 , d ) $ 920.24 , e ) $ 926.24

b

add(multiply(add(multiply(divide(3, const_100), add(multiply(divide(2, const_100), power(const_100, const_2)), power(const_100, const_2))), add(multiply(divide(2, const_100), power(const_100, const_2)), power(const_100, const_2))), divide(4, const_100)), multiply(divide(3, const_100), add(multiply(divide(2, const_100), power(const_100, const_2)), power(const_100, const_2))))

divide(n1,const_100)|divide(n3,const_100)|divide(n5,const_100)|power(const_100,const_2)|multiply(#0,#3)|add(#4,#3)|multiply(#5,#1)|add(#5,#6)|multiply(#7,#2)|add(#8,#6)

gain

walking 4 / 3 of his usual rate , a boy reaches his school 4 min early . find his usual time to reach the school ?

"speed ratio = 1 : 4 / 3 = 3 : 4 time ratio = 4 : 3 1 - - - - - - - - 4 4 - - - - - - - - - ? 16 m . answer : b"

a ) 22 , b ) 16 , c ) 27 , d ) 28 , e ) 20

b

multiply(4, 4)

multiply(n0,n2)|

gain

which is the least number that must be subtracted from 1256 so that the remainder when divided by 7 , 12 , 16 is 4 ?

"first we need to figure out what numbers are exactly divisible by 7 , 12,16 . this will be the set { lcm , lcmx 2 , lcmx 3 , . . . } lcm ( 7 , 12,16 ) = 48 * 7 = 336 the numbers which will leave remainder 4 will be { 336 + 4 , 336 x 2 + 4 , 336 x 3 + 4 , . . . } the largest such number less than or equal to 1256 is 336 x 3 + 4 or 1012 to obtain this you need to subtract 244 . e"

a ) 242 , b ) 232 , c ) 236 , d ) 240 , e ) 244

e

subtract(1256, add(4, multiply(gcd(1256, lcm(lcm(7, 12), 16)), lcm(lcm(7, 12), 16))))

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

general

if n is the smallest integer such that 108 times n is the square of an integer , what is the value of n ?

"108 can written as = 2 * 2 * 3 * 3 * 3 - - > 2 ^ 2 * 3 ^ 3 - - - ( 1 ) so for 108 * n to be a square of an integer , the integer should have even powers to the prime numbers it composed of . here 2 already has even power - > so n has to be 2 to make the power of 2 in ( 1 ) even option a is correct"

a ) 2 , b ) 3 , c ) 6 , d ) 12 , e ) 24

a

divide(divide(divide(divide(divide(divide(108, const_2), const_2), const_2), const_2), const_3), const_3)

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

geometry

of the 500 employees in a certain company , 25 percent will be relocated to city x and the remaining 75 percent will be relocated to city y . however , 40 percent of the employees prefer city y and 60 percent prefer city x . what is the highest possible number of employees who will be relocated to the city they prefer ?

"300 prefer x ( group 1 ) ; 200 prefer y ( group 2 ) . city y needs 375 people : letall 200 who prefer y ( entire group 2 ) be relocated there , the rest 175 will be those who prefer x from group 1 ; city x needs 125 people : 300 - 175 = 125 from group 1 will be relocated to x , which they prefer . so , the highest possible number of employees who will be relocated to the city they prefer is 200 + 125 = 325 . answer : e ."

a ) 65 , b ) 100 , c ) 115 , d ) 130 , e ) 325

e

add(multiply(40, const_2), 60)

multiply(n3,const_2)|add(n4,#0)|

gain

the average of 1 st 3 of 4 numbers is 6 and of the last 3 are 5 . if the sum of the first and the last number is 11 . what is the last numbers ?

"a + b + c = 18 b + c + d = 15 a + d = 11 a – d = 3 a + d = 11 2 d = 8 d = 4 answer : a"

a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8

a

subtract(subtract(multiply(3, 6), add(subtract(11, 6), 3)), 6)

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

general

if the cost price of 50 articles is equal to the selling price of 35 articles , then the gain or loss percent is ?

"percentage of profit = 15 / 35 * 100 = 43 % answer : e"

a ) 16 , b ) 127 , c ) 12 , d ) 18 , e ) 43

e

multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 35), 50)), divide(multiply(const_100, 35), 50)))

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

gain

a batch of cookies was divided amomg 3 tins : 3 / 4 of all the cookies were placed in either the blue or the green tin , and the rest were placed in the red tin . if 1 / 4 of all the cookies were placed in the blue tin , what fraction of the cookies that were placed in the other tins were placed in the green tin

"this will help reduce the number of variables you have to deal with : g + b = 3 / 4 r = 1 / 3 b = 1 / 4 we can solve for g which is 1 / 2 what fraction ( let it equal x ) of the cookies that were placed in the other tins were placed in the green tin ? so . . x * ( g + r ) = g x * ( 1 / 2 + 1 / 3 ) = 1 / 2 x = 3 / 5 answer : d"

a ) 15 / 2 , b ) 9 / 4 , c ) 5 / 9 , d ) 3 / 5 , e ) 9 / 7

d

add(subtract(1, divide(3, 4)), subtract(divide(3, 4), divide(1, 4)))

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

general

10 different biology books and 8 different chemistry books lie on a shelf . in how many ways can a student pick 2 books of each type ?

"no . of ways of picking 2 biology books ( from 10 books ) = 10 c 2 = ( 10 * 9 ) / 2 = 45 no . of ways of picking 2 chemistry books ( from 8 books ) = 8 c 2 = ( 8 * 7 ) / 2 = 28 total ways of picking 2 books of each type = 45 * 28 = 1260 ( option e )"

a ) 80 , b ) 160 , c ) 720 , d ) 1100 , e ) 1260

e

multiply(divide(divide(factorial(10), factorial(subtract(10, 2))), 2), divide(divide(factorial(8), factorial(subtract(8, 2))), 2))

factorial(n0)|factorial(n1)|subtract(n0,n2)|subtract(n1,n2)|factorial(#2)|factorial(#3)|divide(#0,#4)|divide(#1,#5)|divide(#6,n2)|divide(#7,n2)|multiply(#8,#9)|

other

a grocer has a sale of rs . 6400 , rs . 7000 , rs . 6800 , rs . 7200 and rs . 6500 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . 6500 ?

"total sale for 5 months = rs . ( 6400 + 7000 + 6800 + 7200 + 6500 ) = rs . 33900 required sale = rs . [ ( 6500 x 6 ) - 34009 ] = rs . ( 39000 - 33900 ) = rs . 5100 answer : e"

a ) rs . 4500 , b ) rs . 4700 , c ) rs . 4800 , d ) rs . 5000 , e ) rs . 5100

e

subtract(multiply(add(5, const_1), 6500), add(add(add(add(6400, 7000), 6800), 7200), 6500))

add(n5,const_1)|add(n0,n1)|add(n2,#1)|multiply(n6,#0)|add(n3,#2)|add(n4,#4)|subtract(#3,#5)|

general

given that a is the average ( arithmetic mean ) of the first 5 positive multiples of 6 and b is the median of the first 12 positive multiples of 6 , what is the ratio of a to b ?

the first nine positive multiples of six are { 6 , 12 , 18 , 24,30 } the first twelve positive multiples of six are { 6 , 12 , 18 , 24 , 30 , 36,42 , 48 , 54 , 60 , 66 , 72 } both sets are evenly spaced , thus their median = mean : a = 18 and b = ( 36 + 42 ) / 2 = 39 - - > a / b = 18 / 39 = 6 / 13 . answer : b .

