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what is the remainder when 17 ^ 200 is divided by 18 ?

"( 17 ^ 200 - 1 ^ 200 ) is completely divisible by ( 17 + 1 ) as 200 is even . = > ( 17 ^ 200 - 1 ) is completely divisible by 18 . hence , when 17 ^ 200 is divided by 18 , we will get 1 as remainder . answer is b"

a ) 2 , b ) 14 , c ) 3 , d ) 1 , e ) 42

d

subtract(divide(18, const_2), multiply(17, 17))

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

general

a block of wood has dimensions 10 cm x 10 cm x 90 cm . the block is painted red and then cut evenly at the 45 cm mark , parallel to the sides , to form two rectangular solids of equal volume . what percentage of the surface area of each of the new solids is not painted red ?

"the area of each half is 100 + 4 ( 450 ) + 100 = 2000 the area that is not painted is 100 . the fraction that is not painted is 100 / 2000 = 1 / 20 = 5 % the answer is a ."

a ) 60 % , b ) 5 % , c ) 27 , d ) 60 , e ) one

b

multiply(divide(const_100, add(add(multiply(multiply(const_4, const_100), const_4), const_100), const_100)), const_100)

multiply(const_100,const_4)|multiply(#0,const_4)|add(#1,const_100)|add(#2,const_100)|divide(const_100,#3)|multiply(#4,const_100)|

geometry

after 6 games , team b had an average of 75 points per game . if it got only 47 points in game 7 , how many more points does it need to score to get its total above 500 ?

"( 6 * 75 ) + 47 + x > 500 450 + 47 + x > 500 497 + x > 500 = > x > 3 option d"

a ) 7 , b ) 1000 , c ) 3 , d ) 10750 , e ) 21 years

c

subtract(500, add(multiply(6, 75), 47))

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

general

if both 5 ^ 2 and 3 ^ 3 are factors of n x ( 2 ^ 5 ) x ( 6 ) x ( 7 ^ 3 ) , what is the smallest possible positive value of n ?

( 2 ^ 5 ) x ( 6 ) x ( 7 ^ 3 ) has one appearance of 3 ( in the 6 ) and no appearances of 5 . thus n must include at least 3 ^ 2 * 5 ^ 2 = 9 * 25 = 225 the answer is e .

a ) 136.8 , b ) 3400 , c ) 52500 , d ) $ 216 , e ) 225

e

add(add(add(add(add(multiply(multiply(5, 7), 2), multiply(multiply(5, 7), 2)), multiply(multiply(5, 7), 2)), 7), const_4), const_4)

multiply(n0,n7)|multiply(n1,#0)|add(#1,#1)|add(#2,#1)|add(n7,#3)|add(#4,const_4)|add(#5,const_4)

other

a can finish a work in 36 days , b in 9 days and c in 2 days , b and c start the work but are forced to leave after 3 days . the remaining work was done by a in ?

"b + c 1 day work = 1 / 9 + 1 / 12 = 7 / 36 work done by b and c in 3 days = 7 / 36 * 3 = 7 / 12 remaining work = 1 - 7 / 12 = 5 / 12 1 / 36 work is done by a in 1 day 5 / 12 work is done by a in 36 * 5 / 12 = 15 days answer is a"

a ) 40 % , b ) 150 , c ) 15 days , d ) 22 , e ) 80 %', '

c

multiply(divide(const_1, add(divide(const_1, 9), divide(const_1, 2))), 3)

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

physics

there are 20 poles with a constant distance between each pole . a car takes 26 second to reach the 12 th pole . how much will it take to reach the last pole .

"assuming the car starts at the first pole . to reach the 12 th pole , the car need to travel 11 poles ( the first pole does n ' t count , as the car is already there ) . 11 poles 26 seconds 1 pole ( 26 / 11 ) seconds to reach the last ( 20 th ) pole , the car needs to travel 19 poles . 19 pole 19 x ( 26 / 11 ) seconds = 44.9091 seconds answer : b"

a ) 1 : 49 , b ) 45 , c ) 2 / 3 , d ) 0.05 , e ) 44.9091

e

multiply(divide(26, 12), 20)

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

physics

p has $ 21 more than what q and r together would have had if both b and c had 1 / 5 of what p has . how much does p have ?

"p = ( 2 / 5 ) * p + 21 ( 3 / 5 ) * p = 21 p = 35 the answer is a ."

a ) 30 days , b ) 1856 , c ) 31 , d ) $ 35 , e ) 0.036

d

divide(21, subtract(1, multiply(divide(1, 5), const_2)))

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

general

a large box contains 18 small boxes and each small box contains 25 chocolate bars . how many chocolate bars are in the large box ?

"the number of chocolate bars is equal to 18 ? 25 = 450 correct answer c"

a ) 314.3 m , b ) 108 , c ) 450 , d ) 77 , e ) 8

c

multiply(18, 25)

multiply(n0,n1)|

general

because he ’ s taxed by his home planet , mork pays a tax rate of 40 % on his income , while mindy pays a rate of only 30 % on hers . if mindy earned 5 times as much as mork did , what was their combined tax rate ?

"say morks income is - 100 so tax paid will be 40 say mindys income is 5 * 100 = 500 so tax paid is 30 % * 500 = 150 total tax paid = 40 + 150 = 190 . combined tax % will be 190 / 100 + 500 = 31.67 %"

a ) 17 % , b ) 19.2 , c ) 7 , d ) 31.67 % , e ) 25

d

multiply(const_100, divide(add(divide(40, const_100), multiply(5, divide(30, const_100))), add(const_1, 5)))

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

gain

if p ( a ) = 0.4 , p ( b ) = 0.6 and p ( a ∪ b ) = 0.8 . what is the value of p ( a ∩ b ' ) = ?

"solution : p ( a ∪ b ) = p ( a ) + p ( b ) - p ( a ∩ b ' ) = > 0.8 = 0.4 - p ( a ∩ b ) = > p ( a ∩ b ) = 0.2 p ( a ∩ b ' ) = p ( a ) - p ( a ∩ b ) = 0.4 - 0.2 = 0.2 answer b"

a ) 0.2 , b ) 4 , c ) s . 800 , d ) 19 , e ) 49

a

multiply(multiply(0.4, 0.8), const_10)

multiply(n0,n2)|multiply(#0,const_10)|

general

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 . 26000 , then the share of b is ?

"ratio of investments of a and b is ( 70000 * 12 ) : ( 120000 * 6 ) = 7 : 6 total profit = rs . 26000 share of b = 6 / 13 ( 26000 ) = rs . 12000 answer : b"

a ) 20 % , b ) 200 sq feet , c ) 58 , d ) 5 , e ) s . 12000

e

subtract(26000, multiply(const_60, const_100))

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

gain

find the number which is nearest to 3105 and is exactly divisible by 21 .

"sol . on dividing 3105 by 21 , we get 18 as remainder . number to be added to 3105 = ( 21 - 18 ) - 3 . hence , required number = 3105 + 3 = 3108 . option b"

a ) 300 , b ) 11.1 % , c ) 3108 , d ) 18 , e ) 81

c

add(3105, subtract(21, reminder(3105, 21)))

reminder(n0,n1)|subtract(n1,#0)|add(n0,#1)|

general

evaluate : 980 x 436 + 980 x 764

"980 x 436 + 980 x 764 = 986 x ( 436 + 664 ) = 986 x 1200 = 117600 . answer is a ."

a ) 1176000 , b ) $ 576 , c ) 4 , d ) 400 % , e ) 55 cm 2

a

subtract(980, multiply(multiply(436, 980), 764))

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

general

how many terminating zeroes r does 200 ! have ?

you have 40 multiples of 5 , 8 of 25 and 1 of 125 . this will give 49 zeros . c

a ) 6 , b ) 1200 , c ) 49 , d ) 1 km , e ) 4 / 15

c

add(divide(200, add(const_4, const_1)), divide(200, multiply(add(const_4, const_1), add(const_4, const_1))))

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

other

a girl walking at the rate of 9 km per hour crosses a square field diagonally in 12 seconds . the area of the field is :

distance covered in ( 9 × 1000 ) / ( 3600 ) × 12 = 30 m diagonal of squarre field = 30 m . area of square field = 30 ( power ) 2 / 2 = 900 / 2 = 450 sq . m answer is c .

a ) 80 % , b ) 450 sq . m', ' , c ) 10 , d ) 5 kmph , e ) 11.1 %

b

divide(multiply(multiply(12, divide(multiply(9, const_1000), multiply(const_360, const_10))), multiply(12, divide(multiply(9, const_1000), multiply(const_360, const_10)))), const_2)

multiply(n0,const_1000)|multiply(const_10,const_360)|divide(#0,#1)|multiply(n1,#2)|multiply(#3,#3)|divide(#4,const_2)

geometry

x , y , and z are all unique numbers . if x is chosen randomly from the set { 6 , 7 , 8 , 9 , 10 , 11 } and y and z are chosen randomly from the set { 20 , 21 , 22 , 23 } , what is the probability that x and y are prime and z is not ?

