Datasets:

math_qa

like 67

8 people decided to split the restaurant bill evenly . if the bill was $ 214.15 dollars , how much money did they 1 cent is the smallest unit ?

"if the last three digits of a whole number are divisible by 8 , then the entire number is divisible by 8 the last 3 digit 415 not divisible by a hence , we need to add 1 to this number for it to be divisible by 8 correct option : a"

a ) $ 214.16 , b ) $ 214.17 , c ) $ 214.18 , d ) $ 214.19 , e ) $ 214.20

a

add(214.15, divide(const_3, const_100))

divide(const_3,const_100)|add(n1,#0)|

general

in a fuel station the service costs $ 1.50 per car , every liter of fuel costs 0.35 $ . assuming that you own 3 limos and 2 fleet vans and all fuel tanks are empty . how much will it cost to fuel all cars together if a limo tank is 32 liters and an fleet van tank is 75 % bigger ?

"lots of calculations . 1.50 * 4 + 3 * . 35 * 32 + 2 * ( 7 / 4 ) * 32 * . 35 answer = $ 78.80 the correct option is a"

a ) $ 78.80 , b ) $ 79.80 , c ) $ 78.90 , d ) $ 79.90 , e ) $ 77.80

a

multiply(multiply(0.35, 2), 3)

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

general

a soccer store typically sells replica jerseys at a discount of 30 percent to 50 percent off list price . during the annual summer sale , everything in the store is an additional 20 percent off the original list price . if a replica jersey ' s list price is $ 80 , approximately what y percent of the list price is the lowest possible sale price ?

"let the list price be 2 x for min sale price , the first discount given should be 50 % , 2 x becomes x here now , during summer sale additional 20 % off is given ie sale price becomes 0.8 x it is given lise price is $ 80 = > 2 x = 80 = > x = 40 and 0.8 x = 32 so lowest sale price is 32 , which y is 40 % of 80 hence , d is the answer"

a ) 20 , b ) 25 , c ) 30 , d ) 40 , e ) 50

d

divide(80, const_2)

divide(n3,const_2)|

general

each year for 4 years , a farmer increased the number of trees in a certain orchard by 1 / 4 of the number of trees in the orchard of the preceding year . if all of the trees thrived and there were 12500 trees in the orchard at the end of 4 year period , how many trees were in the orchard at the beginning of the 4 year period .

"trees increase by 1 / 4 the number of trees in preceding year . hence , correct answer must be divisible by 4 . based on divisibility rules , if last 2 digits are divisible by 4 then the number is divisible by 4 . thus , we can eliminate a , b , d , e the answer to be c again , trees increase by 1 / 4 the number of trees in preceding year . hence , the number of trees increase by 5 / 4 times the number of trees the preceding year . if x = initial number of trees = 5120 year 1 = 5 / 4 x year 2 = ( 5 / 4 ) ( 5 / 4 ) x year 3 = ( 5 / 4 ) ( 5 / 4 ) ( 5 / 4 ) x year 4 = ( 5 / 4 ) ( 5 / 4 ) ( 5 / 4 ) ( 5 / 4 ) x only for answer d : ( 5 / 4 ) ( 5 / 4 ) ( 5 / 4 ) ( 5 / 4 ) 5120 = 12500 hence , correct answer = c"

a ) 5113 , b ) 5117 , c ) 5120 , d ) 8119 , e ) 10115

c

divide(12500, power(add(divide(1, 4), 1), 4))

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

general

a and b began business with rs . 3000 and rs . 4000 after 8 months , a withdraws rs . 1000 and b advances rs . 1000.00001 more . at the end of the year , their profits amounted to rs . 714 find the share of a .

explanation : ( 3 * 8 + 2 * 4 ) : ( 4 * 8 + 5 * 4 ) 8 : 13 8 / 21 * 714 = 272 answer : b

a ) 240 , b ) 272 , c ) 379 , d ) 277 , e ) 122

b

multiply(divide(714, add(add(multiply(3000, 8), multiply(subtract(3000, 1000), subtract(const_12, 8))), add(multiply(4000, 8), multiply(add(4000, 1000), subtract(const_12, 8))))), add(multiply(3000, 8), multiply(subtract(3000, 1000), subtract(const_12, 8))))

add(n1,n3)|multiply(n0,n2)|multiply(n1,n2)|subtract(n0,n3)|subtract(const_12,n2)|multiply(#3,#4)|multiply(#0,#4)|add(#1,#5)|add(#2,#6)|add(#7,#8)|divide(n5,#9)|multiply(#7,#10)

gain

a train 120 m long running at 60 kmph crosses a platform in 35 sec . what is the length of the platform ?

"d = 60 * 5 / 18 = 35 = 583 – 120 = 463 answer : c"

a ) 338 , b ) 277 , c ) 463 , d ) 456 , e ) 271

c

subtract(multiply(35, multiply(60, const_0_2778)), 120)

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

physics

jar a has 6 % more marbles than jar b . what percent of marbles from jar a need to be moved into jar b so that both jars have equal marbles ?

"an easy way to solve this question is by number plugging . assume there are 100 marbles in jar b then in jar a there will be 106 marbles . now , for both jars to have equal marbles we should move 3 marbles from a to b , which is 3 / 106 = ~ 2.8 % of a . answer : a ."

a ) 2.8 % , b ) 3.0 % , c ) 3.2 % , d ) 3.4 % , e ) 3.6 %

a

multiply(divide(divide(6, const_2), add(6, const_100)), const_100)

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

gain

for any integer n greater than 1 , n * denotes the product of all the integers from 1 to n , inclusive . how many prime numbers t are there between 6 * + 2 and 6 * + 6 , inclusive ?

"given that n * denotes the product of all the integers from 1 to n , inclusive so , 6 * + 2 = 6 ! + 2 and 6 * + 6 = 6 ! + 6 . now , notice that we can factor out 2 our of 6 ! + 2 so it can not be a prime number , we can factor out 3 our of 6 ! + 3 so it can not be a prime number , we can factor out 4 our of 6 ! + 4 so it can not be a prime number , . . . the same way for all numbers between 6 * + 2 = 6 ! + 2 and 6 * + 6 = 6 ! + 6 , inclusive . which means that there are no primes t in this range . answer : a ."

a ) none , b ) one , c ) two , d ) three , e ) four

a

divide(add(factorial(6), 6), add(factorial(6), 6))

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

general

shawn invested one half of his savings in a bond that paid simple interest for 2 years and received $ 400 as interest . he invested the remaining in a bond that paid compound interest , interest being compounded annually , for the same 2 years at the same rate of interest and received $ 605 as interest . what was the value of his total savings before investing in these two bonds ?

"so , we know that shawn received 20 % of the amount he invested in a year . we also know that in one year shawn received $ 200 , thus 0.2 x = $ 200 - - > x = $ 1,000 . since , he invested equal sums in his 2 bonds , then his total savings before investing was 2 * $ 1,000 = $ 2,000 . answer : c"

a ) 3000 , b ) 5000 , c ) 2000 , d ) 4000 , e ) 6000

c

multiply(divide(multiply(divide(400, 2), divide(400, 2)), subtract(605, 400)), 2)

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

gain

a tank is filled to one quarter of its capacity with a mixture consisting of water and sodium chloride . the proportion of sodium chloride in the tank is 40 % by volume and the capacity of the tank is 24 gallons . if the water evaporates from the tank at the rate of 0.5 gallons per hour , and the amount of sodium chloride stays the same , what will be the concentration of water in the mixture in 4 hours ?

"the number of gallons in the tank is ( 1 / 4 ) 24 = 6 gallons the amount of sodium chloride is 0.4 ( 6 ) = 2.4 gallons at the start , the amount of water is 0.6 ( 6 ) = 3.6 gallons after 4 hours , the amount of water is 3.6 - 0.5 ( 4 ) = 1.6 gallons the concentration of water is 1.6 / ( 2.4 + 1.6 ) = 40 % the answer is a ."

a ) 40 % , b ) 44 % , c ) 48 % , d ) 52 % , e ) 56 %

a

multiply(divide(subtract(divide(multiply(4, subtract(const_100, 40)), const_100), multiply(0.5, 4)), subtract(4, multiply(0.5, 4))), const_100)

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

gain

what number has a 5 : 1 ratio to the number 10 ?

"answer : option b explanation : 5 : 1 = x : 10 x = 50 answer : option b"

a ) 74 , b ) 50 , c ) 94 , d ) 59 , e ) 48

b

multiply(10, 5)

multiply(n0,n2)|

other

a 125 meter long train crosses a man standing on the platform in 5 sec . what is the speed of the train ?

