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shangdatalab-ucsd

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mathqa_choice_confusion

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how many boxes do we need if we have to carry 250 apples into boxes that each hold 25 apples ?

sol . apples 250 each carries 25 = 250 / 25 = 10 answer : d

a ) 10 , b ) 3 , 012,121 , c ) 4 , d ) 40.2 degrees c , e ) 1040

a

divide(250, 25)

divide(n0,n1)

general

how many seconds does sandy take to cover a distance of 600 meters , if sandy runs at a speed of 15 km / hr ?

"15 km / hr = 15000 m / 3600 s = ( 150 / 36 ) m / s = ( 25 / 6 ) m / s time = 600 / ( 25 / 6 ) = 144 seconds the answer is c ."

a ) 21.48 hours , b ) 144 , c ) t = 32 , d ) 9.5 gallons , e ) 10

b

divide(600, multiply(15, const_0_2778))

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

physics

if shares of two persons in profits are rs . 600 and rs . 300 then ratio of their capitals is

"total profit = 1000 ratio = 600 / 300 = 2 : 1 answer : e"

a ) - 5 , b ) 1200 , c ) 2 : 1 , d ) 30 , e ) 6

c

divide(600, 300)

divide(n0,n1)|

other

the mean of 50 observations was 36 . it was found later that an observation 48 was wrongly taken as 23 . the corrected new mean is :

"correct sum = ( 36 * 50 + 48 - 23 ) = 1825 . correct mean = 1825 / 50 = 36.5 answer a"

a ) 50 , b ) 5.5 % , c ) 320 km , d ) 0.5 , e ) 36.5

e

divide(add(multiply(36, 50), subtract(subtract(50, const_2), 23)), 50)

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

general

a school has received 60 % of the amount it needs for a new building by receiving a donation of $ 500 each from people already solicited . people already solicited represent 50 % of the people from whom the school will solicit donations . how much average contribution is requited from the remaining targeted people to complete the fund raising exercise ?

"let us suppose there are 100 people . 50 % of them donated $ 25000 ( 500 * 50 ) $ 25000 is 60 % of total amount . so total amount = 25000 * 100 / 60 remaining amount is 40 % of total amount . 40 % of total amount = 25000 * ( 100 / 60 ) * ( 40 / 100 ) = 50000 / 3 this amount has to be divided by 50 ( remaining people are 50 ) so per head amount is 50000 / 3 / 50 = 32000 / 180 = 333.33 ; answer : b"

a ) 16 / 625 , b ) 450 , c ) 3 / 4000 , d ) 300 , e ) $ 333.33

e

divide(multiply(divide(multiply(divide(50, const_100), 500), divide(60, const_100)), divide(50, const_100)), divide(60, const_100))

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

general

a certain bacteria colony doubles in size every day for 19 days , at which point it reaches the limit of its habitat and can no longer grow . if two bacteria colonies start growing simultaneously , how many days will it take them to reach the habitat ’ s limit ?

"if there is one bacteria colony , then it will reach the limit of its habitat in 20 days . if there are two bacteria colonies , then in order to reach the limit of habitat they would need to double one time less than in case with one colony . thus colonies need to double 18 times . answer : d . similar questions to practice : hope it helps ."

a ) 41 , b ) 18 , c ) 5 , d ) 7.9 % , e ) 80 %

b

subtract(19, divide(19, 19))

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

physics

the diameter of a circle is 4 / π . find the circumference of the circle .

circumference = 2 * pi * r = 2 * pi * 4 / pi = > 8 a

a ) 49995 , b ) 60 , c ) 15 , d ) 10 , e ) 8', '

e

circumface(divide(4, const_pi))

divide(n0,const_pi)|circumface(#0)

geometry

simplify : 0.3 * 0.3 + 0.3 * 0.3

"given exp . = 0.3 * 0.3 + ( 0.3 * 0.3 ) = 0.09 + 0.09 = 0.18 answer is c ."

a ) 0.18 , b ) 29 , c ) 1 / 6 , d ) 27.38 % , e ) 45 %

a

add(multiply(0.3, 0.3), multiply(0.3, 0.3))

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

general

how many prime numbers exist between 260 and 280 ?

straight a way lets exclude all the even numbers between 260 and 280 . so now the number starts from 261 to 279 ( only odd ) 261 is a divisible of 3 and next odd divisible by 3 will be 261 + 6 = 267 + 6 = 273 + 6 = 279 . also we can eliminate numbers ending with ' 5 ' so in odd , the excluded numbers are 261 , 265,267 , 273,279 , which leave us with 263,269 , 271,277 . checked the above listed four numbers are divisible by any numbers till 20 . answer : e

a ) 10 , b ) 44 % , c ) 90 , d ) 760 , e ) four

e

subtract(divide(subtract(280, 260), const_4), const_1)

subtract(n1,n0)|divide(#0,const_4)|subtract(#1,const_1)

general

one robot builds a robot in 1 hour , another one builds a robot in 2 hours . the 2 robots work together and when a new robot is complete , it joins the others , working at a constant rate of one robot in 2 hours . how much time will it take until there are 8 robots altogether , if the robots build only one robot at a time ?

"first the 2 robots work at the rate of 1 + 1 / 2 = 3 / 2 so they complete one robot in 2 / 3 rd of an hour = 40 minutes - ( 1 ) now the 3 robots work together at the rate of 1 + 1 / 2 + 1 / 2 = 4 / 2 = 2 / 1 so they complete one robot in 1 / 2 an hour , i . e 30 minutes - ( 2 ) now the 4 robots work together at the rate of 1 + 1 / 2 + 1 / 2 + 1 / 2 = 5 / 2 so they complete one robot in 2 / 5 th of an hour - ( 3 ) now the 5 robots work together at the rate of 1 + 1 / 2 + 1 / 2 + 1 / 2 + 1 / 2 = 3 so they complete one robot in 1 / 3 th of an hour - ( 4 ) now the 6 robots work together at the rate of 1 + 1 / 2 + 1 / 2 + 1 / 2 + 1 / 2 + 1 / 2 = 7 / 2 so they complete one robot in 2 / 7 th of an hour - ( 5 ) now the 7 robots work together at the rate of 1 + 1 / 2 + 1 / 2 + 1 / 2 + 1 / 2 + 1 / 2 + 1 / 2 = 9 / 2 so they complete one robot in 2 / 9 th of an hour - ( 6 ) and now we have 8 robots so total = ( 1 ) + ( 2 ) + ( 3 ) + ( 4 ) + ( 5 ) + ( 6 ) = 146 1 / 7 minutes answer - e"

a ) 3 : 5 , b ) 1 / 12 , c ) 146 1 / 7 min , d ) $ 14.00 , e ) 3.33

c

add(inverse(add(add(inverse(1), inverse(2)), inverse(2))), inverse(add(inverse(1), inverse(2))))

inverse(n0)|inverse(n1)|add(#0,#1)|add(#2,#1)|inverse(#2)|inverse(#3)|add(#5,#4)|

physics

{ - 10 , - 6 , - 5 , - 4 , - 2.5 , - 1 , 0 , 2.5 , 4 , 6 , 7 , 10 } a number is to be selected at random from the set above . what is the probability that the number will be a solution to the equation ( x - 4 ) ( x + 9 ) ( 2 x + 5 ) = 0 ?

x = - 2.5 prob = 1 / 12 answer - a

a ) 1 / 12 , b ) 1764713 , c ) 50 . , d ) 120 , e ) 12', '

a

divide(1, multiply(6, 2))

multiply(n1,n14)|divide(n5,#0)

general

a crow leaves its nest , and flies back and forth from its nest to a nearby ditch to gather worms . the distance between the nest and the ditch is 300 meters . in one and a half hours , the crow manages to bring worms to its nest 15 times . what is the speed of the crow in kilometers per hour ?

"the distance between the nest and the ditch is 300 meters . 15 times mean = a crow leaves its nest , and flies back ( going and coming back ) i . e . 2 times we get total 30 rounds . so the distance is 30 * 300 = 9000 . d = st 9000 / 1.5 = t , i think we can take 9000 meters as 9 km , then only we get t = 6 . ( 1000 meters = 1 km ) d )"

a ) 39 , b ) 378 , c ) 6 , d ) 9.2 miles , e ) 184

c

divide(divide(multiply(300, multiply(15, const_2)), const_1000), divide(15, const_10))

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

physics

a and b are two multiples of 36 , and q is the set of consecutive integers between a and b , inclusive . if q contains 9 multiples of 9 , how many multiples of 4 are there in q ?