"p ( x is prime ) = 1 / 3 p ( y is prime ) = 1 / 4 if y is prime , then z is not prime since y and z are unique . then the probability is 1 / 3 * 1 / 4 = 1 / 12 the answer is d ."

a ) 7 , b ) 24.78 , c ) 1 / 12 , d ) 55 , e ) 14

c

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

divide(const_1,const_2)|divide(const_1,const_4)|multiply(#0,#1)|

probability

what is the total number of integers between 20 and 100 that are divisible by 9 ?

"27 , 36 , 45 , . . . , 90,99 this is an equally spaced list ; you can use the formula : n = ( largest - smallest ) / ( ' space ' ) + 1 = ( 99 - 27 ) / ( 9 ) + 1 = 8 + 1 = 9 answer is e"

a ) 15 , b ) 25 % , c ) 9 , d ) 16 , e ) 5 inches', '

c

add(divide(subtract(100, 20), 9), const_1)

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

general

the perimeter of a triangle is 40 cm and the inradius of the triangle is 2.5 cm . what is the area of the triangle

"explanation : area of a triangle = r * s where r is the inradius and s is the semi perimeter of the triangle . area of triangle = 2.5 * 40 / 2 = 50 cm 2 answer : option c"

a ) 45', ' , b ) 7.2 hr , c ) 13.9 , d ) 50 , e ) 4691530800

d

triangle_area(2.5, 40)

triangle_area(n0,n1)|

geometry

the price of an article is cut by 10 % . to restore it to the former value . the new price must be increased by ?

answer let original price = rs . 100 . then , new price = rs . 90 . ∴ increased on rs . 90 = rs . 10 required increase % = ( 10 x 100 ) / 90 % = 111 / 9 % correct option : c

a ) $ 154.1 , b ) 600 , c ) 2 , d ) 14 % , e ) 11 1 / 9

e

add(subtract(const_100, subtract(const_100, 10)), const_2)

subtract(const_100,n0)|subtract(const_100,#0)|add(#1,const_2)

gain

the probability that event b occurs is 0.6 , and the probability that events a and b both occur is 0.25 . if the probability that either event a or event b occurs is 0.4 , what is the probability that event a will occur ?

p ( a or b ) = p ( a ) + p ( b ) - p ( a n b ) 0.4 = 0.6 + p ( a ) - 0.25 p ( a ) = 0.05 ans : a

a ) 10 / 3 , b ) 1700 , c ) 0.05 , d ) 35 , e ) 9.87 %

c

subtract(add(0.25, 0.4), 0.6)

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

other

if x and y are both odd prime numbers and x < y , how many distinct positive integer e factors does 2 xy have ?

since 2 xy prime e factors are x ^ 1 * y ^ 1 * 2 ^ 1 , its total number or factors must be ( 1 + 1 ) ( 1 + 1 ) ( 1 + 1 ) = 2 ^ 3 = 8 . thus , i think d would be the correct answer .

a ) 1260 , b ) 8 , c ) 2 6 / 7 % , d ) 6 days , e ) 4000

b

multiply(multiply(2, add(const_1, const_1)), add(const_1, const_1))

add(const_1,const_1)|multiply(n0,#0)|multiply(#0,#1)|

other

in a certain group of 10 developers , 4 developers code only in python and the rest program in either ruby on rails or php - but not both . if a developer organization is to choose a 3 - member team , which must have at least 1 developer who codes in python , how many different programming teams can be chosen ?

two ways . . . 1 ) total ways = 10 c 3 = 10 ! / 7 ! 3 ! = 120 . . ways without python developer = 6 c 3 = 6 ! / 3 ! 3 ! = 20 . . ways of at least one python developer = 120 - 20 = 100 . . 2 ) ways of selecting only one = 4 * 6 c 2 = 4 * 15 = 60 . . ways of selecting only two = 4 c 2 * 6 c 1 = 6 * 6 = 36 . . ways of selecting all three = 4 c 3 = 4 = 4 . . total = 60 + 36 + 4 = 100 . . . answer : a

a ) 28 , b ) 100 , c ) 135 , d ) 1 and 10 , e ) 7600

b

subtract(divide(factorial(10), multiply(factorial(subtract(10, 3)), factorial(3))), divide(factorial(subtract(10, 4)), multiply(factorial(3), factorial(3))))

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

other

a dishonest shopkeeper professes to sell pulses at the cost price , but he uses a false weight of 920 gm . for a kg . his gain is … % .

"his percentage gain is 100 * 80 / 920 as he is gaining 80 units for his purchase of 920 units . so 8.69 % . answer : e"

a ) 0 , b ) 46 , c ) 400 m , d ) 8.69 % , e ) 225

d

multiply(subtract(inverse(divide(920, multiply(multiply(add(const_4, const_1), const_2), const_100))), const_1), const_100)

add(const_1,const_4)|multiply(#0,const_2)|multiply(#1,const_100)|divide(n0,#2)|inverse(#3)|subtract(#4,const_1)|multiply(#5,const_100)|

gain

5.40 can be expressed in terms of percentage as

"explanation : while calculation in terms of percentage we need to multiply by 100 , so 5.40 * 100 = 540 answer : option d"

a ) 25 % , b ) 70 , c ) 5 / 4 , d ) 0.3 % , e ) 540 %

e

multiply(5.40, const_100)

multiply(n0,const_100)|

general

a pet store regularly sells pet food at a discount of 10 percent to 30 percent from the manufacturer ’ s suggested retail price . if during a sale , the store discounts an additional 20 percent from the discount price , what would be the lowest possible price of a container of pet food that had a manufacturer ’ s suggested retail price o f $ 25.00 ?

"for retail price = $ 25 first maximum discounted price = 25 - 30 % of 25 = 25 - 7.5 = 17.5 price after additional discount of 20 % = 17.5 - 20 % of 17.5 = 17.5 - 3.5 = 14 answer : option a"

a ) 94 , b ) 20.5 , c ) 6 , d ) $ 14.00 , e ) 2 / 9

d

multiply(divide(subtract(const_100, 20), const_100), multiply(divide(subtract(const_100, 30), const_100), 25.00))

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

gain

a number is mistakenly divided by 5 instead of being multiplied by 5 . find the percentage change in the result due t this mistake .

lets take a number 20 20 / 5 = 4 20 * 5 = 100 diff = 100 - 4 = 96 % answer : a

a ) 32 kmph , b ) 50 days , c ) 1.745 % , d ) 96 % , e ) 2

d

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

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

general

the first , second and third terms of the proportion are 56 , 16 , 49 . find the fourth term .

explanation : let the fourth term be x . thus 56 , 16 , 49 , x are in proportion . product of extreme terms = 56 x product of mean terms = 16 x 49 since , the numbers make up a proportion therefore , 56 x = 16 49 or , x = ( 16 49 ) / 56 or , x = 14 therefore , the fourth term of the proportion is 14 . answer : b

a ) 14 , b ) 555681 , c ) 0 , d ) 600 , e ) 180

a

divide(multiply(49, 16), 56)

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

physics

during a sale , the price of a pair of shoes is marked down 10 % from the regular price . after the sale ends , the price goes back to the original price . what is the percent of increase to the nearest percent from the sale price back to the regular price for the shoes ?