a ) 3 : 4 , b ) 6 : 13 , c ) 5 : 6 , d ) 13 : 10 , e ) 4 : 3

b

divide(multiply(divide(add(5, const_1), const_2), 6), multiply(divide(add(12, const_1), const_2), 6))

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

general

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

"explanation : amount = [ 7500 × ( 1 + 4100 ) 2 ] = ( 7500 × 2625 × 2625 ) = 8112 so compound interest = ( 8112 - 7500 ) = 612 answer : b"

a ) rs . 610 , b ) rs . 612 , c ) rs . 614 , d ) rs . 616 , e ) none of these

b

subtract(multiply(power(add(const_1, divide(divide(4, const_4), const_100)), const_3), multiply(multiply(multiply(const_4, const_4), const_100), sqrt(const_100))), multiply(multiply(multiply(const_4, const_4), const_100), sqrt(const_100)))

divide(n1,const_4)|multiply(const_4,const_4)|sqrt(const_100)|divide(#0,const_100)|multiply(#1,const_100)|add(#3,const_1)|multiply(#4,#2)|power(#5,const_3)|multiply(#6,#7)|subtract(#8,#6)|

gain

how much greater is the combined area in square inches of the front and back of a rectangular sheet of paper measuring 11 inches by 13 inches than that of a rectangular sheet of paper measuring 6.5 inches by 11 inches ?

let ' s just look at the dimensions ( no calculation needed ) . with dimension 11 the same , the other dimension 13 is twice 6.5 then the area will be double which means 100 % greater . the answer is c .

['a ) 50 %', 'b ) 87 %', 'c ) 100 %', 'd ) 187 %', 'e ) 200 %']

c

multiply(divide(subtract(multiply(rectangle_area(11, 13), const_2), multiply(rectangle_area(6.5, 11), const_2)), rectangle_area(11, 13)), const_100)

rectangle_area(n0,n1)|rectangle_area(n0,n2)|multiply(#0,const_2)|multiply(#1,const_2)|subtract(#2,#3)|divide(#4,#0)|multiply(#5,const_100)

geometry

in the xy - plane , a triangle has vertices ( 0,0 ) , ( 4,0 ) and ( 4,9 ) . if a point ( a , b ) is selected at random from the triangular region , what is the probability that a - b > 0 ?

"the area of the right triangle is ( 1 / 2 ) * 4 * 9 = 18 . only the points ( a , b ) below the line y = x satisfy a - b > 0 . the part of the triangle which is below the line y = x has an area of ( 1 / 2 ) ( 4 ) ( 4 ) = 8 . p ( a - b > 0 ) = 8 / 18 = 2 / 9 the answer is d ."

a ) 1 / 5 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 9 , e ) 4 / 5

d

divide(const_4, add(0,0, const_10))

add(n0,const_10)|divide(const_4,#0)|

geometry

if a sum of money doubles itself in 5 years at simple interest , the ratepercent per annum is

"explanation : let sum = x then simple interest = x rate = ( 100 * x ) / ( x * 5 ) = 20 option c"

a ) 12 , b ) 12.5 , c ) 20 , d ) 13.5 , e ) 14

c

divide(divide(const_2, divide(5, const_100)), const_2)

divide(n0,const_100)|divide(const_2,#0)|divide(#1,const_2)|

gain

elvin ' s monthly telephone bill is the sum of the charge for the calls he made during the month and a fixed monthly charge for internet service . elvin ' s total telephone bill for january was $ 52 and elvin ' s total telephone bill for february was 76 $ . if elvin ' s charge for the calls he made in february was twice the charge for the calls he made in january , what is elvin ' s fixed monthly charge for internet service ?

"bill = fixed charge + charge of calls made in jan , bill = fixed charge ( let , y ) + charge of calls made in jan ( let , x ) = $ 52 in feb , bill = fixed charge ( let , y ) + charge of calls made in feb ( then , 2 x ) = $ 76 i . e . x + y = 52 and 2 x + y = 76 take the difference if two equations i . e . ( 2 x + y ) - ( x + y ) = 76 - 52 i . e . x = 24 i . e . fixed monthly charge , y = 28 answer : option d"

a ) $ 5 , b ) $ 10 , c ) $ 14 , d ) $ 28 , e ) $ 29

d

subtract(multiply(52, const_2), 76)

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

general

a started a business with an investment of rs . 70000 and after 9 months b joined him investing rs . 120000 . if the profit at the end of a year is rs . 76000 , then the share of b is ?

"ratio of investments of a and b is ( 70000 * 12 ) : ( 120000 * 3 ) = 7 : 12 total profit = rs . 76000 share of b = 12 / 19 ( 76000 ) = rs . 48000 answer : d"

a ) 33008 , b ) 24000 , c ) 28000 , d ) 48000 , e ) 81122

d

subtract(76000, multiply(const_60, const_100))

multiply(const_100,const_60)|subtract(n3,#0)|

gain

how many 7 in between 1 to 110 ?

"7 , 17,27 , 37,47 , 57,67 , 70,71 , 72,73 , 74,75 , 76,77 ( two 7 ' s ) , 78 , 79,87 , 97,107 21 7 ' s between 1 to 110 answer : b"

a ) 18 , b ) 21 , c ) 22 , d ) 23 , e ) 24

b

divide(110, multiply(7, const_3.0))

multiply(n0,const_3.0)|divide(n2,#0)|

general

a alone can finish a work in 10 days which b alone can finish in 15 days . if they work together and finish it , then out of a total wages of rs . 3000 , a will get :

"explanation : ratio of working days of a : b = 10 : 15 therefore , their wages ratio = reverse ratio = 15 : 10 therefore , a will get 15 units of ratio total ratio = 25 1 unit of ratio = 3000 / 25 = 120 so , a ’ s amount = 120 × 15 = rs . 1800 . answer : option c"

a ) rs . 1200 , b ) rs . 1500 , c ) rs . 1800 , d ) rs . 2000 , e ) none of these

c

multiply(divide(divide(multiply(15, const_2), 10), add(divide(multiply(15, const_2), 10), divide(multiply(15, const_2), 15))), 3000)

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

physics

a started a business with an investment of rs . 70000 and after 6 months b joined him investing rs . 120000 . if the profit at the end of a year is rs . 78000 , then the share of b is ?

"ratio of investments of a and b is ( 70000 * 12 ) : ( 120000 * 6 ) = 7 : 6 total profit = rs . 78000 share of b = 6 / 13 ( 78000 ) = rs . 36000 answer : c"

a ) a ) 34500 , b ) b ) 36099 , c ) c ) 36000 , d ) d ) 38007 , e ) e ) 42000

c

subtract(78000, multiply(const_60, const_100))

multiply(const_100,const_60)|subtract(n3,#0)|

gain

if an object travels 90 feet in 3 seconds , what is the object ’ s approximate speed in miles per hour ? ( note : 1 mile = 5280 feet )

"90 feet / 3 seconds = 30 feet / second ( 30 feet / second ) * ( 3600 seconds / hour ) * ( 1 mile / 5280 feet ) = 20.45 miles / hour ( approximately ) the answer is b ."

a ) 17.36 , b ) 20.45 , c ) 23.87 , d ) 26.92 , e ) 29.56

b

divide(divide(90, 5280), multiply(3, divide(1, const_3600)))

divide(n0,n3)|divide(n2,const_3600)|multiply(n1,#1)|divide(#0,#2)|

physics

if 2 x + y = 7 and x + 2 y = 5 , then xy / 3 =

"2 * ( x + 2 y = 5 ) equals 2 x + 4 y = 10 2 x + 4 y = 10 - 2 x + y = 7 = 3 y = 3 therefore y = 1 plug and solve . . . 2 x + 1 = 7 2 x = 6 x = 3 xy / 3 = 3 * 1 / 3 = 1 a"

a ) 1 , b ) 4 / 3 , c ) 17 / 5 , d ) 18 / 5 , e ) 4

a

divide(add(divide(subtract(multiply(7, 2), 5), subtract(multiply(2, 2), const_1)), subtract(7, multiply(2, divide(subtract(multiply(7, 2), 5), subtract(multiply(2, 2), const_1))))), 3)

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

general

there are 21 students in a class . in how many different ways can a committee of 3 students be formed ?