"now given that q is set the consecutive integers between a and b . and q contains 9 multiples of 9 let take a as 36 . then 36 45 54 63 72 81 90 99 108 . . . so b will 108 . now let ' s check the multiples of 4 among this set 108 - 36 / 4 + 1 = > 18 + 1 = > 19 ans option b ."

a ) 0.7 , b ) 7 2 / 3 , c ) 7 , d ) 18911 , e ) 19

e

subtract(multiply(9, const_2), const_1)

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

physics

find the value of ( 20 c 18 ) * ( 20 c 20 )

"20 c 20 = 1 ( 20 c 2 ) * ( 20 c 20 ) = 20 ! * 1 / 18 ! = 20 * 19 * 18 ! / 18 ! = 20 * 19 * 1 = 380 answer : b"

a ) 1.33 , b ) 380 , c ) 45 , d ) 7.94 , e ) 32

b

multiply(add(divide(18, 20), 20), 20)

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

general

if 30 % of a class averages 95 % on a test , 50 % of the class averages 79 % on the test , and the remainder of the class averages 60 % on the test , what is the overall class average ? ( round final answer to the nearest percent ) .

"this question is a weighted average question with a series of dependent variables . the remaining portion of the class represents 100 % - 30 % - 50 % = 20 % of the class converting the portions of the class population to decimal weights , we find : class average = 0.30 x 95 + 0.50 x 79 + 0.20 x 60 = 80 the class average ( rounded ) is 80 % final answer e ) 80 %"

a ) 36.92 mph , b ) 6 , c ) 80 % , d ) 8925 , e ) $ 600

c

divide(add(add(multiply(30, 95), multiply(50, 79)), multiply(subtract(const_100, add(30, 50)), 60)), const_100)

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

gain

two kinds of vodka are mixed in the ratio 2 : 6 and 6 : 3 and they are sold fetching the profit 10 % and 20 % respectively . if the vodkas are mixed in equal ratio and the individual profit percent on them are increased by 4 / 3 and 5 / 3 times respectively , then the mixture will fetch the profit of

"the profit on the first kind of vodka = x % ; the profit on the second kind of vodka = y % . when they are mixed in the ratio 1 : 2 ( total of 3 parts ) the average profit is 10 % : ( x + 2 y ) / 3 = 10 . when they are mixed in the ratio 2 : 1 ( total of 3 parts ) the average profit is 20 % : ( 2 x + y ) / 3 = 20 . solving gives : x = 30 % and y = 0 % . after the individual profit percent on them areincreased by 4 / 3 and 5 / 3 times respectively the profit becomes 40 % and 0 % , on the first and te second kinds of vodka , respectively . if they are mixed in equal ratio ( 1 : 1 ) , then the mixture will fetch the profit of ( 40 + 0 ) / 2 = 20 % . answer : a"

a ) 22.5 % , b ) 14 , c ) 40 % , d ) 5 , e ) 20 %

e

add(divide(multiply(10, 4), 6), add(10, 5))

add(n4,n8)|multiply(n4,n6)|divide(#1,const_2.0)|add(#0,#2)|

general

if the tens digit of positive integers m , y are 6 , how many values of the tens digit of 2 ( m + y ) can be there ?

if the tens digit of positive integers m , y are 6 , how many values of the tens digit of 2 ( m + y ) can be there ? a . 2 b . 3 c . 4 d . 5 e . 6 - > if m = y = 60 , 2 ( m + y ) = 240 is derived . if m = y = 69 , 2 ( m + y ) = 276 is derived , which makes 4,5 , 6,7 possible for the tens digit . therefore , the answer is c .

a ) 12 / 5 , b ) $ 700 , c ) 9 : 1 , d ) 4 , e ) rs . 2700

d

subtract(6, 2)

subtract(n0,n1)

physics

the maximum number of students among them 848 pens and 630 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 = 848 number of pencils = 630 required number of students = h . c . f . of 848 and 630 = 2 answer is c"

a ) 2 , b ) $ 0.002 , c ) 66 , d ) 5.5 , e ) s . 20

a

gcd(848, 630)

gcd(n0,n1)|

general

find the ratio of the curved surfaces of two cylinders of same heights if their radii are in the ratio 1 : 2 ?

1 : 2 answer : a

a ) 7 , b ) 59 , c ) 0.1 , d ) 4608 , e ) 1 : 2', '

e

divide(1, 2)

divide(n0,n1)

geometry

n and m are each 3 - digit integers . each of the numbers 1 , 2 , 3 , 7 , 8 and 9 is a digit of either n or m . what is the smallest possible positive difference between n and m ?

"you have 6 digits : 1 , 2 , 3 , 7 , 8 , 9 each digit needs to be used to make two 3 digit numbers . this means that we will use each of the digits only once and in only one of the numbers . the numbers need to be as close to each other as possible . the numbers can not be equal so the greater number needs to be as small as possible and the smaller number needs to be as large as possible to be close to each other . the first digit ( hundreds digit ) of both numbers should be consecutive integers now let ' s think about the next digit ( the tens digit ) . to minimize the difference between the numbers , the tens digit of the greater number should be as small as possible and the tens digit of the smaller number should be as large as possible . so let ' s not use 1 and 9 in the hundreds places and reserve them for the tens places . now what are the options ? try and make a pair with ( 2 * * and 3 * * ) . make the 2 * * number as large as possible and make the 3 * * number as small as possible . 298 and 317 ( difference is 19 ) or try and make a pair with ( 7 * * and 8 * * ) . make the 7 * * number as large as possible and make the 8 * * number as small as possible . we get 793 and 812 ( difference is 19 ) a"

a ) 625 , b ) 1.52 % , c ) 19 , d ) 3 / 7 , e ) 2400

c

subtract(subtract(const_100, multiply(subtract(8, 1), const_10)), const_1)

subtract(n5,n1)|multiply(#0,const_10)|subtract(const_100,#1)|subtract(#2,const_1)|

general

the captain of a cricket team of 11 members is 29 years old and the wicket keeper is 3 years older . if the ages of these two are excluded , the average age of the remaining players is one year less than the average age of the whole team . what is the average age of the team ?

"explanation : let the average age of the whole team by x years . 11 x â € “ ( 29 + 32 ) = 9 ( x - 1 ) 11 x â € “ 9 x = 52 2 x = 52 x = 26 . so , average age of the team is 26 years . answer e"

a ) 8 / 9 , b ) $ 300 , c ) 12 , d ) 26 years , e ) 145

d

divide(subtract(add(29, add(29, 3)), multiply(3, 3)), const_2)

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

general

if the sides of a triangle are 31 cm , 29 cm and 15 cm , what is its area ?

"the triangle with sides 31 cm , 29 cm and 15 cm is right angled , where the hypotenuse is 31 cm . area of the triangle = 1 / 2 * 29 * 15 = 217.5 cm 2 answer : e"

a ) 128 , b ) 1000 , c ) 217.5 cm 2 , d ) 12 , e ) 40

c

divide(multiply(29, 15), const_2)

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

geometry

in what ratio should a variety of rice costing rs . 6.5 per kg be mixed with another variety of rice costing rs . 8.75 per kg to obtain a mixture costing rs . 7.50 per kg ?

"let us say the ratio of the quantities of cheaper and dearer varieties = x : y by the rule of allegation , x / y = ( 8.75 - 7.50 ) / ( 7.50 - 6.5 ) = 5 / 4 answer : c"

a ) 4 / 5 , b ) 10 % , c ) 71 , d ) 25 % , e ) 5 / 4

e

divide(divide(subtract(8.75, 7.50), subtract(8.75, 6.5)), subtract(const_1, divide(subtract(8.75, 7.50), subtract(8.75, 6.5))))

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

other

a bag contains 7 green and 8 white balls . if two balls are drawn simultaneously , the probability that both are of the same colour is

"drawing two balls of same color from seven green balls can be done in ⁷ c ₂ ways . similarly from eight white balls two can be drawn in ⁸ c ₂ ways . p = ⁷ c ₂ / ¹ ⁵ c ₂ + ⁸ c ₂ / ¹ ⁵ c ₂ = 7 / 15 answer : e"

a ) 131.6 , b ) 1 , c ) 7 / 15 , d ) s . 295.45 , e ) 5

c

add(multiply(divide(8, add(7, 8)), divide(subtract(8, const_1), subtract(add(7, 8), const_1))), multiply(divide(7, add(7, 8)), divide(subtract(7, const_1), subtract(add(7, 8), const_1))))

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

other

the average monthly salary of 20 employees in an organisation is rs . 1400 . if the manager ' s salary is added , then the average salary increases by rs . 100 . what is the manager ' s monthly salary ?

"explanation : manager ' s monthly salary rs . ( 1500 * 21 - 1400 * 20 ) = rs . 3500 . answer : e"

a ) $ 20,800 , b ) 9 : 16 , c ) 0.3571 , d ) 3500 , e ) 318 $

d

subtract(multiply(add(1400, 100), add(20, const_1)), multiply(1400, 20))

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

general

4 men and 6 women can complete a work in 8 days , while 3 men and 7 women can complete it in 10 days . in how many days will 10 women complete it ?