"assume the price = 100 price during sale = 90 price after sale = 100 percent increase = 10 / 90 * 100 = 11 % approx . correct option : c"

a ) 11 % , b ) 10 , c ) 1.46 % , d ) 23 , e ) 5

a

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

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

gain

sheila works 8 hours per day on monday , wednesday and friday , and 6 hours per day on tuesday and thursday . she does not work on saturday and sunday . she earns $ 324 per week . how much does she earn in dollars per hour ?

"explanation : total hours worked = 8 x 3 + 6 x 2 = 36 total earned = 324 . hourly wage = 324 / 36 = 9 answer : c ) 9"

a ) 9 , b ) 50 m . , c ) 8 days , d ) 8 / 3 , e ) 35

a

divide(324, add(multiply(8, const_3), multiply(6, const_2)))

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

physics

determine the value of 3 * 27 / 31 + 81 / 93

solution : both fractions should be reduced before performing arithmetic operations . we get 3 * 27 / 31 + 3.27 / 3.31 = 3 * 27 / 31 + 27 / 31 = 4 * 27 / 31 = 151 / 31 answer d

a ) 144 , b ) 9 days , c ) 151 / 31 , d ) $ 864 , e ) 67

c

divide(add(subtract(add(81, multiply(27, 3)), subtract(93, 81)), const_1), 31)

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

general

in goshawk - eurasian nature reserve 30 percent of the birds are hawks , and 40 percent of the non - hawks are paddyfield - warblers . if there are 25 percent as many kingfishers as paddyfield - warblers in the reserve , then what percent of the birds e in the nature reserve are not hawks , paddyfield - warblers , or kingfishers ?

"1 . we are given the following percentages : 30 ( 70 ) , 40 ( 60 ) , 25 ( 75 ) . there are two threads from here . first starts at 30 % and finishes there . second one starts at 70 , then 40 , and then 25 . we need a value that is divisible by 7 , 2 , and 5 at least once . lets pick a number now , say 700 . so say if non hawks are 700 ( this is 70 % of the total , so total = 1000 ) , then paddy warbs are 2 / 5 x 700 = 1400 / 5 = 280 . kingfishers , therefore , are 280 / 4 = 70 . lets add them up . 300 hawks + 280 peddy warbs + 70 kingsifhers = 650 . so all others are 1000 - 650 = 350 or 35 % of total birds . the main job here to to identify the smart number to start the question with . this can be time consuming , but once identified , this question can be solved fairly quickly . 2 . another method : if x is total - - > non hawks = 0.7 x - - > warbs = 0.4 ( 0.7 x ) - - > kfs = 0.25 ( 0.4 ( 0.7 x ) ) . our job is to find out e : ( 0.3 x + 0.28 x + 0.07 x ) / x . or 0.65 x / x = 0.65 . we need to find 1 - 0.65 = 0.35 or 35 % . b"

a ) 35 % , b ) 10 years , c ) 879 , d ) 29 , e ) 43

a

add(const_10, divide(add(25, 25), const_2))

add(n2,n2)|divide(#0,const_2)|add(#1,const_10)|

general

what is the smallest integer t greater than 1 that leaves a remainder of 1 when divided by any of the integers 6 , 8 , and 10 ?

or u can just use the answer choices here . since the answers are already arranged in ascending order , the first number which gives remainder t as 1 for all three is the correct answer . in the given question , the first number which gives a remainder of 1 for 6,8 and 10 is 121 . c

a ) t = 121 , b ) 1000 , c ) 15 , d ) 1717.85 , e ) 50

a

add(lcm(lcm(6, 8), 10), 1)

lcm(n2,n3)|lcm(n4,#0)|add(n0,#1)

general

a train 150 m long running at 72 kmph crosses a platform in 20 sec . what is the length of the platform ?

"e 250 e = 72 * 5 / 18 = 20 = 400 â € “ 150 = 250"

a ) 250 m , b ) rs . 1058 , c ) 26 ° , d ) 3 , e ) 6

a

subtract(multiply(20, multiply(72, const_0_2778)), 150)

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

physics

in bangalore there is a well known science institute . during a visit i asked two of the men to tell me their ages . one replied , ' one of our ages subtracted from the other ' s equal 30 . ' then the other man spoke . ' our ages multiplied together equal 1624 . ' what were their ages ?

e their ages were respectively 58 and 28

a ) 28 , b ) 14 : 00 , c ) 21.5 sec , d ) 21 % , e ) 58 and 28

e

divide(divide(multiply(1624, 30), const_4), const_2)

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

general

the smallest number when increased by ` ` 1 ` ` is exactly divisible by 2 , 8 , 24 , 36 is :

lcm = 72 72 - 1 = 71 answer : a

a ) s . 4076 , b ) 1200 , c ) 4966 , d ) 71 , e ) 1235

d

subtract(lcm(24, 36), 1)

lcm(n3,n4)|subtract(#0,n0)

general

a person bought 135 glass bowls at a rate of rs . 15 per bowl . he sold 115 of them at rs . 18 and the remaining broke . what is the percentage gain for a ?

"cp = 135 * 15 = 2025 and sp = 115 * 18 = 2070 gain % = 100 * ( 2070 - 2025 ) / 2025 = 20 / 9 answer : c"

a ) 5625 , b ) 130 , c ) 20 / 9 , d ) 0.2 , e ) 257

c

multiply(divide(subtract(multiply(115, 18), multiply(135, 15)), multiply(135, 15)), const_100)

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

gain

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

"let initial price be 100 price in day 1 after 3 % discount = 97 price in day 2 after 3 % discount = 94.09 price in day 3 after 10 % discount = 84.68 so , price in day 3 as percentage of the sale price on day 1 will be = 84.68 / 97 * 100 = > 87.3 % answer will definitely be ( c )"

a ) 87.3 % , b ) 88 , c ) 5 , d ) 4 , e ) 123

a

add(multiply(divide(divide(10, const_100), subtract(1, divide(1, 3))), const_100), 2)

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

gain

the maximum number of students among them 1200 pens and 820 pencils can be distributed in such a way that each student get the same number of pens and same number of pencils ?

"number of pens = 1200 number of pencils = 820 required number of students = h . c . f . of 1200 and 820 = 20 answer is b"

a ) 8 , b ) 20 , c ) 248 , d ) 20 years , e ) 70400 yards

b

gcd(1200, 820)

gcd(n0,n1)|

general

in a certain pond , 80 fish were caught , tagged , and returned to the pond . a few days later , 50 fish were caught again , of which 2 were found to have been tagged . if the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond , what is the approximate number of fish in the pond ?

this is a rather straight forward ratio problem . 1 . 80 fish tagged 2 . 2 out of the 50 fish caught were tagged thus 2 / 50 2 / 50 = 80 / x thus , x = 2000 think of the analogy : 2 fish is to 50 fish as 50 fish is to . . . ? you ' ve tagged 50 fish and you need to find what that comprises as a percentage of the total fish population - we have that information with the ratio of the second catch . d

a ) 6 , b ) 50 , c ) 500 m , d ) 2,000 , e ) 300

d

divide(80, divide(2, 50))

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

gain

r is the set of positive odd integers less than 100 , and s is the set of the squares of the integers in r . how many elements does the intersection of r and s contain ?