"21 c 3 = 21 * 20 * 19 / 6 = 1330 the answer is b ."

a ) 1180 , b ) 1330 , c ) 1540 , d ) 1760 , e ) 1920

b

multiply(subtract(const_1, divide(3, 21)), 21)

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

probability

the average age of boys in a class is 16 years and that of the girls is 15 years . the average age for the whole class is

"clearly , to find the average , we ought to know the numbers of boys , girls or students in the class , neither of which has been given . so the data provided is inadequate . answer : e"

a ) 15 years , b ) 15.5 years , c ) 16 years , d ) 17 years , e ) can not be computed

e

subtract(16, 15)

subtract(n0,n1)|

general

of 800 surveyed students , 20 % of those who read book a also read book b and 25 % of those who read book b also read book a . if each student read at least one of the books , what is the difference between the number of students who read only book a and the number of students who read only book b ?

"say the number of students who read book a is a and the number of students who read book b is b . given that 20 % of those who read book a also read book b and 25 % of those who read book b also read book a , so the number of students who read both books is 0.2 a = 0.25 b - - > a = 1.25 b . since each student read at least one of the books then { total } = { a } + { b } - { both } - - > 800 = 1.25 b + b - 0.25 b - - > b = 400 , a = 1.25 b = 500 and { both } = 0.25 b = 100 . the number of students who read only book a is { a } - { both } = 500 - 100 = 400 ; the number of students who read only book b is { b } - { both } = 400 - 100 = 300 ; the difference is 400 - 300 = 100 . answer : e ."

a ) 20 , b ) 25 , c ) 30 , d ) 55 , e ) 100

e

subtract(multiply(multiply(divide(800, subtract(add(divide(divide(25, const_100), divide(20, const_100)), const_1), divide(25, const_100))), divide(divide(25, const_100), divide(20, const_100))), subtract(const_1, divide(20, const_100))), multiply(divide(800, subtract(add(divide(divide(25, const_100), divide(20, const_100)), const_1), divide(25, const_100))), subtract(const_1, divide(25, const_100))))

divide(n2,const_100)|divide(n1,const_100)|divide(#0,#1)|subtract(const_1,#1)|subtract(const_1,#0)|add(#2,const_1)|subtract(#5,#0)|divide(n0,#6)|multiply(#7,#2)|multiply(#7,#4)|multiply(#8,#3)|subtract(#10,#9)|

gain

what annual payment will discharge a debt of rs . 1090 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 = 1090 820 x + 1090 * 441 x = 586.21 so , value of each installment = rs . 586.21 answer : option b

a ) 993.2 , b ) 586.21 , c ) 534.33 , d ) 543.33 , e ) 646.33

b

divide(multiply(power(add(divide(5, const_100), const_1), 2), 1090), 2)

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

gain

a man sells a car to his friend at 15 % loss . if the friend sells it for rs . 54000 and gains 20 % , the original c . p . of the car was :

"explanation : s . p = rs . 54,000 . gain earned = 20 % c . p = rs . [ 100 / 120 ã — 54000 ] = rs . 45000 this is the price the first person sold to the second at at loss of 15 % . now s . p = rs . 45000 and loss = 15 % c . p . rs . [ 100 / 85 ã — 45000 ] = rs . 52941.18 correct option : c"

a ) rs . 22941.18 , b ) rs . 32941.18 , c ) rs . 52941.18 , d ) rs . 62941.18 , e ) none of these

c

divide(multiply(divide(multiply(54000, const_100), add(const_100, 20)), const_100), subtract(const_100, 15))

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

gain

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

"answer : option c explanation : a 2 = 3136 = > a = 56 56 * 4 * 3 = 672 â € “ 6 = 666 * 1 = 666 answer : c"

a ) 399 , b ) 272 , c ) 666 , d ) 277 , e ) 311

c

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

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

physics

how many 0 ' s are there in the binary form of 6 * 1024 + 4 * 64 + 2

6 * 1024 + 4 * 64 + 2 = 6144 + 256 + 2 = 6402 in binary code 6402 base 10 = 11001 000000 10 in base 2 . so 9 zeros are there . answer : e

a ) 6 , b ) 7 , c ) 8 , d ) 10 , e ) 9

e

subtract(add(6, 4), const_1)

add(n1,n3)|subtract(#0,const_1)

general

what is 7 1 / 4 - 4 2 / 3 divided by 7 / 8 - 3 / 4 ?

"7 1 / 4 - 4 2 / 3 = 29 / 4 - 14 / 3 = ( 87 - 56 ) / 12 = 31 / 12 7 / 8 - 3 / 4 = ( 7 - 6 ) / 8 = 1 / 8 so 31 / 12 / 1 / 8 = 31 / 12 * 8 = 62 / 3 answer - d"

a ) 87 / 36 , b ) 36 / 17 , c ) 7 / 6 , d ) 62 / 3 , e ) 5 / 4

d

subtract(divide(add(multiply(const_10, 7), 7), 4), divide(add(const_10, 4), 3))

add(n3,const_10)|multiply(const_10,n0)|add(n0,#1)|divide(#0,n5)|divide(#2,n2)|subtract(#4,#3)|

general

the mean daily profit made by a shopkeeper in a month of 30 days was rs . 350 . if the mean profit for the first fifteen days was rs . 225 , then the mean profit for the last 15 days would be

"average would be : 350 = ( 225 + x ) / 2 on solving , x = 475 . answer : c"

a ) rs . 200 , b ) rs . 350 , c ) rs . 475 , d ) rs . 425 , e ) none of these

c

divide(subtract(multiply(30, 350), multiply(15, 225)), 15)

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

gain

find out the c . i on rs . 8000 at 4 % p . a . compound half - yearly for 1 1 / 2 years

"a = 8000 ( 51 / 50 ) 3 = 8489.66 8000 - - - - - - - - - - - 489.66 answer : a"

a ) 489.66 , b ) 406.07 , c ) 406.04 , d ) 306.03 , e ) 306.01

a

subtract(multiply(8000, power(add(1, divide(4, const_100)), divide(const_3, 2))), 8000)

divide(n1,const_100)|divide(const_3,n4)|add(#0,n2)|power(#2,#1)|multiply(n0,#3)|subtract(#4,n0)|

general

jayes can eat 25 marshmallows is 20 minutes . dylan can eat 25 in one hour . in how much time will the two eat 150 marshmallows ?

rate = output / time jayes rate = 25 / 20 = 5 / 4 dylan rate = 25 / 60 = 5 / 12 combined rate = 5 / 4 + 5 / 12 = 20 / 12 combinedrate * combinedtime = combinedoutput 20 / 12 * t = 150 t = 90 mins = > 1 hr 30 min

a ) 40 minutes . , b ) 1 hour and 30 minutes . , c ) 1 hour . , d ) 1 hour and 40 minutes . , e ) 2 hours and 15 minutes .

c

divide(20, 20)

divide(n1,n1)

physics

how many even number in the range between 10 to 120 inclusive are not divisible by 3