"s = 125 / 5 * 18 / 5 = 90 kmph answer : e"

a ) 229 , b ) 108 , c ) 278 , d ) 126 , e ) 90

e

multiply(divide(125, 5), const_3_6)

divide(n0,n1)|multiply(#0,const_3_6)|

physics

excluding stoppages , the speed of a bus is 80 km / hr and including stoppages , it is 70 km / hr . for how many minutes does the bus stop per hour ?

"due to stoppages , it covers 10 km less . time taken to cover 10 km = 10 / 80 * 60 = 8 min . answer : c"

a ) 11 min , b ) 10 min , c ) 8 min , d ) 6 min , e ) 5 min

c

multiply(const_60, divide(subtract(80, 70), 80))

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

physics

a , b and c can do a piece of work in 7 days , 14 days and 28 days respectively . how long will they take to finish the work , if all the three work together ?

"1 / 7 + 1 / 14 + 1 / 28 = 7 / 28 = 1 / 4 all three can finish the work in 4 days answer : a"

a ) 4 , b ) 9 , c ) 2 , d ) 11 , e ) none

a

inverse(add(inverse(28), add(inverse(7), inverse(14))))

inverse(n0)|inverse(n1)|inverse(n2)|add(#0,#1)|add(#3,#2)|inverse(#4)|

physics

in the youth summer village there are 200 people , 100 of them are not working , 75 of them have families and 125 of them like to sing in the shower . what is the largest possible number of people in the village , which are working , that do n ' t have families and that are singing in the shower ?

total = 200 not working = 100 having family = 75 like to sing in shower = 125 working = 200 - 100 = 100 not having family = 200 - 75 = 125 like to sing in shower = 125 largest possible number is the lowest possible among the above thus 100 c

a ) 125 , b ) 150 , c ) 100 , d ) 130 , e ) 140

c

subtract(add(add(100, 75), 125), 200)

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

general

nitin ranks 18 th in a class of 49 students . what is rank from the last ?

"explanation : number students behind the nitin in rank = ( 49 - 18 ) = 31 nitin is 32 nd from the last answer : c ) 32"

a ) 33 , b ) 38 , c ) 32 , d ) 28 , e ) 19

c

subtract(49, 18)

subtract(n1,n0)|

other

lamp a flashes every 6 seconds , lamp b flashes every 8 seconds , lamp c flashes every 10 seconds . at a certain instant of time all 3 lamps flash simultaneously . during the period of 6 minutes after that how many times will exactly two lamps flash ? ( please include any flash of exactly two lights which occurs at the 6 minute mark . )

6 minutes is 360 seconds . lamp a and lamp b will flash together every 24 seconds . 360 / 24 = 15 . in the time period , lamp a and lamp b will flash together 15 times . lamp a and lamp c will flash together every 30 seconds . 360 / 30 = 12 . in the time period , lamp a and lamp c will flash together 12 times . lamp b and lamp c will flash together every 40 seconds . 360 / 40 = 9 . in the time period , lamp b and lamp c will flash together 9 times . all three lights will flash together every 2 * 2 * 2 * 3 * 5 = 120 seconds . 360 / 120 = 3 . we have counted these triple flashes three times , so we need to subtract three times the number of times that all three lights flash together . the number of times that exactly two lights flash together is 15 + 12 + 9 - 9 = 27 times . the answer is d .

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

d

subtract(add(add(divide(multiply(6, const_60), lcm(6, 8)), divide(multiply(6, const_60), lcm(6, 10))), divide(multiply(6, const_60), lcm(8, 10))), multiply(divide(multiply(6, const_60), lcm(lcm(6, 8), 10)), 3))

lcm(n0,n1)|lcm(n0,n2)|lcm(n1,n2)|multiply(n0,const_60)|divide(#3,#0)|divide(#3,#1)|divide(#3,#2)|lcm(n2,#0)|add(#4,#5)|divide(#3,#7)|add(#8,#6)|multiply(n3,#9)|subtract(#10,#11)

physics

when positive integer x is divided by positive integer y , the result is 59.32 . what is the sum q of all possible 2 - digit remainders for x / y ?

"ans b 616 . . . remainders = . 32 = 32 / 100 = 8 / 25 = 16 / 50 and so on . . so two digit remainders are 16 + 24 + 32 + . . . . + 96 . . q = 8 ( 2 + 3 + 4 . . . . + 12 ) = 616"

a ) 560 , b ) 616 , c ) 672 , d ) 728 , e ) 784

b

divide(59.32, subtract(2, floor(2)))

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

general

a man walking at a rate of 10 km / hr crosses a bridge in 12 minutes . the length of the bridge is ?

speed = 10 * 5 / 18 = 50 / 18 m / sec distance covered in 10 minutes = 50 / 18 * 12 * 60 = 2000 m answer is a

a ) 2000 , b ) 1492 , c ) 1667 , d ) 1254 , e ) 1112

a

multiply(divide(multiply(10, const_1000), const_60), 12)

multiply(n0,const_1000)|divide(#0,const_60)|multiply(n1,#1)

gain

there are 1000 students in a school and among them 30 % of them attends chess class . 10 % of the students who are in the chess class are also enrolled for swimming . no other students are interested in swimming so how many will attend the swimming class if all enrolled attends ?

30 % of 1000 gives 300 . so 300 attends chess and 10 % of 300 gives 30 . so 30 enrolled for swimming answer : d

a ) 1 , b ) 10 , c ) 100 , d ) 30 , e ) 20

d

divide(multiply(divide(multiply(30, 1000), const_100), 10), const_100)

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

gain

a computer store offers employees a 10 % discount off the retail price . if the store purchased a computer from the manufacturer for $ 800 dollars and marked up the price 10 % to the final retail price , how much would an employee save if he purchased the computer at the employee discount ( 10 % off retail price ) as opposed to the final retail price .

"cost price = 800 profit = 10 % = 10 % of 800 = 80 selling price = cp + profit sp = 880 a discount of 10 % to employees means 10 % off on 880 so 10 % of 880 = 88 ans b"

a ) 86 , b ) 88 , c ) 90 , d ) 92 , e ) 94

b

divide(add(divide(multiply(800, 10), const_100), 800), multiply(divide(800, const_100), const_2))

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

gain

exactly 15 % of the reporters for a certain wire service cover local politics in country x . if 25 % of the reporters who cover politics for the wire service do not cover local politics in country x , what percent of the reporters for the wire service do not cover politics ?

"let ' s assume there are 100 reporters - - > 15 reporters cover local politics . now , as 25 % of the reporters who cover all politics do not cover local politics then the rest 75 % of the reporters who cover politics do cover local politics , so if there are x reporters who cover politics then 75 % of them equal to 15 ( # of reporters who cover local politics ) : 0.75 x = 15 - - > x = 20 , hence 20 reporters cover politics and the rest 100 - 20 = 80 reporters do not cover politics at all . answer : d ."

a ) 20 % , b ) 42 % , c ) 44 % , d ) 80 % , e ) 84 %

d

multiply(subtract(const_1, divide(15, subtract(const_100, 25))), const_100)

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

gain

two numbers are respectively 50 % and 20 % more than a third number . the percentage that is first of the second is ?

"i ii iii 150 120 100 120 - - - - - - - - - - 150 100 - - - - - - - - - - - ? = > 125 % answer : a"

a ) 125 % , b ) 97 % , c ) 118 % , d ) 52 % , e ) 83 %

a

subtract(const_100, multiply(divide(add(50, const_100), add(20, const_100)), const_100))

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

gain

of 70 players on a football team , 43 are throwers . the rest of the team is divided so one third are left - handed and the rest are right handed . assuming that all throwers are right handed , how many right - handed players are there total ?

"total = 70 thrower = 43 rest = 70 - 43 = 27 left handed = 27 / 3 = 9 right handed = 18 if all thrower are right handed then total right handed is 43 + 18 = 61 so c . 61 is the right answer"

a ) 54 , b ) 59 , c ) 61 , d ) 71 , e ) 92

c

add(multiply(subtract(const_1, divide(const_1, const_3)), subtract(70, 43)), 43)

divide(const_1,const_3)|subtract(n0,n1)|subtract(const_1,#0)|multiply(#2,#1)|add(n1,#3)|

general

in how many seconds will a train 100 meters long pass an oak tree , if the speed of the train is 36 km / hr ?

"speed = 36 * 5 / 18 = 10 m / s time = 100 / 10 = 10 seconds the answer is c ."

a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 14

c

divide(100, multiply(const_0_2778, 36))

multiply(n1,const_0_2778)|divide(n0,#0)|

physics

a pair of articles was bought for $ 1000 at a discount of 10 % . what must be the marked price of each of the article ?