"r is the set of positive odd integers less than 100 , and s is the set of the squares of the integers in r . how many elements does the intersection of r and s contain ? r = 1,3 , 5,7 , 9,11 , 13,15 . . . s = 1 , 9,25 , 49,81 . . . numbers : 1 , 9 , 25 , 49 , and 81 are odd integers ( less than 100 ) that are in both sets . solution : five answer : d"

a ) five , b ) w = 480 , c ) 21 : 124 , d ) 9 / 19 , e ) 130 cm

a

subtract(subtract(100, const_4), const_4)

subtract(n0,const_4)|subtract(#0,const_4)|

physics

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 w 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 w is 40 % of 80 hence , d is the answer"

a ) 427.5 , b ) 1954404 , c ) 3 , d ) 3.33 , e ) 40

e

divide(80, const_2)

divide(n3,const_2)|

general

a brick measures 20 cm * 10 cm * 7.5 cm how many bricks will be required for a wall 28 m * 2 m * 0.75 m ?

"28 * 2 * 0.75 = 20 / 100 * 10 / 100 * 7.5 / 100 * x 28 = 1 / 100 * x = > x = 28000 answer : a"

a ) 49 . , b ) 19 , c ) 7 hours , d ) 28000 , e ) 555681

d

divide(divide(divide(multiply(multiply(multiply(28, const_100), multiply(2, const_100)), multiply(0.75, const_100)), 20), 10), 7.5)

multiply(n3,const_100)|multiply(n4,const_100)|multiply(n5,const_100)|multiply(#0,#1)|multiply(#3,#2)|divide(#4,n0)|divide(#5,n1)|divide(#6,n2)|

physics

an amount at compound interest sums to rs . 17640 / - in 2 years and to rs . 20286 / - in 3 years at the same rate of interest . find the rate percentage ?

"explanation : the difference of two successive amounts must be the simple interest in 1 year on the lower amount of money . s . i = 20286 / - - 17640 / - = rs . 2646 / - rate of interest = ( 2646 / 17640 ) × ( 100 / 1 ) = > 15 % answer : option d"

a ) - 3 , b ) 64 , c ) 15 % , d ) 4 , e ) 2599980

c

multiply(divide(subtract(20286, 17640), 17640), const_100)

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

general

david works at a science lab that conducts experiments on bacteria . the population of the bacteria multiplies at a constant rate , and his job is to notate the population of a certain group of bacteria each hour . at 1 p . m . on a certain day , he noted that the population was 600 and then he left the lab . he returned in time to take a reading at 4 p . m . , by which point the population had grown to 4,800 . now he has to fill in the missing data for 2 p . m . and 3 p . m . what was the population at 3 p . m . ?

let the rate be x , then population of the bacteria after each hour can be given as 600,600 x , 600 ( x ^ 2 ) , 600 ( x ^ 3 ) now population at 4 pm = 4800 thus we have 600 ( x ^ 3 ) = 4800 = 8 thus x = 2 therefore population at 3 pm = 600 ( 4 ) = 2400 answer : a

a ) 30 kmph , b ) 2400 , c ) 153600 m 2', ' , d ) 38 , e ) 20 % loss

b

multiply(multiply(power(divide(multiply(multiply(2, 4), 600), 600), const_0_33), 600), power(divide(multiply(multiply(2, 4), 600), 600), const_0_33))

multiply(n2,n4)|multiply(n1,#0)|divide(#1,n1)|power(#2,const_0_33)|multiply(n1,#3)|multiply(#4,#3)

physics

5 years ago , the average age of a , b , c and d was 45 years . with e joining them now , the average of all the 5 is 50 years . the age of e is ?

solution 5 years ago average age of a , b , c , d = 45 years = > 5 years ago total age of a , b , c , d = 45 x 4 = 180 years = > total present age of a , b , c , d = 180 + 5 x 4 = 200 years if e ' s present age is x years = 200 + x / 5 = 50 x = 50 years . answer a

a ) 8 , b ) 131.95 , c ) 5 / 3 , d ) 50 , e ) 98

d

subtract(multiply(50, 5), add(multiply(45, multiply(const_2, const_2)), multiply(5, const_4)))

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

general

sandy bought 65 books for $ 1180 from one shop and 55 books for $ 860 from another shop . what is the average price that sandy paid per book ?

"average price per book = ( 1180 + 860 ) / ( 65 + 55 ) = 2040 / 120 = $ 17 the answer is c ."

a ) 89 , b ) $ 17 , c ) 11190 , d ) 8 , e ) 20

b

divide(add(1180, 860), add(65, 55))

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

general

there are 15 slate rocks , 20 pumice rocks , and 10 granite rocks randomly distributed in a certain field . if 2 rocks are to be chosen at random and without replacement , what is the probability that both rocks will be slate rocks ?

"total no of rocks = 45 probability of choosing 1 st slate rock = 15 / 45 probability of choosing 2 nd slate rock = 14 / 44 ( without replacement ) so combined probability = 15 / 45 * 14 / 44 = 7 / 66 so , answer d ."

a ) 266 cm 2 , b ) 7 / 66 , c ) 137 / 216', ' , d ) 45 , e ) 30

b

multiply(divide(15, add(add(15, 20), 10)), divide(subtract(15, const_1), subtract(add(add(15, 20), 10), const_1)))

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

other

a cistern can be filled by a tap in 4 hours while it can be emptied by another tap in 9 hours . if both the taps are opened simultaneously then after how much time will the cistern get filled ?

"net part filled in 1 hour 1 / 4 - 1 / 9 = 5 / 36 the cistern will be filled in 36 / 5 hr = 7.2 hr answer is d"

a ) 52 % , b ) 7.2 hr , c ) 272 , d ) 64 , e ) 250

b

divide(const_1, subtract(divide(const_1, 4), divide(const_1, 9)))

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

physics

approximately how many cubic feet of water are needed to fill a circular swimming pool that is 40 feet across and 7 feet deep ?

"answer should be e . v = \ pir ^ 2 h = \ pi * 20 ^ 2 * 7 = approximately 9000"

a ) 12 , b ) $ 2.25 , c ) 180 m , d ) 30 , e ) 9000

e

volume_cylinder(divide(40, const_2), 7)

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

geometry

in a graduate physics course , 70 percent of the students are male and 40 percent of the students are married . if two - sevenths of the male students are married , what fraction of the female students is single ?

"let assume there are 100 students of which 70 are male and 30 are females if 40 are married then 60 will be single . now its given that two - sevenths of the male students are married that means 2 / 7 of 70 = 20 males are married if 40 is the total number of students who are married and out of that 20 are males then the remaining 20 will be females who are married . total females = 30 married females = 20 then single females = 30 - 20 = 10 we need to find the fraction of female students who are single i . e single female students / total female student = 10 / 30 = 1 / 3 [ e ]"

a ) 120 , b ) 1 / 3 , c ) s . 4400 , d ) 84 , e ) 25300

b

divide(const_10, 40)

divide(const_10,n1)|

gain

( 3 x + 2 ) ( 2 x - 1 ) = ax ^ 2 + kx + n . what is the value of a - n + k ?

"expanding we have 6 x ^ 2 - 3 x + 4 x - 2 6 x ^ 2 + x - 2 taking coefficients , a = 6 , k = 1 , n = - 2 therefore a - n + k = 6 - ( - 2 ) + 1 = 8 + 1 = 9 the answer is c ."

a ) 3.2 . , b ) 15552 , c ) 9 , d ) 315 , e ) 18 days

c

add(add(multiply(3, 2), multiply(1, 2)), subtract(multiply(2, 2), multiply(1, 3)))

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

general

a fruit seller had some oranges . he sells 40 % oranges and still has 600 oranges . how many oranges he had originally ?

"60 % of oranges = 600 100 % of oranges = ( 600 × 100 ) / 6 = 1000 total oranges = 1000 answer : c"

a ) - 3 , b ) 11 and 9 , c ) 1000 , d ) 48 , e ) 3

c

add(600, multiply(600, divide(40, const_100)))

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

gain

if the sides of a triangle are 196 cm , 81 cm and 277 cm , what is its area ?