"we have to find the number of terms that are divisible by 2 but not by 6 ( as the question asks for the even numbers only which are not divisible by 3 ) for 2 , 10 , 12,14 . . . 120 using ap formula , we can say 120 = 10 + ( n - 1 ) * 2 or n = 56 . for 6 , 12,18 , . . . 120 using ap formula , we can say 120 = 12 + ( n - 1 ) * 6 or n = 19 . hence , only divisible by 2 but not 3 = 56 - 19 = 37 . hence , answer d"

a ) 15 , b ) 30 , c ) 31 , d ) 37 , e ) 46

d

subtract(divide(subtract(subtract(120, 10), const_2), const_2), divide(divide(subtract(subtract(subtract(subtract(120, const_2), multiply(3, const_4)), 3), 3), 3), const_2))

multiply(n2,const_4)|subtract(n1,n0)|subtract(n1,const_2)|subtract(#1,const_2)|subtract(#2,#0)|divide(#3,const_2)|subtract(#4,n2)|subtract(#6,n2)|divide(#7,n2)|divide(#8,const_2)|subtract(#5,#9)|

general

in the biology lab of ` ` jefferson ' ' high school there are 0.037 * 10 ^ 5 germs , equally divided among 74000 * 10 ^ ( - 3 ) petri dishes . how many germs live happily in a single dish ?

0.037 * 10 ^ 5 can be written as 3700 74000 * 10 ^ ( - 3 ) can be written as 74 required = 3700 / 74 = 50 answer : e

a ) 10 , b ) 20 , c ) 30 , d ) 40 , e ) 50

e

divide(multiply(multiply(const_1000, const_100), 0.037), divide(74000, const_1000))

divide(n3,const_1000)|multiply(const_100,const_1000)|multiply(n0,#1)|divide(#2,#0)

general

the largest 4 digit number exactly divisible by 44 is ?

"largest 4 - digit number = 9999 44 ) 9999 ( 227 9988 largest number : 9988 answer : a"

a ) 9988 , b ) 9939 , c ) 9944 , d ) 9954 , e ) 9960

a

square_area(const_pi)

square_area(const_pi)|

general

if a certain number is divided by 2 , the quotient , dividend , and divisor , added together , will amount to 62 . what is the number ?

"let x = the number sought . then x / 2 + x + 2 = 62 . x = 40 . c"

a ) 18 , b ) 28 , c ) 40 , d ) 38 , e ) 59

c

divide(multiply(subtract(62, 2), 2), add(2, const_1))

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

general

a and b can do a work in 10 days and 15 days respectively . a starts the work and b joins him after 5 days . in how many days can they complete the remaining work ?

"work done by a in 5 days = 5 / 10 = 1 / 2 remaining work = 1 / 2 work done by both a and b in one day = 1 / 10 + 1 / 15 = 5 / 30 = 1 / 6 remaining work = 1 / 2 * 6 / 1 = 3 days . answer : e"

a ) 6 days , b ) 2 days , c ) 8 days , d ) 4 days , e ) 3 days

e

subtract(add(inverse(add(inverse(15), inverse(10))), 10), const_3)

inverse(n1)|inverse(n0)|add(#0,#1)|inverse(#2)|add(n0,#3)|subtract(#4,const_3)|

physics

for the symbol , m ” n = n ^ 2 − m for all values of m and n . what is the value of 4 ” 2 ?

"4 ” 2 = 4 - 4 = 0 answer : e"

a ) 5 , b ) 3 , c ) 2 , d ) 1 , e ) 0

e

subtract(power(2, 2), 4)

power(n2,n0)|subtract(#0,n1)|

general

in an office in singapore there are 60 % female employees . 50 % of all the male employees are computer literate . if there are total 62 % employees computer literate out of total 1100 employees , then the no . of female employees who are computer literate ?

"solution : total employees , = 1100 female employees , 60 % of 1100 . = ( 60 * 1100 ) / 100 = 660 . then male employees , = 440 50 % of male are computer literate , = 220 male computer literate . 62 % of total employees are computer literate , = ( 62 * 1100 ) / 100 = 682 computer literate . thus , female computer literate = 682 - 220 = 462 . answer : option a"

a ) 462 , b ) 674 , c ) 672 , d ) 960 , e ) none

a

multiply(subtract(divide(62, const_100), multiply(subtract(const_1, divide(60, const_100)), divide(50, const_100))), 1100)

divide(n2,const_100)|divide(n1,const_100)|divide(n0,const_100)|subtract(const_1,#2)|multiply(#1,#3)|subtract(#0,#4)|multiply(n3,#5)|

gain

total 48 matches are conducted in knockout match type . how many players will be participated in that tournament ?

"47 players answer : e"

a ) 48 , b ) 47 , c ) 40 , d ) 42 , e ) 47

e

add(48, const_1)

add(n0,const_1)|

general

in 130 m race , a covers the distance in 36 seconds and b in 45 seconds . in this race a beats b by :

"distance covered by b in 9 sec . = 130 / 45 x 9 m = 26 m . a beats b by 26 metres . answer : option d"

a ) 20 m , b ) 25 m , c ) 22.5 m , d ) 26 m , e ) 12 m

d

multiply(divide(130, 45), subtract(45, 36))

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

physics

at a certain university , 69 % of the professors are women , and 70 % of the professors are tenured . if 90 % of the professors are women , tenured , or both , then what percent of the men are tenured ?

"answer is 75 % total women = 69 % total men = 40 % total tenured = 70 % ( both men and women ) therefore , women tenured + women professors + men tenured = 90 % men tenured = 21 % but question wants to know the percent of men that are tenured 21 % / 40 % = 52.5 % d"

a ) 25 , b ) 37.5 , c ) 50 , d ) 52.5 , e ) 75

d

add(subtract(const_100, 69), subtract(90, 69))

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

gain

in a dairy farm , 26 cows eat 26 bags of husk in 26 days . in how many days one cow will eat one bag of husk ?

"explanation : one bag of husk = 26 cows per day ⇒ 26 × 1 × 26 = 1 × 26 × x for one cow = 26 days answer : d"

a ) 1 , b ) 40 , c ) 20 , d ) 26 , e ) 30

d

multiply(divide(26, 26), 26)

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

physics

the average age of a group of n people is 14 years old . one more person aged 34 joins the group and the new average is 16 years old . what is the value of n ?

"14 n + 34 = 16 ( n + 1 ) 2 n = 18 n = 9 the answer is c ."

a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11

c

divide(subtract(34, 16), subtract(16, 14))

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

general

the number 0.650 is how much greater than 1 / 8 ?

"let x be the difference then . 65 - 1 / 8 = x 65 / 100 - 1 / 8 = x x = 21 / 40 ans b"

a ) ½ , b ) 21 / 40 , c ) 1 / 50 , d ) 1 / 500 , e ) 2 / 500

b

subtract(0.650, divide(1, 8))

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

general

if $ 1092 are divided between worker a and worker b in the ratio 5 : 8 , what is the share that worker b will get ?

"worker b will get 8 / 13 = 61.54 % the answer is b ."

a ) 60.32 % , b ) 61.54 % , c ) 62.21 % , d ) 62.76 % , e ) 63.87 %

b

divide(1092, add(5, 8))

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

other

a card is drawn from a pack of 52 cards the probability of getting queen of club or aking of a heart ?