"s . p . of each of the article = 1000 / 2 = $ 500 let m . p = $ x 90 % of x = 500 x = 500 * 100 / 90 = $ 555.55 answer is a"

a ) $ 555.55 , b ) $ 500 , c ) $ 350 , d ) $ 400 , e ) $ 600

a

divide(multiply(subtract(const_100, 10), divide(1000, const_2)), const_100)

divide(n0,const_2)|subtract(const_100,n1)|multiply(#0,#1)|divide(#2,const_100)|

gain

what is 15 % of 2 / 3 of 0.5 ?

"the best way to solve these questions is to convert every term into fraction ( 15 / 100 ) * ( 2 / 3 ) * ( 5 / 10 ) = 150 / 3000 = 0.05 option a"

a ) 0.05 , b ) 0.9 , c ) 9 , d ) 90 , e ) none of the above

a

divide(multiply(15, 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

the bus fare for two persons for travelling between agra and aligarh id 4 - thirds the train fare between the same places for one person . the total fare paid by 6 persons travelling by bus and 8 persons travelling by train between the two places is rs . 1512 . find the train fare between the two places for one person ?

let the train fare between the two places for one person be rs . t bus fare between the two places for two persons rs . 4 / 3 t = > 6 / 2 ( 4 / 3 t ) + 8 ( t ) = 1512 = > 12 t = 1512 = > t = 126 . answer : a

a ) rs . 126 , b ) rs . 132 , c ) rs . 120 , d ) rs . 114 , e ) none of these

a

divide(1512, add(multiply(divide(6, const_2), divide(4, const_3)), 8))

divide(n1,const_2)|divide(n0,const_3)|multiply(#0,#1)|add(n2,#2)|divide(n3,#3)

general

a monkey ascends a greased pole 17 metres high . he ascends 2 metres in first minute and slips down 1 metre in the alternate minute . in which minute , he reaches the top ?

"in 2 minutes , he ascends = 1 metre â ˆ ´ 15 metres , he ascends in 30 minutes . â ˆ ´ he reaches the top in 31 st minute . answer a"

a ) 31 st , b ) 22 nd , c ) 23 rd , d ) 24 th , e ) none of these

a

subtract(multiply(2, 17), 1)

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

physics

a goods bullet train runs at the speed of 72 km / hr and crosses a 250 m long platform in 26 seconds . what is the length of the goods bullet train ?

e 270 m

a ) 220 m , b ) 250 m , c ) 280 m , d ) 210 m , e ) 270 m

e

subtract(multiply(multiply(72, const_0_2778), 26), 250)

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

physics

a shopkeeper labeled the price of his articles so as to earn a profit of 40 % on the cost price . he then sold the articles by offering a discount of 10 % on the labeled price . what is the actual percent profit earned in the deal ?

"explanation : let the cp of the article = rs . 100 . then labeled price = rs . 140 . sp = rs . 140 - 10 % of 140 = rs . 140 - 14 = rs . 126 . gain = rs . 126 â € “ rs . 100 = rs . 26 therefore , gain / profit percent = 26 % . answer : option a"

a ) 26 % , b ) 20 % , c ) 17 % , d ) 18 % , e ) none of these

a

subtract(subtract(add(const_100, 40), multiply(add(const_100, 40), divide(10, const_100))), const_100)

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

gain

what is the tens digit of 36 ^ 5 ?

"36 ^ 5 = 6 ^ 7 ( 6 ^ 2 ) = 6 * 6 = 36 ( 6 ^ 3 ) = 36 * 6 = . 16 ( 6 ^ 4 ) = . 16 * 6 = . . 96 ( 6 ^ 5 ) = . . 96 * 6 = . . 76 ( 6 ^ 6 ) = . . 76 * 6 = . . . 56 ( 6 ^ 7 ) = . . . . 56 * 6 = . . . . 36 if you see there is a pattern here in tens digits 3 , 1,9 , 7,5 , 3,1 and so on . . . continue the pattern up to 6 ^ 7 ( dont actually calculate full values ) and answer is d : 3"

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

d

floor(divide(reminder(power(36, reminder(5, add(const_4, const_1))), const_100), const_10))

add(const_1,const_4)|reminder(n1,#0)|power(n0,#1)|reminder(#2,const_100)|divide(#3,const_10)|floor(#4)|

general

evaluate : | 6 - 8 ( 3 - 12 ) | - | 5 - 11 | = ?

"according to order of operations , inner brackets first . hence | 6 - 8 ( 3 - 12 ) | - | 5 - 11 | = | 6 - 8 * ( - 9 ) | - | 5 - 11 | according to order of operations , multiplication within absolute value signs ( which may be considered as brackets when it comes to order of operations ) next . hence = | 6 + 72 | - | 5 - 11 | = | 78 | - | - 6 | = 78 - 6 = 72 correct answer c ) 72"

a ) 40 , b ) 50 , c ) 72 , d ) 70 , e ) 80

c

subtract(subtract(6, multiply(8, subtract(3, 12))), negate(subtract(5, 11)))

subtract(n2,n3)|subtract(n4,n5)|multiply(n1,#0)|negate(#1)|subtract(n0,#2)|subtract(#4,#3)|

general

find the simple interest on rs . 500 for 9 months at 6 paisa per month ?

"i = ( 500 * 9 * 6 ) / 100 = 270 answer : b"

a ) 287 , b ) 270 , c ) 276 , d ) 129 , e ) 211

b

multiply(500, divide(9, const_100))

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

gain

evaluate : 60 - 12 ÷ 4 × 2 =

"according to order of operations , 12 ÷ 4 × 2 ( division and multiplication ) is done first from left to right 12 ÷ 4 × 2 = 3 × 2 = 6 hence 60 - 12 ÷ 4 × 2 = 60 - 6 = 54 correct answer is b ) 54"

a ) a ) 45 , b ) b ) 54 , c ) c ) 63 , d ) d ) 72 , e ) e ) 81

b

subtract(60, multiply(multiply(12, 4), 2))

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

general

on a trip , a cyclist averaged 11 miles per hour for the first 22 miles and 10 miles per hour for the remaining 20 miles . if the cyclist returned immediately via the same route and took a total of 9 hours for the round trip , what was the average speed ( in miles per hour ) for the return trip ?

the time to go 42 miles was 22 / 11 + 20 / 10 = 2 + 2 = 4 hours . the average speed for the return trip was 42 miles / 5 hours = 8.4 mph . the answer is e .

a ) 7.6 , b ) 7.8 , c ) 8 , d ) 8.2 , e ) 8.4

e

divide(add(22, 20), subtract(9, add(divide(22, 11), divide(20, 10))))

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

physics

on increasing the number of lines in a page by 100 , they become 240 . what is the % of increase in the no . of lines in the page ?

"explanation : number of pages increased = 100 now , the number of pages of book = 240 number of pages of the books before increase = 240 – 100 = 140 % increase in the number of pages in the book = 100 / 140 x 100 % = 71.4 % d"

a ) 20 % , b ) 305 , c ) 50 % , d ) 71.4 % , e ) 60 %

d

subtract(multiply(divide(240, subtract(240, 100)), const_100), const_100)

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

general

find the sum 3 / 10 + 5 / 100 + 8 / 1000 in decimal form ?

answer 3 / 10 + 5 / 100 + 8 / 1000 = 0.3 + 0.05 + 0.008 = 0.358 correct option : b

a ) 0.853 , b ) 0.358 , c ) 3.58 , d ) 8.35 , e ) none

b

add(divide(8, 1000), add(divide(3, 10), divide(5, 100)))

divide(n0,n1)|divide(n2,n3)|divide(n4,n5)|add(#0,#1)|add(#3,#2)

general

3 distinct single digit no a , b , c are in g . p . if abs ( x ) for real x is the absolute value of x ( x if x is positive or zero , and x if x is negative ) , then the no . of different possible values of abs ( a + b - c ) is

if a = 1 , r = 2 then a = 1 , b = 2 , c = 4 then abs ( a + b - c ) = 1 if a = 1 , r = 3 then a = 1 , a = 3 , a = 9 then abs ( 1 + 3 - 9 ) = 5 if a = 2 , r = 2 , then a = 2 , b = 4 , c = 8 then abs ( 2 + 4 - 8 ) = 2 if a = 1 , r = - 2 then a = 1 , b = - 2 , c = 4 the abs ( 1 - 2 - 4 ) = 5 if a = 1 , r = - 3 then a = 1 , b = - 3 , c = 9 then abs ( 1 - 3 - 9 ) = 11 if a = 2 , r = - 2 then a = 2 , b = - 4 , c = - 8 then abs ( 2 - 4 - 8 ) = 10 so total 5 abs ( ) values answer : d

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

d

add(3, const_2)

add(n0,const_2)

general

sum of 19 odd numbers is ?