"let 1 man ' s 1 day work = x and 1 woman ' s 1 day work = y . then , 4 x + 6 y = 1 / 8 and 3 x + 7 y = 1 / 10 solving these two equations , we get : x = 11 / 400 and y = 1 / 400 1 woman ' s 1 day work = ( 1 / 400 * 10 ) = 1 / 40 . hence , 10 women will complete the work in 40 days . answer : b"

a ) 200 , b ) 60 , c ) 35 , d ) 40 days , e ) 8

d

inverse(multiply(divide(subtract(divide(const_1, 10), multiply(3, divide(subtract(divide(const_1, 8), multiply(divide(6, 7), divide(const_1, 10))), subtract(4, multiply(3, divide(6, 7)))))), 7), 8))

divide(const_1,n5)|divide(const_1,n2)|divide(n1,n4)|multiply(#2,#0)|multiply(n3,#2)|subtract(#1,#3)|subtract(n0,#4)|divide(#5,#6)|multiply(n3,#7)|subtract(#0,#8)|divide(#9,n4)|multiply(n2,#10)|inverse(#11)|

physics

a furniture dealer purchased a desk for $ 180 and then set the selling price equal to the purchase price plus a markup that was 40 % of the selling price . if the dealer sold the desk at the selling price , what was the amount of the dealer ' s gross profit from the purchase and the sale of the desk ?

"purchase price = 180 selling price = x 180 + 0.4 * x = x 0.6 * x = 180 x = 300 profit = 300 - 180 = 120 answer : d"

a ) 4 / 12 , b ) 60 , c ) 41 , d ) $ 120 , e ) 5

d

divide(multiply(subtract(divide(180, subtract(const_1, divide(40, const_100))), 180), const_100), 180)

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

gain

a train 400 m long can cross an electric pole in 10 sec and then find the speed of the train ?

"length = speed * time speed = l / t s = 400 / 10 s = 40 m / sec speed = 40 * 18 / 5 ( to convert m / sec in to kmph multiply by 18 / 5 ) speed = 144 kmph answer : b"

a ) 1.8 % , b ) 6 , c ) 144 kmph , d ) 3.75 , e ) 4 cm', '

c

divide(divide(400, const_1000), divide(10, const_3600))

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

physics

a man can row his boat with the stream at 30 km / h and against the stream in 14 km / h . the man ' s rate is ?

"ds = 30 us = 14 s = ? s = ( 30 - 14 ) / 2 = 8 kmph answer : e"

a ) 18 , b ) 480 , c ) 60 , d ) 8 kmph , e ) 11 sec

d

divide(subtract(30, 14), const_2)

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

gain

3 men and 7 women can complete a work in 10 days . but 4 men and 6 women need 8 days to complete the same work . in how many days will 10 women complete the same work ?

explanation : work done by 4 men and 6 women in 1 day = 1 / 8 work done by 3 men and 7 women in 1 day = 1 / 10 let 1 man does m work in 1 day and 1 woman does w work in 1 day . the above equations can be written as 4 m + 6 w = 1 / 8 - - - ( 1 ) 3 m + 7 w = 1 / 10 - - - ( 2 ) solving equation ( 1 ) and ( 2 ) , we get m = 11 / 400 and w = 1 / 400 amount of work 10 women can do in a day = 10 × ( 1 / 400 ) = 1 / 40 ie , 10 women can complete the work in 40 days answer : option b

a ) 7 days , b ) 9216 , c ) 40 , d ) 0 , e ) 5.6

c

inverse(multiply(divide(subtract(divide(const_1, 8), multiply(4, divide(subtract(divide(const_1, 10), multiply(divide(7, 6), divide(const_1, 8))), subtract(3, multiply(4, divide(7, 6)))))), 6), 10))

divide(const_1,n5)|divide(const_1,n2)|divide(n1,n4)|multiply(#2,#0)|multiply(n3,#2)|subtract(#1,#3)|subtract(n0,#4)|divide(#5,#6)|multiply(n3,#7)|subtract(#0,#8)|divide(#9,n4)|multiply(n2,#10)|inverse(#11)

physics

what is the characteristic of the logarithm of 0.0000134 ?

log ( 0.0000134 ) . since there are four zeros between the decimal point and the first significant digit , the characteristic is – 5 . answer : b

a ) 20 , b ) - 5 , c ) 2011 . , d ) 4.8 , e ) 50 m .

b

floor(divide(log(0.0000134), log(const_10)))

log(n0)|log(const_10)|divide(#0,#1)|floor(#2)

other

in the game of dubblefud , red chips , blue chips and green chips are each worth 2 , 4 and 5 points respectively . in a certain selection of chips , the product of the point values of the chips is 16000 . if the number of blue chips in this selection doubles the number of green chips , how many red chips are in the selection ?

this is equivalent to : - 2 x * 4 y * 5 z = 16000 y / 2 = z ( given ) 2 x * 4 y * 5 y / 2 = 16000 2 x * y ^ 2 = 16000 / 10 2 x * y ^ 2 = 1600 now from options given we will figure out which number will divide 800 and gives us a perfect square : - which gives us x = 2 as 2 * 2 * y ^ 2 = 1600 y ^ 2 = 400 y = 20 number of red chips = 2 hence b

a ) 1040 , b ) $ 920.24 , c ) 125 , d ) 2 , e ) 12

d

divide(multiply(multiply(power(2, 4), power(2, const_3)), power(5, const_3)), multiply(power(const_2, multiply(2, const_3)), power(5, const_3)))

multiply(n0,const_3)|power(n0,n1)|power(n0,const_3)|power(n2,const_3)|multiply(#1,#2)|power(const_2,#0)|multiply(#4,#3)|multiply(#5,#3)|divide(#6,#7)

general

compound interest of rs . 2000 at 10 % per annum for 1 1 / 2 years will be ( interest compounded half yearly ) .

"10 % interest per annum will be 5 % interest half yearly for 3 terms ( 1 1 / 2 years ) so compound interest = 2000 [ 1 + ( 5 / 100 ) ] ^ 3 - 2000 = 2000 [ ( 21 / 20 ) ^ 3 - 1 ] = 2000 ( 9261 - 8000 ) / 8000 = 2 * 1261 / 8 = 315 answer : d"

a ) 2.9 , b ) rs . 315 , c ) 24 , d ) 4 cm', ' , e ) 1.8

b

subtract(multiply(2000, power(add(1, divide(divide(10, 2), const_100)), multiply(add(1, divide(1, 2)), 2))), 2000)

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

gain

the radius of a cone is 49 m , slant height is 35 m . find the curved surface area ?

"cone curved surface area = ï € rl 22 / 7 ã — 49 ã — 35 = 154 ã — 35 = 5390 m ( power 2 ) answer is b ."

a ) $ 840 , b ) 1014 , c ) 72 sec , d ) 5390 , e ) 50000

d

volume_cone(49, 35)

volume_cone(n0,n1)|

geometry

the ratio between the length and the breadth of a rectangular park is 3 : 2 . if a man cycling along the boundary of the park at the speed of 12 km / hr completes one round in 10 minutes , then the area of the park ( in sq . m ) is :

"perimeter = distance covered in 10 min . = ( 12000 / 60 ) x 10 m = 2000 m . let length = 3 x metres and breadth = 2 x metres . then , 2 ( 3 x + 2 x ) = 2000 or x = 200 . length = 600 m and breadth = 400 m . area = ( 600 x 400 ) m 2 = 240000 m 2 . answer : e"

a ) 240000 , b ) rs . 15,000 , c ) 12 hours , d ) 123 , e ) 5 / 4

a

rectangle_area(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 10), const_1000), add(3, 2)), const_2), multiply(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 10), const_1000), add(3, 2)), const_2), 2))

add(n0,n1)|multiply(const_2,const_3)|multiply(#1,const_10)|divide(n2,#2)|multiply(n3,#3)|multiply(#4,const_1000)|divide(#5,#0)|divide(#6,const_2)|multiply(n1,#7)|rectangle_area(#7,#8)|

physics

a train covers a distance of 11 km in 10 min . if it takes 6 sec to pass a telegraph post , then the length of the train is ?

"speed = ( 11 / 10 * 60 ) km / hr = ( 66 * 5 / 18 ) m / sec = 55 / 3 m / sec . length of the train = 55 / 3 * 6 = 110 m . answer : c"

a ) 1.1 inches', ' , b ) 17 % , c ) 90 o , d ) 27 / 20 , e ) m

e

divide(11, subtract(divide(11, 10), 6))

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

physics

on average , the boys in the class have 20 pencils and the girls have 38 pencils . if the overall class average is 30 pencils , what is the ratio of boys to girls in the class ?

( 38 g + 20 b ) / ( g + b ) = 30 38 g + 20 b = 30 ( g + b ) 8 g = 10 b b / g = 4 / 5 the answer is d .

a ) 24 , b ) 4 / 5 , c ) 15 days , d ) 49 hr , e ) 3

b

divide(30, 38)

divide(n2,n1)

general

in a certain village , 200 litres of water are required per household per month . at this rate , if there are 5 households in the village , how long ( in months ) will 2000 litres of water last ?