"the triangle with sides 196 cm , 81 cm and 277 cm is right angled , where the hypotenuse is 277 cm . area of the triangle = 1 / 2 * 81 * 196 = 7938 cm 2 answer : option e"

a ) 28.57 % , b ) 31 st , c ) rs . 740 , d ) 6 , e ) 7938

e

divide(multiply(81, 277), const_2)

multiply(n1,n2)|divide(#0,const_2)|

geometry

what is the average ( arithmetic mean ) of 10 , 2030 , 4050 , 6070 , 8090 ?

so addition of all term - 10 , 20 , 30 , . . . . . . . 90 so average = ( 10 + 20 + 30 + 40 + 50 + 60 + 70 + 80 + 90 ) / 9 = ( 450 ) / 9 = 50 hence , the correct answer is e .

a ) 192 kmph , b ) 50 , c ) 44 % , d ) 13 days , e ) 44.9091

b

subtract(divide(add(add(add(add(10, 2030), 4050), 6070), 8090), add(const_4, const_1)), multiply(multiply(const_100, const_10), const_4))

add(n0,n1)|add(const_1,const_4)|multiply(const_10,const_100)|add(n2,#0)|multiply(#2,const_4)|add(n3,#3)|add(n4,#5)|divide(#6,#1)|subtract(#7,#4)

general

a man swims downstream 100 km and upstream 30 km taking 10 hours each time ; what is the speed of the current ?

"100 - - - 10 ds = 10 ? - - - - 1 30 - - - - 10 us = 3 ? - - - - 1 s = ? s = ( 10 - 3 ) / 2 = 3.5 answer : b"

a ) 2 , b ) 7 / 25 , c ) $ 81 , d ) 3.5 , e ) 4 cm

d

divide(add(divide(30, 10), divide(100, 10)), const_2)

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

physics

the length of a rectangular plot is thrice its breadth . if the area of the rectangular plot is 1323 sq m , then what is the breadth of the rectangular plot ?

"let the breadth of the plot be b m . length of the plot = 3 b m ( 3 b ) ( b ) = 1323 3 b 2 = 1323 b 2 = 441 = 21 ( b > 0 ) b = 21 m . answer : d"

a ) 6 , b ) 21 , c ) 13 , d ) 14 , e ) 6.5

b

sqrt(divide(1323, const_3))

divide(n0,const_3)|sqrt(#0)|

geometry

after an ice began to melt out from the freezer , in the first hour lost 3 / 4 , in the second hour lost 3 / 4 of its remaining . if after two hours , the volume is 0.3 cubic inches , what is the original volume of the cubic ice , in cubic inches ?

"let initial volume of ice be = x ice remaining after 1 hour = x - 0.75 x = 0.25 x ice remaining after 2 hour = ( 1 / 4 ) x - ( 3 / 4 * 1 / 4 * x ) = ( 1 / 16 ) x ( 1 / 16 ) x = 0.3 x = 4.8 alternate solution : try to backsolve . initial volume = 4.8 after one hour - - > ( 1 / 4 ) 4.8 = 1.2 after two hours - - > ( 1 / 4 ) 1.2 = 0.3 answer : c"

a ) 4.8 , b ) 9.2 days , c ) $ 0.40 , d ) 6 , e ) 0.3 %

a

divide(divide(0.3, const_0_25), const_0_25)

divide(n4,const_0_25)|divide(#0,const_0_25)|

physics

the perimeter of a triangle is 22 cm and the inradius of the triangle is 3.5 cm . what is the area of the triangle ?

"area of a triangle = r * s where r is the inradius and s is the semi perimeter of the triangle . area of triangle = 3.5 * 22 / 2 = 38.5 cm 2 answer : e"

a ) 4.0 , b ) 12 % , c ) 1.33 , d ) 12 , e ) 38

e

triangle_area(3.5, 22)

triangle_area(n0,n1)|

geometry

for a group of n people , k of whom are of the same sex , the ( n - k ) / n expression yields an index for a certain phenomenon in group dynamics for members of that sex . for a group that consists of 20 people , 6 of whom are females , by how much does the index for the females exceed the index for the males in the group ?

"index for females = ( 20 - 6 ) / 20 = 7 / 10 = 0.7 index for males = ( 20 - 14 / 20 = 3 / 10 = 0.3 index for females exceeds males by 0.7 - 0.3 = 0.4 answer : a"

a ) 35 , b ) 60 km , c ) 4 : 5 , d ) 466 , e ) 0.4

e

subtract(divide(subtract(20, 6), 20), divide(6, 20))

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

general

x does a work in 40 days . y does the same work in 60 days . in how many days they together will do the same work ?

"x ' s 1 day ' s work = 1 / 40 y ' s 1 day ' s work = 1 / 60 ( x + y ) ' s 1 day ' s work = ( 1 / 40 + 1 / 60 ) = 1 / 24 both together will finish the work in 24 days . correct option is c"

a ) 17 , b ) - 5 , c ) 24 , d ) 9.5 gallons , e ) 10 kmph

c

inverse(add(divide(const_1, 40), divide(const_1, 60)))

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

physics

the difference of 2 digit number & the number obtained by interchanging the digits is 36 . what is the sum and the number if the ratio between the digits of the number is 1 : 2 ?

"let the number be xy . given xy – yx = 36 . this means the number is greater is than the number got on reversing the digits . this shows that the ten ’ s digit x > unit digit y . also given ratio between digits is 1 : 2 = > x = 2 y ( 10 x + y ) – ( 10 y + x ) = 36 = > x – y = 4 = > 2 y – y = 4 . hence , ( x + y ) – ( x – y ) = 3 y – y = 2 y = 8 b"

a ) 7.5 sec , b ) 19 , c ) 35 % , d ) 1764713 , e ) 8

e

multiply(divide(36, subtract(multiply(subtract(const_10, 1), multiply(2, 1)), subtract(const_10, 1))), 2)

multiply(n0,n2)|subtract(const_10,n2)|multiply(#0,#1)|subtract(#2,#1)|divide(n1,#3)|multiply(#4,n0)|

general

7 carpet - weavers can weave 7 carpets in 7 days . at the same rate , how many carpets would be woven by 14 carpet - weavers in 14 days ?

explanation : solution : let the required number of carpets be x . more weavers , more carpets ( direct proportion ) more days , more carpets ( direct proportion ) weavers 7 : 14 } : : 7 : x days 7 : 14 . ' . 7 * 7 * x = 14 * 14 * 7 < = > x = 14 * 14 * 7 / 7 * 7 = 28 . answer : b

a ) 8 , b ) 28 , c ) 16 % , d ) 425 , e ) 5.5 %

b

add(14, add(7, 7))

add(n0,n0)|add(n3,#0)

gain

what is the speed of the stream if a canoe rows upstream at 6 km / hr and downstream at 12 km / hr

"sol . speed of stream = 1 / 2 ( 12 - 6 ) kmph = 3 kmph . answer c"

a ) 3 kmph , b ) 11190 , c ) 1 ⁄ 10 , d ) 9600 , e ) 10

a

divide(subtract(12, 6), const_2)

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

physics

a man saves a certain portion of his income during a year and spends the remaining portion on his personal expenses . next year his income increases by 40 % but his savings increase by 100 % . if his total expenditure in 2 years is double his expenditure in 1 st year , what % age of his income in the first year did he save ?

i year best is to give a number to his income , say 100 . . and let saving be x . . so expenditure = 100 - x next year - income = 140 savings = 2 x expenditure = 140 - 2 x . . now 140 - 2 x + 100 - x = 2 ( 100 - x ) . . . 240 - 3 x = 200 - 2 x . . . . . . . . . . . . . . . . x = 40 . . . saving % = 40 / 100 * 100 = 40 % answer : b

a ) 19 , b ) 40 % , c ) 45 , d ) 16 % , e ) 245

b

multiply(divide(subtract(add(add(100, 40), 100), multiply(2, 100)), const_100), const_100)

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

general

a train passes a station platform in 40 sec and a man standing on the platform in 20 sec . if the speed of the train is 54 km / hr . what is the length of the platform ?