"sum of 1 st n odd no . s = 1 + 3 + 5 + 7 + . . . = n ^ 2 so , sum of 1 st 19 odd numbers = 19 ^ 2 = 361 answer : c"

a ) 341 , b ) 351 , c ) 361 , d ) 371 , e ) 381

c

multiply(multiply(19, const_2), divide(19, const_2))

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

general

an engine moves at the speed of 90 kmph without any coaches attached to it . speed of the train reduces at the rate that varies directly as the square root of the number of coaches attached . when 9 coaches are attached speed decreases to 78 kmph . what will be the speed of train when 25 coaches are attached .

"1 . no . of coaches = 9 sqr root = 3 speed decreases by 12 12 = k * 3 k = 4 no . of coaches = 25 swr root = 5 decrease = 5 * 4 = 20 new speed = 90 - 20 = 70 e"

a ) 90 , b ) 85 , c ) 80 , d ) 60 , e ) 70

e

subtract(90, multiply(sqrt(25), divide(subtract(90, 78), sqrt(9))))

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

physics

a train passes a man standing on a platform in 8 seconds and also crosses the platform which is 264 metres long in 20 seconds . the length of the train ( in metres ) is :

"explanation : let the length of train be l m . acc . to question ( 264 + l ) / 20 = l / 8 2112 + 8 l = 20 l l = 2112 / 12 = 176 m answer b"

a ) 188 , b ) 176 , c ) 175 , d ) 96 , e ) none of these

b

multiply(divide(264, subtract(20, 8)), 8)

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

physics

a certain car traveled twice as many miles from town a to town b as it did from town b to town c . from town a to town b , the car averaged 10 miles per gallon , and from town b to town c , the car averaged 12 miles per gallon . what is the average miles per gallon that the car achieved on its trip from town a through town b to town c ?

"step 1 ) took lcm of 10 and 12 . . came as 30 . just multiplied by 10 . . . ( to make easy calculation ) step 2 ) 300 distance between b to c . . . do 300 / 12 hence 25 gallons used step 3 ) twice distance . . hence 300 * 2 = 600 . . . do as above . . 600 / 10 = 60 gallons used step 4 ) total gallons . . 25 + 60 = 85 gallons step ) total miles = 300 + 600 = 900 miles hence . . average of whole journey = 900 / 85 which comes to 10.6 answer : d"

a ) 11.5 , b ) 9.5 , c ) 13.5 , d ) 10.6 , e ) 14.5

d

divide(add(multiply(12, const_10), divide(multiply(12, const_10), const_2)), add(divide(multiply(12, const_10), 10), divide(divide(multiply(12, const_10), const_2), 12)))

multiply(n1,const_10)|divide(#0,const_2)|divide(#0,n0)|add(#1,#0)|divide(#1,n1)|add(#2,#4)|divide(#3,#5)|

general

a shopkeeper buys mangoes at the rate of 4 a rupee and sells them at 3 a rupee . find his net profit or loss percent ?

"explanation : the total number of mangoes bought by the shopkeeper be 12 . if he buys 4 a rupee , his cp = 3 he selling at 3 a rupee , his sp = 4 profit = sp - cp = 4 - 3 = 1 profit percent = 1 / 3 * 100 = 33 1 / 3 % answer : c"

a ) 73 1 / 3 % , b ) 13 1 / 3 % , c ) 33 1 / 3 % , d ) 23 1 / 3 % , e ) 93 1 / 3 %

c

divide(multiply(3, const_100), 4)

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

gain

a train consists of 12 boggies , each boggy 15 metres long . the train crosses a telegraph post in 18 seconds . due to some problem , two boggies were detached . the train now crosses a telegraph post in

"length of train = 12 × 15 = 180 m . then , speed of train = 180 ⁄ 18 = 10 m / s now , length of train = 10 × 15 = 150 m ∴ required time = 150 ⁄ 10 = 15 sec . answer c"

a ) 18 sec , b ) 12 sec , c ) 15 sec , d ) 20 sec , e ) none of these

c

divide(subtract(multiply(12, 15), 15), divide(multiply(12, 15), 18))

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

physics

on sunday , bill ran 4 more miles than he ran on saturday . julia did not run on saturday , but she ran twice the number of miles on sunday that bill ran on sunday . if bill and julia ran a total of 16 miles on saturday and sunday , how many miles did bill run on sunday ?

"let bill run x on saturday , so he will run x + 4 on sunday . . julia will run 2 * ( x + 4 ) on sunday . . totai = x + x + 4 + 2 x + 8 = 16 . . 4 x + 12 = 16 . . x = 1 . . ans = x + 4 = 1 + 4 = 5 answer a"

a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9

a

add(divide(subtract(16, add(4, multiply(const_2, 4))), 4), 4)

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

general

prints a page 40 pg per min . if the printed for 2 hours except 20 min . where there was an paper jam , how many page did it print

40 pages - - - - - - - > 1 min 2 hrs except 20 mints means = 2 * 60 = 120 - 20 = 100 mints i . e . , 100 * 40 = 4,000 pages printed . answer : a

a ) 4,000 , b ) 12,880 , c ) 14,880 , d ) 8,880 , e ) 18,880

a

divide(multiply(subtract(multiply(2, const_60), 20), 40), multiply(const_10, const_100))

multiply(n1,const_60)|multiply(const_10,const_100)|subtract(#0,n2)|multiply(n0,#2)|divide(#3,#1)

general

a train 450 m long is running at a speed of 68 kmph . how long does it take to pass a man who is running at 8 kmph in the same direction as the train ?

"speed of the train relative to man = ( 68 - 8 ) kmph = ( 60 * 5 / 18 ) m / sec = ( 50 / 3 ) m / sec time taken by the train to cross the man = time taken by it to cover 450 m at 50 / 3 m / sec = 450 * 3 / 50 sec = 27 sec answer : c ."

a ) 5 sec , b ) 39 sec , c ) 27 sec , d ) 15 sec , e ) 18 sec

c

divide(450, multiply(subtract(68, 8), const_0_2778))

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

physics

during a thanksgiving weekend , a car rental company rented 6 - tenths of their vehicles , including two - fifths of the 4 wds that it had . if 40 % of the vehicles are 4 wds , then what percent of the vehicles that were not rented were not 4 wds ?

4 / 10 of all the vehicles were not rented . ( 3 / 5 ) ( 2 / 5 ) = 6 / 25 of all the vehicles are 4 wds that were not rented . ( 6 / 25 ) / ( 4 / 10 ) = 3 / 5 is the fraction of non - rented vehicles that were 4 wds 1 - 3 / 5 = 40 % of non - rented vehicles were not 4 wds . the answer is c .

a ) 20 % , b ) 30 % , c ) 40 % , d ) 50 % , e ) 60 %

c

multiply(divide(divide(multiply(const_2, 40), add(const_3, const_2)), 40), const_100)

add(const_2,const_3)|multiply(n2,const_2)|divide(#1,#0)|divide(#2,n2)|multiply(#3,const_100)

gain

if x < y < z and y - x > 7 , where x is an even integer and y and z are odd integers , what is the least possible value of z - x ?

"we have : 1 ) x < y < z 2 ) y - x > 5 3 ) x = 2 k ( x is an even number ) 4 ) y = 2 n + 1 ( y is an odd number ) 5 ) z = 2 p + 1 ( z is an odd number ) 6 ) z - x = ? least value z - x = 2 p + 1 - 2 k = 2 p - 2 k + 1 = 2 ( p - k ) + 1 - that means that z - x must be an odd number . we can eliminate answer choices a , c and e we are asked to find the least value , so we have to pick the least numbers since y is odd and x is even , y - x must be odd . since y - x > 7 the least value for y - x must be 11 , the least value for x must be 2 , and , thus , the least possible value for y must be 11 ( y - 2 = 9 , y = 11 ) 2 < 11 < z , since z is odd , the least possible value for z is 13 z - x = 13 - 2 = 11 answer c"

a ) 6 , b ) 7 , c ) 11 , d ) 8 , e ) 10

c

add(add(7, const_2), const_2)

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

general

a contest will consist of n questions , each of which is to be answered eithertrueorfalse . anyone who answers all n questions correctly will be a winner . what is the least value of n for which the probability is less than 1 / 100000 that a person who randomly guesses the answer to each question will be a winner ?