"i find it much easier to understand with real numbers , so choose ( almost ) any numbers to replace m , n and p : in a certain village , m 200 litres of water are required per household per month . at this rate , if there aren 5 households in the village , how long ( in months ) willp 2000 litres of water last ? water required is 200 * 5 = 1000 ( m * n ) water available is 2000 ( p ) it will last 2 months ( p / m * n ) ans : d"

a ) 106 , b ) 8 , c ) 2 , d ) 30 days . , e ) 200 %

c

divide(2000, multiply(200, 5))

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

gain

sum of two numbers prime to each other is 20 and their l . c . m . is 99 . what are the numbers ?

"as two numbers are prime , only two options satisfy ie option c and d . but option c will not make the product of numbers i . e 99 answer : d"

a ) 11 , b ) 137 / 216', ' , c ) 11 and 9 , d ) 33 % , e ) 3.1 sec

c

add(99, 20)

add(n0,n1)|

physics

a train 180 m in length crosses a telegraph post in 20 seconds . the speed of the train is ?

"s = 180 / 20 * 18 / 5 = 32 kmph answer : c"

a ) 5.33 kmph , b ) 3 , c ) 32 kmph , d ) 3 / 7 , e ) 36

c

multiply(const_3_6, divide(180, 20))

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

physics

rs . 925 becomes rs . 956 in 3 years at a certain rate of simple interest . if the rate of interest is increased by 4 % , what amount will rs . 925 become in 3 years ?

"solution s . i . = rs . ( 956 - 925 ) = rs . 31 rate = ( 100 x 31 / 925 x 3 ) = 124 / 111 % new rate = ( 124 / 111 + 4 ) % = 568 / 111 % new s . i . = rs . ( 925 x 568 / 111 x 3 / 100 ) rs . 142 ∴ new amount = rs . ( 925 + 142 ) = rs . 1067 . answer c"

a ) 55522 feet', ' , b ) 3 / 13 , c ) rs . 1067 , d ) 3 / 4 , e ) 192

c

add(925, divide(multiply(multiply(925, add(divide(multiply(subtract(956, 925), const_100), multiply(925, 3)), 4)), 3), const_100))

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

gain

exactly 3 / 7 of the people in the room are under the age of 21 , and exactly 5 / 12 of the people in the room are over the age of 65 . if the total number of the people in the room is greater than 50 and less than 100 , how many people in the room are under the age of 21 ?

"the total number of the people in the room must be a multiple of both 7 and 12 ( in order 3 / 7 and 5 / 12 of the number to be an integer ) , thus the total number of the people must be a multiple of lcm of 7 and 12 , which is 84 . since , the total number of the people in the room is greater than 50 and less than 100 , then there are 84 people in the room . therefore there are 3 / 7 * 84 = 36 people in the room under the age of 21 . answer : c ."

a ) 36 , b ) 38 % , c ) 9 , d ) 2 : 3 , e ) 2

a

divide(multiply(multiply(7, 12), 3), 7)

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

general

sides of a rectangular park are in the ratio 3 : 2 and its area is 2460 sq m , the cost of fencing it at 50 ps per meter is ?

"3 x * 2 x = 2460 = > x = 20.24 2 ( 79.76 + 50 ) = 259.52 m 259.52 * 1 / 2 = rs . 129.76 answer : c"

a ) 1 / 81 , b ) 2991 , c ) s . 129.76 , d ) 7 , e ) 45

c

divide(multiply(50, rectangle_perimeter(sqrt(divide(multiply(2460, 2), 3)), divide(2460, sqrt(divide(multiply(2460, 2), 3))))), const_100)

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

physics

what is the sum of all the odd numbers between 24 and 50 , inclusive ?

"to solve this problem , all you have to do is take every even number between 24 and 50 and add them together . so we have 25 + 27 + 29 + 31 + 33 + 35 + 37 + 39 + 41 + 43 + 45 + 47 + 49 , which is 481 . final answer : b"

a ) 80 , b ) 2 : 1 , c ) 481 , d ) 40 , e ) 25', '

c

add(add(add(add(add(add(const_12, const_2), const_1), add(add(const_12, const_2), add(add(add(add(add(const_2, const_4), const_4), subtract(const_10, const_1)), add(add(const_2, const_4), const_4)), add(const_10, const_2)))), add(add(add(const_12, const_2), const_1), const_1)), 24), add(const_2, const_4))

add(const_12,const_2)|add(const_2,const_4)|add(const_10,const_2)|subtract(const_10,const_1)|add(#0,const_1)|add(#1,const_4)|add(#5,#3)|add(#4,const_1)|add(#6,#5)|add(#8,#2)|add(#0,#9)|add(#4,#10)|add(#11,#7)|add(n0,#12)|add(#13,#1)|

general

if the average of r , b , c , 14 and 15 is 12 . what is the average value of r , b , c and 29

r + b + c + 14 + 15 = 12 * 5 = 60 = > r + b + c = 60 - 29 = 31 r + b + c + 29 = 31 + 29 = 60 average = 60 / 4 = 15 answer d

a ) 15 , b ) 5 , c ) 18 m , d ) 82 , e ) 1083875

a

divide(add(subtract(multiply(add(const_4, const_1), 12), add(14, 15)), 29), const_4)

add(const_1,const_4)|add(n0,n1)|multiply(n2,#0)|subtract(#2,#1)|add(n3,#3)|divide(#4,const_4)

general

if 15 % of a is the same as 30 % of b , then a : b is :

expl : 15 % of a i = 30 % of b = 15 a / 100 = 30 b / 100 = 2 / 1 = 2 : 1 answer : e

a ) 2 : 1 , b ) 21.21 , c ) 8 , d ) 400 % , e ) 6

a

divide(divide(30, const_100), divide(15, const_100))

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

gain

let a be the event that a randomly selected two digit number is divisible by 2 and let b be the event that a randomly selected two digit number is divisible by 5 . what is p ( a and b ) ?

"p ( a and b ) = 1 / 2 * 1 / 5 = 1 / 10 the answer is b ."

a ) 9 , b ) 5 / 8 , c ) 12 , d ) 75 kg , e ) 1 / 10

e

multiply(divide(subtract(2, const_1), multiply(subtract(2, const_1), 2)), divide(multiply(subtract(2, const_1), const_2), multiply(subtract(2, const_1), 2)))

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

physics

what is the angle between the 2 hands of the clock at 8 : 24 pm ?

"required angle = 240 – 24 × ( 11 / 2 ) = 240 – 132 = 108 ° answer d"

a ) 5 / 4 , b ) 19619 , c ) 108 ° , d ) 4200 , e ) 9990

c

subtract(multiply(8, multiply(const_3, const_2)), 2)

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

geometry

two friends c and d leave point c and point d simultaneously and travel towards point d and point c on the same route at their respective constant speeds . they meet along the route and immediately proceed to their respective destinations in 32 minutes and 50 minutes respectively . how long will d take to cover the entire journey between point d and point c ?

let x per minute be the speed of c and y per minute be the speed of d . after meeting at a point , c travels for 32 mins and d travels for 50 mins . so distance covered by each of them post point of crossing c = 32 x and d = 50 y the distance covered by c and d before they cross each would be distance covered by d and c post crossing respectively . therefore distance covered by d before he meets c = 32 x time taken by d cover 32 x distance = 32 x / y mins therefore total time taken by d = 32 x / y + 50 mins . . . . . . . . . . . . . . . . . i we need to find value of x in terms of y to arrive at final answer . total distance = 32 x + 50 y combined speed of c and d = x + y therefore time taken before c and d meet en - route = ( 32 x + 50 y ) / ( x + y ) time taken by d reach destination after meeting c = 50 mins total travel time for d = [ ( 32 x + 50 y ) / ( x + y ) ] + 50 mins . . . . . . . . . . . . . . . . . . . ii equate i and ii 32 x / y + 50 = [ ( 32 x + 50 y ) / ( x + y ) ] + 50 ( 32 x + 50 y ) / y = ( 82 x + 100 y ) / ( x + y ) 32 x ^ 2 + 50 xy + 32 xy + 50 y ^ 2 = 82 xy + 100 y ^ 2 32 x ^ 2 + 82 xy - 82 xy + 50 y ^ 2 - 100 y ^ 2 = 0 32 x ^ 2 - 50 y ^ 2 = 0 32 x ^ 2 = 50 y ^ 2 16 x ^ 2 = 25 y ^ 2 taking square root . . ( since x and y denote speed , square root ca n ' t be negative ) 4 x = 5 y y = 4 x / 5 . . . . . . . . . . . . iii substitute in i = 32 x / ( 4 x / 5 ) + 50 = 32 x * 5 / 4 x + 50 = 40 + 50 = 90 mins a

a ) 4 th , b ) 90 , c ) 305 , d ) 0.2 , e ) 380

b

add(sqrt(multiply(50, 32)), 50)

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

physics

a sells a cricket bat to b at a profit of 20 % . b sells it to c at a profit of 25 % . if c pays $ 225 for it , the cost price of the cricket bat for a is :

"a 150 125 % of 120 % of a = 225 125 / 100 * 120 / 100 * a = 225 a = 225 * 2 / 3 = 150 ."

a ) 84 % , b ) 22 , c ) 150 , d ) 273309 , e ) 20 %

c

divide(225, multiply(add(const_1, divide(20, const_100)), add(const_1, divide(25, const_100))))

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

gain

the workforce of company x is 50 % female . the company hired 20 additional male workers , and as a result , the percent of female workers dropped to 50 % . how many employees did the company have after hiring the additional male workers ?