"here n ( s ) = 52 let e be the event of getting queen of the club or king of kings n ( e ) = 2 p ( e ) = n ( e ) / n ( s ) = 2 / 52 = 1 / 26 answer ( c )"

a ) 8 / 58 , b ) 9 / 25 , c ) 1 / 26 , d ) 6 / 25 , e ) 8 / 45

c

divide(subtract(52, multiply(const_4, const_4)), 52)

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

probability

1000 men have provisions for 17 days . if 500 more men join them , for how many days will the provisions last now ?

"1000 * 17 = 1500 * x x = 11.3 answer : c"

a ) 12.9 , b ) 12.5 , c ) 11.3 , d ) 12.2 , e ) 12.1

c

divide(multiply(17, 1000), add(1000, 500))

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

physics

a person travels equal distances with speeds of 3 km / hr , 4 km / hr and 5 km / hr and takes a total time of 47 minutes . the total distance is ?

"c 3 km let the total distance be 3 x km . then , x / 3 + x / 4 + x / 5 = 47 / 60 47 x / 60 = 47 / 60 = > x = 1 . total distance = 3 * 1 = 3 km ."

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

c

multiply(multiply(divide(divide(47, const_60), add(add(divide(const_1, 3), divide(const_1, 4)), divide(const_1, 5))), const_3), const_1000)

divide(n3,const_60)|divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#1,#2)|add(#4,#3)|divide(#0,#5)|multiply(#6,const_3)|multiply(#7,const_1000)|

physics

if x is an integer such that 0 < x < 7 , 0 < x < 15 , 5 > x > – 1 , 3 > x > 0 , and x + 2 < 4 , then x is

"0 < x < 7 , 0 < x < 15 , 5 > x > – 1 3 > x > 0 x < 2 from above : 0 < x < 2 - - > x = 1 . answer : a ."

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

a

subtract(subtract(5, 2), 2)

subtract(n4,n8)|subtract(#0,n8)|

general

an error 5 % in excess is made while measuring the side of a square . the percentage of error in the calculated area of the square is :

"explanation : 100 cm is read as 105 cm . a 1 = ( 100 × 100 ) cm 2 = 10000 and a 2 = ( 105 × 105 ) cm 2 = 10816 ( a 2 - a 1 ) = 11025 - 10000 = 1025 = > 1025 / 10000 * 100 = 10.25 answer : c"

a ) 10.01 , b ) 9.25 , c ) 10.25 , d ) 8.25 , e ) 6.25

c

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

1 + 1

e

a ) 9 , b ) 8 , c ) 3 , d ) 0 , e ) 2

e

multiply(divide(1, 1), const_100)

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

general

rahul can do a work in 3 days while rajesh can do the same work in 2 days . both of them finish the work together and get $ 150 . what is the share of rahul ?

"rahul ' s wages : rajesh ' s wages = 1 / 3 : 1 / 2 = 2 : 3 rahul ' s share = 150 * 2 / 5 = $ 60 answer is c"

a ) $ 50 , b ) $ 40 , c ) $ 60 , d ) $ 100 , e ) $ 90

c

multiply(divide(2, add(3, 2)), 150)

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

physics

a boat m leaves shore a and at the same time boat b leaves shore b . they move across the river . they met at 500 yards away from a and after that they met 300 yards away from shore b without halting at shores . find the distance between the shore a & b

if x is the distance , a is speed of a and b is speed of b , then ; 500 / a = ( x - 500 ) / b and ( x + 300 ) / a = ( 2 x - 300 ) / b , solving it , we get x = 1200 yards answer : b

a ) 1000 yards , b ) 1200 yards , c ) 1400 yards , d ) 1600 yards , e ) 1800 yards

b

multiply(300, const_4)

multiply(n1,const_4)

physics

a pineapple costs rs 7 each and a watermelon costs rs . 5 each . if i spend rs 38 on total what is the number of pineapple i purchased ?

explanation : the equation for this problem can be made as : 5 x + 7 y = 38 where x is the number of watermelons and y is the number of pineapples . now test for 2 , 3 and 4 : for y = 2 5 x + 14 = 38 x is not an integer for y = 3 5 x = 17 x not an integer for y = 4 x = 2 so 4 pineapples and 2 watermelons can be bought by 38 rs . answer : d

a ) 7 , b ) 6 , c ) 5 , d ) 2 , e ) 3

d

divide(subtract(38, multiply(const_4, 7)), 5)

multiply(n0,const_4)|subtract(n2,#0)|divide(#1,n1)

general

the sub - duplicate ratio of 25 : 16 is

"root ( 25 ) : root ( 16 ) = 5 : 4 answer : d"

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

d

divide(sqrt(25), sqrt(16))

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

other

in δ pqs above , if pq = 7 and ps = 8 , then

"there are two ways to calculate area of pqs . area remains same , so both are equal . 7 * 8 / 2 = pr * 9 / 2 pr = 56 / 9 c"

a ) 9 / 4 , b ) 12 / 5 , c ) 56 / 9 , d ) 15 / 4 , e ) 20 / 3

c

divide(8, 7)

divide(n1,n0)|

general

ajay can ride 50 km in 1 hour . in how many hours he can ride 1250 km ?

"1 hour he ride 50 km he ride 1250 km in = 1250 / 50 * 1 = 25 hours answer is d"

a ) 10 hrs , b ) 15 hrs , c ) 20 hrs , d ) 25 hrs , e ) 18 hrs

d

divide(1250, 50)

divide(n2,n0)|

physics

a boat covers a certain distance downstream in 1 hour , while it comes back in 11 ⠁ „ 2 hours . if the speed of the stream be 3 kmph , what is the speed of the boat in still water ?

"let the speed of the water in still water = x given that speed of the stream = 3 kmph speed downstream = ( x + 3 ) kmph speed upstream = ( x â ˆ ’ 3 ) kmph he travels a certain distance downstream in 1 hour and come back in 11 ⠁ „ 2 hour . i . e . , distance travelled downstream in 1 hour = distance travelled upstream in 11 ⠁ „ 2 hour since distance = speed ã — time , we have ( x + 3 ) ã — 1 = ( x - 3 ) 3 / 2 2 ( x + 3 ) = 3 ( x - 3 ) 2 x + 6 = 3 x - 9 x = 6 + 9 = 15 kmph answer : d"

a ) 31 kmph , b ) 16 kmph , c ) 19 kmph , d ) 15 kmph , e ) 14 kmph

d

divide(add(multiply(divide(const_3, const_2), const_3.0), 2), subtract(divide(const_3, const_2), 1))

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

physics

in a division , a student took 87 as divisor instead of 36 . his answer was 24 . the correct answer is -

x / 87 = 24 . x = 24 * 87 . so correct answer would be , ( 24 * 87 ) / 36 = 58 . answer : e

a ) 42 , b ) 32 , c ) 48 , d ) 28 , e ) 58

e

divide(multiply(87, 24), 36)

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

general

what is the sum of all digits for the number 10 ^ 26 - 51 ?

"10 ^ 26 is a 27 - digit number : 1 followed by 26 zeros . 10 ^ 26 - 51 is a 26 - digit number : 24 9 ' s and 49 at the end . the sum of the digits is 24 * 9 + 4 + 9 = 229 . the answer is d ."

a ) 199 , b ) 209 , c ) 219 , d ) 229 , e ) 239

d

multiply(add(divide(subtract(subtract(26, 10), const_2), const_2), 10), divide(add(subtract(26, 10), const_2), const_2))

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

general

what is the measure of the angle w made by the diagonals of the any adjacent sides of a cube .