"a contest will consist of n questions , each of which is to be answered eithertrueorfalse . anyone who answers all n questions correctly will be a winner . what is the least value of n for which the probability is less than 1 / 1000 that a person who randomly guesses the answer to each question will be a winner ? a . 5 b . 10 c . 50 d . 100 e . 1000 soln : ans is b probability that one question is answered right is 1 / 2 . now for minimum number of questions needed to take probability less than 1 / 1000 is = > ( 1 / 2 ) ^ n < 1 / 100000 n = 1000 satisfies this . e"

a ) 5 , b ) 10 , c ) 50 , d ) 100 , e ) 1000

e

multiply(const_1000, divide(1, 100000))

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

general

jim drove 1096 miles of a 1200 miles journey . how many more miles does he need to drive to finish his journey ?

"the number of miles to drive to finish his journey is given by 1200 - 1096 = 104 miles correct answer a"

a ) 104 miles , b ) 432 miles , c ) 456 miles , d ) 887 miles , e ) 767 miles

a

subtract(1200, 1096)

subtract(n1,n0)|

physics

ram , who is half as efficient as krish , will take 18 days to complete a task if he worked alone . if ram and krish worked together , how long will they take to complete the task ?

"number of days taken by ram to complete task = 18 since ram is half as efficient as krish , amount of work done by krish in 1 day = amount of work done by ram in 2 days if total work done by ram in 18 days is 18 w amount of work done by ram in 1 day = w amount of work done by krish in 1 day = 2 w total amount of work done by krish and ram in a day = 3 w total amount of time needed by krish and ram to complete task = 18 w / 3 w = 6 days answer d"

a ) 16 days , b ) 12 days , c ) 8 days , d ) 6 days , e ) 18 days

d

inverse(add(divide(const_1, 18), divide(const_1, divide(18, const_2))))

divide(const_1,n0)|divide(n0,const_2)|divide(const_1,#1)|add(#0,#2)|inverse(#3)|

physics

a hollow iron pipe is 21 cm long and its external diameter is 8 cm . if the thickness of the pipe is 1 cm and iron weighs 8 g / cm ^ 3 , then the weight of the pipe is :

"external radius = 4 cm , internal radius = 3 cm . volume of iron = ( 22 / 7 x [ ( 4 ) ^ 2 - ( 3 ) ^ 2 ] x 21 ) cm ^ 3 ( 22 / 7 x 7 x 1 x 21 ) cm ^ 3 462 cm ^ 3 . weight of iron = ( 462 x 8 ) gm = 3696 gm = 3.696 kg . answer b"

a ) 3.6 kg , b ) 3.696 kg , c ) 36 kg , d ) 36.9 kg , e ) 3.06 kg

b

divide(multiply(subtract(volume_cylinder(divide(8, const_2), 21), volume_cylinder(subtract(divide(8, const_2), 1), 21)), 8), const_1000)

divide(n1,const_2)|subtract(#0,n2)|volume_cylinder(#0,n0)|volume_cylinder(#1,n0)|subtract(#2,#3)|multiply(n1,#4)|divide(#5,const_1000)|

general

the average weight of 7 persons increases by 1.5 kg . if a person weighing 65 kg is replaced by a new person , what could be the weight of the new person ?

"total weight increases = 7 × 1.5 = 10.5 kg so the weight of new person = 65 + 10.5 = 75.5 kg answer c"

a ) 76 kg , b ) 77 kg , c ) 75.5 kg , d ) data inadequate , e ) none of these

c

add(65, multiply(7, 1.5))

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

general

mr . jones gave 40 % of the money he had to his wife . he also gave 20 % of the remaining amount to his 3 sons . and half of the amount now left was spent on miscellaneous items and the remaining amount of rs . 12000 was deposited in the bank . how much money did mr . jones have initially ?

explanation : let the initial amount be x , amount given to his wife = ( 40 / 100 ) x = 2 x / 5 balance = ( x - ( 2 x / 5 ) ) = 3 x / 5 amount given to his wife = ( 20 / 100 ) * ( 3 x / 5 ) = 3 x / 25 balance = 3 x / 5 - 3 x / 25 = 12 x / 25 amountt spent on miscellaneous items = ( 1 / 2 ) * ( 12 x / 25 ) = 6 x / 25 which is equal to 12000 hence , = > 6 x / 25 = 12000 = > x = 50000 answer : c

a ) 40000 , b ) 45000 , c ) 50000 , d ) 62000 , e ) none of these

c

divide(12000, multiply(divide(divide(const_100, const_2), const_100), multiply(subtract(const_1, divide(40, const_100)), subtract(const_1, divide(20, const_100)))))

divide(const_100,const_2)|divide(n0,const_100)|divide(n1,const_100)|divide(#0,const_100)|subtract(const_1,#1)|subtract(const_1,#2)|multiply(#4,#5)|multiply(#3,#6)|divide(n3,#7)

gain

he total marks obtained by a student in physics , chemistry and mathematics is 170 more than the marks obtained by him in physics . what is the average mark obtained by him in chemistry and mathematics ?

"let the marks obtained by the student in physics , chemistry and mathematics be p , c and m respectively . p + c + m = 170 + p c + m = 170 average mark obtained by the student in chemistry and mathematics = ( c + m ) / 2 = 170 / 2 = 85 . answer : d"

a ) 55 , b ) 65 , c ) 75 , d ) 85 , e ) 95

d

divide(170, const_2)

divide(n0,const_2)|

general

find the compound interest on rs . 10000 at 12 % rate of interest for 1 year , compounded half - yearly

"amount with ci = 10000 [ 1 + ( 12 / 2 * 100 ) ] 2 = rs . 11236 therefore , ci = 11236 – 10000 = rs . 1236 answer : b"

a ) rs . 1036 , b ) rs . 1236 , c ) rs . 1186 , d ) rs . 1206 , e ) rs . 1226

b

subtract(multiply(power(add(const_1, divide(divide(12, 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

25 onions on a scale weigh 5.12 kg . when 3 onions are removed from the scale , the average weight of the 22 onions is 200 grams . what is the average weight ( in grams ) of the 3 onions which were removed ?

22 * 200 = 4400 . the other 3 onions weigh a total of 720 grams . the average weight is 720 / 3 = 240 grams . the answer is c .

a ) 200 , b ) 220 , c ) 240 , d ) 260 , e ) 280

c

divide(subtract(multiply(5.12, const_1000), multiply(22, 200)), 3)

multiply(n1,const_1000)|multiply(n3,n4)|subtract(#0,#1)|divide(#2,n2)

general

a profit of rs . 600 is divided between x and y in the ratio of 1 / 2 : 1 / 3 . what is the difference between their profit shares ?

"a profit of rs . 600 is divided between x and y in the ratio of 1 / 2 : 1 / 3 or 3 : 2 . so profits are 360 and 240 . difference in profit share = 360 - 240 = 120 answer : b"

a ) s . 220 , b ) s . 120 , c ) s . 320 , d ) s . 50 , e ) s . 90

b

subtract(divide(divide(600, add(divide(1, 2), divide(1, 3))), 2), divide(divide(600, add(divide(1, 2), divide(1, 3))), 3))

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

general

a rectangular photograph is surrounded by a border that is 1 inch wide on each side . the total area of the photograph and the border is m square inches . if the border had been 5 inches wide on each side , the total area would have been ( m + 144 ) square inches . what is the perimeter of the photograph , in inches ?

"let x and y be the width and length of the photograph . ( x + 2 ) ( y + 2 ) = m and so ( 1 ) xy + 2 x + 2 y + 4 = m ( x + 10 ) ( y + 10 ) = m and so ( 2 ) xy + 10 x + 10 y + 100 = m + 144 let ' s subtract equation ( 1 ) from equation ( 2 ) . 8 x + 8 y + 96 = 144 2 x + 2 y = 12 , which is the perimeter of the photograph . the answer is b ."

a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18

b

divide(subtract(144, subtract(power(multiply(5, const_2), const_2), power(multiply(1, const_2), const_2))), const_2)

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

geometry

3 candidates contested an election and received 1000 , 2000 and 4000 votes respectively . what percentage of the total votes did the winning candidate got ?

total number of votes polled = ( 1000 + 2000 + 4000 ) = 7000 required percentage = 4000 / 7000 * 100 = 57 % ( approximately ) answer : option c

a ) 30 % , b ) 50 % , c ) 57 % , d ) 62 % , e ) 75 %

c

multiply(divide(4000, add(add(1000, 2000), 4000)), const_100)

add(n1,n2)|add(n3,#0)|divide(n3,#1)|multiply(#2,const_100)

general

the area of a triangle will be when a = 3 m , b = 5 m , c = 10 m , a , b , c being lengths of respective sides ?

s = ( 3 + 5 + 10 ) / 2 = 9 answer : d

a ) 3 , b ) 6 , c ) 4 , d ) 9 , e ) 1

d

divide(add(add(3, 5), 10), 5)

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

geometry

what is the sum of all 5 digit numbers which can be formed with the digits 0 , 1,2 , 3,4 without repetation ?