"let x be the total worker then 0.5 x = female worker and 0.5 x is male worker then 20 male worker added 05 x / ( 0.5 x + 20 ) = 50 / 100 or 50 x = 50 * ( 0.5 x + 100 ) = 25 x + 5000 or 25 x = 5000 , x = 5000 / 25 = 200 total worker = 200 + 20 = 220 b"

a ) - 1 , b ) 3 / 50 , c ) 405 , d ) 10 years , e ) 220

e

add(divide(multiply(divide(50, const_100), 20), subtract(divide(const_60.0, const_100), divide(50, const_100))), 20)

divide(n2,const_100)|divide(const_60.0,const_100)|multiply(n1,#0)|subtract(#1,#0)|divide(#2,#3)|add(n1,#4)|

gain

x varies inversely as square of y . given that y = 3 for x = 1 . the value of x for y = 5 will be equal to :

"explanation : solution : given x = k / y ^ 2 , where k is constant . now , y = 3 and x = 1 gives k = 9 . . ' . x = 9 / y ^ 2 = > x = 9 / 5 ^ 2 = 9 / 25 answer : e"

a ) 257 , b ) 9 / 25 , c ) 8 min , d ) 50 , e ) 16

b

divide(multiply(1, power(3, const_2)), power(5, const_2))

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

general

in a class of 50 students , 20 play baseball , 15 play cricket and 11 play football . 7 play both baseball and cricket , 4 play cricket and football and 5 play baseball and football . if 18 students do not play any of these given sports , how many students play exactly two of these sports ?

notice that 7 play both baseball and cricket does not mean that out of those 7 , some does not play football too . the same for cricket / football and baseball / football . [ color = # ffff 00 ] { total } = { baseball } + { cricket } + { football } - { hc + ch + hf } + { all three } + { neither } for more checkadvanced overlapping sets problems [ / color ] 50 = 20 + 15 + 11 - ( 7 + 4 + 5 ) + { all three } + 18 - - > { all three } = 2 ; those who play only baseball and cricket are 7 - 2 = 5 ; those who play only cricket and football are 4 - 2 = 2 ; those who play only baseball and football are 5 - 2 = 3 ; hence , 5 + 2 + 3 = 10 students play exactly two of these sports . answer : a .

a ) 150 , b ) 10 , c ) 24 metres , d ) 16 , e ) 0.6

b

add(subtract(5, subtract(50, add(subtract(add(add(20, 15), 11), add(add(7, 4), 5)), 18))), add(subtract(7, subtract(50, add(subtract(add(add(20, 15), 11), add(add(7, 4), 5)), 18))), subtract(4, subtract(50, add(subtract(add(add(20, 15), 11), add(add(7, 4), 5)), 18)))))

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

other

the speed of a train is 90 kmph . what is the distance covered by it in 10 minutes ?

"90 * 10 / 60 = 15 kmph answer : a"

a ) 15 / 11 , b ) 21 hours , c ) 14 m , d ) 665 , e ) 15

e

multiply(divide(10, const_60), 90)

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

physics

in one hour , a boat goes 11 km / hr along the stream and 5 km / hr against the stream . the speed of the boat in still water ( in km / hr ) is :

"speed in still water = ( 11 + 5 ) / 2 = 8 kmph ans - c"

a ) 6 , b ) 195 cm 2 , c ) 8 kmph , d ) 4 , e ) 1.33

c

stream_speed(11, 5)

stream_speed(n0,n1)|

physics

a positive number x is multiplied by 5 , and this product is then divided by 3 . if the positive square root of the result of these two operations equals x , what is the value of x ?

"sq rt ( 5 x / 3 ) = x = > 5 x / 3 = x ^ 2 = > x = 5 / 3 ans - d"

a ) 50 , b ) 5 / 3 , c ) 31 , d ) 9', ' , e ) 605.03

b

divide(5, 3)

divide(n0,n1)|

general

the price of commodity x increases by 40 paise every year , while the price of commodity y increases by 15 paise every year . if in 2001 , the price of commodity x was rs . 4.20 and that of y was rs . 6.30 , in which year commodity x will cost 40 paise more than the commodity y ?

"suppose commodity x will cost 40 paise more than y after z years . then , ( 4.20 + 0.40 z ) - ( 6.30 + 0.15 z ) = 0.40 0.25 z = 0.40 + 2.10 z = 2.50 / 0.25 = 250 / 25 = 10 . therefore , x will cost 40 paise more than y 10 years after 2001 i . e . , 2011 . answer is d ."

a ) 0.6 , b ) 2011 , c ) 561 , d ) 2 , e ) 32.8 %

b

add(2001, divide(add(divide(40, const_100), subtract(6.30, 4.20)), subtract(divide(40, const_100), divide(15, const_100))))

divide(n0,const_100)|divide(n1,const_100)|subtract(n4,n3)|add(#0,#2)|subtract(#0,#1)|divide(#3,#4)|add(n2,#5)|

general

the sum of the present ages of two persons a and b is 60 . if the age of a is twice that of b , find the sum of their ages 8 years hence ?

"a + b = 60 , a = 2 b 2 b + b = 60 = > b = 20 then a = 40 . 5 years , their ages will be 48 and 28 . sum of their ages = 48 + 28 = 76 . answer : d"

a ) 76 , b ) 10.28 % , c ) 10 years , d ) none of these , e ) 2.6 units

a

add(add(multiply(divide(60, 8), const_2), 8), add(divide(60, 8), 8))

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

general

a money lender finds that due to a fall in the annual rate of interest from 8 % to 7 2 / 7 % his yearly income diminishes by rs . 61.50 . his capital is

"explanation : capital = rs . x , then 5 / 7 x = 61.5 x = 87.86 answer : b ) rs . 87.86"

a ) 324 sq m , b ) 38 % , c ) 20 , d ) 0.55 , e ) 87.86

e

divide(61.50, divide(const_4, 7))

divide(const_4,n3)|divide(n4,#0)|

gain

the product of two numbers is 468 and the sum of their squares is 289 . the sum of the number is ?

"let the numbers be x and y . then , xy = 468 and x 2 + y 2 = 289 . ( x + y ) 2 = x 2 + y 2 + 2 xy = 289 + ( 2 x 468 ) = 1225 x + y = 35 . option e"

a ) 14 , b ) q = 35 , c ) 35 , d ) 5 / 8 , e ) 12

c

sqrt(add(power(sqrt(subtract(289, multiply(const_2, 468))), const_2), multiply(const_4, 468)))

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

general

the ratio , by volume , of soap to alcohol to water in a certain solution is 4 : 20 : 60 . the solution will be altered so that the ratio of soap to alcohol is doubled while the ratio of soap to water is halved . if the altered solution will contain 100 cubic centimeters of alcohol , how many cubic centimeters of water will it contain ?

"soap : alcohol initial ratio soap : alcohol : water - - > 4 : 20 : 60 initial soap : alcohol = 4 / 20 = 4 : 20 after doubled soap : alcohol = 2 * 4 / 20 = 8 : 20 initial soap : water = 4 / 60 = 4 : 60 after halved soap : water : 1 / 2 * 4 / 60 = 2 / 60 = 2 : 60 after soap : alcohol : water - - > 8 : 20 : 240 - - > 2 : 5 : 60 given alcohol 100 cubic centimeter . ratio is 40 : 100 : 1200 ( 2 : 5 : 60 ) for 100 cubic centimeter of alcohol - - - 1200 cubic cm water is required ."

a ) 1200 , b ) 62 , c ) 3 , d ) 36 , e ) 3 , 012,121

a

divide(divide(divide(divide(divide(volume_rectangular_prism(100, 60, 20), const_3), const_2), 4), 4), 4)

volume_rectangular_prism(n1,n2,n3)|divide(#0,const_3)|divide(#1,const_2)|divide(#2,n0)|divide(#3,n0)|divide(#4,n0)|

geometry

if n is an integer , f ( n ) = f ( n - 1 ) - n and f ( 4 ) = 13 . what is the value of f ( 6 ) ?