"speed = 54 * 5 / 18 = 15 m / sec . length of the train = 15 * 20 = 300 m . let the length of the platform be x m . then , ( x + 300 ) / 40 = 15 = > x = 180 m answer : c"

a ) $ 1,354 , b ) 33 / 44 , c ) 60 , d ) 4096 , e ) 180 m

e

multiply(20, multiply(54, const_0_2778))

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

physics

speed of a boat in standing water is 10 kmph and speed of the stream is 1.5 kmph . a man can rows to a place at a distance of 105 km and comes back to the starting point . the total time taken by him is ?

"speed upstream = 8.5 kmph speed downstream = 11.5 kmph total time taken = 105 / 8.5 + 105 / 11.5 = 21.48 hours answer is b"

a ) 81 , b ) 60 , c ) 21.48 hours , d ) 6 , e ) 1 / 24

c

add(multiply(add(add(10, 1.5), subtract(10, 1.5)), 105), multiply(subtract(add(divide(105, add(10, 1.5)), divide(105, subtract(10, 1.5))), add(add(10, 1.5), subtract(10, 1.5))), const_60))

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

physics

a girl scout was selling boxes of cookies . in a month , she sold both boxes of chocolate chip cookies ( $ 1.25 each ) and boxes of plain cookies ( $ 0.75 each ) . altogether , she sold 1,585 boxes for a combined value of $ 1 , 587.75 . how many boxes of plain cookies did she sell ?

"let # plain cookies sold be x then # chocolate cookies = ( total cookies - x ) equating for x ( 0.75 ) * x + ( 1.25 ) * ( 1585 - x ) = 1587.75 = > x = 787 e"

a ) 7.05 , b ) 787 , c ) 25 , d ) 3576 , e ) 6.7 kg .

b

divide(add(const_1000, 587.75), const_2)

add(n4,const_1000)|divide(#0,const_2)|

other

find large number from below question the difference of two numbers is 1365 . on dividing the larger number by the smaller , we get 6 as quotient and the 10 as remainder

"let the smaller number be x . then larger number = ( x + 1365 ) . x + 1365 = 6 x + 10 5 x = 1355 x = 271 large number = 271 + 1365 = 1636 a"

a ) 4 / 7 , b ) 25 , c ) 126 , d ) 10 , e ) 1636

e

multiply(divide(subtract(1365, 10), subtract(6, const_1)), 6)

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

general

two assembly line inspectors , lauren and steven , inspect widgets as they come off the assembly line . if lauren inspects every fifth widget , starting with the fifth , and steven inspects every fourth , starting with the fourth , how many of the 98 widgets produced in the first hour of operation are not inspected by either inspector ?

widgets inspected by lauren : ( ( 95 - 5 ) / 5 ) + 1 = 18 + 1 = 19 widgets inspected by steven : ( ( 96 - 4 ) / 4 ) + 1 = 23 + 1 = 24 widgets inspected by both : ( ( 96 / 12 ) + 1 = 9 total : 19 + 24 - 9 = 34 hence , widgets not inspected : 98 - 34 = 64 option d

a ) 22.4 hours , b ) 192 kmph , c ) 12 , d ) 1700 , e ) 64

e

subtract(98, subtract(add(floor(divide(98, add(const_4, const_1))), floor(divide(98, const_4))), floor(divide(98, add(const_10, add(const_4, const_1))))))

add(const_1,const_4)|divide(n0,const_4)|add(#0,const_10)|divide(n0,#0)|floor(#1)|divide(n0,#2)|floor(#3)|add(#6,#4)|floor(#5)|subtract(#7,#8)|subtract(n0,#9)

other

what least number must be added to 3000 to obtain a number exactly divisible by 19 ?

"on dividing 3000 by 19 , we get 17 as remainder .  number to be added = ( 19 - 17 ) = 2 . answer a 2"

a ) 14 , b ) 0.15 , c ) 100 , d ) 158 , e ) 2

e

subtract(multiply(add(multiply(const_4, const_10), const_2), 19), 3000)

multiply(const_10,const_4)|add(#0,const_2)|multiply(n1,#1)|subtract(#2,n0)|

general

the probability that event a occurs is 0.4 , and the probability that events a and b both occur is 0.45 . if the probability that either event a or event b occurs is 0.6 , what is the probability that event b will occur ?

"p ( a or b ) = p ( a ) + p ( b ) - p ( a n b ) 0.6 = 0.4 + p ( b ) - 0.45 p ( b ) = 0.55 ans : e"

a ) 55 , b ) 4.3 days , c ) 0.55 , d ) 1 , e ) 14

c

subtract(add(0.6, 0.45), 0.4)

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

other

the length of the bridge , which a train 130 metres long and travelling at 36 km / hr can cross in 45 seconds , is :

"speed = [ 36 x 5 / 18 ] m / sec = 10 m / sec time = 45 sec let the length of bridge be x metres . then , ( 130 + x ) / 45 = 10 = > 130 + x = 450 = > x = 320 m . answer : a"

a ) 68 kmph , b ) 672 , c ) 16 , d ) 320 m , e ) 11 sec

d

subtract(multiply(divide(multiply(36, speed(const_1000, const_1)), speed(const_3600, const_1)), 45), 130)

speed(const_1000,const_1)|speed(const_3600,const_1)|multiply(n1,#0)|divide(#2,#1)|multiply(n2,#3)|subtract(#4,n0)|

physics

the speed of a car increases by 2 kms after every one hour . if the distance travelled in the first one hour was 35 kms , what was the total distance travelled in 12 hours ?

"total distance travelled in 12 hours = ( 35 + 37 + 39 + . . . upto 12 terms ) . this is an a . p . with first term , a = 35 , number of terms , n = 12 , common difference d = 2 required distance = 12 / 2 ( 2 * 35 + ( 12 - 1 ) * 2 ) = 6 ( 70 + 22 ) = 552 km . correct option : c"

a ) 552 kms , b ) 375 , c ) 80 minutes , d ) 89 % , e ) 100

a

multiply(add(multiply(2, 35), multiply(subtract(12, const_1), 2)), divide(12, 2))

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

physics

evaluate : 11110 + 24 * 3 * 10 = ?

"according to order of operations , 24 ? 3 ? 10 ( division and multiplication ) is done first from left to right 24 / 2 = 8 * 10 = 80 hence 11110 + 24 * 3 * 10 = 11110 + 80 = 11190 correct answer c"

a ) 49.9 kg , b ) 1 : 4', ' , c ) 18 , d ) 642 , e ) 11190

e

subtract(11110, multiply(multiply(24, 3), 10))

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

general

mariah has decided to hire three workers . to determine whom she will hire , she has selected a group of 15 candidates . she plans to have one working interview with 3 of the 15 candidates every day to see how well they work together . how many days will it take her to have working interviews with all the different combinations of job candidates ?

"360 . answer d"

a ) 1 / 27 , b ) $ 78.80 , c ) 54 , d ) 360 , e ) 274

d

subtract(subtract(subtract(divide(divide(divide(factorial(15), factorial(subtract(15, 3))), factorial(3)), const_2), 15), 15), const_10)

factorial(n0)|factorial(n1)|subtract(n0,n1)|factorial(#2)|divide(#0,#3)|divide(#4,#1)|divide(#5,const_2)|subtract(#6,n0)|subtract(#7,n0)|subtract(#8,const_10)|

physics

what is the difference between the largest number and the least number written with the digits 6 , 3 , 2 , 5 ?