"10 ( 11111 ) * 4 ! - 10 ( 1111 ) 3 ! = 2599980 answer : a"

a ) 2599980 , b ) 235500 , c ) 923580 , d ) 765432 , e ) 765434

a

multiply(divide(add(divide(subtract(subtract(const_1000, 5), add(add(multiply(multiply(5, 5), const_10), multiply(5, 5)), 5)), 1,2), const_1), 0), add(subtract(const_1000, 5), add(add(multiply(multiply(5, 5), const_10), multiply(5, 5)), 5)))

multiply(n0,n0)|subtract(const_1000,n0)|multiply(#0,const_10)|add(#2,#0)|add(n0,#3)|add(#4,#1)|subtract(#1,#4)|divide(#6,n2)|add(#7,const_1)|divide(#8,n1)|multiply(#5,#9)|

probability

an old man distributed all the gold coins he had to his two sons into two different numbers such that the difference between the squares of the two numbers is 64 times the difference between the two numbers . how many coins did the old man have ?

"let the number of coins one son got be x and the number of coins another got be y . total = x + y . x ^ 2 - y ^ 2 = 64 ( x - y ) - - > x + y = 64 . answer : d ."

a ) 24 , b ) 26 , c ) 30 , d ) 64 , e ) 40

d

floor(64)

floor(n0)|

general

20 litres of mixture contains 40 % alcohol and the rest water . if 8 litres of water be mixed with it , the percentage of alcohol in the new mixture would be ?

"alcohol in the 20 litres of mix . = 40 % of 20 litres = ( 40 * 20 / 100 ) = 8 litres water in it = 20 - 8 = 12 litres new quantity of mix . = 20 + 8 = 28 litres quantity of alcohol in it = 8 litres percentage of alcohol in new mix . = 8 * 100 / 28 = 50 / 3 = 28.57 % answer is c"

a ) 26.32 % , b ) 35.14 % , c ) 28.57 % , d ) 25 % , e ) 31.14 %

c

multiply(divide(subtract(add(20, 8), add(multiply(divide(subtract(const_100, 40), const_100), 20), 8)), add(20, 8)), const_100)

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

gain

jane makes toy bears . when she works with an assistant , she makes 90 percent more bears per week and works 10 percent fewer hours each week . having an assistant increases jane ’ s output of toy bears per hour by what percent ?

"we can use fractional equivalents here to solve the problem 80 % = 4 / 5 ; this means that in 1 st case if she prepares 5 bears , in 2 nd case she prepares 9 bears 10 % = 1 / 10 ; this means that in 1 st case if she needs 10 hours , in 2 nd case she needs 9 hours now we come to productivity based on above fractional values the productivity in 1 st case is 0.5 bears / hour and in the 2 nd case it is 1 bear / hour hence the productivity is double with the assistant i . e . the increase in productivity is 120 % d"

a ) 20 % , b ) 80 % , c ) 100 % , d ) 120 % , e ) 200 %

d

multiply(divide(10, subtract(subtract(const_100, 90), 10)), const_100)

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

physics

12.036 divided by 0.04 gives :

"= 12.036 / 0.04 = 1203.6 / 4 = 300.9 answer is b ."

a ) 30.09 , b ) 300.9 , c ) 30.06 , d ) 100.9 , e ) 300.6

b

divide(12.036, 0.04)

divide(n0,n1)|

general

if 20 liters of chemical x are added to 80 liters of a mixture that is 5 % chemical x and 95 % chemical y , then what percentage of the resulting mixture is chemical x ?

"the amount of chemical x in the solution is 20 + 0.05 ( 80 ) = 24 liters . 24 liters / 100 liters = 24 % the answer is a ."

a ) 24 % , b ) 26 % , c ) 28 % , d ) 30 % , e ) 32 %

a

add(20, multiply(divide(5, const_100), 80))

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

general

the average speed of a car decreased by 3 miles per hour every successive 8 - minutes interval . if the car traveled 4.4 miles in the sixth 8 - minute interval , what was the average speed of the car , in miles per hour , in the first 8 minute interval ?

"( 4.4 miles / 8 minutes ) * 60 minutes / hour = 33 mph let x be the original speed . x - 5 ( 3 ) = 33 x = 48 mph the answer is c ."

a ) 35 , b ) 40 , c ) 48 , d ) 51 , e ) 54

c

add(add(add(add(divide(4.4, divide(8, const_60)), 3), 3), 3), 3)

divide(n1,const_60)|divide(n2,#0)|add(n0,#1)|add(n0,#2)|add(n0,#3)|add(n0,#4)|

general

for a certain exam , a score of 58 was 2 standard deviations below mean and a score of 98 was 3 standard deviations above mean . what was the mean score r for the exam ?

"a score of 58 was 2 standard deviations below the mean - - > 58 = mean - 2 d a score of 98 was 3 standard deviations above the mean - - > 98 = mean + 3 d solving above for mean r = 74 . answer : a ."

a ) 74 , b ) 76 , c ) 78 , d ) 80 , e ) 82

a

divide(add(multiply(58, 3), multiply(98, 2)), add(2, 3))

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

general

a train is 360 meter long is running at a speed of 45 km / hour . in what time will it pass a bridge of 240 meter length ?

"speed = 45 km / hr = 45 * ( 5 / 18 ) m / sec = 25 / 2 m / sec total distance = 360 + 240 = 600 meter time = distance / speed = 600 * ( 2 / 25 ) = 48 seconds answer : e"

a ) 65 seconds , b ) 46 seconds , c ) 40 seconds , d ) 97 seconds , e ) 48 seconds

e

divide(add(360, 240), divide(multiply(45, const_1000), const_3600))

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

physics

in one year , the population , of a village increased by 10 % and in the next year , it decreased by 10 % . if at the end of 2 nd year , the population was 7920 , what was it in the beginning ?

"x * 110 / 100 * 90 / 100 = 7920 x * 0.99 = 7920 x = 7920 / 0.99 = > 8000 answer : b"

a ) 8008 , b ) 8000 , c ) 8022 , d ) 8021 , e ) 8022

b

divide(divide(7920, subtract(const_1, divide(10, const_100))), add(const_1, divide(10, const_100)))

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

general

the profit earned by selling an article for $ 832 is equal to the loss incurred when the same article is sold for $ 448 . what should be the sale price for making 35 % profit ?

"let c . p . = $ x . then , 832 - x = x - 448 2 x = 1280 = > x = 640 required s . p . = 135 % of $ 640 = $ 864 d"

a ) $ 480 , b ) $ 450 , c ) $ 960 , d ) $ 864 , e ) $ 660

d

add(divide(multiply(divide(add(832, 448), const_2), 35), const_100), divide(add(832, 448), const_2))

add(n0,n1)|divide(#0,const_2)|multiply(n2,#1)|divide(#2,const_100)|add(#3,#1)|

gain

if 4 ( p ' s capital ) = 6 ( q ' s capital ) = 10 ( r ' s capital ) , then out of the total profit of rs 3720 , r will receive

"explanation : let p ' s capital = p , q ' s capital = q and r ' s capital = r then 4 p = 6 q = 10 r = > 2 p = 3 q = 5 r = > q = 2 p / 3 r = 2 p / 5 p : q : r = p : 2 p / 3 : 2 p / 5 = 15 : 10 : 6 r ' s share = 3720 * ( 6 / 31 ) = 120 * 6 = 720 . answer : option a"

a ) 720 , b ) 700 , c ) 800 , d ) 900 , e ) none of these

a

multiply(3720, divide(6, add(add(add(10, add(4, const_1)), 10), 6)))

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

general

it costs $ 2 for the first 15 minutes to use the bumper cars at a fair ground . after the first 15 minutes it costs $ 6 per hour . if a certain customer uses the bumper cars for 3 hours and 25 minutes , how much will it cost him ?