"c . . 60 degrees all the diagonals are equal . if we take 3 touching sides and connect their diagonals , we form an equilateral triangle . therefore , each angle would be 60 . c"

a ) 30 , b ) 45 , c ) 60 , d ) 75 , e ) 90

c

divide(const_180, const_3)

divide(const_180,const_3)|

geometry

what is the compound interest paid on a sum of rs . 6000 for the period of 2 years at 10 % per annum .

solution = interest % for 1 st year = 10 interest % for 2 nd year = 10 + 10 % of 10 = 10 + 10 * 10 / 100 = 11 total % of interest = 10 + 11 = 21 total interest = 21 % 6000 = 6000 * ( 21 / 100 ) = 1260 answer a

a ) 1260 , b ) 1320 , c ) 1200 , d ) 1250 , e ) none of these

a

subtract(multiply(6000, power(add(const_1, divide(10, const_100)), const_2)), 6000)

divide(n2,const_100)|add(#0,const_1)|power(#1,const_2)|multiply(n0,#2)|subtract(#3,n0)

gain

a fair 2 sided coin is flipped 8 times . what is the probability that tails will be the result at least twice , but not more than 8 times ?

"at least twice , but not more than 8 timesmeans exactly 2 times , 3 times , 4 times , 5 times , 6 times , 7 times , 8 times the probability of getting exactly k results out of n flips is nck / 2 ^ n 8 c 2 / 2 ^ 8 + 8 c 3 / 2 ^ 8 + 8 c 4 / 2 ^ 8 + 8 c 5 / 2 ^ 8 + 8 c 6 / 2 ^ 8 + 8 c 7 / 2 ^ 8 = 143 / 128 option : e"

a ) 5 / 8 , b ) 3 / 4 , c ) 7 / 8 , d ) 157 / 64 , e ) 143 / 128

e

subtract(const_1, add(multiply(inverse(power(2, 8)), 8), add(inverse(power(2, 8)), inverse(power(2, 8)))))

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

general

the difference between c . i . and s . i . on an amount of rs . 15,000 for 2 years is rs . 54 . what is the rate of interest per annum ?

"explanation : [ 15000 * ( 1 + r / 100 ) 2 - 15000 ] - ( 15000 * r * 2 ) / 100 = 54 15000 [ ( 1 + r / 100 ) 2 - 1 - 2 r / 100 ] = 54 15000 [ ( 100 + r ) 2 - 10000 - 200 r ] / 10000 = 54 r 2 = ( 54 * 2 ) / 3 = 36 = > r = 6 rate = 6 % answer : option e"

a ) 8 , b ) 2 , c ) 9 , d ) 4 , e ) 6

e

sqrt(54)

sqrt(n2)|

gain

find the compound ratio of ( 1 : 2 ) , ( 2 : 3 ) and ( 3 : 4 ) is

"required ratio = 1 / 2 * 2 / 3 * 3 / 4 = 1 / 4 = 1 : 4 answer is d"

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

d

multiply(divide(1, 2), multiply(divide(1, 2), divide(2, 2)))

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

other

if 1 is added to the denominator of fraction , the fraction becomes 1 / 2 . if 1 is added to the numerator , the fraction becomes 1 . the fraction is

explanation : let the required fraction be a / b . then ⇒ a / b + 1 = 1 / 2 ⇒ 2 a – b = 1 - - - - ( 1 ) ⇒ a + 1 / b = 1 ⇒ a – b = – 1 ⇒ a – b = – 1 - - - - - - - ( 2 ) solving ( 1 ) & ( 2 ) we get a = 2 , b = 3 fraction = a / b = 2 / 3 correct option : c

a ) 4 / 7 , b ) 5 / 9 , c ) 2 / 3 , d ) 10 / 11 , e ) none of these

c

divide(add(1, 1), add(2, 1))

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

general

if the ratio of two number is 2 : 3 and lcm of the number is 120 then what is the number .

"product of two no = lcm * hcf 2 x * 3 x = 120 * x x = 20 answer : b"

a ) 15 , b ) 20 , c ) 25 , d ) 30 , e ) 35

b

divide(120, multiply(2, 3))

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

other

find out the c . i on rs . 4000 at 4 % p . a . compound half - yearly for 1 1 / 2 years

"a = 4000 ( 51 / 50 ) 3 = 4244.832 4000 - - - - - - - - - - - 244.83 answer : a"

a ) 244.83 , b ) 306.07 , c ) 306.04 , d ) 306.03 , e ) 306.01

a

subtract(multiply(4000, power(add(1, divide(4, const_100)), divide(const_3, 2))), 4000)

divide(n1,const_100)|divide(const_3,n4)|add(#0,n2)|power(#2,#1)|multiply(n0,#3)|subtract(#4,n0)|

general

bert and rebecca were looking at the price of a condominium . the price of the condominium was 40 % more than bert had in savings , and separately , the same price was also 75 % more than rebecca had in savings . what is the ratio of what bert has in savings to what rebecca has in savings .

suppose bert had 100 so price becomes 140 , this 140 = 1.75 times r ' s saving . . so r ' s saving becomes 80 so required ratio is 100 : 80 = 5 : 4 answer : e

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

e

divide(divide(const_100, add(const_100, 40)), divide(const_100, add(const_100, 75)))

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

gain

walking at 5 / 6 of her normal speed , a worker is 12 minutes later than usual in reaching her office . the usual time ( in minutes ) taken by her to cover the distance between her home and her office is

"let v be her normal speed and let t be her normal time . d = ( 5 / 6 ) v * ( t + 12 ) since the distance is the same we can equate this to a regular day which is d = v * t v * t = ( 5 / 6 ) v * ( t + 12 ) t / 6 = 10 t = 60 the answer is e ."

a ) 36 , b ) 42 , c ) 48 , d ) 54 , e ) 60

e

multiply(5, 12)

multiply(n0,n2)|

physics

if 12 lions can kill 12 deers in 12 minutes how long will it take 100 lions to kill 100 deers ?

"we can try the logic of time and work , our work is to kill the deers so 12 ( lions ) * 12 ( min ) / 12 ( deers ) = 100 ( lions ) * x ( min ) / 100 ( deers ) hence answer is x = 12 answer : b"

a ) 1 minutes , b ) 12 minute , c ) 120 minutes , d ) 10000 minutes , e ) 1000 minutes

b

multiply(divide(12, 12), 12)

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

physics

a man cycling along the road noticed that every 18 minutes a bus overtakes him and every 6 minutes he meets an oncoming bus . if all buses and the cyclist move at a constant speed , what is the time interval between consecutive buses ?

let ' s say the distance between the buses is d . we want to determine interval = \ frac { d } { b } , where b is the speed of bus . let the speed of cyclist be c . every 18 minutes a bus overtakes cyclist : \ frac { d } { b - c } = 18 , d = 18 b - 18 c ; every 6 minutes cyclist meets an oncoming bus : \ frac { d } { b + c } = 6 , d = 6 b + 6 c ; d = 18 b - 18 c = 6 b + 6 c , - - > b = 2 c , - - > d = 18 b - 9 b = 9 b . interval = \ frac { d } { b } = \ frac { 9 b } { b } = 9 answer : d ( 9 minutes ) .

a ) 5 minutes , b ) 6 minutes , c ) 8 minutes , d ) 9 minutes , e ) 10 minutes

d

divide(subtract(18, divide(18, divide(add(6, 18), subtract(18, 6)))), const_1)

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

physics

x starts a business with rs . 45000 . y joins in the business after 3 months with rs . 27000 . what will be the ratio in which they should share the profit at the end of the year ?