"explanation : 2356 6532 - - - - - - - - - - - - 4176 answer : b"

a ) 1225 , b ) 38.9 % , c ) 4176 , d ) 5.6 , e ) 665

c

subtract(add(add(add(multiply(multiply(6, const_100), const_10), multiply(5, const_100)), multiply(3, const_10)), 2), add(add(add(const_1000, multiply(3, const_100)), multiply(5, const_10)), 6))

multiply(n0,const_100)|multiply(n3,const_100)|multiply(n1,const_10)|multiply(n1,const_100)|multiply(n3,const_10)|add(#3,const_1000)|multiply(#0,const_10)|add(#6,#1)|add(#5,#4)|add(#7,#2)|add(n0,#8)|add(n2,#9)|subtract(#11,#10)|

general

a money lender lent rs . 1000 at 4 % per year and rs . 1400 at 5 % per year . the amount should be returned to him when the total interest comes to rs . 350 . find the number of years .

( 1000 xtx 4 / 100 ) + ( 1400 xtx 5 / 100 ) = 350 â † ’ t = 3.2 answer a

a ) 44 . , b ) 5 , c ) 3.2 , d ) 43 % , e ) 72

c

divide(350, add(divide(multiply(4, 1000), const_100), divide(multiply(1400, 5), const_100)))

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

gain

a group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 5 with 4 students left over . what is the sum of the two smallest possible values of n ?

"n = 4 k + 1 = 5 j + 4 let ' s start at 1 = 4 ( 0 ) + 1 and keep adding 4 until we find a number in the form 5 j + 4 . 1 , 5 , 9 = 5 ( 1 ) + 4 the next such number is 9 + 4 * 5 = 29 . 9 + 29 = 38 the answer is c ."

a ) 16 , b ) rs 66.66 , c ) 38 , d ) 570.07 , e ) 36 metre

c

add(add(multiply(5, const_2), 4), add(multiply(5, multiply(const_2, 4)), 4))

multiply(n2,const_2)|multiply(const_2,n3)|add(n3,#0)|multiply(n2,#1)|add(n3,#3)|add(#2,#4)|

general

in a division sum , the remainder is 6 and the divisor is 5 times the quotient and is obtained by adding 7 to the thrice of the remainder . the dividend is

divisor = ( 6 * 3 ) + 7 = 25 5 * quotient = 25 quotient = 5 . dividend = ( divisor * quotient ) + remainder dividend = ( 20 * 5 ) + 6 = 106 . e )

a ) 6666 , b ) 6 cm', ' , c ) 106 , d ) 15 sec , e ) 12

c

add(multiply(add(multiply(6, const_3), 7), divide(add(multiply(6, const_3), 7), 5)), 6)

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

general

a and b together can do a piece of work in 8 days . if a alone can do the same work in 20 days , then b alone can do the same work in ?

"b = 1 / 8 – 1 / 22 = 0.075 days answer : a"

a ) 4749 , b ) 0 , c ) 35 , d ) 0.075 days , e ) 30

d

inverse(subtract(inverse(8), inverse(20)))

inverse(n0)|inverse(n1)|subtract(#0,#1)|inverse(#2)|

physics

circular gears l and r start to rotate at the same time at the same rate . gear l makes 20 complete revolutions per minute and gear r makes 60 revolutions per minute . how many seconds after the gears start to rotate will gear r have made exactly 8 more revolutions than gear l ?

"gear l - - 20 rotations per 60 seconds - - 2 rotation per 6 seconds . gear r - - 60 rotations per 60 seconds - - 6 rotations per 6 seconds . first 6 seconds - - gear l makes 1 rotation . - - gear r makes 4 rotations - - net difference - - 4 rotations hence every 6 seconds the difference between the number of rotations of r and l gear is 4 units . required net difference should be 8 rotations = > 2 ( 6 seconds later ) = = > 12 seconds . answer : e"

a ) 12 , b ) 12 / 7 , c ) 4327 , d ) $ 1.50 , e ) 80

a

divide(divide(8, subtract(divide(60, const_60), divide(20, const_60))), const_3)

divide(n1,const_60)|divide(n0,const_60)|subtract(#0,#1)|divide(n2,#2)|divide(#3,const_3)|

physics

for all positive integers m and v , the expression m θ v represents the remainder when m is divided by v . what is the value of ( ( 90 θ 33 ) θ 17 ) - ( 97 θ ( 33 θ 17 ) ) ?

( ( 90 θ 33 ) θ 17 ) the remainder of 90 divided by 33 is 24 ; the remainder of 24 divided by 17 is 7 ; ( 97 θ ( 33 θ 17 ) ) the remainder of 33 divided by 17 is 16 ; the remainder of 97 divided by 16 is 1 . 7 - 1 = 6 . answer : d .

a ) 6 , b ) 35 , c ) 257 , d ) 8 / 3 , e ) 605.03

a

subtract(reminder(reminder(90, 33), 17), reminder(97, reminder(33, 17)))

reminder(n0,n1)|reminder(n1,n2)|reminder(#0,n2)|reminder(n3,#1)|subtract(#2,#3)

general

what is the range of all the roots of | x ^ 2 - 3 | = x ?

"we get 2 quadratic equations here . . 1 ) x ^ 2 - x - 3 = 0 . . . . . . . roots 2 , - 1 2 ) x ^ 2 + x - 3 = 0 . . . . . . . . roots - 2 , 1 inserting each root in given equation , it can be seen that - 1 and - 2 do not satisfy the equations . so value of x for given equation . . . . x = 3 or x = 1 i guess range is 3 - 1 = 2 c"

a ) 6 / 25 , b ) 480 , c ) 324 , d ) 9 , e ) 2

e

sqrt(3)

sqrt(n1)|

general

if 1,000 microns = 1 decimeter , and 1,000 , 000,000 angstroms = 1 decimeter , how many angstroms equal 1 micron ?

"given that 1,000 microns = 1 decimeter = 1,000 , 000,000 angstroms so , 1 micron = 1,000 , 000,000 / 1,000 = 1 , 000,000 answer : c"

a ) 50 % , b ) 31 % , c ) 245 , d ) 1 , 000,000 , e ) 205

d

multiply(divide(1, multiply(const_100, const_100)), multiply(const_100, const_100))

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

general

if the sales tax reduced from 3 1 / 2 % to 3 1 / 3 % , then what difference does it make to a person who purchases an article with market price of $ 8400 ?

"required difference = [ 3 1 / 2 % of $ 8400 ] – [ 3 1 / 3 % of $ 8400 ] = [ ( 7 / 20 ) - ( 10 / 3 ) ] % of $ 8400 = 1 / 6 % of $ 8400 = $ [ ( 1 / 6 ) * ( 1 / 100 ) * 8400 ] = $ 14 . answer a ."

a ) 40 , b ) 14 , c ) 1 / 5 , d ) 15 , e ) 7

b

divide(multiply(subtract(add(divide(1, 2), 3), add(divide(1, 3), 3)), 8400), const_100)

divide(n1,n2)|divide(n1,n0)|add(n0,#0)|add(n0,#1)|subtract(#2,#3)|multiply(n6,#4)|divide(#5,const_100)|

general

the sum of three consecutive numbers is 63 . the greatest among these three number is :

"let the numbers be x , x + 1 and x + 2 then , x + ( x + 1 ) + ( x + 2 ) = 63 3 x = 60 x = 20 greatest number , ( x + 2 ) = 22 . answer : d"

a ) 22 , b ) 25 % , c ) 1800 , d ) d , e ) 1 / 26

a

divide(add(63, const_1), const_2)

add(n0,const_1)|divide(#0,const_2)|

physics

machine p and machine q are each used to manufacture 770 sprockets . it takes machine p 10 hours longer to produce 770 sprockets than machine q . machine q produces 10 % more sprockets per hour than machine a . how many sprockets per hour does machine a produce ?