"since f ( n ) = f ( n - 1 ) - n then : f ( 6 ) = f ( 5 ) - 6 and f ( 5 ) = f ( 4 ) - 5 . as given that f ( 4 ) = 13 then f ( 5 ) = 13 - 5 = 8 - - > substitute the value of f ( 5 ) back into the first equation : f ( 6 ) = f ( 5 ) - 6 = 8 - 6 = 2 . answer : d . questions on funtions to practice :"

a ) 30 , b ) 2 , c ) 157 / 160 , d ) 20 , e ) 100

b

subtract(subtract(13, add(1, 4)), 6)

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

general

what will be the area of a semi - circle of 14 metres diameter ?

area of semicircle = ½ π r 2 = ½ × 22 ⁄ 7 × 7 × 7 = 77 m 2 answer b

a ) 81 , b ) 900 , c ) 500 , d ) 77 sq metres', ' , e ) 200 liters

d

divide(circle_area(divide(14, const_2)), const_2)

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

geometry

the owner of a furniture shop charges his customer 10 % more than the cost price . if a customer paid rs . 2200 for a computer table , then what was the cost price of the computer table ?

"cp = sp * ( 100 / ( 100 + profit % ) ) = 2200 ( 100 / 110 ) = rs . 2000 answer : b"

a ) 10 min , b ) 64 : 27', ' , c ) 8 am , d ) 2000 , e ) 30 %

d

divide(2200, add(const_1, divide(10, const_100)))

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

gain

one fourth of a solution that was 10 % salt by weight was replaced by a second solution resulting in a solution that was 16 percent sugar by weight . the second solution was what percent salt by weight ?

"say the second solution ( which was 1 / 4 th of total ) was x % salt , then 3 / 4 * 0.1 + 1 / 4 * x = 1 * 0.16 - - > x = 0.34 . alternately you can consider total solution to be 100 liters and in this case you ' ll have : 75 * 0.1 + 25 * x = 100 * 0.16 - - > x = 0.34 . answer : b ."

a ) 34 % , b ) 70 m , c ) 64 % , d ) 6.5 , e ) $ 20

a

multiply(subtract(multiply(divide(16, const_100), const_4), subtract(multiply(divide(10, const_100), const_4), divide(10, const_100))), const_100)

divide(n1,const_100)|divide(n0,const_100)|multiply(#0,const_4)|multiply(#1,const_4)|subtract(#3,#1)|subtract(#2,#4)|multiply(#5,const_100)|

gain

if the average ( arithmetic mean ) of a and b is 45 and the average of b and c is 85 , what is the value of c â ˆ ’ a ?

"the arithmetic mean of a and b = ( a + b ) / 2 = 45 - - a + b = 90 - - 1 similarly for b + c = 170 - - 2 subtracting 1 from 2 we have c - a = 80 ; answer : b"

a ) 90 , b ) 36 , c ) 80 , d ) 440 , e ) 15 sec

c

subtract(multiply(85, const_2), multiply(45, const_2))

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

general

3 pumps , working 8 hours a day , can empty a tank in 2 days . how many hours a day must 4 pumps work to empty the tank in 1 day ?

the required number of working hours per day x , more pumps , less working hours per day ( indirect ) less days , more working hours per day ( indirect ) pumps 4 : 3 , days 1 : 2 } : : 8 : x therefore 4 * 1 * x = 3 * 2 * 8 , x = ( 3 * 2 * 8 ) / 4 x = 12 correct answer ( d )

a ) 7600 , b ) 34 % , c ) 12 , d ) 400 , e ) 15

c

divide(multiply(multiply(3, 8), 2), 4)

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

physics

peter invests a sum of money and gets back an amount of $ 810 in 3 years . david invests an equal amount of money and gets an amount of $ 854 in 4 years . if both amounts were invested at the same rate ( simple interest ) what was the sum of money invested ?

since both peter and david invested the same amount of money at the same rate , they would earn same interest per year . david invested for one year more than peter and hence he got interest amount for one more year . interest earned per year = amount received by david - amount received by peter = 854 - 810 = 44 interest earned for 3 years = 44 * 3 = 132 amount invested = 815 - 132 = 683 answer : b

a ) - 39 , b ) 88 % , c ) 25 , d ) 683 , e ) 120

d

subtract(810, multiply(divide(subtract(854, 810), subtract(divide(4, const_100), divide(3, const_100))), divide(3, const_100)))

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

gain

a envelop weight 9.2 gm , if 800 of these envelop are sent with an advertisement mail . how much wieght ?

"800 * 9.2 7360.0 gm 7.36 kg answer : e"

a ) rs . 440 , b ) 50 , c ) 80 % , d ) 7.36 kg , e ) 13.75

d

divide(multiply(9.2, 800), const_1000)

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

general

what least number should be added to 1022 , so that the sum is completely divisible by 25 ?

"1022 ã · 25 = 40 with remainder = 22 22 + 3 = 25 . hence 3 should be added to 1022 so that the sum will be divisible by 25 answer : option b"

a ) 8.4 , b ) 3 , c ) 351 , d ) 250 , e ) 20

b

subtract(25, reminder(1022, 25))

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

general

a hostel had provisions for 250 men for 44 days . if 50 men left the hostel , how long will the food last at the same rate ?

"a hostel had provisions for 250 men for 44 days if 50 men leaves the hostel , remaining men = 250 - 50 = 200 we need to find out how long the food will last for these 200 men . let the required number of days = x days more men , less days ( indirect proportion ) ( men ) 250 : 200 : : x : 44 250 × 44 = 200 x 5 × 44 = 4 x x = 5 × 11 = 55 answer a"

a ) 1840 , b ) 55 , c ) 4 , d ) 30 , e ) 150

b

divide(multiply(250, 44), subtract(250, 50))

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

gain

the owner of a furniture shop charges his customer 24 % more than the cost price . if a customer paid rs . 8339 for a computer table , then what was the cost price of the computer table ?

"cp = sp * ( 100 / ( 100 + profit % ) ) = 8339 ( 100 / 124 ) = rs . 6725 . answer : c"

a ) 14 , b ) 4 and 12 , c ) rs . 6725 . , d ) $ 0.50 , e ) 15

c

divide(8339, add(const_1, divide(24, const_100)))

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

gain

63 + 5 * 12 / ( 180 / 3 ) = ?

"63 + 5 * 12 / ( 180 / 3 ) = 63 + 5 * 12 / ( 60 ) = 63 + ( 5 * 12 ) / 60 = 63 + 1 = 64 . answer : d"

a ) 64 , b ) 50 litres , c ) rs . 98.56 , d ) 9.9 % , e ) 4

a

add(63, divide(multiply(5, 12), divide(180, 3)))

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

general

in a fuel station the service costs $ 1.75 per car , every liter of fuel costs 0.45 $ . assuming that a company owns 12 cars and that every fuel tank contains 55 liters and they are all empty , how much money total will it cost to fuel all cars ?

12 * 1.75 + 0.45 * 12 * 55 = 318 hence - a

a ) 1.01 , b ) 63 , c ) 318 $ , d ) 315 , e ) 4327

c

multiply(multiply(0.45, 55), 12)

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

general

if ( - 6 ) ^ ( 2 x ) = 6 ^ ( 7 + x ) and x is an integer , what is the value of x ?

"since x is an integer , ( - 6 ) ^ 2 x is always positive . so , 6 ^ 2 x = 6 ^ ( 7 + x ) 2 x = 7 + x x = 7 answer : e"

a ) 4 , b ) 2 , c ) 1984 , d ) 21 , e ) 7

e

divide(7, 6)

divide(n3,n0)|

general

a hen leaps 3 leaps for every 2 leaps of a duck , but 4 leaps of the duck are equal to 3 leaps of the hen . what is the ratio of the speed of the hen to that of the duck ?

"given ; 4 duck = 3 hen ; or , duck / hen = 3 / 4 ; let hen ' s 1 leap = 4 meter and ducks 1 leap = 3 meter . then , ratio of speed of hen and duck = 4 * 3 / 3 * 2 = 2 : 1 ' ' answer : 2 : 1 ;"

a ) 10 / 21 , b ) 6400 , c ) 8100 , d ) 4 , e ) 2 : 1

e

divide(divide(3, 2), divide(3, 4))

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

other

praveen starts business with rs . 3640 and after 5 months , hari joins with praveen as his partner . after a year , the profit is divided in the ratio 2 : 3 . what is hari ’ s contribution in the capital ?

"let hari ’ s capital be rs . x . then , 3640 * 12 / 7 x = 2 / 3 = > 14 x = 131040 = > x = 9360 . answer : e"

a ) s . 9360 , b ) 982.14', ' , c ) 10 % , d ) 480 , e ) 270

a

divide(divide(3640, subtract(const_1, divide(5, const_12))), divide(2, 3))

divide(n1,const_12)|divide(n2,n3)|subtract(const_1,#0)|divide(n0,#2)|divide(#3,#1)|

other

salad dressing p is made up of 30 % vinegar and 70 % oil , and salad dressing q contains 10 % vinegar and 90 % oil . if the two dressings are combined to produce a salad dressing that is 20 % vinegar , dressing p comprises what percentage of the new dressing ?