"ratio in which they should share the profit = ratio of the investments multiplied by the time period = 45000 × 12 : 27000 × 9 = 45 × 12 : 27 × 9 = 15 × 12 : 9 × 9 = 20 : 9 answer is a"

a ) 20 : 9 , b ) 2 : 1 , c ) 3 : 2 , d ) 2 : 3 , e ) 5 : 3

a

divide(multiply(45000, const_12), multiply(27000, add(const_4, const_3)))

add(const_3,const_4)|multiply(n0,const_12)|multiply(n2,#0)|divide(#1,#2)|

other

the hiker walking at a constant rate of 4 miles per hour is passed by a cyclist traveling in the same direction along the same path at 15 miles per hour . the cyclist stops to wait for the hiker 5 minutes after passing her , while the hiker continues to walk at her constant rate , how many minutes must the cyclist wait until the hiker catches up ?

"after passing the hiker the cyclist travels for 5 minutes at a rate of 15 miles / hour . in those 5 mins the cyclist travels a distance of 5 / 4 miles . in those 5 mins the hiker travels a distance of 1 / 3 miles . so the hiker still has to cover 11 / 12 miles to meet the waiting cyclist . the hiker will need 55 / 4 mins to cover the remaining 11 / 12 miles . so the answer is b ."

a ) 20 , b ) 55 / 4 , c ) 25 , d ) 14 , e ) 13

b

multiply(divide(multiply(subtract(15, 4), divide(5, const_60)), 4), const_60)

divide(n2,const_60)|subtract(n1,n0)|multiply(#0,#1)|divide(#2,n0)|multiply(#3,const_60)|

physics

if x + y = 10 , x - y = 18 , for integers of x and y , x = ?

x + y = 10 x - y = 18 2 x = 28 x = 14 answer is a

a ) 14 , b ) 15 , c ) 25 , d ) 13 , e ) 42

a

divide(add(10, 18), const_2)

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

general

wink , inc . follows a certain procedure that requires two tasks to be finished independently in order for a job to be done . on any given day , there is a 2 / 3 probability that task 1 will be completed on time , and a 3 / 5 probability that task 2 will be completed on time . on a certain day , what is the probability that task 1 will be completed on time , but task 2 will not ?

"p ( 1 and not 2 ) = 2 / 3 * ( 1 - 3 / 5 ) = 4 / 15 . answer : a ."

a ) 4 / 15 , b ) 3 / 40 , c ) 13 / 40 , d ) 7 / 20 , e ) 13 / 22

a

multiply(divide(2, 3), subtract(const_1, divide(3, 2)))

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

probability

find the least number must be subtracted from 964807 so that remaining no . is divisible by 8 ?

"on dividing 964807 by 15 we get the remainder 7 , so 7 should be subtracted a"

a ) 7 , b ) 6 , c ) 5 , d ) 4 , e ) 3

a

subtract(964807, multiply(floor(divide(964807, 8)), 8))

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

general

if 8 people take an hour to complete a piece of work , then how long should 16 people will take to complete the same piece of work ?

if 8 people take an hour to complete a piece of work , then 16 people will take 8 * 60 / 16 = 30 mins to complete the same piece of work . answer : b

a ) 15 min , b ) 30 min , c ) 26 min , d ) 28 min , e ) 18 min

b

multiply(divide(8, 16), const_60)

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

physics

find the area of a parallelogram with base 25 cm and height 15 cm ?

"area of a parallelogram = base * height = 25 * 15 = 375 cm 2 answer : d"

a ) 295 cm 2 , b ) 385 cm 2 , c ) 275 cm 2 , d ) 375 cm 2 , e ) 285 cm 2

d

multiply(25, 15)

multiply(n0,n1)|

geometry

when x is even , [ x ] = x / 2 + 1 , when x is odd [ x ] = 2 x + 1 then [ 6 ] * [ 4 ] = ?

"[ 6 ] * [ 4 ] = ( 6 / 2 + 1 ) ( 4 / 2 + 1 ) = [ 12 ] . ans - a"

a ) [ 12 ] , b ) [ 44 ] , c ) [ 45 ] , d ) [ 88 ] , e ) [ 90 ]

a

multiply(add(divide(6, 2), 1), add(multiply(2, 4), 1))

divide(n4,n0)|multiply(n0,n5)|add(#0,n1)|add(#1,n1)|multiply(#2,#3)|

general

in a certain company , a third of the workers do not have a retirement plan . 60 % of the workers who do not have a retirement plan are women , and 40 % of the workers who do have a retirement plan are men . if 120 of the workers of that company are men , how many of the workers are women ?

"set up equation : x = total number of workers 120 = 0,4 * 2 / 3 * x + 0,4 * 1 / 3 * x 120 = 12 / 30 x x = 300 300 - 120 = 180 answer c"

a ) 80 , b ) 95 , c ) 180 , d ) 120 , e ) 210

c

multiply(divide(120, add(subtract(divide(const_1, const_3), multiply(divide(const_1, const_3), divide(60, const_100))), multiply(subtract(const_1, divide(const_1, const_3)), divide(40, const_100)))), add(multiply(divide(const_1, const_3), divide(60, const_100)), subtract(subtract(const_1, divide(const_1, const_3)), multiply(subtract(const_1, divide(const_1, const_3)), divide(40, const_100)))))

divide(const_1,const_3)|divide(n0,const_100)|divide(n1,const_100)|multiply(#0,#1)|subtract(const_1,#0)|multiply(#2,#4)|subtract(#0,#3)|add(#5,#6)|subtract(#4,#5)|add(#3,#8)|divide(n2,#7)|multiply(#9,#10)|

gain

a bus started its journey from mumbai and reached pune in 65 min with its average speed of 60 km / hr . if the average speed of the bus is increased by 5 km / hr , how much time will it take to cover the same distance ?

sol . distance between ramgarh and devgarh = ( 60 * 65 ) / 60 = 60 average speed of the bus is increased by 5 km / hr then the speed of the bus = 60 km / hr required time = 60 / 60 = 1 hr c

a ) 2 hrs , b ) 3 hrs , c ) 1 hrs , d ) 1 1 / 2 hrs , e ) 2 1 / 2 hrs

c

divide(multiply(60, divide(65, 60)), add(60, 5))

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

general

the mean of 50 observations is 200 . but later he found that there is decrements of 47 from each observations . what is the the updated mean is ?

"153 answer is c"

a ) 165 , b ) 185 , c ) 153 , d ) 198 , e ) 199

c

subtract(200, 47)

subtract(n1,n2)|

general

a bag contains 4 red , 6 blue and 3 green balls . if 3 balls are picked at random , what is the probability that both are blue ?

"p ( both are red ) , = 6 c 3 / 13 c 3 = 10 / 143 c"

a ) 11 / 144 , b ) 12 / 455 , c ) 10 / 143 , d ) 11 / 125 , e ) 14 / 368

c

divide(choose(4, const_2.0), choose(add(add(4, 6), const_4.0), const_2.0))

add(n0,n1)|choose(n0,const_2.0)|add(const_4.0,#0)|choose(#2,const_2.0)|divide(#1,#3)|

other

what is 120 % of 13 / 24 of 360 ?

"120 % * 13 / 24 * 360 = 1.2 * 13 * 15 = 234 the answer is c ."

a ) 52 , b ) 117 , c ) 234 , d ) 312 , e ) 576

c

divide(multiply(120, add(add(multiply(multiply(add(const_3, const_2), const_2), multiply(multiply(const_3, const_4), const_100)), multiply(multiply(add(const_3, const_4), add(const_3, const_2)), multiply(add(const_3, const_2), const_2))), add(const_3, const_3))), const_100)

add(const_2,const_3)|add(const_3,const_4)|add(const_3,const_3)|multiply(const_3,const_4)|multiply(#0,const_2)|multiply(#3,const_100)|multiply(#1,#0)|multiply(#4,#5)|multiply(#6,#4)|add(#7,#8)|add(#9,#2)|multiply(n0,#10)|divide(#11,const_100)|

gain

if a farmer sells 10 of his chickens , his stock of feed will last for 4 more days than planned , but if he buys 15 more chickens , he will run out of feed 3 days earlier than planned . if no chickens are sold or bought , the farmer will be exactly on schedule . how many chickens does the farmer have ?