"p makes x sprockets per hour . then q makes 1.1 x sprockets per hour . 770 / x = 770 / 1.1 x + 10 1.1 ( 770 ) = 770 + 11 x 11 x = 77 x = 7 the answer is c ."

a ) 28 , b ) 5 : 8 , c ) 9.1 litres , d ) 84.6 % , e ) 7

e

divide(subtract(770, divide(770, add(divide(10, const_100), const_1))), 10)

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

gain

the c . p of 20 books is equal to the s . p of 30 books . find his gain % or loss % ?

explanation : 20 cp = 30 sp 30 - - - 10 cp loss 100 - - - ? = > 33.33 % loss answer : c

a ) 28 , b ) 33.33 % , c ) 90.25 sq cm , d ) 1 : 88 , e ) 0

b

multiply(subtract(const_1, divide(20, 30)), const_100)

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

gain

weights of two friends ram and shyam are in the ratio 3 : 5 . if ram ' s weight is increased by 10 % and total weight of ram and shyam become 82.8 kg , with an increases of 15 % . by what percent did the weight of shyam has to be increased ?

"solution : given ratio of ram and shayam ' s weight = 3 : 5 hence , ( x - 15 ) / ( 15 - 10 ) = 3 / 5 or , x = 18 % . answer : option a"

a ) 18 % , b ) 23 , c ) 166.6 , d ) s . 4076 , e ) 10 sec

a

add(15, multiply(subtract(15, 10), divide(3, 5)))

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

gain

in the manufacture of a certain product , 7 percent of the units produced are defective and 4 percent of the defective units are shipped for sale . what percent of the units produced are defective units that are shipped for sale ?

"percent of defective produced = 7 % percent of the defective units that are shipped for sale = 4 % percent of units produced are defective units that are shipped for sale = ( 4 / 100 ) * ( 7 / 100 ) * 100 % = ( 28 / 10000 ) * 100 % = ( 28 / 100 ) % = . 28 % answer b"

a ) 0.28 % , b ) 1076 , c ) 9 / 14 , d ) 72 , e ) 150

a

multiply(7, divide(4, const_100))

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

gain

the average age of 35 students in a class is 16 years . the average age of 21 students is 14 . what is the average age of remaining 7 students ?

"solution sum of the ages of 14 students = ( 16 x 35 ) - ( 14 x 21 ) = 560 - 294 . = 266 . ∴ required average = 266 / 7 = 38 years . answer d"

a ) 16 , b ) 2.9 , c ) 180 km , d ) 38 years , e ) $ 44.33

d

subtract(add(add(multiply(35, 16), 21), 35), multiply(35, 16))

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

general

a trader bought a car at 30 % discount on its original price . he sold it at a 80 % increase on the price he bought it . what percent of profit did he make on the original price ?

"original price = 100 cp = 70 s = 70 * ( 180 / 100 ) = 126 100 - 126 = 26 % answer : b"

a ) 870 , b ) $ 1260 , c ) 1260 , d ) 26 % , e ) s . 240

d

multiply(subtract(divide(divide(multiply(subtract(const_100, 30), add(const_100, 80)), const_100), const_100), const_1), const_100)

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

gain

a boy goes to his school from his house at a speed of 3 km / hr and return at a speed of 2 km / hr . if he takes 5 hours in going and coming , the distance between his house and school is ?

average speed = 2 * 3 * 2 / 3 + 2 = 12 / 5 km / hr distance traveled = 12 / 5 * 5 = 12 km distance between house and school = 12 / 2 = 6 km answer is b

a ) 6 km , b ) rs . 1350 , c ) 98 m , d ) 25 , e ) 86400

a

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

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

physics

how many seconds will a 600 meter long train take to cross a man walking with a speed of 3 km / hr in the direction of the moving train if the speed of the train is 63 km / hr ?

let length of tunnel is x meter distance = 600 + x meter time = 1 minute = 60 seconds speed = 78 km / hr = 78 * 5 / 18 m / s = 65 / 3 m / s distance = speed * time 600 + x = ( 65 / 3 ) * 60 600 + x = 20 * 65 = 1300 x = 1300 - 600 = 700 meters answer : a

a ) 787 , b ) 2013 , c ) 50 litres , d ) 700 , e ) 982.14', '

d

multiply(multiply(subtract(divide(600, multiply(subtract(63, 3), const_0_2778)), const_1), const_10), const_2)

subtract(n2,n1)|multiply(#0,const_0_2778)|divide(n0,#1)|subtract(#2,const_1)|multiply(#3,const_10)|multiply(#4,const_2)

physics

a can do a piece of work in 4 hours ; b and c together can do it in 3 hours , which a and b together can do it in 2 hours . how long will c alone take to do it ?

a ' s 1 hour work = 1 / 4 ; ( b + c ) ' s 1 hour work = 1 / 3 ; ( a + b ) ' s 1 hour work = 1 / 2 ( a + b + c ) ' s 1 hour work = ( 1 / 4 + 1 / 3 ) = 7 / 12 c ' s 1 hour work = ( 7 / 12 - 1 / 2 ) = 1 / 12 c alone will take 12 hours to do the work . answer : a

a ) 12 hours , b ) 1 / 32 , c ) 3 , d ) 23 / 29 , e ) $ 333.33

a

divide(const_1, subtract(divide(const_1, 3), subtract(divide(const_1, 2), divide(const_1, 4))))

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

physics

a particular library has 75 books in a special collection , all of which were in the library at the beginning of the month . these book are occasionally loaned out through an inter - library program . if , by the end of the month , 60 percent of books that were loaned out are returned and there are 65 books in the special collection at that time , how many books of the special collection were loaned out during that month ?

"total = 75 books . 60 % of books that were loaned out are returned - - > 100 % - 60 % = 40 % of books that were loaned out are not returned . now , there are 68 books , thus 75 - 65 = 10 books are not returned . { loaned out } * 0.4 = 10 - - > { loaned out } = 25 . answer : b ."

a ) 120 , b ) 25 , c ) $ 370,000 , d ) 40 , e ) 2520

b

divide(subtract(75, 65), subtract(const_1, divide(60, const_100)))

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

gain

cost is expressed by the formula tb ^ 4 . if b is doubled , the new cost q is what percent of the original cost ?

"original cost c 1 = t 1 * b 1 ^ 4 new cost c 2 = t 2 * b 2 ^ 4 . . . . only b is doubled so t 2 = t 1 and b 2 = 2 b 1 c 2 = t 2 * ( 2 b 1 ) ^ 4 = 16 ( t 1 * b 1 ^ 4 ) = 16 c 1 16 times c 1 = > 1600 % of c 1 ans d = 1600"

a ) q = 1600 , b ) 252 sec , c ) 6666 , d ) 33 % , e ) 94 kmph

a

multiply(power(const_2, 4), const_100)

power(const_2,n0)|multiply(#0,const_100)|

general

albert is 2 times mary ’ s age and 4 times as old as betty . mary is 12 years younger than albert . how old is betty ?

"a = 2 m = m + 12 m = 12 a = 24 a = 4 b , and so b = 6 the answer is a ."

a ) 2 , b ) 90 o , c ) 28 , d ) 6 , e ) 5 / 8

d

divide(multiply(2, 12), 4)

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

general

what is the next number : 2 , 10 , 82 , __

"3 ^ 0 + 1 = 2 3 ^ 2 + 1 = 10 3 ^ 4 + 1 = 82 3 ^ 6 + 1 = 730 the answer is b ."

a ) 612 , b ) 4.3 days , c ) 2250 , d ) 2011 , e ) 730

e

subtract(subtract(subtract(multiply(82, const_10), const_100), 2), 2)

multiply(n2,const_10)|subtract(#0,const_100)|subtract(#1,n0)|subtract(#2,n0)|

general

the wages earned by robin is 40 % more than that earned by erica . the wages earned by charles is 60 % more than that earned by erica . how much % is the wages earned by charles more than that earned by robin ?

"explanatory answer let the wages earned by erica be $ 100 then , wages earned by robin and charles will be $ 140 and $ 160 respectively . charles earns $ 40 more than robin who earns $ 140 . therefore , charles ' wage is 40 / 140 * 100 = 28.57 % . the correct choice is ( c )"

a ) 73 , b ) 24 , c ) 70 , d ) 6 / 7 , e ) 28.57 %

e

multiply(divide(subtract(add(const_100, 60), add(const_100, 40)), add(const_100, 40)), const_100)

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

general