"let x be the percentage of dressing p in the new dressing . 0.3 x + 0.1 ( 1 - x ) = 0.20 0.2 x = 0.10 x = 0.5 = 50 % the answer is b ."

a ) 50 % , b ) 1 / 3 , c ) 18 , d ) 1.11 % , e ) 1984

a

divide(subtract(30, 10), subtract(20, 10))

subtract(n0,n2)|subtract(n4,n2)|divide(#0,#1)|

gain

how many positive integers q between 200 and 300 ( both inclusive ) are not divisible by 2 , 3 or 5 ?

1 ) i figured there are 101 integers ( 300 - 200 + 1 = 101 ) . since the set begins with an even and ends with an even , there are 51 evens . 2 ) question says integers are not divisible by 2 , leaving all of the odds ( 101 - 51 = 50 integers ) . 3 ) question says integers are not divisible by 5 , removing all the integers ending in 5 ( already took out those ending in 0 ) . take out 10 integers ( 2 ? 5 , ? = 0 to 9 ) , leaving us with 40 integers . 4 ) now the painstaking part . we have to remove the remaining numbers that are multiples of 3 . those are 201 , 207 , 213 , 219 , 231 , 237 , 243 , 249 , 261 , 267 , 273 , 279 , 291 , and 297 . . . a total of 14 numbers . 26 numbers left ! 6 ) answer choice e .

a ) 26 , b ) 64 , c ) 0.0012 , d ) $ 6.80 , e ) 756

a

subtract(subtract(subtract(add(subtract(300, 200), const_1), add(subtract(divide(300, 2), divide(200, 2)), const_1)), floor(add(subtract(add(subtract(divide(300, 3), divide(200, 3)), const_1), add(add(const_10, 5), 2)), const_1))), subtract(add(subtract(divide(300, 5), divide(200, 5)), const_1), add(const_10, 5)))

add(n4,const_10)|divide(n1,n2)|divide(n0,n2)|divide(n1,n3)|divide(n0,n3)|divide(n1,n4)|divide(n0,n4)|subtract(n1,n0)|add(#7,const_1)|add(n2,#0)|subtract(#1,#2)|subtract(#3,#4)|subtract(#5,#6)|add(#10,const_1)|add(#11,const_1)|add(#12,const_1)|subtract(#8,#13)|subtract(#14,#9)|subtract(#15,#0)|add(#17,const_1)|floor(#19)|subtract(#16,#20)|subtract(#21,#18)

other

rs . 850 becomes rs . 956 in 3 years at a certain rate of simple interest . if the rate of interest is increased by 4 % , what amount will rs . 850 become in 3 years ?

"solution s . i . = rs . ( 956 - 850 ) = rs . 106 rate = ( 100 x 106 / 850 x 3 ) = 212 / 51 % new rate = ( 212 / 51 + 4 ) % = 416 / 51 % new s . i . = rs . ( 850 x 416 / 51 x 3 / 100 ) rs . 208 . ∴ new amount = rs . ( 850 + 208 ) = rs . 1058 . answer c"

a ) $ 1600 , b ) 10749 , c ) 83 % , d ) 25400 , e ) rs . 1058

e

add(850, divide(multiply(multiply(850, add(divide(multiply(subtract(956, 850), const_100), multiply(850, 3)), 4)), 3), const_100))

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

gain

a specialized type of sand consists of 40 % mineral x by volume and 60 % mineral y by volume . if mineral x weighs 2.5 grams per cubic centimeter and mineral y weighs 4 grams per cubic centimeter , how many grams does a cubic meter of specialized sand combination weigh ? ( 1 meter = 100 centimeters )

"let the volume be 1 m ^ 3 = 1 m * 1 m * 1 m = 100 cm * 100 cm * 100 cm = 1 , 000,000 cm ^ 3 by volume 40 % is x = 400,000 cm ^ 3 60 % is y = 600,000 cm ^ 3 by weight , in 1 cm ^ 3 , x is 2.5 gms in 400,000 cm ^ 3 , x = 2.5 * 400,000 = 1 , 000,000 grams in 1 cm ^ 3 , y is 4 gms in 600,000 cm ^ 3 , y = 4 * 600,000 = 2 , 400,000 gms total gms in 1 m ^ 3 = 1 , 000,000 + 2 , 400,000 = 3 , 400,000 answer : a"

a ) 25 , b ) 3 , 400,000 , c ) 700 , d ) 46080 , e ) 7 / 15

b

subtract(add(multiply(multiply(divide(volume_cube(100), const_10), 2.5), 2.5), multiply(multiply(divide(volume_cube(100), const_10), multiply(const_2, 4)), 4)), volume_cube(100))

multiply(const_2,n3)|volume_cube(n5)|divide(#1,const_10)|multiply(#2,n2)|multiply(#2,#0)|multiply(#3,n2)|multiply(#4,n3)|add(#5,#6)|subtract(#7,#1)|

geometry

it takes ten minutes to load a certain video on a cellphone , and fifteen seconds to load that same video on a laptop . if the two devices were connected so that they operated in concert at their respective rates , how many seconds would it take them to load the video , rounded to the nearest hundredth ?

"the laptop can load the video at a rate of 1 / 15 of the video per second . the phone can load the video at a rate of 1 / ( 60 * 10 ) = 1 / 600 of the video per second . the combined rate is 1 / 15 + 1 / 600 = 41 / 600 of the video per second . the time required to load the video is 600 / 41 = 14.63 seconds . the answer is d ."

a ) $ 94,000 , b ) 14.63 , c ) 14 , d ) 4 / 7 , e ) 7 days

b

subtract(inverse(add(inverse(multiply(add(add(const_2, const_3), const_4), const_60)), inverse(add(multiply(const_3, const_4), const_3)))), divide(subtract(multiply(multiply(const_4, const_4), const_3), const_2), multiply(const_100, const_100)))

add(const_2,const_3)|multiply(const_3,const_4)|multiply(const_4,const_4)|multiply(const_100,const_100)|add(#0,const_4)|add(#1,const_3)|multiply(#2,const_3)|inverse(#5)|multiply(#4,const_60)|subtract(#6,const_2)|divide(#9,#3)|inverse(#8)|add(#11,#7)|inverse(#12)|subtract(#13,#10)|

physics

a certain scholarship committee awarded scholarships in the amounts of $ 1250 , $ 2500 and $ 4000 . the committee awarded twice as many $ 2500 scholarships as $ 4000 and it awarded 3 times as many $ 1250 scholarships as $ 2500 scholarships . if the total of $ 75000 was awarded in $ 1250 scholarships , how many $ 4000 scholarships were awarded ?

since the starting point is given as the $ 4000 scholarship , assume $ 4000 scholarships to be x by the given information , $ 2500 scholarships = 2 x and $ 1250 scholarships = 6 x gievn : total $ 1250 scholarships = $ 75000 6 x * 1250 = 75000 solve for x = 10 option d

a ) 45 , b ) 8 : 5 , c ) 10 , d ) 36 , e ) 10 v 2', '

c

divide(divide(75000, 1250), multiply(const_2, 3))

divide(n8,n0)|multiply(n5,const_2)|divide(#0,#1)

general

in a market , a dozen eggs cost as much as a pound of rice , and a half - liter of kerosene costs as much as 6 eggs . if the cost of each pound of rice is $ 0.33 , then how many cents does a liter of kerosene cost ? [ one dollar has 100 cents . ]

"a dozen eggs cost as much as a pound of rice - - > 12 eggs = 1 pound of rice = 33 cents ; a half - liter of kerosene costs as much as 6 eggs - - > 6 eggs = 1 / 2 liters of kerosene . how many cents does a liter of kerosene cost - - > 1 liter of kerosene = 12 eggs = 12 / 12 * 33 = 33 cents . answer : c ."

a ) 33 , b ) 70400 yards , c ) 270 , d ) t = 121 , e ) 296

a

multiply(divide(divide(6, divide(const_1, const_2)), const_12), multiply(0.33, 100))

divide(const_1,const_2)|multiply(n1,n2)|divide(n0,#0)|divide(#2,const_12)|multiply(#3,#1)|

general

a pupil ' s marks were wrongly entered as 73 instead of 40 . due to the average marks for the class got increased by half . the number of pupils in the class is ?

"let there be x pupils in the class . total increase in marks = ( x * 1 / 2 ) = x / 2 x / 2 = ( 73 - 40 ) = > x / 2 = 33 = > x = 66 . answer : c"

a ) 66 , b ) 10750 , c ) 4 , d ) 14 , e ) 996004

a

multiply(subtract(73, 40), const_2)

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

general

a baseball card decreased in value 25 % in its first year and 10 % in its second year . what was the total percent decrease of the card ' s value over the two years ?

"let the initial value of baseball card = 100 after first year , value of baseball card = ( 1 - 25 / 100 ) * 100 = 75 after second year , value of baseball card = ( 1 - 10 / 100 ) * 75 = 67.5 total percent decrease of the card ' s value over the two years = ( 100 - 67.5 ) / 100 * 100 % = 31.5 % answer c"

a ) 3720 , b ) 12 % , c ) 32.5 % , d ) $ 400 , e ) 20

c

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

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

gain

if an integer n is to be chosen at random from the integers 1 to 96 , inclusive , what is the probability that n ( n + 1 ) ( n + 2 ) will be divisible by 8 ?