3 hrs 25 min = 205 min first 15 min - - - - - - > $ 2 time left is 190 min . . . now , 60 min costs $ 6 1 min costs $ 6 / 60 190 min costs $ 6 / 60 * 190 = > $ 19 so , total cost will be $ 19 + $ 2 = > $ 21 the answer will be ( d ) $ 21

a ) $ 22 , b ) $ 3 , c ) $ 15 , d ) $ 21 , e ) $ 30

d

add(multiply(divide(6, const_60), subtract(add(multiply(3, const_60), 25), 15)), 2)

divide(n3,const_60)|multiply(n4,const_60)|add(n5,#1)|subtract(#2,n1)|multiply(#0,#3)|add(n0,#4)

physics

a rectangular grassy plot 110 m . by 65 m has a gravel path 2.5 m wide all round it on the inside . find the cost of gravelling the path at 30 paise per sq . metre

"area of the plot = 110 m * 65 m = 7150 sq . m area of plot excluding gravel = 105 m * 60 m = 6300 sq . m area of gravel = 7150 sq . m - 6300 sq . m = 850 sq . m cost of building it = 850 sq . m * 30 = 25500 p in rs = 25500 / 100 = rs 255 answer : a"

a ) s 255 , b ) s 780 , c ) s 880 , d ) s 480 , e ) s 980

a

divide(multiply(subtract(multiply(110, 65), multiply(subtract(110, multiply(2.5, const_2)), subtract(65, multiply(2.5, const_2)))), 30), const_100)

multiply(n0,n1)|multiply(n2,const_2)|subtract(n0,#1)|subtract(n1,#1)|multiply(#2,#3)|subtract(#0,#4)|multiply(n3,#5)|divide(#6,const_100)|

physics

john and andrew can finish the work 9 days if they work together . they worked together for 6 days and then andrew left . john finished the remaining work in another 6 days . in how many days john alone can finish the work ?

amount of work done by john and andrew in 1 day = 1 / 9 amount of work done by john and andrew in 6 days = 6 ã — ( 1 / 9 ) = 2 / 3 remaining work â € “ 1 â € “ 2 / 3 = 1 / 3 john completes 1 / 3 work in 6 days amount of work john can do in 1 day = ( 1 / 3 ) / 6 = 1 / 18 = > john can complete the work in 18 days answer : c

a ) 30 days , b ) 60 days , c ) 18 days , d ) 80 days , e ) 90 days

c

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

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

physics

of the 14,210 employees of the anvil factory , 3 / 7 are journeymen . if half of the journeymen were laid off , what percentage of the total remaining employees would be journeymen ?

"the exam gives us a number that is easily divisible by 7 to pique our curiosity and tempt us into calculating actual numbers ( also because otherwise the ratio would be incorrect ) . since the question is about percentages , the actual numbers will be meaningless , as only the ratio of that number versus others will be meaningful . nonetheless , for those who are curious , each 1 / 7 portion represents ( 14210 / 7 ) 2,030 employees . this in turn means that 4,060 employees are journeymen and the remaining 10,150 are full time workers . if half the journeymen were laid off , that would mean 1 / 7 of the total current workforce would be removed . this statistic is what leads many students to think that since half the journeymen are left , the remaining journeymen would represent half of what they used to be , which means 1 / 7 of the total workforce . if 1 / 7 of the workforce is journeymen , and 1 / 7 is roughly 14.3 % , then answer choice a should be the right answer . in this case , though , it is merely the tempting trap answer choice . what changed between the initial statement and the final tally ? well , you let go of 1 / 7 of the workforce , so the total number of workers went down . the remaining workers are still 1 / 7 of the initial workers , but the group has changed . the new workforce is smaller than the original group , specifically 6 / 7 of it because 1 / 7 was eliminated . the remaining workers now account for 1 / 7 out of 6 / 7 of the force , which if we multiply by 7 gives us 1 out of 6 . this number as a percentage is answer choice b , 14.3 % . using the absolute numbers we calculated before , there were 4,060 journeymen employees out of 14,210 total . if 2,030 of them are laid off , then there are 2,030 journeyman employees left , but now out of a total of ( 14,210 - 2,030 ) 12,180 employees . 2,030 / 12,180 is exactly 1 / 6 , or 16.67 % . the answer will work with either percentages or absolute numbers , but the percentage calculation will be significantly faster and applicable to any similar situation . the underlying principle of percentages ( and , on a related note , ratios ) can be summed up in the brainteaser i like to ask my students : if you ’ re running a race and you overtake the 2 nd place runner just before the end , what position do you end up in ? the correct answer is 2 nd place . percentages , like ratios and other concepts of relative math , depend entirely on the context . whether 100 % more of something is better than 50 % more of something else depends on the context much more than the percentages quoted . when it comes to percentages on the gmat , the goal is to understand them enough to instinctively not fall into the traps laid out for you . a"

a ) 14.3 % , b ) 16.67 % , c ) 33 % , d ) 28.6 % , e ) 49.67 %

a

multiply(multiply(divide(divide(divide(3, 7), 3), add(divide(divide(3, 7), 3), subtract(const_1, divide(3, 7)))), const_100), const_3)

divide(n1,n2)|divide(#0,n1)|subtract(const_1,#0)|add(#1,#2)|divide(#1,#3)|multiply(#4,const_100)|multiply(#5,const_3)|

general

what percent is 400 gm of 1 kg ?

"1 kg = 1000 gm 400 / 1000 ã — 100 = 40000 / 1000 = 40 % answer is b"

a ) 25 % , b ) 40 % , c ) 10 % , d ) 8 % , e ) 12 %

b

multiply(divide(400, 1), const_100)

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

gain

the batting average of a particular batsman is 60 runs in 46 innings . if the difference in his highest and lowest score is 160 runs and his average excluding these two innings is 58 runs , find his highest score .

"explanation : total runs scored by the batsman = 60 * 46 = 2760 runs now excluding the two innings the runs scored = 58 * 44 = 2552 runs hence the runs scored in the two innings = 2760 â € “ 2552 = 208 runs . let the highest score be x , hence the lowest score = x â € “ 160 x + ( x - 160 ) = 208 2 x = 368 x = 184 runs answer : a"

a ) 184 , b ) 367 , c ) 269 , d ) 177 , e ) 191

a

divide(add(160, subtract(multiply(60, 46), multiply(58, subtract(46, const_2)))), const_2)

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

general

the total age of a and b is 14 years more than the total age of b and c . c is how many year younger than a

"explanation : given that a + b = 14 + b + c = > a ? c = 14 + b ? b = 14 = > c is younger than a by 14 years answer : option d"

a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15

d

multiply(14, const_1)

multiply(n0,const_1)|

general

a certain car ' s price decreased by 2.5 % ( from the original price ) each year from 1996 to 2002 , during that time the owner of the car invested in a new carburetor and a new audio system for the car , which increased car ' s price by $ 3,500 . if the price of the car in 1996 was $ 22,000 , what is the car ' s price in 2002 ?

"price in 96 = 22000 price decrease each year = 2.5 / 100 * 22000 = 550 price in 97 = 22000 - 550 price in 98 = 22000 - 2 * 550 price in 99 = 22000 - 3 * 550 price in 00 = 22000 - 4 * 550 price in 01 = 22000 - 5 * 550 price in 02 = 22000 - 6 * 550 = 18700 investment in the car = 3500 net price of the car in 02 = 18700 + 3500 = $ 22200 correct option : c"

a ) $ 18,400 , b ) $ 19,500 , c ) $ 22,200 , d ) $ 20,400 , e ) $ 21,100

c

multiply(const_2, const_10)

multiply(const_10,const_2)|

gain

if the difference between the length and breadth of a rectangle is 23 m and its perimeter is 206 m , what is its area ?

l - b = 23 . . . ( 1 ) perimeter = 206 2 ( l = b ) = 206 l + b = 103 . . . ( 2 ) ( 1 ) + ( 2 ) 2 l = 23 + 103 = 126 l = 126 / 2 = 63 metre substituting the value of l in ( 1 ) , we get 63 - b = 23 b = 63 - 23 = 40 metre area = lb = 63 ã — 40 = 2520 m 2 answer : a

['a ) 2520', 'b ) 2510', 'c ) 2525', 'd ) 2025', 'e ) 2020']

a

rectangle_area(add(divide(subtract(206, multiply(const_2, 23)), const_4), 23), divide(subtract(206, multiply(const_2, 23)), const_4))

multiply(n0,const_2)|subtract(n1,#0)|divide(#1,const_4)|add(n0,#2)|rectangle_area(#3,#2)

geometry

a boat having a length 5 m and breadth 2 m is floating on a lake . the boat sinks by 1 cm when a man gets on it . the mass of the man is :

"explanation : volume of water displaced = ( 5 x 2 x 0.01 ) m 3 = 0.10 m 3 . ∴ mass of man = volume of water displaced x density of water = ( 0.10 x 1000 ) kg = 100 kg . answer : a"

a ) 100 kg , b ) 60 kg , c ) 72 kg , d ) 96 kg , e ) none of these

a

multiply(multiply(multiply(5, 2), divide(1, const_100)), const_1000)

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

physics

if 100 cats kill 100 mice in 100 days , then 4 cats would kill 4 mice in how many days ?

as 100 cats kill 100 mice in 100 days 1 cats kill 1 mouse in 100 days then 4 cats kill 4 mice in 100 days answer : d

a ) 1 day , b ) 4 days , c ) 40 days , d ) 100 days , e ) 50 days

d

divide(multiply(multiply(4, 100), 100), multiply(100, 4))

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

physics

what is the unit digit in 7105 ?