"say farmer has n chicken and he is good for d days . : - we have 3 equations given in question : - ( n - 10 ) * d + 4 = ( n + 15 ) * ( d - 3 ) = n * d solving these : ( you can solve 1 st and 3 rd and 2 nd and 3 rd together ) we get : 15 d - 3 n = 45 4 n - 10 d = 40 = > n = 35 ans d it is !"

a ) 12 , b ) 24 , c ) 48 , d ) 35 , e ) 60

d

multiply(add(const_2, 4), multiply(3, 4))

add(const_2,n1)|multiply(n1,n3)|multiply(#0,#1)|

general

how many 1 / 2 s are there in 37 1 / 2 ?

"required number = ( 75 / 2 ) / ( 1 / 2 ) = ( 75 / 2 x 2 / 1 ) = 75 answer : a"

a ) 75 , b ) 150 , c ) 300 , d ) 600 , e ) 700

a

divide(add(37, divide(1, 2)), divide(1, 2))

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

general

a , b , c rent a pasture . if a puts 10 oxen for 7 months , b puts 12 oxen for 5 months and c puts 15 oxen for 3 months for grazing and the rent of the pasture is rs . 175 , then how much amount should c pay as his share of rent ?

a : b : c = 10 * 7 : 12 * 5 : 15 * 3 = 2 * 7 : 12 * 1 : 3 * 3 = 14 : 12 : 9 amount that c should pay = 175 * ( 9 / 35 ) = 5 * 9 = 45 answer is b

a ) 23 , b ) 45 , c ) 15 , d ) 28 , e ) 18

b

multiply(3, 15)

multiply(n4,n5)

general

in a box of 12 pens , a total of 6 are defective . if a customer buys 2 pens selected at random from the box , what is the probability that neither pen will be defective ?

"method - 1 there are 9 fine pieces of pen and 6 defective in a lot of 12 pens i . e . probability of first pen not being defective = ( 6 / 12 ) i . e . probability of second pen not being defective = ( 5 / 11 ) [ 11 pen remaining with 5 defective remaining considering that first was defective ] probability of both pen being non - defective = ( 6 / 12 ) * ( 5 / 11 ) = 5 / 22 answer : option b"

a ) 1 / 6 , b ) 5 / 22 , c ) 6 / 11 , d ) 9 / 16 , e ) 3 / 4

b

multiply(divide(subtract(12, 6), 12), divide(subtract(subtract(12, 6), const_1), subtract(12, const_1)))

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

general

a part of certain sum of money is invested at 6 % per annum and the rest at 15 % per annum , if the interest earned in each case for the same period is equal , then ratio of the sums invested is ?

"15 : 6 = 5 : 2 answer : d"

a ) 4 : 6 , b ) 4 : 9 , c ) 4 : 3 , d ) 5 : 2 , e ) 4 : 2

d

multiply(divide(15, const_100), 6)

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

gain

if a randomly selected non - negative single digit integer is added to { 2 , 3 , 5 , 8 } . what is the probability that the median of the set will increase but the range still remains the same ?

we are selecting from non - negative single digit integers , so from { 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 } . these 10 digits represent the total number of outcomes . hence , the total number of outcomes is 10 . we need to find the probability that the median of the set will increase but the range still remains the same . the median of the set is ( 3 + 5 ) / 2 = 4 , thus the number selected must be 5 or greater . for the range to remain the same , the number must be between 2 and 8 inclusive . to satisfy both conditions , the number selected must be 5 , 6 , 7 , or 8 . the probability is 4 / 10 = 0.4 the answer is c .

a ) 0.2 , b ) 0.3 , c ) 0.4 , d ) 0.5 , e ) 0.6

c

divide(const_4, const_10)

divide(const_4,const_10)

general

on a sum of money , the s . i . for 2 years is $ 600 , while the c . i . is $ 630 , the rate of interest being the same in both the cases . the rate of interest is ?

"difference in c . i . and s . i for 2 years = $ 630 - $ 600 = $ 30 s . i for one year = $ 300 s . i . on $ 300 for 1 year = $ 30 rate = ( 100 * 30 ) / ( 300 ) = 10 % the answer is b ."

a ) 8 % , b ) 10 % , c ) 12 % , d ) 14 % , e ) 16 %

b

divide(multiply(const_100, subtract(630, 600)), divide(600, 2))

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

gain

there are two circles of different radii . the are of a square is 784 sq cm and its side is twice the radius of the larger circle . the radius of the larger circle is seven - third that of the smaller circle . find the circumference of the smaller circle .

"let the radii of the larger and the smaller circles be l cm and s cm respectively . let the side of the square be a cm . a 2 = 784 = ( 4 ) ( 196 ) = ( 22 ) . ( 142 ) a = ( 2 ) ( 14 ) = 28 a = 2 l , l = a / 2 = 14 l = ( 7 / 3 ) s therefore s = ( 3 / 7 ) ( l ) = 6 circumference of the smaller circle = 2 ∏ s = 12 ∏ cm . answer : c"

a ) 165 ∏ cm , b ) 65 ∏ cm , c ) 12 ∏ cm , d ) 14 ∏ cm , e ) 16 ∏ cm

c

add(divide(divide(square_edge_by_area(784), const_2), divide(add(const_3, const_4), const_3)), const_2)

add(const_3,const_4)|square_edge_by_area(n0)|divide(#1,const_2)|divide(#0,const_3)|divide(#2,#3)|add(#4,const_2)|

geometry

y and z walk around a circular track . they start at 5 a . m from the same point in the opposite directions . y and z walk at a speed of 3 rounds per hour and 4 rounds per hour respectively . how many times shall they cross each other before 7 a . m

explanation : relative speed = ( 3 + 4 ) = 9 rounds per hour so , they cross each other 9 times in an hour hence , they cross 18 times before 7 a . m answer : option e

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

e

divide(const_60, add(divide(divide(const_60, 4), const_10), divide(divide(const_60, 3), const_10)))

divide(const_60,n2)|divide(const_60,n1)|divide(#0,const_10)|divide(#1,const_10)|add(#2,#3)|divide(const_60,#4)

physics

a box contains 10 black , 5 red and 4 green marbles . 3 marbles are drawn from the box at random . what is the probability that all the three marbles are of the same color ?

"explanation : total marbles in a box = 10 black + 5 red + 4 green marbles = 19 marbles 3 marbles are drawn from 19 marbles at random . therefore , n ( s ) = 19 c 3 = 969 ways let a be the event that 2 marbles drawn at random are of the same color . number of cases favorable to the event a is n ( a ) = 10 c 3 + 5 c 3 + 4 c 3 = 120 + 10 + 4 = 134 therefore , by definition of probability of event a , p ( a ) = n ( a ) / n ( s ) = 134 / 969 answer : b"

a ) 143 / 969 , b ) 134 / 969 , c ) 120 / 969 , d ) 19 / 969 , e ) 120 / 134

b

divide(add(add(5, 4), add(const_10, add(5, 3))), multiply(add(const_10, 5), 4))

add(n1,n2)|add(n1,n3)|add(const_10,n1)|add(#1,const_10)|multiply(n2,#2)|add(#0,#3)|divide(#5,#4)|

probability