"total nos of ways in which we can choose n = 96 n ( n + 1 ) ( n + 2 ) will be divisible by 8 ? case 1 : n = odd then n + 2 = odd & n + 1 will be even i . e this needs get divided by 8 , hence is a multiple of 8 so we have 8 . . 96 = 12 multiples to fill the n + 1 pos hence 12 ways case 2 : n is even then n + 2 will be even & the product will be divisible by 24 & thus 8 so nos of values that can be used for n = 2 . . . . 96 ( all even nos ) i . e 48 nos total = 48 + 12 = 60 ways so reqd p = 60 / 96 = 5 / 8 ; answer : d"

a ) 5 / 8 , b ) 9 / 14 , c ) 10 , d ) 56 , e ) 25

a

divide(add(divide(96, 2), divide(96, 8)), 96)

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

general

a car going at 40 miles per hour set out on an 60 - mile trip at 9 : 00 a . m . exactly 10 minutes later , a second car left from the same place and followed the same route . how fast , in miles per hour , was the second car going if it caught up with the first car at 10 : 30 a . m . ?

"let car a = car that starts at 9 am car b = car that starts at 9 : 10 am time for which car a travels at speed of 40 m per hour = 1.5 hours distance travelled by car a = 40 * 1.5 = 60 miles since car b catches up car a at 10 : 30 , time = 60 mins = 1 hour speed of car b = 60 / ( 1 ) = 60 miles per hour answer b"

a ) $ 214.16 , b ) 1 : 2 , c ) 10.5 , d ) 60 , e ) 2400

d

divide(60, divide(add(multiply(subtract(10, 9), const_60), subtract(40, 10)), const_60))

subtract(n4,n2)|subtract(n0,n4)|multiply(#0,const_60)|add(#2,#1)|divide(#3,const_60)|divide(n1,#4)|

physics

1 = 5,2 = 25,3 = 253,4 = 150,5 = 225 then 150 = ?

"1 = 5,2 = 25,3 = 253,4 = 150,5 = 225 then 150 = ? 150 = 4 check the fourth eqn . answer : c"

a ) $ 1,240 , b ) 14 % , c ) 4 , d ) 23 / 29 , e ) 270

c

divide(subtract(subtract(225, multiply(multiply(add(const_4, const_2), add(const_4, const_2)), const_10)), 1), const_2)

add(const_2,const_4)|multiply(#0,#0)|multiply(#1,const_10)|subtract(n5,#2)|subtract(#3,n0)|divide(#4,const_2)|

general

a dealer offers a cash discount of 16 % and still makes a profit of 25 % when he further allows 60 articles to be sold at the cost price of 50 articles to a particular sticky bargainer . how much percent above the cost price were his articles listed ?

"given cash discount - 16 % profit - 25 % items sold - 60 price sold at = list price of 50 assume list price = $ 10 total invoice = $ 500 - 16 % cash discount = $ 420 let cost price of 60 items be x so total cost = 60 * x given the shopkeeper had a profit of 25 % 60 * x * 125 / 100 = 420 or x = $ 7 * 4 / 5 = $ 28 / 5 which means his products were listed at $ 10 which is a 78 + ( 4 / 7 ) % markup over $ 28 / 5 answer e"

a ) 8 , b ) 48 - 6 π', ' , c ) 27 / 64 , d ) $ 6 , e ) 78 + ( 4 / , 7 ) 2220

e

multiply(subtract(divide(divide(divide(add(const_100, 25), const_100), subtract(const_1, divide(subtract(60, 50), 60))), divide(subtract(const_100, 16), const_100)), const_1), const_100)

add(n1,const_100)|subtract(n2,n3)|subtract(const_100,n0)|divide(#0,const_100)|divide(#1,n2)|divide(#2,const_100)|subtract(const_1,#4)|divide(#3,#6)|divide(#7,#5)|subtract(#8,const_1)|multiply(#9,const_100)|

gain

if n divided by 8 has a remainder of 1 , what is the remainder when 3 times n is divided by 8 ?

as per question = > n = 8 p + 1 for some integer p hence 3 n = > 24 q + 3 = > remainder = > 3 for some integer q hence b

a ) 3 , b ) 9 , c ) 36 , d ) 360 , e ) 3 / 4000

a

multiply(3, 1)

multiply(n1,n2)

general

the volume of a cube is 2197 cc . find its surface .

a 3 = 2197 = > a = 13 6 a 2 = 6 * 13 * 13 = 1014 answer : b

a ) 4 % , b ) 4 years , c ) 60 cm , d ) 0 , e ) 1014

e

surface_cube(cube_edge_by_volume(2197))

cube_edge_by_volume(n0)|surface_cube(#0)|

geometry

15 business executives and 3 chairmen meet at a conference . if each business executive shakes the hand of every other business executive and every chairman once , and each chairman shakes the hand of each of the business executives but not the other chairmen , how many handshakes would take place ?

"there are 15 business exec and in each handshake 2 business execs are involved . hence 15 c 2 = 105 also , each of 15 exec will shake hand with every 3 other chairmen for total of 45 handshake . total = 45 + 105 = 150 ans : a"

a ) 844.03 m , b ) 5 , c ) 36.3 , d ) 40.6 , e ) 150

e

add(divide(multiply(15, subtract(15, const_1)), const_2), multiply(15, 3))

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

geometry

if pintu is coded as 79523 in a certain code language , how would you code mumbo in the same language ?

"1 2 3 4 5 6 7 8 9 a b c d e f g h i j k l m n o p q r s t u v w x y z sooo . . . mumbo is 43426 . . . answer : a"

a ) 9 , b ) 46 , c ) 43426 , d ) 44 % , e ) 528

c

divide(79523, add(const_3, const_3))

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

general

a is twice as fast as b . if b alone can do a piece of work in 18 days , in what time can a and b together complete the work ?

"a can do the work in 18 / 2 i . e . , 9 days . a and b ' s one day ' s work = 1 / 9 + 1 / 18 = ( 2 + 1 ) / 18 = 1 / 6 so a and b together can do the work in 6 days . answer : d"

a ) 30 minutes , b ) 5 / 6 , c ) 6 , d ) 8 , e ) 48

c

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

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

physics

the area of a triangle is with base 5.5 m and height 6 m ?

"1 / 2 * 5.5 * 6 = 16.5 m 2 answer : b"

a ) 40 days , b ) 16.5 m 2 , c ) 0.01616 , d ) $ 1200 , e ) 340

b

triangle_area(5.5, 6)

triangle_area(n0,n1)|

geometry

working together , printer a and printer b would finish the task in 15 minutes . printer a alone would finish the task in 45 minutes . how many pages does the task contain if printer b prints 3 pages a minute more than printer a ?

"15 * a + 15 * b = x pages in 15 mins printer a will print = 15 / 45 * x pages = 1 / 3 * x pages thus in 15 mins printer printer b will print x - 1 / 3 * x = 2 / 3 * x pages also it is given that printer b prints 3 more pages per min that printer a . in 15 mins printer b will print 45 more pages than printer a thus 2 / 3 * x - 1 / 3 * x = 45 = > x = 135 pages answer : b"

a ) 5 / 4 , b ) 135 , c ) 180 m , d ) 12 hours , e ) 5

b

multiply(divide(3, subtract(divide(45, 15), const_1)), 45)

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

physics

the owner of a furniture shop charges his customer 25 % more than the cost price . if a customer paid rs . 8400 for a computer table , then what was the cost price of the computer table ?

"cp = sp * ( 100 / ( 100 + profit % ) ) = 8400 ( 100 / 125 ) = rs . 6720 . answer : d"

a ) rs . 6720 , b ) 4 , c ) 90 , d ) 16 , e ) rs . 600

a

divide(8400, add(const_1, divide(25, const_100)))

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

gain

if you write down all the numbers from 1 to 100 , then how many times do you write 3 ?

"explanation : explanation : clearly , from 1 to 100 , there are ten numbers with 3 as the unit ' s digit - 3 , 13 , 23 , 33 , 43 , 53 , 63 , 73 , 83 , 93 ; and ten numbers with 3 as the ten ' s digit - 30 , 31 , 32 , 33 , 34 , 35 , 36 , 37 , 38 , 39 . so , required number = 10 + 10 = 20 . answer : c"

a ) 15 days , b ) 75 , c ) 176 , d ) 7 , e ) 20

e

divide(subtract(100, 1), 3)

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

general

the h . c . f . of two numbers is 23 and the other two factors of their l . c . m . are 10 and 11 . the larger of the two numbers is :

"clearly , the numbers are ( 23 x 10 ) and ( 23 x 11 ) . larger number = ( 23 x 11 ) = 253 . answer : option b"

a ) $ 12.48 , b ) 4 , c ) 60 , d ) 253 , e ) 61 cm

d

multiply(23, 11)

multiply(n0,n2)|

other