"unit digit in 7105 = unit digit in [ ( 74 ) 26 * 7 ] but , unit digit in ( 74 ) 26 = 1 unit digit in 7105 = ( 1 * 7 ) = 7 answer : c"

a ) 1 , b ) 5 , c ) 7 , d ) 9 , e ) 11

c

circle_area(divide(7105, multiply(const_2, const_pi)))

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

general

of the 55 cars on a car lot , 40 have air - conditioning , 25 have power windows , and 12 have both air - conditioning and power windows . how many of the cars on the lot have neither air - conditioning nor power windows ?

total - neither = all air conditioning + all power windows - both or 55 - neither = 40 + 25 - 12 = 53 . = > neither = 2 , hence d . answer : d

a ) 15 , b ) 8 , c ) 10 , d ) 2 , e ) 18

d

subtract(55, subtract(add(40, 25), 12))

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

other

what is the remainder when 49 ^ 74 - 5 ^ 74 is divided by 24 ?

"easiest way for me : 49 ^ 74 - 5 ^ 74 = ( 49 ) ^ 37 - 25 ^ 37 = ( 24 * 2 + 1 ) ^ 37 - ( 24 + 1 ) ^ 37 - > remainder is 1 ^ 37 - 1 ^ 37 = 0 ans : c"

a ) 2 , b ) 1 , c ) 0 , d ) 3 , e ) none of these

c

reminder(multiply(74, 49), 5)

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

general

a sum of 13,400 amounts to 14,400 in 2 years at the rate of simple interest . what is the rate of interest ?

"d 4 % s . i . = ( 14400 - 13400 ) = 1000 . rate = ( 100 x 1000 ) / ( 13400 x 2 ) % = 4 %"

a ) 2 % , b ) 1 % , c ) 6 % , d ) 4 % , e ) 8 %

d

multiply(divide(divide(const_3, 2), add(multiply(const_3, 2), add(const_0_25, const_0_25))), const_100)

add(const_0_25,const_0_25)|divide(const_3,n2)|multiply(const_3,n2)|add(#0,#2)|divide(#1,#3)|multiply(#4,const_100)|

gain

a person lent a certain sum of money at 5 % per annum at simple interest and in 8 years the interest amounted to $ 420 less than the sum lent . what was the sum lent ?

"p - 420 = ( p * 5 * 8 ) / 100 p = 700 the answer is c ."

a ) 500 , b ) 600 , c ) 700 , d ) 800 , e ) 900

c

divide(420, subtract(const_1, divide(multiply(5, 8), const_100)))

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

gain

30 men can complete a piece of work in 18 days . in how many days will 27 men complete the same work ?

"explanation : less men , means more days { indirect proportion } let the number of days be x then , 27 : 30 : : 18 : x [ please pay attention , we have written 27 : 30 rather than 30 : 27 , in indirect proportion , if you get it then chain rule is clear to you : ) ] { \ color { blue } x = \ frac { 30 \ times 18 } { 27 } } x = 20 so 20 days will be required to get work done by 27 men . answer : a"

a ) 20 , b ) 77 , c ) 36 , d ) 25 , e ) 13

a

divide(multiply(18, 30), 27)

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

physics

if n is a positive integer and n ^ 2 is divisible by 200 , then what is the largest positive integer that must divide n ?

200 = 2 ^ 3 * 5 ^ 2 if 200 divides n ^ 2 , then n must be divisible by 2 ^ 2 * 5 = 20 the answer is c .

a ) 10 , b ) 15 , c ) 20 , d ) 36 , e ) 50

c

multiply(sqrt(divide(200, 2)), 2)

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

general

john paid a sum of money for purchasing 20 pens , which he recovered in full when he sold 10 of them . what was his percentage of profit or loss per pen ?

a 100 % if the sum he paid whilst purchasing 20 pens = a , then the cost price of each pen = a / 20 . since the amount he got whilst selling 10 pens is also = a then the selling price of each pen = a / 10 . since selling price > cost price , he made a profit . profit per pen = selling price - cost price = a / 10 - a / 20 = a / 20 . profit percentage per pen = profit per pen / cost per pen x 100 = ( a / 20 ) / ( a / 20 ) x 100 = 100 %

a ) 100 % , b ) 150 % , c ) 90 % , d ) 80 % , e ) 95 %

a

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

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

gain

if x is a positive number and 1 / 3 the square root of x is equal to 3 x , then x =

"1 / 3 of sqrt ( x ) = 3 x , which means that sqrt ( x ) = 9 x or x = 81 x ^ 2 - > divide by x 1 = 81 x x = 1 / 81 c ."

a ) 1 / 3 , b ) 1 / 9 , c ) 1 / 81 , d ) 1 , e ) 81

c

power(3, multiply(3, const_3))

multiply(n1,const_3)|power(n1,#0)|

general

find the smallest number of 6 digits exactly divisible by 25 , 3545 and 15 .

smallest number of six digits is 100000 . required number must be divisible by l . c . m . of 25,35 , 45,15 i . e 1575 , on dividing 100000 by 1575 , we get 800 as remainder . therefore , required number = 100000 + ( 1575 â € “ 800 ) = 100775 . answer is b .

a ) 100555 , b ) 100775 , c ) 100885 , d ) 100995 , e ) 100665

b

multiply(power(const_100, const_2), const_10)

power(const_100,const_2)|multiply(#0,const_10)

general

two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively . if they cross each other in 21 seconds , what is the ratio of their speeds ?

"let the speed of the trains be x and y respectively length of train 1 = 27 x length of train 2 = 17 y relative speed = x + y time taken to cross each other = 21 s = ( 27 x + 17 y ) / ( x + y ) = 21 = ( 27 x + 17 y ) / = 21 ( x + y ) = 6 x = 4 y = x / y = 4 / 6 = 2 / 3 i . e 2 : 3 answer : c"

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

c

divide(subtract(27, 21), subtract(21, 17))

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

physics

in a certain apartment building , there are one - bedroom and two - bedroom apartments . the rental prices of the apartment depend on a number of factors , but on average , two - bedroom apartments have higher rental prices than do one - bedroom apartments . let m be the average rental price for all apartments in the building . if m is $ 700 higher than the average rental price for all one - bedroom apartments , and if the average rental price for all two - bedroom apartments is $ 2100 higher that m , then what percentage of apartments in the building are two - bedroom apartments ?

ratio of 2 bedroom apartment : 1 bedroom apartment = 700 : 2100 - - - - - > 1 : 3 let total number of apartments be x no . of 2 bedroom apartment = ( 1 / 4 ) * x percentage of apartments in the building are two - bedroom apartments - - - - > ( 1 / 4 ) * 100 - - - > 25 % answer : a

a ) 25 % , b ) 15 % , c ) 20 % , d ) 40 % , e ) 45 %

a

multiply(divide(const_1, add(const_3, const_1)), const_100)

add(const_1,const_3)|divide(const_1,#0)|multiply(#1,const_100)

general

if two - third of a bucket is filled in 6 minute then the time taken to fill the bucket completely will be .

"2 / 3 filled in 6 mint 1 / 3 filled in 3 mint thn 2 / 3 + 1 / 3 = 6 + 3 = 9 minutes answer : d"

a ) 90 seconds , b ) 70 seconds , c ) 60 seconds , d ) 9 minutes , e ) 120 seconds

d

multiply(divide(6, const_2), const_3)

divide(n0,const_2)|multiply(#0,const_3)|

physics

a batsman makes a score of 76 runs in the 17 th inning and thus increases his average by 3 . find his average after 17 th inning .

"let the average after 17 th inning = x . then , average after 16 th inning = ( x – 3 ) . ∴ 16 ( x – 3 ) + 76 = 17 x or x = ( 76 – 48 ) = 28 . answer b"

a ) 36 , b ) 28 , c ) 42 , d ) 45 , e ) none of the above

b

add(subtract(76, multiply(17, 3)), 3)

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

general